Transport Processes And Unit Operations

Transport Processes

and Unit Operations

CHRISTIE

J.

GEANKOPLIS

University of Minnesota

Transport Processes and Unit Operations Third Edition

Prentice-Hall International, Inc.

r"

ISBN 0-13-045253-X

This edition may be sold only in those countries to which it is consigned by Prentice-Hall International. It is not to be re-exported and it is not for sale in the U.S.A., Mexico, or Canada.

©

1993, 1983, 1978

by P

TR

Prentice-Hall, Inc.

Simon & Schuster Company Englewood Cliffs, New Jersey 07632

A

All rights reserved. No part of this book may be reproduced, in any form or by any means, without permission in writing from the publisher.

Printed in the United States of America 10

9

ISBN -13-DMSES3-X

Prentice-Hall International

(UK) Limited, London

Prentice-Hall of Australia Pty. Limited, Sydney

Prentice-Hall

Canada

Inc.,

Toronto

Prentice-Hall Hispanoamericana, S.A., Prentice-Hall of India Private Limited,

Mexico

New

Delhi

Prentice-Hall of Japan, Inc., Tokyo

& Schuster Asia Pte. Ltd., Singapore Editora Prentice-Hall do Brasil, Ltda., Rio de Janeiro Prentice-Hall, Inc., Englewood Cliffs, New Jersey

Simon

Dedicated to the

memory of my beloved mother, Helen,

for her love and encouragement

Contents xi

Preface

PART 1 TRANSPORT PROCESSES: MOMENTUM, HEAT, AND MASS Chapter!

Introduction to Engineering Principles and Units

and Transport Processes

1.1

Classification of Unit Operations

1.2

SI System of Basic Units

1.3

Methods of Expressing Temperatures and Compositions Gas Laws and Vapor Pressure Conservation of Mass and Material Balances

1.4 1.5 1.6 1.7 1.8

Used

in

1

This Text and Other Systems

5

7 9

Energy and Heat Units Conservation of Energy and Heat Balances Graphical, Numerical, and Mathematical Methods

Chapter 2

Principles of Momentum Transfer

1

3

14 19

23

and Overall Balances

31

2.1

Introduction

31

2.2

Fluid Statics

32

2.3

General Molecular Transport Equation

for

Momentum,

Heat, and

Mass Transfer

39

2.4

Viscosity of Fluids

43

2.5

Types of Fluid Flow and Reynolds Number Overall Mass Balance and Continuity Equation Overall Energy Balance Overall Momentum Balance Shell Momentum Balance and Velocity Profile in Laminar Flow Design Equations for Laminar and Turbulent Flow in Pipes " Compressible Flow of Gases

47

2.6

2.7 2.8 2.9

2.10 2.11

Chapter 3

Principles of

Momentum

Transfer and Applications

3.5

Flow Past Immersed Objects and Packed and Fluidized Beds Measurement of Flow of Fluids Pumps and Gas-Moving Equipment Agitation and Mixing of Fluids and Power Requirements Non-Newtonian Fluids

3.6

Differential

3.1

3.2 3.3

3.4

3.7 3.8 3.9

3.10 3.11

50 56

69 78 83 101

114 114

-127 133

140 '53

164

Equations of Continuity Differential Equations of Momentum Transfer or Motion Use of Differential Equations of Continuity and Motion Other Methods for Solution of Differential Equations of Motion

184

Boundary-Layer Flow and Turbulence Dimensional Analysis in Momentum Transfer

202

170 175

190

vii

Chapter 4

Principles of Steady-State

Heat Transfer

214

4.1

Introduction and Mechanisms of Heat Transfer

214

4.2

220

4.7

Conduction Heat Transfer Conduction Through Solids in Series Steady-State Conduction and Shape Factors Forced Convection Heat Transfer Inside Pipes Heat Transfer Outside Various Geometries in Forced Convection Natural Convection Heat Transfer

4.8

Boiling and Condensation

4.9

Heat Exchangers

259 267 276

43 4.4

45 4.6

4.10

Introduction to Radiation Heat Transfer

223 233

236 247 253

4.11

Advanced Radiation Heat-Transfer

4.12

Heat Transfer of Non-Newtonian Fluids

297

4.13

Special Heat-Transfer Coefficients

300

4.14

Dimensional Analysis in Heat Transfer Numerical Methods for Steady-State Conduction

4.15

281

Principles

308 in

Two

310

Dimensions

Chapters

Principles of Unsteady-State

Heat Transfer

330

5.1

Derivation of Basic Equation

330

5.2

Simplified Case for Systems with Negligible Internal Resistance

332

5.3

Unsteady-State Heat Conduction in Various Geometries Numerical Finite-Difference Methods for Unsteady-State Conduction Chilling and Freezing of Food and Biological Materials Differential Equation of Energy Change Boundary-Layer Flow and Turbulence in Heat Transfer

334

5.4 5.5

5.6 5.7

Chapter 6

Principles of

Mass Transfer

Introduction to

6.2

Molecular Diffusion

in

6.3

Molecular Diffusion in Liquids

Molecular Diffusion

6.5

Molecular Diffusion

in Solids

Methods Two Dimensions

for Steady-State

6.6 ...Numerical

Chapter 7

381

385

Gases

6.4

365

370

381

Mass Transfer and Diffusion

6.1

350 360

397

in Biological Solutions

403

and Gels

408

Molecular Diffusion

in

413

Principles of Unsteady-State and Convective

Mass Transfer

426

7.1

Unsteady-State Diffusion

426

7.2

Convective Mass-Transfer Coefficients

432

73

Mass-Transfer Coefficients for Various Geometries

437

7.4

Mass Transfer

450

to

Suspensions of Small Particles

7.5

Molecular Diffusion Plus Convection and Chemical Reaction

453

7.6

Diffusion of Gases in Porous Solids and Capillaries

462

7.7

Numerical Methods

7.8

Dimensional Analysis in Mass Transfer Boundary-Layer Flow and Turbulence in Mass Transfer

7.9

viii

for

Unsteady-State Molecular Diffusion

468

474 475

Contents

PART 2 UNIT OPERATIONS Chapter 8

Evaporation

489 489

8.1

Introduction

8.2

Types

83

Overall Heat-Transfer Coefficients in Evaporators

495

8.4

Calculation Methods for Single-Effect Evaporators

8.5

Calculation Methods for Multiple-Effect Evaporators

496 502

8.6

Condensers for Evaporators Evaporation of Biological Materials Evaporation Using Vapor Recompression

8.7

8.8

of Evaporation

Equipment and Operation Methods

491

511

513

514

Drying of Process Materials

520

9.1

Introduction and Methods of Drying

520

9.2

Equipment

93

Vapor Pressure

Chapter 9

9.4

93 9.6

9.7 9.8

9.9

9.10 9.11

9.12

10.2 10.3

10.4

103 10.6 10.7

10.8

521

Drying of Water

and Humidity Equilibrium Moisture Content of Materials Rate of Drying Curves Calculation Methods for Constant-Rate Drying Period Calculation Methods for Falling-Rate Drying Period Combined Convection, Radiation, and Conduction Heat Transfer in Constant-Rate Period Drying in Falling-Rate Period by Diffusion and Capillary Flow Equations for Various Types of Dryers Freeze Drying of Biological Materials Unsteady-State Thermal Processing and Sterilization

525

of Biological Materials

569

Chapter 10 10.1

for

533 536 540 545 548 551

556 566

Stage and Continuous Gas— Liquid Separation Processes

Types of Separation Processes and Methods Equilibrium Relations Between Phases

584'

and Multiple Equilibrium Contact Stages Mass Transfer Between Phases Continuous Humidification Processes Absorption in Plate and Packed Towers Absorption of Concentrated Mixtures in Packed Towers Estimation of Mass Transfer Coefficients for Packed Towers

587

586

Single

Chapter 11

113 11.6

Fractional Distillation Using Enthalpy-Concentration

11.7

Distillation of

11.2

113 11.4

Contents

Multicomponent Mixtures

594

602

610 627

632 640

Vapor-Liquid Separation Processes

Vapor-Liquid Equilibrium Relations Single-Stage Equilibrium Contact for Vapor-Liquid System Simple Distillation Methods Distillation with Reflux and McCabe-Thiele Method Distillation and Absorption Tray Efficiencies

11.1

584

Method

640

642

644 649 666 669

679

ix

Liquid-Liquid and Fluid—Solid Separation Processes

Chapter 12

697

12.1

Introduction to Adsorption Processes

697

12.2

Batch Adsorption Design of Fixed-Bed Adsorption Columns Ion-Exchange Processes Single-Stage Liquid— Liquid Extraction Processes Equipment for Liquid-Liquid Extraction

700

716

12.11

Continuous Multistage Countercurrent Extraction Introduction and Equipment for Liquid-Solid Leaching Equilibrium Relations and Single-Stage Leaching Countercurrent Multistage Leaching Introduction and Equipment for Crystallization

12.12

Crystallization

12.3 12.4 12.5 12.6 12.7 12.8

12.9

12.10

708

709 715

Theory

Membrane

Chapter 13

701

723

729 733 737 743

Separation Process

754

13.4

Membrane Separation Processes Membrane Processes or Dialysis Gas Permeation Membrane Processes Complete-Mixing Model for Gas Separation by Membranes

764

13.5

Complete -Mixing Model for Multicomponent Mixtures

769

13.6

772

13.10

Cross-Flow Model for Gas Separation by Membranes Countercurrent-Flow Model for Gas Separation by Membranes Effects of Processing Variables on Gas Separation by Membranes Reverse-Osmosis Membrane Processes Applications, Equipment, and Models for Reverse Osmosis

13.11

Ultrafiltration

13.1

13.2

13.3

13.7 13.8 13.9

Introduction and Types of

754

Liquid Permeation

755

Chapter 14 14.1

Membrane Processes

759

778

780 782 788 791

Mechanical-Physical Separation Processes

800

Introduction and Classification of Mechanical-Physical Separation

Processes

800

14.2

Filtration in Solid-Liquid Separation

14.3

Settling

14.4

Centrifugal Separation Processes

828

14.5

Mechanical Size Reduction

840

and Sedimentation

in Particle-Fluid

801

Separation

815

Appendix

Appendix Appendix Appendix Appendix Appendix

A. 2

Fundamental Constants and Conversion Factors Physical Properties of Water

A.3

Physical Properties of Inorganic and Organic

A. 4 A.5

Physical Properties of Foods and Biological Materials

889

Properties of Pipes, Tubes, and Screens

892

A.l

Compounds

850 854 864

Notation

895

Index

905

x

Contents

Preface

In this third edition, the main objectives and the format of the first and second editions remain the same. The sections on momentum, transfer have been greatly expanded, especially in the sections covering differential equations of momentum transfer. This now allows full coverage of the transport processes of momentum, heat, and mass transfer. Also, a section on adsorption and an expanded chapter on membrane processes have been added to the unit operations sections.

The

field

of chemical engineering involved with physical and physical-chemical

changes of inorganic and organic materials, and to some extent biological materials, overlapping more and more with the other process engineering

fields of

is

ceramic engin-

eering, process metallurgy, agricultural food engineering, wastewater treatment (civil)

engineering, and bioengineering.

and

The

principles of

momentum,

heat,

and mass transport

the unit operations are used in these processing fields.

The engineers.

momentum

principles of

The study

However, engineers and solids.

of

in

mass

transfer

and heat

transfer

have been taught to

all

transfer has been limited primarily to chemical engineers.

other fields have

become more

interested in

mass

transfer in gases,

liquids,

many topics today, a momentum, heat, and mass

Since chemical and other engineering students must study so

more

unified introduction to the transport processes of

transfer

and

to the applications of unit operations

of the transport processes are covered this,

the text

is

PART 1

:

first,

provided. In this text the principles

is

and then the unit operations. To accomplish

divided into two main parts.

Transport Processes: Momentum, Heat, and

Mass

This part, dealing with fundamental principles, includes the following chapters: Introduction to Engineering Principles and Units;

and Overall Balances; ciples of Steady-State

3.

Principles of

Heat Transfer;

Momentum

5.

2.

Principles of

Momentum

Transfer and Applications;

1.

Transfer 4.

Prin-

Principles of Unsteady-State Heat Transfer;

6.

xi

Principles of

Mass

Transfer; and

Principles of Unsteady-State

7.

and Convective Mass

Transfer.

PART 2:

Unit Operations

This part, on applications, covers the following unit operations: 8. Evaporation; 9. Drying of Process Materials; 10. Stage and Continuous Gas-Liquid Separation Processes (humidification, absorption); 1 1. Vapor-Liquid Separation Processes (distillation); 12. Liquid-Liquid and Fluid-Solid Separation Processes (adsorption, ion exchange, extraction, leaching, crystallization); 13. Membrane Separation Processes (dialysis, gas separation, reverse osmosis, ultrafiltration); 14. Mechanical-Physical Separation Processes (filtration, settling, centrifugal separation, mechanical size reduction). In Chapter

1

elementary principles of mathematical and graphical methods, laws of

chemistry and physics, material balances, and heat balances are reviewed. Many, es-

may

pecially chemical engineers, all

be familiar with most of these principles and

may

omit

or parts of this chapter.

A few topics, involved primarily with the processing of biological materials, may be omitted at the discretion of the reader or instructor: Sections 5.5, 6.4, 8.7, 9.11, and 9.12.

Over 230 example or sample problems and over 500 homework problems on

topics are included in the text.

Some

biological systems, for those readers

This text

may be used

plans. In all plans, 1.

Chapter

for a 1

homework problems

of the

who

are especially interested in that area.

course of study using any of the following

may

may not

or

all

are concerned with

five

suggested

be included.

Study of transport processes of momentum, heat, and mass and unit operations. In most of the complete text covering the principles of the transport processes in

this plan,

Part

1

and the

unit operations in Part 2 are

primarily to chemical engineering

and

and one-half year course of study

covered. This plan could be applicable one

also to other process engineering fields in a

at the junior and/or senior level.

Study of transport processes of momentum, heat, and mass and selected operations. Only the elementary sections of Part 1 (the principles chapters 2, 3, 4, 2.

and

7)

—

are covered,

unit 5, 6,

plus selected unit operations topics in Part 2 applicable to a

particular field in a two-semester or three-quarter course.

Those

in

wastewater treatment

engineering, food process engineering, and process metallurgy could follow this plan.

The purpose of this 3. Study of transport processes of momentum, heat, and mass. plan in a two-quarter or two-semester course is to obtain a basic understanding of the transport processes of momentum, heat, and mass transfer. This involves studying sections of the principles chapters

—

2, 3, 4, 5, 6,

and 7

in Part

1

—and omitting Part

2, the

applied chapters on unit operations.

Study of unit operations. If the reader has had courses in the transport processes heat, and mass, Chapters 2 through 7 can be omitted and only the unit operations chapters in Part 2 studied in a one-semester or two-quarter course. This plan 4.

of

momentum,

could be used by chemical and certain other engineers.

For those such as chemical or mechanical engineers who who desire only a background in mass transfer in a one-quarter or one-semester course, Chapters 6, 7, and 10 would be covered. Chapters 9, 11, 12, and 13 might be covered optionally, depending on the needs of the 5.

Study of mass

have had

transfer.

momentum and

heat transfer, or those

reader.

xii

Preface

The

SI (Systeme International d'Unites) system of units has been adopted by the community. Because of this, the SI system of units has been adopted in this text use in the equations, example problems, and homework problems. However, the most

scientific

for

important equations derived English,

when

different.

in the text are also given in a dual set of units, SI

Many example and homework problems

and

are also given using

English units. Christie J. Geankoplis

Preface

xiii

PART

1

Transport Processes

Momentum, Heat, and Mass

CHAPTER

1

Introduction to

Engineering Principles

and Units

CLASSIFICATION OF UNIT OPERATIONS AND TRANSPORT PROCESSES

1.1

1.1A

Introduction

and other physical processing industries and the food and biological many similarities exist in the manner in which the entering feed materials are modified- or processed into final materials of chemical and biological In the chemical

processing industries,

products.

We

can take these seemingly different chemical, physical, or biological pro-

and break them down into a series of separate and distinct steps called unit operations. These unit operations are common to all types of diverse process industries.

cesses

For example, the other foods

is

unit operation distillation

is

used to purify or separate alcohol in

petroleum industry. Drying of grain and similar to drying of lumber, filtered precipitates, and rayon yarn. The unit

the beverage industry

and hydrocarbons

in the

operation absorption occurs in absorption of oxygen from air in a fermentation process or in a sewage treatment plant and

hydrogenation of

oil.

in

Evaporation of

absorption of hydrogen gas salt solutions in the

in a

process for liquid

chemical industry

is

similar to

evaporation of sugar solutions in the food industry. Settling and sedimentation of

suspended solids

hydrocarbons

sewage and the mining industries are similar. Flow of liquid petroleum refinery and flow of milk in a dairy plant are carried out

in the

in the

in a similar fashion.

The

unit operations deal mainly with the transfer and change of energy and the and change of materials primarily by physical means but also by physicalchemical means. The important unit operations, which can be combined in various sequences in a process and which are covered in Part 2, of this text, are described next. transfer

1.1B

Classification of Unit Operations

1.

Fluid flow. This concerns the principles that determine the flow or transportation of

2.

any fluid from one point to another. Heat transfer. This unit operation deals with the principles that govern accumulation and transfer of heat and energy from one place to another.

1

3.

Evaporation. This

a special case of heat transfer, which deals with the evaporation

is

of a volatile solvent such as water from a nonvolatile solute such as salt or

any other

material in solution. 4.

Drying. In this operation volatile liquids, usually water, are removed from solid materials.

5.

Distillation.

This

an operation whereby components of a liquid mixture are

is

separated by boiling because of their differences in vapor pressure. 6.

Absorption. In this process a component

is

removed from a gas stream by treatment

with a liquid. 7.

Membrane membrane

8.

separation. This process involves the separation of a solute from a

by diffusion of

fluid

this solute

from a liquid or gas through a semipermeable

barrier to another fluid.

Liquid-liquid extraction. In this case a solute in a liquid solution

contacting with another liquid solvent which

is

is

removed by

relatively immiscible with the solu-

tion. 9.

Adsorption. In

this

and adsorbed by a

process a component of a gas or a liquid stream

10. Liquid-solid leaching.

dissolves out

removed

This involves treating a finely divided solid with a liquid that

and removes a solute contained

11. Crystallization.

is

solid adsorbent.

in the solid.

This concerns the removal of a solute such as a

salt

from a solution

by precipitating the solute from the solution. 12.

M echanical-physical separations.

These involve separation of solids, liquids, or gases by mechanical means, such as nitration, settling, and size reduction, which are often classified as separate unit operations.

Many

of these unit operations have certain fundamental

occurs

in

and basic principles or

common. For example, the mechanism of diffusion or mass transfer drying, membrane separation, absorption, distillation, and crystallization.

mechanisms

in

Heat transfer occurs

in

drying, distillation, evaporation,

more fundamental nature

following classification of a

is

and so on. Hence, the

often

made

into transfer or

transport processes.

1.1

1.

C

Fundamental Transport Processes

Momentum in

transfer.

This

moving media, such

concerned with the transfer of

is

momentum which

occurs

as in the unit operations of fluid flow, sedimentation,

and

mixing. 2.

Heat transfer. In this fundamental process, we are concerned with the transfer of heat from one place to another; it occurs in the unit operations heat transfer, drying, evaporation, distillation, and others.

3.

Mass

transfer.

Here mass

is

phase; the basic mechanism

This includes

1.1

distillation,

being transferred from one phase to another distinct is

the

tion, adsorption,

and leaching.

D

in

Parts 1 and 2

in

two parts:

Arrangement

This text

is

arranged

same whether

the phases are gas, solid, or liquid.

absorption, liquid-liquid extraction,

Part 1: Transport Processes:

Momentum, Heat, and Mass.

principles are covered extensively in Chapters

1

membrane

separa-

These fundamental

to 7 to provide the basis for study of unit

operations.

2

Chap.

I

Introduction to Engineering Principles and Units

The various

Part 2: Unit Operations. process areas are studied

unit operations and their applications to

Part 2 of this text.

in

There are a number of elementary engineering principles, mathematical techniques, and laws of physics and chemistry that are basic to a study of the principles of momentum, heat, and mass transfer and the unit operations. These are reviewed for the reader in this

Some

chapter.

first

readers, especially chemical engineers, agricultural

and chemists, may be familiar with many of these principles and techniques and may wish to omit all or parts of this chapter. Homework problems at the end of each chapter are arranged in different sections, engineers, civil engineers,

each corresponding to the number of a given section

in the chapter.

SI SYSTEM OF BASIC UNITS USED IN THIS TEXT AND OTHER SYSTEMS

1.2

There are three main systems of basic units employed at present in engineering and science. The first and most important of these is the SI (Systeme International d'Unites) system, which has as its three basic units the meter (m), the kilogram (kg), and the second (s).

The

others are the English foot (ft)-pound (lb)-second

centimeter (cm)-gram (g)-second

(s),

(s),

or fps, system and the

or cgs, system.

At present the SI system has been adopted officially for use exclusively in engineerand science, but the older English and cgs systems will still be used for some time. Much of the physical and chemical data and empirical equations are given in these latter

ing

two systems. Hence, the engineer should not only be proficient also be able to use the other two systems to a limited extent. 1.2A

The

in

the SI system but

must

SI System of Units basic quantities used in the SI system are as follows: the unit of length

is

the meter

mass is the kilogram (kg); the unit of temperature is the kelvin (K); and the unit of an element is the kilogram mole (kg mol). The other standard units are derived from these basic quantities. The basic unit of force is the newton (N), defined as

(m); the unit of time

is

the second

basic unit of work, energy, or heat 1

Power is measured

joule

(J)

in joules/s

=

unit of pressure

1

is

newton

(N m) •

=

thenewton/m or pascal

1

=

1

kg

m 2 /s 2

•

)

1

pascal (Pa)

not a standard SI unit but

is

The standard

watt (W)

(Pa).

newton/m 2 (N/m 2 =

acceleration of gravity 1

(^i)

m

joule/s (J/s)

[Pressure in atmospheres (atm)

(G)

kg-m/s 2

the newton-meter, or joule(J).

2

1

transition period.]

is

1

or watts (W). 1

The

the unit of

newton (N) =

1

The

(s);

g

=

9^80665 m/s

is

is

being used during the

defined as

2

A few of the standard prefixes for multiples of the basic units are as follows: giga = 10 9 mega (M) = 10 5 kilo (k) = 10 3 centi (c) = 10~ 2 milli (m) = 1(T 3 micro = 10~ 6 and nano (n) = 10~ 9 The prefix c is not a preferred prefix.

Sec. 1.2

,

,

,

,

,

,

.

SI System of Basic Units Used in This Text and Other Systems

3

Temperatures are defined in kelvin (K) as the preferred unit in the SI system. in practice, wide use is made of the degree Celsius (°C) scale, which is

However, defined by

= T(K) -

t°C

Note

=

that 1°C

K and

1

that in the case of temperature difference,

= AT K

At"C

The standard

preferred unit of time

of minutes (min), hours (h), or days

1.2B

CGS System

The cgs system

is

the second

is

(s),

but time can be

in

nondecimal units

(d).

of Units

related to the SI system as follows

=

1

g mass

1

cm =

1

dyne (dyn) erg

1

The standard

273.15

1

=

1

(g)

x 10" 3 kg mass

1

2

x 10~

=

1

m g-cm/s 2

=

dyn cm -

acceleration of gravity g

(kg)

1

=

x 10"

1

x 10~

7

joule

5

newton (N)

(J)

is

=

980.665 cm/s

2

English fps System of Units

1.2C

The English system is

'

related to the SI system as follows:

=

mass (lbj

1

lb

1

ft

0.30480

m

1

lb force (lb f )

=

1

ft

1

psia

=

lb f

•

1.8°F

g

=

= =

=

4.4482 newtons (N)

1.35582 newton

1

K=

In

(N m) •

=

1.35582 joules

(N/m 2

(J)

)

2

ft/s

factor for

Newton's law

=

A.l, convenient

is

32.174 ft-lb^lbfS

g c in SI units and cgs units

Appendix

m

1°C (centigrade or Celsius)

gc

The factor

•

2 6.89476 x 10 3 newton/m

32.174

The proportionality

0.45359 kg

is

1.0

and

is

2

omitted.

conversion factors for

all

three systems are tabulated.

Further discussions and use of these relationships are given

in

various sections of the

text.

This text uses the SI system as the primary set of units

in the equations,

sample

problems, and homework problems. However, the important equations derived in the text are given in a dual set of units, SI

4

Chap.

I

and English, when these equations

differ.

Introduction to Engineering Principles

Some

and Units

example problems and homework problems are also given using English units. In some cases, intermediate steps and/or answers in example problems are also stated in English units.

Dimensionally Homogeneous Equations

1.2D

and Consistent Units

A

dimensionally homogeneous equation

units.

is one in which all the terms have the same These units can be the base units or derived ones (for example, kg/s 2 or Pa). •

m

Such an equation can be used with any system of units provided that the same base or derived units are used throughout the equation. No conversion factors are needed when consistent units are used.

The reader should be careful in using any equation and always check it for dimenTo do this, a system of units (SI, English, etc.) is first selected. Then units are substituted for each term in the equation and like units in each term canceled

sional homogeneity.

out.

METHODS OF EXPRESSING TEMPERATURES AND COMPOSITIONS

1.3

13A

Temperature

in common use in the chemical and biological indusThese are degrees Fahrenheit (abbreviated °F) and Celsius (°C). It is often necessary to convert from one scale to the other. Both use the freezing point and boiling point of water at 1 atmosphere pressure as base points. Often temperatures are expressed as absolute degrees K (SI standard) or degrees Rankine (°R) instead of °C or °F. Table 1.3-1 shows the equivalences of the four temperature scales.

There are two temperature scales tries.

The

difference between the boiling point of water

100°C or 180°F. Thus, a 1.8°F

—

273.15"C

is

change

and melting point of ice

at

1

atm

is

equal to a 1°C change. Usually, the value of

is

rounded to -273.2°Cand -459.7°F to -460°F. The following equations

can be used to convert from

one scale

to another.

°F=

32

+

1

°C = - -

(1.3-1)

1.8(°C)

(°F

-

(1.3-2)

32)

1.5

°R

=

T+460

K = °C+

Table

1.3-1.

Boiling water

Melting

ice

Absolute zero

Sec. 1.3

(1.3-3)

273.15

(1.3-4)

Temperature Scales and Equivalents Centigrade

Fahrenheit

100°C

212°F 32°F -459.7°F

0°C -273.15°C

Kelvin

Rankine

Celsius

R

373.15

K

67

273.15

K

491.7°R

100°C o°c

0°R

-273.15°C

0

K

Methods of Expressing Temperatures and Compositions

1.7°

5

amounts of various gases may be compared, standard conditions of

In order that

STP

temperature and pressure (abbreviated

atm) abs and 273.15

(1.0

K (0°C). Under

volume of

1.0

volume of

volume of

SC) are arbitrarily defined as 101.325 kPa volumes are as follows:

or

these conditions the

kg mol (SC)

=

22.414

m

3

g mol (SC)

=

22.414

L

(liter)

=

22414 cm 3

=

359.05

1.0

mol (SC)

1.0 lb

3 ft

EXA MPLE 1.4-1.

Gas-Law Constant Calculate the value of the gas-law constant R when the pressure is in psia, moles in lb mol, volume in ft 3 and temperature in °R. Repeat for SI units. ,

At standard conditions, p = 14.7 psia, V = 359 ft 3 and 32 = 492°R (273.15 K). Substituting into Eq. (1.4-1)

Solution:

460

4-

=

n

,

_?V _ nT

A

3

P siaX359 ft ) lb mol)(492°R)

(14.7 (1.0

(

nT

also at conditions p 2

.

,

m

m

3 )

~

3

Pa

kg mol -K

(1.4-1) for n

moles of gas

at

conditions

Substituting into Eq. (1.4-1),

Pl K,

=nRT

V2

=nRT

x

2

gives

W PiVi

=

T,

(L4" 2)

T '2

Pi v 2

1.4C

-psia

lbmol-°R

from Eq.

V2 T2

,

p2

Combining

ft

kg molX273.15 K)

(1.0

useful relation can be obtained

Vu Tu and

3

_

x 1QS Pa)(22.414

_ L01325

R

p lt

for

mol and solving for R,

1.0 lb

R

T=

Gas Mixtures

Ideal

Dalton's law for mixtures of ideal gases states that the total pressure of a gas mixture

equal to the

sum

P = where P

is

A, B, C,

... in

Pa

and p A p B p c

total pressure

,

,

+ ,

+

Pb ...

fraction of a

component

represented

fraction in

(1-4-3)

are the partial pressures of the

components

is

is

proportional to

its

partial pressure, the

=

—" =

(1.4-4)

Pa

+

Pb

+

Pc

equal to the mole fraction.

+

•

•

Gas

mixtures, are almost always

terms of mole fractions and not weight fractions. For engineering pur-

poses, Dalton's law

is

a few atmospheres or

8

+

is

xA

The volume

pc

the mixture.

Since the number of moles of a component

mole

is

of the individual partial pressures:

sufficiently accurate to use for actual

mixtures at total pressures of

less.

Chap.

1

Introduction to Engineering Principles

and Units

EXAMPLE

1.4-2. Composition of a Gas Mixture gas mixture contains the following components and partial pressures: 75 Hg; 2 595 Hg; CO, 50 Hg; 2 26 Hg. 2 Calculate the total pressure and the composition in mole fraction.

A

C0

mm

mm

,

N

mm

,

P = Pa + The mole

+

Pb

fraction of

Pc

+

Pd

=

+

75

50

and 0.035,

CO,

N2

Vapor Pressure and Boiling Point of Liquids

When

a liquid

The

746

mm

Hg

,

and

02

are calculated as

placed in a sealed container, molecules of liquid will evaporate into the

space above the liquid and will exert

=

respectively.

1.4D

is

26

^ = — = 0.101

In like manner, the mole fractions of 0.067, 0.797,

+ 595 +

CO z is obtained by using Eq. (1.4-4). x,(C0 2 ) =

the liquid.

mm

,

Substituting into Eq. (1.4-3),

Solution:

vapor

0

fill

it

completely. After a time, equilibrium

we

a pressure just like a gas and

value of the vapor pressure

container as long as

some

is

is

call this

is

reached. This

pressure the vapor pressure of

independent of the amount of liquid

in the

present.

an inert gas such as air is also present in the vapor space, it will have very little on the vapor pressure. In general, the effect of total pressure on vapor pressure can

If effect

be considered as negligible for pressures of a few atmospheres or

The vapor

less.

pressure of a liquid increases markedly with temperature. For example,

mm

from Appendix A.2 for water, the vapor pressure at 50°C is 12.333 kPa (92.51 Hg). At 100°C the vapor pressure has increased greatly to 101.325 kPa (760 mm Hg). The boiling point of a liquid is defined as the temperature at which the vapor pressure of a liquid equals the total pressure. Hence,

760

mm

pressure

A line is

Hg, water is

considerably

if the atmospheric total pressure is top of a high mountain, where the total boil at temperatures below 100°C.

boil at 100°C.

will

water

less,

plot of vapor pressure

PA

will

On

of a liquid versus temperature does not yield a straight

but a curve. However, for moderate temperature ranges, a plot of log

PA

versus 1/T

a reasonably straight line, as follows.

(1-4-5)

where

1.5

1.5A

One

m

is

the slope, b

a constant for the liquid A,

is

and

T is the temperature in

K.

CONSERVATION OF MASS AND MATERIAL BALANCES Conservation of

Mass

of the basic laws of physical science

is

the law of conservation of mass. This law,

mass cannot be created or destroyed (excluding, of course, nuclear or atomic reactions). Hence, the total mass (or weight) of all materials entering any process must equal the total mass of all materials leaving plus the mass of any

stated simply, says that

materials accumulating or

left

in the process.

input

Sec. 1.5

=

output

4-

accumulation

Conservation of Mass and Material Balances

(1-5-1)

9

In the' majority of cases there will be

no accumulation of materials

in a process,

and

then the input will simply equal the output. Stated in other words, "what goes in must

We call

come out."

this

type of process a steady-state process.

=

input

1.5B

output (steady state)

(1-5-2)

Simple Material Balances

In this section

we do simple

material (weight or mass) balances in various processes at

We can use units of kg, lb m lb mol, cautioned to be consistent and not to mix

steady state with no chemical reaction occurring. g,

kg mol,

etc., in

our balances. The reader

several units in a balance.

Section

When chemical

one should use kg mol

1.5D),

To

3.6, differential

,

reactions occur in the balances (as discussed in

chemical equations relate moles

units, since

mass balances

reacting. In Section 2.6, overall

Section

is

be covered in more detail and in

will

mass balances.

solve a material-balance problem

it is

advisable to proceed by a series of definite

steps, as listed below. 1.

Sketch a simple diagram of the process. This can be a simple box diagram showing each stream entering by an arrow pointing in and each stream leaving by an arrow pointing out. Include on each arrow the compositions, amounts, temperatures, and so on, of that stream. All pertinent data should be

2.

Write the chemical equations involved

3.

Select a basis for calculation. In

amount 4.

of

one of the streams

on

this

diagram.

(if any).

most cases the problem

in the process,

which

is

is

concerned with a

specific

selected as the basis.

Make a material balance. The arrows into the process will be input items and the arrows going out output items. The balance can be a total material balance in Eq. (1.5-2) or a balance on each component present (if no chemical reaction occurs). Typical processes that do not undergo chemical reactions are drying, evaporation,

dilution of solutions, distillation, extraction,

material balances containing

EXAMPLE 1.5-1.

and so on. These can be solved by setting up these equations for the unknowns.

unknowns and solving

Concentration of Orange Juice

In the concentration of orange juice a fresh extracted

and strained

juice

%

containing 7.08 wt solids is fed to a vacuum evaporator. In the evaporator, water is removed and the solids content increased to 58 wt % solids. For 1000 kg/h entering, calculate the amounts of the outlet streams of concentrated juice and water. Solution:

diagram

Following the four steps outlined, we make a process flow

(step

1) in

Fig. 1.5-1.

Note

that the letter

W kg/h

W represents the unknown

water

1000 kg/h juice evaporator

7.08%

solids

C

kg/h concentrated juice

58% Figure

10

1.5-1.

solids

Process flow diagram for Example 15-1.

Chap.

1

Introduction to Engineering Principles

and Units

amount of water and C reactions are given (step

amount

the

2).

To make the material made using Eq. (1.5-2).

of concentrated juice.

Basis: 10O0 kg/h entering juice (step

balances (step

.

=

1000

No

chemical

3).

a total material balance will be

4),

W+C

(1 .5-3)

This gives one equation and two unknowns. Hence, a on solids will be made.

component balance

'°HwH°> +c(l>) To

two equations, we solve Eq.

solve these

out.

We get C =

(1.5-4) first for

W drops

C since

122.1 kg/h concentrated juice.

C into

Substituting the value of

1000

Eq.

(1.5-3),

=^+122.1

W

= 877.9 kg/h water". and we obtain As a check on our calculations, we can write a balance on the water component.

Solving,

=

929.2

In

Example

processes

1.5-1

in series

877.9

+

51.3

=

929.2

only one unit or separate process was involved. Often, a

are involved.

Then we have a choice

number of

of making a separate balance over

each separate process and/or a balance around the complete overall process.

1.5C

Material Balances and Recycle

Processes that have a recycle or feedback of part of the product into the entering feed are

sometimes encountered. For example, sludge from a sedimentation tank treated. In

in a

sewage treatment plant, part of the activated

recycled back to the aeration tank where the liquid

is

some food-drying operations, the humidity

recirculating part of the hot

wet

of the entering air

air that leaves the dryer.

is

is

controlled by

In chemical reactions, the

material that did not react in the reactor can be separated from the final product and fed

back to the reactor.

EXAMPLE

1.5-2.

Crystallization

In a process producing

KN0

ofKN0 3 and Recycle

1000 kg/h of a feed solution containing 20 wt % an evaporator, which evaporates some water at 3 is fed to 422 K to produce a 50 wt % 3 solution. This is then fed to a crystallizer at 31 1 K, where crystals containing 96 wt % KNOj are removed. The saturated solution containing 37.5 wt 3 is recycled to the evaporator. Calculate the amount of recycle stream R in kg/h and the product stream of crystals Pin kg/h. 3 salt,

KN0

KN0

% KN0

Solution:

Figure

1

.5-2 gives the

use 1000 kg/h of fresh feed.

Sec. 1.5

No

process flow diagram. As a basis chemical reactions are occurring.

Conservation of Mass and Material Balances

we

shall

We

can

11

W kg/h

water,

feed, 1000 kg/h

evaporator

20% KN0 3

422 K

'

5 kg/h

crystallizer

311

50% KNO3

K

R kg/h 37.5% KNO3 recycle,

Figure

make an

crystals,

4%H 2 0

Process flow diagram for Example

1.5-2.

overall balance

on

P kg/h

the entire process for

KN0

1.5-2.

and solve

3

F

for

directly.

= W(0) +

1000(0.20)

F =

F(0.96)

(1.5-6)

208.3 kg crystals/h

we can make a balance around the Using a balance on the crystallizer since it now includes only two unknowns, S and R, we get for a total balance,

To

calculate the recycle stream,

evaporator or the

crystallizer.

= R +

S

For a

KNO3

208.3

(1.5-7)

balance on the crystallizer, S(0.50)

=

F(0.375)

+

208.3(0.96)

Substituting S from Eq. (1.5-7) into Eq. (1.5-8) recycle/h

1.5D In

and S

=

(1.5-8)

R =

and solving,

766.6 kg

974.9 kg/h.

Material Balances and Chemical Reaction

many

cases the materials entering a process

undergo chemical reactions

in the process

so that the materials leaving are different from those entering. In these cases

convenient as

N

kg mol

to

H

2

make

it

is

usually

molar and not a weight balance on an individual component such + or kg atom H, kg mol COJ ion, kg mol CaC0 3 kg atom Na kg mol a

,

,

and so on. For example, in 2 on kg mol of H 2 C, 0 2 or N 2 ,

EXAMPLE

1.5-3.

combustion of CH 4 with

air,

balances can be

made

-

,

,

the

Combustion of Fuel Gas

%

C0

H 2 27.2% CO, 5.6% 0.5% 0 2 and gas containing 3.1 mol 2 63.6% 2 is burned with 20% excess air (i.e., the air over and above that necessary for complete combustion to 2 and 2 0). The combustion of is only 98% complete. For 100 kg mol of fuel gas, calculate the moles of A

fuel

,

,

,

N

H

C0

CO

each component Solution:

12

in the exit flue gas.

First, the

process flow diagram

Chap.

1

is

drawn

(Fig.

1.5-3).

On

Introduction to Engineering Principles

the

and Units

A kg mol

air

F kg mol H20 CO

burner

100 kg mol fuel gas

3.1% H 2 27.2% CO 5.6% C0 2 0.5% 0 2 63.6% N 2

flue gas

co 2 02 N2

100.0 Figure

Process flow diagram for Example

1.5-3.

1.5-3.

diagram the components in the flue gas are shown. Let A be moles of and F be moles of flue gas. Next the chemical reactions are given.

CO + ^0 2 ^C0 2 H + K> -»H 2 0 2

An accounting

0

mol

in fuel

2

all

the

H

for

=

(1.5-10)

0

in

2

the fuel gas

+ 5.6(C0 2 +

(|)27.2(C0)

is

=

0.5(O 2 )

)

as follows:

02

mol

19.7

burned to H 2 0, we need from Eq. (1.5-10) mol H 2 or 3.1(^-) = 1.55 total mol 0 2 For completely burning the CO from Eq. (1.5-9), we need 27.2(j) = 13.6 mol 0 2 Hence, the amount of 0 2 we must add is, theoretically, as follows:

For

\ mol

0

gas

^(1.5-9)

2

of the total moles of

air

2

2

to be completely 1

.

.

0

mol

2

theoretically needed

=

1.55

=

14.65

+

—

13.6

mol

0

0.5 (in fuel gas)

2

we add 1.2(14.65), or 17.58 mol 0 2 Since air contains 79 amount of N 2 added is (79/2 1)(1 7.58), or 66.1 mol N 2 To calculate the moles in the final flue gas, all theH 2 gives 2 0, or 3.1 mol H 2 0. For CO, 2.0% does not react. Hence, 0.02(27.2), or 0.54, mol CO

20%

For mol

a

will

be unburned.

%N

A mol C. 32.8

-

2

excess,

.

the

,

.

carbon balance

total

is

as follows: inlet moles

In the outlet flue gas, 0.54 0.54, or 32.26,

mol as

C0

2

in

=

will

0

we make an

be as

=

C=

27.2

and

the

remainder of

02

balance.

+

5.6

32.8

.

For calculating the outlet mol

O,

CO

mol

19.7 (in fuel gas)

2

+

,

17.58 (in air)

overall

=

0

37.28 mol

2

0 For the N Equating inlet 0 to outlet, the free remaining 0 = 3.2 mol 0 0

2

out

=

(3.1/2) (in

H 2 0) +

CO) +

(0.54/2) (in

=

63.6 (in fuel gas)

outlet flue gas contains 3.10

mol

O z ,and

129.7

mol

N2

mol

for

it

amount

1

.5

is

the

2

2

66.1 (in air), or 129.70

amount of

free

.

2

2

molN 2 The

0, 0.54 mol CO, 32.26 mol

is

.

C0

reactants, the limiting reactant

present in an

amount

to react stoichiometrically with the other reactants.

a reaction

Sec.

compound which

H

+

2)

2

,

3.20

.

In chemical reactions with several

defined as that

C0 +

2

2

balance, the outlet

32.26 (in

this limiting

less

component

is

than the amount necessary

Then

the percent completion of

reactant actually converted, divided by the

originally present, times 100.

Conservation of Mass and Material Balances

13

ENERGY AND HEAT UNITS

1.6

Joule, Calorie, and Btu

1.6A

manner similar to that used in making material balances on chemical and biological we can also make energy balances on a process. Often a large portion of the

In a

processes,

energy entering or leaving a system balances are made,

is

In the SI system energy

form of heat. Before such energy or heat

in the

we must understand

the various types of energy

given in joules

is

expressed in btu (British thermal unit) or cal (calorie). defined as the Also,

amount of

kcal (kilocalorie)

1

raise 1.0 lb

and heat units. (kJ). Energy is

or kilojoules

(J)

The g

also

calorie (abbreviated, cal)

is

heat needed to heat 1.0 g water 1.0°C (from 14.5°C to 15.5°C).

=

1000

The

cal.

btu

defined as the

is

amount of heat needed

to

water 1°F. Hence, from Appendix A.l, btu

1

=

=

252.16 cal

is

defined as the

1.05506 kJ

(1.6-1)

Heat Capacity

1.6B

The heat capacity

of a substance

the temperature by

amount

can be expressed for

of heat necessary to increase

lb, 1 g mol, 1 kg mol, or mol of the substance. For example, a heat capacity is expressed in SI units as J/kg mol K; in other units as cal/g °C. cal/g mol °C, kcal/kg mol °C, btu/lb m °F, or 1

degree.

1

It

g,

1

1

lb

-

•

•

•

•

btu/lbmol-°F. It

can be shown that the actual numerical value of a heat capacity in molar units. That is,

is

the

same

in

mass units or

1.0

cal/g

1.0 cal/g

For example, to prove btu/lb m

°F.

•

453.6 g for

1

lb m

,

mol-°C =

1.0 btu/lb m

1.0 btu/lb

•

°F

(1.6-2)

mol-°F

(1.6-3)

suppose that a substance has a heat capacity of 0.8 made using 1.8 = F for 1°C or 1 K, 252.16 cal for 1 btu, and

this,

The conversion

°C =

is

as follows

heat capacity

(-—]

=

\&-°Cj

btU (0.8

)(

lb m

V

252.16

-°FA

—Y

Y

1.8

btuA453.6 g/lbJV

— °Cj

cal

= The

0.8

g°C

heat capacities of gases (sometimes called specific heat) at constant pressure c p

are functions of temperature

and

for

engineering purposes can be assumed to be indepen-

dent of pressure up to several atmospheres. In most process engineering calculations,

one

is

usually interested in the

amount

of heat needed to heat a gas from one temperature

temperature, an integration must be 2 p performed or a suitable mean c pm used. These mean values for gases have been obtained for T, of 298 K or 25°C (77°F) and various T2 values, and are tabulated in Table 1.6-1 at t,

to

another

101.325

kPa

at

r

Since the

.

pressure or

c

less as c pm in

varies with

kJ/kg mol

EXAMPLE 1.6-1. Heating o/N Gas N 2 at atm pressure absolute is

•

K at various

values of T2 in

K or °C.

2

The gas

Calculate the

14

1

amount

needed

of heat

Chap.

1

being heated in a heat exchanger.

in J

to

heat 3.0 g mol

N

2

in

Introduction to Engineering Principles

the

and Units

Table

Mean Molar Heat

1.6-1.

101.325

at

T{K)

T(°C)

298 373

25 100

473 573

If H

kPa

Capacities of Gases Between 298 and

or Less (SI Units: c

p

ri U

Air

"2

2

=

ft f\ tt

2

28.86 28.99

29.14 29.19

29.16'

29.24

29.19 29.29

29.38 29.66

200

29.13

29.29

29.38

29.40

300

29.18

29.46

29.60

29.61

673

400

29.23

29.68

29.88

29.94

773

500

29.29

29.97

30.19

873

600

29.35

30.27

30.52

973

700

29.44

30.56

1073

800

29.56

173

900

1273

CU r*r\

37.20 38.73

30.07

34.24

30.53

34.39

31.01

30.25

30.56

30.84

30.87

30.85

31.16

29.63

31.16

1000

29.84

1473

1200

1673

1400

35.8 37.6

39.9 41.2

40.62

40.3

42.9

42.32

43.1

44.5

35.21

43.80

45.9

45.8

3 1.46

35.75

45.12

48.8

47.0

31.89

36.33

46.28

51.4

47.9

32.26

36.91

47.32

54.0

48.8

31.18

32.62

37.53

48.27

56.4

49.6

31.49

31.48

32.97

38.14

49.15

58.8

50.3

31.43

31.77

31.79

33.25

38.71

49.91

61.0

50.9

30.18

31.97

32.30

32.32

33.78

39.88

51.29

64.9.

51.9

30.51

32.40

32.73

32.76

34.19

40.90

52.34

Mean Molar Heat (English Units: c p

Capacities of Gases Between 25 and

=

btu/lb

Air

02

NO

6.972

6.965 6.983

6.972 6.996

7.017 7.083

6.957

6.996

7.017

7.021

6.970

7.036

7.07O

7.073

6.982

7.089

7.136

500

6.995

7.159

600

7.011

7.229

N

6.894 6.924

200 300

400

25 100

2

6.961

T°C

at I

atm Pressure or Less

CH,

S0 2 C 2 H, S0 3 C 2 H 6

mol °F)

CO

H2

T(°C)

S0 2

2

33.59 33.85

1

TK (25 and T°C)

kJ/kg mol K)

H20

C0 2

HCl Cl 2

7.134 7.144

8.024 8.084

8.884 9.251

6.96 6.97

8.12 8.24

7.181

7.224

8.177

9.701

6.98

7.293

7.252

8.215

10.108

7.00

7.152

7.406

7.301

8.409

10.462

7.210

7.225

7.515

7.389

8.539

7.289

7.299

7.616

7.470

8.55 8.98

9.54 9.85

10.45 11.35

12.84

12.63 13.76

8.37

9.62

10.25

12.53

13.74

15.27

8.48

10.29

10.62

13.65

14.54

16.72

7.02

8.55

10.97

10.94

14.67

15.22

18.11

10.776

7.06

8.61

11.65

11.22

15.60

15.82

19.39

8.678

11.053

7.10

8.66

12.27

11.45

16.45

16.33

20.58

12.11

700

7.032

7.298

7.365

7.374

7.706

7.549

8.816

11.303

7.15

8.70

12.90

11.66

17.22

16.77

21.68

SOO

7.060

7.369

7.443

7.447

7.792

7.630

8.963

11.53

7.21

8.73

13.48

11.84

17.95

17.17

22.72

900

7.076

7.443

7.521

7.520

7.874

7.708

9.109

11.74

7.27

8.77

14.04

12.01

18.63

17.52

23.69

1000

7.128

7.507

7.587

7.593

7.941

7.773

9.246

11.92

7.33

8.80

14.56

12.15

19.23

17.86

24.56

100

7.169

7.574

7.653

7.660

8.009

7.839

9.389

12.10

7.39

8.82

15.04

12.28

19.81

18.17

25.40

1200

7.209

7.635

7.714

7.719

8.068

7.898

9.524

12.25

7.45

8.94

15.49

12.39

20.33

18.44

26.15

1300

7.252

7.692

7.772

7.778

8.123

7.952

9.66

12.39

1

1400

7^288

7.738

7.818

7.824

8.166

7.994

9.77

12.50

1500

7.326

7.786

7.866

7.873

8.203

8.039

9.89

12.69

1600

7.386

7.844

7.922

7.929

8.269

8.092

9.95

12.75

1700

7.421

7.879

7.958

7.965

8.305

8.124

10.13

12.70

1800

7.467

7.924

8.001

8.010

8.349

8.164

10.24

12.94

1900

7.505

7.957

8.033

8.043

8.383

8.192

10.34

13.01

2000

7.548

7.994

8.069

8.081

8.423

8.225

10.43

13.10

2100

7.588

8.028

8.101

8.115

8.460

8.255

10.52

13.17

2200

7.624

8.054

8.127

8.144

8.491

8.277

10.61

13.24

Source : O. A. Hougen, K.. W. Walson, and R. A. Ragatz, Chemical Process Principles, Part John Wiley & Sons, Inc., 1954. Wilh permission.

I,

2nd ed.

New

York:

following temperature ranges: (a) (b) (c)

Sec. 1 .6

298-673 K(25-400°C) 298-1123 K(25-850°C) 673-1123 K (40O-850°C)

Energy and Heal Units

15

For case (a), Table 1.6-1 gives c pm values at 1 atm pressure or less and can be used up to several atm pressures. For N 2 at 673 K, c pm = 29.68

Solution:

kJ/kg mol-K or 29.68 J/g mol-K.. This range 298-673 K. heat required

known

Substituting the

M

=

is

the

mean

^

mol (xpm -

g

heat capacity for the

(T;

-

TJK.

(1.6-4)

values,

heat required

= (3.0X29.68X673 -

=

298)

33 390

J

For case (b), the c pm at 1123 K (obtained by linear interpolation between 1073 and 1 173 K) is 31.00 J/g mol K. •

heat required

=

3.0(31.00X1123

-

=

298)

76 725

J

For case (c), there is no mean heat capacity for the interval 673-1 123 K. However, we can use the heat required to heat the gas from 298 to 673 K in case (a) and subtract it from case (b), which includes the heat to go from 298 to 673 K plus 673 to 1123 K. heat required (673-1123

K)

=

heat required (298-1123 K)

-

heat required (298-673)

(1.6-5)

Substituting the proper values into Eq. (1.6-5),

On

= 76 725 -

33 390

=

heating a gas mixture, the total heat required

is

heat required

43 335

J

determined by

component and then adding

the heat required for each individual

first

calculating

the results to obtain

the total.

heat capacities of solids and liquids are also functions of temperature and

The

independent of pressure. Data are given

in

Appendix

A.2, Physical Properties of Water;

Compounds; and A. 4, Physical More data are available in (PI).

A. 3, Physical Properties of Inorganic and Organic

Properties of Foods and Biological Materials.

EXA MPLE 1.6-2.

Heating of Milk Rich cows' milk (4536 kg/h) at 4.4°C is being heated 54.4°C by hot water. How much heat is needed?

Solution:

From Appendix A.4

3.85 kJ/kg

is

K. Temperature

=

heat required

The

heat exchanger to

in a

the average heat capacity of rich cows' milk

rise,

AT =

(4536 kg/hX3.85 kJ/kg-

(54.4

-

4.4)°C

KX 1/3600

=

50 K.

h/sX50 K)

enthalpy, H, of a substance in J/kg represents the

sum

=

242.5

kW

of the internal energy

For no reaction and a constant-pressure process with a temperature, the heat change as computed from Eq. (1.6-4) is the difference in

plus the pressure-volume term.

change

in

enthalpy, units,

1.6C

H=

AH, of

the substance relative to a given

temperature or base point. In other

btu/lb m or cal/g.

Latent Heat and Steam Tables

Whenever

a substance undergoes a change of phase, relatively large

changes are involved

at a

pressure can absorb 6013.4 kJ/kg mol. This enthalpy change fusion.

16

Data

for other

amounts of heat 1 atm

constant temperature. For example, ice at 0°C and

compounds

are available in various

Chap.

I

is

called the latent heat of

handbooks

(PI, Wl).

Introduction to Engineering Principles and Units

When a

liquid phase vaporizes to a

amount

temperature, an

vapor phase under

vapor pressure

its

at

constant

of heat called the latent heat of vaporization must be added.

Tabulations oflatent heats of vaporization are given in various handbooks. For water

25°C and 760

mm

mm Hg, the latent heat

a pressure of 23.75

Hg, 44045 kJ/kg mol. Hence, the

engineering calculations. However, there

is

on

heat of water. Also, the effect of pressure

44020 kJ/kg mol, and

is

at

at

25°C and

of pressure can be neglected in

effect

a large effect of temperature on the latent the heat capacity of liquid water

is

small and

can be neglected. Since water

been compiled

in

a very

is

common

chemical, the thermodynamic properties of

steam tables and are given

Appendix A.2

in

in SI

and

in

it

have

English units.

EXA MPLE 1.6-3.

Use of Steam Tables Find the enthalpy change (i.e., how much heat must be added) for each of the following cases using SI and English units. (a) Heating 1 kg (lbj water from 21.11°C (70°F) to 60°C (140°F) at 101.325 kPa (1 atm) pressure. (b) Heating 1 kg {lbj water from 21.11°C (70°F) to 115.6°C (240°F) (c)

and vaporizing at 1 72.2 kPa (24.97 psia). Vaporizing 1 kg (lb J water at 1 15.6°C (240°F) and

172.2

kPa

(24.97

psia).

For part (a), the effect of pressure on the enthalpy From Appendix A.2,

Solution: is

of liquid water

negligible.

Hat

88.60 kJ/kg

or

at

70° F

=

38.09 btu/lb m

60°C = 251.13 kJ/kg

or

at

140°F

=

107.96 btu/lb m

21.1

H

at

1°C=

change

In part

(b),

the saturated

in

H = AH =

251.13

-

88.60

=

162.53 kJ/kg

=

107.96

-

38.09

=

69.87 btu/lb m

the enthalpy at 115.6°C (240°F)

vapor

change

2699.9 kJ/kg or

is

H = AH =

in

= The

water

at

2699.9

-

latent heat of

1

160.7

-

1

and 172.2 kPa (24.97

160.7 btu/lb m

2699.9

-

88.60

=

261

160.7

-

38.09

=

1122.6 btu/lb m

1

1 1

5.6°C (240° F)

psia) of

.

1.3

in part (c)

484.9

=

2215.0 kJ/kg

208.44

=

952.26 btu/lb m

kJ/kg

is

1.6D

Heat of Reaction

When

chemical reactions occur, heat effects always accompany these reactions. This

area where energy changes occur

HC1

is

neutralized with

absorbed

in

NaOH,

is

heat

often called thermochemistry. For example, is

given off and the reaction

an endothermic reaction. This heat of reaction

is

is

when

exothermic. Heat

is

dependent on the chemical

nature of each reacting material and product and on their physical states.

For purposes of organizing data we define a standard heat of reaction change

Sec.

1 .6

in

enthalpy when

1

kg mol

Energy and Heat Units

reacts

under

a

AH 0

as the

pressure of 101.325 kPa at a temper-

17

K (25°C). For example, for the reaction

ature of 298

H2 the

AH 0

is

+

(ff)

\QM- H

2 O(0

(1.6-6)

-285.840 x 10 3 kJ/kg mol or -68.317 kcal/g mol. The reaction

mic and the value reacts with the

Special

02

is

exother-

H2

gas

When

the

negative since the reaction loses enthalpy. In this case, the

is

gas to give liquid water,

names are given

AH 0

to

298

at

all

K (25°C).

depending upon the type of reaction.

0 formed from the elements, as in Eq. (1.6-6), we call theAH heat offormation of the product water, AH° For the combustion of CH„. toformC0 2 andH 2 0, we call it 0 0 heat of combustion, AH Data are given in Appendix A.3 for various values of AH

product

is

,

.

.

.

EXAMPLE 1.6-4. A

Combustion of Carbon g mol of carbon graphite is burned in a calorimeter held at and 1 atm. The combustion is incomplete and 90% of the C goes to and 10% to CO. What is the total enthalpy change in kJ and kcal?

total of 10.0

298

K

C0 2

From Appendix A.3

Solution:

A/f° for carbon going

the

— 393.513 x 10 kJ/kg mol or —94.0518 kcal/g mol, and 3 to CO is - 110.523 x 10 kJ/kg mol or -26.4157 kcal/g 3

C0 2

and

1

AH =

9(- 393.5 13) + 1(-

=

9(

-94.05

of the reaction,

AH

,

C0 2

is

mol. Since 9 mol

-

,

1

= -3652

10.523)

=

1(- 26.41 57)

compounds

of

kJ

-872.9 kcal

is

available, the standard heat

can be calculated by

AH° = In

+

18)

AH 0

a table of heats of formation, 0

to

carbon going

CO are formed,

mol

total

If

for

Appencix A.3, a short

£ AH°

table of

(producls)

- I AH? (reactants)

some values of AHf

is

(1.6-7)

given. Other data are also

available (HI, PI, SI).

EXAMPLE 1.6-5.

Reaction of Methane

For the following reaction of

1

kg mol of CH 4

CUM + H

2

0(Q-

calculate the standard heat of reaction

Solution:

From Appendix

are obtained at 298

at 101.32

+ 3H

CO(g)

AH

0

at

298

A.3, the following

H AH°

that the

standard heats of formation

of

-

110.523 x 10

3

0

2 (9)

elements

all

rnol)

is,

by definition, zero. Substituting into

(1.6-7),

AH 0 = [ -

1

10.523 x 10

= +250.165 18

in kJ.

-74.848 x 10 3 -285.840 x 10 3

H 2 O(0 CO{g)

Eq.

2 (g)

K:

AH° [U/kg

Note

K

kPa and 298 K,

x 10

3

3

-

3(0)]

- -

kJ/kg mol

Chap.

I

(

74.848 x 10

3

-

285.840 x 10

3 )

(endothermic)

Introduction to Engineering Principles and Units

CONSERVATION OF ENERGY AND HEAT BALANCES

1.7

Conservation of Energy

1.7A

making material balances we used the law of conservation of mass, which states that mass entering is equal to the mass leaving plus the mass left in the process. In a similar manner, we can state the law of conservation of energy, which says that all energy In

the

entering a process

is

equal to that leaving plus that

elementary heat balances

will

sidered in Sections 2.7 and

5.6.

Energy can appear electrical energy,

energy, work, In

many

many

in

chemical energy

and heat

More

be made.

Some

forms. (in

terms of

of the

AH

in the process. In this section

common

will

be con-

forms are enthalpy,

reaction), kinetic energy, potential

inflow.

cases in process engineering,

which often takes place

be neglected. Then only the enthalpy of the materials

chemical reaction energy

(AH 0

)

(at

constant pressure), the standard

and the heat added or removed must be taken

at 25°C,

into account in the energy balance. This

at constant pressure,

and work either are not present or can

electrical energy, kinetic energy, potential energy,

generally called a heat balance.

is

Heat Balances

1.7B In

left

elaborate energy balances

making a

we use methods similar to those used in making The energy or heat coming into a process in the inlet materials plus

heat balance at steady state

a material balance.

any net energy added

to the process

equal to the energy leaving

is

in

the materials.

Expressed mathematically,

EH where

£ HR

is

sum

the

temperature

reaction at 298

negative of

Hp = sum 298

K

E« p

(1-7-1)

materials entering the reaction process relative

standard heat of reaction

K

is

taken

to

be positive input heat for an exothermic reaction, q

system.

to the

of enthalpies of

all

If

=

net

heat leaves the system, this item will be negative.

leaving materials referred to the standard reference state

(25°C).

Note negative.

=

at 298 K and 101.32 kPa. If the above 298 K, this sum will be positive. A/-/°9 8 = standard heat of the and 101.32 kPa. The reaction contributes heat to the process, so the

energy or heat added Y,

all

4

is

AH° 98

at

+(-Ar/° 95 ) +

of enthalpies of

to the reference state for the inlet

R

that

if

the materials

coming

Care must be taken not

into a process are

below 298 K,

will

to confuse the signs of the items in Eq. (1.7-1). If

be no

is occurring. Use For convenience it is common Eq. (1.7-1) input items, and those on the

chemical reaction occurs, then simple heating, cooling, or phase change of Eq. (1-7-1) will be illustrated by several examples. practice to call the terms right,

on

the left-hand side of

output items.

EXAMPLE A

1.7-1.

Heating of Fermentation

medium

Medium

30°C is pumped at a rate of 2000 kg/h through a heater, where it is heated to 70°C under pressure. The waste heat water used to heat this medium enters at 95°C and leaves at 85°C. The average heat capacity of the fermentation medium is 4.06 kJ/kg K, and that for water is 4.21 kJ/kg K (Appendix A. 2). The fermentation stream and the wastewater stream are separated by a metal surface through which heat is transferred and do not physically mix with each other. Make a complete heat, balance on the system. Calculate the water flow and the amount of heat added to the fermentation medium assuming no heat losses. The liquid fermentation

process flow

Sec. 1.7

is

given

at

in Fig. 1.7-1.

Conservation of Energy and Heal Balances

19

q heat added

2000 kg/h

2000 kg/h

liquid

70°C

30°C

Wkg/h

W kg/h

85°C

95°C

Figure

Solution:

It

is

water

Process flow diagram for Example

1.7-1.

£ Hr

(25°C) (note that At

From

°f tne enthalpies of the two streams relative to 298 30 - 25°C = 5°C = 5 K):

=

(2000 kg/h)(4.06 kJ/kg- K)(5 K)

=

4.060 x 10

=

W(4.21)(95

8)

=

0

(since there

q

=

0

(there are

H(water)

— AH%

4

kJ/h

-

//(liquid)

=

2000(4.06X70

//(water)

=

1^(4.21X85

Equating input

to

4.060 x 10*

The amount

output

+

25)

= is

2.947 x 10

W

2

(W =

kJ/h

kg/h)

no chemical reaction)

no heat losses or additions)

-

-

25)

25)

=

=

3.65

x

10

2

2

2.526 x 10

Eq. (1.7-1) and solving for

5

kJ/h

W kJ/h

W,

W

=

3.654 x 10

W

=

7720 kg/h water flow

5

+

2.526 x 10

added to the fermentation medium and inlet liquid enthalpies.

of heat

//(outlet liquid)

this

in

2.947 x 10

difference of the outlet

in

K

£ Hp of the two streams relative to 298 K (25°C):

Output items.

Note

K

Eq. (1.7-1) the

=

H(liquid)

(

1.7-1.

convenient to use the standard reference state of 298

(25°C) as the datum to calculate the various enthalpies. input items are as follows.

Input items.

liquid

-

//(inlet liquid)

is

W

2

simply the

=

3.654 x 10

s

-

=

3.248 x 10

s

kJ/h (90.25

4.060 x 10

4

kW)

example that since the heat capacities were assumed

constant, a simpler balance could have been written as follows:

heat gained by liquid 20OO(4.06)(70

Then, solving,

-

30)

W = 7720 kg/h. This

constant. However,

when

=

heat lost by water

=

W(A.7\\95

-

85)

simple balance works well when c p

is

and the material is (25°C) and t K and the

the c p varies with temperature

a gas, c pm values are only available between 298 K simple method cannot be used without obtaining

new

c pm

values over

different temperature ranges.

20

Chap.

1

Introduction to Engineering Principles and Units

EXAMPLE

1.7-2 Heat and Material Balance in Combustion The waste gas from a process of 1000 g mol/h of CO at 473 K is burned at 1 atm pressure in a furnace using air at 373 K. The combustion is complete and 90% excess air is used. The flue gas leaves the furnace at 1273 K.

Calculate the heat removed in the furnace. First the process flow

Solution:

material balance

is

diagram

drawn

is

and then a

in Fig. 1.7-2

made.

CO(g)+}0 Atf° 98

=

2(

5 )->C0 2 (g)

-282.989 x 10 3 kJ/kg mol (from Appendix A.3)

CO =

mol

= mol

0

2

theoretically required

0

mol

mol

02

= i(1.00) =

2

added

=

0.950

~

air

added

=

0.950

+

gas

in outlet flue

2

0.500 kg mol/h

0.500(1.9)

N

C0

moles

kg mol/h

=

0.950

-

=

0.950 kg mol/h

3

3.570

= added —

=

=

-

570 k § mol fa

=

4.520 kg mol/h

=

0.450 kg mol/h

=A

used 0.500

C0

2

in outlet flue

gas

=

1.00

N

2

in outlet flue

gas

=

3.570 kg mol/h

For the heat balance Eq.

1.00

added

actually

2

=

1000 g mol/h

kg mol/h

relative to the

standard state

at

298 K, we follow

(1.7-1).

Input items

H

(CO) =

(The c pm of from Table

H

1.0O(c

CO

p J(473

-

298)

=

of 29.38 kJ/kg mol

•

1.00(29.38X473

K

-

298)

=

between 298 and 473

5142 kJ/h

K

is

obtained

1.6-1.)

-

(air)

=

4.520(c pm )(373

q

=

heat added, kJ/h

298)

=

4.520(29.29X373

-

298)

=

9929 kJ/h

lOOOg mol/h CO flue gas

473 K

A

g

mol/h

furnace

1273 K

air

373 K heat

removed Figure 17-2.

Sec. 1.7

(- q)

Process/low diagram for Example

Conservation of Energy and Heat Balances

1

.7-2.

21

(This will give a negative value here, indicating that heat was removed.)

-&H°29S =

-(-282.989 x 10 3 kJ/kg molXl.OO kg mol/h)

=

282990 kJ/h

00(^1273 -

=

48 660 kJ/h

Output items

H(C0 2 =

1.

)

298)

=

1.00(49.91X1273

-

298)

H(0 2 =

0.450(c pm X1273

-

298)

= 0.450(33.25X1273 -

298)

=

H(N 2 ) =

3.570(c

-

298)

=

-

298)

= 109400

)

(

Equating input 5142

to

+

,

mX1273

3.570(31.43X1273

output and solving for

9929

+ q+282990=

Hence, heat

is

removed:

66O+

14

-125 411

590+ 109400

kJ/h

—34 837 W.

Often when chemical reactions occur

process and the heat capacities vary

in the

with temperature, the solution in a heat balance can be

temperature

trial

and error

if

the final

unknown.

the

is

kJ/h

q,

48

q=

14590 kJ/h

EXAMPLE 1.7-3. Oxidation of Lactose In many biochemical processes, lactose oxidized as follows:

used as a nutrient, which

is

is

C 12 H 22 0 11 (5) + 120 2 (g)^ 12C0 2 fo) + llH 2 0(f) The

&H°

3

Appendix A.3 at 25°C is -5648.8. x 10 J/g heat of complete oxidation (combustion) at 37°C, which

heat of combustion

in

mol. Calculate the is the temperature of 1.20 J/g

•

many biochemical reactions. The c pm of solid lactose K, and the molecular weight is 342.3 g mass/g mol.

is

Solution: This can be treated as an ordinary heat-balance problem. First, the process flow diagram is drawn in Fig. 1.7-3. Next, the datum tempera-

25°C

ture of

is

selected

and the input and output enthalpies calculated. The

temperature difference At

=

(37

-

25)°C

=

(37

-

25) K.

•A//-37°C

1

g

mol

lactose

(s)

37°C 12 g

combustion

mol 0 2 (g)

37°C,

1

12 g

atm

Figure

22

11 g

1.7-3.

mol H 2 0(/), 37°C mol C0 2 (g), 37°C

Process flow diagram for Example

Chap.

I

1.7-3.

Introduction to Engineering Principles and Units

Input items H(Iactose)

= (342.3 =

H(Q 2

gas)

4929

= (12 =

(The c pm of

0

2

c pa

g)(

—

—

)(37

K=

25)

342.3(1.20X37

-

25)

J

g mol)( c pm

12(29.38X37

-

g mol 25)

=

-

K )(37

•

25)

K

4230 J

was obtained from Table

1.6-1.)

-AH°2i = -(-5648.8

x 10 3 )

Output items

H{U 2 0

liquid)

=

— J

11(18.02

g)(

c pm

J(37

— = 11(18.02X4.18X37 - 25) =

(The c pm of liquid water was obtained from Appendix

H(C0 2

(Thec^

of

C0 2

is

Setting input

4929

+

4230

gas)

(12 g moI)( c pm

=

12(37.45X37

obtained from Table

= +

-

gmo ,, K j(37 = 5393

25)

25)

9943

K J

A.2.)

J

=

-

- 25) K

J

1.6-1.)

output and solving, 5648.8 x 10

3

=

AH }rc =

9943

+

5393

- AH lTC

-5642.6 x 10 3 J/g mol =

AH 3I01

GRAPHICAL, NUMERICAL, AND MATHEMATICAL

1.8

METHODS Graphical Integration

1.8A

Often the mathematical function f{x) to integrate

it

analytically.

Or

in

to be integrated

some

is

too complex and we are not able

cases the function

from experimental data, and no mathematical equation

is

is

one that has been obtained

available to represent the data

so that they can be integrated analytically. In these cases, we can use graphical integration.

Integration between the limits x

shown is

Here a function y between the limits x = a to x = b is equal to the the sum of the areas of the rectangles as follows.

in Fig. 1.8-1.

the curve y

=

f{ x)

then equal to

x

=

integral.

This area

b

f{x) dx

Sec. 1.8

= a to x = b can be represented graphically as = f(x) has been plotted versus x. The area under

= A, + A 2 + A 3 + A 4 + A 5

Graphical, Numerical, and Mathematical

Methods

(1.8-1)

23

1.8B

Numerical Integration and Simpson's Rule

Often

it

is

desired or necessary to perform a numerical integration by

computing the

value of a definite integral from a set of numerical values of the integrand /(x). This, of course, can be done graphically, but

methods

The

suitable for the digital

if

data are available

computer are

integral to be evaluated

is

in large quantities,

numerical

desired.

as follows: •i = t

f(x) dx Jx —

where the interval

is

b

—

a.

The most

rule often called Simpson's rule. This

number

(1.8-2)

a

method

generally used numerical

method

divides the total interval b

is

—

the parabolic a into an even

of subintervals m, where

m =

-^ b

(1.8-3)

h

The value of h, a constant, is the spacing in x parabola on each subinterval, Simpson's rule is

/(x) dx Jx =

=

a

- {Jo

+

4(7,

^

+h +

where f0 is the value of/(x) at x = aj x = b. The reader should note that l

at

evenly spaced. This method

24

is

used. Then,

+/, +

2(/2

+h

+/„_,)

+fe + ••

+fm -

2)

+/J

(1-8-4)

= x ...Jm the value of/(x) must be an even number and the increments

the value of/(x) atx

m

approximating f(x) by a

L

,

well suited for digital computation.

Chap.

1

Introduction to Engineering Principles

and Units

PROBLEMS 1.2-1.

Temperature of a Chemical Process. The temperature of a chemical reaction was found to be 353.2 K. What is the temperature in °F, °C, and °R? o Ans. 176 F,80°C,636°R

1.2- 2.

Temperature for Smokehouse Processing of Meat. In smokehouse processing of sausage meat, a final temperature of 155°F inside the sausage is often used. Calculate this temperature in °C, K, and °R.

1.3- 1.

Molecular Weight of Air. For purposes of most engineering calculations, air is oxygen and 79 mol % nitrogen. Calculate assumed to be composed of 21 mol the average molecular weight. Ans. 28.9 g mass/g mol, lb mass/lb mol, or kg mass/kg mol

%

CO and Mole Units. The gas CO is being oxidized by0 2 to form How many kg of C0 2 will be formed from 56 kg of CO? Also, calculate

Oxidation of

1.3-2.

C0

2

.

O

this reaction. (Hint: First write the balanced the kg of z theoretically needed for chemical equation to obtain the mol 0 2 needed for 1.0 kg mol CO. Then calculate the kg mol of in 56 kg CO.) 88.0kgCO 2 32.0 kg 2 Ans.

CO

0

,

1.3-3.

Composition of a Gas Mixture. A gaseous mixture contains 20 g ofN 2 83 g of Calculate the composition in mole fraction and the average z and 45 g of CO z molecular weight of the mixture. Average mol wt = 34.2 g mass/g mol, 34.2 kg mass/kg mol Ans. ,

O 1.3-4.

.

,

%

of a Composition of a Protein Solution. A liquid solution contains 1.15 wt wt KC1, and the remainder water. The average molecular weight of the protein by gel permeation is 525 000 g mass/g mol. Calculate the mole fraction of each component in solution.

%

protein, 0.27

1.3- 5.

Concentration of

NaCl

%

centration of 24.0 wt

Solution.

An aqueous

NaCl with

solution of

a density of 1.178

g/cm 3

NaCl has at

a con25°C. Calculate

the following. (a)

Mole

fraction of NaCl

and water. 3

3 Concentration of NaCl as g mol/liter, lb^/ft lb^/gal, and kg/m Conversion of Pressure Measurements in Freeze Drying. In the experimental measurement of freeze drying of beef, an absolute pressure of 2.4 mm Hg was held in the chamber. Convert this pressure to atm, in. of water at ^C, /jm of Hg, and Pa. (Hint: See Appendix A.l for conversion factors.) 3 Ans. 3.16 x 10" atm, 1.285 in. H 2 0, 2400 /mi Hg, 320 Pa

(b) 1.4- 1.

1.4-2.

Compression and Cooling of Nitrogen Gas. A volume of 65.0ft 3 ofN 2 gasat90°F and 29.0 psig is compressed to 75 psig and cooled to 65°F. Calculate the final

volume

in

3 ft

and

pressures to psia (1.4-1) to 1.4-3.

.

,

obtain

the final density in Ib^/ft 3

first

n, lb

and then

.

\_Hint:

Be sure

to convert

all

to atm. Substitute original conditions into Eq.

mol.]

Gas Composition and Volume. A gas mixture of 0.13 g mol NH 3 1.27 g mol N 2 and 0.025 g mol H 2 0 vapor is contained at a total pressure of 830 mm Hg and 323 K. Calculate the following. (a) Mole fraction of each component. (b) Partial pressure of each component in mm Hg. 3 3 (c) Total volume of mixture in m and ft ,

,

.

1.4-4.

Evaporation of a Heat-Sensitive Organic Liquid. An organic liquid is being evaporated from a liquid solution containing a few percent nonvolatile dissolved solids. Since it is heat-sensitive and may discolor at high temperatures, it will be evaporated under vacuum. If the lowest absolute pressure that can be obtained in the apparatus is 12.0 Hg, what will be the temperature of evaporation in K? It will be assumed that the small amount of solids does not affect the vapor

mm

Chap.

I

Problems

25

pressure, which

is

given as follows:

log

where P A

is

in

PA =

-225^j + 9.05

mm Hg and T in K. T=

Ans. 1.5-1.

282.3

K or9.1°C

Evaporation of Cane Sugar Solutions. An evaporator is used to concentrate cane sugar solutions. A feed of 10000 kg/d of a solution containing 38 wt % sugar is evaporated, producing a 74 wt solution. Calculate the weight of solution produced and amount of water removed. Ans. 5135 kg/d of 74 wt solution, 4865 kg/d water

%

%

1.5-2.

Processing of Fish Meal. Fish are processed into fish meal and used as a supplementary protein food. In the processing the oil is first extracted to produce wet fish cake containing 80 wt water and 20 wt bone-dry cake. This wet cake feed is dried in rotary drum dryers to give a "dry" fish cake product containing water. Finally, the product is finely ground and packed. Calculate the 40 wt

%

%

%

kg/h of wet cake feed needed to produce 1000 kg/h of "dry" Ans. 1.5-3.

fish cake product. 3000 kg/h wet cake feed

Drying of Lumber.

%

is

What

A batch of 100 kg of wet lumber containing 1 1 wt dried to a water content of 6.38 kg water/1.0 kg bone-dry lumber. weight of "dried " lumber and the amount of water removed?

moisture is

the

A wet paper pulp contains 68 wt % water. After the pulp was dried, it was found that 55% of the original water in the wet pulp was removed. Calculate the composition of the "dried" pulp and its weight for a feed of 1000 kg/min of wet pulp.

13-4. Processing of Paper Pulp.

1.5-5.

Production of Jam from Crushed Fruit in Two Stages. In a process producing jam (CI), crushed fruit containing 14 wt% soluble solids is mixed in a mixer with sugar (1.22 kg sugar/1.00 kg crushed fruit) and pectin (0.0025 kg pectin/1.00 kg crushed fruit). The resultant mixture is then evaporated in a kettle to produce a jam containing 67 wt% soluble solids. For a feed of 1000 kg crushed fruit, calculate the kg mixture from the mixer, kg water evaporated, and kg jam produced. Ans. 2222.5 kg mixture, 189 kg water, 2033.5 kgjam

1.5-6.

Drying of Cassava (Tapioca) Root. Tapioca flour is used in many countries for bread and similar products. The flour is made by drying coarse granules of the cassava root containing 66 wt % moisture to 5% moisture and then grinding to produce a flour. How many kg of granules must be dried and how much water removed to produce 5000 kg/h of flour?

1.5-7.

Processing of Soybeans in Three Stages. A feed of 10000 kg of soybeans is processed in a sequence of three stages or steps (El). The feed contains 35 wt moisture, protein, 27.1 wt carbohydrate, 9.4 wt fiber and ash, 10.5 wt

%

%

and oil,

18.0

wt

%

oil.

%

%

In the

first

stage the beans are crushed and pressed to remove

giving an expressed oil stream and a stream of pressed beans containing

Assume no

6%

second step the pressed beans are extracted with hexane to produce an extracted meal stream containing 0.5 wt % oil and a hexane-oil stream. Assume no hexane in the extracted meal. Finally, in the last step the extracted meal is dried to give a dried meal of 8 wt moisture. Calculate: (a) Kg of pressed beans from the first stage. (b) Kg of extracted meal from stage 2. (c) protein in the dried meal. Kg of final dried meal and the wt protein Ans. (a) 8723 kg, (b) 8241 kg, (c) 78 16 kg, 44.8 wt oil.

loss of other constituents with the oil stream. In the

%

%

%

26

Chap.

I

Problems

%

Recycle in a Dryer. A solid material containing 15.0 wt moisture is dried so water by blowing fresh warm air mixed with recycled that it contains 7.0 wt air over the solid in the dryer. The inlet fresh air has a humidity of 0.01 kg water/kg dry air, the air from the drier that is recycled has a humidity of 0.1 kg water/kg dry air, and the mixed air to the dryer, 0.03 kg water/kg dry air. For a feed of 100 kg solid/h fed to the dryer, calculate the kg dry air/h in the fresh air, the kg dry air/h in the recycle air, and the kg/h of "dried" product. Ans. 95.6 kg/h dry air in fresh air, 27.3 kg/h dry air in recycle air, and 91.4 kg/h "dried" product

1.5-8.

%

1.5-9. Crystallization

12H 2 0

impurity.

The

and Recycle. It is desired to produce 1000 kg/h of Na 3 P0 4 from a feed solution containing 5.6 wt % Na 3 P04. and traces of -

crystals

original solution

is first

evaporated

in

an evaporator

to

a 35

wt%

and then cooled to 293 K in a crystallizer, where the hydrated crystals and a mother liquor solution are removed. One out of every 10 kg of mother liquor is discarded to waste to get rid of the impurities, and the remaining mother liquor is recycled to the evaporator. The solubility ofNa 3 P0 4 at 293 K is 9.91 wt %. Calculate the kg/h of feed solution and kg/h of water evaporated.

Na 3 P0 4 solution

7771 kg/h feed, 6739 kg/h water

Ans.

1.5-10. Evaporation and Bypass in Orange Juice Concentration. In a process for concentrating 1000 kg of freshly extracted orange juice (CI) containing 12.5 wt solids, the juice is strained, yielding 800 kg of strained juice and 200 kg of pulpy juice. The strained juice is concentrated in a vacuum evaporator to give an evaporated juice of 58% solids. The 200 kg of pulpy juice is bypassed around the evaporator and mixed with the evaporated juice in a mixer to improve the flavor. This final concentrated juice contains 42 wt solids. Calculate the concentration of solids in the strained juice, the kg of final concentrated juice, and the concentration of solids in the pulpy juice bypassed. (Hint: First, make a total balance and then a solids balance on the overall process. Next, make a balance on the evaporator. Finally, make a balance on the mixer.)

%

%

'._

1.5-11.

%

34.2 wt

solids in

pulpy juice

3

Manufacture of Acetylene. For the making of 6000 ft of acetylene (CHCH) gas at 70°F and 750 mm Hg, solid calcium carbide (CaC 2 ) which contains 97 wt % CaC 2 and 3 wt % solid inerts is used along with water. The reaction is

CaC 2 + 2H 2 0 -> The

final lime slurry

the total wt

%

added

be

C2H 2 CaC 2

,

15.30 lb

feed to lb

then the

member

Ca(OH) 2

is

Ca(OH) 2

1

andCa(OH) 2 lime. In this slurry How many lb of water must

20%.

many

how

and

CHCH +

contains water, solid inerts,

solids of inerts plus

produced? lb of final lime slurry is and convert to lb mol. This gives 15.30 lb mol mol Ca(OH) 2 and 15.30 lb mol CaC 2 added. Convert lb mol and calculate lb inerts added. The total lb solids in the slurry is 3

[Hint: Use a basis of 6000

ft

,

sum of the Ca(OH) 2 some is consumed

that

5200

Ans. 1.5-12.

.Ans.

lb

plus inerts. In calculating the water added, rein the reaction.]

water added (2359 kg), 5815 lb lime slurry (2638 kg)

%

Combustion of Solid Fuel. A fuel analyzes 74.0 wt C and 12.0% ash (inert). Air 1.2% CO, 5.7% is added to burn the fuel, producing a flue gas of 12.4% C0 2 0 2 and 80.7% N 2 Calculate the kg of fuel used for 100 kg mol of outlet flue gas and the kg mol of air used. (Hint: First calculate the mol 0 2 added in the air, using the fact that the N 2 in the flue gas equals the N 2 added in the air. Then make a carbon balance to obtain the total moles of C added.) ,

.

,

1.5-13.

Burning of Coke.

A

the rest inert ash.

needed to burn

components

Chap.

1

Problems

%

furnace burns a coke containing 81.0 wt C, 0.8% H, and uses 60% excess air (air over and above that

The furnace

all

C

to

in the flue gas

C0 if

2

only

and

H

95%

of the carbon goes

to

H 2 0).

Calculate the moles of

toC0 2 and 5%

to

all

CO.

27

of Formaldehyde. Formaldehyde (CH 2 0) is made by the catalytic oxidation of pure methanol vapor and air in a reactor. The moles from this

1.5-14. Production

reactor are 63.1 N 2 13.4 The reaction is ,

0

,

2

H 2 0,

5.9

CH 2 0,

4.1

CH 3 OH,

12.3

and

1.2

HCOOH.

CH 3 OH + |-0 A

side reaction occurring

CH 0 + H 2 0

->

2

2

is

CH 0 + i0 2

2

-^

HCOOH

Calculate the mol methanol feed, mol air feed, and percent conversion of methanol to formaldehyde.

Ans. 1.6-1.

Heating 101.32

o/C0 2

kPa

Gas.

A

17.6

total of

molCHjOH,

79.8

mol

250 g of C0 2 gas at 373

total pressure. Calculate the

23.3% conversion

air,

K

is

K at

heated to 623

amount of heat needed

in cal, btu,

and

kJ.

Ans.

1

5

050

cal, 59.7 btu,

62.98 kJ

1.6-2.

Heating a Gas Mixture. A mixture of 25 lb molN 2 and 75 lb.molCH 4 is being heated from 400°F to 800°F at 1 atm pressure. Calculate the total amount of heat needed in btu.

1.6-3.

Final Temperature in Heating Applesauce. A mixture of 454 kg of applesauce at 10°C is heated in a heat exchanger by adding 121 300 kJ. Calculate the outlet temperature of the applesauce. (Hint: In Appendix A. 4 a heat capacity for

applesauce average c pm

given at 32.8°C.

is

Assume

that this

is

constant and use

this as the

.)

Ans. 1.6-4. -

.

76.5°C

Use of Steam Tables. Using the steam tables, determine the enthalpy change for lib water for each of the following cases. (a) Heating liquid water from 40°F to 240°F at 30 psia. (Note that the effect of total pressure on the enthalpy of liquid water can be neglected.) (b) Heating liquid water from 40°F to 240°F and vaporizing at 240°F and 24.97 psia. (c)

Cooling and condensing a saturated vapor

at

212°F and

1

atm abs

to a liquid

at60°F. (d)

Condensing a saturated vapor Ans.

(a)

(d)

1.6-5.

1.6-6.

at

212°F and

200.42 btu/lb m - 970.3 btu/lb m ,

(b) ,

-

1 1

1

atm

abs.

52.7 btu/lb m

,

(c)

-

1

122.4 btu/lb m

,

2256.9 kJ/kg

Heating and Vaporization Using Steam Tables. A flow rate of 1000 kg/h of water at 21. 1°C is heated to 1 10°C when the total pressure is 244.2 kPa in the first stage of a process. In the second stage at the same pressure the water is heated further, until it is all vaporized at its boiling point. Calculate the total enthalpy change in the first stage and in both stages.

Combustion of CH4 and H 2 For 100 g mol of a gas mixture of 75 mol % CH 4 and 25% H 2 calculate the total heat of combustion of the mixture at 298 K and 101.32 kPa, assuming that combustion is complete. ,

1.6- 7.

Heat of Reaction from Heats of Formation. For the reaction

4NH

3 (£)

calculate the heat of reaction,

+

50 2 (g)^

AH,

at

4NO( 3 )

298

K

4-

6H 2 0(g)

and 101.32 kPa

for

4 g mol of

NH 3

reacting.

Ans. 1.7- 1.

28

AH, heat of reaction

=

—904.7 kJ

Heat Balance and Cooling of Milk. In the processing of rich cows' milk, 4540 kg/h of milk is cooled from 60°C to 4.44°C by a refrigerant. Calculate the heat removed from the milk. Ans. Heat removed = 269.6 kW Chap.

I

Problems

'

Air. A flow of 2200 lbjh of hydrocarbon oil at 100°F enters a heat exchanger, where it is heated to 150°F by hot air. The hot air enters at 300°F and is to leave at 200°F. Calculate the total lb mol air/h needed. The mean heat capacity of the oil is 0.45 btu/lb m °F.

Heating of Oil by

1.7-2.

Ans.

70.1 lb

mol

air/h, 31.8

kg mol/h

Combustion of Methane in a Furnace. A gas stream of 10 000 kg mol/h of CH 4 at 101.32 kPa and 373 K is burned in a furnace using air at 313 K. The combustion is complete and 50% excess air is used. The flue gas leaves the furnace at 673 K. Calculate the heat removed in the furnace. (Hint: Use a datum of 298 and liquid water at 298 K. The input items will be the following: the enthalpy of CH 4 at 373 K referred to 298 K; the enthalpy of the air at 313 K referred to 298 which is referred to liquid K; -AH°, the heat of combustion ofCH 4 at 298 water; and q, the heat added. The output items will include: the enthalpies of C0 2 0 2 N 2 and H z O gases at 673 K referred to 298 K; and the latent heat of H z O vapor at 298 K and 101.32 kPa from Appendix A.2. It is necessary to include this latent heat since the basis of the calculation and of the AH° is liquid

1.7-3.

K

K

,

,

,

water.) 1.7-4.

Preheating Air by Steam for Use in a Dryer. An air stream at 32.2°C is to be used in a dryer and is first preheated in a steam heater, where it is heated to 65.5°C. The air flow is 1000 kg mol/h. The steam enters the heater saturated at 148.9°C, is condensed and cooled, and leaves as a liquid at 137.8°C. Calculate the amount of steam used in kg/h. Ans. 450kgsteam/h

1.7- 5.

Cooling of Cans of Potato Soup After Thermal Processing. A total of 1500 cans of potato soup undergo thermal processing in a retort at 240°F. The cans are then cooled to 100°F in the retort before being removed from the retort by cooling water, which enters at 75°F and leaves at 85° F. Calculate the lb of cooling water needed. Each can of soup contains 1.0 lb of liquid soup and the empty metal can weighs 0. 1 6 lb. The mean heat capacity of the soup is 0.94 btu/lb m °F and that of the metal can is 0.12 btu/lb m °F. A metal rack or basket which is used to hold the cans in the retort weighs 350 lb and has a heat capacity of 0.12 btu/lb m °F. Assume that the metal rack is cooled from 240°F to 85°F, the temperature of the outlet water. The amount of heat removed from the retort walls in cooling from 240 to 100°F is 10 000 btu. Radiation loss from the retort during cooling is estimated as 5000 btu. Ans. 21 320 lb water, 9670 kg •

•

•

1.8- 1.

Graphical Integration and Numerical Integration Using Simpson's Method. following experimental data of y = f(x) were obtained.

X

X

/M

100

0.4

53

0.1

75

0.5

60

0.2

60.5

0.6

72.5

0.3

53.5

0

It is

m

The

desired to determine the integral C

x

=

0.6

A = x

(a)

(b)

=0

Do this by a graphical integration. Repeat using Simpson's numerical method. Ans.

Chap.

I

Problems

(a)

A =

38.55 ;(b)/l

=

38.45

29

1.8-2.

Graphical and Numerical Integration to Obtain Wastewater Flow. The rate of flow of wastewater in an open channel has been measured and the following data obtained:

Time (min)

Flow (m i /min)

Time (min)

(m 3 /min)

0

655

70

10

705

80

800 725

20

780 830

90

670

30

100

40

870

110

640 620

50

890 870

120

610

60

(a)

(b)

Flow

Determine the total flow in m 3 for the first 60 min and also the total for 120 min by graphical integration. Determine the flow for 120 min using Simpson's numerical method. 3 3 (a) 48 460 m for 60 min, 90 390 m for 120 Ans.

m

REFERENCES (CI)

Charm, S. E.-The Fundamentals of Food Engineering, 2nd ed. Westport, Conn.: Avi Publishing Co., Inc., 1971.

(El)

Earle, R. L. Unit Operations

in

Food

Processing. Oxford:

Pergamon

Press, Inc.,

1966.

(HI)

Hougen, O. Part

I,

2nd

A.,

ed.

Watson, K. M., and Ragatz,

New York John :

Wiley

R. A.

& Sons, Inc.,

Chemical Process Principles,

1954.

(01)

Okos, M. R. M.S.

(PI)

Perry, R. H., and Green, D. Perry's Chemical Engineers' Handbook, 6th ed. New York: McGraw-Hill Book Company, 1984.

(SI)

Sober, H. A. Handbook of Biochemistry, Selected Data for Molecular Biology, 2nd ed. Boca Raton, Fla.: Chemical Rubber Co., Inc., 1970.

(Wl)

Weast, R. C, and Selby, S. M. Handbook of Chemistry and Physics, 48th Raton, Fla.: Chemical Rubber Co., Inc., 1967-1968.

30

thesis.

Ohio

State University,

Columbus, Ohio,

Chap.

1972.

I

ed.

Boca

References

CHAPTER

2

Principles of

Momentum

Transfer

and Overall Balances

INTRODUCTION

2.1

The flow and behavior of

A

engineering.

fluid

may

fluids

is

important

in

many

of the unit operations in process

be defined as a substance that does not permanently resist

distortion and, hence, will change

and vapors are same laws. In the process industries, many of the materials are in fluid form and must be stored, handled, pumped, and processed, so it is necessary that we become familiar with the principles that govern the flow of fluids and also with the equipment used. Typical fluids encountered include water, air, C0 2 oil, slurries, and thick syrups. its

shape. In

this text gases, liquids,

considered to have the characteristics of fluids and to obey

many

of the

,

If

a fluid

Most

is

inappreciably affected by changes

in pressure,

it is

said to be incompres-

Gases are considered to be compressible fluids. However, if gases are subjected to small percentage changes in pressure and temperature, their density changes will be small and they can be considered to be incompressible.

sible.

Like

liquids are incompressible.

all

physical matter, a fluid

molecules per unit volume.

mechanics

treats the

A

composed

is

motions of molecules

of individual molecules. In engineering

macroscopic behavior of a

number of

of an extremely large

theory such as the kinetic theory of gases or

fluid rather

in

we

statistical

terms of statistical groups and not

in

terms

are mainly concerned with the bulk or

than with the individual molecular or microscopic

behavior. In

momentum

transfer

we

treat the fluid as a

a "continuum". This treatment as a fluid

contains a large enough

continuous distribution of matter or as

continuum

number

is

valid

when

the smallest

volume of

of molecules so that a statistical average

meaningful and the macroscopic properties of the

fluid

is

such as density, pressure, and so

on, vary smoothly or continuously from point to point.

The study

of

momentum

transfer,

or fluid mechanics as

divided into two branches: fluid statics, or fluids at

motion. In Section 2.2 we treat fluid in

Chapter

term

3,

fluid

"momentum

transfer

is

dynamics. Since

in

transfer" or "transport"

related to heat

and mass

it

is

often called, can be

and fluid dynamics, or fluids in statics; in the remaining sections of Chapter 2 and fluid dynamics momentum is being transferred, the is

rest,

usually used. In Section 2.3

momentum

transfer.

31

FLUID STATICS

2.2

Force, Units, and Dimensions

2.2A

In a static fluid an important property

by a

a surface force exerted

any point

volume of a

in a

is

the pressure in the fluid. Pressure

fluid against the walls of its container. Also,

first

familiar as

fluid.

In order to understand pressure, which

must

is

pressure exists at

defined as force exerted per unit area,

is

we

discuss a basic law of Newton's. This equation for calculation of the force

exerted by a mass under the influence of gravity

F = mg

is

(SI units) (2.2-1)

(English units)

where

in

SI units

F

the force exerted in newtons

is

the standard acceleration of gravity, 9.80665

F

In English units,

conversion factor)

is

is

in lb r ,

32.174 Ib m

m

in lb m 2

•

ft/lb f

s

.

m/s ,

g

N(kg m/s 2 ), m the mass

in kg,

and g

2 .

is

The use

2

and gc (a gravitational of the conversion factory means that 32.1740

ft/s

,

g/g c has a value of 1.0 lb f /lb m and that 1 lb m conveniently gives a force equal to 1 lb f 2 Often when units of pressure are given, the word "force" is omitted, such as in lb/in.

.

(psi)

cra/s

instead of lb r /in. 2 ,

=

and g z

2 .

When

the mass

m

is

given in g mass, 2

980.665 g mass-cm/g force -s

.

F

is

g force, g

=

980.665

However, the units g force are seldom

used.

Another system of units sometimes used in Eq. (2.2-1) is that where the#c is omitted 2 and the force ( F = mg) is given as lb m ft/s which is called poundals. Then I lb m acted on 2 by gravity will give a force of 32.174 poundals (lb m ft/s ). Or if 1 g mass is used, the force 2 (F = mg) is expressed in terms of dynes (g-cm/s ). This is the centimeter-gram-second ,

(cgs)

systems of units.

Conversion factors

for different units of force

and of

force per unit area (pressure) are

given in Appendix A.l. Note that always in the SI system, and usually in the cgs system, the term g c

is

not used.

EXAMPLE 2.2-1.

Units

and Dimensions of Force

Calculate the force exerted by 3 lb mass in terms of the following. (a)

Lb

(b)

Dynes

(c)

Newtons

Solution:

force (English units). (cgs units). (SI units).

For part

(a),

using Eq. (2.2-1),

F (force) = m

= lb.

lb f -s

For part

2

(b),

F = mg =

=

32

3 lb force (tb f)

-ft

32.174

(3 lb

J

453.59

1.332 x 10'

Chap. 2

Principles

980.665

-f-

=

1.332

—

x 10 6 dyn

of Momentum Transfer and Overall Balances

As an alternative method 1

dyn

=

F =

To

calculate

2.2481 x 10"

6

lb r _1

(3 lbf)(

newtons

F

from Appendix A.l,

for part (b),

m -_6 lb /dyn =

,,, 01 ., ,2.2481 x.lO

in part

.

6l

j

1-332 x 10

31b

=

dyn

(c),

^K ^2l^)H

=

6

f

13.32

13.32

65

?)

N

As an alternative method, using values from Appendix A.l,

i^ (dyn)=10 -5^

1

F=

2.2B

Pressure

in a

(1.332 x 10* dyn)

tfr

I

5

(newton)

^^U

13.32

dyn

N

Fluid

Since Eq. (2.2-1) gives the force exerted by a mass under the influence of gravity, the force exerted by a mass of fluid on a supporting area or force/unit area (pressure) also follows

from

this

fluid

The

P 0 N/m 2

is

column of fluid of height/j 2 m and constant where A = A 0 = /t, = A 2 is shown. The pressure above the this could be the pressure of the atmosphere above the fluid.

equation. In Fig. 2.2-1 a stationary

cross-sectional area

A

that

;

m

2 ,

is,

,

above it. It can be shown that must be the same in all directions. Also, for a fluid at rest, the force/unit area or pressure is the same at all points with the same elevation. For example, at/i, m from the top, the pressure is the same at all points shown on the cross-sectional area A any point, say

fluid at

the forces at

,

any given point

must support

in a

points

use of Eq. (2.2-1) will be

in Fig. 2.2-

1

.

The

total

FlGUKE

2.2-

Sec. 2.2

1

.

Pressure

total

kg

shown

=

in a static fluid.

Fluid Statics

in

mass of fluid

fluid

the fluid

[h

2

static fluid

.

,

The

all

nonmoving or

calculating the pressure at different vertical

for

mX/l

h2

m height

and density p kg/m 3

m )^p^j = 2

h

1

Ap kg

is

(2.2-2)

P0

33

Substituting into Eq.

(2.2-2),

the total force

F

of the fluid

on area/1, due

to the fluid

only

is

F= The pressure P

is

is

Ap kgX3 m/s 2 =

=~=

(h 2

the pressure

on A 2 due

P 2 on A 2

the pressure

pressure

,

depth.

(2.2-5)

To

is

calculate

Apg)^ =

P 0 on

*

m (N)

(2.2-3)

h2

fluid

or

Pa

(2.2-4)

above it. However, to must be added.

get the total

the top of the fluid

+ P 0 N/m 2

pg

Pa

or

(2.2-5)

the fundamental equation to calculate the pressure in a fluid at

any

P lt P,

The pressure

ks

pgN/m 2

h2

mass of the

to the

Pi = Equation

Apg

h2

)

defined as force/unit area:

P This

(h 2

=

difference between points 2

Pi-Pi=(h 2 pg + P 0 )-(h

i

pg

h lPg

and

+ P0

(2.2-6)

1 is

+ P0 = )

(h 2

-h

1

)pg

(SI units)

(12_7)

P2 — Since

it

P,

=

(h 2

—

h t )p

at the

bottom of all

fluid,

the

For example, in Fig. 2.2-2, the pressure the same and equal to h v pg + P 0

affect the pressure.

three vessels

EXAMPLE 7. 1-2. A

(English units)

the vertical height of a fluid that determines the pressure in a

is

shape of the vessel does not Pj

—g

is

.

Pressure in Storage Tank

large storage tank contains oil having a density of 917

kg/m 3 (0.917

The tank is 3.66 m (12.0 ft) tall and is vented (open) to the atmos1 atm abs at the top. The tank is filled with oil to a depth of 3.05 m (10 ft) and also contains 0.61 m (2.0 ft) of water in the bottom of the tank. Calculate the pressure in Pa and psia 3.05 m from the top of the tank and at g/cm

3

).

phere of

the bottom. Also calculate the gage pressure at the tank bottom.

Solution:

pressure

First a sketch

P0 =

1

is

made

atm abs = 14.696

P0 =

Figure

34

2.2-2.

Chap. 2

shown in Fig. Appendix A.l). Also,

of the tank, as

psia (from

2.2-3.

The

s 1.01325 x 10 Pa

Pressure

in

vessels

of various shapes.

Principles of Momentum Transfer

and Overall Balances

Figure

Storage tank

2.2-3.

in

Example

2.2-2.

P0 =

atm abs

1

\ 1

0 ft = h

,

1

2U

water

= h2 f

t

Pi

From

Eq. (2.2-6) using English and then SI units,

=

F,

f

+ P0 =

(10 ft)(0.917 x 62.43

^)(l.O

c

+

A= =

14.696 lb f/in.

h lPoH g

+ P0

1.287 x 10

To calculate P 2 Pz

s

are

newtons/m

2

is

18.68 psia

bottom

P^„ - +

=

h2

=

19.55 psia

=

h lPvlMcr g

=

1.347

+

of the tank,p walcr

=

Pi

P,

=

=

1.00

s

g/cm 3 and

(2.0X1.00 x 62.43Xl.0Xrk)

(0.61X1000X9.8066)

bottom

at the

is

19.55 psia

,

in

given in

as given in

many

+

+

18-68

5

1.287 x 10

equal to the absolute pressure P 2 minus

-

different

Appendix

terms of head in

m

14.696 psia

=

1

4.85 psig

Eq.

(2.2-4),

which

relates pressure

A.l.

sets

of units, such

However,

a

as

psia,

common method

dyn/cm

2

same pressure

P and

as the pressures

it

h,

m

or

Using

represents.

height h of a fluid and solving for

and

,

of expressing

or feet of a particular fluid. This height or head in

of the given fluid will exert the

in

1.0132 x 10

x 10 5 Pa

feet

head

+

^§^9.8066

of a Fluid

Pressures

pressures

^Jj^J^)

Pa

P tlsc =

Head

=

=(3.05 m)^917

at the

The gage pressure atm pressure.

2.2C

2

which

is

the

m,

m

/i(head)

(SI)

pg

(2.2-8)

PQj:

h

ft

(English)

PQ

Sec. 2.2

Fluid Statics

35

EXAMPLE 22-3.

Conversion of Pressure to Head of a Fluid 2 1 standard atm as 101.325 kN/m (Appendix A.l), do

Given the pressure of as follows.

Convert Convert

(a)

(b)

this pressure to this pressure to

head in head in

m water at 4°C. m Hg at 0°C.

For part (a), the density of water at 4°C in Appendix A.2 is 1.000 g/cm 3 From A.l, a density of 1.000 g/cm 3 equals 1000 kg/m 3 Substituting

Solution: .

.

these values into Eq. (2.2-8),

P _ pg~

Ca

(

= For part

(b),

equal pressures

P

101.325 x 10

(1000X9.80665)

10.33

m

Hg

in

the density of

from different

p = Pu i

fluids,

K

i

of water at 4°C

Appendix A.l

13.5955 g/cm Eq. (2.2-8) can be rewritten as

=

9

/

Pjj q

2.2D

Devices

In chemical

to

=

h HlQ

1

.

For

(2-2-9)

known

values,

AAA \ k 10 33 ) =

= (133955

3

is

pH 2 o hH 2o 9

Solving for h Hg in Eq. (2.2-9) and substituting

/zHg(head)

3

-

°- 760

m H8

Measure Pressure and Pressure Differences

and other industrial processing plants

it

is

often important to

measure and

control the pressure in a vessel or process and/or the liquid level in a vessel. Also, since

many

fluids are flowing in a pipe

the fluid

is

flowing.

Many

or conduit,

it is

necessary to measure the rate at which

of these flow meters depend

Some common

pressure or pressure difference.

upon devices

to

measure a

devices are considered in the following

paragraphs. /.

36

Simple U-tube manometer.

The U-tube manometer

Chap. 2

is

shown

Principles of Momentum Transfer

in

Fig. 2.2-4a.

The

and Overall Balances

U

2

pressure p a N/m is exerted on one arm of the tube and p b on the other arm. Both pressures p a and p b could be pressure taps from a fluid meter, or p a could be a pressure tap and p b the atmospheric pressure. The top of the manometer is filled with liquid B,

having a density of p B kg/m 3 , and the bottom with a more dense fluid A, having a density 3 Liquid A is immiscible with B. To derive the relationship betweenp a andp b °f Pa kg/m .

p„

is

,

the pressure at point

1

and p b Pi

where

R

is

the reading of the

The pressure

at point 5.

=

+

Pa

point 2

is

+ R)p B g N/m 2

(Z

manometer

at

(2.2-10)

m. The pressure

in

at point 3

must be equal

to

that at 2 by the principles of hydrostatics.

=

P3

The pressure at

point

equals the following:

3 also

Pi

Equating Eq.

The

(2-2-11)

Pi

(2.2-10) to (2.2-12)

=

and

+ Zp B g + Rp A g

Pb

(2.2-12)

solving,

=

Pa

+

(Z

+

R)p B g

Pa

-

Pb

=

&(Pa

-

p B )g

Pa

-

Pb

=

R(Pa

~

Pb)

reader should note that the distance

+ Zp B g + Rp A g

Pb

(2.2-13)

(SI)

(2-2-14)

—g

(English)

Z does

not enter into the

final result

nor do the

tube dimensions, provided that p a and p b are measured in the same horizontal plane.

EXAMPLE

23-4.

A manometer,

Manometer

Pressure Difference in a

shown

being used to measure the head or pressure drop across a flow meter. The heavier fluid is mercury, with a 3 3 density of 13.6 g/cm and the top fluid is water, with a density of 1.00g/cm The reading on the manometer is R — 32.7 cm. Calculate the pressure as

in Fig. 2.2-4a,

is

,

difference in

Solution:

N/m 2

.

using SI units.

Converting R to m,

K

32.7

Also converting p A and p B to kg/m Pa

- Pb =

R(Pa

~

Ps)9

=

= 2.

U

Two-fluid

3

and substituting

(0.327 m)[(13.6

4.040 x 10

4

-

N/m

In Fig. 2.2-4b a two-fluid

tube.

m

a327 =ioo- =

into Eq. (2.2-14),

3 1.0X1000 kg/m )](9.8066 m/s 2

tubes forming the U. Proceeding and

Pawhere R 0 heavier

Sec. 2.2

is

the reading

fluid,

and p B

is

Fluid Statics

Pb

= {R-

m2

making

Ro)(pa

when pa = p b R ,

~

is

)

(5.85 psia)

U

tube

is

device to measure small heads or pressure differences. Let

area of each of the large reservoirs and a

2

shown, which

A

m

2

is

a sensitive

be the cross-sectional

be the cross-sectional area of each of the a pressure balance as for the

Pb

+ ~ Pb - ^PcJ

the actual reading,

U tube, (2.2-15)

9

pA

the density of the lighter fluid. Usually, a/A

is

is

the density of the

made

sufficiently

37

R0

small to be negligible, and also

If p A

and p B are close

EXAMPLE

is

=

often adjusted to zero

- Pb)9

Pa

-

Pa

-P^R(PA-P B L

Pb

R(Pa

2.2-5.

then

(SI)

R

<

(English)

)

to each other, the reading

;

is

12- 16 )

magnified.

Pressure Measurement in a Vessel

The U-tube manometer

used to measure the pressurep^ in a Derive the equation relating the pressure p A and the reading on the manometer as shown. in Fig. 2.2-5a

is

vessel containing a liquid with a density p A

At point 2 the pressure

Solution:

Pi

At point

1

the pressure

is

= Pa.m +

h2

=

pB 9

Wm

2

(2.2-17)

is

Pi=P A + Equating p l

.

h pA g

(2.2-18)

l

p 2 by the principles of hydrostatics and rearranging,

Pa

=

P alm

+

h iP B g

-h lPA g

(2.2-19)

Another example of a U-tube manometer is shown in Fig. 2.2-5b. This device case to measure the pressure difference between two vessels.

is

used

in this

Bourdon pressure gage. Although manometers are used to measure pressures, the most common pressure-measuring device is the mechanical Bourdon-tube pressure gage. A coiled hollow tube in the gage tends to straighten out when subjected to internal pressure, and the degree of straightening depends on the pressure difference between the inside and outside pressures. The tube is connected to a pointer on a calibrated dial.

3.

4.

Gravity separator for two immiscible

liquids.

In Fig. 2.2-6 a continuous gravity

shown for the separation of two immiscible liquids A (heavy liquid) and B (light liquid). The feed mixture of the two liquids enters at one end of the separator vessel and the liquids flow slowly to the other end and separate into two distinct layers. Each liquid flows through a separate overflow line as shown. Assuming

separator (decanter)

is

(b)

(a)

Figure

38

2.2-5.

Measurements of pressure in vessels : (a) measurement of pressure vessel, (b) measurement of differential pressure.

Chap. 2

Principles

in

a

of Momentum Transfer and Overall Balances

vent

light liquid

feed

heavy liquid Figure

A

overflow

Continuous atmospheric gravity separator for immiscible

2.2-6.

the frictional resistance to the flow of the liquids fluid statics

essentially negligible, the principles of

is

A is h A m and that of B is h B The and is fixed by position of the overflow line for B. The heavy discharges through an overflow leg h A2 m above the vessel bottom. The vessel

In Fig. 2.2-6, the depth of the layer of heavy liquid

liquid

A

liquids.

can be used to analyze the performance.

=

total depth h T

and

B

overflow

h Al

+

the overflow lines are vented to the atmosphere.

KPbQ + Substituting h B

=

hT

—

,

.

hB

h Al into

h Al p A

g=

A hydrostatic balance gives

h A1 p A g

Eq. (2.2-20) and solving for h Al

h Al

=

hA i 1

(2.2-20)

,

- h T p B/p A - PbIPa

(2.2-21)

This shows that the position of the interface or height h Al depends on the ratio of the densities of the

two liquids and on the elevations h A2 and h T of the two overflow is movable so that the interface level can be adjusted.

lines.

Usually, the height h A2

GENERAL MOLECULAR TRANSPORT EQUATION FOR MOMENTUM, HEAT, AND MASS TRANSFER

2.3

2.3A

General Molecular Transport Equation and General Property Balance

Introduction to transport processes.

1.

In molecular transport processes in general

we

movement of a given property or entity by molecular movement through a system or medium which can be a fluid (gas or liquid) or a solid. This property that is being transferred can be mass, therma[energy (heat), or momentum.

are concerned with the transfer or

Each molecule of a system has a given quantity of the property mass, thermal energy, or

momentum for

associated with

it.

When

a difference of concentration of the property exists

any of these properties from one region to an adjacent region, a net transport of

this

property occurs. In dilute fluids such as gases where the molecules are relatively far apart, the rate of transport of the property should be relatively fast since few molecules

are present to block the transport or interact. In dense fluids such as liquids the

molecules are close together and transport or diffusion procedes more slowly. molecules is

in solids

are even

more close-packed than

in liquids

The

and molecular migration

even more restricted.

Sec. 2.3

General Molecular Transport Equation

39

2.

of

General molecular transport equation.

momentum,

All three of the

molecular transport processes

heat or thermal energy, and mass are characterized in the elementary

we

sense by the same general type of transport equation. First

start

by noting the

following:

rate of a transfer process

=

driving force

(23-1) resistance

This states what in

is

quite obvious

—that we need a driving force

order to transport a property. This

rate of flow of electricity

is

overcome a

to

Ohm's law

similar to

proportional to the voltage drop (driving force)

is

resistance

where the and inversely

in electricity,

proportional to the resistance.

We

can formalize Eq.

by writing an equation as follows

(2.3-1)

molecular

for

transport or diffusion of a property:

4>t

where

\j/

z

is

=

S dr

(23-2)

dz

defined as the flux of the property as

amount

of property being transferred

per unit time per unit cross-sectional area perpendicular to the z direction of flow in

amount

of property/s

•

m2

<5

,

is

a proportionality constant called diffusivity

concentration of the property in amount of property/m

3 ,

and

m 2/s, T

in

is

z is the distance in the

direction of flow in m. If the process

at

is

steady state, then the

flux^

is

constant. Rearranging Eq. (2.3-2)

and integrating, rr 2

dz

= -5

dr

(23-3)

<5(r\-r 2 ) (2.3-4)

A plot

of the concentration

the flux

is

in the

direction

and the negative sign

in

1

F

versus

z is

shown

in Fig. 2.3-1 a

and

a straight

is

to 2 of decreasing concentration, the slope dF/dz

Eq. (2.3-2) gives a positive flux in the direction

2.3B the specialized equations for

be the same as Eq. (2.3-4)

for the

momentum,

heat,

and mass

1

line. is

to 2.

transfer will be

Since

negative

In Section

shown

to

general property transfer.

unit area

in

=

T z z

t

+

•Az

2.3-1.

|z

+ Az

Az

»-]

(b)

(a)

Figure

= ^z

Molecular transport of a property : (a) plot of concentration versus distance for steady state, (b) unsteady-state general property balance.

40

Chap. 2

Principles of Momentum Transfer

and Overall Balances

EXAMPLE

23-1. Molecular Transport of a Property at Steady State property is being transported by diffusion through a fluid at steady state. 2 At a given point 1 the concentration is 1.37 x 10" amount of property/m 3 2 and 0.72 x 10" at point 2 at a distance z 2 = 0.40 m. The diffusivity 2 5 = 0.013 /s and the cross-sectional area is constant.

A

m

Calculate the

(a) (b) (c)

flux.

Derive the equation for T as a function of distance. Calculate T at the midpoint of the path:

For part

Solution:

^

~ =

For part

(b),

flr, z2

Eq.

(a) substituting into

- T2 _ - z, ~ )

2.113 x 10~*

(2.3-4),

m2

of property/s

integrating Eq. (2.3-2) between

2 x 10" )

-0

0.40

amount

- 0.72

2

10~ (0.013X1-37 x

^

and T and

and

z,

z

and

rearranging,

''

-

dz

•.

dT

j

(2,3-5!

J

r = r, + For

part

(c),

using the midpoint z

=

0.20

r=i,7x,o- + =

3.

1.045 x 10~

2

^

- z)

(23-6)

m and substituting into Eq. (2.3-6),

Mil^ „_ (

amount

General property balance for unsteady state.

system using the molecular transport equation

amount

(z,

0

.

2)

of property/m

3

In calculating the rates of transport in a (2.3-2),

it

necessary to account for the

is

of this property being transported in the entire system. This

general property balance or conservation equation for the property energy, or mass) at unsteady state. only,

which accounts for

all

We

start

volume Az(l) m 3

fixed in space.

(rate of

f rate of generation^

property in/

by writing an equation

\o{ property

shown

in Fig. 2.3-lb,

\

property out/

The |

+

term

which

is

an element of

+

r3tC °^

accum ~

(

\

(23-7)

\ulation of property/

is W^J* 1 amount of property/s and the rate of output is 2 where the cross-sectional area is 1.0 The rate of generation of the K(Az 1), where R is rate of generation of property/s m 3 The accumulation

rate of input

Az )- 1,

property

for the z direction

/ (rate of

(iA z I

done by writing a

the property entering by molecular transport, leaving, being

generated, and accumulating in a system

\

is

(momentum, thermal

is

m

.

•

•

.

is

rate of

accumulation of property

=

—

(Az

•

1)

(23-8)

dt

Sec. 2.3

General Molecular Transport Equation

41

Substituting the various terms into Eq. (2.3-7),

(*,,,)

1

+R(Az-l) =

(^ l|l+4 ,)-l

+^(Az-l)

(23-9)

Dividing by Az and letting Az go to zero,

dr

Substituting Eq. (2.3-2) for

dip.

into (2.3-10)

and assuming that 5

is

constant,

— ~ S^j s— = R r

For the case where no generation

is

(2.3-11)

dz

dt

present,

— -5 = — —£

(2.3-12)

<5

dz

dt

This

final

equation relates the concentration of the property

Equations

(2.3-11)

and

mentum, thermal energy,

T to position z and

time

or mass

and

be used

will

many

in

t.

of mo-

(2.3-12) are general equations for the conservation

sections of this text.

The

equations consider here only molecular transport occurring and the equations do not consider other transport mechanisms such as convection, and so on, which will be

considered when the specific conservation equations are derived text for

of this

in later sections

or mass.

Introduction to Molecular Transport

2.3B

The

momentum, energy,

kinetic theory of gases gives us a

individual molecules

random movement,

in fluids.

random movements if

flux of the property

of molecules

energy the molecules are

in rapid

there

momentum,

mass occurs in a fluid because of Each individual molecule containing all directions and there are fluxes in all

heat, or

of individual molecules.

the property being transferred directions. Hence,

their kinetic

often colliding with each other. Molecular transport or molecular

diffusion of a property such as

these

good physical interpretation of the motion of

Because of

is

moves randomly

in

a concentration gradient of the property, there will be a net

from high to low concentration. This occurs because equal numbers

diffuse

in

each

between

direction

and low-

high-concentration

the

concentration regions.

/.

Momentum

When

transport and Newton's law.

a fluid

direction decreases as

momentum and

its

we approach the

concentration

is

vx

direction

(

+ z and — z

p

in the

The

fluid

x direction vx

in

the

x

has x-directed

momentum/m 3 where the momentum has units 3 (kg m/s)/m By random diffusion of molecules ,

•

.

an equal number moving

in each

directions) between the faster-moving layer of molecules

and the

z

direction,

slower adjacent layer. Hence, the x-directed

momentum

direction from the faster- to the slower-moving layer.

42

flowing

surface in the z direction.

of kg m/s. Hence, the units of v x p are there is an exchange of molecules in the •

is

where the velocity

parallel to a solid surface, a velocity gradient exists

Chap. 2

Principles

has been transferred in the z

The equation

for this transport of

of Momentum Transfer and Overall Balances

momentum

similar to

is

Eq

;

(2.3-2)

and

Newton's law of viscosity written as follows

is

for

constant density p:

r_.

where

x, x

is

momentum density

2.

in

flux of x-directed

m

diffusivity in

kg/m 3 and ;

p. is

2

x

=-v-^

momentum

in the z direction,

/s; z is the direction of

the viscosity in

Heat transport and Fourier's

(23-13)

kg/m

(kg m/s)/s

•

m

2

v is p/p, the

;

transport or diffusion in

m; p

where

qJA

is

m

2 ,

at

is

the thermal diffusivity

concentration of heat or thermal energy in J/m

numbers

colder region. In this

3.

Mass

3 .

When

way energy

is

in

kg mol A/s-m 2

A in B in m /s, and c A to momentum and heat

similar fluid,

A

the flux of 2

is

is

the

a temperature gradient

in

between the hot and the

Fick's law for molecular transport of

J*A:=~D AB

molecule

is

inm 2 /s, andpc p T

transferred in the z direction.

transport and Fick's law.

is

there

of molecules diffuse in each direction

or solid for constant total concentration in the fluid

where JJ.

p and

.

the heat flux in J/s-

a fluid, equal

the

Fourier's law for molecular transport of heat or

law.

heat conduction in a fluid or solid can be written as follows for constant density heat capacity c p

is

-s.

,

equal numbers of molecules diffuse

D AB

when in

in a fluid

(23-15)

^f is

the concentration of

transport,

mass

is

there

the molecular diffusivity of the

A is

in

kg mol/t/m

3 .

In a

manner

a concentration gradient in a

each direction between the high- and the

low-concentration region and a net flux of mass occurs.

Hence, Eqs. all

(2.3-13), (2.3-14),

similar to each other

and

and

(2.3-15) for

to the general

momentum,

heat,

and mass

transfer are

molecular transport equation

equations have a flux on the left-hand side of each equation, a diffusivity

in

m

(2.3-2). All 2

/s,

and the

derivative of the concentration with respect to distance. All three of the molecular

transport equations are mathematically identical. Thus, .

similarity

among

them.

It

we

state

we have an analogy or

should be emphasized, however, that even though there

mathematical analogy, the actual physical mechanisms occurring can

is

a

be totally different.

mass transfer two components are often being transported by relative motion through one another. In heat transport in a solid, the molecules are relatively stationary and the transport is done mainly by the electrons. Transport of momentum can occur by several types of mechanisms. More detailed considerations of each of the For example,

in

transport processes of

momentum, energy, and mass

are presented in this and succeeding

chapters.

2.4

VISCOSITY OF FLUIDS

2.4A

Newton's Law and Viscosity

When

a fluid

is

plates, either of

Sec. 2.4

flowing through a closed channel such as a pipe or between two flat two types of flow may occur, depending on the velocity of this fluid. At

Viscosity

of Fluids

43

low velocities the fluid tends to flow without lateral mixing, and adjacent layers slide past one another like playing cards. There are no cross currents perpendicular to the direction of flow, nor eddies or swirls of fluid. This regime or type of flow is called laminar flow. At higher velocities eddies form, which leads to lateral mixing. This is called turbulent flow.

The

discussion in this section

A

fluid

limited to laminar flow.

is

can be distinguished from a solid

in

this

discussion of viscosity by

behavior when subjected to a stress (force per unit area) or applied force.

deforms by an amount proportional to the applied

stress.

subjected to a similar applied stress will continue to deform, increases with increasing stress.

property of a fluid which gives layers in the fluid.

A

i.e.,

An

However, a

its

elastic solid fluid

when

to flow at a velocity that

fluid exhibits resistance to this stress. Viscosity is that

rise to forces that resist

These viscous forces

arise

the relative

movement

of adjacent

from forces existing between the molecules

and are of similar character as the shear forces in solids. above can be clarified by a more quantitative discussion of viscosity. In Fig. 2.4-1 a fluid is contained between two infinite (very long and very wide) parallel plates. Suppose that the bottom plate is moving parallel to the top plate and at a constant velocity Au z m/s faster relative to the top plate because of a steady force F newtons being applied. This force is called the viscous drag, and it arises from the viscous forces in the fluid. The plates are Ay m apart. Each layer of liquid moves in the z direction. The layer immediately adjacent to the bottom plate is carried along at the velocity of this plate. The layer just above is at a slightly slower velocity, each layer moving at a slower velocity as we go up in the y direction. This velocity profile is linear, with y direction as shown in Fig. 2.4-1. An analogy to a fluid is a deck of playing cards, where, if the bottom card is moved, all the other cards above will slide to some extent. in the fluid

The

ideas

has been found experimentally for

It

many

fluids that the force

directly proportional to the velocity Ay. in m/s, to the area

inversely proportional to the distance

when

the flow

is

Ay

in

,4

in

m2

F

in

newtons

is

of the plate used, and

m. Or, as given by Newton's law of

viscosity

laminar,

F

Av z

A

Ay

(2.4-1)

where

kg/m

•

is

/i

If

s.

a proportionality constant called

we

let

Ay approach

V= where

x

the cgs

44

= F/A and is system, F is in

y

,

the viscosity of the fluid, in

dv z

p.

Chap. 2

in

g/cm-s,

(2.4-2)

(SI units)

-H~r_

the shear stress or force per unit area in

dynes,

Pa-s or

zero, then, using the definition of the derivative,

incm/s, and y in cm.

vz

Principles

2 newtons/m 2 (N/m

We

).

In

can also write

of Momentum Transfer and Overall Balances

Eq. (2.2-2) as dv T

(English units)

*y ,9c=

where

2

of lb f/ft

x yz is'in units

.

The. units of viscosity in the cgs system are g/cm the SI system, viscosity

1

cp

=

is

given in

cp

1

cp

1

Other conversion cosity

is

given as

Pa

•

=

x 1(T 3 kg/m-s

1

=

s

called poise or centipoise (cp). In

(N s/m or kg/m x 1(T 3

1

Pa

10~ 4

6.7197 x

=

s

= 0.01

0.01 poise

=

s,

•

2

•

kinematic viscosity,

in

m

2

•

s).

1

x 10~ 3 N-s/m 2

g/cm

(SI)

-s

lb m /ft-s

Appendix

factors for viscosity are given in

fi/p,

(2.4-3)

Sometimes

A.l.

orcm 2 /s, where p

/s

is

the vis-

the density of the

fluid.

EXAMPLE 2.4-1.

Calculation of Shear Stress in a Liquid

Referring to Fig. 2.4-1, the distance between plates

and the (0.0 177 g/cm -s). 10 cm/s,

(a)

fluid

is

Calculate the shear stress

dvjdy using cgs (b) (c)

x yi

is

Ay

K having a

ethyl alcohol at 273

and the

= 0.5

cm, Av z

viscosity of 1.77

= cp

velocity gradient or shear rate

units.

Repeat, using lb force, s, and Repeat, using SI units.

ft

units (English units).

We

can substitute directly into Eq. (2.4-1) or we can integrate Using the latter method, rearranging Eq. (2.4-2), calling the bottom plate point 1, and integrating,

Solution:

Eq.

(2.4-2).

y2 =

0.5

=0

1)2

dv z yi

=0

v,

hz Substituting the

known

y2

To calculate

— y,

0 .35 4

(2.4-5)

yi-yi

=

i

0.0177

\

_ cm

(10 — 0)cm/s ——g— N ~r~; ~~

=

dy

For part (b), using from Appendix A.l, p.

Viscosity of Fluids

dv z

=

•

— 0) cm

j

cm-sy

(0.5

djm 0 354

cm

the shear ratedy./c/y, since the velocity

shear rate

Sec. 2.4

p^—^

values,

—

=

=

(2.4-4)

= 10

Av, (10-0) —= ^—

Ay

lb force units

change

is

nn = ^20.0

s

cm/s

— 0) cm

(0.5

and

1.77 cp(6.7197

=

1.77(6.7197 x lO"

1

^

,„ A (2.4-7)

the viscosity conversion factor

4 x 10"

=

linear with y,

4 )

lb m /ft

lb m

/ft

•

•

s)/cp

s

45

Integrating Eq. (2.4-3),

T >'-=

/*lb m /ft-sK

-» 2 )ft/s

lb m "ft

-y,)ft

(y 2

(Z4- 8)

known values into Eq. (2.4-8) and converting Ai> r 10- lb /ft 2 .Also,rfu / i>; = 20sFor part (c), Ay = 0.5/100 = 0.005 m, Av z = 10/100 = 0.1

Substituting

=

toft, TyI

1.77

x 10

_3

The shear

1

=

1.77

Momentum

x 10~

(1.77 x 10

same

rate will be the

Transfer

in a

(

J

r

kg/ms = x yz

2.4B

l

7.39 x

3

>

and Ay

_3

=

)(0.10)/0.005 at 20.0s~

0.0354

\i

=

N/m 2

1

Fluid

momentum are mass times xyz

ft/s

m/s, and Pa-s. Substituting into Eq. (2-4-5),

The shear stress x y , in Eqs. (2.4-I)-(2.4-3) can also be momentum in the y direction, which is the rate of flow units of

to

'

.

velocity in kg

=

—m

kg -m/s

=

interpreted as a flux of z-directed of

momentum

The

m/s.

per unit area.

momentum -

m

2

s

2

The

shear stress can be written

(2.4-9)

-s

This gives an amount of momentum transferred per second per unit area. This can be shown by considering the interaction between two adjacent layers of a fluid in Fig. 2.4-1

direction.

which have

different velocities,

The random motions of

and hence

the molecules into the slower-moving layer,

layer.

same

momentum,

in the z

some

of

where they collide with the slower-moving

molecules and tend to speed them up or increase their Also, in the

different

the molecules in the faster-moving layer send

momentum

in the z direction.

fashion, molecules in the slower layer tend to retard those in the faster

This exchange of molecules between layers produces a transfer or flux of z-directed

momentum from high-velocity to low-velocity layers. The negative sign in Eq. (2.4-2) indicates that momentum is transferred down the gradient from high- to low-velocity regions. This

2.4C

is

similar to the transfer of heat from high- to low-temperature regions.

Viscosities of

Newtonian Fluids

Fluids that follow Newton's law of viscosity, Eqs. (2.4-i)-{2.4-3), are called Newtonian

For a Newtonian fluid, there is a linear relation between the shear stress tyz and dvjdy (rate of shear). This means that the viscosity n is a constant and independent of the rate of shear. For non-Newtonian fluids, the relation betweenr yl and dvjdy is not linear; i.e., the viscosity does not remain constant but is a function of shear rate. Certain liquids do not obey this simple Newton's law. These are primarily pastes, slurries, high polymers, and emulsions. The science of the flow and deformation of fluids is often called rheology. A discussion of non-Newtonian fluids will not be given

fluids.

the velocity gradient

j.i

here but will be included in Section

The is

viscosity of gases,

3.5.

which are Newtonian

approximately independent of pressure up

fluids,

increases with temperature and

to a pressure

of about 1000 kPa. At higher

pressures, the viscosity of gases increases with increase in pressure. viscosity of 5

N

2

gas at 298

x 10* kPa (Rl). In

K.

approximately doubles

liquids, the viscosity decreases

liquids are essentially incompressible, the viscosity

46

Chap. 2

Principles of

is

in

For example, the

going from 100 kPa

to

about

with increasing temperature. Since not affected by pressure.

Momentum

Transfer and Overall Balances

Table

2.4-1.

Viscosities

of Some Gases and Liquids at 10132 kPa Pressure

Gases

Liquids

Viscosity

K

Substance

Viscosity

(Pa-sJIO 3 or

Temp.,

fcg/m s) •

f

Temp.,

W

Kej.

Substance

Water

Air

293

0.01813

Nl

co 2

273

0.01370

373

0.01828

CH 4

293

0.01089

Rl Rl Rl

SO 2

373

0.01630

Rl

some experimental

3

Ref.

293

1.0019

SI

373

0.2821

SI

Benzene

278

0.826

Rl

Glycerol

293

Hg

293

Olive

In Table 2.4-1

K

(Pa-sJIO or (kglm-s) 10

oil

LI

1069

303

viscosity data are given for

R2

1.55

El

84

some

typical pure,

kPa. The viscosities for gases are the lowest and do not differ markedly -5 from gas to gas, being about 5 x 10" 6 to 3 x 10 Pa-s. The viscosities for liquids are much greater. The value for water at 293 K is about 1 x 10 -3 and for glycerol fluids at 101.32

1.069 Pa-s. Hence, there is a great difference between viscosities of liquids. More complete tables of viscosities are given for water in Appendix A.2, for inorganic and organic liquids and gases in Appendix A.3, and for biological and food liquids in

Appendix

A.4. Extensive data are available in other references (PI, Rl,

Wl,

Me-

LI).

thods of estimating viscosities of gases and liquids when experimental data are not available are summarized elsewhere (Rl). These estimation

sures below 100

methods

for gases at pres-

±5%,

but the

for liquids are often quite inaccurate.

Introduction and Types of Fluid Flow

2.5A

On

methods

are reasonably accurate, with an error within about

TYPES OF FLUID FLOW AND REYNOLDS NUMBER

2.5

The

kPa

principles of the statics of fluids, treated in Section 2.2, are almost an exact science.

the other hand, the principles of the motions of fluids are quite complex.

The

basic

relations describing the motions of a fluid are the equations for the overall balances of

momentum, which will be covered in the following sections. These overall or macroscopic balances will be applied to a finite enclosure or

mass, energy, and

control volume fixed in space.

We

we wish to describe The changes inside the enclosure are deterstreams entering and leaving and the exchanges of

use the term "overall" because

these balances from outside the enclosure.

mined

in

terms of the properties of the

energy between the enclosure and

When making

its

surroundings.

on mass, energy, and momentum we are not interested in the details of what occurs inside the enclosure. For example, in an overall balance average inlet and outlet velocities are considered. However, in a differential balance the velocity distribution inside an enclosure can be obtained with the use of Newton's law of overall balances

viscosity.

In this section

we

first

discuss the two types of fluid flow that can occur: laminar

turbulent flow. Also, the Reynolds considered.

Then

number used

in Sections 2.6, 2.7,

and

to characterize the

2.8 the overall

Types of Fluid Flow and Reynolds

Number

is

mass balance, energy balance,

and momentum balance are covered together with a number of applications.

Sec. 2.5

and

regimes of flow

Finally, a

47

discussion to

is

given in Section 2.9 on the methods of

making

a shell balance on an element

obtain the velocity distribution in the element and pressure drop.

2.5B

Laminar and Turbulent Flow

The type of flow occurring in a fluid in a channel problems. When fluids move through a closed channel can be observed according

distinct types of flow

types of flow can be of flow

is

commonly

seen

in

to the

observed

is

in

important

any cross

in

fluid

conditions present.

when

dynamics two These two

section, either of

a flowing open stream or river.

slow, the flow patterns are smooth. However,

unstable pattern

is

of

When

the velocity

an which eddies or small packets of fluid particles are present all angles to the normal line of flow.

moving in all directions and at The first type of flow at low

velocities

the velocity

where the layers of

another without eddies or swirls being present viscosity holds, as discussed in Section 2.4A.

is

fluid

called laminar flow

The second type

is

quite high,

seem to slide by one and Newton's law of

of flow at higher velocities

where eddies are present giving the fluid a fluctuating nature is called turbulent flow. The existence of laminar and turbulent flow is most easily visualized by the experiments of Reynolds. His experiments are at

shown

in Fig. 2.5-1.

Water was allowed

to flow

steady state through a transparent pipe with the flow rate controlled by a valve at the

end of the pipe. as

shown and

regular

A

its

fine

and formed a

There was no

steady stream of dye-colored water was introduced from a fine jet

flow pattern observed. At low rates of water flow, the dye pattern was

lateral

fluid,

and

flowed

it

in

streamlines

putting in additional jets at other points in the pipe cross section,

was no mixing type of flow

is

in

shown in Fig. 2.5- la. down the tube. By it was shown that there

single line or stream similar to a thread, as

mixing of the

any parts of the tube and the

fluid

flowed in straight parallel

lines.

This

called laminar or viscous flow.

-dye

Ln

water

water -dye streamline

(a)

water

dye

(b)

FIGURE

2.5-1.

Reynolds' experiment for different types offlow

:

( a)

laminar flow, (b)

turbulent flow.

48

Chap. 2

Principles of

Momentum

Transfer

and Overall Balances

As the velocity was increased, it was found that at a definite velocity the thread of became dispersed and the pattern was very erratic, as shown in Fig. 2.5-lb. This type dye is as turbulent flow. The velocity at which the flow changes is known as flow known of the critical velocity.

Number

Reynolds

2.5C

shown

Studies have

from laminar

that the transition

turbulent flow in tubes

to

is

not only

a function of velocity but also of density and viscosity of the fluid and the tube diameter.

These variables are combined into the Reynolds number, which

dimensionless.

— Dvp

N Rc = N Kc

is

(15-1)

number, D the diameter in m, p the fluid density inkg/m 3 jj the fluid viscosity in Pa s, and v the average velocity of the fluid in m/s (where average velocity is defined as the volumetric rate of flow divided by the cross-sectional area of the 3 pipe). Units in the cgs system are D in cm, p in g/cm \i in g/cm s, and v in cm/s. In the 3 English system D is in ft, p in lb^/ft p. in lb m /ft s, and v in ft/s.

where

is

the Reynolds

,

•

•

,

•

,

The

instability of the flow that leads to disturbed or turbulent flow

is

determined by

the ratio of the kinetic or inertial forces to the viscous forces in the fluid stream.

pv

2

/(fiv/D)

sionless

numbers

For a the flow in

is

is

The

2

and the viscous forces to pv/D, and the ratio the Reynolds number Dup/pi. Further explanation and derivation of dimen-

inertial forces are

proportional

to

pu

given in Section 3.11.

is

straight circular pipe

always laminar.

when

When

the value of the Reynolds

the value

very special cases. In between, which

viscous or turbulent, depending

EXA MPLE 23-1 Water at 303 K

is

and SI

in.

at

than 2100,

called the transition region, the flow can be

is

in

a Pipe

the rate of 10 gal/min in a pipe having an inside

Calculate the Reynolds

number

using both Eng-

units.

From Appendix

Solution:

is less

be turbulent, except

will

the apparatus details, which cannot be predicted.

Reynolds Number flowing

diameter (ID) of 2.067 lish units

upon

number

over 4000, the flow

is

A.l, 7.481 gal

=

3 ft

1

.

The

flow rate

is

calculated

as

flow rate

=

f 10.0

mm J

V

pipe diameter,

D=

2.067

^

= ~^~] (V L^) 60 / V7.481

(

gal/

=0.172

r

,

.

velocity in pipe,

v

iiD = —^— =

=

=

-

(^0.0223

A.2 for water at 303

density, p

viscosity,

Sec. 2.5

n

=

/s

2

ti(0.172)

^

From Appendix

3

ft

ft

2

.

cross-sectional area of pipe

0.0223

s

0.0233

^

r

,

ft

=

0.957

ft/s

K (30°C),

0.996(62.43) lb m /ft

3

=

(0.8007 cp) (6.7197 x 10" V

=

4 5.38 x 10" lb ra /ft

Types of Fluid Flow and Reynolds

Number

4

lb -/ ft

"

S

CP

s

49

Substituting into Eq.

(2.5-1),

Dvp

(0.172 ftXO.957 ft/sX0.996

= Hence, the flow

p

=

/ft

0.0525

m

)

lb m /ft-s

= 996 kg/m 3

(2.067 in.)(l ft/12 in.Xl m/3.2808

0.957

-

)

(1

m/3.2808

ft)

cp

=

ft)

=

0.2917 m/s

m

/

=

2.6

3

lb m

4

3 (0.996X100 kg/m )

(

x 62.43

turbulent. Using SI units,

is

D= v

x 10

1.905

=

x 10"

5.38

ix

4

8.007

x 10"

4

•

s

Pa-s

OVERALL MASS BALANCE AND CONTINUITY EQUATION Introduction and Simple

2.6A

Mass Balances

moved from place to place by means of mechanical devices such as pumps or blowers, by gravity head, or by pressure, and flow through systems of piping and/or process equipment. The first step in the In fluid dynamics fluids are in motion. Generally, they are

is generally to apply the principles of the conservation of mass whole system or to any part of the system. First, we will consider an elementary balance on a simple geometry, and later we shall derive the general mass-balance

solution of flow problems to the

equation.

Simple mass or material balances were introduced input Since, in fluid flow,

we

=

output

+

in

Section

is

zero and

rate of input

where

accumulation

are usually working with rates of flow

the rate of accumulation

1.5,

(1.5-1)

and usually

at steady state,

we obtain

=

rate of

In Fig. 2.6-1 a simple flow system

is

output (steady

shown where

state)

(2.6-1)

fluid enters section

1

with an

A

A2 "2

pro cess

Pi

Pi

Figure

50

2.6-1.

Chap. 2

Mass balance on flow system.

Principles

of Momentum Transfer and Overall Balances

average velocity

m/s and density p x kg/m

vl

leaves section 2 with average velocity v 2

l

in

m=

kg/s-m

lb m /s-ft

vp

kg/s. Often, 2 .

is

1

The

v1

cross-sectional area

is in ft/s,

p

is

A

x

m

2 .

The

fluid

balance, Eq. (2.6-1), becomes

= p2 A 2 v2

G=

expressed as

In English units, v

.

The mass

m=p A where

3

vp,

(2.6-2)

where

in lb m /ft

3

G is mass velocity or mass flux A in ft 2 m in lb m /s, and G in ,

,

2 .

•

EXA MPLE 2.6-1.

Flow of Crude Oil and Mass Balance 3 petroleum crude oil having a density of 892 kg/m is flowing through the 3 3 piping arrangement shown in Fig. 2.6-2 at a total rate of 1.388 x 10~ /s

A

m

entering pipe

The 40 pipe

1.

flow divides equally in each of pipes

Appendix A. 5

(see

The

3.

steel pipes are

schedule

for actual dimensions). Calculate the following

using SI units. (a)

(b) (c)

The total mass flow rate m in pipe The average velocity v in 1 and 3. The mass velocity G in 1.

Solution: 2-in. pipe

:

From Appendix A.5, the dimensions of D (ID) = 2.067 in., cross-sectional area

D 3 (ID) =

l|-in. pipe:

1.610 ft

mass flow

is

rate

= (1.388

0.02330(0.0929)

in.,

=

2

A^ = 0.01414

m,

=

2 ft

,

total

the pipes are as follows:

the

x 10

For part

(b),

1

3

(892

l

JH2_ = Pi A)

3

m

3

m2

/^^V 0

2

0.01414(0.0929)

same through -3

m

3

=

1.313 x 10"

pipes

/sX892 kg/m

and 2 and

1

3 )

-

is

1.238 kg/s

/'

3,

v,

1.238 kg/s

_ Pl A

Figure

Overall

2.165 x 10~

using Eq. (2.6-2) and solving for

_

»

=

cross-sectional area

Since the flow divides equally in each of pipes

Sec. 2.6

3.

,

A = 0.02330

The

and pipes

1

kg/m 3 X2.165 x 10" 3

m

2 )

9^19 (892)(1.313 x 10"

2.6-2.

3

1 )

Piping arrangement for Example 2.6-1

Mass Balance and Continuity Equation

51

For part

(c),

G,

=

v1

p

=

1

—

1.238 :

2.165 x 10

33 =

572

s-nr

Control Volume for Balances

2.6B

The laws

conservation of mass, energy, and

for the

momentum are

system, and these laws give the interaction of a system with

all

stated in terms of a

surroundings.

its

A

system

is

defined as a collection of fluid of fixed identity. However, in flow of fluids, individual particles are not easily identifiable.

through which the

which

is

more convenient,

through which the

As a

fluid flows rather is

result, attention is

focused on a given space

than to a given mass of

to select a control volume,

which

fluid. is

used,

fluid flows.

In Fig. 2.6-3 the case of a fluid flowing through a conduit

shown

The method

a region fixed in space

dashed

is

shown. The control

the surface surrounding the control volume. In

most problems part of the control surface will coincide with some boundary, such as the wall of the conduit. The remaining part of the control surface is a hypothetical surface

surface

as a

through which the

line

can flow, shown as point

fluid

control-volume representation

2.6C

is

is

1

and point 2

in Fig. 2.6-3.

The

analogous to the open system of thermodynamics.

Overall Mass-Balance Equation

In deriving the general equation for the overall balance of the property mass, the law of

conservation of mass

may

be stated as follows for a control volume where no mass

is

being generated. rate of

mass output \

/ rate of mass input

from control volume/ rate + (

\from control volume

of mass accumulation\

.

\m

,

control volume

=0

(rate of

mass generation)

/ (2.6-3)

We now

consider the general control volume fixed in space and located in a fluid

For a small element of area dA m 2 on the control surface, the rate of mass efflux from this element = (pv\dA cos a), where (dA cos a) is the area dA projected in a direction normal to the velocity vector v, a is the angle between the velocity vector v and the outward-directed unit normal vector n to dA, and p is the flow

field,

as

shown

in Fig. 2.6-4.

Figure

52

2.6-3.

Chap. 2

Control volume for flow through a conduit.

Principles of Momentum Transfer

and

Overall Balances

density in

kg/m 3 The quantity pv has

kg/s-m 2 and

units of

.

called a flux or

is

mass

velocity G.

From

we

vector algebra

now

p(v-n) dA. If we

recognize that (pv)(dA cos a)

mass across the control entire control volume V: net outflow of

net

mass

mass

surface, or the net

dA

vp cos a

-jj p(v-n)

A

should note that

is

the

from the

mass

if

is

negative. Hence, there

is

dA

(2.6-4)

A

entering the control volume,

across the control surface, the net efflux of mass in Eq. (2.6-4)

and cos a

A we have

efflux in kg/s

efflux

from control volume

We

the scalar or dot product

is

integrate this quantity over the entire control surface

a net influx of mass. If a

<

is

i.e.,

flowing inward > 90°

negative since a

90°, there

is

a net efflux of

mass.

The

rate of

accumulation of mass within the control volume

V can

be expressed as

follows. rate of

mass accumulation

volume

in control

where (2.6-3),

M

is

the mass of fluid

we obtain

in

the general

d_

pdV =

d_M_

(2.6-5)

dt

dt

the volume in kg. Substituting Eqs. (2.6-4) and (2.6-5) into

form of the overall mass balance.

p(s • n)

dA

p

H

dV =

0

(2.6-6)

dt

The use

of Eq. (2.6-6) can be

one-dimensional flow, where

A2

,

as

shown

all

When

in Fig. 2.6-3.

shown

the flow

for a

inward

is

common normal

a!

is

=

1-0-

180°(cosa,

Where

=

v

l

is

directed inward, a l

-1.0). Since a 2

vp cos

is

adA

Overall

0°

and a[

vp cos a 2

v2

Sec. 2.6

situation for steady-state

A and outward normal l

Mass Balance and

p2 A 2

-

is

>

normal loA 2

to

the the direction of the velocity is 0°

the velocity v 2 leaving (Fig. 2.6-3)

angle a 2 between the normal to the control surface and

and cos a 2

to

n/2,

and

is

,

for the case in Fig. 2.6-3,

180°, using Eq. (2.6-4),

dA +

ViPi

A

l

Continuity Equation

vp cos a!

dA

(2.6-7)

53

For steady

state,

dM/dt = 0

in

Eq.

m= which

is

Eq.

(2.6-2),

pt

v

Ai

t

=

(2.6-6)

p2

becomes

A2

v2

(2.6-2)

derived earlier.

we were

In Fig. 2.6-3 and Eqs. (2.6-3)-(2.6-7)

any

and Eq.

(2.6-5),

not concerned with the composition of

can easily be extended to represent an overall mass multicomponent system. For the case shown in Fig. 2.6-3 (2.6-6), and (2.6-7), add a generation term, and obtain

of the streams. These equations

balance for component

i

we combine Eqs. (2.6-5),

in a

± dM

m

p

<

(2.6-8)

component leaving the control volume and R- is the component i in the control volume in kg per unit time. (Diffusion fluxes are neglected here or are assumed negligible.) In some cases, of course, R = 0 for no generation. Often it is more convenient to use Eq. (2.6-8) written in molar units. where

;i

is

the mass flow rate of

i

t

rate of generation of

(

EXA MPLE 2.6-2.

Overall Mass Balance in Stirred Tank a tank contains 500 kg of salt solution containing 10% salt. At point (1) in the control volume in Fig. 2.6-5, a stream enters at a constant flow rate of 10 kg/h containing 20% salt. A stream leaves at point (2) at a constant rate of 5 kg/h. The tank is well stirred. Derive an equation relating Initially,

the weight fraction

Solution: total

mass

Eq.

total

r

in hours.

mass balance using Eq.

(2.6-7) for the net

from the control volume.

vp cos a

From

of the salt in the tank at any time

we make a

First efflux

wA

(2.6-5),

dA

where

= m2 — m = t

M

is

total

—

5

= —5

10

kg solution/h

kg of solution in control volume

d

p

dV =

(2.6-9)

at time

t,

dM (2.6-5)

dt

dt

Substituting Eqs. (2.6-5) and (2.6-9) into (2.6-6),

5+

dM

=

0

dM =

5

and then

integrating,

dt

(2.6-10)

500

dt

Ji=0

M = 5r + 500 Equation

(2.6-1

1)

relates the total

mass

M

in

the tank at any time

initial

10 kg/h

(20%

(2.6-11)

500 kg

solution

salt)

(/

=

t.

salt

0,

10%

salt)

J

control

volume

(2) 5

Figure

54

2.6-5.

Control volume for flow

Chap. 2

Principles

in

kg/h

a stirred tank for Example 2.6-2.

of Momentum Transfer and Overall Balances

Next, making a component A salt balance, let w A = weight fraction of tank at time t and also the concentration in the stream m 2 leaving at time t. Again using Eq. (2.6-7) but for a salt balance,

salt in

I Using Eq.

vp cos a

dA =

(5)yv<

-

10(0.20)

=

-

5w A

2

kg

salt/h

(2.6-12)

^L

kg

S alt/h

(2.6-13)

(2.6-5) for a salt balance, d_

I

dt

dV=j (Mw A

p

=

)

+ WA

t

Substituting Eqs. (2.6-12) and (2.6-13) into (2.6-6),

5w A

-2+

M

from Eq. and solving for w A

Substituting the value for variables, integrating,

5w A -

2

+

(500

+

-

5w A

A

~t = 0

(2.6-14)

(2.6-11) into (2.6-14), separating

,

+ wM

5f)

2

M^f+w

+

(500

+

=

0

5w A =

0

-5

'-

dt

dw. -~ +

5f)

dt

w^

1

=

o.io 2

-

10w„

,

500

=0

+

500

lOv

5t

+

5f

(2.6-15)

10

500

1

w A = -0.1

100

J00 +

+

0.20

£

(2.6-16)

Note that Eq. balance with

2.6D

R

(2.6-8) for {

=

component

Average Velocity

Use

to

in

In solving the case in Eq. (2.6-7)

constant

v 2 at

section

i"

could have been used for the

salt

0 (no generation).

2. If

we assumed a constant

the velocity

an average or bulk velocity

is

Mass Balance

Overall

is

velocity u, at section

1

and

not constant but varies across the surface area,

defined by 1

v

for a surface

over which

v is

normal

to

A and

dA

the density p

(2.6-17)

is

assumed constant.

EXA MPLE 2.6-3.

Variation of Velocity A cross Control Surface and Average Velocity For the case of imcompressible flow (p is constant) through a circular pipe of radius R, the velocity profile is parabolic for laminar flow as follows

(2.6-18)

Sec. 2.6

Overall

Mass Balance and

Continuity Equation

55

u mal is the maximum velocity at the center where r =.0 and v is the velocity at a radial distance r from the center. Derive an expression for the

where

average or bulk velocity u av to use

in the overall

mass-balance equation.

The average velocity is represented by Eq. (2.6-17). In Cartesian coordinates dA is dx dy. However, using polar coordinates which are more appropriate for a pipe, dA = r dr d8, where 8 is the angle in polar coordi1 nates. Substituting Eq. (2.6-18), dA = r dr dd, and A = nR into Eq. (2.6-17)

Solution:

and

integrating,

(2.6-19)

(2.6-20)

mass balances were made because we wish In this section on overall mass balances, some of the equations presented may have seemed quite obvious. However, the purpose was to develop the methods which should be helpful in the next sections. Overall balances will also be made on energy and momentum in the next sections. These overall balances do not tell us the details of what happens inside. However, in Section 2.9 a shell momentum balance will be made to obtain these details, which will give us the velocity distribution and pressure drop. To further study these details of the processes occurring inside the enclosure, differential balances rather than shell balances can be written and these are discussed in other later Sections 3.6 to 3.9 on differential equations of continuity and momentum transfer, Sections 5.6 and 5.7 on differential equations of energy change and boundary-layer flow, and Section 7.5B on differential equations In this discussion overall or macroscopic

to describe these balances

from outside

the enclosure.

of continuity for a binary mixture.

OVERALL ENERGY BALANCE

2.7

2.7A

Introduction

The second property energy.

We

shall

fixed in space in to

to be considered in the overall balances

on a control volume

is

apply the principle of the conservation of energy to a control volume

much

the

same manner as the principle of conservation of mass was used The energy-conservation equation will then be

obtain the overall mass balance.

combined with

the

first

law of thermodynamics to obtain the

final overall

energy-balance

equation.

We can write the

first

law of thermodynamics as

AE = Q -

W

(2.7-1)

where £ is the total energy per unit mass of fluid, Q is the heat absorbed per unit mass of fluid, and is the work of all kinds done per unit mass of fluid upon the surroundings.

W

56

Chap. 2

Principles of Momentum Transfer

and Overall Balances

must be expressed

In the calculations, each term in the equation as J/kg (SI), btu/lb B

,

or

ft

•

Since mass carries with

we

physical state,

we can

associated energy because of

it

will find that

balance. In addition,

in the

same

units,

such

lb f /lb m (English).

each of these types of energy

its

will

position, motion, or

appear

in the

energy

boundary of the system

also transport energy across the

without transferring mass.

Derivation of Overall Energy-Balance Equation

2.7B

The

entity balance for a conserved quantity such as energy

similar to Eq. (2.6-3) and

is

is

as follows for a control volume.

rate of entity output

The energy E 1.

—

rate of entity input

+

rate of entity

mass

in

a gravitational

The

reference plane.

z

zg/g c in

is

at a given point.

lb f /lb OT

ft

v

ft-lb f/lb m

of a unit

mass of a

and vibrational energy

in

.

the velocity in m/s relative to the

is

Again

J/kg. In the English system the kinetic energy

U

.

the energy present because of ti anslational

is

or rotational motion of the mass, where

boundary of the system

2

•

•

2

Internal energy

m

m/s Multiplying and can be expressed as (kg m/s 2 ) (m/kg), or J/kg. In

Kinetic energy v /2 of a unit mass of fluid

rotational

(2.7-2)

the relative height in meters from a

is

units for zg for the SI system are

English units the potential energy

3.

0

the energy present because of the position

is

where

field g,

dividing by kg mass, the units

2.

=

present within a system can be classified in three ways.

Potential energy zg of a unit mass of fluid of the

accumulation

v

is

system the units o(v 2 /2 are

in the SI 2

/2g c in ft lb f /lb m other energy present, such as •

.

fluid is all of the

chemical bonds. Again the units are

in

J/kg or

.

The total energy

of the fluid per unit

mass

is

then

U+- + zg

(SI)

(2.7-3)

£=[/+

—+— zg

v

2<7 C

The

rate of

(English)

gc

V

accumulation of energy within the control volume f rate of energy accumulation^

c

\m

ct

control volume

/

in Fig. 2.6-4

—+

U +

zg)p

dV

is

(2.7-4)

Next we consider the rate of energy input and output associated with mass in the control volume. The mass added or removed from the system carries internal, kinetic, and potential energy. In addition, energy is transferred when mass flows into and out of the control volume. Net

work

is

done by the

fluid as

volume. This pressure-volume work per unit mass

work

is

usually neglected.

it

flows into and out of the control

fluid is

The pV term and U term

are

pV The .

contribution of shear

combined using

the definition of

enthalpy, H.

H = U + pV Hence, the

Sec. 2.7

total

energy carried with a unit mass

Overall Energy Balance

is

(H +

(2.7-5) v

2

/2

+

zg).

57

dA on

For a small area

+

2

+

the control surface in Fig. 2.6-4, the rate of energy efflux

is

where (dA cos a) is the area dA projected in a direction normal to the velocity vector v and a is the angle between the velocity vector v and the outward-directed unit normal vector a. We now integrate this quantity over the entire

(H

v

/2

zg)(pv)(dA cos

a),

control surface to obtain net energy efflux

\

H +—+

from control volume/

zg

dA

pv) cos a

(2.7-6)

Now we have accounted for all energy associated with mass in the system and moving across the boundary in the entity balance, Eq. (2.7-2). Next we take into account heat and work energy which transfers across the boundary and is not associated with mass. The term q is the heat transferred per unit time across the boundary to the fluid because of a temperature gradient. Heat absorbed by the system is positive by convention.

The work W, which is energy per unit time, can be divided into H^, purely mechanical shaft work identified with a rotating shaft crossing the control surface, and work, which has been included

the pressure-volume

By convention, work done by the system,

is

To

in the

H in Eq. (2.7-6).

enthalpy term

upon the surroundings,

fluid

work out of

i.e.,

obtain the overall energy balance,

H

+

iT — + zg

2

v

we

substitute Eqs. (2.7-4)

and equate the resulting equation tog

entity balance Eq. (2.7-2)

6

\

cos x

}(pv)

dA

u

-i

ot

J

W

—

s

and

pdV =

+1 + zg

(2.7-6) into the

.

q-W

(2.7-7)

s

Overall Energy Balance for Steady-State Flow System

2.7C

A common

special case

of the overall or

macroscopic energy balance

is

that of a

steady-state system with one-dimensional flow across the boundaries, a single single outlet, inlet

and

negligible variation of height

or outlet area. This

(2.7-7)

shown

is

z,

density p, and enthalpy

in Fig. 2.7-1. Setting the

inlet,

a

H across either

accumulation term

in Eq.

equal to zero and integrating.

H m 2

2

+

f/,m,

—

—

2f,

For steady is

the

positive.

state,

on a unit mass

m,

=

z

3 (i-

,

x

z

= m2 =

m. Dividing through by

)

+

x

=

q

-

W

m so

s

(2.7-8)

that the equation

(SI)

3V /(2i av )

and

is

is

can be replaced by v\J2i, where a

equal to f* v /(tf 3 ), v

\ for laminar flow

2.7D.) Hence, Eq. (2.7-9)

«2

A

- gm

1 s

correction factor and flows in pipes

av

z,

basis,

H -H The term

p,

+ gm 2

2u lav

3v

.

The term

and close to

is

(2.7-9)

the kinetic-energy velocity

i has been evaluated for various

1.0 for turbulent flow. (See Section

becomes

-//,+-

(c| av

1-x

-

v\

J + g(z

2

-

z.)

=Q

(SI)

(2.7-10) "?..)

58

Chap. 2

+ -(z J

Principles of

-z = G- ws )

(English)

Momentum

Transfer and Overall Balances

1

Figure

Some

useful conversion factors to be used are as follows

btu

=

778.17

ft

hp

=

550

lb f /s

lb f /lb m

=

2.9890 J/kg

1

1

1

ft

•

1

2.7D

J.

Steady-state flow system for a fluid.

2.7-1.

J=

1

ft

•

•

=

lb f

N-m =

=

1055.06 J

=

from Appendix A.l 1.05506 kJ

kW

0.7457

kg-m 2/s 2

1

Kinetic-Energy Velocity Correction Factor a

Introduction.

In obtaining Eq. (2.7-8)

was necessary

it

to integrate the kinetic-energy

term,

kinetic energy

(pv) cos a

dA

(2.7-11)

which appeared in Eq. (2.7-7). To do this we first take p as a constant and Then multiplying the numerator and denominator by v 3y A, where u av is average velocity and noting that m = pv zv A, Eq. (2.7-1 1) becomes

(u

3 )

dA =

(v

3 )

dA =

2u,„A

Dividing through by

m

so that Eq. (2.7-12)

-J 1 2d

where a

is

A

(u 2i>„„

(u

is

3 )

on

a unit

A

mass

3 )

dA

cos a

=

1.0.

the bulk or

(2.7-12)

basis,

dA =

(2.7-13)

2y„

2a

defined as

(2.7-14)

Sec. 2.7

Overall Energy Balance

59

and

(u

3 is

) a¥

defined as follows: 3 (v

The

)„

=

(v

3 )

dA

(2.7-15)

local velocity v varies across the cross-sectional area of a pipe.

uasa

hence, the value of a, we must have an equation relating

To

evaluate(u

3 )

av

function of position

and,

in the

cross-sectional area.

2.

Laminar flow.

In order to determine the value of a for laminar flow,

Eqs. (2.6-18) and (2.6-20) for laminar flow to obtain

v

=

2t)

v

2.6-3),

r.

(2.7-16)

Hi)"]

Substituting Eq. (2.7-16) into (2.7-15) .and noting that

Example

we first combine

as a function of position

A = n R 2 and dA =

r

dr d9 (see

Eq. (2.7-15) becomes rR

*2k

3

r

Jo (27r)2

3

3 t;

dr dO

j0

«

a

2

(R

-

r

2 3 )

r dr

=

(R

2

-

r

2 3 )

(2.7-17)

r rfr

Integrating Eq. (2.7-17) and rearranging,

(Av =

~ [V

6

-

3r

2

K* + 3r*R 2

-

6

r )r rfr

2vl

(2.7-18)

Substituting Eq. (2.7-1 8) into (2.7-14),

3

(v )„

Hence,

for

laminar flow the value of

at

3

(2.7-19)

0.50

2v „

term of Eq.

to use in the kinetic-energy

(2.7-10)

is

0.50.

3.

Turbulent flow.

For turbulent flow a relationship

is

needed between

v

and

position.

This can be approximated by the following expression

(2.7-20)

where

r

(2.7-15)

is

the radial distance from the center. This Eq. (2.7-20)

and the

resultant integrated to obtain the value of (u

substituted into Eq.

is

3 ) av

.

Next, Eq. (2.7-20)

is

3 and this equation integrated to obtain v iy and (u av) 3 3 Combining the results for (u av and(u av into Eq. (2.7-14X the value of a is 0.945. (See Problem 2.7-1 for solution.) The value of a for turbulent flow varies from about 0.90 to 0.99. In most cases (except for precise work) the value of a is taken to be 1.0.

substituted into Eq. (2.6-17)

.

)

60

Chap. 2

)

Principles of

Momentum

Transfer

and Overall Balances

2.7E

The

Applications of Overall Energy-Balance Equation total

energy balance, Eq.

(2.7-10), in the

form given

is

not often used when appreci-

added (or subtracted) since the kinetic- and potential-energy terms are usually small and can be neglected. As a result, when appreciable heat is added or 'subtracted or large enthalpy changes occur, the methods of doing heat balances described in Section 1.7 are generally used. Examples will be given to illustrate this and other cases.

able enthalpy changes occur or appreciable heat

EXAMPLE

is

Energy Balance on Steam Boiler and 137.9 kPa through a pipe at an average above the liquid inlet velocity of 1.52 m/s. Exit steam at a height of 15.2 leaves at 137.9 kPa, 148.9°C, and 9.14 m/s in the outlet line. At steady state how much heat must be added per kg mass of steam? The flow in the two 2.7-1.

Water enters a

boiler at 18.33°C

m

pipes

is

turbulent.

The

Solution:

process flow diagram

Eq. (2.7-10) and setting a -

shown

is

in Fig. 2.7-2.

Rearranging

and

Ws

(no external

+ (H 2 -

Hi)

for turbulent flow

1

= 0

work),

Q= To

(z 2

-

+

z,)g

(17-21)

solve for the kinetic-energy terms, v\

(1-52)

_

1.115 J/kg

2

2

v\

(9.14)

2

41.77 J/kg

~ 2 Taking

the

datum

From Appendix

A.2,

g

=

2

,

=

steam

z2

=

15.2

=

m.Then,

149.1 J/kg

H t at 18.33°C 2771.4 kJ/kg, and

tables in SI units,

148.9°C

-

2771.4

1,

(15.2X9.80665)

H 2 of superheated steam at

H -

2

height z, at point z2

2

76.97

=

=

2694.4 kJ/kg

=

=

2.694 x 10

76.97 kJ/kg,

6

J/kg

Substituting these values into Eq. (2.7-21),

Q

=(149.1

Q =

-

189.75

0)

+

+

6 1.115)+ 2.694 x 10

-

(41.77

2.694 x 10

6

=

2.6942 x 10

6

J/kg

Hence, the kinetic-energy and potential-energy terms totaling 189.75 6 J/kg are negligible compared to the enthalpy change of 2.694 x 10 J/kg. This

189.75 J/kg

would

raise

the

temperature of liquid water about

0.0453°C, a negligible amount.

steam

~1

Q

15.2

water d,

m

v2

=9.14 m/s

148.9°C, 137.9 kPa

= 1.52 m/s

18.3°c, 137.9 kPa Figure

Sec. 2.7

2.7-2.

Overall Energy Balance

Process flow diagram for Example 2.7- J.

61

cooler

m

20

Q

Figure

EXAMPLE Water

Energy Balance on a Flow System with a Pump

2.7-2.

85.0°C

at

Process flow diagram for energy balance for Example 2.7-2.

2.7-3.

being stored

is

shown

in

a large, insulated tank at atmospheric

pumped

steady state from this The motor driving the pump supplies energy at the rate of 7.45 kW. The water passes through a of heat. The cooled water is then heat exchanger, where it gives up 1 408 pressure as

tank

at

point

1

in Fig. 2.7-3. It

pump

by a

is

being

at the rate of

0.567

at

m 3 /min.

kW

delivered to a second, large open tank at point first

which

2,

is

second tank. Neglect any kinetic-energy changes since the velocities in the tanks are essentially zero.

From Appendix

Solution:

J/kg,p,

=

1/0.0010325

Also, Z[

=

work

H

tables, for

above the

.

=

(0.567X968.5X^o)

initial

vl)/2

=

x 10

is

also negative since

3

9.152 kg/s

it

fluid

gives

= -

J/sXl/9.152 kg/s)

final

x

is

W

s

-(7.45 x 10 3 J/s)(l/9.152 kg/s) = -0.8140 x

3

and

(85°C) = 355.90 x 10 steady state,

W

Q = -(1408 Setting^ -

steam

kg/m 3 Then,

= 20 m. The work done by the done on the fluid and s is negative.

heat added to the fluid

H2 -

A.2,

968.5

0 and z 2 is

Ws = The

=

= m2 =

mi

case

m

20

tank. Calculate the final temperature of the water delivered to the

,

10

but in this

3

J/kg

up heat and

153.8 x 10

3

is

J/kg

0 and substituting into Eq. (2.7-10),

355.90 x 10

3

+ 0 +

9.80665(20

= (-

-

0)

153.8 x 10

H2 =

3 )

-

(-0.814 x 10 3

)

3 Solving, 202.71 x 10 J/kg. From the steam tables this corresponds to r 2 = 48.41'C. Note that in this example, s and g{z 2 — z,) are very small

compared

EXAMPLE A

W

to Q.

Energy Balance

2.7-3.

flow calorimeter

calorimeter, which

is is

in Flow Calorimeter being used to measure the enthalpy of steam. The a horizontal insulated pipe, consists of an electric

heater immersed in a fluid flowing at steady state. Liquid water at

0°C

at a

kg/min enters the calorimeter at point 1. The liquid is vaporized completely by the heater, where 19.63 kW is added and steam leaves point 2 at 250°C and 150 kPa absolute. Calculate the exit enthalpy H 2 of the steam if the liquid enthalpy at 0°C is set arbitrarily as 0. The rate of 0.3964

62

Chap. 2

Principles of

Momentum

Transfer

and Overall Balances

kinetic-energy changes are small and can be neglected. that pressure has a negligible effect

For

Solution:

and

1

mi

=

this case,

m = 0.3964/60 = 2

W

s

=

0 since there

v\/2a)

is

no

shaft

=

will

be

assumed

liquid.)

work between points For steady state,

0 and g(z 2 — z,) = 0. -3 6.607 x 10 kg/s. Since heat

—

Also, (i>|/2a

2.

(It

on the enthalpy of the

is

added to the

system,

Q^ The value

of//!

=

0.

+

Equation // 2

The

final

"'""{I, kg/s

6.607 x 10

(2.7-10)

I

H2 =

'

becomes

-// +0 +

equation for the calorimeter

=2971 kJ/kg5

3

0

=

Q- 0

is

Q +

H

(2.7-22)

l

Q = 297 1 kJ/kg and H x = 0 into Eq. (2.7-22), // 2 = 2971 kJ/kg 250°C and 150 kPa, which is close to the value from the steam table of

Substituting at

2972.7 kJ/kg.

2.7F

Overall Mechanical-Energy Balance

A more

useful type of energy balance for flowing fluids, especially liquids,

is

a modifi-

cation of the total energy balance to deal with mechanical energy. Engineers are often

concerned with

work

this special

type of energy, called mechanical energy, which includes the

term, kinetic energy, potential energy, and the flow

work part

of the enthalpy term.

Mechanical energy is a form of energy that is either work or a form that can be directly converted into work. The other terms in the energy-balance equation (2.7-10), heat terms

and internal energy, do not permit simple conversion into work because of the second law of thermodynamics and the efficiency of conversion, which depends on the temperatures. Mechanical-energy terms have no such limitation and can be converted almost completely into work. Energy converted to heat or internal energy is lost work or a loss in mechanical energy which is caused by frictional resistance to flow. It is convenient to write an energy balance in terms of this loss, £ F, which is the

sum

For the case of steady-state flow, when a unit outlet, the batch work done by the fluid, W, is

of all frictional losses per unit mass.

mass of

fluid passes

from

inlet

to

expressed as

W This work

W

differs

from the

pdV-J^F

W

potential-energy effects. Writing the

(XF>0)

of Eq. (2.7-1), first

(2.7-23)

which also includes kinetic- and this case, where AE

law of thermodynamics for

becomes AU,

AU = Q The equation

(2.7-24)

defining enthalpy, Eq. (2.7-5), can be written as

AH = AU + ApV = AU +

Sec. 2.7

W

Overall Energy Balance

pdV+\

r

V dp

(2.7-25)

63

Substituting Eq. (2.7-23) into (2.7-24) and then combining the resultant with Eq. (2.7-25),

we obtain

AH = Q + Yj F + we

Finally,

substitute Eq. (2.7-26) into (2.7-10)

V

dp

(2.7-26)

and 1/p

for

V, to obtain the overall

mechanical-energy-balance equation

^-[fLv-«l,v] + 2a

For English units the

byg c

+

f +

Y

j

w

and potential-energy terms of Eq.

kinetic-

=

s

o

(2.7-27)

(2.7-27) are divided

-

The

value of the integral in Eq. (2.7-27) depends on the equation of state of the fluid

and the path of the process. (p 2

dp

+

ff(Z2-Zl)

—

Pi)/p and Eq. (2.7-27)

_1_

If

the fluid

v]j +

2a

is

an incompressible

liquid, the integral

g{z 2

~

zi)

+

Pi

~P

-

+

Y,f+ws = o

EXAMPLE 2.7-4.

Mechanical-Energy Balance on Pumping kg/m 3 is flowing at a steady through a uniform-diameter pipe. The entrance pressure 2 68.9 kN/m abs in the pipe, which connects to a pump

Water with

a density of 998

supplies 155.4 J/kg of fluid flowing in the pipe. is

the

same diameter

becomes

becomes

as the inlet pipe.

The

The

exit

(2.7-28)

System mass flow

rate

of the fluid

is

which actually pipe from the pump

exit section of the pipe is 3.05

m

2 higher than the entrance, and the exit pressure is 137.8 kN/m abs. The Reynolds number in the pipe is above 4000 in the system. Calculate the frictional loss F in the pipe system.

£

Solution:

First a flow

diagram is drawn of the system (Fig. 2.7-4), with added to the fluid. Hence, s = — 155.4, since

W

155.4 J/kg mechanical energy the

work done by Setting the

the fluid

is

positive.

datum height

zl

=

0,

z2

=

3.05 m. Since the pipe

is

of

u2

= Pi 137.8

3.05

Pi =

68.9

m

"

kN/m 2

Figure 2.7-4.

64

kN/m 2

Chap. 2

Process flow diagram for Example 2.7-4.

Principles of Momentum Transfer

and Overall Balances

=

constant diameter, u t

z2

v2

g=

.

Also, for turbulent flow a

(3.05

2 mX9.806 m/s )

=

=

1.0

and

29.9 J/kg

Since the liquid can be considered incompressible, Eq. (2.7-28) Pl

68.9 x 1000

p

998

£2

p

Using Eq.

(2.7-28)

_ ~

137.8 x 1000

and solving

for

X

values,

=

„2)

138.0 J/kg

56.5 J/kg

_

+^

j

+

(2.7.29)

p

and solving

F = -(-155.4) +

=

69.0 J/kg

£ F, the frictional losses,

2

known

used.

998

£ F = _ w + J(p _ za Substituting the

=

is

0

-

for the frictional losses,

+

29.9

69.0

-

138.0

(l8.9^

EXAMPLE 2.7-5. Pump Horsepower in Flow System A pump

draws 69.1 gal/min of a liquid solution having a density of 3 from an open storage feed tank of large cross-sectional area through a 3.068-in.-ID suction line. The pump discharges its flow through a 2.067-in.-ID line to an open overhead tank. The end of the discharge line is 50 ft above the level of the liquid in the feed tank. The friction losses in the F = 10.0 ft-lb force/lb mass. What pressure must the piping system are pump develop and what is the horsepower of the pump if its efficiency is 114.8 lbjjft

£

65%

(r\

=

Solution: 5).

0.65)? First,

Equation

The

flow

a

flow

(2.7-28) will

is

turbulent.

diagram of the system

be used. The term

W

5

in

is

drawn

(Fig.

Ws =- n W„ where

Sec. 2.7

— Ws =

mechanical energy actually delivered to the

Overall Energy Balance

2.7-

Eq. (2.7-28) becomes (2.7-30) fluid

by the

65

pump or net mechanical work, or shaft

work delivered

From Appendix 0.05134

v 1

l

=

2 ft

and of the

flow rate

=

V2

=

atm and p 2

=

1

^

M =

atm. Also, a

The flow

Using the datum of z, =

Z2

is

the energy

,

t

p,

=

turbulent. Hence,

_ 067g

ft-lb f

lb.

'

-

0

32.174

c (\ rw

9c

+

The pressure

„„ft-lb ^32^=-5a °lo7

=

9c

ft3 ft> /s

we have

9

50.0

2

2(32.174)

0,

£

1539 ,539

°0

ft/s

== 0.

is

is

p

(2.4-28), solving for

-

W

rate is

^) ==

1.0 since the flow

(6.6 1)

_

2g c

0

^ .

very large. Then v]/2g c

is

v\

=

2

ft

(0.1539^(3^) = 6.61

p

Using Eq.

and

A.5, the cross-sectional area of the 3.068-in. pipe

tank

0, since the

fractional efficiency,

2.067-in. pipe, 0.0233

69,

(

—

t]

pump.

to the

Ws

f

and substituting the known

,

^9c

^9c

0.678

+

values,

P

0

-

10

=

-60.678

^r^1 Ib m

Using Eq.

(2.7-30)

and solving

Ws _

=

To

rate

p

,

60.678 0.65

1

mass flow

W

for

=

(

=

3.00 hp

0.1539

ft

-lb f

_

ft-

lb m

114.8

lb f

lb m

'

^

=

17.65

^

pump must develop, Eq. (2.7-28) must be between points 3 and 4 as shown on the

calculate the pressure the

written over the

pump

itself

diagram.

^^(^^tXcIo^)vi

—

Vi

=

3

-

00

^

6.61 ft/s

Since the difference in level between z andz^ of the pump itself is negligible, 3 it will be neglected. Rewriting Eq. (2.7-28) between points 3 and 4 and

66

Chap. 2

Principles

of Momentum Transfer and Overall Balances

substituting

E±^£i =

known

Z3

P 0

(£

F=

2 9c

9c

0

this

is

for the pipingsystem),

-0 + -

0

+

^--^ 0.140

(2 . 7 . 31)

2 9c

+

60.678

-0

2(32.174)

2(32.174)

=

0 since

£- ZA £. + A_A_ Ws _ lF 9c

.

values

- 0.678 +

=

60.678

60.14

lb„

Pa,

- Pz = 48.0 lb force/in.

2

developed by pump)

(psia pressure

(331

kPa)

Bernoulli Equation for Mechanical-Energy Balance

2.7G

In the special case where no mechanical energy

QT-F

=

flow,

which

0),

is

(Ws =

added

0)

and

for

no

friction

then Eq. (2.7-28) becomes the Bernoulli equation, Eq. (2.7-32), for turbulent is

of sufficient importance to deserve further discussion.

z

This equation covers

many

l

g+—2 + — =z p

2

g

+

— +— 2

situations of practical

(2.7-32)

p importance and

is

often used in

conjunction with the mass-balance equation (2.6-2) for steady state.

m=p A l

Several examples of

its

i>i

= p 2 A 2 v2

vl

(2.6-2)

use will be given.

EXAMPLE 2.7-6. A

l

Rate of Flow from Pressure Measurements

liquid with a constant density

3 p kg/m

is

flowing at an

m/s through a horizontal pipe of cross-sectional area

N/m 2

unknown

A m2 v

velocity

at a pressure

and then it passes to a section of the pipe in which the area is 2 reduced gradually to A 2 m and the pressure is p 2 Assuming no friction losses, calculate the velocity v and v 2 if the pressure difference (p, — p 2 ) is measured.

Pi

,

.

y

diagram is shown with pressure taps to and p 2 From the mass-balance continuity equation (2.6-2), for constant p where Pi = p 2 = p, Solution:

measure

In Fig. 2.7-6, the flow

p^

.

v2

=

V

-^ A

(2.7-33)

2

v2

A,

m

2

J/

d,

1

Figure

Sec. 2.7

2.7-6.

Overall Energy Balance

m/s

m/s

A2 m

Process flow diagram for Example 2.7-6.

67

the items in the Bernoulli equation (2.7-32), for a horizontal pipe,

For

=

Zi

Then Eq. (2.7-32) becomes,

Eq. (2.7-33) for v 2

after substituting

+ Hi +

0

=0

z2

2

= 0- +

£i P

^M 2

,

+-

(2.7-34)

p

Rearranging,

Pi

p^[(/l,//l 2 )

— p2 = Dl

=

-

2

~

/

J

Pi

~

1]

(2.7-35)

2

P2

aw-i]

p

(SI)

(2.7-36) I

Pi-

Pi

29c

Performing the same derivation but

terms of v 2

in

,

~Pi

Pi

=

v2

(English)

2 L(AJA 2 ) -12

P

(2.7-37)

_ {A J Ay)

1

EXAMPLE 2.7-7.

Rate of Flow from a Nozzle in a Tank nozzle of cross-sectional area A 2 is discharging to the atmosphere and is located in the side of a large tank, in which the open surface of the liquid in

A

isHm above the center line of the nozzle. Calculate the velocity v 2 nozzle and the volumetric rate of discharge if no friction losses are

the tank in the

assumed. Solution:

The process flow

is

shown

liquid at the entrance to the nozzle

with point

in Fig. 2.7-7,

and point 2

1

taken

in

the

of the nozzle. Since A^ is very-large compared to A 2 u, = 0. The pressure p y is 2 greater than 1 atm (101.3 kN/m ) by the head of fluid of m. The pressure = 0 exit, which is at the nozzle is at 1 atm. Using point 2 as a datum, 2 p2 and z ( = 0 m. Rearranging Eq. (2.7-32), at the exit

,

H

,

Zi3 Substituting the

known

+

^T 2

+

=

z2 3

—

+

(2.7-38)

2

p

values, 2

0

+

0

+

Pl

-

Pl

=

P Solving for

v2

0

^2

(2.7-39)

m/s

(2-7-40)

+

,

2(Pl

=

Pz)

P

Figure

2.7-7.

Nozzle flow diagram for Example

2.7-7.

Hm JL "»

68

Chap. 2

r

Principles of Momentum Transfer

and Overall Balances

Since Pi

— p 3 = Hpg and p 3 =

p 2 (both

at 1 atm),

H = Pi ~ Pl

m

(2.7-41)

pg where

H

is

the head of liquid with density p.

rate

becomes (2.7-42)

is

=

flow rate

To

(2.4-40)

= JlgH

v2

The volumetric flow

Then Eq.

m 3 /s

A2

v2

(2.7-43)

points can be used in the balance,

illustrate the fact that different

points 3 and 2 will be used. Writing Eq. (2.7-32),

Z2g

V + l + Rl^2± =

I

Since p 2

=

p3

=

1

atm,

=

y3

0,

2.8

2.8A

OVERALL

and

z2

(2.7-44)

L

= 0,

=

v2

v

+ J-

Z3 g

p

= JlgH

(2.7-45)

MOMENTUM BALANCE

Derivation of General Equation

A momentum balance can be written for the control volume shown in Fig. 2.6-3, which is somewhat similar to the overall mass-balance equation. Momentum, in contrast to mass and energy, is a vector quantity. The total linear momentum vector P of the total mass of a moving fluid having a velocity of v is

M

P = Mv

(2.8-1)

M

My

is the momentum of this moving mass enclosed at a particular instant in volume shown in Fig. 2.6-4. The units of Mv are kg m/s in the SI system. Starting with Newton's second law we will develop the integral momentum-balance equation for linear momentum. Angular momentum will not be considered here. Newton 's law may be stated The time rate of change of momentum of a system is equal to the summation of all forces acting on the system and takes place in the direction of the

The term

the control

:

net force.

I where F

is

force. In the SI

the SI system

gc

is

not needed, but

The equation can be written

sum

system F

=

—

(2.8-2)

newtons (N) and 1 N = 1 kg m/s needed in the English system.

is in

it is

for the conservation of

•

momentum

2 .

Note

that in

with respect to a control volume

as follows:

of forces acting\

on control volume /

/ rate of

Overall

momentum

\

/ rate of

\out of control volume/

+

Sec. 2.8

F

/ rate of accumulation of .

.

control volume

momentum\

,„

„

„

(2.8-3)

.

\in control volume

Momentum Balance

\J nto

momentum

/

69

This

is

in the

same form

as the general mass-balance equation (2.6-3), with the

momentum

forces as the generation rate term. Hence,

generated by external forces on the system.

If

sum

of the

not conserved, since

is

external forces are absent,

is

it

momentum

is

conserved.

Using the general control volume shown in Fig. 2.6-4, we shall evaluate the various terms in Eq. (2.8-3), using methods very similar to the development of the general mass balance. For a small element of area dA on the control surface, we write rate of

Note

momentum

that the rate of mass efflux

is

=

efflux

{pv\dA cos

y{pv)(dA cos a)

Also, note that {dA cos a)

a).

projected in a direction normal to the velocity vector v and a velocity vector v

product

(2-8-4)

and the outward-directed-normal vector becomes

n.

is

is

the area

dA

the angle between the

From

vector algebra the

in Eq. (2.8-4)

\{pv)(dA cos a)

=

dA

pv(v-n)

(2.8-5)

Integrating over the entire control surface A, I net

momentum

efflux^

=

\from control volume

The

net efflux represents the

first

v(pu)cos

a.

dA =

jj

pv(v-n)

two terms on the right-hand side of Eq.

Similarly to Eq. (2.6-5), the rate of accumulation of linear

control volume

V

accumulation of

momentumA

Substituting Equations

d

balance for

a

/

within the

(2.8-2), (2.8-6),

and

dV

(2.8-7)

dt

(2.8-7) into (2.8-3), the overall linear

We should note that £

F

in

general

dA +

—

may have

a

pv

dV

component

(2.8-8)

in

any direction, and the

the force the surroundings exert on the control-volume fluid. Since Eq. (2.8-8)

vector equation,

mo-

control volume becomes

pv(v- n)

is

(2.8-3).

momentum

py

volume

in control

F

(2.8-6)

is

rate of

mentum

dA

J"

we may

write the

component

scalar equations for the x, y,

is

a

and z

directions.

vx

pu cos a dA

+

—

pv x

dV

(SI)

ct

(2.8-9)

T.FX =

ux

— v cos

a dA

+

v

y

—

—u

r

dV

(English)

ct

9c

pv cos a dA

+

pudV

(2.8-10)

dV

(2.8-11)

ci

•

v.

pv cos a dA

+

*

—

f

pv :

ct

70

Chap. 2

Principles of

Momentum

Transfer

and Overall Balances

The

£F

force term

Eq. (2.8-9)

x in

composed

is

of the

sum

of several forces. These

are given as follows. 1.

mass

total is

2.

The body

Body force.

Fxg is the x-directed

force

force caused by gravity acting on the

M in the control volume. This force, Fxg

,

Mg x

is

zero

It is

.

the

if

x direction

horizontal.

The

Pressure force.

force

F xp

the x-directed force

is

acting on the surface of the fluid system.

the pressure

cases part of the control surface

the control surface.

caused by the pressure forces

the control surface cuts through the

taken to be directed inward and perpendicular to the surface. In

fluid,

some

is

When

Then

there

outside of this wall, which

is

is

may be

a solid,

and

this wall

a contribution to

Fxp

from

typically

atmospheric pressure.

If

is

included inside

the pressure

gage pressure

on the used,

is

the integral of the constant external pressure over the entire outer surface can be

automatically ignored. 3.

When

Friction force. present,

which

between the

the fluid

flowing,

is

an x-directed shear or

exerted on the fluid by a solid wall

is

fluid

and the solid

wall. In

which

,

is

includes a section of pipe

The

component of the

the x

and

the fluid

may

it

is

resultant of the forces acting on the

contains. This

when

the control

volume

the force exerted by the solid

is

force terms of Eq. (2.8-9) can then be represented as

x

F xg + F xp + F xs + R x

Similar equations can be written.for the y and z directions.

x

this frictional force

fluid.

EF = the

is

neglected.

control volume at these points. This occurs in typical cases surface on the

F xs

In cases where the control surface cuts through a solid, there

4. Solid surface force.

Rx

is

friction force

the control surface cuts

some or many cases

be negligible compared to the other forces and present force

when

(2.8-12)

Then Eq.

(2.8-9)

becomes

for

direction,

£

F x = F *9 + F x P + F xs + R X vr

pv cos

a.

dA

d

pv x

H

dV

(2.8-13)

dt

Overall

2.8B

in

A

quite

One

Momentum

common

flowing

at

in

Flow System

application of the overall

steady state in

Equation

its

f

=F

Integrating with cos a

=

+ F

becomes as follows since v =

+ F +R =

+1.0 andp/1

= m/v av

F *s + F x P + F xs + R X =

Sec. 2.8

Overall

is

the case of a

axis in the

(2.8-13) for the x direction

V

momentum-balance equation

x direction. The fluid will be assumed to be the control volume shown in Fig. 2.6-3 and also shown in Fig.

section of a conduit with

2.8-1.

Balance

Direction

Momentum Balance

»>

cos a

dA

vx

.

(2.8-14)

,

— ^ -m

(2.8-15)

71

Figure 2.8-1.

where

if

the velocity

is

Flow through a horizontal nozzle

in the

x direction

only.

not constant and varies across the surface area,

dA

vl

(2.8-16)

A

The

ratio (u^) av /u iav

is

replaced by u xav//f, where

/?,

which

the

is

momentum

velocity

and I for laminar flow. For most applications in turbulent Rov/,(v x)lJv x „ is replaced by u iav the average bulk velocity. Note that the subscript x on vx andF^ can be dropped since v x = v andF^ = F

correction factor, has a value of 0.95 to 0.99 for turbulent flow

,

for

one directional flow. The term F xp which ,

control volume,

the force caused

is

by the pressures acting on the surface of the

is

-

Pi^i

The

F xs =

0.

acting only in the y direction. Substituting

F

friction force will be neglected in

since gravity

is

(2.8-17)

(2.8-15), replacing (v x ) z Jv xz ,

by

v/fi

Eq. (2.8-15), so

=

(whereyJ[av

v),

setting/}

=

The body

force

from Eq. 1.0,

F

=

0

(2.8-17) into

and solving fovR x

in

Eq. (2.8-15),

R x = mv 2 — wu, + where

Rx

is

the force exerted by the solid

(reaction force)

is

the negative of this or

EXAMPLE

on

— Rx

p2

A2 -

the fluid.

(2.8-18)

Pi/lj

The

force of the fluid on the solid

.

Momentum

Velocity Correction Factor p for Laminar Flow The momentum velocity correction factor /? is defined as follows for flow one direction where the subscript x is dropped. 2.8-1.

in

V.

av

(2.8-19)

v.av

P = Determine p Solution:

for

laminar flow

(2.8-20)

in a tube.

UsingEq.(2.8-16),

v

2

dA

(2.8-21)

A

72

Chap. 2

Principles of Momentum Transfer

and Overall Balances

Substituting Eq. (2.7-16) for laminar flow into Eq. (2.8-21) A = iiR 2 anddA = r dr d9, we obtain (see Example 2.6-3)

~^— -^l (2n)2

= Integrating Eq. (2.8-22)

2

vj f" (R 2

r

and noting

that

2 2 )

rdr

(2.8-22)

and rearranging, (2.8-23)

Substituting Eq. (2.8-23) into (2.8-20), P

EXAMPLE 2 JI-2. Momentum Balance for Horizontal Nozzle flowing at a rate of 0.03154 m /s through a horizontal Water 3

nozzle

is

and discharges to the atmosphere at point 2. The nozzle is attached at the upstream end at point 1 and frictional forces are considered negligible. The upstream ID is 0.0635 m and the downstream 0.O286 m. Calculate the resultant force on the nozzle. The density of the 3 water is 1000 kg/m

shown

in Fig. 2.8-1

.

mass flow and average or bulk

Solution:

First, the

and 2 are

calculated.

x 10"

3

m 2 andX 2 = m,

The v2

=

velocities at points

area at point 1 is /t, = (7r/4X0.0635) 2 2 Then, (tt/4)(0.0286) = 6.424 x 10'*

The

m

= m2 = m =

=

(0.03154X1000)

=

I

3.167

.

31.54 kg/s

point 1 is t>, = 0.03154/(3.167 x 10~ 0.03154/(6.424 x 10~") = 49.1 m/s. velocity

2

at

3 )

= 9.96

m/s and

To evaluate the upstream pressure p t we use the mechanical-energy balance equation (2.7-28) assuming no frictional losses and turbulent flow. (This can be checked by calculating the Reynolds number.) This equation then becomes, for a = 1.0, l±

El

+

=

Et

+

El

1

=

Setting p 2

=

0 gage pressure, p

(2.8-24)

P

1000 kg/m

3

1.156 x 10

6

,

vt

=

9.96 m/s,

v2

=

49.1 m/s,

and solving for p lt 2

(1000X49.1

-

9.96

2 )

Pi

For

the

x direction, the

Substituting the

Rx =

known

=

momentum

N/m 2

(gage pressure)

balance equation (2.8-18)

is

used.

values and solving forK^,

31.54(49.10

-

9.96)

= -2427 N(-546

+

0

-

(1.156

x 10 6 X3.167 x 10" 3 )

lb f )

Since the force is negative, it is acting in the negative x direction or to the left. This is the force of the nozzle on the fluid. The force of the fluid on the solid is —R. or +2427 N.

2.8C

Overall

Momentum

Balance

Another application of the overall system with

Sec. 2.8

fluid entering

Overall

in

Two

Directions

momentum

a conduit at point

Momentum Balance

1

balance

is

shown

inclined at

in Fig. 2.8-2 for a

an angle of ct 1

flow

relative to the

73

Pi

Figure

Overall

2.8-2.

momentum balance for flow system

and leaving

at

with fluid entering at point

horizontal x direction and leaving a conduit at point 2 at an angle

assumed

be flowing at steady state and the (actional force

to

1

2.

F X5

tx

The

.

z

will

fluid will

be

be neglected. Then

Eq. (2.8-13) for the x direction becomes as follows for no accumulation:

+

Fx,

+

F* P

K

vx

dA

pv cos a

(2.8-25)

A

Integrating the surface (area) integral,

F„ + F xp The term the term

2

(u

Fxp

) av

/t> av

Rx

4-

=m

can again be replaced by

cos

o>

2

-m

with

u av //?

f}

cos a,

being

(2.8-26)

From

set at 1.0.

Fig. 2.8-2,

is

F %p — Pi^i Then Eq.

—

(2.8-26)

becomes as follows

R x = nw 2

cos a 2

—

cos a i

~~

Pi &i cos a 2

after solving for

mu cos t

a,

+

p2

A2

Rx

(2.8-27)

:

cos a 2

—

p

l

A

l

cos a,

(2.8-28)

The term F xg = 0 in this case. For R y the body force F is in the negative y direction and F yg = —rn,g, where m, is the total mass fluid in the control volume. Replacing cos a by sin a, the equation for the y direction becomes

R y = mv 2

sin a 2

—

wu,

sin a,

+

p2

A2

sin

a2

—

PiA

{

sin a!

+

m,g

(2.8-29)

EXAMPLE 2.8-3. Momentum Balance in a Pipe Bend Fluid

is

flowing at steady state through a reducing pipe bend, as shown in Turbulent flow will be assumed with frictional forces negligible.

Fig. 2.8-3.

The volumetric flow rate of the liquid and the pressure p 2 at point '2 are known as are the pipe diameters at both ends. Derive the equations to calculate the forces on the bend. Assume that the density p is constant. The velocities u, and v 2 can be obtained from the volumetric flow rate and the areas. Also, m = p v A = p 2 v 2 A 2 As in Example 2.8-2, the mechanical-energy balance equation (2.8-24) is used to obtain the upstream pressure, p,. For the x direction Eq. (2.8-28) is used for the mo-

Solution:

l

74

Chap. 2

l

.

l

Principles of

Momentum

Transfer and Overall Balances

-

l

2

.

P2

y

Figure

mentum

Flow through a reducing bend

2.8-3.

balance. Since a,

R x = mv 2

cos

a.

2

=0°, cos

— mv + x

A2

p2

a,

=

cos a 2

in

Example

Equation

1.0.

— PiA

2.8-3.

becomes

(2.8-28) (SI)

t

(2.8-30)

Rx =

m

—

cos a 2

v2

—

9C

m

—

t)j

For the y direction the sin

=

oti

Pi

Az

cos a 2

—

(English)

P\A^

momentum

balance Eq. (2.8-29)

is

is

mass

total

+

sin a 2

p2 A 2

sin a 2

fluid in the pipe bend.

+

The

m,g

(SI)

(2.8-31)

pressures at points

are gage pressures since the atmospheric pressures acting on cancel.

used where

0.

R = mv 2 where m,

+

9c

The magnitude

control volume fluid

this

1

and 2

surfaces

of the resultant force of the bend acting on the

is

|R|

The angle

all

=JR1 +R

(2.8-32)

makes with the vertical is~ 8 = arctan (RJR t Often the is small compared to the other terms in Eq. (2.8-31) and is ).

Fw

gravity force

-

neglected.

EXAMPLE 2.8-4.

Sudden Enlargement a fluid flows from a small pipe to a large pipe through an abrupt expansion, as shown in Fig. 2.8-4. Use the momentum balance and mechanical-energy balance to obtain an expression for the loss for a liquid. (Hint: Assume that p 0 = p and v 0 = v Make a mechanical-energy balance between points 0 and 2 and a momentum and 2. It will be assumed that p and p 2 are balance between points Friction Loss in a

A mechanical-energy

loss occurs

when

l

t

1

t

uniform over the cross-sectional area.)

Figure

2.8-4.

Losses

in

expansion flow.

1

® )

1

—

—1

"o

1

1

^= 1

1

Sec. 2.8

Overall

Momentum Balance

X

.

The control volume is selected so that R x drops out. The boundaries selected

Solution:

pipe wall, so

it

does not include the

are points

1

and

2.

The

flow through plane 1 occurs only through an area of A 0 The frictional drag force will be neglected and all the loss is assumed to be from eddies in this volume. Making a momentum balance between points 1 and 2 using .

Eq. (2.8-18) and noting that p 0

=

pu

PiA 2

—

p2

=

vx

v0

,

and A^

A 2 = mv 2 — mv

= A2

(2.8-33)

l

rate is m = v 0 pA 0 and v 2 = (AJA 2 )v 0 terms into Eq. (2.8-33) and rearranging gives us

The mass-flow

A2 V

A 2J

Finally, combining Eqs. (2.8-34)

2.8D

Overall

Momentum

a free

=

(2.7-28) to points

^

1

and

2,

(2-8-35)

(2.8-35),

Balance for Free Jet

impinges on a fixed vane as

jet

Substituting these

Vane

Striking a Fixed

When

and

.

p

Applying the mechanical-energy-balance equation

^-ZF

,

in Fig. 2.8-5 the overall

momentum

balance

can be applied to determine the force on the smooth vane. Since there are no changes

in

and after impact, there is no loss in energy and application of the Bernoulli equation shows that the magnitude of the velocity is unchanged. Losses due to impact are neglected. The frictional resistance between the jet and the smooth vane is also neglected. The velocity is assumed to be uniform throughout the jet upstream and downstream. Since the jet is open to the atmosphere, the pressure is the same at all ends elevation or pressure before

of the vane.

making

In

a

momentum

balance for the control volume shown for the curved vane

in Fig. 2.8-5a, Eq. (2.8-28) is written as follows for

are zero, v,

=

v2

,A

l

= A 2 and m = ,

R x = mv 2 Using Eq.

cos

<x

(2.8-29) for the y direction

R y - mv 2 Hence,

The

2

steady state, where the pressure terms

A p =

v2

A2

— mv +

0

=

v-l

i

l

\

p2

:

mu,(cos a 2

and neglecting the body sin a 2

-

0

= mv

x

—

(2.8-37)

1)

force,

sin a 2

(2.8-38)

R x and R y

force

are the force components of the vane on the control volume components on the vane are — R x and — R y

fluid.

.

EXAMPLE 2JI-5. is

Force of Free Jet on a Curved, Fixed Fane m/s and a diameter of 2.54 x 10" 2 deflected by a smooth, curved vane as shown in Fig. 2.8-5a, where

a2

=

A jet

60°.

What

3 1000 kg/m

76

m

of water having a velocity of 30.5

is

the force of the jet

on the vane? Assume

that p

=

.

Chap. 2

Principles of Momentum Transfer

and

Overall Balances

p

FIGURE

Free jet impinging on a fixed vane: (a) smooth, curved vane, (b) smooth, flat vane.

2.8-5.

The

Solution:

cross-sectional

of

area

the

jet

A =

is

t

jr(2.54

4 2 2 2 4 x l0- ) /4 = 5.067 x 10" m Then, m = M1P1 = 30.5 x 5.067 x 10" = x 1000 15.45 kg/s. Substituting into Eqs. (2.8-37) and (2.8-38), .

The

Rx =

15.45 x- 30.5 (cos 60°

Ry =

15.45 x 30.5 sin 60°

force

on the vane

resultant force

is

- Rx =

is

=

= -235.6 N(- 52.97

1)

408.1 N(91.74 lb f)

+235.6

In Fig. 2.8-5b a free jet at velocity u, strikes a

no

loss in energy.

It

there

is

no tangential

must equal

is

No

parallel to the plate.

the final

and

m

3

-R =

and

-408.1 N. The

y

is

momentum

exerted on the fluid by the

Then, the

force.

smooth, inclined

velocities are all equal (u,

initial

balance

flat

where m,

is

=

and

v 3)

in the

the flow

since there

p direction

plate in this direction;

this direction.

in

plate

v2

momentum component

momentum component to Eq. (2.8-26),

flat

=

This means

kg/s entering at

i.e.,

the p direction

in

1

and

£ F p = Q. m2

leaves

at 3, »

By

whose

convenient to make a

force

Writing an equation similar at 2

N

calculated using Eq. (2.8-32).

divides into two separate streams is

lb ( )

X FP =

'

0

=

0

= m2

n»2 v 2

D!

~m

1

—m

l

v

v

1

l

cos a 2

— m3 u3

cos

— m 3 u,

cc

2

(2.8-39)

the continuity equation,

m,

Combining and

m, —

m,

(1

+ cos

a,),

m 3 = -y

resultant force exerted by the plate on the fluid

resultant force

Sec. 2.8

is

(2.8-40)

solving,

m2 = The

= m2 + m3

simply

Overall

m^,

Momentum

sin a 2

.

(1

-

cos a 2 )

must be normal

to

Alternatively, the resultant force

Balance

(2.8-41)

it.

This means the

on

the fluid can be

77

calculated by determining

The

(2.8-32).

2.9

force

R x and R y

on the bend

from Eqs.

(2.8-28)

and

2.8-29)

and then using Eq.

the opposite of this.

is

SHELL MOMENTUM BALANCE AND VELOCITY PROFILE IN LAMINAR FLOW^

2.9A

Introduction

we analyzed momentum balances using an overall, macroscopic control this we obtained the total or overall changes in momentum crossing the control surface. This overall momentum balance did not tell us the details of what happens inside the control volume. In the present section we analyze a small control volume and then shrink this control volume to differential size. In doing this we make a shell momentum balance using the momentum-balance concepts of the preceding section, and then, using the equation for the definition of viscosity, we obtain an expression In Section 2.8

From

volume.

for the velocity profile inside the

enclosure and the pressure drop.

The equations

are

derived for Mo w systems of simple geometry in laminar flow at steady state. In

many

engineering problems a knowledge of the complete velocity profile

maximum

needed, but a knowledge of the

on a surface

is

needed. In this section

is

not

velocity, average velocity, or the shear stress

we show how

to

obtain these quantities from the

velocity profiles.

2.9B

Shell

Momentum

Balance Inside a Pipe

Engineers often deal with the flow of fluids inside a circular conduit or pipe. In Fig. 2.9-1

we have a

horizontal section of pipe in which an incompressible Newtonian fluid

flowing in one-dimensional, steady-state, laminar is

flow..

The flow

is

fully

developed;

is

i.e., it

not influenced by entrance effects and the velocity profile does not vary along the axis

of flow in the x direction.

The

cylindrical control

volume

is

a

shell with

length Ax. At steady state the conservation of follows:

sum

momentum

of forces acting

into volume.

The

pressure forces

on control volume

an inside radius

momentum,

=

rate of

Eq.

r,

and becomes as

thickness Ar,

(2.8-3),

momentum

out

—

rate of

pressure forces become, from Eq. (2.8-17),

= pA

\

x

—

pA

\

x + ^x

=

p(2nr Ar)

\

x

—

p(2nr Ar)

(2.9-1)

\

r

x

Ax Figure

2.9-1.

—

Control volume for shell

momentum balance on

a fluid flowing in a

circular tube.

78

Chap. 2

Principles of

Momentum

Transfer and Overall Balances

The shear stress

momentum mentum

in

and

momentum

net convective

since the flow .x is

is

the shear

can also be considered as the rate of

this

efflux

is

the rate of

momentum

out

—

mo-

rate of

is

net efflux

u x at

cylindrical surface at the radius r

flow into the cylindrical surface of the shell as described by Eq. (2.4-9).

Hence, the net rate of

The

on the

force or drag force acting

r„ times the area 2nr Ax. However,

is

momentum

fully

equal to vx at

Equating Eq.

=

(x rx

— {x rx 2%r

2nr Ax)| r + A,

flux across the

Ax)| r

annular surface

at

(2.9-2)

x and x + Ax

developed and the terms are independent of .x

4-

x.

This

is

is

zero,

true since

Ax.

(2.9-1) to (2.9-2)

and rearranging,

KJUa, -KJIr

r(p\ x

Ar

-p\ Ax

(2.9-3)

In fully developed flow, the pressure gradient (Ap/Ax) is constant and becomes (Ap/L), where Ap is the pressure drop for a pipe of length L. Letting Ar approach zero, we obtain d(rx r (2.9-4)

dr

Separating variables and integrating,

Ap (2-9-5)

2

The constant

of integration

C must t

be zero

if

r

the

momentum

flux

is

not infinite at

r

=

0.

Hence,

Ap\

Po

- Pl (2.9-6)

momentum flux varies linearly with maximum value occurs at r = R at the wall.

This means that the 2.9-2,

and

the

the radius, as

shown

in Fig.

Substituting Newton's law of viscosity, dv x

(2.9-7)

Ir

vx

= 0

v

imix

•parabolic velocity profile

I

momentum flux profile

r rx

Figure

Sec. 2.9

Shell

2.9-2.

Velocity and

0

^rxmax

momentum flux

Momentum Balance and

profiles for laminar flow in a pipe.

Velocity Profile in

Laminar Flow

79

into Eq. (2.9-6),

we obtain

the following differential equation for the velocity:

dv x

Po-Pl

dr

2pL

(2.9-8)

=

Integrating using the boundary condition that at the wall,i; x

0 at

r

=

R, we obtain the

equation for the velocity distribution.

1- -

R'

This result shows us that the velocity distribution

The average velocities

(2.9-9)

0]

is

parabolic as

velocity u XJV for a cross section

shown in Fig. 2.9-2. summing up all

found by

is

over the cross section and dividing by the cross-sectional area as

Following the procedure given

in

Example

2.6-3,

where dA

=r

dr dd and

in

the

Eq. (2.6-17).

A = nR 2

,

R V

-=

Combining Eqs.

1

v^dA =

vr x

7lR*

~A

(2.9-9)

and

o

J

1

=

dr dO

vx

:

nR

,

2nr dr

(2.9-10)

and integrating,

(2.9-10)

-

(Po

Pi)R

2

(Po

-

Pi)D

J

(2.9-11)

where diameter

D=

relates the pressure

The maximum

2R. Hence, Eq.

(2.9-11),

drop and average velocity velocity for a pipe

is

which

for

found from Eq.

Eqs. (2.9-1

1)

and

(2.9-12),

we

the Hagen-Poiseuille equation,

(2.9-9)

and occurs

at

r

=

R

4^L Combining

is

laminar flow in a horizontal pipe. 0.

(2.9-12)

find that

(2.9-13)

2.9C

Shell

We now

Momentum

Balance

for Falling

Film

use an approach similar to that used for laminar flow inside a pipe for the case of

down

flow of a fluid as a film in laminar flow

used to study various phenomena

The control volume for the considered is Ax thick and sufficiently far

in

mass

falling film

is

shown

has a length of

from the entrance and

a vertical surface. Falling films have been

on surfaces, and so on. 2.9-3a, where the shell of fluid

transfer, coatings

L

in Fig.

in the vertical z direction.

exit regions

so that the flow

is

This region

is

not affected by these

means the velocity y.( x) does not depend on position z. we set up a momentum balance in the z direction over a system Ax thick, in bounded in the z direction by the planes z = 0 and z = L, and extending a distance the y direction. First, we consider the momentum flux due to molecular transport. The rate of momentum out-rate of momentum in is the momentum flux at point x Ax minus that at x times the area LW. regions. This

To

start

W

-I-

net efflux

The 80

net convective

momentum Chap. 2

= LW(x x: )\ x + &x - LW(x X! )\ x

flux

is

the rate of

Principles

momentum

entering the area

(2.9-14)

AxH/

at

of Momentum Transfer and Overall Balances

momentum

by

in

convection

- gravity force

momentum in

velocity

by

profile

momentum

out by molecular transport

molecular transport

7xx

M

momentum flux profile

momentum

out by convection (b)

(a)

Figure

Vertical laminar flow of a liquid film : (a) shell momentum balance for thick; (b) velocity and momentum flux profiles.

2.9-3.

a control volume

z v.

= L minus at

z

=

L

This net efflux

is

equal

to

0 since

v z at z

=

0

is

equal to

each value of x. net efflux

The

= 0.

that leaving at z

for

Ax

gravity force acting

on

= AxWv z (pv z )\

the fluid

(2.8-3) for the

— &xWv

z

(pv z )\ zs=0

=

0

= AxWL(pg)

conservation of

momentum

(2.9-16)

at

steady state,

AxWL(pg) = LW(r x:)\ x + ^ x - LW{r xz )\ x + 0 Rearranging Eq. (2.9-17) and

letting

(2.9-15)

is

gravity force

Then using Eq.

= L

z

Ax—

(2.9-17)

0,

Ax

<-xz\x

(2.9-18)

99

Ax t-x,

= 99

(2.9-19)

tlx

Integrating using the boundary conditions at x at

x

=

x, r xz

= t„

This means the momentum-flux profile

Shell

0, r x ,

—

0 at the free liquid surface and

,

t xz

Sec. 2.9

=

Momentum Balance and

= pgx is

linear

(2.9-20)

as

shown

Velocity Profile in

in

Fig. 2.9-3b

Laminar Flow

and the

81

maximum

value

is at

For a Newtonian

the wall.

T

Combining Eqs.

and

(2.9-20)

dv z

=

~

(2.9-21)

Tx

we obtain

(2.9-21)

using Newton's law of viscosity,

fluid

the following differential equation for the

velocity:

d

P9

=

-2i

(2.9-22)

dx Separating variables and integrating gives

(2.9-23)

Using the boundary condition that distribution equation

v2

=

0 at x

=

C =

5,

l

2

(pg/2p)5

Hence, the velocity

.

becomes (2.9-24)

l

in This means the velocity profile velocity occurs at x

=

is

_

(2.9-25)

(2.6-17).

1

1

v.

The maximum

2

pgs

by using Eq.

velocity can be found

Fig. 2.9-3b.

in

is

max

The average

shown

parabolic as

0 in Eq. (2.9-24) and

dA

dx dy

v,

=

W uT

Wb

dx

(2.9-26)

Substituting Eq. (2.9-24) into (2.9-26) and integrating, 2

pg» (2.9-27)

Combining rate q

is

Eqs. (2.9-25)

and

we obtain

(2.9-27),

u :av

=

(2/3)«. mai

.

The volumetric

flow

obtained by multiplying the average velocity u 2av times the cross-sectional area

dW. q

—

=

m

3

(2.9-28)

/s

3^ Often in falling as

films, the

mass

rate of flow per unit width of wall

r = p5v. 3V and a Reynolds number

is

T

in kg/s

_ 4 P 5p r.. Wv Re _ — IE —

for /V Re

1200.

Laminar flow with rippling present occurs above a

of25.

EXAMPLE 2.9-1. An is

82

<

defined

P

P

Laminar flow occurs

is

(2.9-29)

'

]V R£

m

defined as

oil is

flowing

820 kg/m

3

Falling Film Velocity

down

and Thickness

a vertical wall as a film

and the viscosity

Chap. 2

is

0.20

Pas.

1.7

mm

thick.

The

oil

density

Calculate the mass flow rate per

Principles of Momentum Transfer

and Overall Balances

unit width of wall, T, needed and the Reynolds number. Also calculate the average velocity.

The

Solution:

film thickness

=

5

is

0.0017 m. Substituting Eq. (2.9-27) into

the definition of T, 2

(p5)pg6

3

2

p 5 g

* 2

= Using Eq.

=

JTo5o~

°- 053

" kg/s m '

(2.9-29),

Hence, the film

laminar flow. Using Eq.

is in

pg5

2

(2.9-27),

3 2 820(9.806X1.7 x lO" )

=

0.03873 m/s

3(0.20)

3fi

DESIGN EQUATIONS FOR LAMINAR

2.10

AND TURBULENT FLOW 2.1

(2.9-30)

x 1Q- 3 ) 3 (9.8Q6)

(820) (1.7

OA

One

Velocity Profiles

in

IN PIPES

Pipes

of the most important applications of fluid flow

pipes,

and

tubes.

Appendix A. 5 gives

sizes of

flow inside circular conduits,

is

commercial standard

Schedule

steel pipe.

40 pipe in the different sizes is the standard usually used. Schedule 80 has a thicker wall and will withstand about twice the pressure of schedule 40 pipe. Both have the same outside diameter so that they will

same outside diameters

as

steel

the

fit

pipe

same to

Pipes of other metals have the

fittings.

permit

interchanging parts of a

piping

system. Sizes of tubing are generally given by the outside diameter and wall thickness. Perry

When

and Green (PI) give detailed tables of various types of tubing and pipes. and the velocities are measured at different

fluid is flowing in a circular pipe

distances from the pipe wall to the center of the pipe,

laminar and turbulent flow, the fluid

near the walls. These measurements are

entrance to the pipe. Figure 2.10-1 pipe versus the fraction of

given position and u max the

In

is

velocity v'/v m2X

where

v

a true parabola, as derived

velocity u max

is

local velocity at the

in

Eq. (2.9-9).

The

velocity

relationship between

and

v m3X

measured values of Dv z ,pl p and Do max p/n. tally

The average

maximum

is

useful, since in

some

u av

can be used

v a Jv max

velocity over the

to

determine

v zv

On

.

In Fig.

2.

whole cross section of the pipe

is

and the average

velocity

is

Design Equations for Laminar and Turbulent Flow in Pipes

numbers

precisely 0.5 times

momentum

balance

the other hand, for turbulent flow, the curve

flattened in the center (see Fig. 2.10-1)

this

10-2 experimen-

are plotted as a function of the Reynolds

velocity at the center as given by the shell

(2.9-13) for laminar flow.

u av in a

cases only the v mix at the

measured. Hence, from only one point measurement

is

Sec. 2.10

,

velocity at the center of the pipe. For viscous or

engineering applications the relation between the average velocity

maximum

center point of the tube

the

at a reasonable distance from the

zero.

many

pipe and the

is

made

is

a plot of the relative distance from the center of the

maximum maximum

laminar flow, the velocity profile at the wall

is

shown that in both moving faster than the

has been

it

fluid in the center of the pipe

is

in

Eq.

somewhat

about 0.8 times the

83

u o c a C >
~0

0.2

FIGURE

2.10-1.

0.6

0.4

Fraction of

maximum

0.8

1.0

velocity (v'/v mix

)

Velocity distribution of a fluid across a pipe.

maximum. This value of 0.8 varies slightly, depending upon the Reynolds number, as shown in the correlation in Fig. 2.10-2. (Note See Problem 2.6-3, where a value of 0.817 :

is

derived using the-y-power law.)

2.1QB

Pressure

Drop and

Laminar Flow

Friction Loss in

/.

Pressure drop and loss due to friction.

in

a pipe, then for a

Newtonian

fluid the

When

the fluid

shear stress

is

is in

steady-state laminar flow

given by Eq.

(2.4-2),

which

is

rewritten for change in radius dr rather than distance dy, as follows.

(2.10-1)

dr

Using

this relationship

and making

a shell

momentum

balance on the

fluid

over a

Du au p/fi

10

10-

Figure

84

2.10-2.

U

Ratio u av /u m „ as a function of Reynolds number for pipes.

Chap. 2

Principles of

Momentum

Transfer

and Overall Balances

cylindrical shell, the Hagen-Poiseuille equation (2.9-11) for laminar flow of a liquid in

circular tubes

was obtained.

momentum balance. This

A derivation

A/V =

where p,

is

tube, m.

[p\

-

For English

The quantity

D

is

=

p 2)f

upstream pressure at point

velocity in tube, m/s;

also given in Section 3.6 using a differential

is

can be written as

1,

inside diameter,

—

p 2 ) f or Ap f

constant p, the friction loss

Ff

f _

L,)

L

(2.10-2)

g2

N/m 2

p2

;

m; and

units, the right-hand side

(p l

-

32pv(L 2

,

(L 2

pressure at point 2; u

— L,)

of Eq. (2.10-2)

AL

or

is

divided

is

average

length of straight

is

bygc

.

the pressure loss due to skin friction. Then, for

is

is

(Pi

—

Pi)f

N -m

_

J

or

(SI)

kg

kg

P

(2.10-3)

F/

This is

is

1?

(En 8 lish)

the mechanical-energy loss due to skin friction for the pipe in

part of the

(2.7-28).

~

=

£F

This

term for frictional losses

term

(Pi—p 2 )f

for

skin

N

•

m/kg

of fluid

and

mechanical-energy-balance equation

in the

loss

friction

different

is

from

the

owing to velocity head or potential head changes in Eq. (2.7-28). That part of £ F which arises from friction within the channel itself by laminar or turbulent flow is discussed in Sections 2.10B and in 2. 10C. The part of friction loss due to fittings (valves, elbows, etc.), bends, and the like, which sometimes constitute a large part of the friction, is discussed in Section 2. 10F. Note that if Eq. (2.7-28) is applied to steady flow in (Pi

Pi) term,

a straight, horizontal tube,

One

we obtain (p, — p 2 )l p ~Yj^-

is in the experimental measurement of the viscosity of by measuring the pressure drop and volumetric flow rate through a tube of known length and diameter. Slight corrections for kinetic energy and entrance effects are

a

of the uses of Eq. (2.10-2)

fluid

usually necessary in practice. Also, Eq. (2.10-2)

is

often used in the metering of small

liquid flows.

EXA MPLE, 2.10-1. Metering of Small Liquid Flows A small capillary with an inside diameter of 2.22 x

m

10" 3 and a length used continuously measure the flow rate of a liquid being to 0.317 m 3 3 Pas. The pressurehaving a density of 875 kg/m and p. = 1.13 x 10~ drop reading across the capillary during flow is 0.0655 m water (density 996 3 /s if end-effect corrections are neglected? kg/m 3 ). What is the flow rate in is

m

Solution:

Assuming

that the flow

is

laminar, Eq. (2.10-2) will be used. First, water to a pressure drop using Eq.

m

to convert the height h of 0.0655 (2.2-4),

A P/ = hpg =

m)^996

(0.0655

= 640 kg m/s 2 •

^^9.80665 P

m = 2

640

N/m 2

Substituting the following values into Eq. (2.10-2) of p.

=

1.13

Design Equations for Laminar and Turbulent Flow

in

Pipes

Sec. 2.10

x 10" 3 Pa-s,

85

L 2 - L, for

=

0.317 m,

D=

3

2.22 x 10"

Ap f = 640 N/m 2 and

m, and

,

solving

v,

AP/

=

32/u
-L

n2 D

t )

"

(2-10-2) 3

MU "

32(1.13 x i0- Xt>X0-317) 3 2 (2.22 x 10 )

v

The volumetric

rate

is

=

0.275 m/s

then 2

.

volumetric

a

now

=

rate

tm

D = —

10" 0.275(^X2.22 x

4

will

it

was assumed

that laminar flow

be calculated to check

-6

3

/s

occurring, the Reynolds

is

3 * 1Q" X0-275X875) 3 1.13 x 10-

(

is

m

number

this.

Z22 m <~ Dv P Nr p ~ Hence, the flow

)

4

.= 1.066 x 10 Since

3 2

"

473

laminar as assumed.

Use offriction factor for friction loss in laminar flow. A common parameter used in laminar and especially in turbulent flow is the Fanning friction factor, f which is defined

2.

as the drag force per wetted surface unit area (shear stress t, at the surface) divided by the

product of density times velocity head ov^pv

2

The force isApy times the cross-sectional and the wetted surface area is 27T.R AL. Hence, the relation between the pressure drop due to friction and/is as follows for laminar and turbulent flow. area tiR

.

2

Ap f nR 2

Rearranging,

this

\p_l_

2nR Al\

pv /2

becomes

AL — 2

v

Ap/ = 4fp

(SI)

(2.10-5)

AL — &Pf = 4/p D

F ff =

-

2

v

-

(English)

2g c

Ap AL -^ = 4f r

D

p

v

2

(SI)

2 (2.10-6)

f

AL

2

^ 7^ v

(En * llsh)

4/

For laminar flow only, combining Eqs.

(2.10-2)

and

(2.10-5),

16

16

N Rc

Dvp/p

Equations (2.10-2), (2.10-5), (2.10-6), and (2.10-7) for laminar flow hold up to a Reynolds number of 2100. Beyond that at an iV Re value above 2100, Eqs. (2.10-2) and (2.10-7) do not hold for turbulent flow. For turbulent flow Eqs. (2.10-5) and (2.10-6),

86

Chap. 2

Principles

of Momentum Transfer and Overall Balances

however, are used extensively along with empirical methods of predicting the

friction

factor /, as discussed in the next section.

EXAMPLE

2.10-2. Use of Friction Factor in Laminar Flow same known conditions as in Example 2.10-1 except that the velocity of 0.275 m/s is known and the pressure drop Ap is to be predicted. f Use the Fanning friction factor method.

Assume

the

Solution:

The Reynolds number Dvp

Rc_ ii

From

Eq.

(2.

(2.22 x IP"

~

is,

3

as before, 3

mXO.275 m/sX875 kg/m 3

10-

1.13 x

)

kg/m-

10-7) the friction factor/is

= -^r =

/=

AL =

Using Eq. (2.10-5) with 3 p = 875 kg/m

0.0338

0.317 m, v

(dimensionless)

= 0.275

D=

m/s,

2.22 x 10"

3

m,

,

ALv 2 - 4(0.0338X875X0.3 17)(0.275) , An „, N/ m ^ =640 2

Ar Ap,-4fp

,

^lo^)

T1

Example

This, of course, checks the value in

2.10-1.

When the fluid is a gas and not a liquid, the Hagen-Poiseuille equation be written as follows for laminar flow: m=

nD*M{p\ -

(2.10-2)

can

p\)

m{2RT)ii{L 2 -L,)

(

'

(2.10-8)

m=

nD*g e M{p\ T28(2RT), L 2 (

pi)

- L

(Enghsh) {

)

W

= molecular weight in kg/kg mol, 7 = absolute temperature where m = kg/s, and R = 8314.3 N m/kg mol K. In English units, R = 1545.3 ft lb r/lb mol °R. •

2.10C

in

K,

•

Pressure Drop and Friction Factor

in

Turbulent Flow

In turbulent flow, as in laminar flow, the friction factor also depends on the Reynolds

number. However,

not possible to predict theoretically the Fanning friction factor was done for laminar flow. The friction factor must be determined empirically (experimentally) and it not only depends upon the Reynolds number but also it is

for turbulent flow as

it

on surface roughness

of the pipe. In laminar flow the roughness has essentially no effect. Dimensional analysis also shows the dependence of the friction factor on these factors. In Sections 3.11 and 4. 14 methods of obtaining the dimensionless numbers and their

importance are discussed.

A

large

number

of experimental data

on

friction factors of

smooth pipe and pipes

of

varying degrees of equivalent roughness have been obtained and the data correlated. For design purposes to predict the friction factor / and, hence, the frictional pressure drop of

round

pipe, the friction factor chart in Fig. 2.10-3

Sec. 2.10

can be used.

Design Equations for Laminar and Turbulent Flow

in

It is a

Pipes

log-log plot of/

87

88

versus

N Re

.

friction loss

This friction factor/is then used

Ap } or

Ff

in Eqs. (2.10-5)

and

(2.10-6) to predict the

.

AL

2

v

= 4 /p^-y

A P/

(2.10-5)

—— AL D

A P/ = 4/p

F< =

(Si)

Ap r

2

v

AL

= 4f

f

(English)

2 9c v

2

TT

(SI)

(2.10-6)

AL

2

v

For the region with a Reynolds,.number below 2100, the line is the same as Eq. For a Reynolds number above 4000 for turbulent flow, the lowest line in Fig. 2.10-3 represents the friction factor line for smooth pipes and tubes, such as glass tubes, and drawn copper and brass tubes. The other lines for higher friction factors represent lines for different relative roughness factors, a/D, where D is the inside pipe diameter in m and £ is a roughness parameter, which represents the average height in m of roughness

(2.10-7).

On

projections from the wall (Ml).

new e

=

pipes are given (Ml).

pipe,

m(1.5 x 10 *ft). The reader should be cautioned on using friction factor of/in

may be 4

commercial

steel,

roughness for

has a roughness of

-

5 4.6 x 10-

Fanning that

Fig. 2.10-3 values of the equivalent

The most common

Eq. (2.10-6)

is

friction factors /from other sources.

The

the one used here. Others use a friction factor

times larger.

EXAMPLE

2.10-3. Use of Friction Factor in Turbulent Flow flowing through a horizontal straight commercial steel pipe at 4.57 'm/s. The pipe used is commercial steel, schedule 40, 2-in. nominal diameter. The viscosity of the liquid is 4.46 cp and the density 801 kg/m\ Calculate the mechanical-energy friction loss F f in J/kg for a 36.6-m section

A

liquid

is

of pipe.

The

Solution:

0.0525 m,

v

=

following

4.57 m/s, p

data

=

801

are

p =(4.46 cpXl x 10

The Reynolds number

is

From Appendix

given:

kg/m\ AL = _3 )

=

36.6 m,

4.46 x 10

0.0525(4.57X801)

is

turbulent.

2.10-3, the equivalent

4.46 x 10"

D

N Rc

3

kg/m-s

=

4

For commercial steel pipe from the -5 is 4.6 x 10 m. 5 x 10~

0.0525

m

m=

F = 4/ '

ALv 2 =

~FJ

table in Fig.

0.00088

of 4.310 x 10*, the friction factor from Fig. 2.10-3

Substituting into Eq. (2.10-6), the friction loss

Sec. 2.10

-3

roughness

_£ = 4.6 For an

D =

calculated as

p Hence, the flow

A. 5,

and

4(0.O060X36.6X4.57)

is/=

0.0060.

is

2

=

174

-

8

(0.0525X2)

Design Equations for Laminar and Turbulent Flow

kg"

in

Pipes

89

In problems involving the as in

Example

friction loss

F{

Ff

in pipes,

usually the

is

unknown

and pipe length AL known. Then a direct solution However, in some cases, the friction lossFy is already

the diameter D, velocity 2.10-3.

v,

Then

is

with

possible

set

by the

and pipe length are set, the unknown to be calculated is the diameter. This solution is by trial and error since the velocity v appears in both N Rc and /, which are unknown. In another case, with the F f being again already set, the diameter and pipe length are specified. This is also by trial and error, to calculate the velocity. Example 2.10-4 indicates the method to be used to calculate the pipe diameter with F f set. Others (M2) give a convenient chart to aid in available head of liquid.

the volumetric flow rate

if

these types of calculations.

EXAMPLE

2.10-4. Trial-and-Error Solution to Calculate Pipe Diameter 4.4°C is to flow through a horizontal commercial steel pipe having at the rate of 150 gal/min. A head of water of 6.1 m is a length of 305 available to overcome the friction loss Ff Calculate the pipe diameter.

Water

at

m

.

From Appendix

Solution: cosity

=

A.2 the density p

1000 kg/m 3 and the

vis-

p. is

=

fi

friction loss

(1.55

cpXl x 10~

Ff = (6.1m)j =

—

m

(D

is

=

m

velocity

=

solution 0.089

is

by

(6.1X9.80665)

3

area of pipe

D =

x 1CT 3 kg/m-s

1.55

=

59.82 J/kg

J

9.46 x 10"

The

)

SX7sn5X ^ > UBM17 m>m

=

v

=

3

(9.46

trial

2

3

/s

unknown)

x 1(T 3 m^s)

and error

!

since

v

=

^^

appears in

/V Re

m /s

and /. Assume

that

m for first trial. Dl'P

N /v Rc

^

For commercial

_

001204(1000) lOOZ0) (0.089) (Q 0g9)2(i

and using

steel pipe

^

x

0

Fig. 2.10-3, £

8 73Oxl0 x 10 8.730

-

_ ,

<

3)

=

4.6

x 10"

5

m.Then,

"

e

4.6

D From

Fig. 2.10-3 for /V Rc

x 10" 0.089

=

0.00052

m

8.730 x 10*

and e/D

=

0.00052,

/= 0.0051.

Substituting into Eq. (2.10-6),

r F/

=

59.82

Af = 4/

— -= ALvl

—

4(0.0051X305) (0.01204)

2

Solving for D, D = 0.0945 m. This does not check the assumed value of 0.089 m. For the second trial D will be assumed as 0.0945 m.

90

Chap. 2

Principles of

Momentum

Transfer and Overall Balances

From with

2.10-3,/= 0.0052.

Fig.

N Rc in

_

Ff Solving,

It

can be seen that

/ does

not change

much

the turbulent region.

D=

m

0.0954

59.82

-

or 3.75

—

2

^—

4(0.0052X305) (0.01204)

Hence, the solution checks the assumed

in.

value of D closely.

2.1

0D

Drop and

Pressure

Flow of Gases

Friction Factor in

The equations and methods discussed in this section for turbulent flow in pipes hold for incompressible liquids. They also hold for a gas if the density (or the pressure) changes 3 less than 10%. Then an average density, p av in kg/m should be used and the errors ,

than the uncertainty limits in the friction factor /. For gases, Eq. (2.10-5) can be rewritten as follows for laminar and turbulent flow: involved will be

less

~

(Px

Pi) f

4f ALG

=

2

(2.10-9)

where p av is the density at p av = (Pi + p 2 )/2. Also, the N Kc used is DG/p, where G is kg/m 2 s and is a constant independent of the density and velocity variation for the gas. Equation (2.10-5) can also be written for gases as •

RT

4/ ALG

,

DAT-

<

SI > (2.10-10)

4 / ALG RT -rr: Pi-Pi = 2

,

gc

where

R

8314.3 J/kg

is

The

mol

K or

•

1545.3

ft

—

(English)

DM

•

lb r/lb

mol "R and

M

is

molecular weight.

derivation of Eqs. (2.10-9) and (2.10-10) applies only to cases with gases where

is small enough so that large changes in velocity do not becomes large, the kinetic-energy term, which has been omitted, becomes important. For pressure changes above about 10%, compressible flow is occurring and the reader should refer to Section 2.11. In adiabatic flow in a uniform pipe, the velocity in the pipe cannot exceed the velocity of sound.

the relative pressure change occur.

If

the exit velocity

EXAMPLE

Flow of Gas in Line and Pressure Drop flowing in a smooth tube having an inside diameter 2 of 0.010 m at the rate of 9.0 kg/s m The tube is 200 m long and the flow can be assumed to be isothermal. The pressure at the entrance to the tube is 2.10-5.

Nitrogen gas

at

25°C

is

•

.

5 2.0265 x 10 Pa. Calculate the outlet pressure.

Solution:

G= R =

The

viscosity

of the gas

from Appendix A. 3

7 = 298.15 K. Inlet gas pressure kg/s-m 2 D = 0.010 m, M = 28.02 kg/kg

10~ 5 Pa-s

at

pj

=

is

p

=

1.77 x

2.0265 x 10

s

Pa,

mol, AL = 200 m, and 8314.3 J/kg mol K. Assuming that Eq. (2.10-10) holds for this case and that the pressure drop is less than 10%, the Reynolds number is 9.0

,

•

=

DG _ — = ~ p

Hence, the flow

Sec. 2.10

is

0.010(9.0) 1.77

turbulent. Using Fig.

,.-s x 10" ..

2.

1

0-3,

=

/=

5085 0.0090 for a smooth tube.

Design Equations for Laminar and Turbulent Flow

in

Pipes

91

Substituting into Eq. (2.10-10), ,

2

Pl_Pz=

RT

DM 2

- Pz2 _ -

4(0.0090X200X9.0) (83 1 4.3X298. 15)

10

- pi =

10 0.5160 x 10

4.1067 x 10

Solving, p 2 = 1.895 x 10 pressure drop is less than

The

2

rw« xv 1^2 (2.0265 10 )

n

2.10E

4 / ALG

Effect of

5

0.010(28.02)

Pa. Hence, Eq. (2.10-10) can be used since the

10%.

Heat Transfer on Friction Factor

friction factor /given in Fig. 2.10-3 is given for

When

a fluid

is

isothermal flow,

physical properties of the fluid, especially the viscosity.

following

no heat

i.e.,

transfer.

being heated or cooled, the temperature gradient will cause a change

method

of Sieder

in

For engineering practice the

and Tate (PI, S3) can be used to predict the

friction factor for

nonisothermal flow for liquids and gases. 1.

Calculate the

mean bulk temperature t a

and

as the average of the inlet

outlet bulk fluid

temperatures.

3.

N Rr using the viscosity jia at t a and use Fig. 2.10-3 to obtain /. Using the tube wall temperature t w determine /i w at t„

4.

Calculate

2.

Calculate the

,

\p

occurring below.

for the case

^={~j

5.

The

reverse occurs

OF

=

ij,

=

\jf

=

final friction factor

Hence, when the liquid

2.1

ift

is

is

(heating)

N Re > 2100

(cooling) JV Re

>

2100

(2.10-12)

(^j

(heating)

/V~

<

2100

(2.10-13)

(^j

(cooling)

N Ke <

2100

(2.10-14)

(yj

obtained by dividing the

being heated,

on cooling the

(2.10-11)

i/'

is

Rc

/ from

greater than 1.0

step 2 by the

\]/

from step

final / decreases.

and the

4.

The

liquid.

Friction Losses in Expansion, Contraction,

and Pipe Fittings Skin friction losses in flow through straight pipe are calculated by using the Fanning friction factor.

However,

if

the velocity of the fluid

is

changed

in direction or

magnitude,

additional friction losses occur. This results from additional turbulence which develops

because of vortices and other factors. Methods to estimate these losses are discussed below.

/.

Sudden enlargement

little

92

losses.

If

the cross section of a pipe enlarges very gradually, very

or no extra losses are incurred. If the

Chap. 2

change

is

sudden,

it

results in additional losses

Principles of Momentum Transfer

and Overall Balances

due to eddies formed by the jet expanding in the enlarged section. This friction loss can be calculated by the following for turbulent flow in both sections. This Eq. (2.8-36) was derived in Example 2.8-4. (*i

-

v 2)

2

2a where h cx

is

A 2J

V

the friction loss in J/kg,

K cx

2a kg

2a

the expansion-loss coefficient and

is

= (1 —

2

i>i is the upstream velocity in the smaller area in m/s, v 2 is the downstream and a = 1.0. If the flow is laminar in both sections, the factor a in the equation becomes \. For English units the right-hand side of Eq. (2.10-15) is divided by g c Also,

Ai/A 2 )

,

velocity,

.

fc

2.

= ft-lb /lb m f

.

Sudden contraction

losses.

When

the cross section of the pipe

is

suddenly reduced,

the stream cannot follow around the sharp corner, and additional frictional losses due to

eddies occur. For turbulent flow, this

where h c

is

the friction loss, a

=

given by

is

1.0 for

turbulent flow,

v2

is

the average velocity in the

and K c is the contraction-loss coefficient (PI) and — approximately equals 0.55 (1 A 2 /A ). For laminar flow, the same equation can be used = with a j(S2). For English units the right side is divided by g c

smaller or

downstream

section,

1

.

Table

2.10-1.

Friction Loss for Turbulent

Flow Through

Valves and Fittings Frictional Loss,

Frictional Loss,

Number of Type of

Fitting or Valve

Elbow, 45° Elbow, 90°

Heads,

Velocity

Equivalent Length of Straight Pipe in Pipe

Kf

Diameters,

0.35

17

0.75

35

Tee

1

50

Return bend

1.5

75

Coupling

0.04

2

Union Gate valve Wide open Half open Globe valve Wide open Half open Angle valve, wide open Check valve

0.04

2

Ball

Swing Water meter, disk

0.17

LJD

9

225

4.5

6.0

300

9.5

475

2.0

100

70.0

3500

2.0

100

7.0

350

Source: R. H. Perry and C. H. Chilton, Chemical Engineers' Handbook, 5th cd. York: McGraw-Hill Book Company, 1973. With permission.

Sec. 2.10

Design Equations for Laminar and Turbulent Flow

in

Pipes

New

93

Losses in fittings and valves.

3.

lines in

Pipe

fittings

friction loss

for fittings

from these

and valves

fittings

is

fitting.

the fitting or valve

2.

an equivalent pipe length

(2.10-17)

and

Kf

vx

is

the average velocity in the pipe

are given in Table 2.10-1 for turbulent

having the same frictional loss as the

and references (Bl) give data

texts

is

the equivalent length of straight pipe in

fitting,-

and

D

is

the inside pipe diameter in

values in Eqs. (2.10-15) and (2.10-16) can be converted to

K

the

by 50

The L e

(PI).

for losses in

pipe diameters. These data, also given in Table

in

LJD, where Le

2.10-1, are presented as

K.

the

10-2 for laminar flow.

As an alternative method, some fittings as

fittings,

friction loss

f= KS^

Experimental values for

flow (PI) and in Table

The

given by the following equation:

K s is the loss factor for

leading to the

many

could be greater than in the straight pipe.

h

where

and valves also disturb the normal flow

a pipe and cause additional friction losses. In a short pipe with

LJD

m

m. The

values by multiplying

values for\he fittings are simply added to the length of the

straight pipe to get the total length of equivalent straight pipe to use in Eq. (2.10-6).

4.

Frictional losses in mechanical-energy-balance equation.

The

frictional losses

from

the friction in the straight pipe (Fanning friction), enlargement losses, contraction losses,

and losses

in fittings

and

valves are

all

incorporated

in the

£ F term of Eq. (2.7-28) for the

mechanical-energy balance, so that

£

F = 4/

T

and

are the same, then factoring, Eq. (2.10-18) becomes, for

If all the velocities, v, o l3

v2

,

+

2

" 2 + Kc

i + Kf

(2-10-18)

2

this special case,

= £F The will

4/ (

T + K" +

Kc

+ Kf )

(2 " 10~ 19)

7

use of the mechanical-energy-balance equation (2.7-28) along with Eq. (2.10-18)

be shown

following examples.

in the

EXAMPLE

Friction Losses

2.10-6.

and Mechanical- Energy Balance

An It

elevated storage tank contains water at 82.2°C as shown in Fig. 2.10-4. 3 is desired to have a discharge rate at point 2 of 0.223 ft /s. What must be

the height

H

Table

in

ft

of the surface of the water in the tank relative to the

2.10-2.

Friction Loss for

Laminar Flow Through Valves

and Fittings (Kl)

Number of Velocity Heads, Reynolds Number

Frictional Loss,

TyP e °f

Kf

Fitting or

50

100

Elbow, 90° Tee Globe valve Check valve,

17

7

2.5

4.8

3.0

swing

94

200

Valve

9

28

22

17

55

17

9

Chap. 2

400

1000 Turbulent

1.2

0.85

2.0

1.4

14

5.8

10

3.2

Principles of Momentum Transfer

0.75 1.0

6.0

2.0

and Overall Balances

FIGURE

Process flow diagram for Example 2.10-6.

2.10-4.

The pipe used

discharge point?

-

125 ft—

\*

commercial steel pipe, schedule shown.

is

40,

and

the lengths of the straight portions of pipe are

The mechanical-energy-balance equation

Solution:

tween points

Zi

and

1

L + jLJpi_Ei)_ 2ag c

9c

(2.7-28)

is

written be-

2.

\p

Ws =

Z2

£+ gc

p 2J

1

± + Zf

(1,0.20,

:

2ag c

3 A.2, for water, p = 0.970(62.43) = 60.52 lbjft and 4 4 0.347(6.7197 x 10" ) = 2.33 x 10~ lbjft-s. The diam-

From Appendix p

=

0.347

cp

=

eters of the pipes are

The

For

4-in. pipe:

D3 =

For

2-in. pipe:

DA =

velocities in the 4-in.

"3

"*

The

Y,

F term

=

and

0-223

2 067

0 223

=

0.3353

ft;

A3 =

0.1722

ft;

A±

2 ft

0.02330

2 ft

/s 2

=

1523

(4 " ln

ft/S

"

PIPC)

ft

=

9-57

ft/s

0 02330 f° r frictional losses in

contraction loss at tank

=

0.0884

2-in. pipe are

3 ft

0.0884

=

4.026 -— -=

(2 " in

pipe) -

the system includes the following:

the 4-in. straight pipe, (3) contraction loss from 4-in. to 2-in. pipe, (5) friction in the 2-in. straight pipe, and (6) friction in the two 2-in. elbows. Calculations for the six items are as follows.

(1)

friction in 4-in.

/.

Sec. 2.10

elbow,

exit, (2) friction in

(4)

Contraction loss at tank

exit.

From

Eq. (2.10-16) for contraction from

Design Equations for Laminar and Turbulent Flow

in

Pipes

95

A

l

to

A3

cross-sectional area since

K = c

Hence, the 4 (1.5 x 10"

D 3 ^p

2.33

ft-

to

lb f/lb ra

is

-^ xlu 9|Q ,

2.10-3,

Fig.

compared

°- 55

0.054

xl0~*

From

very large

is

~ 0) =

[

_ 0.3353(2.523X60.52)

turbulent.

is

°- 55 (

gg^ =

0-55

^

flow

of the tank

The Reynolds number

2. Friction in the 4-in. pipe.

~

{

=

0.55^1

=

NRe

A

n3

,

—

b

x

4.6

10

-5

m

ft).

-

Then, for

N Rc =

219 300, the Fanning

AL =

tuting to Eq. (2.10-6) for

20.0

ft

friction factor

4(00047)-^f -4f^^_4{a(X}47) ^~ 4/ D 2^ 0.3353 3. Friction in 4-in.

From Table

elbow.

(2 523)2

4.

'

2.10-1,

-K f - -0.75

Contraction loss from 4- to

contraction from

J.

A3

AA

lb m

Kf =

0.75.

0.074

—

The Fanning

AL =

125

F tf

+

Using Eq. (2.10-16) again

for

is

0.00087

0.1722

from

friction factor

10

pipe.

The Reynolds number

^.

«

2-in.

Then, substitut-

cross-sectional area,

Friction in the 2-in. pipe.

D

6.

to

Substi-

_0 1U -QUI

-

2(32.174)

ing into Eq. (2.10-17),

hf

/= 0.0047.

of 4-in. pipe,

+

50

=

185

ft.

Fig. 2.10-3 is/

- ^-Ows) -4^^1 ~* 4(00048) J ~ D

=

0.0048.

2g c

Friction in the two 2-in. elbows.

185(957)2 (0.1722X2X32.174)

For a

=

0.75

~

/

= 2K

,

/^-

Chap. 2

2(0.75X9.57)^

2(32.174)

Principles

=

ft

2 136 -

294

total length

^ lb m

and two elbows,

2 fc

The

Substituting into Eq. (2.10-6),

-

lb r

lb7

of Momentum Transfer and Overall Balances

The

total frictional loss

X

F=

0.054

+

=

32.35

ft

Using as exists, at

1

a

=

£F

1.0.

datum

a

Also, v x

is

+

0.111 •

of items

0.074

+

(1)

through

+

0.575

29.4

(6).

+

2.136

lb f/lb m

level z 2

=

sum

the

z

,

l

=

0 and u 2

atm abs pressure and p 1 = p 2

= i>

4

H =

h, z 2

9.57

=

0. Since turbulent flow Since p t and p 2 are both

ft/s.

,

^1-^ = 0 P Also, since

no pump

is

used,

P

W

s

=

0.

Substituting these values into Eq.

(2.10-20),

+ = 33.77 ft Ib f/lb m (100.9 J/kg) and height of water level above the discharge outlet.

Solving, H(g/g c )

EXAMPLE

-

2.10-7.

32.35

2(32.174)

gc

H

is

33.77

(10.3

ft

m)

Pump in Mechanical-Energy

Friction Losses with

Balance at 20°C is being pumped from a tank to an elevated tank at the rate of x 10" 3 m 3 /s. All of the piping in Fig. 2.10-5 is 4-in. schedule 40 pipe. The pump has an efficiency of 65%. Calculate the kW power needed for the

Water

5.0

pump.

The mechanical-energy-balance equation

Solution:

tween points

1

and

2,

with point

Pi

1

2a

"

' '

"

(2.7-28)

is

written be-

=

0

being the reference plane.

1

"

•

-I

+

p

XF + W

s

(2.7-28)

= 998.2 kg/m 3 = 1.005 x 10~ 3 Pa s. 3 For 4-in. pipe from Appendix A.5, D 0.1023 m and A = 8.219 x 10~ m The velocity in the pipe is v = 5.0 x 10~7(8.219 x 10^ = 0.6083 m/s. The From Appendix

A.2 for water, p

,

/j

2

.

3

)

Reynolds number

is

Dvp

N Re

0.

1.005 x 10"

/j.

Hence, the flow

is

1023(0. 6083)(998. 2)

— = 6.181

x I0 4

turbulent.

100

m

3 £ 15

m 50

m 4-in. pipe

2

C

pump Figure

Sec. 2.10

2.10-5.

Process flow diagram for Example 2.10-7.

Design Equations for Laminar and Turbulent Flow

in

Pipes

The

£

F term

for frictional losses includes the following: (1) contrac-

tion loss at tank exit,

elbows, and 1.

(4)

(2) friction in

expansion

Contraction loss at tank large

A

a small

to

x

A2

Friction

e/D

=

—

1

J

=

x 10- /0.1023 5

= 0.0051.

the

two

Eq. (2.10-16) for contraction from a

= 0.55(1 -0) = 0.55

From

the straight pipe.

in

4.6

From

exit.

(3) friction in

tank entrance.

,

0.55(

2.

the straight pipe,

loss at the

=

Fig. 2.10-3, e

Then

0.00045.

N Rc =

for

Substituting into Eq. (2.10-6) for

x 10

4.6

AL =

+

5

-5

m

6.181 x 10

50

+

15

+

and 4 ,

100

/

=

170 m,

3.

Friction

the two elbows.

in

tuting into Eq.

(2.

10-7) for

From Table two elbows,

A,= 2K,± 4.

Expansion

K" =

K The

=

2(0.75)

=

~ ^2) =

l

{

v

F =

0.102

is

+

1.0

Solving,

%=

10" 3 (998.2)

=

+

—153.93

(2.10-15),

^=L

°

2

£ F.

6.272

9.806(15.0

4.991 kg/s.

Then, substi-

^—^ = 0.185 J/kg

+

0.278

Substituting into Eq. (2.7-28), where {v\

0

~

(0.6083)

K cx — =

total frictional loss

X

2

(1

0.75.

0.278 J/kg

Using Eq.

loss at the tank entrance.

Kf =

2.10-1,

- 0) +

0.185

— v\) =

0

mass

=

6.837 J/kg

and (p 2

—

W

0

+

0 + 6.837

The

J/kg.

Using Eq.

+

=

s

flow

Pi)

rate

is

=

0,

m=

5.0 x

(2.7-30),

Ws =-nW

r

-153.93 = -0.65 551W, Solving,

W

p

=

236.8 J/kg.

The pump

pump kW =

2.1

0G

The

mW

p

=

kW power 4,

"'^6

'

is

8)

=

1.182

1000

kW

Friction Loss in Noncircular Conduits

friction loss in

long straight channels or conduits of noncircular cross section can be

estimated by using the same equations employed for circular pipes

Reynolds number and diameter.

98

in the friction

The equivalent diameter

Chap. 2

factor equation (2.10-6)

is

if

the diameter in the

taken as the equivalent

D is denned as four times the hydraulic radius r„ The .

Principles of Momentum Transfer

and Overall Balances

hydraulic radius

is

defined as the ratio of the cross-sectional area of the channel to the

wetted perimeter of the channel for turbulent flow only. Hence,

D= For example,

for

4r H

=

cross-sectional area of channel

4

(2.10-21)

wetted perimeter of channel

a circular tube,

D

=—4(?tD

-

4frP?/4

For a rectangular duct of sides a and

—= D /4)

nD

For an annular space with outside diameter

D_

2

D and x

inside

ttP|/4)

D2

,

D2

(2.10-22)

b' ft,

D

4{ab)

lab

+

a+b

2a

2b

(2.10-23)

For open channels and partly filled ducts in turbulent flow, the equivalent diameter and Eq. (2.10-6) are also used (PI). For a rectangle with depth of liquid y and width b,

D= b

For a wide, shallow stream of depth

in ducts

(2.10-24)

2y

y,

D = For laminar flow

+

running

full

4y

(2.10-25)

and

in

open channels with various

cross-

sectional shapes other than circular, equations are given elsewhere (PI).

2.10H If the

Entrance Section of a Pipe velocity profile at the entrance region of a tube

is flat,

a certain length of the tube

is

necessary for the velocity profile to be fully established. This length for the establishment of

fully

developed flow

is

called the transition length or entry length. This

2.10-6 for laminar flow. At the entrance the velocity profile

same

is

flat;

i.e.,

is

shown

in

the velocity

is

Fig.

the

As the fluid progresses down the tube, the boundary-layer thickness finally they meet at the center of the pipe and the parabolic velocity profile

at all positions.

increases until is fully

established.

The approximate entry length L e

of a pipe having a diameter of

D

for a fully

^^^^^^^^^^^^^^^^^^^^^^^^^^ velocity profile'"

Figure

Sec. 2.10

2.10-6.

^

boundary layer

Velocity profiles near a pipe entrance for laminar flow.

Design Equations for Laminar and Turbulent Flow

in Pipes

99

developed velocity

profile to be

formed

in

laminar flow

^ = 0.0575N For turbulent

no relation

flow,

developed turbulent velocity

(L2)

is

(2.10-26)

Rc

available to predict the entry length for a fully

is

As an approximation, number and is fully developed

profile to form.

nearly independent of the Reynolds

the entry length

is

after 50 diameters

downstream.

EXAMPLE

Length for a Fluid in a Pipe flowing through a tube having a diameter of 0.010 velocity of 0.10 m/s. (a) Calculate the entry length. (b) Calculate the entry length for turbulent flow.

Water

2.10-8. Entry

20°C

at

is

=

For part (a), from Appendix A. 2, p 10~ 3 Pa-s. The Reynolds number is

Solution:

N "~ Using Eq. (2.10-26)

Dvp n

L = c

0.010(0.10X998.2)

~

1.005 x 10-

"

3

kg/m 3 p = ,

at a

1.005 x

993 2 -

for laminar flow,

^

0.0575(993.2)

=

57.1

part(b),L c ='50(0.01)

=

0.50 m.

^= Hence,

998.2

m

=

0.571 m.

For turbulent flow

The pressure drop

in

or friction factor in the entry length

developed flow. For laminar flow the

friction factor

is

is

greater than in fully

highest at the entrance (L2)

and

then decreases smoothly to the fully developed flow value. For turbulent flow there will

be some portion of the entrance over which the boundary layer friction factor profile

is

difficult to express.

As an approximation

is

laminar and the

the friction factor for the

entry length can be taken as two to three times the value of the friction factor in

fully

developed flow.

2.101

Selection of Pipe Sizes

complex process piping systems, the optimum size of pipe to use for a specific upon the relative costs of capital investment, power, maintenance, and so on. Charts are available for determining these optimum sizes (P 1). However, for small installations approximations are usually sufficiently accurate. A table of representative values of ranges of velocity in pipes are shown in Table 2.10-3. In large or

situation depends

Table

Representative Ranges of Velocities

2.10-3.

in Steel

Pipes Velocity

Type of

Type of Flow

Fluid

Nonviscous

liquid

Inlet to

pump

Process line or

Viscous liquid

Inlet to

2-3

pump

discharge

pump

Process line or

pump

Gas

discharge

Chap. 2

Principles

m/s 0.6-0.9

5-8

1.5-2.5

0.2-0.8

0.06-O.25

0.5-2

0.15-0.6

30-120 30-75

Steam

100

ft/s

9-36 9-23

of Momentum Transfer and Overall Balances

COMPRESSIBLE FLOW OF GASES

2.11

2.11

A

Introduction and Basic Equation for Flow in Pipes

When

pressure changes in gases occur which are greater than about 10%, the friction-

equations (2.10-9) and (2.10-10)

loss

Then

may

the density or specific

be in error since compressible flow

is

occurring.

more complicated because of the variation of volume with changes in pressure. The field of compressible flow is

the solution of the energy balance

is

very large and covers a very wide range of variations in geometry, pressure, velocity, and

temperature. In this section we restrict our discussion to isothermal and adiabatic flow in uniform, straight pipes detail in other references

and do not cover flow

(M2,

in nozzles,

which

is

discussed in

some

PI).

The general mechanical energy-balance equation Assuming turbulent flow, so that a = 1.0; no

point.

(2.7-27)

dp — + dF = 0

+

g dz

Ws = 0;

and

becomes

writing the equation for a differential length dL, Eq. (2.7-27)

vdv +

can be used as a starting

shaft work, so that

(2.11-1)

P

=

For a horizontal duct, dz

Using only the wall shear

0.

frictional

term for

dF and

writing Eq. (2.10-6) in differential form,

dv

v

where V =

1/p.

+V

dp

+

2

dL =0 2D

4fv —

(2.1 1-2)

Assuming steady-state flow and a uniform pipe diameter, G

is

constant

and

G=vp=^

(2.11-3)

dv=GdV. Substituting Eqs.

(2.1 1-3)

and

(2.1 1-4) into (2.11-2)

G

,

dV

V

V

and rearranging,

2fG 2 dL = D

dp

+

(2.11-4)

+

(111 " 5)

°

is the basic differential equation that is to be integrated. To do this the relation between V and p must be known so that the integral of dp/V can be evaluated. This integral depends upon the nature of the flow and two important conditions used are

This

isothermal and adiabatic flow

in pipes.

2.1

IB

Isothermal Compressible Flow

To

integrate Eq. (2.1 1-5) for isothermal flow,

an ideal gas

will

be assumed where

pV = ^-RT

(2.11-6)

M

Solving for

assuming

/

V is

in

Eq. (2.11-6) and substituting

it

into

Eq. (2.11-5), and integrating

constant, 2

G2

f

dV_

+

V

_M_

RT

pdp +

dL = 0

2f—

K, M 2 -p 2 G2 G 2 \n^ + ——(p AL = 2 V 2RT U ^ rw)+2f D l

—

0

(2.11-7)

(2.11-8)

x

Sec. 2.11

Compressible Flow of Gases

101

Substituting

V2 /V

for

pjp 2

and rearranging,

l

,

,

where

T=

M — molecular

weight

in

4fALG 2 RT

2G 2 RT

p,

DM

M

p2

kg mass/kg mol,

RT/M =

temperature K. The quantity

R=

8314.34

where p av

N-m/kg mol-K, and

= (Pi +

p av /p av p 2 )/2 and pav is and p av In English units, R = 1545.3 ft lb f /lb mol °R and the right-hand terms are divided by g c Equation (2.11-9) then becomes

T

the average density at

,

-

•

.

.

ALG G 2 p, 4/ J = ln^ + zz -P2)f 2

(p l

2£>Pav

The

—

(2.11-10)

Pi

Pav

term on the right of Eqs. (2.11-9) and (2.11-10) represents the frictional loss as given by Eqs. (2.10-9) and (2.10-10). The last term in both equations is generally first

negligible in ducts of appreciable lengths unless the pressure

drop

very large.

is

EXAMPLE 2.11-1. Compressible Flow of a Gas in a Pipe Line Natural gas, which is essentially methane, is being pumped through a 5 1.016-m-ID pipeline for a distance of 1.609 x 10 (Dl) at a rate of 2.077 kg mol/s. It can be assumed that the line is isothermal at 288.8 K. The 3 pressure p 2 at the discharge end of the line is 170.3 x 10 Pa absolute. Calculate the pressure p l at the inlet of the line. The viscosity of methane at 5 288.8 K is 1.04 x 10~ Pas.

m

D=

Solution:

G =

2.

s*

= nD 2 /4 =

10-3,

=

e

x lO-

0.8107

m2

Then,

.

J^~m )J = 41.00 s-m

_4 5-4 -^

4.6 x 10

5

2

2

X 1U 6

m.

4.6 x 10"

D friction

=

( 16.0 kg mol/ ( K \0.8107

J \

e

The

2

n( 1.01 6) /4

'-O'^l.OO) 1.04

m

Fig.

*8™f)

( 2.077 V

_gg-~ N Nr <~ From

m, A

1.016

5

= 0.0000453

1.016

factor/ = 0.0027.

in Eq. (2.11-9), trial and error must be used. Estimating p, at 620.5 x 10 3 Pa, R = 8314.34 N m/kg mol K, and AL = 5 1.609 x 10 m. Substituting into Eq. (2. 11-9),

In order to solve for

•

2

_

2

_

Pl_P2_

2

5

4(0.0027X1.609 x 10 )(41.00) (83 14.34X288.8) 1.016(16.0) 2

2(4 ,00) (83 14.34X288.8) 1

+

(16.0)

=

•

4.375 x 10

1

+

3

620.5 x 10 " 170.3 x 10 3

0.00652 x 10

1

=

4.382 x 10

1

(Pa)

2

Now, P 2 = 170.3 x 10 3 Pa. Substituting this into the above and solving for 3 Pi,Pi= 683.5 x 10 Pa. Substituting this new value of p^ into Eq. (2.1 1-9)

again and solving for p u the final result is p, = 683.5 x 10 the last term in Eq. (2.1 1-9) in this case is almost negligible.

When

=

102

0,

G =

0.

Pa.

upstream pressure p, remains constant, the mass flow

downstream pressure p 2

the

p2

the

3

is

varied.

From

Eq. (2.11-9),

whenpi =

This indicates that at some intermediate value of p 2

Chap. 2

,

Principles of Momentum Transfer

p2

Note

rate ,

the

G

that

changes as

G = Oand when flow G must be a

and Overall Balances

maximum. This means differentiation

on Eq.

that the flow

(2.1 1-9) for

is

G Using Eqs.

(2.1 1-3)

and

a

maximum when dG/dp 2 = 0. Performing and / and solving for G,

constant p

this

l

—=

1

^

(2.11-11)

RT

(2.1 1-6),

(2.11-12)

This

is

the equation for the velocity of

sound

in the fluid at the

flow. Thus, for isothermal compressible flow there

conditions for isothermal

maximum

a

is

flow for a given

upstream p and further reduction of p 2 will not give any further increase in flow. Further details as to the length of pipe and the pressure at the maximum flow l

conditions are discussed elsewhere (Dl; M2, Pl).

EXAMPLE 2.11-2. Maximum Flow for Compressible Flow of a Gas For the conditions of Example 2.1 1-1, calculate the maximum velocity that can be obtained and the velocity of sound at these conditions. Compare with Example Solution:

2.1 1-1.

Using Eq.

(2.1 1-12)

and

»

the conditions in

83

1

This

is

the

maximum

Example

2.11-1,

=

387.4 m/s

is

decreased. This

6.0

velocity obtainable

if

p2

also the

is

sound in the fluid at the conditions for isothermal flow. To compare with Example 2.11-1, the actual velocity at the exit pressure p 2 is obtained by combining Eqs. (2.1 1-3) and (2.1 1-6) to give velocity of

,2

RTG =— =

(2.11-13)

831434(288. 8)(4 1.00)

=

36

heat transfer through the wall of the pipe

is

(170.3x10^)16.0 Adiabatic Compressible Flow

2.11C

When

compressible flow (2.1 1-5)

Pl).

13m/s ,

"

in

adiabatic. Equation

is

has been integrated for adiabatic flow and details are given elsewhere (Dl,

Convenient charts

to solve this case are also available (Pl).

batic flow often deviate very

little

from isothermal flow, especially

The

isothermal, but the

maximum

possible difference

is

of about 1000 diameters or longer, the difference

about is

20%

is

Ml,

results for adia-

in long lines.

short pipes and relatively large pressure drops, the adiabatic flow rate

(2.1 1-8)

flow of gas in

the

negligible,

a straight pipe of constant cross section

For very

greater than the

(Dl). For pipes of length

generally less than

can also be used when the temperature change over the conduit

is

5%. Equation small by using

an arithmetic average temperature.

Using the same procedures isothermal case, the the pipe

Sec. 2.1 1

is

maximum

for finding a

flow occurs

maximum

when

the sonic velocity for adiabatic flow. This

Compressible Flow of Gases

flow that were used in the

the velocity at the

downstream end of

is

103

where, y

=

c /c v p

may

flow

For

the ratio of heat capacities.

,

velocity for adiabatic flow

20%

about

is

=

air, y

Hence, the

1.4.

maximum

The

greater than for isothermal flow.

rate of

not be limited by the flow conditions in the pipe, in practice, but by the

development of sonic velocity in a fitting or valve in the pipe. Hence, care should be used in selection of fittings in such pipes for compressible flow. Further details as to the length of pipe

and pressure

A

number, u max

,

at the

maximum flow conditions

convenient parameter often used

N Ma

,

which

is

defined as the ratio of

is

the

PI).

Mach

the speed of the fluid in the conduit, to

v,

the speed of sound in the fluid at the actual flow conditions.

—

N Ma = At a

M2,

are given elsewhere (Dl,

compressible flow equations

in

Mach number

and supersonic

at

of

1.0,

the flow

a number above

sonic.

is

(2.11-15)

At a value

less

than

the flow

1.0,

is

subsonic,

1.0.

PROBLEMS a Spherical Tank. Calculate the pressure in psia and kN/m 2 in a bottom of the tank filled with oil having a diameter of The top of the tank is vented to the atmosphere having a pressure of 14.72

2.2-1. Pressure in

spherical tank at the 8.0

ft.

psia.

The

density of the oil

is

0.922 g/cm

3 .

Ans. 2.2-2. Pressure with

Two

Liquids,

the bottom with 12.1

cm

Hg and of

Water.

Hg and

5.6

17.921b f

/in.

2

(psia),

123.5

kN/m 2

An open test tube at 293 K is filled at cm of water is placed above the Hg.

Calculate the pressure at the bottom of the test tube if the atmospheric pressure 3 3 is 756 Hg. Use a density of 13.55 g/cm for Hg and 0.998 g/cm for water. 2 2 Give the answer in terms of dyn/cm psia, and kN/m See Appendix A.l for

mm

.

,

conversion factors. Ans. 2.2-3.

Head of a

1.175 x 10

6

dyn/cm 2

,

17.0 psia, 2.3 psig, 117.5

and Pressure. The pressure The depth of liquid in the tank is

Fluid of Jet Fuel

fuel is 180.6

kN/m

2 .

3

825 kg/m Calculate the head of the liquid absolute pressure at the bottom of the tank. fuel

2.2-4.

is

.

m

in

kN/m 2

at the top of a tank of jet

m. The density of the which corresponds to the 6.4

Measurement of Pressure. An open U-tube manometer similar

to Fig. 2.2-4a

is

being used to measure the absolute pressure p a in a vessel containing air. The pressure p b is atmospheric pressure, which is 754 Hg. The liquid in the manometer is water having a density of 1000 kg/m 3 Assume that the density p B

mm

.

3

1.30 kg/m and that the distance Calculate p a in psia and kPa.

is

Z

is

very small.

The reading R

Ans. 2.2-5.

Measurement of Small Pressure

Differences.

The

pa

=

is

0.415 m.

15.17 psia, 104.6

two-fluid U-tube

kPa

manometer

is being used to measure the difference in pressure at two points in a line containing air at 1 atm abs pressure. The value of R 0 — 0 for equal pressures. The lighter fluid is a hydrocarbon with a density of 812 kg/m 3 and the heavier 3 water has a density of 998 kg/m The inside diameters of the U tube and reservoir are 3.2 and 54.2 mm, respectively. The reading R of the manometer is 1 17.2 mm. Calculate the pressure difference in Hg and pascal. .

mm

mm

Sea Lab. A sea lab 5.0 m high is to be designed to withstand submersion to 150 m, measured from the sea level to the top of the sea lab. Calculate the pressure on top of the sea lab and also the pressure variation on the side of the container measured as the distance x in m from the top of the sea 3 lab downward. The density of seawater is 1020 kg/m 2 Ans. p = 10.00(150 + x) kN/m

2.2-6. Pressure in a

.

104

Chap. 2

Problems

2.2-7.

Measurement of Pressure Difference in Vessels. In Fig. 2.2-5b the differential manometer is used to measure the pressure difference between two vessels. Derive the equation for the pressure difference p A

and

heights 2.2- 8.

— pB

in terms of the liquid

densities.

Design of Settler and Separator for Immiscible Liquids. A vertical cylindrical 3 settler-separator is to be designed for separating a mixture flowing at 20.0 /h and containing equal volumes of a light petroleum liquid (p B = 875 kg/m 3 ) and 3 a dilute solution of wash water (p A = 1050 kg/m ). Laboratory experiments indicate a settling time of 15 min is needed to adequately separate the two phases. For design purposes use a 25-min settling time and calculate the size of the vessel needed, the liquid levels of the light and heavy liquids in the vessel, and the height h A2 of the heavy liquid overflow. Assume that the ends of the vessel are approximately flat, that the vessel diameter equals its height, and that one-third of the volume is vapor space vented to the atmosphere. Use the nomenclature given in Fig. 2.2-6.

m

Ans. 2.3- 1.

=

h A2

1.537

m

Molecular Transport of a Property with a Variable Diffusivity. A property is being transported through a fluid at steady state through a constant cross2 sectional area. At point 1 the concentration F is 2.78 x 10" amount of 3 2 10" at point 2 at a distance of 2.0 m away. The property/m and 1.50 x diffusivity depends on concentration T as follows. 1

5 (a)

= A + BT =

0.150

+

1.65r

Derive the integrated equation for the flux in terms of

F and f 2 Then, .

x

calculate the flux. (b)

Calculate

T

at z

=

m and plot

1.0

Ans.

(a)

T

^=

versus z for the three points.

-T

[/UT,

2 )

+

(B/2XF

- T 2 )]/(z 2 -

2

z,

of General Property Equation for Steady State. Integrate the general property equation (2.3-11) for steady state and no generation between the points

2.3- 2. Integration

F, at

2.4- 1.

z

and f 2 atz 2 The .

1

Shear Stress

Soybean

final

equation should relate F to z. Ans. r = (r 2 -r,Xz-z

l

-z

)/(z 2

1

+

)

r,

Using Fig. 2.4-1, the distance between the two parallel plates is 0.00914 m and the lower plate is being pulled at a relative velocity of 0.366 m/s greater than the top plate. The fluid used is soybean oil 2 with viscosity of 4 x 10" Pa s at 303 K (Appendix A.4). (a) Calculate the shear stress t and the shear rate using lb force, ft, and s units. in

Oil.

•

(b)

Repeat, using SI units.

glycerol at 293 K having a viscosity of 1.069 kg/ms is used instead of soybean oil, what relative velocity in m/s is needed using the same distance between plates so that the same shear stress is obtained as in part (a)? Also, what is the new shear rate? -1 2 2 Ans. (a) Shear stress = 3.34 x 10~ lb f /ft shear rate = 40.0 s (c)

If

,

(b)

2.4-2.

1.60

N/m 2

;

(c)

relative velocity

Shear Stress and Shear Rate

in Fluids.

=

;

0.01369 m/s, shear rate

Using

Fig. 2.4-1, the

=

lower plate

pulled at a relative velocity of 0.40 m/s greater than the top plate.

The

1.50 s" is

1

being used

fluid

water at 24°C. How far apart should the two plates be placed so that the shear stress r is 2 0.30 N/m ? Also, calculate the shear rate. 2 (b) If oil with a viscosity of 2.0 x 10" Pas is used instead at the same plate spacing and velocity as in part (a), what is the shear stress and the shear is

(a)

rate?

K

Number for Milk Flow. Whole milk at 293 having a density of 3 1030 kg/m and viscosity of 2.12 cp is flowing at the rate of 0.605 kg/s in a glass pipe having a diameter of 63.5 mm.

25-1. Reynolds

Chap. 2

Problems

(a)

(b)

Calculate the Reynolds number. Is this turbulent flow? 3 Calculate the flow rate needed in /s for a Reynolds number of 2100 and

m

the velocity in m/s.

Ans.

(a)

N Rc =

5723, turbulent flow

25-2. Pipe Diameter and Reynolds Number. An oil is being pumped inside a 10.0-mmdiameter pipe at a Reynolds number of 2100. The oil density is 855kg/m 3 and _2 Pa-s. the viscosity is 2.1 x 10 (a) What is the velocity in the pipe? (b) It is desired to maintain the same Reynolds number of 2100 and the same 3 velocity as in part (a) using a second fluid with a density of 925kg/m and a pipe viscosity of 1.5 x 10 Pa s. What diameter should be used? •

2.6-1.

Average Velocity for Mass Balance

in

Flow Down Vertical Plate. For a layer of

liquid flowing in laminar flow in the z direction

the velocity profile

where

5

liquid

toward

the thickness of the layer, x

is

down a

vertical plate or surface,

is

the plate,

and

is

the distance from the free surface of the

v, is the velocity at

a distance x from the free

surface.

maximum velocity o lmax ?

(a)

What

(b)

Derive the expression for the average velocity

is

the

Ans 2.6-2.

-

(

a)

u :av

o* max

and

also relate 2

=

Pff<5

it

/2^, (b) v z av

to v zm3X

=

.

§y z max

in a Pipe and Mass Balance. A hydrocarbon liquid enters a simple flow system shown in Fig. 2.6-1 at an average velocity of 1.282 m/s, where 2 3 and p, = 902 kg/m 3 The liquid is heated in the process /I, = 4.33 x 10" and the exit density is 875 kg/m 3 The cross-sectional area at point 2 is 2 3 5.26 x 10" The process is steady state. (a) Calculate the mass flow rate m at the entrance and exit. (b) Calculate the average velocity v in 2 and the mass velocity G in 1. 2 (a)m, = m 2 = 5.007 kg/s, (b) G = 1156kg/s-m Ans.

Flow of Liquid

m

.

.

m

.

,

,

2.6-3.

Average Velocity for Mass Balance

smooth

in

Turbulent Flow. For turbulent flow in a

circular tube with a radius of R, the velocity profile varies according to

the following expression at a Reynolds

number

R — R where

r is

the radial distance from the center

A

of about 10

5 :

111

and

» max the

maximum

velocity at

the center. Derive the equation relating the average velocity (bulk velocity) u av to v m3X for

an incompressible

substituting z for

R —

fluid. (Hint:

The

integration can be simplified by

r.)

Ans.

2.6-4.

vI ,

= l^\v mix =

0.Snv„

Bulk Velocity for Flow Between Parallel Plates. A fluid flowing in laminar flow in the x direction between two parallel plates has a velocity profile given by the following.

where 2y 0

106

is

the distance between the plates, y

is

the distance from the center

Chap. 2

Problems

and

line,

vx

is

the velocity in the

x

direction at position y. Derive an equation

relating v xay (bulk or average velocity) with vx max 2.6-5.

Mass Balance for

Overall

Dilution Process.

A

.

well-stirred storage vessel con-

methanol solution (wA = 0.05 mass fraction alcohol). A constant flow of 500 kg/min of pure water is suddenly introduced into the tank and a constant rate of withdrawal of 500 kg/min of solution is started. These two flows are continued and remain constant. Assuming that the densities of the solutions are the same and that the total contents of the tank remain constant at 10000 kg of solution, calculate the time for the alcohol content to drop to 1.0 wt %. Ans. 32.2 min tains

2.6-6.

10000 kg of solution of a

Overall stirred

Mass Balance

dilute

for Unsteady-State Process.

A

storage vessel

and contains 500 kg of total solution with a concentration of

5.0

is

well

%

salt.

%

A constant flow rate of 900 kg/h of salt solution containing 16.67 salt is suddenly introduced into the'tank and a constant withdrawal rate of 600 kg/h is also started. These two flows remain constant thereafter. Derive an equation relating the outlet withdrawal concentration as a function of time. Also, calculate the

2.6- 7

concentration after 2.0

h.

Mass Balance for Flow of Sucrose Solution. A 20 wt % sucrose (sugar) solution 3 having a density of 1074 kg/m is flowing through the same piping system as Example 2.6-1 (Fig. 2.6-2). The flow rate entering pipe is 1.892 m 3 /h. The flow 1

divides equally in each of pipes (a)

(b)

The velocity in m/s in pipes The mass velocity G kg/m z

3.

•

Calculate the following:

2

and

s

in pipes 2

3.

and

3.

2.7- 1. Kinetic-Energy Velocity Correction Factor for Turbulent Flow.

Derive the equa-

tion to determine the value of a, the kinetic-energy velocity correction factor, for

turbulent flow. Use Eq. (2.7-20) to approximate the velocity profile and substitute this into Eq. (2.7-15) to obtain (u (2.7- 14) to

obtain

3 )

aV Then use Eqs.

(2.7-20), (2.6-17),

Ans. 2.7-2.

and

a.

a

=

0.9448

Flow Between Parallel Plates and Kinetic-Energy Correction Factor. The equation for the velocity profile for a fluid flowing in laminar flow between two parallel plates is given in Problem 2.6-4. Derive the equation to determine the value of the kinetic-energy velocity correction factor a. [Hint: First derive an equation relating

v to u av

.

Then

derive the equation for(u

3 )

av

and, finally, relate

these results to a.] 2.7-3.

Temperature Drop in Throttling Valve and Energy Balance. Steam is flowing through an adiabatic throttling valve (no heat loss or external work). Steam Renters point 1 upstream of the valve at 689 kPa abs and 171.TC and leaves the valve (point 2) at 359 kPa. Calculate the temperature t 2 at the outlet. [Hint: Use Eq. (2.7-21) for the energy balance and neglect the kinetic-energy and potential-energy terms as shown in Example 2.7-1. Obtain the enthalpy H from Appendix A. 2, steam tables. For H 2 linear interpolation of the values in the table will have to be done to obtain f 2 .] Use SI units. Ans. t 2 = 160.6°C l

,

2.7-4.

Chap. 2

Energy Balance on a Heat Exchanger and a Pump. Water at 93.3°C is being pumped from a large storage tank at 1 atm abs at a rate of 0.189m 3 /min by a pump. The motor that drives the pump supplies energy to the pump at the rate of 1.49 kW. The water is pumped through a heat exchanger, where it gives up 704 kW of heat and is then delivered to a large open storage tank at an elevation of 15.24 m above the first tank. What is the final temperature of the water to the second tank? Also, what is the gain in enthalpy of the water due to the work

Problems

ion

input? (Hint : Be sure and use the steam tables for the enthalpy of the water. Neglect any kinetic-energy changes, but not potential-energy changes.) Ans. t 2 = 38.2°C, work input gain = 0.491 kJ/kg 2.7-5.

Steam

kPa

Boiler

and Overall Energy Balance. Liquid water under pressure

at

150

24°C through a pipe at an average velocity of 3.5 m/s in above the liquid inlet at turbulent flow. The exit steam leaves at a height of 25 150°C and 150 kPa absolute and the velocity in the outlet line is 12.5 m/s in turbulent flow. The process is steady state. How much heat must be added per kg of steam? enters a boiler at

m

2.7-6.

Energy Balance on a Flow System with a Pump and Heat Exchanger. Water stored in a large, well-insulated storage tank at 21.0°C and atmospheric pressure is being pumped at steady state from this tank by a pump at the rate of 40 m 3 /h. The motor driving the pump supplies energy at the rate of 8.5 kW. The water is used as a cooling medium and passes through a heat exchanger where 255 kW of heat is added to the water. The heated water then flows to a second, large vented tank, which is 25 m above the first tank. Determine the final temperature of the water delivered to the second tank.

2.7-7.

Mechanical-Energy Balance in Pumping Soybean Oil. Soybean oil is being pumped through a uniform-diameter pipe at a steady mass-flow rate. A pump supplies 209.2 J/kg mass of fluid flowing. The entrance abs pressure in the inlet 2 pipe to the pump is 103.4 kN/m The exit section of the pipe downstream from above the entrance and the exit pressure is 172.4 kN/m 2 the pump is 3.35 .

m

.

and entrance pipes are the same diameter. The fluid is in turbulent flow. Calculate the friction loss in the system. See Appendix A.4 for the physical properties of soybean oil. The temperature is 303 K. Exit

£F=

Ans. 2.7-8.

Pump Horsepower in

Brine System.

A pump pumps 0.200

3 ft

/s of

101.3 J/kg

brine solution

g/cm 3 from an open feed tank having a large crosssectional area. The suction line has an inside diameter of 3.548 in. and the discharge line from the pump a diameter of 2.067 in. The discharge flow goes to an open overhead tank and the open end of this line is 75 ft above the liquid level in the feed tank. If the friction losses in the piping system are 18.0 ftlb /lb m what pressure must the pump develop and what is the horsepower of having a density of

,

f

the 2.7-9.

1.15

pump

if

the efficiency

is

70% ? The

flow

is

turbulent.

3 Pressure Measurements from Flows. Water having a density of 998 kg/m is flowing at the rate of 1.676 m/s in a 3. 068-in. -diameter horizontal pipe at a pressure pi of 68.9 kPa abs. It then passes to a pipe having an inside diameter of

2.067

in.

Calculate the new pressure p 2 in the 2.067-in. pipe.

(a)

Assume no

friction

losses. If

(b)

p2

the piping

is

vertical

The pressure tap

and the flow

for p 2

is

0.457

is

upward, calculate the new pressure

m above the tap for p

Ans.

(a)p 2

=

63.5

l

.

kPa;(b)p 2

=

59.1

kPa

Cotton Seed Oil from a Tank. A cylindrical tank 1.52 m in diameter 3 and 7.62 m high contains cotton seed oil having a density of 917 kg/m The tank is open to the atmosphere. A discharge nozzle of inside diameter 15.8 and cross-sectional area A 2 is located near the bottom of the tank. The surface of the liquid is located at H = 6.1 m above the center line of the nozzle. The discharge nozzle is opened, draining the liquid level from H = 6. 1 m to H = 4.57 m. Calculate the time in seconds to do this. [Hint : The velocity on the surface of the reservoir is small and can be neglected. The velocity v 2 m/s in the nozzle can be calculated for a given H by Eq. (2.7-36). However, H, and hence are varying. Set up an unsteady-state mass balance as follows. The voluv2 metric flow rate in the tank is (A, dH)/dt, where A, is the tank cross section in m 2

2.7-10. Draining

.

mm

,

108

Chap. 2

Problems

dH is the m 3

and A,

dH

since

H=

6.1

is

must equal the negative of The negative sign is present

liquid flowing in dt s. This rate

the volumetric rate in the nozzle, or the negative of v 2

matt = Oand

H=

.

— A 2 v 2 m 3 /s.

Rearrange

4.57

matt =

this

equation and integrate between

r f .]

Ans.

=

1380

s

Turbine Water Power System. Water is stored in an elevated reservoir. To generate power, water flows from this reservoir down through a large conduit to a turbine and then through a similar-sized conduit. At a point

2.7-11. Friction •

tF

Loss

in

conduit 89.5 m above the turbine, the pressure is 172.4 kPa and at a level below the turbine, the pressure is 89.6 kPa. The water flow rate is 3 0.800 m /s. The output of the shaft of the turbine is 658 kW. The water density in the

5

m

3 1000 kg/m If the efficiency of the turbine in converting the mechanical = 0.89), calculate the energy given up by the fluid to the turbine shaft is 89% friction loss in the turbine in J/kg. Note that in the mechanical-energy-balance equation, the 5 is equal to the output of the shaft of the turbine over/;,. is

.

W

£F = 85.3 J/kg

Ans.

Pumping of Oil. A pipeline laid cross country carries oil at the rate of 3 795 m /d. The pressure of the oil is 1793 kPg gage leaving pumping station 1. The pressure is 862 kPa gage at the inlet to the next pumping station, 2. The second station is 17.4 m higher than the first station. Calculate the lost work 3 friction loss) in J/kg mass oil. The oil density is 769 kg/m (Y, F

2.7-12. Pipeline

.

2.7-13.

Test of Centrifugal

Pump and Mechanical-Energy

Balance.

A

centrifugal

pump

being tested for performance and during the test the pressure reading in the 0.305-m-diameter suction line just adjacent to the pump casing is —20.7 kPa (vacuum below atmospheric pressure). In the discharge line with a diameter of 0.254 at a point 2.53 m above the suction line, the pressure is 289.6 kPa gage. The flow of water from the pump is measured asO.l 133m 3 /s. (The density can be is

m

assumed

as 1000

kg/m 3

.)

Calculate the

kW input of the pump. Ans.

38.11

kW

Pump and Flow

System. Water at 20°C is pumped from the bottom of a large storage tank where the pressure is 310.3 kPa gage to a nozzle which is 15.25 m above the tank bottom and discharges to the atmosphere with a velocity in the nozzle of 19.81 m/s. The water flow rate is 45.4 kg/s. The efficiency of the pump is 80% and 7.5 kW are furnished to the pump shaft. Calculate the following.

2.7-14. Friction Loss in

(a)

(b)

2.7-15.

The The

friction loss in the

pump.

friction loss in the rest of the process.

Power for Pumping

in

Flow System. Water

is

pumped from an open water open storage tank 1500 m away.

being

reservoir at the rate of 2.0 kg/s at 10°C to an

The pipe used is schedule 40 3{-in. pipe and the frictional losses in the system are 625 J/kg. The surface of the water reservoir is 20 m above the level of the storage tank. The pump has an efficiency of 75%. (a) What is the kW power required for the pump? (b)

If

the

pump

is

not present

in

the system, will there be a flow?

Ans. 2.8-1.

(a) 1.143

kW

Momentum

Balance in a Reducing Bend. Water is flowing at steady state through the reducing bend in Fig. 2.8-3. The angle a 2 = 90° (a right-angle bend). The pressure at point 2 is 1.0 atm abs. The flow rate is 0.020 m 3 /s and the diameters at points 1 and 2 are 0.050 and 0.030 m, respectively. Neglect frictional and gravitational forces. Calculate the resultant forces on the bend in newtons and lb force. Use p = 1000 kg/m 3 = + 450.0 N, -R, = -565.8 N. Ans. x

m .

-R

2.8-2.

Forces on Reducing Bend. Water is flowing at steady state and 363 K at a rate 3 /s through a 60" reducing bend (a 2 = 60°) in Fig. 2.8-3. The inlet of 0.0566

m

Chap. 2

Problems

109

is 0.1016 m and the outlet 0.0762 m. The friction loss in the pipe bend can be estimated as vf/5. Neglect gravity forces. The exit pressure 2 p 2 = 1 1 1-5 kN/m gage. Calculate the forces on the bend in newtons. Ans. -R x = +1344 N, ~R y = -1026 N

pipe diameter

'

Force of Water Stream on a Wall. Water at 298 K. discharges from a nozzle and travels horizontally hitting a flat vertical wall. The nozzle has a diameter of 12 and the water leaves the nozzle with a flat velocity profile at a velocity of 6.0 m/s. Neglecting frictional resistance of the air on the jet, calculate the force in

2.8-3.

mm

newtons on the

wall.

—R x = 4.059 N Water at a steady-state rate of 0.050 m /s Ans.

3

Flow Through art Expanding Bend. is flowing through an expanding bend that changes direction by 120°. The upstream diameter is 0.0762 and downstream is 0.2112 m. The upstream pressure is 68.94 kPa gage. Neglect energy losses within the elbow and calculate the downstream pressure at 298 K. Also calculate R x and R f

2.8-4.

m

.

Force of Stream on a Wall.

2.8-5.

Assume no the plate

Repeat Problem 2.8-3 for the same conditions

inclined 45° with the- vertical.

The flow is frictionless. The amount of fluid splitting in each direction along can be determined by using the continuity equation and a momentum

except that the wall

is

loss in energy.

balance. Calculate this flow division and the force on the wall.

Ans.

m2 =

2.8-6.

m3 =

0.5774 kg/s,

-R y =

-2.030

N

(force

0.09907 kg/s,

on

-Rx =

2.030 N,

wall).

Momentum

Balance for Free Jet on a Curved, Fixed Vane. A free jet having a 2 m/s and a diameter of 5.08 x 10" m is deflected by a curved, fixed vane as in Fig. 2.8-5a. However, the vane is curved downward at an angle of 60° instead of upward. Calculate the force of the jet on the vane. The density 3 is 1000 kg/m Ans. x = 942.8 N, —R y = 1633

velocity of 30.5

.

N

—R

2.8-7.

Balance for Free Jet on a U-Type, Fixed Vane. A free jet having a 2 velocity of 30.5 m/s and a diameter of 1.0 x 10~ m is deflected by a smooth, fixed vane as in Fig. 2.8-5a. However, the vane is in the form of a U so that the

Momentum

exit jet travels in

force of the jet

2.8-8.

a direction exactly opposite to the entering

on the vane. Use p

=

jet.

Calculate che

3

1000 kg/m Ans. .

-R x =

146.1 N,

-R y

=

0

Momentum Balance

on Reducing Elbow and Friction Losses. Water at 20°C is flowing through a reducing bend, where cc 2 (see Fig. 2.8-3) is 120°. The inlet pipe 3 diameter is 1.829 m, the outlet is 1.219 m, and the flow rate is8.50 /s. The exit point z 2 is 3.05 m above the inlet and the inlet pressure is 276 kPa gage. Friction losses are estimated as 0.5u 2 /2 and the mass of water in the

m

•

elbow

is

R x and R y and

8500 kg. Calculate the forces

'

control volume fluid.

2.8- 9.

