Principles Of Mass Transfer Operations - I Volume I_nodrm.pdf

A TEXT BOOK OF

PRINCIPLES OF MASS TRANSFER OPERATIONS - I Volume I Enlarged and Revised Fourth Edition

For Degree Courses in Chemical Engineering Covering the Revised Syllabus of : Pune, Mumbai, Shivaji, North Maharashtra, North Gujarat Dr. Babasaheb Ambedekar Technological University, (BATU) Lonere (Dist-Raigad), Jawaharlal Nehru Technological University, (JNTU) Hydrabad, and University of Banglore. Also for Degree Courses in Polymer, Petroleum and Petrochemical Engineering, University of Pune And also for Degree courses in Biotechnology

By

DR. KIRAN D. PATIL M.E. (Chemical Engg.), Ph.D (Chemical Engg.) Professor in Chemical Engineering, Department of Petrochemical and Petroleum Engineering, Maharashtra Institute of Technology, Pune - 411 038 (MS)

N1013

Principles of Mass Transfer Operations - I (Vol. I)

ISBN 978-81-90693-56-1

Fourth Edition : September 2008 Reprint : September 2009, October 2010, January 2012, September 2015, August 2016, September 2017 © : Author The text of this publication, or any part thereof, should not be reproduced or transmitted in any form or stored in any computer storage system or device for distribution including photocopy, recording, taping or information retrieval system or reproduced on any disc, tape, perforated media or other information storage device etc., without the written permission of Author with whom the rights are reserved. Breach of this condition is liable for legal action. Every effort has been made to avoid errors or omissions in this publication. In spite of this, errors may have crept in. Any mistake, error or discrepancy so noted and shall be brought to our notice shall be taken care of in the next edition. It is notified that neither the publisher nor the author or seller shall be responsible for any damage or loss of action to any one, of any kind, in any manner, therefrom.

Published By :

(−Ve)

NIRALI PRAKASHAN Abhyudaya Pragati, 1312, Shivaji Nagar, Off J.M. Road, PUNE – 411005 Tel - (020) 25512336/37/39, Fax - (020) 25511379 Email : [email protected]


DISTRIBUTION CENTRES PUNE Nirali Prakashan : 119, Budhwar Peth, Jogeshwari Mandir Lane, Pune 411002, Maharashtra (For orders within Pune)

Nirali Prakashan (For orders outside Pune)

Tel : (020) 2445 2044, 66022708, Fax : (020) 2445 1538; Mobile : 9657703145 Email : [email protected] : S. No. 28/27, Dhayari, Near Asian College Pune 411041 Tel : (020) 24690204 Fax : (020) 24690316; Mobile : 9657703143 Email : [email protected]

MUMBAI Nirali Prakashan : 385, S.V.P. Road, Rasdhara Co-op. Hsg. Society Ltd., Girgaum, Mumbai 400004, Maharashtra; Mobile : 9320129587 Tel : (022) 2385 6339 / 2386 9976, Fax : (022) 2386 9976 Email : [email protected]


DISTRIBUTION BRANCHES JALGAON Nirali Prakashan : 34, V. V. Golani Market, Navi Peth, Jalgaon 425001, Maharashtra, Tel : (0257) 222 0395, Mob : 94234 91860; Email : [email protected]

KOLHAPUR Nirali Prakashan : New Mahadvar Road, Kedar Plaza, 1st Floor Opp. IDBI Bank, Kolhapur 416 012 Maharashtra. Mob : 9850046155; Email : [email protected]

NAGPUR Nirali Prakashan : Above Maratha Mandir, Shop No. 3, First Floor, Rani Jhanshi Square, Sitabuldi, Nagpur 440012, Maharashtra Tel : (0712) 254 7129; Email : [email protected]

DELHI Nirali Prakashan : 4593/15, Basement, Agarwal Lane, Ansari Road, Daryaganj Near Times of India Building, New Delhi 110002 Mob : 08505972553 Email : [email protected]

BENGALURU Nirali Prakashan : Maitri Ground Floor, Jaya Apartments, No. 99, 6th Cross, 6th Main, Malleswaram, Bengaluru 560003, Karnataka; Mob : 9449043034 Email: [email protected] Other Branches : Hyderabad, Chennai

Note: Every possible effort has been made to avoid errors or omissions in this book. In spite this, errors may have crept in. Any type of error or mistake so noted, and shall be brought to our notice, shall be taken care of in the next edition. It is notified that neither the publisher, nor the author or book seller shall be responsible for any damage or loss of action to any one of any kind, in any manner, therefrom. The reader must cross check all the facts and contents with original Government notification or publications. [email protected] Also find us on

| www.pragationline.com

www.facebook.com/niralibooks

The subject of ‘Mass Transfer’ forms an essential part of the syllabi of undergraduate courses in Chemical Engineering and other diversified courses such as Polymer Engineering, Petroleum Engineering and Petrochemical Engineering. A textbook on ‘ Mass Transfer’ is therefore always welcome, if it is comprehensive and yet easy to understand. Fundamental concepts in mass transfer and their mathematical relationships are difficult to understand for an average student. However without the understanding of these basic concepts, it is difficult to appreciate their applications in chemical engineering unit operations, processes, including their design and maintenance. Only a few books on “Mass Transfer” are available in the Indian market. Also these available books are costly and their orientation and presentation may not always suit the needs of Indian students community. The author of present book, Mr. Kiran Patil has nicely explained each concept in most logical and systematic manner. The mathematical derivations of several equations have been presented most rigorously and without any ambiguity. This may particularly be seen in chapters 2, 3 and chapter 5, wherein the topics on Diffusion mass transfer, Mass transfer coefficient and Gas absorption have been explained in detail. I appreciate the efforts of author to include a large number of problems with solutions and open-ended exercises with hints. The author has taken care to incorporate every concept in logical sequence and developed it step by step to its final form. This will facilitate self-learning by the reader and satisfy the long felt need of the students for such a standard textbook. Mr. Kiran Patil is a senior faculty member of Chemical Engineering Division of the Petrochemical Engineering Department of MAEER’s MIT, Pune (M.S.), India .His industrial and teaching experience and ability to effectively communicate with the student is amply revealed in every chapter of the book. I congratulate him for his untiring efforts and dedications in preparing this textbook. I congratulate Mr. Patil for his untiring efforts and dedications dedication in preparing this textbook. I hope and feel confident that this book will prove to be quite helpful to the engineering students community at large in long run and it will also be quite useful as a noteworthy standard textbook on the subject of “ Mass Transfer-I” for chemical engineering courses in the various Universities of India.

(Prof. Dr. Vishwanath D. Karad) Executive President and Director General, MAEER’S Maharashtra Institute of Technology, Pune-411 038, Maharashtra, India

I am happy to present the Fourth edition of the book, “ Principles and Fundamentals of Mass Transfer Operations - I” (VOLUME I) for students of degree courses of Chemical, Petrochemical, Petroleum, and Polymer Engineering and degree courses in Biotechnology of major Universities of India. My purpose in presenting the fourth edition of this book continues to be that of the previous edition: “ to provide a vehicle for teaching, either through a formal course or through self study, the techniques of, principles of equipment design for, the mass transfer operations of chemical engineering”. The enthusiastic response to the third edition of this book has encouraged me to revise this edition, which contains many new features. In this edition simple ideas are treated first, and are then extended to the more complex. Regarding this new edition: first of all I should say that in spirit it follows the earlier ones, and I try to keep things simple. In fact, I have reorganized materials as per the requirement of topic. I have taken this opportunity to improve and modernize many of the explanations, to modernize the design data, and to lighten the writing as best I could. I feel that problem-solving –the process of applying concepts to new situations –is essential to learning. Consequently this edition includes over 250 solved problems and over 200 homework problems with answers to help the student learn and understand the concepts being taught. The Fourth edition of book is thoroughly revised and split into two volumes : 1. Principles and Fundamentals of Mass Transfer Operations-I (Volume I) 2. Principles and Fundamentals of Mass Transfer Operations-II (Including Separation Processes) (Volume-II) Volume – I consists of following chapters :

1) Overview of Chemical Engineering, Profession, Separation Processes and Mass Transfer Operations 2) Diffusion Mass Transfer 3) Interphase Mass Transfer 4) Convective Mass Transfer 5) Absorption and Stripping Operations 6) Equipment for Gas-Liquid Mass Transfer Operations 7) Humidification and Dehumidification Operations 8) Drying Operations

Volume –II consists of following chapters:

1) 2) 3) 4) 5) 6) 7)

Crystallization Equilibrium Stage Operations Distillation Extraction and Leaching Operations Adsorption and Ion Exchange Membrane Separation Processes Mass Transfer with Chemical Reactions

I am grateful to Prof. (Dr.) V.D. Karad, Executive President and Director General, MAEER’s, Maharashtra Institute of Technology (MIT), Pune for his constant encouragement and moral support during the preparation of this book. My overwhelming debt is to my colleagues and students in the department. Malik types most of the book, and Angha provided valuable editorial help. I express my appreciation my publisher, Shri Dinesh Furia and Shri. Jignesh Furia for positive attitude for bringing out the Fourth edition of this book with great care and in a very attractive manner. Finally may I again thank our readers who, in the past, have made such helpful suggestions and have drawn to our attention errors, many of which would never have been spotted by the author. Would they please continue their good work! I am confident that this book will be definitely useful to the teachers and students of Chemical Engineering of major Universities in India.

Pune 5th September 2008 Teachers Day

Kiran D. Patil E-mail: [email protected]

I am happy to present the third edition of the book, “Principles of Mass Transfer Operations” for students of degree courses of Chemical, Petrochemical, Petroleum, and Polymer Engineering of major Universities of India. The enthusiastic response to the second edition of this book encouraged me to write this edition, which contains many new features: addition of new chapters (chapter 11, 12, 13 and 14) on Distillation, Extraction and leaching operations, Adsorption and ion exchange and membrane Separation Processes. The new material has been included in keeping with current demands of students. While Mass Transfer Operations is a required course in every undergraduate chemical engineering program in the world, there does not exist a comprehensive text on the subject that is specifically tailored to the undergraduate reader. Principles of Mass Transfer Operations responds to this need, providing a thorough, accessible text that presents the latest advances in the science as well as sets of targeted questions that challenge students’ knowledge. This book is designed to equip the reader with sufficient knowledge on mass transfer operations and face the challenges ahead. The focus throughout author’s peerless study is on making the student consider computation from the start of a mass transfer dilemma. Twenty-five to thirty problems at the end of each chapter ensure that readers will remain actively engaged with the material. The objective of this comprehensive and up-to-date textbook is to teach chemical engineering students the principles involved in analyzing a process and apply the desired mass transfer operation to separate the components involved. Chapters encompass : • Fundamentals of mass transfer • Interphase mass transfer • Convective mass transfer • Absorption And Stripping • Equipment For Gas-Liquid Mass Transfer Operations • Humidification And Dehumidification Operations • Drying Operations • Crystallization • Mass Transfer With Chemical Reactions • Distillation • Extraction And Leaching Operations • Adsorption and Ion Exchange • Membrane Separation Processes I express my appreciation and gratefulness to my publisher, Shri Dinesh Furia and Shri. Jignesh Furia and his entire team of Nirali Prakashan, Pune for most co-operative and painstaking attitude with untiring efforts for bringing out the third edition of this book with great care and in a very attractive manner. Suggestions for improvement in the style and content of the textbook, correction of errors if any will be greatly appreciated from the readers and will be incorporated in future editions.

Pune 30th March 2006 Gudi Padva

Mr. Kiran D. Patil E-mail: [email protected]

preface to the second edition I am happy to present the second edition of the book, “A Text book of Mass Transfer - I” for students of degree courses in Chemical, Petroleum, Petrochemical and Polymer Engineering of major Universities of India. The enthusiastic response to the first edition of this book and its several reprints encouraged me to write this edition, which contains many new features. The significant additions to chapter 2,3, 5 and 9 in greater detail. A new chapter on Mass transfer with chemical reaction is added. The book is written in simple and easy to follow language to understand the basic principles, fundamental concepts of mass transfer operations for students of chemical engineering. The course contents have been planned in such a way that the general requirements of engineering students are fulfilled. The new material has been included in keeping with current requirements of students. Over 350 problems (250 solved and 100 unsolved) in this revised edition have been carefully chosen to illustrate the fundamental concepts to help the students to bridge the gap between theory and application of mass transfer. More than 150 quiz questions, 250 study questions from various University question papers and summary notes are given for quick revision of the subject. Additional materials, charts, and data tables have been added, which are equally useful to practicing engineers. I am thankful to my colleagues, friends and my students for creating the atmosphere of academic excellence, which encouraged me to write this book. I am grateful to those observant readers who drew my attention to errors in the first edition. I express my appreciation and gratefulness to my publisher, Mr. Jignesh Furia and his team of Nirali Prakashan for most co-operative and painstaking attitude with untiring efforts for bringing out the second edition of this book with great care and in a very attractive manner. Suggestions for improvement from the readers in the style and content of the textbook, correction of errors if any will be greatly appreciated and will be incorporated in future editions. I am confident that this book will be definitely useful to the teachers and students of Chemical Engineering of major Universities in India.

Pune 18th September 2004 Ganesh Chaturthi

Mr. Kiran D. Patil E-mail: [email protected]

In view of revised syllabus of Chemical Engineering in all the universities in Maharashtra, Gujarat and other states, there is an urgent need to have standard textbooks for various chemical engineering subjects. When I set to write this book, only a very limited number of books on mass transfer were available in the Indian market. Also based on the market survey of available books in Chemical Engineering, it is observed that the books are written by foreign authors (with a few exceptions), and are often very costly. These are generally written for diploma students. Therefore it is not possible for undergraduate students to understand the basic theory and principles of various chemical engineering subjects to the depth they should. As a result their preparation is not up to the mark. This I have observed, often, in oral examinations. Thus there is a need of good quality undergraduate textbooks in Chemical Engineering subjects which: (i) cover topics in the syllabus up to appropriate details, (ii) discuss adequate problems, (iii) useful to all chemical engineering students in Maharashtra, and (iv) are inexpensive. I hope that the students studying chemical engineering will find the book useful for self-study. This book presents an elementary treatment of the principles of Mass transfer. A background in ordinary differential equations is helpful for proper understanding of the material. Presentation of the subject follows classical lines of separate discussions for principles of mass transfer operations, Diffusion mass transfer, Mass transfer coefficients, Interphase mass transfer, Gas Absorption, Humidification and dehumidification operations, Equipment for gas-liquid operations, Drying operations and Crystallization. Throughout the book emphasis has been placed on visualization of physical processes while, at the same time, relying on meaningful experimental data in those circumstances that do not permit a simple analytical solution. Theory alone won’t do, particularly from the point of view of examination in Indian universities and Engineering colleges. Hence emphasis is given on numerical examples in this book. A liberal number of numerical examples are given which include Diffusion mass transfer, Mass transfer coefficient, Gas absorption. A detailed introduction to diffusion and mass transfer is presented in order to acquaint the reader with these processes and to establish more firmly the important analogies between heat, mass, and momentum transfer. Problems are included at the end of each chapter. Some of these problems are of a routine nature to familiarize the student with the numerical manipulations and orders of magnitude of various parameters that occur in the subject of mass transfer. Other problems extend the subject matter by requiring students to apply the basic principles to new situations and develop their own equations. There is also a section at the end of each problem set designated as Practice exercise. The problems in these sections typically are open-ended and do not result in a unique answer. In some cases they are rather extended in length and require judgment decisions during the solution process. Over 100 such problems are included in the text. The subject of mass transfer is not static. New developments occur quite regularly, and better analytical solutions and empirical data are continuously made available to the professional in the field. Because of the huge amount of information that is available in

the research literature, the beginning student could easily be overwhelmed if too many of the nuances of the subject were displayed and expanded. The book is designed to serve as an elementary text, so the author has assumed a role of interpreter of the literature with those findings and equations being presented which can be of immediate utility to the reader. It is hoped that the student's attention is called to more extensive works in a sufficient number of instances to emphasize the depth that is available on most of the subjects of mass transfer. A serious student will do well to go through the end-of-chapter reference I am grateful to Prof. (Dr.) V.D. Karad, Director, Maharashtra Institute of Technology (MIT), Pune for his constant encouragement and moral support during the preparation of this book. I am very grateful to Dr. B.D. Kulkarni, Head, Chemical Engineering Division, National Chemical Laboratory, Pune for writing kind words regarding my book. I am very thankful to Prof. (Dr.) L.K. Kshirsagar for his guidance and suggestions during the preparation of this book. Thanks are due to Prof. Datta. B. Dandge for constructive criticism of the text. I am very thankful to my publisher M/s Nirali Publishers, Pune and especially to Mr. D.K. Furia for giving the present form to the book in minimum possible time. I am also thankful to most efficient and professional staff of M/s Nirali Publishers, Pune. I am immensely grateful to the members of my family who were encouraging, tolerant, totally supportive and generous. I am overwhelmingly happy to acknowledge, how much I owe them and how inadequate this sounds from the heart. I would express my sincere gratitude to my parents, who have always been a major source of encouragement in all my academic pursuits. Finally, my wife Varsha gives me a wonderful rich life. Finally to say, though I have made possible attempts to make this book flawless, yet some errors may surprisingly creep into. Suggestions from the readers regarding improvement of the utility of this book will be gratefully acknowledged and acted upon. I am confident that this book will be definitely useful to the teachers and students of Chemical Engineering. My efforts will be successful only to that extent to which it receives patronage from the wider section of the students and teachers of Chemical Engineering.

Pune 15th September 2002

Mr. Kiran D. Patil E-mail: [email protected]

1. Overview of Chemical Engineering Profession, Separation Processes and Mass Transfer Operations 1.1

1.2

1.3

1.4 1.5 1.6 1.7 1.8

1.1 − 1.30

Overview of Chemical Engineering Profession 1.1.1 Historical Background 1.1.2 Definition of Chemical Engineering 1.1.3 Disciplinary Definition 1.1.4 Occupational Definition 1.1.5 Domain of Chemical Engineering 1.1.6 Where do Chemical Engineering Work ? 1.1.7 The Chemical Engineering Industry 1.1.8 Job Titles 1.1.9 Other Career Opportunities 1.1.10 Job Descriptions Overview of Chemical Engineering Separation Processes 1.2.1 Background 1.2.2 Unit Operations 1.2.3 Examples of early separation techniques 1.2.4 Critical separations in human body 1.2.5 Mechanism of Separation Selection of Separation Processes 1.3.1 Basic Principle 1.3.2 General Rules 1.3.3 Other Considerations Selection of Feasible Separation Process Strengths and Weaknesses of Distillation and other Vapour-Liquid Separation Techniques Factors which favour Separation Processes for Liquid Mixtures Procedure for Separation Process Selection Classification of Separation Processes

1.2 1.2 1.4 1.4 1.4 1.4 1.5 1.6 1.7 1.8 1.8 1.10 1.10 1.11 1.11 1.12 1.12 1.13 1.13 1.13 1.13 1.14 1.15 1.16 1.16 1.16

1.9

Classes of Separation Operations

1.17

1.10

Mass Transfer of Cooking-Chemical Engineering through Cooking

1.19

General Overview

1.19

1.11.1

1.20

1.11 1.12

1.13 1.14

1.15 1.16

Introduction to Mass Transfer Operations

What are the benefits of Mass Transfer Knowledge ? 1.12.1 Where and why Mass Transfer is important ? 1.12.2 Benefits that an Engineer can achieve by learning Mass Transfer General Principles of Mass Transfer 1.13.1 Importance of Mass Transfer Operations Classification of Mass Transfer Operations 1.14.1 Similarities between the Operations 1.14.2 Differences between the Operations Separation Processes Choice of Separation Method

1.23 1.23 1.23 1.23 1.24 1.24 1.25 1.25 1.26 1.28

1.17

1.18

Methods of Conducting the Mass Transfer Operations 1.17.1 Solute Recovery and Fractionation 1.17.2 Unsteady-State Operation 1.17.3 Steady-State Operation 1.17.4 Stagewise Operation 1.17.5 Continuous-Contact Operation Design Principles References

2. Fundamentals of Diffusion Mass Transfer 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10

2.11

2.12

2.13

2.14 2.15

Introduction Kinetic Theory of Gases Molecular Diffusion Moleuclar Diffusion Vs Eddy Diffuion Molar Flux Molecular Diffusion in Fluids and Diffusion Coefficient or Diffusivity Diffusion in Binary Solution Steady-state Diffusion in Fluids at Rest and Laminar Flow Principles of Mass Transfer Properties of Mixtures 2.10.1 Concentration of Species 2.10.2 Mass Averaged Velocity Diffusion Flux 2.11.1 Fick's Law 2.11.2 Relation among Molar Fluxes Diffusivity : 2.12.1 Diffusivity in Gases 2.12.2 Diffusivity in Liquids 2.12.3 Diffusivity in Solids Steady state Diffusion 2.13.1 Diffusion through a Stagnant Gas Film 2.13.2 Pseudo – Steady – State diffusion through a Stagnant Gas Film 2.13.3 Equimolar Counter Diffusion 2.13.4 Diffusion into an Infinite Stagnant Medium 2.13.5 Diffusion in Liquids 2.13.6 Mass Diffusion with Homogeneous Chemical Reaction 2.13.7 Diffusion in Solids Transient Diffusion Differential Equations of Mass Transfer Solved Problems Exercise for Practice Nomenclature References

3. Mass Transfer Coefficients 3.1 Introduction 3.2 Types of Mass Transfer Coefficients 3.2.1 Definition of Mass Transfer Coefficient 3.2.2 Mass Transfer Coefficient for Equimolar Counter Diffusion 3.2.3 Mass transfer Coefficient for A diffusing through stagnant, non-diffusing B

1.28 1.28 1.29 1.29 1.29 1.29 1.29 1.30

2.1 − 2.152 2.2 2.2 2.4 2.5 2.6 2.6 2.7 2.8 2.9 2.10 2.10 2.11 2.11 2.12 2.12 2.13 2.13 2.14 2.15 2.16 2.17 2.19 2.20 2.21 2.24 2.25 2.27 2.33 2.34 2.41 2.140 2.150 2.152

3.1 − 3.88 3.2 3.2 3.2 3.3 3.3

3.3 3.4 3.5 3.6 3.7 3.8 3.9 3.10

Local Mass Transfer Coefficient Mass Transfer Coefficient in Turbulent Flow Eddy/Turbulent Diffusion Theories of Mass Transfer Introduction to Convective Mass Transfer Convective Mass Transfer Coefficient Significant Parameters in Convective Mass Transfer Application of Dimensionless Analysis to Mass Transfer 3.10.1 Transfer into a stream flowing under forced convection 3.10.2 Transfer into a phase whose motion is due to natural convection 3.11 Analogies Among Mass, Heat and Momentum Transfer 3.11.1 Reynolds Analogy 3.11.2 Chilton – Colburn Analogy 3.11.3 Taylor – Prandtl Analogy 3.12 Convective Mass Transfer Correlations

3.4 3.5 3.5 3.6 3.10 3.11 3.11 3.12 3.13 3.15 3.16 3.17 3.17 3.18 3.18

3.12.1

For Flow Around Flat Plat

3.18

3.12.2

For Flow Around Single Sphere

3.19

3.12.3

For Flow Around Single Cylinder

3.19

3.12.4

For Flow Through Pipes

3.19

3.12.5

Mass Transfer to Suspension of Small Particles

3.20

3.12.6

Mass Transfer in Packed Beds

3.20

3.12.7

Mass Transfer in Boundary Layer

3.21

3.13 Mass Transfer between Phases

3.21

3.14 Simultaneous Heat and Mass Transfer

3.22

3.14.1

Condensation of Vapour on Cold Surface

3.23

3.14.2

Wet Bulb Thermometer

3.24

Solved Problems

3.25

Exercise for Practice

3.78

Nomenclature

3.86

References

3.88

4. Interphase Mass Transfer 4.1 4.2 4.3 4.4

4.5 4.6 4.7 4.8 4.9

4.10 4.11

Introduction Concept of Equilibrium Diffusion between phases Film concept in mass transfer 4.4.1 Two-film theory of mass transfer 4.4.2 Two-film theory and equilibrium solubility curve 4.4.3 Analysis of mass transfer process using two-film theory Local two-phase mass transfer Two resistance concept Resistance to mass transfer Overall mass transfer coefficient Material balances for : (a) Steady-state concurrent operations (b) Steady-state counercurrent operations (c) Cascades Stages and stagewise operations Kremser Equation

4.1 − 4.48 4.1 4.2 4.3 4.3 4.4 4.5 4.6 4.7 4.8 4.9 4.10 4.11 4.11 4.11 4.16 4.17

Solved Problems Exercises for Practice Nomenclature References

5. Gas Absorption

4.19 4.44 4.46 4.48

5.1 − 5.98

5.1

Background

5.1

5.2

Physical Vs. Chemical Absorption

5.5

5.3

Method of Operation

5.5

5.4

Henry's Law for Gas Absorption

5.9

5.5

Minimum Liquid/Gas Ratio for Absorption

5.9

5.6

Choice of Solvent for Gas Absorption

5.11

5.7

HETP (Height Equivalent to a Theoretical Plate)

5.12

5.8

Packing Height : The Method of Transfer Units

5.12

5.9

Number of Theoretical Trays

5.16

5.10

The Kremser-Brown-Souders (KBS) Equation

5.17

5.11

Introduction

5.18

5.12

Equipment for Absorption/Stripping

5.18

5.13

Equilibrium Curve

5.20

5.14

Interfacial Mass Transfer : Review

5.23

5.15

Definition of Transfer Coefficients

5.24

5.16

Height of Packed Tower

5.28

5.17

Transfer Unit

5.31

5.18

Evaluation of NTU's Integral

5.33

5.19

Concept of HETP

5.36

5.20 5.21

Analogy with Double Pipe Heat Exchanger Relationship among Hox, Hoy, Hx, Hy

5.36 5.37

5.22

Pressure Drop in Packed Beds

5.38

5.23

Tower Diameter

5.41

5.24

Absorption with Chemical Reaction

5.42

5.25

General Procedure for Design of Packed Absorption Columns

5.43

5.26

Design Considerations

5.44

5.27

Typical Absorber Design Problem

5.45

Solved Problems

5.52

Exercise for Practice

5.92

Nomenclature

5.96

References

6. Humidification and Dehumidification Operations

5.98

6.1 − 6.46

6.1 Introduction

6.2

6.2 General Principles of Humidification and Dehumidification

6.2

6.3 Vapour Gas Mixture

6.2

6.4 Definitions of Humidity Terms

6.3

6.5 Adiabatic Saturation Process

6.5

6.6 Theory of Wet Bulb Temperature

6.6

6.7 Humidity Chart

6.7

6.8 The Lewis Relation

6.7

6.9 Gas Liquid Contact Operations

6.7

6.10 Adiabatic Operations

6.10

6.11 Humidification and Dehumidification Equipments

6.15

6.11.1

Cooling Tower Overview

6.15

6.11.2

Operating Principle of Cooling Towers

6.15

6.11.3

Cooling Tower Selection

6.16

6.11.4

Hot Water Distribution Systems

6.17

6.11.5

Air Flow Distribution Systems

6.17

6.12 Cooling towers : Design and Operational Considerations

6.19

6.12.1

Types of Cooling Towers

6.19

6.12.2

Cooling Tower Theory

6.20

6.12.3

Design Considerations

6.24

6.13 Height of Cooling Tower

6.25

Solved Problem

6.27

Exercise for Practice

6.41

Nomenclature

6.44

References

6.46

7. Equipment for Gas-Liquid Operations

7.1 − 7.34

7.1 Introduction

7.1

7.2 Sparged Vessels (Bubble Columns)

7.2

7.3 Mechanically Agitated Vessels

7.5

7.3.1

Mechanical Agitation of Single-Phase Liquids

7.5

7.3.2

Vortex Formation and Prevention

7.7

7.3.3

Similarity considerations in Agitated Vessels

7.7

7.4 Tray Towers

7.8

7.4.1

General Characteristics

7.4.2

Bubble Cap Trays

7.11

7.4.3

Proprietary Trays : Linde Trays, Valve Trays and Counterflow Trays

7.12

7.4.4

Tray Efficiency

7.13

7.4.5

Design Considerations

7.14

7.5 Packed Towers

7.9

7.15

7.5.1

Introduction

7.15

7.5.2

Construction Details

7.15

7.5.3

Gas and Liquid Phase Coefficients

7.20

7.5.4

Design Criteria for Packed Towers

7.21

7.5.5

Comparision between Tray Towers and Packed Columns

7.23

7.6 Ventrui Scrubbers

7.23

7.7 Wetted Wall Tower

7.24

7.8 Spray Tower

7.25

Solved Problems

7.26

Nomenclature

7.31

References

7.34

8. Drying Operations

8.1 − 8.56

8.1

Introduction

8.1

8.2

Expression of Moisture Content

8.2

8.3

Equilibrium in Drying

8.3

8.4

Types of Moisture

8.3

8.5

Mechanism of Batch Drying

8.3

8.6

Drying Tests

8.4

8.7

Rate of Drying Curve

8.4

8.8

Time required for Drying

8.5

8.9

Drying Time for Droplets

8.8

8.9.1

Constant Rate Period

8.8

8.9.2

Falling Rate Period

8.9

8.9.3

Droplet Trajectory

8.10

8.10

Heat Transfer in Dryers

8.10

8.11

Mechanism of Moisture Movement within the Solid

8.12

8.12

Equipments for Drying

8.13

8.13

Selection, Sizing and Costs

8.18

8.14

Efficient Energy Utilization in Drying

8.18

Solved Problems

8.19

Exercise For Practice

8.54

Nomenclature

8.56

References

8.56

Appendix A : SI Systems, Fundamental Constants and Conversion Units A.1 Appendix B : A Few Useful Observations B.1 Appendix C : Humidification and Water Cooling in a Packed Column C.1 Appendix D : Dimensionless Groups and Correlations in Heat and Mass Transfer D.1 Appendix E : Physical Properties E.1 Appendix F : Determining Mass Transfer Coefficients F.1 Appendix G : Algebraic Solution of Equilibrium Stage Problems : The Kremser Equation G.1 Study Questions S.1




− A.7 − B.2 − C.5 − D.4 − E.12 − F.3 − G.5 − S.10

1

CHAPTER

OVERVIEW OF CHEMICAL ENGINEERING PROFESSION, SEPARATION PROCESSES AND MASS TRANSFER OPERATIONS 1.1

Overview of Chemical Engineering Profession 1.1.1

Historical Background

1.1.2

Definition of Chemical Engineering

1.1.3

Disciplinary Definition

1.1.4

Occupational Definition

1.1.5

Domain of Chemical Engineering

1.1.6

Where do Chemical Engineering Work ?

1.1.7

The Chemical Engineering Industry

1.1.8

Job Titles

1.1.9

Other Career Opportunities

1.1.10 Job Descriptions 1.2

1.3

Overview of Chemical Engineering Separation Processes 1.2.1

Background

1.2.2

Unit Operations

1.2.3

Examples of early separation techniques

1.2.4

Critical separations in human body

1.2.5

Mechanism of Separation

Selection of Separation Processes 1.3.1

Basic Principle

1.3.2

General Rules

1.3.3

Other Considerations

(1.1)

Principles of Mass Transfer Operations − I (Vol. − I)

1.2

Overview of Chemical Engineering Profession, ……

1.4

Selection of Feasible Separation Process

1.5

Strengths and Weaknesses of Distillation and other Vapour-Liquid Separation Techniques

1.6

Factors which favour Separation Processes for Liquid Mixtures

1.7

Procedure for Separation Process Selection

1.8

Classification of Separation Processes

1.9

Classes of Separation Operations

1.10

Mass Transfer of Cooking-Chemical Engineering through Cooking

1.11

General Overview 1.11.1

1.12

1.13

What are the benefits of Mass Transfer Knowledge ? 1.12.1

Where and why Mass Transfer is important ?

1.12.2

Benefits that an Engineer can achieve by learning Mass Transfer

General Principles of Mass Transfer 1.13.1

1.14

Introduction to Mass Transfer Operations

Importance of Mass Transfer Operations

Classification of Mass Transfer Operations 1.14.1

Similarities between the Operations

1.14.2

Differences between the Operations

1.15

Separation Processes

1.16

Choice of Separation Method

1.17

Methods of Conducting the Mass Transfer Operations

1.18

1.17.1

Solute Recovery and Fractionation

1.17.2

Unsteady-State Operation

1.17.3

Steady-State Operation

1.17.4

Stagewise Operation

1.17.5

Continuous-Contact Operation

Design Principles References

1.1 OVERVIEW OF CHEMICAL ENGINEERING PROFESSION 1.1.1 Historical Background Chemical Engineering . . . Takes chemistry out of the lab and into the world Chemical engineers apply the principles of chemistry, maths, and physics to the design and operation of large-scale chemical manufacturing processes. They translate processes developed in the lab into practical applications for the production of products such as plastics, medicines, detergents, and fuels; design plants to maximize productivity and minimize costs; and evaluate plant operations for performance and product quality.

Principles of Mass Transfer Operations − I (Vol. − I)

1.3

Overview of Chemical Engineering Profession, ……

Chemical engineers are employed by almost all companies in the chemical process industry. Their work also extends to processes in nuclear energy, materials science, food production, the development of new sources of energy, and even medicine. In addition to process and product development and design, chemical engineers work in areas such as production, research, environmental studies, market analysis, data processing, sales, and management. They affect or control at some stage the materials or production of almost every article manufactured on an industrial scale. . . . Is the broadest branch of engineering ? Chemical engineering is broader in scope than the other branches of engineering because it draws on the three main engineering foundations : maths, physics, and chemistry - whereas the other branches are based on only the first two. A specific interest in chemistry combined an aptitude for maths and science attracts individuals to the profession. The curriculum of study for chemical engineering is similar to that for chemistry but includes course work in engineeringrelated areas such as thermodynamics, fluid dynamics, process design, and control and electronics. Once processes and equipment are designed, chemical engineers remain on hand at a production facility to solve problems that occur as the processes continue. When changes occur that upset a running system, chemical engineers analyze samples from the system, looking at parameters such as temperatures, pressures, and flow rates to determine where the problem exists. They also work on expanding projects, evaluating new equipment, and improving existing equipment and processes. Meeting safety, health, and environmental regulations is also a large part of a chemical engineer’s work life. As the Industrial Revolution (18th Century to the present) steamed along certain basic chemicals quickly became necessary to sustain growth. Sulfuric acid was first among these "industrial chemicals". For all intents and purposes the chemical engineering profession began in 1888. An effort in 1880, by George Davis to unite these varied professionals through a "Society of Chemical Engineers" proved unsuccessful. However, this muddled state of affairs was changed in 1888, when Professor Lewis Norton of the Massachusetts Institute of Technology introduced "Course X" (ten), thereby uniting chemical engineers through a formal degree. Other schools, such as the University of Pennsylvania and Tulane University, quickly followed suit adding their own four-year chemical engineering programs in 1892 and 1894 respectively. A Few Dates in the History of Chemical Engineering : 1880 : George Davis proposed a "Society of Chemical Engineers" in England. 1888 : The Massachusetts Institute of Technology (MIT) began "Course X" (ten), the first fouryear chemical engineering program in the United States. 1908 : The American Institute of Chemical Engineers (AIChE) was founded.

Principles of Mass Transfer Operations − I (Vol. − I)

1.4

Overview of Chemical Engineering Profession, ……

1915 : The Unit Operations Concept – That a chemical process could be broken down into a series of steps that might include : heat transfer, transportation of solids and liquids, crystallization, distillation, evaporation etc. – was stated by Arthur Little. 1925 : The AIChE began accrediting chemical engineering programs. 1.1.2 Definition of Chemical Engineering Chemical engineering is one of the broadest fields of engineering. This breadth stems from the fact that the discipline is founded on mathematics and on all the basic sciences, namely, chemistry, physics, as well as biology, making it a truely interdisciplinary field of study. In comparison, the other major engineering disciplines are founded mainly on maths and physics. Thus, there are two alternate definitions of chemical engineering : Chemical Engineering

Physics Bilology

Maths Chemistry

Fig. 1.1 : Foundation of Chemical Engineering Science

1.1.3 Disciplinary Definition Chemical Engineering is the profession in which knowledge of mathematics, physics, chemistry and biology, gained by study, experience, and practice, is applied with judgment to develop economic and safe ways of using materials to benefit mankind. 1.1.4 Occupational Definition Chemical Engineering is a broad discipline dealing with processes (industrial and natural) involving the transformation (chemical, biological, or physical) of matter or energy into forms useful for mankind, economically and without compromising environment, safety, or finite resources. 1.1.5 Domain of Chemical Engineering The molecular basis of the chemical and physical transformation of matter that Chemical Engineers are concerned with, coupled with the global principles underlying the discipline (conservation of mass, energy, and momentum, the notion of unit operations, and reaction kinetics) allows them to work from the nano-scale (design of catalysts, or molecular design of drugs) to the meso-scale (petroleum refinery) to the global-scale (air pollution modeling and control). This is unique to Chemical Engineering and is depicted in the figure as shown on next page.

Principles of Mass Transfer Operations − I (Vol. − I)

1.5

Overview of Chemical Engineering Profession, ……

Time scale Month

Enterprise

Week Site day Plants h

Process units min

Single and multiphase systems

s

Particles, thin films

ms

Small Chemical scale

Molecule cluster

Intermediate Large

ns

Molecules ps 1 pm

1 nm

1 mm

1 mm

1m

1 km

Length scale

Fig. 1.2 : Domain of Chemical Engineering in Time and Length Scale

1.1.6 Where Do Chemical Engineers Work ? Chemical Engineers can do just about anything ! The broad training receive as a chemical engineering student equips to not only have highly rewarding careers in the chemical industry, but to pursue further education and careers in just about any other field including business management, banking and finance, law, teaching, and medicine etc. Now-a-days chemical engineering graduates have entered just about every profession imaginable. For this reason, chemical engineering is sometimes called the "liberal arts of engineering". Chemical or process engineers turn great ideas discovered in laboratories into practical devices and processes that : •

Improve our quality of life;

•

Protect the environment;

•

Ensure products and services we purchase are cheaper and of better quality; and

•

That industry increases competitiveness, thereby protecting and creating jobs and wealth for communities.

Chemical engineers do this using a combination of biology, biochemistry and/or chemistry with mathematics (as well as a bit of economics and finance) to predict how these ideas will work on a larger-scale outside the laboratory in the real world, and then building and operating the equipment needed to bring these ideas to life. Chemical engineers have helped do this by performing "research and development" or by "design and operation" of many chemical manufacturing processes.

Principles of Mass Transfer Operations − I (Vol. − I)

1.6

Overview of Chemical Engineering Profession, ……

Work Description : Chemical engineers design and operate plants and processes for large-scale production of chemical products. They use chemistry, physics, and mathematical equations to solve real problems and design ways to produce products safely and economically. Working Conditions : Chemical engineers typically work in manufacturing plants, research laboratories, or pilot plant facilities. They work around large-scale production equipment that is housed both indoors and outdoors. Often they are required to wear safety protective equipment, such as hard hats, goggles, and steel-toe shoes. Workdays may involve of moving from place to place within a facility. Chemical engineers also work in business and management offices; these positions, however, often require visiting research and production facilities. Interaction with other people who are part of a team is critical to the success of projects. Personal Characteristics : A strong interest in chemistry, maths, and physics is vital to success in this field because chemical engineering draws on all three disciplines. Chemical engineers are trained to apply lab processes to large-scale production, monitor processes, and understand highly technical material. As a result, thinking analytically, solving problems, and being creative are essential. Because projects often involve complex processes and problems that require teamwork and the preparation of reports, good interpersonal, oral, and written communication skills are highly desirable. 1.1.7 The Chemical Engineering Industry The following is a listing of the traditional chemical industry subdisciplines along with examples of some companies employing chemical engineers : •

Fuels and Energy Petroleum (ONGC, DGH, OIL, IOCL, GSPC, RELIANCE, CAIRN ENERGY) Natural Gas/Utilities (ONGC, PETRONET, GAIL) Hydrogen (BOC) Batteries (EXIDE) Fuel Cells

•

Commodity Chemicals Agricultural Chemicals (HOCL, INDOFIL, HIL) Plastics (GE Plastics) Rubber (CEAT Tyres)

•

Specialty/Consumer Chemicals Adhesives (Fevicol) Specialty Chemicals (Kodak) Paints, Varnishes, Inks (Asian Paints) Soaps, Detergents (Proctor and Gamble, Godrej Soaps) Cosmetics, Perfumes (Clairol)

Principles of Mass Transfer Operations − I (Vol. − I) •

1.7

Overview of Chemical Engineering Profession, ……

Advanced Materials Glass Ceramics (Kajaria) Composites Polymers (Reliance) Metals (Relience, Pearl) Catalysts

•

Textiles Petrochemicals (IPCL, Reliance) Air Chemicals (Air Products, Praxair)

•

Medicine Biotechnology (BICON) Biomedical Devices

•

Transportation Auto (BAJAJ) Aerospace Microelectronics (Intel, IBM, GE) Pharmaceutical (Pfizer) Process Control (Fisher, Foxboro, Tata Honeywell) Process Design (Aspen Plus) Food and Beverages (Coca-Cola, National Starch) Pulp and Paper (Ballarpur Paper Industries) Design and construction (UHDE, SULZER, CHEMTEX, PRAJ etc.) Environmental, safety, and health

1.1.8 Job Titles Within the subdisciplines listed above, chemical engineers work as : •

Production Engineer

•

Process Engineer

•

Production Manager

• • • • • • • • • • •

Maintenance Engineer Process Control Engineer Sales and Marketing Engineer Design Engineer Environmental Engineer Manufacturing Engineer Research Engineer Development Engineer Technical Service Engineer Quality Control Engineer Chief Executive Officer

Principles of Mass Transfer Operations − I (Vol. − I)

1.8

Overview of Chemical Engineering Profession, ……

1.1.9 Other Career Opportunities In addition to traditional engineering jobs, many chemical engineers work in the following areas : • Consultant • Environmental Consultant • Business Management • Law • Patent Attorney • Medicine • Government • Military • Finance • University Professor • Entrepreneur 1.1.10 Job Descriptions The specific responsibilities of chemical engineers, though varying among industries and even within the same company, can be categorized in general terms. Titles such as "process design engineer" and "project engineer" will describe positions in most industries, whatever the type of work, process, equipment, and product that is involved. Job Function

Description

Process Design Engineer

Designs manufacturing facilities and the equipment and material used inside. Process design engineers work with teams of engineers to develop new or improved processes to meet a company’s production needs.

Environmental Engineer

Develops techniques to reduce and recover usable materials from waste created during manufacture of a product. Designs waste storage and treatment facilities, as well as pollution control strategies for plant operations. Environmental engineers may be responsible for monitoring all systems in a facility for compliance with government environmental regulations.

Plant Process Engineer

Provides technical support to staff and troubleshoots processes in a production facility to keep a plant running efficiently. Plant process engineers work closely with equipment operators to get feedback on the operations of each process and determine how to avoid shutdowns. They may also be involved with some design work for improvement projects.

Process Safety Engineer

Designs and maintains plants and processes that are safer for workers and communities. Process safety engineers may conduct safety analyses of new and existing equipment, and train employees on how to safely operate a new piece of equipment.

Project Engineer

Oversees the design and construction of specific processes in a facility. After construction, they may assist in equipment testing, operator training, and plant start-up. Project engineers may be responsible for the design, and start-up, of a specific process in a facility.

Principles of Mass Transfer Operations − I (Vol. − I) Job Function

1.9

Overview of Chemical Engineering Profession, …… Description

Consultant

Works for many different customers and brings specialized knowledge to individual projects. Consultants in a construction company may work with teams of engineers to design and construct an expansion project for a pharmaceutical company.

Product Engineer

Follows the production cycle of a particular product to ensure that it is being produced according to specification. Product engineers may work with marketing and R&D to ensure that a product will meet the needs of customers, then sees the product through production. They may work on new products or special variations of existing products.

Manufacturing Production

Responsible for the day-to-day operation of a specific manufacturing

Engineer

process. Manufacturing production engineers work directly with operators to ensure that a particular product is made according to specifications.

Research & Development Engineer

Seeks out new and more efficient ways of using and producing existing products. Explores and develops new processes and products and determines their usefulness and applicability. Chemical engineers working in research and development may work with chemists and other engineers to develop a new process or new product that will better meet customer needs.

Project Manager

Oversees the overall design and construction of a facility, and then manages ongoing operations. Project managers may manage a group of project engineers during the design and construction of a new facility.

Attorney

Specializes in intellectual property law, patent law, technology transfer, environmental compliance, and safety issues. Patent attorneys obtain patents for clients and monitor the marketplace for possible patent infringements.

Biomedical Specialist

Works alongside physicians to develop systems that track critical chemical processes in the body. Biomedical specialists may be involved in the design of artificial organs, such as hearts and lungs.

Computer Applications and Technology Engineer

Designs instrumentation and programs systems monitor and control certain processes. Automation engineers may design systems to monitor a series of processes in a chemical, petroleum or biotechnology facility.

Technical Manager

Responsible for the engineering staff and programs at a facility. Manages people, research programs, and daily operations of the engineering functions. Technical managers may oversees the R&D program and work with plant managers to plan and implement the funding programs and expansion necessary to develop a new product.

Business Co-ordinator

Develops budgets and capital projections for a facility or process. Business co-ordinators work closely with production and design team members to ascertain the exact needs of a new process then plans the capital needs necessary to implement the program.

Principles of Mass Transfer Operations − I (Vol. − I) Job Function

1.10

Overview of Chemical Engineering Profession, …… Description

Professor

Instructs students in the field of chemical engineering and conducts research in pertinent areas. Professors may teach several classes in chemical engineering, be members on university committees, and conduct research using funding from government, corporate, or private grants.

Quality Control Engineer

Monitors the manufacture of a product to ensure that it meets specifications. Also, tests materials to determine how they perform over time. Quality control engineers may bring samples of a product in from a field test, or from a normal application, and test them to determine how specific properties, such as strength, color, and weatherability, change over time.

Regulatory Affairs Engineer

Researches, develops, and monitors policies and procedures that companies must follow to ensure the proper handling of chemicals and chemical components. Chemical engineers in regulatory affairs may be government employees, who study the environmental impact of a new chemical, then recommend appropriate guidelines for the chemical’s use.

Technical Services Engineer

Works with customers, usually on-site, to solve production problems caused by a specific process or machine. Chemical engineers working in technical services may represent the manufacturer of a specific machine to determine why it is not performing as designed. They often must understand the other steps in the production process to determine if there is a breakdown in another area.

Sales and Marketing Engineer

Assists customers in solving production and process problems by providing products and services to meet their specific needs. Chemical engineers in sales use their technical knowledge to sell chemicals, equipment, and other products, and provide follow-up services and training where needed.

1.2 OVERVIEW OF CHEMICAL ENGINEERING SEPARATION PROCESSES 1.2.1 Background The traditional concept of unit operations has been a major factor in the phenomenal success of chemical engineers and chemical engineering in last fifty years or so. Unit operations is concerned with those separation processes that depend upon the differences in the physical properties, rather than chemical behaviour. Such processes depend either upon a difference in composition of phases at equilibrium or upon a difference in the rate of mass transfer of constituents of mixture. A typical process, which a chemical engineer might work with, is production of gasoline from crude oil. Process = Sequence of “ unit operations” (physical changes) + Chemical reactors. In this process chemical engineers realized that the many process involved the same physical and chemical operations. Examples of this include : filtration, drying, distillation, crystallization, grinding, sedimentation, combustion, catalysis, heat exchange, extrusion, coating, and so on. These "Unit Operations" repeatedly find their way into industrial chemical practice, and became a convenient manner of organizing chemical engineering knowledge.

Principles of Mass Transfer Operations − I (Vol. − I)

1.11

Overview of Chemical Engineering Profession, ……

Methane-rich gas

Ethane

Reboiled absorption Absorber Wet natural gas Distillation

Propane

Isobutane

Distillation

Distillation

Distillation

Normal

Natural

Butane

Gasoline

Fig. 1.3 : A Typical Chemical Engineering Separation Process

1.2.2 Unit Operations • • • • • •

Separations Mixing Heating/cooling Pumps Drying Humidification

• • • • • • • •

Distillation Leaching Extraction Absorption/Stripping Membrane Separations Mechanical Separations Crystallization Adsorption in fixed-bed separations Distillation Gas absorption

Use

Crystallization Absorption Membranes Chromatography Technology maturity

Fig. 1.4 : Technology Maturity and Use for Unit Operations

1.2.3 Examples for early separation techniques • • •

Salt by evaporating seawater, Metals extracted from ores, Perfumes extracted from plants,

Principles of Mass Transfer Operations − I (Vol. − I) • •

1.12

Overview of Chemical Engineering Profession, ……

Various drinks, medicines extracted from plants, Liquors distilled form plants.

1.2.4 Critical separations in human body •

Separation of oxygen from air and of CO2 from blood in the lungs.

•

Selective removal of water and waste products of metabolism from blood in the kidneys.

•

Selective absorption of nutrients in the intestines.

In chemical industry, separation techniques are mainly used for : •

Isolating and manufacturing valuable chemicals from mixtures,

•

Removing impurities from raw materials,

•

Purification of products,

•

Removal of contaminants (Environmental Protection).

and

valuable

constituents

from

effluent

streams

1.2.5 Mechanism of Separation Mixing : spontaneous, natural process accompanied by increase in entropy and randomness. Second Law of Thermodynamics states that all natural processes take place so as to increase the entropy or randomness of the universe. Separation : Reverse of mixing, demixing. Equivalent of thermodynamic work has to supplied to cause the separation to occur. Separation process is the technology of unmixing and isolating the wanted product or products economically, feasibly and without harming the environment. To accomplish a separation there are usually several competing techniques available. Each technique has to be analyzed in terms of : 1.

Economic conditions

2.

Customer requirements

3.

Applicable official regulations

In order to affect a separation, separating agents are needed in the form of either : •

Energy input (heat, pressure, electricity, magnetism, kinetic or potential energy).

•

Withdrawal of energy (cooling, freezing).

• Matter (filter, membrane, chemicals). A separation process is an operation carried out in a special separation device which transforms a mixture into at least two product streams which are different in composition. In the separation device, separation takes place due to an imposed gradient such as temperature, concentration, pressure or electrical field. Two important elements of separation are : 1. Separating agent used (heat, pressure, solvent, matter such as resins, filters, adsorbents etc.). 2. Principle of separation used, separation gradient applied (temperature, concentration, chemical potential, magnetic field etc.).

Principles of Mass Transfer Operations − I (Vol. − I)

1.13

Overview of Chemical Engineering Profession, ……

1.3 SELECTION OF SEPARATION PROCESSES 1.3.1 Basic Principle Examine physical properties and exploit differences in physical properties between components being separated. 1.3.2 General Rules 1. Consider processes that exploit vapor-liquid equilibrium first (e.g., distillation) since these processes tend to be energy efficient and simple to design, build, and operate. Difference in boiling points between components being separated usually gives a good indication of the feasibility : • A boiling point difference (∆TBoiling) of greater than 100°C usually indicates a single stage process is feasible. • A boiling point difference between 10 and 100°C usually indicates that a multistage distillation process is feasible. • A boiling point difference less than 10°C usually means distillation is unlikely to work since a very large number of stages is required. In general, a boiling point difference of 10°C corresponds to a relative volatility of 1.2 between components. In close cases, estimate the relative volatility and use it as a criterion since a relative volatility greater than 1.2 is usually needed for multistage distillation to be successful. 2.

Don't use distillation if :

•

Boiling points are high (e.g., greater than 200°C).

•

Components decompose when heated.

•

Boiling point difference is less than 10°C (see above).

3. If vapor-liquid equilibrium processes are not promising, consider liquid-liquid extraction next. Solubilities in various solvents are the key physical property. •

For physically interacting solvents (e.g., no hydrogen bonding) the "cohesive energy density" (which is also related to the solubility parameter in regular solution theory) is the key property. Try to select a solvent so that the solubility parameter of the solvent is within 3 (cal/cm3)2 of the solubility parameter of one of the components to be separated and more than 3 units from the other component.

For chemically interacting solvents, try to exploit hydrogen bonding, acid-base interactions, or other "specific" chemical interactions. 4. Consider membrane processes next. A key consideration is whether an appropriate membrane material exists. 5. Consider adsorption processes next. •

1.3.3 Other Considerations •

Favour processes that remove a minor component from a major component.

•

Adsorption processes become favourable as the solute concentration becomes very small since the fact these processes are batch process becomes less of a disadvantage under these conditions. This is because an adsorption bed can be left on stream for long time between regeneration cycles when the solute concentration is dilute.

Principles of Mass Transfer Operations − I (Vol. − I)

1.14

Overview of Chemical Engineering Profession, ……

1.4 SELECTION OF FEASIBLE SEPARATION PROCESS The selection of a best separation process must be frequently made from among a number of feasible candidates. When the feed mixture is to be separated into more than two products, or a combination of two or more operations may be best. Even when only two products are to be produced, a hybrid process of two or more operations may be most economical. The important factors in the selection of feasible separation operations are listed in Table. These factors have to do with feed and product conditions, property differences.

Table 1.1 : Factors That Influences The Selection of Feasible Separation Operations

A. Feed Conditions : 1.

Composition, particularly concentration of species to be recovered or separated

2.

Flow rate

3.

Temperature

4.

Pressure

5.

Phase state (Solid, liquid, and/or gas)

B. Product Conditions : 1.

Required purities of products

2.

Temperature

3.

Pressure

4.

Phase States

C. Property differences that may be exploited : 1.

Molecular

2.

Thermodynamic

3.

Transport

D. Characteristics of separation operation : 1.

Ease of scale-up

2.

Ease of staging

3.

Temperature, pressure, and phase-state requirements

4.

Physical size requirements

5.

Energy requirements

Principles of Mass Transfer Operations − I (Vol. − I)

1.15

Overview of Chemical Engineering Profession, ……

Table 1.2 : Examples of Separation Processes and Separating Agents Process Absorption

Separating agent(s) Solvent

Application(s) Removal of carbon dioxide and hydrogen sulfide from natural gas with amine solvents.

Adsorption and ion Adsorbent/resin

Separation of meta- and paraxylene, air separation, water

exchange

demineralisation.

Chromatography

Adsorbent

Separation of sugars.

Crystallisation

Heat removal

Production of beverages such as "ice" beer.

Distillation

Heat

Propylene/propane separation, production of gasoline from crude oil, air separation.

Drying

Heat

Drying of ceramics, plastics and foods.

Electrodialysis

Membrane

Water desalination.

Evaporation

Heat

Water desalination, sugar manufacture.

Extraction

Solvent

Recovery of benzene/toluene/xylenes from gasoline reformate, removal of caffeine from coffee.

Membranes

Membrane

Separation of hydrogen from hydrocadrbons, concentration of fruit juices, water desalination.

Stripping

Stripping gas

Removal of benzene from wastewaters.

1.5 STRENGTHS AND WEAKNESSES OF DISTILLATION AND OTHER VAPOURLIQUID SEPARATION TECHNIQUES •

Economically, if a stream can be easily vaporized or condensed, distillation or a related vapour-liquid seaparation technique is generally preferred. Reasons of this choice are :

1.

Large number of stages at reasonable equipment costs. High degrees of separation even with low relative volatilities, down to 1.2.

2.

Process equipment requirements in terms of size of units needed are small compared to other separation techniques involving mass separting agent (MSA’s) Throughput/unit volume of equipment is highest for distillation.

3.

Low scale-up factors. Doubling a distillation column’s capacity increases the capital investment by abt. 1.5 times. Higher scale-up factors for other techniques often lead to multiple units arranged in parallel, resulting in higher capital costs.

4.

Energy costs usually run second in impact to capital costs. Substitution of a more complex, higher-investment but lower-energy usage process for an existing distillation usually does not pay off in reduced energy costs. Besides, distillation is suitable for heat integration to cut energy costs.

Principles of Mass Transfer Operations − I (Vol. − I) 1.6 FACTORS WHICH MIXTURES

FAVOUR

1.16

Overview of Chemical Engineering Profession, ……

SEPARATION

PROCESSES

FOR

LIQUID

• Distillation – Relative volatility a greater than 1.2 – Products thermally stable – Rate greater stable – No corrosion, precipitation or explosion problems •

Azeotropic/Extractive Distillation • – Systems normally contain azeotropes – A in solvent greater than for distillation – Solvent thermally stable easily regenerable – Solvent commercially available (at a resonable cost)

Extraction – Solvent selectivity greater than for distillation and greater than 1.5 – 2.0 – Solvent selective for low-concentration – Energy costs high – Easy solvent recovery

•

Adsorption • – Adsorbent selectivity greater than 2 for bulk separations and greater than 10 – 100 for purifications – High percentage solute removal – Acceptable delta and loadings – Adsorbent not susceptible to rapid fouling – Bed(s) easily regenerable – Clean air/water projects

Membranes – Membrane selectivity greater than 10 (except – Bulk separation, clean air/water projects and some trace removals – Acceptable fluxes – Membrane chemically stable – Membrane not susceptible to rapid fouling – Low to moderate feed rates.

1.7 PROCEDURE FOR SEPARATION PROCESS SELECTION 1.

Lay out the nature of the separation task; listing feed rate and its composition and different product streams required and their purities.

2.

Apply first guides for deciding on alternative separation techniques taking into account various factors regarding physical, chemical properties of the feed components, and the production rate of the process.

3.

Compare economic analyses for alternative procedures that use the selected separation techniques.

4.

Evaluate possible column/unit sequences and determine the best flowsheet based on the lowest overall capital plus operating costs.

1.8 CLASSIFICATION OF SEPARATION PROCESSES Several ways to classify separation processes : • Mechanical (heterogeneous-feed) Vs. Diffusional (homogeneous-feed) techniques. • Equilibration processes vs. rate-governed techniques. • Energy-separating agent vs. mass separating agent techniques.

Principles of Mass Transfer Operations − I (Vol. − I)

1.17

Overview of Chemical Engineering Profession, ……

1.9 CLASSES OF SEPARATION OPERATIONS 1.

Equilibration processes : Equilibration of two immiscible phases, which have different compositions at equilibrium.

2.

Rate governed processes : Differences in transport rate through some medium under the influence of an imposed force, resulting from a gradient in pressure, temperature, composition, electric potential, or the like.

Mechanical processes : Heterogeneous feed consisting of more than one phase of matter. These simply serve to separate phases from each other. Name Feed Separating Agent Products Principle of Separation Equilibrium Separation Processes : 1. Evaporation Liquid Head Liquid + Vapour Difference in volatilities (vapour pressure) 2. Flash expansion Liquid Pressure reduction Liquid + Vapour Same (energy) 3. Distillation Liquid and/or Heat Liquid + Vapour Same (repeated internally) Vapour 4. Stripping Liquid Non-condensable gas Liquid + Vapour Same 5. Adsorption Gas Non-volatile Liquid Liquid + Vapour Preferential Solubility 6. Extraction Liquid Immiscible Liquid Two Liquids Different solubilities of different species in the two liquid phases 7. Gel filtration Liquid Solid gel (e.g. crossGel phase and Difference in molecular size linked dextran) liquid and hence in penetrate swollen gel matrix 8. Dual-temperature Fluid Heating and cooling Two fluids Difference in reaction exchange reactions equilibrium constant at two different temperatures 9. Zone melting Solid Heat Solid of non- Same as crystallization uniform composition 10. Osmosis Salt solution More concentrated Two liquids Tendency to achieve salt solution : uniform osmotic pressures membrane removes water from more dilute solution 11. Bubble fractionation; Liquid Rising air bubbles; Two liquids Tendency of surfactant Liquid foam sometimes also molecules to accumulate at fractionation complexing gas-liquid interface and rise surfactants with air bubbles 12. Flotation Mixed Added surfactants; Two solids Tendency of surfactants to powdered rising air bubbles adsorb preferentially on solids one solid species 13. Magnetic separation Mixed Magnetic field Two solids Attraction of materials in powdered magnetic field solids 14. Paper Liquid Capillarity; paper or Regions of Preferential solubilities and chromatography gel phrase moistened paper adsorption potentials in two phrases 3.

Principles of Mass Transfer Operations − I (Vol. − I) 15. Freeze drying

1.18

Overview of Chemical Engineering Profession, ……

16. Desublimation

Frozen water- Heat containing solid Vapour Cooling

Solid and vapour

17. Dialysis

Liquid

Selective membrane; solvent

Liquids

18. Electrodialysis

Liquid

Liquids

19. Gas permeation

Gas

Anionic and cationic membranes; electric field Selective membrane; pressure gradient

20. Electrophoresis

Liquid Electric field containing Liquid mixtures Heat and vacuum

21. Molecular distillation

Dry solid and Sublimation of water water vapour

Gases

Liquids

Preferential condensation (desublimation); preferential participation in crystal structure Different rates of diffusional transport through membrane (no bulk flow) Tendency of anionic membranes to pass only anions, etc. Different solubilities and transport rates through membrane Different ionic mobilities of

Liquid and vapour

Difference in kinetic theory maximum rate of vapourization. Proportional to vapour pressure (molecular weight)1/2

Liquid + Solid

Size of solid greater than pore size of filter medium.

Mechanical Separation Processes : 1.

2. 3.

4.

5. 6. 7.

Filtration

Liquid + Solid

Pressure reduction (energy); filter medium Mesh demister Gas + Solid or Pressure reduction Liquid (energy); wire mesh Settling Liquid + Solid Gravity or another immiscible liquid Centrifuge Liquid + Solid Centrifugal force (semdimentation or another type) immiscible liquid Centrifuge (filtration Liquid + Solid Centrifugal force type) Cyclone Gas + Solid or Flow (inertia) Liquid Electrostatic Gas + Fine Electric field precipitation Solids

Gas + Solid or Same Liquid Liquid + Solid or Density difference another immiscible liquid Liquid + Solid or Density difference another immiscible liquid Liquid + Solid Gas + Solid or Liquid Gas + Fine Solids

Size of solid greater than pore size of filter medium Density difference Charge on fine solid particles.

Principles of Mass Transfer Operations − I (Vol. − I)

1.19

Overview of Chemical Engineering Profession, ……

1.10 MASS TRANSFER OF COOKING-CHEMICAL ENGINEERING THROUGH COOKING Mass transfer is important concept in the field of chemical engineering and science in general. Mass transfer is the movement of something that has mass. Since all matter has mass, then mass transfer just refers to the movement of all things. Mass has the tendency to move from an area of high concentration to an area of low concentration. What we mean by that is, it moves from an area where it is crowded to a more open space. This is just how people act in a crowd. To avoid feeling claustrophobic, many people will move from the center of a crowd to the outskirts to get more breathing room. Everything in nature does this, which is the basis behind mass transfer. Mass transfer is often seen in many day-to-day situations. Have you ever sprayed perfume in one corner of the room and smelled it a few seconds later in another corner of the room ? The perfume moved from the area where it was sprayed (high concentration) to an area where there was less perfume (low concentration). Mass transfer can also be seen when a glass of water is left on the table overnight. The next morning the glass is found empty. What happened to the water ? Was it spilled ? Probably not. The water probably turned into water vapor and spread into the room where there was a low concentration of water vapour. How many of you have ever poured bubble bath into a tub of running water ? The bubbles don't just stay in the area where you poured them. Instead, they move throughout the bathtub evenly. This is just yet another example of mass transfer. The types of mass transfer discussed above are examples of diffusion. The bath bubbles diffuse through the bathtub and the perfume diffuses through the air in the room. Diffusion can also be related to the example of making Brewing Coffee. Have you ever heard a coffee drinker say, "Coffee is an acquired taste" ? Tea, coffee is the most popular drink in the world. It must be a pretty easy taste to acquire. Often times, it is not the flavour of coffee that is desired, but its caffeine. Now, there are several ways of preparing coffee, that is, several ways of extracting caffeine from the coffee bean. The most often used method is to allow hot water to strain through coffee grounds, picking up caffeine and the flavourful tannins as it goes. The coffee is then strained out of the grounds into a collection container. Percolation is a similar method in which the coffee is allowed to circulate through the grounds several times. Other methods, like steeping and vacuum brewing let the boiling water stay in contact with the grounds throughout the entire brewing process. Another method of making coffee is through the use of instant coffees. Instant coffee is actually coffee brewed through a traditional method and then dried or freeze dried. So exactly how do we get the caffeine out of the coffee beans ? This process can be described using mass transfer equations used in chemical engineering. The specific equation, which describes this movement, is called Fick's Law. 1.11 GENERAL OVERVIEW In the heart of all industrial processes where raw materials are processed, separated or purified into useful products resides the science of chemical engineering. since a chemical process may be visualised as a collections or co-ordinated steps or operations – called unit operations – involving, changes in chemical composition or physical properties of materials being prepared, processed, separated or purified, so the fundamental duty of a chemical engineer is to choose proper raw materials, select appropriate steps/operations in appropriate sequence and specify the exact conditions under which each such step is to be carried out. He/she must see that his plants are run efficiently, safely and economically and that the end products meet the standards required by the customers.

Principles of Mass Transfer Operations − I (Vol. − I)

1.20

Overview of Chemical Engineering Profession, ……

Formal theory alone cannot help a chemical engineer to solve his/her process problems. He/she must coin valuable informations from operators who have observed process and learned methods of detail control. The most capable engineer is he/she who gives proper weightage to all the available facts and data regardless of their resource. Mass transfer phenomena are to be found everywhere in nature and are important in all branches of science and engineering. The phrase "mass transfer", which has come into common use only in recent years, refers to the motion of molecules or fluid elements caused by some form of potential or "driving force". It includes not only molecular diffusion but also transport by convection and sometimes simple mixing – not the convenience of a material, as in the flow of a fluid in a pipe. Mass transfer is involved wherever a chemical reaction takes place, whether in an industrial reactor, a biological system, or a research laboratory. The general subject of mass transfer may be divided into four broad areas of particular interest and importance : • molecular diffusion in stagnant media • molecular diffusion in fluids in laminar flow • eddy diffusion or mixing in a free turbulent stream and mass transfer between two phases. The first has been subject of much study by scientists for more than a century. The theory is in good shape for diffusion in gases, though not for diffusion in dense fluids. The second is application of first and is treated by a mathematical manipulation, often difficult, of what is known about molecular diffusion in situations where the flow field can be described or calculated. Eddy diffusion in a free stream away from a phase boundary is the process by which gases leaving a stack are dispersed into the atmosphere and by which mixing occurs in many situations, as in turbulent jets. Transfer between two phases, across an interface, is of particular importance in engineering largely as it is involved in most separation processes, as in the recovery of a pure product from a mixture. Evapouration from a reservoir, oxygenations of blood, removal of pollutants from the atmosphere by rain, chemical reaction at the surface of a solid catalysts or within the porous structure, deposition by electrolysis or electrophoresis, drying of wood are all examples of mass transfer between phases. The chemical engineer's interest in mass transfer stems primarily from his traditional role as a specialist in the design of separation processes. The materials fed to a chemical process are purified by separation or concentration of the reactants and the valuable products must be separated from the stream leaving the reactor. Though the separation equipment is ancillary to the reactor, its cost is often major part of the investment in the plant. 1.11.1 Introduction to Mass Transfer Operations Absorption, Distillation, Stripping, Drying, Extraction are mass transfer operations. These operations are widely used in various (petro-) chemical separation processes. During such an operation either one (or more) component(s) in the vapour phase is transferred to the liquid phase and/or from the liquid phase to the vapour phase. Some examples are : • Absorption is used for removal of gaseous hydrogen sulfide and/or carbon dioxide and/or mercaptans from Natural Gas, Synthesis Gas etc. (any process gas stream, really) by dissolving them in a (reacting) liquid stream. • Distillation is used in every oil refinery and also in many chemical manufacturing plants in the separation and purification of the desired products. Distillation is the most important separation technique, in general. Distillation columns are very visible in the skyline of any refinery and many chemical plants.

Principles of Mass Transfer Operations − I (Vol. − I)

1.21

Overview of Chemical Engineering Profession, ……

•

Stripping is the reverse of Absorption. Stripping of dissolved and contaminating volatile organic components (VOC's) from (ground) water, is a good and actual example. • Drying (removal of water vapour) from Natural Gas by dehydrating liquids (i.e. Triethylene Glycol) is an actual process example. • Extraction is a liquid/liquid contacting operation, and is used in the petroleum industry (to separate aromatics and aliphatic species) and in the pharmaceutical industry (to recover penicillin). These operations are usually performed in cylindrical columns. These columns come in a wide range of sizes, their diameter ranges from 0.05 m (for a typical laboratory scale column) to about 10 to 12 m (for the largest industrial columns) and their height ranges from about 0.5 m to about 100 m. The required contacting needed for the separation(s) is provided by filling these columns with packings and/or trays. Packings (either structured or random) come in many types and sizes and so do trays. Commonly used tray types are : Bubble cap trays, Sieve tray and Valve trays etc. To illustrate the conceptualization of mass transfer processes and provide a few simple applications, we will start by considering two different "thought experiments" that will help us define what mass transfer means. Experiment 1 – Mass exchange between two tanks : Consider two containers connected by a pipe that has a valve, as shown in Fig. 1.5. Both tanks have the same volume V, and they are at the same temperature and pressure (T, P). Tank A contains 1 mole of gaseous oxygen whereas tank B contains 1 mole of gaseoous nitrogen. P, V, T

P, V, T

1 mole N2

1 mole O2

Initial state of the system

Closed valve Tank A Tank B Fig. 1.5 : Two tanks of equal volume at the same pressure and temperature contain initially different ideal gases. The tanks are connected by a pipe that has an initially closed valve. Note that, since P, V and T are the same, the number of moles in both tanks must be the same according to the ideal gas equation of state, PV = nRT

We now open the valve connecting the tanks. What happens ? Fluid Mechanics tells us that, since the pressures at both ends of the pipe are the same, there will be not net flow through the pipe. However, if we let time pass, we observe that N2 starts to appear in tank A, and O2 in tank B. After a long time an equilibrium state will be reached, in which there will be 0.5 moles of each gas in each tank (See Fig. 1.6). P, V, T

P, V, T

0.5 moles O2 0.5 moles N2

0.5 moles O2 0.5 moles N2

Open valve

Equilibrium state

Tank A Tank B Fig. 1.6 : After a long time has passed, the system in Fig. 1.1 reaches equilibrium The two gases are uniformly distributed between the two tanks

Principles of Mass Transfer Operations − I (Vol. − I)

1.22

Overview of Chemical Engineering Profession, ……

Experimental observation tells us that, at equilibrium, the concentration of each chemical species will be uniform throughout the system. In fact, knowing this and applying the principle of mass conservation allows us to know exactly how many holes of each species are in each tank after the equilibrium state is reached. What has happened in this experiment ? We know that, in gas, individual molecules are in continuous motion at all times. This molecular motion implies that the gas will tend to distribute itself uniformly (from a statistical point of view) over all the space available to it. At the final equilibrium stage, both O2 and N2 are uniformly distributed throughout the two tanks. Notice that, if we perform the experiment with two liquids, the final outcome would be the same since the liquid will also try to occupy all the space available, but, since the molecular motion in the liquid phase is much slower than in the gas phase, it would take longer to reach the final equilibrium state. In experiments like this, the use of principle of conservation of mass allows us to predict the final outcome of the experiment. However, if we are interested in determining how fast the material exchange between the two tank is, we would need to use other physical principles, which relate to what we call mass transfer. Experiment 2 – Dissolution of sugar in water : If we put a certain amount of sugar crystals in water at ambient conditions, we will observe that it will slowly dissolve until all the crystals disappear (provided that we do not saturate the solution). After a long time there will be no solid left and, furthermore, the concentration of sugar in the solution will be uniform, i.e., if we take a sample of the solution from any point within the reservoir, we would observe the same sugar concentration. (See Fig. 1.7). Initial state

Final state

Water

Long time

Sugar solution

Sugar crystals

Fig. 1.7 : Solid sugar eventually dissolves in water to form a solution with uniform concentration

In this experiment, there is a phase change : the sugar goes from the solid phase into the liquid phase, and the continuous motion of the dissolved sugar molecules eventually leads to their uniform distribution in the liquid. Once again, if we know the amount of sugar that we added, we can calculate the concentration of sugar in water at the end of the experiment. But if we want to know how fast the system reaches the final equilibrium state, we need to use other physical principles, which fall under what we will call mass transfer. It is interesting to mention that we know how to make the process faster : everyday experience tells us that if we stir the solution, we will achieve the final state faster. Therefore, we already know some mass transfer !

Principles of Mass Transfer Operations − I (Vol. − I)

1.23

Overview of Chemical Engineering Profession, ……

1.12 WHAT ARE THE BENEFITS OF MASS TRANSFER KNOWLEDGE ? In general, most of chemical engineers work : (1) At a producing plant of any of process industries : (a) At producing plant, being a person handling daily plant operation. (b) At laboratory, examining processes involving mass transfer. (2) In engineering departments of process company. (3) In an engineering company working closely with the industry. (4) As consultant to process industry. 1.12.1 Where and Why Mass Transfer is Important ? Mass Transfer is exchange of material between phases : (a) Vapour-Liquid mass transfer occurs in distillation and absorption, including reactive forms of these operations. (b) Liquid-liquid mass transfer occurs in liquid extraction and extractive distillation, as well as in chemical reaction between two liquid phases. (c) Vapour-solid mass transfer occurs in adsorption. (d) Liquid-solid mass transfer occurs in dissolving, crystallization and leaching process (solid-liquid extraction). 1.12.2 Benefits that An Engineer can Achieve By Learning Mass Transfer Today, theory and methods related to mass transfer are well developed and have been implemented in the advanced process simulators. First reason to learn mass transfer theory and methods is to finding out that valuable calculations can be done by average engineer. Secondary, your education might never include sufficient level of mass transfer knowledge (a mechanical engineer or physical chemist working in the process industry). Theory and techniques could have been taught at level existing years ago, when only first attempts to implement theory in production were made. Some faculties, such as mechanical engineering still do not teach their students mass transfer. Other faculties, as chemistry, may give students knowledge about mass transfer, but the knowledge is not always targeted at industrial applications. Thirdly, becoming experienced in mass transfer techniques, you will be able to help your company (engineering or producing) to solve equipment and process related problems. Although there are several processes (such as Distillation) where mass transfer can be calculated by already collected empirical data, many other processes need to be examined in laboratory. For instance, by applying theory and a process simulator, you will be able to study and determine kinetics of mass transfer, e.g., for extraction, leaching or dissolving/ crystallisation processes. These data will be critical information for production engineers. 1.13 GENERAL PRINCIPLES OF MASS TRANSFER In modern chemical process industries, a large number of unit operations of chemical engineering are carried out with the problem of changing the composition of solution and mixtures without involving any chemical reactions. Usually these operations are directed toward separating into its component parts. Thus the operations, which involve changes in composition or concentration of solution, are called mass transfer operations.

Principles of Mass Transfer Operations − I (Vol. − I)

1.24

Overview of Chemical Engineering Profession, ……

The process of mass transfer deals with the flow of constituents in presence of concentration gradient. When two phases of different compositions are brought into contact, a transfer of components may occur from one phase to another phase, and vice versa. This is the physical basis of mass transfer operations. If the two phases are allowed to remain in contact for sufficient time, they will reach equilibrium condition where there is no further net transfer of components between phases. In most cases of interest of mass transfer operations the two phases are only partially miscible, so that at equilibrium there still exist two phases that can be separated from each other. Usually these two phases have compositions different from each other and also different from compositions of the two phases that were initially contacted. As a result, the relative amounts of components transferred between the phases are different so that separations are achieved. Example : When water evapourates from a pool into an air-stream following over water surface, molecules of water vapour diffuses through those of the air at the surface into the main portion of the air-stream where they are carried away.

In this case, the mass transfer is a result of

concentration difference or gradient, the diffusing substance moving from a place of high concentration to low concentration. 1.13.1 Importance of Mass Transfer Operations (i)

It doesn’t require a preliminary purification of raw materials or final separation of products from byproducts.

(ii) In petroleum refinery, large number of processes are to be carried out, in each of which mass transfer operations are frequently carried out for separation or purification. The cost of separation or purification depends directly on the ratio of final initial concentration of the separated substances. Note that if this ratio is large, the product costs are also large. Example : Sulphuric acid is relatively low – priced product as sulphur is found naturally in relatively pure state, where as pure Uranium is expensive because of low concentration in which it is found in the nature. 1.14 CLASSIFICATION OF THE MASS TRANSFER OPERATIONS The classification of mass transfer operations is complicated and may be done in various ways. Mass transfer may occur : (a)

In one direction.

(b) In opposite direction. (c)

With the exchange of a single component.

(d) With the exchange of multicomponent.

Principles of Mass Transfer Operations − I (Vol. − I)

1.25

(e)

With a simultaneous chemical reaction.

(f)

With a simultaneous heat transfer.

Overview of Chemical Engineering Profession, ……

(g) Isothermally and (h) Non-Isothermally. The following phenomena must exist in a mass transfer operations (i)

Two or more phases must come in contact with each other.

(ii)

Materials should flow from one phase to the other.

(iii)

A part of the total flow must be by molecular motions or molecular diffusion. Since similarities between the various mass transfer operations, the mass transfer operations have been classified according to the phase contact as shown in Table 1.13. Table 1.3

No.

Phases in Contact

Mass Transfer Operations

(1)

Liquid – Liquid

Extraction, Liquid thermal diffusion

(2)

Liquid – Solid

Dissolving, Crystallization

(3)

Liquid – Vapour

Distillation

(4)

Liquid – Gas

Gas Absorption

(5)

Solid – Vapour

Sublimation, Adsorption

(6)

Solid – Solid

Solid Diffusion

(7)

Gas – Gas

Gas diffusion, Thermal diffusion

(8)

Solid – Liquid – Solid

Leaching

(9)

Solid – Liquid – Vapour

Adsorption

1.14.1 Similarities between the Operations There are many similarities between the various mass transfer operations. They are : (i)

Phase equilibrium is reached after a sufficiently long period of contact.

(ii)

Rate of transfer is calculated by deviation from equilibrium concentration.

(iii)

Equilibrium exists at phase interface or there is no resistance to mass transfer at the interface, with some exceptions.

(iv)

Material transfer is due to the combined effect of molecular diffusion and turbulence.

1.14.2 Differences between the Operations (a)

Number of components : In distillations, thermal diffusion, adsorption gas diffusion and leaching there are normally two active components, though three, or more components may also be present. Extraction and gas absorption almost always involve three or more components.

Principles of Mass Transfer Operations − I (Vol. − I)

1.26

Overview of Chemical Engineering Profession, ……

(b) Distribution of components between phases : In leaching and adsorption the solid present is more or less inert, in other words, the component appears only in one phase. In distillation, gas diffusion and thermal diffusion all the components are usually well – distributed between the phases, crystallization and sublimation may be either way. (c)

Temperature : Some operations are isothermal. In others, temperature gradients are produced incidentally. There are still others, which depend on difference in temperature.

1.15 SEPARATION PROCESS Modes of Separation : Continuous (as in packed beds) or stagewise (as in tray columns). Classification of Separation Processes : Equilibrium Separation •

Absorption

•

Adsorption

•

Distillation

•

Evapouration

•

Crystallisation

•

Leaching

Rate-governed Separation •

Gas diffusion

•

Reverse osmosis

•

Dialysis

•

Electrophoresis

Mechanical Separation •

Filtration

•

Settling

•

Centrifuge

Examples of Separation Processes : Gas mixture separation Absorption (Principle : Solubility of solute gas in a liquid solvent) : SO2 scrubbing with water. Adsorption (Principle : Adsorption affinity of gas/liquid on a solid adsorbent) : Air separation. Cryogenics (Principle : Condensation of gas to a liquid form) : Air separation.

Principles of Mass Transfer Operations − I (Vol. − I)

1.27

Overview of Chemical Engineering Profession, ……

Liquid mixture separation Distillation (Principle : Boiling point difference) : Crude oil fractionation. Solvent extraction (Principle : Solubility of a liquid solute) : Lube oil/furfural solvent. Diffusional extraction (Principle : Diffusion) : Fish oil removal with alcohol. Liquid-solid separation Drying (Principle : Mass transfer by thermal input) : Paper manufacture. Leaching (Principle : Solubility of a solid solute) : Copper from ore/H2SO4 solvent. Gas/Liquid-solvent separation Cyclones (Principle : Density difference) : Dust removal from flue gas. Settler (Principle : Density difference) : Waste water treatment. Solid-solid separation Screening (Principle : Size difference) : Mineral dressing. Electromagnetic (Principle : Electromagnetic affinity) : Fe removal from non-ferrous matter. Separation System Design Guidelines : Practical guidelines when deciding on separation systems to be used : •

In chemical process, differences in composition dominate the design of the process, therefore, select separation tasks early on.

•

If a feed impurity is inert, then remove it if the quantity is significant; leave it, if the quantity is insignificant.

•

If a feed impurity is not inert, remove it, else, it may lead to raw material losses and need for a complex separation process to recover additional byproducts.

•

If a feed impurity is inert, but is easier to separate from the product than the feed, then it is better to process the feed without separation.

•

If a feed impurity is a catalyst poison or corrosive, remove it.

•

Save difficult separations for last.

Factors to consider in designing separation systems : •

Production rate.

•

Phases present : Liquid gas, solid.

•

Operational issues.

•

Conditions : Temperature, pressure, etc.

•

Product distribution.

Principles of Mass Transfer Operations − I (Vol. − I)

1.28

Overview of Chemical Engineering Profession, ……

1.16 CHOICE OF SEPARATION METHOD The chemical engineer faced with the problem of separating the components of a solution must ordinarily choose from several possible methods. While the choice is usually limited by the peculiar physical characteristics of the material to be handled, the necessity for making a decision nevertheless almost always exists. (i) Choice between using a mass transfer operation of purely mechanical separation method : e.g. In the separation of a desired mineral from its ore, it may be possible to use either the mass transfer operations of leaching with a solvent or the purely mechanical methods of flotation. Vegetable oils can be separated from the seeds in which they occur by extraction, or by leaching with a solvent. A vapour can be removed from a mixture with a permanent gas by the mechanical operation of compression or by the mass transfer operation of gas absorption or adsorption. Some times both mechanical and mass transfer operations are used; where the former is incomplete, as in processes for recovery of vegetable oils wherein extraction is followed by leaching. (ii) Choice between using mass transfer operation and chemical reaction or combination of both method : For example, water can be removed from an ethanol-water solution either by reacting with unslaked lime or by special methods of distillation. H2S can be separated from other gases either by absorption in a liquid solvent with or without simultaneous chemical reaction or by chemical reaction with ferric oxide. (iii) There are also choices to be made within the mass transfer operation : e.g. a gaseous mixture of oxygen and nitrogen may be separated by adsorption of the oxygen on activated carbon, by adsorption, by distillation or by gaseous effusion. The principle basis for choice in any case is cost. The method, which costs the least, is normally the one to be used. In addition, other factors such as simplest operation, favourable previous experience with one method may be given strong considerations. 1.17 METHODS OF CONDUCTING THE MASS TRANSFER OPERATIONS 1.17.1 Solute Recovery and Fractionation If the components of a solution fall into two distinct groups of quite different properties, so that one can imagine that one group of component constitutes the solvent and the other group the solute, separation according to the groups is usually relatively easy and amounts to solute recovery or solute removal operation. e.g. a gas consisting of methane, pentane and hexane can be imagined to consist of methane as solvent with pentane plus hexane as solute, the solvent and solute in this case differing considerably in at least one property. Whether a solute-recovery or fractionation procedure is used may depend upon the property chosen to be exploited. e.g. the separation of propanol from butanol requires a fraction technique (fractional extraction or fractional distillation) as all their properties are very similar.

Principles of Mass Transfer Operations − I (Vol. − I)

1.29

Overview of Chemical Engineering Profession, ……

1.17.2 Unsteady-state Operation It is characteristic of unsteady-state operation that concentration at any point in the apparatus changes with time. This may results from changes in concentrations of feed materials, flow rates or condition of temperature or pressure. In any case batch operation are always of the unsteadystate operations. In purely batch operations, all the phases are stationary from a point of view outside the apparatus. i.e. no flow in or out, even though there may be relative motion within. e.g. the laboratory extraction procedure of shaking a solution with an immiscible solvent. In semibatch operation, one phase is stationary while the other flows continuously in and out of the apparatus. e.g. in case of drier where a quantity of wet solid is contacted continuously with fresh air, which carries away the vapourized moisture until the solid is dry. 1.17.3 Steady-state Operation It is characteristic of steady-state operation that concentration of any position in the apparatus remains constant with passage of time. This requires continuous invariable flow of all phases into and out of the apparatus, constant concentration of the feed streams and unchanging condition of temperature and pressure. 1.17.4 Stagewise Operation If two insoluble phases are first allowed to come into contact so that the various diffusing substances can distribute themselves between the phases, and if the phases are then mechanically separated, the entire operation and equipment required to carry it out are said to constitutes a stage. e.g. laboratory batch extraction in a separatory funnel. The operation can be carried on in continuous fashion (steady state) or batch wise fashion. For separation requiring greater concentration changes, a series of stages can be arranged so that the phases flow through the assembled stages from one to the other. e.g. in countercurrent flow such as assembled is called as cascade. 1.17.5 Continuous-contact (Differential Contact) Operation In these operations, the departure from equilibrium is deliberately maintained, and the diffusional flow between the phases may continue without interruption. A high stage efficiency can mean a relatively inexpensive plant and one whose performance can be reliabely predicted. A low stage efficiency, may make the continuous-contact methods more desirable for reasons of cost and certainty. 1.18 DESIGN PRINCIPLES Following are four factors to be established in the design of any plant involving the diffusional operations : (1) Number of Equilibrium Stages : In order to determine the number of equilibrium stages required in a cascade to bring about a specified degree of separation, or the equivalent quantity for a continuous-contact device, the equilibrium characteristics of the system and the material balance calculations are required.

Principles of Mass Transfer Operations − I (Vol. − I) (2)

1.30

Overview of Chemical Engineering Profession, ……

The Time of Phase Contact Required : In stagewise operations the time of contact is

intimately connected with stage efficiency, whereas for continuous-contact equipment the time leads ultimately to the volume or length of the required device. Material balances permit calculation of the relative quantities required of the various phases. The equilibrium characteristics of the system establish the ultimate concentrations possible and the rate of transfer of material between phases depends upon the departure from equilibrium which is maintained. In addition, the rate of transfer depends upon the physical properties of the phases as well as the flow regime within the equipment. (3)

The Permissible Rate of Flow : This factor enters into consideration of semi-batch and

steady-state operations, where it leads to the determination of cross-sectional area of equipment. Considerations of fluid dynamics establish the permissible flow rate and material balances determine the absolute quantity of each of the streams required. (4)

Energy Requirements : In general heat and mechanical energies are normally required

to carry out the diffusional operations. Heat is necessary for the production of any temperature changes, for the creation of new phases (such as – vapourisation of liquid), and for overcoming heat-of-solution effects. Mechanical energy is required for fluid and solid transport, for dispersing liquids and gases, and for operating moving parts of machinery.




REFERENCES 1.

Pafko, Wayne. "Chemical Engineering Then and Now." Chemistry In Australia, Royal Australian Chemical Institute. Volume 67, Number 6. July 2000. (p. 17-22).

2.

Pafko, Wayne. "What is a Chemical Engineer?" CGP Reprint R-135, Chronicle Guidance Publications. December 1998. (4 pages).

3.

Farhad Sharifi, “Chemical Engineering : Professionally Ignored?” Proceedings of the 2002 American Society for Engineering Education Annual Conference and Exposition, American Society for Engineering Education.

4.

William F. Furter, “History of Chemical Engineering”: Based on a symposium cosponsored by the ACS Divisions of History of Chemistry and Industrial and Engineering Chemistry at the ACS/CSJ Chemical Congress, Honolulu, Hawaii, April 2-6, 1979, Washington, American Chemical Society, 1980.

5.

Clive Cohen,“ The Early History of Chemical Engineering : A reassessment”, British Journal for the History of Science, 1996, Volume 29, Number 101, pp. 171-194.

6.

King C.J., “ Separation Process Principles”, Second Edition, Tata McGraw Hill, 1988.

7.

R.E. Treybal, "Mass Transfer Operations", Third Edition, McGraw Hill, 1981.

,,,

2

CHAPTER

FUNDAMENTALS OF DIFFUSION MASS TRANSFER 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10

2.11

2.12

2.13

2.14 2.15

Introduction Kinetic Theory of Gases Molecular Diffusion Molecular Diffusion Vs Eddy Diffusion Molar Flux Molecular Diffusion in Fluids and Diffusion Coefficient or Diffusivity Diffusion in Binary Solution Steady-state Diffusion in Fluids at Rest and Laminar Flow Principles of Mass Transfer Properties of Mixtures 2.10.1 Concentration of Species 2.10.2 Mass Averaged Velocity Diffusion Flux 2.11.1 Fick's Law 2.11.2 Relation among Molar Fluxes Diffusivity : 2.12.1 Diffusivity in Gases 2.12.2 Diffusivity in Liquids 2.12.3 Diffusivity in Solids Steady state Diffusion 2.13.1 Diffusion through a Stagnant Gas Film 2.13.2 Pseudo – Steady – State diffusion through a Stagnant Gas Film 2.13.3 Equimolar Counter Diffusion 2.13.4 Diffusion into an Infinite Stagnant Medium 2.13.5 Diffusion in Liquids 2.13.6 Mass Diffusion with Homogeneous Chemical Reaction 2.13.7 Diffusion in Solids Transient Diffusion Differential Equations of Mass Transfer Solved Problems Exercise for Practice Nomenclature References

(2.1)

Principles of Mass Transfer Operations − I (Vol. − I)

2.2

Fundamentals of Diffusion Mass Transfer

2.1 INTRODUCTION Molecular transport of mass is called as molecular diffusion. Molecular transport of mass, heat and momentum occurs in gases, liquids and solids. It is the basic physical mechanism underlying many important unit operations. Molecular transport occurs in any homogenous material, it may be gas, liquid or solid. When a system contains two or more components whose concentrations vary from point to point, there is a tendency for mass to be transferred to minimize the concentration differences. There are two mass transfer mechanisms of interest : •

Molecular scale mass diffusion (conduction).

•

Bulk mass transport (convection).

The factors affecting mass transfer includes vapour pressure, solubility and diffusivity. In unit operations, mainly concerned with the transfer of mass between three phases : solid, liquid and gaseous : (i) Distillation : Separation of miscible liquids. (ii) Absorption : Soluble vapour absorbed in liquid. (iii) Desorption-Stripping : Solute moving from liquid to gas. (iv) Dehumidification : Pure liquid condensed from inert gas. (v) Evapouration : Liquid converted to gas. (vi) Liquid extractions : Solid-liquid, liquid-liquid. (vii) Crystallization. (viii) Membrane separations. (ix) Adsorption. (x) Leaching. (xi) Drying. Molecular diffusion is concerned with the movement of individual molecules through a substance by virtue of their thermal energy. An understanding of the molecular transport mechanism may be understood from a study of the kinetic theory of gases and liquids or from consideration of solid-state physics. The kinetic theory of gases provides a means of visualising what occurs and quantitatively describes the diffusional phenomenon. 2.2 THE KINETIC-MOLECULAR THEORY OF GASES The ideal gas law tells us how gases behave, but not why they behave that way. The molecular basis for this law is provided by the kinetic-molecular theory of gases. 2.2.1 Summary of Model 1.

Gases consist of large numbers of molecules that are in continuous, random motion.

2.

The volume of all molecules is negligible.

3.

Attractive and repulsive forces between molecules are negligible.

4.

Collisions are elastic. Energy can be transferred between molecules during collisions, but the average kinetic energy does not change with time, at constant temperature.

Principles of Mass Transfer Operations − I (Vol. − I) 5.

2.3

Fundamentals of Diffusion Mass Transfer

The average kinetic energy of the molecules is proportional to the absolute temperature.

It is the concept of average kinetic energy that is the most important. For a given distribution of molecules with different kinetic energies, and average kinetic energy changes in a predictable and specific way as a function of temperature. Namely as the temperature increases, this distribution shifts towards higher average kinetic energies. For the kinetic energy of a given molecule, 1 2 K.E. = 2 mv For the average kinetic energy, 1 2 ∈ = 2 mu where u is the root-mean-square (r.m.s.) speed. Since the average kinetic energy increases with temperature, so too does the r.m.s. speed of the molecules. 2.2.2 Using Kinetic Theory to Understand the Gas Laws 1.

Kinetic Theory and Boyle’s Law : At constant T, the average kinetic energy and hence, r.m.s. speed remains the same. If the volume increases, however, the molecules must travel a longer distance between collisions, and hence there are fewer collisions per unit time with the container walls. Consequently, as predicted by Boyles Law, as the volume increases, the pressure decreases 2. Kinetic Theory and Charles Law When the T increases, the average kinetic energy increases. Assuming no change in volume, at higher T the molecules will travel faster and thus will hit the walls more often. This results in an increase in the pressure. (Charles’s Law) Molecular Effusion and Diffusion : According to kinetic theory, 1 2 ∈ = 2 mu where, m is the mass of the molecules u is the r.m.s. speed of the molecules. Note that the larger the mass the lower the r.m.s. speed. In a related way, it can be shown that 3RT (MW) Two consequences of this result are predictions for the properties of two phenomenon. Effusion – The escape of gas molecules through a hole. Diffusion – The spread of one substance throughout a space or throughout a second substance. Diffusion : Its easy to see how the rate of diffusion would be dependent on the r.m.s. speed. The faster the molecules are going the faster it will diffuse. u =

Principles of Mass Transfer Operations − I (Vol. − I)

2.4

Fundamentals of Diffusion Mass Transfer

From the equation for the r.m.s. speed, 3RT (MW) Its easy to see that the lighter the gas molecule, the faster the diffusion rate under similar conditions. One qualification, however, is that the rate of diffusion is that these rates are dependent on the density of the medium. u =

At lower pressures, the rates will be faster due to fewer collisions. At higher pressures, the rates will be slower due to too many collisions. In case of a simplified kinetic theory, a molecule is imagined to travel in straight line at uniform velocity until it colloids with another molecule, whereupon its velocity changes both in magnitude and direction. The average distance of molecule velocity is dependent upon the temperature. The molecule thus travels highly zig-zag path, the net distance in one direction, which it moves in a given time, the rate of diffusion, being only a small fraction of the length of its actual path. For this reason the diffusion rate is very slow. Although we can expect it to increase with decreasing pressure, which reduces the number of collisions, and with increased temperature, which increases the molecular velocity. From kinetic theory, it is estimated that the rate of evapouration of water at 25°C into complete vacuum is approximately 3.3 kg/sec per m2 of water surface. But if we place a layer of stagnant air at 1 std. atm. pressure and only 0.1 mm thick above the water surface, reduces the rate by a factor of about 600. Note that for gases, for higher molecular concentration, we find slower rate of diffusion. 2.3 MOLECULAR DIFFUSION Definition : The mass transfer of the species from region of higher concentration to region of lower concentration is accomplished by the actual migration of molecules. This phenomenon is called as molecular diffusion. Molecular diffusion or molecular transport can be defined as the transfer or movement of individual molecules through a fluid by means of random, individual movements of the molecules. The molecules travel only in straight lines and in the process, may collide with other molecules in their path. The molecules then change direction (still in a straight line) after the collision. This is sometimes referred to as a random-walk process as shown in the Fig. 2.1 as follows. A

A B

B

A B

A

B

B

A A

B

A

B A

A A

B

B A

A

B A

Fig. 2.1 : Random walk process

Principles of Mass Transfer Operations − I (Vol. − I)

2.5

Fundamentals of Diffusion Mass Transfer

Now consider a container with a mixture of two components A and B at constant pressure P and constant temperature T as shown in the figure below. A fictitious partition c-c separates the container into two sections. The L.H.S. contains more molecules of component A than the R.H.S., and the reverse is true for the B. Molecule of A Molecule of B

Partition C-C

Fig. 2.2 : Diffusion in Binary mixture

Next consider what happens when the partition c-c is removed. The molecules will move around in all directions in a random manner. But since most of the A-molecules are to the left of c-c, more of the A-molecules will travel from the L.H.S. and to the R.H.S. than in the opposite direction. This is the molecular diffusion often A in the direction of decreasing concentration (i.e. from the region of high concentration to the region of low concentration). At the same time there is also a net diffusion of B from the R.H.S. to the L.H.S. This diffusion continues until the concentrations of A and B is uniform throughout. Note : The rate of diffusion is much greater in gases than in liquids while, diffusion becomes extremely slow process in solids for same difference in concentration. 2.4 MOLECULAR DIFFUSION VERSUS EDDY/TURBULENT DIFFUSION Molecular diffusion is a slow process and with rapid mixing which can be brought about by mechanical stirring and convective movement of fluids. Consider a tank as shown in Fig. 2.3. Here a 0.75 m – deep layer of pure water has been carefully placed over the brine solution without disturbing the brine solution in any way. Water

H = 0.75m Brine Solution Tank

D = 1.5m

Fig. 2.3 : Molecular Diffusion

If the contents of the tank are left completely, undisturbed by molecular diffusion, the salt will completely permeate the liquid, ultimately coming everywhere to one-half its concentration in the original brine. But this process is very slow, and it is estimated that the salt concentration at the top surface will be only 87.5% of its final value after 10 years and will reach 99% of its final value only after 28 years. Now consider Fig. 2.4.

Principles of Mass Transfer Operations − I (Vol. − I)

2.6

Fundamentals of Diffusion Mass Transfer

22 rpm

Paddle Type Agitator Water

H = 0.75m Brine Solution Tank D = 1.5m

Fig. 2.4 : Eddy Diffusion

If we place a simple paddle type agitator rotating in the same tank as shown in Fig. 2.4 at 22 rpm will bring complete uniformity in about 1 min. The mechanical agitation has produced rapid movement of relatively large chunks or eddies, of fluid characteristic of turbulent motion, which has carried the salt with them. This method of solute transfer is called as eddy or turbulent diffusion. 2.5 MOLAR FLUX Rate of diffusion are most conveniently expressed in terms of molecular flux. (mole) (Area) (Time)

i.e.

It may be described by two terms : (a) Molar flux (N) : It is defined w.r.t. fixed locations in space.

 mole  m2.sec. (b) Molar flux (J) : It is defined w.r.t. the relative velocity of all components.

 mole  m2.sec. 2.6

MOLECULAR DIFFUSION IN FLUIDS AND DIFFUSION COEFFICIENT OR DIFFUSIVITY

The diffusivity, DAB, of component A is solution in B which is a measure of its diffusive ∂C A mobility is defined as the ratio of its flux JA to the concentration gradient, ∂Z ∂C JAZ = – DAB

A

∂Z

… (2.1)

This is FICK’S FIRST LAW OF DIFFUSION in Z-direction. where; JAZ = molar flux of A in Z-direction relative to the molar average velocity. 2

(mole/m .sec.).

Principles of Mass Transfer Operations − I (Vol. − I) ∂C

A

∂Z ∂X

A

∂Z

2.7

Fundamentals of Diffusion Mass Transfer 3

= Concentration gradient of A in Z-direction (k mole/m .m)

= Mole fraction gradient of A in Z-direction (1/m)

DAB = Mass diffusivity or diffusion coefficient for A diffusing through 2

component B. (m /s). Note : (1) Negative sign indicates that diffusion occurs in the direction of drop in concentration. (2) Rate of transfer of A in a mixture of two components A and B will be determined not only by the rate of diffusion of A but also behaviour of B. (3) DAB is the diffusivity of A and B, a physical property of both components. (4) It is a characteristic of the constituent and its environment (pressure, temperature, concentration whether in liquid, gas or solid solution and nature of other constituents). 2.7 DIFFUSION IN BINARY SOLUTION P

(i) Box with partition P.

I

(ii) Add H2O / Ethanol.

II

(iii) Remove Partition P. Fig. 2.5 : Illustration of Diffusion in Binary Solution

Consider the box as shown in Fig. 2.5 which is separated into two parts by the partition P. Into section I, 1 kg water (A) is placed and into section II 1 kg ethanol (B) (the densities of the liquids in each section are the same) Imagine the partition to be carefully removed, thus allowing diffusion of both liquids to occur. When diffusion stops, the concentration will be uniform throughout at 50 mass% of each constituents and the masses and moles of each constituent in the two regions will be as indicated in the figure, it is clear that while the water has diffused to the right and the ethanol to the left, there has been a net movement to the right, so that if the box has originally been balanced on a knife edge, at the end of process it would have tipped downward to the right. If the direction to the right is taken as positive, the flux NA of A relative to the fixed position P has been positive and the flux NB of B has been negative. For steady state, the net flux is NA + NB = N

… (2.2)

The movement of A is made up of two parts : (a) That resulting from bulk motion N and fraction XA of N that is A. (b) That resulting from molecular diffusion JA of A itself, w.r.t, the average molal velocity. ∴ NA = NXA + JA … (2.3) From equation (2.2) and (2.3) we get, ∂C CA A NA = (NA + NB) – DAB … (2.4) C ∂z

Principles of Mass Transfer Operations − I (Vol. − I)

2.8

Fundamentals of Diffusion Mass Transfer

Similarly equation for B is, NB = (NA + NB)

∂C CA B – DBA C ∂z

… (2.5)

Adding equation (2.4) and (2.5) we get, – DAB

∂CA ∂z

∂C = DAB

B

… (2.6)

∂z

Adding equation (2.4) and (2.5) we get JA = – JB

i.e.

And if CA + CB = Constant then it follows that DAB = DBA at prevailing concentation and temperature. 2.8

STEADY STATE MOLECULAR DIFFUSION IN FLUIDS AT REST AND IN LAMINAR FLOW Consider equation (2.4).

∂C CA A – DAB C ∂z Applying this equation to the case of diffusion only in Z-direction, with NA and NB both constant (steady state) and separating variables and assuming DAB is constant, we get on integration, NA = (NA + NB)

CA

2

⌠ ⌡ CA

– ∂CA NAC – CA (NA + NB)

1

1 = CDAB

Z2

⌠ ⌡

dZ

… (2.5)

Z1

Where 1 indicates the beginning of diffusion path (CA high) and 2, the end of diffusion path (CA low). Let, Z2 – Z1 = Z then we get, NAC – CA C (NA + NB) 1 Z 2 ln = … (2.6) NA + NB NAC – CA (NA + NB) CDAB 1

NA or,

NA =

1

NA + NB

NA/(NA + NB) – CA /C DAB C 2 ln Z NA/(NA + NB) CA /C

… (2.7)

1

Note : (1) Integration under steady state condition where flux NA is not constant is also possible. (2) To use these equations, NA, NB must be known. Molecular Diffusion in Gases : When the ideal gas law can be applied equation (2.6) can be written in more convenient form for use with gases as follows : CA C

–

=

pA = yA pt

… (2.8)

Principles of Mass Transfer Operations − I (Vol. − I) where,

2.9

Fundamentals of Diffusion Mass Transfer

–

p A = Partial pressure of component A pt = Total pressure. yA = Mole fraction (concentration)

Further,

C=

n V

=

pt RT

… (2.9)

So equation (2.7) becomes, NA

or,

[NA/(NA + NB)] [NA/(NA + NB)]

NA DAB pt = · · ln NA + NB RTZ

NA =

NA DAB pt · · ln NA + NB RTZ

–

pt – pA ·

2

–

pt – pA

… (2.10)

1

NA/(NA + NB) – yA2 N /(N + N ) – y  B A1  A A

… (2.11)

Example : (a) Presence of chemical reactions : Stiochiometric considerations. Methane is being cracked on a catalyst as per following reactions : CH4 → C + 2H2 (A)

(B)

CH4 (A) diffuses to the cracking surface and H2 (B) diffuses back, ∴ By reaction stiochiometry, we have NB = – 2NA ∴

NA NA + NB

=

NA = –1 NA – 2NA

… (2.12)

(b) Absence of chemical reactions : In this case, ratio can be fixed by enthalpy considerations. 2.9 PRINCIPLES OF MASS TRANSFER When a system contains two or more components whose concentrations vary from point to point, there is a natural tendency for mass to be transferred, minimizing the concentration differences within a system. The transport of one constituent from a region of higher concentration to that of a lower concentration is called mass transfer. The transfer of mass within a fluid mixture or across a phase boundary is a process that plays a major role in many industrial processes. Examples of such processes are : (i) Dispersion of gases from stacks (ii) Removal of pollutants from plant discharge streams by absorption (iii) Stripping of gases from waste water (iv) Neutron diffusion within nuclear reactors (v) Air conditioning Many of our day-by-day experiences also involve mass transfer, for example : (i) A lump of sugar added to a cup of coffee eventually dissolves and then eventually diffuses to make the concentration uniform.