Momentum momentum

the resultant force on the

Velocity Correction Factor p for Turbulent Flow. Determine the (i for turbulent flow in a tube. Use Eq.

velocity correction factor

(2.7-20) for the relationship

between

v

and

position.

of Water on Wetted-Wall Tower. Pure water at 20°C is flowing down a column at a rate of 0.124 kg/s m. Calculate the film thickness and the average velocity. Ans. <5 = 3.370 x 0~* m,v „ = 0.3687 m/s

2.9- 1. Film

vertical wetted-wall

•

1

2.9-2. Shell

density

is

Balance for Flow Between Parallel Plates. A fluid of constant flowing in laminar flow at steady state in the horizontal x direction

between two vertical

flat

and

y direction

is

parallel plates.

2y 0

.

Using a

The

shell

for the velocity profile within this fluid

110

z

Momentum

distance between the two plates in the

momentum balance, derive the equation and the maximum velocity for a distance Chap.

2

Problems

Lm

in

the

2.9-3.

the method used in Section 2.9B to derive Eq. used \sdv x /dy = 0 at y = 0.]

x direction. [Hint : See

One boundary condition

(2.9-9).

Non-Newtonian

Velocity Profile for

Newtonian

fluid

is

Fluid.

The

stress rate of shear for a

non-

given by

<--* (-£)" where

K

and n are constants. Find

the relation

between velocity and radial

incompressible fluid at steady state. [Hint: Combine the equation given here with Eq. (2.9-6). Then raise both sides of the resulting equation to the \/n power and integrate.] position

r for this

^-M^T^

Hi)

Momentum Balance for Flow Down an Inclined Plane. Consider the case of a Newtonian fluid in steady-state laminar flow down an inclined plane surface that makes an angle 9 with the horizontal. Using a shell momentum balance, find the equation for the velocity profile within the liquid layer having a

2.9- 4. Shell

maximum velocity of the free surface. (Hint : The convecterms cancel for fully developed flow and the pressure-force terms also cancel, because of the presence of a free surface. Note that there is a

L and momentum

the

thickness tive

gravity force

on the

fluid.)

Ans. 2.10-1.

Viscosity

Measurement of a

One

Liquid.

= pgL 2

u, max

sin 9/2 fx

use of the Hagen-Poiseuille equation

is in determining the viscosity of a liquid by measuring the pressure drop and velocity of the liquid in a capillary of known dimensions. The liquid 3 used has a density of 9^2 kg/m and the capillary has a diameter of 2.222 mm 7 3 and a length of 0.1585 m. The measured flow rate was 5.33 x I0" m /s of 3 liquid and the pressure drop 131 mm of water (density 996 kg/m ). Neglecting end effects, calculate the viscosity of the liquid in Pa s.

(2.10-2)

Ans.

"

,

2.10-2.

Frictional Pressure

drop ...

Drop

in

is

1.22 m/s.

Use

Loss

m and

a length of 76.2 m.

3

3

Pa

s

Is

The

velocity of the fluid

the flow laminar or turbulent?

A. 4.

and

in Straight Pipe

density of 801 kg/m

9.06 x 10"

Oil. Calculate the frictional pressure flowing through a commercial pipe having

K

the friction factor method.

Use physical data from Appendix 2.10-3. Frictional

=

Flow of Olive

in pascal for olive oil at 293

an inside diameter of 0.0525

fi

Effect

and a viscosity of

of Type of Pipe. 10"

A

liquid

having a

3

Pa s is flowing through The commercial steel pipe

1.49 x

a

horizontal straight pipe at a velocity of 4.57 m/s. is ly-in. nominal pipe size, schedule 40. For a length of pipe of 61 m, do as follows. (a) Calculate the friction loss F f .

(b)

For a smooth tube of the same

What

is

Ans. 2.10- 4.

inside diameter, calculate the friction loss.

the percent reduction of the (a)

Ff

for the

348.9 J/kg;(b) 274.2 J/kg(9

Trial-and-Error Solution for Hydraulic Drainage. iron pipe having an inside diameter of 0.156

drain wastewater at 293 K.

Chap. 2

smooth tube?

1.7

Problems

The

m

available head

ft

•

lb / /lb m

),

21.4% reduction

In a hydraulic project a cast

and a 305-m length is

4.57

m

is

used to

of water. Neglecting

111

any

losses in fittings

(Hint

Assume

:

and joints

the physical properties of pure water.

error since the velocity appears in

As a

factor.

2.10-5.

first trial,

flow rate in

in the pipe, calculate the

N Kc

assume that

v

,

=

which

is

The

solution

m 3 /s.

is trial

needed to determine the

and

friction

1.7 m/s.)

Mechanical-Energy Balance and Friction Losses. Hot water is being discharged 3 from a storage tank at the rate of 0.223 ft /s. The process flow diagram and conditions are the same as given in Example 2.10-6, except for different nominal pipe sizes of schedule 40 steel pipe as follows. The 20-ft-long outlet pipe from the storage tank is ly-in. pipe instead of 4-in. pipe. The other piping, which was 2-in. pipe, is now 2.5-in. pipe. Note that now a sudden expansion occurs after the elbow in the 1-j-in. pipe to a 2^-in pipe.

and Pump Horsepower. Hot water in an open storage tank at 3 being pumped at the rate of 0.379 /min from this storage tank. The line from the storage tank to the pump suction is 6.1 m of 2-in. schedule 40 steel pipe and it contains three elbows. The discharge line after the pump is 61 of 2-in. pipe and contains two elbows' The water discharges to the atmosphere at a height of 6. 1 m above the water level in the storage tank.

2.10-6. Friction Losses

82.2°C

m

is

m

(a)

Calculate

(b)

Make a What is

(c)

all frictional

losses

£ F.

mechanical-energy balance and calculate the

kW power of the pump

if its

Ans. 2.10-7. Pressure

Drop of

efficiency

W

s

is

of the

pump

(a)Xf = 122.8 J/kg. (b) Ws = - 186.9 J/kg,

a Flowing Gas. Nitrogen gas

in J/kg.

75%? (c)

schedule 40 commerical steel pipe at kg/s and the flow can be assumed as isothermal. The pipe the inlet pressure is 200 kPa. Calculate the outlet pressure.

is

Ans,

3000 p2

m =

Length for Flow in a Pipe. Air at 10°C and 1.0 atm abs pressure a velocity of 2.0 m/s inside a tube having a diameter of 0.012 m.

2.10-8. Entry at

1.527

kW

flowing through a 4-in. -2 298 K. The total flow rate is 7.40 x 10 is

(a)

Calculate the entry length.

(b)

Calculate the entry length for water at 10°C and the

same

long and 188.5 is

kPa

flowing

velocity.

Pumping Oil to Pressurized Tank. An oil having a density of 833 x 10~ 3 Pas is pumped from an open tank to a pressurized tank held at 345 kPa gage. The oil is pumped from an inlet at the side of the open tank through a line of commercial steel pipe having an inside 3 diameter of 0.07792 m at the rate of 3.494 x 10" m 3 /s. The length of straight pipe is 122 m and the pipe contains two elbows (90°) and a globe valve half open. The level of the liquid in the open tank is 20 m above the liquid level in the pressurized tank. The pump efficiency is 65%. Calculate the kW power of the pump.

2.10-9. Friction Loss in

kg/m 3 and

2.10- 10-...

a viscosity of 3.3

an Annulus and Pressure Drop. Water flows in the annulus of a horiand is being heated from 40°C to 50°C in the exchanger which has a length of 30 of equivalent straight pipe. The flow 3 3 rate of the water is 2.90 x 10" m /s. The inner pipe is 1-in. schedule 40 and the outer is 2-in. schedule 40. What is the pressure drop? Use an average temperature of 45°C for bulk physical properties. Assume that the wall temperature is an average of 4°C higher than the average bulk temperature so that a correction can be made for the effect of heat transfer on the friction factor.

Flow

in

zontal, concentric-pipe heat exchanger

m

2.11- 1. Derivation of Maximum Velocity for Isothermal Compressible Flow. Starting with Eq. (2.11-9), derive Eqs. (2.11-11) and (2.11-12) for the maximum velocity in

isothermal compressible flow.

Drop in Compressible Flow. Methane gas is being pumped through a 305-m length of 52.5-mm-ID steel pipe at the rate of 41.0 kg/m 2 -s. The inlet pressure is p = 345 kPa abs. Assume isothermal flow at 288.8 K.

2.11-2. Pressure

{

112

Chap.

2

Problems

(a)

Calculate the pressure p 2 at the end of the pipe. Pa-s.

(b)

Calculate the

maximum

and compare with

The

viscosity

is

1.04

x 10

5

velocity that can be attained at these conditions

the velocity in part

Ans.

(a)

(a).

p2 v2

= =

298.4 kPa, (b)

i>

max

=

387.4 m/s,

20.62 m/s

K

Isothermal Compressible Flow. Air at 288 and 275 kPa abs is flowing in isothermal compressible flow in a commercial pipe having an ID of 0.080 m. The length of the pipe is 60 m. The mass velocity 2 at the entrance to the pipe is 165.5 kg/m -s. Assume 29 for the molecular weight of air. Calculate the pressure at the exit. Also, calculate the maximum

2.11-3. Pressure

Drop

in

enters a pipe and

allowable velocity that can be attained and compare with the actual.

REFERENCES

(El)

Bennett, C. O., and Meyers, J. E. Momentum, Heat and Mass Transfer, 3rd ed. New York: McGraw-Hill Book Company, 1982. Dodge, B. F. Chemical Engineering Thermodynamics. New York: McGraw-Hill Book Company, 1944. Earle, R. L./Unit Operations in Food Processing. Oxford: Pergamon Press, Inc.,

(Kl)

Kittridge, C.

(LI)

New York: McGraw-Hill Book Company, Langha ar, H. L. Trans. A.S.M.E, 64, A-55 ( 942). Moody, L. F. Trans. A.S.M.E., 66, 67 (1944); Mech Eng., 69, 1005 (1947). McCabe, W.L., Smith, J. C, and Harriott, P. Unit Operations of Chemical Engineering, 4th ed. New York: McGraw-Hill Book Company, 1985. National Bureau of Standards. Tables of Thermal Properties of Gases, Circular

(Bl)

(Dl)

1966.

(L2)

(Ml) (M2) (Nl)

Lange, N.

P.,

and Rowley, D.

S.

Trans. A.S.M.E.,19, 1759(1957).

A. Handbook of Chemistry, 10th ed. 1967. 1

1

464(1955). Perry, R. H., and Green, D. Perry's Chemical Engineers' Handbook, 6th ed. New York: McGraw-Hill Book Company, 1984.

(PI)

(R2)

Reid, R. C, Prausnitz, J. M., and Sherwood, T. K. The Properties of Gases and Liquids, 3rd ed. New York: McGraw-Hill Book Company, 1977. Reactor Handbook, vol. 2, AECD-3646. Washington D.C.: Atomic Energy Com-

(51)

Swindells,

(Rl)

mission,

1

May

1955.

J. F.,

Coe,

J.

R.

Jr.,

and Godfrey, T.

B. J. Res. Nat. Bur. Standards, 48,

(1952).

Sklelland, A. H. P. Non-Newtonian Flow and Heat Transfer. Wiley Sons, Inc., 1967.

(52)

New York: John

&

(53)

Sieder, E. N., and Tate, G. E. Ind. Eng. Chem., 28, 1429 (1936).

(Wl)

Weast, R. C. Handbook of Chemistry and Physics, 48th Chemical Rubber Co., Inc., 1967-1968.

Chap.

2

References

ed.

Boca Raton,

Fla.:

113

CHAPTER

3

Principles of

Momentum

Transfer

and Applications

FLOW PAST IMMERSED OBJECTS AND PACKED AND FLUIDIZED BEDS

3.1

.

3.1A

Definition of

Drag

Coefficient for

Flow Past Immersed

Objects /.

Introduction and types of drag. In Chapter 2 we were concerned primarily with the transfer and the factional losses for flow of fluids inside conduits or pipes.

momentum

In this section

we

consider

in

some

detail the flow

of

fluids

around

solid,

immersed

objects.

The flow of fluids outside immersed bodies appears in many chemical engineering and other processing applications. These occur, for example, in flow past spheres in settling, flow through packed beds in drying and filtration, flow past tubes in applications

heat exchangers, and so on. the force on the

It is

useful to be able to predict the frictional losses and/or

submerged objects

in these various applications.

In the examples of fluid friction inside conduits that transfer of

momentum

drag on the smooth surface parallel fluid

on the

is

fluid,

is

skin friction will

in

in a tangential

to the direction of flow.

solid in^the direction of flow

contact with a flowing

we considered

perpendicular to the surface resulted

Chapter

the

This force exerted by the

called skin or wall drag.

For any surface

In addition to skin friction,

exist.

2,

shear stress or

if

in

the fluid

not flowing parallel to the surface but must change directions to pass around a solid

body such

as a sphere, significant additional frictional losses will

occur and

this

is

called

form drag. In Fig.

3.

1 - 1

and the force F

is parallel to the smooth surface of the flat, solid plate, newtons on an element of area dA m 2 of the plate is the wall shear

a the flow of fluid in

stress t„ times the

area

dA

or

tw

dA. The total force

is

the

sum

of the integrals of these

quantities evaluated over the entire area of the plate. Here the transfer of the surface results in a tangential stress or skin drag

In

many

cases,

however, the immersed body

is

various angles to the direction of the fluid flow. As velocity

114

is

v0

and

is

on

momentum

to

the surface.

a blunt-shaped solid

shown

which presents

in Fig. 3.1-lb, the free-stream

uniform on approaching the blunt-shaped body suspended

in a

very

large duct. Lines called streamlines represent the path of fluid elements around the suspended body. The thin boundary layer adjacent to the solid surface is shown as a dashed line and at the edge of this layer the velocity is essentially the same as the bulk fluid velocity

adjacent to

it.

At the front center of the body, called the stagnation point,

the fluid velocity will be zero and boundary-layer growth begins at this point and it separates. The tangential stress on the body because of boundary layer is the skin friction. Outside the boundary changes direction to pass around the solid and also accelerates near the

continues over the surface until the velocity gradient layer the fluid

in the

and then decelerates. Because of these effects, an additional force is exerted by the on the body. This phenomenon, called form drag, is in addition to the skin drag in the boundary layer. In Fig. 3.1-lb, as shown, separation of the boundary layer occurs and a wake, covering the entire rear of the object, occurs where large eddies are present and contribute to the form drag. The point of separation depends on the shape of the particle, Reynolds number, and so on, and is discussed in detail elsewhere (S3). Form drag for bluff bodies can be minimized by streamlining the body (Fig. 3.1- 1c), which forces the separation point toward the rear of the body, which greatly reduces the size of the wake. Additional discussion of turbulence and boundary layers is given in front fluid

Section 3.10.

Drag coefficient. From the previous discussions it is evident that the geometry of the immersed solid is a main factor in determining the amount of total drag force exerted on the body. Correlations of the geometry and flow characteristics for solid objects suspended or held in a free stream (immersed objects) are similar in concept and form to the

2.

friction

Sec. 3.1

factor-Reynolds number correlation given for flow inside conduits. In flow

Flow Past Immersed Objects and Packed and Fluidized Beds

115

through conduits, the

was defined as the

friction factor

ratio of the

drag force per unit

area (shear stress) to the product of density times velocity head as given in Eq. (2.10-4). In a similar

manner

as the ratio of the total

immersed

for flow past

objects, the

drag coefficient C D

is

defined

drag force per unit area to ptfo/2.

^

CD =

'

(SI)

pu 0/2

(3.1-1)

F

Cd = where F D

is

P

(English)

°ln

Ap

the total drag force in N,

is

an area

in

m2 C D ,

is

dimensionless,

v0

is

and p is density of fluid in kg/m 3 In English units, F D is in 3 and A p in ft 2 The area A p used is the area obtained by lb f v 0 is in ft/s, p is in lb^ft projecting the body on a plane perpendicular to the line of flow. For a sphere, A = p nD p /4, where D p is sphere diameter; for a cylinder whose axis is perpendicular to the flow direction, A = LD p> where L = cylinder length. Solving Eq. (3.1-1) for the total drag p free-stream velocity in m/s,

.

.

,

,

force,

F0 = C D The Reynolds number

for a given solid

pA p

immersed

(3.1-2)

in

a flowing liquid

is

fj.

fj.

where

G0 =

3.1B

Flow Past Sphere, Long Cylinder, and Disk

v 0 p.

For each particular shape of object and orientation of the object with the direction of flow, a different relation of C D versus N Rc exists. Correlations of drag coefficient versus Reynolds number are shown in Fig. 3.1-2 for spheres, long cylinders, and disks. The face of the disk and the axis of the cylinder are perpendicular to the direction of flow. These curves have been determined experimentally. However, in the laminar region for low Reynolds numbers

same

less

than about

as the theoretical Stokes'

1.0,

the experimental drag force for a sphere

F D = 3npD p v 0 Combining Eqs. Stokes' law

(3.1-2)

and

the

(3.1-4)

and solving

for

(3.1-4)

CD

,

the drag coefficient predicted by

is

CD = The

is

law equation as follows.

variation of

CD

with

JV Rc

24

D p v 0 p/p

(Fig.

=

~ N

(3.1-5)

Rc

3.1-2)

is

quite complicated because of the

and form drag. For a sphere as the Stokes' law range, separation occurs and a

interaction of the factors that control skin drag

Reynolds number

wake jV Re

=

is

3

is

increased

beyond

formed. Further increases in

x 10 5 the sudden drop

in

the

N Re CD

cause

is

shifts in the

the result of the

separation point. At about

boundary

layer

becoming

completely turbulent and the point of separation moving downstream. In the region of

N Rc of about

x 10 3

x 10 5 the drag

coefficient is approximately constant for each shape and C D = 0.44 for a sphere. Above a Rc of about 5 x 10 5 the drag coefficients are again approximately constant with C D for a sphere being 0.13, 0.33 for a cylinder, and 1

to 2

N

116

Chop. 3

Principles

of Momentum Transfer and Applications

0.001

0.01

0.1

10

1.0

10

3.1-2.

Drag

coefficients for

10

3

10

4

10

s

10

6

=

Reynolds number, N„

Figure

2

flow past immersed spheres, long

cylinders,

and

(Reprinted with permission from C. E. Lapple and C. B. Shepherd, Ind. Eng. Chem., 32, 606 (1940). Copyright by the American disks.

Chemical Society.)

1.12 for a disk. Additional discussions

and theory on flow past spheres are given

in

Section 3.9E.

For derivations of theory and detailed discussions of the drag force for flow parallel on boundary-layer flow and turbulence should be consulted.

to a flat plate, Section 3.10

The flow

different geometries.

spacings,

and number of

of fluids normal to banks of cylinders or tubes occurs in heat exchangers

other processing applications.

Because of the

many

of tubes can be arranged in a

possible geometric tube configurations and

not possible to have one correlation of the data on pressure drop and

is

it

The banks

friction factors.

many correlations

Details of the

available are given elsewhere (PI).

EXAMPLE

3.1-1. Force on Submerged Sphere Air at 37.STC and 101.3 kPa absolute pressure flows past a sphere having a diameter of 42 at a velocity of 23 m/s. What is the drag coefficient C D and the force on the sphere?

mm

1.90 x

1

= A.3 for air at 37.8°C, p = 1.137 kg/m /j = 0.042 and v 0 = 23.0 m/s. Using Eq. (3.1-3), 3

From Appendix

Solution:

(T

5

Pa-

Also,

s.

Dp

_ FJ^p _ ~ pl

,

m

0.042(23.0X1.137) 1.90

From Fig. 3.1-2 for a sphere, C D = A p = nD p2 /4 for a sphere,

xlO"

Water

at

Sec. 3.1

1

37M

^

where

- 0.1958 N

Force on a Cylinder in a Tunnel flowing past a long cylinder at a velocity of 1.0 m/s in a axis of the cylinder is perpendicular to the direction of

3.1-2.

24°C

large tunnel.

-^ /Slxlu

0.47. Substituting into Eq. (3.1-2),

F.-C.*,A,- (0.47) 2f£ EXAMPLE

5

is

The

Flow Past Immersed Objects and Packed and Fluidized Beds

117

The diameter of

flow.

the cylinder

What

0.090 m.

is

is

the force per meter

length on the cylinder?

From Appendix

Solution:

H = 0.9142 x 10~ Using Eq. (3.1-3),

3

Pa-s. Also,

Re

Dp =

Fig. 3.1-2 for a long cylinder,

where

/I.

1.0(0.090)

=

L=

0.090 m,

0.9142 x 10

u

From

= LD. =

CD =

0.090

m2

1.0

C

1.

Introduction.

Flow

in

(1.4)

997.2

m, and v 0

=

kg/m 3

,

1.0 m/s.

3

1.4.

Substituting into Eq. (3.1-2),

,

V

f d = C d ^- pA p = 3.1

—

A.2 for water at 24°C, p

(997.2X0.09)

=

62.82

N

Packed Beds

A system

engineering fields

of considerable importance in chemical and other process

column which

the packed bed or packed

is

catalytic reactor, adsorption of a solute, absorption, filter bed,

is

used

for

a fixed-bed

and so on. The packing

in the bed may be spheres, irregular particles, cylinders, or various kinds of commercial packings. In the discussion to follow it is assumed that the packing is everywhere uniform and that little or no channeling occurs. The ratio of diameter of the tower to the packing diameter should be a minimum of 8 1 to 10 1 for wall effects to be small. In the theoretical approach used, the packed column is regarded as a bundle of

material

:

:

crooked tubes of varying cross-sectional area. The theory developed in Chapter 2 single straight tubes is used to develop the results for the bundle of crooked tubes. 2.

Laminar flow

in

packed beds.

Certain geometric relations for particles in packed beds

are used in the derivations for flow.

The

void fraction e

volume

m

_1

a.

=

the surface area of a particle in

m

specific surface of a particle a v in

where S p

is

of voids in

in

a packed bed

is

is

defined as

bed

volume of bed (voids plus

total

The

for

solids)

defined as

^ 2

(3.1-7)

and

v

p

the

volume of a

particle in

m

3 .

For

a spherical particle, fl„

where D p

is

diameter

particle diameter

Dp

is

in

=

m. For a packed bed of nonspherical

(1

—

e) is

the

volume

is

(3.1-9)

a.

=

a v {[

-e) =

— ^

(1

- e)

inm"

(3.1-10)

p

the ratio of total surface area in the bed to total

plus particle volume)

118

-

fraction of particles in the bed,

a

where a

particles, the effective

defined as

Dp = Since

(3-1-8)

J"

volume of bed (void volume

1 .

Chap. 3

Principles

of Momentum Transfer and Applications

EXAMPLE

Surface Area in Packed Bed of Cylinders composed of cylinders having a diameter D = 0.02 m and a length h — D. The bulk density of the overall packed bed is 962 kg/m 3 and the density of the solid cylinders is 1600 kg/m 3

A

3.1-3.

packed bed

is

.

Calculate the void fraction a. (b) Calculate the effective" diameter D of the particles. p (c) Calculate the value of a in Eq. (3.1-10).

(a)

Solution:

mass

For part

m

taking 1.00

(a),

kg/m

3

3

of packed bed as a basis, the total

m = 962 3

kg. This mass of 962 kg is ) mass of the solid cylinders. Hence, volume of cylinders = 962 kg/ 3 3 (1600 kg/m ) = 0.601 m Using Eq. (3.1-6),

of the bed

(962

is

(1.00

)

also the

.

e

= volume total

For the

of voids in bed

effective particle

diameter

the surface area of a particle

Sp

The volume

=

Dp

=

0 399

in part (b), for a cylinder

where h

=

D,

is

—

v of a particle

0.601

1.000

nD 2

(2)

—

1.000

volume of bed

(ends)

nD{D)

4-

(sides)

= | nD 2

is

= -£) 2 (Z)) =

—

~

" D

Up

Substituting into Eq. (3.1-7),

G "

v

~

^TtD

p

3

Finally, substituting into Eq. (3.1-9),

DP =

—6 = aa

—=D= 6

0.02

6/D

Hence, the effective diameter to use

is

Dp = D =

m

0.02 m. For part

(c),

using

Eq. (3.1-10),

a

=

J-

(1

-

e)

=

^

(1

-

0.399)

=

180.3 rrT

The average interstitial velocity in the bed is v m/s and it is v' based on the cross section of the empty container by

1

related to the superficial

velocity

v'

The hydraulic

=

(3.1-11)

zv

radius r„ for flow defined in Eq. (2.10-21)

is

modified as follows (B2).

(cross-sectional area\ rH

=

available for flow

/

(wetted perimeter)

void volume available total

for flow

wetted surface of solids

volume of voids/volume of bed

£

wetted surface/volume of bed

a

(3.1-12)

Sec. 3.1

Flow Past Immersed Objects and Packed and Fluidized Beds

119

Combining Eqs.

(3.1-10)

and

(3.1-12),

Since the equivalent diameter

packed bed

is

D

channel

for a

as follows using Eq. (3.1-13)

and

D =

=

ev.

v'

— e)

6(1

p.

(3.1-13)

is

v'

4e

(4r H )vp

Dp

-e)

6(1

£

4r,{

the Reynolds

,

D„

4

p

—

6(1

ix

For packed beds Ergun (El) defined the Reynolds number

number

for a

v'p

n

£)

as

above but without the 4/6

(2.10-2)

can be combined with Eq.

term.

-

(1

where

e)n

-

(1

G = v'p.

For laminar flow, the Hagen-Poiseuille equation (3.1-13) for r H and Eq. (3.1-1 1) to give

^— The

e)/i

true

AL

is

AL

^

32j#) AL

(72)^' AL(1

-li^F""

and use

larger because of the tortuous path

-

1

e)

(3 -

M6)

of the hydraulic radius

Experimental data show that the constant should be 150, which gives the Blake-Kozeny equation for laminar flow, void fractions less than 0.5, effective predicts too large a v

particle diameter

Dp

,

.

and

N Rc <

10.

AL

150/n/

-

(1

2 e)

Dlp J.

Turbulent flow

in

packed beds. For turbulent flow we use the same procedure by and substituting Eqs. (3.1-1 1) and (3.1-13) into this equation to

starting with Eq. (2.10-5)

obtain

Ap =

3fp(v)

2

AL

—

—

1

e

(3.1-18)

3

Dp

£

For highly turbulent flow the friction factor should approach a constant value. Also, it is assumed that all packed beds should have the same relative roughness. Experimental data indicated that 1000,

which

is

3/=

called the

1.75.

Hence, the

equation

final

for turbulent

flow for

W Rc >

Burke-Plummer equation, becomes

Ap =

1.751(10*

AL

1

—

c

p-

(3.1-19)

Adding Eq. (3.1-17) for laminar flow and Eq. (3.1-19) for turbulent flow, Ergun (El) proposed the following general equation for low, intermediate, and high Reynolds numbers which has been tested experimentally. a = Ap

l50 ^ v

'

AL

~5l

(1

-

2 £)

7~ +

l.lSpjv)

2

AL

1

F,

-

E

7~

(3 - 1 " 20)

Rewriting Eq. (3.1-20) in terms of dimensionless groups,

App D p (G')

120

2

£

3

ALl-£

Chap. 3

150

N Rc

,

+

(3.1-21)

1.75

p

Principles of

Momentum

Transfer

and Applications

See also Eq. (3.1-33) for another form of Eq. (3.1-21). The Ergun equation (3.1-21) can be used for gases by using the density p of the gas as the arithmetic average of the inlet and outlet pressures. The velocity v' changes throughout the bed for a compressible fluid, but G' is a constant. At high values of Eqs. (3.1-20) and (3.1-21) reduce R<., P to Eq. (3.1-19) and to Eq. (3.1-17) for low values. For large pressure drops with gases, Eq. (3.1-20) can be written in differential form (PI).

N

EXAMPLE

,

Pressure Drop and Flow

3.1-4.

of Gases in Packed Bed Air at 31

1

K

is

12.7

mm. The

0.61

m

and

flowing through a packed bed of spheres having a diameter of void fraction e of the bed is 0.38 and the bed has a diameter of

The

a height of 2.44 m.

bed drop of the

air enters the

rate of 0.358 kg/s. Calculate the pressure

The average molecular weight

of air

is

at

1.

10

atm abs

air in the

at the

packed bed.

28.97.

-3 311 K,p = 1.90 x 10 Pa- s. The 2 2 2 cross-sectional area of the bed is A = {n/A)D = (tc/4X0.61) = 0.2922 2 Hence, G' — 0.358/0.2922 = 1.225 kg/m s (based on empty cross section of m, AL = 2.44 m, inlet pressure container or bed). D p = 0.0127 5 5 = = x x 10 1.115 10 1.1(1.01325 Pa. ) p, From Eq. (3.1-15),

From Appendix

Solution:

A.3 for

air at

m

.

•

N Rc

- D" G -

P

'

0-0127(1.225)

-

(l-e)/i

(1

5 -0.38X1.90 x 10" )

To

use Eq. (3.1-21) for gases, the density p to use is the average at the p t and outlet p 2 pressures or at (p t + p 2 )/2. This is trial and error since 5 5 p 2 is unknown. Assuming that Ap = 0.05 x 10 Pa, p 2 = 1.115 x 10 — s 5 5 0.05 x 10 = 1.065 x 10 Pa. The average pressure is p av = (1.115 x 10 + 5 5 1.065 x 10 )/2 = 1.090 x 10 Pa. The average density to use is inlet

=

Pav

(3.1-22)

Pnv

28.97(1.090 x 10

5 )

=

1.221

kg/m 3

8314.34(311) Substituting into Eq. (3.1-21) and solving for Ap,

Ap(1.221) 0.0127 2

2.44

(1.225)

Solving,

Ap = 0.0497 x

so a second

4.

Shape factors.

trial is

10

Many

as this particle.

1

-

Pa. This

is

3

150

close

+

1.75

1321

0.38

enough

to the

assumed value,

not needed.

particles in

equivalent diameter of a particle

volume

s

(0.38)

The

is

packed beds are often irregular

in

The same

shape.

defined as the diameter of a sphere having the

(p s of a particle is the ratio of the as the particle to the actual surface

sphericity shape factor

surface area of this sphere having the

same volume

area of the particle. For a sphere, the surface area S = -rrDp and the volume is p v p = TrDp/6. Hence, for any particle, 4> s = -nDplS where S is the actual surface

area of the particle and the

same volume

Dp

p

is

as the particle.

p

S

Then

p -^=

Sec. 3.1

,

the diameter (equivalent diameter) of the sphere having

irDl/cps

V^ =

6

Flow Past Immersed Objects and Packed and Fluidized Beds

(3.1-23)

121

From Eq.

(3.1-7),

Sp

Then Eq.

(3.1-10)

6

becomes 6

a

=—-(1-6) 4>S

=

(3.1-25)

Dp

For a cylinder where the diameter = length, s is s is calculated as 0.806. For granular materials it is difficult to measure the actual volume and surface area to obtain the equivalent diameter. Hence, D p is usually taken to be the nominal size from a screen analysis or visual length measurements. The surface area is determined by adsorption measurements or measurement of the pressure drop in a bed of particles. Then Eq. (3.1-23) is used to calculate 4> s (Table 3.1-1). Typical values for many crushed materials are between 0.6 and 0.7. For convenience for the cylinder and the cube, the nominal diameter is sometimes used (instead of the equivalent diameter) which then gives a For a sphere,



s

1.0.



calculated to be 0.874 and for a cube,

shape factor of 5.

1

(f>

.0.

Mixtures of particles. For mixtures of particles of various sizes specific surface a vm as

we can

define a

mean

tf„

where x

;

is

volume

fraction.

m

~

and

(3.1-26), 1

X xtfltsDJ

mean diameter

the effective

(3.1-24)

(3.1-26)

6

_6_

D pm is

Z x ;^i

Combining Eqs.

Dpm ~ a vm where

=

for the

~~

X xM s D p

(3 " 1 " 27) .)

mixture.

EXAMPLE 3.1-5. Mean Diameter for a Particle Mixture A mixture contains three sizes of particles: 25% by volume of 25 mm size, 40% of 50 mm, and 35% of 75 mm. The sphericity is 0.68. Calculate the effective

mean diameter.

Table

Shape Factors (Sphericity) of Some Materials

3.1-1.

Material

Spheres

s

Reference

0.81

Cylinders,

122

cj>

1.0

Cubes

Dp =

Shape Factor,

0.87

h (length)

Berl saddles

0.3

(B4)

Raschig rings

0.3

(C2)

Coal dust, pulverized

0.73

(C2)

Sand, average

0.75

(C2)

Crushed

0.65

(C2)

glass

Chap. 3

Principles of Momentum Transfer

and Applications

Solution:

D p2 =

0.40,

The following data are given: x 1 = 0.25, D pl = 25 mm; x 2 = 50; x 3 = 0.35, D„ 3 = 75; s = 0.68. Substituting into Eq. (3.1<j>

27),

pm

0.25/(0.68 x 25)

=

30.0

+

0.40/(0.68 x 50)

beds shows that the flow rate viscosity

/i

0.35/(0.68 x 75)

mm

Darcy's empirical law for laminarflow.

6.

+

and length AL. This

Equation

proportional to

is

is

(3.1-17) for laminar flow in

Ap and

packed

inversely proportional to the

the basis for Darcy's law as follows for purely viscous

flow in consolidated porous media.

k

q'

»

where

v is superficial velocity

is

is

,

pressure drop).

The

millidarcy (1/1000 darcy). Hence,

flow

cm

is

length. This equation

at

if 1

)i is-

in

viscosity in cp, A/7

(cm 3 flow/s)

k of

units used for

fluid of 1-cp viscosity will

(3-1-28)

based on the empty cross section in cm/s,

empty cross section in cm 2 length in cm, and k is permeability

cm 3 /s, A

Ap

= 7 = " - 77 A n AL

cm 2

cp/s

-

per

1

cm 2

•

flow rate

q' is

pressure drop in atm,

AL

(cm length)/(cm 2 area) -(atm

atm are

medium

a porous

cm 3 /s

-(cp)

is

often given in darcy or in

has a permeability of

cross section with a

often used in measuring permeabilities of

Ap

of

1 1

darcy, a

atm per

underground

oil

reservoirs.

3.1

D

Flow

in Fluidized

Beds

Minimum velocity and porosity for fluidization. When a fluid flows upward through a packed bed of particles at low velocities, the particles remain stationary. As the fluid J.

Ergun equation where the force of the pressure drop times the cross-sectional area just equals the gravitational force on the mass of particles. Then the particles just begin to move, and this is the onset of fiuidization or minimum fiuidization. The fluid velocity at which fiuidization begins is the minimum fiuidization velocity v'mf in m/s based on the empty cross section of the tower increased, the pressure drop increases according to the

velocity

is

(3.1-20).

Upon

further increases in velocity, conditions finally occur

(superficial velocity).

when true fluidization occurs is the minimum porosity for and is e m/ Some typical values ofe mJ for various materials are given in Table The bed expands to this voidage or porosity before particle motion appears. This

The

porosity of the bed

fiuidization 3.1-2.

-

.

minimum voidage can

be experimentally determined by subjecting the bed to a rising gas

stream and measuring the height of the bed L mf in m. Generally, it appears best to use gas as the fluid rather than a liquid since liquids give somewhat higher values ofe m/ .

As stated onset of

earlier, the

minimum

drop decreases very

pressure drop increases as the gas velocity

fluidization. slightly

Then

as the velocity

is

is

increased until the

further increased, the pressure

and then remains practically unchanged as the bed continues

expand or increase in porosity with increases in velocity. The bed resembles a boiling liquid. As the bed expands with increase in velocity, the bed continues to retain its top to

horizontal surface. Eventually, as the velocity particles

The

increased

much

further,

from the actual fluidized bed becomes appreciable. relation between bed height L and porosity £ is as follows

uniform cross-sectional area A. Since the

Sec. 3.1

is

volume LA{{ —

e) is

for

entrainment of a bed having a

equal to the total volume

Flow Past Immersed Objects and Packed and Fluidized Beds

123

Table

Minimum

Void Fraction, z m{ at

3.1-2.

,

Fluidization

Conditions (L2) Particle Size,

Type of Particles

0.06

0.10

Dp

(nun)

0.20

0.4O

Void fraction,

Sharp sand (

of solids

if

(

=

0.63)

l

l)

0.49

0.43

(0.42)

0.61

0.60

0.56

0.52

is

(3.1-30)

t

height of bed with porosity £ t and

L2

height with porosity

is

As a

Pressure drop and minimum fluidizing velocity.

2.

(3.1-29)

)

l-£ 2 l-e

L2 t

0.53

0.48

= L 2 A(l-e 2

L,

L

0.58

0.53

they existed as one piece,

L A(l-E

where

0.60

first

s2

.

approximation, the pressure

The force obtained from drop times the cross-sectional area must equal the gravitational force exerted by the mass of the particles minus the buoyant force of the displaced fluid. drop

at the start of fluidization

can be determined as follows.

the pressure

ApA = L mf A(l-e mf )(p p -p)g

(3.1-31)

Hence,

Ap

=

(1

-

(1

- <WXP P -

£m/ Xp p

-

(SI)

P )g

(3.1-32)

=

P) —

(English)

Often we have irregular-shaped particles

in the bed, and it is more convenient to use and shape factor in the equations. First we substitute for the effective mean diameter D p the term (p s D P where D P now represents the particle size of a sphere having the same volume as the particle and 4>s tne shape factor. Often, the value of D P

the particle size

is

approximated by using the nominal size from a sieve analysis. Then Eq. (3.1-20) for

pressure drop in a packed bed becomes

Ap

=

L where

AL =

L,

Equation calculate the for

v',

bed length (3.1-33)

minimum

e m/ for £,

and

L mf

cpjDj

DF

3

e

now be used by a

small extrapolation for packed beds to which fluidization begins by substituting v'mf and combining the result with Eq. (3.1-32) to give

fluid velocity v'm{ at

for 2

124

UW^l-a

3

£

m.

in

can

L

L15Dj{v'mf ) p
150^(1-^

2

150(1

P-

-e mf)D P v'mfP _ D P p( Pp S

Chap. 3

E m /t*

P )g

P-

Principles of Momentum Transfer

and Applications

Defining a Reynolds number as ^Rc. m /

^^ DP

=

P

»'mf

(3.1-35)

P Eq. (3.1-34) becomes

l-75(N Rc m/ ) ~ 5

—

When when

2

150(1

,

-smf){N Rc

,

Dlp{p p -p)g

mf )

^

h

N Rc mf < 20 (small particles), the first term of Eq. (3.1-36) N Rc m/ > 1000 (large particles), the second term drops out.

=

0

(3.1-36)

can be dropped and

_

If

the terms e m/ and/or

(j>

s

are not known,

Wen

and Yu (W4) found

for a variety of

systems that

«M^ = i 14

!Z = TTT 4>s

J

W

(3.1-37)

11

Substituting into Eq. (3.1-36), the following simplified equation

"(33.7,1

+

0 0408

DI*P, -P)9'

is

obtained.

1/2

33.7

(3.1-38)

P This equation holds deviation of

+ 25%.

Reynolds number range of 0.001 to 4000 with an average

for a

Alternative equations are available in the literature (Kl, W4).

EXAMPLE 3.1-6.

Minimum

Velocity for Fluidization

Solid particles having a size of 0.12 density of 1000

kg/m 3

The voidage

minimum

(a)

(b) (c)

(d)

at

mm,

a shape factor

are to be fiuidized using air at 2.0 fluidizing conditions

empty bed

is

0.42.

m2

and the bed contains 300 kg of solid, calculate the minimum height of the fiuidized bed. Calculate the pressure drop at minimum fluidizing conditions.

If

the cross section of the

is

0.30

Calculate the minimum velocity for fluidization. Use Eq. (3.1-38) to calculate v'mf assuming that data for



s

and z mf

are not available.

volume of solids = 300 kg/(1000 kg/m ) = 0.300 m The height the solids would occupy in the bed if e, = 0 is L = 3 2 0.300 m /(0.30 m cross section) = 1.00 m. Using Eq. (3.1-30) and calling Lmf = L 2 and £ m/ = e 2 For part

Solution:

(a),

3

the

3

.

t

,

111 = I;.f

Solving,

The

L mf =

1.724

1-0.42 ~

1-0

m.

physical properties of air at 2.0

H=

1.845 x 10"

Pa.

For

5

Pa-s,

the particle,

E "/

!

1.00

Lmf

~

1

DP

=

atm and 25°C (Appendix A.3) are

= 2.374 kg/m 3 3 p F = 1000 kg/m

1.187 x 2

p = 0.00012 m,

= 2.0265 = 0.88,

p

,

,

x 10 i mS

5

=

0.42.

For part A/>

Sec. 3.1

(b)

using Eq. (3.1-32) to calculate Ap,

= L m/ (l -e m/ XP P -P)5 = 1.724(1 - 0.42)(1000 -

2.374X9.80665)

=

0.O978 x 10

Flow Past Immersed Objects and Packed and Fiuidized Beds

5

Pa

125

To

calculate

mf for part

v'

l-75(N Re mr )

(c),

2

,

Eq. (3.1-36)

used.

is

-0-42KN Re

150(1 ,

2

3

,

mr)

3

(0.88) (0.42)

'

(0.88X0.42)

-

2.374(1000

-(0.00012)

2.374X9.80665)

(2.845 x 10-5)2

Solving,

N Re R

-

m/

=

=

0.07764

= /i

=

v'

m/

2

= Solving,

4)

s 5

1.845 x 10

0.005029 m/s

Using the simplified Eq.

N Re. mf

°Z\^3

(3.1-38) for part (d),

0.0408(0.00012)

+

3

(2

.

-

374X1000

(1.845 x 10

-5

1/2

2.374X9.80665)

2

-

33.7

)

0.07129

v'

m/

=

0.004618 m/s.

Expansion of fluidized beds. For the case of small particles and where N Re / = D F v'p/fi < 20, we can estimate the variation of porosity or bed height L as follows. We assume that Eq. (3.1-36) applies over the whole range of fluid velocities with the first term 3.

being neglected. Then, solving for

v',

D F {p B 2

3

150/i

We v'.

find that all terms except

£

e

e

p)g(pj

—

1

£

1

of clumping and other factors, errors can occur

flow rate in a fluidized bed

the other

—

£

are constant for the particular system

This equation can be used with liquids to estimate

The

3

is

£

with

£

<

when used for gases. one hand by the minimum v'mf and on

limited on

by entrainment of solids from the bed proper. This approximated as the terminal settling velocity v\ of the

velocity

is

13.3 for

methods

to calculate this settling velocity.)

the operating range are as follows (P2).

and s depends upon However, because

0.80.

For

Approximate equations

fine solids

and

N Rt

s

—=— v,

maximum

<

allowable

particles. (See Section to calculate

0.4,

90

(3.1-40)

U n,f

For large .solids and

N Rz

f

>

1000, 9

(3.1-41) v'

mJ

1

EXAMPLE 3.1-7.

Expansion of Fluidized Bed Using the data from Example 3.1-6, estimate the maximum allowable velocity v\ Using an operating velocity of 3.0 times the minimum, estimate the .

voidage of the bed. Solution: E m/

=

0.42.

From Example Using Eq. v\

126

3.1-6,

(3.1-40), the

= 90(^ y = )

Chap. 3

N Rc

mf

=

maximum

90(0.005029)

Principles

= 0.005029 m/s, 0.07764, allowable velocity is =

0.4526 m/s

of Momentum Transfer and Applications

Using an operating velocity v'

To

K

3.0{v'mf )

of 3.0 times the

=

3.0(0.005029)

t

Solving,

known

0.01509 m/s

new velocity, we substitute into Eq. minimum fluidizing conditions to deter-

0.005029

=

.

K = t

0.03938.

Then using

Solving, the voidage of the bed e

It is

=

at this

0.01509

3.2

minimum,

values at

determine the voidage

(3.1-39) using the

mine

=

v'

K

l

J

the operating velocity in Eq. (3.1-39),

= (0.03938) =

0.555 at the operating velocity.

MEASUREMENT OF FLOW OF FLUIDS important to be able to measure and control the amount of material entering and

leaving a chemical and other processing plants. Since

form of

fluids,

they are flowing in pipes or conduits.

many

Many

of the materials are

in the

different types of devices are

The most simple are those that measure directly the volume of the fluids, such as ordinary gas and water meters and positive-displacement pumps. Current meters make use of an element, such as a propeller or cups on a rotating arm, which rotates at a speed determined by the velocity of the fluid passing through it. Very widely used for fluid metering are the pitot tube, venturi meter, orifice meter, and used to measure the flow of fluids.

open-channel weirs.

3.2A

Pitot

Tube

The

pitot tube is used to measure the local velocity at a given point in the flow stream and not the average velocity in the pipe or conduit. In Fig. 3.2-la a sketch of this simple device is shown. One tube, the impact tube, has its opening normal to the direction of flow and the static tube has its opening parallel to the direction of flow.

(a)

Figure

3.2-1.

Diagram of pitot tube: (a) simple

(b)

lube, (b) tube with static pressure

holes.

Sec. 3.2

Measurement of Flow of Fluids

127

The

and then remains The difference in the stagnation pressure at this point 2 and the static pressure measured by the static tube represents the pressure rise associated with the deceleration of the fluid. The manometer measures this fluid flows into the

opening at point

2,

pressure builds up,

stationary at this point, called the stagnation point.

small pressure

rise. If

the fluid

Bi

2

Setting v 2

=

0 and solving for

v is

+

write the Bernoulli equation

undisturbed before the fluid decelerates,

u, is

ELZH =

o

(3.2.!)

2 v

v

where

we can

incompressible,

is

between point 1, where the velocity and point 2, where the velocity v 2 is zero.

(2.7-32)

it

2iP2

= C.

the velocity v l in the tube at point

1

~ Pl)

in

(3-2-2)

m/s, p 2

is

the stagnation pressure, p

is

the density of the flowing fluid at the static pressure p 1; and C is a dimensionless coefficient to take into account deviations from Eq. (3.2-1) and generally varies between

about 0.98 to

1.0.

For accurate

use, the coefficient

should be determined by calibration

of the pitot tube. This equation applies to incompressible fluids but can be used to

approximate the flow of gases at moderate velocities and pressure changes of about 10% or less of the total pressure. For gases the pressure change is often quite low and, hence, accurate measurement of velocities

is

difficult.

value of the pressure drop p 2 — Pi or Ap in the manometer, by Eq. (2.2-14) as follows:

The

Pa

is

related to

Aft,

the reading

Ap = Ah(p A - p)g where p A

is

the density of the fluid in the

on

(3.2-3) 3

manometer inkg/m and Aft is the manometer is shown with concentric tubes. In

reading in m. In Fig. 3.2-lb, a more compact design

the outer tube, static pressure holes are parallel to the direction of flow. Further details

are given elsewhere (PI).

Since the pitot tube measures velocity at one point only in the flow, several methods

can be used to obtain the average velocity

measured

in the pipe.

first method the Then by using Fig.

In the

at the exact center of the tube to obtain u max

.

velocity

is

2.10-2 the

u av can be obtained. Care should be taken to have the pitot tube at least 100 diameters downstream from any pipe obstruction. In the second method, readings are taken at several known positions in the pipe cross section and then using Eq. (2.6-17), a graphical

or numerical integration

is

performed to obtain

v iv

.

EXAMPLE 3.2-1.''

Flow Measurement Using a Pitot Tube is used to measure the airflow in a circular duct 600 in diameter. The flowing air temperature is 65.6°C. The pitot tube is placed at the center of the duct and the reading Aft on the manometer is 10.7 of water. A static pressure measurement obtained at the pitot tube position is 205 mm of water above atmospheric. The pitot tube coeffi-

A

pitot tube similar to Fig. 3.2-la

mm

mm

cient

Cp =

(a)

(b)

Calculate the velocity at the center and the average velocity. Calculate the volumetric flow rate of the flowing air in the duct.

For part 5 2.03 x 10"

Solution: are p

=

0.98.

(a),

the absolute static pressure, the

128

from Appendix A.3 kg/m 3 (at 101.325 kPa). To calculate manometer reading Aft = 0.205 m of water

the properties of air at 65.6°C

Pas, p =

Chap. 3

1.043

Principles

of Momentum Transfer and Applications

above 1 atm abs. Using Eq. (2.2-14), the water density kg/m 3 and assuming 1.043 kg/m 3 as the air density,

indicates the pressure as 1000

,

=

Ap-

Then 10

5

=

absolute

the

1.0333

x

10

5

-

0.205(1000

The

=

2008 Pa

s p l = 1.01325 x 10

pressure

static

Pa.

1.043)9.80665

+

0.02008 x

correct air density in the flowing air

is (1.0333 5 5 1.063 kg/m 3 This correct value when used x 10 /1.01325 x 10 X1.043) instead of 1.043 would have a negligible effect on the recalculation ofp,.

=

.

To calculate the Ap for the pitot tube, Eq. (3.2-3) is used. Ap = Ah(p A Using Eq.

(3.2-2), the

p)g

=

(1000

-

1000

maximum

velocity at the center

The Reynolds number using the maximum

Dv^j> _ ~ p.

_ Nk <~

1.063X9.80665)

velocity

0.600(13.76X1.063)

2.03x10- 5

=

104.8

Pa

is

is

- 4 -^ ixlu

From Fig. 2.10-2, vjv mlll = 0.85. Then, av = 0.85(13.76) = 1.70 m/s. To calculate the' flow rate for part (b), the cross-sectional area of 2 2 the duct, A = (jr/4)(0.600) = 0.2827 m The flow rate = 0.2827(11.70) = 1

i>

.

3.308

3.2B

A

m

3

/s.

Venturi Meter

venturi meter

is

shown

in Fig. 3.2-2

manometer or other device

and

is

usually inserted directly into a pipeline.

A

two pressure taps shown and measures the pressure difference p — p 2 between points 1 and 2. The average velocity at point 1 where the diameter is D, m is u, m/s, and at point 2 or the throat the velocity is v 2 and diameter D 2 Since the narrowing down from D, to D 2 and the expansion from D 2 back to !>! is gradual, little frictional loss due to contraction and expansion is incurred. To derive the equation for the venturi meter, friction is neglected and the pipe is assumed horizontal. Assuming turbulent flow and writing the mechanical-energybalance equation (2.7-28) between points 1 and 2 for an incompressible fluid, is

connected to

the'

l

(3.2-4)

2

The

continuity equation for constant p

p

is 2

7t£>f

Figure

Sec. 3.2

3.2-2.

TzD 2

(3.2-5)

Venturi flow meter.

Measurement of Flow of Fluids

129

Combining Eqs.

(3.2-4)

and

v2

To account

and eliminating

(3.2-5)

~ gzj

=•

for the small friction loss

p2

-

an experimental

coefficient

-

(3.2-6)

C„

is

introduced to give

(SI)

,

(3-2-7) »2

-

C» /

,

2g ^'

~

P2)

(English)

For many meters and a Reynolds number >10 4 at point l,C v is about 0.98 for pipe diameters below 0.2 m and 0.99 for larger sizes. However, these coefficients can vary and individual calibration

To

is

recommended

if

the manufacturer's calibration

calculate the volumetric flow rate, the velocity v 2

is

not available.

multiplied by the area

is

A2

.

\2

flow rate

For

the

=

v2

m 3/s

(3.2-8)

measurement .of compressible flow of gases, the adiabatic expansion fromp, must be allowed for in Eq. (3.2-7). A similar equation and the same

to p 2 pressure

coefficient

(shown

Cv

are used along with the dimensionless expansion correction factor

in Fig. 3.2-3 for air) as

C„A 2 Y -

yi A2

is

m

-p

2)

is

(3.2-9)

Pi

(d 2 /£>,)

flow rate in kg/s, p is density of the fluid upstream at point 2 cross-sectional area at point 2 in

where

Y

follows:

t

m

1

in

kg/m 3 and ,

.

The pressure difference pi — p 2 occurs because the velocity is increased from v to However, farther down the tube the velocity returns to its original value ofi?! for — p 2 is not fully liquids. Because of some frictional losses, some of the difference t

v2

.

130

Chap. 3

Principles

of Momentum Transfer and Applications

Figure

Orifice flow meter.

3.2-4.

recovered. In a properly designed venturi meter, the permanent loss differential p

—

l

p2

,

and

measure flows

in large lines,

3.2C

Meter

Orifice

For ordinary It if

installations in a process plant the venturi

is

about 10% of the

venturi meter

often used to

is

is

meter has several disadvantages.

expensive. Also, the throat diameter

is

fixed so that

changed considerably, inaccurate pressure differences may result. meter overcomes these objections but at the price of a much larger permanent

the flow-rate range orifice

head or power

A

A

loss.

such as city water systems.

occupies considerable space and

The

power

this represents

is

loss.

is shown in Fig. 3.2-4. A machined and drilled plate mounted between two flanges in a pipe of diameter Z),. Pressure taps at point 1 upstream and 2 downstream measure p — p 2 The exact positions of the two taps are somewhat arbitrary, and in one type of meter the taps are installed about 1 pipe diameter upstream and 0.3 to 0.8 pipe diameter downstream. The

typical sharp-edged orifice

having a hole of diameter

D0

is

.

l

fluid

stream, once past the orifice plate, forms a vena contracta or free-flowing jet.

The equation

Eq.

for the orifice is similar to

(3.2-7).

-

2(Pi

Pi)

(3.2-10)

7l -(D 0 /D,) 4 where

the velocity in the orifice in m/s,

u 0 is

dimensionless orifice coefficient. imentally.

value of

If

C0

is

the

N Re

The

at the orifice is

a correlation for

As orifice,

in the

C0

is

is

the orifice diameter in m, and

above 20 000

approximately constant and

design for liquids (M2, Pl). Below

D0

C 0 is and D 0/D

orifice coefficient

20000

it

area of the

than about

less

the coefficient rises sharply

case of the venturi, for the

Y given

is

the

0.5,

the

adequate for

and then drops and

measurement of compressible flow of gases

in Fig. 3.2-3 for air

'l-(ZV0,r is

is

given elsewhere (Pl).

a correction factor

m

is

has the value of 0.61, which

C0 A 0 Y

where

l

C0

always determined exper-

flow rate in kg/s,p,

is

n/ 2 Oi

is

in

an

used as follows.

-

(3.2-11)

Pi)Pl

upstream density inkg/m 3 and/l 0 ,

is

the cross-sectional

orifice.

The permanent pressure loss is much higher than for a venturi because of the eddies formed when the jet expands below the vena contracta. This loss depends on D D /D and 1

Sec. 3.2

Measurement of Flow of Fluids

131

is

73%

of pi

-p 2

for £>„/£>,

=

56%

0.5,

D 0/D =

for

and 38%

0.65,

v

for

D 0/D =0.8 l

(PI).

EXA MPLE 32-2.

Metering Oil Flow by an Orifice sharp-edged orifice having a diameter of 0.0566 m is installed in a 0.1541-m pipe through which oil having a density of 878 kg/m 3 and a viscosity of 4.1 cp is flowing. The measured pressure difference across the 2 3 Calculate the volumetric flow rate in orifice is 93.2 kN/m /s. Assume

A

m

.

that

C0 =

0.61.

Solution:

Pi-p 2 = D, =0.1541

kN/m 2 =

93.2

D0 =

m

0.0566

N/m 2

9.32 x 10*

= 0.368 ^2=^^ D, 0.1541

m

Substituting into Eq. (3.2-10),

v0

C0

—

/2(Pi

Vl -(Do/DO

v 0 '= 8.97

volumetric flow rate

=

„0

(0.368)

N Kc

=

4.1

is

calculated to see

x

1

x 10~

3

=

^

=

M0.0566)

.

4

'

0.02257

m

3

kg/m-s

=

3

/s (0.797

4.1

above 2 x 10 4

is

3

greater than 2 x 10

x 10~

Hence, the Reynolds number

)

'878

V

97) (8 v

4

if it is

4.1

4

4

m/s

= The

p2)

/2(9.32 x 10

0.61

71 -

-

4

4

for

x 10~

ft

/s)

C0 = 3

0.61.

Pa-s

.

Other measuring devices for flow in closed conduits, such as rotameters, flow and so on, are discussed elsewhere (PI).

nozzles,

Flow

3.2D In

many

in

Open Channels and Weirs

instances in process engineering and in agriculture, liquids are flowing in

To measure

channels rather than closed conduits. used.

A

weir

is

a

dam

over which the liquid flows.

rectangular weir and the triangular weir weir

and

the height

/j

0

(weir head) in

m

shown. This head should be measured by a

shown

is

at

open

the flow rates, weir devices are often

The two main types of weirs are The liquid flows over

in Fig. 3.2-5.

measured above the

a distance of about

fiat

3/i 0

m

the

the

base or the notch as

upstream of

the weir

level or float gage.

The equation

volumetric flow rate q

for the

q

=

0.41 5(L

-

in

m 3 /s l

0.2h o )h o

-

5

for a rectangular weir

jig

is

given

by

(3.2-12)

2 m, g = 9.80665 m/s and h 0 = weir head in m. This is called the modified Francis weir formula and it agrees with experimental values within 3% if

where L =

132

crest length in

,

Chap. 3

Principles of

Momentum

Transfer and Applications

L

(b)

(a)

Figure

L>

2h 0

,

3.2-5.

Types of weirs : (a) rectangular, (b)

velocity upstream

the bottom of the channel

g

=

32.174

For

is

is

<0.6 m/s, >3/i 0

.

h0

>

0.09

triangular.

m, and the height of the crest above L and h are in ft, q in ft 3 /s, and

In English units

2

ft/s

.

the triangular notch weir, ,2.5

(3.2-13)

Both Eqs.

(3.2-12)

and

(3.2-13) apply only to water.

For other

liquids, see

data given

elsewhere (PI).

3.3

PUMPS AND GAS-MOVING EQUIPMENT

3.3A

Introduction

In order to

make

a fluid flow from one point to another in a closed conduit or pipe,

necessary to have a driving force. Sometimes this force

is

differences in elevation occur. Usually, the energy or driving force

mechanical device such the fluid. This energy

as a

may

pump

it is

supplied by gravity, where is

supplied by a

or blower, which increases the mechanical energy of

be used to increase the velocity (move the

fluid),

the pressure,

or the elevation of the fluid, as seen in the mechanical-energy-balance equation (2.7-28),

which

relates

v,

p, p,

and work. The most

common methods

of adding energy are by

positive displacement or centrifugal action.

word "pump" designates a machine or device for moving an incomand compressors are devices for moving gas (usually air). Fans discharge large volumes of gas at low pressures of the order of several hundred mm of water. Blowers and compressors discharge gases at higher pressures. In pumps and fans the density of the fluid does not change appreciably, and incompressible flow can be assumed. Compressible flow theory is used for blowers and compressors. Generally, the

pressible liquid. Fans, blowers,

3.3B

Pumps

Power and work required. Using the total mechanical-energy-balance equation (2.7on a pump and piping system, the actual or theoretical mechanical energy Ws J/kg added to the fluid by the pump can be calculated. Example 2.7-5 shows such a case. Iff/ the shaft work delivered to the pump, Eq. (2.7-30) gives is the fractional efficiency and p /.

28)

W

(3-3-1)

Sec. 3.3

Pumps and Gas-Moving Equipment

133

The

power of a pump

actual or brake

brakekW =

is

as follows.

^

_J^

=

1000

brake hp v

W

where

J/kg,

is

English units,

m is

W

s is

in

Wm

= —-B— = 550

s

r\

•

lb f/lb m

power

theoretical

The mechanical energy

W

s in

.

5

is

the conversion factor

The theoretical

=

,

(English) 1

x 550

m in Ibjs.

and

,

and 1000

the flow rate in kg/s, ft

(33-2)

Wm

'

(si) v

x 1000

tj

or fluid power

W/kW.

(brake kWX?/)

J/kg added to the

fluid

is

In

is

(33-3)

often expressed as the developed

H of the pump in m of fluid being pumped, where

head

-W

= Hg-

s

(SI)

(33-4)

-W

=

S

To

calculate the

hundred the fan

is

power

of a fan

H~

(English)

where the pressure difference

mm of water, a linear average density of the gas between used to calculate Ws and brake kW or horsepower.

is

of the order of a few

the inlet and outlet of

most pumps are driven by electric motors, the efficiency of the electric motor must be taken into account to determine the total electric power input to the motor. Typical efficiencies rj e of electric motors are as follows: 75% for^-kW motors, 80% for 2 kW, 84% for 5 kW, 87% for 15 kW, and about 93% for over 150-kW motors. Hence, the total electric power input equals the brake power divided by the electric motor drive Since

efficiency

r/

e

.

electric

power input (kW)

=

Suction

lift.

The power

s

(3.3-5)

We'lOOO

>le

2.

-W m—

kW =

brake

calculated by Eq. (2.7-3) depends on the differences in

pressures and not on the actual pressures being above or below atmospheric pressure.

However, the lower fixed

limit of the absolute pressure in the suction (inlet) line to the

by the vapor pressure of the liquid

at the

temperature of the liquid

pump

is

in the suction

on the liquid in the suction line drops to the vapor pressure, some of vapor (cavitation). Then no liquid can be drawn into the pump. For the special case where the liquid is nonvolatile, the friction in the suction line to the pump is negligible, and the liquid is being pumped from an open reservoir, the maximum possible vertical suction lift which the pump can perform occurs. For cold water this would be about 10.4 m of water. Practically, however, because of friction, vapor pressure, dissolved gases, and the entrance loss, the actual value is much less. For

line. If

the pressure

the liquid flashes into

details, see references

3.

elsewhere (PI, M2).

Centrifugal pumps.

Process industries

available in sizes of about 0.004 to 380

m

3

commonly /min

(1

use centrifugal pumps.

100000 gal/min) and

to

They

are

for discharge

m of head to 5000 kPa or so. A centrifugal pump in its simplest form an impeller rotating inside a casing. Figure 3.3-1 shows a schematic diagram

pressures from a few consists of

of a simple centrifugal

The

pump.

liquid enters the

pump

axially at point

rotating eye of the impeller, where

134

it

1

in

the suction line and then enters the

spreads out radially.

Chap. 3

Principles of

On

spreading radially

Momentum

it

enters

Transfer and Applications

outlet-

suction inlet

power shaft

4

N

'

Figure

the channels

3.3-1.

casing

Simple centrifugal pump.

between the vanes

pump

out the

discharge at

5.

point 2 and flows through these channels to point 3 at

at

From

the periphery of the impeller.

The

here

collected in the volute

it is

chamber 4 and flows

rotation of the impeller imparts a high-velocity head to

the fluid, which is changed to a pressure head as the liquid passes into the volute chamber and out the discharge. Some pumps are also made as two-stage or even multistage pumps. Many complicating factors determine the actual efficiency and performance characteristics of a pump. Hence, the actual experimental performance of the pump is usually employed. The performance is usually expressed by the pump manufacturer by means of curves called characteristic curves and are usually for water. The head H in m produced will be the same for any liquid of the same viscosity. The pressure produced, which is p = Hpg, will be in proportion to the density. Viscosities of less than 0.05 Pa-s (50 cp) have little effect on the head produced. The brake kW varies directly as the density. As rough approximations, the following can be used for a given pump. The capacity

q x in

m

3

directly proportional to the

/s is

rpm

rV,,

or

ii

N

92

H

The head

,

is

proportional to q

,

(3.3-6) 2

or

si

(3.3-7)

Hz The power consumed

W

is

l

proportional to the product of//^,, or

W

2

In

most pumps, the speed

is

H2 q

(3.3-8) 2

N\

generally not varied. Characteristic curves for a typical

Most pumps are usually rated on the basis of head and capacity at the point of peak efficiency. The efficiency reaches a peak at about 50 gal/min flow rate. As the discharge rate in gal/min increases, the developed head drops. The brake hp increases, as expected, with single-stage centrifugal

pump

operating at a constant speed are given in Fig. 3.3-2.

flow rate.

EXA MPLE33-1. In order to see at

Sec. 3.3

how

40 gal/min flow

Calculation of Brake Horsepower of a Pump is determined, calculate the brake hp

the brake-hp curve

rate for the

pump

in Fig. 3.3-2.

Pumps and Gas-Moving Equipment

135

Discharge (U.S. gal/min) Figure

Characteristic curves for a single-stage centrifugal

3.3-2.

(From W.

pump

with water.

Badger and J. T. Banchero, Introduction to Chemical Engineering. New York: McGraw-Hill Book Company, 1955. With L.

permission.)

the head

efficiency n from the curve is about 60% and flow rate of 40 gal/min of water with a density of

At 40 gal/min, the

Solution:

H

is

62.4 lb mass/ft

38.5 3

m = The work

W

s

A

ft.

is

is

f

40^^ ntM /

.O

/

3

\ 60 s/min / \7.481 galy \

ft

ib.

/

s

as follows, from Eq. (3.3-4):

^=_H^=-38.5^ The brake hp from Eq. brake hp

=

(3.3-2) is

-Ws m = 38.5(5.56) =

This value checks the value on the curve

0.65

hp

(0.48

kW)

in Fig. 3.3-2.

Positive-displacement pumps. In this class of pumps a definite volume of liquid is drawn into a chamber and then forced out of the chamber at a higher pressure. There are two main types of positive-displacement pumps. In the reciprocating pump the chamber is a stationary cylinder and liquid is drawn into the cylinder by withdrawal of a piston in the cylinder. Then the liquid is forced out by the piston on the return stroke. In the rotary pump the chamber moves from inlet to discharge and back again. In a gear rotary pump two intermeshing gears rotate and liquid is trapped in the spaces between the teeth and forced out the discharge. Reciprocating and rotary pumps can be used to very high pressures, whereas centrifugal pumps are limited in their head and are used for lower pressures. Centrifugal pumps deliver liquid at uniform pressure without shocks or pulsations and can handle liquids with large amounts of suspended solids. In general, in chemical and biological

4.

processing plants, centrifugal

Equations

(3.3-1)

pumps

through

are primarily used.

(3.3-5)

hold for calculation of the power of positive

displacement pumps. At a constant speed, the flow capacity

136

Chap. 3

Principles of

will

Momentum

remain constant with

Transfer and Applications

different liquids. In general, the discharge rate will be directly

The power as the

dependent upon the speed.

increases directly as the head, and the discharge rate remains nearly constant

head increases.

33C Gas-Moving

Machinery

Gas-moving machinery comprises mechanical devices used for compressing and moving gases.

They are

often classified or considered from the standpoint of the pressure heads

produced and are fans

for

low pressures, blowers

for intermediate pressures,

and com-

pressors for high pressures. 1. Fans. The commonest method for moving small volumes of gas at low pressures is by means of a fan. Large fans are usually centrifugal and their operating principle is similar to that of centrifugal pumps. The discharge heads are low, from about 0.1 m to

1.5

mH

2 0.

However,

velocity energy

in

some

cases

and a small amount

much

of the

added energy of the fan

is

converted to

to pressure head.

produced by the rotor produces

In a centrifugal fan, the centrifugal force

a

com-

pression of the gas, called the static pressure head. Also, since the velocity of the gas increased, a velocity head

is

produced. Both the static-pressure-head increase and

is

velocity-head increase must be included in estimating efficiency and power. Operating efficiencies are in the

range 40 to 70%. The operating pressure of a fan

as inches of water gage and

is

the

sum

of the velocity

head and

leaving the fan. Incompressible flow theory can be used to calculate the

EXA MPLE 33-2.

Brake-k

is

generally given

static pressure of the gas

power of fans.

W Power of a Centrifugal Fan

m 3 /min

0 f air (metered at a pressure of 101.3 kPa a process. This amount of air, which is at rest, enters the fan suction at a pressure of 741.7 Hg and a temperature of 366.3 K and is discharged at a pressure of 769.6 Hg and a velocity of 45.7 m/s. A It is

desired to use 28.32

and 294.1 K)

in

mm

mm

centrifugal fan having a fan efficiency of

60%

is

to be used. Calculate the

brake-kW power needed. Incompressible flow can be assumed, since the pressure drop is only (27.9/741.7)100, or 3.8% of the upstream pressure. The average density of the flowing gas can be used in the mechanical-energy-balance equation. Solution:

The

density at the suction, point k

„ Pl

air

V

1, is

k

mo1

-

s g (>* 07 kg moiy V22.414 V

=

0.940

1

m

3

213

A

14[

A

^ ( ( J \3663j { 760 J

kg/m 3

3 (The molecular weight of 28.97 for air, the volume of 22.414 m /kg mol at 101.3 kPa, and 273.2 K were obtained from Appendix A.l.) The density at

the discharge, point

2, is

p2

The average density

^_ Sec. 3.3

=

(0.940)

of the gas

^

=

0.975 kg/m

3

is

^,

0.940

+

Pumps and Gas-Moving Equipment

0.975

= 0958

kg/m]

137

The mass flow

rate of the gas

is

kg

m

kg mol

=

0.5663 kg/s

The developed pressure head I>2

;

(769.6

Pi

-

760

p av

=

is

mm

7 41.7)

2

Hg

f{ \

mm/atm

Qm5

^ {q5

~N/m \

1

/

atm / \0.958 kg/nr

3883 J/kg

The developed

velocity head for v

y

=

0

is

2

v\

(45.7)

1044 J/kg 2

2

Writing-the mechanical-energy-balance equation (2.7-28),

Zig Setting Z[

=

0,

z2

+

=

»1 2

0,t>,

+ =

Z±-Ws = z p

0,

2

and^F =

- W - ELZJl +

g

+

0,

v A = 3883

§2 + ^P + ZF

and solving for

+

1044

=

Ws

,

4927 J/kg

Substituting into Eq. (3.3-2),

2.

Blowers and compressors.

fans, several distinct

For handling gas volumes

at higher pressure rises than

types of equipment are used. Turboblowers or centrifugal com-

move large volumes of gas for pressure rises from about 5 kPa thousand kPa. The principles of operation of a turboblower are the same as those of a centrifugal pump. The turboblower resembles the centrifugal pump in ap-

pressors are widely used to to several

pearance, the main difference being that the gas of the turboblower, as in a centrifugal

pump,

in

the blower

is

is

compressible.

independent of the

fluid

The head handled.

Multistage turboblowers are often used to go to the higher pressures.

Rotary blowers and compressors are machines of the positive-displacement type and are essentially constant-volume flow-rate machines with variable discharge pressure. Changing the speed will change the volume flow rate. Details of construction of the various types (PI) vary considerably and pressures up to about 1000 kPa can be obtained, depending on the type. Reciprocating compressers which are of the positive displacement type using pistons are available for higher pressures. Multistage machines are also available for pressures

up

to

3.3D

lOOOOkPa

or more.

Equations for Compression of Gases

In blowers

and compressors pressure changes are large and compressible flow occurs.

Since the density changes markedly, the mechanical-energy-balance equation must be written in differential form

138

and then integrated

Chap. 3

to obtain the

Principles

work of compression.

In

of Momentum Transfer and Applications

compression of gases the static-head terms, the velocity-head terms, and the friction terms are dropped and only the work term dW and the dp/p term remain in the differential form of the mechanical-energy equation; or,

dW = ^

(33-9)

Integration between the suction pressure p, and discharge pressure p 2 gives the

work

of

compression.

>2 dp

W= To

(33-10)

com-

integrate Eq. (3.3-10) for a perfect gas, either adiabatic or isothermal

assumed. For isothermal compression, where the gas is cooled on com= 8314.3 J/kg mol K in SI units and pression, pjp is a constant equal to RT/M, where

pression

is

R

1545.3

ft

•

mol °R

lb r/lb

Solving for p in Eq. compression is

(3.3-11)

Ws

=

= T2

Ei

[

since the process

,

El

P

Pi

P

(3.3-11)

and substituting

PIdP

=

Ei

p

Pi J,, Also, T,

•

English units. Then,

in

is

ln

Pi

Eq.

in

it

=

El

2.3026RT! :

Pi

work

(3.3-10), the

,

for

isothermal

Pi

(5.3-12)

log

M

isothermal.

For adiabatic compression, the

fluid follows

an isentropic path and

El

(3.3-13)

P

where

y

= cjc a

,

the ratio of heat capacities.

By combining

Eqs. (3.3-10) and (3.3-13) and

integrating,

(y- Dly

-W* y

The adiabatic temperatures are

—

1

M

-

related by

ZWpA' T

l

To

calculate the brake

(3.3-14)

LVPi

power when the

brake

7

"

1 '''

(33-15)

\pj efficiency is

rj,

-W m s

kW

(3.3-16)

foXiooo)

where

m =

The

kg gas/s and

and 1.40 compression in Eq. for ethane,

(3.3-14).

Sec. 3.3

W

s

=

J/kg.

values of y are approximately 1.40 for for

N2

(PI).

(3.3-12)

Hence, cooling

is

is

air, 1.31 for

For a given compression less

methane,

ratio, the

1.29

work

forS0 2 in

than the work for adiabatic compression

sometimes used

in

Pumps and Gas-Moving Equipment

,

1.20

isothermal in

Eq.

compressors.

139

EXAMPLE

Compression of Methane is to compress 7.56 x

3J-3.

A single-stage compressor

10"" 3

kg mol/s of methane

gas at 26.7°C and 137.9 kPa abs to 551.6 kPa abs. (a) Calculate the power required if the mechanical efficiency

and the compression

is

is

80%

adiabatic.

Repeat, but for isothermal compression.

(b)

For part mass/kg mol, and Tj

Solution:

(a),

=

=

p,

273.2

+

M

= 16.0 kg kPa, p 2 = 551.6 kPa, 299.9 K. The mass flow rate per sec

137.9

26.7

=

is

m =

(7.56 x 10"

Substituting

p 2 /Pl

=

3

Eq.

into

=

551.6/137.9

-

1.31

1.31

kg 0.121

methane

for

and

Dh

-

1

4 \(1.31-1W1.3I

-

1

16.0

1

256300 J/kg

(3.3-16),

brake DraK (b)

kw

___ZU^___i_ = 38.74 kW ^-1000

(52.0 hp)

0.80(1000)

using Eq. (3.3-12) for isothermal compression,

_

„, h,s

=

=

2.3026KT, p _____ log ,

—

2.3026(8314.3X299.9)

2

_r

I™

,

4

log

i

216 000 J/kg

Hence, isothermal compression uses 15.8%

3.4

=

y

A 8314.3(299.9)

1.31

For part

(/-

M

y—l

=

for

(3.3-14)

=

4.0/1,

RT,

W*

Using Eq.

kg mol/sX16.0 kg/kg mol)

less

power.

AGITATION AND MIXING OF FLUIDS

AND POWER REQUIREMENTS 3.4A

Purposes of Agitation

In the chemical

great extent

on

and other processing industries, many operations are dependent to a and mixing of fluids. Generally, agitation refers to

effective agitation

forcing a fluid by mechanical

means

to

flow

in a

circulatory or other pattern inside a

Mixing usually implies the taking of two or more separate phases, such as a fluid and a powdered solid, or two fluids, and causing them to be randomly distributed through one another. vessel.

140

Chap. 3

Principles of

Momentum

Transfer and Applications

There are

number

a

of purposes for agitating fluids

and some of these are

briefly

summarized. 1.

2.

3.

Blending of two miscible liquids, such as ethyl alcohol and water. Dissolving solids in liquids, such as salt in water. Dispersing a gas in a liquid as fine bubbles, such as oxygen from air

microorganisms for fermentation or

for the activated

in

a suspension of

sludge process in waste treat-

ment. 4.

Suspending of fine solid particles in a liquid, such as in the catalytic hydrogenation of a liquid where solid catalyst particles and hydrogen bubbles are dispersed in the liquid.

5.

Agitation of the fluid to increase heat transfer between the fluid and a coil or jacket in the vessel wall.

3.4B

Equipment

for Agitation

Generally, liquids are agitated in a cylindrical vessel which can be closed or open to the

The height of mounted on a shaft

liquid

air.

is

is

approximately equal

driven by an electric motor.

to

An

the tank diameter.

A typical

agitator assembly

is

impeller

shown

in

Fig. 3.4-1.

1.

Three-blade propeller agitator.

A common

shown

type,

There are several types of agitators commonly used.

in Fig. 3.4-1, is a

three-bladed marine-type propeller similar to

The propeller can be a side-entering type in a an open vessel in an off-center position. These

the propeller blade used in driving boats.

tank or be clamped on the side of

propellers turn at high speeds of 400 to 1750 for liquids of

low

on the center of

the

flow since the fluid flows axially sides of the tank as

2.

rpm

(revolutions per minute) and are used

The flow pattern in a baffled tank with tank is shown in Fig. 3.4-1. This type of flow

down

shown

Various types of paddle agitators are often used

in Fig. 3.4-2a.

Figure

3.4-1.

is

called axial

and up on the

shown.

Puddle agitators.

the tank diameter

pattern

the center axis or propeller shaft

between about 20 and 200 rpm. Two-bladed and four-bladed as

a propeller positioned

viscosity.

The

total length of the

and the width of the blade

Baffled tank

flat

paddle impeller

5 to jo of

its

at

low speeds

paddles are often used, is

usually 60 to

80%

of

length. At low speeds mild

and ihree-blade propeller agitator with axial-flow pattern

(a) side view, (b) bottom view.

Sec. 3.4

Agitation

and Mixing of Fluids and Power Requirements

141

Figure

Various types of agitators: (a) four-blade paddle, fb) gate or anchor

3.4-2.

paddle, (c) six-blade open turbine, (d) pitched-blade (45°) turbine.

agitation

without agitator

obtained

is

in

baffles, the liquid

ineffective for

is

vertical or axial flow.

an unbaffled vessel. At higher speeds baffles are used, since,

simply swirled around with

is

actual mixing.

little

suspending solids since good radial flow

An anchor

or gate paddle,

shown

The paddle

present but

is

in Fig. 3.4-2b,

is

little

often used.

It

sweeps or scrapes the tank walls and sometimes the tank bottom. It is used with viscous liquids where deposits on walls can occur and to improve heat transfer to the walls.

However,

it

is

a poor mixer. These are often used to process starch pastes, paints,

adhesives, and cosmetics.

3.

Turbines that resemble multibladed paddle agitators with shorter

Turbine agitators.

blades are used

at

The

high speeds for liquids with a very wide range of viscosities.

normally between 30 and 50% of the tank diameter. Normally, the turbines have four or six blades. Figure 3.4-3 shows a flat six-blade turbine agitator diameter of a turbine

is

with disk. In Fig. 3.4-2c a

flat,

six-blade open turbine

shown

is

They

shown. The turbines with

flat

good gas dispersion where the gas is introduced just below the impeller at its axis and is drawn up to the blades and chopped into fine bubbles. In the pitched-blade turbine shown in blades give radial flow, as

Fig. 3.4-2d

axial

flow

4.

with the blades at 45°,

and radial flow

downward and

"is

and

is

at

some

3.4-3.

Fig.

axial flow

present. This type

is

imparted so that a combination of

is

useful in

are also useful for

suspending solids since the currents

then sweep up the solids.

Helical-ribbon agitators.

and operates

in

a low

This type of agitator

is

RPM in the laminar region. The

attached to a central shaft.

The

liquid

moves

used

in highly

ribbon

is

viscous solutions

formed

in a helical

in a tortuous flow path

path

down

the

center and up along the sides in a twisting motion. Similar types are the double helical

ribbon and the helical ribbon with a screw. 5.

Agitator selection and viscosity ranges.

The

viscosity of the fluid

is

one of several

factors affecting the selection of the type of agitator. Indications of the viscosity ranges of

these agitators are as follows. Propellers are used for viscosities of the fluid below about

142

Chap. 3

Principles

of Momentum Transfer and Applications

3 Pa s (3000 cp); turbines can be used below about 100 Pa-s (100000 cp); modified paddles such as anchor agitators can be used above 50 Pa s to about 500 Pa s (500000 -

-

and ribbon-type agitators are often used above this range to about 1000 Pa s and have been used up to 25 000 Pa-s. For viscosities greater than about 2.5 to 5 Pa-s

cp); helical

•

(5000 cp) and above, bafTles are not needed since

little

swirling

is

present above these

viscosities.

3.4C

Flow Patterns

The flow patterns

in

in

Agitation

an agitated tank depend upon the

the tank, types of baffles in the tank,

agitator

mounted

is

fluid properties, the

and the agitator

vertically in the center of a tank with

pattern usually develops. Generally, this

is

no

geometry of

a propeller or other

baffles, a swirling

flow

undesirable, because of excessive air en-

trainment, development of a large vortex, surging, and the

To

If

itself.

like,

especially at high speeds.

an angular off-center position can be used with propellors with small horsepower. However, for vigorous agitation at higher power, unbalanced forces can become severe and limit the use of higher power. For vigorous agitation with vertical agitators, baffles are generally used to reduce swirling and still promote good mixing. Baffles installed vertically on the walls of the prevent

this,

tank are shown in Fig. 3.4-3. Usually four baffles are sufficient, with their width being about jj of the tank diameter for turbines and for propellers. The turbine impeller drives the liquid radially against the wall, where it divides, with one portion flowing upward near the surface and back to the impeller from above and the other flowing

Sometimes,

in

tanks with large liquid depths

three impellers are

bottom impeller

Sec. 3.4

is

mounted on

about

Agitation

1

.0

the

same

much shaft,

downward.

greater than the tank diameter, two or

each acting as a separate mixer. The

impeller diameter above the tank bottom.

and Mixing of Fluids and Power Requirements

143

volume flow

In an agitation system, the

rate of fluid

moved by

the impeller, or

important to sweep out the whole volume of the mixer in a reasonable time. Also, turbulence in the moving stream is important for mixing, since it entrains the material from the bulk liquid in the tank into the flowing stream. Some agitation circulation rate,

is

systems require high turbulence with low circulation rates, and others low turbulence

and high circulation rates. This often depends on the types of fluids being mixed and on the amount of mixing needed.

Typical "Standard" Design of Turbine

3.4D

The turbine

agitator

shown

in Fig. 3.4-3 is the

most commonly used agitator

in the

process industries. For design of an ordinary agitation system, this type of agitator

The geometric proportions

often used in the initial design. are considered as a typical

"standard" design are given

is

of the agitation system which in

Table

3.4-1.

These

relative

proportions are the basis of the major correlations of agitator performance in numerous publications. (See Fig. 3.4-3c for nomenclature.)

some cases W/D a = 3 for agitator correlations. The number of baffles is 4 in most The clearance or gap between the baffles and the wall is usually 0.10 to 0.15 J to

In uses.

ensure that liquid does not form stagnant pockets next to the baffle and wall. In a few correlations the ratio of baffle to tank diameter is J ID , = ^ instead of ^.

Power Used

3.4E

In the design of

in

Agitated Vessels

an agitated

impeller. Since the

vessel,

an important factor

power required

for a given

is

the

power required

empirical correlations have been developed to predict the

power

required.

The

presence

Reynolds number N'Re

or absence of turbulence can be correlated with the impeller

denned

to drive the

system cannot be predicted theoretically,

,

as

(3.4-D

where

D

a

is

density in

N'Ke

<

10,

the impeller (agitator) diameter in m,

kg/m 3 and ,

p.

is

turbulent for N'Rc

kg/m

viscosity in

>

10

4 ,

and

for a

transitional, being turbulent at the impeller

Power consumption

is

•

s.

N

is

rotational speed in rev/s, p

The flow

is

laminar

range between 10 and

and laminar

in

in

is

fluid

the tank for

10*,

remote parts of the

the flow

is

vessel.

related to fluid density p, fluid viscosity p, rotational speed N,

Table

3.4-1.

Geometric Proportions for

"Standard"

a

Agitation

System

H 0.3 to 0.5

W

144

1

Dd

2

5

Da

3

Chap. 3

C 1

D,~

D,

L

Principles

D, 1

J

4

D,

1

~

3

1

"

12

of Momentum Transfer and Applications

and impeller diameter D a by plots of power number N p versus N'Rc The power number .

N =

P

N" =

^—

is

(SI)

(3.4-2)

P = power in J/s

where

Figure 3.4-4

=

In English units, P

W.

or

(English)

pN'Dt

tank,

and impeller

same

impellers in unbaffled tanks

vessel.

for

when

JV'Re is

300 or

an unbaffled vessel

less (B3, Rl).

considerably

is

baffle,

also be used for the

When N'Re

is

above

than for a baffled

less

for other impellers are also available (B3, Rl).

EXAMPLE A.

may

given in Fig. 3.4-3c. These curves

sizes are

power consumption

Curves

lb f/s.

•

Dimensional measurements of

liquids contained in baffled, cylindrical vessels.

300, the

ft

a correlation (B3, Rl) for frequently used impellers with Newtonian

is

Power Consumption

3.4-1.

an Agitator

in

flat-blade turbine agitator with disk having six blades

similar to Fig. 3.4-3.

100 60 40

Q

10

%

m,

1.83

is

installed in a tank

the turbine diameter

Da

A ,3

-

1

—

6

4 2 :

ii

is

i

:

20 a.

The tank diameter D,

V

s i

—

i i

1

0.6 0.4 0.2 0.1

i

2

in 4

i

i

4

2

10

i

i

i

10

i

2 2

"Re Figure

3.4-4.

.

1

.

2.

3.

4.

5.

DJJ =

io

s

D a2 Np baffles (see Fig. 3.4-3c for

W

;

D JW =

5

12.

each

DJJ =

DJW

=

8; four baffles each

DJW

= 2D a ; four baffles each DJJ angular off-center position with no baffles.

Propeller (like Fig. 3.4-1); pitch

same propeller

Propeller: pitch

1, 2,

;

=

8; four

12.

in

= Da

;

four baffles each

propeller in angular off-center position with

[Curves

4

4 2

12.

holds for

Curve

0

Six-blade open turbine but blades at 45° (like Fig. 3.4-2d)

baffles

Curve

4 1

Flat six-blade open turbine (like Fig. 3.4-2c);

DJJ = Curve

iii

i

3 2

Flat six-blade turbine with disk (like Fig. 3.4-3 but six blades)

four baffles each

Curve

10

Power correlations for various impellers and D a ,D,, J, and ).

dimensions

Curve

=

in 4

no

DJJ =

=

10; also

10; also holds for same

baffles.

and 3 reprinted with permission from R.

L. Bales, P. L.

Fondy, and R. R.

1963). Copyright by the American Rushton, E. W. Costich, and H. J. Everett,

Corpstein, lnd. Eng. Chem. Proc. Des. Dev., 2, 310

(

Chemical Society. Curves 4 and 5 from J. 11. Chem. Eng. Progr., 46, 395, 467 (1950). With permission.']

Sec. 3.4

Agitation

and Mixing of Fluids and Power Requirements

145

W

is 0.122 m. The tank contains four m, D, = H, and the width each having a width J of 0.15 m. The turbine is operated at 90 rpm and the liquid in the tank has a viscosity of 10 cp and a density of 929kg/m 3 (a) Calculate the required kW of the mixer. (b) For the same conditions, except for the solution having a viscosity is

0.61

baffles,

.

kW.

of 100000 cp, calculate the required

For part m, D, = and

Solution:

W = 0.122

929 kg/m 3

,

^ Using Eq.

=

(a)

1.83

the Reynolds

c

=

number

-^- =

0.01

m

•

0.01

Pa

s

s

is

=

= 5 and DJJ 12, F = 5 for 1 in Fig. 3.4-4 since 5.185 x 10*. Solving for P in Eq. (3.4-2) and substituting known

DJW

Using curve

Nr =

m,

0.15

3 (10.0cpXl x 10" )

(3.4-1),

data are given: D a = 0.61 m, N = 90/60 = 1.50 rev/s, p =

following

the

m, J =

N

values,

P = N F pN 3 = For part

1324

= =

J/s

3

5(929X1. 50) (0.61)

kW

1.324

(1.77 hp)

(b),

-

=

3 H = 100000(1 x 10~ )

_ N* is

in the

=

P Hence, a

_

,

m 100 ,

kg

,

m

•

s

(0.61)^(1.50)929

m

c

This

5

From

laminar flow region. 3

14(929)(1.50) (0.61)

-5.185.

5

=

Fig. 3.4-4,

3707

J/s

=

NP =

3.71

kW

14.

(4.98 hp)

10 000-fold increase in viscosity only increases the

1.324 to 3.71

power from

kW.

Variations of various geometric ratios from the "standard" design can have different effects

on

the

power number N P

in the turbulent

region of the various turbine agitators as

follows (B3). 1.

2.

For the

flat

six-blade open turbine,

N F-<± [W/D

l

°.

a)

For the flat, six-blade open turbine, varying DJD, from 0.25 to 0.50 has practically no effect on N F With two six-blade open turbines installed on the same shaft and the spacing between the two impellors (vertical distance between the bottom edges of the two turbines) being at least equal to D a the total power is 1.9 times a single flat-blade impeller. .

3.

,

For two

six-blade pitched-blade (45°) turbines, the

power

is

also about 1.9 times that

of a single pitched-blade impeller. 4.

A

baffled vertical square tank or a horizontal cylindrical tank has the

number

as a vertical cylindrical tank.

However, marked changes

in the

same power flow patterns

occur.

The power number

146

for a plain

Chap. 3

anchor-type agitator similar to Fig. 3.4-2b but

Principles

of Momentum Transfer and Applications

without the two horizontal cross bars

NP = where DJD,

= 0.90,

=

WjD,

The power number

N'Ke <

The

is

215(NL)-°-

and C/D,

0.10,

=

N'Rc

= 186(N'Re

Np

= 290(N Re

100 (H2):

955

(3.4-3)

0.05.

for a helical-ribbon agitator for very viscous liquids for

1

(agitator pitch/tank diameter

)

_1

=

(agitator pitch/tank diameter

)

typical dimensional ratios used are

DJD, =

=

.0)

(3.4-4)

0.5)

(3.4-5)

1

0.95, with some ratios as low as

and W/D, = 0.095.

Agitator Scale-Up

3.4F

Introduction.

In the process industries experimental data are often available

laboratory-size or pilot-unit-size agitation system and to design a full-scale unit. Since there single

method can handle

of

is,

Kinematic similarity can be denned

Dynamic

much

is

it is

no and many approaches to course, important and simplest to achieve. diversity in processes to be scaled up,

terms of ratios of velocities or of times (R2).

in

similarity requires fixed ratios of viscous, inertial, or gravitational forces.

geometric similarity

on a

desired to scale-up the results

types of scale-up problems,

all

scale-up exist. Geometric similarity

if

<

as follows (H2, P3).

Np

0. 75.

1.

20

as follows for

is

Even

achieved, dynamic and kinematic similarity cannot often be

is

obtained at the same time. Hence,

it is

often

up

to the designer to rely

on judgment and

experience in the scale-up.

many

an agitation process are as where the liquid motion or corresponding velocities are approximately the same in both cases; equal suspension of solids, where the levels of suspension are the same; and equal rates of mass transfer, where mass transfer is occurring between a liquid and a solid phase, liquid-liquid phases, and so on, and the rates are the same. In

cases, the

main

objectives usually present in

follows: equal liquid motion, such as in liquid. blending,

2.

A

Scale-up procedure.

suggested step-by-step procedure to follow

detailed as follows for scaling up from the initial

given in Table 3.4-1 are

D al D Tl ,

,

Hu W

x

,

in the

scale-up

is

conditions where the geometric sizes

and so on,

to the final conditions of£> a2

,

D T2

,

and so on. 1.

Calculate the scale-up ratio R. Assuming that the original vessel

with

D Tl = H u

the

volume V

x

is

a standard cylinder

is

(3.4-6)

Then

the ratio of the volumes

is

V _

nPy*

V,

nD 3Tl /4

2

The scale-up

ratio

is

_ D\T2 D\Tl

(3.4-7)

then

(3.4-8)

Sec. 3.4

Agitation and Mixing of Fluids and

Power Requirements

147

2.

Using

this

value of R, apply

to all of the dimensions in

it

Table

3.4-1 to calculate the

new dimensions. For example,

=RD al

D a2 3.

Then

=RJ U

J2

,

must be selected and applied

a scale-up rule

to use to duplicate the small-scale results using

=

where n

equal liquid motion, n

1 for

for equal rates of

This value of n 4.

Knowing N 2

.

is

mass transfer (which

= is

N

t

(3.4-9)

N2

determine the agitator speed

to

.

--

This equation

is

as follows (R2):

^ for equal suspension of

solids,

and n

=§

equivalent to equal power per unit volume).

based on empirical and theoretical considerations.

tne

power required can be determined using Eq.

(3.4-2)

and

Fig. 3.4-4.

EXAMPLE

3.4-2. Derivation of Scale-Up Rule Exponent For the scale-up rule exponent n in Eq. (3.4-10), show the following for

turbulent agitation. (a)

That when n

=

power per

§ , the

volume

unit

constant in the

is

scale-up. (b)

That when n

=

1.0,

the tip speed

For part (a), from Fig. Then, from Eq. (3.4-2),

Solution: region.

P Then

for equal

power

is

constant in the scale-up.

NF

3.4-4,

=N r pN]D

1

per unit volume,

constant for the turbulent

is

5

(3.4-11)

al

P /V l

l

= PilV 2

,

or using Eq.

(3.4-6),

ll = K,

P = El = _Jj_ nD 3T1 /4 V2 nD 3T2 /4

a n) (3 1 K

i

P from Eq. (3.4-1 1) and also a similar equation for Eq. (3.4-12) and combining with Eq. (3.4-8),

Substituting

,

'

'

P2

into

3

N = N (j^' 2

For part by

(b),

(3.4-13)

l

using Eq. (3.4-10) with n

=

1.0,

rearranging, and multiplying

7t,

N^N^y nD T2 N 2 = nD Tl N where rnD T2

To

N

2 is

(3.4-14)

(3.4-15)

l

the tip speed in m/s.

new agitation systems and to serve as a guide for evaluating some approximate guidelines are given as follows for liquids of normal 3 (M2): for mild agitation and blending, 0.1 to 0.2 kW/m of fluid (0.0005 to

aid the designer of

existing systems, viscosities

0.001 hp/gal); for vigorous agitation, 0.4 to 0.6

agitation or

where mass

kW

transfer

is

kW/m 3

important, 0.8 to

(0.002 to 0.003 hp/gal); for intense

2.0kW/m 3

(0.004 to 0.010 hp/gal).

power delivered to the fluid as given in Fig. 3.4-4 and Eq. (3.4-2). This does not include the power used in the gear boxes and bearings. Typical efficiencies of electric motors are given in Section 3.3B. As an approximation the power lost in the gear boxes, bearings, and in inefficiency of the electric motor is about 30 to 40% of?, the actual power input to the fluid. This power

148

in

is

the actual

Chap. 3

Principles

of Momentum Transfer and Applications

EXA MPLE

Scale- Up of Turbine Agitation System is the same as given in Example 3.4-la for a flat-blade turbine with a disk and six blades. The given conditions and sizes

An

3.4-3.

existing agitation system

W

= 0.122 m, 7, =0.15 m, D Tl = 1.83 m, D al =0.61 m, t Ni = 90/60 = 1.50 rev/s, p = 929 kg/m\ and p. = 0.01 Pa s. It is desired to scale up these results for a vessel whose volume is 3.0 times as large. Do this for the following two process objectives. (a) Where equal rate of mass transfer is desired. (b) Where equal liquid motion is needed. are

•

Solution:

H = DT =

Since

l

m

14.44

Eq.

3

3

7r(1.83) /4

=

in

the

m3

4.813

Following the steps

.

m,

1.83

,

t

V = {nD T2 JtyHJ =

volume

tank

original

V2 =

Volume

.

3.0(4.813)

=

procedure, and using

the scale-up

(3.4-8),

The dimensions

of the larger agitation system

D T2 =

are as follows:

RD Ti = 1.442(1.83) = 2.64 m, D a2 = 1.442(0.61) = 0.880 m, W2 = 1.442(0.122) = 0.176 m, and J 2 = 1.442(0.15) = 0.216 m. For part (a) for equal mass transfer, n = | in Eq. (3.4-10). 3

2/3

= (L50)

^ = N i(jf' Using Eq.

^J

(3.4-1),

NP =

P2 =

-

^ 8453

fp

m)

><104

0.01

per unit

3

5.0(929X1.175) (0.880)

volume P,

1.324

V

4.813

P2 _ 3.977

V2 ~ value of 0.2752

For part

(b) for

=

3977

J/s

=

3.977

kW

=

0.2752

kW/m 3

=

0.2752

kW/m 3

14.44

kW/m 3

guidelines of 0.8 to 2.0 for

5

is

l

The

rev / s (70 5

5.0 in Eq. (3.4-2),

5 = N e pN\D a2

The power

L175

0.880mi75)929

fi

Using

=

(ii4^)

mass

somewhat lower than

is

the approximate

transfer.

equal liquid motion, n

=

1.0.

1.0

^2=

(1-50)1

r-^l

=

1.040 rev/s

1.442,

P2 =

P2 V2

3.4G

2.757 ~~

=

0.1909

5

=

2757

J/s

=

2.757

kW

kW/m 3

14.44

Mixing Times of Miscible Liquids

one method used amount of HC1 acid In

Sec. 3.4

3

5.0(929X1.040) (0.880)

Agitation

mixing time of two miscible liquids, an added to an equivalent of NaOH and the time required for the

to study the blending or is

and Mixing of Fluids and Power Requirements

149

is noted. This is a measure of molecule-molecule mixing. Other experimental methods are also used. Rapid mixing takes place near the impeller, with slower mixing, which depends on the pumping circulation rate, in the outer zones. In Fig. 3.4-5, a correlation of mixing time is given for a turbine agitator (B5, M5, Nl). The dimensionless mixing factor /, is. denned as

indicator to change color

V

(ND a2 ) 2l f<

*T

l6

D an l

H U2D V2

(34 .j 6)

For /v"Re > 1000, since the /, is approximately For some other mixers it has been shown that t T N is approximately constant. For scaling up from vessel 1 to another size vessel 2 with similar geometry and with the same power/unit volume in the turbulent region, the mixing times are related by

where t T

is

the mixing time in seconds.

constant, then

t

T

N

113

is

constant.

11/18

\D.

'r,

(3.4-17)

!

Hence, the mixing time increases for the larger vessel. For scaling up keeping the same mixing time, the power/unit volume P/V increases markedly. 1

1/4

(p 2 /v 2 ) (3.4-18)

Usually,

in

scaling

that the power/unit

10

up to large-size vessels, a somewhat larger mixing time volume does not increase markedly.

is

used so

J

10' bo

"a:

10

i

i

i

!

1

1

1

i

i

i

Inn

1

1

1

lull

2

10

10

Re =

Figure

3.4-5.

i

10

J

i

i

Inn 10'

1

1

1

lllll

1

1

s

10

1

lllll

10°

NDip

Correlation of mixing time for miscible liquids using a turbine in a ( for a plain turbine, turbine with disk, and pitched-

baffled tank

blade turbine). [From "Flow Patterns and Mixing Rates in Agitated Vessels" by K. W. Norwood and A. B. Metzner, A.I.Ch.E. J., 6, 432 (I960). Reproduced by permission of the

American

150

Chap. 3

Institute

of Chemical Engineers, I960.

Principles of Momentum Transfer

and Applications

The mixing time

Nt T =

126

Nt T = 90

for a helical-ribbon agitator

as follows for

is

N'Re < 20

(H2).

(agitator pitch/tank diameter

=

1.0)

(3.4-19)

(agitator pitch/tank diameter

=

0.5)

(3.4-20)

liquids the helical-ribbon mixer gives a much smaller mixing time than a turbine for the same power/unit volume (M5). For nonviscous liquids, however,

For very viscous

gives longer times.

it

For a propellor

agitator in a baffled tank, a mixing-time correlation

Biggs (B5) and that for an unbaffled tank by

Flow Number and Circulation Rate

3.4H

An

agitator acts like a centrifugal

pump

is

given by

(Fl).

in Agitation

impeller without a casing and gives a flow at

a certain pressure head. This circulation is

Fox and Gex

rate

Q

in

m

3

/s

from the edge of the impeller

the flow rate perpendicular to the impeller discharge area. Fluid velocities have been

measured

in

mixers and have been used to calculate the circulation rates. Data for have been correlated using the dimensionless flow number Nq (Ul).

baffled vessels

N •

Nq =

0.5

marine propeller (pitch = diameter)

NQ =

0.75

6 blade turbine with disk

(W/D a =

NQ =

0.5

6 blade turbine with disk

(W/D a =

Nq =

0.75

pitched-blade turbine

(W/D a =

'

4 - 21)

l

j)

Special Agitation Systems

3.41

/.

(3

°'jk

Suspension of solids.

liquid.

In

some

Examples are where a

agitation systems a solid

finely dispersed solid is to

is

suspended

in the agitated

be dissolved in the liquid,

microorganisms are suspended in fermentation, a homogeneous liquid-solid mixture is to be produced for feed to a process, and a suspended solid is used as a catalyst to speed

up a

reaction.

The suspension

of solids

is

somewhat

similar to a fiuidized bed. In the

The amount and type of agitation needed depends mainly on the terminal settling velocity of the particles, which can be calculated using the equations in Section 14.3. Empirical equations to predict the power required to suspend particles are given in references

agitated system circulation currents of the liquid keep the particles in suspension.

(M2, Wl). 2. Dispersion of gases

and

liquids in liquids.

In gas-liquid dispersion processes, the gas

introduced below the impeller, which chops the gas into very fine

is

bubbles. The type and

degree of agitation affects the size of the bubbles and the total interfacial area. Typical of

such processes are aeration

hydrogen gas

Sec. 3.4

in

sewage treatment plants, hydrogenation of liquids by a catalyst, absorption of a solute from the gas by the

in the presence of

Agitation and Mixing of Fluids and

Power Requirements

151

and fermentation. Correlations are available to predict the bubble size, holdup, power needed (C3, LI, Zl). For liquids dispersed in inmiscible liquids, see reference (Tl). The power required for the agitator in gas-liquid dispersion systems

liquid,

and

kW

can be as much as 10 to 3.

Motionless mixers.

50%

less than that required

Mixing of two

fluids

when no

gas

is

present (C3, T2).

can also be accomplished

in

motionless

mixers with no moving parts. In such commercial devices stationary elements inside a pipe successively divide portions of the stream and then recombine these portions. In

one type a short helical element divides the stream in two and rotates it 180°. The second element set at 90° to the first again divides the stream in two. For each element there are 2 divisions and recombinations, or 2" for n elements in series. For 20 elements about 10

6

divisions occur.

Other types are available which consist of bars or

flat

sheets placed lengthwise

in

a pipe. Low-pressure drops are characteristic of all of these types of mixers. Mixing of

even highly viscous materials

is

quite

good

in these mixers.

Mixing of Powders, Viscous Material, and Pastes

3.4J

In mixing of solid particles or powders it is necessary to displace parts of powder mixture with respect to other parts. The simplest class of devices suitable for gentle blending is the tumbler. However, it is not usually used for breaking up agglomerates. A common type of tumbler used is the double-cone blender, in which two cones are mounted with their open ends fastened together and rotated as shown in Fig. 3.4-6a. /.

Powders.

the

Baffles can also be

used internally.

If

an internal rotating device

is

also used in the

double cone, agglomerates can also be broken up. Other geometries used are a drical

drum with

internal baffles or twin-shell

V

type.

Tumblers used

cylin-

specifically for

breaking up agglomerates are a rotating cylindrical or conical shell charged with metal or porcelain steel balls or rods.

Another container

is

class of devices for solids

stationary

blending

is

the stationary shell device, in which the

and the material displacement

is

accomplished by single or two open

multiple rotating inner devices. In the ribbon mixer in Fig. 3.4-6b, a shaft with

1 and 2 attached to it rotates. One screw is left-handed and one As the shaft rotates, sections of powder move in opposite directions and mixing occurs. Other types of internal rotating devices are available for special situations (PI). Also, in some devices both the shell and the internal device rotate.

helical

screws numbers

right-handed.

In the mixing of dough, pastes, and viscous amounts of power are required so that the material is divided, folded, or recombined, and also different parts of the material should be displaced relative to each other so that fresh surfaces recombine as often as possible. Some machines may require jacketed cooling to remove the heat generated. The first class of device is somewhat similar to those for agitating fluids, with an impeller slowly rotating in a tank. The impeller can be a close-fitting anchor agitator as in Fig. 3.4-6b, where the outer sweep assembly may have scraper blades. A gate impeller can also be used which has horizontal and vertical bars that cut the paste at various levels and at the wall, which may have stationary bars. A modified gate mixer is the 2.

Dough, pastes, and viscous materials.

materials, large

shear-bar mixer, which contains vertical rotating bars or paddles passing between vertical stationary fingers.

Other modifications of these types are those where the can or

container will rotate as well as the bars and scrapers. These are called change-can mixers.

The most commonly used mixer

152

Chap. 3

for

heavy pastes and dough

Principles of

Momentum

is

the double-arm

Transfer and Applications

(a)

(b)

(c)

FIGURE

3.4-6.

Mixers for powders and pastes: (a) double-cone powder mixer, powder mixer with two ribbons, (c) kneader mixer for

(b) ribbon

pastes.

.. .

kneader mixer. The mixing action

bulk movement, smearing, stretching, dividing,

is

and recombining. The most widely used design employs two contrarotating arms of sigmoid shape which may rotate at different speeds, as shown in Fig. 3.4-6c. folding,

3.5

NON-NEWTONIAN FLUIDS Types of Non-Newtonian Fluids

3.5A

As discussed

in

Section

2.4,

Newtonian

fluids are

those which follow Newton's law, Eq.

(3.5-1).

dv

T= -IX-

(SI)

dr (3.5-1)

p dv (English) 9c

where

p. is

the viscosity and

shown of shear

is

dr

a constant independent of shear

stress r versus shear rate

-dv/dr.

The

rate. In Fig. 3.5-1 a plot is

line for a

Newtonian

fluid

is

straight, the slope being p. If

a fluid does not follow Eq.

versus —dv/dr

is

(3.5-1),

it is

a

non-Newtonian fluid. Then a plot of t Non-Newtonian fluids can

not linear through the origin for these fluids.

be divided into two broad categories on the basis of their shear stress/shear rate behavior: those whose shear stress

independent of time or duration of shear (timeis dependent on time or duration of shear (time-dependent). In addition to unusual shear-stress behavior, some non-Newtonian fluids also exhibit elastic (rubberlike) behavior which is a function of time and results in is

independent) and those whose shear^stress

Sec. 3.5

Non-Newtonian Fluids

153

being called viscoelastic fluids. These fluids exhibit normal stresses perpendicular to

their

Most of the emphasis which includes the majority of non-Newtonian

the direction of flow in addition to the usual tangential stresses. will

be put on the time-independent

class,

fluids.

Time-Independent Fluids

3.5B

Bingham plastic fluids. These are the simplest because, as shown in Fig. 3.5-1, they from Newtonian only in that the linear relationship does not go through the origin. A finite shear stress t y (called yield stress) in N/m 2 is needed to initiate flow. Some fluids have a finite yield (shear) stress x y but the plot of z versus —dv/dr is curved upward or downward. However, this departure from exact Bingham plasticity is often small. Examples of fluids with a yield stress are drilling muds, peat slurries, margarine, chocolate mixtures, greases, soap, grain-water suspensions, toothpaste, paper pulp, and sewage /.

differ

,

sludge.

2.

The majority of non-Newtonian

Pseudoplastic fluids.

fluids are in this category

and

include polymer solutions or melts, greases, starch suspensions, mayonnaise, biological fluids,

detergent slurries, dispersion media in certain pharmaceuticals, and paints.

shape of the flow curve

is

shown

The

and it generally can be represented by a Ostwald-deWaele equation).

in Fig. 3.5-1,

power-law equation (sometimes called

(3.5-2)

where

K

is

the consistency index in

index, dimensionless. is

fj.

a

= K(dv/dr)"~

l

The apparent

N-s'/m 2

or \b f

2

and n is the flow behavior from Eqs. (3.5-1) and (3.5-2),

-'s"/(t

viscosity, obtained

,

and decreases with increasing shear rate.

Bingham

plastic

^— dilatant pseudoplastic

Newtonian

Shear rate, - dv/dr

Figure

3.5-1.

Shear diagram for Newtonian and time-independent non-Newtonian fluids.

154

Chap. 3

Principles

of Momentum Transfer and Applications

3.

These

Dilatant fluids.

fluids are far less

common

than pseudoplastic

fluids

and

their

flow behavior in Fig. 3.5-1 shows an increase in apparent viscosity with increasing shear

The power law equation (3.5-2)

rate.

=K

X

For a Newtonian

n

fluid,

=

often applicable, but with n

is

1.

(n>l)

K~7r)

(3,5" 3)

Solutions showing dilatancy are

1.

>

some corn

solutions, wet beach sand, starch in water, potassium silicate in water, tions containing high concentrations of

1.

in water.

Time-Dependent Fluids

3.5C

at

powder

flour-sugar

and some solu-

These

Thixotropic fluids.

fluids exhibit

a reversible decrease

in

shear stress with time

a constant rate of shear. This shear stress approaches a limiting value that depends on

the shear rate. als,

and

Examples include some polymer solutions, shortening, some food materi-

paints.

The theory

time-dependent fluids at present

for

is still

not completely

developed.

2.

These

Rheopectic fluids.

fluids are quite rare in

occurrence and exhibit a reversible

Examples are bentonite clay gypsum suspensions. In design procedures for thixotropic

increase in shear stress with time at a constant rate of shear.

suspensions, certain sols, and

and rheopectic

fluids for steady flow in pipes, the limiting

flow-property values at a

constant rate of shear are sometimes used (S2, W3). /

3.5D

•

Viscoelasric Fluids

Viscoelastic fluids exhibit elastic recovery from the deformations that occur during flow.

They show both viscous and elastic properties. Part of the deformation is recovered upon removal of the stress. Examples are flour dough, napalm, polymer melts, and bitumens.

Laminar Flow of Time-Independent Non-Newtonian Fluids

3.5E

In determining the flow properties of a time-independent 1. Flow properties of a fluid. non-Newtonian fluid, a capillary-tube viscometer is often used. The pressure drop AP N/m 2 for a given flow rate q m 3 /s is measured in a straight tube of length L rh and

D

repeated for different flow rates or average velocities

V

m/s.

time-independent, these flow data can be used to predict the flow

in

any other

diameter fluid

pipe

is

is

If

the

size.

A which fluid

m. This

plot of is

D Ap/4L, which

is

t„, the shear stress at the wall in

proportional to the shear rate at the wall,

following Eq.

Ap

•

Sec. 3.5

plastic

if

versus 8K/D,

power-law

\D

on logarithmic coordinates and Newtonian for ri < 1, pseudoplastic, or curve does not go through the origin; and for ri > 1, dilatant. The

the slope of the line

K' has units of N s"'/m

Bingham

,

in Fig. 3.5-2 for a

/8V\

4L ri is

shown

(3.5-4).

D where

is

N/m 2

the

2

.

For

when

ri

Non-Newtonian Fluids

=

the data are plotted

I,

the fluid

is

;

155

Figure

General flow curve for a power-law fluid in laminar flow in a tube.

3.5-2.

K', the consistency index in Eq. (3,5-4),

rate at the wall,

{-dvldr) w

,

the.value of

is

D Ap/4L

for

ZV/D =

1.

The shear

is

(3.5-5)

—

Also, K'

/x

Equation applied

for

Newtonian

(3.5-4)

is

fluids.

simply another statement of the power-law model of Eq. (3.5-2)

flow in round tubes, and

to

more convenient

is

to

use for pipe-flow

situations (D2). Hence, Eq. (3.5-4) defines the flow characteristics just as completely as

Eq. (3.5-2).

It

has been found experimentally (M3) that for most fluids K' and

constant over wide ranges of

K' and

ri vary.

Then

%VID orD AF/4L

ZVjD

or

D

Ap/4L. For some fluids

the particular values of K'

with which one

flow in a pipe or tube

is

dealing

is

this

is

ri

are

not the case, and

and

in

ri used must be valid for the actual a design problem. This method using

often used to determine the flow properties of a

non-Newtonian

fluid.

In

many

A

flow properties

K

and n

in

discussion of the rotational viscometer

When for

cases the flow properties of a fluid are determined using a rotational

The

viscometer.

many

Eq. (3.5-2) are determined

is

in this

the flow properties are constant over a range of shear stresses

fluids, the

manner.

given in Section 3.51.

which occurs

following equations hold (M3): ri

=

(3.5-6)

n

(iri

+ lV' (3-5-7)

Often a generalized viscosity coefficient

7 is

y

=

X'8"'-

y

=

g c K'8"'-

defined as

1

(SI)

(3.5-8)

2 where y has units of N s"7m orlb^/ft •

i

(English)

2 ""' •

s

Typical flow-property constants (rheological constants) for

some

fluids

are given in

Some

data give y values instead of K' values, but Eq. (3.5-8) can be used to convert these values if necessary. In some cases in the literature, K or K' values are given

Table 3.5-1.

156

Chap. 3

Principles of

Momentum

Transfer and Applications

Table

Flow-Property Constants for Non-Newtonian Fluids

3.5-1.

Flow-Property Constants

1.5% carboxymethylcellulose 3.0% CMC in water 4% paper pulp in water 14.3% clay in water 10% napalm in kerosene

25%

in water

clay in water

Applesauce, brand

=

A (297

K),

g/cm 3 Banana puree, brand A (297 K), density = 0.977 g/cm 3 Honey (297 K) Cream, 30% fat (276 K) density

Tomato

concentrate,

dyn s"7cm 2 or •

lb f

•

s"'/ft

1

Equations for flow

laminar flow

2

in

9.12

(Al)

0.350

0.0512

(W2)

0.520

1.756

0.185

0.3036

(SI)

(SI)

(W2)

0.645

0.500

(CI)

0.458

6.51

(CI)

in

is

5.61

(CI)

1.0

0.01379

(M4)

0.59

0.2226

(HI)

the conversion factors are

=

47.880 N-s"'/m

dyn-s n '/cm 2

=

2.0886 x 10"

n '/ft

3

2

lb f -s"'/ft

2

In order to predict the frictional pressure

a tube.

drop Ap

in

solved for Ap.

is

=

—

(-)

(3.5-9)

desired, Eq. (3.5-4) can be rearranged to give

D V=8

equations are desired

substituted into (3.5-9)

1.00

2

lb f -s

a tube, Eq. (3.5-4)

the average velocity

If the

(SI)

0.575

From Appendix A.l,

.

Ap If

4.17

K)

total solids (305

1

2.

1.369

0.566

1.10

5.8%

as

0.554

and

in

( AdDY'"'

K instead of K' Eqs. (3.5-6) and (3.5-7) can be The flow must be laminar and the generalized

terms of

(3.5-10).

(33-10)

\K'4LJ ,

Reynolds number has been defined as

-

—V~ D"V 2 -"'P

=

=

D"V 2 '"p

D"'V 2 -"'p

=

Jln+f

(SI)

(3 -5_11>

An

3.

Friction factor method.

Alternatively, using the

in Eqs. (2.10-5) to (2.10-7) for

Newtonian

fluids,

Fanning

friction factor

method given

but using the generalized Reynolds

numbers,

Sec. 3.5

Non-Newtonian Fluids

157

(3

'-AT" '

^ 12)

V Re. gen

L V2

-y

Ap = 4/p

(SI)

(33-13)

— 2

L V

— £ 2g

Ap = 4/p

(English)

c

EXA MPLE 35-1.

Pressure Drop of Power-Law Fluid in Laminar Flow 3 power-law fluid having a density of 1041 kg/m is flowing through 14.9 m of a tubing having an inside diameter of 0.0524 m at an average velocity of 0.0728 m/s. The rheological or flow properties of the fluid are K' = 15.23 N s"'/m 2 (0.318 lb f s"'/ft 2 ) and ri = 0.40. (a) Calculate the pressure drop and friction loss using Eq. (3.5-9) for laminar flow. Check the generalized Reynolds number to make

A

•

•

is laminar. but use the friction factor method.

sure that the flow (b)

Repeat part

Solution:

(a)

known

The

data

0.0524 m, V = 0.0728 m/s, using Eq. (3.5-9),

D= (a),

V

\DJ

D

K'

are

as

follows:

L =

14.9

m, and p

=

0.0524

J

0.0524

V

=

1041

15.23,

ri

=

kg/m 3 For .

0.40,

part

«/m

Also, to calculate the friction loss,

An

Using Eq.

=

W

-

o

4O

45 390

43.60 J/kg

(3.5-11),

D"'K ^Re.

Hence, the flow

For part

is

(b),

2

^ gn

8 en-

-"'p ,

-

(0.0524)

-

,

-

(0.0728)'-

6o

(1041)

~

15.23(8)-°- 6

'

laminar.

using Eq. (3.5-12),

/=-^ = ^=1444 44 N 14

}

"

1.106

Rc g£ „ .

Substituting into Eq. (3.5-13),

Ap = 4/p

=

L -

D

45.39

V = 4 2

4(14.44 K X 1041)

^(946 m 2

V

To methods

158

Chap. 3

0.0524

2

^ 2

ft

calculate the pressure drop for a are available for laminar flow

^(^™?

Bingham

plastic fluid with a yield stress,

and are discussed

Principles

in detail

elsewhere (CI, PI, S2).

of Momentum Transfer and Applications

Friction Losses in Contractions, Expansions,

3.5F

and Fittings

in

Laminar Flow

Since non-Newtonian power-law fluids flowing in conduits are often

in

laminar flow

because of their usually high effective viscosity, losses in sudden changes of velocity and fittings are

1.

in is

important in laminar

flow.'

In application of the total mechanical-energy balance

Kinetic energy in laminar flow.

Eq. (2.7-28), the average kinetic energy per unit mass of fluid

is

needed. For

fluids, this

(S2) 2

.

.

V —

=

average kinetic energy/kg

(3.5-14)

2a

For Newtonian

fluids,

a

=

For power-law non-Newtonian

\ for laminar flow.

+

, (2n

a=

lX5n

3(3,

= 0.50, a = 0.585. If n = and non-Newtonian flow, a — 1.0(D1). For example,

if

n

+

3)

(3 -5" 15)

1)*

=

a

1.00,

+

fluids,

For turbulent flow

\.

for

Newtonian

Skelland (S2) and Dodge and Metzner (D2) state and flows through a sudden contraction to a pipe of diameter D 2 or flows from a pipe of diameter D t through a sudden contraction to a pipe of D 2 a vena contracta is usually formed downstream from the contraction. General indications are that the frictional pressure losses for pseudoplastic and Bingham plastic fluids are very similar to those for Newtonian fluids at the same generalized Reynolds numbers in laminar and turbulent flow for contractions and also for fittings and valves. For contraction losses, Eq. (2.10-16) can be used where a = 1.0 for turbulent flow and for laminar flow Eq. (3.5-15) can be used to determine a, since n is not 1.00. For fittings and valves, frictional losses should be determined using Eq. (2.10-17) and values from Table 2.10-1. 2.

Losses

that

contractions and fittings.

in

when

a fluid leaves a tank

,

3.

Losses

in

For the

sudden expansion.

frictional loss for a

laminar flow through a sudden expansion from 3"

" where

is

In

+ +

1

n

7

+

1

2(Sn

1

(

3

+

3)

D x

to

t

£> 2

(

\

\DJ

non-Newtonian

DA

\DJ

3(> +

1)

+

3)

2(5n

the frictional loss in J/kg. In English units Eq. (3.5-16)

/i„isinft-lb f /lb m

is

(3.5-16)

divided by g c and

.

=

Newtonian fluid gives values For turbulent flow 1 (Newtonian fluid). can be approximated by Eq. (2.10-15), with a = 1.0 for non-Newtonian

Equation (2.10-15)

for

laminar flow with a

reasonably close to those of Eq. (3.5-16) for n the frictional loss

fluid in

diameter, Skelland (S2) gives

\ for a

=

fluids (S2).

3.5G

Turbulent Flow and Generalized Friction Factors

number at which turbulent non-Newtonian fluid. Dodge and

In turbulent flow of time-independent fluids the Reynolds flow occurs varies with the flow properties of the

Metzner (D2)

in

a comprehensive study derived a theoretical equation for turbulent flow

of non-Newtonian fluids through Fig. 3.5-3,

Sec. 3.5

where the Fanning

smooth round

tubes.

The

final

equation

is

friction factor is plotted versus the generalized

Non-Newtonian Fluids

plotted in

Reynolds

159

FIGURE

Fanning friction factor versus generalized Reynolds number for timeindependent non-Newtonian and Newtonian fluids flowing in smooth tubes. {From D. W. Dodge and A. B. Metzner, A.I.Ch.E. J., 5, 189

3.5-3.

(1959). With permission.']

number, N Rc gen given in Eq. (3.5-11). Power-law fluids with flow-behavior indexes ri between 0.36 and 1.0 were experimentally studied at Reynolds numbers up to 3.5 x 10* and confirmed the derivation. The curves for different ri values break off from the laminar line at different Reynolds numbers to enter the transition region. For ri = 1.0 (Newtonian), the transition region starts at /V Re 8en = 2100. Since many non-Newtonian power-law fluids ,

have high effective viscosities, they are often in laminar flow. The correlation for a smooth tube also holds for a rough pipe in laminar flow. For rough commercial pipes with various values of roughness e/D, Fig. 3.5-3 cannot be used for turbulent flow, since it is derived for smooth tubes. The functional dependence of the roughness values e/D on ri requires experimental data which are not yet available. Metzner and Reed (M3, S3) recommend use of the existing relationship, Fig. 2.10-3, for Newtonian fluids in rough tubes using the generalized Reynolds number N Rc en This is somewhat conservative since preliminary data indicate that friction factors for pseudoplastic fluids may be slightly smaller than for Newtonian fluids. This is also consistent with Fig. 3.5-3 for smooth tubes that indicate lower / values for fluids .

with

ri

below

1.0 (S2).

EXA MPLE 3J-2.

Turbulent Flow of Power-Law Fluid

A

pseudoplastic fluid that follows the power law, having a density of 961 kg/m 3 is flowing through a smooth circular tube having an inside diameter ,

of 0.0508

m

fluid are ri

at

=

an average velocity of 6.10 m/s. The flow properties of the

0.30 and K'

sure drop for a tubing 30.5 Solution:

V=

160

The data

6.10 m/s, p

=

=

2.744

N

s"'/m

2 .

Calculate the frictional pres-

m long.

are as follows: K"

961 kg/m

3 ,

and L =

Chap. 3

=

= 0.30, D = 0.0508- m, Using the general Reynolds-

2.744, ri

30.5 m.

Principles of Momentum Transfer

and Applications

number equation

(3.5-1

1),

^Rc.gcn-

= Hence, the flow «'

=

0.30,

/=

is

D n.y2- n, p K 8 n,-1 ~

(Q.05Q 8

)O-3

7

(6 1Q)'- (961)

-0

>

2.744I8

-

7 )

1.328 x 10*

turbulent.

Using

Fig. 3.5-3 for

N Rc

gen

=

1.328 x 10

4

and

0.0032.

Substituting into Eq. (3.5-13),

=

kN/m 2

137.4

(2870 lb f /ft

2 )

Velocity Profiles for Non-Newtonian Fluids

3.5H

Starting with Eq. (3.5-2) written as

dv x \

K

n

(33-17)

dr the following equation can be derived relating the velocity v x with the radial position

which

is

the distance from the center. (See

vr

At

r

=

0, v x

=

v x max

=

- Pl 2KL

Problem

(n+

Po

n+l

\

and Eq.

(3.5-18)

=

l)ln


(33-18)

(Ro)'

becomes (n

v xr

r,

2.9-3 for this derivation.)

+

1

)/n

(33-19)

v x max

non-Newtonian fluid to show Newtonian fluid given in Eq. (2.9-9) can differ greatly from that of a non-Newtonian fluid. For pseudoplastic fluids (n < 1), a relatively flat velocity profile is obtained compared to the parabolic profile for a Newtonian fluid. For n = 0, rodlike flow is obtained. For dilitant fluids (n > 1), a much sharper profile is obtained and for n = oo, the velocity is a linear function of the radius.

The

velocity profile can be calculated for laminar flow of a

that the velocity profile for a

3.51

Determination of Flow Properties of Non-Newtonian Fluids Using Rotational

Viscometer

The flow-property

or rheological constants of

using pipe flow as discussed

measuring flow properties

was

first

is

in

non-Newtonian

Section 3.5E. Another,

fluids

can be measured

more important method

for

by use of a rotating concentric-cylinder viscometer. This

described by Couette in 1890. In this device a concentric rotating cylinder

speed inside another cylinder. Generally, there is filled with the fluid. The torque needed to maintain this constant rotation rate of the inner spindle is measured by a torsion wire from which the spindle is suspended. A typical commercial instrument of

(spindle) spins at a constant rotational is

a very small gap between the walls. This annulus

this

type

Sec. 3.5

is

the Brookfield viscometer.

Non-Newtonian Fluids

Some

types rotate the outer cylinder.

161

The shear

stress at the wall

of the bob or spindle

is

given by

(3.5-20)

l

2irR b L

where r w

kg-m

2

/s

spindle,

is

2 ;

N/m 2

Rb

is

The shear

rate at

< R b lR c <

Rc

is

•

0.99:

(3.5-21)

- (R b /R c ) 2/n l

«[1

where

or kg/s

the radius of the spindle,

m. Note that

(M6) for 0.5

2

m; T is the measured torque, m; and L is the effective length of the Eq. (3.5-20) holds for Newtonian and non-Newtonian fluids. the surface of the spindle for non-Newtonian fluids is as follows

the shear stress at the wall,

the radius of the outer cylinder or container,

velocity of the spindle,

rad/s.

Also,

co

= 2vN/6Q, when

m; and w is

the

is

calculated using Eq. (3.5-21) give values very close to those using the

cated equation of Krieger and

The power-law equation

Maron

is

the angular

RPM.

(K2), also given in (P4, S2).

given as

;= where

K

Results

more compli-

= N-s"/m 2 kg-s" _2 /m. ,

(3-5-2)

Substituting Eqs. (3.5-20) and (3.5-21) into (3.5-2)

gives

T=

IttRiLK n[l

(3.5-22)

- (R b /R c ) 2ln ]

Or,

Acj"

(3.5-23)

where,

A

= IttRILK nil

-{R b /R c

(3.5-24)

1/n )

]

Experimental data are obtained by measuring the torque T a given fluid.

The

The parameter,

plotting log

and the intercept from Eq. (3.5-24). Various special cases can be derived for Eq. (3.5-21).

log

a>.

consistency factor

1.

at different values of

may be evaluated by

flow property constants

K

Newtonian fluid.

n, is the slope of the straight line

is

now

is

T

a>

for

versus

Jog A. The

easily evaluated

(/i=l). 2a> 1

2.

< 0.1). This is the case of a spindle immersed Equation (3.5-21) becomes

Very large gap (R b /R c

beaker of

test fluid.

(3.5-25)

- (R b /R c ) 2

in a large

(3.5-26)

162

Chap,

i

Principles of Momentum Transfer

and Applications

Substituting Eqs. (3.5-20) and (3.5-26) into (3.5-2)

2

T=

wn

2irR b LK\^j

(3.5-27)

Again, as before, the flow property constants can be evaluated by plotting log versus log co. 3.

Very narrow gap (R b /R c > 0.99). This is similar Taking the shear rate at radius (R b + R C )I2,

dr)

\

This equation, then,

the

is

Av

dv\

I

same

between

parallel plates.

2
Ar

w

to flow

T

1

(3.5-28)

- (R b /R c ) 2

as Eq. (3.5-25).

Power Requirements in Agitation and Mixing of Non-Newtonian Fluids

3.5J

correlating the power requirements in agitation and mixing of non-Newtonian fluids power number N P is defined by Eq. (3.4-2), which is also the same equation used for Newtonian fluids. However, the definition of the Reynolds number is much more complicated than for Newtonian fluids since the apparent viscosity is not constant for non-Newtonian fluids and varies with the shear rates or velocity gradients in the vessel. Several investigators (Gl, Ml) have used an average apparent viscosity /z a which is used in the Reynolds number as follows:

For the

,

*lu..-^ The average apparent

viscosity

(3-5-29)

can be related to the average shear rate or average For a power-law fluid,

velocity gradient by the following method.

t

For

a

Newtonian

=

K\

-

—

(3.5-30)

|

fluid,

0.5-3.,

Combining Eqs.

(3.5-30)

and

(3.5-31)

ft.

=

(3.5-32)

K^—J

Metzner and others (Gl, Ml) found experimentally that the average shear for pseudoplastic liquids (n

<

1)

rate (dv/dy)^

varies approximately as follows with the rotational

speed:

dA Hence, combining Eqs.

(3.5-32)

=UN

and (3.5-33)

= (\\N)- K l

Ha

Sec. 3.5

Non-Newtonian Fluids

(3-5-33)

(3.5-34)

163

non-Newtonian

Newtonian

Figure

Power

3.5-4.

correlation in agitation for a flat, six-blade turbine with disk in

pseudoplastic non-Newtonian and Newtonian fluids (Gl,

DJW = 5,L/W = 5/4,DJJ

=

Ml, Rl):

10.

Substituting into Eq. (3.5-29),

N'Rc Equation (3.5-35) has been used

„

=

— 11" —

r-1-

(3.5-35)

jFC

to correlate data for

a

flat

six-blade turbine with

and the dashed curve in Fig. 3.5-4 shows the correlation (Ml). The solid curve applies to Newtonian fluids (Rl): Both sets of data were obtained for four baffles with DJJ = 10, DJW = 5, and L/W = 5/4. However, since it has been shown that the difference in results for DJJ = 10 and DJJ — 12 is very slight (Rl), this Newtonian line can be considered the same as curve 1, Fig. 3.4-4. The curves in Fig. 3.5-4 show that the results are identical for the Reynolds number range 1 to 2000 except that they differ only in the Reynolds number range 10 to 100, where the pseudoplastic fluids use less power than the Newtonian fluids. The flow patterns for the pseudoplastic fluids show much greater velocity gradient changes than do the Newtonian fluids in the agitator. The fluid far from the impeller may be moving in slow laminar flow with a high apparent viscosity. Data for fan turbines and propellers are also available (Ml). disk in pseudoplastic liquids,

DIFFERENTIAL EQUATIONS OF CONTINUITY

3.6

3.6A

Introduction

In Sections 2.6, 2.7,

and

2.8 overall

mass, energy, and

momentum

balances allowed us to

These balances were done on an arbitrary finite volume sometimes called a control volume. In these total energy, mechanical energy, and momentum balances, we only needed to know the state of the inlet and outlet solve

many elementary problems on

fluid flow.

streams and the exchanges with the surroundings.

These overall balances were powerful tools in solving various flow problems because knowledge of what goes on inside the finite control volume. Also, in

they did not require

the simple shell

164

momentum

balances

Chap. 3

made

in

Section

Principles of

2.9,

expressions were obtained for

Momentum

Transfer

and Applications

the velocity distribution and pressure drop. However, to advance in our study of these

flow systems,

we must

investigate in greater detail

what goes on inside

this finite control

we now use a differential element for a control volume. The differential balances will be somewhat similar to the overall and shell balances, but now we shall make the balance in a single phase and integrate to the phase boundary using the boundary conditions. In these balances done earlier, a balance was made for each new system studied. It is not necessary to formulate new balances for each new flow

To do

volume.

problem.

It is

this,

often easier to start with the differential equations of the conservation of

mass (equation of continuity) and the conservation of momentum in general form. Then these equations are simplified by discarding unneeded terms for each particular problem. For nonisothermal systems a general differential equation of conservation of energy will be considered in Chapter 5. Also in Chapter 7 a general differential equation of continuity for a binary mixture will be derived.

The differential-momentum-balance

based on Newton's second law and allows us to determine the way velocity varies with position and time and the pressure drop in laminar flow. The equation of momentum balance can be used for turbulent flow with certain modificaequation to be derived

is

tions.

Often these conservation equations are called equations of change, since they describe the variations in the properties of the fluid with respect to position

we

Before

respect to time which occur in these equations will

time.

and a

brief description of vector notation

be given.

Types of Time Derivatives and Vector Notation

3.6B

/.

and

derive these equations, a brief review of the different types of derivatives with

Various types of time derivatives are used

Partial time derivative.

in the derivations

The most common type of derivative is the partial time derivative. For 3 example, suppose that we are interested in the mass concentration or density p inkg/m in a flowing stream as a function of position x, y, z and time t. The partial time derivative to follow.

of p

2.

is

dp/dt. This

is

the local change of density with time at a fixed point x, y,

z.

Suppose that we want to measure the density in the stream moving about in the stream with velocities in the x, y, and z directions of and dz/dt, respectively. The total derivative dp/dt is

Total time derivative.

we

while

are

dx/dt, dy/dt

dp dp dp dx — =—+ This means that the density

and dz/dt

at

is

dp dy H

dy

dt

a function of

which the observer

Substantial time derivative.

dx

dt

dt

3.

and

t

and

+

dt

dp dz (

3.0- 1

dz dt

of the velocity

components

dx/dt, dy/dt,

moving.

is

Another useful type of time derivative

is

obtained

if

the

observer floats along with the velocity v of the flowing stream and notes the change in density with respect to time. This

called the derivative that follows the motion, or the

is

substantial time derivative, Dp/Dt.

where

vx

,

and

u

v,

are the velocity

vector. This substantial derivative

term

(v

Sec. 3.6

•

Vp)

will

be discussed

is

components of

in part 6 of

Differential Equations

the stream velocity

v,

which

applied to both scalar and vector variables.

is

a

The

Section 3.6B.

of Continuity

165

4.

The

Scalars.

can be placed

momentum,

physical properties encountered in

and

in several categories: scalars, vectors,

heat,

and mass

transfer

tensors. Scalars are quantities

such as concentration, temperature, length, volume, time, and energy. They have magnitude but no direction

and are considered

to be zero-order tensors.

The common

mathematical algebraic laws hold for the algebra of scalars. For example, be Hcd) 5.

=

Velocity, force,

Vectors.

momentum, and They

they have magnitude and direction.

B + C by

vector

B = i, j,

The addition of

and are the two

parallelogram construction and the subtraction of two vectors

B — C is shown in Fig. 3.6-1. The and B z on the x, y, and z axes a nd

where

cd,

acceleration are considered vectors since

are regarded as first-order tensors

written in boldface letters in this text, such as v for velocity. vectors

=

and so on.

{bc)d,

B

iB x

represented by

is

+ }B r +

and k are unit vectors along the axes

x, y,

its

three projections B^,

kB,

By

,

(3.6-3)

and

z,

respectively.

In multiplying a scalar quantity r or s by a vector B, the following hold.

The following

rB

=

Br

(3.6-4)

(rs)B

=

r(sB)

(3.6-5)

rB + sB

=

(r

=

(C

B (C + D) =

(B

•

C)

•

(B

•

6.

(B

•

D)

(3.6-8)

= BC

(3.6-10)

•

C)

cos is

<

180°.

momentum

in

transfer

and have nine com-

are discussed elsewhere (B2).

and vectors.

The gradient or "grad"

of a scalar

is

Vp =

i^ + j^ + k^ ox

where p

is

oy

(3.6-11)

oz

a scalar such as density.

Figure

3.6-1.

Addition and subtraction of vectors (b ) subtraction of vectors,

166

+

C)

•

the angle between two vectors and

They

(3.6-7)

(3.6-9)

Differential operations with scalars

field

B)

•

•

Second-order tensors t arise primarily ponents.

(3.6-6)

C)D + B(C D)

(B

s)B

also hold:

(B

where

+

Chap.

3

B -

:

(a) addition of vectors,

B + C;

C.

Principles of

Momentum

Transfer and Applications

The divergence or "div"

of a vector v

,„

,

is

dv x

dv

dv z

ox

oy

oz

(3.6-12)

where

v

is

a function of v z , v y

The Laplacian of a

,

and

vf.

scalar field

is 2

Other operations that

may

2

d p

d

3.t

dy

2

d p

p

dz

(3.6-13)

z

be useful are Vrs

(V-sv)

_

= rVs + sVr

(3.6-14)

= (Vs-v) + s(V-v)

(3.6-15)

3s

.

3s

3s (3.6-16)

Differential Equation of Continuity

3.6C

1.

Derivation of equation of continuity.

A

mass balance

The mass balance

3.6-2.

(rate of

mass

for the fluid with a

in)

— (rate

of

mass

will

Ax Ay Az which

flowing through a stationary volume element

be is

made

concentration ofp kg/m out)

=

(rate of

for

a pure

3

is

mass accumulation)

(3.6-17)

m2

In the x direction the rate of mass entering the face at x having an area of Ay Az {pv x ) x

Ay Az

kg/s and that leaving at x

m2

flux in kg/s

shown

.

Mass entering and

+ Ax

is{pv x ) x + &x

that leaving in the

Ay

Az.

fluid

fixed in space as in Fig.

The term(puj

y and the z directions

is

a

is

mass

are also

in Fig. 3.6-2.

Figure

3.6-2.

Mass balance for in

Sec. 3.6

a pure fluid flowing through a fixed volume

Ax Ay Az

space.

Differential

Equations of Continuity

167

The

rate of

mass accumulation

volume Ax Ay Az

in the

= Ax Ay Az

mass accumulation

rate of

is

dp —

(3.6-18)

dt

Substituting

Ax Ay

and dividing both

these expressions into Eq. (3.6-17)

all

by

sides

Az, [(P"x)x

-(PPx)x + Ax3

Uj>V z\

-(P"y) y + Ay]

l(PVy)y |

Ax Taking the

limit as

Ay

^

dt

fluid.

d_p

d{pv x )

d(pv y)

dt

dx

dy

The vector notation on

d{pv x )

can convert Eq.

= -(V-

how

mass

pv)

(3.6-20)

dz

.

the right side of Eq. (3.6-20)

(3.6-20) tells us

resulting from the changes in the

We

_ dp

+ Az]

Az

(

Equation

z)z

Ax, Ay, and Az approach zero, we obtain the equation of continuity

or conservation of mass for a pure

vector..

~ (j>V

|

density

comes from the

fact that v

is

a

p changes with time at a fixed point

velocity vector pv.

another form by carrying out the actual partial

(3.6-20) into

differentiation.

dp

Tt Rearranging Eq.

dv y

dv

= - pf x + {jx'

dp

v

y

dp

+

V

*Tx

dp

+

V

>Ty

The

left-hand side of Eq. (3.6-22) Hence, Eq. (3.6-22) becomes

Dp Dt

is

I

=

dp

+

V

>Ty

+

dp v

'-Tz

dv r

dv,.

as the substantial derivative in Eq. (3.6-2).

dv,\ 1

= -P( v

-v)

(3.6-23)

Often in engineering with liquids that are

Then p remains constant moves along a path following the fluid motion, or Dp/Dt = 0. becomes for a fluid of constant density at steady or unsteady state,

element as

Hence, Eq. (3.6-23)

z

+ T-pM+T^ dy dz V ox

Equation of continuity for constant density.

p

essentially constant.

is

it

(V

.

v)

=

^ + ^3 = dx

At steady

dp v

- p fdv x + dir + dv \Tx Yy Tz

-Tz =

same

the

relatively incompressible, the density for a fluid

dp

(

(3.6-21),

Tt

2.

dv,\

-

T + -dz-)-{ *Tx +

state, dp/dt

=

dy

0

(3.6-24)

oz

0 in Eq. (3.6-22).

EXAMPLE 3.6-1.

Flow over a Flat Plate one side of a flat plate. The flow in the x direction is parallel to the flat plate. At the leading edge of the plate the flow There is no velocity in the z is uniform at the free stream velocity v x0 direction. The y direction is the perpendicular distance from the plate. Analyze this case using the equation of continuity.

An

incompressible

fluid flows past

.

Solution:

For

this case

where p

is

constant, Eq. (3.6-24) holds.

^ +^+^= ox

168

Chap. 3

dy

0

(3.6-24)

dz

Principles of Momentum Transfer

and Applications

Since there

is

no velocity

we obtain

in the z direction,

-

(3

-~ty

^25)

At a given small value of y close to the plate, the value of v x must decrease from its free stream velocity'v x0 as it passes the leading edge in the x direction because of fluid friction. Hence, dvjdx is negative. Then from Eq. (3.6-25), dv y/dy is positive and there is a component of velocity away from the plate.

3.

Continuity equation in cylindrical and spherical coordinates.

It is

use cylindrical coordinates to solve the equation of continuity

if

often convenient to

fluid

is

flowing in a

The coordinate system as related to rectangular coordinates is shown The relations between rectangular x, y, z and cylindrical r, 0, z coordinates

cylinder.

in Fig.

3.6-3a.

are

x

=

y =

cos 9

r

9

r sin

=

z

z

(3.6-26)

= + Jx + 2

r

Using the relations from Eq. cylindrical coordinates

y

2

x

dp

1

d(prv r )

at

r

or

sin 9 cos

r

1

1

+r

y



2

Jx +y +

= +

r

equation of continuity

in

d(pv 9 )

d[pv x )

+ -^f 3d r,

and

9,

=

0

(3.6-27)

dz cf>

are related to x, y, z by the

in Fig. 3.6-3b.

=

x

-

(3.6-26) with Eq. (3.6-20), the

For spherical coordinates the variables

shown

tan"

is

-£'+ -

following as

=

9

1

2

=

0

z

=

r

sin 0 sin

tan"'

1

z

(f>

=

r

cos 0 (3.6-28)

^

1

^

y 4>

=

tan

z

The equation

of continuity in spherical coordinates

dp

2

+

dt

}_ 1

r

d{pr v r )

d{pv 0 sin

1

becomes 9)

1

djpvj

_ (

dr

r sin

9

dO

r

sin 9

d(f>

y

(b)

(a)

Figure

3.6-3.

Curvilinear coordinate systems

:

( a)

cylindrical coordinates, (b) spheri-

cal coordinates.

Sec. 3.6

Differential Equations

of Continuity

169

DIFFERENTIAL EQUATIONS OF

3.7

Derivation of Equations of

3.7A

The equation of motion equation (2.8-3), which / rate of

Transfer

conservation-of-momentum

can write as

/ rate of

\

\momentum

Momentum

really the equation for the

is

we

MOMENTUM TRANSFER OR MOTION

\momentum

in/

out

(sum

\

of forces

f rate of

momentum (3.7-1)

acting on system/

We

make

will

component

a balance

on an element as

of each term in Eq. (3.6-30).

\accumulation

in Fig. 3.6-2. First

The y and

z

we

shall consider

only the x

components can be described

in

an

analogous manner.

which the x component of momentum enters the face at x in the x by convection is (pvx v x ) x Ay Az, and the rate at which it leaves at x + Ax is 3 (pv x v x ) x+Ax Ay Az. The quantity {pv x ) is the concentration in momentum/m or (kg 3 m/s)/m and it is multiplied by v x to give the momentum flux asmomentum/s- m 2 The x component of momentum entering the face at y is (pv y v x ) y Ax Az, and leaving at y + Ay is (pv v x) y+&y Ax Az. For the face at z we have (pv z v x ) z Ax Ay entering, and at y z + Az we have {pv z v x ) z + Az Ax Ay leaving. Hence, the net convective x momentum flow into the volume element Ax Ay Az is

The

rate at

direction

•

,

.

Lipvx"*)*- (P»x»x)x +

Momentum

JAy Az

+

[{pv v x ) y y

-

+

[(pu z v x ) z

-(pv z v x

(pv y v x ) + Ay ]Ax y )z

+ Az ]Ax

Az

Ay

(3.7-2)

and out of the volume element by the mechanisms of convecin Eq. (3.7-2) and also by molecular transfer (by virtue of the velocity gradients in laminar flow). The rate at which the x component of momentum enters the face at x by molecular transfer is (x xx) x Ay Az, and the rate at which it leaves the surface at x + Ax is {x xx ) x + Ax Ay Az. The rate at which it enters the face at y is and it leaves at y + Ay at a rate of (x yx) y + Ay Ax Az. Note that {x yx ) Ax Az, y tion or

x yx

is

flows in

bulk flow as given

the flux of x

momentum

through the face perpendicular to the y

similar equation for the remaining faces the net x transfer

axis.

Writing a

component of momentum by molecular

is

[(***)*

-

(0

I+

A

JAy

Az

+

\_{x

-

yx ) y

{x yx ) + Ay ]Ax y

Az +

[_(x

zx ) z

-

{x zx ) z +

A f]Ax

Ay (3.7-3)

These molecular fluxes of momentum may be considered as shear stresses and normal stresses. Hence, x yx is the x direction shear stress on the y face and x zx the shear stress on the z face. Also, x xx is the normal stress on the x face. The net fluid pressure force acting on the element in the x direction is the difference between the force acting at x and x + Ax.

-

(p x

The

gravitational force g x acting to give

on a

P;t

unit

+ Ax

)AyAz

mass

in

(3.7-4)

the x direction

is

multiplied by the

mass of the element

pg x Ax Ay Az where g x 170

is

the

x component

of the gravitational vector

Chap. 3

(3.7-5) g.

Principles of Momentum Transfer

and Applications

The

momentum

rate of accumulation of x

Ax Ay Az

in

the element

is

->

d{pv x ) (3.7-6) dt

Substituting Eqs. (3.7-2M3.7-6) into' (3.7-1), dividing limit as A x, Ay, and Az approach zero, we obtain the

by Ax Ay Az, and taking the x component of the differential

equation of motion.

d{pv x )

d(pWx u J

dt

dx dr xx

+

z

components of the

d(pv z v x )

dy

dz

dr arJ dp — —+ -—+ pg yx

+

\ dx

The y and

d{pv y v x )

dy

dz

d{pV x V y )

at

dx dr xy

^

d(pV z V y )

dy

dz

dr yy

dr zy

dy

dz

d(pv z )

d(pv x v z )

dt

dx dr xz

d(pVyVy) ^

dx

^

dy

(3.7-7)

x

dp

\

+ P9y d(pv z v z )

dy

dz

dT zz\ dz

(3.7-8)

3y

/

d(pv y v z )

dTyz

dx

dx

equation of motion are, respectively,

differential

d(pv y )

J

dP

7" + P9 Z

(3.7-9)

ay

/

We

can use Eq. (3.6-20), which is the continuity equation, and Eq. (3.7-7) and obtain an equation of motion for the x component and also do the same for the y and z

components as follows: dv x ,

dv x

dv x b

v

H

T

y

dx

dt

vv

dv x \ H

dy

v,z

dz

dT xx dx

+

dr dr^ — —+

dp

yx

dy

+ P9 X

dx

dz

(3.7-10)

dVy

dVy

+

vz

3'V y\

dVy H

vv

f-

dr xy

dr yy

d~-y.

dz

v,

dt

dx

dy

dz

dx

dy

dv.

dv,

dv.

dv 7

dr xz

3t v ,

dp

+ P9 y -

dx

(3.7-11)

dt

'

dx

dy

dz

3x

+ _2£ +

dp_

d-

dy

+ P9 Z ~ dz

dx (3.7-12)

Adding

vectorially,

we

obtain an equation of motion for a pure

fluid.

Dv P

We

— = -(V-t)- Vp+pg

(3.7-13)

should note that Eqs. (3.7-7) to (3.7-13) are valid for any continuous medium.

Sec. 3.7

Differential

Equations of Momentum Transfer or Motion

171

3.7B

Equations of Motion for Newtonian Fluids with Varying Density and Viscosity

In order to use Eqs. (3.7-7) to (3.7-13) to determine velocity distributions, expressions

must be used for the various stresses in terms of velocity gradients and fluid properties. For Newtonian fluids the expressions for the stresses txx ryx t„, and so on, have been related to the velocity gradients and the fluid viscosity m (Bl, B2, Dl) and are as ,

,

follows. /.

Shear-stress components for Newtonian fluids

— dv x

-2m

in

rectangular coordinates

2

+-/x(V-v)

dx

(3.7-14)

3

dVy

2

dy

3

dv z

2

dz

3

(3.7-15)

(3.7-16)

(dv x

dVy]

+

(3.7-17)

Jy~

~dx~j

'dv y

dv z \ (3.7-18)

dz

dy

dv z

dv x \

K

I

j

(3.7-19)

\dx

(V-v)

2.

=

dz

dv x

dv y

dvA

dx

dy

dz

i

(3.7-20)

Shear-stress components for Newtonian fluids in cylindrical coordinates

dv r 2

,

Teo =

l\ dv e

_M

172

'

(3.7-21)

V)

(V-v)

dv z

- r

= —

(3.7-22)

I

2

-""(V-v)

~ T 6r - - M

T ie

(V

vr\

\r dd

2

T r0

2

T~~T dr 3

d{v e /r)

(3.7-23)

1

dv r

r

38

?

dr

dv e

1

dv z

dz

r

dd

(3.7-24)

(3.7-25)

Chap. 3

Principles of Momentum Transfer

and Applications

dv g

dv r

+

(3.7-26) dz

dr

1

=

(V v) •

d(rv r ) dr

r 3.

dv e

1

+

+

36

r

dv z (3.7-27)

dz

Shear-stress components for Newtonian fluids in spherical coordinates

Trr

dv 2— --(V-v) 2

r

=

dr

Tee

1

=

dVg

V r\

2

--

+ .—

2 \r 66

r

3u a

1

\r sin

~

9

+

v v — + r

Or

(3.7-29)

9

cot

- (V

1

du,

r

dd

+ -

dr

e<}>

~

T e

r sin 0

sin 0

(3.7-33)

r 8/-

d

2

3(r v r )

d

d{vJr)

dv r

+

(V-V)

fi

(3.7-32)

1

1

dt/

1

d9

/-

=-r

(3.7-30)

v)

(3.7-31)

sin 9 3(v^/sin 6)

T

•

V

3

r

dcf)

d(v e /r) T

(V-v)

3

/

4>4>

T re

(3.7-28)

3

sin 0)

1

8v

1

+

3r

(3.7-34) /-

r sin 0

sin 0

d<£

Equation of Motion for Newtonian fluids with varying density and viscosity After Eqs. (3.7-14)-(3.7-20) for shear-stress components are substituted into Eq. (3.7-10) for the x-component of momentum, we obtain the general equation of motion for a

4.

Newtonian

fluid

with varying density and viscosity. 3v x

Dt

/i(V-

dx

3*

— dz

/*

— dx

v)

By

3

dv t \

dv z

d

+

ldv x

2

2/i

-

+ dz

\

3p — + 99 dx

dx

dx

(3.7-35)

x

Similar equations are obtained for the y and z components of

3.7C

dVy

+

momentum.

Equations of Motion for Newtonian Fluids with Constant Density and Viscosity

When the density p and = where (V-v) /a are constant 0, the equations are simplified and we obtain the equations of motion for Newtonian fluids. These equations are also called

The equations above

are seldom used in their complete forms.

the viscosity

the Navier-Stokes equations.

Sec. 3.7

Differential Equations of Momentum Transfer or

Motion

173

Equation of motion in rectangular coordinates. For Newtonian fluids for constant and p for the x component, y component, and z component we obtain, respectively, p 1

.

dv x

+ l

Vr

dy

H

=

Vz

1"

dx

dt

2

'd

Vy

\~

dz

vx

2

+

2

d vx

dx'

By

d v x\

+

2

dz

dp

1

dx

+ 99 x (3.7-36)

I

3Vy

dVy

dVy

^Vy\

2

/3

= M

dz

uv

2

+

3

2

uv

3 w v^

d_p

f

dx'

dy'

dy

dz'

P9y (3.7-37)

'dv z

dv z

+

vv

:

dx

dt

3\ + 3

'3^

3w z

dv z

+

v,

3z

dx

i

2

dy

2

£

vz

dz

dp

)

2 j

~Tz

+ p9 < (3.7-38)

Combining the three equations for the three components, we obtain

—=

P

2.

- Vp + pg + M V-v These equations are and z components,

Equation of motion in cylindrical coordinates. fluids for constant p and p. for the r,

Newtonian

dv r

(dv r

v

dr

e

dv r

vB

dd

r

r

9,

dr

dr \r 1

dVg

dv 8

+ ,

Vr

+

dr

dt

d

+

\

\dt

vr

r

+ dr

3

1

dr

dr

r

Vg dv — r

r

dd

r

dO

d~v

(3.7-40)

P9r

(

+ P9o

39

(3.7-41)

dz'

dv

_d_P

dz

dz

d"v z

vz

2

+ P9 Z

2'

"dl)

p and U0

p.

for the

dv r

(3.7-42)

dz'

in spherical coordinates.

v]

The equations

r, 9,

and


+

for Newtonian fluids components, respectively.

dp_

+ dr

r sin 9 d(p

2

p\V 2 v

r

r

174

r

dp

+

'

dv r v.

39

dv r

.2L

l

e

dz 2

+

2

+ dz'

v,

dv.

3

Equation of motion

dt

+

L

dd

r

are given below for constant

+

z

V.z

d vg

1

p.

r

ldv r

r

+

vg dv —

dr

1

3.

dO

dr

h

+

d vr

+

dd

r~

dv

V rV g

2

2 dv e

dd'

h

d(rv e )\

dv z h

L

Vg dVg

p.

idv z

+

r

r

dr\r

P

2

d vr

1

+

respectively.

dr

J

'

l\ 3(rv r

d

as follows for

dp

dVr) ..dz

(3.7-39)

T 2

dv g

dd

Chap. 3

2 X Vg COt 9 2

r

2

dVfi

pg r

,

r

2

sin 6

(3.7-43)

dcp

Principles of Momentum Transfer

and Applications

(dvg

dvg

+

v.

+

v e dv e

dr

dt

v&

+

36

r

dv e

;

6 d
r sin

2 dv r

V z v 69 + -2? r

(

1-

vr

r

+ a V 2 vj,9

dd

|

\

where

in

1

V2 =

-j2 r

2

r

r sin d

—rS

o 2

sin

6

V

r

2

6

r

sin 6

d

r

dd

(3.7-44) '

J

\

=

•

dv 9

2 cos 6

dcf>

r

j

—

1

dp

+ HP9e9

-\ d

r

dv r

1

BvA

cot 6

r

— —2 i 2

sin

t 2

v 9v $

.

1

J

1

d

=

cos 6

7 2

vr

H

1

dr

\ dt

r

2

1.2 r sin 2

dd

COt d\

r

dv$

1-

Va,

v9

-

VgdVf

dv$

(dv& pi

v rv e

+

-',

1

2

2

sin

6

d

the three equations above. I —d [r —d \ + -y— 1

2

dr

\

dr)

Significant advantages

r

2

sin 6

9/ sin 6

dd \

and uses arise

—d\ + dd

in the

J

( d

1

r

= 2

,

sin

? 2 6>

2

W

\

(3.7-46)

j 2 /

transformation from rectangular coordi-

nates to cylindrical coordinates. For example, in Eq. (3.7-40) the term pv\lr

is

the

from the motion of the fluid in the 6 direction. Note that this term is obtained automatically from the transformation from rectangular to cylindrical coordinates. It does not have to be added to the equation on physical grounds. The Coriolis force pv r v e lr also arises automatically in the transformation of coordinates in Eq. (3.7-41). It is the effective force in the Q direction when there is flow in both the r and the 6 directions, such as in the case of flow near a rotating disk.

centrifugal force. This gives the force in the r direction (radial) resulting

3.8

3.8A

USE OF DIFFERENTIAL EQUATIONS OF CONTINUITY AND MOTION Introduction

The purpose and uses of

the differential equations of motion and continuity, as mentioned previously, are to apply these equations to any viscous-flow problem. For a given specific problem, the terms that are zero or near zero are simply discarded and the remaining equations used in the solution to solve for the velocity, density, and pressure distributions. Of course, it is necessary to know the initial conditions and the

boundary conditions to solve the equations. Several examples the general methods used.

We will consider cases for flow in

will

specific geometries that

be given

to illustrate

can easily be described

mathematically, such as for flow between parallel plates and in cylinders.

3.8B

Differential Equations of Continuity

and Motion

for

Flow between

Parallel

Plates

Two examples

will

be considered, one for horizontal plates and one for vertical plates.

EXAMPLE 3.8-1. Laminar Flow Between Horizontal Parallel Plates Derive the equation giving the velocity distribution at steady state for laminar flow of a constant-density fluid with constant viscosity which is flowing between two flat and parallel plates. The velocity profile desired is Sec. 3.8

Use of Differential Equations of Continuity and Motion

175

at a point far from the inlet or outlet of the channel. The two plates will be considered to be fixed and of infinite width, with the flow driven by the pressure gradient in the x direction.

Solution:

Assuming

that the channel

horizontal, Fig. 3.8-1

is

shows the

axes selected with flow in the x direction and the width in the z direction. The velocities vy and v z are then zero. The plates are a distance 2y 0 apart. The continuity equation (3.6-24) for constant density is dv x

+ ~T ox Since

v

y

and

dv v

dy

dv

+ ~Tz =

_

n

0

^ . „

(3.6-24)

dz

are zero, Eq. (3.6-24) becomes

v,

^=

(3.8-1)

0

dx

The Navier-Stokes equation

^

( dv x

+v

dv x

+v

*Tx

^ dv x

+v

for the x

component 2

(

dv x \

vx

is

2

1

d vx

^)=i^ w +

r+

d vx \

dp

^)-d-x +p9

*

(3.7-36)

Also, dvjdt

We d

2

=

0 for steady

2

=

0.

state, v

=

y

=

0, v x

0,

dvjdx

= 0,

d

vjdx 2

=

0.

no change of v x with z. Then Making these substitutions into Eq. (3.7-36), we obtain

can see that

v x /dz

dvjdz =

2

there

0, ,since

^ 2

dp

£-P9* = we

is

d ur

(3-8-2)

be concerned with gravitational force g x which is g, the gravitational force, in 2 m/s We shall combine the static pressure p and the gravitational force and call them simply p, as follows (note that g x = 0 for the present case of a horizontal pipe but is not zero for the general case of a nonhorizontal pipe): In fluid-flow problems

only

will

in the vertical direction for

,

.

p=p + pgh

FIGURE

176

3.8-1.

(3.8-3)

Flow between two parallel plates

Chap. 3

Principles

in

Example 3 .8-1

of Momentum Transfer and Applications

where h

is

the distance

upward from any chosen reference plane Then Eq. (3.8-2) becomes

(h is in the

direction opposed to gravity).

A

= T ox

We small, p

can see that p is

is

(3 - 8 " 4)

dy

not a function of

not a function of

(Some

y.

Also, assuming that 2y 0

z.

references avoid this

simply use p as a dynamic pressure, which

dynamic pressure gradients cause gradient

is

is

is

problem and

rigorously correct since

flow. In a fluid at rest the total pressure

and the dynamic pressure

the hydrostatic pressure gradient

dp/dx is a constant in this problem since v x is not a function of x. Then Eq. (3.8-4) becomes an ordinary differential equation.

gradient

is

zero.) Also,

dh^^ldp ~ dy

=

1 2

-

-jz fidx

=

const

(3.8-5)

Integrating Eq. (3.8-5) once using the condition for

dv x ldy = 0

at y

=

0

symmetry,

Integrating again using v x

=

dv x

_(ldp

dy

\fi

(3.8-6)

0 at y

=

dx

y0

,

d v

The maximum

*=h £

velocity

Eq.

in

Vxm "

Combining Eqs.

(3.8-7)

and

=

{y2

- yl)

(3.8-7)

1 dP 2y\dx

,

{

(3 - 8 - 7)

occurs

when

y

=

0, giving

~ yo) 2>

a8

"

8)

(3.8-8),

y

>

2

(3.8-9)

Jo Hence, a parabolic velocity in

profile

Eq. (2.9-9) when using a shell

The

results obtained in

force balance

on a

Example

differential

is

obtained. This result was also obtained

momentum

balance.

3.8-1 could also

have been obtained by making a

element of fluid and using the symmetry of the system to

omit certain terms.

EXAMPLE

3.8-2.

Laminar Flow Between

Vertical Plates with

One

Plate

Moving A Newtonian

fluid is confined between two parallel and vertical plates as shown in Fig. 3.8-2 (W6). The surface on the left is stationary and the other is moving vertically at a constant velocity v 0 Assuming that the flow is .

laminar, solve for the velocity profile.

Sec. 3.8

Use of Differential Equations of Continuity and Motion

177

Figure

3.8-2.

Flow between

vertical parallel plates in

The equation to use coordinate, Eq. (3.7-37).

Solution:

idVy

dv y

dv y

2

2

ld v y

d

is

Example

3.8-2.

the Navier-Stokes equation for the y

dv y \

2

vy

d

dp

v y\

state, dv /dt = 0. The velocities v x and v z = 0. Also, y 0 from the continuity equation, dv y /.dz = 0, and pg y = — pg. partial derivatives become derivatives and Eq. (3.7-37) becomes

At steady

=

dvyl dy

The

d zv y

dp

—--r ~P9 = P— ax" ay This

is

dpldy

Example

similar to Eq. (3.8-2) in

is

(3.8-10)

0

3.8-1.

The pressure

gradient

constant. Integrating Eq. (3.8-10) once yields

dv v

x

dp — +pg \dy I

'

p

dx

\

=C,

(3.8-11)

Integrating again gives 2

vy

dp x -— — + pg] = C x + C I

Y

\

2p \fy

(3.8-12)

2

conditions are at x = 0, v y = 0 and at x = H, v y Solving for the constants, C| = v 0 /H - (H/2p)(dp/dy + pg) and 0. Hence, Eq. (3.8-12) becomes

The boundary

1

v,=

178

(dp

Chap. 3

v0

C2

.

=

x

\

-^[^ + P9]( Hx - x

=

)

+ wo

-

Principles of Momentum Transfer

(3.8-13)

and Applications

3.8C

Differential Equations of Continuity

and Motion for Flow

in

Stationary and

Rotating Cylinders

Several examples will be given for flow in stationary and rotating cylinders.

EXAMPLE 3.8-3.

Laminar Flow

in a Circular

Tube

Derive the equation for steady-state viscous flow in a horizontal tube of radius r 0 , where the fluid is far from the tube inlet. The fluid is incompressible and fi is a constant. The flow is driven in one direction by a constantpressure gradient. Solution:

shown

The

fluid will be

assumed

The y

to flow in the z direction in the tube, as

is vertical and the x direction horand v y are zero, the continuity equation becomes dvjdz = 0. For steady state dv z /dt = 0. Then substituting into Eq. (3.7-38) for the z component, we obtain

in Fig. 3.8-3.

direction

izontal. Since v x

dp

To

solve Eq. (3.8-14)

2

fd

we

v,

+

d

2

v\ (3 - 8 - 14)

can use cylindrical coordinates from Eq.

(3.6-26), giving

z=z

=r

x

cos 9

=

y

r sin,

0 (3.6-26)

r= + Jx 2 +

2

0

y

=

tan

_1

y X

Substituting these into Eq. (3.8-14), d ji

dz

2

v.

dr

^

2

~Idv,

2

d v,

1

+

30

or

r

(3.8-15)

2

The flow is symmetrical about the z axis so d 2 v z /3 0 2 is zero (3.8-15). As before, dp/dz is a constant, so Eq. (3.8-15) becomes

-1 dp = -f H dz

2

const

=

d — -f v.

dr

\

[dv£ _\i r

dr

*~

r

d_

dr

\

dv, -rdr

z

Sec. 3.8

3.8-3. Horizontal flow in a tube in

Example

Use of Differential Equations of Continuity and Motion

Eq.

(3.8-16)

of fluid

Figure

in

3.8-3.

Alternatively, Eq. (3.7-42) in cylindrical coordinates can be used for

component and the terms

the z

dv z

Idv,

dr

dt

I

V dv — e

+ dr

dr

V

+

dv z

dp

dz

dz

v,

Z

1

7

p.

r

z

36

r

dv

a

1

+

+

that are zero discarded.

r

2

d v7

d8

2

d

2

v,

+ dz'

+ pg z

(3.7-42)

l As before, dvjdt = 0, d l v z ld8 = 0, v r = 0, dvjdd = 0, at/ z /3z = 0. Then Eq. (3.7-42) becomes identical to Eq. (3.8-16). The boundary conditions for the first integration are dvjdr = 0 at r = 0. For the second integration, v z = 0 at r = r 0 (tube radius). The result is

(3.8-17)

Converting to the

maximum

velocity as before,

= Eq. (3.8-17)

If

(3.8-18)

v.

integrated over the pipe cross section using Eq.

is

(2.9-10) to give the average velocity v zav

,

~fj

0.8-19)

z

Integrating to obtain the pressure drop from z P = Pi we obtain

=

0 for p

=

p l to

z

=L

for

'

Pi

where

D =

2r 0

.

This

is

-

H.. 1

32nv zav L

-

P2

g20)

2

r '0

the Hagen-Poiseuille

equation derived previously as

Eq. (2.9-11).

EXAMPLE

Laminar Flow

3.8-4.

in

a Cylindrical Annulus

Derive the equation for steady-state laminar flow inside the annulus between two concentric horizontal pipes. This type of flow occurs often in concentric pipe heat exchangers. In this case Eq. (3.8-16) also still holds. However, the velocity annulus will reach a maximum at some radius r = r max which is between r and r 2 as shown in Fig. 3.8-4. For the first integration of Eq. (3.8-16), the boundary conditions are dv z ldr = 0 at r = r max which gives

Solution: in the

,

{

,

velocity profile

FIGURE

3.8-4.

Flow through a

Chap. 3

Principles

cylindrical annulus.

of Momentum Transfer and Applications

(3.8-21)

\ndz)\2

2

J

Also, for the second integration of Eq. (3.8-21), v z

=

dr

r

Oat the inner wall where

= r u giving Vz

\2fidz)\2

"3

2

Repeating the second integration but for r = r 2 we obtain

vz

=

(3.8-22)

0 at the outer wall where

,

1

K 2p.

Combining Eqs.

(3.8-22)

dp\(r 2

and

i

(3.8-23)

~

r2 -" ,n

~2

dzj\i

r2

(3.8-23) .

and solving for r max

1

(3.8-24)

ir\-r\)/2 In (r 2 /ri

)

In Fig. 3.8-4 the velocity profile predicted

For the case where

r

by Eq.

(3.8-23)

is

= x

0, r m3X in Eq. (3.8-24) becomes zero (3.8-17) for a single circular pipe.

(3.8-23) reduces to

Eq.

EXAMPLE 3.8-5.

Velocity Distribution for

plotted.

and Eq.

Flow Between Coaxial Cylinders

Tangential laminar flow of a Newtonian fluid with constant density is occurring between two vertical coaxial cylinders in which the outer one is rotating (S4) with an angular velocity of w as shown in Fig. 3.8-5. It can be assumed that end effects can be neglected. Determine the velocity and the shear stress distributions for this flow.

On physical grounds the fluid moves in a circular motion and the velocity v r in the radial direction is zero and v z in the axial direction is zero. Also, dp/dt = 0 at steady state. There is no pressure gradient in the 0 direction. The equation of continuity in cylindrical coordinates as

Solution:

derived before

is

dp — + dt

1

r

d{prv r )

1

+

dr

r

d(pv g ) d9

+

d(pv z )

=0

(3.6-27)

dz

outside cylinder rotates

inside cylinder fixed

Figure

Sec. 3.8

3.8-5.

Laminar flow Example 3.8-5.

in

the region between two coaxial cylinders in

Use of Differential Equations of Continuity and Motion

181

All terms in this equation are zero.

The equations of motion (3.7-41),

in cylindrical coordinates,

Eqs. (3.7-40),

and (3.7-42) cedtMe tojrje following, respectively: 2

— P

dp (r-componenf)

:

(3.8-25)

dr

r

_d_(\ ~ dr

d(rvg) \

(^-component) dr

\r

(3.8-26)

)

dp

0=

(z-component)

pg z

1-

dz

(3.8-27)

Integrating Eq. (3.8-26),

ve

= C

x

+

r

(3.8-28)

solve for the integration constants C x and conditions are used: at r = R\, v e = 0; at r equation is

To

ve

C2

,

the following boundary v e = coR 2 The final

= R2

*1

= 2

(R]

.

,

r

(3.8-29)

- Rl)l{R Rl)

r

x

Using the shear-stress component for Newtonian

fluids

in

cylindrical

coordinates,

d(v e /r) r

dr

The

last term in Eq. (3.7-31) and differentiating gives

is

1

dv r

r

50

+

(3.7-31)

zero. Substituting Eq. (3.8-29) into (3.7-31)

R}IR\ TrO

=

-2fJ.O)R 2

\

(?)

i

The torque T

that is necessary to rotate the outer cylinder of the force times the lever arm.

T=

(3.8-30)

- r]ir\ is

the product

(2irR 2 H)(-T re )\ r = R2 (R 2 )

r}ir\ 4ir ij.HcoR 2 i

- r]ir\

(3.8-31)

H

where is the length of the cylinder. This type of device has been used to measure fluid viscosities from observations of angular velocities and torque and also has been used as a model for some friction bearings.

Chap. 3

Principles of Momentum Transfer

and Applications

EXAMPLE 3.8-6.

Rotating Liquid in a Cylindrical Container

A Newtonian fluid of constant density (B2).

At steady

Solution:

Example

state find the

is

of radius

7?

axis at angular velocity

co

in a vertical cylinder

about

(Fig. 3.8-6) with the cylinder rotating

its

shape of the free surface.

The system can be described in 3.8-5, at steady state, v r = v z =

As in The final

cylindrical coordinates.

0 and

gr = g g =

0.

equations in cylindrical coordinates given below are the same as Eqs. (3.8-25) to (3.8-27) for Example 3.8-5 except that g z = -#inEq. (3.8-27).

dp

v]

P— =— dr

(3.8-32)

r

0

a

/

d(rv 0 )\

1

C,-r-— dr

=

dr \r

•

(3.8-33) J

dp — =~P9

(3.8-34)

dz

Integration of Eq. (3.8-33) gives the

vg

= C

same equation

as in

Example

3.8-5.

C2

1

r+—

(3.8-28)

r

The constant C 2 must be zero since v g cannot be infinite the velocity v g = Ru. Hence, C, = w and we obtain

at r

uj=ur Combining Eqs.

(3.8-35)

0

.

At r = R

,

(3.8-35)

and (3.8-32)

dp — = dr

pu>

2

(3.8-36)

r

R-*4

p = p0

at surface

P=P(r,z)

<4> Figure

3.8-6.

Liquid being rotated

Example

Sec. 3.8

in

a container with a free surface

in

3.8-6.

Use of Differential Equations of Continuity and Motion

183

Hence, we see that Eqs. upon r because of the

(3.8-36)

and

(3.8-34)

centrifugal force

show

depends because of the

that pressure

and upon

z

gravitational force.

dp — =-pg

Since the term p of pressure as

is

a function of position

dp =

Combining Eqs.

(3.8-34)

and

(3.8-34)

we can write

the total differential

dp dp — dr + — dz dr dz

(3.8-37)

(3.8-36) with (3.8-37)

—

and integrating,

poj r

P =

P9z+€

(3.8-38)

3

2

The constant of integration can be determined z = z Q The equation becomes

since

p = pa

at r

— 0 and

.

P-Po The

free surface consists of

=

all

^J-+ pgUo-z) points onihis surface

(3.8-39)

at/?=p 0 Hence, .

(3.8-40)

This shows that the free surface

3.9

3.9A

is in

the shape of a parabola.

OTHER METHODS FOR SOLUTION OF DIFFERENTIAL EQUATIONS OF MOTION Introduction

In Section 3.8

we considered examples where

the Navier-Stokes differential equations

of motion could be solved analytically. These cases were used where there was only

one nonvanishing component of the velocity. To solve these equations for flows in two and three directions is quite complex. In this section we will consider some approximations that simplify the differential equations to allow us to obtain analytical solutions. Terms will be omitted which are quite small compared to the' terms retained. Three cases will be considered in this section: inviscid flow, potential flow, and creeping flow. The fourth case, for boundary-layer flow, will be considered in Section 3.10. The solution of these equations may be simplified by using a stream function and/or a velocity potential
,

184

.

Chap. 3

Principles of Momentum Transfer

and Applications

3.9B

Stream Function

The stream function

convenient parameter by which

ip(x, y) is a

we can

two-dimensional, steady, incompressible flow. This stream function, related to the velocity

tp

represent

m 2 /s,

in

is

components v x and v y by

— dtp

=

dtp

wy

dy

=

-— dx

(3-9-1)

differential equation of

in the x and y components of the motion, Eqs. (3.7-36) and (3.7-37), with v z = 0 to obtain a

differential equation for

tp

These

definitions of

v x and v y can then be used that

is

equivalent to the Navier-Stokes equation. Details are

given elsewhere (B2).

The stream

function

is

steady flow lines defined by

very useful because its physical significance is that in = constant are streamlines which are the actual curves

ip

traced out by the particles of

fluid.

A

stream function exists for

all

two-dimensional,

steady, incompressible flow whether viscous or inviscid and whether rotational or irrotational.

EXAMPLE 3,9-1.

Stream Function and Streamlines

The stream function the

components of

4 and

tp

Solution:

=

relationship

is

given as

ip

= xy Find the equations .

velocity. Also plot the streamlines for

a constant

for

ip

=

1.

Using Eq.

(3.9-1),

_

djp

djxy)

_

-

Vy

d

jL-

ip

for

=

1

i//

d{xy)

~

dx

dx

To determine the streamline y = 0.5 and solve for x.

=

dy

dy

= constant =

= xy =

y

1

=

xy, assume that

x(0.5)

Hence, x = 2. Repeating, for y = 1 x = 1; for y = 2, x = 0.5; for y = 5, x = 0.2, etc. Doing the same for tp = constant = 4, the streamlines for tp = and tp = 4 are plotted in Fig. 3.9-1. A possible flow model is flow around a corner. ,

1

3.9C

Differential Equations of

Motion

for Ideal Fluids (Inviscid

Flow)

Special equations for ideal or inviscid fluids can be obtained for a fluid having a constant density and zero viscosity. These are called the Euler equations. Equations

(3.7-36M3.7-39) for the x, y, and z components of

'

Sec. 3.9

momentum become

dVx

dv x \

dp

dy

dz

dx

Other Methods for Solution of Differential Equations of Motion

185

0

1

0

i

i

i

1

2

3

i

i

4

5

x Figure

Plot of streamlines for

3.9-1.

dVy

BVy

(dv z

17

+

dVy

+V

xy for Example

9 Vy\

^ ^ dv z

V

=

if/

dv z

+

dv t \ V

^)

3.9-1.

3/7

dp

= -J- + Z

P

^

(3 - 9 " 4)

At very high Reynolds numbers the viscous forces are quite small compared to the and the viscosity can be assumed as zero. These equations are useful in calculating pressure distribution at the outer edge of the thin boundary layer in flow past immersed bodies. Away from the surface outside the boundary layer this assumption of an ideal fluid is often valid. inertia forces

3.9D

The

Potential

Flow and Velocity Potential

velocity potential or potential function cf>(x,y) in

problems and

is

m 2 /s

is

useful in inviscid flow

defined as

— d<£

vx

=

vy

dx

d
=

vz

~r~ dy

=

— 34>

(3.9-5)

dz

This potential exists only for a flow with zero angular velocity, or irrotationality. This type of flow of an.ideal or inviscid fluid (p = constant, \l = 0) is called potential flow. Additionally, the velocity potential



exists for three-dimensional flows,

whereas the

stream function does not.

The

vorticity of a fluid is defined as follows:

— df v

dv x

dx

dy

L

=2 &

(3.9-6)

>

l

or,

— +— dx

186

Chap. 3

=r 1

dy

T 1

=-2u

7 z

Principles of Momentum Transfer

(3.9-7)

and Applications

where

=

2a>z

is

0, the flow

is

Using Eq.

_1

is angular velocity about the z axis. If 2« z and a potential function exists. the conservation of mass equation for flows in the x and the y

the vorticity and

co

in s

l

irrotational

(3.6-24),

direction is as follows for constant density:

—

dv

<3t)'

dx

Differentiating v x in

Eq.

+

-^v = 0

(3.9-8)

dy

(3.9-5) with respect to

x and v y with respect

y and

to

substituting into Eq. (3.9-8), 2

d

d

4>

2 4>

-T+T-T=0 dx dy This

is

(3-9-9)

Laplace's equation in rectangular coordinates.

boundary conditions

If suitable

or are known, Eq. (3.9-9) can be solved to give
using numerical analysis, conformal mapping, and functions of a

complex variable and

are given elsewhere (B2, S3). Euler's equations can then be used to find the pressure distribution.

When

the flow

is

inviscid

and

irrotational a similar type of

Laplace equation

is

obtained from Eq. (3.9-7) for the stream function. 2

2

—j+ —7=0 d

ijj

dx

1

d


dy

(3.9-10)

1

Lines of constant 4> are called equal potential lines and for potential flow are everywhere perpendicular (orthogonal) to lines- of constant ip. This can be proved as follows. A line of constant ip would be such that the change in ip is zero.

dtp

Then, substituting Eq.

=

— dx + — dtp

dip

dx

dy

dy

=

0

(3.9-11)

above,

(3.9-1) into the

(3.9-12) ip

Also, for lines of constant

constant

x


—

=

dx



— d(p

d<}>

d
dx +

dy = 0

(3.9-13)

11

(3.9-14)

dy

= constant

Hence, 1

(3.9-15) 4,

Sec. 3.9

= constant

(dy/dx) 0 =

constant

Other Methods for Solution of Differential Equations of Motion

187

An example

of the use of the stream function

is in

obtaining the flow pattern for

The

inviscid, irrotational flow past a cylinder of infinite length.

fluid

approaching the

cylinder has a steady and uniform velocity of u„ in the x direction. Laplace's Equation (3.9-10)

can be converted to cylindrical coordinates 2

d

+ dr

2

dip'

1


d

1

ip

+ dr

r

to give

=0

(3.9-16)

-—

(3.9-17)

dO

r

where the velocity components are given by 1

dip

dtl>-

"r=-— r 30

*>,=

dr

Using four boundary conditions which are needed and the method of separation of variables, the stream function

where

R

is

ip is

the cylinder radius.

lines are plotted in Fig. 3.9-2 as

EXAMPLE 3 .9-2 The

velocity

that

it

The

streamlines and the constant velocity potential

a flow net.

Stream Function for a Flow Field

.

components for a flow

=

vx

Prove

obtained. Converting to rectangular coordinates

satisfies

Solution: First

we

a(x

2

—

field

2

y

)

are as follows: vy

= -2axy

the conservation of mass and determine

determine dv x ldx

ip.

= lax and dv y l dy = -lax.

Substituting these values into Eq. (3.6-24), the conservation of

mass for

two-dimensional flow,

dv x

dv y V

dx

=0

or

lax - lax = 0

dy

9

Figure

3.9-2.

{ip — constant) and constant velocity potential lines constant) for the steady and irrotational flow of an inviscid incompressible fluid about an infinite circular cylinder.

Streamlines {4>

=

and

188

= constant

Chap. 3

Principles

of Momentum Transfer and Applications

Then using Eq.

(3.9-1),

dip

dip

= ax 2 — ay 2

=--=-2axy

vy

~dy

(3.9-19)

Integrating Eq. (3.9-19) for v x

= ax 2y

ip

Differentiating

Eq.

(3.9-20)

(3.9-20) with respect to

x and equating

it

to Eq. (3.9-19),

— = laxy - 0+f(x)= +2axy dip

Hence, f(x)=0 and f(x) = C, a constant. Then Eq.


To plot the

= ax 2

(3.9-21)

becomes

(3.9-20)

y-—+C

stream function, the constant

C can be

(3.9-22)

set

equal to zero before

plotting.

In potential flow, the stream function and the potential function are used to

represent the flow

in

the main

satisfy the condition that v x

viscous drag and

we

body of the

= v y — 0 on

fluid.

is

ideal fluid solutions

use boundary-layer theory where

solutions for the velocity profiles in this thin viscosity. This

These

the wall surface.

discussed

in

Section 3.10.

.

Near

we

the wall

have

obtain approximate

boundary layer taking

Then we

do not

we

into account

splice this solution onto the ideal

flow solution that describes flow outside the boundary layer.

Differential Equations of

3.9E

Motion for Creeping Flow

At very low Reynolds numbers below about particles settling

1,

the term creeping flow

very low velocities. This type of flow applies for the

flow at

through a

fluid.

Stokes' law

is

fall

is

used to describe

or settling of small

derived using this type of flow

in

problems of

and sedimentation.

around a sphere, for example, the fluid changes velocity and direction in a complex manner. If the inertia effects in this case were important, it would be necessary to keep all the terms in the three Navier-Stokes equations. Experiments show that at a Reynolds number below about 1, the inertia effects are small and can be omitted. Hence, In flow

the equations of motion, Eqs. (3.7-36)-(3.7-39) for creeping flow of an incompressible fluid,

become (3.9-23)

(3.9-24)

(3.9-25)

Sec. 3.9

Other Methods for Solution of Differentia! Equations of Motion

189

For flow past a sphere the stream function

\f/

can be used

Navier-Stokes

in the

equation in spherical coordinates to obtain the equation for the stream function and the velocity distribution and the pressure distribution over the sphere. Then by integration

over the whole sphere, the form drag, caused by the pressure distribution, and the skin friction or viscous drag, caused by the shear stress at the surface, can be summed to give the total drag.

F D = 3nnD p v 3nnD — —

(SI)

(3.9-26) v

p <=-

FD =

(English)

:

9c

where F D is total drag force in N, D p is particle diameter in m, fluid approaching the sphere in m/s, and }i is viscosity in kg/m

v s.

free

is

This

stream velocity of

is

Stokes' equation

drag force on a sphere. Often Eq. (3.9-26) is rewritten as follows:

for the

V

- PA

F o = CD

(SI)

(3.9-27) v

= C D —- pA

Fd where

CD

a drag coefficient, which

is

projected area of the sphere, which

is

3.1 for

flow past spheres.

3.10

BOUNDARY-LAYER FLOW AND TURBULENCE

3.10A

is

(English)

equal to 24/N Re for Stokes' law, and

nD p /4.

This

is

discussed in

more

A

is

the

detail in Section

Boundary-Layer Flow

In Sections 3.8

and 3.9 the Navier-Stokes equations were used to find relations that flat plates and inside circular tubes, flow of ideal fluids,

described laminar flow between

and creeping flow. in

more

In this section the flow of fluids

detail, with particular attention

around objects

be considered

will

being given to the region close to the solid

surface, called the boundary layer. In the boundary-layer region near the solid, the fluid this solid surface.

In the bulk of the fluid

away from

the

motion is greatly affected by boundary layer the flow can

often be adequately described by the theory of ideal fluids with zero viscosity. in

the thin

boundary

layer, viscosity

important. Since the region

is

is

However,

thin, simplified

solutions can be obtained for the boundary-layer region. Prandtl originally suggested this division

of the problem into two parts, which has been used extensively in fluid

dynamics. In order to help explain in

boundary

the steady-state flow of a fluid past a

layers,

flat

an example of boundary-layer formation

plate is given in Fig. 3.10-1.

The

velocity of the

upstream of the leading edge at x = 0 of the plate is uniform across the entire fluid stream and has the value y x The velocity of the fluid at the interface is zero and the fluid

.

velocity u x in the x direction increases as

one goes farther from the

plate.

The velocity^

approaches asymptotically the velocity v m of the bulk of the stream.

The dashed line L The layer

velocity v^.

boundary

190

layer.

When

is

drawn so

that the velocity at that point

is

99%

of the bulk

or zone between the plate and the dashed line constitutes the

the flow

is

laminar, the thickness 5 of the boundary layer increases

Chap. 3

Principles

of Momentum Transfer and Applications

i;

turbulent

boundary layer viscous sublayer

x - q

laminar

boundary layer FIGURE

Nrc.

=

x

xv m p/n, where x

Reynolds number

is

in the

transition zone

x

plate.

^direction. The Reynolds number

the distance

less than 2

is

L

*

Boundary layer for flow past a flat

3.10-1.

move

with the yjx as we

—

10

5

downstream from

the flow

is

turbulent, a thin viscous sublayer persists next to the plate.

boundary

the viscous shear in the

present for flow past a

The type

is

the

called skin friction

When

the

boundary

The drag caused by

and

it

is

the only drag

flat plate.

when fluid flows by a bluff or blunt shape such as a mostly caused by a pressure difference, is termed form drag. flow past such objects at all except low values of the Reynolds

of drag occurring

sphere or cylinder, which

is

This drag predominates in

numbers, and often a wake past a bluff shape,

is

and the

present. Skin friction

total

drag

is

the

sum

and form drag both occur of the skin friction

in flow

and the form

(See also Section 3.1 A).

drag.'

3.10B

Boundary-Layer Separation and Formation of

Wakes

We discussed Fig. 3.10-2.

plate

layers

When

shown in Fig. 3.10-1. smooth plate occurs in the

,

is

denned as

laminar, as

The transition from laminar to turbulent flow on a 5 6 Reynolds number range 2 x 10 to3 x 10 as shown in Fig. 3.10-1. layer

is

the leading edge.

the

growth of the boundary layer

at the leading

However, some important phenomena

and other

objects.

At the

trailing

layers

Figure

edge or rear edge of the

and bottom sides of the gradually intermingle and disappear.

Sec. 3.10

3.10-2.

Flow perpendicular

to a flat plate

Boundary-Layer Flow and Turbulence

in

also occur at the trailing edge of this

layers are present at the top

boundary

edge of a plate as shown

plate.

fiat plate, the

On

boundary

leaving the plate, the

and boundary-layer

separation.

191

If

the direction of flow

at the

edge of the

plate,

momentum

however, the

is

in Fig. 3.10-2, a

flowing over the upstream face.

in the fluid that is

Once

from making the separates from the plate. A zone of

in the fluid prevents

abrupt turn around the edge of the plate, and decelerated fluid

shown

right angles to the plate as

is at

boundary layer forms as before

it

it

present behind the plate, and large eddies (vortices), called the wake,

are formed in this area.

The eddies consume large amounts of mechanical energy. This when the change in velocity of the fluid flowing by

separation of boundary layers occurs

an object

is

too large in direction or magnitude for the fluid to adhere to the surface.

wake causes

Since formation of a

large losses in mechanical energy,

is

it

often

necessary to minimize or prevent boundary-layer separation by streamlining the objects

or by other means. This

3. 10C

1.

is

also discussed in Section 3.1 A for flow past

objects.

Laminar Flow and Boundary-Layer Theory

When

Boundary-layer equations.

lamirfar flow

thickness of the boundary layer

99%

surface where the velocity reaches relatively thin

boundary layer leads

and can be neglected. away from the stream velocity. The concept of a negligible

arbitrarily taken as the distance

is

<5

occurring in a boundary layer,

is

become

certain terms in the Navier-Stokes equations

The

immersed

of the free

some important

to

simplifications of the Navier-

Stokes equations.

For two-dimensional laminar flow in the x and y directions of a fluid having a constant density, Eqs. (3.7-36) and (3.7-37) become as follows for flow at steady state as

shown

in

Figure 3.10-1 when dv x

+

dx

*T:

The continuity equation

neglect the body forces g x and g y

^-dv x

V

dv

dv v v

we

+

v

I

=

>ty

dp

p

Tx

I

dp

--plTy

+ +

p.

2

I

2

d v

vx

+

A'^ '¥ 2

n (

v

-pW

.

^

+

]

(3 -

]

i0 ~ 2)

two-dimensional flow becomes

for

dv x

dv v

ox

-fdy

^+ In Eq. (3.10-1), the term pi p(d

2

=

0

(3.10-3)

2

is negligible in comparison with the other ) can be shown that all the terms containing v y and its derivatives are small. Hence, the final two boundary-layer equations to be solved are

terms

in the

equation. Also,

v x /dx

it

Eqs. (3.10-3) and (3.10-4). v

dv x —. +

2

dx 2.

dv dp u d —-=-+ -— — 1

x

v

y

p dx

dy

vx

(3.10-4)

p by

An

Solution for laminar boundary layer on a flat plate.

important case

analytical solution has been obtained for the boundary-layer equations

boundary layer on a flat simplification can be made

plate in steady flow, as

The

final

(3.10-4) in that

in Eq. boundary-layer equations reduce

shown

dpldx

to

is

in Fig.

is

which an laminar

A

further

3.10-1.

zero since v„

the equation of

in

for the

is

motion

constant. for the

x

direction and the continuity equation as follows:

Sv x

2

v

dx dv r

192

Chap. 3

y

v

(3.10-5)

p dy

dy

<3i>„

5

-r dx

dv d 1 = p —f x

+

+

-r 1

=

0

(3.10-3)

dy

Principles of Momentum Transfer

and Applications

The boundar^onditions at.y =" 00/

The

solution of this

function of x and

B2, S3).

are

=

vx

v

=

y

0

y

at

=

0

(y

is

distance from plate), and

vx

^

=

•

y was

for laminar flow over a flat plate giving v x a.ndv T as a obtained by Blasius and later elaborated by Howarth (Bl,

problem first

The mathematical details of the solution are The general procedure will be

not be given here.

quite tedious

and complex and

outlined. Blasius reduced the

equations to a single ordinary differential equation which

is

nonlinear.

will

two

The equation

could not be solved to give a closed form but a series solution was obtained.

The

results of the

thickness

<5,

where

vx

=

work by

0.99v x

Blasius are given as follows.

5= 42^= = where

W Rc x =

The drag

The boundary-layer

given approximately by

is

,

JZ-

5.0

(3-10-6)

xv^p/p. Hence, the thickness 5 varies as N/x. flow past a fiat plate consists only of skin friction and

in

=

the shear stress at the surface at y

0 for any

x as

is

calculated from

follows.

d

:\

•

From

the relation of v x as a function of

(3.10-7)

(3.10-7)

x and y obtained from the

series solution,

Eq.

becomes t0

The

^j

3y

total

drag

is

&

= 0.332^

V

(3-10-8)

fx

given by the following for a plate of length

FD =

b

t0

L and width

b

dx

(3.10-9)

Substituting Eq. (3.10-8) into (3.10-9) and integrating,

FD = The drag coefficient C D A = bL is defined as

0.664b Jupvl,

related to the total

L

(3.10-10)

drag on one side of the plate having an area

Fd=C d ^ P A Substituting the value for

A and

Eq. (3.10-10) into (3.10-11),

CD =

1.328

M^ = -^ N^

V Lv^p

N Rc L = Lv^p/p.. A form of Eq. (3.10-11) movement through a fluid. The definition of C D

where

Fanning

friction factor

The equation less

than about

5

/ for .

(3-10-12)

L

is

used

in

in

Section 14.3 for particle

Eq. (3.10-12)

is

similar to the

pipes.

N

C D applies only to the laminar boundary layer for Rc L Also, the results are valid only for positions where x is

derived for

x 10 5

(3.10-11)

from the leading edge so that x or L is much greater than 5. Experimental on the drag coefficient to a flat plate confirm the validity of Eq. (3.10-12).

sufficiently far

results

Boundary-layer flow past

many

other shapes has been successfully analyzed using similar

methods.

Sec. 3.10

Boundary-Layer Flow and Turbulence

193

3.10D

1.

Nature and Intensity of Turbulence

Nature of turbulence.

Since turbulent flow

is

important

the nature of turbulence has been extensively investigated.

in

many areas

of engineering,

Measurements of

the velocity

have helped explain turbulence. For turbulent flow there are no exact solutions of flow problems as there are in laminar flow, since the approximate equations used depend on many assumptions. However, useful relations have been obtained by using a combination of experimental data and theory. Some of these relations will be discussed. Turbulence can be generated by contact of two layers of fluid moving at different velocities or by a flowing stream in contact with a solid boundary, such as a wall or sphere. When a jet of fluid from an orifice flows into a mass of fluid, turbulence can arise. In turbulent flow at a given place and time large eddies are continually being formed which break down into smaller eddies and which finally disappear. Eddies are as small as about 0.1 or 1 mm or so and as large as the smallest dimension of the turbulent stream. Flow inside an eddy is laminar because of its large size. In turbulent flow the velocity is fluctuating in all directions. In Fig. 3.10-3 a typical

fluctuations of the eddies in turbulent flow

plot of the variation of the instantaneous velocity v x in the x direction at a given point in

turbulent flow

is

shown. The velocity

is

v'x

the deviation of the velocity from the

mean

velocity v x in the x-direction of flow of the stream. Similar relations also hold for the

and

y

z directions.

=

v*

+

v

v.

y

=

vf

+

v' y

,

v,

=

vz

+

v'z

(3.10-13)

(3.10-14)

v

where the mean velocity total velocity in the

v x is

the time-averaged velocity for time

x direction, and

v'

x

t, v x the instantaneous the instantaneous deviating or fluctuating velocity

x direction. These fluctuations can also occur in the y and z directions. The value of v'x fluctuates about zero as an average and, hence, the time-averaged values v x = 0, 2 v = 0, v' = 0. However, the values ofv'xl.v'y, and v' will not be zero. Similar expressions z z y in the

can also be written

for pressure,

which also

fluctuates.

Intensity of turbulence. The time average of the fluctuating components vanishes over a time period of a few seconds. However, the time average of the mean square of the

2.

fluctuating

components

have been analyzed by

is

a positive value. Since the fluctuations are

statistical

methods. The

0

FIGURE

194

level or intensity of

random,

the data

turbulence can be

Time 3.10-3.

Velocity fluctuations in turbulent flow.

Chap. J

Principles of

Momentum

Transfer and Applications

sum

related to the square root of the

This intensity of turbulence

boundary

The

mean squares of the

of the

an important parameter

is

fluctuating components.

in testing

of models and theory of

layers.

intensity of turbulence /

can be denned mathematically as

This parameter

quite important.

/ is

? + ff)

=
/

Such

(3 10 -15) .

boundary-layer transition, separ-

factors as

depend upon the intensity of turbulence. Simulation of turbulent flows in testing of models requires that the Reynolds number and the intensity of turbulence be the same. One method used to measure intensity of turbulence is to utilize a hot-wire anemometer.

and heat- and mass-transfer

ation,

coefficients

Turbulent Shear or Reynolds, Stresses

3.10E

In a fluid flowing in turbulent flow shear forces occur wherever., there

much

gradient across a shear plane and these are

flow. The velocity fluctuations in Eq.' (3. 10-13) give The equations of motion and the continuity equation

For an incompressible

fluid

is

a velocity

larger than those occurring in laminar rise to turbulent

are

shear stresses.

valid for turbulent flow.

still

having a constant density p and viscosity

p.,

the continuity

equation (3.6-24) holds. dv r

2

-T

follows

if

component of

.

S{pv x v x )

+

vx

+

0

(3.6-24)

the equation of motion, Eq. (3.7-36), can be written as

——z.

d{pv x tQ 3(pv x u : )

dy

dx

at

We

=

dz

Eq. (3.6-24) holds:

—— + °{pv x )

+

dy

ox Also, the x

dv.

5i>„

+

2

(

=

^

\dx

oz

2

vx

+

z

d vx

2

+

dy

,v by

v'

x

v

y

+ Q]

y

+

v' y

v,

,

dlp(v x

by

+

+

v.

v,

and p by p

,

±

o+^±

dx

dy

v'^v x

+

+

dx

J

(3.10-16)

pg

p'

+

w±v=

+

v'x )(v

y

,

v,

W

we use the are zero),

+

fact that the

y

+

v' )l y

»/,)]

_

,

+

dx

d(pv x v x )

dt

dx

d{pv x

v

d(pv x

t)

dy

+

Sec. 3.10

^E±n + pgx

(3.10-18)

dx

v'x v'

y

is

not zero.

is

zero

Then Eqs.

become

^+^+^= +

_

time-averaged value of the fluctuating velocities

and that the time-averaged product

(3.10-17) and (3.10-18)

d(pv x )

17)

dy

dz

Now

.

{

dx

+

v

ai0

0

dz

°\_p(v x

»*)3

]

at

,

£

dz

can rewrite the continuity equation (3.6-24) and Eq. (3.10-16) by replacing v x by

oLp(v x

(v x

dp __ +

3 v x\

dy

(3.10-19)

0

dz

u~)

dz d(pv'x v x )

dx

,

d(pv'x

v)

d{pv'x

dy

Boundary-Layer Flow and Turbulence

dz

v'z )

pS 2 v x -^- + pg x

(3.10-20)

ox

195

By comparing these two time-smoothed we

equations with Eqs. (3.6-24) and (3.10-16)

see that the time-smoothed values everywhere replace the instantaneous values.

However,

in

Eq. (3.10-20) new terms arise in the set of brackets which are related to we use the notation

turbulent velocity fluctuations. For convenience

f^ =

?« = P«4«£

K^,

i~ = pvW:

momentum

These are the components of the turbulent

(3.10-21)

and are

flux

called Reynolds

stresses.

Prandtl Mixing Length

3.10F

The equations derived for turbulent flow must be solved to obtain velocity profiles. To do this, more simplifications must be made before the expressions for the Reynolds stresses can be evaluated. A number of semiempirical equations have been used and the eddy diffusivity model of Boussinesq is one early attempt to evaluate these stresses. By analogy to the equation

for shear stress in

= — p(dv x fdy),

laminar flow,i p

the turbulent

shear stress can be written as

f

where

n,

a turbulent or eddy viscosity, which

is

r}

U= -

t

(3.10-22)

-3T dy is

a strong function of position and flow.

This equation can also be written as follows

=-/>£,— where

s,

=

n,/p and

e,

is

eddy

(3.10-23)

momentum

diffusivity of

m

in

2

/s

by analogy to the

momentum diffusivity Prandtl stresses

in

p/p for laminar flow. mixing-length model developed an expression to evaluate these

his

by assuming that eddies move

molecules

in a gas.

The

eddies

move

a fluid in a

in

manner

similar to the

a distance called the mixing length

L

movement

of

before they lose

their identity.

Actually, the

However,

assumed

moving eddy or "lump" of

in the definition

to retain

identity or be

its

L

in

will differ in

absorbed

in the

its

identity.

L and

then to lose

is

its

host region.

the y direction and retaining

mean

gradually lose

will

identity while traveling the entire length

Prandtl assumed that the velocity fluctuation distance

fluid

of the Prandtl mixing-length L, this small packet of fluid

u'x is

mean

its

due

to a

velocity.

velocity from the adjacent fluid by v x

\

"lump"

of fluid

moving

At point L, the lump of

y + L

~

v x \y

a

fluid

Then, the value of

v'x\, is

v'x\,

The

length

L

is

small

enough so

=

Vx\, + L

~

(3.10-24)

»x\y

that the velocity difference can be written as

»'x\,

=

"x\y + L

~

»x\,

= L

dv x (3.10-25) ~dy~

Hence, 1. = L

V v' x

196

Chap. 3

—

(3.10-26)

dy

Principles of Momentum Transfer

and Applications

Prandtl also assumed

v'x =s v'

y

.

Then

the time average, v'x x'y ,

is

(3.10-27)

dy

dy

sign and the absolute value were used to make the quantity experimental data. Substituting Eq. (3.10-27) into (3.10-21),

The minus

v'x v'

y

agree with

dv. dv,

(3.10-28)

dy

dy

Comparing with Eq.

(3.10-23), dv.

(3.10-29)

dy

3.10G

To

Universal Velocity Distribution in Turbulent Flow

determine the velocity distribution for turbulent flow at steady state inside a circular

we

tube,

divide the fluid inside the pipe into two regions: a central core where the

Reynolds

stress

approximately equals the shear stress; and a thin, viscous sublayer is due only to viscous shear and the turbulence

adjacent to the wall where the shear stress effects are

both

assumed

negligible. Later

we

include a third region, the buffer zone, where

stresses are important.

Dropping

and

the subscripts

on

superscripts

the shear stresses

and

velocity,

and

considering the thin, viscous sublayer, we can write

where

t0

is

assumed constant

dv -ii -r dy

=

t0

On

in this region. x0

y

=

(3.10-30)

integration,

(3.10-31)

nv

Defining a friction velocity as follows and substituting into Eq. (3.10-31)

(3.10-32)

P v

yv*

v*

nip

(3.10-33)

The dimensionless

velocity ratio

on

the

left

can be written as

7 r

(SI)

0

(3.10-34) V

=

(English)

V

To 9c

The dimensionless number on y

the right can be written as

=

(SI)

(3.10-35) ?Q

gc p

y

Sec. 3.10

Boundary-Layer Flow and Turbulence

(English)

197

where y where r

is

the distance from the wall of the tube.

is

the distance

distribution

from the center. Hence,

For a tube of radius

r0

,

y

=

r0

—

r,

for the viscous sublayer, the velocity

is

v

=y +

+

(3.10-36)

Next, considering the turbulent core where any viscous stresses are neglected, Eq.

becomes

(3.10-28)

= pL 2

x

(3.10-37)

(^J where dv/dy that the

is

always positive and the absolute value sign

mixing length

is

is

dropped. Prandti assumed

proportional to the distance from the wall, or

L = Ky and that r = t0

=

constant. Equation (3.10-37)

=

r0

(3.10-38)

now becomes

pKV^j

(3.10-39)

Hence, dv

= Ky—-

v*

(3.10-40)

dy

Upon

integration,

v* In y

where

K

x

is

(3.10-41)

The constant K, can be found by assuming

a constant.

small value of y, say y 0

= Kv + Ky that v

is

zero at a

.

—=

y

v*

+

Z

= ii n K

(3.10-42)

y0

+

Introducing the variable y by multiplying the numerator and the denominator of the term y/y 0 by v*/v, where v = /j/p, we obtain Wit'

\

(3.10-43)

(3.10-44)

A large amount of velocity distribution data by Nikuradse and others for a range of j/Reynolds numbers of 4000 to 3.2 x 10 6 have been obtained and the data fit Eq. (3.10-36) :

+

+

Eq. (3.10-44) above y of 30 with K and C, being + universal constants. For the region of y from 5 to 30, which is defined as the buffer region, an empirical equation of the form of Eq. (3.10-44) fits the data. In Fig. 3.10-4 the following relations which are valid are plotted to give a universal velocity profile for in

the region

fluids

up toy

of 5 and also

flowing in smooth circular tubes. + v + u

+ v

198

fit

y

=

5.0 In

=

+

+

=

2.5 In

(0
y +

y

Chap. 3

-

3.05

+5.5

Principles

(3.10-45) (5

<

(30

+

<

y

<

+

y

)

30)

(3.10-46) (3.10-47)

of Momentum Transfer and Applications

FIGURE

Three is

Universal velocity profile for turbulent flow

3.10-4.

distinct regions are apparent in Fig. 3.10-4.

in

The

smooth

first

circular tubes.

region next to the wall

by Eq.

the viscous sublayer (historically called "laminar" sublayer), given

where the velocity

(3.10-45),

proportional to the distance from the wall. The second region, called the buffer layer, is given by Eq. (3. 10-46), which is a region of transition between the viscous sublayer with practically no eddy activity and the violent eddy activity in is

by Eq.

the turbulent core region given

Fanning

related to the

used

These equations can then be used and They can also be

solving turbulent boundary-layer problems.

in

3.10H /.

(3. 10-47).

friction factor discussed earlier in the chapter.

Integral

Momentum

Balance for Boundary-Layer Analysis

Introduction and derivation of integral expression.

boundary

layer

on

a flat plate, the Biasius solution

In the solution for the laminar is

quite restrictive, since

it

is

for

Other more complex systems cannot be solved by this method. An approximate method developed by von Karman can be used when the configuration is more complicated or the flow is turbulent. This is an approximate laminar flow over a

momentum

flat plate.

integral analysis

of the

boundary layer using an empirical

or

assumed

velocity distribution.

In order to derive the basic equation for a laminar or turbulent

small control

The depth

volume

in

the boundary layer on a

in the z direction

from the top curved surface

is b.

at

<5.

Flow

An

flat

plate

is

used as shown

only through the surfaces

is

overall integral

boundary

momentum

A

r

layer, a

in Fig. 3. 10-5.

andA 2 and

also

balance using Eq. (2.8-8)

and overall integral mass balance using Eq. (2.6-6) are applied to the control volume inside the boundary layer at steady state and the final integral expression by von

Karman

is

(B2, S3)

d - = -T P

where

t0

is

dx

C

3

u

>=° -

y x)

dy

(3.10-48)

J0

the shear stress at the surface y

=

0 at point x along the plate. Also, 5 andr 0

are functions of x.

Equation

Sec. 3.10

(3

. 1

0-48)

is

an expression whose solution requires knowledge of the velocity

Boundary-Layer Flow and Turbulence

199

v x as a function of the distance

course,

depend on how

2. Integral

from the

closely the

The accuracy

surface, y.

assumed

velocity profile

of the results

approaches the actual

will,

of

profile.

balance for laminar boundary layer. Before we use Eq. boundary layer,. this equation will be applied to the laminar

momentum

(3.10-48) for the turbulent

boundary layer over a Blasius solution

in

flat

plate so that the results can be

compared with the exact

Eqs. (3.10-6H3.10-12).

In this analysis certain

boundary conditions must be v

=0

at

y

=

0

vx

^

at

y

=

<5

at

y

=

6

UQ

dv

dy

The conditions above are

fulfilled in

the following simple,

v„

The shear stress t 0

at

2 5

boundary

layer.

(3.10-49)

assumed

velocity profile.

U

2

.

satisfied in the

(3.10-50)

a given x can be obtained from f

T0

dv\

=^

(3.10-51)

Differentiating Eq. (3.10-50) with respect to y and setting y

'dv\

=

=

0,

3^

,dy) y=0

(3.10-52)

2<5

Substituting Eq. (3.10-52) into (3.10-51),

T

_

°"

3

^

(3.10-53)

25

Substituting Eq. (3.10-50) into Eq. (3.10-48) and integrating between y

y

=

8,

we

=

0

and

obtain

d5

_

dx~ Combining Eqs. (3.10-53) and = 0 and x = L,

280

t0

(3.10-54)

39 vlp

(3.10-54) and integrating

between

5

=

0 and

5=5,

and

x

FIGURE

200

3.10-5. Control

volume for integral analysis of the boundary-layer flow.

Chap. 3

Principles of Momentum Transfer

and Applications

5

where the length of (3.10-12), the

plate

=

LtL

4. 6 4

x = L. Proceeding

is

drag coefficient

(3.10-55)

in a

manner

similar to Eqs. (3.10-6)-

is

Cp=

HL

~=

1.292

^T

(3.10-56)

A comparison

of Eq. (3.10-6) with (3.10-55) and (3.10-12) with (3.10-56) shows the method. Only the numerical constants differ slightly. This method can be used with reasonable accuracy for cases where an exact analysis is not feasible.

success of

3.

this

momentum

Integral

the integral

analysis

turbulent boundary layer on a

flow which layer

on

a

is

valid

up

flat plate,. to

boundary layer. The procedures used for for^ laminar boundary layer can be applied to the

analysis for turbulent

momentum

to a

A

flat plate.

simple empirical velocity distribution for pipe

Reynolds number of 10 s can be adapted for the boundary

become it/7

(3.10-57)

This

the Blasius 7-power law often used. Equation (3.10-57) is substituted into the integral relation equation (3.10-48).

is

yn y

i

d

Tx

is

\vr dy

6

0

=

—

(3.10-58)

P

hold, as y goes to zero at the wall. Another useful which is consistent at the

The power-law equation does not relation

fy

the Blasius correlation for shear stress for pipe flow,

wall for the wall shear stress t 0

.

For boundary-layer flow over a

^

flat plate,

it

becomes

1/4

0.023(^V

=

py„

(3.10-59)

/w

\

Integrating Eq. (3.10-58), combining the result with Eq. (3.10-59), and integrating

between 5 = 0 and

5=5,

and x

=

0.376

0 and x

=

L,

il5 Lv„pY P -

L=

^

0.376L (3-10-60)

L

Integration of the drag force as before gives

Co =

^ 0.072

(3-10-61)

was assumed to extend to x = 0. Actually, a certain length at the front has a laminar boundary layer. Experimental data 5 7 check Eq. (3.10-61) reasonably well from a Reynolds number of 5 x 10 to 10 More accurate results at higher Reynolds numbers can be obtained by using a logarithmic In this development the turbulent

boundary

layer

.

velocity distribution, Eqs. (3.10-45H3. 10-47).

Sec. 3.10

Boundary-Layer Flow and Turbulence

201

DIMENSIONAL ANALYSIS

3.11

MOMENTUM TRANSFER

IN

3.11A

Dimensional Analysis of Differential Equations

we have derived

In this chapter

several differential equations describing various flow

Dimensional homogeneity requires that each term in a given equation have the same units. Then, the ratio of one term in the equation to another term is dimensionless. Knowing the physical meaning of each term in the equation, we are then able to give

situations.

a physical interpretation to each of the dimensionless parameters or numbers formed. These dimensionless numbers, such as the Reynolds number and others, are useful in correlating and predicting transport phenomena in laminar and turbulent flow. Often it is not possible to integrate the differential equation describing a flow situation. However, we can use the equation to find out which dimensionless numbers

can be used

An

in correlating

experimental data for

important example of

this involves the

this physical situation.

use of the Navier-Stokes equation, which

often cannot be integrated for a given physical situation. the

x component

of the Navier-Stokes equation.

—+ —+ —= v

i-V

dx

Each term

in this

each term

in

dy

dz

d gx

p dx

start,

we

state this

2

d

vx

use Eq. (3.7-36) for

becomes

2

d

vx

2

vx

2

or(L/f

2 ).

each term has a physical significance. First we use

and a

single characteristic length

Eq. (3.11-1)

is

as follows.

right-hand terms, respectively, as

g,

L

The left-hand

for all terms.

Then

a single

charac-

the expression of 2

side can be expressed as v /L and the

p/pL, and pv/pl3. _p_

,

p

equation has the units length/time

In this equation teristic velocity v

y

vz

To

At steady

We

then write

PV

+

(3.11-2) _pi3

This expresses a dimensional equality and not a numerical equality. Each term has 2

dimensions L/t

The

.

left-hand term in Eq. (3.11-2) represents the inertia force

and the terms on the

right-hand side represent, respectively, the gravity force, pressure force, and viscous force. 2 Dividing each of the terms in Eq. (3.11-2) by the inertia force [v /L], the following

dimensionless groups or their reciprocals are obtained. 2

lv /L]

inertia force

[
gravity force

v

2

(Froude number) pressure force

IpIpQ 2

[v /L]

inertia force

rV/L]

inertia force

2 -}

[pv/pL

Note that differential

this

viscous force

method not only

P

pv

2

Lvp

=

N Eu

= Nut

(Euler

number)

(3.11-3)

(3.11-4)

(Reynolds number)

(3.11-5)

gives the various dimensionless groups for a

equation but also gives physical meaning to these dimensionless groups. The

length, velocity,

etc.,

to be used in a given case will be that value

For example, the length

may be

which

is

the diameter of a sphere, the length of a

most

significant.

flat plate,

and so

on.

Systems that are geometrically similar are said to be dynamically similar

202

Chap. 3

Principles of Momentum Transfer

if

the

and Applications

parameters representing ratios of forces pertinent to the situation are equal. This means

Froude numbers must be equal between the two systems. is an important requirement in obtaining experimental data on a small model and extending these data to scale up to the large prototype. Since experiments with full-scale prototypes would often be difficult and/or expensive, it is customary to study small models. This is done in the scaleup of chemical process equipment and in the design of ships and airplanes. that the Reynolds, Euler, or

This dynamic similarity

3.1 IB

Dimensional Analysis Using Buckingham Method

The method

of obtaining the important dimensionless

ential equations

is

generally the preferred method. In

numbers from

many

able to formulate a differential equation which clearly applies.

procedure listing

required, which

is

is

known

Then

Buckingham method.

as the

the basic differ-

we are not more general this method the done first. Then

cases, however,

In

of the important variables in the particular physical problem

is

a

we determine the number of dimensionless parameters into which the variables may be combined by using the Buckingham pi theorem. The Buckingham theorem states that the functional relationship among q quantities or variables whose units may be given in terms of u fundamental units or dimensions may be written as is

(q

—

u)

independent dimensionless groups, often called

maximum number

actually the

rt's.

[This quantity u

of these variables which will not form a dimensionless

group. However, only in a few cases

is

this u not

equal to the

number

of fundamental

units (Bl).]

Let us consider the following example, to illustrate the use of this method.

incompressible fluid

is

variables are pressure

density p.

The

total

flowing inside a circular tube of inside diameter D.

drop Ap, velocity

v,

number of variables is q

diameter D, tube length

=

The

An

significant

L, viscosity

p.,

and

6.

The fundamental units or dimensions are u = 3 and are mass M, length L, and time 2 t. The units of the variables are as follows: Ap in M/Lt v in L/t, D in L, L in L, p in 3 M/Lt, and p in M/L The number of dimensionless groups or 7t's is q — u, or 6 — 3 = 3. ,

.

Thus, re,

=f(n 2 n 3 ,

(3.11-6)

)

Next, we must select a core group of u (or

3) variables which will appear in each n group and among them contain all the fundamental dimensions. Also, no two of the variables selected for the core can have the same dimensions. In choosing the core, the variable whose effect one desires to isolate is often excluded (for example, Ap). This leaves us with the variables v, D, p., and p to be used. (L and D have the same dimensions.) We will select D, v, and p to be the core variables common to all three groups. Then the three dimensionless groups are

7i,

=

k2

=

7c

To

3

b

D°v p c Ap

1

DV/L = DVpV

1

(3.11-8) (3.11-9)

be dimensionless, the variables must be raised to certain exponents First

we

consider the

7t,

a, b, c, etc.

group. tc,

Sec. 3.1 1

(3.11-7)

Dimensional Analysis

in

=

b

D°v p c

Momentum

Ap

{

Transfer

(3.11-7)

203

To

we

evaluate these exponents,

write Eq. (3.1 1-7) dimensionally by substituting the

dimensions for each variable. ,o

Next we equate

r

-/AY AfV M

o,o

the exponents of L

on both sides of this equation, of M, and

(L)

0=

(M)

0

=

0,

b

a+'b-3c-l +

c

(3.11-11)

1 ,

0= -b -2

(t)

Solving these equations, a

=

finally off.

= — 2,andc =

—1.

Substituting these values into Eq. (3.11-7),

=

Ul

Repeating

this

procedure for n 2 and n 3

(3.11-12)

,

;r 2

*3

= Ne "

Vp

=^

=

(3.11-13)

= ^Rc

(3.11-14)

/J

Finally, substituting

7r2

-n-,,

and

,

into Eq. (3.11-6),

7r3

Ap

/L

2-/(7;. Combining Eq.

Dvp

shows was shown before in the factor and Reynolds number) and of length/diameter

a function of the

is

empirical correlation of friction In pipes with

»

L/D

1

(3.11-15)

)

(2.10-5) with the left-hand side of Eq. (3.

that the friction factor

ratio.

—

1

1-15),

Reynolds number

the result obtained

(as

or pipes with fully developed flow, the friction factor

is

found to be independent of L/D. This type of analysis tell

is

experimentation, nor does

it

However, it does not must be determined by

useful in empirical correlations of data.

us the importance of each dimensionless group, which select the variables to be used.

PROBLEMS on a Cylinder in a Wind Tunnel. Air at 101.3 flowing at a velocity of 10 m/s in a wind tunnel.

3.1-1. Force

mm

diameter of 90 perpendicular to the

is

placed

air flow.

in the

What

is

kPa

A

Wind Force on

a

CD =

1.3,

FD =

204

6.94

N

Steam Boiler Stack. A cylindrical steam boiler stack has a and is 30.0 m high. It is exposed to a wind at 25°C having a

diameter of 1.0 m velocity of 50 miles/h. Calculate the force exerted on the boiler stack. Ans. C D = 0.33, F D 3.1-3.

is

tunnel and the axis of the cylinder is held the force on the cylinder per meter length?

Ans. 3.1-2.

absolute and 25°C

long cylinder having a

Effect of Velocity on Force

on a Sphere and Stokes' Law.

A

sphere

Chap. 3

is

=

2935

N

held in a

Problems

small wind tunnel where air at 37.8°C and 1 atm abs and various velocities is forced by the sphere having a diameter of 0.042 m. (a) Determine the drag coefficient and force on the sphere for a velocity of 4 2.30 x 10~ m/s. Use Stokes' law here if it is applicable. 3 2 2.30 x 10~ (b) Also determine the force for velocities of 2.30 x 10" 2.30 x 10" \ and 2.30 m/s. Make a'plot of F D versus velocity. ,

3.1-4.

3.1-5.

,

Drag Force on Bridge Pier in River. A cylindrical bridge pier 1.0 m in diameter is submerged to a depth of 10 m. Water in the river at 20°C is flowing past at a velocity of 1.2 m/s. Calculate the force on the pier. Surface Area in a Packed Bed. A packed bed is composed of cubes 0.020 m on a 3 side and the bulk density of the packed bed is 980 kg/m The density of the solid 3 cubes is 1500 kg/m .

.

(a)

Calculate

(b)

Repeat

D=

e,

for

0.02

effective diameter

the

Dp

,

and

a.

same conditions but

m and a length h =

for cylinders

having a diameter of

1.5/).

Ans.

(a) e

= 0.3467, D p =

0.020 m, a

=

196.0

m~

1

for Number of Particles in a Bed of Cylinders. For a packed bed containing cylinders where the diameter D of the cylinders is equal to the length h, do as follows for a bed having a void fraction e.

3.1-6. Derivation

(a)

(b)

Calculate the effective diameter. Calculate the number, «, of cylinders in

1

m

3

of the bed. (a)

Dp = D

of Dimensionless Equation for Packed Bed. Starting with Eq.

(3.1-20),

Ans. 3.1-7. Derivation

derive the dimensionless equation (3.1-21).

Show

all

steps in the derivation.

3.1-8.

Flow and Pressure Drop of Gases in Packed Bed. Air at 394.3 K flows through a packed bed of cylinders having a diameter of 0.0127 m and length the same as the diameter. The bed void fraction is 0.40 and the length of the packed bed is 3.66 m. The air enters the bed at 2.20 atm abs at the rate of 2.45 kg/m 2 s based on the empty cross section of the bed. Calculate the pressure drop of air in the bed. Ans. Ap = 0.1547 x 10 5 Pa

3.1-9.

Flow of Water in a

3.1-10.

Filter Bed. Water at 24°C is flowing by gravity through a filter bed of small particles having an equivalent diameter of 0.0060 m. The void fraction of the bed is measured as 0.42. The packed bed has a depth of 1.50 m. The liquid level of water above the bed is held constant at 0.40 m. What is the water velocity v' based on the empty cross section of the bed ? in Packed Bed. A mixture of particles in a packed bed contains the following volume percent of particles and sizes: 15%, 10 mm; 25%, 20 mm; 40%, 40 mm; 20%, 70 mm. Calculate the effective mean diameter, D pm if the shape factor is 0.74.

Mean Diameter of Particles

,

Ans.

D pm =

18.34

mm

and Darcy's Law. A sample core of a porous rock obtained from an oil reservoir is 8 cm long and has a diameter of 2.0 cm. It is placed in a core holder. With a pressure drop of 1.0 atm, the water flow at 20.2°C through the 3 core was measured as 2.60cm /s. What is the permeability in darcy?

3.1-11. Permeability

3.1-12.

Minimum Fluidization and Expansion of Fluid Bed. Particles having a size of 0.10 mm, a shape factor of 0.86, and a density of 1200 kg/m 3 are to be fluidized using air at 25°C and 202.65 kPa abs pressure. The void fraction at minimum fluidizing conditions

is

0.43.

The bed diameter

is

0.60

m

and the bed contains 350 kg of

solids.

minimum height of the fluidized bed. Calculate the pressure drop at minimum fluidizing conditions. (c) Calculate the minimum velocity for fluidization. (d) Using 4.0 times the minimum velocity, estimate the porosity of the bed. (a)

Calculate the

(b)

Chap. 3

Problems

205

Ans.

3.1-13.

(a)

L„f =

(c)

v'mf

=

Ap = 0.1212 x

1.810 m, (b)

0.004374 m/s,

(d) e

=

10

5

Pa,

0.604

Fluidization Velocity Using a Liquid. A tower having a diameter of 0.1524 is being fluidized with water at 20.2°C. The uniform spherical beads in and a density of 1603 kg/m 3 Estimate the tower bed have a diameter of 4.42 the minimum fluidizing velocity and compare with the experimental value of 0.02307 m/s of Wilhelm and Kwauk (W5).

Minimum

m

mm

3.1-14. Fluidization

of a Sand Bed

To

Filter.

.

clean a sand bed

filter it is

fluidized at

minimum

conditions using water at 24°C. The round sand particles have a 3 density of 2550 kg/m and an average size of 0.40 mm. The sand has the properties given in Table 3.1-2. (a)

The bed diameter

minimum (b)

0.40

is

m

and the desired height of the bed is 1.75 m. Calculate the amount

fluidizing conditions

needed. Calculate the pressure drop at these conditions and the

minimum

at

these

of solids

velocity for

fluidization. (c)

3.2-1.

Using 4.0 times the minimum velocity, estimate the porosity and height of the expanded bed.

Flow Measurement Using a Phot Tube. A pitot tube is used to measure the flow rate of water at 20°C in the center of a pipe having an inside diameter of 102.3 mm. The manometer reading is 78 of carbon tetrachloride at 20°C.

mm

The (a)

(b)

3.2-2.

pitot tube coefficient

is

0-98.

Calculate the velocity at the center and the average velocity. Calculate the volumetric flow rate of the water. -3 Ans. (a) y max = 0.9372 m/s, v, v = 0.773 m/s, (b) 6.35 x 10

m 3 /s

The flow rate of air at 37.8°C is being duct having a diameter of 800 by a pitot tube. The pressure difference reading on the manometer is 12.4 of water. At the pitot tube position, the static pressure reading is 275 of water above one atmosphere absolute. The pitot tube coefficient is 0.97. Calculate the velocity at the center and the volumetric flow rate of the air.

Gas Flow Rate Using a measured

Pitot Tube.

mm mm

at the center of a

mm

3.2-3.

Pitot-Tube Traverse for Flow Rate Measurement. In a pitot tube traverse of a pipe having an inside diameter of 155.4 in which water at 20°C is flowing, the following data were obtained.

mm

Reading

Distance from

(mm

Wall (mm)

The

in

Manometer

of Carbon Tetrachloride)

26.9

122

52.3

142

77.7

157

103.1

137

128.5

112

pitot tube coefficient

is

0.98.

maximum

(a)

Calculate the

(b)

Calculate the average velocity. [Hint:

velocity at the center.

Use Eq.

(2.6-17)

and do a graphical

integration.] 3.2-4.

Metering Flow by a Venturi. A venturi meter having a throat diameter of 38.9 is installed in a line having an inside diameter of 102.3 mm. It meters 3 water having a density of 999 kg/m The measured pressure drop across the

mm

.

206

Chap. 3

Problems

venturi

m

3

is

156.9 kPa.

The venturi

coefficient

Cv

is

Ans. 3.2-5.

and

0.98. Calculate the gal/min

flow rate.

/s

330 gal/min, 0.0208

m

3

/s

Use of a Venturi to Meter Water Flow. Water at 20°C is flowing in a 2-in. schedule 40 steel pipe. Its flow rate is measured by a venturi meter having a throat diameter of 20 mm. The manometer reading is 214 of mercury. The

mm

venturi coefficient 3.2-6.

is

0.98. Calculate the flow rate.

Metering of Oil Flow by an Orifice. A heavy oil at 20°C having a density of 900 kg/m 3 and a viscosity of 6 cp is flowing in a 4-in. schedule 40 steel pipe. When the 3 flow rate is 0.0174 m /s it is desired to have a pressure drop reading across the 3 manometer equivalent to 0.93 x 10 Pa. What size orifice should be used if the

assumed

orifice coefficient is

3.2- 7.

as 0.61 ?

What

is

permanent pressure

loss?

Water Flow Rate in an Irrigation Ditch. Water is flowing in an open channel in an irrigation ditch. A rectangular weir having a crest length L = 1.75 ft is used. The weir head is measured as h0 = 0.47 ft. Calculate the flow rate in ft 3/s and

m

3

/s.

Ans. 3.3- 1.

the

Flow

rate

=

1.776

3 ft

/s,

0.0503

m 3 /s

Brake Horsepower of Centrifugal Pump. Using Fig. 3.3-2 and a flow rate of 60 gal/min, do as follows. 3 (a) Calculate the brake hp of the pump using water with a density of 62.41b m/ft .

Compare with (b)

3.3-2.

Do

the

same

the value from the curve.

for a nonviscous liquid

having a density of 0.85 g/cm 3 Ans. (b) 0.69 brake hp .

k W-Power of a Fan. A centrifugal fan is to be used to take a flue gas velocity) and at a temperature of 352.6 K and a pressure of 749.3

(0.5

1

kW)

at rest (zero

mm Hg and

to

mm

discharge this gas at a pressure of 800.1 Hg and a velocity of 38.1 m/s. The volume flow rate of gas is 56.6 std 3 /min of gas (at 294.3 and 760 Hg). Calculate the brake of the fan if its efficiency is 65% and the gas has a molecular weight of 30.7. Assume incompressible flow.

m

mm

K

kW

Compression of Air. A compressor operating adiabatically is to comm 3 /min of air at 29.4X and 102.7 kN/m 2 to 311.6 kN/m 2 Calculate the power required if the efficiency of the compressor is 75%. Also, calculate the

3.3- 3. Adiabatic

press 2.83

.

outlet temperature. 3.4- 1.

Power for Liquid Agitation. It is desired to agitate a liquid having a viscosity of -3 1.5 x 10 Pa-s and a density of 969 kg/m 3 in a tank having a diameter of 0.91 m. The agitator will be a six-blade open turbine having a diameter of 0.305

m

operating at 180 rpm.

0.076 m. Also,

The tank has

W = 0.0381

four vertical baffles each with a width J of m. Calculate the required kW. Use curve 2, Fig.

3.4-4.

Ans. 3.4-2.

NF =

2.5,

power

=

0.

172

kW (0.23

1

hp)

Power for Agitation and Scale-Up. A turbine agitator having six flat blades and a disk has a diameter of 0.203 m and -is used in a tank having a diameter of 0.61 m = 0.0405 m. Four baffles are used having a and height of 0.61 m. The width width of 0.051 m. The turbine operates at 275 rpm in a liquid having a density of 909 kg/m 3 and viscosity of 0.020 Pa s. 3 (a) Calculate the kW power of the turbine and kW/m of volume. (b) Scale up this system to a vessel having a volume of 100 times the original for the case of equal mass transfer rates.

W

•

Ans.

(a)

(b)

3.4-3.

Chap. 3

3 P = 0.1508 kW, P/V = 0.845 kW/m 3 = = kW/m kW, P2 15.06 PJV2 0.845 ,

Scale-Down of Process Agitation System. An existing agitation process operates using the same agitation system and fluid as described in Example 3.4-la. It is desired to design a small pilot unit with a vessel volume of 2.0 liters so that effects

Problems

207

on the system can be studied in the laboratory. The mass transfer appear to be important in this system, so the scale-down should be on this basis. Design the new system specifying sizes, rpm, and k\V

of various process variables rates of

power. 3.4-4.

Anchor Agitation System. An anchor-type agitator similar to

that described for Eq. (3.4-3) is to be used to agitate a fluid having a viscosity of 100 Pas and a 3 = 0.90 m. The rpm is density of 980 kg/m The vessel size is D, = 0.90 and 50. Calculate the power required.

m

.

3.4-5.

H

Design of Agitation System. An agitation system is to be designed for a fluid having a density of 950 kg/m 3 and viscosity of 0.005 Pa s. The vessel volume is 3 1.50 m and a standard six-blade open turbine with blades at 45° (curve 3, = 8 and DJD, = 0.35. For the preliminary Fig. 3.4-4) is to be used with 3 design a power of 0.5 kW/m volume is to be used. Calculate the dimensions of the agitation system, rpm, and kW power. •

DJW

of Mixing Times for a Turbine. For scaling up a turbine-agitated system, do as follows:

3.4-6. Scale-Up

(a)

(b)

3.4- 7.

Derive Eq. (3.4-17) for the same power/unit volume. Derive Eq. (3.4-18) for the same mixing times.

Mixing Time (a)

(b)

Do

in a Turbine-Agitated System.

as follows:

Predict the time of mixing for the turbine system in Example 3.4-la. Using the same system as part (a) but with a tank having a volume ot 3 and the same power/unit volume, predict the new mixing time. 10.0 Ans. (a) /, = 4.1, t T = 17.7 s

m

Drop of Power-Law Fluid, Banana Puree. A power-law biological fluid, banana puree, is flowing at 23.9°C, with a velocity of 1.01 8 m/s, through a smooth tube 6.10 m long having an inside diameter of 0.01267 m. The flow properties of 0 454 the fluid are K = 6.00 N s /m 2 and n = 0.454. The density of the fluid is 3 976 kg/m (a) Calculate the generalized Reynolds number and also the pressure drop using Eq. (3.5-9). Be sure to convert K to K' first. (b) Repeat part (a), but use the friction factor method.

3.5- 1. Pressure

•

.

Ans. 3.5-2.

.

8C „

=

63.6,

Ap =

245.2

kN/m 2

(5120 lb f/ft

2 )

Pressure Drop of Pseudoplastic Fluid. A pseudoplastic power-law fluid having a 3 density of 63.2 lb„yft is flowing through 100 ft of a pipe having an inside diameter of 2.067 in. at an average velocity of 0.500 ft/s. The flow properties of = 0.280 lb r s"/ft 2 and n = 0.50. Calculate the generalized Reythe fluid are nolds number and also the pressure drop, using the friction factor method.

K

3.5-3.

N Rc

(a)

•

Turbulent Flow of Non-Newtonian Fluid, Applesauce. Applesauce having the flow properties given in Table 3.5-1 is flowing in a smooth tube having an inside at a velocity of 4.57 m/s. diameter of 50.8 and a length of 3.05 (a) Calculate the friction factor and the pressure drop in the smooth tube. (b) Repeat, but for a commercial pipe having the same inside diameter with a

m

mm

roughness of e

=

4.6 x 10"

5

m.

Ans.

(a)

N Re

.

8en

=

4855,

/=

0.0073, (b)

/=

0.01 00

of a Non-Newtonian Liquid. A pseudoplastic liquid having the follow= 0.53, K = 26.49 N s"7m 2 and p = 975 kg/m 3 is being agitated in a system such as in Fig. 3.5-4 where D, = 0.304 m, D a = 0.151 m, and N = 5 rev/s. Calculate pa N'Rc „ and the kW power for this system. Ans. p.a = 4.028 Pa s, N'KCt „ = 27.60, N P = 3.1, P = 0.02966 kW

3.5-4. Agitation

ing properties of n

•

,

,

,

•

208

Chap. 3

Problems

3.5-5.

Flow Properties of a Non-Newtonian Fluid from Rotational Viscometer Data. Following are data obtained on a fluid using a Brookfield rotational viscometer.

RPM

0.5

Torque

5

10

20

50

754

1365

2379

4636

2.5

1 ,

86.2

402.5"

168.9

(dyn-cm)

The diameter of

the inner concentric rotating spindle cylinder diameter is 27.62 mm, and the effective length the flow properties of this non-Newtonian fluid.

is is

25.15 mm, the outer 92.39 mm. Determine

Ans.

/i

= 0.870

3.6-1. Equation of Continuity in a Cylinder. Fluid having a constant density pis flowing in the z direction through a circular pipe with axial symmetry. The radial

direction (a)

(b)

is

designated by

r.,,-

balance with dimensions dr and dz, derive the equation of continuity for this system. Use the equation of continuity in cylindrical coordinates to derive the equa-

Using a cylindrical

shell

tion.

3.6- 2.

Change of Coordinates for

3.7- 1.

Combining Equations of Continuity and Motion. Using the continuity equation and the equations of motion for the*, y, andz components, derive Eq. (3.7-13).

3.8- 1.

Average Velocity

Continuity Equation. Using the general equation of continuity given in rectangular coordinates, convert it to Eq. (3.6-27), which is the equation of continuity in cylindrical coordinates. Use the relationships in

Eq. (3.6-26) to do

this.

in

a Circular Tube. Using Eq. (3.8-17) for the velocity in a

circular tube as a function of radius r,

derive Eq. (3.8-19) for the average velocity.

<

3.8-2.

Laminar Flow

Example

3.8-4

velocity v z av

P = Po

.

-

a Cylindrical Annulus. Derive all the equations given in all the steps. Also, derive the equation for the average Finally, integrate to obtain the pressure drop from z = 0 for in

showing

= L forp = p L

to z

3 8 - i9 >

.

Ans.

„ lav

dp --1

=

ri-r? '1 ' 1

^2 rt+rt2

|_

In (r 2 / ri )

=

Po

-

r

Pl

8pL

2

- r2

In (r 2 /r,)

3.8-3. Velocity Profile in Wetted-Wall Tower. In a vertical wetted-wall tower, the fluid

flows

down

the inside as a thin film 8

m

thick in laminar flow in the vertical z

direction. Derive the equation for the velocity profile v z as a function of*, the

distance from the liquid surface toward the wall. The fluid is at a large distance from the entrance. Also, derive expressions for v z av and v z max (Hint: At x = 5, which is at the wall, v z = 0. At x = 0, the surface of the flowing liquid, .

vz

Chap. 3

=

v z max .)

Problems

Show

all

Ans.

„,

steps.

=

2

(p ? «5 /2p)[l

- (x/<5) 2 ], v

z av

=

pgS 2 /3u,

v z mal

=

2

P g5 /2p

209

3.8-4.

Film and Differential Momentum Balance.

Velocity Profile in Falling

nian liquid

is

flowing as a falling film on an inclined

makes an angle of & with the

vertical.

Assume

flat

A Newto-

The

surface.

surface

that in this case the section

being considered is sufficiently far from both ends that there are no end effects on the velocity profile. The thickness of the film is 8. The apparatus is similar to Fig. 2.9-3 but is not vertical. (a)

(b) (c)

Do

as follows.

Derive the equation for the velocity profile of v z as a function of x in this film using the differential momentum balance equation. What are the maximum velocity and the average velocity? What is the equation for the momentum flux distribution of rTZ ? [Hint: Can Eq. (3.7-19) be used here?] 2 2 Ans. (a) v z = (pgS cos B/2p-)[\ - (x/S) ] (c)

3.8-5.

= pgx cos B

Velocity Profiles for Flow Between Parallel Plates. In Example 3.8-2 a fluid is flowing between vertical parallel plates with one plate moving. Do as follows. (a)

Determine the average velocity and the

(b)

Make

maximum

velocity.

a sketch of the velocity profile for three cases where the surface

moving upward, downward, and 3.8-6.

rxz

is

stationary.

Conversion of Shear Stresses in Terms of Fluid Motion. Starting with the x (3.7-10), which is in terms of shear stresses, convert it to the equation of motion, Eq. (3.7-36), in terms of velocity gradients, for a Newtonian fluid with constant p and pu. Note that (V v) = 0 in this case. Also, use of Eqs. (3.7-14) to (3.7-20) should be considered.

component of motion, Eq.

•

3.8-7. Derivation of Equation

mass balance over

of Continuity in Cylindrical Coordinates. element whose volume is r Ar

a stationary

By means of a A 6 Az, derive

the equation of continuity in cylindrical coordinates. 3.8- 8.

Flow between Two Rotating Coaxial Cylinders. The geometry of two coaxial cylinders is the same as in Example 3.8-5. In this case, however, both cylinders are rotating with the inner rotating with an annular velocity of coj and the outer at
using the differential equation of

momentum.

Ans. v 0 = —2

~2

K2 - K

1

2 ~~

\

"r 1

~

~

~~

/

= <$> for a given flow situation is cb a constant. Check to see if it satisfies Laplace's equation. Determine the velocity components v x and v y Ans. v x = 2Cx, v = — 2Cy(C = constant)

The C{x 2 - y 2 ), where C

3.9- 1. Potential Function.

potential function is

.

y

3.9-2.

Determining the Velocities from the Potential Function. The potential function flow is given as 4> - Ax + By, where A and B are constants. Determine "the velocities v x and v y

Jot

.

3.9-3.

Stream Function and Velocity Vector. Flow of a fluid in two dimensions is given by the stream function ib = Bxy, where 5 = 50 s" and the units of x and y are in cm. Determine the value of v x v y , and the velocity vector at x = 1

,

1

cm and y =

1

cm. Ans. v

3.9-4.

210

= 70.7 cm/s

Stream Function and Potential Function. A liquid is flowing parallel to the x = 0. axis. The flow is uniform and is represented by v x = U and v y (a) Find the stream function ib for this flow field and plot the streamlines. (b) Find the potential function and plot the potential lines. Ans. (a) i)j = Uy + C (C = constant)

Chap. 3

Problems

manner

at

U

=

vx

A liquid is flowing in a uniform an angle of /3 with respect to the x axis. Its velocity components are cos p and v y = U sin /3. Find the stream function and the potential

Components and Stream Function.

3.9-5. Velocity

function.

Ans.

= Uy

i/»

cos

- Ux

/3

sin

/3

+ C (C =

constant)

FW F/eW

h>i//i Concentric Streamlines. The flow of a fluid that has concentric 2 1 streamlines has a stream function represented by ip = \/{x + y ). Find the

3.9-6.

components of velocity v x and v y Also, determine .

if

determine the vorticity, 2cj z

so,

3.9-7. Potential Function

and

were given. Show

Example

Velocity Field. In

if a

if

the flow

is

rotational,

and

.

3.9-2 the velocity

velocity potential exists and,

Ans.

if

components

so, also determine

= ax 3 II - axy 2 + C{C =

4>

<£.

constant)

Equation of Motion for an Ideal Fluid. Using the Euler equations and zero viscosity, obtain the following equation:

3.9- 8. Euler's

(3.9-2}-(3.9-4) for ideal fluids with constant density

Laminar Boundary Layer on Flat Plate.

3.10- 1.

The

at 0.914 m/s.

plate

is

0.305

Water

at

20°C

flowing past a

is

flat

plate

m wide. m

from the leading edge to determine if Calculate the Reynolds number 0.305 the flow is laminar. (b) Calculate the boundary-layer thickness at x = 0.152 and x = 0.305 m from the leading edge. (a)

(c)

Calculate the total drag on the 0.305-m-long plate. 5 Ans. (a) (b) 5 Rc L = 2.77 x 10

N

3.10-2. Air

,

,

=

0.0029

m at x =

0.305

m

K

Flow Past a

Plate. Air at 294.3 and 101 .3 kPa is flowing past a flat plate m/s. Calculate the thickness of the boundary layer at a distance of 0.3 from the leading edge and the total drag for a 0.3-m-wide plate. at 6.

m

1

3.10-3. Boundary-Layer

Do

0.5 m/s. (a)

Flow Past a

Plate.

Water

at

293

K is flowing past a flat plate at

as follows.

Calculate the boundary-layer thickness in

m at a point 0.1 m from the leading

edge. (b)

At the same point, calculate the point shear drag coefficient.

3.10- 4. Transition Point to Turbulent

flowing past a smooth

stress t 0

Boundary Layer. Air

The

.

Also calculate the total

at 101.3

kPa and 293

is

at

(a)

(b)

=

5

x

10

plate at 100

ft/s.

turbulence

in

3 .

Calculate the distance from the leading edge where the transition occurs. Calculate the boundary-layer thickness <5 at a distance of 0.5 ft and 3.0 ft from the leading edge. Also calculate the drag coefficient for both distances

and .

is

such that the transition from a laminar to a turbulent boundary layer occurs /V Re ,i

3.1 1- 1

K

the air stream

flat

3.0

L

= 0.5

ft.

Dimensional Analysis for Flow Past a Body. A fluid is flowing external to a solid body. The force F exerted on the body is a function of the fluid velocity v, fluid density p, fluid viscosity fj., and a dimension of the body L. By dimensional analysis, obtain the dimensionless groups formed from the variables given. 2 Select v, {Note: Use the M, L, t system of units. The units of F are MLIt p, and L as the core variables.) 2 2 Ans. 7r = {F/L )/pv v 2 = nlLvp .

i

Chap. 3

Problems

,

211

Dimensional analysis is to be used data on bubble size with the properties of the liquid when gas bubbles are formed by a gas issuing from a small orifice below the liquid surface. Assume that the significant variables are bubble diameter/)," orifice diameter d, liquid density p, surface tension trin N/m, liquid viscosity p., and g. Select d, p, and g as the core variables.

3.11-2. Dimensional Analysis for Bubble Formation. to correlate

Ans.

=

n,

=

D/d, n 2

2

a/pd gJ ji 3

= p 2 /p 2 d*g

REFERENCES Chalmers Mfg. Co.

Bull. 1659.

(Al)

Allis

(AT)

American Gas Association, "Orifice Metering of Natural Gas," Gas Measurement Kept.

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New York,

1955.

(Bl)

Bennett, C. O., and Myers, J. E. Momentum, Heat, and Mass Transfer, 3rd ed. New York: McGraw-Hill Book Company, 1982.

(B2)

Bird, R.

Stewart, W.

B.,

York: John WileBates, R.

(B3)

Fondy,

L.,

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E.,

& Sons, Inc., P. L.,

E.

N. Transport Phenomena.

New

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1960.

and Corpstejn, R. R. l.E.C. Proc. Des.

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310

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(B4)

Brown, G. G,

(B5)

Biggs, R. D., A.I.Ch.E.J.,

(CI)

Charm,

S. E.

et al. Unit Operations. 9,

New York: John

The Fundamentals of Food

Avi Publishing Co.,

Wiley

& Sons, Inc., 1950.

636 (1963). Engineering.

2nd

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Conn.:

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(D2)

Carman, P. C. Trans. Inst. Chem. Eng. (London), 15, 150 (1937). Calderbank, P. H. In Mixing : Theory and Practice, Vol. 2, V. W. Uhl and J. B. Gray (eds.). New York: Academic Press, Inc., 1967. Drew, T. B., and Hoopes, J. W., Jr. Advances in Chemical Engineering. New York: Academic Press, Inc., 1956. Dodge, D. W., and Metzner, A. B. A.I.Ch.E.J., 5, 189 (1959).

(E 1 )

Ergun,

(C2) (C3)

(Dl)

Chem. Eng. Progr.,

S.

48, 89

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(Fl)

Fox, E. A., and Gex, V. E. A.I.Ch.E.J.,

(Gl)

Godleski,

E. S.,

(HI)

Harper,

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(H2)

Ho,

(K 1)

Kunii, D., and Levenspiel, O. Fluidization Engineering. Sons, Inc., 1969.

(K2)

Krieger,

(Ll) v

(L2)

.

C, and Kwong,

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J.

Food

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539 (1956).

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617 (1962).

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M., and Maron,

Leamy, G. H. Chem. Eng., Oct.

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H. 1

J.

New York: John

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&

Appl. Phys., 25, 72 (1954).

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(Ml)

F.

J.

and Smith, J. C. A.l.Ch.E.

Mines

Metzner, A.l.Ch.E.

Bull.,

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A. B., Feehs, R. H.,

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Ramos, H.

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Otto, R.

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and Tuthill,

J.

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(M2)

McCabe, W. L., Smith, J. C, and Harriott, P. Unit Operations of Chemical Engineering, 4th ed. New York: McGraw-Hill Book Company, 1985.

(M3)

Metzner, A.

(M4)

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(M5)

Moo- Yung, Tichar, M. K.,and Dullien, F.

212

B.,

and Reed,

J.

C. A.l.Ch.E.

J., 1,

434 (1955).

A. L. A.l.Ch.E.

J., 18,

Chap. 3

1,

Part

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References

McKelvey,

(M6)

M. Polymer Processing, New York: John Wiley

J.

&

Sons, Inc.,

1962.

(Nl)

Norwood,

(Pi)

Perry, R. H., and Green, D. Perry's Chemical Engineers' Handbook, 6th ed. New York: McGraw-Hill Book Company, 1984.

(P2)

Pinchbeck,

(P3)

Patterson, W.

(P4)

Perry, R. H., and Chilton, C. H. Chemical Engineers' Handbook, 5th ed. New York: McGraw-Hill Book Company, 1973.

Rushton,

(Rl)

K. W., and Metzner, A. B. A.I.Ch.E.

P. H., I.,

P.

J.,

Sci., 6,

57 (1956).

and Yap, C. Y. A.I.Ch.E.

Costich, D. W., and Everett, H.

H.,

J.

and Popper, F. Chem. Eng.

Carreau,

432 (1960).

J., 6,

J.,

25, 208 (1979).

Chem. Eng.

J.

Progr., 46, 395,

467(1950).

Rautzen, R.

(R2)

R.,

Corpstein, R.

R.,

and Dickey, D.

S.

Chem. Eng., Oct.

25, 119

(1976). (51)

Stevens, W.

(52)

Skelland, A. H. P. Non-Newtonian Flow and Heat Transfer. Wiley & Sons, Inc., 1967.

(53)

Streeter, V.

Company,

Ph.D.

L.

thesis,

Handbook

University of Utah, 1953.

of Fluid Dynamics.

•

Company,

New

New York: John

York: McGraw-Hill Book

1961.

SchliCHTING, H. Boundary Layer Theory.

(54)

(Tl)

E.

New

York: McGraw-Hill Book

1955.

Treybal, R.

E.

Liquid Extraction. 2nd ed.

New

York: McGraw-Hill Book Com-

pany, 1953. (T2)

Treybal, R. E. Mass Transfer Operations, 3rd ed.

New

(Ul)

Book Company, 1980. Uhl, V. W., and Gray,

and Practice,

J.

B. (eds.), Mixing: Theory

York: McGraw-Hill Vol.

I.

New

York: Academic, 1969.

(Wl)

Weisman, J., and Efferding,

(W2)

Winning, M.D.M.Sc.

(W3)

Walters, K. Rheometry. London: Chapman

(W4)

Wen,

(W5)

Wilhelm, R.

(W6)

Welty, J. R., Wicks, C. E., and Wilson, R. E. Fundamentals of Momentum, Heat, and Mass Transfer, 3rd ed. New York: John Wiley & Sons, 1984.

(Zl)

Zlokarnik, M., and Judat, H. Chem. Eng. Tech.,

Chap. 3

thesis,

L. E. A.I.Ch.E. J., 6,

University of Alberta, 1948.

C. Y., and Yu, Y. H. A.I.Ch.E. H.,

References

419 (1960).

J., 12,

& Hall Ltd.,

1975.

610 (1966).

and Kwauk, M. Chem. Eng. Progr., 44, 201

39,

1

(1948).

163 (1967).

213

CHAPTER

4

Principles of

Steady-State

Heat Transfer

INTRODUCTION AND MECHANISMS OF HEAT TRANSFER

4.1

4.1

A

The

Introduction to Steady-State

transfer of energy in the

Heat Transfer

form of heat occurs

in

many

chemical and other types of

processes. Heat transfer often occurs in combination with other unit operations, such as

drying of lumber or foods, alcohol distillation, burning of

fuel,

and evaporation. The

heat transfer occurs because of a temperature difference driving force and heat flows from the high- to the low-temperature region. In Section 2.3

we derived an equation

for a general

property balance of

thermal energy, or mass at unsteady state by writing Eq.

(2.3-7).

momentum,

Writing

a similar

equation but specifically for heat transfer,

( rate of \ \heat iny

Assuming (2.3-14),

/ rate of gener-\ \ation of heat J

rate of

heat out/

the* rate of transfer of heat occurs only

which

is

/ rate of accumu-

\

+

.

(4.1-1)

\lation of heat

by conduction, we can rewrite Eq.

Fourier's law, as

^=-fc— dx A

(4.1-2)

Making an unsteady-state heat balance for the x direction only on the element of volume or control volume in Fig. 4.1-1 by using Eqs. (4.1-1) and (4.1-2) with the cross-sectional 2 area being A

m

,

+ 4(Ax

214

A)

= q x]x+Ax +

pc p

dT —

(Ax A) •

(

4 ,i_ 3 )

Figure

Unsteady-state balance for heat transfer

4.1-1.

in

control volume.

where q is rate of heat generated per unit volume. Assuming no heat generation and also assuming steady-state heat transfer where the rate of accumulation is zero, Eq. (4.1-3) y~ becomes

U* =


This means the rate of heat input by conduction tion

or q x

;

(4.1-4)

+ -v,

=

the rate of heat output by conduc-

a constant with time for steady-state heat transfer.

is

In this chapter tion of heat

is

we are concerned with

zero and

we have

then constant with time,

change with time. expressions

in

To

a control

volume where the

steady-state heat transfer.

and the temperatures

The

rate of accumula-

rate of heat transfer

at various points in the

is

system do not

solve problems in steady-state heat transfer, various mechanistic

the form of differential equations for the different

modes

of heat transfer

such as Fourier's law are integrated. Expressions for the temperature profile and heat flux are then

again

obtained in this chapter.

Chapter

In

when

5 the

the rate of

conservation-of-energy equations (2.7-2) and

accumulation

The mechanistic expression

is

(4.1-3) will

be used

not zero and unsteady-state heat transfer occurs.

for Fourier's law in the

form of a partial

differential

equation

be used where temperature at various points and the rate of heat transfer change

will

with time. In Section 5.6 a general differential equation of energy change will be

derived and integrated for various specific cases to determine the temperature profile

and heat

4.1

B

flux.

Basic

Mechanisms of Heat Transfer

Heat transfer

may

occur by any one or more of the three basic mechanisms of heat

transfer conduction, convection, or radiation. :

1.

In conduction, heat can be conduced through solids, liquids, and gases. conducted by the transfer of the energy of motion between adjacent mole-

Conduction.

The

heat

is

cules. In a gas the

"hotter" molecules, which have greater energy and motions, impart

energy to the adjacent molecules at lower energy

some

levels.

This type of transfer

is

present to

which a temperature gradient exists. In conduction, energy can also be transferred by "free" electrons, which is quite important in metallic solids. Examples of heat transfer mainly by conduction are heat transfer extent in

all

solids, gases, or liquids in

through walls of exchangers or the

2.

ground during

Convection.

the winter,

The

a refrigerator, heat

steel forgings, freezing of

transfer of heat by convection implies the transfer of heat

transport and mixing of macroscopic elements of

Sec. 4.1

treatment of

and so on.

Introduction

by bulk

warmer portions with cooler portions

and Mechanisms of Heal Transfer

215

of a gas or a liquid. It also often involves the energy exchange between a solid surface

A

must be made between forced-convection heat transfer, where a pump, fan, or other mechanical means, and natural or free convection, where warmer or cooler fluid next to the solid surface causes a circulation because of a density difference resulting from the temperature differences in the fluid. Examples of heat transfer by convection are loss of heat from a car radiator where the air is being circulated by a fan, cooking of foods in a vessel being stirred, cooling of a hot cup of coffee by blowing over the surface, and so on.

and a

fluid.

distinction

flow past a solid surface by a

fluid is forced to

3.

Radiation

Radiation.

no physical

medium

is

from heat transfer by conduction and convection

differs

needed

Radiation

for its propagation.

is

in that

the transfer of energy

in much the same way as The same laws which govern the transfer

through space by means of electromagnetic waves

electro-

magnetic

of light

waves transfer

light

light.

govern the radiant transfer of heat. Solids and liquids tend to absorb the radiation being transferred through

or gases.

so that radiation

it,

important primarily

is

The most important example of

radiation

is

through space

in transfer

the transport of heat to the earth

from the sun. Other examples are cooking of food when passed below red-hot heaters, heating of fluids in coils of tubing inside a

4.1

C

Fourier's

As discussed

Law

Heat Conduction

of

in Section 2.3 for the general

types of rate-transfer processes are characterized

molecular transport equation,

—momentum

transfer, heat transfer,

in this category.

This basic equation

=

rate of a transfer process

is

three

all

and mass

by the same general type of equation. The transfer of

can also be included

electric

combustion furnace, and so on.

main

transfer

electric current

as follows:

driving force

,„„,«.

(2.3-1)

:

resistance

know

This equation states what we as heat or

The

mass,

we need

intuitively: that in order to transfer a

overcome a

a driving force to

property such

resistance.

transfer of heat by conduction also follows this basic equation

and

is

written as

Fourier's law for heat conduction in fluids or solids.

A

dx

where q x is the heat-transfer rate in the x direction in watts (W), A is the cross-sectional area normal to the direction of flow of heat inm 2 T is temperature in K, x is distance in m, and k is the thermal conductivity in W/m K. in the SI system. The quantity qJA is ,

called the heat flux in

direction.

The minus

W/m 2

The quantity dT/dx

.

sign in Eq. (4.1-2)

is

is

the temperature gradient in the x

required because

the heat flow

if

is

positive in a

given direction, the temperature decreases in this direction.

The

cm 2

T

,

k

units in Eq. (4.1-2)

in cal/s

in °F,

x in

•

°C cm, T -

ft,

may

in °C,

also be expressed in the cgs system withq^ in cal/s,

and x

in

cm. In the English system, q x

k in btu/h -°F-ft, and

qJA

in

btu/h

2 •

ft

.

is

in btu/h,

From Appendix

A

A

inft

in 2 ,

A.l, the

conversion factors are, for thermal conductivity,

1

1

216

btu/h btu/h

ft-

•

ft

-

°F =4.1365 x 10" °F

=

1

.73073

Chap. 4

3

cal/s

•

cm °C •

W/m K •

Principles of Steady-State

(4.1-5)

(4.1-6)

Heat Transfer

For heat

flux

and power, 1

1

W/m 2

2

=

3.1546

btu/h

=

0.29307

btu/h-ft

(4-1-7)

W

(4.1-8)

Fourier's law, Eq. (4.1-2), can be integrated for the case of steady-state heat transfer

through a point

T2

and

l

rr 2

x2

dx

A assuming that k

dropping the subscript x on q x

at

— x m away.

a distance of x 2

at point 2

9*

Integrating,

where the inside temperature Rearranging Eq. (4.1-2),

wall of constant cross-sectional area A,

flat

is Tj

1

= -k

dT

'

(4.1-9)

constant and does not vary with temperature and

is

for

convenience,

-.

—

='—

A

x2

—

- T2 )

(T\

(4.1-10)

Xj

EXAMPLE 4.1-1.

Heat Loss Through an Insulating Wall m 2 of surface area for an insulating wall composed of 25.4-mm-thick fiber insulating board, where the inside temperature is 352.7 K and the outside temperature is 297.1 K.

Calculate the heat loss per

From Appendix

Solution:

board

is

-

0.048

W/m

A.3, the thermal conductivity of fiber insulating K. The thickness x 2 - x t =0.0254 m. Substituting

into Eq. (4.1-10),

k

q

0.048

/m

=

105.1

=

(105.1 1

D

The

„

,

,

W/m 2

W/m 2 '

4.1

,„

r=

\

) ;

(3.1546

W/m 2 )/(btu/h

2 •

ft

2 33.30 btu/h-ft

)

Thermal Conductivity defining equation for thermal conductivity

definition, experimental

is

given as Eq.

measurements have been made

to

(4.1-2),

and with

this

determine the thermal con-

ductivity of different materials. In Table 4.1-1 thermal conductivities are given for a few

materials for the purpose of comparison. for

Table

4.1-1, gases

values, 1.

More

detailed data are given in

Appendix A.3 As seen in

inorganic and organic materials and A.4 for food and biological materials.

and

Gases.

In gases the

molecules are

energy and

have quite low values of thermal conductivity, liquids intermediate

solid metals very high values.

in

mechanism

of thermal conduction

is

relatively simple.

The

continuous random motion, colliding with one another and exchanging

momentum.

of lower temperature,

If

a molecule

moves from

a high-temperature region to a region

transports kinetic energy to this region and gives

up

this

energy

through collisions with lower-energy molecules. Since smaller molecules move

faster,

it

gases such as hydrogen should have higher thermal conductivities, as

shown

in

Table

4.1-1.

Theories to predict thermal conductivities of gases are reasonably accurate and are given elsewhere (Rl).

The thermal conductivity

root of the absolute temperature and

increases approximately as the square

independent of pressure up to a few atmospheres. At very low pressures (vacuum), however, the thermal conductivity approaches zero.

Sec. 4.1

Introduction and

is

Mechanisms of Heat Transfer

217

2.

The

Liquids.

physical

mechanism

of conduction of energy in liquids

somewhat

is

where higher-energy molecules collide with lower-energy molecules. However, the molecules are packed so closely together that molecular force fields exert a strong effect on the energy exchange. Since an adequate molecular theory of similar to that of gases,

liquids

is

not available, most correlations to predict the thermal conductivities are

empirical. Reid et

al.

(Rl) discuss these in detail.

varies moderately with temperature

k

where a and

The thermal conductivity

and often can be expressed

b are empirical constants.

=

a

of liquids

as a linear variation,

+ bT

(4.1-11)

Thermal conductivities of

liquids are essentially

independent of pressure.

Water has a high thermal conductivity compared to organic-type liquids such as shown in Table 4.1-1, the thermal conductivities of most unfrozen foodstuffs, such as skim milk and applesauce, which contain large amounts of water have thermal benzene. As

conductivities near that of pure water.

The thermal conductivity of homogeneous solids varies quite widely, as may some typical values in Table- 4.1-1. The metallic solids of copper and aluminum have very high thermal conductivities, and some insulating nonmetallic

3.

Solids.

be seen for

materials such as rock wool and corkboard have very low conductivities.

Heat or energy

is

conducted through solids by two mechanisms. In the first, which is conducted by free electrons

applies primarily to metallic solids, heat, like electricity,

which move through the metal lattice. In the second mechanism, present in all solids, heat is conducted by the transmission of energy of vibration between adjacent atoms.

Table

4.1-1.

Thermal Conductivities of Some Materials Pressure

( k in

at JO J. 325

(K)

(K)

k

Ice

273

2.25

(CI)

Fire claybrick

473

1.00

(PI)

0.130

0.043

(Ml) (Ml) (Ml)

0.168

(Ml)

0.029

(Kl)

Substance

Kef.

273

0.0242

(K2)

373

0.0316

273

0.167

(K2)

Paper

273

0.0135

(P2)

Hard rubber Cork board

273

0.151

303

273

0.569

(PI)

Asbestos

311

366

0.680

Rock wool

266

303

0.159

Steel

291

45.3

333

0.151

373

45

2

rt-Butane

Liquids

Water Benzene

(PI)

Copper

Biological materials

and foods oil

293

0.168

373

0.164

(PI)

Lean beef Skim milk

263

1.35

(CI)

275

0.538

(CI)

Applesauce

296

0.692

(CI)

Salmon

277

0.502

(CI)

248

1.30

218

Ref.

Solids

Air

Olive

Atm)

fc

Gases

H

(1

Temp.

Temp. Substance

kPa

Wjm K J

Chap. 4

Aluminum

Principles

273

388

373

377

273

202

(PI)

(PI)

(PI)

of Steady-State Heal Transfer

For heat

flux

and power, 1

1

W/m 2

2

=

3.1546

btu/h

=

0.29307

btu/h-ft

(4-1-7)

W

(4.1-8)

Fourier's law, Eq. (4.1-2), can be integrated for the case of steady-state heat transfer

through a point

T2

and

l

rr 2

x2

dx

A assuming that k

dropping the subscript x on q x

at

— x m away.

a distance of x 2

at point 2

9*

Integrating,

where the inside temperature Rearranging Eq. (4.1-2),

wall of constant cross-sectional area A,

flat

is Tj

1

= -k

dT

'

(4.1-9)

constant and does not vary with temperature and

is

for

convenience,

-.

—

='—

A

x2

—

- T2 )

(T\

(4.1-10)

Xj

EXAMPLE 4.1-1.

Heat Loss Through an Insulating Wall m 2 of surface area for an insulating wall composed of 25.4-mm-thick fiber insulating board, where the inside temperature is 352.7 K and the outside temperature is 297.1 K.

Calculate the heat loss per

From Appendix

Solution:

board

is

-

0.048

W/m

A.3, the thermal conductivity of fiber insulating K. The thickness x 2 - x t =0.0254 m. Substituting

into Eq. (4.1-10),

k

q

0.048

/m

=

105.1

=

(105.1 1

D

The

„

,

,

W/m 2

W/m 2 '

4.1

,„

r=

\

) ;

(3.1546

W/m 2 )/(btu/h

2 •

ft

2 33.30 btu/h-ft

)

Thermal Conductivity defining equation for thermal conductivity

definition, experimental

is

given as Eq.

measurements have been made

to

(4.1-2),

and with

this

determine the thermal con-

ductivity of different materials. In Table 4.1-1 thermal conductivities are given for a few

materials for the purpose of comparison. for

Table

4.1-1, gases

values, 1.

More

detailed data are given in

Appendix A.3 As seen in

inorganic and organic materials and A.4 for food and biological materials.

and

Gases.

In gases the

molecules are

energy and

have quite low values of thermal conductivity, liquids intermediate

solid metals very high values.

in

mechanism

of thermal conduction

is

relatively simple.

The

continuous random motion, colliding with one another and exchanging

momentum.

of lower temperature,

If

a molecule

moves from

a high-temperature region to a region

transports kinetic energy to this region and gives

up

this

energy

through collisions with lower-energy molecules. Since smaller molecules move

faster,

it

gases such as hydrogen should have higher thermal conductivities, as

shown

in

Table

4.1-1.

Theories to predict thermal conductivities of gases are reasonably accurate and are given elsewhere (Rl).

The thermal conductivity

root of the absolute temperature and

increases approximately as the square

independent of pressure up to a few atmospheres. At very low pressures (vacuum), however, the thermal conductivity approaches zero.

Sec. 4.1

Introduction and

is

Mechanisms of Heat Transfer

217

Thermal conductivities of insulating materials such as rock wool approach that of air since the insulating materials contain large amounts of air trapped in void spaces. Superinsulations to insulate cryogenic materials such as liquid hydrogen are composed of multiple layers of highly reflective materials separated by evacuated insulating spacers.

Values of thermal conductivity are cpnsiderably lower than for Ice has a thermal conductivity

much

air alone.

greater than water. Hence, the thermal con-

ductivities of frozen foods such as lean beef

and salmon given

in

Table 4.1-1 are much

higher than for unfrozen foods.

4.1 E It is

Conveclive-Heat-Transfer Coefficient

known

well

the object.

that a hot piece of material will cool faster

When

the fluid outside the solid surface

when

air is

blown or forced by

in forced or natural

is

motion, we express the rate of heat transfer from the solid to the

fluid,

convective

or vice versa, by

the following equation:

= hA(Tw

q

where q

is

the heat-transfer rate in

solid surface in

h

K, Tf

is

W, A

-Tf

the area in

is

btu/h

2 •

ft

•

The

,

is

the temperature of the

W/m 2

•

K. In English

units,

h

is

in

°F.

coefficient h

is

a function of the system geometry, fluid properties, flow velocity,

and temperature difference. In dict this coefficient, since

when a

m 2 Tw

the average or bulk temperature of the fluid flowing by in K, and

the convective heat-transfer coefficient in

is

(4.1-12)

)

it

many

cases, empirical correlations are available to pre-

often cannot be predicted theoretically. Since

we know

that

by a surface there is a thin, almost stationary layer or film of fluid adjacent to the wall which presents most of the resistance to heat transfer, we often call fluid flows

the coefficient h a film coefficient.

In Table 4.1-2

nisms of given.

some order-of-magnitude values of

free or natural

Water

h for different convective

mecha-

convection, forced convection, boiling, and condensation are

gives the highest values of the heat-transfer coefficients.

To convert

the heat-transfer coefficient h from English to SI units,

1

Table 4.1-2.

btu/h

2 ft

•

°F

=

5.6783

W/m 2 K •

Approximate Magnitude of Some Heat-Transfer Coefficients Range of Values of h

Mechanism

Condensing steam Condensing organics Boiling liquids

Moving water Moving hydrocarbons

Sec. 4.1

Introduction

2

-ft

°F

1000-5000 200-500

300-5000 50-3000 10-300 0.5-4

Still air

Moving

btu/h

air

2-10

and Mechanisms of Heat Transfer

W/m 2 K 5700-28 000

1100-2800 1700-28 000 280-17 000 55-1700 2.8-23

11.3-55

219

4.2

CONDUCTION HEAT TRANSFER Conduction Through a Flat Slab or Wall

4.2A

In this section Fourier's equation (4.1-2)

be used to obtain equations for one-

,will

dimensional steady-state conduction of heat through some simple geometries. For a flat slab or wall where the cross-sectional area A and k in Eq. (4.1-2) are constant, we obtained Eq. (4.1-10), which we rewrite as

A This

shown

is

substituted for

i.

in Fig. 4.2-1,

T2

and x

for

(T,

—

2

—

- T2 = )

(Tj

- T2

(4.2-1)

)

IAJC

where Ax = x 2 — x Equation (4.2-1) indicates that if T is x 2 the temperature varies linearly with distance as shown in t.

,

Fig. 4.2- lb. If

the thermal conductivity

substituting Eq. (4.1-1

1)

is

not constant but varies linearly with temperature, then

and integrating,

into Eq. (4.1-2) T,

a

+

b

+ T2

1 A

(Tl

Ax

^ = Ax

~

(Tl

~

(4.2-2)

Tl)

where

k=a + b This means that the

mean

value of k

(i.e.,

evaluated at the linear average of T, and

As

T2

7,

+ T2 (4.2-3)

k m ) to use in Eq. (4.2-2)

is

the value of k

.

stated in the introduction in Eq. (2.3-1), the rate of a transfer process equals the

driving force over the resistance. Equation (4.2-1) can be rewritten in that form.

q

where R

= Ax/kA

and

is

=

T - T2 l

_T -T

2

X

driving force

_

(4.2-4)

R

Ax/kA

the resistance in

resistance

K/W or h

•

°F/btu.

—

*«

3 0>

a,

E

Figure

4.2-1.

Heat conduction

in

a

flat wall: {a)

geometry of

wall, (b)

temperature

plot.

220

Chap. 4

Principles

of Steady-State Heat Transfer

A U Figure

in

a cylinder.

Conduction Through a Hollow Cylinder

4.2B In

Heat conduction

4.2-2.

many

instances in the process industries, heat

a thick-walled cylinder as in a pipe that

hollow cylinder

in Fig. 4.2-2

being transferred through the walls of

may

or

not be insulated. Consider the

with an inside radius of r„ where the temperature

outside radius of r 2 having a temperature of radially

is

may

T2 and

a length of

,

L m. Heat

from the inside surface to the outside. Rewriting Fourier's law, Eq.

is

is

T1}

an

flowing

(4.1-2),

with

distance dr instead of dx,

(4.2-5)

A The

dr

cross-sectional area normal to the heat flow

A= Substituting Eq. (4.2-6) into

is

2nrL

(4.2-6)

rearranging, and integrating,

(4.2-5),

p dT

dr — =-k

g 2tiL

r

(4.2-7)

Jr,

2tiL

- T2

(7\

(4.2-8)

)

In (r 2 /r,)

Multiplying numerator and denominator by(r 2

= kA

q

T,

-T

r->

-

r,

2

(r 2

r.

—

rj,

- T2 (4.2-9)

- rJ/MJ

R

where (2nLr 2 )

—

A2

(2-nLri)

-

r2

=

J?

r,

lm

The

log

area of (A i

mean

area

+ A 2 )/2

is

substituted for r 2 and instead

A ]m

.

In (r 2 /rj)

T

for

,

the temperature

in Eq. (4.1-10),

it

flat

(4.2-10)

(4.2-11)

In engineering practice,

T2

1

2nkL

within 1.5% of the log

of r as in the case of a

temperature as still

is

~A

ln(/4 2 //},)

In (27tLr 2 /27tLr,)

wall.

mean is

If the

if^//^ < 1.5/1, the linear mean area. From Eq. (4.2-8), if r is

seen to be a linear function of In r

thermal conductivity varies with

can be shown that the

mean

value to use in a cylinder

is

k m ofEq. (4.2-3).

Sec. 4.2

Conduction Heat Transfer

221

EXA MPLE 4.2-1. A 5

Length of Tubing for Cooling Coil

thick-walled cylindrical tubing of hard rubber having an inside radius of

mm

coil

a bath. Ice water

temperature of 14.65

is

274.9 K.

is

The

mm

is being used as a temporary cooling flowing rapidly inside and the inside wall

and an outside radius of 20

in

outside surface temperature

W must be removed from the bath by the cooling

is

A total How many m

297.1 K.

coil.

of tubing are needed?

From Appendix

Solution: k

=

0.151

value will

W/m

A.3, the thermal conductivity at

0°C (273 K)

K. Since data at other temperatures are not available, be used for the range of 274.9 to 297. 1 K. •

5 1

The calculation

=

0,005

m

'

1000

will

—20— = 0.02 m

=

r,

is

this

1000

be done first for a length of 1.0 in Eq. (4.2-10),

m

of tubing. Solving for

A 2 and A lm

the areas

,

A =

2nLr

y

,m

l

=

A 2 - A, ~ In (AJAJ

~

=

2n(1.0)(0.005)

-

0.1257

m2

0.0314

A2 =

0.125?

_

0.0314

2.303 log (0.1257/0.0314)

m2

m2

"

'

Substituting into Eq. (4.2-9) and solving,

q

=

uJ^= —

2

= -15.2 The negative on the

inside.

i

Since 15.2

W

is

removed

is

1-m

for a

W

14.65

=

0.02

0.005

from

r2

on the outside tor!

length, the

needed length

is

. = nn a964A m

mwM

that the thermal conductivity of rubber

quite small. Generally, metal cooling

is

thermal conductivity of metals

resistances in this case are quite small

4.2C

297.1

-

W (51.9 btu/h)

length

Note

v

!

sign indicates that the heat flow

coils are used, since the

-

^74.9

0.151(0.0682)

is

quite high.

The

liquid film

and are neglected.

Conduction Through a Hollow Sphere

Heat conduction through a hollow sphere is another case of one-dimensional conducUsing Fourier's law for constant thermal conductivity with distance dr, where r is

tion.

the radius of the sphere,

dT -=-k— dr A a

The cross-sectional area normal

to the heat flow

A =

47tr

(4.2-5)

is

2

(4.2-12)

Substituting Eq. (4.2-12) into (4.2-5), rearranging, and integrating, T2

dr 4:t 4« JL I

AukiT,

" 222

r

- T2 1A 2

k

~

2

dt

(4.2-13)

Jr,

T,

)

(1/r,

Chap. 4

-

- Ta

(4.M4)

l/r 2 )/4«*

Principles of Steady-State

Heat Transfer

A

7V

B

C

Q

Figure 43-1.

It

Heat flow through a multilayer

wall.

can be easily shown that the temperature varies hyperbolically with the radius. (See

Problem

4.3

4.2-5.)

CONDUCTION THROUGH SOLIDS Plane Walls

4.3A

Series

in

more than one material present as shown in The temperature profiles in the three materials A, B, heat flow q must be the same in each layer, we can write

In the case

where there

is

Fig. 4.3-1,

we proceed

as follows.

and

C

IN SERIES

a multilayer wall of

are shown. Since the

Fourier's equation for each layer as kA

A

)

Ax, Solving each equation

7,

for

kB

T3 =

(7-,

A

A

k

c T3 = -fAx r

(T2

Ax*

)

(T3

(4.3-1)

7i)

AT,

Ax j

- T2

kA

T2

A

-T

3

Ax c

=

(4.3-2)

k„A

Adding the equations for Tj — T2 T2 — T3 and T3 and T3 drop out and the final rearranged equation is ,

,

Ax^/(^

i4)

+ Ax s/(/c

fl

T4

+ Ax c /(/( c /I)

/I)

,

the internal temperatures

+

(43-3)

RB +

/? c

where the resistance R A = AxJk A A, and so on. Hence, the final equation is in terms of the overall temperature dropTi total resistance,

RA + RB

EXAMPLE 43-1. A

+ Rc

T2

— T4

and the

.

Heat Flow Through an Insulated Wall of a Cold Room

room

mm

constructed of an inner layer of 12.7 of pine, a of cork board, and an outer layer of 76.2 of concrete. The wall surface temperature is 255.4 K inside the cold room and 297.1 K at the outside surface of the concrete. Use conductivities from Appendix A.3 for pine, 0.151 for cork board, 0.0433; and for concrete, 0.762 2 K. Calculate the heat loss in for 1 and the temperature at the cold-storage

middle layer of 101.6

is

mm

mm

;

W/m

interface between the

Sec. 4.3

W

•

wood and cork

Conduction Through Solids

in

m

board.

Series

223

T =

Calling

Solution:

and concrete

x

255.4,

T4 =

K, pine as material A, cork and dimensions

297.1

C, a tabulation of the properties

as

as B, is

as

follows:

=

kA

The 1

m

0.

1

5

kB

1

Ax A =

0.0127

Ax B =

0.1016

Ax c =

0.0762

=

0.0433

kc

=

0.762

m m m from Eq.

resistances for each material are,

(4.3-3), for

an area of

2 ,

Ax A A

0.0127

kA

RB =

0.151(1)

Ax — = A B

4

kB

0.1016 X^T^TTTT

=

2.346

0.0433(1)

Ax 0.0762 R c = 7—cT = -~ = 0.100 kc A 0.762(1) Substituting into Eq.

(4.3-3),

- T4 R A + RB + Rc

255.4

T,

^

Since the answer

To

0.0841

T^ = -

is

negative, heat flows in

-

calculate the temperature

T2

297.1

2.346

W (- 56 23

=

16 48

+

-

-

+

0.100

btu /h)

from the outside. between the pine wood

at the interface

and cork,

Ra Substituting the

known

- 16 48 =

values and solving,

25

-

72

n noTi 0.084

and

T2 =

256.79

K

at the interface

alternative procedure to use to calculate T2 is to use the temperature drop is proportional to the resistance.

An

^ T^ R A+ RR

T

A B

+ RC

fact that the

^- T^

(4

^

}

Substituting,

255.4

Hence,

4.3B

T2 =

- T2 =

0.0841(255.4^ 297.1)

256.79 K, as calculated before.

Multilayer Cylinders

In the process industries, heat transfer often occurs for

224

= _ 139K

example when heat

is

through multilayers of cylinders, as

being transferred through the walls of an insulated pipe. Figure

Chap. 4

Principles of Steady-Stale

Heal Transfer

Figure

Radial heat flow through multi-

4.3-2.

ple cylinders in series.

shows a pipe with two

4.3-2

layers of insulation

hollow cylinders. The temperature drop

and Tj

—

The

is

around

— T2

T,

a

it, i.e.,

total of three concentric

across material A,

T2 — T3

across B,

7^ across C. heat-transfer rate q will, of course, be the

same

for

each

layer, since

we are

at

steady state. Writing an equation similar to Eq. (4.2-9) for each concentric cylinder,

-t

r, ('2

-r

t2 - r3

2

l

)/(k A

A A] J

-

(r 3

r3 -

A B]m )

r 2 )/(k B

(r 4

t„4

(43-5)

- rJ/(k c A Ci J

where

Al

A

~

di

^3

d

ln(^ 2 M,)

~ ^2

In (/1 3 //1

^4-^3

.

MJ/4 3 )

In

2)

T2

Using the same method to combine the equations to eliminate

done

for the flat walls in series, the final

and

T3

as

was

equations are

T —T 9

q

" =

(/2

-

r l )/(k A

RA

+R B +R C

EXAMPLE 43-2.

is

-

(r 3

r 2 )/(fc a /I,

,

J+

(r 4

- r3

)/(/c

c

/l c

J

T, - T = —= 4

T,-T4 •

Hence, the overall resistance

A

A A ,J +

(4.3-8)

Y.R again the

sum

of the individual resistances in series.

Heat Loss from an Insulated Pipe

thick-walled tube of stainless steel (A) having a k

m

=

21.63

W/m-K.

with

OD

0.0508 m is covered with a 0.0254-m layer of asbestos (B) insulation, k = 0.2423 W/m K. The inside wall temperature of the pipe is 81 1 K. and the outside surface of the insulation is at 310.8

dimensions of 0.0254

ID and

K. For a 0.305-m length of pipe, calculate the heat loss and also the temperature at the interface between the metal and the insulation. Solution:

Calling T,

=

811 K,

T2

the interface,

and

T3 =

310.8 K, the

dimensions are

^ = 0_25_ = Q0127 m The areas

are as follows for

L=

=

°

8

=

0.0254

2

m

2tz(0.305X0.0127)

=

0.0243

A 2 = 2nLr 2 =

2tz(0.305)(0.0254)

=

0.0487

=

2nLr

x

Conduction Through Solids

r3

=

0.0508

m

0.305 m.

=

A,

Sec. 4.3

r2

in Series

m2 m 2

225

A 3 = 2nLr 3 =

From Eq.

log

(4.3-6), the

2ji(O.3O5XO.05O8)

mean

=

0.0974

m2

areas for the stainless steel

(y4)

and asbestos

(£)are

A Aim _

Ab

,ra

A

mAl

From Eq. (4.3-7)

(A 3 /A 2 )

-

0.0974

0.0487

In (0.0974/0.0487)

kA

AA

°-° 7 ° 3

m2

kD

A D\vr,

0.2423(0.0703) is

811-

_ RA + RB

310.8

0.01673

+

calculate the temperature

T2

-

T\

=

q

K/w

°- 01673

21.63(0.0351)

]m

Hence, the heat-transfer rate

To

_ "

the resistances are

^rr'Sii1

_ QQ351m 2

In (0.0487/0.0243)

~ _- A * A 2 _ ~ In

-0.0243

0-0487

_

\n(A 2 /A,)

%

or

~R~T

w

= 33L?

132 btu/h)

(1

1.491

,

33L7=

811

- T2

aoT673-

Solving, 811 — T2 = 5.5 K and T2 = 805.5 K. Only a small temperature drop occurs across the metal wall because of its high thermal conductivity.

Conduction Through Materials

4.3C

Suppose

that

two plane

direction of heat flow

Then

is

the total heat flow

solids

in Parallel

A and B

are placed side by side in parallel, and the

perpendicular to the plane of the exposed surface of each the

is

sum

of the heat flow through solid

A

solid.

plus that through B.

Writing Fourier's equation for each solid and summing,

Ax B

Ax A where q T is total heat flow, A T3 and T4 for solid B.

and

T2

are the front

and rear surface temperatures of solid

,

;

If

we assume

that Tj

= T3

(front temperatures the

same

for

A and

(T,

- T2

B) and

T2 = T4

(equal rear temperatures),

qT

=

T,1

„

T,2

AxJk A A A

An example would members

—

+

T,1

— T2 ,

Ax B/k B A B

=

—+— \R /

1

1

A

\

be an insulated wall {A) of a brick oven where

(B) are in parallel

and penetrate the

wall.

(4.3-10)

)

R-bJ steel reinforcing

Even though the area>l B of

the steel

would be small compared to the insulated brick area A A the higher conductivity of the metal (which could be several hundred times larger than that of the brick) could allow a large portion of the heat lost to be conducted by the steel. Another example is a method of increasing heat conduction to accelerate the freeze drying of meat. Spikes of metal in the frozen meat conduct heat more rapidly into the ,

insides of the meat.

226

Chap. 4

Principles of Steady-State

Heat Transfer

should be mentioned that

It

occur

if

results using Eq. (4.3-10)

cases

some two-dimensional heat flow can Then the

would be

affected

somewhat.

Combined Convection and Conduction and Overall Coefficients

4.3D

In

some

in

the thermal conductivities of the materials in parallel differ markedly.

many

practical situations the surface temperatures (or

surface) are not

known, but

there

is

a

fluid

on both

boundary conditions

the plane wall in Fig. 4.3-3a with a hot fluid at temperature Tj

cold fluid at

and

hj

on the

T4

on

the outside surface.

inside.

(Methods

The

at the

sides of the solid surfaces. Consider

on the

inside surface

outside convective coefficient

is

h0

and a

W/m 2 K •

to predict the convective h will be given later in Section 4.4

of this chapter.)

The

heat-transfer rate using Eqs. (4.1-12) and (4.3-1)

q

Expressing \/h ; A,

=

h,

A(Tj

- TJ =

Ax A /k A A, and

^ Ax A

(T2

is

given as

- T3 = K A(T3 - T4 )

l/h a~A as resistances

)

(4.3-1 1)

and combining the equations

as

before,

\'lh

The

overall heat transfer by

i

r'- T* A+AxJk A A+\lh 0

- T>~ T>

(43-12)

A

combined conduction and convection U defined by

is

often expressed in

terms of an overall heat-transfer coefficient

q=UAAT where

Arovcrall = T - TA

and

U=

.„.

i

U

1//),.

(43-13)

„ rM

is

W /

1

...

„

+ AxJk A + .

0

l/h 0

~r^7 m2 K •

btu

,22

,

\h

•

ft

•

\

„J °F

(4^-14)

A more important application is heat transfer from a fluid outside a cylinder, through a metal wall, and to a fluid inside the tube, as often occurs in heat exchangers. In Fig. 4.3-3b, such a case is shown.

Sec.

43

Conduction Through Solids

in Series

227

Using the same procedure as before, the overall heat-transfer rate through the cylinder

is

q

yh A +

/4,-

represents

the metal tube; and

The area

A

{

2nLr it

A0

(r

i

i

where

- T4

T)

=

-r )/k A A A]ia +l/h A

0

0

i

tube;^^

the inside area of the metal

(43_15)

£tf ?

the log

mean

area of

the outside area.

U

overall heat-transfer coefficient

or the outside area

q

— T4

T,

=

=

Aa

- T4 =

[/.^(T,

U =

yh +

'

i

U =

A

0

be based on the inside

- T4 =

{T,

i

i

(43-16)

)

-r )AJkA A Aim +AJA

(r 0

i

U„

)

A 0 /A h +(r 0

°

may

for the cylinder

of the tube. Hence,

(43_17) 0

h0

(43_18)

-r )Ao/l< A A Alm +

l/h.

i

EXAMPLE 43-3.