Principles of Mass Transfer Operations − I (Vol. − I)

2.10

Fundamentals of Diffusion Mass Transfer

(ii) Water evapourates from ponds to increase the humidity of passing-air-stream (iii) Perfumes present a pleasant fragrance, which is imparted throughout the surrounding atmosphere. The mechanism of mass transfer involves both molecular diffusion and convection. 2.10 PROPERTIES OF MIXTURES Mass transfer always involves mixtures. Consequently, we must account for the variation of physical properties, which normally exist, in a given system. When a system contains three or more components, as many industrial fluid streams do, the problem becomes unwidely very quickly. The conventional engineering approach to problems of multicomponent system is to attempt to reduce them to representative binary (i.e., two component) systems. In order to understand the future discussions, let us first consider definitions and relations, which are often used to explain the role of components within a mixture. 2.10.1 Concentration of Species Concentration of species in multicomponent mixture can be expressed in many ways. For species A, mass concentration denoted by ρA is defined as the mass of A, mA per unit volume of the mixture. mA ρA = … (2.13) V The total mass concentration density ρ is the sum of the total mass of the mixture in unit volume : n

ρ

=

∑

ρi

i

where ρi is the concentration of species i in the mixture. Molar concentration of, A, CA is defined as the number of moles of A present per unit volume of the mixture. By definition, mass of A Number of moles = molecular weight of A mA nA = … (2.14) MA Therefore from equation (2.13) and (2.14), nA ρA CA = = V MA For ideal gas mixtures, pA V [From Ideal gas law, PV = nRT] RT nA pA CA = = V RT where pA is the partial pressure of species A in the mixture. V is the volume of gas, T is the absolute temperature, and R is the universal gas constant. The total molar concentration or molar density of the mixture is given by, nA

=

C

=

n

∑ i=1

Ci

… (2.15)

Principles of Mass Transfer Operations − I (Vol. − I)

2.11

Fundamentals of Diffusion Mass Transfer

2.10.2 Mass-averaged Velocities In a multicomponent system, the various species will normally move at different velocities; and evaluation of velocity of mixture requires the averaging of the velocities of each species present. If v1 is the velocity of species i with respect to stationary fixed co-ordinates, then mass-average velocity for a multicomponent mixture defined in terms of mass concentration is, ∑ ρi vi ∑ ρi vi i i v= = … (2.16) ∑ ρi ρ i By similar way, molar-average velocity of the mixture v* is, ∑ C i Vi i v* = … (2.17) C For most engineering problems, there will be little difference in v* and v and so the mass average velocity, v, will be used in all further discussions. The velocity of a particular species relative to the mass average or molar average velocity is termed as diffusion velocity i.e. Diffusion velocity = vi – v. The mole fraction for liquid and solid mixture, xA and for gaseous mixtures, yA, are the molar concentration of species A divided by the molar density of the mixtures. CA (Liquids and solids) … (2.18) xA = C CA yA = (Gases) … (2.19) C The sum of the mole fractions, by definition must equal to 1; ∑ xi = 1 i i.e. ∑ yi = 1 … (2.20) i by similar way, mass fraction of A in mixture is; ρA wA = ρ 2.11 DIFFUSION FLUX Just as momentum and energy (heat) transfer have two mechanisms for transportmolecular and convective, so does mass transfer. However, there are convective fluxes in mass transfer, even on a molecular level. The reason for this is that in mass transfer, whenever there is a driving force, there is always a net movement of the mass of a particular species which results in a bulk motion of molecules. Of course, there can also be convective mass transport due to macroscopic fluid motion. In this chapter the focus is on molecular mass transfer. The mass (or molar) flux of a given species is a vector quantity denoting the amount of the particular species, in either mass or molar units, that passes per given increment of time through a unit area normal to the vector. The flux of species defined with reference to fixed spatial co-ordinates, NA is : N A = CA v A

… (2.22)

This could be written in terms of diffusion velocity of A, (i.e., vA – v) and average velocity of mixture, v, as NA = CA (vA – v) + CA v … (2.23)

Principles of Mass Transfer Operations − I (Vol. − I)

2.12

Fundamentals of Diffusion Mass Transfer

∑ Ci v i i By definition, v = v* = C Therefore, equation (2.22) becomes, CA ∑ C v C i i i NA = CA (vA – v) + yA ∑ Ci vi i For systems containing two components A and B, NA = CA (vA – v) + yA (CA vA + CB vB) NA = CA (vA – v) +

= CA (vA – v) + yA (NA + NB) NA = CA (vA – v) + yA N

… (2.24)

The first term on the right hand side of this equation is diffusional molar flux of A, and the second term is flux due to bulk motion. 2.11.1 Fick’s Law An empirical relation for the diffusional molar flux, first postulated by Fick and, accordingly, often referred to as Fick’s first law, defines the diffusion of component A in an isothermal, isobaric system. For diffusion in only the Z direction, the Fick’s rate equation is dCA JA = – DAB dZ Where DAB is diffusivity or diffusion coefficient for component A diffusing through component B, and dCA/dZ is the concentration gradient in the Z-direction. A more general flux relation which is not restricted to isothermal, isobaric system could be written as, dyA … (2.25) JA = – CDAB dZ Using this expression, equation (2.24) could be written as dyA NA = – CDAB + yA N … (2.26) dZ 2.11.2 Relation Among Molar Fluxes For a binary system containing A and B, from equation (2.26), NA = JA + yA N or JA = NA + yA N … (2.27) … (2.28) Similarly, JB = NB + yB N Addition of Equation (2.27) and (2.28) gives, JA + JB = NA + NB – (yA + yB) N … (2.29) By definition N = NA + NB and yA + yB = 1. Therefore equation (2.29) becomes, JA + JB = 0 JA = –JB dyA dyB CDAB = – CDBA … (2.30) dZ dZ From, yA + yB = 1 dyA = – dyB

Principles of Mass Transfer Operations − I (Vol. − I)

2.13

Fundamentals of Diffusion Mass Transfer

Therefore Equation (2.30) becomes, DAB = DBA … (2.31) This leads to the conclusion that diffusivity of A in B is equal to diffusivity of B in A. 2.12 DIFFUSIVITY Fick’s law proportionality, DAB, is known as mass diffusivity (simply as diffusivity) or as the diffusion coefficient. DAB has the dimension of L2/t, identical to the fundamental dimensions of the other transport properties : Kinematic viscosity, ν = (µ/ρ) in momentum K transfer, and thermal diffusivity, α = in heat transfer. ρ · Cρ 2

2

Diffusivity is normally reported in cm /sec; the SI unit being m /sec. Diffusivity depends on pressure, temperature, and composition of the system. In table (2.1), some values of DAB are given for a few gas, liquid, and solid systems. Diffusivities of gases at low density are almost composition independent, incease with the temperature and vary inversely with pressure. Liquid and solid diffusivities are strongly concentration dependent and increase with temperature. Table 2.1 : General range of values of diffusivity – Gases : 1 × 10–5 m2/sec. 5 × 10–6 –6 – Liquids : 10 10–9 m2/sec. – Solids : 1 × 10–10 m2/sec. 5 × 10–14 In the absence of experimental data, semi theoretical expressions have been developed which give approximation, sometimes as valid as experimental values, due to the difficulties encountered in experimental measurements. 2.12.1 Diffusivity in Gases Pressure dependence of diffusivity is given by : 1 DAB ∝ (for moderate ranges of pressures, upto 25 atm). p And temperature dependency is according to, DAB ∝ T3/2 Diffusivity of a component in a mixture of components can be calculated using the diffusivities for the various binary pairs involved in the mixture. The relation given by Wilke is, 1 … (2.32) D1 – mixture = yn' y'2 y'3 + +…… + D1 – 2 D1 – 3 D1 – n Where D1-mixture is the diffusivity for component 1 in the gas mixture; D1-n is the diffusivity for the binary pair, component 1 diffusing through component n; and yn' is the mole fraction of component n in the gas mixture evaluated on a component –1 – free basis, that is, y2 ' y2 = … (2.33) y2 + y3 + … … yn (i) Gilliland's corelation for estimation of diffusivity of gases : DAB = 0.043

T3/2



1 3

PT VA + VB

2

 

1 3

1 1 + MA MB

Principles of Mass Transfer Operations − I (Vol. − I) where,

2.14

Fundamentals of Diffusion Mass Transfer

DAB = diffusivity, cm2/sec. MA and MB = Molecular weight of solute and solvent respectively VA and VB = Molecular volume of solute and solvent respectively, PT

cm3/gm. mole = Total Pressure (atm.)

(ii) Fuller et. al for estimations of diffusivity of gases : Empirical method of Fuller et al. is useful for moderate temperature ranges and can be used for mixtures of non-polar gases and for a polar and non-polar mixture. DAB =

0.001 T1.75 1/MA + 1/MB P [(∑vA)1/3 + (∑vB)1/3]2

∑vA = Sum of structural volume increments P is in atm., T is in K, D is in cm2/s. Table 2.2 : Atomic Diffusion Volumes for use with the Fuller, Schettler, and Giddings Method A. Atomic and Structural Diffusion Volume Increments, v C 16.5 (Cl) 19.5 H 1.98 (S) 17.0 O 5.48 Aromatic ring – 20.2 (N) 5.69 Heterocyclic ring – 20.2 B. Diffusion Volumes for Simple Molecules, ∑ v H2 18.9 7.07 CO CO D2 26.9 6.70 2 N2O 35.9 2.88 He N2 14.9 17.9 NH3 12.7 16.6 O2 H2O 114.8 20.1 (CCl2F2) Air 16.1 (SF6) Ar 69.7 22.8 (Cl2) Kr 37.7 37.9 (Xe) 67.2 (Br2) Ne 41.1 5.59 (SO2 Note : Parenthesis indicate that the value listed is based on only a few data points. Source : From E. N. Fuller, P. D. Schettler and J. C. Giddings, Ind. Eng. Chem. 58,, 19 (1966).

2.12.2 Diffusivity in Liquids Diffusivity in liquid are exemplified by the values given in Table 2.1. Most of these values -5 2 are nearer to 10 cm /sec, and about ten thousand times lower than those in dilute gases. This characteristic of liquid diffusion often limits the overall rate of processes accruing in liquids (such as reaction between two components in liquids). In chemistry, diffusivity limits the rate of acid-base reactions; in the chemical industry, diffusion is responsible for the rates of liquid-liquid extraction. Diffusion in liquids is important because it is slow. Certain molecules diffuse as molecules, while others which are designated as electrolytes ionize in solutions and diffuse as ions. For example, sodium chloride (NaCl), diffuses in + – water as ions Na and Cl . Though each ions has a different mobility, the electrical neutrality of the solution indicates the ions must diffuse at the same rate; accordingly it is possible to speak of a diffusion coefficient for molecular electrolytes such as NaCl. However, if several

Principles of Mass Transfer Operations − I (Vol. − I)

2.15

Fundamentals of Diffusion Mass Transfer

ions are present, the diffusion rates of the individual cations and anions must be considered, and molecular diffusion coefficients have no meaning. Diffusivity varies inversely with viscosity when the ratio of solute to solvent ratio exceeds five. In extremely high viscosity materials, diffusion becomes independent of viscosity. Estimation of Diffusion Coefficient using Wilke-Chang (1958) Equation : The diffusion coefficient or diffusivity of liquids may be estimated by Wilke-Chang correlation : o

–18

DAB = (117.3 × 10

) (φ MB)

1 2

T 0.6

… (1)

µ B VA

This is Wilke-Chang equation and this equation is good only for dilute solutions of non-dissociating solutes upto about ± 10%. where, o

DAB MB T µ VA

= Diffusivity of A in very dilute solutions in solvent B, m2/sec. = = = =

Molecular weight of solvent, kg/kmole Absolute temperature, K Viscosity of solution, kg/m. sec. Solute molar volume at normal boiling temperature, m3/ k mole.

= 0.0756 m3/k mole for water as solute φ = Association factor for solvent = 2.26 for water as solvent. = 1.9 for Methanol as solvent. = 1.5 for Ethanol as solvent. = 1.0 for unassociated solvents such as benzene and ethyl ether. Temperature Dependency of Diffusivity (DAB) : In equation, we can write as T DAB ∝ µ T1 (DAB)T1  µ1  T1 µ2 Hence, = = ×  (DAB)T2 T2  µ1  T2  µ2  ∴ where,

T1 µ2 (DAB)T1 = (DAB)T2 ⋅  ⋅   µ1 T2 (DAB)T1 = Diffusivity of A in B at temperature T1

… (2.)

(DAB)T2 = Diffusivity of A in B at temperature T2 2.12.3 Diffusivity in Solids Typical values for diffusivity in solids are shown in Table '2.3'. One outstanding characteristic of these values is their small size, usually thousands of time less than those in a liquid, which are inturn 10,000 times less than those in a gas. Diffusion plays a major role in catalysis and is important to the chemical engineers. For metallurgists, diffusion of atoms within the solids is of more importance.

Principles of Mass Transfer Operations − I (Vol. − I)

2.16

Fundamentals of Diffusion Mass Transfer

Table 2.3 : Molecular Diffusitivites of Gases at 1 Atm. Abs Pressure System Temperature °C Diffusivity, cm2/sec. Air – NH3 0 0.198 Air – H2O 0 0.220 42 0.288 Air – CO2 3 0.142 44 0.177 Air – ethanol 42 0.145 25 0.135 Air – acetic acid 0 0.106 Air – n-hexane 21 0.080 Air – toulene 25.9 0.086 59.0 0.104 Air – hydrogen 0 0.611 Air – n-butanol 0 0.0703 25.9 0.087 59.0 0.104 Air – n-pentane 21 0.071 H2 – Ar 22.4 0.83 175 1.76 796 8.10 He – Ar 25 0.729 225 1.728 CH4 – Ar 25 0.202 CH4 – He 25 0.675 N2 – He 25 0.687 CH4 – H2 0 0.625 N2 – NH3 25 0.230 85 0.328 H2 – NH3 25 0.783 85 1.093 H2 – N2 25 0.784 85 1.052 H2O – N2 34.4 0.256 55.4 0.303 H2O – CO2 34.3 0.202 55.4 0.211 SO2 – CO2 343°K 0.108 C2H5OH – CO2 67 0.106 (C2H5)2O – air 19.9 0.0896 (CH3)2O – SO2 30 0.0672 (C9H5)2O – NH2 26.5 0.1078 2.13 STEADY STATE DIFFUSION In this section, steady-state molecular mass transfer through simple systems in which the concentration and molar flux are functions of a single space co-ordinate will be

Principles of Mass Transfer Operations − I (Vol. − I)

2.17

Fundamentals of Diffusion Mass Transfer

considered. In a binary system, containing A and B, this molar flux in the direction of z, is given by, dyA NA = – CDAB + yA (NA + NB) … (2.34) dZ 2.13.1 Diffusion Through a Stagnant Gas Film The diffusivity or diffusion coefficient for a gas can be measured, experimentally using Arnold Flow of gas B z = z2 diffusion cell. This cell is illustrated schematically in NAz|z+Dz figure. The narrow tube of uniform cross-section, Dz which is partially filled with pure liquid A, is NAz|z maintained at a constant temperature and pressure. z = z1 Gas B, which flows across the open end of the tube, has a negligible solubility in liquid A, and is also chemically inert to A. (i.e. no reaction between Pure liquid A A and B). Component A vapourizes and diffuses into Fig. 2.6 : Arnold Diffusion Cell the gas phase; the rate of vapourization may be physically measured and may also be mathematically expressed in terms of the molar flux. Consider the control volume S ∆ z, where, S is the cross-sectional area of the tube. Mass balance on A over this control volume for a steady-state operation yields, [Moles of A leaving at z + ∆z] – [Moles of A entering at z] = 0. i.e. SNA Z + ∆Z – SNA Z = 0 … (2.35) Dividing through by the volume, S∆Z, and evaluating in the limit as ∆Z approaches zero, we obtain the differential equation, dNA = 0 … (2.36) dZ This relation stipulates a constant molar flux of A throughout the gas phase from Z1 to Z2. A similar differential equation could also be written for component B as, dNB = 0 dZ and accordingly, the molar flux of B is also constant over the entire diffusion path from z1 and z2. Considering only at plane z1, and since the gas B is insoluble is liquid A, we realize that NB, the net flux of B, is zero throughout the diffusion path; accordingly B is a stagnant gas. We know, dyA NA = – CDAB + yA (NA + NB) dZ Since, NB = 0 dyA NA = – CDAB + yA NA dZ Rearranging, – CDAB dyA … (2.37) NA = 1 – yA dZ

Principles of Mass Transfer Operations − I (Vol. − I) Since,

2.18

Fundamentals of Diffusion Mass Transfer

NB = 0 dyA + yA NA dZ

NA = –CDAB Rearranging, NA =

–CDAB dyA 1 – yA dZ

… (2.37)

This equation may be integrated between the two boundary conditions :

And

at z = z1

YA = YA

at z = z2

YA = yA

1

2

Assuming the diffusivity is to be independent of concentration, and realizing that NA is constant along the diffusion path, by integrating equation (2.37) we obtain, yA 2

Z2

NA

–dy

A ⌠ ⌡ dZ = CDAB ⌠ ⌡ 1 – yA Z1

yA 1

NA =

CDAB Z2 – Z1

1 – yA2 1 – y  A1 

ln

… (2.38)

The log mean average concentration of component B is defined as, yB – yB 2 1 yB, lm = y  B2 ln   yB





1

Since, y B = 1 – y A, (1 – yA ) – (1 – yA ) yA – yA 2 1 1 2 yB, lm = = y y  A2  A2 ln   ln   yA yA





1

Substituting from equation (2.39) in equation (2.38), CDAB (yA1 – yA2) NA = Z2 – Z1 yB, lm For an ideal gas, C =

n V

=

p and RT

for mixture of ideal gases, pA P Therefore, for an ideal gas mixture equation, (2.40) becomes, pA – pA DAB Pt 1 2 NA = – RT (z2 – z1) p B, lm yA =



… (2.39)



1

… (2.40)

Principles of Mass Transfer Operations − I (Vol. − I)

2.19

Fundamentals of Diffusion Mass Transfer

This is the equation of molar flux for steady-state diffusion of one gas through a second stagnant gas. Many mass transfer operations involve the diffusion of one gas component through another non-diffusing component; absorption and humidification are typical operations defined by these equation. The concentration profile (pA vs. z) for this type of diffusion is shown in Fig. 2.7.

p1

Pressure

pt

A

pB2

pB pB1 pA1

z1

pA

Distance, z

pA2 z2

Fig. 2.7 : Diffusion of A through stagnant B

2.13.2 Psuedo–Steady–State Diffusion Through a Stagnant Film In many mass transfer operations, one of the boundaries may move with time. If the length of the diffusion path changes a small amount over a long period of time, a pseudo steady state diffusion model may be used. When this condition exists, the equation of steady state diffusion through stagnant gas’ can be used to find the flux. If the difference in the level of liquid A over the time interval considered is only a small fraction of the total diffusion path, and t0 – t is relatively long period of time, at any given instant in that period, the molar flux in the gas phase may be evaluated by, Flow of gas B z = z2 NAz|z+Dz Dz z = z1 at t1

NAz|z

z = z1 at t0

Pure liquid A

Fig. 2.8 : Arnold diffusion cell with moving liquid interface

CDAB (yA – yA ) NA =

1

2

zyB‚ lm

where z equals z2 – z1, the length of the diffusion path at time t. The molar flux NA is related to the amount of A leaving the liquid by, ρA‚ L dZ NA = MA dt Where

… (2.41)

… (2.42)

ρA‚ L is the molar density of A in the liquid phase. Under Psuedo steady state MA

conditions, equations (2.41) and (2.42) can be equated to give : CDAB (yA – yA ) ρA‚ L dz 1 2 = MA dt zyB‚ lm

… (2.43)

Principles of Mass Transfer Operations − I (Vol. − I)

2.20

Fundamentals of Diffusion Mass Transfer

Equation (2.43) May be integrated from t = 0 to t and from z = z t0 to z = zt as : t

⌠ ⌡ t=0

ρA‚ L yB‚ lm/MA dt = CDAB (yA – yA ) 1

2

Zt

⌠ ⌡ z dz Zt0

Yielding, ρA‚ L yB‚ lm/MA t = CDAB (yA – yA ) 1

2

z2 – z2   t t0  2 

… (2.44)

This shall be rearranged to evaluate diffusivity DAB as, DAB

ρA‚ L yB‚ lm = MA C (yA – yA ) t 1

2

z2– z2   t t0  2 

… (2.45)

2.13.3 Equimolar Counter Diffusion A physical situation which is encountered in the distillation of two constituents whose molar latent heats of vapourization are essentially equal, stipulates that the flux of one gaseous component is equal to but acting in the opposite direction from the other gaseous component; that is, NA = – NB. The molar flux NA, for a binary system at constant temperature and pressure is described by, dyA NA = – CDAB + yA (NA + NB) dz dCA or NA = – DAB + yA (NA + NB) … (2.46) dz With the substitution of NB = –NA, Equation (2.46) becomes, dCA NA = – DAB … (2.47) dz For steady state diffusion Equation. (2.47) may be integrated, using the boundary conditions : at z = z1 CA = CA and z = z2 CA = C A 1

2

Giving, NA

Z2

CA

⌠ ⌡ dz = – DAB

⌠ ⌡

Z1

CA

2

d CA

1

from which NA =

DAB (CA – CA ) z2 – z1 1 2

CA =

nA V

… (2.48)

For ideal gases, =

pA RT

Therefore equation (2.48) becomes, NA =

DAB RT (z2 – z1) (PA – PA ) 1

2

… (2.49)

Principles of Mass Transfer Operations − I (Vol. − I)

2.21

Fundamentals of Diffusion Mass Transfer

This is the equation of molar flux for steady-state equimolar counter diffusion. Concentration profile in this equimolar counter diffusion may be obtained from, d (NA) dz And from equation (2.47) Therefore,

d dz

= 0 (Since NA is constant over the diffusion path).

NA = – DAB

dCA dz

dCA  – DAB = 0 dz   d 2 CA

or

= 0 dz2 This equation may be solved using the boundary conditions to give, CA – CA z – z1 1 = CA – CA z1 – z2 1

… (2.50)

2

Equation, (2.50) indicates a linear concentration profile for equimolar counter diffusion. 2.13.4 Diffusion Into An Infinite Standard Medium Here we will discuss problems involving diffusion from a spherical particle into an infinite body of stagnant gas. The purpose in doing this is to demonstrate how to set up differential equations that describe the diffusion in these processes. The solutions, obtained are only of academic interest because a large body of gas in which there are no convection currents is unlikely to be found in practice. However, the solutions developed here for these problems actually represent a special case of the more common situation involving both molecular diffusion and convective mass transfer. (a) Evaporation of a spherical Droplet : As an example of such problems, we shall consider the evaporation of spherical droplet such as a raindrop or sublimation of naphthalene ball. The vapour formed at the surface of the droplet is assumed to diffuse by molecular motions into the large body of stagnant gas that surrounds the droplet. r+dr

dr

r r0

Fig. 2.9 : Evaporation of a Raindrop

Consider a raindrop, as shown in figure. At any moment, when the radius of the drop is r0, the flux of water vapour at any distance r from the center is given by, NA = – CDAB

dyA + yA (NA + NB) dr

Here NB = 0 (since air is assumed to be stagnant)

Principles of Mass Transfer Operations − I (Vol. − I)

2.22

Therefore,

NA = – CDAB

Rearranging,

NA =

–CDAB 1 – yA

Fundamentals of Diffusion Mass Transfer dyA + yA NA dr dyA dr

… (2.51)

The flux NA is not constant, because of the spherical geometry; decreases as the distance from the center of sphere increases. But the molar flow rate at r and r + δr are the same. This could be written as, ANA r = ANA r + δr where A = surface area of sphere at r or r + δr. 2 Substituting for A = 4 π r in equation (2.52), 4 πr2 NA

or

lim δr → 0

r + δr

r2 N A

– 4Wr2 NA

r + δr

– r2NA

r

… (2.52)

= 0

r

= 0

δr

d (r2 NA) = 0 dr r2 NA = constant

Integrating,

… (2.53) … (2.54)

2

r2 NA = r0NA

0

From equation (2.54), Substituting for NA from equation (2.51), –r2 CDAB dyA 2 = r0NA 1 – yA dr 0 2

r0NA

dr

⌠ 2 0⌡ r

dy

A ⌠ ⌡ 1 – yA

= – CDAB

Boundary condition :

At r = r0

yA = yAS

And

At r → ∞

yA = yA∞

… (2.55)

Therefore equation (2.55) becomes, r0NA Simplifying,

yA∞

∞

2 0

– 1  r r0 = [CDAB ln (1 – yA)]yAS NA

0

=

CDAB 1 – yA∞ ln   r0 1 – yAS

… (2.56)

Time required for complete evapouration of the droplet may be evaluated from making mass balance. In – Out = Accumulation 2 3 ρL  d 4 0 – 4 π r0NA = π r0  dt 3 MA 0 2

or,

= – 4 π r0

ρL dr0 MA dt

… (2.57)

Principles of Mass Transfer Operations − I (Vol. − I)

2.23

Fundamentals of Diffusion Mass Transfer

Substituting for NA from equation (2.56) in equation (2.57), 0

–ρL dr0  1 – yA  1 – y  = M dt  AS A

CDAB ln r0

… (2.58)

Initial condition : When t = 0, r0 = r1 Integrating equation (2.58) with these initial condition, t

⌠ ⌡ 0

–ρL 1 dt = MA CDAB

1 1  – yA∞ ln   1 – yAS

ρL 1 t = MA 2 CDAB

r1

0

⌠ ⌡ r0 dr0 r1

2

… (2.59)

1 – yA∞ ln   1 – yAS

Equation (2.59) gives the total time t required for complete evapouration of spherical droplet of initial radius r1. (b) Combustion of a coal particle : The problem of combustion of spherical coal particle is similar to evapouration of a drop with the exception that chemical reaction (combustions) occurs at the surface of the particle. During combustion of coal, the reaction, C + O2 → CO2 occurs. According to this reaction for every mole of oxygen that diffuses to the surface of coal (maximum of carbon), react with 1 mole of carbon, releases 1 mole of carbon dioxide, which must diffuse away from this surface. This is a case of Equimolar counter diffusion of CO2 and O2. Normally air (a mixture of N 2 and O2) is used for combustion, and in this case N2 does

(

)

not takes part in the reaction, and its flux is zero. i.e. NN = 0 . 2

The molar flux of O2 could be written as, dyO NO

= – CDO

NO

= – CDO

2

2

For steady state conditions, d (r2 NO ) dr 2 r2 N O

Integrating,

2

2 – gas

2 – gas

2

+ yO

dr dyO

2

(N

O2

+ NCO + NN 2

2

) … (2.60)

2

… (2.61)

dr

= 0

… (2.62) 2

= constant = r0 NO

… (2.63)

2S

where r0 is the radius of coal particle at any instant, and NO

2S

in the flux of O2 at the

surface of the particle. Substituting NO from equation (2.61) in equation (2.63). 2

–r2 CDO

dyO

2

2

= r0 N O S dr 2 Boundary condition : r = r0 , yO = yO S 2 – gas

2

2

… (2.65)

Principles of Mass Transfer Operations − I (Vol. − I)

2.24

Fundamentals of Diffusion Mass Transfer

And At, r → ∞, yO = yO 2

2∞

With this boundary condition, equation (2.65) becomes, yO ∞ ∞ 2 2 dr r0 NA ⌠ 2 = –CDO – gas ⌠ dyO ⌡ 0 ⌡ r 2 2 r0

yO S 2

CDO which yields,

NO

2S

2 – gas

=

r0

(y

O2S –

yO

)

… (2.66)

2∞

From fast reaction of O2 with coal, the mole fraction of O2 at the surface of particle is zero. yO S = 0 2

And also at some distance away from the surface of the particle yO = yO 2

(because air is a mixture of 21 mole % O2 and 79 mole % N2) With these conditions, equation (2.66) becomes, 0.21 CDO – gas 2 NO S = r0 2

2∞

= 0.21

… (2.67)

Dr

R

r

N02r NCOr

Fig. 2.10 : Combustion of a particle of Coal

2.13.5 Diffusion in Liquids Equation derived for diffusion in gases equally applies to diffusion in liquids with some modifications. Mole fraction in liquid phases is normally written as ‘x’ (in gases as y). The ρ concentration term ‘C’ is replaced by average molar density,   av M (a) For steady – state diffusion of A through non-diffusing B : NA = constant, NB = 0 NA =

DAB zxBM

 ρ  (x – x ) Mav A1 A2

Where Z = Z 2 – Z 1, the length of diffusion path; and XB – XB 2 1 XBM = XB2 ln  XB 





1

… (2.68)

… (2.69)

Principles of Mass Transfer Operations − I (Vol. − I)

2.25

Fundamentals of Diffusion Mass Transfer

(b) For steady – state equimolar counter diffusion : NA = – NB = constant Na =

DAB Z

(

CA – CA 1

)

=

2

DAB Z

ρ Mav (xA1 – xA2)

…(2.70)

2.13.6 Mass Diffusion with Homogeneous Chemical Reaction Absorption operations involves contact of a gas mixture with a liquid and preferential dissolution of a component in the contacting liquid. Depending on the chemical nature of the involved molecules, the absorption may or may not involve chemical reaction. The following analysis illustrates the diffusion of a component from the gas phase into the liquid phase accompanied by a chemical reaction in the liquid phase. Consider a layer of absorbing medium (liquid) as shown in diagram. Gas mixture (A and inert gas) NAz|z

z z=0

Liquid surface Dz NAz|z+Dz

Liquid B

z=d

Fig. 2.11 : Absorption with homogeneous chemical reaction

At the surface of the liquid, the composition of A is CA 0. The thickness of the film, δ is so defined, that beyond this film the concentration of A is always zero; that is CAδ = 0. If there is very little fluid motion within the film, NA = – DAB

dCA Ca + (NA + NB) dz C

… (2.71)

If concentration of A in the film, CA is assumed small, equation (2.71) becomes, NA = –DAB

dCA dz

… (2.72)

The molar flux NA changes along the diffusion path. This change is due to the reaction that takes place in the liquid film. These changes could be written as, d (NA) – rA = 0 dz

… (2.73)

where –rA is the rate of disappearance of A. For a first order reaction, k >B A –rA = k CA With the substitution from equation (2.74) and (2.72) in equation (2.73), dCA –d  D = kCA = 0 dz  AB dz 

… (2.74)

Principles of Mass Transfer Operations − I (Vol. − I)

2.26

Fundamentals of Diffusion Mass Transfer

For constant Diffusivity, –DAB

d2CA dz2

+ kCA = 0

… (2.75)

Which is a second order ordinary differential equation. The general solution to this equation is,

 

k   z + C2 sin h  DAB  

CA = C1 cos h 

k  z DAB 

… (2.76)

The constants of this equation can be evaluated from the boundary conditions : at

Z=0

CA = CA0

And at

Z=δ

CA = 0 –CA

The constant C1 is equal to CA , and C2 is equal to

0

 tan h  

0

k  δ DAB 

with this

substitution equation (2.76) becomes,

 

CA = CA cos h  0

k  z – DAB 

 

CA sin h  0

 tan h  

k  z DAB  k  z DAB 

… (2.77)

This equation gives the variation of concentration of A with z (i.e concentration profile of A in the liquid). The molar flux at the liquid surface can be determined by differentiating equation (2.77), and evaluating the derivative, dCA at z = 0 dz Differentiating CA with respect to z, dCA = CA dz 0

k DAB

 

sin h 

k  z – DAB 

CA

0

k  cos h  DAB 

 tan h  

k  z DAB 

k  δ DAB 

… (2.78)

Substituting z = 0 in equation (2.78) and from equation (2.72), DAB CA NA

Z=0

=

0

δ

k  δ D  tan h  Dk   AB

AB

   δ 

… (2.79)

For absorption with no chemical reaction, the flux of A is obtained from equation (2.72) as DAB CA NA =

δ

0

… (2.80)

Principles of Mass Transfer Operations − I (Vol. − I)

2.27

Fundamentals of Diffusion Mass Transfer

Which is constant throughout the film of liquid. On comparison of equation (2.79) and k δ DAB (2.80), it is apparent that the term k    tan h  D δ   AB 

   

   

shows the influence of the chemical reactions. This term is a dimensionless quantity, is often called as Hatta Number. 2.13.7 Diffusion in Solids In certain unit operation of chemical engineering such as in drying or in absorption, mass transfer takes place between a solid and a fluid phase. If the transferred species is distributed uniformly in the solid phase and forms a homogeneous medium, the diffusion of the species in the solid phase is said to be structure independent. In this cases diffusivity or diffusion coefficient is direction – independent. At steady state, and for mass diffusion, which is independent of the solid matrix structure, the molar flux in the z-direction is : dCA = constant, as given by Fick’s law.… (2.81) NA = – DAB dz Integrating the above equation, DAB (CA – CA ) 1 2 … (2.82) NA = z Which is similar to the expression obtained for diffusion in a stagnant fluid with no bulk motion (i.e. N = 0). (A) Fick's Law of Diffusion : When the concentration gradient remains unchanged with passage of time, so that the rate of diffusion is constant. Fick's law of diffusion can be applied when, (i) Diffusivity is independent of concentration, (ii) There is no bulk flow. Thus, dCA NA = – DA … (2.83) dz where, DA = Diffusivity of A through the solid (a) If DA = constant, then equation can be integrated (for diffusion through a flat slab of thickness z) give : DA (CA – CA ) NA =

1

2

… (2.84)

Z

which is similar expression obtained for fluids for similar situations. Here CA and CA 1

2

are the concentration at opposite sides of the slab. (b) For other solid shape, the rate is given by, DA Sav (CA – CA ) W = NA Sav =

1

Z

2

… (2.85)

with appropriate values of the average cross-section for diffusion Sav to be applied.

Principles of Mass Transfer Operations − I (Vol. − I)

2.28

Fundamentals of Diffusion Mass Transfer

Example : (i) Radial Diffusion Through a Solid Cylinder : Let,

inner radii = a1 outer radii = a2

so,

length = l 2π l (a2 – a1) Sav = a2 ln   a1

… (2.86)

and z = a2 – a1 … (2.87) (ii) Radial Diffusion Through a Spherical Shell : Let, inner radii = a1 outer radii = a2 … (2.88) So, Sav = 4π a1a2 and z = a2 – a1 … (2.89) Types of Solid Diffusion : Diffusion in polymers is of interest to chemical engineers because thin "permselective" polymer membranes may be employed in separation process and because processes for the manufacture of polymers offers involve diffusion of reactants or products to or from the site of polymerisation reaction. The separation process include hydrocarbon separation dialysis, reverse osmosis, blood oxygenators and artificial kidneys. Diffusion of water and other solvents presents problems in the manufacture of commercial polymers and the spinning of these to produce textile fibers; low water permeability is required in polymer films used to package the food. Separation by electrodialysis or by ion-exchange equipment depends on ion diffusion in polymers. The structure of the solid and its interaction with the diffusion substance have a profound influence on how diffusion occurs and on the rate of transport. (i) Diffusion Through Polymers : Imagine two bodies of a gas (e.g. H2) at different pressures separated by a polymeric membrane (e.g. Polyethylene). The gas dissolves in the solid at the faces exposed to the gas to an extent usually described by Henry's law, concentration directly proportional to pressure. The gas then diffuses from the high-to low-pressure side in a manner usually described as activated : the polymeric chains are in a state of constant thermal motion, and the diffusing molecules jump from one position to another over a potential barrier. A commercial application of these principles has been made for separating hydrogen from waste refinery gases in shell-and-tube devices which resemble in part the common heat exchanger. However, in this use the polymeric fiber tubes are only 30 µm OD, and there are 50 million of them in a shell roughly 0.4 m in diameter. (ii) Diffusion through a solid membrane : Consider the mass transfer process that occurs when a gas (A) diffuses through a solid membrane (B) : the membrane separates two gases in which the partial pressure of species A is different (Figure 2.12). We will assume that the process is at steady state, and that the specified partial pressures of A in the gases (PA1 and PA2 ) have been measured in the immediate vicinity of the membrane.

Principles of Mass Transfer Operations − I (Vol. − I) Gas

2.29

Fundamentals of Diffusion Mass Transfer

Soild (B)

Gas

cA1 PA1

cA2

PA2

NA (PA1 > PA2) x

x=0

x=L

Fig. 2.12 : A solid membrane made of a specific component (B) separates two gases with different partial pressures of a species (A) that is soluble in the membrane material. Because of the partial pressure difference, A will diffuse across the membrane. Other compounds might be present in the gases on each side of the membrane.

One important characteristic of this example, which will appear in a large number of mass transfer applications, is the presence of phase interfaces. In this case, there are two solid/gas interfaces (at x = 0 and x = L). The molar concentration of a chemical species suffers a discontinuity across a phase interface. For the case under consideration, molecules of A are not subjected to appreciable molecular interactions in the gas phase, whereas in the solid phase, they would have to "find their way" through a more compact molecular structure in which interactions with molecules of B will occur. The amount of A that the solid phase can "accommodate" in terms of moles of A per unit volume of material is then expected to be much less than the concentration of A in the gas, and it will depend on molecular compatibility between A and B. This distinguishes concentration profiles from temperature profiles : in heat transfer processes; generally it is considered that the temperature is a continuous function of position, even across a phase interface. However, the concentrations on both sides of the interface can be related by assuming that the phases are at equilibrium across the interface. This condition of local equilibrium is usually assumed to be valid. The rationale behind it is based on the hypothesis that the first few layers of molecules on both sides of the interface reach equilibrium in a characteristic time that is much faster than characteristic times of the mass transfer process. Under the local equilibrium hypothesis, the concentrations on both sides of the interface are related by a mathematical expression that can be determined in independent experiments involving equilibrium contacting between the phases. Going back to the process depicted in Figure 2.12, at the solid/gas interfaces, we can assume that equilibrium between the solid and gas is achieved instantly. Therefore, the concentration of A in the solid that is in contact with the gas is the solubility of A in the solid, when it is exposed to the corresponding partial pressure of A; i.e., referring to the concentrations inside the membrane CA1 and CA2 (Figure 2.12) , we have : CA1 : solubility of A in B exposed to PA1 CA2 : solubility of A in B exposed to PA2 In many cases the solubility will be directly proportional to the partial pressure : CA1 = KPA1 CA2

= KPA2

where K is a constant that depends on temperature and total pressure and can be measured independently.

Principles of Mass Transfer Operations − I (Vol. − I)

2.30

Fundamentals of Diffusion Mass Transfer

We are interested in analyzing the diffusion of A through the solid B. For this particular example, species A will diffuse in the positive x direction since CA1 > CA2 . We start by considering the flux equation for species A. In this case, we expect A to be present at relatively low concentrations in the solid, so that we can consider c to be uniform in the solid. We can write : dcA NA = xA (NA + NB) – DAB dx … (2.89) The solid is stationary and, therefore, NB = 0. This equation can be rearranged to give, dcA … (2.90) (1 – xA) NA = – DAB dx The condition that A is present at low concentrations in the solid (dilute system), can be expressed mathematically by stating that the mole fraction of A everywhere is very small compared to 1 : XA << 1, which implies : 1 – XA = 1, which in turn yields : dcA NA = – DAB dx

… (2.91)

This equation is valid for 0 < x < L. The diffusivity will be considered uniform. On the other hand, if NA were to vary with x, there would be accumulation of A in the membrane. This is not possible under steady state conditions by virtue of the principle of mass conservation. Therefore, we conclude : NA is uniform (independent of x) due to mass conservation. Since NA is constant, we can separate variables and integrate equation (2.91) through the membrane, as follows : cA2

L

NA ⌠ ⌡

dx = – DAB

0

⌠ ⌡

dcA

… (2.92)

cA1

Which leads to, NA = DAB

(cA1 – cA2) L

… (2.93)

This equation can be expressed in terms. of partial pressures as follows, NA = KDAB

(PA1 – PA2) L

… (2.94)

We can then use this equation to calculate the flux of A through the membrane. Equation (2.91) can also be integrated to determined the concentration profile of A within the membrane by making one of the limits of integration a generic co-ordinate x : cA (x)

x

NA ⌠ ⌡ 0

dx = – DAB

⌠ dcA ⌡

… (2.95)

cA1

Integrating, we find that the concentration of A is a linear function of x : NAx = – DAB (cA – cA1 )

… (2.96)

Principles of Mass Transfer Operations − I (Vol. − I)

2.31

Fundamentals of Diffusion Mass Transfer

Also, substituting equation (2.93) and rearranging leads to : x cA = cA1 (cA1 – cA2) L

… (2.97)

i.e., the concentration profile is a straight line joining the two boundary values as shown in Fig. 2.13. Gas

Soild (B)

cA1

Gas

cA(x) PA1

PA2 cA2

x

x=0

x=L

Fig. 2.13 : Concentration profile for the system shown in Figure 2.12

As a practical example of gas diffusion through a solid membrane, consider the following problem : we are interested in setting the shelf like of a plastic soda bottle, knowing that the process to consider is the loss of CO2 by diffusion through the walls. The physical situation is illustrated in Figure 2.14. cA1

Cola

Plastic wall (B)

Liquid with CO2

cA2 CO2(A) diffusing

atmospheric air PA – 0

x

Fig. 2.14 : The loss of CO2 from a carbonated beverage through the plastic wall of the bottle is due to the diffusion of CO2 through the wall. Inside the bottle, the liquid has a certain concentration of CO2, whereas the concentration of CO2 in the atmospheric air outside the bottle is negligible.

The information necessary to solve the problem will be made available as required, and it will be highlighted. This problem can be treated with the diffusion equations developed above, provided that we make the following assumptions : (1) The wall is so thin compared to the bottle curvature that it can be considered flat (diffusion is one-dimensional in the x-direction). (2) The diffusion process through the wall can be considered pseudo-steady; i.e., even though the CO2 content in the liquid will change with time, we will consider that these changes are slow with respect to the time that it takes the diffusion process to become steady. (3) We neglect CO2 concentration profiles in the liquid and the atmosphere.

Principles of Mass Transfer Operations − I (Vol. − I)

2.32

Fundamentals of Diffusion Mass Transfer

If these assumptions hold, then the flux, which will allow us to calculate the loss of CO2, is given by : (cA1 – cA2) … (2.98) NA = DAB L (iii) Surface Diffusion : It is phenomenon accompanying adsorption of solutions onto the surface of the pores of the solid. It is an activated diffusion involving jumping of adsorbed molecules from one adsorption site to another. In general, surface diffusion can be described by a two dimensional analog of Ficks' law. Features of surface diffusion : (a) Surface concentration expressed as mole/area instead of mole/volume. (b) Surface diffusivities range in m2/sec. (1) 10–7 × 10–9 m2/sec at ordinary temperature for physically adsorbed gases. (2) 10–12 m2/sec for liquid solution in adsorbent resin particles. The term "activated" refers to the temperature dependence of the diffusivity, which follows an Arrhenius-type expression : DA = Do e– HD/RT

… (2.99)

where, HD = Energy of activation Do = A constant For simple gases, DA is usually reasonably independent of concentration. It may, however, be a strong function of pressure of molding the polymer. For the permanent gases, diffusivities may be of the order of 10–10 m2/sec. Due to complicated dimensions and awkward units, it has become the practice to describe the diffusional characteristics in terms of quantity P, the permeability. Since at the two faces of a membrane the equilibrium solubility of the gas in the polymer is directly proportional to pressure, equation (2.84) can be converted into : DA SA VA =

(p–

–

A1

– pA

)

2

z

where, VA = Diffusional flux, cm3 gas (STP)/cm2. s DA = Diffusivity of A, cm2/s –

p A = Partial pressure of diffusing gas, cm Hg SA = Solubility coefficient or Henry's law constant, cm3 gas (STP)/(cm3.solid). cm. Hg z = Thickness of polymeric membrane, cm Permeability, P is then defined as, P = DA SA where,

cm Hg P = Permeability, cm3.gas (STP)/cm2.s   cm 

… (2.100)

Principles of Mass Transfer Operations − I (Vol. − I)

2.33

Fundamentals of Diffusion Mass Transfer

Diffusion in Porous Solids : In some chemical operations, such as heterogeneous catalysis, an important factor, affecting the rate of reaction is the diffusions of the gaseous component through a porous solid. The effective diffusivity in the solid is reduced below what it could be in a free fluid, for two reasons. First, the tortuous nature of the path increases the distance, which a molecule must travel to advance a given distance in the solid. Second, the free cross-sectional area is restricted. For many catalyst pellets, the effective diffusivity of a gaseous component is of the order of one tenth of its value in a free gas. If the pressure is low enough and the pores are small enough, the gas molecules will collide with the walls more frequently than with each other. This is known as Knudsen flow or Knudsen diffusion. Upon hitting the wall, the molecules are momentarily absorbed and then given off in random directions. The gas flux is reduced by the wall collisions. By use of the kinetic flux is the concentration gradient is independent of pressure; whereas the proportionality constant for molecular diffusion in gases (i.e. Diffusivity) is inversely proportional to pressure. Knudsen diffusion occurs when the size of the pore is of the order of the mean free path of the diffusing molecule. 2.14 TRANSIENT DIFFUSION Transient processes, in which the concentration at a given point varies with time, are referred to as unsteady state processes or time – dependent processes. This variation in concentration is associated with a variation in the mass flux. These generally fall into two categories : (i) The process which is in an unsteady state only during its initial startup, and (ii) The process, which is in a batch operation throughout its operation. In unsteady state processes there are three variables-concentration, time and position. Therefore the diffusion process must be described by partial rather than ordinary differential equations. Although the differential equations for unsteady state diffusion are easy to establish, most solutions to these equations have been limited to situations involving simple geometries and boundary conditions, and a constant diffusion coefficient. Many solutions are for one-directional mass transfer as defined by Fick’s second law of diffusion : ∂ CA ∂2CA = DAB ∂t ∂z2

… (2.101)

This partial differential equation describes a physical situation in which there is no bulk– motion contribution, and there is no chemical reaction. This situation is encountered when the diffusion takes place in solids, in stationary liquids, or in system having equimolar counter diffusion. Due to the extremely slow rate of diffusion within liquids, the bulk motion contribution of flux equation (i.e., y A Σ Ni) approaches the value of zero for dilute solutions; accordingly this system also satisfies Fick’s second law of diffusion. The solution to Fick’s second law usually has one of the two standard forms. It may appear in the form of a trigonometric series, which converges for large values of time, or it may involve series of error functions or related integrals, which are most suitable for numerical evaluation at small values of time. These solutions are commonly obtained by using the mathematical techniques of separation of variables or Laplace transforms.

Principles of Mass Transfer Operations − I (Vol. − I)

2.34

Fundamentals of Diffusion Mass Transfer

2.15 DIFFERENTIAL EQUATIONS OF MASS TRANSFER 2.15.1 The Differential Equation for Mass Transfer Consider the control volume, ∆x ∆y ∆z, through which a mixture including component A is flowing, as shown in Fig. 2.15. The control volume expression for the conservation of mass is, ∂ ⌠ ⌡ ⌠ ⌡ ρ (v · n) dA + ∂t ⌠ ⌡ ⌠ ⌡ ⌠ ⌡ ρ dV = 0 c.s.

… (2.101)

c.v.

which may be stated in words as Net rate of mass efflux  Net rate of accumulation of    +   = 0 from control volume   mass within control volume

If we consider the conservation of a given species A, this relation should also include a term which accounts for the production or disappearance of A by chemical reaction within the volume. The general relation for a mass balance of species A for our control volume may be stated as

Net rate of mass Net rate of accumulation Rate of chemical production  efflux of A from  +  of A within control  –  of A within the control  = 0  control volume      volume volume … (2.102) y

Dy Dz Dx

x

z

Fig. 2.15 : A differential control volume

The individual terms will be evaluated for constituent A, and a discussion of their meanings will be given below. The net rate of mass efflux from the control volume may be evaluated by considering the mass transferred across control surface. For example, the mass of A transferred across the area ∆y ∆z at x will be ρA υA, x ∆y ∆z |x, or in terms of the flux vector, nA = ρA vA, it would be nA, x ∆y ∆z |x. The net rate of mass efflux of constituent A will be, in the x-direction :

nA, x ∆y ∆z |x + ∆x – nA, x ∆y ∆z |x

in the y-direction :

nA, y ∆x ∆z |y + ∆y – nA, y ∆x ∆z |y

in the z-direction :

nA, z ∆x ∆y |z + ∆z – nA, z ∆x ∆y |z

and

Principles of Mass Transfer Operations − I (Vol. − I)

2.35

Fundamentals of Diffusion Mass Transfer

The rate of accumulation of A in the control volume is, ∂ρA ∆x ∆y ∆z ∂t If A is produced within the control volume by a chemical reaction at a rate rA, where rA has the units (mass of A produced) / (volume) (time), the rate of production of A is, rA ∆x ∆y ∆z This production term is analogous to the energy generation term which appears in the differential equation for energy transfer. Substituting each term in equation (2.102), we obtain nA, x ∆y ∆z |x + ∆x – nA, x ∆y ∆z |x + nA, y ∆x ∆z |y + ∆y – nA, y ∆x ∆z |y + nA, z ∆x ∆y |z + ∆z – nA, z ∆x ∆y |z +

∂ρA ∆x ∆y ∆z – rA ∆x ∆y ∆z = 0 ∂t

… (2.103)

Dividing through by the volume, ∆x ∆y ∆z, and cancelling terms, we have nA‚ x |x + ∆x – nA‚ x |x nA‚ y |y + ∆y – nA‚ y |y nA‚ z |z + ∆z – nA‚ z |z ∂ρA + + + – rA= 0 ∂t ∆x ∆y ∆z … (2.104) Evaluated in the limit as ∆x, ∆y and ∆z approach zero, this yields ∂ρA ∂ ∂ ∂ n + n + n + – rA = 0 ∂x A, x ∂y A‚ y ∂z A‚ z ∂t

… (2.105)

Equation (2.105) is the equation of continuity for component A. Since nA‚ x, nA‚ y, and nA‚ z are the rectangular components of the mass flux vector, nA, equation (2.105) may be written, ∇ · nA +

∂ρA – rA = 0 ∂t

… (2.106)

A similar equation of continuity may be developed for a second constitutent B in the same manner. The differnetial equations are ∂ρB ∂ ∂ ∂ n B, x + n B, y + n B, z + – rB = 0 ∂t ∂x ∂y ∂z

… (2.107)

and ∇ · nB +

∂ρB – rB = 0 ∂t

… (2.108)

where rB is the rate at which B will be produced within the control volume by a chemical reaction. Adding equations (2.106) and (2.107), we obtain, ∇ · (nA + nB) +

∂ (ρA + ρB) – (rA + rB) = 0 ∂t

For a binary mixture of A and B, we have nA + nB = ρA vA + ρB vB = ρv ρA + ρB = ρ and rA = – rB

… (2.109)

Principles of Mass Transfer Operations − I (Vol. − I)

2.36

Fundamentals of Diffusion Mass Transfer

by the law of conservation of mass. Substituting these relations into (2.109), we obtain, ∇ · ρv +

∂ρ ∂t

= 0

… (2.110)

This is the equation of continuity for the mixture. Equation (2.110) is identical to the equation of continuity for a homogeneous fluid. The equation of continuity for the mixture and for a given species can be written in terms of the substantial derivative. The continuity equation for the mixture can be rearranged and written, Dρ + ρ∇ · v = 0 Dt

… (2.111)

Though similar mathematical manipulations, the equation of continuity for species A in terms of the substantial derivative may be derived. This equation is, ρDωA + ∇ · j A – rA = 0 Dt

… (2.112)

We could follow the same development in terms of molar units. If RA represents the rate of molar production of A per unit volume and RB represents the rate of molar production of B per unit volume, the molar-equivalent equations are For component A, ∂cA ∇ · NA + – RA = 0 … (2.113) ∂t For component B, ∂cB ∇ · NB + – RB = 0 … (2.114) ∂t And for the mixture, ∂ (cA + cB) ∇ · (NA + NB) + – (RA + RB) = 0 … (2.115) ∂t For the binary mixture of A and B, we have, NA + NB = cA vA + cB vB = cV and cA + cB = c However, only when the stoichiometry of the reaction is A

B

which stipulates that one molecule of B is produced for each mole of A disappearing, can we stipulate that RA = – RB. In general, the equation of continuity for the mixture in molar units is ∇ · cV +

∂c – (RA + RB) = 0 ∂t

… (2.116)

2.15.2 Special Forms of the Differential Mass Transfer Equation Special forms of the equation of continuity applicable to commonly encountered situations follow. In order to use the equations for evaluating the concentration profiles, we

Principles of Mass Transfer Operations − I (Vol. − I)

2.37

Fundamentals of Diffusion Mass Transfer

replace the fluxes, nA and NA, by the appropriate expressions developed. These expressions are – NA = – cDAB ∇yA + yA (NA + NB)

… (2.117)

or its equivalent, NA = – cDAB ∇yA + cA V and nA = – ρDAB ∇ ωA + ωA (nA + nB)

… (2.118)

or its equivalent, nA = – ρDAB ∇ ωA + ρA v Substituting equation (2.118) into equation (2.113), we obtain, – ∇ · ρDAB ∇ ωA + ∇ · ρA v +

∂ρA – rA = 0 ∂t

… (2.119)

and substituting equation (2.117) into equation (2.105), we obtain, ∂cA – RA = 0 … (2.120) ∂t Either equation (2.119) or equation (2.120) may be used to describe concentration profiles within a diffusing system. Both equations are completely general; however, they are relatively unwidely. These equations can be simplified by making restrictive assumptions. Important forms of the equation of continuity, with their qualifying assumptions, include : (i) If the density, ρ, and the diffusion coefficient, DAB, can be assumed constant, equation (2.119) becomes, ∂ρA – DAB ∇2 ρA + ρA ∇ · v + v · ∇ ρA + – rA = 0 ∂t Dividing each term by the molecular weight of A and rearranging, we obtain ∂cA v · ∇cA + = DAB ∇2 cA + RA … (2.121) ∂t (ii) If there is no production term, RA = 0, and if the density and diffusion coefficient are assumed constant, equation (2.121) reduces to ∂cA + v · ∇cA = DAB ∇2 cA … (2.122) ∂t We recognize that (∂cA/∂t) + v · ∇cA is the substantial derivative of cA; rewriting the lefthand side of equation (2.123), we obtain, DcA = DAB ∇2 cA … (2.123) Dt which is analogous to equation from heat transfer, k DT = ∇2 T … (2.124) Dt ρcp – ∇ · cDAB ∇yA + ∇ · cA V +

DT = α ∇2 T Dt where, α is the thermal diffusitivity. The similarity between these two equations is the basis for the analogies drawn between heat and mass transfer. or

Principles of Mass Transfer Operations − I (Vol. − I)

2.38

Fundamentals of Diffusion Mass Transfer

(iii) In a situation in which there is no fluid motion, v = 0, no production term, RA = 0, and no variation in the diffusivity or density, equation (2.122) reduces to, ∂cA = DAB ∇2 cA ∂t

… (2.125)

Equation (2.125) is commonly referred to as Fick's second "law" of diffusion. The assumption of no fluid motion restricts its applicability to diffusion in solids, or stationary liquids, and for binary systems of gases or liquids, where NA is equal in magnitude, but acting in the opposite direction to NB; that is, the case of equimolar counter diffusion. Equation (2.125) is analogous to Fourier's second "law" of heat conduction. ∂T ∂t

= α ∇2 T

… (2.126)

(iv) Equations (2.121), (2.122), and (2.123) may be simplified further when the process to be defined is a steady-state process; that is, ∂cA/∂t = 0. For constant density and a constant diffusion coefficient the equation becomes, v · ∇ cA = DAB ∇2 cA + RA

… (2.127)

For constant density, constant diffusivity and no chemical production, RA = 0, we obtain, v · ∇ cA = DAB ∇2 cA

… (2.128)

If additionally, v = 0, the equation reduces to ∇2 cA = 0

… (2.129)

Equation (2.129) is the Laplace equation in terms of molar concentration. Each of the equations (2.119) through (2.125) has been written in vector form, thus each applies to any orthogonal co-ordinate system. By writing the Laplacian operator, ∇2, in the appropriate form, the transformation of the equation to the desired co-ordinate system is accomplished. Fick's second "law" of diffusion written is rectangular co-ordinates is 2 2 2 ∂cA ∂ cA + ∂ cA + ∂ cA = DAB … (2.130) 2 2 ∂t ∂y ∂z2   ∂x in cylindrical co-ordinates is 2 2 2 ∂cA ∂ cA + 1 ∂cA + 1 ∂ cA + ∂ cA … (2.131) = DAB ∂t ∂z2   ∂r2 r ∂r r2 ∂θ2 and in spherical co-ordinates is ∂cA ∂cA ∂2 c A  ∂  1  1 ∂ r2 ∂cA + 1 = DAB 2 sin θ + 2 … (2.132) 2 ∂t ∂θ  r sin θ ∂φ2  r ∂r  ∂r  r sin θ ∂θ  The general differential equation for mass transfer of component A, or the equation of continuity of A, written in rectangular co-ordinates is ∂cA ∂NA‚ x ∂NA‚ y ∂NA‚ z  + … (2.133) ∂t  ∂x + ∂y + ∂z  = RA in cylindrical co-ordinates is ∂cA 1 ∂ 1 NA‚ θ ∂NA‚ z  + (rNA‚ r) + + ∂t r ∂θ ∂z   r ∂r and in spherical co-ordinates is

= RA

… (2.134)

Principles of Mass Transfer Operations − I (Vol. − I)

2.39

Fundamentals of Diffusion Mass Transfer

∂cA  1 ∂ 1 ∂ 1 ∂NA‚ φ  + 2 (r2 NA‚ r) + (NA‚ θ sin θ) + = RA ∂t r sin θ ∂θ r sin θ ∂φ  r ∂r

… (2.135)

Example 1 : In a cylindrical nuclear fuel rod which contains fissionable material, the rate of production of neutrons is proportional to the neutron concentration. Use one of the general differential equations for mass transfer to write the differential equation which describes the mass transfer process. List any obvious boundary conditions. For component A, equation (2.113) stipulates ∇ · NA +

∂cA – RA = 0 ∂t

Since the rate of production is proportional to the neutron concentration, RA = kcA. For diffusion in solids, where the bulk motion contribution is zero, NA = – DAB ∇ cA Upon substituting these relations into equation (2.113), we obtain

∂ cA + 1 ∂cA + 1 ∂ cA + ∂ cA + kc A ∂z2   ∂r2 r ∂r r2 ∂θ2 2

∂cA ∂t

= DAB

2

2

If the cylinder is relatively long compared to the radius, ∂2 cA/∂z2 = 0 and if the concentration does not vary with the angle θ, ∂2 cA/∂θ2 = 0; the equation reduces to ∂cA ∂t

= DAB

2 ∂ cA + 1 ∂cA + kc A  ∂r2 r ∂r 

The only obvious boundary condition is, ∂cA ∂r r = 0

= 0

which requires the concentration of the diffusing species to be finite at the center of the rod. Example 2 : In a hot combustion chamber, oxygen diffuses through an air film to a carbon surface where it reacts according to the following equation : 2CO + CO2

3C + 2O2 O2

CO

CO2 z=d

z=0

Fig. 2.16

(a) With the assumption that the carbon surface is flat, reduce the general differential equation for mass transfer to write the specific differential equation that describes this steady-state process.

Principles of Mass Transfer Operations − I (Vol. − I)

2.40

Fundamentals of Diffusion Mass Transfer

For oxygen, equation (2.113) stipulates, ∇ · NO2 +

∂cO2 ∂t

– RO2

= 0

Since this is a steady-state process ∂cO2 /∂t = 0. The reaction occurs at one of the boundaries and not uniformly along the diffusion path; accordingly, RO2 = 0. With diffusion in only the z-direction, ∇ · NO2 reduces to dNO2‚ z /dz and the differential equation becomes, dNO2‚ z dz

= 0

(b) Write Fick's law in terms of only oxygen. Fick's law for molecular diffusion stipulates NO2‚ z

dyO2 = – cDO2 – mixture + yO2 (NO2‚ z + NCO‚ z + NCO2‚ z + NN2‚ z) dz

According to the reaction, two moles of oxyen enter the film and two moles of carbon monoxide leave; accordingly, NO2‚ z = – NCO. The direction also predicts that as the two moles of oxygen enter, one mole of carbon dioxide leave; this stipulates

1 N = – NCO2‚ z . The 2 O2‚ z

net flux of nitrogen is zero. With these stipulations, the bulk contribution term becomes, yO2 (NO2‚ z + NCO, z + NCO2‚ z + NN2‚ z) = yO2 (NO2‚ z – NO2‚ z – = –

yO2 2

1 N + 0) 2 O2‚ z

NO2‚ z

and Fick's law reduces to NO2‚ z

dyO2 yO2 = – cDO2 – mixture – NO2‚ z dz 2

or NO2 , z = –

cDO2 – mixture dyO2 1+

yO2

dz

2

This equation can either be substituted into the differential equation that is defined in part (a) or integrated directly in order to obtain the flux of O2 in the z-direction.

Principles of Mass Transfer Operations − I (Vol. − I)

2.41

Fundamentals of Diffusion Mass Transfer

SOLVED PROBLEMS (1) In an oxygen-nitrogen mixture at 10 atmosphere and 25oC, the concentrations of oxygen at two places of 0.2 cm a part are 10 and 20 volume percent respectively. Calculate the rate of diffusion of oxygen expressed as gm/cm2 – hr for the case of unicomponent diffusion (nitrogen to non-diffusing). Data :

Value of diffusivity between oxygen – nitrogen = 0.181 cm2/sec. atm. cm3 R = 82.06 gm.mole K

Sol. :Given Data : DAB = 0.181 cm2/sec Pt = 10 atm T = 25 + 273 = 298 K Z = 0.2 cm NA = ? gm/cm2.hr. This is the case of steady-state diffusion of gas A (O2) through non-diffusing gas B (N2). For this case, we have equation : NA =

DAB Pt

p– – p–   A1 A2

–

RTZ PB, M

… (A)

Since Volume % = mole % We have, Partial pressure = Mole fraction × total pressure –

pA

1

–

pA

2

Also,

–

–

pA + pB 1

1

–

pB

∴

1

= 0.2 × 10 = 2 atm. = 0.1 × 10 = 1 atm. –

–

= pA + pB = pt = 10 atm. 2

2

–

= 8 atm, p B = 9 atm. 2

–

Thus,

–

pB – PB

–

2

PB, M =

1

p  p–    –

ln

B2

=

9–8 9 = 8.49 atm. ln 8  

B1

Thus, equation (A) becomes, NA =

0.181 × 10 × (2 – 1) 82.06 × 298 × 0.2 × 8.49

NA = 4.36 × 10–5 gm.mole/sec. cm2 = 4.36 × 10–5 × 32 × 3600 = 5.022 gm/cm2. hr.

(Ans.)

Principles of Mass Transfer Operations − I (Vol. − I)

2.42

Fundamentals of Diffusion Mass Transfer

(2) Ammonia gas is diffusing at constant rate through a layer of stagnant air 1 mm thick. Conditions are fixed so that gas contains 50% by volume of ammonia at one boundary of the stagnant layer. The ammonia diffusing to the other boundary is quickly absorbed and the concentration is negligible at that plane. The temperature is 298 K and 1 atm. pressure and under this condition the diffusivity of ammonia in air is 0.18 × 10–4 m2/sec. Calculate the rate of diffusion of ammonia through the layer. Take R = 8.314 kJ/k mole oK. Sol. : Given data : Z = 1 mm = 1 × 10–3 m DAB = 0.18 × 10–4 m2/sec Pt = 101.3 kN/m2 R = 8.314 kJ/k mol K T = 25 + 273 = 298 oK –

pA

1

–

pA

2

= 0.5 × 101.3 = 50.65 kN/m2 = 0

This is case of equimolar counter diffusion so we have equation, DAB · Pt – – NA = RTZ pA – pA   1 2 –4 0.18 × 10 × 101.3 (50.65 – 0) = 8.314 × 298 × 1 × 10–3 = 3.76 × 10–3 k mole/m2. sec. (Ans.) (3) In an O2 – N2 gas mixture at 1 atm. and 25o C, the concentration of O2 at two planes 3 mm apart are 10 and 20 volume percent respectively. Calculate the rate of diffusion kg mole O2 expressed as for the case where there is equimolar counter diffusion takes m2. sec. place. Data : DAB = 0.206 × 10–4 m2/sec. R = 0.08206 atm. m3/kg.mole K Sol. : Given Data : DAB = 0.206 × 10–4 m2/sec. R Pt T Z –

pA

1

–

pA

2

= = = =

0.08206 atm/m3/kg.mole K 1 atm. 25 + 273 = 298 K 3 mm = 3 × 10–3 m

= 0.10 × 1 = 0.1 atm. = 0.20 × 1 = 0.2 atm.

For equimolar counter diffusion, we have, DAB Pt – – NA = RTZ pA – pA   1 2 0.206 × 10–4 × 1 × (0.1 – 0.2) = 0.08206 × 298 × 3 × 10–3 kg mole O2 = 2.85 × 10–5 m2 . sec.

(Ans.)

Principles of Mass Transfer Operations − I (Vol. − I)

2.43

Fundamentals of Diffusion Mass Transfer

(4) Calculate the rate of diffusion of water vapour from a thin layer of water at the bottom of a well 6 m in height to dry air flowing over the top of the well. Assume the entire system is at 298 K and atmosphere pressure. If the well diameter is 3 m, find out the total weight of water diffused per second from the surface of the water in the well. The diffusion coefficient of water vapour in dry air at 298 K and atmospheric pressure is 0.256 × 10–4 m2/sec. The partial pressure of water vapour at 298 K is 0.0323 × 10–4 kg/m2. Sol. : Data given :

–

pA

1

= Partial pressure of water vapour = 0.0323 × 10–4 kg/m2 = 0.0323 kg/cm2 = 0.031 atm.

–

pA

2

= 0, Z = 6m

DAB = 0.256 × 10–4 m2/sec. T = 25 + 273 = 298 K –

pB

Thus,

1

= 1 – 0.031 = 0.969 atm.

–

pB

2

= 1 atm. –

So,

–

P B, M

–

pB – pB 2

=

1

  p–    –

=

pB

ln

2

1 – 0.969 1 ln 0.969  

= 0.984 atm.

B1

We have equation, NA =

DAB Pt –

RTZ PB, M

p– – p–   A1 A2

0.256 × 10–4 × 1 (0.031) 0.08206 × 298 × 6 = 5.496 × 10–9 k mole/m2.sec. Rate of diffusion of water to dry air, = NA × Area × Mol. wt. of water π = 5.496 × 10–9 × 4 D2 × 18 π = 5.496 × 10–9 × 4 × (3)2 × 18 =

= 7 × 10–7 kg/sec.

(Ans.)

(5) A volatile organic compound benzene costing Rs. 45 per kg, is stored in a tank 10 m diameter and open at top. A stagnant air film 10 mm thick is covering the surface of the compound beyond which the compound is absent. If the atmospheric temperature is 25°°C. Vapour pressure of the compound is 150 mm Hg and its molar diffusivity is 0.02 m2/hour, calculate the loss of benzene is Rs./day.

Principles of Mass Transfer Operations − I (Vol. − I)

2.44

Fundamentals of Diffusion Mass Transfer

T = 25 + 273 = 298 K Z = 10 mm = 0.01 m – 150 pA = 150 mm Hg = 760 = 0.197 atm. 1

Sol. : Data Given :

–

pA

2

–

–

pB = 1 – p A 1

1

–

–

pB = 1 – PA 2

2

= 0 = 1 – 0.197 = 0.803 atm. = 1 – 0 = 1 atm. –

PB, M =

So,

–

pB – pB

–

2

1

p  ln  –  p  –

B2

=

1 – 0.803 = 0.898 atm. 1 ln 0.803  

B1

We have,

NA = =

DAB Pt –

RTZ PB, M

P– – P–  A2  A1

0.02 × 1 (0.197 – 0) 0.082 × 298 × 0.01 × 0.898

= 0.018 kg mole/hr. m2 π Area of the tank = 4 (10)2 = 78.5 m2 ∴ Loss of benzene from the tank in one day (24 hours) is given by : Loss in Rs./day = NA × A × hours × Molecular weight × Cost of benzene = 0.018 × 78.5 × 24 × 78 × 45 = 11,19,115 (Ans.) (6) A narrow tube is partially filled with liquid and maintained at a constant temperature. A gentle stream of a gas is passed across the open end of the tube. As the liquid evapourates, the level drops slowly. At a given time t, this level in the tube is Z from the top. Derive an expression to calculate the value of diffusivity of liquid vapour in the gas. Sol. : Since the liquid level drops slowly, we assume that Pseudo steady state conditions prevail in this case. For steady state diffusion; DAB Pt Gas Stream P– – P–  NA = … (1) A A –  1 2 RTZ PB‚ M Let 'S' be the cross sectional area of the tube and dZ be the fall in the liquid level in time dt. Therefore, volume of liquid evapourated is, V = dZ × S

dz

At t = t, z = z At t = dt, z = dz

in time dt.

If ρA is the density of the liquid, then the mass of liquid evapourated is given by, M = dZ × S × ρA

Fig. 2.17 : Diffusivity of liquid vapour in the gas from narrow tube

Principles of Mass Transfer Operations − I (Vol. − I)

2.45

Fundamentals of Diffusion Mass Transfer

∴ Moles of liquid evapourated per unit area per unit time is given by : ρA · dZ · S ρA · dZ NA = = M · dt MA × dt × S A where, MA = Molecular weight of liquid A. So equation (1) becomes, ρA · dZ DAB Pt p– – p–  – MA · dt =  A1 A2 RTZ PB, M –

–

pB – pB

–

2

PB, M =

Since,

… (2)

1

p  p–    –

n

B2 B1

p  P ln  –  p  –

DAB ρA · dZ MA dt

We have,

Also since,

–

–

pA – pA 1

2

=

B2

t

B1

× pA – pA



RTZ pB – pB   2 1 –

–

–

–

–

1

2

 

… (3)

–

= pB – pB 2

1

p  · P ln  –  p  –

DAB we have,

ρA dZ MA · dt

B2

t

B1

=

… (4)

RTZ

Integrating equations (4) between the limits of Z = Z0, when t = 0 Z = Z, when t = t1, we get

p  P ln  –  p  –

DAB We get,

ρA Z1 – Z02 MA  2 

=

B2

t

B1

RT 2

or

DAB =

· ln (t1) 2

RT · ρA (Z1 – Z0)

  p–    –

… (5)

pB

MA · t1 · 2 Pt · ln

2

B1

Equation (5) is called as Stefan's equation. (7) A tube of small diameter was filled with acetone having density 790 kg/m3 upto 1.1 × 10–2 m from top and maintained at a temperature of 20oC in a gentle current of air. After 5 hours, the level of the liquid fall to 2.05 × 10–2 m from the top. Calculate the

Principles of Mass Transfer Operations − I (Vol. − I)

2.46

Fundamentals of Diffusion Mass Transfer

value of diffusivity of acetone in air if the barometric pressure was 750 mm Hg. Vapour pressure of acetone at 20oC is 180 mm Hg. atm. m3 R = 0.08206 k mol K . Sol. : Data Given : Partial Pressure = Vapour pressure of acetone –

= pA = 180 mm Hg = 0.237 atm. 1

Total pressure = Pt

750 = 760 = 0.987 atm.

–

pA = 0, Z1 = 2.05 × 10–2 m, Z2 = 1.1 × 10–2 m, t1 = 5 hours = 18000 sec. 2

Since,

–

–

pA – pA 1

2

–

pB

∴

1

–

pB

2

–

–

= pB – pB = Pt = 0.987 atm. 1

2

= 0.987 – 0.237 = 0.75 atm. = 0.987 atm. –

So,

–

pB – pB

–

2

P B, M =

1

p  p–    –

ln

=

B2

0.987 – 0.75 0.987 ln  0.75   

B1

= 0.863 atm. We have Stefan's equation as, 2

DAB =

2

ρA RT (Z1 – Z2)

p  ln  –  p  –

2 MA Pt t1

B2 B1

DAB =

790 × 0.08206 × 293 [(2.05 × 10–2)2 – (1.1 × 10–2)2] 0.987 2 × 58 × 0.987 × 18000 · ln  0.75   

DAB = 0.10 × 10–4 m2/sec.

(Ans.)

(8) Calculate the amount of diffusion of acetic acid (A) in 2 hours across a film on nondiffusing water (B) solution 1 mm thick at 17oC when the concentration on opposite side of the film are 9 and 3 weight % acid respectively. The diffusivity of acetic acid in solution is 0.95 × 10–9 m2/sec. Data : At 17o C, Density of 9% solution = 1012 kg/m3, Density of 3% solution = 1003 kg/m3. Molecular weight of acetic acid = 60, Molecular weight of water = 18.

Principles of Mass Transfer Operations − I (Vol. − I)

2.47

Fundamentals of Diffusion Mass Transfer

Sol. : Mole fractions of 9% and 3% acetic acid : 9 60 xA = 9 91 = 0.0288 1 60 + 8 3 60 xA = 3 97 = 0.0092 2 60 + 18 xB = 1 – xA = 0.9712 1

xB

2

1

= 1 – xA = 0.9908 2

xB – xB So,

xB, M =

2

1

= 0.981

xB

 ln x   B1 2

Now, Average molecular weight : For 9% Solution : M1 = xA × MCH 1

3COOH

+ xB · M H 1

2

O

= 0.0288 × 60 + 0.9712 × 18 = 19.23 For 3% solution : M2 = xA × MCH 2

3COOH

+ xB × M H 2

2

O

= 0.0092 × 60 + 0.9908 × 18 = 18.41 Thus, for 9% solution :

 ρ  = 1012 = 52.60 19.23 M 1 for 3% solution :

 ρ  = 1003.2 = 54.55 18.41 M 2 52.60 + 54.55 ρ 2 M avg =

Thus,

So,

NA

= 53.58 kg/m3 DAB  ρ  (x – x ) = Z·x A1 A2 B‚ M Mavg =

0.95 × 10–9 × 53.58 × (0.0288 – 0.0092) 0.07707 × 1 × 10–3

= 1.017 × 10–6 kg.mol/m2.sec. ∴ Amount of diffusion of acetic acid in 2 hours, = 1.017 × 10–6 × 3600 × 2 = 7.32 × 10–3 kgmole/m2

(Ans.)

Principles of Mass Transfer Operations − I (Vol. − I)

2.48

Fundamentals of Diffusion Mass Transfer

(9) Calculate the amount of oxygen (A) diffused in one hour under steady state conditions through a non-diffusing gas mixture of methane (B) and hydrogen (C) in the volume ratio of 2 : 1. The diffusivities one estimated to be : DO2 – H2 = 6.99 × 10–5 m2/s DO2 – CH4 = 1.86 × 10–5 m2/sec The total pressure is 1 × 105 N/m2 and temperature is 0oC. The partial pressure of oxygen at two planes 2 mm apart one respectively 13000 and 6500 N/m2. Data : R = 8314 Nm/(k mole) (K). Sol. : Let, us denote e : A = Oxygen B = Methane C = Hydrogen Let, YB = mole fraction of component in the air mixture 2 YB = 2 + 1 = 0.667 Volume ratio of methane to hydrogen = 2 : 1 1 ∴ YC = mole fraction of component C in air free mixture = 2 + 1 = 0.333 Effective diffusivity may be calculated as : 1 DE = y  B  +  yC  D  D   AB  AC =

=

1

   xH2  4 D – D  + D   O2 CH4  O2 – H2 xCH

1 0.667   +  0.333  1.86 × 10–5 6.99 × 10–5

= 2.465 × 10–5 m2/sec. T = 0oC = 273 K

Next,

Z = 2 mm = 2 × 10–3 m –

pA

1

–

pB

2

= 13000 N/m2 = 0.13 × 105 N/m2 = 6500 N/m2 = 0.065 × 105 N/m2

Pt = 1 × 105 N/m2 –

–

pA + pB 1

1

–

∴

pB

∴

pB

1

–

2

–

–

= p A + pB = Pt 2

2

= 87000 N/m2 = 93500 N/m2

Principles of Mass Transfer Operations − I (Vol. − I)

2.49

–

–

pB – pB

–

∴

Fundamentals of Diffusion Mass Transfer

PB

2

=

M

1

p  p–    –

ln

=

B2

93500 – 87000 = 90200 N/m2 93500  ln 8700  

B1

Thus,

NA

= 0.902 × 105 N/m2 DAB Pt p– – p–  = –  A1 A2 RTZ PB, M =

2.46 × 10–5 × 1 × 105 8314 × 273 × 2 × 10–3 × 0.902 × 105 (0.13 × 105 – 0.065 × 105)

= 3.095 × 10–5 kmol/s.m2 = 1.406 × 10–1 kmol/hr. m2 (Ans.) (10) Determine the diffusivity of CO2 (1), O2 (2) and N2 (3) in a gas mixture having the composition : CO2 : 28.5 %, O2 : 15%, N2 : 56.5%, 5

The gas mixture is at 273 K and 1.2 × 10 Pa. The binary diffusivity values are given at 273 K as : 2

D12 P = 1.874 m Pa/sec 2

D13 P = 1.945 m Pa/sec 2

D23 P = 1.834 m Pa/sec Sol. : Diffusivity of CO2 in mixture 1 D1m = y' y'2 2 D12 + D13

where,

y'2

y2 = y +y 2 3

0.15 = 0.15 + 0.565 = 0.21

y'2

y3 = y +y 2 3

0.565 = 0.15 + 0.565 = 0.79

1 D1m P = 0.21 0.79 1.874 + 1.945 Therefore, Since,

= 1.93 m2.Pa/sec. P = 1.2 × 105 Pa, 1.93 D1m = = 1.61 × 10–5 m2/sec. 1.2 × 105

Diffusivity of O2 in the mixture, 1 D2m = y' y'2 1 D21 + D23

Principles of Mass Transfer Operations − I (Vol. − I)

2.50

Fundamentals of Diffusion Mass Transfer

y1 0.285 = y + y = 0.285 + 0.565 = 0.335 1 3

y'1

where, (mole fraction on – 3 free basis). y3 0.565 y'2 = y + y = 0.285 + 0.565 = 0.665 1 3 2

and D21 P = D 12 P = 1.874 m .Pa/sec Therefore

1 D2m P = 0.335 0.6655 1.874 + 1.834 D2m =

= 1.847 m2.Pa/sec.

1.847 = 1.539 × 10–5 m2/sec. 1.2 × 105

By similar calculations diffusivity of N2 in the mixture can be calculated, and is found to be, D3m = 1.588 × 10–5 m2/sec.

(Ans.)

(11) Calculate diffusivity of dry H2 in air at 303 K and 1 atm. The molecular volume of air and hydrogen are : VH

2

= 14.3 cm3/gm. mole

VAir = 29.9 cm3/gm. mole Sol. : Gilliland's correlation is,

DH

2 – air

0.0043 T

=

 

3 2

1 3

Pt VH + Vair

=

  

1 3 2

2

0.0043 (300)

1 1 MH + Mair

×

2

3 2

1 1 2  3 3  1 (14.3) + (29.9) 

×

= 0.544 cm2/sec

1 1 2 + 29 (Ans.)

(12) Use Wilke-Chang correlation to estimate the diffusivity of ethyl benzene into water at 293 oK. The viscosity of water at this temperature may be 1 cP and density of ethyl benzene at normal boiling temperature (409.3 K) is 0.761 gm/cc. Assume φ = 2.6. Compare this value with experimental at value of DAB = 0.81 × 10–5 cm2/sec. Sol. : Wilke–Chang correlation is, 0

DAB Data Given :

=

(117.3 × 10–18) (φ MB) 0.6

µ VA

φ = 2.6 B = H2O;

MB

= 18

1 2

·T

Principles of Mass Transfer Operations − I (Vol. − I)

2.51

Fundamentals of Diffusion Mass Transfer

A = Ethyl benzene (C6H3C2H5); MA = 105 VA = 139.5 cm3/mol T = 293 oK 0

∴

DAB

=

(117.3 × 10–18) (2.6 × 18)0.5 × 293 1 × (139.5)0.6

= 1.215 × 10–14 cm2/sec.

(Ans.)

The deviation of calculated value from its experimental value of diffusivity of ethyl benzene in water is given by : =

(Calculated Value) – (Experimental Value) × 100 (Calculated Value)

=

1.215 × 10–5 – 0.81 × 10–5 × 100 = + 32.92% 1.215 × 10–5

(Ans.)

(13) Mannitol (CH2OH (CHOH)4 · CH2OH) diffusivity in dilute solution in water at 293 ok m2 has been reported to be 5.6 × 10–10 sec.. Compare the observed value with the calculated value. Base your calculations on the following data : φ = 2.26 Dynamic viscosity of dilute solution = 1005 × 10–6 pa. sec. VA = 0.185 m3/k mol Sol. : Wilke–Chang correlation is, o

DAB o

DAB

1 2

=

(117.3 × 10–18) (φ MB) T 0.6

µ VA

1 2

(117.3 × 10–18) (2.26 × 18) × 293 = 1005 × 10–6 × (0.185)0.6 = 6.003 × 10–10 m2/sec.

(Ans.)

The deviation of calculated value from its experimental value of diffusivity of mannitol in its dilute solution in water is given by : =

(Calculated value) – (Experimental value) × 100 (Calculated value)

=

6.003 × 10–10 – 5.6 × 10–10 × 100 6.003 × 10–10

= + 6.66%

(Ans.)

(14) The value of DAB for a dilute solution of methanol in water at 288 K is given by cm2 1.28 × 10−5 s . Estimate DAB for same solution for 373 K, using Wilke-Chang correlation.

Principles of Mass Transfer Operations − I (Vol. − I) Data :

2.52

Fundamentals of Diffusion Mass Transfer

µ1 at 288 K = 1.14 Cp µ2 at 373 K = 0.284 Cp

Sol. : Wilke-Change equation DAB = . . .

0.6

µ ⋅ VA

T µ T1 µ1  T1 µ2 = T = ×T   2 µ1 2 µ2 

DAB ∝ (DAB)T1 = 288 K (DAB)T2 = 373 K

∴

117.3 × 10−18 (φMB)1/2 ⋅ T

µ1 T2 (DAB)T2 = 373 K = (DAB)T1  ⋅ T  µ2 1 1.14 373 = 1.28 × 10−5 0.284 × 288   (DAB)T2 = 373 = 6.6 × 10−5 cm2/sec.

(Ans.)

(15) Estimate DAB for a dilute solution TNT in benzene at 15°°C. Data : µ = 0.705 Cp (Assume pure benzene in solution) VA = 140 cm3/gm.mole (for TNT) φB = 1.0 for benzene MB = 78 for benzene Sol. : Wilke-Chang equation is, 117.3 × 10−8 (φB ⋅ MB)1/2 ⋅ T DAB = µ (VA)0.6 Substituting all values from given data in above equation. 7.4 × 10−8 (1 × 78)1/2 × 288 DAB = 0.705 × (140)0.6 DAB = 1.38 × 10−5 cm2/sec. (Ans.) (16) The air pressure in a tyre reduces from 2 bars to 1.99 bars in five days. The volume of air in the tube is 0.0025 m3, the surface area is 0.5 m2 and wall thickness 0.01 m. The kg mole air . Estimate the diffusivity of air in solubility of air in rubber is 0.00286 m3 rubber rubber. Assume temperature 25o C. Sol. : This is a case of diffusion in solid. We assume that air behave ideally. Equation for the case is, DAB NA = Z (CA – CA ) 1 2 DAB S (CA – CA ) = Z 1 2 DAB · S and W = (CA – CA ) Z 1 2 where W = Loss, kg mole/hour

Principles of Mass Transfer Operations − I (Vol. − I)

2.53

Fundamentals of Diffusion Mass Transfer

CA = Solubility of air in the rubber = 0.00286

Data Given :

1

kg mole air m3 rubber

CA = 0 (since other end is at far surface) 2

Z = 0.01 m S = 0.5 m2 Use ideal gas law to find initial mole of air in the tyre i.e. P1V1 = n1RT1 P1V1 So, n1 = RT 1 2 × 0.025 0.08206 × 298 = 0.002022 kg mole P2 V2 1.99 × 0.025 RT1 = 0.083 × 298 = 0.002011 kg mole = n1 – n2 = (0.002022 – 0.002011) = 0.000011 n1 – n2 = 24 × 5 0.000011 = 120 = 9.167 × 10–8 k mole/hour DAB · S = (CA – CA ) Z 1 2 0.5 = DAB × 0.01 (0.00286 – 0) =

Mole of air after diffusion = n2 = Mole diffused in 5 days,

Loss per hour W W Thus,

W 9.167 × 10–8

Solving for DAB, we get, DAB = 0.64 × 10–6 m2/hour (17)

(Ans.)

o

If an O2 -N2 gas mixture at 1 atm. and 25 C, the concentration of O2 at two planes, 2 mm apart are 10 and 20% (v/v) respectively. Calculate the flux of diffusion of O2 for the case where 1. the N2 is non-diffusing. 2.

there is equimolar counter-diffusion of the gases.

Data : DO

2–N2

o

at 0 C = 1.86 × 10–5 m2/sec.

Solution : We know the variation of diffusivity with temperature as DAB ∝ T1.5 o

We need to calculate DAB at 25 C. ∴

DO

o

2–N2

o T2981.5 at 0 C × 2–N2 T273 1.5 298 = 1.81 × 10–5 273  

at 25 C = DO

Principles of Mass Transfer Operations − I (Vol. − I)

2.54

Fundamentals of Diffusion Mass Transfer

= 2.064 × 10–5 m2/sec. ∴ 1.

o

DAB at 25 C = 2.064 × 10–5 m2/sec. pA

= 0.1 × 105 N/m2

pB

= 1 × 105 – 0.1 × 105

pA

= 0.9 × 105 N/m2 = 0.2 × 105 N/m2

pB

= 1 × 105 – 0.2 × 105

1

1

2

2

= 0.8 × 105 N/m2 5

PT = 1 × 10 N/m2 pB – pB 1 2 (PB)lm = p



   B2

ln p

=

B1

0.9 × 105 – 0.8 × 105 0.9 ln 0.8  

= 0.849 × 105 N/m2 The flux of diffusion of O2 through non-diffusing N2 gas is NA = =

DAB · PT R.T ∆z (PB)lm

(p

A2

– pA

2

)

2.064 × 10–5 × 105 × (– 0.1 + 0.2) × 105 8314 × 298 × 2 × 10–3 × 0.849 × 105

= 4.91 × 10–5 kmole/m2 sec ∴ 2.

NA = 4.91 × 10–5 kmole/m2.sec For equimolar counter diffusion : + DAB NA = pA – pA RT ∆z 1 2

(

)

–5

=

2.064 × 10 (0.1 – 0.2) × 105 8314 × 298 × 2 × 10–3

= – 4.2 × 10–5 kmole/m2.sec

(Ans.)

Negative sign shows that the diffusion is in opposite direction. (18) The O2 (A) is diffusing through CO (B) under standard conditions with CO o

non- diffusing. The total pressure is 1 × 105 N/m2 and temperature is 0 C. The partial pressure of O2 at two planes, 2.0 mm apart is 1300 and 6500 N/m2. The diffusivity of oxygen in CO is DAB = 1.87 × 10–5 m2/sec. Calculate the rate of diffusion of O2 in kmole/m2.sec.

Principles of Mass Transfer Operations − I (Vol. − I)

2.55

Fundamentals of Diffusion Mass Transfer

Solution : Given Data : Let

A → O2 B → CO DAB = 1.87 × 10–5 m2/sec PT = 1 × 105 N/m2 T = 273 K p p

A1 A2

= 1300 N/m2 = 6500 N/m2

∆z = 2 × 10–3 m ∴

p

∴

p

A1 A2

= 13000 N/m2

,

p

= 6500 N/m2

,

p

B1 B2

= 1 × 105 – 13000 = 87000 = 1 × 105 – 6500 = 93500

The flux of diffusion of O2 through a non-diffusing CO is given by,

(

DAB · PT · p NA =

A1

–p

R.T. (∆z) (PB)

R.T. (∆z)

( (p

A1

B1

=

=

∴

NA

–p –p

) )

A2

B2

pB1 ln p   B2

(

DAB · PT · p =

)

lm

DAB · PT · p =

A2

A1

–p

A2

)

(PT – pA1) – (PT – pA2)   RT (∆z) p  B1 ln p   B2 pB2 DAB · PT · ln p   B1 RT ∆z 93500 1.87 × 10–5 × 1 × 105 ln 87000





8314 × 2 × 10–3 × 273

= 2.9 × 10–5 kmole/m2.s = 2.9 × 10–5 kmole/m2.s

(Ans.)

Principles of Mass Transfer Operations − I (Vol. − I)

2.56

Fundamentals of Diffusion Mass Transfer o

(19) Calculate the mass flux of benzene through a layer of air 10 mm thickness at 25 C and 200 kN/m2 (total pressure), partial pressure of benzene is 6 × 103 N/m2, at the left side of the layer and 1 kN/m2 at the right side. The mass diffusivity at this temperature and pressure is 4.4 × 10–6 m2/sec. Solution : Given Data : ∆z = 10 mm = 10 × 10–3 m T = 298 K PT = 200 × 103 N/m2 p p

A1 A2

= 6 × 103 N/m2 = 1 × 103 N/m2

DAB = 4.4 × 10–6 m2/sec For diffusion of A through non-diffusing B, we have mass flux given by DAB · PT NA = p –p … (1) A1 A2 R.T. (PB) · ∆z lm

p

(PB)lm p p

B1 B2

=

B2

–p

p

(

)

B1

… (2)

 ln p   B1 B2

= PT – p = PT – p

A1

= 200 × 103 – 6 × 103 = 194 × 103 N/m2

A2

= 200 × 103 – 1 × 103 = 199 × 103 N/m2

From equation (2), 3

(PB)

lm

=

199 × 10 – 194 × 103 = 196.5 × 103 N/m2 199  ln 194  

From equation (1), NA =

4.4 × 10–6 × 200 × 103 (6 × 103 – 1 × 103) 8.314 × 103 × 298 × 196.5 × 103 × 10–2

NA = 9.037 × 10–9 kmole/m2.sec We know molecular weight of benzene. M = 78 kg/kmole ∴

NA = 9.037 × 10–9 × 78 kg/m2.sec = 704.88 × 10–9 kg/sec.m2

∴

NA = 7.048 × 10−7 kg/m2.sec

(Ans.)

Principles of Mass Transfer Operations − I (Vol. − I)

2.57

Fundamentals of Diffusion Mass Transfer

(20) A Well located in the desert is 10 m deep to the water level and 1.0 m in diameter. o

The stagnant air and water in the Well are at 30 C and normal atmospheric pressure. A slight breeze of dry air is blowing across the top of Well. Calculate the rate of steady-state diffusion of water vapour in the Well. (Assume partial pressure of water o

vapour in the air = vapour pressure of water at 30 C). Assume diffusivity of water o

vapour in air at 30 C is DAB

= 2.6 × 10–5 m2/sec. Vapour pressure of water at

o

30 C = 4.112 kN/m2. Solution : Given Data : p = 4.112 kN/m2 A1

p

A2

p p

B1

B2

= 0 = 1.013 × 105 – 4.112 = 97.215 × 103 N/m2 = 1.013 × 105 N/m2

log mean pressure difference is given as p –p B2 B1 p = p



   B1

B, lm

ln p =

p

B, lm

B2

1.013 × 105 – 97.215 × 103 101.3 ln  97.2   

= 99.77 kN/m2

For one component diffusing, we have rate of diffusion NA =

DAB . PT RT ∆z p

B, lm

=

(p

A1

–p

A2

2.6 × 10–5 × 1.013 × 105 × (4.112 – 0) 8314 × 303 × 10 × 0.9977 × 105

= 4 × 10–9 kmole/m2.sec = 4 × 10–9 × 18 kg/m2.sec = 72 × 10–9 kg/m2.sec ∴ ∴

)

NA = 72 × 10–9 kg/m2.sec Steady-state diffusion of water in kg/s = 72 × 10–9 A kg/s

Principles of Mass Transfer Operations − I (Vol. − I) = 72 × 10–9 ×

2.58

Fundamentals of Diffusion Mass Transfer

πD2 4 kg/s

π = 72 × 10–9 × 4 (1) 2 = 56.54 × 10–9 kg/sec ∴

Steady-state diffusion of water vapour = 56.54 × 10–9 kg/sec.

(Ans.)

(21) Water in the bottom of a narrow metal tube is held at a constant temperature of 293 K. The total pressure of air (assumed dry) is 1.01325 × 105 Pa and the temperature o

is 293 K (20 C). Water evaporates and diffuses through air in the tube, and the diffusion path ∆z is 0.1524 m long. Calculate the steady-state rate of evaporation in kmole/m2.s. The diffusivity of water vapour at 293 K and 1 atm pressure is 0.25 × 10–4 m2/s. Assume the system is isothermal. Vapour pressure of water at o 20 C = 17.54 mm Hg. Solution : Given Data : DAB = 0.25 × 10–4 m2/s where

A → for water vapour B → for dry air 17.54 = 760 × 1.013 × 105 p A1 p

A2

p

p

B1

B2

(pB)lm

= 2.33 × 103 N/m2 = 0 (pure air) = 1.013 × 105 – 2.33 × 103 = 98970 N/m2 = 1.013 × 105 – 0 = 1.013 × 105 N/m2 p –p B2 B1 = p

  

ln p

=

  B1  B2

1.013 × 105 – 98970 1.013 × 105 ln  98970   

= 1 × 105 N/m2

pB1 + pB2  If p is close to p , then linear mean  B1 B2  2  could be used.

Principles of Mass Transfer Operations − I (Vol. − I)

2.59

Fundamentals of Diffusion Mass Transfer

The flux of diffusion of water vapour into air is DAB · PT p –p NA = A1 A2 RT ∆z (PB)lm

(

)

–4

=

5

0.25 × 10 × 1.013 × 10 × (2.33 × 103 – 0) 8314 × 293 × 0.1524 × 105

= 1.595 × 10–7 kmole/s.m2 ∴

NA = 1.595 × 10–7 kmole /s.m2

(Ans.)

(22) A sphere of naphthalene having a radius of 2.0 mm is suspended in a large volume of still air at 318 K and 1.013 × 105 Pa. The surface temperature of the naphthalene can be assumed to be at 318 K and its vapour pressure at 318 K is 0.555 mm Hg. The DAB of naphthalene in air at 318 K is 6.92 × 10–6 m2/s. Calculate the rate of evaporation of naphthalene from the surface. Solution : Given Data : DAB = 6.92 × 10–6 m2/s p p

A1 A2

=

0.555 5 2 760 × 1.013 × 10 = 74 N/m

= 0 (air is pure)

2 r1 = 1000 = 2 × 10–3 m p

B1

= 1.013 × 105 – p

A1

5

= 1.013 × 10 – 74 = 1.012 × 105 N/m2 p

B2

Since values of p

B1

= 1.013 × 105 – 0 = 1.013 × 105 N/m2 and p are close, we take their linear mean. B2

p

(pB)lm

= =

B1

+p

B2

2 (1.012 + 1.013) × 105 2

= 1.0129 N/m2 The flux of diffusion of naphthalene into stagnant air is DAB · PT NA = RT r1 p p –p

( B)lm (

=

A1

A2

)

6.92 × 10–6 × 1.013 × 105 × (74 – 0) 8314 × 318 × 2 × 10–3 × 1.012 × 105

= 9.68 × 10–8 kmole/s.m2

Principles of Mass Transfer Operations − I (Vol. − I)

2.60

Fundamentals of Diffusion Mass Transfer

Rate of evaporation of naphthalene NA × 4πr2 = 9.68 × 10–8 × 4 × π × (2 × 10–3)

2

= 4.86 × 10–12 kmole/sec. NA = 4.761 × 10–14 kmole/m2.sec

∴

Helium loss from 0.1 m2 surface area of diffusion = 4.761 × 10–15 kmole/sec We know, molecular weight of He = 4 kg/kmole ∴

Helium loss = 4.761 × 10–15 × 4 kg/sec = 19.04 × 10–15 kg/sec

Total mass loss of helium = 6.85 × 10–11 kg/hr (23)

(Ans.)

Find the rate of evaporation of chloropicrin into air (considered here as a pure o

substance) at 25 C ? Total pressure

770 mm Hg

Diffusivity Vapour pressure Distance from liquid level to top of the tube

0.088 cm2/sec 23.81 mm Hg 11.14 cm

Density of chloropicrin

1.65 g cm–3

Surface area of liquid exposed for evaporation

2.29 cm2

Solution : Given Data : PT = 770 mm Hg p p

A1 A2

p p

B1 B2

= 23.81 mm Hg

ρ = 1650 kg/m3

= 0

A = 2.29 × 10–4 m2

= 760 – 23.81 = 736.19 mm Hg = 760 – 0 = 760 mm Hg p

∴

( B)lm p

=

B2

–p

B1

pB2 ln p   

B1

 

760 – 736.19 = 748.03 mm Hg 760 ln 736.19   748.03 = 760 × 1.013 × 105

=

∴

DAB = 0.088 × 10–4 m2/s

(pB)lm (pB)lm =

0.997 × 105 N/m2

Principles of Mass Transfer Operations − I (Vol. − I)

2.61

Fundamentals of Diffusion Mass Transfer

For diffusion of A through non-diffusing B, we have mass flux given by DAB · PT p –p NA = A1 A2 R.T p · ∆z

(

( B)lm

)

770 23.81 0.088 × 10–4 × 760 × 1.013 × 105 × 760 × 1.013 × 105 = 8314 × 298 × 0.997 × 105 × 10 × 10–2 0.088 × 10–4 × 1.026 × 105 × 0.0317 × 105 = 2.47 × 1010 = 1.15 × 10–7 kmole/m2.sec NA = 1.15 × 10–7 kmole/m2.sec

∴ ∴

Rate of evaporation = 1.15 × 10–7 × A = 1.15 × 10–7 × 2.29 × 10–4 kmole/sec = 2.63 × 10–11 kmole/sec Molecular weight of chloropicrin = 164.5 Rate of evaporation = 2.63 × 10–11 × 164.5 kg/sec

∴

= 1.55 × 10–4 kg/hr = 0.155 gm/hr

(Ans.)

(24) The diffusivity of the gas-pair oxygen-carbon tetrachloride determined by observing the steady-state evaporation of CCl4 into a tube containing O2. The distance between CCl4 liquid level and the top of the tube is 17.1 cm. The total pressure on the system o

is 100.63 kN/m2 and the temperature is 0 C. The vapour pressure of CCl4 at that temperature is 4.39 kN/m2. The cross-sectional area of the diffusion tube is 0.82 cm2. If it is found that 0.0208 cm3 of CCl4 evaporate in 10 hours period after steady state has been attained, what is the diffusivity of the gas-pair CCl4 – O2 ? ρ = 1590 kg/m3, M = 154 kg/kmole. CCl 4

Solution : Given Data : PT = 100.63 kN/m2 p = 4.39 kN/m2 A1

p

A2

p

∴

p

B1 B2

= 0 = 101.3 – 4.39 = 96.91 kN/m2 = 101.3 – 0 = 101.3 kN/m2 p

∴

( B)lm p

=

B2

–p

p

B1

   B  1

ln p

B2

We need to calculate the flux NAB. A → CCl4 and B → O2

=

101.3 – 96.91 = 99.08 kN/m2 101.3 ln 96.91  

Principles of Mass Transfer Operations − I (Vol. − I)

2.62

1 NAB = 0.0208 × 10–6 × 1590 × 154 ×

∴

Fundamentals of Diffusion Mass Transfer 1 1 × 0.82 × 10–4 10 × 3600

= 7.2 × 10–8 kmole/m2.sec The same flux can be given as PAB · PT × p

(

NAB =

7.2 × 10–8

RT ∆z

A1

–p

A2

)

(pB)lm

DAB × 100.63 × 103 × 4.39 × 103 = 8314 × 273 × 17.1 × 10–2 × 99.08 × 103 = DAB × 0.0114

DAB = 631.57 × 10–8 = 6.315 × 10–6 m2/s (Ans.) ∴ DAB = 6.315 × 10−6 m2/s (25) An open cylindrical tank is filled within 2 m to the top with pure CH3OH. The tank is tapered as shown in Fig. 2.18. The air within the tank is stationary but circulation of air immediately above the tank is adequate to assume negligible concentration of o

CH3OH at this point. The tank and air space are at 77 C and 1 atm. The diffusivity of o

CH3OH in air at 77 C and 1 atm. is 0.6 m2/sec. Calculate the rate of loss of methanol from the tank at steady state. 4m Stationary Air

2m

Methanol

6m

Fig. 2.18 o

Assume vapour pressure of CH3OH at 77 C = 135 mm Hg. Solution : The general equation for diffusion plus convective mass transfer is dxA NA = – C DAB dz + xA (NA + NB) But here, NB = 0, air is stationary ∴ ∴ where,

dxA NA = – C DAB dz + xA NA dxA NA (1 – xA) = – C DAB dz C → total concentration of methanol and air A → methanol B → air

Principles of Mass Transfer Operations − I (Vol. − I) NA =

∴

2.63

Fundamentals of Diffusion Mass Transfer

DAB dxA – C 1 – x dz A

Here, the geometry of the tank is variable, hence rate of loss of methanol depends upon area. ' NA

=

– C DAB dxA × (Area of tank)  1 – xA dz 

=

DAB dxA – C 1 – x dz × A A

π Geometric consideration shows 4 (6 – z) 2 as a variable area when z = 0, z = 2,

area =

at level of CH3OH

π area = 4 42 at top of tank ' NA

∴ ∴

π 62 4

' NA

= – C DAB

π 2 dxA 4 (6 – z) (1 – xA) · dz

dz (6 – z)2

dxA π = – C DAB (1 – x )  4  A  

dz (6 – z)2

=

– C DAB  N' A 

π dxA 4  (1 – xA)



The above equation is integrated as xA 2

z2

⌠ dz ⌠ dxA   ⌡ (6 – z)2 = α ⌡ (1 – xA) z1

xA 1

– C DAB π  4  α =   = constant ' N A  

where 2

0

⌠ dz ⌠ dxA   ⌡ (6 – z)2 = α ⌡ (1 – xA) 0

xA 1 2

∴

0

–1 (– ) (6 – z)  0 = α (–1) [ln (1 – xA)]xA1 2

0

 1  = – α [ln (1 – xA)]x A1 (6 – z)0 ∴

1 – 1 = – α ln (1) – ln (1 – xA ) [ 1 ] 4 6

Principles of Mass Transfer Operations − I (Vol. − I)

2.64

Fundamentals of Diffusion Mass Transfer

 2  = – α ln  1   1 – xA  24 1   1 12

= α ln (1 – xA1) p

xA1 =

A1

P 135 = 760

= 0.1776 ∴

∴ We know,

1 12

= α ln (1 – 0.1776)

= – 0.1955 α α = – 0.427 π – C · DAB · 4 α = ' NA

∴

∴

∴

π – C · DAB · 4 ' NA ' NA

= 0.427 π C · DAB · 4 = 0.427 P DAB · π = RT 1.708 1.013 × 105 0.6 × π = × 1.708 8314 × (77 + 273)

= 0.0383 kmole/sec ' NA = 0.0383 kmole/sec

(Ans.)

(26) Oxygen is diffusing in a mixture of oxygen-nitrogen at 1 std. atm., 25°°C. Concentration of oxygen at planes 2 mm apart are 10 ad 20 volume % respectively. Nitrogen is non-diffusing (a) Derive the appropriate expression to calculate the flux of oxygen. Define units of each term clearly. (b) Calculate the flux of oxygen. Diffusivity of oxygen in nitrogen = 1.89 × 10−5 m2/sec. Sol. : Let us denote as A and nitrogen as B. Flux of A i.e. NA is made up of two components, namely that resulting from the bulk motion of A i.e. NxA and that resulting from molecular diffusion JA : NA = NxA + JA (1) From Fick’s law of diffusion, ∂CA JA = −DAB (2) ∂z

Principles of Mass Transfer Operations − I (Vol. − I)

2.65

Fundamentals of Diffusion Mass Transfer

From one dimensional diffusion equation (2), it becomes, dCA JA = −DAB dz Substituting this in equation (2), dCA NA = NxA − DAB dz Since N = NA + NB, and xA = CA/C equation (3) becomes, CA dCA NA = (NA + NB) − C − DAB dz Rearranging the terms and integrating between the planes between 1 and 2,

… (3)

CA2

⌠ ⌡

dz cDAB = −

dCA ⌠ ⌡ NAC − CA (NA + NB)

… (4)

CA1

Since B is non-diffusing NB = 0. Also, the total concentration C remains constant. Therefore, equation (4) becomes, CA2

z CDAB = −

dCA ⌠ ⌡ N AC − N AC A CA1

1 C − CA2 = N ln C − C A A1 CDAB C − CA2 … (5) Therefore, NA = z ln C − CA1 Replacing concentration in terms of pressures using Ideal gas law, equation (5) becomes, DABPt Pt − PA2 NA = RTz ln P − P … (6) t A1 where, DAB = Molecular diffusivity of A in B PT = Total pressure of system R = Universal gas constant T = Temperature of system in absolute scale z = Distance between two planes across the direction of diffusion PA1 = Partial pressure of A at plane 1, and PA2 = Partial pressure of A at plane 2 Given : DAB = 1.89 × 10−5 m2/sec PT = 1 atm. = 1.01325 × 105 N/m2 T = 25°C = 273 + 25 = 298 K z = 2 mm = 0.002 m PA1 = 0.2 × 1 = 0.2 atm. (From Ideal gas law and additive pressure rule) PA2 = 0.1 × 1 = 0.1 atm. Substituting these equation (6), (1.89 × 10−5) (1.01325 × 105) 1 − 0.1 NA = ln 1 − 0.2 (8314) (298) (0.002)   = 4.552 × 10−5 kmol/m2.sec. (Ans.)

Principles of Mass Transfer Operations − I (Vol. − I)

2.66

Fundamentals of Diffusion Mass Transfer

(27) CO2 is diffusing through a membrane (1.25 mm) thick at 298 oK. The partial pressure of CO2 on one side of the membrane is 5 cm of Hg and other side is zero. The solubility coefficient of CO2 at 298 K is 0.9 m3 gas (NTP) (m3 · rubber) · (atm.) The diffusivity of CO2 in the rubber is 1.1 × 10–5 m2/sec. Determine the permeability of SA =

membrane for CO2 at 298 oK and rate of diffusion through the membrane. Sol. : Diffusivity of CO2 in membrane at 298 K, DA = 1.1 × 10–10 m2/sec Solubility constant, Permeability,

0.9 m3 · gas (NTP) (m3 rubber) · (atm.) P = DA · SA

SA =

m3 gas (NTP) = 9.9 × 10–11 S (m rubber) (atm.) – – 5 ∆P = pA – pA = 5 – 0 = 5 cm = 76 = 0.06578 atm. 1 2 Z = 1.25 × 10–3 m P = 9.9 × 10–11 m3 gas (NTP) sec. (m. rubber) (atm.)

DA · SA PA – PA



VA =

–

–

1

2

 

Z 9.9 × 10–11 × 0.06578 = 1.25 × 10–3 m3 gas (NTP) = 5.210 × 10–9 (Ans.) (s) (m2. rubber) (28) A polyethylene film 0.15 mm thick is being considered for use in packaging a pharmaceutical product at 30oC. If the partial pressure of O2 outside is 0.21 atm. and inside the package it is 0.01 atm., calculate the diffusion flux of O2 at steady state. Data : PM = 4.17 × 10–12 m3 . Solute (STP)/(S.m2.atm./m.) PM pA – pA Sol. : We have equation,



–

–

1

2

 

NA = 22.414 (Z – Z ) 2 1

4.17 × 10–12 (0.21 – 0.01) 22.414 (0.00015 – 0) = 2.480 × 10–10 kg mole/s.m2 Note that a film made of nylon has a much smaller value of permeability PM for O2 and would make a more suitable barrier. (29) H2 gas flows through a tube of neoprene rubber having ID = 30 mm and OD = 55 mm. The pressure and temperature of the gas are 3 std. atm. pressure and 298 K m3 (NTP) respectively. If the solubility of hydrogen in rubber is S = 55 × 10–3 3 m .rubber.atm. –10 2 and diffusivity of H2 through rubber DAB = 1.8 × 10 m /sec. Calculate rate of H2 loss per unit length of tube due to diffusion. =

Principles of Mass Transfer Operations − I (Vol. − I)

2.67

Fundamentals of Diffusion Mass Transfer

Sol. :

H2 H2

CA,2

H2 H2 H2

CA,1 H2

H2 H2

Fig. 2.19 : Diffusion of H2 gas through a tube of neoprene rubber

To calculate the concentration difference (∆ CA) : ∆ CA = CA – CA 1

2

At 3 standard atmosphere pressure, the solubility of H2 in rubber is equal to 3 times solubility (S). m3 (NTP) ∴ 3 (55 × 10–3) = 0.165 3 m rubber ∴ Concentration at inner surface of tube is, k mole H2 0.165 CA = 22.4 = 7.3 × 10–3 3 1 m rubber CA

= concentration of H2 at the outer surface of the tube = 0

∴

∆ CA = 7.3 × 10–3

2

k mole H2 m3 rubber

Molar flux of H2 diffusion : NA = Z = = = ∴

NA =

DA · ∆ CA Z 1 1 2 (55) – 2 (30) 12.5 mm 12.5 × 10–3 m 1.8 × 10–10 × 7.3 × 10–3 12.5 × 10–3

= 1.022 × 10–10

k mole H2 S.m2. rubber

k mole H2 S.m2.rubber

Diffusion Surface : Since diffusion is taking radialy; 2 π l (r2 – r1) Sav = r2 ln r   1 Sav l

=

2 π (r2 – r1) π (d2 – d1) π (25) = = 0.606 r2 d2 ln r  ln d   1  1

= 0.12960 m2/m. length of rubber tube.

Principles of Mass Transfer Operations − I (Vol. − I)

2.68

Fundamentals of Diffusion Mass Transfer

H2 loss through unit length : WA = NA · Sav = 1.022 × 10–10 × 0.12960 k mole H2 = 1.325 × 10–11 s.m. m. length of tube (Ans.) (30) A open bowl 0.3 m in diameter contains H2O at 350 K evapourating into the atmosphere. If the air currents are sufficiently strong to remove into water vapour as it is formed and it its resistance to its mass transfer in air is equivalent to 1 mm layer for conditions of molecular diffusion. What will be the rate of cooling due to evapouration ? The water can be considered as well mixed and the water equivalent of system is 10 kg. The diffusivity of water vapour in air may be taken as 0.2 cm2/sec. and kg molecules volume at NTP condition is 22.4 m3. Sol. : Let '1' refers to water side of stagnant air and '2' refers to air water side of stagnant layer. Let,

A = Water vapour B = Air

Then the rate of diffusion through a stagnant layer is, NA = – –

where,

pA

1

DAB Pt –

RTZ PB‚ M

p– – p–   A1 A2

= Vapour pressure of water at 350 oK = 41.8 kN/m2

–

pA = 0 (Since the air current removes the vapour as it is formed) 2

Pt = 101.3 kN/m2 –

pB

1

–

= Pt – p A

1

= 101.3 – 41.8 = 59.5 kN/m2 –

pB

2

= 101.3 kN/m2 –

So,

–

pB – pB

–

2

PB, M =

1

p  p–    –

ln

=

B2

101.3 – 59.5 101.3 ln  59.5   

B1

= 78.55 kN/m2 R = 8.314 J/kg.mole K. 0.2 × 10–4 × 101.3 × 4.18 8.314 × 350 × 1 × 10–3 × 78.55 = – 3.725 × 10–4 k mole/m2.sec. = – 6.66 × 10–3 kg water/m2. sec.

NA = – NA

Principles of Mass Transfer Operations − I (Vol. − I)

2.69

Fundamentals of Diffusion Mass Transfer

π π Area of the bowl = 4 × D2 = 4 × (0.3)2 = 0.0706 m2 ∴ Rate of evapouration = NA × Area kg. water = 4.7077 × 10–4 sec. We have, Latent heat of vapourisation = 2318 kJ/kg. Specific heat of water = 4.187 kJ/kg K. ∴ Rate of heat removal = (rate of evapouration of water) × Latent heat of vapourisation. ∴ Heat removal = 1.10 kW dQ If rate of cooling = dt k/sec. dQ = (water equivalent) (Special Heat) × dt = 1.10 dθ Solving for dt , we get, dθ (Ans.) dt = 0.026 kg/sec. (31) NH3 absorbed in water from a mixture with air using a column operating at 1 atm and 295 K. The resistance to transfer can be regarded as lying entirely within the gas phase. At a point in the column, the partial pressure of ammonia is 6.6 kN/m2. The back pressure at water interface is negligible and resistance to transfer can be regarded as lying in a stationary gas film 1 mm thick. If the diffusivity of NH3 in air is 0.236 m2/s. What is transfer rate per unit area in that column ? If the gas were compressed to 200 kN/m2, how could the transfer rate be altered ? Sol. : If '1' and '2' refers to the water and air side of stagnant air film and subscripts 'A' and 'B' refers to NH3 and air respectively. Then the rate of diffusion through a stagnant layer is : DAB · Pt p– – p–  NA =  A1 A2 – RTZ PB, M

(

–

pA

Data :

1

–

pA

2

)

= 0 = 6.6 kN/m2

Pt = 101.3 kN/m2 –

pB

So,

1

–

pB

2

–

∴

Thus,

= 101.3 kN/m2 = 94.6 kN/m2

PB, M =

NA

94.6 – 101.3 94.7 ln 101.3  

= 97.96 kN/m2 0.236 × 10–4 × 101.3 × (–6.6) = 8.314 × 295 × 1 × 10–3 × 97.96 = – 6.567 × 10–5 k mole/m2.sec.

Principles of Mass Transfer Operations − I (Vol. − I)

2.70

Fundamentals of Diffusion Mass Transfer

–

–

If pressure is increased to 200 kN/m2, pA is still zero. But pA is doubled. 1

–

2

–

∴ pB and pB becomes, 1

2

–

pB

1

–

pB

2

= 200 kN/m2 = 94.7 × 2 = 189.4 kN/m2

–

PB, M =

So,

189.4 – 200 2 189.4 = 193.3 kN/m  ln 200  

Since diffusivity is inversely proportional to pressure. New value of diffusivity is given (Pt)Initial by, (DAB)New = (DAB)Initial × (P ) t Final 101.3 = 0.236 × 10–4 × 200 = 0.119 × 10–4 m2/s Thus,

NA =

0.119 × 10–4 × 200 (–6.6 × 2) 8.314 × 295 × 10–3 × 193.3

NA = 6.67 × 10–5 k mole/m2.sec. Thus, if pressure is doubled, the rate of diffusion is unaltered.

(Ans.)

(32) By what % would the rate of absorption be increased or decreased by increasing the total pressure from 100 – 200 kN/m2 in the following cases : (a)

The absorption of NH3 from a mixture of NH3 and air containing 10% NH3 by volume using pure water as a solvent. Assume that other resistance to mass transfer lies within the gas phase.

(b) The same condition as (a) but the absorbing solution exerts a partial vapour pressure of NH3 of 5 kN/m2 . The diffusivity can be assumed to be inversely proportional to absolute temperature. Sol. : (a) The rate of diffusion for this process is given by : DAB Pt p– – p–  NA = –  A1 A2 RTZ PB, M –

Data Given :

pA

= 0

–

10 = 100 × 100 = 10 kN/m2

1

pA

2

Pt = 100 kN/m2 –

pB = 100 kN/m2 ; 1

–

pB = 90 kN/m2 2

Principles of Mass Transfer Operations − I (Vol. − I)

2.71

–

2

PB, M =

∴

–

pB – pB

–

1

p  p–    –

ln

Fundamentals of Diffusion Mass Transfer

= 94.91 kN/m2

B2 B1

So,

Pt

= 1.054

–

PB‚ M –10.54 DAB … (1) RTZ 2 If pressure is doubled i.e. to 200 kN/m , the diffusivity becomes half to it's original value. So, in this case, So,

NA =

–

pA

= 0

–

10 = 100 × 200 = 20 kN/m2

1

pA

2

–

pB

1

–

pB

2

= 200 kN/m2 = 180 kN/m2

–

PB, M =

So,

Pt

180 – 200 2 180 = 189.82 kN/m  ln 200  

= 1.054

–

PB‚ M –10.54 · DAB RTZ Comparing equation (1) and (2) we can say that the rate of absorption is unchanged. NA =

∴

… (2)

(b) If the absorbing solution now exerts a partial vapour pressure of NH3 of 5 kw/m2, then at total pressure of 100 kN/m2, –

pA

1

–

pA

2

= 5 kN/m2 = 10 kN/m2 –

∴

–

pB – pB

–

2

PB, M =

1

p  p–    –

ln

B2 B1

–

pB

1

= 95 kN/m2

Principles of Mass Transfer Operations − I (Vol. − I) = –

pB

2

Pt

So,

2.72

Fundamentals of Diffusion Mass Transfer

90 – 95 2 90 = 93.47 kN/m  ln 95  

= 90 kN/m2 = 1.0913

–

PB‚ M NA =

∴

–5.406 DAB RTZ

… (3)

Now, Pressure = 200 kN/m2, D = 0.5 DAB –

pA

1

–

pA

2

–

pB

1

–

pB

2

= 5 kN/m2 = 20 kN/m2 = 195 kN/m2 = 180 kN/m2

–

∴

PB, M =

So,

–

Pt

180 – 195 180 ln 195  

= 187.4 kN/m2

= 1.057

PB‚ M NA =

and thus,

– 8.004 DAB RTZ

… (4)

So, the rate of diffusion is increased by : 8 – 5.406 5.406 × 100 = 47.98%

(Ans.)

(33) Diffusivity of vapour of volatile liquid in air can be conveniently determined by Winklemann's method in which liquid is contained in a narrow vertical tube maintained at constant temperature. Air stream is passed over the top of tube rapidly to measure the partial pressure of vapour remains approximately zero. On the assumption, vapour is transferred from surface of liquid to an stream by molecular diffusion. Calculate the diffusivity of CCl4 vapour in air at 321 oK and 1 atm. pressure from following experimental data. Derive the equation you have used in this problem.

Principles of Mass Transfer Operations − I (Vol. − I)

2.73

Fundamentals of Diffusion Mass Transfer

Time from common cement of experiment (ks)

Liquid Level (cm)

(mm)

0

0

0

1.6

0.25

0.025

11.1

1.29

0.129

27.4

2.32

0.232

80.2

4.39

0.439

117.2

5.47

0.547

168.6

6.70

0.670

199.7

7.38

0.738

289.3

9.03

0.903

10.48

1.048

383.1 Data : Vapour pressure of CCl4 at 321 K is 37.6 kN/m 3

2

and density of liquid is 1540

3

Distance over which diffusion occurs

kg/m . The kilogram molecular volume is 22.4 m . Gas Stream

Liquid

Fig. 2.20 : Estimation of diffusivity of vapours by Winklemann's method

Sol. : Rate of mass transfer is given by, NA = D · where,

CA·  CT   L  CB‚ M

… (1)

CA = Saturation concentration at the interface L = Effective distance through which the mass transfer is taking place. CT = Total concentration of system. CB, M

= Logarithmic

concentration

differential

considering

the

evapouration of liquid, the rate of mass transfer is given by, ρL dL NA = M · dt … (2)   where

ρL = Density of liquid

Principles of Mass Transfer Operations − I (Vol. − I)

2.74

Fundamentals of Diffusion Mass Transfer

Thus, at any time these two rates are equal. Equating equations (1) and equation (2), CA CT ρL dL M · dt = D ·  L  · C  … (3)      BM Integrating equation (3) and putting, t=0 L = Lo , at 2 2 MD CA CTt we get, L2 – L0 =      ρL   CBM  Lo will not be measured accurately nor is the effective distance for diffusion at time t. Most of the practical situations accurate values of (L – Lo) are available. 2 MD CA CTt (L – Lo) (L – Lo + 2 Lo) =  … (4)     ρL   CBM  Equation (4) may be written in the form,  t  =  ρL   CBM  (L – L ) +  ρL – CBM  L o 2 MD CA CT MD CA CT o L – Lo t Plot a graph of L – L  Vs o 

L – Lo

t/(L – L0) (ks/mm)

4

× ×

3

×

×

× 2

× ×

1

× ×

0×

20

40 60 80 100 120 (L – L0) (mm)

Fig. 2.21 : Plot t/(L – Lo) Versus (L – Lo)

From graph,

Now, where, ∴

Slope = S = 0.0310 ks/mm2 = 3.1 × 107 s/m2 1 To CT = 22.4 T     To = reference temperature = 273 K 1 273 CT = 22.4 321     CT = 0.0376 k mole/m3 M = 154 kg/k mole Vapour pressure CA = Total pressure × CT 37.6 = 101.3 × 0.0376

… (5)

Principles of Mass Transfer Operations − I (Vol. − I)

2.75

Fundamentals of Diffusion Mass Transfer

= 0.01396 k mole/m3 ρL = 1540 kg/m3 In this problem, we assume, CT ≅ CB = 0.0376 1

and

CB

2

k mole m3

=

Pt– Vapour Pressure C   T Pt  

=

101.3 – 37.6 × 0.0376  101.3 

= 0.0238 k mole/m3 CB – CB CBM =

∴

2

1

CB

   1

=

2

ln C B

0.0238 – 0.0376 0.4573

= 0.0303 From graph,

(ρL/CBM) S = (2MDC C ) = 3.1 × 107 S/mm2 A T

∴

D =

1540 × 0.0303  ρL · CBM  2M C C S = 2 × 154 × 0.01396 × 0.0376 × 3.1 × 107 A T  

D = 9.2 × 10–6 m2/sec.

(Ans.)

(34) A binary mixture of methane and hydrogen gas is contained in a tube at 101.32 kPa and 273 K. The tube connects two large, well-mixed gas reservoirs of A and B. At one point of partial pressure of methane is PA = 60.79 kPa and at a point 0.02 m away 1

PA = 20.26 kPa. If the total pressure is constant throughout the tube and reservoirs, 2

calculate the flux of methane in the tube, assuming steady-state, equimolar counter diffusion. Data : DAB = 6.25 × 10–5 m2/s. Sol. : Let, CH4 = A , H2 = B P = 101.32 kPa

Data given :

T = 273 K –

pA

1

–

pA

2

= 60.79 kPa = 20.26 kPa

L = 0.02 m DAB = 6.25 × 10–5 m2/s

Principles of Mass Transfer Operations − I (Vol. − I)

2.76 1

Fundamentals of Diffusion Mass Transfer 2

A

B

z Fig. 2.22 : Diffusion of Methane and Hydrogen Gas

This is the case of steady-state, equimolar counter diffusion, DAB pA – pA NA, Z =

So,



–

–

1

2

 

RT (Z2 – Z1)

6.25 × 10–5 m  (60.79 × 103 Pa – 20.26 × 103 Pa) s   = 3 8.314 Pa.m  (273 k) (0.02) mole·K  2

(Ans.) = 5.58 × 10–3 mole A/m2 s. (35) A flat rubber plug (30 mm thick, area 4.0 × 10–4 m2) is used to seal CO2 gas (T = 25 oC, P = 2 atm.) inside a bottle. Assuming steady-state diffusion through the plug, what is the rate of CO2 leakage through the plug ? Data for the system : Solubility (S), CO2 in rubber (T = 25, C = P = 1 atm.) = 40.2 mole CO2/m3 rubber-atm. DCO

2–air

= 0.11 × 10–9 m2/s. You may assume that the partial pressure of CO2 outside the

bottle and plug is zero. Sol. : Data given :

L = 0.030 m A = 4 × 10–4 m2 T = 298 k –

PA

1

Let, CO2 = A, Air = B

= 2 atm.

S = 40.2 mole CO2/m3 atm. DAB = 0.11 × 10–9 m2/s L CO2 CA1 pA2 = 0 CA2 pA1

Z1

Z2

Fig. 2.23 : Steady-state diffusion, through rubber plug

Principles of Mass Transfer Operations − I (Vol. − I) From Fick's law,

NA

1, Z

2.77

Fundamentals of Diffusion Mass Transfer

dCA = – C DAB dZ + CA (NA, Z + NB, Z)

Assume CO2 is very dilute (CA ≅ 0) dCA = – C DAB dZ After integration we get, NA

1, Z

dCA = – DAB dZ DAB (CA – CA )

NA

1, Z

1

=

2

(Z2 – Z1)

We must find CA and CA . 1

2

CA = S · PA 1

1

=

40.2 mole CO2 (2 atm.)   m3 atm.  

= 80.4 CA

2

mole CO2

= S · PA

2

m3 =0

0.11 × 10–9 m  80.4 mole CO2 S   m3 atm.  2

NA

∴

1, Z

=

= 2.95 × 10–7

(0.030 m) mole CO2

m2. s Now, to find rate of leakage of CO2 through the plug : mole CO2  WA , Z = NA Z · A = 2.95 × 10–7  (4 × 10–4 m2) 1 1 m2 · s   mole CO2 (Ans.) = 1.18 × 10–10 s (36) Ammonia (NH3) is selectively removed from an air – NH3 mixture by absorption into water. In this process, ammonia is transferred by molecular diffusion through a stagnant gas layer 2 cm thick and then through a stagnant water layer 1 cm thick. The concentration of ammonia at the outer boundary of the gas layer is 3.42 mole% and the concentration of ammonia at the lower boundary of the water layer is essentially zero. Other useful data for the system (T = 15 °C, P = 1 atm.) : DNH –air = 0.215 cm2/s, DNH –H O = 1.77 × 10–5 cm2/s 3

3

2

Equilibrium data for ammonia in air over aqueous solutions : 5 10 15 PNH in air (mm Hg) 3

CNH in water (mole/cm3 × 106)

6.1

3

Air film

11.9

20.0

z = 0, y NH

20

25

30

32.1

53.6

84.8

= 0.0342 3

z = 2 cm, interface z = 3 cm, x NH = 0 Water film

3

Principles of Mass Transfer Operations − I (Vol. − I)

2.78

Fundamentals of Diffusion Mass Transfer

Assuming steady-state operation, what is the rate of ammonia absorption into the water from the air ? Sol. : Let A = NH3, B = Air and C = H2O T = 15 + 273 = 288 K

Data Given :

P = 760 mm Hg DAB = 0.215 cm2/s DAC = 1.77 × 10–5 cm2/s We must use stagnant film conditions. At steady state, fluxes of NH3 for liquid and for gas must be equal at Air – H2O interface.

P – p  P – p–    –

∴ NA

Pi DAB = RT (Z – Z ) · ln 2 1

t

A2

t

→ for Air film.

A1

2

=

(760 mm Hg) (0.215 cm /s) 3 lit. 62.36 mm Hg 1000 cm  (288 K) (2 cm) mole k   1L  

760 – p–  A2   × ln 760 – 30  

760 – p–  A2   = [4.55 × 10–6 mole/cm2 S] · ln  730  Also,

… (1)

C DAC 1 – XA3 NA = (Z – Z ) · ln 1 – X  → for H2O film 3 2 A



CH

2O

=

NA =

2



55.5 mole  1L 3 = 0.0555 mole/cm3 L  1000 cm   0.0555 mole  (1.77 × 10–5 cm2/s) cm3   (1 cm)

1–0 · ln 1 – X A



1 = (9.82 × 10–7 mole/cm2.s) · ln 1 – X A



2

  

  2 … (2)

There are three unknowns NA pA XA . But we have only two equations. The third



–

1‚

2‚

equation comes from equilibrium data.

2



Principles of Mass Transfer Operations − I (Vol. − I)

2.79

Fundamentals of Diffusion Mass Transfer

100 90 80

CNH3 in Water 3 6 50 (mol./cm × 10 )

20 10 0

5

10

15

20

25

30

PNH3 in Air (mm Hg) Fig. 2.24 : Ammonia absorption into water from air

From above plot, CA × 106 = XA (C × 106) = 3.9765 exp (0.1037 pA ) 2

2

XA

2

=

2

3.9765 e (0.1037 pA ) (0.0555 × 106) xp 2

= 7.16 × 10–5 exp (0.1037 pA ) 2

… (3)

Combining equation (1) and (2), 760 – pA  2  = (4.55 × 10–6) · × ln    730   A2

1 (9.82 × 10–7) · ln 1 – X



760 – pA  2  = 4.63 ln    730   2

1 ln 1 – X A



1 1 – XA

2

760 – pA2 =  730   

4.63

… (4)

Solving for XA , we get, 2

XA

= 1.57 × 10–3

PA

= 29.8 mm Hg

2 2

From either equation (1) or equation (2), we find that, NA =

9.55 × 106 mole  ln 760 – 24.8 2 cm . S    730 

NA = 1.25 × 10–9

mole NH3 cm2 sec.

(Ans.)

Principles of Mass Transfer Operations − I (Vol. − I)

2.80

Fundamentals of Diffusion Mass Transfer

(37) A tank containing water (T = 310 K) has its top open to the air (P = 1 atm.). The tank is cylindrical with a diameter of 1 m. The liquid level is maintained at a level 1 m below to top of the tank as shown below. The vapour pressure of water at 310 K is 47.1 mm Hg. (a) How many moles of water are lost per hour if dry air at 310 K is blow across the top of the tank ? (b) It is proposed to add a tapered top to this water tank as shown in sketch (b). (c) What will be the loss per hour in this case if dry air at 310 K is blown across the top of the tank ? 0.5m

1m

1m

1m

1m

(a)

(b)

Fig. 2.25

Data :

DAB

298

= 2.61 × 10–5 m2/s

T = 310 K

Sol. : Data given :

PVap = 47.1 mm Hg → for H2O (a)

Let, H2O = A and Air = B DAB

298 K

= 2.61 × 10–5 m2/s

To correct for temperature DAB, 298 K =

310 K 298 K

3/2

Air P = 1 atm D=1m

H2O

Fig. 2.26

DAB, 310 oK

310 K 3/2 = 298 K (2.61 × 10–5 m2/s)   = 2.77 × 10–5 m2/s

Principles of Mass Transfer Operations − I (Vol. − I) NA, Z =

∴

2.81

Fundamentals of Diffusion Mass Transfer

(1.01 × 105 P) (2.77 × 10–5 m2/s) 1–0 × ln 1 – 47.1 3 8.314 pa.m (310 K) (1m)  760     mole K 

NA, Z = 6.94 × 10–5 mole/m2.sec. Rate of H2O loss = NA

1, Z

Area

=

6.94 × 10–5 mole  (π) 1m 2 2 m .S  

=

5.45 × 10–5 mole  3600 s  S  1 hour 

= 0.196

2

mole H2O Hour

(Ans.)

(b) r2 = 0.25 m

z

z h=1m

r

z' l = 0.25 m

r1 = 0.5 m Fig. 2.27

From equivalence of triangles (Geometry) 1 h l = 0.25

Z Z Z = r → r =4→r=4

How does r vary with Z ? Z r (Z) = 0.25 + 4 Flippinal direction of Z, Z' = 1 – Z → Z = 1 – Z' r (Z') = 0.25 +

 

(1 – Z') 4

A (Z') = πr2 = π 0.25 +

(1 – Z') 2 4 

0 From Fick's law,

NA

1

, Z'

= –C D AB ·

d XA + XA N A Z' + N D Z' 1

dXA (1 – XA) · NA Z' dZ' = – C DAB . (1 – X ) 1 A

[

B Z' 1

]

Principles of Mass Transfer Operations − I (Vol. − I)

2.82

Fundamentals of Diffusion Mass Transfer

dXA NA dZ' = – CDAB · (1 – X ) 1 A WA Z2 We know that,

NA

1, Z

WA Z' · dZ' 1

∴

π 0.25 +



WA · Z' π

2

(1 – Z') 4 

1

=

1

A (Z)

dXA = – C DAB (1 – X ) A 0

dZ'

1

WA‚ Z' π

0

1 (4) 1 Z' = C·DAB [ln (1 – XA)] XA0 2 – 4   0

4 WA

1‚Z'

π

(4 – 2)

= C DAB

WA , Z' = 1

P C = RT

∴

dXA

⌠ 1 Z' 2 = – C DAB ⌠ (1 – X ) ⌡  ⌡ A  0 Z – 4  X A0

=

(– ln (1 – X )) A0

– π C DAB ln (1 – XA ) 8 0 1.01 × 105 pA 3 8.314 Pa.m  (310 K) mole.K 

= 39.2

mole m3

DAB = 2.77 × 10–5 m2/s 47.1 mm Hg xA = 760 mm Hg = 6.20 × 10–2 0 – π (39.2 mole/m3) (2.77 × 10–5 m2/s) WA , Z = 1 8 m–1 × ln (1 – 6.20 × 10–2) mole 3600 sec = (2.73 × 10–5) sec  hour    WA

1‚ Z

= 9.80 × 10–2 mole H2O/hour

[from part (a)]

(Ans.)

(38) Oxygen is diffusing in a mixture of oxygen-nitrogen at 1 std. atm., 25°°C. Concentrations of oxygen at planes 2 mm-apart are 10 and 20-volume % respectively. Nitrogen is non-diffusing. (a) Derive the appropriate expression to calculate the flux oxygen. Define units of each term clearly. (b) Calculate the flux of oxygen. Diffusivity of oxygen in nitrogen = 1.89 × –5 2 10 m /sec. Sol. : Let us denote oxygen as A and nitrogen as B. Flux of A (i.e.) NA is made up of two components, namely that resulting from the bulk motion of A (i.e.), NxA and that resulting from molecular diffusion JA : NA = NxA + JA

… (1)

Principles of Mass Transfer Operations − I (Vol. − I)

2.83

Fundamentals of Diffusion Mass Transfer

From Fick’s law of diffusion, dCA JA = – DAB dZ Substituting this equation (1),

… (2)

dCA NA = NxA – DAB dZ CA Since, N = NA + NB and xA = C equation (3) becomes, CA dCA NA = (NA + NB) C – DAB dZ Rearranging the terms and integrating between the planes between 1 and 2,

… (3)

CA

⌠ ⌡

dZ CDAB

2

= –

dCA

⌠ N C–N C ⌡ A A A CA 1

C – CA 1 2 = N ln C – C A A 1

Therefore, NA

C – CA CDAB 2 = ln C – C Z A

… (5)

1

Replacing concentration in terms of pressures using ideal gas law, equation (5) becomes, Pt – pA DAB Pt 2 … (6) NA = RTZ ln P – p t

where,

= = = = = =

pA

= Partial pressure of A at plane 2

1 2

Given :

A1

DAB PT R T z pA

Molecular diffusivity of A in B Total pressure of system Universal gas constant Temperature of system in absolute scale Distance between two planes across the direction of diffusion Partial pressure of A at plane 1, and

DAB = 1.89 × 10-5 m2/sec Pt = 1 atm = 1.01325 × 105 N/m2 T = 25 oC = 273 + 25 = 298 K z = 2 mm = 0.002 m pA = 0.2 × 1 = 0.2 atm (From ideal gas law and additive pressure rule) 1

pA

2

= 0.1 × 1 = 0.1 atm

Substituting these in equation (6) NA =

(1.89 × 10–5) (1.01325 × 105) 1 – 0.1 ⋅ ln 1 – 0.2 (8314) (298) (0.002)  

= 4.55 × 10–5 k mole/m2.sec.

(Ans.)

Principles of Mass Transfer Operations − I (Vol. − I)

2.84

Fundamentals of Diffusion Mass Transfer

(39) A vertical glass tube 3 mm in diameter is filled with liquid toluene to a depth of 20 mm from the top opened. After 275 hrs at 39.4°°C and a total pressure of 760 mm Hg the level has dropped to 80 mm from the top. Calculate the value of diffusivity. 2

Data : Vapour pressure of toluene at 39.4°°C = 7.64 kN / m , 3

Density of liquid toluene = 850 kg/m Molecular weight of toluene, (C6 H6 CH3) = 92

Z – Zt02   t   2  2

DAB

Sol. :

ρA‚ L yBlm = M C (y – y ) A A A 1

2

yB – yB 2

yB‚ lm =

1

yB

   B1

ln y

2

where yB = 1 – yA ; yB = 1 – yA 2

2

1

1

pA yA

1

=

1

P

7.64 = 101.3

(760 mm Hg = 101.3 kN/m2)

= 0.0754 yB

1

= 1 – 0.0754 = 0.9246

yA = 0 ; yB = 1 – yA = 1 2

2

yB‚ lm =

Therefore,

P C = RT

=

1 – 0.9246 = 0.9618 1 ln 0.9246   1.01325 × 105 8314 × (273 + 39.4)

= 0.039 k mole/m3 Therefore,

DAB =

2 2 850 × 0.9618 0.08 – 0.02  2 92 × 0.039 × (0.0754 – 0) × 275 × 3600  

= 1.5262 × 10–3 (0.082 – 0.022) = 9.1572 × 10–6 m2/sec.

(Ans.)

(28) Methane diffuses at steady state through a tube containing helium. At point 1 the partial pressure of methane is pA = 55 kPa and at point 2, 0.03 m apart pA = 15 kPa. 1

2

The total pressure is 101.32 kPa, and the temperature is 298 K. At this pressure and –5 2 temperature, the value of diffusivity is 6.75 × 10 m /sec. (i) Calculate the flux of CH 4 at steady state for equimolar counter diffusion. (ii) Calculate the partial pressure at a point 0.02 m apart from point 1. Sol. : For steady state equimolar counter diffusion, molar flux is given by; DAB – – NA = RTz (pA – pA ) 1 2

… (1)

Principles of Mass Transfer Operations − I (Vol. − I)

2.85

Fundamentals of Diffusion Mass Transfer

Therefore, 6.75 × 10–5 k mole (55 – 15) 2 9.314 × 298 × 0.03 m . sec k mole = 3.633 × 10–5 2 m sec And from equation (1), partial pressure at 0.02 m from point 1 is : 6.75 × 10–5 3.633 × 10–5 = (55 – pA) 8.314 × 298 × 0.02 PA = 28.33 kPa NA =

(Ans.)

(41) In a gas mixture of hydrogen and oxygen, steady state equimolar counter diffusion is occurring at a total pressure of 100 kPa and temperature of 20°°C. If the partial pressures of oxygen at two planes 0.01 m apart, and perpendicular to the direction of diffusion are 15 kPa and 5 kPa, respectively and the mass diffusion flux of oxygen in –5 2 the mixture is 1.6 × 10 k mole/m .sec, calculate the molecular diffusivity for the system. Sol. : For Equimolar counter current diffusion : DAB – – NA = RTz (pA – pA ) … (1) 1 2 where, NA = Molar flux of A (1.6 × 10–5) k mole/m2. sec) DAB = Molecular diffusivity of A in B R = Universal gas constant (9.314 kJ/k mole. K) T = Temperature in absolute scale (273 + 20 = 293 K) z = Distance between two measurement planes 1 and 2 (0.01 m) pA = Partial pressure of A at plane 1 (15 kPa); and 1

pA

2

= Partial pressure of A at plane 2 (5 kPa)

Substituting these in equation (1) DAB 1.6 × 10–5 = (8.314) (293) (0.01) (15 – 5) Therefore, –5

2

(Ans.) DAB = 3.898 × 10 m /sec (42) A tube 1 cm in inside diameter that is 20 cm long is filled with CO2 and H2 at a total pressure of 2 atm at 0°°C. The diffusion coefficient of the CO2 – H2 system under these 2

conditions is 0.275 cm /sec. If the partial pressure of CO2 is 1.5 atm at one end of the tube and 0.5 atm at the other end, find the rate of diffusion for : (i) Steady state Equimolar counter diffusion (N A = – N B) (ii) Steady state counter diffusion where N B = – 0.75 N A, and (iii) Steady state diffusion of CO2 through stagnant H2 (NB = 0) Sol. : (i) Given :

NB = –NA

dyA dCA NA = – CDAB dz = –DAB dz pA (For ideal gas mixture CA = RT where, pA is the partial pressure of A; such that pA + pB = P. Therefore,

Principles of Mass Transfer Operations − I (Vol. − I) Therefore

2.86

NA = – DAB

Fundamentals of Diffusion Mass Transfer

d (pA/RT) dz

For isothermal system, T is constant Therefore,

NA =

– DAB RT

dpA dz –

Z2

NA

⌠ ⌡

DAB dz = – RT

Z1

pA 2

⌠ ⌡

dpA

–

pA 1

DAB NA = RTz (pA – pA ) 1 2

i.e.

… (1)

Z = Z2 – Z1

where

2

Given : D AB = 0.275 cm /sec = 0.275 × 10 NA =

–4

2

m /sec ; T = 0 oC = 273 K

0.275 × 10–4 (1.5 × 1.01325 × 105 – 0.5 × 1.01325 × 105) 8314 × 273 × 0.2

= 6.1328 × 10–6

k mole m2 sec

Rate of diffusion = NA ⋅ S where S is surface area. –6

Therefore rate of diffusion = 6.138 × 10 × π r = 6.138 × 10

–6

= 4.821 × 10

–10

= 1.735 × 10

–3

2 –2 2

× π (0.5 × 10 ) k mol/sec

mol/hr.

(ii)

dyA NA = – CDAB dz + yA (NA + NB)

Given :

NB = – 0.75 NA

Therefore,

dyA NA = – CDAB dz + yA (NA – 0.75 NA) dyA = – CDAB dz + 0.25 yA NA dyA NA – 0.25 yA NA = – CDAB dz dyA NA dz = – CDAB 1 – 0.25 y

A

(Ans.)

Principles of Mass Transfer Operations − I (Vol. − I)

2.87

Fundamentals of Diffusion Mass Transfer

For constant NA and C, NA

Z2

yA 2

⌠ dz = – CDAB ⌡

⌠ ⌡

Z1

yA 1

dyA 1 – 0.25 yA

⌠ dx = 1 ln (a + bx) ⌡ a + bx b    y A2

NA z =

(–CDAB)

NA = – Given :

 –1  ln (1 – 0.25 y ) A ] 0.25 [ y

A1

4 CDAB 1 – 0.25 yA2 ln 1 – 0.25 y  z A



1



… (2)

2 × 1.01325 × 105 p = 0.0893 k mole/m3 C = RT = 8314 × 273 pA yA

=

yA

=

1

pA 2

1.5 = 2 = 0.75

1

P 2

P

0.5 = 2 = 0.25

Substituting these in equation (2), 4 × 0.0893 × 0.275 × 10–4 NA = 0.2 k mole = 7.028 × 10–6 2 m sec –6

ln 1 – 0.25 × 0.25  1 – 0.25 × 0.75 –2 2

Rate of diffusion = NA S = 7.028 × 10 × π × (0.5 × 10 ) = 5.52 × 10

–10

k mole/sec

–3

(iii) Given : Therefore

= 1.987 × 10 mole/hr. NB = 0 dyA NA = – CDAB dz + yA NA NA =

= =

Z2

yA 2

⌠ dz = –CDAB ⌡

⌠ ⌡

Z1

yA 1

CDAB Z

(Ans.)

dyA 1 – yA

1 – yA2 ln 1 – y  A 

1

0.0893 × 0.275 × 10 0.2

= 1.349 × 10–5



–4

k mole m2 . sec.

ln 1 – 0.25  1 – 0.75

Principles of Mass Transfer Operations − I (Vol. − I)

2.88

Fundamentals of Diffusion Mass Transfer

Rate of diffusion = 1.349 × 10–5 × π × (0.5 × 10–2)2 = 1.059 k mole/sec = 3.814 mole/hr.

(Ans.)

(43) A sphere of naphthalene having a radius of 2 mm is suspended in a large volume of shell air at 318 K and 1 atm. The surface pressure of the naphthalene can be assumed to be at 318 K is 0.555 mm Hg. The DAB of naphthalene in air at 318 K is –6

2

6.92 × 10 m /sec. Calculate the rate of evaporation of naphthalene from the surface. Sol. : Steady state mass balance over a element of radius r and r + δr leads to SNA

r

– SNA

r + δr

= 0

… (1)

2

Where, S is the surface are (= 4 π r ) Dividing (1) by Sδr, and taking the limit as δr approaches zero, gives : d (r2 NA) dr

= 0

2

2

Integrating r NA = constant (or) 4 π r NA = constant We can assume that there is a film of naphthalene – vapour/air film around naphthalene through which molecular diffusion occurs. Diffusion of naphthalene vapour across this film could be written as, dyA NA = – CDAB dr + yA (NA + NB) NB = 0 (since air is assumed to be stagnant in the film) dyA NA = – CDAB dr + yA NA d  yA  NA = – CDAB dr 1 – y  A  NA = CDAB

d [ln (1 – yA)] dr

WA = Rate of evapouration = 4πr2 NA WA = WA ⌠

dr

⌡ r2

4πr2 CDAB d (ln (1 – yA)) dr

= 4π DAB ⌠ C d ln (1 – yA)

⌡

Boundary condition : yA = At r = R ln (1 – yA) At r = ∞

0.555 –4 760 = 7.303 × 10

= –7.3 × 10–4 yA = 0 ln (1 – yA) = 0

R

= constant

Principles of Mass Transfer Operations − I (Vol. − I) ∞

Therefore,

WA

⌠ ⌡

dr r2

2.89

Fundamentals of Diffusion Mass Transfer 0

⌠ [ln (1 – yA)] ⌡

= 4π DAB C

R

– 7.3 × 10

–4 0

∞

–1 WA  r   R

= 4 π DAB C [ln (1 – yA)]

–7.3 × 10

–4

1 WA 0 + R



–4  = 4π DAB C [0 + 7.3 × 10 ]

WA = 4π RDAB C × 7.3 × 10–4 C=

P Gas constant × T

1.01325 × 105 8314 × 318 = 0.0383 k mole/m3 =

Initial rate of evaporation : Therefore, WA = 4 × 3.142 × 2 × 10–3 × 6.92 × 10–6 × 0.0383 × 7.3 × 10–4 = 4.863 × 10–12 k mole/sec. = 1.751 × 10–5 mole/hr. (Ans.) (44) Calculate the rate of diffusion of butanol at 20°°C under unidirectional steady state conditions through a 0.1 cm thick film of water when the concentrations of butanol at the opposite sides of the film are, respectively 10% and 4% butanol by weight. The –6 2 diffusivity of butanol in water solution is 5.9 × 10 cm /sec. The densities of 10% and 4% butanol solutions at 20°°C may be taken as 0.971 and 0.992 g/cc respectively. Molecular weight of Butanol (C4 H 9 OH) is 74, and that of water 18. Sol. : For steady state unidirectional diffusion, NA =

DAB z

(xA – xA ) C

1

2

xB‚ lm

ρ where C is the average molar density = M avg

 

Conversion from weight fraction the Mole fraction : (0.1/74) xA = (0.1/74 + 0.9/18) = 0.026 1 (0.04/74) xA = (0.04/74 + 0.96/18) = 0.010 2 Average molecular weight at 1 and 2 : 1 M1 = (0.1/74 + 0.8/18) = 19.47 kg/k mole 1 M2 = (0.04/74 + 0.96/18) = 18.56 kg/k mole (ρ1/M1 + ρ2/M2) ρ 2 Mavg = =

0.971/19.47 + 0.992/18.56 2

= 0.0517 g mole/cm3 = 51.7 k mole/m3

Principles of Mass Transfer Operations − I (Vol. − I)

2.90

Fundamentals of Diffusion Mass Transfer

xB – xB (1 – xA ) – (1 – xA ) 2 1 2 1 xB,lm = ln (x /x ) = 1 – x B2 B1 A2  ln 1 – x 



xB,lm =

A1

(1 – 0.01) – (1 – 0.026) 1 – 0.01 ln 1 – 0.026  

i.e.

0.016 = 0.163 = 0.982

Therefore,

DAB = 2

NA

=



ρ M avg

(xA – xA ) 1

2

xB, lm

5.9 × 10–6 × 10–4 × 51.7 (0.026 – 0.010) × 0.982 0.1 × 10–2

= 4.97 × 10–7 = 1.789

g mole m2 . hr.

= 1.789 × 74 = 132.4

k mole m2 sec

g m2.hr.

g m2.hr.

(Ans.)

(45) Carbon monoxide gas (CO) is diffusing at steady state through a tube 0.15 m long with a 0.015 m diameter containing nitrogen at 100oC. The total pressure is constant at 101.32 kPa. The partial pressure of CO at one end is 655 mm Hg and 85 mm Hg at the other. Calculate mass flow rate of CO through the tube. –4

2

Sol. : From literature, 8 × 10 m /s at 100°C and 101.32 kPa o

Temperature T = 373 K Convert pressures into S.I. Units by dividing by 760 mm Hg and multiplying by 101325 Pascals. Use formula for ideal gases in equimolar counter-diffusion : * JAz =

DAB (pA – pA ) 1

2

RT (z2 – z1)

=

101325 3.18 × 10–5 (655 – 85) × 760 8314 × 373 (0.15)

= 5.1951 × 10–6 Net cross-sectional area of tube for molar flowrate : π 2 π AX = 4 di = 4 (0.015)2 = 1.76715 × 10–4 m2 Net molecular weight of carbon monoxide for mass flowrate : 12 + 16 = 28 kg/k mole Mass flowrate = 5.1951 × 10 CO = 2.57 × 10–8 kg/s

–6

× 1.76715 × 10

–4

× 28 = 2.5705 × 10

–8

so mass flow rate of (Ans.)

Principles of Mass Transfer Operations − I (Vol. − I)

2.91

Fundamentals of Diffusion Mass Transfer

(46) Helium and air are contained in a conduit 7 mm in diameter and 0.08 m long at 44°C and atmospheric pressure. The partial pressure of helium at one end of the tube is 0.075 atm and 0.030 atm at the other. Calculate the following for steady-state Equimolal counter diffusion : (a) Molar flux of helium; (b) Molar flux of air; (c) Partial pressure of helium at the halfway point of the conduit. –4

2

Data : From literature, DAB = 0.765 × 10 m /s at 44°C and 101.32 kPa, Sol. : Temperature, T = 317 K. Convert pressures into SI Units by multiplying by 101325 Pascals. Use formula for ideal gases in equimolar conterdiffusion for helium flux : *

JAz =

DAB (pA – pA ) 1

2

= 101324 ×

RT (z2 – z1)

7.65 × 10–5 (0.075 – 0.030) = 1.654 × 10–4 8314 × 317 (0.08)

–6

2

So Molar flux of helium = 1.65 × 10 k mole/m s

(Ans.)

Air is the only other gas present, so pB1 = Ptotal – pA1 = 1 – 0.075 = 0.925 atm. Similarly, pB2 = Ptotal – pA = 1 – 0.030 = 0.970 atm. Use formula for ideal gases in 2

Equimolar conterdiffusion for air, remembering that DBA = DAB : *

JBz =

DBA (pB – pB ) 1

RT (z2 – z1)

2

= 101325 ×

7.65 × 10–5 (0.925 – 0.970) = –1.654 × 10–6 8314 × 317 (0.08)

–6

2

So Molar flux of helium = –1.65 ×10 k mole/m s Negative value indicates opposite direction to helium flux.

(Ans.)

For partial pressure, re-arrange the equimolar conter-diffusion formula, for (z2 – z1) = 0.04 m and pA = 0.075 atm. : 1

*

pA

2

= pA – 1

JAz RT (z2 – z1) DAB

= (0.075 × 101325) –

1.654 × 10–6 × 8314 × 317 (0.04) 7.65 × 10–5

= 7599.375 Thus partial pressure of helium at the halfway point of the conduit = 0.052 atm. (Ans.) (47) Carbon dioxide is diffusing through air under steady-state conditions with air non-diffusing since it is insoluble in one boundary. The total pressure is 101 320 Pa and the temperature is 3°C. The partial pressure of CO2 at one point is 15 625 Pa and at the other point 3 cm away it is 5978 Pa. (a)

Calculate the flux of CO2.

(b) Do the same as (a) but assume that N2 also diffuses; i.e., both boundaries are permeable to both gases and the flux is Equimolal counter diffusion. In which case is the flux greater ?

Principles of Mass Transfer Operations − I (Vol. − I)

2.92

Fundamentals of Diffusion Mass Transfer

–4

2

Data : From literature, DAB = 0.142 × 10 m /s at 3°C and 101.32 kPa Sol. : Use formula for one ideal gas diffusing through a stagnant layer of a second : DAB Ptotal (pA – pA ) pB – pB 1 2 1 2 where, P = NA = BM RT (z – z ) P p 2

1



   B2

BM

ln p

B1

Temperature T = 276 K. Converting pressures into SI Units : pA = 15 625, pA = 5 978 Pascals 1

2

Air is the only other gas present, so pB1 = Ptotal – pA = 101 320 – 15 625 = 85 695 Pa. 1

Similarly pB2 = Ptotal – pA = 101 320 - 5 978 = 95342 Pa. 2

85695 – 95342 PBM = ln (85695/95342) = 90432.758 DAB Ptotal (pA – pA ) NA =

=

1

2

– RT (z2 – z1) PBM 1.42 × 10–5 × 101320 (15625 – 5978) 8314 × 276 (0.03) 90432.758

= 2.2295113 × 10–6 –6

2

Thus, flux of CO2 through air stagnant layer is 2.2295 × 10 k mole/m s

(Ans.)

Using formula for ideal gases in Equimolar conterdiffusion : ' = JAz

DAB (pA – pA ) 1

RT (z2 – z1)

2

=

1.42 × 10–5 (15625 – 5978) = 1.9899413 × 10–6 8314 × 276 (0.03) –6

2

i.e. Equimolar conter-diffusion of CO2 through air is 1.9899 × 10 k mole/m s, LOWER. (48) Water is flowing in a covered irrigation ditch below ground. The ditch is vented to the atmosphere by vertical pipes 0.0254 m long. The outside air can be assumed to be dry and of 101 325 Pascals absolute pressure. Assume that the vapour partial pressure at the water surface is saturated at the air temperature. Calculate the how much greater the evapouration loss of water is on a summer day at 25°C than for a winter day at 0°C. Sol. : Using experimental data for diffusivity, the two different values for the summer's day –4

2

–4

2

and the winter's day are easily found : DAB = 0.260 × 10 m /s, DAB = 0.220 × 10 m /s. 1

2

Using steam tables, the saturated vapour pressure of water for days (1) and (2) are also easily found : pA = 3166, pA = 611.2 Pascals 11

12

Using the formula for one ideal gas diffusing through a stagnant layer of a second : DAB Ptotal Ptotal – pA2 NA = RT (z – z ) ln P  2 1 total – pA



1



Principles of Mass Transfer Operations − I (Vol. − I)

NA

1

∴

NA

2

2.93

Fundamentals of Diffusion Mass Transfer

 Ptotal – pA2  P – p   total A11

DAB Ptotal 1

= RT (z – z ) ln D 1 2 1 AB Ptotal 2

 Ptotal – pA2  ln P –p   total A12

RT2 (z2 – z1)

 Ptotal – pA2  P – p  DAB T2  total A11 1 = D ln AB2 T1  Ptotal – pA2  ln P –p   total A12 Inserting Ptotal = 101 325 Pa, PA = 0, and the other information : 2

 101325  0.26 × 10 × 273 101325 – 3166 = ln = 5.68 –4 101325 0.22 × 10 × 298   ln 101325 – 611.2  

NA

–4

1

NA

2

(Ans.)

Thus, the evaporation loss of water is 5.68 times greater for summer day than winter day. (49) A drop of liquid toluene is kept at a uniform temperature of 298.9 K and is suspended in air by a fine wire. The initial radius of the drop is 3.00 mm and the density of 3

liquid toluene is 866 kg/m . Antoine coefficient A B C Toluene 6.95464 1424.255 219.482 (a) Derive an equation to predict the time tF for the drop to evapourate completely in a large volume of still air. Show all steps. (b) Calculate the time in seconds for complete evaporation. – Sol. : We must return to the Fick's Law formula for a constant number of moles N A of A from a 2

sphere (area = 4pr ) through stagnant B pA  –DAB dpA  NA 1 – P  = RT dr total  –

where,

NA =

NA 4r2 –

– RTNA dr dPA ∴ = Ptotal P 2 4 π DAB r total – A Integrating with limits of pA at r2 and pA at r1 gives : 2

–

–RTNA –1   4π DAB  r r –

∴

RTNA 4π DAB Ptotal

1

pA 2

r2

= Ptotal 1

[–ln (P

total

P –p  1 – 1  = ln  total A2 P –p r1 r2  total A1

]

– pA)

pA 1

Principles of Mass Transfer Operations − I (Vol. − I)

2.94

Fundamentals of Diffusion Mass Transfer

As r1 << r2, then 1/r2 >> 0 : –

pA – pA Ptotal – pA2 1 2 = ln P  = – – p  total A1 PBM

RTNA 4π1 DAB Ptotal –

DAB Ptotal (pA – pA )

NA

∴

1

=

2

2

– RTPBM1

4r1

= NA

1

the flux at the surface. Now consider the rate of change of the number of molecules in the sphere (evapouration) to be equal to their diffusion into the outside world : In time dt, radius of sphere changes by – dr, 2

volume of sphere changes by – 4p r .dr 2

mass of sphere changes by – 4prAr .dr flux of solute to air = NA 1

2

molecular flow to air = 4p r NA

1

2

mass flow to air = 4p r NA MA = evapouration rate 1

4 r2 M A

∴

DAB Ptotal (pA – pA ) 1

2

RTPBM1

dr = – 4 π θA r2 dt

MA DAB Ptotal (pA – pA ) 1

∴

2

ρA RTPBM

dt = –r · dr

This time, the limits are r = r1 at t = 0 and r = 0 at t = tF : tF

MA DAB Ptotal (pA – pA ) 1

∴

0

⌠ dt = – ⌠ r · dr ⌡ ⌡

2

ρA RTPBM

0

2

MA DAB Ptotal (pA – pA ) 1

∴

r1

0 r1 2 r2 –1 tF = –  2  = 2 0 – r1 = 2  r1

2

ρA RTPBM 2

∴ tF

=

r1 ρA RTPBM 2MA DAB Ptotal

(p

A1

– pA

2

)

Using the Antoine data for toluene : B log10 (pA ) = A – T + C 1

where pA is vapour pressure in mm of mercury and T is temperature in oC. 1

1424.255 log10 (pA ) = 6.95464 – (25.9 + 219.482) = 1.1504

∴

1

pA = 1885 Pa 1

pB = 101325 – 1885 = 99 440 Pa 1

pA = 0 (from data) 2

(Ans.)

Principles of Mass Transfer Operations − I (Vol. − I)

2.95

Fundamentals of Diffusion Mass Transfer

pB = 101325 – 0 = 101 325 Pa 2

MA = 6 (12) + 8 (1) = 80 kg/k mole, 3

ρA = 866 kg/m (from data) 2

r1 ρA RTPBM

(0.003)2 866 × 8314 × 298.9 × 100379.56 = tF = 2M D P 2 × 80 × 8.6 × 10–6 × 101325 (1885 – 0) A AB total (pA1 – pA2) = 7398 seconds. Therefore time for complete evaporation = 7398 seconds (Ans.) (50) A pool of liquid benzene diffuses at steady state through a stagnant layer of air in a conduit 2.00 m long at 25°C and a total absolute pressure of 101 320 Pa. The air is saturated with benzene at the base and is assumed benzene-free at the top. The conduit has an equilateral triangular cross-section, the side uniformly tapering from 1.00 meters at the base to 0.500 m at the top. Calculate rate of benzene loss to the air. Antoine coefficient A B C Benzene 6.90565 1211.033 220.790 Sol. : For an equilateral triangle, the formula for area (where u = length of one side) is : 1 u2 A=2 ×u×u = 2 Returning to the Fick's Law formula for a constant number of moles .A of A through a triangle of stagnant B : pA  –DAB dpA  NA 1 – P  = RT dz total  –

where

NA =

2 NA u2

–

dpA – RTNA 2 dz = P ∴ 2 total DAB u Ptotal – pA Before limits are imposed, it must be remembered that u is a function of z, as the size of triangle uniformly tapers with distance along the duct : u2 – u1 u = u1 + z – z (z – z1) 2 1 u2 – u1 u2 – u1 = u1 + z – z z – z – z z 1 2 1 2 1 du dz

∴

u2 – u1 u2 – u1 u2 – u1 = 0 + z – z (1) – z – z (0) = z – z 2 1 2 1 2 1

z2 – z1 dz = u – u du 2 1

∴

We can now substitute z for u in the expression, and integrate with limits of pA at 1

triangle of side u1 and pA when the triangle is of side u2 : 2

–

∴

RTNA z2 – z1 DAB u2 – u1

u2

⌠ ⌡ u1

2 u2 du = Ptotal

pA 2

dpA

⌠ P –p ⌡ total A pA 1

Principles of Mass Transfer Operations − I (Vol. − I) –

Fundamentals of Diffusion Mass Transfer

u

RTNA z2 – z1 –2 2   DAB Ptotal u2 – u1  u u

∴

2.96

1

–

∴

2RTNA z2 – z1 DAB Ptotal u2 – u1

∴

NA

–

∴

pA 2

= – [ln (Ptotal – pA)]p

A1

P –p  1 – 1  = ln  total A1 P –p u2 u1  total A2 P –p u1 – u2 = DAB Ptotal u2 – u1 ln  total A1 u u  P – p  2RT z2 – z1  1 2  total A2 DAB Ptotal u1u2 Ptotal – pA2 – NA = ln P – p  2RT z2 – z1  total A1

Using the Antoine data for benzene : B log10 (pA ) = A – T + C 1

where pA is vapour pressure in mm of mercury and T is temperature in oC 1

1211.033 log10 (pA ) = 6.90565 – (25 + 220.790) = 1.9785456

∴

1

pA = 12 689.619 Pa 1

pB = 101320 – 12689.619 = 88 630.381 Pa 1

pA = 0 (from data) 2

pB = 101320 – 0 = 101 320 Pa 2

–

NA =

DAB Ptotal u1 u2 2RT z2 – z1

Ptotal – pA2 ln P  total – pA 

 0.0962 × 10 × 101320 1 × 0.5  101320  = 2 ln 88630.381 2 × 8314 × 298 1

–4

= 6.58 × 10–9 k mole/m2 s –9

2

Thus rate of benzene loss to the air = 6.58 × 10 k mole/m s

(Ans.)

(51) A solution of ammonia in water at 5°C and 2.5 mm thick is in contact at one surface with an organic liquid at this interface. The concentrations of ammonia in the organic and bulk water phase are both held constant, and aqueous concentrations of ammonia are 10% and 2.0% w/w at the respective interfaces (giving solution densities 3

of 991.7 and 961.7 kg/m respectively). Water is soluble in the organic phase. The –9

2

diffusion coefficient of NH3 in water is 1.24 × 10 m /s. Calculate the steady state flux of ammonia. Sol. : As water can travel through the film as well as ammonia (and in the opposite direction) we have the rare (but not unknown) phenomena of Equimolar conterdiffusion in the liquid phase. *

JAz

DAB (CA – CA ) =

1

z2 – z1

2

Principles of Mass Transfer Operations − I (Vol. − I)

2.97

Fundamentals of Diffusion Mass Transfer

To find the concentrations, we must take the weight fractions and densities from the question, and also the molecular weights (MA = 17, MW = 18 kg/k mole) : CA

1

CA

2

*

JAz

∴

(0.1/17) 91.7 = (0.1/17) + (0.9/18) × 17 = 6.1405573 k mole/m3 (0.02/17) 961.7 = (0.02/17) + (0.98/18) × 17 = 1.196 k mole/m3 1.24 × 10–9 (6.1405573 – 1.1965577) = 0.0025 –6 = 2.452 × 10 k mole/m2 s. –6

2

(Ans.) Thus flux the steady state flux of ammonia is 2.452 × 10 k mole/m s. (52) Molecules of hydrogen chloride (HCl) diffuse through a thin, static film of water 4.0 mm thick at 10°C. The concentration of HCl at one boundary of the film is 12.0% 3

w/w HCl (density = 1060.7 kg/m ), and at the other boundary is 6.0% w/w HCl (density 3

–9

2

= 1030.3 kg/m ). The diffusion coefficient of HCl in water is 2.5 × 10 m /s. Assuming steady state and one boundary impermeable to water, calculate the flux of HCl using both exact and dilute formulae. Does the dilute formula approximate well to the exact one here ? Sol. : The exact formula for diffusion of A through a stagnant layer of liquid B in mole fractions is : DAB Cav NA = x (z – z ) (xA – xA ) 1 2 BM 2 1 xB – xB where,

xBM =

2

1

xB2 ln x  

B1

ρ2  1  ρ1 and Cav = 2 M + M  2  1



The molecular weights are MHCl = 35.6 and Mwater = 18 kg/k mole (0.12/36.5) xA = (0/12/36.5) + (0.88/18) 1 = 0.0630 ∴ xB = 1 – 0.0630 = 0.936 1

xA

2

∴

xB

2

(0.06/36.5) = (0.06/36.5) + (0.94/18) = 0.0305 = 1 – 0.0305 = 0.9694

0.9694 – 0.9369895 xBM = ln (0.9694/0.9369) = 0.9531 The molecular weights of the two solutions can now be found as follows : 1 k mole of solution at point 1 has (0.0630 × 36.5) = 2.2998 kg of HCl 1 k mole of solution at point 1 has (0.9369 × 18) = 16.865 kg of H2O 1 k mole at point 1 weights (16.86581 1+ 2.299) = 19.165 kg M1 = 19.165 kg/K mole 1 k mole of solution at point 2 has (0.0305170 × 36.5) = 1.1138705 kg of HCl 1 k mole of solution at point 2 has (0.9694 × 18) = 17.45 kg of H2O 1 k mole at point 2 weights (17.450694 + 1.1138705) = 18.56 kg

Principles of Mass Transfer Operations − I (Vol. − I)

2.98

Fundamentals of Diffusion Mass Transfer

M2 = 18.564565 kg/k mole 1 1060.7 1030.3 Cav = 2 190165694 + 18.56  = 55.42 k mole/m3   –9 2.5 × 10 × 55.420 NA = (0.0630 – 0.030) 0.953 (0.004) –6 = 1.1808423 × 10 k mole/m3 s

∴ –6

2

i.e. flux is 1.181 × 10 k mole/m s Now compare this result to that of the simplified formula DAB NA = z – z (CA – CA ) 1 2 2 1 3

CA = 0.0630 × 1060.7 = 66.83 k mole/m 1

3

CA = 0.0305 × 1030.3 = 31.44 k mole/m 2

NA =

∴ -6

2.5 × 10–9 (66.835 – 31.44) 0.004 = 2.212 × 10–6 k mole/m2 s

2

i.e. flux is 2.212 ×10 k mole/m s, nearly TWICE the flux if calculated accurately. (Ans.) (53) Predict the diffusivity of methanol in water at 20°C by : (a) Using the Wilke-Chang equation; (b) Extrapolating existing literature values to the required temperature. Sol. : (a) First the Wilke-Chang formula is : Molecular weight of water MB = 18 kg/k mole –6

From steam tables, water viscosity @ 20°C = 1002 × 10 kg/m.s Absolute temperature of water = 20 + 273 = 293 K –3

3

For water : φ = 2.6 VA = [(14.8) + 4(3.7) + 7.4] × 10 = 0.037 m /k mole 293 DAB = 1.173 × 10–18 2.6 × 18 1.002 × 10–3 (0.037)0.6 = 1.6962 × 10–9 m2/s –9

2

Thus diffusivity of methanol in water at 20°C = 1.70 × 10 m /s Using the reduced version of the Wilke-Chang formula : T DAB ∝ µ B DAB ∴

DAB

T 2 µ1 = µ T 2 1

DAB

=

2 1

∴

2

µ1  µ   2

(Ans.)

T2 D T  AB1  1 –9

2

Literature gives the diffusivity of methanol in water as 1.26 x 10 m /s at 288 K. –6

Water viscosity at 288 K (15°C) is 1136 x 10 kg/m.s (from steam tables) –3 1.136 × 10  293 1.26 × 10–9 DAB =   2 1.002 × 10–3 288 = 1.4533 × 10–9 m2/s o –9 2 Thus diffusivity of methanol in water at 20 C = 1.45 × 10 m /s

(Ans.)

Principles of Mass Transfer Operations − I (Vol. − I)

2.99

Fundamentals of Diffusion Mass Transfer

(54) Use the Wilke-Chang equation to predict the diffusivity of dilute acetic acid (CH3COOH) in water at 9.7 and 25°C, and compare predictions with literature values. Sol. : First the Wilke-Chang formula is : T DAB = 1.173 × 10–18 µMB 0.6 µB V A Molecular weight of water MB = 18 kg/k mole Absolute temperature of water = 9.7 + 273 = 282.7 K For water : φ = 2.6 –3

3

VA = [2 (14.8) + 5 (3.7) + 7.4 + 12.0] × 10 = 0.0675 m /k mole Value for viscosity at 9.7 oC needs to be interpolated from values for 5 and 10oC (1501 –6

and 1300 ×10 kg/m.s respectively) : 1300 – 1501 µwater =  10 – 5 (9.7 – 5) + 1501 × 10–6   = 1.312 × 10–6 kg/m.s DAB = 1.173 × 10–16 2.6 × 18 282.7 = 8.7138 × 10–10 m2/s 1.31206 × 10–3 (0.0675)0.6 o

Thus the diffusivity of dilute acetic acid in water at 9.7 C = 8.71 × 10 Using the reduced version of the Wilke-Chang formula : T DAB ∝ µ B DAB ∴

DAB

T 2 µ1 = µ T 2 1

DAB

=

2 1

∴

2

µ1 µ   2

–10

2

m /s

(Ans.)

T2 D T  AB1  1 –6

Water viscosity at 298 K (25°C) is 890 × 10 kg/m.s (from steam tables)

1.31206 × 10   298  8.713 × 10–10  890 × 10–6  282.7   –3

DAB

2

=

= 1.354 × 10–9 m2/s o

–9

2

(Ans.) Thus the diffusivity of dilute acetic acid in water at 25 C = 1.35 × 10 m /s (55) A 7 mm thick layer of 10.0% w/w gelatin in water at 5°C separates two urea solutions. At the gelatine surface, one solution has a concentration of 18 grammes per liter, and 2

the other 2.0 g/l. What is the steady state flux of urea in k mole/m s ? Sol. : Urea has a molecular weight of 60.1 kg/k mole, and a diffusivity of 0.542 × 10 10% gelatin in water at 5°C. Concentrations are converted : 0.018 CA = = 0.2995 k mole/m2s 0.001 × 60.1 1 CA

2

=

0.002 = 0.0332 k mole/m2s 0.001 × 60.1

–9

2

m /s in

Principles of Mass Transfer Operations − I (Vol. − I)

2.100

Fundamentals of Diffusion Mass Transfer

As urea is dilute, xBM >> 1, so we use equimolar counterdiffusion formula : DAB (CA – CA ) 1 2 NA = z –z 2

=

1

(0.542 × 10–9) (0.2995 – 0.0332) 0.007

= 2.061 × 10–8 k mole/m2s –8

2

Thus the steady state flux of urea = is 2.06 × 10 k mole/m s (Ans.) (56) A flat circular vulcanized rubber plug of equal diameter and thickness of 25.4 mm seals the top of a container holding carbon dioxide gas at 1.6 atmospheres and 298 K. Calculate the rate at which the CO2 leaks through the seal to the atmosphere, assuming that the air outside has a negligible CO2 concentration. Sol. : Use the derivation of Fick's Law for diffusion through an impermeable solid slab : –

NA A where, ∴ ∴

DAB (CA – CA ) 2

z2 – z1

SpA CA = 22.414 and pA is in atmosphere 0.90 × 1.6 CA = 22.414 = 0.0642455 k mole/m3 1 CA = 0 2

2

DAB (CA – CA )

–

∴

1

= NA =

NA = =

1

2

z2 – z1

r1

× 4

0.1 × 10–9 (0.0642455 – 0) π (0.0254)2 × 0.254 4

= 1.409 × 10–13 k mole/s i.e. leakage is 1.410 × 10

–13

k mole CO2 per second.

(Ans.)

(57) A loosely packed bed of sand 0.9144 meters thick separates an otherwise open water stream at 25°C and the atmosphere at a total pressure of 101320 Pascals. Assuming that atmospheric air is dry and the air at the water's surface is saturated, what is the steady state rate of diffusion through the bed if its void fraction e is 0.50 ? 0.6 Void fraction ∈ 0.2 0.4 2.0 1.75 1.65 Tortuosity τ Sol. : Use the derivation of Fick's Law for diffusion through a porous solid : DAB (pA – pA ) 1 2 NA = RT (z – z ) 2

1

Interpolated tortuosity gives τ = 1.60 for voidage ∈ = 0.5, At 298 K (25oC), the diffusivity –4

2

of water in air is 0.260 × 10 m /s and its saturated vapour pressure is 0.03166 bar : ∴

NA =

0.5 [0.260 × 10–4] (3166 – 0) 1.60 × 8314 × 298 (0.9144)

= 1.1354 × 10–8 k mole/m2s

(Ans.)

Principles of Mass Transfer Operations

2.101

Diffusion Mass Transfer

(58) A binary system consisting of carbon dioxide and nitrogen at 100 kPa and 0°C undergoes equimolar counter diffusion. The mole fraction of nitrogen at point A is 0.98 and that at point B, which is 4 meters away from point A is 0.8. (a)

What is the molar flux of nitrogen in kmol m–2 h–1 ?

(b)

What is the net mass flux, kg m–2 h–1 ?

(c)

At what velocity would an observer have to move from one point to the other so that the net mass flux, relative to the observer, would be zero ?

(d)

At what velocity would the observer have to move so that, relative to the observer, the nitrogen is stationary ?

(e)

What would be the molar flux of carbon dioxide relative to the observer under condition (d) ?

Sol. : Subject : Equimolar diffusion (EMD) of CO2 and N2 at 100 kPa and 0°C. Given : Compositions of mixture at specified locations. To find : Molar flux of N2, net mass flux, molar flux of CO2, velocities of an observer. Let nitrogen be species A and carbon dioxide be species B. DAB = 0.001 (273.15)1.75 (1/44 + 1/28)0.5/(26.71/3 + 18.51/3)2 cm2/s = 0.14 cm2/s = 0.14 × 10–4 m2/s Molar density of mixture = ρ = 1 kmol/22.71 m3 = 44.03 mol/m3. (a) The molar flux of nitrogen under EMD conditions is given by, NA = JA = – DAB ρ (dy/dz) = – 0.14 × 10–4 × 44.03 × (0.8 – 0.98)/4 = 0.2774 × 10–4 mol m–2 s-1 = 0.9981 × 10–4 kmol m–2 h–1

(Ans.)

(b) The mass flux of nitrogen is equal to 28 × 0.9981 × 10–4 kg m–2 h–1 (from A to B). The mass flux of carbon dioxide is equal to 44 × 0.9981 × 10–4 kg m–2 h–1 (from B to A). The net mass flux is therefore equal to 15.97 × 10–4 kg m–2 h–1 (from B to A). (Ans.) (c) As there is a composition gradient in the system, the velocities depend on position. The velocities of the species and the mixture are given by the following equations : Molar flux of species i = (Velocity of species i) × (Molar concentration of i) Molar flux of mixture = (Molar average velocity) × (Molar concentration of mixture) Mass flux of mixture = (Mass average velocity) × (Mass concentration of mixture) Molar average velocity = ∑i (Mole fraction of species i) × (Velocity of species i) Mass average velocity = ∑i (Mass fraction of species i) × (Velocity of species i)

Principles of Mass Transfer Operations

2.102

Diffusion Mass Transfer

A : Nitrogen, B = Carbon dioxide Point A Distance from point A, m

Point B

0

Mole fraction, yA

1

2

3

4

0.980

0.935

0.890

0.845

0.800

Mass fraction, yAM

0.9689

0.9015

0.8374

0.7762

0.7179

Molar concentration of A, kmol/m3

0.0431

0.0412

0.0392

0.0372

0.0352

Mass concentration of A, kmol/m3

1.2082

1.1527

1.0972

1.0417

0.9863

Velocity of A, m/h

0.0023

0.0024

0.0025

0.0027

0.0028

Molar concentration of B, kmol/m3

0.0009

0.0029

0.0048

0.0068

0.0088

Mass concentration of B, kmol/m3

0.0387

0.1259

0.2131

0.3003

0.3875

– 0.1133

– 0.0349

– 0.0206

– 0.0146

–0.0113

Molar average velocity, m/h

0

0

0

0

0

Mass average velocity, m/h

– 1.281E-03

– 1.249E-03

– 1.219E-03

– 1.190E-03

– 1.162E-03

(c)

Molar flux of B relative to A, kmol m–2 h–1

– 1.018E-04

– 1.067E-04

– 1.121E-04

– 1.181E-04

– 1.248E-04

(e)

Velocity of B, m/h

(d)

The vectors in the direction from A to B are positive and those in the direction from B to A are negative.

(59) Diffusivities for vapours in air can be determined by measuring the rate of evaporation of a liquid contained in a vertical glass tube. For a tube 0.25 cm in diameter filled with n-octane at 20°C, estimate the expected rate of decrease of the liquid level when the meniscus is 1 cm from the top. The published diffusivity for n-octane in air at 0°C is 0.0506 cm2 s–1. The vapour pressure and density of n-octane at 20°C are 1.395 kPa and 0.7022 g cm–3, respectively. Would there be any advantage in using a tube with larger diameter ? Sol. : Subject : Evapouration of n-octane from a small-diameter vertical tube into air at 20°C. Given : Tube diameter, diffusivity for n-octane air at 0°C, density and vapour pressure of n-octane at 20°C. To find : Rate of evapouration (in terms of the rate of decrease of the liquid level). Assumptions : Unimolecular diffusion in ideal gas at 101.3 kPa, the mole fraction of n-octane in the air at the liquid-air interface is given by Raoult's law, and the mole fraction of n-octane at the top of the tube is zero. We can use the following equation, dz/dt = (DAB ρ/z) (ML/ρL) ln [(1 – yA0)/(1 – yAZ)] where, DAB = 0.0506 × (293.15/273.15) 1.75 = 0.0573 cm2/s = 5.73 × 10–6 m2/s ρ = 1.013 × 105/(8.314 × 293.15) = 41.57 mol/m3 z = 0.01 m ML = 114 g/mol ρL = 702.2 kg/m3

Principles of Mass Transfer Operations

2.103

Diffusion Mass Transfer

yA0 = 0 yAZ = 1.395/101.325 = 0.01377 Thus, inserting known values, we get, dz/dt = 0.0536 × 10–6 m/s = 1.93 × 10–2 cm/h Thus, expected rate of decrease of the liquid level = 1.93 × 10–2 cm/h

(Ans.)

(60) An open tank, 4 m in diameter and containing toluene at 25°C, is exposed to air at 100 kPa in such a manner that the surface of the liquid is covered with a stagnant air film estimated to be 5 mm thick. The concentration of toulene beyond the stagnant film is negligible. The vapour pressure and density of liquid toulene at 25°C are 3.8 kPa and 862 kg/m3, respectively. The diffusivity of toulene in air at 0°C is 0.071 cm2/s. Estimate the loss of toulene per day from this tank. Sol. : Subject : Evapouration of toulene (A) at 25°C and 100 kPa from an open tank. The molecules must travel through a stagnant air layer of constant thickness. Given : Tank diameter, thickness of stagnant layer of air, density and vapour pressure of toulene, diffusivity of toulene in air at 0°C. To find : Loss of toulene by evapouration. Assumptions : Steady-stadte diffusion in ideal gas. All mass transfer resistance is in the stagnant layer of air. Raoult's law gives the mole fraction of toulene in the stagnant air at the vapour-liquid interface. The mole fraction of toulene in the air on the other side of the stagnant layer is zero. The molar flux of toulene through a stagnant layer is given by NA = – DAB ρ (dyA/dz)/(1 – yA) Upon substituting the boundary conditions into the integrated form of the equation, we obtain for steady-state Unimolecular diffusion. NA = (DAB ρ/L) ln [(1 – yAL)/(1 – yA0)] The numerical values of the variables in the above equation are given below : DAB = 0.071 × (298.15/273.15)1.75 = 0.0828 cm2/s = 8.28 × 10–6 m2/s ρ = P/(RT) = 105/(8.314 × 298.15) = 40.34 mol/m3 yA0 = 3.8/10 = 0.038 yAL = 0 L = 0.005 m Thus, we obtain, NA = 8.28 × 10–6 × 40.34 × ln (1/0.962)/0.005 = 0.2587 × 10–2 mol m–2 s–1 The cross-sectional area of the tank = 3.14 × (4 m)2/4 = 12.57 m2 The rate of toulene evapouration = 3.25 × 10-2 mol/s = 2810 mol/day = 258821 g/day = 259 kg/day = (259/862) m3/day = 0.3 m3/day So, Loss of toulene = 0.3 m3/day (Ans.) (61) A binary system consisting of carbon dioxide and nitrogen undergoes equimolar counter diffusion. The mole fraction of nitrogen at point A is 0.95 and that at point B, which is 2 meters away from point A, is 0.85. The diffusivity for the nitrogen-carbon dioxide system is 0.14 × 10–4 m2/s and the molar density of the system is 44 mol/m3.

Principles of Mass Transfer Operations

2.104

Diffusion Mass Transfer

What is the molar flux of nitrogen in kmol m–2 h–1 ? What is the net mass flux in kg m-2 h–1 ? What is the velocity of the nitrogen at the mid-point between point A and point B? (d) What is the velocity of the net mass flux at the mid-point between point A and point B ? (e) What is the velocity of the carbon dioxide with reference to the nitrogen at the mid-point between point A and point B ? Sol. : (a) Let A stands for nitrogen and B stands for carbon dioxide. For Equimolar diffusion, the molar flux of A is given by, (a) (b) (c)

NA

dyA = – DAB ρ dz

where, dyA 0.85 – 0.95 DAB = 0.14 × 10–4 m2/s, ρ = 44 mol/m3, dz = 2m – 0 = – 0.05/m Thus, N A = – N B = 0.308 × 10–4 mol/(m2s) = 1.1088 × 10–4 kmol/(m2h)

(Ans.)

(b) The mass flux of nitrogen is NA = 28 × 1.1088 × 10–4 kg/(m2h) = 3.105 × 10–3 kg/(m2h) The mass flux of carbon dioxide is NB = – 44 × 1.1088 × 10–4 kg/(m2h) = – 4.879 × 10–3 kg/(m2h) The net mass flux is N = NA + NB = – 1.774 × 10–3 kg/(m2h)

(Ans.)

(c) At the mid-point between point A and point B, i.e. z = 1 m, the mole fraction of nitrogen is the average of the mole fractions at the two end points, yA = (0.95 + 0.85)/2 = 0.90 The mass fraction of nitrogen at the mid-point between point A and point B is given by, yA = (28 × 0.90)/(28 × 0.90 + 44 × 0.10) = 0.8514 The molar density (i.e. molar concentration) of nitrogen at the mid-point between point A and point B is given by, ρA = ρyA = 44 × 0.9 mol/m3 = 39.6 mol/m3 The molar density (i.e. molar concentration) of carbon dioxide at the mid-point between point A and point B is given by, ρB = ρyB = 44 × 0.1 mol/m3 = 4.4 mol/m3 The velocities of the nitrogen and the carbon dioxide are given by, uA = NA/ρA = 0.0028 m/h uB = NB/ρB = – 0.0252 m/h

(Ans.)

Principles of Mass Transfer Operations

2.105

Diffusion Mass Transfer

(d) The mass average velocity is given by u = yA uA + yB uB = [0.0028 × 0.8514 + (– 0.0252 × 0.1486)] m/h = – 0.001361 m/h (e) The velocity of carbon dioxide relative to nitrogen is given by, uB – uA = [– 0.0252 – (0.0028)] m/h = – 0.0280 m/h (Ans.) (62) A perfectly spherical naphthalene ball having a diameter of 0.01 m is suspended in air at 25°C. The molecular weight and density of naphthalene are 128.17 g/mol and 1150 kg/m3, respectively. The vapour pressure of naphthalene at 25°C is 12.7 Pa and the diffusivity of naphthalene vapour in air at 25°C is 0.06 cm2/s. (a) How many days will it take for this naphthalene ball to complete disappear if the naphthalene evapourates at a steady rate of 7.5 × 10–10 mol/s ? (b) How would you solve this problem if the steady rate of evapouration was not known and you had to rely on the properties given above to get an answer ? Derive the differential equation that is consistent with Fick's law. Sol. : (a) The number of mole of naphthalene contained in the sphere is given by, n = (4πr3/3) ρ/M = (4 × 3.1416 × 0.0053/3) (1150)/128.17 kmol = 4.698 × 10–6 kmol = 4.698 × 10–3 mol At a steady vapourization rate of 7.5 × 10–10 mol/s, the sphere will disappear in t days, where, t = (4.698 ×10–3)/(7.5 × 10–10 × 60 × 60 × 24) = 72.5 days (Ans.) (b) If the vapourization rate is constant, the diffusion flux cannot be constant. This is because the area normal to the diffusion is a function of the distane from the origin, i.e. area = 4πr2, where r is the radius measured from the centre of the sphere. The flux equation for the specified situation (i.e. Unimolecular Diffusion of naphthalene vapour into air) can be written as dn –  dt  – DAB ρ dyA   = (1 – y )  dr  2 4πr  A  dn where, –  dt  is a positive number representing the constant rate of vapourization, its   value can be determined upon integrating the differential equation with the following integration limits : yA = 12.7/101325 at r = 0.005 m (Ans.) yA = 0 at r = ∞ (63) A binary gas mixture at a total pressure of 1.0 atm and 21°C has a molar composition of 30% CO and 70% CO2. The absolute velocities of CO and CO2 are 6.0 m/s and 3 m/s, respectively, all in the direction of the z-axis. (a) Calculate the mass average velocity, v, and the molar average velocity, V, for the mixture. (b) Determine the four fluxes : jCO, z, nCO, z, JCO, z, NCO, z Data : R = 8.2 × 10–5 m3.atm/(mol.K)

Principles of Mass Transfer Operations Sol. : (a)

V =

2.106

Diffusion Mass Transfer

∑ ci vi = ∑ yi vi = (0.3) (6.0) + (0.7) (3.0) = 3.9 m/s c

∑ ρi vi = ∑ wi vi = (0.214) (6.0) + (0.786) (3.0) = 3.642 m/s (Ans.) ρ Moles M Mass (g) Mass fraction CO 0.3 28 8.4 0.214 CO2 0.7 44 30.8 0.786 Total 1.0 – 39.2 1.000 n P 1 atm (b) c = V = RT = = 41.48 mol/m3 (8.2 × 10–5 m3 atm/mol K) (294 K) ρ = cMavg = (41.48 mol/m3) (39.2 g/mol) = 1626.02 g/m3 jCO, z = wCO ρ (vCO – v) = (0.214) (1626.02) (6 – 3.642) = 820.5 g/m2.s JCO, z = yCO c (vCO – V) = (0.3) (41.48) (6 – 3.9) = 26.1 g/m2.s nCO, z = wCO ρ vCO = (0.214) (1626.02) (6) = 2087.8 g/m2.s NCO, z = yCO cvCO = (0.3) (41.48) (6) = 74.7 g/m2.s (Ans.) (64) Ammonia, NH3 (g), diffuses through a 10 cm long tube containing N2 (g) at 1.0 atm and 298 K. The partial pressures of NH3 at the entrance and exit of the tube are 0.1 atm and 0.05 atm, respectively. Calculate DNH3 , N2 and JNH3 . v=

JNH3 , as given by Fick's law, JNH3

Data : MNH3 = 17, MN2 Sol. :

= – DNH3 , N2

dCNH3

dz , can be assumed constant. = 28, R = 82.06 cm3 atm/mol K

Fick's law of diffusion : JNH3

= – DNH3 , N2

dCNH3 dz

Assuming ideal gas law : CNH3 Therefore :

JNH3

=

nNH3

= –

PNH3 = V RT DNH3‚ N2 dPNH3 RT

dz

If JNH3 can be assumed constant : JNH3 (PNH3)2 – (PNH3)1 z2 – z1

= =

DNH3‚ N2 RT DNH3‚ N2 (PNH3)1 – (PNH3)2 RT

z2 – z1

Fuller et al. equation : 10–3 T1.75 M DNH3 ‚ N2

=

1/2

  NH N 3 2  1/3 2  1/3 P (∑v)  NH3 + (∑v)N2   1

1 +M

= 0.2372 cm2/s

MNH3 = 17, MN2 = 28, ∑vNH3 = 14.9, ∑vN2 = 17.9, T = 298 K

(Ans.)

Principles of Mass Transfer Operations

2.107 P =

Diffusion Mass Transfer

(1 + 0.1) + (1 + 0.05) = 1.085 atm 2

(0.2372 cm2/s) (0.1 – 0.05) atm 3 82.06 cm atm (298 K) 10 cm mol K   mol = 4.85 × 10–8 (Ans.) cm2s (65) A flat plug 30 mm thick having an area of 4.0 × 10–4 m2 and made of vulcanized rubber is used for closing an opening in a container. The gas CO2, at 25°C and 2.0 atm pressure is inside the container. Calculate the total leakage of CO2 through the plug to the outside in kgmol CO2/s at steady state. Assume that the partial pressure of CO2 outside the container is zero. Data : For CO2 in vulcanized rubber S = 0.90 m2 STP/m3.atm and DCO2 = 0.11 × 10–9 m2/s Consequently :

Formula : NA = –

JNH3

=

PM ∆ PA , where PM is the permeability of gas A in the solid. 22.4 ∆z 3

3

mSTP  mSTP m2   = 9.9 × 10–11 Sol. : PM = DCO2 S = 0.11 × 10–9 s  0.9 3 2 s m atm/m    m atm 3

 mSTP  9.9 × 10–11  PM ∆PA s m atm (– 2 atm) kgmol  NA = – =– = 2.95 × 10–10 3 m2s 22.4 ∆z  m  22.4 STP  (30 × 10–3 m) kgmol  kgmol kgmol Leakage Rate = NA A = 2.95 × 10–10 (4 × 10–4 m2) = 1.18 × 10–13 s 2s m   (66) Starting with Fick's equation for the diffusion of A through a binary mixture of A and B. NA = – cDAB ∇yA + yA (NA + NB) Derive the following relations, starting the assumptions made in the derivations : (a)

nA = – DAB ∇ρA + wA (nA + nB)

(b)

JA = – DAB ∇cA

(c)

jA = –ρ ρDAB ∇wA (It is important to realize that jA doesn't equal JAMA; why ?)

Sol. : (a) Demonstrate :

nA = – DAB ∇ρA + wA (nA + nB)

Multily each term by MA and assume c is constant : NAMA = – DAB ∇CA MA + yA (NA + NB) MB NAMA = nA CAMA = ρA wA =

ρA cAMA yAMA yAMA = c M +c M =y M +y M = M ρA + ρB A A B B A A B B avg

yAMA = wAMavg yAMA (NA + NB) = wAMavg (NA + NB) = wA (nA + nB)

Principles of Mass Transfer Operations

2.108

Diffusion Mass Transfer

Therefore, nA = – DAB ∇ρA + wA (nA + nB) (b) Demonstrate :

(Ans.)

JA = – DAB ∇cA

Assume constant c : NA = – DAB ∇cA + yA (NA + NB) cA cAvA = – DAB ∇cA + c + c (cAvA + cBvB) A B cAvA = – DAB ∇cA + cAV cA (vA – V) = – DAB ∇cA = JA (c) Demonstrate : From (a) :

(Ans.)

jA = – ρDAB ∇wA nA = – DAB ∇ρA + wA (nA + nB) ρA ρAvA = – ρDAB ∇wA + (ρAvA + ρBvB) ρ ρAvA = – ρDAB ∇wA + ρAv

ρA (vA – v) = – ρDAB ∇wA = jA jA ≠ JAMA because they are measured with respect to different average velocities. (Ans.) (67) Starting with Fick's equation for the diffusion of A through a binary mixture of A and B. Prove : (a) DAB = DBA (b)

JA + JB = 0

(c) NA + NB = cV = DBA Prove that DAB JA = – DAB ∇CA = – cDAB ∇xA JB = – DBA ∇CB = – cDBA ∇xB However, JA + JB = 0, as will be proven in 2.b. Therefore, JA = – JB and – DAB ∇xA = DBA ∇xB. Since, xB = (1 – xA) then dxB = – dxA and ∇xA = – ∇xB. Consequently : DAB ∇xA = DBA ∇xA ⇒ DAB = DBA (b) Demonstrate : JA + JB = 0

Sol. : (a)

(Ans.)

cA (vA – V) + cB (vB – V) = 0 cA [vA – (xAvA + xBvB) + cB [vB – (xAvA + xBvB) = 0 cAvA + cBvB – xAvA (cA + cB) – xBvB (cA + cB) = 0 cAvA + cBvB – xAvAc – xBvBc = 0 cAvA + cBvB – cAvA – cBvB = 0 0 = 0, i.e. JA + JB = 0 (c)

Demonstrate :

(Ans.)

NA + NB = cV NA + NB = cAvA + cBvB = c

cAvA + cBvB = cV c

(Ans.)

Principles of Mass Transfer Operations

2.109

Diffusion Mass Transfer

(68) A gas mixture at a total pressure of 1.5 × 105 Pa and 295 K contains 20% H2, 40% O2, and 40% H2O by volume. The absolute velocities of each species are – 10 m/s, – 2 m/s, and 12 m/s, respectively, all in the direction of the z-axis. (a) Calculate the mass average velocity, v, and the molar average velocity, V, for the mixture. (b) Determine the four fluxes : jO2‚ z , nO2‚ z , JO2‚ z , NO2‚ z . Sol. : (a)

(b)

Moles H2 0.20 O2 0.40 H 2O 0.40 Total 1.00 ∑ ci vi V = = c ∑ ρi vi = v = ρ = Mavg =

M 2.0 32.0 18.0

Mass (g) 0.4 12.4 7.2 20.4

Mass fraction 0.020 0.627 0.353 1.000

∑ yi vi = (0.2) (– 10) + (0.4) (– 2) + (0.4) (12) = 2.0 m/s ∑ wi vi = (0.02) (– 10) + (0.627) (– 2) + (0.353) (12) 2.783 m/s (0.2) (2) + (0.4) (32) + (0.4) (18) = 20.4 g/mol

(Ans.)

P 1.5 × 105 Pa n = 61.16 mol/m3 c = V = RT = (8.314 Pa.m3/mol.K) (295 K) ρ = c.Mavg = (61.16 mol/m3) (20.4 g/mol) = 1247.6 g/m3 jO2‚ z

= ρO2‚ z (vO2 – v) = wO2 ρ (vO2 – v) = (0.627) (1247.6 g/m3)

JO2‚ z

(– 2 – 2.782 m/s) = – 3740.7 g/m2 . s = cO2‚ z (vO2 – V) = yO2 c (vO2 – V) = (0.4) (61.16 mol/m3)

nO2

(– 2 – 2 m/s) = – 97.86 g/m2 . s = ρO2 vO2 = wO2 ρ vO2 = (0.627) (1247.6 g/m3)

NO 2

(– 2 m/s) (kg/1000 g) = – 1.561 kg/m2s = cO2 vO2 = yO2 cvO2 = (0.4) (61.16 mol/m3) (– 2 m/s)

= – 48.93 mol/m2 . s (Ans.) (69) The diffusion coefficient of CO2 in N2 at 25°C and 1 atm is 1.67 × 10–5 m2/s. Use this value to calculate the diffusivity for the system at 125°C and 10 atm using Fueller et. al. correlation. Sol. : Fuller et. al. correlation : 1 1 1/2 10–3 T1.75 M + M  B T1.75  A ⇒ D∝ P DAB = 1/3 1/3 P [(∑ v) + (∑ v) ]2 A

B

1.75

T2 (DAB)2 = (DAB)1 T   1

= 2.77 × 10–6 m2/s

1.75

P1 = 1.67 × 10–5 398 298 P2

1 10 (Ans.)

Principles of Mass Transfer Operations

2.110

Diffusion Mass Transfer

(70) An absorption tower has been proposed to remove selectively two pollutants, hydrogen sulfide and sulfur dioxide, from an exhaust gas stream with molar composition 2.0% H2S, 4.0% SO2 and 94.0% air. The gas mixture is 373 K and 1 atm. using an appropriate empirical correlation, calculate the diffusivity of : (a) Hydrogen sulfide in the gas mixture, (b) Sulfur dioxide in the gas mixture. Sol. : Let A = H2S, B = SO2 and C = air. The correlation of Fuller et al. will be used to estimate DAB. Fuller et. al. correlation is given by : 1 1 1/2 10–3 T1.75 M + M  B  A DAB = 1/3 1/3 P [(∑ v)A + (∑ v)B ]2

(a)

(∑v)A = 2 (1.98) + 17.0 = 20.96

MA = 34 g/mol

(∑v)B = 41.1

MB = 64 g/mol

MC = 29 g/mol (∑v)C = 20.1 The diffusivity of H2S in the gaseous mixture is given by : 1 – yA DA, m = y yC B DAB + DAC

From Fuller et. al. equation, we can calculate as DAB = 0.174 cm2/s, DAC = 0.267 cm2/s DA, m = (b)

1 – 0.02 = 0.261 cm2/s 0.94 0.04 + 0.174 cm2/s 0.267 cm2/s

(Ans.)

The diffusivity of SO2 in the gaseous mixture is given by : DB, m =

1 – yB yC yA DAB + DBC

From Fuller et. al correlation as above we can calculate, DAB = 0.174 cm2/s, DAC = 0.183 cm2/s. 1 – 0.04 DB‚ m = = 0.183 cm2/s 0.94 0.02 + 0.174 cm2/s 0.183 cm2/s

(Ans.)

(71) The solubility of a specific molecule in a polymeric membrane has been measured. The experimental data were obtained by taking a piece of film 1.0 cm × 5.0 cm of thickness 0.0127 cm and exposing it at 30°C to the pure vapour of this solute. The weight gain, measured in equilibrium, was 4.76 × 10–3 g. The solute has a molecular weight of 75.0 g/mol. At 30°C, the vapour pressure of the solute is 76.0 mm Hg. The film weight, free of solute, is 0.0572 g. The film density, free of solute, is 0.9 g/cm3. Sol. : (a) Find the solubility coefficient of the soute in the film. At equilibrium, the solute concentration in the film is : mA/MA (4.76 × 10–3 g)/(75 g/mol) 3 = CA = (1 cm) (5 cm) (0.0127 cm) = 0.001 gmol/cm V CA S' = P A

=

0.001 gmol/cm3 = 0.01 gmol/cm3 atm 0.1 atm

(Ans.)

Principles of Mass Transfer Operations

2.111

Diffusion Mass Transfer

(b) The permeability of this polymeric film to the solute is 3

PM = 7.6 × 10–5 cmSTP / (cm2.s.atm/cm) at 30°C. Calculate DAB at 30°C. 3

DAB

PM = S

=

7.6 ×

PM 3

 cm  2.24 × 104 STP × S' gmol  

=

10–5

cmSTP cm2s atm/cm

3

 cm   2.24 × 104 STP 0.01 gmol gmol   cm3 atm 

= 3.4 × 10–7 cm2/s (Ans.) (c) The solute sits at the bottom of a glass tube at 30°C. To retard evapouration, a polymeric membrane is placed on top of the tube. If the thickness of the polymer membrane is 0.0254 cm and the cross-sectional area is 0.2 cm2, what is the time required for 10 mg of the solute to pass through the barrier ? DABS' ∆P (3.4 × 10–7 cm2/s) (0.01 gmol/cm3atm) (0.1 atm) NA = – = 0.0254 cm ∆z = 1.34 × 10–8 gmol/cm2s Rate = MA ANA = (75 g/mol) (0.2 cm2) (1.34 × 10–8 gmol/cm2s) = 2.7 × 10–7 g/s 0.01 g = 5 × 104 s = 13.9 h (Ans.) t = 2 × 10–7 g/s (72) A 0.15-m-long, 0.015-m-diameter test tube containing ethanol is left open in the laboratory. The level of ethanol is initially 0.1 m below the top. The temperature in the laboratory is 26°C and the atmospheric pressure is 0.987 atm. The vapour pressure of ethanol is 0.08 atm. If the concentration of ethanol in the air outside the test tube is negligible and the concentration of ethanol near the liquid surface can be calculated using Dalton's law, determine : (a) An expression for the concentration profile of ethanol in air inside the test tube if the liquid level is held constant. (b) An expression for the instantaneous molar flux of ethanol. (c) The time required for the level of ethanol to decrease by 0.005 m if the evapouration rate does not change with time. Remember that the decrease in ethanol level will be equal to its rate of transfer to the gas phase : dz = CAvA = CA dt NA | interface Data : Density of ethanol = 784 kg/m3 Sol. : 0.15m

0.1m 0.15m

Ethanol is evaporating

Fig. 2.28 : Evaporation of Ethanol from test tube

Principles of Mass Transfer Operations

2.112

Diffusion Mass Transfer

T = 26°C z

P = 0.987 atm v

PEtoh = 0.08 atm (a)

x

Writing balance in the gas phase,

Assume Air (B) is not soluble in ethanol (A). NA, Z

∴

dyA = – C DAB dz + yA (NAz + NBz ) C · DAB dyA NAZ = – 1 – y dz A

∂cA ∂t

+

∂NA‚ x + ∂NA‚ y + ∂NA‚ z = R A ∂y ∂z   ∂x

Steady state, no reaction, diffusion in z-direction only, dNA‚ z = 0 dz NA‚ z = k1 ∴

Equation (2) becomes, C · DAB dyA – 1–y · dz A

= k1

y

Fig. 2.29

… (2) … (3)

… (4) … (5)

… (6)

k1 (1 – yA) C · DAB

∴

dyA – dz

=

or

dyA 1 – yA

k1 ' = –C·D · dz = k1 dz AB '

– ln (1 – yA) = k1 z + k1

∴

Boundry Conditions : v

PA

(1)

0.08 Z = 0, yA = v = 0.08 + 0.987 = 0.075 PA + Pair

∴

– ln (1 – 0.075) = k2

∴

K2 = 0.0779

(2) z = 0.1, yA = 0 ∴ ∴

'

– ln (1) = k1 (0.1) + 0.0779 '

k1

= – 0.779

Therefore, – ln (1 – yA) = 0.779z + 0.0779 ∴

1 – yA = exp (0.779z – 0.0779) = 0.925 e0.779z

∴

yA = 1 – 0.925 e– 0.779z

(Ans.)

Principles of Mass Transfer Operations

2.113

Diffusion Mass Transfer

(b) The instantaneous molar flux of ethanol at the gas/liquid interface can be obtained as : NA |

z=0

dyA dz z = 0

∴

Therefore, (c)

yB2 = 1 – yA2

C · DAB dyA = – 1–y · dz A z=0 = – (0.925) (0.779)0.779z |

z=0

= – 0.7206 (0.08 + 0.987) P C = RT = (0.08206) (26 + 273) = 0.044 kgmole/m3 DAB = 1.36 × 10–5 m2/s – (0.044) (1.36 × 10–5) = (– 0.7206) NA | 1 – 0.075 z=0 –7 = 4.662 × 10 kgmole/m2 · s = 1; yB1 = 1 – yA1 = 0.925 PA CAL = M A

784 = 46 = 17.04 kgmole/m3 t

0.105

⌠ ⌡

∴

(Ans.)

– – z dz

= 2.738 × 10–9 ⌠ ⌡ dt

0.1

0

Solving for t, we get, t = 1.87 × 105 seconds ≅ 52 hours (Ans.) (73) A 0.20-m-long test tube was used to study the diffusion process in which liquid A diffuses into gas B. In one study the level of liquid A was initially 0.1 m below the top of the tube. The temperature was 25°C and the total pressure was maintained at 1 atm. The molar flux of component A at the top of the test tube was found to be 1.6 × 10–3 kgmol/m2.h. Find the diffusion coefficient for A into gas B. Assume that gas B is insoluble in liquid A. The partial pressure of A at the surface of the liquid was 0.06 atm. Sol. : Data given : T = 25°C P = 1 atm NA |z = 1.6 × 10–3 kgmole/m2.hr 2

DAB = ? B is insoluble in A PA |z = 0.06 atm 1

B (g)

0.1m 0.2m A(l)

Fig. 2.30 : Test tube to study diffusion process

Principles of Mass Transfer Operations

2.114

Diffusion Mass Transfer

This problem is analogous to problem 61. '

∴ – ln (1 – yA) = k1 z + k2 Solving for the generic boundary conditions, yA = yA1 @ z = z1 yA = yA2 @ z = z2 One obtains : 1 – yA ln 1 – y

A1

1 – yA2 z – z1 = z – z ln 1 – y 2 1 A1

For our problem, @

PA 0.06 z = z1 = 0, yA1 = P = 1 = 0.06

@

z = z2 = 0.01, yA2 = 0

Therefore, 1 – yA ln 1 – 0.06 yA

∴

z 1–0 = 0.1 ln 1 – 0.06 = 1 – exp (0.6187z – 0.0619)

= 1 – 0.94 e0.6187z C · DAB dyA NA‚ z = – 1 – y · dz A

Since, dyA dz

= – (0.94) (0.6187) e(0.6187) (0.1) z = 0.1

= – 0.6187 P C = RT ∴

1.6 × 10–3

∴

DAB

(1) = (0.08206) (25 + 273) = 0.041 kgmole/m3 – (0.041) · DAB = (– 0.6187) 1–0 = 0.063 m2/hr

(Ans.)

(74) Gas A diffuses through two immiscible liquids contained in a capaillary tube as shown below. The concentration of A at the bottom of the capillary tube is maintained at a constant value. The partial pressure of A in the gas is 0.05 atm. The equilibrium relationships for A in the two liquids are : CA, I = 200 PA CA, I = 2 CA, II where, PA has units of atm and CA has units of kgmol/m3. The concentration of A at Z2 is 3.0 kgmol/m3. The diffusivities of A in liquids I and II are 1.5 × 10–9 m2/s and 7.5 × 10–9 m2/s respectively. The thickness of the liquid I layer (Z0 – Z1) is 0.02 m. Assume that convective effects at the bottom of the capillary tube are negligible and that liquid I has a very low pressure. Calculate the molar flux A dissolving into liquid II and plot the concentration of A in liquids I and II as a function of Z.

Principles of Mass Transfer Operations

2.115

Diffusion Mass Transfer

Gas Z Liquid I Z1 Liquid II Z2

Fig. 2.31 : Diffusion of gas from capaillary tube

Sol. : Molar balance in Liquid I : ∂CA ∂NAx ∂NAy ∂NAz ∂t + ∂x + ∂y + ∂z = RA dNAz dz

= 0

dyA NAz = – C DAB dz + yA (NAz +NBz)

(stagnant B)

C DAB dyA dyA NAz = – 1 – y dz – C DAB dz A Assuming 1 – yA – 1 ⇒ yA << 1 (gas in liquid solution) Therefore, dyA d  dz – C DAB dz  = 0 I

d 2 CA dz

= 0, Since C can be considered constant.

Integrating twice : I

CA @ z = 0,

I CA

= k1 z + k2

= 200 PA = 200 (0.05) = 10 kgmol/m3 I

Therefore,

CA

= 10 + k1 z

k1 cannot be calculated yet because the concentration of A at z1 is not known. Balance for liquid II : II

d 2C A dz2 II

CA

= 0 = k3 z + k4

II

@ z = z2, CA = 3.0 kmol/m3 I

II

@ z = z 1, C A = 2 C A Therefore, 3 = 0.07 k3 + k4

Principles of Mass Transfer Operations

2.116

Diffusion Mass Transfer

I

CA = 0.02 k3 + k4 2 There are 2 equations and 3 unknowns. The missing equation is obtained by noticing that, at the interface the fluxes of A must be equal : NAI

= NAII @ z = z1

I

DAI dCA dz DAI (k1)

II

dCA dz

= DAII

= DAII (k3)

DAII k1 = D k3 = 5 k3 AI Therefore, k1, k3 and k4 are obtained by solving the system of equations : I

CA

= 10 + k1 z = 10 + 0.02 k1

3 = 0.07 k3 + k4 I

CA 2

= 0.02 k3 + k4

k1 = 5 k3 k1 = – 100

To obtain

k3 = – 20 k4 = 4.4 II

Therefore,

CA

= – 20 z + 4.4

Molar flux of A dissolving in II II

NA = – DAII = – (7.5 ×

dCA dz 10–9)

z = 0.02

(– 20) = 1.5 × 10–7 kgmol/m2 . s

12

CA (kgmol/m^3)

10

C A (I )

8 6 4

CA (II)

2 0 0

0.01

0.02

0.03

0.04

0.05

0.06

z (m)

Fig. 2.32 : Concentration profile for Ex. 62

0.07

(Ans.)

Principles of Mass Transfer Operations

2.117

Diffusion Mass Transfer

(75) Gas A is diffusing from a gas stream at point 1 to a catalyst surface at point 2 and reacts instantaneously and irreversible as follows : 2A → B Gas B diffuses back to the gas stream. Derive the final equation for NA at constant pressure P and steady-state in terms of partial pressure. DAB P dx Sol. : NA = RT dz + xA (NA + NB) From stoichiometry : – NA = 2NB DAB P dxA NA ∴ NA = – RT dz + xA NA – 2    DAB P dxA NA NA = – RT dz + xA 2

1 – xA N = – DAB P dxA dz RT  2 A xA

Z2

2

DAB P ⌠ dxA NA ⌠ ⌡ dz = – RT ⌡  xA Z1 xA 1 – 2  1 xA2

∴

. . .

NA (z2 – z1) =

xA1

=

1 – 2  2D P ln  x  RT 1 – 2  AB

A1

PA1

PA2 ; x = A 2 P P PA2

NA (z2 – z1) =

2DAB P RT

1 – 2P  ln  P  1 – 2P   A1

PA2

1 – 2P  P ln –z )  P  1 – 2P  

2DAB NA = RT (z 2

1

A1

(Ans.)