Heat Loss by Convection and Conduction and Overall U Saturated steam at 267°F is flowing inside af-in. steel pipe having an ID of of 1.050 in. The pipe is insulated with 1.5 in. of 0.824 in. and an insulation on the outside. The convective coefficient for the inside steam 2 surface of the pipe is estimated as h = 1000 btu/h ft °F, and the convective coefficient on the outside of the lagging is estimated as h 0 = 2 btu/h

OD

•

•

t

2 ft

The mean thermal conductivity

°F.

btu/h

•

ft

(a)

(b)

•

°F and 0.064

W/m K

Calling r0

r;

0.525

r

W/m K

45

is

•

or 26

for the insulation. if

the

2.025

r

,

of pipe, the areas are as follows.

i

'0412 2ti(1)(

!

r

Aj^ZhLtj =

2ti(1)(

2nLr 0 -

2n(l)l

Eq. (4.3-6) the log

mean

A A ]m ~~

,

228

°F

the inside radius of the steel pipe, r, the outside radius

A = InLr, =

From

•

the outside radius of the lagging, then

0.412

ft

ft

t

of the pipe, and

1

•

Calculate the heat loss for 1 ft of pipe using resistances surrounding air is at 80°F. Repeat using the overall U based on the inside area /!,-

Solution:

For

of the metal

or 0.037 btu/h

•

In

x

0 525

12

0.2157

=

0.2750

2 ft

N j

2mt>

=

j=

2 ft

2

1.060

ft

areas for the steel (A) pipe and lagging (B) are

-A

0.2750- 0.2157

t

~ (AJAd

_ A o~A _ In (A JA J x

~

In (0.2750/0.2157)

1.060- 0.2750 In (1.060/0.2750)

Chap. 4

Principles

_

^

of Steady-State Heat Transfer

*

From

Eq. (4.3-15) the various resistances are

——

=

R,

h

'

{

A

————

=

-

r.-r, _. (0.525

R RA

-

-k A A Alm

P Kfl

r

_

~r B ^ fllm »

fc

^

'

HzZ

R;

0.412)/12

2

-

025

~

°- 525

V 12 _

<

on

0 472 -

(4.3-1 5),

267

=

+ RA + RB + R

- a °° 148

0.037(0.583)

Using an equation similar to Eq.

=

0.00464

26(0.245)

«-^ = 2ak = , H

=

-

1000(0.2157)

t

0:00464

0

+

-

80

+

0.00148

5.80

+

;

0.472

^

'

6.278

For part

U

the equation relating

(b),

equated to Eq.

to q

x

Eq. (4.3-16), which can be

is

(4.3-19).

q=U,MT,-

=

T„)

(43-20)

Solving for U,,

known

Substituting

values,

=

V,

=

b,U 0.738

0.2157(6.278)

Then

=

U,yl

(

(7;

ft

°F

•

- r0) =

-

0.738(0.2157X267

80)

=

29.8 btu/h (8.73

W)

Conduction with Internal Heat Generation

In certain systems heat

distributed heat source

nuclear fuel rods. Also, of reaction

heaps

2 •

to calculate g,

g

4.3E

h

"

in

is

given

off.

is

is if

generated inside the conducting medium;

Examples of

present.

a chemical reaction

is

activity

occurring

is

a uniformly

and medium, a heat compost heaps and trash

occurring uniformly

In the agricultural and sanitation

which biological

i.e.,

this are electric resistance heaters

will

fields,

have heat given

in

a

off.

Other important examples are in food processing, where the heat of respiration of fresh fruits and vegetables is present. These heats of generation can be as high as 0.3 to

W/kgor0.5to

0.6

/.

1

Heat generation

generation. Heat to be insulated.

is

btu/h -lb m in

.

plane wall.

k

W/m

Sec. 4.3

•

in

is

the one

x

direction.

with internal heat walls are

assumed

K at x = L and x = — L held constant. The 3 q W/m and the thermal conductivity of the medium

The temperature Tw

volumetric rate of heat generation is

is shown The other

In Fig. 4.3-4 a plane wall

conducted only

in

is

K.

Conduction Through Solids

in Series

229

To Eq.

(4.

1

derive the equation for this case of heat generation at steady state, -3)

«,|»

where A.x

A

we

start with

but drop the accumulation term.

is

+

4(&x A)

=

q x]x + Ax

+

(43-22)

0

the cross-sectional area of the pjate. Rearranging, dividing by Ax, and letting

approach zero,

-da

*2

+

dx Substituting Eq. (4.1-2) for q x

A = 0

q

(4.3-23)

,

d2 T

q

(43- 24)

Integration gives the following for q constant

s

T= -

~x

1

+ C,x + C 2

(4.3-25)

where C, and C 2 are integration constants. The boundary condition's are at* = T = Tw and at x = 0, T = T0 (center temperature). Then, the temperature ,

,

£ The center temperature

from the two faces

lost

at

qT

where

2.

A

is

Heat generation

in cylinder.

R

-L,

(43-26)

^-+Tw

(4.3-27)

steady state

=

is

equal to the total heat generated,

q(2LA)

the cross-sectional area (surface area at

cylinder of radius

or

profile is

is

T0 = The total heat g T ,inW.

+ T0

L

In a similar

(4.3-28)

Tw

)

of the plate.

manner an equation can be derived

for a

with uniformly distributed heat sources and constant thermal

-x FIGURE

230

4.3-4.

Plane wall with internal heal generation

Chap. 4

Principles

al

steady state.

of Steady-State Heat Transfer

The heat is assumed to flow only radially; The final equation for the temperature profile is

conductivity. insulated.

{Rl

7= where

r is

distance from the center.

4k

The

~

r2)

+

i.e.,

Tw

(4

center temperature

T0 =

the ends are neglected or

"

3" 29)

7^, is

qR 2

~j-+Tw

(4.3-30)

EXAMPLE 43-4. An

Heat Generation in a Cylinder 200 A is passed through a stainless steel wire having a of 0.001268 m. The wire is L = 0.91 m long and has a resistance R

electric current of

R

radius

The outer surface temperature Tv is held at 422.1 K. The average thermal conductivity is k = 22.5 W/m K. Calculate the center temperature. of 0.126 Q.

where

must be

First the value of q

Solution:

current in

/ is

amps and R 2

I

known

Substituting

is

calculated. Since

power =

2

I R,

ohms,

resistance in

R'= watts = qnR L 2

(4.3-31)

values and solving, (200) (0.126)

=

q7r(0.00 1268) (0.91)

q

=

1.096 x 10

2

2

Substituting into Eq. (4.3-30) and solving,

T0 =

W/m

9

3

441.7 K.

Critical Thickness of Insulation for a Cylinder

4.3F

In Fig. 4.3-5 a layer of insulation

radius

rl

is

fixed with a length L.

inner temperature T, at point

where the cylinder insulation at occurs.

It is

T2

is

is

installed

is

The

around the outside of

outside the cylinder

a metal pipe with saturated

exposed to an environment

not obvious

a cylinder

whose

cylinder has a high thermal conductivity and the

if adding more

is

steam at

T0

fixed.

inside.

An example is the case The outer surface of the

where convective heat

transfer

insulation with a thermal conductivity of k will

decrease the heat transfer rate.

At steady state the heat-transfer rate q through the cylinder and the insulation equals the rate of convection from the surface. q

As insulation

Sec. 4.3

is

=

ho

A(T2

added, the outside area, which

Conduction Through Solids

in Series

-T

(4.3-32)

0)

is

A =

2nr 2 L, increases but

T2

decreases.

231

However,

not apparent whether q increases or decreases.

is

it

To

determine

this,

an

equation similar to Eq. (4.3-15) with the resistance of the insulation represented by Eq. (4.2-1 1) is written

using the two resistances.

- T0

2%UJ,

)

(43-33) I" (r 2 /r.)

1 |

K

r2

To determine respect to

r2

on

the effect of the thickness of insulation

equate

,

this result to zero,

and obtain

- 2nUJ -

dq

ToXlA-2 k

x

dr

q,

we

take the derivative of q with

the following for

-

\/r\ ?

maximum heat

flow.

h)

—=0.

(43-34)

In (r 2 /r,) r2

K

Solving,

=


where

(r 2 ) cr is

Hence,

the value of the critical radius

the outer radius

if

r2

(4-3-35)

f

when

the heat-transfer rate

less-'than the critical value,

is

actually increase the heat-transfer rate q. Also,

if

a

is

maximum.

adding more insulation

the outer radius

is

will

greater than the

adding more insulation will decrease the heat-transfer rate. Using typical values and h 0 often encountered, the critical radius is only a few mm. As a result, adding insulation on small electrical wires could increase the heat loss. Adding insulation to

critical,

of k

large pipes decreases the heat-transfer rate.

EXAMPLE 43-5.

and Critical Radius and covered with a plastic diameter of 1.5 insulation (thickness = 2.5 mm) is exposed to air at 300 K and h 0 = 20 W/m 2 K. The insulation has a k of 0.4 W/m K. It is assumed that the wire surface temperature is constant at 400 K and is not affected by the

An

electric wire

Insulating an Electrical Wire

having

mm

a

•

covering. (a)

(b) (c)

Calculate the value of the critical radius. Calculate the heat loss per of wire length with no insulation. Repeat (b) for the insulation present.

m

Solution:

For part

(a)

using Eq. (4.3-35),

(r 2 ) cr

For part

(b),

L=

1.0

m,

04 —=— = h~ 20 0.020 m _/c__

= r2

=

1.5/(2

x 1000)

=

=

20

mm

0.75 x 10

-3

m,

A =

2nr 2 L.

Substituting into Eq. (4.3-32), q

For (2.5

=

h a A(t 2

part

+

- T0 = )

3 (20X2tt x 0.75 x 10" x 1X400

300)

=

9.42

W

-3

(c)

with insulation, r, = 1.5/(2 x 1000) = 0.75 x 10 m, = 3.25 x 10" 3 m. Substituting into Eq. (4.3-33),

r2

=

1.5/2)/1000

^°X

In (3.25 x

2 400 300) 10-70.75 x 10" 3 )

=

-

0.4

(3.25

Chap. 4

32.98

W

1

x 10" 3 X20)

Hence, adding insulation greatly increases the heat

232

-

loss.

Principles of Steady-State

Heat Transfer

Contact Resistance at an Interface

4.3G

In the equations derived in this section for conduction through solids in series (see Fig. 4.3-1)

ature;

has been assumed that the adjacent touching surfaces are at the same temper-

it

i.e.,

completely perfect contact

is

ing designs in industry, this assumption

power

in nuclear

be present at the

when

resistance, occurs

stagnant

peaks

fluid is

is

the surfaces. For

reasonably accurate. However,

many

engineer-

in cases

such as

plants where very high heat fluxes are present, a significant drop in

may

temperature

made between

the

two

interface. This interface resistance, called contact

solids

do not

fit

tightly together

and a thin layer of

trapped between the two surfaces. At some points the solids touch at

in the surfaces

and

at

other points the fluid occupies the open space.

This interface resistance

is

a complex function of the roughness of the two surfaces,

the pressure applied to hold the surfaces in contact, the interface temperature, and the interface fluid.

Heat

transfer takes place

by conduction, radiation, and convection across

the trapped fluid and also by conduction through the points of contact of the solids.

No

completely reliable empirical correlations or theories are available to predict contact resistances for

all

types of materials. See references (C7, R2) for detailed discussions.

The equation

for the contact resistance

q H

=

hc '

is

often given as follows:

AT = AT A AT = \/h A R

where

hc

is

the contact resistance coefficient in

across the contact resistance in K, and

R

Rc

can be added with the other resistances

c

(4.3-36)

c

c

W/m

2 -

K,

AT

the contact resistance. in

the temperature drop

The contact

resistance

Eq. (4.3-3) to include this effect for solids in

For contact between two ground metal surfaces h c values of the order of mag4 4 2 tol x 10 W/m K have been obtained. An approximation of the maximum contact resistance can be obtained if the maximum gap Ax between the surfaces can be estimated. Then, assuming that the heat transfer across the gap is by conduction only through the stagnant fluid, h c is estimated

series.

nitude of about 0.2 x 10

•

as

K= If

(4.3-37)

-JAx

any actual convection, radiation, or point-to-point contact assumed resistance.

is

present, this will reduce

this

4.4

STEADY-STATE CONDUCTION AND SHAPE FACTORS

4.4A

Introduction and Graphical

Method

for

Two-Dimensional

Conduction In previous sections of this chapter direction. In

many

cases,

we discussed steady-state heat conduction

however, steady-state heat conduction

is

in

one

occurring in two

i.e., two-dimensional conduction is occurring. The two-dimensional solutions more involved and in most cases analytical solutions are not available. One important approximate method to solve such problems is to use a numerical method discussed in detail in Section 4.15. Another important approximate method is the graphical method, which is a simple method that can provide reasonably accurate answers for the heat-transfer rate. This method is particularly applicable to systems having

directions; are

isothermal boundaries.

Sec. 4.4

Steady-State Conduction and Shape Factors

233

method we

In the graphical

through a

flat

first

note that for one-dimensionaJ heat conduction

slab (see Fig. 4.2-1) the direction of the heat flux or flux lines

is

always

perpendicular to the isotherms. The graphical method for two-dimensional conduction also based

on

is

the requirement that the heat flux lines and the isotherm lines intersect

each other at right angles while forming a network of curvilinear squares. This means, as

shown

in Fig. 4.4-1, that

we can sketch

the isotherms

intersect at right angles (are perpendicular to

can obtain reasonably accurate

results.

and also the

flux lines until they

each other). With care and experience we

General steps to use

in this

graphical

method

are

as follows. 1.

Draw

a

model

to scale of the two-dimensional solid. Lable the isothermal boundaries.

and T2 are isothermal boundaries. number N that is the number of equal temperature subdivisions between

In Fig. 4.4-1, Tj 2.

Select a

isothermal boundaries. In Fig. 4.4-1, in the

N=4

subdivisions between T, and

isotherm lines and the heat flow or flux

lines

T2

.

the

Sketch

so that they are perpendicular to

each other at the intersections. Note that isotherms are perpendicular to adiabatic

and also lines of symmetry. Keep adjusting the isotherm and flux lines until

(insulated) boundaries 3.

condition

Ax = Ay

is

for each curvilinear square the

satisfied.

In order to calculate the heat flux using the results of the graphical plot,

assume

shown

unit depth of the material,

a'

lane. Since

q' will

Ax =

heat flow

=

dT = k(Ax -kA— dy

is

the

(4.4-1)

Ay

be the same through each curvilinear square within

Ay, each temperature subdivision

number

AT

is

this heat-flow

equal. This temperature sub-

T — T2 t

and N,

of equal subdivisions.

AT =

234

first

AT 1)

division can be expressed in terms of the overall temperature difference

which

we

through the curvilinear section

q'

given by Fourier's law.

in Fig. 4.4-1 is

This heat flow

The

T — T2 1

Chap. 4

Principles of Steady-State

(4.4-2)

Heat Transfer

through each lane

Also, the heat flow q

and

in

Eq.

same

the

is

since

Hence, the total heat transfer q through

(4.4-1).

Ax = Ay

all

in the

of the lanes

q=Mq' = Mk AT where

M

is

the total

number of heat-flow

construction

is

(4.4-3)

lanes as determined by the graphical procedure.

Substituting Eq. (4.4-2) into (4.4-3),

=^

q

- T2 )

fc(T,

(4.4-4)

EXAMPLE

4.4-1. Two-Dimensional Conduction by Graphical Procedure Determine the total heat transfer through the walls of the flue shown in Fig. 4.4-1 if T, = 600 K, T2 = 400 K,k = 0.90 W/m K, and L (length of flue) = •

5

m.

Solution:

The

In Fig. 4.4-1,

or length

L q

of 5

= 9.25. temperature subdivisions and through the four identical sections with a depth

m is obtained by using Eq. (4.4-4).

=

4

=

8325

Shape Factors

4.4B

77

in

In Eq. (4.4-4) the factor

kL(T,

- 73 =

M/N is called

the conduction shape factor S, where

S

=

M —

q

=

fcS(T,

m and

is

(4.4-5)

- T2

(4.4-6)

)

used in two-dimensional heat conduction where

The shape

factors for a number of geometries have Table 4.4-1. three-dimensional geometry such as a furnace, separate shape factors are used

been obtained and some are given a

^— (0.9X5.0X600-400)

Conduction

only two temperatures are involved.

For

4

W

This shape factor S has units of

to

M

N=4

total heat-transfer rate

in

obtain the heat flow through the edge and corner sections.

dimensions for a

is

uniform wall thickness

S^ n where A

is

When

each of the interior

greater than one-fifth of the wall thickness, the shape factors are as follows

Tw

=Y

:

S edge

=

the inside area of wall and

S corner

0.54L

L

= 0.15TW

(4.4-7)

the length of inside edge.

For a completely

enclosed geometry, there are 6 wall sections, 12 edges, and 8 corners. Note that for a single

fiat

wall, q

= kSwlll (r, - T2 ) = KAJTJJ^ - T2 ), which

is

the

same

as Eq. (4.2-1)

for conduction through a single flat slab.

For

a long

hollow cylinder of length

S

For a hollow sphere from Eq.

=

as that in Fig. 4.2-2,

rrr^ (r^r,)

(44- 8)

In

(4.2-14),

S

Sec. 4.4

L such

Steady-State Conduction

=

^1

and Shape Factors

(4.4-9)

235

Table

4.4-1.

Conduction Shape Factors for q

Cylinder of length L in a square

=

kS^ — T )* 2

3

7\ 7-nL

S=

(

ln(0.54 fl/r,)

r

k -«

*~

a

7nL

H

Horizontal Varied cylinder of length L

Two

i

(//>3r,) ln(2///r,)

2ttZ

parallel

5 =

cylinders of length

£'

H

~

r

\-

cosh" 2r r 2 l

4irr,

S=

Sphere buried

1

*

The thermal condocliviiy

medium

is k.

FORCED CONVECTION HEAT TRANSFER INSIDE

43 4.5A In

of ihe

Introduction and Dimensionless

most

PIPES

Numbers

situations involving a liquid or a gas in heat transfer, convective heat transfer

usually occurs as well as conduction. In most industrial processes

occurring, heat

is

In Fig. 4.5-1 heat

being transformed from one is

fluid

wall in the thin viscous sublayer

the fluid

is

in

fluid to the

turbulent flow,

where turbulence

is

transfer

is

cold flowing

Principles

is

fluid.

fluid.

very steep next to the

absent. Here the heat transfer

mainly by conduction with a large temperature difference of

Chap. 4

where heat

through a solid wall to a second

being transferred from the hot flowing

The temperature profile is shown. The velocity gradient, when

236

-r l /2H

T2 — T3

in the

warm

is

fluid.

of Steady-State Heat Transfer

r

7

metal wall

Figure

Temperature

4.5-1.

another.

profile for heat transfer '

by convection from one fluid to

As we move farther away from the wall, we approach the turbulent region, where rapidly moving eddies tend to equalize the temperature. Hence, the temperature gradient is less and the difference Tj — T2 is small. The average temperature of fluid A is slightly less

T A similar

than the peak value

.

x

explanation can be given for the temperature profile

The convective

is

the fluid in K, and q 2 •

ft

°F,

A

in

is

2

ft

,

fluid in

T

and

given by

K,

Tw in

(4^1)

W/m K, A is the area in m T is the bulk or Tw is the temperature of the wall in contact with 2

•

,

W.

In English units, q

is

in btu/h, h in

°F.

The type of fluid flow, whether laminar or turbulent, great effect on the heat-transfer coefficient h, which is often most of

is

2

the heat-transfer rate in

and

fluid

= kA(T ~ TJ

the convective coefficient in

average temperature of the btu/h

through a

coefficient for heat transfer

q

where h

in

/

the cold fluid.

the resistance to heat transfer

is

of the individual fluid has a called a film coefficient, since

in a thin film close to the wall.

The more

turbulent the flow, the greater the heat-transfer coefficient.

There are two main

classifications of convective heat transfer.

The

or

first is free

natural convection, where the motion of the fluid results from the density changes in heat transfer.

The buoyant

effect

produces a natural circulation of the

fluid,

the solid surface. In the second type, forced convection, the fluid

is

so

it

moves past

forced to flow by

pressure differences, a pump, a fan, and so on.

Most and are

of the correlations for predicting film coefficients h are semiempirical in nature

affected

by the physical properties of the

fluid, the

type and velocity of flow, the

temperature difference^ and by the geometry of the specific physical system.

approximate values of convective

coefficients

were presented

in

Table

Some

4.1-2. In

the

following correlations, SI or English units can be used since the equations are dimensionless.

To

correlate these data for heat-transfer coefficients, dimensionless

the Reynolds and Prandtl numbers are used.

component

of diffusivity for

momentum

The Prandtl number

is

numbers such

as

the ratio of the shear

p/p to the diffusivity for heat k/pc p and physihydrodynamic layer and thermal boundary

cally relates the relative thickness of the layer.

Sec. 4.5

Forced Convection Heat Transfer Inside Pipes

237

N Pr

Values of the

Appendix A. 3 and range from about

for gases are given in

Values for liquids range from about 2 to well over 10

number,

N Uu

,

is

4

0.5 to

The dimensionless

.

1.0.

Nusselt

used to relate data for the heat-transfer coefficient h to the thermal

conductivity k of the fluid and a characteristic dimension D.

hD —

N Nu = For example,

for flow inside a pipe,

D is

(43-3)

k

the diameter.

Heat-Transfer Coefficient for Laminar Flow Inside a Pipe

4.5B

Certainly, the most important convective heat-transfer process industrially

that of

is

cooling or heating a fluid flowing inside a closed circular conduit or pipe. Different types of correlations for the convective coefficient are needed for laminar flow (jV Rc below

(N Re above

2100), for fully turbulent flow

6000), and for the transition region (jV Re

between 2100 and 6000). For laminar flow of fluids inside horizontal tubes or Sieder and Tate (SI) can be used for N Kc < 2100: (N Nu ) a = -7-.= where /j t

=

D =

pipe diameter in m,

=

L

V

heat capacity

in

average heat-transfer coefficient

in

=

J/kg K, k •

W/m

2

fi„

=

N Nu =

in the

pipe

in

m,

viscosity at the wall

thermal conductivity

K, and

•

in

W/m

•

K,

ha

=

dimensionless Nusselt number.

All the-physical properties are evaluated at the bulk fluid

Reynolds number

(43-4)

pipe length before mixing occurs

bulk average temperature in Pa-s,

fluid viscosity at

temperature, c p

L =

p0

N Re /V P -

1.86

k

pipes, the following equation of

The

temperature except

is

N Re =

Dvp —

^

(43-5)

and the Prandtl number,

N ?r=-f

(43-6)

This equation holds for (N Rc N Pr D/ L) > 100. If used down to a(jV jV D/L) > 10, it Rc Pr holds to ±20% (Bl). For(iV Re Pr ~D/L) < 100, another expression is available (PI).

still

N

In laminar flow the average coefficient h depends strongly on heated length. The a average (arithmetic mean) temperature drop AT is used in the equation to calculate the a

heat-transfer rate

q.

q

=

ha

AAT = h a

where Tw is the wall temperature in K, outlet bulk fluid temperature.

a

A

Tbi

(T

"

~ TJ +

(T

"

~ TJ

(43-7)

the inlet bulk fluid temperature,

and

the

For large pipe diameters and large temperature differences AT between pipe wall and bulk fluid, natural convection effects can increase h(Pl). Equations are also available for laminar flow in vertical tubes.

238

Chap. 4

Principles of Steady-State

Heat Transfer

Heat-Transfer Coefficient for Turbulent Flow

4.5C

Inside a Pipe

When

number

the Reynolds

transfer

above 2100, the flow

is

many

greater in the turbulent region,

is

turbulent. Since the rate of heat

is

industrial heat-transfer processes are in

the turbulent region.

The following equation has been found

A'nu

where h L

temperature.

mean

=

h D I \ -~ = 0.027
The

If the

bulk

fluid

on the log mean driving

for

is

mean bulk

To

used.

inlet to the outlet of the pipe,

correct for an

an abrupt contraction, an approximate correction

is

force A7J m (see

n w are evaluated at the

temperature varies from the

of the inlet and outlet temperatures

the entry

(4.5-8)

(

properties except

fluid

014

3

the heat-transfer coefficient based

is

Section 4.5H).

the

to hold for tubes but is also used for 6000, a 7V Pr between 0.7 and 16 000, and LID > 60.

N Re >

pipes. This holds for a

is

L/D <

60,

where

to multiply the

right-hand side of Eq. (4.5-8) by a correction factor given in Section 4.5F.

The evaluate

use of Eq. (4.5-8)

Tw

may

and hence

,

be

trial

and

L because of heat

increases or decreases in the tube length at length

and

L must

be estimated

in

must be known to mean bulk temperature

error, since the value of h L

at the wall temperature. Also,

order to have a

if

the

transfer, the bulk

mean bulk temperature

temperature

of the entrance

exit to use.

The for a

heat-transfer coefficient for turbulent flow

smooth

tube. This effect

much

is

is

somewhat

greater for a pipe than

neglected in calculations. Also, for liquid metals that have Prandtl correlations

must be used

and it numbers

less than in fluid friction,

is

«

usually, 1,

other

to predict the heat-transfer coefficient. (See Section 4.5G.)

For

shapes of tubes other than circular, the equivalent diameter can be used as discussed

in

Section 4.5E.

For

air at

atm

1

total pressure, the

following simplified equation holds for turbulent

flow in pipes.

3.52^

h

0 8 "

L—^T-

(SI)

(4.5-9)

= 77^51 (D')°

hL

where

D

is in

h L in btu/h

Water

m, 2

ft

is

v in

T

in

m/s,

and

W/m

h L in

(English)

2 •

K

for SI units;

often used in heat-transfer equipment.

temperature range of

and

D'

is

in in.,v s in ft/s,

and

English units.

T=

4 to

105°C (40

- 220°F)

A

simplified equation to use for a

is

„0.8

hL

=

+

1429(1

0.01

46T°C)-^

(SI)

(4.5-10) hL

A

=

150(1

+

0.01

1T°F)

(English)

very simplified equation for organic liquids to use for approximations

is

as follows

(P3): I'l

= 423

(SI)

(4.5-11) hL

Sec. 4.5

=

60 7±tT2

(English)

(DT

Forced Convection Heat Transfer Inside Pipes

239

For flow inside

and

helical coils

above 10\ the predicted

jV Rc

be increased by the factor

straight pipes should

EXAMPLE

45-1.

Air at 206.8

kPa and an average

Heating of Air

mm 488.7 K

film coefficient for

3.5D/D coi] ).

Turbulent Flow

K

of 477.6

being heated as

is

flows

it

inside diameter at a velocity of 7.62 m/s.

through a tube of 25.4

medium

in

+

( 1

The

steam condensing on the outside of the tube. Since the heat-transfer coefficient of condensing steam is several thousand W/m 2 K and the resistance of the metal wall is very small, it will be assumed that the surface wall temperature of the metal in contact with the air is 488.7 K. Calculate the heat-transfer coefficient for an L/D > 60 and heating

is

•

also the heat-transfer flux q/A.

A. 3 for physical properties of air at 477.6 K Pa-s, k = 0.03894 W/m, N Pr = 0.686. At 5 2.64 x 10" Pa-s.

Solution:

From Appendix

(204.4°C),

^=2.60x10"

488.7

K (21 5.5°C), H„

=

fi

w

=

5

3

2.60 x 10"

Pa

s

The Reynolds number calculated

=

bulk

at the

fluid

_Ogg_ 6.0254(7.62X1.509) Nrc

Hence, the flow

~

is

n

~

x 10- 5

2.6

5

2.60 x 10"

turbulent and Eq. (4.5-8)

_ _

will

kg/m

•

s

temperature of 477.6

U22

K is

X 10

be used. Substituting into

Eq. (4.5-8),

A Nu = f

^-= 0.0277V»

8 e

M0.0254) — = 0.027(1.122 x 0.03894

10

... 0 8

4

)

(0.686)

n /0.0260\

0 14 '

, l/3

\0.0264 z

Solving, h L = 63.2 W/m To solve for the flux q/A,

K

(11.13 btu/h

2 •

ft

"¥).

- = hAT w - T) = 63.2(488.7 - 477.6) A = 701.1 W/m 2 (222.2 btu/h -ft 2 )

4.5D

Heat-Transfer Coefficient for Transition Flow Inside a Pipe

In the transition region for a 7V Re

between 2100 and 6000, the empirical equations are

not well defined just as in the case of fluid friction factors.

No

simple equation exists

smooth transition from heat transfer in laminar flow to turbulent a transition from Eq. (4.5-4) at a /V Re = 2100 to Eq. (4.5-8) at a 7V Re = 6000.

for accomplishing a

flow,

i.e.,

The

plot in Fig. 4.5-2 represents an

approximate relationship to use between the

number between 2100 and 6000. For below ayV Re of 2100, the curves represent Eq. (4.5-4) and above 10 4 Eq. (4.5-8). The mean AT a of Eq. (4.5-7) should be used with the h a in Fig. 4.5-2. various heat-transfer parameters and the Reynolds

,

240

Chap. 4

Principles of Steady-State

Heat Transfer

Figure

4.5E

A

Correlation of heat-transfer parameters for transition region for Reynolds numbers between 2100 and 6000. (From R. H. Perry and C. H. Chilton, Chemical Engineers' Handbook, 5th ed. New York: McGraw-Hill Book Company, 1973. With permission.)

4.5-2.

Heat-Transfer Coefficient for Noncircular Conduits

heat-transfer system often used

concentric pipes. predicted

The in

that in

same equations

using the

diameter defined

is

which

fluids flow at different

as for circular pipes.

However, the equivalent

Section 2.10G must be used. For an annular space,

minus the OD of the inner pipe equivalent diameter can also be used. the outer pipe

temperatures in

heat-transfer coefficient of the fluid in the annular space can be

£>,

D2

£>

is

cq

the

ID

of

For other geometries, an

.

EXAMPLE

4.5-2. Water Heated by Steam and Trial-and-Error Solution flowing in a horizontal 1 -in. schedule 40 steel pipe at an average temperature of 65.6°C and a velocity of 2.44 m/s. It is being heated by condensing steam at 107. 8°C on the outside of the pipe wall. The steam side

Water

is

coefficient has (a)

(b) (c)

been estimated as h a

=

W/m 2

10 500

•

K.

Calculate the convective coefficient h for water inside the pipe. Calculate the overall coefficient U based on the inside surface area. {

,

m

Calculate the heat-transfer rate q for 0.305 at an average temperature of 65.6°C.

of pipe with the water

From Appendix A. 5 the various dimensions are D-, = 0.0266 m and D a = 0.0334 m. For water at a bulk average temperature of 65.6°C k = from Appendix A.2, N Pr = 2.72, p = 0.980(1000) = 980 kg/m 3 _4 4 kg/m-s. 0.633 W/m K, and \i = 4.32 x 10~ Pa s = 4.32 x 10 The temperature of the inside metal wall is needed and will be assumed as about one third the way between 65.6 and 107.8 or 80°C = T„ for the first

Solution:

,

•

trial.

Hence, First,

•

/j„ at 80°

the

C=

4 3.56 x 10" Pa-s.

Reynolds number of the water

is

calculated at the bulk

average temperature.

D R<

Hence, the flow

~

is

;

0.0266(2.44)(980)

vP

p

~

4.32 x 10"

4

"

turbulent. Using Eq. (4.5-8)

and substituting known

values,

h,D '

Sec. 4.5

= 0.027N° e8

<^

0.14

3

Forced Convection Heal Transfer Inside Pipes

241

A L 0.0266)

-

z 4 \3.56 x 10" /

0.663

=

Solving, h L

For part

=

hi

13 324

tcD.

(

A ]m = n

=

A. for stee!

is

45.0

K.

L =

7i(0.0266)(0.305)

+

(0.0266)

W/m

K.

The

=

0.0320

m

0.0255

=

0.0287

2

m

2

2

rn

resistances are

1

=

= 0.002943

Mi

(13

_r.-r,

=

0.0334)0.305

7i(0.0334)(0.305)

1

R: =

W/m 2

the various areas are as follows for 0.305-m pipe.

(b),

A =

The k

x 10" 4 \°

,n/ 4 32 5, n9 ,/3 8 10 )°' (2.72)

= 0.027(1.473 x

=

324)0.0255

0.0334

-

0.0266

1

=

2

"•""^T

0 -°° 26?3

45.0(0.0287)

= a °02976 (10 500X0.0320)

2

= 0.002943 + 0.002633 + 0.002976 = 0.008552

The overall temperature difference is (107.8 The temperature drop across the water film is R;

temperature drop

^— 2?R

=

65.6)°C

=

=

42.2°C

42.2 K.

/0.002943\ (42.2)

=

=

(42.2)

14.5

K=

14.5°C

\0.008852j

Hence, T w = 65.6 + 14.5 = 80.1°C. This estimate of 80°C. estimate would be trial is

is quite close to the original physical property changing in the second This will have a negligible effect on h; and a second

The only pb w

.

'

not necessary.

For part

(b),

Ui

=

the overall coefficient

is,

by Eq.

(4.3-

1

6),

^—R = 0.0255(0.008552) = 4586 W/m Af 1

1

2 •

K.

2^

For part

(c),

with the water at an average temperature of 65.6°C,

Ta - T,= q

4.5F

= UiA,(T0 -

T;)

107.8

-

65.6

=

42.2°C

=

42.2

K

= 4586(0. 0255)(42. 2) = 4935

W

Entrance-Region EfTect on Heat-Transfer Coefficient

Near the entrance of a pipe where the fluid is being heated, the temperature profile is not developed and the local coefficient h is greater than the fully developed heat-transfer coefficient h L for turbulent flow. At the entrance itself where no temperature gradient has been established, the value of h is infinite. The value of h drops rapidly and is approximately the same as h L at L/D = 60, where L is the entrance length. These relations for turbulent flow inside a pipe are as follows where the entrance is an abrupt contraction. fully

242

Chap. 4

Principles

of Steady-State Heat Transfer

<-<

2

20

L

20

where h

is

<-<

the average value for a tube of finite length

(43-12)

.

60

L and

(4.5-13)

hL

is

the value for a very

long tube.

4.5G

Liquid-Metals Heat-Transfer Coefficient

Liquid metals are sometimes used as a heat-transfer over a wide temperature range in

fluid in cases where a fluid is needed low pressures. Liquid metals are often used

at relatively

nuclear reactors and have high heat-transfer coefficients as well as a high heat capacity

The high

per unit volume.

heat-trarisfer coefficients are

due to the very high thermal

conductivities and, hence, low Prandtl numbers. In liquid metals in pipes, the heat transfer by conduction

is

thermal conductivity and

For

fully

very important in the entire turbulent core because of the high often

is

more important than the convection

equation can be used (LI):

where the Peclet number 10*.

For constant

^

N Nu =

/

and

effects.

developed turbulent flow in tubes with uniform heat flux the following

0.625A&4

=

(43-14)

k

N Pc = N Re N

?r

.

This holds

for

L/D > 60 and

N Pc

between 100

wall temperatures,

rV Nu

^

=

=

5.0

+

0.025/V° c

8

(4315)

k

for

N Pc >

L/D > 60 and

100. All physical properties are

evaluated

at the

average bulk

temperature.

EXAMPLE

Liquid-Metal Heat Transfer Inside a Tube having an inside diameter of 0.05 m. The liquid enters at 500 and is heated to 505 K in the tube. The tube wall is maintained at a temperature of 30 K above the fluid bulk temperature and constant heat flux is maintained. Calculate the required tube length. The average physical properties are as follows: /i = 4 7.1 x 10" Pa-s,p = 7400 kg/m\ c p = 120J/kg-K,/c = 13W/mK.

A

4.5-3.

liquid metal flows at a rate of 4.00 kg/s through a tube

K

Solution:

G=

The area

4.0/1.963 x 10"

A = nD 2 /4 =

is 3

=

3 2.038 x 10

2

ti(0.05) /4

kg/m 2

-s.

_DG_ 0.05(2.038 R<

~

»

7.1

cB n = N = -f£ ?:

Using Eq. hL

120(7.1 x

k

1.963

x 10"

3

m2

.

xl0 3 _ 10" 4 " L

Then is

)

IP"

IU

5

4 )

=

0.00655

13

(4.5-14),

=~ =

Sec. 4.5

x

=

The Reynolds number

(0.625)Af? e4

2512

=

^

(0.625X1.435 x 10

s

x 0.00655) 0

4

W/m -K 2

Forced Convection Heat Transfer Inside Pipes

243

Using

a

heat balance, q

= mc p AT =

-

4.00(120X505

=

500)

2400

(4.5-16)

W

Substituting into Eq. (4.5-1),

— 2400

~q = Hence,/1

=

=

=

2400/75 360

A = Solving,

h L (Tw

L=

-

T)

=

3.185 x 10~

3.185 x 10

-2

2512(30) 2

m2

=

75 360

W/m 2

Then,

.

= nDL =

ji(0.05XL)

0.203 m.

Log Mean Temperature Difference and Varying

4.5H

/

Temperature Drop

s'

Equations is

and

(4.5-1)

constant for

(4.3-12) as written apply only

= ViAiW-T.) =

U.A„{T,

only holds at one point in the apparatus

However, as the and both T and (

and some mean

fluids travel

T

0

ATm

the temperature

which

reverse direction)

is

when

A

;

;

is

heated from

at the outlet of the

AT

in

T2

is

a suitable

heat balance on the hot

rn is

ATj

varies with position,

cooled from T\ to T'2 by a

to Tj as

The

in Fig. 4.5-3a.

A

(in

the

AT shown

is

goes from 0 at the

fluids as in Fig. 4.5-3a, the heat-transfer rate

= UA ATm

mean temperature and

shown

exchanger. is

(4.5-18)

difference to be determined.

For a dA area, a

the cold fluids gives

dq

where

is

AT

Eq. (4.5-17) varies as the area

q

ATm

(4.5-17)

flowing on the outside in a double pipe countercurrently

For countercurrent flow of the two

where

0)

the fluids are being heated or cooled.

T and T0 vary. Then (T — T0 ) or must be used over the whole apparatus.

or either

and

— T

through the heat exchanger, they become heated or cooled

varying with distance. Hence, inlet to

;

- Ta) = UA(AT)

In a typical heat exchanger a hot fluid inside a pipe cold fluid

drop(T

of the heating surface. Hence, the equation

all parts

q

when

flow rate in kg/s.

=

-m'c'

p

dT'

The values of m,

= mc p dT m', c

c

(4.5-19)

and

U

are

assumed constant.

n

^

T'

AT

t

AT2

AT,

AT

AT,

T2

Figure

244

Distance

Distance

(a)

(b)

4.5-3.

Temperature profiles for one-pass double-pipe heat exchangers countercurrent flow; (b) cocurrent or parallel flow.

Chap. 4

Principles of Steady-State

:

(a)

Heat Transfer

Also,

= U(T -

dq

From Eq.

(4.5-19),

dT

and dT =

dq/m'c'p

--

dT

-clT = d(T

T)

dA

dqjmc p Then,

— 1

= -dq

T)

(4.5-20)

+

(43-21)

dA

(4.5-22)

Substituting Eq. (4.5-20) into (4.5-21),

J(T-

-

T)

r -t Integrating between points

1

and

mc'

+ mc,

2,

,'Ti-T2

,

1

-U

=

1

-UA

In

(43-23) mc,

Making

a heat balance

between the q

Solving for m'c'p and

mc p

in

=

m'c'

$

difference

(T;-TJ = mc p (T2

.

Hence,

-T

l

(4.5-24)

)

in

(4.5-23),

UA[(Tj-T2 )-(T; -TM In [(T2 - T2 )/(T; - T,)]

(4.5-25)

'

we

Eqs. (4.5-18) and (4.5-25),

ATlm

outlet,

Eq. (4.5-24) and substituting into Eq.

_ Comparing

p

and

inlet

see that

the case where

AT

OT

is

the log

mean temperature

U

the overall heat-transfer coefficient

constant throughout the equipment and the heat capacity of each fluid

proper temperature driving force to use over the entire apparatus

is

is

the log

is

constant, the

mean

driving

force,

= UAATlm

q

(4.5-26)

where,

—

AT,

ATlm =

AT, (4.5-27)

In

(AT/AT,)

can be also shown that for parallel flow as pictured

It

in Fig. 4.5-3b, the

temperature difference should be used. In some cases where steam

T2 may '

be the same.

The equations

still

hold for

this case.

is

When U

log

condensing,

and

varies with distance

or other complicating factors occur, other references should be consulted (B2, P3,

EXAMPLE

mean

Tj'

Wl).

Heat-Transfer Area and Log Mean Temperature Difference A heavy hydrocarbon oil which has a c pm = 2.30 kj/kg K is being cooled in a heat exchanger from 371.9 K to 349.7 K and flows inside the tube at a rate of 3630 kg/h. A flow of 1450 kg water/h enters at 288.6 K for cooling and 4.5-4.

•

flows outside the tube. (a) Calculate the water outlet temperature and heat-transfer area (b)

U =

Repeat

for parallel flow.

;

340

if

the

W/m 2 K and the streams are countercurrent.

overall

•

Assume a c pm = 4.187 kj/kg K for water. The water inlet T2 = 288.6 K, outlet = T,; oil inlet T[ = 371.9, outlet T2 = 349.7 K. Calculating the heat lost by the oil,

Solution:

-

q

=

= Sec. 4.5

(

3630

^Y

2.30

185400 kJ/h

r-^— kg-K or

)(371.9

51 490

-

349.7)K

W (175 700 btu/h)

Forced Convection Heat Transfer Inside Pipes

245

By

a

heat balance, the q must also equal the heat gained by the water.

=

q

=

Solving, T,

To 349.7

185400 kJ/h = ^1450

288.6

=

~

k^K^ 71

'

319.1 K.

288

'

6)

K

..

mean temperature

solve for the log

-

4 187

y)(

=

AT,

61.1 K,

T[

-

=

T,

difference,

371.9

-

AT = 2

=

319.1

T{

— T2 =

52.8 K. Substi

tutinginto Eq. (4.5-27),

AT.-AT, ATlm _ Using Eq.

=

/!,-

For part 4.5-3b,

"

"

In (61.1/52.8)

(4.5-26),

490

51

Solving,

61.1-52.8 _

(ATj/ATj)

In

AT = 2

2.66

m2

(b),

=

.

the water outlet

371.9

340(/l,X56.9)

-

288.6

=

T =

still

is

K

83.3

l

319.1 K. Referring to Fig.

and AT, = 349.7

-

319.1

=

30.6 K.

Again, using Eq. (4.5-27) and solving, ATj m = 52.7 K. Substituting into Eq. 2 This is a larger area than for counterflow. This (4.5-26), A = 2.87 occurs because counterflow gives larger temperature driving forces and is

m

;

.

usually preferred over parallel flow for this reason.

EXAMPLE

45-5. Laminar Heat Transfer and Trial and Error C hydrocarbon oil at 150 F enters inside a pipe with an inside diameter of 0.0303 ft and a length of 15 ft with a flow rate of 80 Ibjh. The inside pipe surface is assumed constant at 350°F since steam is condensing outside the pipe wall and has a very large heat-transfer coefficient. The properties of the oil are c pm = 0.50 btu/lb m ~F and k m = 0.083 btu/h ft °F. The viscosity of

A

•

•

the oil varies with temperature as follows: 150°F, 6.50 cp; 200"F, 5.05 cp;

250°F, 3.80 cp; 300°F, 2.82 cp; 350°F, 1.95 cp. Predict the heat-transfer and the oil outlet temperature, Tbo

coefficient

.

This

Solution:

is

a trial-and-error solution since the outlet temperature of

unknown. The value of Tbo = 250°F will be assumed and checked later. The bulk mean temperature of the oil to use for the physical properties is (150 + 250)/2 or 200°F. The viscosity at 200°F is the oil

Tbo

is

=

/ib

At

12.23

^

4.72

^

the wall temperature of 350T,

ti

The

w

=

1.95(2.4191)

A

cross-section area of the pipe

4

Rc

_D

;

vp

n

0.000722ft'

2

Dj(j

_

-111 000

lb "

~-

ft

ft

at the bulk

_

is

4

0.000722

The Reynolds number

=

«^ M,

X=^ = G^A

246

=

5.05(2.4191)

2 •

h

mean temperature

0.0303(1

1

1

is

000)

12.23

n

Chap. 4

Principles

of Steady-State Heat Transfer

The Prandtl number

is

yVpr

050(1123)

_ "

_ " fe

~

0.083

J

'

N Re is below 2100, the flow is in the laminar region and Eq. (4.5-4) be used. Even at the outlet temperature of 250°F, the flow is still laminar. Substituting, Since the will

(Nn„).

KD

=

1/3

h a (0.0303)

12.23

1.86

4.72

0.083

W/m

2

Solving, h, = 20. 1 btu/h ft °F (1 14 Next, making a heat balance on the •

=

1

Using Eq.

-

mc(T

b0

- Tbi) =

2 •

K).

oil,

80.0(0.50)(Tbo

-

(4.5-28)

150)

(4.5-7),

q

=

ha

AATa

(4.5-7)

For AT. (T,

- Tbi) +

(350

= Equating Eq. (4.5-28) 80.0(0. 50)(

275

to (4.5-7)

Tb0 -

150)

-

150)

+

- TJ

(350

- TJ

0.5Tto

and substituting,

= MAT; =

20.1[7t(0.0303K15)](275

-

0.5

TJ

Tb0 =

Solving,

This

is

255°F. higher than the assumed value of 250°F. For the second trial bulk temperature of the oil would be (150 + 255)/2 or

mean The new

the

-

(T„

202. 5°F.

viscosity

is

5.0

mate. This only affects the(^ t //j J 0

(N Rc )(N ft )

effect in the

'

cp compared with 5.05 for the first esti14 factor in Eq. (4.5-4), since the viscosity

factor cancels out.

The

heat-transfer coefficient will

change by less than 0.2%, which is negligible. Hence, the outlet temperature of T, = 255°F(i23.9°C) is correct.

HEAT TRANSFER OUTSIDE VARIOUS GEOMETRIES IN FORCED CONVECTION

4.6

4.6A In

Introduction

many

cases a fluid

is

flowing over completely immersed bodies such as spheres, tubes,

occurring between the

plates,

and so

Many

of these shapes are of practical interest in process engineering. The sphere,

cylinder,

and

these surfaces

When Sec. 4.6

on,

and heat transfer

is

flat

plate are perhaps of greatest

and

a

moving

fluid frequently

fluid

and the

solid only.

importance with heat transfer between

encountered.

heat transfer occurs during immersed flow, the flux

Heal Transfer Outside Various Geometries

in

is

dependent on the

Forced Convection

247

geometry of the body, the position on the body

(front, side,

other bodies, the flow rate, and the fluid properties.

The average

over the body.

heat-transfer coefficient

The

back,

etc.),

the proximity of

heat-transfer coefficient varies

given in the empirical relation-

is

ships to be discussed in the following sections.

In general, the average heat-transfer coefficient on immersed bodies

is

given by

N Ha = CNi N}P

(4.6-1)

t

where

C

m

and

depend on the various configurations. The fluid temperature 7} = (Tw + Tb )/2, where Tw is the surface

are constants that

properties are evaluated at the film

Tb

or wall temperature and

the average bulk fluid temperature.

The

velocity in

theN Rc

is

the undisturbed free stream velocity v of the fluid approaching the object.

4.6B

Flow

When

the fluid

the 3

Parallel to Flat Plate

flat plate and heat transfer is occurring between and the fluid, the 7V Nu is as follows for a N Rc L below the laminar region and a7V Pr > 0.7.

flowing parallel to a

is

L

whole plate of length

x 10 5

in

m

N Nu = where

=

jV Rc l

<

0.664N° C5 L

/3

(4.6-2)

Lvp/p.

,

For the completely turbulent region

N Rc

a

at

above

L

3

x 10 5 (Kl, K3) and

N Pr >0.7, =

N^ 3

0.0366iV° t8 L .

(4.6-3)

However, turbulence can start at a N Ke L below 3 x 10 5 if the plate is rough (K3) and then Eq. (4.6-3) will hold and give aN Nu greater than by Eq. (4.6-2). Below about aN Rti L 4 of 2 x 10 Eq. (4.6-2) gives the larger value of N Nu .

,

EXAMPLE 4.6-1. A

smooth,

Cooling a Copper Fin

mm square. Its temperature at

1

mm

from a tube is 51 by 51 approximately uniform at 82.2°C. Cooling air

thin fin of copper extending out

flat,

is

1 atm abs flows parallel to the fin at a velocity of 12.2 m/s. For laminar flow, calculate the heat-transfer coefficient, h. If the leading edge of the fin is rough so that all of the boundary

5.6°C and (a)

(b)

layer or film next to the fin

The

Solution:

Tf =

(Tw

+

fluid properties will

completely turbulent, calculate

be evaluated at the film temperature

=

L±J± =

82-2+15.6

=

48.9°C (322.1 K)

physical properties of air at 48.9°C from

W/m nolds

•

h.

73/2.

Tf

The

is

K, p

=

number

1.097 is,

for

kg/m 3 p =

1.95

,

L =

x 10"

5

Appendix A.3 are k = 0.0280 Pa -s, N Pr = 0.704. The Rey-

0.051 m,

Lvp

(0.051X12.2X1.097)

~ ix

1.95

x 10" 5

~

^

^

10

Substituting into Eq. (4.6-2),

N Nu

=y= 0.664N^ =

Solving, h

248

=

60.7

Np 3 ;

0.664(3.49 x

W/m 2 K (10.7 btu/h •

L

10T-

2 •

Chap. 4

ft

•

5

(0.704)"

3

°F).

Principles of Steady-State

Heal Transfer

For 77.2

4.6C

W/m

part 2 •

K

(b),

substituting

(13.6 btu/h

2 •

ft

Often a cylinder containing a fluid inside

m

and

(4.6-3)

h

solving,

=

Cylinder with Axis Perpendicular to Flow

perpendicular to ficient

Eq.

into

°F).

•

its

axis.

The equation

is

being heated or cooled by a fluid flowing

of the outside of the cylinder for gases and liquids

as given in

Table

4.6-1.

The

7V Rc

=

average heat-transfer coef-

for predicting the

Dvp/p, where

D

is

is

C and

(K3, P3) Eq. (4.6-1) with

the outside tube diameter and

physical properties are evaluated at the film temperature 7}.

The

velocity

is

all

the undis-

turbed free stream velocity approaching the cylinder.

4.6D

Flow Past Single Sphere

When

a single sphere

is

being heated or cooled by a fluid flowing past

equation can be used to predict the average heat-transfer coefficient for 1

to 70 000

and a

N

Pr

the following

= Dvp/p

of

of 0.6 to 400.

=

/V Nu

The fluid

it,

a.-N Rc

2.0

+

0.60/Vg;

5

A^ /3

(4.6-4)

r

properties are evaluated at the film temperature 7} A somewhat more accurate is available for a 7V Rc range 1-17 000 by others (S2), which takes into account .

correlation

the effects of natural convection at these lower

Reynolds numbers.

EXAMPLE 4.6-2.

Cooling of a Sphere Using the same conditions as Example 4.6-1, where air at 1 atm abs pressure and 15.6°C is flowing at a velocity of 12.2 m/s, predict the average heattransfer coefficient for air flowing by a sphere having a diameter of 51 and an average surface temperature of 82.2°C. Compare this with the value

mm

of/i

=

77.2

W/m -K for the flat plate in 2

turbulent flow.

The physical properties at the average film temperature of 48.9°C are the same as for Example 4.6-1. The N Rc is

Solution:

N Rt

Dvp _ =^ = ~

(0.05 1)( 12.2X1 .097)1 1

p.

Table

4.6-1.

.I?.*-, x 10"

=

3-49 x 10*

1.95

Constants for Use

in

Eq. (4.6-1) for Heat

Transfer to Cylinders with Axis Perpendicular to

Sec. 4.6

Flow (N ?r

>

0.6)

N Re

m

C

1-4

0.330

0.989

4-40

0.385

0.911

40-4 x 10 3 3 4 x 10 -4 x 10* 4 s 4 x 10 -2.5 x 10

0.466

0.683

0.618

0.193

0.805

0.0266

Heat Transfer Outside Various Geometries

in

Forced Convection

249

Substituting into Eq. (4.6-4) for a sphere,

hD _

/i(0.051

= =

Solving, h

56.1

=

0.60N°: 5 Nl'

+

2.0

3

0.0280

k

2.0

W/m K 2

•

smaller than the value of h

+

(0.60X3.49 x 10

(9.88 btu/h

=

W/m

77.2

2 •

ft

2 •

•

K

4 )°- 5

°F).

(0.704)

1/3

This value

(13.6 btu/h

2 •

ft

is

somewhat

°F) for a

flat

plate.

4.6E

Flow Past Banks of Tubes or Cylinders

Many

types of commercial heat exchangers are constructed with multiple rows of tubes,

where the fluid flows at right angles to the bank of tubes. An example is a gas heater in which a hot fluid inside the tubes heats a gas passing over the outside of the tubes.

Another example

is

a cold liquid stream inside the tubes being heated by a hot fluid on

the outside.

Figure 4.6-1 shows the arrangement for banks of tubes in-line and banks of tubes staggered where

D

OD

tube

is

m

in

(ft),

S„

is

distance

tubes normal to the flow, and S p parallel to the flow. tubes

is

-

(S n

C and m

D) and (S p

-

D),

and

for staggered tubes

m

between the centers of the

(ft)

The open area it is

(S„

-

£>)

to flow for in-line

and(S'p

-

D).

Values

Reynolds number range of 2000 to 40000 for heat transfer to banks of tubes containing more than 10 transverse rows in the direction of flow are given in Table 4.6-2. Fdr less than 10 rows, Table 4.6-3 gives correction of

to be used in Eq. (4.6-1) for a

factors.

For

where

cases

SJD

consult Grimison (Gl) for

leakage where

all

and

SJD

more

are not equal to each other, the reader should

where there

data. In baffled exchangers

obtained should be multiplied by about 0.6 (P3). The Reynolds using the

minimum

evaluated at

Tf

is

normal

the fluid does not flow normal to the tubes, the average values of h

number

is

calculated

area open to flow for the velocity. All physical properties are

.

EXA MPLE 4.6-3.

Heating

A ir by a Bank of Tubes

15.6°C and 1 atm abs flows across a bank of tubes containing four transverse rows in the direction of flow and 10 rows normal to the flow at a Air

at

flow

Sn

-D

(a)

Figure

4.6-1.

Nomenclature for banks of tubes (b)

250

in

Table 4.6-2

:

(a) in-line

tube rows,

staggered tube rows.

Chap. 4

Principles

of Steady-State Heat Transfer

Table

Values of C and m To Be Used in Eq. (4.6-1) for Heat Transfer to Banks of Tubes Containing More Than

4.6-2.

10 Transverse s„

s„

D

D

_2

E

=

Rows =

1.25

D

1.50

D

D

=

2.0

D

C

m

C

m

C

m

In-line

0.386

0.592

0.278

0.620

0.254

0.632

Staggered

0.575

0.556

0.511

0.562

0.535

0.556

Arrangement

Source

:

ASM E, 59, 583 (1937).

D. Grimison, Trans.

E.

m/s as the

velocity of 7.62

air

approaches the bank of tubes. The tube The outside diameter of the tubes is

surfaces are maintained at 57.2°C.

mm and the tubes are in-line to the flow. The spacing S„ of the tubes normal to the flow is 38.1 mm' and also S p is 38.1 mm parallel to the flow. For a 0.305 m length of the tube bank, calculate the heat-transfer rate. 25.4

Solution:

Referring to Fig. 4.6- la,

S

_38J_L5

2

D ~

25 A

'

Sp

38.1

D ~

1

_

25.4

1.5

~Y

air is heated in passing through the four transverse rows, an outlet bulk temperature of 21.1°C will be assumed. The average bulk temperature is then

Since the

Tl= ii6±iilw The average

film

temperature

7/ /

From Appendix

=

24^7.2 -M8.3 = 37 rc 2

p

=

=

0.02700

1.0048 li

Ratio ofh for (for

2

0

c

4.6-3.

is

A. 3 for air at 37.7 C,

k

Table

l8 .3.C

N

•

1.90 x

Transverse

,

p=

kJ/kg-K

=

NP =

W/m K 10"

5

0.705

1.137

Pa

Rows Deep

kg/m 3

-s

to h for

10 Transverse Rows Deep

Use with Table 4.6-2)

N Ratio for

123456789

10

0.68

0.75

0.83

0.89

0.92

0.95

0.97

0.98

0.99

1.00

0.64

0.80

0.87

0.90

0.92

0.94

0.96

0.98

0.99

1.00

staggered tubes

Ratio

for

in-line tubes Source: W.

Sec. 4.6

M.Kays and

R. K. Lo, Stanford Univ. Tech. Kept. 15,

Navy Contract N6-ONR-251

Heat Transfer Outside Various Geometries

in

Forced Convection

T.O.6, 1952.

251

The (S n

—

of the minimum-flow area to the The maximum velocity in the tube banks

ratio

D)/S n

.

uS„

7.62(0.0381)

s„-D Do m „p H For and

SJD =

=

Sp/D

0.0254(22.86)(1.137)

~

values of

1.5/1, the

-

xlO" 5

1.90

C and m

is

then

is

2Zg6m/s

- 0.0254)

(0.0381

area

frontal

.total

j

-

4/xlu

from Table 4.6-2 are 0.278 and solving for h,

0.620, respectively. Substituting into Eq. (4.6-1)

=|

h

=

CNINU 3

=

(j^j

4 3 (0.278X3.47 x 10 )°-"(0.705)"

W/m 2 K

171.8

This h is for 10 rows. For only four rows in the transverse direction, the h must be multiplied by 0.90, as given in Table 4.6-3. Since there are 10 x 4 or 40 tubes, the total heat-transfer area per 0.305

m length

is

A = 40kDL = The

total

40ti(0.0254)(0.305)

=

m2

0.973

heat-transfer rate q using an arithmetic average temperature and the bulk fluid is

difference between the wall

= hA(Tw - Tb =

q

)

18.3)

=

5852

W

Next, a heat balance on the air is made to calculate its temperature rise using the calculated q. First the mass flow rate of air m must be

AT

calculated.

The

total frontal area of the tube

m long

tubes each 0.305

=

A,

The density of rate

-

(0.90 x 171.8X0.973X57.2

m

bank assembly of

10

rows of

is

10S„(1.0)

=

10(0.0381X0.305)

the entering air at 15.6°C

is

p

=

=

0.1162

m

2

kg/m 3 The mass flow

1.224

.

is

m =

up/4,(3600)

For the heat balance

the

=

=

7.62(1.224X0.1162)

mean

c

p

of air

at

18.3°C

1.084 kg/s is

1.0048

kJ/kg-K and

then q

Solving,

AT =

=

5852

= mc p AT =

1.084(1.0048 x 10

3 )

AT

5.37°C.

Hence, the calculated outlet bulk gas temperature is 15.6 + 5.37 = 20.97°C, which is close to the assumed value of 21.1°C. If a second trial were to be made, the new average Tb to use would be (15.6 + 20.97)/2 or 18.28°C.

4.6F

Heat Transfer

for

Flow

in

Packed Beds

Correlations for heat-transfer coefficients for packed beds are useful in designing fixed-

bed systems such as catalytic reactors, dryers for

solids,

and pebble-bed heat exchangers.

3.1C the pressure drop in packed beds was considered and discussions of the geometry factors in these beds were given. For determining the rate of heat transfer in In Section

packed beds

for

a differential length dz

dq

252

=

in

h(a

m, S dzXTt

Chap. 4

- T2

)

Principles

(4.6-5)

of Steady-State Heat Transfer

where a

the solid particle surface area per unit volume of bed in

is

cross-sectional area of bed in

m\

m -1

S the empty

,

T

the bulk gas temperature in K, and

Tj

2

the solid

surface temperature.

For

the heat transfer of gases in beds of spheres (G2,

G3) and a Reynolds number

"

range of 10-10000,

-

(4.6-6)

where

the superficial velocity based on the cross section of the empty container in

is

v'

m/s [see Eq. superficial film

(3.1-11)], e

mass

temperature with others

for a fluidized bed.

An

N Rc = D P G'j\i s

the void fraction,

is

kg/m 2

velocity in

•

s.

at the

The

,

and G'

=

v'p

the

is

subscript /indicates properties evaluated at the

bulk temperature. This correlation can also be used

alternate equation to use in place of Eq. (4.6-6) for fixed and

is Eq. (7.3-36) for a Reynolds number range of 10-4000. The term J H is Colburn J factor and is defined as in Eq. (4.6-6) in terms of/;. Equations for heat transfer to noncircular cylinders such as a hexagon, etc., are given elsewhere (HI, Jl, P3).

fluidized beds

called the

4.7

NATURAL CONVECTION HEAT TRANSFER Introduction

4.7A

Natural convection heat transfer occurs when a solid surface liquid

which

arising

is

at a different

is

in contact with a

temperature from the surface. Density differences

from the heating process provide the buoyancy force required is observed as a result of the motion of the

Free or natural convection

of heat transfer by natural convection

encountering the radiator forces.

The

theoretical

is

heat

is

move the fluid. An example

fluid.

a hot radiator used for heating a room. Cold air

heated and rises in natural convection because of buoyancy

derivation of equations for natural convection heat-transfer

coefficients requires the solution of

An

is

to

gas or

in the fluid

motion and energy equations.

important heat-transfer system occurring

in

process engineering

is

which

that in

being transferred from a hot vertical plate to a gas or liquid adjacent to

natural convection.

The

fluid

free convection. In Fig. 4.7-1

boundary

layer

is

is

not

moving by

the vertical

ary layer

is

plate

is

zero and also

layer since the free-stream velocity

initially

by

heated and the free-convection

formed. The velocity profile differs from that in a forced-convection

system in that the velocity at the wall

boundary

flat

it

forced convection but only by natural or

is

is

is

zero at the other edge of the

zero for natural convection.

The bound-

laminar as shown, but at some distance from the leading edge

it

starts

become turbulent. The wall temperature is T„ K and the bulk temperature Tb The differential momentum balance equation is written for the x and y directions for the control volume (dx dy 1). The driving force is the buoyancy force in the gravitational field and is due to the density difference of the fluid. The momentum balance becomes to

.

(4.7-1)

where p b

is

the density at the bulk temperature

be expressed

in

T

b

and p

at T.

The

terms of the volumetric coefficient of expansion

/?

density difference can

and substituted back

into Eq. (4.7-1).

P

Sec. 4.7

= Pb~ P

Natural Convection Heat Transfer

(4.7-2)

253

For

gases,

=

[i

1

/T.

The energy-balance equation can be expressed

dT

dT\ y

dx

The

d

2

as follows

T

k

dy

dy J

solutions of these equations have been obtained by using integral methods of

analysis discussed in Section 3.10. Results for a vertical plate have

been obtained, which most simple case and serves to introduce the dimensionless Grashof number discussed below. However, in other physical geometries the relations are too complex and empirical correlations have been obtained. These are discussed in the following sections. is

the

Natural Convection from Various Geometries

4.7B

1.

Natural convection from vertical planes and cylinders. For an isothermal vertical L less than m (P3), the average natural convection

surface or plate with height

1

heat-transfer coefficient can be expressed by the following general equation

hL

where a and

in

viscosity in

capacity

in

c„u\ m

fluid or vice

J/kg K, •

/?

versa in K, k the thermal conductivity in

W/m

•

K,

c

p

the heat

the volumetric coefficient of expansion of the fluid in 1/K [for

and g is 9.80665 m/s 2 All the physical properties are evaluated at temperature Tf = (Tw + Tb )/2. In general, for a vertical cylinder with length L m,

gases P film

AT

are constants from Table 4.7-1, /V Gr the Grashof number, p density in kg/ms, A 7' the positive temperature difference between the wall

kg/m 3 fi and bulk ,

{L3 p 2 g8

is

1/(7} K)],

.

the the

same equations can be used as for a vertical plate. In English units /? is 1/(7}°F + 460) in 2 1/°R and g is 32.174 x (3600) 2 ft/h The Grashof number can be interpreted physically as a dimensionless number that represents the ratio of the buoyancy forces to the viscous forces in free convection and plays a role similar to that of the Reynolds number in forced convection. .

EXAMPLE 4.7-1.

Natural Convection from Vertical Wall of an Oven 1.0 ft (0.305 m) high of an oven for baking food with the surface at 450°F (505.4 K) is in contact with air at 100°F (311 K). Calculate the heat-transfer coefficient and the heat transfer/ft (0.305 m) width of wall. Note that heat transfer for radiation will not be considered. Use English and SI units.

A

FIGURE

heated vertical wall

4.7-1.

Boundary-layer file for natural

velocity

pro-

convection heat

transfer from a heated, vertical

u» 0

plate.

x

I 254

Chap. 4

Principles of Steady-State

Heat Transfer

Table

Constants for Use with Eq. (4.7-4) for Natural Convection

4.7-1.

Physical Geometry

[vertical height

L<

1

m

a

m (3

Kef.

ft)]

<10 4 4

10 -10

^6

1

9

l

fPT>

5 l

Z

>10 9

U.I 5

l

(Mi;

I

(Ml)

n

(W)

Horizontal cylinders

[diameter

D

used for

L and D <

m

0.20

(0.66

ft)]

<10" 5 10" 5 -10- 3 10" 3 -l ./'

4

1

i

/ i

HQ

1

1-10

n 1

U.

C\Q .uv

4 9 10 -10

i

T5 l

I l

4

>10 9

n U.

t 1

i J

1

(P3)

(Ml) (ST J)

Horizontal plates 5 1 10 -2 x 10 7 2 x 10 -3 x 10'°

Upper surface of heated plates or lower surface

0.54 0.14

1

4 1

(Ml)

(Ml)

of cooled plates

Lower surface of heated

10

5

-10 n

0.58

1

7

(Fl)

upper surface

plates or

of cooled plates

The

Solution:

T/=

film

temperature

is

Ld^ = 450 + 100 = 2?50p =

505.4 4-311

=

^

R

275°F are = 0.0198 btu/h ft °F, 0.0343 K; p = 0.0541 lbjft 3 0.867 kg/m 3 /V Pr = 0.690; y. = (0.0232 cp) x 5 3 (2.4191) = 0.0562 lbjft h = 2.32 x 10" Pa- s; /? = 1/408.2 = 2.45 x 10~ 3 10" AT = Tw - Tb = 450 K"\ p = 1/(460 + 275)= 1.36 x °R~ 100 = 350°F (194.4 K). The Grashof number is, in English units,

The

physical properties of air at

W/m

•

fc

,

;

•

1

;

2 2 3 (1.0) (0.0541) (32.174X3600) (1.36

L3 p 2 gPAT '

Gr

~

=

~

2 /i

1.84 x 10

(0.0562)

x 10' 3 X350)

2

8

In SI units, 3

/V

_

\v.ovi) \v~>yJJ/ (0.305) (0.86 7)

2

x 10" 3 X194.4) ^.ouua^-' (9.806X2.4 5 * AjfZiV =

(2.32

The Grashof numbers be the same as shown.

x 10"

84 x 10 8

calculated using English and SI units must, of course,

N c ,N = Pr

3

(1.84 x 10 X0.69O)

Hence, from Table 4.7-1, a

Sec. 4.7

i

=

0.59

Natural Convection Heat Transfer

and

=

8 1.270 x 10

m =\

for use in Eq. (4.7-4).

255

Solving for h

a(N Cr

known

Eq. (4.7-4) and substituting

in

N

0.0198 Pr

T

8 (0.59X1.270 x 10 )

values,

1 '

4

=

1.24 btu/h-

2 ft

•

°F

1.0

0.0343

x 10 8 )" 4 =-7.03

(0.59)(1.27

W/m 2 K •

0.305

For a

1

A

-ft

width of wall, A q

= hA{Tw -

q

=

= 1x1 =

T„)

7.03(0.305

=

1.0

2 ft

(0.305 x 0.305

-

(1.24X1.0)(450

x 0.305X194.4) =

considerable amount of heat in Section 4.10.

127.1

will also

=

100)

m2

).

Then

433 btu/h

W

be lost by radiation. This

will

be considered

Simplified equations for the natural convection heat transfer from air to vertical

planes and cylinders at

atm abs pressure are given in Table 4.7-2/- In SI units the 4 of 10 to 10 9 is the one usually encountered and this 3 3 (L AT) values below about 4.7 m K and film temperatures between 255 and 1

equation for the range of holds for

533 K.

To

N G! N P

,

•

correct the value of h to pressures other than

Table

4.7-2.

1

atm, the values of h

Simplified Equations for Natural Convection

in

Table

from Various

Surfaces

Equation

=

h

btu/h ft

L =fi,

= W/m 2 K L = m, AT = K D=m

°F °F

h

D=ft

Physical Geometry

Air at 101.32 Vertical planes

2

AT =

kPa

>10

cylinders

Horizontal cylinders

atm) abs pressure

(1

4 9 10 -10

and

h

9

Kef.

h

3 9 10 -10

h

>10 9

h

= = = =

0.28(AT/L) 1/4 h 0.18(AT)

I/3

h

0.27(A T/D) 1/4

h

0.18(AT) I/3

h

0.27(AT/L) 1/4

h

h

= =

x 10 5 -3 x 10 10 h

=

0.12(AT/L) 1/4

= = = =

1.37(AT/L)" 1.24

AT

1 '

1.32(AT/Z)) 1.24

AT

1 '

4

3

(PI) (PI)

1/4

3

(Ml) (Ml)

Horizontal plates 5 7 10 -2 x 10

Heated plate facing upward or cooled

7

x 10 -3 x 10

h 10

0.22(AT)

1/3

1.32(AT/L) 1/4 (Ml)

h

= =

h

=

0.59(AT/L) 1/4 (Ml)

1.52

AT

1

'

3

(Ml)

plate facing

downward Heated plate facing

downward

3

or

cooled plate facing

upward Water

Vertical planes

and

4 10 -10 9

at

70°F (294 K) 1/4 h = 26(AT/L)

h

127(AT/L) 1/4

(PI)

h

59(AT/L) 1/4

(PI)

cylinders

Organic liquids Vertical planes

and

4 10 -10 9

at

h

70°F (294 K) 12(AT/L) 1/4

=

cylinders

256

Chap. 4

Principles of Steady-State

Heat Transfer

4.7-2 can be multiplied

by (p/101.32) 1/2

N Cr N Pr >

=

10

4

to 10

10

9

is

9 ,

where p

(I?

N Gl N ?

kN/m

pressure in

encountered when

for

AT)

is

2

4

to 10

9

and by

(p/101.32)

In English units the range of

.

less

10

,

2/3

N Gr jV

for

Pr

of

than about 300ft 3 °F. The value of h can

atm abs by multiplying the h at 1 atm byp 1 2 for N Gr N Pr of 10 to 10 and by p for N Gr N Pr above 10 9 where p = atm abs pressure. Simplified equations are also given for water and organic liquids. be corrected to pressures other than 4

9

'

1.0

2/3

,

EXAMPLE 4.7-2. Repeat Example Solution:

The

Natural Convections and Simplified Equation

4.7-1 but use the simplified equation-.

film

temperature of 408.2

L AT = 3

This

is

slightly greater than

maximum

is

below

=

= hA(T„ - Tb ) =

This value 2.

is

10?-,*

6.88

heat-transfer rate q

q

(

K

the range 255-533 K. Also,

in

is

=

194.4)

5.5

the value of 4.7 given as the

for use of the simplified equation.

value of N Gr iV Pr will be used.

The

(0.305)

3

However,

in

approximate

Example

4.7-1 the

so the simplified equation from Table 4.7-2

W/m 2 -K

2

(1.21 btu/h

ft

-

°F)

is

=

6.88(0.305 x 0.305X194.4)

reasonably close to the value of 127.1

124.4

W (424 btu/h)

W for Example 4.7-1.

Natural convection from horizontal cylinders. For a horizontal cylinder with an D m, Eq. (4.7-4) is used with the constants being given in Table 4.7-1.

outside diameter of

The diameter D is used for L in the equation. Simplified equations are given 4 4.7-2. The usual case for pipes is for the Nq t N'pt range 10 to 10 9 (Ml). 3.

Natural convection from horizontal

used with the constants given

in

plates.

Natural convection

number

in

enclosed spaces.

of processing applications.

plates Eq. (4.7-4)

Table

is

also

Table

4.7-2.

mean

of the

the diameter of a circular disk.

Free convection

One example

inside these enclosed spaces are

in

side of a square plate, the linear

is

in

enclosed spaces occurs in a

an enclosed double window in for energy conservation. The flow

in

which two layers of glass are separated by a layer of air

phenomena

flat

Table 4.7-1 and simplified equations

The dimension L to be used is the length of a two dimensions for a rectangle, and 0.9 times 4.

For horizontal

in

complex since a number of different types of is mainly by conduc-

flow patterns can occur. At low Grashof numbers the heat transfer tion across the fluid layer.

As the Grashof number

is

increased, different flow regimes are

encountered.

The system for two vertical plates of height L m containing the fluid with a gap of shown in Fig. 4.7-2, where the plate surfaces are at T, and T2 temperatures. The Grashof number is defined as <5

m

is

*pW,-T,) „ a< _ The Nusselt number

is

defined as

N Nu .,=j

Sec. 4.7

(4 ,. 5)

Natural Convection Heat Transfer

(4.7-6)

257

The heat

flux

calculated from

is

= MT, - T2 The physical properties

are

evaluated

all

at the

(4.7-7)

)

mean temperature between

the two

plates.

For gases enclosed between

JV N „

.

^Nu.a

=

hS — =1.0

=

0.20

N

1 '

0.073

a

(H

3

1,

J1,K1, PI),

x 10 3 )

(4.7-8)

4

l

,™ 9

;'

< N Cz } N P! <

x 10 3

(6

,

(Nc/pJ" (L/.5)

For liquids

<2

(N Cl .»N Pt

r

and L/3 >

(L/<5)"

N Nu = .

(N °

vertical plates

3 2 x 10 )

(4.7-9)

3

- (2

1/9

x 10

5

<

jV Gr

a

7V Pr

<

2 x 10

7

(4.7-10)

)

in vertical plates,

N Nu = — =1.0 /i<5

.

(N Gri(S /V Pr

4

<

1

x 10 3 )

(4.7-11)

k

N Nu = .

4

C

0.28

d

;;

(1

,J;i

x 10 3

-

< N Gr } N Pr <

1

x 10 7 )

(4.7-12)

{L/b)

For gases or liquids

For gases

annulus, the same equations hold as for vertical plates.

in a vertical

in horizontal plates

with the lower plate hotter than the upper,

Nnu.^-O-ZKNg,.^,,)" 4

N Nu ., = For liquids /V Nu

,

a

0.061(/V G ,

4

(7

N Pr )"

3

x 10 3

<

/V Gr

(N Cr ,,N Pr >

,

a

N <3 Pr

3 x 10

x 10 5 )

(4.7-13)

5

(4.7-14)

)

in

horizontal plates with the lower plate hotter than the upper (G5),

=

0.069(yV Gr

,

iS

N Pr

1/3 )

N°

074 r

(1.5

x 10 5

<

/V Gr ,,N Pr

<

1

x 10

9 )

(4.7-15)

EXAMPLE 4.7-3.

Natural Convection in Enclosed Vertical Space atm abs pressure is enclosed between two vertical plates where L = 0.6 m and 5 = 30 mm. The plates are 0.4 m wide. The plate temperatures are T, = 394.3 K and T2 = 366.5 K. Calculate the heat-transfer rate Air at

1

across the

Figure

air

4.7-2.

gap.

Natural convection

in enclosed

////// /.

vertical space.

7%

/77T77/

258

Chap. 4

Principles of Steady-Stale

Heat Transfer

The mean temperature between

Solution:

=

physical properties. 7}

=

5

(T,

+ T2 )/2 =

the plates

(394.3

30/1000= 0.030 m. From Appendix Pa-s, /c = 0.03219 -3 _1 2.629 x 10 K

x 10"

2.21

=

1/380.4

W/m

5

.

•

+

=

JV Pr

=

G '-

"

Also, /V Gr ,,iV Pr

3.423 x 10

=

~S

=

q

{L/5)

(0.6

366.5)

)

Np

3 1

'

=

2.372 x 10* Using Eq.

4 4 0.03219(0.20X2.352 x 1Q )"

1/4 r)

~

9

(4.7-9),

0.030(0.6/0.030)

1/9

W/m 2 -K

1.909

The area A =

,

-

5 2

4

x 10 4 )0.693

(3.423

(0.20)(jV Gf

=

x 10"

(2.21

=

=

,

.

~

i

p.

0=1/7} =

0.693,

3 2 3 (0.030) (0.9295) (9.806X2.629 x 10~ X394.3 1

evaluate the

to

380.4 K. Also,

p = 0.9295 kg/m\

A.3,

K,

used

is

366.5)/2

x 0.4)

=

0.24

= T2 =

hA(l\

)

m

2

Substituting into Eq.

.

1.909(0.24)(394.3

-

366.5)

=

(4.7-7),

12.74

W

For spheres, blocks, and other types of enKl, Ml, PI, P3) should be consulted. In forced over a heated surface at low velocity in the laminar

Natural convection from other shapes.

5.

closed air spaces, references elsewhere (HI

some cases when a fluid is region, combined forced-convection further discussion of this, see (HI,

plus natural-convection heat transfer occurs. For

Kl, Ml).

BOILING AND CONDENSATION

4.8

4.8A

Boiling

Mechanisms of boiling. and

1.

,

ation and distillation

Heat

transfer to a boiling liquid

also in other kinds of chemical

is

very important in evapor-

and biological processing, such

as petroleum processing, control of the temperature of chemical reactions, evaporation of liquid foods,

and so on. The boiling liquid

usually contained in a vessel with a heating

is

surface of tubes or vertical or horizontal plates

which supply the heat

for boiling.

The

heating surfaces can be heated electrically or by a hot or condensing fluid on the other side of the heated surface.

In boiling the temperature of the liquid

The heated

pressure in the equipment.

is

surface

the boiling point of this liquid at the is,

of course, at a temperature

above

the boiling point. Bubbles of vapor are generated at the heated surface and rise through the

mass

of liquid.

The vapor accumulates

in a

vapor space above the liquid

level

and

is

withdrawn. Boiling

is

a

complex phenomenon. Suppose we consider a small heated horizontal

tube or wire immersed flux

h

is

is

q/A

in a vessel

W/m AT = 2

,

T„

-

containing water boiling at 373.2

373.2 K, where

the heat-transfer coefficient in

values are measured. This

shown

in Fig. 4.8-1 plotted as

In

the

first

region

mechanism

of boiling

Sec. 4.8

Boiling

is

A

W/m

repeated

is

at

2 -

Tw

is

K

(100°C).

The

heat

the tube or wire wall temperature and

K. Starting with a low AT, the q/A and h AT and the data obtained are

higher values of

q/A versus AT.

of the plot in Fig. 4.8-1, at low temperature drops, the

essentially that of heat transfer to a liquid in natural convection.

and Condensation

259

AT 0 25

The variation of h with

approximately the same as that

is

for natural

convection

The very few bubbles formed are released from the and do not disturb appreciably the normal natural

to horizontal plates or cylinders.

surface of the metal

and

rise

convection. In the region

B of nucleate

boiling for

aAT'of about

5

—

25

K (9 —

45°F), the rate of

bubble production increases so that the velocity of circulation of the liquid increases. The heat-transfer coefficient h increases rapidly and

is

proportional to

AT 2

to

AT 3

in this

region. In the region

C

many bubbles

of transition boiling,

are formed so quickly that they

tend to coalesce and form a layer of insulating vapor. Increasing the thickness of this layer and the heat flux and h drop as film boiling,

bubbles detach themselves regularly and

AT

rise

is

AT

increases the

increased. In region

upward. At higher

AT

radiation through the vapor layer next to the surface helps increase the q/A and

The curve

of h versus

AT

or

h.

has approximately the same shape as Fig. 4.8-1. The

values of h are quite large. At the beginning'of region

W/m 2

B

in Fig. 4.8-1 for nucleate boiling,

K, or 1 000-2000 btu/h ft 2 °F, and at the end 2 2 region h has a peak value of almost 57 000 W/m K, or 10 000 btu/hr ft °F.

h has a value of about 5700-1

of this

D

values

1

400

•

•

•

-

-

These values are quite high, and

in

most cases

the percent resistance of the boiling film

is

only a few percent of the overall resistance to heat transfer.

The regions

of commercial interest are the nucleate and film-boiling regions (P3).

Nucleate boiling occurs

in kettle-type

and natural-circulation

reboilers.

In the nucleate boiling region the heat flux is affected by AT, and geometry of the surface and system, and physical properties of the vapor and liquid. Equations have been derived by Rohesenow et al. (PI). They apply to single tubes or flat surfaces and arc quite complex.

2.

Nucleate boiling.

pressure, nature

Simplified empirical equations to estimate the boiling heat-trarisfer coefficients for

water boiling on the outside of submerged surfaces at

developed

260

1.0

atm abs pressure have been

(J2).

Chap. 4

Principles of Steady-State

Heat Transfer

For

a horizontal surface (SI 2

btu/h

h,

W/m 2 K =

h,

btu/h

h,

W/m K =

ft

•

•

units),

1/3

= 151(AT°F)

°F

h,

and English

q/A, btu/h

2 •

ft

,

< 5000 (4.8-1)

1043(ATK) 1/3

•

-ft

2

°F = 0.168(A7°F) 3

-

q/A,

kW/m 2 <

5000

<

16

,

2

q/A, btu/h

•

ft

,

< 75000 (4.8-2)

For

2

5.56(ATK)

•

3

16

<

kW/m < 2

q/A,

240

,

a vertical surface, 2

btu/h

h,

W/m

h,

btu/h

h,

W/m 2 K =

ft

•

= 87(AT

°F

/;,

£>

F)

1 '

7

q/A, btu/h

2 •

ft

,

<

1000 (4.8-3)

2 ft

•

537(ATK) 1/7

'

T„

o

3

,

1000 < q/A, btu/h

:i

2 •

ft

,

< 20000 (4.8-4)

7.95(ATK) 3

<

3

kW/m 2 <

q/A,

,

63

-

sal

p

is

atm abs, the values of h at 1 atm given above are multiplied by and (4.8-3) are in the natural convection region.

(4.8-1)

For forced convection boiling used

kW/m 2 <

q/A,

0.240(AT F)

- T K or °F.

Equations

.

=

°F

•

pressure

If the 0 4

(p/1)

•

•

AT =

where

K=

2

inside tubes, the following simplified relation can be

(J3).

V

55

W/m 2 K

h

=

h

= 0.077(AT°F)V/225

2.55(ATK)

1

(SI)

•

(4.8-5)

where p 3.

in this case is in

Film boiling.

kPa

(SI units)

btu/h

2 ft

•

°F

and psia (English

(English)

units).

In the film-boiling region the heat-transfer rate

drop used, which

large temperature

is

subjected to considerable theoretical analysis.

Bromley (B3) gives

to predict the heat-transfer coefficient in the film-boiling region

h

=

0.62

is

low

in

view of the

not utilized effectively. Film boiling has been the following equation

on a horizontal tube.

klpM - pMhj, + 0-4^ A 'H ] Dp AT

1

/4

g_ 6)

v

where in

kv

kg/m 3

AT =

Tf

=

.

,

T„

p,

the density of the liquid in

— Tsal

T

sal

,

vapor

kg/m 3 h fg ,

W/m

in

•

K, p v the density of the vapor

the latent heat of vaporization in J/kg,

the temperature of saturated vapor in K,

D

the outside tube

m, p v the viscosity of the vapor in Pa s, and g the acceleration of gravity in The physical properties of the vapor are evaluated at the film temperature of

diameter

m/s 2

the thermal conductivity of the

is

(T„

in

•

+T

quite high,

sal )/2

some

and h fg

at the saturation temperature. If the

temperature difference

is

additional heat transfer occurs by radiation (HI).

EXAMPLE

4.8-1.

Rate of Heat Transfer

in a

Jacketed Kettle

atm abs pressure in a jacketed kettle with steam condensing in the jacket at 115.6°C. The inside diameter of the kettle is 0.656 m and the height is 0.984 m. The bottom is slightly curved but it will be assumed to be flat. Both the bottom and the sides up to a height of 0.656 m are jacketed. The kettle surface for heat transfer is 3.2-mm stainless steel with a k of 16.27 W/m-K. The condensing steam coefficient h inside the Water

is

being boiled

at

1

;

jacket has been estimated as 10 200 transfer coefficient h 0 for the

Solution:

Sec. 4.8

A

W/m 2

•

K. Predict the boiling heat-

bottom surface of the

diagram of the

Boiling and Condensation

kettle

is

shown

kettle.

in Fig. 4.8-2.

The

simplified

261

equations

will

be used for the boiling coefficient /i 0 The solution is trial and temperature Tw is unknown. Assuming .

error, since the inside metal surface

that

Tw =

110°C,

AT =

T„

- Tsal =

1

10

-

=

100

=

10°C

K

10

Substituting into Eq. (4.8-2),

=

ht

5.56{AT)

3

=

1 = hAT = A

5.56(10)

3

W/m 2 K

5560

=

55 600

5560(10)

=

•

W/m 2

assumed T„, the resistances R of the condensing steam, R v of and R 0 of the boiling liquid must be calculated. Assuming 2 equal areas of the resistances for A = m then by Eq. (4.3-12),

To check

the

t

the metal wall,

1

R,

=

'

—= M

*° 5]

R =

9.80 x 10"

=

3.2/1000

/^; 5560(1) 1

9.66 x 10

The temperature drop across

AT =

Hence, T„

=

+

=

5.9

-5

19

-

66x

5

17

+

17.98 x 10"

'

98X

is

10_5

is

=

5

47.44 x 10"

5

then

Hf^<

105. 9°C. This

,

10

^

the boiling film

^,n,6-100> =

100

9.80 x 10"

10 200(1)

^ = Tlko) =

+

5

^— =

-

A.x

^• =

,

15 .6>

= 5.rC

lower than the assumed value of

110°C. C second trial Tw — 108. 3 C will be used. Then, AT = = 8.3°C and, from Eq. (4.8-2), the new h 0 = 3180. Calculating new R 0 = 31.44 x 10" 5 and

For

108.3 the

-

the

100

,

'

AT =

31.44 x 1Q- 5X

-

10 °)

108.

rc

m-5 K 115 6 .60.90 x 10" -

I

=

8

-

10C

and T,

=

100

+

8.1

=

jacket

steam

115.6°C

FIGURE

4.8-2.

Steam-jacketed kettle and boiling water for Example 4.8-1.

Chap. 4

Principles

of Steady-State Heat Transfer

This value is reasonably close to the assumed value of 108.3°C, so no further trials will be made.

Condensation

4.8B

Condensation of a vapor to a liquid and vaporization change of phase of a fluid with large heat-transfer coefficients. Condensation occurs when a saturated vapor such as steam comes in contact with a solid whose surface temperature is below the saturation temperature, to form a /.

Mechanisms of condensation.

of a liquid to a vapor both involve a

liquid such as water.

Normally, when a vapor condenses on a surface such as a vertical or horizontal tube or other surfaces, a film of condensate

by the action of gravity.

It is

is

formed on the surface and flows over the surface between the surface and the vapor that

this film of liquid

forms the main resistance to heat transfer. This

is

called film-type condensation.

Another type of condensation, dropwise condensation, can occur where small drops are formed on the surface. These drops grow, coalesce, and the liquid flows from the surface.

During

this

condensation, large areas of tube are devoid of any liquid and are

Very high rates of heat transfer occur on these bare areas. 2 1 10000 W/m K (20000btu/h ft 2 °F), which 10 times larger than film-type coefficients. Condensing film coefficients are is 5 to normally much greater than those in forced convection and are of the order of magnitude 2 of several thousand W/m K or more. Dropwise condensation occurs on contaminated surfaces and when impurities are present. Film-type condensation is more dependable and more common. Hence, for normal design purposes, film-type condensation is assumed.

exposed directly

The average

2.

to the vapor.

coefficient

can be as high as

•

Film-condensation coefficients for vertical surfaces.

vertical wall or

condensate film

Film-type condensation on a

tube can be analyzed analytically by assuming laminar flow of the

down

the wall.

The

film thickness

is

zero at the top of the wall or tube

and increases in thickness as it flows downward because of condensation. Nusselt (HI, Wl) assumed that the heat transfer from the condensing vapor at T;at K, through this liquid film, and to the wall at Tw K was by conduction. Equating this heat transfer by conduction to that from condensation of the vapor, a final expression can be obtained for the average heat-transfer coefficient over the whole surface. In Fig. 4.8-3a vapor at 7^, is condensing on a wall whose temperature is T„ K. The condensate is flowing downward in laminar flow. Assuming unit thickness, the mass of the element with liquid density p, in Fig. 4.8-3b

on (p,

this

—

element

p„)g

the gravitational force

is

where p v

is

is

(5

—

y)(dx-

l)p,.

minus the buoyancy

The downward

force or (5

the density of the saturated vapor. This force

viscous-shear force at the plane y of p,(dv/dy) (dx-

1).

Equating these

is

—

force

y){dx) x

balanced by the

forces,

(4.8-7)

Integrating and using the

boundary condition

that v

=

0 at y

=

0,

(4.8-8)

The mass flow

rate of film

condensate at any point x for unit depth

is

(4.8-9)

Sec. 4.8

Boiling

and Condensation

263

Integrating,

m=

(4.8-10)

3* At the wall

for

area (dx

temperature distribution

is

l)m 2

assumed

the rate of heat transfer

,

in the liquid

dT

=

qx

•

=

-k,(dx-\)

.

dm = d

Pi9(Pi

~

T"' — T"

dx

k,

dx distance, the rate of heat transfer is q x Also, mass from condensation is dm. Using Eq. (4.8-10), In a

-

in this

as-

follows

if

a linear

(4.8-11)

dx distance, the increase

3

Pi9(Pi-Pv)5

Pf)^

is

between the wall and the vapor:

7

in

db (4.8-12)

Mi

Making

a heat balance for dx distance^the

must equal the

qx

from Eq.

(4.8-1

rate

dm

times the latent heath /}

1).

Pt9(Pi- Pv)&

,

h

mass flow

2

d5

fg

T

=

k,

dx

=

x,

"

sal

-Tw

(4.8-13)

Mi

Integrating with 6

=

0

at

x

= 0 and 5

=

5

=

5 at

.x

'4ft,k,x(T„,

gh/gPiiPi

Using the

'

- Tj - P.) .

1/4

(4.8-14)

local heat-transfer coefficient h x at x, a heat balance gives

hx

(dx-l)(Tssl

-TJ =

T — T k,(dx-l)

(4.8-15)

This gives

h

264

=

Chap. 4

(4.8-16)

Principles

of Steady-State Heat Transfer

Combining

Eqs. (4.8-14) and (4.8-16), Pi (Pi

-

Pv)gh fg kf

x(Tsa

By

1/4

(4.8-17)

- TJ

,

integrating over the total length L, the average value of h

is

obtained as follows.

rL 1

.

(4.8-18)

L _ ~

h

=

Pi (Pi

0.943

Pi

However,

for

-

1/4

P v )9h fg kf

(4.8-19)

W»x - TJ

laminar flow, experimental data show that the data are about

20% above

Eq. (4.8-19).

Hence, the

final

recommended expression

for vertical surfaces in

laminar flow

is

(Ml)

N Nu = — =

3

P,(p,-p v )gh fa L H,k,AT

hL

where

p,

is

the density of liquid in

1.13

kg/m 3 and p v

the vertical height of the surface or tube in m, liquid

W/m

thermal conductivity,- in

condensation

in

J/kg at

Tsal

.

•

K,

^' (4.8-20)

that of the vapor, g

/i,

is

9.8066 m/s

the viscosity of liquid in

AT = T — Tw sal

in

temperature 7}

=

4m

s,

,

k,

L

is

the

K, and h fg latent heat of

All physical properties of the liquid except h

(Tsal + T„)/2. For long vertical surfaces bottom can be turbulent. The Reynolds number is defined as

at the film

Pa

2

fg

are evaluated

the flow at the

4r (vertical tube,

diameter D)

(4.8-21)

Pi

4m

4T (vertical plate,

width W)

(4.8-22)

Pi

where m is total kg mass/s of condensate at tube or plate bottom and T — m/nD or mjW. The N Rc should be below about 800 for Eq. (4.8-20) to hold. The reader should note that some references define N Rc as Then this N Re should be below 450. For turbulent flow for N Rc > 1800 (Ml), 1

hL 0.0077

(4.8-23)

\

Solution of this equation to calculate

by

is

trial

and error

Pi

since a value ofjV Re

must

first

be assumed

h.

EXAMPLE

Condensation on a Vertical Tube kPa (10 psia) is condensing on a vertical tube 0.305 m (1.0 ft) long having an of 0.0254 m (1.0 in.) and a surface temperature of 86.1 1°C (187°F). Calculate the average heat-transfer coefficient using English and SI units. 4.8-2.

Steam saturated

at 68.9

OD

Solution:

From Appendix

T = sal

193

T„

Sec. 4.8

Boiling

C

F

A.2,

T„

(89.44°C)

+ Tsal

and Condensation

187

+

193

=

187°F

=

190°F (87.8°C)

(86.1 1°C)

265

latent heat h fg

1143.3

-

161.0

=

982.3 btu/lb m

=

2657.8

-

374.6

=

2283.2 kJ/kg

3

=

60.3(16.018)

1

=

P,

=

=

60.3 lbjft

1

=

L =

1

Assuming

kg/m 3

966.7

0.0244 lbjft

ft

=

°F

0.390 btu/ft h

=

3

=

kg/m 3

0.391

40.95

(0.324 cpX2.4191)

=

k,

=

2.283 x 10 6 J/kg

0.01657

P»

//,

=

=

=

193

sa ,

4 3.24 x 10"

=

(0.390)(1.7307)

AT = T - Tw =

m

0.305

0.784 lbjft-h

1

a laminar film, using Eq. (4.8-20) in English

neglecting p v as

N Nu =

compared

to p,

W/m K

0.675

-

87

Pas •

=

and

6°F

(3.33

K)

also SI units,

and

,

pMiAY P.k.AT

1.13

J

"

2

2

(60.3) (32.174)(3600) (982.3)(1.0)

=

1/4

3

=

1.13

6040

(0.784X0.390)(6)

x 10 5 X0.305) 3

2

'(966.7) (9.806)(2.283

W Nu =

1.13

x 10

(3.24

_hL '"No

~

= 2350

6040

)(0.675)(3.33)

h(\.Q)

SI un tS

6040

•

i

:

^1 =

6040

0.675

0.390

k,

1/4

_4

2

=

°F

W/m

2

K. Next, -the N Kc will be calculated to see if laminar flow occurs as assumed. To calculate the total heat transferred for a tube of area

Solving, h

A = tiDL =

btu/h

•

ft

•

7i(l/12X1.0)

=

7i/

q

However,

this

1

3

350

A =

2

12

ft

•

,

71(0.0254X0.305)

m2

= hA AT

(4.8-24)

q must also equal that obtained by condensation

ofm lb^h

or kg/s. Hence,

= hA AT =

q

h fg

m

(4.8-25)

Substituting the values given and solving for m,

-

2350(ji/12X193 13.35O(7rX0.O254XO.305X3.33)

187)

=

=

982.3(m)

2.284 x

6 1

(m)

m=

3.77

m=

4 4.74 x 10" kg/s

lbjh

Substituting into Eq. (4.8-21),

4m Nrc

nDn,

4(3.77) ti(

Hence, the flow

3.

=

73.5

N = R

1/1 2)(0.784)

is

4(4.74 x 10

-4 )

4 7t(O.0254X3.24 x 10" )

=

73.5

laminar as assumed.

Film-condensation coefficients outside horizontal cylinders.

The

analysis of Nusselt

can also be extended to the practical case of condensation outside a horizontal tube.

For

a single

tube the film starts out with zero thickness at the top of the tube and

increases in thickness as

266

it

flows around to the bottom and then drips

Chap. 4

Principles

off. If

there

is

a

of Steady-State Heat Transfer

bank of horizontal tubes, the condensate from the top tube drips onto the one below; and so on.

For

a vertical tier of TV horizontal

tubes placed one below the other with outside

D (M 1),

tube diameter

(4.8-26)

In

most practical applications, the flow

is

in the

laminar region and Eq. (4.8-26) holds

(C3, Ml).

HEAT EXCHANGERS

4.9

4. 9

/.

A

Types of Exchangers In the process industries the transfer of heat between two fluids

Introduction.

generally done in heat exchangers. the cold fluid

do not come

wall or a

or curved surface.

flat

The most common

type

is

one

in

is

which the hot and

into direct contact with each other but are separated

The

transfer of heat

the wall or tube surface by convection,

then by convection to the cold

is

by a tube accomplished from the hot fluid to

through the tube wall or plate by conduction, and

fluid. In the

preceding sections of

discussed the calculation procedures for these various steps.

this

chapter we have

Now we will

discuss

some

of

equipment used and overall thermal analyses of exchangers. Complete detailed design methods have been highly developed and will not be considered here. the types of

The simplest exchanger is the double-pipe or concenshown in Fig. 4.9-1, where the one fluid flows inside one pipe and the other fluid in the annular space between the two pipes. The fluids can be in cocurrent or countercurrent flow. The exchanger can be made from a pair of single lengths of pipe with fittings at the ends or from a number of pairs interconnected in series. 2.

Double-pipe heat exchanger.

tric-pipe

exchanger. This

This type of exchanger

3.

is

is

mainly

useful

Shell-and-tube exchanger.

used, which

is

the

If

for small flow rates.

larger flows are involved, a shell and tube exchanger

most important type of exchanger

these exchangers the flows are continuous. fluid flows inside these tubes. shell

and the other

fluid flows

The

Many

tubes

is

use in the process industries. In in parallel are

used where one

tubes, arranged in a bundle, are enclosed in a single

outside the tubes in the shell side.

shown in Fig. 4.9-2a for counterflow exchanger. The cold fluid enters tube exchanger

in

is

1

shell pass

and

The 1

simplest shell and

tube pass, or a 1-1

and- flows inside through

all

the tubes in

cold fluid in

hot fluid in

hot fluid out

cold fluid out

Figure

Sec. 4.9.

4.9-1.

Heat Exchangers

Flow

in

a double-pipe heat exchanger.

267

hot fluid out

cold fluid in ill

ill

w

XL;

hot fluid in

cold fluid out

(a)

hot fluid in

cold fluid in

!lt_

III

\\r

i(i

coid fluid out

hot fluid out (b)

Figure

4.9-2.

Shell-and-tube heal exchangers: {a) 1 shell pass and I tube pass exchanger); (b) 1 shell pass and 2 lube passes (1-2 exchanger). (l-l

parallel in

one

pass.

The hot

fluid enters at the

the outside of the tubes. Cross baffles

other end and flows counterflow across

are used so that the fluid

perpendicular across the tube bank rather than parallel with

it.

is

forced to flow

This added turbulence

generated by this cross flow increases the shell-side heat-transfer coefficient. In Fig. 4.9-2b a 1-2 parailel-counterfiow exchanger

tube side flows in two passes as

shown and

first

pass of the tube side the cold fluid

and

in the

hot

fluid.

is

is

shown. The

liquid

on the

the shell-side liquid flows in one pass. In the

flowing counterflow to the hot shell-side

fluid,

second pass of the tube side the cold fluid flows in parallel (cocurrent) with the Another type of exchanger has 2 shell-side passes and 4 tube passes. Other

combinations of number of passes are also used sometimes, with the 1-2 and 2^t types being the most

common.

When a gas such as Cross-flow exchanger. device used is the cross-flow heat exchanger 4)

which

is

air is

being heated or cooled, a

shown

in Fig. 4.9-3a.

One

common

of the fluids,

a liquid, flows inside through the tubes and the exterior gas flows across the tube

bundle by forced or sometimes natural convection. The sidered to be

unmixed

flow outside the tubes

since is

it is

mixed

fluid inside the tubes

confined and cannot mix with any other stream.

since

it

is

con-

The gas

can move about freely between the tubes and there

in the direction normal to the flow. For the unmixed fluid inside the tubes there will be a temperature gradient both parallel and normal to the direction of flow. A second type of cross-flow heat exchanger shown in Fig. 4.9-3b is used typically in

will

268

be a tendency for the gas temperature to equalize

Chap. 4

Principles

of Steady-State Heat Transfer

heating or cooling fluid

'

gas flow

gas flow

(a)

Figure

4.9-3.

(b)

Flow patterns of cross-flow heat exchangers: (a) one gas) and one fluid unmixed; {b) both, fluids unmixed.

fluid

mixed

(

air-conditioning and space-heating applications. In this type the gas flows across a

unmixed since it is confined in separate flow channels between The fluid in the tubes is unmixed. Discussions of other types of specialized heat-transfer equipment is deferred to Section 4.13. The remainder of this section deals primarily with a shell-and-tube and finned-tube bundle and the fins as

it

is

passes over the tubes.

cross-flow heat exchangers.

Log Mean Temperature Difference Correction Factors

4.913

In Section

4.5H

it

was shown

that

when

the hot

and cold

fluids in a heat

true countercurrent flow or in cocurrent (parallel) flow,

the log

exchanger are

in

mean temperature

difference should be used.

-

AT*, In

where

AT2

is

AT,

(AT2 /AT,)

the temperature difference at one end of the exchanger

end. This AT" m holds for a double-pipe heat exchanger

pass and

1

and

andATj

at the other

a 1-1 exchanger with

1

shell

tube pass in parallel or counterflow.

In the cases

where a multiple-pass heat exchanger

obtain a different expression for the the arrangement of the shell

two-tube-pass exchanger counterflow with the hot

in

mean temperature

and tube

is

involved,

it

is

necessary to

difference to use, depending

passes. Considering

first

Fig. 4.9-2b, the cold fluid in the first tube pass

fluid.

on

the one-shell-pass,

In the second tube pass the cold fluid

is

is

in

in parallel flow

Hence, the log mean temperature difference, which applies to either parallel or counterflow but not to a mixture of both types, as in a 1-2 exchanger, cannot

with the hot

fluid.

mean temperature drop without a correction. The mathematical derivation of the equation for the proper mean temperature to use is quite complex. The usual procedure is to use a correction factor F T which is so defined that when it is multiplied by the ATj m the product is the correct mean temperbe used to calculate the true

,

ature drop

warmer

Sec. 4.9.

ATm

to use. In using the correction factors

fluid flows

through the tubes or

Heat Exchangers

shell (Kl).

FT

The

it

is

factor

immaterial whether the

FT

has been calculated

269

(B4) for a 1-2 exchanger and

is

shown

in Fig. 4.9-4a.

Two dimensionless ratios are

used

as follows: (4.9-2)

T —T Y = where

T = hi

inlet

cold fluid, and

temperature of hot

Tc0 =

fluid in

K

T"

(4.9-3)

'ci

(°F),

Th0 =

outlet of hot fluid,

T

ci

inlet of

outlet of cold fluid.

FT

In Fig. 4.9-4b the factor

recommended

T" — 'hi

to use a heat

(B4) for a 2-4 exchanger

is

shown. In general,

exchanger for conditions under which F T

<

0.75.

it is

not

Another

(b)

Figure

4.9-4.

F r to log mean temperature difference: (a) 1-2 2-4 exchangers. [From R. A. Bowman, A. C. Mueller, and W. M. Nagle, Trans. A.S.M.E., 62, 284, 285 (1940). With per-

Correction factor exchangers,

(b)

mission.']

270

Chap. 4

Principles of Steady-State

Heat Transfer

shell and tube arrangement should be used. Correction factors for two types of cross-flow exchangers are given in Fig. 4.9-5. Other types are available elsewhere (B4, PI).

Using the nomenclature of Eqs.

(4.9-2)

and

A7 lm

(4.9-3), the

of Eq. (4.9-1) can be

written as

ATlm Then

~ Tco — (Tho — T [(T„ - TJ/(ThB - T

C^ii

"

In

the equation for an exchanger

)

ci ) "

"

'

ci )]

is

q=U

i

A ATm = U 0 A 0 ATm

(4.9-5)

i

where

ATm = F r A7] m EXAMPLE 4.9-1. A

(4.9-6)

Temperature Correction Factor for a Heat Exchanger

1-2 heat exchanger containing one shell pass and two tube passes heats

2.52 kg/s of water from 21.1 to 54.4°C by using hot water under pressure

entering at 115.6 and leaving at 48.9°C. tubes in the exchanger (a)

Calculate the

(b)

and For

the

First

m2

The

outside surface area of the

.

mean temperature

difference

The temperatures

making

ATm

in the

exchanger

.

2-4 exchanger, what would

are as follows.

Th0 =

T =

4S.9°C

ci

Tco =

21.1°C

54.4°C

a heat balance on the cold water assuming a c pm of water ci

= mc pm (T - T =

4.9-5.

a

U0

?

J/kg-KandTco - T = q

FIGURE

9.30

same temperatures but using

TH =115.6°C of 4187

Aa =

the overall heat-transfer coefficient

betheATm Solution:

is

C0

ci )

-

21.1)°C

(2.52X4187X54.4

Correction factor

exchangers [Z

(54.4

=

=

-

33.3°C 21.1)

=

=

33.3

348 200

K,

W

F T to log mean temperature difference for cross-flow (TM - TJ/(Tco - Tei )] : (a) single pass, shell fluid

mixed, other fluid unmixed, (b) single pass, both fluids unmixed. [From R. A. Bowman, A. C. Mueller, and W. M. Nagle, Trans. A.S.M.E., 62,

288,289(1940). With permission.]

Sec. 4.9.

Heat Exchangers

111

The

mean temperature

log

-

(115.6

A7ta

Next

"

In [(1 15.6

difference using Eq. (4.9-4)

- 21.1) 54.4)/(48.9 - 21.1)] -

54.4)

-

(48.9

and

substituting into Eqs. (4.9-2)

T„-T»„

Z= y

TM

Fig. 4.9-4a,

21.1

-

21.1

Then, by Eq. 0.74(42.3)

Rearranging Eq. (4.9-5) to solve for

U0

„ K

^ n

3

=

2.00

(4.9-2)

=

0.352

(4.9-3)

115.6-21.1

ci

F T = 0.74. ATm = F T ATlm =

From

54.454.4

-T

t

4

~~

(4.9-3)

115.6-48.9

T — TT -T =

C

3

~~

is

=

(4.9-6),

=

31.3°C

K

31.3

(4.9-6)

and substituting the known values,

we have 348200

U =

A 0 ATm

For part Then,

(b),

(211 btu/h

•

2

,°F)

ft :

FT

using a 2-4 exchanger and Fig. 4.9-4b,

ATm = F T Hence,

= U96 W/m 2 -K

(9.30X31.3)

in this case the

AT, m

=

0.94(42.3)

2-4 exchanger

=

utilizes

=

39.8°C

more

0.94.

K

39.8

of the available temper-

ature driving force.

Heat-Exchanger Effectiveness

4.9C

/.

ki tbe preceding section the log

Introduction,

in the

when

equation q the inlet

= UA ATlm

in the

mean temperature

and outlet temperatures of the two

fluids are

known

by a heat balance. Then the surface area can be determined

when

difference was used

design of heat exchangers. This form

if

is

convenient

or can be determined

U

is

known. However,

known and a given necessary. To solve these

the temperatures of the fluids leaving the exchanger are not

exchanger cases, a

is

to be used, a tedious trial-and-error

method

procedure

called the heat-exchanger effectiveness e

is

is

used which does not involve

any of the outlet temperatures.

The heat-exchanger

effectiveness

transfer in a given exchanger to the infinite heat-transfer

exchanger

is

shown

is

defined as the ratio of the actual rate of heat

maximum

area were available.

amount

possible

The temperature

of heat transfer

if

an

profile for a counterflow heat

in Fig. 4.9-6.

'Ha

(cH

>cc

)

'a Distance

Figure

272

4.9-6.

Temperature

profile for countercurrent heat exchanger.

Chap. 4

Principles

of Steady-State Heat Transfer

2.

The heat balance

Derivation of effectiveness equation.

and the hot (H)

for the cold (C)

fluids is

q Calling

(mcp ) H =

CH

=

-

(mc p ) H (Tm

and (mcp ) c

= Cc

T„„)

,

=

then

in Fig. 4.9-6,

undergoes a greater temperature change than the hot

C min

or

transfer,

minimum heat capacity. Then, TCo = TH1 Then the effectiveness .

Cj/(Tw; C;/(T Wi

=

f.

there

if

fluid,

and the cold

Hence, we designate

fluid

Cc

as

infinite area available for heat

Cmzx(T ——m —=— T

—T T„ —HoB

) )

.

Tc d — Ta cd

,

T —Ho — Tc— d

,,,

Cm\n(TH

rHo = Tci

Cn(T,ii

CH > C c

fluid.

an

is

(4.9-7)

)

e is

cd

l

- Tci

(mc p ) c (TCo

)

i

and ~~ T cd ~ — 7^,) Tc d

Cmax(Tco

C ^-min(^//i min (Twi

)

(4.9-8)

x In

both equations the denominators are the same and the numerator gives the actual

heat transfer. q

Note

= sC m .JTln - Tc d

that Eq. (4.9-10) uses only inlet temperatures,

temperatures are

known and

it

is

(4.9-10)

which

is

an advantage when

inlet

desired to predict the outlet temperatures for a given

existing exchanger.

For the case of a

single-pass, counterflow exchanger,

combining Eqs.

(4.9-8)

and

(4.9-9),

- Tc d — Ta

_ C n(Tm - THo = C c{TCo )

c

We

consider

first

Cm

i

n (^Hi

—

C mm (T

Tcd

}li

(4.9-11) )

minimum

the case of the cold fluid to be the

Rewriting Eq.

fluid.

(4.5-25) using the present nomenclature,

<7

Combining Eq.

——

——

In L(

Tcdl(Tin

= C c{TCo - Ta = UA )

(4.9-7)

with the

left

side of Eq. (4.9-1

Tm = Ta + Subtracting

TCo

from both

(4.9-7) for

- (TCo

1)

- Ta

—

(4.9-12)

- TCo )\

and solving for T Hi

,

(4.9-13)

)

sides,

Tm - TCo = Ta From Eq.

tho ~

[

-TCo + - (TCo

C min = C c

and

- TC! = )

C max = C H

THo = Tm -

(TCo

0 - 1^(T

C„

- TCl

)

(4.9-14)

,

- Tc d

(4-9-15)

This can be rearranged to give the following:

T»o

- Ta = Tm - T„ -

^

(TCo

- 7Ci

)

(4.9-16)

''max

Sec. 4.9.

Heat Exchangers

273

Substituting Eq. (4.9-13) into (4.9-16),

TH . - TCi = -

(TCo

- Tc d -

(TCo

Finally, substituting Eqs. (4.9-14)

and (4.9-17Tmto and solving for £,

antilbg of both sides,

1

We define NTU

as the

UA

- exp

1

-

1

number of transfer

(4.9-17)

(4.9-12), rearranging, taking the

(4.9-18)

UA C

exp

- Tci)

Cr 1

units as follows

UA NTU = r

(4.9-19)

.

The same For

result

would have been obtained

parallel flow

we

CH = Cn

if

obtain

I

£

—

UA exp

=

1

C„

+

C„

c„ (4.9-20)

1

+

In Fig. 4.9-7, Eqs. (4.9-18) and (4.9-20) have been plotted

in

convenient graphical form.

Additional charts are available for different shell-and-tube and cross-flow arrange-

ments (Kl).

Number

Number of

of transfer units,

NTU

=

NTU

UA/C. mm '

4.9-7.

transfer units,

UA/C. mm (b)

(a)

Figure

=

Heat-exchanger effectiveness e: (a) counterftow exchanger,

(b) parallel flow

exchanger.

274

Chap. 4

Principles

of Steady-State Heat Transfer

EXAMPLE 2.85 kg/s (c

A =

p

at a rate

.

of 0.667 kg/s enters a countercurrent heat exchanger

K

heated by an oil stream entering at 383 at a rate of kJ/kg -K). The overall V = 300 W/m 2 -K and the area Calculate the heat-transfer rate and the exit water temperis

=

m2

15.0

Exchanger

Effectiveness of Heat

4.9-2.

Water flowing at 308 K and

1.89

ature.

Assuming

that the exit water temperature is about 370 K, thec p average temperature of (308 + 370)/2 = 339 K is 3 4.192 kJ/kg-K (Appendix A.2). Then, (mc p ) H = C„ = 2.85(1.89 x 10 ) = 3 = = = = 5387 W/K and (mc p ) c C c 0.667(4.192 x 10 ) 2796 W/K C min Since C c is the minimum, C min/C m „ = 2796/5387 = 0.519. = UA/C m]n = 300(15. 0)/2796 = 1.607. Using Eq. (4.9-19), Using Fig. (4.9-7a) for a counterfiow exchanger, s = 0.71. Substituting

Solution:

water

for

an

at

.

NTU

into

Eq.

(4.9-10),

= tC min (Tm - Tci =

q

)

Using Eq.

4.9D

-

308)

-

308)

=

148 900

W

(4.9-7),

q= Solving,

0.71(2796X383

TCo =

148 900

=

2796(TCo

361.3 K.

Fouling Factors and Typical Overall

U

Values

do not remain

In actual practice, heat-transfer surfaces

clean. Dirt, soot, scale,

and other

and on other heattransfer surfaces. These deposits form additional resistances to the flow of heat and reduce the overall heat-transfer coefficient U. In petroleum processes coke and other substances can deposit. Silting and deposits of mud and other materials can occur. Corrosion products may form on the surfaces which could form a serious resistance to heat transfer. Biological growth such as algae can occur with cooling water and in the

deposits form on one or both sides of the tubes of an exchanger

biological industries.

To avoid

or lessen these fouling problems chemical inhibitors are often added to

minimize corrosion,

salt deposition,

and algae growth. Water

deposition of solids on surfaces and should be avoided

The

effect of

such deposits and fouling

is

if

above

velocities

generally used to help reduce fouling. Large temperature differences

may

1

m/s are

cause excessive

possible.

usually taken care of in design by adding a

term for the resistance of the fouling on the inside and the outside of the tube

in

Eq.

(4.3-17) as follows.

where h di

is

+

+

'

(r*

-

r

i

)A /k A i

A A lm + AJA B h„ + AjA 0 hdo

the fouling coefficient for the inside

outside of the tube in

W/m 2

•

K.

A

and

h do the fouling coefficient for the

similar expression can be written for

Ua

using Eq.

(4.3-18).

Fouling coefficients recommended for use available in coefficients

many is

references (P3, Nl).

A

in

designing heat-transfer equipment are

short tabulation of

some

In order to

do preliminary estimating or

sizes of shell-a^nd-tube heat exchangers,

typical values of overall heat-transfer coefficients are given in

Table 4.9-2. These values

should be useful as a check on the results of the design methods described

Sec. 4.9.

typical fouling

given in Table 4.9-1.

Heat Exchangers

in this chapter.

275

Table

4.9-1

Typical Fouling Coefficients (P3, Nl)

.

K

K

(W/m 2 : K) Distilled

and seawater

11

City water

Muddy

1990-2840

water

2

2000 1000

350-500

Gases

2840

Vaporizing liquids

2840

500

1990

350

oils

-°F)

500

INTRODUCTION TO RADIATION HEAT TRANSFER

4.10A J.

350

5680

Vegetable and gas

4.10

(btu/h-ft

Introduction and Basic Equation for Radiation

Nature of radiant heat

transfer.

In the preceding sections of this chapter

studied conduction and convection heat transfer. In conduction heat

is

we have

transferred from

one part of a body to another, and the intervening material is heated. In convection the heat is transferred by the actual mixing of materials and by conduction. In radiant heat transfer the medium through which the heat is transferred usually is not heated. Radiation heat transfer

is

the transfer of heat by electromagnetic radiation.

Thermal radiation is a form of electromagnetic radiation similar to x rays, light waves, gamma rays, and so on, differing only in wavelength. It obeys the same laws as light: travels in straight lines, can be transmitted through space and vacuum, and so on. It is an important mode of heat transfer and is especially important where large

Table

4.9-2.

Typical Values of Overall Heat-Transfer Coefficients Shell-and-Tube Exchangers {HI, P3, Wl)

in

V (W/m 2 -K) Water Water Water Water Water Water Water

Gas

to

1140-1700

570-1140 570-1140 1420-2270

to organic liquids to

condensing steam

340-570 140-340 110-285 110-285 1420-2270 110-230 230-425 55-230

to gasoline to gas oil

to vegetable oil

oil

Steam Water

water

to brine

to gas oil

to boiling water to air (finned tube)

Light organics to light organics

Heavy organics

276

to

heavy organics

Chap. 4

Principles

V (blu/h-ft

2

-°F)

200-300 100-200 100-200 250-400 60-100 25-60 20-50 20-50 250-400 20-40 40-75 10-40

of Steady-State Heat Transfer

temperature differences occur,

as, for

example,

furnace with boiler tubes, in radiant

in a

and in an oven baking food. Radiation often occurs in combination with conduction and convection. An elementary discussion of radiant heat transfer will be given here, with a more advanced and comprehensive discussion being given in Section 4.1 1.

dryers,

In an elementary sense the

mechanism

of radiant heat transfer

is

composed

of three

distinct steps or phases: 1.

The thermal energy

of a hot source, such as the wall of a furnace atT,,

is

converted

into the energy of electromagnetic radiation waves. 2.

These waves object at

3.

T2

travel

through the intervening space

in straight lines

and strike a cold

such as a furnace tube containing water to be heated.

The electromagnetic waves

that strike the

body are absorbed by the body and

converted back to thermal energy or heat. 2.

Absorptivity and black bodies.

body, part

is

When

absorbed by the body

thermal radiation

in the

(like light

form of heat, part

is

waves)

reflected

falls

upon

a

back into space,

and part may be actually transmitted through the body. Foremost cases in process engineering, bodies are opaque to transmission, so this will 6e neglected. Hence, for opaque bodies, 3t

where a

A

is

absorptivity or fraction absorbed and p

black body

Hence, p

+ p=1.0

is

0 and a

is

=

The

enters the hole

is

defined as one that absorbs 1.0 for a

is

inside surface of the hollow

The

reflected rays

continues. Hence, essentially

radiant energy and reflects none.

a small hole in a hollow body, as

body

is

and impinges on the rear wall; part

in all directions.

reflectivity or fraction reflected.

all

black body. Actually, in practice there are no perfect black

bodies, but a close approximation to this Fig. 4.10-1.

(4.10-1)

all

hole acts as a perfect black body.

shown

in

blackened by charcoal. The radiation is

absorbed there and part is reflected is absorbed, and the process

impinge again, part

is absorbed and the area of the surface of the inside walls are " rough " and rays are

of the energy entering

The

all directions, unlike a mirror, where they are reflected at a definite angle. As stated previously, a black body absorbs all radiant energy falling on it and reflects none. Such a black body also emits radiation, depending on its temperature, and does not reflect any. The ratio of the emissive power of a surface to that of a black body is c&Wed-emissioity e and it is 1.0 for a black body. Kirchhoffs law states that at the same temperature Tj, a, and e of a given surface are the same or

scattered in

]

a,

=

e,

(4.10-2)

Equation (4.10-2) holds for any black or nonblack solid surface.

Sec. 4.10

Introduction to Radiation

Heal Transfer

277

3.

The

Radiation from a body and emissivity.

basic equation

radiation from a perfect black body with an emissivity q

where q 2

W/m

is

heat flow in

4

(0.1714 x 10"

K

•

W, A 8

is

m2

1.0

for

heat transfer by

is

= AoT*

(4.10-3)

surface area of body, a a constant 5.676 x 10"

2

btu/h-ft

=

£

°R

4

and

),

T is

temperature of the black body

in

8

K

(°R).

For a body that is not power is reduced by e, or

a black

body and has an

q

Substances that have emissivities of emissivity

is

emissivity e

<

1.0,

the emissive

= AeoT 4 than

less

independent of the wavelength. All

(4.10-4)

1.0 are called

real materials

gray bodies when the

have an emissivity

e

<

1.

and absorptivity^ of a body are equal at the same temperature, the emissivity, like absorptivity, is low for polished metal surfaces and high for oxidized metal surfaces. Typical values are given in Table 4.10-1 but do vary some with Since the emissivity

temperature.

e

Most nonmetallic substances have high

values. Additional data are tabu-

lated in

Appendix

4.10B

Radiation to a Small Object from Surroundings

When we have

A. 3.

A m2

the case of a small gray object of area

T2

i

at

temperature T,

in a large

The body emits an amount of radiation to the enclosure given by Eq. (4.10-4) of 4 /IjEjffT The emissivity £, of this body is taken at T,. The small body also absorbs enclosure at a higher temperature

,

there

a net radiation to the small object.

is

small

.

energy from the surroundings

body same

1

at

T2

given by /l,a 12 aT\. Thea !2 at T2 The value of a 12

from the enclosure

for radiation

body

as the emissivity of this

at

.

T2

.

The

is

the absorptivity of

is

approximately the

net heat of absorption

is

then, by the

Stefan-Boltzmann equation, q

A

=A

1

e1

aT* -

further simplification of Eq.

using one emissivity of the small

A^ 12 aT\ (4. 10-5) is

body q

Table

/l^T? - a 12 T$)

usually

= A to{T\- T 4 i

made

T2

temperature

4.10-1. Total Emissivity, e,

.

for engineering purposes

of Various Surfaces

Polished aluminum

500

440

0.039

1070

0.057

Polished iron

850 450

350

0.052

Oxidized iron

373

212

0.74

Polished copper

353

176

0.018

Asbestos board

296 373 273

74

212

Chap. 4

Principles

Water

colors

by

(4.10-6)

)

T(K)

all

(4.10-5)

Thus,

Surface

Oil paints,

278

at

=

T(°F)

32

Emissivity,

c.

0.96

0.92-0.96 0.95

of Steady-State Heat Transfer

EXAMPLE

Radiation to a Metal Tube

4.10-1.

OD

A

m

of 0.0254 small oxidized horizontal metal tube with an (1 in.) and being 0.61 (2 ft) long with a surface temperature at 588 K (600°F) is in a

m

very large furnace enclosure with fire-brick walls and the surrounding air at and 0.46 1088 K (1500°F). The emissivity of the metal tube is 0.60 at 1088

K

588 K. Calculate the heat transfer to the tube by radiation using SI and English units.

at

Since the large furnace surroundings are very large compared with the small enclosed tube, the surroundings, even if gray, when viewed from the position of the small body appear black and Eq. (4.10-6) is applicable. Substituting given values into Eq. (4. 10-6) with an e of 0.6 at 1088 K

Solution:

A, q

= nDL = = A

x

Ea{T\

= -2130 =

m2 =

7t(0.0254X0.61)

-

=

Tj)

tt(1/1

2 ft

2X2.0)

8 4 [ti(O.0254XO.61)](0.6)(5.676 x 10" )[(588)

-

4

(1088) ]

W 10" 8 )[(1060) 4

[7r(l/12)(2)](0.6)(0.1714 x

Other examples of small objects

in large

- (I960) 4 ] = -7270

btu/h

enclosures occurring in the process indus-

are a loaf of bread in an oven receiving radiation from the walls around

it, a package meat or food radiating heat to the walls of a freezing enclosure, a hot ingot of solid iron cooling and radiating heat in a large room, and a thermometer measuring the

tries

of

temperature

4.10C

When

in

a large duct.

Combined Radiation and Convection Heat Transfer radiation heat transfer occurs from a surface

convective heat transfer unless the surface at

we can

a uniform temperature,

is

As discussed

sum

is

usually accompanied by the radiating surface

is

in

The

the previous sections of this chapter.

Then

the

of convection plus radiation.

by convection and the convective

before, the heat-transfer rate

coef-

given by 9co n v

where q conv

the heat-transfer rate

is

convection coefficient

temperature of the

W/m 2 K can

in

W/m 2

•

= M,(Ti - T2

by convection in W, h c the natural or forced the temperature of the surface, and T2 the t

and the enclosure.

air

A

radiation heat-transfer coefficient h r in

be defined as

= Ml(Tl

where g ra(1 is the heat-transfer rate by radiation of Eqs. (4.10-7) and (4.10-8), <7

=

4co„v

obtain an expression for h r

,

+

q,«

we

1

m(T ~ T ^ — —= — h

~T

(4.10-8)

2)

in

W. The

total heat transfer

is

= (K + K)A i(T, - T2 )

the

sum

(4.10-9)

equate Eq. (4.10-6) to (4.10-8) and solve for h r

.

(T./ioor-qyioo) 4

*

K=

(4.10-7)

)

T

K,

9rad

To

it

vacuum. When

calculated by the Stefan-Boltzmann equation (4.10-6).

total rate of heat transfer is the

ficient are

in a

calculate the heat transfer for natural or forced

convection using the methods described radiation heat transfer

is

£(5.676)

'2

'1

—

(SI)

'2

(4.10-10)

hr

Sec. 4.10

=

s(0.

1714)

h~

Introduction to Radiation

(English) '2

Heat Transfer

279

A convenient chart giving values of in English units calculated from Eq. (4. 10-10) e= 1.0 is given in Fig. 4.10-2. To use values from this figure, the value obtained from /i

with

r

e to give the value of h r touse in Eq. (4.10-9). If the same as T 2 of the enclosure, Eqs. (4.10-7) and (4.10-8) must be used separately and not combined together as in (4.10-9). this figure air

should be multiplied by

temperature

is

not the

EXAMPLE 4.10-2. Combined Convection Plus Radiation from a Tube Recalculate Example 4.10-1 for combined radiation plus natural convection to the horizontal 0.0254-m tube.

The

A

=

=

m

2

For the natural convection coefficient to the 0.0254-m horizontal tube, the simplified equation from Table 4.7-2 will be used as an approximation even though the

Solution:

film

area

temperature

is

of the tube

0.0487

.

quite high.

*•

Substituting the

ti(0.0254X0.61)

known

1


values,

Using Eq. (4.10-10) and

e

=

0.6,

Temperature of one surface (°F) FIGURE

280

4.10-2.

Radiation heat-transfer coefficient as a function of temperature. (From R. H. Perry and C. H. Chilton, Chemical Engineers' Handbook, 5th ed. New York: McGraw-Hill Book Company, 1973. With permission.)

Chap. 4

Principles

of Steady-State Heat Transfer

Substituting into Eq. (4.10-9), q

=

=

(h c

+

h r )Ai(T

x

-2507

- T2 = )

+

(15.64

87.3)(0.0487)(588

-

1088)

W

W

Hence, the heat loss of —2130 for radiation is increased to only — 2507 when natural convection is also considered. In this case, because of the large temperature difference, radiation is the most important factor.

W

Perry and Green (P3, p. 10-14) give a convenient table of natural convection plus + h r ) from single horizontal oxidized steel pipes as a function of

radiation coefficients {h c

the outside diameter and temperature difference.

The

coefficients for insulated pipes are

about the same as those for a bare pipe (except that lower surface temperatures are involved for the insulated pipes), since the emissivity of cloth insulation wrapping

about that of oxidized

be given

will

in

steel,

approximately

A more

0.8.

is

detailed discussion of radiation

Section 4.11.

ADVANCED RADIATION HEAT-TRANSFER

4.11

PRINCIPLES

4.11 A

Introduction and Radiation Spectrum

will cover some basic principles and also some advanced were not covered in Section 4.10. The exchange of radiation between two surfaces depends upon the size, shape, and relative orientation of these two surfaces and also upon their emissivities and absorptivities. In the cases to be considered the surfaces are separated by nonabsorbing media such as air. When gases such as C0 2 and H 2 0 vapor are present, some absorption by the gases occurs, which is not con/.

Introduction.

This section

topics on radiation that

sidered until later

2.

in this section.

Radiation spectrum and thermal radiation.

Energy can be transported

electromagnetic waves and these waves travel

many forms

of radiant energy, such as

on. In fact, there

is

gamma

at the

speed of

light.

in the

Bodies

form of

may

rays, thermal energy, radio waves,

emit

and so

a continuous spectrum of electromagnetic radiation. This electro-

magnetic spectrum is divided into a number of wavelength ranges such as cosmic rays 7 13 13 10 (A < 10" m), gamma rays (A, 10" m), thermal radiation(;., 10" to 10"* m), to 10"

and so on. The electromagnetic radiation produced solely because of the temperature of 7 is called thermal radiation and exists between the wavelengths of 10" and 4 10~ m. This portion of the electromagnetic spectrum is of importance in radiant 7 thermal heat transfer. Electromagnetic waves having wavelengths between 3.8 x 10" 7 and 7.6 x 10" m, called visible radiation, can be detected by the human eye. This visible

the emitter

radiation

lies

When

within the thermal radiation range.

same temperature, they do not all emit or A bddy that absorbs and emits the maximum amount of energy at a given temperature is called a black body. A black body is a standard to which other bodies can be compared. different surfaces are heated to the

absorb the same amount of thermal radiant energy.

Planck's law and emissive power. When a black body is heated to a temperature T, photons are emitted from the surface which have a definite distribution of energy. 3.

Sec. 4.11

Advanced Radiation Heat-Transfer

Principles

281

monochromatic emissive power E BX wavelength X in m.

Planck's equation relates the ature

T in K

and a

plot of Eq. (4.1 1-1)

is

^5|" e 1.4388xlO-JMT

_

(4.11-1)

shows that the energy given off increases power reaches a maximum value at a wave-

that decreases as the temperature

in

the low / region.

spectrum straddles

temper-

j"|

T

At a given temperature the

increases.

radiation emitted extends over a spectrum of wavelengths.

occurs

at a

given in Fig. 4.1 1-1 and

with T. Also, for a given T, the emissive length

W/m 3

16 3.7418 x 1(T

^ BX =

A

in

The sun has

The

visible light

a temperature of about 5800

K

spectrum

and the

solar

this visible range.

For a given temperature, the wavelength

which the black-body emissive power

at

maximum can be determined by differentiating Eq. (4.11-1) with respect to X constant T and setting the result equal to zero. The result is as follows and is known a

VVien

's

4.

locus of the

maximum

values

The

Stefan-Boltzmann law.

T = is

2.898 x 1CT

shown

3

m-K

black body, the total emissive all

power

total emissive

power

is

(4.11-2)

in Fig. 4.1 1-1.

is

energy per unit area leaving a surface with temperature over

as

displacement law: X m3X

The

is

at

the total

T

over

amount of

all

given by the integral of Eq. (4.11-1)

wavelengths or the area under the curve

in

radiation

wavelengths. For a at a

given

T

Fig. 4.1 1-1.

(4.11-3)

0

2

4

6

Wavelength, A Figure

282

4.1 1-1.

8

10

12

(m X 10 6 )

Spectral distribution of total energy emitted by a black body at various temperatures of the black body.

Chap. 4

Principles of Steady-State

Heat Transfer

This gives

£B =

crT

5.676 x 10"

8

2

W/rrr

•

K4

.

The

units of

-

.

An

Emissivity and Kirchhojfs law.

The

of a surface.

(4.11-4)

=

The result is the Stefan-Boltzmann law with a ' E B are W/m 2 5.

4

important property

emissivity s of a surface

is

in

radiation

the emissivity

is

defined as the total emitted energy of the

body

surface divided by the total emitted energy of a black

at the

same temperature.

£

body emits

Since a black

We

the

maximum amount

of radiation,

t is

always

1.0.

and allowing

material by placing this material in an isothermal enclosure

same temperature at thermal equilibrium'. on the body, the energy absorbed must equal the energy emitted.

enclosure to reach the

a If this

<

can derive a relationship between the absorptivity a, and emissitivy

body

is

1

G=£

G

is

£

t

of a

body and

the irradiation

(4.11-6)

1

removed and replaced by a black body of equal

Dividing Eq. (4.11-6) by

If

the

size,

then at equilibrium,

ajG = EB

(4.11-7)

- = 7£,i

(4.11-8)

(4.11-7), a,

Buta 2 =

1.0 for a

black body. Hence, since £,/£„

a

This

body

is is

its

£ 1;

,=§i=£,

KirchhofTs law, which states that not at equilibrium with

=

at

(4.11-9)

thermal equilibrium a

surroundings, the result

is

not

=

£

of a body.

When

a

valid.

Concept of gray body. A gray body is defined as a surface for which the monochromatic properties are constant over all wavelengths. For a gray surface,

6.

£x

Hence, the

total absorptivity a

are equal, as are £ and

ex

=

ax

const.,

=

const.

and the monochromatic absorptivity a x of

(4.11-10) a gray surface

.

a

Applying KirchhofTs law

to a gray

=

OL

body

x

£

,

ax

=

e.

a

=

£

x

=

£,

(4.11-11)

and (4.11-12)

As a result, the total absorptivity and emissivity are equal for a gray body even if the body is not in thermal equilibrium with its surroundings. Gray bodies do not exist in practice and the concept of a gray body is an idealized one.

The

absorptivity of a surface actually varies with the wavelength of the incident

radiation. Engineering calculations can often be based

Sec. 4.11

Advanced Radiation Heat-Transfer

Principles

on the assumption of a gray body

283

The a

with reasonable accuracy. incident radiation. Also,

in

assumed constant even with a variation

is

actual systems, various surfaces

atures. In these cases, a for a surface

is

may be

X of the

in

at different temper-

evaluated by determining the emissivity not

at the

actual surface temperature but at the temperature of the source of the other radiating surface or emitter since this

is

the temperature the absorbing surface

would reach

if

the

absorber and emitter were at thermal equilibrium. The temperature of the absorber has only a slight effect on the absorptivity.

4.11B

Derivation of View Factors

definitions presented in Section 4.11 A

The concepts and

Introduction.

/.

ficient

Radiation

in

Various Geometries

for

form a

suf-

foundation so that the net radiant exchange between surfaces can be determined.

two surfaces are arranged so that radiant energy can be exchanged, a net flow of energy occur from the hotter surface to the colder surface. The size, shape, and orientation of two radiating surfaces or a system of surfaces are factors in determining the net heat-flow rate between them. To simplify the discussion we assume that the surfaces are separated by a nonabsorbing medium such as air. This assumption is adequate for many engineering applications. However, in cases such as a furnace, the presence of C0 2 and H 2 0 vapor make such a simplification impossible because of their high absorptivities. If

will

The

simplest geometrical configuration will be considered

exchange between edge effects

parallel, infinite planes.

case of

in the

finite surfaces. First,

the surfaces are black bodies

first,

that of radiation

This assumption implies that there are no the simplest case will be treated in which

and then more complicated geometries and gray bodies

will

be treated.

View factor for infinite parallel black planes. If two parallel and infinite black planes emits aT\ radiation to plane 2, and T2 are radiating toward each other, plane which is all absorbed. Also, plane 2 emits aT\ radiation to plane 1, which is all absorbed.

2.

-

at T,

1-

Then

for plane

1,

the net radiation

is

from plane

q l2

In this case

1

that

is

to 2

1

is

2.

-

,

is 1.0.

2; that is,

The

the fraction of

factor

Fn

fraction of radiation leaving surface

in all

1

is

called

(4.11-14)

directions which

is

intercepted

Also, -

=F 2i A

92i

In the case for parallel plates

3.

intercepted by

Fn

is

=F i2 AMT\-T\)

q l2

by surface

is

(4.11-13)

which Hence,

intercepted by 2

the geometric view factor or view factor.

where F 12

2,

= AMTt-T$)

the radiation from

all

radiation leaving

to

1

View factor for

F i2 = F 2l =

infinite parallel

are gray with emissivities

and

2

o(T*-Tt)

1.0

(4.11-15)

and the geometric factor

gray planes.

If

absorptivities ofgj

is

simply omitted.

both of the parallel plates

=

ande 2

= a2

,

A

x

respectively,

and A 2

we can

proceed as follows. Since each surface has an unobstructed view of each other, the view factor e2

is 1.0.

(where a 2

In unit time surface

=

£2)

is

A

x

emits

absorbed by

284

e^CTT'J

radiation to/l 2

.

Of this,

the fraction

absorbed

A2 =

Chap. 4

A oT*)

(4.11-16)

Principles

of Steady-State Heat Transfer

e 2 {e

i

i

(1 — £ 2 ) or the amount (1 — amount /l, reflects back to A 2 a fraction(l — The surface A 2 absorbs the fraction e 2 or

Also, the fraction

£ 2 Xe,/l

,
is

reflected

—

or an amount(l

£,)

back toA v Of this

—

£,X1

£ 2K E

i^ l^^t)-

,

absorbed by

e-

-

2 (1

A from A 2 is of this and reflects back to A 2 The surface A 2 then absorbs

The amount absorbs

A^=

e,

(e,/4,(7Tf).

back to

reflected

e 2 (1

- s^A.aT^

— £ 2 )(1 — £iXl —

(1

x

=

absorbed by A 2

e,)(l

an amount

- e,)(l -

e 2 )(1

-

and

A2

1,-2

= A aT*Ze l

The

result

is

l

B2

2

+

-£ 2 +

-£,)(1

e l e 2 {l

A,oT\-

ElE

-

? 2 -i

(1

-

is

=

1.

2

-

sum

Sl ) (l

2 £ 2)

1

e 2)

=A,oT\

+

the difference of Eqs. (4.1 1-20)

at

\ l/e 2

and

= AMTt-Tt) u 1/e, +

q 12

e2

the

= A aTt u 1/e, +

-

-

\ e,)(1

1/e,

=

-

£,£ 2 (l

)

Repeating the above for the amount absorbed

Ife,

- e^A^T^)

is

Then A l

— E iXl

x

£ 2)

(4.11-18)

of Eqs. (4.11-16),

so on.

1

net radiation

lO'^t)-

+

-

•

(4.11-19)

-]

a geometric series (Ml).

q^ =

The

e 2 K e i^4

—e,X1 —£ 2 X1

(1

e,)(1

This continues and the total amount absorbed at (4.11-17), (4.11-18),

(4.11-17)

0 for black bodies, Eq.

(4. 11

-22)

A

]

,

j

l/e 2

-

(4.11-20) 1

which comes from

A2

,

(4.11-21)

-

1

(4.1 1-21).

)

l/e 2

becomes Eq.

-

(4.11-22) 1

(4.11-13).

EXAMPLE 4.11-1.

Radiation Between Parallel Planes which are very large have emissivities of £, = 0.8 and e 2 = 0.7 and surface 1 is at 1100°F (866.5 K) and surface 2 at 600°F (588.8 K). Use English and SI units for the following. (a) What is the net radiation from 1 to 2? (b) If the surfaces are both black, what is the net radiation?

Two

parallel gray planes

Solution:

For

part

(a),

using Eq. (4.11-22) and substituting the

known

values,

g l2

A

x

o{T\-T\) (ai/14x "1/e, + l/ E2 -l-

= 4750

For black

btu/h

Note than

Sec. 4.11

•

460)

4

-

(600

+

460)*

1/0.8+1/0.7-1

ft

=

(b),

using Eq.

7960 btu/h

(4.1 1-13),

2 •

ft

the large reduction in radiation 1.0

+

2

surfaces in part

q

.-_„. (1100 1U ]

or

25

when

1

10

W/m 2

surfaces with emissivities less

are used.

Advanced Radiation Heat-Transfer

Principles

285

Example

4.11-1

radiation. This fact

is

as a radiation shield.

the interchange

is,

shows the large influence

that emissivities less than 1.0 have

For example,

by Eq.

for

two

andT2

parallel surfaces of emissivity e atT\

,

(4.1 1-22),

5^.f(lLl2 A The

on

used to reduce radiation loss or gain from a surface by using planes

—

Ijz

(4

„.23)

1

no planes in between the two surfaces. Suppose more radiation planes between the original surfaces. Then it or

subscript 0 indicates that there are

that we now insert one can be shown that

foi2)«

N+ where

N

the

is

— T\) 2/e -


1

1

(4.11-24)

1

of radiation planes or shields between the original surfaces.

number

Hence, a great reduction in radiation heat loss

is

obtained by using these shields.

Suppose that we Derivation of general equation for view factor between black bodies. consider radiation between two parallel black planes of finite size as in Fig. 4.1 l-2a.

4.

Since the planes are not infinite in strike surface 2, lost to the

which

is

and vice

fraction of radiation leaving surface

differential surface

we can

of the radiation from surface

The

surroundings.

Before

some

Hence, the net radiation interchange

intercepted by surface 2

by taking

size,

versa.

is

called

F 12 and must

is 1

1

does not

since

less

in all

some

is

directions

be determined for each geometry

elements and integrating over the entire surfaces.

derive a general relationship for the view factor between two

finite

bodies we must consider and discuss two quantities, a solid angle and the intensity of radiation.

A

solid angle

w

is

a dimensionless quantity which

is

solid geometry. In Fig. 4.1 l-3a the differential solid angle dco

projection of

dA 2

dA 2

a measure of an angle in 1

is

equal to the normal

divided by the square of the distance between the point

P

and area

.

dco,

=

——

dA-,2 cos 0 27 -

(4.11-25)

7

r

The

units of a solid angle are steradian or

subtended by

The

this surface

is

sr.

For a hemisphere

the

number

of sr

In,

intensity of radiation for a black body,/ B , is the rate of radiation emitted per normal to the surface and per unit solid angle in a

unit area projected in a direction

'////////////////A

'//////////////A (a)

Figure

286

4.1 1-2.

(b)

Radiation between two black surfaces: (a) two planes alone, planes connected by refractory reradiating walls.

Chap. 4

Principles

(fc)

two

of Steady-State Heat Transfer

dA^

normal to area dA^'cos 8^

normal to

cos $2

du>

dA

/-»

(b)

(a)

Figure

4.1 1-3.-

Geometry for a geometry,

specified direction as

centers

dA

is

cos



shown

and intensity of radiation : (a) solid-angle of radiation from emitting area dA.

solid angle

(b) intensity

in Fig. 4.

1

The

l-3b.

projection of

dA on

the line between

0.

dq

dA

W

(4.11-26)

cos dco

and I B is in W/m 2 sr. We assume that the black body is a diffuse surface where q is in which emits with equal intensity in all directions, i.e., / = constant. The emissive power E B which leaves a black-body plane surface is determined by integrating Eq. (4.11-26) over all solid angles subtended by a hemisphere covering the surface. The final result is as follows. [See references (C3, H 1, Kl) for details.] •

(4.11-27)

where E B

is

in

W/m

2 .

In order to determine the radiation heat-transfer rates between

we must determine

surface and arrives on a second surface. Using only black surfaces,

shown in Fig. 4.11-4, and dA 2 The line r is .

the normals to

we

consider the case

exchanged between area elements dA the distance between the areas and the angles between this line and the two surfaces are 0 and 9 2 The rate of radiant energy that leaves dA in

which radiant energy

is

l

.

l

i

Figure

Sec. 4.11

two black surfaces

the general case for the fraction of the total radiant heat that leaves a

4.1 1-4.

Area elements for radiation shape factor.

Advanced Radiation Heat-Transfer

Principles

287

in the direction given

arrives

on dA 2

by the angle 0

dq ^ 2 x

where Eqs.

doa

dA

I Bl

is

X

=

dA

I BX

and

(4.1 1-25)

dA

x

and

dA 2

From Eq. (4.1 1-27), I BX = E BX /u.

=

as seen

from dA

Combining

,.

Substituting ffT* for

=

£ B1 and aT 2

cos 0

2

for I BX into Eq. (4.1 1-29),

dA dA 2

cos 9 2

X

x

(4.11-30)

2

dA

at

—

is

,

cos 0 2 cos 0[ J/l 2

—5

=

£ fl2

for

dA

' x

(4.11-31)

from Eq.

(4.11-4)

and taking

the difference of

net heat flow,

cos 0

=

dA

cos 0 2

X

E BX /n

x

(4.1 1-31) for the

dq l2

cos 9

x

(4.11-29)

E Bi

The energy leaving dA 2 and arriving dq 2 ^

dA

/ fll

Substituting

a<7i- 2

and

(4.11-28)

(4.11-28),

^1-2 =

Eqs. (4.1 1-30)

rate that leaves

cos 9 X d(a l

the solid angle subtended by^the-area

is x

The

cos 0,.

given by Eq. (4.1 1-28).

is

a[T*'- 71)

X

dA dA-

cos 0 2

x

(4.11-32)

Performing the double integrations over surfaces A

x

and A 2

will yield the total net

heat flow between the finite areas.

q X2

cos 0

= o(Tl-Ti)

dA dA 2

cos 0 2

X

x

(4.11-33) itr

Equation

(4.

1

1-33) can also be written as

q X2

where

F X2

is

x

T\)

= A 2 F 2I «x(T? -

a geometric shape factor or view factor

total radiation leaving

which strikes

= A F l2 o(T* -

A

x

.

A which x

strikes

A 2 and F 2]

T\)

(4.11-34)

and designates the

fraction of the

represents the fraction leaving

A F 12 = A 2 F 21

(4.11-35)

1

which

is

A2

Also, the following relation exists.

valid for black surfaces

and also nonblack cos 0

1

A

i

X

The view

surfaces.

cos 0 2

factor

F X2

is

then

dA dA 2 x

(4.11-36)

J

Values of the view factor can be calculated for a number of geometrical arrangements.

5.

View factors between black bodies for various geometries.

A number

of basic relation-

ships between view factors are given below.

The

reciprocity relationship given by Eq.

^1^12

(4.

1

1-35)

is

= ^2^21

This relationship can be applied to any two surfaces

(4.11-35) i

and j.

A F = A J FH i

288

ij

Chap. 4

Principles of Steady-Slate

(4.11-37)

Heat Transfer

Figure

Radiant exchange between a and a hemisphere

4.1 1-5.

flat surface

for Example 4.1 1-2.

If

If

F 12 = 1.0. surfaces A 2 A it

A can only see surface A 2 surface A sees a number of

surface

then

,

l

F,, If

A

the surface

cannot see

x

itself

and

,

x

enclosure then the enclosure relationship 4-

+ F 13 +••• =

12

(surface

all

form an

the surfaces

is

is fiat

1.0

(4.11-38)

or convex), F,

,

=

0.

EXAMPLE

4.11-2. View Factor from a Plane to a Hemisphere Determine the view factors between a plane A{ covered by a hemisphere A 2

as

shown

in Fig. 4.

Solution:

Eq.

1-5.

1

Since surface

/I,

sees only

=A

/t,F 12

The for

A2

,

the view factor

F 12 =

1.0.

Using

(4.1 1-35),

area

F 21

A = nR

2

x

,

A 2 = 2kR

2

2

F 21

Substituting into Eq. (4.11-35) and solving

.

,

4i_M m 1

Using Eq. writing Eq.

(4.11-38) for surface

^

>2

F 22 F 2 , =

MPLE

,

1.0

-F

21

1

=

1.0

- F 12 =

1.0

-

1.0

=

0.

Also,

,

^22+^21 = 10 Solving for

_

2

Au Fu

surface A 2

(4.1 1-38) for

(4.11-35)

(4.11-39)

= 1.0-! = i

Radiation Between Parallel Disks A is parallel to a large disk of area A 2 and /I, is centered directly below A 2 The distance between the centers of the disks is R and the radius of A 2 is a. Determine the view factor for radiant

fA"/!

4.11-3.

In Fig. 4.11-6 a small disk of area

x

.

heat transfer from

Sec. 4.11

A

x

to

A2

.

Advanced Radiation Heat-Transfer

Principles

289

Solution:

The

x so that dA 2

A2

differential area for

= 2nx

dx.

taken as the circular ring of radius

is

The angle 0, = 0 2 Using Eq. .

(4.1 1-36),

dA^nx

cos 0, cos 0j

dx)

JA2

In this case the area

A

x

compared

very small

is

to

A2

so

,

dA

t

can be

and the other terms inside the integral can be assumed 2 2 2 2 constant. From the geometry shown, r = (R + x ) ' cos 0, = R/{R + 2 112 Making these substitutions into the equation for Fj 2 x ) integrated to

^4,

1

,

.

,

2

2R x dx

F 12 =

+ x2

(K 2

o

2 )

Integrating,

.R

The

+

ofF 12 q i2

where F 12

2

for

numerous geometrical

= F i2 A

i

-

a(T\

T$)

= F 2i A 2 a{T\ - T 2

the fraction of the radiation leaving/1, which

is

becomes Eq.

A 2 ..Since

the flux from

1

to

is

(4.11-34)

)

intercepted

byA 2 andF 21

2 must equal that from 2

A2

is

1.0,

factors

since

all

for a

small surface

the radiation leaving

F 12 between

.

290

4.11-7.

^4,

is

A

,

easily.

For

completely enclosed by a large surface

l

intercepted by

parallel planes are given,

Ratio of Figure

1,

(4.11-35)

Hence, one selects the surface whose view factor can be determined most

F 12

to

as given previously.

(4.1 1-35)

A,F l2 = A 2 F 2i example, the view factor

configur-

tabulated. Then,

the fraction reaching A^ from (4.1 1-34)

a

done

integration of Eq. (4.11-36) has been

ations and values

Eq.

2

and

A2

.

In Fig. 4.11-7 the view

in Fig. 4.11-8 the

view factors

for

smaller side or diameter ;

distance between planes

View factor between parallel planes directly opposed. (From W. H. Mc Adams, Heat Transmission, 3rd ed. New York: McGraw-Hill Book Company, 1954. With permission.)

Chap. 4

Principles of Steady-State

Heat Transfer

dimension

Figure

4.1 1-8.

ratio,

7=

0.1

View factor for adjacent perpendicular rectangles. [From H. Mech. Eng., 52, 699 (1930). With permission.']

C.

Hottel,

adjacent perpendicular rectangles.

View

factors for other geometries are given elsewhere

(HI, K1.P3, Wl).

4.1

1C

When

View Factors

Surfaces Are Connected

by Reradiating Walls

and A 2 are connected by nonconducting (refractory) a larger fraction of the radiation from surface 1 is intercepted by 2. This view factor is called F 12 The case of two surfaces connected by the walls of an enclosure such as a furnace is a common example of this. The general equation for this case assuming a uniform refractory temperature has been derived (M 1, C3) for two radiant sources /I, and A 2 which are not concave, so they do not see If

the

two black-body surfaces A

but reradiating walls as

l

in Fig. 4.1 l-2b,

.

,

themselves. 2 2 \-(AJA A 12 -A,F 17 2 )F 12 L_i?2 = A + A 2 — 2A F 12 AJA 2 + 1 -2(AJA 2 )F 12 .

F 12

v

'

(4.11-40)

1

y

Also, as before,

A,F X2 = A 2 F 2 q 12

The

factor

F )2

= F i2 A i0-(Tt-n)

a wall as in a furnace

is given in Fig. 4.1 1-7 and for other geometries can be For view factorsF 12 andF 12 for parallel tubes adjacent to

and also for variation in refractory wall temperature, see elsewhere no reradiating walls,

If there are

F 12

Sec. 4.11

(4.11-42)

for parallel planes

calculated from Eq. (4.11-36).

(Ml, P3).

(4.11-41)

,

=F 12

Advanced Radiation Heat-Transfer

Principles

(4.11-43)

291

4.1

A

ID

View Factors and Gray Bodies

more practical case, which is the same as and A 2 being gray with emissivities e, and

genera] and

surfaces

A

y

for Eq. (4.11-40) but with the e 2) will

conducting reradiating walls are present as before. Since the there will be

some

reflection of radiation

between the surfaces below that

which

will decrease the net radiant

for black surfaces.

q i2

be considered. Non-

two surfaces are now

The final equations

gray,

exchange

for this case are

= J 12 AMT*i-Tt)

(4.11-44)

(4.11-45)

where

J i2

new view

the

is

factor for

two gray surfaces A and A 2 which cannot see If no refractory walls are present, F 12 1

themselves and are connected by reradiating walls. is

used

in place of

F 12

Eq.

in

(4.1 1-41).

A

Again, l

:T l2

= A 2 J 21

"

(4.11-46)

EXAMPLE

4.11-4. Radiation Between Infinite Parallel Gray Planes Derive Eq. (4.11-22) by starting with the general equation for radiation between two gray bodies A and y4 2 which are infinite parallel planes having ,

and

emissivities a,

Solution:

comes F 12 surface

2,

£2

,

respectively.

Since there are no reradiating walls, by Eq. (4.11-43), F, 2 beAlso, since all the radiation from surface 1 is intercepted by

.

F 12 =

1.0.

Substituting into Eq.

(4.1 1-45),

noting that

AJA 2 = 1.0,

1

1

£2

£,

Then using Eq. q 12

(4.1 1-44),

=

J X2 AMT\ -

71)

=A

1

a{T* 1

-

T\) -

—+

1

£2

£,

This

is

identical to Eq. (4.1 1-22).

EXAMPLE

Complex View Factor for Perpendicular Rectangles for the configuration shown in Fig. 4.11-9 of the rectangle with area A 2 displaced from the common edge of rectangle^, and perpendicular to /I,. The temperature of A l is T, and that of A 2 and A 3 is 4.11-5.

Find the view factor F 12

T2

.

Solution:

area

A2

The

plus

A3

area

A 3 is

as/i l23)

.

a fictitious area between areas

The view factor F 1(23

)

for areas

A 2 and A Call the A and/i (23) can be }

.

1

obtained from Fig. 4.11-8 for adjacent perpendicular rectangles. Also, F 13 can be obtained from Fig. 4.11-8. The radiation interchange between A 1 and A {23) is equal to that intercepted by A 2 and by A 2 .

A.F^aiT* 292

T$)

=

A,¥, 2 a{J\

- T 2 ) + A,F„(T\ - T 2

Chap. 4

Principles

)

(4.11-47)

of Sieady-Siate Heat Transfer

Figure

Configuration

4.1 1-9.

Example

for

4.11-5.

Hence, J

Solving for

F 12

— A F l2 + A F i3

^1(23)

l

,

F 12 ~ Methods

(4.11-48)

l

-^1(23) ~~

example can be employed

similar to those used in this

factors for a general orientation of

(4.11-49)

Fl3

two rectangles

shape

to find the

perpendicular planes or parallel

in

rectangles (C3,H1,K1).

EXAMPLE

4.11-6. Radiation to a Small Package small cold package having an area/1 and emissivity £ is at temperature T1 It is placed in a warm room with the walls at T2 and an emissivity £, Derive the view factor for this using Eq. (4.1 1-45), and the equation for the

A

1

t

.

radiation heat transfer.

For the small surface A completely enclosed by the enclosure A 2 F 12 = F l2 by Eq. (4.11-43), since there are no reradiating (refractory) walls. Also, F 2 = 1.0, since all the radiation from A is intercepted by the enclosure A 2 because A does not have any concave surfaces and cannot "see" itself. Since A 2 is very large compared \o A A /A 2 = 0. Substituting

Solution:

x

,

i

x

x

x ,

x

into Eq. (4.1 1-45), 1

F,12,

A-,2 n

\£, \t 2

Substituting into Eq. q l2

This

is

the

For methods

same

1

Y1

\£

//

+

o(\E,

= +

£i

1

El

(4.1 1-44),

= ? 12 A

l

o(T* l

-Tt) =

e

l

A o(T+-T$) 1

as Eq. (4.10-6) derived previously.

to solve

complicated radiation problems involving more than four or

five heat-transfer surfaces,

matrix methods to solve these problems have been developed

and

are discussed in detail elsewhere (HI, Kl).

4.1

IE

7.

Introduction to absorbing gases in radiation.

Radiation

in

Absorbing Gases

As discussed in this section, However, most gases that

liquids emit radiation over a continuous spectrum.

atomic or diatomic, such as He, Ar, radiation

Sec. 4.11

;

i.e.,

H 02 2

,

,

and

N2

,

and

mono-

are virtually transparent to thermal

they emit practically no radiation and also

Advanced Radiation Heat-Transfer

solids

are

Principles

do not absorb

radiation. Gases

293

moment and higher polyatomic gases emit significant amounts of radiation and also absorb radiant energy within the same bands in which they emit radiation. with a dipole

NH

and organic vapors. C0 2 H 2 0, CO, S0 2 3 For a particular gas, the width of the absorption or emission bands depends on the pressure and also the temperature. If an absorbing gas is heated, it radiates energy to the

These gases include

,

The

cooler surroundings.

,

,

net radiation heat-transfer rate between surfaces

some

these cases because the gas absorbs

is

decreased

in

of the radiant energy being transported

between the surfaces.

The absorption

Absorption of radiation by a gas.

2.

of radiation in a gas layer can be

number of

described analytically since the absorption by a given gas depends on the

molecules

in

the path of radiation. Increasing the partial pressure of the absorbing gas or

amount

the path length increases the

same wavelength

after

decrease dl x

where

J x is

in

W/m 2

.

it

in

define I x0 as the intensity of and 1 XL as the intensity at the

enters the gas

having traveled a distance of L

a gas layer of thickness dL, the

We

of absorption.

radiation at a particular wavelength before

in the gas. If

intensity,^

=-a

x lx

,

is

the

beam impinges on I x and dL.

proportional to

dh

(4.11-50)

Integrating,

The constant a x depends on

the particular gas,

radiation. This equation

called Beer's law.

is

its partial pressure, and the wavelength of Gases frequently absorb only in narrow-

wavelength bands.

mean beam length of absorbing gas. The calculation methods for gas For the purpose of engineering calculations, Hottel (Ml) has presented approximate methods to calculate radiaton and absorption when gases such as CO z and water vapor are present. Thick layers of a gas absorb more energy than do thin layers. Hence, in addition to specifying the pressure and temperature of a gas, we must specify a characteristic length (mean beam length) of a gas mass to determine the emissivity and absorptivity of a gas. The mean beam length L depends on Characteristic

3.

radiation are quite complicated.

the specific geometry.

For a

a

black differential receiving surface area

dA

located

in

the center of the base of

hemisphere of radius L containing a radiating gas, the mean beam length

mean beam

length has been evaluated for various geometries and

For other shapes,

L can

is

V

is

The

L.

1-1.

be approximated by

L = 3.6A where

is

given in Table 4.1

volume of the gas

in

m3 A ,

(4.11-52)

the surface area of the enclosure in

m

2 ,

and

L

is

in

m. Emissivity, absorptivity, and radiation of a gas. Gas emissivities have been correlated and Fig. 4.11-10 gives the gas emissivity e Q ofC0 2 at a total pressure of the system of 1.0 atm abs. The pG is the partial pressure of C0 2 in atm and the mean beam length L in m. The emissivity e G is defined as the ratio of the rate of energy transfer from the hemispherical body of gas to a surface element at the midpoint divided by the rate of

4.

energy transfer from a black hemisphere surface of radius

L and

temperature

TG

to the

same element.

294

Chap. 4

Principles

of Steady-State Heat Transfer

Table

Mean Beam Length for Gas Radiation

4.1 1-1.

Entire Enclosure Surface

,

Mean Beam

Geometry of Enclosure

Sphere, diameter

to

(Ml R2, P3)

D

Length,

L

0.65D

Infinite cylinder,

diameter

Cylinder, length

=

D

0.95D

D

diameter

0.60D 1.8D

Infinite parallel plates, separation

D

distance

R

Hemisphere, radiation to element in base, radius

R

Cube, radiation to any face, side D Volume surrounding bank of long tubes

0.60D 2.80

with centers on equilateral triangle, clearance

The

=

tube diameter

rate of radiation emitted

element, where

eg

is

evaluated at

D

-

from the gas

TG

.

is

TG

az G

If the surface

in

W/m 2

of receiving surface

element at the midpoint

at 7i

is

beaa G T\, where a G is surface atTj. The cc G of C0 2

radiating heat back to the gas, the absorption rate of the gas will the absorptivity of the gas for blackbody radiation from the is

determined from Fig. 4.11-10 at Ti but instead of using the parameter of p G L, the

parameter p G

L(TJTG)

is

used.

The

resulting value from the chart

is

then multiplied by

0.004

0.003

250 500

FIGURE

Sec. 4.11

4.1 1-10.

1250 1000

I

2750

1750

2000 Temperature (K) 1500

2500

Total emissivity of the gas carbon dioxide at a total pressure of 1.0 atm. {From W. H. Mc Adams, Heat Transmission, 3rd ed. New York : McGraw-Hill Book Company, 1954. With permission.)

Advanced Radiation Heat-Transfer

Principles

295

(T0 /T,) 0

-

65

to give a G

surface of finite area

The

.

A

the total pressure

T\)

(4.11-53)

not 1.0 atm, a correction chart

is

is

available to correct the

C0 2 Also, charts are available for water vapor (HI, Kl, Ml, P3). When C0 2 and H 0 are present the total radiation reduced somewhat, since each gas

emissivity of

both

and a black

then

is

q=oA(z c T G -a c When

TG

net rate of radiant transfer .between a gas at

at Tj

,

.

is

2

somewhat opaque to radiation from available (HI, Kl, Ml, P3).

EXAMPLE

4.11-7.

A

in the

is

the other gas. Charts for these interactions are also

Gas Radiation to a Furnace Enclosure form of a cube 0.3 m on a side inside, and these interior walls can be approximated as black surfaces. The gas inside at 1.0 atm total pressure and 1 100 K contains 10 mole % C0 2 and the rest is 0 2 and N 2 The small amount of water vapor present will be neglected. The walls of the furnace are maintained at 600 K by external cooling. Calculate the total heat transfer to the walls neglecting heat transfer by convection. furnace

is

.

From Table

Solution:

cube face is

Pq

From

=

is

L=

0.60

0.10(100)

=

Fig. 4.1 1-10,

To

(0.0180)(6O0/l 100)

0.60(0.30)

0.10

=

eg

obtain a G

mean beam length for radiation to a 0.180 m. The partial pressure of C0 2 atm. Then p G L = 0.10(0.180) = 0.0180 atm-m.

4.11-1, the

D=

0.064 at

=

=

aG

Substituting into Eq.

For

six sides,

A =

100 K.

1

=

q

and p G L(TJTG )

=

by the correction factor(TG /T

1

0 65

=

-

)

,

the

0.0712

(4.1 1-53),

4.795 x 10

x

K

600

Fig. 4.11-10, the uncorrected 0 65

r?)

x 10- 8 )[0.064(1100) 4

(5.676

6(0.3

this

0.048(1 100/600)

t% - * G

\= =

=

we evaluate a G at T, 0.00982 atm-m. From

,

value of a G = 0.048. Multiplying final correction value is

=

T"G

=

3

0.3)

=

W/m = 2

=

0.540

4.795

m

4.795(0.540)

2

=

.

-

0.0712(600)

kW/m

4 ]

2

Then, 2.589

kW

For the case where the walls of the enclosure are not black, some of the radiation is reflected back to the other walls and into the gas. As an approximation when the emissivity of the walls is greater than 0.7, an effective emissivity e' can be striking the walls

used.

F,

.

£

where

e is

'

=

^

(4.11-54)

the actual emissivity of the enclosure walls.

Then Eq.

(4.1 1-53) is

modified to

give the following (Ml):

Q

=

oAz'{e g

Tg —

u G T*)

(4.11-55)

Other approximate methods are available for gases containing suspended luminous flames, clouds of nonblack particles, refractory walls and absorbing gases present, and so

on(Ml,P3).

2%

Chap. 4

Principles

of Steady-State Heat Transfer

HEAT TRANSFER OF NON-NEWTONIAN FLUIDS

4.12

4.12A

Most

Introduction of the studies

However,

a

on heat

transfer, with fluids

wide variety of non-Newtonian

have been done with Newtonian

fluids are

chemical, biological, and food processing industries.

encountered

To

in

fluids.

the industrial

design equipment to handle

must be available or must be measured experimentally. Section 3.5 gave a detailed discussion of rheological constants for non-Newtonian fluids. Since many non-Newtonian fluids have high effective viscosities, they are often in laminar flow. Since the majority of non-Newtonian fluids are pseudoplastic fluids, which can usually be represented by the power law, Eq. these fluids, the flow property constants (rheological constants)

(3.5-2), the

discussion will be concerned with such fluids.

For other

fluids,

the reader

is

referred to Skelland (S3).

Heat Transfer

4.12B

Inside

Tubes

1. Laminar flow in tubes. A large portion of the experimental investigations have been concerned with heat transfer of non-Newtonian fluids in laminar flow through cylindrical

tubes.

The

physical properties that are needed for heat transfer coefficients are density,

heat capacity, thermal conductivity, and the rheological constants K' and In heat transfer in a fluid in laminar flow, the

mechanism

is

ri

or

K and n.

one of primarily

conduction. However, for low flow rates and low viscosities, natural convection effects can be present. Since many non-Newtonian fluids are quite "viscous," natural convection effects are reduced substantially. fluids, the

For laminar flow inside

circular tubes of

power-law

equation of Metzner and Gluck (M2) can be used with highly "viscous"

non-Newtonian fluids with negligible natural convection Graetz number N Gx > 20 and ri > 0. 10.

for horizontal or vertical tubes

for the

(N Nu )„ =

^ k

A

= U55^(N G J^ (f)° \y

*

(4.12-1)

where,

c

3

N- = The

=

——+ 3n'

n Dvp c„u

mc„

(4.12-2)

D

iV

l£- 4 ~V

viscosity coefficients y b at temperature

1

Tb

and y„

D

n

"" N *Z 4 at

Tw

„

, ,

„N

(4 12 " 3) -

are defined as

K '£"" = ^± = ^l 1

y±

=

(4.12-4)

kg/m 3 flow rate m in kg/s, length of heated section of tube L in m, inside diameter Bin m, the mean coefficient 2 h a in W/m K, and K and ri rheological constants (see Section 3.5). The physical properties and K b are all evaluated at the mean bulk temperature Tb and K w at the

The nomenclature

is

as follows: k in

W/m

•

K, c p

in

J/kg K, p

in

,

•

average wall temperature

Sec. 4.12

Tw

.

Heat Transfer of Non-Newlonian Fluids

297

found to not vary appreciably However, the rheological constant K' or K has been found to vary appreciably. A plot of log K' versus 1/Tabs (CI) or versus T°C (S3) can often be approximated by a straight line. Often data for the temperature effect on K are not available. Since the ratio KJK„ is taken to the 0.14 power, this factor can sometimes be neglected without causing large errors. For a value of the ratio of 2 1, the error is only about 10%. A plot of log viscosity versus 1/Jfor Newtonian fluids is also often a straight line. The value of h a obtained from Eq. (4.12-1) is the mean value to use over the

The value of

the rheological constant ri or n has been

over wide temperature ranges

(S3).

:

tube length

L with

the arithmetic temperature difference AT„.

a

when T„

—^———

—

AT =

whole tube and

the average wall temperature for the

is

temperature and

Tbo

the outlet bulk temperature.

q

EXAMPLE

=

ha

AAT

a

=

(412-5)

The

heat flux q

h a (jtDL)

AT

T

bi

the inlet bulk

'

-

0

is

is

(4.12-6)

Heating a Non-Newtonian Fluid in Laminar Flow flowing at a rate of 7.56 x 10" 2 kg/s inside a 25.4-mm-ID tube is being heated by steam condensing outside the tube. The fluid enters the heating section of the tube, which is 1.524 m long, at a temperature of 37.8°C. The inside wall temperature T„ is constant at 93.3°C. 4.12-1.

A non-Newtonian

fluid

=

kg/m 3

c pm = 2.093 a power-law fluid having the following flow property (rheological) constants: n = ri = 0.40, which is approximately constant over the temperature range encountered, and

The mean

physical properties of the fluid are p

kJ/kg K, and

k

=

W/m

1.212

•

K. The

fluid

2

•

is

temperature of the Solution:

The

fluid

if it is

solution

is

For

this fluid a plot of log

Calculate the outlet bulk

laminar flow.

in

trial

,

is

K = 39.9 N s"Vm at 37.8°C and 62.5 at 93.3°C. K versus T°C approximately a straight line. 1

1041

and error since the outlet bulk temperature

Tbo of the fluid must be known to calculate ha from Eq. (4.12-2). Assuming Tbo = 54.4 = C for the first trial, the mean bulk temperature Tb is (54.4 + 37.8)/2, or 46.1 °C. Plotting the two values of

K

given

at 37.8

and 93.3°C as log

K

versus

T°C and drawing a straight line throuah these two points, a value of 123.5 at Tb = 46.1°C is read from the plot. At Tw = 93.3°C, K w = 62.5. Next,

K

b

of

calculated using Eq. (4.12-2).

6 is

3w'+ 4ri

1

=

3(0.40)

+

1

=

4(0.40)

Substituting into Eq' (4.12-3),

_mcJL _

'

From

c *~

(7.56 x 1Q-

kL~

3 X2.093 x 10 )

1.212(1.524)

Eq. (4.12-4), y„

yw

298

2

_ Kb =

K„

Chap. 4

123.5

62.5

Principles of Steady-State

Heat Transfer

Then

substituting into Eq. (4.12-1),

h.D

(0.0254)

/i.

^'

W

1.75(1. 375)

Solving, h a

By

=

448.3

W/m 2

•

K.

a heat balance, the value of q in

q

This

equated

is

to

u/

tnesll3lA ,

=

1.212

1/3

W

/y

t

Vy

(85.7)

is

1/3

123.5' /123 5\ 014 {

(4.12-1)

-^J

V 62

-

5

as follows:

= mc pm (Tb0 - T

(4.12-7)

bt )

Eq. (4.12-6) to obtain q

= mc pm (Tbo - T6i = h.frDL) ATa

(4.12-8)

)

The arithmetic mean temperature (T„

difference

AT

by Eq. (4.12-5)

0

- T + (X ~ TJ bi )

SsJ

=

2

74.4 '"'

-

0.5 7L *"

"~

Substituting the known' values in Eq. (4.12-8) and solving for 3

(7.56 x 10" X2.093 x 10 )(T> 0 2

= T = fco

is

448.3(71

-

Tba

,

37.8)

x 0.0254 x 1.524)(74.4

- 0.5 TJ

54.1°C

This value of 54.1°C is close enough to the assumed value of 54.5°C so that a second trial is not needed. Only the value of K b would be affected. Known values can be substituted into Eq. (3.5-1 1) for the Reynolds number to show that it is less than 2100 and is laminar flow.

For vection

"viscous" non-Newtonian power-law fluids

less

may

affect the heat-transfer rates.

in

laminar flow, natural con-

Metzner and Gluck (M2) recommend use of an

empirical correction to Eq. (4.12-1) for horizontal tubes.

2.

Turbulent flow

in

lubes.

For turbulent flow of power-law

fluids

through tubes, Clapp

(C4) presents the following empirical equation for heat transfer:

W Nu

=^ =

0.004

UN,!

0 9' -

(4.12-9) _

where log

N Rt>gcn

is

defined by Eq. (3.5-1

mean temperature

driving force.

1)

and h L

The

is

k

the heat-transfer coefficient based on the

fluid properties are

evaluated

at

the bulk

mean

temperature. Metzner and Friend (M3) also give equations for turbulent heat transfer.

4.1

2C

Natural Convection

Acrivos (Al, S3) gives relationships for natural convection heat transfer to power-law fluids

from various geometries of surfaces such as spheres, cylinders, and

Sec. 4.12

Heat Transfer of Non-Newtonian Fluids

plates.

299

SPECIAL HEAT-TRANSFER COEFFICIENTS

4.13

4.1 3A

/.

Heat Transfer

in

Many

Introduction.

Agitated Vessels

chemical and biological processes are often carried out in agi-

As discussed

tated vessels.

in Section 3.4, the liquids are generally agitated in cylindrical

an impeller mounted on a shaft and driven by an electric motor. Typical

vessels with

agitators and vessel assemblies have

been shown

in Figs. 3.4-1

and

3.4-3.

Often

necessary to cool or heat the contents of the vessel during the agitation. This

done by heat-transfer

surfaces,

is

is

it

usually

which may be in the form of cooling or heating jackets immersed in the liquid.

in

In Fig. 4.13-la a vessel with a cooling or heating jacket

is

the wall of the vessel or coils of pipe

2.

Vessel with heating jacket.

When

shown.

heating, the fluid entering

and leaves at the bottom. The vessel

is

often steam, which condenses inside the jacket

equipped with an agitator and

is

in

most cases

also

with baffles (not shown).

Correlations for the heat-transfer coefficient from the agitated Newtonian liquid inside the vessel to the jacket walls of the vessel

have the following form:

(4.13-1)

where h

W/m 2 Da

is

•

the heat-transfer coefficient for the agitated liquid to the inner wall in

is

K, D,

is

the inside diameter of the tank in

diameter of agitator

density in

kg/m 3 and ,

p. is

in

m,

N

is

m, k

is

thermal conductivity in

rotational speed in revolutions per sec,

range(NL

Tw

=

.

is

•

K,

fluid

liquid viscosity in Pa-s. All the liquid physical properties are

which

evaluated at the bulk liquid temperature except

temperature

W/m p

Below are

listed

some

is

evaluated at the wall

correlations available and the Reynolds

number

£>>p/K).

heating fluid

—

heating

(b)

(a)

Figure

4.13-1.

Heat

transfer in agitated vessels: (a) vessel with heating jacket,

(b) vessel with heating coils.

300

Chap. 4

Principles of Steady-State

Heat Transfer

Paddle agitator with no

1.

a 2.

=

=

b

0.54,

Anchor

=

=i

m=

'

agitator with

no

jVr c

=

300 to 3 x 10 5

=

N'Rc

0.14,

3 30 to 3 x 10

Some

U

m=

=i

b

no b

0,633,

typical overall

m=

b=\,

Helical ribbon agitator with

=

0.14.

N'Rc

= 500

to 3

x 10 5

baffles (Ul)

a =0.36,

a

m=

=i

b

0.74,

a =1.0,

5.

0.21,

Flat-blade turbine agitator with baffles (B4, B5) a

4.

m=

=i

b

0.36,

Ul)

Flat-blade turbine agitator with no baffles (B4) a

3.

baffles (C5,

baffles

=

},

N'Rc

0.18,

0.18,

N'Rc

=

=

10 to 300

300 to 4 x 10*

(G4) wi

=

=

0.18,

8 to'lO

3

values for jacketed vessels for various process applications

are tabulated in Table 4.13-1.

EXAMPLE

Heat-Transfer Coefficient in Agitated Vessel with Jacket A jacketed 1.83-m-diameter agitated vessel with baffles is being used to heat a liquid which is at 300 K. The agitator is 0.61 m in diameter and is a flat-blade turbine rotating at 100 rpm. Hot water is in the heating jacket. The wall surface temperature is constant at 355.4 K. The liquid has the 3 following bulk physical properties: p = 961 kg/m , c p = 2500 J/kg K, 4.13-1.

•

k

=

0.173

W/m

K, and

/i

=

1.00

Pa

•

s at

300

K

and 0.084 Pa

s

at 355.4

K. Calculate the heat-transfer coefficient to the wall of the jacket.

Table

4.13-1.

Typical Overall Heat-Transfer Coefficients

in

Jacketed Vessels

U Fluid in

Fluid

in

Wall

btu

w

2 hft -°F

m 2 -K

Jacket

Vessel

M aterial

Agitation

Steam

Water

Copper

None

150

852

Simple

250

1420

125

710

(PI)

Ref.

(PI)

stirring

Steam

Paste

Cast iron

Double scrapers

Steam

Boiling

Copper

None

250

1420

(PI)

Enameled

None

200

1135

(PI)

Stirring

300

1700

None

70

398

(PI)

Agitation

30

170

(CI)

water

Steam

Milk

cast iron

Hot water

Steam

Cold water

Tomato

Enameled cast iron

Metal

puree

Sec. 4.13

Special Heat-Transfer Coefficients

301

Solution

:

The following are given D,

=

1.83

m K)

=

1.00

/U355.4 K)

=

0.084 Pa

/i(300

First, calculating the

N *<

Pa

Reynolds number 2 a

s

•

•

at

=

1.00

=

s

100/60 rev/s

kg/m

0.084

s

•

kg/m

s

300 K,

2

D Np ==

N=

D a = 0.61m

(0.61) (100/60X961)

=

-1T

.

=

596

m=

0.14

loo

The Prandtl number is

Using Eq.

(4.13-1) with a

k

Substituting and solving for

2/3,

and

h,

0.74( 5 96)- (14

=

170.6

450)-3(l000)

W/m 2 K

with a turbine agitator

Vessel with heating coils.

cooling coil

is

(30.0 btu/h

2 •

a paddle agitator with

k

This holds for a Reynolds

For a

When

is

ft

°F)

power-law non-Newtonian

also available elsewhere (C6).

In Fig. 4.13-lb an agitated vessel with a helical heating or

shown. Correlations

for the heat-transfer coefficient to the outside surface

of the coils in agitated vessels are listed

For

14

(4.13-1)

correlation to predict the heat-transfer coefficient of a

fluid in a jacketed vessel

-

014

^§ .

3.

=

= 0.74(N^ 3 (iV Pr )" 3 (ii-j

-

A

0.74, b

( u \0

) /iD,

/,

=

no

'

below

for various types of agitators.

baffles (C5),

\

n

number range

\ k

J

of 300 to4 x 10

5 .

flat-blade turbine agitator with baffles, see (Ol).

the heating or cooling coil

is

flat-blade turbine, the following correlation

in

the

form of

vertical

tube baffles with a

can be used (Dl).

D 0 is the outside diameter of the coil tube in m,n b is the number of vertical and n f is the viscosity at the mean film temperature. Perry and Green (P3) give typical values of overall heat-transfer coefficients

where

baffle

tubes,

coils

immersed

4.13B

in

U for

various liquids in agitated and nonagitated vessels.

Scraped Surface Heat Exchangers

Liquid-solid suspensions, viscous aqueous and organic solutions, and numerous food products, such as margarine and orange juice concentrate, are often cooled or heated in a

scraped-surface exchanger. This consists of a double-pipe heat exchanger with a jacketed

302

Chap. 4

Principles of Steady-State

Heat Transfer

r

Figure

4.

Scraped surface heat exchanger.

3-2.

1

cylinder containing steam or cooling liquid and an internal shaft rotating and fitted with

wiper blades, as shown

The viscous the rotating shaft

in Fig. 4.

liquid

1

3-2.

product flows

and the inner

pipe.

at

low velocity through the central tube between

The

rotating scrapers or wiper blades continually

scrape the surface of liquid, preventing localized overheating and giving rapid heat

This device

transfer.

Skelland

in

some cases

et al. (S4)

also called a votator heat exchanger.

is

give the following equation to predict the inside heat-transfer

coefficient for the votator.

HW(™Hv)°W'W" where

D =

a

=

0.014

p

=

0.96

for viscous liquids

a

=

0.039

P

=

0.70

for nonviscous liquids

Ds =

diameter of rotating shaft

diameter of vessel

velocity of liquid in m/s, agitator.

Data cover

a

N

in

=

m,

agitator speed

in

rev/s,

and

nB

in

m,

,.3-,

v

= number

—

axial flow

of blades on

region of axial flow velocities of 0.076 to 0.38

m/min and

rotational speeds of 100 to 750 rpm.

Typical overall heat-transfer coefficients

(300 btu/h

2 •

ft

•

in

°F) for cooling margarine with

NH

3

,

with steam, 1420 (250) for chilling shortening with

cream with water

4.13G 1

.

1700 W/m K 2270 (400) for heating applesauce

food applications are

NH

3

,

2

U=

and 2270

(400) for cooling

(B6).

Extended Surface or Finned Exchangers

Introduction.

The

on the outside of a heat exchanger

use of fins or extended surfaces

pipe wall to give relatively high heat-transfer coefficients in the exchanger

common. An automobile

radiator

is

through a bank of tubes and loses heat to the surfaces receive heat from the tube walls

Two common wall

and the direction of gas flow

is

number

quite

On

the outside of the tubes, extended

it

to the air

by forced convection.

shown

in Fig.

of longitudinal fins spaced around the tube

parallel to the axis of the tube. In Fig. 4.13-3b the gas

flows normal to the tubes containing

Sec. 4.13

air.

and transmit

types of fins attached to the outside of a tube wall are

4.13-3. In Fig. 4.13-3a there are a

is

such a device, where hot water passes inside

many circular or

Special Heat-Transfer Coefficients

transverse

fins.

303

(a)

Figure

(b)

Two common

4.13-3.

types

on

of fins

a

section

circular

of

tube:

(a) longitudinal fin, {b) circular or transverse fin.

The

qualitative effect of using extended surfaces can be

(4.13-5) for a fluid inside a tube coefficient of h D

The

resistance

;

of the wall can often be neglected.

-R me a) ,

the tube.

A 0 and

For example,

The presence of

hence reduces the resistance l/h 0 A B of the

if

we have

outside the tube, which

is

condensing steam, which

h-t for

fluid

is

the fins on the on the outside of

very large, and/i D for

quite small, increasing A„ greatly reduces l/h 0

AB

.

This in

turn greatly reduces the total resistance, which increases the heat-transfer rate. positions of the

two

fluids are reversed with air inside

heat transfer could be obtained by using

Equation

(4.13-5)

is

surface of the bare tube resistance to heat flow

area of fin.

A

fin

in Eq.

.

outside increases

air

shown approximately

having a heat-transfer coefficient of h and an outside

surface

is

fin efficiency n

is

little

If

the

increase in

fins.

only an approximation, since the temperature on the outside not the

same

as that at the

by conduction from the

not as f

and steam outside,

efficient as a unit

end of the

fin tip to

fin

because of the added

the base of the

fin.

Hence, a unit

area of bare tube surface at the base of the

has been mathematically derived for various geometries of fins.

Derivation of equation for fin efficiency. We will consider a one-dimensional fin exposed to a surrounding fluid at temperature Tx as shown in Fig. 4.13-4. At the base of the^fin. the temperature is 7^, and at point x it is T. At steady state, the rate of heat

2.

conducted

in to the

element at x

is

q^

and

is

equal to the rate of heat conducted out plus

the rate of heat lost by convection.

1x\x=

1x\x + *x

+

(4.13-6)

1c

Substituting Fourier's equation for conduction and the convection equation,

-kA

— dx

-kA

— dx

+

h(P

AxXT - TJ

(4.13-7)

where A is the cross-sectional area of the fin in m 2 P the perimeter of the fin (P Ax) the area for convection. Rearranging Eq. (4.13-7), dividing by Ax, and ,

304

Chap. 4

Principles of Steady-State

in

m, and

letting

Ax

Heat Transfer

Figure

Heat balance for one-dimensional conduction and convection

4.13-4.

a

in

rectangular Jin with constant cross-sectional area.

approach zero, d

2

T

hP

^ (T - TJ =

dx 2 Letting 0

= T - Tm

,

(4 - 13- 8)

°

Eq. (4.13-8) becomes 2

hP

d 0

dV^u 0 = 0

(4

-

13- 9)

The first boundary condition is that 0 = 9 0 = T0 — Tm at x = 0. For the second boundary condition needed to integrate Eq. (4.13-9), several cases can be considered, depending upon the physical conditions at x = L. In the first case, the end of the fin is insulated and dO/dx = 0 at x = L. In the second case the fin loses heat by convection from the tip surface so that —k(dT/dx) L = h{TL — TJ. The solution using case 2 is quite involved and will not be considered here. Using the first case where the tip is insulated, integration of Eq.

(4.

1

3-9) gives

cosh 0 -=

m =

The

(hP/kA)

1 '

—

x)]

*

cosh

uQ

where

\m{L L

(4.13-10)

mL

2 .

heat lost by the

fin is

expressed as

dT ^~ kA Tx Differentiating Eq. (4.13-10) with respect to q

=

(hPkA)

ll2

(4.13-11)

x and combining

(Ta

- TJ

tanh

it

with Eq. (4.13-1

mL

1),

(4.13-12)

In the actual fin the temperature T in the fin decreases as the tip of the fin is approached. Hence, the rate of heat transfer per unit area decreases as the distance from the tube base efficiency

r\

transferred

s if

is

is

increased.

indicate this effectiveness of the fin to transfer heat, the

—

=

- TJ tanh mL kpixt0 -tj

(hPkAy i2 (T0

T0

— —zr

the entire fin were at the base temperature

n*

where

To

defined as the ratio of the actual heat transferred from the

fin

fin

to the heat

.

tanh

mL (413- 13)

PL is the entire surface area of fin.

Sec. 4.13

Special Heat-Transfer Coefficients

305

The expression

for

mL is mL =

For

fins

which are

thin, 2t

small

is

U)

compared

for

length of the fin

by

t/2,

a

fin

1 '

2

J

L

with an insulated

fin loses

1/2

2w and

to

;2h\

Equation (4.13-15) holds

2t)

(4.13-14)

'

hold for the case where the

+

h(2w

L

heat from

(4.13-15)

tip.

its tip.

where the corrected length

L

c

This equation can be modified to

This can be done by extending the to use in Eqs. (4.13-13) to (4.13-15)

is

Lc = L +

The

fin efficiency

(4.13-16)

calculated from Eq. (4.13-13) for a longitudinal fin

4.13-5a. In Fig. 4.13-5b the fin efficiency for a circular fin

abscissa

on the curves

is

Lc {h/kt) 1/2 and

EXAMPLE 4.13.-2.

L

not

Fin Efficiency

c

{2h/kt)

112

is

is

presented.

shown in Fig. Note that the

as in Eq. (4.13-15).

and Heat Loss from Fin

in Fig. 4.13-3b (/c = 222 W/m-K) is attached to a copper tube having an outside radius of 0.04 m. The length of and the thickness is 2 mm. The outside wall or tube base is the fin is 0.04 and the external surrounding air at 343.2 has a convective at 523.2

A

circular

aluminum

fin

as

shown

m

K

30

from the

fin.

loss

306

K

coefficient of

W/m 2

•

K. Calculate the

Chap. 4

fin efficiency

Principles

and

the rate of heat

of Steady-State Heat Transfer

The

Solution: r,

=

16),

0.04 m,

Lc = L +

given data are

=

t

=

0.002 m, k

=

t/2

+

0.040

y/2

T0 =

222

Tm = 343.2 K, L = 0.04 m, K, h = 30 W/m 2 K. By Eq. (4.130.041 m. Then,

523.2 K,

W/m =

0.002/2

•

1/2

30

=

=

(0.041)

0.337

222(0.002)

+

Also, (L c

The

0.89.

=

r, )/r,

(0.041

+

=

0.040)/0.040

heat transfer from the q/

2.025.

Using

=n / hA J{T0 -T

is the outside surface area (annulus) of the following for both sides of the fin

+

2tt[(Lc

r,)

2

-

2

(longitudinal

0.040)

given by the

fin) .'

.

+

is

(circular fin)

(r,) ]

Hence, 2n[(0.041

and

fin

(4.1 3-1 8)

A f = 2n(Lc xw)

As =

^=

(4.13-17)

a) )

where A {

As =

Fig. 4.13-5b,

fin itself is

2

-

=

2

(0.040) ]

3.118 x 10

-2

m

2

Substituting into Eq. (4.13-17),

=

qf

3.

10" 2 X523.2 0.89(30X3.1 18 x

-

343.2)

We

Overall heat-transfer coefficient for finned tubes.

=

149.9

W

consider here the general case

where heat transfer occurs from a fluid inside a cylinder or tube, through the cylinder metal wall A of thickness Ax^ and then to the fluid outside the tube, where the tube has fins on the outside. The heat is transferred through a series of resistances. The total heat q leaving the outside of the tube is the sum of heat loss by convection from the base of the bare tube q, and the loss by convection from the fins,^. similar to Fig. 4.3-3b,

,

=

q

+

q,

qf

=

K a,(t0 - rj +

h0

as ^t0 - rj

This can be written as follows as a resistance since the paths are

q

=

(h 0

+

A,

h0

r°~

A j n s ){T, - TJ =

h 0 (A

where A,

is

t

Tco

+A f n f

(4.13-19)

in parallel.

=

(4.13-20)

)

and/i„ the fins, A f the area of the fins, resistance in Eq. (4.3-20) can be substituted for the

the area of the bare tube between the

outside convective coefficient. resistance {\/h 0

A0

)

in

The

Eq. (4.3-15) for a bare tube to give the overall equation for a finned

tube exchanger. 9

where

T4

is

~

A,

l/h,

+ Ax A /k A A A lm +

l/h 0{A,

+A f n f )~ l

the temperature of the fluid inside the tube and T,

temperature. Writing Eq. (4.13-21)

based on the inside area

U,

T,)

coefficient l/

;

and

=

of fins

the outside fluid

form of an overall heat-transfer

A i? q= U-,AITA —

(4.13-22) l/h-,

The presence

in the

R

+ Ax A AJk A A A lm + AJh 0 {A, + A { nf)

on the outside of the tube changes the

characteristics of the fluid

flowing by the tube (either flowing parallel to the longitudinal finned tube or transverse

Sec. 4.13

Special Heat-Transfer Coefficients

307

to the circular finned tube). Hence, the correlations for fluid flow parallel to or transverse to bare tubes

cannot be used

con vective

to predict the outside

are available in the literature (K4,

Ml

,

coefficient h 0

Correlations

.

PI P3) for heat transfer to various types of fins. ,

DIMENSIONAL ANALYSIS IN HEAT TRANSFER

4.14

4.14A

Introduction

As seen

in

many

and heat transfer, many dimensionless number and Prandtl number, occur in these correlations.

of the correlations for fluid flow

groups, such as the Reynolds

Dimensional analysis

is

group the variables

often used to

into dimensionless parameters or

numbers which can be

in a

given physical situation

and

useful in experimentation

correlating data.

An important way

of obtaining these dimensionless groups

is

to use

analysis of differential equations described in Section 3.11. Another useful

dimensionaJ

method

is

the

Buckingham method, in which the listing of the significant variables in the particular physical problem is done first. Then we determine the number of dimensionless paramwhich the variables

eters into

4.14B 1.

may

be combined.

Buckingham Method

Heat

transfer inside a pipe.

The Buckingham theorem, given in Section 3.11 among q quantities or variables whose units may

states that the function relationship

be given

in

may be

terms of u fundamental units or dimensions

—

written as (q

u)

dimensionless groups.

As an additional example flowing

in

method,

to illustrate the use of this

turbulent flow at velocity v inside a pipe of diameter

transfer to the wall.

We

The fundamental

and undergoing heat

wish to predict the dimensionless groups relating the heatk,

and

units or dimensions are u

=4

transfer coefficient h to the variables D, p, p,c p

and temperature T. The

us consider a fluid

let

D

,

v.

The

total

number

of variables

is

and are mass M, length L, time /, fundamental units are as

units of the variables in terms of these

follows:

f

M

M

M

3

L

Lt

T

Hence, the number of dimensionless groups or 3.

ML

13 "

n's

2 t

3

T

t

T

T

t

can be assumed to be 7

—

4,

or

Then

We less

will

choose the four variables D,

and

t>

to be

common

to

all

the dimension-

d

7t,

= D°k bp

7t

2

= D ek fpg vh cp

(4.14-3)

3

=

(4.14-4)

tt

308

k, p,

groups. Then the three dimensionless groups are c

v

p

k

D'k J p v'h

Chap. 4

Principles

(4.14-2)

of Steady-Stale Heat Transfer

For

7t„ substituting the actual dimensions.

Summing

each exponent,

for

0=a + 6- c+
(L)

+c+

(M)

0

=

(0

0

= -

b

0=

(T)

3b

1

(4.14-6)

-

c

-

d

b

=

0,

c

-£>

=

Solving these equations simultaneously, a

1,

= —

1,
=

1.

Substituting these values into Eq. (4.14-2),

71,=

Repeating for

7t

2

and

rc

3

—

= N Rt

(4.14-7)

and substituting the actual dimensions, C

Substituting

This

is

tt,, 7r

in the

2

,

and

tt

3

n2

= -^ = N Pt

(4.14-8)

"3

=

Y=N

(4.14-9)

k

into Eq. (4.14-1)

N„

and rearranging,

form of the familiar equation for heat transfer inside pipes, Eq.

This type of analysis

is

(4.5-8).

useful in empirical correlations of heat-transfer data.

The

importance of each dimensionless group, however, must be determined by experimentation (Bl.Ml). 2.

Natural convection heat transfer outside a vertical plane.

In the case of natural-

L to an adjacent fluid, groups should be expected when compared to forced convection

convection heat transfer from a vertical plane wall of length different dimensionless

inside a pipe since velocity

is

not a variable. The buoyant force due to the difference

density between the cold and heated fluid should be a factor. As seen (4.7-2), the

in

in

Eqs. (4.7-1) and

buoyant force depends upon the variables /?, g, p, and AT. Hence, the and their fundamental units are as follows:

list

of

variables to be considered

M

=

M c

Tt

L

,

Sec. 4.14

9. 7t 3

Dimensional Analysis

Since u ,

tt 4

in

,

L2 = 13 2 t

= h=-jh

=

4,

the

„

1

T

M

AT = T

The number of variables is q = is 9 — 4, or 5. Then 7t, = f(n 2

p

=

ML —

7f

k

number

of dimensionless groups or

7t's

7T ). 5

Heal Transfer

309

We less

will

choose the four variables, L,

and g

to be

common

p

n3

groups. Tt,

7i

For

^, k,

7i,,

4

= -

L°p.

b

c

d

k g p

n

Hrfi

k°g p

AT

n2

=

L'p.

ns

=

Ufi'k'g'h

f k'g h c

=

to

all

^

the dimension-

substituting the dimensions,

=

Solving for the exponents as before, a

b b

b;

1=^7:

= — 1, c =

f,b

0,

(4.14-n)

=

d

3.

Then

7t,

becomes (4.14-12)

Taking

the square of

both sides

to eliminate fractional exponents,

TT,

Repeating

for the

——

c

n

(4.14-13)

other n equations,

—— 2

Tt,

Combining

=

=

I?P 9

Equation (4.14-14)

is

the

7t

p ~j-

k

kAT

hL

the dimensionless groups

7t,7r 3

=

n2

fi

4

=

L3 p 2 3

7t,, 7t 3

L^g/?

,

fc

= N Pr

n3

k

follows,

AT

2

3

=

j

Grashof group given

in

L

Eq.

,

4.15A

LfigfJ

and 7t 4 as p

g/?

AT

(4.7-4).

^N U =/(N Gr N Pr

4.15

—

=

= WG

r

(4.14-14)

Hence, (4.14-15)

)

NUMERICAL METHODS FOR STEADY-STATE CONDUCTION IN TWO DIMENSIONS Analytical Equation for Conduction

In Section 4.4

we discussed methods

for solving two-dimensional heat-conduction prob-

lems using graphical procedures and shape factors. In

and numerical methods. The equation for conduction

x direction

in the

is

this section

we consider

analytical

as follows:

dT

=-kA—-

qx

Now we

shall derive

an equation

for steady-state

Referring to Fig. 4.15-1, a rectangular block input to the block

is

conduction

Ax by Ay by L

in is

two directions x and y. shown. The total heat

equal to the output. Rx\x

310

(4.15-1)

ox

+

Ry\,

=


+ Ax

Chap. 4

+

1y)y + *,

Principles

(4.15-2)

of Steady-State Heat Transfer

Figure

Sieady-siaie conduction in two directions

4.15-1.

Now, from

y\y +

Ay

Eq. (4.15-1),

qx

dT

—k(AyL) —

$=

(4.15-3)

dx

Writing similar equations for the other three terms and substituting into Eq. (4.15-2)

- k(AyL)

dT

-

k(AxL)

-

Ax AyL and

2

T

dx is

k{AxL)

2

called the Laplace equation.

method

dT — dy

Ax and Ay approach

two

in

d

solve this equation. In the

dT — dx

letting

equation for steady-state conduction

This

k(AyL)

dy

dx

Dividing through by

dT —

zero,

(4.15-4) y

+ Ay

we obtain

the final

directions. d

2

T

dy

(4.15-5)

2

There are a number of analytical methods to

of separation of variables, the final solution

We

is

shown in Fig. 4.15-2. The solid is called a semi-infinite solid since one of its dimensions is oo. The two edges or boundaries at x = 0 and x = L are held constant at T, K. The edge at y = 0 is held at T2 And at y = co, T = 7,. The solution relating T to position y and x is expressed as an

infinite

Fourier series (HI, G2, Kl).

consider the case

.

T - T, T -T,

Itix

3llX

1

sin

sin

+

(4.15-6)

2

Other analytical methods are available and are discussed in many texts (C2, HI G2, Kl). A large number of such analytical solutions have been given in the literature. However, there are many practical situations where the geometry or boundary con-

complex for analytical solutions, so that methods are used. These are discussed in the next section.

ditions are too

Figure

4.15-2.

Steady-slate heal conduction in

two directions

in

a semiat

infinite plate.

Sec. 4.15

finite-difference numerical

Numerical Methods for Steady-Stale Conduction

in

y =

°°

Two Dimensions

311

4.15B

J.

Finite-Difference Numerical

Methods

Since the advent of the fast digital computers, solutions to

Derivation of the method.

many complex two-dimensional heat-conduction problems by numerical methods we can

readily possible. In deriving the equations

equation

(4.15-5). Setting

up

the finite difference of d

t+ 1

d

2

T _

d(dT/3x)

1

ox

n

—T

m

n,

_

_

m

~

Tn,m

Tn —

1

.

m

Ax

+

7/1-1,

T

on

(4.15-7)

2

stands for a given value of the

x

y,

m+

divided into squares.

The this

stands for y

1

This

scale.

concentrated at the center of the square, and

is

,

Ax

index indicating the position of

two-dimensional solid

T/dx 2 1

m

Ax

(Ax)

m

,

ox ^n+l.m

where the index

1

2

are

start with the partial differential

is

+

shown

1

solid inside a square

concentrated mass

Ay, and n

is

Fig. 4.15-3.

in

is

the

The

imagined to be

Each node is imagined to be connected to the adjacent nodes by a small conducting rod as shown. The finite difference oid 2 T/dy 2 is written in a similar manner. S

2

T_

5y

T„.

m+1

2

-2T

n,

(Ay)

+

m

m+1

+

T„.

m _,

+ Tn+Um +

a "node."

T„,

(4-15-8)

2

Substituting Eqs. (4.15-7) and (4.15-8) into Eq. T„.

is

(4.

T„_

15-5)

1

.

and

setting

m -4T„, m

=

Ax =

0

Ay, "(4.15-9)

n, in

n +

\,m

Lffl-1

n, rn

Figure

4.15-3.

-

1

Temperatures and arrangement of nodes for two-dimensional steadystale heat conduction.

312

Chap. 4

Principles of Steady-State

Heat Transfer

This equation states that the net heat flow into any point or node

The shaded area

in Fig. 4.15-3 represents the

Alternatively, Eq. (4.15-9) can be derived by

The k

total heat in for unit thickness

Ay (T„- l.m

'

Ax

~

m) H

{T„ +

I

Ax

+

i.ct

—

making

a heat balance on this shaded area.

^i.m)

— k

zero at steady state.

is

Ay

k ^B.

is

area on which the heat balance was made.

Ax (TBtIB+i Ay

k

Ax

,„ (T -T..J+— Ay

1

flt

-T„.J =

0

(4.15-10)

Rearranging, this becomes Eq. (4.15-9). In Fig. 4.15-3 the rods connecting the nodes act as fictitious heat conducting rods.

To

N

for

use the numerical method, Eq. (4.15-9)

unknown

nodes,

equations solved

N

for the

is

written for each node or point. Hence,

linear algebraic equations

must be written and r^he system of

various node temperatures. For a hand calculation using a

modest number of nodes, the

iteration

method can be used

to solve the system of

equations.

2. (4.

Iteration 1

5-9)

is

method of solution.

set

equal

<7n.

Since q„ m

=

In using the iteration method, the right-hand side of Eq.

to a residual q„

.

= T„A_ m

m

n,

0 at steady state, solving for T„ m

in

m—

1

-4T

(4.15-11)

Eq. (4.15-11) or (4.15-9),

(4.15-12)

Equations (4.15-11) and (4.15-12) are the

final

equations to be used. Their use

is

illus-

trated in the following example.

EXAMPLE

4.15-1. Steady-State Heat Conduction in Two Directions Figure 4.15-4 shows a cross section of a hollow rectangular chamber with the inside dimensions 4 x 2 and outside dimensions m. The

m

chamber

0,1

I

is

0,2

20

m

long.

The

8x8

inside walls are held at

600

K

and the outside

at

0,3

600 K 300

1,1

1,2

1,3

2,1

2,2

2,3

2,4

3,1

3,2

3,3

3,4

K

3,5

I

3,6

I

I

4,1

4,2

4,3

4,4

4,5"

5,1

5,2

5,3

5,4

5,5

~J !

4,6 5,6

^— 300 K Figure

Sec. 4.15

4.15-4.

Square grid pattern for Example 4.15-1

Numerical Methods for Steady-State Conduction

in

Two Dimensions

313

300 K. The k

W/m

1.5

is

K. For steady-state conditions find the heat loss

Use

per unit chamber length.

grids

1

x

m.

1

Since the chamber is symmetrical, one-fourth of the chamber (shaded part) will be used. Preliminary estimates will be made for the first approximation. These are for node T, 2 = 450 K, T2 2 = 400, T3 2 = 400, Solution:

'

T3> 3 = 400, T 4 = 450, 73> 5 = 500, T4> 2 = 325, 74> 3 = 350, T4 4 = 375, and T4-5 = 400. Note- that T0 2 = T2 2 T3>6 = T3>4 and T4 6 = T4 4 by '

3i

,

,

symmetry.

To

start the calculation,

one can

any

select

T

usually better to start near a boundary. Using

2

x

,

interior point, but

we calculate

it

is

the residual

q u2 by Eq. (4.15-11). qt

HenceJpF] value of rj

by Eq.

we

set

2

to 0

and calculate a new

"

(4.15-12).

<

T\ /,, 2

.

not at steady state. Next,

2 is 2

.i=Tul + TU3 + T0-2 + T2 2 -4Tli2 = 300 + 600 + 400 + 400 - 4(450) = -100

+ T

i

-

i

+

3

T„

2

+ T2

2

+

300

-

600

+ -

400

+

400

-425

4

K

This new value of T, 2 of 425 will replace the old one of 450 and be used to calculate the other nodes. Next,

= T2i + T2> 3 +


,

= Setting q 2 •

T /2

2

- -=2

Continuing

+ T

2

,

=

(4.15-12),

3

+-

T, 2

+

400

Tx2 =

364,

300

3

=

364

3

=

441

<7j.4

=

441

T3

.

4

=

479

<7 3

5

=

479

.

5

=

489

2

=

300

74> 2 =

329

<7

3

.

T3

.

T3
4

.

4

3

=

329

.

T4

3

=

.

361

4 4

=

361

4

=

385

4 5

=

385

T4> 5 =

390

<7

<7

,

T4 cj

314

+

425

+

+ T3

2

400

+

431

300

-

for all the rest of the interior
Using Eq.

600

2

-

4(400)

=

125

zero and using Eq. (4.15-12),

to

T

2

+

300

+ T3- - 4T2i 2

T,, 2

+

600

+ -

425

+

400

^, - 431

nodes,

+

325

- 4(400) = - 144

+

450

+

600

+

350

-

=

164

+

500

+

600

+

375

- 4(450) =

116

+

479

+

600

+

400

-

4{50O)

= -42

+

350

+

364

+

300

-

4(325)

=

14

+

375

+

441

+

3 00

-

4{350)

=

45

+

400

+

479

+

300

-

4(375)

=

40

+

385

+

489

+

399

-

4(400)

= -41

Chap. 4

4(400)

Principles of Steady-State

Heat Transfer

Having completed one sweep across the grid map, we can start a second approximation, using, of course, the new values calculated. We can start again with T, 2 or we can select the node with the largest residual. Starting with T,

2

again, qK

2

= 300 + 600 + 431 + 431 -

T,,

2

=

2

= 300 + 600 + 440 + 364 -

2

=

T2 This

,

4{425)

=

4(431)

= -20

62

440

426

continued until the residuals are as small as desired. The

is

final

values

are as follows: ^.. 2

=

T3

=

5

441, 490,

T2

2

=

,

T4

2

=

,

.To calculate the

we use

length,

432, 340,-

=

T3

.

2

T4

,

3

=

.

744 =

372,

T3

=461,

3

.

=

4

T4-5 =

387,

485,

391

from the chamber'per unit chamber T2 4 to T3 4 with Ax = Ay and 1 m

total heat loss

Fig. 4.15-5.

T3

384,

For node

deep,

<7

=

—Yx—

kA AT Ax =

/c[A.x(l)]

2

^

'

~

4

3'

=

4)

2 4

_

3 - 4^

'

'

(4,15~ 13)

heat flux for node T2 5 to T3 5 and for T, 3 to Tu2 should be multiplied by \ because of symmetry. The total heat conducted is the sum of the five paths for \ of the solid. For four duplicate parts,

The

q,

-T + + (T2 4 - T3 4 + |(T2> - r3 )] = 4(1.5)^(600 - 441) + (600 - 432) +

=

4/c[i(T,.

3

- Tu 2 +

(T2

)

)

+

(600

W

= 3340

-

485)

per 1.0

2

3

5

,

.

.

+

^(600

m

-

,

,

2)

(T2

,

3

- T3

,

3)

(4.15-14)

5

(600

-

461)

490)]

deep

Also, the total heat conducted can be calculated using the nodes at the outside, as

value

shown

<j av

FIGURE

4.15-5.

Sec. 4.15

in Fig. 4.15-5.

= 3430 W. The

This gives q n

average

is

=

3340

+

3430

=

W ...

3385

per

1.0

m

deep

Drawing for calculation of total heat conduction.

Numerical Methods for Steady-Slate Conduction

in

Two Dimensions

315

If

a larger

number of nodes,

i.e.,

a smaller grid size,

can be obtained. Using a grid size of 0.5

W

3250

digital

is

obtained.

If

a very

computer would be needed

for

method used here

iteration

is

is

instead of 1.0

used, a

more accurate

solution

m for Example 4.15-1, a q,

v

of

more accuracy can be obtained but a the large number of calculations. Matrix methods of simultaneous equations on a computer. The

fine grid

are also available for solving a set

m is

used,

method. Conte (C7) gives an

often called the Gauss-Seidel

actual subroutine to solve such a system of equations. Most computers have standard

subroutines for solving these equations (G2, Kl).

3.

Equations for other boundary conditions.

In

boundaries were such that the node points were there

is

in Fig. 4.15-6a

Ay

Ax

as follows,

is

k

Ax

2

Ay

where heat

(T„.

Ay

2

Ax =

in

Ax

k

+ r-rSetting

4.15-1 the conditions at the

constant. For the case where

convection at the boundary to a constant temperature

node n,m k

Example

known and

m-

,

=

Tm

,

a heat balance on the

heat out (Kl):

-T

)

=

- 7J

h Ay(7„. m

=

Ay, rearranging, and setting the resultant equation

(4.15-15)

q nm residual, the

following results. (a)

For convection

—

Ax

h

7aj +

-

at

a boundary,

—

(h |(27„_

1 ,

+

m

7„,

m+1

+

T„.

m _,)-

T„J

Ax

+

2

n,

m+

5

(4.15-16)

1

insulated

surface

(b)

'n-l,m

n,

m+

1

' '

4

Ay \

n+

1

1 1

<

T

1,

m

-*Ax*~ n - l,m (c)

(d)

FIGURE

4.15-6.

Other types of boundary conditions: (a) convection at a boundary, boundary, (c) exterior corner with convective boundary, (d) interior corner with convective boundary. (b) insulated

316

Chap. 4

Principles of Steady-State

Heat Transfer

manner for the cases in Fig. 4.15-6: For an insulated boundary,

In a similar (b)

?(Tn m+1 ,

(c)

— h

Ax

.

m_

l

+ Tn - Um

)

For an exterior corner with convection

^ (d)

+ Tn

For an

Tx +

T„ + i(T„_ 1>m

+

T„.

„_

,)

~2Tn m = q ntn

at the

interior corner with convection at the

+

T„_

+ i(Tn+

1

=

(4.15-18)

g„, m

boundary,

+ Tnin,_ ,) -

,

T„. m +

boundary,

(~ +

-

(4.15-17)

,

— h

/ I

3

+

Ax\

JT^ =

(4.15-19)

4„.„

For curved boundaries and other types of boundaries, see (C3, Kl). To use Eqs. (4.15-16H4.15-19), the residual q„ m is first obtained using the proper equation. Theng, m is

set

T

equal to zero and

n

m solved for in the resultant equation.

PROBLEMS Cold Room. Calculate the heat loss per m 2 of surface area for a temporary insulating wall of a food cold storage room where the outside temperature is 299.9 K and the inside temperature 276.5 K. The wall is composed of 25.4 mm of corkboard having a k of 0.0433 W/m- K.

4.1-1. Insulation in a

Ans. 4.1- 2.

39.9

W/m 2

Determination of Thermal Conductivity. In determining the thermal conductivity of an insulating material, the temperatures were measured on both sides of a flat slab of 25 of the material and were 318.4 and 303.2 K. The heat 2 flux was measured as 35.1 W/m Calculate the thermal conductivity in btu/

mm

.

h-ft-°FandinW/m-K. Mean Thermal Conductivity

in a Cylinder. Prove that if the thermal conducwith temperature as in Eq. (4.1-11), the proper mean value k m to use in the cylindrical equation is given by Eq. (4.2-3) as in a slab. 4.2-2. Heat Removal of a Cooling Coil. A cooling coil of 1.0 ft of 304 stainless steel tubing having an inside diameter of 0.25 in. and an outside diameter of 0.40 in. is

4.2- 1.

tivity varies linearly

being used to remove heat from a bath. The temperature at the inside surface of the tube is 40°F and 80°F on the outside. The thermal conductivity of 304 stainless steel

is

a function of temperature. k

where k is and watts.

in

btu/h

•

ft

•

=

7.75

°F and

T

is

+

7.78 x 10~

3

T

in °F. Calculate the heat

Ans. 4.2-3.

4.2-4.

and

W

•

k a, b,

1.225 btu/s, 1292

a Bath.

Variation of Thermal Conductivity. A maintained at T, and the other at according to temperature as

where

in btu/s

Repeat Problem 4.2-2 but for a cooling coil made having an average thermal conductivity of 15.23 W/m K.

Removal of Heat from of 308 stainless steel

removal

flat

T

2

.

plane of thickness Ax has one surface If the thermal conductivity varies

= A + bT + cT 3

c are constants, derive

an expression for the one-dimensional

heat flux q/A.

Chap. 4

Problems

317

4.2-5.

Temperature Distribution in a Hollow Sphere. Derive Eq. (4.2-14) for the steadyconduction of heat in a hollow sphere. Also, derive an equation which shows that the temperature varies hyperbolically with the radius r. state

Ans.

Neededfor Food Cold Storage Room. A food cold storage room is to mm of pine wood, a middle layer of cork mm of concrete. The inside wall surface temperature is — 17.8°C and the outside surface temperature is 29.4°C at the outer concrete surface. The mean conductivities are for pine, 0.151; cork, 0.0433; and concrete, 0.762 W/m K. The total inside surface area of the room

4.3-1. Insulation

be constructed of an inner layer of 19.1 board, and an outer layer of 50.8

•

to use in the calculation is

m

effects).

to

approximately 39 What thickness of cork board is needed

2

(neglecting corner and end

keep the heat Ans.

of a Furnace.

4.3-2. Insulation

A

0.

586 W? m thickness

loss to 1

28

m thick constructed of W/m K. The wall will be average k of 0.346 W/m K, so

wall of a furnace 0.244

material having a thermal conductivity of 1.30

is

•

on the outside with material having an from the furnace will be equal to or less than 1830W/m 2 The inner surface temperature is 1588 K and the outer 299 K. Calculate the thickness of insulated

•

the heat loss

.

insulation required.

Ans. 4.3-3.

0.179

m

Heat Loss Through Thermopane Double Window. A double window called thermopane is one in which two layers of glass are used separated by a layer of dry stagnant air. In a given window, each of the glass layers is 6.35 mm thick separated by a 6.35 mm space of stagnant air. The thermal conductivity of the glass is 0.869 W/m K and that of air is 0.026 over the temperature range used. For a temperature drop of 27.8 K over the system, calculate the heat loss for a window 0.914 m x 1.83 m. (Note: This calculation neglects the effect of the convective coefficient on the one outside surface of one side of the window, the convective coefficient on the other outside surface, and convection inside the •

window.) 4.3-4.

Heat Loss from Steam Pipeline. A steel pipeline, 2-in. schedule 40 pipe, contains saturated steam at 121. 1°C. The line is insulated with 25.4 mm of asbestos. Assuming that the inside surface temperature of the metal wall is at 121. 1°C and the outer surface of the insulation

is

at 26.7°C, calculate the heat loss for 30.5

m

kg of steam condensed per hour in the pipe due to the heat loss. The average k for steel from Appendix A. 3 is 45 W/m K and by linear interpolation for an average temperature of (121.1 + 26.7)/2 or 73.9°C, the k for

of pipe. Also, calculate the

•

asbestos

is

0.182.

Ans. 4.3-5.

5384

W,

8.8

1

kg steam/h

Heat Loss with Trial-and-Error Solution. The exhaust duct from

a heater has an with ceramic walls 6.4 thick. The average thick is k = 1.52 W/m K. Outside this wall, an insulation of rock wool 102 installed. The thermal conductivity of the rock wool is k = 0.046 + 1.56 x -4 10 T°C(W/m K). The inside surface temperature of the ceramic is = 588.7 K, and the outside surface temperature of the insulation is T3 = 311 Ti K. Calculate the heat loss for 1.5 m of duct and the interface temperature T2 between the ceramic and the insulation. [H/"/ir ; The correct value of k m for the insulation is that evaluated at the mean temperature of (T2 + T3 )/2. Hence, for the first trial assume a mean temperature of, say, 448 K. Then calculate the heat loss and T2 Using this new T2 calculate a new mean temperature and proceed

inside diameter of 114.3

mm

mm

mm

•

•

.

,

as before.] 4.3-6.

Heat Loss by Convection and Conduction. 0.557

318

m2

is

installed in the

wooden

A

glass

window with an area of The wall dimensions

outside wall of a room.

Chap. 4

Problems

are 2.44 x 3.05 m.

The

The wood has

mm

a k of 0.1505

and has

W/m-K

and

is

mm

25.4

room temperature is 299.9 K (26.7°C) and the outside air temperature is 266.5 K. The convection coefficient /i on the inside wall of the glass and of the wood is thick.

glass

is

3.18

thick

a

k of 0.692. The inside

(

estimated as 8.5 W/m 2 K and the outside h 0 also as 8.5 for both surfaces. Calculate the heat loss through the wooden wall, through the glass, and the -

total.

Ans.

569.2 646.8

W (wood) (1942 btu/h); 77.6 W (glass) (265 btu/h); W (2207 btu/h) (total)

Conduction, and Overall U. A gas at 450 K is flowing inside a 2-in. The pipe is insulated with 51 of lagging having a mean k = 0.0623 W/m K. The convective heat-transfer coefficient of the gas 2 K and the convective coefficient on the outside of inside the pipe is 30.7 W/m the lagging is 10.8. The air is at a temperature of 300 K. of pipe using resistances. (a) Calculate the heat loss per unit length of 1 (b) Repeat using the overalL[/ 0 based on the outside area A 0

4.3-7. Convection, steel pipe,

mm

schedule 40.

•

•

m

.

4.3-8.

in Steam Heater. Water at an average of 70°F is flowing in a 2-in. schedule 40. Steam at 220°F is condensing on the outside of the pipe.The convective coefficient for the water inside the pipe is h = 500 btu/h ft 2 °F and the condensing steam coefficient on the outside is h — 1500.

Heat Transfer steel pipe,

•

(b)

Calculate the heat loss per unit length of 1 ft of pipe using resistances. Repeat using the overall U, based on the inside area/},-.

(c)

Repeat using

(a)

[/„..

Ans.

(a)

(b) (c)

4.3-9.

= 26 7 10 btu/h (7.828 kW), U,= 329.1 btu/h- 2 -°F (1869 W/m 2 -K), U 0 = 286.4 btu/h 2 °F (1626 W/m 2 K)

q

ft

•

ft

•

•

A steel pipe carrying steam has an lagged with 76 of insulation having an Two thermocouples, one located at the interface between the pipe wall and the insulation and the other at the outer surface of the insulation, give temperatures of 115°C and 32°C, respectively. Calculate the

Heat Loss from Temperature Measurements. outside diameter of 89 mm. average k = 0.043 W/m K.

heat loss in

mm

It is

W per m of pipe.

of Convective Coefficients on Heat Loss in Double Window. Repeat Problem 4.3-3 for heat loss in the double window. However, include a convec2 tive coefficient of h = 1 1.35 W/m K on the one outside surface of one side of the window and an h of 1 1.35 on the other outside surface. Also calculate the

4.3-10. Effect

•

overall U.

Ans. 4.3-11.

q

=

106.7

W, U =

2.29

W/m 2 K •

Uniform Chemical Heat Generation. Heat is being generated uniformly by a chemical reaction in a long cylinder of radius 91.4 mm. The generation rate is constant at 46.6 W/m 3 The walls of the cylinder are cooled so that the wall temperature is held at 311.0 K. The thermal conductivity is 0.865 W/m-K. Calculate the center-line temperature at steady state. Ans. Tg = 311.112 K .

4.3-12.

Heat of Respiration of a Food Product. A fresh food product is held in cold storage at 278.0 K. It is packed in a container in the shape of a flat slab with all faces insulated except for the top flat surface, which is exposed to the air at 278.0 K. For estimation purposes the surface temperature will be assumed to be 2 278 K. The slab is 152.4 thick and the exposed surface area is 0.186m The 3 density of the foodstuff is 641 kg/m The heat of respiration is 0.070 kJ/kg-h and the thermal conductivity is 0.346 W/m K. Calculate the maximum temperature in the food product at steady state and the total heat given off in W.

mm

.

.

•

(Note

Chap. 4

:

It is

Problems

assumed

in this

problem that there

is

no

air circulation inside the

319

foodstuff. Hence, the

results

will

be conservative, since circulation during

respiration will reduce the temperature.)

Ans: 4.3-13.

278.42 K, 0.353

W (1.22 btu/h)

Heating Wire. A current of 250 A is passing through a having a diameter of 5.08 mm. The wire is 2.44 m long and has a resistance of 0.0843 CI. The outer surface is held constant at 427.6 K. The thermal conductivity is k = 22.5 W/m K. Calculate the center-line temperature Temperature Rise

in

stainless steel wire

•

at steady state.

A metal steam pipe having an outside diameter of has a surface temperature of 400 K and is to be insulated with an and a k of 0.08 W/m K. The pipe is insulation having a thickness of 20 and a convection coefficient of 30 W/m 2 K. exposed to air at 300

4.3-14. Critical Radius for Insulation.

30

mm

mm

K

(a)

Calculate the critical radius and the heat loss per

m

of length for the bare

pipe. (b)

Calculate the heat loss for the insulated pipe assuming that the surface temperature of the pipe remains constant.

=

Ans.^. (b) q 4.4-1. Curvilinear-Squares

W

54.4

Graphical Method. Repeat "Example 4.4-1 but with the

following changes. (a)

(b)

number of equal temperature subdivisions between the isothermal boundaries to be five instead of four. Draw in the curvilinear squares and determine the total heat flux. Also calculate the shape factor S. Label each isotherm with the actual temperature. Repeat part (a), but in this case the thermal conductivity is not constant but k = 0.85 (1 + 0.00040T), where T is temperature in K. [Note : To calculate the overall q, the mean value of k at the mean temperature is used. The spacing of the isotherms is independent of how k varies with T (Ml). However, the temperatures corresponding to the individual isotherms are a function of how the value of k depends upon T. Write the equation for q' for a given curvilinear section using the mean value of k over the temperature interval. Equate this to the overall value of q divided by or q/M. Then solve for the isotherm temperature.] Select the

M

4.4-2.

A rectangular furnace with inside dimensions of has a wall thickness of 0.20 m. The k of the walls is 0.95 W/m K. The inside of the furnace is held at 800 K and the outside at 350 K. Calculate the total heat loss from the furnace. Ans. q= 25 081

Heat Loss from a Furnace. 1.0

x

1.0

x 2.0

m

•

W

4.4- 3.

Heat Loss from a Buried Pipe. A water pipe whose wall temperature is 300 K has a diameter of 150 mm and a length of 10 m. It is buried horizontally in the ground at a depth of 0.40 m measured to the center line of the pipe. The ground surface temperature is 280 K and k = 0.85 W/m K. Calculate the loss of heat from the pipe. •

Ans. 4.5- L.

Heating Air by Condensing Steam. Air

q

=

451.2

W

flowing through a tube having an average temperature of 449.9 K, and pressure of 138 kPa. The inside wall temperature is held constant at 204.4°C (477.6 K) by steam condensing outside the tube wall. Calculate the heat-transfer coefficient for a long tube and the heat-transfer flux. inside diameter of 38.1

mm

is

at a velocity of 6.71 m/s,

Ans. h

=

39.27

W/m 2 K •

(6.91 btu/h

•

ft

2 •

°F)

45-2. Trial-and-Error Solution for Heating Water. Water is flowing inside a horizontal 1^-in. schedule 40 steel pipe at 37.8°C and a velocity of 1.52 m/s. Steam at 108.3°C is condensing on the outside of the pipe wall and the steam coefficient is

320

assumed constant

at

9100 W/m

2 •

K.

Chap. 4

Problems

Calculate the convective coefficient h for the water. (Note that this is trial and error. A wall temperature on the inside must be assumed first.) (b) Calculate the overall coefficient 17,- based on the inside area and the heat (a)

t

transfer flux qjA in ;

4.5-3.

W/m 2

.

Heat-Transfer Area and Use of Log Mean Temperature Difference. A reaction mixture having a c pm = 2.85 kJ/kg is flowing at a rate of 7260 kg/h and is to be cooled from 377.6 K to 344.3 K. Cooling water at 288.8 K is available and 2 K. the flow rate is 4536 kg/h. The overall U a is 653 W/m (a) For counterflow, calculate the outlet water temperature and the area A 0 of the exchanger. (b) Repeat for cocurrent flow. 2 2 Ans. (a) T, = 325.2 K, A 0 = 5.43 m (b) A 0 = 6.46 m -

K

•

,

4.5-4.

Heating Water with Hot Gases and Heat-Transfer Area. A water flow rate of 13.85 kg/s is to be heated from 54.5 to 87.8°C in a heat exchanger by 54430 kg/h of hot gas flowing counterflow and entering at 427°C (c pm = 2 K. Calculate the exit-gas tem1.005 kJ/kg K). The overall- U„ = 69.1 W/m •

•

perature and the heat-transfer area. Ans. 4.5-5.

T=

299.5°C

and Overall U. Oil flowing at the rate of 7258 kg/h with a c pm = 2.01 kJ/kg-K is cooled from 394.3 K to 338.9 K in a counterflow heat exchanger by water entering at 294.3 K and leaving at 305.4 K. Calculate the flow 2 rate of the water and the overall U, if the A is 5.11 m 2 Ans. 17 420 kg/h, 17,= 686 W/m K Cooling

OH

.

t

.

.

4.5-6.

Laminar Flow and Heating of Oil. A hydrocarbon oil having the same physical properties as the oil in Example 4.5-5 enters at 175°F inside a pipe having an inside diameter of 0.0303 ft and a length of 15 ft. The inside pipe surface temperature is constant at 325°F. The oil is to be heated to 250°F in the pipe. How many lb^yh oil can be heated? (Hint: This solution is trial and error. One method is to assume a flow rate of say m = 75 lb mass/h. Calculate the;V Re and the value of h a Then make a heat balance to solve for q in terms of m. Equate this q to the q from the equation q = h a A AT0 Solve for m. This is the .

.

new

m

to use for the

second

trial.)

Ans. 4.5-7.

m=

84.2

lbjh

(38.2 kg/h)

Heating Air by Condensing Steam. Air at a pressure of 101.3 kPa and 288.8 K enters inside a tube having an inside diameter of 12.7 and a length of 1.52 m with a velocity of 24.4 m/s. Condensing steam on the outside of the tube maintains the inside wall temperature at 372.1 K. Calculate the convection coefficient of the air. (Note : This solution is trial and error. First assume an outlet temperature of the air.)

mm

4.5- 8.

Heat Transfer with a Liquid Metal. The liquid metal bismuth at a flow rate of 2.00 kg/s enters a tube having an inside diameter of 35 mm at 425°C and is heated to 430°C in the tube. The tube wall is maintained at a temperature of 25°C above the liquid bulk temperature. Calculate the tube length required. The physical properties are as follows (HI): k = 15.6 W/m K, c p = 149 J/kg- K, •

H= 4.6- 1.

1.34 x 10"

3

Pa

s.

Flat Plate. Air at a pressure of 101.3 kPa and a temperflowing over a thin, smooth flat plate at 3.05 m/s. The plate length in the direction of flow is 0.305 m and is at 333.2 K. Calculate the

Heat Transfer from a ature of 288.8

K

is

heat-transfer coefficient assuming laminar flow.

Ans,

h

=

12.35

W/m 2 K (2.18 btu/h •

2 •

ft

-

°F)

—

Frozen Meat. Cold air at 28.9°C and 1 atm is recirculated at a velocity of 0.61 m/s over the exposed top flat surface of a piece of frozen meat. The sides and bottom of this rectangular slab of meat are insulated and the top surface is 254 by 254 square. If the surface of the meat is at — 6.7°C,

4.6-2. Chilling

mm

Chap. 4

Problems

mm

321

As an approxican be used, depending on the

predict the average heat-transfer coefficient to the surface.

mation, assume that either Eq. (4.6-2) or

(4.6-3)

n *
Ans.

/i

=

6.05

W/m 2 K •

Heat Transfer to an Apple. It is desired to predict the heat-transfer coefficient blown by an apple lying'on a screen with large openings. The air velocity is 0.61 m/s at 101.32 kPa pressure and 316.5 K. The surface of the apple is at 277.6 K and its average diameter is 1 14 mm. Assume that it is a sphere. for air being

4.6- 4.

Heating Air by a Steam Heater. A total of 13 610 kg/h of air at 1 atm abs pressure and 15.6°C is to be heated by passing over a bank of tubes in which steam at 100°C is condensing. The tubes are 12.7 OD, 0.61 long, and arranged in-line in a square pattern with S p = S„ = 19.05 mm. The bank of tubes contains 6 transverse rows in the direction of flow and 19 rows normal to the flow. Assume that the tube surface temperature is constant at 93.33°C.

m

mm

Calculate the outlet air temperature. 4.7- 1.

Natural Convection from an Oven Wall. The oven wall in Example 4.7-1 is insulated so that the surface temperature is 366.5 K instead of 505.4 K. Calculate the natural convection heat-transfer coefficient and the heat-transfer rate of width. Use both Eq. (4.7-4) and the simplified equation. per (Note : Radiation is being neglected in this calculation.) Use both SI and English units.

m

mm

in by Natural Convection from a Cylinder. A vertical cylinder 76.2 diameter and 121.9 high is maintained at 397.1 K at its surface. It loses heat by natural convection to air at 294.3 K. Heat is lost from the cylindrical side and the flat circular end at the top. Calculate the heat loss neglecting radiation losses. Use the simplified equations of Table 4.7-2 and those equations for the lowest range o( N Gr The equivalent L to use for the top flat surface is 0.9 Pr

4.7-2. Losses

mm

N

.

times the diameter.

Ans. 4.7-3.

g

= 26.0W

Heat Loss from a Horizontal Tube. A horizontal tube carrying hot water has a K and an outside diameter of 25.4 mm. The tube is

surface temperature of 355.4

exposed to room air 1-m length of pipe? 4.7-4.

at 294.3

K.

What

is

the natural convection heat loss for a

Natural Convection Cooling of an Orange. An orange 102 mm in diameter having a surface temperature of 21.1°C is placed on an open shelf in a refrigerator held at 4.4°C. Calculate the heat loss by natural convection, neglecting radiation. As an approximation, the simplified equation for vertical planes can be used with L replaced by the radius of the sphere (Ml). For a more accurate correlation, see (S2).

4.7-5.

Natural Convection the case

in

Enclosed Horizontal Space. Repeat Example 4.7-3 but for the bottom plate is hotter than

where the two plates are horizontal and

the upper plate.

Compare

the results.

Ans. 4.7-6.

q=

12.54

W

Natural Convection Heat Loss in Double Window. A vertical double plate-glass window has an enclosed air-gap space of 10 mm. The window is 2.0 high by 1.2 m wide. One window surface is at 25°C and the other at 10°C. Calculate the free-convection heat-transfer rate through the air gap.

m

Heat Loss for Water in Vertical Plates. Two vertical square metal plates having dimensions of 0.40 x 0.40 are separated by a gap of 12 and this enclosed space is filled with water. The average surface temperature of one plate is 65.6°C and the other plate is at 37.8°C. Calculate the heat-transfer rate through this gap.

4.7-7. Natural Convection

m

mm 4.7-8.

0.8

322

Two horizontal metal plates having dimensions of comprise the top of a furnace and are separated by a distance of

Heat Loss from a Furnace. x

1.0

m

Chap. 4

Problems

1

mm. The lower plate is at 400°C and the upper at 1 00°C and air at 1 atm abs enclosed in the gap. Calculate the heat-transfer rate between the plates.

5

is

Jacketed Kettle, Predict the boiling heat-transfer coefjacketed sides of the kettle given in Example 4.8-1. Then, using this coefficient for the sides and the coefficient from Example 4.8-1 for the bottom, predict the total heat transfer.

4.8-1. Boiling Coefficient in a ficient for the vertical

Tw =

Ans. 4.8-2. Boiling

107.65°C,

AT =

7.65 K,

and ^vertical)

=

3560

W/m 2

K.

•

Coefficient on a Horizontal Tube. Predict the boiling heat-transfer

under pressure boiling at 250°F for a horizontal surface of having a k of 9.4 btu/h ft °F. The heating medium on the other side of this surface is a hot fluid at 290°F having an h of 275 btu/h ft 2 °F. Use the simplified equations. Be sure and correct this h

coefficient for water

yg-in. -thick stainless steel

•

•

-

•

value for the effect of pressure. 4.8-3.

Condensation on a Vertical Tube. Repeat Example 4.8-2 but for a vertical tube 1 .22 (4.0 ft) high instead of 0.305 m (1.0 ft) high. Use SI and English units. 2 2 K, 1663 btu/h ft °F; Kc = 207.2 (laminar flow) Ans. h = 9438 W/m

m

-

4.8-4.

•

N

Condensation of Steam on Vertical Tubes. Steam at 1 atm pressure abs and high and 100°C is condensing on a bank of five vertical tubes each 0.305 having an of 25.4 mm. The tubes are arranged in a bundle spaced far enough apart so that they do not interfere with each other. The surface temperature of the tubes is 97.78°C. Calculate the average heat-transfer coefficient and the total kg condensate per hour.

m

OD

Ans.

h

=

15 240

W/m 2 -K

Condensation on a Bank of Horizontal Tubes. Steam at 1 atm abs pressure and 100°C is condensing on a horizontal tube bank with five layers of tubes (N = 5) placed one below the other. Each layer has four tubes (total tubes = 4 x 5 = 20) and the of each tube is 19.1 mm. The tubes are each 0.61 m long and the tube surface temperature is 97.78°C Calculate the average heat-transfer coefficient and the kg condensate per second for the whole condenser. Make a sketch of the tube bank.

4.8- 5.

OD

Temperature Difference in an Exchanger. A 1-2 exchanger with one shell pass and two tube passes is used to heat a cold fluid from 37.8°C to 121. 1°C by using a hot fluid entering at 315.6°C and leaving at 148.9°C Calculate theAT; m

Mean

4.9- 1.

and the mean temperature difference ATm

in

K.

Ans.

ATlm =

148.9

K;

ATm =

131.8

K

Water in an Exchanger. Oil flowing at the rate of 5.04 kg/s (c pm = 2.09 kJ/kg K) is cooled in a 1-2 heat exchanger from 366.5 K to 344.3 K by 2.02 kg/s of water entering at 283.2 K. The overall heat-transfer coefficient U B is 340 W/m 2 K. Calculate the area required. (Hint : A heat balance must first be made to determine the outlet water temperature.)

4.9-2. Cooling Oil by

•

•

'

4.9-3.

Heat Exchange Between

Oil

and Water. Water

is

flowing at the rate of 1.13 kg/s

1-2 shell-and-tube heat exchanger and is heated from 45°C to 85°C by an oil having a heat capacity of 1.95 kJ/kg K. The oil enters at 120°C and leaves at 85°C. Calculate the area of the exchanger if the overall heat-transfer coefficient in a

•

is

300

W/m 2

•

K.

and Effectiveness of an Exchanger. Hot oil at a flow rate of kJ/kg-K) enters an existing counterflow exchanger at p 400 K and is cooled by water entering at 325 K (under pressure) and flowing at a 2 2 rate of 0.70 kg/s. The overall U = 350W/m -K and A = 12.9 m Calculate the heat-transfer rate and the exit oil temperature. Radiation to a Tube from a Large Enclosure. Repeat Example 4.10-1 but use the slightly more accurate Eq. (4.10-5) with two different emissivities. Ans. q = -2171 (-7410 btu/h)

4.9-4. Outlet Temperature

3.00 kg/s

(c

=

1.92

.

4.10- 1.

W

Chap. 4

Problems

323

Loaf of Bread in an Oven. A loaf of bread having a surface temperK is being baked in an oven whose walls and the air are at 477.4 K. The bread moves continuously through the large oven on an open chain belt conveyor. The emissivity of the bread is estimated as 0.85 and the loaf can be assumed a rectangular solid 114.3 mm high x 114.3 mm wide x 330 mm long.

4.10-2. Baking a

ature of 373

Calculate the radiation heat-transfer rate to the bread, assuming that to the oven and neglecting natural convection heat transfer.

compared

Ans.

and Convection from a Steam Pipe.

4.10-3. Radiation

OD

carrying steam and having an

=

278.4

small

W (950 btu/h)

horizontal oxidized steel pipe

m

has a surface temperature of in a large enclosure. Calculate the heat exposed to air at 297.1 for 0.305 m of pipe from natural convection plus radiation. For the steel

374.9 loss

A

q

it is

K and

of 0.1683

K

is

pipe, use an a of 0.79.

4.10-4. Radiation

Ans_

and Convection

to a

Loaf of Bread.

q

=

163 3

w (557 btu/h)

Calculate the total heat-transfer

rate to the loaf of bread in Problem 4.10-2, including the radiation plus natural convection heat transfer. For radiation first calculate a value of/i r For natural convection, use the simplified equations for the lower N Gl Pl range. For the four vertical sides, the equation for vertical planes can be used with an L of 1 14.3 mm. For the top surface, use the equation for a cooled plate facing upward and for the bottom, a cooled plate facing downward. The characteristic L for a horizontal rectangular plate is the linear mean of the two dimensions. .

N

4.10- 5.

Heat Loss from a Pipe. A bare stainless steel tube having an outside diameter of 76.2 mm and an £ of 0.55 is placed horizontally in air at 294.2 K. The pipe surface temperature plus radiation

is

366.4 K. Calculate the value of h c

and the heat

4.11- 1. Radiation Shielding.

of 0.7. Surface

1

is

loss for 3

m

+

h r for convection

of pipe.

Two very large and parallel planes each have an emissivity at 866.5

K and

surface 2

is at

588.8 K.

Use

SI and English

units. (a)

(b)

What is the net To reduce this

radiation loss of surface

1

?

two additional radiation shields also having an emisare placed between the original surfaces. What is the new

sivity of 0.7

loss,

radiation loss?

4300 btu/h ft 2 (b) 4521 W/m 2 1433 btu/h -ft 2 4.11-2. Radiation from a Craft in Space. A space satellite in the shape of a sphere is traveling in outer space, where its surface temperature is held at 283.2 K. The sphere "sees" only outer space, which can be considered as a black body with a temperature of 0 K. The polished surface of the sphere has an 2 emissivity of 0.1. Calculate the heat loss per m by radiation. 2 Ans. q 12 /A = 36.5 W/m Ans.

(a)

13 565

W/m 2

,

,

;

t

4.11-3. Radiation

uration

and Complex View Factor. Find the view factor

shown

in Fig. P4.11-3.

Figure P4.1

324

1-3.

The areas j4 4 andj4 3

Fn

for the config-

are fictitious areas (C3).

Geometric configuration for Problem 4.1 1-3.

Chap. 4

Problems

The areaA 2 + A 4

is

called

A (24) and A) + A 3

is

called

A

(

i

Areas A (24) and A (13

3 ).

)

are perpendicular to each other. [Hint: Follow the methods in Example 4.11-5. First, write an equation similar to Eq. (4.11-48) which relates the interchange

A (24) Then A (13) and A 4 AlK- A 2 =

between Ay and Finally, relate

.

.]

relate the interchange

between A( 13 and )

A (24)

.

' ,

^4(1 3)

^(13X24)

+

•^3-^34

—^

3

— ^(13) -^(13)4

-^3(24)

Between Parallel Surfaces. Two parallel surfaces each 1 .83 X 1.83 m square are spaced 0.91 m apart. The surface temperature of A is 811 K and that of A 2 is 533 K. Both are black surfaces. (a) Calculate the radiant heat transfer between the two surfaces. (b) Do the same as for part (a), but for the case where the two surfaces are connected by nonconducting reradiating walls. (c) Repeat part (b), but A has an emissivity of 0.8 and A 2 an emissivity of 0.7.

4.11-4. Radiation

,

j

Between Adjacent Perpendicular Plates. Two adjacent rectangles are perpendicular to each other.,The first rectangle is 1.52 x 2.44 m and the second is 1.83 x 2.44 m with the 2.44-m side common to both. The temperature of the first surface is 699 K and that of the second is 478 K. Both surfaces are black. Calculate the radiant heat transfer between the two

4.11-5. Radiation

surfaces. 4.11-6.

View Factor for Complex Geometry. Using the dimensions given P4.11-3, calculate the individual view factors and also n

F

4.11-7.

in

Fig:

-

Radiatwn from a Surface to the Sky. A plane surface having an area of 1.0 m 2 is insulated on the bottom side and is placed on the ground exposed to the atmosphere at night. The upper surface is exposed to air at 290 K and the convective heat-transfer coefficient from the air to the plane is 12 W/m 2 K. The plane radiates to the clear sky. The effective radiation temperature of the sky can be assumed as 80 K. If the plane is a black body, calculate the •

temperature of the plane

at

equilibrium.

T = 266.5 K = -6.7°C Ans. and Heating of Planes. Two plane disks each 1.25 m in diameter are parallel and directly opposed to each other. They are separated by a distance of 0.5 m. Disk 1 is heated by electrical resistance to 833.3 K. Both disks are insulated on all faces except the two faces directly opposed to each other. Assume that the surroundings emit no radiation and that the disks are in space. Calculate the temperature of disk 2 at steady state and also the electrical energy input to disk 1. (Hint : The fraction of heat lost from area number 1 to

4.11-8. Radiation

space

is 1

— F i2

.)

F 12 =

Ans. 4.11-9. Radiation by Disks to

Each Other and to Surroundings.

Two

0.45,

T2 =

K m in

682.5

disks each 2.0

diameter are parallel and directly opposite each other and separated by a distance of 2.0 m. Disk 1 is held at 1000 by electric heating and disk 2 at 400 K by cooling water in a jacket at the rear of the disk. The disks radiate only to each other and to the surrounding space at 300 K. Calculate the electric heat input and also the heat removed by the cooling water.

K

A small black disk is vertical having an area of 2 0.002 and radiates to a vertical black plane surface that is 0.03 wide and 2.0 m high and is opposite and parallel to the small disk. The disk source is 2.0 away from the vertical plane and placed opposite the bottom of the plane. Determine F 12 by integration of the view-factor equation.

4.11-10. View Factor by Integration.

m

m

m

Ans. 4.11-11.

Gas Radiation

to

Gray Enclosure. Repeat Example

4.

1

1-7 but

F 12 =

0.00307

with the follow-

ing changes (a)

The

interior walls are not black surfaces but

gray surfaces with an emissivity

of 0.75.

Chap. 4

Problems

325

(b)

The same conditions

as part (a) with gray walls, but in addition heat

convective coefficient of 8.0

W/m 2

is

Assume an average

transferred by natural convection to the interior walls.

K.

•

Ans.

(b)

^(convection

+

radiation)

=

4.426

W

and Convection to a Stack'. A furnace discharges hot flue gas at 1000 K and atm abs pressure containing 5% C0 2 into a stack having an inside diameter of 0.50 m. The inside walls of the refractory lining are at 900 K and the emissivity of the lining is 0.75. The convective heat-transfer 2 coefficient of the gas has been estimated as 10 W/m K. Calculate the rate of heat transfer qIA from the gas by radiation plus convection.

4.11-12. Gas Radiation

1

•

4.12-1.

Laminar Heat Transfer of a Power-Law Fluid. A non-Newtonian power-law fluid banana puree flowing at a rate of 300 lb m /h inside a l.O-in.-ID tube is being heated by a hot fluid flowing outside the tube. The banana puree enters the heating section of the tube, which is 5 ft long, at a temperature of 60°F.

The

inside wall temperature is constant by Charm (CI) are p = 69.91b m /ft\ c p

at

=

180°F.

The

fluid properties as given

0.875 btu/lb m -°F,and k

= 0.320

has the following Theological constants: n = n' = 2 0.458, which can be assumed constant and K = 0. 146 lb f s" -ft" at70°F and 0.0417 at 190°F. A plot of log K versus T°F can be assumed to be a straight line. Calculate the outlet bulk temperature of the fluid in laminar btu/h-ft-°F.

The

fluid

•

flow.

Power-Law Fluid in Laminar Flow. A non-Newtonian power-law having the same physical properties and Theological constants as the 2 kg/s fluid in Example 4.12-1 is flowing in laminar flow at a rate of 6.30 x 10 inside a 25.4 mm-ID tube. It is being heated by a hot fluid outside the tube.

4.12- 2. Heating a fluid

D

C and leaves the heating section at an outlet bulk temperature of 46.1°C. The inside wall temperature is constant at 82.2°C. Calculate the length of tube needed in m. (Note: In this case the unknown tube length L appears in the equation for h a The

fluid enters the heating section of the tube at 26.7

and

in

the heat-balance equation.) Ans.

4.13- 1.

L=

1.722

m

Jacketed Vessel with a Paddle Agitator. A vessel with a paddle agitator and no baffles is used to heat a liquid at 37.8°C in this vessel. A steam-heated jacket furnishes the heat. The vessel inside diameter is 1.22 m and the agitator diameter is 0.406 m and is rotating at 150 rpm. The wall

Heat Transfer

in a

is 93.3°C. The physical properties of the liquid are 3 = c 2.72 kJ/kg-K, k = 0.346 W/m-K, and p. = 0.100 977 kg/m p kg/m-s at 37.8°C and 7.5 x 10" 3 at 93.3°C. Calculate the heat-transfer coefficient to the wall of the jacket.

surface temperature

p

4.13-2.

=

,

Heat Loss from Circular Fins. Use the same data and conditions from Example 4.13-2 and calculate the fin efficiency and rate of heat loss from the following different (a)

(b)

fin

materials.

= 44 W/m K). Stainless steel (k = 17.9 W/m JC). Carbon

steel (k

(a)^ A longitudinal aluminum Ans.

4.13-3.

=

0.66,q

=

111.1

W

Heat Loss from Longitudinal Fin. fin as shown in Fig. 4.13-3a (k = 230 W/m-K) is attached to a copper tube having an outside radius of 0.04 m. The length of the fin is 0.080 m and the thickness is 3 mm. The tube base is held at 450 K and the external surrounding air at 300 K has 2 a convective coefficient of 25 W/m K. Calculate the fin efficiency and the heat loss from the fin per 1.0 m of length. •

4.13-4.

326

Heat Transfer is Finned Tube Exchanger. Air at an average temperature of 50°C is being heated by flowing outside a steel tube (k = 45.1 W/m-K) having an inside diameter of 35 mm and a wall thickness of 3 mm. The outside

Chap. 4

Problems

covered with

of the

tube

L=

mm and a thickness of

13

is

16

=

/

longitudinal .0

1

steel

with

fins

mm. Condensing steam

a

length

inside at 120°C

has a coefficient of 7000 W/m 2 K. The outside coefficient of the air has been 2 estimated as 30 W/m K. Neglecting fouling factors and using a tube 1.0 m long, calculate the overall heat-transfer coefficient U based on the inside •

-

;

area

A,-.

4.14-1. Dimensional Analysis for Natural Convection.

Repeat the dimensional analysis

for natural convection heat transfer to a vertical plate as given in Section 4.14.

However, do (a)

(b)

as follows.

the detailed steps solving for all the exponents in the 7t's. Repeat, but in this case select the four variables L, fi, c p and g to be common to all the dimensionless groups.

Carry out

all

,

For unsteady-state conduction in a solid the following variables are involved: p, c p L (dimension of solid), k, and z (location in solid). Determine the dimensionless groups

4.14- 2. Dimensional Analysis for Unsteady-State Conduction.

,

relating the variables.

Ans

n

-

^ i

=—f2^2 pc L p

4.15- 1. Temperatures in a Semi-Infinite Plate.

A

semi-infinite plate

is

=T L

similar to that in

At the surfaces x = 0 and x = L, the temperature is held constant at 200 K. At the surface y = 0, the temperature is held at 400 K. If L = .0 m, calculate the temperature at the point y = 0.5 m and x = 0.5 m at Fig. 4.15-2.

1

steady state.

Heat Conduction

in a Two-Dimensional Solid. For two-dimensional heat conduction as given in Example 4. 15-1 derive the equation to calculate the total heat loss from the chamber per unit length using the nodes at the outside. There should be eight paths for one-fourth of the chamber. Substitute the actual temperatures into the equation and obtain the heat loss.

4.15-2.

,

Ans. 4.15-3. Steady-State

q

=

3426

W

A chamber that is in the shape

Heat Loss from a Rectangular Duct.

m

and of a long hollow rectangular duct has outside dimensions of 3 x 4 thick. The inside surface inside dimensions of 1 x 2 m. The walls are 1 temperature is constant at 800 K and the outside constant at 200 K. The k = 1 .4 W/m K. Calculate the steady-state heat loss per unit m of length of duct. Use a grid size of A x = Ay = 0.5 m. Also, use the outside nodes to calculate

m

•

the total heat conduction.

Ans. 4.15-4. Two-Dimensional Heat Conduction

q

=

7428

W

and Different Boundary Conditions. Avery

long solid piece of material 1 by 1 m square has its top face maintained at a constant temperature of 1000 K and its left face at 200 K. The bottom face and right face are exposed to an environment at 200 K and have a convection 2 coefficient of h = 10 W/m -K. The k W/m-K. Use a grid size of = Ay = and calculate the steady-state temperatures of the various nodes.

=10

|m

Nodal Point

4.15-5.

at Exterior

Corner Between Insulated Surfaces. Derive the

difference equation for the case of the nodal point

between insulated surfaces. The diagram the two boundaries are insulated.

is

Tn m

similar to Fig. 4.

A °S-

q n m =1 l(Tn -).m+ .

finite-

corner 15-6c except that

at an exterior

Tn.n,-l)-T„.m

REFERENCES (A 1)

Acrivos, A. A.l.Ch.E.

(Bl)

Bird, R.

B.,

York John Wiley :

Chap. 4

J., 6,

Stewart, W.

References

584 (1960). E.,

and Lightfoot, E. N. Transport Phenomena.

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(B2)

Badger, W. L., and Banchero, J. T. Introduction York: McGraw-Hill Book Company, 1955.

(B3)

Bromley,

L.

(B4)

Bowman,

R. A.,

Chem. Eng. Progr.,

A.

to

New

Chemical Engineering.

46, 221 (1950).

Mueller, A. C, and Nagle, W. M. Trans. A.S.M.E.,

283

62,

(1940).

and Su, G. Chem. Eng. Progr., 55, 54

(B5)

Brooks,

(B6)

Bolanowski,

J. P.,

(CI)

Charm,

The Fundamentals of Food Engineering, 2nd

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S. E.

Avi Publishing Co., (C2)

and Lineberry, D. D. Inc.,

(1959).

Ind. Eng. Chem., 44,

657 (1952).

ed.

Westport, Conn.:

1971.

Carslaw, H. S., and Jaeger, J. E. Conduction of Heat York Oxford University Press, Inc., 1959.

in Solids,

2nd

New

ed.

:

New

(C3)

Chapman,

(C4)

Clapp, R. M. International Developments in Heat Transfer, Part American Society of Mechanical Engineers, 1961.

(C5)

Chilton, T.

(C6)

Carreau,

(C7)

Clausing, A. M.

(Dl)

Dunlap,

I.

(Fl)

Fujii, T.,

and Imura, H.

(Gl)

Grimison,

(G2)

Geankoplis, C.

A.

Heat Transfer.

J.

Drew, T.

H.,

York: Macmillan Publishing Co.,

and Jebens, R. H.

B.,

Charest, G., and Corneille,

P.,

R.,

Int. J.

Heat Mass Transfer,

and Rushton,

J.

Can.

York:

44, 3 (1966).

791 (1966).

Heat Mass Transfer,

Int. J.

Chem. Eng.,

J.

H. Chem. Eng. Progr. Symp.,

E. D. Trans. A.S.M.E., 59, J.

9,

Inc., 1960.

New

Ind. Eng. Chem., 36, 510 (1944).

L.

J.

III.

49(5), 137 (1953).

755 (1972).

15,

583 (1937).

Mass Transport Phenomena. Columbus, Ohio: Ohio

State

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Gupta,

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Gluz, M.

D.,

(G5)

Globe,

and Dropkin, D.

(HI)

Holman,

A.

S.,

S.,

J.

Chaube, R.

B.,

and Upadhyay,

and Pavlushenko,

P.

J.

S.

L. S. J.Appl.

N. Chem. Eng.

Heat Transfer, 81,24

Heat Transfer, 4th

ed.

New

Sci., 29,

839 (1974).

Chem., U.S.S.R., 39, 2323 (1966). (1959).

York: McGraw-Hill Book Company,

1976.

M.Heat

New York

John Wiley

& Sons, Inc.,

(Jl)

Jacob,

(J2)

Jacob, M., and Hawkins, G. Elements of Heat Transfer, 3rd ed. Wiley & Sons, Inc., 1957.

(J3)

Jacob, M. Heat Transfer, Vol.

(Kl)

Kreith,

F.,

Transfer,Vo\.

and Black, W.

1.

2.

New York

Z. Basic

:

John Wiley

:

Heat Transfer.

New

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York: John 1957.

&

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York:

York: Harper

Publishers, 1980.

(K2)

Keyes,F.G. Trans. A.S.M.E.,

(K3)

Knudsen, J. G., and Katz, D. McGraw-Hill Book Company,

(K4)

73, 590

( 1

L. Fluid

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Dynamics and Heat Transfer.

1958.

Kern, D. Q. Process Heal Transfer.

New

York: McGraw-Hill Book Company,

1950.

(Ml)

B., and Kaufman, S. J. NACA Tech. Note No. 3336 (1955). McAdams, W. H. Heat Transmission, 3rd ed. New York: McGraw-Hill Book Company, 1954.

(M2)

Metzner,

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(Nl)

Nelson, W.

(LI)

Lubarsky,

L.

Gluck, D.

and Friend,

F.

Chem. Eng.

185 (1960).

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P. S. Ind. Eng. Chem., 51, 879 (1959).

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Perry, R. H., and Chilton, C. H. Chemical Engineers' Handbook, 5th York: McGraw-Hill Book Company, 1973.

328

J.

Y.,

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I.

Chem. Eng. Progr.,

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New

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Perry,

J.

H. Chemical Engineers' Handbook, 4th ed.

Book Company, (P3)

New

York: McGraw-Hill

1963.

Perry, R. H., and Green, D. Perry's Chemical Engineers' Handbook, 6th ed. York: McGraw-Hill Book Company, 1984.

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Reid, R.

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J.

M.-;

and Sherwood,

T. K.

The

Properties of Gases and 1977.

New York: McGraw-Hill Book Company,

(R2)

Rohesenow, W. M., and Hartnett, J. P., eds. Handbook of Heat Transfer. York: McGraw-Hill Book Company, 1973.

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Steinbercer, R. L, and Treybal, R. E. A.I.Ch.E.

(53)

Skelland, A. H. P. Non-Newtonian Flow and Heat Transfer. Wiley & Sons, Inc., 1967.

(54)

Skelland, A. H.

P.,

Oliver, D.

R.,

J., 6,

and Tooke,

S.

227

Brit.

New

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New

Chem.

York: John

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(Ul)

Uhl,

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Welty,

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W. Chem. Eng. J.

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Transfer, 3rd ed.

References

329

CHAPTER

5

Principles of

Unsteady-State

Heat Transfer

DERIVATION OF BASIC EQUATION

5.1

5.1

A

Introduction

we considered various heat-transfer systems in which the temperature at any given point and the heat flux were always constant with time, i.e., in steady state. In the present chapter we will study processes in which the temperature at any given point in the system changes with time, i.e., heat transfer is unsteady state or transient.

In Chapter 4

Before steady-state conditions can be reached after

the heat-transfer process

disappear. For example,

steady state.

We

in

in a process,

some

time must elapse

initiated to allow the unsteady-state conditions to

is

Section 4.2A

we determined

the heat flux through a wall at

did not consider the period during which the one side of the wall was

being heated up and the temperatures were increasing.

Unsteady-state heat transfer

and cooling problems occurring predict cooling

is

important because of the large number of heating

industrially. In metallurgical processes

and heating rates of various geometries of metals

in

it

time required to reach certain temperatures. In food processing, such as industry, perishable canned foods are heated by

immersion

in

cold water. In the paper industry

immersion

wood

in the

steam baths or

canning

chilled

by

immersed in steam baths suddenly immersed into a

is

of higher or lower temperature.

5.1

B

To

derive the equation for unsteady-state condition

Derivation of Unsteady-State Conduction Equation

Fig. 5.1-1.

Heat

conduction

is

in the

being conducted

in the

in

x direction

one direction in a solid, we refer to cube Ax, Ay, Az in size. For

in the

x direction, we write

q*

The term dT/dx means

330

necessary to

logs are

before processing. In most of these processes the material fluid

in

is

order to predict the

dT =~kA— ox

the partial or derivative of

(5.1-1)

T

with respect to x with the other

variables, y,

z,

and time

t,

being held constant. Next, making a heat balance on the cube,

we can write

+

rate of heat input

=

rate of heat generation

rate of heat output

+ The

rate of heat input to the

cube

rate of heat accumulation

(5.1-2)

is

=

rate of heat input

dT

= — k(Ay

qx x ,

Az)

(5.1-3)

dx

Also,

=

rate of heat output

qx

\

dT

= — k{Ay

x + Ax

Az)

(5.1-4) ~dx~

The

rate of accumulation of heat in the

rate of heat

The

rate of heat generation in

volume Ax Ay Az

=

accumulation

volume Ax Ay Az

(5.

1

-3>— {5. 1 -6) into (5.1-2)

Ax approach

Letting

on the

left

zero,

+

\ox

we have

k/pc

= (Ax Ay

Az)q

(5.1-6)

and dividing by Ax Ay Az,

x+

dT

Ax)

(5.1-7)

the second partial of

T

2

with respect to x ord T/dx

2

Then, rearranging,

side.

di

is

(5.1-5)

Ax

dT __k_

where a

dT —

dT

(oT q

(Ax Ay Az)pc

is

is

rate of heat generation

Substituting Eqs.

time dt

in

p

,

d

2

T

pc. dx

2

— 2

d

q

T

ox 2

pc„

— q

+

(5.1-8)

pc p

thermal diffusivity. This derivation assumes constant

k, p,

andCj,. In SI

= m 2 /s, T = K, = s, k = W/m K, p = kg/m 3 q = W/m 3 and c p = J/kg K. In 2 3 3 English units, a = and /h, T = °F, t = h, k = btu/h ft °F, p = lbjft q = btu/h ft = btu/lb c °F. p units,

i

f

,

,

•

ft

•

,

,

ra

For conduction

Figure

5.1-1.

Sec. 5.1

in

three dimensions, a similar derivation gives 2

T

d_T_

d

~d~i

2+ lix

Unsteady-stale conduction one direction.

Derivation of Basic Equation

d

2

T

2+ ~dy

d

2

T

~dz

2

+

(5.1-9)

P cp

in

331

In

many

generation

is

conduction

cases, unsteady-state heat

zero.

Then Eqs.

(5.

-8)

1

dT _ =

and

(5.

occurring but the rate of heat

is

become

-9)

1

2

a—T 8

(5.1-10)

dT _ (d 2 T =a + 17 \~d7

d

2

T

2

d

lT +

T

~dl

y

T

Equations (5.1-10) and (5.1-11) relate the temperature

with position

x, y,

and

The solutions of Eqs. (5.1-10) and (5.1-1 1) for certain specific cases as well time the more general cases are considered in much of the remainder of this chapter. t.

z

and

as for

SIMPLIFIED CASE FOR SYSTEMS WITH NEGLIGIBLE

5.2

INTERNAL RESISTANCE Basic Equation

5.2A

We

begin our treatment of transient heat conduction by analyzing a simplified case. In

this situation

we consider

a solid which has a very high thermal conductivity or very low

internal conductive resistance tion occurs is

from the external

compared

to the external surface resistance, where convec-

fluid to the surface of the solid.

very small, the temperature within the solid

An example would immersed

T0 K

be a small, hot cube of steel at

into a large bath of cold water at

that the heat-transfer coefficient h in

Tm

W/m

which

K

2

Since the internal resistance

essentially uniform at

is

is

is

any given time. t = 0, suddenly

at time

Assume Making a heat

held constant with time.

constant with time.

balance on the solid object for a small time interval of time

dt

s,

the heat transfer from the

bath to the object must equal the change in internal energy of the object.

hA{Tx where A

is

time

s,

-T)dt=c pP V dT

m T the average temperature of the object at in kg/m and V the volume in m Rearranging the 2

the surface area of the object in

,

3

p the density of the object equation and integrating between the limits of ;

in

(5.2-1)

3

.

,

T = T0 when = t

dT :ro

hA

T.

T = T when = I

c pP

'

f

=

V

f,

=

dt

T„-T T—

,

0 and

(5.2-2)

o

e-WpfiV)'

(5.2-3)

This equation describes the time-temperature history of the solid object. The term c

pV p

is

often called the lumped thermal capacitance of the system. This type of analysis

often called the lumped capacity

5.2B

Equation for Different Geometries

In using Eq. (5.2-3) the surface/volume ratio of the object

assumption of negligible internal resistance was is

reasonably accurate

made

must be known. The basic

in the derivation. This assumption

when

N Bi =

332

is

method or Newtonian heating or cooling method.

hx ^<0A

Chap. 5

Principles of Unsteady-State

(5.2-4)

Heat Transfer

where hx Jk is called the Biot number N Bi which is dimensionless, andx, is a character= V/A. The Biot number compares the istic dimension of the body obtained from x relative values of internal conduction resistance and surface convective resistance to heat ,

l

transfer.

For a sphere, 3

V

4nr /3

r

*,=-7 = -r-r = 7 For

(

5 2-5) -

a long cylinder,

kD 2 L/4 nuL

V

A

D

r

4

2

For a long square rod, X

V

= >

=

A

(2x)

2

L

x

=

thickness)^

4^x)L

(5.2-7)

2

EXAMPLE

5.2-1. Cooling of a Steel Ball having a radius of 1.0 in. (25.4 mm) is at a uniform temperature of 800°F (699.9 K). It is suddenly plunged into a medium whose temperature is held constant at 250°F (394.3 K). Assuming a convective coefficient 2 2 of h = 2.0 btu/h-ft -°F (11.36 W/m K), calculate the temperature of the ball after 1 h (3600 s). The average physical properties are k = 25 btu/ 3 3 Ji-ft-°F (43.3 W/m-K), p = 490 lbjft (7849 kg/m ), and c p = 0.11 btu/ lb m °F (0.4606 kJ/kg K). Use SI and English units.

A

steel ball

•

•

Solution:

For

a sphere

from Eq.

Xl==

7

(5.2-5),

=

=

T

3

=

75ft

25.4

1000 x 3

From

Eq. (5.2-4) for the Biot number,

N Bi =

^=^=

0.00222

N.- il*Sgp3 - 0.00222 This value

is

hA

<0.1 hence, the lumped capacity method can be used. Then, ;

2

c

pP V

c

pP V

=

1.335 h"

hA

11.36

(0.4606 x 1000X7849X8.47 x 10

Substituting into Eq. (5.2-3) for

T—T T0 — Tg,

T-

250°F

800

-

T699.9

Sec. 5.2

1

0.11(490X^)

Simplified

e

=

1

.0

3.71

x 10"* s"

1

(1.335

h" 1 )

)

h and solving for T,

-{hAlc pP VM

_

e

-
-jr

_ 395°p

250

394.3

-

t

=

-3

K

T=

474.9

394.3

Case for Systems With Negligible Internal Resistance

K 333

Amount of Heat Transferred

Total

5.2C

The temperature t,

of the solid at any time

t

the instantaneous rate of heat transfer

can be calculated from Eq.

W from

q(t) in

At any time

(5.2-3).

the solid of negligible internal

resistance can be calculated from q(t)

= hA(T — TJ

Substituting the instantaneous temperature q(t)

To time

t

0 to

r

=

I,

we can

(5.2-3) into

Eq.

q(t)dt

=

Q

W

in

•

s

(5.2-9)

or J transferred from the solid from

TJe-^ Alc '" y)

M'o-

Q = c pP V(T0 - TJ[1

-

-

e

(h

'

dt

(5.2-10)

«]

(5.2-11)

Amount of Heat in Cooling Example 5.2-1, calculate the

Total

5.2-2.

For the conditions removed up to time

in £

= 3600

total

amount of heat

s.

From Example 5.2-1, hA/c p pV = 3.71 x 10" 4 s" = 4(tiX0.0254) 3 /3 = 6.864 x 10- 5 m 3 Substituting into

Solution: 3

47tr /3

.

Q =

(0.4606 x 10O0X7849)(6.864 x 10 |"|

.

=

(5.2-8),

integrate Eq. (5.2-9).

Q =

EXAMPLE

from Eq.

= hA(T0 - TJe-^'"^'

determine the total amount of heat

=

T

(5.2-8)

_

g-(3.71

5.589 x 10

4

_5

X699.9

-

1 .

Also,

V =

Eq. (5.2-1

1),

394.3)

x 10-->)(3600)-j

J

UNSTEADY-STATE HEAT CONDUCTION

5.3

IN VARIOUS GEOMETRIES

Introduction and Analytical

5.3A

In Section 5.2

we considered

Methods

a simplified case of negligible internal resistance

object has a very high thermal conductivity. situation

where the internal resistance

constant in the solid. convective resistance

is

The

first

is

Now

we

will consider the

where the

more

general

not small, and hence the temperature

is

not

we shall consider is one where the surface compared to the internal resistance. This could occur

case that

negligible

because of a very large heat-transfer coefficient

at

the surface or because of a relatively

large conductive resistance injhe object.

To

illustrate

an analytical method of solving this

first

case,

equation for unsteady-state conduction in the x direction only in a

2H

as

shown

uniform

and held

at

in Fig. 5.3-1.

T = T0

.

The

At time

there. Since there

is

t

=

initial profile

flat

will

derive the

plate of thickness

of the temperature in the plate at

/

=

0

is

is suddenly changed to T, the temperature of the surface is

0, the ambient temperature

no convection resistance,

also held constant at T,. Since this

is

conduction

« dt

334

we

Chap. 5

in the

x direction, Eq.

(5.1-10) holds.

—

(5.1-10)

dx

Principles of Unsteady-State

Heat Transfer

The

initial

and boundary conditions are

T = T0

,

T = Tu T = Generally,

between 0 and

it is

1.

t

= =o, 0,

X x

= xX

I

=

t,

x

=0

«

=

t,

x

= 2H

(53-1)

convenient to define a dimensionless temperature

Y

so that

it

varies

Hence,

T.-T

Y =

(53-2) J

J

1

o

Substituting Eq. (5.3-2) into (5.1-10),

dY = ~

— d

a

initial

T\

Tx T\

Y =

— r,

A

convenient procedure

variables,

which leads

to a

Y

(53-3)

T~ ox

ot

Redefining the boundary and

2

conditions,

- ^0 - T0

1

,

- r, - T0

0,

- r, = - r To 0

0,

to use to solve

•

t

=

0,

x = x

t

=

t,

x = 0

t

=

t,

x = 2H

Eq. (5.3-3)

e

- a2

«(-A

where A and B are constants and a

is

cos ax

+ B

a parameter.

conditions of Eq. (5.3-4) to solve for these constants

Fourier series (Gl).

FIGURE

5.3-1.

the

is

method

of separation of

product solution

y =

infinite

(53-4)

V nsteady-stale '

conduction

in

sin ax)

(53-5)

Applying the boundary and in

Eq.

(5.3-5), the final

initial

solution

is

an

a

2H-

flat plate with negligible sur-

face resistance.

T0 T

at

at

/

t

H

Sec. 5.3

Unsteady-State Heat Conduction

in

= 0 =

t

7H

Various Geometries

335

T, T,

- T — T0

-l 2

4/1

-

- exp \l

7i

—

2

4tf

-5 2

2 7r

5

Itcx

.

sin

1

h

2//

at

7i

- exp 3

—

3tix

.

sin

r

2W

\

5tcx

.

-17^

exp

-3 2 af 4H 2 2

at

=

1

+

2 7t

+ -'j

sin

-i77

Hence from Eq. (5.3-6), the temperature T at any position x and time can be determined. However, these types of equations are very time consuming to use, and convenient charts have been prepared which are discussed in Sections 5.3B, 5.3C, 5. 3D, and t

where

5.3E,

a surface resistance

is

present.

Unsteady-State Conduction

5.3B

In Fig. 5.3-2 a semiinfinite solid

conduction occurs only

uniform

T0

at

of ambient

W/m

2 •

K

.

At time

fluid at

or btu/h

t

Hence, the temperature

The solution

=

0,

°F

ft

is

Semiinfinite Solid

shown

+x

that extends to oo in the

x direction. Originally, the temperature the solid is suddenly exposed to or immersed

in the

is

Ts at

direction.

in a large

1

;

the surface

- Y=

—

mass

T,,

is

not the same as T,.

of Eq. (5.1-10) for these conditions has been obtained (SI) and

=

Heat

in the solid is

which is constant. The convection coefficient h in present and is constant i.e., a surface resistance is present.

temperature 2

•

in a

is

erfc

x exp

h^/at

+

x

h

erfc

(5.3-7)

at

where x

is

the distance into the solid

2 a = k/pc p in m /s. In English — erf), where erf is the error (1

units,

from the surface in SI units in m, t = time in = h, and a = ft 2 /h. The function erfc x = ft, t

function and numerical values are tabulated in standard

tables

and texts (Gl, PI, SI), Y is fraction of unaccomplished change

and

— 7 is fraction

1

Figure

s,

is

(7^

-

T)I{T\

—To),

of change.

5.3-3, calculated

using Eq. (5.3-7),

is

a convenient plot used for unsteady-state

heat conduction into a semiinfinite solid with surface convection. If conduction into the solid

is

slow enough or h

EXAMPLE The depth

53-1.

is

very large, the top line with hy/at/k

Freezing Temperature

in the

=

00

is

used.

Ground

which freezing temperatures penetrate is often of importance in agriculture and construction. During a certain fall day, the temperature in the earth is constant at 15.6°C (60°F) to a depth of several meters. A cold wave suddenly reduces the air temperature from in the soil of the earth at

15.6 to — 17.8°C (0°F). The convective coefficient W/m 2 K (2 btu/h 2 °F). The soil properties can •

Figure

5.3-2.

ft

•

Unsteady-slate conduction

in

above the soil is 11.36 be. assumed as a = 4.65

a

semiinfinite solid.

T0

336

Chap. 5

at

Principles of Unsteady-State

t

= 0

Heal Transfer

m

x 10" 7

2

/s (0.018

any latent heat (b)

This

Solution:

x a

= -

and k = 0.865 W/m K Use SI and English units.

/h)

btu/h

(0.5

•

ft

What is the surface temperature after 5 h? To what depth in the soil will the freezing temperature penetrate in 5h?

(a)

For part

solid.

2 ft

effects.

is

(a),

°F).

Neglect

of 0°C (32°F)

a case of unsteady-state conduction in a semiinfinite

the value of x which

the distance from the surface

is

is

=

0 m. Then the value of x/2 N/af is calculated as follows for t 5 h, x 10- 7 m 2 /s, k = 0.865 W/m °C, and A = 1 1.36 W/m 2 °C. Using

4.65

•

•

SI and English units,

x

~ ijal

0 _7 2^/(4.65 x 10 )(5

x

0 ~~

x 3600)

iJoTt

2^0.018(5)

Also, 7 11.36^/(4.65 x 10" X5 x 3600)

hy/ai

~

k

=

at

2^/0.018(5)

0.865

0.5

=

1.2

1.2

Fig. 5.3-3, for x/2 N/af = 0 and hy/at/k = 1.2, the value of — Y = 0.63 is read off the curve. Converting temperatures to K, T0 = 15.6°C + 273.2 = 288.8 K (60°F) and T, = - 17.8°C + 273.2 = 255.4 K

Using 1

(0°F).Then 1

-

Y =

T - T0

„ ,„

0.63

=

T— 255.4

288.8

-

288.8

i

i

i

Figure

5.3-3.

Unsteady-slate heat conducted

in

a semiinfinite solid with surface con-

vection. Calculated from Eq. (5.3-7){SI).

Sec. 5.3

Unsteady-Stale Heat Conduction

in

Various Geometries

337

Solving for

T

at the surface after 5 h,

T = For part

K

273.2

K

T=

(b),

Substituting the

267.76

known

-5.44°C

or

and

or 0°C,

255.4 -

288.8

273.2

- T0 )/(T, - T0 =

Fig. 5.3-3 for

0.16

read off the curve for.x/2 N/af. Hence,

)

= Solving for

=

Unsteady-State Conduction

5.3C

A geometry

0.16 '

0.0293

=

1.2,

a value of

— ^= = ^i==0.16 zjat 2^/0.018(5)

temperature penetrates in

x, the distance the freezing

x

=

0.467 and h^/of/A:

7 2^/(4.65 x 10" X5 x 3600)

2701

unknown.

is

288.8

From

(7

the distance x

values,

T — T0 = T, - T0 is

(22.2°F)

m

(0.096

5 h,

ft)

Large Flat Plate

in a

that often occurs in heat-conduction

problems

is

a flat plate of thickness 2x 1

in the x direction and having large or infinite dimensions in the y and z directions, as

shown in the

in Fig. 5.3-4. Heat is being conducted only from the two flat and parallel surfaces x direction. The original uniform temperature of the plate is T0 and at time = 0, t

,

the solid

occurs.

is

exposed to an environment

A surface resistance

is

at

temperature

Tj

and unsteady-state conduction

present.

The numerical results of this case are presented graphically in Figs. 5.3-5 and 5.3-6. Figure 5.3-5 by Gurney and Lurie (G2) is a convenient chart for determining the temperatures at any position in the plate and at any time The dimensionless pat.

rameters used in these and subsequent unsteady-state charts

Table

5.3-1 (x is the distance

one half the thickness of the

from the surface

When

n

=

fiat plate,

cylinder, or sphere, x,

plate radius of cylinder, or radius of sphere,

x

=

is

distance

for a semiinfinite solid.)

the position

0,

temperature history for

from the center of the

flat

in this section are given in

is

at the center

the center of the plate in Fig. 5.3-5. Often the

at

quite important.

of the plate

is

determining only the center temperature

is

A more

given in Fig. 5.3-6

accurate chart

in the Heisler

(HI)

chart. Heisler (HI) has also prepared multiple charts for determining the temperatures at

other positions.

EXAMPLE A

5.3-2.

Heat Conduction

rectangular slab of butter which

of 277.6 297.1

K

K

(4.4°C) in a cooler

(23.9°C).

The

sides

in a is

Slab of Butter

46.2

mm

thick at a temperature

removed and placed in an environment at and bottom of the butter container can be is

y Figure

338

5.3-4.

Unsteady-state conduction

Chap. 5

in

a large flat

plate.

Principles of Unsteady-Slate

Heat Transfer

considered to be insulated by the container side walls. The flat top surface of the butter is exposed to the environment. The convective coefficient is

constant and

W/m 2

8.52

is

K. Calculate the temperature

•

mm

below the surface, and the insulated bottom after 5 h of exposure.

surface, at 25.4

The

Solution:

butter can be considered as a large

tion vertically in the x direction. Since heat

and the bottom

mm

at 46.2

face

is

insulated, the 46.2

plate with thickness x,

=

mm.

46.2

=

x

in Fig. 5.3-4, the center at

in the

butter at the

below the surface

fiat

at

plate with conduc-

entering only at the top face of butter is equivalent to a half

is

mm

In a plate with two exposed surfaces as

0 acts as an insulated surface and both halves

are mirror images of each other.

The

Appendix A.4 are k = 0.197 2300 J/kg K, and p = 998 kg/m 3 The ther-

physical properties of butter from

W/m K,cp =

2.30 kJ/kg

mal

is

diffusivity

=

Also, x,

46.2/1000

=

K=

•

•

.

0.0462 m. for use

The parameters needed

m=

For

the top surface

hx

~x\~

Table

2

(0.0462)

=

where x = Xi x^

_ "

_U/

0.0462 m,

_

0.0462

0.0462

x,

~

'

Fig. 5.3-5,

Y Solving,

o 50

x 10" 8 X5 x 3600)

(8.58

"

Then using

=

8.52(0.0462)

t

at

X

0197

JL =

are

in Fig. 5.3-5

T=

=

0.25

- T 7,-70

297.1

-

277.6

292.2 K(19.0°C).

Dimensionless Parameters for Use

5.3-1.

- 7

297.1

T,

=

in

Unsteady-State

Conduction Charts

7,

- 7

k

-T0 7 -T

/ix.

7,

0

7,

-

=

n

—X x,

To

at

A SI units: a

=

m

2

English units a :

h

Cgs

btu/h-

units: a

h

Sec. 5.3

=

=

7=

/s,

=

2 ft

2 ft

/h,

K,

7=

cal/s-

cm

=

s,

°F,

x t

= m, x, = m, k = W/m-K,/i = W/m 2 K = h, x = x = = btu/h °F, ft,

,

ft

ft, /c

-°F

= cm 2 /s, 7 = 2

t

°C,

t

=

s,

x

=

cm,x,

=

cm, k

=

cal/s

•

cm

•

"C,

°C

Unsteady-State Heat Conduction in Various Geometries

339

340

Chap. 5

Principles of Unsteady-Stale

Heat Transfer

341

mm from the top surface or 20.8 mm from the center,

At the point 25.4 x

=

0.0208

m.and n

From

Fig. 5.3-5,

T=

°-

ft

^ T ^ 7 ._ QA5= iT, - T0 .

Y = Solving,

x 0208 « — =^ = 0.45 0.0462 x,

=

K

287.4

for the center point,

5.3D

277.6

0 and

—=—=0 0

n

5.3-5,

Y = Solving,

-

m from the top, x =

x

=

n

T=

297.1

—^ T

K (15.1 °C).

288.3

For the bottom point or 0.0462

Then, from Fig.

,

297.1

Y

x

-T

291

T,

- T0

297.1

= T

0.50

-

277.6

(14.2°C). Alternatively, using Fig. 5.3-6, which = 0.53 and T = 286.8 (1 3.6°C).

is

only

K

Unsteady-State Conduction

Long Cylinder

a

in

Here we consider unsteady-state conduction

in a

long cylinder where conduction occurs

long so that conduction at the ends can be

only in the radial direction. The cylinder is neglected or the ends are insulated. Charts for

this

and

determining the temperatures at any position

case are presented in Fig. 5.3-7 for

Fig. 5.3-8 for the center temperature

only.

EXAMPLE

Transient Heat Conduction in a Can of Pea Puree can of pea puree (C2) has a diameter of 68.1 and a height of 101.6 and is initially at a uniform temperature of 29.4°C. The cans are stacked vertically in a retort and steam at 115.6°C is admitted. For a heating time of 0.75 h at 115.6°C, calculate the temperature at the center of the can. Assume that the can is in the center of a vertical stack of cans and that it is insulated on its two ends by the other cans. The heat capacity of the

A

5J-3.

mm

cylindrical

mm

metal wall of the can will be neglected. The heat-transfer coefficient of the steam is estimated as 4540 W/m 2 K. Physical properties of puree are 7 2 k = 0.830 W/m K and a = 2.007 x 10" /s.

m

•

Since the can long cylinder. The radius x = 0, Solution:

we can consider it as a 0.03405 m. For the center with

is

insulated at the two ends

is

x,

=

0.0681/2

=

n

=

—x = —0 = 0n

Also,

m=

8

it,

=

X =-=

=

454(So .03405) (2-007

x 1Q- 7 X075 x 3600)

x\

(0.03405)

Using Fig. 5.3-8 by Heisler

Y =

342

T=

2

=

^

for the center temperature,

0.13

=

T, T,

Solving,

°-°0537

-T - T0

115.6

—T

115.6-29.4

104.4°C.

Chap. 5

Principles of Unsteady-Slate

Heat Transfer

5.3E

Unsteady-State Conduction

In Fig. 5.3-9 a chart

any position

is

given by

in

a Sphere

Gurney and Lurie

in a sphere. In Fig. 5.3-10

for

determining the temperatures at

a chart by Heisler

is

given for determining the

center temperature only in a sphere.

Sec. 5.3

Unsteady-State Heat Conduction in Various Geometries

343

344

Figure

5.3F

5.3-9.

Unsteady-state heat conduction in a sphere. [From H. P. Gurney and Lurie, ind. Eng. Chem., 15, 1170(1923).']

Unsteady-State Conduction

in

J.

Two-

and Three-Dimensional Systems

The heat-conduction problems considered so far have been limited to one dimension. However, many practical problems are involved with simultaneous unsteady-state conduction in two and three directions. We shall illustrate how to combine one-dimensional solutions to yield solutions for several-dimensional systems.

Newman Sec. 5.3

(Nl) used the principle of superposition and showed mathematically

Unsteady-State Heal Conduction

in

Various Geometries

how 345

o °^ -

346

l

J-

„ 'Or <

o

o

o

Figure

5.3-1

Unsteady-state conduction in

1.

three

directions

a

in

rect-

angular block.

2*i

2x^~y to

combine the solutions

one-dimensional heat conduction in the

for

direction into an overall solution for simultaneous conduction in

x, the y,

all

For example, a rectangular block with dimensions 2x,, 2y 1> and 2z, 5.3-1 1. For the Y value in the x direction, as before, T,

-

and the

is

shown

the temperature at time

is

before. Also, n

=

x/x„

m=

k/hx

1

,

f

and

(53-8)

and'position x distance from the center

Zx =

in Fig.

71

>;=^-^? where 7^

z

three directions.

ar/xj, as before.

Then

for the

line,

as

y direction, (53-9)

and n

=

m=

y/y u

k/hy^

and

X = y

ar/j^. Similarly, for the z direction,

T —T n = rr-^r Then,

for the

simultaneous transfer in

Yx y „ ,

where

Tx

block.

The value

y

conduction

,

=(YJYy

all

(53-10)

three directions,

m VV* — =

J i

J

(53-11)

o

from the center of the rectangular two parallel faces is obtained from Figs. 5.3-5 and 5.3-6 for The values of Yy and Yz are similarly obtained from the same

the temperature at the point x, y, z

is

of

Yx

for the

in a flat plate.

charts.

For

a short cylinder with radius x y

cylinder.

and length 2y v the following procedure is is obtained from the figures for a long parallel planes is obtained from Fig. 5.3-5

Yx for the radial conduction Then Yy for conduction between two

followed. First

or 5.3-6 for conduction in a

flat plate.

y .„ = J

Then,

(OT =

7

i'~ M ~ t" 0

(5 -3_12)

J

EXAMPLE

53-4. Two-Dimensional Conduction in a Short Cylinder Repeat Example 5.3-3 for transient conduction in a can of pea puree but assume that conduction also occurs from the two flat ends.

The can, which has a diameter shown in Fig. 53-12. The given

Solution:

mm,

Sec. 5.3

is

Unsteady-State Heal Conduction

in

mm

and a height of 101.6 values from Example 5.3-3 are

of 68.1

Various Geometries

347

Figure

Two

5.3-12.

in

dimensional conduction

Exam-

a short cylinder in

ple 5.3-4.

2yi

x,

=

10"

7

= 0.1016/2 = 0.0508 m, k = 0.830 W/m K, a = 2.007 x 0.03405 m, 2 2 and t = 0.75(3600) = 2700 s. /s, h = 4540 W/m For conduction in the x (radial) direction as calculated previously, •

m

K

^" x

"

=

0

=

x]

From

0.830

k

n

m= kx~r

°'

(0.03405)

=

45^40(0.03405)

^

nm „„

2

Fig. 5.3-8 for the center temperature,

^ For conduction

in the

y

=

0.13

(axial) direction for the center

y

o

y,

0.0508

temperature,

0.830 hyi

X = «2 y

Using

~

0.00360 4540(0.0508)

J 2m ];ZJ (0.0508)

21 °°

2

=

0-210

Fig. 5.3-6 for the center of a large plate (two parallel

Yy =

opposed

planes),

0.80

Substituting into Eq. (5.3-12),

Yxy = (Yx XYy = )

0.13(0.80)

=

0.104

Then,

T ~ T*-v = Tj - T0 i

Txy =

115 6 -

~ T*.y =

115.6-29.4

a 104

106.6°C

This compares with 104.4°C obtained in Example 5.3-3 for only radial conduction.

53G If

Charts for Average Temperature in a Plate, Cylinder, and Sphere with Negligible Surface Resistance

the surface resistance

fraction of

348

is

negligible, the curves given in Fig. 5.3-13 will give the total

unaccomplished change, E, for slabs, cylinders, or spheres

Chap. 5

for unsteady-state

Principles of Unsteady-State

Heat Transfer

conduction.

The

value of E

is

(53-13)

where

T0

t

T

the original uniform temperature,

is

which the solid

is

suddenly subjected, and

Tav

is

x

the environment temperature to

is

the average temperature of the solid after

hours.

The

values of

faces as in a plate.

E a E b and E c ,

,

For example,

are each used for conduction between a pair of parallel

for

conduction

in the

a

and

b directions in a rectangular

bar,

E = Ea E b For conduction from

all

(53-14)

three sets of faces,

E - E a E b Ec For conduction

in a short cylinder 2c

(53-15)

long and radius

a,

E = Ec E r

(5.3-16)

0.006 0.004 0.003 0.002

°- 00

0

0.1

0.2

0.4

0.3

af at a

FIGURE

5.3-13.

o

0.5

0.6

0.7

at c

Unsteady-state conduction and average temperatures for negligible Mass Transfer Operations,

surface resistance. {From R. E. Treybal,

2nd

ed.

New York : McGraw-Hill Book Company,

1968. With per-

mission.)

Sec. 5.3

Unsteady-State Heat Conduction in Various Geometries

349

NUMERICAL FINITE-DIFFERENCE METHODS FOR UNSTEADY-STATE CONDUCTION

5.4

Unsteady-State Conduction

5.4A

1.

As discussed

introduction.

in

in

a Slab

previous sections of this chapter, the partial differential

equations for unsteady-state conduction analytically

if

the

various simple geometries can be solved

in

boundary conditions are constant

solutions the initial profile of the temperature at

unsteady-state charts used also have these dition. initial

T=

at t

=

0

T,

is

with time. Also,

uniform

at

T=

same boundary conditions and

T0

in the .

initial

The con-

However, when the boundary conditions are not constant with time and/or the conditions are not constant with position, numerical methods must be used.

Numerical calculation methods

conduction are similar to

for unsteady-state heat

numerical methods for steady state discussed in Section 4.15. The solid is subdivided into sections or slabs of equal length and a fictitious node is placed at the center of each

Then a heat balance is made for each node. This method differs from the method in that we have heat accumulation in a node for unsteady-state

section.

steady-state

conduction.

2.

The unsteady-state equation

Equations for a slab.

in a slab

for

conduction

x

in the

direction

is

dT _

d

2

T (5.1-10)

This can be set up for a numerical solution by expressing each partial derivative as an

AT, At, and Ax. However, an alternative method will be used to making a heat balance. Figure 5.4-1 shows a slab centered at position n, represented by the shaded area. The slab has a width of Ax m and a 2 cross-sectional area of A m The node at position n having a temperature ofT„ is placed at the center of the shaded section and this node represents the total mass and heat capacity of the section or slab. Each node is imagined to be connected to the adjacent actual finite difference in

derive the final result by

.

node by

a fictitious, small conducting rod. (See Fig. 4.15-3 for

The

an example.)

shows the temperature profile at a given instant of time t heat balance on this node or slab, the rate of heat in — the rate of heat out figure

heat accumulation in At

kA — where

,T„

timet-)-

1

At

later.

—

- T„)- kA l

(,Tn -,T„ +i )=

Rearranging and solving

,

Making

a

the rate of

s.

the temperature at point n at time

is 1

(,Tn -

s.

=

+ A ,T„

=

-Jt

(A Ax)pc„ P P

J

and, + A ,T„

t

for

,

l.Tjt + (M

{, +

is

*Tn -,TJ

(5.4-1)

the temperature at point n at

+ A ,T„

—

2),T„

+

,Tn -J

(5-4-2)

where

MJ^L-

(5.4-3)

a At

Note time

t

+

time. This

350

that in Eq. (5.4-2) the temperature

At

is

is

l

+ A ,T„ at position or

calculated from the three points which are

known

node

called the explicit method, because the temperature at a

Chap. 5

Principles

n

at time

and t,

at a

new

the starting

new time can be

of Unsteady-State Heat Transfer

calculated explicitly from the temperatures at the previous time. In this

method

the

calculation proceeds directly from one time increment to the next until the final temper-

ature distribution

Of

calculated at the desired final time.

is

course, the temperature

and the boundary conditions must be known. Once the value of Ax has been selected, then from Eq. (5.4-3) a value of M or the time increment At may be picked. For a given value of M, smaller values of Ax mean must be as follows: smaller values of At. The value of

distribution at the initial time

M

M>2 M

(5.4-4)

2, the second law of thermodynamics is violated. It also can be shown must be > 2. and convergence of the finite-difference solution Stability means the errors in the solution do not grow exponentially as the solution proceeds but damp out. Convergence means that the solution of the difference equation approaches the exact solution of the partial differential equation as At and Ax go to zero with fixed. Using smaller sizes of At and Ax increases the accuracy in general but greatly increases the number of calculations required. Hence, a digital computer is often If

less

is

than

M

that for stability

M

ideally suited for this type of calculation.

method for a slab. If the value of Eq. (5.4-2) occurs, giving the Schmidt method.

Simplified Schmidt

3.

plification of

,

This means that t

At

4-

n

—

1

5.4B

is

when

a time

1

+ a,

K=

T

4-

time

2,

then a great sim-

T (3.4-5)

^

At has elapsed, the new temperature

the arithmetic average of the temperatures at the

at the original

M=

at a given point n at

two adjacent nodes

n

+

1

and

t.

Boundary Conditions

for

Numerical Method for a Slab

For the case where there is a finite convective resistance boundary and the temperature of the environment or fluid outside is suddenly changed to Ta we can derive the following for a slab. Referring to Fig. 5.4-1, we make a /.

Convection at the boundary.

at the

,

Figure

Sec. 5.4

5.4-

1

.

Unsteady-state conduction

in

a slab.

Numerical Finite-Difference Methods for Unsteady-State Conduction

351

heat balance on the outside \ element. heat out by conduction

-

hA(,Ta where ,T12 s

>

=

The

by convection

the rate of heat accumulations in At

— ^— (,Ti —

,7,)

Ax

=

,T2 )

{

T

(

A/

—

the rate of

s.

AM9££z ,

the temperature at the

s

rate of heat in

midpoint of the 0.5

Ax

T

(5.4.6)

)

outside slab. As an

approximation, the temperature Ti at the surface can be used to replace that ofTj Rearranging,

,

=

+ a,^

-j-

[2N,Ta

+ [M —

(2N +

+

2)] ,7;

2,T2 ]

25

.

(5.4-7)

where

N=

—Ax— /i

(5.4-8)

k

Note that

the value of

M must be such that

M > 2N + 2 2.

Insulated boundary condition.

N

In the case for the

made on

boundary condition where

the rear

Axslabjust as on the front^ Ax slab Fig. 5.4-1. The resulting equation is the same as Eqs. (5.4-6) and (5.4-7), but h = 0 or = 0 and ,Tf _ = ,7} + because of symmetry.

face in

(5.4-9)

is

insulated, a heat balance

i

is

the rear|

t

,

+ *,Tf

=

-

^

2),7>

+ 2Js .

(5.4-10)

-\

x

Alternative convective condition. To use the equations above for a given problem, the same values of M, Ax, and At must be used. If N gets too large, so that may be inconveniently too large, another form of Eq. (5.4-7) can be derived. By neglecting the i.

M

heat accumulation in the front half-slab in Eq. (5.4-6),

i

Here the value of well

when

neglected

4.

M

is

small

—

N ^~

j

t

+

AiTa

not restricted by the

number compared

a large

is

+ ai^i

of increments

in

1 ,

j

+ & T2

value. This

Ax

(5.4-11)

t

approximation works

are used so that the

amount

fairly

of heat

to the total.

Procedures for use of initial boundary temperature. When the temperature of the is suddenly changed to T the following procedures should be used. a

environment outside 1.

N

+ —

M

=

,

hand calculation of a limited number of increments is used, a and (5.4-7) or (5.4-11). For the first time increment one should use an average value for Ta of (Ta + 0 Ti)/2, where 0 Tj is the initial temperature at point 1. For all succeeding Ar values, the value of Ta should

Whejn-

2 and a

special procedure should be used in Eqs. (5.4-5)

x

be used (Dl, Kl). This special procedure

for the

value of T„ to use for the

first

time

increment increases the accuracy of the numerical method, especially after a few time

3.

T

varies with^ime r, a new value can be used for each At interval. and many time increments are used with a digital computer, this special procedure is not needed and the one value of Ta is used for all time increments. = 3 or more and a hand calculation of a limited number of increments or a When intervals. If

2.

When

352

M

=

a

2

M

Chap. 5

Principles

of Unsteady-State Heat Transfer

digital

computer calculation of many increments

used for

time increments. Note that

all

compared

are needed

when

M

=

4,

which

to the case for

is

M

when

M

=

=

used, only the one value of

is

The most accurate

2.

Ta

is

more, many more calculations

3 or

results are

obtained

the preferred method, with slightly less accurate results for

M = 3(D1,K1,K2). EXAMPLE 5.4-1.

Unsteady-State Conduction and the Schmidt Numerical Method A slab of material 1.00 m thick is at a uniform temperature of 100°C. The front surface is suddenly exposed to a constant environmental temperature of 0°C. The convective resistance is zero (h = co). The back surface of the 3 2 slab is insulated. The thermal diffusivity a = 2.00 x 10" /s. Using five = 2.0, slices each 0.20 m thick and the Schmidt numerical method with calculate the temperature profile at = 6000 s. Use the special procedure for

m

M

£

the

first

time increment.

Figure 5.4-2 shows the temperature profile at r = 0 and the environmental temperature of T„ = 0°C with five slices used. For the = 2. Substituting into Eq. (5.4-3) with a = 2.00 x 10" 5 Schmidt method,

Solution:

M

and Ax

=

0.20 and solving for At, 2

M = 2 = (Ax) aAr

(0.20)

(2.00

2

At

x lO" 5 ) Ar

=

1000

s

(5.4-3)

This means that (6000 s)/(1000 s)/increment), or six time increments must be used to reach 6000 s. For the front surface where n = 1, the temperature Ta to use for the first At time increment, as stated previously, is x

i

where

0

T

X

is

'

i

the initial temperature at point

-t

n

1.

=

(5.4-12)

1

For

the remaining time

=0

insulated wall

Figure

Sec. 5.4

5.4-2.

Temperature for numerical method. Example

5.4-1.

Numerical Finite-Difference Methods for Unsteady-State Conduction

353

increments,

To calculate using Eq.

n=\

= Ta

Ty

the temperatures for

time increments for the slabs n

all

,Jn

+

=

'

T"-' +

2 to

5,

Tn+l

'

n

=

5

2, 3, 4,

(5.4-14)

2

For the insulated end for all time increments and /= 6 into Eq. (5.4-12),

For the first time increment of 1 by Eq. (5.4-12),

+

f

at n

=

6,

substituting

M=2

Af and calculating the temperature

=

Ta +

T For n

=

2,

0+100

oTl

using Eq. (5.4-14),

50

+

100

100

+

100

—

100

—

,7y+,T3 Continuing for n =

3, 4, 5,

we have

+

,T2

For

=

(5.4-5),

,

atn

(5.4-13)

n

=

6,

1

+ 41

I

+ Al

r

+ Ar

3

T4 —

+

,T3

i

J

5

,T4 ~~

2

—

,TS

2

+

,T4

,T6

2"

—

—

2

100

+ 2

100

+

100 ~~

j

using Eq. (5.4-15), r+Al

T6 =

,T5

For 2 Ai using Eq. (5.4-13) for n using Eqs. (5.4-14) and (5.4-15), l+ 2 Al^l

= Ta = l

/

=

= 1

100

and continuing

for n

=

2 to

6,

0

+ Ar^l

+ 2 Al *2

+t-..r, + Al^3 !

o+ioo 5Q

2 l

+ Ai^

1+ 2 Al^3

f

354

+ 2

Al^5

+

Al^i

2

+

Al^

Ai^4

+ Al^4

+

t

=

+ Al^5

2 1

1

1

87.5

2 l

1+ 2

+

f

A:?4

+

=

100

=

,„„ 100

2

l

+ Al^6

+

100

2

= l

+ Al^s

100

CTzo;?.

5

Principles of Unsteady-State

Heat Transfer

For

3 Af,

1+3

= 1 MT

0

1

0 J

+

87.5

+ 3 41^2

=

43.75

2

50+

100

1+ 3 41^3

+

87.5

=

75

100 93.75

1+3 AtT*

+

100

100

=

1+ 3 4/^5

1+3 41^6

For 4

=

100

Af,

r

+ 4 4(^1

1

+ 4 41^2

1

+ 4 4/^3

1

+ 4 41^4

:

0 0

—

+

75

+

37.5

93.75

=

68.75

100 87.5

+

93.75 1+4 4r^5

100

=

96.88

2

100

1

+ 4 4<^6

/

+5

/

+ 5 41^2

+ 5 41^3

=

!

+ 5 41^4

=

1

l

+ 5

5 Af, 41

0

^1

0

+

37.5

68.75

+

=

34.38

87.5

68.75

+

+

96.88

100

4i^5

I+5A ,T6

=

62.50

87.5

For

=

+

43.75

75

For

100

=

=

82.81

93.75

96.88

6 Af (final time),

f+64l

0 0

+

62.5

31.25

(+6 41^2 34.38

Si

1+6 4r^3

+

82.81

58.59

V Sec. 5.4

3

Numerical Finite-Difference Methods for Unsteady-State Conduction

355

„ ea ,T«

=93.75

The temperature profiles for 3 At increments and the final time of 6 At increments are plotted in Fig. 5.4-2. This example shows how a hand calculation can be done. To increase the accuracy, more slab increments and more time increments are required. This, then, is ideally suited for computation using the digital computer.

EXAMPLE 5.4-2.

Unsteady-State Conduction Using the Digital Computer Repeat Example 5.4-1 using the digital computer. Use a Ax = 0.05 m. Write the computer program and compare the final temperatures with Example = 2. Although not needed 5.4-1. Use the explicit method of Schmidt for for many time increments using the digital computer, use the special procedure for the value of fTa for the first time increment. Hence, a direct comparison of the effect of the number of increments on the results can be

M

made

with

Example

5.4-1.

The number

Solution:

of slabs to use

Substituting into Eq. (5.4-3) with a

M=

and solving

2,

Hence, (6000/62.5) from n = to 21.

is

1.00 m/(0.05 m/slab) or

=

2.00 x 10"

-

(

5

m

2

/s,

Ax =

20

0.05

slabs.

m, and

for At,

M

=

At

=

=

2

=

(Ax)2

°- 05)2

(2.00 x 10"

a At

5

XAO

62.5 s

96 time increments to be used. The value of n goes

1

The equations to use to calculate the temperatures are again Eqs. However, the only differences are that in Eq. (5.4-14) n goes from 2 to 20, and in Eq. (5.4-15) n = 21, so that r+A ,T21 = ,T20 The computer program for these equations is easily written and is left up to the reader. The results are tabulated in Table 5.4-1 for comparison with Example 5.4-1, where only 5 slices were used. The table shows that the (5.4-12}-<5.4-15).

.

Table

5.4-1

.

Comparison of Results for Examples 5.4-1 and 5. 4-2 Results Using

Ax = 0.20

m

Results Using

Ax =

0.05

m

Distance from

Front Face

m

356

Temperature

Temperature n

°C

n

"C

0

1

0.0

1

0.0

0.20

2

31.25

5

31.65

0.40

3

58.59

9

58.47

0.60

4

78.13

13

77.55

0.80

5

89.84

17

88.41

1.00

6

93.75

21

91.87

Chap. 5

Principles of Unsteady-Slate

Heat Transfer

reasonably close to those for 20 both cases deviating by 2% or less from each other. results for 5 slices are

As a rule-of-thumb guide at

hand

for

with values from

slices

minimum

calculations, using a

of five slices

and

least 8 to 10 time increments should give sufficient accuracy for most purposes. Only

when the

very high accuracy

problem using

known with

desired or several cases are to be solved

is

accuracy to justify a computer solution.

sufficient

EXAMPLE 5.4-3.

Unsteady-State Conduction with Convective Boundary Condition

Use the same conditions h k

desirable to solve

is it

computer. In some cases the physical properties are not

a digital

= 25.0 W/m 2 K = lO.OW/m-K.

Example 5.4-1, but a convective coefficient of present at the surface. The thermal conductivity

as

now

is

and

can be used for convection at the h Ax/k = 25.0(0.20)/10.0 = 0.50. Then 2N + 2 = 2(0.50) + 2 = 3.0. However, by Eq. (5.4-9), the value of must = 2 be equal to or greater than 2N 4- 2. This means that a value of = 4.0. [Ancannot be used. We will select the preferred method where Equations

Solution:

From

surface.

(5.4-7)

Eq.

(5.4-8)

N—

(5.4-8),

M

M

M

other less accurate alternative is to use Eq. (5.4-11) for convection and then the value of value.] is not restricted by the Substituting into Eq. (5.4-3) and solving for At,

M

N

M=4

^ 2

(Ax)

(0.20)

=

2

=

Af

(2.00xlO-S)(AO

5 °° S

Hence, 6000/500 = 12 time increments must be used. For the first At time increment and for all time increments, the value of the environmental temperature Ta to use is Ta = 0°C since > 3. For = 4 and convection at the node or point n = 1 we use Eq. (5.4-7), where

M

M

N=

0.50.

,

For n =

2, 3, 4, 5,

,

For

n

=

+ i ,T,

+ *,Tn

=

i[2(0.5)Ta

=

0.25TD

we

+

+(4-2)„7; + ,Tn _

=

0.25,T„ +

1

+

0.50, 7„

+

„T6 =

+ a ,T,

2, 3, 4, 5

,

T6 =

+

2),T,

=

n

2,T2 ] (5.4-16)

1

]

0.25 r Tn _,

=

=

n

use Eq. (5.4-10) and

5 (5.4-17)

2, 3, 4,

/=

6.

i[(4-2),T6 + 2,r5 ] 0.50,T6

+

0.50,T5

n

increment of temperature at node 1,

+

0.25(0)

0.25(100)

+

t 4-

=

6

At,

0.50(100)

=

(5.4-18)

Ta =

0.

Using Eq.

75.0

using Eq. (5.4-17),

+ A ,T2

=

0.25,T3

=

0.25(100)

+ 0.50,T2 +

Also, in a similar calculation,

Sec. 5.4

we

+

1

At, for the first time

I

(5.4-18),

+

0.50,T2

KX+i

(5.4-16) to calculate the

=

0.5

use Eq. (5.4-2),

=

For n

x

(2

+

0.25,7,

6 (insulated boundary),

1

-

[4

=

,

For

+

+

0.25,7,

0.50(100)

+

0.25(100)

T3 T4 and 75 = ,

,

100.0.

=

100.0

For n =

6,

using Eq.

100.0.

Numerical Finite-Difference Methods for Unsteady-State Conduction

357

For

2 At,

T4

Also,

Using Eq.

0.

=

A ,T,

+2

,

Using Eq.

Ta =

0.25(0)

=

(5.4-1 7) for n

,

+

2i ,T2

,

+

2/^3

and

T5 =

(5.4-16),

+

0.25(75.0)

+

0.50(100)

=

68.75

2, 3, 4, 5,

==

0.25(100)

+

0.50(100)

4-

0.25(75.00)

=

0.25(100)

+

0.50(100)

+

0.25(100)

+

0.50(93.75)

=

=

93.75

100.0

100.0.

For n = 6, using Eq. (5.4- 8), T6 = 1 00.0. For 3 Af, T0 = 0. Using Eq. (5.4-16), 1

,

Using Eq.

Also,

=

+ 3^7,

0.25(0)

=

(5.4-17) for n

0.25(68.75)

0.25(100)

+

0.50(93.75)

A ,T3

=

0.25(100)

+

0.50(100)

+

0.25(93.75)

3A ,r4

=

0.25(100)+ 0.50(100)

+

0.25(100)

+3

a.

,

+

3

,

+

100.0 and

In a similar

T6 =

manner

=

64.07

2,3, 4,5,

T2 =

,

T5 =

+

+

=

0.25(68.75)

=

=

89.07

98.44

100.0

100.0.

the calculations can be continued for the remaining

time until a total of 12 Ar increments have been used.

5.4C. J.

Other Numerical Methods

Unsteady-state conduction

unsteady-state conduction out. In a cylinder

it

changes

used where the cylinder thick.

Assuming

rate of heat in

—

in

is

in

a

for

Unsteady-State Conduction

a cylinder.

flat slab,

radially.

To

In deriving the numerical equations for

the cross-sectional area

was constant through-

derive the equation for a cylinder, Fig. 5.4-3

divided into concentric hollow cylinders whose walls are

a cylinder

1

is

Ax m

m long and making a heat balance on the slab at point n, the

rate of heat out

=

rate of heat accumulation.

(5.4-19)

Figure

358

5.4-3.

Unsteady-stale conduction in a cylinder.

Chap. 5

Principles of Unsteady-State

Heat Transfer

Rearranging, the final equation

is

+

2n

^

1

+(M-2),T„ +

,T„ +1

2n-l -— -,T

where

M

2

=

(A,x) /(a At) as before.

n

2n

2/i

Also, at the center

where

n

=

_

(5.4-20)

1

0,

M-4

4

(5.4-21)

To

use Equations (5.4-20)

and

(5.4-21),

M Equations

for

2n

-

,

+

1

the temperature at the surface

is

(5.4-22) If

neglected,

is

nN

T„

4

convection at the outer surface of the cylinder have been derived (Dl).

the heat capacity of the outer 7 slab

where

>

^

(2/i

+

,„ 2/j

and T„_

-

-

at

!

, 1

l)/2

.^T...

+

(5.4-23)

/iN

a position in the solid

1

Ax below

the surface.

Equations for numerical methods for two-dimensional unsteady-state conduction

have been derived and are available

a

in

number of references (Dl,

K2).

In some practical problems > 2 by stability requirements may prove imposed on the value inconvenient. Also, to minimize the stability problems, implicit methods using different finite-differer.ee formulas have been developed. An important one of these formulas is the Crank-Nicolson method, which will be considered here. In deriving Eqs. (5.4-1) and (5.4-2), the rate at which heat entered the slab in Fig. 5.4-1 was taken to be the rate at time t.

2.

Unsteady-state conduction and implicit numerical method.

M

the restrictions

kA Rate of heat

It /

A

in at

t

Ax

(

T

X)

(5.4-24)

was then assumed that this rate could be used during the whole interval from / to At. However, this is an approximation since the rate changes during this At interval.

+

better value

would be the average value

averace rate of heat

in

of the rate at

+ i ,r„

+1

-(2M +

This means that time

l

2), +

now

A ,Tn a

+

calculated simultaneously.

1+Ar

all

To do

at

£

+

At, or

(5.4-25)

is

used.

The

final

equation

is

Tn _ = -,Tn+l +(2-2M)X-,T„-i

new value

as in Eq. (5.4-2) but that

and

kA — Ax

=

Also, for the heat leaving, a similar type of average

,

t

1

of, + 4 ,7^

the

this

(5.4-26)

cannot be calculated only from values at T at t + At at all points must be

new values of an equation

is

written similar to Eq. (5.4-26) for

each of the internal points. Each of these equations and the boundary equations are linear algebraic equations.

methods used, such and so on (Gl, Kl).

These then can be solved simultaneously by the standard

as the Gauss-Seidel iteration technique, matrix inversion technique,

An important advantage

Sec. 5.4

of Eq. (5.4-26)

is

that the stability

and convergence

Numerical Finite-Difference Methods for Unsteady-State Conduction

criteria

359

are satisfied for

A

positive values of M. This

all

disadvantage of the implicit method

means

that

the larger

is

M can have values

number

than

2.0.

needed

for

less

of calculations

each time step. Explicit methods are simpler to use but because of stability considerations, especially in

complex situations, implicit methods are often used.

CHILLING AND FREEZING OF FOOD AND BIOLOGICAL MATERIALS

5.5

Introduction

5.5A

Unlike

many

inorganic and organic materials which are relatively stable, food and other

and deteriorate more or less rapidly with time at room due to a number of factors. Tissues of foods such as fruits continue to undergo metabolic respiration and ripen vegetables, after harvesting, and and eventually spoil. Enzymes of the dead tissues of meats and fish remain active and induce oxidation and other deteriorating effects. Microorganisms attack all types of foods by decomposing the foods so that spoilage occurs; also, chemical reactions occur, biological materials decay

temperature. This spoilage

is

such as the oxidation of fats.

At low temperatures the growth rate of microorganisms temperature

is

below that which

is

optimum

for

will

be slowed

if

the

growth. Also, enzyme activity and

chemical reaction rates are reduced at low temperatures. The rates of most chemical and biological reactions in storage of chilled or frozen foods

and biological materials are

reduced by factors of 5 to^ for each 10 K (10°C) drop in temperature. Water plays an important part in these rates of deterioration, and substantial percentage in

most of these

for

most biological

rates to

materials.

To

it is

present to a

reach a low enough temperature

approximately cease, most of the water must be frozen.

Materials such as food do not freeze at 0°C (32°F) as pure water does but at a range of

temperatures below 0°C. However, because of some of the physical effects of

and other

effects,

ice crystals

such as concentrating of solutions, chilling of biological materials

is

often used instead of freezing for preservation.

Chilling of materials involves removing the sensible heat and heat of metabolism and reducing the temperature usually to a range of 4.4°C (40°F) to just above freezing. Essentially no latent heat of freezing is involved. The materials can be stored for a week or so up to a few months, depending on the product stored and the gaseous atmosphere. Each material has its optimum chill storage temperature. In the freezing of food and biological materials, the temperature is reduced so that most of the water is frozen to ice. Depending on the final storage temperatures of down to

— 30°C,

many

the materials can be stored for

frozen foods, they are

5.5B

Chilling of

Food and

first

up

to

1

year or so. Often

in

the production of

treated by blanching or scalding to destroy enzymes.

Biological Materials

In the chilling of food and biological materials, the temperature of the materials is reduced to the desired chill storage temperature, which can be about — 1.1°C (30°F) to 4.4°C (40°F). For example, after slaughter, beef has a temperature of 37.8°C (100°F) to

40°C (104 chilled

C

F),

and

it

is

often cooled to about 4.4°C (40° FY. Milk from

quickly to temperatures just above freezing.

packing are

at a

Some

cows must be

fish fillets at

the time of

temperature of 7.2°C (45°F) to 10°C (50°F) and are chilled to close to

0°C.

These rates of

chilling or cooling are

conduction discussed

360

governed by the laws of unsteady-state heat

in Sections 5.1 to 5.4.

Chap. 5

The heat

is

removed by convection

Principles of Unsteady-State

at the

Heat Transfer

and by unsteady-state conduction in is used to remove

surface of the material

the material.

outside the foodstuff or biological materials

The

and

this heat,

fluid

many

in

it is air. The air has previously been cooled by refrigeration to — 1.1°C to +4.4°C, depending on the material and other conditions. The convective heat-transfer coefficients, which usually include radiation effects, can also be predicted by the methods in

cases

Chapter 7 btu/h

In

•

and

4, 2 ft

°F),

•

some

depending primarily on

cases the fluid used for chilling

Then

with the material.

(1.5 to

a liquid flowing over the surface

and the

40

to

8.5

-

air velocity. is

values of h can vary from about 280 to 1700

other cases, a contact or plate cooler

W/m 2 K

from about

for air the coefficient varies

W/m 2 K

(50-300 btu/h

used where chilled plates are

is

the temperature of the surface of the material

is

2 ft

•

°F). Also, in

in direct

Contact freezers are used

to be equal to or close to that of the contact plates.

contact

usually assumed for freezing

biological materials.

Where

the food

packaged

is

in

boxes or where the material

add the resistance

of the

tightly

is

covered by a

must be considered. One method to do

film of plastic, this additional resistance

package covering to that of the convective

R T = RF + Rc where R P

is

and R T the

the resistance of covering,

Then,

total resistance.

for

Rc

(5.5-1)

the resistance of the outside convective film,

each resistance,

Kc =

—

(5-5-2)

RP =

Ax —

(5.5-3)

hc

A

Kr = rr. hA where h c

is

covering, k

The

(5.5-4)

A

the convective gas or liquid coefficient, is

is

the area,

the thermal conductivity of the covering, and h

overall coefficient h

is

this is to

film:

Ax is

is

the thickness of the

the overall coefficient.

the one to use in the unsteady-state charts. This assumes a

negligible heat capacity of the covering,

which

usually the case. Also,

is

the covering closely touches the food material so there

is

no

it

assumes that

resistance between the

covering and the food.

The major

sources of error

in

using the unsteady-state charts are the inadequate

data on the density, heat capacity, and thermal conductivity of the foods and the prediction of the convective coefficient. stances,

Food

and the physical properties are often

water occurs on

chilling, latent heat losses

EXAMPLE 5J-1.

can

Chilling Dressed

materials are irregular anisotropic sub-

difficult to evaluate. Also,

affect the

accuracy of the

if

evaporation of

results.

Beef

Hodgson (H2) gives physical properties of beef carcasses during chilling of 3 p = 1073 kg/m c p = 3.48 kJ/kg K, and k = 0.498 W/m K. A large slab of •

,

m

beef 0.203 thick and initially at a uniform temperature of 37.8°C is to be cooled so that the center temperature is 10°C. Chilled air at 1.7°C (assumed 2 constant) and having an h = 39.7 W/m is used. Calculate the time needed. •

Solution:

The thermal diffusivity

Sec. 5

J

Chilling

=

^

is

0 498

k

"

a

"

K

(1073X3.48 x 1000)

= 1334 X

10

"m

and Freezing of Food and Biological Materials

'

/s

361

Then

for the half-thickness x, of the slab,

x,=^ = 0.1015m For the center of the

slab,

x__0_ 0

n

x

x

i

i

Also,

0498 0.123 (39.7)(0.1015)

/ix,

T,

= l.TC +

273.2

=

K

274.9

T=

10

+

T0 =

273.2

7-,-T

274. 9

r,-T

274.9

0

Using Fig. 5.3-6 for the center of a large at

pop Y -0.90-^-

=

-

283.2

=

6.95 x 10

5.5C

Freezing of

Food and

1.

first

latent heat of freezing

total heat

s

removed on

273.2

=

K

311.0

K

311.0

flat plate,

0-334 x

10^X0

(Q1015)2

(19.3 h).

Biological Materials

In the freezing of food

Introduction.

sensible heat in chilling

The

4

+

283.2

A"

Solving,!

37.8

and other biological materials,

the removal of

occurs and then the removal of the latent heat of freezing.

water of 335 kJ/kg (144 btu/lb m ) is a substantial portion of the Other slight effects, such as the heats of solution of salts,

freezing.

and so on, may be present but are quite small. Actually, when materials such as meats are frozen to -29°C, only about 90% of the water is frozen to ice, with the rest thought to be

bound water

(Dl).

Riedel (Rl) gives enthalpy-temperature-composition charts for the freezing of many different foods.

These charts show that freezing does not occur

at a given

extends over a range of several degrees. As a consequence, there

is

temperature but

no one

freezing point

with a single latent heat of freezing.

Since the latent heat of freezing

is

present in the unsteady-state process of freezing,

the standard unsteady-state conduction equations and charts given in this chapter

cannot be used

for prediction of freezing times.

and biological materials

freezing of food

physical properties withjernperature, the

and other 2.

factors.

An approximate

is

temperature but

is

is

analytical solution of the rate of

difficult

is

often used.

Plank (P2) has derived an approximate

often sufficient for engineering purposes.

derivation are as follows. Initially,

unfrozen.

The thermal

because of the variation of

amount of freezing varying with temperature,

Approximate solution of Plank for freezing. in the

full

solution by Plank

solution for the time of freezing which

assumptions

A

very

all

the food

conductivity, of the frozen part

in the frozen layer

occurs slowly enough so that

it is

The

at the freezing is

constant. All

The heat

transfer by under pseudo-steady-

the material freezes at the freezing point, with a constant latent heat.

conduction

is

state conditions.

362

Chap. 5

Principles

of Unsteady-State Heat Transfer

frozen

Figure

Temperature

5.5-1.

m

In Fig. 5.5-1 a slab of thickness a

given time

s,

t

temperature of the environment constant at

The

m

a thickness of x

.

heat leaving at time

q

t is

W. Since we

is

is

is

is

present.

are at pseudo-steady state, at time

the

t,

is

(53-5)

x

is

kA — (T x

=

q

k

The

the freezing temperature

the surface area. Also, the heat being conducted through the frozen layer of x

thickness at steady state

where

x

= hA(Ts - T )

q

A

T K and

the center at Ty

heat leaving by convection on the outside

where

cooled from both sides by convection. At a

of frozen layer has formed on both sides.

constant'at

is

Tf An unfrozen layer in

is

profile during freezing.

f

- Ts)

(53-6)

the thermal conductivity of the frozen material. In a given time dt

Then multiplying A times dx times p latent heat X in J/kg and dividing by dt,

thick of material freezes.

Multiplying

this

by the

A

gives the

s,

a layer

kg mass

dx

frozen.

dx

dx pX

(5.5-7)

where p

is

the density of the unfrozen material.

Next, to eliminate

Ts

from Eqs.

(5.5-5)

and

(5.5-6),

Eq.

(5.5-5) is solved for

Ts

and

substituted into Eq. (5.5-6), giving

..

Equating Eq.

..

(53-8)

(5.5-8) to (5.5-7),

(Tf

-T )A

x/k

Rearranging and integrating between

(Tf

Sec. 5.5

- TAA x/k + l/h

(7>

_

q

-

Chilling

-

dx

X

+ /

l/h

~

= 0 and

P x

=

(53-9) dt 0,

to

t

=

t

and x

T,)

and Freezing of Food and Biological Materials

= a/2, (53-10)

363

Integrating and solving for

f,

-

t

To

(5.5-11)

-

7}

\2h

T,

8/c

generalize the equation for other shapes,

(5.5-12)

Tf -T,\h where a

is

the thickness of an infinite slab (as in Fig. 5.5-1), diameter of a sphere, diameter

of a long cylinder, or the smallest dimension of a rectangular block or brick. Also,

P=

\ for infinite slab, \ for sphere, \ for infinite cylinder

R=

| for infinite slab, jz for sphere,

for infinite cylinder

For a rectangular brick having dimensions a by p a by fi 2 a, where a is the shortest side, Ede (B 1) has prepared a chart to determine the values of P and R to he used to calculate t in Eq. (5.5-12). Equation (5.5-11) can also be used for calculation of thawing times by replacing the k of the frozen material by the k of the thawed material. x

EXAMPLE 55-2.

Freezing of Meat thick are to be frozen in an air-blast freezer at

m

Slabs of meat 0.0635

K - 28.9°C). The meat

is initially at the freezing temperature of 270.4 meat contains 75% moisture. The heat-transfer coefficient 2 3 K. The physical properties are p = 1057 kg/m for the unfrozen meat and k = 1.038 W/m-K for the frozen meat. Calculate the

244.3

(

K — 2.8°C). The is h = 17.0 W/m (

•

freezing time.

Solution:

Since the latent heat of fusion of water to meat with 75% water,

ice is

335 kJ/kg(144

btu/lb m ), for

k

The other given 3 p = 1057 kg/m

=

0.75(335)

=

h

17.0

251.2 kJ/kg

0.0635 m, 7} = 270.4 K, 1\ = 244.3 K, K, k = 1.038 W/m K. Substituting into

=

variables are a ,

=

W/m 2

•

Eq.(5.5-ll), " ' '

Xp _ ~ 7} -

=

3.

fa_ T,

\2h

2.395 x 10

+

a*\

_

5

(2.512 x 10 )1057 \

270.4

8JcJ~~

4 s

-

244.3

f 0.0635

+

(O.0635)

J [2(17.0)

2

8(1.038)

(6.65 h)

Neumann (CI, C2) has derived a compliHe assumes the following conditions. The surface

Other methods to calculate freezing times.

cated equation for freezing in a slab.

temperature of freezing coefficient

the

is is

the

same

as the environment,

method

constant. This

cannot be used

method does include

no surface resistance. The temperature

from

this

limitation

that a convection

assumes no surface resistance. However, of cooling from an original temperature, which may be

at the surface, since

the effect

i.e.,

suffers it

above the freezing point. Plank's equation does not

make

provision for an original temperature, which

An approximate method from temperature T0 down to the

be above the freezing point.

may

to calculate the additional time

freezing point 7} is as follows. Calculate by means of the unsteady-state charts the time for the average temperature in

necessary to cool

no freezing occurs using the physical properties of no surface resistance, Fig. 5.3-13 can be used directly for

the material to reach 7} assuming that the unfrozen material. If there

364

is

Chap. 5

Principles of Unsteady-State

Heat Transfer

is present, the temperature at several points in the material will have from the unsteady-state charts and the average temperature calculated from these point temperatures. This may be partial trial and error since the time is unknown, which must be assumed. If the average temperature calculated is not at the

this. If

a resistance

to be obtained

new time must be assumed. This

freezing point, a

is

an approximate method since some

material will actually freeze.

DIFFERENTIAL EQUATION OF

5.6

ENERGY CHANGE Introduction

5.6A

In Sections 3.6 and 3.7

equation

we

of momentum

derived a differential equation of continuity and a differential

momentum

parts of Chapter 2 did not

However,

is

it

balances

us what goes on inside a control volume. In the over-

tell

new

balances performed, a

all

These equations were derived because made on a finite volume in the earlier

transfer for a pure fluid.

overall mass, energy, and

made

balance was

new system

for each

studied.

often easier to start with the differential equations of continuity and

momentum transfer in general form and then to simplify the equations by discarding unneeded terms for each specific problem. In Chapter 4 on steady-state heat transfer and Chapter 5 on unsteady-state heat transfer new overall energy balances were made on a finite control volume for each new situation. To advance further in our study of heat or energy transfer in flow and nonflow systems we must use a differential volume to investigate in greater detail what goes on inside this volume. The balance will be made on a single phase and the boundary conditions at the phase boundary will be used for integration. In the next section

we

derive a general differential equation of energy change: the

conservation-of-energy equation. Then this equation

is

modified for certain special cases

that occur frequently. Finally, applications of the uses of these equations are given. Cases

both

for

and

steady-state

unsteady-state

conservation-of-energy equation, which

energy

are

transfer

perfectly general

is

studied

and holds

using

this

for steady- or

unsteady-state conditions.

5.6B

Derivation of Differential Equation of Energy Change

As

the derivation of the differential equation of

in

momentum

balance on an element of volume of size Ax, Ay, and Az which write the law of conservation of energy, which

for a control

volume element at any volume given in Section 2.7.

(,

—

for the fluid in this

rate of

1

energy in/

/ rate

\

/

\

\

/ rate of

\

—

\ en ergy out/

is

time.

really the first

The following

\

transfer,

write a

We

then

law of thermodynamics

is

same

the

as Eq. (2.7-7)

»

external

\ system

work done by

1

I

on surroundings/ /

-

rate o(

^1 accumulation \ energy

is

we

stationary.

\

of

/ (

is

\

of 1(5.6-1) /

As in momentum transfer, the transfer of energy into and out of the volume element by convection and molecular transport or conduction. There are two kinds of energy

being transferred. The

Sec. 5.6

first is

internal energy

Differential Equation

U

in

of Energy Change

J/kg(btu/lb m ) or any other set of units.

365

This

is

random translational and internal motions of The second is kinetic energy pv 2 /2, which is

the energy associated with

molecules plus molecular interactions.

energy associated with the bulk fluid motion, where v

Hence, the energy

total

energy per unit volume

volume element

in the

in

m

3 (ft

3

is

)

Ax Ay Az The at

x

total energy

+ Ax

coming

in

the local fluid velocity, m/s

is

(pU + pv 2 /2). The them

by convection

+

pv —

in the

the

(ft/s).

rate of accumulation of

is

I —d IpU

the

(5.6-2)

x direction at x minus that leaving

is

pv

Ay Az

pV

Ay Az

Similar equations can be written for the y and

z

I

(5.6-3)

-[

directions using velocities v y and

vz

,

respectively.

The

net rate of energy into the element by conduction in the x direction

-(q x)x + Ax]

AyAz[_(q x) x

Similar equations can be written for the y and

components of convenient

The

(5-6-4)

directions where q x q y and q z are the 2 2 is in (btu/s-ft ) or any other

z

set of units.

work done by the system on its surroundings For the net work done against

net

- p Ax Ay is

gravitational force.

The

net

is

N/m 2 (lb /ft 2 f

)

is

the

sum

of the following

the gravitational force,

Az{v x g x )

(5.6-5)

work done against

Ay Azl(pv x )x + Ax ~{pv x where p

,

,

W/m

the heat flux vector q, which

three parts for the x direction.

where g x

is

or any other convenient

set

the static pressure p

is

(5.6-6)

) x]

of units. For the net

work against

the

viscous forces,

(Ay Az)[(x xx

vx

+

v

t

4- x X2

-

v z) x+ *x

»*

In Section 3.7 these viscous forces are discussed in

Writing equations similar to (5.6-3)-{5.6-7) equations and Eq. (5.6-2) into

Az approach

zero,

(5.6-1); dividing

vJ

pU +

dx dq x

+

|

dx

dy

—

(r xx v x

Fz

pv — dq z

dq y ]

366

more

v,

+

(5.6-7)

r xz u T)J

detail.

in all three directions; substituting these

by Ax, Ay, and Az; and

letting Ax, Ay,

and

we obtain

pv

For further

+

(r zx v x

+

dy

x xy v y

y

+

Piv x g x

+

x xz v z )

dz

+

pV —

+ — v [pU + r

+— dz

+

+

v

y

—

gy

+

vz

{x yx v x

vA P U

+

pv

gz)

+

x

yy

v

y

+

x yz v z )

x zy

(5.6-8)

details of this derivation, see (B2).

Chap. 5

Principles of Unsteady-State

Heat Transfer

Equation

However,

(5.6-8) is the final

equation of energy change relative to a stationary point.

We first

not in a convenient form.

it is

combine Eq.

(5.6-8) with the

equation of

continuity, Eq. (3.6-23), with the equation of motion, Eq. (3.7-13), and express the internal

energy

terms of

in

DT =

pc

This equation

utilizes

fluid

kV

d7

"

and heat capacity. Then writing the

with constant thermal conductivity

fdp\ (V-v) + \dTj

,

k,

we obtain

^

(5 " 6' 9)

Fourier's second law in three directions, where

W 2T = The

T

temperature

fluid

Newtonian

resultant equation for a

viscous dissipation term p
velocity gradients exist.

It will

change

the equation of energy

is

—

'd *(

2

T

dx

2

T

2

d

+

2

d

T

+

TT TT dz dy 2

(5-6-10)

)

generally negligible except where extremely large

be omitted in the discussions to follow. Equation (5.6-9)

Newtonian

for a

fluid

with constant k

in

terms of the

is

fluid

temperature T.

5.6C

Special Cases of the Equation of Energy

The following

forms of Eq.

special

conductivity are

(5.6-9) for

commonly encountered.

Change

a Newtonian fluid with constant thermal

First,

Eq. (5.6-9)

be written in rectan-

will

gular coordinates without the \x§ term.

dT —

—

H vx

at

dT —

h v,

d

2

dz

T

d

2

T

d

2

T\

( dp\

k

dv r

\dT/ p

+

dv„ -rI

dy

\ dx

The equations below can

Fluid at constant pressure.

fluids as well as for

dT —

oy

ox

=

/.

dT —

V v

dv.

+~r)

(5.6-11)

be used for constant-density

constant pressure.

DT (5.6-12)

In rectangular coordinates,

dT

dT

+ '

dt

v '*x

—

1-

dx

v

d

2

T

'

y

dx

Jy~

2

+

d

2

T

dy

d

2

T

+

2

(5.6-13) dz'

In cylindrical coordinates,

dT

dT

v0

dT

dT\

dt

dr

r

dO

oz J

(>c ,

d

2

T

1

dT

d

1

2

T

d

2

T\

f

,

..

,

,

In spherical coordinates,

{dT pCp

\dt

+

dT

vg

+

v "

~d7

=

7

dT

ae

dj>

k J__3 r

Sec. 5.6

dT ^ _i +

2

dr

77in e

dT dr

Differential Equation

1

+ 1

r

2

sin 0 sin d

dO

of Energy Change

— d6

d

i

+ r

2

sin

2

0

2

T

J^

2

(5.6-15)

367

For definitions of cylindrical and spherical coordinates, see Section zero, DT/Dt becomes dT/dt.

3.6.

the velocity v

If

is

2.

Fluid at constant density

DT pc B y

Dt

"

Note 3.

that this

is

is

=

constant and v

8T

0.

~—=kN T

pc p

2

(5.6-17)

~dt

often referred to as Fourier's second law of heat conduction. This also holds for a

This

is

fluid

with zero velocity at constant pressure.

4.

(5.6-16)

identical to Eq. (5.6-12) for constant pressure.

Here we consider p

Solid.

= k\ 2 T

Heat generation.

there

If

is

heat generation in the fluid by electrical or chemical

means, then q can be added to the right side of Eq.

(5.6-17).

57 (5.6-18)

where q

is

the rate of heat generation in

dissipation

is

also a heat source, but

W/m

3

3 ft )

(btu/h

or other suitable

units.

Viscous

inclusion greatly complicates problem solving

its

because the equations for energy and motion are then coupled.

5.

Other coordinate systems.

Fourier's second law of unsteady-state heat conduction

can be written as follows.

For rectangular coordinates,

dT_k is

k/pc p and

is

2

T

sc

\dx

pc p

dt

where a

fd

V2T =

2

+

d

2

T

d

2

t (5.6-19)

dy

inm 2 /s (ft 2 /h).

thermal diffusivity

For cylindrical coordinates,

8T

=

d

2

T

a

1

dT

~r

Ik

+

dt

+ j_ 1 r

8

T

W

+

°

T (5.6-20)

Ih 2

For spherical coordinates,

dT

lis ! 9

dt

5.6D

i

r

2

sin 0

30

d

1

s>n0-| + r

2

s\n

2

2

T 2

(5.6-21)

Olub

Uses of Equation of Energy Change

In Section 3.8 fluid flow

we used

problems.

We

the differential equations of continuity and of motion to set up did this by discarding the terms that are zero or near zero and

using the remaining equations to solve for the velocity and pressure distributions. This

was done instead of making new mass and momentum balances for each new situation. In a similar manner, to solve problems of heat transfer, the differential equations of

368

Chap. 5

Principles

of Unsteady-Stale Heat Transfer

unneeded terms being discarded. methods used.

continuity, motion, and energy will be used with the

Several examples will be given to illustrate the general

EXAMPLE 5.6-1.

Temperature Profile with Heat Generation which heat generation is occurring uniformly asg

A

solid cylinder in

is

insulated on the ends.

only at

The temperature

TW K. The

held constant at

if

be used

(5.6-20) will

dT =

-T-

tion

2

state

=0

(d 2 T

k

I

dr pc p \ TT

dT/dt

and

d

2

=

T/d0

0. 2

dT

r

Also

to the right side, giving d

1

+ -T+ r dr

2

dt

d T/dz

is

=R

m. Heat flows the temperature profile

is

for cylindrical coordinates.

added

the term q/pc p for generation will be

For steady

3

the solid has a constant thermal conductivity.

Equation

Solution:

2

of the surface of the cylinder

radius of the cylinder

the radial direction. Derive the equation for

in

steady state

W/m

2

T

2 2 -2-5HT r 80

2

T\ TT+— d

+

8z

q

2

J

(5.6-22)

pc p

Also, for conduction only in the radial direction

=0. This

gives the following differential equa-

:

d

2

T

dr

2

+

}_dT_ _q

~

dr

r

(5.6-23)

k

This can be rewritten as

d T

Note

that Eq. (5.6-24) can

2

T .dT _ +

dr 2


dr

k

be rewritten as follows d_

dr

(

dT\

_ ~ \ dr)~

qr^

(5.6-25)

k

Integrating Eq. (5.6-25) once,

(5.6-26)

where

K

,

is

a constant. Integrating again,

7= where

K

2 is

+ K

l

In r

The boundary conditions are when r = when r = R,T = Tw The final equation is

a constant.

(by symmetry); and

the

is

same

EXAMPLE

as Eq. (4.3-29),

5.6-2.

(5.6-27)

0,

dT/dr

=

0

.

T= This

+ K2

+ Tw

(5.6-28)

which was obtained by another method.

Laminar Flow and Heat Transfer Using Equation of Energy Change

differentia] equation of energy change, derive the partial differenequation and boundary conditions needed for the case of laminar flow of a constant density fluid in a horizontal tube which is being heated. The fluid

Using the tial

flowing at a constant velocity u. At the wall of the pipe where the radius r 0 the heat flux is constant at q 0 The process is at steady state and it is assumed at z = 0 at the inlet that the velocity profile is established. Constant is

r

.

=

-

.

,

physical properties will be assumed.

Sec. 5.6

Differential Equation

of Energy Change

369

From Example 3.8-3, the equation of continuity gives dvjdz = Solution 0. of the equation of motion for steady state using cylindrical coordinates gives the parabolic velocity profile. Solution:

2

(5.6-29)

Since the fluid has a constant density, Eq. (5.6-14) in cylindrical coordinates will be used for the equation of energy change. For this case v = 0 r

and v„ = 0. Since this will be symmetrical dT/dO and d 2 T/d0 2 For steady state, dT/dt = 0. Hence, Eq. (5.6-14) reduces to

dT

k

Yd 2 T

dT

1

B

2

will

be zero.

T\

2

T/dz 2 term) is small compared to dT/dz and can be dropped. Finally, substituting Eq.

Usually conduction the convective term

vz

(5.6-29) into (5.6-30),

we obtain

in the z direction (d

The boundary conditions are

For

5.7

details

At

z

=

0,

At

r

=

0,

At

r

=

r0

T = r0 T — finite = —

q0

,

(all r)

—-

k

(constant)

on the actual solution of this equation,

see Siegel et

BOUNDARY-LAYER FLOW AND TURBULENCE HEAT TRANSFER Laminar Flow and Boundary-Layer Theory

5.7A

in

al. (S2).

IN

Heat

Transfer

10C an exact solution was obtained for the velocity profile for isothermal flat plate. The solution of Blasius can be extended to include the convective heat-transfer problem for the same geometry and laminar flow. In Fig. 5.7-1 the thermal boundary layer is shown. The temperature of the fluid approaching the plate is Tx and that of the plate is T at the surface. s In section 3.

laminar flow past a

We start by writing the differential energy balance, Eq. (5.6-13). dT

3T

dT _

dT

flow

is

in the

x and y directions,

neglected in the x and z directions, so d

y direction.

The

result

vz 2

=

0.

T/dx

2

At steady

=

d

2

pcp \ dx

dz If the

fd 2 T

k

2

d T/dz

2

T

dy state,

2

=

0.

2

d

2

dz

t 2

(5.7-1)

= 0. Conduction is Conduction occurs in the

dT/dt

is

dT

dT

k

d

2

T (5.7-2)

pc p dy

370

Chap.

2

Principles of Unsteady-State

Heat Transfer

The

simplified

vation

is

momentum

very similar and

balance equation used

in the velocity

boundary-layer deri-

is

— + — - - —j-

The continuity equation used

2

dvx

dv x

vx

v

dx

previously

y

fi

d vx

inin (3.10-5) ,,

2 p By

8y

is

(3.10-3)

dx

dy

Equations (3.10-5) and (3.10-3) were used by Blasius for solving the case for laminar boundary-layer flow. The boundary conditions used were

^=^=0

at

y

=

0

=

1

at

y

=

co

^=

1

at

x

=

0

(5.7-3)

The

similarity between Eqs. (3.10-5) and (5.7-2) is obvious. Hence, the Blasius solution can be applied if klpc p = /x/p. This means the Prandtl number c p fi/k = 1 Also, the boundary conditions must be the same. This is done by replacing the temperature T in .

Eq.

(5.7-2)

conditions

by the dimensionless variable (T

- TS )/{T„ - Ts ). The boundary

become

T

= Ji "CO

T«

"CO

-T -T s

=

at

y

=

0

1

at

y

=

oo

1

at

x =0

0

s

-Ts = -T T -T = T -Ts T

T 1

CO

*

CO

s

We

see that the equations

(5.7-4)

s

and boundary conditions are

identical for the temper-

ature profile and the velocity profile. Hence, for any point x, y in the flow system, the

dimensionless velocity variables vjv w and (T — T^/iT^ — Ts ) are equal. profile solution is the same as the temperature-profile solution.

The

velocity-

edge of thermal

boundary layer

^^^^^^^^^^^^^^ \

Ts

x

x = 0 Figure

Sec. 5.7

5.7-1.

Laminar flow offluid past a flat plate and thermal boundary

Boundary-Layer Flow and Turbulence

in

Heat Transfer

layer.

371

This means that the transfer of momentum and heat are directly analogous and the

boundary-layer thickness

5 for the velocity profile

the thermal boundary-layer thickness

Prandtl numbers are close to

By combining

<5

T are equal.

1.

Eqs. (3.10-7) and (3.10-8), the velocity gradient at the surface 9

=

-f). where

N Rcx =

xv m

0.332

= TV

(5.7-5)

and

SyJ y =o The convective equation can be in J/s

or

(5.7-6)

- Tx

/0.332

{T„

is

(5.7-5)

e

(5.7-6),

dT\

where q y

^'%

is

Also,

v„

Combining Eqs.

(hydrodynamic boundary layer) and is important for gases, where the

This

- Ts)[

,,,

N£

,

(5.7-7)

x

related to the Fourier equation by the following,

W (btu/h). ^=h

{Ts

x

-Tj=

-k[<—)

(5.7-8)

y=0

Combining Eqs.

(5.7-7)

and

(5.7-8)

^=N N Nu

= 0.332N^ x

NUiI

(5.7-9)

number and h x is the local heat-transfer on the plate. Pohlhausen (Kl) was able to show that the relation between the hydrodynamic and thermal boundary layers for fluids with Prandtl number >0.6 gives approximately

where

x

is

the dimensionless Nusselt

coefficient at point x

f = NH As

a result, the

3

(5-7-10)

equation for the local heat-transfer coefficient

is

k h

x

= 0.332 - N\g x Wpr3

(5.7-11)

x

Also,

^ The equation plate of width b

for the

mean

= N Nu

.

x

=

% NU

0.332/V''e

3

heat-transfer coefficient h from x

(5.7-12)

=

0 to x

= L

is

for

a

and area bL,

dx A.

Jo

l

-

372

0.332Jfc

f—Y'Vr'

Chap. 5

L 3

dx (5.7-13)

Principles of Unsteady-State Heat Transfer

Integrating,

h= 0.644 ^Nk' 2 L NH 3

(5.7-14)

e,

~=W

Nu

= 0.644JV# L Np/ 3

(5.7-15)

As pointed out previously, this laminar boundary layer on smooth plates holds up to a Reynolds number of about 5 x 10 s In using the results above, the fluid properties are .

usually evaluated at the film temperature 7}

= (Ts + Tm )/2.

5.7B Approximate Integral Analysis of the Thermal Boundary

Layer

hydrodynamic boundary layer, the Blasius solution is more complex systems cannot be solved by this method. The approximate integral analysis was used by von Karman to calculate the hydrodynamic boundary layer and was covered in Section 3.10. This approach can be used to analyze the thermal boundary layer. This method will be outlined briefly. First, a control volume, as previously given in Fig. 3.10-5, is used to derive the final energy integral expression.

As discussed

in the analysis of the

accurate but limited in

its

scope. Other

This equation

momentum

\(TM - T) dy

dx

P C p\ d yJy=0

(5.7-16)

o

analogous to Eq. (3.10-48) combined with Eq. (3.10-51) for the

is

analysis, giving

" Equation

(5.7-16)

can be solved

known. The assumed velocity

if

profile

v x)

dy

(5.7-17)

both a velocity profile and temperature used is Eq. (3.10-50).

profile are

3

(3.10-50)

v„

The same form

of temperature profile

T-

2 is

T,

2

<5

assumed. 3

v

1

/

V

V

j^i-tAi)

(5 7 - |8) '

Substituting Eqs. (3.10-50) and (5.7-18) into the integral expression and solving,

N Na x = 036N»

2

x

.

Nl>

3

(5.7-19)

is only about 8% greater than the exact result in Eq. (5.7-1 1), which indicates that approximate integral method can be used with confidence in cases where exact solutions cannot be obtained.

This this

In a similar fashion, the integral

hydrodynamic boundary layer in turbulent flow.

These give Section

Sec. 5.7

in

momentum

Again, the Blasius y-power law

results

analysis

method used

for the turbulent

Section 3.10 can be used for the thermal boundary layer is

used for the temperature distribution.

that are quite similar to the experimental equations as given in

4.6.

Boundary-Layer Flow and Turbulence

in

Heal Transfer

373

5.7C Prandtl Mixing Length aod Eddy Thermal Diffusivity

1.

Eddy momentum

f

summed

butions are

In Section 3.10F the total shear stress

diffusivity in turbulent flow.

for turbulent flow

was written as follows when

the molecular

and turbulent

contri-

together:

(5.7-20)

dy

The molecular momentum

diffusivity p./p

inm 2 /s

is

a function only of the fluid molecular

However, the turbulent momentum eddy diffusivity £, depends on the fluid motion. In Eq. (3.10-29) we related e, to the Prandtl mixing length L as follows: properties.

=

e,

13

(3.10-29)

dy

We

Prandtl mixing length and eddy thermal diffusivity.

2.

the

eddy thermal

fluid are

transported a distance

differs in

mean

velocity

can derive

diffusivity a, for turbulent heat transfer as follows.

velocity

in

a similar

manner

Eddies or clumps of

y direction. At this point L the clump of fluid fluid by the velocity^, which is the fluctuating Section 3.10F. Energy is also transported the distance L

L

in the

from the adjacent

component discussed

in

y direction together with the mass being transported. The instantaneous temperature of the fluid is T = T' + f, where T is the mean value and is similar to the fluctuating the deviation from the mean value. This fluctuating velocity v'x The mixing length is small enough so that the temperature difference can be

with a velocity

v' t

in the

T

T

.

written as

dT

T = L Ty The

(5.7-21)

isq^A and

rate of energy transported per unit area

equal to the mass flux

is

in

the y direction times the heat capacity times the temperature difference.

(5.7-22)

A In Section 3.10F

we assumed

dy

v'x s= v'

v'x

=

and

v' y

that

=L

dv\ (5.7-23)

dy

Substituting Eq. (5.7-23) into (5.7-22),

-pc p L2

A The term is

in

dv.

(5.7-24)

dy

L 2 \dvx ldy\ by Eq. (3.10-29) is the momentum eddy diffusivity s When this term r

the turbulent heat-transfer equation (5.7-24),

Then Eq.

dT

(5.7-24)

it is

called

a,

,

.

eddy thermal

diffusivity.

becomes

dT (5.7-25)

A Combining

374

this

dy

with the Fourier equation written in terms of the molecular thermal

Chap. 5

Principles

of Unsteady-State Heat Transfer

difTusivity a,

^=

-/7c p (a

+ a,)—

i.

among momentum,

Similarities

to Eq. (5.7-20) for total

eddy

momentum

difTusivity

e,

and mass transport. Equation (5.7-26) is similar The eddy thermal difTusivity a, and the

heat,

momentum

(5.7-26)

try

/I

transport.

have been assumed equal

Experimental

in the derivations.

An eddy mass

data show that

this equality is

transfer has also

been defined in a similar manner using the Prandtl mixing length theory

and

is

assumed equal

to a,

and

e,

only approximate.

difTusivity for

mass

.

PROBLEMS 5.2-1.

Temperature Response in Cooling a Wire. A small copper wire with a diameter of and initially at 366.5 K is suddenly immersed into a liquid held constant at 311 K. The convection coefficient- h = 85.2 W/m 2 -K. The physical properties can be assumed constant and are k = 374 W/m K, c = 0.389 p 3 kJ/kg K, and p = 8890 kg/m (a) Determine the time in seconds for the average temperature of the wire to drop to 338.8 K (one half the initial temperature difference). 2 K. (b) Do the same but for h= 1 1.36 W/m (c) For part (b), calculate the total amount of heat removed for a wire 1.0 long.

mm

0.792

•

•

.

•

m

Ans. 5.2-2.

(a)

t

=

5.66

s

mm

Quenching Lead Shot in a Bath. Lead shot having an average diameter of 5.1 is at an initial temperature of 204.4°C. To quench the shot it is added to a quenching oil bath held at 32.2°C and falls to the bottom. The time of fall is 15 s. Assuming an average convection coefficient of h — 199 W/m 2 K, what will be 3 the temperature of the shot after the fall? For lead, p = 1 1 370 kg/m and c = 0.138 kJ/kg-K. p •

5.2- 3.

m

3 Unsteady-State Heating of a Stirred Tank. A vessel is filled with 0.0283 of water initially at 288.8 K. The vessel, which is well stirred, is suddenly immersed into a steam bath held at 377.6 K. The overall heat-transfer coefficient U be2 tween the steam and water is 1 1 36 W/m 2 K and the area is 0.372 Neglecting the heat capacity of the walls and agitator, calculate the time in hours to heat the water to 338.7 K. [Hint : Since the water is well stirred, its temperature is uniform. Show that Eq. (5.2-3) holds by starting with Eq. (5.2-1).]

m

-

5.3- 1.

Temperature

in

a Refractory Lining.

.

A combustion chamber has a To predict the thermal

refractory lining to protect the outer shell.

2-in. -thick

stresses at

is needed 1 min after startup. temperature T0 = lOOT, the hot gas 2 temperature T, = 3000° F, h = 40 btu/h -ft °F, k = 0.6 btu/h- ft -°F, and 2 a = 0.020 ft /h. Calculate the temperature at a 0.2 in. depth and at a 0.6 in. depth. Use Fig. 5.3-3 and justify its use by seeing if the lining acts as a semiinfinite solid during this 1-min period. Ans. For x = 0.2 in., (T - T0 )/(T, - T0 ) = 0.28 and T = 912°F (489°C); o for x = 0.6 in., (T - T0 )/(T, - T0 ) = 0.02 and T = 158 F(70°C)

startup, the temperature 0.2

The following data are

in.

below the surface

available.

The

initial

•

53-2. Freezing Temperature in the Soil. The average temperature of the soil to a considerable depth is approximately 277.6 K (40°F) during a winter day. If the outside air temperature suddenly drops to 255.4 K (0°F) and stays there, how long will it take for a pipe 3.05 (10 ft) below the surface to reach 273.2 2 (32°F)? The convective coefficient is h = 8.52 W/m K (1.5 btu/h ft 2 °F). The -7 2 2 soil physical properties can be taken as 5.16 x 10 /s (0.02 ft /h) for the

K

m

•

•

m

Chap. 5

Problems

375

thermal difTusivity and 1.384 W/m K (0.8 btu/h ft °F) for the thermal conductivity. (Note: The solution is trial and error, since the unknown time appears twice in the graph for a semiinfinite solid.) •

•

53-3. Cooling a Slab of Aluminum. A large piece of aluminum that can be considered a semiinfinite solid initially has a uniform temperature of 505.4 K. The surface is with a surface convection suddenly exposed to an environment at 338.8

K

W/m

2

K. Calculate the time in hours for the temperature to reach 388.8 K at a depth of 25.4 mm. The average physical properties are 2 a = 0.340 m /h and k = 208 W/m K.

coefficient of

455

•

5.3-4.

m

Transient Heating of a Concrete Wall. A wall made of concrete 0.305 thick insulated on the rear side. The wall at a uniform temperature of 10°C (283.2 K)

is is

exposed on the front side to a gas at 843°C (1 116.2 K). The convection coefficient 2 3 2 is 28.4 W/m /h, and the thermal K, the thermal difTusivity is 1.74 x 10~ conductivity is 0.935 W/m K. (a) Calculate the time for the temperature at the insulated face to reach 232°C

m

•

(505.2 K). (b)

m

Calculate the temperature at a point 0.152

below the surface

at this

same

time.

Ans. 5.3-5.

(a)

at/xl

=

0.25,1

=

13.4 h

mm

Cooking a Slab of Meat. A slab of meat 25.4 thick originally at a uniform temperature of 10°C is to be cooked from both sides until the center reaches 12TC in an oven at 177°C The convection coefficient can be assumed constant 2 at 25.6 W/m K. Neglect any latent heat changes and calculate the time required. The thermal conductivity is 0.69 W/m K and the thermal difTusivity 4 2 5.85 x 10" m /h. Use the Heisler chart. •

Ans. 5.3-6.

Unsteady-State Conduction

in a

Brick Wall.

A

flat

brick wall 1.0

0.80 h (2880 ft

thick

is

s)

the

on one side of a furnace. If the wall is at a uniform temperature of 100°F and the one side is suddenly exposed to a gas at 1 100°F, calculate the time for the furnace wall at a point 0.5 ft from the surface to reach 500°F. The rear side of the 2 wall is insulated. The convection coefficient is 2.6 btu/h ft °F and the physical

lining

properties of the brick are/c 5.3-7.

=

0.65 btu/h

•

•

ft

°F and a

=

2

0.02ft /h.

rod 0.305 m in diameter is initially at a in an oil bath maintained at 311 K. The 2 surface convective coefficient is 125 W/m K. Calculate the temperature at the center of the rod after 1 h. The average physical properties of the steel are k = 38 W/m K and a = 0.0381 m 2 /h. Ans. T = 391 K Cooling a Steel Rod. temperature of 588 K.

A

long

It is

steel

immersed

•

•

of Size on Heat-Processing Meat. An autoclave held at 121. 1°C is being used to process sausage meat 101.6 in diameter and 0.61 m long which is originally at 21.1°C. After 2 h the temperature at the center is 98.9°C. If the diameter is increased to 139.7 mm, how long will it take for the center to reach 2 98.9°C? The heat transfer coefficient to the surface is h = 1100 W/m K, which is very large so that the surface resistance can be considered negligible. (Show this.) Neglect the heat transfer from the ends of the cylinder. The thermal conductivity k = 0.485 W/m K. 3.78 h Ans,

5.3-8. Effect

mm

•

•

5.3-9.

Temperature of Oranges on Trees During Freezing Weather. In orange-growing on the trees during cold nights is economically important. If the oranges are initially at a temperature of 21.1°C, calculate the center temperature of the orange if exposed to air at — 3.9°C for 6 h. The oranges in diameter and the convective coefficient is estimated as 11.4 are 102 W/m 2 -K. The thermal conductivity k is 0.431 W/m-K and a is 4.65 x 10 _1 2 /h. Neglect any latent heat effects. Ans. (T, - T)/(Ti - T0 ) = 0.05, T = — 2.65°C

areas, the freezing of the oranges

mm

m 376

Chap. 5

Problems

5.3-10.

To harden a steel sphere having a diameter of 50.8 heated to 1033 K and then dunked into a large water bath at 300 K. Determine the time for the center of the sphere to reach 366.5 K. The surface

Hardening a Steel Sphere.

mm,

it is

coefficient can be

m

2

assumed as 710

W/m

2

K, k

=

45

W/m

•

K, and

a.

=

0.0325

/h.

Conduction in a Short Cylinder. An aluminum cylinder is initially heated so it is at a uniform temperature of 204.4°C. Then it is plunged into a 2 large bath held at 93.3°C, where h = 568 W/m K. The cylinder has a diameter of 50.8 and is 101.6 long. Calculate the center temperature after 60 s. The 5 2 physical properties area = 9.44 x 10" m /s and k = 207.7 W/m K.

5.3-11. Unsteady-State

•

mm

mm

•

Three Dimensions in a Rectangular Block. A rectangular steel is initially at 315. 6°C. It is suddenly by 0.457 m by 0.61 immersed into an environment at 93.3°C. Determine the temperature at the 2 center of the block after 1 h. The surface convection coefficient is 34 W/m K. 2 = = The physical properties are k 38 W/m K and a 0.0379 m /h.

5.3-12. Conduction in

m

m

block 0.305

•

•

5.4-1.

Schmidt Numerical Method for Unsteady-State Conduction. A material in the form of an infinite plate 0.762 thick is at an initial uniform temperature of 366.53 K. The rear face of the plate is insulated. The front face is suddenly exposed to a temperature of 533.2 K. The convective resistance at this face can be assumed as zero. Calculate the temperature profile after 0.875 h using the = 2 and slabs 0.1524 m thick. The thermal Schmidt numerical method with

m

M

diffusivity

is

0.0929

m

2

/h.

Ans. 5.4-2.

At

=

0.125

h,

seven time increments needed

Unsteady-State Conduction with Nonuniform Initial Temperature Profile. Use the same conditions as Problem 5.4-1 but with the following change. The initial temperature profile is not uniform but is 366.53 Kat the front face and 422.1 K at the rear face with a linear variation between the two faces.

Unsteady-State Conduction Using the Digital Computer. Repeat Problem 5.4-2 but use the digital computer and write the Fortran program. Use slabs 0.03048

5.4-3.

m

thick

M

and

5.4-4. Chilling

=

2.0.

Calculate the temperature profile after 0.875

Meat Using Numerical Methods. A

slab of beef 45.7

h.

mm

thick

and

uniform temperature of 283 K is being chilled by a surface contact cooler at 274.7 K on the front face. The rear face of the meat is insulated. Assume that the convection resistance at the front surface is zero. Using five slices and = 2, calculate the temperature profile after 0.54 h. The thermal diffusivity is initially at a

M

4.64 x 10

-4

m

2

/h.

Ans. 5.4-5.

Af

=

0.090 h, six time increments

mm

Cooling Beef with Convective Resistance. A large slab of beef is 45.7 thick and is at an initial uniform temperature of 37.78°C. It is being chilled at the front surface in a chilled air blast at — 1.1 1°C with a convective heat-transfer coef2 ficient of h = 38.0 W/m K. The rear face of the meat is insulated. The thermal -4 2 conductivity of the beef is k = 0.498 W/m and a = 4.64 x 10 /h. Using a = 4.0, calculate the temperature profile numerical method with five slices and after 0.27 h. [Hint : Since there is a convective resistance, the value of must be calculated. Also, Eq. (5.4-7) should be used to calculate the surface temperature •

•

K

m

M

N

i

5.4-6.

+ Ai^i-]

Cooling Beef Using the Digital Computer. Repeat Problem 5.4-5 using the digital = 4.0. Write the Fortran program. computer. Use 20 slices and

M

5.4-7.

Convection and Unsteady-State Conduction. For the conditions of Example 5.4-3, continue the calculations for a total of 12 time increments. Plot the temperature profile.

5.4-8. Alternative Convective

ample use

Chap. 5

M

Boundary Condition for Numerical Method. Repeat Exboundary condition, Eq. (5.4-1 1). Also,

5.4-3 but instead use the alternative

=

4.

Calculate the profile for the

Proble ms

full

12 time increments.

377

5.4- 9.

Numerical Method for Semi-infinite Solid and Convection. A semiinfinite solid a uniform temperature of 200°C is cooled at its surface by convection. The cooling fluid at a constant temperature of 100°C has a convective coefficient 2 K. The physical properties of the solid are k = 20 W/m and of h = 250 W/m 5 2 10" = 4.0, /s. Using a numerical method with Ax = 0.040 m and a = 4 x calculate the temperature profile after 50 s total time. Ans. T, = 157.72, T2 = 181.84, T3 = 194.44, TA = 198.93, T5 = 199.90°C initially at

K

M

m

of Beef. Repeat Example 5.5-

5.5- 1. Chilling Slab

1 ,

10°C at the center but use air of 0°C

to

W/m 2

•

where the slab of beef a

at

is

cooled

= 22.7

lower value of h

K. Ans.

Cod

(T,

-

T)/(T,

- T0 = )

0.265,

X=

0.92,

t

=

19.74 h

10°C are packed to a thickness of 102 mm. Ice is packed on both sides of the fillets and wet-strength paper separates the ice and fillets. The surface temperature of the fish can be assumed as essentially 0°C. Calculate the time for the center of the filets to reach 2.22°C and the temperature at this time at a distance of 25.4 from the surface. Also, plot temperature versus position for the slab. 3 The physical properties are (Bl) k = 0.571 W/m-K, p = 1052 kg/m and = 4.02 kJ/kg-K. c p

5.5-2. Chilling Fish Fillets.

fish fillets originally at

mm

,

5.5-3.

Average Temperature in Chilling Fish. Fish fillets having the same physical properties given in Problem 5.5-2 are originally at 10°C. They are packed to a thickness of 102 with ice on each side. Assuming that the surface temperature of the fillets is 0°C, calculate the time for the average temperature to reach 1.39°C. (Note: This is a case where the surface resistance is zero. Can Fig. 5.3-13 be used for this case?)

mm

5.5-4.

to Freeze a Slab of Meat. Repeat Example 5.5-2 using the same conditions except that a plate or contact freezer is used where the surface coefficient can be assumed as h = 142 W/m 2 -K.

Time

Ans.

t

=

2.00 h

A

package of meat containing 75% moisture and in the form of a long cylinder 5 in. in diameter is to be frozen in an air-blast freezer at -25°F. The meat is initially at the freezing temperature of 27°F. The

5.5- 5. Freezing a Cylinder

of Meat.

heat-transfer coefficient

is

h = 3.5

btu/h

•

2

ft

•

°F.

The

physical properties are

3 p = 64 lb m /ft for the unfrozen meat and k = 0.60 btu/h meat. Calculate the freezing time.

5.6- 1.

Heat Generation Using Equation of Energy Change.

W/m

internal heat generation of q conduction only in the x direction.

temperature constant at

3

is

ft

•

°F for the frozen

plane wall with uniform

insulated at four surfaces with heat

The wall has a thickness of 2L m. The one wall atx = +L and at the other wall at x = —L is held T w K. Using the differential equation of energy change, Eq.

at the

(5.6-18), derive the equation for the final

5.6-2.

A

•

temperature

profile.

Heat Transfer in a Solid Using Equation of Energy Change. A solid of thickness L is at a uniform temperature of T 0 K. Suddenly the front surface temperature of the solid at z = 0 m is raised to T\ at t = 0 and held there and at z = L at the rear to T 2 and held constant. Heat transfer occurs only in the z direction. For constant physical properties and using the differential equation of energy change, do as follows. (a) Derive the partial differential equation and the boundary conditions for unsteady-state energy transfer.

Do the same for steady state and integrate the final equation. Ans. (a) dT/dt = a d 2 T/8z 2 B.C.(l): t = 0, z = z, T = T0 B.C.(2): = T = T B.C.(3): t = t, z = L, T = T2 (b) T = (T2 - T,)z/L + T, (b)

378

;

f

;

;

1

I,

z

=

0,

;

Chap. 5

Problems

Temperature Profile Using the Equation of Energy Change. Radial heat is occurring by conduction through a long hollow cylinder of length L with the ends insulated. (a) What is the final differential equation for steady-state conduction? Start with Fourier's second law in cylindrical coordinates, Eq. (5.6-20). (b) Solve the equation for the temperature profile from part (a) for the boundary

5.6-3. Radial

transfer

(c)

5.6-4.

conditions given as follows: T = 7] for r = r,, T = T0 forr = Using part (b), derive an expression for the heat flow q in W.

r0

.

Heat Conduction

in a Sphere. Radial energy flow is occurring in a hollow sphere with an inside radius of r ; and an outside radius of r a At steady state the inside surface temperature is constant at 7" ; and constant at T Q on the outside surface. .

Using the

(a)

differential

temperature

equation of energy change, solve the equation

for the

profile.

(a), derive an expression for the heat flow in W. Heat Generation and Equation of Energy Change. A plane wall is insulated so that conduction occurs only in the x direction. The boundary conditions which apply at steady state are T = T 0 at x = 0 and T = T L at x = L. Internal heat generation per unit volume is occurring and varies as xlL where q 0 and B are constants. Solve the general differential q = q 0 e~^ equation of energy change for the temperature profile. Thermal and Hydrodynamic Boundary Layer Thicknesses. Air at 294.3 K and 101.3 kPa with a free stream velocity of 12.2 m/s is flowing parallel to a smooth flat plate held at a surface temperature of 383 K. Do the following. 5 calculate the critical length x = L of the (a) At the critical N Ke L = 5 x 10 plate, the thickness <5 of the hydrodynamic boundary layer, and the thickness <5 T of the thermal boundary layer. Note that the Prandtl number is not 1.0.

Using part

(b)

5.6- 5. Variable

,

,

5.7- 1.

,

Calculate the average heat-transfer coefficient over the plate covered by the laminar boundary layer.

(b)

5.7-2.

atm abs Boundary-Layer Thicknesses and Heat Transfer. Air at 37.8°C and flows at a velocity of 3.05 m/s parallel to a flat plate held at 93.3°C. The plate is m wide. Calculate the following at a position 0.61 m from the leading edge. (a) The thermal boundary-layer thickness 5 T and the hydrodynamic boundary1

1

layer thickness 5. (b)

Total heat transfer from the

plate.

REFERENCES (Bl)

Blakebrough, N. Biochemical and York Academic Press, Inc., 1968.

2.

New

Phenomena.

New

Biological Engineering Science, Vol.

:

(B2)

B., Stewart, W. York John Wiley & Sons,

Bird, R. :

E.,

(CI)

Carslaw, H. S., and Jaeger, don Press, 1959.

(C2)

Charm,

S. E.

E. N. Transport

C. Conduction of Heat

in Solids.

Oxford: Claren-

Engineering, 2nd ed. Westport, Conn.:

Inc., 1971.

Dusinberre, G. M. Heat Transfer Calculations by Finite Differences. Scranton, Pa.: International

Chap. 5

J.

The Fundamentals of Food

Avi Publishing Co., (Dl)

and Lightfoot,

Inc., 1960.

References

Textbook

Co., Inc., 1961.

379

(Gl)

Geankoplis, C.

J.

Mass Transport Phenomena. Columbus, Ohio: Ohio

State

University Bookstores, 1972.

(G2)

Gurney, H.

(HI)

Heisler, H. P. Trans. A.S.M.E., 69, 227 (1947).

(H2)

Hodgson,

P.,

and Lurie, J. Ind. Eng. Chem.,

15,

1

170 (1923).

T. Fd. Inds. S. Afr., 16, 41 (1964); Int. Inst. Refrig. Annexe, 1966, 633

(1966).

(Kl)

Kreith,

F.

(K2)

Kreith,

Heat Transfer, 2nd

Principles of

Textbook Company,

ed.

Scranton, Pa.: International

1965.

&

Row,

ed.

New

R. Z. Ges. Kalteind., 20, 109 (1913); Z. Ges. Kalteind. Bieh. Reih, 10

(3), 1

F.,

and Black, W.

Z. Basic

Heat Transfer.

New

York: Harper

Publishers, 1980.

(Nl)

Newman,

(PI)

Perry, R. H., and Chilton, C. H. Chemical Engineers' Handbook, 5th York: McGraw-Hill Book Company, 1973.

(P2)

Plank,

A. H. Ind. Eng. Chem., 28, 545 (1936).

(1941).

(Rl)

Riedel, L. Kaltetchnik,

(51)

Schneider, P.

(52)

380

J.

8,

374 (1956);

9,

38 (1957); 11,41 (1959);

12,

4 (1960).

Conduction Heat Transfer. Reading, Mass.: Addison-Wesley

Publishing

Company,

Siegel, R.,

Sparrow,

Inc.,

1955.

E. M.,

and Hallman,

T.

M.

Appl.

Sci. Res.,

A7, 386 (1958).

Chap. 5

References

CHAPTER

6

Principles of

Mass

Transfer

INTRODUCTION TO MASS TRANSFER AND DIFFUSION

6.1

Similarity of Mass, Heat, and

6.1A

Momentum

Transfer Processes

/.

Introduction.

In

classified into three

Chapter

1

heat transfer, and mass transfer. in

we noted

that the various unit operations could be

fundamental transfer (or "transport") processes:

The fundamental process

of

such unit operations as fluid flow, mixing, sedimentation, and

occurs

in

momentum

momentum

filtration.

conductive and convective transfer of heat, evaporation,

transfer,

transfer occurs

Heat

transfer

distillation,

and

drying.

The

third fundamental transfer process,

mass

transfer, occurs in distillation,

tion, drying, liquid-liquid extraction, adsorption,

mass

is

being transferred from one distinct phase to another or through a single phase,

the basic

was

absorp-

and membrane processes. When

also

mechanisms are the same whether the phase is a gas, liquid, or solid. This shown in heat transfer, where the transfer of heat by conduction followed

Fourier's law in a gas, solid, or liquid.

2.

General molecular transport equation.

of

momentum,

given

in

heat,

All three of the molecular transport processes and mass are characterized by the same general type of equation

Section 2.3A. rate of a transfer process

=

driving force

(23-1) resistance

This can be written as follows for molecular diffusion of the property

momentum,

heat,

and mass: (2.3-2)

3.

Molecular diffusion equations for momentum, heat, and mass transfer. Newton's equamomentum transfer for constant density can be written as follows in a manner

tion for

similar to Eq. (2.3-2): Aj

d(v x p)

p

dz

(6.1-1)

381

where in

x JX is

momentum transferred/s

•

3 m, and v x p is rnomentum/m where ,

m2

kinematic viscosity

\i/p is

,

the

momentum has

in

m 2/s,

z

is

distance

units of kg m/s. -

Fourier's law for heat conduction can be written as follows for constant p

andc p

^=-«^ A

:

(6.1-2)

dz

where qJA

is

heat flux in

The equation (2.3-2).

It is

for

W/m 2

,

a

the thermal diffusivity

is

molecular diffusion of mass

inm 2 /s, andpcp T isJ/m 3

Fick's law

is

and

is

.

also similar to Eq.

written as follows for constant total concentration in a fluid

=-Dab^

J*Ai

(6.1-3)

where J* z is the molar flux of component A in the z direction due to molecular diffusion 2 in kg mol A/s m D AB the molecular diffusivity of the molecule A in B in m 2 /s, c A the ,

kg mol/m 3 and z the distance of diffusion in m. In cgs units J Az is D AB is cm 2 /s, and c A is g mol A/cm 3 In English units, J*, is lb mol/h ft 2

concentration of A 2

in

,

g mol A/s cm D AB is ft 2 /h, and c A is lb mol/ft 3 The similarity of Eqs. (6.1-1), ,

•

.

,

.

and (6.1-3) for momentum, heat, and mass on the left-hand side of the three equations have as units transfer of a quantity of momentum, heat, or mass per unit time per unit area. The transport properties u/p, a, and D AB all have units of m 2 /s, and the concentrations (6.1-2),

transfer should be obvious. All the fluxes

are represented as

momentum/m 3 J/m 3 ,

3 or kgmol/rn

,

.

Turbulent diffusion equations for momentum, heat, and mass transfer. In Section 5.7C equations were given discussing the similarities among momentum, heat, and mass 4.

transfer in turbulent transfer.

For turbulent momentum

T„ = - - + For turbulent heat

transfer for constant

A For turbulent mass

= — (a +

J*Az In these equations

x

transfer for constant

e,

is

=

m 2 /s.

(6.1-4)

,

—

a,)

~

"

dz

(6.1-5)

c,

-(^ B +£ M )^

the turbulent or eddy

turbulent or eddy thermal diffusivity in diffusivity in

-

dz

J

p and c

and constant density,

——

E,

\P

transfer

m

2

/s,

(6-1-6)

momentum and

e

M

diffusivity in

the turbulent or

Again, these equations are quite similar to each other.

theoretical equations

and empirical correlations

for

m

2

/s,

a,

the

eddy mass

Many

of the

turbulent transport to various

geometries are also quite similar.

6.1

B

Mass

Examples of Mass-Transfer Processes transfer

is

important

in

many

areas of science and engineering.

Mass

transfer

occurs when a component in a mixture migrates in the same phase or from phase to

phase because of a difference

phenomena involve mass air

382

in

transfer.

concentration between two points.

Liquid

in

an open

pail of

Many

familiar

water evaporates into

still

because of the difference in concentration of water vapor at the water surface and the

Chap. 6

Principles of Mass Transfer

surrounding

air.

There

a "driving force" from the surface to the

is

air.

A

piece of sugar

added to a cup of coffee eventually dissolves by itself and diffuses to the surrounding solution. When newly cut and moist green timber is exposed to the atmosphere, the wood will

dry partially when water in the timber diffuses through the wood, to the surface, and

then to the atmosphere. In a fermentation process nutrients and oxygen dissolved in the solution diffuse to the microorganisms. In a catalytic reaction the reactants diffuse from the surrounding

Many nium

medium

to the catalyst surface,

purification processes involve

mass

where reaction occurs. transfer. In uranium processing,

a ura-

by an organic solvent. Distillation to separate alcohol from water involves mass transfer. Removal of S0 2 from flue gas is done by absorption salt in solution is extracted

in a basic liquid solution.

We

can treat mass transfer

in a

manner somewhat

similar to that used in heat

However, an important difference is that in molecular mass transfer one or more of the components of the medium is moving. In heat transfer by conduction the medium is usually stationary and only energy in the form of heat is being transported. This introduces some differences between heat and mass transfer with Fourier's law of conduction.

transfer that will be discussed in this chapter.

6.1

C

Fick's

Law

for

Molecular Diffusion

Molecular diffusion or molecular transport can be defined as the transfer or movement of individual molecules through 'a fluid by means of the random, individual movements of the molecules. We can imagine the molecules traveling only in straight lines and changing direction by bouncing off other molecules after collisions. Since the molecules travel in a random path, molecular diffusion is often called a random-walk process.

In Fig. 6.1-1 the molecular diffusion process that molecule

shown.

If

A might

there are a greater

molecules diffuse randomly (2)

is

take in diffusing through

number in

of

A

shown

B

schematically.

molecules near point

both directions, more

A random

molecules from point

A

(1)

molecules

than at

(1)

(2),

will diffuse

path

to (2)

is

then, since

from

(1)

to

than from (2) to (1). The net diffusion of A is from high- to low-concentration regions. As another example, a drop of blue liquid dye is added to a cup of water. The dye

molecules

will diffuse

slowly by molecular diffusion to

parts of the water.

all

To

increase

mixing of the dye, the liquid can be mechanically agitated by a spoon and convective mass transfer will occur. The two modes of heat transfer, conduction and convective heat transfer, are analogous to molecular diffusion and convective mass this rate of

transfer. First, we moving but is

will

FIGURE

Schematic aiugrum ocntmaiic diagram of molecu oj motecu-

6.1-1.

consider the diffusion of molecules

stationary. Diffusion of the molecules

is

when

and Diffusion

not

\^ }

& Transfer

is

\

)

.

Mass

fluid

concentration gradient. »

'

^

B

Introduction to

a

j

lar diffusion process.

Sec. 6.1

whole bulk

the

due to

J®

383

A and

Fick's law equation can be written as follows for a binary mixture of

The general B:

where

B in kg mol A + B/m 3 and x A

concentration of A and

c is total

of A in the mixture of

A and

B. If c c

Substituting into Eq. (6.1-7)

,

=

dx A

we obtain

varies

the

is

is

is

the

mole

fraction

,

(6.1-8)

Eq. (6.1-3) for constant total concentration.

~ dc

-D AB

more commonly used one

some, an average value

cx A

= dc A

d(cx A )

J A ,= This equation

=

constant, then sincec^

is

in

(6.1-3)

many molecular

diffusion processes.

If

c

often used with Eq. (6.1-3).

EXAMPLE

6.1-1. Molecular Diffusion of Helium in Nitrogen mixture of He and N 2 gas is contained in a pipe at 298 and 1 atm total pressure which is constant throughout. At one end of the pipe at point 1 the partial pressure p Ai of He is 0.60 atm and at the other end 0.2 (20 cm) p A2 = 0.20 atm. Calculate the flux of He at steady state if D AB of the He-N 2 4 mixture is 0.687 x 10~ m 2 /s (0.687 cm 2 /s). Use SI and cgs units.

K

A

m

Solution:

Since total pressure

as follows for a gas

P

is

constant, then c

is

constant, where c

PV = nRT

where

n

8314.3

m

is

is

kg mol

3

(6.1-9)

kg mol A plus B, V is volume inm 3 T Pa/kg mol K or R is 82.057 x 10" 3 ,

is 3

m

A

plus

B/m 3

.

In cgs units,

R

is

82.057cm 3

in K, R is atm/kg mol K, and c atm/g mol K.

temperature

is

•

•

For steady gas

is

from the perfect gas law.

•

state the flux J* Az in Eq. (6.1-3) is constant. Also, constant. Rearranging Eq. (6.1-3) and integrating,

/*

dz

-D AB

=

j

D AB

for a

"dc A

leu

j a2 Also, from the perfect gas law,p^,

= D ab(c a1 z2 -

V =

nA

c a2 )

(6in)

z,

RT, and

Substituting Eq. (6.1-12) into (6.1-11), ,*

_

E>ab(Pai

A>

This

is

and p A2

=

0.2

atm

=

0.2

- Paz) - 2l

(6.1-13)

)

in the form easily used for gases. x 1.01325 x 10 5 = 6.08 x 10 4 Pa 4 x 1.01325 x 10 5 = 2.027 x 10 Pa. Then, using SI

the final equation to use, which

Partial pressures are p Al

384

RT(z 2

=

0.6

atm

=

is

0.6

Chap. 6

Principles of Mass Transfer

4

units,

JJ* Az

—

(0.687 x 1(T X6.08 x 10

4

8314(298X0.20

= If

atm

pressures in /* Az

—

5.63

are used with SI units,

3 (82.06 x 10" X298X0.20

„

x 10 4 )

x 10~ 6 kg mol A/s-m 2

4 (0.687 x 10" X0.60-0.20)

For cgs

- 2.027 - 0)

-

=

6 5.63 x 10"

kg mol A/s-m 2

0)

units, substituting into Eq. (6.1-13),

0.687(0.60-

=

0.20)

82.06(298)(20-

=

5 63 "

]n _ 7 X 10

*

0)

Cm 2

mo1 "

Other driving forces (besides concentration differences) for diffusion also occur electrical potential, and other gradients. Details are

because of temperature, pressure, given elsewhere (B3). 6.1

D

Convective Mass-Transfer Coefficient

When

a fluid

is

flowing outside a solid surface in forced convection motion, we can

express the rate of convective mass transfer from the surface to the

fluid,

or vice versa, by

the following equation

N A =k

y

c

(c Ll

-cLi

(6.1-14)

)

where k c is a mass-transfer coefficient in m/s, c L1 the bulk fluid concentration in kg mol /1/m 3 and c u the concentration in the fluid next to the surface of the solid. This mass-transfer coefficient is very similar to the heat-transfer coefficient h and is a function of the system geometry, fluid properties, and flow velocity. In Chapter 7 we consider convective mass transfer in detail. ,

6.2

6.2A

MOLECULAR DIFFUSION Equimolar Counterdiffusion

in

IN GASES Gases

is given of two gases A and B at constant total pressure P in two chambers connected by a tube where molecular diffusion at steady state is oc-

In Fig. 6.2-1 a diagram large

Pa

i

Pbi

P

2

1

PB2

P

J*A J *3 «-

2

B2

P A ,PB ,oiP

A2

Figure

Sec. 6.2

6.2-1.

Molecular Diffusion

Equimolar counterdiffusion of gases

in

Gases

A and B. 385

chamber keeps the concentrations in each chamber uniform. The p A2 and p B2 > p B1 Molecules of A diffuse to the right and B to the

curring. Stirring in each partial pressure

p A1

>

.

Since the total pressure

left.

P

is

constant throughout, the net moles of

must equal the net moles of B to the not remain constant. This means that right

for

subscript z

constant

often

is

If this is

A diffusing

to the

not so, the total pressure would

= -Jl

J*a,

The

left.

(6.2-1)

dropped when the direction

is

obvious. Writing Fick's law for

B

c,

=-D BA

JB

Now since P =

+

pA

pB

=

d

^

(6.2-2)

dz

constant, then (6.2-3)

Differentiating both sides,

= -dc B

dc A

Equating Eq.

(6.2-4)

(6.1-3) to (6.2-2),

Ja

=

= -Dam

~J*b

^

= ~(-Pb,

(6.2-5)

Substituting Eq. (6.2-4) into (6.2-5) and canceling like terms,

D AB = D BA

(6.2-6)

This shows that for a binary gas mixture of A and B the difTusivity coefficient for

A

diffusing in

B

EXAMPLE

is

the

6.2-1.

same

as

D BA

for

B

D AB

diffusing into A.

Equimolar Counterdijfusion

m

Ammonia

gas (A) is diffusing through a uniform tube 0.10 long containing 5 Pa press and 298 K. The diagram is similar to 2 gas (B) at 1.0132 x 10 4 Fig. 6.2-1. At point 1, p Al = 1.013 x 10 Pa and at point 2, p A2 =0.507 4 x 10 Pa. The diffusivityD = 0.230 x 10" 4 m 2 /s.

N

/4B

(a)

Calculate the flux J* at steady state.

(b)

Repeat

Solution: z2

-z, = ,*

A

for J%.

Equation 0.10 m, and

5 can be used where P = 1.0132 x 10 Pa, Substituting into for part 298 K. Eq. (6.1-13) (a),

(6.1-13)

T=

= DabIPai-Pai) = RT{z 2

=

(0-23 x

1Q-*X1.013 x 10

-iA

4.70 x 10"

7

4

8314(298X0.10

- 0.507 - 0)

x 10

4 )

kg mol A/s-m 2

Rewriting Eq. (6.1-13) for component B for part (b) and noting that p Bl 4 4 Pa and P - p Al = 1.0132 x 10 5 - 1.013 x 10 = 9.1 19 x 10 p B1

P

-p A2 =

5 1.0132 x 10

-

0.507 x i0

~ Pbi) _ ,* _ Dab(Pbi ""~ RT(z 2 ~zA ~ =

=

9.625 x 10

4

Pa,

4 4 (0-23 x 10' )(9.119 x 10

8314(298X0.10

- 9.625 - 0)

x 10

4 )

-4.70 x 10" 7 kg mol B/s-m 2

The negative value

386

4

= =

for J B

means

the flux goes from point 2 to

Chc'.p. 6

1

Principles of Mass Transfer

A

General Case for Diffusion of Gases

6.2B

and B Plus

Convection

Up

to

now we have

considered Fick's law for diffusion in a stationary

fluid;

i.e.,

there has

been no net movement or convective flow of the entire phase of the binary mixture A and B. The diffusion flux J* occurred because of the concentration gradient. The rate at

which moles of A passed a fixed point to the is

m2

J* kg mol A/s

which

right,

be taken as a positive

will

flux,

This flux can be converted to a velocity of diffusion of A to the right

.

by J* (kg mol A/s

where

v Ad is

Now

the diffusion velocity of

let

stationary point

kg mol

~

)

in

A\

3

(6.2-7)

j

m/s.

what happens when the whole fluid is moving in bulk or the right. The molar average velocity of the whole fluid relative to a

is

v

diffusion velocity v Ad

moving

A

m 2 = vM c A I-

us consider

convective flow to

is

(m

-

faster

m/s.

M is

Component A

measured

is

moving

than the bulk of the phase, since

of the bulk phase

v

stationary point

the

is

M

diffusing to the right, but

still

relative to the

fluid.

To

diffusion velocity v Ad

its

Expressed mathematically, the velocity of

.

sum

now

its

a stationary observer

A

is

added

A

to that

the

relative to

of the diffusion velocity and the average or convective

velocity.

=

VA

where v A

is

V AJ

+ vm

(6.2-8)

the velocity of A relative to a stationary point. Expressed pictorially,

V

A

M

uAd

Multiplying Eq. (6.2-8) by c A

,

c A "a

Each of the three terms represents a 2 This is the flux N A kg mol A/s m .

second term

is

JA

,

U

=

<m »Ad

flux.

The

+

first

total flux of

(6-2-9)

ca vM

term.c,, u A

A

,

can be represented by the

The

relative to the stationary point.

the diffusion flux relative to the

moving

fluid.

The

third

term

is

the

convective flux of A relative to the stationary point. Hence, Eq. (6.2-9) becomes

N A = J*A + c A v M Let

N

(6.2-10)

be the total convective flux of the whole stream relative to the stationary

point. Then,

N = Or, solving for

v

M

= NA + NB

cv M

(6.2-11)

,

Vm

=

Na + Nb

(fi

2 _, 2)

c

Substituting Eq. (6.2-12) into (6.2-10),

NA = Sec. 6.2

Molecular Diffusion

in

J*

Gases

+

— (N A + N B

)

(6.2-13)

387

Since J A

is

Fick's law, Eq. (6.1-7),

^

Na = - cD AB Equation

(6.2-14)

NA be written for N B when

the flux

is

is

A

+

/V B)

(6.2-14)

the final general equation for diffusion plus convection to use

used, which

is

relative to

a stationary point.

A

similar equation can

.

N„ = - cD BA To

^ (N

+

— (N A + N B

+ dz

(6.2-15)

)

c

solve Eq. (6.2-14) or (6.2-15), the relation between the flux

NA

and

jV B

must be known.

Equations (6.2-14) and (6.2-15) hold for diffusion in a gas, liquid, or solid. For equimolar counterdiffusion, N A = — B and the convective term

N

becomes

6.2C

zero. Then,

Special Case for

Nondiff using

The case

in

Eq. (6.2-14)

N A =J*=— N B =—J^. A

Diffusing Through Stagnant,

B

of diffusion of A through stagnant or nondiffusing

one boundary

B

at steady state often occurs.

end of the diffusion path is impermeable to component B, so it cannot pass through. One example shown in Fig. 6.2-2a is in evaporation of a pure liquid such as benzene (A) at the bottom of a narrow tube, where a large amount of inert or nondiffusing air (S) is passed over the top. The benzene vapor (A) diffuses through the air (B) in the tube. The boundary at the liquid surface at point 1 is impermeable to air, since air is insoluble in benzene liquid. Hence, air (B) cannot diffuse into or away from the surface. At point 2 the partial pressure p A2 = 0, since a large volume of air is passing by. Another example shown in Fig. 6.2-2b occurs in the absorption of NH 3 (A) vapor which is in air {B) by water. The water surface is impermeable to the air, since air is only

In this case

at the

very slightly soluble in water. Thus, since

To

derive the case for

A

diffuse, jV B = 0. stagnant, nondiffusing

B cannot

diffusing in

B,

NB

=

0

is

substituted into the general Eq. (6.2-14).

NA

= - cD AD

air

Figure

6.2.-2.

d

^+-(N

dz

A

+0)

(6.2-16)

c

{B)

A through stagnant, nondiffusing B: (a) benzene evaporating into air, (b) ammonia in air being absorbed into

Diffusion of

water.

388

Chap.

6'

Principles

of Mass Transfer

Keeping

the

P

pressure

total

constant,

substituting

= P/RT,

c

pA

=x A P,

and

cjc = P A/P into Eq. (6.2- 16), d

N A = - ~^ -~- + RT dz

~^ P

(6.2-17)

A

Rearranging and integrating,

i

A,

l

d _ EA = _ 5d£ lA Pj RT dz

(6.2-18)

PA1

NA

'dz

|

NA = .

ever, is

Equation

(6.2-20)

defined as follows.

P~

PA2

~^—\J—^

(6.2-20)

HowB = P — p AU and p B2 =

the final equation to be used to calculate the flux of A.

is

Since

A

form as follows.

often written in another

is

it

(6.2-19)

RT

[

P =

p A1

~

Pbi

+

=

p Bi

+

p A2

mean

log

p B2

p Bi

,

value of the inert

i

PSM

_ ~

Pb2

~

Pa\

1" (P B2 /P B >)

~

1" [(P

-

,

P A2 )/(P

-

~

,

2J\

p Al )l

Substituting Eq. (6.2-21) into (6.2-20),

N* = EXAMPLE

^ P"

PT(z 2 -

P

^

1 z,)p flM

"

>

(

6 2 " 22 > -

Diffusion of Water Through Stagnant, Nondiffusing Air bottom of a narrow metal tube is held at a constant temper5 ature of 293 K. The total pressure of air (assumed dry) is 1.01325 x 10 Pa (1.0 atm) and the temperature is 293 K (20°C). Water evaporates and diffuses through the air in the tube and the diffusion path z 2 — z, is 0.1524 m (0.5 ft) long. The diagram is similar to Fig. 6.2-2a. Calculate the rate of evapor2 2 ation at steady state in lb mol/h ft and kg mol/s-m The diffusivity of 4 10~ water vapor at 293 K and 1 atm pressure is 0.250 x m 2 /s. Assume that the system is isothermal. Use SI and English units.

Water

6.2-2.

in the

-

Solution:

The

factor from

Appendix

diffusivity

D AB =

A.

.

converted to

is

2 ft

/h by using the conversion

1

4 4 0.250 x 10" (3.875 x 10 )

=

0.969

2 ft

/h

mm

From Appendix A. 2 the vapor pressure of water at 20°C is 17.54 5 3 or p Ai = 17.54/760 = 0.0231 atm = 0.0231(1.01325 x 10 ) = 2.341 x 10 = P a > Paj the temperature is 20°C (68°F), 0 (pure air). Since

T =

460

+

68

= 528°R =

atm/lb mol °R. •

p Bt

= p-

To

Pai

293

=

1.00

= IS1ZLM. =

Since p B1

In (p B2 /p B1 ) is

-

0.0231

0

=

1.00

A.l,

R =

0.730

3 ft

•

0.9769 atm

atm

6*

=

0.988

atm

=

1.001

x 10* Pa

In (1.00/0.9769)

close to p B2 the linear close to p BM

Molecular Diffusion

=

^-^l

,

would be very

Sec. 6.2

From Appendix

calculate the value ofp BM from Eq. (6.2-21),

P B2 = P ~ Pa2 = 100 p BM

K.

mean

(p B1

+

p B2 )/2 could be used and

.

in

Gases

389

-

Substituting into Eq. (6.2-22) with z 2

2abP

NA =

-

RT(z 2

z,)p BM

~

(Pai

Pat)

=

z,

0.5

ft

(0.1524 m),

0.969(1.0X0.0231

=

-

0)

0.730(528X0.5X0.988)

4 3 1.175 x 10" lb mol/h-ft 3 4 5 (0.250 x 10~ X1.01325 x 10 X2.341 x 10

NA =

8314(293X0.1524X1.001 x 10

=

1.595 x 10"

-

0)

5 )

kg mol/s-m 2

7

EXAMPLE

6.2-3. Diffusion in a Tube with Change in Path Length Diffusion of water vapor in a narrow tube is occurring as in Example 6.2-2 under the same conditions. However, as shown in Fig. 6.2-2a, at a given time

m

the level is z from the top. As diffusion proceeds, the level drops slowly. Derive the equation for the time t F for the level to drop from a starting point

f,

m

of z 0

at

=

r

0 toz F at

We

Solution:

—

t

t

F s as

shown.

assume a pseudo-steady-state condition since the

level

drops

very slowly. As time progresses, the path length z increases. At any time Eq. (6.2-22) holds; but the path length follows where

NA

and

z are

now

is

and Eq.

z

(6.2-22)

variables.

(6.2-23)

Pai)

Assuming a cross-sectional area of 1 m 2 the level drops dz m p A (dz \)/M A is the kg mol of A that have left and diffused. Then, ,

N A -l = Equating Eq. (6.2-24) to

=

limits of z

z0

when

t

=

Pa

Solving

for

t

F

zF

(6.2-24)

M A dt when

E>abP{Pax

t

=

-

and

and integrating between the t

F

,

Pai)

(6.2-25)

dt

RTp BM

in

=

PaSA ~ z o)R t Pbm - p A2 A D AB P{p Ai

™

Example

6.2-3 has

(6.2-26)

been used to experimentally determine the is measured at £ = 0 and

In this experiment the starting path length z 0

.

also the final z F at t F

6.2D

=

in dt s

,

The method shown

D AB

z

dz

tr

diffusivity

PaVz-1)

(6.2-23), rearranging,

0 and

£,

becomes as

Diffusion

.

Then Eq.

Through

a

(6.2-26)

is

used to calculate

D AB

.

Varying Cross-Sectional Area

we have considered N A and J* as constants in the A m 2 through which the diffusion been constant with varying distance z. In some situations the area A may

In the cases so far at steady state

integrations. In these cases the cross-sectional area

occurs has vary.

Then

it is

convenient to define

N A as Na =

where

NA

is

A

kg moles of A diffusing per second or kg mol/s. At steady

constant but not

390

(6-2-27)

~~T state,

NA

will

be

A for a varying area. Chap. 6

Principles of Mass Transfer

from a sphere. To illustrate from a sphere in a gas will such cases as the evaporation of a drop thalene, and the diffusion of nutrients to Diffusion

the use of Eq. (6.2-27), the important case of

diffusion to or

be considered. This situation appears often in

].

Fig. 6.2-3a

shown

is

of liquid, the evaporation of a ball of napha spherical-like microorganism in a liquid. In

a sphere of fixed radius

r

m

t

in

an

infinite gas

medium. Component

pressure p Al at the surface is diffusing into the surrounding stagnant where p A2 = 0 at some large distance away. Steady-state diffusion will be

(A) at partial

medium

(B),

assumed.

The 4nr 2

at

flux

point

NA r

can be represented by Eq.

where

(6.2-27),

A

distance from the center of the sphere. Also,

is

the cross-sectional area

NA

is

a constant at steady

state.

N,=£^ Since this

used

a case of

is

A

in its differential

diffusing through stagnant, nondiffusing

NA

form and

Note

that dr r2

was substituted

will

be equated to Eq.

NA = _ 2 =

Ma =

point

4nr

,

Eq. (6.2-18) will be

^Va (6.2-29)

T, RT{l-p A /P)dr

Rearranging and integrating between

for dz.

rid

rt

and some

dp.

L= 2

[

JL

RT

r

NA :

£d£^

RT

4tc

r2

Dab

B

(6.2-28), giving

a large distance away,

"a

Since

(6.2-28)

>

r,, l/r 2

s

0.

,

N A1 =

n

L^EJI

(6.2-31)

P-p A1

77^7-

RTr

This equation can be simplified further.

If

t

p Al

_

—

(6.2-32)

p BM is

small compared to

P

(a dilute

gas

(b)

(a)

6.2-3.

„

Substituting p BM from Eq. (6.2-21) into Eq. (6.2-31),

4n 2

FIGURE

i

(6.2-30)

d-pJP)

Diffusion through a varying cross-sectional area: (a) from a sphere to a surrounding medium, (b) through a circular conduit that is tapered uniformally.

Sec. 6.2

Molecular Diffusion

in

Gases

391

=

phase), p BM

P. Also, setting 2r,

= D u diameter, and c A = ,

NA =

(

\

This equation can also be used

for liquids,

p A1 /RT, we obtain

— c ai)

c ai

where D AB

(6-2-33)

the diffusivity of

is

A

in

the liquid.

EXAMPLE 6.2-4.

Evaporation of a Naphthalene Sphere is suspended in a large sphere of naphthalene having a radius of 2.0 volume of still air at 318 K and 1.01325 x 10 5 Pa (1 atm). The surface temperature of the naphthalene can be assumed to be at 318 K and its vapor pressure at 318 K is 0.555 Hg. TheD^ of naphthalene in air at 318 K is 6 6.92 x 10~ m 2 /s. Calculate the rate of evaporation of naphthalene from

mm

A

mm

the surface.

The

Solution:

m2 /s,

=

,R

=

Pai

D AB =

similar to Fig. 6.2-3a.

is

= 74.0 Pa, p A2 = 0, = P - p A1 = 1.01325

s (0.555/760)(1.01325 x 10 )

m

8314

flow diagram

1.01251 x 10

3 -

5

Pa/kg

Pa, p B2

mol K,

p Bi

•

=P-

p A2

=

1.01325 x 10

- 0.

5

r,

6 6.92 x 10"

=

2/1000 m,

-

x 10 5

74.0

=

Since the values of

p B1 and p B2 are close to each other,

=

PBM =

(1 -°

325)xl ° 5

125+1

=

f

Pa

1.0129 x 10'

Substituting into Eq. (6.2-32),

Dab PiP Ai ~ Pai) _ v A1 _ RTr p BM

6 (6-92 x 1Q- X1.01325 x

l

= If

9.68 x 10"

8

the sphere in Fig. 6.2-3a

10^74.0

-

0)

5 8314(318X2/1000X1.0129 x 10 )

'

kg mol Ajs-m 1 is

evaporating, the radius

r

of the sphere decreases slowly

with time. The equation for the time for the sphere to evaporate completely can be derived by assuming pseudo-steady state and by equating the diffusion flux equation

where

(6.2-32),

r is

now a

variable, to the

moles of

solid

material-balance

method

is

similar to

Example

A evaporated Problem

unit area as calculated from a material balance. (See

The

6.2-3.

final

per dt time and per

6.2-9 for this case.)

equation

PAr\RTp BM

2M A D AB P(p Al -p A2 where

r

i

is

The

is

(62_ 34) )

the original sphere radius, p A the density of the sphere,

and

MA

the molecular

weight.

2.

Diffusion through a conduit of nonuniform cross-sectional area.

nent as

A

is

diffusion at steady state through a circular conduit

shown. At point 1 the radius is r and at point 2 A diffusing through stagnant, nondiffusing B, i

it is

r2

In Fig. 6.2-3b compo-

which is tapered uniformally At position z in the conduit,

for

NA =

M A~ — = nr

2

dPA - DAs RT(l-pJP)dz

Using the geometry shown, the variable radius

r

(6.2-35)

can be related to position z in the path

as follows:

(z 2

392

- zJ

Z

+

r

'

Chap. 6

(6.2-36)

Principles of Mass Transfer

Thisjvalue of

r is

then substituted into Eq. (6.2-35) to eliminate

r

and the equation

infegrated.

dz

Dab z

A case similar to this

1.

„

r.

1

-Pa/P

given in Problem 6.2-10.

Diffusion Coefficients for Gases

6.2E

tal

is

+

(6.2-37)

RT

Experimental determination of diffusion coefficients. A number of different experimenmethods have been used to determine the molecular diffusivity for binary gas

methods are

mixtures. Several of the important

One method

as follows.

pure liquid in a narrow tube with a gas passed over the top as

shown

is

to evaporate a

in Fig. 6.2-2a.

The

measured with time and the diffusivity calculated from Eq. (6.2-26). In another method, two pure gases having equal pressures are placed in separate sections of a long tube separated by a partition. The partition is slowly removed and diffusion proceeds. After a given time the partition is reinserted and the gas in each section analyzed. The diffusivities of the vapor of solids such as naphthalene, iodine, and benzoic acid in a gas have been obtained by measuring the rate of evaporation of a sphere. Equation (6.2-32) can be used. See Problem 6.2-9 for an example of this. A useful method often used is the two-bulb method (Nl). The apparatus consists of two glass bulbs with volumes V and V2 m 3 connected by a capillary of cross-sectional 2 area A m and length L whose volume is small compared to K, and V2 as shown in Fig. 6.2-4. Pure gas A is added to K, and pure B to V2 at the same pressures. The valve is opened, diffusion proceeds for a given time, and then the valve is closed and the mixed contents of each chamber are sampled separately. The equations can be derived by neglecting the capillary volume and assuming each bulb is always of a uniform concentration. Assuming quasi-steady-state diffusion in the fall in

liquid level

is

l

,

capillary,

where

c 2 is

going to V2

D

d

J*=-D AB — = A

the concentration of is

in

V2

AJ*

D AB(C2 ~

=

\

(6.2-38)

at time

equal to the rate of accumulation

_

,

f

in

and

V2

c

l

in V^.

The

rate of diffusion of

A

.

C l) A

dc, (6.2-39)

dt

The average

value c av at equilibrium can be calculated by a material balance from the

starting compositions c°

and c°

at

(Vx

Figure

6.2-4.

f

=

0.

+ V2 )c„=V c°+V2C°2

measurement of gases by the iwo-bulb method.

(6.2-40)

l

Diffusivity

v2

valve

Vi

rxi L

-

Cl

z~*

Sec. 6.2

Molecular Diffusion

in

Cases

393

A

similar balance at time

t

gives

+ y2 )c„'=V c + V2 c

(yx

Substituting c

and

t

=

t,

from Eq. (6.2-41) into

x

the final equation

l

(6.2-41)

2

and integrating between

(6.2-39), rearranging,

t

=

0

is

-

c„ If c 2 is

l

c

=

o

ex P {LI

obtained by sampling at

t,

D AB

(6.2-42)

AW

2

V,)

can be calculated.

Some typical data are given in Table 6.2-1. Other data and Green (PI) and Reid et al. (Rl). The values range from -4 about 0.05 x 10" 4 m 2 /s, where a large molecule is present, to about 1.0 x 10 m 2 /s, 2 where H 2 is present at room temperatures. The relation between diffusivity inm /s ar>d Experimental

2.

diffusivity data.

are tabulated in Perry

2 ft

3.

/h

is

1

m

2

/s

=

3.875 x 10

4

2 ft

/h.

The

Prediction of diffusivity for gases.

gas region,

i.e.,

at

theory of gases.

diffusivity of a binary gas

mixture

in the dilute

low pressures near atmospheric, can be predicted using the kinetic

The

gas

assumed

is

to consist of rigid spherical particles that are

completely elastic on collision with another molecule, which implies that

momentum

is

conserved. In a simplified treatment

assumed that there

it is

free

distance that a molecule has traveled between collisions.

D AB = where

i7

and X into Eq.

attractive or repulsive forces

path

The

final

X,

which

equation

is

the average

is

\uX

(6.2-43)

The

the average velocity of the molecules.

is

no

are

between the molecules. The derivation uses the mean

equation obtained

final

after

approximately correct, since it correctly predicts D AB proportional to 1/pressure and approximately predicts the temsubstituting expressions for

perature

i7

(6.2-43)

is

effect.

A more accurate and rigorous treatment must consider the intermolecular forces of attraction and repulsion between molecules and also the different sizes of molecules A and B. Chapman and Enskog (H3) solved the Boltzmann equation, which does not utilize the

mean

free

path X but uses a distribution function.

To

solve the equation, a

relation between the attractive

and repulsive forces between a given pair of molecules nonpolar molecules a reasonable approximation to the forces

must be used. For a pair of the Lennard-Jones function.

is

The

final relation to

predict the diffusivity of a binary gas pair of

A

and B molecules

is

D AB = where

D AB

is

the diffusivity in

kg mass/kg mol,

1.8583 x 10-

-j-r

m

2

/s,

T

7

T 3/2

/

1

1

77-

temperature

in

Y

+ XT K,

MA

/2

(S- 2- 44 )

I

molecular weight of

A

in

M B molecular weight of B, and P absolute pressure in atm. The terma^

an "average collision diameter" and Q D AB Lennard-Jones potential. Values of a A and a B and

is

a collision integral based on the

is

fJ D

AB can be obtained from a

number

of sources (B3, G2, H3, Rl).

The

collision integral Cl D AB

compared

interactions.

394

is

a ratio giving the deviation of a gas with interactions

to a gas of rigid, elastic spheres. This value

Equation

would be

(6.2-44) predicts diffusivities with

1.0 for a gas with

no

an average deviation of about

Chap. 6

Principles of Mass Transfer

Table

Diffusion Coefficients of Gases at

6.2-1.

101.32 kPa Pressure Temperature

Diffusivity 2

[(m /s)I0*

0

273

0 VA

0

273

VAX, jL\J

25

298

42

315

VAX,

Air-H 111 Air-C H 1

Air-CH V— 1

/All

L

-j

OH rOOH

V-/I 1

A r — /IM-hpY^np All 11 »» A 0. V, 1

1

1

3

276

0 VA

44

317

VA 1

/

0

273

0 f\ VAU

1 1 I 1

25

298

n JJ U.l

42

315

0

273

VA

294

0 0R0 VAV/OVJ

1

1


n n

25.9

298.9

VAUOU

/All

fl-UUlcillUI

273

fi

25.9

298.9

0 0R7

11 2

^4

25

298

n 79^ U, / ZD

"2

1N 2

25

298

n 7Sd va / 04

85

358

n2

uenzcnc Ar

38.1

311.1

11

1

070^

fR

'\

1

0S9 JZ .U

o dfM

22.4

295.4

25

298

n 7sn VA / OJ

50

323

o6

67

340

o U. JOU

25

298

0 79Q / zy u.

150

423 317

0 JO SS7/ VA

44 25

298

n 67s

v_.

IN /J

UU

fnliipriP LUiUCllC

Hp-PH no n

\

ns

A 11 >r i\

n-riii n r\! ULlLclIlUl

1

/

298

OH

>

I'M (iVIL)

14?

21

0

fWII I" X J (Li)

0 988 OO

25

T-I P I1C

(tsij 1 [t>i)

0 76S VA / U J

(T|)

1

J

^ J

T-Ip-N J V, N2

25

298

o 6R7

"2 V_ rA\ r-PH ^

25

298

0 79Q va / Ly

25

298

H 909 u.zuz

C0 2 -N 2 co 2 -o N 2 -n-butane

25

298

0.167

20

293

0.153

(W3) (W4)

25

298

0.0960

(B2)

H 2 0-C0 2

34.3

307.3

0.202

(S3)

100

373

0.318

(Al)

30

303

0.0693

(C3)

26.5

299.5

0.1078

(S4)

J

1

110 1

l l

2

CO-N 2 CH 3 C1-S0 (C 2 H 2 0-NH 3 2

5)

up to about 1000

K (Rl).

the correct force constant

is

the potential-energy function effect of

For a polar-nonpolar gas mixture Eq. (6.2-44) can be used for the polar gas (Ml, M2). For polar-polar gas pairs

commonly used

is

B in

in

maximum

In most cases the effect

V*—'J

used

concentration of A

gases with interactions the

Sec. 6.2

98 sO

Kpn 7p n1C f» UC11Z.C1

Hp— Ar

(G2).

1 1

Ref.

0 290

A1r /All

H n 2 — in n 3 n 2 0^2 H P H

The

cm 2 /s]

Air-H 2 0

r\l

if

or

Air-NH 3

Air- C0 2

8%

K

"C

System

is

effect of

considerably

Molecular Diffusion in Gases

the Stockmayer potential (M2).

Eq. (6.2-44)

is

not included. However, for real

concentration on diffusivity

less,

and hence

it is

is

about

4%

usually neglected.

395

Equation

(6.2-44)

relatively

is

such as a AB are not available or

complicated to use and often some of the constants

difficult to estimate.

Hence, the semiempirical method of

much more convenient to use, is often utilized. The equation was obtained by correlating many recent data and uses atomic volumes from Table 6.2-2, which are summed for each gas molecule. The equation is Fuller

which

et al. (Fl),

is

lo-^-^i/M, + i/M B y 2 p[(Y,v A y» + &v B y»y

LOO x

n where

£

vA

= sum

is

-

of structural volume increments, Table 6.2-2, and

method can be used accuracy

(6 2" 45)

for mixtures of

nonpolar gases or

for a

D AB —

m

2

/s.

This

polar-nonpolar mixture.

Its

not quite as good as that of Eq. (6.2-44).

Table

Atomic Diffusion Volumes for Use with the Fuller, Schettler, and

6.2-2.

Giddings Method*

Atomic and

C

structural diffusion

volume increments,

(CI)

16.5

H O

1.98

(S)

5.48

Aromatic ring

(N)

5.69

Heterocyclic ring

17.0

—20.2 -20.2

Diffusion volumes for simple molecules,

H2 D2

£v

16.6

CO co N20 NH H 20

Air

20.1

(CC1 2 F 2 )

Ar Kr

16.1

(SF 6 )

22.8

(Cl 2 )

37.7

(Xe)

37.9

(Br 2 )

67.2

(SOJ

41.1

7.07

6.70

He

17.9

o

2

Ne *

18.9

26.9

2

2.88

35.9 14.9

3

5.59

v

19.5

12.7

114.8 69.7

Parenlheses indicate that the value listed

is

based on only a few data

points.

Source: Reprinted with permission from E. N. Fuller, P. D. Schettler, and J. C. Giddings, Ind. Eng. Chem., 58, 19(1966). Copyright by the

American Chemical

The equation shows value of

D AB

another

T

and

relationship

4.

is

to l/P it is

and to

T 115

.

If

oc

is

D AB

T and P

at

by the

T uli /P.

Schmidt number of gases.

dimensionless and

an experimental

desired to have a value of

one should correct the experimental value to the new

P,

D AB

D AB is proportional a given T and P and

that

available at

Society.

The Schmidt number

of a gas mixture of dilute

A

in

B

is

defined as

(6-2-46)

P&AB

396

Chap. 6

Principles of Mass Transfer

which is viscosity of B for a dilute mixture in Pa s and p is the density of the mixture in kg/m 3 For a gas the Schmidt number can be assumed independent of temperature over moderate 5 ranges and independent of pressure up to about 10 atm or 10 x 10 Pa. The Schmidt number is the dimensionless ratio of the molecular momentum diffusivity nip to the molecular mass diffusivity D AB Values of the Schmidt number for gases range from about 0.5 to 2. For liquids Schmidt numbers range from about 100 to over 1 0 000 for viscous liquids.

where

or

/i

viscosity of the gas mixture,

is

kg/m

•

D AB is

s,

m

diffusivity in

2

/s,

.

.

EXAMPLE 6.2-5. Normal butanol Fuller et

Estimation of Diffusivity of a Gas Mixture

(A)

diffusing through air (B) at

is

method, estimate the

al.

diffusivity

D AB

1

atm

abs.

Using the

for the following temper-

compare with the experimental data. For0°C. For 25.9°C. For 0°C and 2.0 atm abs.

atures and (a)

(b) (c)

For part

Solution: ol)

=

74.1,

MB

£>„ = X vB =

P = 1.00 atm, T = 273 + From Table 6.2-2,

(a),

=

(air)

29.

4(16.5)

+

10(1.98)

+

1(5.48)

=

0

= 273 K,

MA

(butan-

91.28 (butanol)

20.1 (air)

Substituting into Eq. (6.2-45), 1.0

AB

x lQ-

~

7

(273)'-

1.0[(91.28)

=

6 7.73 x 10"

+10%

This value deviates by 2 /s from Table 6.2-1.

m

2

75

1/3

(l/74.1

+

+

I/3

(20.1)

l/29)' /2 2

]

/s

from the experimental value of 7.03 x 10

-6

m

For part

d ab =

tal of 8.70

in

(b),

x

-6

T=

m 10" 6 m

9.05 x 10

2

273 + 25.9 = 298.9. Substituting into Eq. (6.2-45), This value deviates by +4% from the experimen-

/s.

2

/s.

For part (c), the total pressure P = part (a) and correcting for pressure, 0" 6 .0/2.0) Dab = 7 -73 x 1

2.0

=

( 1

MOLECULAR DIFFUSION

6.3

6.3A

atm. Using the value predicted

3.865 x 10"

6

m

2

/s

IN LIQUIDS

Introduction

Diffusion of solutes

in liquids is

very important in

many

industrial processes, especially in

such separation operations as liquid-liquid extraction or solvent extraction, gas absorption, and distillation. Diffusion in liquids also occurs in many situations in nature, such as oxygenation of rivers It

and lakes by the

air

slower than

in gases.

The molecules

in

will

diffusion of salts in blood.

and

diffuse

more slowly than

in liquids is

considerably

a liquid are very close together compared to a gas.

Hence, the molecules of the diffusing solute often

and

should be apparent that the rate of molecular diffusion

A

in gases.

will collide

with molecules of liquid

B more

In general, the diffusion coefficient in a gas

be of the order of magnitude of about 10 3 times greater than in a liquid. However,

the flux in a gas

concentrations

Sec. 6.3

is

not that

in liquids

much

greater, being only

about 100 times

faster, since

the

are considerably higher than in gases.

Molecular Diffusion

in Liquids

397

Equations for Diffusion

6.3B

Liquids

in

Since the molecules in a liquid are packed together

much more closely than in gases, the much greater. Also, because of this

density and the resistance to diffusion in a liquid are closer spacing of the molecules, the attractive forces

between molecules play an import-

ant role in diffusion. Since the kinetic theory of liquids

we

only partially developed,

is

write the equations for diffusion in liquids similar to those for gases.

an important difference from diffusion in gases is that the dependent on the concentration of the diffusing components.

In diffusion in liquids diffusivities are often quite

Equimolar counterdiffusion.

1.

Starting with the general equation (6.2-14),

N A = — NB

obtain for equimolal counterdiffusion where (6.1-1 1) for

gases at steady state.

NA =

D AB (c A —

NA

is

c av

A

in

A

D AB c

=

y (x A1

s,

—

x A2 )

t£ii\

(6.3-1)

Zl-Zl

kg mol A/s m 2 kg mol A/m 3 at point

the flux of

concentration of

c A2 )

,

-Z,

Z2

where

we can

an equation similar to Eq.

,

in

,

D AB

the diffusivity of

A

Bin

in

x A1 the mole fraction of

1,

A

m

at

2

/s, c A1

point

1,

the

and

defined by

where

c 3V

is

A +B

the average total concentration of

molecular weight of the solution at point of the solution in

Equation

kg/m 3

at

point

(6.3-1) uses the

is

used as

kg mol/m 3

in

,

M,

the average

kg mass/kg mol, andp, the average density

1.

D AB

average value of

centration and the average value of the linear average of c

1

in

Eq.

in

(6.3-2).

which may vary some with con-

,

may

which also

c,

vary with concentration. Usually,

The case

of

equimolar counterdiffusion

in

Eq. (6.3-1) occurs only very infrequently in liquids.

2.

The most important case of diffusion in liquids and solvent B is stagnant or nondiffusing. An example

Diffusion of A through nondiffusing B.

where solute A

is

that

is

a dilute solution of

toluene.

Only

is

diffusing

propionic acid (A)

in a

water (B) solution being contacted with

the propionic acid (A) diffuses through the water phase, to the boundary,

and then into the toluene phase. The toluene-water interface is a barrier to diffusion of B and N B = 0. Such cases often occur in industry (T2). If Eq. (6.2-22) is rewritten in terms of concentrations

by substituting

obtain the equation for liquids

at

c av

= P/RT,

c A1

=

p A1 /RT, and x BM

{Xai

_

Xai)

=

p BM /P,

we

steady state.

D ** c «

Na =

z l) x BM

(Z 2

(6.3-3)

where

*bm Note

that

x A1

+

x By

essentially constant.

= x A2 + Then

x B2

=

=

1.0.

—— ;

7—,

For

dilute solutions

x BM

is

close to 1.0

and

c is

Eq. (6.3-3) simplifies to

N A = DACai ~ Cai) 398

(6-3-4)

:

In (x B2 /x Bl )

Chap. 6

(63-5)

Principles of Mass Transfer

EXAMPLE 63-1.

Diffusion ofElhanol (A ) Through Water (B) ethanol (/l)-water (B) solution in the form of a stagnant film 2.0 thick at 293 is in contact at one surface with an organic solvent in which

mm

An

K

is soluble and water is insoluble. Hence, N B = 0. At point 1 the and the solution density is p = 972.8 concentration of ethanol is 16.8 wt kg/m 3 At point 2 the concentration of ethanol is 6.8 wt and p 2 = 988.1 kg/m 3 (PI). The diffusivity of ethanol is 0.740 x 10" 9 2 /s (T2). Calculate

ethanol

%

,

%

.

m

the steady-state flux

NA

.

The diffusivity is D AB = 0.740 x 10~ 9 m 2 /s. The molecular weights or/1 and B areM^ = 46.05 and of 6.8, the B = 18.02. For a wt mole fraction of ethanol (A) is as follows when using 100 kg of solution. Solution:

M

6.8/46.05

Xa1

+

6.8/46.05

Then x B2 = 1 — 0.0277 = x Ai = 0.0732 and x B1 = 1 weight

M

2 at

point

%

0.1477

_

+

0.1477

93.2/18.02

_

5.17

0.9723. Calculating x Ai in a similar manner, 0.0732 = 0.9268. To calculate the molecular

-

2.

100 ke

+

(0.1477 Similarly, Af,

=

c 3v

To

=

p,/M,

20.07.

From

+p 2 /M 2 =

5.17)

Eq.

kg mol

x

18-75 kg/kg

mol

(6.3-2),

+

972.8/20.07

calculate x BM from Eq. (6.3-4),

and x B2 are close

=

988.1/18.75

we can use

=

50.6

the linear

kg mol/m 3

mean

since

fll

to each other.

09268

+

w-

+

°- 9123

2

=

0-949

Substituting into Eq. (6.3-3) and solving,

D^c,, a

~(z 2 -Zi)x bm

=

8.99 x 10"

7

_

(Xai

9 (0.740 x 1Q- X50.6)(0.0732

V ' 12

"~"

'

0.0277)

(2/1000)0.949

kg mol/s-m 2

Diffusion Coefficients for Liquids

6.3C

Experimental determination of diffusivities. 1. determine diffusion coefficients experimentally diffusion in a long capillary tube

concentration profile. is

-

D AB

.

If

is

one method unsteady-state

carried out and the diffusivity determined from the

A

the solute

Several different methods are used to in liquids. In

Also, the value of diffusivity

is

diffusing in B, the diffusion coefficient determined

is

often very dependent

the diffusing solute A. Unlike gases, the diffusivity

common method

D AB

upon the concentration of

does not equal

D BA

for liquids.

and a slightly more concentrated solution are placed in chambers on opposite sides of a porous membrane of sintered glass as shown in Fig. 6.3-1. Molecular diffusion takes place through the narrow passageways of the pores in the sintered glass while the two compartments are stirred. The effective diffusion length is K 5, where K, > 1 is a constant and corrects for the fact the path is actually greater than 5 cm. In this method, discussed by Bidstrup and Geankoplis (B4), the effective diffusion length is obtained by calibrating with a solute such as having a known diffusivity. In a relatively

a relatively dilute solution

t

KG

Sec. 6.3

Molecular Diffusion

in

Liquids

399

V' stirrer

—

C

X porous

glass

-)

1

>-

5

y

,

,-

|

z

-)V 1

V c

Figure

To

6.3-

1

.

Diffusion cell for determination of diffusivity in a liquid.

derive the equation, quasi-steady-state diffusion in the

NA =

membrane

is

assumed.

C

-pA. O

eD AB

(6.3-6)

[

where

c

is

the upper,

solute

A

making

the concentration

and

in

£ is

in the

lower chamber

the fraction of area of the glass

the upper chamber, where the rate in

a similar balance

and integrating the

final

a time

at

open

=

t, c'

is

to diffusion.

rate out

+

on the lower chamber, using volume V

equation

the concentration in

Making a balance on

rate of accumulation,

=

V, and combining

is

where 7zAjK bV is a cell constant that can be determined using a solute of known diffusivity, such as KC1. The values c 0 and c'0 are initial concentrations and c and c' final x

concentrations.

2.

Experimental

Experimental

liquid diffusivity data.

diffusivity

data for binary mix-

tures in the liquid phase are given in Table 6.3-1. All the data are for dilute solutions of

the diffusing solute in the solvent. In liquids the diffusivities often vary quite markedly

with concentration. Hence, the values in Table 6.3-1 should be used with

when

solutes in solution are given

values are quite small and relatively

63D

nonviscous

in

As noted

the next section.

in the range of about 0.5 X

liquids. Diffusivities in

in the table, the diffusivity

10~ 9 to 5 x 10~ 9

gases are larger by a factor of 10

m 4

2

/s

for

to 10

5 .

Prediction of Diffusivities in Liquids

The equations

for predicting diffusivities of dilute solutes in liquids are

semiempirical, since the theory for diffusion

Stokes-Einstein equation, one of the

molecule (A) diffusing

assuming that

all

first

in liquids is

theories,

in a liquid solvent (B) of

describe the drag on the

moving

was derived

Then

yet.

The

for a very large spherical

was used

to

the equation was modified by

molecules are alike and arranged in a cubic

D AB =

by necessity

not well established as

small molecules. Stokes' law

solute molecule.

the molecular radius in terms of the molar

400

some caution

outside the dilute range. Additional data are given in (PI). Values for biological

lattice

and by expressing

volume (W5),

9.96 x 10

_16

r (6-3-8)

j^p Chap. 6

Principles

of Mass Transfer

Table

Diffusion Coefficients for Dilute Liquid Solutions

6.3-1.

Temperature

Diffusivity 2

l(m ls)10 Solute

in

Solvent

n3

1

t-tnyi ajconoi

12

285

(cm

i

Formic

(Mil (1NZ) 1. 77 1 1 1

QR 1.70

(INZ)

25

298

9 At

[VI)

w ater

25

298

z.uu

(VI)

WI n tar water

25

298

^.o

mil

288 281

96 I .ZD

water

acid

15

10

I

nil t J1 )

/TIN

1A

1

U.o

?R? 7

u.

(J 1

/

J

/

I.jZ

9 .7. 7

T1

f

I

288 25

WF o tar w ater

Acetic acid

Ref.

291

w aiei

ropy djconoi

or

/s)./0 ]

288

298 /i-r

9 3

2

18

Wq tpr w a ter

n 1 f r\ r\ t~\ aiconui

\/t of V\ im ivicLiiyi

K

15

w a ter

^2

°C

(154)

/oy

(»4J

l.ZO

(#4)

w ater

25

298

Water

10 10

283 283

Z.J

Dciizuic aciu

w aier

25

298

i1

91 .Zl

l*-4J

Acetone

Water Benzene

25

1.28

(A2)

25

298 298

2.09

(C5)

Urea Water

Ethanol

12

285

0.54

(N2)

Ethanol

25

298

1.13

(H4)

KCI KC1

Water

25

1.870

(P2)

Ethylene

25

298 298

0.119

(P2)

Propionic acid rii^j \y

g moi/mer)

(Z.j

g mol/iiter)

Acetic acid

l.UI

(B4J t~\Tl\

(NZ) (N2)

glycol

where or

D AB is

kg/m

s,

diffusivity in

and VA

is

m 2 /s, T is

the solute

temperature in K, ^ is viscosity of solution in Pa-s molar volume at its normal boiling point inm 3 /kg mol.

This equation applies very well to very large unhydrated and spherical-like solute

VA

molecules of about 1000 molecular weight or greater (Rl), or where the about 0.500

m

3

is

above

/kg mol (W5) in aqueous solution.

For smaller solute molar volumes, Eq. etical derivations

(6.3-8) does not hold. Several other theorhave been attempted, but the equations do not predict diffusivities very

accurately. Hence, a number of semitheoretical expressions have been developed (Rl). The Wilke-Chang (T3, W5) correlation can be used for most general purposes where the

solute (A)

is

dilute in the solvent (B).

D AB = where

VA

MB

is

l6 (
the molecular weight of solvent B,/j b

is

m —L-6

the viscosity of B in

(

Pa

•

s

or

6

3" 9>

-

kg/m

-s,

molar volume at the boiling point (L2) that can be obtained from Table 6.3-2, and q> is an "association parameter" of the solvent, where q> is 2.6 for water, 1.9 methanol, 1.5 ethanol, 1.0 benzene, 1.0 ether, 1.0 heptane, and 1.0 other unassociated 3 3 solvents. When values of VA are above 0.500 m /kg mol (500 cm /g mol), Eq. (6.3-8) is

the solute

should be used.

Sec. 6.3

Molecular Diffusion

in

Liquids

401

Table

Atomic and Molar Volumes at the Normal Boiling Point

6.3-2.

Material

Atomic Volume 3 {m 3 /kg mol) 10

M oterial

Atomic Volume 3 {m /kg mol) 10 3

14.8

Ring, 3-membered

-6

3.7

as in ethylene

c H

0

(except as below)

7.4

Doubly bound

7.4

as

oxide

carbonyl

Coupled

to

two

other elements

In aldehydes, ketones

7.4

In methyl esters

9.1

In methyl ethers

9.9

In ethyl esters

9.9

In ethyl ethers

9.9

-8.5 -11.5

Anthracene ring

-47.5

-15 -30

Molecular Volume (m 3 /kg mol) JO 2

In higher esters

11.0

In higher ethers

11.0

Air

29.9

(—OH)

12.0

o2

25.6

N

31.2

In acids

Joined to

N

S, P,

8.3

N Doubly bonded

15.6

In primary amines

10.5

In secondary

amines

CI in

RCHC1R'

24.6

RC1

21.6

(terminal)

53.2

Cl 2

48.4 30.7 34.0

2

H2 H zO H S

27.0 in

2

Br 2

CO co

12.0

Br CI

4-membered 5-membered 6-membered Naphthalene ring

14.3 18.8

32.9

2

I

37.0

NH NO

S

25.6

N 0

36.4

P

27.0

so 2

44.8

F

8.7

25.8

3

23.6

2

Source: G. Le Bas, The Molecular Volumes of Liquid Chemical Compounds.

New York: David McKay

Co., Inc.,

1915.

When

water

is

the solute, values

from Eq.

(6.3-9)

should be multiplied by a factor of

mean deviation of 10-15% for nonaqueous solutions. Outside the range 278313 K, the equation should be used with caution. For water as the diffusing solute, an equation by Reddy and Doraiswamy is preferred (R2). Skelland (S5) summarizes the correlations available for binary systems. Geankoplis (G2) discusses and gives an equation to predict diffusion in a ternary system, where a dilute solute A is diffusing in a Equation (6.3-9) predicts aqueous solutions and about 25% 1/2.3 (Rl).

mixture of B and

C solvents. This

EXAMPLE

63-2.

diffusivities

with a

in

case

is

often approximated in industrial processes.

Prediction of Liquid Diffusivity

Predict the diffusion coefficient of acetone (CH 3

COCH

3)

in

water

50°C using the Wilke-Chang equation. The experimental value _9 2 10 m /sat25°C(298 K). Solution:

From Appendix A.2

3 0.8937 x 10"

402

Pa

s

and

the viscosity of water at 3 at 50°C, 0.5494 x 1QFrom .

Chap. 6

at is

25° and 1.28 x

25.0°C is /j b = Table 6.3-2 for

Principles of Mass Transfer

CH COCH 3

3

VA =

+

with 3 carbons 3(0.0148)

+

6

hydrogens

6(0.0037)

+

+

1

oxygen,

=

1(0.0074)

0.0740

m 3 /kg

mol

M

For water the association parameter (p = 2.6 and B = 18.02 kg mass/kg mol. For 25°C, T = 298 K. Substituting into Eq. (6.3-9),

D AB =

^

(1.173 x 10-

Mb)

i/2_1_ hB* A

Q-' 6 (1.173 x ! X2.6 x 10- 3 x (0.8937

18.02)

1/2

(298)

6

X0.0740)°'

=

1.277 x 10"

9

m

2

/s

For50°Cor T = 323 K, 16 (1.173 x 1Q)(2.6

AB

~ =

(0.5494 x 10 9 2.251 x 10"

ICQ

Electrolytes in solution such as

more

x

18.02)

1/2

_3

(323)

6

X0.0740)°-

m 2/s

dissociate into cations and anions and diffuse

rapidly than the undissociated molecule because of their small size. Diffusion

coefficients can be estimated using ionic conductance at infinite dilution in water. Equations and data are given elsewhere (S5, T2). Both the negatively and positively charged ions diffuse at the same rate so that electrical neutrality is preserved.

MOLECULAR DIFFUSION IN BIOLOGICAL SOLUTIONS AND GELS

6.4

6.4A 1.

Diffusion of Biological Solutes in Liquids

Introduction.

(e.g.,

proteins) in

biological systems

processing

is

The

diffusion of small solute molecules and especially macromolecules

aqueous solutions are important

and

in the life

in

the processing

and storing of

processes of microorganisms, animals, and plants.

an important area where diffusion plays an important

liquid solutions of fruit juice, coffee,

and

tea,

role. In the

Food

drying of

water and often volatile flavor or aroma

constituents are removed. These constituents diffuse through the liquid during evaporation.

In fermentation processes, nutrients, sugars, oxygen, and so on, diffuse to the microorganisms and waste products and at times enzymes diffuse away. In the artificial kidney machine various waste products diffuse through the blood solution to a membrane and then through the membrane to an aqueous solution.

Macromolecules in solution having molecular weights of tens of thousands or more were often called colloids, but now we know they generally form true solutions. The macromolecules in solution is affected by their large sizes and shapes, which can be random coils, rodlike, or globular (spheres or ellipsoids). Also,

diffusion behavior of protein

interactions of the large molecules with the small solvent and/or solute molecules affect

the diffusion of the macromolecules

and also of the small solute molecules.

Besides the Fickian diffusion to be discussed here, mediated transport often occurs in biological

systems where chemical interactions occur. This

latter

type of transport will

not be discussed here.

Sec. 6.4

Molecular Diffusion

in

Biological Solutions

and Gels

403

2.

Protein macromolecules are very large com-

Interaction and "binding" in diffusion.

pared to small solute molecules such as urea, KC1, and sodium caprylate, and often have

number

a

example

of sites for interaction or "binding" of the solute or ligand molecules.

the binding of oxygen to

is

hemoglobin

in the

Human

blood.

An

serum albumin

protein binds most of the free fatty acids in the blood and increases their apparent

Bovine serum albumin, which

solubility.

is in

mol sodium caprylate/mol

milk, binds 23

3 albumin when the albumin concentration is 30kg/m solution and the sodium caprylate is about 0.05 molar (G6). Hence, Fickian-type diffusion of macromolecules and small

solute molecules can be greatly affected by the presence together of both types of

molecules, even in dilute solutions.

3.

Experimental methods to determine

Methods

diffusiuity.

to determine the diffusivity

of biological solutes are similar to those discussed previously in Section 6.3 with modifications. In the

diaphragm diffusion

cell

shown

in

Fig. 6.3-1, the

chamber

some

made

is

of Lucite or Teflon instead of glass, since protein molecules bind to glass. Also, the

porous membrane through which the molecular diffusion occurs acetate or other polymers (G5, G6,

4.

K

Experimental data for biological solutes.

on protein

ture

diffusivities

is

composed

of cellulose

1).

Most

of the experimental data in the litera-

have been extrapolated to zero concentration since the

diffusivity is often a function of concentration.

A

tabulation of diffusivities of a few

proteins and also of small solutes often present in biological systems

is

given in Table

6.4-1.

The

diffusion coefficients for the large protein molecules are of the order of

nitude of 5 x 10

_u m 2 /s compared

solutes in Table 6.4-1. This

slow

as small solute

Table

to the values of about

means macromolecules

x 10~ 9

1

diffuse at a rate

mag-

for the small

about 20 times as

molecules for the same concentration differences.

Diffusion Coefficients for Dilute Biological Solutes in

6.4-1.

m 2 /s

Aqueous Solution

Temperature Molecular Weight

Diffusivity

Solute

Urea

°C

K

(m J /s)

20

293

9 1.20 x 10~

25

298

1.378

9 0.825 x 10" 9 1.055 x 10~

Glycerol

20

293

Glycine

25

298

Sodium caprylate

25

298

Bovine serum albumin Urease

25

60.1

x lO" 9

Kef.

(N2) (G5)

92.1

(G3)

75.1

(L3)

166.2

(G6)

298

10 8.78 x 10" -11 6.81 x 10

67 500

(C6)

25

298

11 4.01 x 10"

482 700

(C7)

20

293

3.46 x 10"

Soybean protein

20

293

11 2.91 x 10"

Lipoxidase

20

293

5.59 x 10"

human

20

293

serum albumin

20

293

human

20

293

x 10" 5.93 x 10" 4.00 x 10-

Creatinine

37

Sucrose

37

310 310

20

293

Fibrinogen,

Human

y-Globulin,

404

1.98

11

(S6)

97 440

(S6)

11

339 700

(S6)

11

72 300

(S6)

11

153100

(S6)

113.1

(C8)

342.3

(C8)

u

9 1.08 x 10" 9 0.697 x 10" 0.460 x 10" 9

Chap. 6

(S6)

361800

(P3)

Principles

of Mass Transfer

When coefficient

the concentration of

would be expected

macromolecules such as proteins

increases, the diffusion

to decrease, since the diffusivity of small solute

molecules

show

decreases with increasing concentration. However, experimental data (G4, C7)

some

the diffusivity of macromolecules such as proteins decreases in in

that

cases and increases

other cases as protein concentration increases. Surface charges on the molecules

appear to play a role

When

in these

phenomena.

small solutes such as urea, KC1, and sodium caprylate, which are often

present with protein macromolecules in solution diffuse through these protein solutions,

polymer concentration (C7, G5, G6, N3). Experi-

the diffusivity decreases with increasing

mental data for the diffusivity of the solute sodium caprylate {A) diffusing through bovine

serum albumin

(P) solution

show

D AP

that the diffusivity

reduced as the protein (P) concentration reduction is due to the binding of A to due to blockage by the large molecules.

P

is

of

A

through

increased (G5, G6).

so that there

is

less free

A

A

P

markedly

is

large part of the

to diffuse.

The

rest

is

For predicting the diffusivity of small aqueous solution with molecular weights less than about 1000 or solute" 3 molar volumes less than about 0.500 m /kg mol, Eq. (6.3-9) should be used. For larger solutes the equations to be used are not as accurate. As an approximation the Stokes5.

Prediction of diffusivities for biological solutes.

solutes alone in

Einstein equation (6.3-8) can be used.

D AB =

—7^

6 9.96 x 1Q-' T

-77T73

<

6

^

8)

Probably a better approximate equation to use is the semiempirical equation of Poison which is recommended for a molecular weight above 1000. A modification of his

(P3),

equation to take into account different temperatures

is

as follows for dilute aqueous

solutions.

9.40 x io-'

where

MA

is

5

r

the molecular weight of the large molecule A.

When

the shape of the

molecule deviates greatly from a sphere, the equation should be used with caution.

EXAMPLE

Prediction of Diffusivity of Albumin

6.4-1.

Predict the diffusivity of bovine

serum albumin

at

298

K in

solution using the modified Poison equation (6.4-1) and

experimental value

is

•

s

and

T=

298 K. Substituting into Eq. 5

_9.40xl0-' T AB

lAMJ

= This value

6.

the

6.4-1

The molecular weight of bovine serum albumin (A) from Table = 67 500 kg/kg mol. The viscosity of water at 25°C is 0.8937 x

M

10" 3 Pa

Table

A

Solution: 6.4-1

in

water as a dilute

compare with

is 1

1%

7.70 x

1 '3

10-"

(6.4-1),

x 10~ 15 )298 3 3 (0.8937 x 10- X6750O)" (9.40

m 2 /s

higher than the experimental value of 6.81 x 10"

Prediction of diffusivity of small solutes

in

protein solution.

When

11

m 2 /s.

a small solute (A)

through a macromolecule (P) protein solution, Eq. (6.3-9) cannot be used for prediction for the small solute because of blockage to diffusion by the large molecules.

diffuses

The data needed

Sec. 6.4

to predict these effects are the diffusivity

Molecular Diffusion

in Biological Solutions

D AB

and Gels

of solute

A

in

water alone,

405

on

the water of hydration

the protein, and an obstruction factor.

equation that can be used to approximate the diffusivity

P

protein

solutions

G6) and no binding

c

p

= D ab(1

.

D NA = where c A1 7.

is

concentration of A

in

^~ ^ c

kg mol /1/m 3

is

(6.4-3)

.

When A

is

Methods

where

A

DP

=

is

D AB (\

-

1.81

x l(T 3 c p )l "

Then Eq.

concentration of A

6.4B

P and

the solution

in

to predict this flux are available

% A\ — ——) +

%

ioo

»

Diffusion

in

in

(6.4-3)

is

is

is

bound A\ (6.4-4)

£>

iocT

\

the diffusivity of the protein alone in the solution,

not bound to the protein which

coefficient.

A

binding data have been experimentally obtained. The equation used

free

D AP

a protein solution

in

equal to the flux of unbound solute

plus the flux of the protein-solute complex.

when

(6.4-2)

is

Prediction of diffusivity with binding present.

binds to P, the diffusion flux of A

(G5, G6)

considered (C8, G5,

-1-81 x l(T 3 c p )

3 kg P/m Then the diffusion equation

=

is

A semi-theoretical A in giobular-type

present I>ap

where

of

as follows, where only the blockage effect

is is

D AP

m

2

/s;

and

free

A

is

that

determined from the experimental binding

used to calculate the flux where

cA

is

the total

the solution.

Biological Gels

Gels can be looked upon as semisolid materials which are "porous." They are composed of macromolecules which are usually

few wt

in dilute

aqueous solution with the

gel

comprising a

% of the water solution. The "pores" or open spaces in the gel structure are filled

with water.

The

rates of diffusion of small solutes in the gels are

somewhat

less

than

in

aqueous solution. The main effect of the gel structure is to increase the path length for diffusion, assuming no electrical-type effects (S7). Recent studies by electron microscopy (L4) have shown that the macromolecules of gel agarose (a major constituent of agar) exist as long and relatively straight threads. the This suggests a gel structure of loosely interwoven, extensively hydrogen-bonded polysaccharide macromolecules. Some typical gels are agarose, agar, and gelatin. A number of organic polymers exist as gels in various types of solutions. To measure the diffusivity of solutes in gels, unsteady-state methods are used. In one method the gel is melted and poured into a narrow tube open at one end. After solidification, the tube is placed in an agitated bath containing the solute for diffusion. The solute leaves the solution at the gel boundary and diffuses through the gel itself. After a period of time the amount diffusing in the gel is determined to give the diffusion coefficient of the solute in the gel. A few typical values of diffusivities of some solutes in various gels are given in Table 6.4-2. In some cases the diffusivity of the solute in pure water is given so that the decrease in diffusivity due to the gel can be seen. For example, from Table 6.4-2 at 278 K, urea in water has a diffusivity of 0.880 x 10" 9 m 2 /s and in 2.9 wt % gelatin, has a value of _9 2 0.640 x 10 m /s,adecreaseof27%.

406

Chap. 6

Principles of Mass Transfer

Table

Typical Diffusivities of Solutes

6.4-2.

in Dilute Biological

Gels

in

Aqueous Solution Temperature

Wt % Gel

Solute

Sucrose

in

Diffusivity

K

Solution

(m 2 /s)

"C

278

0

Gelatin

Urea

Gel

Ref.

0.285 x 10"

5

9

10~~ 9

3.8

278

5

0.209 x

10.35

278

5

9 0.107 x 10"

5.1

293

20

0.252 x 10"

9

0

278

5

0.880 x 10"

9

Gelatin

5.1

278

5

0.644 x 10" 9 9 0.609 x 10"

10.0

278

5

0.542 x 10"

5.1

293

20

2.9

278

5

9

9 0.859 x 10"

Methanol

Gelatin

3.8

278

5

0.626 x 10"

9

Urea

Agar

1.05

278

5

0.727 x 10"

9

3.16

278

5

0.591 x 10"

9

5.15

278

5

0.472 x 10"

9

(F2) (F2) (F2)

(F3) (F2)

(F2) (F3) (F2)

(F3) (F3) (F3)

(F3) (F3)

Glycerin

Agar

2.06

278

5

6.02

278

5

9 0.297 x 10" 9 10" 0.199 x

Dextrose

Agar Agar Agar Agarose

0.79

278

5

9 0.327 x 10"

(F3)

0.79

278

5

9 0.247 x 10"

(F3)

5.15

278

5

0.393 x 10"

0

298

25

1.511 x 10

2

298

25

1.398 x 10"

Sucrose

Ethanol

NaCl

M)

(0.05

In both agar linearly with

and

or batches of the

-9 9

(F3)

(F3) (S7)

(S7)

gelatin the diffusivity of a given solute decreases approximately

an increase

smaller than that

9

(F3)

in

wt

%

gel.

However, extrapolation to

0%

gel gives a value

shown for pure water. It should be noted that in different preparations same type of gel, the diffusivities can vary by as much as 10 to 20%.

EXAMPLE

6.4-2.

Diffusion

of Urea

in

Agar

tube or bridge of a gel solution of 1.05 wt agar in water at 278 K is 0.04 long and connects two agitated solutions of urea in water. The urea concentration in the first solution is 0.2 g mol urea per liter solution and 0 in

%

A

m

the other. Calculate the flux of urea in kg mol/s

•

m2

at

steady state.

From„Table 6.4-2 for the solute urea at 278 K, D AB = 0.727 x 10" 9 m 2 /s. For urea diffusing through stagnant water in the gel, Eq. (6.3-3) can be used. However, since the value ofx^ is less than about 0.01, the solution is quite dilute and x BM = 1.00. Hence, Eq. (6.3-5) can be used. The 3 concentrations are c A1 = 0.20/1000 = 0.0002 g mol/cm 3 = 0.20 kg mol/m and c A1 = 0. Substituting into Eq. (6.3-5),

Solution:

D ^(.c AX - c A1 _ NA = " z 2 -z, )

=

Sec. 6.4

3.63

Molecular Diffusion

9 0.727 x 10- (0.20

-

0)

0.04-0

x 10" 9 kg mol/s-m 2

in

Biological Solutions

and Gels

407

MOLECULAR DIFFUSION

6.5

IN SOLIDS

Introduction and Types of Diffusion in Solids

6.5A

and solids in solids are generally slower mass transfer in solids is quite important in chemical and biological processing. Some examples are leaching of foods, such as soybeans, and of metal ores; drying of timber, salts, and foods; diffusion and catalytic reaction in solid catalysts; separation of fluids by membranes; diffusion of gases through polymer films Even though

than rates

used

in

rates of diffusion of gases, liquids,

in liquids

and

gases,

packaging; and treating of metals

We can

broadly classify transport

at

high temperatures by gases.

in solids

into

two types of diffusion: diffusion

that

can be considered to follow Fick's law and does not depend primarily on the actual structure of the solid, and diffusion in porous solids where the actual structure and void

channels are important. These two broad types of diffusion will be considered.

1.

Law

Diffusion in Solids Following Fick's

6.5B

This type of diffusion

Derivation of equations.

dissolved leaching,

through

The

in solids

does not depend on the actual

when the fluid or solute diffusing is actually in the solid to form a more or less homogeneous solution for example, in where the solid contains a large amount of water and a solute is diffusing

structure of the solid.

diffusion occurs

—

this solution,

or in the diffusion of zinc through copper, where solid solutions are

hydrogen through rubber, or

present. Also, the diffusion of nitrogen or

in

some

cases

diffusion of water in foodstuffs can be classified here, since equations of similar type can

be used. Generally, simplified equations are used. Using the general Eq. (6.2-14) for binary diffusion,

dx

*

c

N A = -cD AB -+ + ^(N A + N B the bulk flow term,

quite small. Hence,

{c Afc){N A it

is

+

NB

),

even

if

neglected. Also, c

(6.2-14)

)

c

clz

present,

is

usually small, since

assumed constant giving

is

cjc orx A

is

for diffusion in

solids,

NA = where

D AB

is

diffusivity

independent of pressure

in

m 2 /s

for solids.

of

D

.„

dc (6-5-1)

~j dz

A through B and usually that D AB + D BA in solids.

is

assumed constant

Note

Integration of Eq. (6.5-1) for a solid slab at steady state gives

NA =

gdgifdlUfdij z2

For the case of diffusion and length L,

radially

(63.2)

- z,

through a cylinder wall of inner radius

r,

and outer

r2

N A =D AB (c A1 -c A2 408

2tiL

(6.54)

)

In (r 2 /r,)

Chap.

6

Principles

of Mass Transfer

This case

is

similar to conduction heat transfer radially through a hollow cylinder in Fig.

4.3-2.

.The diffusion coefficient D AB in the solid as stated above is not dependent upon the pressure of the gas or liquid on the outside of the solid. For example, ifC0 2 gas is outside a slab of rubber pA is

,

and

the partial pressure of

diffusing through the rubber,

is

C0

directly proportional to p A

2 .

at the surface.

This

is

The solubility

The

solubility of a solute gas {A) in a solid

of 0°C and

(STP)/atm

•

cm 3

1

3

solid per

solid in the cgs system.

3 kg mol /1/m using SI

c A.

C0

be independent of

2 in the solid,

however,

02

To

is

atm

02

in water

by Henry's law. 3 usually expressed as 5 in m solute in the air

partial pressure of (A). Also, S

convert this toe,, concentration

(at

= cm 3

in the solid in

units,

m 3 (STP)/m 3 = S r -

Using cgs

m

atm) per

of

similar to the case of the solubility of

being directly proportional to the partial pressure of

STP

D AB would

m

22.414

3

atm

solid

(STP)/kg mol

atm

p,, A

A

Sp A

=

22.414

kg mol

m

; 3

A (6.5-5)

solid

units,

Ca=

2^14 cm 3

(6

^

6)

solid

EXAMPLE 6.5-1. Diffusion of H 2 Through Neoprene Membrane The gas hydrogen at 17°C and 0.010 atm partial pressure is diffusing through a membrane of vulcanized neoprene rubber 0.5 mm thick. The pressure of H 2 on the other side of the neoprene is zero. Calculate the steady-state flux, assuming that the only resistance to diffusion

is

in the

membrane. The solubility S of H 2 gas in neoprene at 17°C is 0.051 m 3 (at -1 ° STP of 0°C and atm)/m 3 solid -atm and the diffusivity is 1.03 x 10 1

m

2

/satl7°C.

A sketch showing the concentration is shown in Fig. 6.5-1. The equilibrium concentration c at the inside surface of the rubber is, from Eq. Solution:

(6-5-5),

•"'

22.414

= 2 28xlO ^ i= 2^Mi0) 22.414 '

Al

-

.

Since p A2 at the other side

is 0, c A2

=

0.

5

kg mol

H 2 /m 3

solid

Substituting into Eq. (6.5-2) and

solving,

NA _ =

FIGURE

6.5-1.

°ab(c ai z2 4.69

-

-c A2 _ )

z,

x 10"

Concentrations

(L03 x 10-'°K2.28x 10-

for

(0.5 12

kg mol

Example

H 2 /s-m

PA

j

-

5

-0)

oyiooo

2

'A1

6.5-1.

Sec. 6.5

Molecular Diffusion

in Solids

409

2.

Permeability equations for diffusion in

solids.

many cases

In

the experimental data for

and solubilities but. as per3 meabilities, P M in m of solute gas A at STP (0°C and 1 atm press) diffusing per second 2 per m cross-sectional area through a solid 1 m thick under a pressure difference of 1 atm

diffusion of gases in solids are not given as diffusivities ,

pressure. This can be related to Fick's equation (6.5-2) as follows.

N A = gigifdi-Zf^ From

Eq.

(6.5-2)

(6.5-5),

s Pai Al

SPa (65-7)

AZ

22.414

22.414

Substituting Eq. (6.5-7) into (6.5-2),

DabS(Pa\ — Pai) = Pm{Pai-Pai), -kgmol/s-m N * = ~^r777, T ^T77^ ,,

22.414(z 2

where the permeability F M

-

m 3 (STP) '

is

is

(6.5-9)

; -atm/m .

C.S.

s- rn

Permeability

stcot (65-8)

2

z,)

is

Pm = Dab S

the permeability

-

22.414(z 2

Z()

also given in the literature in several other ways.

given as P'Si cc(STP)/(s ,

•

cm 2 C.S.

atm/cm). This

is

For

the cgs system,

related to

P M by

Pm = W~*P'm In

some

cm Hg/mm

cases

in

(6-5-10)

the literature the permeability

thickness). This

is

is

given as P"M cc(STP)/(s ,

cm 2 C.S.

P M by

related to

P M =7.60 x;irr 4 P«'

When

there are several solids

(65-11)

series

in

2, 3,

1,

•

and L,, L 2

,

represent the

thickness of each, then Eq. (6.5-8) becomes

N =

Pai

~

Pa2

,

LJP UI +LJP U2 +

22.414

where p A 3.

,

—

p A2

Experimental

is

the overall partial pressure difference.

diffusivities, solubilities,

fusivities in solids

(6.5-12)

---

is

Accurate prediction of

and permeabilities.

dif-

generally not possible because of the lack of knowledge of the theory

of the solid state. Hence, experimental values are needed. diffusivities, solubilities,

and permeabilities are given

and solids diffusing in solids. For the simple gases-such as He,

in

Some

experimental data for

Table

6.5-1 for gases diffusing in

C0 2

with gas pressures up to

solids

H2 02 N2 ,

,

,

and

,

1

or 2 atm, the solubility in solids such as polymers and glasses generally follows Henry's

law and Eq.

(6.5-5) holds. Also, for these gases the diffusivity

independent of concentration, and hence pressure. For the the In

H2

,

is

PM

and permeability are

effect of

temperature

T

in

K,

approximately a linear function of 1/T. Also, the diffusion of one gas, say approximately independent of the other gases present, such as 0 2 and 2 is

For metals such as Ni, Cd, and

N

Pt,

where gases such as H 2 and

02

are diffusing,

.

it is

found experimentally that the flux is approximately proportional to( v/p^ — ^/p A2 ), so Eq. (6.5-8) does not hold (B5). When water is diffusing through polymers, unlike the

410

Chap. 6

Principles of Mass Transfer

Table

6.5-1.

Diffusivities

and Permeabilities

D AB

in Solids

,

Solubility. S Vm^soluteiSTP)!

Diffusion Coefficient

Solute

W H

2

[m 2 /s-]

T(K)

Solid (B)

Vulcanized

L

-9

m 3 solid -atm

Permeability, 3

PM

V m solute{STP)~\

J

[_

s

m2

atm/m J

Ref.

10

)

0.040

0.342(10"

)

(B5)

)

0.070

0.152(10-'°)

(B5)

)

0.035

0.054(10-'°)

(B5)

)

0.90

1.01(10-'°)

(B5)

298

0.85(10

298

0.21(10~

9

298

0.15(10"

9

298

o.ikio"

9

290

0.103(10~ 9 )

300

0.180(10-

rubber

o2 N2 C0 2 H2

Vulcanized

0.051

(B5)

neoprene

H2 o2

Polyethylene

N,

o2 N2

Nylon

Air

English

9

0.053

)

(B5)

298

6.53(10-

12

303

4.17(10"

12

303

1.52(10" 12 )

303

0.029(10"

303

0.0152(10"

298

4 0.15-0.68 x 10"

(B5)

(B5)

)

(R3)

)

(R3)

(R3)

12

(R3)

)

12

(R3)

)

leather

H20

Wax

306

0.16(10"'°)

H2 0

Cellophane

311

0.91-1.82(10"

He

Pyrex glass

293

4.86(10"

15

373

20.1(10"

15

He

293

2.4-5.5(10"

H2

Si0 2 Fe

293

2.59(10-

Al

Cu

293

1.3(10"

simple gases,

14

EXAMPLE

in

(B5)

)

)

(B5)

)

(B5)

0.01

)

(B5)

13

(B5)

)

34

(B5)

)

P M may depend somewhat on

Further data are available

10

the relative pressure difference (C9, B5).

monographs by Crank and Park (C9) and Barrer

(B5).

Packaging Film Using Permeability A polyethylene film 0.00015 m (0.15 mm) thick is being considered for use in packaging a pharmaceutical product at 30°C. If the partial pressure 0fO 2 outside is 0.21 atm and inside the package it is 0.01 atm, calculate the 6.5-1. diffusion flux of 2 at steady state. Use permeability data from Table Assume that the resistances to diffusion outside the film and inside are negligible compared to the resistance of the film. 6.5-2.

Diffusion Through a

0

Solution:

From Table

6.5-1

atm/m). Substituting into Eq. xt

A

_

Note for

Sec. 6.5

that a film

-

2.480 x 10"

made

z

x

10

4-17(10"

)

Solids

12

kg mol/s-m

of nylon has a

in

)

m3

)(0-21

22.414(0.00015

much

O z and would make a more suitable Molecular Diffusion

12

4.17(10"

solute(STP)/(s

m2

•

(6.5-8),

P M {p Ai ~Pa2) _ 22.414(z 2

=

PM =

-0.01)

-

0)

2

smaller value of permeability P M

barrier.

411

4.

Membrane separation processes. In Chapter membrane separation processes of gas

the various

13 a detailed discussion

is

given of

separation by membranes, dialysis,

reverse osmosis, and ultrafiltration.

6.5C

Depends on Structure

Diffusion in Porous Solids That

porous solids. In Section 6.5B we used Fick's law and treated homogeneous-like material with an experimental diffusivity D Ag In this section we are concerned with porous solids that have pores or interconnected voids in the solid which affect the diffusion. A cross section of such a typical porous solid 1.

Diffusion of liquids

in

the solid as a uniform

is

shown

.

in Fig. 6.5-2.

For the situation where the voids are filled completely with liquid water, the concentration of salt in water at boundary 1 is c Ai and at point 2 isc^j. The salt in diffusing through the water in the void volume takes a tortuous path which is unknown and greater than (z 2 — zj by a factor t, called tortuosity. Diffusion does not occur in the inert solid. For a dilute solution using Eq. (6.3-5) for diffusion of salt in water at steady state,

NA =

eD

^-°^ -

(63-13)

z,)

t(z 2

where e is the open void fraction, D AB is the diffusivity of salt in water, and x is a factor which corrects for the path longer than (z 2 — zj. For inert-type solids t can vary from about 1.5 to 5. Often the terms are combined into an effective diffusivity.

D Ac „ = ~D AB

m 2 /s

(6.5-14)

T

EXA MPLE A

6J-3. Diffusion sintered solid of silica 2.0

and

ofKCl in Porous Silica

mm thick

a tortuosity x of 4.0.

The pores

face the concentration of

KC1

by the other

face.

is

porous with a void fraction

e

of 0.30

are filled with water at 298 K. At one

held at 0.10 g mol/liter, and fresh water Neglecting any other resistances but that in the porous solid, calculate the diffusion of KC1 at steady state. flows rapidly

The

Solution: 1.87 x

10" 9

kg mol/m 3

,

m

in

water from Table 6.3-1

Also, c Ai = 0.10/1000 = 1.0 x 10~* g and c A2 = 0. Substituting into Eq. (6.5-13), /s.

iDab(c a1 x(z 2

= 6.5-2.

KC1

2

NA

Figure

of

diffusivity

is

Sketch

7.01 x

of a

-

-

c A2 )

z,)

9 0.30(1.870 x 1Q- X0.10

4.0(0.002

-

is

D AB =

mol/cm 3 =

-

0.10

0)

0)

10- 9 kg mol KCl/s-m 2

typical

porous

solid.

z,

412

Chap. 6

z2

Principles of Mass Transfer

2.

Diffusion of gases in porous solids.

gases, then a

somewhat

diffusion occurs only

shown

If the voids

in Fig. 6.5-2 are filled with

similar situation exists. If the pores are very large so that

by Fickian-type diffusion, then Eq. £ D AB (c Ai KJ = N A -

C A2 )

=

2 i)

t(z 2

becomes, for gases,

(6.5-13)

—tKT^-z,) — — — £D AB (j) Ai

p A2 )

Ciri (63-15) ,/r

Again the value of the tortuosity must be determined experimentally. Diffusion

assumed

to occur only through the voids or pores

and

is

through the actual solid

riot

particles.

A

versus the void fraction of various unconsolidated

correlation of tortuosity

porous media of beds of glass spheres, sand,

approximate values of E

=

= 1.65. When the pores

t for

salt, talc,

different values of

e: s

and so on

=

0.2,

t

(S8), gives the

=

2.0; s

=

0.4,

following t

=

1.75;

0.6, t

are quite small in size and of the order of magnitude of the

free path of the gas, other types of diffusion occur,

6.6

which are discussed

in

mean

Section 7.6.

NUMERICAL METHODS FOR STEADY-STATE

MOLECULAR DIFFUSION

IN

TWO DIMENSIONS Derivation of Equations for Numerical Method

6.6A /.

Derivation of method for steady state.

with unit thickness

is

In Fig. 6.6-1 a two-dimensional solid

divided into squares.

The numerical methods

shown

for steady-state

molecular diffusion are very similar to those for steady-state heat conduction discussed in Section 4.15. Hence, only a brief summary will be given here. The solid inside of a is imagined to be concentrated at the center of the square at c n m and is called "node," which is connected to the adjacent nodes by connecting rods through which the mass diffuses. A tota mass balance is made at steady state by stating that the sum of the molecular

square

a

1

diffusion to the

shaded area for unit thickness must equal zero.

U-n-l.m

^x

L n. mi

^

<

+

D

>

R

L n.m)

V-n+l.m

Ax (c„,

+

m+1 -c„.J

n,

m

+

1

D AR Ax

^r

-(c„, m .

m+

m-

Figure

Sec. 6.6

6.6-1:

1

-c„.J = 0

(6.6-1)

1

1

Concentrations and spacing of nodes for two-dimensional steady-state molecular diffusion.

Numerical Methods for Steady-Stale Molecular Diffusion

in

Two Dimensions 413

where

c nm

concentration of

is

A

node n,m

at

in

A/m 3

kg mol

Setting

.

Ax =

Ay, and

rearranging,

+

+c„. m _!

2. Iteration

equation

unknown

method of numerical written for each

is

c n+lim

solution.

unknown

+

c B _ lim

-4c n

m

.

=

0

In order to solve Eq.

(6.6-2)

(6.6-2),

For a hand calculation using a modest number of nodes, the iteration

points.

method can be used to solve the equations, where the right-hand side of Eq. equal to a residual

/V„

c„,

m

m+

+

i

c„.

(6.6-3)

and

c„,

m _,

Example

Cn

=

m

+

-i.m

c„

(Ax

=

Cn

Um -

4c„ im

for steady state

+l.m

+

C n.m+1

conduction

4.15-1 for steady-state heat

+

=

N„ m

and c„

m

is

(6.6-3)

calculated by

. , , „ (6.6-4)

^n.m-l

all

the

illustrates the detailed steps for the

to those for steady-state diffusion.

the concentrations have been calculated, the flux can be calculated for each 6.6-1, the flux for the

node

or element c„ m toc„ „_

Ay)

N = ^(c

-c

= D AB {c n

-c n<m _

where the area A

is

Ax

_

m

times

1

Equations

for Special

=

)

m deep

E^fflgg

,

-

c

c

)

(6.6-5)

1 )

N is kg mol

and

sum

other appropriate elements and the

6.6B

0

c„_

the final equations to be used to calculate

element as follows. Referring to Fig. is

=

+

state.

method, which are identical

Once

+

+1>m

'

(6.6-4) are

concentrations at steady

iteration

(6.6-2) is set

.

Setting the equation equal to zero, JV„ m

Equations

a separate

point giving TV linear algebraic equations for JV

A/s.

Equations are written

for the

of the fluxes calculated.

Boundary Conditions

for

Numerical

Method /.

Equations for boundary conditions.

When one

boundary where convective mass transfer the bulk fluid

shown

balance on the node

DAB Ay '

x

(c c n-l.m V

-cc

n.

a different equation must be derived.

in Fig. 6.6-2a

n,

m, where mass in

ml)

+ '

Dab AX 0 A 2 Ay

(c L n, \

is

Ax = Ay,

a mass

-cL n,m)

Ay

)

K m-l -Cn.m) = K ^n.

the convective mass-transfer coefficient in

Setting

Making

= mass out at steady state,

m+1

2

where k c

is

of the nodal points c„ m is at a occurring to a constant concentration c m in

m/s defined by Eq.

rearranging, and setting the resultant equation

m

~

O

(6.6-6)

(6.1-14).

=

JV„_

m , the

re-

sidual, the following results.

414

Chap. 6

Principles of Mass Transfer

(a)

Figure

(b)

6.6-2.

Different at a tive

1.

For convection k

Ax cx

This equation

at a

+ is

boundary conditions for steady-state

diffusion: (a) convection

boundary, (b) insulated boundary, (c) exterior corner with convecboundary, (d) interior corner with convective boundary.

boundary

{{2c n _

Um +

(Fig. 6.6-2a),

c„,

m+

+

,

c„.

m_

-

,)

Ax

(k

c„,

J -|— +

\ 2

=

7V„.

m

(6.6-7)

and convection with

similar to Eq. (4.15-16) for heat conduction

Ax/D AB being used in place of h Ax/k. Similarly, Eqs. (6.6-8)-(6.6-10) have been derived for the other boundary conditions shown in Fig. 6.6-2. For an insulated boundary (Fig. 6.6-2b), kc

2.

i(c„. 3.

+c„. m -

i

1

For an exterior corner with convection

^

cx

U AB

4.

For an

kc

Ax

+

j(c _ rt

+

)

+

c„_

at the

c„.

m _,)

1

m

.

_

Km +

c„,

m+1

+

\(c n+Um

+

cn

,

n.

m

-

+

-

/ 3

+

Ax\ —— k

shown

K=

Sec. 6.6

is

(6.6-9)

-

kr

When

in Fig. 6.6-3

m

(Fig. 6.6- 2d),

Boundary conditions with distribution coefficient. K between the liquid and the

distribution coefficient

n_

/

boundary

m _,)

lV. = N

AB

\

The

(6.6-8)

7V„. m

(Fig. 6.6-2c),

AB

distribution coefficient as

=

boundary

,

n

-2c

\

interior corner with convection at the

c„+c -j— U 2.

m+

L> AB

m

=

N„, m

(6.6-10)

J

Eq. (6.6-7) was derived, the

solid at the surface interface

was

1.0.

defined as

°-^±

Numerical Methods for Steady-State Molecular Diffusion

(6.6-11)

in

Tno Dimensions 415

Figure

Interface concentrations for conveciive mass transfer at a solid surface

6.6-3.

and an equilibrium distribution

coefficient

K=

c„_

m Jcntm

.

where c„ mL is the concentration in the liquid adjacent to the surface and c n m is the concentration in the solid adjacent to the surface. Then in deriving Eq. (6.6-6), the

—

right-hand side k c Ay(c„ m

c^) becomes

K &Ac where

c ro is

- ^W* Ay(^ K) K

c

Hence, whenever k c appears as

cJK should

For convection

KK

Ax\

-cJ

mL

(6-6-12)

Kcn

m for c„ mL from Eq.

- '-A

(6.6-13)

and multiplying and dividing by K,

Kk

1.

.

the concentration in the bulk fluid. Substituting

(6.6-1 l)into(6. 6-12)

appears,

n

be used.

at a

in

Eq.

(6.6-7),

Then Eq.

boundary

(6.6-7)

Ay(c „ J n m "•

e

Kk

K_

,

\

'

c

should be substituted and when c c

becomes

as follows.

(Fig. 6.6-2a),

c„

v + D AB J K i

2

(2c„ -

1.

m

+

c„

m+

+

|

c„ m _

J

(Kk Ax

-c„.^-^- +

\ 2j

=

N„, m

(6.6-14)

(6.6-9) and (6.6-10) can be rewritten in a similar manner as follows, For an exterior corner with convection at the boundary (Fig. 6.6-2c),

Equations 2.

Kk Ax\ c a AB J K r

3.

For an

KK

Ax\

(Kk Ax

\

c

,

D AB

\

interior corner with convection at the

boundary

j (Fig. 6.6-2d),

c„

DXS

"t"

Cn -

1 ,

m

"1"

C n,

+

EXAMPLE

m+

1

ita, +

,

.

m

+

c..

m.

, )

-

/ (

3

+

K/c c

Ax\

km=

JV,,

m

(6.6-16)

Numerical Method for Convection and Steady-State Diffusion For the two-dimensional hollow solid chamber shown in Fig. 6.6-4, determine the concentrations at the nodes as shown at steady state. At the inside surfaces the concentrations remain constant at 6.00 x 10~ 3 kgmol/m 3 At 6.6-1.

.

the outside surfaces the convection coefficient k c

2.00 x 10"

416

3

kg mol/m 3 The .

=

diffusivity in the solid

Chap. 6

is

-7

m/s and c m = D AS = 1.0 x 10" 9

2 x 10

Principles

of Mass Transfer

-3 .-6.00 X 10"

©

A

1x

2

1

^»

1,3

1,4

^

\

2,2

,1

c

=2.00 X 10"

J§—

Figure

m

2

/s.

The

3,2

grid size

To

Solution:

i

2,3

i

i

2 ,5

—

(

3,4

3,3

I—

_1_ \«Ax+\

3,5

Concentrations for hollow chamber for Example 6.6- J.

6.6-4.

per 1.0-m depth.

£


2,4

—X

3,1

V»

v./

Ax = Ay =

is

The

0.005 m. Also, determine the diffusion rates

K=

distribution coefficient

simplify the calculations,

1.0.

the concentrations will be multi-

all

by 10 3 Since the chamber is symmetrical, we do the calculations on the I shaded portion shown. The fixed known values arec! 3 = 6.00, c 14 = = 2.00. Because of symmetry, c, 2 = c 2 3 c 2> 5 = c 2 3 c 2 = c 3 2 6.00, c 3. 3 = c 3. 5 To speed up the calculations, we will make estimates of the plied

.

,

_

,

,

t

•

unknown concentrations c3

,

=

as follows: c 2

=

=

,

.

=

2

=

"

2.50, c 3 2 2.70, c 3 3 3.00, c 3>4 For the interior points c 2 2 , c 23

,

3.80, c 2

3

=

4.20, c 2 4

=

4.40,

3.20.

and

c

21 we

use Eqs. (6.6-3) and

Eq. (6.6-9); for the other the corner convection point c 3 i convection points c 3 2 c 3 3 c 3 4 Eq. (6.6-7). The term k c Ax/D AB = (2 7 x 10~ )(0.005)/(1.0 x lO *) = 1.00. (6.6-4);

for

,

,

,

First approximation.

residual

N2

Starting with c 2

+ Q.2 + +

4.20

Hence, c 2

_

c 2>

2

_ _

2

we

use Eq. (6.6-3)

and

calculate the

2 is

2.70

+

C 2,

2.70

+

1

+

-

C 2-3

-

4.20

not at steady state. Next 2 from Eq. (6.6-4).

4C 2

,

2

= N2

2

4(3.80)

= - 1.40

N

to zero

we

set

4.20

+

2. 1

and

calculate a

of c 2

—

C| 2

new value Forc 2 3

This

,

2

Cj.2

new value

,

+

c3

+

2

c2

+

t

c2

3

-

2.70

2

2.70

+

4.20

_ -

4

4 of c 2

4-

replaces the old value.

,

+c 2 .4

C2.2

3.45

+

4.40

_ c 2 2 +c 2 -4 +c, C2.3--" t

Sec. 6.6

T

-

+ 3

'

+c,. 3 6.00

+c 3

+c 3

+ 3

,

3.00

3

-4c 2-3 =

N 2>3

- 4{4.20) = 0.05

3.45

+

4.40

+

6.00

+

3.00

4

Numerical Methods for Steady-State Molecular Diffusion

in

-4.21

Two Dimensions 417

For

c,

+

c 2 .3

+

4.21

+

For c 3

we

,,

c2

+

5

,

+

4.21

+

6.00

-

3.20

+

c 2|5

^2,4 = ^2,4

+ C 3-

c,, 4

+

4.21

+

l(c 2

+

, .

+

(1.0)2.00

c3

+

£(2.70

-

2)

.

+

(1.0

-

2.70)

+

(1.0)2.00 2

,

we

+

(1.0)c m

+

6.00

|(2.70

+

3.20

=

,

3

,

=

-0.30

,

=0

x

c3

=

,

2.35

_

use Eq. (6.6-7).

i(2 x

c2

.

2

+

c3

, .

+

c3

.

3)

-

(1.0

+

2)c 3>

2

= N 3> 2

+

^(2 x 3.45

+

2.35

+

3.00)

-

(3.0)2.70

=

0.03

(1.0)2.00

+

\{2 x 3.45

+

2.35

+

3.00)

-

(3.0)c 3- 2

=

0

3

4.41

,

(1.0)2.00

Forc 3

=

,,

- (2.0)c 3

2.70)

l)c 3

2.0(2.50)

Setting Eq. (6.6-9) to zero and solving for c 3

c3

+

4.21

0.02

use Eq. (6.6-9): (1.0)c m

For

=

4(4.40)

c3

_

=

2

2.71

,

(1.0)c m

+

(1.0)2.00

(1.0)2.00

(3.0)c 3 3-

= Nx 3

2.71

J+ 3.20) -

(3.0)3.00

=

0.17

+

2.71

+

3.20)

-

(3.0)c 3i 3

=

0

+

c3 3

+

c3

_

(3.0)c 3> 4

=

tf 3 4

1(2 x 4.41

+

3.06

+

3.06)

-

(3.0)3.20

=

-0.13

\(2 x 4.41

+

3.06

+

3.06)

-

(3.0)c 3 4

=

0

i

3

+

Cji 2

+

i(2 x 4.21

+

+

\{2 x 4.21

(1.0)c ro

+

|(2 x c 2

(1.0)2.00

+

(1.0)2.00

+

(2

x c2

,

+

c3

,

,

c3

.

3

=

3.06

F ° rC 3.4, ,

4

,

5)

,

3.16

Having completed one sweep across the grid map, we can start a second approximation using the new values calculated and starting with c 2 2 or any other node. This is continued until all the residuals are as small as desired. The final values after three approximations are c 2 2 = 3.47, c 2 3 = 4.24, _

c2 4

=

4.41,

To

=

3-1

,

2.36,c 3i2

=

2.72,c 3-3

calculate the diffusion rates

diffusion rate, leaving the

we

bottom surface

=

3.06,c 3t4

=

3.16.

calculate the total convective

first

at nodes c 3

„ c3

2

,

c3

3

,

and

c3 4

for 1.0-m depth.

N=

k c (Ax

1)

Hr^ +

(c 3>2

"2.36 (2 x

+ = Note

10-°X0.005 x

(3.06

-

2.540 x 10-

that the

first

12

-

1)

3.16 2.00)

- cj +

-

+

2.00

2.00

+

(c 3 3 .

- cj +

C co

£i 4 .

2

J

(2.72-2.00)

x 10"

kgmol/s

and fourth paths include only \ of a

Chap. 6

surface.

Next we

Principles of Mass Transfer

calculate the total diffusion rate in the solid entering the top surface inside,

using an equation similar to Eq.

N

+

-1. 3

Ay

x 10~ 9 X0.005 x

(1.0

(6.6-5).

1)

-

(6.00

4.24)

+

6.00-4.41

x 10"

0.005

=

2.555 x 10"

kg mol/s

At steady state the diffusion rate leaving by convection should equal that entering by diffusion. These results indicate a reasonable check. Using smaller grids would give even more accuracy. Note that the results for the

whole chamber.

diffusion rate should be multiplied by 8.0 for the

PROBLEMS of Methane Through Helium. A gas of CH 4 and He is contained in a kPa pressure and 298 K. At one point the partial pressure of methane is p Al = 60.79 kPa and at a point 0.02 m distance away, p A2 = 20.26 kPa. If the total pressure is constant throughout the tube, calculate the flux of CH 4 (methane) at steady-state for equimolar counterdiffusion. 2 5 2 6 Ans. J*, = 5.52 x 10" kg mol A/s m (5.52 x 10" g mol A/s cm )

6.1-1. Diffusion

tube

at

101.32

C0 2

Binary Gas Mixture. The gas CO, is diffusing at steady m long having a diameter of 0.01 m and containingN 2 at 298 K. The total pressure is constant at 101.32 kPa. The partial pressure of Hg at the other end. The diffusivity C0 2 at one end is 456 Hg and 76 D AB is 1.67 x 10" 5 m 2 /s at 298 K. Calculate the flux of 2 in cgs and SI units for equimolar counterdiffusion.

6.1- 2. Diffusion

of

in a

state through a tube 0.20

mm

mm

C0

6.2- 1.

Equimolar Counterdiffusion of a Binary Gas Mixture. Helium and nitrogen gas are contained in a conduit 5 in diameter and 0.1 m long at 298 K and a uniform constant pressure of 1.0 atm abs. The partial pressure of He at one end of the tube is 0.060 atm and 0.020 atm at the other end. The diffusivity can be obtained from Table 6.2-1. Calculate the following for steady-state equimolar

mm

counterdiffusion.

6.2-2.

He

kg mol/s

(b)

Flux of Flux of

(c)

Partial Pressure of

(a)

N

2

in

•

m2

and

g mol/s

•

cm 2

.

.

He at

a point 0.05

NH

m from

N

either end.

Steady State. Ammonia gas {A) 3 2 and nitrogen gas (B) are diffusing in counterdiffusion through a straight glass tube 2.0 ft (0.610 m) long with an inside diameter of 0.080 ft (24.4 mm) at 298 K and 101.32 kPa. Both ends of the tube are connected to large mixed chambers at 101.32 kPa. The partial pressure of kPa 3 in one chamber is constant at 20.0 and 6.666 kPa in the other chamber. The diffusivity at' 298 K and 101.32 kPa is 2 2.30 x 10- 5 m /s. Equimolar Counterdiffusion of

and

at

NH

(a)

(b) (c)

Calculate the diffusion ofNH 3 in lb mol/h and kg mol/s. Calculate the diffusion ofN 2 Calculate the partial pressures at a point 1.0 ft (0.305 m) in the tube and plot p A ,p B and P versus distance z. = 7.52 x 10" 7 lb mol A/\ (a) Diffusion of Ans. .

,

NH

9.48 x 10" (c)

Chap. 6

= B

3

kg mol A/s\

1.333 x 10

4

Pa

and Effect of Type of Boundary on A Through Stagnant Ammonia gas is diffusing through N 2 under steady-state conditions with

6.2-3. Diffusion

Flux.

pA

11

of

Problems

419

N

nondiffusing since

2

1.013 x 10

point

5

Pa and

1.333 x 10

is

it

is

4

Pa and

D AB for the mixture at

The

insoluble in one boundary.

the temperature

(a)

Calculate the flux of

(b)

Do

at the

is

other point 20

1.013 x 10

NH 3 in

K.The

298

5

Pa and 298

kg mol/s

•

m2

The

total pressure

partial pressure

mm

away

K is 2.30

it

ofNH 3

one

3

Pa.

is6.666 x 10

x 10"

5

m

is

at

2

/s.

.

assume that N 2 also diffuses; i.e., both boundaries are permeable to both gases and the flux is equimolar counterdiffusion. In which

same

the

case

is

as (a) but

the flux greater? "

Ans.

(a)A^

=

6 3.44 x 10"

kgmol/s

•

m

2

of Methane Through Nondiffusing Helium. Methane gas is diffusing in tube 0.1 m long containing helium at 298 K and a total pressure of 4 5 1.01325 x 10 Pa. The partial pressure of CH 4 at one end is 1.400 x 10 Pa and 3 1.333 x 10 Pa at the other end. Helium is insoluble in one boundary, and hence is nondiffusing or stagnant. The diffusivity is given in Table 6.2-1. Calculate the 2 flux of methane in kgmol/s m at steady state.

6,2-4.) Diffusion

a straight

•

6.2-5.

Mass

Transfer from a Naphthalene Sphere to Air. Mass transfer is occurring from a sphere of naphthalene having a radius of 10 mm. The sphere is in a large volume of still air at 52.6 C and 1 atm abs pressure. The vapor pressure of naphthalene at 52.6°C is 1.0 Hg. The diffusivity of naphthalene in air at 0°C 6 is 5.16 x 10~ m 2 /s. Calculate the rate of evaporation of naphthalene from the 2 surface in kg mol/s m [Note : The diffusivity can be corrected for temperature cr

mm

-

.

using the temperature-correction factor of the Fuller et

al.

Eq. (6.2-45).]

of Diffusivity of a Binary Gas. For a mixture of ethanol (CH 3 CH 2 OH) vapor and methane (CH 4 ), predict the diffusivity using the

6.2-6. Estimation

method of Fuller et al. 5 (a) At 1.0132 x 10 Pa and 298 and 373 K. 5 (b) At 2.0265 x 10 Pa and 298 K. Ans.

(a)

D AB =

1.43 x 10"

5

m 2 /s (298 K)

Flux and Effect of Temperature and Pressure. Equimolar counterdiffusoccurring at steady state in a tube 0.11 long containingN 2 and gases at a total pressure of 1.0 atm abs. The partial pressure ofN 2 is 80 Hg at one end and 10 at the other end. Predict the D AB by the method of Fuller et al. 2 (a) Calculate the flux in kg mol/s at 298 K for 2

6.2-7. Diffusion

ion

m

is

CO

mm

mm

•

(b) (c)

m

N

.

Repeat at 473 K. Does the flux increase? Repeat at 298 K but for a total pressure of 3.0 atm abs. The partial pressure of N 2 remains at 80 and 10 Hg, as in part (a). Does the flux change? -5 Ans. (a) D AB = 2.05 x 10 m 2 /s, N A = 7.02 x 10" 7 kg mol/s -m 2 7 (b) N A = 9.92 x 10" kg mol/s -m 2 7 10" (c) N kg mol/s -m 2 A = 2.34 x

mm

;

;

r \

~"~\

6.2-8J Evaporation Losses of Water in Irrigation Ditch. Water at 25°C is flowing in a covered irrigation ditch below ground. Every 100 ft there is a vent line 1.0 in. inside diameter and 1.0 ft long to the outside atmosphere at 25°C. There are 10 vents in the 1000-ft ditch. The outside air can be assumed to be dry. Calculate the

V

y

evaporation loss of water inlb^d. Assume that the partial pressure of water vapor at-the surface of the water is the vapor pressure, 23.76 Hg at 25°C. Use the diffusivity from Table 6.2-1. total

mm

6.2-9.

Time to Completely Evaporate a Sphere. A drop of liquid toluene is kept at a uniform temperature of 25.9°C and is suspended in air by a fine wire. The initial radius r, = 2.00 mm. The vapor pressure of toluence at 25.9°C isf^, = 3.84 kPa and the density of liquid toluene is 866 kg/m 3 (a) Derive Eq. (6.2-34) to predict the timef f for the drop to evaporate completely .

volume of still air. Show all steps. Calculate the time in seconds for complete evaporation.

in a large (b)

Ans.

420

(b)i f

Chap. 6

=

1388

s

Problems

Nonuniform Cross-Sectional Area. The gas ammonia (A) is diffusN 2 (B) by equimolar counterdiffusion in a conduit 1.22 m long at 25°C and a total pressure of 101.32 kPa abs. The partial pressure of ammonia at the left end is 25.33 kPa and 5.066 kPa at the other end. The cross section of the conduit is in the shape of an equilateral triangle, the length of each side of the triangle being 0.0610 m at the left end and tapering uniformally to 0.0305 m at the right end. Calculate the molar flux of ammonia. The diffusivity is D AB = 0.230 x 10 -4 m 2 /s.

6.2-10. Diffusion in a

ing at steady state through

of A Through Stagnant B in a Liquid. The solute HC1 (A) is diffusing thick at 283 K. The concentration of HC1 through a thin film of water (B) 2.0 at point 1 at one boundary of the film is 12.0 wt % HC1 (density p x = 1060.7 HC1 (p 2 = 1030.3 kg/m 3 ), and at the other boundary at point 2 it is 6.0 wt 9 in water is 2.5 x 10~ m 2 /s. Assuming kg/m 3 ). The diffusion coefficient of steady state and one boundary impermeable to water, calculate the flux of HC1 in

6.3-1. Diffusion

mm

%

HO

kgmol/s

•

m2

.

NA

Ans. 6.3-2. Diffusion

of Ammonia K and 4.0

in

an Aqueous Solution.

mm thick

= 2.372 x 10 6 kg mol/s m 2 An ammonia (/l)-water (B) solu•

contact at one surface with an organic liquid of ammonia in the organic phase is held constant and is such that the equilibrium concentration of ammonia in the water at this surface is 2.0 wt ammonia (density of aqueous solution is 991.7 kg/m 3 ) and the concentration of ammonia in water at the other end of the film 4.0 tion at 278

is

in

The concentration

at this interface.

%

mm

away

is

10 wt

%

(density of 961.7

kg/m 3 Water and

each other. The diffusion coefficient ofNH 3 in water (a) At steady state, calculate the flux N A in kg mol/s (b) Calculate the flux B Explain.

N

the organic are insoluble in

).

is

1.24

m

m

x 10~ 9

2 /s.

2 .

.

63-3. Estimation of Liquid Diffusivity.

It is

desired to predict the diffusion coefficient of

(CH 3 COOH) in water at 282.9 K and at 298 Wilke-Chang method. Compare the predicted values with the dilute acetic acid

K using the experimental

values in Table 6.3-1.

D AB =

Ans.

9 0.897 x 10"

m 2 /s (282.9

K);

D AB =

1.396

x 10" 9

m

2

/s

(298 K)

H

of Diffusivity of Methanol in 2 0. The diffusivity of dilute methanol 2 9 in water has been determined experimentally to be 1.26 x 10" m /s at 288 K. (a) Estimate the diffusivity at 293 K using the Wilke-Chang equation. (b) Estimate the diffusivity at 293 K by correcting the experimental value at 288 K to 293 K. (Hint : Do this by using the relationship/}^ oc T/n B .)

6.3- 4. Estimation

of Diffusivity of Enzyme Urease in Solution. Predict the diffusivity of enzyme urease in a dilute solution in water at 298 K using the modified Poison equation and compare the result with the experimental value in Table

6.4- 1. Prediction

the

Ans. Predicted 6.4-2. Diffusion

of Sucrose

%

in Gelatin.

A

D AB =

3.995 x 10"

layer of gelatin in water 5

mm

11

m

2

/s

thick contain-

K

separates two solutions of sucrose. The concentration of sucrose in the solution at one surface of the gelatin is constant at 2.0 solution and 0.2 g/100 at the other surface. Calculate the g sucrose/100 ing 5.1 wt

gelatin at 293

mL

flux of sucrose in 6.4-3. Diffusivity

kg sucrose/s

of Oxygen

in

•

mL m 2 through the gel at steady state.

Protein Solution.

of bovine serum albumin (BSA) at 298 K.

BSA. Predict protein/100

D

AF of oxygen in a protein solution containing 11 g in solution. (Note: See Table 6.3-1 for the diffusivity of 2

the diffusivity

mL

Oxygen is diffusing through a solution Oxygen has been shown to not bind to

0

water.)

Ans.

D AP =

1.930

x 10" 9

m

2

6.4-4. Diffusion of Uric Acid in Protein Solution and Binding. Uric acid (A) at 37°C diffusing in an aqueous solution of proteins (P) containing 8.2 g protein/100

/s

is

mL

Chap.

6

Problems

421

solution. Uric acid binds to the proteins

and over the range of concentrations

present, 1.0 g mol of acid binds to the proteins for every 3.0 g mol of total acid present in the solution. The diffusivity AB of uric acid in water is

D

x 10" 5 cm 2 /s and

1.21

cm 2 /s.

5 0.091 x 10" binding, predict the ratio

Assuming no

(a)

DP

=

D AP ID AB

due only

to

blockage

effects.

Assuming blockage plus binding effects, predict the ratio D AP /D AB Compare this with the experimental value for D AP ID AB of 0.616 (C8).

(b)

(c)

.

2

Predict the flux in g uric acid/s • cm for a concentration of acid of 0.05 g /L at point (1) and 0 g/L at point (2) a distance 1.5 pm away.

Ans.

NA

(c)

= 2.392 x 10~ 6

g/s

•

cm 2

/"""~~'~~\ .

Through Rubber. A flat plug 30 mm thick having an area of and made of vulcanized rubber is used for closing an opening in a container. The gas C0 2 at 25°C and 2.0 atm pressure is inside the container. Calculate the total leakage or diffusion of C0 2 through the plug to the outside in kg mol C0 2 /s at steady state. Assume that the partial pressure ofC0 2 3 outside is zero. From Barrer (B5) the solubility oftheC'0 2 gas is 0.90 m gas (at 3 STP of 0°C and 1 atm) per m rubber per atm pressure of C0 2 The diffusivity is

6.5-1. j Diffusion

of 10" 4

C0 2

m

y 4.0 x

2

.

0.11 x 10-

9

m

2

/s.

Ans. 6.5-2.

1.178

x 10"

13

kgmolC0 2 /s

Leakage of Hydrogen Through Neoprene Rubber. Pure hydrogen gas at 2.0 atm abs pressure and 27°C is flowing past a vulcanized neoprene rubber slab 5 mm thick. Using the data from Table 6.5-1, calculate the diffusion flux in kg mol/sm 2 at steady state. Assume no resistance to diffusion outside the slab and zero partial pressure of H 2 on the outside. Between Diffusivity and Permeability. The gas hydrogen is diffusing thick at 25°C. The partial pressure through a sheet of vulcanized rubber 20 of H 2 inside is 1.5 atm and 0 outside. Using the data from Table 6.5-1, calculate

6.5-3. Relation

mm

the following. (a)

The

diffusivity

with the value (b)

The

NA

flux

from the permeability Table 6.5-1.

solubility

S and compare

of H 2 at steady state.

Ans. 6.5-4.

P M and

D AB in

(b)N A

=

10 1.144 x 10"

kg mol/s

•

m

2

Loss from a Tube of Neoprene. Hydrogen gas at 2.0 atm and 27°C is flowing in a neoprene tube 3.0 outside diameter. Calculate inside diameter and 11 state. the leakage of H 2 through a tube 1.0 m long in kg mol 2 /s at steady

mm

mm

H

Through Membranes in Series. Nitrogen gas at 2.0 atm and 30°C is diffusing through a membrane of nylon 1.0 thick and polyethylene 8.0 thick in series. The partial pressure at the other side of the two films is 0 atm. Assuming no other resistances, calculate the flux N A at steady state.

6.5-5. Diffusion

mm

6.5-6. Diffusion

diffusion

276

K

of

C0 2

ofC0 2

and

a Packed Bed of Sand.

3

It

t is 0.30.

Pa and 0 Pa

The

x 10 5 Pa. The bed depth

CO z

partial pressure of

at the

bottom. Use a Ans. A

x

is

1.25

and the bed is

1.609 x 10"

9

kg mol

CO z /s m

2

to Keep Food Moist. Cellophane is being used to keep food moist at 38°C. Calculate the loss of water vapor in g/d at steady state for a wrapping 0.10 2 thick and an area of 0.200 when the vapor pressure of water vapor inside is 10 Hg and the air outside contains water vapor at a pressure of 5 Hg. Use the larger permeability in Table 6.5-1.

Packaging

mm

m

mm

mm

Ans.

422

m

at the top of the

of 1.87.

N =

6.5-7.

desired to calculate the rate of

is

gas in air at steady state through a loosely packed bed of sand at

a total pressure of 1.013

void fraction 2.026 x 10

in

mm

0.1667

Chap.

6

gH 2 Q/day. Problems

mm

4 65-8. Loss of Helium and Permeability. A window of Si0 2 2.0 thick and 1.0 x 10~ 2 area is used to view the contents in a metal vessel at 20°C. Helium gas at 14 202.6 kPa is contained in the vessel. To be conservative use D AB = 5.5 x 10"

m

m

2

/s

from Table

6.5-1.

Calculate the loss of He in kg mol/h at steady state. (b) Calculate the permeability P M and P"M Ans. (a)Loss = 8.833 x 1CT (a)

.

6.6-1.

kgmolHe/h from Example

Numerical Method for Steady-State Diffusion. Using the results 6.6-1, calculate the total diffusion rate in the solid using the bottom nodes and paths of c 2> 2 to c 3i 2 c 2 3 to c 3 3 and so on. Compare with the other diffusion ,

.

Example

rates in

,

6.6-1.

N=

Ans. 6.6-2.

15

2.555 x 10"

Numerical Methodfor Steady-State Diffusion with Distribution

1

Coefficient.

kg mol/s

Use

the

conditions given in Example 6.6-1 except that the distribution coefficient defined by Eq. (6.6-11) between the concentration in the liquid adjacent to the external surface and the concentration in the solid adjacent to the external surface

is

K

=

and the

1.2. Calculate the steady-state concentrations

diffusion

rates.

for Steady-State Diffusion. Use the conditions given in Example 6.6-1 but instead of using Ax = 0.005 m, use Ax = Ay = 0.001 m. The overall dimensions of the hollow chamber remain as in Example 6.6-1. The only difference is that many more nodes will be used. Write the computer program and solve for the steady-state concentrations using the numerical method.. Also, calculate the diffusion rates and compare with Example 6.6-1. Numerical Method with Fixed Surface Concentrations. Steady-state diffusion is occurring in a two-dimensional solid as shown in Fig. 6.6-4. The grid A x = Ay = 0.010 m. The diffusivity D AB = 2.00 x 10"' m 2 /s. At the inside of 3 kg the chamber the surface concentration is held constant at 2.00 x 10~

6.6-3. Digital Solution

6.6-4.

mol/m 3 At

the outside surfaces, the concentration

.

Calculate the steady-state concentrations and

3

is

constant at 8.00 x 10" of depth.

the diffusion rates per

.

m

REFERENCES (A

1)

Amdur,

I.,

and Shuler,

(A2)

Anderson, D.

(Bl)

Bunde, R.

(B2)

Boyd, C.

K.,

L.

M.

J.

J.

R.,

and Babb, A.

Hall.

E. Univ. Wisconsin

A., Stein, N.,

Chem.

Phys., 38, 188 (1963).

404

L. J. Phys. Chem., 62,

(1958).

Naval Res. Lab. Kepi. No. CM-850, August 1955.

Steingrimisson,

V.,

and Rumpel, W.

Chem. Phys.,

F. J.

19,

548(1951). (B3)

Bird, R. B., Stewart, W. York John Wiley & Sons, :

E.,

and Lightfoot,

N. Transport Phenomena.

E.

and Geankoplis, C.

(B4)

Bidstrup, D.

(B5)

Barrer, R. M. Diffusion

(CI)

Carmichael,

(C2)

Carswell,

(C3)

Chakraborti,

(C4) (C5)

Chang, S. Y. M. S. thesis, Massachusetts Institute of Technology, Chang, Pin, and Wilke, C R. J. Phys. Chem., 59, 592 (1955).

(C6)

Charlwood,

(C7)

Cameron,

(C8)

Colton,

E.,

New

Inc., 1960.

in

J. J.

and Through

Chem. Eng. Data, Solids.

8,

170 (1963).

London: Cambridge University

Press, 1941. L. T., Sage, B. H.,

AVJ.,

J.

and Lacey, W. N.

and Stryland, J. C. Can.

P.

K, and Gray,

P. Trans.

J.

A.I.Ch.E.J.,

1,

385 (1955).

Phys., 41, 708 (1963).

Faraday Soc,

62, 333

1

(1961).

1959.

P. A. J. Phys. Chem., 57, 125 (1953).

R.

M.

S. thesis,

Ohio State

C. K., Smith, K. A.,

Merrill,

University, 1973. E. W.,

and Reece,

J.

M. Chem. Eng.

Progr.

Symp., 66(99), 85(1970).

Chap. 6

References

423

(C9)

Crank,

and Park, G.

J.,

S. Diffusion in

New

Polymers.

York: Academic Press,

Inc., 1968.

(Fl)

Fuller,

Schettler,

E. N.,

and Giddings,

P. D.,

C. Ind. Eng. Chem., 58, 19

J.

(1966)

and Kramer,

(F2)

Friedman,

L.,

(F3)

Friedman,

L.

(Gl)

Gilliland,

(G2)

Geankoplis, C.

J.'

0.

E.

Am. Chem. Soc,

E. R.lnd. J.

Am. Chem. Soc,

J.

52, 1298 (1930).

52, 1305, 1311 (1930).

Eng. Chem., 26, 681 (1934)

Mass Transport Phenomena. Columbus, Ohio: Ohio

State

University Bookstores, 1972.

(G3)

Garner, G.

and Marchant, P.

H.,

J.

M. Trans.

Inst.

Chem. Eng. (London),

39,

397(1961).

(G4)

Gosting, L.

Advances

S.

Protein Chemistry, Vol.

in

1.

New York:

Academic

Press,

Inc., 1956.

(G5)

Geankoplis, C. (1978)

(G6)

Geankoplis, C. (1979)

J.,

Okos, M.

and Grulke, E.

R.,

A. J.

Chem. Eng. Data,

23,

40

.

J.,

Grulke,

E. A.,

and Okos, M. R. Ind. Eng. Chem. Fund.,

18,

233

.

(HI)

Holsen,

(H2)

Hudson, G.

J.

and Strunk, M.

N.,

H.,

McCoubrey,

R. Ind. Eng. J.

Chem. Fund.,

C, and Uddelohde,

143 (1964).

3,

A. R. Trans. Faraday Soc.,

56, 1144(1960).

(H3)

Hirschfelder, J. O., Curtiss, C. F„ and Bird, R. B. Molecular Theory of Gases and Liquids. New York: John Wiley & Sons, Inc., 1954.

(H4)

Hammond,

(Jl)

Johnson,

(Kl)

Keller, K.

(LI)

Lee, C. Y., and Wilke, C. R. Ind. Eng. Chem.,46, 2381 (1954).

(L2)

Le Bas, G. The Molecular Volumes of Liquid Chemical Compounds. David McKay Co., Inc., 1915.

(L3)

(M2)

Longsworth, L. G. J. Phys. Chem., 58, 770 (1954). Langdon, A. G., and Thomas, H. C. J. Phys. Chem., 75, 1821 (1971). Mason, E. A., and Monchick, L. J. Chem. Phys., 36, 2746 (1962). Monchick, L, and Mason, E. A. J. Chem. Phys.,35, 1676 (1961).

(Nl)

Ney,

(N2)

National Research Council, International Critical Tables, Vol. V.

(N3)

McGraw-Hill Book Company, 1929. Narvari, R. M., Gainer, J. L, and Hall, K.

(L4)

(Ml)

(PI)

B. R.,

P. A.,

E. P.,

and Stokes, R. H. Trans. Faraday Soc,

and Babb, A.

L.

Chem.

and Friedlander,

H.,

and Armistead,

F. C.

J.

890 (1953).

387 (1956).

Revs., 56,

K.

S.

49,

Gen. Physiol, 49, 68 (1966).

New

York:

New

York:

Phys. Rev., 71, 14 (1947).

R. A.I.Ch.E.J., 17, 1028 (1971).

Perry, R. H., and Green, D. Perry's Chemical Engineers' Handbook, 6th ed. York: McGraw-Hill Book Company, 1984.

New (P2)

Perkins, L.

(P3)

PolsoN, A.

(Rl)

Reid, R.

R., J.

and Geankoplis, C.

J.

Chem. Eng.

Sci.,

24, 1035 (1969).

Phys. Colloid Chem., 54, 649 (1950).

C, Prausnitz,

Liquids, 3rd ed.

M., and

J.

Sherwood,

(R2)

Reddy, K.

(R3)

Rogers, C.

(51)

Schafer, K. L. Z. Elecktrochem., 63,

(52)

Seager,

A.,

T. K.

The

Properties of Gases and 1977.

New York: McGraw-Hill Book Company,

and Doraiswamy,

E. Engineering

L. K. Ind. Eng.

Design for Plastics.

Chem. Fund.,

New

6,

77 (1967).

York: Reinhold Publishing

Co., Inc., 1964.

S. L.,

Geertson,

L. R.,

1 1 1

(1959).

and Giddings,

J.

C. J. Chem. Eng. Data,

8,

168

(1963).

(53)

424

Schwertz,

F. A.,

and Brow,

J.

E. J.

Chem. Phys.,

19,

640

(1951).

Chap.

6

References

N, and Srivastava,

Chem. Phys., 38,

(54)

Srivastava,

(55)

Skelland, A. H.

(56)

Sorber, H. A. Handbook of Biochemistry, Selected Data for Molecular Biology. Cleveland: Chemical Rubber Co., Inc., 1968.

(57)

Spalding, G.

(58)

Satterfteld, C. N.

Company,

B.

P. Diffusional

I.

B.

Mass

J.

Transfer.

New

1

183 (1963).

York: McGraw-Hill Book

1974.

E. J.

MIT

Mass: The

Phys. Chem., 73, 3380 (1969).

Mass

Transfer

Heterogeneous Catalysis. Cambridge,

in

Press, 1978.

(Tl)

Trautz, M., and Muller, W. Ann. Physik,

(T2)

Treybal, R.

2nd

E. Liquid Extraction,

ed.

22,

333 (1935).

New

York: McGraw-Hill Book Com-

pany, 1963. (T3)

Treybal, R. E. Mass Transfer Operations, 3rd

Book Company,

and King, C.

(VI)

Vivian,

(Wl)

Wintergerst, V.

(W2)

Westenberg, A.

(W3)

(W4)

Walker, Walker,

(W5)

Wilke, C.

Chap. 6

J. E.,

R.

E.,

New

York: McGraw-Hill

E.

A.,

J.

AA.Ch.EJ.,

Ann. Physik,

4,

10,

220 (1964).

323, (1930).

and Frazier, G.

J.

Chem. Phys.,

and Westenberg, A. A.

J.

and Westenberg, A. A.

J.

36,

Chem. Phys.,

3499 (1962).

29,

1

139 (1958).

Chem. Phys., 32, 436 and Chang, Pin. A.l.Ch.EJ., 1, 264 (1955).

R. E., R.,

ed.

1980.

References

(1960).

425

CHAPTER

7

Principles of

Unsteady- State and

Mass

Convective

Transfer

UNSTEADY-STATE DIFFUSION

7.1

Derivation of Basic Equation

7.1A

In Chapter 6

we considered various mass-transfer systems where

partial pressure at

steady

state.

process

the concentration or

any point and the diffusion flux were constant with time, hence

at

Before steady state can be reached, time must elapse after the mass-transfer

initiated for the unsteady-state conditions to disappear.

is

In Section 2.3 a general property balance for unsteady-state molecular diffusion was

made

for the properties

momentum,

heat,

and mass. For no generation

present, this

was

(2-13-12)

oz~

at

In Section

5.1

an unsteady-state equation for heat conduction was derived,

£01

The transfer 7.1-1

(5-1-10)

ox

derivation of the unsteady-state diffusion equation in one direction for mass

is

similar to that

where mass

is

done

obtaining Eq.

for heat transfer in

diffusing in the x direction in a cube

gas, or stagnant liquid

direction

«

(5.1-10).

composed of

and having dimensions Ax, Ay, and Az. For

We

refer to Fig.

a solid, stagnant

diffusion in the x

we write

N Ax

=-D AB

^

(7.1-1)

ox

The term dcjdx means the partial of c A with respect to x or the rate of change ofc,, with x when the other variable time is kept constant. Next we make a mass balance on component A in terms of moles for no generation. t

rate of input

426

=

rate of output

+

rate of

accumulation

(7.1-2)

rate of output

NAx\x + Ar

Figure

The

rate of input

7.1-1.

Unsteady-state diffusion

and rate of output

in

rate of input

rate of output

The

accumulation

rate of

is

one direction.

kg mo] A/s are

= N Ax

= N Ax

x

,

= -D AB

(7.1-3)

dx

= —D AB

x + Ax

,

as follows for the

rate of

in

(7.1-4)

dx

volume Ax Ay

=

accumulation

x

+ tix

Azm 3.

(Ax Ay Az)

dc A (7.1-5) ot

Substituting Eqs. (7.1-3), (7.1-4),

and

dc A

dc A

dx

-D AB Letting

Ax approach

Equation all

dx

x

dc A

x + Ax

(7.1-6)

'

Ax

dt

zero, 2

dc A

The above holds

and dividing by Ax Ay Az,

(7.1-5) into (7.1-2)

for a

=

constant diffusivity D^,,

(7.1-7) relates the

d cA

D,

(7.1-7)

IfD^

.

is

dc A

d(D AB dc A /dx)

dt

ox

a variable,

(7-1-8)

concentration c A with position x and time

/.

For diffusion

in

three directions a similar derivation gives 2

dt

In the

remainder of

= D AB\

1

d c, -,

OX

2

this section, the

+

df

dz

(7.1-9)

2

solutions of Eqs. (7.1-7) and (7.1-9) will be

considered. Note the mathematical similarity between the equation for heat conduction,

dT _

lh~ and Eq.

(7.1-7) for diffusion.

Because of

for solution of the unsteady-state

state

mass

Sec. 7.1

transfer.

This

is

2

T T

(5.1-6)

~dx

this similarity, the

mathematical methods used

heat-conduction equation can be used for unsteady-

discussed

Unsteady-State Diffusion

d

a

more

fully in

Sections

7.

IB, 7.1C,

and

7.7.

421

7.1

B

Diffusion

in a

Flat Plate with Negligible

Surface Resistance

To

illustrate

an analytical method of solving Eq.

(7.1-7),

we

will derive the solution for

unsteady-state diffusion in the x direction for a plate of thickness 2x u as 7.1-2.

For

one

diffusion in

= D AB the subscripts

A and B

for convenience,

3c

_ D

Ji' The

initial profile

in Fig.

(7.1-7)

dx 2

dt

Dropping

shown

direction,

2

d c (7.1-10)

lx~2

of the concentration in the plate at

t

=

0

is

uniform at

c

=

c 0 at

x values, as shown in Fig. 7.1-2. At time t = 0 the concentration of the fluid in the environment outside is suddenly changed to c,. For a very high mass-transfer coefficient outside the surface resistance will be negligible and the concentration at the surface will be equal to that in the fluid, which is c The initial and boundary conditions are all

.

l

c

=

cn

t

= 0,

x

=

x,

c

=

c,

t

=

x

=

0,

c

=

c,

t

Redefining the concentration so

t,

=

x

t,

it

=

Y=

i

— —

2x u

goes between 0 and

Y =

=

1

=

0

=

0

(7.1-11)

c0 c

i

cn

1,

c,-c (7.1-12) c0

c,

8

2

Y (7.1-13)

dt

The

solution of Eq. (7.1-13)

is

an

ox infinite

Fourier series and

is

identical to the

solution of Eq. (5.1-6) for heat transfer.

Figure

7.1-2.

Unsteady-state diffusion in a plate with negligible sur-

flat

face resistance.

428

Chap. 7

Principles of Unsteady-Stale

and Convective Mass Transfer

/

1

3

2

n 2 X\

.

3tix

2

5

1

2 ti

X

5nx

.

sin

-

—+

(7.1-14)

2x,

where,

X= =

c

1

Dt/x], dimensionless

concentration at point x and time

Y =

(c

— 7=

(c

l

c)/(cj

—

c0 )

=

fraction of

unaccomplished change, dimensionless

c 0 )/(cj

—

c0 )

=

fraction of

change

— -

in slab

t

Solution of equations similar to Eq. (7.1-14) are time consuming; convenient charts for various geometries are available and

7.1

C

]

Convection and boundary conditions

.

Unsteady-State Diffusion

tive resistance at the surface.

convective mass transfer ficient k c

,

is

in

,

at the surface.

However,

occurring

in

many

when

A

at the surface.

=Mc tl

coefficient k c

is

is

next section.

a fluid

is

was no convec-

outside the solid,

convective mass-transfer coefis

defined as follows

-Cid

a mass-transfer coefficient in m/s, c L1 c Li

in the

In Fig. 7.1-2 there

cases

similar to a convective heat-transfer coefficient,

is

A/m 3 and

be discussed

Various Geometries

NA where k c

will

is

(7-1-15)

an empirical coefficient and

will

be discussed more

fully in

Section

In Fig. 7.1 -3a the case for a mass-transfer coefficient being present at the

shown. The concentration drop across the solid

c,

at the surface

Figure

7.1-3.

is

in

fluid is c L1

equilibrium with c u

— c u The .

The

7.2.

boundary

is

concentration in the

.

Interface conditions for convective mass transfer and an equilibrium distribution coeffcienl K = c vJc { : (a) > 1, (c) < 1, 1, (b)

K=

(d)K >

Sec. 7.1

kg mol

the bulk fluid concentration in

the concentration in the fluid just adjacent to the surface of the solid.

1

andk c =

Unsteady-Slate Diffusion

K

K

oo.

429

and c in the However, unlike heat transfer, where the temperatures are equal, the concentrations are in equilibrium and are related by In Fig. 7.1 -3a the concentration c Li in the liquid adjacent to the solid

solid at the surface are in equilibrium

and also

K =— where

K

is

;

equal.

(7.1-16)

the equilibrium distribution coefficient (similar to Henry's law coefficient for a

gas and liquid).

The value of K

in Fig.

7.1-3a

is 1.0.

In Fig. 7.1-3b the distribution coefficient in equilibrium.

Other cases are shown

K is >

1

in Fig. 7.1-3c

andc t; > c even though they are and d. This was also discussed in ; ,

Section 6.6B.

2.

Relation between mass- and heat-transfer parameters.

In order to use the unsteady-

Chapter 5 for solving unsteady-state diffusion problems, the dimensionless variables or parameters for heat transfer must be related to those for mass transfer. In Table 7.1-1 the relations between these variables are tabulated. For K ^ 1-0, whenever k c appears, it is given nsKk c and whenever c, appears, it is given as

state heat-conduction charts in

,

cJK. Tadle

7.1-1.

Relation Between Mass- and

Heat-Transfer Parameters for Unsteady-State Diffusion*

Mass

K = cjc=

Ileal Transfer

T,

—T

T,

— T0

Cj

T — 70 T — T0

—c — c0

K = cjc+

cJK — c

/K

:

cJK —

«'

D AB

D AB

*7

xi

Y,

y

y

t

c

—

—

c — c0 c, — c 0

-

1

Cj

1.0

Transfer

1.0

c

c0

c0 c0

t

~

2j^t

'

h K

n,

x

D AB

_±_

m

*

2j~D A B~t

hxy

k c x,

,-

kc

Kk

c

Kk

c

XXX ,

AB

— X, is

Xy ,

AB

—

— X,

X,

the distance from the center of the slab, cylinder, or sphere; for

a semiinfinite slab, x

is

uniform concentration

the distance

from the surface. c 0

in the solid, c, the

is

the original

concentration

in the fluid

outside the slab, and c the concentration in the solid at position x

and lime

430

t.

Chap. 7

Principles of Unsteady-State

and Convective Mass Transfer

3.

Charts for diffusion

The various

various geometries.

in

heat-transfer

charts for

unsteady-state conduction can be used for unsteady-state diffusion and are as follows. 1.

Semiinfinite solid, Fig. 5.3-3.

2.

Flat plate, Figs. 5.3-5

3.

Long

4.

Sphere, Figs. 5.3-9 and 5.3-10.

5.

Average concentrations, zero convective resistance, Fig. 5.3-13.

and

5.3-6.

and

cylinder, Figs. 5.3-7

EXAMPLE

5.3-8.

Unsteady-State Diffusion in a Slab of A gar Gel thick and contains a % agar gel at 278 K is 10.16 3 uniform concentration of urea of 0.1 kg mol/m Diffusion is only in the x direction through two parallel flat surfaces 10.16 apart. The slab is suddenly immersed in pure turbulent water so that the surface resistance can be assumed to be negligible; i.e., the convective coefficient k c is very large. The diffusi vity of urea in the agar from Table 6.4-2 is 4.72 x 10 _10 2 /s. (a) Calculate the concentration at the midpoint of the slab (5.08

A

7.1-1.

solid slab of 5.15

mm

wt

.

mm

m

mm

from the surface) and 2.54 (b) If the

mm from the surface after 10 h.

thickness of the slab

concentration

in 10

is

halved, what would be the midpoint

h?

=

kg mol/m 3 c = 0 for pure water, and c = concentration at distance x from center line and time t s. The equilibrium distribution coefficient K in Eq, (7.1-16) can be assumed to be 1.0, since the water in the aqueous solution in the gel and outside should be very For part

Solution:

(a),

similar in properties.

c0

0.10

From Table __

t

,

7.1-1,

cJK -

c

cJK-c 0

_

0/1.0

-

c

0/1.0-0.10

3 Also, x, = 10.16/(1000 x 2) = 5.08 x 10" m (half-slab thickness), x = 0 _10 3 2 = D AB t/xj = (4.72 x 10 (center), )<10 x 360O)/(5.08 x 10" ) = 0.658. The relative position n = x/Xj = 0/5.08 x 10~ 3 = 0 and relative resistance

X

m = D AB/Kk x, = 0, since k From Fig. 5.3-5 for X =

c

c

is

very large (zero resistance). = 0, and n = 0,

0.658,

m

0

Y = 0.275 0

-

-

c

0.10

= 0.0275 kg mol/m for x = 0. For the point 2.54 from center, from the surface or 2.54 = 0.658, m = 0, n = x/x, = 2.54 x x = 2.54/1000 = 2.54 x 10" 3 m, 10" 3 /5.08 x 10" 3 = 0.5. Then from 5.3-5, Fig. 7 = 0.172. Solving, 3 c = 0.0172 kg mol/m For part (b) and half the thickness, X = 0.658/(0.5) 2 = 2.632, n = 0, and m = 0. Hence, Y = 0.0020 and c = 2.0 x 10" 4 kg mol/m 3 3

Solving, c

mm

mm

X

.

.

EXA MPLE

7.1-2. Unsteady-State Diffusion in a Semiinfinite Slab very thick slab has a uniform concentration of solute A of c 0 = 1.0 x 10" 2 kg mol A/m 3 Suddenly, the front face of the slab is exposed to a

A

.

flowing fluid having a concentration c, = 0.10 kg mol yl/m 3 and a convec7 tive coefficient k c = 2 x 10" m/s. The equilibrium distribution coefficient

K =

c L Jc = 2.0. Assuming that the slab is a semiinfinite solid, calculate the concentration in the solid at the surface (x = 0) and x = 0.01 m from the 4 9 surface after / = 3 x 10 s. The diffusivity in the solid is D AB = 4 x 10" ;

m 2 /s.

Sec. 7.1

Unsteady-State Diffusion

431

To

Solution:

use Fig. 5.3-3,

V^I = jfc For x

=

2 X

2

7)

"

X 1Q 9X3 X 1Q4)

-4

x lQ°^

=

LQ9S

m from the surface in the solid,

0.01

x

0.01

=

2y(4lTl0- 9 X3 x

2^D^~t

10

0.457

4 )

the chart, 1 — Y = 0.26. Then, substituting into the equation for 7) from Table 7.1-1 and solving,

From

—

(1

c

For x

=

0

c°

cJK-c 0

=

m

~

c

_y=

1

2.04 x 10

~

c

=

1

x 10

2 (10 x 10~ )/2

-2

~2

-(1 x

kg mol/m 3

x

(for

10

=

=

-2

o 26

)

0.01

m)

at the surface of the solid),

(i.e.,

x

=

0

,t

From same

the chart, as c i( as

1

— Y=

shown

=

0.62. Solving, c

3.48 x 10"

To calculate

in Fig. 7.1-3b.

2 .

This value

is

the

the concentration c u in the

liquid at the interface,

C Li = K Ci =

A 4.

2.0(3.48 x 10"

2 )

=

6.96 x 10"

2

kg mol/m 3

plot of these values will be similar to Fig. 7.1-3b.

Unsteady- state diffusion

in

more than one

In Section 5.3F a

direction.

method was

given for unsteady-state heat conduction to combine the one-dimensional solutions to yield solutions for several-dimensional systems.

unsteady-state diffusion in in a

more than one

rectangular block in the x,

where c x value of

is y x

Yx for

the x direction.

y,

and

The same method can be used

z directions,

the concentration at the point x, y, the two parallel faces

The values

of

for

direction. Rewriting Eq. (5.3-11) for diffusion

is

Yy and Y

z

z

from the center of the block. The

obtained from Fig. 5.3-5 or 5.3-6 for a are similarly obtained from the

short cylinder, an equation similar to Eq. (5.3-12)

is

same

flat plate in

charts.

For a

used, and for average concentrations,

ones similar to Eqs. (5.3-14), (5.3-15), and (5.3-16) are used.

CONVECTIVE MASS-TRANSFER COEFFICIENTS

7.2

7.2A

Introduction to Convective

Mass Transfer

In the previous sections of this chapter and Chapter 6 diffusion in stagnant fluids or fluids in laminar flow. In

and more rapid transfer is turbulent mass transfer occurs.

slow,

To have

desired.

this,

the rate of diffusion

the fluid velocity

is

is

increased until

a fluid in convective flow usually requires the fluid to be flowing by another

immiscible fluid or by a solid surface.

432

To do

we have emphasized molecular

many cases

Chap. 7

An example

Principles

is

a fluid flowing in a pipe, where part

of Unsteady-State and Convective Mass Transfer

of the pipe wall

is

made by

benzoic acid dissolves and

When

a fluid

is

in

a slightly dissolving solid material

is

such as benzoic acid. The

transported perpendicular to the main stream from the wall.

turbulent flow and

is

flowing past a surface, the actual velocity of

small particles of fluid cannot be described clearly as in laminar flow. In laminar flow the

and its behavior can usually be described mathematically. However, in turbulent motion there are no streamlines, but there are large eddies or "chunks" of fluid moving rapidly in seemingly random fashion. When a solute A is dissolving from a solid surface there is a high concentration of this solute in the fluid at the surface, and its concentration, in general, decreases as the distance from the wall increases. However, minute samples of fluid adjacent to each other fluid flows in streamlines

do not always have concentrations close to each other. This occurs because eddies having solute in them move rapidly from one part of the fluid to another, transferring relatively large

amounts

or eddy transfer

of solute. This turbulent diffusion

comparison to molecular transfer. Three regions of mass transfer can be visualized. surface, a thin viscous sublayer film

is

molecular diffusion, since few or no eddies are

In the first,

Most of

present.

present".

A

which

mass

the

quite fast in

is

is

adjacent to the

by drop occurs

transfer occurs

large concentration

across this film as a result of the slow diffusion rate.

The

transition or buffer region

present and the mass transfer

gradual transition

in this

is

sum

the

is

adjacent to the

first

at the

turbulent diffusion, with a small

A

.

is

eddies are is

a

other end.

In the turbulent region adjacent to the buffer region,

decrease

Some

region from the transfer by mainly molecular diffusion at the

one end to mainly turbulent

A more

region.

of turbulent and molecular diffusion. There

amount by molecular

most of the transfer is by The concentration

diffusion.

very small here since the eddies tend to keep the fluid concentration uniform.

detailed discussion of these three regions

is

given

in

Section 3.10G.

from a surface to a turbulent fluid in a conduit is given in Fig. 7.2-1. The concentration drop fromc^,, adjacent to the surface is very abrupt close to the surface and then levels off. This curve is very similar to the shapes found for heat and momentum transfer. The average or mixed concentration c A is shown and is slightly greater than the minimum c A2 typical plot for the

mass

transfer of a dissolving solid

.

7.2B /.

Types of Mass-Transfer Coefficients

Definition of mass-transfer coefficient.

incomplete, to that for

Figure

we attempt

Since our understanding of turbulent flow

to write the equations for turbulent diffusion in a

molecular diffusion. For turbulent mass transfer

7.2-1.

Concentration profile lent

mass

transfer

in

for

constant

manner c,

is

similar

Eq. (6.1-6)

is

turbu-

from a

sur-

face to a fluid.

z

0

Distance from surface

Sec. 7.2

Convective Mass-Transfer Coefficients

433

written as

dz

where

D AB

The value

of

of £ w

is

?,

M

is

a variable

and

is

not generally

We

=

-

_

=

^

(7 2 2)

is

often not

-

may

known. Hence, Eq.

vary.

(7.2-2)

is

.

c A2 )

(7.2-3)

A from thfe surface A relative to the whole bulk phase, k' is an experimental mass-transfer coefficient in kg mol/s m 2 (kg 3 or simplified as m/s, and is the concentration at point 2 in kg mol A/m or

where J*Al (Dab

k't {c Al

2,

since the cross-sectional area

i

/s.

since the variation

written using a convective mass-transfer coefficient k'c

is

J*ai

+

c

Zj

2

and increases as

and

1

m

difTusivity in

M

£

between points

(7.2-1)

The flux J*! is based on the surface area A The value of z 2 — z,, the distance of the path, and

mass eddy

the

at the interface or surface

D AB + e M Z2

simplified

is

then use an average value

known. Integrating Eq. jm

and e M

near zero

from the wall increases.

the distance

m 2 /s

the molecular diffusivity in

is

the flux of

is

£ w)/( z 2

mol/m 3 ) more usually

—

c

l

z i)

•

the average bulk concentration c A2 This defining of a convective massis quite similar to the convective heat-transfer coefficient h. .

transfer coefficient k'c

2.

Mass-transfer coefficient for equimolar counter diffusion.

in

NA

which

the flux of

,

is

A

relative to stationary coordinates.

For

can

similar to that for molecular diffusion but the term£ M

N A = ~c(D AB + £m)^- + the case of

steady state, calling

Equation

=

+ £ M )/(z 2 - Zj), N A = K(c A1 -c A2

(D AB

terms of partial pressure is

if

mole

added.

is

NB

(7.2-4)

)

NA — — Ng

and integrating

,

a gas. Hence,

we can

at

(7.2-5)

)

terms of mole fraction

in

interested

with the following,

start

equation for the mass-transfer

(7.2-5) is the defining

however, we define the concentration several ways. If y A

*a(N a +

equimolar counterdiffusion, where

k'c

we are

Generally,

We

if

coefficient. Often,

a liquid or gas or in

define the mass-transfer coefficient in

fraction in a gas phase andx,., in a liquid phase, then Eq. (7.2-5)

can be written as follows for equimolar counterdiffusion

Gases:

NA =

k'c (c Al

- c A2 =

k'0 (p Ai

Liquids:

NA =

k'c (c Al

-

=

k'L (c Al

All of these

)

c A2 )

-

-

p A2 )

=

c A2 )

=

k' (y Ai y

k'x (x A1

-

y A2 )

(7.2-6)

~

x A2 )

(7.2-7)

mass-transfer coefficients can be related to each other. For example, using

Eq. (7.2-6) and substituting y A1

NA =

K(c Ai

-

c A2 )

=

=

c Al /c

k' (y Al y

and y A2

-

y A2 )

=

=

c A2 /c into the equation,

k'

y

^-f

-

C

~fj

=^

(c Al

-

c A2 )

(7.2-8)

Hence,

K = -zc These relations among mass-transfer given in Table 7.2-1.

434

Chap. 7

coefficients,

Principles

(7.2-9)

and the various

flux equations, are

of Unsteady-Slate and Convective Mass Transfer

Table

Flux Equations and Mass-Transfer Coefficients

7.2-1.

Flux equations for equimolar counter diffusion

NA =

Gases:

k'c {c Al

NA =

Liquids':

NA =

Gases: ..

NA =

Liquids:

-

k'c (c Al

A

Flux equations for

-

c A2 )

=

c A2 )

=

k'G (p Al

- p A2 =

k'L (c Al

-

y {y Al

=

c A2 )

-

k'

)

y A2 )

- xA2

k'x (x A1

B

diffusing through stagnant, nondiffusing

k c {c Al k e (c Al

-

=

c A2 )

-

=

c A2 )

k G (p Al

- p A2 =

k L {c Al

- c A2 =

)

)

)

k y (y Al

-

k x (x Al

- x A2

y A1 ) )

Conversions between mass-transfer coefficients

Gases:

Kc = K

~

-~ = K

=

=

k 'aP

=

ka PBM

ky

=

y BM

k' y

=

k c y BM c

=

k G y BM

P

Liquids: k'c c

(where p

=

=

k'L c

k L x Bht c

M

density of liquid and

is

is

= k'L plM =

=

k'z

k x x BM

molecular weight)

Units of mass-transfer coefficients SI Units kc

kL

,

kx

k

,

y

,

,

,

k x1 ky

m

s

kc

English Units

cm/s

ft/h

m/s

k'c , k'L

kg mol

g

2

2

-molfrac

cm

s

kg mol

kg mol kG

Cgs Units

•

mol

h

frac

mol

g

s-m 2 -Pa s-m 2 -atm

mol •

mol

2

mol

ft

lb

s-cm 2 -atm

lb

frac

mol

hTt 2 -atm

(preferred)

3.

Mass-transfer coefficient for

A

diffusing through stagnant, nondiffusing D.

diffusing through stagnant, nondiffusing

NA = where the x BM and

its

transfer coefficient for

A

— X BM

B

(c Al

where

-

c A2 )

NB =

=

k c (c A

,

-

=

——

X B2

~~

X Bl

7—,

In (x B2 /x Bl

y bm

i )

A

(7.2-10)

c A2 )

counterpart y BM are similar to Eq. (6.2-21) and k c diffusing through stagnant B. Also,

x bm

For

0, Eq. (7.2-4) gives for steady state

)>B2

== ,

~

;

—

yB ,

is

the mass-

/-nlt\ (7.2-11)

l

:

In (y B2 /y Bi )

Rewriting Eq.(7.2-10) using other units,

Again

Sec. 7.2

(c A1

-

c A2 )

=

k G (p Ai

-

p A2 )

=

k y (y Al

- y A2

)

(7.2-12)

k c {c A1

-

c A2 )

=

k L (c Al

-

c A2 )

=

k x (x Al

- x A1

)

(7.2-13)

(Gases):

NA = k

(Liquids):

NA =

all

c

the mass-transfer coefficients can be related to each other

Convective Mass-Transfer Coefficients

and are given

in

435

Table

7.2-

For example,

L.

setting Eq. (7.2- LO) equal to (7.2-L3),

"a = -rX

(c Al

-

= Ux, - x A2 = k/^di _ £dij

c A2 )

(7.2-14)

)

t

BM

c

\

Hence,

K

K

EXAMPLE A

Vaporizing

7.2-1.

volume of pure gas B

large

which pure

A

A and Convective Mass

Transfer flowing over a surface from completely wets the surface, which

atm pressure

at 2

The

vaporizing.

is

(7.2-15)

A

liquid

is

a blotting paper. Hence, the partial pressure of A at the surface is the vapor pressure of A at 298 K, which is 0.20 atm. The k'y has been estimated -5 2 to be 6.78 x 10 mol frac. Calculate A the vaporization kg mol/srate, and also the value of k and k G is

m

N

•

,

.

This is the case of A diffusing through B, where the flux of B normal to the surface is zero, since B is not soluble in liquid A. p Al = 0.20 atm and p A2 = 0 in the pure gas B. Also, y Al = p Al /P = 0.20/2.0 = 0.10 and

Solution:

)>A2

=

0-

We can use Eq. (7.2-

1

with mole fractions.

2)

N A = ky(y Al -y A2 However, we have a value

for k'

y

which

is

The term y BM

similar to

is

x BM and

is,

Substituting into Eq. (7.2-1

0.90

(W^bi) y B2 =

1.0-0-90

ky

=

"

—

y BM

I

1),

-y Al

=

-0=

1

1.0

_ nQ ,

in (1.0/0.90)

(7.2-16),

6.78 x 10 _ — = ~ 0.95

k'

y -^t-

(7.2-1

1),

^ BM

Then, from Eq.

by

(7.2-16)

k' y

from Eq.

In

^1 = 1-^1 = 1-0.10 =

related lok y from Table 7.2-1

=

ky yB „

(7.2-12)

)

5

=

5

7.138 x 10

m2

kg mol/s

•

mol

frac

Also, from Table 7.2-1,

kG yB

uP = KVbm

(7.2-17)

Hence, solving for k G and substituting knowns,

°

=^ = 2

kG

7 =^= F

k

For the

NA = 436

k y (y Al

X 10

l

l* 2.0

flux

-

Pa =

x L01325

atm

=

^

3.569 x 10"

5

X 10 "

k

'°

kg mol/s

•

m °'/S

«

m

2 •

atm

using Eq. (7.2-12),

y AI )= 7.138 x 10

Chap. 7

_5

(0.10

-

0)

=

7.138 x 10"

Principles of Unsieady-Siaie

6

2 kg mol/s-m

and Convective Mass Transfer

Also,

p Al

Using Eq.

=

=

0.20 atra

0.20(1.01325 x 10 5 )

=

2.026 x 10*

Pa

(7.2-12) again,

NA =

*o(Pai

- p A2 = )

= HA =

k G (p Al

10 4 3.522 x 10(2.026 x 10

0)

7.138 x 10" 6 kg mol/s-m 2

-p A2 =

5 3.569 x 10" (0.20-0)

=

6 7.138 x 10~ kg mol/s

)

-

Note that in this case, since the concentrations were 1.0 and k and k' differ very little. y y

m

2

dilute, y BSf

is

close to

7.2C Methods to Determine Mass-Transfer Coefficients

Many

different experimental

methods have been employed

to experimentally obtain

mass-transfer coefficients. In determining the mass-transfer coefficient to a sphere, Steele

and Geankoplis

(S3) used a solid

pipe. Before the run the

sphere of benzoic acid held rigidly by a rear support

sphere was weighed. After flow of the

fluid for a

in

a

timed interval,

was removed, dried, and weighed again to give the amount of mass transwhich was small compared to the weight of the sphere. From the mass transferred

the sphere ferred,

and the area of the sphere, the flux was used to calculate k L where c AS ,

NA is

was

calculated.

Then

the driving force (c AS

—

0)

the solubility and the water contained no benzoic

acid.

Another method used is to flow gases over various geometries wet with evaporating For mass transfer from a flat plate, a porous blotter wet with the liquid serves as

liquids.

the plate.

7.3

7.3A

MASS-TRANSFER COEFFICIENTS FOR VARIOUS GEOMETRIES Dimensionless Numbers Used to Correlate Data

The experimental data

for mass-transfer coefficients

fluids, different velocities,

numbers similar

and

obtained using various kinds of

different geometries are correlated using dimensionless

to those of heat

and momentum

transfer.

Methods

of dimensional

analysis are discussed in Sections 3.11,4. 14, and 7.8.

The most important dimensionless number

is

the

Reynolds number

yV Re

,

which

indicates degree of turbulence.

N*

e

-^—

(7.3-1)

\>-

where L

is

velocity v v'

in

the

diameter

is

D p for a

sphere, diameter

the mass average velocity

empty

cross section

is

D

for

a pipe, or length

L

for a flat plate.

The

packed bed the superficial velocity often used or sometimes v = v'/e is used, where v is if

in a pipe.

In a

and e void fraction of bed. The Schmidt number is

interstitial velocity

/V Sc

Sec. 7.3

= -JL-

(7.3-2)

pD AB

Mass-Transfer Coefficients for Various Geometries

437

and density p used are the actual flowing mixture of solute A and is dilute, properties of the pure fluid B can be used. The Prandtl number c p p/k for heat transfer is analogous to the Schmidt number for mass transfer. The Schmidt number is the ratio of the shear component for diffusivity p/p to the diffusivity for mass transfer D AB and it physically relates the relative thickness of the hydrodynamic layer and mass-transfer boundary layer. The Sherwood number, which is dimensionless, is

The

viscosity

fluid B. If the

p.

mixture

,

— =lj,v,- -L

^=

t;

L

=

Other substitutions from Table 7.2-1 can be made The Stanton number occurs often and is k ~ =

ky

c

Again, substitution for

k'c

==

N Sh

and

JD For heat

=

- (^sc)

2/3

=

~

2' 3

(Wsc)

J H factor

c

1.

tions of

we can say

Among Mass,

(7.3-3).

c

in

=

(^p

G

p

-

•

" '

•

(7.3-4)

•

vc.

JD

= N s J(N Rc N^)

2'3

factor which

is

(73-5)

(7-3-6)

r)

Momentum

Heat, and

which were pointed out

Newton

k P ~ =

(7.3-3)

as follows:

is

In molecular transport of

Introduction.

similarities,

Eq.

for k'c in

-

as follows.

transfer a dimensionless

Analogies

—

=

correlated as a dimensionless

is

Jh =

7.3B

Dab

G M = vp/M av =

can be made.

Often the mass-transfer coefficient related to k'c

L

c

Dab

ab

^si

k'

Transfer

momentum,

Chapters 2 to

6.

heat, or

mass there are many

The molecular

diffusion equa-

momentum, Fourier for heat, and Fick for mass are very similar and we have analogies among these three molecular transport processes.

for

that

There are also similarities in turbulent transport, as discussed in Sections 5.7C and 6.1A, where the flux equations were written using the turbulent eddy momentum diffusivity g, the turbulent

eddy thermal

However, these

more

difficult to relate to

A great among these

each other.

deal of effort has been devoted in the literature to developing analogies three transport processes for turbulent transfer so as to allow prediction of

one from any of the others. 2.

Reynolds analogy.

and

We discuss several

Reynolds was the

next.

to note similarities in transport processes

first

momentum and heat transfer. Since then, mass transfer has also momentum and heat transfer. We derive this analogy from Eqs. (6.1-4)—

relate turbulent

been related to

(6.1-6) for turbulent transport.

the wall, Eq. (6.1-5)

becomes as

For

fluid flow in a pipe for

follows, where

q

\= A 438

and the turbulent eddy mass diffusivity e,v , mathematically or physically and are

diffusivity a,,

similarities are not as well defined

Chap. 7

_p C

Principles

z is

heat transfer from the fluid to

distance from the wall:

dT

(a

+ a,)—

(73-7)

dz

of Unsteady-State and Convective Mass Transfer

For momentum

becomes

transfer, Eq. (6.1-4)

+ Next we assume a and

/i/p

and

are negligible

C734)

that a,

=

e(

.

Then dividing Eq.

(7.3-7)

by

(7.3-8),

dT =dv

c.

If

we assume

that heat flux q/A in a turbulent system

the ratio r/(q/A) must be constant for

conditions

same

at the wall

T= T

where

;

and

written as t s

.

=

v

analogous to

We now

0 to some point

T

at this

point

at the wall,

momentum

in the fluid is

the

as

is

flux

r,

integrate between

T

where

sameas

v ay

,

the

is

the bulk

the shear- at the wall,

Hence,

c

q/A Also, substituting q/A

=

h(T

-

p

-

(T

h

2

In a similar

Eq. (7.3-8)

manner using Eq. for

momentum

7D

= > a2v

7])andt s

f = -

this to

is

radial positions.

all

and assume that the velocity Also, q/A is understood to be the flux

as the bulk

velocity.

(73-9)

c

p

y av

=

». ¥

-

0

(7.3-10)

p/2 from Eq. (2.10-4) into Eq. (7.3-10),

=—G h

p

c

(7.3-11)

p

J* and also J* = k'L (c A — c Ai ), we can Then, the complete Reynolds analogy is

(6.1-6) for

transfer. J f = -

— =-±

2

c

h

k'

G p

v 3V

relate

(7.3-12)

Experimental data for gas streams agree approximately with Eq. (7.3-12) if the Schmidt and Prandtl numbers are near 1.0 and only skin friction is present in flow past a flat plate or inside a pipe. When liquids are present and/or form drag is present, the analogy is not valid. Other analogies. The Reynolds analogy assumes that the turbulent difTusivities a,, and e M are all equal and that the molecular difTusivities /i/p, a, and D AB are negligible compared to the turbulent difTusivities. When the Prandtl number (;Vp)/a 1S 10. tnen n/p = a; also, for N Sc = 1.0, /i/p = D AB Then,(p./p + £,) = (a + a,) = (D AB + e M and the Reynolds analogy can be obtained with the molecular terms present. However, the analogy breaks down when the viscous sublayer becomes important since the eddy difTusivities diminish to zero and the molecular difTusivities become important. Prandtl modified the Reynolds analogy by writing the regular molecular diffusion equation for the viscous sublayer and a Reynolds-analogy equation for the turbulent core region. Then since these processes are in series, these equations were combined to produce an overall equation (Gl). The results also are poor for fluids where the Prandtl and Schmidt numbers differ from 1.0.

3.

r

.

Von Karman in

)

further modified the Prandtl analogy by considering the buffer region

addition to the viscous sublayer and the turbulent core. These three regions are

in the

universal velocity profile in Fig. 3.10-4. Again, an equation

is

in

for the turbulent core. Both the molecular

an equation for the buffer

Sec. 7.3

layer,

where the velocity

and eddy

in this layer

Mass-Transfer Coefficients for Various Geometries

shown

written for molecular

diffusion in the viscous sublayer using only the molecular diffusivity

analogy equation

,

is

and a Reynolds

diffusivity are

used

used to obtain an

439

equation for the eddy diffusivity. These three equations are then combined to give the von KaYm&n analogy. Since then, numerous other analogies have appeared (PI, S4).

4.

The most

Chilton and Colburn J-factor analogy.

analogy

experimental data for

and

is

written as follows:

72 = Although flow,

and most widely used

successful

and Colburn J-factor analogy (C2). This analogy is based on gases and liquids in both the laminar and turbulent flow regions

the Chilton

is

it

this is

can be

J« = cp

(^p r )

G

2/3

= Jo =

-

2/3

(73-13)

(^sc)

y av

an equation based on experimental data for both laminar and turbulent

shown

to satisfy the exact solution derived from laminar flow over a flat

plate in Sections 3.10

and

5.7.

Equation (7.3-13) has been shown to be quite useful in correlating momentum, heat, and mass transfer data. It permits the prediction of an unknown transfer coefficient when

one of

the other coefficients

obtained for the

due

is

known.

friction loss,

In

momentum

transfer the friction factor

which includes form drag or

momentum

is

losses

and also skin friction. For flow past a flat plate or in a pipe where no = J H = J D When form drag is present, such as in flow in packed other blunt objects J/2 is greater than J H or J D and J H = J D

to blunt objects

form drag

is

beds or past

present,//2

.

.

Derivation of Mass-Transfer Coefficients

7.3C /.

drag or

total

When

Introduction.

a fluid

is

flowing

in

Laminar Flow

laminar flow and mass transfer by molecular

in

by and mass transfer are not always completely analogous since in mass transfer several components may be diffusing. Also, the flux of mass perpendicular to the direction of the flow must be small so as not to diffusion

is

conduction

occurring, the equations are very similar to those for heat transfer

in

laminar flow.

The phenomena

of heat

distort the laminar velocity profile.

In theory

it

is

not necessary to have experimental mass-transfer coefficients for

laminar flow, since the equations

However,

in

many

for geometries,

actual cases

it

for is

momentum

such as flow past a cylinder or

2.

Mass

two cases

in

in

and

mass-transfer coefficients are often obtained derivation will be given for

and

transfer

for diffusion

packed bed. Hence, experimental

a

correlated.

A

simplified theoretical

laminar flow.

transfer in laminar flow in a tube.

We

consider the case of mass transfer from a

tube wall to a fluid inside in laminar flow, where, for example, the wall

benzoic acid which the flowing fluid

is

dissolving in water. This

where natural convection

is

is

negligible.

= vx

is

is

made

of solid

similar to heat transfer from a wall to

parabolic velocity derived as Eqs. (2.6-18) and (2.6-20)

where

can be solved.

describe mathematically the laminar flow

difficult to

2v

For

fully

developed flow, the

is

1

-

m

1

'

(7.3-14)

from the center. For mass balance can be made on a differential element by convection plus diffusion equals the rate out radially by diffusion to

the velocity in

the x direction at the distance r

steady-state diffusion in a cylinder, a

where the rate

in

give

dc A

440

Chap.

7

fldc A

2

d c

2

d cA \

Principles of Unsteady-State

and Convective Mass Transfer

Then, d 2 c A /dx 2

=

0

the diffusion in the x direction

if

Combining Eqs.

convection.

and

(7.3-14)

Graetz solution for heat assumed that the velocity profile

series similar to the If

it is

easily obtained (SI).

A

obtained, where there

transfer is flat

negligible

compared

and a parabolic

is

by complex

to that

a

velocity profile.

as in rodlike flow, the solution

approximate Leveque

third solution, called the

is

is

(7.3-15), the final solution (SI)

solution,

is

more

has been

a linear velocity profile near the wall and the solute diffuses only

a short distance from the wall into the fluid. This

similar to the parabolic velocity

is

Experimental design' equations are presented

profile solution at high flow rates.

in

Section 7.3D for this case.

3.

In Section 2.9C we derived the equation for the

Diffusion in a laminar falling film.

velocity profile in a falling film

A

solute

shown

in Fig. 7.3-la.

which

into a laminar falling film,

is

developing theories to explain mass transfer

The

A

We

will

consider mass transfer of

important in wetted-wall columns,

in

and

in

stagnant pockets of

in

fluids,

and then diffuses a distance into the liquid so that it has not penetrated the whole distance x = 5 at the wall. At steady state the inlet concentration c A = 0. At a point z distance from the turbulent mass transfer.

inlet the

A

concentration profile ofc^

mass balance

=

rate of input

will

be

is

made on

in the gas

shown

absorbed

is

at the interface

in Fig. 7.3-la.

the element

shown

in Fig. 7.3-lb.

For steady

state,

rate of output.

N AAx (\ For a

solute

Az)

+ JV^l

dilute solution the diffusion

Ax)

= N Axlx + &x(l

equation for

dc N Ax = -D AB — A

A

Az)

in the

h zero

+ N Az]: + Az (l

x direction

Ax)

(7.3-16)

is

convection

(7.3-17)

ox

For the

z direction the diffusion is negligible.

N A: =

0 +

(a)

Figure

Sec. 7.3

7.3-1.

(7.3-18)

cA v2

(b)

A in a laminar falling film : (a) velocity profile and concentration profile, (b) small element for mass balance.

Diffusion of solute

Mass-Transfer Coefficients for Various Geometries

441

Dividing Eq. (7.3-16) by (7.3-17)

and

Ax

Ax and Az approach

Az, letting

(7.3-18) into the result,

we

5V

8c A

dx

From

—

(7.3-19)

2

Eqs. (2.9-24) and (2.9-25), the velocity profile

=

2

and substituting Eqs.

zero,

obtain

parabolic and

is

vz

is

=

Also, v z mal (3/2)u z av If the solute has penetrated only a short distance into the fluid, i.e., short contact times of t seconds equals z/i> max then the A that i)

!mlI [1

(x/<5) ].

.

,

has diffused has been carried along at the velocity

!>.

or v mll

max

if

the subscript z

is

dropped. Then Eq. (7.3-19) becomes dc.

dx

Using the boundary conditions of c A = 0 at x = oo, we can integrate Eq. (7.3-20) to obtain

—= where

erf

y

is

the error function

z

=

(7.3-20)

2

=

0, c A

c A0

at

x =

0,

and

0 at

(7.3-21)

erfc

and

=

cA

erfc

y

=

1

—

Values of erf y are standard

erf y.

tabulated functions.

To

determine the local molar flux

entrance,

we

at the surface

x

=

0 at position z from the top

write (Bl)

= -D A

dc.

(7.3-22)

ox

The z

=

moles of A transferred per second to the liquid over the entire length where the vertical surface is unit width is

total

L,

JV„(L-1)=(1)

z

=

0 to

(N o

H2

= (D 4D.,

(L-l)c A0

The term L/v m3X is t L time of exposure of means the rate of mass transfer is proportional ,

(7.3-23)

nL

V'

the liquid to the solute to

D° s5 and

l/t°"

5 .

This

A is

in

the gas. This

the basis for the

penetration theory in turbulent mass transfer where pockets of liquid are exposed to unsteady-state diffusion (penetration) for short contact times.

7.3D /.

Mass Transfer

Mass

for

Flow

Inside Pipes

transfer for laminar flow inside pipes.

pipe and the Reynolds

number Dvp/u

is

When

a liquid or a gas

is

flowing inside a

below 2100, laminar flow occurs. Experimental

data obtained for mass transfer from the walls for gases (G2, LI) are plotted in Fig. 7.3-2 cA

is

face

W/D AB pL

less

than about

between the wall and the

N Re N Sc (D/L)(n/4), where W 442

The ordinate

—

—

c A0 ), where the exit concentration, c A0 inlet concentration, and c Ai concentration at the inter-

for values of

is

Chap. 7

gas.

70.

is (c^,

c A0 )/(c Ai

W/D AB pL or is length of mass-transfer section in m.

The dimensionless abscissa

flow in kg/s and

L

is

Principles of Unsteady-State

and Convectivc Mass Transfer

Graetz^\\ parabolic

^

^"^-^^-v.

10"

parabolic flow io"

s

approximate
10"

i

i

10

10

i

2

i

i

10

10

D

orN„Re"'Sc N.

DAB» L Data for

i

10

w 7.3-2.

flow

2 -

10"

Figure

^^rodlike

10"

10

7

7T

~4

~l

diffusion in a fluid in streamline

s

flow inside a pipe:

filled

vaporization data of Gilliland and Sherwood; open circles, dissolving-solids data of Linton and Sherwood. [From W. H. Linton circles,

and T. K. Sherwood, Chem. Eng. Progr.,

46,

255 (1950). With per-

mission.']

Since the experimental data follow the rodlike plot, that line should be used. profile

is

For is

assumed

fully

liquids that have small values of D AB

as follows for

Mass

J;

-

=

,

data follow the parabolic flow

W

-

line,

which

2/3

(73-24)

5.5

c.

transfer/or turbulent flow inside pipes.

for gases

velocity

W/D AB pL over 400.

c

2.

The

developed to parabolic form at the entrance.

For turbulent flow

for

Dvp/p above 2100

or liquids flowing inside a pipe,

= The equation holds

D .

kc

D = KPbm =

D

—

{Dvp\° o.023

D,

1 I-

(7.3-25)

pD AB

J

3000 (G2, LI). Note that the Sc for gases is in the above 100 in general. Equation (7.3-25) for mass transfer and Eq. (4.5-8) for heat transfer inside a pipe are similar to each other. for
range 0.5-3 and for liquids

3.

Mass

is

transfer for flow inside wetted-wall towers.

of a wetted-wall tower the

same correlations

laminar or turbulent flow

in a

(7.3-25)

When

a gas

that are used for

pipe are applicable. This

can be used to predict mass transfer

for the gas.

is

flowing inside the core

mass

means

transfer of a gas in

that Eqs. (7.3-24)

For the mass

and

transfer in the

down the wetted-wall tower, Eqs. (7.3-22) and (7.3-23) can be used for Reynolds numbers of 4T/p as defined by Eq. (2.9-29) up to about 1200, and the theoretically predicted values should be multiplied by about 1.5 because of ripples and other factors. These equations hold for short contact times or Reynolds numbers above about 100 (SI). liquid film flowing

EXAMPLE 7.3-1. A

tube

of 20

Sec. 7.3

is

mm

Mass Transfer Inside a Tube coated on the inside with naphthalene and has an inside diameter and a length of 1.10 m. Air at 318 K and an average pressure of

Mass-Transfer Coefficients for Various Geometries

443

101.3

kPa

pressure remains

Example

6.2-4.

Solution:

From Example p Ai = 74.0 Pa or

=

p

Use the physical properties given

D AB =

6.2-4, c Ai

air.

that

calculate the con-

constant,

essentially

centration of naphthalene in the exit

pressure _i 10 kg

Assuming

flows through this pipe at a velocity of 0.80 m/s.

the absolute

6.92 x 10

= pJRT =

-5

m

2

in

and the vapor

/s

= 2.799 x x 10~ 5 Pas,

74.0/(8314.3 x 318)

mol/m 3 For air from Appendix 3 1.114 kg/m The Schmidt number is

A.3,

.

=

1.932

=

2 506

p.

.

Nsc

p.

=

=

JdZ

The Reynolds number

1.932 x 10"

is

Pop R

Hence, the flow

is

5

6 1-114 x 6.92 x 10-

<~

0.020(0.80X1.114)

~

n

1.932 x 10-

1

"

*

laminar. Then,

NuN*^ = 922.6(2.506)^ = = 33.02 Using

Fig. 7.3-2

Also,

c x0 (inlet)

Mass Transfer

/.

Mass

transfer in

from

liquids

for

-

Then,

0.

=

c^exit concentration)

7.3E

—

the rodlike flow line, (c A — c A0 )/(c Ai c A0 ) = 0.55. 5 {c A - 0)/(2.799 x 10" Solving, 0) = 0.55.

and

=

-

1.539 x 10

5

kg mol/m 3

.

Flow Outside Solid Surfaces

flow parallel to

flat plates.

a plate or flat surface to a

The mass

flowing stream

and vaporization of

transfer

is

of interest in the drying of

inorganic and biological materials, in evaporation of solvents from paints, for plates in wind tunnels, and in flow channels in chemical process equipment. When the fluid flows past a plate in a free stream in an open space the boundary layer is not fully developed. For gases or evaporation of liquids in the gas phase and for the laminar region of

±25%

N Rc

L

=

Lvp/pi less than 15 000, the data can be represented within

by the equation (S4) JD

Writing Eq. (7.3-26)

in

U AB

is

=

J i,

=

iV Sh

=

(7.3-26) /V Sh

,

5

0.664iV°; L

(7.3-27)

the length of plate in the direction of flow. Also, J D = J „ = f/2 for this Rc L of 1 5 000-300 000, the data are represented within + 30%

geometry. For gases and

by J D

0.6647V- °l

terms of the Sherwood number

^ where L

=

= f/2

N

as

JD

Experimental data for a /V Re L of

=

-

O.036/V R e° L

2

for liquids are correlated within

-

about

(7.3-28)

+ 40%

by the following

600-50 000 (L2): JD

=

-°

5

0.99iV R e L

(7.3-29)

EXA MPLE 73-2. Mass A

solid

444

Transfer from a Flat Plate volume of pure water at 26.1°C is flowing parallel benzoic acid, where L = 0.244 m in the direction of

large

Chap. 7

Principles of Unsteady-State

to

a

flat

flow.

plate of

The water

and Connective Mass Transfer

velocity

is

0.061 m/s.

mol/m 3 The

The

solubility of benzoic acid in water

diffusivity of

.

benzoic acid

mass-transfer coefficient k L and the flux

x 10~ 9

1.245

is

NA

m

2

/s.

is 0.02948 kg Calculate the

.

Since the solution is quite dilute, the physical properties of water 1°C from Appendix A.2 can be used.

Solution: at 26.

^

=

p

= 996 kg/m 3

D AB =

10~*Pa-s

8.71 x

1.245 x 10~ 9

m

2

/s

The Schmidt number is x 10"*

8.71

"

Nsc

The Reynolds number

is

\ _^_ ~ ~

WX

l

10

(7.3-29),

jD

=

=

5

%

0.99Ar R C

definition of J D

from Eq.

0.99(1.700 x lO

(7.3-5)

=

JD Solving for

k'c

,

k'c

K =

= J D v(N Sc )~ 2/i

.

is

| (Wsc)

)

=

0.00758

2'3

(7.3-5)

Substituting

A

for

-05

4

is

0.00758(0.0610X702)-

In this case, diffusion (7.2-10)

x lO" 4

8.71

p.

The

_ ~

0-244(0.0610X996)

"*<- L

Using Eq.

702

9 996(1.245 x 10~ )

2'3

=

known

values

6 5.85 x 10~

and

solving,

m/s

through nondifFusing B, so k c

in

Eq.

should be used.

N a = ~~~ {c Al ~ X BM

=

c A2 )

-

k c (c A1

c A2 )

(7.2-10)

Since the solution is very dilute, x BSI = 1.0 and k'c = k c Also, c A = 2.948 x 10" 2 kg mol/m 3 (solubility) and c A2 = 0 (large volume of fresh water). Substituting into Eq. (7.2-10), .

t

NA = Mass

(5.85

x 10" 6 X0.02948

-

0)

=

1.726 x 10~

7

kg mol/s

•

m

2

For flow past single spheres and for very where v is the average velocity in the empty test section before the sphere, the Sherwood number, which is k'c D pJD A8 should approach a value of 2.0. This can be shown from Eq. (6.2-33), which was derived for a stagnant medium. Rewriting Eq. (6.2-33) as follows, where D is the sphere diameter, p 2.

low

transfer for flow past single spheres.

N Rc = D p vp/p,

,

N A = ^f(c Al - c A2

)

=

k c (c A

,

-

(73-30)

c A2 )

p

The mass-transfer

coefficient k c

,

which

is k'c

k'c

Sec. 7.3

=

for a dilute solution,

^

Mass-Transfer Coefficients for Various Geometries

is

then

(7.3-31)

445

Rearranging,

-j—z = Of course,

N Sh =

(73-32)

2.0

natural convection effects could increase

k'c

.

Schmidt number range of 0.6-2.7 and a Reynolds number range of 1-48 000, a modified equation (G 1) can be used. For gases

for a

N Sh = 2 + 0.552iV° 53 e

^

3

(7.3-33)

number

This equation also holds for heat transfer where the Prandtl

replaces the

Schmidt number and the Nusselt number hDJk replaces the Sherwood number. For liquids (G3) and a Reynolds number range of 2 to about 2000, the following can be used.

N Sh = 2 + 0.95 N^Nli For

liquids

3

(73-34)

and a Reynolds number of 2000-17 000, the following can be used

(S5).

N sh = 0341N™ 2 Nlt 3

(73-35)

EXAMPLE

7J-3. Mass Transfer from a Sphere Calculate the value of the mass-transfer coefficient and the flux for mass transfer from a sphere of napthalehe to air at 45°C and 1 atm abs flowing at

The diameter of the sphere is 25.4 mm. The diffunaphthalene in air at 45°C is 6.92 x 10~ 6 m 2 /s and the vapor pressure of solid naphthalene is 0.555 Hg. Use English and SI units.

a velocity of 0.305 m/s. sivity of

mm

D AB = 6.92 x 10" 6 (3.875 x m = 0.0254(3.2808) = 0.0833

In English units

Solution:

The diameter D p = 0.0254

10

4

ft.

= 0.2682 ft 2/h. From Appendix )

A. 3 the physical properties of air will be used since the concentration of

naphthalene fi

=

is

low.

10"

1.93 x

p

=

5

Pas =

1.93 x

kg/m 3 =

1.113

10" 5 (2.4191 x i0 3 )

-±^- =

=

0.0695 lbjft

0.0467 lbjft-h

3

5 v

=

0.305 m/s

The Schmidt number

=

0.305(3600 x 3.2808)

0 0461

_

t

pD AB

"

The Reynolds number

_

446

2 5o 5

"

6

2

505

)

is

Nrc_

+

=

5

1.113(6.92 x lO"

_ D p vp _ ~

2

ft/h

0.0695(0.2682)

1.93 x 10

Ns <

N Sh =

3600

is

N Sc

Equation

=

0.0833(3600)(0.0695)

0.0467

" 446

"

0.0254(0.3048X1.113) 1.93 x

10"

5

(7.3-33) for gases will be used.

0.552(N Rc )°-

Chap. 7

53

(/v"

1/3

Sc )

=

2

+

0.552(446)°'

Principles of Unsteady-Stale

53

(2.505)"

3

=

21.0

and Convective Mass Transfer

From

Eq.

(7.3-3),

—

=K

Wsh

=

k'

^AB

knowns and

Substituting the

r D AB

solving,

6 6.92 x 10"

From Table

7.2-1,

Hence, for

T= k'G

+

45

=

Since the gas

= RT ~

0.555/760

" is

A

for

(7.2-12)

The

=

318(1.8)

=

574°R,

=

""

2 163 X 10 "

9

kg

-

x 10

=

Pai) 4

lb

mol/h

-ft

-

=

2.163 x 10~ (74-0

area of the sphere

0)

1.599 x 10

^

'

Pa

ti(0.O833)

xl0~

packed beds.

-

0)

-7

kg mol/s-m 2

is

2

2

=

2.18 x 10"

)(^)^ N

transfer to

'

2

Mass

transfer to

2

2

ft

2.025

Total amount evaporated = A A = (1.18 x _7 6 x 10" lb mol/h = (1.599 x 10 )(2.025 x 10"

Mass

atm

.

4 0.1616(7.303 x 10"

9

A = nDp =

3.

•

= 1.0 and k'G = k G Substituting into Eq. through stagnant B and noting that p Al = atm = 74.0 Pa and p A2 = 0 (pure air),

*

1.180 x 10~

(2->8

ft

m ° 1/S

very dilute, y gM diffusing

=

2

0.1616 lb mol/h

(0.730X573)

x i0" 3

Na = MP,. -

=

K=

67.6

""8314(378)

= 7.303 =

318

k'c

5.72

k '°

=

273

a

RT

c

xlO-m

10~" 4 3 )

=

2

2

10~ )= 2.572 X2.18 x 10" 10 kg mol/s x 3.238

and from packed beds occurs often

in

processing operations, including drying operations, adsorption or desorption of gases or liquids

by

contained

and mass transfer of gases and liquids to Using a packed bed a large amount of mass-transfer area can be

solid particles such as charcoal,

catalyst particles. in

a relatively small volume.-

The void

fraction in a bed

of void space plus solid.

The

is £,

m

3

volume void space divided by the

m

3

total

volume

values range from 0.3 to 0.5 in general. Because of flow

channeling, nonuniform packing,

etc.,

accurate experimental data are difficult to obtain

and data from different investigators can deviate considerably. For a Reynolds number range of 10-10000 for gases in a packed bed of spheres (D4), the recommended correlation with an average deviation of about +20% and a

maximum of about +50%

is

jD

Sec. 7.3

=

jH

=

^2 0 4548

Af-o.*o69

Mass-Transfer Coefficients for Various Geometries

(7 .3.36)

447

It

has been shown (G4, G5) that J D and J„ are approximately equal. The Reynolds

number

is

defined as

iV Re

= D p v'p/n, where D p

is

diameter of the spheres and

empty tube without packing. For Eqs.

superficial mass average velocity in the

the

v' is

(7.3-36)—

and Eqs. (7.3-5H7.3-6), y' is used. For mass transfer of liquids in packed beds, the correlations of Wilson and Geankoplis (Wl) should be used. For a Reynolds number D p v'p/p. range of 0.0016-55 and a Schmidt number range of 165-70600, the equation to use is

(7.3-39)

For

liquids

N R" 2

=

JD

£

'3

(7.3-37)

and a Reynolds number range of 55-1500 and a Schmidt number range

of 165-10690,

JD

=

—

N R- 031

(73-38)

e

Or, as an alternate, Eq. (7.3-36) can be used for liquids for a Reynolds number range of 10-1500.

For fluidized beds of spheres, Eq. (7.3-36) can be used for gases and liquids and a Reynolds number range of 10-4000. For liquids in a fluidized bed and a Reynolds

number range of 1-10 (D4),

UD =

1.1068

N^

12

(73-39)

If packed beds of solids other than spheres are used, approximate correction factors can be used with Eqs. (7.3-36)-(7.3-38) for spheres.This is done, for example, for a given

nonspherical particle as follows. The particle diameter to use in the equations to predict

JD

is

same surface area

the diameter of a sphere with the

flux to these particles in the

bed

alternative approximate procedure to use

4.

obtained and then k c

in

is

calculated using Eqs. (7.3-40)

and

An

calculate the total flux in a packed bed,7 D is Then knowing the total volume Vb m 3 of

m/s from the J D

.

the bed (void plus solids), the total external surface area transfer

The

given elsewhere (G6).

is

To

Calculation method for packed beds.

first

as the given solid particle.

then calculated using the area of the given particles.

is

A

m2

of the solids for

mass

(7.3-41).

(73-40)

where a

is

the

m 2 surface area/m 3

total

volume of bed when the

solids are spheres.

A = aVb To

calculate the mass-transfer rate the log

(7.3-41)

mean

driving force

at the inlet

and

outlet

of the bed should be used. '

kf *, N A A = Ak

(cAi

a

~

C4l)

~

(C

-

4

'

~

Cai) (7.3-42)

c

In

c Ai

where the

final

of the solid,

in

term

is

the log

kg mol/m 3

;

c Al

mean is

driving force: c Ai

the bulk stream

NA A = Chap. 7

cA2 is

the concentration at the surface

the inlet bulk fluid concentration;

The material-balance equation on

448

-

Principles

V(c A2

andc^ 2

is

tne outlet.

is

-c Al

)

(7.3-43)

of Unsteady-State and Convective Mass Transfer

V

inm 3 /s.

Equations (7.3-42) and (7.3-43) must both be satisfied. The use of these two equations is similar to the use of the log mean temperature difference and heat balance in heat exchangers. These two equations can also be used for a fluid flowing in a pipe or past a flat plate, where A is the pipe wall area

where

is

volumetric flow rate of fluid entering

or plate area.

EXAMPLE 73-4.

Mass Transfer of a Liquid in a Packed Bed 7 3 Pure water at 26.TC flows at the rate of 5.514 x 1CT m /s through a packed bed of benzoic acid spheres having a diameter of 6.375 mm. The 2 total surface area of the spheres in the bed is 0.01198 m and the void fraction is 0.436. The tower diameter is 0.0667 m. The solubility of benzoic 2 3 acid in water is 2.948 x 10" kgmol/m .

Compare

(a)

Predict the mass-transfer coefficient k c

(b)

mental value of 4.665 x 10" 6 m/s by Wilson and Geankoplis (Wl). Using the experimental value of k c predict the outlet concentration of benzoic acid in the water.

.

with the experi-

,

Since the solution is dilute, the physical properties of water will 3 be used at 26.1°C from Appendix A.2. At 26A°C,n = 0.8718 x 10" Pa-s, 3 3 10" Pa s and from Table 6.3-1, p = 996.7 kg/m At 25.0°C,/i = 0.8940 x D A g= 1.21 x 10" 9 m 2 /s. To correct D AB to 26. using Eq. (6.3-9), D Ag oz T/fi. Hence, Solution:

•

.

PC

D, B (26.1°C)

The tower

v= (5.514

o x 10- 9

(1.21

=

9 1.254 x 10"

=

cross-sectional area 7

x 10" )/(3.494 x 10"

=

/V, Sc

—DH— = P

The Reynolds number

AB

ix

Then, using Eq.

=

1

1

The

predicted

=

=

3.494 x 10"

0.8718x 10" 3 s9 996.7(1.245 x 10" )

09

and

=

3

m

2 .

Then

=

702.6

_

for dilute solutions,

k'c

09

1

2'3

=^|(1.150)-

=

2'3

2.277

solving,

= j(N^ k'c

2

1 1.578 x 10" m/s. Then,

=

)

k

JD

/s

4 0.006375(1.578 x 10" X996.7) 3 100.8718 x

f(^r

(7.3-5)

2

(tt/4X0.0667)

3

Using Eq. (7.3-37) and assuming k c

/o

m

is

ZVp _ ~

_ N *<~

/299.1V0.8940 x 10" 3 \ )(^-j( 08718 xlo - 3 j

=

2.277

= L578

^

10

-, (702.6)*'

4.447 x 10~ 6 m/s. This compares with the experimental

value of 4.665 x 10" 5 m/s. For part (b), using Eqs. (7.3-42) and (7.3-43), (

Ak c

c Ai

~~

~

c Al) ,

c Ai

(

c Ai

C A\

~

c Al)

=V(c A2 -c M

(73-44)

)

In

C Ai

The

A = Sec. 7.3

~

C A2

values to substitute into Eq. (7.3-44) are c Ai 0.01198,

V =

7 5.514 x 10"

=

2.948 x 10"

2 ,

c A1

=0,

.

Mass-Transfer Coefficients for Various Geometries

449

0.01198(4.665 x 10"

^2 -

0)

7

(5.514 x i
-0 - c A2

2 2.948 x 10~

" 2.948 x 10 _ Solving, c A2

=

2.842 x 10"

3

kgmol/m 3

- 0)

.

Mass

Experimental data have been obtained transfer for flow past single cylinders. mass transfer from single cylinders when the flow is perpendicular to the cylinder. The cylinders are long and mass transfer to the ends of the cylinder is not considered. For the

5.

for

Schmidt number range of 0.6

to 2.6 for gases

and 1000

to 3000 for liquids

number range of 50 to 50000, data of many references plotted and the correlation to use is as follows: JD

The data

scatter considerably

transfer with J D 6.

= JH

=

0.600(N Re )-

(B3, LI,

Ml,

and a Reynolds

S4, VI) have been

0 487 -

(7.3-45)

by up to ±30%. This correlation can also be used

for heat

.

Liquid metals mass transfer.

coefficients of liquid metals

In recent years several correlations for mass-transfer

have appeared

in the literature.

It

has been found (Gl) that

with moderate safety factors, the correlations for nonliquid metals mass transfer

may be

used for liquid metals mass transfer. Care must be taken to ensure that the solid surface wetted. Also,

if

the solid

is

an

alloy, there

may

is

exist a resistance to diffusion in the solid

phase.

MASS TRANSFER TO SUSPENSIONS OF SMALL

7.4

PARTICLES Introduction

7.4A

Mass transfer from or to small suspended particles in an agitated solution occurs in a number of process applications. In liquid-phase hydrogenation, hydrogen diffuses from gas bubbles, through an organic liquid, and then to small suspended catalyst particles. In fermentations, oxygen diffuses from small gas bubbles, through the aqueous medium, and then to small suspended microorganisms.

For a liquid-solid dispersion, increased agitation over and above that necessary freely

suspend very small particles has very

to the particle (B2).

When

on the mass-transfer

little effect

to

coefficient k L

the particles in a mixing vessel are just completely suspended,

turbulence forces balance those due to gravity, and the mass-transfer rates are the same as for particles freely

which

so,

is

moving under

the size of

particles, their size

is

gravity.

With very small

many microorganisms

in

particles of say a few

fermentations and

smaller than eddies, which are about 100 /im or so

increased agitation will have

little effect

on mass

some in size.

or

catalyst

Hence,

transfer except at very high agitation.

For a gas-liquid-solid dispersion, such as in fermentation, the same principles hold. However, increased agitation increases the number of gas bubbles and hence the interfacial area. The mass-transfer coefficients from the gas bubble to the liquid and from the liquid to the solid are relatively unaffected.

7.4B

1.

Equations for

Mass

Mass Transfer

transfer to small particles

to

Small Particles

<0.6 mm.

Equations to predict mass transfer

to

small particles in suspension have been developed which cover three size ranges of

450

Chap. 7

Principles of Unsteady-State

and Conveclive Mass Transfer

The equation for particles <0.6 mm (600 pm) is discussed first. The following equation has been shown to hold to predict mass-transfer

particles.

from small gas bubbles such as oxygen or

air to the liquid

coefficients

phase or from the liquid phase

to the surface of small catalyst particles, microorganisms, other solids, or liquid drops

(B2.C3).

^ ^ + 03W^(^Y 2

k

m

(7.4-1)

2

/s, D is the diameter of the p is the bubble or the solid in m, viscosity of the gas particle solution in kg/m-s, pc 2 g = 9.80665 m/s Ap = {p c — p p ) or {p p — p c ), p c is the density of the continuous phase in

where D AB

is

the diffusivity of the solute

A

in solution in

,

kg/m 3 and p p is the density of the gas or solid particle. The value of Ap is always positive. The first term on the right in Eq. (7.4-1) is the molecular diffusion term, and the second term is that due to free fall or rise of the sphere by gravitational forces. This ,

equation has been experimentally checked for dispersions of low-density solids

and

tated dispersions

for

small gas bubbles

EXAMPLE 7.4-1.

Mass

in agi-

in agitated systems.

Transfer from Air Bubbles

Fermentation fermenter from air 2 bubbles at 1 atm abs pressure having diameters of 100 ^im at 37°C into water having a zero concentration of dissolved 0 2 The solubility of 0 2 7 from air in water at 37°C is 2.26 x 10" gmol0 2 /cm 3 liquid or 2.26 x 10"* 3 kg mol 0 /m The diffusivity of 0 in water at 37°C is 3.25 x 10~ 9 m 2 /s.

Calculate the

maximum

rate of absorption of

0

in

in a

.

.

2

Agitation

is

2

used to produce the air bubbles.

The mass-transfer

resistance inside the gas bubble to the outside bubble can be neglected since it is negligible (B2). Hence, the mass-transfer coefficient k'L outside the bubble is needed. The given data are

Solution:

interface of the

Dp =

=

100 fim

x 10

1

-4

D AB =

m

3.25 x 10"

9

m

2

/s

At 37X, /^(water)

=

6.947 x 10

p c (water)

=

Nl>

3

=

(215)

2/3

s

6.947 x 10

p„(air)

-4

kg/m-s

=1.13 kg/m 3

10-

10" 9 ) (994X3.25 x

- pp =

Ap = p c

35.9

=

6.947 x

_

p c D AB

=

Pa

994 kg/m 3 fe

Sc

~ 4

994

-

1.13

= 993 kg/m 3

Substituting into Eq. (7.4-1),

x IP" 4 x 10"

2(3.25 1

= The

flux

is

6.50 x 10"

as follows

NA = = Sec. 7.4

5

9 )

0.31

-

993 x 6.947 x 1Q- 4 x 9.806

35.9

+

c A2 )

5.18 x 10"

(994)

16.40 x 10"

assumingfc L

k L (c Al

Mass Transfer

+

8

=

=

k'L

5

=

2.290 x 10

-4

m/s

for dilute solutions.

4 4 2.290 x 10" (2.26 x 10"

kg mol

1/3

2

0

2

-

0)

/s-m 2

to Suspensions of Smalt Particles

451

Knowing

the total

number of bubbles and

possible rate of transfer of

0

their area, the

to the fermentation liquid

2

In Example 7.4-1, k L was small. For mass transfer of

microorganism with

Dp =

1

/im, the

0

2D AB/D p would be 100

term

maximum

can be calculated.

2

in

a solution

times larger.

to

Note

a

that

in Eq. (7.4-1) becomes small and the mass-transfer becomes essentially independent of size D p In agitated vessels with gas introduced below the agitator in aqueous solutions, or when liquids are aerated with

diameters the second term

at large

coefficient k L

.

sintered plates, the gas bubbles are often in the size range covered by Eq. (7.4-1) (B2, C3, Tl).

In aerated mixing vessels the mass-transfer coefficients are essentially independent

of the power input. However, as the power

is increased, the bubble size decreases and the mass transfer coefficient continues to follow Eq. (7.4-1). The dispersions include those in which the solid particles are just completely suspended in mixing vessels. Increase in agitation intensity above the level needed for complete suspension of these small particles

results in only a small increase in k L (C3).

Equation

(7.4-1)

has also been

shown

to

apply to heat transfer and can be written as

follows (B2, C3):

(7.4-2)

2.

Mass

>

transfer to large gas bubbles

mm, the

2.5

>

2.5

mm.

For large gas bubbles or

drops

liquid

mass-transfer coefficient can be predicted by

(7.4-3)

Large gas bubbles are produced when pure liquids are aerated

in mixing vessels and columns (CI). In this case the mass-transfer coefficient k'L or k L is independent bubble size and is constant for a given set of physical properties. For the same

sieve-plate

of the

physical properties the large bubble Eq. (7.4-3) gives values of k L about three to four

times larger than Eq. (7.4-1) for small particles. Again, Eq. (7.4-3) shows that the k L essentially independent of agitation intensity in an agitated vessel

and gas velocity

in

is

a

sieve-tray tower.

3.

Mass

in the size

transfer in the transition

range 0.6 to 2.5

can be -approximated by assuming that

coefficient

mass

In

transfer to particles in transition region.

region between small and large bubbles

mm,

the mass-transfer

increases linearly with bubble

it

diameter (B2, C3).

4.

Mass

In the preceding three regions,

transfer to particles in highly turbulent mixers.

the density difference between phases

is

sufficiently large to

cause the force of gravity to

primarily determine the mass-transfer coefficient. This also includes solids just pletely

suspended

needed

for

in

mixing

vessels.

When

agitation

power

is

com-

increased beyond that

suspension of solid or liquid particles and the turbulence forces become larger

than the gravitational forces, Eq. (7.4-1)

where small increases

in k'L

is

not followed and Eq. (7.4-4) should be used

are observed (B2, C3).

(7.4-4)

where PjV

452

is

power input per

Chap.

7

unit

volume defined

Principles

in

Section

3.4.

The data deviate

of Unsteady-State and Convective Mass Transfer

substantially by is

up to 60% from

this correlation. In the case of gas-liquid dispersions

it

quite impractical to exceed gravitational forces by agitation systems.

The experimental data easily

suspended and

coefficient will

if

are complicated by the fact that very small particles are

their size

is

of the order of the smallest eddies, the mass-transfer

remain constant until a large increase

power input

in

added above that

is

required for suspension.

MOLECULAR DIFFUSION PLUS CONVECTION AND CHEMICAL REACTION

7.5

Different Types of Fluxes and Fick's

7.5A

Law

was defined as the molar flux of A in kg raoM/s m 2 relative to the molar average velocity v M of the whole or bulk stream. Also, N A was defined as the molar flux of A relative to stationary coordinates. Fluxes and velocities can also be defined in other ways. Table 7.5-1 lists the different types of fluxes and velocities often In Section 6.2B the flux J*

used

in

Table

-

binary systems.

7.5-1.

Types of Fluxes and Velocities

Different

in

Binary Systems

Mass Flux A/s-m 2

(kg

Relative to fixed coordinates

'U

Relative to molar average velocity v M

j*

Relative to mass average velocity v

Ja

= = =

Pa

Pa("a

NA

+

J* J'a

+ NB =

=0

J*

+ 7b =

JA

is

Forms

= Ja/ M a

of Fick's

mass average

the

Law jA

~ ~

J*

=

C A (v A

Ja

=

c a(v a

»m) ")

N A =J A +

nJM A

= ~ cD ab dxjdz

J*

velocity v

=-

)

- vM - v)

)

Above

N A = -J*a+c a v m

0

Different

The

NA

cu M

m1

N A = Ca "a

"a

Pa(v a

Relations Between Fluxes

Molar Flux [kg mol A/s

)

CA V

= pv

"A

+

nA

=j A + Pa"

for Diffusion

nB

Flux

= -pD AS dwjdz

velocity of the stream relative to stationary

coordinates and can be obtained by actually weighing the flow for a timed increment. is

related to the velocity v A

and

v

vB

= wa

by

»a

+ w b vb =

— Pa

va

+

P where w A velocity of

m/s

is

is

A

pjp,

It

the weight fraction of A;

wB

,

Pb —

vs

the weight fraction of B;

relative to stationary coordinates in m/s.

rn z i\ ('•S-l)

P

The molar average

and v A is the velocity v M in

relative to stationary coordinates.

v

Sec. 7.5

m = xa

va

+ xb vB =

— vA + — c

vB

(7.5-2)

c

Molecular Diffusion Plus Convection and Chemical Reaction

453

The molar

diffusion flux relative to the

molar average velocity v M defined previously

is

=

J*a

The molar diffusion

flux

JA

- vM

c A {v A

relative to the

(73-3)

)

mass average

Ja =

c a ("a

-

velocity v

is

(73-4)

v)

Fick's law from Table 7.5-1 as given previously

is

relative

lov^ and

is

J*a=-cD ab -^ Fick's law can also be defined in terms of a

mass

(7.5-5)

flux relative to v as given in

Table

7.5-1.

~

jA=-pD AB

dw.

(73-6)

EXAMPLE 75-1. Proof of Moss Flux Equation Table 7.5-1 gives the following relation: ;„+J"b

Prove

this relationship

=0

(7.5-7)

using the definitions of the fluxes in terms of velo-

cities.

From Table and rearranging,

Solution: for j B

,

7.5-1 substituting

Pa v a

Pa

va

Substituting Eq. (7.5-1) for

+

pB

vB

and p

pB

-

for

vb

i\p A

pA

+

—

-

pB v

= 0

(7.5-8)

+

pB)

=

(7.5-9)

v) for j A

and pg(v B

0

p B the identity ,

is

v)

proved.

Equation of Continuity for a Binary Mixture

7.5B

A

v

- pA v +

—

p A {v A

general equation can be derived for a binary mixture of

A and B

for diffusion

and

convection that also includes the terms for unsteady-state diffusion and chemical reac-

Wc

tion.

shall

make

a

space as shown in Fig. rate of

\

mass A in/

mass balance on component A on an element Ax Ay Az The general mass balance on A is

fixed in

7.5-1.

/rate of

\mass A out \

/ rate of

+

,

^generation of mass A J

=

/ rate of

\ ,

(7.5-10)

,

^accumulation of mass A J

rate of mass A entering in the direction relative to stationary coordinates is Az kg A/s and leaving is (n A x\x + Ax)&y Similar terms can be written for the y and z directions. The rate of chemical production of A is r A kg A generated/s m 3 volume and the total rate generated is r A (Ax Ay Az) kg A/s. The rate of accumulation of A is (dp A /dt)Ax Ay Az. Substituting into Eq. (7.5-10) and letting Ax, Ay, and Az approach

The

(n Ax]x )Ay

•

zero,

454

Chap. 7

Principles of Unsteady-Slate

and Convective Mass Transfer

z

Figure

Mass balance for A

7.5-1.

in

a binary mixture.

In vector notation,

(7.5-12) dt

Dividing both sides of Eq.

dN Ax

+ dt

where R A

is

MA

by

(7.5-1 1)

dN Ay

,

m3

•

.