(76) An inert fluid B occupies the space between concentric spheres with radii r1 and r2. The inner sphere is porous and is supplied with fluid A at a rate of Q moles/sec. Fluid A diffuses through B to the outer sphere, which is catalytic and which catalyzes the instantaneous reaction A = 2C. Sol. : (a)

Derive a relationship for the steady-state concentration, xA1 , of A at the surface of the inner sphere. Be sure to state all assumptions.

(b) What is the rate of production of C ?

Principles of Mass Transfer Operations (c)

2.118

Diffusion Mass Transfer

Show that the result of the molar flux of A is the same as that for diffusion between two flat parallel plates separated by a distance δ = r2 – r1, if the transport area is based on the geometric mean radius, rGM, where, rGM =

r1 r2

Assumptions : (1) Steady-state; (2) No homogeneous reaction; (3) Spherical symmetry – radial transport only. A → 2C NC = – 2NA, NB = 0; N = NA + NB + NC = – NA (4) Mixture is mostly B. Therefore, use DAB even though not a binary system. ∂xA NA = – CDAB ∂r + xA (NA + NB + NC) ∂xA – CDAB ∂r ∂xA ∴ NA = – CDAB ∂r – xA NA ⇒ NA = 1 + x A ∂CA 1 ∂ (r2 NA) = – 2 + RA ∂r ∂t r ∂CA = 0; At steady-state, ∂t RA = 0 because there is no homogeneous reaction in gap 1 ∂ (r2 NA‚ r) ∴ 0 = – 2 ∂r r ∂xA – CDAB r2 ∂r Q 2 r2 NA‚ r = = r1 NA‚ r = 1 + xA 4π 1 Recall that we're at steady-state and flux at (r) = Flux at surface r2 0 ∂xA Q ⌠ ⌠ ∂r ⌡ 1 + xA = – 4πCDAB ⌡ r2 xA‚ 1 r1 Note here at xA‚ 2 = 0 since catalytic reaction is instantaneous. ln (1 + xA‚ 1) =

Q 1 – 1 r 2 4πCDAB r1

Production of C = 2Q Using the flat plate geometry

NA

∂xA – CDAB ∂z = 1 + xA ;

As before, steady-state, no homogeneous reaction ∂NA‚ z = 0 ∂z ∴ NA‚ z = constant

(Ans.) (Ans.)

Principles of Mass Transfer Operations

2.119

Diffusion Mass Transfer r2

0

⌠ ⌡ xA‚ 1

∂xA 1 + xA

NA = – CD AB

⌠ ∂z; ⌡ r1

Remember here that we are looking at a separation δ = r2 – r1. NA‚ z ln (1 + xA‚ 1) = CD (r2 – r1) AB Now in the spherical case, we can define a mean flux as Q Q NAm = A = 2 M 4πr M

Remember – in the spherical case, we said that (r2 NA‚ r) was constant 2

∴

4πrM NAM 4πCDAB

 1 – 1  = ln (1 + xA‚ 1) = NA‚ z (r2 – r1) CDAB r1 r2 2

For

NAM 2

rM

rM 1 1 1 = NA‚ z ⇒ CD r – r  = CD (r2 – r1) AB  1 2 AB

r2 – r1 = (r – r ) 2 1  r 1r 2 

∴ rM = r1r2 = rGM (Ans.) (77) A rubber basketball is initially filled with helium and left in the middle of the air business centre. Assuming the size of the basketball does not charge and neglecting the flux of air into the basketball, how long will it take for the pressure inside the basketball to decrease to 75% of its initial value ? Show all calculations and clearly state all assumptions necessary to solve this problem. DHe/rubber = 22 × 10–6 cm2/s at 25°C Radius of basketball = 20 cm Thickness of basketball wall = 0.5 cm Sol. : This could be considered an instantaneous mass flux problem but even though the concentration gradient through the basketball is linear as for steady-state diffusion, the concentration inside will change over time as the pressure decreases. The problem should be set up first as an instantaneous mass flux problem that is then integrated over pressure on one side and over time on the other side. dC J = – D dx Using the ideal gas law PV = nRT P dC 1 n C = V = RT ⇒ dP = RT dC dC dP 1 dP ∴ J = – D dx = – D dP dx = – D RT dx dn We also know that JA = – dt i.e. the flux of helium out of the basketball dn 1 dP – dt = – D RT dx · A, where A is the surface area of the basketball

Principles of Mass Transfer Operations

2.120

Diffusion Mass Transfer

dn – dt

D dP = – RT dx (4πr2) PHe – outside – PHe – inside dP Now let's assume we can represent dx = ∆x and that PHe – outside ≈ 0 i.e. very low concentration of He in air dn D dP D (– PHe – inside) 2 ∴ (4πr2) dt = – RT dx (4πr = – RT ∆x

∴

–P

∴

dn

He

– inside

PV From the ideal gas law n = RT dP V – RT (P He – inside) dP – (P He –inside)

∴ ∴

– (P

∴

He

0.75 PHe

–

dP – inside)

– inside

t

dP

⌠ ⌡ (PHe – inside)

PHe – inside

∴

D (4πr2) = RT dt ∆x V ⇒ dn = RT dP D (4πr2) = RT dt ∆x D (4πr2) dP D (4πr2) = v dt ⇒ – (P = dt ∆x He – inside) 4 πr3 ∆x 3  3D = dt r∆x

0.75PHe – inside – ln  P   He – inside 

3D = r∆x =

⌠ dt ⌡ 0

3D 3 (22 × 10–6) t ⇒ – ln (0.75) = (20) (0.5) t r∆x

∴ t = 43588s = 12.1 hrs. Note that here we assumed that the OUTER radius of the basketball was 20 cm. If you interpreted the question as that was the INNER radius, then the time increases to 12.4 hrs. (Ans.) (78) Two large storage tanks containing natural gas mixtures, each maintained at a pressure of 10 atmosphere and 298 K. They are connected by a pipeline 1 inch in diameter and 5 feet. In one tank, it cntains mainly H2S separate from the natural gas : 90% H2S and 10% CH4 by volume. The second tank contains clean natural gas : 100% CH4. Accidently, the valve is left open by an operator. At the initial tank conditions, DH2 S–CH4 = 2.02 × 10–6 m2/s. Assume the mixtures behave like ideal gases. Find. (a)

What is the steady rate of transfer (pseudo-steady state initially) of Methane, – NCH4 , between the two tanks ?

(b) What is the mass velocity of Methane with respect to a stationary co-ordinate ? (c)

What is the molar velocity of Methane with respect to Hydrogen Sulfide ?

(d) What are the mass concentrations in the two tanks ? Sol. : Equimolar counter diffusion : N = 0 Let A = CH4; B = H2S.

Principles of Mass Transfer Operations

2.121

Diffusion Mass Transfer

yA2 – yA1 dyA * NA = JA = – C · DAB dz = – C · DAB · ∆z y – y A1 A2 – N A = A · NA = A · C · DAB ∆z

∴

10 atm 298 K

10 atm 298 K

Storage Tank 1

Storage Tank 2

Fig. 2.33 : Storage task containing natural gas

10 0.1 – 0 π = 4 (2.54 × 10–2)2 ⋅ × 2.02 × 10–6 · 0.08206 × 298 5 × 0.3048 = – 2.472 × 10–10 kmol/sec. (Ans.) = – 2.472 × 10–7 mole/sec. mole 16.04 gm/mole · (b) m A = – 2.472 × 10–7 sec · π –2 2 2 4 (2.54 × 10 ) m = – 7.826 × 10–6 kg/(m2 · sec) (Ans.) Negative sign indicating flux transfer from tank 2 to tank 1. (c) NA – NB ⇒ NA + NB = 0 = N – 2.472 × 10-7 mole/sec. ∴ NA – NB = 2NA = 2 × π –2 2 2 4 (2.54 × 10 ) m mole = – 9.757 × 10–4 2 m . sec (d) ρA = MA · CA = MA yA · C P 10 kmol kmol . . C = RT = = 0.40895 . 3 m m3 0.08206 × 298 MA = 16.04 kg/kmol; MB = 2 × 1.0079 + 32.06 kg/kmol = 34.08 kg/kmol Tank – 1 : ρA1 = 16.04 × 0.40895 × 0.1 = 0.65596 kg/m3 (a)

– NCH4

ρB 1

= 34.08 × 0.40895 × 0.9 = 12.543 kg/m3

Principles of Mass Transfer Operations

2.122

Diffusion Mass Transfer

Tank – 2 : ρA2

= 16.04 × 0.40895 × 1 = 6.5596 kg/m3

ρB2

= 0

(Ans.)

(79) Water lost due to evaporation : A large water storage tank is filled with water. The tank has a vent line open to the atmosphere. The vent line is 1 m long and 1 inch ID How many kg of water are lost per hour if the atmosphere is dry air at 40°C ? Assume that the liquid level remains constant. The vapour pressure of water at 40°C is 55.3 kPa. Dwater-air = 2.77 × 10–4 m2/s. Sol. : A schematic diagram of the problem is as shown in figure. Air

2 : Vent line tip Air z

L

A = Water vapor B = Dry air

1 : Liquid water surface Water Water

Water Water

Fig. 2.34 : Water storage tank

Total molecular flux moving upward (z-direction) inside the ventral is given by : N = NA + NB Since the liquid level is not changing (assumption in the problem), the dry air remains stationary inside the ventral line, thus, NB = 0 The general diffusion equations : N = NA + NB *

or

∴ ∴

⇒

NA = NyA + JA C · dyA * and Fick's law : JA = – DAB · dz CA DAB dyA NA = C N – C · dz P, T = constant pA DAB dpA NA = P · NA – RT · dz pA DAB dpA NA 1 – P  = – RT dz   dpA RT pA = – DAB · NA dz 1– P P – pA1 NA RT = – P · D · ∆z ln P – p A2

AB

Principles of Mass Transfer Operations ⇒

2.123

Diffusion Mass Transfer

P – pA1 P · ∆z NA = – RT · DAB · ln P – p A2 pA1

= 55.3 kPa, pA2 = 0

P = 101.325 kPa 101.325 × 103 × 1 101.325 – 55.3 ∴ NA = × 2.77 × 10–4 × ln 101.325 – 0 8314 × (273.15 + 40) kmol = 8.5073 × 10–6 2 m · sec Kg of water lost per hour is given by : – G A = M A · NA · A π = 18.015 × 8.5073 × 10–6 × 4 (2.54 × 10–2) 2 = 7.7658 × 10–8 kg/sec. (Ans.) = 2.7957 × 10–4kg/hr (80) Transport of A in a binary system is given by CA dCA NA = C (NA + NB) – DAB dz where, C is the total concentration. (a) Write down the counter-part of the above eqution for component B. (b) Show that DAB = DBA (c) It is generally true for a binary system that NA = k NB with k being a constant (independent of location and concentration). Show that CA2 NA – NA + NB C NA CDAB NA = N + N ln CA1 ∆ z NA A B – NA + NB C CA dCA Sol. : NA = C (NA + NB) – DAB · dz CB dCB (a) NB = C (NA + NB) – DBA · dz (Ans.) CA + CB dCA dCB (b) NA + NB = (NA + NB) – DAB · dz – DBA dz C dCA dCB ⇒ DAB dz + dBA dz = 0 dCA dCB dC . . C = CA + CB ⇒ dz = 0 = dz + dz . dCA dCA i.e. DAB dz + DBA – dz  = 0   dCA ∴ ≠0 dz i.e. DAB – DBA = 0 ∴ DAB = DBA (Ans.)

Principles of Mass Transfer Operations (c) ⇒

⇒

2.124

CA dCA NA = C (NA + NB) – DAB dz NA CA dCA (NA + NB) · N + N – C  = – DAB dz B  A  (NA + NB) dCA = – dz NA CA DAB NA + NB – C Integrate ;

⌠ ⌡ CA1

C ln

(Ans.)

z = 0 ∆z CA = CA1 CA2

CA2

⇒

Diffusion Mass Transfer

∆z

dCA CA NA – NA + NB C CA2 NA – N +N C A

B

CA1 NA – NA + NB C

⌠ ⌡

=

–

0

= –

NA + NB dz DAB

NA + NB ∆z DAB

CDAB 1 1 = N +N ∆z A B

∴

CA2 NA – NA + NB C ln CA1 NA N +N – C A

NA CDAB NA = N + N ∆z A B

CA2 NA – NA + NB C ln CA1 NA – N +N C A

B

(Ans.)

B

(81) A crystal of copper sulphate CuSO4 · 5H2O falls through a large tank of pure water at 20°C. Estimate the rate at which the crystal dissolves by calculating the flux of CuSO4 from the crystal surface to the bulk solution. Repeat by calculating the flux of water. Data and assumptions : Molecular diffusion occurs through a film of water uniformly 0.0305 mm thick, surrounding the crystal. At the inner side of the film, adjacent to the crystal surface, the concentration of copper sulphate is its solubility value, 0.0229 mole fraction CuSO4 (solution density = 1193 kg/m3). The outer surface of the film is pure water. The diffusivity of CuSO4 in water is 7.29 × 10–10 m2/s. Sol. : We can set this up a straight forward flux expression for diffusion through a thin film : m2 7.29 × 10–10 s D N = k∆C = ∆C = (Csat – 0) ∆z 0.0305 × 10–3 m Here we note that the surface concentration = Csat and that the concentration on the other side of the film (i.e. across the diffusion pathway) = 0. m N = 2.4 × 10–5 s (Csat)

Principles of Mass Transfer Operations

2.125

Diffusion Mass Transfer

Csat can be calculated from the fact that it is equal to 0.0229 mole fraction of CuSO4 mole 1 fraction of CuSO4in CuSO4 · 5H2O = 6 ≈ 0.17. MWCuSO4 159.6 Mass fraction of CuSO4 in CuSO4 · 5kH2O = MW = 249.6 = 0.64 CuSO4 · 5H2O

(Assuming 1 mole of CuSO4 · 5H2O). kg m3 mol . . . . [CuSO · 5H O] = Solution density . . 4 2 g = 4.8 m3 249.6 mol mol mol [CuSO4] = 0.17 × 4.8 3 = 0.8 3 m m mol mol [CuSO4]solubility = 0.0229 × 0.8 3 = 0.018 3 m m m mol mol ∴ N = 2.4 × 10–5 s 0.018 3  = 4.4 × 10–7 2 (Ans.) m  m ·s  (82) Derive the differential equation describing the concentration (in the convenient units - mole fraction, mass fraction, molar concentration, mass concentration) of species A as a function of position z from the continuity equations in the following situations. Ensure that you define the applicable boundary conditions. Assume steady-state and that, T, P and DAB are constants. A horizontal solid surface of A is sublimating slowly into a stagnant air film of thickness L. Air is completely insoluble in the solid. Let z = L be located at the interface between the gas and solid. At z = 0, yA = yA0 kg = 1193 3 m

1193

Sol. : From continuity (mass balance) equations : From continuity (mass balance) dNAZ dNBZ = 0 and = 0 dz dz ∴ NAZ and NBZ are both constants for 0 < z < L. . . NBZ = 0 at z = L. ∴ NBZ = 0 for 0 < z < L. . From Fick's law, dxA NAZ – xA (NAZ + NBZ) = – DAB C dz . . NBZ = 0 for 0 < z < L . dxA NAZ – xA (NAZ) = – DAB C dz DABC dxA NAZ = – (1 – x ) dz A dNAZ d  DABC dxA ∴ = dz dz – (1 – xA) dz  = 0 dxA d  1 ∴ (Ans.) dz (1 – xA) dz  = 0 (83) 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 H2 on the other side of the neoprene is 0. Calculate the steady-state flux, assuming that the only

Principles of Mass Transfer Operations

2.126

Diffusion Mass Transfer

resistance of diffusion is in the membrane. The solubility S of H2 gas in neoprene at 17°C is 0.051 m3/m3 solid – atm, and the diffusivity DAB is 1.03 × 10–10 m2/s. Sol. : The key to solving this problem is to convert all our units to concentration terms S (0.051) (0.010) cA = 22.414 PA = = 2.28 × 10–5 kgmol H2/m3 solid 22.414 Here, the solubility is that of gas A in a solid and is defined in terms of m3/m3 solid/atm. partial pressure of A. The 22.414 arises from an ideal gas concentration. DAB (CA‚ 2 – CA‚ 1) 1.03 × 10–10 (2.28 × 10–5 kgmol H2/m3 solid – 0) = ∴ NA = z2 – z1 (0.5 – 0) 1000 ∴ NA = 4.69 × 10–12 kgmol H2/s-m2 (Ans.) (84) Water evaporating from a pond does so as if it were diffusing across an air film 0.12 cm thick. The diffusion coefficient of water in 20°C air is about 0.25 cm2/s. If the air out of the film is 50% saturated, how fast will the water level drop in one day ? Sol. : Since this is a diffusion question, we can use Fick's law as follows : dCH2O JH2 O = – DH2 O dz – in this cases, we can tacitly assuming that we are using a one-dimensional model, and as such, diffusion only occurs in the z-direction. Known information : Path length (z) = 0.12 cm; DH2 O = 0.25 cm2/s CH2 O in order to get this value, we have to resort to literature references, so from the literature and partial pressure values. At the stated temperature of 20°C, PH2 O = 2.34 kPa. We are also told that the air is 50% saturated on the other side of the stagnant film. ∴ PH2 O |z = 0 = 2.34 kPa; PH2 O |z = 0.12 cm = 1.17 kPa; JH2 O = – DH2 O

dCH2O

=

– DH2O dPH2O

dz RT dz – 0.25 cm2/s (1.17 – 2.34) kPa = (8.314 J/mol · K) (293.15 K) 0.12 cm mol · cm L × (be careful here - look at units) JH2 O = 1 × 10–3 L · s 1000 cm3 mol JH2 O = 1 × 10–6 cm2 · s Now to calculate the rate at which the pond level drops, we simply convert the molar flux rate into a volumetric flux rate, using the density of water. g cm3 cm3 ρH2 O = 18 mol × 1 g = 18 mol mol cm3 cm cm × 18 mol = 1.8 × 10–6 s = 1.8 × 10–6 s JH2 O = 1 × 10–6 2 cm · s 3600 × 24 s × day cm JH2 O = 1.56 day (Ans.)

Principles of Mass Transfer Operations

2.127

Diffusion Mass Transfer

(85) A conical vulcanized rubber plug is used to close an opening of a container. The gas CO2 at 25°C and 2.0 atm is inside the container. The solubility of CO2 in the rubber is 0.90 Nm3/(m3-rubber atm-CO2). The plug is 30 mm thick and has a cross-sesctional area of 3.00 × 10–4 m2 facing inside the container and 4.00 × 10–4 m2 exposing outside. The CO2 content in the atmosphere is negligible. The diffusivity of CO2 in the rubber is 0.11 × 10–9 m2/s. Calculate the total leakage of CO2 through the rubber in kmol/s. Sol. : 2 30 mm

Z

1

Z Fig. 2.35 : Conical Valcanized rubber plug

This is the case of steady state diffusion through solids. A1 = 3.00 × 10–4 m2, A2 = 4.00 × 10–4 m2 A = αZ2, A1 = αZ2, A2 = α (Z1 + 0.03)2 0.03 ⇒ Z1 = m = 0.193923 m, Z2 = 0.223923 m 1/2 4   –1 3 α = 7.97742 × 10–3 dCA – dCA NAZ = – D dZ , N AZ = A · NAZ = – A · D dZ Z2

– – NAZ = constant ⇒ N AZ

⌠ dZ = ⌠ – D dCA ⌡ A ⌡ Z1

∴ ∴

1 1 1 – NAZ · – Z – Z  α  2 1

CA2

CA1

= –D (CA2 – CA1 )

Dα (CA1 – CA 2) – NA, Z = Z1 Z2 , CA = SPA/VA0 Z2 – Z1 Dα S (pA1 – pA2) = V (Z – Z ) A0

=

0.11 ×

2

10–9

1

× 7.97742 × 10–3 × 0.9 × (2.0 – 0) 22.414 × 0.03 × 0.193923 × 0.223923 kmol/s

– NA, Z = 1.02003 × 10–13 kmol/s

(Ans.)

(86) Mass transfer is occuring to a spherical water droplet having a radius of 1 mm. The droplet is falling in the sky toward the ground. Along the path, the air is saturated with water vapour and the water droplet is assumed not to grow on the water vapour. It takes 10 seconds to pass through a layer of still air containing 1% (in volume

Principles of Mass Transfer Operations

2.128

Diffusion Mass Transfer

fraction) acid gas (SO2) at 5°C and 0.8 atm before reaching the ground. Assume that the acid gas reacts very fast with water droplet in the liquid phase. Thus, the vapour pressure of the SO2 + H2O → H2SO3 acid gas may be assumed zero on the water droplet. Furthermore, assume that the water droplet is isolated, i.e. only one in the sky. The diffusivity of air in sulfur dioxide is 1.4 × 10–5 m2/s at 20°C and atmospheric conditions. (a) Estimate the diffusivity of SO2 in air at 5°C and 0.8 atmosphere. (b) Calculate the diffusion flux of SO2 to the water droplet surface in kmol/(m2 · s). (c) What is the concentration of H2SO3 in the rain water ? '

(d) What is the mass transfer coefficient, kc , based on your calculations ? (e) Sol. : (a) ∴

What is percentage of mass transfer resistance in the liquid film ? T1.75 DAB ∝ P , DAB, Z 0 °C and 1 atm = 1.4 × 10–5 m2/s (2.73.5 + 5)1.75 1 DAB = (273.15 + 20) × 0.8 × 10–5 m2/s = 1.596 × 10–5 m2/s

(b)

r dr NAr

Fig. 2.36 : Illustration of spherical water droplet in Ex. 74

⇒ – NAr NAr = 4πr2

dyA NAr = (NAr + NBr) yA – C DAB dr C DAB dyA NAr = 1 – y dr A

– NAr C DAB ⇒ dr = 1 – y dyA 4πr2 A r = r0, yA = 0 = y A0

r → ∞, yA = 0.9 = y

A∞

∴

– NAr – 4πr

y

∞

= – C DAB ln (1 – yA) r0

– NAr 4πr0

y

A∞

A0

1 – yA0 = C DArs ln 1 – y

NAr0 =

A∞

– NAr 2 4πr0

=

1 – yA0 C DAB ln r0 1 – yA∞

Principles of Mass Transfer Operations

2.129

NAr0

Diffusion Mass Transfer

1 – yA0 P DAB 0.8 × 101325 × 1.596 × 10–5 = RT r ln 1 – y = 8314 × 278.15 × 10–3 0 A∞ kmol (1 – 0) × ln (1 – 0.9) 2 m · sec. = – 5.6225 × 10–6 kmol/m2 · s

(Ans.)

– 2 NAr = 4πr0 ⋅ NAr0 = 4 × 3.141592654 × (10–3)2

(c)

× (– 5.6225 × 10–6) kmol/s 10–11

nA CA

– NAr0

(d)

kmol/s = – 7.0654 × – = – N Ar · t = 7.0654 × 10–10 kmol nA nA 7.0654 × 10–10 kmol = V = 4 = 4 m3 3 –3)3 π r × π × (10 0 3 3 = 0.16867 kmol/m3 = 0.16867 M = kc (CA∞– CA0 ) = kc C (yA∞– yA0 )

(Ans.)

1 – yA0 ' = kc · Cln 1 – y A∞ DAB 1.596 × 10–5 = m/s = 0.01596 m/s (Ans.) r0 10–3 (e) The percentage of mass transfer resistance in the liquid film = 0. (Ans.) (87) A binary gas mixture containing 50% A and 50% B (by volume) at 388 K and 1 atm is present in the bulk to which a narrow pore is opened. The pore is cylindrical in shape and the bottom is restrictive. On the bottom of the pore, reaction : A → 2B takes place. The reaction rate is given by – rA = k' CA with k' = 0.1 m/s. Assume that the pore is 10 µm in diameter and 0.2 mm deep. The diffusion coefficient measured at 25°C and 1 atmosphere is DBA = 2.0 × 10–5 m/s. Find. '

kc

∴

(a)

=

What is the value of diffusivity DAB at 388 K and 1 atm ?

(b) What is the molar flux ratio,

NA + NB NB , inside the pore ? Is it constant

everywhere ? (c)

What is the mole fraction of A at the bottom of the pore ?

(d) What is the molar transfer rate of A into the pore ? (e) What is the molar rate of B coming out of the pore ? Sol. : Z2 Z Z1 a

b

Fig. 2.37 : Diffusion of binary gas mixture

Principles of Mass Transfer Operations

2.130

Diffusion Mass Transfer

Ac = a · b o

pA1

= PA = 2.487 kPa

pA2

1 o = 0.25 PA = 4 PA1

Let A = Water vapour, B = Air T = 21°C = 294.15 K, P = 1 atm = 101.325 kPa, PL = 918.0 kg/m3, NBZ = 0, Z2 – Z1 = 0.02 m, DAB ℑ (a) ∴

25°C

= 2.60 × 10–5 m2/s.

T1.75 We know, DAB ∝ P T 1.75 DAB = T  · DAB, 1 =  1

1.75

274.15 298.15

× 2.60 × 10–5 m2/s = 2.539 × 10–5 m2/s (Ans.)

NBZ = 0

(b)

(Ans.)

(c) AC = ab = 10 km × 0.3 m = 3000 m2 (d) General diffusion equation : dyA NAZ = yA (NAZ + NBZ) – C DAB dZ AC = Constant ⇒ NAZ = Constant dyA NAZ (1 – yA) = – C DAB dZ

⇒

yA2

Z2

⌠ ⌡

NAZ dZ = –

Z1

⌠ C DAB dyA; C and DAB = Constants ⌡ 1 – yA yA1

1 – yA2 NAZ (Z2 – Z1) = C DAB ln 1 – y

∴

A1

Ideal gas law :

PV = nRT pA1 pA2 P 1 pA1 C = RT , yA1 = P , P = 4 P = yA2 P – pA2 P DAB ∴ NAZ = (Z – Z ) RT ln P – p 2 1 A1 (e)

(Ans.)

P – pA2 P DAB AC – N AZ = NAZ · AC = (Z – Z ) RT ln P – p 2 1 A1 =

10–5

101325 × 2.539 × × 3000 ⋅ ln 0.02 × 8314 × 294.15

1 101325 – 4 × 2.487 101.325 – 2.487

kmol s

Principles of Mass Transfer Operations

2.131

Diffusion Mass Transfer

= 2.9501 × 10–3 kmol/s · – m A = MA N AZ = 2.9501 × 10–3 kmol/s × 18.015 kg/kmol = 0.053146 kg/s = 4591.82 kg/day · mA 4591.81885 = m3/day = 4.601 m3/day (Ans.) QA = 998.0 ρL (88) Mass Transfer from a Naphthalene Sphere to Air : Mass transfer is occuring 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 vapour pressure of naphthalene at 52.6°C is 1.0 mm Hg. The diffusivity of naphthalene in air at 0°C is 5.16 × 10–6 m2/s. Calculate the flux of sublimation of naphthalene from the surface in mol/(s·m2). Note : The diffusivity can be corrected for temperature using the temperature-correction factor from the Fuller et al. equation. 1 Sol. : KA1 = 760 = 1.316 × 10–3 atm kA2 = 0 T = 52.6 + 273 = 325.6 K P = 1.00 atm = 1.0132 × 105 Pa DAB = 5.16 × 10–6 m2/k at 52.6°C and 1.0 atm r1 = 10 mm = 0.010 m pB2 – pB1 kBM = ln (p /p ) B2 B1 – pB1 – pB2

= 1.00 – 1.316 × 10–3 = 0.9987 atm.

325.6 1.75 DAB = 5.16 × 10–6  273   

= 1.00 – 0 = 1.00 atm

DAB = 7.024 × 10–6 m2/s

1.00 – 0.9987 – PB, M = ln 1.00/0.9987 = 0.9993 atm We know, NA1

=

=

DAB P (k1 – pA2) – RT r1 PBM (7.024 × 10–6) (1.0132 × 10–5) (1.316 × 10–3) (1.0132 × 105) (8314) (325.6) (0.010) (0.9993 × 1.0132 × 105)

NA1 = 3.46 × 10–8 kgmol/s · m2

(Ans.)

(89) A flat plug 30 mm thick having an area of 4.0 × 10–4 m2 and made of vulcanized rubber is used for closing an opening in a container. The gas CO2 at 25°C and 2.0 atm pressure is inside the container. Calculate the total leakage or diffusion of CO2 through the plug to the outside in kmol/s at steady state. Assume that the partial pressure of CO2 outside is zero. The solubility of CO2 gas is 0.90 m3-STP/(m3-rubber · atm-CO2). The diffusivity = DAB = 0.11 × 10–9 m2/s.

Principles of Mass Transfer Operations

2.132 A = 4.0 × 10–4 m2

Sol. : L = 30/1000 = 0.030 m T = 25°C S=

kA1

0.90 m3 (STP) m3 rubber · atm

Diffusion Mass Transfer

= 2.0 atm

DAB = 0.11 × 10–9 m2/s

Concentration in solids : S kA1 cA1 = 22.414

=

0.90 (2.0) 22.414

= 8.031 × 10–2 kgmol/m3 rubber Diffusion through solids : NA = NAA =

DAB (CA1 – CA2) z2 – z1 A DAB (CA1 – CA2) z2 – z1 L

cA1 CO2 kA1 cA2

kA1 = 0

x z1

z2

Fig. 2.38 : Concentration profile for diffusion of CO2 through plug

4 × 10–4 (0.11 × 10–9) (8.031 × 10–2 – 0) (0.030 – 0) –13 NAA = 1.178 × 10 kgmol/sec. (Ans.) (90) Due to valve leakage, water has spilled on the floor on an industrial complex. Determine the period the water will take to evapourate out in the adjacent stagnant air of the surroundings. Data : The water layer = 1.25 mm thick The temperature of water = 297 K The surrounding air temperature = 297 K The surrounding air pressure = 1 std. atm. Absolute humidity of air = 2 × 10–3 kg of water/kg of dry air. Assume steady-state molecular diffusion taking place through a still air film 5 mm thick during evapouration process. You can take diffusivity of water in air = 2.6 × 10–5 m2/s at 298 K and 1 std. atm. Sol. : Basis : 1 m2 of evapourative surface area considered. ∴ Volume of water evapourated = Area × thickness of film = 1 × 1.25 × 10–3 = 1.25 × 10–3 m3 =

Principles of Mass Transfer Operations

2.133

Diffusion Mass Transfer

So, mass of water evapourated = 1.25 × 10–3 × 1000 = 1.25 kg 2

2

Stagnant Air

Z-direction

1

5mm

1

Water film

1.25mm

Fig. 2.39 : Water molecule diffusing into the stagnant air above the water film

∴

1.25 18 = 0.0694 kmoles. Next, molar rate of evapouration of water is given by : DAB · Pt YA1 – YA2 NA, Z = ZRT  Y  ; kmol/m2 · s  B‚ M  Kilomoles of water evapourated =

Let A = water; B = Air Now, diffusivity of water in air at 298 K. DAB, T1 = 298 K = 2.6 × 10–5 m2/s at 1 std. atm. DAB, T2

= 297 K

T2 3/2 P1 = DAB, T1 T  · P  1  2 297 3/2 1 = 2.6 × 10–5 298 1





 

= 2.587 × 10–5 m2/s at 297 K and 1 std. atm. At (1) – (1) surface, air is saturated with water. At (2) – (2) surface, air is partly saturated with water. Therefore, the molecular diffusion of water will take place from (1) – (1) plane to (2) – (2) plane. kg water At 297 K, the saturated humidity is 1.89 × 10–3 kg · dry air . kg · water kmol · water 29 kg dry air H = 1.89 × 10–3 kg · dry air · 18 kg · dry air ·  kmol dry air        kmol · water = 30.45 × 10–3 kmol · dry air i.e. 1 kmol dry air is associated with 30.45 × 10–3 kmol water. 30.45 × 10–3 ∴ YA1 = = 0.0295 (1 + 30.45 × 10–3) The humidity in the bulk air stream is given by kg · water 2 × 10–3 kg · dry air

Principles of Mass Transfer Operations

2.134

Diffusion Mass Transfer

1

= [2 × 10–3

18 kmol water ] 1  kmol air 29  

= 0.00322 kmol water i.e. 1 kmol of dry air is associated with 0.00322 kmol of water vapour. 0.00322 ∴ YA2 = (1 + 0.00322) = 0.003211 ∴ YB1 = 1 – YA1 = 1 – 0.0295 = 0.970 and

YB2

= 1 – YA2 = 1 – 0.003211 = 0.9968

YB1 – YB2 0.970 – 0.9968 YB, M = ln (Y /Y ) = = 0.9835 0.970 B1 B2 ln 0.9968   –5 2.5869 × 10 101325 (0.02955 – 0.003211) ∴ NA, Z = (5/1000) × 8314 × 297 × 0.9835 –6 2 NA, Z = 5.685 × 10 kmol/m · s We have 0.06944 kmol H2O/m2 of water surface to be evapourated. kmol H2O/m2 0.06944 ∴ Required time = –6 5.685 × 10 kmol H2O/m2 · s = 12213.8 s = 3.39 hours (91) The diffusion coefficient relative to the volume average velocity is defined by ∇ c1 – j1 = D∇ That relative to the molar average velocity is defined by ∴

*

– j1

(Ans.)

= D* c∇ ∇y1

Show that the diffusion coefficients in these two definitions are equal, even if the molar concentration c is not constant. Sol. : Use the definition of the volume-based diffusion flux j1 … (1) j1 = c1 (v1 – v0) = – D∇c1 *

Use the definition of the molar-based diffusion flux j1 *

j1 = c1 (v1 – v*) = – D* c∇y1 Assuming subscript 1 for the solute and 2 for the solvent and volume-based flux : 1 (v1 – v0) = – c D∇c1 = – D∇ln c1 1 1 (v2 – v0) = – c D∇c2 = – D∇ ln c2 2 And from 3 and 4 : c1 c · y1 y1 (v1 – v2) = – D∇ ln c = – D∇ ln c · y = – D∇ ln y 2 2 2 Similarly, for the molar-based flux : 1 1 1 (v1 – v*) = – c D* c∇y1 = – (c /c) D* ∇y1 = – y D* ∇y1 = – D* ∇ ln y1 1 1 1

… (2) … (3) … (4)

… (5)

… (6)

Principles of Mass Transfer Operations

2.135

Diffusion Mass Transfer

1 (v2 – v*) = – c D* c∇y2 = D*∇ ln y2 … (7) 2 And from 6 and 7 : y1 (v1 – v2) = – D* ∇ ln y … (8) 2 Comparing equations 5 and 8 we see that D = D* (92) The following irreversible and very fast cracking reaction A → 2B is taking place on the surface of a spherical catalyst particle (radius Rcat = 0.2 mm). Both species (A and B) behave like ideal gases. The temperature in the reactor is 573 K and the total pressure is 2 bar. Very far from the surface (at r/Rcat → infinity), the partial pressure of species A is 0.5 bar. The diffusion coefficient DAB at the given conditions is approximately 0.2 cm2/sec. Calculate the total molar flux of species A at the surface of the catalyst particle. Hint : Use that the total molar flow rate (moles/sec.) moving towards the catalyst is constant. Sol. : r

yA,¥ Catalyst

A yA,0 R Reaction A 2B

2B

Fig. 2.40 : Illustration cracking reaction on spherical catalyst particle

From the general mass balance equation for spherical co-ordinates (see Appendix K). ∂cA 1 ∂ 2 1 ∂ 1 ∂nA‚ θ + rA ∂t = – r2 ∂r (r nA, r) – r sin θ ∂θ (nA, θ sin θ) – r sin θ ∂φ Assumption : ∂cA (i) Steady state : ∂t = 0 ∂nA‚ θ ∂ (ii) Symmetry with respect to θ and φ = (nA, θ sin θ) = =0 ∂θ ∂φ (iii) No chemical reaction in the diffusion path : rA = 0. Then, you can get the simplified governing equation as follows : 1 ∂ – 2 ∂r (r2 nA, r) = 0 → r2 · nA = constant … (1) r The total molar flux, dyA nA = – Dc dr = cA · v* where the molar average velocity, v* is defined as follows : nA + nB v* = c

Principles of Mass Transfer Operations

2.136

Diffusion Mass Transfer

The mass flux of species B is always 2 times of species A because one mole of A produces 2 moles of B on the catalyst surface. nB = – 2nA Therefore, dyA nA = – Dc dr – yAnA dyA (1 + yA) nA = – Dc dr dyA Dc nA = – (1 + y ) dr … (2) A According to equation (1), the total molar flow rate [r2 · nA, mol/s] at any r-position is constant. Let us refer to the molar flux [mol/(m2s] at the catalyst surface (r = R) as NA. Then, we get the following relation from the equation (1). 2

r2nA = Rcat · NA … (3) Substitute the equation (2) into (3). dyA 2 Dc – (1 + y ) r2 dr = Rcat · NA … (4) A Boundary conditions (because A disappears fast by the chemical reaction) At r = R, yA = 0 r= ∞ yA = yA, ∞ = 0.5/2 = 1/4 The solution of equation (4) is given by the following : 2

Rcat NA 1 ln (1 + yA) = D · c r + b By the boundary condition at r = Rcat Rcat NA b = – D·c 2

Rcat NA 1 1 ln (1 + yA) = D · c  r – R  cat 

… (5)

By the boundary condition at r = ∞ 2

Rcat NA 1 ln (1 + yA, ∞) = D · c – R   cat D·c NA = – R ln (1 + yA, ∞) cat

… (6)

where, D = 0.2 × 10–4 m2/sec P 2 · 105 (Pa) c = RT = = 42 moles/m3 m3 · Pa 8.314 mol · K 573 K Rcat = 0.0002 m y∞ = 0.25 Total molar flux of species A at the surface NA = – 0.937 moles/m2-sec (The negative sign is because the diffusion of species A is taken place toward the surface of catalyst).

Principles of Mass Transfer Operations

2.137

Diffusion Mass Transfer

(93) A vessel filled with the water stands in a hood where dry air is circulating. To reduce evapouration, it is covered with a sponge that is floating on the water surfae (Fig. 2.41). The cavities of the sponge can be modeled as cylindrical capillaries. The capillaries are partially filled with water, with a length h above the water level. The temperature of air and water are constant and identical. The diffusion coefficient and saturation pressure of water vapour in air are given below. (1)

Calculate the ratio of the total flux at 60°C to the total flux at 10°C in steady state.

(2)

How large (in percent) is the contribution of convection to the total flux at 60°C. (a)

at the water surface.

(b)

halfway up the capillary (z = h/2)

(c)

at the top of the capillary ?

Data : At 60°C : D = 0.305 cm2/s, vapour pressure of water = 0.199 bar At 10°C : D = 0.240 cm2/s, vapour pressure of water = 0.0123 bar 1 atm = 1.013 bar Dry air Dry air ch h z

csat

Water Water

Fig. 2.41 : Water vessel covered with a sponge

Sol. : (1)

nH2O (60°C) Wanted : n (10°C) H2 O

We now rename the species : 1 = H2O, 2 = air The concentration profile within such a capillary was already is given by : z

1 – y1  1 – y1l  l = 1 – y  1 – y10 10  The air that is blown over the top of the tube has a water vapour content of y1l = 0

… (1) … (2)

So we can rearrange equation (1) and get for the concentration profile : 1–

y1 = 1 – (1 – y10)

z l

… (3)

Principles of Mass Transfer Operations

2.138

Diffusion Mass Transfer

The total flux of H2O within the capillary is independent of z. The total flux was also calculated by equation : 1 Dc  1 – y1l  Dc n1 = l ln 1 – y  = l ln 1 – y  … (4) 10 10   At 60°C, D = 0.305 cm2/s, y10 = 0.199/1.013 = 0.196, c = p/R (273.15 + 60) = p/(333.15R) At 10°C, D = 0.240 cm2/s, y10 = 0.0123/1.013 = 0.0121, c = p/R (273.15 + 10) = p/(283.15R) Therefore, 1 D (60°C) c (60°C)  p ln 1 – y (60°C) D (60°C) 333.15 R ln (1 – y (60°C)) l n (60°C) 10     10 = n1 (10°C) = D (10°C) c (10°C)  p ln (1 – y10 (10°C)) 1    D (10°C) 283.15 R ln 1 – y (10°C) l   10   0.305 283.15 ln (1 – 0.196) = 0.240 333.15 ln (1 – 0.0121) = 19.4     (2) The total fux was calculated above (Equation 4). It is also given by : n1 = j1 + c1 v*

… (5)

The diffusive flux (Fick's 1st law), that is part of the flux, is depending on z. z

Dc  1 – y1l  l  1 – y1l  j1 = l (1 – y10) 1 – y  ln 1 – y  10 10   With this, the contribution of convection to the total flux can be obtained :

… (6)

z

j1 c1 v*  1 – y1l  l = 1 – n = 1 – (1 – y10) 1 – y  … (7) n1 1 10  This allows to calculate the contribution of convection at 60°C at the three positions : (a) At z = 0 : c1 v* = 1 – (1 – y10) = y10 n1 = 0.196 = 19.6% (b) At z = h/2 :

c1 v* n1

1 2 1 = 1 – (1 – y10) 1 – y  = 1 – (1 – y10)1/2 10 

= 0.103 = 10.3 % c1 v* (c) At z = h : = 0 n1 (94) Ammonia gas (A) and nitrogen (B) are stored in 2 large storage tanks respectively. The tanks are maintained at constant pressure of 1.0132 × 105 Pa pressure and temperature of 298 K. A uniform tube 0.1 m long connects the 2 tanks, as shown below :

Ammonia Tank (A)

Nitrogen Tank (B)

(Z2 – Z1) 1

2

Fig. 2.42

Principles of Mass Transfer Operations

2.139

Diffusion Mass Transfer

The partial pressure of A at point 1 is pA1 = 1.013 × 104 Pa and at point 2, pA2 = 0.507 × 104 Pa. The diffusivity DAB = 0.230 × 10−4 m2/s. R = 8314 m3.Pa/kg-mole.K. (a) Calculate the flux JA at steady state. (b) Repeat for JB. Comments on your results. Sol. : Given :

Total pressure PT = 1.0132 × 105 Pa (constant) Temperature = 298 K DAB = 0.230 × 10−4 m2/s R = 8314 m3.Pa/kg-mole.K At point 1, pA1 = 1.013 × 104 Pa At point 2, pA2 = 0.507 × 104 Pa Diffusion path = (z2 − z1) = 0.1 m

We can use the following equation for flux of A in a binary mixture with B : JA =

DAB (pA1 − pA2) RT (z2 − z1)

Component – A is diffusing from point 1 to point 2, as its partial pressure is higher at point 1. Putting in the number and check for the appropriate units :

0.230 × 10−4 m  (1.013 × 104 − 0.507 × 104 Pa) s  = 8314 m2 ⋅ Pa ⋅ 298 K (0.1 m) kg-mole.K   2

JA

JA = 4.697 × 10−7 kg-mole/m2.s We can use the Dalton’s Law of partial pressures to determine the partial pressures of component-B at points 1 and 2 : pT = pA + pB At point 1 : pB1 = PT − pA1 PB1 = 1.0132 × 105 − 1.013 × 104 = 91,190 Pa At point 2 : pB2 = PT − pA2 PB2 = 1.0132 × 105 − 0.507 × 104 = 96,250 Pa Component-B is diffusing in the opposite direction to component-A : from point 2 to point 1, as the partial pressure B at point 1 is higher. We can calculate the flux of B in A using : JB =

DAB (pB1 − pB2) RT (z2 − z1)

Principles of Mass Transfer Operations

2.140

Diffusion Mass Transfer

0.230 × 10−4 m  (91‚190 − 96‚25 Pa) s  = 3.Pa m 8314  kg-mole.K ⋅ 298 K (0.1)  2

JB

Thus, JB = 4.697 × 10−7 kg-mole/m2.s (Ans.) The flux for component-B is the same as the flux for component-A, but with a negative sign, indicating that it is in the opposite direction. The 2 components are diffusing in opposite direction. This is the concept of Equimolar Counter-Diffusion. (95) One Component Diffusing in another Stagnant Component : Oxygen (A) is diffusing through carbon dioxide (B) under steady-state conditions, with the CO2 non-diffusing. The total pressure is 1 × 105 N/m2, and the temperature 0°°C. The diffusion path is 2.0 mm. The partial pressures of oxygen at the 2 ends are 13,000 and 6,500 N/m2 respectively. The diffusivity of the mixture is 1.87 × 10−5 m2/s. Calculate the molar flux of O2 in the mixture. Given : R = 8314 (m3.Pa)/(kg-mole.K). Sol. : This is a case of component-A diffusing in another non-diffusing component-B. The equation to be used is : DAB P P − pA2 NA = RT (z − z ) ln P − p  2 1  A1 Given : P = 1 × 105 N/m2 (i.e. Pa) T = 0°C = 273 K DAB = 1.87 × 10−5 m2/s R = 8314 m3.Pa/kg-mole.K (z2 − z1) = 2.0 mm = 2 × 10−3 m pA2 = 6,500 N/m2 (i.e. Pa) pA2 = 13,500 N/m2 (i.e. Pa) Putting in all the numbers and simplifying the units :

1.87 × 10−5 m  (1.0 × 105 Pa) s   1.0 × 105 − 6‚500  = ln 3 8314 m .Pa  (273 K) (2 × 10−3 m) 1.0 × 105 − 13‚500 kg-mole.K  2

NA

NA = 2.97 × 10−5

kg-mole m 2⋅ s

(Ans.)

EXERCISE FOR PRACTICE (1) Ammonia gas (A) and nitrogen gas (B) are diffusing in counter diffusion through a straight glass tube 0.61 m long with an ID of 24.4 mm at 298 k and 101.32 kPa. Both the ends of tube are connected to large mixed chambers at 101.32 kPa. The partial pressure of NH3 in one chamber is constant at 20 kPa and 6.666 kPa in the other chamber. The diffusivity at 298 k and 101.32 kPa is 2.3 × 10–5 m2/sec. (a) Calculate the diffusion of NH3 in kg mole/sec. (b) Calculate the diffusion of N2. (c)

Calculate the partial pressures at a point 0.305 m in the tube.

(Ans. : (a) Diffusion of NH = 9.48 × 10 3

–11

–

kg mole/sec‚ (b) PA = 1.333 × 104 Pa

)

Principles of Mass Transfer Operations

2.141

Diffusion Mass Transfer

(2) NH3 gas is diffusing through N2 under steady-state conditions with N2 non-diffusing since it is insoluble in one boundary. The total pressure is 1.013 × 105 Pa and the temperature is 298 k. The partial pressure of NH3 at one point is 1.333 × 104 Pa and other point 20 mm away it is 6.666 × 103 pa. The DAB for the mixture at 1.013 × 105 Pa and 298 K is 2.30 × 10–5 m2/s. Calculate the flux of NH3 is kg mole/s.m2.

(Ans. : NA = 3.44 × 10–6 kg mole/s.m2) (3) An ethanol (A) – water (B) solution in the form of a stagnant film 2 mm thick at 293 K is in constant at one surface with an organic solvent in which ethanol is soluble and water is insoluble. (Hence NB = 0). At point 1, the concentration of ethanol is 16.8 weight % and solution density. ρ1 = 972.8 kg/m3. At point 2, the concentration of ethanol is 6.8 weight % and ρ2 = 988.1 kg/m3. The diffusivity of ethanol is 0.740 × 10–9 m2/sec. Calculate the steady-state flux NA.

(Ans. : NA = 8.99 × 10–7 kg mole/s.m2) (4) The gas hydrogen at 17o C and 0.010 atm. pressure is diffusing through a membrane of vulcanised neoprene rubber 0.5 mm thick. The pressure of H2 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 H2 gas in neoprene at 17o C is 0.051 m3 (at STP of 0o C and 1 atm)/m3. solid.atm. and diffusivity DAB = 1.03 × 10–10 m2/s at 17o C.

(Ans. : NA = 4.69 × 10–12 kg mole H2/s.m2) (5) Predict the diffusion coefficient of acetone [CH3 CO CH3] in water at 25 o C and

50 oC

using Wilke-Chang equation. The experimental value is 1.28 × 10–9 m2/S at 25o C. Data : Viscosity of water at 25 o C = µB = 0.893 × 10–3 Pa·sand at 50 oC = 0.5494 × 10–3 Pas. VA = 0.0740 m3/kg mole φ = 2.6 MB = 18

(Ans. : D

AB50oC

= 2.251 × 10–9 m2/sec‚ DAB

25oC

= 1.277 × 10–9 m2/sec

)

(6) A gas of carbon dioxide and air is contained in a tube at 101.32 kPa pressure and 44°C. At one point the partial pressure of CO2 is PA1 = 58.21 kPa and at a point 0.035 m distance away, PA2 = 18.76 kPa. If the total pressure is constant throughout the tube, calculate the flux of CO2 at steady state.

–6

2

(Ans : flux of CO2 = 7.57 × 10 k mole/m .s)

(7) Gaseous ammonia and air are diffusing in counter diffusion through a straight glass tube 0.6096 m long with an inside diameter of 25.4 mm at 0oC and 101.32 kPa. Both ends of the tube are connected to large mixed chambers at 101.32 kPa. The partial pressures of NH3 are kept constant at 22.1 kPa in one chamber and 7.52 kPa in the other. (a)

Calculate the diffusion of ammonia in k mole/s.

(b) Calculate the diffusion of air.

Principles of Mass Transfer Operations (c)

2.142

Diffusion Mass Transfer

Calculate the partial pressures halfway along the tube and plot PA, PB and P against distance z.

(Ans. : (a) Diffusion of ammonia = 1.06 × 10

–8

k mole/s; (b) Diffusion of air = –1.65 ×

–6

10 k mole/s; (c) Partial pressure of ammonia halfway along the tube = 14.81 kPa) (8) Oxygen is diffusing in a straight tube 0.2 m long containing helium at 298 K and a total pressure of 101 320 Pa. The partial pressure of O2 at one end is 11 500 Pa and 1215 Pa at the other. Helium is insoluble in one boundary, and hence is non-diffusing or stagnant. Calculate the steady state flux of oxygen. –4

2

(Ans. : Steady state flux of O2 = 7.04 × 10 k mole/m s) (9) For a mixture of methane gas and ethanol vapour, predict the diffusivity : (a)

At atmospheric pressure and 15 and 200 oC;

(b)

At 3.0 atmospheres and 100 oC. –5

2

–6

2

(Ans. : (a) DAB = 3.20 × 10 m /s; (b) DAB = 7.04 × 10 m /s) (10) A tube 11 cm long contains carbon monoxide and nitrogen gases at a total pressure of 101.32 kPa. The partial pressure of N2 is 80 torr at one end and 10 torr at the other. Using experimental data as much as possible, calculate the steady state Equimolal counter diffusion flux of nitrogen : (a) At 15°C; (b) At 200°C; (c) At 47°C and a total pressure of 200 kPa, with the same N2 partial pressures as before. –7

2

–6

2

–6

2

(Ans : (a) 7.35 × 10 k mole/m s; (b) 1.04 × 10 k mole/m s; (c) 3.93 × 10 k mole/m s) (11) 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 325.6 K and 101 325 Pascals absolute pressure. The vapour pressure of naphthalene at 325.6 K is 1.0 torr. The diffusivity of naphthalene in air at 0°C is 5.16 × 10 evapourating.

–6

2

m /s. Calculate the rate at which the naphthalene is currently –6

2

(Ans. : Rate of the naphthalene evapourating = NA = 3.46 × 10 k mole/m .s) (12) Oxygen diffuses through a solution of bovine serum albumin (BSA) at 18°C, to which it is shown that oxygen does not bind. Predict the diffusivity of oxygen through a protein solution containing 0.96 g protein/10 ml solution. –9

2

(Ans. : Diffusivity = 1.64 × 10 m /s) (13) Hydrogen gas at 1.4 atmospheres absolute pressure and 27°C flows inside a 6 foot long vulcanized neoprene tube with a 0.5 inch outer and 0.25 inch inner diameter. Calculate the diffusion rate through the tube walls, assuming no resistance to diffusion outside the slab and zero partial pressure of H2 on the outside. (Conversion factors : 1 foot = 0.3048 meters, 12 inches = 1 foot). –12

(Ans. : NA = 5.43 × 10 k mole H2 per second) (14) Oxygen gas at 2.0 atm and 27°C is flowing in a vulcanized rubber tube 4.5 mm inside diameter and 9.5 mm outside diameter. Calculate the leakage of O2 to the atmosphere through the walls of tube 5.0 meters long. (Ans. : Leakage of O2 = 5.13 × 10

–11

k mole O2 per second)

Principles of Mass Transfer Operations

2.143

Diffusion Mass Transfer

(15) Hydrogen gas diffuses through a 2 mm thick polyethylene sheet at 25°C. The partial pressure of H2 inside is 1.2 atm. and zero outside. Calculate the steady state flux of H2. (Ans. : Steady state flux of H2 = 1.748 × 10

–10

2

k mole/m s)

(16) Oxygen gas at 1.75 atm and 30°C diffuses through 1.5 mm nylon and a 6.5 mm vulcanized rubber membrane in series to the atmosphere. Assuming that there are no other resistances to diffusion, calculate the steady state flux of O2. (Ans. : Steady state flux of O2 = 7.004 × 10

-14

2

k mole/m s)

(17) The diffusivity of methanol in air at 298 K and 1 atm. was measured to be 0.16 cm2/s in an Arnold diffusion cell. The cell has a 0.8 cm2 cross-sectional area and a 20 cm height. The initial liquid height is 16 cm. At 298 K, the vapour pressure of methanol is 1.7 × 10–4 Pa and its density is 0.7914 g/cm3. Calculate the liquid height after 10 hours. (18) Determine the time for one mole of ethanol to diffuse through a 4 µm thick stagnant film of water if the concentrations at the two planes of the film are 0.1 and 0 kg mole/m3, respectively. The area for mass transfer is 100 cm2 and the diffusivity of ethanol in water is 0.84 × 10–5 cm2/s. (19) Ammonia, NH3 is selectively removed from an air NH3 mixture by absorption into water. In this steady-state process, ammonia is transferred by molecular diffusion through a stagnant gas layer 2 cm thick and then through a stagnant water layer 1 cm thick. The concentration of ammonia at the outer boundary of the gas layer is 3.42 mole percent and the concentration at the lower boundary of the water layer is essentially zero. The temperature of the system is 15 oC and the total pressure is 1 atm. At 15 oC, the diffusivity of ammonia in air is 0.215 cm2/s and in liquid water it is 1.77 × 10–5 cm2/s. The concentration at the interface between the gas and the liquid phases is given by the following equilibrium data : Ammonia Partial Pressure (mm Hg)

Ammonia Concentration (mole/cm3) × 106

5

6.1

10

11.9

15

20.0

20

32.1

25

53.6

30 84.8 (a) Calculate the mole fractions of ammonia at the gas-liquid interface. (b) Determine the total flux of ammonia. (20) A 2 mm deep pool of ethyl alcohol is exposed to air at 23 oC and 1 atm. This pool is evapourating through a resistance equivalent to a 2 mm thick layer of stagnant air. Ethanol properties at 23 oC are as follows : Density 0.79 g/cm3, diffusivity in air 0.132 cm2/sec, vapour pressure 53 mm Hg. (a) Calculate the time required for complete evapouration of the alcohol pool. (b) For the above calculation you presumably took the alcohol surface temperature to be 23 oC. What may cause it to be different ? How ? (c) If air is moving over the surface of the pool, approximately how would you expect the evapouration time to change with air velocity ? Why ?

Principles of Mass Transfer Operations

2.144

Diffusion Mass Transfer

(d) Would you expect the alcohol surface temperature to depend on air velocity ? Yes ? No ? Are both answers possible ? Explain. (21) Ammonia gas (A) and nitrogen (B) are stored in 2 large storage tanks respectively. The tanks are maintained at constant pressure of 1.0132 × 105 Pa pressure and temperature of 298 K. A uniform tube 0.1 m long connects the 2 tanks. The partial pressure of A at point 1 is pA1 = 1.013 × 104 Pa and at point 2, pA2 = 0.507 × 104 Pa. The diffusivity DAB = 0.0230 × 10–4 m2/s. R = 8314 m3.Pa/kg-mole.K. (a)

Calculate the flux JA at steady state.

(b)

Repeat for JB.

Comments on your results. (Ans. : JA = 4.697 × 10–7 kg moles/m2.s, JB = – 4.697 × 10–7 kg moles/m2.s, both fluxes JA and JB are same, but with negative sign, indicating that it is in the opposite direction). (22) The gas CO2 (MW = 44) is diffusing at steady state through a tube 20 cm long having a diameter of 1.0 cm and containing N2 (MW = 28) at 298 K. The total pressure is constant at 101.32 kPa. The partial pressure of CO2 at one end is 456 mm Hg and 76 mm Hg at the other end. The diffusivity DAB is 0.167 cm2/s at 298 K. Calculate the flux of CO2 in N2. Repeat your calculations in the diffusion is between (a) H2 (MW = 2) and N2, where the diffusivity DAB is 0.784 cm2/s at 298 K, and (b) NH3 (MW = 17) and N2, where the diffusivity DAB is 0.230 cm2/s at 298 K. Discuss your results. (Ans. : 1.707 × 10–6 kg-mole/m2.s) (23) What equation would you use to estimate the diffusivity of O2 in liquid water at 25°C ? Which of the necessary constants do you know (can you calculate) ? What are their values ? You do not need to solve the equation. (24) A method for separating He from natural gas was proposed. It had been reported that Pyrex glass is impermeable to all gases except for He. The diffusion coefficient for He is about 25 times that for H2. Assuming you have a Pyrex tub with an inner diameter of R1 and an outer diameter of R2. Obtain an expression for the rate of He leakage through the tube in terms of the diffusivity of He in Pyrex (DHe/Pyrex), the interfacial concentration of He in the Pyrex and the tube dimensions. (25) A small quantity of water (component – A, MW = 18) is located at the bottom of a narrow metal tube. The tube is exposed to dry air (component – B, MW = 29). The total air pressure is 1.01325 × 105 Pa and the temperature is 293 K. The water is held at a constant temperature of 293 K and it slowly evapourates and diffuses through the air in the tube. The diffusion path (from the surface of water to the open end of the tube) is 0.1524 m long. Make a sketch of the above system. Given that R = 8314 m3.Pa/kg-mole.K and that the system can be assumed isothermal, calculate the rate of evapouration of water at steady-state in kg-mole/s.m2. The diffusivity of water vapour at 293 K and 1 atm pressure is 0.250 × 10–4 m2/s. Vapour pressure of water at 293 K is 0.0231 atm. Repeat your calculations if component – A is (a) Benzene, MW = 78, (b) Ethanol, MW = 46. The diffusivities of benzene and ethanol are 9.62 × 10–6 m2/s and 1.35 × 10–5 m2/s respectively.

Principles of Mass Transfer Operations

2.145

Diffusion Mass Transfer

The vapour pressure of pure component can be estimated from the following Antoine Equations : (where Pvp is in kPa, T is in K). 2948.78 Benzene : ln (Pvp) = 14.1603 – T – 44.5633 3423.53 Ethanol : ln (Pvp) = 16.1963 – T – 55.7152 (Ans. : NA = 1.595 × 10–7 kg-mole/s.m2) (26) The solute HCl (component–A) is diffusing through a thin film of water (component-B) of thickness 2.0 mm at 298 K. The concentration of HCl at point 1 at the boundary of the film is 12.0 wt% HCl (density = 1060.7 kg/m3), and at the other boundary at point 2 it is 6.0 wt% HCl (density = 1030.3 kg/m3). The diffusion coefficient of HCl in water DAB is 2.5 × 10–9 m2/s. Assuming steady-state and that water is non-diffusing, calculate the flux of HCl in kg-mole/m2.s. (Ans. : 2.372 × 10–6 kmol/m2.s) (27) An ethanol-water solution in the form of a stagnant film 2.0 mm thick at 293 K is in contact with an organic solvent in which ethanol is soluble and water is not. Hence, ethanol is diffusing through the solution into the solvent. At point 1 in the solution the concentration of ethanol is 16.8 wt% and the solution density is 972.8 kg/m3. At point 2 the concentration of ethanol is 6.8 wt% and 988.1 kg/m3. The diffusivity of ethanol in water DAB = 0.740 × 10–9 m2/s at 293 K. Also given MW of ethanol = 46, MW of water = 18. Determine the steady-state flux NA (in kg-mole/m2.s) between point 1 and point 2 using the following equation : DAB Cavg 1 – XA2 NA = (z – z ) ln 1 – X  2 1 A1  where, Cave is the average total concentration (ethanol and water) in the solution (kg-mole/m3). Cave = (C1 + C2)/2; CA1 is the total concentration in point 1 and C2 is the total concentration in point 2. xA1 is the mole fraction of A in point 1. xA2 is the mole fraction of A in point 2. (Ans. : 8.967 × 10–7 kg-mole/m2.s) (28) Using the general equation for molecular flux given below : dCA CA NA = DAB  dz  + C (NA +NB)   Derive the following equation for the case of diffusion in liquid whereby one component is diffusing another component is stagnant : DAB C C – CA2 NA = Z – Z ln C – C  2 1 A1  The symbols have their usual meanings.

Principles of Mass Transfer Operations

2.146

Diffusion Mass Transfer

(29) A flat plug 30 mm thick having an area of 4.0 × 10–4 m2 and made of vulcanized rubber is used for closing an opening in a container. The gas CO2 at 25°C and 2.0 atm pressures is inside the container. Calculate the total leakage or diffusion of CO2 through the plug to the outside in kmol/s at steady state. Assume that the partial pressure of CO2 outside is zero. The solubility of CO2 is 0.90 m3 STP/(m3-rubber·atm-CO2). The diffusivity is 0.11 × 10–9 m2/s. (Ans. : NA = 1.44 × 10–10 kgmole/m2.s) (30) An absorption tower has been proposed to selectively remove the pollutant sulfur dioxide (SO2) from an exhaust gas stream. One of the design parameters, the Schmidt number, requires a value of the gas diffusivity of the diffusing species in the surrounding air. Estimate the diffusivity of SO2 in air at 1 atm and 200°C using the correlation of Fuller et al. How well do your estimates compared with the experimental value measured at 273 K : DAB = 0.122 cm2/s ? (Ans. : 0.282 cm3/s, to compare with the experimental result, correct it to 474 K assuming the temperature dependency given by Fuller et al. (473)1.75 = (0.122 cm2/s) (2.617) = 0.319 cm2/s. Therefore, Fuller et. (273)1.75 al. correlation differs from the extrapolated experimental value by approximately 10%)).

(DAB)473 = (DAB)273

(31) Dry ice (solid carbon dioxide : CO2) is placed in the bottom of a capillary tube (see schematic below) of height L. Pure air (CO2 free) is blown across the top of the tube. The capillary tube is axisymmetric with circular cross-sections. The radius of the capillary tube at the top is rL, while at the bottom r0 and at any location rL − r0 z : r = (r0 + B ⋅ z) = r0 + L ⋅ z.   (a) Calculate the molar flux (mol/s/m2) of gaseous carbon dioxide, at z = L/2 at steady-state, assuming slow sublimation of dry ice (no convection effects). (c) Using the above results, what percentage of the CO2 molar concentration at z = 0 is the molar concentration of CO2 at z = L/2 at steady state if r0 = 1.5 rL ? rL

z

ro

L

r

Fig. 2.43

Principles of Mass Transfer Operations

2.147

Diffusion Mass Transfer

(32) The bottom of a deep cylindrical vessel (30 cm diameter) is covered with a thin slab of salt. At time t = 0, the vessel is filled with pure water. The salt dissolves in the water, maintaining a fixed molar concentration (mol/m3) at the water-salt interface. If the molar concentration of the solid salt in solution at the water-salt interface is approximately 7 Kmol/m3 what is the mass of salt dissolved into water after 5 h ? Justify your approach. Additional Information : The salt-water diffusion coefficient is approximately 10−9 m2/s, while the salt’s molecular weight is MWA = 58 g/mol. (33) A cylindrical 10 cm long glass tube with 0.5 cm in diameter contains 1 mm deep water. Air at 10% relative humidity gently flows outside the tube, (a) calculate the water concentration profile at steady state inside the tube and (b) how long will it take to dry up the test tube ? Take any physical properties you need from problem 31). (34) Water evaporates from an open tank into absolutely still air at 20°C and fifty percent saturated. How much will the water drop in a day (a) when the diffusion film thickness is l = 20 mm and (b) when there is no such film (l = ∞) ? Data : Saturation pressure of water vapour, Psat (H2O : 20°C) = 2.34 kPa. Gas constant, R = 8.314 J/(mol.K) Diffusivity of water, d = 0.25 cm2/sec. (35) In many catalytic chemical process, the mass transfer of reactant or product controls the reaction rate. Here, we consider an infinitely fast and irreversible chemical reaction at steady state. 2A → B that is taking place on a flat catalyst surface. Both species (A) and (B) behave like ideal gases. Close to the surface, they are diffusing through a thin unstirred film of thickness l = 0.1 mm. Outside of this film, the gases are ideally mixed and the partial pressures of both species are constant (partial pressure of (A) outside the film : 1 bar). The temperature in the reactor is 600 K and the total pressure 4 bar. The diffusion coefficient DAB = D at these conditions is about 0.3 cm2/sec. (a)

Calculate the total flux of species (A).

(b)

Derive the concentration profile of species (A) in the thin film. (1 bar = 105 Pa).

(c)

What is the molar fraction of species (B) at z = 0 (catalyst surface) if at z = l yB1 = 0.5 ?

(36) Dry ice (solid carbon dioxide : CO2) is placed in the bottom of a capillary tube (see schematic below) of height L. Pure air (CO2 free) is blown across the top of the tube. The capillary tube is axisymmetric with circular cross-sections. The radius of the capillary tube at the top is rL, while at the bottom r0. (a)

What is the profile of the molar fraction of CO2 along the capillary tube assuming fast sublimation (concentrated solution) of dry ice, ideal gases, isothermal conditions and constant total pressure ?

(b)

What is the molar fraction of CO2 at z = L/2 if r0 = 2 rL and yCO2 (z = 0) = yAo = 0.4 ?

Principles of Mass Transfer Operations

2.148

Diffusion Mass Transfer

rL

z

ro

L

r

Fig. 2.44

(37) In some laboratory plant, polymer tubes are used as ducts for hydrogen (H2) gas. In total, the plant contains 10 meters of tubes, with an inner radius Ri = 1.0 cm and an outer radius Ro = 1.3 cm. The tubes are slightly permeable for H2 at 25°C, the partition coefficient in the wall is 4 ⋅ 10−6 mol H2/cm3⋅atm and the diffusion coefficient in the wall is 2.5⋅10−6 cm2/sec. The environment around the plant is well ventilated and practically free of H2. The hydrogen pressure in the tubes is 1.2 atm. everywhere, the temperatures is 25°C. How much H2 is diffusing out of the tubes during 1 day ? (38) During a summer thunderstorms, a house owner forgets to close the skylight. When the thunderstorm is over, a 1.235 m2 water slop6 of uniform thickness of 0.3 cm spreads over the floor. Water evaporating from this slop does so as if it were diffusing across an air film, which is 0.1 cm thick. The diffusion coefficient of water in 30°C air is about 0.28 cm2/sec. If the air outside the film is 55% saturated, how long will it take until the floor is dry again ? Additional data : Gas constant R = 8.314 J/(mol⋅K) Vapour pressure of water at 30°C : pH2O = 4.35 kPa Molar mass of water MW = 18 g/mol Density of water pH2O = 1000 kg/m3 (39) The gas CO2 (MW = 44) is diffusing at steady state through a tube 20 cm long having a diameter of 1.0 cm and containing N2 (MW = 28) at 298 K. The total pressure is constant at 101.32 kPa. The partial pressure of CO2 at one end is 456 mm-Hg and 76 mm-Hg at the other end. The diffusivity DAB is 0.167 cm2/s at 298 K. Calculate the flux of CO2 in N2. Repeat your calculations in the diffusion is between (a) H2 (MW = 2) and N2, where the diffusivity DAB is 0.784 cm2/s at 298 K, and (b) NH3 (MW = 17) and N2, where the diffusivity DAB is 0.230 cm2/s at 298 K. Discuss your results. (Ans. : 1.707 × 10−6 kg-mole/m2⋅s)

Principles of Mass Transfer Operations

2.149

Diffusion Mass Transfer

(40) Ammonia gas (component-A) and nitrogen (component-B) are diffusing in counterdiffusion through a straight glass tube 0.610 m long with an inside diameter of 24.4 mm at 298 K and 101.32 kPa. Both ends of the tube are connected to a large mixed chambers at 101.32 kPa. The partial pressure of NH3 in one chamber is constant at 20.0 kPa and 6.666 kPa in the other chamber. The diffusivity at 298 K and 101.32 kPa is 2.30 ×10−5 m2/s. (a) Calculate the diffusion of NH3 in kg-mole/s. (b) Calculate the diffusion of N2 in kg-mole/s. (c) Calculate the partial pressures at a point 0.305 m in the tube. (Ans. : (a) 9.48 × 10−11 kg-mole/s; (c) pNH3 = 1.333 × 104 Pa) (41) A small quantity of water (component-A, MW = 18) in located at the bottom of a narrow metal tube. The tube is exposed to dry air (component-B, MW = 29). The total air pressure is 1.01325 × 105 Pa and the temperature is 293 K. The water is held at a constant temperature of 293 K and it slowly evaporates and diffuses through the air in the tube. The diffusion path (from the surface of water to the open end of the tube) is 0.1524 m long. Make a sketch of the above system. Given that R = 8314 m2.Pa/kg-mole.K and that the system can be assumed isothermal, calculate the rate of evaporation of water at steady-state in kg-mole/s.m2. The diffusivity of water vapor at 293 K and 1 atm pressure is 0.250 × 10−4 m2/s. Vapor pressure of water at 293 K is 0.0231 atm. Repeat your calculations if component-A is (a) Benzene, MW = 78, (b) Ethanol, MW = 46. The diffusivities of benzene and ethanol are 9.62 × 10−6 m2/s and 1.35 × 10−5 m2/s respectively. The vapor pressure of pure component can be estimated from the following Antoine equations : (where PVP is in kPa, T is in K). 2948.78 Benzene : ln (PVP) = 14.1603 − T − 44.5633 3423.53 Ethanol : ln (PVP) = 16.1962 − T − 55.7152 (Ans. NA = 1.595 × 10−1 kg-mole/sm2) (42) Using the general equation for molecular flux given below : dCA CA NA = −DAB  dz  + C (NA + NB)   Derive the following equation for the case of diffusion in liquid whereby one component is diffusing while another component is stagnant : DAB C C − CA2 NA = z − z ln C − C  2 1  A1 The symbols have their usual meanings. Analyze the situation below : The solute HCl (component-A) is diffusing through a thin film of water (component-B) of thickness 2.0 mm at 298 K. The concentration of HCl at point 1 at the boundary of the film is 12.0 wt.% HCl (density = 1060.7 kg/m3), and at the boundary at point 2 it is 6.0 wt.% HCl (density = 1030.3 kg/m3). The diffusion coefficient of HCl in water DAB is 2.5 × 10−9 m2/s. Assuming steady-state and that water is non-diffusing, calculate the flux of HCl in kg-mole/m2.s.

Principles of Mass Transfer Operations

2.150

Diffusion Mass Transfer

(43) An ethanol-water solution in the form of a stagnant film 2.0 mm thick at 293 K is in contact with an organic solvent in which ethanol is soluble and water is not. Hence, ethanol is diffusing through the solution into the solvent. At point 1 in the solution the concentration of ethanol is 16.8 wt% and the solution density is 972.8 kg/m3. At point 2 the concentration of ethanol is 6.8 wt% and 988.1 kg/m3. The diffusivity of ethanol in water DAB is 0.740 × 10−9 m2/s at 293 K. Also given MW of ethanol = 46, MW of water = 18. Determine the steady-state flux NA (in kg-mole/m2.s) between point 1 and point 2 using the following equation : DAB Cave 1 − xA2 NA = (z − z ) ln 1 − x  2 1  A1 where, Cave is the average total concentration (ethanol and water) in the solution (kg-mole/m3) Cave = (C1 + C2)/2; CA1 is the total concentration in point 1 and C2 is the total concentration in point 1. xA1 is the mole fraction of A in point 1. xA2 is the mole fraction of A in point 2. (Ans. : 8.967 × 10−7 kg-mole/m2.s) NOMENCLAUTRE Any consistent set of units may be used, except as noted. Symbols Meaning a One-half thickness; radius, m b One-half width, m c Concentration, mole/volume, mole L3 d Differential operator, also pore diameter, m D Diffusivity or diffusivity coefficient, m2/s; Dff, effective diffusivity; DK, Do E f, f', f'' HD J k ln M n N p –

p P

Knudsen diffusivity. Diffusivity for a solute at infinite dilution, m2/s. Fraction of solute removal, dimensionless. Functions Energy of activation, Flux of diffusion relative to the molar average velocity, mole/m2s Constant Natural logarithm Molecular weight, M/mole Number of moles, dimensionless Molar flux relative to a fixed surface, mole/m2. s Vapour pressure, kN/m2 Partial pressure, kN/m2 Permeability, [cm2 gas (STP)/cm2. s.(cm Hg/cm)]

Principles of Mass Transfer Operations Symbols Pt R s S T Tb u –

u v V w x xi y yi y'i z Greek Letters : α ∂ ∆ θ λ µ γ ρ φ π Subscripts : A B i n M x y z 1 2 av. 0 ∞

2.151

Diffusion Mass Transfer Meaning

2

Total Pressure, kN/m Universal gas law constant Solubility coefficient or Henry's law constant [cm3 gas (STP) × cm3 . (cm Hg) Cross-sectional area, m2 Absolute temperature, K Normal boiling point, K Linear velocity m/s Mean molecular velocity, m/s Liquid molar volume, m3/kmol Volume, m3 Rate of diffusion (with no subscript) in the x-direction, m Mole fraction concentration of component i in a liquid. (with no subscript) Distance in the y-direction, m Mole-fraction concentration of component i, in a gas Mole-fraction concentration of component i, diffusing-solute-free basis Distance in z-direction, m; also thickness of membrane or pellet, m; also distance in direction of diffusion, m Thermal diffusivity, m2/s Partial differential operator Difference Time, sec Mean free path of a molecule, m Viscosity, kg/m.s Kinematics viscosity or momentum diffusivity = µ/ρ , m2/s Density, kg/m3 Dissociation factor for a solvent, dimensionless 3.1416 Component A Component B Component i The last of n components Mean In the x-direction In the y-direction In the z-direction Beginning of diffusion path End of diffusion path Average Initial (at time zero) At time ∞; at equilibrium

Principles of Mass Transfer Operations

2.152

Diffusion Mass Transfer

REFERENCES 1.

R.E. Treybal, "Mass-Transfer Operations", Third Edition, McGraw-Hill, 1981.

2.

J.M. Coulson and J.F. Richardson, "Chemical Engineering” Volume 1, Third Edition, Pergamon Press, 1986.

3.

E.L. Cussler, “Diffusion : Mass Transfer in Fluid Systems”, Second Edition, Cambridge University Press, 1998.

4.

P. Chattopadhya, “Unit Operations “(Vol-I), Khanna Publishers, New Delhi, 1996.

5.

G.K.Roy, “Fundamentals of Heat and Mass Transfer”, Second Edition, Khanna Publishers, New Delhi, 1990

6.

A.S. Foust, L.A. Wenzel, C.W. Clump, L. Maus and L.B. Andersen, "Principles of Unit Operations", Second Edition, John Wiley and Sons, 1980.

7.

W.L. McCabe and J.C. smith, “Unit Operations in Chemical Engineering”, Fifth Edition, McGraw Hill, New York, 1993.

8.

A.H.P.Skelland, “Diffusional Mass Transfer”, Kriegar, Malbar FL, 1985.

9.

C.J. Geankoplis, "Transport Processes and Unit Operations", Fourth Edition, Prentice Hall, 2003

10. R.B. Bird et al, “Transport Phenomena”, John Wiley and Sons, 1960. 11. C.J. Geankoplis, “Mass Transport Phenomena”, Columbus, Chio, 1972 12. R.H. Perry and D. Green, "Perry's Chemical Engineers' Handbook", Seventh Edition, McGraw-Hill, 1997. 13. T.K.Shrewood and R.L. Pigford, “Mass Transfer”, McGraw Hill, 1975. 14. C.R. Wilke and J. Crank, “The Mathematics of Diffusion”, Oxford University Press, 1956. 15.

J. Crank and G. Park, “Diffusion in Polymers”, Academic Press, 1968.

,,,

3 CHAPTER

MASS TRANSFER COEFFICIENTS 3.1 3.2

3.3 3.4 3.5 3.6 3.7 3.8 3.9 3.10

3.11

3.12

3.13 3.14

Introduction Types of Mass Transfer Coefficients 3.2.1 Definition of Mass Transfer Coefficient 3.2.2 Mass Transfer Coefficient for Equimolar Counter Diffusion 3.2.3 Mass transfer Coefficient for A diffusing through stagnant, non-diffusing B Local Mass Transfer Coefficient Mass Transfer Coefficient in Turbulent Flow Eddy/Turbulent Diffusion Theories of Mass Transfer Introduction to Convective Mass Transfer Convective Mass Transfer Coefficient Significant Parameters in Convective Mass Transfer Application of Dimensionless Analysis to Mass Transfer 3.10.1 Transfer into a stream flowing under forced convection 3.10.2 Transfer into a phase whose motion is due to natural convection Analogies Among Mass, Heat and Momentum Transfer 3.11.1 Reynolds Analogy 3.11.2 Chilton – Colburn Analogy 3.11.3 Taylor – Prandtl Analogy Convective Mass Transfer Correlations 3.12.1 For Flow Around Flat Plat 3.12.2 For Flow Around Single Sphere 3.12.3 For Flow Around Single Cylinder 3.12.4 For Flow Through Pipes 3.12.5 Mass Transfer to Suspension of Small Particles 3.12.6 Mass Transfer in Packed Beds 3.12.7 Mass Transfer in Boundary Layer Mass Transfer between Phases Simultaneous Heat and Mass Transfer 3.14.1 Condensation of Vapour on Cold Surface 3.14.2 Wet Bulb Thermometer Solved Problems Exercise for Practice Nomenclature References

(3.1)

Principles of Mass Transfer Operations − I (Vol. − I) 3.1

3.2

Mass Transfer Coefficients

INTRODUCTION

Imagine we are interested in transfer of mass from some interface into a well mixed solution. We expect that the amount of mass transferred is directly proportional to the concentration difference and the interfacial area. Thus, (Rate of mass transfer) = k (Interfacial area) (Concentration difference) … (3.1) where the proportional constant is denoted by k and is called as mass transfer coefficient. If we divide both sides of equation (3.1) by the area, we can write the equation in more convenient form as : … (3.2) N1 = k (C1, i – C1) where,

N1 = Flux at interface C1, i = Concentration at the interface C1 = Concentration at bulk of the solution.

Remarks : (i) The flux N1 includes both diffusion and convection, it is like total flux n1, except that it is located at the interface. (ii) The concentration C1, i is at interface but in the same fluid as the bulk concentration C1. (iii) The flux equation (3.2) makes some practical sense. It says that concentration is doubled, the flux also doubles. (iv) It also suggests that if the area is doubled, the total amount of mass transfer will also double but flux per unit area will not change. 3.2

TYPES OF MASS TRANSFER COEFFICIENTS

3.2.1 Definition of Mass Transfer Coefficient Since our understanding of turbulent flow is incomplete, we write the equations for turbulent diffusion in a manner similar to that for molecular diffusion. For turbulent mass transfer we can write, dC * A … (3.3) JA = – (DAB + EM) dz DAB = Molecular diffusivity in m2/sec.

where,

EM = Mass eddy diffusivity in m2/sec. The value of EM is a variable and is near zero at interface or surface and increases as the –

distance from the wall increases. We use an average value E M since the variation of EM is not generally known. Integrating equation (3.3) between points 1 and 2, –

* JA1

DAB + EM = z –z (CA – CA ) 1 2 2 1

… (3.4)

*

The flux JA1 is based on the surface area A1 since the cross-sectional area may vary. The value of z2 – z1, the distance of the path, is often not known. Hence equation (3.4) is written using a '

convective mass transfer coefficient kC as : *

JA1

'

= kC (CA – CA ) 1

2

… (3.5)

Principles of Mass Transfer Operations − I (Vol. − I)

3.3

Mass Transfer Coefficients

'

'

where JA is the flux of A from the surface A1 relative to the whole bulk phase, kC is 1

(DAB + EM) (z2 – z1) an experimental mass transfer coefficient in m/s and CA2 is the concentration at point 2 in kg mole A/m3. '

Note : This definition of a convective mass transfer coefficient kC is similar to the convective heat-transfer coefficient h. 3.2.2 Mass Transfer Coefficient for Equimolar Counter Diffusion The flux of A relative to stationary co-ordinates is given by : dxA NA = – C (DAB + EM) dz + xA (NA + NB) … (3.6) For the case of equimolar counter diffusion, where, NA = – NB so integrating equation (3.6) and calling, (DAB + EM) ' kC = (z – z ) 2 1 We get,

NA = kC' (CA – CA ) 1

… (3.7)

2

Equation (3.7) is the defining equation for the mass transfer coefficient. Often, we define the concentration in terms of mole fraction if a liquid or gas or in terms of partial pressure of gas. Thus, we can define the mass transfer coefficient is several ways as : '

Gases :

NA = kc (CA – CA )

Liquids :

NA = kc (CA – CA )

1

'

1

'

1

'

= kG (pA – pA ) = ky (yA – yA )

2

2

1

'

… (3.8)

2

'

= kL (CA – CA ) = kx (xA – xA )

2

1

2

1

… (3.9)

2

yA = Mole fraction in gas phase

where,

xA = Mole fraction in liquid phase 3.2.3 Mass Transfer Coefficient for a Diffusing Through Stagnant, Non-diffusing B For A diffusing through stagnant, non-diffusing B where NB = 0, equation (3.6) gives for '

steady state,

NA

kc = x (CA – CA ) = kC (CA – CA ) 1 2 1 2 BM

… (3.10)

'

where, kC = mass transfer coefficient for A diffusing through stagnant B. xB – xB 2 1 xB, M = x

 B2 ln x   B1

… (3.11)

Expressing equation (3.10) using other units : Gases :

NA = kc (CA – CA ) = kG (pA – pA ) = ky (yA – yA )

… (3.12)

Liquids :

NA = kc (CA – CA ) = kL (CA – CA ) = kx (xA – xA )

… (3.13)

1 1

2 2

1 1

2

2

1

1

2

2

The relations among mass transfer coefficients and the various flux equations, are given in Table (3.1).

Principles of Mass Transfer Operations − I (Vol. − I)

3.4

Mass Transfer Coefficients

Flux equations for equimolar counter diffusion : Gases : Liquids :

'

'

'

NA = kc (cA – cA ) = kG (pA – pA ) = ky (yA – yA ) 1

2

1

'

2

'

1

2

'

NA = kc (cA – cA ) = kL (cA – cA ) = kkx (xA – xA ) 1

2

1

2

1

2

Flux equations for A diffusing through stagnant, non-diffusing B : Gases : NA = kc (cA – cA ) = kG (pA – pA ) = ky (yA – yA ) 1

Liquids :

2

1

2

1

2

NA = kc (cA – cA ) = kL (cA – cA ) = kx (xA – xA ) 1

2

1

2

1

2

Conversions between mass transfer coefficients : PBM P Gases : kC' = kC' RT = kc RT = kG' P = kG pPM = ky yBM = k'y = kc yBM c = kG yBM P Liquids :

kC' c = kL' c = kL xB, M c = kL' ρ/M = k'x = kx xB, M

(where ρ is density of liquid and M is molecular weight) Types of MTC kc, kL, k'c , kL'

Table 3.1 : Units of mass transfer coefficients SI Units CGS Units m/s cm/s

kx, ky, k'x , k'y kG, kG'

English Units ft/h

kg mol s·m2·mol frac

g mol s·cm ·mol fract

lb mol h·ft2·mol frac

kg mol kg mol s·m2·Pa s·m2·atm

g mol s·cm2·atm

lb mol h·ft2·atm

2

(preferred) 3.3 LOCAL MASS TRANSFER COEFFICIENT

Gas

z

z=Z x

CA1 (in gas) Evaporating liquid

Fig. 3.1 : Mass Transfer to a Confined Liquid

For a fluid flowing past a solid surface or gas flowing over a liquid surface, the rate of mass transfer through the phase interface (or phase boundary) is given by, NA

NA/(NA + NB) – CA /C NA 2 = N + N · F N /(N + N ) – C /C A B A A B A

2

(k mole/m .sec.) … (3.14)

1

where, A and B are two components. A is the solute diffusing. i.e. it enters or leaves the phase. CA = Concentration of A at the beginning of transfer path, (k mole/m3). 1

CA = Concentration of A at the end of transfer path (k mole/m3). 2

The local mass transfer coefficient (F) for the process is defined as : C ∴ F = DAB · Z k mole/m2. sec.

… (3.15)

Principles of Mass Transfer Operations − I (Vol. − I)

3.5

Mass Transfer Coefficients

Note : (i) "F" is defined for a particular location at the phase boundary. (ii) It may vary along this interface depending on the nature of fluid motion. (iii) Flux = (Coefficient) × (Concentration difference). 3.4 MASS TRANSFER COEFFICIENT IN TURBULENT FLOW (i) In case of most practically useful situations involving turbulent flow, it is not possible to compute the mass transfer coefficient, because the flow cannot be described mathematically. (ii) Instead, we depend principally on experimental data, which are limited in scope w.r.t. the circumstances, situations as well as fluid properties. (iii) Hence, it is important to be able to extend their application to situations not covered experimentally and to draw upon the knowledge about other transport properties. (e.g. Heat transfer or fluid mechanics) for help. 3.5 EDDY/TURBULENT DIFFUSION Turbulence :

x (1) z

(2)

l

ux + Dux, CA+ DCA, t + Dt

u'z

ux, CA, t

Fig. 3.2 : Eddy Diffusion

It is characterized by motion of the fluid particles, which is irregular w.r.t. both direction and time. For fluid flowing turbulently in a duct, the flow is in the net, in the axial (X) direction. At any location (2) within the content region of the cross section, the time average of the velocity may be 4x, but at any instant the velocity will actually be 4x + 4x where uix is deviating or fluctuating velocity. The values of u’is will vary with time through a range of (+ve) and (–ve) value, the time average being zero, through Ux1 =

(uix1) 2 will be finite.

… (3.16)

Although, the time average of u Z = 0, since the net flow is axially directed, the deviating 1 velocity in the Z – direction (i.e. ui Z) is non-zero at any instant. In case of fluid in turbulence, eddy diffusion takes place in addition to molecular diffusion and overall rate is given by :

where,

∂C A NA = – (DB + ED) ∂Z DB = Molecular diffusivity. ED

… (3.17)

= Eddy Diffusivity.

Note : (i) The eddy diffusivity depends on flow pattern and the position (Z) in addition to physical properties of the fluid.

Principles of Mass Transfer Operations − I (Vol. − I)

3.6

Mass Transfer Coefficients

(ii) For this reason, simplified methods have been developed to evaluate mass transfer rates in case of turbulent flow. (iii) Analogy between momentum, mass and heat also provides an alternative method for estimation of mass transfer coefficient and their by the transfer rate in case of turbulent condition. 3.6 MODELS FOR MASS TRANSFER AT A FLUID-FLUID INTERFACE (THEORIES OF MASS TRANSFER) The mechanism of transfer can be explained by various theories, important one are as under : (a) Whitman’s two-film theory (1923) (b) Higbie’s penetration theory (1935) (c) Danckwert’s surface renewal theory (1951) (d) Toor and Marchello’s film penetration theory (1958) Theories of Mass Transfer : Of greater interest in separation processes is mass transfer across an interface between a gas and a liquid or between two liquid phases. Such interfaces exist in absorption, distillation, extraction and stripping. pA

Bulk Liquid

pA

Liquid film cAi

cAi

Gas

Gas cAb

z=0

cAb

Well-mixed bulk region at cAb

Interfacial region

z = dL

(a) (b) Fig. 3.3 : Theories for mass transfer from a fluid-fluid interface into a liquid : (a) Film theory; (b) Penetration and surface-renewal theories

At fluid-fluid interfaces, turbulence may persist to the interface. The following theoretical models have been developed to describe mass transfer from a fluid to such an interface. 1.

Film Theory :

A simple theoretical model for turbulent mass transfer to or from a fluid-phase boundary was suggested in 1904 by Nernst, who postulated that the entire resistance to mass transfer in a given turbulent phase is in a thin, stagnant region of that phase at the interface, called a film. This is shown schematically in Fig. 3.3 (a) for the case of a gas-liquid interface, where the gas is pure component A, which diffuses into non-volatile liquid B. Thus, a process of absorption of A into liquid B takes place, without desorption of B into gaseous A. Because the gas is pure A at total

Principles of Mass Transfer Operations − I (Vol. − I)

3.7

Mass Transfer Coefficients

pressure P = pA, there is no resistance to mass transfer in the gas phase. At the gas-liquid interface, equilibrium is assumed so the concentration of A, cA, is related to the partial pressure of A, pA, by some form of Henry's law, for example, cAi = HApA. In the thin, stagnant liquid film of thickness δ, molecular diffusion only occurs with a driving force of cAi – cAb . Since the film is assumed to be very thin, all of the diffusing A passes through the film and into the bulk liquid. If, in addition, bulk flow of A is neglected, the concentration gradient is linear as in Fig. 3.3 (a). Accordingly, Fick's first law, for the diffusion flux integrates to JA =

DAB (cAi – cAb) δ

=

cDAB (xAi – xAb) δ

… (3.18)

If the liquid phase is dilute in A, the bulk-flow effect can be neglected and (3.18) applies to the total flux : DAB cDAB NA = (cAi – cAb) = (xAi – xAb) … (3.19) δ δ If the bulk-flow effect is not negligible, then, we have, cDAB cDAB 1 – xAb ln  1 – x  = (x – xAb) … (3.20) NA = δ δ (1 – xA)LM Ai Ai   xAi – xAb (1 – xA)LM = ln [(1 – x )/(1 – x )] = (xB)LM Ab Ai

… (3.21)

In practice, the ratios DAB/δ in (3.19) and DAB/δ(1 – xA)LM in (3.20) are replaced by mass '

transfer coefficients kc and kc , respectively, because the film thickness, δ, which depends on the flow conditions, is not known. The film theory, which is easy to understand and apply, is often criticized because it appears to predict that the rate of mass transfer is directly proportional to the molecular diffusivity. This dependency is at odds with experimental data, which indicate a dependency of Dn, where n ranges from about 0.5 to 0.75. However, if DAB/δ is replaced with kc, which is then estimated from the 2/3

Chilton-Colburn analogy, we obtain kc proportional to DA/B , which is in better agreement with experimental data. The theory has been and continues to be widely used in the design of mass transfer separation equipment. 2.

Penetration Theory – (Higbie, 1935) :

A more realistic physical model of mass transfer from a fluid-fluid interface into a bulk liquid stream is provided by the penetration theory of Higbie, shown schematically in Fig. 3.3 (b). The stagnant-film concept is replaced by Boussinesq eddies that, during a cycle, (1) move from the bulk to the interface; (2) stay at the interface for a short, fixed period of time during which they remain static so that molecular diffusion takes place in a directional normal to the interface; and (3) leave the interface to mix with the bulk stream. When an eddy moves to the interface, it replaces another static eddy. Thus, the eddies are intermittenty static and moving. Turbulence extends to the interface. In the penetration theory, unsteady-state diffusion takes place at the interface during the time eddy is static. This process is governed by Fick's second law, with boundary conditions. cA = cAb at t = 0 for 0 ≤ z ≤ ×; cA = cAi at z = 0 for t > 0; and cA = cAb at z = × for t > 0.

Principles of Mass Transfer Operations − I (Vol. − I)

3.8

Mass Transfer Coefficients

These are the same boundary conditions as in unsteady-state diffusion in a semi-infinite medium. Thus, the solution can be written as cAi – cA  z  … (3.22) cAi – cAb = erf 2 DAB tc   where, tc = "contact time" of the static eddy at the interface during one cycle. The corresponding average mass transfer flux of A in the absence of bulk flow is given by the following form of equation : DAB NA = 2 (cAi – cAb) … (3.23) πtc … (3.24) or NA = kc (cAi – cAb ) Thus, the penetration theory gives DAB πtc

kc = 2

… (3.25)

which predicts that kc is proportional to the square root of the molecular diffusivity, which is at the lower limit of experimental data. The penetration theory is most useful when mass transfer involves bubbles or droplets, or flow over random packing. For bubbles, the contact time, tc, of the liquid surrounding the bubble is taken as the ratio of bubble diameter to bubble rise velocity. For example, an air bubble of 0.4 cm diameter rises through water at a velocity of about 20 cm/s. Thus, the estimated contact time, tc, is 0.4/20 = 0.02s. 3. Surface Renewal Theory (Danckwerts, 1951) : The penetration theory is not satisfying because the assumption of a constant contact time for all eddies that temporarily reside at the surface is not reasonable, especially for stirred tanks, contactors with random packings, and bubble and spray columns where the bubbles and droplets cover a wide range of sizes. In 1951, Danckwerts suggested an improvement to the penetration theory that involves the replacement of the constant eddy contact time with the assumption of a residence-time distribution, wherein the probability of an eddy at the surface being replaced by a fresh eddy is independent of the age of the surface eddy. The instantaneous mass transfer rate for an eddy with an age t is given by (3.23) for the penetration theory in flux form as, DAB NAt = (cAi – cAb) … (3.26) πt The integrated average is ∞

(NA)avg =

⌠ φ {t} NA dt ⌡ t

… (3.27)

0

Danckwerts selected the model for his surface renewal theory, using the corresponding φ {t} function : … (3.28) φ {t} = se–st 1 where, s = t = fractional rate of surface renewal Combining (3.26), (3.27) and (3.28) and integrating : (NA)avg = Thus,

kc =

DABs (cAi – cAb)

DABs

… (3.29) … (3.30)

Principles of Mass Transfer Operations − I (Vol. − I)

3.9

Mass Transfer Coefficients

4.

Film-Penetration Theory (Toor and Marchello, 1958) : Toor and Marchello in 1958, combined features of the film, penetration and surface renewal theories to develop a film-penetration model, which predicts a dependency of the mass transfer coefficient kc, on the diffusivity, that varies from DAB to DAB. Their theory assumes that the entire resistance to mass transfer resides in a film of fixed thickness δ. Eddies move to and from the bulk fluid and this film. Age distributions for time spent in the film are of the Higbie or Danckwerts type. Fick's second law, still applies, but the boundary conditions are now cA = cAb at t = 0 for 0 ≤ z ≤ ×. cA = cAi at z = 0 for t > 0; and cA = cAb at z = δ for t > 0

Infinite-series solutions are obtained by the method of Laplace transforms. The rate of mass transfer is then obtained in the usual manner by applying Fick's first law at the fluid-fluid interface. For small t, the solution, given as, x DAB1/2 1 + 2 ∑ exp – n2δ2  At = (cAi – cAb )  πt    DABt n=1

… (3.31)

converges rapidly. For large t, NAt

x DABt  DAB  ∑ = (cAi – cAb )  1 + 2 exp – n2π2 2   δ   δ   n=1

… (3.32)

Equation with φ {t} from (3.28) can then be used to obtain average rates of mass transfer. Again, we can write two equivalent series solutions, which converage at different rates. Equations (3.29) and (3.30) become, respectively, NAavg = kc (cAi – cAb ) = (cAi NAavg = kc (cAi – cAb ) = (cAi

x   ∑ – cAb ) exp – 2nδ 1 + 2  n=1  ∞  DAB  1 – cAb )  1+2 ∑  D   δ   n = 1 1 + n2π2 AB sδ2  

(sDAB)1/2

s  DAB



… (3.33) … (3.34)

In the limit, for a high rate of surface renewal, sδ2/DAB, (3.33) reduces to the surface renewal theory, (3.29). For low rates of renewal, (3.34) reduces to the film theory, (3.29). At conditions in n

between, kc is proportional to DAB , where n is in the range of 0.5 to 1.0. The application of the film-penetration theory is difficult because of lack of data on δ and s, but the predicted effect of the molecular diffusivity brackets experimental data. Salient Features : (a) Two Film Theory : Salient points relating to the two-film theory are : (i) Resistance to transfer in the two phases on both sides of the interface is replaced by the two fictitious films one on each side close to the interface. (ii) The transfer in these films is by a steady process of molecular diffusion. (iii) The concentration gradient is assumed to be linear in these films and zero outside. (iv) The time taken for concentration gradient to establish is small compared to the time of transfer or in other words, the film capacity is negligible.

Principles of Mass Transfer Operations − I (Vol. − I)

3.10

Mass Transfer Coefficients

(v) The theory assumes that the turbulence in the bulk fluids dies at the interface. (vi) According to this theory, the mass transfer coefficients for different solvents transferred under identical fluid flow conditions are proportional to their diffusivities. (b) Penetration Theory : Main features of this theory are : (i) The time of exposure of a fluid to mass transfer is generally being short, establishment of concentration gradient of film theory, characteristics of steady state, is not possible. (ii) The transfer is largely due to fresh material brought to the interface by the eddies. (iii) A process of unsteady transfer takes place for a fixed period at the freshly exposed surface. (iv) Each fluid element (eddy) resides for the same time interval at the surface. (v) The mass transfer coefficient is proportional to the square root of diffusivity according to this theory. (c) Surface Renewal Theory : This theory states that : (i) Eddies of the fluid at the surface do not have constant exposure time, but are exposed to varying lengths of time. (ii) On the basis of exposure history, an age distribution for the surface elements is calculated. (iii) Mass transfer is proportional to the square root of the product of diffusivity and surface renewal rate, is generally defined as the rate of production of fresh surface per unit total area of surface. (d) Film Penetration Theory : This theory incorporates some salient features of both the two-film theory and the penetration theory. The outlines of the theory are : (i) Complete resistance to transfer is regarded as lying within a finite laminar film at the interface. (ii) The transfer process is regarded as an unsteady process. (iii) Fresh surface is formed at the intervals from the fluid, which is brought to the interface from the bulk by the eddies. (iv) The transfer then takes place as in the penetration theory and material traversing the film mixes with the bulk of the fluid immediately. (v) For short exposure time, the process is identical to the postulation made in case of penetration theory, whereas for long exposure time with a steady state concentration gradient, conditions similar to two-film theory exist. 3.7 INTRODUCTION TO CONVECTIVE MASS TRANSFER Our discussion of mass transfer in the previous chapter was limited to molecular diffusion, which is a process resulting from a concentration gradient. In system involving liquids or gases, however, it is very difficult to eliminate convection from the overall mass transfer process. Mass transfer by convection involves the transport of material between a boundary surface (such as

Principles of Mass Transfer Operations − I (Vol. − I)

3.11

Mass Transfer Coefficients

solid or liquid surface) and a moving fluid or between two relatively immiscible, moving fluids. There are two different cases of convective mass transfer : (1) Mass transfer takes place only in a single phase either to or from a phase boundary, as in sublimation of naphthalene (solid form) into the moving air. (2) Mass transfer takes place in the two contacting phases as in extraction and absorption. In the first few section we will see equation governing convective mass transfer in a single fluid phase. 3.8 CONVECTIVE MASS TRANSFER COEFFICIENT In the study of convective heat transfer, the heat flux is connected to heat transfer coefficient as, Q … (3.35) A = q = h (ts – tm) The analogous situation in mass transfer is handled by an equation of the form … (3.36) NA = kc (CAs – CA) The molar flux NA is measured relative to a set of axis fixed in space. The driving force is the difference between the concentration at the phase boundary, CAS (a solid surface or a fluid interface) and the concentration at some arbitrarily defined point in the fluid medium, CA. The convective mass transfer coefficient kC is a function of geometry of the system and the velocity and properties of the fluid similar to the heat transfer coefficient, h. 3.9 SIGNIFICANT PARAMETERS IN CONVECTIVE MASS TRANSFER Dimensionless parameters are often used to correlate convective transfer data. In momentum transfer Reynolds number and friction factor play a major role. In the correlation of convective heat transfer data, Prandtl and Nusselt numbers are important. Some of the same parameters, along with some newly defined dimensionless numbers, will be useful in the correlation of convective mass transfer data. The molecular diffusivities of the three-transport process (momentum, heat and mass) have been defined as : µ … (3.37) Momentum diffusivity, v = ρ k Thermal diffusivity, α = … (3.38) ρ Cp … (3.39)

and Mass diffusivity DAB 2

It can be shown that each of the diffusivities has the dimensions of L / t, hence, a ratio of any of the two of these must be dimensionless. The ratio of the molecular diffusivity of momentum to the molecular diffusivity of heat (thermal diffusivity) is designated as the Prandtl Number, Cp µ Momentum Diffusivity v = Pr = = … (3.40) Thermal Diffusivity K α The analogous number in mass transfer is Schmidt number given as, Momentum diffusivity v µ = Sc = D = Mass diffusivity ρDAB AB

… (3.41)

The ratio of the molecular diffusivity of heat to the molecular diffusivity of mass is designated the Lewis Number, and is given by, Thermal diffusivity k α … (3.42) Mass diffusivity = Le = DAB = ρ Cp DAB

Principles of Mass Transfer Operations − I (Vol. − I)

3.12

Mass Transfer Coefficients

Lewis number is encountered in processes involving simultaneous convective transfer of mass and energy.

n¥

cAS – cA¥

y

cAS – cA = [cAS – cA] (y)

n = n(y) x

Fig. 3.4 : Concentration and Velocity Profiles for a Fluid Flowing Past a Solid Surface

Let us consider the mass transfer of solute A from a solid to a fluid flowing past the surface of the solid. The concentration and velocity profile is depicted in Fig. (3.3). For such a case, the mass transfer between the solid surface and the fluid may be written as, … (3.43) NA = kc (CAs – CA∞) Since the mass transfer at the surface is by molecular diffusion, the mass transfer may also described by, NA = – DAB

dCA dy

… (3.44)

y=0

When the boundary concentration, CAs is constant, equation (3.44) may be written as, NA = – DAB

d (CA – CAs) dy

y=0

… (3.45)

Equation (3.43) and (3.44) may be equated, since they define the same flux of component A leaving the surface and entering the fluid d kc (CAs – CA∞) = – DAB dy (CA – CAs) … (3.46) y=0 This relation may be rearranged into the following form : kc DAB

= –

d (CA – CAs)/dy

(CA – CA∞)

… (3.47) y=0

Multiplying both sides of equation (3.47) by a characteristic length, L we obtain the following dimensionless expression : kc L DAB

d (CA – CAs)/dy = –

(CAS – CA∞)/L

y=0

… (3.48)

The right hand side of equation (3.48) is the ratio of the concentration gradient at the surface to an overall or reference concentration gradient; accordingly, it may be considered as the ratio of molecular mass transport resistance to the convective mass transport resistance of the fluid. This ratio is generally known as the Sherwood number, Sh and analogous to the Nusselt number Nu, in heat transfer. 3.10 APPLICATION OF DIMENSIONLESS ANALYSIS TO MASS TRANSFER One of the method of obtaining equations for predicting mass transfer coefficients is the use of dimensionless analysis. Dimensional analysis predicts the various dimensionless parameters, which are helpful in correlating experimental data. There are two important mass transfer

Principles of Mass Transfer Operations − I (Vol. − I)

3.13

Mass Transfer Coefficients

processes, which we shall consider, the transfer of mass into a steam flowing under forced convection and the transfer of mass into a phase which is moving as the result of natural convection associated with density gradients. 3.10.1 Transfer Into a Stream Flowing Under Forced Convection Consider the transfer of mass from the walls of a circular conduit to a fluid flowing through the conduit. The mass transfer is due to the concentration driving force CAs – CA. The important variables, their symbols and their dimensions are listed in the Table (3.2). Table 3.2 Variable

Symbol

Dimensions

tube diameter

D

L

fluid density

ρ

M/L3

fluid viscosity

µ

M/Lt

fluid velocity

v

L/t

DAB

L2/t

kc

L/t

fluid diffusivity mass transfer coefficient

These variables include terms descriptive of the system geometry, the flow and fluid properties and the quantity of importance, kc. Example 1 : Application of the Buckingham Pi Theorem : By the Buckingham method of grouping the variables, the number of dimensionless π groups is equal to the number of variables minus the number of fundamental dimensions. Hence the number of dimensionless group for this problem will be three. With DAB, ρ and D as the core variables, the three π groups to be formed are, a

π1 = DAB ρb Dc kc d

π2 = DAB ρe Df v and

… (3.49) … (3.50)

g

π3 = DAB ρh Di µ

… (3.51)

Substituting the dimensions for π, a

π1 = DAB ρb Dc kc 2

1 =

a

… (3.52)

b

L  M (L)c L t  t  L3

… (3.53)

Equating the exponents of the fundamental dimensions on both sides of the equation, we have, L : 0 = 2a – 3b + c + 1 t : 0=–a–1 M : 0=b Solving these equations, a = –1, b = 0 and c = 1

Principles of Mass Transfer Operations − I (Vol. − I) kc D Thus, π1 = D AB

3.14

Mass Transfer Coefficients

which is the Sherwood number.

The other two π groups could be determined in the same manner, yielding Dv π2 = D AB and

π3 =

µ = Sc ρ DAB

… (3.54) … (3.55)

which is termed as Schmidt Number Dividing π2 by π3, we get,

π2 π3

 Dv  Dv ρ DAB = = µ = Re µ   ρ D   AB

… (3.56)

which is the Reynolds Number. The result of the dimensional analysis of mass transfer by forced convection in a circular conduit indicates that a correlating relation could be of the form, Sh = f (Re, Sc)

… (3.57)

Which is analogous to the heat transfer correlation Nu = f (Re, Pr)

… (3.58)

Example 2 : Application of the Buckingham π-Theorem : To relate mass transfer coefficient to known variables : kc = f (L, ρ µ, v, DAB) The total number of variables, Nv (including kc) = 6. In terms of basic dimensions M (mass), L (length), T (time) : kc = L/T; ρ = M/L3; µ = M/L; v = L/T; DAB = L2/T The total number of basic variables (M, L and T), Nb = 3. Then according to the Buckingham Pi theorem on dimensional analysis, the number of dimensionless groups are Nv – Nb = 6 – 3 = 3. We choose the variables L, ρ and DAB to be the variables common and form the 3 dimensionless groups : P1 = (DAB)a (ρ)b (L)c Kc P2 = (DAB)a (ρ)b (L)c v P3 = (DAB)a (ρ)b (L)c µ In P1 (which has a dimension of 1), let us substitute the actual dimensionless as follows : 1 = (L2/T)a (M/L3)b (L)c (L/T) Summing up each exponent, L : 0 = 2a – 3b + c + 1 M : 0=b T : 0=–a–1 Solving these : a = – 1, b = 0, c = 1. Thus, P1 = kcL/DAB

Principles of Mass Transfer Operations − I (Vol. − I)

3.15

Mass Transfer Coefficients

Repeating these operations for P2 and P3, and rearranging the groups, we get : Sh = f (Re, Sc) which is the most common way mass transfer correlations are presented. 3.10.2 Transfer Into a Phase Whose Motion is Due to Natural Convection Natural convection currents develop if there exists any variation in density within the fluid phase. The density variation may be due to temperature differences or to relatively large concentration differences. In the case of natural convection involving mass transfer from a vertical plane wall to an adjacent fluid, the variables of importance are listed in the Table 3.3. Table 3.3 Variable

Symbol

Dimensions

L

L

DAB

L2/t

fluid density

ρ

M/L3

fluid viscosity

µ

M/LT

byoyant force

g ∆ρA

M/L2t2

kc

L/t

characteristic length fluid diffusivity

mass transfer coefficient

According to Buckingham theorem, there will be three dimensionless groups. Choosing DAB, L and µ as the core variables, the π groups to be formed are, a

π1 = DAB Lb µc kc d

π2 = DAB Le µf ρ and

g

π3 = DAB Lh µi g ∆ ρ A

Solving for the dimensionless groups, we obtain, kc L π1 = D = Nu, the Nusselt number AB

and

π2 =

ρ DAB 1 = S , the reciprocal of Schmidt number µ c

π3 =

L3 g ∆ ρ A µ DAB

… (3.59) … (3.60) … (3.61)

… (3.62)

… (3.63)

With the multiplication of π2 and π3, we obtain a dimensionless parameter analogous to the Grash of number in heat transfer by natural convection, 3 ρ DAB L g ∆ ρA π2 π3 =  µ   µD     AB  L3 ρ g ∆ ρ A = GrAB … (3.64) µ2 The result of the dimensional analysis of mass transfer by natural convection indicates that a correlating relation could be of the form, Sh = f (GrAB, Sc) … (3.65) =

Principles of Mass Transfer Operations − I (Vol. − I) 3.11

3.16

Mass Transfer Coefficients

ANALOGIES AMONG MASS, HEAT AND MOMENTUM TRANSFER

Analogies among mass, heat and momentum transfer have their origin either in the mathematical description of the effects or in the physical parameters used for quantitative description. To explore those analogies, it could be understood that the diffusion of mass and conduction of heat obey very similar equations. In particular, the Fick’s Law as describes diffusion in one dimension. dCA JA = – DAB dz

… (3.66)

Similarly, Fourier's law as describes heat conduction dT q = – K dz

… (3.67)

Where K is the thermal conductivity. The similar equation describing momentum transfer as given by Newton's law is, dv τ = µ dz

… (3.68)

Where τ is the momentum flux (or shear stress) and µ is the viscosity of fluid. At this point it has become conventional to draw an analogy among mass, heat and momentum transfer. Each process uses a simple law combined with a mass or energy or momentum balance. In this section, we shall consider several analogies among transfer phenomenon, which has been proposed because of the similarity in their mechanisms. The analogies are useful in understanding the transfer phenomena and as a satisfactory means for predicting behaviour of systems for which limited quantitative data are available. The similarity among the transfer phenomena and accordingly the existence of the analogies require that the following five conditions exist within the system. (i) The physical properties are constant. (ii) There is no mass or energy produced within the system. This implies that there is no chemical reaction within the system. (iii) There is no emission or absorption of radiant energy. (iv) There is no viscous dissipation of energy. (v) The mass transfer does not affect the velocity profile. This implies there should be a low rate of mass transfer. Heat, Mass and Momentum Transfer Analogies : •

The molecular diffusion equations are quite similar for fluid flow (Newton's law), for heat conduction (Fourier's law) and for mass diffusion (Fick's law). Thus, we have analogies (or similarities) among these three molecular transport processes.

•

For convection, the analogies are not easily defined as turbulent flow is difficult to analyse. Over the years several analogies have been proposed. Among the most obvious being : the eddy diffusivities in heat, mass and momentum transfer are equal. However, in practice this analogy does not work and in general none of the eddy diffusivities are known.

Principles of Mass Transfer Operations − I (Vol. − I) •

3.17

Mass Transfer Coefficients

The most successful and most widely used analogy is the Chilton-Colburn J-factor analogy. The analogy is based on experiment data for gases and liquids in both laminar and turbulent flow and is given by : f/2 = JH = JM

where,

f = Fanning friction factor JH = (StH) (Pr)2/3 JM = (StM) (Sc)2/3

where,

StH = Stanton number for heat transfer = Nu/(Re.Pr) Nu = Nusselt number, hL/k Pr = Prandtl number, Cpµ/k

•

The Chilton-Chilton analogy entirely satisfies the exact solution derived theoretically for laminar flow over flat plate.

•

The equation for transport analog is valuable in estimating convective mass transfer coefficient when friction factor or heat transfer coefficient for a particular case is known and vice-versa. Since instruments for concentration measurements are expensive, the above method of using analogies to derive mass transfer correlation is widely used.

3.11.1 Reynolds Analogy Osborne Reynolds reported the first recognition of the analogous behaviour of mass; heat and momentum transfer in 1874. Although his analogy is limited in application, it served as the base for seeking better analogies. Reynolds postulated that the mechanisms for transfer of momentum, energy and mass are identical. Accordingly, kc v∞

=

h f =2 ρv∞ Cp

… (3.69)

Here h is heat transfer coefficient, f is friction factor, v∞ is velocity of free stream. The Reynolds analogy is interesting because it suggests a very simple relation between different transport phenomena. This relation is found to be accurate when Prandtl and Schmidt numbers are equal to one. This is applicable for mass transfer by means of turbulent eddies in gases. In this situation, we can estimate mass transfer coefficients from heat transfer coefficients or from friction factors. 3.11.2 Chilton – Colburn Analogy Because the Reynold’s analogy was practically useful, many authors tried to extend it to liquids. Chilton and Colburn, using experimental data, sought modifications to the Reynold’s analogy that would not have the restrictions that Prandtl and Schmidt numbers must be equal to one. They defined for the j factor for mass transfer as, kc jD = v (Sc) 2/3 ∞

… (3.70)

The analogous factor for heat transfer is, jH = St pr2/3

… (3.71)

Principles of Mass Transfer Operations − I (Vol. − I) where,

3.18

Mass Transfer Coefficients

Nu h St is Stanton number = Re Pr = ρ v∞ C p

Based on data collected in both laminar and turbulent flow regimes, they found f jD = jH = 2 … (3.72) This analogy is valid for gases and liquids within the range of 0.6 < Sc < 2500 and 0.6 < Pr < 100. The Chilton-Colburn analogy has been observed to hold for many different geometries for example, flow over flat plates, flow in pipes, and flow around cylinders. 3.11.3 Taylor – Prandtl Analogy It is modification over Reynolds analogy and presence of the laminar sub-layer in the path of transfer is considered. The expressions developed in this analogy are : (i) For Heat-Transfer : f 2 h … (3.73) CP G (Stanton No.) = f 1+ 2 (Pr – 1) hD (ii) For Mass Transfer : U = Stanaton number for mass transfer f 2 … (3.74) = f 1+ 2 (SC – 1) 3.12 CONVECTIVE MASS TRANSFER CORRELATIONS Extensive data have been obtained for the transfer of mass between a moving fluid and certain shapes, such as flat plates, spheres and cylinders. The techniques include sublimation of a solid, vapourization of a liquid into a moving stream of air and the dissolution of a solid into water. These data have been correlated in terms of dimensionless parameters and the equations obtained are used to estimate the mass transfer coefficients in other moving fluids and geometrically similar surfaces. 3.12.1 Flat Plate From the experimental measurements of rate of evapouration from a liquid surface or from the sublimation rate of a volatile solid surface into a controlled air-stream, several correlations are available. These correlation have been found to satisfy the equations obtained by theoretical analysis on boundary layers, 1/2

Sh = 0.664 ReL Sc1/3 (laminar) ReL < 0.3 × 105 0.8

Sh = 0.036 ReL Sc1/3 (turbulent) ReL > 3 × 105

… (3.75) … (3.76)

Using the definition of j factor for mass transfer on equation (3.75) and (3.76) we obtain, –1/2

(laminar) ReL < 3 × 105

… (3.77)

JD = 0.037 ReL (turbulent) ReL > 3 × 105

… (3.78)

jD = 0.664 ReL

0.2

Principles of Mass Transfer Operations − I (Vol. − I)

3.19

Mass Transfer Coefficients

These equations may be used if the Schmidt number in the range 0.6 < Sc < 2500. 3.12.2

For Flow Around Single Sphere

Correlations for mass transfer from single spheres are represented as addition of terms representing transfer by purely molecular diffusion and transfer by forced convection, in the form Sh = Sho + CRem + Scn

… (3.79)

Where C, m and n are constants, the value of n is normally taken as 1/3. For very low Reynold’s number, the Sherwood number should approach a value of 2. This value has been derived in earlier sections by theoretical consideration of molecular diffusion from a sphere into a large volume of stagnant fluid. Therefore the generalized equation becomes, Sh = 2 + CRem Sc1/3

… (3.80)

For mass transfer into liquid streams, the equation given by Brain and Hales : Sh =

2/3

4 + 1.21 Pe   AB 

1/2

… (3.81)

Correlates the data that are obtained when the mass transfer Peclet number, PeAB is less than 10,000. This Peclet number is equal to the product of Reynolds and Schmidt numbers, PeAB = Re Sc

… (3.82)

For Peclet numbers greater than 10,000, the relation given by Levich is useful, 1/3

Sh = 1.01 PeAB

… (3.83)

The relation given by Froessling, … (3.84) Sh = 2 + 0.552 Re1/2 Sc1/3 Correlates the data for mass transfer into gases for at Reynold’s numbers ranging from 2 to 800 and Schmidt number ranging 0.6 to 2.7. For natural convection mass transfer the relation given by Schutz : Sh = 2 + 0.59 (GrAB Sc) 1/4 … (3.85) 8

is useful over the range 2 × 10 < Gr AB Sc < 1.5 × 10

10

3.12.3 For Flow Around Single Cylinder Several investigators have studied the rate of sublimation from a solid cylinder into air flowing normal to its axis. Bedingfield and Drew correlated the available data in the form, kG P·Sc0.56 = 0.281 (Re')–0.4 Gm Which is valid for 400 < Re' < 25000 … (3.86) and 0.6 < Sc < 2.6 Where Re' is the Reynold’s number in terms of the diameter of the cylinder, Gm is the molar mass velocity of gas and P is the pressure. 3.12.4 Flow Through Pipes Mass transfer from the inner wall of a tube to a moving fluid has been studied extensively. Gilliland and Sherwood, based on the study of rate of vapourization of nine different liquids into air given the correlation, pB‚ lm Sh P = 0.023 Re0.83 Sc0.44 … (3.87)

Principles of Mass Transfer Operations − I (Vol. − I)

3.20

Mass Transfer Coefficients

Where pB, lm is the log mean composition of the carrier gas, evaluated between the surface and bulk stream composition. P is the total pressure. This expression has been found to be valid over the range, 2000 < Re < 35000, 0.6 < Sc < 2.5. Linton and Sherwood modified the above relation making it suitable for large ranges of Schmidt number. Their relation is given as, … (3.88) Sh = 0.023 Re0.83 Sc1/3 and found to be valid for 2000 < Re < 70000 and 1000 < Sc < 2260. 3.12.5 Mass Transfer to Suspension of Small Particles Mass transfer from or to small suspended particles in an agitated solution occurs in number of process applications. For example, (i) In liquid phase hydrogenation, H2 diffuses from the gas bubbles through an organic liquid and then to small suspended particles. (ii) In fermentation process O2 diffuse from small gas bubbles through the aqueous medium and then to a small suspended micro-organisms. To predict mass transfer coefficient from small gas bubbles such as O2/air to the liquid phase, or from liquid phase to the surface of small catalyst particles, other solids or liquid drops, following correlation is used : 2 DAB –0.67 ∆ ρ µC g 0.33 ' kL = D + 0.31 NSC  … (3.89) 2  P ρ  C  where,

DAB = Diffusivity of A in solution, m2/s DP = Diameter of gas bubble or solid particle, m µC = Viscosity of solution, kg/m. sec. g = Acceleration due to gravity = 9.80 m/s2 ∆ρ = ρC – ρP or ρP – ρC, kg/m3

where,

ρC = Density of continuous phase, kg/m3

ρP = Density of gas or solid particle, kg/m3 Note that ∆ρ is always positive. 3.12.6 Mass Transfer in Packed Beds The mass transfer to and from packed beds occurs often in processing operations and many other mass transfer operations of gases or liquids by solid particles such as charcoal and mass transfer of gases and liquids to catalyst particles. Using a packed bed, a large amount of mass transfer area can be contained in a relatively small volume. The void fraction, ∈ in a packed bed is given by : Volume of void m3 ∈ = Total volume , 3 m The value of ∈ varies from 0.3 to 0.5. For mass transfer of liquid in packed bed, the correlation is given by : Wilson–Geankoplis (1966) correlation : –0.67 1.09 (i) JD = · NRe … (3.90) ∈ for NRe ⇒ 0.0016 – 55 NSC ⇒ 165 – 70,600 0.250 –0.31 (ii) JD = NRe … (3.91) ∈

Principles of Mass Transfer Operations − I (Vol. − I)

3.21

Mass Transfer Coefficients

NRe ⇒ 55 – 1500

for

NSC ⇒ 165 – 10,690 kC' JD = V (NSC) 0.67

(iii)

… (3.92)

3.12.7 Mass Transfer in Boundry Layer In a turbulent boundry layer zone, the velocity of flow is sufficiently large in which case … (3.93) Reynolds number is, Rex > 3 × 106 (i) Laminar boundry wall thickness at a distance x from leading edge of plate is given (x) … (3.94) by : δ = 4.64 (Rex)0.5 (ii) Boundry layer wall thickness (δ) for turbulent flow at a distance x from leading edge of (x) plate is given by : δ = 0.376 … (3.95) (Rex)0.5 3.13 MASS TRANSFER BETWEEN PHASES Instead of a fluid in contact with a solid, suppose we now consider two immiscible fluids, designated 1 and 2, in contact with each other. If fluid 1 has dissolved in it a substance A that is also soluble in fluid 2, then as soon as the two fluids are brought together, substance A will begin to diffuse into fluid 2. As long as the two phases remain in contact, the transport of A will continue until a condition of equilibrium is reached. The situation discussed here occurs in a variety of engineering processes such as gas absorption, stripping, and in liquid – liquid extraction. In all these separation processes, two immiscible fluids are brought into contact and one or more components are transferred from one fluid phase to the other. In the system of fluids 1 and 2 with A, the transported component, the concentration gradients in the region of the interface between the two fluids are illustrated in Fig. 3.5. Fluid 1

Fluid 2 CA,i

C A1

C A2

CA,i C*A

Fig. 3.5 : Concentration gradients near the interface between immiscible fluids 1 and 2

Concentration CA1 and CA2 are the bulk phase concentrations. in fluids 1 and 2, respectively, CAi is the concentration of A at the interface, and NA is the molar flux of A. For steady state conditions, we can define the flux of A as, NA = kC (CA – CA ) = kc (CA – CA ) = KC (CA – CA ) … (3.96) 1

1

i

2

i

2

i

2

where kC = Individual mass transfer coefficient defined in terms of the concentration difference in a single phase. KC = Overall mass transfer coefficient defined in terms of the overall difference in composition.

Principles of Mass Transfer Operations − I (Vol. − I)

3.22

Mass Transfer Coefficients

Equation (3.96) is analogous to that in heat transfer, where the individual coefficients h are related to the overall coefficient U. From equation (3.96), 1 1 1 = K … (3.97) kc + kc C 1

2

In equation (3.96), the potential for mass transfer is exposed in terms of composition. However, this is not always the most convenient way to express it. For example, if fluid 1 is a gas and fluid 2 a liquid, as in gas absorption, the potential in gas phase is often expressed in terms of partial pressures, while that in the liquid phase may be expressed in terms of concentrations. The expression for the molar flux is then written for the individual phases as : … (3.98) NA = KP (PAG – PAi) = KC (CAi – CA ) L

where,

KP = Individual mass transfer coefficient for the gas phase with the potential defined in terms of partial pressure. , C = Partial pressure and concentration of A in the bulk gas and liquid PAG A L

phases, respectively. PAi, CAi = Partial pressure and concentration of A, respectively, at the interface. At interface, it is usually assumed that the two phases are in equilibrium and equilibrium relationship is given by, … (3.99) PAi = H · CAi The flux NA can also be expressed in terms of overall mass transfer coefficient as : NA = KP (PAG – PAE) = Kc (CAE – CAL) … (3.100) where, Kp = Overall mass transfer coefficient with the overall potential defined in terms of partial pressures. Kc = Overall mass transfer coefficient with the overall potential defined in terms of concentrations. PAE , CAE = Equilibrium composition. PAE is related to the bulk liquid composition CAL AS … (3.101) PAE = HCAL PAG Similarly, CAE = H … (3.102) The relationship between the individual and overall coefficients is readily obtained through the use of equations (3.98) to (3.102) as, 1 Kp

1 H H = k +k =K p c c

… (3.103)

In many system, mass transfer resistance is mainly in one phase. For example, gases such as nitrogen and oxygen do not dissolve much in liquids. Their Henry’s law constant H is very large, thus K c ≈ k c is a good approximation. In this case, the liquid phase controls the mass transfer press since mass transfer is slowest there. 3.14 SIMULTANEOUS HEAT AND MASS TRANSFER Diffusional mass transfer is generally accompanied by the transport of energy, even with in an isothermal system. Since each diffusing constituent carries its own individual enthalpy, the heat flux at a given plane is expressed as : q =

∑ i

–

N i Hi

… (3.104)

Principles of Mass Transfer Operations − I (Vol. − I)

3.23

Mass Transfer Coefficients –

where q is the heat flux due to diffusion of mass past the given plane, and Hi is the partial molar enthalpy of constituent i in the mixture. When there is a temperature difference, energy transfer also occurs by one of the three heat transfer mechanisms (conduction, convection, radiation); for example, the equation for energy transport by convection and molecular diffusion becomes, –

q = h ∆ T + ∑ N i Hi

… (3.105)

i

If the heat transfer is by conduction, the first term on the right hand side of equation (3.105) k ∆T becomes L where L is the thickness of the phase through which conduction takes place. The most common examples of processes involving heat and mass transfer are condensation of mist on a cold surface and in wet bulb thermometer. There are a number of such processes involving simultaneous heat and mass transfer such as in formation of fog, and in cooling towers. 3.14.1 Condensation of Vapour On Cold Surface Condensate liquid film

A process important in many engineering processes as well as in day-to-day events involve the condensation of a vapour upon a cold surface. Examples of this process include "sweating" on cold water pipes and the condensation of moist vapour on a cold surface. Figure illustrates the process which involves a film of condensed liquid following down a cold surface and a film of gas through which the condensate is transferred by molecular diffusion. This process involves the simultaneous transfer of mass and energy. The heat flux passing through the liquid film is given by, q = h1 (T2 – T3)

Boundary of gas film

T = T(z)

T1

yA = yA (z)

yA1

T2

T3

yA2

… (3.106)

This flux is also equal to the total energy transported by convection and molecular diffusion in the gas film, i.e., q = hc (T1 – T2) + NA MA (H1 – H2)

… (3.107)

z3

where MA is the molecular weight of the diffusing constituent A. H 1 and H 2 are enthalpies of the vapour at plane 1 and liquid at plane 2.

z2

z1

Fig. 3.6 : Vapour Condensation on a cold surface

From equation (3.87) and (3.88), q = h1 (T2 – T3) = hc (T1 – T2) + NA MA (H1 – H2)

… (3.108)

The molar flux NA is calculated by diffusion through stagnant gas model as, – CDAB dyA NA = 1 – y dZ A Substituting the appropriate limits, the integral form of equation is : (CDAB)avg NA =

(y

A1

– yA

(Z2 – Z1) yB‚ lm

2

)

… (3.109)

Principles of Mass Transfer Operations − I (Vol. − I)

3.24

Mass Transfer Coefficients

3.14.2 The Wet Bulb Thermometer The another example of simultaneous heat and mass transfer is that taking place in wet bulb thermometer. This convenient device for measuring relative humidity of air consists of two conventional thermometers, one of which is clad in a cloth nick wet with water. The unclad dry-bulb thermometer measures the air’s temperature. The clad wet-bulb thermometer measures the colder temperature caused by evapouration of the water. We want to use this measured temperature difference to calculate the relative humidity in air. This relative humidity is defined as the amount of water actually in the air divided by the amount at saturation at the dry-bulb temperature. To find this humidity, we can write equation for the mass and energy fluxes as : NA = kc (CA – CA) = ky (yA – yA) i

and

i

Air Flow Wick

Liquid Reservoir

Fig. 3.7 : Concept of wet bulb thermometer

… (3.110) … (3.111)

q = h (Ti – T)

where CA and CA are the concentrations of water vapour at the wet bulb’s surface and in the i

bulk of air, yA and yA are the corresponding mole fractions; Ti is the wet-bulb temperature, and T i

is the dry bulb temperature. It can be noted that yAi is the value at saturation at Ti. In the air-film surroundings the wet bulb, the mass and energy fluxes are coupled as : … (3.112) NA λ = – q where λ is the latent heat of vapourization of water Thus, ky (yA – yA) λ = h (T – Ti) i

Rearranging,

λky Ti = T – h (yA – yA) i

… (3.113)

From Chilton – colburn analogy, jH = jD or

kc h (Pr) 2/3 = v (Sc) 2/3 2 ρ v Cp α

… (3.114)

For gas Pr ≈ 1 and Sc ≈ 1. Therefore equation (3.114) becomes, kc 1 (as ky C ≈ ky ρ = kc) h = Cp Therefore equation (3.113) becomes, λ Ti = T – C

p

(yAi – yA)

Where Cp is the humid heat of air. By similar method, the other industrial processes of importance involving simultaneous heat and mass transfer such as humidification and drying can be analyzed.

Principles of Mass Transfer Operations − I (Vol. − I)

3.25

Mass Transfer Coefficients

SOLVED PROBLEMS (1)

A stream of air at 100-kPa pressure and 300 K is flowing on the top surface of a thin flat sheet of solid naphthalene of length 0.2 m with a velocity of 20 m/sec. The other data are : –6 2 Mass diffusivity of naphthalene vapour in air = 6 × 10 m /sec –5 2 Kinematic viscosity of air = 1.5 × 10 m .sc Concentration of naphthalene at the air-solid naphthalene interface –5 3 = 1 × 10 k mole/m Calculate : (a) The overage mass transfer coefficient over the flat plate. (b) The rate of loss of naphthalene from the surface per unit width. Note : For heat transfer over a flat plate, convective heat transfer coefficient for laminar flow can be calculated by the equation. 1/2

Pr1/3

Nu = 0.664 ReL

You may use analogy between mass and heat transfer. Sol. : Given : Correlation for heat transfer. 1/2

Pr1/3

Nu = 0.664 ReL The analogous relation for mass transfer is,

1/2

Sh = 0.664 ReL where

Sc1/3

… (1)

Sh = Sherwood number = kL/DAB ReL = Reynolds number = Lvρ/µ Sc = Schmidt number = µ / (ρDAB) k = Overall mass transfer coefficient L = Length of sheet DAB = Diffusivity of A in B v = Velocity of air

µ = Viscosity of air ρ = Density of air, and µ/ρ = Kinematic viscosity of air. Substituting for the known quantities in equation (1) k (0.2) 6 × 10–6

= 0.664

(0.2) (0.20)   1.5 × 10–5 

1/2

–5

1.5 × 10  1/3  6 × 10–6   

k = 0.014 m/sec

(Ans.)

Rate of loss of naphthalene = k (CAi – CA∞) = 0.014 (1 × 10–5 – 0) = 1.4024 × 10–7 k mole/m2 sec Rate of loss per meter width = (1.4024 × 10–7) (0.2) = 2.8048 × 10–8 k mole/m.sec = 0.101 g mole/m. hr.

(Ans.)

Principles of Mass Transfer Operations − I (Vol. − I)

3.26

Mass Transfer Coefficients

(2) If the local Nusselt number for the laminar boundary layer that is formed over a flat 1/2

Nux = 0.332 Rex Sc1/3

plate is

Obtain an expression for the average film-transfer coefficient kc, when the Reynolds number for the plate is : (a) ReL = 100000, (b) ReL = 1500 000 5

The transition from laminar to turbulent flow occurs at Rex = 3 × 10 . L

⌠ kc dx ⌡

–

Sol. : (a)

kc =

0 L

⌠ dx ⌡ 0

kc x By definition : Nux = D AB

xvρ µ ; Rex = µ ; Sc = and ρDAB

For ReL = 100 000; (which is less than the Reynolds number corresponding to Transition 5

value of 3 × 10 ) L

1

1

DAB xvρ ⌠ 0.332  µ 2 (Sc)3 x dx   ⌡

–

kc =

0

0.332 (Sc)1/3 =

L 1/2 vρ µ

L

L

DAB

dx

⌠ 1/2 ⌡ x 0

vρ 1/2 0.332 Sc1/3  µ  DAB = 1   L 2

L

[x1/2]0

–

kc L DAB Lt

x

⌠ 0.332 Re1/2 Sc ⌡ –

(Ans.)

ReL = 1500000 (> 3 × 105)

(b) For

kc = DAB

1/2

= 0.664 ReL Sc1/3

0

1/3

dx x +

L

4/5

dx

⌠ 0.0292 Rex Sc1/3 x ⌡ Lt

L where Lt is the distance from the leading edge of the plane to the transition point where Rex = 3 × 105.

–

k c = DAB

Lt L 1/2 4/5  1/3 vρ 0.332 Sc1/3 vρ ⌠ dx ⌠ + 0.0292 Sc  µ  ⌡ x1/2 µ ⌡   0 Lt

L

dx x1/5

   

Principles of Mass Transfer Operations − I (Vol. − I)

3.27

–

kc L DAB

Mass Transfer Coefficients

1/2

L 0.0292 Vρ 4/5 Sx1/3 + 4/5 Sc1/3 [x4/5]L  µ  t  

1/2

Sc1/3 + 0.0365 Sc1/3 ReL – Ret

= 0.664 Ret = 0.664 Ret

4/5

4/5

 

–

kc L DAB where,

1/2

= 0.664 Ret

4/5

4/5

Sc1/3 + 0.0365 ReL Sc1/3 – 0.0365 Ret

Ret = 3 × 105

Sc1/3 (Ans.)

(3) The mass flux from a 5 cm diameter naphthalene ball placed in stagnant air at 40°°C –3 2 and atmospheric pressure, is 1.47 × 10 mol/m . sec. Assume the vapour pressure of naphthalene to be 0.15 atm at 40οC and negligible bulk concentration of naphthalene in air. If air starts blowing across the surface of naphthalene ball at 3 m/s by what factor will the mass transfer rate increase, all other conditions remaining the same ? For spheres :

Sh = 2.0 + 0.6 (Re)

0.5

(Sc)

0.33

Where Sh is the Sherwood number and Sc is the Schmidt number. The viscosity and –5 3 density of air are 1.8 × 10 kg/m.s and 1.123 kg/m , respectively and the gas constant is 3 82.06 cm . atm/mole.K. kc L Sol. : Sh = D AB where L is the characteristic dimension for sphere L = Diameter. µ Sc = ρ DAB Rc = Mass flux,

Dvρ µ

NA = Kc ∆C

… (1)

Sh = 2.0 + 0.6 (Re)0.5 (Sc)0.33 kc D µ  0.33 0.5  = 2.0 + 0.6 (DVρ)   DAB ρDAB

… (2)

–

N = KG ∆ p A kc RT = KG

also Therefore

N = 1.47 × 10–3

Given : kc RT

Kc mole – = RT ∆ p A 2 m .sec

0.15 – 0 = 1.47 × 10–3 × 10–4 mole  1  cm2.sec kc =

1.47 × 10–7 × 82.06 × (273 + 40) 0.15

= 0.0252 cm/sec kc = 2.517 × 10–4 m/sec.

… (3)

Principles of Mass Transfer Operations − I (Vol. − I)

3.28

Mass Transfer Coefficients

Estimation of DAB : From equation (2), 2.517 × 10–4 × 5 × 10–2 = 2 DAB

(since v = 0)

DAB = 6.2925 × 10–6 m2/sec.

Therefore, And kc × 5 × 10–2

–2 5 × 10 × 3 × 1.1230.5 = 2 + 0.6   6.2925 × 10–6 1.8 × 10–5  

–5

1.8 × 10  0.33 1.123 × 6.2925 × 10–6  

7946 kc = 2 + 0.6 × (96.74) × (1.361) kc = 0.0102 m/sec. (4) (3)

⇒

NA

2

NA

1

=

… (4)

0.0102 = 40.5 2.517 × 10–4

Therefore, rate of mass transfer increases by 40.5 times the initial conditions.

(Ans.)

o

(4) A solid disc of benzoic acid 3 cm in diameter is spin at 20 rpm and 25 C. Calculate the rate of dissolution in a large volume of water. Diffusivity of benzoic acid in water is –5 2 1.0 × 10 cm /sec, and solubility is 0.003 g/cc. The following mass transfer correlation is applicable : _

Sh = 0.62 Re1/2 Sc Where Re = Sol. : Where

1/3

D2 ω ρ and ω is the angular speed in radians/time. µ Dissolution rate = N A S NA = mass flux, and

… (1)

S = surface area for mass transfer … (2) NA = kc (CAs – CA∞) Where CAs is the concentration of benzoic and at in water at the surface of the dose. CA∞ is the concentration benzoic acid in water for and from the surface of the disc. Given : (i.e.)

_

Sh = 0.62 Re1/2 Sc kc D DAB

1/3 1

1

D2 ω ρ 2 µ 3 = 0.62  µ     ρ DAB

… (3)

1 rotation = 2 π radian Therefore 20 rotation per minute = 20 × 2 π radian/min 20 = 60 × 2 π radian/sec For water ρ = 1 g/cm3 µ = 1 centipoise = 0.01 g/cm. sec. 1

From equation (3),

kc

1

µ 3 ωρ 2 = 0.62 DAB  µ     ρ DAB 1

(40 π/60) × 12 = 0.62 × 1.0 × 10 ×  0.01   5

Principles of Mass Transfer Operations − I (Vol. − I)

3.29

Mass Transfer Coefficients 1

0.01  3 –5 1 × 1.0 × 10  From equation (2), From equation (1),

kc = 8.973 × 10–4 cm/sec. NA = 8.973 × 10u–4 (0.003 – 0) = 2.692 × 10–6 g/cm2. sec NAS = NA × (2πr2)

= 2.692 × 10–6 × (2π × 1.52) = 3.805 × 10–5 g/sec = 0.137 g/hr. (Ans.) (5) Air at 1 atm is blown past the bulb of a mercury thermometer. The bulb is covered with a wick. The wick is immersed in an organic liquid (molecular weight = 58). The reading of the thermometer is 7.6°°C. At this temperature, the vapour pressure of the liquid is 5 kPa. Find the air temperature, given that the ratio for heat transfer coefficient to the mass transfer coefficient (psychrometric ratio) is 2 kJ/kg. Assume that the air, which is blown, is free from the organic vapour. Sol. : For simultaneous mass and heat transfer, heat flux q and mass flux NA are related as : q = NA λ

… (1)

where λ is the latent heat of vapourization Mass Flux is given by, '

NA = kY (Yω – Y') where,

… (2)

kY = Mass transfer coefficient

Yω' = Mass ratio of vapour in surrounding air at saturation; and Y' = Mass ratio of vapour in surrounding air. Convective heat flux is given by, q = h (T – Tω) where,

Substituting for NA

… (3)

h = heat transfer coefficient; Tω = wet bulb temperature of air; and T = dry bulb temperature of air. and q from equation (2) and equation (3) in equation (1), h (T – Tω' )

= ky

T – Tω =

(T' – Y') λ ω

λ (Yω' – Y') h/kY

Given : Y' = 0; λ = 360 kJ/kg; h/kY = 2 kJ/kg.K; and Tω = 7.6o C kg organic vapour at saturation Yω' = kg dry air 5 58 = 101.3 – 5 29 = 0.1038 Substituting these in equation (4), (360) (0.1038 – 0) T – 7.6 = = 18.69 2 T = 18.69 + 7.6 = 26.29oC Temperature of air = 26.29oC

… (4)

(Ans.)

Principles of Mass Transfer Operations − I (Vol. − I)

3.30

Mass Transfer Coefficients

(6) Imagine that water is evaporating into initially dry air in the closed vessel as shown in Fig. 3.8. Air c1(t) Water

Fig. 3.8 : Evaporation of water in closed vessel

The vessel is isothermal at 25o C, so the water's vapour pressure is 23.8 mm Hg. This vessel has 0.8 liter of water with 150 cm2 of surface area in a total volume of 19.2 litres. After 3 minutes, the air is five percent saturated. (a) What is mass transfer coefficient ? (b) How long will it take to reach ninety percent saturation ? Sol. : (a) The flux N1 at 3 minutes can be found from following expression. (Vapour Concentration) (Air Volume) (Liquid Area) (Time) 23.8 1 mol 273 0.05  760   22.4 liter 298 (18.4 liters)     = (150 cm2) (180 second)

N1 =

N1 = 4.39 × 10–8 mol/cm2 sec. The concentration difference is that at the water's surface minus that in the bulk solution. That at the water's surface is the value at saturation, that in bulk at short times is essentially zero. Thus, mass transfer coefficient (k) is calculated as : N1 = k (C1i – C1) In this case, C1 = 0, since at water surface, the concentration value at saturation = 0 23.8 1 mol 273  ∴ 4.39 × 10–8 = k  760 × 3 3 × 298 – 0 22.4 × 10 cm   So, k = 3.427 × 10–2 cm/sec. (b) The time required for 90% saturation can be determined from a simple mass balance, Accumulation = (Rate of evaporation) in gas phase  d dt VC1 = N1.A ∴ = A k [C1 (sat.) – C1] It is given that the air is initially dry, So, at t = 0, C1 = 0 We use this condition to integrate the mass balance, d dt VC1 = A. k [C1 (sat)] VC 1 d ⌠ C (sat.) = A k ⌠ dt

⌡

or,

1

⌡

VC ln C1 (sat.) = A. k. t 1

Principles of Mass Transfer Operations − I (Vol. − I) or,

C1 C1 (sat.)

3.31

Mass Transfer Coefficients

= 1 – e–(kA/V). t

Rearranging above equation, we get, V t = – kA =

C1   ln 1 – C (sat.) 1  

18.4 × 103 cm3 3.4 × 10–2 cm × (150 cm2) sec 

ln (1 – 0.90)

= 8.3 × 103 seconds = 2.3 hours (Ans.) (7) A bubble of O2 originally 0.1 cm in diameter is injected into excess stirred water as shown in figure 3.9. After 7 minutes, the bubble is 0.054 cm in diameter. What is the mass transfer coefficient ?

Size = f(t)

Fig. 3.9 : Illustration gas bubble in Ex. 7

Data : O2 concentration = 1.5 × 10–3 mole/liter. Sol. : We write mass balance equation not on surrounding solution but on bubble itself. So, writing mass balance, d  4 3 dt C1 3 π r  = A . N1 Where, N1 = – k [C1 (sat.) – 0]

… (1) … (2)

2

… (3) A = 4πr Putting equation (2) and (3) in equation (1) we get, d  4 3 2 C . … (4) 1 dt  3 πr  = – 4πr . k [C1 (sat.) – 0] Where, C1 = O2 concentration in the bubble at standard condition 1 mole = 22.4 lit C1 (sat.) = O2 concentration at saturation in water So,

dr dt

= 1.5 × 10–3 mole /lit. C1 (sat.) = –k C 1 = –

(given)

k (1.5 × 10–3)  1  22.4

dr dt = – 0.0336 k Integrating equation (5) using following boundary condition, t = 0, r = 0.05 cm r = 0.05 – 0.034 kt

… (5)

Principles of Mass Transfer Operations − I (Vol. − I)

3.32

Mass Transfer Coefficients

To find mass transfer coefficient k, put t = 420 sec. and r =

0.054 = 0.027 cm 2

0.054 = 0.05 – 0.034 kt × k (420) 2 k = 1.63 × 10 –3 cm/sec. (Ans.) (8) Bromine is being rapidly dissolved in water as shown in Fig. 3.10. Its concentration is about half saturated in 3 minutes. What is the mass transfer coefficient ? So,

c1(t)

Fig. 3.10 : Illustration of Bromine Liquid Drops in Ex. 8

Sol. : Mass balance equation may be written as, d dt VC1 = AN1 = A.k [C1 (sat.) – C1] Equation (1) may be written in terms of interfacial area as, dC1 = ka [C1 (sat.) – C1] dt A Where, a = V = Interfacial area per unit volume of aqueous solution. (m2/m3) If we assume water initially contains no bromine. at t = 0, C1 = 0 Using this boundary condition and integrating equation (2), we get, dC1 = k. a. dt [C1 (sat.) – C1] ln, (C1 (sat.) – C1)

= k. a.t

or

C1 (sat.) ln  C 

= k.a.t

or

C1 (sat) C1

= ekat

C1 C1 (sat.)

= 1 – e–kat



1



… (1)

… (2)

… (3)

Rearranging equation (3) we get, C1  1  k.a = – t ln 1 – C (sat.) 1   1 = – 3 min. ln (1 – 0.5) (Ans.) k.a = 3.9 × 10–3 sec –1 (9) Assume that 0.2 cm. diameter spheres of benzoic acid are backed into a bed like that shown in Fig. 3.11.

Principles of Mass Transfer Operations − I (Vol. − I)

3.33

Mass Transfer Coefficients

The spheres have 23 cm2 surface per 1 cm3 of bed. Pure water flowing at a superficial velocity of 5 cm/sec. into the bed is 62 % saturated with benzoic acid after it has passed through 100 cm of bed. Predict the mass transfer coefficient.

z z + Dz

z

c1(Z)

Fig. 3.11 : Packed Bed of benzoic acid spheres in Ex. 9

Sol. : To solve this problem, we depends on concentration difference used in definition of mass transfer coefficient. Another way is to use empirical equations to find out the mass transfer coefficient. We choose this difference as the value at the sphere's surface minus that in the solution. However, we can define different mass transfer coefficient by choosing the concentration difference at various position in the bed. For example, we can choose the concentration difference at the entrance of packed bed. The flux N1 is given by : N1 = k (C1 (sat.) – 0) =

… (1)

0.62 C1 (sat.) (5 cm/sec.) A

(23 cm2/cm3) (100 cm) A … (2) = k C1 (sat.) Where 'A' is the bed's cross-section, Thus, k = 1.3 × 10–3 cm2/sec. Alternatively, we can choose as our concentration difference that at a position Z in the bed and write a mass balance on a differential volume A ∆ Z at this position : Amount of (accumulation) = (flow in – flow out) + dissolution   0 0 = A C1νo – C1 (A ∆ Z) a N1 



Z

Z + δZ



Where a is the sphere surface area per bed volume. Substituting for N1 as N1 = k (C1i – C1) and dividing by A∆Z, and taking the limit as ∆Z goes to zero, we get dC1 ka dZ = νo [C1 (sat.) – C1] This is subject to the initial condition that, Z = 0, C1 = 0 Integrating, we obtain an exponential of the form C1 – (ka/νo) Z C1 (sat.) = 1 – e Rearranging the equation and putting the known values, we get, C1  νo  k = aZ ln 1 – C (sat.)    1  – 5 cm/sec. ln (1 – 0.62) (23 cm2/cm3) (100 cm) = 2.1 × 10–3 cm/sec.

=

(Ans.)

Principles of Mass Transfer Operations − I (Vol. − I)

3.34

Mass Transfer Coefficients

(10) Sherwood Number for Eqimolar Counter Diffusion : For equimolar counter-diffusion from a sphere to a surrounding stationary infinite medium, the mass flux NAi of the diffusing component A at the interface is given by NAi = DA (CAi − CAb)/R where DA is the diffusivity, R the radius of the sphere and CAi and CAb the molar concentrations of A at the interface and at a point far away from the sphere. Show that the Sherwood number, based on the diameter of the sphere, is equal to 2. kcD Sol. : Sherwood number SH = D A Here, kc = DA/R (since, NAi = DA (CAi − CAb)/R = kc (CAi − CAb)). Therefore, DA D Sh = R D A D = R 2R = R =2 Hence proved. (11) Heat and Mass Transfer Analogy : A stream of air at 100 kPa pressure and 300 K is flowing on the top surface of a thin flat sheet of solid naphthalene of length 0.2 m with a velocity of 20 m/sec. The other data are : Mass diffusivity of naphthalene vapor in air = 6 × 10−6 m2/sec. Kinematic viscosity of air = 1.5 × 10−5 m2/sec. Concentration of naphthalene at the air-solid naphthalene interface = 1 × 10−5 kmol/m3. Calculate : (a) The average mass transfer coefficient over the flat plate. (b) The rate of loss of naphthalene from the surface per unit width. Note : For heat transfer over a flat plate, convective heat transfer coefficient for laminar flow can be calculated by the equation : 1/2

Nu = 0.664 ReL Pr1/3 You may use analogy between mass and heat transfer. Sol. : Given : Correlation for heat transfer 1/2

Nu = 0.664 ReL Pr1/3 The analogous relation for mass transfer is, 1/2

Sh = 0.664 ReL Sc1/3 where, Sh ReL Sc DAB k L ν µ ρ µ/ρ

= = = = = = = = = =

Sherwood number = kL/DAB Reynolds number = Lνρ/µ Schmidt number = µ/(ρDAB) Diffusivity of A in B Overall mass transfer coefficient Length of sheet Velocity of air Viscosity of air Density o air, and Kinematic viscosity of air

… (1)

Principles of Mass Transfer Operations − I (Vol. − I)

3.35

Mass Transfer Coefficients

Substituting for the known quantities in equation (1), 1/2 −5 1/3 k(0.2)  (0.2) (20)  1.5 × 10  = 0.664 6 × 10−6 1.5 × 10−5  6 × 10−6  Rate of loss of naphthalene = k (CAi − CA∞) kmol = 0.014 (1 × 10−5 − 0) = 1.4024 × 10−7 m2.sec kmol Rate of loss per meter width = (1.4024 × 10−7) (0.2) = 2.8048 × 10−8 m.sec. gmol = 0.101 m.hr. (12) Liquid A evaporates and diffuses through a stagnant gas B as shown in Fig. 3.12. Assume that process is carried out isothermally at 25oC and total pressure 1 atm. Calculate the molar flux and convective mass transfer coefficient when the level of liquid A is 0.05 m below the top of the container. Assume that DAB = 1 × 10–5 m2/sec. The more fraction of component A at gas liquid interface is 0.2. At the top of the container, the mole fraction is 0.001. Gas B

Z = Z, CA = CA Gas B

Z = 0, CA = CA 1

Gas A

Fig. 3.12 : Evaporation of Single Component

Sol. : (i) Molar flux may be calculated as, NA = –

Pt. DAB –

RTZ (PB,M)

(P–

A1

–

– PA2

)

–

PA = yA. Pt ; PB = yB Pt –

PA1 = 0.2 atm –

–

PB1 = Pt – PA1 = 1 – 0.2 = 0.8 atm –

PA2 = 1 × 10–3 atm –

PB2 = 0.999 atm –

So,

–

PB,M =

–

PB2 – PB1

–

PB2 ln  –  P   B1

=

0.999 – 0.8 0.999 ln 0.8   

–

PB,M = 0.8958 atm atm. m3 R = 0.08206 kg mole K

Principles of Mass Transfer Operations − I (Vol. − I)

3.36

Mass Transfer Coefficients

Z = 0.05 m. 1 × 1 × 10–5 (0.2 – 0.001) NA = 0.08206 × 298 × 0.05(0.8958)

∴

NA = 1.816 × 10–6

kg mole m2sec

(Ans.)

(ii) Convective mass transfer coefficient, kg is given by kg =

=

Pt. DAB –

RTZ (PB,M) 1 × 1 × 10–5 0.08206 × 298 × 0.05 (0.8958)

kg = 9.13 × 10–6

kg mole m2 sec. atm

(Ans.)

(13) Air containing water soluble vapour is flowing up and water is flowing down in a experimental column as shown in Fig. 3.13. Almost Pure Air

The water flow in 0.07 cm thick film is 3 cm/sec. The column diameter is 10 cm and the air is essentially well mixed. The diffusing coefficient in water of absorbed vapour is 1.8 × 10–5 cm2/sec. How long a column is required to reach a gas concentration in water having 10% saturation ?

Pure Water z Dz

Air & Water Soluble Vapor

Data : You may use following mass transfer relation. Dν νo 0.5 k = 0.69  Z   

Gas Dissolved in Water

Fig. 3.13 : Gas Scrubbing in a Wetted-Wall Column

Sol. : We have to write a mass balance as the water in a differential column of height ∆ Z So,

(Accumulation) = (Flow in – Flow out) + (Absorption) 0 =

Where,

[π dl νo C1_] Z – [π dl νo C1] Z + ∆Z

+ πd∆zk

[C1 (sat.) – C1]

… (1)

d = Column diameter = 10 cm. l = Film thickness = 0.07 cm νo = Velocity = 3 cm/sec. C1 = Vapour concentration in water.

Therefore, equation (1), can be written as, dC1 0 = l νo dZ + k [C1 (sat.) – C1]

… (2)

Principles of Mass Transfer Operations − I (Vol. − I)

3.37

Mass Transfer Coefficients

k is given by Dνo 0.5 k = 0.69  Z    Dνo 0.5 dC1 ∴ 0 = – lν3o dZ + 0.69  Z  [C1 (sat.) – C1]     It is given that the entering water is pure so, Z = 0, C1 = 0 Therefore, integrating equation (3), we get, dC1 = 0.69 Dνo 0.5 C (sat.) – C lνo dZ 1]    Z  [ 1 lνo⌠ dC1

⌡

[C1 (sat.) – C1] l νo ln {C1 (sat.) – C1}

… (3)

OR

Dνo 0/5 ⌠ = 0.69  Z    ⌡ dz 1  Z3/2  [0.069 (Dν0) 0.05/lν0]

= – 0.69 (Dν0)0.5 .

C1 C1 (sat.)

= e

C1 C1 (sat.)

= 1–e

OR

– 1.38 (DZ l2 ν0)0.5

Inserting the known values, we get, 2  l ν  ln 1 – C1 2 1.90 D   C1(sat.)  2  (0.07 cm) (3 cm/sec.)  [ln (1 – 0.1)]2 =   (1.90) × 1.8 × 10–5cm2/sec.

Z =

Z = 4.182 cm. (Ans.) (14) A solid disc of benzoic acid 2.5 cm in diameter is spinning at 20 r.p.m. and 25oC. How fast will it dissolve in a large volume of water ? How fast will it dissolve in a large volume of air ? The diffusion coefficients are 1 × 10–5 cm2/sec. in water and 0.233 cm/sec. in air. The solubility of water is 0.003 gm/cm3, its equilibrium vapour pressure is 0.30 mm Hg. You may use following correlation to find mass transfer coefficient : ω 1/2 V 1/3 k = 0.62 D   D ν    Sol. : The dissolution rate is given by N1 = kC1 (sat) Where, C1 (sat) is the concentration at equilibrium For water, the mass transfer coefficient is, cm2 (20.60) (2π,/sec) 1/2 k = 0.62 1 × 10–5 sec      0.01 cm2/sec.  2

×

 0.01 cm /sec  1/3 1 × 10–5 cm2/sec  

k = 0.90 × 10–3 cm/sec

Principles of Mass Transfer Operations − I (Vol. − I)

3.38

Mass Transfer Coefficients

Thus, the flux is, cm N1 = (0.90 × 10–3) sec × (0.003) gm/cm3 = 2.7 × 10–6 gm / cm2. sec For air, the mass transfer coefficient is, cm2 k = 0.62 0.233 sec   

2

(20/60) (2π/sec) 1/2 ×  0.15 cm /sec  1/3  0.15 cm2/sec  0.233 cm2/sec

k = 0.47 cm/ sec. The flux is, N1 =

1 mol  × 273 122 gm 0.47 cm    0.3 mm Hg   3 3 sec.  760 mm Hg 22.4 ∞ 10 cm  298  mol  

N1 = 0.9 × 10–6 gm/cm2 sec. 1rd The flux in air is about 3 of that in water even though the mass transfer coefficient in air is about 500 times larger than that in the water. (Ans.) (15) Calculate convective mass transfer coefficient in a wetted wall tower for the transfer of benzene into CO2 at OoC. The gas velocity through the column is 1.25 m/s. The diameter of the column is 0.16 m. Assume that benzene concentration in CO2 is very low. Date : ρCO = 1.966 kg /m3 2

µCO2 = 1.35 × 10–5 kg/m.sec. DAB = 5.28 × 10–6m2/sec. Sol. : It is given that benzene concentration in CO2 is very low, hence for dilute solution, –

PB,M = P –

So, Sherwood Number

ShM ShM

kcd PB,M = P = kC d µ 1.35 × 10–5 = 1.966 ρ = 6.8 × 10–6 m2/sec

ν =

Schmidt Number,

Reynolds Number, So,

SC =

µ n 6.8 × 10–6 =D = ρDAB 5.28 × 10–6 AB

= 1.2878 × 103 dν dνρ = Re = µ = ν

0.16 × 1.25 = 29.411 × 103 6.8 × 10–6

ShM = 0.023 Re0.82 Sc0.44 = 0.023 (29.411 × 103) = 118.642

0.82

(1.2878)0.44

Principles of Mass Transfer Operations − I (Vol. − I) and,

Mass Transfer Coefficients

kcd ShM = D AB kC =

∴

3.39

118.642 × 5.28 × 10–6 0.16

kC = 0.003915 m/s

(Ans.)

(16) Air flow through cylindrical tube made of naphthalene at velocity of 5 m/s. The diameter of tube is 0.1 m and temperature of air is 293 K. Using correlation proposed by Gilliand and Sherwood, calculate the mass transfer coefficient for transfer of naphthalene to air. Data : Viscosity of air = µ (air) = 1.8 × 10–5 kg/ms Density of air = ρ(air) = 1.2 kg/m3 Diffusivity = DAB = 4.24 × 10–6m2/s µ 1.8 × 10–5 = = 1.5 × 10–5m2/sec 1.2 ρ 1.5 × 10–5 ν SC = D = = 3.54 4.24 × 10–6 AB ν =

Sol. :

Re =

0.1 × 5 DV = = 33.33 × 105 ν 1.5 × 10–5

So, we use Gilliland-Shrewood correlation, ShM = 0.023 Re0.83 . SC0.44 = 0.23 (33.33 × 105)

0.83

× (3.54)0.44

ShM = 198.10 So,

kC =

ShM × DAB 198.10 × 4.24 × 10–6 = d 0.1

kC = 8.39 × 10–3 m/sec.

(Ans.)

(17) Pure water at 26.1oC flows at the rate of 5.514 × 10–7 m3 /sec through a packed bed of benzoic acid spheres having a diameter of 6.375 mm. The total surface area of the spheres in the bed is 0.01198 m2 and the void fraction is 0.436. The tower diameter is kgmol 0.0667 m. The solubility of Benzonic acid in water at 2.498 × 10–2 m3 (a) Predict the mass transfer coefficient kC. (b)

Using the value of kC calculated in (a), predict the outlet concentration of benzoic acid.

Data : Physical properties of water at 26.1oC Viscosity = µ = 0.8718 × 10–3 Pa. sec. Density = ρ = 996.7 kg/m3 Diffusivity = DAB = 1.254 × 10–9 m2/sec.

Principles of Mass Transfer Operations − I (Vol. − I)

3.40

Mass Transfer Coefficients

Sol. : π (a) A = Tower cross-sectional area = 4 × D2 2 π = 4 × (6.635 × 10–3) = 3.45 × 10–5 m2 Q 5.514 × 10–7 V = velocity in the tower= A = = 0.015 m/s 3.45 × 10–5 µ SC = ρ.DAB = Re = = JD = =

0.8718 × 10–3 = 697.52 996.7 × 1.254 × 10–9 ρVD µ (996.7) (0.015) (6.635 × 10–3) = 113.78 (0.8718 × 10–3) 1.09 (Re)–0.67 ∈ 1.09 –0.67 = 0.4768 4.36 (113.78)

'

Assume kC = kC for dilute solution, '

(NSc) 0.67 ν' JD.ν' 0.4768 × 0.015 = 0.67 = (NSc) (697.52)0.67

JD =

Thus,

'

kC

kC

'

(Ans.) kC = 4.447 × 10–6 m/s (b) To calculate outlet concentration of benzoic acid in water, we may use of following expressions. (CAi – CA ) – (CAi – CA ) 1 2 NA · A = AkC C –C



Ai



Ai – CA

ln C

  2

A1

Where, CAi = concentration at the surface of the solid (k mole/m3) CA1 = Inlet bulk fluid concentration and CA2 is the outlet material balance equation for the bulk stream NA · A = V (CA – CA ) 2

1

m3 where V = volumetric flow rate of above two fluid entering, sec Thus, we can write from equations, (CAi – CA ) – (CAi – CA ) V 1 2 A kC = (C – C ) C – C A2 A1  Ai A1 ln C – C 



Ai

A2



Principles of Mass Transfer Operations − I (Vol. − I)

3.41

Mass Transfer Coefficients

Here, CA = 2.948 × 10–2 1

CA = 0 1

A = 0.01198 m2 V = 5.514 × 10–7 m3/s Putting all these values in above equation, we get, 0.01198 (4.665 × 10–6) (CA – 0) 2 = (5.514 × 10–7) (CA – 0) 2 2.948 × 10–2 – 0   ln  2.948 × 10–2– CA 



2



kg mole (Ans.) m3 (18) Ammonia is absorbed from an air stream into a water film flowing down an inclined plane at 20o C and 1 atm. as shown in fig. The flow rate of the water is 5 × 10–3 kg/sec per meter width of the plate. If the laminar flow is assumed, the thickness of the film can be calculated from the expression : 3µΓ Γ  1/3 δ =  2 ρ g cos β  Where, Γ is the mass rate of flow per unit width of the plate. Assuming that the film thickness calculated from above expression above is equal to the thickness of the stagnant film for the film theory, estimate the mass transfer coefficient for the absorption of ammonia into the water. Solving for CA , we get,

CA

2

2

= 2.842 × 10–3

z x C A1 C A2

o

6 b=

0

Fig. 3.14 : Absorption of Ammonia

Data : Physical properties at 20°°C Viscosity of water = µH O 2

= 10.5 × 10–4

kg –3 ms , Γ = 5 × 10 kg/sec

Density of water = ρH

2O

D NH

3 – H2O

= 998 kg/m3

= 1.76 × 10–9 m2/sec.

Principles of Mass Transfer Operations − I (Vol. − I) δ =

Sol. :

3.42

Mass Transfer Coefficients

 3µΓ  1/3 ρ2g cos β –4

=

(3 × 10.5 × 10 ) × (5 × 10  (998)2 (9.8) (cos 60) 

–3

)1/3

 

δ = 1.39 × 10–4 m From the films theory, ' kC

=

DAB

=

D NH

2 – H2O

δ δ 1.76 × 10–9 m2/s = 1.39 × 10–4m

(Ans. ) = 1.27 × 10–5 m/s (19) NH3 is absorbed at 1 bar from an NH3 air stream by passing it a vertical tube down which dilute H2SO4 is flowing. The following laboratory test data are available : Length of the tube = 825 mm Diameter of tube = 15 mm Partial pressure of NH3 at inlet = 7.5 kN/m2 Partial pressure of NH3 at outlet = 2 kN/m2 The amount of NH3 absorbed at this condition is 1.12 × 10–6 k mole/sec. Determine the overall transfer coefficient kG based on gas phase. Sol. : Driving force at inlet = 7500 N/m2 Driving force at outlet = 2000 N/m2 So,

Mean driving force =

7500 – 2000 7500 ln 2000  

= 4161 N/m2 Wetted surface = π D. L = π × 5 × 10–3 × 825 × 10–3 = 0.038 m2 The overall transfer coefficient KG is given by KG

k mol (NH3 absorbed) sec = (wetted surface) m2 × mean driving force N/m2 =

1.12 × 10–6 0.038 × 4161

KG = 6.93 × 10–9 k mole/N·sec.

(Ans.) –8

(20) Pure water at 30°°C is flowing through a packed bed at the rate of 6.9 × 10 m3/sec through a packed bed of β -napthol spheres having a diameter of 10 mm. The total surface area of sphere in bed is 0.015 m2 and the void fraction is 0.35. The tower diameter is 0.0950 m. The solubility of β -Napthol in water is 5.3 × 10–2 kg mol/m3. Calculate the mass transfer coefficient kC.

Principles of Mass Transfer Operations − I (Vol. − I)

3.43

Mass Transfer Coefficients

Data : The physical properties of H2O at 30oC Viscosity = µ = 1 × 10–3 Pa. sec Density = ρ = 1000 kg / m3 Diffusivity = DAB = 2.60 × 10–10 m2/sec. Sol. :

Tower cross-sectional area

Velocity in tower = (V)

=

π A = 4 (D2) π = 4 (0.950) 2 = 7.0882 × 10–3 m2 Q 6.9 × 10–8 = A 7.0882 × 10–3 = 9.734 × 10–6 m/s µ SC = ρDAB = Re

= =

JD

=

JD

=

1 × 10–3 = 3846.154 1000 × 2.60 × 10–10 ρVD 1000 × 9.734 × 10–6 × 10 × 10–3 µ = 1 × 10–3 0.09734 1.09 1.09 (Re) –0.67 = 0.35 (0.09734)–0.67 ∈ 14.886 '

and,

JD '

kC '

kC

kC

= V (Sc) –0.67 JD.V 14.866 × 9.734 × 10–6 = 0.67 = (3846.154)0.67 (NSc) = 6 × 10–7 m/s

(Ans.)

(21) Calculate the maximum rate of absorption of O2 in a fermenter from air bubbles at 1 atm having diameter of 100 µm at 37oC into water having a zero concentration of dissolved oxygen. The solubility of O2 from an air in water at 37oC is equal to 2.26 × 10–4 kg mole O2/ m3. Data : Diffusivity of O2 in H2O = 3.25 × 10–9 m2/s Viscosity of water (µc) = 6.947 × 10–4 kg/ms Density of Water (ρ ρc) = 994 kg/m3 Density of air particle (ρ ρp) = 1.13 kg/m3 Assume that agitation is used to produce air bubbles. Determine, '

(a) Connective mass transfer coefficient, kL (b) Mass transfer rate. Sol. : (a) In this case, agitation is used to produce air bubbles, hence the mass transfer resistance in gas bubble to outside interface of the bubble can be neglected. µ 6.947 × 10–4 NSc = = = 215.044 ρDAB 994 × 3.25 × 10–9

Principles of Mass Transfer Operations − I (Vol. − I) ∆ρ

= ρc– ρp

=

3.44

994 – 1.13

Mass Transfer Coefficients 992.87 kg/m3

=

'

Here we use following correlation to find kL '

kL

=

2 DAB ∆ρ µc.g 1/3 –0.67  + 0.31 (N )  2  Sc DP  (ρc) 

=

–4 2 × 3.25 × 10–9 992.87 × 6.947 × 10 × 9.81/3 + 0.31 (215.044)–0.67   –6 2 (994) 100 × 10  

'

(b) CA

2

kL = 2.25 × 10–4 m/s = 0 (Since O2 is consumed faster than it supplied so CA2 at interface is zero) CA

1

= 2.26 × 10–4 k mol/sec. '

NA = kL (CA – CA ) = 2.25 × 10–4 (2.26 × 10–4 – 0) 1

2

k mol (Ans.) NA = 5.08 × 10 m2 sec (22) Water at 293 K is flowing over a horizontal plate at a velocity of 3 m/s. Determine the thickness of boundary layer at a distance of 50 mm from leading edge. Data : Kinematic viscosity = 1 × 10–6 m2/s x.V 50 × 10–3 × 3 = Sol. : Rex = ν 1 × 10–6 = 1,50,000 < 3 × 105 The flow is laminar, over this length of plane. So we are have for laminar flow 4.64 (x) 4.64 (50 × 10–3) δ = = (Rex)0.5 1500000.5 –8

(Ans.) δ = 6.008 × 10–4 mm = 0.6 mm (23) A napthalene 75 mm diameter and 60 mm length is exposed to a stream of pure CO2 flowing perpendicular direction to the cylinder with gas velocity of 6 m/s. The temperature of CO2 is 373 K and pressure of CO2 is 101.32 kN/m2. Data : Viscosity of CO2 at 373 k = 19 × 10–6 pa. sec Density of CO2 at 373 K = 1.449 kg/m3 Vapour pressure of napthalene at 373 k = 1.33

kN m2

DNaphthalene – CO2 = 8.224 × 10–6 m2/sec. (a) (b)

–

Determine mass transfer coefficient, F Determine the rate of sublimation. –

You may use following correlation for estimation of mass transfer coefficient, F ––

Sh ––

= 0.43 + 0.532 × Re0.5 × Sc0.3 –

F.dp where , Sh = C.D AB dp = Diameter of naphthalene cylinder C = Molar density at 373 K = 0.03267 kg/m3 ρVD 1.449 × 6 × 75 × 10–3 Sol. : (a) NRe = µ = 19 × 10–6

Principles of Mass Transfer Operations − I (Vol. − I)

NSc

3.45

Mass Transfer Coefficients

= 34.318 × 103 µ 19 × 10–6 = = ρ.DAB 1.449 × 8.224 × 10–6 = 1.5944

––

So,

Sh ––

So,

Sh

= 0.43 + 0.532 (34.318 × 103)0.5 (1.5944)0.31 = 114.318 –

F. dp = C.D AB –

F × 75 × 10–3 360.649 = 0.03267 × 8.22 × 10–6 –

F

= 409.529

× 10–6 k mol/m2 sec.

(Ans.)

(b) Here, NB = 0 NA NA + NB

∴

= 1

– (1 – CA2/C) NA = F ln (1 – C /C) A1  

∴ ...

CA2 = 0,

CA2 C =0 CA1 P = Pt C

1.33 × 103 = 0.0131 101.32 × 103 1–0 ∴ NA = 409.529 × 10–6 ln 1 – 0.013   = 5.411 × 10–6 k mole naphthalene/m2 sec This is based on total area of cylinder which is equal to, π π π × dp × h + 2 4 × d2 = π (0.075) (0.6) + 4 (0.075)2 = 0.1502 m2   Hence, the rate of sublimation of naphthalene cylinder : kg = (5.411 × 10–6) (128) (0.1502) (3600) hr = 0.3745 kg/hr (Ans.) (24) Air at 75oC flows through a 0.05 m long packed tube of naphthalene spheres at an interstitial velocity of 0.6 m/s as shown in Fig. 3.15. and,

o

Air Temp. = 75 C

V = 0.6 m/s

=

Î = 0.5

dp = 0.01 m NA = ? mass – transfer coefficient = 0.057 m/s CA

Z = 0.05 m Fig. 3.15 : Convection mass transfer coefficient for packed tube of napthalene spheres

The convective mass transfer coefficient for this process was found to be 0.057 m/s. (a) Calculate the bulk concentration of naphthalene in air at the exist of the tube.

Principles of Mass Transfer Operations − I (Vol. − I)

3.46

Mass Transfer Coefficients

(b) Determine the percent saturation of the exiting air assuming the entering air is free of naphthalene. The vapour pressure of naphthalene at 75oC is 7.47 × 102 N/m2 Sol. : (a) A shell balance over a differential section of the packed bed is given by dNA * dZ = kCa (CAi – CA) Where, CAi = Solute concentration at the solid interface.

… (1)

CA = Solute concentration in the bulk fluid at any position in the tube. Since, mass transfer in the axial direction is due to bulk flow, the molar flux can be written as : … (2) NA ≅ CAU If we assume plug flow in the packed bed and U is constant, then dCA * U dz = kC a (CA‚i – CA) Where, a = Interfacial area (m2/m3) Separating the variables and integrating over the tube length gives, CA2 Z * kC. a dCA ⌠ C – C = ⌠ U dZ ⌡ Ai A ⌡ CA1 0 CA2

[ln (CAi– CA)]CA

=

kC . a

.Z

U

* kC a.Z CAi – CA1  ln C – C  = U  Ai A2

Where,

… (5)

U = ∈ × Interstitial Velocity = 0.5 × 0.6 = 0.3 m/s a = Interfacial area = a =

6 (1 – ∈) dp 6 (1 – 0.5) 2 3 0.01 = 300 m /m

CA1 = 0 (... CO2 entering air free from naphthalene) PAi CAi = RT

=

7.5 × 102 8314 × 348

= 2.59 × 10-4 k mol. m3 Put all these values in equation (5) –4  2.59 × 10 – 0  ln   –4 2.59 × 10 – CA2

… (4)

*

1

or,

… (3)

=

0.057 × 300 × 0.05 0.3

Principles of Mass Transfer Operations − I (Vol. − I)

3.47

Mass Transfer Coefficients

Solving for CA2 gives, CA2 = 2.44 × 10–4 kg mole/m3

(Ans.)

(b) % saturation is related to the concentration at the solid interface. Thus, CA2 % Saturation = C × 100 Ai =

2.44 × 10–3 × 100 2.59 × 10–4

= 94 % (Ans.) (25) A wetted wall column of diameter 50 mm is used to strip CO2 from an aqueous solution flowing in a thin wall while an air stream is flowing up through the column. The concentration of CO2 in the air stream is 1.5 mole % while at that same point, the concentration of CO2 in water is 0.5 mole %. Calculate the gas mass transfer coefficient and molar flux at this point in the column. The operating pressure and temperature of column are 10 atm and 298 K. At 298 K, Henry's law constant for CO2 in the water is equal to 1.6 × 10–2 per mole of CO2 in the solution. Data : DCO2 air at 298 K = 1.55 × 10–6 m2/sec Viscosity = µ = 18.3 × 10–6 kg/m.sec Density of air = ρair = 11.86 kg/m3 You may use the following expression to find at the mass transfer coefficient. ––

Sh

= 0.02 (Re)0.83 (Sc)0.44 –

––

Where, Sol. :

Sh

PBM kc = D . D. P AB

18.3 × 10–6 11.86 × 1.55 × 10–6 = 0.995 ρVD 11.86 × 15 × 10–3 × 75 × 10–2 Re = µ = 18.3 × 10–6 = 2430. 27 Partial pressure at interface : Let A = CO2, B = Air SC =

Henry's law, *

Where, yA

µ ρ. DAB

* yA

=

PA* = P = H. xA

= Mole fraction of solute gas (A) in gas phase in solution.

* PA

= Equilibrium partial pressure (A) P = Total pressure XA = Mole fraction of solute (A) in solution

So,

*

PCO

2

= H. xA 0.5 = 100 × 1.64 × 10–2

*

PCO

2

= 0.82 atm.

Principles of Mass Transfer Operations − I (Vol. − I)

3.48

Mass Transfer Coefficients

1 std. atm. → 0.82 ∴

*

10 std atm. → 8.2 = PCO

2

*

P*Air = 10 – PCO

2

= 10 – 8.2 = 1.8 atm Sh = 0.023 (Re)0.33 (Sc)0.44 = 0.023 (2430.27)0.33 (0.995)0.44 = 100 Log mean partial pressure of non-diffusing air : Partial Pressure of CO2

Position

PA,1 = 8.2 atm

Interface

Partial Pressure of Air –

–

PB1 = Pt – PA1 = 10 – 8.2 = 1.8 atm.

–

So,

–

PA,2 = yA,2 × P 1.5 = 100 × 100 = 0.15 atm

Bulk gas

–

PB,M =

= 1 – 0.15 = 9.85 atm.

–

PB2 – PB1 –

–

PB2 = Pt – PA2

–

ln (PB2/PB1 )

=

9.85 – 1.8 9.85 ln  1.8   

–

PB,M = 4.736 atm –

So,

––

Sh

PB‚M kC = D .D P AB 100 × 1.55 × 10–6 × 10 50 × 10–3 × 4.736 = 6.56 × 10–3 m/s

kC = kC

(Ans.)

Molar Flux, NA = kC (CA,i – CA, ∞) kc = RT (PA,i – PA, ∞) Where, PA,i = Partial pressure of CO2 at the interface = 8.2 atm PA, ∈ = Partial pressure of CO2 at the bulk stream = 0.15 atm So,

6.562 × 10–3 (m/s) (8.2 – 0.15) (atm) atm.m3 0.08205  k mol.k × 298 (k)   k mol = 2.1604 × 10–3 2 (Ans.) m sec

NA =

NA

Principles of Mass Transfer Operations − I (Vol. − I)

3.49

Mass Transfer Coefficients

(26) H2O containing dissolved air is to be deoxygenated by contact with N2 in an agitated vessel 1 m in diameter. The H2O will enter counter currently at bottom of the vessel and flow through an overflow pipe set in air side of vessel at an appropriate height. Neglecting gas density, estimate, mass transfer coefficient on liquid side if temperature is 298 K. Use the following equations to solve this problem : ShL = 2.0 + 0.31 Ra1/3 FL dp Where, ShL = C. D L and Ra = Rayleigh Number and is given by : dp3 (ρ ρL– ρo)g Ra = DLµL Where, dp = Average bubble diameter in m ρL = Density of liquid in kg/m3 ρg

= Density of gas in kg/m3

DL = Diffusivity of N2 in water in m2/sec. µL = Viscosity in kg/m. sec The value of dp may be calculated by : 0.4 d p = 2.88 × 10–3 φ G Where φG = Gas hold up and is given by relation : φG =

7.067 × 10–3 Vt0.833

Where, Vt = Terminal settling velocity of bubble. Data at 298 K : Vt = 0.25 m/s DL = 1.9 × 10–9 m2/s µL = 8.9 × 10–4 kg/m.s ρL = 1000 kg/m3 FL.dp ShL = C.D L

Sol. : Where,

FL = Mass transfer coefficient on liquid side, k mol/m2 sec ShL = Sherewood Number for liquid dp = Average bubble diameter in m 1000 C = Concentration of liquid = 18 = 55.5 k mol/m3 DL = Diffusitivity of N2 in water φG

So,

= 1.9 × 10–9 m2/s = 0.0224

dp = 2.88 × 10–3 (φG)0.4 = 2.88 × 10–3 (0.0224)0.4

Principles of Mass Transfer Operations − I (Vol. − I)

3.50

Mass Transfer Coefficients

dp = 6.302 × 10–4 m and,

Ra =

dp3 (ρL – ρo) g DL µL

(6.302 × 10–4)3 (1000) (9.81) 1.9 × 10–9 × 8.9 × 10–4 Ra = 1.452 × 106 =

So, ShL = 2.0 + 0.31 (Ra)1/3

Hence,

= 2.0 + 0.31 (1.452 × 106)1/3 ShL = 37.103 FL. dp Sh = C. D L 37.103 =

FL × 6.302 × 10–4

1.9 × 10–9 × 55.5 k mol ∴ FL = 6.208 × 10–3 2 (Ans.) m sec (27) Benzene vapours from a dilute mixture with inert gas is to be absorbed in packed bed with Berl Saddles. The circumstances are as follows : (i) For gas side : Average Mol. Weight = 11 Viscosity = 10 × 10–5 kg/m. sec. Temperature = 270C Pressure = 107 kN/m2 Diffusivity = DG = 1.3 × 10–5m2/s (ii)

Flow rate = GL = 0.716 kg/m3 sec. For liquid side : Average Mol. Weight = 2 6 0 Viscosity = 2 × 10–3 kg/m. sec Density = 840 kg/m3 Diffisivity = DL = 4.71 × 10–10 m2/sec

(1) (2) (3) (4)

Flow Rate = LL = 2.71 kg/m3. sec Additional Data : Fractional void volume in a dry packed bed = ∈ = 0.75 Liquid hold up in a packed tower = φt = 0.0333 m2/m3 Diameter of sphere of same surfaces as a single particle = ds = 0.0472 m Specific interfacial surface = aA = 37.4 m2/m3 Compute the volumetric mass transfer coefficient using following equations : For Berl saddle the gas phase coefficient Fa is given by FG. SCG2/3 G

l  ds G –0.36 = 1.195   µG(1 – ∈ Lo)

Principles of Mass Transfer Operations − I (Vol. − I)

3.51

Mass Transfer Coefficients

Where, ∈Lo = Operating void space in packing = ∈ – φLt SCG = Schmidt number for gas G = gas flow rate, k mol/m2 sec. The liquid coefficient kL is given by – kLds DL

l ds.L  0.45 . Sc 0.5 = 25.1  µ  L  L 

Sol. : (a) for gas side : Assume that ideal gas behaviour ∴ PV = nRT P Mavg and hence density of gas ρG = RT 107 × 103 × 11 = 0.08205 × 305 101.3 × 103 ρG = 0.472 kg/m3 =

Next, ScG

µG ρG.DG =

∈ Lo

10 × 10–5 0.472 × 1.3 × 10–5

= 1.63 = ∈ – φLt

= 0.75 – 0.033 = 0.716 0.716 G = 11 2/3 G

FG × SC ∴ ∴

G

–0.36 FG × (1.63)2/3  0.0472 × 0.716  = 1.195 –5 10 × (1 – 0.716) 0.716  11 

FG = 1.913 × 10–3 k mol/m2 sec.

∴ (b)

For liquid side : SC

L

kL.ds DL

So, ∴

l  ds. G –0.36 = 1.195   µG(1 – ∈Lo)

kL ×

0.0472 4.71 × 10–10

=

µL ρLDL

=

2 × 10–3 = 5055.1 840 × 4.71 × 10–10

l ds. L  0.45 (S )0.5 = 25.1  µ  CL  L 

0.0472 × 2.71 0.45 = 25.1  [5055.1]0.5  2 × 10–3 

kL = 1.164 × 10–4k mol/m2 sec.

Principles of Mass Transfer Operations − I (Vol. − I)

3.52

Mass Transfer Coefficients

ρ C = M

(c) For liquid side

k mol 840 = 260 = 3.23 m3 FL = kL.C = 1.164 × 10–4 × 3.23 = 3.76 × 10–4 k mol/m2 sec. and, volumetric coefficients are : (i)

For gas side, FG. aA = 1.913 × 10–3 × 37.4 = 0.071 k mol/m2 sec.

(ii)

(Ans.)

For liquid side, 2

FL. G A = 3.76 × 10–4 × 37.4 (Ans.) = 0.014 k mol/m3 sec. (28) Prove or show the following relationships starting with the flux equations : (a) Convert k'C to ky and kG; (b) Convert kL to kx and k'x; (c) Convert kG to ky and kC. Sol. : Remembering that dashes mean equimolar counter diffusion and a lack of them diffusion through a stagnant layer : ky kG (a) NA = k'c (CA – CA ) y (yA – yA ) = P (PA – PA ) 1 2 1 2 1 2 BM BM ∴

ky kG = P (PA – PA ) = P y 1 2 total BM BM

k'c RT (PA1 – PA2) '

kc = P

∴

(PA – PA ) 1

2

RT RT ky = P kG y total BM BM

(Ans.)

k'x (xA – xA ) NA = kL (CA – CA ) = kx (xA – xA ) = x

(b)

1

∴ ∴

2

1

2

1

BM

2

k'x kL C (xA – xA ) = kx (xA – xA ) = x (xA – xA ) 1

2

1

2

BM

1

2

kx k'x kL = C = Cx BM

(Ans.)

NA = kG (PA – PA ) = ky (yA – yA ) = kC (CA – CA )

(c)

1

∴ ∴

2

1

2

1

2

ky kc kG (PA – PA ) = P (PA – PA ) = RT (PA – PA ) 1 2 1 2 1 2 total ky kc kG = P = RT total

(Ans.)

Principles of Mass Transfer Operations − I (Vol. − I)

3.53

Mass Transfer Coefficients

' (29) Experimental analysis found a convective mass transfer coefficient (kG ) of −8

2

1.486 × 10 k mole/m s. Pa for Equimolal counter diffusion of species A through B at atmospheric pressure. Calculate the corresponding value (kG) and the resulting mass flux for diffusion of A through a stagnant layer of B for the same conditions and concentrations, namely partial pressures PA = 5000 and PA = 2400 Pa. 1

2

'

Sol. : Find conversion between kG and kG : – ' kGPtotal = kG P BM '

Ptotal kG = – PBM

kG

∴

From a total pressure of 101325 Pascals : pA

= 5000 Pa/PB = 101325 – 5000 = 96325 Pa

pA

= 2400 Pa/PB = 101325 – 2400 = 98925 Pa

1

1

2

2

96325 – 98925 – PBM = ln (96325/98925) = 97619.225 kG =

∴

(1.486 × 10–8) × 101325 97619.229

= 1.5424108 × 10–8 k mole/m2 s. Pa For Stagnant layer, NA = kG (pA – pA ) = 1.5424108 × 10–8 (5000 – 2400) 1

2

= 4.010268 × 10–5 k mole/m2 s. i.e. flux is 4.010 × 10–5 k mole/m2 s, (Ans.) mass transfer coefficient 1.542 × 10–8 k mole/m2 s. Pa (30) An air stream at 52.6 oC and 2 atmospheres (abs.) flows through a duct at 1.524 m/s past samples of solid naphthalene. Naphthalene diffusivity at 0 oC and 101.32 kPa is –6

2

5.16 × 10 m /s. Estimate mass transfer coefficient k'C for flow past the following shapes : (a) Parallel to a flat plate 152 mm long; (b) A lone sphere of 12.7 mm diameter. Sol. : Data from literature at 52.6 oC (425.6 K) : µ = 1.9682734 × 10–5 kg/m.s ρ = 1.0876331 kg/m3 For the Schmidt number, we must use Fuller et al. equation to predict the diffusivity of naphthalene in air under these conditions, knowing that it is 5.16 × 10–6 m2/s at 101.32 kPa and 273 K; DAB = DAB 2

1

T2 1.75 P1 = 5.16 × 10–6 325.6 1.75 101.32 = 3.5125175 × 10–6 m2/s T  P2  273   202.6   1 µ Sc = D

AB

=

1.9682734 × 10–5 = 5.1521026 1.0876331 (3.5125175 × 10–6)

Principles of Mass Transfer Operations − I (Vol. − I)

Mass Transfer Coefficients

ρVxd 1.0876331 × 1.524 × 0.152 = = 12800.5 µ 1.9682734 × 10–5

Re =

For flat plate :

3.54

0.5

'

i.e. less than 15,000 so,

kc

0.664 DAB Re Sc0.33 DAB = L Sh = L =

0.664 (3.512175 × 10–6) (12800.5)0.5 (5.1521026)0.33 0.152

= 2.9983611 × 10–3 m/s Re =

For single sphere :

(Ans.)

ρV∞d 1.0876331 × 1.524 × 0.0127 = 1069.512 µ = 1.9682734 × 10–5

i.e. a gas between 1 and 48,000 although Schmidt falls outside 0.6 to 2.7 range, the best we DAB DAB ' can do is : kc = L Sh = L (2 + 0.552 Re0.53 Sc0.33) 3.5125175 × 10–6 [2 + 0.552 (1069.512)0.53 (5.1521026)0.33] 0.0127 = 0.0111835 m/s (Ans.)

=

3

(31) A micro-organism batch of density 1082 kg/m and spherical diameter 0.75 mm weighing 3.6 grams is added to an aqueous solution of oxygen at 37 oC (oxygen –4

3

concentration = 2.26 × 10 k mole/m ) in a shaker flask for a fermentation. Air can enter through a porous stopper. If oxygen transfer to surface is rate-determining step, calculate : (a) The mass O2 consumption rate of oxygen micro-organism is the only consumer of oxygen in a saturated solution; (b) The percentage saturation of the solution if the micro-organisms only consumed 0.5 grams per second and other nutrients accounted for the rest. Sol. : The prescribed correlation for particle suspensions of less than 0.6 mm is : 2DAB g µ | ∆ ρ| ' –0.67  kL = d + 0.31 Sc ρ2   Before we calculate the Schmidt number, remember that the particle and the water are of the same density, i.e. Dr = 0, so the second half of the correlation disappears. 2DAB 2 (3.27 × 10–9) ' kL = = = 3.0293144 × 10–3 m/s d 1.08 × 10–6 NA = k'L(CA – CA ) = 3.0293144 × 10 1

i.e. 6.846 × 10

2

–7

–3

–4

[(2.26 × 10 ) – 0] = 6.8462505 x 10

–7

2

k mole/m s.

When we talk of a RATE DETERMINING STEP, this means that the transfer through the porous stopper is so fast as to be considered instantaneous compared to the transfer to the organism surface, and that the absorption takes place so quickly that the effective surface oxygen concentration is zero. The prescribed correlation for particle suspensions of less than 0.6 mm is : '

kL =

2DAB d

+ 0.31 Sc–0.67

g µ | ∆ ρ |  ρ2  

0.33

Principles of Mass Transfer Operations − I (Vol. − I)

3.55

Mass Transfer Coefficients

From literature data for water at 37 oC : –6

m = 695 × 10 kg/m.s ρ = 993.04866 kg/m3 Using existing data for a known temperature and the corresponding viscosity, extrapolate the diffusivity for this system using Wilke-Chang : T 2 µ1 310 890 × 10–6 –9 2 DAB = DAB T µ = 2.41 × 10–9 298    695 × 10–6 = 3.2716595 × 10 m /s 2 1 1 2 µ Sc = D

AB

∴

=

695 × 10–6 = 213.91743 993.04866 (3.2716595 × 10–9)

∆ρ = 1082 – 993.04866 = 88.951341 kg/m3 9.81 × 0.000695 × 88.95130.33 2 (3.27166 × 10–9) ' + 0.31 (213.917)–0.67  kL = –7 7.5 × 10 (993.049)2   -3

= 8.7981258 × 10 m/s The flux of oxygen to the organism is therefore : '

NA = kL (CA – CA ) = 8.7981258 × 10 1

2

–3

–4

[(2.26 × 10 ) – 0] = 1.988377 × 10

–6

2

k mole/m s

With mass M of organisms comprising a total number of particles N, the total surface area A is given by : A = Np d

∴ ∴

2

π M = Nρ 6 d3 6M N = π ρ d3 6 M 6M 6 × 0.0036 A =π πd2 = = = 26.617375 m2 ρd ρd3 1082 × (7.5 × 10–7) –6

molar uptake = 26.617375 × 1.988377 × 10 = 5.2925377 × 10 With a molecular weight of 32 kg/k mole : –5

–3

–5

k mole/s

Mass uptake = 5.2925377 × 10 × 32 = 1.6936121 × 10 kg/s. (Ans.) (32) A circular metal pipe was used to transport oil sands. After a long time of service, the inner surface was coated with a thin layer of bitumen. The inner diameter is D and the tube length is L. It was decided to use a solvent to clean the tube turbulen flow conditions. Assume a dilute system. What is the exit bitumen concentration if (a) tripling the solvent flow rate. (b) increasing the temperature from 20°C to 30°C. The solvent viscosity dropped by 20% from 20°C to 30°C. For simplicity, assume that the solubility of bitumen in the solvent is unaffected by temperature. (c) comment on (b) if solubility change is accounted for. Assume that the bitumen solubility in the solvent is 2.5 mol/m3 and prior to any changes the effluent concentration of bitumen was 1.2 mole/m3. No mass transfer experiment was performed, but the heat transfer is known to obey 0.83

Sol. : CA1 = 0, CA2 ⇒

Nu = 0.023 Re = CA, Z1 = 0, Z2 = L. ln C

CAs As – CA

=

kc πD L Q

1/3

·Pr

Principles of Mass Transfer Operations − I (Vol. − I)

CA

∴ (a)

0.83

Nu = 0.023 Re

1/3

Pr

3.56

Mass Transfer Coefficients

kc πD  – Q L   = CAs 1 – e  0.83

⇒ Sh = 0.023 Re

1/3

Sc

QII = 3QI ⇒ ReII = 3ReI , uII = 3uI kcD Sh = D

AB

ln CAs   CAs – CAS  II

∴

ln CAs   CAs – CA  I

⇒ kcII /kcI =

ReII0.83  Re  = 30.83  I

kcII QI = k Q = 30.83 – 1 = 3– 0.17 = 0.859639 cI II

ln CAs  = 0.829639 × ln 2.5  CAs – CA 2.5 – 1.2 = 0.542523  II

∴

= CAs (1 – e– 0.542523) = 1.0468 mole/m3 (b) TII = 30 + 273.15 K = 303.15 K TI = 20 + 273.15 K = 293.15 K DABII TII µI T 303.15 ∞ 1 = T µ = = 1.29264 DAB ∝ µ ⇒ D 293.15 ∞ (1 – 0.2) ABI I II CAII

µ Dρu ,R = µ ρDAB e ScII µII DABI TI 1 = 0.8 , S = µ D = T cI I ABII II

0.83

Sh = 0.023 Re ReII

µI ReI = µII

∴

ln CAs   CAs – CA  II ln CAs   CAs – CA  I

(Ans.)

1/3

Sc , Sc =

kcII ReII0.83 = k = R  cI  eI 

TII µI 293.15 1/3 ScII1/3 DABII Sc  DAB = 0.8–0.83 × 303.15 × TI µII I  I

= 1.53835691

l CAs   CAs – CA  

= 1.00597 ⇒ CAII = 1.5858 mole/m3

(Ans.)

II

(c)

As CAs increases, if temperature is increased.

(Ans.)

Therefore, CA II will be slightly higher. (33) Pure water at a velocity of 0.11 m/s is flowing at 26.1°C past a flat plate of solid benzoic acid where L = 0.40 m. The solubility of benzoic acid in water is 0.02948 kg mol/m3. The diffusivity of benzoic acid is 1.245 × 10–9 m2/s. (a) What is the corresponding mass transfer coefficient ? (b) What is the equivalent film thickness ? Sol. : To solve this, we need to look at the Re value at the end of the plate. Lvρ (0.40) (0.11) (998.2) Re µ = = 44364 9.93 × 10–4

Principles of Mass Transfer Operations − I (Vol. − I)

3.57

Mass Transfer Coefficients

Using a transition Re of 200000, we recognize this is a laminar flow situation here. The overall mass transfer coefficient can be calculated as follows : kcL = 0.6654 (ReL )1/2 Sc1/3 ShL = D AB kc (0.40) µ 1/3 = 0.664 (44364)1/2  –9 (1.245 × 10 ) ρDAB 1/3 kc (0.40) 9.93 × 10–4  = 0.664 (44364)1/2  –9 -9 (1.245 × 10 ) (998.2) (1.245 × 10 ) kc (0.40) = 0.664 (44364)1/2 (799)1/3 = 1297.8 (1.245 × 10–9) kc = 4.04 × 10–6 m/s (Ans.) (b) The equivalent film thickness is given by DAB m 1.245 × 10–9 kc = 4.04 × 10–6 s = = δ δ m2 1.245 × 10–9 s δ = (Ans.) m = 0.3 mm 4.04 × 10–6 s (35) Water flows at 10 cm/s over a sharp-edged plate of benzoic acid. The dissolution of benzoic acid is diffusion-controlled, with a diffusion coefficient of 1.00 × 10–5 cm2/s. Find the following : (a) The distance at which the laminar boundary layer ends; (b) The thickness of the flow and concentration boundary layer at the point; (c) The local transfer coefficients at the edge (kc (edge), and at the position of transition (kc (transition)), as well as the average mass transfer coefficient over this length (kc (average). Sol. : (a) To solve for this, we need to identify the point at which the Re exceeds the laminar-turbulent flow transition. Now, assuming that 300,000 is the critical value, cm g  x 10 s  1 3 cm   xvρ   = = = 3,00,000 ⇒ x = 300 cm (Ans.) µ g 0.01  cm · s  (b) Recall that for laminar flow, 1/3 µ 1/3 993 × 10–6 Pa · s δ  ≈ 10 =  =  Sc1/3 = 2 2  kg cm m δc ρDAB 998.2 m3 1 × 10–5 s × 1002 cm2   Now, we need to solve for either the velocity (flow) or concentration boundary layer and then we can get the other value. From fluid mechanics, we remember that the boundary layer for laminar flow over a 5 δ plate is ~ x = – this is derived from the Blausius solution to the laminar flow Rex profile over a flat plate. In this case, x = 300 cm. 5 (300 cm) = 2.7 cm ∴ δ = 300000 ∴ δc = 0.27 cm

(Ans.)

Principles of Mass Transfer Operations − I (Vol. − I)

3.58

Mass Transfer Coefficients

(c)

Now we can use the standard correlation to calculate the local mass transfer coefficients as follows : 0.332 DAB (Re)1/2 Sc1/3 kcx 1/2 Sc1/3 ⇒ k = = 0.332 (R ) ex c x DAB at the leading edge, x = 0, ∴ kc = ∞ cm2 0.332 1 × 10–5 s  (300000)1/2 (10)1/3   at the transition, x = 300, ∴ kc = 300 cm cm = 1.3 × 10-5 s The average value for the convective mass transfer coefficient can be determined from – kcL cm – 1/2 1/3 –5 DAB = 0.664 (ReL ) Sc so, k c = 2kc = 2.61 × 10 s . Note here that this is ONLY for laminar flow over a flat plate. (Ans.) (35) Water flows through a thin tube, the walls of which are lightly coated with benzoic acid. The benzoic acid is dissolved very rapidly, and so is saturated at the pipe's wall. The water flows slowly, at room temperature and 0.1 cm/s. The pipe is 1 cm in diameter. Under these conditions, the mass transfer coefficient varies along the pipe : 1

kx D

1

xvρ ρ 2 µ 3 = 0.3  µ      ρD

where, x is the distance along the pipe, and v is the average velocity in the pipe. What is the average concentration of benzoic acid in the water after 2 m of pipe ? Some useful information : Dbenzoic/water : 1 × 10–5 cm2/s Sc =

Sol. :

µ ρDAB

=

(993 × 10–6 Pa · s) = 995 2 2  998 kg3 1 × 10–5 cm × m s 1002 cm2  m 

cm g  (1 cm) 0.1 s  0.998 cm3   Re = = 10 9.93 × 10–3 g  cm · s 

∴

∴

kx DAB

=

 0.3  

k =

cm g  x 0.1 s  0.998 cm3   9.93 × 10–3 g  cm · s 

DAB

x Now let's set up a mass balance.

  

=

1 2

1 3

(995) 0.3

x

(10)1/2

(995)

1 3

9.47 × 10–5 cm s x

Rate of accumulation in fluid = Transfer from pipe – over the exposed area of the pipe in – out = Accumulated; Note however, that the areas are different – one is the pipe πd2 cross-section  4  , and one is the area of interfacial transport (πd ∆x).  

Principles of Mass Transfer Operations − I (Vol. − I)

3.59

Mass Transfer Coefficients

πd2 v  4  [C |x – C |x + ∆x = k (πd ∆x) (Csat – C)   Now, we can divide thisby the volume of the cylindrical element and take the limit, 0

∆x → πd2

v 4 

 [C | – C | x + ∆x] k (πd ∆x) (Csat – C)  x = 2 2 πd  ∆x πd  ∆x  4   4  =

∂C v ∂x

= 4k (Csat – C)

∂C (Csat – C)

= 4

9.47 × 10–4 ∂x x

– ln (Csat – C) = 2 (9.47 × 10-4)

∴

C 1–C

sat

∴

k (4) (Csat – C) d

∂C v ∂x

1 – e

– 2 (9.47 × 10–4)

200

= e

– 2(9.47 × 10–4)

x x

∴

1 – e

– 2 (9.47 × 10–4)

x

C =C

C = C sat

(Ans.)

C Csat = 0.02654

∴

sat

(Ans.)

(36) A tube is coated in the inside with naphthalene (A) and has an inside diameter of 20 mm and a length of 1.1 m. Air at 318 K and an average pressure of 101.3 kPa flows through this pipe at a velocity of 0.8 m/s. Assuming that the absolute pressure remains essentially constant, calculate the concentration of naphthalene in the exit air. The following constants may be helpful : DAB = 6.92 × 10–6 m2/s; PAi = 74 Pa Air : µ = 1.932 × 10–5 Pa-s, ρ = 1.114 kg/m3 µ 1.932 × 10–5 Sc = = = 2.506 ρDAB (1.114) (6.92 × 10–6)

Sol. :

Re = Flow is laminar …… ∴

Dvρ µ

(0.020) (0.80) (1.114) = 922.6 (1.932 × 10–5) We can use the appropriate correlation equation – =

1 3 D Sh = 1.86  L Sc Re  

∴

Sh = 1.86 (4.2)

1 3

DAB kL = 3 L

1

1

3 3 D 0.002 = 1.86  L Sc Re = 1.86  1.1 (2.506) (922.6)    

kLL = 3 = D AB = 3

(6.92 × 10–6) m = 1.898 × 10–5 s 1.1

Principles of Mass Transfer Operations − I (Vol. − I) Csat =

0.074 kPa mol = 0.026 1 RT

3.60

= 2.6 × 10–5

Mass Transfer Coefficients mol m3

Since, 4 (1.1) 1.898 × 10–5 (CAs – AA0) 4L KL ln (C – C ) = D  D  = (0.002)  (0.80)  = 0.052195      As AL  –5 (2.6 × 10 )  ∴ ln   = 0.052195 (2.6 × 10–5 – CAL) ∴

(2.6 × 10-5) = 1.0535 (2.6 × 10–5 – CAL)

mol (Ans.) m3 (37) Jasmone (C11H6O) is a valubale material in the perfume industry, used in many soaps and cosmetics. Suppose we are recovering this material from a water suspension of jasmine flowers by an extraction with benzene. The aqueous phase is continuous; the mass transfer coefficient in the benzene drops is 3 × 10–4 cm/s; the mass transfer coefficient in the aqueous phase is 2.4 × 10–3 cm/s. Jasmone is about 170 times more soluble in benzene than in the suspension. What is the overall mass transfer coefficient ? Sol. : If we write the overall flux as ' ' N = k (C10 – C1i) = k' (C1i – C10 ) ∴

CAL = 1.322 × 10–6

(i.e. flux from the water side = flux into benzene side) We need to solve for the various mass transfer coefficients. In this case, we recognize that the interfacial concentrations are in equilibrium or '

C1i = HC1i – Here the trick, as it were, is to recognize that the equilibrium constant between the two phases is given by that assertion that jasmone is 170 times more soluble in benzene than in the suspension. 1 ' ∴ N = 1 H (HC10 – C10) , k' + k 1 K' = = 1.3 × 10–5 cm/s (Ans.) 1 170 + 3.0 × 10–4 cm/s 2.4 × 10–3 cm/s (38) A part of the manufacture of microelectronic circuits, silicon wafers are partially coated with a 5400. A film of a polymerized organic film called a "photoresist". The density of this polymer is 0.96 g/cm3. After the wafers are etched, this photoresist must be removed. To do so, the wafers are placed in groups of 20 in an inert "boat", which in turn is immersed in strong organic solvent. The solubility of the photoresist is 2.23 × 10–3 g/cm3. If the photoresist dissolves in 10 min, what is its mass transfer coefficient ? Sol. :

This is a relatively simple calculation. Given that the silicon wafers are uniformly coated with 5400 A° of photoresist and the density of the resist of 0.96 g/cm3.

Principles of Mass Transfer Operations − I (Vol. − I)

3.61

Mass Transfer Coefficients

On a unit area basis, there are 0.96 g/cm3 × 5400 × 10–8 cm = 5.184 × 10–5 g/cm2 If this amount of material is removed in 10 minutes, then the mass flux is 5.184 × 10–5 g/cm2 = 8.64 × 10–8 g/cm2 · s 10 × 60 s N = 8.64 × 10–8 g/cm2 · s = k∆C In this case, the ∆C = (solubility – 0) 8.64 × 10–8 g/cm2 · s k = = 3.8 × 10–5 cm/s (Ans.) 2.23 × 10–3 g/cm3 (39) Air flows through a cylindrical tube made of naphthalene at a velocity of 5 m/s. The diameter of the tube is 0.1 m and the temperature of the air is 20°C. (a) Using the correlation proposed by Linton and Sherwood given below, calculate the mass transfer coefficient for the transfer of naphthalene to air. ∴

0.83

1/3

Sh = 0.023 Re Sc kcd ρdv v Sh = D , Re = µ , Sc = D AB AB

Compare this result with the values obtained using the analogies of (b) Reynolds (c) Prandtl (e) Von Kármán (e) Chilton-Colburn Note that the friction factor for this system, Cf can be estimated using the equation Cf –2 2 = (2.236 ln Re – 4.639) Data : µair = 1.8 × 10–5 kg/m.s ρair = 1.2 kg/m3 DAB = 4.24 × 10–6 m2/s Sol. : µair = 1.8 × 10–5 kg/m.s µ = 5 m/s ρair = 1.2 kg/m3 DAB = 4.24 × v µ = Sc = D ρD AB AB

=

10–6

D = 1m m2/s

1.8 × 10–5 = 3.54 (4.24 × 10-6) (1.2)

ρDµ (1.2) (0.1) (5) = = 3.33 × 104 µ 1.8 × 10–5 (a) Linton-Sherwood Correlation : kcD 0.83 1/3 Sc DAB = Sh = 0.023 Re 4.24 × 10–6 kc = (0.023) (3.33 × 104)0.83 (3.54)1/3 0.1 kc = 8.43 × 10–3 m/s (b) Reynolds Analogy : kc f –2 vm = 2 = (2.236 ln Re – 4.639) Re =

kc = (5) [2.236 ln (3.33 × 104) – 4.639]–2 f kc = 1.44 × 10–2 , 2 = 2.876 × 10–3

(Ans.)

(Ans.)

Principles of Mass Transfer Operations − I (Vol. − I) (c)

3.62

Mass Transfer Coefficients

Prandtl Analogy : kc f vm = 2

=

  1 + 5 

1   f  (S – 1) c 2 

kc = 5 (0.002876)

1+5 kc = 8.55 × 10–3 m/s

1 0.002876 (3.54 – 1) (Ans.)

(d) Von-Kármán Analogy : kc vm

(e)

f/2

=

f  5Sc + 1  2 Sc + ln  6  – 1 5 (0.002876) kc = 5 (3.54) + 1  1 + 5 0.002876 3.54 + ln  6    – 1 kc = 7.24 × 10–3 m/s Chilton-Colburn j-factor Analogy : kc 2/3 f = 2 vm S c 1+5

(Ans.)

kc = (0.002876) (5)/(3.54)2/3 = 6.19 × 10–3 m/s (Ans.) (40) The average heat transfer coefficient for natural convection from a single sphere in a large body of fluid is given by hd β∆T β∆  cpµ1/3 d3ρ2gβ∆ = 2 + 0.6 for Gr1/4 Pr1/3 < 200 2 k  µ  k  where, d is the diameter of the sphere and the fluid properties are evaluated at the mean temperature of the sphere and bulk fluid. Using the analogy between mass and heat transfer, calculate the instantaneous rate of sublimation at the surface of a naphthalene sphere in air at 145.5°C and 1 atm. Explain the analogy between Nu = Sh, Sc = Pr, Gr = GrAB. vap

Data : Pnaphthalene = 0.13 atm, DAB = 5.85 × 10–6 m2/s, d = 7.5 × 10–2m ρair = 0.839 kg/m3, µair = 2.3 × 10–5 kg/m.s 1/4 1/3 hmd d2ρ2gβ∆T Cpµ Sol. : = 2.0 + 0.6 2 k  µ   k  Nu = 2.0 + 0.6 Gr1/4 Pr1/3 The equivalent expression for mass transfer is : 1/4

Sh = 2.0 + 0.6 GrAB Sc1/3 or

kc d DAB

d3ρ2g∆x1/4  µ 1/3 = 2.0 + 0.6   µ2  ρDAB

Considering only the properties of air (naphthalene has a very small vapour pressure). ρ = 0.839 kg/m3 µ = 2.3 × 10–5 kg/m.s CAs = 0.13/1 = 0.13 (A = naphthalene) ∆xA = 0.13 – 0 = 0.13

Principles of Mass Transfer Operations − I (Vol. − I)

3.63

Mass Transfer Coefficients

1/4 –2 3 2 (7.5 × 10 ) (0.839) (9.8) (0.13) (2.3 × 10–5)2   –5 2.3 × 10   × = 31.20 (0.839) (5.85 × 10–6)

Sh = 2.0 + 0.6

Sh DAB (31.20) (5.85 × 10–6) = d 75 × 10–2 –3 kc = 2.43 × 10 m/s 0.13 NA = kc ∆C = (2.43 × 10–3) 0.082 × 418.5 NA = 9.21 × 10–6 kmol/m2 · s (Ans.) (41) A "cooling bag", commonly used for storing water in deserts, is made of porous canvas. A small amount of water diffuses through the canvas and evapourates from the surface. The evapouration of the water cools the surface of the bag and a temperature driving force is established. Determine the temperature of the ambient air if the following values are assumed : Surface temperature of the bag 66 °F Pr 0.72 Sc 0.61 Density of air 0.072 lb/ft3 Viscosity of air 0.018 cp Thermal conductivity of air 0.014 Btu/hr.ft2 (°F/ft) Specific heat of air 0.24 Btu/lb.°F v∞ 0.5 mph 1056 Btu/lb Latent heat of vapourization of H2O Vapour pressure of H2O 0.31 psi Partial pressure of H2O in air 8 mmHg R 0.7302 ft3.atm/lbmole/°R kc =

'

Sol. :

NA

' = kc (cs – C∞)

kc Ps P∞  = R T – T  s ∞ 

Q = NA λH2 O = h (T∞ – Ts)

… (1) … (2)

Combining equations (1) and (2) '

kc Ps P∞   R Ts – T∞ λH2 O = h (T∞ – Ts) Rearranging equation (3)

… (3)

'

kc Ps P∞ λH2O   = (T∞ – Ts) h Ts – T∞ R Using Chilton-Colburn analogy

… (4)

'

kc

h Sc2/3 = Pr2/3 jD = U ρC m p Um '

kc h

=

1 Pr2/3 ρCp Sc

… (5)

Principles of Mass Transfer Operations − I (Vol. − I)

3.64

Mass Transfer Coefficients

Substituting equation (5) in equation (4) 1 Pr2/3 Ps P∞ λH2O Ps – P∞ = (T – T ) – = a   Ts T∞ ∞ s ρCp Sc Ts T∞ R   Consequently Ps 2  T∞ – T∞ Ts + a T  + aP∞ = 0 s





… (6)

where, a = =

1 Pr2/3 λH2O R ρCp Sc (0.072

lb/ft3)

2/3 (18 lb/lbmol) 1 0.72 (1056 Btu/lb) (0.24 Btu/lb. °F) 0.61 0.7302 ft3 atm/lbmol. °R

= 1682490.5 °R2/atm and Ps = 0.31 psi/(14.7 psi/atm) = 0.021 atm and P∞ = 8 mmHg/(760 mmHg/atm) = 0.0105 atm Therefore, equation (6) becomes 2

T∞ – 592.3 T∞ + 17666.2 = 0 (Ans.) Solving the quadratic equation for T∞ we get : T∞ = 560.8 °R = 101.1 °F (42) Naphthalene is sublimating from a flat plate into a turbulent air stream flowing past the plate. It is found that the thickness of the naphthalene coating at a point decreases 0.5 mm in one hour. Surface temperature at that point is 78°F. Using data given below, calculate the gas-phase mass transfer coefficient, kG in gmol/cm2 sec mmHg and the thickness of a fictitious laminar film (cm) having a resistance to mass transfer equal to that in the boundary layer. Properties of naphthalene : Vapor pressure at 78°F = 0.0895 mmHg Density (cast material) = 1.08 g/cm3 Molecular weight = 128.16 Diffusivity in air (at 20°C) = 0.0611 cm2/sec. Sol. : We can model the flux as N = k∆C Since we know that the thickness decreased by 0.5 mm/hr = 1.4 × 10–5 cm/s and we know that the bulk density was 1.08 g/cm3. ∴ On a unit area basis (cm2), the flux of naphthalene was = 1.5 × 10–5 g/s = 1.18 × 10–7 mol/cm2 · s

∴

k =

1.18 × 10–7 mol/cm2 · s 1.18 × 10–7 mol/cm2 · s = 1 ∆C RT (∆P)

k =

1.7 × 10–7 mol/cm2 · s 1 cm3 0.0895  8.314 (298 K)  760 101.325 × 1000 L

k = 3.54 × 10–5 cm/s

Principles of Mass Transfer Operations − I (Vol. − I)

3.65

Mass Transfer Coefficients

To calculate the fictitious film thickness, we can use the relationship k =

D 0.0611 cm2/s D ⇒ δ= k = = 1731 cm 3.54 × 10–5 cm/s cm/s δ

Note : This is based on film theory.