3N Az

+ ,

dy

dz

Substituting

NA

dx

V

kg mol A generated/s

,

N = -cD A

dx t

+

A

dz

and writing the equation

oc.

is

7.5C

/.

(

(7.5-14)

cA U vM

becomes

cD AB Vx A = )

RA

(7.5-15)

Special Cases of the Equation of Continuity

Equation for constant c and

D AB

= P/RT,

the general equation (7.5-15)

+

c

In diffusion with gases the total pressure

.

and Eq.

(7.5-

1

5)

is

c A (V

•

v

M + )

(v

Vc A

M

For the

2

dt

is

T

ox

)

- D AB V 2 c A = R A

special case of

=

(7.5-16)

equimolar counterdiffus-

constant, v M

=

0,

2

2

d cA -

dy

1 2

D AB =

constant,

d\, A

„

dz

(7.1-9)

2

Eq. (7.1-9) derived previously and this equation

unsteady-state diffusion of a dilute solute

Sec. 7.5

often

becomes S cA

This equation

is

and substituting Vx A = Vc A /c, we obtain

Equimolar counterdijfusion for gases.

0,

P

constant for constant temperature T. Starting with

ion of gases at constant pressure and no reaction, c

RA =

7.5-1,

the final general equation.

constant. Then, since c

2.

+ V -^ V ")-(V

(7.5-13)

and Fick's law from Table

for all three directions, Eq. (7.5-13)

-^f This

R.

A

in a solid or a liquid

is

whenD^

Molecular Diffusion Plus Convection and Chemical Reaction

also is

used for

constant.

455

Equation for constant p and D AB (liquids). In dilute liquid solutions the mass density p D AB can often be considered constant. Starting with Eq. (7.5-12) we substitute

3.

and

= ~pDAB VwA + pA \ from Table 7.5-1

nA

Then

into this equation.

using the fact that for

constant p, Vw A = Vp^/p and also that (V • v) = 0, substituting these into the resulting equation, and dividing both sides by A we obtain

M

?5±

+

(

V

,

2 VcJ - D AB V cA = R A

-

at

(7.5-17)

Special Cas«s of the General Diffusion

7.5D

Equation at Steady State

The

Introduction and physical conditions at the boundaries.

/.

diffusion

and convection

of a binary mixture in

general equation for

one direction with no chemical reaction

has been given previously.

To

integrate this equation

conditions at is

z,

and

at z 2

d

.

C

^+ -^(N

N A = -cD AB

-

dz

steady state

at

Often

in

many

A

+N B

(6.2-14)

)

c

it

is

necessary to specify the boundary

mass-transfer problems the molar ratio

N JN B

determined by the physical conditions occurring at the two boundaries.

As an example, one boundary of the diffusion path may be impermeable to species B B is insoluble in the phase at this boundary. Diffusion of ammonia (A) and nitrogen (B) through a gas phase to a water phase at the boundary is such a case since nitrogen is essentially insoluble in water. Hence, N B = 0, since at steady state N B must have the same value at all points in the path of z 2 — z,. In some cases, a heat balance in the adjacent phase at the boundary can determine the flux ratios. For example, if component A condenses at a boundary and releases its latent heat to component B, which vaporizes and diffuses back, the ratios of the latent heats determine the flux ratio. In another example, the boundary concentration can be fixed by having a large volume of a phase flowing rapidly by with a given concentration x Al In some cases the concentration x A may be set by an equilibrium condition, whereby x A is in equilibrium with some fixed composition at the boundary. Chemical reactions can also influence the rates of diffusion and the boundary conditions. because

.

,

2.

y

Equimolar counterdiffusion.

N A = —N B

,

For the special case of equimolar counterdiffusion where

Eq. (6.2-14) becomes, as

shown

previously, for steady state and constant

N A = J*=-cD AB -— = dz

3. Diffusion

of

A

—

(7.5-18) z

t

For gas A diffusing through and integration of Eq. (6.2-14) gives Eq. (6.2-22).

through stagnant, nondiffusing B.

stagnant nondiffusing gas B,

NB

NA = Several other

z2

c,

=

0,

—

Da " P

RT(z 2

more complicated

-

(pA

t

-p A2

)

(6.2-22)

z x )p BM

cases of integration of Eq. (6.2-14) are considered

next.

4.

Diffusion

and B

456

at a boundary. Often in catalytic reactions where A from a catalyst surface, the relation between the fluxes N A and

and chemical reaction

are diffusing to'and

Chap. 7

Principles of Unsteady-State

and Convective Mass Transfer

NB

at

steady state

A

is

controlled by the stoichiometry of a reaction at a boundary.

from the bulk gas phase to the catalyst surface, where instantaneously and irreversibly in a heterogeneous reaction as follows: example

is

gas

diffusing

it

A->2B Gas B

then diffuses back, as

At steady state or

N B = — 2N A

.

1

shown

is

mol of A

The

An

reacts

(7.5-19)

in Fig. 7.5-2.

diffuses to the catalyst for every 2

mol

of B diffusing away,

negative sign indicates that the fluxes are in opposite directions.

Rewriting Eq. (6.2-14) in terms of mole fractions,

NA Next, substituting

=- cD AB

+ x A (N A + N B

N B = — 2N A into Eq. (7.5-20),

N A = -cD AB ~^ + x A (N A -2N A Rearranging and integrating with constant 1.

(13-20)

)

Instantaneous surface reaction

c

[P

=

constant),

r Z2 =

r *" 2

i

dz

1

Na =

—r ^ 5

catalyst surface.

instantaneous, x A2

Equation

we obtain

(75-21)

the following:

dx

= -cD

11=0

is

..

:

NA

Since the reaction

)

=

0,

1

+

(73-22)

+X

-

(75-23)

x A2

because no

A can

exist next

to the

(7.5-23) describes the overall rate of the process of diffusion

plus instantaneous chemical reaction. 2.

Slow surface reaction. If the heterogeneous reaction at the surface is not instantaneous but slow for the reaction A -» 25, and the reaction is first order,

N Az = = }

where k\ (7.5-23)

k\c A

=k\cx A

(7.5-24)

the first-order heterogeneous reaction velocity constant in m/s. Equation

is

still

holds for this case, but the boundary condition x A2 at

z

=

8

is

obtained by

catalyst surface

NR

Figure

Sec. 7.5

7.5-2.

Diffusion of A and heterogeneous reaction at a surface.

Molecular Diffusion Plus Convection and Chemical Reaction

457

xA

solving for

Eq. (7.5-24),

in

—

= Xa2=-T, = 77k c k^c

Xa

(73-25)

l

For steady

The

NA

state,

,

=

= NA

6

rate in Eq. (7.5-26)

equation

is

+

1

than in Eq.

+

is 1

Diffusion

from point

denominator

at

a Boundary

at a partial pressure of 101.32

1

in the latter

N Jk\c.

and Chemical Reaction

7J-2.

diffuses

Pure gas A

(7.5-23), since the

=1+0 and in the former

x A2

EXAMPLE

less

is

Substituting Eq. (7.5-25) into (7.5-23),

.

kPa

to point

mm

away. At point 2 it undergoes a chemical reaction at the catalyst surface and A —* 2B. Component B diffuses back at steady state. The total pressure is P = 101.32 kPa. The temperature is 300 K and D AB = 2 4 0.15 x 10" m s. 2 a distance 2.00

For instantaneous rate of reaction, calculate x A2 and N A For a slow reaction where k\ =5.63 x 10" 3 m/s, calculate x A2 and

(a)

.

(b)

NA

.

For part

Solution:

(a),

5

=

4.062 x 10"

3

m, T = kg mol/m 3

2.00 x 10" 2

=

=

0 since no A can exist next to the Eq. (7.5-23) will be used as follows: 300 K, c = P/R.T = 101.32 x 10 3 /(8314 x 300) = 3 3 x Al = p Al /P = 101.32 x 10 /101.32 x 10 =

p A2

x A2

N B = —2N A

catalyst surface. Since

,

,

1.00.

A

_cD i? ~

=

5

1

1

+

2 4 (4.062 x 1Q- X0.15 x IP" )

x Al

~

+x A2

2.112 x 10"-* kg

2.00 x 10-

mol

/1/s- ni

=

(4.062

x 10-

2

)(0.15

4 x 10"

)

N Jk\c = N A /{5.63

+

x 4.062 x 10" 2 )

3

4

N

Even though

5.

is

in part (a) of

diffusion controlled.

Diffusion and

Example

2 )

kg mol A/s

=

m2

7.5-2 the rate of reaction

in a

phase.

Then.x^ =

.

0.4390.

is

As the reaction rate slows, the fluxA/^

homogeneous reaction

x

1.00

AV(5.63 x 10"

Solving by trial and error, A = 1.004 x 10 4 3 x 4.062 x 10" (1.004 x 10" )/(5.63 x 10"

NA

3

ln

1

flux

x 10"

,

iooTTo^ +

1.00

1+0

2

For part (b), from Eq. (7.5-25), x A2 = 2 4.062 x 10" ). Substituting into Eq. (7.5-26),

1

+

1

3

Equation

(7.5-23)

instantaneous, the

is

decreased also.

was derived

for the

some cases homogeneous phase B

case of chemical reaction of A at the boundary on a catalyst surface. In

component A undergoes an while diffusing as follows,

irreversible chemical reaction in the

A—

»

C.

Assume

component A

that

is

which can be a gas or a liquid. Then at steady state the equation follows where the bulk-flow term is dropped.

N Az = -D AB 458

Chap. 7

^+

0

Principles of Unsteady-Stale

very dilute in phase B, for diffusion of

A

is

as

(7.5-27)

and Convective Mass Transfer

Figure

Homogeneous chemical

7.5-3.

tion

and diffusion

in

reac-

a fluid.

TV 1

Az

I

I

|

I

!

I

I

I

\z

in

1

Writing a material balance on A shown

TV

Az\z + hz

.

Az

i

out

r

in Fig. 7.5-3 for the

Az element

for steady

state,

The

( rate of\

/ rate of

\A

^generation of

in

of

rate

first-order reaction rate of

A

A

per

A

m3

+

k'

is

= — k'c A

the reaction velocity constant in s"

cross-sectional area of

1

m

2

A

is

rate of generation

where

(73-28)

\ accumulation of

out

volume

/ rate of

1 .

(7.5-29)

Substituting into Eq. (7.5-28) for a

with the rate of accumulation being 0 at steady state,

N Azi tf) -

*'cm(1XAz)

Next we divide through by Az and

=

N Ml + AJU) + 0

Az approach

let

dN

=

(73-30)

zero.

k'c.

(7.5-31)

Substituting Eq. (7.5-27) into (7.5-31), d*c<

dz

The boundary conditions

are c A

sinh cA

=

(7.5-32)

2

=

c AX for z

z

1

+

0 and c A

=

c A2 for z

=

L. Solving,

(L-z)

c Al sinh

=

(7.5-33) k'

sinh

D7

f

This equation can be used at steady state to calculate^ at any reaction in gases, liquids, or even solids, where the solute

As an alternative derivation of Eq.

and

(7.5-32),

A

we can

is

z

and can be used

for

dilute.

use Eq. (7.5-17) for constant p

Z)„

dc A

D AB V 2 c A = R A

(7.5-17)

We set the first term dcjdt = 0 for steady state. Since we are assuming dilute solutions and neglecting the bulk flow term, v = 0, making the second term in Eq. (7.5-17) zero. For 3 a first-order reaction of A where A disappears, R A = —k'c A kg mol A generated/s m .

Sec. 7.5

Molecular Diffusion Plus Convection and Chemical Reaction

459

— D AB W

Writing the diffusion term

2

c A for only the

D AB which

=

-j^r

z

we obtain

direction,

k cA

(73-34)

of course, identical to Eq. (7.5-32).

is,

Unsteady-State Diffusion and Reaction

7.5E

Medium

in a Semiinfinite

Here we consider a case where dilute A is absorbed at the surface of a solid or stagnant fluid phase and then unsteady-state diffusion and reaction occur in the phase. The fluid or solid phase of c A

is

B

is

considered semiinfinite. At the surface where

kept constant

at c A0

The dilute solute A

.

z

=

0,

the concentration

mechanism

reacts by a first-order

A + B—y C and the

rate of generation

(73-35)

—k'c A The same diagram

is

Using Eq.

accumulation, dc

-

JV^ |:(1)

as in Fig. 7.5-3 holds.

.

(7.5-30) but substituting (dc A /dt){Az)(l) for the rate of

k'c A (l)(Az)

A = N A:iz + ^(l) +[-£)

(AzXl)

(7.5-36)

This becomes 8c A ,

01

The

initial

—

t

=

z

=

z

=

2

cA

=

0,

cA

=

oo,

cA

=

0,

amount Q

S= where

g

equation

is

kg mol

is

~ k'c A

useful

/i

of

for z

>

0

c A0

for

t

>

0

0

for

I

>

0

0

J k'/D AB

)

i exp(z /fc7£) /la ) x

/I

erfcf

/D AB/kWt +

c AOS

2 .

(

Many actual

and

to

+

J

^

(7.5-39)

)

is

i)erf s//c'<

where absorption occurs

of dissolved gases,

!

erfcf

absorbed up to time

absorbed/m

(7.5-38)

—-L= - V^f

+ V^e"*"]

(7-5-40)

cases are approximated by this case.

at the surface of

and unsteady-state diffusion and reaction occurs in used to measure the diffusivity of a gas in a solution, k'

(7.5-37)

is

cxp(- z

4-

total

oz

and boundary conditions are

The solution by Danckwerts (Dl)

The

2

d cA

n

Dab -zrf z

the solid or fluid. to

The

a stagnant fluid or a solid

The

results can be

determine reaction rate constants

determine solubilities of gas

in liquids

with which they react.

Details are given elsewhere (D3).

EXAMPLE 7J-3. Pure C0 gas at

Reaction and Unsteady-State Diffusion kPa pressure is absorbed into a dilute alkaline buffer solution containing a catalyst. The dilute, absorbed solute 2 undergoes a first-order reaction with k' = 35 s _I and D AB = 1.5 x 10~ 9 2

101.32

C0

460

Chap. 7

Principles

of Unsteady-State and Conveciive Mass Transfer

m

2

The

/s.

surface

CO

solubility of

2.961 x 10"

is

z

exposed to the gas for 0.010

is

absorbed/m 2

s.

7

kg mol/m 3 -Pa (D3). The Calculate the kg mol CO z

surface.

=

Solution: For use in Eq. (7.5-40), k't 35(0.01) = 0.350. Also, c A0 = 7 3 3 2 2.961 x 10" (kg mol/m -PaX101.32 x 10 Pa) 3.00 x 10" kg mol

=

S0 2 /m 3 Q=

.

(3.00 x 10

=

_2

_

7

1.458 x 10"

+

)yi.5 x 10 735[(0.35

^035 + yO^/Tce" 0 35 ] -

^)erf

CO z /m 2

kg mol

Multicomponent Diffusion of Gases

7.5F

The equations derived

in this

chapter have been for a binary system of A and B, which

is

probably the most important and most useful one. However, multicomponent diffusion

sometimes occurs where three or more components A,B,C, case

is

for diffusion of

A

NB =

Hence,

at constant total pressure.

the Stefan-Maxwell

method (Gl)

is

the log

mean ofp n = P

D Am = where x'B

= mol

B/mol

The

simplest

0,

inerts

=

N =

The

0,

c

final

equation derived using

for steady-state diffusion is

-

RT(z 2 where p iM

are present.

a gas through a stagnant nondiffusing mixture of B, C,D,

in

—

z,)p iM

p A1 and p i2

— —

p A2 Also, .

1

-77^ + x'c/D AC +

(7-5-42>

:

x'B/D AB

x B /(l

- P—

xj,

x'c

=

x c /(l

—

x^),

—

EXA MPLE 7.5-4. Diffusion of A Through Nondiffusing B and C At 298 K and atm total pressure, methane {A) is diffusing at steady state through nondiffusing argon (B) and helium (C). At z = 0, the partial pressures in atm are p Al = 0.4, p B1 = 0.4, p ci = 0.2 and at z 2 = 0.005 m, p A2 = 0.1, p B2 = 0.6, and p C2 = 0.3. The binary diffusivities from Table 6.2-1 5 are D AB = 2.02 x 10" m 2 /s, D AC = 6.75 x 10" 5 m 2 /s, and D BC = 7.29 x 1

t

10"

m 2 /s.

5

Calculate N A

=

- x A = 0.4/(1 - 0.4) = 0.667. At point 2, 1, x'B = x s/(l = 0.667. The value of x'B is constant throughout the path. - xj = 0.2/(1 - 0.4) = 0.333.

At point

Solution: x'B

.

0.6/(1

—

)

0.1)

= x c/(l Substituting into Eq. (7.5-42),

Also,Xc

D Am x'b/Dab

=

-

0.1

=

x'c/

2.635 x 10"

For calculating 1.0

+

w pn Then,

p,

0.90.

,

5

m

(p,- 2

Sec. 7.5

p Al

=

p,, 2

=

p A1

=

1.0

!;2

In (0.90/0.60) ,

0.333/6.75 x 10"

+

5

/s

= P—

= rr=fi In /Pn)

5

2

ft

=

0.667/2.02 x 10"

d ac

0.4(1.01325 x 10

5

0.1(1.01325 x 10

5

-

—

0.4

°- 740

=

0.6

atm

=

4.053 x 10

4

)

Pa

)

=

1.013 x 10*

Pa

atm, p i2

-

7

-

=P—

p A2

=

496 x 104 Pa

Molecular Diffusion Plus Convection and Chemical Reaction

461

Substituting into Eq. (7.5-41),

RT(z 2

-

z Y )p

m

5 5 _ (2.635 x 10" X1.01325 x 10 X4.053 - 1.013X10*) 4 (8314X298X0.OO5 - 0X7.496 x 10 )

=

8.74 x 10

Using atm pressure

RT{z 2

=

8.74

A number

-

x 10

^ (

r,)p,

_

2 kg mol /1/s-m

units,

°A " P

NA

-5

v

_

5

(2.635 x 10

X1 -0X0.4 (82.06 x 10- X298X0.OO5

Pa2>

kg mol /1/s-m

3

of analytical solutions have been obtained for other cases such as for

stagnant C, and the general case of two or

reader

7.6

is

0.1)

2

B through multicomponwith examples by Geankoplis (Gl) and the

equimolar diffusion of three components, diffusion of components ent mixture.

-

- 0X0.740)

These are discussed

more components

in detail

A

and

diffusing in a

referred there for further details*

DIFFUSION OF GASES IN POROUS SOLIDS AND CAPILLARIES

7.6A

Introduction

in porous solids that depends on structure was discussed for For gases it was assumed that the pores were very large and Fickian-type diffusion occured. However, often the pores are small in diameter and the mechanism of diffusion is basically changed. Diffusion of gases in small pores occurs often in heterogeneous catalysis where gases diffuse through very small pores to react on the surface of the catalyst. In freeze drying of foods such as turkey meat, gaseous H 2 0 diffuses through very fine pores of the porous

In Section

liquids

and

6.5C diffusion for gases.

structure.

Since the pores or capillaries of porous solids are often small, the diffusion of gases

may depend upon

the diameter of the pores.

We

first

the average distance a gas molecule travels before

it

define a

mean

free

)J*L f*L 2kM P

path X; which

is

collides with another gas molecule.

(7.6-1)

V

where X is in m, ji is viscosity in Pa -'s, P is pressure in N/m 2 T is temperature in K, = molecular weight in kg/kg mol, and R = 8.3143 x 10 3 N m/kg mol K. Note that ,

M

•

low pressures give large values of

X.

For

liquids, since X

is

so small, diffusion follows

Fick's law. In the next sections

we

shall consider

what happens

diffusion in gases as the relative value of the

mean

diameter varies. The total pressure P in the system of A and

462

B may

will

free

to the basic

mechanisms of

path compared to the pore

be constant, but partial pressures

be different.

Chap. 7

Principles of Unsteady-State

and Convective Mass Transfer

Knudsen Diffusion of Gases

7.6B

A

In Fig. 7.6-1 a a gas molecule

at partial pressure p A1 at the entrance to a capillary

m. The

diffusing through the capillary having a diameter of d

throughout.

The mean

free

path A

is

large

compared

to the

total pressure

diameter

P is

As a

d.

is

constant

result, the

molecule collides with the wall and molecule-wall collisions are important. This type called

Knudsen

is

diffusion.

The Knudsen diffusi vity

is

independent of pressure P and

D KA =

is

calculated from

\rv A

(7.6-2)

where D KA is diffusivity in m 2 /s, f is average pore radius in m, and v A is the average molecular velocity for component A in m/s. Using the kinetic theory of gases to evaluate v A the final

equation for

D KA

is

1/2

7

where

MA

is

molecular weight of

EXAMPLE 7.6-1.

A

in

Knudsen

kg/kg mol and

T

is

temperature

in

K.

of Hydrogen

Diffusivity

A H 2 (/1)-C 2 H 6 (B)

gas mixture is diffusing in a pore of a nickel catalyst used 5 for hydrogenation at 1.01325 x 10 Pa pressure and 373 K. The pore radius is 60 A (angstrom). Calculate the Knudsen diffusivity D KA of H 2 .

Solution:

Substituting into Eq. (7.6-3) for

r

=

6.0

x 10~

9

m,

MA

=

2.016,

and T = 373 K, /

D KA =

97.0ri

J — M

\l/2 )

/

=

9 97.0(6.0 x 10~ )

\ 2.016

'

AJ

= The

flux

7.92 x 10

z,

=

0,

NA = diffusion of

Sec. 7.6

m m 22
equation for Knudsen diffusion

Integrating between

The

-66

/t

for

pA

=

373 \I/2

"

p A1 and

z2

in a

=

L,

pore

pA

=

is

p A1

,

- x A2 ) = ^'(P-i. - ^2)

ICnudsen diffusion

is

completely independent of B, since

Diffusion of Gases in Porous Solids

and

Capillaries

(7-6-5)

A

collides

463

A

with the walls of the pore and not with B.

component

B.

When the Knudsen number A^,,

defined as

N Kn = is

>

similar equation can be written for

~

(7.6-6)

Knudsen and Eq.

10/1, the diffusion is primarily

about a 10% error. As N Kn gets larger, proaches the Knudsen type.

(7.6-5) predicts the flux to within

this error decreases, since the diffusion

ap-

Molecular Diffusion of Gases

7.6C

As shown

when

in Fig. 7.6-lb

diameter d or where

N Kn <

mean

the

1/100,

free path X is small compared to the pore molecule-molecule collisions predominate and

molecule-wall collisions are few. Ordinary molecular or Fickian diffusion holds and Fick's law predicts the diffusion to within about

smaller since the diffusion approaches

more

The equation for molecular diffusion

D

given in previous sections

P dx A

AB N A = - -jgr

A

flux ratio factor a

the diffusion

is

gets

+ *a(N a +

is

N„)

(7.6-7)

=

1

~

4-

(7.6-8)

Eqs. (7.6-7) and (7.6-8) and integrating for a path length of

NA =

« If

asN Kn

can be denned as

a

Combining

10%. The error diminishes

closely the Fickian type.

equimolar,

molecular diffusivity

D AB

D

,

ocR

n 1

P

L

NA = — NB

1

In 1

— —

L cm,

axj,

^

(7.6-9)

ax Al

and Eq.

(7.6-7)

becomes

Fick's law.

The

inversely proportional to the total pressure P.

is

Transition-Region Diffusion of Gases

7.6D

As shown in Fig. 7.6- lc, when the mean free path X and pore diameter are intermediate in between the two limits given for Knudsen and molecular diffusion, transition-type

size

diffusion occurs where molecule-molecule

and molecule-wall

collisions are important in

diffusion.

loss

The

transition-region diffusion equation can be derived by adding the

due

to molecule-wall collisions in Eq. (7.6-4)

collisions in Eq. (7.6-7) final differential

on

equation

a slice of capillary.

is

and that due

No ^chemical

momentum

to molecule— molecule

reactions are occurring.

The

(Gl)

RT

dz

where

D NA =

464

Chap. 7

(l-ax A )/D AB +l/D KA

Principles of Unsteady-State

(7.6-11)

and Convective Mass Transfer

This transition region diffusivity D N/4 depends slightly on concentrationx^,. Integrating Eq. (7.6-10),

N — ,

*RTL

In

(7.6-12)

l-axAl + D AB/D KA

This equation has been shown experimentally to be valid over the entire transition region (Rl). It reduces to the

equation

Knudsen equation

high pressures.

at

An

at

low pressures and

to the

molecular diffusion

equation similar to Eq. (7.6-12) can also be written for

component B. The term D AB/D KA is proportional to l/P. Hence, the term D AB /D KA becomes very small and N A in Eq.

as the total pressure (7.6-12)

P

increases,

becomes independent of

pressure since D AB P is independent of P. At low total pressures Eq. (7.6-12) becomes the Knudsen diffusion equation (7.6-5) and the flux N A becomes directly proportional to P for constant x Al and x A2 This is illustrated in Fig. 7.6-2 for a fixed capillary diameter where the flux increases as total pressure increases and then levels off at high pressure. The relative position of the curve depends, of course, on the capillary diameter and the molecular and Knudsen diffusivities. Using only a smaller diameter, D KA would be smaller, and the Knudsen flux line would be parallel to the existing line at low pressures. At high pressures the flux line would asymptotically approach the existing horizontal line since molecular diffusion is total

-

independent of capillary diameter. If

that

A

is

A —*

from Eq.

diffusing in a catalytic pore

and

reacts at the surface at the

B, then at steady state, equimolar counterdiffusion occurs or (7.6-8), a

=

1

-

1

=

0.

The effective diffusivity£> Wi4 from

end of the pore so

NA

N

R

.

Then

Eq. (7.6-11) becomes

1

+

l/D AB

The

diffusivity

is

(7.6-13)

\/D KA

then independent of concentration and

is

constant. Integration of

Knudsen diffusion equation (7.6-5) 10"

molecular diffusion equation (7.6-9)

10"

5?

transition region

10

equation (7.6-12)

-6

10" 10'

10

J

10" Pressure,

Figure

7.6-2.

10

P

Effect of total pressure

J

10

c

(Pa)

P

on the diffusion flux

NA

in

the transition

region.

Sec. 7.6

Diffusion of Gases in Porous Solids

and

Capillaries

465

(7.6-10) then gives

N A = ^{x Al -x A2 = -jj£±(p Al -p A2 )

(7.6-14)

)

This simplified diffusivity D'NA is often used in diffusion in porous catalysts even when equimolar counterdiffusion is not occurring. This greatly simplifies the equations for diffusion

An

and

by using

reaction

this simplified diffusivity.

alternative simplified diffusivity to use

is

to use

an average value ofx A

in

Eq.

(7.6-11), to give

-

(1

=

where x Alv

+ x A2 )/2.

(x Al

+

ax A „)/D AB

This diffusivity

more

is

l/D KA accurate than D'NA

.

Integration of

Eq. (7.6-10) gives

NA = Flux Ratios

7.6E

/.

Diffusion

reaction

in

D

for Diffusion of

open system.

~

(*ai

£jl

Gases

=

*ai)

{Pai

~ Pai)

(7.6-16)

in Capillaries

If diffusion in

porous solids or channels with no chemical P remains constant, then for an open

occurring where the total pressure

is

binary counterdiffusing system, the ratio

regimes and

is

o(NJN B

is

constant

in all

of the three diffusion

(Gl)

Nb Na

(7.6-17)

Hence,

(7.6-18)

In this case, gas flows by the

two open ends of the system. However, when chemical N B/N A and not Eq. (7.6-17).

reaction occurs, stoichiometry determines the ratio

2.

Diffusion

shown

in

When

closed system.

molecular diffusion

is

occurring

in

a closed system

constant total pressure P, equimolar counterdiffusion occurs.

in Fig. 6.2-1 at

EXAMPLE

N

A

K

7.6-2. Transition-Region Diffusion of He and 2 gas mixture at a total pressure of 0.10 atm abs and 298 is composed of (A) and He The diffusing an (S). mixture is through open capillary 0.010 2 m long having a diameter of 5 x 10" 6 m. The mole fraction ofN 2 at one end

N >

s

is

= 0-8 and at the other end is x A2 0.2. The molecular diffusivity D AB 5 2 6.98 x 10" /s at 1 atm, which is an average value by several investi-

=

xa i

m

gators.

Calculate the flux

(b)

Use the approximate equations

The

Solution:

x 10" 6 x Ai =0.8, 2.5

are

466

NA

(a)

MA =

at

given values are

m,

=

L =

m,

0.01

0.2, .C^g

=

28.02 kg/kg mol,

Chap. 7

steady state.

B

(7.6-16), for this case.

T = 273 + 25 = 298 K, f = 5 x 10" 6 /2 = P = 0.1(1.01325 x 10 5 ) = 1.013 x 10 4 Pa,

6.98 x 10"

M

and

(7.6-14)

=

5

m 2 /sat

1

atm. Other values needed

4.003.

Principles of Unsteady-State

and Convective Mass Transfer

The molecular 10" 4

m2

/s.

D KA = FromEq.

Eq.

is

5 6.98 x 10~ /0.1 = Knudsen diffusivity,

D AB =

x 10~ 6 )V298/28.02

97.0(2.5

=

x 10

7.91

-4

m

2

6.98

x

/s

(7.6-17),

Nn NA From

aim

diffusivity at 0.1

Substituting into Eq. (7.6-3) for the

\M A

M

v

/28.02

=

—2.645

sj 4.003

R

(7.6-8),

2.645 = l+^=lM

=

o

Substituting into Eq. (7.6-12) for part

~

A

=

(-1.645X8314X298X0.01)" -5

6.40 x 10

For part

(b),

kg mol/s-m

1

"

1

+ +

1.645(0.2)

~ U d ab +

(x

1

"rT~L

=

9.10 x 10"

~ 5

x Xa2))

-

6.98/7.91

1.645(0.8)+ 6.98/7.91

1

3) is

used.

1

~

1/6-98 x 10

=

3.708 x 10

-4

-4

+

m

Substituting into Eq. (7.6-14), the approximate flux

D '»' P N ~

+

2

the approximate equation (7.61

NA

(a),

4 x 1Q-*)(1.013 x 10 )

(6.98

- 1.645

A

.

1/7.91

2

x 10" 4

/s

is

4 4 (3-708 x 10- X1.013 x 10 )

,

8314(298)(0.01)

( °' 8

~ °" 2)

2 kg mol/s-m

Hence, the calculated flux is approximately 40% high when using the approximation of equimolar counterdiffusion (a = 0). The more accurate approximate equation (7.6-15) is used next. The average concentration is x A av = (x Ai + x A2 )/2 = (0.8 + 0.2)/2 = 0.50.

D"V!

(l-ccx A

J/D AB +

l/D KA 1

(1

=

+

4 1.645 x 0.5)/(6.98 x 10" )

2.581 x 10

-4

m

+

1/(7.91

4 x 10" )

2

/s

Substituting into Eq. (7.6-16),

4 4 x 10" X1.013 x 10 )

(2.581

(0.8

- 0.2)

8314(298X0.01)

In this case the flux

Sec. 7.6

=

6.33 x 10~

is

only

—

5

kg mol/s-m 2

1.1% low.

Diffusion of Gases in Porous Solids

and

Capillaries

467

Diffusion of Gases in Porous Solids

7.6F

In actual diffusion in porous solids the pores are not straight and cylindrical but are irregular.

Hence, the equations for diffusion in pores must be modified somewhat for The problem is further complicated by the fact that the pore

actual porous solids.

diameters vary and the Knudsen diffusivity

DA

is a function of pore diameter. As a result of these complications, investigators often measure effective in porous media, where c((

Na = If

a tortuosity factor t

side

is

is

~rtl"

used to correct the length

multiplied by the void fraction

Na ^ Comparing

e,

a6~ 19)

~ Xai)

{Xai

L

in

diffusivities

Eq. (7.6-16), and the right-hand

Eq. (7.6-16) becomes

£

^TWl

(Xai

~

(7 - 6~ 20)

Xa2)

Eqs. (7.6-19) and (7.6-20),

D At(f =

A

(7.6-21)

—f

some cases investigators measure D Acl[ but use D'NA instead of the more accurate inEq. (7.6-21). Experimental data (C4, S2, S6) show that r varies from about 1.5 to over 10. A reasonable range for many commercial porous solids is about 2-6 (S2). If the porous solid consists of a bidispersed system of micropores and macropores instead of a monodispersed pore system, the approach above should be modified (C4, S6). Discussions and references for diffusion in porous inorganic-type solids, organic solids, and freeze-dried foods such as meat and fruit are given elsewhere (S2, S6). In

Another type

of diffusion that

may occur

is

surface diffusion.

When

a molecular

layer of absorption occurs on the solid, the molecules can migrate on the surface. Details are given elsewhere (S2, S6).

NUMERICAL METHODS FOR UNSTEADY-STATE

7.7

MOLECULAR DIFFUSION 7.7A

Introduction

Unsteady-state diffusion often occurs in inorganic, organic, and biological solid materials. If

boundary conditions are constant with time, if they are the same on all sides or and if the initial concentration profile is uniform throughout the the methods described in Section 7.1 can be used. However, these conditions are

the

surfaces of the solid, solid,

not always

7.7B

1.

fulfilled.

Hence, numerical methods must be used.

Unsteady-State Numerical Methods for Diffusion

Derivation for unsteady state for a slab.

For unsteady-state diffusion

in

one direction,

Eq. (7.1-9) becomes dc. '

dt

468

Chap. 7

2

d c

D AB

(7.7-1)

dx

Principles of Unsteady-Slate

and Convective Mass Transfer

Since this equation

is

identical mathematically to the unsteady-state heat-conduction Eq.

(5.10-10),

= ~a

dt

identical

mathematical methods can be used

(5-10-10)

dx both diffusion and conduction

for solving

numerically.

Figure 7.7-1 shows a slab with width

shaded in

—

Making

area.

=

rate out

rate of accumulation in At

D AB A -^-(,c„-i where A

is

D

fin)

cross-sectional area

(

centered at point n represented by the this slab at the

time

t

when

the rate

s,

A

AB - —^T

Ax

A on

a mole balance of

Cn

and I+A,c„

~ is

1)

=

{A Ax) (

,

+M cH

- ,c„)

(7.7-2)

concentration at point n one At

later.

Rearranging,

,

where

M

is

+ a,c„

=

+ (M —

C,c„ +1

in

+

,c„_

J

(7.7-3)

a constant.

(Ax)

M= As

2),c„

heat conduction,

M>

2

(7.7-4)

2.

the concentration + Al c„ at position n and the new time t + At is calculated explicitly from the known three points at t. In this calculation method, starting

In using Eq.

with the

known

(7.7-3),

concentrations at

increment to the next

2.

Simplified Schmidt

l

t

until the final

=

0, the calculations

time

is

method for a slab. Schmidt method.

proceed directly from one time

reached.

If the

value of

M=

2,

a simplification of Eq.

(7.7-3) occurs, giving the

Sec. 7.7

Numerical Methods for Unsteady-State Molecular Diffusion

469

Boundary Conditions for Numerical Method

7.7C

for

a Slab

For the case where convection occurs outside in the fluid we can make a mass is suddenly changed to c„ balance on the outside \- slab in Fig. 7.7-1. Following the methods used for heat transfer to derive Eq. (5.4-7), we write rate of mass entering by convection — rate of mass leaving 1.

Convection at a boundary.

and the concentration of the

=

by diffusion

D AB A

^—

,Ci)

(,c,

in

At hours.

- ,c 2 = )

——— {A Ax/2)

(

I+Al c,. 2S

Ax

the concentration at the midpoint of the 0.5

is

,c l25

,

mass accumulation

rate of

K A{fi, where

fluid outside

approximation using

,c i

for,^

,+ A.Ci

=

25

and rearranging Eq.

(7.7-6),

+ [M -

2)], Cl

Ta PN,c.

M

(2N

+

+

-

125 )

(7.7-6)

outside slab.

As an

,c

(7.7-7)

2,c 2 ]

(7.7-8)

CAB where

M

fe

the

is

c

> {IN +

convective

mass-transfer

2.

Insulated boundary condition.

kc

= 0(N =

0) in Eq. (7.7-7),

,

J.

N

coefficient

in

m/s.

note

Again,

that

2).

For the insulated boundary

at fin Fig. 7.7-1, setting

we obtain

+

ucr =

^ [(M -

2),c

r

Alternative convective equation at the boundary.

+

2, C/ _ t ]

(7.7-9)

Another form of Eq.

can be obtained by neglecting the accumulation

gets too large

(7.7-7) to use if

in the front half-slab

of

Eq. (7.7-6) to give

N

1

-j-

The value

of

increments

M

in

is

Ax

1

+ 4 ,c 0

+

^—

(7.7-10)

+ A ,c 2

j-,

not restricted by the

N

value in this equation.

When

amount

of

mass neglected

compared

are used, the

is

small

a large

number of

to the total.

Procedure for use of initial boundary concentration. For the first time increment we should use an average value for c a of(c„ + 0 Ci)/2, where 0 c is the initial concentration 4.

i

point

at

1.

procedure after a

x

For succeeding times, the

for the

full

value of c a should be used. This special

value of c a increases the accuracy of the numerical method, especially

few time intervals.

In Section 5.4B for heat-transfer

on the best value

of

M to use in

Eq.

numerical methods, a detailed discussion (7.7-3).

The most accurate

results are

is

given

obtained for

M =4. 5.

Boundary conditions with distribution

in

Eqs. (7.7-7) and (7.7-10) were derived for the distribution coefficient

470

Chap. 7

coefficient.

Equations for boundary conditions

Principles of Unsteady-State

K

given in Eq.

and Convective Mass Transfer

(7.7-7)

being

1.0.

When K

is

not

1.0,

as in the boundary conditions for steady state,

Kk

c

should be substituted for kc in Eq. (7.7-8) to become as follows. (See also Sections 6.6B

and

7.1C.)

—

Kk Ax tf=-7T c

(7-7-11)

and (7.7-10), the termc^X should be substituted forc„. Other cases such as for diffusion between dissimilar slabs in series, resistance between slabs in series, and so on, are covered in detail elsewhere (Gl), with actual numerical examples being given. Also, in reference (Gl) the implicit numerical method is

Also, in Eqs. (7.7-7)

discussed.

EXAMPLE 7.7-1. A A

Numerical Solution for Unsteady-State Diffusion with a Distribution Coefficient slab of material 0.004 m thick has an initial concentration profile of solute as follows, where x is distance in m from the exposed surface:

Concentration {kg mol

0 0.002

x 10 3 1.25 x 10" 3 10~ 1.5 x

0.003

1.75

0.004

)

Position, n

-3

1.0

0.001

The

Aim 1

x(m)

x 10~ 2.0 x 10-

3

3

1

(exposed)

2 3

4 5 (insulated)

D AB = 1.0 x 10~ 9 m 2 /s. Suddenly, the top surface is exposed having a constant concentration c a = 6 x 10 3 kg mol A/m 3 The

diffusivity

to a fluid

.

distribution coefficient

K

= cjc n =

1.50.

The

rear surface

is

insulated

and

occurring only in the x direction. Calculate the concentration profile after 2500 s. The convective mass-transfer coefficient = 2.0. k c can be assumed as infinite. Use Ax = 0.001 m and unsteady-state diffusion

is

M

lated

n Figure

7.7-2.

Concentrations for numerical method for unsteady-state diffusion. Ex-

ample

Sec. 7.7

7.7-1.

Numerical Methods for Unsteady-State Molecular Diffusion

471

Figure 7.7-2 shows the initial concentration profile for four slices = 2, substituting into Eq. (7.7-4) v/ith Since

Solution:

=

3 6 x 1CT

and

ca

Ax =

0.001 m,

M

.

and solving

for At,

D AB =

At

500

=

time increment, as stated previously,

i

where

CjK

=

ca

time increments are needed. the concentration to use for the

five

where n

front surface

x 10^ 9 XA0

(1

s

Hence, 2500 s/(500 s/increment) or

For the

At

1,

Cl ^° =

1

c1

=

the initial concentration at n

is

first

is

(n

=

1.

For the remaining time

(7.7-12)

l)

increments, C

c,=

To

calculate the concentrations for

using Eq. (7.7-5) for

M=

end

the insulated

all

(7.7-13)

1)

time increments for slabs n

=

2, 3, 4,

2,

'

= For

=

{n

f

C

,C "

""'

+

'

—

at n

substituting

5,

=

("

2

2 3 4) >

<

>

7 7 - 14 ) '

M = 2 and f = n = 5 into

Eq.

n1

.„

(7.7-9),

,

For

At or

1

=

centration for n

+

1

by Eq.

c + ai

For n

=

c,

=

—

i

2, 3,

,c,-i

4,

—

l

r

+ Al c 4

For n

=

fii

3 6 x 10- /1.5

+rC n+1 _

+

fix

+

5,

472

2ai c

_ og

X

3

+

1.5

x 10' 3

_

_

iQ

3

3

+

x 10-

1.75

3

=

x l0

_3

fis

=

1.5

x IP"

3

+

2 x 10-

3

=

j

?5 x 1Q

-3

2

using Eq. (7.7-15),

= i

con-

2

+ a,c 5

=

,c 4

For 2 At using Eq. (7.7-13) for n using Eq. (7.7-15) for n = 5, +

(7.7-15)

5)

2

1

,

x IP" 3

1

x 10~

2.5

,c 3

10-

1-25 x

2

=

+

2

+ £4 =

=

("

,

2

2

£3

,c 4

using Eq. (7.7-14),

2

+ A( C 3

=

(7.7-12),

2

and

2,c 4

Af, the first time increment, calculating the

c^K+qC, !

+

2),c 5

^

t

1

-

(2

=

+ a,c 5

c

t7

K

=

6

x 10~ 3

tj

=

=

1.75

=

1,

4.0 x 10~

3

x 10

-3

using Eq. (7.7-14) for n

=

2-4,

and

(constant for rest of time)

1.5

Chap. 7

Principles of Unsteady-State

and Convective Mass Transfer

=

.

,

+a( -3

—

=

~

i

+ A( C 2

+

[

2 x 10~ 3

+ Al C 4

+

2 I+AI C 3

=

+ 2Ai c 4

=

+ AI C 4-

I

2 75 x 1Q

-3

KT 3

1.75 x

1.875 x 1CT

3

2

+=

l+Al C 5

2

I+2AI C 5

For

r

=

2

2 2 Al C 3

3

=

x 10- 3

1.5

=

+= 1.75

x 1(T 3

2

=

I- 625

x 10

1-75 X 10

3 At, u c,

=

4 x 10 4

" Cz

~

1A ,c 3

=

-3

x IP" 3

+

1.875 x IP'

3

=

2.938 x

2.75 x 10-

+

3

1.625 x

KT

1.875

xlO- 3 +

1+3 Al -*

+

3A ,c 5

3

3

= 2.188x10

^ 1.75

x IP' 3

= L813xlQ ,3

1

,

KT

2

2

=

1.625 x 10-

3

For 4 At, ,

+ 4a,Ci

1

+ 4 Al L 2

=

4 x 10

3

4xlQ-' +

2.1 8

8xlQ-'

2 2.938 x 10

-3

+

1.813 x lO

-3 3

2.376 x 10

2

,

l

For

+4

aA =

+4

AI

L

2.188 x 10

-3

1.625 x 10

-3

=

^

1

906 x 10"

3

2

1.813 x 10"

5

+

3

5 At,

r+5A,Cl

=

4 x 10

-

«»">" +."*»">. 3.188

3

x 10-

2 3.094 x IP' 1

+

5 AI<-3

1

+

Ld 4 5 Al c

The

final

=

1.906 x 10"

3

1.813 x 10~

3

=

3 2.500 x 10"

2.376 x 10"

3

-

3

+

=

„

concentration profile

is

^

tn 2.095 x 10

plotted in Fig. 7.7-2.

more slab increments and more time increments

type of calculation

Sec. 7.7

+ 2

1.906 x 10

accuracy,

3

is

suitable for

a

digital

To

. 3

increase the

are needed. This

computer.

Numerical Methods for Unsteady-State Molecular Diffusion

473

7.8

DIMENSIONAL ANALYSIS IN MASS TRANSFER

7.8A

The

Introduction use of dimensional analysis enables us to predict the various dimensional groups

which are very helpful flow and

experimental mass-transfer data. As

in correlating

we saw

in fluid

heat transfer, the Reynolds number, the Prandtl number, the Grashof

in

number, and the Nusselt number were often used in correlating experimental data. The Buckingham theorem discussed in Section 3.11 and Section 4.14 states that the functional relationship among q quantities or variables whose units may be given in terms of u fundamental units or dimensions may be written as {q - u) dimensionless groups.

Dimensional Analysis for Convective Mass Transfer

7.8B

We

mass

consider, a case of convective

convection in a pipe and mass transfer flows at a velocity

v

is

and

v,

fi,

fluid

is

flowing by forced

occurring from the wall to the

inside a pipe of diameter

coefficient k'c to the variables D, p,

where a

transfer

D and we

fluid.

fluid

D AB The total number of variables is q = 6. = 3 and are mass M, length L, and time .

The fundamental units or dimensions are u

The

The

wish to relate the mass-transfer

f.

units of the variables are

kc

L —— t

The number

We

p

=

M —

=

u

3

L

M — Lt

L

o——

D,ABR

t

D AB

-

3,

or

3.

Then,

*i

=/("2. * 3)

p,

and D to be the variables

,

D= L

t

of dimensionless groups or n' are then 6

choose the variables

1} =—

(7-8-1) .

common

to all the

dimensionless groups, which are

= D AB p b Dc k'

(7.8-2)

7i,

= D AB p c D r v

(7.8-3)

K3

= D AB p D^

(7.8-4)

n

For

7t,

we

Summing

i

c

h

substitute the actual dimensions as follows:

for

each exponent,

=

(L)

0

(M)

0=6

(t)

0

=

2a

a

Solving these equation simultaneously, a

-

3b

+

c

+

1

(7.8-6)

-

1

= —

1,

b

=

0, c

=

1.

Substituting these values

into Eq. (7.8-2),

k'D

Ki=77-=Nsh

(7-8-7)

AB

474

Chap. 7

Principles of Unsteady-State

and Convective Mass Transfer

Repeating for k 2 and

rc

3

,

vD (7.8-8)

(7.8-9)

If

we

divide n 2

byn 3 we

obtain the Reynolds number. 7t

vD

2

Hence, substituting into Eq.

If

p

Dvp

\

(7.8-10)

d abI \PD AB )

n3

H

(7.8-1),

Nsh =f(N Rc ,NSc

7.9

7.9A

(7.8-11)

)

BOUNDARY-LAYER FLOW AND TURBULENCE MASS TRANSFER Laminar Flow and Boundary-Layer Theory

in

IN

Mass

Transfer In Section 3. 10C

an exact solution was obtained for the hydrodynamic boundary layer for

isothermal laminar flow past a plate and in Section 5.7A an extension of the Blasius solution was also used to derive an expression for convective heat transfer. In

analogous manner we use the Blasius solution for convective mass transfer

an same shown

for the

geometry and laminar flow. In Fig. 7.9-1 the concentration boundary layer is where the concentration of the fluid approaching the plate is c Am and c AS in the

fluid

adjacent to the surface.

We

start

by using the

steady state where dcjdt

=

neglecting diffusion in the x

differential 0,

RA =

and

0,

mass balance, Eq. (7.5-17), and simplifying it for in the x and y directions, so v, = 0, and

flow only

z directions to give

dc A

j_

dc A

n dy

(7.9-1)

2

edge of concentration C

A

boundary

layer

x = 0 Figure

7.9-1.

Laminar flow of fluid past aflat plate and concentration boundary layer.

Sec. 7.9

Boundary-Layer Flow and Turbulence

in

Mass Transfer

475

The momentum boundary-layer equation

very similar

is

a d dv dv — — - + - = v

dx

The thermal boundary-layer equation

is

y

v

dy

p dy

;'

also similar

dT dT _ T — + — = — —T dy k

vx

The continuity equation used

2

r

r

v

dx

previously

y

dy

d

2

(5.7-2)

pc p

is

^ + —-=0 dv r

dv„

dx

dy

(3.10-3)

The dimensionless concentration boundary conditions

T — Ts

vx

Ts

v co

The is

similarity

obvious, as

is

oo

between the three

the similarity

c AS

cA

=

0

y

at

=

0

c AS

c Aco

T - Ts = = T — *S

y co

-

cA

are

(7.9-2)

-

c AS

=

y

at

1

=

oo

c Aao ~~ C AS

equations (7.9-1), (3.10-5), and (5.7-2)

differential

among the

three sets of boundary conditions in Eq. (7.9-2).

In Section 5.7A the Blasius solution was applied to convective heat transfer {(i/p)/a

= N Pr =

1.0.

We

use the

=N

same type of solution

=

transfer vihcn(p/p)/D AB 1.0. Sc The velocity gradient at the surface

3v. '

where

N Rz x =

when

laminar convective mass

for

was derived previously.

= 1

V

0.332

-^N R 2 x l'

(5.7-5)

c

y=0

xv x p/p. Also, from Eq.

(7.9-2),

-

CA

"x

CAS (7.9-3)

'Att

Differentiating Eq. (7.9-3)

and combining the

<-AS

result with Eq. (5.7-5),

/0.332

dc A \

Rc, x

dyj y=0 The

convective mass-transfer equation can be written as follows and also related with

Fick's equation for dilute solutions:

N Ay = Combining

K(c AS

U AB

This relationship

476

dc

c Aao )

= — D AB l—^A dy

(7.9-5)

j

y

=0

Eqs. (7.9-4) and (7.9-5),

^ The

-

is

= N Sh ., = 0.332/^,

restricted to gases with a

relationship

N Sc =

1

(7.9-6)

.0.

between the thickness 5 of the hydrodynamic and 5 C of the

Chap. 7

Principles

of Unsteady-State and Convective Mass Transfer

number

concentration boundary layers where the Schmidt

is

not 1.0

is

(7.9-7)

As

a result, the equation for the local convective mass-transfer coefficient

^=N U AB

We can obtain the equation for a plate of width b

for the

result

mass-transfer coefficient^ from x

— b

f

0 to x

=L

L

K

I

dx

(7.9-9)

= N Sh = 0.664^£ L N S

1'3

similar to the heat-transfer equation for a

is

=

is

^ This

(7.9-8)

by integrating as follows:

K= The

= 0.332iV^^ 3

Sh . x

mean

is

flat plate,

the experimental mass-transfer equation (7.3-27) for a

(7.9-10)

Eq. (5.7-15), and also checks

flat plate.

In Section 3.10 an approximate integral analysis was made for the laminar hydrodynamic and also for the turbulent hydrodynamic boundary layer. This was also done in Section 5.7 for the thermal boundary layer. This approximate integral analysis can also be done in exactly the same manner for the laminar and turbulent concentration boundary layers.

Prandtl Mixing Length and Turbulent Eddy

7.9B

Mass

Diffusivity

In

many

applications the flow in mass transfer

turbulent flow of a fluid

is

and

quite complex

is

The random eddy occurring, we refer to

turbulent and not laminar.

the fluid undergoes a series of

movements throughout the turbulent core. When mass transfer is this as eddy mass diffusion. In Sections 3.10 and 5.7 we derived equations for turbulent eddy thermal diffusivity and momentum diffusivity using the Prandtl mixing length theory. In a similar £

M

.

manner we can

derive a relation for the turbulent eddy mass diffusivity,

Eddies are transported a distance L, called the Prandtl mixing length,

direction.

At

velocity v'x

,

this

in the y from the adjacent fluid by the the fluctuating velocity component given in Section 3.10F. The

point

which

is

L

the fluid

eddy

differs in velocity

instantaneous rate of mass transfer of A at a velocity^ for a distance

L

in the

y direction

is

J* y

where

c'A

is

the

instantaneous

centration of the fluid

is

cA

=

c'A

= c>;

fluctuating

+

cA

where

from the mean value. The mixing length L

is

(7.9-11)

concentration. cA

is

the

mean

The instantaneous convalue and c'A the deviation

small enough so the concentration difference

is

cA

Sec. 7.9

=L

^

(7-9-12)

dy

Boundary-Layer Flow and Turbulence

in

Mass Transfer

477

The

rate of

mass transported per unit area

J*y Combining Eqs.

is

.

and

(7.9-11)

(7.9-12),

dc.

v'L "~ From

(7.9-13)

dy

Eq. (5.7-23), V'y

=

V'x

do.

=L

(7.9-14)

dy

Substituting Eq. (7.9-14) into (7.9-13), dv. (7.9-15)

dy

dy

The term l3\dvjdy (7.9-

1

5)

\

is

called the turbulent

J*Ay

The transfer

7.9C

1.

eddy mass

with the diffusion equation in terms of D AB

similarities

between Eq. (7.9-16)

have been pointed out

Models

for

d£A e«)

(7.9-16)

dy

mass

for

Combining Eq.

.

is

and heat and momentum

transfer

in detail in Section 6.1 A.

Mass-Transfer Coefficients

For many years mass-transfer

Introduction.

on empirical

= -(Dab +

diffusivity e w

the total flux

,

coefficients,

which were based primarily

correlations, have been used in the design of process equipment.

A

better

needed before we can give a theoretical explanation of convective-mass-transfer coefficients. Some theories of convective mass understanding of the mechanisms of turbulence transfer,

is

such as the eddy diffusivity theory, have been presented

following sections

we present

briefly

some

of these theories

in this chapter. In the

and also discuss how they can

be used to extend empirical correlations.

2.

Film

mass-transfer

theory.

The

This

film,

where only molecular diffusion

resistance to regions.

mass

Then

The

is

c A2 )

is

the

simplest

laminar film next

to the

and most boundary.

assumed to be occurring, has the same and turbulent core

=

k'c is

~

(c A1

-

(7.9-17)

c A2 )

is

(7.9-18)

proportional to D AB However, since we have shown that V1 then k' oc D ' [p./ pD AB ) c A B Hence, the film theory is .

.

,

great advantage of the film theory

Penetration theory.

related to this film thickness 5 f by

=

proportional to

complex situations such 3.

= K(Cax -

mass-transfer coefficient

Eq. (7.3-13), J D

not correct.

is

the actual mass transfer coefficient

K in

fictitious

transfer as actually exists in the viscous, transition,

J*a

The

which

theory,

film

elementary theory, assumes the presence of a

as simultaneous diffusion

is its

simplicity where

it

can be used

in

and chemical reaction.

The penetration theory derived by Higbie and modified by

Danckwerts (D3) was derived

for diffusion or penetration into a laminar falling film for

short contact times in Eq. (7.3-23) and

is

as follows

K

*Dab

(7.9-19)

nt,

478

Chap. 7

Principles

of Unsteady-State and Convective Mass Transfer

where t L is the time of penetration of the solute in seconds. This was extended by Danckwerts. He modified this for turbulent mass transfer and postulated that a fluid

eddy has a uniform concentration in the turbulent core and is swept to the surface and undergoes unsteady-state diffusion.^Then the eddy is swept away to the eddy core and other eddies are swept to the surface and stay for a random amount of time. A mean surface renewal factor s in s"

1

defined as follows:

is

K=

(7-9-20)

The mass-transfer coefficient k'c is proportional to D^J In some systems, such as where liquid flows over packing and semistagnant pockets occur where the surface is being renewed, the results approximately follow Eq. (7.9-20). The value of s must be .

obtained experimentally. Others (D3, T2) have derived more complex combination

change of the exponent on D AB from depending on turbulence and other factors. Penetration theories have been cases where diffusion and chemical reaction are occurring (D3).

film-surface renewal theories predicting a gradual 0.5 to 1.0

used

4.

in

The boundary-layer theory has been

Boundary-layer theory.

Section 7.9 and surfaces.

is

useful in predicting

and correlating data

For laminar flow and turbulent flow the mass-transfer

This has been experimentally verified for

many

discussed in detail in

for fluids

flowing past solid

coefficient k'c oc

D%g

.

cases.

PROBLEMS Unsteady-State Diffusion in a Thick Slab. Repeat Example 7.1-2 but use a K = 0.50 instead of 2.0. Plot the data. 2 2 Ans. (x = 0), c = 2.78 x 10' (x = 0.01 m), c = c = 5.75 x 10"

7.1-1.

distribution coefficient

;

c Li

=

2.87 x 10"

2

kg mol/m 3

Plot of Concentration Profile in Unsteady-State Diffusion. Using the same conExample 7.1-2, calculate the concentration at the points x — 0,

7.1-2.

ditions as in

0.005, 0.01, 0.015,

and 0.02

m

from the

surface. Also calculate c Li in the liquid at

the interface. Plot the concentrations in a

manner

similar to Fig. 7.1-3b, showing

interface concentrations.

Unsteady-State Diffusion in Several Directions. Use the same conditions as in 7.1-1 except that the solid is a rectangular block 10.16 thick in the x

7.1-3.

mm

Example and

mm

mm

thick in the y direction, and 10.16 thick in the z direction, diffusion occurs at all six faces. Calculate the concentration at the midpoint

direction, 7.62

after 10 h.

Ans. 7.1-4

c

=

4 6.20 x 10~

kg mol/m 3

Drying of Moist Clay. A very thick slab of clay has an initial moisture content of c 0 = 14 wt %. Air is passed over the top surface to dry the clay. Assume a relative resistance of the gas at the surface of zero. The equilibrium moisture content at the surface is constant at Cj = 3.0 wt %. The diffusion of the moisture in the clay 2 can be approximated by a diffusivity of D AB = 1.29 x 10" 8 m /s. After 1.0 h of drying, calculate the concentration of water at points 0.005, 0.01,

and

0.02

m

below the surface. Assume that the clay is a semiinfinite solid and that the Y 3 value can be represented using concentrations of wt rather than kg mol/m

%

Plot the values versus 7.1-5.

Unsteady-State Diffusion in a Cylinder of Agar Gel. A wet cylinder of agar gel at 3 278 K containing a uniform concentration of urea of 0.1 kg mol/m has a diameter of 30.48 and is 38. 1 long with flat parallel ends. The diffusivity 10 is 4.72 x 10" m 2 /s. Calculate the concentration at the midpoint of the cylinder

mm

Chap. 7

.

x.

Problems

mm

479

100 h for the following cases

after

the cylinder

if

suddenly immersed

is

in

turbulent pure water. (a)

For

(b)

Diffusion occurs radially

radial diffusion only.

and

axially.

mm

Drying of Wood. A flat slab of Douglas fir wood 50.8 thick containing 30 wt moisture is being dried from both sides (neglecting ends and edges). The equilibrium moisture content at the surface of the wood due to the drying air moisture. The drying can be assumed to be blown over it is held at 5 wt 6 2 represented by a diffusivity of3. 72 x 10~ m /h. Calculate the time for the center

7.1- 6.

%

%

to reach

10%

moisture.

Flux and Conversion of Mass-Transfer Coefficient. A value of k G was experimen2 tally determined to be 1.08 lb mol/h ft atm for A diffusing through stagnant B. For the same flow and concentrations it is desired to predict k'G and the flux of A for equimolar counterdiffusion. The partial pressures arep <
7.2- 1.

•

•

•

m

•

,

show

Conversion of Mass-Transfer Coefficients. Prove or

7.2-2.

the following relation-

ships starting with the flux equations.

Convert k[ to k y and k G Convert k L to k x and k'x (c) Convert k G to k and k c y Absorption of 2 S by Water. In a wetted-wall tower an air-H 2 S mixture is flowing by a film of water which is flowing as a thin film down a vertical plate. The H 2 S is being absorbed from the air to the water at a total pressure of 4 1.50 atm abs and 30°C. The value of k'c of 9.567 x 10~ m/s has been predicted for the gas-phase mass-transfer coefficient. At a given point the mole fraction of H 2 S in the liquid at the liquid-gas interface is 2.0(10" 5 ) and p A ofH 2 S in the gas is 0.05 atm. The Henry's law equilibrium relation is p A (atm) = 609x_4 (mole

(a)

.

(b)

.

.

H

7.2- 3.

fraction in liquid). Calculate the rate of absorption of

the interface the given x A

and point

2 the gas phase.

The value

of p A2

.

is

Then

Mass

from a Flat Plate to Example 7.3-2 calculate

and a plate length of calculate 7.3-2.

/

its

L=

[Hint: Call point

1

from Henry's law and

0.137 m.

NA = —

1

.485 x

1

0

~ 6

a Liquid. Using the data

Transfer

properties of

,

0.05 atm.)

Ans. 7.3- 1.

H 2 S.

calculate p A

kg mol/s

m

2

and physical

the flux for a water velocity of 0.152 m/s

Do

not assume that x BM

=

1.0,

but actually

value.

Mass

Transfer from a Pipe Wall. Pure water at 26.1°C is flowing at a velocity of 0.0305 m/s in a tube having an inside diameter of 6.35 mm. The tube is 1.829 long with the last 1.22 having the walls coated with benzoic acid. Assuming that the velocity profile is fully developed, calculate the average concentration of

m

m

benzoic acid at the outlet. Use the physical property data of Example 7.3-2. [Hint: First calculate the Reynolds number Dvp/fj.. Then calculate Re Sc

N N

(

D/L)( 7t/4), which

Ans. 7.3-3.

is

the

[c A

same

as

W/D AB pL.~]

-c A0 )/[c Ai -c A0

)

=

=0.0744, c A

2.193 x I0-

3

kgmol/m 3

Mass-Transfer Coefficient for Various Geometries. It is desired to estimate the mass-transfer coefficient k G in kg mol/s m 2 Pa for water vapor in air at 338.6 K and 101.32 kPa flowing in a large duct past different geometry solids. The velocity in the duct is 3.66 m/s. The water vapor concentration in the air is small, so the physical properties of air can be used. Water vapor is being •

Do this for the following geometries. A single 25.4-mm-diameter sphere. A packed bed of 25.4-mm spheres with £ = 0.35.

transferred to the solids. (a)

(b)

Ans.

480

(a)Jt G

=1.98x

10 -

8

kg mol/s

•

m

2 -

Pa

(1.48 lb

mol/h

Chap. 7

2 ft

atm)

Problems

[7.3-4 J

Mass Transfer to

^

ficient in a

—

Definite Shapes. Estimate the value of the mass-transfer coefstream of air at 325.6 K flowing in a duct by the following shapes made of solid naphthalene. The velocity of the air is 1.524 m/s at 325.6 K and 6 2 202.6 kPa. The D AB of naphthalene in air is 5.16 x 10~ m /s at 273 K and 101.3 kPa. (a) For air flowing parallel to a flat plate 0. 152 m in length. (b) For air flowing by a single sphere 12.7 in diameter.

mm

73-5.

Mass

Transfer to

Packed Bed and Driving Force. Pure water

at

26.1°C

is

flowing

3

/h through a packed bed of 0.251-in. benzoic acid spheres having a total surface area of 0.129 ft 2 The solubility of benzoic acid in water is

at a rate of 0.0701

ft

.

0.00184 lb mol benzoic acid/ft 3 solution. The outlet concentration c A2 10"* lb mol/ft 3 Calculate the mass-transfer coefficient k c 7.3-6.

is

1.80

x

.

.

Mass

Transfer in Liquid Metals. Mercury at 26.5°C is flowing through a packed bed of lead spheres having a diameter of 2.096 with a void fraction of 0.499. The superficial velocity is 0.02198 m/s. The solubility of lead in mercury is 1.721; -3 wt %, the Schmidt number is 124.1, the viscosity of the solution is 1.577 x 10 3 Pa s, and the density is 13 530 kg/m (a) Predict the value of J D Use Eq. (7.3-38) if applicable. Compare with the experimental of J D = 0.076 (D2). (b) Predict the value of k c for the case of A diffusing through nondiffusing B. Ans. (a) J D = 0.0784, (b) k c = 6.986 x 10" 5 m/s

mm

-

.

.

7.3-7.

Transfer from a Pipe and Log Mean Driving Force. Use the same physical conditions as Problem 7.3-2, but the velocity in the pipe is now 3.05 m/s. Do as

Mass

follows. (a)

(b)

(c)

Predict the mass-transfer coefficient k'c (Is this turbulent flow?) Calculate the average benzoic acid concentration at the outlet. [Note: In this case, Eqs. (7.3-42) and (7.3-43) must be used with the log mean driving force, where A is the surface area of the pipe.] .

Calculate the total kg mol of benzoic acid dissolved per second.

of Relation Between J D and-Nsh Equation (7.3-3) defines the SherEq. (7.3-5) defines the J D factor. Derive the relation between and J D in terms of N Rc and N Sc

7.3-8. Derivation

.

wood number and

N Sh

.

N .N^

N

7.3- 9.

Ans. K Sh = J D Driving Force to Use in Mass Transfer. Derive Eq. (7.3-42) for the log mean driving force to use for a fluid flowing in a packed bed or in a tube. (Hint : Start by making a mass balance and a diffusion rate balance over a differential area dA as follows:

N A dA = where V = 7.4- 1.

m

3

/s

flow rate.

Assume

-

c,)

dA = V

dc A

dilute solutions.)

Maximum Oxygen Uptake

of a Microorganism. Calculate the maximum possible rate of oxygen uptake at 37°C of microorganisms having a diameter off suspended in an agitated aqueous solution. It is assumed that the surrounding liquid is saturated with 0 2 from air at 1 atm abs pressure. It will be assumed that the microorganism can utilize the oxygen much faster than it can diffuse to it. The micoorganism has a density very close to that of water. Use physical property data from Example 7.4-1. (Hint : Since the oxygen is consumed faster than it is supplied, the concentration c A1 at the surface is zero. The concentration c Al in the solution

is

Ans. 7.4-2.

k c (c At

at saturation.)

kc

=

9.75 x 10~

3

m/s,

NA =

2.20 x 10

-6

kg mol

0

2 /s

•

m

2

Mass

Transfer of0 2 in Fermentation Process. A total of 5.0 g of wet microorganisms having a density of 1 100 kg/m 3 and a diameter of 0.667 /im are added to 0.100 L of aqueous solution at 37°C in a shaker flask for a fermentation. Air can enter through a porous stopper.

Chap. 7

Problems

Use physical property data from Example

7.4-1.

481

maximum

Calculate the

(a)

0 2 /s

mass transfer of oxygen in kg mol microorganisms assuming that the solution is

rate possible of

to the surface of the

kPa abs

saturated with air at 101.32

pressure.

By material balances on other nutrients, the actual utilization of 0 2 by the What would be the actual microorganisms is 6.30 x 10~ 6 kg mol

(b)

concentration of

0 2 in

OA

the solution as percent saturation during the fermen-

tation?

Ans.

(a) k'L

x 10

9.82

-3

x 10" 5 kg mol "OA

Sum

of Molar Fluxes. Prove the following equation using the definitions in Table 7.5-1. ...

HA + NB = 7.5-2.

N A A = 9.07

m/s,

6.95% saturation

(b)

7.5-1.

=

cv M

Proof of Derived Relation. Using the definitions from Table 7.5-1, prove the following:

= »a- w a("a +

Ja 7.5-3. Different

(Hint

dx A

Forms ofFick's Law. Using Eq.

First relate

:

Form of

7.5-4. Other

wA

M=

Finally, use

.

= - PD AB

h

=--M M p

to x^.

B

^

D AB

(2)

dz

differentiate this equation to relate

+ x B M B to simplify.) Law. Show that the following form of

xA

Fick's

(2).

(1)

A

Then

prove Eq.

(1),

Ja

n B)

MA

dw A and

Fick's law

is

valid:

c(v A

(Hint 7.5-1

Start with

:

and

7.5-5. Different

NA =

+

cD AB dx A

-v B )=

cA v

M

.

x A x B dz

Substitute the expression for

from Table

simplify.)

Form of Equation of Continuity.

^

+

(V

•

Starting with Eq. (7.5-12),

nj =

r„

(7.5-12)

convert this to the following for constant p:

% + (v-VpJ-(V.D =

VpJ =

r

(1)

1

A + p A v into Eq. (7.5-12). Note that Then substitute Fick's law in terms of j A .] Diffusion and Reaction at a Surface. Gas A is diffusing from a gas stream at point to a catalyst surface at point 2 and reacts instantaneously and irreversibly as

{Hint

(V

7.5-6.

From Table

4B

.

:

v)

=

0

for

7.5-1, substitute n A

constant

'}

p.

1

follows:

2A—B Gas B

diffuses

back to the gas stream. Derive the final equation P and steady state in terms of partial pressures.

for

NA

at

constant pressure

=

AnS "

482

*

JftJT1^) Chap. 7

T^~p~J2P Problems

7.5-7.

and Reaction. Solute A is diffusing at unsteady state of pure B and undergoes a first order reaction with B. Solute A is dilute. Calculate the concentration c A at points z = 0, 4, and 10 mm from the surface for t = 1 x 10 5 s. Physical property data are D AB = 1 x 10" 9 m 2 /s, k' = 1 x 10~ 4 s"\ c A0 = 1.0 kg mol/m 3 Also calculate the kg mol absorbed/m 2 Unsteady-State Diffusion

into a semiinfinite

medium

.

.

7.5-8.

Multicomponent Diffusion. At a total pressure of 202.6 kPa and 358 K, ammonia gas (A) is diffusing at steady state through an inert nondiffusing mixture of nitrogen (B) and hydrogen (C). The mole fractions at z, = 0 are x Al = 0.8, x BX =0.15, and x cl = 0.05; and at z 2 = 4.0 mm, x A i = 0.2, x^ = 0.6, and =3.28 x 10" 5 m 2 /s xc2 = 0.2. The diffusivities at 358 K and 101.3 kPa are and D AC = 1.093 x 10" 4 m 2 /s. Calculate the flux of ammonia. = 4.69 x 10~ 4 kg mol /1/s m 2 Ans. t>' •

A

7.5-9. Diffusion in Liquid Metals and Variable Diffusivity. The diffusion of tin (A) in liquid lead (B) at 510°C was carried out by using a 10.O-mm-long capillary tube 1

and maintaining the mole fraction of tin at x A at the left end and x A2 at the right end of the tube. In the range of concentrations of 0.2 < x A < 0.4 the diffusivity of tin in lead has been found to be a linear function of x A (S7). t

D AB = A + Bx A where (a)

A and B

and

are constants

D AB

is

in

m 2 /s.

molar density to be constant at c = c A + c B equation for the flux N A assuming steady diffuses through stagnant B. For this experiment, A = 4.8 x 10" 9 B = -6.5 x 1CT 9 c av x Al = 0.4, x A2 = 0.2. Calculate N A

Assuming

the

final integrated

(b)

,

,

=

c av

state

=

,

derive the

and

A

that

3 50 kg mol/m

,

.

Ans.

NA =

(b)

4.055 x 10"

5

kg mol

/1/s

mz

•

and Chemical Reaction of Molten Iron in Process Metallurgy. In a steelmaking process using molten pig iron containing carbon, a spray of molten "iron particles containing 4.0 wt carbon fall through a pure oxygen atmosphere. The carbon diffuses through the molten iron to the surface of the drop, where it is assumed that it reacts instantly at the surface, because of the high temperature, as follows by a first-order reaction:

7.5-10. Diffusion

%

C + Calculate the

i0

maximum drop

2

(g)^CO(g)

size allowable so that the final

drop

after

a

%

contains on an average 0.1 wt carbon. Assume that the mass transfer rate of gases at the surface is very great, so there is no outside resistance. Assume no internal circulation of the liquid. Hence, the decarburization rate is controlled

2.0-s fall

by the rate of diffusion of carbon to the surface of the droplet. The diffusivity of carbon in iron is 7.5 x 10"" 9 m 2 /s (S7). {Hint: Can Fig. 5.3-13 be used for this case?)

radius

Ans. 7.5-11. Effect

catalyst surface at point 2,

where

it

A

=

0.217

mm

from point 1 to a reacts as follows: 2A -» B. Gas B

of Slow Reaction Rate on Diffusion. Gas

diffuses

back a distance S to point 1. Derive the equation for N A for a very fact reaction using mole fraction units x A [, and so on. -4 For D AB = 0.2 x 10 m 2 /s, x Al = 0.97, P = 101.32 kPa, <5 = 1.30 mm, and T = 298 K, solve for N A Do the same as part (a) but for a slow first-order reaction where k\ is the

diffuses (a)

(b)

.

(c)

(d)

reaction velocity constant. Calculate A and x A2 for part

N

(c)

where k\ Ans.

Chap. 7

Problems

=

(b)

0.53 x 10"

NA =

2

m/s.

4 8.35 x 10"

kg mol/s

•

m2 483

m

and Heterogeneous Reaction on Surface. In a tube of radius R filled with a liquid dilute component A is diffusing in the nonflowing liquid phase represented by

7.5-12. Diffusion

where z

distance along the tube axis.

is

The

inside wall of the tube exerts a

and decomposes A so that the heterogeneous rate of decomposition on the wall in kg mol A/s is equal to kc A A w where k is a first-order constant 2 and A w is the wall area in m Neglect any radial gradients (this means a uniform

catalytic effect

,

.

radial concentration).

Derive the differential equation for unsteady state for diffusion and reaction : First make a mass balance for A for a Az length of tube as follows: rate of input (diffusion) + rate of generation (heterogeneous) = rate of output (diffusion) + rate of accumulation.] for this system. [Hint

7.6-1.

Knudsen

Pa

Diffusivities.

pressure

total

A mixture of He(A) and and 298 K through a

Ar(B)

2

—-D -^--c dc A

Ans.

2k

d cA

AB

is

diffusing at

capillary

having

a

1

.013

x

A

10

5

of

radius

100 A.

Knudsen Knudsen

(a)

Calculate the

(b)

Calculate the

(c)

Compare with

diffusing through

7.29 x 10~

Predict the flux

(c)

0.2.

5

m

x 10 5

Pa.

mm

7.29 x 10"

5

m K

2

/s

is (B) at 298 long with a radius of 1000 A.

The molecular

diffusivity

D AB

at 1.013

2

/s.

diffusivity of He (A).'using Eq. (7.6-18) and Eq. (7.6-12)

Knudsen

Calculate the

(b)

=

1.013

is

(a)

x A2

.

(a)

an open capillary 15

total pressure is

{A). (B).

D AB D KA = 8.37 x 1CT 6 m 2 /s;(c) D AB = Diffusion. A mixture of He (A) and Ar

7.6-2. Transition-Region

x 10 5 Pa

He Ar

the molecular diffusivity

Ans.

The

diffusivity of

diffusivity of

NA

if

x Al =0.8 and

Assume steady state.

Predict the flux

NA

using the approximate Eqs. (7.6-14) and (7.6-16).

Pore in the Transition Region. Pure H2 gas (A) at one end of a noncatalytic pore of radius 50 A and length 1.0 mm (x Ai = 1.0) is diffusing through this pore with pure C 2 H 6 gas (B) at the other end at x A2 = 0. The total pressure is constant at 1013.2 kPa. The predicted molecular diffusivity -5 of H 2 -C 2 H 6 is 8.60 x 10 m 2 /s at 101.32- kPa and 373 K. Calculate the Knudsen diffusivity ofH 2 and flux A/ ^ ofH 2 in the mixture at 373 K and steady

7.6-3. Diffusion in a

state.

D KA =6.60

Ans.

x 10~ 6

m 2 /s, N A =

1.472 x 10"

3

kg mol A/s

m

2

Region Diffusion in Capillary. A mixture of nitrogen gas (A) and helium (B) at 298 is diffusing through a capillary 0.10 m long in an open system with a diameter of 10 ^m. The mole fractions are constant at x Al = 1.0 ..and x A2 = 0. See Example 7.6-2 for physical properties.

7.6- 4. Transition

K

(a)

Calculate the 0.1,

and

Knudsen

diffusivity

D KA andD Kg

at the total pressures

of 0.001,

10.0 atm.

(b)

Calculate the flux

(c)

Plot

NA

steady state at these pressures. log-log paper. What are the limiting lines at lower pressures and very high pressures? Calculate and plot these lines.

NA

versus

at

P on

Method for Unsteady-State Diffusion. A solid slab 0.01 m thick has uniform concentration of solute A of 1.00 kg mol/m 3 The diffusivity of A in the solid is D AB = 1.0 x 10" 10 m 2 /s. All surfaces- of the slab are insulated except the top surface. The surface concentration is suddenly

7.7- 1. Numerical

an

484

initial

.

Chap. 7

Problems

dropped to zero concentration and held there. Unsteady-state diffusion occurs in the one x direction with the rear surface insulated. Using a numerical method, = 2.0. determine the concentrations after 12 x 10* s. Use Ax = 0.002 m and

M

The

value of

K

is 1.0.

Ans.

cl c2 c3

c4 c5

= 0( front surface, x = 0 m), = 0.3125 kg mol/m 3 (x = 0.002 m) = 0.5859 (x = 0.004 m), = 0.7813 (x = 0.006 m) = 0.8984 (x = 0.008 m), = 0.9375 (insulated surface, x = 0.01

c6 m) Computer and Unsteady-State Diffusion. Using the conditions of Problem 7.7-1, solve that problem by the digital computer. Use A x = 0.0005 m. Write the computer program and plot the final concentrations. Use the explicit = 2. method, Numerical Method and Different Boundary Condition. Use the same conditions as in Example 7.7-1, but in this new case the rear surface is not insulated. At time t = 0 the concentration at the rear surface is also suddenly changed to c 5 = 0 and held there. Calculate the concentration profile after 2500 s. Plot the initial and final concentration profiles and compare with the final profile of Example 7.7-1.

7.7-2. Digital

M

7.7- 3.

7.8- 1.

Dimensional Analysis in Mass Transfer. A fluid is flowing in a vertical pipe and mass transfer is occurring from the pipe wall to the fluid. Relate the convective mass-transfer coefficient k'c to the variables D, p, p., v,D AB g, and Ap, where D is pipe diameter, L is pipe length, and Ap is the density difference. ,

7.9- 1.

Mass Transfer and Turbulence Models. Pure water flowing at 26. 1°C past a

flat

at

a velocity of 0.11 m/s is where L = 0.40 m. Do

plate of solid benzoic acid

as follows. (a)

Assuming

dilute solutions, calculate the mass-transfer coefficient k c

physical property data from

Example

(c)

Using the film model, calculate the equivalent film thickness. Using the penetration model, calculate the time of penetration.

(d)

Calculate the

(b)

mean

Use

.

7.3-2.

surface renewal factor using the modified penetration

model. Ans.

(b)

5f

=

0.2031

mm,

=

(d) 5

3.019 x 10

-2

s"

1

REFERENCES (Bl)

Bird, R.

B.,

Stewart, W.

York: John Wiley (B2)

E.,

and Lightfoot,

& Sons, Inc.,

Blakebrough, N. Biochemical and York: Academic Press,

Inc., 1967.

(B3)

Beddington, C.

and Drew,

(Cl)

Carmichael, L.T., Sage,

(C2)

Chilton, T.

(C3) (C4)

(Dl) (D2)

H.,

Jr.,

B. H.,

E. N. Transport

Phenomena.

New

1960.

Biological Engineering Science, Vol.

T. B. Ind. Eng. Chem., 42,

and Lacey, W. N. A.I.Ch.E.

1

1.

New

164 (1950).

J., 1,

385 (1955).

83 H., and Colburn, A. P. Ind. Eng. Chem., 26, 934). Calderbank, P. H., and Moo- Young, M. B. Chem. Eng. Sci., 16, 39 (1961). Cunningham, R. S., and Geankoplis, C. J. Ind. Eng. Chem. Fund., 1, 535 (1968). Danckwerts, P. V. Trans. Faraday Soc, 46, 300(1950).

Dunn, W. E, Bonilla,

1

C.

F.,

1

Ferstenberg, C, and Gross,

( 1

B. A.I.Ch.E.

J., 2,

184

(1956).

Chap. 7

References

485

(D3)

Danckwerts,

Gas-Liquid Reactions.

P. V.

New York: McGraw-Hill Book Com-

pany, 1970.

(D4)

Dwtvedi,

P. N.,

and Upadhyay,

N. Ind. Eng. Chem., Proc. Des. Dev.,

S.

16,

157

(1977).

(Gl)

Geankoplis, C.

J.

Mass Transport Phenomena. Columbus, Ohio: Ohio

State

University Bookstores, 1972.

and Sherwood, T. K.

(G2)

Gilliland, E.

(G3)

Garner,

(G4)

Gupta, A.

S.,

and Thodos, G.

(G5)

Gupta,

A.

S.,

and Thodos, G. A.I.Ch.E.

(G6)

Gupta, A.

S.,

and Thodos, G. Chem. Eng. Progr., 58

(LI)

Linton, W. H.,

(L2)

Litt,

(Ml)

McAdams, W. H. Heat Transmission, 3rd ed. New York: McGraw-Hill Book Company, 1954. Perry, R. H., and Green, D. Perry's Chemical Engineers' Handbook, 6th ed. New York: McGraw-Hill Book Company, 1984.

(PI)

R.,

M, and

(Rl)

Remick, R.

(51)

Seager,

Jr.,

S. L.,

Ind.

J., 4,

1

Chem. Eng. Fund., J., 8,

14 (1958). 3,

218

(1964).

609 (1962). (7),

58 (1962).

and Sherwood, T. K. Chem. Eng. Progr., 46, 258

Friedlander,

R.,

Ind. Eng. Chem., 26, 516 (1934).

and Suckling, R. D. A.I.Ch.E.

F. H.,

S.

K. A.I.Ch.E.

J., 5,

and Geankoplis,

C.

Geertson,

and Giddings,

L. R.,

J.

Ind. Eng.

(1950).

483 (1959).

Chem. Fund., J.

C. J.

12, 214(1973).

Chem. Eng. Data,

8,

168

(1963).

(52)

Satterfield, C. N.

The MIT

Mass

Transfer in Heterogeneous Catalysis. Cambridge, Mass.:

Press, 1970.

(53)

Steele, L. R., and Geankoplis, C.

(54)

Sherwood, T. K., Pigford, R. McGraw-Hill Book Company,

J.

A.I.Ch.E.

J., 5,

and Wilke, C.

L.,

178 (1959). R.

Mass

Transfer.

New

York:

1975.

(57)

W. L., and Treybal, R. E. A.I.Ch.E. J., 6, 227 (1960). M. Chemical Engineering Kinetics, 2nd ed. New York: McGraw-Hill Book Company, 1970. Szekely, J., and Themelis, N. Rare Phenomena in Process Metallurgy. New York

(Tl)

Treybal, R. E. Mass Transfer Operations, 3rd

(55)

Steinberger,

(56)

Smith,

J.

Wiley-Interscience, 1971.

Company,

M.

(T2)

Toor, H. L, and Marchello,

(VI)

Vogtlander,

P. H.,

(Wl)

Wilson,

and Geankoplis, C.

486

E.

ed.

New York: McGraw-Hill Book

1980.

J.,

J.

A.I.Ch.E.

and Bakker, C. A. J.

P.

J., 1,

97 (1958).

Chem. Eng.

Ind. Eng.

Sci., 18,

583 (1963).

Chem. Fund.,5, 9

(1966).

Chap. 7

References

1

PART

2

Unit Operations

i

CHAPTER

8

Evaporation

INTRODUCTION

8.1

8.1

A

Purpose

In Section 4.8

we

discussed the case of heat transfer to a boiling liquid.

An important

instance of this type of heat transfer occurs quite often in the process industries

name

given the general

and

is

evaporation. In evaporation the vapor from a boiling liquid

removed and a more concentrated solution remains. In the majority of cases removal of water from an aqueous solution. Typical examples of evaporation are concentration of aqueous solutions of sugar, sodium chloride, sodium hydroxide, glycerol, glue, milk, and orange juice. In these cases the concentrated solution is the desired product and the evaporated water is normally

solution

is

the unit operation evaporation refers to the

discarded. In a few cases, water, which has a small

amount

of minerals,

is

evaporated to

which is used as boiler feed, for special chemical processes, or for other purposes. Evaporation processes to evaporate seawater to provide drinking water give a solids-free water

have been developed and used. In some cases, the primary purpose of evaporation

is

to

concentrate the solution so that upon cooling, salt crystals will form and be separated.

This special evaporation process, termed crystallization,

8.1B

The

is

discussed in Chapter

12.

Processing Factors

and chemical properties

physical

of the solution being concentrated

and of the vapor

being removed have a great effect on the type of evaporator used and on the pressure and

temperature of the process.

Some

of these properties

which

afTect the processing

methods

are discussed next.

J.

Concentration

dilute, so

its

in

the liquid.

viscosity

are obtained.

is

Usually, the liquid feed to an evaporator

low, similar to water, and

As evaporation proceeds,

is

relatively

relatively high heat-transfer coefficients

the solution

may become

very concentrated and

drop markedly. Adequate circucoefficient from becoming too low.

quite viscous, causing the heat-transfer coefficient to lation and/or turbulence

must be present

to

keep the

489

As solutions

2. Solubility.

are heated

and concentration of

the solubility limit of the material in solution

may

This

limit the

maximum

evaporation. In Fig. 8.1-1

may

the solute or salt increases,

be exceeded and crystals

may

form.

concentration in solution which can be obtained by

some

solubilities of typical salts in

water are shown as a

function of temperature. In most cases the solubility of the salt increases with temperature. This

room 3.

means

Temperature

when a hot concentrated

may

sensitivity of materials.

logical materials, after

that

temperature, crystallization

may

solution from an evaporator

is

cooled to

occur.

Many

products, especially food and other bio-

be temperature-sensitive and degrade at higher temperatures or

prolonged heating. Such products are pharmaceutical products; food products such orange juice, and vegetable extracts; and fine organic chemicals. The amount of

as milk,

degradation

4.

is

a function of the temperature and the length of time.

some cases materials composed of caustic solutions, food and some fatty acid solutions form a foam or froth during This foam accompanies the vapor coming out of the evaporator and entrainment

Foaming

or frothing.

In

solutions such as skim milk, boiling.

losses occur.

5.

The

Pressure and temperature.

of the system.

temperature

The higher

boiling point of the solution

concentration of the dissolved material

at boiling. Also, as the

increases by evaporation, the temperature of boiling

and

boiling-point rise or elevation

low in heat-sensitive materials, under vacuum. 6.

is

it is

discussed

in

may

rise.

Section

8.4.

This

Some

in

phenomenon

To keep

often necessary to operate under

Scale deposition and materials of construction.

called scale

related to the pressure

is

the operating pressure of the evaporator, the higher the

1

solution is

called

the temperatures

atm pressure,

i.e.,

solutions deposit solid materials

on the heating surfaces. These could be formed by decomposition products or

The result is that the overall heat-transfer coefficient decreases and must eventually be cleaned. The materials of construction of the evaporator are important to minimize corrosion.

solubility decreases.

the evaporator

0,

0

50

100

Temperature (°C) Figure

490

8.1-1.

Solubility curves for

some

typical sails in waier.

Chap. 8

Evaporation

TYPES OF EVAPORATION EQUIPMENT

8.2

AND OPERATION METHODS General Types of Evaporators

8.2A

In evaporation, heat

water.

The heat

is

added to a solution

is

to vaporize the solvent,

which

is

usually

generally provided by the condensation of a vapor such as steam

on

one side of a metal surface with the evaporating liquid on the other side. The type of equipment used depends primarily on the configuration of the heat-transfer surface and on the means employed to provide agitation or circulation of

These general

the liquid.

types are discussed below.

Open kettle or pan. The simplest form of evaporator consists of an open pan or kettle which the liquid is boiled. The heat is supplied by condensation of steam in a jacket or in coils immersed in the liquid. In some cases the kettle is direct-fired. These evaporators are inexpensive and simple to operate, but the heat economy is poor. In some cases, /.

in

paddles or scrapers for agitation are used.

The horizontal-tube natural circuThe horizontal bundle of heating tubes is similar to the bundle of tubes in a heat exchanger. The steam enters into the tubes, where it condenses. The steam condensate leaves at the other end of the tubes. The boilingliquid solution covers the tubes. The vapor leaves the liquid surface, often goes through some deentraining device such as a baffle to prevent carryover of liquid droplets, and leaves out the top. This type is relatively cheap and is used for nonviscous liquids having high heat-transfer coefficients and liquids that do not deposit scale. Since liquid circuHorizontal-tube natural circulation evaporator.

2.

lation evaporator

lation

is

shown

is

in Fig. 8.2-la.

poor, they are unsuitable for viscous liquids. In almost

all

cases, this

and the types discussed below are operated continuously, where the constant rate and the concentrate leaves at a constant rate. Vertical-type natural circulation evaporator.

3.

In this

rather than horizontal tubes are used, and the liquid

is

type of evaporator, vertical

inside the tubes

condenses outside the tubes. Because of boiling and decreases in

evaporator

feed enters at a

and the steam

in density, the liquid rises

shown in Fig. 8.2- lb and flows downward through downcomer. This natural circulation increases the heat-

the tubes by natural circulation as

a large central open space or transfer coefficient.

It

short-tube evaporator.

but the heating element as the

ator,

variation of this is

which has

is

is

often called the

the basket type, where vertical tubes are used,

body so there is an annular open space from the vertical natural circulation evapor-

held suspended in the

downcomer. The basket type a central instead of

widely used in the sugar, 4.

not used with viscous liquids. This type

is

A

salt,

differs

annular open space as the downcomer. This type

and caustic soda

Long-tube vertical-type evaporator.

compared

is

industries.

Since the heat-transfer coefficient on the steam

on the evaporating liquid side, high liquid velocities are desirable. In a long-tube vertical-type evaporator shown in Fig. 8.2- lc, the liquid is inside the tubes. The tubes are 3 to 10 m long and the formation of vapor bubbles inside

side

is

very high

the tubes causes a

to that

pumping

action giving quite high liquid velocities. Generally, the

liquid passes through the tubes only once

quite low in this type. In

some

cases, as

and

when

is

not recirculated. Contact times can be

the ratio of feed to evaporation rate

natural recirculation of the product through the evaporator

is

pipe connection between the outlet concentrate line and the feed for

is

low,

done by adding a large line.

This

is

widely used

producing condensed milk.

Sec. 8.2

Types of Evaporation Equipment and Operation Methods

491

J.

A

Falling-film-type evaporator.

evaporator, wherein the liquid

is

variation of the long-tube type

fed to the

is

the falling-film

top of the tubes and flows down the walls as a

thin film. Vapor-liquid separation usually takes place at the bottom. This type

used for concentrating heat-sensitive materials such as orange juice and other

because the holdup time

is

very small (5 to 10

s

is

widely

fruit juices,

or more) and the heat-transfer coefficients

are high

6.

Forced-circulation-type evaporator.

The

liquid-film heat-transfer coefficient can be

pumping to cause forced circulation of the liquid inside the tubes. This could be done in the long-tube vertical type in Fig. 8.2-lc by adding a pipe connection with a pump between the outlet concentrate line and the feed line. However, usually in a forced-circulation type, the vertical tubes are shorter than in the long-tube type, and this increased by

vapor vapor

steam -

steam concentrate

condensate feed

condensate concentrate feed (c)

FIGURE

8.2-1.

Different types of evaporators

tube type,

492

(c)

:

(a)

horizontal-tube type,

(b) vertical-

long-tube vertical type, (d) forced-circulation type.

Chap. 8

Evaporation

type

is

shown

exchanger

7.

falling-film

One way

very useful for viscous liquids.

is

In an evaporator the main resistance to heat transfer

Agitated-film evaporator.

the liquid side. coefficient, is

by actual mechanical agitation of this

liquid film. This

by the

top of the tube and as

vertical agitator blades.

done

it

flows downward,

The concentrated

bottom and vapor leaves through a separator and out the is

is

a modified

in

evaporator with only a single large jacketed tube containing an internal

into a turbulent film

ator

on

is

to increase turbulence in this film, and hence the heat-transfer

agitator. Liquid enters at the

the

other cases a separate and external horizontal heat

in Fig. 8.2-ld. Also, in

used. This type

is

it is

spread out

solution leaves at

This type of evapor-

top.

very useful with highly viscous materials, since the heat-transfer coefficient

greater than in forced-circulation evaporators.

It

is

used with heat-sensitive viscous

is

and fruit juices. However, it has a high For interested readers, Perry and Green (P2) give more

materials such as rubber latex, gelatin, antibiotics,

cost and small capacity. detailed discussions

8.

in

and descriptions of evaporation equipment.

A

Open-pan solar evaporator. open pans. Salt water is put

in

slowly in the sun to crystallize the

still

used process

is

solar evaporation

salt.

Methods of Operation of Evaporators

8.2B

1.

very old but yet

shallow open pans or troughs and allowed to evaporate

A simplified diagram of a single-stage or single-effect evap-

Single-effect evaporators.

orator

is

The feed enters at TF K and saturated steam at Ts enters the Condensed steam leaves as condensate or drips. Since the solu-

given in Fig. 8.2-2.

heat-exchange section. tion in the evaporator

assumed

is

the solution in the evaporator

to be completely mixed, the concentrated product and have the same composition and temperature T,, which is

The temperature The pressure

the boiling point of the solution.

of the vapor

equilibrium with the boiling solution. the solution at If

is

Pj, which

is is

also

7*,,

since

it is

in

the vapor pressure of

7*,.

the solution to be

steam condensing

will

evaporated

entering has a temperature

The concept

is

assumed

evaporate approximately

TF

to 1

be dilute and

rate of heat transfer in an evaporator.

q

=

If

The

A

vapor

Tp

steam,

7$

will

then

hold

if

1

kg of

the feed

near the boiling point.

of an overall heat-transfer coefficient

feed,

like water,

kg of vapor. This is

used

in the

calculation of the

general equation can be written

AT = UA(TS T

Tj)

(8.2-1)

to condenser

heat-exchange tubes

1

condensate concentrated product Figure

Sec. 8.2

8.2-2.

Simplified diagram of single-effect evaporator.

Types of Evaporation Equipment and Operation Methods

493

W (btu/h), U is the overall heat-transfer coefficient A is the heat-transfer area in m 2 (ft 2 Ts is the temperature of the condensing steam in K (°F), and T is the boiling point of the liquid in K (°F).

where q in

is

the rate of heat transfer in

W/m 2 K (btu/h

2

•

ft

•

°F),

),

l

Single-effect evaporators are often used relatively small cost.

However,

when

the required capacity of operation

is

cheap compared to the evaporator large-capacity operation, using more than one effect will markedly

and/or the cost of steam for

relatively

is

reduce steam costs.

2.

A

Forward-feed multiple-effect evaporators.

8.2-2

is

much

discarded. However,

evaporator

ration system

is

shown

the feed to the

kg of steam

A

diagram of a forward-feed

simplified

as

is

shown

in Fig.

not used but

is

and reused by employing

of this latent heat can be recovered

multiple-effect evaporators.

If

single-effect

wasteful of energy since the latent heat of the vapor leaving

triple-effect

evapo-

Fig. 8.2-3.

in

first

effect

is

near the boiling point at the pressure

evaporate almost

kg of water. The

in the first effect,

effect operates

at a high-enough temperature so that the evaporated water serves as the heating medium to 1

will

1

first

kg of water is evaporated, which can be As a very rough approximation, almost 3 kg of water will be evaporated for 1 kg of steam for a three-effect evaporator. Hence, the steam economy, which is kg vapor evaporated/kg steam used, is increased. This also approximately holds for a number of effects over three. However, this increased steam the second effect. Here, again, almost another

used as the heating

economy of a

medium

to the third effect.

multiple-effect evaporator

is

gained at the expense of the original

first

cost

of these evaporators.

shown in Fig. 8.2-3, the fresh feed is added to the first same direction as the vapor flow. This method of operation is used when the feed is hot or when the final concentrated product might be damaged at high temperatures. The boiling temperatures decrease from effect to effect. This means that if the first effect is at P, = 1 atm abs pressure, the last effect will be under vacuum at a pressure P 3 In forward-feed operation as

effect

and flows

to the next in the

.

3.

Backward-feed multiple-effect evaporators.

Fig. 8.2-4 for

a

triple-effect

In the backward-feed operation

evaporator, the fresh feed enters the

continues on until the concentrated product leaves the feed

is

advantageous when the fresh feed

heated to the higher temperatures

I

concentrate

from effect

Figure

494

8.2-3.

first

first effect.

cold, since a smaller

the second

vapor

vapor -

in

is

i

and

T2

last

first effects.

concentrate from second

f

shown in and

coldest effect

This method of reverse

amount

of liquid must be However, liquid pumps

vapor I

and

T3

to

vacuum

""condenser

concentrated product

effect Simplified diagram offorward-feed triple-effect evaporator.

Chap. 8

Evaporation

vapor

vapor

vapor

I

product Figure

8.2-4.

Simplified diagram of backward-feed triple-effect evaporator.

are used in each effect, since the flow

used

when

effects

the concentrated product

is

is

from low to high pressure. This method

highly viscous.

also

is

high temperatures in the early

reduce the viscosity and give reasonable heat-transfer coefficients. Parallel feed in multiple-effect evaporators

Parallel-feed multiple-effect evaporators.

4.

The

involves the adding of fresh feed and the withdrawal of concentrated product from each

The vapor from each effect is still used to heat the next effect. This method of is mainly used when the feed is almost saturated and solid crystals are the product, as in the evaporation of brine to make salt.

effect.

operation

OVERALL HEAT-TRANSFER COEFFICIENTS

8.3

IN EVAPORATORS The

U in an evaporator is composed of the steam-side 2 which has a value of about 5700 W/m 2 K (1000 btu/h ft °F);

overall heat-transfer coefficient

condensing

coefficient,

resistance; the resistance of the scale

which

is

•

•

the metal wall, which has a high thermal conductivity

on the

liquid side;

•

and usually has a negligible

and the liquid

film coefficient,

usually inside the tubes.

The steam-side condensing coefficient outside the tubes can be estimated using Eqs. The resistance due to scale formation usually cannot be predicted.

(4.8-20)-(4-8-26).

Increasing the velocity of the liquid in the tubes greatly decreases the rate of scale

formation. This

can be

salts,

an increase

if

is

one important advantage of forced-circulation evaporators. The scale

such as calcium sulfate and sodium sulfate, which decrease

in

in solubility

with

temperature and hence tend to deposit on the hot tubes.

For forced-circulation evaporators the coefficient h inside the tubes can be predicted is little or no vaporization inside the tube. The liquid hydrostatic head in the

there

tubes prevents most boiling in the tubes.

The standard equations

for predicting the h

value of liquids inside tubes can be used. Velocities used often range from 2 to 5 m/s (7 to 15

ft/s).

The

heat-transfer coefficient can be predicted from Eq. (4.5-8), but using a

constant of 0.028 instead of 0.027 (B all

1).

If there is

of the tubes, the use of the equation assuming

some

or appreciable boiling in part or

no boiling

will give

conservative safe

results (PI).

For long-tube vertical natural circulation evaporators the heat-transfer coefficient is more difficult to predict, since there is a nonboiling zone in the bottom of the tubes and a boiling zone in the top. The length of the nonboiling zone depends on the heat transfer in the two zones and the pressure drop in the boiling two-phase zone.

Sec. 8.3

Overall Heat-Transfer Coefficients in Evaporators

The

film heat-transfer

495

Table

Typical Heat-Transfer Coefficients for Various Evaporators*

8.3-1.

(B3, B4, LI, P2)

Overall

Wlm 2 K

Type of Evaporator Short-tube

vertical,

natural circulation

Horizontal-tube, natural circulation

Long-tube

vertical,

natural circulation

Long-tube

vertical,

forced circulation

Agitated film *

U Btu/h-ft

2

-°F

1100-2800 1100-2800 1100-4000

200-500 200-500 200-700

2300-11000 680-2300

400-2000 120-400

Generally, nonviscous liquids have the higher coefficients and viscous liquids the lower coefficients

in the

ranges given.

zone can be estimated using Eq. (4.5-8) with a constant of For the boiling two-phase zone, a number of equations are given by Perry and

coefficient in the nonboiling"

0.028.

Green (P2). For short-tube vertical evaporators the heat-transfer coefficients can be estimated by using the same methods as for the long-tube vertical natural circulation evaporators. Horizontal-tube evaporators have heat-transfer coefficients of the same order of magnitude as the short-tube vertical evaporators.

For the agitated-film evaporator, the heat-transfer

coefficient

may

be estimated

using Eq. (4.13-4) for a scraped surface heat exchanger.

The methods given above are

useful for actual evaporator design and/or for evalu-

ating the effects of changes in operating conditions on

preliminary designs or cost estimates, coefficients usually

encountered

in

it is

the coefficients. In

making

helpful to have available overall heat-transfer

commercial practice. Some preliminary values and

ranges of values for various types of evaporators are given in Table 8.3-1.

CALCULATION METHODS FOR

8.4

SINGLE-EFFECT EVAPORATORS Heat and Material Balances

8.4A

The

for Evaporators

basic equation for sol ving for the capacity of a single-effect evaporator

Eq.

is

(8.2-1),

which can be written as q

where .

AT K (°F)

is

= UA AT

(8.4-1)

the difference in temperature between the condensing steam and the

boiling liquid in the evaporator. In order to solve Eq. (8.4-1) the value of q in

W (btu/h)

must be determined by making a heat and material balance on the evaporator shown in Fig. 8.4-1. The feed to the evaporator is F kg/h(lb m/h) having a solids content ofx F mass fraction,

temperature

concentrated liquid

TF

L

enthalpy h L The vapor .

,

and enthalpy h F J/kg (btu/lb m

kg/h

V T

(lb m/h)

).

Coming

having a solids content of x L

kg/h {Vo^Jh)

is

,

out as a liquid

temperature

is

Tj,

the

and

given off as pure solvent having a solids content

and enthalpy H v Saturated steam entering is S kg/h (Vojh) and has a temperature of 7^ and enthalpy of Hs The condensed steam leaving of S kg/h This is assumed usually to be at 7^, the saturation temperature, with an enthalpy o(h s

of y Y

=

0,

temperature

1;

.

.

.

496

Chap. 8

Evaporation

means

that the

steam gives off only

its

latent heat,

A Since the vapor

V

is

in

at its boiling

point

P

l

=

(8.4-2)

assumes no boiling-point

rise.)

=

steady state, the rate of mass in

at

rate of

(8.4-3)

Fx F = Lx L

(8.4-4)

the solute (solids) alone,

For the heat balance, since the 4-

hs

L+V

F =

heat in feed

where

X,

the saturation vapor pressure of the liquid of

is

Tj. (This

For the material balance since we are mass out. Then, for a total balance,

For a balance on

is

equilibrium with the liquid L, the temperatures of vapor and

liquid are the same. Also, the pressure

composition x L

= Hs -

which

total heat entering

=

total heat leaving,

heat in steam

heat in concentrated liquid

4-

+

heat in vapor

heat in condensed steam

This assumes no heat lost by radiation or convection. Substituting into Eq.

Fh F

+ SHS =

(8.4-5)

(8.4-5),

VH V + Sh s

Lh L +

(8.4-6)

Substituting Eq. (8.4-2) into (8.4-6),

Fh F The heat q

+ SX=

transferred in the evaporator

q In Eq. (8.4-7) the latent heat

is

Lh L +

VH V

(8.4-7)

SI

(8.4-8)

then

= S(H S -

hs )

=

A of steam at the saturation temperature

Ts

can be

obtained from the steam tables in Appendix A. 2. However, the enthalpies of the feed and products are often not available. These enthalpy-concentration data are available for only a few substances in solution. Hence,

make 1.

It 1

some approximations

temperature of the boiling solution

T,

order to

(exposed surface temperature)

rather than the equilibrium temperature for pure water at If

in

can be demonstrated as an approximation that the latent heat of evaporation of kg mass of the water from an aqueous solution can be obtained from the steam

tables using the

2.

made

are

a heat balance. These are as follows.

the heat capacities c pF of the liquid feed

and

c pL of the

P

.

i

product are known, they can

be used to calculate the enthalpies. (This neglects heats of dilution, which in most cases are not known.)

Figure

8.4-1.

Heat and mass

vapor

balance for

single-effect evaporator.

feed

F

F'

xc

Ti ,

F'

h„ F

,

V

y v,

Hv

Pi

7

condensate S

Ts>

steam S

Ts-

Hs

Calculation Methods for Single-Effect Evaporators

s

concentrated liquid L

Ti,x L Sec. 8.4

h

,

hL

497

EXAMPLE 8.4-1.

Heat-Transfer Area in Single-Effect Evaporator continuous single-effect evaporator concentrates 9072 kg/h of a 1.0 wt salt solution entering at 311.0 K. (37.8°C) to a final concentration of 1.5 wt %. The vapor space of the evaporator is at 101.325 kPa (1.0 atm abs) and the steam supplied is saturated at 143.3 kPa. The overall coefficient U = 1704 W/m 2 K. Calculate the amounts of vapor and liquid product and

%

A

-

Assume

the heat-transfer area required.

that, since

it

the solution

is dilute,

has the same boiling point as water.

The

Solution:

flow diagram

same

the

is

as that in Fig. 8.4-1.

For the

material balance, substituting into Eq. (8.4-3),

F=L+V —L+ V

9072 Substituting into Eq. (8.4-4)

and

(8.4-3)

solving,

Fx F = Lx L

=

9072(0.01)

(8.4-4)

L(0.015)

L = 6048 kg/h

of liquid

Substituting into Eq. (8.4-3) and solving,

V = 3024 kg/h

*

of vapor

=

4.14 kJ/kg- K. The heat capacity of the feed is assumed to be cpF (Often, for feeds of inorganic salts in water, the c p can be assumed approxi-

mately as that of water alone.) To make a heat balance using Eq. (8.4-7), it is convenient to select the boiling point of the dilute solution in the evaporator, which is assumed to be that of water at 101.32 kPa, Tj = 373.2 K (100°C), as the datum temperature. Then y is simply the latent heat of

H

water at 373.2 K, which from the steam tables in Appendix A.2

is 2257 kJ/kg steam at 143.3 kPa [saturation temperature Ts = 383.2 K (230°F)] is 2230 kJ/kg (958.8 btu/lbj. The enthalpy of the feed can be calculated from

(970.3 btu/lbj.

The

latent heat A of the

hF

=

(TF - TJ

=

Substituting into Eq. (8.4-7) with h L 9072(4.14)(311.0

-

373.2)

S

The

=

(8.4-9)

c pF

+

0,

since

S(2230)

=

it

is

at the

6048(0)

datum of 373.2 K,

+

3024(2257)

4108 kg steam/h

heat q transferred through the heating surface area

A

is,

from Eq.

(8.4-8),

q

q

=

Solving,

8.4B

A =

m

= 2544000

W

AT = Ts — T

l7

= 2544000 = UA AT = 149.3

(8.4-8)

S(a)

4108(2230X1000/3600)

Substituting into Eq. (8.4-1), where q

=

1704(^X383.2

-

373.2)

2 .

Effects of Processing Variables on

Evaporator Operation

/. Effect offeed temperature. The inlet temperature of the feed has a large effect on the operation of the evaporator. In Example 8.4-1 the feed entering was at a cold temper-

498

Chap. 8

Evaporation

ature of 31 1.0

K

compared

temperature of 373.2 K. About £ of the steam

to the boiling

used for heating was used to heat the cold feed to the boiling point. Hence, only about

| of the steam was left for vaporization of the feed. If the feed is under pressure and enters the evaporator at a temperature above the boiling point in the evaporator, additional vaporization

obtained by the flashing of part of the entering hot

is

feed.

Preheating the

feed can reduce the size of evaporator heat-transfer area needed.

In Example 8.4-1 the pressure of 101.32 kPa abs was used in the Effect of pressure. vapor space of the evaporator. This set the boiling point of the solution at 373.2 K and gave a AT for use in Eq. (8.4-1) of 383.2 — 373.2, or 10 K. In many cases a larger AT is 2.

AT

desirable, since, as the

decrease.

To

and vacuum

pump can

be used. For example,

the boiling point of water

A

33.3 K.

A

increases, the heating-surface area

reduce the pressure below 101.32 kPa, if

i.e.,

and cost of evaporator vacuum, a condenser

to be under

the pressure were reduced to 41.4 kPa,

K

and the new AT would be 383.2 heating-surface area would be obtained.

would be 349.9

large decrease in

—

349.9, or

Using higher pressure, saturated steam increases the AT, and cost of the evaporator. However, high-pressure steam is more costly and also is often more valuable as a source of power elsewhere. Hence, overall economic balances are really needed to determine the optimum steam pressures.

3.

Effect of steam pressure.

which decreases the

8.4C

size

Boiling-Point Rise of Solutions

In the majority of cases in evaporation the solutions are not such dilute solutions as

those considered in

being evaporated solutions are high

Example

may

differ

8.4-1. In

most

thermal properties of the solution

cases, the

considerably from those of water. The concentrations of the

enough so

that the heat capacity

and boiling point are quite

different

from that of water. For strong solutions of dissolved solutes the boiling-point rise due to the solutes the solution usually cannot be predicted. However, a useful empirical law known Diihring's rule can be used. In this rule a straight line

solution in °C or °F for a given

is

obtained

is

if

the boiling point of a

plotted against the boiling point of pure water at the

concentration

at different pressures.

A

different straight line

each given concentration. In Fig. 8.4-2 such a Diihring

sodium hydroxide in water. It is necessary at only two pressures to determine a line.

to

know

line chart

is

in

as

same

is

pressure

obtained

for

given for solutions of

the boiling point of a given solution

EXAMPLE 8.4-2.

Use of Diihring Chart for Boiling-Point Rise As an example of use of the chart, the pressure in an evaporator is given as 25.6 kPa (3.72 psia) and a solution of 30% NaOH is being boiled. Determine the boiling temperature of the NaOH solution and the boiling-point rise

BPR of the solution over that Solution:

of water at the

From the steam tables kPa is 65.6°C. From

water at 25.6

NaOH,

the boiling point of the

boiling-point rise In Perry

number addition

is

79.5

-

65.6

and Green (P2) a chart

= is

in

same

Appendix

pressure. A.2, the boiling point of

Fig. 8.4-2 for 65.6°C (150°F)

NaOH

solution

and 30%

79.5°C (175°F). The

13.9°C (25°F).

given to estimate the boiling-point rise of a large

common aqueous solutions used in chemical and biological processes. In to the common salts and solutes, such as NaN0 3 NaOH, NaCl, and H 2 S0 4

of

,

,

the biological solutes sucrose, citric acid, kraft solution,

Sec. 8.4

is

and glycerol are

Calculation Methods for Single-Effect Evaporators

given.

These

499

Boiling point of water 0

50

25

50

0

(

75

150

100

C) 100

125

200'.

250

300

Boiling point of water (°F) Figure

8.4-2.

biological solutes have

common

Diihring lines for aqueous solutions of sodium hydroxide.

quite small

boiling-point-rise values

compared

to those

of

salts.

Enthalpy-Concentration Charts of Solutions

8.4D If the

large,

heat of solution of the aqueous solution being concentrated in the evaporator neglecting

it

is

could cause errors in the heat balances. This heat-of-solution

phenomenon can be explained as follows. If pellets of NaOH are dissolved in a given amount of water, \[ is found that a considerable temperature rise occurs; i.e., heat is evolved, called heat of solution. The amount of heat evolved depends on the type of substance and on the amount of water used. Also, if a strong solution of NaOH is diluted to a

lower concentration, heat

is

liberated. Conversely,

if

a solution

is

concentrated from

a low to a high concentration, heat must be added.

In Fig. 8.4-3 an enthalpy-concentration chart for

enthalpy

is

in

weight fraction

made

500

kJ/kg (btu/lb m ) solution, temperature

NaOH

for solutions

in solution.

NaOH

in

°C

is

(°F),

given (Ml) where the

and concentration

Such enthalpy-concentration charts

in

are usually not

having negligible heats of solution, since the heat capacities can be

Chap. 8

Evaporation

easily

used to calculate enthalpies. Also, such charts are available for only a few

solutions.

liquid water in Fig. 8.4-3 is referred to the same datum or steam tables, i.e., liquid water at 0°C (273 K). This means that enthalpies from the figure can be used with those in the steam tables. In Eq. (8.4-7) values for h F and h L can be taken from Fig. 8.4-3 and values for X andH v from the steam tables. The uses of Fig. 8.4-3 will be best understood in the following example.

The enthalpy of the

reference state as in the

Evaporation of an NaOH Solution used to concentrate 4536 kg/h (10000 lb^/h) of a 20% solution of NaOH in water entering at 60°C (140°F) to a product of 50% solids. The pressure of the saturated steam used is 172.4 kPa (25 psia) and the pressure in the vapor space of the evaporator is 11.7 kPa (1.7 psia). The 2 2 overall heat-transfer coefficient is 1 560 W/m °F). Calcu(275 btu/h ft late the steam used, the steam economy in kg vaporized/kg steam used, and 2 the heating surface area in m

EXA MPLE 8.4-3. An

evaporator

is

•

K

•

•

.

The process flow diagram and nomenclature are the same as The given variables are F = 4536 kg/h, x F = 0.20 wt fraction, TF = 60°C, P = 11.7 kPa, steam pressure = 172.4 kPa, and x L = 0.50 wt fraction. For the overall material balance, substituting into Eq.

Solution:

.

given in Fig. 8.4-1.

{

(8.4-3),

F=

4536

=L+ V

Concentration (wt fraction Figure

Sec. 8.4

8.4-3.

(8.4-3)

NaOH)

Enthalpy-concentration chart for the system NaOH-water. [Reference state liquid water at 0°C {273 K) or 32°F.~] [From W, L. McCabe, Trans. A.l.Ch.E.,31, 129(1935). With permission.}

Calculation

Methods for

Single-Effect Evaporators

501

Substituting into Eq. (8.4-4)

and solving (8.4-3) and

(8.4-4) simultaneously,

Fx F = Lx L 4536(0.20)

=

L(0.50)

K =

L=1814kg/h

(8.4-4)

2722 kg/h

To determine the boiling point Tx of the 50% concentrated solution, we obtain the boiling point of pure water at 1 1.7 kPa from the steam tables, Appendix A.2, as 48.9°C (120°F). From the Diihring chart, Fig. 8.4-2, for a first

boiling point of water of 48.9°C

solution

is

=

T,

boiling-point rise

From

and

50% NaOH,

the boiling point of the

89.5°C (193°F). Hence,

=

T,

- 48.9 =

89.5

- 48.9 =

40.6°C (73°F)

the enthalpy-concentration chart (Fig. 8.4-3), for

20% NaOH

at

60°C (140°F), /I, = 214 kJ/kg (92 btu/lbj. For 50% NaOH at 89.5°C (193°F), h L = 505kJ/kg(217btu/lbJ. For the superheated vapor V at 89.5°C (193°F) and 11.7 kPa [superheated 40.6°C (73°F) since the boiling point of water is 48.9°C (120°F) at 11.7 kPa], from the steam tables, v = 2667 kJ/kg (1147 btu/lbj. An alternative method to use to calculate the v is to first obtain the enthalpy of saturated vapor at 48.9°C (120°F) and 11.7 kPa of 2590 kJ/kg (1113.5

H

H

btu/lb m ). Then using a heat capacity of 1.884 kJ/kg-K for superheated steam with the superheat of (89.5 - 48.9)°C = (89.5 - 48.9) K,

Hv =

2590

+

1.884(89.5

-

=

48.9)

2667 kJ/kg

For the saturated steam the steam tables

at 172.4 kPa, the saturation temperature from 115.6°C (240°F) and the latent heat is X = 2214 kJ/kg

is

(952 btu/lbj. Substituting into Eq. (8.4-7) and solving for S,

VH y

Fh f + SX = Lh L + 4535(214)

+

S(2214)

S

=

=

1814(505)

+

(8.4-7)

2722(2667)

3255 kg steam/h

Substituting into Eq. (8.4-8),

= SX =

q

=

3255(2214>

2002

kW

Substituting into Eq. (8.4-1) and solving,

2002(1000)= Hence,

8.5

8.5A

A =

49.2

m

2 .

Also,

1560(/1)(1 15.6

steam economy

=

-

89.5)

2722/3255

0.836.

CALCULATION METHODS FOR MULTIPLE-EFFECT EVAPORATORS Introduction

In evaporation of solutions in a single-effect evaporator, a

steam used

502

=

to

evaporate the water.

A

single-effect

major cost

evaporator

is

is

the cost of the

wasteful of steam costs,

Chap. 8

Evaporation

vapor leaving the evaporator

since the latent heat of the

usually not used. However, to

is

reduce this cost, multiple-effect evaporators are used which recover the latent heat of the vapor leaving and reuse it.

A three-effect evaporator, discussed briefly in Section 8.2B, this

system each

steam

is

effect in itself acts as

used as the heating

and pressure condensing

medium

is

shown in

a single-effect evaporator. In the

to this first effect,

which

Fig. 8.2-3. In

raw

first effect

boiling at temperature T,

is

The vapor removed from the first effect is used as the heating medium, second effect and vaporizing water at temperature T2 and pressure F 2

F,.

in the

To

from the condensing vapor to the boiling liquid in this T2 must be less than the condensing temperature. This means that the pressure P 2 in the second effect is lower than F, in the first effect. In a similar manner, vapor from the second effect is condensed in heating the third effect. Hence, pressure F 3 is less than F 2 If the first effect is operating at 1 atm abs pressure, the second and third effects will be under vacuum. In the first effect, raw dilute feed is added and it is partly concentrated. Then this partly concentrated liquid (Fig. 8.2-3) flows to the second evaporator in series, where it is

in this effect.

second

transfer heat

boiling temperature

effect, the

.

further concentrated. This liquid from the second effect flows to the third effect for final

concentration.

When

a multiple-effect

evaporator

is

at steady-state operation, the

rate of evaporation in each effect are constant.

The

flow rates and

pressures, temperatures,

and

internal

flow rates are automatically kept constant by the steady-state operation of the process itself.

To change

the concentration in the final effect, the feed rate to the

first effect

must

be changed. The overall material balance made over the whole system and over each evaporator increased,

itself

and

must be

satisfied. If the final

vice versa.

Then

solution

is

too concentrated, the feed rate

the final solution will reach a

new steady

is

state at the

desired concentration.

Temperature Drops and Capacity

8.5B

of Multiple-Effect Evaporators

The amount of heat transferred per /. Temperature drops in multiple-effect evaporators. hour in the first effect of a triple-effect evaporator with forward feed as in Fig. 8.2-3 will be qx

where AT, liquid,

is

= U A l

l

AT,

(8.5-1)

steam and the boiling point of the no boiling-point rise and no heat of

the difference between the condensing

Ts — T

l

Assuming

.

that the solutions have

solution and neglecting the sensible heat necessary to heat the feed to the boiling point,

approximately

all

the latent heat of the condensing steam appears as latent heat in the

vapor. This vapor then condenses

amount

in the

second

effect,

up approximately the same

giving

of heat. q2

This same reasoning holds

for

L/,/t,AT, Usually,

in

q3

.

= U 2 A 2 AT2 Then

= U2 A2

since q l

(8.5-2)

=

=

q2

AT = U A 3

commercial practice the areas

2

3

q3

,

then, approximately,

AT,

in all effects are

(83-3)

equal and

(8-5-4)

Sec. 8.5

Calculation Methods for Multiple-Effect Evaporators

503

AT

Hence, the temperature drops

a multiple-effect evaporator are approximately

in

inversely proportional to the values of U. Calling'

£ AT

as follows for

no boiling-point

rise,

X AT = AT

Note

that

!/[/,,

then

= AT

°C

K,

AT

2

+ AT2 + AT, = Ts - T3

AT, °C

Similar equations can be written for

= AT2

AT

2

(83-5)

K, and so on. Since ATj

and AT3

proportional to

is

.

Capacity of multiple-effect evaporators. A rough estimate of the capacity of a threeevaporator compared to a single effect can be obtained by adding the value of q for

2.

effect

each evaporator. q If

we make

=

+


<\i

+

<7

3

=

V\A.\

AT + ^2^2 AT + U 3 A AT3

assumption that the value of (8.5-7) becomes

the

U

(8.5-7)

3

i

is

same

the

in

each

effect

and the values of

A are equal, Eq.

= UA(ATi + AT, +

q

where If

AT = £ AT =

is

(8.5-8)

3)

+ AT2 + AT3 = Ts - T3

ATj

a single-effect evaporator

same

the

AT = UA AT .

used with the same area A, the same value of U, and

temperature drop AT, then

total

q

= UA AT

(8.5-9)

same capacity as for the multiple-effect evaporators. Hence, the economy obtained by using multiple-effect evaporators is obtained at

This, of course, gives the

increase in steam

the expense of reduced capacity.

Calculations for Multiple-Efrect Evaporators

8.5C In

doing calculations

evaporator system, the values

for a multiple-effect

to

be obtained

are usually the area of the heating surface in each effect, the kg of steam per hour to be

supplied, and the

amount of vapor leaving each

or

known

in

vapor space of the

values are usually as follows:

concentration

last effect, (3)

in the liquid

(I)

effect, especially the last effect.

steam pressure to

feed conditions

The given

first effect, (2) final

and flow

to first effect, (4)

pressure the final

leaving the last effect, (5) physical properties such as en(6) the overall heat-transfer

thalpies and/or heat capacities of the liquid and vapors, and coefficients in

The

equations q trial

8.5D

and

each

effect.

calculations are

= UA AT

error.

The

each

effect.

A

convenient

way

balances, and the capacity

to solve these

equations

is

by

Step-by-Step Calculation Methods

From

the

known

boiling point

in

from a Duhring

504

done using material balances, heat

for

basic steps to follow are given as follows for a triple-effect evaporator.

for Triple-EfFect 1.

Usually, the areas of each effect are assumed equal.

Evaporators

outlet concentration

and pressure

the last effect. (If a boiling-point rise

is

in the last effect,

determine the

present, this can be determined

line plot.)

Chap. 8

Evaporation

2.

Determine the

total

amount

of vapor evaporated by an overall material balance.

For

among the three effects. (Usually, equal vapor produced in each effect, so that V = V2 = K3 is assumed for the first trial.) Make a total material balance on effects I, 2, and 3 to obtain L lt L 2 and L 3 Then this first trial

apportion

this total

amount

of

vapor

l

.

,

on each effect. and AT3 in the three

calculate the solids concentration in each effect by a solids balance 3.

Using Eq. effects!

(8.5-6),

Any

estimate the temperature drops A7"i, AT2 , has an extra heating load, such as a cold feed, requires a

effect that

proportionately larger

AT. Then

calculate the boiling point in each effect.

a boiling-point rise (BPR) in °C

[If

and 2 and determine the estimate

is

needed since

is

present, estimate the pressure in effects

available for heat transfer without the superheat three

all

BPRs from

5.

the overall

AT of Ts — T3

is

obtained by subtracting the

(saturation).

Using Eq.

(8.5-6),

sum

of

estimate

AT2 ,and AT3 Then calculate the boiling point in each effect.] Using heat and material balances in each effect, calculate the amount vaporized and the flows of liquid in each effect. If the amounts vaporized differ appreciably from those assumed in step 2, then steps 2, 3, and 4 can be repeated using the amounts of

AT,, 4.

1

BPR in each of the three effects. Only a crude pressure BPR is almost independent of pressure. Then the £ AT

.

evaporation just calculated. (In step 2 only the solids balance is repeated.) Calculate the value of q transferred in each effect. Using the rate equation q = UA AT for each effect, calculate the areas A u A 2 and/l 3 Then calculate the average value .

,

4. by

Am = If

A.

+ A 2 + A 31

(8.5-10)

f

these areas are reasonably close to each other, the calculations are complete and a

second

trial is

not needed.

If

these areas are not nearly equal, a second

trial

should be

performed as follows. 6.

start trial 2, use the new values of L„ L, L 3 Vu V2 and K3 calculated by the heat balances in step 4 and calculate the new solids concentrations in each effect by a solids

To

,

balance on each 7.

,

,

effect.

Obtain new values of AT',, AT'2 and AT'3 from ,

AT',

ATM, —

=

-

AT'j

=

—— AT,

/I,

AT'3 =

AT — — 3 /1 3

(8.5-11)

The sum AT', + AT', + AT'3 must equal the original £ AT. If not, proportionately readjust all AT' values so that this is so. Then calculate the boiling point in each effect. [If a boiling point rise is present, then using the new concentrations from step 6, determine the new BPRs in the three effects. This gives a new value of £ AT available for heat transfer by subtracting the sum of all three BPRs from the overall AT. Calculate the new values of AT' by Eq. (8.5-11). The sum of the AT' values just calculated must be readjusted to this new £ AT value. Then calculate the boiling point in each effect.] Step 7

is

essentially a repeat of step 3 but using Eq. (8.5-1

1)

to

obtain a better estimate of the AT' values. 8.

Using the new AT' values from step

Two

trials

7,

repeat the calculations starting with step 4.

are usually sufficient so that the areas are reasonably close to being equal.

EXAMPLE 8J-1.

Evaporation of Sugar Solution in a Triple-Effect Evaporator A triple-effect forward-feed evaporator is being used to evaporate a sugar solution containing 10 wt solids to a concentrated solution of 50%. The boiling-point rise of the solutions (independent of pressure) can be estimated

%

Sec. 8.5

Calculation Methods for Multiple-Effect Evaporators

505

from

BPR°C =

+

l.78x

6.22x

2

(BPR°F =

3.2x

+

1.2x

1

steam

fraction of sugar in solution (Kl). Saturated

2

where x

),

at 205.5

kPa

wt

is.

(29.8 psia)

[121. 1°C (250°F) saturation temperature] is being used. The pressure in the vapor space of the third effect is 13.4 kPa (1.94 psia). The feed rate is 22 680 kg/h (50000 Ibjh) at 26.7°C (80°F). The heat capacity of the liquid solutions = 4.19 - 2.35x kJAg" K (1.0 - 0.56x btu/lb m °F). The heat of is (Kl) c solution is considered to be negligible. The coefficients of heat transfer have been estimated as U l = 3123, U 2 = 1987, and U 3 = 1136 W/m 2 -K or 550, 2 350, and 200 btu/h ft °F. If each effect has the same surface area, calculate the area, the steam rate used, and the steam economy. -

•

The

Solution:

process flow diagram

is

given in Fig.

8.5-1.

Following the

eight steps outlined, the calculations are as follows. 1. For 13.4 kPa (1.94 psia), the saturation temperature is 51.67°C (125°F) from the steam tables. Using the equation for BPR for evaporator number 3 with x = 0.5,

Step

BPR 3 = T3 = Step

1.78x

51.67

+ +

Making an

2.

2.45

=

2

6.22x

=

1.78(0.5)

+

6.22(0.5)

=

2

2.45°C (4.4°F)

54.12°C (129.4°F)

overall

and a

solids balance to calculate the total

+ K3 )andL 3 F = 22680 = L 3 + (V, + V + K3

amount vaporized (K; + V2

,

2

Fx F = 22 680(0.1) = L 3 (0.5) +

(K,

)

+ V2 + K3 X0)

L 3 = 4536 kg/h (10000 Ibjh) total

vaporized

=

+ V2 + K3 ) =

{V,

Assuming equal amount vaporized (13 333 \bj\s).

Making

in

18 144 kg/h (40 000 lbjh)

each

effect,

a total material balance

V = V2 = K3 = 6048

kg/h

l

on

effects

and

2,

1,

3

and

solving, (1)

F = 22 680 =

K,

+ L

t

6048

+ L

1;

6048

+ L

2

(2)

L,

=

16 632

= V2 + L 2

(3)

L2 =

10 584

= K + L 3 = 6048 + L 3

Making a

3

solids balance

on

effects

1, 2,

L2 =

,

3

and solving

= L [Xl =

16632(x,),

x,

=

0.136

(2)

16632(0.136)

= L2 x2 =

10 584{x 2 ),

x2

=

0.214

(3)

10584(0.214)

= L3 x3 =

4536(x 3 ),

x

667^^)

10 584(23 334)

22 680(0.1)

x

0.500

V2 =£,

-

for x,

(check balance)

L2

V3

=L 2

~ 4536

22 680

xir = 0.1,7> = 26.7°C kPa = 121. 1°C

S, 205.5

TSi

Figure

506

16 632 kg/h (33

L 3 = 4536(10000)

,

and

=

(1)

V = 22,680- L

F=

Lj

8.5-1.

13.7 kPa (4)

Flow diagram for

(2)

triple-effect

(3)

evaporation for Example 85-1

Chap. 8

Evaporation

Step (1)

BPR, =

(2)

BPR 2 =

(3)

BPR in

The

3.

+

+

1.78(0.5)

=

1.78(0.136)

6.22(0.214)

=

2

6.22(0.5)

=

2

+

= 0.36°C

2

6.22(0. 136)

(0.7°F)

0.65°C (1.2°F)

2.45°C (4.4°F)

= TS1 - T3 (saturation) - (BPR, + BPR + BPR 3 = 121.1 - 51.67 - (0.36 + 0.65 + 2.45) = 65.97°C (1 18.7°F) 2

AT, and similar equations

(8.5-6) for

l/^i

AT = L V AT 1

AT,

effect is calculated as follows

6.22xj

1.78(0.214)

available

Using Eq.

+

1.78jc,

BPR 3 =

£ AT

each

1/1/,

=

+

l/U 3

AT2 =

12.40°C

2

and

AT3

,

(65.97X1/3123)

= +

l/U 2

AT

for

)

+

(1/3123) +(1/1987)

AT =

19.50°C

3

(1/1136)

34.07°C

However, since a cold feed enters effect number 1, this effect requires more heat. Increasing AT, and lowering AT2 and AT3 proportionately as a first estimate,

=

AT,

=

15.56°C

AT =

=

AT,

=

32.07°C

3

To calculate the actual T, = Ts - AT, (1)

K

15.56

=

18.34°C

32.07

18.34

K

K

boiling point of the solution in each effect,

,

=

Ts = T =

T,

2

=

T,

=

= 105.54°C

(condensing temperature of saturated steam to

effect 1)

— BPR, - AT,

105.54

TS2 =

(3)

15.56

121. 1°C

,

(2)

-

121.1

-

0.36

-

- BPR, =

18.34

=

-

105.54

86.84°C

0.36

(condensing temperature of steam to

105.18°C

effect 2)

T3 = T2 - BPR - AT, 2

=

-

86.84

0.65

-

32.07

=

86.19°C

=

54.1

2°C

TS3 = T - BPR 2 2

=

—

86.84

0.65

(condensing temperature of steam to

effect 3)

The temperatures

in the

three effects are as follows: Effect 2

Effect I

Tsl = T,

Step

4.

TS2 =

121.1°C

1

The heat capacity

=

4.19

—

F:

CP

L,:

C

C

Calculation

P

P

P

=

To 4

=

51.67

T3 = 54.12—

1

is

calculated from the

.

=

4.19

=

4.19

-

=

4.19

=

4.19

Methods for

Condenser

3

86.19

of the liquid in each effect

2.35x.

C

Sec. 8.5

-To,

T2 = 86.84—

= 105.54-

equation c p

Effect

105.18

2.35(0.1)

=

3.955

kJ/kg-K

2.35(0.136)

=

3.869

-

2.35(0.214)

=

3.684

-

2.35(0.5)

=

3.015

Multiple-Effect Evaporators

507

The values of the enthalpy H of the various vapor streams relative at 0°C as a datum are obtained from the steam table as follows: Effect 7,

=

1

TS2 =

105.54°C,

H = H si = Sl

=

+

2684

— H SI

105.18 (221.3°F),

=

1.884(0.36)

BPR, =

TS2 +

(saturation enthalpy at

i

/.

to water

)

-

508)

=

121.1 (250°F)

1.884 (0.36°C superheat)

2685 kJ/kg

—

(vapor saturation enthalpy)

(2708

Tsl =

0.36,

h si (liquid enthalpy at

TS1

)

2200 kJ/kg latent heat of condensation

Effect 2:

T2 = #2 = W S3 + X S2

86.84°C,

TS3 =

1.884(0.65)

= 2654 +

= H -h S2 = l

2685

-

BPR 2 =

86.19,

=

441

0.65

=

1.884(0.65)

2655 kJ/kg

2244 kJ/kg

Effect 3:

=

T3

0

TS4 =

54.12 C,

BPR 3 =

51.67,

2.45

H = H Si + 1.884(2.45) = 2595 + 1.884(2.45) = = 2655 - 361 = 2294 kJ/kg S3 =H 2 - h S3

2600 kJ/kg

3

;.

(Note that the superheat corrections in this example are small and could possibly have been neglected. However, the corrections were used to demonstrate the method of calculation.) Flow relations to be used in heat balances are K,

= 22680 —

K2 =

L,,

L,-L

2

K3 = L 2 - 4536,

,

L 3 = 4536

Write a heat balance on each effect. Use 0°C as a datum since the values of // of the vapors are relative to 0°C (32°F) and note that (7> - 0)°C = (7> - 0) K and (T, - 0)°C = (T, - 0) K, (1)

Fc p (TF

Substituting the

known

22680(3.955)(26.7

-

-

+

0)

= L.cJJ, -

S/. S1

+

0)

L, c,(T,

X

values,

=

S(2200)

—

L,(3.869X105.54

+

0)

-

0)

K,

+

;.

S2

(22 680

-

-

0)

L,X2685)

= L 2 c p(T2 -

680

(22

-

L,(3.869X105.54

+ (2)

+ V H,

0)

+ V H2

0)

2

L 2 (3.684X86.84 -

L,X2244) =

+ L2

(3)

c

p

(T2

-0)+V

L 2 (3.684X86.84 -

2

0)

+

(L,

;.

S3

= L3

c,(T3

-

- L 2 X2294) =

last

tuting into the

L,

S

two equations simultaneously

first

-L

2

X2655)

+ K3 H 3

0)

4536(3.015X54.12

+ Solving the

(L,

for

0)

(L 2

L

l

-

-

0)

4536X2600)

and L 2 and

substi-

equation,

= 17078

= 8936

kg/h

Vv = 5602

L2 =

1 1

068

V2 = 6010

L 3 = 4536 K3 = 6532 Chap. 8

Evaporation

calculated values of Vlt V2 and K3 are close enough to the assumed values so that steps 2, 3, and 4 do not need to be repeated. If the calculation were repeated, the calculated values of K„ V2 and K3 would be used starting

The

,

,

with step 2 and a solids balance

Step

each

in

effect

would be made.

Solving for the values of q in each effect and area,

5.

=

g,

=

SX SI

q2

=

K,;. S2

q3

= V2

s,

/t,

=

= .

at,

Aj

=

6 x * 10

6

1L-. = — U AT

6

3.492 x 10 " 1" 1987(18.34)

2

3.830 x 10


= 1830

100°)

W

460 * 106

W

6 3.492 x 10

=

_

-

-

1P4 m

=

95.8

.«

x 10 6

W

2

m2

=.105.1

m

2

1136(32.07)

3

3

5

3123(15.56)

— 3i— = ? U AT 2

=

x 1000)

[^wp 294 x 5.460 - 1460

<71

rj,

2200 X 100 °)

= (^)(2244

/.

_

,

^j^)(

The average area /4 m = 104.4 m 2 The areas differ from the average value by less than 10% and a second trial is not really necessary. However, a second trial will be made starting with step 6 to indicate the calculation methods .

used.

Step L,

6.

=

Making a new solids balance on effects 1, 2, and L2 = 068, and L 3 = 4536 and solving for x,

17 078,

(1)

22680(0.1)

=

17078(x,),

x,

=

0.133

(2)

17 078(0.133)

=

11

068(x 2 ),

x2

=

0.205

(3)

1 1

Step

7.

068(0.205)

= 4536(x 3

The new BPR

in

(1)

BPR, =

(2)

BPR 2 =

1.78(0.205)

(3)

BPR 3 =

1.78(0.5)

X AT The new

3 using the

new

1 1

1.78X!

available

values for

+

=

each

6.22x

+

+

2

x 3 = 0.500 (check balance)

),

effect is then

=

6.22(0.205)

6.22(0.5)

121.1

-

+

1.78(0133)

2

=

51.67

2

=

6.22(0.133)

=

0.35°C

0.63°C

2.45°C

-

(0.35

+

0.63

AT are obtained using Eq. (8.5-1 AT, A,

2

+

2.45)

= 66.00°C

1).

15.56(112.4)

104.4

AT

AT2 A

AT'3

X AT = Sec. 8.5

2

2

18.34(95.8)

104.4

AT3 A

3

32.07(105.1) 104.4

16.77

+

16.86

+

32.34

=

65.97°C

Calculation Methods for Multiple-Effect Evaporators

509

These AT' values are readjusted so that AT\ = 16.77, AT'2 = 16.87, AT'3 = 32.36, and £ AT = 16.77 + 16.87 + 32.36 = 66.00°C. To calculate the actual boiling point of the solution in each effect,

= TS1 - AT; =

(1)

Ti

(2)

T2 =

T,

TS2 =

T,

-

121.1

1

7S3 = T2 - BPR 2 = Step

87.1

Following step

8.

- 0.63 =

1

heat

the

4,

-

0.35

121. 1°C

16.87

=

1°C

87.1

103.98°C

- 0.63 -

87.11

TSI =

104.33°C,

- BPR; - AT'2 = 104.33 - BPR = 104.33 - 0.35 =

T3 = T2 - BPR 2 - AT'3 =

(3)

=

16.77

32.36

=

54.12°C

86.48°C capacity

of

the

liquid

c

is

p

=

4.19 -2.35.x.

F:

c„

=

3.955

L,:

c

=

4.19

-

2.35(0.133)

=

3.877

L2

:

c

=

4.19

-

2.35(0.205)

=

3.708

:

c

=

3.015

L3

p

p p

H are as follows in each effect.

The new values of the enthalpy (1)

kJ/kg-K

H = H S2 + 1.884(°C superheat) = 2682 + 1.884(0.35) = 2683 = 2708 - 508 = 2200 kJ/kg S1 =H sl -h sl H = H S3 + 1.884(0.63) = 2654 + 1.884(0.63) = 2655 kJ/kg

kJ/kg

x

;.

(2)

2

;.

(3)

S2

= H -h S2 =

tf 3

= H Si +

;.

=H

S3

-

2

-

2683

l

1.884(2.45)

=

h S3

2655

440

=

2595

-

362

Writing a heat balance on each (1)

22 680(3.955X26.7

-

0)

=

+

2243 kJ/kg

+

=

-

L,(3.877XI04.33

0)

+

2600 kJ/kg

2293 kJ/kg

effect,

=

5(2200)

1^(3.877X104.33

+ (2)

=

1.884(2.45)

(22 680

-

680

(22

-

L 2 (3.708X87.1

1

-

+

0)

(L,

- L 2 )(2293) =

L.X2683)

= L 2 (3.708)(87.1 -

L,)(2243)

1

+ (3)

- 0)

(L,

- L 2 X2655)

4536(3.015X54.12

+

(L 2

-

0)

-

0)

4536)(2600)

Solving,

L,

=

K,

Note

L 2 =10952

17005 kg/h

=

S = 8960 (steam used)

K3 = 6416

K2 = 6053

5675

5,

from trial 2 differ very little from and solving for q in each effect and A,

=

SX SI

that these values

Following step

q

x

=

^

= V^sz =

510

L 3 = 4536

(2200 x 1000)

(

=

2243 x 100 °)

the trial

6 5.476 x 10

=

3

-

539 x 106

1

results.

W W

Chap. 8

Evaporation

93

=

^2

^"S3

— 3600 6 5.476 x 10

9i 1

C/, 17,

AT', AT,

3123(16.77) s 6 3.539 x 10

A,

2 = - g92 ^f- = ^-^rr^r = t/ AT' 1987(16.87)

3.855 x 10

93

The average area

m

105.0

2

is

105.6

m

, 2

104.9

m

2

2

2

s

=

U3

AT'3

1136(32.36)

/l m

=

m

105.0

z

to use in each effect.

quite close to the average value of

steam economy

K V V = — + ^2 + 3i = t

104.4m

5675

+

that this value of

from the

first trial.

Me —— + 6416 = 2.025

6053

„

8960

o

8.6

Note 2

CONDENSERS FOR EVAPORATORS

8.6A

Introduction

In multiple-effect evaporators the vapors

vacuum,

i.e.,

from the

discharged as a liquid at atmospheric pressure. This using cooling water.

usually leaving under

last effect are

atmospheric pressure. These vapors must be condensed and

at less than

The condenser can be

is

done by condensing the vapors

a surface condenser, where the vapor to be

condensed and the cooling liquid are separated by a metal wall, or a direct contact condenser, where the vapor and cooling liquid are mixed directly.

8.6B

Surface Condensers

Surface condensers are employed where actual mixing of the condensate with condenser

cooling water

on the

not desired. In general, they are shell and tube condensers with the vapor

is

shell side

and cooling water

in multipass flow

on the tube

side.

gases are usually present in the vapor stream. These can be air,CO z

which

may have

decomposition

Noncondensable

N2

,

or another gas

entered as dissolved gases in the liquid feed or occur because of

in the solutions.

well-cooled point

,

in

These noncondensable gases may be vented from any is below atmospheric

the condenser. If the vapor being condensed

pressure, the condensed liquid leaving the surface condenser can be

removed by pumping

noncondensable gases by a vacuum pump. Surface condensers are much more expensive and use more cooling water, so they are usually not used in cases where a

and

the

direct-contact condenser

8.6C

is

suitable.

Direct-Contact Condensers

In direct-contact condensers cooling water directly contacts the vapors

vapors.

One

of the most

common

current barometric condenser

condensed by

rising

shown

discharge point in the

tail

tail

pipe.

The vapor

the counteris

The condenser

is

of cooling water droplets.

The condenser

is

enters the condenser and

is

high enough above the

pipe so that the water column established in the pipe more

than compensates for the difference

Sec. 8.6

in Fig. 8.6-1.

upward against a shower

located on top of a long discharge

and condenses the

types of direct-contact condensers

in

pressure between the low absolute pressure in the

Condensers for Evaporators

511

cold water

noncondensables

vapor

inlet

tailpipe

warm

water, 7t3?7

Figure

Schematic of barometric condenser.

8.6-1.

condenser and the atmosphere. The water can then discharge by gravity through a seal pot at the bottom.

A

height of about 10.4

m (34

used.

ft) is

The barometric condenser is inexpensive and economical of water consumption. It can maintain a vacuum corresponding to a saturated vapor temperature within about 2.8

K (5°F)

water

is

10.1 k

Pa

of the water temperature leaving the condenser. For example, 316.5

at

K

(1

10°F), the pressure

corresponding

The water consumption can be estimated by is

W

derivation

If

the vapor flow to the condenser

kg/h is

if

the discharge

K

2.8 or 319.3

is

at

a simple heat balance for a barometric

V kg/'h at temperature Ts and the water T and a leaving temperature ofT the

is

an entering temperature of

2

l

,

as follows.

VH S

4-

Wc p(T x

273.2)

=(K+

W)c p (T2

H s is the enthalpy from the steam tables of the vapor vapor stream. Solving,

where the

+

(1.47 psia).

condenser. flow

to 316.5

W=— kg water H — =— V kg

s ^

vapor

-

273.2)

at

Ts K and

(8.6-1)

the pressure in

—

- 273.2) - cJT 2 p c (T - T p 2

(8.6-2)

t )

The noncondensable gases can be removed from the condenser by a vacuum pump. The vacuum pump used can be a mechanical pump or a steam-jet ejector. In the ejector high-pressure steam enters a nozzle at high speed and entrains the noncondensable gases from the space under vacuum.

Another type of direct-contact condenser

is

the jet barometric condenser. High-

velocity jets of water act both as a vapor condenser

condensables out of the

tail

pipe. Jet

and as an entrainer of the non-

condensers usually require more water than the

more common barometric condensers and are more

difficult to throttle at

low vapor

rates.

512

Chap. 8

Evaporation

EVAPORATION OF BIOLOGICAL MATERIALS

8.7

Introduction and Properties of Biological Materials

8.7A

The evaporation

many biological materials often differs from evaporation of inorganic NaCl and NaOH and organic materials such as ethanol and acetic

of

materials such as

acid. Biological materials

such as pharmaceuticals, milk, citrus juices, and vegetable

and often contain

extracts are usually quite heat-sensitive

matter

must be designed

for

fine particles of

Also, because of problems, due to bacteria growth, the

in the solution.

easy cleaning.

Many

suspended equipment

biological materials in solution exhibit only a

when concentrated. This is due to the fact that suspended solids dispersed form and dissolved solutes of large molecular weight contribute little

small boiling-point rise in

a fine

to this rise.

The amount

of degradation of biological materials on evaporation is a function of and the length of time. To keep the temperature low, the evaporation must be done under vacuum, which reduces the boiling point of the solution. To keep the time of contact low, the equipment must provide for a low holdup time (contact time) of

the temperature

equipment used and some biological

the material being evaporated. Typical types of

materials processed are given below. Detailed discussions of the equipment are given in

Section

8.2.

Condensed

1.

Long-tube

2.

Falling-film evaporator. Fruit juices.

3.

Agitated-film (wiped-film) evaporator.

4.

Heat-pump

8.7B

vertical evaporator.

milk.

Rubber

latex, gelatin, antibiotics, fruit juices.

cycle evaporator. Fruit juices, milk, pharmaceuticals.

Fruit Juices

In evaporation of fruit juices

such as orange juice the problems are quite different from

evaporation of a typical

such as NaCl. The

salt

fruit

juices are heat-sensitive

viscosity increases greatly as concentration increases. Also, solid fruit juices

and the

suspended matter

in

has a tendency to cling to the heating surface and thus causes overheating,

leading to burning and spoilage of the matter (B2).

To

reduce

this

circulation over the sensitive,

tendency to stick and to reduce residence time, high is also necessary. Hence, a and not a multiple evaporation

low-temperature operation

employs a

plant usually

single

fruit

unit.

of

rates

heat-transfer surface are necessary. Since the material

is

heat-

juice concentration

Vacuum

is

used to

reduce the temperature of evaporation.

A typical fruit juice evaporation system using the heat pump cycle is shown (PI, CI), which uses low-temperature ammonia as the heating fluid. A frozen concentrated citrus juice process

evaporator.

is

A

described by

major

volatile constituents

fault of

Charm

during evaporation.

juice bypasses the evaporation cycle

8.7C

(CI).

The process

concentrated orange juice

and

To overcome

is

is

uses a multistage falling-film

a

flat

this, a

flavor

due

to the loss of

portion of the fresh pulpy

blended with the evaporated concentrate.

Sugar Solutions

Sugar (sucrose)

is

obtained primarily from the sugar cane and sugar beet. Sugar tends to high temperatures for long periods (B2).

caramelize

if

Sec. 8.7

Evaporation of Biological Materials

kept

at

The

general tendency

is

to

513

use short-tube evaporators of the natural circulation type. In the evaporation process of sugar solutions, the clear solution of sugar having a concentration of 10-13° Brix (1013 wt %) is evaporated to 40-60° Brix (Kl, SI).

The

feed

is first

preheated by exhaust steam and then typically enters a six-effect

The first effect operates at a pressure in the vapor space kPa (30 psia) [121. 1°C (250°F) saturation temperature] under vacuum at about 24 kPa (63.9°C saturation). Examples of the

forward-feed evaporator system. of the evaporator of about 207

and the

last effect

relatively small boiling-point rise of

Example

sugar solutions and the heat capacity are given in

8.5-1.

Paper-Pulp Waste Liquors

8.7D

In the making of paper pulp in the sulfate process, wood chips are digested or cooked and spent black liquor is obtained after washing the pulp. This solution contains primarily sodium carbonate and organic sulfide compounds. This solution is concentrated by evaporation in a six-effect system (Kl, SI).

EVAPORATION USING VAPOR RECOMPRESSION

8.8.

8.8A

Introduction

In the single-effect evaporator the vapor from the unit

is

generally

condensed and

discarded. In the multiple-effect evaporator, the pressure in each succeeding effect

is

lowered so that the boiling point of the liquid is lowered in each effect. Hence, there is a temperature difference created for the vapor from one effect to condense in the next effect and boil the liquid to form vapor. In a single-effect vapor recompression (sometimes called vapor compression)

evaporator the vapor

is

compressed so

increased. This compressed vapor

is

that its

condensing or saturation temperature

is

returned back to the heater of steam chest and

condenses so that vapor is formed in the evaporator (B5, Wl, Zl). In this manner the latent heat of the vapor is used and not discarded. The two types of vapor recompression units are the mechanical and the thermal type. Mechanical Vapor Recompression Evaporator.

8.8B

vapor recompression evaporator, a conventional single effect evapois used and is shown in Fig. 8.8-1. The cold feed is preheated by exchange with the hot outlet liquid product and then flows to the unit. The vapor coming overhead does not go to a condenser but is sent to a centrifugal or positive displacement compressor driven by an electric motor or steam. This compressed vapor or steam is "sent back to the heat exchanger or steam chest. The compressed vapor condenses at a higher temperature than the boiling point of the hot liquid in the effect and a temperature difference is set up. Vapor is again generated and In a mechanical

rator similar to that in Fig. 8.2-2

the cycle repeated.

Sometimes

it

is

necessary to add a small amount of makeup steam to the vapor

may be added compressed vapor to remove any superheat, if present. Vapor recompression units generally operate at low optimum temperature differences' of 5 to 10°C. Hence, large heat transfer areas are needed. These units usually have higher capital costs than multiple-effect units because of the larger area and the line

before the compressor (B5, K2). Also, a small amount of condensate

to the

514

Chap. 8

Evaporation

makeup steam

concentrated

condensate

product

(for desuperheating)

feed heater

Figure

8.8-1.

Simplified process flow for mechanical vapor recompression evaporator.

costs of the relatively expensive compressor and drive unit.

vapor recompression units the

power

effect

to drive the

cold feed •

is in

The main advantage of

the lower energy costs. Using the steam equivalent of

compressor, the steam economy

is

equivalent to a multiple-

evaporator of up to 10 or more units (Zl).

Some

typical applications of mechanical

tion of sea

water to give

distilled water,

vapor recompression

units are evapora-

evaporation of kraft black liquor

industry (L2), evaporation of heat-sensitive

materials

crystallizing of salts having inverse solubility curves

such as

where the

the paper

in

fruit juices,

solubility

and

decreases

with increasing temperature (K2, M3). Falling-film evaporators are well suited for vapor recompression systems (Wl) because they operate at low-temperature-difference values and have very little

entrained liquid which can cause problems

has been used

in distillation

in

the compressor.

towers where the overhead vapor

Vapor recompression is recompressed and

used in the reboiler as the heating medium (M2).

Thermal Vapor Recompression Evaporator

8.8C

A steam jet can unit.

also be used to

The main disadvantages

removal of

this

compress the vapor

in a

thermal vapor recompression

are the low efficiency of the steam jet, necessitating the

excess heat, and the lack of flexibility to changes in process variables more durable than mechanical compressors and can

(M3). Steam jets are cheaper and

more

easily handle large

volumes of low-pressure vapors.

PROBLEMS 8.4-1.

Heat-Transfer Coefficient in Single-Effect Evaporator. A feed of 4535 kg/h of a 2.0 wt salt solution at 31 1 K enters continuously a single-effect evaporator and is being concentrated to 3.0%. The evaporation is at atmospheric pressure and z the area of the evaporator is 69.7 m Saturated steam at 383.2 K is supplied for heating. Since the solution is dilute, it can be assumed to have the same boiling point as water. The heat capacity of the feed can be taken asc = 4.10 kJ/kg K. p Calculate the amounts of vapor and liquid product and the overall heat-transfer

%

.

•

coefficient U.

Ans. 8.4-2. Effects

of Increased Feed Rate

in

U =

Problems

W/m 2 K •

Evaporator. Using the same area, value of U,

steam pressure, evaporator pressure, and feed temperature as

Chap. 8

1823

in

Problem

8.4-1,

515

calculate the

centration

amounts of

the feed rate

if

is

liquid

and vapor leaving and the

liquid outlet con-

increased to 6804 kg/h.

V=

Ans.

1256 kg/h,

L= 5548 kg/h, x L = 2.45%

of Evaporator Pressure on Capacity and Product Composition. Recalculate Example 8.4-1 but use an evaporator pressure of 41.4 kPa instead of 101.32 kPa abs. Use the same steam pressure, area A, and heat-transfer coefficient U in the

8.4-3. Effect

calculations. (a)

(b)

Do this to obtain the new capacity or feed rate under these new conditions. The composition of the liquid product will be the same as before. Do this to obtain the new product composition if the feed rate is increased to 18 144 kg/h.

m

2 evaporator having an area of 83.6 and a U = 2270 W/m K is used to produce distilled water for a boiler feed. Tap water having 400 ppm of dissolved solids at 15.6°C is fed to the evaporator operating at 1 atm pressure abs. Saturated steam at 1 15.6°C is available for use. Calculate the amount of distilled water produced per hour if the outlet liquid contains 800 ppm

8.4-4. Production

An

of Distilled Water. 2

•

solids.

8.4-5. Boiling-Point Rise

of NaOH Solutions. Determine the boiling temperature of the

solution and the boiling-point rise for the following cases. (a)

A 30%

NaOH

solution boiling in an evaporator at a pressure of 172.4

kPa

solution boiling in an evaporator at a pressure of 3.45

kPa

(25 psia). (b)

A 60%

NaOH

(0.50 psia).

Ans. 8.4-6. Boiling-Point Rise rise for

(a)

Boiling point

=

130.6°C, boiling point rise

=

15°C

of Biological Solutes in Solution. Determine the boiling-point

the following solutions of biological solutes in water.

Use the

figure in

(PI), p. 11-31. (a)

(b)

A 30 A 40

wt wt

% solution of citric acid in water boiling at 220°F (104.4°C). % solution of sucrose in water boiling at 220°F (104.4°C). Ans.

(a)

Boiling-point

rise

=

2.2°F (1.22°C)

of Feed Temperature on Evaporating an NaOH Solution. A single-effect evaporator is concentrating a feed of 9072 kg/h of a 10 wt % solution of NaOH in water to a product of 50% solids. The pressure of the saturated steam used is 42 kPa (gage) and the pressure in the vapor space of the evaporator is 20 kPa 2 (abs). The overall heat-transfer coefficient is 1988 W/m K. Calculate the steam used, the steam economy in kg vaporized/kg steam, and the area for the following

8.4-7. Effect

•

feed conditions. (a)

(b)

Feed temperature of 288.8 Feed temperature of 322.1

K (15.6°C). K (48.9°C). Ans.

8.4-8.

(a)

S

= 8959 kg/h of steam, A = 295.4 m NaOH. In order to concentrate 2

Heat-Transfer Coefficient to Evaporate 4536 kg/h of an NaOH solution containing 10 wt % NaOH to a 20 wt % 2 solution, a single-effect evaporator is being used with an area of 37.6m The feed enters at 21.TC (294.3 K). Saturated steam at 1 10°C (383.2 K) is used for heating and the pressure in the vapor space of the evaporator is 51.7 kPa. Calculate the kg/h of steam used and the overall heat-transfer coefficient. .

8.4-9.

Throughput of a Single-Effect Evaporator. An evaporator is concentrating at 311 K of a 20 wt solution of NaOH to 50%. The saturated steam used for heating is at 399.3 K. The pressure in the vapor space of the evaporator 2 2 is 13.3 kPa abs. The overall coefficient is 1420 W/m and the area is 86.4m

%

F kg/h

-

Calculate the feed rate

F

K

.

of the evaporator.

Ans. 8.4-10. Surface

rator

516

is

F = 9072 kg/h

Area and Steam Consumption of an Evaporator. A single-effect evapoconcentrating a feed solution of organic colloids from 5 to 50 wt %. The

Chap. 8

Problems

solution has a negligible boiling-point elevation. The heat capacity of the feed is c = 4.06 kJ/kg " (0.97 btu/lb m °F) and the feed enters at 1 5.6°C (60°F). Satu-

K

p

rated steam at 101.32

kPa

is

available for heating, and the pressure in the vapor

space of the evaporator is 15.3 kPa. A total of 4536 kg/h (lOOOOlbnyh) of water is 2 K (350 to be evaporated. The overall heat-transfer coefficient is 1988 W/m 2 2 and the steam consumpbtu/h ft °F). What is the required surface area in •

•

m

•

tion?

of Tomato Juice Under Vacuum. Tomato juice having a con-

8.4-11. Evaporation

centration of 12 wt

evaporator.

%

solids

25%

being concentrated to

is

The maximum allowable temperature

solids in a film-type

tomato juice

for the

is

135°F,

The feed enters at 100°F. Saturated steam at 25 psia is used for heating. The overall heat-transfer coef2 2 ficient f / is 600 btu/h ft °F and the area A is 50 ft The heat capacity of the which

will

be the temperature •

of the product.

•

.

estimated as 0.95 btu/lb m °F. Neglect any boiling-point rise p Calculate the feed rate of tomato juice to the evaporator.

feed c

8.4- 12.

is

if

•

present.

Concentration of Cane Sugar Solution. A single-effect evaporator is being used to concentrate a feed of 1 0 000 lbjjli of a cane sugar solution at 80°F and containing sugar) to 30° Brix for use in a a sugar content of 15° Brix (degrees Brix is wt

%

food process. Saturated steam at 240°F is available for heating. The vapor space 2 °F in the evaporator will be at 1 atm abs pressure. The overall U = 350 btu/n ft •

and the heat capacity of the feed is c p = 0.91 btu/lb m °F. The boiling-point rise can be estimated from Example 8.5-1. The heat of solution can be considered negligible and neglected. Calculate the area required for the evaporator and the amount of steam used per hour. Ans.

Boiling-point rise

8.5-1. Boiling Points in a Triple-Effect Evaporator.

point rise at 121.

The

is

=

2.0°F(l.l°C),/4

=

667

ft

2

(62.0

m2

)

A solution with a negligible boiling-

being evaporated in a triple-effect evaporator using saturated steam The pressure in the vapor of the last eflect is 25.6 kPa abs.

1°C (394.3 K).

U

= 1988, and = 1420 heat-transfer coefficients are [/, = 2840, 2 3 • and the areas are equal. Estimate the boiling point in each of the

U

W/m 2 K

evaporators. Ans.

of Sugar Solution evaporator with forward feed

8.5-2. Evaporation

T,

=

108.6°C (381.8

in a Multiple-Effect Evaporator. is

A

K)

triple-effect

evaporating a sugar solution with negligible

K, which will be neglected) and containing Saturated steam at 205 kPa abs is being used. The

boiling-point rise (less than 1.0 5

wt

% solids to 25% solids.

pressure in the vapor space of the third effect is 13.65 kPa. The feed rate is 22 680 kg/h and the temperature 299.9 K. The liquid heat capacity is c = 4.19 p - 2.35 x, where c is in kJ/kg and x in wt fraction (Kl). The heat-transfer p coefficients are =3123, U 2 = 1987, and U 3 = 1136 W/m 2 -K. Calculate the surface area of each effect if each effect has the same area, and the steam rate. 2 Ans. Area A = 99.1 steam rate S = 8972 kg/h •

K

m

,

Double-Effect Reverse-Feed Evaporators. A feed containing dissolved organic solids in water is fed to a double-effect evaporator 2 wt with reverse feed. The feed enters at 100°F and is concentrated to 25% solids.

8.5- 3. Evaporation

in

%

The

be considered negligible as well as the heat of Each evaporator has a 1000-ft 2 surface area and the heat-transfer 2 °F. The feed enters coefficients are [/, = 500 and U 2 = 700 btu/h ft evaporator number 2 and steam at 100 psia is fed to number The pressure in the vapor space of evaporator number 2 is 0.98 psia. Assume that the heat boiling-point rise can

solution.

•

•

1

.

capacity of aU liquid solutions is that of liquid water. Calculate the feed rate F and the product rate L, of a solution containing 25% solids. [Hint: Assume a feed rate of, say, F = 1000 lb m /h. Calculate the area. Then calculate the actual feed rate

Chap. 8

by multiplying 1000 by 1000/calculated area.) Ans. F= 133 800 Vojh (60 691 kg/h), L, = 10 700 lbjh (4853 kg/h)

Problems

517

A forced-circulation

8.5-4. Concentration oj'NaOH Solution in Triple-Effect Evaporator.

evaporator using forward feed is to be used to concentrate a 10 wt NaOH solution entering at 37.8°C to 50%. The steam used enters at 58.6 kPa gage. The absolute pressure in the vapor space of the third effect is 6.76 kPa. The feed rate is 13 608 kg/h. The heat-transfer coefficients are U\ — 6246, U 2 = 3407, and U 3 = 2271 W/m 2 K. All effects have the same area. Calculate the surface area and steam consumption.

triple-effect

%

•

Ans. 8.5-5.

A=

97.3

m 2 ,S =

5284 kg steam/h

Evaporator with Reverse Feed. A feed rate of 20 410 kg/h of solution at 48.9°C is being concentrated in a triple-effect reverse-feed evaporator to produce a 50% solution. Saturated steam at 178. 3°C is fed to the first evaporator and the pressure in the third effect is 10.34 kPa abs. The heat-transfer coefficient for each effect is assumed to be 2840 W/m 2 K. Calculate the heat-transfer area and the steam consumption rate. Triple-Effect

10

wt

% NaOH

•

of Sugar Solution in Double Effect Evaporator. A double-effect evaporator with reverse feed is used to concentrate 4536 kg/h of a 10 wt sugar solution to 50%. The feed enters the second effect at 37.8°C. Saturated steam at 1 15.6°C enters the first effect and the vapor from this effect is used to heat the second effect. The absolute pressure in the second effect is 13.65 kPa abs. The 2 overall coefficients are U\ = 2270 and U 2 = 1705 W/m K. The heating areas for both effects are equal. Use boiling-point-rise and heat-capacity data from Example 8.5-1. Calculate the area and steam consumption. Water Consumption and Pressure in Barometric Condenser. The concentration of NaOH solution leaving the third effect of a triple-effect evaporator is 50 wt %. The vapor flow rate leaving is 5670 kg/h and this vapor goes to a barometric condenser. The discharge water from the condenser leaves at 40.5°C. Assuming that the condenser can maintain a vacuum in the third effect corresponding to a saturated vapor pressure of 2.78°C above 40.5°C, calculate the pressure in the third effect and the cooling water flow to the condenser. The cooling water enters at 29.5°C. (Note: The vapor leaving the evaporator will be superheated because of the boiling-point rise.) = 306 200 kg water/h Ans. Pressure = 8.80 k Pa abs,

8.5- 6. Evaporation

%

•

8.6- 1.

W

REFERENCES (Bl)

Badger, W. L., and BancherO, J. T. Introduction York: McGraw-Hill Book Company, 1955.

(B2)

BLAKEBROUGH, N. Biochemical and York: Academic Press, Inc., 1968.

(B3)

to

Chemical Engineering.

Biological Engineering Science, Vol.

2.

New New

Badger, W. L., and McCabe, W. L. Elements of Chemical Engineering, 2nd York: McGraw-Hill Book Company, 1936.

ed.

New

New

& Sons, Inc.,

(B4)

Brown,

(B5) (CI)

Beesley, A. H., and Rhinesmith, R. D. Chem. Eng. Progr., 76(8), 37 (1980). Charm, S. E. The Fundamentals of Food Engineering, 2nd ed. Westport, Conn.:

(Kl)

Kern, D. Q. Process Heat Transfer.

G. G., et

al.

Avi Publishing Co.,

Unit Operations.

York: John Wiley

1950.

Inc., 1971.

New

York: McGraw-Hill Book Company,

1950.

(K2)

King, R.

J.

Chem. Eng. Progr.,

(LI)

Lindsey, E. Chem. Eng., 60

(L2)

Logsdon,

(Ml)

McCabe, W. L.

518

J.

(4),

80(7), 63 (1984).

227 (1953).

D. Chem. Eng. Progr., 79(9), 36 (1983). Trans. A.I.Ch.E., 31, 129 (1935).

Chap. 8

References

Chem. Eng., 94 (Feb.

(M2)

Meili, A., and Stuecheli, A.

(M3)

Mehra, D. K. Chem. Eng., 93(Feb.

(PI)

Perry, R. H., and Chilton, C. H. Chemical Engineers Handbook, 5th ed. New York: McGraw-Hill Book Company, 1973.

(P2)

Perry, R. H., and Green, D. Perry's Chemical Engineers' Handbook, 6th ed. New York: McGraw-Hill Book Company, 1984.

(SI)

Shreve, R. N., and Brink, J. A., Jr. Chemical Process Industries, 4th ed. York: McGraw-Hill Book Company, 1977.

(Wl)

Weimer, L. D., Dolf, H. R., and Austin, D. A. Chem. Eng. Progr.,76

3),

133 (1987).

16),

56 (1986). 1

New

(11),

70

(1980).

(Zl)

Zimmer, A. Chem. Eng. Progr.,

Chap. 8

References

76(9), 37 (1980).

519

CHAPTER

9

Drying of Process Materials

INTRODUCTION AND METHODS OF DRYING

9.1

9.1

A

The

Purposes of Drying discussions of drying in this chapter are concerned with the removal of water from

process materials and other substances.

The term

drying

is

also used to refer to removal

of other organic liquids, such as benzene or organic solvents, from solids. types of equipment and calculation

methods discussed

for

Many

of the

removal of water can also be

used for removal of organic liquids.

Drying,

in

general, usually

material. Evaporation refers

means removal removal of

to

material. In evaporation the water

water

is

usually

removed

as a

is

removed

vapor by

amounts of water from amounts of water from

of relatively small relatively

large

as vapor at

its

boiling point. In drying the

air.

may be removed mechanically from solid materials by presses, and other methods. This is cheaper than drying by thermal means for removal of water, which will be discussed here. The moisture content of the final dried product varies depending upon the type of product. Dried salt contains about 0.5% In

some

cases water

centrifuging,

water, coal about

4%, and many food products about 5%. Drying is usually the final makes many materials, such as soap powders and

processing step before packaging and

more

dyestufTs,

suitable for handling.

Drying or dehydration of biological materials, especially foods, is used as a preservation technique. Microorganisms that cause food spoilage and decay cannot grow and multiply

in

the absence of water. Also,

many enzymes

that cause chemical changes in

food affd other biological materials cansjot function without water.

is

When

the water

reduced below about 10 wt %, the microorganisms are not active. However, it usually necessary to lower the moisture content below 5 wt in foods to preserve

content

is

%

flavor

and

Some

nutrition. Dried foods

biological materials

ordinary drying,

may

for

extended periods of time.

and pharmaceuticals, which may not be heated

be freeze -dried as discussed

method often employed

in

Section 9.11. Also,

and other biological materials to preserve such materials.

9.12, sterilization of foods

520

can be stored

is

discussed, which

for

in

Section

is

another

General Methods of Drying

9.1B

Drying methods and processes can be classified in several different ways. Drying prowhere the material is inserted into the drying equipment and drying proceeds for a given period of time, or as continuous, where the material is continuously added to the dryer and dried material continuously removed. Drying processes can also be categorized according to the physical conditions used cesses can be classified as batch,

add heat and remove water vapor:

added by direct is removed by the air (2) in vacuum drying, the evaporation of water proceeds more rapidly at low pressures, and the heat is added indirectly by contact with a metal wall or by radiation (low temperatures can also be used under vacuum for certain materials that may discolor or decompose at higher temperatures); and (3) in freeze drying, water is sublimed from to

(1) in the first

category, heat

is

contact with heated air at atmospheric pressure, and the water vapor formed ;

the frozen material.

9.2

EQUIPMENT FOR DRYING Tray Dryer

9.2A

In tray dryers, which are also called shelf, cabinet, or

which

may

be a lumpy solid or a pasty solid,

depth of 10 to 100

mm. Such

a typical tray dryer,

compartment

dryers, the material,

spread uniformly on a metal tray to a

is

shown

in Fig. 9.2-1,

contains removable

trays loaded in a cabinet.

Steam-heated

air

is

recirculated by a fan over

and

parallel to the surface of the trays.

low heating loads. About 10 to 20% of the air passing over the trays is fresh air, the remainder being recirculated air. After drying, the cabinet is opened and the trays are replaced with a new batch of trays. A modification of this type is the tray-truck type, where trays are loaded on trucks Electrical heat

is

also used, especially for

which are pushed into the dryer. This saves considerable time, since the trucks can be loaded and unloaded outside the dryer. In the case of granular materials, the material can be loaded on screens which are the

bottom of each

tray.

Then

in

this

through-circulation dryer, heated air passes

through the permeable bed, giving shorter drying times because of the greater surface area exposed to the

9.2B

air.

Vacuum-Shelf Indirect Dryers

Vacuum-shelf dryers are indirectly heated batch dryers similar to tray dryers. Such a dryer consists of a cabinet

made

of cast-iron or steel plates fitted with tightly fitted doors

trays

adjustable louvers

/

r air

m-

-air

out

fan

heater

26

Figure

Sec. 9.2

Equipment For Drying

9.2-

1

.

Tray or

shelf dryer.

521

it can be operated under vacuum. Hollow shelves of steel are fastened permanently inside the chamber and are connected in parallel to inlet and outlet steam headers. The trays containing the solids to be dried rest upon the hollow shelves. The heat is conducted through the metal walls and added by radiation from the shelf above. For low-temperature operation, circulating warm water is used instead of steam for

so that

furnishing the heat to vaporize the moisture.

The vapors

usually pass to a condenser.

These dryers are used to dry expensive, or temperature-sensitive, or easily oxidizable materials. They are useful for handling materials with toxic or valuable solvents.

9.2C

Continuous Tunnel Dryers

Continuous tunnel dryers are often batch truck or tray compartments operated as

shown

in

Fig. 9.2-2a.

The

on

solids are placed

trays or

in series,

on trucks which move

continuously through a tunnel with hot gases passing over the surface of each tray. flow can be countercurrent, cocurrent, or a combination.

hot

air

this

way.

Many

The

foods are dried in

When granular particles of solids are to be dried, perforated or screen-belt continuous conveyors are often used, as in Fig. 9.2-2b. The wet granular solids are conveyed as a layer 25 to about 150

upward through

mm deep on a screen or perforated apron while heated air

the bed, or

each with a fan and heating fan. In

some cases pasty

downward. The dryer

coils.

A portion of the

is

blown

consists of several sections in series,

exhausted to the atmosphere by a

air is

materials can be preformed into cylinders

and placed on the bed

to be dried.

blower louvers fresh air in

wet material

dry material

trucks enter -air

moving trucks

out

trucks leave

(a)

air

air

granulai feed

It!

r 0*0

I

I

^

fan i

o|o|o 1

1

1

1

flow

— steam

heaters

0J00

H

•oofofoto

-air

ii

Jp

J. screen belt

dry product

(b)

FIGURE

9.2-2.

Continuous tunnel dryers:

[a)

tunnel dryer trucks with countercurrent

air flow, (b) through-circulation screen

522

conveyor dryer.

Chap.

9

Drying of Process Materials

air -<

Figure

9.2D

A

Schematic drawing of a direct-heat rotary dryer.

9.2-3.

Rotary Dryers

rotary dryer consists of a hollow cylinder which

toward the

The wet granular

outlet.

and move through the

shell as

it

gases in countercurrent flow. In

is

rotated and usually slightly inclined

solids are fed at the high

rotates.

some

The heating shown

cases the heating

is

is

end as shown in Fig. 9.2-3 by direct contact with hot

by indirect contact through the

heated wall of the cylinder.

The granular

move forward

particles

slowly a short distance before they are

showered downward through the hot gases as shown.

Many

other variations of this

rotary dryer are available, and these are discussed elsewhere (PI).

9.2E

Drum

A drum

Dryers

dryer consists of a heated metal

thin layer of liquid or slurry roll,

which

and

for solutions.

is

Drum

is

roll

shown

in Fig. 9.2-4,

evaporated to dryness. The

final

on

the outside of which a

dry solid

is

scraped off the

revolving slowly.

dryers are suitable for handling slurries or pastes of solids in fine suspension

The drum functions

and also as a dryer. Other drums with dip feeding or with top using drum dryers, to give potato flakes.

partly as an evaporator

variations of the single-drum type are twin rotating

feeding to the two drums. Potato slurry

9.2F

is

dried

Spray Dryers

In a spray dryer a liquid or slurry solution a mist of fine droplets.

The water

is

is

sprayed into a hot gas stream

in the

form of

rapidly vaporized from the droplets, leaving particles

internally steam-

film

heated

drum

spreader dried material liquid or slurry feed

•knife scraper

FIGURE

Sec. 9.2

Equipment For Drying

9.2-4.

Rotary-drum dryer.

523

liquid feed

spray chamber

heated

exhaust gas

air

t

cyclone separator

droplets

hopper

dry product

screw conveyor FIGURE

9.2-5.

Process flow diagram of spray-drying apparatus.

of dry solid which are separated from the gas stream.

spray chamber

The

may be

The

flow of gas and liquid in the

countercurrent, cocurrent, or a combination.

fine droplets are

formed from the liquid feed by spray nozzles or high-speed

rotating spray disks inside a cylindrical chamber, as in Fig. 9.2-5.

It is

necessary to ensure

do not strike and stick to solid surfaces before Hence, large chambers are used. The dried solids leave at the

that the droplets or wet particles of solid

drying has taken place.

bottom of the chamber through a screw conveyor. The exhaust gases flow through a cyclone separator to remove any fines. The particles produced are usually light and quite porous. Dried milk powder is made from spray-drying milk.

9.2G

Drying of Crops and Grains

from a harvest, the grain contains about 30 to 35% moisture and about 1 year should be dried to about 13 wt % moisture (HI). A

In the drying of grain for safe storage for

typical continuous-flow dryer layer of grain

is

0.5

m

is

shown

in Fig. 9.2-6. In the

drying bin the thickness of the

or less, through which the hot air passes.

bottom section cools the dry grain before storage bins are described by Hall (H 1).

it

leaves.

Unheated

air in

the

Other types of crop dryers and

grain inlet

»-heated air for drying wire screen »-

cooling

air

dry grain Figure

524

9.2-6.

Vertical continuous-flow grain dryer.

Chap.

9

Drying of Process Materials

VAPOR PRESSURE OF WATER AND HUMIDITY

93

93A /.

Vapor Pressure of Water

Introduction.

number

In a

make

necessary to

of the unit operations and transport processes

These calculations involve knowledge of the concentration of water vapor

air.

it

calculations involving the properties of mixtures of water vapor

is

and

in air

under various conditions of temperature and pressure, the thermal properties of these mixtures, and the changes occurring

when

this

mixture

is

brought into contact with

water or with wet solids in drying. Humidification involves the transfer of water from the liquid phase into a gaseous

mixture of air and water vapor. Dehumidification involves the reverse transfer, whereby

water vapor

is

transferred from the vapor state to the liquid state. Humidification

and

dehumidification can also refer to vapor mixtures of materials such as benzene, but most practical applications occur with water.

To

better understand humidity,

it is

first

neces-

sary to discuss the vapor pressure of water.

2.

Vapor pressure of water and physical

physical states: solid

ice, liquid,

states.

Pure water can

and vapor. The physical

state in

exist in three different

which

it

exists

depends

on the pressure and temperature. Figure 9.3-1 illustrates the various physical states of water and the pressuresolid, liquid, and AB, the phases liquid and vapor coexist. Along coexist. Along line AD, ice and vapor coexist. If ice at

temperature relationships at equilibrium. In Fig. 9.3-1 the regions of the

vapor line

AC,

point is

shown. Along the

states are

and

the phases ice

(1) is

heated

shown moving

at

liquid

line

rises and the physical condition As the line crosses AC, the solid melts, and on crossing AB Moving from point (3) to (4), ice sublimes (vaporizes) to a vapor

constant pressure, the temperature

horizontally.

the liquid vaporizes.

without becoming a liquid. Liquid and vapor coexist in equilibrium along the line AB, which

when

is

the vapor-

vapor pressure of the water is equal to the total pressure above the water surface. For example, at 100°C (212°F) the vapor pressure of water is 101.3 kPa (1.0 atm), and hence it will boil at 1 atm pressure. At 65.6°C (150°F), from the steam tables in Appendix A. 2, the vapor pressure of water is 25.7 kPa (3.72 psia). pressure line of water. Boiling occurs

Hence,

at 25.7

If a

pan

kPa and

of

water

the

65.6°C, water will boil. is

held at 65.6°C in a

room

at 101.3

kPa abs

pressure, the vapor

pressure of water will again be 25.7 kPa. This illustrates an important property of the

liquid region

solid

vapor region.

D Temperature Figure

Sec. 9.3

9.3-1.

Phase diagram for water.

Vapor Pressure of Water and Humidity

525

vapor pressure of water, which air;

i.e.,

is

not influenced by the presence of an inert gas sucrf a's

the vapor pressure of water

essentially independent of the total pressure of the

is

system.

Humidity and Humidity Chart

9.3B

1.

H of an

The humidity

Definition of humidity.

air-water vapor mixture

is

defined as

kg of dry air. The humidity so denned depends only on the partial pressure p A of water vapor in the air and on the total pressure P (assumed throughout this chapter to be 101.325 kPa, 1.0 aim abs, or 760 mm Hg). Using the molecular weight of water (A) as 18.02 and of air as 28.97, the humdiity H in kg H 2 0/kg dry air or in English units as lbH 2 0/lb dry air is as follows: the kg of water vapor contained in

H 0 = ~

kg

H

kg dry

kg mol

pA

7

-

—

P

air

1

H20

kg mol

pA

kg mol

air

Saturated

air

18.02

pA

P-p A

which the water vapor

air in

is

is

H

air

equilibrium with liquid water at

this

mixture the partial pressure of

Hs

is

p AS

H F is denned

The percentage humidity

Percentage humidity.

humidity

mol

equal to the vapor pressure p AS of pure water

18.02

2.

in

is

Hence, the saturation humidity

at the given temperature.

28.97 kg air/kg

(9.3-1)

and temperature. In

the water vapor in the air-water mixture

1

x

2

28.97

H

the given conditions of pressure

H 20 H 0

18.02 kg

x

of the air divided by the humidity

Hs

if

as 100 times the actual

the air were saturated at the

same

temperature and pressure.

H P =100—

(9.3-3)

"s 3.

Percentage relative humidity.

ture

The amount

also given as percentage relative

is

of saturation of an air-water vapor mix-

HK

humidity

HR

= 100

using partial pressures.

—

(9.3-4)

Pas

Note

that

(9.3-2),

and

HR # HP (9.3-3)

HF =

since

,

HF

100

H = — H

is

in partial

—

18.02 (100)

28.97

s

This, of course,

expressed

pressures by combining Eqs. (9.3-1),

is

p.

P - pj

not the same as Eq.

EXAMPLE

93-1.

The

room

/

18.02

/

28.97

D.c v_as_

P-

p AS

=

Pa Ya_ p AS

P~ Pas ^noo) P -

K

(9.3-5)

pA

(9.3-4).

Humidity from Vapor-Pressure Data at 26.7°C (80°F) and a pressure of 101.325 kPa and contains water vapor with a partial pressure p A = 2.76 kPa. Calculate the air in a

is

following. (a)

Humidity,

(b)

Saturation humidity, s and percentage humidity, Percentage relative humidity, R

(c)

526

//.

H

,

H

Hr

.

.

Chap. 9

Drying of Process Materials

From the steam tables at 26.7°C, the vapor pressure of water kPa (0.507 psia). Also, p A = 2.76 kPa and P = 101.3 kPa

Solution: is

p AS

=

3.50

For part

(14.7 psia).

18.02

r,

"

18.02(2.76)

p.

= 28^7

For part

T=JA =

using Eq.

(b),

using Eq. (9.3-1),

(a),

from Eq.

Dew

^„

TT

-

a,r

is

100(0.01742)

(9.3-4), the

percentage relative humidity

is

3.50

Pas 4.

nntn „,kg

°-° 1742

(9.3-3), is

H (c),

=

2.76)

saturation humidity

(9.3-2), the

The percentage humidity, from Eq.

For part

-

28.97(101:3

The temperature

point of an air-water vapor mixture.

at

which a given mixture

and water vapor would be saturated is called the dew-point temperature or simply the dew point. For example, at 26.7°C (80°F), the saturation vapor pressure of water is Pas = 3-50 kPa (0.507 psia). Hence, the dew point of a mixture containing water vapor having a partial pressure of 3.50 kPa is 26.7°C. If an air-water vapor mixture is at 37.8°C of

air

bulb temperature, since this is the actual temperature a dry thermometer bulb would indicate in this mixture) and contains water vapor ofp^ = 3.50 kPa, the mixture would not be saturated. On cooling to 26.7°C, the air would be

(often called the dry

saturated,

i.e.,

dew

at the

point.

On

some water vapor would condense,

further cooling,

since the partial pressure cannot be greater than the saturation vapor pressure.

5.

Humid

present by

over the 1.88

The humid heat

heat of an air-water vapor mixture.

in J (or kJ)

required to raise the temperature of

K

kg of dry

1

cs

is

the

amount of heat

air plus the

water vapor

The heat capacity of air and water vapor can be assumed constant temperature ranges usually encountered at 1.005 kJ/kg dry air-K and 1

or

1

°C.

kJ/kg water vapor K, respectively. Hence, for SI and English units, •

kJ/kg dry

cs

air

•

K =

+

1.005

1.88//

(SI)

(9.3-6)

c s btu/lb m dry air

[Insomecases 6.

c s will

Humid volume

volume

in

m3

of

•

=

°F

be given as (1.005

+

0.24

+ 0.45H

1.88//)10

3

kg of dry

vapor

air plus the

J/kg-K.]

The humid volume v H is the total kPa (1.0 atm) abs

of an air-water vapor mixture. 1

(English)

contains at 101.325

it

pressure and the given gas temperature. Using the ideal gas law,

v„

mVkg

dry

air

=

22.41

f

T K

v., ft

"

3

/lb m dry y air

=

=

—

Sec. 9.3

+

18

-

H 02

4.56 x 10"

\ /

3

H)T

K ( 9 -3-7)

—— + —— H] l

T°R

492

=

3

1

+

-

(2.83 x 10-

'

I

V 28 97

273

(0.0252

[

V 28 97 -

18

-

02

/

+ 0.0405H)T°R

Vapor Pressure of Water and Humidity

527

For a saturated air-water vapor mixture,

is

the saturated volume.

H

water vapor

components, the

total

enthalpy

is

the sensible heat of the air-water vapor mixture plus

the latent heat A 0 in J/kg or kJ/kg

- r0 )°C = (T - T0 )

(T

and v H

,

Total enthalpy of an air-water vapor mixture. The total enthalpy of 1 kg of air plus is J/kg or kJ/kg dry air. If T0 is the datum temperature chosen for both y

7.

its

H = Hs

K

and

water vapor of the water vapor

that this enthalpy

air

=

c s (T

- T0 +

Hy

btu/lb m dry air

=

(0.24

+ 0A5H)(T - T0 °F) +

for

Hy becomes

H

v

kJ/kg dry

)

H/. 0

+

(1.005

at

T0 Note .

that

referred to liquid water.

is

- T0 °C) + HX 0

1.88//XT

(93-8)

If

H

the total enthalpy

referred to a base temperature

is

air

=

(1.005

btu/lb m dry air

=

(0.24

kJ/kg dry

v

H/. 0

+

1.88H) (T°C

- 0) +

T0

of

0°C

(32° F), the equation

2501.4H

(SI)

1075.4H

(English)

(9.3-9)

H 8.

+

0.45H) (T°F

-

32)

Humidity chart of air-water vapor mixtures.

+

A convenient

air-water vapor mixtures at 1.0 atm abs pressure this figure

the humidity

H

is

is

chart of the properties of

the humidity chart in Fig. 9.3-2. In

plotted versus the actual temperature of the air-water vapor

mixture (dry bulb temperature).

The

Hs

as

curve marked

H 2 0/kg

0.02226 kg 9.3-2,

100% running upward to the right gives the saturation humidity Example 9.3-1, for 26.7°C H s was calculated as

a function of temperature. In

it

falls

on

the

air.

100%

Plotting this point of 26.7°C (80°F) and

saturated

Hs = 0.02226

on Fig.

line.

Any point below the saturation line represents unsaturated air-water vapor mixThe curved lines below the 100% saturation line and running upward to the right represent unsaturated mixtures of definite percentage humidity H P Going downward vertically from the saturation line at a given temperature, the line between 100% tures.

.

saturation and zero humidity

increments of

10%

H

(the

bottom horizontal

line)

is

divided evenly into 10

each.

All the percentage humidity lines

HP

mentioned and

the saturation humidity line

Hs

can be calculated from the data of vapor pressure of water.

MPLE

93-2. Use of Humidity Chart EXA Air entering a dryer has a temperature (dry bulb temperature) of 60°C

dew point of 26.7°C (80°F). Using the humidity chart, determine the actual humidity H, percentage humidity H P humid heatc s and the humid volume v„ in SI and English units.

(140°F) and a

,

,

The dew point of 26.7°C is the temperature when the given 100% saturation. Starting at 26.7°C, Fig. 9.3-2, and drawing a vertical line until it intersects the line for 100% humidity, a humidity of H = 0.0225 kg H 2 0/kg dry air is read off the plot. This is the actual

Solution:

mixture

is

at

humidity of the air at 60° C. Stated in another way, if air at 60°C and having a humidity H = 0.0225 is cooled, its dew point will be 26.7°C. In English units, H = 0.0225 lb H 2 0/lb dry air. Locating this point of H = 0.0225 and r = 60°C on the chart, the percentage humidity H ? is found to be 14%, by linear interpolation vertically between the 10 and 20% lines. The humid heat for = 0.0225 is, from

H

528

Chap.

9

Drying of Process Materials

Sec. 9.3

Vapor Pressure of Water and Humidity

529

Eq.

(9.3-6),

c5

=

1.005

=

1.047 kJ/kg dry air

=

0.24

=

0.250 btu/lb m dry air °F

c5

1.88(0.0225)

4-

+

3 1.047 x 10 J/kg

or

(2.83

=

K

(English)

•

=

•

0.45(0.0225)

The humid volume v„

K

•

at

60°C (140°F), from Eq.

x 10" 3

m

0.977

3

+

4.56 x 10"

3

(9.3-7), is

x 0.0225X60

273)

4-

/kg dry air

In English units, vH

9.3C

=

(0.0252

0.0405 x 0.0225)(460

4-

4-

140)

=

3

15.67

ft

/lb m dry air

Adiabatic Saturation Temperatures

Consider the process shown mixture

is

in

where the entering gas of air-water vapor

Fig. 9.3-3,

contacted with a spray of liquid water.

humidity and temperature and the process

some makeup water added. The temperature of the water being

is

recirculated reaches a steady-state temperature

called the adiabatic saturation temperature,

H

having a humidity of

is

Ts

not saturated,

saturated at

is

Ts

will

the entering gas and the spray of droplets

equilibrium, the leaving air

The gas leaves having a different The water is recirculated, with

adiabatic.

Ts

is

,

.

If

the entering gas at temperature

be lower than T.

enough

If the

to bring the gas

having a humidity

Hs

T

contact between

and liquid to

.

Writing an enthalpy balance (heat balance) over the process, a datum of 7^ is used. The enthalpy of the makeup H 2 0 is then zero. This means that the total enthalpy of the entering gas mixture

=

enthalpy of the leaving gas mixture,

~ Ts

c s (T

)

4-

Or, rearranging, and using Eq. (9.3-6)

H-H T—

HX S = for c s

c s (Ts

)

4-

using Eq.

(9.3-8),

H S XS

(9.3-10)

,

1.005

s

- Ts

or,

4-

1.88W (SI)

Tc

(9.3-11)

H - //, T — 's

0.24

T<;

Equation

(9.3-1 1)

is

0.45

4-

H (English)

As

the equation of an adiabatic humidification curve

Hs

T

makeup H 2 O

Figure

530

plotted

outlet gas.

inlet gas

H,

when

9.3-3.

.

Ts

7777777777771

Adiabatic air-waier vapor saturator.

Chap. 9

Drying of Process Materials

on Fig. 9.3-2, which passes through and other points of H and T. These

Hs

the point

and Ts on the 100% saturation curve running upward to the left, are called

series of lines,

adiabatic humidification lines or adiabatic saturation

lines.

Since c s contains the term H,

when plotted on the humidity chart. If a given gas mixture at 7\ and Hj is contacted for a sufficiently long time in an adiabatic saturator, it will leave saturated at H SI and T^. The values ofH sl andT^ are

the adiabatic lines are not quite straight

determined by following the adiabatic saturation intersects the at a

100% saturation

percentage saturation

less

line. If

contact

is

line

not

going through point

sufficient, the

than 100 but on the same

Tu

until

it

leaving mixture will be

line.

EXAMPLE 93-3.

Adiabatic Saturation of Air = 0.030 kg H 2 0/kg dry air is stream at 87.8°C having a humidity contacted in an adiabatic saturator with water. It is cooled and humidified

An to

H

air

90% (a)

(b)

saturation.

What are the final values of// and T? For 100% saturation, what would be the

Solution:

For part

the humidity chart.

(a),

H

until

it

intersects the

90%

2 0/kg dry air. (b), the same line

is

followed to

100%

followed upward to the

and T?

= 0.030 and T = 87.8°C is located on adiabatic saturation curve through this point is

the point

The

values of//

left

line at

42.5°C and

H = 0.0500 kg H T= 9.3D

The large

For part 40.5°C and

Wet

H=

0.0505 kg

saturation, where

H z O/kg dry air.

Bulb Temperature

adiabatic saturation temperature

amount

of water

is

is

the steady-state temperature attained

when

contacted with the entering gas. The wet bulb temperature

steady-state nonequilibrium temperature reached

when

a small

amount

is

a

the

of water

is

contacted under adiabatic conditions by a continuous stream of gas. Since the amount of liquid

is

and humidity of the gas are not changed, contrary to where the temperature and humidity of the gas are

small, the temperature

the adiabatic saturation case,

changed.

The method used to measure the wet bulb temperature is illustrated in Fig. 9.3-4, where a thermometer is covered by a wick or cloth. The wick is kept wet by water and is immersed in a flowing stream of air-water vapor having a temperature of T (dry bulb temperature) and humidity H. At steady state, water

The wick and water heat of evaporation

stream at

T

are cooled to is

Tw and

is

evaporating to the gas stream.

stay at this constant temperature.

The

latent

exactly balanced by the convective heat flowing from the gas

to the wick at a lower temperature

Tw

.

thermometer reads

Tw

makeup water

\ Figure

Sec. 9.3

9.3-4.

•wick

Measurement of wet bulb temperature.

Vapor Pressure of Water and Humidity

531

A

taken at

Tw

.

is

MA

where q

is

kW(kJ/s),

m2 A

is

surface area

In English units, q

is

,

(9.3-12)

N

and X w

btu/h,N A

is

is

molH 2 0

A is kg the latent heat of vaporization at

molecular weight of water,

is

m2

MA NA X w A

=

q

,

is

of heat lost by vaporization, neglecting the small sensible heat change of-the

vaporized liquid and radiation,

s

The datum temperature

heat balance on the wick can be made.

The amount

2

lbmol/h-

NA =

ft

,

-

(y w

=

k y (y w

-

evaporating/

in

H 2 0.

kJ/kg

H z O. The fluxN^

aridX w isbtu/lb m

y)

Tw

-

is

(93-13)

y)

where k'y is the mass-transfer coefficient in kg mol/s m 2 mol frac, x BM is the log mean inert mole fraction of the air, y w is the mole fraction of water vapor in the gas at the surface, and y is the mole fraction in the gas. For a dilute mixture x BM ^ 1.0 andfc^ ^ k y The relation between H and y is. •

.

H/M A HjM A

(93-14)

l/M„ + where

MB

is

the molecular weight of air

and

MA

=

HM —5

the molecular weight

ofH 2 0.

Since

H is

small, as an approximation,

y

B

(93-15)

Substituting Eq. (9.3-15) into (9.3-13) and then substituting the resultant into Eq. (9-3-12),

q

The

=

MB k

rate of convective heat transfer

q

=

y

Xw

[H w

— H)A

(93-16)

from the gas stream

h(T

H-H w T-Tw

1),

2 •

,

lines

for others,

is

•

°F).

(9.3-18)

called the psychrometric ratio,

approximately 0.96-1.005. Since

show

that

this value

is

approximately 1.005, Eqs. (9.3-18) and (9.3-11) are

almost the same. This means that the adiabatic saturation

bulb

ft

h/M B k y

Experimental data on the value of h/M B k y close to the value of c s in Eq. (9.3-1

Tw

(93-17)

where h is the heat-transfer coefficient in kW/m K (btu/h Equating Eq. (9.3-16) to (9.3-17) and rearranging,

is

the wick at

- TW )A 2

for water vapor-air mixtures, the value

T to

at

with reasonable accuracy. (Note that this

lines

can also be used

for

wet

only true for water vapor and not

is

such as benzene.) Hence, the wet bulb determination

is

often used to determine

the humidity of an air-water vapor mixture.

EXAMPLE 93-4.

Wet Bulb Temperature and Humidity vapor-air mixture having a dry bulb temperature of T = 60°C is passed over a wet bulb as shown in Fig. 9.3-4, and the wet bulb temperature obtained is Tw = 29.5°C. What is the humidity of the mixture?

A water

Solution:

same

The wet bulb temperature

of 29.5°C can be

as the adiabatic saturation temperature

Ts

the adiabatic saturation curve of 29.5°C until

temperature of 60°C, the humidity

532

is

H=

0.0135

,

assumed

to be the

as discussed. Following it

reaches the dry bulb

kgH z O/kg

Chap. 9

dry

air.

Drying of Process Materials

EQUILIBRIUM MOISTURE CONTENT OF MATERIALS

9.4

9.4A

Introduction

As in other transfer processes, such as mass transfer, the process of drying of materials must be approached from the viewpoint of the equilibrium relationships and also the rate relationships. In most of the drying apparatus discussed in Section 9.2, material is dried in contact with an air-water vapor mixture. The equilibrium relationships between the air-water vapor and the solid material will be discussed in this section.

An

important variable in the drying of materials

is

the humidity of the air in contact

with a solid of given moisture content. Suppose that a wet solid containing moisture

brought into contact with a stream of air having a constant humidity

A

large excess of air

is

used, so

its

is

H and temperature.

conditions remain constant. Eventually, after exposure

of the solid sufficiently long for equilibrium to be reached, the solid will attain a definite

moisture content. This

is

known

as the equilibrium moisture content of the material

under the specified humidity and temperature of the air. The moisture content is usually expressed on a dry basis as kg of water per kg of moisture-free (bone-dry) solid or kg H 2 O/100 kg dry solid; in English units as lb H 2 O/100 lb dry solid.

For some

solids the value of the equilibrium moisture content

direction from which equilibrium

moisture content

is

approached.

obtained according

is

A

whether a wet sample

to

depends on the

different value of the equilibrium is

allowed to dry by

desorption or whether a dry sample adsorbs moisture by adsorption. For drying calculations

it is

the desorption equilibrium that

the larger value

and

is

of particular interest.

Experimental Data of Equilibrium Moisture Content

9.4B

for Inorganic

/.

is

and Biological Materials

Typical data for various materials.

If

the material contains

more moisture than

equilibrium value in contact with a gas of a given humidity and temperature, until

the material contains less moisture than

its

air

0%

equilibrium value.

its

equilibrium value,

having

its

dry

will

adsorb water until it reaches its equilibrium value. For humidity, the equilibrium moisture value of all materials is zero.

reaches

it

it

it

If

will

The equilibrium moisture content varies greatly with the type of material for any given percent relative humidity, as shown in Fig. 9.4-1 for some typical materials at room temperature. Nonporous insoluble solids tend to have equilibrium moisture contents which are quite low, as shown for glass wool and kaolin. Certain spongy, cellular materials of organic and biological origin generally show large equilibrium moisture contents. Examples of this in Fig. 9.4-1 are wool, leather, 2.

Typical food materials.

and wood.

In Fig. 9.4-2 the equilibrium moisture contents of

some

typical food materials are plotted versus percent relative humidity. These biological

materials also in'

show

large values of equilibrium moisture contents.

Fig. 9.4-1 for biological materials

about 60

to

80%,

show

Data

in this figure

and

that at high percent relative humidities of

the equilibrium moisture content increases very rapidly with increases

of relative humidity. In general, at low relative humidities the equilibrium moisture content

is

greatest for

food materials high in protein, starch, or other high-molecular-weight polymers

lower

for

food

materials high in soluble solids. Crystalline salts

generally adsorb small 3.

amounts

Effect of temperature.

Sec. 9.4

and sugars and

and

also fats

of water.

The equilibrium moisture content

Equilibrium Moisture Content of Materials

of a solid decreases

some-

533

Relative humidity (%)

Figure

9.4-1.

Typical equilibrium moisture contents of some solids at approximately K (25°C). [From National Research Council, International Criti-

298

New York : McGraw-Hill Book Company, 1929. Reproduced with permission of the National Academy of Sciences.]

cal Tables, Vol. 11.

what with an increase of

50%,

temperature. For example, for raw cotton at a relative humidity

in

the equilibrium moisture content decreased

37.8°C (311 K) to about

5.3 at

from

7.3

kg

H 2 O/100

kg dry

solid at

93.3°C (366.5 K), a decrease of about 25%. Often for

moderate temperature ranges, the equilibrium moisture content

when experimental data are not available

at different

will

be assumed constant

temperatures.

At present, theoretical understanding of the structure of solids and surface pheno-

mena does not enable

us to predict the variation of equilibrium moisture content of

various materials from

first

principles.

However, by using models such

as those used for

adsorption isotherms of multilayers of molecules and others, attempts have been

made

to

Henderson (H2) gives an empirical relationship between equilibrium moisture content and percent relative humidity for some agricultural materials. In general, empirical relationships are not available for most materials, and equilibrium moisture contents must be determined experimentally. Also, equilibrium moisture relationships often vary from sample to sample of the same kind of material.

correlate experimental data.

9.4C

Bound and Unbound Water

In Fig. 9.4-1,

if

the equilibrium moisture content of a given material

intersection with the

534

in Solids

100% humidity

line, the

moisture

is

called

Chap. 9

is

continued to

its

bound water. This water

Drying of Process Materials

20

y

-

c o c o

-t-»

15

y

//

r

/

o

£ O

//

>>

K o ED * ^1 Ic O O £

10

/

/

'3 cr

W

40

20

80

60

100

Relative humidity (%)

Figure

Typical equilibrium moisture contents of some food materials at approximately 298 K (2rC): (/) macaroni, (2) flour, (3) bread;{4) crackers, (5) egg albumin. [Curue (5) from ref. {El). Curves (7) to (4) from

9.4-2.

National Research Council, International Critical Tables,

New York : McGraw-Hill Book Company, permission of the National

in the solid exerts If

II.

Academy of Sciences.]

a vapor pressure less than that of liquid water at the

more water than

such a material contains

Vol.

1929. Reproduced with

same temperature. 100%

indicated by intersection with the

line, it can still exert only a vapor pressure as high as that of ordinary water at same temperature. This excess moisture content is called unbound water, and it is held primarily in the voids of the solid. Substances containing bound water are often called

humidity the

hygroscopic materials.

As an example, consider curve 10 for wood in Fig. 9.4-1. This intersects the curve for at about 30 kg H 2 O/100 kg dry solid. Any sample of wood containing than 30 kg H 2 O/100 kg dry solid contains only bound water. If the wood sample

100% humidity less

H 2 O/100 kg dry solid, 4 kg H z O would be unbound and 30 kg H 2 0 bound per 100 kg dry solid. The bound water in a substance may exist under several different conditions. Moisture in cell or fiber walls may have solids dissolved in it and have a lower vapor

contained 34 kg

pressure. Liquid water in capillaries of very small diameter will exert a lowered

pressure because of the concave curvature of the surface. materials

is

in

vapor

natural organic

chemical and physical-chemical combination.

in

Free and Equilibrium Moisture of

9.4D

Water

Free moisture content in a sample content. This free moisture

is

is

a

Substance

the moisture

above the equilibrium moisture

the moisture that can be removed by drying under the given

For example, in Fig. 9.4-1 silk has an equilibrium moisture kg dry material in contact with air of 50% relative humidity O/100 2 and 25°C. If a sample contains 10 kg H 2 O/100 kg dry material, only 10.0 - 8.5, or 1.5, kg H 2 O/100 kg dry material is removable by drying, and this is the free moisture of the percent relative humidity.

content of 8.5 kg

H

sample under these drying conditions. In

many

texts

dry basis. This

Sec. 9.4

is

and

references, the moisture content

exactly the

same

as the

kg

H 2 O/100

Equilibrium Moisture Content of Materials

is

given as percent moisture on a

kg dry material multiplied by

100.

535

RATE OF DRYING CURVES

9.5

Introduction and Experimental

9.5A 1.

Methods

In the drying of various types of process materials from one moisture

Introduction.

content to another,

it

usually desired to estimate the size of dryer needed, the various

is

operating conditions of humidity and temperature for the air used, and the time needed to perform the

amount

of drying required.

As discussed

in Section 9.4, equilibrium

moisture contents of various materials cannot be predicted and must be determined experimentally. Similarly, since our knowledge of the basic is

quite incomplete,

ments of drying

it

is

necessary

in

most cases

to obtain

mechanisms of rates of drying some experimental measure-

rates.

Experimental determination of rate of drying. To experimentally determine the rate of drying for a given material, a sample is usually placed on a tray. If it is a solid material it 2.

should

fill

the tray so that only the top surface

suspending the tray from a balance

in

is

exposed to the drying

air stream.

a cabinet or duct through which the

air

is

By

flowing,

the loss in weight of moisture during drying can be determined at different intervals

without interrupting the operation.

/

In doing batch-drying experiments, qertain precautions should be observed to

obtain usable data under conditions that closely resemble those to be used in the large-scale operations.

The sample should not be too small

in

weight and should be

a tray or frame similar to the large-scale one. The ratio of drying to nondrying surface (insulated surface) and the bed depth should be similar. The velocity,

supported

in

humidity, temperature, and direction of the air should be the

same and constant

to

simulate drying under constant drying conditions.

Rate of Drying Curves for Constant-Drying Conditions

9.5B

Conversion of data to rate.- of-drying curve. Data obtained from a batch-drying experiment are usually obtained as total weight of the wet solid (dry solid plus moisture) 1.

W

at different

times

drying data

t

in the

hours

in the

drying period. These data can be converted to rate-of-

following ways. First, the data are recalculated.

the wet solid in kg total water plus dry solid

and

W

s

is

If

W

is

the weight of

the weight of the dry solid in kg,

(9.5-1)

For the given constant drying conditions, the equilibrium moisture content is determined. Then the free moisture content

equilibrium moisture/kg dry solid free

water/kg dry solid

is

=

Using the data calculated from Eq. /

in h

is

made

as in Fig. 9.5-la.

slopes of the tangents

kg

calculated for each value of X,

X time

X* kg

X in

drawn

To

(9.5-2)

X,

(9.5-2),

a plot of free moisture content

to the curve in Fig. 9.5-la

values oidX/dt at given values of

t.

The

X versus

obtain the rate-of-drying curve from this plot, the

rate

R

is

can be measured, which give

calculated for each point by

(9^-3)

536

Chap. 9

Drying of Process Materials

where

R

is

drying rate in kg

m

area for drying in 2 ft

.

2

is

•

,

In English units,

.

For obtaining R from

drying-rate curve

H 2 0/h m 2 L s R

is

kg of dry lb m

H

Fig. 9.5-la, a value of

then obtained by plotting

R

solid used,

2 0/h

2 •

ft

Ls/A

,

Ls

is

of 21.5

and

A

lb m

dry

exposed surface solid, and A is

kg/m 2 was

used.

The

versus the moisture content, as in Fig.

9.5-lb.

loss

Another method to obtain the rate-of-drying curve is to first calculate the weight for a Ar time. For example, if = 0.350 at a timer, = 1.68 h andx 2 = 0.325 at

AX

a time

t2

=

2.04

h,

AX/At =

(0.350

- 0.325)/(2.04 -

1.68).

Then, using Eq.

(9.5-4)

and

0.5 T3

o X MM

•-=1

o

A 0.4

>>

E Time

t

10

12

0.5

0.6

14

(h)

(a)

'0

0.1

0.3

0.2

Free moisture

X

(kg

0.4

H 2 0/kg

dry solid)

(b)

Figure

Sec. 9.5

9.5-1.

Typical drying-rate curve for constant drying conditions : (a) plot of data as free moisture versus time, (b) rate of drying curve as rate versus free moisture content.

Rate of Drying Curves

537

L s/A =

21.5,

0.350 2.04

This rate

R

-

0.325

=

and should be plotted

the average over the period 1.68 to 2.04 h

is

1.493

1.68 at the

X = (0.350 + 0.325)/2 = 0.338.

average concentration

2. Plot of rate-of-drying curve. In Fig. 9.5-lb the rate-of-drying curve for constantdrying conditions is shown. At zero time the initial free moisture content is shown at

point A. In the beginning the solid

is

usually at a colder temperature than

temperature

may

start at point A'.

usually quite short and

From

equilibrium value. Alternatively,

rises to its

with, the rate

B

point

constant during

to

it is

C in

This

initial

if

the solid

is

ultimate

its

B

temperature, and the evaporation rate will increase. Eventually at point

the surface

quite hot to start

unsteady-state adjustment period

is

often ignored in the analysis of times of drying.

Fig. 9.5-la the line

this period.

is

straight,

and hence the slope and

This constant-rate-of-drying period

shown

is

rate are

as line

BC

in

Fig. 9.5-lb.

At point until

9.5-lb

C on

both

is

drying rate starts to decrease

plots, the

reaches point D. In this

it

in the falling-rate period

shown

falling-rate period, the rate

first

as line

CD

in Fig.

often linear.

At point

D

the rate of drying falls even

the equilibrium moisture content dried, the region

more

rapidly, until

X* and X = X* — X* =

is

CD may be missing completely

or

it

may

it

0.

reaches point £, where

In

some

constitute

all

materials being

of the falling-rate

period.

Drying

9.5C

in

the

Constant-Rate Period

Drying of different solids under different constant conditions of drying

will often give

curves of different shapes in the falling-rate period, but in general the two major portions of the drying-rate curve

—constant-rate period and

falling-rate period

In the constant-rate drying period, the surface of the solid

continuous

film of

the given air conditions

is

a free liquid surface.

higher rates than from a

if

the solid were not present.

independent of the solid and

is

The

—are present.

initially

water exists on the drying surface. This water

water and the water acts as

from

is

is

rate of

very wet and a

evaporation under

essentially the

same

Increased roughness of the solid surface, however,

flat

unbound

entirely

as the rate

may

lead to

surface.

porous, most of the water evaporated in the constant-rate period

is

supplied from the interior of the solid. This period continues only as long as the water

is

the solid

If

is

supplied to the surface as fast as

it

is

evaporated. Evaporation during

this

period

is

wet bulb temperature, and in the absence of heat transfer by radiation or conduction, the surface temperature is approximately that of the

similar to that in determining the

wet bulb temperature.

Drying

9.5D Point

C

in

in Fig. 9.5-lb

insufficient water

surface

the Falling-Rate Period

is

is

at

the critical free moisture content

on the surface

to

Xc

.

At

this

point there

no longer wetted, and the wetted area continually decreases

falling-rate period until the surface

is

is

maintain a continuous film of water. The entire

completely dry at point

D

in

this

first

in Fig. 9.5-1 b.

The second falling-rate period begins at point D when the surface is completely dry. The plane of evaporation slowly recedes from the surface. Heat for the evaporation is

538

Chap.

9

Drying of Process Materials

transferred

through the solid -tegabe zone of vaporization. Vaporized water moves

through the solid into the air.stream. In some cases no sharp discontinuity occurs

change

but the time required

may

in the falling-rate period

be long. This can be seen

H

reduction of 0.21 kg

2

0/kg dry

The

X

from 0.40

falling-rate period

CE

relatively small

The period

BC

for

about 0.19, a about 9.0 h and

to

lasts

X only from 0.19 to 0.

Moisture Movements

in Solids

During Drying

Falling-Rate Period

in the

When

solid.

may be

in Fig. 9.5-1.

and reduces

constant-rate drying lasts for about 3.0 h

9.5E

change from

the

so gradual that no sharp

is

detectable.

is

The amount of moisture removed

reduces

and

at point D,

partially wetted to completely dry conditions at the surface

drying occurs by evaporation of moisture from the exposed surface of a solid,

move from

moisture must

movement theories

affect the

advanced

the depths of the solid to the surface.

The mechanisms of the

drying during the constant-rate and falling-rate periods.

to

Some of the

explain the various types of falling-rate curves will be briefly

reviewed.

/.

is

In this theory diffusion of liquid moisture occurs

Liquid diffusion theory.

when there method

a concentration difference between the depths of the solid and the surface. This

of transport

of moisture

is

nonporous

usually found in

solids

where single-phase

solutions are formed with the moisture, such as in paste, soap, gelatin, and glue. This also found in drying the last portions of moisture starches,

and

textiles.

The shapes 7.

is

wood, leather, paper, movement of water in the

clay, flour,

food materials, the

by diffusion.

of the moisture distribution curves in the solid at given times are

qualitatively consistent with

Chapter

many

In drying

falling-rate period occurs

from

The moisture

use of the unsteady-state diffusion equations given in

diffusivity

D AB

usually decreases with decreased moisture

content, so that the diffusivities are usually average values over the range of con-

centrations used. Materials drying in this

although the actual mechanisms

from the surface rate

is

quite

through the solid

fast,

may

i.e.,

way

are usually said to be drying by diffusion,

be quite complicated. Since the rate of evaporation

the resistance

is

quite low,

in the falling-rate period, the

compared

moisture content

to

the diffusion

at the surface is at

the equilibrium value.

The shape

of a diffusion-controlled curve in the falling-rate period

9.5-2a. If the initial constant-rate

drying

is

quite high, the

first

is

similar to Fig.

falling-rate period of

a no •

C —

Q Free moisture,

X

(a)

Figure

9.5-2.

Typical drying-rate curves: (b)

Sec. 9.5

Free moisture,

X

(b) (a) diffusion-controlled falling-rate period,

capillary-controlled falling-rale period in a fine porous solid.

Rate of Drying Curves

539

unsaturated surface evaporation

may

not appear.

the period of unsaturated surface evaporation

9.5-lb and the diffusion-controlled curve

drying

in this

theory are given in

and Problem 7.1-6 the Chapter 7 Problems.

Capillary movement

sand,

soil,

in

the constant-rate drying

is

usually present in region

DE. Equations

region

quite low,

CD

in

porous

solids.

drying of

for the

When granular and

paint pigments, and minerals are being dried,

wood

porous

unbound or

Problem

using diffusion

solids free

in Fig.

for calculating

period where diffusion controls are given in Section 9.9. Also,

7.1-4 for the drying of clay

2.

is

If

is

such as clays,

moisture moves

through the capillaries and voids of the solids by capillary action, not by diffusion. This mechanism, involving surface tension, is similar to the movement of oil in a lamp wick. A porous solid contains interconnecting pores and channels of varying pore sizes. As water is evaporated, a meniscus of liquid water is formed across each pore in the depths of the solid. This sets up capillary forces by the interfacial tension between the water and solid. These capillary forces provide the driving force for moving water through the pores to the surface. Small pores develop greater forces than those developed by large pores. At the beginning of the falling-rate period at point C in Fig. 9.5-lb, the water is being brought to the surface by capillary action, but the surface layer of water starts to recede below the surface. Air rushes in to fill the voids. As the water is continuously removed, a point is reached where there is insufficient water left to maintain continuous films across the pores, and the rate of drying suddenly decreases at the start of the second falling-rate period at point D. Then the rate of diffusion of water vapor in the pores and rate of conduction of heat in the solid may be the main factors in drying. In fine pores in solids, the rate-of-drying curve in the second falling-rate period may conform to the diffusion law and the curve is concave upward, as shown in Fig. 9.5-2b. For very porous solids, such as a bed of sand, where the pores are large, the rate-ofdrying curve in the second falling-rate period is often straight, and hence the diffusion equations do not apply. 3.

A

Effect of shrinkage.

the solid as moisture

is

factor often greatly affecting the drying rate

is

the shrinkage of

removed. Rigid solids do not shrink appreciably, but colloidal

and fibrous materials such as vegetables and other foodstuffs do undergo shrinkage. The most serious effect is that there may be developed a hard layer on the surface which is impervious to the flow of liquid or vapor moisture and slows the drying rate; examples are clay and soap. In many foodstuffs, if drying occurs at too high a temperature, a layer of closely packed shrunken cells, which are sealed together, forms at the surface. This presents a barrier to moisture migration and is known as case hardening. Another effect of shrinkage is to cause the material to warp and change its structure. This can happen in drying wood.

Sometimes

to decrease these effects of shrinkage,

it is

desirable to dry with moist

air.

This decreases the rate of drying so that the effects of shrinkage on warping or hardening at the surface are greatly

CALCULATION METHODS FOR CONSTANT-RATE DRYING PERIOD

9.6

9.6A /.

reduced.

Method Using Experimental Drying Curve

Introduction.

Probably the most important factor

in

drying calculations

of time required to dry a material from a given initial free moisture content

moisture content

540

X2

.

For drying

in the constant-rate period,

Chap. 9

we can

the length

is

X

t

to

a final

estimate the time

Drying of Process Materials

needed by using experimental batch drying curves or by using predicted mass- and heat-transfer coefficients.

2.

Method

To

using drying curve.

method

material, the best

is

estimate the time of drying for a given batch of

based on actual experimental data obtained under con-

ditions where the feed material, relative exposed surface area, gas velocity, temperature,

and humidity

same

are essentially the

as in the final drier.

Then

the time required for the

constant-rate period can be determined directly from the drying curve of free moisture

content versus time.

EXAMPLE

Time of Drying from Drying Curve

9.6-1.

A solid whose drying curve is represented by Fig. 9.5-la is free moisture content X = 0.38 kg H 2 0/kg dry solid i

H 2 0/kg dry solid. Solution:

From

X = 0.25,

t2

2

—

to

to

be dried from a X 2 = 0.25 kg

Estimate the time required.

X =

Fig. 9.5-la for

0.38,

x

f

is

j

3.08 h. Hence, the time required t

=

t2

-t = 1

3.08

-

=

1.28

read off as 1.28

h.

For

is

1.80 h.

Method using rate-of-drying curve for constant-rate period. Instead of using the drying curve, the rate-of-drying curve can be used. The drying rate R is defined by Eq. 3.

(9.5-3) as

R

—

Lc dX = - tS-

A

(9.5-3)

dt

This can be rearranged and integrated over the time interval to dry from

X

2

at( 2

=

t

If

X

att {

l

=

0

to

t.

=

dt=^-

f

Xl

dX R

A

(9.6-1)

X, and

the drying takes place within the constant-rate period so that both

Xc

greater than the critical moisture content

R = constant = R c

then

,

.

X

2

are

Integrating Eq.

(9.6-1) for the constant-rate period,

f=^f-(X AR r

EXAMPLE

9.6-2.

Repeat Example Solution: Fig. 9.5-lb

1

-X

(9.6-2)

2)

Drying Time from Rate-of-Drying Curve

9.6-1 but use Eq. (9.6-2)

and

Fig. 9.5-lb.

As given previously, a value of 21.5 for L s /A was used to prepare 2 from 9.5-la. From Fig. 9.5-lb, R c = 1.51 kg H 2 0/h m Substi.

tuting into Eq. (9.6-2),

1

This

9.6B

is

=

jtc {Xl ~ Xl) = TIT (038

close to the value ofl.80 h of

Method Using Predicted Transfer for

Example

-

a25)

=

9.6-1.

Coefficients

Constant-Rate Period In the constant-rate period of drying, the surfaces of the grains of solid

/.

Introduction.

in

contact with the drying air flow remain completely wetted.

Sec. 9.6

L85 h

Calculation

Methods for Constant-Rate Drying Period

As

stated previously, the

541

under a given set of

rate of evaporation of moisture

type of solid and

is

essentially the

same conditions.

surface under the

air

conditions

independent of the

is

same as the rate of evaporation from a However, surface roughness may increase

free liquid

the rate of

evaporation.

During were not solid.

this

there.

The

constant-rate period, the solid

The water evaporated from

rate of evaporation

that occurring at a wet bulb

2.

so wet that the water acts as

is

the surface

if

the solid

supplied from the interior of the

from a porous material occurs by the same mechanism as

thermometer, which

is

essentially constant-rate drying.

Drying of a material occurs by mass

Equations for predicting constant-rate drying.

transfer of water vapor

is

from the saturated surface of the material through an

air film to

The rate of moisture movement within the solid saturated. The rate of removal of the water vapor (drying)

the bulk gas phase or environment. sufficient to

keep the surface

is is

controlled by the rate of heat transfer to the evaporating surface, which furnishes the latent heat of evaporation for the liquid. At steady state, the rate of

mass

transfer

balances the rate of heat transfer.

To

we

derive the equation for drying,

neglect heat transfer by radiation to the solid

surface and also assume no heat transfer by conduction from metal pans or surfaces. In

Section 9.8, convection and radiation will also be considered.— Assuming only heat

from the hot gas

transfer to the solid surface by convection

and mass transfer from the surface which are

The

the

same as those

to the hot

for deriving the

h

is

same

m

2

2

(ft

as Eq. (9.3-1

3)

).

and

=

h(T

The equation

in

write equations

Eq. (9.3-18).

btu/h) from the gas at

- TW )A

W/m

2

K

•

fc,

needed to vaporize

the small sensible heat changes,

is

k (y w y

(9.3-1 5)

NA =

).

is

the latent heat at

Figure

542

(°F) to

(9.6-3)

(btu/h

2 •

ft

and A

°F)

is

the exposed is

the

the

same

(9.6-4)

and substituting

^

NA

-y)

(H H

.

-

kg mol/s

into Eq. (9.6-4),

H)

-

m2

(9.6-5)

(lb

mol/h

2 •

ft

)

9.6-1.

Tw

in

water, neglecting

as Eq. (9.3-12).

q=M A N A A w A where w

T°C

is

is

Using the approximation from Eq.

of heat

Tw

of the flux of water vapor from the surface

NA =

The amount

(J/s,

)

the heat-transfer coefficient in in

W

Tw °C, where (T — TW )°C = (T — Tw K q

drying area

we can

wet bulb temperature

rate of convective heat transfer q in

the surface of the solid at

where

to the surface of the solid

gas (Fig. 9.6-1),

(9.6-6)

J/kg (btu/lb m ).

Heat and mass transfer

in

constant-rate drying.

Chap. 9

Drying of Process Materials

Equating Eqs.

and

(9.6-3)

Rc = Equation

(9.6-7)

and substituting Eq.

(9.6-6)

q

AA W

- Tw

h(T

=

(9.6-5) for jV^

,

~ = k M B (H w -H) )

(9.6-7)

y

Aw

identical to Eq. (9.3-18) for the wet bulb temperature.

is

and

the absence of heat transfer by conduction

Hence,

radiation, the temperature of the solid

in is

wet bulb temperature of the air during the constant-rate drying period. Hence, the

at the

R c can be calculated using the heat-transfer equation h(T — Tw )/X w or the mass-transfer equation k y B {H w — H). However, it has been found more reliable to use the heat-transfer equation (9.6-8), since an error in determining the interface temperature rate of drying

M

Tw at

the surface affects the driving force

R c kg

H

2

=

0/h -m 2

(T

— Tw much less than )

— (T — Tw °CX360O)

it

affects

—

(H w

H).

1^

(SI)

(9.6-8)

Rc

lb m

H

2

=

2

0/h

ft

•

/.

To

predict

Rc

case where the air

w

(English)

known. For

the

flowing parallel to the drying surface, Eq. (4.6-3) can be used for

air.

in

is

- Tw °F)

(T

-A-

Eq.

the heat-transfer coefficient must be

(9.6-8),

However, because the shape of the leading edge of the drying surface causes more turbulence, the following can be used for an air temperature of 45-150°C and a mass 2 2 velocity G of 2450-29 300 kg/h m (500-6000 lbjh ft ) or a velocity of 0.61-7.6 m/s •

(2-25

ft/s).

=

h

0.0204G

0 8 '

(SI)

(9.6-9)

=0.0128G°-

h

where

and

in SI units

h in btu/h

19 500 kg/h

-

m2

Equations

G

is

2 ft

•

vp kg/h

°F.

•

When

8

(English)

m2

and

is

in

air

flows perpendicular to the surface for a

G

h

is

W/m 2

or a velocity of 0.9-4.6 m/s (3-15

•

K. In English units,

= 1.17C 037

(SI)

h

= 0.37G 037

(English)

(9.6-8) to (9.6-10)

of

2 •

ft

3900-

(9.6-10)

can be used

when

lb^h

ft/s),

h

constant-rate period. However,

G

to estimate the rate

possible, experimental

of drying during the

measurements of

the drying

rate are preferred.

To

estimate the time of drying during the constant-rate period, substituting Eq.

(9.6-7) into (9.6-2),

Ls^wWi — Ah(T

EXAMPLE

%i)

- Tw

)

(9 6-11)

Prediction of Constant-Rate Drying is dried in a pan 0.457 x 0.457 deep. The material is 25.4 deep in the pan, and 25.4 9.6-3.

An

insoluble wet" granular material

(1.5

x

1.5 ft)

^s(^j ~ jj) ~ Ak M B (H w — H) y

mm

m

mm

and the sides and bottom can be considered to be insulated. Heat transfer is by convection from an air stream flowing parallel to the surface at a velocity of 6.1 m/s (20 ft/s). The air is at 65.6°C (150°F) and has a humidity of 0.010 kg H 2 0/kg dry air. Estimate the rate of drying for the constant-rate period using SI and English units. Solution:

Sec. 9.6

For a humidity

Calculation

H—

0.010 and dry bulb temperature of 65.6°C

Methods for Constant-Rate Drying Period

543

and using the humidity chart, Fig. found as 28.9

:>

ration line (the

C

wet bulb temperature Tw is by following the adiabatic satuto the saturated humidity. Using

9.3-2, the

H w = 0.026

and

(84°F)

same as the wet bulb line) humid volume,

Eq. (9.3-7) to calculate the vH

The

= =

(2.83

=

0.974

density for

1.0

P

The mass

Using Eq.

At

x 10~ 3

(2.83

x 10"

+

LQ

=

3

m 3 /kg

kg dry

+

air

+

4.56 x 1(T

3

x 10"

3

4.56

H)7

+

x 0.01X273

65.6)

dry air

+ 0.010kgH 2 O

" ° 10

=

1-037

is

kg/m 3

3

(0.0647 lbjft

)

0 9 4

velocity

G is

G =

vp

=

6.1(3600X1.037)

G=

vp

=

20(3600)(0.0647)

= 22770 kg/h-m 2 -

4660 lbjh

2 •

ft

(9.6-9), 0 8

h

=

h

= 0.0128G 0

0.0204G

= 0.0204(22 770) 0 = 62.45 W/m 2 K '

'

8

•

=

8

'

0.0128(4660)°-

8

=

2

11.01 btu/h-ft

-°F

Tw =

28.9°C (84°F), k w = 2433 kJ/kg (1046 btu/lbj from steam tables. Substituting into Eq. (9.6-8), noting that (65.6 - 28.9)°C = (65.6 28.9)

-

K,

R< =

=

Tw (T " T^

total

(150

-

84)

= RC A =

=

As

(65 6 "

2433 x 1000

28 9X3600)

-

-

=

2

0.695 lbjh-ft

evaporation rate for a surface area of 0.457 x 0.457

total rate

9.6C

-

kg/h-m 2

3.39

*c=^ The

62 4 3600)

Effect of Process Variables

stated previously, experimental

over using the equations

3.39(0.457 x 0.457) 0.695(1.5 x 1.5)

=

=

H

2

is

H 2 0/h

0.708 kg

1.564 lb m

m

2

0/h

on Constant-Rate Period

measurements of the drying rate are usually preferred However, these equations are quite helpful to

for prediction.

predict the effect of changing the drying process variables

when

limited experimental

data are available.

/.

When

Effect of air velocity.

the rate

Rc

conduction and radiation heat transfer are not present,

of drying in the constant-rate region

is

proportional to h and hence to G

given by Eq. (9.6-9) for air flow parallel to the surface.

The

effect of

gas velocity

0 8 '

is

as

less

important when radiation and conduction are present.

2.

Effect of gas humidity.

If

the gas humidity

H

is

decreased for a given

then from the humidity chart the wet bulb temperature (9.6-7),

544

Rc

will increase.

For example,

if

Tw

will decrease.

the original conditions

Chap. 9

areK cl T,, ,

T

of the gas,

Then using Eq.

TWI H ,

lt

and

Drying of Process Materials

H W1

,

then

if

H

l

is

changed

to

R C2 = However, since X WI

=

J.

W2

H 2 and H WI

o/" gas

r = Ra n l

temperature.

If

"J** wi

w\

T is

the gas temperature in T.

Hence,

Rc

R C 2 becomes

v ':

~ "2 n

(9-6-13)

1

increased, Tjy

is

also increased

increases as follows:

n wi ~ "I

'Wl

'l

Rc

,

(9.6-12)

T~T

Ra T

some, but not as much as the increase

4.

H W2

,

'

£/fect

to

7^ = *c,

He,

R C2 =

J.

changed

is

For heat transfer by convection only, the rate solid. However, the time t for drying between be directly proportional to the thickness x This

Effect of thickness of solid being dried. is

independent of the thickness x, of the

X and X 2 will shown by Eq. (9.6-2), where increasing increase the amount of L s kg dry solid. fixed moisture contents

y

is

l

the thickness with a constant

A

.

will directly

Experimental effect of process variables. Experimental data tend to bear out the conclusions reached on the effects of material thickness, humidity, air velocity, and

5.

T-Tw 9.7

.

CALCULATION METHODS FOR FALLING-RATE DRYING PERIOD Method Using, Graphical

9.7A

Integration

In the falling-rate drying period as

shown

in Fig. 9.5-

constant but decreases when drying proceeds past the

When

the free moisture content

The time

for

the rate

If

is

X

is

1

b, the rate

critical free

of drying

R

is

moisture content

not

Xc

.

zero, the rate drops to zero.

drying for any region between

X

t

and X 2 has been given by Eq.

(9.6-1).

constant, Eq. (9.6-1) can be integrated to give Eq. (9.6-2). However, in

the falling-rate period,

R

varies.

For any shape

can and determining the area under the

of falling-rate drying curve, Eq. (9.6-1)

be graphically integrated by plotting l/R versus

X

curve.

EXAMPLE

9.7-1. Graphical Integration in Falling-Rate Drying Period batch of.wet solid whose drying-rate curve is represented by Fig. 9.5-lb is to be dried from a free moisture content of X, = 0.38 kg H 2 0/kg dry solid to X 2 = 0.04 kg H 2 0/kg dry solid. The weight of the dry solid is L s = 399 kg dry solid and A = 18.58 2 of top drying surface. Calculate the time for drying. Note that Ls/A = 399/18.58 = 21.5 kg/m 2

A

m

.

Solution:

kg

H

From

2 0/kg dry

Fig. 9.5-lb, the critical free

solid.

Hence, the drying

is

moisture content isX c

in the constant-rate

and

=

0.195

falling-

rate periods.

Sec. 9.7

Calculation Methods for Falling-Rate Drying Period

545

For the constant-rate period, X = 0.38 and X 2 = X c = 0.195. From = 1.51 kg H z O/h m 2 Substituting into Eq. (9.6-2), {

Fig.9.5-lb,R c C

For the

from

.

=

Ls

7^

(

R

area

= 163

h

various values of

for

X

R

l/R

0.195

1.51

0.663

0.065

0.71

1.41

0.150

1.21

0.826

0.050

0.37

2.70

0.100

0.90

1.11

0.040

0.27

3.70

=0.195

X

=

A,

=

0.189

X2 =

(point C) to

+ A2 + A 3 =

is

made and

0.040

is

+

x 0.024)

(2.5

X

prepared:

is

1/R

In Fig. 9.7-1 a plot of l/R versus {

R

falling-rate period, reading values of

X

X

0.195)

(18.58X1.51)

Fig. 9.5- lb, the following table

from

-

399(0.38

*'~ X2)=

the area under the curve

determined: (1.18 x 0.056)

+

(0.84

+

0.075)

Substituting into Eq. (9.6-1), *'

dX

399

R

18.58

x2

The

9.7

B

total

time

is

2.63

+

(0.189)

4.06

=

6.69

h.

Falling-Rate Region

In certain special cases in the falling-rate region, the

/.

4.06 h

Calculation Methods for Special Cases in

(9.6-1),

=

equation

for the

time for drying, Eq.

can be integrated analytically.

Rate

linear in

is

a linear function of X.

If

both

X

l

X2

and

are less than

Xc

R = aX + b where a is the slope of the line and b is a dX. Substituting this into Eq. (9.6-1),

R

is

=

,

b

and R 2

>

dR L — =— — aA R R

= aX 2 +

j,

Eq.

(9.7-1) givesrfR

R,

s

In

aA

= aX +

(9.7-1)

a constant. Differentiating

R t

Since R,

and the rate

X over this region,

(9.7-2)

2

b,

x

l

-x

(9.7-3) 2

Substituting Eq. (9.7-3) into (9.7-2),

L S {X — X 2 ) L

A(R,-R

546

2)

R^

"r

(9.7-4) 2

Chap. 9

Drying of Process Materials

2.

Rate

is

a linear function through origin.

some

In

cases a straight line from the critical

moisture content passing through the origin adequately represents the whole falling-rate period. In Fig. 9.5- lb this

would be a

lack of

more

origin,

where the rate of drying

is

from C to £ at the origin. Often for made. Then, for a straight line through the

straight line

assumption

detailed data, this

is

directly proportional to the free moisture content,

R = aX Differentiating,

dX =

dR/a. Substituting into Eq.

of the line

is

R c/X c and ,

R

aA

R2

X = Xc

at

R = Rc

Rl

for

,

Ls

Xc

AR C = X c/X 2

Noting also that R c /R 2

(9.6-1),

""^ = ^,n^

aA The slope a

(9.7-5)

(9.7-6)

[

R — R c

In

(9.7-7)

,

AR r

X,

or

R = Rc

(9.7-9) Ac-

example rate

R

versus

Approximation of Straight Line for Falling-Rate Period but as an approximation assume a straight line of the through the origin from point X c to X = 0 for the falling-

9.7-2.

Repeat Example

X

9.7-1,

rate period.

Solution:

Sec. 9.7

Rc =

Calculation

1.51

kg

H 2 0/h- m and X c = 2

0.195.

Methods for Falling-Rate Drying Period

Drying

in the falling-

547

rate region

is

from

Xc 1

X = 0.040. Substituting into Eq. (9.7-8), LS X C X c 399(0.195) 0.195

to

2

0

~ AR C =

X2 ~

n

0.040

18.58(1.51)

4.39 h

This value of 4.39 h compares with the value of 4.06 h obtained 9.7-1 by graphical integration.

9.8

in

Example

COMBINED CONVECTION, RADIATION, AND CONDUCTION HEAT TRANSFER IN CONSTANT-RATE PERIOD

9.8A

Introduction

In Section 9.6B an equation was derived for predicting the rate of drying in the constant-rate period. Equation (9.6-7) was derived assuming heat transfer to the solid by

convection only from the surrounding

air to the drying surface. Often the drying is done an enclosure, where the enclosure surface radiates heat to the drying solid. Also, in some cases the solid may be resting on a metal tray, and heat transfer by conduction through the metal to the bottom of the solid may occur.

in

Derivation of Equation for Convection, Conduction,

9.8B

and Radiation In Fig. 9.8-1 a solid material being dried by a stream of air heat transfer to the drying surface

<1

where q c in

W

is

=

1c

+

<7k

+

is

total rate of

(9-8-1)

Ik

the convective heat transfer from the gas at

(J/s),

shown. The

is

is

the radiant heat transfer

T°C to the solid surface at TS °C TR to Ts in W (J/s), andg K is

from the surface at

qR the rate of heat transfer by conduction from the

bottom

in

W. The

rate of convective

-hot radiating surface iqji radiant

heat

gas

-

— H>y -

T

>

i

Qq convective

TR

KNA

drying surface

i

heat

ys.

gas

t7h.

HS

(surface)

nondrying surface ~y

conduction heat Figure

548

TS.

9.8-1.

Heat and mass transfer

in

drying a solid from the top surface.

Chap. 9

Drying of Process Materials

heat transfer

where A

is

where h R

is

similar to Eq. (9.6-3)

and

is

qc

=

the exposed surface area in

m

qK

=

is

as follows, where (T

hc 2 .

— TS)°C = (T — Ts

(T-Ts )A heat transfer

is

- TS)A

(9.8-3)

denned by Eq.

the radiant-heat-transfer coefficient

K, (9.8-2)

The radiant

h R (TR

)

(4.10-10).

(AY-

WOO/ = £(5.676) ^ L

K

V 100;

'r

TR

—

)—*-

(4.10-10)

's

Ts are in K.

For the heat transfer by conduction from by convection from the gas to the metal plate, then by conduction through the metal, and finally by conduction through the solid. Radiation to the bottom of the tray is often small if the tray is placed above another tray, and it will be neglected here. Also, if the gas temperatures are not too high, radiation from the top surface to the tray will be small. Hence, the heat by radiation should not be overemphasized. The heat by conduction is

Note

that in Eq.

(4.

10-10)

the bottom, the heat transfer

and

is first

qK

=U K (T-Ts)A

(9.8-4)

1

UK =

(9.8-5) l

+

/h c

Z A// fc

M +

z sl k s in

W/m- K,z s

m, and k s the solid thermal conductivity. The value (9.8-4) is assumed to be the same as in Eq. (9.8-2). The equation for the rate of mass transfer is similar to Eq. (9.6-5) and is

of/i c in Eq.

where

z

M

is

the metal thickness in

m, k M the metal thermal conductivity

the solid thickness in

NA =

ky

^ M

(Hs

- H)

(9.8-6)

A

(9.8-7)

A

Also, rewriting Eq. (9.6-6), q

Combining Eqs.

=

MA NA X

s

(9.8-1), (9.8-2), (9.8-3), (9.8-4), (9.8-6),

= ^ = JL aa

+ U K )(T-Ts] +

(»c

;.

s

hR[ TR

and

(9.8-7),

-T

s)

^_

=

s

This equation can be compared to Eq.

(9.6-7), which gives the wet bulb temperature and conduction are absent. Equation (9.8-8) gives surface temperature 7^ greater than the wet bulb temperature Tw Equation (9.8-8) must also intersect the saturated humidity line at 7^ and H s and Ts > Tw and H s > H w The equation must be solved by trial and error.

Tw when

radiation

.

.

,

To

facilitate solution of

(h s h c /k

The

ratio h c /k y

mately c s

in

Eq.

y

Eq.

m MB

a \

M B was shown

in the

+

it

can be rearranged (Tl) to the following:

uA (T _ hc

h, Ts]

+

{Tr hc

J

_

(9>g. 9)

7s)

wet bulb derivation of Eq. (9.3-18)

to

be approxi-

(9.3-6).

cs

Sec. 9.8

s

(9.8-8),

= (1.005 +

1.88//)10

3

J/kg

K

Combined Convection, Radiation, and Conduction Heat Transfer

(93-6)

549

EXAMPLE An

Constant-Rate Drying When Radiation and Convection Are Present

9£-l.

insoluble granular material wet with water

0.457 x 0.457

m

and 25.4

mm

being dried

is

The material

deep.

mm

25.4

is

pan

in a

deep

in the

mm

metal pan, which has a metal bottom with thickness z M = 0.610 having a thermal conductivity k M = 43.3 W/m K. The thermal conductivity of the solid can be assumed as k s = 0.865 W/m K. Heat transfer is by convection from an air stream flowing parallel to the top drying surface and the bottom •

•

metal surface at a velocity of

and humidity

H = 0.010

2

from steam-heated pipes whose surface temperature TR of the solid is e = 0.92. Estimate the rate of drying

direct radiation

93.3°C.

m/s and having a temperature of 65.6°C 0/kg dry air. The top surface also receives

6.1

H

kg

The emissivity

= for

the constant-rate period.

Some

Solution:

T =

of the given values are as follows

65.6°C, zM

zs

=

=

0.0254 m,

m

0.00061

e

fe

=

M=

ks

43.3,

H=

0.92,

=

0.865

0.010

velocity, temperature, and humidity of air are the same as Example and the convective coefficient was predicted as h c = 62.45 W/m 2 K. The solution of Eq. (9.8-9) is by trial and error. The temperature Ts will be above the wet bulb temperature of Tw = 28.9°C and will be estimated as 7^ = 32.2°C. Then l s = 2424 kJ/kg from the steam tables. To predict h R from Eq. (4.10-10) for e = 0.92, T, = 93.3 + 273.2 = 366.5 K, and T2 = 32.2 + 273.2 = 305.4 K,

The

9.6-3

•

K - (0.92X5.676) Using Eq.

(366 5/ -

^

)

t

;:

3 0 S 4/100 ''

30 s

- 7.96 W/m K 1

4

(9.8-5),

1

1

+ zJk M +

= From Eq.

22.04

W/m

2 •

z s /k s

1/62.45

+

0.00061/43.3

+

0.0254/0.865

K

(9.3-6), cs

=

(1.005

+

1.88HJ10

3

This can be substituted for {hJk y other knowns,

^^

=

+

=

+

(1

=

(1.005

=

1.024 x 10

MB

)

+

1.88

J/kg-K

into Eq. (9.8-9). Also, substituting

22.04/62.45X65.6

(7.96/62.45X93.3

1.353(65.6

3

x 0.010)10 3

- Ts

- Ts + )

- Ts

)

)

0.1275(93.3

- Ts

)

(9.8-10)

Ts assumed as 32.2°C, A s = 2424 x 10 J/kg. From the humidity Ts = 32.2°C, the saturation humidity H s = 0.031. Substituting into Eq. (9.8-10) and solving for Ts 3

For

chart for

,

(0.031- 0.010)(2424 x 10 3 ) 1

To

550

=

1.353(65.6

=

34.4°C

Q24 x 1Q3

- Ts>.+_ )

Chap. 9

0.1275(93.3

- r) Ts )

Drying of Process Materials

For the second trial, assuming that Ts = 32.5°C, Xs = 2423 x 10 3 and H s from the humidity chart at saturation is 0.032. Substituting into Eq. (9.8-10) while assuming that h R does not change appreciably, a value of

Ts =

32.8°C

is

obtained. Hence, the final value

is

32.8°C. This

greater than the wet bulb temperature of 28.9°C in

Example

is

3.9°C

9.6-3,

where

radiation and conduction were absent.

Using Eq.

(9.8-8),

(h c

+U K W-T

+

s)

hR

-T

(TR

5)

(3600)

J*

+

(62.45

-

22.04X65.6

32.8)

+

7.96(93.3

-

32.8)

(3600)

2423 x 10 3

=

kg/h-m 2

4.83

This compares with 3.39 conduction.

9.9

9.9A

kg/h-m 2

for

Example

no radiation or

9.6-3 for

DRYING IN FALLING-RATE PERIOD BY DD7FUSION AND CAPILLARY FLOW Introduction

no longer completely methods time of drying. In one method the actual rate of drying curve was

In the falling-rate period, the surface of the solid being dried

wetted, and the rate of drying steadily

were used to predict the

falls -with

is

time. In Section 9.7 empirical

graphically integrated to determine the time of drying.

In another

content

method an approximate

to the origin at

straight line

from the

critical

free

moisture

zero free moisture was assumed. Here the rate of drying

assumed to be a linear function of the defined by Eq. (9.5-3).

free

moisture content. The rate of drying

*=-^ A

was

R

is

(9.5-3)

at

When R

is

a linear function of

X in the falling-rate period, R = aX

where a

is

(9.7-5)

a constant. Equating Eq. (9.7-5) to Eq. (9.5-3),

R= —

—A — = aX

(9.9-1)

dt

Rearranging,

(9.9-2)

Ls

dt In

many instances, however, as mentioned briefly in movement in the falling-rate period is governed by

moisture

ment of

the liquid

by

moisture movement

liquid diffusion or will

be considered

by in

Section 9.5E, the rate of

the rate of internal movemovement. These two methods of more detail and the theories related to

capillary

experimental data in the falling-rate region.

Sec. 9.9

Drying

in

Falling-Rate Period by Diffusion

and

Capillary

Flow

SSI

9.9B

Liquid Diffusion of Moisture in Drying

When

liquid diffusion of moisture controls the rate of drying in the falling-rate period,

the equations for diffusion described as

X

kg

second law

in

Chapter 7 can be used. Using the concentrations

3 moisture/kg dry solid instead of concentrations kg mol moisture/m Fick's

free

,

for unsteady-state diffusion,

Eq. (7.10-10), can be written as

dX d X — — = D L —j 2

where D L

the liquid diffusion coefficient

is

This type of diffusion

(9.9-3)

dx 2

dt

'

inm 2 /h and x

is

distance in the solid in m.

often characteristic of relatively slow drying in nongranular

is

materials such as soap, gelatin, and glue, and in the later stages of drying of in clay,

A

wood, major

distribution

and other hydrophilic

paper, foods, starches,

textiles, leather,

analyzing diffusion drying data

difficulty in

not uniform throughout the solid at the start

is

a drying period at constant

During diffusion-type drying, the resistance to mass is usually very small, and the diffusion in the

rate precedes this falling-rate period.

transfer of water vapor

that the initial moisture

is

if

bound water

solids.

from the surface

Then

solid controls the rate of drying.

the moisture content at the surface

equilibrium value X*. This means that the free moisture content

X

is

at the

at the surface

is

essentially zero.

Assuming

that the initial moisture distribution

be integrated by the methods

—

X ~ X* = =— X — X* X '

f y

at

t

=

X= 0,

average

X* =

e

uniform

is

-D L l(7r/2x

2 t

+

)

^c

-9D L ,M2x

l

^ e -25D

+

)2

free

moisture content at time

equilibrium free moisture content,

h,

£

x

l

X = ^

=\

/

=

0,

Eq. (9.9-3)

L .(,/2x 1 )2

+

.

.

."J

may

(9,9.4)

initial free

drying only from the top

moisture content

the thickness of the slab

drying occurs'from the top and the bottom parallel faces, andxj if

at

Chapter 7 to give the following:

71

j

where

^

in

=

when

total thickness of slab

face.

assumes that D L is constant, but D L is rarely constant; it varies with moisture content, temperature, and humidity. For long drying times, only the first term

Equation

in Eq. (9.9-4)

is

(9.9-4)

significant,

and

becomes

the equation *

X

e

-DL.W2x,)i

(9.9.5)

K

[

Solving for the time of drying,

t

= -r±2 n

In this equation

if

the diffusion

In

DL

(9.9-6)

mechanism starts at AT = and rearranging,

Xc

,

then.^ =

Xc

.

Differen

tiating Eq. (9.9-6) with respect to time

dX _ It ~ Multiplying both sides by

n

2

DL X

(9.9-7)

4x1

— L s/A, L dX — — = n 4xL AD, x A 2

R = -

s

s

dt

Hence, Eqs.

(9.9-7)

times, the rate of drying diffusivity

552

and

and is

(9.9-8) state that

when

internal diffusion controls for long

directly proportional to the free

that the rate of drying

is

(9.9-8)

2

moisture

X

and the liquid

inversely proportional to the thickness squared.

Chap. 9

Drying of Process Materials

Or, stated as the time of drying between fixed moisture limits, the time varies directly as

The

the square of the thickness.

drying rate should be independent of gas velocity and

humidity.

EXAMPLE

Drying Slabs of Wood When Diffusion of Moisture Controls

9.9-1.

diffusion coefficient of moisture in a given wood is x 10" 6 m 2 /h (3.20 x 10" 5 ft 2 /h). Large planks of wood 25.4 thick are dried from both sides by air having a humidity such that the equilibrium moisture content in the wood is X* = 0.04 kgH z O/kg dry wood. The wood is to be dried from a total average moisture content of n = 0.29 to X, = 0.09. Calculate the time needed.

The experimental average

mm

2.97

X

The

Solution:

moisture

free

content

x 1000)

25.4/(2

=

The

half-slab

thickness x,

=

0.0127 m. Substituting into Eq. (9.9-6),

4x\ 1

X = X n — X* = 0.29 — 0.04 = i

X = X, - X* = 0.09 - 0.04 = 0.05.

0.25,

~

n

=

30.8 h

2

4(0.0127)

8*i

DL

n

2

X

"

2 7t

2

8

x

0.25

6 " n 2 x 0.05 (2.97 x 10" )

Alternatively, Fig. 5.3-13 for the average concentration in a slab can be

The ordinate E a = X/X t

used.

= D L t/x 2

value of 0.56

t

9.9C

Capillary

=

Movement

,

=

=

0.05/0.25

substituting,

0.20.

and solving

for

Reading off the plot a t,

2

x?(0.56)

=

~^T

(0.0127) (0.56)

2.97

x 10" 6

=m4h

of Moisture in Drying

Water can flow from regions of high concentrations to those of low concentrations as a result of capillary action rather than by diffusion if the pore sizes of granular materials are suitable.

The

capillary theory (PI) assumes that a

a void space between the spheres called pores. set

up by

As water

the interfacial tension between the water

driving force for

A

packed bed of nonporous spheres contains

moving

is

and

evaporated, capillary forces are solid.

These forces provide the

the water through the pores to the drying surface.

modified form of Poiseuille's equation for laminar flow can be used

in

with the capillary-force equation to derive an equation for the rate of drying

by capillary movement. the rate of drying

R

this period

same

is

the

will

If

the moisture

movement

conjunction

when

vary linearly with X. Since the mechanism of evaporation during

same as for the constant-rate drying period. The defining equation for the rate of drying

R _ For the

rate

R

is

as in the constant-rate period, the effects of the variables of the

drying gas of gas velocity, temperature of the gas, humidity of the gas, and so on, the

flow

follows the capillary-flow equations,

varying linearly with

hi

will

be

is

d

JL

(9.5-3)

~~A~dt

X given

previously,

R = Rc

X — X

(9.7-9)

c

t

Sec. 9.9

Drying

in

= ^s*c AR C

,

In

X — X

Falling-Rate Period by Diffusion

c

(9.7-8)

2

and

Capillary

Flow

553

We define

as the time

t

when X

= X2

and

L s — x^Ap s

=

where p s

solid density

kg dry solid/m

3

(9.9-9)

Substituting Eq. (9.9-9) and

.

X = X2

into Eq.

(9.7-8),

= -^7^

t

Substituting Eq. (9.6-7) for

Rc

In

(9.9-10)

,

=

t

(9.9-11)

Hence, Eqs. (9.9-10) and (9.9-11) state that when capillary flow controls falling-rate period, the rate of drying

of drying between fixed moisture limits varies directly as the thickness the gas velocity, temperature,

the

and humidity.

Comparison of Liquid Diffusion and Capillary Flow

9.9D

To

in

The time and depends upon

inversely proportional to the thickness.

is

determine the mechanism of drying

obtained of moisture content

experimental data

in the falling-rate period, the

various times using constant drying conditions are often

at

analyzed as follows. The unaccomplished moisture change, defined as the ratio of free moisture present

in

the solid after drying for

t

paper.

a straight line

If

hours to the

X/X c

present at the start of the falling-rate period,

B

obtained, such as curve

is

,

is

total free

moisture content

plotted versus time

in Fig. 9:9-1

on semilog

using the upper scale for

the abscissa, then either Eqs. (9.9-4)-(9.9-6) for diffusion are applicable or Eqs. (9.9-10)

and

(9.9-1 1) for capillary flow are applicable.

flow applies, the slope of the falling-rate drying line

If the relation for capillary

Fig. 9.9-1

of R c

which contains the constant drying

related to Eq. (9.9-10),

is

agrees with the experimental value of

value of

Rc

If the

of line

B

,

in

movement

the moisture

values o(

R c do

Fig. 9.9-1

in the

DL

(9.9-6)

(X/X c

shows for

a

)

is

plotted versus

curvature

X/X c < 0.6. When the

in

line,

which

movement

should equal

moisture contents, and an average value of D L In

B

in

value

is

X

by diffusion and the slope

is

— n 2 DJ4x\.

In actual practice,

usually less at small moisture contents than at large

is

the moisture range of interest.

— R c/x

.

by capillary flow.

not agree, the moisture

from Eq.

however, the diffusivity

is

Rc

R c The

l ps c> and if it constant drying period or the predicted

calculated from the measured slope of the

is

rate

A

is

usually determined experimentally over is shown as same plot as

plot of Eq. (9.9-4)

D L t/x\.

This

is

the

the line for values of X/X c between 1.0

show

experimental data

that

the

line A,

where ln(X/X,) or

Fig. 5.3-13 for a slab

and

and

line

movement

0.6

and a straight

of moisture follows the

diffusion law, the average experimental diffusivities can be calculated as follows for different concentration ranges.

A

value

oiX/X c

is

chosen

at 0.4, for

example.

experimental plot similar to curve B, Fig. 9.9-1, the experimental value of

From

curve

A

at

substituting the

X/X c = known

0.4, the theoretical

values of

value of D L over the range

X/X c =

t

and

.x,

1.0-0.4

is

value of (D L t/x]) lhcol

is

t

is

From an obtained.

obtained. Then, by

into Eq. (9.9-12), the experimental average

obtained.

(9.9-12)

554

Chap. 9

Drying of Process Materials

This

is

repeated for various values

olX/X c Values ofD L obtained [oxX/X c > .

0.6 are in

error because of the curvature of line A.

EXAMPLE 9.9-2. Diffusion Coefficient in the Tapioca Root Tapioca flour is obtained from drying and then milling the tapioca root. Experimental data on drying thin slices of the tapioca root 3 mm thick on both sides in the falling-rate period under constant drying conditions are tabulated below. The time t = 0 is the start of the falling-rate period. X/X c

t

(h)

X/X c

t

XjX c

(h)

t

(h)

0

0.55

0.40

0.23

0.94

0.80

0.15

0.40

0.60

0.18

1.07

0.63

0.27

0.30

0.80

1.0

It has been determined that the data do not follow the capillary-flow equation but appear to follow the diffusion equation. Plot the data as X/X c versus t on semilog coordinates and determine the average diffusivity of the moisture up to a value o(X/X c = 0.20.

Solution: £

In Fig. 9.9-2 the data are plotted as

X/Xc

on the log

on a linear scale and a smooth curve drawn through the data.

Figure

scale versus

AtX/X c =

Plot of equations for falling-rate period : (A) Eq. (9.9-4) for moisture diffusion, (B) Eq. (9.9-10) for moisture movement by

9.9-1.

movement by

(From R. H. Perry and C. H. Chilton, Chemical EnHandbook, 5th ed. New York: McGraw-Hill Book Company, 1973. With permission.)

capillary flow.

gineers

Sec. 9.9

Drying

in

Falling-Rate Period by Diffusion and Capillary Flow

555

O.L'0

oil

0.4

0.6

Time Figure

0.20, a value of

L5

m

t

=

0.8

1.2

(h)

t

Plot of drying data for

9.9-2.

1.0

Example

The

1.02 h is read off the plot.

9.9-2.

value ofx 1

=

3

mm/2 =

mm for drying from both sides. From Fig. 9.9-1, line A, for X/X c = 0.20,

L f/x

2 ) lhcor

=

0.56.

Then

substituting into Eq. (9.9-12),

EQUATIONS FOR VARIOUS TYPES OF DRYERS

9.10

9.10A

Through Circulation Drying

in

Packed Beds

For through circulation drying where the drying gas passes upward or downward through a bed of wet granular period of drying

may

both a constant-rate period and a falling-rate

solids,

Often the granular solids are arranged on

result.

the gas passes through the screen

a screen

and through the open spaces or voids between

so that

the solid

particles.

/.

Derivation of equations.

assumed, so the system granular solids. gas flow of

We

G kg dry

is

To

derive the equations for this case,

The drying

adiabatic.

will be for

no heat

losses will

unbound moisture

in the

bed of uniform cross-sectional area A m 2 where a cross section enters with a humidity of ff L By a material

shall consider a

gas/h

m

2

be

wet

,

.

balance on the gas at a given time, the gas leaves the bed with a humidity

amount of water removed from

the

bed by the gas

is

H2

.

The

equal to the rate of drying at this

time.

R = G(H 2 - HJ where R

=

kg

H 2 0/h m 2 •

cross section and

In Fig. 9.10-1 the gas enters at

temperature

556

T and humidity

H both

T

x

(9.10-1)

G =

kg dry air/h

H

and leaves

and

l

vary through the bed.

Chap.

9

•

m2 at

cross section.

T2 and H 2

Making

.

Hence, the

a heat balance over

Drying of Process Materials

the short section dz

m of the bed,

dq=-Gcs AdT where A

=m

humid heat in kg/s

•

m

2

cross-sectional area, q

and

a

the heat-transfer rate in

is

of the air-water vapor mixture in Eq. (9.3-6).

2 .

The

Note

W

that

G

and cs

(J/s),

in this

the

is

equation

is

heat-transfer equation gives

dq where

(9.10-2)

=

ha A dz{T

- Tw

(9.10-3)

)

Tw = wet bulb temperature of solid, h is the heat-transfer coefficient in W/m 2 K, 2 3 bed volume. Equating Eq. (9.10-2) to (9.10-3), is m surface area of solids/m •

rearranging, and integrating,

ha

Gcs

dT

dz= -

(9.10-4)

T — T,w

0

^ = ln^

(9.10-5)

where z = bed thickness = m. For the constant-rate period of drying by air flowing parallel to (9.6-11) was derived. Ls {X — X 2 ) L S X W {X — X 2 ) i

l

Ah(T-Tw Using Eq.

(9.9-9)

a surface,

~ Ak y M B (H w —

)

(9 6-11)

H)

and the definition of a, we obtain

^7

A

=^

(9.10-6)

a

X

X

Xc

= c for drying to 2 the equation for through circulation drying in the constant-rate period. Substituting Eq. (9.10-6) into (9.6-11) and setting

t

=

Ps

~ Xc = ah(T-Tw

manner, Eq.

p s (X

)

)

In a similar

Eq.

aky

1

-X c

,

we obtain

)

(9

M B (H W — H)

(9.7-8) for the falling-rate period,

which assumes that

R

is

proportional to X, becomes, for through circulation drying,

£

Figure

9.10-1.

=

Ps l w

X c In (XJX) = PsX c

ah{T-Tw

)

ak

Heat and material balances

in

a through circulation dryer

in

M

B

\n

(X c/X)

{H w

—

H)

T2 #2 ,

a packed bed.

T+dT,

H+dH

dz T,

T

Sec. 9.10

Equations for Various Types of Dryers

H

557

Both Eqs.

and

(9.10-7)

however, hold only for one point

(9.10-8),

T

9.10-1, since the temperature

mean temperature

similar to the derivation in heat transfer, a log as

an approximation

(T

- TW

)

for the

whole bed

In

in place of

T — Tw

- Tw - (T - Tw [(T, - TW )/{T2 - T„)]

(T\

=

LM

)

2

T2 from Eq. (9.10-5) (T

and

2

- TW )/(T2 - Tw

In [(T,

(9.10-8).

~T

7*1

)

in Fig.

manner

difference can be used

in Eqs. (9.10-7)

Substituting Eq. (9.10-5) for the denominator of Eq. (9.10-9)

value of

bed

in the

of the gas varies throughout the bed. Hence, in a

(9.10-9))~]

and also substituting

the

into (9.10-9),

- TW

)

= (7\

LM

- Tw)0. -

e

- hatlGcs )

(9.10-10)

haz/Gc s

Substituting Eq. (9.10-10) into (9.10-7) for the constant-rate period and setting

x

l

=

z,

t

Ps ^-w x i(X

=

Gcs {Tx Similarly, for the falling-rate period

— %c - e -haxilGcs'

i

- Tw \\

an approximate equation

~ Gc (T! - T w \\ s A

major

with the use of Eq. (9.10-12)

difficulty

easily estimated. Different

is

e

is

obtained.

(9.10-12)

- haxllGcs )

that the critical moisture content

is

not

forms of Eqs. (9.10-11) and (9.10-12) can also be derived, using

humidity instead of temperature

2.

(9.10-11)

(Tl).

For through circulation drying, where the gases pass

Heat-transfer coefficients.

through a bed of wet granular

following equations for estimating h for

solids, the

adiabatic evaporation of water can be used (Gl, Wl).

h

=

0.151

(SI)

D n G, 0.11

h

h

=

=

D o. S

(9.10-13)

<

(9.10-14)

(English)

D o.*i

0.214

> 350

(SI) .

DP

G,

350

(English)

0.15 D'l

where h

is

in

W/m 2

K,

Dp

the particle in the bed, G, viscosity in

kg/m

-

h.

is

is

diameter

in

m

of a sphere having the same surface area as

the total mass velocity entering the bed in kg/h

In English units, h

is

btu/h

2 •

ft

°F,

Dp

is ft,

G

t

is

lb^h

m2

,

2 ft

,

and and

p. is

y. is

lbjffh. Geometry factors in a bed. To determine the value of
6(1

in

e)

(9.10-15)

D„

558

Chap. 9

Drying of Process Materials

where

e is the

void fraction in the bed. For cylindrical particles,

4(i-

where

Dc is

diameter of cylinder in

+ asp c )

txft

m and h

is

length of cylinder in m.

use in Eqs. (9.10-13) and (9.10-14) for a cylinder

same surface area

The value of

D

to

the diameter of a sphere having the

as the cylinder, as follows

D p = {D 4.

is

c

+

h

0.5Dl)

112

(9.10-17)

Equations for very fine particles. The equations derived for the constant- and fallingpacked beds hold for particles of about 3-19 in diameter in shallow

mm

rate periods in

beds about 10-65

mm) and

mm

thick (Tl, Ml).

bed depth greater than

1

1

For very

mm,

fine particles of

10-200 mesh (1.66-0.079

the interfacial area a varies with the moisture

content. Empirical expressions are available to estimate a and the mass-transfer coefficient (Tl,

A 1).

EXAMPLE

9.10-1. Through Circulation Drying in a Bed granular paste material is extruded into cylinders with a diameter of 6.35 and length of 25.4 mm. The initial total moisture content X n = 1.0 kg

A

mm H

and the equilibrium moisture is X* = 0.01. The density of 1602 kg/m 3 (100 lb^/ft 3 ). The cylinders are packed on a screen to a depth of x t = 50.8 mm. The bulk density of the dry solid in the = 0.04 kgH z O/kg bed is p s = 641 kg/m 3 The inlet air has a humidity dry air and a temperature Tt = 121. 1°C. The gas superficial velocity is 0.81 m/s and the gas passes through the bed. The total critical moisture content is X lC = 0.50. Calculate the total time to dry the solids to X, = 0.10 kg H 2 0/kg dry solid. 2

0/kg dry

solid

the dry solid

is

.

For the

Solution:

solid,

X = X n - X* = 1.00 - 0.01 = 0.99 X c = X lC - X* = 0.50 - 0.01 = 0.49 X = X, - X* = 0.10 - 0.01 = 0.09 x

kg

H

z

O/kg dry

solid

and tf, = 0.04 kg H z O/kg dry air. The wet 47.2°C and H w = 0.074. The solid temperature is at Tw if radiation and conduction are neglected. The density of the entering air at 121.1°C and 1 atm is as follows. For the

gas, T,

bulb temperature

vH

P

The mass

=

=

(2.83

=

1.187

=

'^

x 10" 3 3

4.56 x 10"

ft04

=

x 0.04X273

0.876 kg dry air

+

121.1)

(93-7)

+

H 0/m 3 2

is

= 0.811(3600X0.876)^^ =

2 2459 kg dry air/h-m

H = 0.040 and l

H

the outlet will be less than 0.074, an approxiof 0.05 will be used to calculate the total average mass

The approximate average G,

3

/kg dry air

velocity of the dry air

Since the inlet mate average

Sec. 9.10

m

+

87

G = pp( +° 10 004)

velocity.

121. 1°C

Tw =

=

2459

+

2459(0.05)

G,

=

is

2582 kg

Equations for Various Types of Dryers

air

+

H 2 0/h m 2 •

559

.

For the packed bed, the void fraction

e

is

calculated as follows for

1

m3

A total of 641 kg dry solid is present. The 1602 kg dry solid/m 3 solid. The volume of the 3 3 solid. Hence, solids in 1 m of bed is then 641/1602, or 0.40 e = 1 - 0.40 = 0.60. The solid cylinder length h — 0.0254 m. The diameter D c = 0.00635 m. Substituting into Eq. (9.10-16), of bed containing solids plus voids.

density of the dry solid

is

m

_ ~

a

= To

-

4(1

E)(h

+

D

h

m

283.5

c

2

_ ~

0.5£ rel="nofollow"> c )

Dp

0.6)[0.0254

+

0.5(0.00635)]

0.00635(0.0254)

surface area/m

calculate the diameter

-

4(1

3

bed volume

of a sphere with the

same area

as the cylinder

using Eq. (9.10-17),

D p = {Dc :h + 0.5D 2 )" 2 =

+

[0.00635 x 0.0254

0.5(0.00635)

2

]" 2

m

= 0.0135

The bed thickness x, = 50.8 mm = 0.0505 m. To calculate the heat-transfer coefficient, the Reynolds number is first calculated. Assuming an approximate average air temperature of 93.3°C, the -5 viscosity of air is n = 2.15 x 10 kg/nvs = 2.15 x 10~ 5 (3600) = 7.74 x " 2 kg/m h. The Reynolds number is 1 0 •

/VRe

Using Eq.

(9.

D p G,

~

u

10- 13),

«m^8T\o.si 0.151(2582)°

-0.59

k

For

Tw =

steam cs

= =

To (9.10-1

t

=

=

151

°-

r,

=

-

i

(0.0135)°

-

=

1.005

+

1.88H

1.099 x 10

3

=

+

1.005

1.88(0.05)

=

air-K

J/kg-K

and

G = 2459/3600 =

p s X w x {X Gcs (T, - r„Xl l

0.6831 kg/s

•

m

2 ,

- Xc -e'"-^) )

l

6

641(2.389 x 10 )(0.0508)(0.99

(0.683X1.099 x 10 X121.1 s

=

-

47.2)[1

-

e"'

90

9 "

283

-

0.49)

5 x

*

»-o»»" 103

>]

0.236 h

For the time of drying

=

1.099 kJ/kg dry

calculate the time of drying for the constant-rate period using Eq.

1)

850

'

>.

3

=

W/m2 K

9 °- 9

6 w = 2389 kJ/kg, or 2.389 x 10 J/kg (1027btu/lbJ, from The average humid heat, from Eq. (9.3-6), is

47.2°C,

tables.

0.0135(2582)

450 "7.74 x 10- 2 ~

Ps'-wX^Xc

In

GcsiT.-T^l -

for the falling-rate period, using Eq. (9.10-12),

(XJX) e-"°*"^)

6 641(2.389 x 10 X0.0508X0.49) 3

(0.6831X1.099 X 10 X121.1 -47.2)[1

=

1412

s

=

e

In (0.49/0.09) -(90.9K 283.Sx 0.0508)/(0.683x 1.099

x 103)-]

0.392 h total

560

-

time

t

=

0.236

+

0.392

=

0.628 h

Chap. 9

Drying of Process Materials

9.10B

Tray Drying with Varying Air Conditions

For drying

in a

compartment or tray dryer where

the air passes in parallel flow over the

Heat and material balances must be made to determine the exit-gas temper-

surface of the tray, the air conditions do not remain constant. similar to those for through circulation

ature and humidity. In Fig. 9.10-2 air T[

and humidity

dry

air

flow

is

shown passing over a tray. It enters having a temperature of at T2 and H 2 The spacing between the trays is b m and

is

and leaves

G kg dry

air/s

length dL, of tray for a section

•

1

m 2 cross-sectional m wide, dq

The

heat-transfer equation

= Gc s (l

x

b)

area. Writing a heat balance over

dT

a

(9.10-18)

is

dq

=

x dL,)(T

h{l

- Tw

(9.10-19)

)

Rearranging and integrating, hL,

=

In

Gc*b

T7

(9.10-20)

-Tw

Denning a log mean temperature difference similar to Eq. (9.10-10) and substituting and (9.7-8), we obtain the following. For the constant-rate period,

into Eqs. (9.6-11)

t

x iPs

=

Gcs b(T, For the

falling-rate period,

Xc -<:-"«)

L X w (X — 7^X1

an approximate equation

t

=

Gcs 6(T, - T^Xl

9.1

OC

/.

Simple heat and material balances.

)

l

t

-

-e

is

(9.10-21)

obtained.

(9.10-22)

-hLt/Gcsb

Material and Heat Balances for Continuous Dryers

In

Fig. 9.10-3 a flow

diagram

is

given for a

continuous-type dryer where the drying gas flows countercurrently to the solids flow.

The

solid enters at a rate of

Figure

Sec. 9.10

L s kg

9.10-2.

dry solid/h, having a

Heat and material balances

Equations for Various Types of Dryers

free

in

moisture content

X

l

and a

a tray dryer.

561

Figure

Process flow for a countercurrent continuous dryer.

9.10-3.

X2

Ts2.

leaves at X 2 and TS2 The gas enters at a rate G kg dry air/h, having dry air and a temperature of TG2 The gas leaves at Tcl andH 0/kg 2 2 material balance on the moisture,

temperature

humidity

H

For a

TS1 kg

.

It

a

.

H

.

t

GH 2 + LS X^ = GH + L s Xj

.

(9.10-23)

l

For a heat balance a datum of T0 °C is selected. A convenient temperature is 0°C The enthalpy of the wet solid is composed of the enthalpy of the dry solid plus

(32°F).

The

that of the liquid as free moisture.

of the gas H'G in kJ/kg dry air

H'G

where the

?.

0 is the latent

humid

=

heat of water at

cs

c pS

is

(TG

-T )+

usually neglected.

is

The enthalpy

=

Hl 0

0

T0 °C,

(9.10-24)

2501 kJ/kg (1075.4 btu/lbj

=

1.005

+

c„s(Ts

in

0°C, andc s

1.88//

is

(9.3-6)

where (Ts

- T0 )°C

- T0 + XcpA (Ts - T0 )

kJ/kg

H z O-K.

=

(Ts

- T0

)

K,

is

(9.10-25)

)

K and c A is the heat The. heat of wetting or adsorption is

the heat capacity of the dry solid in kJ/kg dry solid

capacity of liquid moisture

at

K.

of the wet solid H's in kJ/kg dry solid, H's

where

cs

heat, given as kJ/kg dry air

The enthalpy

heat of wetting

is

•

neglected.

A heat balance on the dryer

is

GH'G2 + Ls H'Si = GH'01 + L s H'S2 + Q where Q added, Q

is is

the heat loss in the dryer in kJ/h.

(9.10-26)

For an adiabatic process

Q =

0,

and

if

heat

is

negative.

EXAMPLE

9.10-2. Heat Balance on a Dryer continuous countercurrent dryer is being used to dry 453.6 kg dry solid/h containing 0.04 kg total moisture/kg dry solid to a value of 0.002 kg total moisture/kg dry solid. The granular solid enters at 26.7°C and is to be discharged at 62.8°C. The dry solid has a heat capacity of 1.465 kJ/kg K, which is assumed constant. Heating air enters at 93.3°C, having a humidity of 0.010 kg H 2 0/kg dry air, and is to leave at 37.8°C. Calculate the air flow rate and the outlet humidity, assuming no heat losses in the dryer.

A

Solution:

The flow diagram

453.6 kg/h dry^solid, c pS

=

is

given in Fig. 9.10-3. For the solid,

1.465 kJ/kg dry solid

•

K,

Ls =

X = 0.040

kg H 2 0/kg 62.8X, X 2 =

1

dry solid, c pA =4.187 kJ/kg H 2 0-K, TS1 = 26.7°C, Ts2 = 0.002. (Note that X values used are X, values.) For the gas, TC2 = 93.3°C, H 2 = 0.0 10 kg H,0/kg dry air, and TC1 = 37.8°C. Making a material balance on the moisture using Eq. (9.10-23),

GH + LS X, = GH + LS X + 453.6(0.040) = GH + 453.6(0.002) 2

C(0.010)

562

l

l

Chap. 9

2

(9.10-27)

Drying of Process Materials

For the heat balance, the enthalpy of the entering gas at 93.3°C using 0°C as a datum is, by Eq. (9.10-24), AT°C = AT K, and X 0 = 2501 kJ/kg, from the steam tables, ^"02

For the

cs

=

[1.005

=

120.5 kJ/kg dry air

(TC2

)

+

-

1.88(0.010)](93.3

+

0)

0.010(2501)

exit gas,

#Gi

For

— T0 + H 2 X 0

=

=c s (TG1

-T

=

+

(1.005

0)

+

// 1 A 0

1.88// 1 K37.8

- 0) +

77,(2501)

=

37.99

4-

2572/7,

the entering solid using Eq. (9.10-25),

H'si

H'S2

- T0 + X lCpA (Tsl - T0

=

c pS

=

1.465(26.7

=

c pS {TS2

=

1.465(62.8

(Tsl

)

- 0) +

)

0.040(4.187X26.7

- T0 + X 2 c pA (TS2 - T0 )

- 0) +

-

0)

= 43.59

-

0)

=

kJ/kg dry solid

)

0.002(4.187X62.8

92.53 kJ/kg

Substituting into Eq. (9.10-26) for the heat balance with

Q=

0 for no heat

loss,

G( 120.5)

+

.453:6(43.59)= G(37.99

+

2572/7,)

+

453.6(92.53)

+

0

(9.10-28)

Solving Eqs. (9.10-27) and (9.10-28) simultaneously,

G = 2.

1

Air recirculation

166 kg dry air/h

many

In

in dryers.

77,

=

0.0248 kg

dryers

it

is

H 2 0/kg

dry air

desired to control the wet bulb

temperature at which the drying of the solid occurs. Also, since steam costs are often

important

in

heating the drying

air,

recirculation of the drying air

reduce costs and control humidity. Part of the moist hot

air

is

sometimes used to

leaving the dryer

is

and combined with the fresh air. This is shown in Fig. 9.10-4. Fresh air having a temperature Tc and humidity 77, is mixed with recirculated air atTC2 and H 2 to give air at T03 and 77 3 This mixture is heated to TG4 with 77 4 = 77 3 After recirculated (recycled)

,

.

.

TG2

and a higher humidity 77 2 The following material balances on the water car. be made. For a water balance on

drying, the air leaves at a lower temperature

.

recirculated air (6)

wet solid

dry solid

x l' TS1 Figure

Sec. 9.10

9. 10-4.

Process flow for air recirculation

Equations For Various Types of Dryers

in

drying.

563

the heater, noting that

H6 = H = H s

G,H + G 6 H 2 = l

Making a water balance on

+ G 6 )tf 4

(G,

+ G 6 )H 2 + L S C 2

manner heat balances can be made on

In a similar

(9.10-29)

the dryer,

+ G 6 )H A + LS X,=

(G,

(G 1

(9.10-30)

the heater and dryer

and on the

overall system.

Continuous Countercurrent Drying

9.10D

/.

Drying continuously

Introduction and temperature profiles.

offers a

number of ad-

vantages over batch drying. Smaller sizes of equipment can often be used and the product has a

more uniform moisture

content. In a continuous dryer the solid

the dryer while in contact with a

moving gas stream

may

that

is

moved through

flow parallel or counter-

current to the solid. In countercurrent adiabatic operation, the entering hot gas contacts the leaving solid, which has been dried. In parallel adiabatic operation, the entering hot

gas contacts the entering wet solid. In Fig. 9.10-5 typical temperature profiles of the gas

TG

and the

solid

Ts

are

shown

continuous countercurrent dryer. In the preheat zone, the solid is heated up to the wet bulb or adiabatic saturation temperature. Little evaporation occurs here, and for for a

low-temperature drying

zone

this

usually ignored. In the constant-rate zone,

is

bound and surface moisture are evaporated and

essentially constant at the adiabatic saturation temperature

The

convection.

rate of drying

I,

un-

the temperature of the solid remains if

heat

is

transferred by

would be constant here but the gas temperature

is

X

changing and also the humidity. The moisture content falls to the critical value c at the end of this period. In zone II, unsaturated surface and bound moisture are evaporated and the solid is value X 2 The humidity of the entering gas The material-balance equation (9.10-23) may

dried to

its final

rises to

Hc

.

.

entering zone

II is

H

2

be used to calculate

and

Hc

it

as

follows.

L S{X C where L s

is

kg dry solid/h and G

is

zone

-X

2

)= G(H C - H 2

(9.10-31)

)

kg dry gas/h.

zone

I,

constant rate

II,

falling rate

c o

TSl Xy Ts ,

(solid)

T G 7,

H2

TS7<

X2

Jsi. *c

Distance through dryer FIGURE

564

9.10-5.

Temperature

profiles for a continuous countercurrent dryer.

Chap. 9

Drying of Process Materials

2.

The

Equation for constant-rate period.

zone

I

in this

rate of drying in the constant-rate region in

would be constant if it were not for the varying gas conditions. The section is given by an equation similar to Eq. (9.6-7).

R= The time

for

drying

is

ky

M£H W -H) = ^-(TG - Tw

m 2 /kg

where A/Ls is the exposed drying surface (9. 10-33) and (G/L s ) dH for dX,

t=°(h LS \A

G = kg

dry

Ls = kg dry

air/h,

M

and

Xc

.

(9.10-33)

dH Hw — H

B j He

solid/h,

x

dry solid. Substituting Eq. (9.10-32) into

1

ky

X

dX R

Xc

where

(9.10-32)

)

given by Eq. (9.6-1) using limits between '*«

rate of drying

(9.10-34)

= m 2 /kg

and A/Ls

dry solid. This can be

integrated graphically.

For

where

the case

Tw

Hw

or

integrated.

constant for adiabatic drying, Eq. (9.10-34) can be

is

G hi L S \A

The above can be modified by use

H w — Hr H w -H

1

ky

of a log

MB

In

mean humidity

— H c — {H w — H l(H w - HC)/(H W - HS

(H w In

)

x

(9.10-35)

l

)

difference.

H — Hc [_{H W H C)/(H W - H l

In

Substituting Eq. (9.10-36) into (9.10-35), an alternative equation

t=

— — ky

From

Eq.

(9.

10-3

1),

Hw

is

(9.10-37)

H c can be calculated as follows.

Equation for falling-rate period.

occurs,

t

M B AH LM

Hc = H 2 + ^ 3.

obtained.

H - Hc

1

\

is

(9.10-36) t )]

(X c

- X2

(9.10-38)

)

For the situation where unsaturated surface drying is directly dependent upon

constant for adiabatic drying, the rate of drying

X as in Eq. (9.7-9), and Eq. (9.10-32) applies. R = Rc

—=

ky

M B{H w

H)

(9.10-39)

Substituting Eq. (9.10-39) into (9.6-1), Xc

KM B Substituting

G dH/L s

G

for

dX and(H -

H

2

=

Again, to calculate

Sec. 9.10

G Ls

L

\Ajk ,

(

-

X y

M B (H w - H

(9.10-40)

Hw

H)X

+ X2

HC

fL

Hc

x2

)G/L S

-y-'B J h x

t

dX

mm

for

X,

dH - h 2 )g/ls + * j

1

2

)G/LS +

Xz

In

(9.10-41)

H X {H W - H c %c(H w ~ 2

2)

(9.10-42)

)

Eq. (9.10-38) can be used.

Equations For Various Types of Dryers

565

These equations for the two periods can also be derived using the and temperatures instead of humidities.

part of Eq.

last

(9.10-32)

FREEZE DRYING OF BIOLOGICAL MATERIALS

9.11

9.11

A

Introduction

Certain foodstuffs, pharmaceuticals, and biological materials, which

even to moderate temperatures

in

may

ordinary drying,

usually frozen by exposure to very cold

be dried

is

removed

as a

it is

not be heated

In freeze drying the water

air.

is

a-vacuum chamber After removed by mechanieal^iJuum pumps or steam

vapor by sublimation from the frozen material

the moisture sublimes to a vapor,

may

be freeze-dried. The substance to

in

jet ejectors.

As a

rule, freeze

drying method.

drying produces the highest-quality food product obtainable by any

A prominent

factor

is

the structural rigidity afforded by the frozen

substance when sublimation occurs. This prevents collapse of the remaining porous

When

structure after drying. its

water

is

added

rehydrated product retains

later, the

original structural form. Freeze drying of biological

advantage of little

loss

of flavor and aroma.

much

in

of

also has the

The low temperatures involved minimize

degradative reactions which normally occur freeze drying

and food materials

the

ordinary drying processes. However,

an expensive form of dehydration

for foods because of the slow drying vacuum. Since the vapor pressure of ice is very small, freeze drying requires very low pressures or high vacuum. If the water were in a pure state, freeze drying at or near 0°C (273 K) at a pressure of 4580 /im (4.58 mm Hg abs) could be performed. (See Appendix A. 2 for the properties of ice.) However, since the water usually exists in a solution or a combined state, the material must be cooled below 0°C to keep the water in the solid phase. Most freeze drying is done at — 10°C (263 K) or lower at pressures of about 2000

rate

/zm or

9.1

is

and the use

IB

of

less.

Derivation of Equations for Freeze Drying

In the freeze-drying process the original material

is

composed

of a frozen core of

As the ice sublimes, the plane of sublimation, which started at the outside surface, recedes and a porous shell of material already dried remains. The heat for the latent heat of sublimation of 2838 kJ/kg (1220 btu/lb m ice is usually conducted inward through the layer of dried material. In some cases it is also conducted through the frozen material.

)

layer

from the

rear.

The vaporized water vapor

material. Hence, heat

and mass

In Fig. 9.11-1 a material being freeze-dried tion,

transferred through the layer of dried

is

transfer are occurring simultaneously.

pictured. Heat

is

by conduction to the

ice layer. In

some

cases heat

may

frozen material to reach the sublimation front or plane.

long enough so that the final moisture content

degradation of the

final

dried food and reached

material on storage. in

the frozen food

is

is

then transferred

also be conducted through the

The

total drying time

below about 5 wt

%

to

must be prevent

The maximum temperatures reached

must be low enough

minimum. The most widely used freeze-drying process

566

by conduction, convec-

and/or radiation from the gas phase reaches the dried surface and

is

to

in the

keep degradation to a

based upon the heat of sublimation

Chap. 9

Drying of Process Materials

conduction

conduction

ice front

Figure

Heat and mass

9.1 1-1.

transfer in freeze drying.

being supplied from the surrounding gases to the sample surface.

Then

transferred by conduction through the dried material to the ice surface.

model by Sandall

The

et al. (SI) is

shown

the heat

A

is

simplified

in Fig. 9.1 1-2.

heat flux to the surface of the material in Fig. 9.1 1-2 occurs by convection and in

the dry solid

by conduction to the sublimation

surface.

The heat

flux to the surface

is

equal to that conducted through the dry solid, assuming pseudo steady state.

q

where q

heat flux in

is

W

=

(J/s),

h(Te

h

is

- T =

T

is 3

temperature of the sublimation front or in

- Tf)°C =

W/m

{Ts

fC,

- Tf

In a similar

)

and

AL

is

NA

is

(9.11-1)

W/m 2

Te

is

surface temperature of the dry solid in °C, 7}

is

ice layer in °C,

k

is

•

K,

thermal conductivity of the

the thickness of the dry layer in m.

Note

that

manner, the mass flux of the water vapor from the sublimation front

flux of

{Ts

K.

NA = where

- 7»

external heat-transfer coefficient in

external temperature of the gas in °C,

dry solid

[TM

s)

water vapor

(p fw

in

kgmol/s

•

ps J

m2

,

=

kg

k (p s „ g

is

-

Pe

J

is

(9.11-2)

external mass-transfer coefficient in

L 2

FIGURE

Sec. 9.1!

9.1 1-2.

Model for uniformly

retreating ice from in freeze drying.

Freeze Drying of Biological Materials

567

kg mol/s

•

m2

•

atm, p sw

temperature

and p f „

in the

partial pressure of water

is

partial pressure of water

vapor

dry layer, D'

is

vapor

at the surface in atm,

bulk gas phase

in the external

in

an average effective diffusivity

atm,

in the

T

is

pew

is

the average

dry layer in

m 2 /s,

the partial pressure of water vapor in equilibrium with the sublimation ice

is

front in atm.

Equation

(9.1 1-1)

can be rearranged to give

*~whm< T'- T

(<m - 3)

')

Also, Eq. (9.11-2) can be rearranged to give

The

and k g

by the gas~veIocitieS%Td c!5ara5ferTstics of the Te and p ew are set by the external operating conditions. The values of k and D' are determined by the nature of the dried material. The heat flux and mass flux at pseudo steady state are related by coefficients h

are determined

The values of

dryer and hence are constant.

= AH,N A

q

where

AH

S

is

determined by

(9.11-5)

the latent heat of sublimation of ice in J/kg mol. Also, p fw is uniquely T since it is the equilibrium vapor pressure of ice at that temperature; or f

,

Pf„=AT,) Substituting Eqs.

(9.1 1-3)

and

TJhTKLTk Also, substituting Eqs.

First, the

{T <

~ T'> = AH

<

and

into (9.1

^-T

As Te and, hence, Ts be reached.

(9.11-4) into (9.1 1-5),

(9.1 1-1)

AlA

(9.11-6)

'>

(9.1 1-4)

=

^W 9

~

+ RT AL/D'

l/fc.

iP '~

are raised to increase the rate of drying, 7^

(9 -

U

-

7)

(9

n

-

8)

1-5),

+ RT AL/D'

outer surface temperature

^

'

P<J

two

limits

may

'

possibly

cannot go too high because of thermal

damage. Second, the temperature Tf must be kept well below the melting point. For the where k/AL is small compared to kg and D'/RT AL, the outer-surface temperature limit will be encountered first as Ts is raised. To further increase the drying rate, k must be raised. Hence, the process is considered to be heat-transfer-controlled. Most commercial freeze-drying processes are heat-transfer-controlled (Kl).

situation

In order to solve the given equations, free

AL

is

related to x, the fraction of the original

moisture remaining.