(Ans.)

(43) A particular implant has been designed to release a drug at constant rate of 0.5 µg/cm2-s. The drug has a solubility in blood of 10 g/l and a diffusivity of 10–6 cm2/s. The effective therapeutic concentration is one-tenth of the solubility. What is the effective film thickness in the blood ? Sol. : We can view this as follows : N A

D∆C = z fm

where, zfm = approximate film thickness, and N A = flux per unit area cm2 g g  10–6 s 10–2 – 10–3 cm3 cm3 D∆C  zfm = N = g 0.5 × 10–6 A cm2 · s

∴

zfm = 0.18 mm

∴

(Ans.)

(44) A fluid is flowing in a vertical pipe and mass transfer is occuring from the pipe wall to '

the fluid. Relate the convective mass transfer coefficient kc to the variables d, L, µ, v, DAB, G, ∆ρ, ∆ρ where, D is the pipe diameter, L is pipe length, and ∆ρ is the density difference. L M M L L2 L2 M ' Sol. : kc =  t  , D = [L], ρ=  3 , µ = Lt , v =  t  , DAB =  t  , g =  t  , ∆ρ =  3           L  L  We have 8 variables and 3 parameters. Therefore, we need to come up with 5 equations. So using the Buckingham π approach, and selecting D, ρ, µ as the variables common to all sets. '

π1 : Da ρb µc kc M b M c L [L]a  3 Lt  t  L      L : a – 3b – c + 1 = 0; M : b + c = 0; t : – c – 1 = 0 c = – 1; b = 1; a = 1 Dρ ' π1 : µ kc π2 : Da ρb µc DAB M b M c L2 [L]a  3 Lt  t  L     

Principles of Mass Transfer Operations − I (Vol. − I)

3.66

Mass Transfer Coefficients

L : a – 3b – c + 2 = 0; M : b + c = 0; t : – c – 1 = 0 c = – 1; b = 1, a = 0 ρDAB π2 : µ π3 : Da ρb µc v M b M c L [L]a  3 Lt  t  L      Dρ π3 : µ v π4 : Da ρb µc ∆ρ M b M c M [L]a  3 Lt  3 L    L  L : a – 3b – c – 3 = 0; M : b + c + 1 = 0; t : – c = 0 c = 0, b = – 1, a = 0 ∆ρ π4 : ρ π5 : Da ρb µc g M b M c L2 [L]a  3 Lt  t  L      L : a – 3b – c + 2 = 0; M : b + c = 0; t : – c – 1 = 0 c = – 1, b = 1, a = 0 ρg π5 : µ ρDAB Dρ ' Dρv ρg ∆ρ ; π5 : µ π1 : µ kc ; π2 : µ ; π3 : µ ; π4 : ρ '

Dkc π1 Dρρv ∆ρ ρg = Sh = π2 DAB = f  µ ‚ ρ ‚ µ 

(Ans.)

(45) Mercury at 26.5 °C is flowing through a packed bed of lead spheres having a diameter of 2.096 mm with a void fraction of 0.499. The superficial velocity is 0.02198 m/s. The solubility of lead in mercury is 1.721 wt%, the Schmidt number is 124.1, the viscosity of the solution is 1.577 × 10–3 Pa·s, and the density is 13530 kg/m3. (a)

Predict the value of jD. Compare with the experimentally determined value of jD = 0.076.

(b) Predict the value of mass transfer coefficient of lead in mercury. Sol. : ε = 0.499, ' = 0.02198 m/e (superficial velocity). NDC = 124.1, T = 26.5°C, ρ = 13530 kg/m3, µ = 1.577 × 10–3 Pa·s. Dp = 2.096 mm = 2.096/1000 = 0.002096 m.

Principles of Mass Transfer Operations − I (Vol. − I)

3.67

Mass Transfer Coefficients

Fig. 3.16 : Packed bed and lead spheres

NR e =

Dk É' ρ (0.002096) (0.02198) (13530) = = 395.3 µ 1.577 × 10–3 JD =

– 0.31 0.250 0.250 NRe = 0.499 (395.3) – 0.31 ε

JD = 0.07843

(a) Let A = B,

∴

MA (Pb) = 207.19

B = Mercury, ∴ MB (Hg) = 200.59

XAi = 0.01721 wt. fraction (Solubility of Pb(A) in Hg (B)

Fig. 3.17

Assume 100 kg saturated solution : 1.721 kg A and 98.279 kg B

XA1 (mol frac.)

1.721 207.19 = 1.721 98.279 = 0.01667 + 207.19 200.59

XB1

= 1.000 – 0.01667 = 0.98333

XB2

= 1.000 – 0 = 1.000

XB2

≅

XB1 + XB2 2

=

0.9833 + 1.000 = 0.9916 2 '

We have correlation,

' kc C

kc

= kc XBM C or kc = X

BM

Also equation for JD is given by, ' kc

2/3

JD = É' NDC

Principles of Mass Transfer Operations − I (Vol. − I)

3.68

Mass Transfer Coefficients

'

kc

0.07843 = 0.02198 (124.1) 2/3 '

kc

= 6.929 × 10–5 m/s

(Ans.)

' kc

6.929 × 10–5 kc = X = kc = 6.986 × 10–5 m/s (Ans.) 0.9916 BM (46) Ether and water are contacted in the Lewis cell shown in the figure. An iodine-like solute is originally present in both phases at 3.10–3 M. However, it is 700 times more soluble in ether. Diffusion coefficients in both phases are around 10–5 cm2/s. Resistance to mass transfer in the ether is across a 10–2 – centimeter thin film; resistance to mass transfer in the water involves a surface renewal time of 10 seconds. What is the solute concentration in the ether after 20 minutes ? (b)

3

20 cm ether Surface area of 12.2 cm

2

3

100 cm H2O Fig. 3.18 : Lewis cell in Ex. 44

Sol. : • • • • •

Given information : Initial iodine-like solute concentrations in both phases : 3 × 10–3 M. Solute is 700 times more soluble in ether. Diffusion coefficients in both phases : 1 × 10–5 cm2/s. Resistance to mass transfer in the ether : across 10–2 cm film. Resistance to mass transfer in the water : a surface renewal time of 10 s. From the given information, the mass transfer coefficients in both phases can be calculated. In the ether phase, film theory can be applied. Therefore, D 10–5 kc = 1 = = 10–3 cm/s … (1) 10–2 In the water phase, surface renewal theory can be applied. Therefore, kw =

D/τ

Ether phase

=

10–5/10 = 10–3 cm/s c1i'

Water phase

Solute flux

c1'

c1i

Fig. 3.19

… (2)

c1

Principles of Mass Transfer Operations − I (Vol. − I)

3.69

Mass Transfer Coefficients

For convenience, let us designate all concentrations in the ether phase with a prime and all those in the water without a prime. Then, the flux of solute is, '

'

N1 = ke (c1i – c1 ) = kw (c1 – c1i)

… (3)

The interfacial concentrations are in equilibrium. Therefore, '

c 1i

= Hc1i

… (4)

Eliminating these interfacial concentrations, we get, N1 =

1   (Hc1 – c' ) 1 + H/k 1/k e w  

… (5)

Therefore, the overall coefficient based on a driving force in ether is 1 Ke = 1/k + H/k  e w 

=

1 = 1.43 × 10–6 cm/s 1/10–3 + 700/10–3

… (6)

From the mass balance in the ether phase, '

dc1 Ve dt

'

= Ke A (Hc1 – c1 )

… (7)

From the mass balance of the solute, '

'

Vtc = Vec1 + Vwc1 ⇒ 120 × 3 × 10–3 = 20c1 + 100c1 '

c1 = 3.6 × 10–3 – 0.2 c1

… (8)

If we plug equation (8) into equation (7), then we obtain, '

dc1 '

2.52 – 141c1

=

KeA Ve dt

… (9)

If we integrate equation (9) from t = 0 to t, then '

 2.52 – 141 c1  1   ln – 141 2.52 – 141 c'   10

=

KeAt Ve

… (10)

'

 2.52 – 141 c1  –6 1   = 1.43 ∞ 10 ∞ 12.2 ∞ 20 ∞ 60 ln –3 20 – 141 2.52 – 141 × 3 × 10  ⇒

'

c1

= 5.0 × 10–3 M

(Ans.)

(47) A thin metal foil is put in a wind tunnel in order to measure the drag force exerted by an air current flowing parallel to the foil's surfaces. For a speed of 40 m/s, 1 bar pressure and 25°C, the drag force measured was 0.134 N. The metal foil is now replaced by a naphthalene plate of exactly the same dimensions. (a)

Calculate the diffusion coefficient of naphthalene in air, the Schmidt Number, Sc and the naphthalene molar concentration at the plate's surface for the above conditions. The diffusion coefficient of naphthalene in air at 25°C and 3 bar is 2.064 × 10–6 m2/s, while its saturation pressure is 0.25 Torr. Properties of air at 25°C and 1 bar : kinematic viscosity v = 1.57 · 10–5 m2/s, and density ρ = 1.17 kg/m3.

Principles of Mass Transfer Operations − I (Vol. − I)

3.70

Mass Transfer Coefficients

(b) What is (assuming steady state conditions) the maximum rate of sublimation of naphthalene (in mol/s), at the same air current conditions, using the correlation : f 2

=

k (Sc) 2/3 u0

… (1)

where, f is the friction factor, Sc, the Schmidt Number, K, the average mass transfer coefficient and u0 the air current velocity ? Hint : Remember that the shear stress τ0 on a plate is the force exerted parallel to the plate divided by its surface area. Air current Metal foil/ Naphthalene plate Drag force measuring device

Fig. 3.20 : A thin metal foil is put in a wind tunnel to measure drag force

Sol. : (a) For 1 bar and 25°C :

Sc =

P0 D = D0 P

3 = 2.064 × 10–6 1 m2/s = 6.19 × 10–6 m2/s

µ v = D ρ·D

≈ 2.54

The concentration on the plate's surface is that at saturation : P1i c1i = RT ≈ 1.345 × 10–2 mol/m3 (b) For the maximum sublimation rate : F1 N N1 = k (c1i – c∞i) ≈ k c1i ⇒ k = c = A · c 1i 1i

… (2)

where, A is the total surface of the plate. Using the definition of the friction factor f and equations (1) and (2) : Fdrag F1 τ0 f 1 Sc2/3 = 2 = = A · ⇒ A · c 1i · u0 ρ · (u0)2 ρ · (u0)2

F1 =

Fdrag · c1i Sc2/3 · ρ · u0

P1i  Fdrag · RT   = ≈ 2.07 × 10–5 mol/s Sc2/3 · ρ · u0

(Ans.)

(48) A single potassium chloride (KCl) crystal about 0.063 cm in diameter, which is immersed in a 5.2% supersaturated solution containing about 25 wt% potassium chloride, is growing at a rate of 0.0013 cm/min. If the system is well mixed, this growth is second order, presumably because both potassium and chloride ions are involved. In our case, the solution may not be well mixed; it flows past the crystal at 6 cm/s. The solution's viscosity is about 1.05 · 10–3 Pa · s; its density is 1.2 g/cm3, and the crystal's density is 1.984 g/cm3. Does diffusion influence the rate of crystal growth ? (Additional information : Diffusion coefficient of KCl in water at the given conditions : 1.994 · 10–5 cm2/s)

Principles of Mass Transfer Operations − I (Vol. − I)

3.71

Mass Transfer Coefficients

Sol. : To check if the diffusion influences the rate of the crystal growth you have to calculate the mass transfer coefficient where growing effects take place and compare with the mass transfer coefficient without growing effects. By comparing these two values you can say if the crystal growth is the limiting step or not. The mass transfer coefficient without growing effects can be calculated using equation for forced convection around a solid sphere is given by : 1

1

0 kd dv 2  v 3 = 2.0 + 0.6 … (1) D  v  D where the diffusion coefficient D is 1.994 · 10–5 cm2/s, the relative velocity of the sphere v0 is 6 cm/s, the solution viscosity v is (1.05 · 10–3 Pa · s) (1.2 g/cm3) = 8.75 m2/s, and the diameter of the crystal d is 0.063 cm. The mass transfer coefficient k without growing effects becomes therefore : 1

1

 dv0 2 v 3 D k = 2.0 + 0.6  v  D  · d      m

 6.3 · 10 m · 0.06 s   8.75 · 10 2.0 + 0.6   8.75 · 10 m  1.994 · 10 s    –4

=

2

–7

1 2

m2 s 2 m –9 s

–7

m2 s –4 m

  1.994 · 10   · 6.3 · 10  1 3

–9

… (2) m cm = 1.01 · 10–4 s = 0.0101 s The mass transfer coefficient K including growth effects can be calculated using the information of the growing rate and the concentration of the solution. j = K (c1 – c1, sat) … (3) The flux j can be calculated with the information of the growing rate of 0.0013 cm/min, and the density of the crystal of 1.984 g/cm3 1 dm ρ dV (π/2) · d2 · dd ρ ρ dd j = A · dt = A · dt = π · d2 · = 2 · dt … (4) dt g cm 1 min g j = 0.5 · 1.984 · 0.0013 min · 60s · = 2.15 · 10–5 cm3 cm2 · s This flux has to be equal to : g g 1.052 · 0.25 1 · 0.25 – 1.052 · 1.2 = K · 1.48 · 10–2 … (5) K (c1 – c1, sat) = K  1.052 3 cm cm3   Combining equation (4) and (5), the mass transfer coefficient K including growth effects becomes : g 2.15 · 10–5 cm2 · s cm K = = 0.00145 s … (6) g 1.48 · 10–2 3 cm Now, we see that when comparing (equation 2) and (equation 6), that the mass transfer which includes the growing effect is only : K … (7) k = 0.143 14.3% of the mass transfer coefficient without growing effects, therefore the crystal growth is the limiting step ! (Ans.)

Principles of Mass Transfer Operations − I (Vol. − I)

3.72

Mass Transfer Coefficients

(49) The mass transfer coefficient in gas-solid fludized beds has been correlated with the dimensionless expression – –0.44  dGM   = 1.77 · RT µ (1 – ε)

kp  µ 2/3 G ρD

where k is the mass transfer coefficient (cm/s), ρ is the mean pressure of the fluidized – gas (atm), G is the total convective flux, d is the particle diameter, M is the molecular weight of the gas, ε is the bed porosity, and R is the gas constant (82.05 cm3 · atm/ (mol · K). You are studying 0.1 centimeter particles of coal burning in a bed with ε = 0.42 fluidized by air at 1250°C. The air flux at 1.7 atm is 380 lb/(ft2 · h). The produced CO2 is caried away by an excess of air. What is the film or unstirred layer thickness for the CO2 diffusion around the coal particles ? Additionally required data : Air viscosity at the given conditions : 550 · 10–6 g/(cm · s); molecular weight of air : 29 g/mol, of CO2 : 44 g/mol, D = 1.467 cm2/s Sol. : If we assume that the surface layer of air around the carbon particles is an unstirred film, we know from the film theory, that the film thickness correlates diffusivity and mass transfer coefficient : D l = k

… (1)

For the calculation of the mass transfer coefficient, we still need the convective flux G in mol/(cm2 · s) (applying the given conversion) and the density of the air (assumed as an ideal gas). The total convective flux is G = 380 = 380

lb ft2 · hr g 1 ft2 1 hr 1 mol lb · 453.59 lb · · 3600 s · 29 g 2 · hr (30.48 cm)

ft2

= 1.777 · 10–3

… (2)

mol cm2s

The density of the gas is : – – Mp ρ = M c = RT

g 29 mol · 1.7 atm

g = 3.95 · 10–4 cm3 cm3atm 82.05 mol K · 1523 K So finally the mass transfer coefficient becomes, – –0.44 dGM G ρD 2/3 air   k = 1.77 · · Rt · p ·  µ    µ (1 – ε) Putting all known parameters, we get, cm = 70.32 s =

… (3)

… (4)

Principles of Mass Transfer Operations − I (Vol. − I)

3.73

Mass Transfer Coefficients

Eventually we get for the film thickness : 1.467 D l = k = 70.32 cm = 0.02 cm

(Ans.)

(50) Find the dissolution rate of a cholesterol gallstone 1 cm in diameter immersed in a solution of bile salts. The solubility of cholesterol in this solution is about 3.5 · 10–3 g/cm3; the kinematic viscosity of this soution is about 0.06 cm2/sec; the diffusion coefficient of cholesterol is 1.8 · 10–6 cm2/sec. Sol. : A schematic of the problem can be seen in Fig. 3.21. Solute flux away

Dr

r

Fig. 3.21 : Steady dissolution of a cholesterol gasstone

The dissolution rate of the gall stone can be defined like : · dm M = dt

= N1 A = kA (csat – c∞)

… (1)

The surface of the sphere is A = πd2 and the concentration far away is c∞ = 0 whereas at the surface of the sphere, the concentration is the saturation concentration, which is the solubility of cholesterol in bile solution and is given. Because of the density difference between saturated bile solution and pure bile solution, we get a free convection around the sphere. For this case, Cussler gives us the mass transfer correlation : kd D

d3 ∆ρg1/4  v 1/3 = 2.0 + 0.6   ρv2  D

… (2)

We can solve this for the mass transfer coefficient k and obtain : D d3 ∆ρg1/4  v 1/3 D k = 2.0 d + 0.6   ρv2  D  d 

… (2a)

For the density of bile solution, we can take the density of water : g ρ = 1 cm3 All other quantities are given in the text or are well-known constants. So we get for the mass transfer coefficient : cm2 1.8 · 10– 6 s 1/4 1/3 2.0 + 0.6 13 · 0.003 · 981  0.06   = 1.89 · 10–4 cm k= 2 –6 1 cm s   1 · 0.06  1.8 · 10  

Principles of Mass Transfer Operations − I (Vol. − I)

3.74

Mass Transfer Coefficients

… (3) Finally, we can calculate the dissolution rate : S = k πd2 csat cm g s · 86400 day … (4) = 1.89 · 10–4 s · π · 1 cm2 · 3.5 · 10–3 3 cm g = 0.18 day How would the solution look like if we had only diffusive flux, but no free convection ? A mass balance on a shell around the sphere has the following form : Solute accumulation = Diffusion into – Diffusion out  within the shell   the shell   of the shell  or in mathematical terms : ∂ 2 2 2 … (5) ∂t (4πr ∆rc) = (4πr j)r – (4πr j)r + ∆r The left-hand side of the above equation is equal to zero, because we can treat the diffusion as in steady-state. We divide both sides of this equation by the spherical shell volume and take the limit ∆r → 0, then we find, 1 d 0 = 2 dr (r2 j) r Combine this with the Fick's law, D d dc 0 = 2 dr r2 dr … (6) r   This differential equation has two boundary conditions d r = R0 = 2 , c = csat r=∞ , c=0 After two integrations and using the boundary condition we get the concentration profile R0 c = csat r … (7) The diffusive dissolution flux j can be found from Fick's law DR0 dc j = – D dr = csat … (8) r2 which, at the sphere's surface, is D 2D j = R csat = d csat 0 The mass flow by diffusion alone then would be : dm' 2D 2 dt = jA = d πd csat

… (9)

… (10)

By comparison of (1) and (10) we see, that 2D k' = d which is the first term of the right hand side of equation (2a), is the mass transfer coefficient for pure diffusion from a solid sphere. The second term in equation (2a) is accounting for the effect of free convection around the sphere. (Ans.)

Principles of Mass Transfer Operations − I (Vol. − I)

3.75

Mass Transfer Coefficients

(51) Consider a laboratory bio-reactor in which micro-organisms need oxygen (at a rate of 0.9 g/h) to work property. This oxygen is continuously supplied by feeding air (20% O2, 80% N2, at a pressure of 1 atm) through a sparger at the bottom of the reactor. The reactor is stirred with a power input of 450 W/m3. By this, air bubbles are generated with an average diameter of 1.2 mm that have an average residence time in the reactor of 25 seconds. Data (at the operating conditions) : Density of the liquid :

1 g/cm3

Kinematic viscosity :

0.1 cm2/s

Diffusion coefficient of O2 :

2.1 · 10–5 cm2/s

Partition coefficient of oxygen in the liquid :

4 · 10–5 g/(atm · cm3)

(a)

Use the appropriate correlation to calculate the mass transfer coefficient in that process.

(b) Calculate the volume rate of air that is needed to keep the oxygen concentration in the liquid at 40% of its saturation concentration on the bubble/liquid interface. It can be assumed that bubble diameter and partial pressure of O2 in the bubbles are constant. Sol. :

Air

Fig. 3.22 : Laboratory bio-reactor

(a) For the stirred tank with gas bubbles, the mass transfer correlation is : 1/3 4 1/4 kd (P/V) d   v  = 0.13 D  ρv3  D This can be solved for kstirred. (P/V)1/4 D2/3 v1/3 kstirred = 0.13   ρv3  ≈ 0.0021 cm/s

(b) Mass balance :  Mass of oxygen  = consumed per time

Mass transfer rate  of oxygen 

Principles of Mass Transfer Operations − I (Vol. − I) or

· m O2

3.76

Mass Transfer Coefficients

= k A (cO2‚ sat – cO2 )

The saturation concentration is cO 2 = H pO2 = H yO2p ‚ sat

The wanted oxygen concentration is cO2 = 0.4 cO2‚ sat So the required exchange area becomes : · mO2 A = 0.6 kc O 2‚

sat

This is also the total surface area of all the bubbles in the liquid : A = N · πd2 Therefore, the number of bubbles in the tank must be : · mO2 1 N = πd2 0.6 k cO2‚ sat From this, the needed air volume rate can be calculated : Vbubble N · V air = τ · mO2 πd3 1 1 = 6 · πd2 0.6 k cO2‚ sat τ · mO2 1 d = 3.6 τ k HyO2 p 0.9/3600 1 0.12 = 3.6 25 0.0021 · 4 · 10–5 · 0.2 · 1 cm3 ≈ 19.8 s

(Ans.)

(52) Water containing 0.1-M benzoic acid flows at 0.1 cm/s through a 1 cm diameter rigid tube of cellulose acetate, the walls of which are permeable to small electrolytes. These walls are 0.01 cm thick; solutes within the walls diffuse as through water. The tube is immersed in a large well-stirred water bath. Under these conditions, the flux of benzoic acid from the bulk to walls can be described by, 1 2 kd d v3 = 1.62 D DL  where, L is the length of the tube and v is the average velocity of the tube. (a) After 50 cm of tube, what fraction of a 0.1-M benzoic acid solution has been removed ?

Remember that there is more than one resistance to mass transfer in this system. (b) How much does the previous answer change if the benzoic acid solution in the tube is in benzene, not water ? Sol. : A schematic of the problem can be seen in Fig. 3.23.

Principles of Mass Transfer Operations − I (Vol. − I)

3.77

Mass Transfer Coefficients Water containing Benzoic acid out

Larger, well stirred water bath Water containing Benzoic acid in

Fig. 3.23 : Schematic of the problem (50)

(a) The diffusion coefficient of benzoic acid at infinite dilution in water at 25°C is D = 1.00 · 10–5 cm2/s. Under the given conditions, the flux of benzoic acid from the bulk to the walls can be described using the given equation in the text. Using this equation it is possible to calculate the mass transfer coefficient k : 1 d 2v 3

k = 1.62  DL 





cm

 (1 cm) 0.1 s  D · d = 1.62   cm 1 · 10 s · 50 cm 2

2

–5

1 3

cm2 1 · 10–5 s cm · 1 cm = 9.47 · 10–5 s

… (1)

The total mass transfer coefficient kT is put together by the mass transfer coefficient resulted form the flux of benzoic acid from the bulk to the walls and from the mass transfer coefficient resulted from diffusion : 1 1 cm kT = 1 = 8.65 · 10–5 s … (2) 1 1 = 0.01 cm + + 2 k cm D cm 9.47 × 10–5 s 1.00 · 10–5 s To calculate the concentration profile of benzoic acid along the tube, a mass balance over a section of the pipe has to be made. πd2 … (3) 4 v∆c = – kT (πd ∆z) (c – csat) Equation (3) can be written as, 4kT dc – dz = dv (c – cext)

… (4)

Considering a infinite small volume. The boundary conditions are : c = c0 at z = 0 cext = 0 for all z (large well-stirred water bath) Solving equation (4) applying the boundary conditions : 4kT c ln c  = – dv L  0

… (5)

Principles of Mass Transfer Operations − I (Vol. − I)

3.78

Mass Transfer Coefficients

The concentration at the end of the pipe is therefore, 4kT c = c0 exp – dv L  

–5

=

cm

 4 · 8.65 · 10 s 0.1 M exp – cm 1 cm · 0.1 s 

  

· 50 cm

= 0.0841 M

… (6)

Therefore, 15.9% of the benzoic acid has been removed. (b) The diffusion coefficient of benzoic acid in benzene is D = 1.38 · 10–5 cm2/s. Under the given conditions, the flux of benzoic acid from the bulk to the walls can be described using the given equation in the text. Using the equation it is possible to calculate the mass transfer coefficient k : 1 cm cm2 1 2 0.1 –5 3 (1 cm) 1.38 · 10 s s d 3v 3 D cm · = 1.17 · 10–4 s k = 1.62  DL  · d = 1.62 2 1 cm cm   1.38 · 10–5 s · 50 cm … (7) The total mass transfer coefficient kT is put together by the mass transfer coefficient resulted form the flux benzoic acid from the bulk to the walls and from the mass transfer coefficient resulted from diffusion from benzoic acid through the walls into water : 1 1 cm = 1.05 · 10–4 s … (8) kT = 1 1 = 1 0.01 cm + + k D cm cm2 1.17 · 10–4 s 1.00 · 10–5 s The concentration at the end of the pipe is therefore : cm 4 · 1.05 · 10–4 s 4kT   c = c0 exp – dv L … (9) cm 50 cm = 0.0811 M   = 0.1 M exp – 1 cm · 0.1 s Therefore, 18.9 % of the benzoic acid has been removed, that means that 3% more is removed when the benzoic acid is solved in benzene. (Ans.)

  

  

  

  

EXERCISE FOR PRACTICE (1) Estimate the overall liquid side mass transfer coefficient at 250C for oxygen from water into air. In this estimate, assume that each individual mass transfer coefficient is : D k = 0.01 cm mol Ans. : k = 2.1 × 10–3cm/sec; kp = 9.4 × 10–4 gm. ; K = 2.1 × 10–3 cm/sec 2 L cm sec. atm L   (2) We are studying gas absorption into water at 2.2 atm. total pressure in a packed bed tower containing Berl saddles. From previous experiments with ammonia and methane, we assume that for both gases, the mass transfer coefficient times the packing area per kg mole tower volume is 18 hr.m3 For the gas side and 530 kg/mole hr. m3 for the liquid side. Henry's law constants for ammonia is 9.6 atm and 41,000 atm for methane. Estimate the overall gas-side mass transfer coefficient for each gas.

(Ans. : kyaammonia = 16 kg mole/hr. m3‚ kyamethane = 0.03 kg mole/hr.m3)

Principles of Mass Transfer Operations − I (Vol. − I)

3.79

Mass Transfer Coefficients

(3) Estimate the average mass transfer coefficient for water evapourating from a film falling at 0.82 cm/sec into air. The air is at 25oC and 2 atm, and the film is 186 cm long. Express your results in cm3 of H2O vapour at STP/(hr. cm2. atm)

(Ans. : 130 in units given) (4) Calculate the convective mass transfer coefficient in a wetted wall column for the transfer of benzene into carbon-dioxide at 0°C. The gas velocity through the column is 1.25 m/s. The diameter of the column is 0.16 m. Assume that the benzene concentration in Carbon dioxide is very low.

(Ans : kC = 4.3 × 10–3 m/s) (5) Determine the film thickness of oil (density = 980 kg/m3) at a distance of 100 mm downstream of the leading edge of a flat surface over which the oil is flowing at a rate of 0.4 m/s. Ans. : Film thickness = 16.57 mm   Displacement thickness = 0.375 m   Boundary wall thickness = δ = 6.21 mm (6) It is desired to estimate the mass transfer coefficient kG in kg mol/s m2 for water vapour 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 vapour concentration in the air is small, so the physical properties of air can be used. Water vapour is being transferred to the solids. Do this for the following geometries. (a) A single 25.4 mm diameter sphere (b) A packed bed of 25.4 mm spheres with ∈ = 0.35 Ans. : (a) k = 1.98 × 10–8 kg mole  G s.m2   (7) Calculate the maximum possible rate of O2 uptake 37oC of micro-organisms having 2 diameter of 3 µm suspended in an agitated aqueous solution. It is assumed that the surrounding liquid is saturated with O2 from air at 1 atm. pressure. It will be assumed that the microorganism can utilize the oxygen much faster than it can diffuse to it. The micro-organism has density very close to that of water. Physical data : g mol O2 (1) The solubility of O2 from air in water at 37oC = 2.26 × 10–7 cm3 liquid (2)

The diffusivity of O2 in water at 37oC = 3.25 × 10–9 m2/sec.

(3)

Viscosity of water at 37oC = 6.94 × 10–4 kg/ms

(4)

Density of water = 994 kg/m3

(5)

Density of Air = 1.13 kg/m3

(Hint : Since O2 is consumed faster than, it is supplied, the concentration CA2 at the surface is zero. The concentration CA1 in the solution is at saturation)

Ans : k = 9.75 × 10–3 m/s‚ N = 2.20 × 10–6 kg mole O2   C A s. m2  

Principles of Mass Transfer Operations − I (Vol. − I)

3.80

Mass Transfer Coefficients

(8) A total of 5 gm of wet micro-organisms having a density of 1100 kg/m3 and a diameter of 0.667 µm are added to 0.1 liter of aqueous solution at 37oC in a shaker flask for a fermentation. Air can enter through a porous stopper. Use physical property data from example (8) (a) Calculate the maximum rate possible of mass transfer of O2 in kg mol/O2. sec. to the surface of micro-organisms assuming that the solution is saturated with air at 101.32 kpa. (b) If the actual utilization of O2 by the micro-organisms is 6.30 × 10–6 kg mol. O2 sec. What would be the actual concentration of O2 in the solution as percent saturation during the fermentation ? –3 –5 Ans : (a) kL = 9.82 × 10 m/s‚ NAA = 9.07 × 10 kg. mol O2/sec  (b) 6.95 % saturation    (9) A stream of air at 100 kPa pressure and 300 K flowing on the top of surface of a thin flat sheet of solid naphthalene of length 0.2 m with a velocity of 20 m/s. The order date are : (i) Mass diffusivity of naphthalene vapour in air = 6 × 10–6 m2/sec. m2 (ii) Kinematics viscosity of air = 1.5 × 10–5 s (iii) Concentration of napthalene of the air solid napthalene interface = 1 × 10–5 k mole/m3 Calculate : (a) The average mass transfer coefficient over the plate. (b) The rate of loss of naphthalene from the surface per unit width.

Ans. : (a) k– = 5.597 ∞ 10–6 k mole/m2.s kpa  G   (b) Rate of loss of naphthalene = 1.1194 × 10–11 k mole/m (10) Air at 25°C and atmospheric pressure passes up a duct at 2.48 m/s. Estimate the mass transfer coefficient kG of water vapour to this air at from a number of different solid objects placed in the duct. As the water vapour concentration in the air is small, use the physical properties of pure air. (a)

A single sphere of 19.05 mm diameter

(b)

A bed of such spheres with a voidage of 50%. -8

(Ans: (a) mass transfer coefficient kG = 1.90 × 10 m/s; (b) mass transfer coefficient (kG = 4.91 × 10

–8

m/s)

(11) A mixture of air and carbon dioxide flows in a wetted-wall tower past a thin film of water flowing down a vertical plate. The CO2 is absorbed into the water from the air at 20°C and 1.20 atm. (abs.), and a value for k'C of 7.289 × 10–4 m/s has been predicted. At a given –5

point the mole fraction of CO2 in the liquid at the liquid-gas interface is 1.8 × 10 and the partial pressure in the bulk of the air phase is 0.03 atm. Calculate the absorption rate of 4

CO2. Henry's Law coefficient of CO2 in air as 0.142 × 10 atmospheres per mole fraction. -7

2

(Ans : Rate of absorption of CO2 = –1.34 × 10 k mole/m .s; i.e. flow is from gas bulk to the liquid).

Principles of Mass Transfer Operations − I (Vol. − I)

3.81

Mass Transfer Coefficients

(12) Large volumes of pure water are made to flow past a piece of solid benzoic acid. The water is at 26.1 oC, at which the solubility of benzoic acid is 0.02948 k mole/m3. Calculate (i) the flux of benzoic acid perpendicular to the acid block, making any approximations to the log mean water concentration if necessary and (ii) the exit concentration (if applicable) for : (a) Water at 0.127 m/s parallel to a flat acid plate of length (parallel to flow) 0.254 m. (b)

Water at 0.04572 m/s through a circular acid tube of length (parallel to flow) 0.9144 m and diameter 9.525 mm. (Ans : (a) NA = 2.45 × 10

–7

2

k mole/m s; (b) Exit concentration= CA = 8.050 × 10 3

k mole/m ; NA = 9.585 × 10

−8

–4

2

k mole/m s)

(13) A micro-organism of diameter 1.08 mm floats suspended in water of similar density. If the surrounding water is saturated with oxygen at 37 oC (oxygen concentration –4

3

= 2.26 × 10 k mole/m ) and the micro-organism absorbs the O2 so quickly on its arrival at the surface that the surface concentration is zero, calculate : (a)

The diffusivity of oxygen in the water;

(b)

The molar flux of oxygen absorption. (Ans : (a) DAB = 3.27 × 10

–9

2

m /s; (b) NA = 6.84 × 10

–7

2

k mole/m .s)

(14) Water, at 25 oC, flows through a 6 ft long 2 in I.D. pipe at a velocity of 1.5 m/s. The pipe is coated on the inside with a thin layer of benzoic acid which slowly dissolved in the water. The solubility of benzoic acid is 0.029 g mole/l. The diffusion coefficient of benzoic acid is 1.06 × 10–4 cm2/sec. (a)

If the concentration of benzoic acid in the exit stream is 1 × 10–4 g mole/l what is the average mass transfer coefficient, kc, cm/sec ?

(b)

What is mass transfer coefficient in units of g mole/[cm2.sec. (mole fract)] ?

(c)

Evaluate the Sherwood and Stanton numbers (using kc from part (a)).

(15) Water at 20 oC flows through a 5 cm ID pipe at an average velocity of 3 m/s. A 60 cm section of the pipe is replaced with a tube made of solid sodium chloride. Compare the mass transfer coefficient for the dissolution of sodium chloride in water using the mass transfer forms of the : (a) Prandtl analogy; (b) Von Karman analogy and (c) Chilton-Colburn analogy. (16) In studying the sublimation of naphthalene into an air stream, an investigator constructed a 3 m long annular duct. The inner pipe was made from 2.5 cm OD solid naphthalene rod; this was surrounded by a 5 cm ID naphthalene pipe. Air at 20 oC and an average pressure of 1 atm flowed through the annular space at a bulk velocity of 15 m/s. At 20 oC, naphthalene has a vapour pressure of 0.039 mm Hg and a diffusivity in air of 0.05 cm2/s. Determine the concentration of naphthalene in the air stream existing from the annular duct. (17) It is known that a thin microlayer of organic matter exists on the surface of the water in the petrochemical complex near Mumbai. Oxygen which is transported from the

Principles of Mass Transfer Operations − I (Vol. − I)

3.82

Mass Transfer Coefficients

atmosphere to the water below must diffuse through the organic microlayer of thickness of 10–4 cm. You are asked to develop appropriate expressions for the overall water-side mass transfer coefficient and the overall air-side mass transfer coefficients for the transport of oxygen from the atmosphere to the water phase through the surface microlayer. You may wish to use the nomenclature below but note that you may not need all of the listed parameters. k

=

air-side mas transfer coefficient (above the microlayer)

kw

=

water-side mass transfer coefficient (below the microlayer)

kow

=

overall water-side mass transfer coefficient

koa

=

overall air-side mass transfer coefficient

kmw

=

mass transfer coefficient in the microlayer for microlayer-water mass transfer

kma

=

mass transfer coefficient in the microlayer for microlayer-air mass transfer

Haw =

dimensionless air-water equilibrium partition coefficient (Haw = Ca/Cw)

Ham =

dimensionless air-microlayer partition coefficient (Ham = Ca/Cm)

Hwm =

dimensionless water-microlayer partition coefficient (Hwm = Cw/Cm)

h

thickness of microlayer.

=

(18) The mass transfer coefficient for mass transfer from (or to) a flat plate is given by the following equations : Laminar conditions (Rez < 3 × 105) : kcz/DAB = 0.332 Rez1/2 Sc1/3 Turbulent conditions (Rez < 3 × 105) : kcz/DAB = 0.0292 Rez4/5 Sc1/3 where Rez = Vzρ/µ. V is the wind velocity and z is the distance along the plate (z = 0 is the position where the air stream first encounters the plate. Note : the above correlation is not applicable when z approaches zero, however, it can be used to average kc over the plate from say z = 0 to z = L, where L is the length of the plate). A 6 m/s wind flows parallel to a pan containing water. The mass diffusivity of water in air at the temperature and pressure of the system is 0.25 cm2/s. Determine : (a)

The value of the mass transfer coefficient at a distance of 1.2 m from the leading edge of the pan.

(b)

The average mass transfer coefficient for the surface between z = 0.25 m and z = 0.45 m.

(c)

The average value of the mass transfer coefficient for the entire pan if the pan length is 1.5 m.

(19) Chlorine water for pulp bleaching is being prepared by absorbing chlorine gas in water within a packed tower operating at 293 K and 1.013 × 105 Pa pressure. At one point in the tower, the chlorine pressure in the gas is 4 × 104 Pa and the concentration in the liquid is 1 kg/m3. Data on the solubility of chlorine in water at 293 K are given below. If the mass transfer coefficient in the gas phase is four times larger than the mass transfer coefficient in the liquid phase, what are the interfacial concentrations ?

Principles of Mass Transfer Operations − I (Vol. − I)

3.83

Mass Transfer Coefficients

Partial pressure of Cl2, mm Hg Solubility of Cl2, g/liter

5

10

30

50

100

150

0.438

0.575

0.937

1.21

1.773

2.27

(20) Estimate the time that a spherical drop of water, originally 2 mm in diameter, must fall in quiet air at 295 K and 1.013 × 105 Pa pressure in order to reduce its volume by 50%. The free-fall terminal velocity of water drops in air was reported by Sherwood and Pigford (1952) as given below : Diameter, mm 0.05 0.2 0.5 1.0 2.0 3.0 Velocity, ft/s 1.18 2.3 7.0 12.7 19.2 23.8 (21) Air is flowing counter-currently to a falling water liquid film in a 3-cm-ID wetted-wall column. The gas is at 298 ok and 1.013 × 105 Pa total pressure. Determine the gas-side mass transfer coefficient, kG, if (a)

At a point of 1 m from the gas inlet, the airflow rate is 9.5 × 10–4 m3/s, the average partial pressure of water vapour in the gas stream is 665 Pa and the average airflow rate is 9.5 × 10–4 m3/s. The kinematic viscosity of the gas stream is 1.7 × 10–5 m2/s.

(b)

The average partial pressure of the water vapour in the gas stream is 665 Pa. The kinematic viscosity of the gas stream is 1.7 × 10–5 m2/s.

(22) Component A is being separated from a gas mixture of A and B in a wetted-wall absorption tower with the liquid mixture flowing downward along the wall. At a certain point in the tower, the bulk gas concentration yA is 0.7 mole fraction and the bulk liquid xA is 0.1 mole fraction. At the temperature 30°C and 1 atm at which the tower is operating, the equilibrium data is given below was mole fractions. yA

xA 0 0.1 0.2 0.25 0.3 0.35

0 0.155 0.340 0.465 0.650 0.905 '

The film mass transfer coefficient for A is kx = 3.1 lb mole/hr-ft2 and '

ky = 2.06 lbmole/ hr-ft2. (a)

Calculate the interfacial compositions yAi and xAi for the case of equimolar counter diffusion of A and B through both the liquid and gas films.

(b)

Calculate the interfacial compositions yAi and xAi for these case of diffusion of A through stagnant B where the liquid is a non-diffusing solvent. (Ans. : (a) yAi = 0.47, xAi = 0.253, (b) yAi = 0.573, xAi = 0.284)

(23) A wetted wall adsorption tower is fed with water as the wall liquid and ammonia-air mixture is passed over the liquid. At a particular level in the tower the ammonia concentration in the bulk gas is 0.7 mole fraction and that in the liquid is 0.05 mole fraction. Compute the lcoal mass transfer flux for the absorption of ammonia ignoring the vaporization of water.

Principles of Mass Transfer Operations − I (Vol. − I)

3.84

Mass Transfer Coefficients

Equilibrium data xA

yA

0.05 0.10 0.25 0.3

0.0707 0.1347 0.591 0.920

When dilute solutions were used, kx = 1.59 gmole/m2s (mole fraction) and ky = 1.47 gmole/m2s (mole fraction) were found to be the local mass transfer coefficient from the correlation. (a) What is the local mass transfer flux for the absorption of ammonia ? (b) What is the error if you assumed that this is an equimolar/dilute case rather than the true diffusion of A through stagnant B ? (24) A perfume scent can be extracted from an aqueous solution using oil drops (diameter : 0.1 cm). The perfume sent is about 200 times more soluble in the oil than in the aqueous solution. The mass transfer coefficient in the aqueous phase (outside the drops) is 8.4 ⋅ 10−3 cm/sec. After 10 minutes, the concentration of the perfume scent in an initially pure oil drop is 100 times larger than in the aqueous phase. Calculate the mass transfer coefficient of the perfume scent inside the oil drops. (25) A gas stream at 75°C flows through a 0.07 m long packed bed of naphthalene spheres at a superficial velocity of 0.32 m/s. The diameter of each sphere is 0.011 m and the bed porosity is 0.4. The average mass transfer coefficient for this steady-state) process was found to be approximately 0.05 m/s. (a) Write the naphthalene molar balance and calculate the bulk concentration of naphthalene in the gas stream at the exit of the tube. (b) Determine the percent saturation of the exiting gas stream. Additional Information : The entering gas stream is free of naphthalene. The vapour pressure of naphthalene at 75°C is 5.61 mmHg. (26) Consider a laboratory bio-reactor in which micro-organisms need oxygen (at a rate of 0.9 g/hr) to work properly. This oxygen is continuously supplied by feeding air (20% O2, 80% N2, at a pressure of 1 atm.) through a sparger at the bottom of the reactor. The reactor is stirred with a power input of 450 W/m3. By this, air bubbles are generated with an average diameter of 1.2 mm that have an average residence time in the reactor of 25 seconds. Data (at the operating conditions) : Density of the liquid :

1 g/cm3

Kinematic viscosity :

0.1 cm2/sec

Diffusion coefficient of O2 :

2.1 ⋅ 10−5 cm2/sec

Partition coefficient of oxygen in the liquid

4⋅10−5 g/(atm.cm3)

(a) Use the appropriate correlation to calculate the mass transfer coefficient in that process. (b) Calculate the volume rate of air that is needed to keep the oxygen concentration in the bulk liquid at 40% of its saturation concentration (on the bubble/liquid interface) ?

Principles of Mass Transfer Operations − I (Vol. − I)

3.85

Mass Transfer Coefficients

(27) In a cylindrical gas absorption column (5 m in height and 1 m in diameter), a solute gas A (with molecular weight of about 50 g/mol) diffuses into solvent liquid droplets within which it also reacts. The solvent liquid droplet diameter is 2 cm and remains constant, while the reaction is zero order with respect to the solute. (a) At a depth L = 0.4 cm below the droplet liquid surface, the concentration of A has fallen to one-third of the value at the surface. What is the solute molar concentration profile inside the droplet at steady-state ? (b) For a given droplet, what is the molar flux (moles/cm2/s) at this depth L at steady-state conditions ? The solute-solvent molecular diffusivity D = 2 × 10−6 cm2/s, the zero-order rate constant k0 = 10−10 mol/cm3/s and the solute concentration CAi inside the droplets at the droplet liquid interface is 4 × 10−4 mol/cm3. (c) If the number concentration of droplets in the tower is approximately 104/m3, what is the total mass depletion rate of the gas in the absorption column ? (28) A spherical balloon is filled with 15 liters of pure Helium. The balloon skin is slightly permeable to Helium and can be considered as a membrane with a constant thickness of 0.015 cm. Outside the balloon, the wind is blowing strongly and leading to a surface renewal time of 0.8 sec at the outer surface of the balloon. The balloon is shrinking, so that the pressure inside the balloon (p = 1.5 bar) is constant. The temperature of balloon and environment is 20°C. The effective diffusion coefficient of Helium (D⋅H) in the membrane is approximately 5⋅10−7 cm2/sec. (a) Calculate the overall mass transfer coefficient. What is the limiting step ? (b) How long does it take until the balloon has lost 20% of its initial volume ? (29) A spherical droplet of water (substance A) is suspended in stagnant air (gas B) at constant temperature and ambient pressure (25°C, 1.013 bar). The droplet radius is R. Around the droplet there is a spherical stagnant air film of radius r2, through which water vapor is diffusing radially. The molar fraction of water in the air is xA1, at R and xA2 at r2. Consider ideal gases. (a) Calculate the flux (mol/(sm2)) of water vapor evaporating from the droplet as function of c (the total concentration), DAB (the diffusion coefficient), R, pB1 (partial pressure of B at r = R) and pB2 pressure of B at r = r2), assuming steady state, constant droplet radius and concentrated solutions. (b) Calculate the flow rate of water at the surface of the droplet when the film becomes infinitely thick. DAB = 0.26 cm2/s; R = 2 mm; xA1 = 0.4. (30) Calculate the sublimation flux (mol/(cm2⋅s) of naphthalene from a cylinder made of naphthalene with a diameter of d = 4 mm and surface temperature of 25°C. The cylinder is submerged in an air stream flowing perpendicularly to it with a velocity of u = 3 m/s, at p = 1 atm, and T = 25°C. Make use of the following correlation for an average heat transfer coefficient h (g/s3⋅K)) to derive the corresponding correlation for the average mass transfer coefficient k (cm/s) by approximately altering the dimensionless groups to account for mass transfer (instead of heat transfer). h ⋅ d d ⋅ u0.5  ν 0.31 = 0.43 + 0.532 ⋅  Nu = 0.43 + 0.532 ⋅ Re0.5 ⋅ Pr0.31 ⇒   λ   ν  ⋅ α where, λ is the thermal conductivity (g⋅cm/(s3⋅K)), ν is the kinematic viscosity (cm2/s), and α is the thermal diffusivity (cm2/s). Note that α = λ/(ρ⋅cp), ρ being the density (g/cm3) and cp the specific heat capacity (cm2/(s2⋅K). The above correlation is valid for 1 < Re < 4000.

Principles of Mass Transfer Operations − I (Vol. − I)

3.86

Mass Transfer Coefficients

The diffusion coefficient of naphthalene in air (at 3 atm and 25°C) is D = 0.206 × 10−5 m2/s, while the saturation pressure of naphthalene at 25°C is 0.25 mmHg. The air kinematic viscosity ν (at 1 atm, 25°C) = 1.57 × 10−5 m2/s, and air density (at 1 atm, 25°C) ρ = 1.17 Kg/m3. The gas constant R = 0.082 L⋅atm/(mol⋅K). (31) A very dilute solution of ethyl chloride in water flows at 298 K with an average velocity of 0.07 cm/s through a 1.5 cm diameter tube (of total length L = 40 cm) the walls of which are permeable to ethyl chloride. These walls are 0.011 cm thick. Solutes within the walls diffuse as through water. The tube is immersed in a large well-stirred tank of water (assume no resistance to mass transfer from the outer walls to the tank). Use the appropriate mass transfer coefficient to calculate what fraction (in % mole) of the ethyl chloride has been removed at the exit of the tube. The viscosity of water at 298 K is 0.008937 g/(cm⋅s). NOMENCLATURE Any consistent set of units may be used, except as noted. Symbols Meaning a Specific surface of a fixed bed of pellets, pellet surface per volume of bed, m2/m3 A Mass transfer surface, m2 b 1, b 2 Constant c Solute concentration (if subscripted), molar density of solution (if not subscripted), mole/m3 – Bulk average concentration, mole/m3 c d dc de dp D ED EH EV f F g G G' Gr h H jD jH

Differential operator; diameter, m Diameter of cylinder, m Equivalent diameter of a non-circular duct = 4 (cross-sectional area)/ Perimeter, m Diameter of a sphere; for a non-spherical particle, diameter of sphere of the same surface as the particle, m Molecular diffusivity, m2/s Eddy mass diffusivity, m2/s Eddy thermal diffusivity, m2/s Eddy momentum diffusivity, m2/s Friction factor, f1, f2, functions Mass transfer coefficient, mole/m2 s Acceleration due to gravity, m2/s Molar mass velocity, kg/S Mass velocity, kg/S Grasph of number, dimensionless Heat transfer coefficient, w/m2.K Enthalpy Mass transfer dimensionless group, St . SC2/3 D

Heat transfer dimensionless group, St Pr2/3 H

Principles of Mass Transfer Operations − I (Vol. − I) Symbols J k kc, kG, kx, ky K l ld le L m n N Nu P –

3.87

Mass Transfer Coefficients

Meaning Mass transfer flux relative to molar average velocity, mole/m2. sec Thermal conductivity, W/m. K Mass transfer coefficients, mole/ m2. sec (concentration difference) Constant Length; Prandtl mixing length; L Length scale of eddies in the universal range, m Length scale of medium sized eddies, m Length of a wetted-wall tower, m Mass, kg A number, a dimensionless Mass transfer flux at, and relative to, a phase boundary, mole/m2.s Nusselt Number, dimensionless Vapour pressure, kN/m2 Partial Pressure kN/m2

P Pe Pr q R Re Re' Re '' Re

Heat transfer flux, w/m2 Universal gas law constant Reynolds number, dimensionless Reynolds number for flow outside a cylinder, dcG'/µ, dimensionless Reynolds number for flow past a sphere, dpG'/µ, dimesionless Reynolds number for flow in a non-circular duct, deG'/µ, dimensionless

Re

Reynolds number computed with x as the length dimension

e

x

s S SC Sh St t u –

u u' u'i

Peclet number, dimensionless Prandtl number, dimensionless

Fractional rate of surface-element replacement, s–1 Cross-sectional area of a duct, m2 Schmidt number, dimensionless Sherwood number, dimensionless Stanton number Temperature, K Linear velocity, m/s Bulk-average velocity, m/s Root-mean–deviating velocity, m/s Instaneous deviating velocity, m/s

ud'

velocity of eddies in the universal range, m/s

x, y, z (no subscript) xA

Distance in the x, y, z direction respectively, m Concentration of component A in a liquid, mole fraction

yA

Concentration of component A in a gas, mole fraction

z Zb

Distance from the centre of a pipe, m Depth of penetration (film and surface-renewal theories), m

ZF

Effective film thickness (film theory), m

Principles of Mass Transfer Operations − I (Vol. − I)

3.88

Mass Transfer Coefficients

Symbols Meaning Greek letters : α Thermal diffusivity = k/ρCp, m2/s β Volumetric coefficient of expansion, s–1 Mass rate of flow/unit width, kg/m.sec ϒϒτ Thickness of a layer, m δ Molar heat evolution on passing through a surface λ µ Viscosity, kg/m.s ρ Density, kg/m3 φ Area of surface elements (surface-renewal theory), m2/m2.sec Subscripts : A Component A B Component B C Concentration D For mass transfer H For heat transfer i Interface; instantaneous when used with velocity M Logarithmic mean r Reference condition s Sensible heat t Total x, y, z in the x, y, z directions, respectively 0 Approach, or initial value 1 At beginning of mass transfer path 2 At the end of mass transfer path.

✱✱✱ REFERENCES 1. 2.

R.E. Treybal, "Mass-Transfer Operations", Third Edition, McGraw-Hill, 1981. W.L. McCabe, J.C. Smith and P. Harriot, "Unit Operations of Chemical Engineering", Fifth Edition, McGraw-Hill, 1993. 3. J.D. Seader and E.J. Henley, "Separation Process Principles", John Wiley and Sons, 1998. 4. A.H.P.Skelland, “ Diffusional Mass Transfer”, Kriegar, Malbar FL, 1985. 5. C.J. Geankoplis, "Transport Processes and Unit Operations", Fourth Edition, Prentice Hall, 2003 6. A.S. Foust, L.A. Wenzel, C.W. Clump, L. Maus and L.B. Andersen, "Principles of Unit Operations", Second Edition, John Wiley and Sons, 1980. 7. P. Chattopadhya, “Unit Operations “(Vol-I), Khanna Publishers, New Delhi, 1996. 8. G.K.Roy, “ Fundamentals of Heat and Mass Transfer”, Second Edition, Khanna Publishers, New Delhi, 1990 9. R.H. Perry and D. Green, "Perry's Chemical Engineers' Handbook", Seventh Edition, McGraw-Hill, 1997. 10. Philip Wankat,” Equilibrium Staged Operations”, McGraw Hill, 1988. 11. A.L.Hines and R.N. Maddox,” Mass Transfer: Fundamentals and Applications”, Prentice Hall, Inc., New Jersey, 1985 12. J.R. Welty, R.E. Wilson and C.E. Pikes, “Fundamentals of Momentum, Heat and Mass Transfer”, Wiley New York, 1980.

,,,

4

CHAPTER

INTERPHASE MASS TRANSFER 4.1 4.2 4.3 4.4

4.5 4.6 4.7 4.8 4.9

4.10 4.11

Introduction Concept of Equilibrium Diffusion between phases Film concept in mass transfer 4.4.1 Two-film theory of mass transfer 4.4.2 Two-film theory and equilibrium solubility curve 4.4.3 Analysis of mass transfer process using two-film theory Local two-phase mass transfer Two resistance concept Resistance to mass transfer Overall mass transfer coefficient Material balances for : (a) Steady-state concurrent operations (b) Steady-state counercurrent operations (c) Cascades Stages and stagewise operations Kremser Equation Solved Problems Exercises for Practice Nomenclature References

4.1 INTRODUCTION Thus so far we have considered only the diffusion of substances within a single phase. In most of the mass transfer operations, however, two insoluble phases are brought into contact in order to permit transfer of constituent substances between them. Therefore we are now concerned with the simultaneous application of the diffusional mechanism for each phase to the combined system. We have seen that the rate of diffusion within each phase is dependent upon the concentration gradient existing within it. At the same time the concentration gradients of the two-phase system are indicative of the departure from equilibrium, which exists between the phases. Should equilibrium be established, the concentration gradients and hence the rate of diffusion will fall to zero. It is therefore, necessary, to consider both the diffusional phenomena and the equilibria in order to describe the various situations fully. Purpose of this Chapter : (A) To show the interrelation of the resistance and driving forces in an interphase mass transfer process at a particular point in a piece of equipment and to integrate the relationships over the entire surface area. (4.1)

Principles of Mass Transfer Operations − I (Vol. − I)

4.2

Interphase Mass Transfer

(B) A design equation relating the rate of transfer to the total required transfer area will be developed. 4.2 CONCEPT OF EQUILIBRIUM

If we now inject additional ammonia into the container, a new set of equilibrium concentrations will eventually be established, with higher concentrations in each phase than were at first obtained. In this manner we can eventually obtain the complete relationship between the equilibrium concentrations in both phases. If the ammonia is designated as substance A, the equilibrium concentrations in the gas and liquid, yA and xA mole fractions, respectively, give rise to an equilibriumdistribution curve of the type shown in Fig. 4.1. This curve results irrespective of the amounts of water and air that we start with and is influenced only by the conditions, such as temperature and pressure, imposed upon the three-component system.

Mole fraction of A in gas yA

It is convenient first to consider the equilibrium characteristics of a particular operation and then to generalise the result for others. As an example, consider the gas-absorption operation, which occurs when ammonia is dissolved from an ammonia-air mixture by liquid water. Suppose a fixed amount of liquid water is placed in a closed container together with a gaseous mixture of ammonia and air, the whole arranged so that the system can be maintained at constant temperature and pressure. Since ammonia is very soluble in water, some ammonia molecules will instantly transfer from the gas into the liquid, crossing the interfacial surface separating the two phases. A portion of the ammonia molecules escapes back into the gas, at a rate proportional to their concentration in the liquid. As more ammonia enters the liquid, with consequent increase in concentration within the liquid, the rate at which returns to the gas increases, until eventually the rate at which it enters the liquid exactly equals that at which it leaves. At the same time, through the mechanism of diffusion, the concentrations throughout each phase become uniform. A dynamic equilibrium now exists, and while ammonia molecules continue to transfer back and forth from one phase to the other, the net transfer falls to zero. The concentrations within each phase no longer change. To the observer who cannot see the individual molecules the diffusion has apparently stopped.

Mole fraction of A in liquid = xA

Fig. 4.1 : Equilibrium Distribution Curve

It is important to note that at equilibrium the concentrations in the two phases are not equal; instead the chemical potential of the ammonia is the same in both phases, and it will be recalled that it is equality of chemical potentials, not concentrations, which causes the net transfer of solute to stop. The curve of Fig. 4.1 does not of course show all the equilibrium concentrations existing within the system. For example, the water will partially vapourize into the gas phase, the components of the air will also dissolve to a small extent in the liquid, and equilibrium concentrations for these substances will also be in the liquid, and equilibrium concentrations for these substances will also be established. For the moment we need not consider these equilibria, since they are of minor importance to the discussion at hand. Obviously also, concentration units other than mole fractions may be used to describe the equilibria.

Principles of Mass Transfer Operations − I (Vol. − I)

4.3

Interphase Mass Transfer

Generally speaking, whenever a substance is distributed between two insoluble phases, a dynamic equilibrium of this type can be established. The various equilibria are peculiar to the particular system considered. For example, replacement of the water in the example considered above with another liquid such as benzene or with a solid adsorbent such as activated carbon or replacement of the ammonia with another solute such as sulfur dioxide will each result in new curves not at all related to the first. The equilibrium resulting for a two-liquid-phase system bears no relation to that for a liquid-solid system. A discussion of the characteristic shapes of the equilibrium curves for the various situations and the influence of conditions such as temperature and pressure must be left for the studies of the individual unit operations. Nevertheless the following principles are common to all systems involving the distribution of substance in the two insoluble phases : 1. At a fixed set of conditions, referring to temperature and pressure, there exists a set of equilibrium relationships, which can be shown graphically in the form of equilibrium – distribution curve for each distributed substance by plotting the equilibrium concentrations of the substance in the two phases against each other. 2. For a system in equilibrium, there is no net diffusion of the components between the phases. 3. For a system not in equilibrium, diffusion of the components between the phases will occur so as to bring the system to a condition of equilibrium. If sufficient time is available, equilibrium concentrations will eventually prevail. 4.3 DIFFUSION BETWEEN PHASES Having established that departure from equilibrium provides the driving force for diffusion, we can now study the rates of diffusion in terms of the driving forces. Many of the mass transfer operations are carried out in steady-flow fashion, with continuous and invariant flow of the contacted phases and under circumstances such that concentrations at any position in the equipment used do not change with time. It will be convenient to use one of these as an example with which to establish the principles and to generalize respecting other operations later. For this purpose, let us consider the absorption of a soluble gas such as ammonia (substance A) from a mixture such as air and ammonia, the wetted-wall tower previously described. The ammonia-air mixture may enter at the bottom and flow upward while the water flows downward around the inside of the pipe. The gas mixture changes its composition from a high to a low solute concentration as it flows upward, while the water dissolves the ammonia and leaves at the bottom as an aqueous ammonia solution. Under steady-state conditions, the concentrations at any point in the apparatus do not change with passage of time. 4.4 FILM CONCEPT IN MASS TRANSFER As previously noted, gas absorption operation involves mass transfer from the gas phase to the liquid phase. That means the gas molecules must diffuse from the main body of the gas phase to the gas-liquid interface, then cross this interface into the liquid side, and finally diffuses from the interface into the main body of the liquid. The Figure 4.2 shows a typical gas-liquid interface. This interface can represent any location in the gas absorption equipment where the gas contacts the liquid.

Principles of Mass Transfer Operations − I (Vol. − I)

4.4

Interphase Mass Transfer

In the gas phase, 3 flow Gas-Liquid interface regimes can be visualized : • Fully developed turbulent region where most of the (a) Laminar Film mass transfer takes place (b) Transition Zone by eddy diffusion. • A transition zone with (c) Fully-developed Gas turbulence Wall some turbulence • A laminar film with molecular diffusion Such phenomena are difficult to analyze. Instead, we (a) (b) (c) will use a simplified TWOFILM THEORY as a basis for analysis as well as Liquid development of various Fig. 4.2 : A typical gas-liquid interface correlations of mass transfer phenomena. 4.4.1 Two-Film Theory of Mass Transfer Once again consider the interface between the gas phase and the liquid phase, now simplified as shown in the Figure 4.3. yAG or pAG

Gas Film

yAi or pAi

Bulk Gas Phase

Liquid Film

Bulk Liquid Phase

XAi or CAi

XAL or CAL

Interface

Liquid Entering

Gas Leaving

Any point in the column

Gas Film

Bulk Gas Phase

Liquid Film

Bulk Liquid Phase

Interface Liquid Leaving

Gas Entering

Fig. 4.3 : The interface between the gas phase and the liquid phase

Principles of Mass Transfer Operations − I (Vol. − I)

4.5

Interphase Mass Transfer

This interface can represent any point in the gas absorption equipment where the gas contacts the liquid. See the Figure below that shows a counter-current gas absorption column. We will study the diffusion of solute A from the gas phase into the liquid phase, for example, NH3 that is diffusing from a gaseous air-NH3 mixture into liquid phase water. Assumptions of Two-Film Theory : • • • •

Steady state : concentrations at any position in the tower do not change with time. Interface between the gas phase and the liquid phase is a sharp boundary. Laminar film exist at the interface on both sides of the interface Equilibrium exists at the interface, thus there is negligible resistance to mass transfer across the interface : (xi, yi) is the equilibrium concentration.

•

No chemical reaction : rate of diffusion across the gas-phase film must equal the rate of diffusion across the liquid-phase film.

4.4.2 Two-Film Theory and Equilibrium Solubility Curve In the analysis of gas absorption, we are interested in the transfer of materials throughout the entire gas absorption equipment, not just a single location in the equipment. Therefore the two-film theory can be analyzed more effectively by using the equilibrium solubility curve. The concentrations at the interface in the gas (yAi) and in the liquid (xAi) is represented as a point M on the equilibrium solubility curve (see Fig. 4.4). Point M thus has the co-ordinates (yAi, xAi). As we move along the column along the continuous interface, we can trace out an equilibrium curve. Very often, the subscript "Ai" is dropped, and the equilibrium curve is simply a relationship between y and x; i.e. y = f(x). The concentrations in the bulk gas phase (yAG) and in the bulk liquid phase (xAL) is represented as a point P above the equilibrium curve. Point P thus has the co-ordinates (xAL, yAG). Point P is located above the equilibrium curve. [Remember that it is the departure from equilibrium that provide the driving force for mass transfer] See the Figure as follows. Slope of PM = – Kx / ky yAG

Equilibrium Curve y = f(x)

P

yAi

M

xAL

xAi

Fig. 4.4 : Equilibrium curve

Principles of Mass Transfer Operations − I (Vol. − I)

4.6

Interphase Mass Transfer

Notation : yAG = composition of A in the bulk gas phase (mole fraction) xAL = composition of A in bulk liquid phase (mole fraction) (xAi, yAi) = equilibrium interface compositions (mole fraction) 4.4.3 Analysis of Mass Transfer Process Using Two-Film Theory In the gas-phase, the concentration falls from yAG in the bulk gas to yAi at the interface. Thus, there is a concentration driving force for mass transfer from the bulk gas to the gas film to the interface. At the interface, the component A crossed the interface and enters the liquid side. In the liquid-phase, the concentration falls from xAi at the interface to xAL in the bulk liquid. Thus, there is a concentration driving force for mass transfer from the interface to the liquid film to the bulk liquid. The mass transfer process can be represented by the line PM. See the Fig. 4.4. NOTE : The bulk concentrations yAG, xAL are not equilibrium values, otherwise there would be no diffusion of A. REMINDER :

pAG Bulk Gas Phase

Gas Film

pAi

Liquid Film

Bulk Liquid Phase

cAi

cAL

Partial pressure of solute in the gas phase, pA

Interface Equilibrium Curve, p = f(c) pAG

P

pAi pA*

D

M C cAL

cAi cA*

Concentraction of solute in the liquid phase, cA

Fig. 4.5 : Two-film theory and equilibrium curve expressed in partial pressures and concentrations

The two-film theory and equilibrium curve can be expressed in other ways, e.g. in terms of partial pressure (for the gas phase) and concentration (for the liquid phase); but the analysis for them is the same as outlined before for mole fractions (x and y). Refer to the Figure 4.5, which showed the two-film theory and equilibrium curve expressed in partial pressures and concentrations.

Principles of Mass Transfer Operations − I (Vol. − I)

4.7

Interphase Mass Transfer

4.5 LOCAL TWO-PHASE MASS TRANSFER Let us investigate the situation at a particular level along the tower, e.g., at a point midway between top and bottom. Since the solute the solute is diffusing from the gas phase into the liquid, there must be a concentration gradient in the direction of mass transfer within each phase. This can be shown graphically in terms of the distance through the phases, as in Fig. 4.6, where a section through the two phases in contact is shown. It will be assumed that no chemical reaction occurs. Gas Concentration of diffusing solute A

The concentration of A in the main body of the gas is yA,G mole fraction, and it falls to yA, at the interface. In the liquid, the concentration falls xA,I at the interface to XA,L in the bulk liquid. The bulk concentrations yA,g and XA,L are clearly not equilibrium values, since otherwise diffusion of the solute would not occur. At the same time, these bulk concentrations cannot be used directly with a mass transfer coefficient to describe the rate of interphase mass transfer, since the two concentrations are differently related to the chemical potential, which is the real “driving force” of mass transfer.

Liquid

yAG

xAi

Interface

yAi

xAL

Distance

Fig. 4.6 : The two-resistance concept

To get around this problem, Lewis and Whitman assumed that the only diffusional resistances are those residing in the fluids themselves. Then there is no resistance to solute transfer across the interface separating the phases, and as a result the concentrations yA,I and XA,I are equilibrium values, given by the system’s equilibrium-distribution curve. The reliability of this theory has been the subject of a great amount of study. A careful review of the results indicates that departure from concentration equilibrium at the interface must be a rarity. It has been shown theoretically that departure from equilibrium can be expected if mass-transfer rates are very high, much higher than are likely to be encountered in any practical situation. Careful measurements of real situations when interfaces are clean and conditions carefully controlled verify the validity of the assumption. Unexpectedly large and small transfer rates between two phases nevertheless occur, and these have frequently been incorrectly attributed to departure from the equilibrium assumption. For example, the heat of transfer of a solute, resulting from a difference in the heats of solution in two phases, will either raise or lower the interface temperature relative to the bulk-phase temperature and the equilibrium distribution at the interface will then differ from that which can be assumed for the bulk-phase temperature. Because of unexpected phenomena near the interface in one or both of the contacted phases (caused by the presence of surfactants and the like) there may be a departure of the transfer rates from the expected values which may improperly be attributed to departure from equilibrium at the interface. These matters will be considered later. Consequently, in ordinary situations the interfacial concentrations of Fig. 4.6 are those corresponding to a point on the equilibrium-distribution curve. Referring again to Fig. 4.6 it is clear that the concentration rise at the interface, from ya,I to xA,I is not a barrier to diffusion in the direction gas to liquid. They are equilibrium concentrations and hence correspond to equal chemical potential of substance A in both phases at the interface.

4.8

Interphase Mass Transfer

The various concentrations can also be shown graphically, as in Fig. 4.7, whose co-ordinates are those of the equilibrium distribution curve. Point P represents the two bulk-phase concentrations and point M those at the interface. For steady-state mass transfer, the rate at which A reaches the interface from the gas must equal that at which it diffuses to the bulk liquid, so that no accumulation or depletion of A at the interface occurs. We can therefore write the flux of A in terms of the mass-transfer coefficients for each phase and the concentration changes appropriate to each (the development will be done in terms of the k-type coefficients, since this is simpler, and the results for F-type coefficients will be indicated later). Thus, when ky and kx are the locally applicable coefficients, NA = ky (yA, G – yA, i) = Kx (xA, i – xA, L)

Concentration of solute in the gas

Principles of Mass Transfer Operations − I (Vol. − I)

Equilibrium Distribution Curve, yAi = f(xAi) yAG

k P Slope = – —x ky

yAi

M

xAL

xAi

Concentration of solute in the liquid

Fig. 4.7 : Departure of bulk-phase concentration form equilibrium

… (4.1)

And the differences in y’s or x’s are considered the driving forces for the mass transfer. Rearrangement as : yA‚G – yA‚ i kx … (4.2) XA‚ L – XA‚ i = – ky Provides the slope of the line PM. If the mass-transfer coefficients are known, the interfacial concentrations and hence the flux NA can be determined, either graphically by plotting the line PM or analytically by solving equation (4.2) with an algebraic expression for the equilibrium distribution curve. … (4.3) YA, i = f (xA, i) 4.6 TWO RESISTANCE CONCEPT

0

Gas Liquid phase phase

CA,i

pA,i CA,L Interface if H < 1 Distance, Z

Concentration of diffusing species A

Concentration of diffusing species A

Gas Liquid phase phase

0

pA,G

pA,i

CA,L Interface if H = 1 Distance, Z

Fig. 4.8 (a) : Steps in Interphase mass transfer process

The interphase mass transfer involves three steps : (i) Transfer of mass from the bulk of one phase to the interface separating the two phases. (ii) Transfer of mass from this interface to the second phase. (iii) Transfer of mass to the bulk of the second phase.

Partial pressure of A in the gas phase

Principles of Mass Transfer Operations − I (Vol. − I)

4.9

Interphase Mass Transfer

kL Slope = – — k

G

pA,G

pA,i

Equilibrium Curve

cA,i

cA,L

Concentration of A in the liquid phase Fig. 4.8 (b) : Equilibrium curve of a component A distributed between the gas and the liquid phase in contact

The interface offers no resistance to mass transfer. The resistance to the interface of diffusing component is offered by the phases on each side of the interface. For steady-state transfer of component A from the gas phase to the liquid phase, the molar rate of diffusion in : (a) Gas Phase : NA, Z = Kg (pA, G – pA, i) … (4.4) (b) Liquid Phase :

NA, Z = Kl (CA, i – CA, L)

Where,

pA, G = Partial pressure of A in the bulk gas, Pa

… (4.5)

pA, i = Partial pressure of A in the interfacial gas, Pa CA, i = Concentrations of A in the intefacial liquid, mol/m3 CA, L = Concentations of A in the bulk liquid, mol/m3 Kg = Convective mass transfer coefficient in the gas phase expressed as, moles of A transferred   (time) (interfacial area) (∆P – Units of concentration) Kl = Convective mass transfer coefficent in the liquid phase expressed as moles of A transferred   (time) (interfacial area) (∆C – Units of concentration) Under steady-state conditions, the molar rate of mass transfer of A in one phase must equal to that of A in the other phase. Therefore, NA, Z = Kg (pA, G – pA, i) = kl (CA, i – CA, L) … (4.6) And equation (4.6) may be written as, pA‚ G – pA‚ i Kl CA‚ L – CA‚ i = – Kg

… (4.7)

4.7 RESISTANCE TO MASS TRANSFER ∆PA‚ gas film 1/Kg Resistance in the gas phase = = 1/K Overall Resistance (i.e. resistance in both phases) ∆PA‚ Total G Resistance in the liquid phase Overall Resistance

=

∆CA‚ gas film ∆CA‚ total

… (4.8) … (4.9)

Principles of Mass Transfer Operations − I (Vol. − I)

4.10

Interphase Mass Transfer

As long as following linear equilibrium relation exists, we can express, pA, i = m·CA, i pA, G = m·CA

… (4.10)

pA = m·CA, L The overall mass transfer coefficient (KG) based on the partial pressure difference, as driving force is given by :

*

NA = KG (pA, G – pA )

… (4.11)

Rearranging above equation, we get,'

or,

*

*

1 KG

pA‚ G – pA pA‚ G – pA‚ i pA, i – pA = +N NA‚ Z = NA‚ Z A, Z

1 KG

=

pA‚ G – pA‚ i m (CA‚ i – CA, L) + NA‚ Z NA‚ Z

1 1 m = + KG Kg Kl

… (4.12)

The overall mass transfer coefficient (KL) based on the concentration difference, as driving force is given by : NA = KL (CA* – CA, L ) … (4.13) Rearranging above equation, we get,

Thus we have,

1 KL

=

CA* – CA‚ L CA. – CA‚ i CA‚ i – CA‚ L = + NA‚ Z NA‚ Z NA‚ Z

1 KL

=

PA‚ G – PA‚ i (CA‚ i – CA‚ L) m·NA‚ Z + NA‚ Z

1 1 1 KL = m·Kg + K1

… (4.14)

Note : If we assume numerical value of KL and KG are approximate the same and m is very small (Equilibrium distribution curve is flat), so that at equilibrium, only small concentration of A m in the gas will be provide a very large concentration in the liquid, the term K in equation (4.12) l 1 becomes minor, the major resistance is represented by K and the rate of mass transfer is g gas-phase controlled. Thus, 1 1 KG ≈ Kg or

*

pA, G – pA

≈ pA, G – pA, i

4.8 OVERALL MASS TRANSFER COEFFICIENT Determination of mass transfer coefficient is necessary to evaluate the rate of transfer between two phases. The two phases are assumed to be in equilibrium at all points on the interface. In a gas liquid operations (e.g. gas absorption), the liquid concentration (Ci) and the

Principles of Mass Transfer Operations − I (Vol. − I)

4.11

Interphase Mass Transfer

partial pressure (pi) at the interface are related by equilibrium conditions. For a liquid-liquid transfer process (e.g. liquid-liquid extraction), the value of Ci will be different in each of the phase, but the two will correspond to equilibrium concentration in the two phases. The extreme difficulty in physical measurement of Partial pressure and concentration at the interface makes it necessary to employ overall mass transfer coefficient. The overall mass transfer coefficient (KG) based on the partial pressure difference, as the driving force is given by : *

NA = KG (pA, G – pA ) Where,

… (4.15)

pA, G = partial pressure of A in the bulk gas (Pa) *

pA

= partial pressure of A i the equilibrium with the bulk composition in the liquid phase (Pa.)

KG accounts for the entire diffusional resistance in the bulk phase. The overall mass transfer coefficient (KL) based on the concentration difference, as driving force is given NA =

by : where,

*

CA

(K

*

L

(CA – CA‚ L)

)

… (4.16)

= Concentration of A in equilibrium with the bulk composition of gas phase. (mol/m3).

CA, L = Concentration of A in the bulk liquid (mol/m3). Equilibrium Curve

pA, G

pA, i

DpA,Total

DpA, G

k Slope = – —L kG

DpA, L p*A cA, L

cA, i DcA, L

c*A DcA, G

Fig. 4.9 : Equilibrium curve of component A

4.9 MATERIAL BALANCES FOR (A) STEADY–STATE CO–CURRENT OPERATIONS (B) STEADY–STATE COUNTERCURRENT OPERATIONS (C) CASCADES

Introduction : The concentration difference – driving force are existing at one position in the equipment used to contact the immiscible phases in batch or steady-state processes. These changes produce corresponding variations in the driving force, and these can be followed with the help of material balance. In other words, all concentrations are the bulk average values for indicated streams.

Principles of Mass Transfer Operations − I (Vol. − I)

4.12

Interphase Mass Transfer

(A) Steady State Co-Current Operations : Consider any mass transfer operation conducted in steady state co-current fashion as shown in Fig. 4.6, which is simply shown by rectangular box. Let the two phases R identified as two insoluble phases in gas and liquid respectively. Consider only single substance A diffuses from phase R to phase E during their contact. The other constituents at the phases, solvents for the diffusing solutes, are considered not to diffuse.

E1 moles total material/time Es moles non-difusing material/time y1 mole fraction solute Y1 mole ratio solute

II

I

E Es y Y

R1 moles total material/time Rs moles non-difusing material/time x1 mole fraction solute X1 mole ratio solute

R Rs x X Fig. 4.10 : Steady-State Concurrent Process

The content of the leaving streams is R2x2 and E2y2 respectively. R1x1 = R2x2 + E2y2 or R1x1 – R2x2 = E2y2 – E2y1 x1 but, R1x1 = RS 1 – x = RS X1 1

E2 Es y2 Y2

R2 Rs x2 X2

… (4.17) … (4.18) … (4.19)

where X1 = Mole ratio concentration of 'A' at the entrance mole A/mole non–A. Equation (4.18) becomes, RS (X1 – X2) = ES (Y2 – Y1) … (4.20) RS This is equation of straight line on X, Y co-ordinates of slope is – E passing through two S points whose co-ordinates are (X1, Y1) and (X2, Y2) respectively. At any section B-B through the apparatus, the mole fractions of A are x and y and the mole ratios X and Y, in phases R and E respectively, and if envelope – II is drawn to include all the device from the entrance to section B-B, the A balance becomes. RS (X1 – X2) = ES (Y – Y1) … (4.21) RS This is also the equation of straight line on x, y co-ordinates of slope – E , through the points S (X1, Y1) and (X, Y). Note : Since the two-straight lines have the same slope and a point in common they are the same straight line and equation (4.21) is therefore a general expression relating the compositions of the phases in the equipment at any distance from the interface. It must be emphasized that the graphical representation of the operating line as a straight line is greatly dependent upon the units in which the concentrations of the material balance are expressed. The representations of Fig. 4.7 and 4.8 are straight lines because the mole-ratio concentration are based on the unchanging quantities ES and RS. If equation (4.18) were plotted

Principles of Mass Transfer Operations − I (Vol. − I)

4.13

Interphase Mass Transfer

y = Conc. in phase E, mole A/mole non-A

on mole-fraction co-ordinates, or if any concentration unit proportional to mole fractions such as partial pressure were used, the nature of the operating curve obtained would be indicated as in Fig. 4.13. In any operation where the total quantities of each of the phases E and R remain constant while the compositions change owing to diffusion of several components a diagram in

K

Equilibrium Curve, Yi = f(xi)

L

Y2 Y1 Y2

T

Operating Line, Slope = – RS/ES

Y1

M

X*2 Xe X2 X1 x = Conc. in phase R, mole A/mole non-A

Fig. 4.11 : Steady-State Co-current process, Transfer of solute from phase R to phase E

terms of mole-fraction will provide a straight line operating line, as equation 4.18 indicates (let E2 = E1 = E; R2 = R1). If all the components diffuse so that the total quantities of each phase do

Y1 Y2 Ye

P

Operating Line, Slope = – RS/ES Q T

Equilibrium Curve, Yi = f(xi)

X1 X2 Xe Concentration in phase R, moles A/mole non-A = X Fig. 4.12 : Steady-State Co-current process Transfer of solute from phase E to phase R

Concentration in phase E, moles A/mole non-A = y

Concentration in phase E, moles A/mole non-A = Y

not remain constant, the operating line will generally be curved.

ye y2

T Q

Equilibrium Curve, Yi = f(xi) Operating Curve

y1

P xe

x2

x1

Concentration in phase R, moles A/mole non-A = x Fig. 4.13 : Steady-State Co-current Process transfer of solute from phase R to Phase E

To sum-up, the cocurrent operating line is a graphical representation of the material balance. A point on the line represents the bulk average concentrations of the stream in contact with each other at any section in the apparatus. Consequently the line passes from point representing the streams entering the apparatus to that representing the effluent streams. (B) Steady-State Countercurrent Operations : If the same process (combined out in co-current fashions) is carried out in countercurrent fashion as shown in Fig. 4.14. Here 1 indicates that end of apparatus where phase R enters and 2 that end where phases R leaves.

Principles of Mass Transfer Operations − I (Vol. − I)

4.14

Interphase Mass Transfer

Material Balance for envelope – I : E2y2 + R = 1x1 = 1y1 + R2x2 and RS (x1 – x2) = ES (y1 – y2)

… (4.22) … (4.23)

Material Balance for envelope – II : Ey + R1x1 = E1y1 + Rx and RS (x1 – x) = ES (y1 – y)

… (4.24) … (4.25)

Equations (4.24) and (4.25) gives general relationship between the concentrations in the phases at any section. Equation (4.22) and (4.23) establish the entire material balance. RS Equation 4.23 is that of a straight line on X, Y co-ordinates of slope E , through points of S co-ordinates (X1, Y1), (X2, Y2), as shown in Fig. 4.11. The line will lie above the equilibrium curve if diffusion proceeds from phase E to phase R (Absorber) and below for diffusion in the opposite direction (stripper) (i.e. phase R to phase E). B

E1 moles total material/time Es moles non-difusing material/time y1 mole fraction solute Y1 mole ratio solute R1 moles total material/time Rs moles non-difusing material/time x1 mole fraction solute X1 mole ratio solute

I

E Es y Y

II

R Rs x X

E2 Es y2 Y2

R2 Rs x2 X2

B Fig. 4.14 : Steady-State Countercurrent Process

The operating lines of Fig. 4.14 are straight because the mole-ratio concentrations X and Y are based on the quantities RS and ES, which are stipulated to be constant. If for this situation mole fractions (or quantities such as partial pressures, which are proportional to mole fractions) are used the operating lines are curved, as indicated in Fig. 4.11. However, for some operations should the total quantity of each of the phases E and R be constant while the compositions change, the mole fraction diagram would provide the straight line operating lines, as equation 4.22 would indicate (let E = E1 = E2; R = R1 = R2). As in the concurrent case, however, the operating line will generally be curved if all the components diffuse so that the total quantities of each phase to not remain constant. The countercurrent operating line is a graphical representation of the material balance, passing from the point representing the streams at one end of the apparatus to the point representing the streams passing each other at any section in the apparatus. The advantage of the countercurrent method over concurrent is that, for the former, the average driving force for given situation will be larger, resulting either smaller equipment for a given set of flow conditions or smaller flows for a given equipment size.

Principles of Mass Transfer Operations − I (Vol. − I)

4.15

Interphase Mass Transfer

Concentration in phase E, moles A/mole non-A = y

Equilibrium Curve, yi = f(xi) y1

Operating Curve

y2 x2

x1

Concentration in phase E, moles A/mole non-A = x

Fig. 4.15 : Steady-State Countercurrent Process, Transfer of Solute from Phase R to Phase E

(C) Cascades : A group of stages interconnected so that the various streams flow from one to the other is called cascade. Its purpose is to increase the extent of mass transfer over and above that which is possible with a single stage. The fractional overall stage efficiency of a cascade is then defined as the number of equilibrium stages to which the cascade is equivalent divided by the number of real stages. ES1 y1 Y1

Concentration in phase E, moles A / mole non-A

RS x0 X0

1 ES1 y0 Y0

ES2 y2 Y2 RS x1 X1

2 ES2 y0 Y0

ES3 y3 Y3 RS x2 X2

Y1 RS – —– ES2

Y2 Y3 Y0

3 ES3 y0 Y0

RS x3 X3

Slope RS = – —– ES1

R – —–S ES3

X3 X2 X1 X0 Concentration in phase R, moles A / mole non-A

Fig. 4.16 : A cross flow cascade of three real stages

Two or more stages connected so that the flow is concurrent between stages will, of course, never be equivalent to more than one equilibrium stage, although the overall stage efficiency can thereby-increased. For effects greater than one equilibrium stage, they can be connected for cross flow or countercurrent flow.

Principles of Mass Transfer Operations − I (Vol. − I)

4.16

Interphase Mass Transfer

In Fig. 4.16, each stage is represented simply by a circle and within each the flow is concurrent. The R-phase flows from one stage to the next, being contacted in each stage by a fresh E phase. There may be different flow rates of the E phase to each stage and each stage may have a different Murphee stage efficiency. The material balances are obviously merely a repetition of that a single stage and the construction on the distribution is obvious. Cross flow is used sometimes in adsorption, leaching, drying and extraction operations. 4.10 STAGES AND STAGESWISE OPERATIONS (1) Stage : Stage may be defined as any device or combination of devices in which two insoluble phases are brought into intimate contact, where mass transfer occurs between the phases tending to bring them to equilibrium, and where the phases are then mechanically separated. A process carried out in this manner is single stage process. (2) Equilibrium/Theoretical/Ideal Stage : In order to establish a standard for the measurement of performance, the equilirbium/ideal/theoretical stage is defined as one where effluent phases are in equilirbium so that a longer time of contact will bring about no additional change of composition. (3) Stage Efficiency : A stage efficiency is defined as the fractional approach to equilibrium which a real stage produces. Referring to Fig. 4.7, that be taken as the fraction which the line QP represents of the line TP, or the ratio of actual solute transfer to that if equilibrium were attained. (4) Murphee Stage Efficiency : The most commonly used expression in design of absorber and striper is the Murphee stage efficiency. It is defined as, the fractional approach of one leaving stream to equilibrium with the actual concentration in the other leaving stream. Again referring to Fig. 4.7, this can be expressed in terms of the concentrations in phase E or in phase R – Y2 – Y1 … (4.26) EME = * Y2 – Y1 X1 – X2 EMR = X – X 1 2

… (4.27)

These definitions are somewhat arbitrary since, in the case of a truly co-current operation such that of Fig. 4.7, it would be impossible to obtain a leaving concentration in phase E higher than Ye or one in phase R lower than Xe. The two Murphee efficiencies are not normally equal for a given stage and they can be simply related only when the equilibrium relation is straight line. Thus for a straight equilibrium line of slope, *

m = it can be shown that :

where,

RS A = mE S

(Y2 – Y2) *

(X2 – X2)

EMR EME = E (1 – S) + S = MR is called the absorption factor.

m·ES 1 S =A = R is the stripping factor. S

EMR 1 1 EMR 1 – A + A  

Principles of Mass Transfer Operations − I (Vol. − I)

4.17

Interphase Mass Transfer

RS = Rate of flow of non-diffusing solvent in phase R. ES = Rate of flow of non-diffusing solvent in phase E. m = Equilibrium curve slope. Comments : (i) A high degree of absorption can be obtained, either by using a large number of plates or RS by using high absorption factor; A = m E S RS Since m is fixed by the system, this means that E must be large if a high degree of S absorption can be obtained, but this will result in a low value of x for the liquid leaving at the bottom. RS (ii) It is important to note that, if m E < 1, then very large number of plates are required to S achieve a high recovery and even an infinite number will not give complete recovery. RS Slope of operating line (iii) Absorption factor = A = mE = Slope of equilibrium curve S RS RS If E < m i.e. m E < 1, then the operating line will never cut the equilibrium curve and S S gas leaving the top of column will not therefore reach equilibrium with entering liquid. 4.11 KREMSER EQUATION (KREMSER - SOUDERS - BROWN EQUATION) (a) For transfer from R to E (stripping of R) : Xo – XN (1/A)NP + 1 – 1/A P A≠1: = … (4.30) Xo – YN + 1/m (1/A)NP + 1 – 1 P

Xo – YNP + 1/m  (1 – A) + A X – Y  NP NP + 1m 

NP = log Xo – XN A=1:

P

Xo – YN + 1/m P

NP = N +1 P

… (4.31)

… (4.32)

Xo – XN P NP = X – Y + 1/m N N P

P

(b) For transfer from E to R (absorption into R) : A similar treatment yields : YN + 1 – Y1 ANP + 1 – A P A≠1 = YN + 1 – mXo ANP + 1 – 1 P

NP = log YN A=1:

P+1

YN

P

– Y1

+ 1 – mXo

YNP + 1 – mXo  1  1   Y – mX  1 – A + A o  1 log A

NP = N +1 P YN

NP =

P+1

… (4.33)

… (4.34) … (4.35)

– Y1

Y1 – mXo

… (4.36)

Principles of Mass Transfer Operations − I (Vol. − I)

4.18

Interphase Mass Transfer

The equations from (4.30) to (4.36) are called as Kremser – Brown – Souders (or simply Kremser equations) where, NP = Total number of stages in a cascade mole A X = Concentration of A in phase R, mole non – A mole Y = Concentration of A in phase E mole non – A   m = Equilibrium curve slope A = Absorption factor The Kremser equations (equation 4.30 to 4.36) are plotted in Fig. 4.17, which becomes very convenient for quick solutions. 1.0 0.8

A / 1/A 0.3

0.6

0.5 0.6

0.4

0.7

0.3

0.8

X0 – YNp+1 / m

Y1 – mX0 YNp+1 – mX0

(absorption) or

YNp – YNp+1 / m

(stripping)

0.2

0.9 0.1 0.08 0.95

0.06 0.04

1.0

0.03 0.02 1.05 1.1

0.01 0.008 0.006

1.2 1.3

0.004

1.6 1.4 1.5 2.0 1.8

0.003 0.002

A (absorption) or 1/A (stripping)

5.0 4.0

3.0

2.5

0.001 0.0008 0.0006 0.0005 1

2

3

4

8 10 20 5 6 Np = Number of theoretical trays

30

40

50

Fig. 4.17 : Number of theoretical stages for counter-current cascades, with Henry's law equilibrium and constant absorption or stripping factors

Principles of Mass Transfer Operations − I (Vol. − I)

4.19

Interphase Mass Transfer

1.0 8 Plates

Y1 – mXo YNP+1 – mXo

0.9 es lat 5 P tes la 4P es lat 3P es lat P 2 es lat 1P

0.5

0.3 0.2 0.1 0

0.2

0.4

0.6

0.8

1.0

1.5

2.0

3.0

Absorption factor (A)

Fig. 4.18 : Graphical representation of the effect of the absorption factor and the number of plates on the degree of absorption

SOLVED PROBLEMS (1) In a mass transfer apparatus operating at 1 atm., the individual mass-transfer coefficients have the following values : Kx = 22 kg mole/m2 hour (∆ ∆x = 1) ∆y = 1) Ky = 1.07 kg mole/m2 hour (∆ The equilibrium compositions of the gaseous and liquid phases are characterised by Henry's law. p* = 0.08 × 106 · x mm Hg (i) Determine the overall mass-transfer coefficient Kx and Ky. (ii) How many times the diffusion resistance of the liquid phase differs that of the gaseous phase ? Sol. : Data given : Kx = 22 kg mole/m2. hour (∆x = 1) = Kl Ky = 1.07 kg mole/m2.hour (∆y = 1) = Kg p* = H.C. = 0.08 × 106 · x mm Hg So, (i)

∴

= 105.26 x atm. H = 105.26 1 1 1 Kx = Kx + HKy

Kx

1 1 = 0.0539 = 22 + 105.26 × 1.07 = 18.55 kg mole/m2. hour

Principles of Mass Transfer Operations − I (Vol. − I)

4.20

Interphase Mass Transfer

1 Ky

1 H = K =K y x

∴

Ky

105.26 1 = 1.07 + 22 = 5.715 = 0.175 kg mole/m2. hour

(ii)

Rx Ry

Similarly,

=

1 Kx

Resistance in liquid phase Resistance in gas phase = 1 Ky

0.0539 = 5.715 = 0.0094 (Ans.) (2) In a wetted wall column, a gas A is being stripped from A – water solution into an air stream. At a certain point in the column, the concentration of component A in liquid is 4.8 k mole/m3 and partial pressure of component A in gas stream is 1 atm. The equilibrium relation for dilute solution of component A in water is given by PA, i = 0.25 CA, i Where, PA, i = Equilibrium partial pressure of component 'A' (atm.) k mole A  CA, i = Equilibrium concentration of A in water  3 m solution The overall liquid coefficient : k mole A KL = 0.0144 k mole A (hr. m2)   m3  Assume that gas phase offers 70% of total resistance to mass transfer. Calculate the following : (a) Convective mass transfer coefficient in liquid phase, Kl. (b) Convective mass transfer coefficient in gas phase, Kg. (c) Overall mass transfer coefficient in gas phase, KG. Sol. : (a) Since 70% of the total resistance to mass transfer offers in the gas phase; hence the resistance in liquid phase is 30%. ∆ CA‚ liquid film Resistance in liquid phase ∴ = Overall resistance ∆ CA‚ Total

=

1 Ki 1 KL

=

0.30 100

=

1 Kl 1 0.0144

k mole A (hour · m2) (k mole . A/m2) (b) Corrective mass transfer coefficient is given by 1 1 1 KL = m · Kg + Kl

So,

Kl = 0.048

Here,

m = 0.25

(Ans.)

Principles of Mass Transfer Operations − I (Vol. − I)

4.21

1 0.0144

∴

=

Interphase Mass Transfer 1 1 + 0.25 × Kg 0.048

k mole A (hour · m2) (atm.) (c) Overall mass transfer coefficient in gas phase, KG is given by 1 m 1 KG = Kg + KL Kg = 0.082

∴

1 KG

∴

(Ans.)

1 0.25 = 0.082 + 0.0144

k mole A (Ans.) (hour · m2) (atm.) (3) In a typical chemical process, component A is desorped from an aqueous solution into an air stream in a mass transfer tower at a certain operating temperature and pressure. At a particular point in the tower, analysis report shows that : PA, G = 12 mm Hg; KG = 0.0338

∴

CA, L = 4 k mole/m3 K mole A The overall mass transfer coefficient, KG = 0.269 2 (m .hr.atm.) If Henry's law is applicable to this system and if 56% of the total mass transfer resistance is encountered in gas film. Calculate : (a) Gas film coefficient, Kg. (b) Liquid film coefficient, Kl. (c)

Molar flux of component A, NA. atm mole A/m3 solution 12 = 12 mm Hg = 760 atm = 0.01578 atm.

Data : Henry's law constant = 7.5 × 10–3 Sol. : Data given :

PA, G

CA, L = 4 k mole/m3 KG = 0.269 k mole A/[(hr) (m2) (atm.)] (a) Since 56% of the total mass transfer resistance in gas film. So, for liquid film the resistance is 44%. We know that, ∆ CA‚ liquid film Resistance in liquid phase = Overall resistance ∆ CA‚ total

∴

∴

56 100

=

1 Kg 1 0.269

Kg = 0.469

(k mole A) (hr. m2) (atm.)

Henry's law constant is given as, m = 7.5 × 10–3

atm. (mole/m3 solution)

Principles of Mass Transfer Operations − I (Vol. − I)

4.22 1 KG

So,

1 0.269

1 m = K + K g l 1 7.5 × 10–3 = 0.469 = Kl

Kl = 4.85

∴ (c)

Interphase Mass Transfer

k mol A (hr. m ) (mole A/m3) 2

*

(

(Ans.)

)

NA = KG pA‚G – pA

Molar flux NA :

* pA

where,

NA

∴

NA

= Partial pressure of component A = m × CA, L = 7.5 × 10–3 × 4 = 0.03 k mole/m3 = 0.269 (0.01578 – 0003) k mole = 3.825 × 10–3 (hour · m2)

(Ans.)

(4) Ammonia is absorbed by water in a weffed-wall column being operated at 20 oC and 1 std. atm. The overall gas coefficient is 1 k mole NH3/[(m2) (std. atm.)]. At one point in the column, the gas contains 10 mole% ammonia and the liquid phase contains 0.155 mole ammonia per m3 of solution. 96% of the total resistance is in the gas phase. Assume Henry's law constant at 293 K = 4.247 × 10–3 std.atm./(mole NH3/m3 solution). Determine the interfacial film coefficient and the interfacial compositions. Sol. : (i) Calculation of gas phase coefficient Kg : It is given that 96% of the total resistance is in the gas phase.

∴

Resistance in the gas phase = Overall resistance

1 Kg

1 = 96% KG

1 Kg ∴

or,

1 KG KG Kg = 0.96

= 0.96 k mole NH3 1 = 0.96 = 1.0416 (hr.) (m2) (std. atm.)

(ii) Calculation of liquid phase coefficient, Kl : We have,

1 KG

1 m = K + K g l 1 4.247 × 10–3 = 1.0416 + Kl

∴

1 1

Solving for Kl, we get,

Kl = 0.1063

k mole NH3 (mole NH3) (hr) (m2) 3 (m solution)

Principles of Mass Transfer Operations − I (Vol. − I)

4.23

Interphase Mass Transfer

*

(iii) Partial pressure, pA in equilibrium with the liquid : At a certain point in the column, the bulk concentration in the gas phase, 10 PA, G = YA · P = 100 (1) = 0.1 std. atm.   CA, L = 0.155 mole NH3/m3 solution *

Since Henry's law is applicable to this system. So pA in equilibrium with the bulk liquid concentration is, mole NH3 std. atm. * pA = H · CA, L = 4.247 × 10–3 × 0.155 3 m solution  mole NH3 

m3 solution   *

pA (iv)

(v)

(vi)

Molar flux, NA :

= 0.658 × 10–3 std. atm.

NA = KG

(p

A‚ G

–

(Ans.)

* pA

)

= 1 (0.1 – 0.658 × 10–3) k mole NH3 = 0.0993 (hr. m2) Partial pressure at the interface, pA, i : NA = Kg (PA, G – PA, i) 0.0993 = 1.0416 (0.1 – PA, i) ⇒ PA, i = 4.66 × 10–3 std. atm. Concentration in the liquid phase at the interface : pA, i = m · CA, i

(Ans.)

(Ans.)

4.66 × 10–3 = 4.247 × 10–3 CA, i mole NH3 (Ans.) ∴ CA, i = 1.097 m3 (5) In an experimental study of the absorption of NH3 by water from an air-ammonia mixture in a wetted-wall tower, the value of overall gas mass transfer coefficient, KG was found to be : 93.25 × 10–2 k mole NH3/[(hr) (m2) (std. atm.)] The operating pressure and temperature of the tower were 2 std. atm and 288 K respectively. For dilute solutions of ammonia in water at 288 K, the equilibrium partial pressure is given by : pA, i = 4 (CA, i) where,

pA, i = equilibrium partial pressure of ammonia, std. atm. CA, i = concentration of ammonia in water, k mole NH3/m3 solution

At the top of the tower, the outlet gas contained ammonia 1% by volume and inlet liquid, which this contacted with pure water. Assuming 85% of the total resistance to mass transfer was contributed by the gas phase, calculate : (i) the gas film coefficient, Kg.

Principles of Mass Transfer Operations − I (Vol. − I)

4.24

Interphase Mass Transfer

(ii)

the liquid film coefficient, Kl.

(iii)

the overall liquid mass transfer coefficient, KL.

(iv)

the interfacial concentrations pA, i and CA, i

Sol. : Given data : Since, So,

KG = 93.25 × 10–2 K mole NH3/[(hr) (m2) (std. atm.)] P = 2 std. atm. volume % = mole %, 1 pA, G = yA. P = 100 × 2 = 0.02 std. atm.

(i) Gas film coefficient, Kg : Since 85% of the total resistance to mass-transfer is in the gas-phase. So, 1 K Resistance in the gas phase g = overall resistance 1 = 85% KG ∴ ∴

KG Kg

= 0.85

Kg =

K mole NH3 93.25 × 10–2 = 1.097 0.85 (hr) (m2) (std. atm) (Ans.)

(ii) Liquid film coefficient, Kl : We have, 1 1 m KG = Kg + Kl Here m = 4 std.atm/(k mole NH3/m3 solution) ∴ ∴

1 1 4 = 1.097 + K 93.25 × 10–2 l Kl = 24.87

k mole NH3 2

(hr) (m ) (k mole NH3/m3 solution)

(Ans.) (iii) Overall liquid mass-transfer coefficient, KL : Since 85% of the total resistance lies in the gas phase, so the liquid phase shares rest 15% of the total resistance, therefore, 1 Kl = 0.15 1 KL ∴

KL Kl

∴

KL = 0.15 (24.87)

= 0.15

= 3.73

k mole NH3 2

(hr.) (m ) (k mole/m3 solution) (Ans.)

Principles of Mass Transfer Operations − I (Vol. − I)

4.25

Interphase Mass Transfer

(iv) Molar flux, NA : We have, *

NA = KL (CA – CA, L) from equilibrium relation, *

PA, G = m · CA , we get k mole NH3 0.02 –3 4 = 5 × 10 m3 solution Since the liquid in contact with the gas = 0 having 1 volume % NH was pure water.   3

*

=

CA CA, L

NA = 3.73 (5 × 10–3 – 0)

∴

= 18.65 × 10–3

k mole NH3

(hr) (m2) (v) Concentration at the interface : For the continuity of mass flow, *

NA = KL (CA – CA, L ) = Kl (CA, i – CA, L) 18.65 × 10–3 = 24.87 (CA, i – 0)

∴ Solving for CA, i we get,

CA, i = 7.5 × 10–4 k mole NH3/m3 solution

and

PA, i = m CA, i = 4 (7.5 × 10–4) = 3 × 10–3 std. atm.

(Ans.)

(6) The following data were obtained from a wetted wall column employing a constant liquid flow rate : Molar gas flow rate, Gm (k mole/sec)

0.01

0.02

0.04

0.06

0.08

0.10

Overall mass transfer coefficient, KG

50.8

67.2

84.0

91.7

93.5

100

2

2

6

(k mole/m sec. (kN/m ) × 10 ) The relationship between the equilibrium vapour pressure PA (kN/m2) and the molar 1

concentration in the liquid phase CA (K mole/m3) is given by 1

PA = 20 CA. For a gas flow rate of 0.05

k mole sec . Calculate overall and individual

mass transfer coefficients. 1 KG

Use the relation :

1 m = K + K g l

where Kg and Kl are individual mass transfer coefficients for gas and liquid phase respectively. KG = Overall mass transfer coefficient for gas phase

kN2  m 

m = Henry's law constant of value 20 k mole    m3  0.8

It is given that : 0.8 Kg ∝ G m , where Gm = gas velocity.

Principles of Mass Transfer Operations − I (Vol. − I)

4.26

Interphase Mass Transfer

PA = 20 CA

Sol. : Given data :

kN2  m 

m = 20 k mole  3   m  1 KG

1 m = kg + K l

… (1)

0.8

Kg ∝ G m

0.8

∴

Kg =

Thus equation (1) becomes,

1 Ka

a · Gm 0.8 , where a = constant 0.8

=

a·

m + K l

0.8 Gm

y = mx + C 0.8 m = slope = a

where,

m C = Intercept = K

l

1 Plot a graph of K

G

1/KG 0.8

1/G m

v/s

1 0.8

Gm

0.0196

0.0149

0.0119

0.0109

0.0106

0.010

39.81

22.86

13.13

9.494

7.542

6.30

0.020 1 KG

–10 0.8 Slope = m = — a = 2.831 ×10

0.015

}

0.01 0.005 0

0

–9 m Intercept = — kl = 8.3 ×10

10

20

1

30

40

0.8 Gm

1 Fig. 4.19 : A Graph of K

G

∴ ∴ from graph,

0.8 a 0.8 a= m

v/s

1 0.8

Gm

in Ex. (6)

= m =

0.8 = 2.82 × 109 2.831 × 10–10

Principles of Mass Transfer Operations − I (Vol. − I)

4.27

m C=K

l

Interphase Mass Transfer

= 8.3 × 10–9

… (2)

0.8

So,

Kg

a · Gm = 0.8 =

2.82 × 109 (0.05)0.8 0.8

Kg = 3.21 × 108 k mole/m2. sec 1 KG

and

(Ans.)

1 m = k +K g l =

1 3.21 × 10

8

+ 8.3 × 10

–9

KG = 8.75 × 107 k mole/m2 sec.

(Ans.)