AL=(l-x)| The

rate of freeze drying can be related

NA = where

568

MA

is

(9.11-9)

loN A by L

\

(

dx

2

M A Vs

\

dt)

molecular weight of water,

Vs

is

(9.11-10)

the

volume of solid material occupied by a

Chap. 9

Drying of Process Materials

unit

dry

kg of water initially (Vs = \/X 0 p s), X 0 is initial and p s is bulk density of dry solid inkg/m 3

solid,

Combining

Eqs. (9.1

1-3), (9-11-5), (9.1 1-9),

L AH, (

and

dx\

free

moisture content in

kgH z O/kg

.

we obtain,

(9.1 1-10),

for heat transfer,

1

iwX-t)- mW^L/ik (T <

Similarly for

mass

/

1

dx\

WJ {"di)~ S

1

l/k g

for the

time of drying to x 2

~

+ RT(l-x)L/2D'

Integrating Eq. (9.11-11) between the limits of

equation

n

(9 -

1W2)

-

n)

transfer,

L 2

(9 -

is

t

=

=

0 atX!

and

1.0

P'

t

J

=

t

atx 2

= x2

,

the

as follows for h being very large (negligible

external resistance):

L2 r

AH,

/

1

%-^M:^w>(

-

Xi

x\ X2

-T

+

xf\ (9 -

Tj

1M3)

AHJM A is heat of sublimation in J/kg H 2 0. For x 2 = 0, the slab is completely dry. Assuming that the physical properties and mass- and heat-transfer coefficients are known, Eq. (9.1 1-8) can be used to calculate the ice sublimation temperature Tf when the environment temperature Te and the environment partial pressure p ew are set. Since h is very large, Tc = T, Then Eq. (9.1 1-8) can be solved for Tf since Tf and p fw are related by where

.

the equilibrium-vapor-pressure relation, Eq. (9.11-6). In Eq. (9.11-8) the value to use for

T can be approximated The uniformly

by (7}

actual freeze-drying data.

of

65-90%

+ T )/2. 5

model was

retreating ice-front

The model

of the total initial water (SI, Kl).

interface did

remain essentially constant

removal of the

last

10-35% of the

tested by Sandall et

al.

(SI) against

satisfactorily predicted the drying times for

as

The temperature 7}

assumed

removal

of the sublimation

However, during markedly and the actual

in the derivation.

water, the drying rate slowed

time was considerably greater than the predicted for this period.

The

effective thermal conductivity

significantly with the total pressure

k

in the dried material

and with the type of gas

material affects the value of k (SI, Kl).

has been found to vary

present. Also, the type of

The effective diffusivity D' of the dried material is Knudsen diffusivity, and molecular diffusivity

a function of the structure of the material,

(Kl).

9.12

UNSTEADY-STATE THERMAL PROCESSING

AND STERILIZATION OF BIOLOGICAL MATERIALS 9.12A

Introduction

Materials of biological origin are usually not as stable as most inorganic and

organic materials. Hence,

it

is

some

necessary to use certain processing methods to preserve

biological materials, especially foods. Physical

and chemical processing methods

for

preservation can be used such as drying, smoking, salting, chilling, freezing, and heating.

Freezing and chilling of foods were discussed in Section 5.5 as methods of slowing the spoilage of biological materials. Also,

in

Section 9.11, freeze drying of biological materials

was discussed.

Sec. 9.12

Unsteady-State Thermal Processing and Sterilization of Biological Materials 569

An important method

is

heat or thermal processing, whereby contaminating micro-

organisms that occur primarily on the outer surface of foods and cause spoilage and health problems are destroyed. This leads to longer storage times of the food biological materials.

A common method

Also, thermal processing

used to

is

for preservation

sterilize

is

and other

to heat seal cans of food.

aqueous fermentation media

to be

used in

fermentation processes so that organisms which do not survive are unable to compete with the organism that

The

is

sterilization of

to be cultured.

food materials by heating destroys bacteria, yeast, molds, and so

and also destroys pathogenic (disease-producing) orwhich may produce ganisms, deadly toxins if not destroyed. The rate of destruction of varies with microorganisms the amount of heating and the type of organism. Some

on, which cause the spoilage

bacteria can exist in a vegetative

spore forms are

much more

in a dormant or spore form. The mechanism of heat resistance is not

growing form and

resistant to heat. This

clear.

For foods it is desired to kill essentially all the spores of Clostridium botulinum, which produces a toxin that is a deadly poison. Complete sterility with respect to this spore

is

the

difficult

to

purpose of thermal processing. Since use, other spores,

CI.

botulinum

is

so dangerous and often

such as Bacillus stearothermophilus, which

is

a non-

pathogenic organism of similar heat resistance, are often used for testing the heattreating processes (A2, CI).

Temperature has a great

effect

on the growth rate of microorganisms, which have no

temperature-regulating mechanism. Each organism has a certain optimal temperature

range in which

temperature

grows

it

best. If

for a sufficient time,

it

any microorganism will

be rendered

heated to a sufficiently high

is

sterile

or killed.

The exact mechanism of thermal death of vegetative bacteria and spores is still somewhat uncertain. It is thought, however, to be due to the breakdown of the enzymes, which are essential to the functioning of the living

9.12B

Thermal Death-Rate

The destruction

of

Kinetics of Microorganisms

microorganisms by heating means

in the physical sense. If

inactivate the

cell (Bl).

cell,

it

is

assumed

and not destruction enzyme in a cell will

loss of viability

that inactivation of a single

then in a suspension of organisms of a single species at a constant

temperature, the death rate can be expressed as a first-order kinetic equation (A2). rate of destruction

(number dying per

unit time)

is

The

proportional to the number of

organisms.

dN — =

-kN

(9.12-1)

at

where

N

is

the

number

of viable organisms at a given time,

reaction velocity constant in

min"

The

1 .

is

t

time

in

reaction velocity constant

min, and k

is

is

a

a function of

temperature and the type of microorganism. Afte_r

rearranging, Eq. (9.12-1) can be integrated as follows: v

'

'No

where

N0

is

the original

number

at

dN — = "

t

called the contamination level (original

570

f

-

Jl

In

—N =

=

0 and

number

kdt

(9.12-2)

=0

(9.12-3)

kt

N

is

the

number

at

time

t.

Often

N0

is

of contaminating microbes before sterili-

Chap. 9

Drying of Process Materials

N the sterility level. Also, Eq. (9.12-3) can be written as

zation) and

N = N 0 e~ in

kt

(9.12-4)

Sometimes microbiologists use the term decimal reduction time D, which is the time min during which the original number of viable microbes is reduced by 1/10. Substitut-

ing into Eq. (9.12-4),

N — =—= N 1

Taking

and solving

the log 10 of both sides

e~ kD

(9.12-5)

10

0

for D,

D =

—~ 2.303

(9.12-6)

k

Combining Eqs.

and

(9.12-3)

(9.12-6),

t

If

the log 10

(N/N 0 )

is

Experimental data bear for

=D

plotted versus

this

t,

log 10

^

(9.12-7)

from Eq. (9.12-3). and approximately for spores. Data

a straight line should result

out for vegetative

cells

the vegetative cell E. coli (Al) at constant temperature follow this logarithmic

death-rate curve. Bacterial spore plots sometimes deviate rate of death, particularly during a short period

However,

is

used.

experimentally measure the microbial death rate, the spore or

a solution

is

usually sealed in a capillary or test tube.

suddenly dipped into a hot bath ately chilled.

temperature

The

the logarithmic

thermal-processing purposes for use with spores such as CI. botulinum, a

for

logarithmic-type curve

To

somewhat from

immediately following exposure to heat.

The number is

for

a given time.

A number

Then

cell

suspension in

of these tubes are then

they are removed and immedi-

of viable organisms' before and after exposure to the high

then usually determined biologically with a plate count.

effect of

temperature on the reaction-rate constant k

may

be expressed by an

Arrhenius-type equation.

=

k

ae- EIRT

(9.12-8)

where a = an empirical constant, R is the gas constant in kJ/g mol K (cal/g mol K), T is absolute temperature in K, and E is the activation energy in kJ/g mol (cal/g mol). The

£ is in the range 210 to about 418 kJ/g mol (50-100 kcal/g mol) and spores (A2) and much less for enzymes and vitamins.

value of cells

for vegetative

Substituting Eq. (9.12-8) into (9.12-2) and integrating,

N

r<

ln^ = N

e~ EIRT dt

a

At constant temperature T, Eq. (9.12-9) becomes temperature, the decimal reduction time D, which function of temperature. Hence,

D

is

(9.12-9)

o

is

(9.12-3).

Since k

a function of

is

related to k by Eq. (9.12-6),

often written as

DT

to

show

that

it is

is

also a

temperature-

dependent.

9.12C

Determination of Thermal Process Time for Sterilization

For canned foods,

CI.

botulinum

has been established that the

Sec. 9.12

is

the primary

minimum

organism

to be reduced in

heating process should reduce the

number (S2). It number of the

Unsteady-Slate Thermal Processing and Sterilization of Biological Materials 571

12 This means that since D is the time to reduce the original 12 10" number by \ substituting N/N 0 = 10" into Eq. (9.12-4) and solving for t,

spores by a factor of 10

.

t

=

12

2.303 — — =12D

(9.12-10)

k

This means that the time

t is equal to 12D (often called the 12D concept). This time in Eq. number by 10" 12 is called the thermal death time. Usually, the a number much less than one organism. These times do not represent

(9.12-10) to reduce the sterility level

complete

N

is

sterilization but

a mathematical concept which has been found empirically to

give effective sterilization.

Experimental data of thermal death rates of CI. botulinum, when plotted as the

T

D T at a given T versus the temperature in °F on a semilog plot, give essentially straight lines over the range of temperatures used in food sterilization

decimal reduction time

(S2).

A

typical thermal destruction curve

Eqs. (9.12-6) and (9.12-8), degrees absolute)

is

it

a straight

obtained when log 10

DT

is

is

shown

in Fig. 9.12-1. Actually,

can be shown that the plot of log 10 line,

.

Since the plot

is

D T2 -

is

T°F or °C.

a straight line, the equation

log 10

by combining (T in

versus 1/T

but over small ranges of temperature a straight line

plotted versus

In Fig. 9.12-1 the term z represents the temperature range in

DT

DT

log 10

°F

for a 10

:

1

change

in

can be represented as

D Tt = - (T, - T2

(9.12-11)

)

z

Letting

T = {

250°F

(121. 1°C),

which

processes are compared, and calling

is

the standard temperature against which thermal

T2 =

T, Eq. (9.12-1

D T = D 250 For the organism

CI.

10

1)

becomes

(250 - TVZ

(9.12-12)

botulinum the experimental value of z

each increase in temperature of 18°F (10°C) This compares with the factor of 2

for

will increase the

many chemical

=

18°F. This

means

that

death rate by a factor of

10.

reactions for an 18°F increase in

temperature.

Using Eq.

(9.12-7),

'

Substituting

572

T = 250°F

= DT

(121. 1°C) as the

log.o

^

(9-12-7)

standard temperature into

Chap. 9

this

equation and

Drying of Process Materials

substituting

F0

forf,

= £250 where the F 0 value of a process

the time

is.

t

degree of sterilization as the given process at (9.12-12),

and

(9.12-13), the

F0

This

is

the

T°F and

F0

min

value in

a given time

t

at 250°F that will produce the same temperature T. Combining Eqs. (9.12-7),

min

in its

=

(rx - niA

10

(^-wn

.

t

^

10

Tis

(SI)

(9.12-14)

(English)

thermal process at a given constant temperature

for the given

in

t

(9.12-13)

of the given process at temperature

F0 = Fo

log, 0 —j-

min. Values for

F 0 and

adequate

z for

sterilization with CI.

botulinum vary somewhat with the type of food. Data are tabulated by

Stumbo

(S2)

and

Charm (C2) for various foods and microorganisms. The

but successive sterilization processes in a given material are

effects of different

additive. Hence, for several different different times

f,,

f

2

.

,

Fo =

EXAMPLE

f

.

,

.

1

the

F0

-10 (r '- 25O)/z

9.12-1.

T T2

temperature stages

x,

and so on, each having

,

values for each stage are added to give thetotal

+

t

2

Sterilization

-10 (72

- 25O)/z

+

---

(English)

F0

.

(9.12-15)

of Cans of Food

a retort for sterilization. TheF 0 for CI. botulinum in this type of food is 2.50 min and z = 18°F. The temperatures in the center of a can (the slowest-heating region) were measured and were

Cans of a given food were heated

in

approximately as follows, where the average temperature during each time period is listed: t, (0-20 min), T, = 160=F; t 2 (20-40 min), T2 = 210°F; = 230°F. Determine if this sterilization process is adet 3 (40-73 min), T3

Use English and SI

quate.

For the three time periods the data are

Solution:

2

-0 = 20 min, = = 40 - 20 = 20 min, T2 =

3

=

=

t,

t

f

units.

20

73

7",

- 40 =

73 =

33 min,

as follows:

160°F (71.1°C),

=

z

210°F (98.9°C)

230°F

10°C)

(1

Substituting into Eq. (9.12-15) and solving using English

F0 =

t,

•

10

(r '- 25O)/l

<16O_:5O)/18

=

(20)10

=

0.0020

=

(20)10

=

2.68

+

+

0.1 199

t

•

2

+

+

172 " 250 '/*

(20)10

2.555

(711 - 1211) "°

min

io

+

+

r

•

3

(21o_25O,/18

=

18°F (10°C)

min

2.68

io<

+

and

SI units,

r >- 250 "'

(33)10

(9.12-15)

(23O_25o)/18

(English)

(98 9 - 1211)/, ° (20)10 -

+

(33)10

(11

°- 1211)/10

(SI)

Hence, this thermal processing is adequate since only 2.50 min is needed for complete sterilization. Note that the time period at 160°F (71.1°C) contributes an insignificant amount to the final F 0 The major contribution is at 230°F (1 10°C), which is the highest temperature. .

In the general case is

when cans

of food are being sterilized in a retort, the temperature

not constant for a given time period but varies continuously with time. Hence, Eq.

(9.12-15) can be modified

Sec. 9.12

and written

for a

continuously varying temperature

T

by

Unsteady-Slate Thermal Processing and Sterilization of Biological Materials 573

taking small time increments of

equation

min

dt

for

T

each value of

and summing. The

final

is

10 1

(7-F-2SW(i-F)

(Eng ii s h)

dt

=0

(9.12-16) rt=i

10

(7-C-121.1,/( Z °Q

dt

(SI)

This equation can be used as follows. Suppose that the temperature of a process

T

varying continuously and a graph or a table of values of calculated by the unsteady-state

graphically integrated

methods given

in

Chapter

5.

versus

The

plotting values of io (r-250)/z versus

by

t

is

known

or

is

is

Eq. (9.12-16) can be

and taking the area

t

under the curve. In

many

cases the temperature of a process that

is

varying continuously with time

is

determined experimentally by measuring the temperature in the slowest-heating region. In cans this

Methods given

the center of the can.

is

heating of short,

fat

in

Chapter 5

for unsteady-state

cylinders by conduction can be used to predict the center temper-

ature of the can as a function of time. However, these predictions can be error, since physical

and often can can

affect the

vary.

somewhat

in

and thermal properties of foods are difficult to measure accurately Also, trapped air in the container and unknown convection effects

accuracy of predictions.

EXAMPLE

Thermal Process Evaluation by Graphical Integration

9.12-2.

In the sterilization of a canned puree, the temperature in the slowest-heating

of the can was measured giving the following timetemperature data for the heating and holding time. The cooling time data region (center)

will be neglected as a safety factor.

The F 0

T

(min)

l

(°F)

t

-

T

(min)

(°F)

0

80

40

225

(107.2°C)

15

165

(73.9)

50

230.5

(1

25

201

(93.9)

64

235

(112.8)

30

212.5 (100.3)

(26.7°C)

value of CI. botulinum

2.45

is

min and

value of the process above and determine

if

z

is

10.3)

18°F. Calculate the

the sterilization

is "

Solution:

In order to use Eq. (9.12-16), the values of lO

calculated for each time.

10

Fort

=

15 min,

For

(r-25O)/r

T=

=

25 min,

T =

574

t

=

0 min,

1()

7 =

(80-250)/18

(16S-25O)/18

80°F, and

=

3

z

=

-250 " 1

must be

18°F,

5 x jq-10

=

0.0000189

(2Ol-25O>/18

=

0.0O189

10 (212.5-25O)/18

=

0.00825

201°F 10

For

=

=

165°F 10

Forf

t

17

F0

adequate.

30 min,

Chap. 9

Drying of Process Materials

For

For

=

t

=

t

40 min,

•

10 U25-250>/18

= 00408

(230.5-250)/18

=

50 min, 10

For

t

—

64 min, 10 (235

- 250)/18

These values are plotted versus rectangles

t

= 0.1465

in Fig. 9.12-2.

Ai

of the various

+ A 2 + A 3 + AA

=

10(0.0026)

=

0.026

+

+

10(0.0233)

0.233

The process value of 2.50 min sterilization

is

Sterilization

+

0.620

is

+

+

10(0.0620)

1.621

=

2.50

14(0.1160)

min

greater than the required 2.45

Methods Using Other Design

min and

the

Criteria

which are not necessarily involved with

foods, other types of design criteria are used. In foods the

number

+

adequate.

In types of thermal processing

reduce the

The areas

shown are

F0 =

9.12D

0.0825

of spores by a factor of 10

-12 i.e.,

,

minimum

iV/JV 0

=

10

sterilization of

heat process should

-12 .

However,

in

other

batch sterilization processes, such as in the sterilization of fermentation media, other criteria are often used.

specific

organism

to

Often the equation

be used,

is

for k, the reaction velocity

constant for the

available.

k

=

ae- EIRT

(9-12-8)

0.16 r

Time FIGURE

Sec. 9.12

9.12-2.

t

(min)

Graphical integration for Example 9.12-2.

Unsteady-State Thermal Processing and Sterilization of Biological Materials 575

Then Eq. (9.12-9) is

written as

e- EIRT dt

where V

temperature -E/RT j

k dt

(9.12-17)

N0

the design criterion. Usually, the contamination level

is

N

either the sterility level

ae

=

is

the

unknown

the

is

is

available

and

or the time of sterilization at a given

unknown. In either case a graphical integration is usually done, where t and the area under the curve is obtained.

plotted versus

s

In sterilization of food

in

a container, the time required to render the material safe

is

calculated at the slowest-heating region of the container (usually the center). Other

and

regions of the container are usually heated to higher temperatures

Hence, another method used

container. These details are given by others (C2, S2). In

short-time, continuous-flow process

9.1 2E

are overtreated.

based on the probability of survival in the whole

is

is

still

another processing method, a

used instead of a batch process

in

a container (B2).

Pasteurization

The term

pasteurization

is

drastic than sterilization.

compared

resistance

used today to apply to mild heat treatment of foods that It is

used to

to those for

kill

is

less

organisms that are relatively low in thermal

which the more drastic

sterilization processes are

designed to eliminate. Pasteurization usually involves killing vegetative microorganisms

and not heat-resistant spores. The most common process is the pasteurization of milk to kill Mycobacterium tuberculosis, which is a non-spore-forming bacterium. This pasteurization does not sterilize the milk but kills the M. tuberculosis and reduces the other bacterial count sufficiently so that the milk can be stored if refrigerated. For the pasteurization of such foods as milk, fruit juices, and beer, the same mathematical and graphical procedures covered for sterilization processes in this section

The much shorter and the temperatures used in pasteurization are much the F 0 value is given at 150°F (65.6°C) or a similar temperature rather

are used to accomplish the degree of sterilization desired in pasteurization (Bl, S2).

times involved are lower. Generally,

than at 250°F as rise in

in sterilization.

temperature of z°F

F 9ls0 means

will

Also, the concept of the

z

value

is

employed,

increase the death rate by a factor of

F value

10.

in

which a

An F 0

value

150°F with a z value of 9°F (S2). In pasteurizing milk, batch and continuous processes are used. The U.S. health regulations specify two equivalent sets of conditions, where in one the milk is held at

written as

the

at

.

145°F (62.8°C)

The

for 30

min and

in

the other at 161°F(71.7°C) for 15

s.

general equations used for pasteurization are similar to sterilization

and can be

written as follows. Rewriting Eq. (9.12-13),

F zTi = D Tl

log I0

(9.12-18)

N

Rewriting Eq. (9.12-14), (9.12-19)

where

7\

is

the standard temperature being used such as 150°F,

for a tenfold increase in

EXAMPLE A

typical

F

exchanger the

576

9.12-3.

rate,

and

T

is

z is

the value of

z in

°F

the temperature of the actual process.

Pasteurization of Milk

value given for the thermal processing of milk in a tubular heat

is

number

death

F' 50

=

9.0

min and

D 150 =

0.6 min. Calculate the reduction in

of viable cells for these conditions.

Chap. 9

Drying of Process Materials

The z value is 9°F (5°C) and the temperature of the process 150°F (65.6°C). Substituting into Eq. (9.12-18) and solving,

Solution:

F? 50

=

N0 N ~ This gives a reduction

9.12F

Effects of

it

log I0

^

1S

10 1

number

of viable cells of 10

15 .

Thermal Processing on Food Constituents

Thermal processing but

in the

= 0.6

9.0

is

is

used

to

cause the death of various undesirable microorganisms,

also causes undesirable effects, such as the reduction of certain nutritional values.

B2

Ascorbic acid (vitamin C) and thiamin and riboflavin (vitamins Bj and destroyed by thermal processing.

The

)

are partially

reduction of these desirable constituents can also

F 0 and z values in the same way as for sterilization and pasteurization. Examples and data are given by Charm (C2). These same kinetic methods of thermal death rates can also be applied to predict the time for detecting a flavor change in a food product. Dietrich et al. (Dl) determined a curve for the number of days to detect a flavor change of frozen spinach versus temperature of storage. The data followed Eq. (9.12-8) and a first-order kinetic relation.

be given kinetic parameters such as

PROBLEMS 9.3-1.

Humidity from Vapor Pressure. The air in a room is at 37.8°C and a total pressure of 101.3 kPa abs containing water vapor with a partial pressure p A = 3.59 kPa. Calculate: (a)

Humidity.

(b)

Saturation humidity and percentage humidity. Percentage relative humidity.

(c)

9.3-2.

Percentage and Relative Humidity.

H

kg

2

0/kg

dry air at 32.2°C and

H

Percentage humidity P Percentage relative humidity

(a)

(b)

1

The air in a room has a humidity kPa abs pressure. Calculate:

of 0.021

.

HR

.

Ans. 9.3-3.

H

01.3

Use of the Humidity Chart. The

(a)

HP =

67.5%

;

(b)

HR =

68.6%

temperature of 65.6°C (150°F) and dew point of 15.6°C (60°F). Using the humidity chart, determine the actual humidity and percentage humidity. Calculate the humid volume of this mixture and also calculate cs using SI and English units. = 0.0113 kg H 2 0/kg dry air, H P = 5.3%, Ans. air entering a dryer has a

H

cs

=

1.026

kJAg

vH

=

0.976

m

3

•

air

btu/lb m °F), water vapor/kg dry air

K. (0.245

+

•

of Air to a Dryer. An air-water vapor mixture going to a drying process has a dry bulb temperature of 57.2°C and a humidity of 0.030 kg

9.3-4. Properties

H 2 0/kg dry air.

Using the humidity chart and appropriate equations, determine

the-percentage humidity, saturation humidity at 57.2°C,

dew

point,

humid

heat,

and humid volume. 93-5. Adiabatic

H= It

leaves at

(a)

(b)

Saturation

0.0655 kg

Temperature. Air

80%

Problems

in

82.2°C and having a humidity an adiabatic saturator with water.

saturation.

What are the final values of H and T°C? For 100% saturation, what would be the Ans.

Chap. 9

at

H 2 0/kg dry air is contacted

values of

(a)H = O.079 kg

H and

T?

H 0/kg dry air, 7 = 2

52.8°C

577

93-6. Adiabatic Saturation of Air. Air enters an adiabatic saturator having a temperature of 76.7°C and a dew-point temperature of 40.6°C. It leaves the saturator

90%

saturated.

What are

the final values of

H and T°C?

93-7. Humidity from Wet and Dry Bulb Temperatures. An air-water vapor mixture has a dry bulb temperature of 65.6°C and a wet bulb temperature of 32.2°C. What is the humidity of the mixture?

Ans.

H=

0.0175 kg

H OAg dry air 2

The humidity of an air-water vapor kg H 2 0/kg dry air. The dry bulb temperature of the

93-8. Humidity and Wet Bulb Temperature.

mixture mixture

H=

0.030

60°C.

What

is

is

9.3-9. Dehumidification

is

the wet bulb temperature?

of Air. Air having a dry bulb temperature of 37.8°C and a wet to be dried by first cooling to 15.6°C to condense water vapor

bulb of 26.7°C is and then heating to 23.9°C. (a) Calculate the initial humidity and percentage humidity. (b) Calculate the final humidity and percentage humidity. [Hint: Locate the initial point on the humidity chart. Then go horizontally (cooling) to the

100%

saturation line. Follow this line to 15.6°C.

Then go horizontally

to the right to 23.9°C]

Ans.

(b)

H = 0.01 15 kg H

2

0/kg dry

air,

HP =

60%

93-10. Cooling and Dehumidifying Air. Air entering an adiabatic cooling chamber has a temperature of 32.2°C and a percentage humidity of 65%. It is cooled by a cold water spray and saturated with water vapor in the chamber. After leaving, it is heated to 23.9°C. The final air has a percentage humidity of 40%. (a) What is the initial humidity of the air? (b) What is the final humidity after heating? 9.6-1.

Time for Drying

in Constant-Rate Period.

A

tray dryer using constant drying conditions

mm. Only

batch of wet solid was dried on a and a thickness of material on the

was exposed. The drying rate during the 2 (0.42 lb m H 2 0/h ft ). The ratio 2 2 used was 24.4 The L s/A kg dry solid/m exposed surface (5.0 lb m dry solid/ft initial free moisture was X = 0.55 and the critical moisture content X c = 0.22 tray of 25.4

the top surface

constant-rate period was

R=

2.05

kgH 2 0/h m 2 -

).

l

kg

free

moisture/kg dry

solid.

X =

X

2

Calculate the time to dry a batch of this material from x = 0.30 using the same drying conditions but a thickness of 50.8

drying from the top and bottom surfaces. [Hint

:

First calculate

L s/A

0.45

mm,

to

with

new

for this

case.)

Ans.

£

=

1.785 h

of Effect of Process Variables on Drying Rate. Using the conditions in Example 9.6-3 for the constant-rate drying period, do as follows. (a) Predict the effect on R c if the air velocity is only 3.05 m/s. (b) Predict the effect if the gas temperature is raised to 76.7°C and H remains the

9.6-2. Prediction

same. (c)

Predict the effect on the time if

t

for drying

the thickness of material dried

drying

is still

is

between moisture contentsX l ioX 2 and the instead of 25.4

38.1

mm

mm

in the constant-rate period.

Ans.

(a)

(b)

R c = 1.947 kg H 2 0/h m 2 R c =4.21 kg H z O/h m 2 •

(0.399 lb m

H

z

O/h

2 •

ft

)

Constant-Rate Drying Region. A granular insoluble solid material wet with water is being dried in the constant-rate period in a pan 0.61 x 0.61 and the depth of material is 25.4 mm. The sides and bottom are insulated. Air flows parallel to the top drying surface at a velocity of 3.05 m/s and has a dry

9.6-3. Prediction in

m

m

bulb temperature of 60°C and wet bulb temperature of 29.4°C. The pan contains 1 1.34 kg of dry solid having a free moisture content of 0.35 kg H solid 2 0/kg dry

578

Chap.

9

Problems

and the material

is

to be dried in the constant-rate'period to 0.22

kgH 2 0/kg

dry

solid. (a)

Predict the drying rate

(b) Predict the

and the time

time needed

if

in

hours needed.

the depth of material

is

mm.

increased to 44.5

Drying a Filter Cake in the Constant-Rate Region. A wet filter cake in a pan 1 ft square and 1 in. thick is dried on the top surface with air at a wet bulb 1 ft x temperature of 80°F and a dry bulb of 120°F flowing parallel to the surface at a 3 velocity of 2.5 ft/s. The dry density of the cake is 120 lb^ft and the critical free moisture content is 0.09 lb H z O/lb dry solid. How long will it take to dry the material from a free moisture content of 0.20 lb H 2 0/lb dry material to the critical moisture content? Ans. t = 13.3 h

9.6-4.

Graphical Integration for Drying in Falling-Rate Region. A wet solid is to be dried in a tray dryer under steady-state conditions from a free moisture content = 0.02 kg H 2 0/kg dry solid. The dry of X l = 0.40 kg H 2 0/kg dry solid to 2 2 solid weight is 99.8 kg dry solid and the top surface area for drying is 4.645 The drying-rate curve can be represented by Fig. 9.5-lb.

9.7- 1.

X

m

(a)

Calculate the time for drying using graphical integration

.

the falling-rate

in

period. (b)

Repeat but use a straight

line

through the origin

drying rate

for the

in the

falling-rate period.

Ans. 9.7-2.

f(constant rate)

(a)

=

=

2.91 h, f(falling rate)

6.36

h, f(total)

=

9.27

h

Drying Tests with a Foodstuff. In order to test the feasibility of drying a certain foodstuff, drying data were obtained in a tray dryer with air flow over the top 2 exposed surface having an area of 0.186 The bone-dry sample weight was 3.765 kg dry solid. At equilibrium after a long period, the wet sample weight was 3.955 kg H 2 0 + solid. Hence, 3.955 - 3.765, or 0.190, kg of equilibrium moisture was present. The following sample weights versus time were obtained in the

m

drying

Time

;

.

test.

Weight

(h)

(kg)

Time

(h)

Weight

Time

(kg)

Weight

(h)

(kg)

0

4.944

2.2

4.554

7.0

4.019

04

4.885

3.0

4.404

9.0

3.978

0.8

4.808

4.2

4.241

12.0

3.955

1.4

4.699

5.0

4.150

X

kg H 2 0/kg dry solid for each data Calculate the free moisture content point and plot versus time. {Hint: For 0 h, 4.944 - 0.190 - 3.765 = 0.989 = 0.989/3.765.) kg free moisture in 3.765 kg dry solid. Hence, 0/h m 2 and plot R (b) Measure the slopes, calculate the drying rates R in kg 2 versus X. (c) Using this drying-rate curve, predict the total time to dry the sample from (a)

X

X

H

X=

0.20 to

What

is

X=

0.04.

the drying rate

Use graphical integration

Rc

,

for the falling-rate period.

in the constant-rate period

Ans.

•

and

Xc 2

R c = 0.996 kg H 2 0/h m 2 X c = 0.12, = 4.1 h (total)

(c)

t

A

material was dried in a tray-type batch dryer using constant drying conditions. When the initial free moisture content was 0.28 kg free moisture/kg dry solid, 6.0 h was required to dry the material to a free

9.7-3. Prediction

of Drying Time.

moisture content of 0.08 kg free moisture/kg dry solid.

Chap.

9

Problems

The

critical free

moisture

579

content

is

0.14.

Assuming a drying

rate in the falling-rate region

where the rate

is

a straight line from the critical point to the origin, predict the time to dry a

sample from a

moisture content of 0.33 to 0.04 kg free moisture/kg dry solid.

free

and Then use

(Hint: First use the analytical equations for the constant-rate falling-rate periods with the

known

total time of 6.0 h.

the linear

the

same

equations for the new conditions.) 9.8-1.

Drying of Biological Material in Tray Dryer. A granular biological material wet with water is being dried in a pan 0.305 x 0.305 m and 38.1 deep. The material is 38.1 deep in the pan, which is insulated on the sides and the bottom. Heat transfer is by convection from an air stream flowing parallel to the top surface at a velocity of 3.05 m/s, having a temperature of 65.6°C and humidity H = 0.010 kg H 2 0/kg dry air. The top surface receives radiation from steam-heated pipes whose surface temperature TR = 93.3°C. The emissivity of the solid is £ = 0.95. It is desired to keep the surface temperature of the solid below 32.2°C so that decomposition will be kept low. Calculate the surface temperature

mm

mm

and the

rate of drying for the constant-rate period.

Ans. 9.8- 2.

Ts =

31.3°C,i? c

=

2.583

kg

H 2 0/h-m 2

Drying When Radiation, Conduction, and Convection Are Present. A material is granular and wet with water and is being dried in a layer 25.4 deep in a batch-tray dryer pan. The pan has a metal bottom having a thermal conductivity of k M = 43.3 W/m K and a thickness of 1.59 mm. The thermal conductivity of the solid is k s = 1.125 W/m-K. The air flows parallel to the top exposed surface and the bottom metal at a velocity of 3.05 m/s and a temperature of 60°C and humidity H = 0.010 kg H 2 0/kg dry solid. Direct radiation heat from steam pipes having a surface temperature of 104.4°C falls on the exposed top surface, whose emissivity is 0.94. Estimate the surface temperature and the drying rate for the constant-rate period.

mm

•

9.9- 1. Diffusion Drying in

Wood. Repeat Example 9.9-1 using the physical properties

given but with the following changes. (a) Calculate the time needed to dry the 0.1 3. (b)

Use

wood from a

total

moisture of 0.22 to

Fig. 5.3- 13.

mm

thick from Calculate the time needed to dry planks of wood 12.7 thickn = 0.29 to X, = 0.09. Compare with the time needed for 25.4

mm

X

ness.

Ans.

(b)t

=

7.60 h (12.7

mm thick)

Drying Tapioca Root. Using the data given in Example 9.9-2, determine the average diffusivity of the moisture up to a value of XIX C =

9.9-2. Diffusivity in

0.50. 9.9-3. Diffusion Coefficient. ical

Experimental drying data of a typical nonporous biolog-

material obtained under constant drying conditions in the falling-rate

region are tabulated below.

x/x c

t(h)

x/x c

t(h)

1.00

0

0.17

11.4

0.65

2.50

0.10"

14.0

0.32

7.00

0.06

16.0

Drying from one side occurs with the material having a thickness of 10.1 mm. The data appear to follow the diffusion equation. Determine the average diffusivity over the range XIX C = 1 .0-0.10. 9.10-1. Drying a

580

Bed

of Solids by Through Circulation. Repeat

Example

Chap. 9

9.10-1 for

Problems

drying of a packed bed of wet cylinders by through circulation of the drying Use the same conditions except that the air velocity is 0.381 m/s.

air.

of Equation for Through Circulation Drying. Different forms of Eqs. and (9.10-12) can be derived using humidity and mass-transfer equations rather than temperature and heat-transfer equations. This can be done by writing a mass-balance equation similar to Eq. (9. 10-2) for a heat balance and a mass-transfer equation similar to Eq. (9.10-3).

9.10-2. Derivation

(9.10-11)

(a)

(b)

Derive the final equation for the time of drying in the constant-rate period using humidity and mass-transfer equations. Repeat for the falling-rate period. „. ,__

Through Circulation Drying in the Constant-Rate Period. Spherical wet catalyst are being dried in a through circulation pellets having a diameter of 12.7 dryer. The pellets are in a bed 63.5 mm thick on a screen. The solids are being dried by air entering with a superficial velocity of 0.914 m/s at 82.2°C and having = 0.0 1 kg H 2 0/kg dry air. The dry solid density is determined as a humidity 3 and the void fraction in the bed is 0.35. The initial free moisture 1522 kg/m content is 0.90 kg H 2 0/kg solid and the solids are to be dried to a free moisture content of 0.45, which is above the critical free moisture content. Calculate the

9.10-3.

mm

H ,

time for drying in this constant-rate period.

and Heat Balances on a Continuous Dryer. Repeat Example 9.10-2, making heat and material balances, but with the following changes. The solid enters at 15.6°C and leaves at 60°C. The gas enters at 87.8°C and leaves at 32.2°C. Heat losses from the dryer are estimated as 2931 W.

9.10-4. Material

A

rate of feed of 700 lb m dry solid/h Tunnel Dryer. .— 0.4133 lb containing a free moisture content of 2 0/lb dry solid is to be = solid in continuous counterflow tunnel dried to 0.0374 lb 0/lb dry a 2 2 = 0.0562 lb flow of 13 280 lb m dry air/h enters at 203°F with an dryer. 2 at the, wet bulb temperature of 119°F and 2 0/lb dry air. The stock enters remains essentially constant in temperature in the dryer. The saturation humidity at 119°F from the humidity chart is w = 0.0786 lb 2 0/lb dry air. The surface area available for drying is (AlL s ) = 0.30 ft 2 /lb m dry solid. A small batch experiment was performed using constant drying conditions,

9.10-5. Drying in a Continuous

X

H

x

H

X

H

A

H

H

H

air velocity, and temperature of the solid approximately the same as in the continuous dryer. The equilibrium critical moisture content was found to be = 0.0959 lb 2 0/\b dry solid, and the experimental value of ky B was found c 2 In the falling-rate period, the drying rate was directly as 30.15 lb m air/h -ft

M

H

X

.

proportional to X. For the continuous dryer, calculate the time in the dryer in the constant-rate zone and in the falling-rate zone. Ans. Hc = 0.0593 lb H 2 0/lb dry air, H, = 0.0760 lb H 2 0/lb dry air, zone t = 4.24 h in the constant-rate zone, £ = 0.47 h in the falling-rate

The wet feed material to a continuous dryer contains 50 wt water on a wet basis and is dried to 27 wt by countercurrent air flow. The dried product leaves at the rate of 907.2 kg/h. Fresh = 0.007 kg H 2 0/kg dry air to the system is at 25.6°C and has a humidity of = 0.020 and part of it is air. The moist air leaves the dryer at 37.8°C and recirculated and mixed with the fresh air before entering a heater. The heated = 0.010. The solid enters at 26.7°C mixed air enters the dryer at 65.6°C and and leaves at 26.7°C. Calculate the fresh-air flow, the percent air leaving the dryer that is recycled, the heat added in the heater, and the heat loss from the

9.10-6. Air Recirculation in a Continuous Dryer.

%

%

H H

H

dryer.

Ans.

Chap.

9

Problems

32 094 kg fresh dry air/h, 23.08% recycled, 440.6

kW in heater

581

9.12-1. Sterilizing

Canned Foods.

In a sterilizing retort, cans of a given food

were heated

and the average temperature in the center of a can is approximately 98.9°C for the first 30 min. The average temperature for the next period is 1 10°C. If the F 0 for the spore organism is 2.50 min and z = 10°C, calculate the time of heating at

1

make

10°C to

the process safe.

Ans. 29.9 min on Decimal Reduction Time. Prove by combining Eqs. and (9.12-8) that a plot of log,„ D T versus 11T (T in degrees

9.12-2. Temperature Effect (9.12-6)

absolute) 9.12-3.

is

a straight line.

Thermal Process Time for Pea Puree. For cans of pea puree, the F 0 = 2.45 min and z = 9.94°C (C2). Neglecting heat-up time, determine the process time for adequate sterilization at 112.8°C at the center of the can.

=

16.76 min Time for Adequate Sterilization. The F 0 value for a given canned food is 2.80 min and z is 18°F (10°C). The center temperatures of a can of this food when heated in a retort were as follows for the time periods given: t\ (0-10

Ans.

t

9.12-4. Process

min),

T =

(10-30 min), T 2 = 185°F; (30-50 min), T 3 = 220°F; f 4 230°F; t 5 (80-100 min), T 5 = 190°F. Determine if adequate obtained.

140°F;

l

(50-80 min), T4 sterilization is

t2

=

Time and Graphical Integration. The following time-temperature data were obtained for the heating, holding, and cooling of a canned food product in a retort, the temperature being measured in the center of the can.

9.12-5. Process

T(°F)

t(min)

110 (43.3°C)

80

232 (111.1)

20

165 (73.9)

90

40

205

225 (107.2) 160 (71.1)

60

228 (108.9)

0

100

(96.1)

The F 0 value used is this

T{°F)

t(min)

2.60

process and determine

min and

if

z is

18°F (10°C). Calculate theF 0 value

the thermal processing

is

for

adequate. Use SI and

English units.

Medium. The aqueous medium in a fermentor being sterilized and the time-temperature data obtained are as follows.

9.12-6. Sterility Level of Fermentation

Time

(min)

Temperature (°C)

The reaction

0

10

20

25

30

35

100

no

120

120

110

100

velocity constant k in

min

1

is

contaminating bacterial spores

for the

can be represented as (A I) k

T=

=

7.94 x 10

3

V- (58

K. The contamination sterility level N at the end and V.

where

9.12-7.

-

level

7x

'°3)/

N0 =

1

.987r

1

x 10 12 spores. Calculate the

Time for Pasteurization of Milk. Calculate the time in min at 62.8°C for pasteurization of milk. The F 0 value to be used at 65.6°C is 9.0 min. The z value is

5°C. Ans.

582

Chap. 9

t

=

32.7

min

Problems

Number of Viable Cells in Pasteurization. In a given pasteurization 15 process the reduction in the number of viable cells used is 10 and the F 0 value used is 9.0 min. If the reduction is to be increased to 10 16 because of increased contamination, what would be the new F 0 value?

9.12-8. Reduction in

REFERENCES (Al)

Allerton,

Brownell,

J.,

L. E.,

and Katz, D.

L.

Chem. Eng. Progr., 45, 619

(1949).

(A2)

Aiba,

S.,

Humphrey,

A. E., and Millis, N. F. Biochemical Engineering,

New York: Academic Press, Inc., (Bl)

Blakebrough, N. Biochemical and York: Academic Press, Inc., 1968.

(B2)

Bateson, R. N. Chem. Eng. Progr. Symp.

(CI)

Cruess,

W.

ed.

New

Biological Engineering Science, Vol. 2.

Ser.,

67 (108), 44 (1971).

New

V. Commercial Fruit and Vegetable Products, 4th ed.

McGraw-Hill Book Company,

2nd

1973.

York:

1958.

(C2)

Charm,

(Dl)

Dietrich, W. C, Boggs, M. M., Nutting, M. D., and Weinstein, N. E. Food Technol., 14,522(1960).

(El)

Earle, R. L. Unit Operations

S. E. The Fundamentals of Food Engineering, 2nd Avi Publishing Co., Inc., 1971.

in

Food

ed.

Westport, Conn.:

Pergamon

Processing. Oxford:

Press, Inc.,

1966.

and Hougen, O. A. Trans. A.l.Ch.E.,

(Gl)

Gamson, B. W., Thodos,

(HI)

Hall, C. W. Processing Equipment for Agricultural Products. Westport, Conn.: Avi Publishing Co., Inc., 1972.

(H2)

Henderson,

(K 1)

King, C.

J.

M.

S.

G.,

Agr. Eng., 33

39,

1

(1943).

29 (1952).

(1),

Freeze Drying of Foods. Boca Raton, Fla. Chemical Rubber Co., :

Inc.,

1971.

(Ml)

Marshall, W.

(PI)

Perry, R. H., and Green, D. Perry's Chemical Engineers' Handbook, 6th ed. New York: McGraw-Hill Book Company, 1984.

(51)

Sandall, O. C, King, C.

R., Jr.,

and Hougen, O. A. Trans. A.l.Ch.E.,

J,

and Wilke,

C. R. A.l.Ch.E.

38, 91 (1942).

J., 13,

428 (1967); Chem.

Eng. Progr., 64(86), 43 (1968). (52)

Stumbo, C. R. Thermobacteriology Academic Press, Inc., 1973.

(Tl)

Treybal, R.

Company, (Wl)

Chap. 9

E.

Mass Transfer

in

Food Processing, 2nd

Operations, 3rd ed.

ed.

New

York:

New York: McGraw-Hill Book

1980.

Wilke, C. R.,and Hougen, O. A. Trans. A.l.Ch.E., 41, 441

References

(1945).

583

CHAPTER

10

Stage and Continuous Gas-Liquid Separation Processes

TYPES OF SEPARATION PROCESSES AND METHODS

10.1

I0.1A

Introduction

Many chemical process materials and biological substances occur as mixtures of different components in the gas, liquid, or solid phase. In order to separate or remove one or more of the components from its original mixture, it must be contacted with another phase. The two phases are brought into more or less intimate contact with each other so that a solute or solutes can diffuse from one to the other. The two bulk phases are usually only somewhat miscible in each other. The two-phase pair can be gas-liquid, gas-solid, liquid-liquid, or liquid-solid. During the contact of the two phases the components of the original mixture redistribute themselves between the two phases. The phases are then separated by simple physical methods. By choosing the proper conditions and phases, one phase is enriched while the other is depleted in one or more components.

10.1B

/.

Types of Separation Processes

When

Absorption.

operation

is

the

two contacting phases are

called absorption.

A

solute

A

a

gas and a liquid, the unit

or several solutes are absorbed from the gas

phase into a liquid phase in absorption. This process involves molecular and turbulent diffusion or

mass transfer of solute A through a stagnant nondiffusing gas B into a An example is absorption of ammonia A from air B by the liquid water

stagnant liquid C.

C. Usually, the exit

ammonia-water solution

distilled

is

to

recover relatively pure

ammonia. Another example

is

absorbing

S0

2

from the

flue gases

by absorption in alkaline hydrogen gas is

solutions. In the hydrogenation of edible oils in the food industry,

bubbled into

oil

and absorbed. The hydrogen

presence of a catalyst.

same

584

theories

The

in

reverse of absorption

solution then reacts with the is

oil in

called stripping or desorption,

and basic principles hold. An example

is

the

and the

the steam stripping of nonvolatile

oils, in oil

which the steam contacts the

oil

and small amounts of volatile components of the

pass out with the steam.

When

the gas

is

pure

and the

air

humidification. Dehumidification involves

2.

liquid is pure water, the process removal of water vapor from air.

is

called

In the distillation process, a volatile vapor phase and a liquid phase that

Distillation.

vaporizes are involved.

An example is

distillation of

an ethanol-water solution, where the

vapor contains a concentration of ethanol greater than in the liquid. Another example is distillation of an ammonia-water solution to produce a vapor richer in ammonia. In the distillation of oils,

3.

crude petroleum, various fractions, such as gasoline, kerosene, and heating

are distilled

off.

Liquid-liquid extraction.

removed from one

are

liquid-liquid extraction.

When

liquid

the

two phases are

liquids,

where a solute or solutes

phase to another liquid phase, the process

One example

is

is

called

extraction of acetic acid from a water solution by

isopropyl ether. In the pharmaceutical industry, antibiotics in an aqueous fermentation solution are sometimes

4.

Leaching.

If

a fluid

Sometimes from solid ores by leaching.

removed by extraction with an organic is

this

being used

a solute from a

solid, the process

is

called

Examples are leaching copper acid and leaching vegetable oils from solid soybeans by

process

sulfuric

to extract

solvent.

is

also called extraction.

organic solvents such as hexane. Vegetable

oils

are also leached from other biological

products, such as peanuts, rape seeds, and sunflower seeds. Soluble sucrose

is

leached by

water extraction from sugar cane and beets.

Membrane processing. Separation of molecules by the use of membranes is a new unit operation and is becoming more important. The relatively thin, solid membrane controls the rate of movement of molecules between two phases. It is used to

5.

relatively

remove 6.

salt

from water, purify gases,

Crystallization.

in

food processing, and so on.

Solute components soluble in a solution can be removed from the

solution by adjusting the conditions, such as temperature or concentration, so that the

one or more solute components is exceeded and they crystallize out as a Examples of this separation process are crystallization of sugar from solution

solubility of solid phase.

and

crystallization of metal salts in the processing of metal ore solutions.

an adsorption process one or more components of a liquid or gas stream are adsorbed on the surface or in the pores of a solid adsorbent and a separation 7.

is

Adsorption.

obtained.

In

Examples include removal of organic compounds from polluted water,

separation of paraffins from aromatics, and removal of solvents from

10.1C

Processing Methods

Several

methods of processing are used

in the

air.

separations discussed above.

phases, such as gas and liquid, or liquid and liquid, can be mixed together

then separated. This

is

a single-stage process. Often the phases are

separated, and then contacted again

The two and

in a vessel

mixed

in

one

stage,

in a multiple- stage process. These two methods can

still another general method, the two phases packed tower. can be contacted continuously in a In this chapter, humidification and absorption will be considered; in Chapter 1, distillation; in Chapter 12, adsorption, liquid-liquid extraction, leaching, and crystal-

be carried out batchwise or continuously. In

1

Sec. 10.1

Types of Separation Processes and Methods

585

and

lization,

Chapter

in

membrane processes.

13,

In-all of these

processes the

equilibrium relations between the two phases being considered must be known. This is discussed for gas-liquid systems in Section 10.2 and for the other systems in

Chapters

and

11, 12,

13.

EQUILIBRIUM RELATIONS BETWEEN PHASES

10.2

Phase Rule and Equilibrium

10.2A

In order to predict the concentration of a solute in each of two phases in equilibrium,

experimental equilibrium data must be available. Also, equilibrium, the rate of mass transfer

departure from equilibrium. In

all

is

if

the

two phases are not

proportional to the driving force, which

cases involving equilibria,

is

at

the

two phases are involved,

such as gas-liquid or liquid-liquid. The important variables affecting the equilibrium of a

and concentration. The equilibrium between two phases in a given

solute are temperature, pressure,

situation

is

restricted

by the phase

rule:

F = C—P+

2

(10.2-1)

where P is the number of phases at equilibrium, C the number of total components in the two phases when no chemical reactions are occurring, and F the number of variants or degrees of freedom of the system. For example, for the gas-liquid system of

C0 2 -air-water, there are

two phases and three components (considering

component). Then, by Eq.

(10.2-1),

air as

one

inert

F=C-P+2=3-2+2=3 This means that there are

3

are set, only one variable

is left

x A of

sition

pressure p A

CO

in

z

degrees of freedom.

(A) in the liquid phase

the gas phase

is

If

is

set,

and the temperature mole fraction compo-

the total pressure

that can be arbitrarily

set. If

the

the mole fraction composition y A or

automatically determined.

The phase rule does not tell us the partial pressure p A in equilibrium with the The value of p A must be determined experimentally. The two phases can, of

selected x A

.

course, be gas-liquid, liquid-solid,

and so on. For example,

the equilibrium distribution

of acetic acid between a water phase and an isopropyl ether phase has been determined

experimentally

various conditions.

Gas-Liquid Equilibrium

10.2B

/.

for

To

Gas-liquid equilibrium data.

equilibrium data, the system

S0 2

air,

,

S0

and water are put

temperature

until

equilibrium

illustrate the

-air-water 2

will

obtaining of experimental gas-liquid be considered.

An amount

of gaseous

in

a closed container and shaken repeatedly at a given

is

reached. Samples of the gas and liquid are analyzed to

give the partial pressure p A in

atm of SO,

(A) in the gas

and mole fraction

x,, in

the

Figure 10.2-1 shows a plot of data from Appendix A. 3 of the partial pressure p A of in the vapor in equilibrium with the mole fraction x A of S0 2 in the liquid at 293 K

liquid.

S0 2

(20°C).

Often the equilibrium relation between p A in the gas phase be expressed by a straight-line Henry's law equation at low concentrations.

2.

Henry's law.

Pa

586

Chap. 10

= Hx A

andx^ can

(10.2-2)

Stage and Continuous Gas-Liquid Separation Processes

where

H

is

the Henry's law constant in

sides of Eq. (10.2-2) are divided

by

atm/mole fraction

total pressure

Va

P in

for the given system. If

both

atm,

= H'x A

(10.2-3)

where H' is the Henry's law constant in mole frac gas/mole frac liquid and is equal to H/P. Note that H' depends on total pressure, whereas H does not. In Fig. 10.2-1 the data follow Henry's law up to a concentration x A of about 0.005, where H = 29.6 atm/mol frac. In general, up to a total pressure of about 5 x 10 s Pa (5 atm) the value of H is independent of P. Data for some common gases with water are given in Appendix A. 3.

EXAMPLE

10.2-1. Dissolved Oxygen Concentration in Water be the concentration of oxygen dissolved in water at 298 when the solution is in equilibrium with air at 1 atm total pressure? The Henry's law constant is 4.38 x 10 4 atm/mol fraction.

What

K

will

Solution:

The

partial pressure

p A of oxygen (A)

in air

is

0.21 atm.

Using

Eq. (10.2-2), 0.21

Solving,

0

2 is

=

6 4.80 x 10~

dissolved in 1.0

= Hx A =

4.38 x

10^

x 10" 6 mol mol water plus oxygen or 0.000853 partOj/lOO parts

mol

fraction. This

means

that 4.80

water.

10.3

10.3A

SINGLE AND MULTIPLE EQUILIBRIUM CONTACT STAGES Single-Stage Equilibrium Contact

In many operations of the chemical and other process industries, the transfer of mass from one phase to another occurs, usually accompanied by a separation of the components of the mixture, since one component will be transferred to a larger extent than will

another component.

0.8 r

Mole fraction S0 2 Figure

Sec. 10.3

Single

10.2-1.

in liquid

Equilibrium plot for

phase, x A

S0 2 -vva(er system

and Multiple Equilibrium Contact Stages

at

293

K

(20°C).

587

Figure

A

Single-stage equilibrium process.

10.3-1.

Kj

one

single-stage process can be defined as

in

which two

different phases are

brought into intimate contact with each other and then are separated. During the time of contact, intimate mixing occurs and the various components diffuse and redistribute themselves between the two phases.

If mixing time is long enough, the components are two phases after separation and the process is considered

essentially at equilibrium in the

a single equilibrium stage.

A

L 0 and V2

,

equilibration occur, other.

can be represented as

single equilibrium stage

phases,

Making a

in Fig. 10.3-1.

known amounts and compositions, enter and the two exit streams L and V leave in x

total

The two entering and

the stage, mixing

of

equilibrium with each

x

mass balance,

L0

+ V2 = L + V = x

x

M

(10.3-1)

M

is kg (lb m ), V is kg, and is total kg. Assuming that three components, A, B, and C, are present making a balance on A and C,

where L

streams and

in the

L 0 x A0 + V2 y A2 = L x AX + V y AX =

Mx AM

(103-2)

LqX C o + V2Yc2 = L x ci + Kyci =

M x CM

(10.3-3)

x

x

i

An equation for B is not needed since x A + x B + x e — 1.0. The mass fraction of A in the L stream is x A andy^ in the V stream. The mass fraction of A in the stream isx AM To solve the three equations, the equilibrium relations between the components

M

.

must be known. In Section 10.3B, this will be done for a gas-liquid system and in Chapter for a vapor-liquid system. Note that Eqs. (10.3-l)~{10.3-3) can also be written mole units, with L and V having units of moles and x A and y A units of mole using 1 1

fraction.

Single-Stage Equilibrium Contact

10.3B

for

Gas-Liquid System

In the usual gas-iiquid system the solute

and

the liquid phase

in

L along with

insoluble in the water phase

phase

is

a

binary

A-B and

and

A

is

inert

in

the gas phase

V along ,

water C. Assuming that

with inert air

B

air is essentially

that water does not vaporize to the gas phase, the gas

the liquid phase

is

a binary A-C. Using moles and mole

fraction units, Eq. (10.3-1) holds for a single-stage process for the total material balance.

Since

component A

balance on

A can be

is

the only

component

that redistributes between the

two phases, a

written as follows.

(10.3-4)

where L

is moles inert water and usually known.

To

C

and

V

is

moles

inert air B.

Both£ and

solve Eq. (10.3-4), the relation between y Al and x Ai

in

V

are constant

equilibrium

is

given by

Henry's law.

588

Chap. 10

Stage and Continuous Gas-Liquid Separation Processes

(103-5) If

the solution

not dilute, equilibrium data in the form of a plot of p A or y A versus

is

x A must be available, such as in Fig. 10.2-1.

EXAMPLE 103-1. A gas mixture at

1.0

Equilibrium Stage Contact for

atm pressure abs containing

air

CO ^-Air-Water and CO z

is

contacted in

a single-stage mixer continuously with pure water at 293 K.. The two exit gas and liquid streams reach equilibrium. The inlet gas flow rate is 100 kg mol/h, = 0.20. The liquid flow rate entering is with a mole fraction of 2 of y A2

C0

300 kg mol water/h. Calculate the amounts and compositions of the two outlet phases. Assume that water does not vaporize to the gas phase. Solution:

The flow diagram

water flow

is

Eq.

the same as given in Fig. 10.3-1. The inert is obtained from mol/h. The inert air flow

is

V

L — L 0 = 300 kg

(10.3-6).

V Hence, the

inert air

flow

is

V=

Substituting into Eq. (10.3-4) to

°

\

+

l-o/'

=

-

V{1

(103-6)

y A)

K2 (l - y A2 ) = 100(1 - 0.20) = make a balance on C0 2 {A),

80 kg mol/h.

JT^)= 300(^) + ^r^-) -^! v-W 0 20 /

V1

'

(103-7)

V1

-

At 293 K, the Henry's law constant from Appendix A. 3 is H = 0.142 x 4 4 4 atm/mol frac. Then H' = H/P = 0.142 x 10 /1.0 = 0.142 x 10 mol 10 frac gas/mol frac liquid. Substituting into Eq. (10.3-5),

y Al

=

4

0.142 x 10 x^,

(103-8)

=

Substituting Eq. (10.3-8) into (10.3-7) and solving, x Ai y Al = 0.20. To calculate the total flow rates leaving,

x 10~ 4 and

10ft

I'

L

1.41

^T~^-l-1.41xl0-^ 300 kgmOl/b "

^r^=T^oIn this case, since the liquid solution

10.3C

/.

100

kgmol/h

so dilute,

is

L0 ^ L

l

.

Countercurrent Multiple-Contact Stages

Derivation of general equation.

transfer the solute say, the

V

l

However,

In Section

10.3A we used single-stage contact to

between the V and L phases. In order

to transfer

more

solute from,

stream, the single-stage contact can be repeated by again contacting the

stream leaving the streams.

A

this is

T°

first

wasteful of the

conserve use of the

L0 L0

L0

in

l

.

l

stream and to get a more concentrated product,

countercurrent multiple-stage contacting countercurrent heat transfer

V

This can be repeated using multiple stages. stream and gives a dilute product in the outlet L

stage with fresh

is

generally used. This

is

somewhat

similar to

a heat exchanger, where the outlet heated stream ap-

proaches more closely the temperature of the

inlet

hot stream.

The process flow diagram for a countercurrent stage process is shown in Fig. 10.3-2. The inlet L stream is L 0 and the inlet V stream is VN+ instead of K2 as for a single-stage in Fig. 10.3-1. The outlet product streams are V and L N and the total number of stages is N. The component A is being exchanged between the V and L streams. The V stream is ,

l

Sec. 10.3

Single

and Multiple Equilibrium Contact Stages

589

* 2

1

' 3

'

'

N

n

2

1

n+l

'

n

L2

L0 Figure

vV

L» _, Countercurrent multiple-stage process.

10.3-2.

composed mainly of component B and the L stream of component C. Components B and C may or may not be somewhat miscible in each other. The two-phase system can be gas-liquid, vapor-liquid, liquid-liquid, or other.

Making a

total overall

balance on

all

L 0 + VN+1 where VN +

stages,

=L N

M

+ V = l

(103-9)

M

mol/h entering, L N is mol/h leaving the process, and is the total flow. that any two streams leaving a stage are in equilibrium with each other. For example, in stage n, V n and L n are in equilibrium. For an overall component

Note

1

is

in Fig. 10.3-2

balance on A, B, or C,

+ viyi = Mx m

L 0 x 0 + JViJV+i = LnXn where used

.x

and y are mole

in these

fractions.

Flows

(103-10)

kg/h (lb^/h) and mass fraction can also be

in

equations.

Making a

total

balance over the

L0

Making a component balance over

first

n stages,

+ Vn+l

the

=L

+ V

n

(103-11)

x

n stages,

first

Lo*o + K+lJWi = L n x n + Solving for y„ +

,

Eq. (10.3-12).

in

L.x.

V This

L0

+

V

x

y

x

(10.3-13)

V

an important material-balance equation, often called an operating

is

the concentration y„ +l in the )),,

(103-12)

,

and x 0

are

V stream

L

with x„ in the

constant and usually

known

stream passing

or can

it.

line. It relates

The terms

K,,

be determined from Eqs.

(10.3-9H10.3-12). 2.

A

V

being transferred occurs when the solvent stream

with no

V

An important

Countercurrent contact with immiscible streams. is

C

and the solvent stream L contains A and

are immiscible in each other with only

plotted

on an xy

since the slope

plot (x A

LJVn+

x

A

C

with no B.

being transferred.

and y A of component A)

of the operating line varies

if

case where the solute

A and B The two streams L and

contains components

When

as in Fig. 10.3-3,

the

Eq. (10.3-13)

it is

L and V streams

is

often curved,

vary from stage

to stage.

In Fig. 10.3-3

is

plotted the equilibrium line that relates the compositions of

streams leaving a stage

in

equilibrium with each other.

To determine

the

number

two

of ideal

stages required to bring about a given separation or reduction of the concentration of

A

from y N + to y u the calculation is often done graphically. Starting at stage l,y l and x 0 are on the operating line, Eq. (10.3-13), plotted in the figure. The vapory; leaving is in t

equilibrium with the leaving x

x

and both compositions are on the equilibrium line. Then is in equilibrium withx 2 and so on. Each

y 2 and x x are on the operating line and y 2

590

Chap. 10

,

Stage and Continuous Gas-Liquid Separation Processes

drawn on Fig. 10.3-3. The steps are continued on the graph we can start at y N+l and draw the steps going toy,. yN + l If the streams L and V are dilute in component A, the streams are approximately constant and the slope LJV„ +l of Eq. (10.3-13) is nearly constant. Hence, the operating line is essentially a straight line on an xy plot. In distillation, where only components A and B are present, Eq. (10.3-13) also holds for the operating line, and this will be covered in Chapter 11. Cases where A, B, and C are appreciably soluble in each other often occur stage

is

represented by a step

until

is

reached. Alternatively,

in liquid-liquid extraction

and

will

be discussed in Chapter

12.

EXAMPLE 103-2. It is

desired to

Absorption of Acetone in a Countercurrent Stage Tower absorb 90% of the acetone in a gas containing 1.0 mol

%

acetone in air in a countercurrent stage tower. The total inlet gas flow to the tower is 30.0 kg mol/h, and the total inlet pure water flow to be used to absorb the acetone is 90 kg mol 2 0/h. The process is to operate isothermally at 300 K and a total pressure of 101.3 kPa. The equilibrium relation for the acetone (A) in the gas-liquid is y A = 2.53x A Determine the number of theoretical stages required for this separation.

H

.

The

Solution:

process flow diagram

is

similar to Fig. 10.3-3.

x A0 = 0, VN+1 = 30.0 kg mol/h, and Making an acetone material balance,

are y AN+

1

amount

=

0.01,

of entering acetone

=

y AN +

entering air

=

(1

=

29.7 kg

=

0.10(0.30)

=

LN =

0.90(0.30)

= 0.27

acetone leaving in Vx acetone leaving in

Figure

Sec. 10.3

10.3-3.

Single

Number of stages

l

VN+1 =

0.01(30.0)

- yAN+ JJ^

in a

mol

L0

,

=

=

Given values kg mol/h.

90.0

0.30 kg mpl/h

=(1-0.01X30.0)

air/h

0.030 kg mol/h

kg mol/h

countercurrent multiple-stage contact process.

and Multiple Equilibrium Contact Stages

591

0.012r

X

XAN

XA0 Mole fraction acetone Figure

10.3-4.

K,

in water,

Theoretical stages for countercurrent absorption

=

29.7

=

0.03

4-

mol

29.73 kg

air

+

Example

in

10.3-2.

acetone/h

0.030

^=2^= a0° LN =

90.0

=

0.27

mol water + acetone/h

90.27 kg

0.27

=

**"

+

101

a0030° 9027 =

Since the flow of liquid varies only slightly from

LN =

L0 =

90.0 at the inlet

LJVn+ of the operating line in Eq. (10.3-13) is essentially constant. This line is plotted in Fig. 10.3-4 and the equilibrium relation y A = 2.53X,, is also plotted. Starting to

90.27 at the outlet and

V from

at point y Al x A0 the stages are stages are required. ,

,

drawn

as

shown. About

,

5.2 theoretical

Analytical Equations for Countercurrent Stage Contact

10.3D

When

30.0 to 29.73, the slope

the flow rates

V and L

in a

operating-line equation (10.3-13)

countercurrent process are essentially constant, the

becomes

straight. If the equilibrium line

is

also a

straight line over the concentration range, simplified analytical expressions can be

derived for the

number

of equilibrium stages in a countercurrent stage process.

Referring again to Fig. 10.3-2, Eq. (10.3-14)

component

is

an overall component balance on

A.

^o*o+ VN+ iy» + =L*x N + V y l

i

l

(10.3-14)

Rearranging,

l n x n ~ VN+iys+i = L 0 *o - Kyi Making

a

component balance

for

A on

the

first

(10.3-15)

n stages,

L 0 x 0 + Vn+i yn+l =L„x„+ Viyi

(103-16)

Rearranging,

L0 x 0 592

Chap. JO

-V y =L ,x l

l

l

n

-V„ +l yn+l

(10.3-17)

Stage and Continuous Gas-Liquid Separation Processes

Equating Eq. (10.3-15)

to (10.3-17),

L„x n — Vn+i y„ + Since

= LN x N — VN +

= LN =

molar flows are constant, L„

the

constant

i

=

V.

l

yN+

constant

(10.3-18)

!

= L and Vn+1 = VN+l

Then Eq. (10.3-18) becomes

Ux„ - x N = V(y n+l - yN+l )

(103-19)

)

Since y n+l and x„ +1 are in equilibrium, and the equilibrium line is straight, y n + = mx n+l Also y N+l = mx N+l Substituting mx n+l fory„ + and calling A = L/mV, Eq. .

.

(10.3-19)

1

becomes n

where A

is

+

Ax„

l

an absorption factor and

All factors

on

=

Ax.

(103-20)

constant.

is

the right-hand side of Eq. (10.3-20) are constant. This equation

and can be solved by

linear first-order difference equation

is

a

the calculus of finite-difference

methods (Gl, Ml). The final derived equations are as follows. For transfer of solute A from phase L to V (stripping),

x0

N+l -(1/A) N+l (l/A)

Xh

Xn

(l/A)

- {y N+ iM

(103-21)

1

x0

log

-

(yN+J m )

x N -{y N+ Jm)

N=

(1

-A) + A (103-22)

log(iM)

When A =

1,

N= For

A from

transfer of solute

-

(y N+l /m)

V

to

L (absorption),

phase

A N+l yN +

l

-mx

A

N

-A

N+l

log

(103-23)

xN

-

y N+

(103-24) 1

i

+

=•

(103-25) log

When A =

A

1,

N=

- yi - mx 0

yN+i yl

(103-26)

Often the term A is called the absorption factor and S the stripping factor, where S = l/A. These equations can be used with any consistent set of units such as mass flow and mass

molar flow and mole fraction. Such series of equations are often Kremser equations and are convenient to use. If A varies slightly from the inlet outlet, the geometric average of the two values can be used, with the value of m dilute end being used for both values of A.

called

fraction or

EXAMPLE 103-3. Repeat Example

Number of Stages by

10.3-2

but

use

the

to the at the

Analytical Equation.

Kremser

analytical

equations for

countercurrent stage processes.

Sec. 10.3

Single

and Multiple Equilibrium Contact Stages

593

At one end of the process at stage

Solution:

y Al =0.00101, L 0 = 90.0, and x^ 0 = y A = 2.53x^ where m = 2.53. Then,

L0

90.0

mV

2.53 x 29.73

_ L

mV

1

At stage N,

VN+

,

=

30.0,

y

y AN +

=

x

0.01,

L"

AN

1,

LN =

90.27,

9Q 27 -

mKN+1

V = x

29.73 kg mol/h,

Also, the equilibrium relation

0.

x

2.53

and x^ N

=

is

0.00300.

=119

30.0

The geometric average A = ^4^4^= ^Zl.20 x 1.19 = 1.195. The acetone solute is transferred from the V to the L phase (absorption). Substituting into

0.01

log

Eq. (10.3-25),

-

2.53(0)

+

,

1

0.00101 -2.53(0)'

N=

I.I95)

I.I95]

=

5.04 stages

log (1.195)

This compares closely with 5.2 stages obtained using the graphical method.

MASS TRANSFER BETWEEN PHASES

10.4

Introduction and Equilibrium Relations

10.4A /.

Introduction to interphase mass transfer.

from a

fluid

In

Chapter 7 we considered mass transfer A was

phase to another phase, which was primarily a solid phase. The solute

mass transfer and through the concerned with the mass transfer of

usually transferred from the fluid phase by convective solid

by

solute

diffusion. In the present section

A

from one

we

shall be

phase by convection and then through a second

fluid

convection. For example, the solute

through and be absorbed

in

may

diffuse

fluid phase by through a gas phase and then diffuse

an adjacent and immiscible liquid phase. This occurs

in the

case of absorption of ammonia from air by water.

The two phases

are in direct contact with each other, such as in a packed, tray, or

spray-type tower, and the interfacial area between the phases In two-phase

mass transfer most cases. 2.

mass

transfer, a

concentration gradient

to occur. At the interface

Equilibrium relations.

between the two

Even when mass

transfer

is

usually not well defined.

is

will exist in

each phase, causing

fluid phases,

equilibrium exists in

occurring equilibrium relations are

important to determine concentration profiles for predicting rates of mass

transfer. In

Section 10.2 the equilibrium relation in a gas-liquid system and Henry's law were discussed. In Section 7.1C a discussion covered equilibrium distribution coefficients

between two phases. These equilibrium relations

will

be used

in

discussion of mass

transfer between phases in this section.

10.4B

Concentration Profiles

in

Interphase

In the majority of mass-transfer systems,

Mass Transfer

two phases, which are

essentially immiscible in

each other, are present and also an interface between these two phases. Assuming the solute

A

is

diffusing

from the bulk gas phase

phase G, through the

594

interface,

Chap. 10

and then

G

to the liquid

into phase

L

phase L,

in series.

A

it

must pass through

concentration gradient

Stage and Continuous Gas-Liquid Separation Processes

Figure

Concentration

10.4-1.

solute

A

liquid -phase solution

of through

profile

diffusing

[

of

A

L

in liquid

gas-phase mixture

|Of^4LngasG

two phases.

I—

»

~N~A

2j

*AL

interface |

distance from interface

cause this mass transfer through the resistances in each phase, as shown in The average or bulk concentration of A in the gas phase in mole fraction units is y AG where y AG = pJP, and x AL in the bulk liquid phase in mole fraction units. The concentration in the bulk gas phase y AG decreases loy Ai at the interface. The

must

exist to

Fig. 10.4-1.

,

x Ai at the interface and falls to x AL At the interface, since would be no resistance to transfer across this interface, y Ai and x M are in equilibrium and are related by the equilibrium distribution relation liquid concentration starts at

.

there

y Ai where y Ai If the

is

=/(*J

(10.4-1)

a function of x Ai They are related by an equilibrium plot such as Fig. 10.1-1. .

system follows Henry's law, y A P or p A and x A are related by Eq. (10.2-2) at the

interface.

Experimentally, the resistance at the interface has been

shown

to be negligible for

most cases of mass transfer where chemical reactions do not occur, such as absorption of common gases from air to water and extraction of organic solutes from one phase to another. However, there are some exceptions. Certain surface-active compounds may concentrate at the interface and cause an "interfacial resistance" that slows down the diffusion of solute molecules. Theories to predict

are

still

10.4C

when

interfacial resistance

may occur

obscure and unreliable.

Mass Transfer Using Film Mass-Transfer

Coefficients

and Interface Concentrations

Equimolar counterdiffusion.

For equimolar counterdiffusion the concentrations of on an xy diagram in Fig. 10.4-2. Point P represents the bulk phase compositions x AG and x AL of the two phases and point the interface concentrations y Ai and x Ai For A diffusing from the gas to liquid and B in equimolar counterdiffusion from liquid to gas, 1.

Fig. 10.4-1 can be plotted

M

.

NA =

K(y AG

-

y Ai )

=

k'x (x Ai

-

x AL)

(10.4-2)

2 mol frac (g mol/s where k'y is the gas-phase mass-transfer coefficient in kg mol/s m 2 2 cm mol frac, lb mol/h ft mol frac) and k'x the liquid-phase mass-transfer coefficient in kg mol/s m 2 mol frac (g mol/s cm 2 mol frac, lb mol/h ft 2 mol frac). Rearranging •

•

•

•

•

•

-

-

Eq. (10.4-2),

-^= kf

yAG

- yAi

X AL

X Ai

(10.4-3)

The driving force in the gas phase is (y AG — y Ai) and in the liquid phase it is (x,,,- — x AL ). The slope of the line PM is — k'Jk'y This means if the two film coefficients k'x and/c, are .

known,

the interface compositions

— k'Jk'y

intersecting the equilibrium line.

Sec. 10.4

can be determined by drawing

Mass Transfer Between Phases

line

PM

with a slope

595

slope

=-k'jk'

y

equilibrium line

AL FIGURE

10.4-2.

Ai

Concentration driving forces and interface concentrations phase mass transfer (equimolar counterdiffusion).

in

inter-

The bulk-phase concentrations y AG and x AL can be determined by simply sampling mixed bulk gas phase and sampling the mixed bulk liquid phase. The interface concentrations are determined by Eq. (10.4-3). the

2. Diffusion

of

A

For the common case of A and then through a stagnant liquid phase, the 10.4-3, where P again represents bulk-phase compo-

through stagnant or nondiffusing B.

diffusing through a stagnant gas phase

concentrations are shown

in Fig.

M

and interface compositions. The equations gas and then through a stagnant liquid are

sitions

NA =

k (y AG y

-

y Ai )

for

=

k x (x Ai

k

=

A

-

diffusing through a stagnant

x A[)

(10.4-4)

Now, k'

slope

=

k'

(10.4-5)

k'x /(\-x A ) iM

yAG

equilibrium

line

xAi FIGURE

596

10.4-3.

Concentration driving forces and interface concentrations phase mass transfer (A diffusing through stagnant B).

Chap. 10

in

inter-

Stage and Continuous Gas— Liquid Separation Processes

where

-^ =

n (1

(i

\

- yJ -

.n [(

(1

~

y AG )

einA

l-^/(l-^)]

^

X

-

(l

=

*A) iM

/Vr^~u,~

"n

(

10 - 4" 7>

Then, ky

"a = Note (1

that

— xA

(1

) iM is

—

-

=

ix

:

M - x AL

(10.4-8)

)

is the same asy BM of Eq. (7.2-11) but is written for the interface, and same as x BM of Eq. (7.2-1 1). Using Eq. (10.4-8) and rearranging,

y A ) iM

the

— xA

~~ k'J{\

The

-

(y AG

:

/ .

PM in Fig.

slope of the line

)

iM

y A )iM

yag

~~

X AL

-

yAi

(10 4-9) *Ai

10.4-3 to obtain the interface compositions

is

given

by the left-hand side of Eq. (10.4-9). This differs from the slope of Eq. (10.4-3) for equimolar counterdiffusion by the terms (1 — y A ) iM and (1 — x A ) iM When A is diffusing through stagnant B and the solutions are dilute,(l — y A ) iM and(l — x A ) s/li are close to 1. .

A

trial-and-error

method

needed to use Eq. (10.4-9) to

is

get the slope, since the

—

left-hand side contains y Ai and x Ai that are being sought. For the first trial (1 y A ) iM and x are assumed and Eq. is used to get the slope andj^,to be 1.0 (10.4-9) andx^ (1 A ) iM

—

values.

Then

for the

second

trial,

these values of y Ai

new

and x Ai are used

to calculate a

new

values of y Ai and x M This is repeated until the interface compositions do not change. Three trials are usually sufficient. slope to get

.

EXAMPLE

10.4-1.

The

A

Interface Compositions in Interphase Mass Transfer being absorbed from a gas mixture of A and B in a wetted-wall tower with the liquid flowing as a film downward along the wall. At a certain point in the tower the bulk gas concentration y AG = 0.380 mol solute

fraction

is

and the bulk liquid concentration is x AL = 0.100. The tower is K and 1.013 x 10 5 Pa and the equilibrium data are as

operating at 298 follows:

XA

The

solute

A

0

0

0.20

0.131

0.05

0.022

0.25

0.187

0.10

0.052

0.30

0.265

0.15

0.087

0.35

0.385

diffuses

through stagnant

B

in the

gas phase and then through

a nondiffusing liquid.

Using correlations for dilute solutions

in

wetted-wall

towers,

the

phase is predicted as k y = 3 2 1.465 x 10" kg mol A/smol frac(1.08 lbmol/h- ft 2 mol frac) and for -3 the liquid phase as k x = 1.967 x 10 kg mol A/s- z mol frac (1.45 lb 2 mol/h ft mol frac). Calculate the interface concentrations y A; and x Ai and

film mass-transfer coefficient for

m

the flux

NA

Solution:

Sec. 10.4

A

in the gas

-

.

Since the correlations are for dilute solutions,

Mass Transfer Between Phases

(1

— yA

) iU

and

597

— xA

are the same as k'y and ) iM are approximately 1.0 and the coefficients The equilibrium data are plotted in Fig. 10.4-4. Point P is plotted at y AG = 0.380 and x AL = 0.100. For the first trial (1 - y A ) M and (1 - x A ) iht are assumed as 1.0 and the slope of line PM is, from Eq. (10.4-9),

(1

k'x

A

.

line

fr'x/U

1-967

x 10-yi.O

K/d-yAv

1.465

xl0- 3 /1.0

through point

P

with a slope of —1.342

intersecting the equilibrium line at

For the second the

new

M„ where y Ai

we use y Ai and

trial

x^,

slope. Substituting into Eqs. (10.4-6)

( 1

-

YaY.m

=

-

(1

In [(1

- yA

m-

0.183)

— 0.

1

(1

83)/(

1

is

=

plotted in Fig. 10.4-4

0.183 a.ndx Ai

from the first and (10.4-7),

-yj-d -y M

(l

in [(l

_

=

trial to

0.247.

calculate

)

y A c)l

- 0.380) - 0.380)]

0.715

(\-x AL )-(l-xJ (1

In

_

[d-xj/d-xj] -

(1

In [(1

.

0.100)

-

-

(1

0.100)/(1

Substituting into Eq.( 10.4-9) to obtain the

-x A iM ^d-yJ.M

k'Jjl

A

line

through point

equilibrium line at

with a slope of

0.825

slope,

-1.163

x 10-V0-715

1.465

M, where yA; =

Using these new values

new

3 1.967 x lQ- /0.825

)

P

- 0.247) = - 0.247)]

—

0.197

1.163

plotted and intersects the

is

andx Ai =

0.257.

for the third trial, the following values are

calculated: (1 1

598

n)iM ~

Chap. 10

-

In [(

1

0.197)

-

0.

1

-

(1

97)/(

1

-

0.380) 0.380)]

-

U

"

IW

Stage and Continuous Gas-Liquid Separation Processes

m U

_

(i

xj« -

- (i - Q.257) _ o.100)/(1 - 0.257)] ~ V * M

Q.ioo)

i

i

n

[(1

_KA^AtL=

3 1.967 x 10- /0.820

yd-yJiM

3 1.465 x 10- /0-709

_

This slope of — 1.160 is essentially the same as the slope of — 1.163 for the second trial. Hence, the final values arey^,- = 0.197 andx Ai = 0.257 and are as point M. To calculate the flux,

shown

= Na =

x 10-* kg

3.78 1

Eq. (10.4-8)

08

07(i

(0-380

k'

mol/sm 2

- 0.197) =

,

(x

,

"-

=

(T=tu

=

4 3.78 x 10" kg

used.

is

0.2785 lb mol/h

=

Xal)

1.967 x 10"

2

3

(0

'

257

0.820

,

-

aiO0)

mol/s-m 2

that the flux N A through each phase which should be the case at steady state.

Note

10.4D

-ft

is

same

the

as in the other phase,

Overall Mass-Transfer Coefficients

and Driving Forces

1.

Introduction.

Film or single-phase mass-transfer coefficients k'y and k'x or k y and k x measure experimentally, except in certain experiments designed so

are often difficult to

that the concentration difference across result, overall

one phase

small and can be neglected.

is

As

a

mass-transfer coefficients K'y and K'x are measured based on the gas phase

method is used in heat transfer, where overall heat-transfer coefmeasured based on inside or outside areas instead of film coefficients.

or liquid phase. This ficients are

The overall mass

transfer K' y

is

defined as

NA = where K'y y*

is

is

-

K'y (y AG

(10.4-10)

y*)

based on the overall gas-phase driving force

the value that would be in equilibrium with x AL as ,

in

kgmol/s

shown

m

2

mol

in Fig. 10.4-2.

frac,

and

A\so,K'x

is

defined as

NA = where K'x

is

-x A[)

K'x (x A

(10.4-11)

based on the overall liquid-phase driving force

would be

equilibrium with y AG

x*

is

2.

Equimolar counter diffusion and/or diffusion

the value that

in

holds for equimolar counterdiffusion, or

when

in

in

kg mol/s

•

m

2 •

mol

frac

and

.

dilute solutions.

Equation (10.4-2)

the solutions are dilute, Eqs. (10.4-8)

and

(10.4-2) are identical.

NA = From

k y (y AG

y Ai )

=

-

yAi)

k'x (x Ai

-

x AI)

(10.4-2)

Fig. 10.4-2,

Vac -y*A

Sec. 10.4

-

=

iy AG

Mass Transfer Between Phases

+

iy

M - y\)

(10.4-12)

599

Between the points £ and

M the slope m' can be given as yA

,

i

-

(10.4-13)

X AL

X Ai Solving Eq.

-

(10.4- 1 3) for (y Ai

and substituting into Eq.

- y* =

yAG

Then on

y*)

iy ag

-

y A i)

'(

x Ai

- *ai)

left-hand side of Eq. (10.4-15)

and equals the gas

In a similar

+

on the

overall gas driving .

y

manner from

Fig. 10.4-2,

~

*a

W"x

x al =

y *°~

m" =

+

a.)

(*Ai

~

x al)

(10.4-16)

yM

~

XA

(10.4-17)

X Ai

as before,

(104- 18)

k"M"k now

Several special cases of Eqs. (10.4-15) and (10.4-18) will

numerical values of k'x and are very important.

If

m

k'

y

are very roughly similar.

hence the term m'/k'x

in

Eq. (10.4- 15)

in

M has

major resistance

moved down

Similarly,

small,

A

is in

is

=

when m" and

is

-

3.

air

is

controlling" and x Ai

by water are similar to Eq.

Diffusion

ofA

^

(10.4-19)

A

very large, the solute

x

"liquid phase

controlling."

The

point

=

(10.4-20)

very insoluble in the liquid, l/(m"k')

is

1

(10.4-21)

tt k'x

x*. Systems for absorption of oxygen

/ 1

I

-

y*A

=

^

1

[y ag

(Vag

:

~

/A)iM

-

Chap. 10

orC0 2

(10.4-21).

For the case of A diffusing through

through stagnant or nondiffusing B.

=

600

is

yAi

nondiffusing B, Eqs. (10.4-8) and (10.4-14) hold and Fig. 10.4-3

yAG

is

in

then very soluble in the liquid phase, and

= yA c -

y*A

1

from

The

or m"

the gas will give a large value of x A

the gas phase, or the "gas phase

— = K The

m

very close to E, so that

yA G

becomes

be discussed.

values of the slopes

very small. Then,

is

7^7

the

The

quite small, so that the equilibrium curve in Fig. 10.4-2

is

almost horizontal, a small value of y A equilibrium in the liquid. The gas solute

and

""M5)

(

K

film resistance l/k' plus the liquid film resistance m'/k'x

x

Proceeding

and canceling outN^

the total resistance based

is

10.4-1 4)

(

m

1

K~K force

+m

(10.4-1 2),

substituting Eqs. (10.4-10) and (10.4-2) into (10.4-14) 1

The

y*

y A i)

=

r. \

+ ™\x Ai -

l

x al)

^X

:

is

(

used.

x a;

-

x al)

(10.4-8)

AliM (i

0.4-14)

Stage and Continuous Gas-Liquid Separation Processes

We must, however, define the equations for K' .(1

-

(yxo

-y A ).M

the flux using overall coefficients as follows:

y*)

=

The bracketed terms are often written as

- yA

(i

follows:

K = ).

(10.4-22)

- XA ).M J

(i

M

(10.4-23)

-

(1

x a)*m

where K y is the overall gas mass-transfer coefficient for A diffusing through stagnant B and K x the overall liquid mass-transfer coefficient. These two coefficients are concentration-dependent. Substituting Eqs. (10.4-8) and (10.4-22) into (10.4-14), we obtain 1

1

- y A\ M

K'y /( 1

+

KK " yJm

k 'J( 1

1

-

(10.4-24)

x a),m

where (l

- yJ.M

(i

In

-y%-(i-y M [(1 -yj)/(l -y AG y] )

(10.4-25)

Similarly, for l

Kic/U

where

-

(1

-

In It

-

m"k'y/(l

)

(1

1

1

" xA M

+ k'J(l

y A ) iM

x AI)

W -x

-

(1

At)/(l

(10.4-26)

- xA M )

- x5) - x*y]

(10.4-27)

should be noted that the relations derived here also hold for any two-phase system

where y stands

for the

one phase and x

EXAMPLE

For example, for the by isopropyl ether (x phase),

for the other phase.

extraction of the solute acetic acid (A) from water (y phase) the same relations will hold.

Overall Mass-Transfer Coefficients from Film Coefficients Using the same data as in Example 10.4-1, calculate the overall masstransfer coefficient K' the flux, and the percent resistance in the gas and r liquid films. Do this for the case of A diffusing through stagnant B. 10.4-2.

From Fig. 10.4-4, y\ = 0.052, which is in equilibrium with the bulk liquid x AL = 0.10. Also, y AG = 0.380. The slope of chord m' between E and from Eq. (10.4-13) is, tory Ai = 0.197 andx^,- = 0.257,

Solution:

M

0.197

yAi

0.257

From Example

-

0.052

=

10.4-1,

1.465 x 10"

-

(i

Using Eq.

yJ.M

0.709

1.967 x 10 (1

-

x A)

M

-

3

0.820

(10.4-25),

(1

Sec. 10.4

0.923

0.100

Mass

-

(1

y„).*

-y AG

)

[(1-^3/(1-^)] (1 - 0.052) -(1 - 0.380) = In [(1 - 0.052)/(l - 0.380)]

In

Transfer Between Phases

0.773

601

Then, using Eq.

(10.4-24),

1

K;/0.773

= ~ =

0.923

1

+

3 1.465 x 10" /0.709

+

484.0

4 8.90 x 19~

384.8

=

3 1.967 x 10- /0.820

868.8

Solving, K'y = The percent resistance in the gas film is (484.0/868.8)100 = 55.7% and 44.3%' in the liquid film. The flux is as follows, using Eq. (10.4-22):

w

.

-(r^u^=

4 3.78 x 10" kg

This, of course,

is

the

same

rt,

=

8

^r

(0380

-

0052)

mol/s-m 2

flux value as

was calculated

in

Example

10.4-1

using the film equations. 4.

Discussion of overall coefficients.

phase as

If

the two-phase system

is

such that the major

mass should be centered on increasing the gas-phase turbulence, not the liquid-phase turbulence. For a two-phase system where the liquid film resistance is

resistance

in the gas

is

in Eq. (10.4-19), then to increase the overall rate of

transfer, efforts

controlling, turbulence should be increased in this phase to increase rates of

mass

transfer.

To

design mass-transfer equipment, the overall mass-transfer coefficient

is

syn-

thesized from the individual film coefficients, as discussed in this section.

CONTINUOUS HUMIDIFICATION PROCESSES

10.5

10.5A

J.

Introduction and Types of Equipment for Humidification

Introduction to gas-liquid contactors.

contacted with gas that

is

unsaturated,

When some

a relatively

of the liquid

warm is

liquid

vaporized.

is

directly

The

liquid

drop mainly because of the latent heat of evaporation. This direct contact of a gas with a pure liquid occurs most often in contacting air with water. This is done for the following purposes: humidifying air for control of the moisture content of air in drying or air conditioning; dehumidifying air, where cold water condenses some water vapor from warm air; and water cooling, where evaporation of water to the air temperature

cools

warm

will

water.

In Chapter 9 the fundamentals of humidity and adiabatic humidification were

and design of continuous air-water contactors on cooling of water, since this is the most important type of process in the process industries. There are many cases in industry in which warm water is discharged from heat exchangers and condensers when it would be more economical to cool and reuse it than to discard it.

discussed. In this section the performance is

2.

considered.

The emphasis

Towers for water

is

cooling.

In a typical

water-cooling tower,

countercurrently to an air stream. Typically, the

tower and cascades

down through

warm

warm

water flows

water enters the top of a packed

the packing, leaving at the bottom. Air enters at the

and flows upward through the descending water. The tower packing often consists of slats of wood or plastic or of a packed bed. The water is distributed by troughs and overflows to cascade over slat gratings or packing that provide large interfacial areas of contact between the water and air in the form of droplets and films

bottom of

602

the tower

Chap. 10

Stage and Continuous Gas-Liquid Separation Processes

of water.

The flow of

warm

of the

air

air in the

upward through the tower can be induced by the buoyancy tower (natural draft) or by the action of a fan. Detailed

descriptions of towers are given in other texts (Bl, Tl). the wet bulb temperature. The driving force for approximately the vapor pressure of the water less the vapor pressure it would have at the wet bulb temperature. The water can be cooled only to the wet bulb temperature, and in practice it is cooled to about 3 K or more above this.

The water cannot be cooled below

the evaporation of the water

is

Only a small amount of water heat of vaporization of water

is

lost by evaporation in cooling water. Since the latent about 2300 kJ/kg, a typical change of about 8 in water

is

K

temperature corresponds to an evaporation loss of about 1.5%. Hence, the total flow of

water

is

usually assumed to be constant in calculations of tower

size.

In humidification and dehumidification, intimate contact between the gas phase and liquid phase

is

needed

for large rates

The gas-phase

of mass transfer and heat transfer.

resistance controls the rate of transfer. Spray or packed towers are used to give large interfacial areas

to

promote turbulence

Theory and Calculation

10.5B /.

and

of

Temperature and concentration

in the

gas phase.

Water-Cooling Towers profiles at interface.

In Fig. 10.5-1 the temperature

and the concentration profile in terms of humidity are shown at the water-gas interface. Water vapor diffuses from the interface to the bulk gas phase with a driving force in the gas phase of (H, — H G ) kg H z O/kg dry air. There is no driving force for mass profile

transfer in the liquid phase, since water

TL — T t

in the liquid

phase and

is

a pure liquid.

— TG K

7]

from the bulk liquid to the interface

The temperature

driving force

in the liquid. Sensible heat also flows

interface to the gas phase. Latent heat also leaves the interface in the

diffusing to the gas phase.

The

is

or °C in the gas phase. Sensible heat flows

from the

water vapor,

sensible heat flow from the liquid to the interface equals

the sensible heat flow in the gas plus the latent heat flow in the gas.

The conditions

occur at the upper part of the cooling tower. In the is higher than the wet

in Fig. 10.5-1

lower part of the cooling tower the temperature of the bulk water bulb temperature of the air but

may

be below the dry bulb temperature. Then the

direction of the sensible heat flow in Fig. 10.5-1

2.

Rate equations for heat and mass

transfer.

is

reversed.

Wc

shall consider a

packed water-cooling

interface

liquid water

Hq

humidity

sensible heat in liquid

sensible heat in gas

FIGURE

Sec. 10.5

10.5-1.

Temperature and concentration

Continuous Humidification Processes

profiles in

upper part of cooling tower.

603

downward in the tower. The and water phases is unknown, since the surface area of the packing is not equal to the interfacial area between the water droplets and the air. 2 Hence, we define a quantity a, defined as m of interfacial area per m 3 volume of 2 3 packed section, or m /m This is combined with the gas-phase mass-transfer coefficient 2 Pa or kg mol/s -m 2 atm to give a volumetric coefficient k G a in k G in kg mol/s -m 3 kg mol/s m volume Pa or kg mol/s m 3 atm (lb mol/h ft 3 atm). The process is carried out adiabatically and the various streams and conditions are shown in Fig. 10.5-2, where tower with

air

flowing upward and water countercurrently

between the

total interfacial area

air

.

-

-

•

L=

TL = G=

•

water flow, kg water/s

dry

air flow, kg/s

temperature of

H=

humidity of

y

m2

(lbjli

temperature of water, °C or

TG =

H =

•

•

m

air,

air,

2

•

•

(lbjh

K

•

2 •

ft

)

(°F) 2

•

ft

)

K (°F)

°C or

kg water/kg dry

water/lb dry

air (lb

enthalpy of air-water vapor mixture, J/kg dry

The enthalpy H y

as given in Eq. (9.3-8)

H = c (TH = c (T y

s

T0 )+ HX 0 =

(1.005

y

s

T0 + HX 0 =

(0.24

)

air)

air (btu/lb m

dry

air)

is

+

1.88/f)10

3

(T

+ 0.45HXT -

-

32)

+

0)

+

2.501 x 10

6

H

1075.4H

(SI)

(English) (9.3-8)

The base temperature

selected

is

0°C or 273

K (32°F). Note that(T - T0 )°C = (T - T0

)

K.

Making a operating line

Figure

is

total heat balance for the dashed-line

box shown

Fig. 10.5-2,

in

an

obtained.

10.5-2.

Continuous count ercurrent adiabatic water cooling.

G2

water

H2 Hyi

G TG +dTG

H+dH Hy +dH

L+dL TL +dTL dz

—

G TG \-

L

H

Li

i

___

J

air

H 604

Chap. 10

Stage and Continuous Gas-Liquid Separation Processes

G(H,-H, This assumes that

L is

l

)

-TLl

= Lc L (TL

(103-1)

)

only a small amount

essentially constant, since

is

evaporated.

The

K

3 heat capacity c L of the liquid is assumed constant at 4.187 x 10 J/kg (1.00 btu/lb m this is a versus Eq. (10.5-1) straight line with a °F). When plotted on a chart of L y •

H

Making an

slope of LcJG.

T

•

,

overall heat balance over both ends of the tower,

-H

G(H y2 Again making a heat balance

yl )

-T

= Lc L (TL2

for the dz

(103-2)

L1 )

column height and neglecting

sensible heat

terms compared to the latent heat,

= G dH y

Lc L dTL

The total

sensible heat transfer from the bulk liquid to the interface

3 ft

•

hL

a

°F) and

GdHy = h L a

dTL =

Lc L

where

(103-3)

dz{TL

-

(refer to Fig. 10.5-1)

is

(103-4)

jQ

the liquid-phase volumetric heat-transfer coefficient in

is

Tj is

W/m -K 3

(btu/h-

the interface temperature.

For adiabatic mass

due

transfer the rate of heat transfer

to the latent

heat in the

water vapor being transferred can be obtained from Eq. (9.3-16) by rearranging and using

a volumetric basis.

=M ^ A where

qJA

is in

W/m 2

(btu/h

2 -

ft

),

kG

B

i

M B = molecular m

mass-transfer coefficient in the gas in kg mol/s latent heat of water in J/kg water, /J,

water/kg dry

dry

air.

The

air,

and

HG

3

q

is

in

W/m 2

W/m K. Now from Fig.

weight of

k G a is a volumetric pressure in Pa, l 0 is the

air,

P = atm

Pa,

the humidity of the gas at the interface in kg

is

rate of sensible heat transfer in the gas

qJA

(103-5)

the humidity of the gas in the bulk gas phase in

is

-f A

where

H G )dz

aPX 0 (H ~

and

li

c

a

is

=

h c a(T

i

kg water/kg

is

-TG )dz

(10.5-6)

a volumetric heat-transfer coefficient in the gas in

3

10.5-1,

GdH y = Equation

Eq. (10.5-4) must equal the

M B k G aP/.

0

{Hi

- HG

)

sum

+

dz

of Eqs. (10.5-5)

h c a(7]

- TG

)

and

dz

(10.5-6).

(10.5-7)

(9.3-18) states that

h° a

MB k Substituting

Pk G a

y

a

=c s

(10.5-8)

for k a, y

=c s

M B Pk G a

(103-9)

Substituting Eq. (10.5-9) into Eq. (10.5-7) and rearranging,

GdH = y

M

B

kG

Adding and subtracting cs T0

G dH f =

Sec.

1

0.5

aP

dz

Uc s

T,

+

\

0

Hi)

-

(c s

TG +

\

0

HG

)]

(103-10)

inside the brackets,

M B k G aP dz{c

s

(T

:

- T0 + H,X 0 )

Continuous Humidification Processes

[c s

(TG

- T0 + H G A 0 ]} )

(10.5-11)

605

The terms

inside the braces

are(H yi

— Hy\ and Eq. (10.5-11)

M B k G aP dz(H - H

G dH y =

yi

becomes (10.5-12)

y)

Integrating, the final equation to use to calculate the tower height

dz

If

Eq. (10.5-4)

is

z

=

G

Hyj

—H

y

(10.5-14)

aM B P

Design of Water-Cooling Tower Using Film Mass-Transfer Coefficients

The tower design The enthalpy in

(10.5-13) y

equated to Eq. (10.5-12) and the result rearranged,

kG

1.

_dHy_

1

M B k G aP Jw,i H yi~ H hL a

10. 5C

f"'

is

Fig.

is

done using

of saturated air

10.5-3.

the following steps.

H

This enthalpy

yi

is

humidity from the humidity chart

is

plotted versus

7]

on an

H versus

T

plot as

shown

calculated with Eq. (9.3-8) using the saturation for a given temperature, with

0°C (273 K)

as a base

temperature. Calculated values are tabulated in Table 10.5-1.

and H lt the enthalpy of this air yl is and TL1 (desired leaving water temperature) is plotted in Fig. 10.5-3 as one point on the operating line. The operating line is plotted with a slope LcJG and ends at point TL2 which is the entering water temperature. This gives H y2 Alternatively, H y2 can be calculated from Eq. (10.5-2). Knowing h L a and k G a, lines with a slope of —h L a/k G aM B P are plotted as shown in

Knowing

2.

the entering air conditions

calculated from Eq. (9.3-8).

The point

H

TGl

Hyl

,

.

3.

Fig. 10.5-3.

point

From

Eq. (10.5-14) point

M represents H

yi

and

7],

P

represents

Hy

and

TL on

the operating line, and

the interface conditions. Hence, line

MS orH - H yi

y

represents the driving force in Eq. (10.5-13).

o a.

1

.a

U >•

3

operating line, slope = Lcl/G

£ oo

T3

slope

00 -x

-

- h^a k G aM B P

Liquid temperature (°C) Figure

10.5-3.

Temperature enthalpy diagram and operating

line

for water-cooling

lower.

606

Chap. 10

Stage and Continuous Gas-Liquid Separation Processes

Table

Enthalpies of Saturated Air-Water Vapor Mixtures (0°C Base Temperature)

10.5-1.

J

btu

°F

°C

60

15.6

80

m dry

95

55.4

The driving force Then by plotting performed

kg dry

air

63.7

148.2 x 10 3

84.0 x 10

105

40.6

74.0

172.1 x 10 3

97.2 x 10

3

110

43.3

84.8

3 197.2 x 10

112.1 x 10

3

115

46.1

96.5

3 224.5 x 10

128.9 x 10

3

140

60.0

198.4

3 461.5 x 10

H —H yi

l/(H yi

dry air

lb

37.8

48.2

35.0

°C

100

41.8

32.2

3

°F

3

43.68 x 10

36.1

29.4

90

kg dry air

air

18.78

26.7

85

4.

lb

J

btu

computed

is

y

—H

y

versus

)

for various values

H

y

from

H yl

to

H y2

of 7^ between TLl andTt2 a graphical integration is .

,

to obtain the value of the integral in Eq. (10.5-13). Finally, the height z

is

calculated from Eq, (10.5-13).

Design of Water-Cooling Tower Using Overall Mass-Transfer Coefficients

10. 5D

Often, only an overall mass-transfer coefficient mol/s

•

m

3 •

atm

is

available,

and Eq.

(10.5-13)

The value of H*

kg mol/s-m 3 -Pa or kg

in

becomes

dH y

h, :

M B K G aP

KG a

H* -

)H

H

(10.5-15) y

H

determined by going vertically from the value of y at point P up H* at point R, as shown in Fig. 10.5-3. In many cases the experimental film coefficients k G a and h L a are not available. The few experimental data available indicate that h L a is quite large and the slope of the lines -h L a! (k G aM B P) in Eq. (10.5-14) would be very large and the value of H yi would approach is

to the equilibrium line to give

that

of//*

in Fig. 10.5-3.

The tower design using

the overall mass-transfer coefficient

is

done using the

following steps. 1.

The enthalpy-temperature data from Table

2.

The operating

10.5-1 are plotted as

shown

in

Fig.

10.5-3. line is calculated as in steps

1

and 2

for the film coefficients

and

plotted in Fig. 10.5-3. 3.

In Fig. 10.5-3 point

represents

H* on

P

represents

represents the driving force 4.

The

driving force

TL2 Then by .

integration

H* -

plotting

Hy

in is

T L on

the operating line and point

H

y

)

for various values of

versus

Hy

from

H y]

performed to obtain the value of the integral is obtained from Eq. (10.5-15).

available, then using

in

RP

or

H*y -

R

Hy

Eq. (10.5-15).

computed

U(H* -

experimental cooling data

Sec. 10.5

and

is

the height z If

Hy

the equilibrium line. Hence, the vertical line

an actual run

in a

in

to

H y2

,

a graphical

Eq. (10.5-15). Finally,

cooling tower with

Eq. (10.5-15), the experimental value of

Continuous Humidification Processes

T L between T u and

known

KGa

height z are

can be obtained.

607

EXAMPLE 10.5-1.

Design of Water-Cooling Tower Using Film Coefficients

A packed G = 1.356

countercurrent water-cooling tower using a gas flow rate of kg dry air/s-m 2 and a water flow rate of L = 1.356 kg 2 is to cool the water from T water/s L2 = 43.3°C (1 10°F) to Ll = 29.4°C (85°F). The entering air at 29.4°C has a wet bulb temperature of 23.9°C. The -7 3 mass-transfer coefficient k G a is estimated as 1.207 x 10 kg mol/s Pa 4 and h L a/k G aM B P as 4.187 x 10 J/kg-K (10.0 btu/lb m -°F). Calculate the 5 height of packed tower z. The tower operates at a pressure of 1.013 x 10 Pa. •

m

T

•

m

-

Following the steps outlined, the enthalpies from the saturated air-water vapor mixtures from Table 10.5-1 are plotted in Fig. 10.5-4. The inlet air at TG1 = 29.4°C has a wet bulb temperature of 23.9°C. The humidity from the humidity chart is H, = 0.0165 kg H 2 0/kg dry air. Substituting

Solution:

-

into Eq. (9.3-8), noting that (29.4

H yl = = The

point

+

(1.005

71.7 x 10

H yl =

into Eq. (10.5-2)

71.7

and

and

129.9 x 10

TL2 =

43.3°C

3

3

x 0.0165)10

3

=

- 0) K,

(29.4

(29.4

-

0)

+

6

2.501 x 10 (0.0165)

J/kg

x 10 3 and

TLl =

29.4°C

is

plotted.

Then

3

-

substituting

solving,

-

1.356(// y2

H y2 =

1.88

0)°C

71.7 x 10

3 )

=

1.356(4.187 x 10 ){43.3

29.4)

H

J/kg dry air (55.8 btu/lbj. The point = 129.9 x 10 3 y2 also plotted, giving the operating line. Lines with slope -41.87 x 10 3 J/kg-K are plotted giving and yi y

is

-h L a/k c aM B P =

H

which are tabulated

H

Table 10.5-2 along with derived values as shown. Values of l/(H yi — H y ) are plotted versus y and the area under the 3 3 curve from H yl =71.7 x 10 to// > 2 = 129.9 x 10 is values,

in

H

,

Tlx Figure

608

10.5-4.

Chap. 10

Liquid temperature (°C)

T L2

Graphical solution of Example 105-1.

Stage and Continuous Gas-Liquid Separation Processes

Table

Enthalpy Values for Solution to

10.5-2.

Example

10.5-1 ( enthalpy in J'/kg dry air)

H. 71.7 x 10

3

83.5 x 10

3

x 10 3 3 108.4 x 10 94.4

22.7 x 10

3

24.9 x 10

3

4.02 x 10"

x KT 5 5 2.83 x 10" _5 2.29 x 10 5 1.82 x 10~

124.4 x 10

3

x 10

3

29.5 x 10

3

141.8 x 10

3

106.5 x 10

3

35.3 x 10

3

162.1 x 10

3

3 118.4 x 10

43.7 x 10

3

184.7 x 10

3

3

54.8 x 10

3

94.9

129.9 x 10

4.41

x 10" 5 5

3.39

Substituting into Eq. (10.5-13), 1.356

M B k c aP =

6.98

m

Minimum Value

10.5E

Often the

shown

flow

air

H

and

yl

TL{

the equilibrium line

is

line at a point farther

value of

G min

is

10. 5F

(22.9

of Air

a

7 29(1.207 x 10" X1.013

y

Flow must be

set for the design of the cooling tower.

minimum value of G,

the operating line

MN

quite curved, line

As

drawn through TL2 point N. If

is

with a slope that touches the equilibrium line at

down

(1.82)

x 10 5 )

ft)

G is not fixed but

in Fig. 10.5-5 for

the point

H

,

MN could become tangent to the equilibrium

the equilibrium line than point

N. For the actual tower,

G greater than G min must be used. Often, a value of G equal to

1.3 to 1.5

a

times

used.

Design of Water-Cooling

Tower Using Height

of a

Transfer Unit

Often another form of the film mass-transfer coefficient ~"> 2 z

=

is

used

in

Eq. (10.5-13):

AIL

Hr

(10.5-16) H,i

"yi

equilibrium line

operating line for slope =Lc L /G mia

G min

,

operating line, slope = Lc L JG

Figure

Sec. 10.5

10.5-5.

Operating-line construction for

Continuous Hurnidification Processes

minimum gas flow.

609

HG = ~

G

M B k G aP~

where

HG

is

number

the

-

(10.5-17)

the height of a gas enthalpy transfer unit in

HG

The term

of transfer units.

is

m and the integral term

often used since

flow rates than.fc 0 a. Often another form of the overall mass-transfer coefficient

•Pa or kg mol/s -m z

3 -

=

atm

is

less

it is

KG a

is

called

dependent upon in

kg mol/s

m3

•

used and Eq. (10.5-15) becomes

-

M K c aP B

where H oa is the height of an overall gas enthalpy transfer unit in m. The value of H* is determined by going vertically from the value of H y up to the equilibrium line as shown in Fig.

10.5-3.

This method should be used only

However, the

straight over the range used. line is

somewhat curved because of

10. 5G

H 0G

when is

the equilibrium line

often used even

if

is

almost

the equilibrium

the lack of film mass-transfer coefficient data.

Temperature and Humidity of Air Stream

in

Tower

The procedures outlined above do not yield any information on the changes in temperature and humidity of the air-water vapor stream through the tower. If this information is of interest, a graphical method by Mickley (M2) is available. The equation used for the graphical method

Gc s dT q and combining

it

dH

'

-7=* dTG

10. 5H

Dehumidification

is

derived by

first

setting

Eq.

(10.5-6) equal to

with Eqs. (10.5-12) and (10.5-9) to yield Eq. (10.5-19).

=

H —M VT;- TG ;

v

(10.5-19)

'

Tower

For the cooling or humidification tower discussed, the operating line lies below the is cooled and air humidified. In a dehumidification tower cool water is used to reduce the humidity and temperature of the air that enters. In this case the operating line is above the equilibrium line. Similar calculation methods are equilibrium line and water

used (Tl).

10.6

ABSORPTION IN PLATE AND PACKED TOWERS

10.6A /.

Equipment

for Absorption

Introduction to absorption.

ma"ss-transfer process in

and Distillation

As discussed

briefly in Section

which a vapor solute A

in a

gas mixture

10. is

IB, absorption

is

a

absorbed by means of

which the solute is more or less soluble. The gas mixture consists mainly of an and the solute. The liquid also is primarily immiscible in the gas phase; i.e., its vaporization into the gas phase is relatively slight. A typical example is absorption of the solute ammonia from an air-ammonia mixture by water. Subsequently, the solute is a liquid

in

inert gas

recovered from the solution by distillation. In the reverse process of desorption or stripping, the

610

same

principles

and equations hold.

Chap. 10

Stage and Continuous Gas-Liquid Separation Processes

Equilibrium relations for gas-liquid systems

in

absorption were discussed in Section

in

and such data are needed for design of absorption towers. Some data are tabulated Appendix A. 3. Other more extensive data are available in Perry and Green (PI).

2.

Various types of tray (plate) towers for absorption and distillation.

10.2,

vapor and liquid

efficiently contact the

are often used.

A

1.

common

very

shown schematically

through the flowing

2.

common

size.

openings

in the tray

Section

1

1.4A for

is

the sieve tray, which

is

distillation.

same type of

tray,

liquid.

mm

The vapor area of the Holes varies between 5 to 15% of the tray area. The liquid is maintained on the tray surface and prevented from flowing down through the holes by the kinetic energy of the gas or vapor. The depth of liquid on the tray is maintained by an overflow, outlet weir. The overflow liquid flows into the downspout to the next tray below. Valve tray. A modification of the sieve tray is the valve tray, which consists of area which at

low vapor

tray

is

and a

lift-valve

cover for each opening, providing a variable open

varied by the vapor flow inhibiting leakage of liquid

is

Hence,

rates.

than the sieve

3.

in

sieve tray is used in gas absorption and in vapor bubbles up through simple holes in the tray Hole sizes range from 3 to 12 in diameter, with 5 mm a

Essentially, the

Sieve tray.

and

In order to

absorption and distillation, tray (plate) towers

type of tray contacting device

in Fig. 10.6-la

distillation. In the sieve

in

tray,

this

down

the

opening

type of tray can operate at a greater range of flow rates

with a cost of only about

20%

higher than a sieve tray.

The

valve

being increasingly used today.

Bubble-cap tray.

Bubble-cap

shown

trays,

in Fig. 10.6- lb,

have been used

for

over

100 years, but since 1950 they have been generally superseded by sieve-type or valve

which is almost double that of sieve-type trays. In the vapor or gas rises through the opening in the tray into the bubble caps. Then the gas flows through slots in the periphery of each cap and bubbles trays because of their cost,

bubble

tray, the

upward through

the flowing liquid. Details

and design procedures

and other types of trays are given elsewhere (B2, PI, Tl). The efficiencies are discussed in Section

1

1

for

many

of these

different types of tray

.5.

I

(b)

(a)

Figure

10.6-1.

Tray contacting devices:

(a) detail

of sieve-tray tower,

(b) detail

of

bubble-cap tower tray.

Sec. 10.6

Absorption

in

Plate

and Packed Towers

611

3.

Packed towers for absorption and

Packed towers are used

distillation.

for

continuous

countercurrent contacting of gas and liquid in absorption and also for vapor-liquid contacting in distillation. The tower

column and distributing the top, a gas outlet at the top, a liquid outlet at the bottom, and a packing or the tower. The entering gas enters the distributing space below the packed

containing a gas device at filling in

section

and

rises

and

inlet

upward through

the openings or interstices in the packing

contact between the liquid and gas different types of

A

same openings.

the descending liquid flowing through the

Many

10.6-2 consists of a cylindrical

in Fig.

distributing space at the bottom, a liquid inlet

and contacts

large area of intimate

provided by the packing.

is

tower packing have been developed and a number are used

commonly. Common types of packing which are dumped at random in the tower shown in Fig. 10.6-3. Such packings and other commercial packings are available in sizes of 3 mm to about 75 mm. Most of the tower packings are made of inert and cheap materials such as clay, porcelain, graphite, or plastic. High void spaces of 60 to 90% are characteristic of good packings. The packings permit relatively large volumes of liquid to

quite

are

pass countercurrently to the gas flow through the openings with relatively low pressure

drops

for the gas.

These same types

of

packing are also used

in vapor-liquid separation

processes of distillation.

Stacked packing having

mm

75

sizes of

and larger

or so

is

also used.

The packing is The

stacked vertically, with open channels running uninterruptedly through the bed.

advantage of the lower pressure drop of the gas is offset in part by the poorer gas-liquid contact in stacked packings. Typical stacked packings are wood grids, drip-point grids, spiral partition rings,

and

others.

gas outlet

liquid inlet

liquid distributor

/ ,

J.n.n.n.n.M ,

I

\

/

V

|

I

\

)

V

«

'

I

'

»

I

I

W

i

\/

I

\

-packing

gas inlet liquid outlet

Figure

612

10.6-2.

Packed tower flows and characteristics for absorption.

Chap. 10

Stage and Continuous Gas-Liquid Separation Processes

(b)

(a)

Figure

(c)

Typical tower packings:

10.6-3.

(a)

Raschig

(d) ring, (b) Lessing ring,

(c)

Berl

saddle, (d) Pall ring:

In a given packed tower with a given type

flow of liquid, there

is

and

size of

packing and with a definite

an upper limit to the rate of gas flow, called the flooding velocity.

Above this gas velocity the tower cannot operate. At low gas velocities the downward through the packing essentially uninfluenced by the upward gas gas flow rate

increased at low gas velocities, the pressure drop

is

flow rate to the

1.8

is

liquid flows flow.

As the

proportional to the

power. At a gas flow rate called the loading point, the gas starts to

hinder the liquid downflow and local accumulations or pools of liquid start to appear the packing.

The pressure drop of the gas

starts to rise at a faster rate.

As the gas flow

in

rate

holdup or accumulation increases. At the flooding point, the down through the packing and is blown out with the gas. In an actual operating tower the gas velocity is well below flooding. The optimum economic gas velocity is about one half or so of the flooding velocity. It depends upon an economic balance between the cost of power and the fixed charges on the equipment cost (SI). Detailed design methods for predicting the pressure drop in various types of packing is

increased, the liquid

liquid can no longer flow

are given elsewhere (PI, LI, Tl).

10.6B

Design of Plate Absorption Towers

A

Operating-line derivation.

1.

diagram as vertical tray

(B)

plate (tray) absorption tower has the

the countercurrent multiple-stage process in Fig. 10.3-2

tower

and then

in Fig. 10.6-4. In the case of solute

diffusing through a stagnant gas

from air (B) by water remain constant throughout the

into a stagnant fluid, as in the absorption of acetone (A)

water, the moles of inert or stagnant air entire tower. If the rates are

kg mol

A

same process flow and is shown as a

inert/s

•

m

2

units (lb

and

inert

V kg mol inert air/s and L kg mol inert solvent water/s or in mol inert/h ft 2 ), an overall material balance on component A

in Fig. 10.6-4 is

(10.6-1)

A

balance around the dashed-line box gives

where x is the mole fraction A in the liquid, y the mole fraction of A in the gas,L„ the moles liquid/s, and V„ + the total moles gas/s. The total flows/s of liquid and of gas vary throughout the tower. Equation (10.6-2) is the material balance or operating line for the absorption tower total

Sec. 10.6

,

Absorption

in Plate

and Packed Towers

613

and

is

similar to Eq. (10.3-13) for a countercurrent-stage process, except that the inert

streams

V

L and

The terms V, 2.

Z,

L and

are used instead of the total flow rates

x0

,

V. Equation (10.6-2)

stream with x„ in the liquid stream passing and y x are constant and usually known or can be determined.

relates the concentration y„ +

,

in the gas

A plot

Graphical determination of the number of trays.

it.

of the operating-line equation

x and y are very dilute, the denominators 1 — x and 1 — y will be close to 1.0, and the line will be approximately straight, with a slope = L/V. The number of theoretical trays are determined by simply stepping off the number of trays, as done in Fig. 10.3-3 for a countercurrent multiple-stage process.

(10.6-2) as y versus x will give a curved line. If

EXAMPLE

o/S0 2 in a Tray Tower designed to absorb S0 2 from an air stream by using pure water at 293 K (68°F). The entering gas contains 20 mol S0 2 and that leaving 2 mol % at a total pressure of 101.3 kPa. The inert air flow 2 rate is 150 kg air/h m and the entering water flow rate is 6000 kg

A

Absorption

10.6-1.

tray tower

is

to be

%

,

water/h

-m 2 Assuming an

etical trays

293

.

and actual

overall tray efficiency of

trays are

25%, how many

theor-

needed? Assume that the tower operates

at

K (20°C).

Solution:

First calculating the

molar flow

V — -— = 5.18 L =

kg mol inert air/h

6000 — — = 333 kg mol 1

rates,

•

m

inert water/h

•

2

m

,

o.O

Referring to Fig. 10.6-4, y N+l = 0.20, y and solving for x N

into Eq. (10.6-1)

l

=

0.02,

and x 0

=

0.

Substituting

,

xN

=

Substituting into Eq. (10.6-2), using

0.00355

V

and £ as kg mol/h

m2

instead of

-i-U- n x n ,

n+l

'

N-

1

N Ln< x n Figure

614

10.6-4.

Chap. 10

Material balance

in

an absorption tray lower.

Stage and Continuous Gas-Liquid Separation Processes

kg mol/s

m

-

2 ,

be

In order to plot the operating line, several intermediate points will calculated. Setting y„ +

1

=

0.07

and substituting

into the operating equation,

»^
Hence, x n y„ +

l

—

=

0.13,

0.000855.

and xn

To

calculate another intermediate point,

we

set

The two end points and

the

two

calculated as 0.00201.

is

intermediate points on the operating line are plotted in Fig. 10.6-5, as are the equilibrium data from Appendix A.3. The operating line is somewhat curved.

The number

of theoretical trays

The

to give 2.4 theoretical trays.

is

determined by stepping off the steps

actual

number

of trays

is

2.4/0.25

=

9.6

trays.

10.6C

Design of Packed Towers for Absorption

/. Operating-line derivation. For the case of solute A diffusing through a stagnant gas and then into a stagnant fluid, an overall material balance on component A in Fig. 10.6-6 for a packed absorption tower is

(10.6-3)

where L

mol

is

kg mol

inert gas/s

-m

2 ,

and y

l

V

kg mol inert liquid/s m 2 is kg mol inert gas/s or kg and x are mole fractions A in gas and liquid, respectively.

inert liquid/s or

•

,

{

operating line

x0

xN

Mole Figure

Sec. 10.6

10.6-5.

fraction, x

Theoretical number of trays for absorption of S0 2 in Example 10.6-1.

Absorption

in

Plate and Packed Towers

615

v2

.

y-i

i dz z

:_~r_ V.y

Vu Figure

The

flows

V

y,

Material balance for a counter current packed absorption tower.

10.6-6.

L and

x

L,

are constant throughout the tower, but the total flows

L and V

are

not constant.

A

balance around the dashed-line box in Fig. 10.6-6 gives the operating-line equa-

tion.

(10.6-4)

This equation, when plotted on yx coordinates,

Equation

10.6-7a.

—

=

y,)

taken as

1

.0

Pi/{P

and Eq.

~ (

10.6-4)

When stripping.

2.

LJV

the solute

its

l

^Lx + x

line, as

shown

in Fig.

line is

LjV

L

below the equilibrium ratios.

composition

V'y

(10.6-5)

line is essentially straight.

being transferred from the

Limiting and optimum

and

V'y

and the operating is

The operating

(Fig. 10.6-6)

curved

becomes

Lx+ This has a slope

will give a

can also be written in terms of partial pressure />, of A, where P\\ and so on. If x and y are very dilute, (1 — x) and (1 — y) can be

(10.6-4)

to the

V stream, the process is shown in Fig. 10.6-7b.

called

line, as

In the absorption process, the inlet gas flow K,

The

y, are generally set.

exit concentration y 2

is

also

and the concentration x 2 of the entering liquid is often fixed by process requirements. Hence, the amount of the entering liquid flowL 2 orL' is open to usually set by the designer

choice.

V and the concentrations y 2 x 2 and y are set. When the minimum slope and touches the equilibrium line at point P, the liquid flow L is a minimum at L'min The value of x, is a maximum atx, max whenL' is a minimum. At point P the driving forces y — y*,y — y- x* — x, orx — x are all zero. To solve for L'mm the values y, and x, m „ are substituted into the operating-line equation. In some cases if the equilibrium line is curved concavely downward, the minimum value of L is reached by the operating line becoming tangent to the equilibrium line instead of In Fig. 10.6-8 the flow

,

x

,

x

operating line has a

.

t

,

;

,

intersecting

it.

The choice

of the

optimum L/V

ratio to use in the design

depends on an economic

balance. In absorption, too high a value requires a large liquid flow,

616

Chap. 10

and hence

a

Stage and Continuous Gas-Liquid Separation Processes

Mole

fraction,

x

Mole fraction, x

(a)

Figure

10.6-7.

the

A small

optimum

A from V

Location of operating lines: (a) for absorption of stream, (b)for stripping of A from LtoV stream.

large-diameter tower.

be high.

(b)

The cost of recovering

to

L

from the liquid by distillation will which is costly. As an approximation, obtained by using a value of about 1.5 for the ratio of the the solute

liquid flow results in a high tower,

liquid flow

is

average slope of the operating factor can vary depending

line to that of the

equilibrium line for absorption. This

on the value of the solute and tower type.

in packed towers. As discussed in Section measure experimentally the interfacial area A m 2 between phases L and V. Also, it is difficult to measure the film coefficients k'x and k'y and the overall coefficients K'x and K' Usually, experimental measurements in a packed tower y yield a volumetric mass-transfer coefficient that combines the interfacial area and mass-

3.

Film and overall mass-transfer coefficients

10.5;

it

is

very

difficult

to

.

transfer coefficient.

Defining a as interfacial area

in

m2

per

m3

volume of packed

section, the

volume of

operating line for actual liquid flow

Figure

Sec. 10.6

Absorption

10.6-8.

in Plate

Minimum

liquid/gas ratio for absorption.

and Packed Towers

617

packing

in a height

m (Fig.

dz

S dz and

10.6-6) is

dA = where S

m2

is

dz

(10.6-6)

The volumetric

cross-sectional area of tower.

film

and overall mass-

transfer coefficients are then defined as

ICy

kg mol

=

,

ky a

s

•

s

•

packing- mol frac

m

packing- mol frac

k1

k§

a-

3

lb

ka=

mol

packing -mol frac

kg s

•

m

3

(SI)

packing mol frac •

mol

lb

ka = ,

,

J

h

packing- mol frac

ft

•

kg mol

s-m 3

-

K'1 a

\

h 4.

a=

,

m3

ft

,_ (English)

packing mol frac

For absorption of A from stagnant For the differential height of tower dz leaving V equal the moles entering L.

Design method for packed towers.

operating-line equation (10.6-4) holds. (10.6-6), the

moles of A

=

d{Vy)

where V

=

kg mol

d(Lx)

B,

the

in Fig.

(10.6-7)

L = kg mol total liquid/s, and d{Vy) = d(Lx) = kg mol A m. The kg mol A transferred/s from Eq. (10.6-7) must equal the

total gas/s,

transferred/s in height dz

kg mol A transferred/s from the mass-transfer equation for the flux N A using the gas-film and liquid-film coefficients.

-

-

NA

.

Equation

(10.4-8) gives

xj M are defined by Eqs. (10.4-6) and (10.4-7). Multiplying the (1 y A ) iM and (1 left-hand side of Eq. (10.4-8) by dA and the two right-side terms by aSdz from

where

;

Eq. (10.6-6),

^ = - -±K-~^

Na where

a

y

-

ii

N A dA =

kg mol A

,

(y (y.4G AC

- yJS y A ds

,

dz

=

—^~

transferred/s in height dz

Equating Eq.(10.6-7)

to (10.6-8)

and usingy MG

k'a \_

^

(x Ai

-

x AL )S dz

(10.6-8)

m(lb mol/h). for the

bulk gas phase a.ndx AL for the

bulk liquid phase,

d(

Vy AG = )

(i

d(Lx AL ) Since

V=

K( 1

—

d(

Substituting

V

y AG ) or

Vy for

M

)

V =

= ijV

K'/(l

-

= -

V'/(l

ky

-

\

(y AC

- yJS

\

(x Ai

-

dz

(10.6-9)

x AL )S dz

(10.6-10)

y A )iM

kx

- y AC

),

1—, yd] = Vd(-^-) =

y AG )

in

(10.6-1

1)

Eq. (10.6-11) and then equating Eq. (10.6-11) to

(10.6-9),

f^ 618

Chap. 10

=

ky

n

\

{yAo-yA^Sdz

(10.6-12)

Stage and Continuous Gas-Liquid Separation Processes

Repeating

for

Eq. (10.6-10) since

L =

L dx A 1

-

— x Ai),

L/(l

k'a

*al

(1

(10.6-13)

-

x a)>m

L and

subscripts A, G, and

Dropping the

integrating, the final equations are as

follows using film coefficients:

dz

=

V

=

z

(10.6-14)

aS

k'

J

dz

=

-(i-yto-yd L dx

=

z

dy

(10.6-15)

k'aS

-

(1

(1

-

- X)

xXx,

x) iM

In a similar manner, the final equations can be derived using overall coefficients.

z

V dy

=

(10.6-16)

(i

ri

-

y).M

L dx

>

(10.6-17)

(1

-

x)

and

In the general case, the equilibrium k'x a, (

k' a, y

M

(I

-xX**-x)

the operating lines are usually curved

K'y a, and K'x a vary somewhat with

gas and liquid flows.

total

must be integrated graphically. The methods to do

10.6- 14)-( 10.6- 17)

centrated mixtures will be discussed in Section 10.7.

Methods

and

Then Eqs.

this for

con-

for dilute gases will

be

considered below.

10.6D

Simplified Design of Dilute

Methods

Gas Mixtures

Absorption

for

Packed Towers

in

Since a considerable percentage of the absorption processes include absorption of a dilute gas A, these cases will be considered using a simplified design procedure.

The concentrations can be considered

dilute for engineering design purposes

when

mole fractions y and x in the gas and liquid streams are less than about 0.10, i.e., 10%. The flows will vary by less than 10% and the mass-transfer coefficients by considerably less than this. As a result, the average values of the flows V and L and the mass-transfer coefficients at the top and bottom of the tower can be taken outside the integral. the

Likewise, the terms

(1

-

y) iM /(l

-

- y),J{l -

y), (1

y), (1

- x) iM /(l -

x),

and

(1

- x).J

— x)

can be taken outside and average values of the values at the top and bottom of the tower used. (Often these terms are close to 1.0 and can be dropped out entirely.) Then

(1

Eqs. ( 10.6- 14H 10.6- 17)

become V

-

(1

_k' aS y

dy (10.6-18)

-y

1

yi

y-yi

~

L [k'x aS

Sec. 10.6

Absorption

in

Plate

- x) iM

dx

-

(1 1

—

X

av

and Packed Towers

,

X2

X,

—

(10.6-19)

X

619

at UJ

K'y I\

1 1

l

r

=

z

*yi

»i (I - - yj.w

V

=

z

av

ay y

» j»i

y

dx

(i-

K'x aS

—

1

X

—X

X*

av „ 12

Since the solutions are dilute, the operating line will be essentially straight*

suming the equilibrium

line

approximately straight over the range of concentriifc

is

— y,) varies linearly with y and also with x.

used, (y

y

where

- y. =

+

%

ky

b

(IE

and b are constants. Therefore, the integral of Eq. (10.6-18) can be

k

integral.-.

give the following.

I dy

y-y>

Jyi

where (y

—

y )M (

is

the log

mean (y

driving force. (yi

-

If

the term

(10.6-18)

and

(1

-

y) iM /(l

doing the

same

-

1

yd*

(y

-

y,)« in C(yi

(y

- yi

y\

— y)

-

(yj

y*) M

in CCv,

- (y - y, - ya)/(> - yi2)l

y.i)

2)

2

- (y -

y*)

2

y?)

- yD/(y 2 - y!)]

considered

is

2

,

then substituting Eq. (10.6-23)ib

1.0,

for Eqs. (10.6-19)—( 10.6-21), the final results are as folksss.

I (yi

5

where the

left

side

is

the right-hand side (

+ V2 )/2

K,

the

is

and of L

is

v

-

y2)

= ^ a
x 2) =

1

i<;

4-

= ^;cz(y-y*) M

(X,

x 2 ) = K'x az{x* -x) M

•

m

2

mass

1.

The

mol/h

transfer.

2

material balance ) by The value of V is the as^ ft

|

below and shown

operating-line equation (10.6-4)

is

in slightly different

ways. Theisfc:

in Fig. 10.6-9.

plotted as in Fig. 10.6-9 as a straig&fc

L„ L 2 and L lv = (L, + iCjE Average experimental values of the film coefficients k'y a and k'x a are available;®: obtained from empirical correlations. The interface compositions y n and x n at& whose slope is calculri^; y,, X! in the tower are determined by plotting line Calculate

2.

(lb

L 2 )/2.

Equations (10.6-26) to (10.6-29) can be used steps to follow are discussed

x) M

y 2)

kg mol absorbed/s

,

-

(y,

the rate equation for

(L

«z(x,.

v

y.)«

Vu V2 and Kav = ,

(K,

+ V2 )/2\

also calculate

,

y

Eq. (10.6-30):

slope

= —

k'x k'

y

620

Chap. 10

kx a

a/(l a/(l

-

y) iM

k a y

Stage and Continuous Gas-Liquid Separation Prtm.

i

(10.6-31)

and (1 — y) iM are used, the procedure is trial and error, as in However, since the solutions are dilute, the terms (1 — and (1 — yj can be used in Eq. (10.6-31) without trial and error and with a small error in the slope. If the coefficients k a and k x a are available for the approximate y concentration range, they can be used, since they include the terms (1 — x) iM and For line P 2 (1 — y) iM 2 at the other end of the tower, values of y i2 and x i2 are determined using Eq. (10.6-30) or (10.6-31) andy 2 andx 2 If

terms

— x) iM

(1

Example

10.4-1.

M

.

.

3. If

the overall coefficient K'y a

10.6-9. If X^.ais 4.

Calculate the log (y

—

'

s

is

being used.y? andyf are determined as shown

in Fig.

used.x* andx* areobtained.

mean

driving force [y

—

by Eq. (10.6-24)

\ik'

y

a

is

used.

ForK^a,

calculated by Eq. (10.6-25). Using the liquid coefficients, the appropriate

driving forces are calculated. 5.

Calculate the column height z

m

by substituting into the appropriate form

of Eqs.

(10.6-26H10.6-29).

EXAMPLE

10.6-2. Absorption of Acetone in a Packed Tower being absorbed by water in a packed tower having a cross2 sectional area of 0.186 at 293 and 101.32 kPa (1 atm). The inlet air contains 2.6 mol acetone and outlet 0.5%. The gas flow is 13.65 kg mol inert air/h (30.1 lb mol/h). The pure water inlet flow is 45.36 kg mol water/h (100 lb mol/h). Film coefficients for the given flows in the tower are 2 k' a = 3.78 x 10" kg mol/s m 3 mol frac (8.50 lb mol/h ft 3 mol frac) and y -2 = 6.16 x 10 kg mol/s -m 3 mol frac (13.85 lb mol/h ft 3 mol frac). k'x a Equilibrium data are given in Appendix A. 3. (a) Calculate the tower height using k' a. y (b) Repeat using k'x a. (c) Calculate K' a and the tower height. y

Acetone

is

m

K

%

•

•

•

-

Solution: frac, p A

rium

Sec. 10.6

=

line

•

•

From Appendix A. 3 for acetone-water and x A = 0.0333 mol = 0.0395 atm or y A = 0.0395 mol frac. Hence, the equilib= m (0.0333). Then, y = 1.186x. This equiis y A = mx A or 0.0395 30/760

Absorption

in

Plate

and Packed Towers

621

is plotted in Fig. 10.6-10. The given data are L = 45.36 kg = 13.65 kg mol/h, y, = 0.026, y 2 =0.005, andx 2 = 0. mol/h, Substituting into Eq. (10.6-3) for an overall material balance using flow

librium line

V

rates as

kg mol/h instead of kg mol/s, 0.026

0

45.361 1

-0 +

13.65;

-

1

= X!

The points y ,x and y 2 x 2 are drawn for the operating line. l

Using Eq. (10.6-31) the approximate slope aty ,x 1

slope

k'z a/(l

=

6.16 x !Q-

-x,)

0.005 13.65; 1

- 0.005

=0.00648

plotted in Fig. 10.6-10

,

1

l-xj +

45.36|

0.026

2

/(l

1

and a

-

is

is

-0.00648)

2 3.78 x 10~ /(1

straight line

= -

1.60

0.026)

Plotting this line through y u x 1( the line intersects the equilibrium line at = 0.0154 and x n = 0.0130. Also, y* = 0.0077. Using Eq. (10.6-30) to

yn

more accurate slope, the preliminary values of y n and x fl will be used in the trial-and-error solution. Substituting into Eq. (10.4-6),

calculate a

-yn)-(i -y.)

(1

(1

-

In [(1

Using Eq.

-

-y„V(l

ln[(l

0.0154)

-

yi )l

-(1

0.0154)/(1

- 0.026) - 0.020)]

0.979

(10.4-7),

(1

-

(1

-x,)-(l

x) iM In [(1 (1

-

In [(1

-x n

)

-x,)/d -x,,)] 0.00648)

-

-(1

0.00648)/(1

- 0.0130) = - 0.0130)]

0.993

slope = -

1

.62

0.014 i

*2

Figure

622

*

W2 10.6-10.

1

Location of interface compositions for Example 10.6-2.

Chap. 10

Stage and Continuous Gas-Liquid Separation Processes

Substituting into Eq. (10.6-30), 2 6.16 x 10- /0.993

k'*a/(l-x),H S10pe

"

-

Jf,a/(1

~~

2 3.78 x 10- /0.929

y)-,M

Hence, the approximate slope and interface values are accurate enough. For the slope at point y 2 x 2 , ,

~ _ ^a/q-^) _ = " Jt; a /(1 -

,

S10PC

The slope changes little 0.0018, and y% = 0.

2 6.16 x 10- /(l-0)

x 10- 2 /(l

3.78

- 0.005)

_ _

=

in the tower. Plotting this line, y i2

0.0020,

xn

=

Substituting into Eq. (10.6-24),

in CCvi

(0.026

"

To calculate

Kav =

K,1

K, '

=

^,i)/(y2

-

In [(0.026

the total

+

-

(a),

-

molar flow rates

3.893 x 10'

3

+

(0.005

kg mol/s,

in

3.811 x 10~

~

=

85'~ x 10

(y>

-

3

=

„

OM

3.852 x 10

1-260 x 10"

substituting into Eq. (10.6-26)

V -f 3

y.-z)]

- 0.0020) = - 0.0154)/(0.005 - 0.0020)] 0.00602

0.0154)

£SI,SL,S^, = For part

-

and

2

3

,

kg mol/s

kg mol/s

solving,

- y^

yi)

=

k az (y

0.005)

=

2 (3.78 x 10' )z(0.00602)

z

=

'y

-3

'

-

(0.0260

tae

0.186

For part

1.911

m

(6.27

ft)

using an equation similar to Eq. (10.6-24),

(b),

>U 1" [(x,i

-

(0.0130

x,)/(x i2

-

In [(0.0130

-

0.00648)

x 2 )]

-

(0.0018

-0)

- 0.00648)/(0.0O18 -

= Q00m

0)]

Substituting into Eq. (10.6-27) and solving,

L26

0 H Tor 0. 1 86

This checks part

(a)

Absorption

in

Sec. 10.6

(0.00648

-

0)

=

(6.16

x 10

z

=

1.936

m

_2

)z(0.00368)

quite closely.

Plate and

Packed Towers

623

For part (i

-

y).M

substituting into Eq. (10.4-25) for point

(c),

- yX - (1 - yj Cd - yf)/(l - j^)]

(1

=

In

The overall

(1

In

CU

t ,

- (1 -

0.0077)

-

_y

0.0077)/(l

-

x it 0.026)

0.983 0.026)]

mass-transfer coefficient K' a at pointy^x, y

is

calculated by

substituting into Eq. (10.4-24). 1

K'a/(l

m

1

- y). u

-

k' o/(l

1

y)

M

1

K/z/0.983

x 10~ 70.979

3.78

=

K'y a

- x) M

k'x a/(l

+

1.186 •

2 6.16 x 10" /0.993

x 10" 2 kg mol/s-

2.183

3 •

mol

frac

Substituting into Eq. (10.6-25),

(y

-

y*) M

- [yj - yt) In - yty(y 2 - y$] (0.0260 - 0.0077) - (0.0050 - 0) ~ In [(0.0260 - 0.0077)/(0.0050 0)] -

Cvi

=

yt)

0.01025

Finally substituting into Eq. (10.6-28),

3.852 x 10

-3 (0.0260

-

0.0050)

=

2 (2.183 x 10" )z(0.01025)

z

=

1.944

0.186

This checks parts

10.6E

(a)

and

m

(b).

Design of Packed Towers Using Transfer Units

Another and

in

some ways a more useful design method of packed towers is the use of the For the most common case of A diffusing through stagnant and

transfer unit concept.

nondiffusing B, Eqs. (10.6-

can be rewritten as

—(10.6-17)

14)

z

=

y)iM

HG

=

"

HL

(1

= Hn

" ,

X2

dx (10.6-33)

-XXX,-X) ( 1

- Hog >,

z

-x),- M

(1

:2

z

(10.6-32)

W-yiy-yb

>,

z

dy

~

y).M dy

(10.6-34)

-y)(y-y*)

(i

(1

-

x).„ dx

(l-xXx*-x)

(10.6-35)

where

V

»0 = k' y

aS

L

k

y

a(l

-

(10.6-36) y) lM

S

L (10.6-37)

624

Chap. 10

Stage and Continuous Gas-Liquid Separation Processes

The The

units of

(ft).

K y a{\-y). M S

K'x aS

K x a(l-x).u S

The H G

For example,

The average

-

x).

y

a

is

,

integrals

N 0> N L ,N OG

(10.6-39)

on the gas

film.

more constant than

K0 7

often proportional to

values of the mass-transfer coefficients,

M must be used

The units

k'

(10.6-38)

the height of a transfer unit based

is

values of the heights of transfer units are

coefficients.

(1

H are in m

HoG ~ K'y aS

—

(1

,

then

y) iM

,

H

the mass-transfer l 0 0 7 0 3 a: V . G oc y /y -

— y). M U —

(1

>

x) lM

an d

,

in Eqs. (10.6-36H10.6-39).

on the right side of Eqs. (10.6-32X10-6-35) are the number of transfer and N 0L respectively. The height of the packed tower is then ,

,

= H G N G = H L N L = H OG N og = H0L N C

z

(10.6-40)

These equations are basically no different than those using mass-transfer coefficients. One still needs k'y a and k'x a to determine interface concentrations. Disregarding (1 — y) iM /(l — y), which is near 1.0 in Eq. (10.6-32), the greater the amount of absorption iy\

~~

y-i)

or tne smaller the driving force (y

—

y), the larger the

number of transfer

units

N G and the taller the tower. (1

When the solutions are dilute with concentrations below 10%, the terms (1 — y) iMl - x)iM/(l - x), (1 - y),J(\ - y), and (1 - x). M/(l - x) cajube taken outside the

-y),{l

integral

and average values used. Often they are quite

close. to

1'

and can be dropped

out.

The equations become z

z

= H L NL =

Hoc.

L "(1

-x) iM l - X av

H,

N nr = .

-y

1

1

Z

-

(1

He

(10.6-41)

_ av

»

'

Xl

dx (10.6-42)

.

dy

y). u

- y

1

z

dy

= H n Nr. = H ( ro -jOuT

(10.6-43)

y

-

y*

dx

= H OL N OL = H 0L

(10.6-44) 1

the operating and equilibrium lines are both straight

If

integral

shown

and

the solutions dilute, the

in Eq. (10.6-23) is valid.

dy

y-yi

_

.

y2

(10.6-23)

(y-

This then can be substituted into Eq. (10.6-41) and similar expressions into Eqs. (10.6-

42H 10.6-44).

EXAMPLE

10.6-3. Use of Transfer Units for Packed Tower Repeat Example 10.6-2 using transfer units and height of a transfer unit as

follows. (a)

(b)

Solution:

Sec. 10.6

H G and N G to calculate tower height. H 0G and N OG to calculate tower height. For part a= 3.78 x 10" 2 kg mol/s-m 3

Use Use

Absorption

(a),

in

Plate

k' y

and Packed Towers

mol

frac

from

625

Example

From

10.6-2.

Eq. (10.6-36),

Hr =

(10.6-36)

.

k'

y

The average V

is

3

3.852 x 10

aS

=

kg mol/s and 5

0.186

m2

.

Substituting

and

solving,

x 10" 3

3.852 0

Since the solution

=

2 (3.78 x 10- X0.186)

number

the

is dilute,

0.548

m from Eq.

of transfer units

(10.6-41)

is

dy (10.6-45)

i-y

L

_ av

t

y-yt

y,

The term y,

in the brackets will be evaluated at point 1 and point =0.026, y n =0.0154 from Example 10.6-2. Also, from Eq.

Example at point

10.6-2, (1

-

=

y) iu

0.979. Also,

- y)«

0-979

l-y (1

point

_

y).

-y=

- 0.026 =

1

At point

(10.4-6) in

0.974.

Then

1,

(1

At

1

2.

M =

=

y2

0.005,

0.997 and

1

=

y i2

-y=

0.002.

1

-

0.005

(l-y) iM - y

1.005

-

y

;)

Eq.

into

Then

(10.4-6),

at point 2,

-

+

y,

y

M =

brackets in Eq. (10.6-45) 1.002

=

is

1.003

is

dy

10.6-2, (y

0.995.

in the

(10.6-23), the integral

From Example

=

1.002

y) iM

y

Substituting

0.995

Hence, the average value of the term

Using Eq.

1.005

0.997

1

d ~

=

0.974

(y

t

-

y2

(10.6-23)

0.OO602. Substituting Eq. (10.6-23) into

Eq. (10.6-45),

Nr.

-y)iM

(i

=

i

-

y

yi

Jav

(y

- yi - y,) M

(10.6-46)

Substituting into Eq.(10.6-46)

NG =

(1.003)

/0 ° 26

^ 0 005 =

3.50 transfer units

0.00602

Finally, substituting into Eq. (10.6-40), z

= HG N G =

(0.548X3.50)

=

1.918

m

For part (b), using K'y a = 2.183 x 10' kg mol/s Example 10.6-2 and substituting into Eq. (10.6-38), 2

H °G

V

K'aS

Chap. 10

3.852 x 10

•

m

3 •

mol

frac

from

-3

2 (2.183 x 10- X0.186)

0.949

m

Stage and Continuous Gas-Liquid Separation Processes

The number

of transfer units in Eq. (10.6-43) becomes as follows is carried out.

when

the

integration similar to Eq. (10.6-23) (i

N 0G =

- y). u i

Substituting (y

knowns

-

=

y*) M

into Eq. (10.6-47)

Note

that the

- 0.005

0.026

number

=

(10.6-47)

- y*) M

Cv

calling the bracketed

10.6-2

term

and the other

1.0,

2.05 transfer units

0.01025 ..

= H OG N OG =

z

,»

from Example

0.01025

and

N oc = (1.0) Finally by Eq. (10.6-40),

y

0.949(2.05)

of transfer units

N OG

=

1.945

of 2.05

is

m

not the same as

NG

of

3.50.

10.7

ABSORPTION OF CONCENTRATED MIXTURES IN PACKED TOWERS

In Section 10. 6D simplified design

methods were given

for absorption of dilute gases in

packed towers when the mole fractions in the gas and liquid streams were less than about 10%. Straight operating lines and approximately straight equilibrium lines are obtained.

and usually the equilibrium line will be and k'y a may vary with total flows. Then the design must be integrated graphically or numerically.

In concentrated gas mixtures the operating line substantially curved

and

k'x

equations (10.6-14)—(10.6-17)

dz

a

=

z

V

=

(10.6-14) k'

y

(1

dz

=

z

=

dz=z =

The The The

L

dx

(1

-xKx, -x)

V

dy

(10.6-15)

(1

=

aS

- y)m. V-y)(y -yd k'x

dz

dy

aS

-x) M

(10.6-16)

K'y aS

{\~y){y -y*)

L dx (10.6-17)

z

K'aS (1 - *).M

(I

-

xXx*

-

x)

detailed general steps to follow are as follows:

and the equilibrium line are plotted. and k'x a are obtained from empirical equations. y 2 and G™, kg total are functions ofGJ, kg total gas/s m

operating-line equation (10.6-4) values of the film coefficients

These two liquids/s

•

film coefficients

m2

,

where n and

equation values, total

tower and converted variation between

k'

y

k'

a

,

m

are in the range 0.2-0.8. Using the operating-line-

V and L are calculated for different values of y and x in the G y and G x Then values of k'y a and k'x a are calculated. If the

to

a or

.

k'x

a at the top and bottom of the tower

is

small,

an average

value can be used.

Sec. 10.7

Absorption of Concentrated Mixtures

in

Packed Towers

627

3.

bottom at point P^y^ xj, the interface compositions yn x ;i by plotting a line P with a slope calculated by Eq. (10.6-30). This

Starting with the tower are determined

,

1

line intersects the

M

l

equilibrium line at the interface concentrations at point Mi

—-

= -—

slope

k,a/(l (1

tively.

This

(1

4.

—

where

—

y

x)

—

—- =

--

(10.6-30)

ky a

y) iU

and (1 — x) iM are determined from Eqs. (10.4-6) and (10.4-7), respecand error. As a first trial, (1 — x ) can be used for (1 — x) jM and — y)iM The values of y n and x n determined in the first trial are used in

y) iM

is trial

for (1

t

.

Eq. (10.6-30) for the second trial. At point P 2 {y 2 x 2) determine a new slope using Eq.

(10.6-30) repeating step

>

for several intermediate points in the tower.

This slope

may

3.

Do

this

vary throughout the

tower. 5.

Using the values ofy, andx determined, graphically integrate Eq. (10.6-14) the tower height by plotting/ (y), where f(y) is as follows

to obtain

(

f{y)

=

—rrz k„ aS

(

y

„ (i versus y between y 2 and y y

.

- y)m >

(i

Then determine

-

yXy

-

10 -7-D

yd

the area under the curve to give the tower

height. If k'x a or other coefficients are used the appropriate functions indicated in Eqs.

15H10.6-17) are plotted. assumed to be 1.0.

(10.6-

EXAMPLE

If

a stream

is

quite dilute,

(1

—

y),

M or(l -

x) ;M can be

Design of an Absorption Tower with a Concentrated Gas Mixture A tower packed with 25.4-mm ceramic rings is to be designed to absorb SO z s from air by using pure water at 293 K and 1.013 x 10 Pa abs pressure. The entering gas contains 20 mol % S0 2 and that leaving 2 mol %. The inert air - 2 kg flow is 6.53 x 10~* kg mol air/s and the inert water flow is 420 x 10 mol water/s. The tower cross-sectional area is 0.0929 m 2 For dilute S0 2 the film mass-transfer coefficients at 293 K are for 25.4-mm (1-in.) rings 10.7-1.

.

,

(Wl), k'

y

a

=

7

0.0594G°- G2'

25

k'x

a

= 0A52G°X

&2

where k'y a is kg mol/s m 3 mol frac, k'x a is kg mol/s m 3 mol frac, and G x and G y are kg total liquid or gas, respectively, per sec per m 2 tower cross section. Calculate the tower height. •

Solution: mol/h),

x2

=

•

The given data

L=

4.20 x 10"

2

•

are

V

=

kg mol/s (333

6.53 x 10"* lb mol/h), y,

kg mol

=

0.20,

(5.18

air/s

y2

=

0.02,

lb

and

0.

Substituting into the overall material-balance equation (10.6-3),

4.

M , 10-^)

+

6.53 ,

.O-^) =

4.20 x

+

628

Chap. 10

6 53 '

10-^) x 10

J

"4

0.02

\

(rrab2j

Stage and Continuous Gas-Liquid Separation Processes

Solving, x,

= 0.00355. The operating line Eq. (10.6-4) is +

4.20 x 10"

0.2

6.53 x 10" 1

= 4.20

-0.2

0.00355

x 10" 1

+

\

- 0.00355 )

6.53 x 10"

l-y

Setting y = 0.04 in the operating-line equation above and solving for x, x = 0.000332. Selecting other values of y and solving for x, points on the

operating line were calculated as shown in Table 10.7-1 and plotted in Fig. 10.7-1 together with the equilibrium data from Appendix A. 3. The total molar flow V is calculated from V = K'/(l - y). At y = 0.20, V = 6.53 x 10~7(1 - 0.2) = 8.16 x 10~ 4 Other values are calculated and 2 tabulated in Table 10.7-1. The total mass flow G y inkg/sis equal to the mass flow of air plus S0 2 divided by the cross-sectional area. .

m

6.53 x 1(T*(29)

G,

kg

air/s

+

6.53 x 10"

= 0.0929

Setting^

=

m

y

l-y

64.1)

kg

S0 2 /s

2

0.20,

4 6.53 x 10" (29)

0.2

+

6.53 x 10

1-0.2

64.1

0.3164 kg/s-m 2

0.0929

y\- -o.2o

/

0.18

\ /

/.

/

JM

0.16

^ 0.14 •2

A-

0.1 2 -

0.10

/

7

'

"o

0.06

n\/ y

/

.

^2--o.o;

i

,

0.004

0.002 I

0.006

I

x2 Mole Figure

Sec. 10.7

10.7-1.

Operating

line

and

fraction, x

interface compositions for

Absorption of Concentrated Mixtures

in

Example

Packed Towers

10.7-1.

629

I

I

m

on NO co CN on r--° no

CN

to to to o NO cn oo o o cn O CN o co o o o d d d

to to U-i oo oo oo ON o\ on oo CO

o o O O d d

o o o oo to ro on on ON d O d o to to ON co cn * o o O o d d d

r-~

oo o r-~

oo CO

dd

to to o .00 d d

NO CO o vi oo to co

O o o o o 8 o o o o o O CO ON 0 CO C^ T to to o co 00 00 OO CO d d 0 d d 00 t co CN NO ON 0 r- ro ON co to tO O co ro co XT xr 0 O O O 0 0 O 0 d d

o o o o

00 r- CN NO co
0 co CN

NO CN cn CN

OO*

(N CO CO

CO CN tO rco r-~ CN CN CO

O O O 0 d g O CO O co 0 to CN CN CN CN CN Tt ^ 0 tr 0 O O 0 0 d d d d *

*

Tf

0— O O 0 0 1

1

1

1

•

X

X X X X

0 O

NO to CN NO CO to NO NO CO cn to co v-i co CO

to 0 to CO 0 O O O 0 O 0 S 0 0 d d d d r— CO 0 0 0 0 CN d d d d d

CN

CTiap. 70

Stage and Continuous Gas••-Liquid Separation Processes

Similarly,

L=

Gy

is

calculated for

points and tabulated. For the liquid flow,

all

— x). Also, for the total liquid mass flow rate,

L'/(l

2 4.20 x 10" (18)

4.20 x 10"

+

Gx =

2 (

——

J64.1

0.0929

Calculated values of L and

Gx

for various values of

x are tabulated

in

Table

10.7-1.

To calculate values =

0.152G°

82

=

o(k'x a, for

=

x

0.152(8.138)

0,

0 82

Gx =

=

8.138 and

0.848

kg mol/s -m 3 mol -

frac

The

rest of these values are calculated and given in Table 10.7-1. For the value of k'y a, for y = 0.02, G y = 0.2130, G x = 8.138, and k' y

a

=

25 0.0594G°- G°7

o 7

=

0.0594(0.2130)

=

0.03398 kg mol/s

-

(8.138)

m3

0 25 -

mol

•

frac

This and other calculated values of k'y are tabulated. Note that these values vary considerably in the tower. Next the interface compositions y and x must be determined for the given y and x values on the operating line. For the point y t = 0.20 and x l = 0.00355 we make a preliminary estimate of (1 — y) iM = 1 — y = 1 — 0.20 =? 0.80. Also, the estimate of (1 - x) iM =1 - x =s 1 - 0.00355 s 0.996. The slope of the line by Eq. (10.6-30) is approximately ;

f

P^,

'

-

Ka/(l

nnP=

,, 0pe S

second

(i

y)

n-^ ;

^

(1

M"vf

=

_

_

0.04496/(0.80)

0.1688

andx = ;

0.00566. Using these values

Eqs. (10.4-6) and (10.4-7),

trial in

-

0.857/(0.996)

_

k y a/{l-y) iM

Plotting this on Fig. 10.7-1, y-, for the

x) iM

-

ln [(1

_

(1

-

ln [(1

0.1688)

_

o

-(1 -

mm

0-00355)

_

-(1

0.20)

o 20)]

-

" no

.

ft

0.00566)

-0.003 55)/(l -0.00566)]

The new slope by Eq. (10.6-30) is (-0.857/0.995)/(0.04496/0.816) = -15.6. Plotting, y,- = 0.1685 and x = 0.00565. This is shown as point v This calculation is repeated until point y 2 x 2 is reached. The slope of Eq. (10.6-30) increases markedly in going up the tower, being —24.6 at the top of the tower. The values of y and x are given in Table 10.7-1.

M

;

,

t

:

In order to integrate Eq. (10.6-14), values of (1 {y is

"

yd a r e needed and are tabulated in Table 10.7-1. calculated from Eq. (10.7-1).

/(,)

This y.

is

The

V -

=

y), (1

— y) ;M and = 0.20, /(y) ,

for y

816 * 1(r '

=

.6.33

repeated for other values of y. Then the function f(y) is then the sum of four rectangles:

is

plotted versus

total area

total area

=

Hence, the tower height

Sec. 10.7

—

Then

0.312 is

+

0.418

+

0.318

+

0.540

=

1.588

equal to the area by Eq. (10.6-14) andz

Absorption of Concentrated Mixtures

in

Packed Towers

=

1.588 m.

631

ESTIMATION OF MASS-TRANSFER COEFFICIENTS FOR

10.8

PACKED TOWERS Experimental Determination of Film Coefficients

10.8A

film mass-transfer coefficients k'y a and k'x a depend generally upon Schmidt number, Reynolds number, and the size and shape of the packing. The interactions among these factors are quite complex. Hence, the correlations for mass-transfer coef-

The individual

ficients are highly empirical.

Deviations of up to

25%

coefficient or resistance

To

ances in series.

reliability

To measure solute

is

negligible or can

systems as

NH

known

this 2

or

02

0

coefficient

two

2

the gas-phase film coefficient

C0

k'x a,

3

data for

2

k' a, y

we

is

used.

The experiment

negligible.

is

desire to use a system

by correcting

from the overall resistance

k'x a)

film resist-

so arranged

a system for absorption or

in water

k'z a

such that the

negligible.

is

have a liquid-phase resistance of about 10%. 3 -air-water

NH

is

By

Most such

subtracting

data for absorption of

in Eq. (10.4-24),

we obtain

the

Details of these are discussed elsewhere(Gl, SI, S2).

k' a. y

Correlations for Film Coefficients

10.8B

The experimental data in

or

liquid phase resistance (obtained to

because an overall

be approximately calculated.

the liquid film mass-transfer coefficient

very soluble in the liquid and the liquid-phase resistance

is

not too satisfactory.

that represents the

K'x a, which equals k'z a, since the gas-phase resistance

gives

is

difficulty arises

obtain the single-phase film coefficient, the experiment

desorption of very insoluble gases such as

C0

of these correlations

uncommon. A main

measured experimentally

is

that the other film resistance

To measure

The

are not

terms of H G

,

for the gas film coefficient in dilute mixtures

have been correlated

where

HG = The empirical equation

is

(10.6-36) k

y

aS

as follows:

H0 = where G y = kg total gas/s m G x = kg for a packing as given in Table 10.8-1 2

•

;

5

zGfGlNi,

(10.8-1)

total liquid/s

(T2).

•

m

2 ;

and

The temperature

a,

ft,

and y are constants which is small, is

effect,

number p-lpD, where p. is the viscosity of the gas mixture in kg/m 3 and D the diffusivity of solute A in the gas in m 2 /s. The can be shown to be independent of pressure.

included in the Schmidt

kg/m

s,

p the density in

and

coefficients k' a y

Equation

Ha

(10.8-1)

,

can be used

to correct existing

gas on a specific packing to absorption of solute

mass-flow

rates.

This

is

done by Eq.

data for absorption of solute

E

same system and

-]

0 5 -

until

A

in

a

same

(10-8-2)

correlations for liquid film coefficients in dilute mixtures

independent of gas rate

the

(10.8-2).

H Gm = n G(A lN SclE)/N SciA) The

the

in

show

that

HL

is

loading occurs, as given by the following:

HL =

N°Sc 5

0

(10.8-3)

(j^J where

632

HL

is

in

m, p L

is

liquid viscosity in

Chap. 10

kg/m

-s,

is

Schmidt number pJpD, p

is

Stage and Continuous Gas-Liquid Separation Processes

Table

Gas Film Height of a Transfer Unit H G

10.8-1.

Meters*

in

Range of Values o

Packing Type

ct

p

0.620

0.45

y

Raschig rings

mm (| in.) 25.4 mm (1 in.) 38.1 mm (1.5 38.1 mm (1.5 50.8 mm (2 in.) 9.5

0.557

0.32

in.)

0.830

0.38

in.)

0.689

0.38

0.894

0.41

-0.47 -0.51 -0.66 -0.40 -0.45

0.271-0.678

0.678-2.034

0.271-0.814

0.678-6.10

0.271-0.950

0.678-2.034

0.271-0.950

2.034-6.10

0.271-1.085

0.678-6.10

-0.74 -0.24 -0.40 -0.45

0.271-0.950

0.678-2.034

Berl saddles

mm (0.5 mm (0.5 25.4 mm (1 38.1 mm (1.5 12.7

in.)

0.541

0.30

12.7

in.)

0.367

0.30

0.461

0.36

0.652

0.32

in.)

HG -

in.)

0.271-0.950

2.034-6.10

0.271-1.085

0.542-6.10

0.271-1.356

0.542-6.10

= n/pD. aG"r G\ N^ s where G = kg total gas/s m 2 G„ = kg total liquid/s m J and y Data from Fellinger (P2) as given by R. E. Treybal, Mass Transfer Operations. New York: McGraw-Hill Book Company, 1955, p. 239. With permission. *

•

•

,

,

,

Source:

kg/m 3 and D

liquid density in

,

is

diffusivity of solute

A

in the liquid

inm 2/s. Data

are

given in Table 10.8-2 for different packings. Equation (10.8-3) can be used to correct existing data

on a given packing and solute

EXAMPLE

10.8-1.

to another solute.

Prediction of Film Coefficients for

Ammonia Absorp-

tion

Predict

solution

HG HL ,

,

and K'y a

NH

for

a packed tower with

in

Table

from water in a dilute absorption of 3 25.4-mm Raschig rings at 303 (86°F) and

K

Liquid Film Height of a Transfer Unit

10.8-2.

HL

Meters*

in

Packing

n

Range of G x

0.46

0.542-20.34

0

Raschig rings 9.5

12.7

25.4 38.1

50.8

mm mm mm mm mm

4 3.21 x 10~

(| in.)

(0.5 in.) (1

in.)

4 7.18 x 10" -3 2.35 x 10

0.35

0.542-20.34

0.22

0.542-20.34

10~ 3

0.22

0.542-20.34

0.22

0.542-20.34

0.28

0.542-20.34

(1.5 in.)

2.61 x

(2 in.)

2.93 x 10"

3

Berl saddles 12.7

25.4 38.1 • fl L

kg/m

•

s, :

1.456 x 10

(0.5 in.) (1 in.)

1.285 x 10

(1.5 in.)

1.366 x 10

= OiGJu^N^ 5

Source

Mass

mm mm mm

and

,

where

Gx =

= nJpD. G f is

less

"

3

-3 -3

kg total liquid/s than loading.

-

0.28

0.542-20.34

0.28

0.542-20.34

m2

,

pL

=

viscosity of liquid in

Based on data by Sherwood and Holloway (S3) as given by R. E. Treybal, New York: McGraw-Hill Book Company, 1955, p. 237.

Transfer Operations.

With permission.

Sec. 10.8

Estimation of Mass-Transfer Coefficients For Packed Towers

633

101.32 kg/s

kPa

m

•

The flow

pressure.

Gx =

rates are

2.543 kg/s

•

m2

and Gy

2

=

0.339

.

From Appendix

Solution:

A. 3 the equilibrium relation in a dilute solution

= 1.20x. Also, from Appendix A.3 for 3 5 10" x kg/m-s. Density p = 1.168 kg/m The diffusivity of H _5 2 /s. Correcting to 303 in air at 273 K from Table 6.2-1 is 1.98 x 10 = 2.379 x 10" 5 m 2/s. Hence, Eq. (6.2-45), D is

0.0151 =m(0.0126) or y

=

1.86

air,

NH K by

.

m

3

AB

1.86

pD ~

x 1Q-

10~ (1.168X2.379 x

5 )

Substituting into Eq. (10.8-1) using data from Table 10.8-1,

HG =

aG^y Gxy N^ 5

=

0.557(0.339)°-

32

(2.543)-°- 51 (0.669)°- 5

=

0.200

m

The viscosity of water = 1.1404 x 10" kg/m-s at 15°C and 0.8007 x 10" 3 at 30°C from Appendix A.2. TheD^ ofNH in water at 288 K(15°C) 3 -9 m 2 /s. Correcting this to 303 (30°C), using from Table 6.3-1 is 1.77 x 10 Eq. (6.3-9), 3

K

°-

-(££^%\'.T> * 10-) x 7\288

2-652 , 10

_

^0.8007

=

Then, using p

10

996 kg/m 3

for water,

0.8007

g

88

- ™vs

xlO- 3

9 (996X2.652 x 10" )

pD

Substituting into Eq. (10.8-3), using data from Table 10.8-2,

„ t . 8 (|)'»SV , (2.35

2 S

0^)( 0 800 7 ^, 0 -, )°"

x

(303,,-

= 0.2412 m Converting

'*

For

k'x a

y

V

=

k a

to k' a using

JTs

Eq. (10.6-36),

0139/29

=

o2oo

= 00584 kg

mol/s

m3 mo1

'

'

frac

using Eq. (10.6-37),

=

k'x a

25

—^— =

8

=

A ^(!,

0.586 kg mol/s

•

m3

•

mol

frac

Substituting into Eq. (10.4-24) for dilute solutions,

IFK a

=

y

ky a

+ IT = kxa K'y a

Note

=

fT7^o7 0.0584

= 1712 + + T^k 0.586

0.0522 kg mol/s

•

m3

•

mol

that the percent resistance in the gas film

is

2 048 -

=

19 168 -

frac

(17.12/19.168X100)

=

89.3%.

PROBLEMS 10.2-1.

Equilibrium and Henry's 1.333 x 10

634

4

Pa and

Law

The

Constant.

the total pressure

is

partial pressure

1.133

ofC0 2

in air is

x 10 5 Pa. The gas phase

Chap. 10

is

in

Problems

equilibrium with a water solution at 303 K. What is the value of x A ofC0 2 in equilibrium in the solution? See Appendix A. 3 for the Henry's law constant.

Ans. x A 10.2-2.

x 10" 5 mol fracC0 2

Aqueous Solution. At 303 K the concentration of C0 2 in 4 kg C0 2Ag water. Using the Henry's law constant from Appendix A. 3, what partial pressure of C0 2 must be kept in the gas to keep the C0 2 from vaporizing from the aqueous solution? Ans. p A = 6.93 x 10 3 Pa (0.0684 atm) Gas

Solubility in

water

10.2- 3.

= 7.07

is

0.90 x lO"

Phase Rule for a Gas-Liquid System. For the system S0 2 -air-water, the total pressure is set at 1 atm abs and the partial pressure ofS0 2 in the vapor is set at 0.20 atm. Calculate the

number

of degrees of freedom, F.

What

variables are

unspecified that can be arbitrarily set?

Stage Contact for Gas-Liquid System. A gas mixture at 2.026 containing air and S0 2 is contacted in a single-stage equilibrium mixer with pure water at 293 K. The partial pressure ofS0 2 in the original gas is 1.52 x 10* Pa. The inlet gas contains 5.70 total kg mol and the inlet water 2.20 total kg mol. The exit gas and liquid leaving are in equilibrium. Calculate the amounts and compositions of the outlet phases. Use equilibrium data from Fig. 10.2-1. Ans. x Al = 0.00495, y Al = 0.0733, L, = 2.21 1 kg mol, V = 5.69 kg mol

10.3- 1. Equilibrium

x 10 s

Pa

total pressure

l

Countercurrent Stage Tower. Repeat Example 10.3-2 using the same conditions but with the following change. Use a pure water flow to the tower of 108 kg mol H 2 0/h, that is, 20% above the 90 used in Example 10.3-2. Determine the number of stages required graphically. Repeat using the analytical Kremser equation.

10.3-2. Absorption in a

from Cream by Steam. Countercurrent stage stripping is to be used to remove a taint from cream. The taint is present in the original cream to the stripper at a concentration of 20 parts per million (ppm). For every 100 kg of cream entering per unit time, 50 kg of steam will be used for stripping. It is desired to reduce the concentration of the taint in the cream to 1 ppm. The equilibrium relation between the taint in the steam vapor and the liquid cream where yA is ppm of taint in the steam and x A ppm in the cream is y A = l0x A (El). Determine the number of theoretical stages needed. [Hint: In this case, for stripping from the liquid (L) stream to the vapor (V) stream the operating line

10.3- 3. Stripping Taint

,

be below the equilibrium line on the y A — x A diagram. It is assumed that none of the steam condenses in the stripping. Use ppm in the material balances.] Ans. Number stages = 1.85 (stepping down starting from the concentrated end) will

Mass-Transfer Coefficient from Film Coefficients. Using the same data 10.4-1, calculate the overall mass-transfer coefficients K'x and K x the flux, and the percent resistance in the gas film. 2 3 = 1.519 x 10" 3 Ans. K'x = 1.173 x 10" kg mol/s-m -mol frac, x 4 2 = 3.78 x 10~ kg mol/s-m 36.7% resistance A

10.4- 1. Overall

as in

Example

,

K

N

10.4- 2.

,

,

Use the Interface Concentrations and Overall Mass-Transfer Coefficients. same equilibrium data and film coefficients k'y and k'x as in Example 10.4-1. However, use bulk concentrations of y AG = 0.25 and x AL = 0.05. Calculate the following. (a) (b) (c)

Interface concentrations y Ai a.ndx Ai and flux Overall mass-transfer coefficients K' and

NA

Overall mass-transfer coefficient K'x and flux

NA

y

K

y

.

and

flux

NA

.

.

Water-Cooling Tower. A forced-draft countercurrent watercooling tower is to cool water from 43.3 to 26.7°C. The air enters the bottom of the tower at 23.9°C with a wet bulb temperature of 21.1°C. The value of Hp for the flow conditions is H a = 0.533 m. The heat-transfer resistance in the liquid

10.5- 1. Countercurrent

Chap. 10

Problems

635

will be neglected; i.e., h L is very large. Hence, values oiH* should be used. Calculate the tower height needed if 1.5 times the minimum air rate is used.

phase 10.5-2.

Minimum Gas Rate and Height of Water-Cooling

Tower. It is planned to cool 10°F to 85°F in a packed countercurrent water-cooling tower using entering air at 85°F with a wet bulb temperature of 75° F. The water flow is 2000 2 2 lb„/h-ft and the air flow is 1400 lb m air/h -ft The overall mass-transfer 3 coefficient is K G a = 6.90 lb mol/h ft atm. (a) Calculate the minimum air rate that can be used. 2 (b) Calculate the tower height needed if the air flow of 1400 lb m air/h ft is

water from

1

.

•

•

-

used, 2

935 lb m air/h ft (4241 kg air/h m ); (b) z = 21.8 ft (6.64 m) 10.5-3. Design of Water-Cooling Tower. Recalculate Example 10.5-1, but calculate the Ans.

(a)

minimum

G mi „ =

air rate

2

-

and use

-

1.75 times the

minimum

air rate.

of Changing Air Conditions on Cooling Tower. For the cooling tower in Example 10.5-1, to what temperature would the water be cooled if the entering air enters at 29.4°C but the wet bulb temperature is 26.7°C? The same gas and

10.5- 4. Effect

The water enters at 43.3°C, as before. (Hint : In this unknown. The tower height is the same as in Example 10.5-1. The operating line is as before. The solution is trial and error. Assume a

liquid flow rates are used.

case

TLl

is

the

slope of the

value of

TL1

that

is

greater than 29.4°C

Do

same height z is obtained.) Amount of Absorption in a Tray Tower. An

the graphical integration to see

if

the 10.6-

1.

existing tower contains the equiva-

being used to absorb S0 2 from air by pure water at 293 K and 1.013 x 10 5 Pa. The entering gas contains 20 mol S0 2 and the inlet air flow rate is 150 kg inert air/h m 2 The entering water rate is 2 6000 kg/h-m Calculate the outlet composition of the gas. (Hint: This is a trial-and-error solution. Assume an outlet gas composition of, say, y 1 = 0.01. Plot the operating line and determine the number of theoretical trays needed. If lent of 3.0 theoretical trays

and

is

%

•

.

.

this

number

is

not 3.0 trays, assume another value ofy 1( and so on.) Ans.

yt

=

0.011

for Number of Trays in Absorption. Use the analytical equations in Section 10.3 for countercurrent stage contact to calculate the

10.6-2. Analytical

number

Method

of theoretical trays needed for

Example

10.6-1.

of Ammonia in a Tray Tower. A tray tower is to be used to remove 99% of the ammonia from an entering air stream containing 6 mol % ammonia 5 at 293 K and 1.013 x 10 Pa. The entering pure water flow rate is 188 kg 2 H 2 0/h m and the inert air flow is 128 kg air/h m 2 Calculate the number of

10.6-3. Absorption

•

•

.

Use equilibrium data from Appendix A.3. For the tower, plot an expanded diagram to step off the number of

theoretical trays needed. dilute end of the trays

more accurately. Ans.

10.6-4.

Minimum

yl Liquid Flow

=

0.000639

(exit),

x N = 0.0260

(exit), 3.8

theoretical trays

The gas stream from a chemical reactor contains 25 mol % ammonia and the rest inert gases. The total flow is 5 181.4 kg mol/h to an absorption tower at 303 K and 1.013 x 10 Pa pressure, where water containing 0.005 mol frac ammonia is the scrubbing liquid. The outlet gas concentration is to be 2.0 mol % ammonia. What is the minimum flow L'min ? Using 1.5 times the minimum plot the equilibrium and operating in

a Packed Tower.

lines.

Ans. 10.6-5.

LV n =

262.6 kg mol/h

Stripping and Number of Trays. A relatively nonvolatile hydrocarbon oil contains 4.0 mol propane and is being stripped by direct superheated steam in a stripping tray tower to reduce the propane content to 0.2%. The temperature is held constant at 422 K by internal heating in the tower at 2.026 x 10 5

Steam

%

636

Chap. 10

Problems

A

1 1.42 kg mol of direct steam is used for 300 kg mol of The vapor-liquid equilibria can be represented by y = 25x, where y is mole fraction propane in the steam and x is mole fraction propane in the oil. Steam ran be considered as an inert gas and will not condense. Plot the operating and equilibrium lines and determine the number of theoretical trays

Pa

pressure.

total of

total entering liquid.

needed.

Ans.

5.6 theoretical trays (stepping

down from the tower

top)

%

of Ammonia in Packed Tower. A gas stream contains 4.0 mol reduced to 0.5 mol % in a packed absorption 3 and its ammonia content is s tower at 293 K and 1.013 x 10 Pa. The inlet pure water flow is 68.0 kg mol/h and the total inlet gas flow is 57.8 kg mol/h. The tower diameter is 0.747 m. The 3 film mass-transfer coefficients are k' a = 0.0739 kg mol/s-m -mol frac and y

10.6-6. Absorption

NH

k'x

a

—

kg mol/s-m 3 -mol do as follows.

0.169

mixtures,

frac.

Using the design methods

Calculate the tower height using k'y a. Calculate the tower height using K'y a.

(a)

(b)

Ans. 10.6-7.

(a) z

=

for dilute gas

2.362

m (7.75

ft)

Repeat Example 10.6-2, using the overall liquid mass-transfer coefficient K'x a to calculate the tower

Tower Height Using Overall Mass-Transfer

Coefficient.

height. 10.6-8.

m

Experimental Overall Mass-Transfer Coefficient. In a tower 0.254 in diameter absorbing acetone from air at 293 and 101.32 kPa using pure water, the following experimental data were obtained. Height of 25.4-mm Raschig = 3.30 kg mol air/h, y = 0.01053 mol frac acetone, y 2 = rings = 4.88 m, 0.00072, L = 9.03 kg mol water/h, x l = 0.00363 mol frac acetone. Calculate the experimental value of K y a.

K

V

x

to Transfer Unit Coefficients from Mass-Transfer CoeffiExperimental data on absorption of dilute acetone in air by water at 80°F and 1 atm abs pressure in a packed tower with 25.4-mm Raschig rings 2 were obtained. The inert gas flow was ,95 lb m air/h ft and the pure water flow 3 2 was 987 Ibjh ft The experimental coefficients are k G a = 4.03 lb mol/h ft 3 3 '= The equilibrium data can be atm and k L a 16.6 lb mol/h ft -lb mol/ft 3 expressed by c A = 1.37p y4 where c A = lb mol/ft and p A = atm partial pressure

10.6- 9. Conversion cients.

•

•

.

;

.

,

of acetone. Calculate the film height of transfer units Calculate OG

(a)

H

(b)

H G and H L Ans.

(b)

H oa =

of Tower Using Transfer Units. Repeat Example and calculate H Ll N L and tower height.

10.6-10. Height

units

.

.

ft

(0.292

m)

,

10.6-11. Experimental Value of 10.6- 8, calculate the

0.957

10.6-2 but use transfer

Hoa

number

.

Using the experimental data given in Problem N O0 and the experimental value of

of transfer units

HogAns,

H oa =

1.265

m

Film Coefficients and Design ofS0 2 Tower. Using the data of Example 10.7- 1, calculate the height of the tower using Eq. (10.6-15), which is based on the liquid film mass-transfer coefficient k'x a. [Note: The interface values X,- have already been obtained. Use a graphical integration of Eq.

10.7- 1. Liquid

(10.6-15).]

Ans.

= 1.586 m of Example

z

10.7-2. Design of S0 2 Tower Using Overall Coefficients. Using the data 10.7-1, calculate the tower height using the overall mass-transfer coefficient K' a. [Hint: Calculate K'a at the top of the tower and at the bottom of the y

tower from the film coefficients. Then use a linear average of the two values for the design. Obtain the values of y* from the operating and equilibrium

Chap. 10

Problems

637

Graphically integrate Eq. (10.6-16), keeping K'y a outside the

plot.

line

integral.]

of Packed Tower Using Transfer Units. For Example 10.7-1 calculate the tower height using the G [Hint: G and the number of transfer units Calculate G at the tower top using Eq. (10.6-36) and at the tower bottom. Use the linear average value for G Calculate the number of transfer units G by graphical integration of the integral of Eq. (10.6-32). Then calculate the tower height.]

10.7-3. Height

N

H

.

H

H

.

N

H

m

Ans. (average value) G = 0.2036 of Absorption Tower Using Transfer Units. The gas S0 2 is being 5 scrubbed from a gas mixture by pure water at 303 and 1 .013 x 10 Pa. The inlet gas contains 6.00 mol S0 2 and the outlet 0.3 mol S0 2 The tower

10.7- 4. Design

K

%

%

cross-sectional area of packing inert air/h

and the

inlet

transfer coefficients are

is

0.426

water flow

HL

=

0.436

is

m

m

2 .

The

inlet

gas flow

.

is

mol

13.65 kg

984 kg mol inert water/h. The mass-7 3 and k G a = 6.06 x 10 kg mol/s m the tower for the given concentration •

Pa and are to be assumed constant in range. Use equilibrium data from Appendix A.3. By graphical integration, determine N G Calculate the tower height. (Note: The equilibrium line is •

.

markedly curved, so graphical integration

is

necessary even for

this dilute

mixture.)

Ans.

NG =

8.47 transfer units, z

=

1.31

m

1

10.8- 1. Prediction of Mass-Transfer Coefficients. Predict the mass-transfer coefficients

H G H i,

k'x a, k' a, and K'x a for absorption of C0 2 from air by water in y same' packing and using 1.6 times the flow rates in Example 10.8-1 at 303 (Hint: Use equilibrium data from Appendix A. 3 for Henry's law constants C0 2 in water. Use diffusivity data for C0 2 in water from Table 6.3-1 and ,

C0 2

in air

from Table

6.2-1.

Correct data to 303 K.) Ans. G = 0.2186 m,

H

HL =

the

K. for for

0.2890

m

NH

Film Mass-Transfer Coefficients. For absorption of dilute 3 from air by water at 20°C and 101.3 kPa abs pressure, experimental values are G = 0.1372 m and L = 0.2103 m for heights of transfer units. The flow 2 rates are G x = 13 770 kg/lv m and G y = 1343 kg/h m 2 For absorption of acetone from air by water under the same conditions in the same packing, predict k' a, k'x a, and K' a. Use equilibrium data from Appendix A. 3. Use y y the diffusivity for acetone in water from Table 6.3-1. The diffusivity of acetone 4 in air at 0°C is 0.109 x 10~ m 2 /s at 101.3 kPa abs pressure.

10.8-2. Correction of

H

H

•

.

REFERENCES Chemical Engineering.

New

Bubble Tray Design Manual. American Institute of Chemical Engineers,

New

Badger, W. L., and BaNCHERO, J. T. Introduction York: McGraw-Hill Book Company, 1955.

to

York, 1958. EaRLE, R.

L.

Unit Operations

in

Food Processing. Oxford: Pergamon

Press, Inc.,

1966.

Geankoplis, C.

J.

Mass Transport Phenomena. Columbus, Ohio: Ohio

State

University Bookstores, 1972.

Leva, M. Tower Packings and Packed Tower Design, 2nd ed. Akron, Ohio: U.S. Stoneware, Inc., 1953.

MiCKLEY, H.

S.,

Sherwood,

Chemical Engineering, 2nd ed. Mick. ley, H.

638

S.

T. K., and Reed, C. E. Applied Mathematics

New

Chem. Eng. Progr.,

in

York: McGraw-Hill Book Company, 1957. 45, 739(1949).

Chap. 10

References

(PI)

(51)

(52) (53)

(Tl)

(T2)

Perry, R. H., and Green, D. Perry's Chemical Engineers' Handbook, 6th ed. New York: McGraw-Hill Book Company, 1984.

Sherwood, T. K., Pigford, R. L., and Wilke, C. R. Mass Transfer. New York: McGraw-Hill Book Company, 1975. Shulman, H. L., and coworkers. A.I.Ch.E. J., 1, 274(1955); 9, 479 (1963). Sherwood, T. K., and Holloway, F. A. L. Trans.. A.I.Ch.E., 30, 39 (1940). Treybal, R. E. Mass Transfer Operations, 3rd ed. New York: McGraw-Hill Book Company, 1980. Treybal, R. E. Mass Transfer Operations. New York: McGraw-Hill Book Company, 1955.

(W 1)

Whitney, R.

Chap. 10

P.,

References

and Vivian,

J.

E.

Chem. Eng. Progr.,

45, 323 (1949).

639

CHAPTER

11

Vapor-Liquid Separation Processes

VAPOR-LIQUID EQUILIBRIUM RELATIONS

11.1

11.1 A

As

Phase Rule and Raoult's

Law

equilibrium in vapor-liquid systems is restricted by the As an example we shall use the ammonia-water, vapor-liquid system. For two components and two phases, F from Eq. (10.2-1) is 2 degrees of freedom. The four variables are temperature, pressure, and the composition y A of NH 3 in the vapor phase and x A in the liquid phase. The composition of water (B) is fixed ify,, orx^ is specified, since y A + y B = 1.0 and x A + x B = 1.0. If the pressure is fixed, only one more variable can be set. If we set the liquid composition, the temperature and vapor compoin the gas-liquid systems, the

phase

rule, Eq. (10.2-1).

sition are automatically set.

An

ideal law, Raoult's law,

can be defined PA

where p A

is

A

in

holds only for ideal

vapor-liquid phases

in

equilibrium.

= PAxA

(11.1-1)

in the vapor in Pa(atm), P A is the vapor mole fraction of A in the liquid. This law solutions, such as benzene-toluene, hexane-heptane, and methyl

the partial pressure of

pressure of pure

for

Pa

(atm),

component A

and x A

is

the

alcohol-ethyl alcohol, which are usually substances very similar to each other.

systems that are ideal or nonideal solutions follow Henry's law

11. IB

Boiling-Point

Diagrams and xy Plots

Often the vapor-liquid equilibrium relations for a binary mixture of

shown in Fig. 11.1-1 kPa. The upper line is

a boiling-point diagram

for the

total pressure of 101.32

the saturated

and the lower is

line

in the region

is

640

it

if

we

are given as

vapor

at

a

line (the dew-point line)

line).

The two-phase

region

lines.

mixture of x Al =0.318 and heat the 98°C (371.2 K) and the composition of the first vapor in

start with a cold liquid

will start to boil at

A and B

system benzene (X)-toluene (B)

the saturated liquid line (the bubble-point

between these two

In Fig. 11.1-1,

mixture,

Many

in dilute solutions.

120 vapor region

xa

yAi

i

Mole fraction benzene in xA or vapor, y A

liquid,

,

Figure

Boiling point diagram for benzene (A)-toluene (B) at 101.325

11.1-1.

(1

equilibrium left

since y A

is is

y A1

=

kPa

atm) total pressure.

0.532.

As we continue

boiling, the

composition x A

will

move

to the

richer in A.

The system benzene-toluene

follows Raoult's law, so the boiling-point

diagram can

be calculated from the pure vapor-pressure data in Table 11.1-1 and the following equations:

Pa+Pb = P

(H.l-2)

P A x A + P B(l-x A ) = P

(11.1-3)

EXAMPLE 11.1-1.

Use of Raoult's Law for Boiling-Point Diagram vapor and liquid compositions in equilibrium at 95°C (368.2 K) for benzene-toluene using the vapor pressure from Table 1 1.1-1 at 101.32 kPa. Calculate the

At 95°C from Table 11.1-1 for benzene,

Solution:

PB =

63.3 kPa. Substituting into Eq. 155.7( X/4 )

Hence, x A

=

0.411

+

- xA =

63.3(1

and x B

=

(1 1.1-3)

)

1

- xA =

101.32 1

PA =

155.7

kPa and

and solving,

kPa

(760

mm

- 0.411 = 0.589.

Hg)

Substituting into

Eq. (11.1-4),

Vj yA

A common method Sec. 11. 1

-

PA x A = P

155.7(0.411)

=

0.632

101.32

of plotting the equilibrium data

Vapor-Liquid Equilibrium Relations

is

shown in

Fig. 11.1-2,

where y A

641

Table

1

1.1-1.

Vapor-Pressure and Equilibrium-Mole-Fraction Data for Benzene-Toluene System Vapor Pressure

Temperature

K

mm Hg

kPa

°C

Mole Fraction Benzene at 101325 kPa

Toluene

Benzene

kPa

mm Hg

XA

yA

353.3

80.1

101.32

760

1.000

1.000

358.2

85

116.9

877

46.0

345

0.780

0.900

363.2

90

135.5

1016

54.0

405

0.581

0.777

368.2

95

155.7

1168

63.3

475

0.411

0.632

373.2

100

179.2

1344

74.3

557

0.258

0.456

378.2

105

204.2

1532

86.0

645

0.130

0.261

383.8

110.6

240.0

1800

101.32

760

0

0

is is

plotted versus x A for the benzene-toluene system. richer in component A than is x A

The

45° line

is

given to

show thaty^

.

The

boiling-point diagram in Fig. 11.1-1

typical of an ideal system following

is

Raoulfs law. Nonideal systems differ considerably. In Fig. ll.l-3a the boiling-point diagram is shown for a maximum-boiling azeotrope. The maximum temperature Tmilx corresponds to a concentration x Az andx^, = y Az at this point. The plot ofy^ versusx^ would show the curve crossing the 45° line at this point. Acetone-chloroform is an

example of such a system. In Fig. 11.1 -3t> a minimum-boiling azeotrope y Az = x Az at Tmin Ethanol-water is such a systcn.

is

shown with

.

11.2

If

SINGLE-STAGE EQUILIBRIUM CONTACT FOR VAPOR-LIQUID SYSTEM

a vapor-liquid system

liquid,

is

being considered, where the stream

and the two streams are contacted

V2

is

a vapor and

in a single equilibrium stage

similar to Fig. 10.3-1, the boiling point or the

which

L0 is

is

a

quite

xy equilibrium diagram must be used is not available. Since we are

because an equilibrium relation similar to Henry's law considering only two components material balances.

If

A and

B, only Eqs. (10.3-1)

and the A condenses, 1 mol

sensible heat effects are small

and

(10.3-2) are

used

latent heats of

for the

both com-

when 1 mol of of B must vaporize. Hence, vapor V2 entering will equal Vl leaving. Also, moles 0 = L,. This case called one of constant molal overflow. An example is the benzene-toluene system.

pounds are

the same, then

the total moles of is

EXA M PLE

1 1.2-1. Equilibrium Contact of Vapor-Liquid Mixture vapor at the dew point and 101.32 kPa containing a mole fraction of 0.40 benzene (A) and 0.60 tolueneJB) and 100 kg mol total is contacted with 1 10 kg mol of a liquid at the boiling point containing a mole fraction of 0.30 benzene and 0.70 toluene. The two streams are contacted in a single stage, and the outlet streams leave in equilibrium with each other. Assume constant molal overflow. Calculate the amounts and compositions of the exit

A

streams.

Solution:

642

The process flow diagram

is

the

same

Chap. II

as in Fig. 10.3-1.

The given

Vapor-Liquid Separation Processes

Mole fraction benzene FIGURE

11.1-2

in liquid,

xA

Equilibrium diagram for system benzene (Aytoluene (B) at 101.32 kPa (1 atm).

V2 = 100 kg mol, yA2 = 0.40, L 0 = 110 kg mol, and x A0 = 0.30. For constant molal overflow, V2 = Vl and L 0 = L t Substituting into Eq. (10.3-2) to make a material balance on component A, values are

.

L 0 x A0 + V2 y A2 = L x Al + V y A l

110(0.30)

To 1

is

solve Eq.

1.1-1

(1 1.2-1),

+

100(0.40)

=

x

110x^,

+ lOOy^

the equilibrium relation between

must be used. This

is

by

trial

(103-2)

,

and error since an

y^

(11.2-1)

and x Al

in Fig.

analytical expression

not available that relates y A and x A First, we assume that x Al =0.20 and substitute into Eq. (11.2-1) to .

solve for y Al

.

110(0.30)

Solving, y Al =0.51. ted in Fig. 11.2-1. It

Sec. 11.2

The is

+

100(0.40)

=

110(0.20)

+ lOOy^

equilibrium relations for benzene-toluene are plot-

evident thaty^,

=

0.51

andx Al =

0.20

do not

Single-Stage Equilibrium Contact For Vapor-Liquid System

fall

on

643

Mole fraction benzene Figure

the curve. This point

and

solving, y Al

=

is

11.2-1.

plotted

Solution to

Example

xA

1 1.2-1.

on the graph. Next, assuming thatx^

This point

0.29.

in liquid,

is

also plotted in Fig. 11.2-1.

= 0.40

Assuming

x Al = 0.30, y Al = 0.40. A straight line is drawn between these three points which represents Eq. (1 1.2-1). At the intersection of this line with the

that

equilibrium curve, y Al

0.455

andx^ =

0.25,

which check Eq.

(1 1.2-1).

SIMPLE DISTILLATION METHODS

11.3

11. 3 A

The

=

Introduction is a method used to separate the components of a liquid upon the distribution of these various components between a phase. All components are present in both phases. The vapor phase is

unit operation distillation

solution, which depends

vapor and a liquid created from the liquid phase by vaporization

The

at the boiling point.

basic requirement for the separation of the

components by

distillation is that

the composition of the vapor be different from the composition of the liquid with which is

in

where

all

components are appreciably

however, of a solution of

salt

such as

volatile,

water solutions, where both components

will

be

and water, the water

in

in the is

it

concerned with solutions ammonia-water or ethanol-

equilibrium at the boiling point of the liquid. Distillation

is

vapor phase. In evaporation,

vaporized but the

salt is not.

The

process of absorption differs from distillation in that one of the components in absorption

from

is

essentially insoluble in the liquid phase.

air

11.3B

by water, where

air is

insoluble in the

An example

is

water-ammonia

absorption of

ammonia

solution.

Relative Volatility of Vapor-Liquid Systems

In Fig. 11.1-2 for the equilibrium diagram for a binary mixture of A

distance between the equilibrium line and the 45°

line,

and

B, the greater the

the greater the difference between

the vapor composition y A and liquid composition x A . Hence, the separation made. A numerical measure of this separation is the relative volatility a. AB

easily

denned

644

as the ratio of the concentration of

A

in the

Chap. 11

is .

more

This

vapor over the concentration of A

is

in

Vapor-Liquid Separation Processes

the liquid divided by the ratio of the concentration of

B

in the

vapor over the con-

centration of B in the liquid.

a AB

where a AB If

is

=

V*b

the relative volatility of

A

(1

- 3^/(1 ~ x A

with respect to

B in

(113-1) )

the binary system.

the system obeys Raoult's law, such as the benzene-toluene system,

Pa* a yA

yB JD

p

=

Prt x t

(113-2)

P

Substituting Eq. (11.3-2) into (11.3-1) for an ideal system,

Pa

Equation

(1 1.3-1)

(113-3)

can be rearranged to give

yA 1

+

(a

-

(11-3-4) l)x A

where a = a AB When the value of a is above 1.0, a separation is possible. The value of oc may change as concentration changes. When binary systems follow Raoult's law, the .

relative volatility often varies

only slightly over a large concentration range at constant

total pressure.

EXAMPLE

113-1. Relative Volatility for Benzene— Toluene System Using the data from Table 11.1-1, calculate the relative volatility for the benzene-toluene system at 85°C (358.2 K) and 105°C (378.2 K).

Solution:

At 85°C, substituting into Eq. (11.3-3)

for a

system following

Raoult's law,

a

=

116.9

2.54

46.0 Similarly at 105°C,

*-w = The

11.

3C

variation in a

is

2 38 -

about 7%.

Equilibrium or Flash Distillation

1. Introduction to distillation methods. Distillation can be carried out by either of two main methods in practice. The first method of distillation involves the production of a vapor by boiling the liquid mixture to be separated in a single stage and recovering and condensing the vapors. No liquid is allowed to return to the single-stage still to contact the rising vapors. The second method of distillation involves the returning of a portion of the condensate to the still. The vapors rise through a series of stages or trays, and part of

the condensate flows

downward through

the vapors. This second

method

is

the series of stages or trays countercurrently to

called fractional distillation, distillation with reflux, or

rectification.

There are three important types of distillation that occur

Sec. 11.3

Simple Distillation Methods

in

a single stage or

still

and

645

that

do not involve

second

is

rectification.

The

first

of these

equilibrium or flash distillation, the

is

simple batch or differential distillation, and the third

is

simple steam

distil-

lation.

2.

In equilibrium or flash distillation, which occurs in a

Equilibrium or flash distillation.

single stage, a liquid mixture

is

partially vaporized.

The vapor

is

allowed to

come

to

equilibrium with the liquid, and the vapor and liquid phases are then separated. This can

be done batchwise or continuously. In Fig.. 11.3-1 a binary mixture of

mol/h into a heater separated.

is

The composition

component

A

of

F

is

L = F —

Then

A and B

flowing at the rate of

F

and is material balance on

the mixture reaches equilibrium

x F mole fraction of A.

A

total

as follows:

is

Fx F Since

components

partially vaporized.

V, Eq.

(1

1.3-5)

=Vy +

Lx

(113-5)

becomes

Fx F

^Vy + (F-

V)x

(113-6)

Usually, the moles per hour of feed F, moles per hour of vapor V, and moles per hour of

L

are

known

or

set.

Hence, there are two unknowns x and y in Eq. (11.3-6). The other is the equilibrium line. A convenient method to

relationship needed to solve Eq. (11.3-6)

use

is

to plot Eq. (11.3-6)

and the equilibrium

shown

11.3D

on the xy equilibrium diagram. The

line

is

the desired solution. This

is

intersection of the equation

similar to

Example

11.2-1

and

in Fig. 11.2-1.

Simple Batch or Differential Distillation

first charged to a heated kettle. The withdrawn as rapidly as they form to a condenser, where the condensed vapor (distillate) is collected. The first portion of vapor condensed will be richest in the more volatile component A. As vaporization proceeds, the vaporized product becomes leaner in A. In Fig. 1 1.3-2 a simpJe still is shown. Originally, a charge ofi^ moles of components A and B with a composition of x, mole fraction of A is placed in the still. At any given time, there are L moles of liquid left in the still with composition x and the composition

In simple batch or differential distillation, liquid

liquid charge

is

is

boiled slowly and the vapors are

is y. A differential amount of dL is vaporized. The composition in the still pot changes with time. For deriving the equation for this process, we assume that a small amount of is vaporized. The composition of the liquid

of the vapor leaving in equilibrium

V,

heater

F,

y

separator

xp

L, x

FIGURE

646

11.3-1.

Equilibrium or flash distillation.

Chap. II

Vapor-Liquid Separation Processes

V moles

vapor to condenser

y

L moles

liquid

x

Figure

Simple batch or differential

11.3-2

distillation.

— dx and the amount of liquid from Ho L — dL. A material balance made where the original amount = the amount left in the liquid + the

changes from x to x

on A can be amount of vapor.

xL = (x- dx)(L-dL) + ydL Multiplying out the right

(113-7)

side,

xL = xL - x dL - L dx + dx dL + y dL

(113-8)

Neglecting the term dx dL and rearranging,

dL

dx

L

y

—

(113-9)

x

Integrating, L '.

dL In

where

L

l

is

the original moles charged,

composition, and x 2 the

The

final

integration of Eq.

—= L2

dx (113-10)

y-x

the moles

left

in the

still,

x,

the original

composition of liquid.

(1 1.3-10)

can be done graphically by plotting

l/(y

—

x) versus

andx 2 The equilibrium curve gives 1.3-10) is known as the Rayleigh equation.

x and getting the area under the curve between x,

.

between y and x. Equation (1 The average composition of total material distilled, y as the relationship

,

can be obtained by a material

balance.

L,x,

= L2

x2

+

(L,

- L 2 )yas

(11.3-11)

EXAMPLE 113-2.

Simple Differential Distillation mixture of 100 mol containing 50 mol n-pentane and 50 mol % n-heptane is distilled under differential conditions at 101.3 kPa until 40 mol is distilled. What is the average composition of the total vapor distilled andthe composition of the liquid left? The equilibrium data are as follows,

%

A

where x and y are mole X

Sec. 11. 3

fractions of n-pentane.

y

X

y

X

y

1.000

1.000

0.398

0.836

0.059

0.271

0.867

0.984

0.254

0.701

0

0

0.594

0.925

0.145

0.521

Simple

Distillation

Methods

647

Solution:

x

{

=

0.50,

to be used in Eq. (11.3-10) are L t = 100 mol, (moles distilled) = 40 mol. Substituting into

The given values L 2 = 60 mol, V

Eq. (11.3-10), = o.s

r Xi

d (113-12)

The unknown

is

x2

the composition of the liquid

,

To do

differential distillation.

versus

x

is

value of y l/(y

— x)=

made

=

L2

at the

end of the

the graphical integration a plot of l/(y

For x

in Fig. 11.3-3 as follows.

=

— x)

0.594, the equilibrium

Then \/{y - x) = 1/(0.925 - 0.594) = 3.02. The point and x = 0.594 is plotted. In a similar manner, other points

0.925.

3.02

are plotted.

To

determine the value of x 2 the area of Eq. (11.3-12) is obtained under x x = 0.5 to x 2 such that the area = 0.510. Hence, x 2 = 0.277. Substituting into Eq. (11.3-11) and solving for the average composition of the 40 mol distilled, ,

the curve from

100(0.50)

=

60(0.277)

ya v

=

0.835

+

40(y av )

Simple Steam Distillation

11.3E

At atmospheric pressure high-boiling liquids cannot be purified by distillation since the

components of the

may decompose

liquid

at the high

temperatures required. Often the

high-boiling substances are essentially insoluble in water, so a separation at lower distillation. This method is often used to component from small amounts of nonvolatile impurities. If a layer of liquid water (A) and an immiscible high-boiling component (B) such as a hydrocarbon are boiled at 101.3 kPa abs pressure, then, by the phase rule, Eq. (10.2-1), for three phases and two components,

temperatures can be obtained by simple steam separate a high-boiling

F= Hence, each

if

the total pressure

will exert its

is

7

—

3

+

2

fixed, the

own vapor

=

1

degree of freedom

system

is

fixed.

Since there are two liquid phases,

pressure at the prevailing temperature and cannot be

influenced by the presence of the other.

When

sum

the

of the separate vapor pressures

equals the total pressure, the mixture boils and

PA + P B = P

(11.3-13)

4r

x 2 =0.277

x

x

=0.5

x Figure

648

11.3-3.

Graphical integration for Example

Chap.

1 1

1 1.3-2.

Vapor-Liquid Separation Processes

where P A is vapor pressure of pure water vapor composition is

yx

=

A

and

y

PB

is

vapor pressure of pure B. Then the

= jr

y„

(H3-i4)

liquid phases are present, the mixture will boil at the same tempervapor of constant composition y A The temperature is found by using the vapor-pressure curves of pure A and pure B.

As long as the two ature, giving a

.

Note that by steam distillation, as long as liquid water is present, the high-boiling component B vaporizes at a temperature well below its normal boiling point without using a vacuum. The vapors of water (A) and high-boiling component (B) are usually condensed in a condenser and the resulting two immiscible liquid phases separated. This method has the disadvantage that large amounts of heat must be used to simultaneously evaporate the water with the high-boiling compound. The ratio moles of B distilled to moles of A distilled is (113-15)

Steam distillation is sometimes used in the food industry for the removal of volatile and flavors from edible fats and oils. In many cases vacuum distillation is used

taints

instead of steam distillation to purify high-boiling materials.

The

total pressure

low so that the vapor pressure of the system reaches the total pressure

is

quite

at relatively

low

temperatures.

Van Winkle amount of a

(VI) derives equations for steam distillation where an appreciable

nonvolatile

component

involves a three-component system.

is

present with the high-boiling component. This

He

also considers other cases for binary batch,

continuous, and multicomponent batch steam distillation.

DISTILLATION WITH REFLUX McCABE-THIELE METHOD

11.4

11.4A

AND

Introduction to Distillation with Reflux

Rectification (fractionation) or stage distillation with reflux, from a simplified point of

view, can be considered to be a process in which a series of flash-vaporization stages are

arranged

in

a series in such a manner that the vapor and liquid products from each stage

flow countercurrently to each other.

The

liquid in a stage

stage below and the vapor from a stage flows

upward

is

conducted or flows

to the

to the stage above. Hence, in each

V and a liquid stream L enter, are mixed and equilibrated, and a vapor and a liquid stream leave in equilibrium. This process flow diagram was shown in Fig. 10.3-1 for a single stage and an example given in Example 11.2-1 for a benzenestage a vapor stream

toluene mixture.

For the countercurrent contact with multiple stages

in Fig. 10.3-2, the material-

balance or operating-line equation (10.3-13) was derived which relates the concentrations of the vapor and liquid streams passing each other in each stage. In a distillation

column

the stages (referred to as sieve plates or trays) in a distillation tower are arranged

shown schematically in Fig. 1 1.4-1. column in Fig. 11.4-1 somewhere in the middle of the column. If liquid, it flows down to a sieve tray or stage. Vapor enters the tray and bubbles

vertically, as

The the feed

feed enters the

is

Sec. 11.4

Distillation With Reflux

and McCabe-Thiele Method

649

through the liquid on stage,

where

it

is

centration of the

The vapor and liquid The vapor continues up to the next tray or

this tray as the entering liquid flows across.

leaving the tray are essentially in equilibrium.

again contacted with a downflowing liquid. In

more

component

volatile

(the lower-boiling

this case the

component A)

is

conbeing

upward and decreased in the liquid from vapor product coming overhead is condensed in a condenser and a portion of the liquid product (distillate) is removed, which contains a high concentration of A. The remaining liquid from the condenser is returned (refluxed) increased in the vapor from each stage going

each stage going downward. The

final

as a liquid to the top tray.

The and

liquid leaving the

The vapor from the shown in the tower

where it is partially vaporized, withdrawn as liquid product. the bottom stage or tray. Only three trays are

bottom tray enters a

the remaining liquid, which reboiler

is

is

sent

lean in

back

A

to

reboiler,

or rich

of Fig. 11.4-1. In most cases the

the sieve tray the vapor enters through an opening give intimate contact of the liquid

and

in B, is

and vapor on the

liquid leaving are in equilibrium.

The

number of trays

is

much

greater. In

and bubbles up through the tray.

liquid to

In a theoretical tray the vapor

reboiler can be considered as a theoretical

stage or tray.

Figure

650

1

1.4-1.

Process flow of a fractionating tower containing sieve trays.

Chap.

II

Vapor-Liquid Separation Processes

11. 4B

McCabe-Thiele Method of Calculation

for

Number

of Theoretical Stages

A

Introduction and assumptions.

mathematical-graphical method

for determining needed for a given separation of a binary mixture of A and B has been developed by McCabe and Thiele. The method uses material balances around certain parts of the tower, which give operating lines somewhat similar to Eq. (10.3-13), and the xy equilibrium curve for the system. The main assumption made in the McCabe—Thiele method is that there must be equimolar overflow through the tower between the feed inlet and the top tray and the feed inlet and bottom tray. This can be shown in Fig. 11.4-2, where liquid and vapor /.

the

number

of theoretical trays or stages

streams enter a tray, are equilibrated, and leave.

A

total material balance gives

Vn + +L„_, = Vn +L„

(11.4-1)

Kn+1 y„ +1 + L„_ 1 x„_ = Vn yn + L n x n

(11.4-2)

l

A component balance on A gives 1

mol/h of vapor from tray n + 1,L„ is mol/h liquid from tray n,y„ +1 is mole Vn+l and so on. The compositions y„ andx„ are in equilibrium and the temperature of the tray n is T„. If Tn is taken as a datum, it can be shown by a heat

where Vn+ , fraction of

is

A

in

,

balance that the sensible heat differences

in

the four streams are quite small

solution are negligible. Hence, only the latent heats in stream

Since molar latent heats for chemically similar

Vn and 2.

L = n

L„_

,.

Therefore,

In Fig.

feed being introduced to the

1

in

at

if

heats of

are important.

=

Vn+l

the tower.

continuous

1.4-3 a

column

V„

are almost the same,

we have constant molal overflow

Equations for enriching section.

shown with

compounds

Vn+l and

distillation

column

is

an intermediate point and an

product and a bottoms product being withdrawn. The upper part of o^ above the feed entrance is called the enriching section, since the entering feed of binary components A and B is enriched in. this section, so that the distillate is richer in A than the feed. The tower is at steady state. An overall material balance around the entire column in Fig. 1 1.4-3 states that the entering feed of F mol/h must equal the distillate D in mol/h plus the bottoms in

overhead

distillate

the tower

W

mol/h.

F = A

total material

D+W

balance on component A gives

Fx F = Dx D +

Figure

Sec. II A

(11.4-3)

11.4-2.

Vapor and

Distillation Willi Reflux

Wx w

liquid flows entering

(11.4-4)

and leaving a

and McCabe-Thiele Method

tray.

651

«

Figure

1

1.4-3.

Distillation

column showing material-balance sections for McCabe-

Thiele method.

In Fig. 11.4-4a the distillation tower section above the feed, the enriching section,

shown

schematically.

The vapor from

is

the top tray having a composition y l passes to the

condensed so that the resulting liquid is at the boiling point. The reflux stream L mol/h and distillate D mol/h have the same composition, so y l = x D Since equimolal overflow is assumed, L, = L 2 = L n and Vv = V2 = Vn = Vn+l

condenser, where

it is

.

.

Making a

total material

balance over the dashed-line section

Va+l -L„ + D

652

Chap.

1 1

in Fig.

1 1

.4-4a,

(11.4-5)

Vapor-Liquid Separation Processes

Making

a balance on component A, (11.4-6)

Solving for yn+l the enriching-section operating line is ,

Dx r

L.

+ "1/ y +

x„ 1/ k

ji

Since

Kn+1 = L„ + D,L„/Kn+1 = R/{R +

+

1)

1

n

and Eq.

R R +

x. 1

+

(11.4-7) 1

(11.4-7)

becomes

xD

R +

(11.4-8) 1

where R = LJD = reflux ratio = constant. Equation (11.4-7) is a straight line on a plot of vapor composition versus liquid composition. It relates the compositions of two streams passing each other and is plotted in Fig. 11.4-4b. The slope is LJV„ +l or R/(R + 1), as given in Eq. (11.4-8). It intersects the y = x line (45° diagonal line) at x = x D The intercept of the operating line at x = 0 is y = Xp/(K + 1). The theoretical stages are determined by starting at x D and stepping off the first plate to x Then y 2 is the composition of the vapor passing the liquid x In a similar manner, the other theoretical trays are stepped off down the tower in the enriching .

t

.

.

t

section to the feed tray.

Making

Equations for stripping section.

3.

a total material balance over the dashed-line

section in Fig. 11.4-5a for the stripping section of the tower

Km+1 Making a balance on component

=Lm - W

(11.4-9)

A,

L„x m -

F,

below the feed entrance,

Wx K

(11.4-10)

Xf

m 1

7

m+

y m +i

N

yN xN

LN W, x

W x w xN

--steam

I

(a)

Figure

Sec. 11.4

11.4-5.

Material balance and operating line for stripping section: matic of tower, (b) operating and equilibrium lines.

Distillation With Reflux

and McCabe-Thiele Method

(a)

sche-

653

Solving for y m+

lt

the stripping-section operating line

is

Wx w

L„

(11.4-11) \

Again, since equimolal flow

Equation (11.4-11)

LJVm+v

slope of

— Wx w/Vm+

is

It

is

Lm = L N =

assumed,

constant and

Vm+ = VN = i

constant.

a straight line when plotted as y versus x in Fig. 11.4-5b, with a intersects the y = x line at x = x w The intercept at x = 0 is y = .

,.

Again the theoretical stages for the stripping section are determined by starting at x w going up to y w and then across to the operating line, etc. ,

,

The condition of the feed stream F entering the tower Effect of feed conditions. determines the relation between the vapor Vm in the stripping section and Vn in the enriching section and also between L m and L„ If the feed is part liquid and part vapor, 4.

.

the

vapor

will

add

Vm

to

to give

Vn

.

For convenience, we represent the condition

of the feed

by the quantity

q,

which

is

defined as

q

If

=

heat needed to vaporize

r~

;

molar

the feed enters at

denominator and q

its

=

mol of

1

is

Equation

1.4-12)

(1

and

superheated vapor q

<

fraction of feed that

liquid.

0,

and

is

the

same

as the

can also be written in terms of enthalpies.

He (1L4- I3)

dew

HL

point,

the enthalpy of the feed at the

H r the enthalpy of the feed at its entrance conditions.

the feed enters as vapor at the

We

(11.4-12)

Jf^t

the enthalpy of the feed at the

boiling point (bubble point),

is

77^

boiling point, the numerator of Eq. (11.4-12),

1.0.

Hy

Hv

:

:

latent heat of vaporization of feed

q== where

feed at entering conditions

;

;

dew

point, q

=

0.

For cold

for the feed being part liquid

liquid feed q

and

>

1.0,

part vapor, q

is

If

for

the

can look at q also as the number of moles of saturated liquid produced on the by each mole of feed added to the tower. In Fig. 1 1.4-6 a diagram shows the

feed plate

relationship between flows above and below the feed extrance.

From

the definition of q,

the following equations hold:

L m = L„ + qF Vn

Fig lire

1

1.4-6.

(11.4-14)

=Vm + (l-c )F

(11.4-15)

l

between flows above and below the feed en-

Relationship trance.

0-Q)F J

F

^ v.

\

V

654

Chap. II

I

Vapor-Liquid Separation Processes

The

point of intersection of the enriching and the stripping operating-line equations

on an xy

plot can be derived as follows. Rewriting Eqs. (11.4-6)

and (11.4-10) as follows

without the tray subscripts:

Vn y=L n x + Dx D

(11.4-16)

Vm y = Lm x-Wx w

(11.4-17)

where the y and x values are the point of intersection of the two operating Subtracting Eq. (11.4-16) from (11.4-17),

~ Vn)y = (L m - K)x -

iYm

Substituting Eqs.

(1 1.4-4), (1 1.4-14),

y

and

(1

= q

This equation

is

+ Wx w

(Dx D

-

and rearranging,

^-

x

-

q

1

the q-line equation and

(11.4-18)

)

1.4-15) into Eq. (11.4-18)

is

lines.

(11.4-19)

1

the locus of the intersection of the

two

Qperating lines. Setting y = x in Eq. (1 1.4-19), the intersection of the q-line equation with the 45° line is y = x = x f where x F is the overall composition of the feed. ,

In Fig.

The

1

1.4-7 the q line

is

plotted for various feed conditions given below the figure.

—

1). For example, for the liquid below the boiling point, shown. The enriching and operating lines are plotted for the case of a feed of part liquid and part vapor and the two lines intersect on the q line. A

q

>

slope of the q line

1,

and the slope

convenient

way

is

is

>

q/(q

1.0, as

to locate the stripping operating line

operating line and the q

line.

Then draw

q line and enriching operating line and the point y

5.

Location of the feed tray

theoretical trays intersect

stepped

needed

in

to first plot the enriching

is

the stripping line between the intersection of the

= x =x B

,.

a tower and number of trays.

in a tower, the stripping

To determine

and operating

the

are

lines

number of drawn to

on the q line as shown in Fig. 11.4-8. Starting at the top atx fl the trays are For trays 2 and 3, the steps can go to the enriching operating line, as shown ,

off.

xw

xF

Mole fraction Figure

1

1.4-7.

in liquid,

x

Location of the q line for various feed conditions point [0

Sec. 11.4

xD

Distillation

<

q

(q

>

1),

<

I),

saturated vapor (q

liquid

at

boiling

=

point

(q

:

=

liquid /),

below boiling + vapor

liquid

0).

With Reflux and McCabe-Thiele Method

655

in

11.4-8a.

Fig.

At step 4 the step goes to the stripping

theoretical steps are needed.

The

For the correct method, Fig. 11.4-8b.

number

the line

A

total of only

of trays to a

the shift

about

first

In Fig. 11.4-8b the feed

and the

plate 2 liquid, all

it

2,

vapor,

it

is

A

total of

2 to the stripping line, as

shown

in

needed with the feed on tray 2. To keep from the enriching to the stripping operating

part liquid

and part vapor since 0 < q < 1. Hence, in is separated and added beneath from above entering tray 2. If the feed is all

the vapor portion of the feed

to the liquid flowing to tray 2

from the tray above.

If

the feed

is

should be added below tray 2 and joins the vapor rising from the plate below.

Since a reboiler

is

considered a theoretical step

when

the

vapory^

is

with x w as in Fig. 11.4-5b, the number of theoretical trays in a tower

number

about 4.6

opportunity after passing the operating-line intersection.

is

added to the liquid

liquid

should be added

line.

3.7 steps are

minimum, the shift

should be made at the

adding the feed to tray

on tray 4. made on step

feed enters

in is

equilibrium

equal to the

of theoretical steps minus one.

EXAMPLE

of a Benzene-Toluene Mixture is to be distilled in a fractionating tower at 101.3 kPa pressure. The feed of 100 kg mol/h is liquid and it contains 45 mol % benzene and 55 mol toluene and enters at 327.6 K (130°F). A distillate containing 95 mol % benzene and 5 mol % toluene and a bottoms containing 10 mol % benzene and 90 mol % toluene are to be „ obtained. The reflux ratio is 4 1 The average heat capacity of the feed is 1 59 kJ/kg mol K (38 btu/lb mol °F) and the average latent heat 32 099 kJ/kg mol (13 800 btu/lb mol). Equilibrium data for this system are given in Table 11.1-1 and in Fig. 1 1.1-1. Calculate the kg moles per hour distillate, kg moles per hour bottoms, and the number of theoretical trays needed.

A

11.4-1.

Rectification

liquid mixture of benzene-toluene

%

:

Solution:

The

=

and

xw

0.10,

.

•

•

F = 100 kg mol/h, x F = 0.45, x D = 0.95, For the overall material balance substituting

given data are

R = LJD =

4.

into Eq. (11.4-3),

F= 100

656

D+W

=£> +

Chap.

(11.4-3)

W

11

Vapor-Liquid Separation Processes

Substituting into Eq. (11.4-4) and solving for

Fx F = Dx D +

=

100(0.45)

D=

R R+

+

D(0.95)

J?

+

-

DX0.10)

W = 58.8 kg mol/h 0 95

4 1

(100

(11.4-4)

using Eq. (11.4-8),

line,

4

1

The equilibrium data from Table above are plotted

Wx w

41.2 kg mol/h

For the enriching operating

D and W,

+

x„+-

4

1

11.1-1

+

r=

0.800x„

+

0.190

1

and the enriching operating

line

in Fig. 11.4-9.

Next, the value of q is calculated. From the boiling-point diagram, Fig. x F = 0.45, the boiling point of the feed is 93.5°C or 366.7 (200.3T). From Eq. (11.4-13),

K

11.1-1, for

=

<7

The value

Hv — HL =

of

Eq. (11.4-13)

Hv - H F Hy ~ H L =

latent heat

(11.4-13)

32 099 kJ/kg mol.

The numerator

of

is

Hy — H F = (Hy — H L ) + (H L —

HF

(11.4-20)

)

Also,

»f =

H,

cATB - TF

(11.4-21)

=

159 kJ/kg mol K, TB = (inlet feed temperature).

)

where the heat capacity of the liquid feed c pL 366.7

K

(boiling point of feed),

Substituting Eqs.

(1

1.4-20)

and (Hy

and

(1

TF =

327.6

•

K

1.4-21) into (1 1.4-13),

-H ) + c vL {TB l

TF

)

(11.4-22)

Hy - H L

o a,

> .5

£

O

jo

O

S

Mole fraction Figure

1

1.4-9.

McCabe-Thiele diagram for

in liquid,

distillation

x

of benzene-toluene for Exam-

ple 11.4-1.

Sec. 11.4

Distillation

With Reflux and McCabe-Thiele Method

657

Substituting the

From

The

9

"

q

=

known

values into Eq. (11.4-22),

-

32099 + 159(366.7 32099 13 800

+

38(200.3

-

^ L195

327.6)

130)

UlA95

13800

Eq. (11.4-19), the slope of the q line

q line

a

1.195

1.195-1

=

6.12

1.4-9 starting at the

1

(Engii5h)

is

g-1 plotted in Fig.

is

(SI)

pointy

=

xF

=

0.45 with a

slope of 6.12.

The

stripping operating line

drawn connecting

is

the point

y= x

=

=

0.10 with the intersection of the q line and the enriching operating line. Starting at the point y = x = x D the theoretical steps are drawn in as

xw

,

shown

in Fig.

11.4-9.

The number

minus a reboiler, which gives tray 5 from the top.

11. 4C

Total and

Minimum

of theoretical steps

6.6 theoretical trays.

The

is

feed

7.6 or 7.6 steps is

introduced on

Reflux Ratio for McCabe-Thiele

Method J.

Total reflux.

A and B

In distillation of a binary mixture

the feed conditions, distillate

composition, and bottoms composition are usually specified and the number of theoretical trays are to be calculated. However, the number of theoretical trays needed depends upon the operating lines. To fix the operating lines, the reflux ratio R = LJD at the top of the column must be set.

One of the limiting values R = LJD and, by Eq. (11.4-5),

of reflux ratio

K„

then

Ln

is

very large, as

is

+1

the vapor flow

=

L„

is

that of total reflux, or

R=

+ D

co.

Since

(11.4-5)

Vn This means .

that the slope

R/(R

+

1)

of the

enriching operating line becomes 1.0 and the operating lines of both sections of the

column coincide with the 45° diagonal line, as shown in Fig. 11.4-10. The number of theoretical trays required is obtained as before by stepping trays

from the

distillate to the

bottoms. This gives the

minimum number

off the

of trays that can

possibly be used to obtain the given separation. In actual practice, this condition can be realized

by returning

all

to the

tower as reflux,

Hence,

all

the overhead condensed vapor V, from the top of the tower back i.e.,

total reflux. Also, all the liquid in the

the products distillate

and bottoms are reduced

bottoms

to zero flow, as

is

is

reboiled.

the fresh feed

to the tower.

This condition of total reflux can also be interpreted as requiring infinite sizes of condenser, reboiler, and tower diameter for a given feed rate. If the relative volatility a

of the binary mixture

following analytical expression by of theoretical steps

M m when

Fenskecan be used

a total condenser

is

is

approximately constant, the

to calculate the

minimum number

used.

logl

Nm =

658

v" :

— -

log

Chap.

ot

'

(11.4-23)

a<

U

Vapor-Liquid Separation Processes

xw

0

xF

Mole Figure

11.4-10.

fraction

xD

A

in liquid,

1

x

Total reflux and minimum number of trays by McCabe-Thiele method.

For small variations in cc, a av = (a a^) 2 where a, is the relative overhead vapor and a w is the relative volatility of the bottoms liquid. 1'

1

2.

Minimum

reflux ratio.

,

The minimum reflux ratio can be defined as number of trays for the given separation

that will require an infinite

volatility

of the

the reflux ratio

desired of x D

Rm

and

x w This corresponds to the minimum vapor flow in the tower, and hence the minimum reboiler and condenser sizes. This case is shown in Fig. 11.4-11. If R is decreased, the slope of the enriching operating line R/(R + 1) is decreased, and the intersection of this line and the stripping line with the q line moves farther from the 45° line and closer to the .

Mole fraction Figure

11.4-11.

Minimum

reflux ratio

and

A

in liquid,

infinite

x

number of trays by McCabe-

Thiele method.

Sec. 11.4

Distillation

With Reflux and McCabe-Thiele Method

659

equilibrium increases.

and

line.

When

As a result, the number of steps required to give a fixed x D and x w the two operating lines touch the equilibrium line, a "pinch point" at y

number of

occurs where the

x'

enriching operating line points

x',

/, andx D

(y

=

is

as follows

steps required

becomes

infinite.

from Fig. 11.4-11, since the

The slope of

line passes

x D ).

—

9

Rm + In

some

cases,

the

through the

where the equilibrium

minimum

11.4-12, the operating line at

(11.4-24)

1

line

has an inflection

in

it

as

shown

in Fig.

reflux will be tangent to the equilibrium line.

For the case of total reflux, the number of plates minimum, but the tower diameter is infinite. This corresponds to an infinite cost of tower and steam and cooling water. This is one limit in the tower operation. Also, for minimum reflux, the number of trays is infinite, which again gives an infinite cost. These are the two limits in operation of the tower. The actual operating reflux ratio to use is in between these two limits. To select the proper value of R requires a complete economic balance on the fixed costs of the tower and operating costs. The optimum reflux ratio to use for lowest total cost per year is 3. is

Operating and optimum reflux ratio.

a

between the minimum R m and total reflux. This has been shown for an operating reflux ratio between L2R m to \.5R m

many

cases to be at

.

EXAMPLE

Minimum

11.4-2.

Reflux Ratio and Total Reflux in Rectifi-

cation

For the being

rectification in

distilled to give

composition of x w (a) (b)

=

Example

11.4-1,

0.10, calculate the following.

Minimum reflux ratio R m Minimum number of theoretical

Solution:

where a benzene-toluene feed is = 0.95 and a bottoms

a distillate composition of x D .

For part

(a)

plates at total reflux.

the equilibrium line

is

plotted in Fig.

1

1.4-13

and

the

x F = 0.45. Using the samex D and x w as in Example 11.4-1, the enriching operating line for minimum reflux is plotted as a dashed line and intersects the equilibrium line at the same point at

q-line equation

is

also

shown

for

which the q

/=

line

Reading

intersects.

Rm

xD

+

R-m

For the case of total

shown

1

x'

=

0.49

and

,

- 0.702 0.95 - 0.49

0.95 ~~

x'

R„ —

values of.

the

and solving for ltm

1.17.

reflux in part (b), the theoretical steps are

The minimum number of

in Fig. 11.4-13.

which gives

- y'

~ xD -

Hence, the minimum reflux ratio

11.4D

off"

0.702, substituting into Eq. (11.4-24),

drawn

theoretical steps

is

as

5.8,

4.8 theoretical trays plus a reboiler.

Cases for Rectification Using McCabe-Thiele

Special

Method I.

Stripping-column distillation.

an intermediate point

shown

in

in Fig. 11.4- 14a.

some

In

a column but

The feed

is

cases the feed to be distilled

added

is

is

not supplied to

to the top of the stripping

column as

usually a saturated liquid at the boiling point and the

VD is the vapor rising from the top plate, which goes to a condenser with no reflux or liquid returned back to the tower.

overhead product

W

The bottoms product usually has a high concentration of the less volatile component B. Hence, the column operates as a stripping tower with the vapor removing the more volatile A from the liquid as it flows downward. Assuming constant molar flow rates, a material balance of the more volatile component A around the dashed line in Fig. I

I. 4-

14a gives, on rearrangement,

Wx w

(11.4-25)

y m +i m+1 This stripping-line equation

is

tower given as Eq. (11.4-11). constant at

LJVm +

the

It

same

as the stripping-line equation for a complete

intersects the

y

= x

line at

x = xw

,

and the slope

is

.

l

enriching operating line for

0

0.2

0.4|

!

0.6

0.8

Rm

!1.0

J

xw

xF x

Mole fraction Figure

1

1.4-13.

Distillation

in liquid,

x

Graphical solution for minimum reflux ratio

Example

Sec. 11.4

xD

Rm

and

total reflux for

1 1.4-2.

With Reflux and McCabe-Thiele Method

661

Figure

1.4-14.

1

Material balance and operating tower,

the feed

If

is

(b)

saturated liquid, then

point, the q line should be used

line

operating and equilibrium

and q

L„ =

>

for stripping tower

;

(a) flows in

line.

F. If the feed

is

cold liquid below the boiling

I.

Lm = qF In Fig.

1

equation (11.4-25)

1.4-14 the stripping operating-line

Eq. (11.4-19),

is

also

shown

for q

=

1.0.

(11.4-26)

Starting at

xf

,

is

and the q line, drawn down the

plotted

the steps are

tower.

EXAMPLE A

11.4-3.

Number of Trays in

liquid feed at the boiling point of

Stripping Tower 400 kg mol/h containing 70 mol

%

benzene (A) and 30 mol % toluene (B) is fed to a stripping tower at 101.3 kPa pressure. The bottoms product flow is to be 60 kg mol/h containing only 10 mol A and.the rest B. Calculate the kg mol/h overhead vapor, its composition, and the number of theoretical steps required.

%

Solution:

known values are F — 400 kg = 0.10. The equilibrium data 11.4-15. Making an overall material

Referring to Fig. 11.4-14a, the

mol/h, x f

=

from Table

0.70,

W = 60

kg mol/h, and x w

11.1-1 are plotted in Fig.

balance,

F=

W

+ VD

=

60

+ VD

400 Solving,

VD = 340 kg

mol/h.

Making a component A balance and

Fx F =

662

solving,

Wx w + VD y D

400(0.70)

=

60(0.10)

yD

=

0.806

Chap.

+

II

3400? D )

Vapor-Liquid Separation Processes

Mole fraction, x Figure

1

Stripping tower for

1.4-15.

Example

11.4-3.

For

a saturated liquid, the q line is vertical and is plotted in Fig. 11.4-15. The operating line is plotted through the point y = x w — 0.10 and the intersection of y D = 0.806 with the q line. Alternatively, Eq. (11.4-25) can be used with a slope of m + = 400/340. Stepping off the trays from the top, 5.3

LJV

l

theoretical steps or 4.3 theoretical trays plus a reboiler are needed.

2.

Enriching-column

distillation.

Enriching towers are also sometimes used, where the

bottom of the tower

as a vapor. The overhead distillate is produced in the same manner as in a complete fractionating tower and is usually quite rich in the more volatile component A. The liquid bottoms is usually comparable to the feed in composition, being slightly leaner in component A. If the feed is saturated vapor, the vapor in

feed enters the

the tower

Vn =

F. Enriching-line equation (11.4-7) holds, as does the q-line equation

(11.4-19).

3. Rectification with direct

steam

injection.

Generally, the heat to a distillation tower

is

applied to one side of a heat exchanger (reboiler) and the steam does not directly contact the boiling solution, as volatile

A and water B

open steam injected

shown is

However, when an aqueous solution of more heat required may be provided by the use of bottom of the tower. The reboiler exchanger is then

in Fig.

being

1

1.4-5.

distilled, the

directly at the

not needed.

The steam

is

in Fig. 11.4-16a. sufficient contact

on A,

injected as small bubbles into the liquid in the tower

The vapor leaving is

obtained.

the liquid

Making an

is

bottom, as shown

then in equilibrium with the liquid

overall balance

on

if

the tower and a balance

....

F+ S =D + Fx F + Sys = Dx D where S

= mol/h

(11.4-27)

+ Wx w

(11.4-28)

and ys = 0 = mole fraction of A the same as for indirect steam.

of steam

operating-line equation

Sec. 11.4

W

Distillation

is

With Reflux and McCabe-Thiele Method

in

steam. The enriching

663

For the

and a balance on component A

stripping-line equation, an overall balance

are as follows:

Lm + L m x„ + Solving for y m+

,

in

Eq.

S=Vm +

S(0)

l

+

W

(11.4-29)

= Vm+i ym+l + Wx w

(11.4-30)

(1 1.4-30),

Lm K m+

,

For saturated steam entering, S

= Vm+

^

xm

(11.4-31)

m+1

and hence, by Eq.

W

i

K

1

ing into Eq. (11.4-31), the stripping operating line

y m+

WXyy -—Z

(11.4-29),

L m = W.

Substitut-

is

W

=y X ™~Y * w

(11.4-32)

x = x w Hence, the stripping line passes through the point y = 0, x = x w and is continued to the x axis. Also, for the intersection of the stripping line with the 45° line, when y = x in Eq. (1 1.4-32),* = Wx w /(W — S). For a given reflux ratio and overhead distillate composition, the use of open steam rather than closed requires an extra fraction of a stage, since the bottom step starts below the y = x line (Fig. 11. 4- 16b). The advantage of open steam lies in simpler construction of

When y = as

shown

0,

the heater, which

4.

.

is

a sparger.

Rectification tower with side stream.

664

,

in Fig. 11.4- 16b,

In certain situations, intermediate product or

Chap. 11

Vapor—Liquid Separation Processes

side streams are

removed from

bottoms. The side stream

may

sections of the tower between the distillate

be vapor or liquid and

may

and the

be removed at a pgint above

the feed entrance or below depending on the composition desired.

The flows for a column with a liquid side stream removed above the feed inlet are shown in Fig. 11.4-17. The top enriching operating line above the liquid side stream and the stripping operating line below the feed are found in the usual way. The equation of the q line is also unaffected by the side stream and is found as before. The liquid side stream

and hence the material balance or operating and liquid side stream plates. material balance on the top portion of the tower as shown

alters the liquid rate

below

it,

line in

the middle portion between the feed

Making

a total

in

the

dashed-linebox in Fig. 11.4-17,

Vs+1 where 0 is

is

=L S +

mol/h saturated liquid removed

0

+D

(11.4-33)

as a side stream. Since the liquid side

stream

saturated,

Ln = Ls + 0

(11.4-34)

= K +l

(11.4-35)

Vs +i

Making

on the most

a balance

volatile

ys+

iy s+

component,

=Ls x s

i

+ Ox 0 + Dx D

(11.4-36)

Solving for ys+ „ the operating line for the region between the side stream and the feed

Ls

>'s+i-

v YS +

xs 1

+

0x 0 + Dx D y yS +

(11.4-37)

1

\D, x D

n+l

W S+

0, i

is

Xc

saturated liquid

I

F,

xp

Vm +

Figure

Sec. 11.4

1

1.4-17.

Distillation

Process flow for a rectification lower with a liquid side stream.

With Reflux and McCabe-Thiele Method

665

Figure

11.4-18.

McCabe-Thiele

plot for a

tower with a liquid stream above the feed.

side

xF

xw

The slope line,

of this line

is

which determines

L sjVs+

,.

The

line

can be located as

the intersection of the stripping line

shown

x0

xd

in Fig. 11.4-18

and Eq.

(1 1.4-37),

or

by the

fixed by the specification ofx 0 the side-stream composition. The step on the Thiele diagram must actually be at the intersection of the two operating lines at x Q ,

q

may be McCabeit

an

in

actual tower. If this does not occur, the reflux ratio can be altered slightly to change the steps.

5.

Partial condensers.

In a few cases

it

may

be desired to

remove

the overhead distillate

product as a vapor instead of a liquid. This can also occur when the low boiling point of

The

the distillate

makes complete condensation

condenser

returned to the tower as reflux and the vapor

is

difficult.

liquid condensate in a partial

removed

as product as

shown

in Fig. 11.4-19. If

partial

the time of contact between the vapor product

condenser

is

a theoretical stage.

Then

the

equilibrium with the vapor composition y D

condenser separation

11.5

is is

liquid

sufficient, the

is

=

where y D

xD

.

If the

cooling

do not reach equilibrium, only

rapid and the vapor and liquid

in

is

in

the

a partial stage

obtained.

DISTILLATION

11.5A

,

and the

composition x R of the liquid reflux

AND ABSORPTION TRAY EFFICIENCIES

Introduction

In all the previous discussions

on

theoretical trays or stages in distillation,

we assumed

vapor leaving a tray was in equilibrium with the liquid leaving. However, if the time of contact and the degree of mixing on the tray is insufficient, the streams will not be that the

equilibrium. As we must use more

in

a result the efficiency of the stage or tray

is

not 100%. This means that

actual trays for a given separation than the theoretical

determined by calculation. The discussions

in this section

number

of trays

apply to both absorption and

distillation tray towers.

Three types of tray or plate efficiency are used: overall tray efficiency E Q Murphree £ Af and point or local tray efficiency E MP (sometimes called Murphree point efficiency). These will be considered individually. ,

tray efficiency

666

,

Chap.

11

Vapor-Liquid Separation Processes

11. 5B /.

Types of Tray Efficiencies

Overall tray efficiency.

tower and

number

The

simple to use but

is

overall tray or plate efficiency

is

the least fundamental.

It is

E 0 concerns

denned

the entire

as the ratio of the

number

of theoretical or ideal trays needed in an entire tower to the

of actual

trays used.

number of number For example,

number trays

is

if

7/0.60, or

Two

is

eight

minus

(115-1)

needed and the overall

eight theoretical steps are

of theoretical trays

ideal trays

of actual trays

a reboiler, or seven trays.

efficiency

The

60%, the number of

is

actual

1.7 trays.

1

empirical correlations for absorption

and

distillation overall tray efficiencies in

commercial towers are available for standard tray designs (Ol). For hydrocarbon distillation these values

from about 10

range from about 50 to

85% and

for

hydrocarbon absorption

50%. These correlations should only be used

to

for

approximate

esti-

mates.

2.

Murphree tray

efficiency.

The

M urphree tray efficiency E M EM =

y -j~^y*

-

y n+

is

defined as follows

(113-2)

i

mixed vapor leaving the tray n shown mixed vapor entering tray n, and y* the concentration of the vapor that would be in equilibrium with the liquid of concentration x n leaving the tray to the downcomer. where y n in Fig.

1

is

the average actual concentration of the

1.5-1, y„ +

Sec. 11 .5

,

the average actual concentration of the

Distillation

and Absorption Tray

Efficiencies

667

The tray,

in the liquid as

it

flows across the tray.

different concentrations,

3.

Point efficiency.

The

point or local efficiency

1

n

travels across the

the tray contacts liquid of

E MP on a y" +

tray

is

defined as

1

(11.5-3)

t

shown

in Fig.

concentration of the vapor entering the plate n at the same point, andy^*

would be

the concentration of the vapor that y'

it

the concentration of the vapor at a specific point in plate n as

1.5-1, y'n+x the

Since

as

a concentration gradient

not be uniform in concentration.

will

y

y' is n

is

The vapor entering

and the outlet vapor

Emf = l~ where

and

liquid entering the tray has a concentration ofx„_,,

concentration drops to x„ at the outlet. Hence, there

its

cannot be greater than

y*

,

equilibrium with

in

x'n at the

same

point.

the local efficiency cannot be greater than 1.00 or

100%. In small-diameter towers the

Then

vapor flow

sufficiently agitates the liquid so that

it is

on Then y'„ = yn y'n+l = yn+l and y'* = y* The point efficiency then equals the Murphree tray efficiency or E M = E MP In large-diameter columns incomplete mixing of the liquid occurs on the trays. Some vapor will contact the entering liquid x„_ which is richer in component A than

uniform on the

tray.

the tray.

,

the concentration of the liquid leaving

is

the

same

as that

.

,

.

l

,

x„. This will give a richer vapor at this point than at the exit point, where x„ leaves.

E M will be greater than the point efficiency E MP The E MF by the integration of E MP over the entire tray.

Hence, the tray efficiency

£ M can 11.5C

The

be related to

.

value of

Relationship Between Efficiencies

relationship between

E MP and E M can

be derived mathematically

if

the

amount of

and the amount of vapor mixing is also set. Derivations for sets different of assumptions three are given by Robinson and Gilliland (Rl). However, experimental data are usually needed to obtain amounts of mixing. Semitheoretical methods to predict E MP and E M are summarized in detail by Van Winkle (VI). When the Murphree tray efficiency E M is known or can be predicted, the overall tray liquid

668

mixing

is

specified

Chap.

II

Vapor-Liquid Separation Processes

equilibrium line

operating line

Figure

is

Use of Murphree

1.5-2.

E Q can be

efficiency

expression slope

1

L/V of the

EM

related to

as follows

when

plate efficiency to determine actual

by several methods. In the

the slope

m

first

of the equilibrium line

is

number of trays.

method an

analytical

constant and also the

operating line: log [1

+£.>K/L-1)]

<»">

io^K/L)

If the equilibrium and operating lines of the tower are not straight, a graphical method in the McCabe-Thieie diagram can be used to determine the actual number of trays when the Murphree tray efficiency is known. In Fig. 1 1.5-2 a diagram is given for an actual plate as compared with an ideal plate. The triangle acd represents an ideal plate; and the smaller triangle abe the actual plate. For the case shown, the Murphree efficiency E M = 0.60 = ba/ca. The dashed line going through point b is drawn so that bajca for each tray is 0.60. The trays are stepped off using this efficiency, and the total number of steps gives the actual number of trays needed. The reboiler is considered to be one

theoretical tray, so the true equilibrium curve 1

1.5-2, 6.0

11.6

is

used for

this tray as

shown. In Fig.

actual trays plus a reboiler are obtained.

FRACTIONAL DISTILLATION USING ENTHALPY-CONCENTRATION

METHOD 11.6A

/.

Enthalpy-Concentration Data

Introduction.

number

In Section

1

McCabe-Thieie method was used

1.4B the

to calculate the

of theoretical steps or trays needed for a given separation of a binary mixture of

A and B

by rectification or fractional

distillation.

The main assumptions

in the

method

are that the latent heats are equal, sensible heat differences are negligible, and constant

molal overflow occurs

in

each section of the

distillation tower. In this section

we

shall

consider fractional distillation using enthalpy-concentration data where the molal overflow rates are not necessarily constant.

The

analysis will be

made

using enthalpy as well

as material balances.

Sec. 11.6

Fractional Distillation Using Enthalpy-Concentration

Method

669

2.

An

Enthalpy-concentration data.

A and B

vapor-liquid mixture of

enthalpy-concentration diagram for a binary

takes into account latent heats, heats of solution or

mixing, and sensible heats of the components of the mixture.

needed to construct such a diagram at a constant pressure: as a function of temperature, composition,

of temperature

and composition;

and pressure;

(1)

The

following data are

heat capacity of the liquid

heat of solution as a function

(2)

heats of vaporization as a function of

(3) latent

composition and pressure or temperature; and

(4)

boiling point as a function of pressure,

composition, and temperature.

The diagram

at a given constant pressure is based on arbitrary reference states of and temperature, such a 273 K (32°F). The saturated liquid line in enthalpy h kJ/kg (btu/lb m ) or kJ/kg mol is calculated by

liquid

h

where x A

is

=

- T0 +

x A c pA (T

wt or mole

)

A T is

fraction

,

(1

-

x A )c pB(T

- T0 + AHxi

(1 1.6-1)

)

boiling point of the mixture in

K (°F)

or °C,

T0

reference temperature, K, c pA is the liquid heat capacity of the component A in (btu/lb m -°F) or kJ/kg mol K, c pB is heat capacity of B, and A// so is heat kJ/kg of solution at T 0 in kJ/kg (btu/lb m ) or kJ/kg mol. If heat is evolved on mixing, the A// so is

K

•

|

|

will

be a negative value

in

Eq. (11.6-1). Often, the heats of solution are small, as in

hydrocarbon mixtures, and are neglected. The saturated vapor enthalpy line of composition y A is calculated by

H = y A [\ A

+

c

pyA

H

{T- T 0 )] +

kJ/kg (btu/lb m ) or kJ/kg mol of a vapor

(1

- y A )[\ g +

c pyB (T

- T 0 )]

(11.6-2)

c py ^ is the vapor heat capacity of A and c py g for B. The latent heats \ A and \ B are the values at the reference temperature T 0 Generally, the latent heat is given as

where

.

.

T bA of the pure component A and X Bb for B. correct this to the reference temperature T 0 to use in Eq. (11.6-2), \ Ab at the normal boiling point

A

= c pA {T bA - T 0 + =

ab

c pB {T bB

In Eq. (11.6-3) the pure liquid

cooled as a vapor to

T0

reference temperature boiling

.

is

T0

is

- T0 +

\ Bb

)

heated from

Similarly,

Eq.

-

k Ab

)

,t

-

T0

c py (J bA c p) {T bB B ,

to

(11 .6-4) also

- T0

(11.6-4)

at T bA and then For convenience, the

T bA vaporized ,

.

to

(11.6-3)

)

- T0 )

holds for A B

Then

,

often taken as equal to the boiling point of the lower-

component A. This means X A = \ Ab Hence, only .

must be corrected

to A B

.

EXAMPLE

11.6-1. Enthalpy-Concentration Plot for Benzene-Toluene atm Prepare an enthalpy-concentration plot for benzene-toluene at pressure. Equilibrium data are given in Table 11.1-1 and Figs. 11.1-1 and 11.1-2. Physical property data are given in Table 11.6-1. 1

Table

1

1.6-1.

Physical Property Data for Benzene and Toluene

Heat Capacity, (Ulkg mol K)

Vapor

Latent Heat of Vaporization {kJIkg mol)

138.2

96.3

30 820

167.5

138.2

33 330

Boiling Point,

670

Component

CQ

Benzene (A) Toluene (B)

80.1

110.6

Liquid

Chap.

II

Vapor-Liquid Separation Processes

A reference temperature of T 0 = 80.1°C will be used for convenience so that the liquid enthalpy of pure benzene (x A = 1 .0) at the boiling point will be zero. For the first point we will select pure toluene {x A = 0). For liquid toluene at the boiling point of 110.6°C using Eq. (11.6-1) with zero heat of solution and data from Table 1 1.6-1,

Solution:

h

= x A c pA (T -

h

=

0

+

(1

-

80. 1)

+

- x A )c pB (T -

(1

0)(167.5)(110.6

-

80.

= 5109

80.1)

+

1)

0

(11.6-5)

kJ/kg mol

For the saturated vapor enthalpy line, Eq. (11.6-2) is used. However, we first must calculate X B at the reference temperature T 0 = 80.1°C using Eq. As

(11.6-4).

c pB {J bB

=

167.5(1 10.6

= 34 224 To

- T0 +

=

calculate

H = y A [X A =

0

X Bb

-

c pyB (T bB

80.1)

+

33 330

)

-

- T0

(11.6-4)

)

- 138.2(110.6 -

80.1)

kJ/kg mol

H, Eq.

+

(11.6-2)

c pyA (T

is

- T0 )] +

used and y A (1

-

y A )[X B

=

0.

+

c pyB (T

- T0 )]

(11.6-2)

+ (1.0 - 0)[34 224 + 138.2(110.6 - 80.1)]

= 38 439

kJ/kg mol

For pure benzene, x A = 1.0 and y A = .0. Using Eq. (1 1.6-5), since 7* = 80.1, h = 0. For the saturated vapor enthalpy, using Eq. 0 (11.6-2) and T = 80.1, 1

T =

H=

1.0[30 820 + 96.3(80.1

Selecting x A

=

- 80.1)] +

0.50, the boiling point

0

=

30 820

T b = 92°C and

the tempera-

for y A =0.50 is 98.8°C from Fig. 11.1-1. (11.6-5) for the saturated liquid enthalpy at the boiling point,

ture of saturated

h

vapor

= 0.5(138.2)(92 -

80.1)

+

(1

- 0.5)(167.5)(92 - 80.1) = 1820

Using Eq. (11.6-2) for>> A = 0.5, the saturated vapor enthalpy

H=

0.5[30 820 + 96.3(98.8

-

Using Eq.

80. 1)]

+

(1

-

+

138.2(98.8

at

98.8°C

is

0.5)[34 224

-

80.1)]

=

34 716

Selecting^ =0.30 and y A = 0.30, h = 2920 and H = 36268. Also, = 0.80 and y A = 0.80, h = 562 and H = 32 380. These values are tabulated in Table 11.6-2 and plotted in Fig. 11.6-1. for x A

Some

properties of the enthalpy-concentration plot are as follows.

The

region in

between the saturated vapor line and the saturated liquid line is the two-phase liquid-vapor region. From Table 11.1-1 for x A = 0.411, the vapor in equilibrium is y A = 0.632. These two points are plotted in Fig. 11.6-1 and this tie line represents the Sec. 11.6

Fractional Distillation Using Enthalpy-Concentration

Method

671

Table

11.6-2.

Enthalpy-Concentration Data for Benzene-Toluene Mixtures at 101.325 kPa (1 atm) Total Pressure

Saturated Vapor

Saturated Liquid

Mole

Enthalpy, h,

Mole

fraction, x A

{kJIkg mol)

fraction, y A

enthalpy,

H,

{kJIkg mol)

0

5109

0

38 439

0.30

2920

0.30

36 268

0.50

1820

0.50

34 716

0.80

562

0.80

32 380

1.00

0

1.00

30 820

enthalpies and compositions of the liquid and vapor phases in equilibrium. Other lines

can be drawn

a similar manner.

in

The

tie

region below the h versus x A line

represents liquid below the boiling point.

11. 6B

Distillation in

Enriching Section of Tower

To analyze the enriching section of a fractionating tower using enthalpy-concentration 1.6-2. data, we make an overall and a component balance in Fig. 1

V n+x -=L n

+D

(11.6-6)

Vn + y n + = L„x n + Dx D

(11.6-7)

\

i

40000, -O

Hvs

1 1

y^4 (saturat ;d

'

vapor)

1 1 1

~~

II

30000

/ 1

J

1 1

1

1

1

1 1 1 1 1

Enthalpy of mixture,

1

1*

Hovh, (kJ/kg

<;

tie line

1

mol

1 1

mixture) 1 1

10000 1 1

/

I

"l

1

1

1

1

1

1

1

II

1

0.2

Mole Figure

11.6-1.

I

I

i

0.4

i

1

1

h v s.

i

i

i

i

xa i

i

i

(satur ated liquid)

i

i

0.6

fraction benzene, y/[ or

i

fr?n~r-t-+4-i-u } 0.8

1.0

x&

Enthalpy-concentration plot for benzene-toluene mixture at 1.0

atm abs.

672

Chap.

11

Vapor-Liquid Separation Processes

FIGURE

Enriching section of distillation tower.

11.6-2.

Equation (11.6-7) can be rearranged to give the enriching-section operating

L„ y n + x=-

xn

+

Dx D -

line.

(11.6-8)

is the same as Eq. (1 1.4-7) for the McCabe-Thiele method, but now the liquid and vapor flow rates V n+ and L n may vary throughout the tower and Eq. (1 1.6-8) will not be a straight line on an xy plot. Making an enthalpy balance,

This

Vn +

l

H n+]

=

where q c is the condenser duty, kJ/h or around the condenser.

L,,h n

kW

+ Dh D +

(btu/h).

An

(11.6-9)

qc

enthalpy balance can be

made

just

qc

By combining

= V

l

H

- Lh D - Dh D

l

(11.6-10)

Eqs. (11.6-9) and (11.6-10) to eliminate q c

an alternative form

,

is

obtained.

V n + ,H n , Substituting the value of

x

=L

L n from Eq.

Vn +

l

H

n+

i

nh n

+V,H,-Lh D

(11.6-11)

(11.6-6) into (11.6-11),

= (V n +

1

Equations (11.6-8) and (11.6-12) are the

-D)h n + V H X

final

X

- Lh D

(11.6-12)

working equations for the enriching

section.

In order to plot the operating line Eq. (11.6-8), the terms

Vn +

l

and L n must be

determined from Eq. (11.6-12). If the reflux ratio is set, V and L are known. The values and h D can be determined by Eqs. (11.6-1) and (11.6-2) or from an x

H

Sec. 11.6

t

Fractional Distillation Using Enthalpy-Concentration

Method

673

enthalpy-concentration plot. If a value of x n to obtain 1.

2.

Hn +

1

since y„+

is

{

not

3.

selected,

it

is

a trial-and-error solution

to follow are given

D

Assume n+1 = t = L + and L n = L. values in Eq. (11.6-8), calculate an approximate value of y„ +1 straight operating line. Select a value of x n

V

.

H

V

.

below.

Then using these This assumes a

Using thisy„ +1 obtain n + and also obtain h„ using x n Substitute these values into Eq. (11.6-12) and solve for V n + Obtain L„ from Eq. (11.6-6). Substitute into Eq. (11.6-8) and solve fory n+1 If the calculated value of y n+1 does not equal the assumed value of y„ +1 repeat steps 2-3. Generally, a second trial is not needed. Assume another value of x„ and ,

.

{

1

4.

is

known. The steps

.

.

,

repeat steps 1-4. 5.

Plot the curved operating line for the enriching section. Generally, only a few values

of the flows L„ and

V n+i

are needed to determine the operating line,

which

is

slightly curved.

11. 6C

Distillation in Stripping Section of

To analyze

Tower an overall and a component

the stripping section of a distillation tower,

material balance are

made on

Fig. 11.4-5a.

Lm

=W+V m+[

L m x m = Wx w + V m + ym +

1

=

-L m *m +

xm

(11.6-13)

]

ym+

(11.6-14)

i

-~Wx w V

!

m+

(11.6-15)

l

Making an enthalpy balance with q R kJ/h or kW(btu/h) entering the reboiler 11.4-5a and substituting (V m+l +W) forL m from Eq. (11.6-13),

V m+l H m+] = (V m+I + W)h m + Making an

- Wh w

(11.6-16)

overall enthalpy balance in Fig. 11.4-3,

qR = The

qR

in Fig.

Dh D + Wh w +

q c - Fh F

(11.6-17)

working equations to use are Eqs. (11. 6-15)-(l 1.6-17). Using a method similar to that for the enriching section to solve the equations, select a value of y m+1 and calculate an approximate value of x m from Eq. (11.6-15) assuming constant molal overflow. Then calculate V m + and L m from Eqs. (11.6-16) final

1

Then use Eq. (1 1.6-15) the ofx m with assumed value.

and

(11.6-13).

EXAMPLE

to

determine x m Compare this calculated value .

Using Enthalpy-Concentration Method benzene-toluene is being distilled using the same conditions as in Example 1 1 .4-1 except that a reflux ratio of 1.5 times the minimum reflux ratio is to be used. The value of R m = 1.17 from Example 1 1 .4-2 will be used. Use enthalpy balances to calculate the flow rates of the liquid and vapor at various points in the tower and plot the curved operating lines. Determine the number of theoretical stages needed.

A

674

11.6-2.

Distillation

liquid mixture of

Chap. II

Vapory-Liquid Separation Processes

Solution: The given data are as follows: F = 100 kg mol/h, x F = 0.45, x D = 0.95, x w = 0.10,/? = 1.5 /?„, = 1.5(1.17) = 1.755, /> = 41.2 = 58.8 kg mol/h. The feed enters at 54.4°C and q= 1.195. kg mol/h, The flows at the top of the tower are calculated as follows.

W

L -=

L=

1.755;

1.755(41.2)

=

V =L+D=

72.3;

{

72 Ji

+ 41.2=113.5 The 0.95

H

{

saturation temperature at the top of the tower for y 82.3°C from Fig. 11.1-1. Using Eq. (11.6-2),

is

= 0.95[30 820 +

96.3(82.3

-

80.1)]

+

(1

-

=

t

jc

d

0.95)[34 224

+ 138.2(82.3 -

80.1)]

=

31

206

This value of 3 1 206 could also have been obtained from the enthalpyconcentration plot, Fig. 11.6-1. The boiling point of the distillate D is obtained from Fig. 11.1-1 and is 81.1°C. Using Eq. (11.6-5),

=

hD

Again

-

0.95(138.2)(81.1

this

80.1)

+

(1

-

0.95)(167.5)(81.1

-

80.1)

=

139

value could have been obtained from Fig. 11.6-1.

Following the procedure outlined for the enriching section for step 1, a value of x n =0.55 is selected. Assuming a straight operating line for Eq. (11.6-8), an approximate value of y n+I is obtained.

y " +1

41.2

72.3

=

TTTT

+

Xn

TTT7

(0,95)

=

°-

637U " + )

°- 345

= 0.637(0.55) + 0.345 = 0.695 x n = 0.55, /;„ = 1590 and 33 240. Substituting into Eq. (11.6-12) and

Starting with step 2 and using Fig. 11.6-1, for

fory n+1 = 0.695,

H n+l

=

solving,

y„ +

I

(33 240)

= (V„ + - 41.2)1590 + 113.5(31 206) - 72.3(139) ,

V„ + = 109.5 1

Using Eq.

(11.6-6),

109.5

For step

3,

= L„ + 41.2

or

Ln =

68.3

substituting into Eq. (11.6-8),

y„" + y

=

41.2

68.3 (0.55)

1

+

(0.95)

= 0.700

109.5

109.5

This calculated value of y„ + =0.700 is sufficiently close to the approximate value of 0.695 so that no further trials are needed. 1

Sec. 11.6

Fractional Distillation Using Enthalpy-Concentration

Method

675

Selecting another value for x n = 0.70 and using Eq. (11.6-8), an approximate value of y n+i is calculated.

— 41.2

72.3

y„ +I = 77^77 (0-70)

+

(0.95)

=

0.791

Using Fig. 11.6-1 for x„ = 0.70, h n = 1000, and for>- n+1 = 0.791, = 32 500. Substituting into Eq. (11.6-12) and solving,

H n+]

V„ + (32 500) 1

= {V H+ i—

Vn+l = Using Eq.

41.2)1000

+

1

13.5(31 206)

-

72.3(139)

110.8

(11.6-6),

Ln =

1

10.8

- 41.2 = 69.6

Substituting into Eq. 11.6-8),

y " +]

=

69.6

no

41.2 (0-70)

+

(0 95) '

TToTs

=

°" 793

In Fig. 1 1 .6-3, the points for the curved operating line in the enriching section are plotted. This line is approximately straight and is very slightly

above

Mole

that for constant molal overflow.

fraction

in vapor, y

Mole FIGURE

676

11.6-3.

fraction in liquid, x

of curved operating lines using enthalpy-concentration method for Example 11.6-2. Solid lines are for enthalpy-concentration method and dashed lines for constant molal overflow.

Plot

Chap.

11

Vapor-Liquid Separation Processes

Using Eq. (11.6-10), the condenser duty

qc

To

=

113.5(31 206)

=

3

-

calculated.

is

72.3(139)

- 41.2(139)

526 100 kJ/h

q R values for h w and h F are needed. Using x w = 0.10, h w = 4350. The feed is at 54.5°C. Using Eq.

obtain the reboiler duty

Fig. 11.6-1 for

,

(11.6-5), /i

F

=

0.45(138.2)(54.5

-

80.1)

+

- 0.45)(167.5)(54.5 -

(1

80.1)

= -3929 Using Eq. (11.6-17), qR

=

41.2(139)

=

4

+ 58.8(4350) +

3

526 100 - 100(

-

3929)

180 500kJ/h

Using Fig. 1 1.4-5 and making a material balance below the bottom around the reboiler,

tray and

LN Rewriting Eq.

(1 1.6-16) for this

=W+V W

(11.6-18)

bottom section,

V W H W = (V w + W)h N + q R -

Wh w

(11.6-19)

the equilibrium diagram, Fig. 1 1.1-2, for x w = 0. 10, y w = 0.207, which is the vapor composition leaving the reboiler. For equimolal overflow in the stripping section using Eqs. (11.4-14) and (11.4-15),

From

= L n + qF =

L„,

72.3 + 1.195(100) = ftl.8

(11.4-14)

V m+l = V n + -(1 - q)F l

=

113.5

-

(1

-

1.195)100 = 133.0

(11.4-15)

Selecting y m + = y w — 0.207, and using Eq. (1 1.6-15), an approximate value of x m = x N is obtained. \

W

L

y m+

i=^m L-x m ~—^ Vm

0.207

v

=

+

+

l

(11.6-15)

\

—

58.8

191.8 133.0

x

(x N )

- (0.

10)

133.0

x N = 0.174. From Fig. (11.6-1) for x N = 0.174, h N = 3800, andfory vv = 0.207, H w = 37 000. Substituting into Eq. (11.6-19),

Solving,

Vw Sec. II .6

(37 000)

= (V w + 58.8)(3800) +

4 180

500 - 58.8(4350)

Fractional Distillation Using Enthalpy-Concentration

Method

677

Vw =

Solving,

Eq.

(11 .6-15)

LN =

125.0. Using Eq. (11.6-18), and solving for x N

183.8. Substituting into

,

°- 207

=

^

xN

=

0.173

58.8

183.8

(

^-Il53

(()

-

10)

This value of 0.173 is quite close to the approximate value of 0.174. Assuming a value of y m+ = 0.55 and using Eq. (11.6-15), an approximate value of x m is obtained. i

=

y m+ i

133.0

xm

From

Hm

(0.10)

(jcJ

133.0

= 0.412

x m = 0.412, h m = 2300, and for y m + 34 400. Substituting into Eq. (11.6-16),

Fig. (11.6-1) for

=

Vm +

58.8

191.8

=

0.55

1

(34 400)

Solving,

= {V m + + 58.8)(2300) + i

V m+l =

=

0.55,

1

500 - 58.8(4350)

4 180

126.5. Using Eq. (11.6-13),

Lm =

W+

V m+ = ,

and solving forx m

Substituting into Eq. (11.6-15)

^- =

0 55 -

58.8 + 126.5

=

=

185.3

,

58.8

185.3

Ii6T5-^-T2^

( °-

1)

x m = 0.407 This value of 0.407 so that no further

is

sufficiently close to the

trials

are needed.

approximate value of 0.412

The two

stripping section are plotted in Fig.

11.6-3.

points calculated for the

This stripping

line

is

also

approximately straight and is very slightly above the operating line for constant molal overflow. Using the operating line for the enthalpy balance method, the number of theoretical steps is 10.4. For the equimolal method 9.9 steps are obtained. This difference would be larger if the reflux ratio of 1.5 times R m were decreased to, say, 1.2 or 1.3. At larger reflux ratios, this difference in number of steps would be less.

Note that in Example from 125.0 to 126.5

slightly

11.6-2 in the stripping section the in

vapor flow increases

going from the reboiler to near the feed tray. These

values are lower than the value of 133.0 obtained assuming equimolal overflow. Similar

conclusions hold for the enriching section. useful in calculating the internal

These data are then used in

vapor and

in sizing

The enthalpy-concentration method liquid flows at

any point

in

is

the column.

the trays. Also, calculations of q c and q R are used method is very applicable for design

designing the condenser and reboiler. This

make tray to tray mass and enthalpy balances for the whole tower. A more restrictive Ponchon-Savarit graphical method for only binary mixtures is available (K3, T2). using a computer solution for binary and multicomponent mixtures to

678

Chap.

II

Vapor-Liquid Separation Processes

DISTILLATION OF MULTICOMPONENT

11.7

MIXTURES 11.7A

Introduction to Multicomponent Distillation

many of the distillation processes involve the separation of more than two components. The general principles of design of multicomponent distillation towers are the same in many respects as those described for binary systems. There is one mass

In industry

balance for each component in the multicomponent mixture. Enthalpy or heat balances are

made which

are similar to those for the binary case. Equilibrium data are used to

calculate boiling points

and dew points. The concepts of minimum

reflux

and

total reflux

as limiting cases are also used.

Number

In binary distillation one tower

was used to components with A in the overhead and B in the bottoms. However, in a multicomponent mixture of n components, n — 1 fractionators will be required for separation. For example, for a threecomponent system of components A, B, and C, where A is the most volatile and C the least volatile, two columns will be needed, as shown in Fig. 1 1.7-1. The feed of A, B, and C is distilled in column 1 and A and B are removed in the overhead and C in the bottoms. Since the separation in this column is between B and C, the bottoms containing C will contain a small amount of B and often a negligible amount of A (often called trace component). The amount of the trace component A in the bottoms can often be neglected if the relative volatilities are reasonably large. In column 2 the feed of A and B is distilled with A in the distillate containing a small amount of component B and a much smaller amount of C. The bottoms containing B will also be contaminated with a small amount of C and A. Alternately, column 1 could be used to remove A overhead with B plus C being fed to column 2 for separation of B and C. /.

of distillation towers needed.

separate the two components

A and B

into relatively pure

'

2.

Design calculation methods.

assumed

In multicomponent distillation, as in binary, ideal stages

Using equilibrium data, equilibrium calculations are used to obtain the boiling point and equilibrium vapor composition from a given liquid composition or the dew point and liquid composition from a given vapor composition. Material balances and heat balances similar to those or trays are

described

in

Section

in the stage-to-stage calculations.

1

1.6 are then

used to calculate the flows to and from the adjacent

These stage-to-stage design calculations involve trial-and-error calculations, and high-speed digital computers are generally used to provide rigorous solutions.

stages.

A,

A

B

feed 2

A, B,

C

C Figure

Sec. 11.7

Distillation

11.7-1.

B

Separation of a ternary system of A, B, and C.

of Multicomponent Mixtures

679

In a design the conditions of the feed are generally

known

or specified (temperature,

most cases, the calculation procedure follows method, the desired separation or split between

pressure, composition, flow rate). Then, in

two general methods. In the first two of the components is specified and the number of theoretical trays are calculated for a selected reflux ratio. It is clear that with more than two components in the feed the complete compositions of the distillate and bottoms are not then known and trial-anderror procedures must be used. In the second method, the number of stages in the enriching section and in the stripping section and the reflux ratio are specified or assumed and the separation of the components is calculated using assumed liquid and vapor flows and temperatures for the first trial. This approach is often preferred for computer calculations (H2, PI). In the trial-and-error procedures, the design method of Thiele and Geddes (PI, SI, Tl), which is a reliable procedure, is often used to calculate resulting distillate and bottoms compositions and tray temperatures and compositions. Various combinations and variations of the above rigorous calculation methods are available in the literature (H2, PI, SI) and are not considered further. The variables in the design of a distillation column are all interrelated, and there are only a certain number of these which may be fixed in the design. For a more detailed

either of

discussion of the specification of these variables, see

3.

Kwauk (K2).

In the remainder of this chapter, shortcut calculation approximate solution of multicomponent distillation are considered.

Shortcut calculation methods.

methods These methods are quite useful to study a large number of cases rapidly to help orient the designer, to determine approximate optimum conditions, or to provide information for a cost estimate. Before discussing these methods, equilibrium relationships and calculation methods of bubble point, dew point, and flash vaporization for multicomponent systems for the

are covered.

11.7B

Equilibrium Data

in

Multicomponent

Distillation

For multicomponent systems which can be considered determine the composition of the vapor

ideal,

Raoult's law can be used to

equilibrium with the liquid. For example, for a

in

system composed of four components, A, B, C, and D,

Pb=P b x b

Pa=Pa*a, Va

=

PB

pa

Pa

T = T Xa

Pc=Pc*c,

,

'

^ =y

Xfi

yc

'

Pd

PC = -yx c

= Pd x d yD

,

=

Pd — XD

(1 1-7-1)

(11.7-2)

In hydrocarbon systems, because of nonidealities, the equilibrium data are often

represented by

yA

where

KA

is

=

K-A*A,

}b=K„x s

yc

,

=K c x c

,

yD

=

KD x D

the vapor-liquid equilibrium constant or distribution coefficient for

nent A. These

K

(11.7-3)

compo-

values for light hydrocarbon systems (methane to decane) have been

determined semiempirically and each

K factor

K

is

a function of temperature and pressure.

and Hadden and Grayson hydrocarbon systems K is generally assumed not to be a function of composition, which is sufficiently accurate for most engineering calculations. Note that for an ideal system, K A = PJP, and so on. As an example, data for the hydrocarbons n-butane, /i-pentane, n-hexane, and /i-heptane are plotted in Fig. 11.7-2 at 405.3 kPa (4.0 atm) absolute (Dl.Hl). Convenient

(HI).

680

For

charts are available by Depriester (Dl)

light

Chap. 11

Vapor-Liquid Separation Processes

FIGURE

Equilibrium

11.7-2.

(4.0

The

relative volatility a (

mixture can be defined in a

=

C

a

11.7C

1

1

will

each individual component

for

D

is

*=lT> a.

aB

=

lF' ^c

kPa

a multicomponent

component

=

-7T-=10, K. c

a

D=T^ A.

(11.7-4)

c

be a stronger function of temperature than thea,- values since the

Boiling Point,

Dew

temperature

C

component,

ac

in a

K

t

similar manner.

Point, and Flash Distillation

At a specified pressure, the boiling point or bubble point of a given satisfy the relation £ y = 1.0. For a mixture of A, B, C,

multicomponent mixture must

Sec. 11.7

in

similar to that for a binary mixture. If

selected as the base

.7-2 all increase with

Boiling point.

values for light hydrocarbon systems at 405.3

manner

c

-

The values of K; lines in Fig.

a

-77-> JS.

/.

in

mixture of A, B, C, and

«.

K

atm) absolute.

Distillation

of Multicomponent Mixtures

t

681

and

D with C as

component,

the base

X y, = X

= Kc

K, x,-

X

a,-

X|

=

1

.0

(1

1.7-5)

The calculation is a trial-and-error process, as follows. First a temperature is assumed and the values of a are calculated from the values of X at this temperature. Then the ;

value of

Kc

of

value

calculated

;

Kc =

calculated from

is

1.0/£ a, *,-. The temperature corresponding to the 1 '.0/2 otiXj. The temperature corresponding to

Kc =

the calculated value of

Kc

is

compared to is used

temperature

Kc

is

K

c = calculated from

calculated from

the liquid composition

is

x,

EXAMPLE 11.7-1. A

is

vf^

£

=

Boiling Point

(n.7-6)

The composition

mol

in

is

and

also trial

{y-Ja^. After the final

error,

temperature

is

known,

a

Jf I (yA-)

(H.7-8)

'

.

of a Multicomponent Liquid

liquid feed to a distillation tower at 405.3

tower.

values

If the

After the final

calculated from

For the dew point calculation, which

Dew point.

The value of

=

assumed temperature.

for the next iteration.

known, the vapor composition

is

yi

2.

the

the calculated temperature

differ,

fractions

is

kPa abs

is

fed to a distillation

as follows: n-butane(x /4

=

0.40),

n-hexane(x c = 0.20), n-heptane(x D = 0.15). Calculate the boiling point and the vapor in equilibrium with the liquid. n-pentane(x B

=

0.25),

First a temperature of 65°C is assumed and the values of K obtained from Fig. 1 1.7-2. Using component C (n-hexane) as the base component, the following values are calculated using Eq. (1 1.7-5) for. the first

Solution:

trial:

Trial

J

(65°C) Trial 3 (70°C) Final

Cornp.

K

x-

Kc

t

'

a,*,

K.,

a.

a,x,

y.,

A B C

0.40

1.68

6.857

2.743

1.86

6.607

2.643

0.748

0.25

0.63

2.571

0.643

0.710

2.522

0.631

0.178

0.20

0.245

1.000

0.200

0.2815

1.000

0.200

0.057

D

0.15

0.093

0.380

0.057

0.110

0.391

0.059

0.017

3.533

1.000

£

1.00

Kc = The

1/1

a.x,

=

trial 3,

is

Z

3.643

= 0.2745

C)

Kc =

a x i

i

=

1/3.533

=

0.2830 (70°C)

Kc

for trial 2, a

the calculations

70°C, which

==

1/3.643

calculated value of

Using 69°C

3.

~ a

is 0.2745, which corresponds to 69°C, Fig. 1 1.7-2. temperature of 70°C was obtained. Using 70°C for

shown

in the table give a final calculated value of

the bubble point. Values of y, are calculated from Eq.

(1

1

.7-6).

Flash distillation of multicomponent mixture.

diagram

682

is

shown

in Fig.

1

1.3-1,

For flash distillation, the process flow Defining/ = V/F as the fraction of the feed vaporized Chap.

J J

Vapor-Liquid Separation Processes

and

—f) = L/F

(1

nent

as the fraction of the feed remaining as liquid

balance as in Eq.

/

(1 1.3-6),

the following

y, is

(11.7-9)

i

the composition of component

in the vapor,

/

in the liquid after vaporization. Also, for equilibrium, y,

KJKC

.

and making a compo-

obtained:

= Lj-x +-^

yi

where

is

which

is

in equilibrium withx,-

= K iXi = K c a, Xi,

where a,

=

Then Eq. (1 1.7-9) becomes y

Solving for x and ;

= Kc x = a-,

t

is

;

^

(11.7-10)

=1 °

(1L7- H)

summing for all components,

^• = £ /(* This equation

x +

;

solved by

trial

When

c

J-l) +

and error by

£

l

assuming a temperature

first

if

the fraction

x values add up to 1.0, the proper temperature has been chosen. The composition of the vapor y can be obtained fromy, = c a,-Xj or by a vaporized has been

set.

the

;

K

;

material balance.

11.7D

Key Components

in

Multicomponent

Fractionation of a multicomponent mixture

Distillation

in

a distillation tower will allow separation

only between two components. For a mixture of A, B, C, D, and so on, a separation in

B and C, and so on. The components more volatile (identified by the subscript L), and the heavy key (H). The components more volatile than the light key are called light components and will be present in the bottoms in small amounts. The components less volatile than the heavy key are called heavy components and are present in the distillate in small amounts. The two key components are present in significant amounts one tower can only be made between

A and

separated are called the light key, which

in

both the

11.7E

/.

distillate

is

B, or

the

and bottoms.

Total Reflux for Multicomponent Distillation

Minimum

Just as in binary distillation, the

stages for total reflux.

Nm

minimum number

can be determined for multicomponent distillation for total reflux. The Fenske equation (11.4-23) also applies to any two components in a multicomponent system. When applied to the heavy key H and the light key L, it of theoretical stages or steps,

,

becomes

N

log [(x

" D/x "° D ^x " w W'Xt w W ^ -

!og

{ct

L

,

J

(117-12)

where x LD is mole fraction of light key in distillate, x LW is mole fraction in bottoms, x [ID is mole fraction of heavy key in distillate, and x„ w is mole fraction in bottoms. The average value of a L of the light key is calculated from thea t0 at the top temperature (dew point) of the tower and a LW at the bottoms temperature. «L.av

Note and not

that the distillate dew-point

= AoCtif

(11.7-13)

and bottoms boiling-point estimation is partially trial components in the distillate and bottoms is

error, since the distribution of the other

known and can

Sec. 11.7

affect these values.

Distillation

of Multicomponent Mixtures

683

2.

To

Distribution of other components.

determine the distribution or concentration of

other components in the distillate and the bottoms at total reflux, Eq. (11.7-12) can be

rearranged and written for any other component

^4" X iW

w =(«.-.. v)

as follows:

i

-?4 HW "

(iL7 - i4 >

These concentrations of the other components determined at total reflux can be used as approximations with finite and minimum reflux ratios. More accurate methods for finite and minimum reflux are available elsewhere (H2, SI, VI).

EXAMPLE ]]..7-2.

Calculation of Top and Bottom

Temperatures and

Total Reflux

mol/h at the boiling point given in Example 11.7-1 is tower at 405.3 kPa and is to be fractionated so that 90% of the H-pentane (B) is recovered in the distillate and 90% of the n-hexane (C) in the bottoms. Calculate the following. (a) Moles per hour and composition of distillate and bottoms. (b) Top temperature (dew point) of distillate and boiling point of bottoms. (c) Minimum stages for total reflux and distribution of other components in the distillate and bottoms.

The

liquid feed of 100

fed to a distillation

For part (a) material balances are made for each component, with component n-pentane (B) being the light key (L) and n-hexane (C) the heavy key (H). For the overall balance,

Solution:

W

F = D + For component

B, the light key,

x BF Since

x BW

90%

W=

of

2.5.

(11,7-15)

B

F =

0.25(100) the

in

is

=

=

25.0

distillate,

y BD

For component C, the heavy x CF

F=

0.20(100)

=

y BD

20.0

D + x BW

D =

W

(0.90X25)

(11.7-16)

=

22.5.

Hence,

key,

=

y CD

D + x cw

W

(11.7-17)

W

= 0.90(20) = 18.0. Then, bottoms and x cw assumed that no component D (heavier y CD than the heavy key C) is in the distillate and no light A in the bottoms. Hence, moles A in distillate = y AD D = 0.40(100) = 40.0. Also, moles D in = 0.15(100) = 15.0. These values are tabulated below. bottoms = x DW 90% of C is D = 2.0. For the

Also,

in

the

first trial, it is

W

Feed.

Comp.

A B

(It

C

(hy key H)

key L)

D

xF

F xF

Distillate,

F

xw

d

xw

W

40.0

0.620

40.0

0

25.0

0.349

22.5

0.070

2.5

0.20

20.0

0.031

2.0

0.507

18.0

0.15

15.0 100.0

0

0

0.423 1.000

W = 35.5

F

=

For the dew point of the

684

x

-

W

0.25

D =

1.000

distillate (top first trial.

Chap.

64.5

0

15.0

temperature) in part (b), a value values are read from Fig.

The K a values calculated. Using Eqs.

of 67°C will be estimated for the

and the

~~

Bottoms,

0.40

1.00

11.7-2

>'d

D

11

(11.7-7)

and

(11.7-8),

the

Vapor-Liquid Separation Processes

following values are calculated:

Comp.

K/67°C)

y iD

x

<*.-

i

A

0.620

1.75

6.73

0.0921

0.351

B(L)

0.349

0.65

2.50

0.1396

0.531

C(H)

0.031

0.26

1.00

0.0310

0.118

0

0.10

0.385

0

D

=

1.000

Kc = The

Kc

is

0.2627, which corresponds very closely to

the final temperature of the

is

1.000

I yA = 0-2627

calculated value of

67°C, which

0

0.2627

dew

point.

For the bubble point of the bottoms, a temperature of 135°C is assumed for trial 1 and Eqs. (11.7-5) and (11.7-6) used for the calculations. A second trial using 132°C gives the final temperature as shown below.

Comp.

x inr

K,

A

y.

0

5.00

4.348

0

0

B(L)

0.070

2.35

2.043

0.1430

0.164

C(H)

0.507

1.15

1.000

0.5070

0.580

D

0.423

0.61

0.530

0.2242 0.8742

0.256 1.000

1.

calculated value of

For part

(c)

=

000

Kc = The

<x,x,

=

1/0.8742

Kc

is

1.144

1.144, which' is close to the value at 132°C.

the proper a values of the light key

L

(n-pentane) to use in

Eq. (11.7-13) are as follows: a LD

=

2.50

(£

= 67°C

LW

=

2.04

(t

= 132°C

=

^oc LD a LW

a.

av

Substituting into Eq.

=

J

2.

at the

column

at the

50(2.04)

top)

column bottom)

=

2.258

(1 1.7-13)

(1 1.7-12),

log [(0.349 x 64.5/0.031 x 64.5X0-507 x 35.5/0.070 x 35.5)]. log (2.258)

= The

5.404 theoretical stages (4.404 theoretical trays)

components can be calculated For component A, the average a value to use is

distribution or compositions of the other

using Eq.

«„.

av

(1 1.7-14).

= J* ad* aw =

Making an

v/6.73 x 4.348

overall balance

1

1.7

Distillation

5.409

on A,

x AF F

Sec.

=

=

40.0

=

x AO D

of Mullicomponent Mixtures

+

x AW

W

(1 1-7-18)

685

Substituting x AD

D = lOHx^

W from Eq.

(11.7-14) into (1 1.7-18)

and

solv-

ing,

W = 0.039,

x AW

For

D,a D

the distribution of component

x DF F

=

15.0

=

x DD D

W

x AD D av

=

=

39.961

^0.385 x 0.530

+ x DlK W = 14.977.

x D0 D = 0.023, jc dh The revised distillate and bottoms compositions are as

Solving,

,

= 0.452.

D

Distillate,

Com p.

follows.

Bottoms,

xD D

xw

W xw

W

A

0.6197

39.961

0.0011

B(L)

0.3489

22.500

0.0704

2.500

C(H)

0.0310

2.000

0.5068

18.000

D

0.0004

0.023

0.4217

14.977

64.484

1.0000

W = 35.516

1.0000

Hence, the moles of bottoms.

D

D=

0.039

in the distillate are quite small, as

are the moles of

A

in the

Using the new distillate composition, a recalculation of the dew point assuming 67°C gives a calculated value of K c = 0.2637. This is very close obtained when the trace amount of D in the distillate was Hence, the dew point is 67°C. Repeating the bubblepoint calculation for the bottoms assuming 132°C, a calculated value of K c = 1.138, which is close to the value at 132°C. Hence, the bubble point remains at 132°C. If either the bubble- or dew-point temperatures had changed, the new values would then be used in a recalculation of N m to that of 0.2627

assumed

as zero.

.

11.7F

Shortcut

Method

component

As

in the case for

will require

an

for

Minimum

Reflux Ratio for Multi-

Distillation

minimum reflux ratio R m is that reflux ratio that of trays for the given separation of the key components.

binary distillation, the

infinite

number

For binary distillation only one "pinch point" occurs where the number of steps become infinite, and this is usually at the feed tray. For multicomponent distillation two in the section above the feed and another below the feed tray. The rigorous plate-by-plate stepwise procedure to calculate is trial and error and can be extremely tedious for hand calculations. Underwood's shortcut method to calculate R m (Ul, U2) uses constant average a values and also assumes constant flows in both sections of the tower. This method

pinch points or zones of constant composition occur: one plate

provides a reasonably accurate value.

minimum

The two equations

1-^ = 686

to be solved to determine the

reflux ratio are

1^—

Chap. 11

(H.7-19)

Vapor-Liquid Separation Processes

Km +

'

The

ofx iC

l=Z^~ —a a

(H.7-20)

;

each component in the

distillate in Eq. (11.7-20) are supposed to be However, as an approximation, the values obtained using the Fenski total reflux equation are used. Since each a may vary with temperature, the average value of a,- to use in the preceeding equations is approximated by using a at the average temperature of the top and bottom of the tower. Some (PI, SI) have used the average a which is used in the Fenske equation or the a at the entering feed temperature. To solve for R m the value of 9 in Eq. (1 1.7-19) is first obtained by trial and error. This value of 0 lies between the a value of the light key and a value of the heavy key, which is 1.0. Using this value of 0 in Eq. (11.7-20), the value of R m is obtained

values

the values at the

for

minimum

reflex.

t

;

,

When distributed components appear between methods described by others (S1,T2, VI) can be used.

directly.

11.7G

Shortcut

Method

for

Number

the key

components, modified

of Stages at

Operating Reflux Ratio

Number of stages at operating reflux ratio. The determination of the minimum /. number of stages for total reflux in Section 11. 7E and the minimum reflux ratio in Section 11. 7F are useful for setting the allowable ranges of number of stages and flow These ranges are helpful for selecting the particular operating conditions for

conditions.

a design calculation.

The

relatively

complex rigorous procedures

for

doing a stage-by-

stage calculation at any operating reflux ratio have been discussed in Section

An

1

1.7A.

important shortcut method to determine the theoretical number of stages

required for an operating reflux ratio (El) given

in

Fig.

11.7-3.

R

is

the empirical correlation of Erbar and

This correlation

is

somewhat

Maddox

similar to a correlation

by

and should be considered as an approximate method. In Fig. 1 1.7-3 the operating reflux ratio R (for flow rates at the column top) is correlated with the minimum R m obtained using the Underwood method, the minimum number of stages N m obtained Gilliland (Gl)

by the Fenske method, and the number of stages

2.

Estimate offeed plate location.

number

to estimate the

Kirkbride

of theoretical stages

to estimate the feed stage location.

log

where

N

e is

the

number

£7 =

N at the operating R.

(K.1)

has devised an approximate method

above and below the

This empirical relation

0.206 log

D

is

feed

which can be used

as follows:

WJ

(11.7-21)

J

of theoretical stages above the feed plate and

N

s

the

number of

theoretical stages below the feed plate.

EXAMPLE

Minimum Reflux Ratio and Number of Stages at Operating Reflux Ratio Using the conditions and results given in Example 11.7-2, calculate the 11.7-3.

following. (a) (b)

Minimum reflux ratio using the Underwood method. Number of theoretical stages at an operating reflux ratio R using the

(c)

Sec.

1

1.7

Erbar-Maddox

of i.5R„

correlation.

Location of feed tray using the method of Kirkbride.

Distillation

of Multicomponent Mixtures

687

For

Solution:

a

;

is

Example

11.7-2)

Fig. 11.7-2

Eqs.

part (a) the temperature to use for determining the values of

67°C and the bottom

the average between the top of

and

is

(67

4-

and the a, values and distillate and and (1 1.7-20) are as follows

of 132°C (from

values obtained from

feed compositions to use in

(1 1.7-19)

Comp.

X iF

X iD

K,.(99.5°C)

X iW

a,(99.5°C)

A

0.40

0.6197

3.12

5.20

0.0011

B(L)

0.25

0.3489

1.38

2.30

0.0704

C(H)

0.20

0.0310

0.60

1.00

0.5068

D

0.15 1.00

0.0004 1.0000

0.28

0.467

0.4217 1.0000

=

Substituting into Eq. (11.7-19) with q

1

_„9 = _ _ 0 1

1.0 for feed at the boiling point,

5 2(X°- 4Q )

2.30(0.25)

-

,

1

5.20

+ 688

The K,

132)/2 or 99.5°C.

-

8

1.00(0.20)

1.00

2.30

-9 +

Chap. II

-

0

0.467(0.15)

0.467

-

(UJ ' Z1)

0

Vapor-Liquid Separation Processes

is trial and error, so a value of 9 = 1.210 will be used for the first trial must be between 2.30 and LOO). This and other trials are shown below.

This

0.575

2.08

9 (Assumed)

-

5.2

9

-

2.3

9

1.210

0.5213

0.5275

1.200

0.5200

0.5227

1.2096

0.5213

0.5273

The

final value of 6

=

R +

1.2096

i

5.20

0.070

1.0-6*

0.467

-0.9524 -1.0000 -0.9542

-0.0942 -0.0955 -0.0943

substituted into Eq.

is

5.2CX0.6197)

_

0.200

-

2.30

Rm =

Solving,

For

RJ(K + 0.49.

Hence,

For

1.7-20) to solve for

R„

1.2096

j

0.467

1.2096

-

1.2096

0.593,

0.395/(0.395

NJN = 0.49

+ =

+

R/(R

1.0)

=

values

=

0.2832.

1.0 (reboiler)

are

calculated.

0.593/(0.593

+

From

11.7-3,

Solving,

5.40//V.

—

1)

Fig.

N=

1.0)

=

0.3723,

NJN

=

11.0 theoretical stages

or 10.0 theoretical trays.

the location of the feed tray in part (c) using Eq. (11.7-21),

log

Hence,

=

This gives 11.0

in the tower.

-

following

the

(b)

1.5(0.395)

=

1)

+ 0.0001

0.395.

part

R = L5R m =

-

+ 0.0022 -0.0528

0.467(0.0004)

1. 00(0.031)

1.00

X (Sum)

2.30(0.3489)

+

1.2096

(1

-9

(0

j£ =

0.206 log

0.20 ^ 35.516 / 0.0704 0.25 J

=

0.07344

64.484 ^0.0310/

NJN, = 1.184. Also, N e + N = 1.184N, + N s = N = 11.0 stages. N s = 5.0 and N e = 6.0. This means that the feed tray is 6.0 trays s

Solving,

from the top.

PROBLEMS 11.1-1.

Phase Rule for a Vapor System. For the system NH 3 -water and only a vapor phase present, calculate the number of degrees of freedom. What variables can be fixed? Ans.

11.1-2. Boiling Point

and Raoult's Law. For

using the data of Table (a)

F =

1

3 degrees of freedom

;

variables T, P, y A

the system benzene-toluene,

do

as follows

1.1-1.

At 378.2 K, calculate y A and x A using Raoult's law. mixture has a composition ofx^ = 0.40 and is at 358.2 K and 101.32 kPa pressure, will it boil? If not, at what temperature will it boil and what will be the composition of the vapor first coming off?

(b) If a

Chap.

II

Problems

689

The vapor-pressure data are given below

11.1-3. Boiling-Point-Diagram Calculation.

for

the system hexane-octane.

Vapor Pressure

n-Hexane T(°F)

155.7

68.7

175

(a)

mm Hg

kPa

T(°C)

79.4

n-Octane

kPa

101.3

760

16.1

136.7

1025

23.1

173

37.1

278

434 760

200

93.3

197.3

1480

225

107.2

284.0

2130

57.9

258.2

125.7

456.0

3420

101.3

Using Raoult's

mm Hg

and

law, calculate

plot the

xy data

121

at

a

total pressure

of

101.32 kPa. (b)

Plot the boiling-point diagram.

of Vapor-Liquid System. A mixture of 100 mol containing n-pentane and 40 mol n-heptane is vaporized at 101.32 kPa abs pressure until 40 mol of vapor and 60 mol of liquid in equilibrium with each other are produced. This occurs in a single-stage system and the vapor and liquid are kept in contact with each other until the vaporization is complete. The equilibrium data are given in Example 11.3-2. Calculate the composition of the

11.2- 1 Single-Stage Contact

%

60 mol

%

vapor and the

liquid.