(7) Show that for the absorption of a soluble gas, from a mixture with a non-soluble gas, 1 H 1 into a liquid : K G = K G + kL where the symbols have their usual meaning. Hence indicate the conditions under which either the gas or liquid films can become the controlling factor in the overall rate. For the absorption of a certain gas from an air stream into water KG = 10–3 m/s and 15% of the total resistance lies in the liquid phase. Estimate kG and kL. T = 300 K Molecular weight of gas = 36 H = 0.2142, where pressure is expressed as kPa and concentration in the liquid phase as kg/m3. Solution : The principles of mass transfer define the following : In the gas phase, 1 KG

=

PG – P* PG – PGi PGi – P* PG – PGi PGi – P* CLi – CL = + N = N +C –C NA NA NA A A Li L

In the gas phase,

1 kG

=

PG – PGi NA

In the liquid phase,

1 kL

=

CLi – CL NA

By substitution,

1 KG

1 1 = k +Hk G L

If H is low or kL high KG approximates to kG. Similarly, when kG is high, KG approximates to kL.

Principles of Mass Transfer Operations − I (Vol. − I) For the given case :

4.28

1 10–3

Interphase Mass Transfer

1 H = k + k = 1000 s/m G L

and we know that 85% of this resistance is in the gas phase. 1 ∴ k = 850 s/m G

⇒ kG = 1.176 × 10–3 m/s

(Ans.)

H ∴ k = 150 s/m L

H 0.2142 ⇒ kL = 150 = 150 = 1.428 × 10–3 Ns/kg

(Ans.)

(8) Given that the solvent liquid to gas ratio is reduced, explain the way this would affect : (a) The slope of the operating line. (b) The number of theoretical plates required. Solution : (a) The operating line in an absorber problem has a gradient of L/G. With increased gas flow G, the gradient will decrease. (b) As the gap between the operating line and the equilibrium curve narrows, the steps between the two become shallower but greater in number, i.e. the number of theoretical stages increases. (9) Define the term mole fraction and mole ratio and derive a relationship between the two. Solution : Denote mole fractions with a lower case "x" and mole ratios with a capital "X" : moles of solute x = moles of mixture

moles of solute = moles of solvent + moles of solute

moles of solute X = moles of solvent 1 ∴ x

=

moles of solvent + moles of solute moles of solvent moles of solute 1 = moles of solute + moles of solute = X + 1 moles of solute

∴

1 x

∴

X x = X+1

=

1+X X

(10) Interfacial Gas Concentration : Consider a system in which component A is being transferred from a gas phase to a liquid phase. The equilibrium relation is given by yA = 0.75xA where yA and xA are mole fraction of A in gas and liquid phase respectively. At one point in the equivalent, the gas contains 10 mole % A and liquid 2 mole % A. Gas film mass transfer coefficient ky at this point is 10 kmol/(hr.m2.∆ ∆yA) and 60% of the resistance is in the gas film. Calculate : (a) The overall mass transfer coefficient in kmol/(hr.m2.∆ ∆yA). (b) Mass flux of A in kmol/(hr.m2). (c) The interfacial gas concentration of A in mole fraction.

Principles of Mass Transfer Operations − I (Vol. − I)

4.29

Interphase Mass Transfer

* Sol. : Equilibrium composition of gas yA corresponding to xA = 0.02 (i.e. 2 mole %). * yA = 0.75 xA × 0.02 = 0.015 Since individual gas film resistance 1/ky is 60% of overall resistance 1/Ky. 1 0.6 K

y

1 = k y

Ky = 0.6 × 10 = 6 kmol/(hr.m2.∆yA) Mass flux in terms of overall mass transfer coefficient : * Mass flux = Ky (yA − yA)

… (1)

= 6 (0.1 − 0.015) = 0.51 kmol/(hr.m2) Mass flux in terms of individual mass transfer coefficient : Mass flux = ky (yA − yAi)

… (2)

where, yAi = Interfacial gas concentration. Substituting for mass flux, ky and yA in equation (2), 10 (0.1 − yAi) = 0.51 yAi = 0.049 (11) Consider a wetted-wall column 117 cm long with an inner diameter of 2.67 cm. In an experiment, water and air streams are allowed to flow co-currently. Under the following experimental conditions, the rate of water evaporation was measured to be 0.234 kg/h. Estimate the average mass transfer coefficient. Data : Top : T = 31.1 C; pw = 2.5 mmHg; pvap = 33.9 mmHg Bottom : T= 26.1C; pw = 19.3 mmHg; pvap = 25.4 mmHg Solution : Consider a wetted-wall column 117 cm long with an inner diameter of 2.67 cm. In an experiment, both water and air streams are allowed to flow concurrently (downwards). Under the following experimental conditions, the rate of evaporation was measured to be 0.234 kg/h. Given :

o

T(top) = 31.1 C o

T(bottom) = 26.1 C pw (inlet) = 2.5 mm Hg pvap = 33.9 mm Hg pw (out) = 19.3 mm Hg pvap = 25.4 mm Hg

Principles of Mass Transfer Operations − I (Vol. − I)

4.30

Interphase Mass Transfer

We have to estimate the average mass transfer coefficient (kc) Air (Pw = 2.5mm Hg) o 31.1 C

0.0267 m

1-17 m

o

26.1 C PW0 = 19.3mm Hg

Fig. 4.20 : Wetted wall column in Ex. (10) –

·

m w = kc A (∆C )L A = πDL → area for mass transfer = π × 0.0287 × 1.17 = 0.0981 m2 ·

mw =

 ∆CL  =  

∆C1 (top) (of water)

0.234 –6 = 3.61 × 10 kmol/s 18 × 3600 (∆C)1 – (∆C)2 ∆C1 ln   ∆C2 

101325 (33.9 – 2.5) ×  760  ∆pw1   = RT = 8314 × (273.15 + 31.1) 1 = 0.00165

∆PAir (∆C2) = RT 2 ∴

( ∆C )

L

kmol m3

101325 760 kmol = = 0.00033 8314 × (273.15 + 26.1) m3 (25.4 – 19.3) ×

=

0.00165 – 0.00033 kmol = 0.00082 0.00165 m3 ln 0.00033   –6

∴

kC =

3.61 × 10 kmol 1 m3 . 2 . kmol s 0.0981 × 0.00082 m

m m kC = 0.045 s or 162 h

(Ans.)

Principles of Mass Transfer Operations − I (Vol. − I)

4.31

Interphase Mass Transfer

(12) A hailstone falling freely through the atmosphere grows by transfer of water vapour from the surrounding air saturated with water vapour. Estimate the mass transfer coefficient (kc) for a 3-cm diameter hailstone falling at a Reynolds number of 4 × 104. Assume air temperature to be – 10C. Use the following mass transfer coefficient correlation for this type of phenomenon : [Sh Sc

–0.33

= 0.51 Re

0.5

+ 0.02235 Re

0.78

].

o

Solution : We need DAB at – 10 C o

From data : DAB for air-water mixture @ 25.9 C –4

∴

D2 D1

∴

D2

= 0.258 × 10 m2/s T2 1.5 = T   1 273.15 – 10 3/2 –4 2 = 273.15 + 25.9   0.258 × 10 m /s = 0.213 × 10

–4

m2/s –6

∴

12.5 × 10 ν o = Schmidt No. at (– 10 C) for air = D –4 = 0.585 AB 0.213 × 10 kC D –0.33 4 0.5 4 0.78 = 0.51 × (4 × 10 ) + 0.02235 (4 × 10 ) DAB (Sc)

(Ans.) ∴ kC = 0.112 m/s or 404 m/h. (13) Air flows through a cylindrical tube made of naphthalene at a velocity of 5 m/s. The tube diameter is 0.1 m, and the air temperature is 20C. Estimate the mass transfer –6

coefficient. Given : Dnaph-air = 4.24 ×10 m2/s. pM 101325 × 28.9 Solution : For air : ρ = RT = = 1.2 kg/m3 8314 × (293.15) µ = 1.8 × 10

–5

pa.s –5

1.8 × 10 µ = Sc = –6 = 3.54 ρ DN–A 1.2 × 4.24 × 10 –

Re =

ρDu 1.2 × 0.1 × 5 4 µ = 1.8 × 10–5 = 3.33 × 10 0.83

Use : ∴

1/3

Sc (Linton – Sherwood correlation) Sh = 0.023 Re 0.83 1/3 DN-A 4 kC = 0.1 × 0.023 (3.33 × 10 ) (3.54) = 0.00843 m/s

Try Colburn jD factor jD = St. Sc Sh =

2/3

Re Sc Sc

2/3

f –0.20 = 2 = 0.023 Re

. 0.023 Re

= 0.023 (Re) ∴

–3

0.8

(Sc)

kc = 6.2 × 10 m/s = 0.0062 m/s

–0.20

1/3

Principles of Mass Transfer Operations − I (Vol. − I)

4.32

Interphase Mass Transfer

(14) In the dilute concentration region, equilibrium data for SO2 distributed between air and water can be approximated by [PA = 25 xA], where the partial pressure of SO2 (A) in the vapour phase is in atmospheres. For an absorption column operating at 10 atm, the bulk vapour and liquid concentrations at one point in the column are : yA = 0.01 and xA = 0.0. The mass transfer coefficients for this process are : kx = 10 kmol/(m2h) and ky = 8 kmol/(m2h). (a) Find the overall mass transfer coefficient, Kx. (b) Determine the interfacial compositions, xAi and yAi. (c) Calculate the molar flux NA. Solution : (a) pA = 25 xA pA 25 yA = P = 10 atm. xA yA = 2.5 xA (a straight line) m' = m" = slope of equilibrium line 1 1 1 Kx = m" ky + kx m" = 2.5 (slope of equilibrium curve, see Fig. 4.21) 1 1 kmol 1 ∴ Kx = 2.5 × 8 + 10 ⇒ Kx = 6.67 (m2.h) 1 1 m' Similarly, Ky = ky + kx 2.5 kmol 1 (Ans.) = 8 + 10 ⇒ Ky = 2.67 (m2·h) yA P 0.01

Equlibrium curve Slope = 2.5

D

M

From Part (b)

0.005

O X 0.005 0.01 C Fig. 4.21 : Equilibrium curve in Ex. (13)

(b) or, ∴

kx yAi – yAG yAi – 0.01 ky = slope = xAi – xAL = xAi – 0 yAi – 0.01 2.5 xAi – 0.01 10 – 8 = = xAi xAL 0.01 xAi = 2.5 + 1.25 = 0.00267 yAi = 2.5 xAi = 0.00669

(Ans.)

Principles of Mass Transfer Operations − I (Vol. − I)

4.33

Interphase Mass Transfer

Equation for line PM yA = – 1.25 XA + 0.01 NA = kx (xAi – xAL) = 10 × (0.00267 – 0) kmol = 0.0267 2 m .h NA = ky (yAG – yAi) = 8 (0.01 – 0.00667) kmol = 0.0267 2 m .h

(c)

or

(

*

NA = Kx XA – XAL

or

)

yAG 0.01 = 2.5 = 2.5 = 0.004 Kx = 6.67 ∴ NA = 6.67 (0.004 – 0.0) kmol (Ans.) NA = 0.0267 m2.h (15) In an apparatus for studying the absorption of SO2 by water, the overall mass transfer coefficient Ky was found to be 0.3 kmol/(m2h). Assuming that 40% of the resistance to mass transfer is in the gas phase, calculate the overall mass transfer coefficient based on the liquid concentration. The system temperature was 30°°C and the pressure was maintained at 1 atm. The equilibrium data for SO2H2O system at 30°°C are : *

XA

(kg SO2)/(100 kg H2O) 0.1 0.3 0.5 0.7 1.0 Solution : Need xA vs. yA data

PSO2, atm 0.0062 0.0259 0.0474 0.0684 0.1039

kg SO2/64.06 xA = kg SO /64.06 + kg H O/18 2 2 PSO 2 yA = p We get : yA

xA 2.81 × 10

–4

0.0062

8.46 × 10

–4

0.0259

14.1 × 10

–4

0.0474

19.8 × 10

–4

0.0684

28.2 × 10

–4

0.1039

Principles of Mass Transfer Operations − I (Vol. − I)

4.34

Interphase Mass Transfer

With the straight line equilibrium relationship, m' = m" the tall one data essentially a straight line

Fig. 4.22 : Operating line in Ex. (14)

m' = m" = Also

0.1039 – 0.0062 28.2 × 10

–4

– 2.81 × 10

–4

= 38.5

 1  = 0.4  1  ky Ky 1 m' = 0.4 k + k   y x kmol (m2.h)

Since

Ky = 0.3

∴

ky = 0.75

or

0.6 0.4 m' = ky kx 1 kx

1 Kx

0.6 = 0.4 × 0.75 × m'

2 m'

1 1 = m" k + k y x =

1 2 + 38.5 38.5 ∞ 0.75

= 11.6 ∴

1 ⇒ k x

0.6 = 0.4 m' k y =

Now

kmol (m2.h)

Kx = 11.6

kmol of SO2 m2.hr kmol (of SO2) (m2.h)

(Ans.)

(16) At one point in an absorption column, the bulk compositions were found to be xA = 0.0 and yA = 0.08. The corresponding interfacial compositions were estimated to be xAi = 0.025 and yAi = 0.04. If the overall mass transfer coefficient for the liquid phase (Kx) is 50 kmol/(m2h), determine the percentage resistance to mass transfer for the gas phase. Assume that the equilibrium relationship for the gas and liquid phases can be described by Henry's law.

Principles of Mass Transfer Operations − I (Vol. − I)

4.35

Interphase Mass Transfer

Solution : Henry's law : ∴ Equilibrium relationship is yA = HxA, a straight line yA varies from 0 to 0.08 xA varies from 0 to 0.025 Slope of line PM (See Fig. 4.23) kx 0.04 – 0.08 = – k = 0.025 – 0 y ∴

kx ky

= 1.6 → kx = 1.6 ky

yA 0.1 0.08

P

D

k Slope = – —x ky

0.06 0.04

Equilibrium

M

0.02 0

C

0.01

0.02

0.03

xA

Fig. 4.23 : Equilibrium curve in Ex. (15)

For liquid phase : 1 Kx

Slope

1 1 = m" k + k y x m" = slope of line MD = m' (straight line) 0.04 – 0 m' = m" = 0.025 – 0 = 1.6

∴

1 50

∴

ky = 62.5 kmol/(m2.h)

where,

1 1 = 1.6 k + 1.6 k y y

kx = 1.6 × 62.5 = 100 kmol/(m2.h) Both resistances are equal. % resistances for the gas phase = 50% Also

1 Ky

1 m' = k + k y x 1 1.6 = 62.5 + 100

∴

Ky = 31.25 kmol/m2.hr.

(Ans.)

Principles of Mass Transfer Operations − I (Vol. − I)

4.36

Interphase Mass Transfer o

(17) Water is used to absorb CO2 from a N2-CO2 mixture in a unit operating at 10 C and 2 MPa. The average gas - and liquid-side mass transfer coefficients are 4.0 and o

48.0 kmol/(m2.h), respectively. Henry's law for the CO2-water system at 10 C is : 5

[p = 1.05 × 10 x], where p is the partial pressure (in kPa) of CO2 in the gas phase and x is the aqueous-phase mole fraction of CO2. Water is assumed to be non-volatile, and the solubility of N2 in water is neglected. (a) Calculate the overall mass transfer coefficients, Kx and Ky. (b) At a certain location in the absorption unit, the mole fractions of CO2 in the liquid and gas phases were measured to be 0.0005 and 0.10, respectively. Estimate the interfacial mole fractions of CO2 at this location. (c) At the location in Part (b), obtain the mass rate of CO2 absorption for an interfacial area of 0.6 m2. o

Solution : Interphase mass transfer @ 10 C and 2 MPa. CO2 diffuses through gas phase into the liquid (water) phase. Let A + CO2, B = N2 or water A diffusing, B non-diffusing Henry's law for CO2 : 5

pA = H xA ⇒ pA = 1.05 × 10 xA 5 pA 1.05 × 10 yA = p = 3 xA t 2 × 10

yA = 52.5 xA

∴

yA P(xAL, yAG)

M(xAi, yAi)

xA

Fig. 4.24 : Graph of xA versus yA in Ex. (16)

ky = 4.0 kmol/(m2.h) kx = 48.0 kmol/(m2.h) (a)

Calculation of Kx and Ky : Here

m' = m" = 52.5 –1

Kx

1 1 = m" k + k  y x 

Principles of Mass Transfer Operations − I (Vol. − I)

4.37

Interphase Mass Transfer

–1

=

1  1 +  52.5 × 4.0 48

Kx = 39.1 kmol/m2.h

(Ans.)

–1

Ky

1 m' = k + k  x  y

–1

=

 1 + 52.5 4.0 48 

Ky = 0.744 kmol/m2.h (b) xAi and yAi when yAG = 0.10, xAL = 0.0005 kx 48.0 –k = – 4.0 ≡ slope of PM y and

(Ans.) yAi – yAG = x –x Ai AL

yAi = 52.5 xAi 52.5 xAi – 0.10 – 12.0 = x – 0.0005 Ai

– 12 xAi + 0.006 = 52.5 xAi – 0.10 yAi = xAi = 0.00164, (c) Absorption rate : mass rate = MCO × NA × a

0.0863

(Ans.)

2

= 44 × 0.6 × NA = 26.4 NA NA = ky (yAG – yAC) = kx (xAC – xAL) *

(

) = Kx (x*A – xAL)

= Ky yAG – yA *

yA

*

xA

= 52.5 × xAL = 52.5 × 0.0005 = 0.02625 yAG 0.10 = 52.5 = 52.5

= 1.905 × 10 Mass rate of CO2 absorption (kg/h)

–3

26.4 (4.0 (0.10 – 0.0863) = 1.45 kg/h 26.4 (4.8 (0.00164 – 0.0005) = 1.44 kg/h 26.4 (0.744 (0.10 – 0.02625) = 1.45 kg/h –3

26.4 (39.1 (1.905 × 10 – 0.0005) = 1.45 kg/h Rate of CO2 absorption = 1.45 kg/h (Ans.) (18) A wetted-wall column, with 20-cm diameter, is used for absorbing a solute 'A' from an 'A'+ air mixture at 100 kPa. The solvent vapour pressure and the solubility of air in the solvent are negligible. The equilibrium data are plotted. At a certain column location, the steady-state liquid- and gas-phase mole fractions of 'A' are 0.02 and 0.07, respectively, and the liquid- and gas-side mass transfer coefficients are estimated to be 70 and 50 kmol/(m2h), respectively. Calculate the following at this location.

Principles of Mass Transfer Operations − I (Vol. − I) (a)

4.38

Interphase Mass Transfer

The equilibrium mole fractions : yAi and xAi.

(b) The mass transfer coefficients : kG, Ky and Kx. (c)

The molar absorption rate of 'A' per unit column height.

Solution : Absorption of 'A' in a non-volatile solvent. Solubility of air in solvent is negligible. ∴ "A" is diffusing, but air (or solvent) is not ! Gas

Liquid

yAG

yAi

xAi

interface

Fig. 4.25 : Illustration of gas-liquid interface in Ex. (17)

(a)

xAi and yAi Given

kz = 70

kmol m2h

ky = 50

kmol m2h

yAG = 0.07, xAL = 0.02 Point P : xAL = 0.02, yAG = 0.07 (See Fig. 4.26) Slope Pm

kx 70 = – k = – 50 = – 1.4 y

From P, draw a line of slope = – 1.4 Locate M on the equilibrium curve Graphically : xAi = 0.05, yAi = 0.027 Find the equation of PM (y – 0.07) = – 1.4 (x – 0.02) y = – 1.4 x + 0.098 (y intercept) xAi = 0.05, yAi = 0.027

(Ans.)

Principles of Mass Transfer Operations − I (Vol. − I)

4.39

Interphase Mass Transfer

yA P(0.02, 0.07)

yAG (0.07)

D

yAi (0.028)

M

yA* (0.01)

C

xAL xAi xA* (0.02) (0.05) (0.083) Fig. 4.26 : Graphical calculations of interfacial concentrations

(b) Calculation of kG, Ky and Kx kG (∆pA) = ky (∆yA) → ∴ kG = kG = 0.5

ky (∆yA) 50 kmol = 100 pt(∆yA) kPa.m2

kmol kmol = 0.0005 2 kPa.m .h Pa.m2.h

∆pA = pt (∆yA) –1

Ky =

 1 + m' ky kx 

0.027 – 0.01 m' (slope of CM ) = 0.05 – 0.02 = 0.57 –1

∴

Ky

1 0.57 = 50 + 70   

= 35.5 kmol/m2.h –1

Kx

1 1 = m" k + k  y x 

m" =

yAG – yAi *

xA – xAi 0.07 – 0.027 = 0.083 – 0.05 = 1.3 –1

∴

kx

1 1 =  + 70 1.3 × 50  

= 33.7 kmol/m2.h Check :

(

*

)

ky (yAG – yAi) = Ky yAG – yA

 yA – y*A    xAi – xAL 

Principles of Mass Transfer Operations − I (Vol. − I) Ky = ky

∴

4.40

Interphase Mass Transfer

(yAG – yAi)

(yAG – y*A)

(0.07 – 0.027) = 50 × (0.07 – 0.01) = 35.8 kmol/m2.h

(*

)

∴

kx (xAi – xAL) = Kx xA – xAL 0.05 – 0.02 Kx = 70 × 0.083 – 0.02 kmol = 33.3 2 m .h

(c)

· A = Ky a yAG – yA m

(

*

)

(*

= Kx a xA – xAL

)

a = πdL

·A m L

∴

= 35.5 × (π × 0.2) × (0.07 – 0.01) kmol = 1.34 m.h

Check (with kx) :

·A m L

= 33.7 × π × 0.2 × (0.083 – 0.02)

kmol (Ans.) = 1.33 m.h ; OK (19) An oil is used for removing cyclohexane (C6H6) from an air stream using a countercurrent packed column. The steady-state operating conditions are 30°°C and 101.3 kPa. At these conditions, the oil has low vapour pressure and the solubility of air in the oil is negligible. The equilibrium relationship (Henry's law) is given by : PA = 16.2 xA, where PA (in kPa) is partial pressure of C6H6. Average liquid- and gas-side mass transfer coefficients are 5.0 and 100.0 kmol/(m2h), respectively. (a)

Estimate the average values of overall mass transfer coefficients, Kx and Ky.

(b) Somewhere in the middle of the column, the liquid and gas phase mole fractions of C6H6 are 0.02 and 0.005, respectively. Estimate the molar flux of C6H6 and the interfacial compositions at this location. (c) For the packed column, the inlet compositions are : yA = 0.01 and xA = 0.005, and the outlet compositions are : yA = 0.002 and xA = 0.04. Use the average values of mass transfer coefficient(s) to calculate the molar absorption rate of C6H6 when the total interfacial area is 50.0 m2. Solution : "A" (C6H6) diffuses into oil ("C"). Air ("B") does not diffuse. "C" is non-volatile pA = 16.2 xA

Principles of Mass Transfer Operations − I (Vol. − I) ∴

yA =

4.41

Interphase Mass Transfer

pA 16.2 pt = 101.3 xA

yA = 0.16 xA yA

0.012 0.008 P

0.004

0.01

0.02

0.03

0.04

Fig. 4.27 : Graphical calculations in Ex. (18)

(a) Calculation of Kx and Ky m' = m" = 0.16 –1

Ky =

–1

 1 + m' ky kx 

= 23.8

=

kmol m2.h –1

Kx

1 1 = m"k + k   y x = 3.8

 1 + 0.16 100.0 5.0 

–1

1 1 =  + 5.0 0.16 × 100.0 

kmol m2.h

(Ans.)

(b) xAL = 0.02, yAG = 0.005 NA = Ky

(yAG – y*A)

= 23.8 (0.005 – 0.16 × 0.02) = 0.043 kmol/m2h or

(*

NA = Kx xA – xAL

)

= 3.8 (0.03125 – 0.02) = 0.043 Equation of PM kx 5.0 slope = – k = – 100.0 = – 0.05 y (yA – y1) = m (xA – x1) → (yA – 0.005) = – 0.05 (xA – 0.02) ∴ yA = – 0.05 xA + 0.006 Point 'm' : Point of intersection with yA = 0.16 xA ∴ 0.16 xA = – 0.05 xA + 0.006

kmol m2.h

Principles of Mass Transfer Operations − I (Vol. − I)

4.42

Interphase Mass Transfer

xAi = 0.0286 yAi = 0.16 × 0.0286 = 0.00457

∴

kmol m2.h xAi = 0.0286 yAi = 0.00457 · m = NA × (area) = Kg  ∆yA

NA = 0.043

∴

(c)



 ∆yA   

lm

=

=

 

lm

× area

(yA2 – y*A2) – (yA1 – y*A1) yA2 – y*A2  ln yA1 – y*  A1  (0.002 – 0.16 × 0.005) – (0.01 – 0.16 × 0.04) 0.002 – 0.16 × 0.005 ln  0.01 – 0.16 × 0.04 

= 2.185 × 10

–3

yA2 = 0.002

xA2 = 0.005

2

1 yA1 = 0.01

xA1 = 0.04

Fig. 4.28 : Counter current packed bed

∴

· m

–3

= 23.8 (2.185 × 10 ) × 50.0 = 2.6 kmol of A/h

With arithmetic mean, ∆yA ∴

· m

= 2.4 × 10

–3

= 23.8 × 2.4 × 10 kmol = 2.86 h

–3

× 50.0

Check with kx : · m

= Kx  ∆xA 

 area 

Principles of Mass Transfer Operations − I (Vol. − I)

 ∆xA  =   =

4.43

Interphase Mass Transfer

(x*A2 – xA2) + (x*A1 – xA1) 2

0.002 – 0.005 + 0.01 + 0.04  0.16  0.16  2

= 0.015 · m

∴

= 3.8 (0.015) 50 = 2.85

kmol h ; OK

(Ans.)

(20) The following schematics illustrate possible concentration profiles near the gas (yb) liquid (xb) interface. For each curve, establish whether mass transfer can occur and if so, in which direction (gas → liquid/liquid → gas). Explain/justify your answer by describing a situation in which such a profile might exist.

Mole fraction

xb

xb yb

yb

Distance

(a)

xb

(b)

xb

xb yb

yb

yb (c)

(d)

(e)

Fig. 4.29 : Concentration profiles for gas-liquid interface

Solution : (a)

Solute will transfer from the gas to the liquid phase. The fact that the interfacial liquid mole fraction is higher than the gas concentration is no impediment. It is merely an indication of high gas solubility.

(b) There is no transfer from gas to liquid nor from liquid to gas. Such a profile (where the interfacial concentration is higher from both sides) only arises in the case where the solute is generated at the gas-liquid interface.

Principles of Mass Transfer Operations − I (Vol. − I) (c)

4.44

Interphase Mass Transfer

The solute will desorb from the liquid to the gas phase. Gas solubility is low because the interfacial concentrations are nearly identical.

(d) The flat liquid phase profile indicates that the liquid phase is well-mixed and shows no mass transfer resistance. Because the gas phase concentration diminishes in the negative direction (away from the interface), the transfer will be from liquid to gas. (e)

Transfer is from gas to liquid. As in (d), the liquid phase is well-mixed.

EXERCISE FOR PRACTICE (1) A gas mixture, A – Air is fed into an absorption tower where absorption of the component A in water is taking place at 288 K and 3 std. atm. pressure. Given :

Kl = 0.152 Kg = 1.22

k mole A (hr. m2) (mole A/m3)

k mole A hr. m2. atm.

The equilibrium partial pressure of the gas A over dilute solution of A in water is given PA, i = 0.25 CA, i

by where, PA, i is the atm.

CA, i is in mole A/m3 Determine the values of the following mass transfer coefficients : (i)

Ky,

(ii) KC for gas film,

(iii)

KG,

(iv) KL.

(Ans. :

(i) Ky = 3.66

k mole A , hr. m2

(ii) KC = 28.83 m/hr, (iii) KG = 0.405

k mole A , hr. m2. atm.

(iv) KL = 0.1014

k mole A (hr. m2) (mole A/m3)

(2) A component A is being transferred from the liquid phase to the gas phase in a mass transfer apparatus such that the following equilibrium relation : yA, i = 0.75 (xA, i) holds good.

Principles of Mass Transfer Operations − I (Vol. − I)

4.45

Interphase Mass Transfer

At one point in the apparatus, the liquid contains 90 mole% of A and the gas in contact with this liquid contains 40 mole % of A. The individual gas-film mass transfer coefficient at the above point is, Ky = 2

k mole (hr) (m2) (∆ yA)

and 72% of the total resistance to mass transfer lies in the gas phase. Determine : (i)

molar flux of A, NA.

(ii)

interfacial concentration of A in both phases.

(iii)

the overall mass transfer coefficient, Ky.

(Ans. :

(i) NA = –0.396 k mole A/(hr.m 2), (ii) yA, i = 0.202, xA, i = 0.269, (iii) Ky = 1.44

k mole A ) (hr) (m2) (∆YA)

(3) In the absorption of component A from an air stream into an aqueous stream, the bulk composition of the two adjacent streams were analyzed to be PA, G equals 0.1 atm and CA, L equals 0.004 g mole/cm3. The Henry's constant for this system is 16.5 atm. cm3/g mole. The overall gas-side mass transfer coefficient KOG was 7.45 × 10–6 g mole/ cm2.s. atm. If 57% of the total resistance to mass transfer is in the gas film, determine (a) the gas-side mass transfer coefficient, kG; (b) the concentration on the liquid side of the interface, CA, i (c) the liquid-side mass transfer coefficient, kL; (d) the mass flux of A. (4) The partial pressure of CO2 in air is 1.333 × 104 Pa and the total pressure in 1.133 × 105 Pa. The gas phase is in equilibrium with a water solution at 303 K. What is the value of xA of CO2 in equilibrium in the solution ? See appendix for the Henry's law constant. (Ans. : xA = 7.07 × 10–5 mole fraction CO2) (5) At 303 K the concentration of CO2 in water is 0.90 × 10–4 kg CO2/kg water. Using the Henry's law constant from Appendix, what partial pressure of CO2 must be kept in the gas to keep the CO2 from vapourising from aqueous solution ? (Ans. : pA = 6.93 × 103 Pa) (6) The solute A is being absorbed form 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

Principles of Mass Transfer Operations − I (Vol. − I)

4.46

Interphase Mass Transfer

tower, the bulk gas concentration yAG = 0.25 and the bulk liquid concentration is xAL = 0.05. The tower is operating at 298 K and 1.013 bar and the equilibrium data can be represented by yA =

0.4749 x 4/3

1 – 2.304 xA

The solute A diffuses through stagnant B in the gas phase and then through a non-diffusing liquid. Using correlations for dilute solutions in a wetted-wall tower, the –3 film mass transfer coefficient for A in the gas phase is predicted as ky = 1.465 × 10 kmolA/(s.m2) and for the liquid phase Calculate. (a)

is predicted as kx = 1.967 × 10

–3

kmol-A(s.m2).

Interface concentrations yAi And xAi and flux NA. Sketch graphically how to find the interface concentration. Obtain the actual numbers through numerical solution.

' (b) Overall mass transfer coefficients Ky and Ky and flux NA. (c)

' Overall mass transfer coefficient Kx and flux NA.

(Ans. :

–4

(a) xAi = 0.16904, yAi = 0.10231, NA = 2.633 × 10 kmol/(cm2.s) ' –3 –3 (b) Ky = 1.003 × 10 kmol/(m2.s), Ky = 1.169 × 10 kmol/(cm2s), –4

NA = 2.633 × 10 kmol/(m2.s) ' –4 –4 (c) Kx = 8.967 × 10 kmol/(m2.s), NA = 2.633 × 10 kmol/(m2.s) NOMENCLATURE Any consistent set of units may be used, except as noted. Symbols

Meaning

A

Substance A, the diffusing solute, Absorptions factor, dimensionless

e

2.7183

E

Phase E; total rate of flow of phase E, mole/s

Es

Rate of flow of non-diffusing solvent in phase E, mole/s

EME

Fractional Murphee stage efficiency of phase E

EMR

Fractional Murphee stage efficiency for phase R

f

Equilibrium-distribution function

FG

Gas-phase mass transfer coefficient, mole/m2.s

FL

Liquid-phase mass transfer coefficient, mole/m2. s

FOG

Overall gas mass transfer coefficient, mole/m2.s

Principles of Mass Transfer Operations − I (Vol. − I)

4.47

Symbols FOL

Interphase Mass Transfer Meaning

Overall liquid mass transfer coefficient, mole/m2.s

i

Generalized concentration in phase R

j

Generalized concentration in phase E

kx

Liquid mass transfer coefficient, mole/m2.s(mole fraction)

ky

Gas mass transfer coefficient, mole/ m2. s (mole fraction)

Kx

overall liquid mass transfer coefficient, mole/m2.s (mole fraction)

Ky

overall gas mass transfer coefficient, mole/m2.s (mole fraction)

m

Equilibrium curves slope, dimensionless

m', m"

slopes of chords of the equilibrium curve, dimensionless

n N Np

Stage n

R Rs

Phase R; total rate of flow of phase R, mole/s Rate of flow of non-diffusing solvent in phase R, mole/sec

S

stripping factor, dimensionless

x

Concentration of A in a liquid or in phase R, mole fraction

X

Concentration of A in phase R, mole A/mole non-A

y

Concentration of A in a gas or in phase E, mole fraction

Y

Concentration of A in phase E, mole A/mole non-A

Mass transfer flux, mole/m2.s Total number of ideal stages in a cascade

Subscripts : A

Substance A

e

Equilibrium

E

Phase E

G

Gas

i

Interface

L

Liquid

n

Stage n

O

Overall

R

Phase R

0

Entering stage 1

1, 2

Positions 1 and 2; stage 1 and 2

Subscripts : *

In equilibrium with bulk concentration in other phase.

✱✱✱

Principles of Mass Transfer Operations − I (Vol. − I)

4.48

Interphase Mass Transfer

REFERENCES 1.

A.H.P.Skelland, “Diffusional Mass Transfer”, Kriegar, Malbar FL, 1985.

2.

C.J. Geankoplis, "Transport Processes and Unit Operations", Fourth Edition, Prentice Hall, 2003

3.

A.S. Foust, L.A. Wenzel, C.W. Clump, L. Maus and L.B. Andersen, "Principles of Unit Operations", Second Edition, John Wiley and Sons, 1980.

4.

R.E. Treybal, "Mass-Transfer Operations", Third Edition, McGraw-Hill, 1981.

5.

W.L. McCabe, J.C. Smith and P. Harriot, "Unit Operations of Chemical Engineering", Fifth Edition, McGraw-Hill, 1993.

6.

J.D. Seader and E.J. Henley, "Separation Process Principles", John Wiley and Sons, 1998.

7.

C.J. Geankoplis, “ Mass Transport Phenomena”, Columbus, Chio, 1972

8.

R.H. Perry and D. Green, "Perry's Chemical Engineers' Handbook", Seventh Edition, McGraw-Hill, 1997.

9.

T.K.Shrewood and R.L. Pigford, ”Mass Transfer”, McGraw Hill, 1975.

10. A.L.Hines and R.N. Maddox, ”Mass Transfer : Fundamentals and Applications”, Prentice Hall, Inc., New Jersey, 1985 11. J.R. Welty, R.E. Wilson and C.E. Pikes, “Fundamentals of Momentum, Heat and Mass Transfer”, Wiley New York, 1980. 12. E.L. Cussler, “Diffusion : Mass Transfer in Fluid Systems”, Second Edition, Cambridge University Press, 1998.

,,,

5 CHAPTER

GAS ABSORPTION 5.1 5.2 5.3 5.4 5.5 5.6 5.7 5.8 5.9 5.10 5.11 5.12 5.13 5.14 5.15 5.16 5.17 5.18 5.19 5.20 5.21

Background Physical Vs. Chemical Absorption Method of Operation Henry's Law for Gas Absorption Minimum Liquid/Gas Ratio for Absorption Choice of Solvent for Gas Absorption HETP (Height Equivalent to a Theoretical Plate) Packing Height : The Method of Transfer Units Number of Theoretical Trays The Kremser-Brown-Souders (KBS) Equation Introduction Equipment for Absorption/Stripping Equilibrium Curve Interfacial Mass Transfer : Review Definition of Transfer Coefficients Height of Packed Tower Transfer Unit Evaluation of NTU's Integral Concept of HETP Analogy with Double Pipe Heat Exchanger Relationship among Hox, Hoy, Hx, Hy

5.22 5.23 5.24 5.25 5.26 5.27

Pressure Drop in Packed Beds Tower Diameter Absorption with Chemical Reaction General Procedure for Design of Packed Absorption Columns Design Considerations Typical Absorber Design Problem Solved Problems Exercise for Practice Nomenclature References

5.1 BACKGROUND Gas absorption (also known as scrubbing) is an operation in which a gas mixture is contacted with a liquid for the purpose of preferentially dissolving one or more components of the gas mixture and to provide a solution of them in the liquid. Therefore we can see that there is a mass transfer of the component of the gas from the gas phase to the liquid phase. The solute so transferred is said to be absorbed by the liquid. (5.1)

Principles of Mass Transfer Operations − I (Vol. − I)

5.2

Gas Absorption

In gas desorption (or stripping), the mass transfer is in the opposite direction, i.e. from the liquid phase to the gas phase. The principles for both systems are the same. We will focus on the analysis for gas absorption, for the simple case whereby only one component of the gas solute is being absorbed. The other components of the gas are assumed to be non-soluble in the liquid (i.e. the other gas components are inert components), and the liquid is non-volatile, which means that there is no transfer of molecules from the liquid to the gas phase. In addition, we assume that there is no chemical reaction in the system and that it is operating at isothermal condition. The process of gas absorption thus involves the diffusion of solute from the gas phase through a stagnant or non-diffusing liquid. Gas absorption and desorption (stripping) can often integrated. Gas absorption is a process in which a gas mixture is in placed in contact with a liquid for the purpose of preferential dissolving of one ore more components of the gas mixture. Examples : •

NH3 is removed from oven gas by water.

•

CO2, H2S are removed from natural gas using water solutions of Alkaline salts.

•

Benzene, toluene are removed from natural gas using hydrocarbon oil.

Absorption : Physics of the process

P, T

Absorption : Physics of process

Fig. 5.1 : Absorption : Physics of process

Imagine a system where we have a binary mixture of two components A and B, one (A) is volatile and the second one (B) is a non-volatile solvent. The mixture coexists with an equilibrium gas of almost pure volatile component, which has pressure P. So if we place a piston on the top of the system, exerting pressure P, the system will be in equilibrium. The concentration of the volatile component A in solvent B is called solubility of A in B at P, T. This characteristics can be measured for a range of binary systems under different P, T conditions. In general solubility decreases with temperature. it is natural to guess that if we heat the system up, the energy supplied in the system will be used to evaporate more of the volatile component A.

Principles of Mass Transfer Operations − I (Vol. − I)

5.3

Gas Absorption

Physics of the process : Multicomponent mixture :

P, T

Physics of process : Multicomponent mixture Fig. 5.2 : Physics of process : Multicomponent mixture

Now we deal with a jar where a number of components are dissolved in the solvent and also forms and equilibrium gas mixture. Each component can be characterized with it’s own solubility under these conditions (composition, P, T). In general these solubilities (and other properties of the systems) must be described using methods of multicomponent phase equilibria (or measured experimentally). In many cases the system can be described with Simplified relations. If ideal gas is in equilibrium with ideal mixture we can apply the following expression to find partial pressures of the components : *

Pi

= Pxi

If ideal gas is in equilibrium with non-ideal mixture, Henry’s law applies *

= mxi Absorption : General design considerations Pi

1. 2. 3.

4. 5. 6. 7. 8.

Entering gas composition flow rate, temperature and pressure Desired degree of separation Choice of solvent : • High gas solubility • Low volatility • Low corrosiveness • Low cost • Low viscosity • Non-toxic, non-explosive Minimum solvent flow rate -> actual solvent flow rate Number of equilibrium stages Heat/cooling requirements Type of absorber Geometrical parameters of the system/economy

Design objectives

Theromophysical Characteristics

Basic Design Elements

Technical Specifications

For example : Integration of absorber and stripper : In practice, absorbers and strippers are often interconnected with one stripper serving several absorbers. In general, operating pressure should be high and temperature low for an absorber, to minimize stage requirements and/or solvent flow rate and to lower the equipment volume required to accommodate the gas flow. Unfortunately, both compression and refrigeration of a

Principles of Mass Transfer Operations − I (Vol. − I)

5.4

Gas Absorption

gas are expensive. Therefore most absorbers are operated at feed gas pressure, which may be higher than ambient pressure. Operating temperature is kept low, but not to condense the feed gas. The Figure shows an amine treating system typical in refineries with one absorber (scrubber) and one regenerator (stripper). Overhead condenser

Acid gas

Sweet gas Lean amine

Outlet drum

Lean amine cooler Overhead accumulator

Lean amine charge pump

Absorber (scrubber)

Reflux water

Reflux pump Lean amine booster pump

Sour gas

Rich amine

Stripper (regenerator)

Reboiler Rich amine Inlet drum

Lean amine storage tank

Steam

Flash gas

Condensate Lean/rich amine heat exchanger

Rich amine flash drum

Lean amine

Carbon filter Flowsheet of Amine treating system Fig. 5.3 : Flowsheet of Amine treating system

In the absorber, H2S is removed from the incoming sour gas by contacting it with liquid amine (DEA or MEA) in a counter-current manner. The amine solution (known as lean amine) enters the column at the top. The gas leaves the absorber as sweet gas with the H2S removed. The amine solution that leaves the bottom of the column is rich in absorbed H2S (known as rich amine) and is feed to the stripper or regenerator. In the regenerator, the opposite process takes place whereby H2S is removed from rich amine by counter-current contact with steam. The amine stripped of H2S leaves the bottom of the regenerator is send back to the scrubber as lean amine for more absorption. The entire process takes place essentially in a closed loop. Note that in this case, the liquid entering the absorber is no longer pure (i.e. x2 is not 0) as it may contain traces of H2S in the entering liquid amine.

Principles of Mass Transfer Operations − I (Vol. − I)

5.5

Gas Absorption

5.2 PHYSICAL VS. CHEMICAL ABSORPTION There are two types of absorption processes : Physical absorption and chemical absorption, depending on whether there is any chemical reaction between the solute and the solvent (absorbent). When water and hydrocarbon oils are used as absorbents, no significant chemical reactions occur between the absorbent and the solute, and the process is commonly referred to as physical absorption. When aqueous sodium hydroxide (a strong base) is used as the absorbent to dissolve an acid gas, absorption is accompanied by a rapid and irreversible neutralization reaction in the liquid phase and the process is referred to as chemical absorption or reactive absorption. More complex examples of chemical absorption are processes for absorbing CO2 and H2S with aqueous solution of monoethanolamine (MEA), diethanolamine (DEA), diethyl-eneglycol (DEG) or triethyleneglycol (TEG), where a reversible chemical reaction takes place in the liquid phase. Chemical reactions can increase the rate of absorption, increase the absorption capacity of the solvent, increase selectivity to preferentially dissolve only certain components of the gas, and convert a hazardous chemical to a safe compound. 5.3 METHOD OF OPERATION Two methods of contacting the gas and liquid are possible : counter-current operation and co-current operation. We will focus principally on the counter-current gas absorption, as it was widely used in the industry. The main differences between the two configurations will be highlighted. Note that for counter-current operation, the gas enters Lin

Gout

Xin

yout

Lout

Gout

Xout

yout

Distance

Lout

Gin

Lin

Gin

Xout

yin

Xin

yin

Gas concentration of solute

Liquid concentration of solute Distance

Concentration

Concentration

Distance

Gas concentration of solute

Liquid concentration of solute Distance

Method of operation Fig. 5.4 : Method of operation for contacting gas and liquid

the column or tower from below as leaves at the top, while liquid enters from the top and flows in opposite direction and exits from the bottom. [Recall the similarity with heat exchanger operation].

Principles of Mass Transfer Operations − I (Vol. − I)

5.6

Gas Absorption

We will be concerned primarily with counter-current gas absorption. As shown in the Figure below, the gas flows upwards while the liquid flow downwards. Inside the column where there is vapor-liquid contact, mass transfer by absorption occurs, i.e. there is a transfer of solute(s) from the gas phase to the liquid phase. Gas out: Inert gas + unabsorbed solute A (trace amount)

Liquid in: Inert liquid + little or no solute A

Mass transfer of solute A from gas into liquid (solute A absorbed into inert liquid)

Solute content in liquidphase increases down the column

Solute content in gas phase decreases up the column

Liquid out: inert liquid + absorbed solute A

Gas in: inert gas + solute A

Flow of gas and liquid in counter-current gas absorption Fig. 5.5 : Flow of gas and liquid in counter-current gas absorption

(a) Counter-Current Gas Absorption : Refer to the Figure below for a dilute system : L2 x2

G2 y2

1 Bottom of column 2

2 Top of column

L x

G y

Envelope 1 1 G1 y1

L1 x1

Counter-current gas absorption Fig. 5.6 : Counter-current gas absorption

Principles of Mass Transfer Operations − I (Vol. − I)

5.7

Gas Absorption

Notations : In terms of mole fraction and total flow rates y : mole fraction of solute A in the gas phase x : mole fraction of solute A in the liquid phase 2

G : total molar flow rate of the gas stream (gas flux), kg-moles/m .s 2

L : total molar flow rate of the liquid stream, kg-moles/m .s 2

Gy and Lx are the molar flow rates of A in the gas and liquid respectively (kg-moles A/m .s) at any point inside the column. Inside the column, mass transfer takes place as the solute (component A) is absorbed by the liquid. The quantities of L and x (for the liquid side) and G and y (for the gas side) vary continuously : as we gradually move up the column, component A is continuously being transferred from the gas phase to the liquid phase. Thus, in going up the column, there is a decrease in the total gas flow rate, and a decrease in the concentration of A in the gas phase. At the same time, in going down the column, there is an increase in the total liquid flow rate, and an increase in the concentration of A in the liquid phase. Thus, G1 > G > G2

L1 > L > L2

y1 > y > y2

x1 > x > x2

… (5.1)

For dilute systems, the solute content is small relative to the non-soluble inerts and non-volatile liquid. Thus, we can assume : G1 = G2 = G = constant

… (5.2)

L1 = L2 = L = constant The relationship between these variables L, x, G and y is the operating line equation. The operating line equation is obtained by material balance around the column (as shown in Envelope 1 of the Figure above). At steady-state : Thus,

IN = OUT

… (5.3)

G y + L 1 x1 = L x + G1 y1

Using the dilute system assumptions, we simply the equation and obtain : G y = L x + G y1 – L x1

… (5.4)

Re-arranging : Gy1 – Lx1 L y = G x +     G 

… (5.5)

Since L and G are assumed to be approximately constant, the operating line is a straight line of the form y = mx + c, with the gradient of L / G, the liquid-to-gas ratio. The operating line connects the two end points - point 1 (x1, y1) that represents conditions at the bottom of the column, and point 2 (x2, y2) that represents conditions at the top of the column. For dilute solution, the equilibrium solubility line is also straight, as represented by Henry's Law, y = mx, where m is the Henry's Law constant which is also the gradient of the line.

Principles of Mass Transfer Operations − I (Vol. − I)

5.8

Gas Absorption

Operating line (bottom)

y1

y

y2

Any point P (x, y) -k /k x y

(top)

x2

x

Mole fraction a decreases up the column

Mole fraction solute A in vapour, y

When these two lines are plotted on mole fraction co-ordinates, we have the following Figure.

Equilibrium line (Henry's law)

x1

Mole fraction A increases down the column

Fig. 5.7 : Concept of operating and equilibrium line

Any point P (x, y) on the operating line represents gas-liquid contact for which the analysis can be carried out using the two-film theory covered in earlier section. The larger the distance between the operating line and equilibrium line, the larger the concentration difference for mass transfer, and thus, the easier the separation. Note : • •

•

Operating line for gas absorption lies above the equilibrium line. Also, in the analysis of gas absorption, we will need to know the minimum liquid rate that can be used for a given separation, i.e. to remove a specified amount of solute from the gas. This is known as the minimum liquid-to-gas ratio. The analysis is applicable to both tray and packed column.

(b) Co-Current Gas Absorption : This mode of operation is seldom used in practice. See the Figure below. The main points to note about this operation are as follow : L1 x1

G1 y1

L x

L2 x2

G y

G2 y2

Mole fraction solute in vapour, y

Operating line, slope = -L/G

Equilibrium line, y=mx

y1

y2 (xe, ye)

x1

x2

Mole fraction of solute a in liquid, x Co-current gas absorption

Fig. 5.8 : Co-current gas absorption

Principles of Mass Transfer Operations − I (Vol. − I)

5.9

Gas Absorption

The operating line has negative slope. There is no minimum liquid-to-gas ratio. To produce an exit liquid and gas streams at equilibrium (xe, ye) on the equilibrium curve, an infinitely tall column must be used. It is less efficient than counter-current operation. 5.4 HENRY'S LAW FOR GAS ABSORPTION When the gas mixture in equilibrium with an ideal liquid solution follows the ideal gas behaviour, we have - as seen previously - the Raoult's Law : p* = Pvp x … (5.6) [at this point, students should revise this concept previously covered in the topic on distillation]. When the solution is non-ideal, Raoult's Law cannot be applied. For non-ideal solution, we must use Henry's Law, which states that : … (5.7) p* = Hx or y* = p* / PT = m x where, y* = Equilibrium mole fraction in gas phase PT = Total system pressure m = Henry's Law Constant In many cases, the superscript '*' is dropped for convenience : pA = H xA

yA = m xA

… (5.8)

Henry's Law is often used to represent equilibrium solubility curves. As we can see, Henry's Law predicts a linear equilibrium relationship. Still, most equilibrium relationships are actually non-linear. Henry's Law is only applicable over a modest liquid concentration range, especially when the solution is dilute. Note : Several other variations of Henry's law are available, depending on applications, e.g. membrane separation. Care must be exercised in the correct unit for the constant. For example, H has the unit of pressure/mole fraction, while m is dimensionless. m is independent of total pressure, whereas H does not. Table below showed the values of Henry's Law constant as a function of temperature for several gases. T °C 0 10 20 30 40

Table 5.1 : Values of Henry’s Law Constant as a function of temperature Gases CO2 C2H6 C2H4 H2 H2S CH4 N2 CO He 0.0728 0.104 0.142 0.186 0.233

3.52 4.42 5.36 6.20 6.96

1.26 1.89 2.63 3.42 4.23

0.552 0.768 1.02 1.27 –

12.9 12.6 12.5 12.4 12.1

5.79 6.36 6.83 7.29 7.51

0.0268 0.0367 0.0483 0.0609 0.0745

2.24 2.97 3.76 4.49 5.20

5.29 6.68 8.04 9.24 10.4

O2 2.55 3.27 4.01 4.75 5.35

5.5 MINIMUM LIQUID/GAS RATIO FOR ABSORPTION The inlet gas has a solute mole fraction of y1. The solute mole fraction is reduced to y2 at the outlet. By material balance for the solute in the gas, the amount to be removed is G (y1 – y2). The least amount of liquid Lmin that can remove this amount of solute is the minimum liquid rate, often expressed in terms of a liquid-to-gas ratio, Lmin/G.

Principles of Mass Transfer Operations − I (Vol. − I)

5.10

Gas Absorption

Understanding the effect of reducing liquid rate requires an analysis of the operating line equation. This is shown in the Fig. 5.9. Mole fraction solute in vapour, y

Operating line at minimum liquid rate (bottom) E

y1

y2

F

Mole fraction solute in liquidr, x

(top)

D x2

Rotation

x1

Equilibrium line

M

x1, max

Effect of reducing liquid rate : Analysis of operating line equation

Fig. 5.9 : effect of reducing liquid rate : Analysis of operating line equation

The condition at the top of the column (point D) is known : x2 the mole fraction of entering liquid is known, and the mole fraction of gas leaving y2 is known. Hence point D is fixed. The mole fraction of gas entering y1 is known. The mole fraction of liquid leaving x1 obviously depends on the liquid rate used. For the same amount of solute to be removed, using a larger quantity of liquid will result in smaller value of x1, and vice versa. Hence, when the liquid rate is changed, the condition at the bottom of the column varies along the horizontal line through y1. Recall that the operating line has a gradient of L/G. By reducing the liquid rate, we are decreasing the slope of the operating line and increasing the exit concentration x1. Therefore the operating line rotates around point D as L is decreased, e.g. from line DE to DF. Notice that the operating line has moved closer to the equilibrium curve. When this happens, the driving force for mass transfer is smaller, i.e. the absorption process becomes more difficult. At point M, the operating line intersects the equilibrium line, and we have a condition of operation at zero driving force. At this point, we cannot reduce the liquid rate anymore. Hence, the liquid rate at this point of equilibrium is known as the minimum liquid rate, Lmin. At minimum liquid, the outlet liquid concentration is a maximum, x1(max). The minimum liquid rate results in infinite column height : infinite number of trays or packed height required for separation (at zero driving force). The minimum liquid rate, Lmin can be calculated from the gradient of the operating line :

Lmin =  y1 – y2   G  x1 (max) – x2

… (5.9)

Note : x1(max) can also be calculated using Henry's Law. The actual liquid rate to be used is specified as multiples of the minimum liquid rate. If the liquid rate for absorption is initially unknown, then one must calculates the minimum liquid rate first.

Principles of Mass Transfer Operations − I (Vol. − I)

5.11

Gas Absorption

5.6 CHOICE OF SOLVENT FOR GAS ABSORPTION If the principal purpose of the absorption operation is to produce a specific solution, as in the manufacture of hydrochloric acid, for example, the solvent is specified by the nature of the product, i.e. water is to be the solvent. If the principal purpose is to remove some components (e.g. impurities) from the gas, some choice is frequently possible. The factors to be considered are : Gas Solubility : The gas solubility should be high, thus increasing the rate of absorption and decreasing the quantity of solvent required. Generally solvent with a chemical nature similar to the solute to be absorbed will provide good solubility. A chemical reaction of the solvent with the solute will frequently result in very high gas solubility, but if the solvent is to be recovered for re-use, the reaction must be reversible. For example, H2S can be removed from gas mixtures using amine solutions since the gas is readily absorbed at low temperatures and easily stripped at high temperatures. Caustic soda absorbs H2S excellently but will not release it in a stripping operation. Volatility : The solvent should have a low vapor pressure to reduce loss of solvent in the gas leaving an absorption column. Corrosiveness : The materials of construction required for the equipment should not be unusual or expensive. Cost : The solvent should be inexpensive, so that losses are not costly, and should be readily available. Viscosity : Low viscosity is preferred for reasons of rapid absorption rates, improved flooding characteristics in packed column, low pressure drops on pumping, and good heat transfer characteristics. Others : The solvent should be non-toxic, non-flammable and chemically stable. Counter-Current Gas Stripping : Important points to note : •

Mass transfer from the liquid-phase to the gas-phase.

•

Sometimes gas stripping units go by the name of regenerators.

•

Analysis is similar to counter-current gas absorption.

•

Important difference : operating line lies below to equilibrium line.

Principles of Mass Transfer Operations − I (Vol. − I)

5.12

Gas Absorption

Mole fraction of solute in gas, y

See the Fig. 5.10.

y1

y2

(Top)

Equilibrium line, y = m x

P Operating line, slope = L/G (Bottom) x1

x2

Mole fraction solute in liauid, x Counter-current gas stripping

Fig. 5.10 : Counter-current gas stripping

5.7 HETP (HEIGHT EQUIVALENT TO A THEORETICAL PLATE) As we have noted, instead of a tray (plate) column, a packed column can be used for various unit operations such as continuous or batch distillation, or gas absorption. With a tray column, the vapours leaving an ideal plate will be richer in the more volatile component than the vapour entering the plate by one equilibrium "step". When packings are used instead of trays, the same enrichment of the vapor will occur over a certain height of packings, and this height is termed the height equivalent to a theoretical plate (HETP). As all sections of the packings are physically the same, it is assumed that one equilibrium (theoretical) plate is represented by a given height of packings. Thus the required height of packings for any desired separation is given by (HETP × No. of ideal trays required). HETP values are complex functions of temperature, pressure, composition, density, viscosity, diffusivity, pressure drop, vapour and/or liquid flow rates, packing characteristics, etc. Empirical correlations, though available to calculate the values of HETP, are restricted to limited applications. The main difficulty lies in the failure to account for the fundamentally different action of tray and packed columns. In industrial practice, the HETP concept is used to convert empirically the number of theoretical trays to packing height. Most data have been derived from small-scale operations and they do not provide a good guide to the values, which will be obtained, on full-scale plant. The Method of Transfer Units had largely replaced this method. 5.8 PACKING HEIGHT : THE METHOD OF TRANSFER UNITS A newer concept in the analysis of packed column centered on the method of transfer units. This method is more appropriate because the changes in compositions of the liquid and vapour phases occur differentially in a packed column rather than in stepwise fashion as in trayed column. In this method, height of packings required can be evaluated either based on the gas-phase or the liquid-phase. The packed height (z) is calculated using the following formula : Z = N×H

Principles of Mass Transfer Operations − I (Vol. − I)

5.13

Gas Absorption

where, N = number of transfer units (NTU) − Dimensionless H = height of transfer units (HTU) − Dimension of length The number of transfer units (NTU) required is a measure of the difficulty of the separation. A single transfer unit gives the change of composition of one of the phases equal to the average driving force producing the change. The NTU is similar to the number of theoretical trays required for trayed column. Hence, a larger number of transfer units will be required for a very high purity product. The height of a transfer unit (HTU) is a measure of the separation effectiveness of the particular packings for a particular separation process. As such, it incorporates the mass transfer coefficient that we have seen earlier. The more efficient the mass transfer (i.e. larger mass transfer coefficient), the smaller the value of HTU. The values of HTU can be estimated from empirical correlations or pilot plant tests, but the applications are rather restricted. The calculation of packing height follows the same nomenclature as before and this is shown in the Fig. 5.11. L2 x2

G2 y2 2 G y

1 Bottom of column 2 Top of column

L x

1 G1 y1

L1 x1

Calculation of packing height : Nomenclature

Fig. 5.11 : Calculation of packing height : Nomenclature

In this Section, we will focus on the applications of the equations rather than any derivation of them. Determination of the packed height can be based on either the gas-phase or the liquid-phase. For the gas-phase, we have :

z = NOG × HOG y1 – y2 NOG = (y – y*) LM

… (5.11)

Principles of Mass Transfer Operations − I (Vol. − I)

5.14

Gas Absorption

*

(y – y*)LM =

HOG =

*

(y1 – y1) – (y2 – y2)

… (5.12)

(y1 – y*1)  ln   (y – y*)   2  G

… (5.13)

*

KY a (1 – y)LM *

*

(1 – y)LM

(1 – y1) – (1 – y1) = (1 – y1) ln  * (1 – y1)

… (5.14)

and KY is the overall gas-phase mass transfer coefficient. "a" is the packing parameter that we had seen earlier (recall the topic on column pressure drop. Normally, packing manufacturers report their data with both KY and "a" combined as a single parameter. Since KY has a unit of mole/(area.time.driving force), and "a" has a unit of (area/volume), the combined parameter KY a will have the unit of mole/(volume.time.driving force), such as kg-mole/(m3.s.mole fraction). As seen earlier, other than mole fraction, driving force can be expressed in partial pressure (kPa, psi, mm-Hg), wt% etc. *

y1 is the mole fraction of solute in vapour that is in equilibrium with the liquid of mole *

fraction x1 and y2 is mole fraction of solute in vapour that is in equilibrium with the liquid of mole fraction x2. *

*

The values of y1 and y2 can be obtained from the equilibrium line as previously covered

Mole fraction solute a in vapour, y

(see section on Two-Film Theory). See Fig. 5.12. Operating line Equilibrium line

(bottom)

y1

P (x, y) y1* y2 (top) y2* x2

x2*

x1

Mole fraction solute in liquid, x

x1* *

*

Fig. 5.12 : Calculation of y1 and y2 from equilibrium line

Principles of Mass Transfer Operations − I (Vol. − I)

5.15

Gas Absorption

*

(y1 – y1 ) is the concentration difference driving force for mass transfer in the gas phase at *

point 1 (bottom of column) and (y2 – y2 ) is the concentration difference driving force for mass transfer in the gas phase at point 2 (top of column). [Point P (x, y) as shown is any point in the column. The concentration difference driving force for mass transfer in the gas phase at point P is (y – y*) as shown previously, this time no subscripts are shown]. Note : •

Both equilibrium line and operating line are straight lines under dilute conditions.

•

Alternatively, equilibrium values y1 and y2 can also be calculated using Henry's Law

*

*

(y = m x, where m is the gradient) which is used to represents the equilibrium relationship at dilute conditions. *

*

Thus, we have : y1 = m x1 ; y2 = m x2

•

Similarly for the liquid-phase we have : z = NOL × HOL x1 – x2 NOL = (x* – x) LM *

(x* – x)LM =

HOL =

… (5.15) *

(x1 – x1) – (x2 – x2)

… (5.16)

(x*1 – x1)  ln  * (x – x2)  2  L

… (5.17)

*

KX a (1 – x)LM *

*

(1 – x)LM

(1 – x1) – (1 – x1) = (1 – x1) ln  * (1 – x1)

… (5.18)

and KX is the overall liquid-phase mass transfer coefficient, and "a" is the packing parameter seen earlier. Again, normally both KX and "a" combined as a single parameter. *

Likewise, x1 is the mole fraction of solute in liquid that is in equilibrium with the vapour of *

mole fraction y1 and x2 is mole fraction of solute in liquid that is in equilibrium with the vapour *

*

of mole fraction y2. Refer to Figure kp for finding values of x1 and x2 from the equilibrium line. *

*

Alternatively, x1 = y1 /m and x2 = y2 /m. *

(x1 – x1) is the concentration difference driving force for mass transfer in the liquid phase at *

point 1 (bottom of column) and (x2 – x2) is the concentration difference driving force for mass transfer in the liquid phase at point 2 (top of column).

Principles of Mass Transfer Operations − I (Vol. − I)

5.16

Gas Absorption

Using either gas-phase or liquid-phase formula should yield the same required packing height : z = N × H = NOG × HOG = NOL × HOL … (5.19) where, N = Number of transfer units (NTU) H = Height of transfer units (HTU) Comparison between number of theoretical trays, HETP and Method of Transfer Units : The Number of Transfer Units (NTU) and Height of Transfer Units (HTU) such as NOG, HOG should not be confused with the number of theoretical trays (N), and the height equivalent to theoretical plate (HETP) respectively. When the operating line and equilibrium line are straight and parallel : NTU = N; and HTU = HETP Otherwise, the NTU can be greater than or less than N as shown in the Figure below : y

y

y Operating line

Operating line

Equilibrium line

Operating line Equilibrium line

Equilibrium line x

x

NTU = N

NTU > N

x NTU < N

Fig. 5.13

When the operating line is straight but not parallel, we have the following relationships : ln (1/A) … (5.20) HETP = HOG (1 – A)/A   ln (1/A) NOG = N (1 – A)/A  

… (5.21)

L where, A = mG is the Absorption Factor. 5.9 NUMBER OF THEORETICAL TRAYS The number of theoretical trays can be determined graphically using a method similar to the McCabe-Thiele Method used in continuous distillation. Refer to the Figure below that shows the graphical solution. Note : We are following the consistent nomenclature of using subscript "1" to refer to conditions at the bottom of the packed column, and subscript "2" to refer to conditions at the top of the packed column. Note the slight difference in the way the triangles are drawn : Start from point 1 and work the way down towards point 2, and draw triangles between the operating line and equilibrium line. In the above example, 5 triangles are drawn. Therefore 5 theoretical trays are required for the separation. For the case of gas absorption, the last triangle represents a theoretical tray, not a reboiler.

Principles of Mass Transfer Operations − I (Vol. − I)

5.17

Gas out G2, y2 Column top 2

Gas Absorption For dilute systems: ~ ~ G1 = G2 = G = constant ~ ~ L1 = L2 = L = constant

Liquid in L2, x2

1

Tray y decreases up the column

y1 (a) yb xa

(b) yc (c)

xb

Operating line e

d

(d) c

(e)

yc

Equilibrium line

b yb GAs in

2

y2

a x increases down the column

G1, y1 Column bottom 1

Liquid out

xa

x2

xb

x1

L1, x1 Fig. 5.14 : Number of trays calculation through graphical solution (similar to McCabe-Thiele Method used in continuous distillation)

Analysis for the changes in gas phase and liquid phase compositions is similar to the distillation process : For example, consider tray (b) where the liquid concentration changed from xa at the inlet to xb at the outlet, and the gas composition changed from yc at the inlet to yb at the outlet. (yc – yb) showed the decreases in gas concentration as it passed through tray (b), and (xb – xa) showed the increase in liquid concentration as it passed through tray (b). The larger the triangle, the more effective the separation. 5.10 THE KREMSER-BROWN-SOUDERS (KBS) EQUATION For dilute system, recall that both the operating line and equilibrium line is straight, even on x-y co-ordinates. In this case, the number of theoretical stages required for a given separation can be calculated using the Kremser-Brown-Souders Equation as shown : • Absorption

N =

 y1 – mx2  log y – mx  (1 – 1/A) + (1/A) 

2

2



log (A)



… (5.22)

Principles of Mass Transfer Operations − I (Vol. − I)

5.18

Gas Absorption

where, A = L /mG is the absorption factor and is assumed constant. • Stripping x2 – (y1/m) log x – (y /m) (1 – A) + A N =



1

1





… (5.23)

log (1/A)

where, the stripping factor S = 1/A, is assumed constant. Note : m = Henry's Law constant = slope of the equilibrium line.

5.11 INTRODUCTION Gas absorption : Two component of a gas are separated by contact with a liquid (in which one component is preferentially soluble).

Stripping : Two components of a liquid are separated by contact with a gas. An example of gas absorption is the removal of ammonia from air by contact with liquid water. Ammonia is very soluble in water whereas air is only slightly soluble. Both gas absorption and stripping involve at least three components. Usually only one of these components crosses the phase boundary. In the example of ammonia and air, ammonia is the component whose molar flow rate changes by the largest percentage of the inlet value. Although some air will also dissolve in water, and some water will evaporate into the air, the molar flow rates of air and water change by negligible fractions : their flows can usually be considered constant. These are the two main differences between them and distillation : (i) at least three component. (ii) By contrast, in distillation, all of the components are present in both phases.

Gas out

Liquid in Liquid distributor Packing restrainer

Shell

Random packing

Liquid re-distributor

5.12 EQUIPMENT FOR ABSORPTION/ STRIPPING Although liquid and gas streams for absorption or stripping could be contacted using a tray column (like that used in distillation), tray columns are seldom used. The reason is that tray efficiencies are generally much lower for absorption and stripping that for distillation (perhaps only 5% instead of 50%). Because of this very low efficiency, very large numbers of trays are required - perhaps 100's or 1000's. Fabrication costs just become prohibitively expensive. Fortunately, a viable alternative exists : the packed tower.

Packing support Gas in Liquid out Fig. 5.15 : Packed Tower

Principles of Mass Transfer Operations − I (Vol. − I)

5.19

Gas Absorption

A packed tower is simply a tube or pipe, which is filled with some sort of "packing." The packing typically consists of particles around an inch in diameter. In commercial packed towers, the usual choices are particles with one of three different shapes.

(a) Rashing Ring

(b) Lessing Ring

(c) Partition (d) Berl saddle (e) Intalox Ring saddle Fig. 5.16 : Types of Packings

(f) Taller ette

(g) Poll Ring

Rasching ring (which is just a piece of pipe which has been cut into segments, whose length and diameter are about the same) 1 1 L ≈ D ≈ 2 to 1 2 inches • Berl saddle • Pall ring Although, in a pinch, almost anything you have lying around would do --Ping-Pong balls golf balls, etc. The purpose of the packing is to promote good contact between the liquid and vapour streams, which are being brought together to permit interfacial mass, transfer. The liquid stream is usually fed into the top of the tower while the vapour is fed into the bottom. Thus we have countercurrent flow of the two streams, which has the same advantages for mass transfer as it did for heat transfer. The packing promotes good contact between the phases by dividing the two feed streams into many parallel-interconnected paths. Ideally, you would like the liquid to flow downward as a thin film over the surface of the packing. This would give the maximum surface area of contact between the gas and liquid.

•

If you just pour the liquid from the end of the pipe onto the top of the packing in tower having' much larger diameter than the pipe, most of the packing will not even be wet. Only some of the channels will be carrying flow. This is called as Channeling-maldistribution of liquid flow. Liquid in

Dry Packing

Iriggated Packing

(a)

(b)

Fig. 5.17 : Liquid Distribution and Packing Irrigation (a) Inadequate, (b) Adequate

Principles of Mass Transfer Operations − I (Vol. − I)

5.20

Gas Absorption

So some sort of device to distribute the flow over the entire cross section of the tower is needed. This device is called a distributor. Even if the flow is evenly distributed at the top of tower, channeling might still develop as the bed trickles down. When two thin films converge they tend for form a thick film and a dry patch, which results in a reduction in contact area. So redistributors are placed every 10-15 feet along the length of tile tower. 5.13 EQUILIBRIUM CURVE In typical problem McCabe gives us equilibrium data in the form of partial pressure of acetone in the gas phase, PA' as a function of the mole fraction of acetone in the liquid, x : 0

PA = P

AY

A

x

… (5.24)

0

o

P A = Vapour pressure of acetone at 27 C = 0.33 atm γA = Activity coefficient for liquid mixture Activity coefficient is a measure of non-ideality of the liquid phase. For ideal solutions the activity coefficient is unity : γA = 1 (ideal solution) 0.18 0.16 0.14 0.12 y

0.10 0.08 0.06 0.04 0.02 0 0.02 0.04 0.06 0.08 0.10 0.12 x Fig. 5.18 : Equilibrium Curve

0

For ideal solutions, equation (5.1), reduces to Raoult's law. Our solution of acetone in water is not ideal, but McCabe kindly gives us a model. lnyA = 1.95 (1 – x)

2

Note that γA → 1 (i.e lnyA → 0) as x → 1. This is a general rule : pure components also behave ideally since non-ideal behaviour results from interactions between different molecules; there are no different molecules when the fluid is pure. We can obtain an expression for the mole fraction of acetone in the gas just by dividing the partial pressure by the total pressure : PA y= P

=

0.33 atm 1.95 (1 – x)2 1 atm × e

Where we use P = 1 atm, Repeating this for different x's to obtain y's up to the feed concentration), we obtain the curve at right, where x = mole fraction of acetone in the liquid (2 y = mole fraction of acetone in the gas (2

nd

nd

component is water)

component is air)

Principles of Mass Transfer Operations − I (Vol. − I)

5.21

Gas Absorption

Operating Line : Suppose we are trying to absorb acetone from a mixture of air and acetone by contacting the air mixture with water. Let, L, V = Total molar flowrate X, Y = Mole fraction of transfer component Where the transferable component is acetone in this example. Performing a component balance on the transferable component about the top section of the tower yields : LaXa + Vy = LX + Vaya … (5.25) Va ya – Laxa L Solving for y : y = V x+ V This relationship y (x) between the mole fractions in the liquid and gas streams is again known as the operating line, since the relationship is imposed by a component mole balance. However as acetone is transferred from the gas to Liquid, L and V' change. In particular, their ratio changes. Thus the operating line is not straight. This makes it difficult to determine the minimum water flow rate. Va Lean gas

La Weak liquor

yb OL's for inc L

L

ya

V Vb Rich gas

Lb Strong liquor

Fig. 5.19 (a) : Packed tower

Equilibrium Curve

Operating Line Lmin

xa

Fig. 5.19 (b) : Equilibrium curve

There are two limiting cases where we can easily predict the change in L and V. Approx. #1 : Only One Transferable Component : It is often possible to assume that only one component is undergoing transfer between the liquid and gas streams. For example, in our problem acetone is being transferred from the air to the water : Transferable : acetone, Non-transferable : air, water Although some of the water will evaporate when it contacts the air and some of the air will dissolve in the water, the molar rates of transfer of these components can often (but not always) be neglected compared to rate of acetone transfer. When you can neglect the molar transfer rate of all but one component, then significant simplification can be made. If we can neglect evaporation of water and dissolution of air, then the moles of water (the non-transferable component) in the liquid stream must be the same at all elevations : L' (1 – x) L = L' constant so, L = 1 – x … (5.26) Similarly, the moles of air in the gas stream must be the same at all elevations. V (1 – y) V = V = constant so, V = 1 – y … (5.27) Put equation (5.26) and (5.27) into equation (5.25). L'Xa V'ya V'y L'x =1–X +1–y = 1–x +1–y a a

Principles of Mass Transfer Operations − I (Vol. − I)

5.22

Gas Absorption

Where L; V' are constants. If we now express concentrations in terms of mole ratios instead of mole fraction : x X = 1–x

moles of A in liquid = moles of non – A in liquid

moles of A in gas y Y = 1 – y = moles of non – A in gas Then this equation becomes very simple : L' Xa + V'Y = L'X + V'Ya L' L' Y = V' x + Ya – V' Xa  

Or,

… (5.28)

Since L'V' = constant, this is the equation of a straight line. Thus for the special case of one-transferable component, the operating line is straight one mole ratio co-ordinates. Note be smaller than unity : 0≤x≤1 But

0 ≤ X≤ ∞ Y (Xb, Yb)

e Op

i rat

ng

e Lin

(Xa, Ya) X

Fig. 5.19 (c) : Concept of operating line for one transferable component

Approx #2 : Dilute Solution : In other problems, the solution might be very dilute. If the solutions are sufficiently dilute, then mole ratios are virtually equal to mole fraction : If Then and

x << 1 1–x ≈ 1 x X = 1–x ≈x

When this is also true for the gas stream, the equation (5.28) can be approximated as L' L' … (5.29) y = V' x + ya – V' xa   in fact, we can also drop the prime, since L' ≈ L : L L y = V x + ya – V xa  

… (5.30)

Thus the operating line will also be linear on mole fraction co-ordinates, if the solutions are sufficiently dilute. This linear operating line on mole fraction co-ordinates will also apply even if we have more than one transferable component provided both phases are sufficiently dilute in all the transferable components.

Principles of Mass Transfer Operations − I (Vol. − I)

5.23

Gas Absorption

5.14 INTERFACIAL MASS TRANSFER : REVIEW An important design parameter is the depth of packing Heat required. In distillation, the tower height was determined by the Th transport number of plates required times the plate separation (which is T i usually 1-2 feet). The operating and equilibrium lines determine Liquid the number of plates required for a given separation. For packed Gas towers, the height also depends on the operating and equilibrium Tc lines, but in addition it is inversely proportional to the mass transfer coefficient, which in distillation plays only minor role in Fig. 5.20 : Interfacial mass determining plate efficiency. transfer Now that we have established the importance of interfacial mass transfer in packed towers, let's discuss about modeling mass transfer across a phase boundary. Let's start by recalling the driving force for heat transfer across an interface. Suppose we contact a hot gas with a cold liquid. The temperature profile near the interface will look something like that show at right. There are two characteristics of this sketch, which are important : (1) Heat flows from high to low temperature. (2) Temperature is continuous across the interface. (3) Interfacial mass transfer is similar to interfacial heat transfer, but it is also different. (4) Mass transfer occurs in the direction from high to low chemical potential (not necessarily from high to low concentration). (5) Chemical potential is continuous across interface (concentration is generally not continuous). The main difference is evident in the sketch of concentration profile near the interface. Note the discontinuity in the mole fraction at the interface. yI ≠ xI Tyi = Txi = Ti The reason for this discontinuity in concentration across the interface has to do with thermodynamic criteria for phase equilibrium. Recall v

= µj

v

= T

Phase equilibrium :

µj

Thermal equilibrium :

T

L

L

Where µj is called the chemical potential, which plays the role of temperature in mass transfer. Unfortunately, there exists no "thermometer" for measuring chemical potential. Instead, we are forced to measure chemical concentration. While chemical potential usually increases with concentration within any given phase (this is why diffusion of a solute occurs from high to low concentration), when comparing chemical potentials between two phases, there is no general correlation between chemical potential and concentration. To illustrate this, consider the simplest case of VLE : an ideal gas mixture in equilibrium with an ideal solution. This lead's to Raoult's law : p

0

yj

= xj P J

Note that :

yi xj

pj = p ≠1

Thus,

yi ≠ xj

0

for VLE

Principles of Mass Transfer Operations − I (Vol. − I)

5.24

Gas Absorption

Even in the simplest case of vapour-liquid equilibrium, the mole fractions are not equal, except the trivial case when we have only one component and the total pressure is the vapour pressure. 5.15 DEFINITIONS OF TRANSFER COEFFICIENTS The local heat flux through the interface can be related to the local temperatures using any one of three types of local heat transfer coefficients : rate of heat transfer = hx (Th – Ti) = hy (Ti – TC) = U (Th – TC) … (5.31) interfacial area Where hx, hy one-phase local heat transfer coefficients U = overall local coefficient. Basically these equations say that the rate of heat transfer is proportional to the driving force (which is the departure from equilibrium) and the proportionality constant is the heat transfer coefficient. Similar definitions of transport coefficients can be made for mass transfer : molar rate of transfer = kx (xi – x) = ky (y – yi) = kx (x* – x) = ky (y – y*) … (5.32) interfacial area Where kx, ty = one-phase local mass transfer coefficients Kx, Ky = overall local mass transfer coefficients Comparing the mass transport expressions in equation (5.32) with, their heat-transfer analogues in equation (5.31), there is a good deal of similarity especially in the one-phase relations. For a single phase, the driving force is the difference between the concentration or temperature in the bulk and in the concentration or temperature at the interface. But the driving force for the overall coefficient look a little different. The overall driving force for heat transfer is just the difference between the temperature of the hot and cold fluid, (Th – Tc). By simple analogy, we expect the overall driving force for mass transfer to be the difference in concentration of the two phases : y – x ≠ overall driving force Instead y – y* and x* – x Appear in equation (5.32) as the overall driving force. Operating line The *'s' are defined as shown on the Fig. 5.21. Basically y – y* and x* – x are two different measures of the distance between the operating line and the equilibrium curve at Equilibrium that elevation in the packed bed where the liquid and gas y Curve concentrations are (x, y). x-y does not represent the overall y* driving force for mass transfer, because the transport rate does not go to zero when, x – y = 0. The transport rate is x x* zero only at equilibrium, and x – y ≠ 0 at equilibrium. Fig. 5.21 : Driving force

Determining the Interfacial Concentrations : (xi, yi) : Although overall mass transfer coefficients are the most easy to use in design, correlation's are usually more available in the form of single-phase coefficients. Then we need to calculate overall coefficients Kx and Ky from the single-phase coefficient kx and ky. As the first step, we will need to evaluate the interfacial concentration xi and yi, which appear in the definitions above. Example #1 : Given : x, y, kx, ky and the equilibrium curve Find :

xi and yi.

Principles of Mass Transfer Operations − I (Vol. − I)

5.25

Gas Absorption

Solution : Consider the rate of interfacial transport at some arbitrary elevation in the absorber, where the local concentrations in the liquid and gas are x and y and (x, y) is a point on the operating line. The profile of concentration of the transferable component near the interface at this location in the absorbed might look like the sketch as shown in Fig. 5.22.

Liquid

xi

x Gas

y

Mass Transport

yi

Fig. 5.22 (a) : Profile of concentration of transferable component near interface

At steady state, the flux of ammonia through the gas film must equal the flux of ammonia through the liquid film : NA = kx (xi – x) = ky (y – yi) … (5.33) If the fluxes were not equal, we would have ammonia building up at the interface. We will denote this interfacial flux by NA. Now the bulk compositions x, y are known, together with the two single-phase mass transfer coefficients. Think of the two interfacial concentrations as two unknowns. We need two equations. One equation is provided by the requirement that (xi, yi) must lie on the equilibrium curve. The second relation is equation (5.33), which can be re-written as a linear relationship between xi and yi . Y (Xb, Yb)

e Op y

ing

e Lin

rat

x

(Xa, Ya) X

Fig. 5.22 (b)

Fig. 5.22 (c)

The intersection of equation (5.34) and the equilibrium curve gives the interfacial kx kx concentrations. yi = k xi + y + k x … (5.34) y y slope

intercept

Example # 2 : Next, let's determine the value of the overall coefficient, which leads to the same flux for these interfacial concentrations. Given : kx and ky Find : ky Solution : Using the definitions of the k's and of Ky' … (5.35) NA = kx (xi – x) = Ky (y – yi) ky (y – y*) The relationship among the concentrations is shown in the Fig. 5.22 (d). Adding and subtracting yi from the overall driving force y – y*; y – y* = y – yi + yi – y* … (5.36) NA ky

NA ky

Principles of Mass Transfer Operations − I (Vol. − I)

5.26

Gas Absorption

Using (5.35) we can assign meaning to two of the three differences appear above, leaving : NA ky

NA = k + yi – y* y

… (5.37)

(xi – x) m

Y k Slope = – —x ky

e Op

y

in rat

gl

ine Equilibrium Curve

yi Slope = m

y*

X x xi Fig. 5.22 (d) : Relation among concentration

The remaining difference yi – y* can be related to xi – x and the local slope of the equilibrium curve m : NA yi – y* (xi – x) m = k m x

… (5.38)

In the second equality above, xi – x is expressed in term of the remaining single-phase mass transfer coefficient using equation (5.33). Substituting equation (5.38) into equation (5.37) and dividing by NA we get, 1 ky where,

1 m = k + k y x

… (5.39)

yi – y* m = x – x Average slope of equilibrium curve i

Of course, if the equilibrium curve is straight (as it will be in dilute solutions), then m is its slope. Comment : This is similar to the expression for overall heat transfer coefficients for a double-pipe heat exchanger : 1 U

1 1 = h +h x y

… (5.40)

We said that 1/U represented the total resistance to heat transfer through the two phases, which is just the sum of the resistances of each phase. The main difference between equation (5.39) and equation (5.40) is the appearance of m in equation (5.39). For heat transfer, the slope of the equilibrium line is unity (m = 1) because at thermal equilibrium Ty = mTx = Tx. In short, (5.39) states that the total resistance to mass transfer equals the sum of the resistances of the gas and liquid phases. We could also have showed, in a similar fashion. That 1 1 1 = + … (5.41) kx mky kx

Principles of Mass Transfer Operations − I (Vol. − I) where,

5.27

Gas Absorption

y – yi m = x* – x i

If the operating line is straight, these two slopes would be they same and they could relate two overall mass transfer coefficient : m 1 Straight Equilibrium Curve : ky = kx Equimolar Counter-Diffusion vs. Diffusion through Stagnant Fluid : In the form most analogous to Fourier's law of heat conduction, Fick's law of diffusion of a binary mixture of components A and B, the flux (moles/area/lime) of component A in the +z direction is given by : dcA J*AZ = –DAB d z Where CA is the molar concentration of component A and DAB is the diffusion coefficient. This gives the flux of A relative to a reference frame, which moves, with the mole-average velocity (i.e. the average of the species velocity of each component) of the mixture (A + B). Less fundamental, but more useful in calculations is the molar flux of A relative to a stationary reference frame is : dcA NAZ + NBZ NAZ = – DAB d + CA … (5.42) C z Where C = CA + CB is the total molar concentration. Case I : If we use a capillary tube to connect two gas reservoirs having the same total pressure but different amounts of gases A and B, we will obtain equimolar counter-diffusion of A and B (i.e. NAZ = –NBZ). In this way, the mole-average velocity V = 0 and the total pressure in both reservoirs remains constant. For NAZ = –NBZ, equation (5.42) gives dcA NAZ = – DAB d = J*AZ z

… (5.43)

Case II : If instead, we have diffusion of benzene vapour above its liquid through air (which is virtually insoluble in the liquid), the air must remain stagnant (i.e. NBZ = 0) since it cannot enter the liquid. This is diffusion of A through stagnant B. For NB = 0, equation (5.42) gives : dcA CA dcA NA = – DAB d + NAZ C or (1 – yA) NA = – DAB d z z Finally we have,

J*AZ NAZ = 1 – y A

… (5.44)

For the same driving force (concentration gradient), J*AZ is the same for both cases, but NAZ are different. Since 1 – yA is always less than one, we see that Equimolar counter-diffusion is slower than diffusion through a stagnant fluid. This can be qualitatively understood as follows. Suppose that to get to class, you need to walk down a corridor that's crowded with other students. If everyone else was standing almost still (stagnant fluid), it would be easier to walk around them than if everyone is walking toward you (counter diffusion). All of the flux expressions above apply locally at every point in the fluid. For one dimensional steady equimolar counter-diffusion (from a reservoir having CA1 to a second reservoir having CA2), equation (5.43) integrates to, DAB cDAB NAZ = J* AZ = Z – Z (CA – CA ) = Z – Z (yA – yA ) = K'y (yA – yA ) … (5.45) 1 2 1 2 1 2 2 1 2 1

Principles of Mass Transfer Operations − I (Vol. − I)

5.28

Gas Absorption

Where Z2 – Z1 is the length of the capillary through which diffusion is occurring and k'y is the single-phase (gas) mass transfer coefficient to be used with mole fractions and equimolar-counter diffusion. The analogous expression for diffusion through stagnant film of B is : cDAB NAZ = (Z – Z ) (1 – y ) (yA – yA ) = Ky (yA – yA ) 1 2 1 2 2 1 A M (1 – yA)M =

… (5.46)

(1 – yA) 1 – (1 – yA)2 (1 – yA)1 ln (1 – y ) A 2

where, (1 – yA)M is the log-mean of yB evaluated at either end of the diffusion path. Comparing equation (5.45) and equation (5.41), we see that k' y Ky = (1 – y ) A M Similar relations exist for diffusion in the liquid phase : k' y Kx = (1 – y ) A M And between the overall mass transfer coefficients for equimolar counter diffusion and diffusion through a stagnant film : Ky =

k'y (1 –

* yA)M

and Kx =

K'x

*

where, (1 –

* yA)M

=

(1 – yA) – (1 – yAG) *

(1 – yA) ln (1 – y ) AG

… (5.47)

*

(1 – xA)M *

and (1 –

* xA)M

=

(1 – XA) – (1 – xAL) *

(1 – xA) ln (1 – x ) AL

… (5.48)

*

and where yAG is the mole fraction in the bulk of the gas, yA is the mole fraction which would *

be in equilibrium with the bulk liquid having a mole fraction of XAL, and XA is mole fraction in the bulk of the liquid which would be in equilibrium with the bulk gas having a mole fraction of yAG. 5.16 HEIGHT OF A PACKED TOWER The analysis, which follows, has as its goal the determination of the height of packing required. The approach is similar to that used in the design of double-pipe heat exchangers in which the goal is the determination of the area of heat-exchange surface required. The main complication arises from the fact that the mass flux is not proportional to y − x; instead, the mass flux is proportional to y – y*. We have to chop up the tower into pieces, which are small, enough so that the driving force is virtually uniform throughout each piece. Since the compositions change only with z, we chop up the tower in such a way that we produce pieces which have z const, which is a thin horizontal slice.

Va

La

ya

xa

Dz ZT Z Vb

Lb

yb

xb

Fig. 5.23 : Height of Packed Tower

Principles of Mass Transfer Operations − I (Vol. − I)

5.29

Gas Absorption

Now let's take a closer look at what happens inside the slice of the tower. The slice contains solid packing as well as liquid and gas streams. In what follows, we will ignore the solid and treat the liquid and gas phases as if they were completely separated, rather than interspersed in each other.

V(z+Dz)

L(z+Dz)

Gas

Liquid

NA

Dz

V(z)

L(z)

Fig. 5.24 (a) : Closed look inside slice of tower

Let's do a mass balance on the acetone in the gas phase only contained within our slice of column. Besides the liquid streams entering and leaving the slice, we have acetone crossing the interface between the gas and liquid phases. The rate of absorption per unit area can be expressed in terms of the local overall mass transfer coefficient : NA =

molar rate of transfer = Ky (y – Y*) interfacial area

… (5.49)

We could also have used one of the other expressions in equation (5.32); this one proves later to be more convenient than some of the others. Just like the overall heat transfer coefficient U0 depends on the flow rates of both fluids being contacted in the heat exchanger, the overall mass transfer coefficient Kv depends on both flow rates, Ky = Ky (L, V) To obtain the rate of transfer across the interface in our slice of column, we need to multiply equation (5.47) by the interfacial area in that slice. In a heat exchanger, the heat transfer surface is fixed by the geometry of the equipment selected : it is just the area of the pipe wall or the tubes. In particular, the heat transfer area does not depend on the flow rates of the hot and cold streams. On the other hand, the boundary between liquid and gas in a packed bed is very complex and very hard to measure directly. Most importantly, the area also depends on the flow rates of the gas and liquid streams (L, V). The interfacial area is usually expressed as a, the interfacial area per unit volume of packing : interfacial area volume of packing πD0 dZ π (D0 2)2

= a = a (L, V)

… (5.50)

4 = D 0

One empirical correlation relating area to flow rates is the Schulman Correlation : α

β

a = mG x Gy Where Gx and Gy are the mass flow rates of the liquid or gas stream divided by the cross-sectional area of the lower. mass/time G = πD2/4 and where m, α, β are constants which depend on the type and size of packing used. Multiplying equation (5.49) by (5.50) we get, … (5.51) dq = U0 d A0 (Th – TC) π D0dz

Principles of Mass Transfer Operations − I (Vol. − I)

5.30

Gas Absorption

If we now multiply by the volume of this slice of tile packed column, we will obtain the rate at which acetone crosses the interface in this slice : Rate of transfer = r S ∆z S = empty tower cross-section 2

πDT = 4 where

S ∆z = volume of slice DT

DZ

Fig. 5.24 (b)

At steady state, the rate of transport of acetone into the vapour must equal the rate out : In = out (Vy)

z

=

(Vy)

z + ∆z +

r S∆z

… (5.52)

mpcp T (z) = mpcp T (z + ∆z) + dq Dividing by –∆z and letting ∆z → 0 equation (5.50) becomes, d (Vy) dz dT mpcp dz

= –rS

… (5.53)

dq = – dz = π U0 D0 (Th – TC)

Now the total molar flowrate (V) will change as the gas phase loses acetone, but the flowrate of acetone-free air (V') does not change with Z : V'y V' V = 1 – y → Vy = 1 – y V"y y d (Vy) = d 1 – y V' d 1 – y     = …V' Equation (5.51) becomes,

V dy 1 – y dz

dy dy =V1–y (1 – y)2

= –rS = –Ky a (y – a*)S

… (5.54)

The second equation is obtained by substituting our expression for the rate of absorption r from equation (5.49). We can now solve this equation for the height dz of the slice : dz =

–Vdy (1 – y) (1 – y*) (Ky a)S

Principles of Mass Transfer Operations − I (Vol. − I)

5.31

Gas Absorption

So, if we know the local mole fractions y and y* and the change in the mole fraction dy of the gas phase which occurred between the top and bottom of this slice, we can calculate the height of the slice. The total height of packing is the sun of the height of each slice : Zr

yb

–Vdy

⌠ dz = ⌠ ⌡ ⌡ (1 – y) (y – y*) (Ky a)S

ZT =

0

ya

Factoring out the S, which is constant along the entire height of the packing, and changing the order of integration (since yb > ya) : 1 ZT = S

yb

⌠ ⌡

Vdy (1 – y) (y – y*) (Ky a)

ya

Now Ky depends on Gx and Gy, which in turn depend on gas composition. For the case of a dilute gas stream, we know that y <<1 1 – y ≈ 1 V ≈ V' = constant Kya ≈ constant After making these approximations, we have =

ZT

V K aS

yb

dy

⌠ y – y* ⌡

… (5.55)

ya N oy

Hoy

Which serves as the basic design equation for packed bed absorbers. For very dilute solutions, this integral can be evaluated analytically (i.e. numerical integration is not required). If both x << 1 and y << 1 hold, then the molar flow rates of the two streams won't change much as the solute is transferred across the phase boundary. In particular, the ratio of the flowrates will be constant : Then

L/V ≈ L'/V' = constant

This means that the operating line will be straight, even on mole fraction co-ordinates : y = (L/V) x + [ya – (L/V) xa] Also, for sufficiently dilute solutions, Henry's law always applies so that y* = mx Will also be a straight line. When both operating and equilibrium curves are straight, then the integration can be done analytically instead of numerically. The result is : Straight Operating Line and Equilibrium Curve : yb – ya Noy = (y – y*) L *

… (5.56 (a)) *

Where (y – y*)L is the jog-mean of yb – yb and ya – ya which represent the driving force at the bottom and top of the tower, respectively. 5.17 TRANSFER UNIT Meaning can be given to the integral. Even if the integrand varies significantly over the domain of integration, we can still pull it outside the integral, provided we replace it with some

Principles of Mass Transfer Operations − I (Vol. − I)

5.32

Gas Absorption

appropriate value, which can be thought of as the mean value for this range. Recall the Mean Value Theorem of Calculus : yb

dy

yb

1

yb – ya

⌠ y – y* = (y – y*) ⌠ dy = (y – y*) ⌡ av ⌡ avg ya

…(5.57 (b))

ya

Of course, it is not usually known what type of average to use, but the precise value is not important. The integral can be thought of as the total change in gas mole fraction divided by the average driving force. (Gas phase) Transfer unit : A slice (not necessarily thin) of an absorber in which the gas undergoes a change in y equal to (y – y*)avg which represents the average driving force over the entire absorber. NTU : The number of these it takes to make the entire absorber (not necessarily an integer). HTU : Height of a slice of an absorber corresponding to one transfer unit. From equation (5.56 (b)), we see that the integral clearly represents the number of transfer units : yb 0

Noy =

dy

⌠ y – y* ⌡

… (5.57)

ya

This is called the number of overall gas phase transfer units since it is based on the overall driving force expressed as the gas-phase mole fraction. Since the product in equation (5.28) represents the total height of packing, the coefficient of this integral must represent the height of a V|S transfer unit : Hoy K a avg  y  So that our design equation can be rewritten as : ZT = HoyNoy Which is called the HTU method for sizing an absorber. Throughout the derivation above we made use of one type of mass transfer coefficient : r = Ky a(y – y*) Of course, other types of mass transfer coefficient are commonly encountered and can also be used to compute the height of packing required to mention a few : r = Kx a(x* – x) = Kx (xi – x) = Ky(y – yi) Of course, there are still more definitions based on driving forces expressed as differences in molar concentration or partial pressure. Any one of these mass transfer coefficients can be used in the HTU method : ZT = HoyNoy = HoxNox = HXNX = HyNy xa

Where,

Nox =

L/S dx ⌠ x* – x ; Hox = K a ⌡  x  avg

xb xa

Nx =

L/S dx ⌠ x – x ; Hx = K a ⌡ i  x  avg

xb yb

Ny =

dy V/S ⌠ y – y ; Hy = K a ⌡ i  y  avg

ya

… (5.58)

Principles of Mass Transfer Operations − I (Vol. − I)

5.33

Gas Absorption

Note that the N's are not equal to each other; thus the number of transfer units depends on which driving force we use. Ny ≠ Noy 5.18 EVALUATION OF NTU INTEGRALS When the equilibrium curve is non-linear, the NTU integral must be evaluated numerically (or graphically); for example : dy

⌠ ⌡ y – yi'

dx

dy

dx

⌠ x – x' ⌠ y – y*' ⌠ x* – x' ⌡ i ⌡ ⌡

… (5.59)

which represent areas under the curves : 1 1 1 1 y – yi vs y, xi – x vs x, y – y* vs y, x* – x vs x

… (5.60)

The last two are particularly simple to do since vertical lines represent y – y* and horizontal lines represent x* – x. The first two can be obtained by noting that : y – yi xi – x

kx = k y

… (5.61)

which is a straight line of slope – kx/ky passing through (x, y) and (xi, yi). The former point lies on the operating line (see point 1 above) and the latter on the equilibrium line (point 2 above). From this it is possible to generate y – yi and xi – x. Operating Line (y, x) y kx ky (yi, xi)

Equilibrium Curve

x

Fig. 5.25 : Evaluation of NTU integrals

When the equilibrium curve is linear (e.g., Henry's law), a special form can be obtained which does not involve any numerical or graphical integration. This same concept applies in countercurrent heat exchangers in heat transfer. The result is : y B

⌠ ⌡

dy (y – yi)

=

yB – yT (∆y)lm

… (5.62)

y T

where (∆y)lm is defined as : and

(∆y)lm =

∆yB – ∆yT

∆yB ln   ∆yT

∆y = y – yi

… (5.63)

Principles of Mass Transfer Operations − I (Vol. − I)

5.34

Gas Absorption

This only requires y and yi at the top and bottom of the tower. If (y – yi)T and (y – yi)B are within a factor of two (2) of one another, then : y B

dy

yB – yT

⌠ (y – y ) = (y – y ) ⌡ i i mean

… (5.64)

y T

where arithmetic mean, ∆ymean = (y – yi)mean, is a good approximation to logarithmic mean. What is a "transfer unit ?" : For example, the average driving force for mass transfer based on the gas phase is defined as : y2

⌠ dy ⌡

y2

(y – y*)mean +

y1

⌠ (y – y*) dy = ⌡

y2

⌠ ⌡

y1

… (5.65)

dy (y – y*)

y1

If the denominator of this expression is set to be one (NTU)OG; y2

i.e.

(NTU)OG +

⌠ ⌡

dy (y – y*)

y1

then equation (5.65) becomes y2

⌠ dy ⌡ y1

(y – y*)mean = 1 (NTU) OG

(y2 – y1) = 1 (NTU) OG

(y2 – y1) (NTU)OG = (y – y*) … (5.66) mean This means that an NTU is defined at the point where the mean driving force over a transfer unit becomes equal to the total change in concentration that occurs. This result leads to the graphical "Baker construction" method for an NTU. This involves drawing in a curve midway between the operating line and the equilibrium curve (when using an overall gas-phase driving force) and performing the construction indicated below : or

Baker Construction y2

Baker Construction 50% Curve (1/2 vertical distance)

Operating Line

y y1

b

a

Dy = y2 –y1 b

a Mean Driving Force, 2a = (y –y*) mean Equilibrium Curve x Fig. 5.26 : Baker construction method for NTU

Principles of Mass Transfer Operations − I (Vol. − I)

5.35

Gas Absorption

By similar triangles : ∆y ∆y a b = 2b ⇒ a = 2 ⇒ 2a = ∆y, and thus ∆y = y2 – y1 = (y – y*)mean; which is an NTU. Cocurrent Operation : Finally, it is pointed out that all of the above applies to countercurrent flow of the gas and liquid streams. Sometimes, however, both streams flow concurrently downward in a tower. In this case the only real change is in the slope of the operating line, such that – Operating Line

Operating Line

y

y

Pinch

(y – y*)

(y – y*) Equilibrium Curve

Equilibrium Curve

x x Countercurrent Cocurrent Fig. 5.27 : Graphical construction of Co-current and Co-current operations

Stripping Operations : Stripping is basically the opposite of gas absorption. The mass transfer process is reversed; that is, the solute is removed from the liquid phase by contact with a gas phase. In stripping operations, the gas phase may be steam or air and in order to effective, the solute should only be sparingly soluble in the liquid. For example, volatile organics contaminants (VOCs) in water. Consequently, mass transfer in stripping operations is typically liquid-film controlling. For dilute systems, the mass balances on the tower still apply as before; that is : (subscript B = bottom) Bottom : Gy + LXB = GyB + Lx or y = (L/G) x + (yB – (L/G) xB) … (5.67) Top : In the same way : y = (L/G) x + (yT – (L/G) xT) (subscript T = top) … (5.68) However, in stripping, the liquid phase solute concentration must always be greater than the corresponding equilibrium value, in order to provide the driving force for mass transfer from the liquid to the gas phase; that is, x > x*. On a y-x diagram, this means that the operating line must always lie below the equilibrium curve, rather than above it, as is the case for gas absorption. Also, now one of the principal operating parameters becomes the gas flow rate, G. As G is decreased, the slope of the operating line increases such that the intersection with the equilibrium curve will occur at the top of the tower for countercurrent operation. This intersection will then define the minimum allowable gas flow rate (G)min, or (L//G)max. Just as for gas absorption, a typical design

Counter Current Equilibrium Curve, Stripping y = Hx

y

(L/G)max or (G/L)min Operating Lines

Decreasing G

XT

XB x

Fig. 5.28 : Graphical analysis for counter current stripping operations

Principles of Mass Transfer Operations − I (Vol. − I)

5.36

Gas Absorption

"rule of thumb" is to operate at a gas flow rate that is about 40% greater than (G)min. All other design calculations are similar to those presented above for gas absorption. 5.19 CONCEPT OF HETP : (HEIGHT EQUIVALENT OF THEORETICAL PLATE) This is the height of packing that will give the same separation as one theoretical plate i.e. section of packing of a height such that the vapour leaving the top of the section will have the same composition as the vapour in equilibrium with the liquid leaving the bottom of the section. The HETP must be an experimental determined quantity characteristics of each packing. It depends not only on the type and size of packing but also with the flow rate of each fluid and concentration. The required height of packing for any desired separation is given by : Z = (HETP) × (Number of theoretical stages required) where, Z = height of packing. This is a simple method representation which has been widely used as a method of design. Murch's correlation for HETP : Murch has considered column from 50 – 750 mm diameter and packed over heights of 0.9 – 3 mm with rings, saddles and other packings. 1 α µL C3 3 C … (5.67) HETP = C1 G' 2 d C Z ρL where,

C1, C2 and C3 vary with type of packing. These values are available in the literature. G' dc Z α µL

= Mass velocity of vapour in kg/m2sec of tower area. = Column diameter in m = Packed height in m = Relative Volatility. = Liquid viscosity, N/S.m2

ρL = Liquid density, kg/m3 Elli's correlation for HETP : Elli derived a general equation which is consistent units for HETP (Zt) of packed column using 25 and 50 mm Rasching rings : G' Zt = 18 dr + 12 m  L' – 1



where



… (5.68)

dr = Diameter of the rings m = Average slope of equilibrium curve G' = Vapour flow rate L' = Liquid flow rate

In industrial practices, the HETP concept is used to convert empirically the number of theoretical stages to packing height. 5.20 ANALOGY WITH DOUBLE PIPE HEAT EXCHANGER Comparing equation (5.56 (a)) with equation (5.56 (b)), we see that for very dilute solutions. The average driving force is the log-mean. This result is quite analogous to our design equation for double-pipe exchangers. To see analogy, let's compute the total rate of ammonia transfer over the entire packed tower : qT = V (yb – ya)

… (5.69)

Recall the design equation : V|S ZT = Hoy Noy = K a y

yb – ya Ky a (y – y*)L

… (5.70)

Principles of Mass Transfer Operations − I (Vol. − I)

5.37

Gas Absorption

r This contains the product that appears in equation (5.42). Solving equation (5.43) for (y – y ) b a . V (yb – ya) = Ky a SZT (y – y*)L total volume of packing

where, SZT is the total volume of packing in the absorber and a is the interfacial area per unit volume of packing. Thus, aSZT = total interfacial area in entire absorber = AT and substituting in equation (5.42) yields : qT = Ky AT (y – y*)L Which is analogous to : qT = UAT ∆TL 1 InyN y yeq 2

1 0 –6

–5

–4

–3 –2 –1 In (y) Fig. 5.29 : Analogy of double pipe heat exchanges

Thus, for dilute solutions, the design equation for an absorber is identical in form to that for heat exchangers : with the overall mass transfer coefficient Ky analogous to the overall heat transfer coefficient U and the log-mean of y – y* at the two ends of the absorber analogous to the log-mean ∆T. 5.21 RELATIONSHIPS AMONG Hox, Hoy, Hx and Hy We know a relationship between the overall mass transfer coefficient and the single-phase mass transfer coefficients. This relationship can be easily modified to relate the corresponding heights of transfer units. Dividing both sides of equation (5.39) by a the (interfacial area per volume : of packing), we obtain : 1 m 1 … (5.71) Kya = Kya + Kxa Where m = (yi – y*) (xi – x) = avg. slope of equilibrium curve Of course, if the equilibrium Curve is straight (as it will be in dilute solutions), then m is its slope. This is similar to the expression for overall heat transfer coefficients for a double-pipe heat exchanger : 1 1 1 … (5.72) U = hx + hy We said that 1/U represented the total resistance to heat transfer through the two phases, which is just the sum of the resistances of each phase. The main difference between

Principles of Mass Transfer Operations − I (Vol. − I)

5.38

Gas Absorption

equation (5.71) and equation (5.72) is the appearance of m in equation (5.71). For heat transfer, the slope of the equilibrium line is unity (m = 1) because at thermal equilibrium Ty = mTx. In short, equation (5.44) states that the total resistance to mass transfer equals the sum of the resistances of the gas and liquid phases. Using equation (5.16), we could also have showed, in a 1 1 1 similar fashion, that Kxa = mkya + kxa Where now :

m = (y – yi) (x* – xi)

If the operating line is straight, these two slopes would be the same and we could relate the two overall mass transfer coefficient : Straight equilibrium curve :

1/ky = mkx

If we now multiply equation (5.42) by V/S we have the height on an overall gas transfer unit on the left-hand side : V/S V/S mV/S Kya = Kya + kxa

V/s VL/S = K a + m LK a y x

Dry

=2 x

0,0 =3 x

G

1.0

G

G

Dp, Inches Water/ft packing

x

=3

6,0

00

00

2.0

5,0 00 Gx = 20,0 00 Gx = 15,0 00 Gx = 12 Gx = ,000 9,00 0 Gx = 6,00 0 Gx = 3,00 0

5.22 PRESSURE DROP IN PACKED BEDS Usually, flow in a packed-bed absorber is countercurrent. Gravity causes the liquid to flow down through the packing, whereas a small pressure drop drives the gas flow upward. This pressure drop plays an important role in packed beds. Without liquid present, the pressure-drop across dry packing increases approximately with the square of the gas flow rate.