Volatility of a Binary System. Using the equilibrium data for the n-pentane-n-heptane system given in Example 11.3-2, calculate the relative volatility for each concentration and plot a versus the liquid composition x A

11.3- 1. Relative

.

11.3-2.

Comparison of Differential and Flash Distillation. A mixture of 100 kg mol which % n-pentane (A) and 40 mol % /i-heptane (B) is vaporized at 101.32 kPa pressure under differential conditions until 40 kg mol are distilled. Use equilibrium data from Example 11.3-2. (a) What is the average composition of the total vapor distilled and the compocontains 60 mol

sition of the liquid left?

same vaporization is done in an equilibrium or flash distillation and 40 kg mol are distilled, what is the composition of the vapor distilled and of the

(b) If this

liquid left?

Ans.

(a)

x2

=

0.405,

y av

=

0.892; (b) x 2

=

0.430,

y 2 = 0.854

%

of Benzene-Toluene. A mixture containing 70 mol is distilled under differential conditions at 101.32 kPa (1 atm). A total of one third of the moles in the feed is vaporized. Calculate the average composition of the distillate and the composition of the remaining liquid. Use equilibrium data in Table 11.1-1.

11.3-3. Differential Distillation

benzene and 30 mol

11.3-4.

Steam

Distillation

%

toluene

of Ethylaniline. A mixture contains 100kgofH 2 O and 100 kg = 121.1 kg/kg mol), which is immiscible with water. A

of ethylaniline (mol wt very slight

amount of nonvolatile impurity

the ethylaniline

it is

is

dissolved in the organic.

steam-distilled by bubbling saturated

To

purify

steam into the mixture

kPa (1 atm). Determine the boiling point of the mixture and the composition of the vapor. The vapor pressure of each of the pure at a total pressure of 101.32

690

Chap.

11

Problems

compounds

is

as follows (Tl):

Temperature

K

°C

PJiethylaniline)

(kPa)

353.8

80.6

48.5

1.33

369.2

96.0

87.7

2.67

98.3

3.04

163.3

5.33

372.3

99.15

386.4

1

P A (water) (kPa)

113.2

Steam Distillation of Benzene. A mixture of 50 g mol of liquid benzene and 50 g mol of water is boiling at 101.32 kPa pressure. Liquid benzene is immiscible in water. Determine the boiling point of the mixture and the composition of the vapor. Which component will first be removed completely from the still? Vaporpressure data of the pure components are as follows:

13-5.

Temperature

P water

p

(mm Hg)

{mm Hg)

,

K

°C

308.5

35.3

benzene

43

150

325.9

52.7

106

345.8

72.6

261

300 600

353.3

80.1

356

760

McCabe-Thiele Method. A rectification column is fed 100 kg mol/h of a mixture of 50 mol % benzene and 50 mol toluene at 101.32 kPa abs pressure. The feed is liquid at the boiling point. The distillate is to contain 90 mol % benzene and the bottoms 10 mol % benzene. The reflux ratio is 4.52 1. Calculate the kg mol/h distillate, kg mol/h bottoms, and the number of theoretical trays needed using the McCabe-Thiele method. = 50 kg mol/h, 4.9 theoretical trays plus reboiler Ans. D = 50 kg mol/h,

11.4-1. Distillation Using

%

:

W

A saturated liquid feed of 200 42 mol heptane and 58% ethyl benzene is to be fractionated at 101.32 kPa abs to give a distillate containing 97 mol heptane and a bottoms containing 1.1 mol heptane. The reflux ratio used is 2.5 1. Calculate the mol/h distillate, mol/h bottoms, theoretical number of trays, and the feed tray number. Equilibrium data are given below at 101.32 kPa abs

11.4-2. Rectification

mol/h

of a Heptane— Ethyl Benzene Mixture.

at the boiling point containing

%

%

%

:

pressure for the mole fraction n-heptane x H and y H

Temperature

Temperature

yH

K

0 0.230

K

°C

409.3

136.1

402.6

129.4

0.08

392.6

119.4

0.250

0.514

XH

Ans.

Chap. II

Problems

.

0

D=

85.3 mol/h,

feed

on

°C

XH

yH

383.8

110.6

0.485

0.730

376.0

102.8

0.790

0.9O4

371.5

98.3

1.000

1.000

W=

114.7 mol/h, 9.5 trays

+

reboiler,

tray 6 from top

691

11.4-3.

Minimum

Graphical Solution for

Problem

tification given in

toluene

is

being distilled

Reflux Ratio and Total Reflux. For the recwhere an equimolar liquid feed of benzene and to give a distillate of composition x D = 0.90 and a 1

1.4-1,

bottoms of composition x M,= methods.

0.10,

calculate the following using graphical

Minimum reflux ratio R m Minimum number of theoretical plates at total reflux. Ads. (a) R m — 0.9 1 (b) 4.0 theoretical trays plus a reboiler Minimum Number of Theoretical Plates and Minimum Reflux Ratio. Determine the minimum reflux ration R n and the minimum number of theoretical plates at (a)

.

(b)

;

11.4-4.

mixture of heptane and ethyl benzene as by the graphical methods of McCabe—Thiele.

total reflux for the rectification of a

given in Problem 11.4-2.

Do

this

Vaporized Feed. A total feed of 200mol/h having % heptane and 58 mol ethyl benzene is to be fractionated at 101.3 kPa pressure to give a distillate containing 97 mol heptane and a bottoms containing 1.1 mol % heptane. The feed enters the tower partially vaporized so that 40 mol is liquid and 60 mol vapor. Equilibrium data are given in Problem 11.4-2. Calculate the following. (a) Moles per hour distillate and bottoms.

11.4-5. Rectification Using a Partially

%

an overall composition of 42 mol

%

%

(b) (c)

(d)

Minimum reflux ratio R m Minimum steps and theoretical

%

.

trays at total reflux.

Theoretical number of trays required for an operating reflux ratio of 2.5 1. Compare with the results of Problem 11.4-2, which uses a saturated liquid :

feed.

11.4-6. Distillation Using

a Vapor Feed. Repeat Problem 11.4-1 but use a feed that dew point. Calculate the following.

is

saturated vapor at the (b)

Minimum reflux ratio R m Minimum number of theoretical

(c)

Theoretical

(a)

.

number

of trays at

plates at total reflux.

an operating

reflux ratio of 1.5(R m

).

Tower for Benzene— Toluene. An enriching tower is fed 100 kg mol/h of a saturated vapor feed containing 40 mol % benzene {A) and 60 mol % toluene benzene. The reflux (B) at 101.32 kPa abs. The distillate is to contain 90 mol and their ratio is set at 4.0 1. Calculate the kg mol/h distillate D and bottoms

11.4-7 Enriching

%

W

:

compositions. Also, calculate the number of theoretical plates required. = 80 kg mol/h, x w Ans. D = 20 kg mol/h,

W

1

1.4-8. Stripping

^-octane

Tower. is

fed at

A

liquid mixture containing 10

its

mol

%

=

0.275

n-heptane and 90 mol

boiling point to the top of a stripping tower at 101.32

%

kPa

The bottoms are to contain 98 mol % n-octane. For every 3 mol of feed, 2 mol of vapor is withdrawn as product. Calculate the composition of the vapor and the number of theoretical plates required. The equilibrium data below are given as mole fraction n-heptane.

abs.

11.4-9. Stripping

X

y

X

y

0.284

0.459

0.039

0.078

0.097

0.184

0.012

0.025

0.067

0.131

Tower and Direct Steam

%

Injection.

A

liquid feed at the boiling point

%

contains 3.3 mol ethanol and 96.7 mol water and enters the top tray of a stripping tower. Saturated steam is injected directly into liquid in the bottom of the tower. The overhead vapor which is withdrawn contains 99% of the alcohol

Assume equimolar overflow for this problem. Equilibrium data mole fraction of alcohol are as follows at 101.32 kPa abs pressure(l atm abs).

in the feed.

692

Chap.

II

for

Problems

(a)

(b)

0

0

0.0080

0.0750

0.020

0.175

0.0296

0.250

0.033

0.270

infinite number of theoretical steps, calculate the minimum moles of steam needed per mole of feed. (Note : Be sure and plot the q line.) Using twice the minimum moles of steam, calculate the number of theoretical

For an

steps needed, the composition of the

overhead vapor, and the bottoms

composition. Ans.

(a)

0.121

mol steam/mol

feed;

xD

(b) 5.0 theoretical steps,

11.5- 1.

=

0.135,

xw

=

0.OOO33

Murphree Efficiency and Actual Number of Trays. For the distillation of heptane and ethyl benzene in Problem 11.4-2, the Murphree tray efficiency is estimated as 0.55. Determine the actual number of trays needed by stepping off the trays using the tray efficiency of 0.55. Also, calculate the overall tray efficiency

Ea

.

of Enthalpy-Concentration Method to Distill an Ethanol-Water Solution.

11.6- 1. Use

mixture of 50 wt

% ethanol and 50 wt % water which

A

saturated liquid at the pressure to give a distillate is

is to be distilled at 101.3 kPa ethanol. The feed ethanol and a bottoms containing 3 wt containing 85 wt rate is 453.6 kg/h and a reflux ratio of 1.5 is to be used. Use equilibrium and enthalpy data from Appendix A. 3. Note that the data are given in wt fraction and kJ/kg. Use these consistent units in plotting the enthalpy-concentration

boiling point

%

%

data and equilibrium data. Do as follows. (a) Calculate the amount of distillate and bottoms. (b) Calculate the number of theoretical trays needed. (c) Calculate the condenser and reboiler loads.

Ans.

W

(a)

D =

(b)

3.9 trays plus a reboiler

(c)

<7 C

260.0 kg/h,

= 698 750

kJ/h,

=

193.6 kg/h

^=704

770 kJ/h

of Ethanol-Water Solution Using Enthalpy-Concentration MethRepeat Problem 11.6-1 but use a reflux ratio of 2.0 instead of 1.5. Ans. 3.6 theoretical trays plus reboiler

Distillation

od.

A

feed of ethanol- water Minimum Reflux and Theoretical Number of Trays. containing 60 wt ethanol is to be distilled at 101.32 kPa pressure to give a ethanol. ethanol and a bottoms containing 2 wt distillate containing 85 wt The feed rate is 10 000 kg/h and its enthalpy is 116.3 kJ/kg (50 btu/lb m ). Use consistent units of kg/h, weight fraction, and kJ/kg. (a) Calculate the amount of distillate and bottoms. (b) Determine the minimum reflux ratio using enthalpy-concentration data

%

%

%

(c)

(d) (e)

from Appendix A. 3. Using 2.0 times the minimum reflux ratio, determine the theoretical number of trays needed. Calculate the condenser and reboiler heat loads. Determine the minimum number of theoretical plates at total reflux. Ans. (b) R m = 0.373 (c)

Chap. 11

Problems

4.4 theoretical trays plus reboiler

=

3

634 kW, q R = 4 096

kW

(d)

qc

(e)

2.8 theoretical trays plus reboiler

693

of Benzene-Toluene Feed Using Enthalpy-Concentration Method. A of 100 kg mol/h of benzene-toluene at the boiling point contains 55 toluene. It is being distilled at 101.32 kPa benzene and 45 mol mol pressure to give a distillate with x D =0.9S and a bottoms of x w = 0.04. Using a reflux ratio of 1.3 times the minimum and the enthalpy-concentration method

11.6-4.. Distillation

liquid feed

%

do (a)

%

as follows.

Determine the theoretical number of trays needed. condenser and reboiler heat loads. Determine the minimum number of theoretical trays

(b) Calculate the (c)

11.6- 5.

Use of Enthalpy-Concentration

at total reflux.

For the system benzene-toluene do as

Plot.

follows. (a)

Plot the enthalpy -concentration data using values from Table 1 1.6-2. For a value of x = 0.60 = y, calculate the saturated liquid enthalpy h and the saturated vapor enthalpy and plot these data on the graph.

H

(b)

A

mixture contains 60 mol of benzene and 40 mol of toluene. This mixture is heated so that 30 mol of vapor are produced. The mixture is in equilibrium. Determine the enthalpy of this overall mixture and plot this point on the enthalpy-concentration diagram.

11.7- 1. Flash Vaporization

tower of Example (a)

Dew

of Multicomponent Feed. For the feed to the

distillation

11.1-1, calculate the following.

point of feed and composition of liquid

in equilibrium.

(Note: The

boiling point of 70°C has already been calculated.) (b)

The temperature and composition of both phases when 40% of vaporized

the feed

is

in a flash distillation.

Ans.

(a)

107°C, x A

(bT 82°C, x A

yA

= 0.610,

=

=

yB

x B = 0.158, x c = 0.281, x D = 0.447; x B = 0.254, x c = 0.262, x D = 0.224;

0.114,

0.260,

=

0.244,

yc

=

0.107, y D

=

0.039

Dew Point, and Flash Vaporization. Following is the composition of a liquid feed in mole fraction: n-butane (x A = 0.35), n-pentane (x B = 0.20), rc-hexane, (x c = 0.25), ^-heptane (x D = 0.20). At a pressure of 405.3 kPa calculate the following.

11.7-2. Boiling Point,

(a)

Boiling point and composition of the vapor in equilibrium.

(b)

Dew point

(c)

The temperature and composition vaporized

and composition of the liquid in equilibrium. of both phases when 60% of the feed

is

in a flash distillation.

Multicomponent Alcohol Mixture. given below for the following alcohols.

11.7-3. Vaporization of

The vapor-pressure data are

Vapor Pressure (mm Hg) T(°C)

694

Methanol

Ethanol

n-Propanol

n-Buianol

50

415

220.0

88.9

33.7

60

629

351.5

148.9

59.2

65

767

438

190.1

77.7

70

929

542

240.6

99.6

75

1119

665

301.9

131.3

80

1339

812

376.0

165.0

85

1593

984

465

206.1

90

1884

1185

571

225.9

100

2598

1706

843

387.6

Chap.

11

Problems

Following is the composition of a liquid alcohol mixture to be fed to a distillation tower at 101.32 kPa: methyl alcohol {x A = 0.30), ethyl alcohol (;c B = 0.20), npropyl alcohol (x c = 0.15), and n-butyl alcohol (x D = 0.35). Calculate the following assuming that the mixture follows Raoult's law. (a) Boiling point and composition of vapor in equilibrium. (b) Dew point and composition of liquid in equilibrium. (c) The temperature and composition of both phases when 40% of the feed is vaporized in a flash distillation. Ans.

(a)

(b)

83°C, y A = 0.589, y B = 0.241, yc = 0.084, y D = 0.086; 100°C, x A = 0.088, x B = 0.089, x c = 0.136, x D = 0.687

Minimum Reflux, Number of Stages. The following feed of 100 the boiling point and 405.3 kPa pressure is fed to a fractionating tower: rc-butane (x A = 0.40), /i-pentane {x B = 0.25), «-hexane (x c = 0.20), n-heptane (x D = 0.15). This feed is distilled so that 95% of the «-pentane is recovered in the distillate and 95% of the n-hexane in the bottoms. Calculate

11.7-4. Total Reflux,

mol/h

at

the following. Moles per hour and composition of distillate and bottoms. (b) Top and bottom temperature of tower. (c) Minimum stages for total reflux and distribution of other components (trace components) in the distillate and bottoms, i.e., moles and mole fractions. [Also correct the compositions and moles in part (a) for the traces.] (a)

(e)

Minimum reflux ratio using the Underwood method. Number of theoretical stages at an operating reflux minimum using the Erbar-Maddox correlation.

(f)

Location of the feed tray using the Kirkbride method.

(d)

D = 64.75

ratio of 1.3 times the

=

0.6178, x BD = 0.3668, x CD = 0.0154, x DD 0; x BW = 0.0355, x cw = 0.5390, x DW = 04255; (b) top, 66°C; bottom, 134°C; m = 7.14 stages; trace compositions,

Ans.

(a)

=

mol/h, x AD

W = 35.25 mol/h, x AW =

0,

N

x AW (d) (e) (f)

11.7-5. Shortcut

=

=

R m = 0.504; N — 16.8 stages; N e = 9.1 stages, N = s

=

(q

to

a

distillation

(x A

=

0.35),

The

0.30)

The

tower.

feed

is

=

-5

;

from top

A

feed of part liquid and rate of 1000 mol/h overall composition of the feed is rc-butane Distillation

kPa

is

n-hexane

0.30),

distilled so

10

7.7 stages, feed 9.1 stages

405.4

at

n-pentane (x B

x

4.0

Design of Multicomponent

part vapor

0.15).

x 10~\ x DD

1.2

that

97%

Tower.

fed

=

(x c

the

at

0.20)

and /i-heptane

of the n-pentane

is

(x D

=

recovered in the

(a)

and 85% of the n-hexane in the bottoms. Calculate the following. Amount and composition of products and top and bottom tower temper-

(b)

Number

distillate

atures.

of stages at total reflux and distribution of other

components

in the

products. (c)

Minimum

reflux ratio,

number

of stages at l.2R m

Multicomponent Alcohol Mixture.

11.7-6. Distillation of

A

,

and feed

tray location.

feed of 30 mol

%

meth-

ethanol (B), 15% rt-propanol (C), and 35% /i-butanol (D) is distilled at 101.32 kPa absolute pressure to give a distillate composition containing 95.0 mole methanol and a residue composition containing 5.0%

anol

(A),

20%

%

methanol and the other components point, so that q

applies

=

1.1.

The operating

as calculated.

reflux ratio

is

The feed is below the boiling Assume that Raoult's law

3.0.

and use vapor-pressure data from Problem

11.7-3.

Calculate

the

following. (a)

(b)

Composition and amounts of distillate and bottoms for a feed of 100 mol/h. Top and bottom temperatures and number of stages at total reflux. (Also, calculate the distribution of the other components.)

Chap. II

Problems

695

(c)

Minimum

number

reflux ratio,

of stages at

R=

3.00,

and the

feed

tray

location.

Ans.

(a) D'

=

=

27.778 mol/h, x AD

W = 72.222 mol/h,

=

0.95,

0.0500,

x BD

x BW

= 0.05, x CD = 0, x DD = 0; = 0.2577, xctr = 0.2077, x DW =

0.4846;

65.5°C top temperature, 94.3°C bottom, N m = 9.21 stages, 10" 7 (trace compositions); 10~ 5 ,x 00 = 8.79 x *cd = 3.04 x

(b)

Rm =

(c)

8.6

N=

2.20,

16.2 stages,

N = 7.6, N =

feed

8.6,

e

s

on stage

from top

Design Method for Distillation of Ternary Mixture. A liquid feed at its bubble point is to be distilled in a tray tower to produce the distillate and bottoms as follows. Feed, x AF = 0.047, x BF = 0.072, x CF = 0.881; distillate, x AD = 0. 1260, x BD = 0.1913, x CD = 0.6827; bottoms, x AW = 0, x bw = 0.001 x cw = 0.999. Average a values to use are a A = 4.19, a B =

11.7-7. Shortcut

,

a c = 1.00. For a feed rate of 100 mol/h,

1.58, (a)

calculate D and W, number of stages and distribution (concentration) of A in the bottoms. Calculate R m and the number of stages at 1.25R m

at total

reflux,

(b)

.

REFERENCES (D 1)

DEPR1ESTER, C.

(El)

Erbar,

(G

Gilliland,

1)

(HI) (H2)

J.

H.,

L.

Chem. Eng. Progr. Symp.

and Maddox,

E. R. Ind.

R.

Ser., 49(7), 1 (1953).

N. Petrol. Refiner,

40(5), 183 (1961).

Eng. Chem., 32, 1220 (1940).

Hadden, S. T., and Grayson, H. G. Petrol. Refiner, 40(9), 207 (1961). Holland, C. D. Multicomponent Distillation. Englewood Cliffs, N.

J.:

Prentice-

Hall, 1963.

(Kl)

Kirkbride, C. G., Petrol. Refiner, 23, 32(1944).

(K2)

Kwauk, M.

(K.3)

King, C.

Company, (Ol) (PI)

(Rl)

A.l.Ch.E.J.,

J.,

2,

240 (1956).

Separation Processes, 2nd ed.

New

York: McGraw-Hill

Book

1980.

O'Connell, H.

E.

Trans.

/l./.C/i.E., 42,

741 (1946).

Perry, R. H., and Green, D. Perry's Chemical Engineers' Handbook, 6th ed. New York: McGraw-Hill Book Company, 1984. Robinson, C. S., and Gilliland, E. R. Elements of Fractional York: McGraw-Hill Book Company, 1950.

Distillation, 4th ed.

New (SI)

Smith, B. D. Design of Equilibrium Stage Processes.

Book Company,

New York: McGraw-Hill

1963.

(Tl)

Thiele, E. W., and Geddes, R. L. ind. Eng. Chem., 25, 289 (1933).

(T2)

Treybal,

(U2)

R. E. Mass Transfer Operations, 3rd ed. New York: McGraw-Hill Book Company, 1980. Underwood, A. J. V. Chem. Eng. Progr., 44, 603 (1948); 45, 609 (1949). Underwood, A. J. V. J. Inst. Petrol., 32, 614 (1946).

(VI)

Van Winkle, M.

(Ul)

696

Distillation.

New York McGraw-Hill Book Company, :

Chap.

II

1967.

References

CHAPTER

.-

12

Liquid-Liquid and Fluid-Solid Separation Processes

INTRODUCTION TO ADSORPTION PROCESSES

12.1

12.1 A

Introduction

one or more components of a gas or liquid stream are adsorbed and a separation is accomplished. In commercial processes, the adsorbent is usually in the form of small particles in a fixed bed. The fluid is passed through the bed and the solid particles adsorb components from the fluid. When the bed is almost saturated, the flow in this bed is stopped and the bed is regenerated thermally or by other methods, so desorption occurs. The adsorbed material (adsorbate) is thus recovered and the solid adsorbent is ready for another In adsorption processes

on the surface of a

solid adsorbent

cycle of adsorption.

Applications of liquid-phase adsorption include removal of organic

compounds

from water or organic solutions, colored impurities from organics, and various fermentation products from fermentor effluents. Separations include paraffins from aromatics and fructose from glucose using zeolites. Applications of gas-phase adsorption include removal of water from hydrocarbon gases, sulfur compounds from natural gas, solvents from air and other gases, and odors

from

air.

12. IB

Many

Physical Properties of Adsorbents

adsorbents have been developed for a wide range of separations. Typically, the

adsorbents are 0. 1

mm to

12

in

the form of small pellets, beads, or granules ranging from about

mm in size with the larger particles being used in packed beds. A particle many fine pores and pore volumes up to volume. The adsorption often occurs as a monolayer on the pores. However, several layers sometimes occur. Physical adsorp-

of adsorbent has a very porous structure with

50%

of

total particle

surface of the fine tion,

or van derWaals adsorption, usually occurs between the adsorbed molecules and

the solid internal pore surface and

The fluid is

is

readily reversible.

overall adsorption process consists of a series of steps in series.

flowing past the particle in a fixed bed, the solute

fluid to the

gross exterior surface of the particle.

Then

first

diffuses

When

the

from the bulk

the solute diffuses inside the

697

pore to the surface of the pore. Finally, the solute the overall adsorption process

is

is

adsorbed on the surface. Hence,

a series of steps.

There are a number of commercial adsorbents and some of the main ones

are.

described below. All are characterized by very large pore surface areas of 100 to over

2000 1.

m 2 /g.

made by thermal decomposiand has surface areas of 300 to 1200 m 2 /g with average pore diameters of 10 to 60 A. Organics are generally adsorbed by

Activated carbon. This

is

a microcrystalline material

tion of wood, vegetable shells, coal, etc.,

activated carbon. 2.

Silica gel. This

then dried.

It

20 to 50 A.

adsorbent

is

made by

acid treatment of

has a surface area of 600 to 800

It is

sodium

silicate solution

m 2 /g and average

primarily used to dehydrate gases and liquids

and

pore diameters of

and

to fractionate

hydrocarbons. 3.

Activated alumina.

by heating

To prepare

areas range from 200 to 500 4.

5.

this material,

hydrated aluminum oxide

is

activated

used mainly to dry gases and liquids. Surface with average pore diameters of 20 to 140 A.

to drive off the water. It is

m 2 /g,

Molecular sieve zeolites. These zeolites are porous crystalline aluminosilicates that form an open crystal lattice containing precisely uniform pores. Hence, the uniform pore size is different from other types of adsorbents which have a range of pore sizes. Different zeolites have pore sizes from about 3 to 10 A. Zeolites are used for drying, separation of hydrocarbons, mixtures, and many other applications. Synthetic polymers or resins. These are made by polymerizing two major types of monomers. Those made from aromatics such as styrene and divinylbenzene are used to adsorb nonpolar organics from aqueous solutions. Those made from acrylic esters are usable with more polar solutes in aqueous solutions.

12.1C

Equilibrium Relations for Adsorbents

The equilibrium between

the concentration of a solute in the fluid phase and

concentration on the solid resembles somewhat the equilibrium solubility of a gas

its

in

a

Data are plotted as adsorption isotherms as shown in Fig. 12.1-1. The concentration in the solid phase is expressed as q, kg adsorbate(solute)/kg adsorbent 3 (solid), and in the fluid phase (gas or liquid) as c, kg adsorbate/m fluid. liquid.

Freundlich,

favorable

Langmuir, strongly favorable

kg adsorbate/

kg adsorbent

c,

FIGURE

698

12.1-1.

kg adsorbate/m 3 fluid

Some common

Chap.

12

types

of adsorption isotherms.

Liquid-Liquid and Fluid-Solid Separation Processes

Data

that follow a linear

law can be expressed by an equation similar to Henry's

law.

q = Kc

where

K

isotherm

not

common,

but in the dilute region

it

m 3 /kg

adsorbent. This linear can be used to approximate data of

a constant determined experimentally,

is

is

(12.1-1)

many systems. The Freundlich isotherm equation, which is empirical, often approximates data for many physical adsorption systems and is particularly useful for liquids.

q=Kc" K and n

(12.1-2)

and must be determined experimentally. If a log-log plot made, the slope is the dimensionless exponent n. The dimensions of K depend on the value of n. This equation is sometimes used to correlate data for hydrocarbon gases on activated carbon. The Langmuir isotherm has a theoretical basis and is given by the following, where q„ and K are empirical constants

where

of q versus c

are constants

is

q =

where q 0

—

(12.1-3)

a constant, kg adsorbate/kg solid; and

is

K

is

a constant,

kg/m 3

.

The

equation was derived assuming that there are only a fixed number of active sites available for adsorption, only a

monolayer

and reaches an equilibrium condition. the intercept

Almost

is all

\lq Q

is

formed, and the adsorption

By plotting

is

l/q versus 1/c, the slope is

reversible

Klq 0 and

.

adsorption systems

show

that as temperature is increased the

adsorbed by the adsorbent decreases strongly. This

is

amount

useful since adsorption

is

normally at room temperatures and desorption can be attained by raising the temperature.

EXAMPLE 12.1-1.

Adsorption Isotherm for Phenol in Wastewater Batch tests were performed in the laboratory using solutions of phenol in water and particles of granular activated carbon (R5). The equilibrium data at room temperature are shown in Table 12.1-1. Determine the isotherm that fits the data.

TABLE

12.1-1

.

Equilibrium Data for Example 12.1-1 (R5) c.

kg phenol \

^m

Sec. 12.1

3

solution

J

(

kg phenol\

\kg carbon

0.322

0.150

0.117

0.122

0.039

0.094

0.0061

0.059

0.0011

0.045

Introduction to Adsorption Processes

699

Plotting the data as \lq versus 1/c, the results are not a straight

Solution:

A

and do not follow the Langmuir equation

plot of log q (12.1-3). versus log c in Fig. 12.1-2 gives a straight line and, hence, follow the Freundlich isotherm Eq. (12.1-2). The slope n is 0.229 and the constant is 0.199, to give

line

K

q

=

0.199c

0229

BATCH ADSORPTION

12.2

Batch adsorption

is

often used to adsorb solutes from liquid solutions

when

the

quantities treated are small in amount, such as in the pharmaceutical industry or other industries.

or

As

many

in

other processes, an equilibrium relation such as the Freundlich

Langmuir isotherms and a

tion

is

material balance are needed.

c F and the final equilibrium concentration

the solute adsorbed on the solid

balance on the adsorbate

is

M

is

the

M+

amount of adsorbent,

When the variable q If the equilibrium

The

feed concentra-

initial

Also, the

initial

final

equilibrium value

qM

+ cS

concentration of

is

q.

The

material

is

qF

where

q F and the

is c.

is

=

and S

kg;

Eq. (12.2-1)

in

isotherm

c FS

is

the

(12.2-1)

volume of feed

plotted versus c, the result

is

also plotted on the

q and

lines gives the final equilibrium values of

same graph, the

is

solution,

a straight

m

3 .

line.

intersection of both

c.

EXAMPLE 12.2-1 Batch Adsorption on Activated Carbon A wastewater solution having a volume of 1.0 m 3 contains phenol/m 3 of solution (0.21 g/L). A total of 1.40 kg of fresh .

0.21

kg

granular

activated carbon is added to the solution, which is then mixed thoroughly to reach equilibrium. Using the isotherm from' Example 12.1-1, what are the final equilibrium values, and what percent of phenol is extracted?

M

=

m

values are c F = 0.21 kg phenol/m 5 = 1.0 , 1.40 kg carbon, and q F is assumed as zero. Substituting into Eq.

Solution:

3

The given

,

3

(12.2-1)

0(1.40) + 0.21(1.0) = 9(1.40)

+

c(1.0)

0.20

0.10 <7>

kg phenol kg adsorbent

0 .05

0.01

0.005

0.001

c,

Figure

700

12.1-2.

Chap. 12

0.05

0.01

0.10

0.5

3

kg phenol/m solution

Plot of data for

Example

12.1-1

Liquid-Liquid and Fluid-Solid Separation Processes

This straight-line equation

Example

is

plotted in Fig. 12.2-1.

%

12.3A

q

-

.

is

extracted

=

cF

-c

=

(100)

0.210-0.062 (100)

= 70.5

0.210

cF

12.3

intersection,

= 0.062 kg phenol/m 3 The percent of

0.106 kg phenol/kg carbon and c phenol extracted

The isotherm from

At the

12.1-1 is also plotted in Fig. 12.2-1.

DESIGN OF FIXED-BED ADSORPTION COLUMNS Introduction and Concentration Profiles

A widely

used method for adsorption of solutes from liquid or gases employs a fixed bed of granular particles. The fluid to be treated is usually passed down through the packed bed at a constant flow rate. The situation is more complex than that for a simple stirred-tank batch process

which reaches equilibrium. Mass-transfer resistances are

important in the fixed-bed process and the process

dynamics of the system determines the

is

unsteady

state.

The

overall

efficiency of the process rather than just the

equilibrium considerations.

The concentrations of the solute in the fluid phase and of the solid adsorbent phase change with time and also with position in the fixed bed as adsorption proceeds. At the inlet to the bed the solid is assumed to contain no solute at the start of the process. As the fluid first contacts the inlet of the bed, most of the mass transfer and adsorption takes place here.

As

the fluid passes thrcugh the bed, the concentration in this fluid

drops very rapidly with distance in the bed to zero reached. The concentration

where the concentration

way

before the end of the bed

shown bed length. The

profile at the start at time fj is

ratio clc a

tion c 0 is the feed concentration

is

plotted versus

and c

is

is

in Fig. 12.3- la, fluid

concentra-

the fluid concentration at a point in the bed.

After a short time, the solid near the entrance to the tower is almost saturated, and most of the mass transfer and adsorption now takes place at a point slightly farther from the inlet. At a later time t 2 the profile or mass-transfer zone where most of the

Sec. 12.3

Design of Fixed-Bed Adsorption Columns

701

FIGURE

12.3-1.

Concentration profiles for adsorption at various positions

and times

a fixed bed: (a) profiles (b) breakthrough

in

the bed,

in

concentration profile in the fluid at outlet of bed.

moved farther down the bed. The concentration shown are for the fluid phase. Concentration profiles for the concentration of adsorbates on the solid would be similar. The solid at the entrance would be nearly saturated and this concentration would remain almost constant down to the masstransfer zone, where it would drop off rapidly to almost zero. The dashed line for time shows the concentration in the fluid phase in equilibrium with the solid. The f 3 concentration change takes place has profiles

difference in concentrations

12. 3B

is

the driving force for

mass

transfer.

Breakthrough Concentration Curve

As seen

Fig. 12.3-la, the major part of the adsorption at any time takes place in a narrow adsorption or mass-transfer zone. As the solution continues to flow, this mass-transfer zone, which is S-shaped, moves down the column. At a given time in Fig. 12.3-la when almost half of the bed is saturated with solute, the outlet f 3 in

relatively

concentration

is

still

approximately zero, as shown

in

Fig.

This outlet

12.3-1 b.

concentration remains near zero until the mass-transfer zone starts to reach the tower outlet at time

/ 4

.

Then

the outlet concentration starts to rise and at

concentration has risen to c b

,

After the break-point time point c d

,

which

is

which is

is

t

5

the outlet

called the break point.

reached, the concentration c rises very rapidly up to

the end of the breakthrough curve

where the bed

is

judged

The break-point concentration represents the maximum that can be discarded and is often taken as 0.01 to 0.05 for c b lc 0 The value c d lc 0 is taken as the

ineffective.

.

point where c d

702

is

approximately equal to c Q

Chap. 12

.

Liquid-Liquid and Fluid—Solid Separation Processes

For a narrow mass-transfer zone, the breakthrough curve is very steep and most is used at the break point. This makes efficient use of the adsorbent

of the bed capacity

and lowers energy costs

for regeneration.

Capacity of Column and Scale-Up Design Method

12.3C

The mass-transfer zone width and shape depends on

the adsorption isotherm, flow

and diffusion in the pores. A number of theoretical methods have been published which predict the mass-transfer zone and concentration profiles in the bed. The predicted results may be inaccurate because of many uncertainties due to flow patterns and correlations to predict diffusion and mass transfer. Hence, experiments in laboratory scale are needed in order to scale up the rate, mass-transfer rate to the particles,

results.

The total or stoichiometric capacity of the packed-bed tower, if the entire bed comes to equilibrium with the feed, can be shown to be proportional to the area between the curve and a line at clc 0 = .0 as shown in Fig. 12.3-2. The total shaded 1

area represents the total or stoichiometric capacity of the bed as follows (R6):

(12.3-1)

where

t,

is

the time equivalent to the total or stoichiometric capacity.

capacity of the bed up to the break-point time

t

b is the

The usable

crosshatched area.

(12.3-2)

where

t

u is

the time equivalent to the usable capacity or the time at which the effluent

concentration reaches close to that of

The

ratio

t

/

b

its

maximum

permissible level.

The value of

f„ is usually

very

.

u lt t is

the fraction of the total bed capacity or length utilized

break point (C3, LI, Ml). Hence, for a bed used up to the break point,

HB

total

=

bed length of

~H T

HT

m,

HB

is

up

to the

the length of

(12.3-3)

1

Sec. 12.3

Design of Fixed-Bed Adsorption Columns

703

The

H UNB in m

length of unused bed

is

=y~ jjH T

Hunb

H UNB

The

velocity and

H UNB

may,

is

then the unused fraction times the total length.

(12.3-4)

represents the mass-transfer section or zone.

therefore, be

measured

depends on the fluid column. The value of

It

essentially independent of total length of the

at the design velocity in

a small-diameter

Then the full-scale adsorber bed can be designed simply by first calculating the length of bed needed to achieve the required usable capacity, H B at the break point. The value of H B is directly proportional to b Then the length H UNB of the mass-transfer section is simply added laboratory column packed with the desired adsorbent.

,

t

to the length

HB

.

needed to obtain the

Hy = This design procedure

is

total length,

H UNB

+

widely used and

its

HT

.

HB

(12.3-5)

depends on the conditions

validity

the laboratory column beilig similar to those for the full-scale unit.

The

in

small-diameter

must be well insulated to be similar to the large-diameter tower, which operates The mass velocity in both units must be the same and the bed of sufficient length to contain a steady-state mass transfer zone (LI). Axial dispersion or axial mixing may not be exactly the same in both towers, but if caution is exercised, this method is a useful design method. unit

adiabatically.

An approximate areas

is

0.5 and

assume

to t

s

.

Then

alternative procedure to use instead of integrating and obtaining

that the breakthrough

the value of

below the curve between

t

b

/, in

and

t

s

curve

is

equal to

in Fig. 12.3-2 is

symmetrical

is

simply

ts

by

a packed bed having a diameter of 4 cm and containing 79.2 g of carbon. The inlet gas stream having a

activated carbon particles

cm

=

.

EXAMPLE 12.3-1. Scale-Up of Laboratory Adsorption Column A waste stream of alcohol vapor in air from a process was adsorbed length of 14

at clc 0

This assumes that the area the area above the curve between t s and

Eq. (12.3-1)

in

3

concentration c a of 600 ppm and a density of 0.00115 g/cm entered the bed at a flow rate of 754 cm'is. Data in Table 12.3-1 give the concentrations of the breakthrough curve. The break-point concentration is set at clc 0 (a)

=

0.01.

as follows.

Determine the break-point time, the fraction of total capacity used up to the break point, and the length of the unused bed. Also determine the saturation loading capacity of the carbon.

Table

704

Do

12.3-1.

Breakthrough Concentration for Example 12.3-1

Time, h

clc 0

0 3

Time, h

clc 0

0

5.5

0.658

0

6.0

0.903

3.5

0.002

6.2

0.933

4

0.030

6.5

0.975

4.5

0.155

6.8

0.993

5

0.396

Chap. 12

Liquid-Liquid and Fluid—Solid Separation Processes

(b) If the break-point time required for

new

total length of the

Solution:

The data from Table

=

a

new column

is

6.0 h, what

is

the

column required? 12.3-1 are plotted in Fig. 12.3-3.

For part

0.01 the break-point time is t b = 3.65 h from the graph. The value of t d is approximately 6.95 h. Graphically integrating, the areas are A) = 3.65 h and A 2 = 1.51 h. Then from Eq. (12.3-1), the time equivalent to the total or stoichiometric capacity of the bed is (a),

for clc 0

,

dt

= A, + A 7 =

3.65

+

1.51

=

5.16 h

The time equivalent to the usable capacity of the bed up to the break-point time is, using Eq. (12.3-2), f/»=

-J.

3.65

c\

r^ /

,=ii=

3.65 h

Hence, the fraction of total capacity used up to the break point is tjt, = 3.65/5.16 = 0.707. From Eq. (12.3-3) the length of the used bed is H B = 0.707(14) = 9.9 cm. To calculate the length of the unused bed from Eq. (12.3-4),

=

(1

- 0.707)14 =

4.1

cm

time, t{h)

Figure

Sec. 12.3

12.3-3.

Breakthrough

cur\'e

Design of Fixed-Bed Adsorption Columns

for Example 12.3-1.

705

For part

(b) for a

new

t

b

of 6.0 h, the

new H B

is

obtained simply from

the ratio of the break-point times multiplied by the old

HB

.

6.0

H B = -—(9.9)

=

cm

16.3

J.OJ

H T = H B + H UNB = We determine Air flow rate

=

cm 3 /s)(3600

(754

=

= Saturation capacity

3

s)(0.00115 g/cm

= 3122 g

)

air/h

/600 g alcohoA (3122 g air/h)(5.16

10" g

y

h)

air

J

9.67 g alcohol

= 9.67

g alcohol/79.2 g carbon

= 0.1220 fraction of the

= 20.4 cm

4.1

the saturation capacity of the carbon.

Total alcohol adsorbed

The

+

16.3

g alcohol/g

new bed used up

carbon

to the break point

is

now

16.3/20.4,

or 0.799.

In the scale-up

it

may be

necessary not only to change the column height, but also,

the actual throughput of fluid might be different from that used in the experimental

laboratory unit. Since the mass velocity

in

must remain constant

the bed

for scale-up,

the diameter of the bed should be adjusted to keep this constant.

Typical gas adsorption systems use heights of fixed beds of about 0.3 with downflow of the gas.

m

to 1.5

Low superficial gas velocities of 15 to 50cm/s (0.5 to

are used. Adsorbent particle sizes range from about 4 to 50

mesh

m

1.7 ft/s)

(0.3 to 5

mm).

Pressure drops are low and are only a few inches of water per foot of bed. The adsorption time in

the bed

12. 3D

is

is

about 0.5 h up to 8

about 0.3 to 0.7 cm/s

h.

For liquids the superficial velocity of the liquid

(4 to 10 gpm/ft

2 ).

Basic Models to Predict Adsorption

most important method used for this process. A fixed filled or packed with the adsorbent Adsorbers are mainly designed using laboratory data and the methods

Adsorption

in fixed

beds

is

the

or packed bed consists of a vertical cylindrical pipe particles.

in Section 12. 3C. In this section the basic equations are described for isothermal adsorption so that the fundamentals involved in this process can be better

described

understood.

An

unsteady-state solute material balance

in the fluid is as

follows for a section dz

length of bed.

e

dc — + dt

706

2

(1

- e)p D p

Chap. 12

dq dc — = -v — + E—

d C x

dt

dz

dz

2

(12.3-6)

Liquid-Liquid and Fluid—Solid Separation Processes

where

e

is

the external void fraction of the bed; v

bed, m/s; p p

is

density of particle, kg/m

3 ;

and

is

superficial velocity in the

empty

E is an axial dispersion coefficient, m 2 /s.

first term represents accumulation of solute in the-liquid. The second term is accumulation of solute in the solid. The third term represents the amount of solute flowing in by convection to the section di of the bed minus that flowing out. The last term represents axial dispersion of solute in the bed, which leads to mixing of the

The

solute and solvent.

The second differential equation needed to describe this process relates the second term of Eq. (12.3-6) for accumulation of solute in the solid to the rate, of external mass transfer of the solute from the bulk solution to the particle and the diffusion and adsorption on the internal surface area. The actual physical adsorption is very rapid. The third equation is the equilibrium isotherm. There are many solutions to these three equations which are nonlinear and coupled. These solutions often do not fit experimental results very well and are not discussed here.

12.3E

Processing Variables and Adsorption Cycles

Large-scale adsorption processes can be divided into two broad classes. The

most important

is

first

and

the cyclic batch system, in which the adsorption fixed bed

is

and then regenerated in a cyclic manner. The second is a continuous flow system which involves a continuous flow of adsorbent countercurrent alternately saturated

to

a flow of feed.

There are four basic methods in common use for the cyclic batch adsorption system using fixed beds. These methods differ from each other mainly in the means used to regenerate the adsorbent after the adsorption cycle. In general, these four basic methods operate with two or sometimes three fixed beds in parallel, one in the adsorption cycle, and the other one or two in a desorbing cycle to provide continuity of flow. After a bed has completed the adsorption cycle, the flow is switched to the second newly regenerated bed for adsorption. The first bed is then regenerated by any of the following methods.

1.

Temperature-swing cycle. This is also called the thermal-swing cycle. The spent adsorption bed is regenerated by heating it with embedded stream coils or with a hot purge gas stream to remove the adsorbate. Finally, the bed must be cooled so that it

can be used for adsorption

in

the next cycle.

The time

for regeneration is generally

a few hours or more. 2.

desorbed by reducing the pressure at low pressure with a small fraction of the product stream. This process for gases uses a very short cycle time for regeneration compared to that for the temperature-swing cycle.

Pressure-swing cycle. In this case the bed

is

essentially constant temperature and then purging the bed at this

3.

Inert-purge gas stripping cycle. In this cycle the adsorbate

is

removed by passing

a nonadsorbing or inert gas through the bed. This lowers the partial pressure or

concentration around the particles and desorption occurs. Regeneration cycle times are usually only a 4.

few minutes.

Displacement-purge cycle. The pressure and temperature are kept essentially constant as in purge-gas stripping, but a gas or liquid is used that is adsorbed more

Sec. 12.3

Design of Fixed-Bed Adsorption Columns

707

strongly than the adsorbate and displaces the adsorbate. Again, cycle times are

usually only a few minutes.

Steam

stripping

often used in regeneration of solvent recovery systems using

is

activated carbon adsorbent. This can be considered as a combination of the temper-

ature-swing cycle and the displacement-purge cycle.

ION-EXCHANGE PROCESSES

12.4

12.1 A

Introduction and Ion-Exchange Materials

Ion-exchange processes are basically chemical reactions between ions

in solution and exchange so closely adsorption that for the majority of engineering purposes ion

The techniques used

ions in an insoluble solid phase.

resemble those used

in

in ion

exchange can be considered as a special case of adsorption. In ion exchange certain ions are removed by the ion-exchange solid. Since electroneutrality must be maintained, the solid releases replacement ions to the solution. The first ion-exchange materials were natural-occurring porous sands called 2+ zeolites and are cation exchangers. Positively charged ions in solution such as Ca + diffuse into the pores of the solid and exchange with the Na ions in the mineral.

Ca 2+ + Na2 fl

^

CaR + 2Na + (12.4-1)

(solution)

R

where the

Almost

left.

all

NaCl

(solution)

(solid)

represents the solid. This

is

the basis for "softening" of water.

To

added which drives the reversible to reaction above of these inorganic ion-exchange solids exchange only cations.

regenerate, a solution of the

(solid)

is

Most present-day ion-exchange

solids are synthetic resins or polymers. Certain

synthetic polymeric resins contain sulfonic, carboxylic, or phenolic groups.

These

anionic groups can exchange cations.

Na +

+

KR

H+

+

^± Nafi

(12.4-2) (solution)

Here the with

H

+

R

(solid)

represents the solid resin.

(solution)

(solid)

The Na +

in

the solid resin can be exchanged

or other cations.

Similar synthetic resins containing amine groups can be used to exchange anions

and

OH"

in solution.

CI"

+

/?NH 3 OH

^

KNH3CI

+

OH" .

(solution)

12. 4B

Equilibrium Relations

(solid)

in

(solid)

Ion Exchange

The ion-exchange isotherms have been developed using

the law of

example, for the case of a simple ion-exchange reaction such as Eq.

708

(12.4-3)

(solution)

Chap. 12

mass

action.

(12.4-2),

For

HR and

Liquid-Liquid and Fluid—Solid Separation Processes

Nai? represent the ion-exchange +

sodium ion Na It is assumed + or Na At equilibrium,

sites

that

.

all

on

the resin filled with a proton

of the fixed

number of sites

H+

and a

are filled with

H+

.

[Na*][H + j

K = rvT

+,

(12.4-4)

„.,

rTr [Na + ][Kfl]

Since the total concentration of the ionic groups [K] on the resin

=

[R]

Combining Eqs.

(12.4-4)

and

constant

=

[Nai?]

K[iR][Na

=

[KLR]

(12.4-5)

is

buffered so [H

+

or adsorption

is

similar to the

Langmuir isotherm.

]

is

+ ]

(12 4 " 6) -

[Hn + ^[Na-]

the solution

constant, the equation above for sodium exchange

Design of Fixed-Bed Ion-Exchange Columns

12.4C

The

fixed (B7),

(12.4-5),

[Na/?]

If

+

is

exchange depends on mass transfer of ions from the bulk solution to the exchange the surface, and diffusion of the exchange ions back to the bulk solution.

rate of ion

particle surface, diffusion of the ions in the pores of the solid to the surface,

of the ions at

This

similar to adsorption.

is

those for adsorption. similar

The

and are described

The

differential equations derived are also

very similar to

design methods used' for ion exchange and adsorption are in

Section 12.3 for adsorption processes.

SINGLE-STAGE LIQUID-LIQUID EXTRACTION PROCESSES

12.5

Introduction to Extraction Processes

12.5A

In order to separate

one or more of the components in a mixture, the mixture is The two-phase pair can be gas-liquid, which was dis-

contacted with another phase. cussed

in

Chapter

10; vapor-liquid,^which

was covered

in

Chapter 11; liquid-liquid; or processes are considered

fluid-solid. In this section liquid— liquid extraction separation first.

Alternative terms are liquid extraction or solvent extraction. In distillation the liquid

vapor.

The

is

partially vaporized to create another phase,

which

is

a

separation of the components depends on the relative vapor pressures of the

The vapor and liquid phases are similar chemically. In liquid-liquid extractwo phases are chemically quite different, which leads to a separation of the components according to physical and chemical properties. Solvent extraction can sometimes be used as an alternative to separation by distillation or evaporation. For example, acetic acid can be removed from water by distillation or by solvent extraction using an organic solvent. The resulting organic solvent-acetic acid solution is then distilled. Choice of distillation or solvent extraction would depend on relative costs (C7). In another example high-molecular-weight fatty acids can be

substances. tion the

Sec. 12.5

Single-Stage Liquid-Liquid Extraction Processes

709

separated from vegetable

which

distillation,

is

oils

by extraction with liquid propane or by high-vacuum

more expensive.

In the pharmaceutical industry products such as penicillin occur in fermentation

mixtures that are quite complex, and liquid extraction can be used to separate the

Many metal separations are being done commercially by extraction of aqueous solutions, such as copper-iron, uranium-vanadium, and tantalum-columbium. penicillin.

Equilibrium Relations in Extraction

12.5B

1.

Phase

rule.

Generally in a liquid-liquid system we have three components, A, B, and

C, and two phases in equilibrium. Substituting into the phase rule, Eq. (10.2-1), the

number of

degrees of freedom

The

3.

is

variables are temperature, pressure,

and four

concentrations. Four concentrations occur because only two of the three mass fraction

concentrations in a phase can be specified. total to

1

.0,

x A + xB + x c

=

1

.0. If

The

third

must make the

total

mass

pressure and temperature are set, which

is

fractions

the usual

case, then, at equilibrium, setting one concentration in either phase fixes the system.

2.

Triangular coordinates and equilibrium data.

Equilateral triangular coordinates are

often used to represent the equilibrium data of a three-component system, since there are

three axes. This

component, A,

is

B,

shown

in Fig. 12.5-1.

or C.

The

point

perpendicular distance from the point of

C

in the

mixture at

distance to base

M,

A common is

shown

Each of

the distance to base

+

xB

the three corners represents a pure

represents a mixture of A, B,

+ xc =

CB

the

mass

fraction

0.40

+

0.20

+

0.40

0

The c

x A of A, and the

=

(12-5-1)

1.0

A

and B are

partially

Typical examples are methyl isobutyl ketone (/l)-water

\o.

1.0. .

C.

of B. Thus,

phase diagram where a pair of components in Fig. 12.5-2.

and

M to the base AB represents the mass fraction x

AC the mass fraction x B xA

miscible

M

0.2

0.4

0.8

0.6

1.0

(B)

Mass fraction

Figure

710

12.5-1.

Chap. 12

B

Coordinates for a triangular diagram.

Liquid-Liquid and Fluid-Solid Separation Processes

(B)-acetone (Q, water (y4)-chloroform (B)-acetone (C), and benzene (/4)-water (B)-acetic acid (C). Referring to Fig. 12.5-2, liquid C dissolves completely in A or in E. Liquid A is

only slightly soluble

in

B and £

slightly soluble in A.

The two-phase

M

lines are also

3.

shown. The two phases are identical

Equilibrium data on rectangular coordinates.

is

included

will

separate

region

below the curved envelope. An original mixture of composition into two phases a and b which are on the equilibrium tie line through point inside

M. Other

tie

at point P, the Plait point.

Since triangular diagrams have

some

disadvantages because of the special coordinates, a more useful method of plotting the three

component data

is

to use rectangular coordinates.

shown in Fig. 12.5-3 for Data are from the partially miscible. The con-

This

is

the system acetic acid (,4)-water (B)-isopropyl ether solvent (C).

Appendix A.3

The solvent pair B and C are component C is plotted on the vertical axis and that of A on the The concentration of component B is obtained by difference from Eqs.

for this system.

centration of the horizontal

axis.

(12.5-2) or (12.5-3).

The two-phase

xB

=

1.0

- xA -

xc

(12.5-2)

yg

=

1.0

- yA -

yc

(12.5-3)

region in Fig. 12.5-3

is

A

and the

ether-rich solvent layer g, called the extract layer.

tie

line gi is

shown connecting

designated by x and the extract by

y.

and the one-phase region

inside the envelope

outside.

the water-rich layer

i,

The

called the raffinate layer, is

C is

designated as y c in construct the tie line gi using the

the extract layer and as x c in the raffinate layer. To equilibrium y A — x A plot below the phase diagram, vertical lines to g

EXAMPLE

composition

raffinate

Hence, the mass fraction of

and

/

are drawn.

Material Balance for Equilibrium Layers kg and containing 30 kg of isopropyl ether (C), 10 kg of acetic acid {A), and 60 kg water [B) is equilibrated and the equilibrium phases separated. What are the compositions of the two equilibrium phases?

An

12J-1.

original mixture weighing 100

The composition of the original mixture is x c = 0.30, x A = 0.10, and x B = 0.60. This composition of x c = 0.30 and x A = 0.10 is plotted as point h on Fig. 12.5-3. The tie line gi is drawn through point h by trial and error. The composition of the extract (ether) layer at g is y A = 0.04, y c = 0.94, and y„ = 1.00 - 0.04 - 0.94 = 0.02 mass fraction. The raffinate — (water) layer composition at is x A = 0.12, x c = 0.02, and x B = 1.00

Solution:

i

0.12

-

0.02

=

0.86.

C

one-phase region equilibrium

tie line

two-phase region

B

A Figure

12.5-2.

Liquid-liquid phase diagram where components

A and B are

partially

miscible.

Sec. J2.5

Single-Stage Liquid-Liquid Extraction Processes

711

C

(ether)

tie line

extract layer,

yc

yA

vs.

two-phase region

one-phase region

raffinate layer,

Xq

vs.

A 0.5

Mass fraction

(acetic acid)

1.6

A

{.x

A

,

yA

)

—

/

A-*

o X

4-

CD

a
15° line

1

o 1

>> _£3 l-i

/

O

%

O,

E (S

/

CO

/

/

/

-

-eq uilibrium

line

1.0

0.5

Water layer (raffinate) composition, x^ Figure

12.5-3.

Acetic acid (A)-water {B)-isopropyl ether (C) liquid-liquid phase

diagram

at

K {20° Q.

293

common type of phase diagram C and also A and C are partially

Another pairs

B and

is

shown

miscible.

in Fig. 12.5-4,

Examples

where the solvent

are the system styrene

and the system chlorobenzene (/4)-methyl

(/4)-ethylbenzene (B)-diethylene glycol (C) ethyl ketone (6>-water(C).

12.5C /

.

Single-Stage Equilibrium Extraction

Derivation of lever-arm rule for graphical addition.

This

will

be derived for use in

the rectangular extraction-phase-diagram charts. In Fig. 12.5-5atwo streams,

V

L kg and

components A, B, and C, are mixed (added) to give a resulting mixture stream kg total mass. Writing an overall mass balance and a balance on A, kg, containing

M

V +

L= M

Vy A + Lx A = where x A!tl

is

the

mass fraction of A

in the

Chap. 12

(123-5)

M stream. Writing a balance for component C,

Vy c + Lx c =

712

Mx AM

(12.5-4)

MxCM

(123-6)

Liquid-Liquid and Fluid—Solid Separation Processes

-one-phase region -tie line

two-phase region

one-phase region

Figure

12.5-4.

Phase diagram where

the solvent pairs

B-C and A-C

are partially

miscible.

Combining Eqs.

Combining Eqs.

(12.5-4)

(12.5-4)

and

and

(12.5-5),

(123-7)

V

XJ W

^

yc

V

X CM

X

A

(12.5-6),

-x -

c

(123-8)

XC

Equating Eqs. (12.5-7) and (12.5-8) and rearranging,

XC

~

XA This shows that points L,

X CM

_ X CM ~

yc

X AM

X AM

~

yA

M, and V must

lie

(123-9)

on a straight

line.

By using

the properties of

similar right triangles,

L

(kg)

V

(kg)

m

a

>

M, x AM X CM L,

VM

C o u

y A yc

v,

_ "

(123-10)

xc X CM

CO

S

yc

xA x c ,

yA X AM X A

(a)

Mass fraction

A

(b)

Figure

12.5-5.

Graphical addition and lever-arm rule

:

(a)

process flow,

{b)

graphical

addition.

Sec. 12.5

Single-Stage Liquid-Liquid Extraction Processes

713

This

V

and states that kg L/kg

the lever-arm rule

is

is

equal to the length of line

KM/length ofline LM. Also, L(kg)

.

=

VM

=

(123-11)

These same equations also hold for kg mol and mol frac,lb m , and so on.

EXA MPLE

Amounts of Phases in Solvent Extraction two equilibrium layers in Example 12.5-1 are for the extract layer (V) y A = 0.04, y B = 0.02, and y c = 0.94, and for the raffinate layer (L) x A =0.12, x B = 0.86, and x c = 0.02. The original mixture contained 100 kg andx,, M = 0.10. Determine the amounts of V and L. 125-2.

The compositions of

Solution:

the

Substituting into Eq. (12.5-4),

M = 100 where M = 100 kg and x^ = 0.10,

V+L= Substituting into Eq. (12.5-5),

K(0.04)

+ U0A2) =

100(0.10)

Solving the two equations simultaneously, L = 75.0 and V = 25.0. Alternatively, using the lever-arm rule, the distance hg in Fig. 12.5-3 is measured as 4.2 units and gi as 5.8 units. Then by Eq. (12.5-1 1),

L _hg

L

M~ Solving,

L =

72.5

kg and V =

100

27.5 kg,

4.2 5.8

^7

which

is

a reasonably close check on

the material-balance method.

2.

Single-stage equilibrium extraction.

We now

study the separation of

A

from a mix-

A and B by a solvent C in a single equilibrium stage. The process is shown in Fig. 12.5-6a, where the solvent, as stream V 2 enters and also the stream L 0 The streams are mixed and equilibrated and the exit streams L, and V] leave in equilibrium with

ture of

,

.

each other.

The equations

714

for this process are the

Chap.

12

same

as those given in Section 10.3 for a single

Liquid-Liquid and Fluid-Solid Separation Processes

equilibrium stage where y represents the composition of the streams. l

M

+

ViYai

L 0 + V2 = L + V = t

=

L 0 xA o + ViYai

V

streams and x the

L

(123-12)

=

Afx^,

(123-13)

U*co + ^y C 2 = -Mci + ^yci = Mx CM

(123-14)

^i*4i

Since x A + x B + x c = 1.0, an equation for B is not needed. To solve the three equations, the equilibrium-phase diagram in Fig. 12.5-6b is used. Since the amounts and compo-

L 0 and V2 are known, we can calculate values of M,x AM andxCM from Eqs. The points L 0 V2 and M can be plotted as shown in Fig. 12.5-6b. Then using trial and error a tie line is drawn through the point M, which locates the compositions of L and V The amounts of L and V can be determined by substitution sitions of

,

(12.5-12)-(12.5-14).

1

{

.

l

l

by using the lever-arm

into Eqs. (12.5-12)—(12.5-14) or

rule.

EQUIPMENT FOR LIQUID-LIQUID EXTRACTION

12.6

12.6A

As

,

,

Introduction and Equipment Types

in the separation processes

liquid extraction

of absorption and distillation, the two phases

must be brought

in

liquid-

into intimate contact with a high degree of turbulence

order to obtain high mass-transfer rates. After this contact of the two phases, they must be separated. In absorption and in distillation, this separation is rapid and easy because of the large difference in density between the gas or vapor phase and the liquid phase. In solvent extraction the density difference between the two phases is not large and separation is more difficult. There are two main classes of solvent-extraction equipment, vessels in which mechanical agitation is provided for mixing, and vessels in which the mixing is done by the flow of the fluids themselves. The extraction equipment can be operated batchwise or operated continuously as in absorption and in distillation. in

12.6B

To

Mixer-Settlers for Extraction

provide

efficient

mass

mechanical mixer

transfer, a

contact of the two liquid phases.

One phase

is

is

often used to provide intimate

usually dispersed into the other in the form

of small droplets. Sufficient time of contact should be provided for the extraction to take place. Small droplets

produce large

interfacial areas

and

faster extraction.

droplets must not be so small that the subsequent settling time in the settler

The design and power requirements of discussed

in detail in

the mixer or agitator

Section is

3.4.

baffled

from the

is

too large.

have been shown, where

agitators or mixers

In Fig. 12.6- la a typical mixer-settler

entirely separate

However, the

settler.

The

feed of

is

aqueous phase and

organic phase are mixed in the mixer, and then the mixed phases are separated in the settler.

In Fig. 12.6- lb a

combined mixer-settler

is

shown, which

is

sometimes used

in

extraction of uranium salts or copper salts from aqueous solutions. Both types of mixer-settlers can be used in series for countercurrent or multiple-stage extraction.

12.6C

Plate and Agitated

As discussed similar

Sec. 12.6

types

in

Tower Contactors

for Extraction

Section 10.6 for plate absorption and distillation towers,

of devices

are

used

for

liquid-liquid

Equipment for Liquid-Liquid Extraction

contacting. In

Fig.

somewhat 12.6-2a a

715

Figure

Typical- mixer-settlers for extraction: (a) separate mixer-settler,

12.6-1.

(b)

combined mixer-settler.

perforated-plate or sieve-tray extraction tower light solvent liquid are dispersed.

are then re-formed

on each

is

shown wherein

The dispersed

the rising droplets of the

tray by passing through the perforations.

liquid flows across each plate,

where

it is

and The heavy aqueous

droplets coalesce below each tray

contacted by the rising droplets and then passes

through the downcomer to the plate below. In Fig. 12.6-2b an agitated extraction tower

mounted on a agitator

is

is

shown.

A

series

of paddle agitators

central rotating shaft provides the agitation for the

two phases. Each

separated from the next agitator by a calming section of wire mesh to

encourage coalescence of the droplets and phase separation. This apparatus is essentially a series of mixer-settlers one above the other (C8, PI, Tl). Another type is the Karr reciprocating-plate column, which contains a series of sieve trays with a large

of

60% where

types of extraction

12.6D

open area

moved up and down (C6, C8, L3). This is one of the few towers that can be scaled up with reasonable accuracy (C6, K3).

the plates are

Packed and Spray Extraction Towers

Packed and spray tower extractors give differential contacts, where mixing and settling proceed continuously and simultaneously (C8). In the plate-type towers or mixersettler contactors, the extraction

and

settling

proceeds

the heavy liquid enters the top of the spray tower,

phase, and flows out through the bottom.

The

in definite stages.

fills

In Fig. 12.6-3

the tower as the continuous

light liquid enters

through a nozzle

bottom which disperses or sprays the droplets upward. The light coalesces at the top and flows out. In some cases the heavy liquid is sprayed

distributor at the liquid

downward

into a rising light continuous phase.

A more intervals

tower is to pack the column with packing such as Raschig which cause the droplets to coalesce and redisperse at frequent

effective type of

rings or Berl saddles,

throughout the tower. Detailed discussions of flooding and construction of

packed towers are given elsewherefTl, PI).

12.7

12.7A

CONTINUOUS MULTISTAGE COUNTERCURRENT EXTRACTION Introduction

was used to transfer the solute A from more solute, the single-stage contact can

In Section 12.5 single-stage equilibrium contact

one liquid

716

to the other liquid phase.

Chap.

12

To

transfer

Liquid-Liquid and Fluid-Solid Separation Processes

heavy liquid

rising

drops

of light solvent

coalesced solvent

m

— (b)

Figure

12.6-2.

Extraction towers:

(a)

perforated-plate or sieve-tray tower, (b) agi-

tated extraction tower.

be repeated by contacting the exit

way a

stream with fresh solvent

greater percentage removal of the solute

A

is

V2

in Fig. 12.5-6. In this

obtained. However, this

the solvent stream and also gives a dilute product of

A

streams. In order to use less solvent and to obtain a

more concentrated

stream, countercurrent multistage contacting

Many

is

is

wasteful of

in the outlet solvent extract

exit extract

often used.

of the fundamental equations of countercurrent gas absorption and of rec-

tification are the

same

or similar to those used in countercurrent extraction. Because of

two liquid phases in each other, the equilibrium more complicated than in absorption and distillation.

the frequently high solubility of the

relationships in extraction are

light liquid outlet

heavy liquid

*

0mB&\ oV ° o o

1

o

o

°o

° o°o ° O o ° O o o

^

O O O O O o o° o oo O Oo°o light liquid

Figure

Sec. 12.7

J

12.6-3.

U U U

-coalesced interface

-rising droplets

-spray nozzle

Spray-type extraction lower.

Continuous Multistage Countercurrent Extraction

717

Continuous Multistage Countercurrent Extraction

12.7B

/.

The process flow for and is shown in Fig.

Countercurrent process and overall balance.

process

is

the

same

as given previously in Fig. 10.3-2

A

stream containing the solute

to be extracted enters at

The

solvent stream enters at the other end.

and the

currently from stage to stage, stage

1

and the

stream

raffinate

Making an

overall balance

all

the

M represents total kg/h (lbjjli) and

the inlet solvent flow rate in kg/h,

^o x co

Combining Eqs.

(12.7-1)

+

and

a constant,

is

V

x

LN

the exit

C,

— L N x CN + V yCi — Mx CM

(12.7-2)

1

and rearranging,

(12.7-2)

i

i

A similar

leaving

flow rate in kg/h,

0 the inlet feed

the exit extract stream, and

L 0 Xco + ^N+iycN+ + Vn +

'cm

x

(12.7-1)

component balance on component

^n + i^cn+i

V

M

l

VN+l

overall

raffinate streams flow counter-

N.

LN + V =

where

Making an

feed

N stages,

L 0 + VN

raffinate stream.

The

one end of the process and the

products are the extract stream

final

L N leaving stage on

and

extract

this extraction

12.7-1.

L N x cs + V^ci L N +V,

(12.7-3)

balance on component A gives

X AKf

—

c /io I-oX,,

+

^n+iYan +

i

L n x an + V y Al i

(12.7-4)

Equations (12.7-3) and (12.7-4) can be used to calculate the coordinates of point

M

on the phase diagram that ties together the two entering streams L 0 and Kv+1 and the two exit streams V and L N Usually, the flows and compositions ofL 0 and VN+ are known and the desired exit composition x AN is set. If we plot pointsL 0 VN+ „ and as in Fig. 12.7-2, a straight line must connect these three points. Then L N M, and V must must also lie on the phase envelope, as shown. These lie on one line. Also, L N and balances also hold for lb ra and mass fraction, kg mol and mol fractions, and so on. ,

.

l

M

,

,

l

EXAMPLE

12.7-1. Material Balance for Countercurrent Stage Process Pure solvent isopropyl ether at the rate of VN+ = 600 kg/h is being used to extract an aqueous solution of L 0 = 200 kg/h containing 30 wt % acetic acid (A) by countercurrent multistage extraction. The desired exit acetic acid concentration in the aqueous phase is 4%. Calculate the compositions and amounts of the ether extract V and the aqueous raffinate L N Use equilibrium data from Appendix A. 3. .

l

The given values

Solution:

=

200,

0.30,

are

xK

VN+ = 600, y AN+ = 0, yCN + = = 0, and X AN = 0.04. In i

t

j

1.0,

L0 =

=

extract "

solvent r

1

r

2

'N

3

n

2

1

VN+l

N

L2

LN

feed

raffinate

Figure

718

12.7-1.

Couniercurrcnl-mullistage-extr action-process flow diagram.

Chap. 12

Liquid-Liquid and Fluid-Solid Separation Processes

Figure

Use of

12.7-2.

the mixture point

M

c m vN+1

for overall material balance in counter current solvent extraction.

XAM xa y a •

and L 0 are at x AN

-

plotted. Also, since

0.04.

For

LN

is

on the phase boundary,

the mixture point

M,

it

can be plotted

substituting into Eqs. (12.7-3) and

(12.7-4),

L 0 xco + VN+l y CN+i

200(0)

+ 600(1.0) + 600

0.75

(12.7-3)

200

L 0 x a0 + VN+ iyAN+ L 0 + VN

AM

,

+ 600(0) = 0.075 + 600

200(0.30)

200

(12.7-4)

M M

Using these coordinates, the point is plotted in Fig. 12.7-3. We locate Vl by drawing a line from L N through and extending it until it intersects the phase boundary. This gives y Al = 0.08 and y ci = 0.90. For L N a value of x CN = 0.017 is obtained. By substituting into Eqs. (12.7-1) and (12.7-2) and solving, L N = 136 kg/h and K, = 664 kg/h. 2.

The next step after an go stage by stage to determine the concentrations at number of stages N needed to reach L N in Fig. 12.7-1.

Stage-to-stage calculations for countercurrent extraction.

overall balance has

been made

each stage and the total

C

is

to

(ether)

X CM

O 6

H

0

B

(water)

Figure

Sec. 12.7

12.7-3.

L0

X AM Method

0.5

1.0

A

(acetic acid)

xa. yA to

perform overall material balance for Example

Continuous Multistage Countercurrent Extraction

1 2.7-1.

719

Making a

balance on stage

total

1,

+V =L +V

L0 Making a

similar balance

on stage

2

l

n,

= L„+

Ln _ + KB+1 l

Rearranging Eq. (12.7-5)

to obtain the difference i-o

This value of A in kg/h

l

A = L0 - V =

L„

x

This also holds

A

(12.7-6)

V„

in flows,

- V = L - V2 = A

(12.7-7)

f

constant and for

is

(12.7-5)

l

all

stages,

- Vn+ = L N - FN+1 =

•

(12.7-8)

•

,

a balance on component A, B, or C.

for

=

Ax A = L 0 x 0 - V y = L„x„ - Vn+l y n+1 = L N x N - FN+1 y„ +1 l

Combining Eqs.

(12.7-8)

where x A

is

and

LoX 0

x *=

l

L ^0

(12.7-9)

~V y l

and solving forx A

=

_ K ¥

——

Lx —

l

L

\

V„+iy„ +

i

_ KK n+1

(12.7-9)

,

= L s x s — VN+l y N+1

,„,,., (12 7 " 10) -

y N+l

N

the x coordinate of point A.

Equations (12.7-7) and

L0 = A

(12.7-8)

V

4-

can be written as L„

l

= A + Vn+

L N = A + VN+l

(12.7-11)

From Eq. (12.7-11), we see that L 0 is on a line through A and V L„ is on a line A and Vn+ ,, and so on. This means A is a point common to all streams passing each. other, such as L 0 and V L n and V„ +l L N and VN + and so on. The coordinates to y

,

through

y,

locate this

points

A

,

l

,

operating point are given for x CA and x^ A in Eq. (12.7-10). Since the end

VN+l L N

or

,

Alternatively, the

L 0 V and L N VN+l

A

Vu and L 0 point

is

are

known, x A can be calculated and point A

located.

located graphically in Fig. 12.7-4 as the intersection of lines

.

x

In order to step off the

the line

L 0 A, which

Figure

12.7-4.

locates

Operating

number of

we

number of stages using Eq.

(12.7-1

1)

V on

Next a

tie

y

point

the phase boundary.

A

start at

line

L 0 and draw

through

K, locates

and

theoretical stages

needed for countercurrent extraction.

720

Chap. 12

Liquid-Liquid and Fluid-Solid Separation Processes

L

l7

which

in

is

equilibrium with

V Then lineL t A

drawn. This stepwise procedure

reached The number of stages

N

y

.

is

drawn giving V2 The .

tie line

V2 L 2 LN

repeated until the desired raffinate composition

is

is is

obtained to perform the extraction.

is

EXAMPLE

12.7-2. Number of Stages in Countercurrent Extraction Pure isopropyl ether of 450 kg/h is being used to extract an aqueous solution of 150 kg/h with 30 wt % acetic acid (A) by countercurrent multistage extraction. The exit acid concentration in the aqueous phase is 1 wt %. Calculate the number of stages required.

values are Fv+1 =450, y AN+ i=0, y C N+i = l-0, = 0.30, x B0 = 0.70, xco = 0, and x AN = 0.10. The points 150, x^o VN+l L0 and L N are plotted in Fig. 12.7-5. For the mixture point M, substituting into Eqs. (12.7-3) and (12.7-4),x CM = 0.75 andx^ M = 0.075. The

The known

Solution:

L0 =

,

,

M

point is plotted and Vx is located at the intersection of lineL^M with the phase boundary to give y A1 = 0.072 and y ci = 0.895. This construction is not shown. (See Example 12.7-1 for construction of lines.)

L 0 V and L N Vs+l are drawn and the intersection is the A as shown. Alternatively, the coordinates of A can be calculated from Eq. (12.7-10) to locate point A. Starting at L 0 we draw line in equilibrium L 0 A, which locates Vj. Then a tie line through V locates with V (The tie-line data are obtained from an enlarged plot such as the bottom of Fig. 12.5-3.) Line L A is next drawn locating V2 A tie line through V2 gives L 2 A line L 2 A gives V3 A final tie line gi vesL 3 which has gone beyond the desired L N Hence, about 2.5 theoretical stages are needed. The

lines

l

operating point

1

l .

.

t

.

.

,

.

3.

Minimum

solvent rate.

If

a solvent rate

VN+

is

,

selected at too

low a value a limiting

xA yA .

Figure

Sec. 12.7

L2.7-5.

Graphical solution for countercurrent extraction

Continuous Multistage Countercurrent Extraction

in

Example

12.7-2.

721

A and a tie line being the same. Then an infinite needed to reach the desired separation. The minimum amount of

case will be reached with a line through

number solvent

of stages will be is

reached. For actual operation a greater

The procedure

L0

point

to obtain this

minimum

(Fig. 12.7-4) to intersect the

is

amount

extension of lineL w

must be used. drawn through Other tie lines to the left

of solvent

as follows.

A

VN +

t

tie line is .

drawn including one through L H to intersect the line L N VN+l The intersection of a tie line on line L N VN+l which is nearest to VN+l represents the A min point for minimum solvent. The actual position of A used must be closer to VN+l than A min for a finite number of stages. This means that more solvent must be used. Usually, the tie line through L 0 represents the A min of this

tie line

are

.

.

Countercurrent-Stage Extraction with Immiscible Liquids

12.7C

the solvent stream VN+l contains components A and C and the feed stream L 0 contains A and B and components B and C are relatively immiscible in each other, the stage calculations are made more easily. The solute A is relatively dilute and is being transIf

ferred

from

L0

to

KN+

,.

and making an

Referring to Fig. 12.7-1

and then over the

first

overall balance for

A

over the whole system

n stages,

(12.7-12)

(12.7-13)

where L = kg inert B/h, V = kg inert C/h, y = mass fraction A in V stream, and = mass fraction A in L stream. This Eq. (12.7-13) is an operating-line equation whose slope = LjV. If y and x are quite dilute, the line will be straight when plotted on an xy x

diagram.

The number of stages

are stepped off as

shown

previously in cases in distillation and

absorption. If

the equilibrium line

is

straight, the analytical Eqs.

calculate the

number

relatively dilute, then since the operating line ( 1

is

essentially

0.3-2 1 )—( 10.3-26) given in Section 10.3D can be used to

of stages.

EXAMPLE

12.7-3. Extraction of Nicotine with Immiscible Liquids. water solution of 100 kg/h containing 0.010 wt fraction nicotine {A) in water is stripped with a kerosene stream of 200 kg/h containing 0.0005 wt fraction nicotine in a countercurrent stage tower. The water and kerosene are essentially immiscible in each other. It is desired to reduce the concentration of the exit water to 0.0010 wt fraction nicotine. Determine the theoretical number of stages needed. The equilibrium data are as follows (C5), with x the weight fraction of nicotine in the water solution and y in the

An

inlet

kerosene.

722

X

y

X

y

0.001010

0.000806

0.00746

0.00682

0.00246

0.001959

0.00988

0.00904

0.00500

0.00454

0.0202

0.0185

Chap. 12

Liquid-Liquid and Fluid-Solid Separation Processes

Solution:

The given

kg/h, yw +

=

1

L = U\ K'

=

-

V{1

Making an

values are

=

0.0005, x N

0.0010.

x)

= L0 (1 -

x 0)

y)

= KN+

-

t

(l

=

100(1

y N+

overall balance

L0 = The

,)

=

100 kg/h, x 0 = 0.010, streams are

VN +

1

=

200

inert

-

0.010)

=

99.0 kg water/hr

- 0.0005) =

200(1

on A using Eq.

199.9

(12.7-12)

and

kg kerosene/hr solving,

y

l

=

0.00497. These end points on the operating line are plotted in Fig. 12.7-6.

Since the solutions are quite dilute, the line is straight. The equilibrium line = 3.8 theois also shown. The number of stages are stepped off, giving

N

retical stages.

12.8

12.8A

/.

INTRODUCTION AND EQUIPMENT FOR LIQUID-SOLID LEACHING Leaching Processes

Introduction.

Many

biological

mixture of different components

and inorganic and organic substances occur in

in

a

a solid. In order to separate the desired solute

component from the solid phase, the solid is The two phases are in intimate contact and the solute or

constituent or remove an undesirable solute

contacted with a liquid phase.

solutes can diffuse from the solid to the liquid phase,

components originally in the leaching. The term extraction

solid. is

This process

2.

which causes a separation of the

called liquid-solid leaching or simply

also used to describe this unit operation, although

refers to liquid-liquid extraction. In

moved from a

is

leaching

solid with water, the process

is

when an

undesirable

component

it

also

is

re-

called washing.

Leaching processes for biological substances. In the biological and food processing many products are separated from their original natural structure by liquid-

industries,

solid leaching.

An important

process

water. In the production of vegetable

Sec. 12.8

Introduction

is

the leaching of sugar from sugar beets with hot

oils,

organic solvents such as hexane, acetone, and

and Equipment for Liquid-Solid Leaching

723

ether are used to extract the

oil

from peanuts, soybeans,

sunflower seeds, cotton seeds, tung meal, and halibut

many

industry,

flax seeds, castor beans,

livers.

In the pharmaceutical

different pharmaceutical products are obtained by leaching plant roots,

and stems. For the production of soluble "instant" coffee, ground roasted coffee is produced by water leaching of tea leaves. Tannin is removed from tree barks by leaching with water. leaves, is

3.

leached with fresh water. Soluble tea

Leaching processes for inorganic and organic materials.

cesses occur in the metals processing industries.

The

Large uses of leaching pro-

useful metals usually occur in

mixtures with very large amounts of undesirable constituents, and leaching

remove the metals

as soluble salts.

Copper

salts are dissolved or

is

used to

leached from ground

ores containing other minerals by sulfuric acid or ammoniacal solutions. Cobalt and sulfuric acid-ammonia-oxygen mixtures. Gold sodium cyanide solution. Sodium hydroxide is leached from a slurry of calcium carbonate and sodium hydroxide prepared by reacting Na 2 C0 3 withCa(OH) 2

nickel salts are leached from their ores is

leached from

its

by

ore using an aqueous

.

Preparation of Solids for Leaching

12.8B

1.

Inorganic and organic materials.

large extent

upon

The method

of preparation of the solid depends to a

the proportion of the soluble constituent present,

its

distribution

—

throughout the original solid, the nature of the solid i.e., whether it is composed of plant cells or whether the soluble material is completely surrounded by a matrix of insoluble matter

—and the

original particle size.

surrounded by a matrix of insoluble matter, the solvent and dissolve the soluble material and then diffuse out. This occurs in many hydrometallurgical processes where metal salts are leached from mineral ores. In these cases crushing and grinding of the ores is used to increase the rate of leaching since the soluble portions are made more accessible to the solvent. If the soluble substance is in solid solution in the solid or is widely distributed throughout the whole If

must

the soluble material

is

diffuse inside to contact

solvent

is

made

then

of the particles the solid.

2.

is

easier,

and grinding

not necessary

if

to very small sizes

the soluble material

Then simple washing can be used

Animal and vegetable

materials.

cell

is

dissolved in solution adhering to

as in washing of chemical precipitates.

Biological materials are cellular in structure and the

soluble constituents are generally found inside the

comparatively slow because the

cells.

is

impractical.

The

rate of leaching

may be

walls provide another resistance to diffusion.

ever, to grind the biological material sufficiently small to

ual cells

The passage of additional may not be needed. Grinding

could form small channels.

solid, the solvent leaching action

How-

expose the contents of individ-

Sugar beets are cut into thin wedge-shaped

slices for

leaching so

that the distance required for the water solvent to diffuse to reach individual cells

reduced.

The

cells

is

of the sugar beet are kept essentially intact so that sugar will diffuse

through the semipermeable

cell walls,

while the undesirable albuminous and colloidal

components cannot pass through the walls. For the leaching of pharmaceutical products from

leaves, stems,

and

roots, drying of

the.material before extraction helps rupture the cell walls. Thus, the solvent can directly dissolve the solute.

The

cell

and many vegetable seeds are largely in size to about 0.1 mm to 0.5 mm by but the walls are ruptured and the vegetable

walls of soybeans

ruptured when the original materials are reduced rolling or flaking. Cells are smaller in size, oil is easily

724

accessible to the solvent.

Chap. 12

Liquid-Liquid and Fluid-Solid Separation Processes

Rates of Leaching

12.8C

Introduction and general steps.

In the leaching of soluble materials from inside a by a solvent, the following general steps can occur in the overall process. The solvent must be transferred from the bulk solvent solution to the surface of the solid. 1.

particle

Next, the solvent must penetrate or diffuse into the solid.

The

solvent.

Finally, the solute

particle.

The

solute dissolves into the

solute then diffuses through the solid solvent mixture to the surface of the is

phenomena encountered make

transferred to the bulk solution. it

The many

different

almost impracticable or impossible to apply any one

theory to the leaching action.

from the bulk solution to

In general, the rate of transfer of the solvent

surface

somewhat rapid or

slow. These are not, in

many

cases, the rate-limiting steps in the

leaching process. This solvent transfer usually occurs

overall

the solid

quite rapid, and the rate of transfer of the solvent into the solid can be

is

particles are first contacted with the solvent.

The

initially

when

the

dissolving of the solute into the

may be simply a physical dissolution process or an actual chemical reaction that frees the solute for dissolution. Our knowledge of the dissolution

solvent inside the solid

process

limited and the

is

The

mechanism may be

different in

each

solid (Kl).

through the solid and solvent to the surface of the solid is often the controlling resistance in the overall leaching process and can depend on a number of different factors. If the solid is made up of an inert porous solid structure rate of diffusion of the solute

with the solute and solvent needed. This

is

described

in

pores in the solid, the diffusion through the porous

in the

can be described by an

solid

The void

effective diffusivity.

fraction

and tortuosity are

Section 6.5C for diffusion in porous solids.

In biological or natural substances, additional complexity occurs because of the cells present. In the leaching of thin sugar beet slices, in the slicing of the beets.

remaining

(Yl). In the

The leaching

cells,

about one-fifth of the

of the sugar

is

cells

are ruptured

then similar to a washing process

sugar must diffuse out through the

The

cell walls.

net result of

the two transfer processes does not follow the simple diffusion law with a constant effective diffusivity.

With soybeans, whole beans cannot be leached

effectively.

the soybeans ruptures cell walls so that the solvent can action.

The

rate of diffusion of the

permit simple interpretation.

soybean

A method

oil

more

The

rolling

and flaking of by capillary

easily penetrate

solute from the soybean flakes does not

to design large-scale extractors

is

given by using

small-scale laboratory experiments (02) with flakes.

The solvent

(01)

resistance to in

is

itself.

mass transfer of the solute from the solid surface

general quite small

compared

Rate of leaching when dissolving a

on

the extraction rate (03, Yl).

When

solid.

solid to the solvent solution, however, the rate of

the liquid is

bulk

This has been found for leaching soybeans where the degree of agitation of the

external solvent has no appreciable effect

2.

to the

to the resistance to diffusion within the solid

is

the controlling factor.

a pure material.

The equation

a material

mass

is

being dissolved from the

transfer from the solid surface to

There is essentially no resistance in the solid phase if it can be derived as follows for a batch system. The

for this

following can also be used for the case

when

diffusion in the solid

is

very rapid

compared

to the diffusion from the particle.

The

m

rate of

mass

transfer of the solute

A

being dissolved to the solution of volume

V

3 is

^A = Sec. 12.8

k L (c AS

-c A

)

Introduction and Equipment for Liquid-Solid Leaching

(12.8-1)

725

N A is kg mol

where kL

is

A

of

a mass-transfer coefficient in m/s, c AS

kg mol/m

is

surface area of particles

inm 2

the saturation solubility of the solid solute

is

,

A

kgmol/m 3 and c A is the concentration of A in the solution at time sec By a material balance, the rate of accumulation of A in the solution is

in the solution in in

A

dissolving to the solution/s,

3 .

t

,

equal to Eq. (12.8-1) times the area A.

V

dc A dt

Integrating from

=

t

0 and c A

=

= N A =Ak L (c AS -c A

c A0 to CA

t

= and c A t

Ak

dc A

1 1

Cas

(12.8-2)

)

~

V

cA

Jr

dt

(12.8-3)

=0

C AS

area

(12.8-4)

The solution approaches a saturated condition exponentially. Often the interfacial A will increase during the extraction if the external surface becomes very irregular. If

the soluble material forms a high proportion of the total solid, disintegration of the

may

particles

occur. If the solid

completely dissolving, the interfacial area changes

is

markedly. Also, the mass-transfer coefficient If

may

then change.

the particles are very small, the mass-transfer coefficient to the particle in

For

agitated system can be predicted by using equations given in Section 7.4.

an

larger

which are usually present in leaching, equations to predict the mass-transfer mixing vessels are given in Section 7.4 and reference (Bl).

particles

coefficient k L in agitated

3.

Rate of leaching when diffusion

diffusion in the solid

In the case where unsteady-state

in solid controls.

the controlling resistance in the leaching of the solute by an

is

external solvent, the following approximations can be used. If the average diffusivity

DA

eff

of the solute

A

is

approximately constant, then for extraction

batch process,

in a

unsteady-state mass-transfer equations can be used as discussed in Section particle

7.1. If

the

approximately spherical, Fig. 5.3-13 can be used.

is

EXAMPLE

Prediction of Time for Batch Leaching

J2JS-1.

Particles having an average diameter of

approximately 2.0

mm

are leached

a batch-type apparatus with a large volume of solvent. The concentration of the solute A in the solvent is kept approximately constant. A time of 3. 1 h is needed to leach 80% of the available solute from the solid. Assuming that diffusion in the solid is controlling and the effective diffusivity is in

constant, calculate the time of leaching

if

the particle size

reduced to

is

1.5

mm. For 80% extraction, the fraction unextracted E s

Solution: Fig.

for a sphere,

5.3-13

obtained, where is

radius in

for

D A efr

mm. For

a different

size.

is

the

for

Es =

the effective diffusivity

same

fraction

f

2

is

Es

,

inmm 2 /s,

the value

0.20.

is

Using

D Ae!r tfa 2 =0.112 t

o(D A

is eff

time t/a

in

1

is

s,

is

and a

constant

Hence,

h = where

a value of

0.20,

^

(12.8-5)

time for leaching with a particle size a 2

.

Substituting into

Eq. (12.8-5), r

726

Chap. 12

mn

(L5/2)2

175h

Liquid-Liquid and Fluid-Solid Separation Processes

4.

Methods of operation

There are a number of general methods of operThe operations can be carried out in batch or unsteady-

in leaching.

ation in the leaching of solids. state conditions as well as in

continuous or steady-state conditions. Both continuous and

stagewise types of equipment are used in steady or unsteady-state operation. In unsteady-state leaching a in-place leaching,

In other cases the leach liquor

ground

level as

it

common method

in the

mineral industries

is

is

is

drains from the heap.

sulfide ores in this

used

allowed to percolate through the actual ore body. pumped over a pile of crushed ore and collected at the

where the solvent

Copper

is

leached by sulfuric acid solutions from

manner.

by percolation through stationary solid beds in a The solids should not be too fine or a high resistance to flow is encountered. Sometimes a number of tanks are used in series, called an extraction battery, and fresh solvent is fed to the solid that is most nearly extracted. The tanks can be open tanks or closed tanks called diffusers. The solvent flows through the tanks in series, being withdrawn from the freshly charged tank. This simulates a continuous countercurrent stage operation. As a tank is completely Crushed

solids are often leached

vessel with a perforated

leached, a fresh charge

bottom

is

tanks do not have to be

to permit drainage of the solvent.

added to the tank

moved

for

at the other end. Multiple piping

countercurrent operation. This

is

is

used so

often called the

Shanks system. It is used widely in leaching sodium nitrate from ore, recovering tannins from barks and woods, in the mineral industries, in the sugar industry, and in other processes.

In

some processes

the crushed solid particles are

type conveyors or a screw conveyor.

The

moved continuously by bucketmoving

solvent flows countercurrently to the

bed.

Finely ground solids

may be

leached in agitated vessels or

in thickeners.

The process

can be unsteady-state batch or the vessels can be arranged in a series to obtain a countercurrent stage process.

12.8D

J.

Types of Equipment for Leaching Fixed-bed leaching

Fixed-bed leaching.

is

used

in the beet

sugar industry and

is

also

used for the extraction of tanning extracts from tanbark, for the extraction of pharmaceuticals from barks and seeds, and beet diffuser or extractor cossettes can be

in

other processes. In Fig. 12.8-1 a typical sugar

shown. The cover

is

removable, so sugar beet

slices called

Heated water at 344 K (7PC) to 350 K (77°C) leach out the sugar. The leached sugar solution flows out the

dumped

flows into the bed to

is

into the bed.

— movable cover hot water

sugar beet slices (cossettes)

movable bottom sugar solution

Figure

Sec. 12.8

12.8-1.

Introduction

Typical fixed-bed apparatus for sugar beet leaching.

and Equipment for Liquid-Solid Leaching

727

bottom onto the next tank in series. Countercurrent operation is used in the Shanks The top and bottom covers are removable so that the leached beets can be removed and a fresh charge added. About 95% of the sugar in the beets is leached to yield an outlet solution from the system of about 12 wt %.

system.

2.

Moving-bed

There are a number of devices for stagewise countercurrent moves instead of being stationary. These are used widely

leaching.

leaching where the bed or stage in extracting oil

from vegetable seeds such as cottonseeds, peanuts, and soybeans. The first, sometimes precooked, often partially dried, and rolled or

seeds are usually dehulled

Sometimes preliminary removal of

flaked.

accomplished by expression. The

oil is

vents are usually petroleum products, such as hexane. solution, called miscella,

may contain some finely

The

final

sol-

solvent-vegetable

divided solids.

an enclosed moving-bed bucket elevator device is shown. This is called the Bollman extractor. Dry flakes or solids are added at the upper right side to a perforated basket or bucket. As the buckets on the right side descend, they are leached by In Fig. 12.8-2a

a dilute solution of

oil in

solvent called half miscella. This liquid percolates

through the moving buckets and miscella.

is

downward

bottom as the strong solution or full

collected at the

The buckets moving upward on the left are leached countercurrently by fresh on the top bucket. The wet flakes are dumped as shown and removed

solvent sprayed

continuously.

The Hildebrandt extractor in Fig. 12.8-2b consists of three screw conveyors arranged U shape. The solids are charged at the top right, conveyed downward, across the bottom, and then up the other leg. The solvent flows countercurrently. in

a

When the solid can be ground fine to about 200 mesh (0.074 can be kept in suspension by small amounts of agitation. Continuous countercurrent leaching can be accomplished by placing a number of agitators in series with

3.

Agitated solid leaching.

mm),

it

settling tanks or thickeners

between each agitator.

pure

dr y

solvent

na kes

Figure

728

12.8-2.

.

.

dry

Equipment for moving-bed leaching : (a) Bollman bucket-type extractor, (b) Hildebrandt screw-conveyor extractor.

Chap. 12

Liquid-Liquid and Fluid-Solid Separation Processes

Sometimes the thickeners themselves are used as combination contactor-agitators

shown in Fig. 12.8-3. In this countercurrent stage system, fresh solvent shown to the first stage thickener. The clear settled liquid leaves and flows from stage to stage. The feed solids enter the last stage, where they are contacted with solvent from the previous stage and then enter the settler. The slowly rotating rake moves the solids to the bottom discharge. The solids with some liquid are pumped as a slurry to the and

settlers as

enters as

next tank. If the contact

is

insufficient,

a mixer can be installed between each

AND

EQUILIBRIUM RELATIONS SINGLE-STAGE LEACHING

12.9

12.9A

Equilibrium Relations

To

settler.

Leaching

in

leaching, an and the equilibrium relations between the two streams are needed as in liquid-liquid extraction. It is assumed that the solute-free solid is insoluble in the solvent. In leaching, assuming there is sufficient /.

Introduction.

analyze

single-stage

and

countercurrent-stage

operating-line equation or material-balance relation

solvent present so that

equilibrium

is

all

the solute in the entering solid can be dissolved into the liquid,

reached when the solute

dissolved in the

first

stage.

is

There usually

dissolved. Hence, is

all

is

completely

occur

in the first

the solute

sufficient time for this to

stage. It

is

also

leaching. This

assumed that there is no adsorption of the solute by the solid in the means that the solution in the liquid phase leaving a stage is the same as

the solution that remains with the solid matrix in the settled slurry leaving the stage. In

the settler in a stage

it is

not possible or feasible to separate

all

solute

is

present. This solid-liquid stream

Consequently, the concentration of to the

concentration of solute

in

oil

is

or solute

from the solid. which dissolved

the liquid

Hence, the settled solid leaving a stage always contains some liquid

in

called the underflow or slurry stream. in the liquid

or overflow stream

accompanying the equilibrium line is on the 45° line.

the liquid solution

is

equal

slurry or under-

on an xy plot the The amount of solution retained with the solids in the settling portion of each stage may depend upon the viscosity and density of the liquid in which the solid is suspended. flow stream. Hence,

Sec. 12.9

Equilibrium Relations and Single-Stage Leaching

729

This, in turn, depends

upon

the concentration of the solute in the solution. Hence,

experimental data showing the variation of the

amount and composition

of solution

retained in the solids as a function of the solute composition are obtained. These data

should be obtained under conditions of concentrations, time, and temperature similar to those in the process for which the stage calculations are to be made.

2.

The equilibrium data can be

Equilibrium diagrams for leaching.

plotted on the

rectangular diagram as wt fraction for the three components: solute (A), inert or leached

and solvent (Q. The two phases are the overflow (liquid) phase and the underflow (slurry) phase. This method is discussed elsewhere (B2). Another convenient

solid (B),

method of plotting the equilibrium data will be used, instead, which is similar to the method discussed in the enthalpy-concentration plots in Section 11.6. The concentration of inert or insoluble solid B in the solution mixture or the slurry mixture can be expressed in kg (lb m )

N=

units.

kg B

=

C

kg A + kg

—

kg B

solid

=

kg solution

lb solid

(12.9-1) lb solution

There will be a value of N for the overflow where N = 0 and for the underflow N will have different values, depending on the solute concentration in the liquid. The compositions of solute A in the liquid will be expressed as wt fractions.

A

kg solute

kg

A + kg C

kg solution

1

A +

kg

=

A

kg

kg

a

(overflow liquid)

i

(12.9-2)

kg solute

=

Z~~7kg C

—

i~T kg solution

i

(

h 9 uld

(12.9-3)

ln sIurr y)

where x A is the wt fraction of solute A in the overflow liquid andy^ is the wt fraction of A on a solid B free basis in the liquid associated with the slurry or underflow. For the entering solid feed to be leached, entering solvent

N=

0 and x A

=

N is

kg

inert solid/kg solute

diagram

In Fig. 12.9-la a typical equilibrium

soluble

in

is

shown where

solvent C, which would occur in the system of soybean

meal (B)-hexane solvent

solid

A and

=

yA

1.0.

For pure

0.

The upper curve of

(C).

N

solute oil

A

is

infinitely

(/i)-soybean inert

versus y A

for the slurry

underflow represents the separated solid under experimental conditions similar to the

N

actual stage process.

The bottom

the overflow liquid

composition where

small

line of

amounts of solid may remain

versusx^, where

all

in the

N = Oon the axis, represents

the solid has been removed. In

The

overflow.

tie lines

some

cases

are vertical, and on a yx

is y A = x A on the 45° line. In Fig. 12.9-lb the tie lines are not vertical, which can result from insufficient contact time, so that all the solute is not

diagram, the equilibrium line

dissolved; adsorption of solute If

the underflow line of

N

A on

the solid; or the solute being soluble in the solid B.

versus y

associated with the solid in the slurry that the underflow liquid rate

overflow stream. This

12.9B In Fig.

is

is

is

is

straight

constant throughout the various stages as well as the

a special case which

is

sometimes approximated

in practice.

Single-Stage Leaching is shown where V is kg/h {\bjb) of L is kg/h of liquid in the slurry solution with B kg/h of dry solute-free solid. The material-

12.9-2a a single-stage leaching process

overflow solution with composition x A and

composition y A based on a given flow rate

730

and horizontal, the amount of liquid all concentrations. This would mean

constant for

Chap. 12

Liquid-Liquid and Fluid—Solid Separation Processes

(b)

(a)

Figure

Several typical equilibrium diagrams:

12.9-1.

yA

=

xA

,

(b)

case where y A

^

balance equations are almost identical to Eqs.

x A for

(

(a)

case for vertical

tie lines

and

lie lines.

L2.5- 1 2)—(12.5- 1 4) for single-stage liquid-

and are as follows for a total solution balance (solute A + solvent component balance on A, and a solids balance on B, respectively. liquid extraction

L0

V2 = L,

-I-

LoYao + VjXai

No Lo +

B = where

M

point

M. A balance on C

shown line.

M

is

L MV shown in

is

l

l

V\

M

=

(12.9-4)

= J- i^i + Kx Al = Mx AM = N L + 0 = NM M l

(12.9-5) (12.9-6)

1

and N M are the coordinates of this + x c = 1.0 and y A + yc = 1.0. As must lie on a straight line and L Q MV2 must also lie on a straight Fig. 12.9-2b. Also, L and V must lie on the vertical tie line. The

the total flow rate in kg

before,

This

point

is

0

-(

C), a

is

A +

C/h and x A

not needed, since x A

l

y

two lines. IfL 0 entering is the fresh solid feed to be present, it would be located above the N versus y line in Fig.

the intersection of the

leached with no solvent

C

12.9-2b.

EXAMPLE

12.9-1.

Single-Stage Leaching of Flaked Soybeans soybean oil from flaked soybeans with hexane,

In a single-stage leaching of

100 kg of soybeans containing 20 wt % oil is leached with 100 kg of fresh hexane solvent. The value of N for the slurry underflow is essentially 1.5 kg insoluble solid/kg solution amounts and compositions of the overflow V and

constant at

l

retained.

Calculate

the underflow slurry

the

L

l

leaving the stage.

Solution:

The known

Sec. 12.9

The

process flow diagram

is

the

same

as given in Fig. 12.9-2a.

process variables are as follows.

Equilibrium Relations and Single-Stage Leaching

731

Figure

Process flow and material balance for single-stage leaching

12.9-2.

:

(a) pro-

cess flow, (b) material balance.

The

V2 = 100

entering solvent flow

kg,

x A1

=

0,

x C2

=

1.0.

For the

entering slurry stream, B = 100(1.0 — 0.2) = 80 kg insoluble solid, L 0 = 100(1.0 - 0.8) = 20 kg A, N 0 = 80/20 = 4.0 kg solid/kg solution, y A0 = 1.0. To calculate the location of M, substituting into Eqs. (12.9-4), (12.9-5),

and

(12.9-6)

and solving,

L 0 + K2 = 20 + LoVao + V2 x A1 Hence, x AM

=

=

NM = point

M

20(1.0)

+

4.0(20)

=

=

120 kg

=

100(0)

=

M

l20x AM

0.167.

B = N 0 L0 =

The

100

is

80

= N M (120)

0.667

plotted in Fig. 12.9-3 along with

V2 andL 0 The .

vertical tie

equilibrium with each other. Then N = 1.5, y Al = 0.167, x Al =0.167. Substituting into Eqs. (12.9-4) and (12.9-6) and solving or using the lever-arm rule, L = 53.3 kg and V = 66.7 line is

drawn

locating

L and t

K,

in

l

1

l

kg-

Va. x a Figure

732

12.9-3.

Graphical solution of single-stage leaching for Example 12.9-1.

Chap. 12

Liquid-Liquid and Fluid-Solid Separation Processes

COUNTERCURRENT MULTISTAGE LEACHING

12.10

12.10A

Introduction and Operating Line for Countercurrent

A is

Leaching

process flow sheet for countercurrent multistage leaching similar to Fig. 12.7-1 for liquid-liquid extraction.

The

The solvent

direction of the solids or underflow stream.

is

shown

and

in Fig. 12.10-1

ideal stages are

numbered

in the

V

phase

(C)-solute (A) phase or

from stage to stage countercurrently to the solid phase, and it dissolves solute as it moves along. The slurry phase L composed of inert solids (B) and a liquid phase of A and C is the continuous underflow from each stage. Note that the composition of the V phase is denoted by x and the composition of the L phase by y, which is the reverse of that for liquid-liquid extraction. It is assumed that the solid B is insoluble and is not lost in the liquid V phase. The flow rate of the solids is constant throughout the cascade of stages. As in the single-stage leaching V is kg/h (lb^/h) of overflow solution and L is kg/h of liquid solution in the " slurry retained by the solid. In order to derive the operating-line equation, an overall balance and a component balance on solute A is made over the first n stages.

is

the liquid phase that overflows continuously

(12.10-1)

Vn+l x n+l + L 0 y 0 = VlXl + L n yn Solving for x n+

and eliminating Vn +

1

l

,

1

+(K

1

The

1

-L

operating-line equation (12.10-3)

terminal points

x,,}',,

and x N+

In the leaching process,

if

,,

yN

(12.10-2)

y„

+

VlXl (12.10-3)

0 )/L„

when

plotted on an

xy

plot passes through the

.

the viscosity

and density of the solution changes appreci-

ably with the solute (A) concentration, the solids from the lower-numbered stages where solute concentrations are high

may

retain

more

liquid solution than the solids from the

higher-numbered stages, where the solute concentration retained in the solids underflow, will vary

and

is

dilute.

Then L„,

the liquid

the slope of Eq. (12.10-3) will vary

from

The overflow will also vary. If the amount of solution Ln retained by the solid is constant and independent of concentration, then constant underflow occurs. This simplifies somewhat stage to stage. This condition of variable underflow will be considered

the stage-to-stage calculations. This case will

be considered

first.

later.

exit

leaching

-overflow

solvent

—

v„

yo,

w

N

2

1

+

XN+ yN>

0

1

nn

LN B

Ln, Br

,

—

L feed

1

solids

Figure

Sec. 12.10

12.10-1.

underflow stream

leachedsolids

Process flow for countercurrent multistage leaching.

Countercurrent Multistage Leaching

733

Variable Underflow in Countercurrent Multistage

12.10B

Leaching

The methods

in this section are very similar to those

current solvent extraction, where the

Making an

L and V

A

overall total solution (solute

4-

used in Section 12.7B for counter-

flow rates varied from stage to stage.

solvent C) balance on the process of Fig.

12.10-1,

L0 where

M

is

M

+ VN+ = L N + V = ;

l

,

the total mixture flow rate in

kg

A +

'

(12.10-4)

-

C/h. Next making a component balance

on A,

Loy A o + yN+i x AN+i Making

a total solids balance

V x Al = x

Mx AM

(12.10-5)

on B,

B =

Nw

=L N y AN +

N0 L0 = N N LN = N M M

(12.10-6)

M

shown in Fig. 12.10-2, which is the shown previously L 0 MVN+l must lie on a straight line and V^ML S must be on a straight line. Usually the flows and compositions of L 0 and Kv+1 are known and the desired exit concentration y AN is set. Then the coordinates N M

where

and x AM are the coordinates of point

operating diagram for the process. As

M

and x AM can be calculated from Eqs. (12.1 0-4)—( 12.10-6) and point plotted. ThenL N M, and V must lie on one line as shown in Fig. 12.10-2. In order to go stage by stage on Fig. 12.10-2, we must derive the operating-point ,

x

Figure

734

12.10-2.

N umber of stages for multistage countercurrent

Chap. 12

leaching.

Liquid-Liquid and Fluid-Solid Separation Processes

Making a

equation.

on stage

total balance

1

and than on stage

n,

L 0 + V2 = L + V t

L„_

+ K +l = Ln +

1

Rearranging Eq. (12.10-7) for the difference flows

L0 - V = Lx 1

This value

A

is

(12.10-8)

kg/h,

- V2 = A

This can also be written

for

X AA~

i

a balance on solute

r

^0

/I

-

_ K'I

r

L-'N

B

N\ is the As shown

where

=

_ A

=•••

(12.10-10)

i

1/

'N+l

balance given on solids gives

—N —L — 0

0

(12.10-12)

N coordinate of the operating point A. in

Section

12. 7B,

A

is

see that

V

x

is

on a

line

A is located L 0 V and L N VN+1 From Eq. V2 is on a line between L and

the operating point. This point

graphically in Fig. 12.10-2 as the intersection of lines

we

for all stages.

to give

the x coordinate of the operating point A.

NA =

(12.10-10)

(12.10-9)

=L„- K + =^n- VN+

y

is

A in

V„

constant and also holds for Eq. (12.10-8) rearranged and

A = L0 - V

where x AA

(12.10-7)

x

between

L 0 and

A,

.

l

x

Vn+ is on a line between L„ and A, and so on. To graphically determine the number of stages, we start atL 0 and draw lineL0 A to locate V A tie line through F, locates L LineL; A is drawn given V2 A tie line gives L 2 This is continued until the desired L N is reached. In Fig. 12.10-2, about 3.5 stages are A,

,

.

t

l

.

.

.

required.

EXAMPLE 12.10-1.

Countercurrent Leaching of Oil from Meal continuous countercurrent multistage system is to be used to leach oil from meal by benzene solvent (B3). The process is to treat 2000 kg/h of inert solid meal (B) containing 800 kg oil (A) and also 50 kg benzene (Q. The inlet flow per hour of fresh solvent mixture contains 1310 kg benzene and 20 kg oil. The leached solids are to contain 120 kg oil. Settling experiments

A

similar to those in the actual extractor

show

that the solution retained

depends upon the concentration of oil in the solution. The data (B3) are tabulated below as N kg inert solid B/kg solution and y A kg oil A/kg solution.

2.00

0

1.82

1.98

0.1

1.75

0.5

1.94

0.2

1.68

0.6

1.89

0.3

1.61

0.7

0.4

Calculate the amounts and concentrations of the stream leaving the process and the number of stages required. Solution:

N

The underflow data from the table are plotted in Fig. 12.10-3 For the inlet solution with the untreated solid, L 0 = 800

versus y A

Sec. 12.10

.

Countercurrent Multistage Leaching

as

+ 735

50 = 850 kg/h, y A0 = 800/(800 + 50) = 0.941, B = 2000 kg/h, N 0 = 2000/(800 + 50) = 2.36. For the inlet leaching solvent, KN+ = 1310 + 20 = 1330 kg/h and x AN + = 20/1330 = 0.015. The points VN+l and L 0 are ,

,

plotted. lies on the N versus y A line in Fig. 12.10r3. Also for this N N/y AN = (kg solid/kg solution)/(kg oil/kg solution) = kg

The point L N point, the ratio

solid/kg

yA

=

oil

0 and

2000/120 = 16.67. Hence, a dashed line through the origin at 0 is plotted with a slope of 16.67, which intersects the N = at L N The coordinates of L s at this intersection are N

=

N=

N

versus y A line 1.95 kg solid/kg solution and y AN .

Making an

=

kg oil/kg

0.1 18

solution.

overall balance by substituting into Eq. (12.10-4) to deter-

mine point M,

L0 +

Kv +

i

= 850 + 1330 = 2180

kg/h

=

M

Substituting into Eq. (12.10-5) and solving,

+ VN+l x AN+l =

MUo

850(0.941)

+

1330(0.015)

=

2180^

*ah = 0 376 Substituting into Eq. (12. 10-6) and solving,

B = 2000 The point

M

Fig. 12.10-3.

is

= NM M

=N M (2 180)

plotted with the coordinates

The

line

VN+l ML 0

the abscissa at point K, where x A

The amounts

is l

0.918

x AM = 0.376 andN M = 0.918 in is line L N M, which intersects

drawn, as

=

0.600.

V and L N

of streams

NM =

l

are calculated by substituting into

Eqs. (12.10-4) and (12.10-5), and solving simultaneously:

L„ + V = l

M

=

2180

L^ah + v^ai = M0- 11 8) + V (0.600) = x

2180(0.376)

= Hence, L N = 1016 kg solution/h in the outlet underflow stream and 1164 kg solution/h in the exit overflow stream. Alternatively, the amounts could have been calculated using the lever-arm rule. The operating point A is obtained as the intersection of lines L 0 Vx and 736

Chap. 12

Liquid— Liquid and Fluid-Solid Separation Processes

L N VN+1 in Fig. 12.10-3. Its coordinates can also be calculated from Eqs. (12.10-11) and (12.10-12). The stages are stepped off as shown. The fourth

L4

stage for

is

slightly past the desired

LN

.

Hence, about

3.9 stages are

required.

Constant Underflow

12.10C

in

Countercurrent Multistage

Leaching In this case the liquid L„ retained in the underflow solids

This means that a plot of N versus y A the operating-line equation (12.10-3) equilibrium line

may

line

is

a straight line

L0

yA

= xA

.

12.10-1).

method used

when

generally not equal to L„, since

is

Then

many

is

it

made on

contains

stage

1

the straight operating line can be used

to step off the

number

N

is

constant.

Then The

plotted asy^ versusx^.

cases the equilibrium

Special treatment must be given the

separate material and equilibrium balance Fig.

constant from stage to stage.

a horizontal straight line and

can also be plotted on the same diagram. In

also be straight with

however, since

is

is

little

or

to obtain

no

first

stage,

solvent.

L and V2 l

A

(see

and the McCabe-Thiele

of stages.

Since this procedure for constant underflow requires almost as

many

calculations as

the general case for variable underflow, the general procedure can be used for constant

underflow by simply using a horizontal off the stages with the

12.11

line

of

N

versus y A in Fig. 12.10-2 and stepping

point.

INTRODUCTION AND EQUIPMENT FOR CRYSTALLIZATION

12.11 A

1.

A

Crystallization and

Introduction.

been treated

Types of Crystals

Separation processes for gas-liquid and liquid-liquid systems have

in this

and previous chapters. Also,

the separation process of leaching

discussed for a solid-liquid system. Crystallization

is

was

also a solid-liquid separation

which mass transfer occurs of a solute from the liquid solution to a pure solid An important example is in the production of sucrose from sugar beet, where the sucrose is crystallized out from an aqueous solution. Crystallization is a process where solid particles are formed from a homogeneous phase. This process can occur in the freezing of water to form ice, in the formation of snow particles from a vapor, in the formation of solid particles from a liquid melt, or in the formation of solid crystals from a liquid solution. The last process mentioned, crystallization from a solution, is the most important one commercially and will be treated in the present discussion. In crystallization the solution is concentrated and usually cooled until the solute concentration becomes greater than its solubility at that temperature. Then the solute comes out of the solution forming crystals of approximately process

in

crystalline phase.

pure solute. In commercial crystallization the yield

and purity of crystals are not only important

but also the sizes and shapes of the crystals. in size.

Size uniformity

is

It is

often desirable that crystals be uniform

desirable to minimize caking in the package, for ease of

pouring, for ease in washing and

filtering,

and

for

uniform behavior when used. Some-

times large crystals are requested by the purchaser, even though smaller crystals are just as useful. Also, crystals of a certain

shape are sometimes required, such

as needles rather

than cubes.

Sec. 12.11

Introduction

and Equipment for

Crystallization

737

2.

A

Types of crystal geometry.

crystal can be defined as a solid

composed of atoms,

which are arranged in an orderly and repetitive manner. It is a highly organized type of matter. The atoms, ions, or molecules are located in three-dimensional arrays or space lattices. The interatomic distances in a crystal between these imaginary ions, or molecules,

planes or space planes.

The

lattices are

measured by x-ray diffraction

pattern or arrangement of these space lattices

as are the angles is

repeated in

all

between these directions.

flat faces and sharp corners. The relative and edges of different crystals of the same material may differ greatly. However, the angles between the corresponding faces of all crystals of the same material

Crystals appear as polyhedrons having

sizes of the faces

are equal

and are characteristic

of that particular material. Crystals are thus classified

on

the basis of these interfacial angles.

There are seven classes of

depending upon the arrangement of the axes

crystals,

to

which the angles are referred 1.

Cubic system. Three equal axes

at right angles to

2.

Tetragonal system. Three axes

at right angles to

each other.

each other, one axis longer than the

other two.

6.

Orthorhombic system. Three axes at right angles to each other, all of different lengths. Hexagonal system. Three equal axes in one plane at 60° to each other, and a fourth axis at right angles to this plane and not necessarily the same length. Monoclinic system. Three unequal axes, two at right angles in a plane and a third at some angle to this plane. Triclinic system. Three unequal axes at unequal angles to each other and not 30, 60, or

7.

Trigonal system. Three equal and equally inclined axes.

3. 4.

5.

90°.

The

relative

development of different types of faces of a crystal may differ for a given Sodium chloride crystallizes from aqueous solutions with cubic faces

solute crystallizing.

only. In another case,

if

sodium chloride

from an aqueous solution with a

crystallizes

given slight impurity present, the crystals will have octahedral faces. Both types of crystals are in the cubic system but differ in crystal habit.

shapes of plates or needles has no relation

depends upon the process conditions under which the

12.11B

Equilibrium Solubility

In crystallization equilibrium

This

is

is

The

crystallization in overall

system and usually grown.

to crystal habit or crystal

crystals are

in Crystallization

when

attained

the solution or

represented by a solubility curve. Solubility

is

mother liquor

is

saturated.

dependent mainly upon temper-

on solubility. Solubility data are given in the form some convenient units are plotted versus temperature. given in many chemical handbooks (PI). Solubility curves for

ature. Pressure has a negligible effect

of curves where solubilities

Tables of solubilities are

some

in

were given in Fig. 8.1-1. In general, the markedly with increasing temperature.

typical salts in water

salts increase slightly or

A

very

common

solubilities

of most

in Fig. 8.1-1 for KN0 3 where the solubility and there are no hydrates. Over the whole range of KN0 3 The solubility of NaCl is marked by its small

type of curve

is

shown

,

increases markedly with temperature

temperatures, the solid phase

is

.

change with temperature. In solubility plots the solubility data are ordinarily given as parts by weight of anhydrous material per 100 parts by weight of total solvent (i.e., water in

many

cases).

In Fig. 12.11-1 the solubility curve

is

shown

for

sodium

thiosulfate,

Na 2 S 2 0 3 The .

solubility increases rapidly with temperature, but there are definite breaks in the curve

738

Chap. 12

Liquid-Liquid and Fluid—Solid Separation Processes

which indicate

different hydrates.

Na 2 S 2 0 3 5H 2 0. •

The

Na 2 S 2 0 3 2H 2 0. A

half-hydrate

•

12.1 1C

to 48.2°C

•

From

solubility line, only a solution exists.

the stable

up

Na 2 S 2 0 3 5H 2 0.

formed are

48.2°C), the solid crystals

is

stable phase

is

the pentahydrate

This means that at concentrations above the solubility line (up to

is

At concentrations below the

48.2 to about 65°C, the stable phase

is

present between 65 to 70°C, and the anhydrous salt

phase above 70°C.

Heat and Material Balances

Yields and

in

Crystallization

In most of the industrial crystallizaand the solid crystals are in contact for a long enough time to reach equilibrium. Hence, the mother liquor is saturated at the final temperature of the process, and the final concentration of the solute in the solution can be obtained from the solubility curve. The yield of crystals from a crystallization process can then be calculated knowing the initial concentration of solute, the final temperature, and the solubility at this temperature. In some instances in commercial crystallization, the rate of crystal growth may be quite slow, due to a very viscous solution or a small surface of crystals exposed to the 1.

Yields and material balances

in

crystallization.

tion processes, the solution (mother liquor)

solution. Hence,

some supersaturation may

exist,

still

giving a lower yield of crystals than

predicted.

making

In

the material balances, the calculations are straightforward

and

solute crystals are anhydrous. Simple water

When

the crystals are hydrated,

some

of the

water

when

solute material balances are in the solution is

the

made.

removed with

the

crystals as a hydrate.

EXAMPLE 12.11-1.

Yield of a Crystallization Process weighing 10000 kg with 30 wt Na 2 CO a is cooled to 293 K (20°C). The salt crystallizes as the decahydrate. What will be the yield of Na 2 C0 3 10H 2 crystals if the solubility is 21.5 kg anhydrous Na 2 CO 3 /100 kg of total water? Do this for the following cases. (a) Assume that no water is evaporated. (b) Assume that 3% of the total weight of the solution is lost by

A

%

salt solution

O

•

evaporation of water

)

1

i

10

i

20

30

in

cooling.

i

40

i

i

i

50

60

70

Temperature Figure

Sec. 12.11

12.1 1-1.

Introduction

(

1

80

1

90

C)

Solubility of sodium thiosulfate, /Va 2 S 2 0 3

and Equipment for

Crystallization

,

in water.

739

The molecular weights are 106.0 for Na 2 C0 3 180.2 for 10H 2 O, Na 2 C0 3 10H 2 O. The process flow diagram is shown in Fig. 12.11-2, with being kg H 2 0 evaporated, S kg solution (mother liquor), and C kg crystals of Na 2 C0 3 10H 2 O. Making a material balance around

Solution:

,

and 286.2

for

•

W

-

box

the dashed-Iine

for water for part (a),

°- 70(10000)

where (180.2)/(286.2)

forNa 2 C0 3

is

= 100^15

W=

+

H

(S)

wt fraction of water

Solving the two crystals

For part comes

in the crystals.

=

looT^B

{S)

+

2H

W

(b),

Eqs. -

= TbT+TiL5

(S)

+

ii

An

(C)

+

0

(12

300

in

salt is

When

in crystallization.

increases as temperature increases dissolves, there

heat of solution.

+

(C)

-

is

a

(12-

the

C=

and (12.11-3) simultaneously, crystals and S = 3070 kg solution.

and heat balances

effects

-

1M

>

Making a balance

(12.11-2)

Na 2 C0 3 10H 2 O Heat

(12

°

n

"

2)

C = 6370 kg of Na 2 C0 3 = 3630 kg solution. = 0.03(10000) = 300 kg H 2 0. Equation (12.11-1) be-

Equation (12.11-2) does not change, since no

2.

+

(C)

equations simultaneously,

and S

a70(10000)

Solving

0,

,

a30(10000)

10H 2 O

where

W

n

"

3)

stream.

6630 kg of

compound whose

solubility

an absorption of heat, called the

when a compound dissolves whose solubility compounds dissolving whose solubility does not no heat evolution on dissolution. Most data on heats of

evolution of heat occurs

decreases as temperature increases. For

change with temperature, there

is

mol (kcal/g mol) of solute occurring kg mol of the solid in a large amount of solvent at essentially

solution are given as the change in enthalpy in kJ/kg

with the dissolution of

1

infinite dilution.

In crystallization the opposite of dissolution occurs. crystallization

is

At equilibrium the heat of

equal to the negative of the heat of solution at the same concentration

solution. If the heat of dilution

from saturation

in the solution to infinite dilution

can be neglected, and the negative of the heat of solution

this

used

for the heat of crystallization.

With many materials

this

at infinite dilution

heat of dilution

in

small,

is

can be

is

small

W kg H 2 O I

I

10 000 kg solution

Cooler and

30% Na 2 C0 3

Crystallizer

I

S kg solution

J21.5

kgNa 2 CO 3 /100

kg

H2 0

I

L

J

I

C kg FIGURE

740

12.11-2.

crystals,

Na 2 C0 3

Process flow for crystallization

Chap. 12

in

1

0

H2 O

Example

12.11-1.

Liquid-Liquid and Fluid-Solid Separation Processes

compared with the heat of solution, and

this

approximation

is

reasonably accurate. Heat

of solution data are available in several references (PI, Nl).

Probably the most satisfactory method of calculating heat effects during a crystalliis to use the enthalpy-concentration chart for the solution and the various

zation process

which are present

solid phases

However, only a few such charts are

for the system.

magnesium

available, including the following systems: calcium chloride-water. (HI),

and

sulfate-water (P2),

following procedure

temperature of the

final

is

ferrous sulfate-water (K2).

The enthalpy H, whereHj is kJ

used.

q

is

positive, heat

kJ

in

is

is

obtained

is

(12.11-4)

)

must be added to the system.

initial

also read off

of the water vapor

= (H 2 + H v - Hi

q If

absorbed q

total heat

available, the

The enthalpy H 2

temperature

final

Hv

occurs, the enthalpy

If

is

of the entering solution at the

mixture of crystals and mother liquor at the

some evaporation from the steam tables. Then the the chart.

such a chart

(btu) for the total feed.

read off the chart,

is

When

If it is negative,

heat

is

evolved or given

off.

EXAMPLE 12.11-2.

Heat Balance

in

Crystallization

A feed solution of 2268 kg at 327.6 K (54.4X) containing 48.2 kg MgSCVlOO kg total water is cooled to 293.2 K (20°C), where MgS0 4 7H 2 0 crystals are removed. The solubility of the salt is 35.5 kg MgSO^/lOO -

kg total water (PI). The average heat capacity of the feed solution can be assumed as 2.93 kJ/kg-K (HI). The heat of solution at 291.2 K (18°Q is - 13.31 x 10 3 kJ/kg mol MgS0 4 7H z O (PI). Calculate the yield of crystals and make a heat balance to determine the total heat absorbed, q, assuming that no water is vaporized.

Making a water balance and

Solution:

a balance for

and (12.11-2) in Example and S = 1651.1 kg solution.

tions similar to (12.11-1)

MgS0 4 -7H 2 0 crystals

MgS0 4

12.11-1,

using equa-

C=

616.9 kg

K

To make a heat balance, a datum of 293.2 (20°C) will be used. The molecular weight of MgS0 4 7H 2 0 is 246.49. The enthalpy of the feed is Hp •

H, = 2268(327.6 The heat

of solution

heat

the

of

54.0(616.9)

=

as at 293.2 K.

q

Since q

12.1

/.

ID

is

293.2X2.93) 3

is

-(13.31 x 10 )/246.49

crystallization

is

—(— 54.0)

=

228 600 kJ

= -54.0

=+ 54.0

kJ/kg crystals. Then kJ/kg crystals, or

33 312 kJ. This assumes that the value at 291.2

The total heat absorbed,

= -228

600

negative, heat

Equipment

-

is

33 312

K

is

the

same

q, is

= -261

912 kJ (-248240 btu)

given off and must be removed.

for Crystallization

Introduction and classification of crystallizers.

Crystallizers

may

be classified accord-

is done for Continuous operation of crystallizers is generally preferred. Crystallization cannot occur without supersaturation. A main function of any cry-

ing to whether they are batch or continuous in operation. Batch operation certain special applications.

stallizer

is

to cause a

supersaturated solution to form.

A

classification of crystallizing

equipment can be made based on the methods used to bring about supersaturation as follows: (1) supersaturation produced by cooling the solution with negligible evaporation tank and batch-type crystallizers; (2) supersaturation produced by evaporation of the solvent with little or no cooling evaporator-crystallizers and crystallizing

—

Sec. 12.11

Introduction

—

and Equipment for

Crystallization

741

evaporators;

(3)

supersaturation by combined cooling and evaporation in adiabatic

—vacuum

evaporator

crystallizers.

In crystallizers producing supersaturation by cooling the substances must have a

markedly with temperature. This occurs for many subused. When the solubility curve changes little with common salt, evaporation of the solvent to produce super-

solubility curve that decreases

and

stances,

this

method

is

temperature, such as for

commonly

Sometimes evaporation with some cooling will also be used. In vacuum, a hot solution is introduced into a vacuum, where the solvent flashes or evaporates and the solution is cooled adiabatically. This method to produce supersaturation is the most important one for large-scale saturation

is

often used.

method of cooling

the

adiabatically in a

production. In another

method of

classification of crystallizers, the

equipment

is

classified

according to the method of suspending the growing product crystals. Examples are crystallizers

or

is

where the suspension

is

agitated in a tank,

is

circulated by a heat exchanger,

circulated in a scraped surface exchanger.

An important difference in many commercial

crystallizers

is

the

manner

in

which the

supersaturated liquid contacts the growing crystals. In one method, called the circulating

magma method,

magma

the entire

of crystals and supersaturated liquid

is

circulated

through both the supersaturation and crystallization steps without separating the solid from the liquid into two streams. Crystallization and supersaturation are occurring together in the presence of the crystals. In the second method, called the circulating liquid

method, a separate stream of supersaturated liquid crystals,

liquid

is

is passed through a fluidized bed of where the crystals grow and new ones form by nucleation. Then the saturated passed through an evaporating or cooling region to produce supersaturation

again for recycling.

2.

Tank

crystallizers.

In tank crystallization, which

is

an old method

still

used in some

open tanks. After a period of time the mother liquor is drained and the crystals removed. Nucleation and the size of crystals are difficult to control. Crystals contain considerable amounts of occluded mother liquor. Labor costs are very high. In some cases the tank is cooled by coils or a jacket and an agitator used to improve the heat-transfer rate. However, crystals often build up on these surfaces. This type has limited application and sometimes is used to specialized cases, hot saturated solutions are allowed to cool in

produce some 3.

fine

chemicals and pharmaceutical products.

Scraped surface

crystallizers.

One

type

of scraped

surface

crystallizer

is

the

Swenson-Walker crystallizer, which consists of an open trough 0.6 m wide with a semicircular bottom having a cooling jacket outside. A slow-speed spiral agitator rotates and suspends the growing crystals on turning. The blades pass close to the wall and break off any deposits of crystals on the cooled wall. The product generally has a

somewhat wide

crystal-size distribution.

In the double-pipe scraped surface crystallizer, cooling water passes in the annular space.

An

internal agitator

is

fitted

with spring-loaded scrapers that wipe the wall and

provide good heat-transfer coefficients. This type crystallizing ice

4.

cream and

plasticizing margarine.

Circulating-liquid evaporator-crystallizer.

shown liquid

heater.

giving

742

in Fig.

12.1 l-3a,

supersaturation

is

is

called a votator

and

is

used in

A sketch is shown in Fig. 4.13-2.

In a

combination evaporator-crystallizer

generated by evaporation. The circulating

drawn by the screw pump down inside the tube side of the condensing steam The heated liquid then flows into the vapor space, where flash evaporation occurs some supersaturation. The vapor leaving is condensed. The supersaturated liquid

is

Chap. 12

Liquid-Liquid and Fluid-Solid Separation Processes

down

the downflow tube and then up through the bed of fluidized and agitated which are growing in size. The leaving saturated liquid then goes back as a recycle stream to the heater, where it is joined by the entering feed. The larger crystals settle out and a slurry of crystals and mother liquor is withdrawn as product. This type is

flows

crystals,

also called the Oslo crystallizer.

5.

Circulating-magma vacuum

stallizer in Fig. 12.

l-3b, the

1

In this circulating-magma vacuum-type cry-

crystallizer.

magma

or suspension of crystals

is

circulated out of the

main body through a circulating pipe by a screw pump. The magma flows through a heater, where its temperature is raised 2 to 6 K.. The heated liquor then mixes with body slurry and boiling occurs at the liquid surface. This causes supersaturation in the swirling liquid near the surface, which causes deposits on the swirling suspended crystals until they leave again via the circulating pipe. ejector provides the

A steam-jet

Introduction and Nucleation Theories

12.12A

Introduction.

phase

grow

leave through the top.

CRYSTALLIZATION THEORY

1112

1.

The vapors

vacuum.

is

is

When crystallization occurs in a homogeneous mixture, a new solid An understanding of the mechanisms by which crystals form and then in designing and operating crystallizers. Much experimental and theo-

created. helpful

work has been done

to help understand crystallization. However, the differences between predicted and actual performance in commercial crystallizers are still often quite retical

large.

The overall process

of crystallization from a supersaturated solution

consist of the basic steps of nucleus formation or nucleation

solution

is

free of all solid particles, foreign

formation must

first

is

occur before crystal growth

starts.

New nuclei may

continue to form

pump

(a) 12.1 1-3.

If the

or of the crystallizing substance, then nucleus

pump

Figure

considered to

and of crystal growth.

(b)

Types of crystallizers: (a) circulating-liquid evaporator-crystallizer, circulating-magma vacuum crystallizer.

(b)

Sec. 12.12

Crystallization Theory

743

The driving force for the nucleation step and the do not occur in a saturated or'undersatu-

while the nuclei present are growing.

growth step

supersaturation. These two steps

is

rated solution.

2.

Primary nucleation

Nucleation theories.

is

a result of rapid local fluctuations on a

molecular scale in a homogeneous phase. Particles or molecules of solute happen to

come

together and form clusters.

clusters so they grow,

The growing

clusters

More solute molecules may be added to one or more may break up and revert to individual molecules.

whereas others

become

and continue

crystals

to

absorb solute molecules from the

solution.

This type of nucleation

is

called

crystal, the smaller its solubility.

range

is

crystals.

homogeneous or primary nucleation. The larger the solubility of small crystals in the micrometer size

greater than that of a large crystal.

Hence,

large crystal

This

The

The ordinary

also present, the larger crystal will

is

solubility data apply to large

a supersaturated solution a small crystal can be in equilibrium.

in

effect of particle size

is

grow and

If

a

the smaller one will dissolve.

an important factor in nucleation. In

magma crystallization,

primary nucleation happens to a small degree.

An

Miers attempts to explain

early qualitative explanation of crystallization by

formation of nuclei and crystals in an unseeded solution. This theory 12.12-1,

where

cooled,

it

AB is

line

the

normal

solubility curve. If a

The sample

crosses the solubility curve.

first

is

shown

sample of solution will

in Fig.

at point

not crystallize until

it

a

is

has

supercooled to some point b where crystallization begins, and the concentration drops to point c

if

no further cooling

is

done.

The curve CD,

called the supersolubility curve,

represents the limit at which nucleus formation starts spontaneously crystallizaton can start.

tendency

is

to

Any

and hence where

crystal in the metastable region will grow.

The

present

regard the supersolubility curve as a zone where the nucleation rate

increases sharply.

However, the value of Miers' explanation

is

that

it

points out that the

greater the degree of supersaturation, the greater the chance of more nuclei forming.

Secondary or contact nucleation, which occurs

when

is

the most effective

walls of the pipe or container. This type of nucleation intensity of agitation. It occurs at

the

optimum

for

good

crystals.

demonstrated experimentally. in

magma

nor

is

method of

nucleation,

crystals collide with each other, with the impellors in mixing, or with the

crystallization.

The

is,

of course, affected by the

low supersaturation, where the crystal growth rate is at This type of contact nucleation has been isolated and

It is

the

precise

most effective and common method of nucleation mechanisms of contact nucleation are not known,

a complete theory available to predict these rates.

"supersolubility" curve labile region

B

£— solubility curve unsaturated region

A

metastable region

Temperature Figure

744

12.12-1

Miers' qualitative explanation of crystallization: solubility curve {AB) and "supersolubility "curve (CD).

Chap. 12

Liquid-Liquid and Fluid-Solid Separation Processes

J.

AL Law

Rate of Crystal Growth and

12.12B

Rate of crystal growth and growth

the distance

moved

growth

crystal

is

The

coefficients.

per unit time in a direction that

growth of a crystal face

rate of

perpendicular to the

is

and since the growth can occur only

a layer-by-layer process,

is

The

face.

at the

outer face of the crystal, the solute material must be transported to that face from the

The solute molecules reach the face by diffusion through the liquid The usual mass-transfer coefficient k y applies in this case. At the surface the

bulk of the solution. phase.

resistance to integration of the molecules into the space lattice at the face

must be

considered. This reaction at the surface occurs at a finite rate, and the overall process consists of two resistances in series.

and

The solution must be supersaturated

for the diffusion

interfacial steps to proceed.

The equation for mass transfer of solute A from the bulk solution of supersaturation concentration^, mole fraction of A, to the surface where the concentration isy^ is ,

^~ = kJy A ~

(12.12-1)

/a)

where k y is the mass-transfer coefficient in kg mol/s m 2 mol frac, N A is rate in kg mol 2 A/s, and A is area in m of surface Assuming that the rate of reaction at the crystal surface is also dependent on the concentration difference, •

•

i.

{

-f =

-

(12.12-2)

y^)

kg mol/s m 2 mol saturation concentration. Combining Eqs. (12.12-1) and (12.12-2), where k

a surface reaction coefficient in

is

s

and y Ae

frac

(m2- 3)

^TiK^r^-^ where

K is the overall

The

transfer coefficient.

mass-transfer coefficient k can be predicted by methods given in Section 7.4 for y

convective mass-transfer coefficients.

velocity of particles can

be used

in the correlation

\/k

y

the mass-transfer coefficient k y negligible. Conversely,

is

diffusional resistance coefficients

is

velocities

all

also

and

is

is

controlling

very small,

overall transfer

is

V 1).

not directly applicable, because the con-

measurement differ greatly from those in a commercial crystallLzer. Also, the and the level of supersaturation in a system are difficult to determine, and vary

The AL law of

The growth crystal.

crystals.

very large, the surface reaction the mass-transfer coefficient

controlling. Surface reaction coefficients

magma in

crystal growth.

geometrically similar and of the

one

is

have been measured and reported on a number of systems (B4, H2, P3,

with position of the circulating

rate.

is

when

of the information in the literature

ditions of

2.

the saturated solution

for the prediction.

When

Much

or the velocity of the solution relative

The Schmidt number of

to the crystals in the suspension.

and

The correlation for mass transfer through fixed and The velocity obtained from the terminal settling

beds of solids can be used.

fluidized

needed

the

is

is

measured

This increase

This increase

is

in

the crystallizer.

McCabe (Ml)

same material

has shown that

in the

as the increase in length

length

is

same AL,

in

all

solution

mm,

crystals that are

grow

in linear

at the

for geometrically corresponding distances

independent of the

initial size

of the

initial crystals,

same

dimension of

on

all

provided that

same environmental conditions. This law follows from where the overall transfer coefficient is the same for each face of all crystals.

the crystals are subject to the

Eq. (12.

12-3),

Sec. 12.12

Crystallization Theory

745

Mathematically,

this

can be written

—= G

(12.12-4)

At

where At is time in h and growth rate G is a constant in mm/h. Hence, dimension of a given crystal at timeti and D 2 at time t 2

if

D

t

is

the linear

,

AL = D 2 The

total

—

growth (D 2

The AL law

or

fails in

AL is

the

£>!

same

=

G(t 2

-

tj

(12.12-5)

for all crystals.

cases where the crystals are given any different treatment based

on size. It has been found to hold for many materials, particularly when the crystals are under 50 mesh in size (0.3 mm). Even though this law is not applicable in all cases, it is reasonably accurate

12.12C

many situations.

in

Particle-Size Distribution of Crystals

An important

factor in the design of crystallization

equipment

size distribution of the crystals obtained. Usually, the

determine the particle

The in

sizes.

The percent

retained

on

is

the expected particle-

dried crystals are screened to

different-sized screens

is

recorded.

sieve or clear openings

whose

screens or sieves used are the Tyler standard screens,

mm are given in Appendix A.5. The data

are plotted as particle diameter (sieve opening in screen) in

cumulative percent retained

at that size

particles

from a typical

data

show an approximate

will

A common variation,

CV,

crystallizer (B5)

straight line for a large portion of the plot.

parameter used to characterize the size distribution

PD 16%

is

the coefficient of

as a percent.

CV = where

mm versus the

on arithmetic probability paper. Data for urea are shown in Fig. 12.12-2. Many types of such

is

100

PD '6%- pP 8«x

(12.12-6)

2PD 50%

the particle diameter at 16 percent retained.

variation

and mean

obtained

if

the line

is

the coefficient of is

approximately straight between 90 and 10%. For a product

removed from a mixed-suspension

G

By giving

particle diameter, a description of the particle-size distribution

crystallizer, the

CV

value

is

about

50%

(Rl). In a

1.20

P 0.80

2

Ph

0.40

i

5

10 20 40 60 80

95

99

99.9

Cumulative percent retained Figure

12.12-2.

Typical particle-size distribution from a crystallizer, [From R. C. Bennett and M. Van Buren, Chem. Eng. Progr. Symp., 65(95), 46 (1969).']

746

Chap. 12

Liquid-Liquid and Fluid-Solid Separation Processes

mixed-suspension system, the crystallizer suspension

magma with

Models

I2.12D

is

steady state and contains a well-mixed-

at

no product classification and no solids entering with the

feed.

for Crystallizers

In order to analyze data from a mixed-suspension crystallizer, an overall theory combining the effects of nucleation rate, growth rate, heat balance, and material balance

needed.

Some

progress has been

Randolph and Larson and

their

made and an

idealized

is

model has been investigated by

co-workers (Rl, R2, R3, M2, S2, PI, P3, C2). Their

equations are rather complicated but allow determination of some fundamental factors of growth rate and nucleation rate from experimental data.

A

crystallizer is first obtained and a screen and retention time in the crystallizer are also needed. By analysis to a population density of crystals of various sizes and

crystal product

analysis run.

The

sample from the actual

slurry density

converting the size

plotting the data, the nucleation rate and the

growth

the actual conditions tested in the crystallizer.

determine the

effects of

calculation for this

rate in

mm/h

can be obtained

operation effects on growth rate and nucleation

method

is

for

Then experiments can be conducted

A

given elsewhere (PI).

rate.

A

to

sample

contact nucleation model for the

design of magma crystallizers has been developed (B6) based on single-particle-contact nucleation experiments (C4, M3). Larson (L2, PI) gives examples of design of crystallizing

systems.

PROBLEMS 12.1-1. Equilibrium Isotherm for Glucose Adsorption. Equilibrium isotherm data for

adsorption of glucose from an aqueous solution to activated alumina are as follows (H3): c (g/cm

3 )

q (g solute/g alumina)

0.0040

0:0087

0.019

0.027

0.094

0.195

0.026

0.053

0.075

0.082

0.123

0.129

Determine the isotherm that

fits

the data and give the constants of the

equation using the given units.

Ans. Langmuir isotherm, q 12.2- 1.

=

0.

145c7(0.0 74 + c) 1

Batch Adsorption for Phenol Solution. A wastewater solution having a volume 3 3 contains 0.25 kg phenol/m of solution. This solution is mixed of 2.5 thoroughly in a batch process with 3.0 kg of granular activated carbon until equilibrium is reached. Use the isotherm from Example 12.2-1 and calculate the final equilibrium values and the percent phenol extracted.

m

Column Data. Using the break-point time from Example 12.3-1, do as follows: The break-point time for a new column is to be 8.5 h. Calculate the new total length of the column required, column diameter, and the fraction of total capacity used up to the break point. The flow rate is to remain

12.3- 1. Scale-Up of Laboratory Adsorption

and other (a)

results

constant (b)

Use to

the

2000

at

754

cm 3 /s.

same conditions as

part (a), but the flow rate

is to

be increased

cm 3 /s. Ans.

(a)

# r = 27.2

cm, 0.849

fraction; (b) £>

= 6.52 cm

and Scale-Up of Column. Using molecular sieves, water vapor was removed from nitrogen gas in a packed bed (C3) at 28.3°C. The column height was 0.268 m, with the bulk density of the solid bed being equal 3 The initial water concentration in the solid was 0.01 kg to 712.8 kg/m water/kg solid and the mass velocity of the nitrogen gas was 4052 kg/m""- h.

12.3-2. Drying of Nitrogen

.

Chap. 12

Problems

747

water concentration in the gas was c o = 926xl0" 6 kg water/kg nitrogen. The breakthrough data are as follows.

The

initial

H z O/kg N 2

c (kg

x

10

6

<0.6

)

c (kg

A

H 2 O/kgN 2 xl0 6

9.6

2.6

0.6

11.5

717

630

418

)

9.2

11.25

10.8

(h)

f

9

0

(h)

t

10

21

91

12.0

12.5

855

10.4

235 12.8

906

926

= 0.02 is desired at the break point. Do as follows. Determine the break-point time, the fraction of total capacity used up to the break point, the length of the unused bed, and the saturation loading

value of clc g

(a)

capacity of the solid. (b)

For a proposed column length

HT

= 0.40 m,

calculate the break-point

time and fraction of total capacity used.

Ans. 12.4-1. Scale-Up of Ion-Exchange

(a)

t

b

= 9.58

h, fraction

used

=

0.878

Column. An ion-exchange process using a resin to

remove copper ions from aqueous solution is conducted in a 1.0-in. -diameter column 1.2 ft high. The flow rate is 1.5 gph and the break point occurred at 7.0 min. Integrating the breakthrough curve gives a ratio of usable capacity

Design a new tower that will be 3.0 ft high and operating at 4.5 gph. Calculate the new tower size and break-point time. to total capacity of 0.60.

Ans.

t

b

=24.5 min, £>= .732 1

m

in. 3

/h of Tower in Ion-Exchange. In a given run using a flow rate of 0.2 an ion-exchange tower with a column height of 0.40 m, the break point occurred at 8.0 min. The ratio of usable capacity to total equilibrium capacity is 0.65. What is the height of a similar column operating for 13.0 min to the break point at the same flow rate?

12.4-2. Height in

12.4-3. Ion

Exchange of Copper

in

An

Column.

ion-exchange column containing

2+ 99.3 g of amberlite ion-exchange resin was used to remove Cu from a solution where c 0 = 0. 18 CuS0 4 The tower height = 30.5 cm and the

M

.

diameter = 2.59 cm. The flow rate was 1.37 breakthrough data are shown below. t

c (g t

420

(s)

mol Cu/L)

(s)

c (g

0

720

mol Cu/L)

0.1433

The concentration desired

480 0.0033

0.0075

0.1722

break point

solution/s to the tower.

600

0.0157

870

810

0.1634

3

540

510

780

at the

cm

is

660

0.0527

0.1063

900

0.1763 clc 0

The

=

0.180

0.01 0. Determine the

break-point time, fraction of total capacity used up to the break point, length

of unused bed, and the saturation loading capacity of the 12.5-1.

solid.

Composition of Two Liquid Phases in Equilibrium. An original mixture weighing 200 kg and containing 50 kg of isopropyl ether, 20 kg of acetic acid, and 130 kg of water is equilibrated in a mixer-settler and the phases separated. Determine the amounts and compositions of the raffinate and extract layers. Use equilibrium data from Appendix A.3.

A single-stage extraction is performed in which 400 kg acetic acid in water is contacted with 400 kg of of a solution containing 35 wt pure isopropyl ether. Calculate the amounts and compositions of the extract and raffinate layers. Solve for the amounts both algebraically and by the

12.5-2. Single-Stage Extraction.

%

748

Chap. 12

Problems

lever-arm rule. What percent of the acetic acid is removed? Use equilibrium data from Appendix A.3. = 358 kg, x B1 = 0.715, x ci = 0.03, K, = 442 kg, Ans. y Al = 0.11, y ci = 0.86, 34.7% removed.

1,

Unknown Composition. A feed mixture weighing 200 kg of unknown composition containing water, acetic acid, and isopropyl ether is contacted in a single stage with 280 kg of a mixture containing 40 wt acetic acid, 10 wt water, and 50 wt isopropyl ether. The resulting raffinate layer weighs 320 kg and contains 29.5 wt % acetic acid, 66.5 wt water, and isopropyl ether. Determine the original composition of the feed 4.0 wt mixture and the composition of the resulting extract layer. Use equilibrium data from Appendix A.3.

12.5-3. Single-Stage Extraction with

%

%

%

%

%

Ans,

x A0

=

0.030,

x B0

=

0.955, y A1

=

0.15

of Acetone in a Single Stage. A mixture weighing 1000 kg contains acetone and 76.5 wt % water and is to be extracted by 500 kg methyl isobutyl ketone in a single-stage extraction. Determine the amounts and compositions of the extract and raffinate phases. Use equilibrium data from Appen-

12.5-4. Extraction

23.5 wt

%

dix A.3.

Fresh Solvent in Each Stage. Pure water is to be used to extract acetic acid from 400 kg of a feed solution containing 25 wt acetic acid in isopropyl ether. Use equilibrium data from Appendix A.3. (a) If 400 kg of water is used, calculate the percent recovery in the water solution in a one-stage process. (b) If a multiple four-stage system is used and 100 kg fresh water is used in each stage, calculate the overall percent recovery of the acid in the total outlet water. (Hint: First, calculate the outlet extract and raffinate streams for the first stage using 400 kg of feed solution and 100 kg of water. For the second stage, 100 kg of water contacts the outlet organic phase from the first stage. For the third stage, 100 kg of water contacts the outlet organic phase from the second stage, and so on.)

12.7-1. Multiple-Stage Extraction with

%

Balance in Countercurrent Stage Extraction. An aqueous feed of 200 kg/h containing 25 wt acetic acid is being extracted by pure isopropyl ether at the rate of 600 kg/h in a countercurrent multistage system. The exit acid acetic acid. Calculate concentration in the aqueous phase is to contain 3.0 wt the compositions and amounts of the exit extract and raffinate streams. Use equilibrium data from Appendix A. 3.

12.7-2. Overall

%

%

12.7-3.

Minimum

Solvent and Countercurrent Extraction of Acetone. An aqueous feed water is acetone and 76.5 wt being extracted in a countercurrent multistage extraction system using pure

solution of 1000 kg/h containing 23.5 wt

%

%

methyl isobutyl ketone solvent at 298-299 K. The outlet water raffinate will contain 2.5 wt acetone. Use equilibrium data from Appendix A. 3. (a) Calculate the minimum solvent that can be used. [Hint: In this case the tie line through the feed L 0 represents the condition for minimum solvent flow to give the rate. This gives K, min Then draw lines L N K, mln and L 0 VN+ mixture point min and the coordinate x AMmin Using Eq. (12.7-4), solve for

%

.

1

M

(b)

.

VN+i min the minimum value of the solvent flow rate VN+ ,.] Using a solvent flow rate of 1.5 times the minimum, calculate the number of theoretical stages.

12.7-4.

Countercurrent Extraction of Acetic Acid and Minimum Solvent. An aqueous feed solution of 1000 kg/h of acetic acid-water solution contains 30.0 wt acetic acid and is to be extracted in a countercurrent multistage process with acid in the pure isopropyl ether to reduce the acid concentration to 2.0 wt final raffinate. Use equilibrium data in Appendix A.3. (a) Calculate the minimum solvent flow rate that can be used. (Him: See

%

%

Problem

Chap. 12

12.7-3 for the

Problems

method

to use.)

749

(b) If

2500 kg/h of ether solvent

stages required. (Note:

used, determine the

is

may be

It

number of

theoretical

necessary to replot on an expanded scale

the concentrations at the dilute end.)

Ans. 12.7-5.

(a)

Number of Stages an

exit acid

Minimum solvent in

flow rate

VN+ =

1630 kg/h;

,

(b) 7.5 stages

Countercurrent Extraction. Repeat Example 12.7-2 but use

concentration in the aqueous phase of 4.0 wt %.

A water solution of 1000 kg/h containing stripped with a kerosene stream of 2000 kg/h

12.7-6. Extraction with Immiscible Solvents. 1.5

%

wt

nicotine in water

is

%

nicotine in a countercurrent stage tower. The exit water is containing 0.05 wt to contain only 10% of the original nicotine, i.e., 90% is removed. Use equilibrium data from Example 12.7-3. Calculate the number of theoretical stages needed. Ans. 3.7 stages 12.7-7. Analytical Equation for

Number of Stages. Example

cible. (

A total

12.7-3 gives data for ex-

two solvents are immisUse the analytical equations number of theoretical stages and compare

traction of nicotine from water by kerosene

where

the

of 3.8 theoretical stages were needed.

10.3-2I)—{10.3-26) to calculate the

with the value obtained graphically. 12.7-8.

Minimum

Solvent Rate with Immiscible Solvents. Determine the

minimum

sol-

Example 12.7-3. Using determine the number of theoretical stages

vent kerosene rate to perform the desired extraction in

times this

1.25

minimum

rate,

needed graphically and also by using Eqs.

(1

0.3-2 1)—{ 1 0.3-26).

A

kerosene flow of 100 kg/h contains 1.4 wt pure water in a countercurrent multistage tower. It is desired to remove 90% of the nicotine. Using a water rate of 1.50 times the minimum, determine the number of theoretical stages required. (Use the equilibrium data from Example 12.7-3.)

12.7- 9. Stripping Nicotine from Kerosene.

%

nicotine and

is

to be stripped with

Example

12.8- 1. Effective Diffusivity in Leaching Particles. In

of the solid particle of

3.1

h

1

is

needed to remove

12.8-1 a time of leaching

80%

Do

of the solute.

the

following calculations. (a)

Using the experimental data, calculate the

(b)

Predict the time to leach

90%

Ans. 12.9- 1. Leaching

effective diffusivity,/)

cff

.

mm particle. mm /s; (b) =

of the solute from the 2.0 (a)

D Ac!! =

5 x 10"

1.0

2

t

5.00 h

of Oil from Soybeans in a Single Stage. Repeat Example 12.9-1 for from soybeans. The 100 kg of soybeans contains 22 and the solvent feed is 80 kg of solvent containing 3 wt % soybean oil.

single-stage leaching of oil

wt

% oil

Ans. 12.9-2.

L,

=

52.0 kg, y A

,

=

0.239,

V = Y

50.0 kg,

xA

,

=

0.239,

N,

=

1.5

Leaching a Soybean Slurry in a Single Stage. A slurry of flaked soybeans weighing a total of 100 kg contains 75 kg of inert solids and 25 kg of solution with 10 wt oil and 90 wt % solvent hexane. This slurry is contacted with 100 kg of pure hexane in a single stage so that the value of N for the outlet underflow is 1.5 kg insoluble solid/kg solution retained. Calculate the amounts and compositions of the overflow Vl and the underflow L, leaving the stage.

%

12.10-1. Constant Underflow in Leaching Oil

from Meal. Use

the

same conditions

given in Example 12.10-1, but assume constant underflow of solid/kg solution. Calculate the exit flows stages required.

Compare with Example

and compositions and

of Less Solvent Flow

as given in

hour

is

Example

in

the

1.85

as

kg

number of

12.10-1.

Ans. 12.10-2. Effect

N=

y AN

=

0.1

1 1,

x A1 =

0.623, 4.3 stages

Leaching Oil from Meal. Use the same conditions

12.10-1, but the inlet fresh solvent mixture flow rate per

decreased by 10%, to of stages needed.

1

179

kg of benzene and 18 kg of oil. Calculate the

number

12.10-3. Countercurrent Multistage

750

Washing of Ore.

A

treated ore containing inert solid

Chap. 12

Problems

gangue and copper sulfate is to be leached in a countercurrent multistage extractor using pure water to leach the CuS0 4 The solid charge rate per hour consists of 10000 kg of inert gangue (£), 1200 kg of CuS0 4 (solute A) ( and 400 kg of water (C). The exit wash solution is to contain 92 wt % water and 8 wt % .

95% of the CuS0 4 in the inlet ore is to be recovered. The = 0.5 kg inert gangue solid/kg aqueous solution. constant at Calculate the number of stages required.

CuS0 4 A

total of

underflow

is

.

N

12.10-4. Countercurrent Multistage

containing 25.7 wt

95%

of the

oil

%

Leaching of Halibut Livers. Fresh halibut

livers

are to be extracted with pure ethyl ether to remove in a countercurrent multistage leaching process. The feed rate is oil

1000 kg of fresh livers per hour. The final exit overflow solution is to contain 70 wt oil. The retention of solution by the inert solids (oil-free liver) of the liver

%

N

where

varies as follows (CI),

is

kg inert solid/kg solution retained andy A

is

kg oil/kg solution.

N

N 4.88

0

1.67

0.6

3.50

0.2

1.39

0.81

2.47

0.4

Calculate the

number

amounts and compositions of

12.10-5. Countercurrent Leaching

%

the exit streams

and

the total

of theoretical stages.

of Flaked Soybeans. Soybean

flakes containing 22

wt

are to be leached in a countercurrent multistage process to contain 0.8 kg oil/100 kg inert solid using fresh and pure hexane solvent. For every 1000 kg oil

soybeans, 1000 kg hexane is used. Experiments (SI) give the following retention of solution with the solids in the underflow, where N is kg inert solid/kg solution retained and is wt fraction of oil in solution.

N 1.73

0

1.52

0.20

1.43

0.30

Calculate the exit flows

and compositions and

the

number

of theoretical

stages needed.

A hot solution of Ba(N0 3 2 from an evaporator Ba(NO 3 ) 2 /100 kg H z O and goes to a crystallizer where the cooled and Ba(N0 3 ) 2 crystallizes. On cooling, 10% of the original

12.11-1. Crystallization

of Ba(N0 3 ) z

.

)

contains 30.6 kg solution

is

water present evaporates. For a feed solution of 100 kg

total, calculate the

following. (a)

The

yield of crystals if the solution

solubility (b)

The

is

yield

8.6 if

kg

is

cooled to 290

Ba(NO 3 ) 2 /100 kg total

and Subsequent

Crystallization.

is

35 wt

Chap. 12

Problems

its

%

KC1

solubility

is

7.0

kg

in

A

(a)

17.47 kg

Ba(N0 3

)2

crystals

batch of 1000 kg of KC1 is solution at 363 K, where the is cooled to 293 K, at which

make a saturated water. The solution

dissolved in sufficient water to

temperature

is

water.

Ans.

solubility

where the

(17°C),

cooled instead to 283 K, where the solubility

Ba(NO 3 yi00 kg total 12.11-2. Dissolving

K

water.

25.4

wt %.

751

(a)

(b)

What is the weight of water required for solution and the weight of crystals of KC1 obtained? What is the weight of crystals obtained if 5% of the original water evaporates

on cooling? Ans.

1857 kg water, 368 kg crystals ;(b) 399 kg crystals

(a)

of MgSO t 7H2 0. A hot solution containing 1000 kg of MgS0 4 and water having a concentration of 30 wt % MgS0 4 is cooled to 288.8 K, where crystals of MgS0 4 7H 0 are precipated. The solubility at 288.8 K is 24.5 wt % anhydrous MgS0 4 in the solution. Calculate the yield of crystals

12.11-3. Crystallization

•

•

2

obtained 12.11-4.

if

5%

of the original water in the system evaporates on cooling.

Heat Balance

in Crystallization.

FeSO 4/10O lb

taining 47.0 lb

total

A

feed solution of 10

water

000

lb m at

130°F con-

whereFeS0 4 7H 2 0 lb FeSO 4 /100 lb total

cooled to 80°F,

is

•

removed. The solubility of the salt is 30.5 water (PI). The average heat capacity of the feed solution is 0.70 btu/lb m °F. The heat of solution at 18°C is -4.4 kcal/g mol(-18.4 kJ/g mol) FeSOv 7H 2 0 (PI). Calculate the yield of crystals and make a heat balance. Assume that no water is vaporized. Ans. 2750 lb m FeS0 4 -7H^O crystals, q = -428 300 btu (-451 900 kJ)

crystals are

of Temperature on Yield and Heat Balance in Crystallization. Use the conditions of Example 12.11-2, but the solution is cooled instead to 283.2 K, where the solubility is 30.9 kg MgSGyiOO kg total water (PI). Calculate the effect on yield and the heat absorbed by using 283.2 K instead of 293.2 for the

12.11-5. Effect

K

crystallization.

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Environmental Engineer-

Publishers, 1982.

(R6)

Ruthven, D. M. Principles of Adsorption and Adsorption Processes, York: John Wiley & Sons, Inc., 1984.

(51)

Smith, C. T. J.

(52)

Saeman, W. C. A.I.Ch.E.

(Tl)

Treybal, R. E. Mass Transfer Operations, 3rd

Am.

Book Company, (VI)

Van Hook, A.

New

Sons, Inc., 1987.

Reynolds, T. D. Unit Operations and Processes ing,

J., 8,

New

Oil Chemists' J., 2,

Soc,

New

28, 274 (1951).

107 (1956). ed.

New

York: McGraw-Hill

1980.

Crystallization,

Theory and Practice,

New York: John

Wiley

&

Sons, Inc., 1951. (Yl)

Yang, H. H., and Brier,

Chap. 12

References

J.

C. A.I.Ch.E.

J., 4,

453 (1958).

753

CHAPTER

13

Membrane Separation Processes

INTRODUCTION AND TYPES OF MEMBRANE SEPARATION PROCESSES

13.1

13. 1A

Introduction

membranes are becoming increasingly important in the process new unit operation, the membrane acts as a semipermeable barrier and separation occurs by the membrane controlling the rate of movement of various molecules between two liquid phases, two gas phases, or a liquid and gas phase. The two fluid phases are usually miscible and the membrane barrier prevents actual, ordinary hydrodynamic flow. A classification of the main types of membrane separation Separations by the use of

industries. In this relatively

is

as follows.

13.

IB

Classification of

Membrane

Processes

/. Gas diffusion in porous solid. In this type a gas phase is present on both sides of the membrane, which is a microporous solid. The rates of molecular diffusion of the various gas molecules depend on the pore sizes and the molecular weights. This type of diffusion in the molecular, transition, and Knudsen regions was discussed in detail in Section 7.6.

2.

Gas permeation

in a

membrane.

The membrane

such as rubber, polyamide, and so on, and dissolves

in

the

membrane and

is

in this process is usually a

polymer

not a porous solid. The solute gas

first

then diffuses in the solid to the other gas phase. This was

discussed in detail in Section 6.5 for solutes following Fick's law and

is

considered again

where resistances are present in Section 13. 3; Examples are hydrogen diffusing through rubber and helium being separated from natural gas by permeation through a fluorocarbon polymer. Separation of a gas mixture occurs since each type of molecule diffuses at a different rate through the membrane. for the case

3.

Liquid permeation or dialysis.

In this case the small solutes in one liquid phase diffuse

readily because of concentration differences liquid phase (or

754

through a porous membrane to the second

vapor phase). Passage of large molecules through the membrane

is

more

membrane

This

difficult.

process has been applied in chemical processing separations

such as separation of H 2 S0 4 from nickel and copper sulfates in aqueous solutions, food processing, and artificial kidneys and is covered in detail in Section 13.2. In electrodialysis, separation of ions

4.

A membrane,

Reverse osmosis.

weight solute,

is

occurs by imposing an emf difference across the membrane.

which impedes the passage of a low-molecular-

placed between a solute-solvent solution and a pure solvent.

diffuses into the solution

imposed which causes the flow of solvent to reverse This process also

is

The solvent

by osmosis. In reverse osmosis a reverse pressure difference

used to separate other low-molecular-weight solutes, such as

in

and simple acids from a solvent (usually Sections 13.9 and 13.10.

5.

Ultrafiltration

sugars,

is

as in the desalination of seawater.

water). This process

is

covered

salts,

in detail

membrane process. In this process, pressure is used to obtain a separby a semipermeable polymeric membrane (M2). The membrane

ation of molecules

on the

discriminates

relatively high

basis of molecular size, shape, or chemical structure

and separates

molecular weight solutes such as proteins, polymers, colloidal materials

such as minerals, and so on. The osmotic pressure is usually negligible because of the high molecular weights. This is covered in Section 13.11. 6.

Gel permeation chromatography.

The porous

molecular-weight solutes. The driving force in

is

gel

retards

diffusion

concentration. This process

of the highis

quite useful

analyzing complex chemical solutions and purification of very specialized and/or

valuable components.

13.2

13. 2A

In

LIQUID PERMEATION MEMBRANE PROCESSES OR DIALYSIS Series Resistances in

membrane

Membrane

Processes

processes with liquids, the solute molecules must

first

be transported or

phase on one side of the solid membrane, through the membrane itself, and then through the film of the second liquid phase. This is shown in Fig. 13.2-la, where Cj is the bulk liquid-phase concentration of the diffusing

diffuse

through the liquid film of the

Sec. 13.2

Liquid Permeation

first

liquid

Membrane Processes

or Dialysis

755

solute solid,

with

A

kg mol

in

and

A/m 3

c lis is the

C u The .

,

cu

the concentration of

is

concentration of

A

A

and is in equilibrium m/s. The equilibrium distri-

mass-transfer coefficients are k cl and k c2 in

bution coefficient K'

adjacent to the

in the fluid just

in the solid at the surface

defined as

is

K'

^=^

^=

=

CL

C

(13.2-1)

C 2t

li

Note that K' is the inverse of K defined in Eq. (7.1-16). The flux equations through each phase are all equal

and

to each other at steady state

are as follows:

N A = Mc, - c u =

(c

)

= K'c u and

Substituting c us

HA =

fc

(c,

cl

- cu =

Dab K (c

u

-

A

in

m 2 /s.

the solid in

Note

from the permeability P M defined

different

Eq.

L

is

In

-

k c2 (c 2;

c2)

the thickness in m, and

Eq.

in

pM

in

(13.2-3)

D AB and

K'

D AB

is

the

Eq. (13.2-4)

(6.5-9). Also, the value ofp M

-c u =

N — A

K ci

c

u

-c = 2i

N — A

c 2i

in

is

is

inversely

two separate

N =— A

-c 2

and

(13.2-5)

k cz

Pm-

the equations, the internal concentrations c u

equation

=

more convenient

Cl

Adding

c 2i)

to determine p M in one separate diffusion experiment. of the parts in Eq. (13.2-3) for the concentration difference,

it is

Solving each

(13.2-2),

that the permeability

proportional to the thickness L. Instead of determining experiments,

(13.2-2)

c 2)

(13.2-4)

the permeability in the solid in m/s,

is

into

-

-~-

Pm =

diffusivity of

i

= p M {c u -

c 2i )

j

where p M

k c2 (c 2i

)

= K'c 2

c 2 is

)

- c 2is =

us

drop

c 2i

out,

and the

final

is

some cases, the resistances in the two liquid films are quite membrane resistance, which controls the permeation rate.

small

compared

to that of

the

EXAMPLE A

13.2-1.

Membrane

Diffusion

A

liquid containing dilute solute

at a

and Liquid Film Resistances concentration c,

=

3 x 10~

2

kg

mol/m 3 is flowing rapidly by a membrane of thickness L = 3.0 x 10" 5 m. The distribution coefficient K' = 1.5 and D AB = 7.0 x 10" 11 m 2 /s in the membrane. The solute diffuses through the membrane and its concentration on the other side is c 2 = 0.50 x 10" 2 kg mol/m 3 The mass-transfer coef.

ficient

10"

5

k cl

is

large

and can be considered

as infinite

=

and k c2

2.02 x

m/s. (a)

Derive the equation

to calculate the steady-state flux

NA

and make

a sketch. (b)

Calculate the flux and the concentrations at the

membrane

inter-

faces.

Solution:

For part

(a)

the sketch

concentration profile on the

7S6

left

is

shown

side

is

in Fig. (13.2-2).

flat (k cl

Chap.

13

=

oo)

and

Membrane

Note C[

=

that the cu

.

The

Separation Processes

Figure

derivation

is

same

the

Concentrations for Example 13.2-1.

13.2-2.

NA = For part

=

+ VPm c,

= calculate c 2i

NA =

7.458

7.0

=

-T~

A

To

c l1

—

Eqs. (13.2-4) and (13.2-7),

x 10"

3.0

l/k e2

1/3.5

x 10~

2

6

-

win° - 6m/ 5xl m/ s,

-

0.5

x 10

-2

1/2.02 x 10"

+

-

c 2)

=

5 2.02 x 10- (c 2f

= 0.869 x 10" 2 kg mol/m 3 Also, .

=

1.5

=

1.304 x 10

-2

=

Cl!S

- 0.5

x 10"

2 )

using Eq. (13.2-1),

0.869 x 10"

kg mol/m 3

2

.

Dialysis Processes

Dialysis uses a semipermeable different diffusion rates in the

membrane

to

separate species by virtue of their

membrane. The feed

solution or dialyzate, which

contains the solutes to be separated, flows on one side of the solvent or diffusate stream the

5

,

c 2i

13. 213

, 3

=

x 10- 8 kg mol/s-m 2

K' Solving, c 2is

(13.2-7)

x 10-"(1.5)

8 7.458.x 10' =/c c2 (C 2i

Solving, c 2i

c

3.0X10" 5

c2

0 to give

,„

,

,

(b) to calculate the flux using

P»

=

as for Eq. (13.2-6) but \lk cX

membrane

in

on the other

side.

Some

solvent

may

membrane and

the

also diffuse across

the opposite direction, which reduces the performance

by

diluting

the dialyzate. In practice

it is

used to separate species that differ appreciably

in size,

which have

a reasonably large difference in diffusion rates. Solute fluxes depend on the concentration gradient in the

comparison

to other

membrane. Hence,

membrane

membrane processes

that

dialysis

is

characterized by low flux rates, in

processes, such as reverse osmosis and ultrafiltration,

depend on applied pressure.

used with aqueous solutions on both sides of the membrane. The film resistances can be appreciable compared to the membrane resistance. Applications include recovery of sodium hydroxide in cellulose processing, recovery In general, dialysis

Sec. 13.2

is

Liquid Permeation

Membrane Processes

or Dialysis

757

of acids from metallurgical liquors, removal of products from a culture solution fermentation, desalting of cheese beer.

Many

whey

solids,

small-scale applications are used in

in

and reduction of alcohol content of the pharmaceutical industry.

Types of Equipment for Dialysis

13. 2C

Various types of geometrical configurations are used

common

in liquid

membrane

processes.

A

one similar to a filter press where the membrane is a fiat plate. Vertical solid membranes are placed in between alternate liquor and solvent feed frames, with the liquor to be dialyzed being fed to the bottom and the solvent to the top of these frames. The dialyzate and the diffusate are removed through channels located at the top and bottom of the frames, respectively. The most important type consists of many small tubes or very fine hollow fibers arranged in a bundle like a heat exchanger. This type of unit has a very high ratio of membrane area to volume of the unit. 13. 2D

An

type

is

Hemodialysis

in Artificial

Kidney

important example of liquid permeation processes

kidney

in the

biomedical

principal solutes

field.

is

dialysis with

In this application for purifying

removed are the small

human

an

artificial

blood, the

solutes urea, uric acid, creatinine, phosphates,

and excess amounts of chloride. A typical membrane used is cellophane about 0.025 mm thick, which allows small solutes to diffuse but retains the large proteins of the blood. During the hemodialysis, blood is passed on one side of the membrane while an aqueous dialyzing fluid flows on the other side. Solutes such as urea, uric acid, NaCl, and so on, which have elevated concentrations in the blood, diffuse across the membrane to the dialyzing aqueous solution, which contains certain concentrations of solutes such as potassium salts, and so on, to ensure that concentrations in the blood do not drop below certain levels. In one configuration the membranes are stacked in the form of a multilayered sandwich with blood flowing by one side of the membrane in a narrow channel and the dialyzing fluid by the other side in alternate channels.

EXAMPLE

13.2-2.

Remove Urea from Blood

Dialysis to

Calculate the flux and the rate of removal of urea at steady state

g/h from

in

cuprophane (cellophane) membrane dialyzer at 37°C. The mem2 The mass-transfer is 0.025 mm thick and has an area of 2.0 m -5 coefficient on the blood side is estimated as/c cl = 1.25 x 10 m/s and that 5 on the aqueous side is 3.33 x 10" m/s. The permeability of the membrane -5 is 8.73 x 10 m/s (B2). The concentration of urea in the blood is 0.02 g urea/100 mL and that in the dialyzing fluid will be assumed as 0. blood brane

in a

.

Solution:

The concentration

and c 2 =

0. Substituting into

NA =

c

l/k cl

+

=0.02/100 = 2.0 x 10" 4 g/mL Eq. (13.2-6),

c,

=

l/ Pu

For a time of

1

h

rate of

758

1/1.25 x 10~ 8.91

3 200 g/m

-

+

\/k c2

200

~

=

5

+

1/8.73-

-

0

x 10

-6

+

1/3.33 x 10

-5

x 10~ 4 g/s-m 2

and an area of 2.0m 2 removal

=

8.91

x

,

4

1

0 " (3600X2.0)

Chap.

13

=

6.42 g urea/h

Membrane

Separation Processes

GAS PERMEATION MEMBRANE PROCESSES

13.3

13.3A

Membrane

Series Resistances in

Processes

membrane processes with two gas phases and a solid membrane, similar equations can be written for the case shown in Fig. 13.2-lb. The equilibrium relation between the solid and gas phases is given by

In

H = —^— = Si = SlS = £25. 22.414

where S

the solubility of

is

A

in

equations

in

m

3

m

(STP)/atm

kg mol/m

the equilibrium relation in

p AU

pA

3

3

solid, as

atm. This

(13.3-1)

p A2i

shown

in

Eq.

(6.5-5),

similar to Henry's law.

is

H

flux

is

each phase are as follows:

~

R~f

=

Pa

s ~~ °

~L =

D AB H

—£— (Pah - PaiO

= ^(Pa2,-Pa2) The permeability Pm

in

kg mol/s

•

m

•

atm

is

this

Pm

differs

from the

PM

(13.3-2)

given by

Pm = D AB H = Note that

and

The

^f

(13.3-3)

4

defined in Eq. (6.5-9)

asD^gS. Eliminating

the

interfacial concentrations as before,

ff

Pai-Pai

= \KkJRT) +

where pure A

In the case

resistance k cl

in

{p Al )

the gas phase

is

and

on the k cl

left

\/{PJL)

+

side of the

(13 (

l/(k c2 /RT)

membrane,

can be considered

4)

is no diffusional Note that k Gl =

there

to be infinite.

/RT.

An example oxygenator cation, pure

Oxygen

of gas permeation in a

for a

02

membrane

is

use of

a

polymeric membrane as an

heart-lung machine to oxygenate blood. In

gas

is

on one side of a

diffuses through the

thin

membrane

this

membrane and blood

into the blood

and

C0

2

is

biomedical appli-

on the other

side.

diffuses in a reverse

direction into the gas stream.

13.3B

Types of Membranes and Permeabilities for Separation of Gases

Types of membranes. Early membranes were limited in their use because of two gases and quite low permeation fluxes. This low-flux problem was due to the fact that the membranes had to be relatively thick (1 mil or 1.

low-selectivities in separating

1/1000 of an inch or greater) in order to avoid tiny holes which reduced the separation by allowing viscous or Knudsen flow of the feed. Development of silicone polymers (1 mil thickness) increased the permeability by factors of 10 to 20 or so. Some newer asymmetric membranes include a very thin but dense skin on one side of the membrane supported by a porous substructure (Rl). The dense skin has a

Sec. 13.3

Gas Permeation Membrane Processes

759

A

and the porous support thickness is about 25-100 fj.m. The is thousands of times higher than the 1 -mil-thick original membranes. Some typical materials of present membranes are a composite of polysulfone coated with silicone rubber, cellulose acetate and modified cellulose acetates, aromatic polyamides or aromatic polyimides, and silicone-polycarbonate copolymer on a porous support.

thickness of about 1000 flux increase of these

2. Permeability

membranes tal

data for

is

membranes

of membranes.

The accurate prediction of permeabilities of gases

in

generally not possible, and experimental values are needed. Experimen-

common

that there are

gases in some typical membranes are given in Table 13.3-1. Note wide differences among the permeabilities of various gases in a given

membrane. Silicone rubber exhibits very high permeabilities for the gases in the table. For the effect of temperature T in K, the In P'A is approximately a linear function of l/T and increases with T. However, operation at high temperatures can often degrade the membranes. When a mixture of gases is present, reductions of permability of an individual component of up to 10% or so can often occur. In a few cases much larger reductions have been observed (Rl). Hence, when using a mixture of gases, experimental data should be obtained to determine if there is any interaction between the gases. The presence of water vapor can also have similar effects on the permeabilities and can also possibly damage the membranes. Types of Equipment for Gas Permeation Membrane Processes

13.3C

membranes are mainly used for experimental use to membrane. The modules are easy to fabricate and use and the areas of the membranes are well defined. In some cases modules are stacked together like a multilayer sandwich or plate-and-frame filter press. The major drawback of this type is the very small membrane area per unit separator volume. I.

membranes.

Flat

Flat

characterize the permeability of the

Table

13.3-1.

Permeabilities of Various Gases

in

Membranes

cm^iSTP) cm Permeability, P'A

x

,

10

10

cm ' cm Hg

s

Temp. Material

(°C)

He

Hi

CH

co

A

2

Oi

Silicone rubber

25

300

550

800

2700

500

Natural rubber

25

31

49

30

131

24

Polycarbonate

25-30

15

12

5.6,10

1.4

0.17

0.034

N

2

250 8.1

Ref.

(S2)

(S2) (S2)

(Lexane)

Nylon 66

25

1.0

Polyester

1.65

0.035

0.31

0.008

(S2)

0.031

(HI)

(Permasep) Silicone-

210

25

160

970

70

(W2)

polycarbonate

copolymer

(57% Teflon

silicone)

FEP

30

62

Ethyl cellulose

30

35.7

49.2

7.47

47.5

Polystyrene

30

40.8

56.0

2.72

23.3

760

2.5

1.4

Chap. 13

11.2

7.47

3.29 2.55

(SI)

(W3) (W3)

Membrane Separation Processes

Small commercial

flat

membranes

are used- for producing oxygen-enriched air for

individual medical applications. 2.

Spiral-wound membranes. This configuration maintains the simplicity of fabricatmembranes while increasing markedly the membrane area per unit separator

ing flat

2 3 ft /ft

volume up to 100

m 2 /m 3

) while decreasing pressure drops (Rl). The assembly consists of a sandwich of four sheets wrapped around a central core of a perforated collecting tube. The four sheets consist of a top sheet of an open separator grid for the feed channel, a membrane, a porous felt backing for the permeate channel, and another membrane as shown in Fig. 13.3-1. The spiral-wound element is 100 to

(328

mm

m

in diameter and is about 1 to 1.5 200 long in the axial direction. The fiat sheets before rolling are about 1 to 1.5 m by about 2 to 2.5 m. The space between the membranes (open grid for feed) is about 1 and the thickness of the porous backing (for permeate) is about 0.2 mm.

mm

The whole spiral-wound element enters at the -

channel

Then

left

end of the

in the axial direction

is

located inside a metal shell.

The feed gas

feed channel, and flows through this

shell, enters the

of the spiral to the right end of the assembly (Fig. 13.3-1).

the exit residue gas leaves the shell at this point.

feed channel, permeates perpendicularly through the

The feed

stream, which

is in

the

membrane. This permeate then

flows through the permeate channel in a direction perpendicular to the feed stream toward the perforated collecting tube, where it leaves the apparatus at one end. This is illustrated in Fig. 13.3-2, where the local gas flow paths are shown for a small element of the assembly. 3.

Hollow-fiber membranes.

hollow

fibers.

The

The membranes

inside diameter of the fibers

are in the shape of very small diameter is in

the range of 100 to 500 /j.m and the

The module resembles a Thousands of fine tubes are bound together at each end

outside 200 to 1000 fxm with the length up to 3 to 5 m. shell-and-tube heat exchanger.

membrane

permeate channel FIGURE

13.3-1.

Spiral-wound elements and assembly. [From R. Eng., 88 (July

Sec. 13.3

13),

I.

Berry,

Chem.

63 (1981). With permission.]

Gas Permeation Membrane Processes

761

surrounded by a metal shell having a diameter of 0.1 to 0.2 m, 2 3 per unit volume is up to 10 000 m /m as in Fig. 13.3-3. Typically, the high-pressure feed enters into the shell side at one end and leaves at the other end. The hollow fibers are closed at one end of the tube bundles. The permeate gas inside the fibers flows countercurrently to the shell-side flow and is

into a tube sheet that

so that the

is

membrane area

collected in a

chamber where the open ends of the

fibers terminate.

Then

the permeate

exits the device.

reject (residue)

4,

sealed

end fiber

permeate

bundle

mm t

feed

7ZZA

0,0

YZZL

permeate Figure

762

13.3-3.

Hollow-fiber separator assembly.

Chap. 13

Membrane

Separation Processes

Introduction to Types of Flow in

13.3D

Gas Permeation

Types offlow and diffusion gradients. In a membrane process high-pressure feed is supplied to one side of the membrane and permeates normal to the membrane. The permeate leaves in a direction normal to the membrane, accumulating on the low-pressure side. Because of the very high diffusion coefficient in gases, concentration gradients in the gas phase in the direction normal to the surface of the membrane are quite small. Hence, gas film resistances compared to the membrane resistance can be neglected. This means that the concentration in the gas phase in a direction perpendicular to the membrane is essentially uniform whether or not the gas stream is

1.

gas

flowing parallel to the surface or

stream

If the gas

is

is

not flowing.

flowing parallel to the

concentration gradient occurs

in this.direction.

membrane

plug flow, a

in essentially

Hence, several cases can occur

in the

membrane module. The permeate side of the membrane can be operated so that the phase is completely mixed (uniform concentration) or where the phase is in plug flow. The high-pressure feed side can also be completely mixed or in plug flow. Countercurrent or cocurrent flow can be used when both sides are in plug flow. Hence, separate theoretical models must be derived for these different types of operation as

operation of a

given in Sections 13.4 to 13.7. 2.

Assumptions used and ideal flow patterns.

In deriving theoretical

models for gas

separation by membranes, isothermal conditions and negligible pressure drop in the feed stream and permeate stream are generally assumed.

It is

assumed that and that the between different

also

the effects of total pressure and/or composition of the gas are negligible

permeability of each component

is

constant

no interactions

(i.e.,

components). Since there are a number of idealized flow patterns, the important types are

summarized in Fig. 13.3-4. In Fig. 13.3-4a complete mixing is assumed for the feed chamber and the permeate chamber. Similar to a continuous-stirred tank, the reject or residue and the product or permeate compositions are equal to their respective uniform compositions

in

the chambers.

MM

permeate

|

|

c3

permeate

y//>////////////////////////, 1-

c3

feed

feed

reject

(b)

(a)

permeate

permeate -«

-x

'///////////////////////////, t-

p.

1_

feed

&~

:

-

>-

>-

feed

(c)

13.3-4.

*

'///////////////////////////,

—— reject

Figure

^-

g>_

reject (d)

Ideal flow patterns in a membrane separator for gases: (a) complete mixing, (b) rossflow, (c) countercurrent flow, (d) <

cocurrent flow.

Sec. 13.3

Gas Permeation Membrane Processes

763

An

ideal cross-flow pattern

is

given in Fig. 13.3-4b, where the feed stream

plug flow and the permeate flows

in

a normal direction

away from

without mixing. Since the feed composition varies along

its

the

is in

membrane

flow path, the local

permeate concentration also varies along the membrane path. In Fig. 13.3-4c both the feed stream and permeate stream are in plug flow countercurrent to each other. The composition of each stream varies along its flow path. Cocurrent flow of the feed and permeate streams

is

shown

in Fig. 13.3-4d.

COMPLETE-MIXING MODEL FOR GAS SEPARATION BY MEMBRANES

13.4

13. 4A

Basic Equations Used

In Fig. 13.4-1 a detailed process flow diagram

separator element

is

is

operated at a low recovery

a small fraction of the entering feed rate), there

Then

shown for complete mixing. When a where the permeate flow rate is

(i.e.,

a minimal change in composition.

is

model provide reasonable mates of permeate purity. This case was derived by Weller and Steiner (W4). The overall material balance (Fig. 13.4-1) is as follows: the results derived using the complete-mixing

qf

=q 0

+

esti-

(13.4-1)

qp

where qf is total feed flow rate in cm 3 (STP)/s; q 0 is outlet reject flow rate, cm 3 (STP)/s; 3 and q p is outlet permeate flow rate, cm (STP)/s. The cut or fraction of feed permeated, d, is

given as

0

=

q — P

(13.4-2)

qf

The

rate of diffusion

or permeation of species

A

(in

a binary of A and B)

by an equation similar to Eq. (6.5-8) but which uses rather than flux in kg mol/s-

_

q_a

An

cm 2

,

cm 3 (STP)/s

is

given below

as rate of

permeation

.

q Py P

(13.4-3)

P,y P )

Am

permeate out yP

1

CO

low-pressure side A

A

p A

t

,

yp A

A

'//////////////////////////////////////////////s

=

qf ,xf »

feed

in

Ph' x o

T Figure

764

13.4-1.

(1-6)9/ >~

high-pressure side

reject out

xo

Process flow for complete mixing case.

Chap. 13

Membrane Separation Processes

permeability of A in the membrane,

cm 3 (STP)

cm 2 cm Hg); q A 2 is membrane is flow rate of A in permeate, cm (STP)/s; A m is membrane area, cm is total thickness, cm;p h is total pressure in the high-pressure (feed) side, cm Hg; pressure in the low-pressure or permeate side, cm Hg; x Q is mole fraction of A in reject side; xj is mole fraction of A in feed; and y p is mole fraction of A in permeate. Note that Pf,x 0 is the partial pressure of A in the reject gas phase. where P'A

is

cm/(s

•

•

•

3

;

A

component B.

similar equation can be written for

f -£p± = =

q

'

-

x„)

where P'B is permeability of B, cm 3 (STP) cm/(s- cm 2 by (13.4-4) •

a*[x a -

yp

l-y p

t

- piX -

•

cm

(13.4-4)

Hg). Dividing Eq. (13.4-3)

[pi/p h )y p ]

-x 0 -{ Pl/p h )(\ -y p

(l

y p)]

)

(13.4-5) )

This equation relates y p the permeate composition, to x a the reject composition, and ,

,

the ideal separation factor a*

defined as

is

(13.4-6)

"B

Making an

on component A

overall material balance

1fXf = q 0 x 0 + and solving

Dividing by

x" =

Substituting q p area, A m

membrane

=

for the outlet reject composition,

xf

- By

o^TT

or

x f - x 0 (l - 6)

=

y"

(13 4 " 8) -

e

Oqj from Eq. (13.4-2) into Eq. (13.4-3) and solving for the

,

Am =

13.4B

—

(13.4-9)

-p,y p

(P'A /t)( Ph x 0

)

Solution of Equations for Design of Complete-Mixing Case

For design of a system there are seven variables in a *> PilPh> an d A m four of which

*/> x o> y P

1.

the complete-mixing

,

<

commonly occurring cases Case

(13.4-7)

p yp

This

is

the simplest case

is

and y p d, and quadratic equation use of the

where Xf, x 0 a* and p,/p h ,

,

By

are given

solved for the permeate composition y p

yp

Sec. 13.4

Two

are considered here.

A,„ are to be determined by solution of the equations.

formula, Eq. (13.4-5)

model (HI),

are independent variables.

=

-b +

2

\jb

in

,

terms of x 0

.

- Aac (13.4-10)

2a

Complete-Mixing Model for Gas Separation by Membranes

765

where

=

a

- a*

1

Ph b=—

- x 0) -

,

,

(I

+ a*

1

Pi

—x

Ph

0

+

a*

Pi

Ph Pi

Hence, to solve this case, y p is first calculated using Eq. (13.4-10). Then the fraction of feed permeated, 9, is calculated using Eq. (13.4-8) and the membrane area, A m from Eq. (13.4-9).

EXAMPLE 13.4-1. Design of a Membrane Unit for Complete Mixing A membrane is to be used to separate a gaseous mixture, of A and B whose is feed flow rate is qj = 1 x 10 cm (STP)/s and feed composition of xj = 0.50 mole fraction. The desired composition of the reject is x a = 3 0.25. The membrane thickness t = 2.54 x 10~ cm, the pressure on the

A

3

p h = 80 cm

and on the permeate side is p = 20 cm Hg. = 50 x 10 -10 cm 3 (STP) cm/(s cm 2 10 10~ Assuming the complete-mixing model, cm Hg) and P'B = 5 x calculate the permeate composition, y p the fraction permeated, 9, and the feed side

The

is

Hg

t

permeabilities are P'A

•

•

•

.

,

membrane

area,

Am

.

Solution: Substituting into Eq. (13.4-6),

P\

10 50 x 10"

P'B

(5

x 10" 10 )

=

10

Using Eq. (13.4-10), a

=

b =

1

- a* =

—

( 1

1

-

- X 0) -

10

1

= -9

—x

+ a*

Pi

=

80 —

(1

-

0.25)

c=a**p>> -x 0 =

-1 +

-10

yP

-b +

/

/80\ 10

1

—

1(0.25)

+

10

= 22.0

80 V(0.25) = -10 —

- 4ac

2

yjb

2a -22.0

+

/(22.0)

N

2

- 4(-9)(-10)

= 0 604 -

v=» 766

+ a*

\20/

Pi

=

0

Pi

Chap. 13

Membrane Separation Processes

Using the material balance equation

x f - 8y D

0.25=

^-J^f; Solving, 6

=

(13.4-8),

_

0.50 - 0(0.604)

0.706. Also, using Eq. (13.4-9),

Hf y

Am

(P'A /t)(p h

P

-p,y p )

x0

4 0.706(1 x 10 )(0.604)

~

=

10 3 [50 x 10" /(2.54 x 10~ )](80 8 2.735 x 10

4 (2.735 x 10

cm 2

x 0.25 - 20 x 0.604)

m2

)

Case 2. In this case xj-, 6, a* and p\lph are given and y p x 0 and A m are to be determined. Equation (13.4-5) cannot be solved fory^ since x 0 is unknown. Hence, x a ,

from Eq. yp

,

,

(13.4-8) is substituted into

Eq. (13.4-5) and the resulting equation solved for

using the quadratic equation to give

yP

b,

+

—

yfb}

=

2a

-

4a, c,

(13.4-n) i

where a,

=

d

Pi

Pi

+

6

Ph

Ph

b\=
1

-

0

- a*6 - a*

Pi — + a* — Pi

Ph Pi

- xf

Pi +—

Ph

0

0

Ph

+ a*0'.+ a*

Ph

Pi

a*

Pi — 6 + a*x f

Ph

Ph

= — a*Xf

After solving tovy p

,

the value of

x0

is

calculated from Eq. (13.4-8) and

Am

from Eq.

(13.4-9).

EXAMPLE

Membrane Design for Separation of Air membrane area needed to separate an air stream using a membrane mil thick with an oxygen permeability of 10 cm 3 (STP)-cm/(s-cm 2 -cmHg). An a* = 10 for P'A = 500 x 10"" oxygen permeability divided by nitrogen permeability (S6) will be used. x 10 6 cm 3 (STP)/s and the fraction cut d = 0.20. The feed rate is qf = The pressure selected for use are p h = 190 cm Hg and p = 19 cm Hg. It is

13.4-2.

desired to determine the 1

1

l

Again, assuming the complete-mixing model, calculate the permeate composition, the reject composition, and the area. Solution:

a,

=

Using Eq. (13.4-11) for a feed composition of Xj- = 0.209,

0+

Pi

Pi 0

Ph

=

0.2

+

19

Pi Pi — + a* — 0

Ph

Ph

Ph 19

(0.2)

190

Sec. 13.4

- a*^ 9 - a*

190

-

10(0.2)

-

/l9\

(19)

+

10

190

10

(0.2) 1

= -2.52

190/

Complete-Mixing Model for Gas Separation by Membranes

767

by

=

=

-

i

X,

ILL

ILL

Ph

Ph

0.2

a*—

Ph

Ph

9

+ a*xf

19

- 0.209

+ 190

+ 10 |-^-| \l90j

Pi

Pi

+ a*6 + a*

19

-

1

-

e

+

(0.2)

10(0.2)

190

10\~ 1(0.2) \

190

+ 10(0.209) = 5.401 c,

= -<x*xf = -10(0.209) = -2.09 b

+

x

- 4a,cj

\jbf

2a

,

- 5.401 +

2

V(5.401) 2(

-

-

4(

-

2.52)(

-

2.09)

2.52)

= 0.5067 Substituting into Eq. (13.4-8),

_

X °

9y p _ 0.209 - 0.2(0.5067)

xf

~~

(1

- 6) ~

(1

-

= 0.1346

0.2)

Finally, using Eq. (13.4-9) to find the area

0q f y p

(P'JtHPhXo - Piy P

)

6

0.2(1 x 10 )(0.5067) _,u /2.54 x (500 x 10

= 3.228 x

Minimum

13. 4C

10

8

]

190 x 0.1346

0

-

19 x 0.5067)

cm 2

Concentration of Reject Stream

and the feed composition xf = y p For all values of 6 < 1, the permeate composition y p > Xf (HI). Substituting the value Xf = y p into Eq. (13.4-5) and solving, the minimum reject composition x oM for a given xj If all

value

of the feed

is

is

permeated, then 9 =

1

.

obtained as xf

x oM -

+ (a* -

1)

—

(1

Ph a*(\

-Xf + )

xf

-xf

)

(13.4-12)

Hence, a feed of Xf concentration cannot be stripped lower than a value of x oM even with an infinitely large membrane area for a completely mixed system. To strip beyond this limiting value a cascade-type system could be used. However, a single unit could be used which is not completely mixed but is designed for plug flow.

768

Chap.

13

Membrane Separation Processes

EXAMPLE 13.4-3.

Effect of Feed Composition on

Minimum Reject Concen-

tration

Calculate the minimum reject concentration for Example 13.4-1 where the feed concentration is Xf = 0.50. Also, what is the effect of raising the feed purity to Xf = 0.65?

= 0.50

Solution: Substituting

XJ

into Eq. (13.4-12),

+(«* -

1

1)

—

-Xf

(1

)

Ph

x oM -

a*(l

0.50

1

-xf)+xf

+(1010(1

-

1)1^1(1 -0.50)

+

0.50)

0.50

= 0.1932 For an xj = 0.65,

0.65

1

x oM -

10(1

-

= 0.2780 0.65)

+ 0.65

COMPLETE-MIXING MODEL FOR MULTICOMPONENT MIXTURES

13.5

13. 5A

Derivation of Equations

When multicomponent is

+ (10- 1)(^)(1 -0.65)

quite useful. This

mixtures are present, the iteration method by Stern et

method

will

and C. The process flow diagram is the same as Fig. composition Xf is xjA XfB and xjC The known values are

xfA

>

The unknown values

xfB, xfc to be

;

qf, 0;

ph

,

Pl

;

P'A

,

P'B

eight

13.4-1,

where the feed

,

P'c

;

and

t

determined are

y P A' y P B< y P c'< x 0 a< x i?b- x <,c'< q p or

These

(SI)

.

,

,

al.

be derived for a ternary mixture of components A, B,

unknowns can be obtained by

<7

0

;

and

A

„,

solving a set of eight simultaneous

equations using an iteration method. Three rate of permeation equations similar to Eq. (13.4-3) are as follows for

components A, B, and

Q Py P A


Sec. 13.5

Py P B

=

=

—A P'a

—A

C

m {p h x oA - p,y pA )

m {p h x oD

- Piy pB )

Complete-Mixing Model for Multicomponent Mixtures

(13.5-1)

(13.5-2)

769

— Am

P'c

=

q Py P c

The three

{p h x oC - Piy pC )

(13.5-3)

material balance equations similar to Eq. (13.4-8) are written for components

A, 5, and C. e

1

1

X/A

-

yPA

(13.5-4)

yPB 7^~8 >p

(13.5-5)

61

-

1

6

1

x oB

~ i

xfB

-

e

8

1

x oC Also, two

final

8

XJfC

:

i-e"

(13.5-6)

yPc

i-9

"

equations can be written as

2

=

yp"

y PA

+ y PB + y P c =

X oA

+ x oB + x oC ~

i-o

(13.5-7)

-0

(13.5-8)

n

2X

=

°"

1

Substituting x oA from Eq. (13.5-4) into Eq. (13.5-1) and solving fof?A

m

,

Q P y P At (13.5-9)

Ph

Pa 1

For component B, Eq.

(13.5-5)

is

(x/a

-

-

Qy P A) - Piy pA

8

substituted into Eq. (13.5-2), giving

q P y P B<

Ph 1

-

U/b - Sy pB - p y pB )

0

Rearranging Eq. (13.5-10) and solving fory pfl

(13.5-10)

t

,

Ph x;bK

\

-

9) (13.5-11)

q p t/(P'B A m ) + 9p h /(l - 8)+p, In a similar

manner Eq.

(13.5-12)

is

derived for y p c-

Ph xfc/(\ yPc

13. 5B

q p t/(P'c

Aj

Iteration Solution Procedure for

The following

iteration or trial-and-error

+

-

9)

9 Ph /(\

- 8) +

(13.5-12) p,

Multicomponent Mixtures procedure can be used

to solve the

equations

above. 1.

A

2.

Using Eq.

3.

The membrane area

770

value of y pA

is

(13.4-2)

assumed where y pA > XfA and the known value of is

.

9,

qp

is

calculated.

calculated from Eq. (13.5-9).

Chap. 13

Membrane Separation Processes

4. 5.

6.

Values of y pB and y pC are calculated from Eqs. (13.5-11) and (13.5-12). is calculated from Eq. (13.5-7). If this sum is not equal to "1 through 5 are repeated until the sum is 1.0.

The sum £„ y pn Finally, x oA

1.0, steps

x oB and x oC are calculated from Eqs. (13.5-4), (13.5-5), and ,

,

EXAMPLE

(13.5-6).

Design of Membrane Unit for Multicomponent Mixture composition of xj-A = 0.25, = = and 0.55, 0.20 to separated by a membrane with a is be XfC ~ XfB 3 thickness of 2.54 x 10 cm using the complete-mixing model. The feed 4 3 flow rate is 1.0 x 10 cm (STP)/s and the permeabilities are P'A = 200 x 10" 10 cm 3 (STP)-cm/(s-cm 2 -cm Hg), P'B = 50 x 10 -10 and P'c = " 10 The pressure on the feed side is 300 cm Hg and 30 cm Hg on 25 x 10 the permeate side. The fraction permeated wiU be 0.25. Calculate the permeate composition, the reject composition, and the membrane area using the complete-mixing model. 13.5-1.

A multicomponent gaseous mixture having a

,

.

Solution: Following the iteration procedure, a value of y pA assumed. Substituting into Eq. (13.4-2) for step 2, 4 4 q p = 6qf = 0.25 x 1.0 x 10 = 0.25 x 10

Using Eq.

(13.5-9), the

membrane area

=

0.50

is

cm 3 (STP)/s

for step (3) is

q P y P At

Ph P'a

- oy P A) -Piy P A

(*/a

4 0.25 x 10 (0.50)(2.54 x 10

200 x 10" 10

-3 )

/

300 1

-

= 4.536 x

10

Following step (13.5-11)

and

q p ll(P'B

0.50)

-

30(0.50)

6

4,

cm 2 the values y pB

and y pC are calculated using Eqs.

(13.5-12).

-

xpsl(\

Ph yPB

- 0.25 x

(0.25

0.25

Am + )

8)

9 Ph /(l

- 0)+p, 300 x 0.55/(1 - 0.25)

0.25 x 10

4

x 2.54 x 10" /(50 x I0~'° x 4.536 x 10 6 ) + 0.25 x 300/(1 - 0.25) + 30 3

= 0.5366 p h x fc l( y P c-

q p l/(P'c

-

\

0)

A m )+ 8p h l{\ -8)+p, 300 x 0.20/(1 - 0.25)

0.25 x 10

4

=

Sec. 13.5

x 2.54 x iO" /(25 x 10"'° x 4.536 x |0 6 ) + 0.25 x 300/(1 - 0.25) + 30 3

0.1 159

Complete-Mixing Model for Multicomponenl Mixtures

771

Substituting into Eq. (13.5-7),

Ey

Pn

=

+ y pB + y P c = o.sooo + 0.5366 +

y PA

o.

1 1

59 =

1

.

1525

n

For the second

iteration,

assuming that y pA

=

0.45, the following

values are calculated: 6 3.546 x 10

Am =

cm 2

y pB

Xy The y B

Pn

= 0.4410

y pC = 0.0922

= 0.9832

A m = 3.536 x 10 6 cm 2 y pA = 0.4555, = = 0.4502, and y pC 0.0943. Substituting into Eqs. (13.5-4),

final iteration

(13,5-5),

and

values are

;

(13.5-6)

=

'oa

r^i *m

>m =

"

rrb?

(0 - 25)

0.25 1

=

- 0.25

(0.4555)

= 0.1815

7^7 */» - rz7 y ^ =

(0 55) "

0.25 1

-

(0.4502)

= 0.5833

(0.0943)

= 0.2352

0.25

0.25 1

13.6.

13. 6A

A

-

0.25

CROSS-FLOW MODEL FOR GAS SEPARATION BY MEMBRANES Derivation of Basic Equations

detailed flow diagram for the cross-flow

(W3, W4)

is

shown

in

Fig.

13.6-1.

model derived by Weller and Steiner

In this case the longitudinal

velocity of the

plug flow and membrane. On the low-pressure side the permeate stream is vacuum, so that the flow is essentially perpendicular to the

high-pressure or reject stream

is

large

enough so

that this gas stream

is in

flows parallel to the

almost pulled into

membrane. This model assumes no mixing in the permeate side and also no mixing on the high-pressure side. Hence, the permeate composition at any point along the membrane is determined by the relative rates of permeation of the feed components at that point. This cross-flow pattern approximates that

772

in

an actual spiral-wound membrane

Chap. 13

Membrane Separation Processes

permeate out

low-pressure side

feed

reject out

in

high-pressure

volume element

plug

side

flow FIGURE

13.6-1.

Process flow diagram for cross-flow model.

separator (Fig. 13.3-1) with a high-flux asymmetric

membrane

resting

on a porous

felt

support (P2, Rl). Referring to Fig. 13.6-1, the local permeation rate over a differential

area

dA m

at

any point

-(1

membrane

in the stage is

-ydq =

—

[p h x

- y)dq =

—

[

where dq is the total flow by (13.6-2) gives

rate

P'a

- p,y]dA r

Ph (\ -

-p,(l-

x)

(13.6-1)

y))dA,

permeating through the area

a*[x -

y

l-y

(1

-x) -

dA m

(13.6-2)

.

Dividing Eq. (13.6-1)

{p,/p h )y]

(13.6-3) (

Pl /p h )(l

-y)

This equation relates the permeate composition y to the reject composition x at a point along the path. It is similar to Eq. (13.4-5) for complete mixing. Hwang and

Kammermeyer

(HI) give a computer program for the solution of the above system of

by numerical methods. Weller and Steiner (W3, W4) used some ingenious transformations and were able

differential equations

to

obtain an analytical solution to the three equations as follows: (1

-

0*)(1

(1

-

fuf - EID\

x)

-*/)

u

- E/D

R

lu f

- a* + F\ s (u f - F\

+

F

T

(13.6-4)

where

1-^ 9/

i

Sec. 13.6

=

Cross-Flow Model for Gas Separation by Membranes

773

= -Di + (D 2 i 2 + 2Ei + F 2 ) 05

u

- a*)p,

(1

D=

•+ a'

0.5

Ph

E=

Z>F 2

-

(1

F=

«*) PJ

-0.5

-

1

Ph 1

2D -

a*(D -

+F (2D - l)(a*/2 - F)

5 =

1

is

1)

1

7= The term wy

1

the value of u at

permeated up to the value of x

-D/

=

if

(F/F)

—

Xfl{\

—

The value of 9* is outlet where x = x 0

Xf).

the fraction

At the the value of 9* is equal to 9, the total fraction permeated. The composition of the exit permeate stream is y p and is calculated from the overall material balance, Eq. (13.4-8). The total membrane area was obtained by Weller and Steiner (W3, W4) using

some

in Fig. 13.6-1.

,

additional transformations of Eqs. (13.6-1) to (13.6-3) to give tq f

Am

(1

fij

-

-

0*)(1

x) di

(13.6-5)

PhP'n Ji

Pi /

1

(/;

" 0 i

+

Ph

«

1

V

+ fi.

where fi

Values of 6*

l = (Di - F) + (D l i' + 2Ei + F 2x0.5 )

in the integral

can be obtained from Eq.

The term

(13.6-4).

The

integral

can be

feed Xf and i 0 is the value of at the outlet x Q A shortcut approximation of the area without using a numerical integration is available by Weller and Steiner (W3) which has a maximum error of

calculated numerically.

if is

the value of

i

at the

.

about 20%. 13. 6B

Procedure for Design of Cross-Flow Case

model there are seven variables and two of the most common cases were discussed in Section 13. 4B. Similarly, for the cross-flow model these same common cases occur. In the design for the complete-mixing

Case

1.

The values of

Xf,

xD

,

a*, and p lp h are given and y p f

,

9,

and

Am

are to be

determined. The value of.0* or 9 can be calculated directly from Eq. (13.6-4) since other values in

this

calculate the area

774

equation are known.

Am

,

Then y p

is

calculated from Eq. (13.4-8).

all

To

a series of values of x less than the feed Xf and greater than the

Chap. 13

Membrane Separation Processes

reject outlet

x 0 are substituted

Eq. (13.6-4) to give a series of Q* values. These

into

values are then used to numerically or graphically integrate Eq. (13.6-5) to obtain the

areaA m Case

.

2.

y p ,~x 0 and A m and error, where values of x„ are substituted into Eq. solve the equation. The membrane area is calculated as in Case 1.

In this case the values of xj, d, a*, and p\lp\, are given and

are to be determined. This (13.6-4) to

,

is trial

EXAMPLE

13.6-1. Design of a Membrane Unit Using Cross-Flow The same conditions for the separation of an air stream as given in Example 13.4-2 for complete mixing are to be used in this example. The process flow streams will be in cross-flow. The given values are Xf — 0.209, 0 = 0.20, a* = 10, p h = 190 cm Hg, p, = 19cmHg,
and (a)

(b)

t

= 2.54 x

10

cm. Do as follows: and A m

Calculate y p x a Compare the results with ,

,

Solution: Since this

used for the

is

first trial

/

•

3

.

Example

same

the

Case

as

13.4-2. 2,

a value of x„

=

0.

1642

will

be

for part (a). Substituting into Eq. (13.6-4)

xf

=

0.209

xf

0.2642

-

1

0.209

0.1642 1

D=

=

- 0.1642

0.1965

- a*)p,

(1

+ a

0.5

Ph

- 10)19

(1

=

+ 10 = 4.550

0.5

190

F=

(1

-0.5

" a*)Pi

-

1

Ph

=

(1

-

10)19

-

-0.5

= 0.950

1

190

a* — - DF = — 10

1

R =

S =

1

2D -

1

«*(£>-

2(4.550) 1)

+

-

= 0.12346 1

F

(2D - l)(o*/2 - F) 10(4.550 (2

Sec. 13.6

4.550(0.950) = 0.6775

x

4.550 -

1)

+ 0.950

l)(10/2

-

=

1.1111

0.950)

Cross-Flow Model for Gas Separation by Membranes

775

1

1= 1

-D -

1

-

(E/F) 1

uf =

=

= -0.2346

- 0.6775/0.950

4.550

-Di + {D 2 i 2 + 2Ei + F 2 ) 05 -(4.550)(0.2642)

+

2

[(4.550) (0.2642)

+ 2(0.6775)(0.2642) +

(0.950)

2

05

2 ]

= 0.4427

=

u

2

-(4.550)(0.1965) + [(4.550) (0.1965)

+

2(0. 6775)(0. 1965)

+ (0.950) 2 ] 0

-

2

5

= 0.5089 (1

-

fl*)(l

(1

- xf)

-

x)

_

(1

-

- 0.1642)

fl*)(l

~ (1

- 0.209)

/0.4427

-

0. 6775/4. 550\

^0.5089

-

0. 6775/4. 550y

0 12346

11,11 /0.4427 - 10 + 0.950\

\0.5089 - 10 + 0.950/ /0.4427 - 0.950\ >0.5089

-

""°' 2346

0.950;

Solving 9* = 0.0992. This value of 0.0992 does not check the given value of 8 = 0.200. However, these values can be used later to solve Eq. (13.6-5).

For the second iteration, a value of x 0 = 0.142 is assumed and it is used again to solve for 6* in Eq. (13.6-4), which results in 9* = 0.1482. For the final iteration, x 0 = 0.1190 and 6* = 6 = 0.2000. Several more values are calculated for later use and are for x a = 0.187, 6* = 0.04876, and {otx 0 = 0.209, 0* = 0. These values are tabulated in Table 13.6-1.

Table

Calculated Values for

13.6-1.

Example x

yp

F,

0

0.209

0.6550

0.6404

0.04876

0.1870

0.6383

0.7192

0.0992

0.1642

0.6158

0.8246

0.1482

0.1420

0.5940

0.9603

0.2000

0.1190

0.5690

1.1520

e*

776

13.6-1

Chap.

13

Membrane Separation Processes

Using the material-balance equation (13.4-8) to calculate y p xf

-x 0 (l -

0.209

0)

-

0.1190(1

calculate

=

givey p

To

- 0.2000) 0.5690

0.2000

9

To

,

=

at 9*

yp

0,

Eqs. (13.6-3) and (13.4-10) must be used and

0.6550.

solve for the area, Eq. (13.6-5) can be written as

-

(1

Am — PhP'B

9*)(l

Pi

1

Ji (/,•

- 0 1

+

-x)

V

Ph

i

PhP'B

1

+fij (13.6-6)

where the function

F

f

is

defined as above. Values of

F,-

will

for different values of i in order to integrate the equation.

x a = 0.119 and fromEq.

be calculated = 0.200,

For 9*

(13.6-4),

0.119

= 0.1351 (1-*) From Eq.

(1-0.119)

(13.6-5),

fi

= (Di - F)+ (D 2 i 2 + 2Ei +

F2

0 5 '

)

= (4.55 x 0.1351 - 0.950) + [(4.55) 2 (0. 135 1) 2 + 2(0.6775)(0.1351) + (0.95) 2 ]

05

= 0.8744 Using the definition of F,- from Eq.

- 0*)U -x)

(1

F,=

Pi

1

(//

(13.6-6),

- 0 i

+

'

PhV+fij (1

-

0.200)(1

-

0.119)

1

19

(0.8744 - 0.1351) 1

+ 0.1351

190

V 1

+ 0.8744,

= 1.1520 Other values of F,- are calculated for the remaining values of 9* and are tabulated in Table 13.6-1. The integral of Eq. (13.6-6) is obtained by using the values from Table 13.6-1 and plotting F,- versus / to give an area of 0.1082. Finally, substituting into Eq. (13.6-6) tq f

PhP'B

2.54 x 10

fif

)i_

Sec. 13.6

( 1

x 10 6 )

u 190(50 x 10"' )/10

'

8 2.893 x 10

-3

cm 2

Cross-Flow Model for Gas Separation by Membranes

777

For 10

8

cm

from Example 13.4-2, y = 0.5067 and A m = 3.228 x Hence, the cross-flow model yields a higher y p of 0.5690

part (b), .

compared

to 0.5067 for the complete-mixing model. Also, the area for the cross-flow model is 10% less than for the complete-mixing model.

COUNTERCURRENT-FLOW MODEL FOR GAS SEPARATION BY MEMBRANES

13.7

13.7A

A

Derivation of Basic Equations

flow diagram for the countercurrent-flow model

streams are in plug flow.

The

is

given in Fig. 13.7-1, where both

derivation follows that given

by Walawender and Stern

(W5). Others (Bl, P4) have also derived equations for this case.

Making a

total

and a component balance

for

A

over the volume element and the

reject,

q

= q0 +

q'

(13.7-1)

qx = q 0 x 0 + q'y

(13.7-2)

Differentiating Eq.(13.7-2)

d(qx)

A balance

for

component

A

= 0 +

(13.7-3)

d(q'y)-

on the high- and low-pressure side of the volume element

gives

qx= Simplifying,

- dq){x -

dx)

+ y dq

(13.7-4)

d(qx)

(13.7-5)

we have y dq

The

(q

=

q dx

+ x dq =

local flux out of the element with area

dA m

is

-y dq = -^{p h x- p,y) dA m

plug

—

q ,y

permeate

-<

Pi

-*

9p = **S

(13.7-6)

-^-j

flow

-<-j j

j

feed in reject out

Ph

r xf

g

q-dq, x-dx

q,x i

plug flow

-tJ

dA m

q Q = (i-e)g;

differential

volume element Figure

778

13.7-1.

Flow diagram for the countercurrent-flow model.

Chap. 13

Membrane Separation Processes

Combining Eqs.

and

(13.7-3), (13.7-5),

(13.7-6)

pi

=~ (p

-d(q'y) = -d(qx)

Similarly, for

(13.7-7)

-d[q{\-x)]=^j{p h {\-x)- Pl {\-y)}dA m

(13.7-1) with (13.7-2) to eliminate q'

Combining Eq.

(~~j(~q

q Q dx =

can be shown

that

Eq. (13.7-10)

-q dx

from Eq.

(13.7-8)

and multiplying by dx,

dx)

(13.7-9)

valid.

is

- q dx = -(1 Substituting

x-p,y) dA m

component B,

-d[q'{\-y)] =

It

k

(13.7-10),

x) d(qx)

+

x

-d{qx) from Eq.

-

x)]

(13.7-10)

(13.7-7),

and d[q(\-x)] from

d[_q{\

Eq. (13.7-8) into Eq. (13.7-9) gives qat

dx

\

\PiP'b

I

dA m

x

\y

-

y

-

xa

{(1

- x)a*(rx

-y)-

x[r(l

-

x)

-

(1

-

(13.7-11)

y)]}

r = Pf,lpj and a* = P\IP'bEquation (13.7-12) can also be derived using the same methods as those used for

where

Eq. (13.7-9).

q 0 dy =

It

can also be shown

that

q'

(13.7-12)

\^—~^(- q 'dy)

Eq. (13.7-13)

is

valid,

which

dy=(\ -y)d(q'y)-y

similar to

is

-

d[q'(\

Eq. (13.7-10). (13.7-13)

y)]

Substituting q' dy from Eq. (13.7-13), d(q'y) from Eq. (13.7-7), and d[q'{

Eq. (13.7-8) into

\p,P'B

dA m

—

y)}

from

Eq. (13.7-12),

(x-y

dy

9oM

1

\x

- x

{(1

-

y) a *{rx

-

y)

-

y[r{\

~

x)

-

(1

-

(13.7-14)

y)]}

(

At the outlet of the residue stream of composition x 0 the permeate y is given as Eq. (13.7-15). ,

=

y and x a (

are related by Eq. (13.4-5), which

a*[x 0 -

y-,

1-y,-

The

(1

solution of this quadratic equation

13.7B

(p,/p ; ,)y,]

-x 0 )-{p,lp h ){\ is

(13.7-15)

-y,)

identical to Eq. (13.4-10).

Solution of Equations for Countercurrent Flow

Equations (13.7-11) and (13.7-14) are solved simultaneously by numerical methods starting at the high-pressure outlet stream of composition x a The area A,n can be .

Sec. 13.7

Counlercurrent-Flow Model for Gas Separation by Membranes

119

equal to zero at this outlet and a negative area will be obtained whose must be reversed. Equation (13.7-14) along with Eq. (13.7-15) is indeterminate at the high-pressure outlet. Using L'Hopital's rule for A m —> 0, arbitrarily set

sign

/

dy

\

\dA,

A. =

0

- yM«* ->,<«*- D] - {Uo - ?)[(«* - DC?,- - rx a - 1) - rM^/^A Ja„ = o (x 0

QoWb) For Eq.

(13.7-16)

(13.7-11),

dx \

dA
-

a*{rx a

P/P'b

\

yj)(x 0

-

y,)

(13.7-17)

Am

=

lot

Q

It is more convenient to solve Eqs. (13.7-11) and (13.7-14) in terms of x as the independent variable. So dividing Eq. (13.7-14) by (13.7-11),

dy

dx

{(1

- y)a*{rx -

y)

-

y[r(\

-

x)

-

(1

-

y)]}

x 0 ) {(1

- x)a*{rx -

y)

-

x[r(\

-

x)

-

(1

-

y)]}

x _ (y- 0 ) " 0c

(13.7-18)

Inverting Eq. (13.7-11)

dAjn

_ J7of ~ dx p,P'B

Since Eq. (13.7-18)

is

-x 0 )/(x-y)] -y)- x[r(l - x) -

i(y {(1

- x)a*(rx

indeterminate

at

(1

-

the high-pressure outlet,

(13.7-19) y)]}

it

can be evaluated

using Eqs. (13.7-16) and (13.7-17) as follows

{dy/dA m ) Am = Am =

Assuming

that the stage cut 6

is

0

~ (dx/dA ) m Am

specified, the

=

Q

(13.7-20)

0

procedure

is trial

and

error. First a

value of x a at the high-pressure outlet is assumed, and by solving the two Eqs. (13.7-18) and (13.7-19) numerically, the value of the area A m and the value of the

permeate^

at the outlet are

obtained. Using thisy^ and the material balance equation

assumed and the calculated values of x a do not agree, assumed and the procedure repeated. Further details and computer programs are given elsewhere (HI, Rl, W5). For cocurrent flow, the equations are quite similar and are given by others (Bl, HI, Rl, W5). (13.4-8),

xa

is

another value

13.8.

13. 8A

calculated. If the is

EFFECTS OF PROCESSING VARIABLES ON GAS SEPARATION BY MEMBRANES Effects of Pressure Ratio

and Separation Factor on Recovery

Using the Weller-Steiner equation (13.4-5) for the compete-mixing model, the effects of pressure ratio, Phlpi, and separation factor, a*, on permeate purity can be determined for a fixed feed composition. Figure 13.8-1 is a plot of this equation for a

780

Chap. 13

Membrane

Separation Processes

20

0

I

I

I

I

1

20

0

I

I

40

1

60

1

80

Separation factor, a* FIGURE

13.8-1.

of separation factor and pressure ratio on permeate (Feed Xf = 030.) [From "Membranes Separate Gases Selectively," by D. J. Stookey, C. J. Pattern, and G. L. Malcolm, Chem. Eng. Progr., 82(11), 36 (1986). Reproduced by permission of the American Institute of Chemical Engineers, Effects

purity.

1986.]

can be expected to provide estimates of product purity and trends for conditions of low to modest recovery in all types of models, such as complete-mixing, cross-flow, and countercurrent. Figure 13.8-1 shows that above an a* of 20, the product purity is not greatly

'feed concentration of 30%j(S7). This equation

affected. Also,

above a pressure

product purity.

Some typical

Table

ratio of

about

6, this ratio has a diminishing effect

13.8-1.

If liquids are

membrane

present in the gas separation process, a liquid film can increase the

resistance markedly. Liquids can also

action or by swelling or softening. If water vapor point

on

separation factors for commercial separators are given in

may be

reached

in

damage

is

present

the in

membrane by chemical

the gas streams, the

dew

the residue product and liquid condensed. Also, condensation

of hydrocarbons must be avoided.

Table

13.8-1.

Typical Separation Factors for

Some

Industrial

Membranes Gases Separated

H 2 0/CH 4 He/CH 4

H /CO H /N 2 H /0 2 H 2 /CH

500 5-44 35-80

2

2 2

Refs.

(M4) (M4), (SI), (W3) (K3), (M4)

3-200

(HI), (K3), (M4), (W2), (W3)

,4-12

(M4), (S2), (W3)

6-200

(HI), (K3),(M4), (W3).

0 2 /N 2

2-12

(M4), (K3), (S2), (W2), (W3)

C0 2 /CH 4 C0 2 /O 2

3-50 3-6

(HI), (M4), (S2), (W3)

4

CH 4 /C H 6 2

Sec. 13.8

Separation factor

2

Effects of Processing Variables

(M4), (S2), (W2), (W3)

(M4)

on Gas Separation by Membranes

781

0.6

0.2

i

I

I

1

I

i

I

i

02

0

i

I

I

i

0.6

0.4

Stage cut, 6

FIGURE

Effect of stage cut and flow pattern on permease purity. Operating conditions for air are as follows: Xf = 0.209, a* = 10 10, p h lpi = 380 cm Hg/76 cm Hg = 5. P'A = 500 x cm 3 (STP) • cmls cm 2 cm Hg. (1) countercurrent flow, (2) cross-flow, (3) cocurrenl flow, (4) complete mixing (W5). [Reprinted from W. P. Walawender and S. A. Stern, Sep. Sci., 7, 553 (1972). By courtesy of Marcel Dekker, Inc.]

13.8-2.

W'

13.8B

Effects of Process

Flow Patterns on Separation and Area

Detailed parametric studies have been done by various investigators (P4, P5,

binary systems.

They compared

W5)

for

the four flow patterns of complete mixing, cross-flow,

cocurrent, and countercurrent flow. In Fig. 13.8-2 (W5) the permeate concentration

shown

plotted versus stage cut,

0, for

a feed of air (xj

= 0.209

for oxygen) with a*

is

=

10 and p h /pi = 5. It is shown that, as expected, the countercurrent flow pattern gives the best separation. The other patterns of cross-flow, cocurrent, and complete mixing give lower separations in descending order.

Note

that

when

8= 0, all flow same permeate same value of y p = 0.209, the stage cut

patterns are equivalent to the complete mixing model and give the

composition. Also,

which

at 0

=

1

.00,

all

patterns again give the

also the feed composition.

is

The required membrane

areas for the same process conditions and air feed versus were also determined (W5). The areas for all four types of flow patterns were shown to be within about 10% of each other. The countercurrent and cross-flow flow stage cut

patterns give the lowest area required. In general,

it

has been concluded by

many parametric

studies that at the

same

operating conditions the countercurrent flow pattern yields the best separation and requires the lowest rent

>

13.9

13.9A /.

cross-flow

membrane

>

cocurrent

area.

>

The order of efficiency

is

as follows: countercur-

complete mixing.

REVERSE-OSMOSIS MEMBRANE PROCESSES Introduction

Introduction.

To be

useful for separation of different species, a

passage of certain molecules and exclude or greatly

restrict

membrane must allow

passage of other molecules. In

osmosis, a spontaneous transport of solvent occurs from a dilute solute or salt solution to

782

Chap. 13

Membrane Separation Processes

a concentrated solute or salt solution across a semipermeable membrane which allows passage of the solvent but impedes passage of the salt solutes. In Fig. 13.9-la the solvent

water normally flows through the semipermeable membrane to the salt solution. The levels of both liquids are the same as shown. The solvent flow can be reduced by exerting a pressure on the salt-solution side and membrane, as shown in Fig. 13.9Tb, until at a certain pressure, called the osmotic pressure n of the salt solution, equilibrium

and the amount of

the solvent passing in opposite directions

potentials of the solvent

on both

sides of the

membrane

is

are equal.

solution determines only the value of the osmotic pressure, not the that

it is

truly

equal.

is

reached

The chemical

The property of the membrane, provided

semipermeable. To reverse the flow of the water so that it flows from the fresh solvent as in Fig. 13.9Tc, the pressure is increased above the

salt solution to the

osmotic pressure on the solution side. This phenomenon, called reverse osmosis,

important commercial use

is in

fresh water. Unlike distillation

osmosis can operate

at

is

used in a number of processes.

the desalination of seawater or brackish water to

and

freezing processes used to

remove

An

produce

solvents, reverse

ambient temperature without phase change. This process

is

quite

and chemically unstable products. Applications include and milk, recovery of protein and sugar from cheese whey,

useful for processing of thermally

concentration of

fruit

juices

and concentration of enzymes. 2.

Osmotic pressure of solutions.

a solution

is

Experimental data show that the osmotic pressure n of

proportional to the concentration of the solute and temperature T. Van't

Hoff originally showed that the relationship For example, for dilute water solutions, x

=

is

similar to that for pressure of an ideal gas.

— RT

(13.9-1)

kg mol of solute, Vm the volume of pure solvent water inm 3 R the gas law constant 82.057 x 10~ 3 m 3 -atm/kg mol K, and T is temperature in K. If a solute exists as two or more ions in solution, n represents the total number of ions. For more concentrated solutions Eq. (13.9-1) is where n

is

the

number

of

associated with n kg mol of solute, •

modified using the osmotic coefficient

cp,

which

is

the ratio of the actual osmotic pressure

solvent

flow fresh

membrane

water solvent

NaCl-water (b)

(a)

FIGURE

13.9-1.

Osmosis and reverse osmosis: (c)

Sec. 13.9

(c) (a)

osmosis,

(b)

osmotic equilibrium,

reverse osmosis.

Reverse-Osmosis Membrane Processes

783

7r

to the ideal n calculated from the equation.

For very

dilute solutions

unity and usually decreases as concentration increases. In Table 13.9-1 tal

values of 7i are given for

NaCl

solutions, sucrose solutions,

has a value of

tf>

some experimen-

and seawater solutions

(S3,

S5).

EXAMPLE

Calculation of Osmotic Pressure of Salt Solution of a solution containing 0.10 g

13.9-1.

mol

Calculate the osmotic pressure

NaCl/1000gH 2 Oat25°C. 3 Then, A.2-3, the density of water = 997.0 kg/m 4 3 10" 10" = 2.00 x n = 2 x 0.10 x kg mol (NaCl gives two ions). Also, the volume of the pure solvent water Vm = 1.00 kg/(997.0 kg/m 3 ). Substituting into Eq. (13.9-1),

From Table

Solution:

.

10-*(82.057 x 1Q- 3 X298.15) ,.000/997.0

" st = 2.00 x n=-RT „

=4

"

88 atm

This compares with the experimental value in Table 13.9-1 of 4.56 atm.

Types of membranes for reverse osmosis. One of the more important reverse-osmosis desalination and many other reverse-osmosis processes

3.

acetate

membrane. The asymmetric membrane

thin dense layer

about

thicker (50 to 125

The

0.1 to 10

um

is

made

as a

membranes

composite film in which a

thick of extremely fine pores supported

um) layer of microporous sponge with

for

the cellulose

is

little

upon a much

resistance to permeation.

dense layer has the ability to block the passage of quite small solute molecules.

thin,

In desalination the

membrane

rejects the salt solute

and allows the solvent water to pass

through. Solutes which are most effectively excluded by the cellulose acetate are the salts NaCl, NaBr,

The main

CaCl 2 and ,

Na 2 SO A

limitations of the cellulose acetate

;

membrane

are that

it

membrane

ammonium

sucrose; and tetralkyl

salts.

can only be used

in aqueous solutions and that it must be used below about 60°C. Another important membrane useful for seawater, wastewater, nickel-plating

mainly

solutions,

made

in

and other solutes the form of very

the synthetic aromatic

is

hollow

fine

fibers (LI, P3). This type of

industrially withstands continued operation at

anisotropic

can be used

membranes have in

pH

membrane used

values of 10 to 11 (S4).

also been synthesized of synthetic polymers,

13.9-1.

Many

some

organic solvents, at higher temperatures, and at high or low

Table

rinse

polyamide membrane "Permasep,"

pH

Osmotic Pressure of Various Aqueous Solutions

other

of which

(M2, Rl).

at

25° C (P1,S3,S5J Sodium Chloride Solutions

Sea Salt Solutions

Osmotic g mol NaCl kg

*

784

H 0

Density

(kg/m')

Pressure (atm)

%

Wt.

Salts

Sucrose Solutions

Osmotic

Solute

Osmotic

Pressure

Mol. Frac. x 10 3

Pressure

(atm)

2

0

0

(atm)

0

997.0

0.01

997.4

0.10

1001.1

4.56

3.45*

25.02

0.50

1017.2

22.55

7.50

58.43

10.69

15.31

1.00

1036.2

45.80

10.00

82.12

17.70

26.33

2.00

1072.3

96.2

0 0.47

1.00

7.10

0 1.798

5.375

0 2.48

7.48

Value for standard seawater.

Chap. 13

Membrane

Separation Processes

Flux Equations for Reverse Osmosis

13.9B

There are two basic types of mass-transport 1. Basic models for membrane processes. mechanisms which can take place in membranes. In the first basic type, using tight membranes, which are capable of retaining solutes of about 10 A in size or less, diffusiontype transport mainly occurs. Both the solute and the solvent migrate by molecular or Fickian diffusion in the polymer, driven by concentration gradients set up in the membrane by the applied pressure difference. In the second basic type, using loose, microporous membranes which retain particles larger than 10 A, a sieve-type mechanism occurs where the solvent moves through the micropores in essentially viscous flow and

enough to pass through the pores are carried by convection second type, see (M2, Wl).

the solute molecules small

with the solvent.

For

details of the

For diffusion-type membranes, the steady-state equations Diffusion-type model. governing the transport of solvent and of solute are to a first approximation as follows (M2, M3). For the diffusion of the solvent through the membrane as shown in Fig.

2.

13.9-2,

Nw = y

li

P„ =

D

(AF c

— An) = A„{AP —

An)

(13.9-2)

V

~f^

(13-9-3)

Aw =

(13.9-4)

where N w is the solvent (water) flux in kg/s m 2 F w the solvent membrane permeability, kg solvent/s m atm; Lm the membrane thickness, m; A„ the solvent permeability 2 constant, kg solvent/s m atm AF = P x — F 2 (hydrostatic pressure difference with F t pressure exerted on feed and F 2 on product solution), atm; An = n t — n 2 (osmotic •

;

•

•

;

— osmotic pressure of product solution), atm; D w is the diffusimembrane, m 2 /s; c w the mean concentration of solvent in membrane, kg solvent/m 3 V„ the molar volume of solvent, m 3 /kg mol solvent; R the gas law constant, 3 3 82.057 x 10" m atm/kg mol K; and T the temperature, K. Note that subscript 1 is the feed or upstream side of the membrane and 2 the product or downstream side of the membrane. pressure of feed solution

vity of solvent in ;

•

membrane product permeate solution

feed concentrate

solution

Figure

Sec. 13.9

13.9-2.

Concentrations and fluxes

Reverse-Osmosis Membrane Processes

in

reverse-osmosis process.

785

For the diffusion of the solute through the membrane, an approximation for the of the solute

is

N = A5 =

N

s

is

membrane,

-

(c,

M

where

=

c 2)

m

/s;

K = cjc

- c2

A,{c t

(13.9-5)

)

5

(13.9-6)

Lm

the solute (salt) flux in kg solute/s-m 2

flux

Ml)

(CI,

2

Ds

;

the diffusivity of solute in

5

brane/concentration of solute

A3

in solution;

the solute permeability constant, m/s; c,

is

the solute concentration in upstream or feed (concentrate) solution, kgsolute/m the solute concentration in

mem-

(distribution coefficient), concentration of solute in

downstream

3

and

;

c2

or product (permeate) solution, kg solute/m

3 .

The distribution coefficient K s is approximately constant over the membrane. Making a material balance at steady state for the solute, the solute diffusing through the membrane must equal the amount of solute leaving in the downstream or product (permeate) solution.

N

N =

c

(13.9-7)

s

where c w2 is the concentration of solvent in stream 2 (permeate), kg solvent/m 3 If the stream 2 is dilute in solute, c„ 2 is approximately the density of the solvent. In reverse .

R

osmosis, the solute rejection

membrane

is

defined as the ratio concentration difference across the

divided by the bulk concentration on the feed or concentrate side (fraction of

solute remaining in the feed stream).

R =

c,

—

c,

=

c,

1

-—

c,

c,

This can be related to the flux equations as follows by

Nw

(13.9-5) into (13.9-7) to eliminate

and

N

s in

Eq.

(13.9-8) •

first

substituting Eqs. (13.9-2)

Then

(13.9-7).

and

solving for c 2 /c, and

substituting this result into Eq. (13.9-8),

B(AP - An)

_ 1

B =

+ B(AP - An)

~— =

Ds Ks c„ 2

(13.9-10)

A s c„ 2

where B is in atm" Note that B is composed of the various physical properties P w D s and K 5 of the membrane and must be determined experimentally for each membrane. 1

.

Usually, the product

many s

•

m

,

Ds K s

is

determined, not the values of

of the data reported in the literature give values

2 •

atm and (D s KJLm )

EXAMPLE

or

A s in

m/s and not separate

D s and K 5

separately. Also,

o((PJL m or A w in kg values of L m P w and so )

,

,

,

solvent/

on.

Experimental Determination of Membrane Permeability Experiments at 25°C were performed to determine the permeabilities of a cellulose acetate membrane (A I, Wl). The laboratory test section shown in -3 2 Fig. 13.9-3 has membrane area A = 2.00 x 10 The inlet feed solution concentration of NaCl is c t = 10.0 kgNaCl/m 3 solution (lO.Og NaCl/L, 3 p[ = 1004 kg solution/m ). The water recovery is assumed low so that the 13.9-2.

m

786

Chap. 13

.

Membrane

Separation Processes

-+~ exit

f^-^

feed solution

TzMz^ZZ^Z^Zz I

c

Figure

feed solution

boundary layer

— membrane

'

product solution

2

Process flow diagram of experimental reverse-osmosis laboratory

13.9-3.

unit.

concentration c, in the entering feed solution flowing past the membrane and the concentration of the exit feed solution are essentially equal.

The product

=

997 (p 2 solution/s.

kg solution/m

A

3

0.39 kg NaCl/m 3 solution measured flow rate is 1.92 x 10" 8 m 3

and

)

its

kPa (54.42 atm) is used. Calculate membrane and the solute rejection R.

pressure differential of 5514

the permeability constants of the

Since c 2

Solution:

=

contains c 2

solution

is

very low (dilute solution), the value of c w2 can be

assumed as the density of water (Table

= 997 kg solvent/m 3 N„, using an area of

13.9-1) or c w2

To

convert the product flow rate to water 3 2 2.00 x 10" m

flux,

.

,

N w = (1.92 =

x 10" 8

9.57 x 10"

3

m

3

3

/sX997 kg solvent/m Xl/2.00 x 10

kg solvent/s-m

-3

m2

)

2

Substituting into Eq. (13.9-7),

Nw c _ 2

N.

(9.51

997

c„ 2

= To determine

x 1Q- 3 X0.39)

3.744 x 1CT

6

kg solute NaCl/s-m

2

the osmotic pressures from Table 13.9-1, the con-

For c u 10 kg NaCl is in 1004 kg 3 3 solution. solution/m {p y = 1004). Then, 1004 - 10 = 994 kg H 2 0 in 1 Hence, in the feed solution where the molecular weight of NaCl = 58.45, (10.00 x 1000)/(994 x 58.45) = 0.1721g mol NaCl/kg H 2 0. From Table 13.9-1, 7r = 7.80 atm by linear interpolation. Substituting into Eq. (13.9-1), the predicted rr = 8.39 atm, which is higher than the experimental value. For the product solution, 997 - 0.39 = 996.6 kg 2 0. Hence,

centrations are converted as follows.

m

j

]

H

x 1000)/(996.6 x 58.45) = 0.00670g mol NaCl/kg H 2 0. From Table 13.9-1 tt2 = 0.32 atm. Then, Att = tt, - tt2 = 7.80 -,0.32 = 7.48 atm and

(0.39

,

AP =

54.42 atm. Substituting into Eq. (13.9-2),

Nw = Solving

9.57

{PJL m

x 10" 3

= -== (AP -

Att)

=

(54.42

= A w = 2.039 x 10" 4 kg solvent/s

)

•

-

7.48)

m2

•

atm. Substi-

tuting into Eq. (13.9-5),

N = s

Solving, (D s

Sec. 13.9

3.744

x 10" 6

KJLm = A = )

s

=

( Cl

-

c2)

=

(10.00

-

0.39)

7 3.896 x 10" m/s.

Reverse-Osmosis Membrane Processes

787

To

calculate the solute rejection

R

by

substituting into Eq. (13.9-8),

10.00

c,

Also substituting into Eq. (13.9-10) and then Eq.

PJLm

(3.896

B(AP - An) + B(AP -An)

_ 1

x 10-, _ -°" Q5249atm 5249atm x 10- 7 )997

2.039

{D,KJLJcw2 D

0.5249(54.42

+

1

- 7.48)

0.5249(54.42

APPLICATIONS, EQUIPMENT, AND

13.10

(13.9-9),

-

=

Q

7.48)

m

MODELS

FOR REVERSE OSMOSIS Effects of Operating Variables

13.10A In

many commercial

units operating pressures in reverse osmosis range

kPa

1035 up to 10 350

Comparison of Eq.

(150 up to 1500 psi).

from about

(13.9-2) for solvent flux

N

and Eq. (13.9-5) for solute flux shows that the solvent flux w depends only on the net pressure difference, while the solute flux N, depends only on the concentration difference. is increased, solvent or water flow through the membrane and the solute flow remains approximately constant, giving lower solute

Hence, as the feed pressure increases

concentration in the product solution.

At a constant applied pressure, increasing the feed solute concentration increases the product solute concentration. This since as

more solvent

solute concentration

is

is

caused by the increase in the feed osmotic pressure,

extracted from the feed solution (as water recovery increases), the

becomes higher and

the water flux decreases. Also, the

amount

of

solute present in the product solution increases because of the higher feed concentration. If

membrane

a reverse-osmosis unit has a large

between the feed

the path

inlet

considerably higher than the

and

outlet

inlet feed c l

feed compared to the inlet (K2).

is

Then

.

Many

area (as in a commercial unit), and

long, the outlet feed concentration

manufacturers use the feed solute or

concentration average between inlet and outlet to calculate the solute or

R

in

Eq.

salt

salt rejection

(13.9-8).

EXAMPLE

Prediction of Performance

13.10-1.

in a

A

can be

the salt flux will be greater at the outlet

Reverse-Osmosis Unit

membrane to be used at 25°C for a NaCl 3 3 p = 999 kg/m ) g NaCl/L (2.5 kg NaCl/m

reverse-osmosis

feed solution

containing 2.5 has a water 4 2 permeability constant A w = 4.81 x 10~~ kg/satm and a solute (NaCl) 7 permeability constant A s = 4.42 x 10" m/s (Al). Calculate the water flux and solute flux through the membrane using a AP — 27.20 atm and the solute rejection R. Also calculate c 2 of the product solution. ,

m

Solution:

In the feed solution, c

l

=

2.5

•

3 kg NaCl/m and p

x

=

999 kg

solution/m 3 Hence, for the feed, 999 - 2.5 = 996.5 kg H 2 0 in 1.0 m 3 solution; also for the feed, (2.50 x 1000)/(996.5 x 58.45) = 0.04292 g mol .

NaCl/kg

H 2 0. From Table

13.9-1

,

tt,

=

1

.97 atm. Since the

product solu-

=0.1 kg NaCl/m 3 will be assumed. 3 Also, since this is quite dilute, pj = 997 kg solution/m and C 2, 3 = 997kg solvent/m Then for the product solution, (0.10 x 1000)/ (996.9 x 58.45) = 0.00172 g mol NaCl/kg H 2 0 and n 2 = 0.08 atm. Also, tion c 2

is

unknown,

a value of c 2

.

Att

788

=

7t[

-

ti

2

=

1.97

-

0.08

=

1.89 atm.

Chap. 13

Membrane Separation Processes

Substituting into Eq. (13.9-2),

N w = A W (AP -

An)

=

=

1.217 x 10-

For calculation ofR, substituting

—A — = — w

B= Next

A,cw2

2

1.89)

kgH 2 O/s-m 2

into Eq. (13.9-10),

first

—7— — = 1.092„ atra x 997

4 4.81 x 10~

4.42

-

x 10- 4 (27.20

4.81

rtn

,

_.1

7

x 10

substituting into Eq. (13.9-9),

B(AP-Att) + B(AP - Arc)

1

Using

this

value of

R

in

1

+

1.092(27.20

=

0.0875 kg

significantly

NaCl/m

3

2.50

on

a

second

trial.

product solution. This

for the

assumed value of

to the

1.89)

=^^ = ^-^ c,

enough

1.89)

-

Eq. (13.9-8),

R=0 .965 Solving, c 2

-

1.092(27.20

_

c2

=

0.10 so that n 2

Hence, the

final

will

value of c 2

is

close

not change is 0.0875 kg

NaCl/m 3 (0.0875

g NaCl/L). Substituting into Eq. (13.9-5),

N = s

13.10B

A;{c y

-

c 2)

= =

7 4.42 x 10" (2.50

-

0.0875)

2 6 1.066 x 10" kg NaCl/s-m

Concentration Polarization in Reverse-Osmosis Diffusion Model

where the The solute accumulates in a relatively stable boundary layer (Fig. 13.9-3) next to the membrane. Concentration polarization, B, is defined as the ratio of the salt concentration at the membrane surface In desalination, localized concentrations of solute build up at the point

solvent leaves the solution and enters the membrane.

to the salt concentration in the bulk feed stream c

Concentration polarization causes

.

j

the water flux to decrease since the osmotic pressure

tt x

increases as the boundary

layer concentration increases and the overall driving force

(AP —

Also, the solute flux increases since the solute concentration

boundary. Hence, often the

AP

must be increased

power costs (K2). The effect of the concentration modifying the value of

A-rr in

polarization

It is

the

can be included approximately by

j3

1

(13.10-1)

tt 2

assumed that the osmotic pressure 7r, is directly proportional to the concentrawhich is approximately correct. Also, Eq. (13.9-5) can be modified as

N, = A 5 {Bc,~

the

at

compensate which gives higher

Eqs. (13.9-2) and (13.9-9) as follows (P6): Att = Btt

tion,

to

Att) decreases.

increases

The usual concentration polarization boundary layer is 1.2 to 2.0 times c l

difficult to predict. In

ratio (K3)

is 1.2

to 2.0,

i.e.,

the concentration in

in the bulk feed solution. This ratio

desalination of seawater using values of about 1000 psia

can be large. Increasing this

Sec. 13.10

(13.10-2)

c 2)

it,

by a factor of

Applications, Equipment,

1.2

is

often

= AP,K

t

to 2.0 can appreciably reduce the

and Models for Reverse Osmosis

789

and using A'P values of atm abs, the value of n is low and concentration polarization is not important. The boundary layer can be reduced by increasing the turbulence using higher feed

solvent flux. For brackish waters containing 2 to 10 g/L 17 to 55

l

solution velocities. However, this extra flow results in a smaller ratio of product solution to feed. Also, screens

13. IOC

can be put

in the flow

path to induce turbulence.

Permeability Constants of Reverse-Osmosis

Membranes

Permeability constants for membranes must be determined experimentally for the particular type of

membrane

water permeability constants

•m 2 -atm

(Al,

to

Aw

be used. For cellulose acetate membranes, typical -4 4 to 5 x 10 1 x 10~ kg solvents/s

range from about

M3, Wl). Values

for other types of

membranes can differ widely. membrane does not depend

Generally, the water permeability constant for a particular

upon the solute present. For the solute permeability constants A s of cellulose acetate membranes, some relative typical values are as follows, assuming a value of A s = -7 7 7 4 x 10" m/s for NaCl: 1.6 x 10~ 7 m/s (BaCl 2 ), 2.2 x 10~ (MgCl 2 ), 2.4 x 10 -7 -7 7 (CaCl 2 ), 4.0 x 10 (Na 2 S0 4 ), 6.0 x 10 (KC1), 6.0 x 10~ (NH 4 C1) (Al). Types of Equipment

13.10D

for Reverse

The equipment for reverse osmosis is membrane processes described in Section plastic support plates with thin

Osmosis quite sknilar to that for gas permeation

13.3C. In the plate-and-frame type unit, thin

grooves are covered on both sides with membranes as

between the closely spaced membranes in the grooves to an outlet. In the tubular-type unit, membranes in the form of tubes are inserted inside poroustube casings, which serve as a pressure vessel. These tubes are then arranged in

in

a

filter

press. Pressurized feed solution flows

(LI). Solvent permeates through the

bundles

like a heat

membrane and flows

exchanger.

membrane is used and a flat, porous support sandwiched between the membranes. Then the membranes, support, and a mesh feed-side spacer are wrapped in a spiral around a tube. In the hollow-fiber type, fibers of 100 to 200 /u.m diameter with walls about 25 /xm thick are arranged in a bundle similar to a heat exchanger (LI, Rl). In the spiral-wound type, a planar

material

is

Complete-Mixing Model

13.10E

The process flow diagram model

is

for Reverse

Osmosis

for the complete-mixing

model

is

shown

in Fig. 13.10-1.

a simplified one for use with low concentrations of salt of about

1% or so

The such

reject out CO

feed in

9f.Cf

(exit feed) >-

TZSZS^SZZSSZS^m^ZS^SZS^ 1 f

if


= (1-8)9

permeate out w ^2~^1f

FIGURE

790

13.10-1.

Process flow for complete-mixing model for reverse osmosis.

Chap. 13

Membrane

Separation Processes

as occur in brackish waters. Also, a relatively low recovery of solvent occurs and the

Since the concentration of the permeate very low, the permeate side acts as though it were completely mixed. For the overall materia] balance for dilute solutions,

effects of concentration polarization are small. is

+ where qf <7i

is

flow rate of residue or exit,

is

m 3 /s; q 2

volumetric flow rate of feed,

m

3 /s.

cf q

s

Denning the cut or

=

(13.10-3)

<7 2

Making a

+

c,
is

flow rate of permeate,

The previous equations derived

=

=

and

q 2 /q/, Eq. (13.10-4) becomes (13.10-5)

and rejection are useful

for the fluxes

/s;

(13.10-4)

c 2q 2

- 0)c, + 6c 2

(1

3

solute balance,

fraction of solvent recovered as 6

cf

m

in this

case and

are as follows:

Nw

=

A 2 (AP -

Ns = A s R=

-

c,

-c 2

(13.9-5)

)

c2

(13.9-8)

B(AP -

R= 1

When

( Cl

(13.9-2)

Att-)

Att) (13.9-9)

+ 5(AP-Att)

is trial and error. Since and c 2 are unknown, a value of c 2 is assumed. Then c, is calculated from Eq. (13.10-5). Next N w is obtained from Eq. (13.9-2) and c 2 from Eqs. (13.9-8) and (13.9-9). If the calculated value of c 2 does not equal the assumed value, the procedure is repeated.

the cut or fraction recovered, 6,

is

specified the solution

the permeate and reject concentrations c

When

t

concentration polarization effects are present an estimated value of

be used to make an approximate correction for

this effect.

to obtain a value of Att for use in Eqs. (13.9-2)

and

This

is

used

(13.9-9). Also,

in

/3

can

Eq. (13.10-1)

Eq. (13.10-2)

will

A

more detailed analysis of this complete mixing model is given by others (HI, Kl) in which the mass transfer coefficient in the concentration polarization boundary layer is used. The cross-flow model for reverse osmosis is similar to that for gas separation by membranes discussed in Section 13.6. Because of the small solute concentration, the permeate side acts as if completely mixed. Hence, even if the module is designed for countercurrent or cocurrent flow, the cross-flow model is valid. This is discussed in replace Eq. (13.9-5).

detail

elsewhere (HI).

13.11

ULTRAFILTRATION MEMBRANE PROCESSES

13.11A

Introduction

Ultrafiltration is a

membrane process

that

is

quite similar to reverse osmosis.

pressure-driven process where the solvent and,

pass through the

Sec. 13.11

membrane and

Ultrafiltration

when

It is

a

present, small solute molecules

are collected as a permeate. Larger solute molecules

Membrane Processes

791

do not pass through the membrane and are recovered in a concentrated solution. The solutes or molecules to be separated generally have molecular weights greater than 500 and up to 1 000 000 or more, such as macromolecules of proteins, polymers, and starches and also colloidal dispersions of clays, latex particles, and microorganisms. Unlike reverse osmosis, ultrafiltration membranes are too porous to be used for desalting. The rejection, often called retention, is also given by Eq. (13.9-8), which is defined for reverse osmosis. Ultrafiltration different molecular weight proteins.

is

also used to separate a mixture of

The molecular weight

membrane

cut-off of the

defined as the molecular weight of globular proteins, which are

90%

retained

is

by the

membrane. used

Ultrafiltration is

in

many

Some

different processes at the present time.

of

these are separation of oil-water emulsions, concentration of latex particles, processing of blood

and plasma, fractionation or separation of proteins, recovery of whey

proteins in cheese manufacturing, removal of bacteria and other particles to sterilize

wine, and clarification of

fruit juices.

Membranes for ultrafiltration are in general similar to those for reverse osmosis and are commonly asymmetric and more porous. The membrane consists of a very thin dense skin supported by a relatively porous layer for strength. Membranes are made from aromatic polyamides, cellulose acetate, cellulose nitrate, polycarbonate, polyimides, polysulfone, etc. (M2, P6, Rl).

13.11B

Types of Equipment for Ultrafiltration

The equipment

for ultrafiltration is similar to that used for reverse osmosis

separation processes described less

prone to foul and

However,

this

Flat sheet

type

is

is

in

more

Sections 13. 3C and

13.

easily cleaned than

10D.

The

and gas

tubular type unit

is

any of the other three types.

relatively costly.

membranes

in

a plate-and-frame unit offer the greatest versatility but

the highest capital cost (P6).

Membranes can

at

be cleaned or replaced by

easily

disassembly of the unit. Spiral-wound modules provide relatively low costs per unit

membrane

area.

These

units are

more prone

to foul

than tubular units but are more

resistant to fouling than hollow-fiber units. Hollow-fiber

to fouling

when compared

configuration has the highest ratio of

However,

membrane area per

13.11C

Flux Equations for Ultrafiltration

The

equation for diffusion of solvent through the

flux

modules are the

to the three other types.

unit

least resistant

the hollow-fiber

volume.

membrane

is

the

same as Eq.

(13.9-2) for reverse osmosis:

Nw = A In ultrafiltration the

1U

(AP - Att)

(13.9-2)

membrane does not allow the passage of the The concentration in moles/liter of

generally a macromolecule.

molecules

Then Eq.

is

usually small. Hence, the osmotic pressure

(13.9-2)

which

is

very low and

is

neglected.

becomes

NW

= A W (&P)

(13.11-1)

about 5 to 00 psi pressure drop compared to 400 to 2000 For low-pressure drops of, say, 5 to 10 psi and dilute solutions of

Ultrafiltration u nits operate at

for reverse osmosis.

792

is

solute,

the large solute

1

Chap. 13

Membrane

Separation Processes

up to

1

wt

%

or so, Eq. (13.11-1) predicts the performance reasonably well for

well-stirred systems.

at

accumulates and starts to build up is increased and/or concentration of

by the membrane, the surface of the membrane. As pressure drop Since the solute

the solute

is

is

rejected

increased, concentration polarization occurs, which

than in reverse osmosis. This

shown

is

in Fig. 13.1 1-la,

3 of the solute in the bulk solution, kg solute/m

solute at the surface of the

As

it

where c

and c s

,

is

is

much more severe

is

x

the concentration

the concentration of the

membrane.

N

w to and through membrane. This gives a higher convective transport of the solute to the membrane, i.e., the solvent carries with it more solute. The concentration c s increases and gives a larger back molecular duTusion of solute from the membrane to the bulk solution. At the pressure drop increases, this increases the solvent flux

the

steady state the convective flux equals the diffusion flux,

Nwc p

where

D AB

N w clp =

is

2

[kgsolvent/(s-

diffusivity

of solute

dc

=-D AB — dx 3

)](kg solute/m )/(kg solvent/m

in solvent,

equation between the limits of x

=

0

m

2

/s;

and c =

(13.11-2) 3 )

= kg

solute/s

•

m2

;

and x is distance, m. Integrating this c s and x = 5 and c = c 1

,

(13.11-3)

where k c

is

the mass-transfer coefficient, m/s. Further increases in pressure drop

increase the value of c s to a limiting concentration where the accumulated solute forms a semisolid gel

Figure

where c s = c g

13.11-1.

,

as

shown

in Fig. 13.11-lb.

Concentration polarization

in ultrafiltration: (a)

concentration

profile before gel formation, (b) concentration profile with

gel layer formed at

Sec. 13.1 1

Ultrafiltration

membrane

Membrane Processes

a

surface.

793

Still

to

further increases in pressure drop

do not change c 3 ,and the membrane

is

said

be "gel polarized." Then Eq. (13.11-3) becomes (PI, P6, Rl)

(13.11-4)

P With increases

in

Vij

pressure drop, the gel layer increases in thickness and this causes the

added gel layer resistance. Finally, the net flux becomes equal to the back diffusion of solute into the bulk solution because of the polarized concentration gradient as given by Eq. (13.1 1-4). The added gel layer resistance next to the membrane causes an increased resistance to solvent flux as given by solvent flux to decrease because of the of solute by convective transfer

AP (13.11-5) l/A w

where \/A v (s

•

m2

•

the

is

membrane resistance and R g is the variable gel The solvent flux in this gel-polarized regime

atm)/kg solvent.

pressure difference and

is

determined by Eq.

data confirm the use of Eq. (13.1

such as proteins,

13.11D

A

+ R,

etc.,

1-4) for

layer resistance, is

independent of

back diffusion. Experimental a large number of macromolecular solutions, (13.

1

1-4) for

and colloidal suspensions, such as latex

particles, etc. (PI, P6).

Effects of Processing Variables in Ultrafiltration

plot of typical experimental data of flux versus pressure difference

is

shown

in Fig.

At low pressure differences and/or low solute concentrations the data typically follow Eq. (13.1 1-1). For a given bulk concentration, c the flux approaches a constant value at high pressure differences as shown in Eq. (13.11-4). Also, more dilute protein concentrations give higher flux rates as expected from Eq. (13.11-4). Most commercial applications are flux limited by concentration polarization and operate in the region where the flux is approximately independent of pressure 13.11-2 (HI, P6).

t

,

difference (Rl).

Eq. (13.11-1)

AP Figure

794

13.11-2.

Effect of pressure difference

Chap.

13.

on solvent flux.

Membrane

Separation Processes

N

Using experimental data, a plot of w lp versus In c, is a straight line with the negative slope of k c , the mass-transfer coefficient, as shown by Eq. (13.11-4). These plots also give the value of'c

the gel concentration.

,

Data

show

(PI)

that the gel

ff

concentration for to

50%. For

many macromolecular

colloidal dispersions

The concentration

it is

solutions is about 25 wt %, with a range of 5 about 65 wt %, with a range of 50 to 75%.

polarization effects for hollow fibers

because of the low solvent

flux.

Hence, Eq.

often quite small

is

(13.11-1) describes the flux. In order to

membrane can be used to sweep away part of the polarized layer, thereby increasing k c in Eq. (13.11-4). Higher velocities and other methods are used to increase turbulence, and hence, k c In increase the ultrafiltration solvent flux, cross-flow of fluid past the

.

mode.

most cases the solvent

flux is too small to operate in a single-pass

to recirculate the feed

by the membrane with recirculation rates of

It is

necessary

10/1 to 100/1 often

used.

Methods

to predict the mass-transfer coefficient k c in Eq. (13.1 1-4) are given

others (PI, P6) for

known geometries

known geometries such

by

as channels, etc. Predictions of flux in

using these methods and experimental values of c g in Eq. (13.1 1-4) regime compare with experimental values for macromolecular

in the gel polarization

30%. However, for colloidal dispersions the experimental by factors of 20 to 30 for laminar flow and 8 to 10 for turbulent flow. Hence, Eq. (13.11-4) is not useful for predicting the solvent flux accurately. Generally, for design of commercial units it is necessary to obtain experimental data on single modules. solutions within about 25 to

flux is higher than the theoretical

PROBLEMS Through Liquids and a Membrane. A membrane process is being designed to recover solute A from a dilute solution where c t = 2.0 x 10" 2 kg mol /l/m 3 by dialysis through a membrane to a solution wherec 2 = 0.3 x 10" 2 The membrane thickness is 1.59 x 10~ 5 m, the distribution coefficient 11 K' = 0.75, D AB = 3.5 x 10" m 2 /s in the membrane, the mass-transfer coef= 3.5 x 10" 5 m/s,and/c = 2.1 x 10" 5 ficient in the dilute solution is k

13.2-1. Diffusion

.

(b)

Calculate the individual resistances, total resistance, and the total percent resistance of the two films. 2 Calculate the flux at steady state and the total area in for a transfer of

m

0.01 (c)

.

c2

cl

(a)

kg mol solute/h.

Increasing the velocity of both liquid phases flowing by the surface of the membrane will increase the mass-transfer coefficients, which are approxi0 6 mately proportional to u where v is velocity. If the velocities are doubled, total percent calculate the resistance of the two films and the percent '

,

increase in flux.

Ans.

(a)

(b)

13.2-2. Suitability

Total resistance = 6.823 x 10 5 s/m, 11.17% resistance, 2 10" 8 kg mol .4/s-m 2 area =111.5 A = 2.492 x

N

,

m

of a Membrane for Hemodialysis. Experiments are being conducted

mm

determine the suitability of a cellophane membrane 0.029 thick for use in artificial kidney device. In an experiment at 37°C using NaCl as the diffusing solute, the membrane separates two components containing stirred aqueous 3 3 solutions of NaCl, where c, = 1.0 x 10~ 4 g mol/cm (100 g mol/m ) and 7 10" The mass-transfer coefficients on either side of the membrane c 2 = 5.0 x have been estimated as k cl = kc2 = 5.24 x 10" 5 m/s. Experimental data obto

an

.

tained gave a flux

NA =

8.11

x 10

-4

g

mol NaCl/s-

m2

at

pseudo-steady-state

conditions. (a)

(b)

Chap. 13

Calculate the permeability p M in m/s and D AB K' inm 2 /s. Calculate the percent resistance to diffusion in the liquid

Problems

films.

795

13.3- 1. Gas-Permeation

Membrane for Oxygenation. To determine the suitability of its use as a membrane for a heart-lung machine to oxygenate

silicone rubber for

blood, an experimental value of the permeability at 30°C of oxygen was ob-

O

7 3 2 tained where P"u = 6.50 x 10 ~ cm cm Hg/mm). z (STP)/(s cm 2 (a) Predict the maximum flux of with an0 2 pressure of 700 2 in kgmol/s Hg on one side of the membrane and an equivalent pressure in the blood film side of 50 mm. The membrane is 0.165 thick. Since the gas •

•

m

0

mm

mm

film

is

pure oxygen, the gas film resistance

zero. Neglect the

is

blood film

resistance in this case. (b)

Assuming a maximum requirement

an adult of 300

0

cm 3

2 (STP) per 2 minute, calculate the membrane surface area required in (Note: The actual area needed should be considerably larger since the blood film resistance, which must be determined by experiment, can be appreciable.)

for

m

.

Ans.

(b) 1.953

m

2

13.4-1. Derivation of Equation for Permeate Concentration. Derive Eq. (13.4-1 1) for Case 2 for complete mixing. Note that x Q from Eq. (13.4-8) must first be substituted

Eq. (13.4-5) before multiplying out the equation and solving for y p Model for Membrane Design. A membrane having a -1 thickness of 2 x 10 ~cm, a permeability P'A = 400 x 10 ° cm 3 (STP)-cm/ 2 = 1 0 is to be used to separate a gas mixture of A and (s-cm -cm Hg), and an a* 3 3 B. The feed flow rate is q f = 2 x 10 cm (STP)/s and its composition is Xf = 0.413. The feed-side pressure is 80 cm Hg and the permeate-side pressure is 20 cm Hg. The reject composition is to be x a = 0.30. Using the complete-mixing model, calculate the permeate composition, the fraction of feed permeated, and into

.

13.4-2. Use of Complete-Mixing _3

the

membrane

area.

Ans. 13.4-3. Design Using Complete-Mixing Model.

A

yp

=

0.678

gaseous feed stream having a compo-

= 0.50 and a flow rate of 2 x 10 3 cm 3 (STP)/s is to be separated in a membrane unit. The feed-side pressure is 40 cm Hg and the permeate is 10 cm Hg. The membrane has a thickness of 1.5 x 10~ 3 cm, a permeability 10 cm 3 (STP)-cm/(s-cm 2 -cm Hg), and an a* = 10. The P'A = 40 x 10" sition

Xf

permeated is 0.529. the complete-mixing model to calculate the permeate composition, the reject composition, and the membrane area.

fraction of feed (a)

Use

(b) Calculate the (c) If

minimum

reject concentration.

the feed composition

minimum

is

increased to Xf

=

0.60, what

is

new

this

reject concentration?

Ans.

(a)

(c)

Am =

5.153 x 10 x oM = 0.2478

7

cm 2

Minimum Reject Concentration. For the conditions of Problem 13.4-2, xr = 0.413, a* = 10, p, = 20 cm Hg, p h = 80 cm Hg, and x 0 = 0.30. Calculate the minimum reject concentration for the following cases.

13.4- 4. Effect of Permeabilities on

Calculate x oM for the given conditions. Calculate the effect on x oM if the permeability of B increases so that a* decreases to 5. (c) Calculate the limiting value of x oM when a* is lowered to its minimum value. Make a plot of x oM versus a* for these three cases. (a)

(b)

13.4-5.

Minimum

Reject Concentration

and Pressure

Effect.

For Example 13.4-2

for

separation of air, do as follows.

minimum reject concentration. pressure on the feed side is reduced by one-half, calculate the effect

(a)

Calculate the

(b)

If the

on x oM

.

Ans.

796

(b)

Chap

.

x oM

13

= 0.0624 Problems

Gas Mixtures. Using the same feed composition and membrane as in Example 13.5-1 do the following using the complete-mixing model. (a) Calculate the permeate composition, the reject composition, and the membrane area for a fraction permeated of 0.50 instead of 0.25. (b) Repeat part (a) but for 9 = 0.90. (c) Make a plot of permeate composition y pA versus 9 and also of area A m versus 9 using the calculated values for 9 = 0.25, 0.50, and 0.90.

13.5-1. Separation of Multicomponent

and flow

rate, pressures,

,

A

13.5- 2. Separation of Helium from Natural Gas. (SI) is 0.5% He (A), 17.0% 2 (B),

N

typical composition of a natural gas

CH 4 (C), and 6.0% higher hydrocarbons (D). The membrane proposed to separate helium has a thickness ~3 of 2. 54 x 10 cm and the permeabilities are P'A = 60 x 10 ~'° cm 3 (STP)-cm/ 2 cm Hg), P'B = 3.0 x 10" 10 ,andP'c = 1.5 x 10 It is assumed (s cm that the higher hydrocarbons are essentially non-permeable (P'D s= 0). The feed 3 5 flow rate is 2.0 x 10 cm (STP)/s. The feed pressure p h = 500 cm Hg and the = permeate pressure p/ 20 cm Hg. (a) For a fraction permeated of 0.2, calculate the permeate composition, the reject composition, and the membrane area using the complete mixing model. (b) Use the permeate from part (a) as feed to a completely mixed second stage. The pressure p h = 500 cm Hg and p = 20 cm. For a fraction permeated of 0.20, calculate the permeate composition and the membrane area. 76.5%

•

•

t

Model for Membrane. Use the same conditions for the separation of an air stream as given in Example 13.6-1. These given values are

13.6- 1. Design Using Cross-Flow

xf

cm t

= 3

0.209, a* (STP)/s, P'A

(a)

6 p h = 190 cm Hg, p, = 19 cm Hg, qf = 1 x 10 -10 2 = 500 x 10 cm^STP) crn/(s crn cm Hg), and

10,

•

•

•

3

=2.54 x 10

(b)

=

cm. Do as follows using the cross-flow model. Calculate y p x 0 and A m for d = 0.40. Calculate y and x a for 6 = 0. p s Ans. (&)y p = 0.452, x 0 = 0.0303, A m = 6.94 x 10 = = (b)y p 0.209 0.655, x 0 ,

,

13.7- 1. Equations for Countercurrent-Flow Model.

cm 2

(S6)

For the derivation of the equations membrane do the following.

for countercurrent flow in a gas separation using a (a)

(b) (c)

Obtain Eq. (13.7-5) from Eq. (13.7-4). Show that Eq. (13.7-10) is valid. Obtain Eq. (13.7-12).

13.7-2. Design Using Countercurrent-Flow

tions as given in

Example

Model for Membrane. Use the same condi-

13.6-1 for the separation of

an

air stream.

The given

=

0.209, a* = 10, p h 190 cm Hg,p, = 19 cm Hg, qf = 1 x 6 10 cm 3 (STP)/s, P'A = 500 x 10~ 10 cm 3 (STP) cm/(s • cm 2 cm Hg), and 3 t = 2.54 x 10 cm. Using the countercurrent-flow model, calculate y p ,x r and A m for 6 = 0.40. (Note that this problem involves a trial-and-error procedure along with the numerical solution of two differential equations.) values are*/

=

•

•

,

13.9-1.

Osmotic Pressure of Salt and Sugar Solutions. Calculate the osmotic pressure of the following solutions at 25°C and compare with the experimental values. (a) Solution of 0.50 g mol NaCl/kg H 2 0. (See Table 13.9-1 for the experimental value.) (b) (c)

Solution of Solution of

1.0

g sucrose/kg

1.0 g

H 2 0.

MgCl 2 /kg H 2 0.

(Experimental value = 0.0714 atm.) (Experimental value = 0.660 atm).

Ans.

(a)

7t

(c)

7c

= =

24.39 atm, (b) n 0.768 atm

=

13.9-2. Determination of Permeability Constants for Reverse Osmosis.

acetate

Chap. 13

membrane

Problems

with an area of 4.0 x 10

-3

m2

is

0.0713 atm,

A

cellulose-

used at 25°C to determine

in

the permeability constants for reverse osmosis of a feed salt solution containing 12.0

3 kg NaCl/m (p

of 0.468 kg 8 3.84 x 10"

=

NaCl/m

m

3

/s

and

kg/m 3 The product solution has a concentration 3 997.3 kg/m ). The measured product flow rate is

1005.5

3

(p

=

).

the pressure difference used

is

56.0 atm. Calculate the

permeability constants and the solute rejection R. Ans. A w = 2.013 x 10" 4 kg solvent/s13.9-3.

m

2 -

atm,

R=

0.961

Performance of a Laboratory Reverse-Osmosis Unit. A feed solution at 25°C 3 contains 3500 mg NaCl/L (p = 999.5 kg/m ). The permeability constant A w = 4 2 10" 3.50 x kg solvent/s m atm and A, = 2.50 x 10~ 7 m/s. Using a AP = 35.50 atm, calculate the fluxes, solute rejection R, and the product solution concentration in mg NaCl/L. Repeat, but using a feed solution of 3500 mg BaCl 2 /L. Use the same value of/l w but /I, = 1.00 x i0" 7 m/s(Al). •

Using the same

13.10-1. Effect of Pressure on Performance of Reverse-Osmosis Unit.

conditions and permeability constants as in Example 13.10-1, calculate the fluxes, solute rejection R, and the product concentration c 2 for AP pressures

of 17.20, 27.20, and 37.20 atm. (Note: The values for 27.20 atm have already been calculated.) Plot the fluxes, R, and c 2 versus the pressure. 13.10-2. Effect of Concentration Polarization on Reverse Osmosis. Repeat Example 13.10-1 but use a concentration polarization of 1.5. (No/e:The flux equations

and the solute rejection R should be calculated using

Nw =

Ans.

c2

=

new value ofc

this

2 1.170 x 10"

0.1361 kg

kg

t

.)

solvent/s

NaCl/m

•

m

2 ,

3

Model for Reverse Osmosis. Use the same Example 13. 10-1. Assume that the cut or fraction recovered of the solvent water will be 0.10 instead of the very low water recovery assumed in Example 13.10-1. Hence, the concentration of the entering feed solution and the exit feed will not be the same. The flow rate q-, of the permeate water solution is 100 gal/h. Calculate c and c 2 in kg NaCl/m and the membrane area in m 2 2 3 3 Ans. c, = 2.767 kg/m c 2 = 0.0973 kg/m area = 8.68 m

13.10- 3. Performance of a Complete-Mixing

feed conditions and pressures given in

,

.

,

,

wt % protein is to undergo ultrafiltration using a pressure difference of 5 psi. The membrane permeability 10~ 2 kg/s- m 2 atm. Assuming no effects of polarization, is A w = 1.37 x 2 predict the flux in kg/s-m and in units of gal/ft day which are often used in

13.11- 1. Flux for Ultrafiltration.

A

solution containing 0.9

•

-

industry.

Ans. 13.11-2.

9.88 gal/ft

2 •

day

Time for

Ultrafiltration Using Recirculation. It is desired to use ultrafiltration for 800 kg of a solution containing 0.05 wt of a protein to obtain a solution of 1.10 wt %. The feed is recirculated by the membrane with a surface area of 2 The permeability of the membrane is A w = 2.50 X 10 2 9.90 m 2 atm. Neglecting the effects of concentration polarization, if any, kg/s-m calculate the final amount of solution and the time to perform this using a pressure difference of 0.50 atm.

%

.

-

REFERENCES (Al)

Agrawal,

(Bl)

Blaisdell, C. T., and Kammermeyer, K. Chem. Eng.

(B2)

Babb, A. L., Maurer, C. J., Fry, D. L., Popovich, R. P., and McKee, R. E. Chem. Eng. Progr. Symp., 64(84), 59 (1968).

(B3)

Berry, R.

798

J. P.,

I.

and Sourirajan, S. Ind. Eng. Chem., 69(11), 62 (1969).

Chem. Eng., 88(July

13),

Sci., 28, 1249 (1973).

63 (1981)

Chap. 13

References

(CI)

(HI)

W.

Clark,

Hwang,

E. Science, 138, 148 (1962).

S. T.,

John Wiley

and Kammermeyer, K. Membranes

& Sons,

and Sourirajan, S. A.I.Ch.E.

(Kl)

Kimura,

(K2)

Kaup, E. C.

Chem. Eng., 80(Apr.

(K3)

Kurz,

and Narayan, R.

S.,

J.

E.,

Membrane Tech nology (LI)

in

Separations.

2),

J., 13,

497 (1967).

46 (1973).

"New Developments

S.,

in

57 (1972).

4),

(Ml)

McCabe, W. L.

(M2)

Michaels, A. S. Chem. Eng. Progr., 64(12), 31 (1968).

(M3)

Merten, U.

Ind. Eng.

(ed.).

and Applications

. '

Lacey, R. E. Chem. Eng., 79(Sept.

MIT Press,

New York:

Inc., 1975.

Chem.,

21, 112 (1929).

Desalination by Reverse Osmosis. Cambridge, Mass.:

The

1966.

(M4)

Mazur, W. H., and Martin, C. C. Chem. Eng.

(PI)

Perry, R. H., and Green, D. Perry's Chemical Engineers' Handbook, 6th ed. New York: McGraw-Hill Book Company, 1984.

(P2)

Pan, C. Y. A.I.Ch.E.

(P3)

Permasep Permeators, E.

(P4)

Pan, C. Y., and Habgood, H.

(P5) (P6)

(51)

I.

545 (1983).

duPont Tech.

Bull., 401, 403,

405 (1972).

W. Ind. Eng. Chem. Fund., 13, 323 (1974). Pan, C. Y., and Habgood, H. W. Can. J. Chem. Eng., 56 197, 210 (1978). Porter, M. C. (ed.). Handbook of Industrial Membrane Technology. Park Ridge, N.J.:

(Rl)

J., 29,

Progr., 78(10), 38 (1982).

Noyes

Rousseau, R. W. York: John Wiley

Publications, 1990.

Handbook of Separation Process Technology. Sons, Inc., 1987.

(ed.).

&

Stern, S. A., Sinclair, T. F., Gareis, P. H. Ind. Eng. Chem., 57, 49 (1965).

P.

J.,

Vahldieck, N.

P.,

New

and Mohr,

.

(52)

Stannett, V. T., Koros, W. J., Paul, D. R., Lonsdale, H. K., and Baker, R. W. Adv. Polym. Sci., 32, 69 (1979).

(53)

Stoughten, R. W., and Lietzke, M. H.

(54)

Schroeder, E. D. Water and Wastewater Treatment.

Book Company,

Chem. Eng. Data,

10,

254 (1965).

New York: McGraw-Hill

1977.

(55)

Sourirajan, S. Reverse Osmosis.

(56)

Stern, S. A., and

(57)

Stookey, D.

J.,

Jr.

New

Walawender, W.

Patton, C.

J.,

York: Academic Press, Inc., 1970.

P. Sep. Sci., 4, 129 (1969).

and Malcolm, G. L. Chem. Eng. Progr.,

82(1

1),

36 (1986).

(Wl) (W2)

Weber, W. J., Jr. Physicochemical Processes for Water Quality Control. York: Wiley-Interscience, 1972.

Ward, W.

J.,

Browal, W.

R.,

and Salemme, R. M.

J.

Membr.

New

Sci., 1,

99

(1976).

(W3)

Weller,

S.,

and Steiner, W. A. Chem. Eng. Progr.,

(W4)

Weller,

S.,

and Steiner, W. A.

(W5)

Walawender, W.

Chap. 13

References

P.,

J. Appl. Phys., 21,

and Stern,

S.

A. Sep.

46, 585 (1950).

279 (1950).

Sci., 7, 553 (1972).

799

CHAPTER

14

Mechanical-Physical Separation Processes

INTRODUCTION AND CLASSIFICATION OF MECHANICAL-PHYSICAL SEPARATION PROCESSES

14.1

14.1

A

Introduction 11

gas-liquid and vapor-liquid separation processes were con-

The separation

processes depended on molecules diffusing or vaporizing from

In Chapters 10 sidered.

and

distinct phase to another phase. In Chapter 12 liquid-liquid separation processes were discussed. The two liquid phases are quite different chemically, which leads to a

one

separation on a molecular scale according to physical-chemical properties. Also, in

Chapter 12 we considered liquid-solid leaching and adsorption separation processes. Again differences in the physical-chemical properties of the molecules lead to separation on a molecular scale. In Chapter 13

we

discussed

membrane

separation

processes where the separation also depends on physical-chemical properties. All the separation processes considered so far

chemical differences In this

way

in the

have been based upon physical-

molecules themselves and on mass transfer of the molecules.

individual molecules were separated into two phases because of these

molecular differences. In the present chapter a group of separation processes considered where the separation the differences

among

is

not accomplished on a molecular scale nor

the various molecules.

The

is it

will

due

be to

separation will be accomplished using

mechanical-physical forces and not molecular or chemical forces and diffusion. These will be acting on particles, liquids, or mixtures of and not necessarily on the individual molecules.

mechanical-physical forces

and

liquids themselves

The mechanical-physical

particles

forces include gravitational and centrifugal, actual me-

chanical, and kinetic forces arising from flow. Particles and/or fluid streams are separated

because of the different 14. IB

effects

produced on them by these

forces.

Classification of Mechanical-Physical

Separation Processes

These mechanical-physical separation processes are considered under the following classifications.

800

1. Filtration. The general problem of the separation of solid particles from liquids can be solved by using a wide variety of methods, depending on the type of solids, the proportion of solid to liquid in the mixture, viscosity of the solution, and other factors. In filtration a pressure difference is set up and causes the fluid to flow through small holes of

a screen or cloth which block the passage of the large solid particles, which,

up on the cloth 2.

3.

in turn, build

porous cake.

In settling and sedimentation the particles are separated by gravitational forces acting on the various size and density particles.

and sedimentation.

Settling

from the

as a

fluid

Centrifugal settling and sedimentation.

In centrifugal separations the particles are

separated from the fluid by centrifugal forces acting on the various size and density particles.

Two

general types of separation processes are used. In the

first

type of process,

centrifugal settling or sedimentation occurs.

4.

In this second type of centrifugal separation process, centrifu-

Centrifugal filtration.

gal filtration occurs which

similar to ordinary nitration where a bed or cake of solids

is

builds up on a screen, but centrifugal force

is

used to cause the flow instead of a pressure

difference.

5.

Mechanical size reduction and separation. In mechanical size reduction the solid broken mechanically into smaller particles and separated according to size.

particles are

FILTRATION IN SOLID-LIQUID SEPARATION

14.2

14.2A

Introduction

In filtration, suspended solid particles in a fluid of liquid or gas are physically or

mechanically removed by using a porous of applications.

very fine

(in

The

fluid

medium

filtrate.

much

the micrometer range) or

spherical or very irregular

valuable product

may be

that retains the particles as a separate

Commerical filtrations cover a very wide range can be a gas or a liquid. The suspended solid particles can be

phase or cake and passes the clear

larger, very

rigid

or plastic particles,

The some

shape, aggregates of particles or individual particles.

in

the clear filtrate

from the

cases complete removal of the solid particles

is

filtration or the solid cake. In

required and in other cases only partial

removal.

The

feed or slurry solution

may

amount. When the concentration of time before the

filter

is

carry a heavy load of solid particles or a very small

very low, the

filters

can operate for very long periods

needs cleaning. Because of the wide diversity of filtration

problems, a multitude of types of filters have been developed. Industrial filtration

the

amount

equipment

differs

from laboratory

filtration

equipment only

of material handled and in the necessity for low-cost operation.

A

in

typical

is shown in Fig. 14.2-1, which is similar to a Biichner The liquid is caused to flow through the filter cloth or paper by a vacuum on the end. The slurry consists of the liquid and the suspended particles. The passage of the

laboratory filtration apparatus funnel. exit

particles

is

blocked by the small openings

relatively large holes

is

used to hold the

form of a porous filter cake as for the suspended particles. As

Sec. 14.2

in

the pores of the

filter cloth.

The

filter

cloth.

A

support with

solid particles build

up

in the

the filtration proceeds. This cake itself also acts as a filter the

cake builds up, resistance to flow also increases.

Filtration in Solid-Liquid Separation

801

slurry solution filter

/

cake

\

/

cloth or paper filter

open support for filter cloth v

filtrate

Figure

14.2-1

Simple laboratory filtration apparatus.

.

In the present section 14.2 the ordinary type of filtration will be considered where a pressure difference

is

used to force the liquid through the

filter

cloth and the

filter

cake

that builds up.

In Section 14.4E centrifugal filtration will be discussed, where centrifugal force

used instead of a pressure difference. In centrifugal

14. 2B

/.

filters

Types of

are often competitive

Filtration

filters.

In one classification

ordinary

filters

is

and

and either type can be used.

There are a number of ways

not possible to

it is

filtration applications,

Equipment

Classification of filters.

equipment, and

many

filters

make

to classify types of filtration

a simple classification that includes

are classified according to whether the

desired product or whether the clarified filtrate or outlet liquid

is

filter

all

types of

cake

is

the

desired. In either case

the slurry can have a relatively large percentage of solids so that a cake

is

formed, or have

just a trace of suspended particles.

can be

Filters

the cake

is

classified

removed

by operating

after a run, or

cycle. Filters

can be operated as batch, where is continuously removed.

continuous, where the cake

filters can be of the gravity type, where the liquid simply flows by a hydrostatic head, or pressure or vacuum can be used to increase the flow rates. An important method of classification depends upon the mechanical arrangement of the filter media. The filter cloth can be in a series arrangement as flat plates in an enclosure, as individual leaves dipped in the slurry, or on rotating-type rolls in the slurry. In the following sections only the most important types of filters will be described. For more

In another classification,

details, see references (B

2.

Bed filters.

1

,

P1

).

The simplest type of

filter is

useful mainly in cases

the bed

where

filter

shown

schematically in Fig.

amounts of solids are to be removed from large amounts of water in clarifying the liquid. Often the bottom layer is composed of coarse pieces of gravel resting on a perforated or slotted plate. Above the gravel is fine sand, which acts as the actual filter medium. Water is introduced at the top onto a baffle which spreads the water out. The clarified liquid is drawn out at the bottom. 14.2-2. This type

The

is

filtration

continues until the precipitate of filtered particles has clogged the sand

so that the flow rate'drops. direction so that

it

filtering

802

Then

the flow

is

stopped and water introduced

in the reverse

flows upward, backwashing the bed and carrying the precipitated

solid away. This apparatus

to the sand

relatively small

and can be

can only be used on precipitates that do not adhere strongly removed by backwashing. Open tank filters are used in

easily

municipal water supplies.

Chap 14 .

Mechanical-Physical Separation Processes

One of the important types of filters is the plate-andwhich is frame filter press, shown diagrammatically in Fig. 14.2-3a. These filters consist of plates and frames assembled alternatively with a filter cloth over each side of the plates. The plates have channels cut in them so that clear filtrate liquid can drain down along each plate. The feed slurry is pumped into the press and flows through the duct into each of the open frames so that slurry fills the frames. The filtrate flows through the filter cloth and the solids build up as a cake on the frame side of the cloth. The filtrate flows between the filter cloth and the face of the plate through the channels to the outlet.

3.

Plate-and-frame filter presses.

The 14.2-3a

filtration

all

proceeds until the frames are completely

the discharge outlets go to a

common

header. In

with solids. In Fig.

filled

many cases

the

filter

press

have a separate discharge to the open for each frame. Then visual inspection can be made to see if the filtrate is running clear. If one is running cloudy because of a break in

will

the filter cloth or other factors,

completely is

full,

the frames

and

reassembled and the cycle If

the cake

is

is

it

and

the frames are

the cake removed.

Then

the

filter

repeated.

be washed, the cake

to

performed, as shown

When

can be shut off separately.

plates are separated

is

left in

the plates

and through washing

is

provided for wash water opening behind inlet. The enters the inlet, which has ports the wash water the filter cloths at every other plate of the filter press. The wash water then flows through the

through It

to

filter

in Fig. 14.2-3b. In this

cloth, through the entire

press a separate channel

is

cake (not half the cake as

in filtration),

of the frames, and out the discharge channel.

the filter cloth at the other side

should be noted that there are two kinds of plates in Fig. 14.2-3b: those having ducts admit wash water behind the filter cloth, alternating with those without such ducts.

The plate-and-frame presses suffer from the disadvantages common to batch proThe cost of labor for removing the cakes and reassembling plus the cost of fixed charges for downtime can be an appreciable part of the total operating cost. Some newer types of plate-and-frame presses have duplicate sets of frames mounted on a rotating shaft. Half of the frames are in use while the others are being cleaned, saving downtime cesses.

and labor Filter

costs.

Other advances

in

automation have been applied

throughput processes. They are simple

and can be used the

4.

filter

to these types of filters.

employed

presses are used in batch processes but cannot be

at

high pressures,

to operate, very versatile

when

necessary,

if

and

for

high-

flexible in operation,

viscous solutions are being used or

cake has a high resistance.

Leaf filters.

The

filter

press

is

useful for

handling large quantities of sludge or for

many purposes but

efficient

is

not economical for

washing with a small amount of wash

inlet liquid

baffle

fine particles

perforated or coarse particles

slotted plate

clarified liquid

Figure

Sec. 14.2

14.2-2.

Bed filter of solid panicles.

Filtration in Solid-Liquid Separation

803

FIGURE

14.2-3.

Diagrams of plate-and-frame filter presses closed delivery,

804

{b)

through washing

Chap. 14

in

:

(a) filtration

of slurry with

a press with open delivery.

Mechanical-Physical Separation Processes

water.

The wash water

and more filter

efficient

and large volumes of wash water may was developed for larger volumes of slurry hollow wire framework covered by a sack of

often channels in the cake

be needed. The leaf filter shown

in Fig. 14.2-4

washing. Each leaf

is

a

cloth.

A number of these tank and

leaves are

hung

in parallel in a closed tank.

forced under pressure through the

The

slurry enters the

where the cake deposits on the outside of the leaf. The filtrate flows inside the hollow framework and out a header. The wash liquid follows the same path as the slurry. Hence, the washing is more efficient than the through washing in plate-and-frame filter presses. To remove the cake, the shell is opened. Sometimes air is blown in the reverse direction into the leaves to help in dislodging the cake. If the solids are not wanted, water jets can be used to simply wash away the cakes without opening the filter. Leaf niters also suffer from the disadvantage of batch operation. They can be automated for the filtering, washing, and cleaning cycle. However, they are still cyclical and are used for batch processes and relatively modest throughput processes. 5.

is

Continuous rotary filters.

common

to

all

cloth,

The plate-and-frame

filters

batch processes and cannot be used

number of continuous-type (a).

filter

filters

suffer

from the disadvantages

for large capacity processes.

Continuous rotary vacuum-drum filter.

This

filter

shown

in Fig. 14.2-5 filters,

washes, and discharges the cake in a continuous repeating sequence. The

with a suitable

filtrate leaves

through the axle of the

The automatic Also,

used the

if

to

drum

is

covered

medium. The drum rotates and an automatic valve in the center the filtering, drying, washing, and cake discharge functions in the cycle.

filtering

serves to activate

The

A

are available as discussed below.

filter.

valve provides separate outlets for the

filtrate

and the wash

liquid.

needed, a connection for compressed air blowback just before discharge can be help in cake removal by the knife scraper.

vacuum

filter is

only

1

atm. Hence, this type

The maximum pressure

is

differential for

not suitable for viscous liquids or for

slurry inlet

FIGURE

Sec. 14.2

14.2-4.

Filtration in Solid-Liquid Separation

Leaffiller.

805

slurry feed

FIGURE

Schematic of continuous rotary-drum filter.

14.2-5.

that must be enclosed. If the drum is enclosed in a atmospheric can be used. However, the cost of a pressure type of a vacuum-type rotary drum filter (P2).

liquids

shell, is

pressures above

about two times that

Modern, high-capacity processes use continuous filters. The important advantages filters are continuous and automatic and labor costs are relatively low. However, the capital cost is relatively high.

are that the

Continuous rotary disk

{b).

This

filter.

filter

consists of concentric vertical disks

mounted on a horizontal rotating shaft. The filter operates on the same principle as the vacuum rotary drum filter. Each disk is hollow and covered with a filter cloth and is partly submerged in the slurry. The cake is washed, dried, and scraped off when the disk is in the upper half of its rotation. Washing is less efficient than with a rotating drum type.

Continuous rotary horizontal filter.

(c).

This type

is

a

vacuum

filter

with the rotat-

As the horizontal filter rotates it and the cake is scraped off. The washing

ing annular filtering surface divided into sectors. successively receives slurry, efficiency

is

is

washed, dried,

better than with the rotary disk

This

filter.

filter

is

widely used in ore

extraction processes, pulp washing, and other large-capacity processes.

14. 2C

Filter

Media and

The

Filter media.

1.

Filter

filter

Aids

medium

requirements. First and foremost,

and give

it

for industrial filtration

must remove the

a clear filtrate. Also, the pores should not

fulfill

a

number of

solids to be filtered from the slurry

become plugged so

becomes too slow. The filter medium must allow the and cleanly. Obviously, it must have sufficient strength

filtration

easily

must

filter

that the rate of

cake to be removed

to not tear

and must be

chemically resistant to the solutions used.

Some

widely used

woven heavy

cloth,

filter

woolen

media are

twill

or duckweave heavy cloth, other types of

cloth, glass cloth, paper, felted

nylon cloth, Dacron cloth, and other synthetic cloths.

pads of

cellulose, metal cloth,

The ragged

fibers of natural

more effective in removing fine particles than the smooth plastic or metal Sometimes the filtrate may come through somewhat cloudy at first before the first

materials are fibers.

layers of particles,

which help

filter

the subsequent slurry, are deposited. This filtrate can

be recycled for refiltration.

2.

Filter aids.

806

Certain

filter

aids

may be

used to aid

Chap. 14

filtration.

These are often incom-

Mechanical-Physical Separation Processes

diatomaceous earth or kieselguhr, which is composed primarily of cellulose, asbestos, and other inert porous solids.

pressible

used are

Also

silica.

wood

These

filter

number of ways. They can be used

aids can be used in a

before the slurry

is filtered.

This

as a precoat

prevent gelatinous-type solids from plugging the

will

medium and

filter

also give a clearer filtrate. They can also be added to. the slurry before This increases the pororsity of the cake and reduces resistance of the cake

filtration.

during

In a rotary

filtration.

subsequently thin

can be applied as a precoat, and

the filter aid

filter

are sliced off with the cake.

slices of this layer

The use of cases

filter aids is usually limited to cases where the cake is discarded or to where the precipitate can be separated chemically from the filter aid.

14.2D

Basic Theory of Filtration

1. Pressure drop offluid through filter cake. Figure 14.2-6 is a section through a filter cake and filter medium at a definite time t s from the start of the flow of filtrate. At this

time the thickness of the cake the linear velocity of the

A

m

Lm

is

(ft).

the

filtrate in

The

filter

L direction

cross-sectional area

is

v m/s

(ft/s)

is

A m2

based on the

2 (ft

filter

),

and

area of

2 .

The

flow of the

filtrate

through the packed bed of cake can be described by an

equation similar to Poiseuille's law, assuming laminar flow occurs

Equation (2.10-2) gives Poiseuille's equation can be written Ap 32uv

for

in the filter channels.

laminar flow in a straight tube, which

(SI)

D

(14.2-1)

where Ap diameter is

is

in

pressure drop in

m

(ft),

L

32.174 lb m -ft/lb f -s

is

Ap

32/jy

L

9^

N/m 2

length in

m

(lb f /ft

(ft), \i is

(English)

2 ),

v is

open-tube, velocity in m/s

viscosity in, Pa

•

s

or kg/m-s

(ft/s),

(lb m /ft-s),

D

is

andgc

2 .

For laminar flow in a packed bed of particles, the Carman-Kozeny relation is similar and to the Blake-Kozeny equation (3.1-17) and has been shown to apply

to Eq. (14.2-1) to filtration.

_

_ k^vjl ~ 3 L e

Ap,

z

e)

Sl

(14.2-2)

filter

filtrate

slurry flow-

Figure

Sec. 14.2

medium

14.2-6.

Section through a filter cake.

Filtration in Solid-Liquid Separation

807

where

a constant and equals 4.17 for

fc, is

viscosity of is

filtrate in

Pa

-

s (lb m /ft

L

void fraction or porosity of cake,

area of particle in pressure drop (14.2-2)

and

is

m

2

in the

(ft

2 )

.

random

N/m 2

The

filter

thickness of cake in m(ft),

is

m3

3

S0

and shape,

area inm/s

p. is

(ft/s), e

specific surface

is

and Apc is ) right-hand For English units, the side of Eq. ). velocity is based on the empty cross-sectional area

(lb f /ft

linear

particles of definite size

ishnear velocity based on

of particle area per

cake in

divided by g c

v

s),

volume of solid

(ft

particle,

2

is

v

=

—A— dV/dt

(14.2-3)'

where A is filter area in m 2 (ft 2 ) and V is total m 3 (ft 3 ) of nitrate collected up to time t The cake thickness L may be related to the volume of filtrate V by a material balance. 3 3 c s is kg solids/m (lb m /ft ) of filtrate, a material balance gives

- e)p p =

LA(l

cs

(V + tLA)

s.

If

(14.2-4)

where p p is density of solid particles in the cake inkg/m 3 (lb m /ft 3 ) solid. The final term of Eq. (14.2-4) is the volume of filtrate held in the cake. This is usually small and will be neglected.

Substituting Eq. (14.2-3) into (14.2-2) and using Eq. (14.2-4) to eliminate L,

we obtain

the final equation as

dV A dt

- Apc - e)Sl pc s V

/cj(l

pp 8

where a

is

the specific cake resistance in

For the

filter

medium

resistance,

pcs

Q

A

Mi

(14.2-5)

V

A

m/kg (ft/lb m ),

=

a

3

- Apc

defined as

-g)S§ (14.2-6)

we can

write, by analogy with Eq. (14.2-5),

-A dV _ ~~ A dt pR where R m

is

the resistance of the

pressure drop.

When R m

is

filter

medium

to filtrate flow

treated as an empirical constant,

flow of the piping leads to and from the

filter

and the

Since the resistances of the cake and the

and

(14.2-7)

(14.2-7)

filter

dV

where Ap = Ap c

+

a

volume of

equal to

The volume

808

)

andAp r

is

the

includes the resistance to

medium resistance. medium are in series, Eqs.

filter

(14.2-5)

dt

Rm

filtrate

(14.2-8)

is

modified as follows:

-Ap_

(14 2 9)

ua.Cr

K)

A is

l

facs V

Apy. Sometimes Eq. (14.2-8)

A is

(ft'

~ Ap

dt

dV

resistance

it

1

can be combined, and become

A

where Ve

inm"

necessary to build up a fictitious

filter

cake whose

.

of filtrate

V can

also be related to

Chap. 14

W,

the kg of accumulated dry cake

Mechanical-Physical Separation Processes

solids, as follows :

W=c

y

V=

s

where cx

and p

is

is

mass fraction of

density of filtrate

2. Specific

e

m

solids in the slurry,

kg/m 3

(lb m /ft

From Eq.

cake resistance.

function of void fraction affect

in

and S 0

.

It

V

-

- mc,

1

is

mass

ratio of wet cake to dry cake,

3 ).

(14.2-6)

we see that the specific cake resistance is a

also

a function of pressure, since pressure can

is

By conducting constant-pressure experiments Ap can be found.

e.

(14.2-10)

at various pressure drops, the

variation of a with

Alternatively, compression-permeability experiments

cake at a low pressure drop and atm pressure with a porous bottom. pressure.

Then

(14.2-9).

This

A

piston

filtrate is fed to is

is

can be performed.

up by gravity

built

is

filtering in

A

filter

a cylinder

loaded on top and the cake compressed to a given

the cake and a

is

determined by a

differential

form of Eq.

then repeated for other compression pressures (Gl).

If a is independent of — Ap, the sludge incompressible. Usually, increases with — Ap, since most cakes are somewhat compressible. An empirical equation often used is is

=

a

where a 0 and

5

times the following

,

and

/?,

(14.2-11)

The compressibility constant The constant s usually falls between 0.1

is

s

is

zero for

to 0.8.

Some-

used.

a a'0

a 0 (-A P r

are empirical constants.

incompressible sludges or cakes.

where

oc

=

+

a'0 [l

/?(

- Apf]

(14.2-12)

are empirical constants. Experimental data for various sludges are

s'

given by Grace (Gl).

The data obtained from filtration experiments often do not have a high degree of The state of agglomeration of the particles in the slurry can vary and

reproducibility.

on the

have an

effect

14.2E

Filtration Equations for Constant-Pressure Filtration

/.

specific

cake resistance.

Basic equations for filtration rate

in

batch process.

Often a

filtration

is

done

for

conditions of constant pressure. Equation (14.2-8) can be inverted and rearranged to give

— = V + R dV A (-Ap) A(-&p) 2

where

K

is

in

s/m 6

(s/ft

6 )

and B

in

s/m 3

m

= Kp V + B

(14.2-13)

3

(s/ft

).

^ = A (-Ap) 2

(SI)

(14.2-14)

K, = B =

Vacs

A,, 2 ATif (-&p)gc

(English)

m (SI)

A(-Ap) (14.2-15)

B =

Sec. 14.2

aR a, a"\ A(-Ap)g c

Filtration in Solid-Liquid Separation

(English)

809

= Kp/2

slope

=B

intercept

0 Filtrate

FIGURE

V (m 3 )

volume,

Determination of constants

14.2-7.

For constant pressure, constant

in

a constant-pressure filtration run.

V

and incompressible cake,

a,

and

variables in Eq. (14.2-13). Integrating" to obtain the time of filtration in

dt

=

t

=

(K p

is

B)dV

(14.2-16)

V 2 + BV

1

Kp V

t

To

t s,

(14.2-17)

V

Dividing by

where V

V+

are the only

f

total

volume of filtrate

evaluate Eq. (14.2-17)

V

using Eq. (14.2-18). Data of

m3

in

it is

3 (ft

)

fall

on the

(14.2-18)

collected to

necessary to

t

s.

know a and R m

collected at different times

experimental data are plotted as t/V versus the graph does not

+B

line

and

is

V

This can be done by

.

Then

are obtained.

the

on and the

as in Fig. 14.2-7. Often, the first point

omitted.

intercept B. Then, using Eqs. (14.2-14)

t

and

The slope of the

line is

Kp Rm l2

a and

(14.2-15), values of

can be

determined.

EXAMPLE Data

14.2-1

.

Evaluation of Filtration Constants for Constant-Pressure Filtration

laboratory filtration of

for the

CaC0

3

slurry in water at 298.2

K 2

(25°C) are reported as follows at a constant pressure ( — Ap) of 338kN/m 2 (7060 lb f /ft ) (Rl, R2, Ml). The filter area of the plate-and-frame press was

A — 0.0439

m

(1.465 lb m /ft given, where

t

4.4

810

2

3 ).

(0.473

2 ft

)

and

Calculate the constants a and

f is

time in

V 0.498 x io-

3

9.5

1.000 x io-

3

16.3

1.501 x IO"

3

24.6

2.000 x 10"

3

s

wasc s

the slurry concentration

and V

is

filtrate

R m from

volume

t

V

34.7

2.498 X io- 3

46.1

3.002

X 10"

59.0

3.506 X 10"

3

Chap. 14

23.47

kg/m 3

the experimental data

collected in

t

3

=

73.6 89.4 107.3

m

3 .

V 3

4.004 X io3 4,502 X 10" 5.009 X IO"

3

Mechanical-Physical Separation Processes

The data are

14.2-1.

data are calculated as t/V and tabulated in Table V in Fig. 14.2-8 and the intercept

First, the

Solution:

plotted as //V versus

K p /2 =

determined as B = 6400 s/m 3 (181 s/ft 3 ) and the slope as 6 6 3.00 x 10 s/m Hence, K p = 6.00 x 10 6 s/m 6 (4820 s/ft 6 ). is

.

-4 At 298.2 K the viscosity of water is 8.937 x 10 Pa s = 8.937 x 10" 4 kg/m- s. Substituting known values into Eq. (14.2-14) and solving -

Kp = a

=

x

6.00

10

6

p.ac 5

=

=

)(a)(23.47) =

=

2

2

3 (0.0439) (338 x 10 )

A (-Ap)

10" m/kg(2.77 x

1.863 x

~4

(8.937 X 10

10

11

ft/lb

J

Substituting into Eq. (14.2-15) and solving,

5 = 6400 =

Rm = EXAMPLE The same

14.2-2.

—txR m

=

(8.937 x 10

10

m"

Time Required

)(/?J

3 0.0439(338 x 10 )

A(-Ap)

10.63 x 10

-4

to

1

x

(3.24

1

0

10

ft"

!

)

Perform a Filtration

Example 14.2-1 is to be filtered in a plate-and-frame 2 2 press having 20 frames and 0.873 m (9.4 ft ) area per frame. The same pressure will be used in constant-pressure filtration. Assuming the same 3 filter cake properties and filter cloth, calculate the time to recover 3.37 m

(119

3 ft

)

filtrate.

Solution:

s/m

Kp

slurry used in

In

Example

6

A = 0.0439 m 2 K p = 6.00 x 10 6 a and R m will be the same as before,

14.2-1, the area

,

and B = 6400 s/m Since the can be corrected. From Eq. (14.2-14), 3

.

,

Table

Kp

is

proportional to 1/A

Sec. 14.2

The

14.2-1

U = s,V = m

KxlO

.

Values of t/V for

14.2-1.

Example

t

2

1

3

)

(t/V) X 10" 3

0

0

4.4

0.498

9.5

1.000

9.50

16.3

1.501

10.86

24.6

2.000

12.30

34.7

2.498

13.89

46.1

3.002

15.36

59.0

3.506

16.83

73.6

4.004

18.38

89.4

4.502

19.86

107.3

5.009

21.42

8.84

Filtration in Solid-Liquid Separation

811

24

slope =

20

KJ2>

16

±xlP

3

12

(s/m 3 )

s 8

=

inter cept

-

i3

4

0

1

1

0

Volume FIGURE

new

area

Kp

is

=

The new

0.873(20)

6.00 x 10

B

is

i

I

l

4

3

5

V x 10 (m 3

of filtrate,

3 )

Determination of constants for Example 14.2-1.

14.2-8.

A =

...

2

1

6

=

m

17.46

(0. 0439/17. 46)

2

2

(188

=

2

ft

).

The new

37.93 s/m

6

Kp

(0.03042

is s/ft

6 )

proportional to 1/A from Eq. (14.2-15).

0.0439

B =

(6400)

= 16.10 s/m 3

(0.456

3

s/ft

)

17.46 Substituting into Eq. (14.2-17),

Kp

37.93 (3.37)

2

2

Using English

=

,

2

+

BV

=

0.03042 2 ( 1

1

+ (0.456)0

9)

19)

= 269.7

The washing

Equations for washing offilter cakes and total cycle time.

2.

s

units,

—V AT „

t

+ (16.10X3.37) = 269.7

s

of a cake after

the filtration cycle takes place by displacement of the filtrate and by diffusion.

amount

of wash liquid should be sufficient to give the desired washing efTed.

washing

rates,

existed at the

when wash In

it is assumed that the conditions during washing are the same as those that end of the filtration. It is assumed that the cake structure is not affected

where the wash

filters,

liquid follows the flow path, similar to that

the final filtering rate gives the predicted

pressure filtration using the is

The

calculate

liquid replaces the slurry liquid in the cake.

filters

as in leaf

To

same pressure

in

washing as

washing

during

rate.

filtration

For constant-

in filtering, the final filtering rate

the reciprocal of Eq. (14.2-13).

gU K dt) f

1

... Vf p

m 3 /s (ft 3/s) and Vf 3 3 the entire period at the end of filtration in m

where (dV/dt) f

=

rate of

washing

in

(ft

812

Chap. 14

(14.2-19)

+Bis

the total

volume of

filtrate for

).

Mechanical-Physical Separation Processes

For plate-and-frame

wash

presses, the

filter

liquid travels

through a cake twice as

thick and an area only half as large as in filtering, so the predicted washing rate

is 3;

of the

final filtration rate.

dV\

1

1

(14 - 2 - 20)

«)r4K&TB In actual experience the washing rate

may

be

than predicted because of cake

less

and formation of cracks. Washing rates in a small plate-andframe filter were found to be from 70 to 92% of that predicted (M 1). After washing is completed, additional time is needed to remove the cake, clean the filter, and reassemble the filter. The total filter cycle time is the sum of the filtration time, plus the washing time, plus the cleaning time. consolidation, channeling,

EXAMPLE

m

3

Rate of Washing and Total Filter Cycle Time

14.2-3.

At the end of the

filtration

cycle in

Example

14.2-2, a total filtrate

volume of

be washed by through washing in the plate-and-frame press using a volume of wash water equal to 10% of the filtrate volume. Calculate the time of washing and the total filter cycle time if cleaning the filter takes 20 min. 3.37

is

For

Solution:

s/m

6

B =

,

—

fdV\ \dt

collected in a total time of 269.7

this filter,

The time

of washing

is

Kp = is

37.93

as follows:

.

=

4 (37.93)(3.37)

is to

,

1

= --

)

The cake

Eq. (14.2-20) holds. Substituting Vf = 3.37 m 3 the washing rate

16.10 s/m-, and 1

s.

+ 16.10

1.737 x

10"" 3

m 3 /s

(0.0613

then as follows for 0.10 (3.37), or 0.337

m

3

3 ft

/s)

of wash

water.

t

;

x 10

1.737

The

total filtration cycle

-3

—m tr 3

= 194.0

/s

is

269.7

+

194.0

+ 20 = 27.73 min

60

3.

m3

0.337

=

60

Equations for continuous filtration.

In a

filter

that

is

continuous in operation, such as

vacuum type, the feed, filtrate, and the cake move at steady, continuous rotary drum the pressure drop is held constant for the filtration. The cake

a rotary-drum rates. In a

formation involves a continual change

ance of the

filter

Eq. (14.2-13),

medium

B =

is

5 = dt

1o

f

f

is

is

So

in

0,

= Kr

V dV

=

than the total cycle time

(14.2-21)

K„— ~2

the time required for formation of the cake. In a rotary-drum

less

tc

(14.2-22)

filter,

the filter

by

t=ft

Sec. 14.2

filtration the resist-

the cake resistance.

o

t

time

continuous

compared with

0.

Integrating Eq. (14.2-13) with

where

in conditions. In

generally negligible

(14.2-23) c

Filtration in Solid-Liquid Separation

813

where /is the fraction of the cycle used for cake formation. In the rotary drum,/ is the fraction submergence of the drum surface in the slurry. Next, substituting Eq. (14.2-14) and Eq. (14.2-23) into (14.2-22) and rearranging, flow rate

=

—=

1/2

(14.2-24)

At, If the specific cake resistance varies with pressure, the constants in Eq. (14.2-1 1) are needed to predict the value of a to be used in Eq. (14.2-24). Experimental verification of

Eq. (14.2-24) shows that the flow rate varies inversely with the square root of the viscosity

and

the cycle time (Nl).

When

short cycle times are used

resistance

is

in

continuous nitration and/or the

B must be

term

relatively large, the filter resistance

medium

filter

and Eq.

included,

becomes

(14.2-13)

t

Then Eq.

(14.2-25)

=

A=

+ BV

K.

(14.2-25)

becomes

-RJt +

V

=

flow rate

t

IRlIt]

+

2c s

«(-A

/

>)//( A1 £ e )]

,>

J

(14.2-26)

XCr

At,

EXAMPLE 14.2-4. Filtration in a Continuous Rotary Drum Filter A rotary vacuum drum filter having a 33% submergence of the drum in the slurry is to be used to filter a CaC0 slurry as given in Example 14.2-1 using 3

The solids concentration in the slurry is cx = 0.191 kg solid/kg slurry and the filter cake is such that the kg wet cake/kg dry cake = m = 2.0. The density and viscosity of the filtrate can be assumed as that of water at 298.2 K. Calculate the filter area needed to filter 0.778 kg slurry/s. The filter cycle time is 250 s. The specific cake resistance 9 03 where — Ap is in Pa and a can be represented by a = (4.37 x 10 ) ( — Ap) a

pressure drop of 67.0 kPa.

in

m/kg.

,

From Appendix A.2 Pa-s. From Eq. (14.2-10),

Solution: 10

_3

cs

=

for water, p

=

996.9 (0.191)

pc -^= Y^TT^M) = 3081 T x

9 3 03 Solving for a,cc = (4.37 x 10 ) (67.0 x 10 ) culate the flow rate of the filtrate,

=

996.9

kg/m 3

p.

,

=

0.8937 x

k S SO " ds / m

fiUrate

1.225 x 10

m/kg-

1

To

cal-

V

-= t

0.778 (c x )/(c s )

c

kg slurry\

/

=

kg

/ '

s

=

solid \ //

4.823 x 10"

4

J\ rn

3

kg slurryy/ \308.1 kg solid/m

4.823 x 10

Hence,

814

-4 2(0.33) (67.0

[250 (0.8937 x 10~

A

At c

A =

6.60

m

3

filtrate y

filtrate/s

Substituting into Eq. (14.2-24), neglecting and setting

V

\

1

0.191

0.778

3 )

B =

x 10

0,

and solving, 1/2

3 )

(1.225 x 10

11 )

(308.1)

2 .

Chap. 14

Mechanical-Physical Separation Processes

Filtration Equations for Constant-Rate Filtration

14.2F In

some

cases filtration runs are

constant pressure. This occurs

pump. Equation

(14.2-8)

made under

conditions of constant rate rather than

by a positive displacement can be rearranged to give the following for a constant rate if

the slurry

is

fed to the filter

(dV/dt)m 3 /s. /iac s

-Ap

dV

iV\_ KyV + C dt)-

V +

Fit

(14.2-27)

where

Kv

_{n<*c s dv

\A

(SI)

2

dt

(14.2-28) /^otcs

dV

\A 2 g c

dt

(English)

——

HR dV A Tt

c=

(SI)

(14.2-29)

c=

Kv

is

in

N/m 5 (lb /ft 5 ) and C is in N/m 2 f

Assuming

that the cake

is

volume

the total

increases

of filtrate collected,

is

(lb f /ft

K

).

The

gives a straight line for a constant rate dV/dt. is

The

C.

and the volume of filtrate collected

any moment during the

2

K v and C are constants characteristic of and so on. Hence, a plot of pressure, — Ap, versus

K v and the intercept

The equations can t

(English)

dt

incompressible,

the slurry, cake, rate of filtrate flow,

slope of the line

—

Ag c

pressure increases as the cake thickness

increases.

also be rearranged in terms of filtration, the total

volume V

is

— Ap

and time

t

At

as variables.

related to the rate

and

total

time

as follows:

K =

f

dV —

(14.2-30)

dt

Substituting Eq. (14.2-30) into Eq. (14.2-27),

-Ap

=

ixac s

fdV

(14.2-31)

A

~dt

For

the case where the specific cake resistance a

(14.2-11), this can

be substituted for a

is

not constant but varies as

Eq. (14.2-27) to obtain a

in

Eq.

final equation.

SETTLING AND SEDIMENTATION IN PARTICLE-FLUID SEPARATION

14.3

14. 3A

Introduction

In filtration the solid particles are filter

in

dt

medium, which blocks

removed from

the slurry by forcing the fluid through a

the passage of the solid particles

and allows the

filtrate to

pass through. In settling and sedimentation the particles are separated from the gravitational forces acting on the particles.

Sec. 14.3

Settling

and Sedimentation

in

Particle-Fluid Separation

fluid

by

815

Applications of settling and sedimentation include removal of solids from liquid sewage wastes, settling of crystals from the mother liquor, separation of liquid-liquid mixture from a solvent-extraction stage in a settler, settling of solid food particles from a liquid food, and settling of a slurry from a soybean leaching process. The particles can be solid particles or liquid drops. The fluid can be a liquid or gas and it may be at rest or in

motion.

some processes

In

from the

particles

fluid

of settling

and sedimentation the purpose

stream so that the

fluid

free of particle

is

is

to

remove the

contaminants. In other

processes the particles are recovered as the product, as in recovery of the dispersed phase in liquid— liquid extraction. In

some

cases the particles are suspended in fluids so that the

particles can be separated into fractions differing in size or in density.

When

a particle

is

other particles so that Interference is

less

When

less

is

than

1

if

1

%

if

from the walls of the container and from

not affected by them, the process

its fall is

is

called free settling.

the ratio of the particle diameter to the container diameter

the particle concentration

is

less

%

than 0.2 vol

in the solution.

and the process is called or suspension by gravity settling into a

the particles are crowded, they settle at a lower rate

The separation

hindered settling. clear fluid

14. 3B

1.

than

200 or

:

at a sufficient distance

and a slurry of higher

of a dilute slurry solids content

is

called sedimentation.

Theory of Particle Movement Through a Fluid

Whenever

Derivation of basic equations for rigid spheres.

through a difference

a

fluid, is

number

needed between the particle and the

needed to impart motion to the particle. the

buoyant force on the particle

will

not

move relative

For a

a particle

is

moving

of forces will be acting on the particle. First, a density

will

fluid.

An

external force of gravity

If the densities of the fluid

and

is

particle are equal,

counterbalance the external force and the particle

to the fluid.

moving in a fluid, there are three forces acting on the body: downward, buoyant force acting upward, and resistance or drag force

rigid particle

gravity acting

acting in opposite direction to the particle motion.

We

will consider a particle of mass m kg falling at a velocity The density of the solid particle is p p kg/m 3 solid and that of liquid. The buoyant force F b in N on the particle is fluid.

F„=

— =V

p

v

m/s

relative to the

the liquid

is

pkg/m 3

(14.3-1)

pg

Pp

where m/p p m/s 2

is

the

volume Vp

in

m

3

of the particle

and g

is

the gravitational acceleration in

.

The

gravitation or external force

Fg

in

N on the particle

is

F g = mg

(14.3-2)

N

force F D on a body in may be derived from the fact that, like in flow of fluids, 2 the drag force or frictional resistance is proportional to the velocity head v /2 of the fluid

The drag

displaced by the

moving body. This must be multiplied by

a significant area A,

such as the projected area of the

the density of the fluid and

particle.

by

This was defined previously

in Eq. (3.1-1).

Fd = C d where the drag

816

coefficient

CD

is

v

2

- pA

the proportionality constant and

Chap. 14

(14.3-3)

is

dimensionless.

Mechanical-Physical Separation Processes

The

resultant force

on the body

F g — Fb — F D

then

is

.

This resultant force must

equal the force due to acceleration.

j=F

m

g

- Fb - F D

(14.3-4)

t

Substituting Eqs. (14.3-1H14.3-3) into (14.3-4),

m

dv

^l dt

CD v2pA

mpg

= m9

(14.3-5)

^

2

pp

we start from the moment the body is released from its position of rest, the falling of the body consists of two periods: the period of accelerated fall and the period of constant velocity fall. The initial acceleration period is usually very short, of the order of If

a tenth of a second or so. Hence, the period of constant velocity

The velocity

To

is

fall is

called the free settling velocity or terminal velocity u t

=

solve for the terminal velocity in Eq. (14.3-5), dvldt

the important one.

.

0 and the equation

becomes

=

v,

For spherical (14.3-6),

we

particles

m =

2.

Drag

shown

p)m (14.3-fi)

A =

nDjiA. Substituting these

m/s(ft/s),

p

is

4 (Pp-P^D„

=

.

kg/m 3 (Ibjft 3 ), g

2 9.80665 m/s

is

The

coefficient for rigid spheres.

to be a function of the

Section

3.

p. is

174 ft/s

2 ),

andD p

is

m

(ft).

sphere and is shown in Fig. law region for /V Re < 1, as discussed

vp/pi of the

is

CD = where

(32.

drag, coefficient for rigid spheres has been

Reynolds number D p

IB, the drag coefficient

(14 3 . 7)

3C D p

V

14.3-1. In the laminar flow region, called the Stokes' in

into Eq.

obtain, for spherical particles

'

v, is

-

2g(p p

irDpp p l6 and

„

where

j

the viscosity of the liquid in

24

Dp Pa

=

vp/n •

s

— N 24

(14.3-8)

Rc

or kg/m s^b^ft

s).

•

Substituting this into

Eq. (14.3-7) for laminar flow,

=

(14-3-9)

18^

For other shapes of particles, drag coefficients wiLl differ from those given in Fig. 14.3-1 and data are given in Fig. 3.1-2 and elsewhere (B2, L2, PI). In the turbulent Newton's law region above a Reynolds

number

approximately constant at C D = Solution of Eq. (14.3-7) the terminal velocity velocity

is

to

and error when the

,

particle

be obtained. This occurs because

the drag coefficient

CD

diameter

known and

is

also depends

upon

the

Brownian motion

is

present. Brownian

movement

surrounding the particle and the

Settling

is

to the particle by collisions between the molecules of the particle.

This movement of the particles

directions tends to suppress the effect of gravity, so settling of the particles

Sec. 14.3

is

v,

random motion imparted

fluid

5

0.44.

is trial

If the particles are quite small,

the

of about 1000 to 2.0 x 10

and Sedimentation

in Particle-Fluid

Separation

in random may occur

817

1000

1 1 1 1

\

—

100

10

Al

24

CD

-

^s \

Mil 10

-3

1

10

II

1

-2

1

1

1

1

1

II

1

10°

10"

1

1

14.3-1

10

Drag

.

1

1

2

10

Reynolds number, FIGURE

II

/V Re

II

1

10

=

II

1

1

1

104

3

II

1

\l

5

l>

10 6

10

•

coefficient for a rigid sphere.

more slowly or not at all. At particle sizes of a few micrometers, the Brownian effect becomes appreciable and at sizes of less than 0.1 ^m, the effect predominates. In very small particles, application of centrifugal force helps reduce the effect of Brownian movement.

EXAMPLE

14.3-1.

Settling Velocity

of Oil Droplets

Oil droplets having a diameter of 20 pm (0.020 mm) are to be settled from air at an air temperature of 37.8°C (311 K) at 101.3 kPa pressure. The density of the

oil is

900 kg/m 3 Calculate the terminal settling velocity of the drops. .

The various knowns are D p = 2.0 x 10" 5 m, p p = 900 kg/m 3 From Appendix A.3 for air at 37.8°C, p = 1.137 kg/m 3 p = 1.90 x 10" 3 Pa s. The drop will be assumed to be a rigid sphere. The solution is trial and error since the velocity is unknown. Hence, C D cannot be directly evaluated. The Reynolds number is as follows

Solution:

.

,

N *< = For the

first trial,

Dp

v.p

(2.0

~*f

assume

=

x 10" 5 Xi>,XU37) 1.90 x 10-'

=

that v,

0.365. Substituting into Eq. (14.3-7)

v,

=

l

4(p p

-

p) 9 D p

3C dP

CD = Using

v,

=

^

/

Then

and solving for

4{900

-

Ll91V

(14 - 3 - 10) -

7V Re

CD

=

1.197(0.305)

=

,

1.137X9.8066X2.0 x

1(F)

(3)C D (1.137)

V

(14.3-11)

C D = 0.2067/(0.305) 2 = 2.22. v, = 0.0305 m/s, N Re = 0.0365 from Eq.

0.305 m/s,

Assuming that 222 from Eq.

CD =

0.00305 m/s,

818

_

0.305 m/s.

=

N Rc =

(14.3-11).

0.00365 and

For the

CD =

Chap. 14

third trial,

(14.3-10)

assuming

and

that v,

=

22200. These three points of7V Re and

Mechanical-Physical Separation Processes

values of

shown

CD

calculated are plotted on a plot similar to Fig.

in Fig. 14.3-2. It

14.3-1

can be shown that the line through these points

The intersection

and is

a

and the drag-coefficient correlation line is the solution to the problem at N Kc — 0.012. The velocity can be calculated from the Reynolds number in Eq. (14.3-10).

straight line.

N Re = v,

=

of this line

=

0.012

1.197u,

0.0100 m/s (0.0328

ft/s)

The particle is in the Reynolds number range less than 1, which is the laminar Stokes' law region. Alternatively, the velocity can be calculated by substituting into Eq. (14.3-9). 9.8066(2.0 x 10" V '

Note

For drag

18(1.90

5 2 )

(900-

was

in the

1.137; "

x 10-*)

that Eq. (14.3-9) could not be

particle fall

particle

=

used

until

nntM " a01 ° 3

it

,

m/S

was determined

that the

laminar region.

particles that are rigid but nonspherical, the

and the orientation of

drag depends upon the shape of the motion. Correlations of

the particle with respect to its

coefficients for particles of different

shapes are given in a number of references (B2,

CI, PI).

Drag

3.

coefficients for nonrigid spheres.

lation inside the particle

and

particle

When

particles are nonrigid, internal circu-

deformation can occur. Both of these effects affect

the drag coefficient and terminal velocity. Drag coefficients for air bubbles rising in water are given in Perry and Green (PI), and for a Reynolds number less than about 50, the curve is the same as for rigid spheres in water.

For is

drops

liquid

in gases, the

same drag relationship

as for solid spherical particles

obtained up to a Reynolds number of about 100 (HI). Large drops

will

deform with an

increase in drag. Small liquid drops in immiscible liquids behave like rigid spheres and the drag coefficient curve follows that for rigid spheres 10.

Above

this

and up

to a

up

to a

Reynolds number of about

Reynolds number of 500, the terminal velocity

is

greater than

that for solids because of internal circulation in the drop.

A^e =0.012 Reynolds number, Figure

Sec. 14.3

14.3-2.

Settling

N Rc

Solution of Example 14.3-1 for settling velocity of a particle.

and Sedimentation

in

Particle-Fluid Separation

819

Hindered Settling

14.3C

For many cases

in settling,

particles interfere with the

a large number of particles are present and the surrounding

motion of individual

particles.

The

velocity gradients sur-

rounding each particle are affected by the close presence of other displace the liquid, and an appreciable

in settling in the liquid

liquid

is

generated. Hence, the velocity of the liquid

apparatus

the particle than with respect to the

The

true drag force

particles

velocity of the

is

appreciably greater with respect to

is less

than would be calculated from Eq.

itself.

For.this hindered flow the settling velocity (14.3-9) for Stokes' law.

The

particles.

upward

greater in the suspension because of the

is

interference of the other particles. This higher effective viscosity of the mixture /i m to the actual viscosity of the liquid

which depends upon

e,

the

itself,

volume

is

equal

divided by an empirical correction factor, \p p

p.,

,

fraction of the slurry mixture occupied by the liquid

(SI).

=

V*

(14.3-12)

Jwhere

\p

p

dimensionless and

is

as follows (SI):

is

_

*P

The which

is

(14.3-13)

t,

density of the fluid phase becomes effectively the bulk density of the slurry

pm

,

as follows:

pm

where p m

= ep+(l -e) Pp

density of slurry in kg solid

is

Pp

The

1

= 10~ ,n..82 (1 -

settling velocity

v,

-P m =

-

PP

+

(14.3-14)

liquid/m 3 The density difference .

l>P

+

(1

- e)P P = ~)

with respect to the apparatus

is £

£(P P

-

is

now (14.3-15)

P)

times the velocity calculated by

Stokes' law.

(p p

Substituting mixture properties of

p.

-

-

pm

)

from Eq.

(14.3-15) for (p p

relative velocity effect,

m from Eq. p),

(14.3-12) for

and multiplying the

Eq. (14.3-9) becomes, for laminar

v,

=

—z-f

(e-i^

p. in

result

Eq. (14.3-9),

by

£

for the

settling,

(14.3-16)

)

18/i

This

is

the velocity calculated from Eq. (14.3-9), multiplied by the correction factor (e 2 ^).

The Reynolds number

is

then based on the velocity relative to the fluid and

= When

the

Reynolds number

Reynolds numbers above spherical particles

is

1.0,

and angular

0^

less

=

than

see (PI).

is

D>g(p,-p)p m «p,

1,

the settling

The

effect of

is

in the Stokes'

concentration

is

law range. For

greater for non-

particles (SI).

EXAMPLE 14.3-2. Hindered Settling of Glass Spheres Calculate the settling velocity of glass spheres having a diameter of 4 1.554 x 1CT (5.10 x 10"* ft) in water at 293.2 K (20°C). The slurry

m

contains 60 wt% solids. 3 (154 lbjft ).

820

The

density of the glass spheres

Chap. 14

is

pp

=

2467 kg/m

3

Mechanical-Physical Separation Processes

3 Density of water p = 998 kg/m 3 (62.3 lb^ft ), and viscosity of 3 4 10" 10" water ^=1.005 x Pas (6.72 x lbjft-s). To calculate the

Solution:

volume

fraction e of the liquid,

1

40/998

40/998

The bulk density of the

-

ep

=

1553 kg/m

+

(1

=

e)p p 3

60/2467

by Eq.

slurry p m

=

Pm

+

(14.3-14) is

0.622(998)

(96.9 lbjft

+

(1

- 0.622X2467)

3 )

Substituting into Eq. (14.3-13),

=

=

jQl.82(l -t)

jq1.82(1 -0.622)

=

0.205

Substituting into Eq. (14.3-16), using SI and English units, 4 2 9.807(1.554 x 10" ) (2467

_ "* ~~

1.525 x i0"

3

4 2 32.174(5.1 x 10" ) (154

5.03

x 10"

The Reynolds number

-

2

^m £

x 0.205)

ft/s

obtained by substituting into Eq. (14.3-17),

_ Dp»,Pm _ D P v,Pn, _ (/*/^>

3 d-554 x 10-^X1-525 x 10' )1553 3 (1.005 x 10" /0.205)0.622

0.121

Hence, the settling

14. 3D

is

3

x 0.205)

)

62.3X0-622 -4 18(6.72 x 10 )

=

2

m/s

~

=

998X0.622 -3

18(1.005 x 10

=

Re

-

is

in

the laminar range.

Wall Effect on Free Settling

When the diameter of D p of the particle becomes appreciable with respect to the diameter D w of the container in which the settling is occurring, a retarding effect known as the wall effect

is

exerted on the particle.

settling in the Stokes'

The

terminal settling velocity

the following to allow for the wall effect (Zl) for D

kw

For

p

<

/D w

reduced. In the case of

0.05.

l -

=

\+2A{D p/D w

(14.3-18) )

the completely turbulent regime, the correction factor

[i

14.3E

is

law regime, the computed terminal velocity can be multiplied by

Differential Settling

is

+(/yzv) 4 ]" 2

and Separation

of Solids in Classification 1.

Sink-and-float methods.

fractions based

Sec. 14.3

upon

Settling

Devices

for

the separation of solid particles into several

their rates of flow or settling

and Sedimentation

in

through

fluids are

Particle-Fluid Separation

known

as classifiers.

821

There are several separation methods to accomplish this by sink-and-float and differential settling. In the sink-and-float method, a liquid is used whose density is intermediate between that of the heavy or high-density material and the light-density material. In this

method

medium, and

the heavy particles will not float but settle from the

the light

particles float. sizes of the particles and depends only upon the two materials. This means liquids used must have densities greater than water, since most solids have high densities. Unfortunately, few such liquids exist that are cheap and noncorrosive. As a result, pseudoliquids are used, consisting of a

This method

is

independent of the

relative densities of the

suspension in water of very fine solid materials with high specific gravities such as galena

and magnetite (specific gravity = 5.17). is used and the bulk density of the medium can be varied widely by varying the amount of the fine solid materials in the medium. Common applications of this technique are concentrating ore materials and cleaning coal. The fine solid

(specific gravity

Hindered

=

7.5)

settling

materials in the

medium

are so small in diameter that their settling velocity

is negligible,

giving a relatively stable suspension.

The separation of solid particles into several size fracupon the settling velocities in a medium is called differential settling or classification. The density of the medium is less than that of either of the two substances 2.

Differential settling methods.

tions based

to be separated. In differential settling both light and

disadvantage of sizes

is

this

method

if

heavy materials

settle

through the medium.

A

the light and heavy materials both have a range of particle

that the smaller, heavy particles settle at the

same terminal

velocity as the larger,

light particles.

Suppose that we consider two

different materials: heavy-density material

A

(such as

galena, with a specific gravity p A = 7.5) and light-density material B (such as quartz, with a specific gravity p B = 2.65). The terminal settling velocity of components A and B from

Eq. (14.3-7) can be written 1/2

pA

(14.3-20)

2C DA p

Ap pB For

particles of equal settling velocities,

(14.3-20) to (14.3-21), canceling terms,

~

{PpA

1/2

v,

p)gD pB

A

=

(14.3-21)

and we obtain, by equating Eq.

v tB

and squaring both sides,

P)D pA

{p pB

pC DA

-

p)D'?b

(14.3-22)

pc c

or

For

D pA

PpB- P

D pb

P pA

particles that are essentially spheres at very

turbulent Newton's law region,

CD

is

constant and

D pA

822

-P

CL C DB

-P PpA - P PpB

Chap. 14

(14.3-23)

high Reynolds numbers

C DA = C DB

,

in the

giving

(14.3-24)

Mechanical-Physical Separation Processes

For laminar Stokes' law settling,

24u

C DA =D pA v, A p Substituting Eq. (14.3-25) into (14.3-23)

_pa

_

D PB

C DB =

24u

and rearranging

i

—-

VpB

\P P A

y

(14.3-25)

D pB v, B p for Stokes'

law settling, where

\

(14.3-26)

Pj

For transition flow between laminar and turbulent flow,

Bid

D pB For

= (Ei^LE)" \P PA -Pj

where

particles settling in the turbulent range,

f< „ <

Eq. (14.3-24) holds for equal

— D pB and

For particles where D pA settling region, combining Eqs. (14.3-20) and (14.3-21),

velocities.

(14.3-27)

i

is in

Pm-P V/2 If

both

A

and

B

particles are settling in the

(14.3-28) can be used to

diameter to

the

A

B

same medium, then Eqs.

(14.3-24)

and

the plots given in Fig. 14.3-3 for the relation of velocity to

a mixture of particles of materials

Dp4 for both types of material.

fraction of substance

range

make

(14.3-28)

for

First,

Dpl

A and B. we consider

settling

the turbulent Newton's law

B

In the size range

A and B

D pl

to

can be obtained since no particles of A

D p2

with a size range of

in Fig. 14.3-3, a

settle as slowly.

pure

In the size

D pi to D pi a pure fraction of A can be obtained since no B particles settle as fast as particles in this size range. In the size range pl to D pi A particles settle as rapidly ,

D

D p2

D p4

,

forming a mixed fraction of A and B. Increasing the density p of the medium in Eq. (14.3-24), the numerator becomes smaller proportionately faster than the denominator, and the spread between D pA and as

particles in the size range

to

D pB is increased. Somewhat similar

Sec. 14.3

Settling

,

curves are obtained in the Stokes' law region.

and Sedimentation

in Particle-Fluid Separation

823

EXAMPLE 14.3-3.

Separation of a Mixture of Silica and Galena and galena (A) solid particles having a size range of 5 is to be separated by hydraulic classifito 2.50 x 10" cation using free settling conditions in water at 293.2 (Bl). The specific gravity of silica is 2.65 and that of galena is 7.5. Calculate the size range of the various fractions obtained in the settling. If the settling is in the laminar region, the drag coefficients will be reasonably close to that for spheres.

A

mixture of 6 5.21 x 10"

silica (B)

m

m

K

The

Solution:

particle-size range

is

D p = 5.21

x 10~ 6

m to 1^ = 2.50 x

10"

5

m. Densities are p A = 7.5(1000) = 7500 kg/m p pB = 2.65{ 1000) = 2650 kg/m 3 p = 998 kg/m 3 for water at 293.2 K (20°C). The water viscosity _3 Pa s = 1.005 x 10~ 3 kg/m-s. fi = 1.005 x 10 Assuming Stokes* law settling, Eq. (14.3-9) becomes as follows: 3

,

,

i£l&*LZJ±

0tA=

The

(.14.3-29)

Reynolds number occurs for the largest particle and the biggest where D pA = 2.50 x 10" 5 m and p pA = 7500. Substituting into Eq.

largest

density,

(14.3-29),

V,j

=

9.807(2.50

x 10

_5

z )

(75O0

-

998)

2.203 x 10"

3 18(1.005 x 10" )

Substituting into the Reynolds

3

m/s

number equation,

D N = P A v,a P

(14.3-30)

V-

(2.50

x 1Q- S X2.203 x 10" 3 )998 nne/ln -0-054/ 3 1.005 x 10-

Hence, the settling is in the Stokes' law region. Referring to Fig. 14.3-3 and using the same nomenclature, the largest = 2.50 x 10~ 5 m. The smallest size is D pl = 5.21 x 10~ 6 m. The size is D p4 pure fraction of A consists ofD,,^ = 2.50 x 10" 5 toD pA3 The particles, _

m

.

having diameters D pA3 and D pBi are related by having equal settling 5 velocities in Eq. (14.3-26). Substituting D pB4 = 2.50 x 10" m into Eq. ,

(14.3-26)

and solving, /2650 - 998Y' 2 \1500 - 998

D pA3 2.50 x 10"

5

D pA3 = The diameter

size

range of pure

D pB2

is

B

related to

1.260 x 10~

fraction

D pA

,

is

3

m

D pB2 ioD pBX =

5.21

=5.21 x 10" 6 by Eq.

x 10~ 6 m. The

(14.3-26) at equal

settling velocities. 5.21 x 10"

6

D pB2 D pB2 = The 1.

The

- 998\ 1/2

V 7500

-" 8

1.033 x 10"

5

m

three fractions recovered are as follows.

size

range of the

D pA3 =

824

/2650

first

fraction of pure

1.260 x 10"

5

m

to

Chap. 14

A

(galena)

D pAi =

is

2.50

as follows:

x 10~ 5

m

Mechanical-Physical Separation Processes

2.

The mixed-fraction

D pB2 = D pAl = 3.

The

range

is

1.033 x 1(T5.21

x 10' 5

as follows:

m

5

m

D pB4 =

to

.

D pA3 =

to

range of the third fraction of pure

size

D p8l =

14.3F

size

5.21

x 10~

6

m

to

2.50 x 10

1.260 x 10"

B (silica) is

D pB2 =

-5 5

m m

as follows:

1.033 x 10~

5

m

Sedimentation and Thickening

Mechanisms of sedimentation. When a dilute slurry is settled by gravity into a clear and a slurry of higher solids concentration, the process is called sedimentation or sometimes thickening. To illustrate the method of determining settling velocities and the mechanisms of settling, a batch settling test is carried out by placing a uniform concentration of the slurry in a graduated cylinder. At the start, as shown in Fig. 14.3-4a, all the particles settle by free settling in suspension zone B. The particles in zone B settle at a uniform rate at the start and a clear liquid zone A appears in Fig. 14.3-4b. The height z drops at a constant rate. Also, zone D begins to appear, which contains the settled particles at the bottom. Zone C is a transition layer whose solids content varies from that in zone B to zone D. After further settling, zones B and C disappear, as shown in Fig. 14.3-4c. Then compression first appears, and this moment is called the critical point. During compression liquid is expelled upward from zone D and the thickness of zone D /.

fluid

decreases.

2.

In Fig. 14.3-4d the height z of the clear liquid

Determination of settling velocity.

interface height

slope of the

line,

is is

plotted versus time. As shown, the velocity of settling, which

constant at

first.

The

critical

point

is

shown

is

the

at point C. Since sludges

vary greatly in their rates, experimental settling rates of each sludge is necessary. Kynch (Kl) and Talmage and Fitch (Tl) describe a method to predict thickener sizes from the

batch settling

Sec. 14.3

test.

Settling

and Sedimentation

in

Particle- Fluid Separation

825

The is

is determined by drawing a tangent to the curve in Fig. and the slope —dzidt = v At this point the height is z and

settling velocity v

14.3-4d at a given time

t

.

1

z.,-

x

x

the intercept of the tangent to the curve. Then,

(14.3-31)

The concentration

cx

is,

therefore, the average concentration of the suspension ifz

height of this slurry. This

is

c^i^CoZq where c 0

is

is ;

the

calculated by

or

cx

= kg/m 3

the original slurry concentration in

(14.3-32)

co

(^fj

at z 0

height and

t

=

0.

repeated for other times and a plot of settling velocity versus concentration

This

is

is

made.

Further details of this and other methods to design the thickener are given elsewhere (CI, Fl, F2, Tl, PI). These

and other methods

should be exereised

in their use.

14. 3G

for Settling

/.

Equipment

in the literature

are highly empirical and care

and Sedimentation

Simple gravity settling tank.

In Fig. 14.3-5a a simple gravity settler is

shown

for

phase from another phase. The velocity horizontally to the right must be slow enough to allow time for the smallest droplets to rise from the bottom to the interface or from the top down to the interface and coalesce. In Fig. 14.3-5b a gravity settling chamber is shown schematically. Dust-laden air

removing by

enters at

settling a dispersed liquid

one end of a large boxlike chamber. Particles settle toward the floor at their The air must remain in the chamber a sufficient length of time

terminal settling velocities.

Knowing

(residence time) so that the particles reach the floor of the chamber.

throughput of the time of the

air in

air

the

stream through the chamber and the chamber

chamber can be

calculated.

The

size,

vertical height of the

the

the residence

chamber must

be small enough so that this height, divided by the settling velocity, gives a time

less

than

the residence time of the air.

2.

Equipment for

classification.

The

simplest type of classifier

is

one

in

which a large

air

dust

FIGURE

826

14.3-5.

Gravity sealing tanks settling chamber.

:

(a) seitler for liquid-liquid dispersion, (b) dust-

Chap. 14

Mechanical-Physical Separation Processes

fluid

slurry in

coarse

intermediate

fine

particles

particles

particles

FIGURE

tank

is

14.3-6.

Simple gravity settling

subdivided into several sections, as shown

out

classifier.

A

in Fig. 14.3-6.

liquid slurry feed

enters the tank containing a size range of solid particles. The larger, faster-settling particles settle to the

to the

bottom close

result of the

bottom

close to the entrance

to the exit.

The

and the slower-settling

particles settle

linear velocity of the entering feed decreases as a

enlargement of the cross-sectional area

at the entrance.

the tank allow for the collection of several fractions.

The

The vertical

baffles in

settling-velocity equations

derived in this section hold.

Another type of gravity settling chamber is the Spitzkasten, which consists of a series of conical vessels of increasing diameter in the direction of flow. The slurry enters the first vessel, where the largest and faster-settling particles are separated. The overflow goes to the next vessel, where another separation occurs. This continues in the succeeding vessel or vessels. In each vessel the velocity of upflowing inlet water is controlled to give the desired size range

3.

Spitzkasten classifier.

shown

in Fig.

14.3-7,

for each vessel.

4.

Sedimentation thickener.

clear fluid

and

The separation of

a dilute slurry by gravity settling into a

a slurry of higher solids concentration

is

called sedimentation. Industrially,

sedimentation operations are often carried out continuously

in

equipment called

thick-

slurry inlet

upward flow of water

water

coarse

intermediate

solids

solids

Figure

Sec. 14.3

Settling

14.3-7.

Spitzkasten gravity settling chamber.

and Sedimentation

in Particle-Fluid

Separation

827

eners.

A

continuous thickener with a slowly revolving rake for removing the sludge or

is shown in Fig. 14.3-8. The slurry in Fig. 14.3-8 is fed at the center of the tank several feet below the surface of the liquid. Around the top edge of the tank is a clear liquid overflow outlet. The rake serves to scrape the sludge toward the center of the bottom for removal. This gentle stirring dds in removing water from the sludge.

thickened slurry

In the thickener the entering slurry spreads radially through the cross section of the

thickener and the liquid flows

zone by

free settling.

Below

upward and out

the overflow.

this dilute settling

zene

is

The

solids settle in the upper

the transition zone, in which the

A

concentration of solids increases rapidly, and then the compression zone. flow can be obtained

minimal terminal

The settling

if

the

upward

velocity of the fluid in the dilute

is

clear over-

less

than the

settling velocity of the solids in this zone.

rates are quite

slow in the thickened zone, which consists of a compress-

ion of the solids with liquid being forced

upward through the

case of hindered settling. Equation (14.3-16) velocities,

zone

may be used

solids. This

is

an extreme

to estimate the settling

but the results can be in considerable error because of agglomeration of

result, laboratory settling or sedimentation data must be used in the design of a thickener as discussed previously in Section 14.3F.

particles.

14.4

14. 4A

As a

CENTRIFUGAL SEPARATION PROCESSES Introduction

In Section 14.3 were discussed the processing and sedimentation where particles are separated from a fluid by gravitational forces acting on the particles. The particles were solid, gas, or liquid and the fluid was a liquid or a gas. In the present section we discuss settling or separation of particles from a fluid by centrifugal forces acting on the particles. Use of centrifuges increases the forces on particles manyfold. Hence, particles that will not settle readily or at all in gravity settlers can often be separated from fluids by centrifugal force. The high settling force means that practical rates of settling can be obtained with much smaller particles than in gravity settlers. These high centrifugal 1.

Centrifugal settling or sedimentation.

methods of

settling

feed

free

thickened sludge underflow

FIGURE

828

14.3-8.

Continuous thickener.

Chap. 14

Mechanical-Physical Separation Processes

do not change

the relative settling velocities of small particles, but these forces

do overcome the disturbing effects of Brownian motion and free convection currents. Sometimes gravity separation may be too slow because of the closeness of the densities of the particle and the fluid, or because of association forces holding the components together, as in emulsions. An example in the dairy industry is the separation of cream from whole milk, giving skim milk. Gravity separation takes hours, while centrifugal separation is accomplished in minutes in a cream separator. Centrifugal settling or separation is employed in many food industries, such as breweries, vegetable oil processing, fish protein concentrate processing, fruit juice processing to remove cellular materials, and so on. Centrifugal separation is also used in drying crystals and for separating emulsions into their constituent liquids or solid-liquid. The principles of centrifugal sedimentation are discussed in Sections 14. 4B and 14. 4C. forces

2.

Centrifugal filtration.

centrifugal force

is

Centrifuges are also used in centrifugal filtration where a

used instead of a pressure difference to cause the flow of slurry

in

a

where a cake of solids builds up on a screen. The cake of granular solids from the slurry is deposited on a filter medium held in a rotating basket, washed, and then spun "dry." Centrifuges and ordinary filters are competitive in most solid-liquid separation problems. The principles of centrifugal filtration are discussed in Section 14. 4E. filter

14. 4B

1.

Forces Developed

in

Centrifugal Separation

make

Centrifugal separators

Introduction.

use of the

common

principle that

an

object whirled about an axis or center point at a constant radial distance from the point is

acted on by a force. The object being whirled about an axis

direction

and

is

centripetal force acts in a direction If

exert

the object being rotated

an equal and opposite

container. This

is

constantly changing

thus accelerating, even though the rotational speed

is

is

is

constant. This

toward the center of rotation.

a cylindrical container, the contents of fluid

force, called centrifugal force,

outward

and

solids

to the walls of the

the force that causes settling or sedimentation of particles through a

layer of liquid or filtration of a liquid through a bed of filter cake held inside a perforated

rotating chamber. In Fig. 14.4-la a cylindrical

bowl

is

shown

rotating with a slurry feed of solid

and liquid being admitted at the center. The feed enters and is immediately thrown outward to the walls of the container, as in Fig. 14.4- 1 b. The liquid and solids are now acted upon by the vertical gravitational force and the horizontal centrifugal force. The centrifugal force is usually so large that the force of gravity may be neglected. The liquid layer then assumes the equilibrium position with the surface almost vertical. The particles settle horizontally outward and press against the vertical bowl wall. In Fig. 14.4-lc two liquids having different densities are being separated by the centrifuge. The more dense fluid will occupy the outer periphery, since the centrifugal force is greater on the more dense fluid. particles

2.

Equations for centrifugal force.

gal force

In circular

motion the acceleration from

ae

=

rco

2

2 where a e is the acceleration from a centrifugal force in m/s 2 (ft/s ), from center of rotation in m (ft), and co is angular velocity in rad/s.

Sec. 14.4

the centrifu-

is

Centrifugal Separation Processes

(14.4-1)

r is

radial distance

829

,

slurry feed

slurry feed

liquid-liquid feed

I

m

II

i /

c

-

h

D

liquid

(a)

heavy

solids

(b)

<

\

U

light

liquid

liquid

fraction

fraction (c)

FIGURE

The

Sketch of centrifugal separation: (a) initial slurry feed entering, settling of solids from a liquid, (c) separation of two liquid fractions.

14.4-1.

centrifugal force

Fc

in

N (lb

f)

acting on the particle

Fc = ma e =

mrco

1

is

(b)

given by

(SI)

(14.4-2)

mrco (English) 9c

where g

c

=

32.174 lb m

Since co

=

v/r,

•

ft/lb f

where

2 •

v is

s

.

the tangential velocity of the particle in

m/s

(ft/s),

(14.4-3)

Often rotational speeds are given as

N

rev/min and

co

2nN

=

(14.4-4)

60

N

(14.4-5)

2nr Substituting Eq. (14.4-4) into Eq. (14.4-2),

F newton = mr

2nN

=

'

0.01097mr/V 2

(SI)

60 (14.4-6)

F

c

lb f

=

mr f2nN\ 2

9c

By Eq.

=

0.00034 ImrN 2

(14.3-2), the gravitational force

on

a particle

F = mg where g

is

the acceleration of gravity and

the centrifugal force

830

is

(English)

\ 60

is

is

(14.3-2)

9.80665 m/s

2 .

In terms of gravitational force,

as follows by combining Eqs. (14.3-2), (14.4-2), and (14.4-3).

Chap. 14

Mechanical-Physical Separation Processes

F = rw —

2

Fg

2

[2izN\ 2

= — = -{——) = v

c

r

9

0.001 11 8rN

2

(SI)

g \ 60 J

rg

— = 0.000341WV

(14.4-7)

2

(English)

F9

Hence, the force developed force.

This

is

often expressed as equivalent to so

EXAMPLE 14.4-1. A

2

2

or v /rg times as large as the gravity

in a centrifuge is ra> /g

Force

many g

forces.

in a Centrifuge

centrifuge having a radius of the bowl of 0.1016

N=

m (0.333

ft)

is

rotating at

lOOOrev/min. in terms of gravity forces. a bowl with a radius of 0.2032

(a)

Calculate the centrifugal force developed

(b)

Compare

rotating at the

For part

Solution:

m

this force to that for

same rev/min.

m and N

= 0.1016

(a), r

=

1000. Substituting into Eq.

(14.4-7),

^ = 0.001118WV =

2

=

0.001 11 8(0. 101 6X1 0OO)

2

(SI)

113.6 gravities or g's

= 0.00O341(0.333X10O0) ^ t

2

=

1

(English)

13.6

9

For part

(b), r

=

0.2032 m. Substituting into Eq. (14.4-7),

F -~ = 0.001

F9

14. 4C

2

8(0.2032X1000)

Equations for Rates of Settling

General equation for

/.

11

If a

settling.

in

=

227.2 gravities or g's

Centrifuges

centrifuge

used for sedimentation (removal of

is

removed from

particles by settling), a particle of a given size can be if

sufficient residence time of the particle in the

For a

the wall.

particle

moving

bowl

is

the liquid in the

bowl

available for the particle to reach

radially at its terminal settling velocity, the diameter

of the smallest particle removed can be calculated. In Fig. 14.4-2 a schematic of a tubular-bowl centrifuge the

bottom and

it is

assumed

all

is shown. The feed enters at moves upward at a uniform velocity, carrying assumed to be moving radially at its terminal

the liquid

The particle is The trajectory or path of the particle is shown in Fig. 14.4-2. A particle of a given size is removed from the liquid if sufficient residence time is available for the particle to reach the wall of the bowl, where it is held. The length of the bowl is b

solid particles with

it.

settling velocity v,.

m.

At

the

distance the fluid.

rB

end of the residence time of the particle

m

from the axis of rotation.

If r B

=

r2

,

it is

If r B

<

r2

,

in the fluid, the particle is at

a

then the particle leaves the bowl with

deposited on the wall of the bowl and effectively removed from

the liquid.

For

settling in the Stokes'

law range, the terminal settling velocity

at

a radius

r is

obtained by substituting Eq. (14.4-1) for the acceleration g into Eq. (14.3-9).

= »'rPfr,

p)

(14 4 . 8)

18/i

Sec. 14.4

Centrifugal Separation Processes

831

liquid

discharge

liquid

surface

particle

trajectory

feed flow-

Figure

where

v

is t

14.4-2.

Particle settling in segmenting tubular-bowl centrifuge.

Dp

settling velocity in the radial direction in m/s,

kg/m 3 p

particle density in

If

hindered settling occurs, the right-hand side of Eq. (14.4-8)

(e

2 i/>

is

,

particle diameter in

is

kg/m 3 and

liquid density in

is

p. is

,

liquid viscosity in

m,p p

Pa

-

s.

multiplied by the factor

is

given in Eq. (14.3-16).

„)

Since

becomes

drldt, then Eq. (14.4-8)

v,

=

dt

Integrating between the limits

=

r

18^

dr

w 2 (p p -p)D p2

r

r, at

=

t

0 and

r

=

(14.4-9)

r2

at

t

=

t

r

.

18/j 2

co (p p

The

residence time

t

T

is

and solving for

m

3

The volume V =

/s.

V m3

2 p)D p/r „

=

(K)

co

2

(p p

-p)Dl

Particles having diameters smaller than that calculated

reach the wall of the bowl and

A

-j

layer pr

the distance between r x

(r 2

-

then between

=

(r l

+

18/i In

Substituting into Eq.

r 2 )/2

at

t

r2

.

1)

will

will

not

reach

can be defined as the diameter of a particle that This particle moves a distance of half the liquid

=

0 and

r

=

2

+

(14.4-1

go out with the exit liquid. Larger particles

and

r 2 at 2

P)D PC

[2r 2 /(r,

from Eq.

(14.4-11)

liquid.

during the time this particle

r,)/2

r

u 2 (p p

832

will

and be removed from the

cut point or critical diameter

reaches

the bowl divided by the

r\).

[#! - rf)]

18;i In (r 2 /r.)

lift In (i-j/r,)

the wall

in

—

nb{r\

q,

= <Ap,

q

(14.4-10)

In

p)D

equal to the volume of liquid

feed volumetric flow rate q in

(14.4-10)

-

£

co (p p

(V) r 2 )J

is

18/i In

Chap. 14

The Then we obtain

in the centrifuge.

=

t

T

.

- p)D

[2r 2 /(r,

pc

+

[jrf>(ri

integration

- rf)]

is

(14.4-12)

r 2 )]

Mechanical-Physical Separation Processes

At

q c particles with a diameter greater than Dpc and most smaller particles will remain in the liquid.

this flow rate

to the wall

,

will

predominantly

settle

For the special case where the thickness of the liquid layer is Special case for settling. small compared to the radius, Eq. (14.4-8) can be written for a constant r = r 2 and

2.

Dp =

as follows:

^Bfrr-p)

(14413)

lap

The time

of settling

t T is

then as follows for the critical

tT

case.

—

-V = — - r,)/2 (r 2

=

(14.4-14)

Substituting Eq. (14.4-13) into (14.4-14) and rearranging,

1c

The volume V can be expressed

*r<

18^[(r 2

-

(14.4-15)

and

(14-4-15)

r,)/2]

as

V = 2nr 2 (r 2 Combining Eqs.

™

P

=

—

(14.4-16)

r x )b

(14.4-16),

9 p. analysis above is somewhat simplified. The pattern of flow of the fluid is more complicated. These equations can also be used for liquid-liquid systems where droplets of liquid migrate according to the equations and coalesce in the other

The

actually

liquid phase.

EXAMPLE

14.4-2. Settling in a Centrifuge J viscous solution containing particles with a density p p =1461 kg/m is to 3 and its be clarified by centrifugation. The solution density p = 801 kg/m viscosity is 100 cp. The centrifuge has a bowl with r 2 = 0.02225 m, r =

A

,

t

0.00716 m, and height b = 0.1970 m. Calculate the critical particle diameter of the largest particles in the exit stream if JV - 23 000 rev/min 3 and the flow rate q = 0.002832 /h.

m

Solution:

Using Eq.

(14.4-4),

2nN -^_i= = c=— 2tt(23 000)

..

2 410 rad/s

The bowl volume V is

V=

=

nb{r\

r\)

7t(0.1970)[(0.02225)

Viscosity qc

-

p.

=

3 100 x 10"

=

-

0.100

Pa

s

4 2.747 x 10"

2

=

=

0.100 kg/m

(0.00716) ]

•

s.

m3

The flow

rate

is

9r

Sec. 14.4

2

=

0.002832

3600

7.87 = nnn

Centrifugal Separation Processes

,

x 10

7

m 33(/s 833

Substituting into Eq. (14.4-12) and solving for

=

qc

7.87 x 10"

-

2

4 801)0^(2.747 x 1(T )

18(0.100) In [2 x 0.02225/(0.00716

m

6 0.746 x 10"

=

,

7

(2410) (1461

~

D pc

or

+

0.02225)]

0.746 p.m

Substituting into Eq. (14.4-13) to obtain v, and then calculating the Reynolds number, the settling is in the Stokes' law range.

Sigma values and scale-up of centrifuges. A useful physical characteristic of a tubularbowl centrifuge can be derived by multiplying and dividing Eq. (14.4-12) by 2g and

3.

then substituting Eq. (14.3-9) written for (p,

U = where

v, is

D pc

- totiU

Eq. (14.4-12)

<*v

18/i

= + r 2 )]

2g In I7rj(r x

to obtain

z

2

(14 . 4 . 18)

the terminal settling velocity of the particle in a gravitational field

2 = is

—Li

= 2g In [2r 2 /(r,

where Z

into

+

—

Ly

1

2g

r 2 )]

In [2r 2 /(r

a physical characteristic of the centrifuge

and

+

1

Q4

4-19)

r 2 )]

and not of the

fluid-particle system

being separated. Using Eq. (14-4-17) for the special case for settling for a thin layer,

Z =

CLp-nb2r\

(14.4-20)

9

The value of Z same sedimentation from a laboratory

m

really the area in

is

2

of a gravitational settler that will have the

characteristics as the centrifuge for the

andZ tog 2

test of q l

(

(for v n

=

same

feed rate.

To

up

v, 2 ),

= Z

scale

(14.4-21)

T

type and geometry centrifuges and if from each other. If different configurations are used, efficiency factors E should be used where q fL E = q 2J~L 1 E 1 These efficiencies are determined experimentally and values for different types of centrifuges are

This scale-up procedure

dependable

is

for similar

the centrifugal forces are within a factor of 2

.

l

l

l

given elsewhere (Fl, PI).

4.

Separation of liquids

in

Liquid-liquid separations in which the liquids

a centrifuge.

are immiscible but finely dispersed as an emulsion are

and other

industries.

An example

is

common

operations in the food

the dairy industry, in which the emulsion of milk

is

separated into skim milk and cream. In these liquid-liquid separations, the position of the outlet overflow weir in the centrifuge

volumetric holdup

V

in the centrifuge

is

quite important, not only

in

controlling the

but also in determining whether a separation

is

actually made. In Fig. 14.4-3

rating

two

light liquid

a tubular-bowl centrifuge is shown in which the centrifuge is sepaone a heavy liquid with density p H kg/m 3 and the second a

liquid phases,

with density p L The distances shown are as follows :r is radius to surface is radius to liquid-liquid interface, and r 4 is radius to surface of .

{

of light liquid layer, r 2

heavy

liquid

To

834

downstream.

locate the interface, a balance

must be made of the pressures

Chap. 14

in

the

two

layers.

Mechanical-Physical Separation Processes

The

force on the fluid at distance

by Eq.

r is,

Fc = The

differentia] force across a thickness dr

(14.4-2),

mrco

2

(14.4-2)

is

dFc = dmrco 2

(14.4-22)

But,

dm =

(14.4-23)

[(2nrb) dr]p

where b is the height of the bowl in m and (2nrb) dr is the volume of fluid. Substituting Eq. (14.4-23) in (14.4-22) and dividing both sides by the area A = lirrb,

dP= where

P is pressure in N/m 2

(lb f /ft

—= dFc A

w 2 pr

dr

(14.4-24)

2 ).

Integrating Eq. (14.4-24) between

r,

and

r2

,

(14.4-25)

Applying Eq. (14.4-25) to Fig. 14.4-3 and equating the pressure exerted by the phase of thickness

r2

— r,

to the pressure exerted

at the liquid-liquid interface at r 2

by the heavy phase of thickness r 2

(i r\

,

rA

-

(14.4-26)

r\)

the interface position, r2

Ph

The

—

,

p„co

Solving for

light

interface at r 2

must be located

at

-

(14.4-27)

Pl

a radius smaller than r 3 in Fig. 14.4-3.

outlets

heavy

liquid,

light liquid,

Ph

pL

liquid— liquid interface

feed

Figure

Sec. 14.4

14.4-3.

Tubular bowl centrifuge for separating two

Centrifugal Separation Processes

liquid phases.

835

EXAMPLE 14.4-3

Location of Interface

.

In a vegetable-oil-refining process, an

in

Centrifuge

aqueous phase

is being separated phase in a centrifuge. The density of the oil is 919.5 kg/m 3 and 3 The radius r x for overflow of the that of the aqueous phase is 980.3 kg/m and the outlet for the heavy liquid at light liquid has been set at 10.160

from the

oil

.

mm

10.414

mm.

Calculate the location of the interface in the centrifuge.

Solution: The densities are p L = 919.5 and into Eq. (14.4-27) and solving for r 2 , 2

980.3(10.414)

"2

~

r2

=

2

980.3 13.75

-

= 980.3 kg/m 3

p tl

919.5(10.160)

.

Substituting

2

919.5

mm

Centrifuge Equipment for Sedimentation

13.4D

A schematic of a tubular bowl centrifuge is shown in Fig. 14.4-3. and has a narrow diameter, 100 to 150 mm. Such centrifuges, known as super centrifuges, develop a force about 13 000 times the force of gravity. Some narrow centrifuges having a diameter of 75 mm and very high speeds of 60000 or so rev/min are known as ultracentrifuges. These supercentrifuges are often used to separate liquid-liquid 1.

Tubular centrifuge.

The bowl

is tall

emulsions.

2.

Disk bowl centrifuge.

liquid-liquid separations. travels

The

upward through

The The

disk

bowl centrifuge shown in Fig. 14.4-4 is often used in compartment at the bottom and

feed enters the actual

vertically

spaced feed holes,

filling

the spaces between the disks.

holes divide the vertical assembly into an inner section, where mostly light liquid

present,

and an outer

section,

where mainly heavy liquid

is

present. This dividing line

is

is

similar to an interface in a tubular centrifuge.

The heavy The

liquid flows beneath the underside of a disk to the periphery of the bowl.

light liquid flows over the

Figure

836

upper side of the disks and toward the inner

14.4-4.

outlet.

Any

Schematic of disk bowl centrifuge.

Chap. 14

Mechanical-Physical Separation Processes

amount of heavy solids is thrown to the outer wall. Periodic cleaning is required to remove solids deposited. Disk bowl centrifuges are used in starch-gluten separation, concentration of rubber latex, and cream separation. Details are given elsewhere (PI, LI). small

14.4E

1

.

Centrifugal FUtration

Theory for

centrifugal filtration.

Theoretical prediction of filtration rates

The

gal filters have not been too successful.

filtration in centrifuges is

in centrifu-

more complicated

than for ordinary filtration using pressure differences, since the area for flow and driving

and the

force increase with distance from the axis

markedly. Centrifuges for

specific

filtering are generally selected

cake resistance may change by scale-up from tests on a

similar-type laboratory centrifuge using the slurry to be processed.

The theory of constant-pressure fied

filtration

discussed

in

Section

14.

2E can

be modi-

and used where the centrifugal force causes the flow instead of the impressed pressure

difference.

The equation will be derived for the case where a cake has already been shown in Fig. 14.4-5. The inside radius of the basket is r 2 r is the inner

deposited as

,

radius of the face of the cake, and

assume that the cake

is

;

the inner radius of the liquid surface.

is

We

will

nearly incompressible so that an average value of a can be used

for the cake. Also, the flow

centrifuge, then the area

r^

A

laminar. If

is

for

flow

we assume

a thin

cake

in a

large-diameter

approximately constant. The velocity of the liquid

is

is

=

v

where q

is

the filtrate flow rate in

m

3

dV

q

l = TZt

(14 - 4 " 28)

and v the velocity. Substituting Eq. (14.4-28)

/s

into (14.2-8),

_A = J7

where

m = c

c s V,

mass of cake

in

g /i|-f r -+-=]

kg deposited on the

Fora hydraulic head ofdzm.the pressure drop dp In a centrifugal field, g

is

replaced by rco

dp

Sec. 14.4

=

=

2

filter. is

pg dz

from Eq. (14.4-1) and dz by

prcv

Centrifugal Separation Processes

(14.4-29)

2

dr

(14.4-30) dr.

Then, (14.4-31)

837

Integrating between

r

t

and

r2

>

-Ap =

pu 2 {r\ -

r\)

(14.4-32) 2

Combining Eqs.

(14.4-29)

and (14.4-32) and solving for q,

<J

For the case where the flow area A

=

P*>\A -

r\)

(14.4-33)

varies considerably with the radius, the following has

been derived (Gl). paj

2

2 (r 2

-

r\)

(14.4-34)

where A 2 = 2nr 2 b (area of filter medium), A L = 2nb(r 1 — r )/ln (r 2 /r ) (logarithmic cake area), and A a = (r,. + r 2 )nb (arithmetic mean cake area). This equation holds for a cake of a given mass at a given time. It is not an integrated equation covering the whole i

;

filtration cycle.

Equipment for centrifugal filtration.

2.

to a rotating basket

cake builds up on the surface of the the end of the filtration cycle, feed

Then

the cake.

motor

is

the

In a centrifugal filter, slurry is fed

which has a perforated wall and

wash

liquid

is

is

filter

medium

is

covered with a

to the desired thickness.

is

is added or sprayed on to spun as dry as possible. The

then shut off or slowed and the basked allowed to rotate while the solids are

Finally, the filtyer

medium is rinsed

in the

basket

m

in

floor.

clean to complete the cycle. Usually, the batch cycle

completely automated. Automatic batch centrifugals have basket sizes up

1.2

The Then at

stopped, and wash liquid

stopped and the cake

discharged by a scraper knife so the solids drop through an opening

is

continuously

filter cloth.

to

about

diameter and usually rotate below 4000 rpm.

up to about 25 000 kg on the filter medium is removed by being pushed toward the discharge end by a pusher, which then retreats again, allowing the cake to build up once more. As the cake is being pushed, it passes through a wash region. The filtrate and wash liquid are kept separate by partitions in the collector. Details of Continuous centrifugal

filters

are available with capacities

solids/h. Intermittently, the cake deposited

different types of centrifugal filters are available (PI).

14. 4F

/.

Gas-Solid Cyclone Separators

For separation of small solid particles or mist from most widely used type of equipment is the cyclone separator, shown in Fig.

Introduction and equipment.

gases, the

14.4-6.

The cyclone

consists of a vertical cylinder with a conical bottom.

The

gas-solid

particle mixture enters in a tangential inlet near the top. This gas-solid mixture enters in

and thewortex formed develops centrifugal force which throws the toward the wall.

a rotating motion, particles radially

On entering, the wall.

When

the air in the cyclone flows

downward

in

the air reaches near the bottom of the cone,

a spiral or vortex adjacent to it

spirals

upward

cone and cylinder. Hence, a double vortex downward and upward spirals are in the same direction.

spiral in the center of the

838

Chap. 14

is

in a smaller

present.

The

Mechanical-Physical Separation Processes

fall downward, leaving out the bottom which the outward force on the particles at high tangential velocities is many times the force of gravity. Hence, cyclones accomplish "" much more effective separation than gravity settling chambers. The centrifugal force in a cyclone ranges from about 5 times gravity in large, low-velocity units to 2500 times gravity in small, high-resistance units. These devices are used often in many applications, such as in spray drying of foods, where the dried particles are removed by cyclones; in cleaning dust-laden air; and in removing mist droplets from gases. Cyclones offer one of the least expensive means of gas-particle separation. They are generally applicable in removing particles over 5 /im in diameter from gases. For particles over 200 psn in size, gravity settling chambers are often used. Wet scrubber cyclones are sometimes used where water is sprayed inside, helping to

The particles

of the cone.

remove 2.

are thrown toward the wall and

A cyclone is a

settling device in

the solids.

Theory for cyclone separators.

It is

assumed

that particles

on entering a cyclone

quickly reach their terminal settling velocities. Particle sizes are usually so small that Stokes' law is considered valid. For centrifugal motion, the terminal radial velocity v tR is

given by Eq. (14.4-8), with v lR being used for v, co

2

rD 2p (p p

-

p)

(14.4-35)

Ify

Since

cu

(14.4-35)

=

v lan /r,

where

v an is tangential velocity of the particle at radius r, Eq. ,

becomes

D P2 g(p P -

p) vi

(14.4-36)

v,

18/;

where

gr

gr

v, is the gravitational terminal settling velocity v, in

Eq.

(14.3-9).

gas out t

gas-solids in

K

—

f>~

}-f-'

I

^ 1-

C

I

i

i

4 -'

(b)

solids out (a)

FIGURE

Sec. 14.4

14.4-6.

Gas-solid cyclone separator :

Centrifugal Separation Processes

(a) side view, (b)

top view.

839

The higher it

the terminal velocity

v,

,

the greater the radial velocity

v, R

and the

easier

should be to "settle" the particle at the walls. However, the evaluation of the radial difficult, since

a function of gravitational terminal velocity, tangential

velocity

is

velocity,

and position radially and

equation

is

is

it

axially in the cyclone. Hence, the following empirical

often used (S2).

,R

= Mfo-p)

(14-4 . 3

iv

where b and n are empirical constants. l

Smaller particles have smaller settling velocities Efficiency of collection of cyclones. by Eq. (14.4-37) and do not have time to reach the wall to be collected. Hence, they leave with the exit air in a cyclone. Larger particles are more readily collected. The efficiency of

3.

separation for a given particle diameter

is

defined as the mass fraction of the size particles

that are collected.

A

typical collection efficiency plot for a cyclone

rapidly with particle

mass of

size.

cut diameter

the entering particles

MECHANICAL

14.5

The

14. 5A

Introduction

Many

solid materials

retained.

in sizes that

Often the solids are reduced

shows that the efficiency rises which one half of the

the diameter for

REDUCTION

SIZE

occur

is

is

are too large to be used

in size so that the

carried out. In general, the terms crushing

and must be reduced.

separation of various ingredients can be

and grinding are used to

signify the subdividing

of large solid particles to smaller particles. In the food processing industry, a large size reduction. Roller mills are

number of food products

are subjected to

used to -grind wheat and rye to flour and corn. Soybeans

and ground to produce oil and flour. Hammer mills are often used to and other flours. Sugar is ground to a finer product. Grinding operations are very extensive in the ore processing and cement industries. Examples are copper ores, nickel and cobalt ores, and iron ores being ground before chemical processing. Limestone, marble, gypsum, and dolomite are ground to use as fillers in paper, paint, and rubber. Raw materials for the cement industry, such as lime, alumina, and silica, are ground on a very large scale. are rolled, pressed,

produce potato

Solids

flour, tapioca,

may

be reduced

in size

by a number of methods; Compression or crushing

generally used for reduction of hard solids to coarse sizes. Impact gives coarse, or fine sizes. Attrition or rubbing yields fine products. Cutting

14. 5B

The

Particle-Size

opening

One common way

in screen) in

(Openings

and the product are defined

mm

to plot particle sizes

arithmetic probability paper is

made,

is

in

terms of the particle-

to plot particle diameter (sieve

Appendix

A. 5.)

size.

Such a plot was given on

in Fig. 12.12-2.

instead, as the cumulative

the stated size versus particle size as

840

sizes.

or fim versus the cumulative percent retained at that

for various screen sizes are given in

Often the plot

used to give definite

Measurement

feed-to-size reduction processes

size distribution.

is

is

medium,

shown

Chap. 14

amount

in Fig. 14.5-la.

as percent smaller than

In Fig. 14.5-lb the

same data

Mechanical-Physical Separation Processes

The ordinate is obtained by taking the slopes and converting to percent by weight per fim. necessary for most comparisons and calculations.

are plotted as a particle distribution curve.

of the 5-fim intervals of Fig. 14.5-la

Complete

is

Energy and Power Required

14. 5C

1.

particle-size analysis

Introduction.

strain the particles is

The

materials are fractured.

The

particles of feed are

and strained by the action of the size-reduction machine. This work to

distorted

force

Reduction

In size reduction of solids, feed materials of solid are reduced to a

smaller size by mechanical action. first

in Size

added

is first

stored temporarily in the solid as strain energy.

to the stressed particles, the strain

As additional

energy exceeds a certain

level,

and the

material fractures into smaller pieces.

When

the material fractures,

surface requires a certain

the is

new

amount

new

surface area

of energy.

surface, but a large portion of

it

Some

is

created.

Each new

of the energy added

is

unit area of

used to create

appears as heat. The energy required for fracture

a complicated function of the type of material, size, hardness, and other factors.

The magnitude

of the mechanical force applied; the duration

;

the type of force, such

and impact; and other factors affect the extent and efficiency of the size-reduction process. The important factors in the size-reduction process are the amount of energy or power used and the particle size and new surface formed. as compression, shear,

Power required in size reduction. The various theories or laws proposed for prepower requirements for size reduction of solids do not apply well in practice. The most important ones will be discussed briefly. Part of the problem in the theories is that of estimating the theoretical amount of energy required to fracture and create new surface area. Approximate calculations give actual efficiencies of about 0.1 to 2%. The theories derived depend upon the assumption that the energy E required to produce a change dX in a particle of size X is a power function of X.

2.

dicting

6

Figure

Particle size (/im)

Particle size iptm)

(a)

(b)

14.5-1

.

Particle-size-distribution curves size, [b)

: (a) cumulative percent versus particle percent by weight versus particle size. [From R. II Perry and

C. H. Chilton, Chemical Engineers'

Handbook, 5th ed. McGraw-Hill Book Company, 1973. With permission)

Sec. 14.5

Mechanical Size Reduction

.

New York:

841

C

dE

(14 - 5 - 1}

!x=-T* X

in mm, and n and C are constants depending upon and type machine. type and size of material of Rittinger proposed a law which states that the work in crushing is proportional to

where

new

the

size or

is

diameter of particle

=

surface created. This leads to a value of n

2

Eq. (14.5-1), since area

in

is

proportional to length squared. Integrating Eq. (14.5-1),

E--£-(-L. ' -±-\ n-[\X"

X

where

is

1

X2

mean diameter of feed and we obtain

(14.5-2)

X\~ J

x [

2

mean diameter

is

of product. Since n

=

2 for

Rittinger's equation,

E= K »(l2 E

where

is

work

to reduce a'unit

mass

to 50

mm as

iromX

of feed

law implies that the same amount of energy

is

(14 - 5 - 3)

-i) loX 2 andK^

a constant.

The

needed to reduce a material from 100

mm

l

needed to reduce the same material from 50

is

mm

is

to 33.3

mm.

has been

It

this law has some validity in grinding fine powders. Kick assumed that the energy required to reduce a material in size was directly proportional to the size-reduction ratio. This implies n = 1 in Eq. (14.5-1), giving

found experimentally that

E = C where

KK

is

£i = K K

(14.5-4)

log

same amount of energy is required to needed to reduce the same material from

a constant. This law implies that the

reduce a material from 100 50

In

mm to

50

mm as

is

mm to 25 mm. Recent data by Bond (B3) on correlating extensive experimental data suggest that

the

work

required using a large-size feed

is

proportional to the square root of the

=

surface/volume ratio of the product. This corresponds to n

KB

1.5 in

Eq.

(14.5-1), giving

-L

(14.5-5)

Fx-,2

where in

kW

K B is •

a constant.

To

Bond proposed

use Eq. (14.4-5),

a

work index £

h/ton required to reduce a unit weight from a very large size to

100-/jm screen. feed with

80%

Bond's

Then

the

work E

passing a diameter

final

is

the gross

work required

to

;

as the

80%

work

passing a

reduce a unit weight of

X F fim to a product with 80% passing X F pm.

equation in terms of English units

£=

1

.46£,

(

is

-)= -

-U

(14.5-6) J

where P is hp, T is feed rate in tons/min,£> f is size of feed in ft, andD,, is product size in ft. Typical values of £ for various types of materials are given in Perry and Green (PI) and ;

by Bond

(B3).

shale (16.4),

Some

typical values are bauxite (£

and granite

(14.39).

;

=

9.45),

coal

(1 1.37),

potash

These values should be multiplied by

salt (8.23),

1.34 for

dry

grinding.

EXAMPLE 14.5-1. It is

842

Power

to

Crush Iron Ore by Bond's Theory

desired to crush 10 ton/h of iron ore hematite.

Chap. 14

The

size of the feed

is

Mechanical-Physical Separation Processes

such that 80% passes a 3-in. (76.2-mm) screen and 80% of the product is to pass a j-in. (3.175-mm) screen. Calculate the gross power required. Use a work index E-, for iron ore hematite of 12.68 (PI).

= 0.250 ft (76.2 mm) and the product is D F = 0.0104 ft (3.175 mm). The feed rate is T = 10/60 = 0.167 ton/min. Substituting into Eq. (14.5-6) and solving for P, The

Solution: size is

DP =

feed size

£/12

=

P 14. 5D

1.

Equipment

way

a colloid

(17.96

kW)

Size-reduction equipment

may

be classified accord-

between two surfaces, as in crushing and impact; and by action of the surrounding medium, as

the forces are applied as follows:

shearing; at one solid surface, as in in

hp

for Size Reduction

Introduction and classification.

ing to the

24.1

mill.

A

more practical classification and cutters.

to divide the

is

equipment into crushers,

grinders, fine grinders,

2.

Jaw

crushers.

Equipment

for

coarse reduction of large amounts of solids consists of

slow-speed machines called crushers. Several types are in

common

use. In the first type, a

jaw crusher, the material is fed between two heavy jaws or flat plates. As shown in the Dodge crusher in Fig. 14.5-2a, one jaw is fixed and the other reciprocating and movable on a pivot point at the bottom. The jaw swings back and forth, pivoting at the bottom of the V. The material is gradually worked down into a narrower space, being crushed as it moves.

The Blake

is more commonly used, and the pivot point is at The reduction ratios average about 8 1 in the Blake crusher. used mainly for primary crushing of hard materials and are usually

crusher in Fig. 14.5-2b

the top of the movable jaw.

Jaw

crushers are

:

followed by other types of crushers.

feed

feed pivot

fixed

movable jaw

jaw

point

fine

pivot

fixed

product

point

jaw (b)

(a)

Figure

Sec. 14.5

14.5-2.

movable jaw

Types ofjaw crushers

Mechanical Size Reduction

:

(a)

Dodge

type, (b) Blake type.

843

The gyratory crusher shown in Fig. 14.5-3a has taken over to a and mineral crushing applications. Basically it is like a mortar-and-pestle crusher. The movable crushing head is shaped like an inverted truncated cone and is inside a truncated cone casing. The crushing head rotates eccentrically and the material being crushed is trapped between the outer fixed cone and the

3.

Gyratory crushers.

large extent in the field of large hard-ore

inner gyrating cone.

4.

In Fig. 14.5-3b a typical

Roll crushers.

rotated toward each other at the

same

smooth

roll

problem. The reduction ratio varies from about 4 used, rotating against a fixed surface,

Many

:

is

shown. The

of the rolls

is

rolls are

a serious

to 2.5:1. Single rolls are often

1

and corrugated and toothed

rolls

are also used.

rolls.

Hammer

Hammer

mill grinders.

mill devices are

used to reduce intermediate-sized

material to small sizes or powder. Often the product from the feed to the

hammer

cylindrical casing. Sets of

The

Wear

food products that are not hard materials, such as flour, soybeans, and starch, are

ground on

5.

crusher

or different speeds.

mill. In

hammers

the

hammer

jaw and gyratory crushers

is

mill a high-speed rotor turns inside a

are attched to pivot points at the outside of the rotor.

and the particles are broken as they fall through the broken by the impact of the hammers and pulverized into powder between the hammers and casing. The powder then passes through a grate or feed enters the top of the casing

cylinder.

The

material

is

screen at the discharge end.

6.

Revolving grinding

For intermediate and

mills.

fine reduction of materials revolving

grinding mills are often used. In such mills a cylindrical or conical shell rotating on a

horizontal axis

is

charged with a grinding

or with steel rods.

The

size

reduction

is

medium such

effected

as steel,

flint,

or porcelain balls,

by the tumbling of the

on up the

balls or rods

the material between them. In the revolving mill the grinding elements are carried

and fall on the particles underneath. These mills may operate wet or dry. Equipment for very fine grinding is very specialized. In some cases two flat disks are used where one or both disks rotate and grind the material caught between the disks (PI).

side of the shell

feed

(a)

FIGURE

14.5-3.

Types of size-reduction equipment

:

(a)

gyratory crusher,

(£>)

roll

crusher.

844

Chap. 14

Mechanical-Physical Separation Processes

PROBLEMS 14.2-1 Constant-Pressure Filtration

CaC0 3

a constant pressure

— Ap)

(

frame press was 0.0439 solid/m s

and Filtration Constants.

slurry in water at 298.2

3

m2

K (25°C) are reported kN/m 2

of 46.2

2

(0.473

ft

m

V x 70 3

V x

i

for the filtration of

The area

Ml)

at

of the plate-and-

and the slurry concentration was 23.47 kg R m Data are given as = time in

)

Calculate the constants a and volume of filtrate collected in 3

filtrate.

and V =

(6.70 psia).

Data

as follows (Rl, R2,

f

.

.

10*

V

i

x

W

1 f

0.5

17.3

1.5

72.0

2.5

152.0

1.0

41.3

2.0

108.3

3.0

201.7

Mb

a = 1.106 x 10 11 m/kg(1.65 x 10" m ), -1 R m = 6.40 x 10 10 m~' (1.95 x 10 10 ft )

Am.

Constants for Constant-Pressure Filtration. Data for constant2 pressure filtration at 194.4 kN/m are reported for the same slurry and press as 3 in Problem 14.2-1 as follows, where / is in s and V in

14.2-2. Filtration

m

V x 10 3

V x

r

10

}

:

V x

r

W

3 t

0.5

6.3

2.5

51.7

4.5

134.0

1.0

14.0

3.0

69.0

5.0

160.0

1.5

24.2

3.5

88.8

2.0

37.0

4.0

110.0

Calculate the constants a and

Rm

.

=

x 10" m/kg cake resistance a from

Ans. a

1.61

of Filter Cake. Use the data for specific and Problems 14.2-1 and 14.2-2 and determine the compressibility constant s in Eq. (14.2-1 1). Plot the In of a versus the In of -Ap and determine the slope s.

14.2-3. Compressibility

Example

14.2-1

14.2-4. Prediction of Filtration Time and Washing Time. The slurry used in Problem 2 14.2-1 is to be filtered in a plate-and-frame press having 30 frames and 0.873

m

2 will be used in constantarea per frame. The same pressure, 46.2 kN/m pressure filtration. Assume the same filter cake properties and filter cloth, and 3 calculate the time to recover 2.26 of filtrate. At the end, using through ,

m

washing and 0.283 filter

cycle time

if

m

3

of

wash water,

calculate the time of washing

filter

34.5

2 press with an area of 0.0929m performed constant-pressure filtration at

kPa

,

of a 13.9 wt

% CaC0

3

w

The mass ratio was 1017 kg/m 3 The

solids in water slurry at 300 K.

of wet cake to dry cake was 1.59. data obtained are as follows, where

The dry cake

W = kg

density

filtrate

and

£

.

=

W

i

W

time

in s:

t

,0.91

24

3.63

244

6.35

1.81

71

4.54

372

7.26

888

2.72

146

5.44

524

8.16

1188

Calculate the values of a and

Chap. 14

total

McMillen and Webber (M2), using

14.2-5. Constants in Constant-Pressure Filtration.

a

and the

cleaning the press takes 30 min.

Problems

690

R

845

14.2-6. Constant-Pressure Filtration filter

and Washing

press having an area of 0.0414

m

2

where (a)

/

is

and V

in s

in

10.25 x 10

m3

6

is

The

slurry at a constant pressure of 267 kPa.

^=

in a

(Rl)

K +

Leaf

used to

filtration

An experimental an aqueous BaC0 3 equation obtained was Filter. filter

x 10 3

3.4

.

and conditions are used in a leaf press having an area of 2 3 6.97 m how long will it take to obtain 1.00m of filtrate? 3 After the filtration the cake is to be washed with 0.100 m of water. If

same

the

slurry

,

(b)

Calculate the time of washing.

Ans. 14.2-7. Constant-Rate Filtration of Incompressible Cake. filtration at a

The

constant pressure of 38.7 psia (266.8 kPa)

£=

6.10 x 10~

5

K

+

(a)

filtration

f

=

381.8

s

equation for

is

0.01

where is in s, —Ap in psia, and V in liters. The specific resistance of the cake is independent of pressure. If the filtration is run at a constant rate of 10 liters/s, t

how

long

will

it

take to reach 50 psia?

14.2-8. Effect of Filter Medium Resistance on Continuous Rotary-Drum Filter. Example 14.2-4 for the continuous rotary-drum vacuum filter but

neglect the constant

Compare with

Rm

results of

,

which

the

is

Example

filter

medium

do not

resistance to flow.

14.2-4.

Ans. 14.2- 9.

Repeat

A =

7.78

m

2

Continuous Rotary Drum Filter. A rotary drum filter having an is to be used to filter the CaC0 3 slurry given in Example 14.2-4. The drum has a 28% submergence and the filter cycle time is 300 s. A pressure 2 drop of 62.0 kN/m is to be used. Calculate the slurry feed rate in kg slurry/s for Throughput

in

area of 2.20

m

2

the following cases. (a)

(b)

Neglect the filter medium resistance. Do not neglect the value of B.

14.3- 1. Settling Velocity

of a Coffee Extract Particle.

Solid spherical particles of coffee

extract (Fl) from a dryer having a diameter of 400 jj.m are falling through air at

temperature of 422 K. The density of the particles is 1030kg/m 3 Calculate the terminal settling velocity and the distance of fall in 5 s. The pressure is 101.32 kPa. Ans. v, = 1.49 m/s, 7.45 m fall a

.

14.3-2. Terminal Settling

Velocity

of Dust Particles.

Calculate the terminal settling Kand 101.32 kPa. The dust particles can be considered spherical with a density of 1280 velocity of dust particles having a diameter of 60 /jm in air at 294.3

kg/m 3

.

Ans. 14.3-3. Settling Velocity

of Liquid Particles.

are settling from

900 kg/m 3 velocity.

number

.

A

still

settling

How

chamber

long will

above about spheres cannot be used.) is

K

air at 294.3

it

is

v,

=

0.1372 m/s

Oil droplets having a diameter of 200

and 101.32 kPa. The density of

0.457

m

^m

the oil

is

high. Calculate the terminal settling

take the particles to settle? (Note : If the Reynolds and form drag correlation for rigid

100, the equations

of Quartz Particles in Water. Solid quartz particles having a pm are settling from water at 294.3 K. The density of the 3 spherical particles is 2650 kg/m Calculate the terminal settling velocity of

14.3-4. Settling Velocity

diameter of 1000

.

these particles.

846

Chap. 14

Problems

14.3-5. Hindered Settling of Solid Particles. Solid spherical particles having a diameter or 0.090 and a solid density of 2002 kg/m 3 are settling in a solution of

mm

water

at 26.7°C.

The volume fraction of the solids and the Reynolds number.

in the

water

is

0.45. Calculate

the settling velocity

of Quartz Particles in Hindered Settling. Particles of quartz having a and a specific gravity of 2.65 are settling in water at diameter of 0.127 293.2 K. The volume fraction of the particles in the slurry mixture of quartz and water is 0.25. Calculate the hindered settling velocity and the Reynolds number.

14.3-6. Settling

mm

14.3-7. Density Effect on Settling Velocity

and Diameter.

Calculate the terminal setdiameter having a density of 2469 kg/m 3 in air at 300 and 101.32 kPa. Also calculate the diameter of a sphalerite sphere having a specific gravity of 4.00 with the same terminal settling velocity.

sphere 0.080

tling velocity of a glass

mm

in

K

of Particles. Repeat Example 14.3-3 for particles having a 2 2 range of 1.27 x 10" to 5.08 x 10~ mm. Calculate the size range of the various fractions obtained using free settling conditions. Also calculate the

14.3-8. Differential Settling

mm

size

value of the largest Reynolds

of 0.075-0.65

(b)

occurring.

is

mixture of galena and silica particles has a size range to be separated by a rising stream of water at 293.2 K.

from Example 14.3-3. To obtain an uncontaminated product of galena, what velocity of water flow is needed and what is the size range of the pure product ? If another liquid, such as benzene, having a specific gravity of 0.85 and a 4 viscosity of 6.50 x 10~ Pa -s is used, what velocity is needed and what is the size range of the pure product?

Use (a)

mm and

number

A

14.3-9. Separation by Settling. specific gravities

14.3-10. Separation by Sink-and-Float Method.

and hematite having a It is

Quartz having a

specific gravity of 5.

desired to separate

1

specific gravity of 2.65

are present in a mixture of particles.

them by a sink-and-float method using a suspension of having a specific gravity of 6.7 in water. At what

fine particles of ferrosilicon

%

ferrosilicon solids consistency in vol tained for the separation?

in

water should the

medium

be main-

and Sedimentation Velocities. A batch settling test on a slurry gave the following results, where the height z in meters between the clear liquid and the suspended solids is given at time hours.

14.3-11. Batch Settling

t

t

The

z(m)

(h)

r(h)

z(m)

r

z (m)

(h)

0

0.360

1.75

0.150

12.0

0.102

0,50

0.285

3.00

0.125

20.0

0.090

1.00

0.211

5.00

0.113

3 is 250 kg/m of slurry. Determine the veloand concentrations and make a plot of velocity versus con-

original slurry concentration

cities of settling

centration. 14.4-1.

Comparison of Forces

Centrifuges.

in

peripheral velocity of 53.34 m/s.

Two

centrifuges rotate at the

The first bowl has

a radius of

r,

=

76.2

same

mm and

the second r 2 = 305 mm. Calculate the rev/min and the centrifugal forces developed in each bowl. Ans. N = 6684 rev/min, N 2 = 1670 rev/min, 3806 g's in bowl 1, 951 g's in bowl 2 i

A

14.4-2. Forces in a Centrifuge.

rev/min.

Chap. 14

What

Problems

radius bowl

is

centrifuge bowl

is

spinning

at a

constant 2000

needed for the following?

847

A A

(a)

(b)

force of 455 g's. force four times that in part

(a).

Ans.

=

r

0.1017

but with the following changes. Reduce the rev/min to 10 000 and double the outer-bowl radius 0.0445 m, keeping/-! = 0.00716 m.

(a)

Keep all

(b)

variables as in

Example

14.4-2 but

Remove Food

A

Particles.

.

rev/min

and

is

=

r2

(b)

Dp =

6 1.747 x 10~

m

dilute slurry contains small solid

mm

which are to be removed by 5 'x 10" 1050 kg/m 3 and the solution density is 1000 The viscosity of the liquid is 1.2 x 10" 3 Pas. A centrifuge at 3000 2

food particles having a diameter of centrifuging.

kg/m 3

to

r2

double the throughput. Ans.

14.4-4. Centrifuging to

m

Repeat Example 14.4-2

of Varying Centrifuge Dimensions and Speed.

14.4-3. Effect

(a)

The

particle density

is

The bowl dimensions are b = 100.1 ram,r, = 5.00 mm, Calculate the expected flow rate in m 3 /s just to remove these

to be used.

30.0

mm.

particles.

14.4-5. Effect

of Oil Density on Interface Location.

case where the vegetable 14.4-6. Interface in

oil

Repeat Example

14.4-3, but for the

density has been decreased to 9 4.7 1

A cream

Cream Separator.

kg/m 3

.

separator centrifuge has an outlet

mm

and outlet radius r4 = 76.2 mm. The density of discharge radius r = 50.8 3 3 the skim milk is 1032 kg/m and that of the cream is 865 kg/m (El). Calculate l

the radius of the interface neutral zone.

Ans. 14.4-7.

Scale-Up and 14.4-2, (a)

(b)

E

r2

For the conditions given

Values of Centrifuges.

=

mm

150

Example

in

do as follows.

Calculate the E value. A new centrifuge having the following dimensions is to be used. r 2 = 0.0445 m, r = 0.01432 m, b = 0.394 m, and N = 26000 rev/min. Calculate the new E value and scale-up the flow rate using the same solution. 2 Ans. (a) E = 196.3 m i

14.4- 8. Centrifugal Filtration Process.

has a bowl height b 25.0°C. The filtrate

= is

0.457

m

A

batch centrifugal

and

r2

= 0381

essentially water.

m

filter

similar to Fig. 14.4-5

and operates

At a given time

in

at 33.33 rev/s at

the cycle the slurry

and cake formed have the following properties. c s = 60.0 kgsolids/m 3 filtrate, = 0.82, p p = 2002 kg solids/m 3 cake thickness = 0.152 m, a = 6.38 x 10 10 m/kg, R m = 8.53 x 10' 0 m "', r^ = 0.2032 m. Calculate the rate of filtrate flow.

£

,

Ans. 14.5- 1.

Change

in

Power Requirements

80%

in

=

6.11

4 x 10"

m 3 /s

In crushing a certain ore, the feed

Crushing.

mm

q

and the product size is such that than 6.35 mm. The power required is 89.5 kW. What will be the power required using the same feed so that 80% is less than 3.18 mm? Use the Bond equation. (Hint : The work index £ is unknown, but it can be determined is

such that

80%

is

is less

than 50.8

in size

less

;

using the original experimental data in terms of T. In the equation for the size, the same unknowns appear. Dividing one equation by the other

new will

eliminate these unknowns.)

Ans. 14.5-2. Crushing

from ^in. (a)

(b)

of Phosphate Rock. where 80% is

a feed size

desired to crush 100 ton/h of phosphate rock

than 4

in.

to a product where

The work index is 10.13 (PI). Calculate the power required. Calculate the power required to crusiwhe product than 1000 /jm.

848

It is

less

kW

146.7

80%

further where

is

less

80%

than

is

less

.'

Chap. 14

Problems

REFERENCES (Bl)

Badger, W. L., and Banchero, J. T. Introduction York: McGraw-Hill Book Company, 1955.

(B2)

Becker, H. A. Can.

(B3)

Bond,

(CI)

Coulson, J. M., and Richardson, York: Pergamon Press, Inc., 1978.

(El)

Earle, R. L. Unit Operations

J.

Chem.

to

Chemical Engineering.

New

Eng., 37, 85 (1951).

484 (1952).

F. C. Trans. A.I.M.E., 193,

in

F. Chemical Engineering, Vol.

J.

Food

Processing. Oxford:

2,

3rd ed.

Pergamon

New

Press, Inc.,

1966.

(Fl)

Foust, A. S.,et Sons,

al.

Principles of Unit Operations,

(F2)

Fitch, B. lnd. Eng. Chem., 58

(Gl)

Grace, H. J.,

ed.

New York: John

Wiley

&

(10),

18(1966).

Chem. Eng. Progr., 46, 467 (1950); 49, 303, 367, 427 (1953); A.l.Ch.E.

P.

2,307(1956).

(HI)

Hughes, R.

(Kl)

K.YNCH, G.

(LI)

2nd

Inc., 1980.

R.,

and Gilliland,E.

Chem. Eng. Progr., 48, 497

R.

(1952).

Trans. Faraday Soc, 48, 166(1952).

J.

LaRiaN, M. G. Fundamentals of Chemical Engineering Operations. Englewood Cliffs,

N.J.

:

Prentice-Hall, Inc., 1958.

and Shepherd, C.

605 (1940).

(L2)

Lapple, C.

(Ml)

McCabe, W. L., and Smith, J. C. Unit Operations of Chemical Engineering, 3rd ed. New York: McGraw-Hill Book Company, 1976. McMillen, E. L., and Webber, H. A. Trans. A.l.Ch.E., 34, 213 (1938). Nickolaus,N., and Dahlstrom, D. A. Chem. Eng. Progr., '52(3), 87M (1956). Perry, R. H., and Green, D. Perry's Chemical Engineers' Handbook, 6th ed. New York: McGraw-Hill Book Company, 1984.

(M2) (Nl) (PI)

(P2)

E.,

B. lnd. Eng. Chem., 32,

Perry, R. H., and Chilton, C. H. Chemical Engineers' Handbook, 5th ed. York: McGraw-Hill Book Company, 1973.

New

and Kempe,

Trans A.l.Ch.E.,

(Rl)

Ruth,

B. F.,

(R2)

Ruth,

B. F. lnd. Eng. Chem., 25, 157 (1933).

(51)

Steinour, H. H. lnd. Eng. Chem., 36, 618, 840(1944).

(52)

Shepherd, C.

B.,

L. L.

and Lapple, C.

'.

34,

213 (1938).

E. lnd. Eng. Chem., 31,

972 (1939); 32, 1246

(1940).

(Tl)

Talmage, W.

(Zl)

Zenz, F. A., and Othmer, D. F. Fluidization and Fluid-Particle Systems. York: Reinhold Publishing Co., Inc., 1960.

Chap. 14

P.,

References

and Fitch,

E. B. lnd.

Eng. Chem., 47(1), 38 (1955).

New

849

APPENDIX

A.l

Fundamental Constants and Conversion Factors

Gas Law Constant R

A. 1-1

Units

Numerical Value

1.9872

g cal/g mol K btu/lb mol-°R

82.057

cm 3 atm/g mol K

8314.34

J/kg mol

1.9872

•

82.057 x 10"

m

3

10.731

2

/s

•

K

2

•

•

f

1545.3

1

atm/kg mol

m

-

•

f

8314.34

1

•

It

•

kg mol K J 3 lb mol °R lb /in. ft 3 -atm/lb mol-°R ft ft-lb /]b mol-°R m 3 Pa/kg mol K

0.7302

1

3

kg

8314.34

A. 1-2

•

•

•

•

Volume and Density

0°C,760mm Hg = 22.4140 liters = 22414 cm 3 3 lb mol ideal gas at 0°C, 760 mm Hg = 359.05 ft kg mol ideal gas at 0°C, 760 mm Hg = 22.414 m 3 gmol

ideal gas at

Density of dry

air at

0°C, 760

mm Hg = =

Molecular weight of 1

1

1

A. 1-3 1

Length in.

100

850

air

=

28.97

lb m /tb

g/cm 3 = 62.43 lbjft 3 = 1000 kg/m 3 g/cm 3 = 8.345 lbJU.S. gal 3 3 lbjft = 16.0185 kg/m

=

2.540

cm =

1

cm

m (meter)

1.2929 g/liter

0.080711 Ibjft 3

mol = 28.97

g/g

mol

=

m=

10" 6

1

micron

1

A (angstrom) =

1

mile

= 5280

1

m=

3.2808

10"

4

10

cm =

10 10" 4 /im

m=

10

mm

=

/jm (micrometer)

1

ft

=

ft

39.37

in.

Mass

A. 1-4

= 453.59 g = 0.45359 kg = 16oz= 7000 grains lb m = 1000 g = 2.2046 lb m kg ton (short) = 2000 !b m

lb m

1

1 1 1

=

2240 lb m 1000 kg ton (metric) ton (long)

1

=

1

Standard Acceleration of Gravity

A. 1-5

g g

g

= = =

9.80665 m/s

z

980.665 cm/s 32.174

ft/s

2

2

g c (gravitational conversion factor)

= =

32.1740

1

bm

•

ft/1

br

980.665 g m -cm/g f

-

s

-s

2

2

Volume

A. 1-6 1

L

(liter)

1

in.

1

ft

3

3

= =

=

16.387

28.317

L

1

m =

1

U.S. gal

3

cm 3 cm 3

1000

(liter)

= 0.028317 m 3 3 = 7.481 U.S. gal ft m 3 = 264.17 U.S. gal

1 1

ft

1000

= = =

1

U.S. gal

1

U.S. gal

1

British gal

1

m3 =

3

1

L

(liter)

4 qt

3.7854 3785.4

=

35.313

L (liter) cm 3

1.20094 U.S. gal ft

3

Force

A.l-7 1

g cm/s 2 (dyn) g cm/s

1

kg m/s

2

1

ib f

•

•

=

1 1

1

1

10"

5

kg-m/s 2 = 10" 5 N(newton)

= 7.2330 x 10" 5 = 1 N (newton)

4.4482

lg-cm/s 2

A. 1-8

=

•

2

1

--

3

=

lb m

ft/s

2

(poundal)

N

6 2.2481 x 10" lb f

Pressure

= psia = psia = psia =

bar

x 10 5

1

1

Pa

lb f /in.

2.0360 2.311

in.

ft

(pascal)

=

1

x 10 5

N/m 2

z

Hg

at

0°C

H O at 70°F z

lpsia= 51.715 mm Hg at 0°C(p Hg = 13.5955 g/cm 3 ) 2 5 1 atm = 14.696 psia = 1.01325 x 10 N/m = 1.01325 bar = = atm 1.01325 x 10 5 Pa 760 mm Hg at 0°C = atm 29.921 in.HgatOX latm = 33.90ftH 2 Oat4°C 1

1

Appendix A.l

= =

psia

1

6.89476 x 10

4

g/cnvs 2 dyn/cm 2

4

psia 6.89476 x 10 dyn/cm 2 = 2.0886 x 10"

1

1

3

2

lb f /ft

= 6.89476 x 10 N/m = 6.89476 x 10 3 Pa 2 2 2 lb /ft = 4.7880 x 10 dyn/cm = 47.880 N/m 2 2 mmHg(0°C) = 1.333224 x 10 N/m = 0.1333224 kPa psia

1

2

3

2

1

f

1

Power

A.l-9

hp = 0.74570 kW hp = 550 ft lb /s hp = 0.7068 btu/s

1

•

1

f

1

A. 1-10

=

2

2

btu

1

btu

= =

1

kcal (thermochemical)

1

cal

1

cal (IT)

1

btu

1

btu

1055.06

=

J

=

= 0.7457 kW-h

ft

778.17

= =

1

btu/h

1

btu/h-

cm 2 /s 2

g

(erg)

4.1840 kJ

ft

lb r

-

1.35582 J

=

2.9890 J/kg

1

1

ft

•

•

°F

=

4.1365 x 10"

°F=

ft

1.73073

3

cm

cal/s

J

C

W/m-K

Heat-Transfer Coefficient

A. 1-1 2

btu/h btu/h

2 •

•

btu/h

1

kcal/h

A. 1-13

ft

2

1

•

ft

•

2 ft •

•

x 10" 4 4 °F = 5.6783 x 10~

°F

=

°F

= 5.6783 W/m K = 0.2048 btu/h ft 2

m2

•

1.3571

cal/s

cm 2 °C

W/cm 2 °C

2

•

°F

•

•

°F

Viscosity

cp

= 10" 2 g/cm

1

cp

1

cp

= = = =

1

7

Thermal Conductivity

A.l-11

1

W (watt) W

2544.5 btu

lb r/lb m

-

1

4.1868 J

hp h hp h

I

10

= 1000 cal = = 4.1840 J

251.996 cal (IT)

ft-lb f

=

252.16 cal (thermochemical)

(thermochemical)

1

14.340 cal/min

0.29307

1.05506 kJ

= =

1

=

2

2

=

1

1

J/s fjoule/s)

J (joule)

m

=

1

/s

kg

/s

btu/h

1

1

1

852

N-m = kg-m

1

watt (W)

1

Work

Heat, Energy,

J

1

=

1

cp

1

cp

1

Pa

•

s

s

•

(poise)

2.4191 lbjft-h

6.7197 x 10

10~ 3 Pa-s

_4

=

2.0886 x 10

:

lb-m /ft -s 10~ 3 kg/m-s

-5

lb r

= IN- s/m 2 =

-

1

s/ft

=

10~ 3

=

1000 cp

N-s/m 2

2

kg/m

s

App. A.]

=

0.67197 lb m /ft-s

Fundamental Constants and Conversion Factors

A.l-14

Diffusivity

cm 2 /s = 3.875 ft 2 /h cm 7s = 10" 4 m7s m 2 /h = 10.764 ft 2/h

1 1 1

1 1

1

m7s =

1

centistoke

2 ft

/h

cm 2 /s

•

•

•

m2

Heat Flux and Heat Flow

= 3.1546 W/nr = 0.29307 W = 1.1622 x 10" 3 W

btu/hft 2

1

btu/h

1

cal/h

Heat Capacity and Enthalpy

A.l-17 1

btu/lb m °F

=

1

btu/lb m -°F

= 1.0OOcal/g°C

=

1

btu/ib m

1

ft-lb f/lb m

1

cal (IT)/g

1

kca!/g

kJ/kg-K

2326.0 J/kg

= •

4.1868

2.9890 J/kg

=

°C

4.1868 kJ/kg

•

K

4.1840 x 10 3 kJ/kg mol

mol =

Mass-Transfer Coefficient

A.l-18

= 10" 2 m/s = 8.4668 x 10" 5

1

k c cm/s

1

kjt/h

1

k x g mol/s

1

10" 2

•

1

1

=

4

cm 2 = 7.3734 x 10 3 Ibjh ft 2 3 2 2 lb mol/h ft g mol/s cm = 7.3734 x 1 2 4 2 g mol/s cm = 10 kg mol/s m = 1 x 10 g mol/s 3 2 2 10" kg mol/s -m lb mol/h -ft = 1.3562 x g/s

A.l-16

1

3.875 x 10

Mass Flux and Molar Flux

A.l-15 1

1

cm 2 mol 2 mol/s cm mol •

m/s

frac

=

10 kg mol/s

•

m2

•

mol

frac

4

x 10 g mol/s- m 2 mol frac 2 3 2 k x lb mol/h -ft -mol frac = 1.3562 x 10" kg mol/s mol frac 3 3 10" mol frac = 4.449 x k x a lb mol/h ft kg mol/s m 3 mol frac kx g

frac

=

1

•

m

•

•

kG

kg mol/s -rrr-atm

1

kG

a kg mol/s

Appendix A.]

m

3 •

=

5 0.98692 x 10" kg mol/s 5 10" atm = 0.98692 x kg mol/s

1

•

•

•

•

•

m m 2

-

•

3

Pa Pa •

APPENDIX

A.

Physical Properties

of Water

Latent Heat of Water at 273.15

A.2-1

Latent heat of fusion

K

(0°C)

=

1436.3 cal/g mol

=

79.724 cal/g

=

2585.3 btu/lb mol

=

6013.4 kJ/kg mol

Source: O. A. Hougen, K. M. Waison, and R. A. Ragatz, Chemical Process Principles, Pari 1,2nd cd. New York: John Wiley & Sons, Inc., 1954.

Latent heat of vaporization Pressure

at

298.15

(mm Hg)

K

(25°C)

Latent Heat

44 020 kJ/kg mol, 10.514 kcal/g mol, 18925 btu/lb mol

23.75

44 045 kJ/kg mol, 10.520 kcal/g mol, 18 936 btu/lb mol

760

Source: National Bureau of Standards, Circular 500.

A .2-2

Vapor Pressure of Water

Temperature

K

Vapor Pressure

kPa

°C

mm Hg

Temperature

X

°C

V apor Pressure kPa

mm Hg

273.15

0

0.611

4.58

323.15

50

12.333

283.15

10

1.228

9.21

333.15

60

19.92

149.4

293.15

20

2.338

17.54

343.15

70

31.16

233.7

298.15

25

3.168

23.76

353.15

80

47.34

355.1

303.15

30-

4.242

31.82

363.15

90

70.10

525.8

313.15

40

7.375

55.32

373.15

100

101.325

760.0

92.51

Source: Physikalish-iechnishe, Reichsansalt, Holborn, Scheel, and Henning, Warmeiabellen. Brunswick, Germany: Friedrich Vicwig and Son, 1909.

854

Density of Liquid Water

A.2-3

Temperature

Temperature

Density

Ais

c

273.15

0

0.99987

999.87

277.15

4

1.00000

1000.00

283.15

10

0.99973

293.15

20

298.15

o /~

kg/m

g/cm

Density

kg/m

C

g/cm

323.15

50

O.98807

988.07

333.15

60

0.98324

983.24

999.73

343.15

70

0.97781

977.81

0.99823

998.23

353.15

80

0.97183

971.83

25

0.99708

997.08

363.15

90

0.96534

965.34

303.15

30

0.99568

995.68

373.15

100

0.95838

958.38

313.15

40

0.99225

992.25

R. H. Perry and C. H. Chilton, Chemical Engineers' Handbook, 5th McGraw-Hill Book Company, 1973. With permission.

Source:

Water

Viscosity ef Liquid

A.2-4

Viscosity

Temperature

[(Pa

•

JO

s)

Viscosity

2 ,

{kglm-s) 10\ or cp]

Temperature

l(Pa [kg/m

s) s)

J0\ 70

K

"C

or cp]

70? /yZ l

323.15

50

0.5494

£779.

325.15

52

0.5315

l<\

327.15

54

0.5146

Z6

329.15

56

0.4985

JoOU

331.15

58

0.4832

Ml

1

333.15

60

0.4688

12

1.2363

335.15

62

0.4550

287.15

14

1.1709

337.15

64

0.4418

289.15

16

1.1111

339.15

66

0.4293

291.15

18

1.0559

341.15

68

0.4174

293.15

20

1.0050

343.15

70

0.4061

293.35

20.2

1

.0000

345.15

72

0.3952

295.15

22

0.9579

347.15

74

0.3849

297.15

24

0.9142

349.15

76

0.3750

298.15

25

0.8937

351.15

78

0.3655

299.15

26

0.8737

353.15

80

0.3565

301.15

28

0.8360

355.15

82

0.3478

303.15

30

0.8007

357.15

84

0.3395

305.15

32

0.7679

359.15

86

0.3315

307.15

34

0.7371

361.15

88

0.3239

309.15

36

0.7085

363.15

90

0.3165

311.15

38

0.6814

365.15

92

O.3095

313.15

40

0.6560

367.15

94

0.3027

315.15

42

0.6321

369.15

96

0.2962

K

"C

273.15

0

1

275.15

2

1

277.15

4

1

.30

279.15

6

1

.4

281.15

8

1

.

283.15

10

I.

285.15

I

1

.

/

317.15

44

0.6097

371.15

98

0.2899

319.15

46

0.5883

373.15

100

0.2838

321.15

48

0.5683

Source

:

Bingham, Fluidity and

pany, 1922. With permission.

Appendix A.

ed.

Plasticity.

New

3

York: McGraw-Hill Book Com-

New

York:

Heat Capacity of Liquid Water

A.2-5

Heat Capacity,

Temperature

K

°c

caljg

°C

c

at

101.325 kPa

Temperature

K

;

C

X

Atm)

Heat Capacity, c p

p

kJ/kg

(1

caljg

-"C

Id I kg -K

0

273.15

1.0080

4.220

50

323.15

0.9992

4.183

10

283.15

1.0019

4.195

60

333.15

1.0001

4.187

20

293.15

0.9995

4.185

70

343.15

1.0013

4.192

25

298.15

0.9989

4.182

80

353.15

1.0029

4.199

30

303.15

0.9987

4.181

90

363.15

1.0050

4.208

40

313.15

0.9987

4.181

100

373.15

1.0O76

4.219

Source

N.

:

S.

Osborne, H.

F.

Stimson, and D. C. Ginnings, Bur. Standards J. Res., 23, 197

(1939).

Thermal Conductivity of Liquid Water

A. 2-6

Thermal Conductivity

Temperature

°c

K

°F

0

32

37.8

100

btu/h

fi'F

Wjm-K

273.15

0.329

0.569

311.0

0.363

0.628

93.3

200

366.5

0.393

0.680

148.9

300

422.1

0.395

0.684

215.6

420

588.8

0.376

0.651

326.7

620

599.9

0.275

0.476

Source: D. L. Timrol and N. B. Vargaflik, J. Tech. Phys, (U.S.S.R.), 10, 1063 (1940); 6lh International Conference on the Properties of Steam, Paris, 1964.

Vapor Pressure of Saturated Ice-Water Vapor and Heat

A.2-7

of Sublimation

K 273.2

°F

°C

32

0

266.5

20

-6.7

261.0

10

255.4

0

249.9

-10 -20 -30 -40

-12.2 -17.8 -23.3 -28.9 -34.4 -40.0

244.3 238.8

233.2 Source:

856

Heat of Sublimation

Vapor Pressure

Temperature

ASHRAE, Handbook

kPa

mm Hg

psia

6.107 X 10-'

8.858 X 10"

2

5.045 X 10"

2

2.128 X 10-'

3.087 X io-

2

1.275 X 10"

1

1.849 X 10"

2

2

1.082 X 10"

2

X 10"

3

3.440 X 10"

3

X 10"

3

3.478 X lO"

7.411 X 10"

1

2

3,820 X io2 2.372 X 10~ 1.283 X IO"

of Fundamentals.

2

New

6.181

1.861

York:

ASHRAE,

btu/lb m

kJ/kg

4.581

1218.6

2834.5

2.609

1219.3

2836.1

1.596

1219.7

2837.0

0.9562

1220.1

2838.0

0.5596

1220.3

2838.4

0.3197

1220.5

2838.9

0.1779

1220.5

2838.9

0.09624

1220.5

2838.9

1972.

App. A.2

Physical Properties of Water

A. 2-8

Heat Capacity of Ice

Temperature

op

c

X

Temperature

p

°F

btu/Ib m

kJ/kg

K

32

273.15

0.500

2.093

20

266.45

0.490

2.052

10

260.95

0.481

2.014

0

255.35

0.472

1.976

Source

A.2-9

Adapted from

ASHRAE, Handbook

Properties of Saturated

c

p

f

K

btu/lb„-°F

-10 -20 -30 -40

249.85

0.461

1.930

244.25

0.452

1.892

238.75

0.442

1.850

233.15

0.433

1.813

=

of Fundamentals.

New

York:

ASHRAE,

kJ/kg

K

1972.

Steam and Water (Steam Table),

SI Units

Specifi c

Temper

Vapor

ature

Pressure

(°C)

(kPa)

Volume

Enthalpy

(m '/kg) Liquid

(kJ/kg)

Sal'd Vapor

Liquid

Sal'd Vapor

Entropy (kJ/kg-K) Liquid

Sal

'd

Vapor

0.01

0.6113

0.0010002

206.136

0.00

2501.4

0.0000

9.1562

3

0.7577

0.0010001

168.132

12.57

2506.9

0.0457

9.0773

6

0.9349

0.0010001

137.734

25.20

2512.4

0.0912

9.0003

9

1.1477

0.0010003

113.386

37.80

2517.9

0.1362

8.9253

12

1.4022

0.0010005

93.784

50.41

2523.4

0.1806

8.8524

15

1.7051

0.0010009

77.926

62.99

2528.9

0.2245

8.7814

18

2.0640

0.0010014

65.038

75.58

2534.4

0.2679

8.7123

21

2.487

0.0010020

54.514

88.14

2539.9

0.3109

8.6450

24

2.985

0.0010027

45.883

100.70

2545.4

0.3534

8.5794

25

3.169

0.0010029

43.360

104.89

2547.2

0.3674

8.5580

27

3.567

0.0010035

38.774

113.25

2550.8

0.3954

8.5156

30

4.246

0.0010043

32.894

125.79

2556.3

0.4369

8.4533

33

5.034

0.0010053

28.011

138.33

2561.7

0.4781

8.3927

36

5.947

0.0010063

23.940

150.86

2567.1

0.5188

8.3336

40

7.384

0.0010078

19.523

167.57

2574.3

0.5725

8.2570

45

9.593

0.0010099

15.258

188.45

2583.2

0.6387

8.1648

50

12.349

0.0010121

12.032

209.33

2592.1

0.7038

8.0763

55

15.758

0.0010146

9.568

230.23

2600.9

0.7679

7.9913

60

19.940

0.0010172

7.671

251.13

2609.6

0.8312

7.9096

65

25.03

0.0010199

6.197

272.06

2618.3

0.8935

7.8310

70

31.19

0.0010228

5.042

292.98

2626.8

0.9549

7.7553

75

38.58

0.0010259

4.131

313.93

2635.3

1.0155

7.6824

80

47.39

0.0010291

3.407

334.91

2643.7

1.0753

7.6122

85

57.83

0.0010325

2.828

355.90

2651.9

1.1343

7.5445

"

90

70.14

0.0010360

2.361

376.92

2660.1

1.1925

7.4791

95

84.55

0.0010397

1.9819

397.96

2668.1

1.2500

7.4159

100

101.35

0.0010435

1.6729

419.04

2676.1

1.3069

7.3549

Appendix A.2

857

A.2-9

SI Units, Continued

Volume

Specific :

Temper-

Vapor

'

ature

Pressure (kPa)

Liquid

(°Q

Entropy {kJ/kg-K)

Enthalpy

{m^/kg)

{kJ/kg)

Sat'd Vapor

Liquid

Sat'd Vapor

Liquid

Sat'd Vapor

105

120.82

0.0010475

1.4194

440.15

2683.8

1.3630

7.2958

110

143.27

0.0010516

1.2102

461.30

2691.5

1.4185

7.2387

115

169.06

0.0010559

1.0366

482.48

2699.0

1.4734

7.1833

120

198.53

0.0010603

0.8919

503.71

2706.3

1.5276

7.1296

0.0010649

0.7706

524.99

2713.5

1.5813

7.0775

546.31

7.0269

125

232.1

130

270.1

0.0010697

0.6685

2720.5

1.6344

135

313.0

0.0010746

0.5822

567.69

2727.3

1.6870

6.9777

140

316.3

0.0010797

0.5089

589.13

2733.9

1.7391

6.9299

145

415.4

0.0010850

0.4463

610.63

2740.3

1.7907

6.8833

150

475:8

0.0010905

0.3928

632.20

2746.5

1.8418

6.8379

155

543.1

0.0010961

0.3468

653.84

2752.4

1.8925

6.7935

160

617.8

0.0011020

0.3071

675.55

2758.1

1.9427

6.7502

165

700.5

0.0011080

0.2727

697.34

2763.5

1.9925

6.7078

170

791.7

0.0011143

0.2428

719.21

2768.7

2.0419

6.6663

175

892.0

0.0011207

0.2168

741.17

2773.6

2.0909

6.6256

180

1002.1

0.0011274

0.19405

763.22

2778.2

2.1396

6.5857

190

1254.4

0.0011414

0.15654

807.62

2786.4

2.2359

6.5079

200

1553.8

0.0011565

0.12736

852.45

2793.2

2.3309

6.4323

225

2548

0.0011992

0.07849

966.78

2803.3

2.5639

6.2503

250

3973

0.0012512

0.05013

1085.36

2801.5

2.7927

6.0730

275

5942

0.0013168

0.03279

1210.07

2785.0

3.0208

5.8938

300

8581

0.0010436

0.02167

1344.0

2749.0

3.2534

5.7045

Source: Abridged from

New

J.

York: John Wiley

A.2-9

H. Keenan, F. G. Keyes,

&

P.

G.

Hill,

and

J.

—

New

G. Moore, Sieam Tables Metric Uniis. & Sons, Inc.

Sons, Inc., 1969. Reprinted by permission of John Wiley

Properties of Saturated Steam and

Water (Steam Table),

English Units

Specific

Volume

3

Entropy

Enthalpy

Temper-

Vapor

ature

Pressure

(°F)

(psia)

32.02 35

0.08866

0.016022

3302

0.00

1075.4

0.000

2.1869

0.09992

0.016021

,2948

3.00

1076.7

0.00607

2.1764

40

0.12166

0.016020

2445

8.02

1078.9

0.01617

2.1592

45

0.14748

0.016021

2037

13.04

1081.1

0.02618

2.1423

50

0.17803

0.016024

1704.2

18.06

1083.3

0.03607

2.1259

55

0.2140

0.016029

1431.4

23.07

1085.5

0.04586

2.1099

(/*

Liquid

/'*>J

Sat'd Vapor

(btu/lbj

Liquid

[btu/lb m

Sat'd Vapor

App. A.2

Liquid

°F)

Sat'd

—

Vapor

Physical Properties of Water

A.2-9

English Units, Continued

Specific 3

Temper-

Vapor

ature

Pressure

(°F)

(psia)

60

0.2563

65

0.305

/0 Ac ID

0.3622

oa

80 oc

85

90 95

Entropy

(WO

{btu/lb m -"F)

Liquid

OA£ a izuo.y

AO AO 28.08

Cil < 1021.D

AA 33.09

0.4300

0.016042 a a rnf 0.016051 a a /ca^ 0.016061

0.50/3 a c c\c a 0.5964

0.0160/3 A A1 £AO < 0.01 6085

632.8

c.

i

a ai o.oi

2.225

140

2.892

150

3.722

160

4.745

170

5.996

180

7.515

190

£nnn ooyy

1

1

1

1

0.016130 a a c 0.01 61 66 a ai a ac 0.016205 1 zr 1 zr

zr

.6945

130

1

a a z; 14 0.0161 a ai tz ia

1.2763 1

i

1

0.9503

120

Enthalpy

a ai cc\i 0.01 6035

/

1

1

Volume

/'U Sat'd Vapor

Liquid

0.6988 a o z!A 0.8162

100 A 110 1

(/>


z^

0.016343 a a / one 0.016395 A A ZT CA 0.016450

1

1

"3 "3

1

86

1 O AA 38.09

/. /

T2.Q /39. 11

A 1

AA

43.09

A O AA 48.09 C 1 AO 53.08

CA 1 543.1 1

a cn n 46/./ AC\A A 404.0

Liquid

Sat'd Vapor

AO1 A 1087.7 1 A O A A 1089.9 1 AAA A 1 092.0 AA A A 1094.2

A AC C C C 0.05555 A AZT C A 0.065 14

A AA A T 2.0943

1

AAZT A

1

AAO

A A AT T 0.09332 A AA C A 0.10252

Sat'd

Vapor

1

1

1096.4 C

1098.6

58.07

1

63.06

1 1

100.7 02.9

2.049/

2.0356 A AA O 2.0218 1

AAO T 2.0083 A

AA C 1 1.9951 AO A A 1.9822 1

1

13.5

0.16465

1.9336

157.17

aa ao 97.98

1 1

17.8

0.18172

1.9109

AA OO

A AzT 107.96

1

121.9

0.19851

1.8892

17.96

1

126.1

1

1

122.88

'"7

137.97

1 1

147.99

1138.2

9.343

A A £ C AA 0.016509 a a c r\ 0.016570

62.02 f A OA 50.20

40.95

158.03

1

142.1

145.9

a

A A A c\n

109.3

AA

1130.1

1

1

1

1

OO

203.0

1105.0

127.96

1

1165 A AAZT O 0.12068

0.1

ATA

1

no AA 78.02

77.23

,4

1

A

2.0791 A f\£ A A 2.0642

88.00

68.05

AC C 1 265.1 AA1 A

1

1

A AT Ad 0.07463 A AO A AA 0.08402

a nA/i 0.12963 A ,mA -0.14730

1 CA A 350.0

96.99

1

1

34.2

1

AA

1

c r\ "1

0.21503 AA A 0.23130 A A A T) A 0.24732 *> 1

1

1.9574

1.8684

*">

1.8484 1.8293

0.26311 A A AO 0.27866 A A A ^ AA 0.29400

1.8109

1.7599

ZT ZT

1.7932

200

1

1.529

0.016634

33.63

168.07

1

210

14.125

27.82

178.14

1149.7

212

14.698

0.016702 A AI £71 / 0.016716

180.16

1

150.5

0.30913 A1 A 1 0.31213

220

17.188

0.01

188.22

1

153.5

0.32406

1.7441

230 240

20.78

AA f AC 0.016845 A A AAA

250

29.82

260 270

35.42

6772 Ci

1

1

Z"

6922 A A 7AA 0.01 700

24.97

0.0 1 1

A £ OA 26.80 AT AC 23.15

1

1

1.7762 A cH A 1.7567 1

1

*7

A A

1

198.32

1

157.1

0.33880

1.7289

208.44

1

160.7

0.35335

1.7143

13.826

218.59

1

164.2

0.36772

1.768 1 A AzT 10.066

228.76

1

167.6

0.38193

238.95

1 1

70.9

0.39597

1.7001 1 CO £. A 1.6864 Z* AO 1 1.6731

19.386 1

Z"

*)

AA

16.327 1

*5

OA £

300

66.98

310

77.64

A A *7 AO A 0.017084 A A "7 HA 0.01 7170 A A lAf 0.017259 A A T) f 1 0.017352 AA A A O 0.01 7448 A A 1CAO 0.017548

320

89.60

0.017652

4.919

290.43

1185.8

0.46400

1.6123

330

103.00

0.017760

4.312

300.84

1188.4

0.47722

1.6010

340

117.93

0.017872

3.792

311.30

1190.8

0.49031

1.5901

350

134.53

0.017988

3.346

321.80

1193.1

0.50329

1.5793

360 370

152.92

0.018108

2.961

332.35

1195.2

0.51617

1.5688

173.23

0.018233

2.628

342.96

1197.2

0.52894.

1.5585

380 390

195.60

0.018363

2.339

353.62

1199.0

0.54163

1.5483

220.2

0.01

8498

2.087

364.34

1200.6

0.55422

1.5383

400

247.1

0.018638

1.8661

375.12

1202.0

0.56672

1.5284

410 450

276.5

0.018784

1.6726

385.97

1203.1

0.57916

1.5187

422.1

0.019433

1.1011

430.2

1205.6

0.6282

1.4806

1

1

41.85

280

49.18

290

57.33 -

J.

New

&

Appendix

A2

ZT

r\

f

a r\r\ r>

1

1

8.650

249.18

1

174.1

0.40986

1.6602

1

7.467

259.44

1

177.2

0.42360

1.6477

6.472

269.73

1

1

80.2

0.43720

1.6356

5.632

280.06

1

183.0

0.45067

1.6238

1

""7

1

Source: Abridged from

York: John Wiley

1

1

H. Keenan, F. G. Keyes, P. G.

—

Hill, and J. G. Moore, Steam Tables English Units. Sons, Inc., 1969. Reprinted by permission of John Wiley & Sons, Inc.

859

Steam (Steam Table), entropy, kJ/kg K)

Properties of Superheated

A.2-10

enthalpy, kJ/kg;

s,

SI Units

specific

(i>,

volume,

m 3 /kg;

H,

•

Absolute Pressure,

kPa Temperature (°Q

(Sat.

Temp.,

°Q

100

150

200

250

300

360

420

500

v

17.196

19.512

21.825

24.136

26.445

29.216

31.986

35.679

10

H

2687.5

2783.0

2977.3

3076.5

3197.6

3320.9

3489.1

(45.81)

s

8.4479

8.6882

2879.5 O HAT O 8.9038

lLXJi

9.2813

9.4821

9.6682

9.8978

9.

3.418

3.889

4.356

4.820

5.284

5.839

6.394

7.134

50

H

2682.5

2780.1

2877.7

29 /o.O

3075.5

3196.8

3320.4

3488.7

(81.33)

5

7.6947

7.9401

8.1580

8.3556

8.5373

8.7385

8.9249

9.1546

v

2.270

2.587

2.900

3.211

3.520

3.891

4.262

4.755

75

H

2679.4

2778.2

2876.5

2975.2

3074.9

3196.4

3320.0

3488.4

(91.78)

s

7.5009

7.7496

7.9690

8.1673

8.3493

8.5508

8.7374

8.9672

v

-

v

1.6958

1.9364

2.172

2.406

2.639

2.917

3.195

3.565

100

H

2672.2

2776.4

2875.3

2974.3

3074.3

3195.9

3319.6

3488.1

(99.63)

s

7.3614

7.6134

7.8343

8.0333

8.2158

8.4175

8.6042

8.8342

v

1.2853

1.4443

1.6012

1.7570

1.9432

2.129

2.376

150

H

2772.6

2872.9

2972.7

3073.1

3195.0

3318.9

3487.6

(111.37)

s

7.4193

7.6433

7.8438

8.0720

8.2293

8.4163

8.6466

v

0.4708

0.5342

0.5951

0.6548

0.7257

0.7960

0.8893

400

H

2752.8

2860.5

2964.2

3066.8

3190.3

3315.3

3484.9

(143.63)

s

6.9299

7.1706

7.3789

7.5662

7.7712

7.9598

8.1913

v

0.2999

0.3363

0.3714

0.4126

0.4533

0.5070

700

H

2844.8

2953.6

3059.1

3184.7

3310.9

3481.7

(164.97)

s

6.8865

7.1053

7.2979

7.5063

7.6968

7.9299

r oocc

v

0.2060

0.2327

0.2579

0.2873

0.3162

0.3541

1000

H

2827.9

2942.6

3051.2

3178.9

3306.5

3478.5

(179.91)

5

6.6940

6.9247

7.1229

7.3349

7.5275

7.7622

v

0.13248

0.15195

0.16966

0.18988

0.2095

0.2352

1500

H

2796.8

2923.3

3037.6

3.1692

3299.1

3473.1

(198.32)

s

6.4546

6.7090

6.9179

7.1363

7.3323

7.5698

0.11144

0.12547

0.14113

0.15616

0.17568

2000

H

2902.5

3023.5

3159.3

3291.6

3467.6

(212.42)

5

6.5453

6.7664

6.9917

7.1915

7.4317

v

v

0.08700

0.09890

0.11186

0.12414

0.13998

2500

H

2880.1

3008.8

3149.1

3284.0

3462.1

(223.99)

s

6.4085

6.6438

6.8767

7.0803

7.3234

v

3000

.

(233.90)

0.07058

0.08114

0.09233

0.10279

0.11619

H

2855.8

2993.5

3138.7

3276.3

3456.5

s

6.2872

6.5390

6.7801

6.9878

7.2338

— Metric

J. H. Keenan, F. G. Keyes, P. G. Hill, and J. G. Moore, Steam Tables Sons, Inc., 1969. Reprinted by permission of John Wile^& Sons, Inc.

Source: Abridged from

Wiley

&

860

App.

A .2

Units.

New York: John

Physical Properties of Water

A.2-10

Properties of Superheated fic

volume,

3 ft

/lb m ;//,

Steam (Steam Table), English Units

enthalpy, btu/lb„;

s,

entropy, btu/lb„

(v, speci-

°F)

-

Absolute Pressure,

Temperature

psia [Sat.

—

(°F)

Temp.,

200

°F)

300

400

500

600

800

700

900

1000

392.5

452.3

511.9

571.5

631.1

690.7

750.3

809.9

869.5

1.0

H

1150.1

1195.7

1241.8

1288.5

1336.1

1384.5

1433.7

1483.8

1534.8

(101.70)

s

2.0508

2.1

7 7 70 Z.Z /VJO 9 7 147

7

7 7Q77

7 47 Q4

150.01

161.94

173.86

1433.5

1483.7

1534.7

v

v

78.15

90.24

102.24

114.20

126.15

1194.8

1241.2

1288.2

1335.8

1.9367

1.9941

2.0458

5.0

H

1148.6

(162.21)

s

1.8715

(193.19)

1384.3 Z.

1

JO

/

Z. 1

/ /

J

Z.Z1

7<\7fl Z.Z3ZU JO 7

38.85

44.99

51.03

57.04

63.03

69.01

74.98

80.95

86.91

1146.6

1193.7

1240.5

1287.7

1335.5

1384.0

1433.3

1483.5

1534.6

s

1.7927

1.8592

1.9171

1.9690 z.u i ty+ Z.UOU 1

v

14.696

138.08

H

v

10.0

150 2.1720 2.2235

H

(211.99) s

7

1

AAQ

z.

1

Jy J

z.l

/

JJ

30.52

34.67

38.77

42.86

46.93

51.00

55.07

59.13

1192.6

1239.9

1287.3

1335.2

1383.8

1433.1

1483.4

1534.5

1.8157

1.8741

1.9263

Q7T7 l.y 15 1

z.ui

z.Ujo4 l.WO

1

/_>

/

13 j\J

z.

22.36

25.43

28.46

31.47

34.77

37.46

40.45

43.44

20.0

H

1191.5

1239.2

1286.8

1334.8

1383.5

1432.9

1483.2

(227.96)

s

1.7805

1.8395

1.8919

l

Q87/1 JH .yo

Z.Uz4 J

Z.UOZ

1534.3 7 flQSQ

7.260

8.353

9.399

10.425

11.440

12.448

13452 14.454

60.0

H

1181.9

1233.5

1283.0

1332.1

1381.4

1431.2

1481.8

(292.73)

s

1.6496

1.7134

1.7678

1

i.oouy

I

v

v

1

Q7q<;

Q

1

.5 10!)

1

I

.vUzz

1

i

/

.y4Uo

1533.2

QT77 l.y 1 1 i 1

4.934

5.587

6.216

6.834

7.445

8.053

8.657

100.0

H

1227.5

1279.1

1.6517

1.7085

1429.6 QAA Q 1 .544

1532.1

s

1379.2 QPH "3 1 .oUjj

1480.5

(327.86)

1329.3 *7 COO 1 / JOZ

v

1

.

1

1

1

.88 Jo

3.221

3.679

4.111

4.531

4.944

5.353

5.759

H

1219.5

1274.1

1325.7

1376.6

1427.5

1530.7

(358.48)

1.5997

1.6598

1

.9

1478.8 Q1Q

2.361

2.724

3.058

3.379

3.693

4.003

4.310

200.0

H

1210.8

1268.8

1322.1

1373.8

1425.3

1477.1

1529.3

(381.86)

s

1.5600

1.6239

1.6767

1.7234

1.7660

1.8055

1.8425

2.150

2.426

2.688

2.943

3.193

3.440

250.0

H

1263.3

1318.3

1371.1

1423.2

1475.3

1527.9

(401.04)

s

1.5948

1.6494

1.6970 1.7401

1.7799

1.8172

v

150.0

v

v

1

.

1/

1

1

1

1

Pi

u

1. /

JOo

1

1

1.0

/JU

1.766

2.004

2.227

2.442

2.653

2.860

300.0

H

1257.5

1314.5

1368.3

1421.0

1473.6

1526.5

(417.43)

s

1.5701

1.6266

1.6751

1.7187

1.7589

1.7964

v

1.2843

1.4760

1.6503

1.8163

1.9776

2.136

400

H

1245.2

1306.6

1362.5

1416.6 1470.1

1523.6

(444.70)

s

1.5282

1.5892

1.6397

1.6884

1.7632

v

New

—

Metric J. H. Keenan, F. G. Keycs, P. G. Hill, and J. G. Moore, Steam Tables York: John Wiley & Sons, Inc., 1969. Reprinted by permission of John Wiley & Sons, Inc.

Source: Abridged from Units.

1.7252

Appendix A.

Heat-Transfer Properties of Liquid Water, SI Units

A.2-11

3

T

T

CO

p (kg/m

(K)

c 1

273.2

OGG £ 777.0

15.6

288.8

ooq n

26.7

299.9

996.4

37.8

311.0

65.6

338.8

93.3

366.5 jyn.j

0

)

p x /O (Pa s, or

(kJ/kg-K) kg/m-s) H.ZZy /I 1 87

k

(W/m-K) N Fr

786

1

1

1

1 1 J.J

P x 10*

\9PP IPx /0~ 8

U/K)

(1/K-m*)

-0.630

111 J

U.Joon

St

01

1.44

10.93

4.183

0.860

0.6109

5.89

2.34

30.70

994.7

4.183

0.682

0.6283

4.51

3.24

68.0

981.9

4.187

0.432

0.6629

2.72

5.04

256.2

962.7

4.229

0.3066

0.6802

1.91

6.66

943.5

4.271

0.2381

0.6836

1.49

O.HU

1

1

.

1

642 1

JUu

148.9

422.1

917.9

4.312

0;1935

0.6836

1.22

10.08

2231

204.4

477.6

858.6

4.522

0.1384

0.6611

0.950

14.04

5308

260.0

533.2

784.9

4.982

0.1042

0.6040

0.859

19.8

11030

315.6

588.8

679.2

6.322

0.0862

0.5071

1.07

31.5

19 260

Heat-Transfer Properties of Liquid Water, English Units

A.2-1 1

p

c

p

p x 10*

k (9

T

f'Jlm\

{ htu \

( l^m)

f

(°F)

U'V

\lb m -'Fj

Xft-sJ

{h-ft-'Fj

btu

\

p x

N Fr

, 0*

(1/°R)

P

l ,o-^ urR-fi

3 )

32

62.4

1.01

1.20

0.329

60

62.3

1.00

0.760

0.340

8.07

0.800

17.2

80

62.2

0.999

0.578

0.353

5.89

1.30

48.3

100

62.1

0.999

0.458

0.363

4.51

1.80

107

150

61.3

1.00

0.290

0.383

2.72

2.80

403

200 250

60.1

1.01

0.206

0.393

1.91

3.70

1010

58.9

1.02

0.160

0.395

1.49

4.70

2045

300 400

57.3

1.03

0.130

0.395

1.22

5.60

3510

53.6

1.08

0.0930

0.382

0.950

7.80

8350

500

49.0

1.19

0.0700

0.349

0.859

11.0

17 350

600

42.4

1.51

0.0579

0.293

1.07

17.5

30 300

862

)

-0.350

13.3

App. A. 2

Physical Properties of Water

Heat-Transfer Properties of Water Vapor (Steam)

A.2-12

kPa

at 101.32

Atm

(1

Abs), SI Units s p x 10

T

T

(°Q

(K)

100.0

373.2

0.596

1.888

148.9

422.1

0.525

204.4

477.6

260.0

3 $ x 70

gPp /p

(W/m-K)N Pr

(1/K)

(//Km 3

1.295

0.02510 0.96

2.68

8 0.557 X 10

1.909

1.488

0.02960 0.95

2.38

8 0.292 X 10

0.461

1.934

1.682

0.03462 0.94

2.09

0.154 X 10

533.2

0.413

1.968

1.883

0.03946 0.94

1.87

8 0.0883 X 10

315.6

588.8

0.373

1.997

2.113

0.04448

0.94

1.70

52.1

X 10 5

371.1

644.3

0.341

2.030

2.314

0.04985 0.93

1.55

33.1

X 10 5

426.7

699.9

0.314

2.068

2.529

0.05556 0.92

1.43

5 21.6 x 10

A.2-12

p (kg/m

)

k

2

2

)

8

Atm

(1

Abs), English Units

p.x I0 5

k

p

flbA }

\ft

kPa

c

p

l°F)

3

Heat-Transfer Properties of Water Vapor (Steam) at 101.32

T

(Pa-s,or p (kJ/kg-K) kg/m s) c

)

f

btu

\

\lb m -'Fj

f

lb

\

f

btu

\

\ffsj \h-ff'Fj

p x

N Pr

y03

(1/°R)

gppl/fi 2

(irR-fi

3 )

212

0.0372

0.451

0.870

0.0145

0.96

1.49

6 0.877 x 10

300

0.0328

0.456

1.000

0.0171

0.95

1.32

0.459 x 10

6

400

0.0288

0.462

1.130

0.0200

0.94

1.16

0.243 x 10

6

500

0.0258

0.470

1.265

0.0228

0.94

1.04

0.139 x 10

6

600

0.0233

0.477

1.420

0.0257

0.94

0.943

82 x 10

3

700

0.0213

0.485

1.555

0.0288

0.93

0.862

52.1

800

0.0196

0.494

1.700

0.0321

0.92

0.794

34.0

x 10 3 x 10 3

Source: D. L. Timrol and N. B. Vargaftik,7. Tech. Phys. (U.S.S.R.), 10, 1063(1940); R. H. Perry and C. H. Chillon, Chemical Engineers' Handbook, 5th ed. New York: McGraw-Hill Book Company, 1973; J. H. Kecnan, F. G. Keyes, P. G. Hill, and J. G. Moore, Sleam Tables. New York: John Wiley & Sons, Inc., 1969; National Research Council, International Critical Tables. New York: McGraw-Hill Book Company, 1929; L. S. Marks, Mechanical Engineers' Handbook, 5th ed. New York: McGraw-Hill Book Company, 1951.

Appendix A.2

863

APPENDIX

A.

Physical Properties

of Inorganic

and Organic Compounds

Standard Heats of Formation

A.3-1

Abs),

=

(c)

crystalline, (g)

=

gas, (/)

=

at

298.15

K

(25°C) and 101.325 kPa

AH° Compound

NH

3 (g)

NO(g) H,0(/)

H 2 0(

f/ )

HCN(y)

H,SO 4 (0

H 3 P0 4 (c) NaCl(c)

NH 4 Cl(c) J.

Company,

New

864

Compound

-46.19

-11.04

CaC0 3 (c)

+

+ 21.600

CaO(c)

-635.5

-285.840 -241.826

-68.3174 -57.7979

CO(

- 110.523

+

+ 31.1

chm

-22.063

C 2 H 6 (g) C 3 H 8 (y)

kcal/g

90.374

130.1

-92.312

HC\(g)

Source:

3

-811.32 -1281.1

-411.003 -315.39

-

193.91

-306.2 -98.232 -75.38

C0

(kJ/kg mo/);0"

-

fl )

2 (g)

CH 3 OH(/) •CH 3 CH 3 OH(0

H. Perry and C. H. Chillon. Chemical Engineers' Handbook, 5lh ed.

1206.87

-393.513 -74.848 -84.667

-

103.847

-238.66 -277.61

3

kcal/g mol

-288.45 -151.9 -26.4157 -94.0518 -17.889 -20.236 -24.820 -57.04 -66.35

New York: McGraw-Hill Book

1973; and O. A. Hougen, K. M. Walson, and R. A. Ragaiz, Chemical Process Principles, Part

York: John Wiley

Atm

AH°f mol

{kJ/kgmoI)10'

(1

liquid

& Sons, Inc.,

1954.

1,

2nd

cd.

K (25°C) and 101.325 kPa

Standard Heats of Combustion at 298.15

A.3-2

(?)

=

gas, (/)

=

liquid, (s)

=

(1

Atm

Abs)

solid

ah; (kj/kg 1

n/u

/"»f>j

i

C nmhti*itinn

yi/l

R pnctinn

krniln mn!

7A 41 ^7

_ 67.6361 — 94.0518

CCl{n\

cm n A9)

74

— 7RS

S7 7Q7Q

741

£.8.

+ 20 2 ( 2C0 {g) + 3H 0(f) 9 + 50 2 (g)-» 3C0 2 9 + 4H 2 O(0

CH A (9)

CH.fe)

C 2 H 6 fo) C3 H 8 9

C2 H 6 CjH 8

(

)

(

(

)

2

)

2

)

(

— 282.989 — 393.513

7

1

R4fl

-890.346

-212.798 -372.820

-

-530.605

-2220.051

-673

-2816

1

559.879

^-Glucose (dextrose)

C 6 H, 2 0 6

(

C 6 H 12 0 6 (5) + 60,( y )- 6CO,(y) + 6H 2 O(0

S)

Lactose (anhydrous)

C 12 H 2 ,0 M (s)

C.zH^O^fs) + 120 2 9 )- I2C0,(g)+ 11H 2 0(/) (

-1350.1

-5648.8

-

-5643.8

Sucrose

C,,H 22 O u

(.s)

C l2 H 22 0, U)+ 120 2 1

( ff

)->

12C0 (g)+ llH 2 O(0 2

1348.9

Source: R. H. Perry and C. H. Chilton, Chemical Engineers' Handbook, 5th ed. New York: McGraw-Hill Book Company, 1973 and O. A. Hougcn, K. M. Watson, and R. A. Ragatz, Chemical Process Principles, Part 1, 2nd cd. New York: John Wiley Sons, Inc., 1954. ;

&

Temperature (°F) Figure A. 3-1.

Mean molar

heat capacities from

7TF

(25°C) to t°F at constant pressure of 101.325

(From O. A. Hougen, K. M. Watson, and R. A. Ragatz, Chemical Process Principles, Part I, 2nd ed. New York : John Wiley & Sons, Inc., 1954. With

kPa

(1

aim

abs).

permission.)

Appendix

A3

865

Physical Properties of Air a 1

A.3-3

1

01 .325

kPa

x JO 1 {Pa s, or

( 1

Atm

Abs), SI Units

p.

T

T

to

(K)

-17.8

255.4

1.379

c p (kg/m 3 ) (kJ/kg-K)

k

x I0 3

(t

kg/ms) [Wjm-K)

2

9Pp /p (1/K-m

2

I

U/K)

1.0048

1.62

0.02250 0.720

3.92

2.79 X 10

!

0

273.2

1.293

1.0048

1.72

0.02423

0.715

3.65

2.04 X 10'

10.0

283.2

1.246

1.0048

1.78

0.02492 0.713

3.53

1.72 X 10'

37.8

311.0

1.137

1.0048

1.90

0.02700 0.705

3.22

1.12 x 10

65.6

338.8

1.043

1.0090

2.03

0.02925 0.702

2.95

0.775 X 10'

93.3

366.5

0.964

1.0090

2.15

0.03115 0.694

2.74

0.534 X

394.3

0.895

1.0132

2.27

0.03323

2.54

0.386 X 10

;

121.1

0.692

:

10'

148.9

422.1

0.838

1.0174

2.37

0.03531

0.689

2.38

0.289 X 10

!

176.7

449.9

0.785

1.0216

2.50

0.03721

0.687

2.21

0.214 x 10

:

204.4

477.6

0.740

1.0258

2.60

0.03894 0.686

0.168 X 10

:

2.09

232.2

505.4

0.700

1.0300

2.71

0.04084 0.684

1.98

0.130 X 10

;

260.0

533.2

0.662

1.0341

2.80

0.04258

1.87

0.104 X 10

A.3-3

Physical Properties of Air at 101.325

kPa

0.680

(1

Atm

Abs), English

Units

P

T C (

F)

flb^\ \f,i)

k

<=,

(

bin

\lb„

c

\

M

biu

(

Fj (cemipoise)\h

Ji

p*10>

\

'F)

N

gpp'/fi

(l/°R-ft

pr

1 2 )

6

0

0.0861

0.240

0.0162

0.0130

0.720

2.18

4.39 X 10

32

0.0807

0.240

0.0172

0.0140

0.715

2.03

3.21

50

0.0778

0.240

0.0178

0.0144

0.713

1.96

2.70 X 10

6

6

X 10 6

100

0.0710

0.240

0.0190

0.0156

0.705

1.79

1.76

x 10

150

0.0651

0.241

0.0203

0.0169

0.702

1.64

1.22

X 10 6

200 250

0.0602

0.241

0.0215

0.0180

0.694

1.52

6 0.840 X 10

0.0559

0.242

0.0227

0.0192

0.692

1.41

0.607 X 10

300

0.0523

0.243

0.0237

0.0204

0.689

1.32

6 0.454 X 10

350

0.0490

0.244

0.0250

0.0215

0.687

1.23

0.336 X 10

6

400 450 500

0.0462

0.245

0.0260

0.0225

0.686

1.16

0.264 X 10

6

0.0437

0.246

0.0271

0.0236

0.674

1.10

0.204 X 10

6

0.0413

0.247

0.0280

0.0246

0.680

1.04

0.163 X 10

6

6

Source: National Bureau of Standards, Circular 461C, 1947; 564, 1955; NBS-NACA, Tables of Thermal Properties of Gases, 1949; F. G. Keyes, Trans. A.S.M.E., 73, 590, 597 (1951); 74, 1303 (1952); D. D. Wagman, Selected Values of Chemical Thermodynamic Properties.

866

Washington, D.C.: National Bureau of Standards, 1953.

App.

A .3

Physical Properties of Inorganic and Organic

Compounds

S

4

A.3-4

Viscosity of Gases at

(Pa

-

10\ (kg/m

s)

•

s)

101325 kPa 10\ orcp)

(1

Atm

Abs) [Viscosity

in

Temperature

K

Of

255.4

0

273.2

32

283.2

50

311.0

CO

co 2

0.0158

0.0156

0.0128

0.0166

0.0165

0.0137

0.0171

0.0169

0.0141

0.0213

0.0183

0.0183

0.0154

0.00960

0.0228

0.0196

0.0195

0.0167

93.3

0.0101

0.0241

0.0208

0.0208

0.0179

121.1

0.0106

0.0256

0.0220

0.0220

0.0191

148.9

0.0111

0.0267

0.0230

0.0231

0.0203

350

176.7

0.0115

0.0282

0.0240

0.0242

0.0215

400 450 500

204.4

0.0119

0.0293

0.0250

0.0251

0.0225

232.2

0.0124

0.0307

0.0260

0.0264

0.0236

260.0

0.0128

0.0315

0.0273

0.0276

0.0247

Hi

o2

17.8

0.00800

0.0181

0

0.00840

0.0192

10.0

0.00862

0.0197

100

37.8

0.00915

338.8

150

65.6

366.5

422.1

200 250 300

449.9 477.6

394.3

505.4 533.2

°C

-

Source: Nalional Bureau of Standards, Circular 461C, 1947; 564, 1955; NBS-NACA, Tables of Thermal Properties of Gases, 1949; F. G. Keyes, Trans. A.S.M.E., 73, 590, 597 (1951): 74, 1303 (1952); D. D. Wagman, Selected Values of Chemical Thermodynamic Properties. Washington, D.C.: Nalional Bureau of Standards, 1953.

Appendix A.

867

no

—

I

tj-

oo

Q O

NO oo (N

O

no oo ci

O — QQ— — n — — .(N OOOOOOOOOOO o o CO oo -

iai

_

.

'

o

3t

OOO O do o o ooo o fN m rN o o no ON o NO oo xr V} CO NO OO o CO V} r- ON (N rN rN m ci ci O o O o o
oo^oooo— '-^Tj-Nnv-)cnrNi/~i rNciciNONor— ooono n m - —
^^^•^"•z? OOOOOOOOOOOO oooo'oo'oo'oooo i

'

'

o

m o ON rNO oo o O m m o m *r oO O OoO O O oo O © O O oO O © oo o o o Os (N on no oo oo ci NO r— fN rl rN rN fN fN rN fN

NO O (N m ci no r— on O rN m rN rN (N rN rN 3O o o o o o O oO o o CD O o © o o o © O o o o rN oo ci

oo ON oo rci oo rN NO oo rN NO CO cn rN rN rN rN ci

m 2 O mO 8 oO O O oo o o o O 3 S d o o d CD o* o d d O o o -o

< s

<

O r- oo o o o o O d d d dd r— On o r— Cs rN fN rN rN rN m O o o o o d d d d d

fN rN

3

•

m o d

-3

a ft. -a:

CO fN rN

o d

IT)

rN

wo

O o o o d d d oo ON

o m m OO O dd d

rN io rN rN rN fN rN

fN CO rN oo

NO OO ON rN CI m m o mO m O O o o d d dd d d ^J-

^1-

o rN rrN Cn NO rN On NO on O. ON o rN rN m NO o O O d d d d dd d d dd d d rN o rN o oo ON m NO CA rN VO m r- OO OO r- CO o rN CI NO rrN rN rN rN rN rN rN d d d d d d O d d d d d r<-)

i^N

3t

3 -a c

o

u

o

o

r-'

CO

o 8 o o O o o 8 o o o o o NO rN rN CI m o oo -o ON rN o O d NO ON rN oo nO * rN d NO m O rN rN rN rN

r»l

r*"i

t~r~-

H

1

fN rN r-i

r"l

r- oo rN CN rN

868

App. A.3

o

oo

ON NO

rN

oo m

NO rN CN r~" NO ON fN r~

m o ci

Physical Properties of Inorganic and Organic

Compounds

\o r~ m rN m m rs O rS cn cn CN rN d o o O do O o Od d d NO Vl XT m xr ON o NO NO r- o CN Vl CO o CM m r- ON g ON O C) CO OO OO CO OO ON ON On o" d d O O d o O o d V) rOn ON ON

CN)

CO"

(

i

cn)

E

r^i


E

o o

xr VI o o — V) Vl VI rN CN O© o d Oo o o Odd d VI On m CO CO m r— rND NO NT xr xr xf m CO Om o O o o o o O V) o o Oo

r- OO CO On ON XT xr xT xr xf Vi >J-> CN rN r-i rs CN rs rs

O

5.

r*l

^

v-j

o

—

-3i

3

r-i

v-i

-«

u, E

m o O vi m V, v> rN CN od oo o doo O ddd r— OO OO CO OO On ON xr xr xr XT rs rN CN CN rN CN|

vi ON OO OO OO OO r- r~ V, V) m o m m o O o o o VI Om OO Oo o c"i

cn

-o ^.

< z

V:

S3

E

Z o E

o d o ood oOo O Vi ON CO NO VI ON m V) CN
o m m d d d

rs r- OO On
Vj t— On ON ON

O dd

g>

3'Q •

NO

t

„-

-

rT

— — *— ^ Urn —— — o>

u-,

r-i

„

nO c~i

d

o CO

ON

r-i

C*i

xr xr xr r-i c-i rn

rn rn

m m O XT

r— On On CN NO

O xr

o

xt-"

xr xf xf

CO

ro

xr ON CN ON XT xr xr

CN V

CN vi

^

-

^

xr

nT

-

-

u

"^-

o o oo oo o o o o Vl o V> o V o o O vi CN m ON o o OO CO xf d o d vi NO O Cl NO xr o CN CN CN XI-

r-4

n-i

r-i

t-~

1

cn

m

v> r-i VI r- CO cs CN r-i

A3

r-i

O^

-

r-i

cn

r-'

Appendix

C

Xf

ro cs

fNl

o

u

—

xr

NO NO NO r-

xr

o

m Nf m NO m On m

CO

V-N

CO

NO'

ON NO xr CN Vl rn ON Cn|

Xf

xr r~

O m Vl Vl

869

A.3-7

Prandtl

Number

of Gases at

101325 kPa

(1

Aim

Abs)

Temperature

°c

°F

K

w2

°2

N

-17.8

2

CO

C0 1 0.775

0

255.4

0.720

0.720

0.720

0.740

0

32

273.2

0.715

0.711

0.720

0.738

0.770

10.0

50

283.2

0.710

0.710

0.717

0.735

0.769

37.8

100

311.0

0.700

0.707

0.710

0.731

0.764

65.6

150

338.8

0.700

0.706

0.700

0.727

0.755 0.752

93.3

200

366.5

0.694

0.703

0.700

0.724

121.1

250

394.3

0.688

0.703

0.696

0.720

0.746

148.9

300

422.1

0.683

0.703

0.690

0.720

0.738

176.6

350

449.9

0.677

0.704

0.689

0.720

0.734

204.4

400 450 500

477.6

0.670

0.706

0.688

0.720

0.725

505.4

0.668

0.702

0.688

0.720

0.716

533.2

0.666

0.700

0.688

0.720

0.702

232.2

260.0

*

: National Bureau of Standards, Circular 46\C, 1947; 564, 1955; NBS-NACA, Tables of Thermal Properties of Oases, 1949; F. G. Keyes, Trans. A.S.M.E., 73, 590, 597 (1951); 74, 1303 (1952); D. D. Wagman, Selected Values of Chemical Thermodynamic Properties. Washington, D.C. National Bureau of Standards, 1953.

Source

;

870

App.

A3

Physical Properties of Inorganic

and Organic Compounds

Temperature (°C)

Viscosity

(°F)

[(kg/m-s)10

-100I

3

or cp]

—

0.1

0.09

— - 100

-

0.08

-

0.07

0.06

0.05

30 100

28

0.04

26 100 — _

200 24 300

-

0.03

—

0.02

22

20 200

400 18

500

16

600

14

700

12

800

10

900

8

300

— 400

500

-0.01

1000 6

1100

600

1200

700 '

800

— 1300 — 1400

—t- 1500

900

1600

-

1700 1800

-

1000 Figure

-

A.3-2.

4

2

-

0.009

-

0.008

-

0.007

0

10

8

x

12

14

16

18

0.006

t- 0.005

Viscosities of gases at 101.325 kPa (1 aim abs). (From R. H. Perry and C. H. Chilton, Chemical Engineers' Handbook, 5th ed. New York: McGraw-Hill Book Company, 1973. Wilh permission.) See Table A. 3-8 for coordinates for use with Fig.

A.3-2.

Appendix A.

871

A.3-8

Viscosities of

Gases (Coordinates

X

Gas

No.

for

y

No.

Use with

Fig. A.3-2)

Acetic acid

7.7

14.3

29

Freon-1 13

2

Acetone

8.9

13.0

30

9.8

14.9

31

1.0

20.0

32

Helium Hexane Hydrogen

1

3

Acetylene

4

Air

1

16.0

33

3H 2 + 1N 2

34

Benzene

8.5

13.2

35

Bromine

8.9

19.2

Butene

9.2

13.7

36 37

7 8

9

14.0

20.5

8.6

22.4

Argon

1.3

1

8.4

Ammonia

y

10.9

1

10.5

5 ,6

X

Gas

1.2

1

1.8

12.4

1.2

17.2

8.8

20.9

chloride

8.8

18.7

cyanide

9.8

14.9

iodide

9.0

21.3

8.6

18.0

1

10

Butylene

8.9

13.0

38

Hydrogen Hydrogen Hydrogen Hydrogen Hydrogen

11

9.5

18.7

39

Iodine

9.0

18.4

8.0

16.0

40

5.3

22.9

9.9

15.5

bromide

sulfide

13

Carbon dioxide Carbon disulfide Carbon monoxide

11.0

20.0

41

14

Chlorine

9.0

18.4

42

Mercury Methane Methyl alcohol

15

Chloroform

8.9

15.7

43

16

Cyanogen

9.2

15.2

17

Cyclohexane

9.2

12.0

18

Ethane

9.1

14.5

19

Ethyl acetate

8.5

13.2

20

Ethyl alcohol

9.2

21

Ethyl chloride

8.5

22

Ethyl ether

23

Ethylene

12

8.5

15.6

Nitric oxide

10.9

20.5

44

Nitrogen

10.6

20.0

45

Nitrosyl chloride

8.0

17.6

Nitrous oxide

8.8

19.0

Oxygen

11.0

21.3

14.2

46 47 48

Pentane

7.0

12.8

15.6

49

Propane

9.7

12.9

8.9

13.0

50

Propyl alcohol

8.4

13.4

9.5

15.1

51

Propylene

9.0

13.8

24

Fluorine

7.3

23.8

52

Sulfur dioxide

9.6

17.0

25

Freon-1

10.6

15.1

53

Toluene

8.6

12.4

26

Freon-12

11.1

16.0

54

2,3,3-Trimethylbutane

9.5

10.5

27

Freon-21

10.8

15.3

55

8.0

16.0

28

Freon-22

10.1

17.0

56

Water Xenon

9.3

23.0

872

App. A.3

Physical Properties of Inorganic and Organic

Compounds

=cal/g.°C = btu/lb m .°F

cp

co

2

4.0

-

o

Temperature (°C) 0

(°F)

—

100

100

—

200

200

——

400

— 300

300 7

500

—

600

3

400 —

700 o nn

48

500

900

5

6

°n

700

900 1000

6>

O o

0I6

15

Ol7 17A 0 17C

1300

17BOo

1500 1600 1700 1800

O 17D

20

18

o

19

22

o

100

1200 1300

2000 2100

O

_j

23o o

y

25 21

1900 1

13

14

1100 1200 1400

800

12

10O o

1000

600

O 80

8

29 30

% 32

31

o

2200 2300 2400 2500

33

o

34 0

1400

35 i; -

0.1

0.09

36

o

0.08

— — t-

Figure

A.3-3.

0.06 0.05

Heal capacity of gases at constant pressure at 101.325 kPa (1 aim abs). (From R. H. Perry and C. H. Chilton, Chemical Engineers' Handbook, 5th ed. New York: McGraw-Hill Book Company, 1973. With permission.) See Table A .3-9 for use with Fig.

Appendix A.

0.07

A .3-3.

873

A.3-9 Heat Capacity of Gases at Constant Pressure (for Use with Fig. A. 3-3)

1

u

Range

Gas

No. TM£ir"*£» A lyiene Ace

l

0

11 Z1 1

1

1 c± y~i c±

Acetylene

1

Air All

U

7 Z

Ammonia Ammonia

U— OUU

R o

t^al DQI1

1

1 1

(OVl Crw

\^d.TDOU UIUaJUC

ZD

I m v in P Laroon munoxiuc *~i

iijonnc

~KA

Chlorine

Af\f\ ^tUU

ft ?AA u— zuu 70TU1 400 — 1 *+u\j zuu

7 J

f n o n y cifianc

Q

Ethane

9 0

Pt n ] np

4

cuiyiene

u— zuu

1

tiinyiene

AAA 7f\A ZUU— OvU

1 1

u— zuu 7AA_AAA zUU— DUU r>00-l 400

l—.

<jYi

J

7ft I/O 1

1

1

i

/~\ r\ i~\

t~ r~\ /~\ y~\

1? JZ

1

1 *-r\J\J

U HUU 40A 400-1I HUU HLAJ

UIOaIUC

94 Z*T

1

C)

U ZUU ?0A__/tAA ZUU —fUU 4AA AOC1

/-^ii»t

A y ACCl j JcjIc 1

(

7P

17A

Frprvn

1

1

^PHPIF

rreon-iij t I

z

JJ

zu

(v^^i 2 r

5

0 1

— c^jr

Hydrogen Hydrogen Hydrogen bromide Hydrogen chloride Hydrogen fluoride Hydrogen iodide nyurogen suinae Methane Methane Methane

1

t

2J

A u— /^AA DUU AAA 14UU /1AA DUU— A 14LKJ AA U— 1

1 /t

A 1400 AA U— A A f\r\ 1 /I

1

A_ 1 4UAJ A AA U— A_~7AA 0— /uu 7AA AA / UU— 1 ^tUU 1

1 /I

A jOO 1 AA 01 00 AA— /oo AA j "7

"7AA /UU—

/tAA 14UU t

0-700

25

Nitric oxide

28

Nitric oxide

26

Nitrogen

23 29

Oxygen Oxygen

33

Sulfur

22

Sulfur dioxide

31

Sulfur dioxide

17

Water

874

cn ft U— 1 JU A_ 1 JU SO U— ca A U— I JU 1

1

riyurogen suinoc Z. 1

4AA

1

rreun-zi ^v-ii^^r Frpr>n

I


(PF1 F^

App.

700-1400 0-1400 0-500 500-1400 300-1400 0-400 400-1400 0-1400

A3

Physical Properties of Inorganic

and Organic Compounds

Thermal Conductivities of Gases and Vapors at 101.325 kPa (1 Atm Abs); k = W/m K)

A.3-10

•

Gas or Vapor

,

Acetone' 11

Ammonia 12

Butane 13

'

'

Carbon monoxide'

Chlorine 14

2

'

'

(1)

k

239

0.0149

273

0.0183

k

273

0.0099

319

0.0130

373

0.0171

373

0.0303

457

0.0254

293

0.0154

273

0.0218

373

0.0215

373

0.0332

273

0.0133

473

0.0484

319

0.0171

273

0.0135

373

0.0227

373

0.0234

273

0.0175

173

0.0152

323

0.0227

273

0.0232

373

0.0279

373

0.0305

27?

0.0125

273

0.00744

293

0.0138

273

0.0087

373

0.0119

Gas or Vapor

Ethane' 5

-

6'

Ethyl alcohol'" Ethyl ether' 1

Ethylene'

'

6'

n-Hexane' 3

'

Sulfur dioxide'

Source:

K

K

Moser, dissertation, Berlin, 1913;

(2)

7'

F. G. Kcyes, Tech. Rept. 37, Project Squid,

1952; (3) W. B. Mann and B. G. Dickens, Proc. Roy. Soc. (London), A134, 77 (1931); (4) International Critical Tables. New York; McGraw-Hill Book Company, 1929; (5) T. H. Chilton and R. P. Genercaux, personal communication, 1946; (6) A. Eucken, Physik, Z., 12,

Apr.

1

1,

101(1911); 14, 324 (1913);

(7) B.

G. Dickens. Proc. Roy. Soc. (London), A143, 517 (1934).

Heat Capacities of Liquids

A. 3-11

Liquid

Acetic acid

Acetone Aniline

Benzene

Butane

(c

p

=

kJ/kg

Liquid

K

Hydrochloric acid (20 mol %)

273

2.43

293

2.474

293

0.01390

293

2.512

313

2.583

283

1.499 1.419

K 1.959

2.240

273

2.119

293

2.210

273

2.001

323

2.181

293

1.700

303

333

1.859

363

1.436

273

2.300

293

3.39

330

3.43

293

1.403

Mercury Methyl alcohol Nitrobenzene

Sodium

chloride

303

2.525

273

2.240

Sulfuric acid

298

2.433

Toluene

273

1.825

Glycerol

p

311

Ethyl alcohol

acid

c

273

/-Butyl alcohol

Formic

K)

289

2.131

288

2.324

305

2.412

o-Xylene

(9.1

(100%)

mol %)

273

1.616

323

1.763

303

1.721

Source : N. A. Lange, Handbook of Chemistry, 10th ed. New York; McGraw-Hill Book Company, ty67; National Research Council, International Critical Tables, Vol. V. New York: McGraw-Hill Book Company, 1929; R. H. Perry and C. H. Chilton, Chemical Engineers' Handbook, 5th ed. New York:

McGraw-Hill Book Company,

Appendix A.3

1973.

875

Temperature Viscosity

(°C)

200

(°F)

[(kg/m

—

•

s)10

390 380 190 370 - 360 180 - 350 340 170 - 330 160 — 320 150

3

-

3

60

50

40

1

jUU 290 140 - - 280 — 270 130 260 Ten

——

90 — 80

-

30

220 210 200

10 9

28

8

7

190

26

6 5

170 1 60

24

150

22

4 3

140

-

50

20

- _ 180

70 — 60

30

Z4U

no 100

or cp)

-100 90 80 70

20

130

—

120

40 -

18

16

100

y 90

30 -

14 -

80

20

-

70 -

10

-

60

0.9 0.8 0.7

10

0.6

50

0

—

- 10

-

40 30

1

12

0.5 8

-

0.4 6

-

4

•

H-

20

0.3

10 0.2

2

0

-20 -

-10 -30

-

4

-20

10

12

14

16

18

20

X

—

I

Figure

876

A.3-4.

0.1

Viscosities of liquids. (From R. H. Perry and C. H. Chilton, Chemical Engineers' Handbook, 5th ed. New York .-'McGraw-Hill Book Company, 1973. With permission.) See Table A. 3-1 2 for use with Fig. A.3A.

App. A.3

Physical Properties of Inorganic and Organic

Compounds

A. 3-12

Viscosities

of

Liquids (Coordinates for

X

Liquid

Acetaldehyde

Use with

y

Fig. A.3-4)

X

Liquid

Y

15.2

4.8

Cyclohexanol

2.9

24.3

12.1

14.2

Cyclohexane

9.8

12.9

9.5

17.0

Dibromomethane

12.7

15.8

Acetic anhydride

12.7

12.8

Dichloroethane

13.2

12.2

Acetone, 100%

14.5

7.2

Dichloromethane

14.6

8.9

7.9

15.0

Diethyl ketone

13.5

9.2

Diethyl oxalate

11.0

16.4

5.0

24.7 18.3

Acetic acid,

Acetic acid,

Acetone,

100% 70%

35%

Acetonitrile

14.4

7.4

Acrylic acid

12.3

13.9

Diethylene glycol

Diphenyl

12.0

Dipropyl ether

13.2

8.6

Dipropyl oxalate

10.3

17.7

Ethyl acetate

13.7

9.1

Ethyl acrylate

12.7

10.4

10.5

13.8

9.8

14.3

6.5

16.6 11.5

Allyl alcohol

10.2

14.3

Ally 1 bromide

14.4

9.6

Allyl iodide

14.0

11.7

Ammonia, 100% Ammonia, 26% Amyl acetate Amyl alcohol

12.6

2.0

10.1

13.9

11.8

12.5

Ethyl alcohol,

100%

7.5

18.4

Ethyl alcohol,

Aniline

8.1

18.7

Ethyl alcohol,

95% 40%

Anisole

12.3

13.5

Ethyl benzene

13.2

Arsenic trichloride

13.9

14.5

Ethyl bromide

14.5

8.1

Benzene

12.5

10.9

2-Ethyl butyl acrylate

11.2

14.0

Brine,

CaCL

25%

6.6

15.9

Ethyl chloride

14.8

6.0

10.2

16.6

Ethyl ether

14.5

5.3

14.2

13.2

Ethyl formate

14.2

8.4

20.0

15.9

2-Ethyl hexyl acrylate

9.0

15.0

Butyl acetate

12.3

11.0

Ethyl iodide

14.7

10.3

Butyl acrylate

11.5

12.6

Ethyl propionate

13.2

9.9

8.6

17.2

Ethyl propyl ether

14.0

7.0

Butyric acid

12.1

15.3

Ethyl sulfide

13.8

8.9

Carbon dioxide Carbon disulfide Carbon tetrachloride

11.6

0.3

Ethylene bromide

11.9

15.7

16.1

7.5

Ethylene chloride

12.7

12.2

12.7

13.1

Ethylene glycol

6.0

23.6

Chlorobenzene Chloroform

12.3

12.4

Ethylidene chloride

14.1

'8.7

14.4

10.2

13.7

10.4

Chlorosulfonic acid

11.2

18.1

Fluorobenzene Formic acid

10.7

15.8

Chlorotoluene, ortho

13.0

13.3

Freon-11

14.4

9.0

Chlorotoluene, meta

n;3

12.5

Freon-12

16.8

15.6

Chlorotoluene, para

13.3

12.5

Freon-21

15.7

7.5

2.5

20.8

Freon-22

17.2

4.7

2

,

NaCl, 25% Bromine Bromotoluene

Brine,

Butyl alcohol

Cresol,

meta

Appendix

A3

A. 3-12

Viscosities of Liquids,

X

Liquid

Freon-1 13 /—^

i

Continued

y

12.5

Liquid

1.4

Octyl alcohol

X

y

14 100%

6.6

21.1

2.0

30.0

Pentachloroethane

10.9

17.3

50%

6.9

19.6

Pentane

14.9

5.2

14.1

O A 8.4

6.9

20.8

13.8

16.7

r\i~\ r% t

Glycerol, Glycerol,

Heptane Hexane

1

A

1

1

Phenol

14.7

7.0

Hydrochloric acid, 31.5%

13.0

16.6

Phosphorus tribromide Phosphorus trichloride

16.2

10.9

~ A« lodobenzene

1

15.9

Propionic acid

12.8

13.8

IT T

J

„1

|

_

•

•

t

1

1

Isobutyl alcohol

CO/

1 O

1

O

(\

7.1

18.0

Propyl acetate

13.1

10.3

12.2

14.4

Propyl alcohol

9.1

16.5

o o 8.2

16.0

Propyl bromide

14.5

9.6

Isopropyl bromide

14.1

9.2

Propyl chloride

14.4

7.5

Isopropyl chloride

13.9

7.1

Propyl formate

13.1

9.7

Isopropyl iodide

13.7

11.2

Propyl iodide

14.1

11.6

Kerosene

10.2

16.9

13.9

27.2

Sodium Sodium hydroxide, 50%

16.4

7.5

3.2

25.8

Mercury

18.4

16.4

Stannic chloride

13.5

12.8

Methanol, 100% Methanol, 90%

12.4

10.5

Succinonitrile

10.1

20.8

12.3

11.8

Sulfur dioxide

7.8

15.5

Sulfuric acid,

14.2

8.2

Sulfuric acid,

13.0

9.5

Sulfuric acid,

12.3

9.7

Sulfuric acid,

13.2

10.3

15.0

Isobutyric acid

Isopropyl alcohol

Linseed

oil,

Methanol,

raw

40%

Methyl acetate Methyl acrylate Methyl i-butyrate Methyl n-butyrate Methyl chloride Methyl ethyl ketone Methyl formate Methyl iodide

15.2

7.1

110% 100%

7.2

27.4

8.0

25.1

98% 60%

7.0

24.8

10.2

21.3

Sulfuryl chloride

15.2

12.4

3.8

Tetrachloroethane

11.9

15.7

13.9

8.6

13.2

11.0

14.2

7.5

14.4

12.3

13.7

10.4

14.3

9.3

Thiophene Titanium tetrachloride Toluene

Methyl propionate Methyl propyl ketone

13.5

9.0

Trichloroethylene

14.8

10.5

14.3

9.5

Triethylene glycol

4.7

24.8

Methyl sulfide Naphthalene

15.3

6.4

Turpentine

11.5

14.9

7.9

18.1

Vinyl acetate

14.0

8.8

12.8

13.8

Vinyl toluene

13.4

12.0 13.0

Nitric acid,

95% 60%

10.8

17.0

Water

10.2

Nitrobenzene

10.6

16.2

Xylene, ortho

13.5

12.1

Nitrogen dioxide

12.9

8.6

Xylene, meta

13.9

10.6

Nitrotoluene

11.0

17.0

Xylene, para

13.9

10.9

Octane

13.7

10.0

Nitric acid,

878

App. A .3

Physical Properties of Inorganic and Organic

-

Compounds

c

p

= cal/g

°C =

•

Temperature •0.2

(°F)

(°C)

1
200No

Acetic Acid

29 32

-350

150

-E

-300

A

37 26 30 23 27 10 49

Al

hi

Orbon

Tetraehior Ide

CMphenyl OxitJe

Dowlherm A

" "

1

7

39

ao 30

120 100

0

200 200

-

5

-30 -100

Chloride Ether iodide Ethylene Glycol

13 36

0 0 so -30 -40

30 20 20

Banzene Bromide

60

60 100 50 25 60 50

-50

100% 95% 50%

0.3

50

30 20

0

Ethyl Acetate Alcohol

4°0 4A

20 1QO 25

-100 10

Dlphenylmelhane

42 46 5o 25

10

-40 0

Dlcnloroathane Olchlorome thane Ol phenyl

5

15 22 16 16 24

2 5

130 60

-4 0

Chlorob«nzane Chloroform Decane

6A

-50

-30

25% C*Cl2 25% NaCI Butyl Alcohol C*rbon DUuHld* Brine, Brln*.

51

44

2)

-200

1

O o o 3A

80 50 50

Q

20

Benzyl AlcohoJ B«nzyl Chloride

3 a

100-

10O%

Acetone

2A 2C-

9

52

2

•250

* n9<

Liquid

-

0

-

-40

•

06A 9

Q

OlO 0.4

n O\oI2 Ol3A O13

25 80 80 ao 100 25 40 25 100 200

22ool7

19

OZ1

Z

024

0.5

O

150

28p

33

41

50-

CD 44°

100

n 43

46°

48

O

O

40

047

Q49 •50

NO

2A 6

Freon-ll(CCl3F)

"

-12(CCI 2 F 2 )

-21(CHCI2F) -22fCHCIF2) -113(CCl2F-CCIF2)

4A 7A 3A

-50 -50-

38 28 35 48 41 43 47 31 40

13A 14 12 34

33 3

ioo

45 20 9 11

23 53 19 18 17

-100-

Glycerol

Beptana Hezane Hydrochloric Acid, lioamyl Alcohol liobutyl Alcohol Uopropyl Alcohol Uopropyl Ether Methyl Alcohol Methyl Chloride Naphthalene Nitrobenzene

Nona no Octane Perchl or ethylene

Propyl Alcohol Pyridine Sulfuric Acid 98% Suilur Dioxide

Toluene Water Xylene Ort? rel="nofollow">D Meta '*

P3r*

-0.7

Range Dog C

Liquid

30%

70 15 70 60 70 20 0 - 60 -80- 20 20 - 100 10 - 100 0 - 100 -20 50 -80 - 20 -40 - 20 ao - 20 90 200 0 100 -50 25 -50 25 -30 140 -20 IOO -50 25 10 45 -20 100 60 0 -20 -40 -20 -20 -20 -40

10 0 0

0

-

O50

51,

-0.8

0.9

200 100 100 100

52

O

53

O

1.0

FIGURE A. 3-5.

Heat capacity of liquids. (From R. H. Perry and C. H. Chilton, Chemical Engineers' Handbook, 5th ed. New York : McGraw-Hill Book Company, 1973. With permission.)

A3

879

Appendix

Thermal Conductivities of Liquids

A. 3-13

Liquid

Ar*f»tif q

nn

= W/m K)* •

Liquid

K

A.

F h vlf*np t

100% S0%

0.171

nivrernl

ol vr*r»l 1

AIS

K

273

0 765

00%

0 ?R4 0

m

A rvi i~i n i o /AlllillUIllct

0 SO?

n-Amyl alcohol Benzene

Carbon

(k

tetrachloride

/i-Decane

Ethyl acetate

303

0.163

Kerosene

1

"{R

jjj

U.l JJ

293

0.149

348

0.140

373

0.154

303

0.159

333

0.151

100%

293

0.215

273

0.185

293

0.329

341

0.163

60% 20%

293

0.492

303

0.147

100%

323

0.197

333

0.144

293

0.175

Methyl alcohol

rj-Octane

NaCl

Ethyl alcohol

303

0.144

333

0.140

brine

100%

293

0.182

25%

303

0.571

60% 20%

293

0.305

12.5%

303

0.589

293

0.486

100%

323

0.151

90% 60%

303

0.364

303

0.433

Vaseline

332

0.183

Sulfuric acid

* A linear variation with temperature may be assumed between the temperature limits given. Source: R. H. Perry and C. H. Chilton, Chemical Engineers' Handbook, 5th ed. New York:

McGraw-Hill Book Company,

880

1973.

App.

With permission.

A3

Physical Properties of Inorganic

and Organic Compounds

3

Heat Capacities of

A.3-14

Solid

Solids (c p

K

r L P

=

kJ/kg

•

K)

Solid

Benzene Benzoic acid

273

1.147

293

1.243

1.05

Camphene

308

1.591

0.92

Caprylic acid

271

2.629

0.829

Dextrin

273

1.218

^ AO 1.248

Formic acid

273

1.800

Cement, portland

0.779

Glycerol

273

1.382

Clay

0.938

Lactose

293

1.202

Concrete

0.63

Oxalic acid

323

1.612

0.167

Tartaric acid

309

1.202

0.84

Urea

293

1.340

Alumina

373

1773 Asbestos Asphalt Brick, fireclay

373 1773

Corkboard

303

Glass

Magnesia

0.84

1

373

0.980

1773

0.787

Oak

1.570

2.39

Pine, yellow

Porcelain

298

293-373

2.81

0.775

Rubber, vulcanized

2.01

Steel

0.50

Wool

1.361

Source: R. H. Perry and C. H. Chilton, Chemical Engineers' Handbook, 5th ed. New York: McGraw-Hill Book Company, 1973; National Research Council, International Critical Tables, Vol. V. New York: McGraw-Hill Book Company, 1929; L. S. Marks, Mechanical Engineers' Handbook, 5lh cd. New York: McGraw-Hill Book Company, 1951; F. Krcith, Principles of Heat Transfer, 2nd ed. Scranton, Pa.: Internationa!

Textbook Co., 1965.

Appendix

A3

Thermal Conductivities of Building and Insulating Materials

A.3-15

P

k(W/m-K)

Material

Asbestos

577

Asbestos sheets

889

Brick, building

0.151 (0°C) 51

0.166

20

0.69 1.00 (200° C)

Brick, fireclay

Clay

soil,

4% H 2 0

Concrete,

1

:4

4.5

dry

Corkboard

30

1.47 (600°C)

1.64 (1000°C)

0.061 (37.8°C)

0.068 (93.3°C)

0.57

0.0433 0.055 (0°C)

80.1

wool

0.19O (93.3°C)

0.762 160.2

Cotton Felt,

1666

0.168 (37.8°C)

330

30

0.052

237

21

0.048

Fiber insulation

board Glass,

window

0.52-1.06

Glass wool

64.1

0

921

Ice

85%

Magnesia,

0.0310 (-6.7°C) 0.0414 (37.8°C) 0.0549 (93.3°C)

30

2.25

271

0.068 (37.8°C)

0.071 (93.3°C)

0.080 (204.4°C)

208

0.059 (37.8°C)

0.062 (93.3°C)

0.066 (148.9°C)

Oak, across grain

825

15

0.208

Pine, across grain

545

15

0.151

Paper

0.130

Rock wool

0.0317 (-6.7°C) 0.0391 (37.8°C) 0.0486 (93.3°C)

192

0.0296 (-6.7°C) 0.0395 (37.8°C) 0.0518 (93.3°C)

128

Rubber, hard

1198

0

1826

4.5

1.51

2

1922

4.5

2.16

Sandstone

2243

40

1.83

559

0

0.47

Sand

0.151

soil

4% H 2 0 10% H 0 Snow Wool

110.5

30

0.036

Room temperature when none is noted. Source: L. S. Marks, Mechanical Engineers' Handbook, 5th ed. New York: McGraw-Hill Book Company, 1951; W. H. McAdams, Heal Transmission, 3rd ed. New York: McGraw-Hill Book Company. 1954; F. H. *

Norton, Refractories. tional Critical Tables. Sia., Bull. 28,

882

New York: McGraw-Hill Book Company, 1949; National Research Council, InternaNew York: McGraw-Hill Book Company, 1929; M. S. Kerslen, Univ. Minn. Eng. Ex.

June 1949; R. H. Heilman,

App.

A3

Ind. Eng. Chem., 28, 782 (1936).

Physical Properties of Inorganic and Organic

Compounds

A. 3-16

Thermal

Conductivities, Densities, and

Heat

Capacities of Metals

kiW/mK)

Material

Aluminum

20

2707

0.896

202 (0°C)

206 (100°C)

215 (200°Q

230 (300°C) Brass (70-30)

20

8522

0.385

97 (0°C)

104 (100°C)

109 (200°C)

Cast iron

20

7593

0.465

55 (0°C)

52 (100°C)

48 (200°C)

Copper Lead

20

8954

0.383

388 (0°C)

377 (100°C)

20

11370

0.130

35 (0°C)

33 (100°C)

372 (200°C) 31 (200°C)

20

7801

0.473

45.3 (18°C)

45 (100°C)

45 (200°C)

Steel

1%C

43 (300°C) 308 stainless

20

7849

0.461

15.2 (100°C)

304 stainless

0

7817 7304

0.461

13.8 (0°C)

16.3 (100°C)

18.9 (300°C)

0.227

62 (0°C)

59 (100°C)

57 (200°C)

Tin

20

21.6 (500° C)

Source: L. S. Marks, Mechanical Engineers' Handbook, 5lh ed. New York: McGraw-Hill Book Company, 1951; E. R. G. Eckert and R. M. Drake, Heat and Mass Transfer, 2nd ed. New York: McGraw-Hill Book Company, 1959; R. H. Perry and C. H. Chilton, Chemical Engineers' Handbook, 5lh ed. New York: McGraw-Hill Book Company, 1973; National Research Council, International Critical Tobies. New York: McGraw-Hill Book Company, 1929.

Appendix A J

883

Normal Total Emmissivities

A. 3- 17

K

Surface

of Surfaces

Surface

E

Aluminum

K

£

Lead, unoxidized

400

0.057

highly oxidized

366

0.20

Nickel, polished

373

0.072

highly polished

500

0.039

Nickel oxide

922

0.59

850

0.057

Oak, planed

294

0.90

550

0.63

Paint

Asbestos board

296

0.96

aluminum

373

0.52

Brass, highly

520

0.028

oil (16 different,

polished

630

0.031

Paper

373

0.075

Roofing paper

294

0.91

Rubber

296

0.94

Aluminum

oxide

all

colors)

373

0.92-0.96

292

0.924

Chromium, polished

Copper

(hard, glossy)

oxidized

298

0.78

polished

390 295

0.023

oxidized at 867

472

0.79

0.94

polished stainless

373

0.074

304 stainless

489

0.44

oxidized

373

0.74

273

0.95

tin-plated

373

0.07

373

0.963

772

0.85

Glass,

smooth

Steel

Iron

Iron oxide

K

Water

Source: R. H. Perry and C. H. Chilton, Chemical Engineers' Handbook, 5th ed. New York." McGraw-Hill Book Company, 1973; W. H. McAdams, Heat Transmission, 3rd ed. New York:

McGraw-Hill Book Company, 1954;

A. 3-18

Law

Henry's

E. Schmidt, Gesundh.-Ing. Beiheft, 20, Reihe

Constants

for

Gases

in

Water {H x 10

1,

1

(1927).

")*

f

K

=

C

CO c 2 w 6

CO,

C2 H A

H

He

2

S

CH i

N2

o2

273.2

0

0.0728

3.52

1.26

0.552 12.9

5.79

0.0268

2.24

5.29

283.2

10

0.104

4.42

1.89

0.768 12.6

6.36

0.0367

2.97

6.68

3.27

293.2

20

0.142

5.36

2.63

1.02

12.5

6.83

0.0483

3.76

8.04

4.01

303.2

30

0.186

6.20

3.42

1.27

12.4

7.29

0.0609

4.49

9.24

4.75

313.2

40

0.233

6.96

4.23

12.1

7.51

0.0745

5.20

, p A = partial pressure of A in gas in atm, x A = mole fraction of A Henry's law constant in atm/mole frac. Source: National Research Council, International Critical Tables, Vol. III. New York:

*

pA

= Hx A

10.4

in

2.55

5.35

liquid,

H= Hill

884

Book Company,

McGraw-

1929.

App.

A3

Physical Properties of Inorganic

and Organic Compounds

Equilibrium

A.3-19

Data

for

S0 2-Water

System

Partial Pressure of S0 2 in

Mole Fraction SO 2

Vapor, p A (mm Hg)

in

Vapor, y A

;

P=

1

Atm

Mole Fraction

S0 2

in Liquid,

xA

20°C (293 K)

0

0.0000562

30°C (303 K)

30°C

20°C

0

0

0

0

0.5

0.6

0.000658

0.000790

0.0O01403

1.2

1.7

0.00158

0.0O223

0.000280

3.2

4.7

0.00421

0.00619

0.000422

5.8

8.1

0.00763

0.01065

0.000564

8.5

11.8

0.01120

0.0155

0.000842

14.1

19.7

0.01855

0.0259

0.001403

26.0

36

0.0342

0.0473

0.001965

39.0

52

0.0513

0.0685

0.OO279

59

79

0.0775

0.1

0.00420

92

125

0.121

0.1645

040

0.OO698

161

216

0.212

0.284

0.01385

336

452

0.443

0.594

0.0206

517

688

0.682

0.905

0.0273

698

Source

:

T. K.

A. 3-20

Sherwood,

lnd.

0.917

Eng. Chem.,

i

7,

745 (1925).

Equilibrium Data for Methanol-Water System Partial Pressure of in

Methanol

Vapor, p A (mm Hg)

Mole Fraciion Methanol

Source

:

in

Liquid, x A

—

39.9°C (313.1 K)

59.4°C (332.6K)

0

0

0

0.05

25.0

50

0.10

46.0

102

0.15

66.5

151

National Research Council, International Critical Tables, Vol.

III.

New York: McGraw-Hill Book Company;i929.

Appendix A.

885

Equilibrium Data for Acetone-Water System

A.3-21

at

20°C (293 K)

Mole Fraction Acetone

in

Liquid,

Partial Pressure of Acetone

xA

in

Vapor, p A (mm Hg)

0

0 0.0333

30.0

0.0720

62.8

0.117

85.4

103

0.171

Source: T. K. Sherwood, Absorption and Extraction. McGraw-Hill Book Company, 1937. With permission.

A. 3-22

Equilibrium Data for

York:

Ammonia-Water System

Partial Pressure of

Vapor, p A

in

New

NH^

Mole Fraction

(mm Hg)

Vapor, y A

;

NH^

P=

/

in

Aim

Mole Fraction in Liquid,

xA

0

20°C (293 K) 0

0.0126

886

0

0

30°C

0 0.0151 0.0201

15.3

0.0208

12

19.3

0.0158

0.0254

0.0258

15

24.4

0.0197

0.0321

0.0309

18.2

29.6

0.0239

0.0390

0.0405

24.9

40.1

0.0328

0.0527

0.0503

31.7

51.0

0.0416

0.0671

0.0737

50.0

79.7

0.0657

0.105

0.0960

69.6

110

0.0915.

0.145

114

179

0.150

0.235

0.175

166

260

0.218

0.342

0.210

227

352

0.298

0.463

0.241

298

454

0.392

0.597

0.297

470

719

0.618

0.945

0.137

1963.

20°C

11.5

0.0167

Source

30°C (303 K)

: J.

H. Perry, Chemical Engineers' Handbook, 4th ed.

New

York: McGraw-Hill Book Company,

With permission.

App. A.3

Physical Properties of Inorganic

and Organic Compounds

o

rn

S/-I

no
c-4

o

m

CO

rsi

(N

o

c S

o

a c

3

On ro6 t—

m

O ro

ON

r— •A oo <* oo vi

o o

D 6

r-'

cn>

a o

a.

o

o

r-'

oo ON ON s/S tN oo >o ro (N

NO NO ON ON OO

a

-a-

C

5

rr NO 3; NO C-l oo ON CO NO VI CN

o

^

E

Cr

s a i-

o

O

o

V) ON O o o m O d d d

pO

vi NO -
o)

s

r-'

(N

5

u

O

o

o

OO rrn CO r-i ON CO oo CO r-

d

m oo"

o

Z r-i oo r-l On oo vn r~ On (N vi r- oo oo On OS ON ON

NT

d d dd d d d 7^ .2 '3 M

o o o o o oo o o o d d d d d d

-<* NO SO r— oo ON ON ON

CO
p

co r- oo

p

r-i

r-i

r-'

NO NT

r-

r- r- r- r- r~ r- r-

p d CO oo

r-i

r->

r-4

Si

o

"1

6

S

"N

K|

J

r->

CO CO CO CO oo r- r- r- r- r- rcri

U s-

O ON — —

r-l

t— r~ VO rn NO r— On r
~

2

O

Cr It}

m d d d d d d d

f «"5 3 « o.

o o o o o o o o o o o o O
r-i

c

& c

sn ^r rs) rsj VI r~ NO ON CO r-i r- ON ON CO oo oo

rs

o O

>, ^>

^

o c '

I 2

*

.

O

»»Z I3i

p

8

CN OO r-i (N r-i rs oo vi ON ON ON oo oo oo oo

887

Acetic Acid-VVater-Isopropyl Ether System,

A.3-24

Liquid-Liquid Equilibria at 293

Water Layer

(wt

K

or

20°C

Isopropyl Ether Layer (wt

%)

%)

/ sopropyl

Acetic Acid

Water

0

Ether

Acetic Acid

Water

0

Ether

98.8

1.2

0.6

99.4

0.69

98.1

1.2

0.18

0.5

99.3

1.41

97.1

1.5

0.37

0.7

98.9

2.89

95.5

1.6

0.79

0.8

98.4

6.42

91.7

1.9

1.93

1.0

97.1

13.30

84.4

2.3

4.82

1.9

93.3

25.50

71.1

3.4

11.40

3.9

84.7

36.70

58.9

4.4

21.60

6.9

71.5

44.30

45.1

10.6

31.10

10.8

58.1

46.40

37.1

16.5

36.20

15.1

48.7

Source

Trans. AJ.Ch.E., 36, 601, 628 (1940). With permission.

:

A. 3-25

Liquid-Liquid Equilibrium Data for

Acetone-Water-Methyl Isobutyl Ketone (MIK) System at 298-299 K or 25-26°C Composition Data

Acetone Distrihut ion Data

(m %)

MIK

Acetone

{wt

Water

Water Phase

%)

MIK

Phase

98.0

0

2.00

2.5

4.5

93.2

4.6

2.33

5.5

10.0

77.3

18.95

3.86

7.5

13.5

71.0

24.4

4.66

10.0

17.5

65.5

28.9

5.53

12.5

21.3

54.7

37.6

7.82

15.5

25.5

46.2

43.2

10.7

17.5

28.2

12.4

42.7

45.0

20.0

31.2

5.01

30.9

64.2

22.5

34.0

3.23

20.9

75.8

25.0

36.5

94.2

26.0

37.5

2.12

2.20

3.73

0

97.8

Source: Reprinted with permission from D. F. Othmer, R. E. White, and E. Trueger, lnd. Eng. Chem., 33, 1240(1941). Copyright by the American'chemical Society.

888

App. A. 3

Physical Properties of Inorganic and Organic

Compounds

APPENDIX

A.

Physical Properties

of Foods and Biological Materials

Heat Capacities of Foods (Average cp 273-373 K or 0-1 OOX)

A.4-1

H20 Material

Apples

cp

%)

(U/kg K)

75-85

3.73-4.02

(wt

4.02*

Apple sauce Asparagus Fresh

93

3.94f

Frozen

93

2.01t

Bacon, lean

51

3.43

Banana puree

3.66§

72 44-45

Beef, lean

Bread, white

3.43

2.72-2.85

Butter

15

Cantaloupe

92.7

2.30H 3.94|

Cheese, Swiss

55

2.68|

Corn, sweet Fresh

3.32f

Frozen

Cream, 45-60%

Cucumber

1.77t fat

57-73 97

3.06-3.27 4.10

Eggs Fresh

3.18f

Frozen

1.68J

Fish,

cod

Fresh

70

Frozen

70 12-13.5

Flour

100

Ice

Appendix

A .4

3.18

1.72J 1.80-1.88

1.958H

889

Continued

A.4-1

H 0 2

Material Ice

(kJ/kg-K)

%)

{wt

cream

Fresh

58-66-

3.27t

Frozen

58-66

1.88J 3.18*

Lamb

70

Macaroni

1.84-1.88

12.5-13.5

Milk, cows'

Whole Skim Olive

3.85

87.5

3.98-4.02.

91

2.01**

oil

Oranges Fresh

87.2

3.77t

Frozen

87.2

1.93J

14

1.84

Fresh

74.3

3.31t

Frozen

74.3

1.76J 4.10

Peas, air-dried

Peas, green

Pea soup

Plums Pork

75-78

3.52

Fresh

60

Frozen

60

1.34 J

75

3.52

Fresh

74

3.31|

Frozen

74

1.55J

Potatoes

2.85|

Poultry

Sausage, franks

Fresh

60

3.60t

Frozen

60

2.35J

Fresh

88.9

3.8 It

Frozen

88.9

1.97J

95 63

3.98t 3.22

100

4.185

String beans

Tomatoes Veal

Water *

32.8°C.

t

Above

freezing.

J

Below

freezing.

§

24.4°C.

1i

4.4°C.

II

-20-C

** 20°C.

Source: W. O. Ordinanz, Food lnd„

18, 101 (1946);

G. A.

Rcidy, Department of versity, 1968; S. E.

eering,

1971;

2nd R.

ed.

L.

Food Science, Michigan Stale UniCharm, The Fundamentals of Food Engin-

Westporl, Conn.: Avi Publishing Co.,

Earle,

Unit -Operations

in

Food

Inc..

Processing.

Oxford: Pergamon Press, Inc, 1966; ASHRAE, Handbook New York: ASHRAE, 1972, 1967; H. C. Mannheim, M. P. Steinberg, and A. I. Nelson, Food

of Fundamentals.

Technol.,9, 556(1955).

890

App. A.4

Physical Properties of Foods and Biological Materials

Thermal

A.4-2

Conductivities,

Densities,

and

Viscosities

of

Foods

Temp-

H 0 2

Material

[wt

%)

Apple sauce

k

(X)

(W/m-K)

295.7 15

Butter

erature

277.6

Cantaloupe

p (kg/m 3)

p.

[(Pa s)J0\ or cp]

0.692

998

0.197 0.571

Fish

Fresh

273.2

Frozen

263.2

Flour, wheat

Honey

Lamb

1.22

0.450

8.8

275.4

0.50

,100

273.2

2.25

100

253.2

2.42

71

278.8

0.415

12.6

Ice

0.431

Milk

Whole Skim

1030

2.12

298.2

1041

1.4

298.2

924

293.2

274.7

0.538

Oil

Cod

liver

Corn

288.2

Olive

293.2

0.168

Peanut

277.1

0.168

Soybean Oranges

921

303.2 61.2

Pears

303.5

0.431

281.9

0.595

275.4

0.460

258.2

3.109

919

84

919

40

Pork, lean

Fresh

74

Frozen Potatoes

Raw

0.554

Frozen

260.4

1.09

977

Salmon Fresh

67

277.1

0.50

Frozen

67

248.2

1.30

80

294.3

74

276.0

0.502

248.2

1.675

0.485

Sucrose solution

1073

1.92

Turkey Fresh

Frozen Veal Fresh

75

335.4

Frozen

75

263.6

1.30

100

293.2

0.602

100

273.2

0.569

Water

Source: R. C. Weasl, Handbook of Chemistry and Physics, 48th ed. Cleveland: Chemical Rubber Co., Inc., 1967; C. P. Lentz, Food Techno!., 15, 243 (1961); G. A. Reidy, Department of

Food

Science, Michigan Stale University, 1968; S. E.

Engineering, 2nd ed. Westport, Conn.; Avi Publishing

Charm, The Fundamentals of Food

Co,

Inc.,

1971

;

R. Earle, Unit Operations

Food Processing, Oxford: Pergamon Press, 1966; R. H. Perry and C. H. Chilton, Chemical Engineers' Handbook, 5th ed. New York: McGraw-Hill Book Company, 1973; V. E. Sweat, J. in

FoodSci., 39, 1080(1974).

Appendix A.4

APPENDIX

A.

Properties of Pipes,

Tubes, and Screens

A.5-1

Dimensions of Standard Steel Pipe

Nominal

Outside

Pipe

Diameter

Size (in.)

1

I

4

3

1 1

2

3

4

1

I*

Wall

Inside

Inside Cross-

Thickness

Diameter

Sectional Area

ule in.

0.405

0.540

0.675

0.840

mm 10.29

13.72

17.15

21.34

mm

mm

m

2

2

Number

in.

40

0.068

1.73

0.269

6.83

0.00040

0.3664

80

0.095

2.41

0.215

5.46

0.00025

0.2341

40

0.088

2.24

0.364

9.25

0.00072

0.6720

80

0.119

3.02

0.302

7.67

0.00050

0.4620

40

0.091

2.31

0.493

12.52

0.00133

1.231

80

0.126

3.20

0.423

10.74

0.00098

0.9059

40

0.109

2.77

0.622

15.80

0.00211

1.961

80

0.147

3.73

0.546

13.87

0.00163

1.511

20.93

0.00371

3.441

2.791

in.

ft

x 70*

1.050

26.67

40

0.113

2.87

0.824

80

0.154

3.91

0.742

18.85

0.00300

1.315

33.40

40

0.133

3.38

1.049

26.64

0.00600

5.574

80

0.179

4.45

0.957

24.31

0.00499

4.641

40

0.140

3.56

1.380

35.05

0.01040

9.648

80

0.191

4.85

1.278

32.46

0.00891

1.610

40.89

0.01414

1.660

42.16

8.275

1.900

48.26

40

0.145

3.68

80

0.200

5.08

1.500

38.10

0.01225

11.40

2

2.375

60.33

40

0.154

3.91

2.067

52.50

0.02330

21.65

80

0.218

5.54

1.939

49.25

0.02050

19.05

2{

2.875

73.03

40

0.203

5.16

2.469

62.71

0.03322

30.89

80

0.276

7.01

2.323

59.00

0.02942

27.30

3

3.500

88.90

40

0.216

5.49

3.068

77.92

0.05130

47.69

3i

4.000

4

4.500

101.6

114.3

13.13

0.300

7.62

2.900

73.66

0.04587

42.61

40

0.226

5.74

3.548

90.12

0.06870

63.79

80

0.318

8.08

3.364

85.45

0.06170

57.35

40

0.237

6.02

4.026

0.08840

82.19

80

0.337

8.56

3.826

102.3

97.18

0.07986

74.17

5

5.563

141.3

40

0.258

6.55

5.047

128.2

0.1390

129.1

80

0.375

9.53

4.813

122.3

0.1263

117.5

6

6.625

168.3

40

0.280

7.11

6.065

154.1

0.2006

186.5

80

0.432

10.97

5.761

146.3

0.1810

168.1

40

0.322

8.18

7.981

202.7

0.3474

322.7

80

0.500

12.70

7.625

193.7

0.3171

294.7

8

892

Sched-

8.625

219.1

App. A. 5

Properties of Pipes, Tubes,

and Screens

A.5-2

Dimensions of Heat-Exchanger Tubes

Outside

Wall

Inside

Inside Cross-

Diameter

Thickness

Diameter

Sectional Area

BWG in.

5 "5

3

4

7

Number

in.

mm

15.88

12

0.109

14

0.083

16

19.05

22.23

25.40

1

,

mm

mm

2.77

0.407

10.33

0.000903

0.8381

2.11

0.459

11.66

0.00115

1.068

0.065

1.65

0.495

12.57

0.00134

1.241

18

0.049

1.25

0.527

13.39

0.00151

1.408

12

0.109

2.77

0.532

13.51

0.00154

1.434

14

0.083

2.11

0.584

14.83

0.00186

1.727

16

0.065

1.65

0.620

15.75

0.00210

1.948

18

0.049

1.25

0.652

16.56

0.00232

2.154

12

0.109

2.77

0.657

16.69

0.00235

2.188

14

0.083

2.11

0.709

18.01

0.00274

2.548

16

0.065

1.65

0.745

18.92

0.00303

2.811

18

0.049

1.25

0.777

19.74

0.00329

3.060

10

0.134

3.40

0.732

18.59

0.00292

2.714

12

0.109

2.77

0.782

19.86

0.00334

3.098

14

0.083

2.11

0.834

21.18

0.00379

3.523

16

0.065

1.65

0.870

22.10

0.00413

3.836

10

0.134

3.40

0.982

24.94

0.00526

4.885

12

0.109

2.77

1.032

26.21

0.00581

5.395

14

0.083

2.11

1.084

27.53

0.00641

5.953

16

0.065

1.65

1.120

28.45

0.00684

6.357

10

0.134

3.40

1.232

31.29

0.00828

7.690

12

0.109

2.77

1.282

32.56

0.00896

8.326

14

0.083

2.11

1.334

33.88

0.00971

9.015

10

0.134

3.40

1.732

43.99

0.0164

15.20

12

0.109

2.77

1.782

45.26

0.0173

16.09

2

fl

m2

x /0*

i

U

31.75

38.10

2

in.

50.80

Appendix A.S

893

A3-3

Tyler Standard Screen Scale

Nominal Wire Diameter

Sieve Opening

in.

in.

Tyler

[approx.

(approx.

Equivalent

equivalents)

Designation

mm

equivalents)

aa LOO

oa n

7

1

O< A 25.4

on

U.l

jjj

I.UjU

in.

U.l

j /o o on

u.ooj

in.

/4z

in.

C\f\

j.oU

ZZ.o

0.O /5

J.JKJ

iy.o A C\ lo.O

A O" C A 0.750 A AO C 0.625

1 3.30

T.A

ft

1

n

AA

A

1

O

~7

1

1

ACTA 0.530

13.5 1 o n 1

1

")

1.2

A C1 9.51

O T7 2.2 / o 2.0 /

m

1

8.00 A n 6.73 "i

6.35

AA

1

n /

0.1

1

1.10

A AAT7 0.0937

1.00

1

1

1

1.00

Ao

1.41

0.0555

1.19

0.0469

1

1

AA

0.841

0.0394 A An 0.0331

0.707

A AO TO 0.0278

1.00

A OA

1

i

0.595 A C AA 0.500 A A "% A 0.420 A CA 0.354

0.0234

0.297

0.01 17 A AAA O

0.0197 A A £C 0.0165 A A 1A 0.0139 1

A AO 15< 0.08 A ATT/C 0.0/ 36 1

A A/C A/C U.U606 A ncifi 0.0539

1.3

0.0787 A f\C£ 0.0661

in.

<.A 1.54

z.3o

aa

0.441

A AiC/C 1 U.0661

z.o3

Z.00

A AO./C < 0.0V65 A AOA/i U.U894

AO

0.132

**)

in.

1

1

o0 1.23

3.36

u.jzd

t

0")

1

1

1

A

AA

A AAA o.yoo A O A 0.810 A TO C 0. /25 A £ CA 0.650 1

0.580 A C A 0.510 A A CA 0.450 1

A AA 0.390 A AA 0.340 ")

")

A A/t OA 0.0484 A f\A "3A 0.0430 A A'JA/I 0.U394 A A1 C ^ 0.0354 A A1 A 0.0319 A AT O C 0.0285 1

0.0256 A A'l O O 0.0228 0.0201 A A T7 0.0177 i

0.0154 A A 1A 0.01 34 1

AA

1

1

A

0.210

0.0083

0.152

0.0060

0.177

0.0070

0.131

0.0052

0.149

0.0059

0.110

0.0043

0.125

0.0049

0.091

0.0036

0.105

0.0041

0.076

0.0030

0.088

0.0035

0.064

0.0025

0.074

0.0029

0.053

0.0021

0.063

0.0025

0.044

0.0017

0.053

0.0021

0.037

0.0015

0.044

0.0017

0.030

0.0012

0.037

0.0015

0.025

0.0010

894

1

1

mesh mesh

3-j

mesh

1

0.0098

0.250

""i

3

2-j

in

0.01 14

1

0.3/1 in.

A AT H o.o

0.290 A AH 0.247 AO C 0.215 A OA 0.180

~}

A AO/1 i« u.oz4 in.

U.IUoj a a^ U. 1U51

1

/

1.00

1

4.00

Q1 0

Q~7

1.82

1

4.76

1

U. 1 J

1

1.0

")

5.66

/I

< /5 O AT 2.6/ O AC 2.45 2.

A CAA 0.500 A A O 0.438 a inc 0.375 A 1 O 0.312 A OA C 0.265 A O CA 0.250 A OO 0.223 A OT 0.187 A CH 0.157

o

1

3.0U

0.

0.0097 A ATVO C 0.0085 A AA*7 0.0O71 1

App. A. 5

4 mesh

mesh mesh 7 mesh O t_ 8 mesh mesh 9 10 mesh 12 mesh 14 mesh £ 16 mesh 20 mesh 24 mesh 28 mesh 32 mesh 35 mesh 42 mesh 48 mesh 60 mesh 65 mesh 80 mesh 100 mesh 115 mesh 1 50 mesh 170 mesh 200 mesh 250 mesh 270 mesh 325 mesh 400 mesh 5

6

1

t_

Properties of Pipes, Tubes,

and Screens

Notation

SI units are given

first

(followed by English and/or cgs units).

m

a

particle radius,

a

area,

ae

acceleration from centrifugal

a0

specific surface area of particle,

A A

m2

2 (ft

);

(ft)

also

cross-sectional

m 2 area/m

area,m

=

absorption factor

2

2 (ft

,

3

volume bed or packing (ft 2 2 force, m/s (ft/s )

m~

cm

2 );

(ft

membrane

A„ A

solvent permeability constant, kg/s

b

length,

B

flow rate of dry solid, kg/h

5

cm

,

m

(ft

)

2

2 (ft

also

)

area,m

filter

2

2 (ft

)

2 ) •

m2

•

atm

solute permeability constant, m/s (ft/h)

(s/ft

B

area,

3

')

also area,m

Am

2

/ft

~

1

L/mV, dimensionless; 2

2

m

(ft,

cm)

(lb^);

also filtration constant,

s/m 3

3 )

physical property of membrane,

atm

1

cs

kg mol/m (lb^ft 3 g mol/cm 3 ) 3 3 concentration of absorbate in fluid, kg/m (lb m /ft ) 3 3 3 concentration of A, kg mol/m (lb mol/ft g mol/cm ) 3 3 break-point concentration, kg/m (lb m /ft ) 3 concentration of solute in gel, kg solute/m 3 (lb m /ft g/cm 3 ) 3 concentration of solute at surface of membrane, kg solute/m 3 3 (lb m /ft ,g/cm ) 3 3 3 concentration of A in solid, kg mol A/m (lb m /ft g mol/cm )

cs

concentration of solids in slurry, kg/m 3 (lb^ft 3 )

concentration,

c

c

cA cb c

g

cs

kg/m

3

3

,

,

,

,

,

K (btu/lbm dry air

cs

humid heat, kJ/kg dry

cx

concentration of solids in slurry, mass frac

c

heat

p

mol

capacity •

K (btu/lb m

at •

air

•

°F)

constant pressure, J/kg-K, kJ/kg-K, kJ/kg

°F, cal/g-°C)

cP

membrane, kg/m 3 (lb^ft 3 ) deviation of concentration from mean concentration mol/m 3 concentration of P, kg P/m 3

c„

heat capacity at constant volume, J/kg

C C C Cp

filtration constant,

c„ c'

A

Notation

mean solvent concentration

fluid heat capacity,

N/m 2 (lb

in

2 f

ft

W/K (btu/h

); •

also,

•

cA

,

kg

K

number

of components

°F)

height of bottom of agitator above tank bottom,

m (ft)

pitot tube coefficient, dimensionless

895

CD C C0 D D

drag coefficient, dimensionless

D AB D D KA D NA D A c[l Dp m Da

molecular diffusivity,

Venturi coefficient, orifice coefficient, dimensionless

molecular diffusivity,

m 2/s

2

(ft

cm 2 /s)

/h,

decimal reduction time, min; also

;

also diameter,

m (ft)

flow rate, kg/h, kg

distillate

mol/h (lbjh) particle diameter,

m

2

2

m (ft)

Knudsen diffusivity

m 2 /s (ft

2

transition-region diffusivity,

m

effective diffusivity, effective

2

diameter of agitator,

critical

E E

activation energy, J/kg

energy

/s (ft /h,

/s (ft /h,

2

cm

2

cm 2 /s)

/s)

for mixture,

in protein solution,

mol

(cal/g

m

(ft)

m 2 /s

mol)

kW-h/ton;

reduction,

size

for

m

m (ft) m (ft)

diameter of tank,

D pc D AP

of A

cm 2/s)

2

m (ft)

D,

diffusivity

/h,

2

mean diameter

diameter,

cm 2 /s)

/s (ft /h,

tray

also

efficiency,

dimensionless

E E E

total energy,

J/kg (ft lb r/lbj

fraction unaccomplished change, dimensionless

emitted radiation energy,

W/m 2 (btu/h

m

2

ft

2 )

2

E E BX

axial dispersion coefficient,

/ /

fraction of feed vaporized; also cycle fraction, dimensionless

F F F F FT F0 F,

monochromatic emissive power,

W/m

3

(btu/h -ft

frictional loss, J/kg

)

(ft

•

lb f /lb m )

flow rate, kg/h, kg mol/h (lbjh) force,

N (lb

r

,

dyn)

correction factor for temperature difference, dimensionless

process time at ,

2

2

1

2

1

.

1

°C (250°F), min

geometric view factor for gray surfaces, dimensionless geometric view factor, dimensionless

g

standard acceleration of gravity (see Appendix A.

gc

gravitional conversion factor (see

G G

h

mass velocity = vp, kg/s m 2 kg/h m 2 (lb„yh growth constant, mm/h mass velocity = v'p, kg/s m 2 kg/h m 2 (lb„yh 2 irradiation on a body, W/m (btu/h ft 2 ) constant spacing in x for Simpson's rule

h

head, J/kg

h

heat-transfer coefficient,

G'

G

h h

3

Fanning friction factor, dimensionless mixing factor, dimensionless number of degrees of freedom

/,

F

/s (ft /h)

fg

•

ft

•

•

ft

)

m (ft)

lb r/lbj; also height of fluid,

W/m 2 K (btu/h

)

2

•

,

2

°F) enthalpy of liquid, J/kg, kJ/kg, kJ/kg mol (btu/lb m ) latent heat of vaporization, J/kg, k J/kg (btu/lb m) •

hc

contact resistance coefficient,

H H H H

distance,

ft

•

•

W/m 2 K (btu/h

2 ft

•

°F)

m (ft)

Henry's law constant, atm/mol frac head, J/kg

(ft

lb f/lb m ); also height of fluid,

enthalpy, J/kg, kJ/kg, kJ/kg

dry

896

2

•

,

(ft

1)

Appendix A. 1)

air (btu/lb m

dry

mol

(btu/lb m

)

;

m (ft) also enthalpy kJ/kg

air)

Notation

H

humidity, kg water vapor/kg dry

H H H

enthalpy of vapor, J/kg, kJ/kg, kJ/kg mol (btu/lb m ) equilibrium relation, kg mol/m 3

bed length,

H'

m

water vapor/lb dry

air (lb

atm

•

m

also effective length of spindle,

(ft);

air)

(ft)

enthalpy, kJ/kg dry solid (btu/lb m ); also enthalpy, kJ/kg dry air (btu/lb m dry air)

HB HG Hl ,

m

length of bed used up to break point,

m

height of transfer unit,

,

(ft)

(ft)

HoG' Hol Hp, H R

percentage humidity, percentage relative humidity,

HT

total

HyNg

length of unused bed,

respectively

m

bed length,

x

(ft)

m

(ft)

i

unit vector along

/

intensity of turbulence, dimensionless

1B

radiation intensity of black body,

axis

Ix

intensity of radiation,

j

unit vector along

J

width of baffle,

W/m

2

(btu/h

•

amp

also current,

;

W/m2

•

sr

2

(btu/h

•

ft

•

sr)

2

ft )

axis

y

m (ft)

mass flux of A relative to mass average velocity, kg/s m 2 mass flux of A relative to molar average velocity, kg/s m 2 molar flux of A relative to mass average velocity, kg mol/s m 2 molar flux vector of A relative to molar average velocity, kg mol/s m 2 (lb mol/h ft 2 g mol/s cm 2 )

jA j*

JA

•

J*

•

•

•

,

mass-transfer and heat-transfer factors, dimensionless

JD JH k ,

unit vector along z axis

reaction velocity constant, h ~

W

k,

thermal conductivity,

k

reaction velocity constant, h"

k\

first-order

,k c ,k G ,k x

,

...

,

~

min

1 ,

W/m K (btu/h

k

k'c

1

ft

•

;

l ,

or

1

s

°F)

min -1 or s"

1

,

heterogeneous reaction velocity constant, m/s (cm/s)

mass-transfer coefficient, kg mol/s

m

2

cone cone diff, g mol/s cm -cone diff); kg mol/s m 2 atm (lb mol/h ft 2 atm). (See Table 7.2-1.) •

mass-transfer coefficient, m/s

kc y

a, ...

k„

(ft/h,

3 ft

•

cone

g mol/s

diff,

•

cm

3 •

ft"

Pa, kg mol/s-

m

3

-

•

;

-

m

2 -

mol

frac

wall constants, dimensionless

K'y K'x

overall

,

2

).

kg mol/s mol frac)

'ass-transfer coefficient,

-

m

2

mol

K

g mol/s cm consistency index, s"/m 2 (lb f

K,

constant in Eq. (3.1-39), m/s

K',K S

equilibriun distribution coefficient, dimensionless

K'

consistency index,

K Kp Kv K c K ex K f

equilibrium distribution coefficient, dimensionless

L L

length,

h

ft

-

mol

2

frac,

•

N

,

•

cone diff (lb 3 cone diff) kg mol/s m Pa

surface reaction coefficient, kg mol/s k'w

,

2

cm/s)

volumetric mass-transfer coefficient, kg mol/s

mol/h k,

•

-

•

k G a,k x a,k

mol/h

diff (lb

m

2

•

,

Notation

N

•

•

s"'/m

2

frac (lb

mol/

2

•

s"/ft )

(ft/s)

(lb,-

2 •

s"'/ft )

6 (s/ft )

filtration constant,

s/m 6

filtration constant,

N/m 5

(lb f /ft

5 )

contraction, expansion, fitting loss coefficient, dimensionless

m (ft); also amount, kg, kg mol (lb m ); alsokg/h m

liquid flow rate, kg/h,

2

•

kg mol/h (lb^h)

897

m (ft) m (ft)

Prandtl mixing length,

mean beam

length,

dry solid weight, kg dry solid (lb ra dry solid) flow rate, kg inert/h, kg mol inert/h slope of equilibrium

line,

(lb m inert/h)

dimensionless

flow rate, kg/s, kg/h (\bjs)

parameter ratio,

m

= k/hx 1 also, position Table 7. 1-1, dimensionless; also position parameter kg wet cake/kg dry cake (lb wet cake/lb dry cake)

dimensionless ratio

;

in

molecular weight, kg/kg mol (lbjlb mol)

kg(lb m ); also parameter

total mass,

=

2 (Ax) /aAt, dimensionless

= (Ax) /D AB At, dimensionless flow rate, kg/h, kg mol/h (\bjh) 2

modulus

amount of adsorbent,

kg(lb m )

exponent, dimensionless; also flow behavior index, dimensionless slope of line for power-law

fluid,

dimensionless

position parameter; also dimensionless ratio total

amount, kg mol

A

flux of

rpm

=

mol, g mol) relative to stationary coordinates, kg/s

or rps also ;

parameter

=

x/x l

(lb

•

m

2

number of radiation shields

h Axjk, dimensionless; also

number of

viable or-

ganisms, dimensionless total

relative

flux

stationary coordinates, kg mol/s

to

2

•

ft

-m 2

(lb

2

g mol/s cm ) modulus = k c Ax/D AB dimensionless; also

mol/h

•

,

,

number of stages

concentration of solid B, kg solid B/kg solution (lb solid

itylb

solution)

number

of equal temperature subdivisions, dimensionless

solute flux, kg/s

-

m2

(lbj'h

•

solvent flux, kg solvent/s-m

ft 2

2 )

(lb^

residual defined by Eq. (6.6-3), kg

molar

flux vector of

m 2 (lb mol/h

A

2 •

ft

)

mol/m 3

(lb

mol/ft

3 )

relative to stationary coordinates,

kg

2 g mol/s cm ) mass transfer of A relative to stationary coordinates, kg mol/s

mol/s

•

2

•

ft

•

,

(lb

mol/h, g mol/s) number of transfer units, dimensionless

Grashof number defined in Eq. (4.7-4), dimensionless Biot number = hxjk, dimensionless Graetz number defined in Eq. (4.12-3), dimensionless Grashof number defined in Eq. (4.7-5), dimensionless Mach number defined by Eq. (2.1 1-15), dimensionless Peclet number = N Kc N P dimensionless Nusselt number = hbjk, dimensionless Froude number = v 2/gL, dimensionless Knudsen number = X/2r, dimensionless Nusselt number = hL/k, dimensionless Nusselt number = h x xjk, dimensionless Prandtl number = c p fi/k, dimensionless Reynolds number = Dvp/fi, dimensionless Reynolds number = D\Npj\i, dimensionless Reynolds number defined by Eq. (3.5-1 1), dimensionless ,

,

Notation

i.

X

p

nr

c.

mf

Reynolds number = Lv^ p/p, dimensionless Reynolds number = xv m p/p, dimensionless Reynolds number defined by Eq. (3.5-20), dimensionless

Reynolds number denned by Eq. (3. 1-15), dimensionless Reynolds number at minimum fluidization denned by Eq.

(3.1-

35), dimensionless

=

Euler number

p/pv

2

dimensionless

,

= p/pD AB dimensionless Sherwood number = k'c D/D AB dimensionless Schmidt number

N Sh NP

,

,

rn

power number defined by Eq. (3.4-2), dimensionless flow number, dimensionless Stanton number = k'Jv, dimensionless number of transfer units, dimensionless flow rate, kg/h, kg mol/h (lb^/h) 2 pressure, N/m Pa (lb f /ft 2 atm, psia, mm Hg)

Pa

partial pressure of A,

Pbm

log

NQ N NTU Si

0

,

mean

(lb f /ft

2 ,

,

N/m 2

Pa(lb f /ft 2 atm,

psia,

,

B

inert partial pressure of

total pressure in

Pm

permeability in solid, m/s

parameter in Eq.

Pa

vapor pressure of pure A,

power,

W

(ft/h)

dimensionless

(5.5-12),

N/m 2 Pa (lb /ft 2 ,

rate, kg/h,

f

mm H)

atm, psia,

,

hp)

(ft- lb f /s,

momentum

,

low-pressure side (permeate)

P P P P P

total pressure,

N/m 2 Pa

cm Hg, Pa (atm) cm Hg, Pa (atm)

pressure in high-pressure side (feed),

total

Pi

mm Hg)

Eq. (6.2-21),

in

mm Hg)

atm, psia,

Ph

flow

,

kg/min; also number of phases kg m/s (lb m ft/s)

vector,

•

permeability of A,

P„

solvent

Pm

permeability, kg mol/s

Pm

permeability,

P'm

permeability,

membrane

3

cm

P'a

at

equilibrium

•

N/m 2 Pa (lb ,

(STP) cm/s

f

•

•

/ft

2 ,

atm, psia,

kg solvent/s

permeability,

mm Hg)

cm 2 -cm Hg •

m

•

atm

(lbn/h

•

ft

•

atm)

m atm (lb mol/h m 3 (STP)/(s m C.S. atm/m) •

ft

•

•

atm)

2

•

•

P"

cm 3 (STP)/(s cm 2 3 2 permeability, cm (STP)/(s cm

q

heat-transfer rate,

atm/cm)

C.S.

•

C.S.

cm Hg/mm)

•

W (btu/h); also net energy added

to system,

W

(btu/h); also J (btu)

m

3

3

i

flow rate,

q

heat flux vector,

/s (ft /s)

W/m 2

W/m

9

rate of heat generation,

9

feed condition defined

9'

flow rate in Darcy's law,

q

kg adsorbate/kg adsorbent

Q\

flow rate of residue,

m

flow rate of permeate,

qA

flow rate of A

cm 3

residual defined

3

)

(btu/h

3 -

ft

)

(1 1.4-12)

(lb m /lb m ) 3

cm 3 /s) 3 /h, cm /s)

/s (ft /h,

m

3

/s (ft

kW

3

cm 3

(STP)/s,

m

3

3

/s (ft /h)

(btu/h)

(STP)/s,

by Eq.

m

3

3

/s (ft /h)

(4.5-11),

K (°F)

cm 3 (STP)/s, m 3 /s (ft 3 /h) 3 3 3 rate, cm (STP)/s, m /s (ft /h)

Qo

reject flow rate,

qP

permeate flow

qR

reboiler duty, kJ/h,

Notation

ft

cm 3 /s

permeate,

condenser duty, kJ/h, feed flow rate,

•

3

by Eq.

Qi

in

2

(btu/h

kW

(btu/h)

899

m 3 /s

circulation rate,

3

/h)

(ft

amount absorbed, kg mol/m 2 absorbed, J/kg (btu/lb m radius,

m

•

also heat loss,

;

W (btu/h); also heat

J

lb r /lb

(ft)

m

hydraulic radius,

3 ft

)

(ft)

m

value of radius,

rate of drying,

m 3 (lb m Afh

kg A/s

rate of generation,

critical

ft

,

kg/h

m

•

2

(ft) 2

(lb m /h

ft

•

)

solute rejection, dimensionless

scaleup ratio, dimensionless radius,

m

K/W (h

also resistance,

(ft);

parameter

Eq.

in

(5.5-12),

°F/btu)

•

dimensionless; also resistance,

ohms

= LJD,

gas constant (see Appendix A.l); also reflux ratio

dimen-

sionless rate of generation,

kg mol/l/s-

radius of spindle,

m

K/W

gel layer resistance, s

resistance of

rate of generation of

moM/h

(lb

3 •

ft

)

•

m

i,

x component of force,

(ft)

(h -"F/btu)

m2

medium,

filter

3

(ft)

radius of outer cylinder,

contact resistance,

m

kg/s

N (lb

atm/kg

•

m"

-1

1

(ft

)

(lb m /h) f

,

dyn)

compressibility constant, dimensionless mean surface renewal factor, s~ 1

m m3

conduction shape factor, solubility

of

a

gas,

(ft)

solute(STP)/m 3

-aim

solid

[cc

sol-

ute(STP)/cc solid atm] •

m

distance between centers,

(ft);

steam flow

also

rate, kg/h,

kg

mol/hflbjh)

volume of feed

solution,

m3

cross-sectional area of tower,

(ft

m

3 )

2

2 (ft

);

also stripping factor,

l/A specific surface

area,m 2 /m 3 volume (ft 2 /ft 3 volume)

surface area of particle, fin

thickness,

time,

s,

m

m2

2 (ft

)

(ft)

min, or h

temperature, K, °C (°F)

membrane

thickness, cm,

m

(ft)

break-point time, h

*

•

time equivalent to total capacity, h

mixing time,

s

time equivalent to usable capacity up to break-point time, h torque, kg

-

m

2

/s

2

temperature, K, °C (°F, °R); also feed deviation of temperature from

average velocity, m/s

rate,

ton/min

mean temperature

T,

K (°F)

(ft/s)

overall heat-transfer coefficient,

W/m 2 K (btu/h •

2 ft

•

°F)

internal energy, J/kg(btu/lb m ) velocity, m/s (ft/s)

velocity vector, m/s

(ft/s)

Notation

vA

velocity of A relative to stationary coordinates,

v Ad

diffusion velocity of

A

relative to

m/s (ft/s, cm/s) molar average velocity, m/s

(ft/s,

cm/s)

vH

molar average velocity of stream relative to stationary coordinates, m/s (ft/s, cm/s) humid volume, m 3 /kg dry air (ft 3 /lb m dry air)

v,

terminal settling velocity, m/s

oM

+

(ft/s)

dimensionless velocity defined by Eq. (3.10-34)

u

by Eq.

(3.10-42), dimensionless

v*

velocity defined

v'

deviation of velocity in

v'

superficial velocity based

x

jc

mean velocity v x m/s (ft/s) on cross section of empty tube, m/s (ft/s,

direction from

,

cm/s)

minimum

velocity at

v'mt

fluidization,

terminal settling velocity, m/s

v'j

m 3 /kg mol 3 flow rate, kg/h, kg mol/h, m /s (lbjh,

VA V V V

(ft/s)

solute molar volume,

volume, velocity,

V

W W W W W W W

m/s

(ft/s)

m

3

3 (ft

m/s

cm

,

(ft/s);

3 );

3 ft

/s)

also specific volume,

m J /kg

3 (ft

/lb

J

amount, kg, kg molObjJ kg mol/h (Ib^yh)

also total

inert flow rate, kg/h, kg/s,

mass fraction of A work done on surroundings, W(ft lb f/s) free water, kg (lb,J mechanical shaft work done on surroundings, flow rate, kg/h, kg mol/h (lbjh) work done on surroundings, J/kg (ft lb f /lb.J weight of wet solid, kg (lb m ) -

s

height or width,

m

also power,

(ft);

W

(ft

•

lb f /s)

W (hp); also mass dry cake, kg

ObJ

W Ws W

width of paddle, m (ft) mechanical shaft work done on surroundings, J/kg (ft shaft

p

work

delivered to

x

distance in x direction,

x

mass free

xA x'B

xj

xa x oM x BM (1 (1

— —

x A ). M x a)im

X X

fraction or

mole

pump, J/kg (ft

m

•

'

lt> /lfc> f

m)

lb f /lb m )

(ft)

fraction; also fraction remaining of original

moisture

mole fraction of A, dimensionless inert mole ratio, mol 5/mol inert mole fraction of A in feed, dimensionless mole fraction of A in reject, dimensionless mole fraction of minimum reject concentration, dimensionless log mean mole fraction of inert or stagnant B given in Eq. (6.3-4) log mean inert mole fraction defined by Eq. (10.4-27) log mean inert mole fraction defined by Eq. (10.4-7) particle size, m (ft); also parameter = at/xj, dimensionless parameter = Dt/x 2 dimensionless; also free moisture, kg water/kg ,

dry solid

(lb

water/lb dry solid)

m (ft); also mole fraction mole fraction of A; also kg A/kg solution

y

distance in y direction,

yA

mass

fraction of A or

(lb

A/lb solution)

y BM +

y

Notation

log

mean mole

fraction of inert or stagnant

B given

in

Eq. (7.2-1

1)

dimensionless number defined by Eq. (3.10-35)

901

y

t

mole fraction of A

in

permeate

in

permeate, dimensionless

at outlet of

residue stream,

dimensionless yp

- yi)n ~ y^iM

(>'

('

mole fraction of A l°g log

mean driving force denned by Eq. (10.6-24) mean inert mole fraction defined by Eq. (10.4-6)

Y Y 7

temperature ratio defined by Eq.

2

distance in

z

temperature range

Z

height,

fraction of unaccomplished

(4.10-2),

dimensionless

change in Table

5.3-1 or Eq. (7.1-12)

expansion correction factor defined in Eqs. (3.2-9) and

Greek

m

direction,

z

also tower height,

(ft);

for 10:1

change

in

m

(3.2-1 1)

(ft)

D T °C (°F) ,

m (ft); also temperature ratio in Eq. (4.10-2)

letters

=

turbulent flow and^, laminar flow

a

correction factor

a

absorptivity, dimensionless; also flux ratio

a

thermal diffusivity

1.0,

relative

= + N B/N A = k/pc p m 2 /s (ft 2 /h, cm 2 /s) volatility of A with respect to B, dimensionless

a

specific

cake resistance, m/kg (ft/lb m )

a

angle, rad

3 AB

ct,

1

,

a

kinetic-energy velocity correction factor, dimensionless

a*

ideal separation factor

aG

gas absorptivity, dimensionless

aT

eddy thermal

p

concentration polarization, ratio of

=

P'A /P'B

m

diffusivity,

brane surface to the

2

,

dimensionless

2

/s (ft /h)

salt

concentration at

mem-

concentration in bulk feed stream, di-

salt

mensionless

momentum

/3

velocity correction factor, dimensionless

volumetric coefficient of expansion, 1/K (1/°R) 2- "') 2

(3

N

y

viscosity coefficient,

y

ratio of heat capacities

r r

flow rate, kg/s

5

molecular

5

boundary-layer thickness,

§

constant

A

difference; also difference operating-point flow rate, kg/h(lb m/h)

ATC AT]m AH

arithmetic temperature drop, K,

•

s"'/m

i)bjii

= cjc v

m (lb.jTi

•

s

dimensionless

,

ft)

concentration of property, amount of property/m 3

log

diffusivity,

in

2

/s

m (ft); also distance, m

(ft)

Eq. (4.12-2), dimensionless

mean temperature

°C

(°F)

driving force, K, °C(°F)

Y

enthalpy change, J/kg, kJ/kg, kJ/kg mol (btu/lb m btu/lb mol) design criterion for sterilization, dimensionless

Ap

pressure drop,

£

roughness parameter,

e

emissivity, dimensionless; also

£ v,

mass eddy diffusivity m 2 /s

e

heat-exchanger effectiveness, dimensionless

£G £,

t mJ r]

f

,

N/m\ Pa (lb /ft 2 f

m

)

or void fraction, dimensionless 2

(ft

volume

/h,

fraction, dimensionless

cm 2 /s)

gas emissivity, dimensionless

momentum eddy diffusivity, m 2 /s (ft 2 /s) void fraction at minimum fluidization, dimensionless fin efficiency,

dimensionless

r/,

turbulent eddy viscosity,

r]

efficiency,

0

angle, rad

902

(ft);

Pa

-

s,

kg/m

s (lb m /ft

s)

dimensionless

Notation

0

parameter

9

cut or fraction of feed permeated, dimensionless

6*

fraction

in

Eq. (11.7-19)

permeated up to a value of x =

1

-

qlqf, dimen-

sionless

wavelength,

latent heat, J/kg, kJ/kg(btu/lb m ); also mean free path, latent heat of at normal boiling point, J/kg, kJ/kg

kg/m s, N s/m 2 (lb^/ft apparent viscosity, Pa s, kg/m s (lb^yft Pa

viscosity,

H \x

(ft)

•

s,

•

•

-

•

•

a

s, •

lb^ft

mol

(btu/lb m )

h, cp)

s)

dimensionless group, dimensionless

,

osmotic pressure, Pa, N/m 2

7t

kg/m

p

density,

a

constant, 4

x,

3

(lb^/ft

);

2

(lb f /ft

atm)

,

also reflectivity, dimensionless

W/m 2 K 4

8

•

also collision diameter, for centrifuge,

(0.1714 x 10

-8

m 2 (ft 2

z direction,

(kg- m/s)/s

t x

tortuosity, dimensionless

4>

velocity potential,

4>

angle, rad; also association parameter, dimensionless

osmotic

•

•

m

l

or

coefficient,

m 2 /s

,

2 (ft

)

/h)

dimensionless

shape factor of particle, dimensionless

s

amount

of property/s

NP.

flux of property,

\p

correction factor, dimensionless

m

2

m

2 2 stream function, /s (ft /h) parameter defined by Eq. (14.3-13), dimensionless

i/r

\\i

ft

)

,

f

(f>

2 •

)

in

N/m 2 (lb /ft 2 dyn/cm 2 2 2 2 shear stress, N/m (lb /ft dyn/cm f

btu/h

A

momentum

flux of x-directed

x

3

5.676 x 10"

°R ); sigma value

E

(j>

(ft)

momentum diffusivity /Vp,m 2/s (ft 2 /s, cm 2/s)

v 7i

m

A

A,lb

p

(o

angular velocity, rad/s

a)

solid angle, sr

Q0

n

m

A A

AB

Notation

collision integral, dimensionless

903

Index

Adiabatic saturation temperature, 530-

A

531

Absorption {see also Humidification processes; Stage processes)

adsorbents

absorption factors, 593

design methods for packed towers for concentrated gas mixtures,

627-628 for dilute gas mixtures,

619-621

equipment

for,

624-625

610-613 586-587

gas-liquid equilibrium,

interface compositions, 595-597

interphase mass transfer, 594 - 602 introduction to, 584-585

Kremser

analytical equations, 592-

593 driving force, 620

by fixed-bed process adsorption cycles, 707-708 basic model, 706-707

breakthrough curves, 702-703 concentration profiles, 701-702

616—

617

exchange

processes) Agitation {see also Mixing)

theoretical trays, 614

operating lines, 613-616 liquid to gas ratio,

baffles for,

143-144

circulation rate in, 151

616-617

packing mass transfer coefficients 595-597, 617-618, 632-633

599-601, 618 stage calculations, 587-592 overall,

stripping,

Freundlich isotherm, 698-699 Langmuir isotherm, 698-699 linear isotherm, 698-699

ion exchange {see Ion

liquid to gas ratio,

number of

film,

types of, 698 by batch process, 700

design method, 703-706

mean minimum

log

optimum

applications of, 697

physical properties of, 697-698

equilibrium in

genera] method, 615—619 transfer unit method,

Adsorbents {see Adsorption) Adsorption

616-617

equipment flow

for,

number

141-145

in, 151

flow patterns

in,

143-144

heat transfer in, 300-302

mixing time

in,

149-151

motionless mixers, 152

Absorptivity, 276-277, 283-284

of non-Newtonian fluids, 163-164

Adiabatic compressible flow, 103-104

power consumption, 144-147

904

Index

purposes scale-up

of,

140-141 147-149

in,

special agitation systems, 151-152

standard agitation design, 144 types of agitators, 141-144

Boiling point rise, 499-500 Bollman extractor, 728 Bond's crushing law, 842 Boundary layers (see also Turbulent flow)

Air, physical properties of, 866

continuity equation for, 192

Analogies

energy equation

for,

laminar flow

192-193

boundary layer, 370-375 between momentum, heat, and mass in

370-373

mass transfer equation

for,

transfer, 42-43, 375, 381-382,

momentum

430, 438-440, 477-478

separation of, 115, 191-192

Arithmetic

mean temperature drop,

238

Azeotrope, 642-643

475-477

equation for, 199-201

theory for mass transfer in, 478-479 wakes in, 115, 191-192 Bound moisture, 534-535 Bourdon gage, 38 Brownian motion, 817-818

Bubble trays

B

efficiencies,

646-647 Bernoulli equation, 67-68 Batch

in,

distillation,

Bingham-plastic

fluids, 154

666-669

types of, 611

Buckingham

pi theorem, 203-204, 308-310, 474-475

Biological diffusion in gels,

406-407 403-406

in solutions,

C

Biological materials chilling of,

360-361

diffusion of,

403-407

equilibrium moisture content, 533-

535

Capillaries (see

Porous

solids)

Capillary flow, in drying, 540, 553-555 Centrifugal filtration

evaporation of fruit juices,

513

paper pulp liquors, 514 sugar solutions, 513-514 freeze drying of, 566-569 freezing of, 362-365

equipment, 838 theory for, 837-838 Centrifugal pumps, 134-136 Centrifugal settling and sedimentation (see also Centrifugal filtration;

Centrifuges; Cyclones)

diameter, 832-833

grain dryer, 524

critical

heat of respiration, 229

equations for centrifugal force, 829-

leaching of, 723-729 physical properties of, 889-891 sterilization of,

569-577

Biot number, 332-333

Blake crusher, 843 Blasius theory, 193, 201

831

purpose

of,

829

separation of liquids, 834-835

of particles, 831-834 Centrifuges (see also Centrifugal settling

settling

and sedimentation)

Blowers, 138-139

disk bowl, 836-837

Boiling

scale-up

film, 261

natural convection, 259-260

nucleate, 260-261

physical mechanisms of, 259-261

temperature

of, 9,

681-683

Boiling point diagrams, 640-642

Index

of,

834

sigma value of, 834 tubular, 836 Cgs system of units, 4

Chapman-Enskog

theory, 394—395

Chemical reaction and diffusion, 456460

905

Chilling of food and biological materials,

in cylindrical coordinates,

360-361

in

Chilton-Colburn analogies, 440

in spherical coordinates,

Circular pipes and tubes

compressible flow friction

in,

101-104

graphical curvilinear square method,

233-235

of,

with heat generation, 229-231

in, 238-243 78-80, 83-87

heat transfer coefficients

laminar flow

mass

in,

hollow sphere, 222-223 through materials in parallel, 226in

440-443

transfer in,

227

turbulent flow in, 83-84, 87-91

mechanism

universal velocity distribution

shape factors

in,

197-199

215 235-236

of,

in,

two dimensions

steady-state in

Classifiers (see also Settling

and

Laplace equation, 311 numerical method, 312-317

sedimentation) centrifugal,

368

Fourier's law, 43, 214, 216-217, 368

892-893 factors in, 86-92

dimensions

368

rectangular coordinates, 368

831-833

with other boundary conditions,

821-823, 826-

differential settling,

316-317 through a wall, 220

827

Comminution

(see

Mechanical size

through walls

reduction)

Compressible flow of gases adiabatic conditions, 103-104 basic differential equation, 101

for

binary mixture, 454-456

for

boundary layer, 192-193

for

pure

isothermal conditions, 101-103

Mach number for, 104 maximum flow conditions,

102-104

equations for, 138-139

equipment, 138

Convection heat transfer (see also Heat transfer coefficients; Natural convection) •

general discussion of, 215-216, 219,

Condensation

227-228 physical

derivation of equation for, 263-267

(see

(see

for evaporation

of,

236-238

Heat transfer

coefficients)

Convective mass transfer coefficients

267 outside vertical surfaces, 263-265

mechanism

Convective heat transfer coefficients

263

outside horizontal cylinders, 266-

Condensers

50-54, 164-165, 167-

169, 176

Concentration units, 7

of,

fluid,

Control volume, 52-54

Compressors

mechanisms

223-225

in series,

Continuity equation

Mass

transfer coefficients)

Conversion factors, 850-853 Cooling towers (see Humidification

calculation methods, 512

processes)

direct-contact type, 511-512

Coriolis force, 175

surface type, 511

Countercurrent processes (see also

Conduction heat transfer (see also

Absorption; Distillation;

Leaching; Liquid-liquid

Unsteady-state heat transfer)

combined conduction and convection, 227-228 contact resistance at interface, 233 critical

thickness of insulation, 231-

232 in cylinders, 221,

224-226

effect of variable thermal

conductivity, 220

equations for

906

extraction; Stage processes) analytical equations, 592-593 in

heat transfer, 244-245

multiple stages, 589-591

Creeping flow, 189-190 Critical diameter, 832-833 Critical

moisture content, 538-539

Critical thickness of insulation,

231—

232

Index

Crushing and grinding {see Mechanical Crystallization {see also Crystallizers) crystal geometry, 737-738

of metals, 883

of solids, 882

crystal particle-size distribution,

of water, 855

Dew

746-747

on solubility, 744 heat balances, 740-741 heat of solution, 740-741 crystal size effect

mass transfer

in,

point temperature, 527, 682

Dialysis

equipment

758

for,

hemodialysis, 758

theory for, 755-756

745

material balances, 739

McCabe AL

Density of foods, 891

size reduction)

use

law, 745-746

of,

754-755, 757-758

Differential operations

Miers' qualitative theory, 744

time derivatives, 165

models

with scalars, 166-167

for,

747

nucleation theories, 744, 747

purpose of, 737 rate of crystal growth, 743-746

with vectors, 166-167 Differential settling, 821-823,

solubility (phase equilibria), 490,

diffusion;.

738-739 supersaturation, 741-744

of

circulating-liquid evaporator-

crystallizer, 743

through stagnant B, 388-390,

456 in biological gels,

tank crystallizer, 742 Crystals {see also Crystallization)

746

rate of growth, 745-746

types of, 738

Cubes, mass transfer to, 448 Curvilinear squares method, 233-235 Cyclones equipment, 838-839 theory, 839-840

Cylinders

and blockage by proteins, 405-406 in capillaries, 462—468 and convection, 387-389 in drying, 539-540, 552-553

and equation of continuity, 454-456 equimolar counterdiffusion, 385-386, 456 Fick's law, 43, 383-384, 453-454, 464 in

gases, 385-397

general case for

A and

Knudsen, 463-464 in

leaching, 725-726

in liquids,

flow across banks of, 250-251

molecular, 464

heat transfer coefficients for, 249-

multicomponent gases, 461-462

251, 559 transfer coefficients for, 450

Cylindrical coordinates, 169, 174, 369

D

B, 387-388

introduction to, 42-43, 381-383

drag coefficient for, 116-117

mass

406-407

with chemical reaction, 456-460

741-743

scraped-surface crystallizer, 742

particle-size distribution,

A

of biological solutes, 403-406

742-743

circulating-magma vacuum classification of,

Unsteady-state

diffusion)

Crystallizers {see also Crystallization)

crystallizer,

826-827

Diffusion {see also Steady-state

liquids,

397-399

402

porous solids, 412-413, 462, 468 similarity of mass, heat, and momentum transfer, 39-43, 381-382

in

in solids

Dalton's law, 8

classification of, 408

Darcy's law, 123

Fick's law, 408-409 permeability equations, 410-411

Dehumidification, 525, 602-603

Index

907

enriching operating line, 651-653

Diffusion {cont.) to a sphere,

enriching tower, 663

391-392

two dimensions,

steady-state, in

transition,

enthalpy-concentration method

benzene-toluene data, 672

413-416

construction of plot for, 669- 672

464-466

with variable cross-sectional area,

390-393, 408-409

enriching equations, 672-674

equilibrium data, 672, 887

example

velocities in, 387

of,

number of

Diffusivity (mass) in biological gels,

406-407

674-678

stages, 676, 678

stripping equations, 674

of biological solutes, 404 - 406

equilibrium or flash, 682-683

effective

feed condition and location, 654-

464- 466

in capillaries, in

porous solids, 412, 468

estimation of

405-406

for biological solutes, in

gases, 394-396

in liquids,

400-402

experimental determination in biological gels,

406 404

in biological solutions, in

399-400

total,

406-407 for biological solutes, 404-405 for gases, 394-395 for liquids, 400-401 for solids, 410-411 Knudsen, 463-464 molecular and eddy, 374-375 transition, 464-466 Diffusivity (momentum), 374-375, 381for biological gels,

382 Diffusivity (thermal), 331, 364-365,

374-375, 381-382

308-310, 474 equations, 202-203

in

heat transfer, 308-310

in

mass

in

momentum

transfer,

single stage contact,

642

stripping operating line, 653-654 stripping tower, 661-662 tray efficiencies

introduction, 666

Murphree, 667-669 overall, 667-669

Disk, drag coefficient for, Distillation (see also

1

Dodge crusher, 843 Drag coefficient

for flat plate, 115-116, 192-193

5

form drag, 115-117, 190

16-1 17

for long cylinder,

Multicomponent

Vapor-liquid

skin drag,

1

1 1

6-1 17

14-116

for sphere, 115-118,

190,816-819,

822-823

equilibrium)

constant molal overflow, 651-652 direct steam injection,

mass

429-430, 470-471

for disk, 116-118

202-204

transfer,

transfer,

definition of, 115-116, 201

474-475

Dimensional homogeneity,

908

simple batch or differential, 646- 647

simple stream, 648- 649

Distribution coefficient, in

theorem, 203-204,

distillation;

,

sidestream, 664 - 665

types of trays, 61 1-612

Dimensional analysis

in differential

658-659, 683 644- 645 681

relative volatility,

point, 668

Dilitant fluids, 155

pi

reflux ratio

operating, 660

experimental values

Buckingham

condensers, 666 Ponchon-Savarit method, 678 partial

minimum, 659- 660, 686- 687

gases, 390, 393-394

in liquids,

656, 687-688 Fenske equation, 658-659, 683 introduction and process flow, 644, 649-650 McCabe-Thiele method, 651-666 packed towers, 612-613

663-664

Dryers (see also Drying) continuous tunnel, 522

Index

drum, 523 fixed bed,

Energy balances for boundary layer, 370-373

556-559

grain, 524

incompressible flow, 101-104 365-368 mechanical, 63-67

rotary, 523

differential,

spray, 523-524 tray, 521, 561

vacuum

shelf,

overall,

521-522

Drying (see also Dryers) air recirculation in, 563—564

Enthalpy, 15, 57 Enthalpy of air-water vapor mixtures,

of biological materials, 533-535, 540

bound moisture, 534-535 capillary

movement

theory, 540,

through circulation, 556-559

544

effect

of

effect

of humidity, 544-545

effect of solid thickness,

545

538-540, 545-

period, 541-545

time for, 540-543

use of drying curve, 540-541

continuous countercurrent, 564-566 moisture, 538

540

equilibrium moisture, 533-535

experimental methods, 536 freeze drying, 566-569

561-562

Equivalent diameter in fluid flow,

98-99

in heat transfer, 241

correlation, 687-688

Euler equations, 185-186

Evaporation {see also Evaporators) of biological materials, 513-514 boiling point rise in,

499-500

capacity

in multiple-effect,

Duhring

lines,

504

499-500

effects of processing variables on,

489-490, 498-499 enthalpy-concentration charts for,

496

methods of operation

introduction to, 520-521 liquid diffusion theory, 539-540,

551-555

backward-feed multiple-effect,

494-495 forward-feed multiple-effect, 494

packed beds, 556-559 with varying air conditions,

parallel-feed multiple-effect, 495 single-effect,

493-494

multiple-effect calculations, 502-505

561

unbound moisture, 534-535 lines, 499-500

Duhring

E

single-effect calculations,

496-497

steam economy, 494 temperature drops in, 503-504 vapor recompression mechanical recompression, 514515 thermal recompression, 515~

Emissivity definition of, 277-278,

Index

Equilibrium moisture, 533-535

heat transfer coefficients for, 495-

heat balance in continuous dryers,

of,

645-

500-501

free moisture, 535

values

distillation,

Euler number, 202

536-538

rate-of-drying curve,

in trays

Equilibrium or flash

Erbar-Maddox

547 prediction for constant-rate

in

Enthalpy-concentration method (see

646

effect of temperature, 545

effect of shrinkage,

669-672

evaporation, 500-501

Distillation)

constant-rate period, 538, 540-545 air velocity,

606-607

in distillation, in

constant-drying conditions

critical

528,

Enthalpy-concentration diagram

553-555

falling-rate period,

56-61

English system of units, 4

884

283-284

Evaporators {see also Evaporation) agitated film, 493

909

Evaporators {cont.) condensers for, 511-512 falling-film,

492

washing, 803-805, 812-813 Finned-surface exchangers efficiency of,

forced-circulation, 492-493

horizontal-tube natural circulation, 491

304-306

overall coefficients for, 307-308

types of, 303-304

law of thermodynamics, 56-57

First

long-tube vertical, 491

Fixed-bed extractor, 727-728

open

Rat

kettle, 491

short-tube, 491 vertical-type natural circulation, 491

Extraction (see Leaching; Liquid-

plate

boundary layer equations heat transfer, 370-373 mass transfer, 475-477

liquid extraction)

total drag,

for

190-193, 199-201

drag coefficient for, 114-117, 193 heat transfer to, 248, 543

F

mass

transfer to, 444

Flooding velocities, 613 Falling film

Flow meters

diffusion in,

441-442

momentum

shell

tube meter; Venturi meter;

balance, 80-82

velocity profile, 82

Weirs)

Flow separation, 191-192

Fans, 137

Fenske equation, 658-659, 683 Fermentation, mass transfer in, 450453

Fluid friction

chart for

453-454 for steady-state, 383-384 for unsteady-state, 426-427 Film temperature, 248 Film theory, 478 Filter aids, 806-807

Newtonian

fluids,

160 effect

of heat transfer on, 92

of,

in

entrance section of pipe, 99-100

in fittings

Filters (see also Filtration)

bed, 802 classification of, 802

continuous rotary, 805-806 leaf, 803-805 plate-and-frame, 803 Filtration (see also Centrifugal filtration; Filters)

basic theory, 809-810

friction factor in pipes,

non-Newtonian

93

Fluidized beds

expansion

of,

126

heat transfer in, 253

mass

transfer in, 448

minimum

fluidization velocity in,

123-125

minimum

porosity

Fluid statics, 32-39

Flux (mass)

filter

filter

media, 806

"

"

conversion factors 466

in,

123-124

for,

853

ratios, 464,

types of, 453-454

pressure drop, 807-808

Foods, physical properties

purpose

Form

specific

of,

801

cake resistance, 808-809

153-161

from sudden contraction, 93 from sudden expansion, 75-76, 92-

continuous, 813-814

806-807 cycle time, 812-813

fluids,

roughness effect on, 89

constant rate, 815 filter aids,

86-92 98-99

for noncircular channels,

compressible cake, 809 constant pressure, 809-810

and valves, 92-94

for flow of gases, 91

for

Filter media, 806

910

88

chart for non-Newtonian fluids, 159-

Fick's law (see also Diffusion)

forms

(see Orifice meter; Pitot

of,

889-891

drag, 114-117, 190

Fouling factors, 275-276

Index

Fourier's law, 43, 214, 216-217, 368,

radiation, 293-296

Gas-solid equilibrium, 409-411

transfer)

General molecular transport equation for heat transfer, 214-216

Fractionation (see Distillation)

Free moisture, 535 Free

Gas

382 (see also Conduction heat

between momentum, heat, and mass, 39-43, 381-382 for steady state, 39-40 for unsteady state, 41-42 General property balance, 39-42 Graphical methods integration by, 23-24 two-dimensional conduction, 233similarity

settling (see Settling

and

sedimentation)

Freeze drying, 566-569 Freezing of food and biological materials, 362-365

Freundlich isotherm (see Adsorption) Friction factor (see Fluid friction)

Froude number, 202 Fuller et

al.

235

equation, 396

Fundamental constants, 850-853

Grashof number, 254 Gravitational constant, 851

Gravity separator, 38-39

Grinding (see Mechanical size

G

reduction)

Gurney-Lurie charts, 340, 343, 345 Gyratory crusher, 844

Gases equations for, 7-9 physical properties of, 864-875,

H

884-886

Gas law constant, R, Gas law, ideal, 7-9

7,

850 Hagen-Poiseuille equation, 80, 87, 180

Heat balances

Gas-liquid equilibrium

740-741

acetone-water, 886

in crystallization,

ammonia-water, 886

in drying,

enthalpy-temperature, 606-607

in

Henry's law, 586-587, 884

principles of, 19-22

rule,

data for foods, 889-890

586

sulfur dioxide-water, 586-587, 885

Gas-liquid separation processes (see

Absorption; Humidification processes; Stage processes)

Gas permeation membrane processes equipment

minimum

for,

evaporation, 496-497, 505

Heat capacity

methanol-water, 885

phase

760-762

data for gases, 16, 866, 869, 873-874 data for liquids, 875, 879 data for solids, 881, 883 data for water, 856-857 discussion of, 14-15

Heat exchangers (see also Heat

reject concentration, 768

permeability

561-562

in,

759-760

transfer coefficients)

cross-flow, 268-269, 271

processing variables for, 780-782

double-pipe, 267

separation factor

effectiveness of, 272-274

in,

765

series resistances in, 759

extended-surface, 303-308

theory of

fouling factors for, 275-276

.

mean temperature

cocurrent model, 780

log

completely-mixed model, 764-768 countercurrent model, 778-780 cross-flow model, 772-775

244-245, 269-271 scraped-surface, 302-303 shell and tube, 267-268

introduction to, 763-764

temperature correction factors, 269-

multicomponent mixture, 769-771 types of membranes, 759-760

Index

difference,

271

Heat of reaction, 17-18 911

equilibrium relations, 606- 607

Heat of solution, 740-741 Heat transfer coefficients for agitated vessels, 300-302 approximate values

of,

equipment

219

calculation

minimum

average coefficient, 238 to

methods

air

Humidity (see also Humidification

evaporators, 495-496

processes) adiabatic saturation temperature,

for finned-surface exchangers,

303-

530-531 chart for, 528-529

308 for flow parallel to flat plate, 248

definition of,

for flow past a cylinder, 249

dew

for fluidized bed, 253

equations for, 526

fouling factors,

for laminar flow in pipes,

238

percentage, 526

for natural convection,

253-259

relative,

non-Newtonian

297-299 260-261

fluids,

253 275-276

for other geometries,

overall, 227-228,

for

526

saturated, 526

noncircular conducts, 241

for nucleate boiling,

526

point temperature, 527

humid heat, 527 humid volume, 527

275-276

for liquid metals, 243

for

603-610

610

for film boiling, 261

in

for,

flow, 609

packed tower height, 607, 609-

237

entrance region effect on, 242-243 in

602-603

operating lines, 604- 605

banks of tubes, 250-251 263-267

for condensation,

definition of, 219,

for,

water-cooling

total enthalpy, 528 wet bulb temperature, 531-532 Humid volume, 527-528

packed beds, 252-253, 447-448 and convection, 279-

for radiation

I

281 for scraped-surface exchangers,

302-

856-857, 889

Ice, properties of,

Ideal fluids, 185-189

303

Ideal gas volume, 850

for spheres, 249

for transition flow

in

pipes, 240-241

Ideal tray (see

Bubble trays) boundary layers

239-240 Heat transfer mechanisms, 215-216 Height of a transfer unit, 609-610, 624-625, 632-633 Heisler charts, 341, 344, 346

Integral analysis of

Hemodialysis, 758 Henry's law

Internal energy, 16, 57

for turbulent flow in pipes,

for energy balance, 373 for

momentum

balance, 199-201

Intensity of turbulence, 194-195

Interface contact resistance, 233

Interphase mass transfer

data for gases, 884

interface compositions, 595-597

equation, 586-587

introduction

Hildebrandt extractor, 728

Hindered

Humid

settling,

820-822

heat, 527

Humidification processes (see also

Humidity) adiabatic saturation temperature,

530-531 definition of, 525

dehumidification, 525, 590-591, 602-

603

912

to,

594-595

use of film coefficients, 595-597,

605-606, 617-621 use of overall coefficients, 599-601, 607, 610

Inverse lever-arm rule, 712-713 Inviscid flow (see Ideal fluids)

Ion exchange processes (see also

Adsorption) design of, 709 equilibrium relations

in,

708-709

Index

16-17

introduction to, 708

discussion

types of, 708

of ice, 854, 856

Isothermal compressible flow, 101-103

of,

of water, 854, 857-859

Leaching countercurrent multistage constant underflow, 737

J

number of 429, 438, 440

J -factor,

operating

stages,

line,

734-735

733

variable underflow, 734-735

equilibrium relations, 729-730

K

equipment for agitated tanks, 728-729

K

factors (distillation), 680-681

Bollman extractor, 728

Kick's crushing law, 842

fixed-bed (Shanks), 727

Kinetic energy

Hildebrandt extractor, 728

velocity correction factor for,

59-

thickeners, 728-729

preparation of solids, 724-725

60, 159

processing methods, 727

definition of, 57

purpose

KirchhofFs law, 277-278, 283 Kirkbride method, 687

of,

723

rates of

when

Knudsen diffusion, 463-464 Knudsen number, 464 Kremser equations for stage

diffusion in solid controls,

726

when

processes, 592-593

dissolving a solid, 725-726

introduction to, 725 single-stage contact, 730-731

washing, 723 Le Bas molar volumes, 401-402

L Laminar flow boundary layer

Lennard-Jones function, 394-395 Lever-arm rule, 712-713 for, 192-193,

199-

definition of,

47-49

on

190-193

fiat

plate,

Hagen-Poiseuille equation for, 80, kinetic energy correction factor for,

transfer in,

440-443

correction factor for,

72-73

{see Adsorption)

Laplace's equation heat transfer, 310-311

for potential flow, 187

for stream function, 187

Latent heat

Index

minimum solvent rate, 721-722 number of stages, 719-722 equilibrium relations acetic acid-water-isopropyl ether,

831, 711-712, 888 acetone-water-methyl isobutyl

ketone, 888

84

Langmuir isotherm

countercurrent multistage

overall material balance, 718

non-Newtonian fluids in, 155-159 pressure drop in, 84-87 Reynolds number for, 49, 86 velocity profile in tubes for, 80, 83-

in

ketone, 888

immiscible liquids, 722

58-60, 159

momentum

711-712, 888

acetone-water-methyl isobutyl Liquid-liquid extraction

87, 180

mass

Liquid-liquid equilibrium acetic acid-water-isopropyl ether,

201

phase

rule,

710

rectangular coordinates, 711 triangular coordinates, 710-711

types of phase diagrams, 710-712 equipment for agitated tower, 715-716

913

experimental determination of, 437,

Liquid-liquid extraction (cont.)

632

Karr column, 716 mixer-settler, 715

for falling film,

packed tower, 716

film

and overall, 595-597, 599- 601, 606-610, 632-633 for flat plate, 444 for fluidized bed, 448 inside pipes, 440-443 introduction to, 385, 432-433

plate tower, 715-716

spray tower, 716

type and classes of operation,

715-716 lever-arm rule, 712-713

purpose

of,

441-442

709-710

for liquid metals,

models

single-stage equilibrium contact,

714-715

for,

450

478-479

packed bed, 447-449, 556-557

for

for packed tower, 632-633

Liquid-metals heat transfer

to small particle suspensions,

coefficients, 243

Liquid-metals mass transfer, 450

450—

453

445-446 433-436

Liquid-solid equilibrium, 729-730

for sphere,

Liquid-solid leaching (see Leaching)

types

of,

area, 225

441-443 453-454 transfer between phases concentration profiles, 594-595 film coefficients, 595-597, 606-607

concentration difference, 448-449,

overall coefficients, 599-601, 607,

for wetted-wall tower,

Liquid-solid separation (see

Mass Mass

Crystallization; Settling and

sedimentation)

Log-mean value

620

transfer fluxes,

610, 617

corrections for heat exchanger, 269-

271

temperature difference, 244-245,

269-271

Mass Mass Mass

transfer models,

478-479

units, 6

velocity, 51

Material balances

Lost work, 63

chemical reaction and, 12-13

Lumped

for crystallization, 739

capacity analysis, 332-334

496-497, 504-505 methods of calculation, 10-11 overall, 50-56 for evaporation,

M

recycle and,

1

McCabe AL Law Mach number,

Manometers, 36-38

Mass balance overall, 50-56 Mass transfer, boundary conditions

in,

456

Mass

A

through stagnant B, 435-436

for crystallization, 745

packed bed, 448 450 for cylinders in packed bed, 448 definition of, 429, 433^434 dimensionless numbers for, 437-438 for equimolar counterdiffusion, 434for

cubes

in

for cylinder,

435

914

McCabe-Thiele method, 651-666 Mean free path, 462 Mechanical energy balance, 63-67 Mechanical-physical separation

transfer coefficients

analogies for, 438-440 for

of crystal growth,

745-746

104

processes (see also Centrifugal filtration;

Centrifugal settling

and sedimentation; Cyclones; Filtration; Mechanical size reduction; Settling and sedimentation)

800-801 methods of separation, 800- 801 Mechanical size reduction equipment for classification of,

Blake crusher, 843

Index

Dodge

crusher, 843

differential

gyratory crusher, 844

jaw crushers, 843

measurement, 840-841

power required

164-165,

ideal fluids, 185-186

for jet striking a vane, 76-78

crusher, 844

particle size

for,

Euler equations, 185-186

revolving grinding mill, 844 roll

equations

170-175

Laplace's equation

in,

187

general theory, 841-842

Newtonian fluids, 172-175 overall, 69-78 between parallel plates, 175-178

Kick's law, 842

potential flow, 186-189

Rittinger's law, 842

in rotating cylinder,

for

for

Bond's law, 842

purpose

840

of,

shell,

Membrane processes (see also Dialysis; Gas permeation membrane processes; Reverse osmosis; Ultrafiltration)

equipment

for, 758,

permeability

410-411, 756, 785-

in,

stream function, 185 transfer of

series resistances in, 755-756, 759

462

of,

883-884

Miers' theory, 744

Mixing (see also Agitation) discussion of, 140-141, 152-153 for,

152-153

equilibrium data for, 680-681

682-683

introduction to, 679-680

key components, 683 minimum reflux (Underwood equation), 686-687

number of

with pastes, 152-153

components,

684 flash distillation,

728-729

Mixer-settlers, 715,

equipment

point, 682

distribution of other

types of, 585, 754-755

Metals, properties

46, 79

Multicomponent distillation boiling point, 681-682

dew

786, 790

momentum,

Motionless mixers (see Agitation) Multicomponent diffusion, 402, 461-

760-762, 790,

792

181-184

78-82

with powders, 152

stages

feed tray location, 687

Mixing time (see Agitation) Moisture (see also Drying; Humidity) bound and unbound, 534-535

by short-cut method, 687-688

number of towers total reflux

in, 679 (Fenske equation), 683

capillary flow of, 540, 553-555 diffusion of, 539-540, 551-555

N

equilibrium, for air-solids, 533-535 free moisture, 535

Molar volumes, 400-402 Molecular diffusion (see Diffusion)

Natural convection heat transfer derivation of equation (vertical

Molecular transport (see Diffusion)

Mole

units,

6

plate),

equations for various geometries,

Momentum

254-259

definition of, 46,

69-70

introduction to, 3 velocity correction factor for, 72-73

Momentum

boundary

layer,

192-193, 199-

201 in circular tube,

179-181

Coriolis force, 175

Index

introduction to, 253

Navier-Stokes equations, 173-175 Newtonian fluids, 46-47, 153-154

Newton's law

balance

applications of, 175-184 for

253-254

of

momentum

transfer,

42-45

second law (momentum balance), 69-70 of settling, 817 of viscosity, 42-45

915

Non-Newtonian

fluids

Overall energy balance {see Energy balances)

agitation of, 163-164

flow-property constants, 156-157 friction loss in fittings, 159

P

heat transfer for, 297-299

laminar flow for, 155-159 rotational viscometer, 161-163

turbulent flow for, 159-160

types

of,

153-155

Packed beds Darcy's law for flow drying

velocity profiles for, 161

253, 447-448, 558-559

Nucleation {see also Crystallization)

mass

mass

secondary, 744 of transfer units

274 609- 610 624 - 625

for heat exchangers, for humidification,

mass transfer, Numerical methods for integration,

by Simpson's

method, 24 for steady-state

conduction

with other boundary conditions,

316-317 in

two dimensions, 310-317

with other boundary conditions,

414-416 two dimensions, 413-414

for unsteady-state heat transfer

boundary conditions, 351-353 in a cylinder, 358-359 implicit method, 359-360 Schmidt method, 351 in a slab, 350-353 for unsteady-state mass transfer boundary conditions, 470-471 Schmidt method, 469-470 in a slab, 468-471

transfer in,

tortuosity

in,

448-449

412, 468

pressure drop in

laminar flow, 118-120, 123 turbulent flow, 120-121

shape factors for particles, 121-122 surface area in, 118, 448, 558-559

Packed towers, 602-603, 612-613, 716 Packing, 612-613 Particle-size measurement in crystallization, 746-747 in mechanical size reduction, 840— 841

for steady-state diffusion

in

transfer coefficients for, 447-

448

primary, 744

for

123

in,

556-559

heat transfer coefficients for, 252-

Notation, 895-903

Number

in,

standard screens for, 894 Pasteurization, 576

Penetration theory, 442, 478-479

Permeability of solids, 410-411, 756,

760 764-765, 785-786, 790 ,

Phase

rule, 586, 640, 648,

710

Physical properties of compounds,

854-891 Pipes

dimensions

of,

892-893

schedule number

of, 83,

892

size selection of, 100 Pitot tube meter, 127-128

Nusselt equation, 263-265

Planck's law, 281-282

Nusselt number, 238

Plate tower, 611, 715-716

Pohlhausen boundary layer

relation,

372 Poiseuille equation {see

O

Hagen-

Poiseuille equation)

Porous solids Operating lines {see Absorption; Distillation; Humidification

processes; Leaching) Orifice meter, 131-132

Osmotic pressure, 783-784

916

effective diffusivity of, 412, 468

introduction to, 408, 412-413, 462

Knudsen

463-464 412-413, 464 transition diffusion in, 464-466 diffusion in,

molecular diffusion

in,

Index

between gray bodies, 292-293

Potential energy, 57 Potential flow, 186-189

Raoult's law, 640-641, 680

Prandtl analogy, 439

Reflux

Prandtl mixing length (see Turbulent

Relative volatility, 644- 645

number

definition of,

658-660, 686- 687

237-238

concentration polarization

of gases, 866, 870

681

in,

introduction to, 782-783

conversion factors for, 851-852

membranes

devices to measure, 36-39

operating variables

head and, 35 units of, 7, 32-34

osmotic pressure

in,

784

788 783-784

in,

in,

solute rejection, 786

Pressure drop (see also Fluid friction)

theory of, 785-786, 789-791 Reynolds analogy, 438-439 Reynolds number

in

compressible flow, 91, 101-104

in

laminar flow, 84-87

in

packed beds, 118-121 84-94

for condensation, 265

turbulent flow, 87-90

definition of, 49, 202, 437

in

in pipes,

agitation/ 144-145

Pseudoplastic fluids, 154

for flat plate, 191, 193

Psychometric

for flow in tube, 49, 238, 437

ratio,

532

Pumps centrifugal,

134-135

developed head of, 134-136 efficiency of, 133-136

for non-Newtonian fluids, 157 Reynolds stresses, 195-196 Rheopectic fluids, 155

Rittinger's crushing law, 842

positive displacement, 136-137

Rotational viscometer, 161-163

power requirements

Roughness

suction

789-

790

Pressure

in

,

Reverse osmosis complete-mixing model, 790-791

flow)

Prandtl

ratio,

lift

of,

for,

133-136

in

pipes,

87-89

134

Schmidt method Radiation heat transfer in

absorbing gases, 293-296

absorptivity, 277

black body, 277-278

for heat transfer, 351 for mass transfer, 469-470 Schmidt number, 396-397, 437-438 Scraped-surface heat exchangers, 302-

combined radiation and convection, 279-281

303

Screen analyses

746-747 measurement, 746

emissivity, 277-278, 283, 884

in crystallization,

gray body, 278, 283-284

particle size

heat transfer coefficient, 279-280 introduction to, 216, 276-278, 281

Tyler screen table, 894 Sedimentation {see Settling and

KirchhofTs law, 277, 283 Planck's law for emissive power,

sedimentation)

Separation (see also Settling and

281-282

286 Stefan-Boltzmann law, 278, 283 to small object, 278-279 view factors between black bodies, 284-291 general equation for, 286-288

shields,

Index

sedimentation)

by

differential settling,

821-823,

826-827 of particles from gases, 838-840 of particles from liquids, 815-823, 831-834 of two liquids, 834-835

917

Separation processes, types

of,

584—

Size reduction (see Mechanical size reduction)

585 Settlers (see also Settling

and

Skin friction, 114-115, 440 Solid-liquid equilibrium, 729-731

sedimentation) gravity chamber, 826

Solids, properties of,

gravity classifier, 826-827

Solubility of gases

gravity tank, 826

in liquids,

Spitzkasten classifier, 827

in solids,

thickener, 718, 728-729, 825-829

and sedimentation (see also Centrifugal settling and

Settling

sedimentation; Settlers;

sodium

738-739

extraction)

821-827 drag coefficient for sphere, 816-819, 822-823 differential settling,

hindered settling, 820, 822

Newton's law, 817 purpose of, 815-816 sedimentation and thickening, 825828

Spheres diffusion to, 391-392

drag coefficient for, 114-117, 190 heat transfer to, 249

mass transfer to, 445-446, 450-453 Newton's law for, 817 settling velocity of,

817

Stoke's law for, 116, 190, 817 Spherical coordinates, 169, 368

Spitzkasten classifier, 827

Stoke's law, 116, 189-190, 817 terminal settling velocity for sphere,

Spray tower, 716 Stage processes (see also Absorption;

817

Distillation; Liquid-liquid

theory for rigid sphere, 816-819

extraction; Leaching)

wall effect, 821 factors, in conduction,

235-236

factors, for particles, 121-122,

absorption, 613-614 analytical equations for, 592-593

countercurrent multistage, 589-591

124

Shear stress (see also

Momentum

definition of,

distillation,

of,

172-173

44-45

liquid-liquid extraction,

185-186

laminar flow on flat plate, 193 laminar flow in tube, 79

715-722

single stage, 587-588, 642, 712-715,

730-731

Euler equations, 185-186 ideal fluids,

649-688

leaching, 730-735, 737

balance)

components

in

thiosulfate,

typical solubility curves, 490

Solvent extraction (see Liquid-liquid

Brownian movement, 817-818 classification, 821-823, 826-827

in

586-587, 884-886 409-411

Solubility of salts

Thickeners)

Shape Shape

881-884

Standard heat of combustion, 18, 865 Standard heat of formation,

18,

Standard screen sizes, 894

normal, 170

Stanton number, 438

potential flow, 186-189

Steady-state diffusion (see also

stream function, 185

Diffusion; Diffusivity;

Sherwood number, 438

Numerical methods) and

Sieve (perforated) plate tower, 611, 667-668, 716-717

in biological solutions

Sigma value

in gases,

(centrifuge), 834

Simple batch or differential distillation, 646- 647 Simpson's numerical integration

method, 24 SI system of units, 3-4

918

864

gels,

403-

407

385-393 397-398

in liquids, in

two dimensions, 413—416 408-413

in solids,

Steady-state heat transfer (see

Conduction heat

transfer;

Index

Convective heat transfer; Numerical methods)

Steam Steam

distillation,

648-649 857-861

table, 16-17,

.

introduction to, 666

Murphree, 667-669 overall, 667-669 point, 668

Stefan-Boltzmann radiation law, 278, 283 Sterilization of biological materials effects on food, 577 introduction to, 569-570

Tray towers, 611, 613-614, 715-716 Triangular coordinates, 710-711 Tubes, sizes

of,

893 (see also Circular

pipes and tubes)

pasteurization, 576

Turboblowers, 138 Turbulence, 194-195

thermal death rate kinetics, 570-571,

Turbulent flow (see also Boundary

575-576

layers)

thermal process time, 571-574

boundary layer separation

Stoke's law, 116, 190, 817

Stream function, 185

191—

boundary layer theory for

Streamlined body, 115

heat transfer, 373

Stripping (see Absorption)

mass

Supersaturation, 741-746 (see also

momentum

transfer,

475-477

transfer, 192-193,

199-201

Crystallization)

Suspensions, mass transfer

in,

192

to,

450-

453

deviating velocities

in,

194-195

discussion of, 48-49, 194-195

Systeme International (SI) system of basic units, 3-4 table of values and conversion factors, 850-853

T

on

flat plate,

maximum

190-191, 199-201

velocity

in,

83-84

Prandtl mixing length in

heat transfer, 374-375

in

mass

in

momentum

transfer,

477-478

transfer, 196-197,

374-375 Reynolds number

for,

49

in tubes, 49, 83-84, 87-89 turbulent shear in, 195-196

Temperature scales, 5 Thermal conductivity definition of, 217-219

turbulent diffusion

in,

of foods, 891 of gases, 217-218 866, 868 875 of liquids, 218, 880 of solids, 218-219, 882-883

velocity profile

83-84, 197-199

,

,

in,

382

Two-dimensional conduction, 233-236 Tyler standard screens, 894

of water, 856, 862-863 Thickeners, 728-729, 825-828 (see also Settling and sedimentation)

Thixotropic

fluids,

Tortuosity factor, 412-413, 468 Total energy, 56-57

Transfer unit method, 609-610, 624625

Transport processes

boundary layers, 370-373

classification of, 2

similarity of,

39-43, 381-382, 426,

430, 438-440

Tray

efficiency

Index

Ultrafiltration

comparison to reverse osmosis, 791— 792 concentration polarization

Transition region diffusion, 464-466

in

U

155

in,

789-

791

equipment

for,

792

introduction to, 791-792

membranes

for,

792

processing variables

in,

794-795

theory of, 792-794

Unbound

moisture, 534-535

919

Underwood, minimum

reflux ratio,

calculations using Raoult's law,

686-687

640-641

Unit operations, classification of, 1-2

enthalpy-concentration, 669-672

Units

ethanol-water, 887

cgs system, 4

n-heptane-ethylbenzene, 691

and dimensions, 5

n-heptane-n-octane, 692

English system, 4

n-hexane-n-octane, 690

SI system, 3-4 table of,

multicomponent, 680-682

850-853

n-pentane-n-heptane, 647

Universal velocity distribution, 197— 199

phase

rule, 640,

648

water-benzene, 691

Unsteady-state diffusion (see also water-ethylaniline, 691

Numerical methods) analytical equations for fiat plate,

xy

428-429

426-427 boundary conditions for, 428-430

Distillation)

basic equation for,

charts for various geometries, 431

Vapor pressure data for organic compounds, 642,

690- 692

and chemical reaction, 460 in leaching, 724-726

data for water, 525-526, 854, 856-

859

mass and heat transfer parameters for,

discussion of, 9

430

in three directions,

Vapor recompression

432

Unsteady-state heat transfer (conduction) (see also

Vectors, 165-167

Numerical methods)

Velocity

334-336 average temperature for, 348-349 in biological materials, 360-365 in cylinder,

342-344

330-332 plate,

average, 55, 80, 82 interstitial,

laminar flow, 80, 82-84

profile in turbulent flow,

relation

334-336, 338-341

437

maximum, 83-84 profile in

derivation of equation for, 214-215,

flat

between

superficial, 119,

430

lumped capacity method

for,

332-

for negligible internal

universal, in pipes, 197-199

Venturi meter, 129-131

336-337 sphere, 343, 345-346 three directions, 345, 347-349

in semiinfinite solid,

in

437

754

resistance, 332-334

in

83-84

types of, in mass transfer, 387, 753-

334

method

83-84

velocities,

representative values in pipes, 100

heat and mass transfer parameters for, 339,

(see

Evaporation)

analytical equations for,

to

641-643

plots,

Vapor-liquid separation processes (see

View

factors in radiation, 284-293

Viscoelastic fluids, 155.

Viscosity discussion of, 43-47

of foods, 891 of gases, 866-867, 871-872

V

of liquids, 876-878

Valve

Newton's law

trays, 611

Vapor-liquid equilibrium

"

azeotropic mixtures, 642

benzene-toluene, 641-642 boiling point diagrams,

920

640-642

of, 43-45,

381-382

of water, 855, 863

Void fraction (packed and beds),

Von Karman

1

fluidized

18-1 19, 123-124, 447

analogy, 439-440

Index

w

Weirs, 132-133

Wet-bulb temperature

Washing {see Water

Filtration;

Leaching)

physical states of, 525 properties of, 854-863 Water cooling {see Humidification

processes)

Index

relation to adiabatic saturation

temperature, 532

theory of, 531-532

Wilke-Change correlation, 401-402

Work, 57-58, 63

Work

index for crushing, 842