0.8 0.6

Column dia. = 30 in. Packing height = 10 ft

0.4

0.2 3

Liquid rate,lb/ft – h as parameter 0.1 100

200

300 400

600

1,000

2,000

5,000

3

Air Mass Velocity, Gx, lb/ft – h

Fig. 5.30 : Pressure drop in packed bed

lb/hr of gas Gy = mass velocity of gas = cross-section of lower

=

MyV S

Principles of Mass Transfer Operations − I (Vol. − I)

5.39

Where My is the average molecular weight of the gas. Once liquid is flowing down the column as gas is flowing up, some of the space between the packing particles is taken up by liquid, leaving less space for the gas to flow. Forcing the same gas flow through a smaller opening will increase the frictional contribution to the pressure drop. Thus the curves for increasing liquid flow are above that wet for dry packing although the curves tend to be parallel. Note that the curves with liquid flow curl up at high gas flow rates. This can be explained as follows : The upward flow of the gas exerts a shear force on the liquid, retarding its downward motion. As the gas flowrate increases so does this shear force. When the shear force becomes comparable to gravity, the liquid flow might slow down and liquid begins to accumulate in the tower.

Gas Absorption

Packing

Gas

Liquid

Fig. 5.31 : Packed bed log (Dp)

Flooding Loading

» 1.8 to 2.0

Constant holdup Wet (Gx > 0)

(Turbulent)

Dry (Gx = 0)

log (Gy)

Fig. 5.32 : Relation between ∆P and Gy

Packing particle

Stagnant air

Packing particle

Flowing air

Fig. 5.33 : Effect of flow as shear force

(i) Liquid Holdup : Holdup refers to the liquid retained in the tower as films wetting the pakcing and as pools caught in the crevices between packing particles. It is found that the total holdup φL is made of two parts, φLt = φLo + φLs where,

φLs = Static holdup (volume liquid/packed volume) φLo = Operating or moving holdup (volume liquid/packed volume) (ii) Interstitial volume : Fraction of interstitial volume occupied by liquid. (iii) Pore Volume : "Empty" space between and inside packing particles; that volume occupied by liquid or gas. (iv) Loading : An increase in liquid holdup caused by an increase in gas flowrate. For given liquid flowrate, there is a maximum flowrate of a gas, which can be forced through the column. If you exceed this maximum the shear forces on the liquid are so high that they exceed the weight to the liquid. Then the net force on the liquid is upward and you blow the liquid back out the top of the column. This is called as loading.

Principles of Mass Transfer Operations − I (Vol. − I)

5.40

Gas Absorption

(v) Flooding : Liquid downflow is essentially stopped by high gas upflow. Since you can't get the liquid through the column, flooding is clearly a condition to be avoided. Generally, to get a high area of contact between liquid and gas you want to operate will below the flooding velocity : At flooding :

∆pf/L ≈ 2 to 3 inch H2O/ft of packing

Loading :

∆pf/L ≈

Normal operating pressure

∆pf/L ≈ 0.25 to 0.5 inch H2O/ft

0.5 inch H2O/ft

Normal operating pressure drops are just below those for which loading begins. Notice Fig. 5.30 that when we increase the liquid flowrate, flooding occurs at lower gas flows. If we increase the size of the packing particle, we can tolerate higher gas flows. While higher gas flows can be obtained with larger packing particles, the amount of interfacial area generally less (i.e. a is decreased). Fig. 5.30 are for a particular packing (1" Intalox saddles) and for a particular temperature and pressure (20 oC and 1 atm). A more general correlation of pressure drop is provided by Fig. 5.34. The size and type of packing is accounted for in this correlation through the parameter. Packing factor (Fp) :Values of the packing factor for various types of packing are given in Table 5.2. The y-co-ordinate of this graph Fig. 5.34, is not dimensionless, so we need to use the units summarized in the table as : Symbol Units Gx Gy 1b/ft2–s µ CP PxPy 1b/ft3 gc 32.2 1bf – ft/b1b – s2 0.60 0.40

J 0.1

C1mL

0.2

G

rG (rL – rG)

0.20

Parameter of curves is pressure drop in inches of water foot of packed height 1.50

0.10 0.060

1.00 0.50

0.040

0.25

0.020

0.010

0.010 0.006 0.004

0.05

0.002 0.001 0.01 0.02 0.040.060.1 0.2 0.4 0.6 1.0 ra L rL G

Fig. 5.34

2.0 4.0 6.0 10.0

Principles of Mass Transfer Operations − I (Vol. − I)

5.41

Gas Absorption

Table 5.2 Type

Rasching rings

Material

Ceramic

Nominal

Bulk density 1b/ft3

Total area ft2/ft3

Porosity

Packing factors Fp

fp

1 2

55 42 43

112 58 37

0.64 0.74 0.73

580 155 95

1.52 ξ

1

41

28

0.74

65

0.92 ξ

30 24 22

63 39 31

0.94 0.95 0.96

56 40 27

1.54 1.36 1.09

5.5 4.8

63 39

0.90 0.91

55 40

1.36 ξ 1.18 ξ

54 45

142 76

0.62 0.68

240 110

40

46

0.71

65

1.58 ξ 1.36 ξ 1.07 ξ

46

190

0.71

200

2.27

42 39 38 36

78 59 36 28

0.73 0.76 0.76 0.79

92 52 40 22

1.54 1.18 1.0 0.64

2 3 1 2

– –

– –

– –

60 30

1.54 1.0

1

–

–

0.97

41

1.74

1 12

–

–

0.98

24

1.37

–

–

0.98

18

1.19

2

–

–

–

–

–

1

19

54

0.96

45

1.54

1 12

–

–

–

29

1.36

2

14

29

0.97

26

1.09

1.36 ξ 1.0

1

12 2 Pall rings

Metal

Plastic

1 1 12 2 1 1 12

Berl saddles

Ceramic

1

12

1 12

Intalox saddles

Ceramic

1 1 12

Super intalox saddles IMTP ®

Hy-Pak

Ceramic

Metal

Metal

5.23 TOWER DIAMETER Once the total molar liquid and gas flowrates (L and V) are known, we can choose the diameter of tower we need. The diameter of the tower is usually chosen such that the pressure drop is some prescribed value below flooding.

Principles of Mass Transfer Operations − I (Vol. − I)

5.42

Gas Absorption

The pressure drop depends on the mass velocities : Choose DT such that :

dp dz

∆p = Z T

∆p ZT

= f (Gy Gx)

≈ 0.25 to 0.5 inch H2O/ft of packing

and where Mx is the average molecular weight of the liquid. M xL M xL mass flowrate of liquid   where,Gx cross-sectional area of column = S = 2   πD /4 T

MyV Similarly, we have, Gy = S So to get certain pressure drop for a certain set of flowrates, we must choose a particular value for the tower diameter. Although it looks like you have to guess the diameter DT in order to calculate Gx and Gy to get ∆p, a trial-and-error can be avoided by noting that the abscissa (x co-ordinate) does not depend on the diameter : MxL/S M xL Gx = = … (5.73) Gy MyV/S MyV Note that the unknown cross-sectional area S conveniently cancels out. What remains is known so the procedure is as follows : Given : L,V Find : DT Procedure : (1) Calculate the abscissa. (2) Locate the point on the curve of Fig. 5.30 which has the desired ∆p and this value of the abscissa. (3) Read the corresponding ordinate of this point from Fig. 5.30. (4) Calculate Gy which gives this value for the ordinate. (5) Calculate the desired tower diameter from 2

πDT 4

MyV = S = G y

5.24 ABSORPTION WITH CHEMICAL REACTION Absorption followed by reaction in the liquid phase is often used to get more complete removal of a solute from a gas mixture. For example, a dilute acid solution can be used to scrub NH3 from gas streams and basic solutions are used to remove CO2 and other acid gases. Reaction in the liquid phase reduces the equilibrium partial pressure of the solute over the solution, which greatly increases the driving force for mass transfer. If the reaction is essentially irreversible at absorption conditions, the equilibrium partial pressure is zero, and N0y can be calculated just from the change in gas composition. For y* = 0, b

N0y =

⌠ ⌡

yb dy y = ln ya

… (5.74)

a

A further advantage of absorption plus reaction is the increase in the mass transfer coefficient. Some of this increase comes from a greater effective interfacial area, since absorption can now take place in the nearly stagnant regions (static holdup) as well as in the dynamic liquid

Principles of Mass Transfer Operations − I (Vol. − I)

5.43

Gas Absorption

holdup. For NH3 absorption in H2SO4 solutions, Kga was 1.5 to 2 times the value for absorption in water. Since the gas-film resistance is controlling, this effect must be due mainly to an increase in effective area. The values of Kga for NH3 absorption in acid solutions were about the same as those for vapourisation of water, where all the interfacial area is also expected to be effective. The factors Kgavap/Kgaabs and Kgareact/Kgaabs decrease with increasing liquid rate and approach unity when the total holdup is much large than the static holdup. The factor Kgavap/Kgaabs also depends on the concentration of reactant and is smaller when only a slight excess of reagent is present in the solution fed to the column. Data on liquid holdup and effective area have been published for Rasching rings and Berl saddles, but similar results for newer pickings are not available. When the liquid-film resistance is dominant, as in the absorption of CO2 or H2S in aqueous solutions, a rapid chemical reaction in the liquid in the liquid can lead to a very large increase in the mass transfer coefficient. When absorption is accompanied by a very slow reaction, the apparent values of Kga may be lower than with absorption alone. An example is the absorption of Cl2 in water, followed by hydrolysis of the dissolved chlorine. The slow hydrolysis reaction essentially controls the overall rate of absorption. 5.25 GENERAL COLUMNS

PROCEDURE

FOR

DESIGN

OF

PACKED

ABSORPTION

It involves three steps : (a)

Overall mole balance : This indicates the overall size of the process to be designed.

(b) Diameter of the column : For rapid mass transfer between the phases, intimate contact between the gas and the liquid is desirable. This is made possible if the gas and liquid flow through a narrow range inside the tower. (c) Height of the column : The solute inside the column must be allowed sufficient time to diffuse from the gas into the liquid. This time depends on the height of the column. Overall mole balance : In the case of dilute gas, if the liquid is non-volatile and the inert gas is practically insoluble in the liquid, we can take the gas and liquid flow rates to be constant throughout the column. The overall mole balance at steady state becomes :

Solute entering – Solute leaving – Solute leaving  – Solute entering  in gas stream  in gas stream  in liquid stream in liquid stream GM (y1 – y2) = LM (x1 – x2)

i.e.

… (5.75)

where GM and LM are the molar flow rates of the gas and liquid respectively (k moles/sec.) y1, y2 = Inlet and outlet concentrations of the solute in the gas, mole fraction. x1, x2 = Outlet and inlet concentration of the solute in the liquid, mole fraction. Diameter of Absorption Column : The diameter of absorber (or stripper) is generally determined from following correlation which consists of a logarithmic plots of :

U2 α ρG 0.2 g · ∈3 ρ µ  Vs L  C 

L G 

ρG  ρL

 

…( 5.76)

Principles of Mass Transfer Operations − I (Vol. − I)

5.44

Gas Absorption

rG rL

m

0.2

0.4 0.2

1200

0.1

800

Approximate flooding

400

0.04

a Î3

200 0.02

V 9c

2

100

0.004

Gas pressure 2 drop, N/m /m

0.001 0.01

0.1 L G

1.0

2

4

10

rG rL

Fig. 5.35 : Determination of diameter of absorber/stripper

where,

µ = Viscosity ρG = Density of gas ρL = Density of liquid ∈ = Porosity operating velocity in the tower is usually 40 – 50% of the flooding velocity, Volumetric gas flow rate Therefore, cross-sectional area = Uopt where

Uopt = Operating velocity

5.26 DESIGN CONSIDERATIONS (1) Height of an absorber/stripper : Z = HTU × NTU Case I : When the controlling resistance is offered by the gas film : GM Then, HTU = K a P G T y1

and

NTU =

dy

y1 – y2

⌠ y – y = (y – y ) ⌡ e e lm

… (5.77)

… (5.78)

… (5.79)

y2

when equilibrium relation is a straight line. (y1 – ye ) – (y2 – ye ) 1 2 (y – ye)lm = y –y

 1 e 1 ln y – y   2 e 2

… (5.80)

Note : When equilibrium relationship is other than straight line, equation (5.52) is evaluated by graphical integration. Case II : When the controlling resistance is offered by the liquid film : LM HTU = K a C … (5.81) L T

Principles of Mass Transfer Operations − I (Vol. − I)

5.45 x1

NTU =

Gas Absorption dx

⌠ x –x ⌡ e

… (5.82)

x2 x1

where,

dx

x1 – x2

⌠ x – x = (x – x) ⌡ e e lm

… (5.83)

x2

In terms of individual film coefficients and overall mass transfer coefficient, the HTU expressions changes as under : GM … (5.84) (HTU)g = K a P g T GM (HTU)OG = K a P G T

… (5.85)

LM (HTU)l = K a C l T

… (5.86)

LM (HTU)OL = K a C L T

… (5.87)

The equations (5.84) to (5.85) are interrelated as : GM (HTU)OG = (HTU)g + m L (HTU)l M

… (5.88)

LM (HTU)OL = (HTU)l + m G

… (5.89)

M

(HTU)OG = where,

(HTU)g

m GM LM (HTU) OL

… (5.90)

GM = Molar mass velocity of inert (MT–1 L–2) LM = Molar mass velocity of absorbing liquid (MT–1L–2) a = Interfacial area per unit volume (L2/L3) m = Slope of equilibrium curve

5.27 TYPICAL ABSORBER DESIGN PROBLEM Designing a randomly packed column is a subtle blend of art and science. Packed columns are most frequently used to remove contaminants from a gas stream (absorption). However, packed columns can also be used to remove volatile components from a liquid stream by contacting it with an inert gas (striping). They are also used in distillation applications where the separation is particularly difficult due to close boiling components. While we will discuss all of these applications, we'll focus on adsorption. However, the design methods are similar for any of the scenarios.

Gas out G out y out

Liquid in L in x in Absorber or Striping Column

Liquid out L out x out

Gas in G in y in

Fig. 5.36 : Absorber or stripping column

Principles of Mass Transfer Operations − I (Vol. − I)

5.46

Gas Absorption

The first step in designing a packed tower is more science than art. The equilibrium data between the contaminant and the solvent (or the distillation components) is needed for the analysis. If tabulated data for your system is unavailable and the total amount of the contaminant is small (as it usually will be), Raoult's Law can be used to estimate the equilibrium data for adsorption or stripping applications. For distillation, equilibrium data can be predicted by selecting the appropriate thermodynamic model. The operating line for the tower is constructed differently depending on whether you are dealing with distillation or adsorption/stripping. Since we are focusing on adsorption, we will use it as an example. In adsorption/stripping, the operating line is constructed differently depending on whether the contaminated stream can be considered "dilute" or if it must be treated as a concentrated stream. Usually, it is safe to treat the stream as dilute if the contaminant makes up less than 10 mole percent of the stream. For streams that can not be considered dilute, the mass transfer coefficients must be evaluated in terms of the gas and liquid flows. Then, graphical evaluation of several integral relationships must be completed. Here we will consider dilute streams which are more common for packed tower adsorption and stripping. Dilute streams allow the column Gas out = 10,000 mol/h L in = 8000 mol/h designer to assume constant mass transfer y outm = 0.005 x in = 0 and the operating line can be constructed in terms of the simplified balance shown below : Lout xout + Gout yout = Lin xin + Gin yin Absorber or This relation is used in the following Striping manner : Suppose you wish to remove Column acetone from a gas stream of 10,000 mole/h in a packed column. The inlet gas constrains 2.6 mole percent acetone and the L out = 8000 mol/h G in = 10,000 mol/h y in = 0.026 x out = ? outlet gas stream can contain no more than 0.5 mole percent acetone. Assume a pure Fig. 5.37 : Absorber or Stripping Column water stream enters the packed tower at a rate of 8,000 mole/h. Lout xout + Gout yout = Lin xin + Gin yin (8000) xout + (10000) (0.005) = (8000) (0) + (10000) (0.026) xout = 0.02625 The equilibrium and operating lines are constructed as follows : 0.03

Tower Bottom Operating Line

y

0.02

Equilibrium Line

0.01

0.01

0.02

0.03

0.04

x Fig. 5.38 : Construction of equilibrium and operating lines

Principles of Mass Transfer Operations − I (Vol. − I)

5.47

Gas Absorption

Just as in the McCabe – Thiele analysis of distillation, the equilibrium stages are stepped-off between the two lines. Note that for stripping, the operating line would be on the other side of the equilibrium line. Once the theoretical number of stages have been determined, you can proceed with the design of the column by following the three steps that we will outline as shown in Fig. 5.38. To best illustrate the other steps in the process, we will use the following example : Specify the packing type and column dimensions for a column that will be used to remove chlorine from a gas stream using an organic solvent. Assume the separation requires 20 theoretical stages. The vapour flow is 7000 kg/h, the average vapour density is 4.8 kg/m3. The liquid flow is 5000 kg/h, the average liquid density is 833 kg/m3. The liquids' kinematic viscosity is 0.48 centistokes (4.8 × 10–7 m2/s). Step 1 : Selecting a Type and Size of Packing : This is where the art of designing packed columns beings. Some people believe that there are stringent rules surrounding the choice between random and structured packing. We can think of random packing as the type that comes in a sack and it is simply dumped into the column. Structured packing may come in bales or intricate designs that are stacked in specific patterns. This is probably one of those areas of engineering where past experience in the application is the best guide. Two "areas of choice" where structured packing is used are in very low pressure drop applications and for increasing the capacity of an existing column. Since we are considering a new design with no serious pressure drop constraint, we will choose the more economical random packing. Below are charts showing both English and Metric unit packing factors. The most common random packing types are shown here : Table 5.3 : Packing Factors for Column Packing Packing Type

Material

Nominal Packing Size, in 1/4

3/8

1/2

5/8

3/4

1

1 1/4

1 1/2

2

3

3 1/2

Hy-Pack

Metal

43

18

Super Intalox Saddles

Ceramic

60

30

Super Intalox Saddles

Plastic

33

21

Pall Rings

Plastic

97

52

40

24

22

16

Pall Rings

Metal

70

48

33

20

37

16

Intalox Saddles

Ceramic

725

330

200

145

92

52

40

Rasching Rings

Ceramic

1600

1000

580

380

255

155

125

95

65

Rasching Rings

Metal, 1/32 in

700

390

300

170

155

115

Rasching Rings

Metal, 1/16 in

410

290

220

137

110

83

57

Berl Saddles

Ceramic

170

110

65

45

Tellerettes

Plastic

Mas Pac

Plastic

900

240

38

15

16

32

19 32

Quartz Rock

20

160

Cross Partition

Ceramic

80

Flexipack

Metal

33

22

16

Interlox

Metal

41

27

18

Chempak

Metal

29

Principles of Mass Transfer Operations − I (Vol. − I)

5.48

Gas Absorption

Table 5.4 : Packing Factors for Column Packing Material Nominal Packing Size, in

Packing Type

6.35

9.53

12.70

15.88

19.05

25.40

31.75

38.10

43

50.80

76.20

Metal

Super Intalox Saddles

Ceramic

60

30

Super Intalox Saddles

Plastic

33

21

Pall Rings

Plastic

97

52

40

24

22

16

Pall Rings

Metal

70

48

33

20

37

16

Intalox Saddles

Ceramic

145

92

52

40

Rasching Rings

Ceramic

125

95

65

Rasching Rings

Metal, mm

.79

Rasching Rings

Metal, mm

1.6

110

83

57

Berl Saddles

Ceramic

65

45

Tellerettes

Plastic

Mas Pac

Plastic

725

330

200

1600

1000

580

380

255

155

700

390

300

170

155

115

410

290

220

137

170

110

900

240

38

18

88.90

Hy-Pack

15 16

32

19 32

Quartz Rock

20

160

Cross Partition

Ceramic

Flexipack

Metal

33

22

80 16

Interlox

Metal

41

27

18

Chempak

Metal

29

Generally, the column diameter to packing size ratio should be greater than 30 for Rasching rings, 15 for ceramic saddles, and 10 for rings or plastic saddles. The geometry of your packing will typically be a function of the needed surface area and/or allowable pressure drop. If several packings meet your requirements, you will typically choose the least expensive so long as it has an acceptable operating life. For our example, we will choose Pall rings (plastic). For columns over 24 inches in diameter, No. 2 or 2 inch packing should be examined first. By looking at our flowrates, the chances of our column having a diameter of at least 24 inches are good, but we will verify this later. For now we will settle on 2 inch plastic Pall rings for our initial analysis. Step 2 : Determine the Column Diameter : Most methods for determining the size of randomly packed towers are derived from the Sherwood correlation. A design gas rate, G, can be determined with the help of the Fig. 5.39 which is based on correlation from the Sherwood equation. Each line on the graph is marked with an acceptable pressure drop in inches of water per foot of packing (number sin parenthesis are in mm of water per meter of packing). Guidelines are as follows : Moderate to high pressure distillation = = Vacuum Distillation = = Absorbers and Strippers =

0.4 to 0.75 in water/ft packing 32 to 63 mm water/m packing 0.1 to 0.2 in water/ft packing 8 to 16 mm water/m packing 0.2 to 0.6 in water/ft packing

= 16 to 48 mm water/m packing

Principles of Mass Transfer Operations − I (Vol. − I)

5.49

Gas Absorption

10.0 6.0 4.0 1.5

(12 5) 10 (83 0.50( ) 42)

2.0

1.0

0.10(8

PV(PL – PV)

2

CG FV

0.1

0.6

)

0.05(4

0.4

)

0.2

0.1 0.06 0.04

0.02

Symbol Property V Gas rate Liquid rate L Gas density PV Liquid density PL Liquid viscosity n Conversion factor C Packing factor F Specific gas rate G

0.01 0.01

0.02 0.04 0.06

British Units lb/n lb/n lb/n3 lb/n3 centistokes 10 – lb/(ft2)(s)

0.1

Metric Units kg/n kg/n kg/n3 kg/n3 centistokes 10.764 – lb/(m2)(s)

0.2 0.4 L rV V rL

0.6

10

20

40

60

100

Fig. 5.39 : Procedure for determination of column diameter

These guidelines are designed around "flooding pressure drops" documented in literature. In other words, for most cases, designing with these pressure drops should help you avoid flooding. In the alter stages of design, we may want to therefrom a through flooding calculation. Perry's Chemical Engineers' Handbook covers this topic well. Since we are designing an absorber, we will design for 42 mm water/m packing (we could design for a lower pressure drop, but the column will increase in diameter and most likely cost). First, we will evaluate the x-axis of the graph above : (L/V) (vapour density/liquid density)0.5 = (5000/7000) (4.2/833)0.5 = 0.0507 Note that 4.2 kg/m3 was used for the vapour density. The average vapour density was given as 4.8 kg/m3. However, at the top of the column, the vapour will be less dense and at its' highest velocity. This is what you should design for. As a rule of thumb, we reduce the average vapour density by about 15% for design, however if you can get real data from a similar tower, certainly do so ! Reading the intersection of the 42 mm water/m packing line and 0.05 on the axis, we find a value of 1.5 for the y-axis.

Principles of Mass Transfer Operations − I (Vol. − I)

5.50

Gas Absorption

10.0 6.0 4.0 1.5

(1 2 5) 10 (83 0.50( ) 42)

2.0

1.0

0.10(8

PV(PL – PV)

2

CG FV

0.1

0.6

)

0.05(4

0.4

)

0.2

0.1 0.06 0.04

0.02

Property Gas rate Liquid rate Gas density Liquid density Liquid viscosity Conversion factor Packing factor Specific gas rate

0.01 0.01

Symbol V L PV PL n C F G

0.02 0.04 0.06

British Units lb/n lb/n lb/n3 lb/n3 centistokes 10 – lb/(ft2)(s)

0.1

Metric Units kg/n kg/n kg/n3 kg/n3 centistokes 10.764 – lb/(m2)(s)

0.2 0.4 L V

0.6

10

20

40

60

100

rV rL

Fig. 5.40 : Procedure for determination of column diameter

From the previous charts, we read a packing factor of 24 for 2 inch plastic Pall rings. All other information is know so we can solve for G as shown on the y-axis of the graph : G = [1.5 [(4.2) (8.44 – 4.2)]/[(10.764) (24) (0.48)0.1]] 0.5 = 4.66 kg/m2 s Now, we solve for the column cross sectional area : Ax = vapour flow/G = 7000 kg/h/](4.66 kg/m2 s) (3600 s/hour)] = 0.42 m2 and the column diameter is calculated by : Diameter = [Ax/(Pi/4)]0.5 = [0.42/(Pi/4)]0.5 = 0.73 m or 2.4 ft So our assumption of at least a 24 in column diameter is accurate. If it had not been accurate, G would be recalculated using a smaller packing which would also correspond to a larger packing factor. Step 3 : Determine the Column Height : Perhaps the most interesting step in designing a packing column is deciding how tall to build it. You should first ask yourself "What stage of the design are we currently working on ?" If the seeing is preliminary, the general HETP (Height Equivalent to a Theoretical Plate) will work well. If the design requires a higher degree of

Principles of Mass Transfer Operations − I (Vol. − I)

5.51

Gas Absorption

accuracy, we recommend consulting the packing manufacturer or a book entitled Distillation Design by Henry Kister (McGraw – Hill. Distillation Design contains an exhaustive list of HETP values based on the components of the system and the type of packing used. As for preliminary estimates, the following HETP values should be used. Table 5.5 : HETP values for various mass transfer operations SETUP Method Distillation

Vacuum Distillation Absorption/Stripping

HETP expressed as ft (meters) Packing Size (in) 1.0

1.5 (0.46)

1.5

2.2 (0.67)

2.0

3.0 (0.91)

1.0

2.0 (0.67)

1.5

2.7 (0.82)

2.0

3.5 (1.06)

All Sizes

6.0 (1.83)

To determine the height of the adsorption tower in our example, we multiple the 20 theoretical stages by 6 ft or 1.83 m. We estimate the height of the tower to be 120 ft or about 37 meters. Short-cut design guidelines for packed bed absorbers : (A) Information available : 1. Feed gas flow rate and composition, V1, y1. 2.

Feed liquid quality (x2).

3.

Pressure at the top of the column (P) and operating temperature (T).

4.

Information on gas-liquid equilibrium (equilibrium relation y vs. x @ P and T).

(B) Information that might be available …… 5. Exit gas composition (y2) (goal of

6.

…… what do to if it is not available Assume 99% recovery of solute :

the separation process)

V1y1 – V2y2 = 0.99 V1y1

Feed liquid flow rate (L2)

Use 1.5 times the minimum : L2 = 1.5 L2, min

7.

Column diameter (D)

Assume gas superficial mass flow rate to be 50 to 80% of the flooding value.

(C) Report design parameters : 1. Use graphical correlation to estimate flooding superficial flow rate (Gy, F) and set Gy (0.5 to 0.8)Gy, F. Calculate column diameter from Gy. 2.

Calculate height of packed bed from appropriate design equation. Estimate height of transfer units using empirical correlations.

3.

Use actual gas and liquid superficial mass flow rates to determine total pressure drop from graphical correlation.

Principles of Mass Transfer Operations − I (Vol. − I)

5.52

Gas Absorption

SOLVED PROBLEMS (1) A mixture of acetone vapour and air containing 5% by volume of acetone is to be freed of its acetone content by scrubbing it with water in a packed bed absorber. The flow rate of the gas mixture is 700 m3/hr of acetone-free air measured at NTP and that of water is 1500 kg/hr. The absorber operates at an average temperature of 20 oC and a pressure of 101 kPa. The scrubber absorbs 98% acetone. The equilibrium relation for the acetone vapour-water system is given by : Y* = 1.68 x where, Y = k mole acetone/k mole dry air X = k mole acetone/k mole water Calculate : (a) Mean driving force for absorption, (b) Mass transfer area if the overall mass transfer coefficient is – kG = 0.4 k mole of acetone/m2. hr (k mole acetone/per k mole dry air) Sol. : Y2 = 0.00105 Gg

LS X2 = 0

Packed Bed Absorber

Y1 = 0.0526 X2= 0.0193 kmol/water kmol dry air Gg = 31.25 LS = 83.33 hour hour Fig. 5.41 : Absorption of acetone vapour and air in packed bed

700 k mole dry air Gas Rate = Gg = 22.4 = 31.25 hour 1500 k mole water Liquid Rate = LS = 18 = 83.33 hour y1 = 0.05 y1 Y1 = 1 – y

So,

1

0.05 = 1 – 0.05 = 0.0526

Y2 = 0.02 y1 = 0.02 × 0.0526 = 0.00105 X2 = 0 since the entering water is pure. Material Balance : Gg (Y1 – Y2) = LS (X1 – X2) ∴

X1 =

Gg (Y – Y ) = 0.0193 L  1 2  S

The quantity of acetone absorbed is, m = Gg (Y1 – Y2) = 1.611

k mole acetone hour

Principles of Mass Transfer Operations − I (Vol. − I)

5.53

Gas Absorption

The driving force for the absorption process is : (i) at the bottom of the tower : ∆Y1 = = = =

*

Y1 – Y1 Y1 – 1.68 X1 0.0526 – 1.68 × 0.0193 0.0202

(ii) at the top of the tower : ∆Y2 = = = = So, the mean driving force is, ∆Y =

*

Y2 – Y2 Y2 – 1.68 Y2 0.00105 – 0 0.00105 ∆Y1 – ∆Y2 ln

∆Y1

=

∆Y   2

0.0202 – 0.00105 0.0202 ln 0.00105  

k mole acetone = 0.0065 k mole dry air and the required mass transfer area is, m 1.611 A = = kG · ∆Y 0.4 × 0.0065

(Ans. : a)

A = 619.6 m2 (Ans. : b) (2) 5000 kg/hr. of a SO2 – air mixture containing 5% by volume of SO2 is to be scrubbed with 2,00,000 kg/hr of water in a packed tower. The exit concentration of SO2 is reduced to 0.15%. The tower operates at 1 atm. The equilibrium relation is given by : Y = 30 X Mole SO2 where Y = Mole air Mole SO2 X = Mole water If the packed height of tower is 0.42 m, calculate the height of transfer unit. Sol. : LS X2 = 0

Y2 = 0.0015 Gg

Packed Bed Absorber

Y1 = 0.0526 kg mole Gg = 154.5 hour

X2= 0.0007 kg mole LS = 11,111 hour

Fig. 5.42 : Absorption of SO2 - air mixture in packed bed tower

Principles of Mass Transfer Operations − I (Vol. − I)

5.54

Gas Absorption

Average molecular weight of SO2 – Air mixture is, (64 × 0.05) – (29 × 0.95) = 30.75 5000 Total gas rate = 30.75 = 162.6 kg mole/hour Inert gas rate = Gg = Total gas rate × 0.95 = 154.5 kg mole/hour 5 Y1 = 100 – 5 = 0.0526 0.15 Y2 = 1 – 0.15 = 0.0015 Amount of water = 2,00,000 kg/hour 2‚00‚000 ∴ LS = = 11,111 kg mole/hour 18 Material Balance : Gg (Y1 – Y2) = LS (X1 – X2) X2 = 0, Since entering pure water to the tower. ∴ 154.5 (0.0526 – 0.0015) = 11,111 X1 X1 = 0.0007 ∴ Ye = 30 X 1

where So,

Ye = equilibrium value Ye = 30 × 0.0007 1

= 0.021 ...

X2 = 0 ,

Ye = 0 2

Y1 – Y2 (Y1 – Ye ) – (Y2 – Ye ) Thus,

NTU =

1

2

(Y1 – Ye ) 1

ln (Y – Y ) 2 e



2



0.0526 – 0.0015 NTU = (0.0526 – 0.021) – (0.0015 – 0) 0.0526 – 0.021 ln   0.0015  NTU = 5.17 Thus, Height of tower = Z = HTU × NTU 0.42 = HTU × 5.17 0.420 HTU = 5.17 = 0.0812 m (Ans.) (3) A packed tower is designed to recover 98% CO2 from a gas mixture containing 10% CO2 and 90% air using water. The equilibrium relation is given as y = 14 X. kg CO2 where, y = kg dry air kg CO2 X = kg water

Principles of Mass Transfer Operations − I (Vol. − I)

5.55

Gas Absorption

The water to gas rate is kept 30% more than the minimum value. Calculate the height of tower if (HTU)0G = 1 m. 10 Y1 = 100 – 10 = 0.1111 10 Y2 = 100 – 10 × 0.02 = 0.0022   X2 = 0 Since entering water is pure, y = 14 X in mole fraction term, X y 14 44 = 14 1 in mole ratio term. 1 29 18 y = 8.69 X (y and X are mole ratios) y1 X1 = 8.69 = 0.0128

Sol. :

On simplification we get, For minimum liquid rate, Material Balance :

G (Y1 – Y2) = Lmin . X1 (... X2 = 0) ∴

L G

=

min.

L G

Actual

Y1 – Y2 X1

=

0.1111 – 0.0022 = 8.5 0.0128

L = 1.3 G  

min.

= 1.3 × 8.5 = 11.05 ∴ Liquid concentration is, Y1 – Y2 L G

'

= X1

Actual '

0.1111 – 0.0022 11.05

X1

=

Ye

= 8.69 X1

0.0098

'

1

= 8.69 × 0.0098 Ye

=

2

= 0.085

= 0, since X2 = 0 Y1 – Y2 (Y1 – Ye ) – (Y2 – Ye )

∴

So, height of packed tower,

NTU =

1

2

Y1 – Ye

 ln Y – Y   2 e 2

Z = HTU × NTU Z = 11.29 m

= 11.29

1

=

1 × 11.29 (Ans.)

Principles of Mass Transfer Operations − I (Vol. − I)

5.56

Gas Absorption

(4) Gas from a petroleum distillation tower has its concentration of H2S reduced from k mole H2S 0.03 k mole inert hydro carbon gas to 1% of this value by scrubbing with triethanolamine water solvent in a counter-current tower operating at 300 K and at 1 atm. H2S is soluble in such solution and equilibrium relation may be taken as, Y = 2X where

k mole H2S Y = k mole inert gas

k mole H2S X = k mole solvent Y2 = 0.0003 The solvent enters the tower free of H2S and leaves having concentration 0.013 k mole of H2S/k mole of solvent. If the flow of inert hydrocarbon gas is 0.015 k mole/m2 sec. of tower cross section and the gas phase resistance controls the process, calculate, (a) the height of absorber necessary. (b) the number of transfer units required. Data : The overall coefficient for absorption KG · a may be taken as 0.04 k mole/s.m3 of tower volume. (Unit driving force in Y). Driving force at the top of the column : = Y2 – Ye

LS X2 = 0

Gg

Packed Tower

Y1 = 0.03 X2= 0.013 k mole Gg = 0.015 LS m2 .s Fig. 5.43 : Absorption of gas from petroleum distillation tower

2

(... X2 = 0, Ye = 0)

= (0.01 × 0.03) – 0 = 0.0003 Driving force at the bottom of the column : = Y1 – Ye

2

1

= 0.03 – 2 × 1 (from equilibrium relation) = 0.03 – 2 × 0.013 = 0.004 (Y1 – Ye ) – (Y2 – Ye ) 1 2 Logarithemic mean driving force = = 0.00143 Y –Y



1



Material Balance : G g (Y 1 – Y 2 ) · A i.e.

  2

e1

ln Y – Y 2 e

= k G · aP (Y – Y e ) ln · A · Z

Gg (Y1 – Y2) = kG· aP (Y – Ye)ln · Z = 0.015 (0.03 – 0.0003) = 0.04 × 0.00143.Z Z = 7.79 ≅ 7.8 m

Principles of Mass Transfer Operations − I (Vol. − I)

5.57

Gas Absorption

Gm = (HTU)oG = k .a G

Height of transfer units

0.015 = 0.04 = 0.375 m So, Number of transfer units, Z (NTU)oG = (HTU) oG

=

7.8 0.375

=

20.8 ≅

21

(Ans.)

(5) The acetone from an acetone-air mixture is scrubbed with water in a packed absorption tower using 1" Rasching rings. The entering gas mixture has 2 mole % of acetone and the gas leaving the tower is expected to have acetone only to the extent of 0.2 mole %. The gas rate and liquid rate are 160 and 290 kg/hr. m2 respectively. The temperature is 25 oC and pressure is 1 atm. The equilibrium relationship is given by : Y = 2.53 X Following expressions are available to calculate the height of transfer units as under (in feet) :

0.5

HG = 6.41 G0.32 L–0.51 (NSC) g

0.22 0.5 L (NSC) l HL = 0.01 µ (2.42)  

where

G = lb/hr. ft2 (gas) L = lb/hr. ft2 (liquid) µ = Viscosity = 0.82 Cp (NSC)g = Schmidt number for gas phase = 1.6 (NSC)l = Schmidt number for liquid phase = 690

Determine the height of pakced tower. 2 Sol. : Y1 = 100 – 2 = 0.0204 0.2 Y2 = 99.8

= 0.002

Y = 2.53 X is the equilibrium relation G = 160 kg/hr. m2 = 32.77 lb/hr. ft2 L = 290 kg/hr. m2 = 59.40 lb/hr. ft2 0.50

HG = 6.41 G0.32 L–0.51 · (NSC)9

= 6.41 (32.77)0.32 (59.4)–0.51 (1.6)0.5 = 3.08 and

0.22 1 0.50 HL = 0.01 µ (2.42) (NSC)L  

59.4 0.22 = 0.01  [690]0.50 0.82 × 2.42 = 0.555 Average molecular weight of gas = 0.02 × 58 + 0.98 × 29 = 29.58

Principles of Mass Transfer Operations − I (Vol. − I) 160 Total gas rate = 29.58

5.58

Gas Absorption

= 5.409 kg mole/hr. m2

Gm = 0.98 × 5.409 = 5.301 kg mole/hr. m2 Liquid rate = 290 kg/hr. m2 = 16.11 kg mole/hr. m2 From solute balance, Gm (Y1 – Y2) = L (X1 – X2) ...

X2 = 0 Gm (Y1 – Y2) = LX1 Gm X1 = L (Y1 – Y2) = = 0.00605 Gm X1 = L Y1 – Y2 = 0.329

∴ So,

5.301 16.11 (0.0204 – 0.002)

from equilibrium relation,

So, ∴

Y = 2.53 X Y = mX + C = 2.53

m

(HTU)G = HG + M Ye

1

Gm H L L

= 3.08 + 0.329 × 2.53 × 0.555 = 2.53 X1

=

3.54 feet = 1.08 m

= 2.53 × 0.00605 = 0.0153 Y1 – Y2 (Y1 – Ye ) – (Y2 – Ye ) 1 2 NTU = Y –Y



1



  2

e1

ln Y – Y 2 e

0.024 – 0.002 = (0.0204 – 0.0153) – (0.002 – 0) 0.0204 – 0.0153 ln  0.002  

=

5.56

So, height of packed tower =

Z = HTU × NTU = 1.08 × 5.56 = 6 m (Ans.) (6) A soluble gas is absorbed from a dilute gas air mixture by counter-current scrubbing with solvent in a pakced tower. If the liquor led to the top of the tower contains no solute, show that number of transfer units required is given by : 1  ln 1 – mGm y1 + mGm N =   Lm  y2 Lm  mG m   1 – L  m   where, Gm and Lm = flow rates of the gas and liquid in k mole/S.m2 of tower area y1, y2 = mole fraction of the gas at the inlet and outlet of the column.

Principles of Mass Transfer Operations − I (Vol. − I)

5.59

Gas Absorption

The equilibrium relation is given as, ye = mx where, ye = mole fraction in the gas is equilibrium with x mole fraction in the liquid. Sol. : By definition, y2

(NTU)0G =

dy

⌠ y–y ⌡ e

…(1)

y1

A mass balance between the top and some plane in the tower where the mole fraction are x and y gives, Gm (y – y2) = Lm (x – x2)

…(2)

If we assume the inlet liquid is solute free, x2 = 0 Gm x = L m

and

(y – y2)

…(3)

If the equilibrium data is represented as, ye = mx Gm ye = m · L (y – y2) m

So,

…(4)

Put equation (4) in equation (1) y1

(NTU)0G =

⌠ ⌡ y2 y1

=

⌠ ⌡ y2

dy m · Gm y– L (y – y2) m dy mG m   mGm y 1 – L  + L y2 m  m 

on integration we get, (NTU)0G =

1  mGm y1 mGm    mGm ln 1 – Lm  y2 + Lm  1 – L  m  

(Ans.)

(7) Some experiments are made on the absorption of CO2 air mixture is 2.5 N caustic soda using a 250 mm diameter tower pakced to a height of 3 m with 19 mm Rasching rings. In one experiment at 1 atm, the following results were obtained : gas rate = G' = 0.34 kg/m2. s liquid rate = L' = 3.94 kg/m2.s The CO2 in the inlet gas is 3.15 PPM and in the exit gas 31 PPM. Estimate the value of overall gas transfer coefficient, kGa.

Principles of Mass Transfer Operations − I (Vol. − I)

5.60

Gas Absorption

G' = 0.34 kg/m2. sec

Sol. :

Gm = ? Y2 = 31 × 10

2

–6

L' = 3.94 kg/m .s X2 = 0

Packed Tower

–6

X1 = 0 Lm = ?

Y1 = 315 × 10 2 Gg = 0.34 kg/m .sec Gm = ?

Fig. 5.44 : Absorption of CO2

Conditions at the bottom of the tower : y1 = 315 × 10–6 G' = 0.34 kg/m2.s 0.34 Gm = Average molecular weight of air 0.34 k mole = 29 = 0.017 m2 .S Conditions at the top of the tower : y2 = 31 × 10–6 x2 = 0 L' = 3.94 kg/m2 sec 2.5 NaOH contains 2.5 × 40 gm/lit. = 100 kg/m3 NaOH (... Molecular weight of NaOH = 40) Mean molecular weight of liquid =

So,

NaOH  + Water  100 × 40 900 × 18

1000 = 20.2 kg/k mole k mole = 0.195 2 m . sec

3.94 Lm = 20.2

(

)

Material balance over the tower ... y ≅ y for dilute gases Gm (y1 – y2) A = kG a P (y – ye)ln · Z · A We assume that all resistance to transfer lies in the gas phase ∴ Driving force at the top of the tower = y2 – ye 2 –6

= 31 × 10 – 9 = 31 × 10

–6

… (1)

(... x2 = 0, ye = 0) 2

Principles of Mass Transfer Operations − I (Vol. − I)

5.61

Driving force at the bottom of the tower, = y1 – ye

∴ Log. mean driving force,

Gas Absorption

1

= 315 × 10–6 – 0 = 315 × 10–6 (315 – 31) × 10–6 = (y – ye)ln = –6 315 × 10  ln  –6   31 × 10 

= 122.5 × 10–6 Put all these values in equation (1), 0.0117 (315 – 31) × 10–6 = kG a × 101.3 × 122.5 × 10–6 × 3 (Ans.) ∴ kG.a = 8.93 × 10–5 k mole/m3.s (kN/m2) (8) Ammonia is to be removed from its mixture with air by scrubbing with water in a packed tower. A gas mixture entering the column contains 6% NH3 (vol.) and rest air (vol.) water free of NH3 enters in the column in counter-current direction. If 90% of the ammonia is to be removed using NH3 free water at the rate of 2 mole water per mole of air. Determine the exit concentration of ammonia. The gas-liquid equilibrium relationship is y = 0.08 x, where y = moles of NH3/mole air, mole of NH3 x = mole H O . 2 6 100 Sol. : Y1 = 6 = 0.06382 1 – 100 0.1 6 × 100 Y2 = = 0.0060362 6 × 0.1 1 – 100 LS X1 = 0 , G = 2 m Material Balance : Gm (Y1 – Y2) = LS (X2 – X1) 1 = 2 (0.06382 – 0.0060362) So, X2 = 0.0288 moles of NH3/mole water ∴ Exit concentration of NH3 in water moles of NH3 (Ans.) = 0.0288 moles of water (9) Equilibrium relationships for the system heptane-oil-air is given by y = 2X (Y and X are kg heptane/kg air and kg heptane/kg oil respectively). Oil containing 0.005 – kg heptane/kg oil is being used as solvent for reducing the heptane content of air from 0.10 to 0.02 kg-heptane per kg air in a continuous counter-current packed bed absorber. What column height is required to treat 1400 kg/(hr) (m2 of empty tower cross-section) of pure air containing heptane if the overall gas mass transfer coefficient is 320 kg/hr) (m3) per unit gradient of y. The oil rate employed is 3100 kg/(hr) (m2) solve analytically.

Principles of Mass Transfer Operations − I (Vol. − I)

5.62

Gas Absorption

Sol. : Air (Y)

dh

Oil (X)

Fig. 5.45 : Continuous countercurrent packed bed absorber

Writing material balances for heptane across the dotted boundary, we get, X (X – 0.005) = Y (Y – 0.02) where X and Y are the compositions of the liquid and air streams, leaving and entering the differential element of height dh. Y X = X (Y – 0.02) + 0.005

∴

X = 0.452 (Y – 0.02) + 0.005

… (1)

We have for the differential element, kg· dH (y – y*) = y dy ∴ Integrating both the sides, we get, Y H = k g

y2

dy

⌠ (y – y*) ⌡ y1

Now, since,

y* = 2x (given), we have, Y H = k g

y2

dy

⌠ y – 2x ⌡ y1

Now put value of x from equation (1) Y = k g

y2

dy

⌠ ⌡ (y – 2) (0.452 (y – 0.002) + 0.005) y1

Y = k g

y2

dy

⌠ ⌡ 0.1 y + 8.1 × 10–2 y1

–2 –2 Y 1400 y2 + 8.1 × 10   0.1 + 8 × 10  = (0.1) k ln  = ln    –2 0.02 + 8 × 10–2 g y1 + 8.1 × 10  (0.1) (320)

∴

= 25.7 m H = 25.7 m

(Ans.)

Principles of Mass Transfer Operations − I (Vol. − I)

5.63

Gas Absorption

(10) Carbon disulphide is to be absorbed from a dilute gas mixture of CS2 – N2 into a pure non-volatile oil at atm. pressure in a counter-current absorber. The mole fraction of CS2 in inlet gas stream is 0.05 and the flow rate of gas stream, G is 1500 k mole/hr. the equilibrium relation is given by : y = 0.5 x where x = mole fraction of CS2 in liquid stream. It is desired to reduce the mole fraction of CS2 in the exit gas stream to 0.005. (a)

L

Calculate the minimum value of G where L is the liquid flow rate in k mole/hr. L

(b) Derive the equation for the operating line if G is equal to 1.5 times the minimum value. Sol. : x2 = 0

y2 = 0.005 Countercurrent Absorber

x1 = ?

y1 = 0.05 Fig. 5.46 : Absorption of Carbon Disulphide in Countercurrent Absorber

G = 1500 k mole/hr.

Data given :

The equilibrium relation, y = 0.5 x Material Balance : Gm (y1 – y2) = LS (x1 – x2) ∴

y1 – y2 x1 – x2

at

L G

min.

=

 LS  G   m

y1 – y2 = y 1 0.5 – x2

 . y = 0.5  ∴ x = y  0.5  . .

Putting values in above equation, we get, 0.05 – 0.005 L = 0.05 G min. 0.5 – 0

L G

= 0.45

(Ans.)

min.

L When G is 1.5 times

L G

= 1.5 × 0.45

min.

= 0.675

(Ans.)

Principles of Mass Transfer Operations − I (Vol. − I)

5.64

Gas Absorption

The equation of the operating line for a dilute mixture is, L (x1 – x2) + y2 Y = 1.5 G  

(Ans.)

min.

(11) A counter-current plate absorber is to be installed for scrubbing of an air mixture containing 5% ammonia by volume. The scrubber is fed with water containing 0.002 mole NH3 per mole of water. The scrubbing water flows at a rate of 1 mole water per mole air. It is necessary to absorb 85% of the ammonia present in the gas by operating the absorber at 20 oC. mole NH3 k = 0.80 mole NH /mole H O 3 2 Calculate the concentration of NH3 in the outgoing liquid and estimate number of stages necessary for this operation. y = 0.80 x

Sol. :

0.05 y2 = 1 – 0.05

y1

L G x1

mole NH3 = 0.0526 mole of air = (1 – 0.85) (0.0526) mole NH3 = 7.89 × 10–3 mole of air mole of water = 1 mole of air mole NH3 = 0.002 mole of H 2

x1 = 0.02 L

y1 = 7.80 × 10 G

–3

1

2 y2 = 0.0526 G1

L x2 = ?

Fig. 5.47 : Counter-current Plate Absorber for Ammonia-Air Mixture

So,

L (x2 – x1) = G (y2 – y1) x2 =

∴

G (y2 – y1) + x1 L

x2 = (0.0526 – 0.00789) = 0.04471

Principles of Mass Transfer Operations − I (Vol. − I) Now

L A=m

G

5.65

Gas Absorption

1 = 0.8 = 1.25 → Absorption factor.

Assuming Kresmer equation to be valid for this dilute system, ln NP =

ln A ln

=

yNP + 1 – mx 1 – 1  + 1   y – mx  A A   1  0.05260 – 0.8 × 0.002 1 – 1  + 1  0.00789 – 0.8 × 0.002  1.25 1.25  ln (1.25) NP = 3.96 ≅ 4

∴ Number of plates = 4

(Ans.)

(12) In an experiment 200 liters of air – SO2 mixture per minute is scrubbed continuously by water in counter-current fashion in a packed tower. The mixture contains 10% SO2 by volume and is admitted at 20oC and 1000 mm Hg pressure into the tower. During absorption the tower temperature is maintained constant at 20 oC by means of cooling arrangement. Determine the tower diameter required for absorbing 95% SO2. It may be assumed that the height of the tower may have no limitation. Vapour pressure of SO2 (gas) over aqueous SO2 solution at 20 oC is given below : % concentration of SO2

in water

0.5

1.0

2.0

3.0

5.0

10.0

26

59

123

191

336

698

(weight %) Partial pressure of SO2 mm Hg

Additional Data : Maximum allowable flow rate of water is 200 litres per hour per m2. Sol. : The equilibrium data are to be converted as : mole SO2 X = mole water mole SO2 Y = mole air

=

0.5 64 99.5 = 0.0014 18

Partial pressure of SO2 = Partial pressure of air 26 = 1000 – 26 = 0.0267

As shown above, the values are converted and the equilibrium relationship is obtained as : x

0.0014

0.0028

0.0057

0.0087

0.0148

0.0313

y

0.0267

0.0627

0.1403

0.2362

0.5061

2.31

Principles of Mass Transfer Operations − I (Vol. − I)

5.66

Gas Absorption

These values are plotted as shown in Fig. 5.48. 0.24 0.20 0.16

Y

0.12 Y1 0.08 0.04

X1 0.03

0.006

0.009

X

Fig. 5.48 : Plot of Y versus X

10 y1 = 100 – 10

= 0.1111

y2 = 0.05, y1 = 0.0055 P = 1000 mm Hg T = 20 + 273 = 293 K PV n = RT

=

0.2 1000 ×  760  0.082 × 293 = 0.011 kg mole/min.

Inert rate = 0.011 × 0.9 ≅ 0.01 kg mole/min. It is given that no limitation for scrubber height, minimum liquid can be used. When minimum liquid is used, the liquid concentration is obtained from the equilibrium plot. X1 = 0.0045 (from Fig. 5.48 corresponding to Y1 = 0.1111)

∴ Overall material balance :

Gm (Y1 – Y2) = Lm X1 (... X2 = 0) ∴

Lm = =

Gm (Y1 – Y2) X1 0.01 (0.1111 – 0.0055) 0.0045

= 0.231 kg mole/min. Maximum flow rate of water = 200 lit./hr. m2 =

200 60 × 18

= 0.185 kg mole/min. m2

Principles of Mass Transfer Operations − I (Vol. − I)

5.67

Gas Absorption

Cross sectional area of scrubber, 0.231 = 0.185

= 1.249 m2

∴

π Scrubber diameter = 4 · D2 = 1.249

∴

D =

1.249 × 4 π

D = 1.25 m

(Ans.)

(13) An effluent gas containing 0.15% by volume of a very soluble gas A is scrubbed counter-currently with pure water in a packed column at 298 K and atmospheric pressure to reduce the soluble A content to 0.01%. Find the number of theoretical plates in the column for a water flow rate of 1.7 times the minimum value. Take equilibrium data as : y* = 0.03 x. Sol. : Assume the molar balance about the column to be DILUTE; i.e. ∴

G1 ≈ G2 ≈ Gand L1 ≈ L2 ≈ L L (x1 – x2) = G (y1 – y2) y1 = 0.0015, y2 = 0.0001

Liquid solvent (water) contains no solute on entry so x2 = 0 Operating line of minimum gradient crosses equilibrium line at either entry or exit 0.0015 * conditions. For this, we see that x1 would be 0.0.3 = 0.05. y1 – y2 L = * G   min x1 – x2 = ∴

0.0015 – 0.001 = 0.028 0.05 – 0

L L = 1.7 G G   actual   min = 1.7 × 0.028 = 0.0476

∴

y1 – y2 0.0476 = x – x 1 2 =

∴

0.0015 – 0.0001 x1 – 0

0.0014 x1 = 0.0476 = 0.0294 mole fraction A

Principles of Mass Transfer Operations − I (Vol. − I)

5.68

Gas Absorption

0.16

Mole ratio A to S in gas phase

0.14 0.12 0.1 0.08 0.06 eqm y(L/G)min y(L/G)act Steps

0.04 0.02 0

0

1

2

3 4 Mole ratio A to S in liquid phase Fig. 5.49 : Graphical Solution for Problem (13)

5

6

The operating line is drawn between (0, 0.0001) and (00294, 0.0015). When drawn on the graph, the gap between this operating line and the equilibrium data needs four steps to pass from the top of the line to below the bottom. (Ans.) (14) A plate absorption column is used to reduce the concentration of a pollutant A in an air stream from 5.4% to 0.3% v/s by counter-current scrubbing with solvent "S". This solvent is fresh on entering the top of column and the gas stream enters at the bottom of the column at a flow rate of 2.4 m3/s at the column operation conditions of 293 K and 1 atmosphere. Equilibrium data is given by : X (k mole A/k mole S) 0 0.005 0.010 0.020 0.030 0.040 0.045 Y (k mole A/k mole air) 0 0.002 0.005 0.015 0.032 0.053 0.065 (a) Determine the minimum flow rate needed for fresh solvent S. (b) If the actual fresh solvent flow rate is 1.1 times the minimum, estimate the number of ideal plates needed. Sol. : (a) Rather than use mole fractions, use mole ratios (equilibrium data is in ratio). Y1 = y1/(1 – y1) = 0.054, (1 – 0.054) = 0.0571 Y2 = y2/(1 – y2) = 0.003, (1 – 0.003) = 0.0030 Mass balance : GS (Y1 – Y2) = LS (X1 – X2) From ideal gas equation : PV' G1 = RT =

(101.325 × 2.4) (8.3144 × 293)

= 0.0988 k mole/s GS = G1 (1 – y1) = 0.0988 (1 – 0.054) = 0.0944 k mole/s air

Principles of Mass Transfer Operations − I (Vol. − I)

5.69

Gas Absorption

Minimum liquid flow occurs when inlet gas and outlet liquid are in equilibrium. From *

graph, this gives x1 = 0.042.  LS  G   S

= min

∆Y 0.0571 – 0.0030 = = 1.288 0.042 – 0 ∆X*

(LS)min = 1.288 × GS = 1.288 × 0.0944 = 0.122 k mole/s  LS   LS  = 1.1 × G  = 1.1 × 1.288 = 1.417 G   S  S

(b)

actual

 LS  G   S

∴

min

= actual

∆Y 0.0541 = X – 0 = 1.417 ∆X 1

Mole ratio A to S in gas phase

X1 = 0.054, 1.417 = 0.038 Operating line links points (0.038, 0.0571) and (0, 0.0030 between which and the equilibrium curve gives 5 ideal plates. (Ans.) 0.07 0.065 0.06 0.055 0.05 0.045 0.04 0.035 0.03 0.025 0.02 0.015 0.01 0.005 0 0

eqm (L/G)minimum (L/G)actual Steps 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 Mole ratio A to S in liquid phase Fig. 5.50 : Graphical Solution for Problem (14)

(15) A gaseous mixture of ammonia and air is scrubbed counter-currently in a plate column with fresh water. The ammonia concentration is reduced from 7.0 mole% to 0.5 mole% and the inlet water and gas rates are respectively 450 and 370 kg/m2s. Calculate using the data below, the number of actual plates required for the absorber. Equilibrium data for the system at 20 oC and 1 atm is : X (k mole NH3/k mole H2O)

0.0050

0.0164

0.0252

0.0349

0.0455

0.0722

Y (k mole NH3/k mole air)

0.0054

0.0210

0.0320

0.0420

0.0533

0.0800

Take molar masses of air, ammonia and water as 29, 17 and 18 kg/k mole respectively. State any assumptions made. Plate efficiency = 67% for all plates in column. Sol. : Equilibrium data for the system at 20 oC and 1 atm is : x (mole % NH3)

0.4675

1.6135

2.4580

3.3723

4.3520

6.7338

y (mole % NH3)

0.5371

2.0568

3.1008

4.0307

5.0603

7.4074

Principles of Mass Transfer Operations − I (Vol. − I)

5.70

Gas Absorption

Assume that the solubility of air in water is negligible compared to the solubility of ammonia. Molecular weight of inlet gas = (0.93/29) + (0.07/17) = 28.16 kg/k mole. G1 = (370/28.16) = 13.14 k mol/m2s, of which 93% (12.22 k mole/m2s)

So, is air and

G2 = (12.22/0.995) = 12.28 k mole/m2s of which 99.5% (12.22 k mole/m2s) is air. The inlet liquid is pure water so L2 = (450/18) = 25 k mole/m2s. L1 = L2 + (G1 – G2) = 25 + (13.14 – 12.28)

So,

= 25.858 k mole/m2s Compare liquid and gas flow rates at both top and bottom before deciding if system is L2 25 dilute or not : G2 = 12.28 L1 G1

= 2.0358 26.858 = 13.14

Mole % ammonia in air

= 1.9679 These values are close enough to justify an approximation of liquid-to-gas ratio of 2.0019 throughout the column. y1 = 7%, y2 = 0.5% Liquid solvent (water) contains no solute on entry so x2 = 0. 7.5 7 6.5 6 5.5 5 4.5 4 3.5 3 2.5 2 1.5 1 0.5 0 0

eqm y(L/G)actual Steps 1

2

3

4

5

6

7

8

Mole % ammonia in water Fig. 5.51 : Graphical Solution for Problem (15)

When drawn on the graph, the gap between this operating line and the equilibrium data gives 3 theoretical plates. As the efficiency for all plates is 67%, this means that 5 actual plates are needed. (Ans.) (16) For a certain soluble gas/water system Henry's Law holds and y* = 0.15 x. Estimate the number of theoretical stages required to reduce the soluble content of a gas stream from 4.8% to 0.5% if the water flow is one fifth of the non-soluble gas flow.

Principles of Mass Transfer Operations − I (Vol. − I)

5.71

Gas Absorption

Sol. : Assume that the system is dilute. i.e. G1 ≈ G2 ≈ and L1 ≈ L2 ≈ L ∴ L (x1 – x2) = G (y1 – y2) y1 = 4.8 , y2 = 0.5 Assume that the water used is fresh so x2 = 0 y1 – y2 L = x1 – x2 G min

=

4.8 – 0.5 4.3 x1 – 0 = x1 = 0.2

x1 = 21.5%

So, 5 4.5

Mole % solute in gas

4 3.5 3 2.5 2 eqm y(L/G)act Step

1.5 1 0.5 0

0

5

10 15 Mole % solute in water Fig. 5.52 : Graphical Solution for Problem (16)

20

25

When we draw on the graph, the gap between this operating line and the equilibrium data gives 4 theoretical plates. (Ans.) (17) An acetone-air mixture containing 0.015-mole fraction of acetone has the mole fraction reduced to 1% of this value by counter-current absorption with water in a packed tower. The gas flow rate G is 1 kg/m2s of air and the water entering is 1.6 kg/m2s. Calculate, using the data below : (a) The number of overall transfer units NOG, (b) The height of packing required. Equilibrium relation : y* = 1.75x where y* is the mole fraction of acetone in vapour in equilibrium with a mole fraction x in the liquid. The overall coefficient for a absorption KGa = 0.06 k mole/m3s (unit mole fraction driving force)–1.

Sol. :

Molar mass of air = 29 kg/k mole Molar mass of water = 18 kg/k mole y1 = 0.015 y2 = 0.015' 0.01 = 0.00015

Principles of Mass Transfer Operations − I (Vol. − I)

5.72

Gas Absorption

x2 = 0 x1 = ? G = 1 kg/m2s = (1, 29) k mole/m2s = 0.0345 k mole/m2s L = 1.6 kg/m2s = (1.6, 18) k mole/m2s = 0.0889 k mole/m2s Overall Mass Balance gives : G (y1 – y2) = L (x1 – x2) 0.0.345 (0.015 – 0.00015) = 0.0889 (x1 – 0) x1 = 0.00576 *

Dy = y1 – 1

y1

= 1.75 x1 = 0.0101

* y2

= 1.75 x2 = 0

* y1

= 0.015 – 0.0101 = 0.0049 top driving force

*

Dy = y2 – y2 2

= 0.00015 bottom driving force

lm (y – y*) = (0.0049 – 0.00015), ln (0.0049, 0.00015) = 0.00136 NOG = (0.015 – 0.00015), 0.00136 = 10.92 Z = (G/KGa) NOG = (0.0345, 0.06) 10.92 = 6.28 m

(Ans. : a) (Ans. : b)

(18) A rotary film washer is used for a the removal of ammonia from coal gas at 20 oC. The washer is made up of eight brush compartments and is designed to treat 3700 m3/hr of gas with 1.5 times the theoretical amount of wash water. If the coal gas contains 1.3% NH3 on entering the scrubber and 0.3% NH3 on leaving, calculate : (a) The mass of water used per hour, (b) The number of theoretical stages, (c) The overall stage efficiency, (d) The NH3 content of the exit water. Equilibrium data at 20 oC : y* = 0.8 x Sol. : Assume that the system is dilute. i.e. ∴

G1 ≈ G2 ≈ Gand L1 ≈ L2 ≈ ≈ L L (x1 – x2) = G (y1 – y2) y1 = 1.3% , y2 = 0.3%

Assume that the water used is fresh so x2 = 0 Operating line of minimum gradient crosses equilibrium line at either entry or exit *

conditions. For this, we see that x1 would be (1.3 ÷ 0.8) = 1.625% y1 – y2 1.3 – 0.3 L = = 1.625 – 0 = 0.615 * Gmin x1 – x2 ∴

L L = 1.5 G G  actual   min = 1.5 × 0.615 = 0.923

Principles of Mass Transfer Operations − I (Vol. − I)

5.73

Gas Absorption

(a) Assuming that the system is dilute, the wash water molecular weight will approximate to 18 kg/k mole. Also, if the gas is assumed to be ideal at atmospheric pressure and 20 oC : pV = nRT p n V = RT

∴

=

101325 = 0.04159 k mole/m3 8314.4 × 293

∴

G = 3700 × 0.04159 = 153.89 k mole/hr

∴

L = 0.923 × 153.89 = 142.06 k mole/hr

(Ans.)

(b) When drawn on the graph, the gap between this operating line and the equilibrium data gives 3 theoretical plates.

(Ans.)

Mole % ammonia in coal gas

1.5 1.4 1.3 1.2 1.1 1 0.9 0.8 0.7 0.6

eqm y(L/G)min y(L/G)act Steps

0.5 0.4 0.3 0.2 0.1 0

0

0.2

0.4

0.6 0.8 1 1.2 1.4 Mole % ammonia in wash water

1.6

1.8

2

Fig. 5.53 : Graphial Solution for Problem (18)

(c) The actual number of plates will give the efficiency : Actual number of plates =

Theoretical number of plates Efficiency

∴

Efficiency =

Theoretical number of plates Acutal number of plates

∴

3 Efficiency = 7 = 0.43

(Ans.)

(d) The liquid-to-gas ratio will give the exit concentration :

 L  = 0.923 = y1 – y2 = 1.3 – 0.3 x1 – x2 x1 – 0 G ∴

x1 =

0.923 10 = 0.923 mole fraction NH3

(Ans.)

Principles of Mass Transfer Operations − I (Vol. − I)

5.74

Gas Absorption

(19) An experiment is carried out on the counter-current absorption of acetone from air at 298 K and 1 atm. using pure water. The following data is obtained : Packed height 4.0 m Gas flow rate 2.0 k mole/hr Water fluorite 5.6 k mole/hr Inlet gas mole fraction acetone 0.01 Outlet gas mole fraction acetone 0.00075 Outlet liquid mole fraction acetone 0.0004 Equilibrium relationship y* = 1.18 x Calculate : (a) The number of transfer units, NOG (b) The height of a transfer unit, HOG (c)

The overall gas phase mass transfer coefficient, KGa.

Sol. : (a) The compositions at entry and exit points will give the number of stages : y1 = 0.01, y2 = 0.00075, x1 = 0.0004, x2 = 0 y1

*

= 1.18 x1

= 0.000472

*

= 1.18 x2

=0

y2 y1

= y1 –

y2

= y2 – y2

* y1 *

0.01 – 0.000472

= 0.009528

Bottom driving force

= 0.00075 – 0

= 0.00075

Top driving force

NOG = (y1 – y2)

(b)

(c) ∴

ln ∆ y1 – ln ∆ y2 ∆ y1 – ∆ y2

(y1 – y2) = lm (y – y*)

ln (0.009528/0.00075) = (0.01 – 0.00075) 0.009528 – 0.00075 = 2.6786 meters z = HOG NOG z 4 HOG = N = 2.6786 = 1.4933 meters (Ans.) OG G (y1 – y2) = Kg aPT lm (y – y*) Az G (y1 – y2) Kga = P lm (y – y*) Az T

2 (0.01 – 0.00075) ln (0.009528/0.00075) × 0.009528 – 0.00075 1 × (π/4) 0.22 × 4 = 42.63 k mole/m3 hr. atm (Ans.) (20) Minimum Liquid Rate for Absorption The carbon dioxide issuing out of a fermenter contains 0.01 mole fraction of ethanol, which has to be reduced to 0.0001 mole fraction by scrubbing with water in a countercurrent packed tower. The gas flow rate is 227.3 kmol/hr and may be assumed constant throughout the tower. The equilibrium mole fraction of ethanol in the gas phase y* is related to that in the liquid x as y* = 1.07 x Determine the minimum liquid rate needed, and the number of overall gas-side transfer units needed at 1.5 times the minimum liquid rate. The entering liquid may be assume to be free of ethanol. =

Principles of Mass Transfer Operations − I (Vol. − I)

5.75

Gas Absorption

Solution : Mole fractions x, y and mole ratios X, Y are related as : X Y and x = 1 + X y = 1+Y y Y = 1–y

In reverse,

x and X = 1 – x

Therefore, equilibrium relation becomes Y* X 1 + Y* = 1.07 1 + X Inversing the above equation, 1 1+X 1 + Y* = 1.07 X Y* 1 Y*

1 1 = 1.07 X + 1.07 – 1

0.9346 – 0.0654 X 0.9346 – 0.0654 X = X X Y* = 0.9346 – 0.0654 X Nature of curve of equation (1) * X Y =

0.1 0.2 0.3

0.108 0.217 0.328

Y

… (1)

y2 G2

x2 L2

y1 G1

x1 L1

Operating line

Equilibrium curve

0

X Fig. 5.54 : Graphical analysis in Problem (20)

These values indicates that, the equilibrium curve in terms of moles ratios Y* vs. X is concave upward. The operation line Y vs. X is straight line. The minimum liquid-gas ratio then corresponds to an exit liquid concentration in equilibrium with the entering gas (i.e., the point of intersection of operating and equilibrium lines).

Principles of Mass Transfer Operations − I (Vol. − I)

5.76

Gas Absorption

Entering mole ratio of gas 0.01 Y1 = 1 – 0.01 = 0.0101 *

Exit liquid concentration X2 in equilibrium with Y1 is obtained from equation (1) as : *

0.0101 = *

0.00944 – 0.00066 X2 *

X2

X2 *

0.9346 – 0.0654 X2 *

= X2 = 0.00943

Minimum liquid rate Lmin : 0.0001 Y2 = 1 – 0.0001 = 0.0001 *

G (Y1 – Y2) = Lmin (X1 – X2) 0.0101 – 0.0001 Lmin = 227.3 0.00943 – 0 = 241.04 kmol/hr Operating liquid rate : L = 1.5 Lmin = (1.5) (241.04) = 361.56 kmol/hr Absorption factor A : Slope of operating line A = Slope of equilibrium curve L = mG 361.56 = (1.07) (227.3) = 1.4455 1 A

= 0.692

For absorption of dilute solutions, overall number of transfer units NtNOG is given by

NtOG =

1 1 y1 – mx2 lny – mx 1 – A + A



2



2





… (2)

1 1–A

Subtituting for the values in equation (2),

NtOG =

0.01 ln 0.0001 (1 – 0.692) + 0.692







1 – 0.692

= 11.2 Number of overall gas-side transfer units = 11.2.

… (Ans. )

Principles of Mass Transfer Operations − I (Vol. − I)

5.77

Gas Absorption

(21) Height of Absorption Column Equilibrium relationship for the system helptane-oil-air is given by Y = 2X (Y and X are kg-heptane/kg-air and kg=heptane/kg-oil respectively). Oil containing 0.005 kg-heptane/kg-oil is being used as solvent for reducing the heptane content of air from 0.10 to 0.02 kg-heptane/kg-air in a continuous countercurrent packed bed absorber. What column height is required to treat 1400 kg/(hr.m2 of empty tower crosssection) of pure air containing heptane if the overall gas mass transfer coefficient is ∆Y). The oil rate employed is 3100 kg/(hr.m2). Solve analytically. 320 kg/ (hr.m3.∆ Solution : Material balance for absorber : Ls (X1 – X2) = Gs (Y1 – Y2) 3100 (X1 – 0.005) = 1400 (0.1 – 0.02) X1 – 0.005 = 0.03613 X1 = 0.04113 Y

Y2 Gs

X2 Ls

Y1 Gs

X1 Ls

0.1 Operating line

Equilibrium curve

0

X

0.04113 Fig. 5.55 : Graphical analysis in Problem (21)

Equation of operating line : Ls Y1 – Y2 = G (X – X2) s 3100 Y – 0.02 = 1400 (X – 0.005) Y = 2.2143 X + 0.0089 Number of overall gas transfer units NtOG is related to gas-phase mole ratios as

… (1)

Y2

NtOG =

⌠ dY 1 + Y2 1  ⌡ Y – Y* + 2 ln 1 + Y1 Y1

where,

Y* = equilibrium gas composition in mole ratio. Given : Y* = 2X

… (2)

Principles of Mass Transfer Operations − I (Vol. − I)

5.78

Gas Absorption

Susbtituting for Y* in equation (2) and from equation (1), equation (2) becomes Y1

NtOG =

⌠ 1 + Y2 dY 1  ⌡ 0.2143X + 0.0089 + 2 ln 1 + Y1

… (3)

Y2

From equation (1), Y = 2.2143X + 0.0089 differentiating, dY = 2.2143 dX. Using this and changing the integration limits in terms of X, equation (3) becomes X2

NtOG =

⌠ 1 + X2 2.2143 dX 1  + ln 1 + X1 ⌡ 0.2143X + 0.0089 2 X1

Integrating,

2.2143 NtOG = 0.2143 ln

0.2143 X1 + 0.0089 + 1 ln 1 + X2 0.2143 X + 0.0089  2 1 + X 2 1  

… (4)

Substituting for X1 = 0.04113 and X2 = 0.005, in equation (4) 2.2143 (0.2143) (0.04113) + 0.0089 1 1 + 0.005 NtOG = 0.2143 ln  (0.2143) (0.005) + 0.0089  + 2 ln 1 + 0.04113   = 5.9375 – 0.0177 = 5.9198 Height of overall gas transfer unit HtOG is given by Gs HtOG = K a (1 – Y) Y *M where,

… (5)

Kya = overall gas phase mass transfer coefficient, and

(1 – Y)*M = mean of logarithmic averages of (1 – Y) and (1 – Y*) between terminals of tower, given as (1 – Y)2 – (1 – Y*)2  1  (1 – Y)1 – (1 – Y*)1 … (6) (1 – Y)*M = 2 ln [(1 – Y) /(1 – Y*) ] + ln[(1 – Y) /(1 – Y*) ]  1 1 2 2   From the diagram shown, it can be visualised that when

and when

*

Y1 = 0.02

Y1

= 0

Y2 = 0.1

* Y2

= 2 × 0.04113 = 0.08226

Substituting these in equation (6), 1 (1 – 0.02) – 1 (1 – 0.1) – (1 – 0.08226) (1 – Y)*M = 2 ln[(1 – 0.02)/(1)] + ln[(1 – 0.1)/(1 – 0.08226)]    = (0.5) + (0.99 + 0.9088) = 0.9494 Then, from equation (5), 1400 HtOG = (320) (0.9494) = 4.6081 Height of column = NtOG HtOG = 5.9198 × 4.6081 = 27.29 m

(Ans. )

Principles of Mass Transfer Operations − I (Vol. − I)

5.79

Gas Absorption o

(22) Experimental data on absorption of dilute acetone in air by water at 80 F and 1 atmosphere in a packed tower with 25.4 mm Raschig rings were obtained. The inert 2

gas flow was 95 lb-air (h.ft ) and the pure water

2

flow was 987 lb/(h.ft ). The 3

experimental mass transfer film coefficients are kG a = 4.03 lb-mol/(h.ft .atm) and –1

kLa = 16.6 h . The equilibrium data can be expressed by CA = 1.37 pA, where pA is in 3

3

atm and CA is in lb-mol/ft . The density of water is 62.216 lb/ft . (a) What are the heights of a gas and a liquid film mass tranfer unit ? (b) What is the height of an overall gas mass transfer unit ? ' ' (c) Calculate kx a and ky a. ' ' (d) How would you expect kxa and kya to vary if the operating temperature is dropped by a few degrees ? Do you expect any changes in the equilibrium line and how if there is ? Solution : 2

2

3

V' = 95 lb-air (h.ft ), L' = 987 lb/(h.ft ), kGa = 4.03 lb-mol/(h.ft .atm.) –1

*

kLa = 16.6 h , CA = 1.37 pA, p = 62.216 lb/ft

3

P 62.216 3 3 P = 1 atm., C = M = 18.015 lb-mol/ft = 3.453566 lb-mol/ft 95 28.966 V V' (a) hG = ' ≈ k aP = ft = 0.88362 ft 4.03 × 1 G t kya L hL = ' kxa

987 18.015 L' ≈ k aC = ft = 0.95567 ft 16.6 × 3.453566 L

1 m' 1 3.453566 1 3 + h.ft /lb-mol (b) K a = k a + k a = 4.03 × 1 1.37 × 16.6 × 3.453566 × 1 y y x 3

= 0.29211 h.ft /lb-mol 3

⇒ Kya = 3.42336 lb-mol/(h.ft ) V 95/28.966 HOG = K a ≈ 3.42336 ft = 0.958037 ft. y ' 3 3 (c) kx a ≈ kxa = 16.6 × 3.453566 lb-mol/(ft .h) = 57.329 lb-mol (h.ft ) ' 3 ky a ≈ kya = 4.03 lb-mol/(h.ft ) ' ' ' ' (d) T↓ , DAB ↓ ⇒ kx a↓, ky a↓ (As 'T' decreases, so DAB decreases, hence kxa and kya also decreases). T↓, p*↓ ⇒ lower vapour composition is expected higher equilibrium liquid composition.

Principles of Mass Transfer Operations − I (Vol. − I)

5.80

Gas Absorption

(23) An absorber, packed with pall rings to the height of 8 meters, is currently being used o

at 20 C and 2 atmospheres to remove a pollutant from an exhaust gas stream. Eight hundred cubic meters per hour of the exhaust gas containing 2.9% (volume) pollutant is fed to the bottom of the tower. By feeding a recycled non-volatile solvent steam (containing 0.4% by mol of pollutant) to the top of tower, the pollutant concentration in the existing gas stream is reduced to 0.15%. The solvent stream leaves the bottom of the tower containing 4.6% pollutant. The equilibrium data (X vs. Y) for the particular o

pollutant-solvent pair at 20 C and 1 atmosphere may be plotted on the graph. Calculate the following : (a)

What is the molar rate of the gas stream entering the tower, in mol/h ?

(b) What is the operating line ? Show that the operating line is given by : L' Y = Y2 + V' , (X – X2) (c)

What is the inert liquid solvent rate, mol/h, through the tower ?

(d) What is the ratio of (L//V/)/(L//V/) min ? (e)

What is the connecting line ? Show that the connecting line is given by ' kxa

1 + Xi ' Yi = (1 + Y) 1 + X kya – 1   (f)

Estimate the number of gas mass transfer units if the liquid film mass transfer resistance is negligible using three integration points.

(g) Calculate a kG' if the column has a diameter of 2 meters. Solution : o

Z = 8 m, T = 20 C, 2

Z

Pt = 2 atm

QG1 = 800 m3/h,

yA1 = 0.029

XA2 = 0.004,

yA2 = 0.0015

XA1 = 0.046

1

Fig. 5.56 : Absorber packed with pall rings in Ex. (23)

(a)

V1 =

QG1P 800 × 2 × 101325 RT = 8314 × 293.15 kmol/h 4

= 6.652 × 10 mol/h

(Ans.)

Principles of Mass Transfer Operations − I (Vol. − I)

5.81

Gas Absorption

(b) Operating line is derived from the mole balanced between the tower end and some middle point in tower. Between z and 2 : In = V'Y + L'X2 Out = V'Y2 + L'X ∴

V'Y + L'X2 = V'Y2 + L'X L' Y = Y2 + V' (X – X2)

⇒

(Ans.)

V' = V1 X (1 – y1)

(c)

4

6.652 × 10 = (1 – 0.029) mol/h 4

= 6.4589 × 10 mol/h L' V'

Y1 – Y2 = X –X 1 2 0.002987 – 0.001502 = 0.04822 – 0.004016 = 0.64175

XA1 X1 = 1 – X A1 = 0.04822 X2 = 0.004016 YA1 Y1 = 1 – Y A1 = 0.02987 Y2 = 0.001502 L' = 0.64175 V'

∴

4

= 4.145 × 10 mol/h o

(Ans.) o

(d) Equilibrium line was for T = 20 C and P = 1 atm. To convert to T = 20 C and P = 2 atm, only the Y scale needs to be changed. o

*

p = 1 atm ,

*

p = 2 atm

y1 atm y2 atm

o

Principles of Mass Transfer Operations − I (Vol. − I)

5.82

Gas Absorption

*

*

y2 atm

y1 atm = 2

y2 atm Y2 atm = 1 – y 2 atm y1 atm 2 = y1 atm 1– 2 y1 atm = 2–y 1 atm Y1 atm 1 + Y1 atm = Y1 atm 2 – 1+Y 1 atm Y1 atm = 2 (1 + Y 1 atm) – Y1 atm Y1 atm Y1 atm = 2–Y ≈ 2 1 atm Operating line

L' Y = V'

(X – X2) + Y2 *

Y1 ⇒ X1 ∴

= 0.05

Y1 – Y2  L'  = * V'min X1 – X2 0.02987 – 0.001502 = 0.0485 – 0.004016 = 0.6177

∴

 L'  V' 0.64175 = 0.6177 L'   V'min

= 1.0063

(Ans.)

Principles of Mass Transfer Operations − I (Vol. − I)

Y

5.83

Gas Absorption

0.06

0.03093 Y1

0.05

0.02564

0.04

0.02041

0.03

0.01523

0.02

0.01010

0.01

0.005025

0 0.00 0.00

0.01

0.02

0.03

0.04

0.05 X1

o

Fig. 5.57 : Graphical solutions for example (23)

(e)

Contacting line is the flux condition of the interface. Mass transfer flux from the gas phase to the interface is the same as the mass transfer flux from the interface to the liquid phase. kxa (xi – x) = kya (y – yi) 1 – yi 1–x ' ' ⇒ kx a ln 1 – x = ky a ln 1 – y i

∴

X Y x = 1+X , y = 1+Y 1 + Xi 1+Y ' ' kx a ln 1 + X = ky a ln 1 + Y i ' k xa

– '

(f)

⇒

1 + Xi kya – 1 Yi = (1 + Y) 1 + X  

(X1 = 0.04822,

Y1 = 0.02987)

(X2 = 0.004016, Y2 = 0.001502) Middle point

X = 0.026118,

Y = 0.015686

(Ans.)

Principles of Mass Transfer Operations − I (Vol. − I)

5.84

Gas Absorption

*

Y1 = 0.02961 ⇒

Y1 = 0.02987,

1 + Y* = – 3846 (Y* – Y) (1 + Y)*m

*

Ymid = 0.015686,

Ymid = 0.0102 ⇒ – 182

Y2 = 0.001502,

Y2 = 0.00065 ⇒ – 1173

*

Y2

NG =

1 + Yi ⌠  dY * ⌡ (Y – Y) (1 + Y) im i Y 1

0.001502 – 0.02987 × (– 3846 – 4 × 182 – 1173) 3 = 54.3 =

(Ans.)

(g) Z = HG NG Z 8 HG = N = 54.3 m = 0.1473 m G V V HG = ' = ' kya Ac kGa Ac P

∴

6.4589 × 10 × 1 + 4

V ' kG a = A P H = c G

∴



0.02987 + 0.001502 2  mol/h

π 2 3 4 × 1 × 2 atm × 0.1473 m

= 283.5 kmol/(h.atm.m3) = 0.07876 kmol/(s.atm.m3)

(Ans.)

(24) Pressure Drop and Tower Diameter : Ammonia is being asborbed in a tower using o

pure water at 25 C and 1.0 atm. abs. pressure. The feed rate is 2000 lbm/h (908 kg/h) and contains 3.0 mol% ammonia in air. The process design specifies a liquid-to-gas mass flow rate ratio GL/GG of 2.2 and flow rates at 60% flooding level. Use 1" Intalox packing. Calculate the (a) Pressure drop, (b) Gas and liquid flows and (c) Tower diameter o

Sol. : GG = 908 kg/h, 25 C, 1 atm, MG = 29.97 × 0.97 + 0.03 × 17 kg/kmol = 28.61 kg/kmol. pG =

PMG 101325 ∞ 28.61 3 3 RT = 8314 ∞ 298.5 kg/m = 1.169 kg/m

µL = 0.8937 CP, ρL = 987.08 kg/m3, FP = 134 m–1 (1" Intalor packing) (a)

–

∆Pflooding Z

= min

{167‚ 417 F0.7 P } mm H O/m 2

= 128.6 mmH2O/m = 1.54 in-H2O/ft GL GG

ρa ρL

= 2.2 ×

1.169 897.08 = 0.07533

Principles of Mass Transfer Operations − I (Vol. − I)

5.85

GG H2 0.05 From graph ⇒ A  p  C  L

Gas Absorption

Fp pG (pL – pG)

(5.41)

≈ 0.46 at flooding

o

Using 60 C flooding ⇒ GG = GG flooding × 0.6 GG µL0.05 AC ρL 

⇒

Fρ = 0.46 × 0.6 = 0.276 ρG (ρL – ρG)

mmH2O ∆p From graph, – Z ≈ 0.25 in H2O/ft = 20.8 m

(Ans.)

(5.41) GG µL–0.05 × AC = 0.276 ρL 

(b)

0.8937 × 10 = 0.276 ×  997.08 

ρG (ρa – ρG) Fρ

  

–2 –0.05

×

1.169 ∞ (987.08 – 1.169) kg 134 m2.s

= 1.4546 kg/(m2.s) GL = 2.2 GG = 1997.6 kg/h GG GL 2 AC = 2.2 AC = 3.2 kg/(m .s)

(Ans.)

C

GG mL rL A

( )

0.05

FP 0.6 –1.05 ,m s rG(rL– rG)

DP — = 2 in H2O/ft Z 0.6 0.5

Flooding

1.5 1.0 0.5

0.4 0.3

0.25 60% flooding

0.2

0.10 0.05

0.1 0.0 0.1 GL rG GG rL

0.01

Fig. 5.58 : Random Packing

(c)

GG = 908 kg/h = 0.2522 kg/s ∴

GG 0.2522 AC = G /A = 1.4546 m2 = 0.1734 m2 G C

1

Principles of Mass Transfer Operations − I (Vol. − I)

5.86

Gas Absorption

π 2 AC = 4 D 4 4 (Ans.) π AC = π × 0.1734 m = 0.4699 m (25) Minimum Liquid Flow in a Packed Tower : The gas stream from a chemical reactor contains 25 mol% ammonia and the rest inert gases. The total flow is 181.4 kmol/h to D =

5

an absorption tower at 303 K and 1.013 × 10 Pa. pressure, where containing 0.005 mol % ammonia is the scrubbing liquid. The outlet gas concentration is to be 2.0 mol % ' ammonia. What is the minimum flow Lmin ? Using 1.5 times the minimum, plot the equilibrium and operating line. Equilibrium Data for Ammonia-Water System Mole fraction of NH3 in Partial Pressure of NH3 in Vapour phase, pA, aqueous phase, xA mmHg o

20 C (293 K) 0 − − 12 15 18.2 24.9 31.7 50.0 69.6 114 166 227 298 470

0 0.0126 0.0167 0.0208 0.0258 0.0309 0.0405 0.0503 0.0737 0.0960 0.137 0.175 0.210 0.241 0.297

o

30 C (303 K) 0 11.5 15.3 19.3 24.4 29.6 40.1 51.0 79.7 110 179 260 352 454 719

Source : J. H. Perry, Chemical Engineers' Handbook, McGraw-Hill Book Company, 1984.

Solution :

V'Y2 + L'X = V'Y + L'X2 Y – Y2 L' ∴ X – X2 = V' = constant operating line is straight on (X, Y) plane. V1

L1

V

L

V2

L2

Fig. 5.59 : Minimum liquid flow in a packed tower

Principles of Mass Transfer Operations − I (Vol. − I)

5.87

⇒

y2 = 0.25

x1 = 0.00005 ⇒ ⇒

y1 = 0.02

Gas Absorption

0.25 Y2 = 1 – 0.25 = 0.333 0.00005 X1 = 1 – 0.00005 = 0.0000500025 0.02 Y1 = 1 – 0.02 = 0.020408

The equilibrium data is first converted to X and Y yA PA o Y = 1 – y = P – p , at 30 C A t A

XA X = 1–X , A XA

0

0.0126

0.0167

0.0208

0.0258

0.0308

0.0405

0.0503

……

X

0

0.01276

0.016 98

0.02124

0.02648

0.031 80

0.04221

0.052%

……

PA,

0

11.5

15.3

18.3

24.4

28.6

40.1

51.0

……

0

0.01536

0.02054

0.02606

0.03317

0.04052

0.05570

0.07193

……

mmHg Y

(A) Plot the equilibrium line on (X, Y) plane. (B) Locate the tower Top operating point (X1, Y1) (C) Draw a line (horizontal) at Y2 = 0.333 (D) The pinch line (minimum slope or maximum gas rate) is the line pass (X1, Y1) and intercept with the equilibrium line only once between Y1 < Y < Y2. As shown by the long dashed line. (See Fig. 5.60).

(

*

(E) Read the pinch point : X2,Y2 *

X2

∴

' Lmin V'

)

= 0.1664 =

Y1 – Y2 *

X1– XL

0.333 – 0.02041 = 0.1664 – 0.00503 = 1.9392 ⇒

' Lmin

= 1.9392 V' = 1.9392 × 181.4 (1 – 0.25) km = 263.8 kmol/h ' Lmin

L' V'

= 1.5 V'

L' V'

Y1 – Y2 = X –X 1 2

= 2.90875

Principles of Mass Transfer Operations − I (Vol. − I)

5.88

Gas Absorption

V' X2 = X1 – L' (Y1 – Y2)

⇒

= 0.00503 +

0.333 – 0.02041 2.90875

= 0.1126 Draw the operating line by connecting the two end points = Top (X1, Y1) and bottom (X2, Y2). 0.35

Bottom

0.30

0.25

0.20 Y 0.15

0.10

0.05

0.00 0.00

0.10 X

0.05

0.15

0.20

Fig. 5.60 : Graphical analysis for example (25)

(26) Design of Absorption Tower Using Transfer Units. The gas SO2 is being scrubbed from a gas mixture by pure water at 303 K and 1.013 × 5

10 Pa. The inlet gas contains 6.00 mol % SO2 and the outlet 0.3 mol% SO2. The tower cross-sectional area of the packing is 0.426 m2. The inlet gas flow is 13.65 kmol inert air/h and the inlet water flow is 984 kmol inert water/h. The mass transfer coefficients are HL = 0.436 m and kGa = 6.06 × 10

–7

kmol/(m3.s.Pa) and are to be assumed constant

in the tower of the given concentration range. Calculate the member gas transfer units NG and the tower height.

Principles of Mass Transfer Operations − I (Vol. − I)

5.89

Gas Absorption

Equilibrium Data for SO2-Water System

Mole fraction of

Partial Pressure of SO2 in vapour, mmHg

SO2 in liquid

20°°C (293 K)

0

30°°C (303 K)

0

0

0.0000562

0.5

0.6

0.0001403

1.2

1.7

0.000280

3.2

4.7

0.000422

5.8

8.1

0.000564

8.5

11.8

0.000842

14.1

19.7

0.001403

26.0

36

0.001965

39.0

52

0.00279

59

79

0.00420

92

125

0.00698

161

216

0.01385

336

452

0.0206

517

688

0.0273

698

–

(Source : T. K. Sherwood, Ind. Eng. Chem., 17, 745 (1925)]

Sol. :

5

A = SO2,

T = 303 K,

P = 1.013 × 10 Pa

y2 = 0.06,

Y2 = 0.06383,

y1 = 0.003,

Y1 = 0.003009

Ac = 0.426 m2,

V' = 13.65 kmol/h

x1 = X1 = 0

' –7 L' = 984 kmol/h, HL = 0.436 m, kG a = 6.06 × 10 kmol/(m3.s.Pa) HL =

L ' ' kya = kGa Pt ' kxa Ac 1 V

L 2

Fig. 5.61 : Design of Absorption tower using transfer units

Principles of Mass Transfer Operations − I (Vol. − I)

5.90

Gas Absorption

Mole balance in the tower Y'V' + L'X2 = X.L' + V'Y2 L' ⇒ Y = Y2 + V' (X – X2) straight line on (X, Y) plane V' X2 = X1 + L' (Y2 – Y1) 13.65 = 0 + 984 × (0.06383 – 0.003009) = 0.0008437 ' ' –7 5 ⇒ kya = kGa Pt = 6.06 × 10 × 1.013 × 10 kmol/(m3.s) –2

= 6.139 × 10 kmol/(m2.s) L 984 × (1 + 0.0008437/2) ' kmol/(m3.s) kx a = H A = 0.436 × 3600 × 0.426 L C = 1.4722 kmol/(m3.s) V HG = ' kya Ac 13.65 × [1 + (0.003 + 0.06)/2]/3600 m = 0.1496 m –2 6.383 × 10 × 0.426 (A) Convert the equilibrium data to (X, Y) =

XA

0

0.0000562

0.0001403

0.000280

0.000422

0.000564

0.000842

…

XA X=1–X A

0

0.0000562

0.0001403

0.0002801

0.0004222

0.0005643

0.0008427

…

pA, mmHg

0

0.6

1.7

4.7

8.1

11.8

18.7

…

PA Y=P –P t A

0

0.0007901

0.002242

0.006223

0.01077

0.01577

0.02661

…

(B) Plot equilibrium line. (C) Plot operating line by connecting (X1, Y1) and (X2, Y2). (D) Choose three points for integration. Plot the connecting line passing (X1, Y1), X1 + X2 , Y1 + Y2  to intercept with the equilibrium line. The connecting line (X2, Y2) and 2  2  is specified by the flux condition : kxa (X1 – X) = NAa ℑ through liquid film = NAaℑthrough gas film = kya (y – yi) ⇒

1 – xi ' kx a ln 1 – x

1 – yi ' = ky a ln 1 – y

⇒

1 + Xi ' kx a ln 1 + X

1+Y ' = ky a ln 1 + Y

i

i ' kxa

⇒

1 + Xi– ' Yi = – 1 + (1 + Y) 1 + X kya   1 + Xi239 = – 1 + (1 – Y) 1 + X  

Principles of Mass Transfer Operations − I (Vol. − I)

5.91

Gas Absorption

0.07 Bottom 0.06

0.05

Y 0.04

0.03

0.02

0.01 Top 0.00 0.0000

0.0005

0.0010 X Fig. 5.62 : Graphical solution for example (26)

0.0015

(E) Tabulate the operating and interface conditions X

0

0.0008437

0.00042185

Y

0.003009

0.06383

0.033419

Yi

0.00123

0.04985

0.02449

1 + Yi (Y – Yi) (1 + Y)lm

562

71.06

111.5

Y2

NG =

1 + Yi ⌠  ⌡ (Y – Yi) (1 + Y)lm dY Y1

=

0.06383 – 0 × (71.06 + 4 × 111.5 + 562) 3×2

= 11.48 Z = N G HG = 11.48 × 0.1496 m = 1.72 m

(Ans.)

Principles of Mass Transfer Operations − I (Vol. − I)

5.92

Gas Absorption

EXERCISE FOR PRACTICE (1) A mixture containing 10 mole % solute and rest inert is fed to a packed tower in which 90% of the solute is absorbed. Solute-free water used for absorption contains 5% (mole) solute when it leaves the tower at the bottom. If the equilibrium relationship is Ye = 0.05 Xe and Hy = 0.5 m, Hx = 0.4 m, determine the height of packed section. (Ans. : NTU = 2.34, (HTU)G = 0.51, Height (Z) = 1.19 m) (2) A gas mixture containing 31 weight % ammonia and 69 weight % air is absorbed in a water in a counter current packed tower. The gas and liquid flow rates on solute free basis are 2150 moles/hr. m2 and 5276 moles/hr. m2 respectively. The inlet water is pure and the exit gas contains 1.5% of the ammonia entering the tower. The equilibrium relation is, Y* = 1.6 X where

Y* =

kg ammonia kg air

X =

kg ammonia kg water

Compute the number of overall gas phase transfer units.

(Ans. : NTU = 6.14)

(3) You are testing a new packed tower to strip oxygen from water using excess nitrogen. The oxygen-free water is to be used in microelectric manufacture. Your tower is small, about 2 m high and 0.6 m in diameter, filled with 1 inch Hy-pack rings. you expect the value of mG for oxygen is large and the dominant transfer coefficient in the liquid will be 2.2 × 10–3 cm/sec. The water flow rate is to be 300 cm3/sec. Estimate the value of oxygen that can be removed from this tower. (Ans. : 98% of the oxygen) 3

m (4) A gas absorber has to be designed to handle 900 hr of coal gas containing 2% (v/v) benzene. Coal gas enters at a temperature of 300 K and 805 mm Hg. 95% of benzene should be recovered by the solvent. The solvent enters at 300 k containing 0.005 mole fraction of benzene and has an average molecular weight of 260. Calculate the circulation rate of solvent per second if the column is to be operated at 1.5 times the minimum LS. Equilibrium data : where,

Y 1+Y

X = 0.125 1 + X

Y = Mole ratio of benzene to dry gas X = Mole ratio of benzene to solvent (Ans. : Circulation rate = 0.43 kg/sec.)

(5) Acetone is to be recovered from an air stream by scrubbing counter-currently with pure water in a plate column. The air stream has a flow rate of 0.1 kg/s and contains 2 mole % by volume acetone. The liquid flow rate is 1.4 times the minimum liquid flow rate. Dilute mixtures of acetone in water obey Henry's Law, y* = 1.2x, where y and x are the mole fractions of acetone in the gas and liquid phases respectively. The column has to recover 98% of the acetone from the gas stream. If the overall plate efficiency is 60%, estimate : (a)

The required water flow rate in kg/s;

Principles of Mass Transfer Operations − I (Vol. − I) (b)

5.93

Gas Absorption

The actual number of plates required in the column.

(Ans : (a) water flow rate = 0.099 kg/s; (b) Actual number of plates required in the column = 15) (6) An effluent gas, containing 0.15% SO2 is scrubbed with water in a packed column at 25°C to reduce the SO2 content to 100 PPM. Find the number of theoretical stages in the column for a water flow 1.7 times the minimum value. Equilibrium data : y = x/36 (Ans : The number of theoretical stages in the column = 4) (7) Sulphur dioxide is to be removed from an air stream in which it is present at a concentration of 2.96 %v/v. A packed absorption tower will be used of such a diameter 2

that the gas flow is 0.00922 kmol/m s. Pure water will be fed to the top of the tower and 2

it will descend at 0.437 kmol/m s counter-current to the gas. At 293 K and 1 atm., the 3

following data are available : KGa = 0.019 kmol/m s.atm y* = 29.6x. Calculate the height of the packing needed to remove 95% of the entering SO2 under these conditions. (Ans : Height of packing = 2.7089 meters) (8) An absorber of packed height 10.7 m is used to reduce the ammonia concentration in an air stream from 4.93% v/v to 0.2% by counter-current absorption with fresh water. The 2

inlet gas rate is 0.136 kmol/m s, based on the tower cross-sectional area, and the inlet water flow rate is 1.4 times the minimum required for the separation. The operating conditions are 298 K, 1 atmosphere and equilibrium is given by y* = 1.08x. Determine : (a) The absorption rate of ammonia; (b) The concentration of ammonia in the exit liquid stream; (c) The overall mass transfer coefficient (KGa) under these conditions. (Ans : (a) 6.43 x 10

−3

2

kmol/m .s; (b) x1 = 3.26% ; (c) KGa = 0.0519 kmol/m3.s.atm.)

(9) An air stream that is 40% saturated with acetone at 20°C and one atmosphere total pressure is to be treated with fresh water in an absorption column in order to recover 90% of the acetone. For water flow 1.8 times the minimum value, calculate the number of overall gas phase mass transfer units required. Saturated vapour pressure of acetone @ 20°C = 48.3 mm Hg. Equilibrium data y* = 1.75x, where y* is the mole fraction of acetone in the gas stream in equilibrium with mole fraction x in the liquid stream. (Ans : The number of overall gas phase mass transfer units required = NOG = 3.897) (10) A packed column is used to remove ammonia from the gas purged from an ammonia plant in order to reduce the effluent to 0.10 mole% NH3. The gas to be treated contains 2.0-mole% ammonia, 32-mole% nitrogen and 66-mole% hydrogen. It will be scrubbed with pure water in a counter-current absorber packed with 25 mm Berl saddles using an 2

inlet gas rate of 1.13 kg/m s and a pure water flow rate of 1.6 times the minimum value. Assume isothermal operation at 293 K and 1 atmosphere, since heat of solution and cooling due to water evaporation offset each other at this NH3 concentration. Equilibrium can be approximated to y* = x under these conditions. Neglect small 3

absorption of nitrogen and hydrogen. The value of KGa is 7.9 10–4 mol/m s. Pa. Find : (a) How many equilibrium stages (theoretical plates) are required ? (b) How many overall transfer units, NOG is required ? (c) How many meters of packing are required ?

Principles of Mass Transfer Operations − I (Vol. − I)

5.94

Gas Absorption

Hint : calculate average molecular weight of inlet gas mixture and use this to calculate 2

the inlet gas flow rate in kmol/m s. For this dilute mixture the gas flow rate can be assumed constant throughout column in mass balance. (Ans : (a) 5 equilibrium stages; (b) NOG = 5.889; (c) z = 7.829 meters) (11) An absorber, packed to the height of 2.4 m, is employed to reduce the solute (A) concentration in a gas stream from 6.5 to 1.0%. An aqueous stream, initially containing 1.0% A, is fed to the top of the tower. The feed the tower is twenty thousand cubic meter per hour of the gas mixture at 60 oC and 1 atm. The equilibrium data for this system are : XA

0.00

0.01

0.03

0.03

0.04

0.05

0.06

0.07

YA

0.00

0.002

0.005

0.010

0.021

0.036

0.055

0.079

3

(12) 5.66 m per minute of an air-SO2 mixture containing 10% SO2 by volume at 287K and 1 atm are to be scrubbed with water in a counter current packed tower for the purpose of recovering 95% of the SO2. Cooling coils will be installed in the tower to maintain the liquid temperature constant at 287 K. Determine the minimum water rate at which the tower would operate as planned, assuming there is no limitation on the height of packing 3 available. Assume density of SO2 = 2.94 kg/m . Vapour pressure of SO2 over aqueous SO2 solution at 287K is given below Concentration kg SO2/ 100 kg water

0.5

1.0

2.0

3.0

5.0

10.0

Partial Pressure SO2, mm Hg

26

59

123

191

336

698

[Solution notes : Step 1 :

Determine partial pressure of SO2 for the feed mixture.

Step 2 :

Determine the amount of SO2 to be absorbed in mass units (gm) = A.

Step 3 :

At minimum water rate, the gas and solution will be in equilibrium at the bottom of the column.

Given the SO2 partial pressure (of the gas), determine from the equilibrium data the corresponding solution concentration = B. Minimum amount of water = 100A/B.] (13) Vapour-liquid equilibrium data in mole fraction for the system acetone-air-water at 1 atm are as follows : y, acetone in air

0.004

0.008

0.014

0.017

0.019

0.0200

x, acetone in water

0.002

0.004

0.006

0.008

0.010

0.0120

(a) Plot the data (1) as graph of moles of acetone per mole air versus moles of acetone per mole water. (2) Partial pressure of acetone versus gm acetone per gm water. (3) y versus x. (b) If 20 moles of gas containing 0.015 mole fraction acetone is brought into contact with 15 moles of water in an equilibrium stage, what would be composition of the discharge 3 3 streams ? Assume density of air = 1.225 kg/m and density of acetone = 790 kg/m .

[Solution notes : For part (a), convert the supplied data to the variables needed for the plots in 1 and 2. For part (b), following steps are needed Step 1 : Use plot (1) to solve the problem graphically. Step 2 : Locate feed point F in the figure from step (1).

Principles of Mass Transfer Operations − I (Vol. − I)

5.95

Gas Absorption

Step 3 : Determine the slope of the operating line = – L/G. Step 4 : Extend the line starting from point F to meet the equilibrium curve at point A. Step 5 : For point A, read The x-axis value = A moles acetone/mole water. The y-axis value = B moles acetone /mole gas. Convert the product compositions to mole fractions.] (14) Ninety-five percent of the acetone vapour in an 85-vol % air stream is to be absorbed by counter-current contact with pure water in a value tray column with an expected overall tray efficiency of 50%. The column will operate essentially at 20°C and 1 atm. Equilibrium data for acetone-water at these conditions are : Mole % acetone in water

3.30

7.20

11.7

17.1

Acetone partial pressure, (mm Hg)

30

62.8

85.4

103.0

Calculate (use 100 kmol/h of feed mixture as a basis) (a)

The minimum value of L/G (the ratio of moles of water per mole of air).

(b) The number of equilibrium stages required using a value of L/G of 1.25 times the minimum. (c) The number of stages by using the Kremser method. (d) The concentration of acetone in the exit water stream. [Solution notes : (1) Convert the equilibrium data to X-Y form. (2) Locate the column top condition - point D (for gas-stream 95% of acetone has been removed; liquid stream is pure water). (3)

Draw the operating line joining point D and intersecting the equilibrium curve (problem a).

(4)

Draw the operating line for the specified slope (L/G is 1.25 times the minimum) and mark out the number of stages (problem b).

(5)

To use the Kremser method, compute Ai and fraction solute recovered. By trial and error, find the number of stages.

Fraction of solute absorbed = (6)

(Ai (N + 1) − Ai)/(Ai (N + 1) – 1)

Convert the X-value (obtained from the graphical solution in b) at the bottom of the column to mole fractions to define the Concentration of acetone in the exit water stream.]

(15) Gas, from a petroleum distillation column, has its concentration of H2S reduced from 0.03 kmol H2S per kmole of inert hydrocarbon gas to 1% of this value by scrubbing with a triethanolamine-water solvent in a counter-current tower operating at 300K and 1 atm. The equilibrium relation may be assumed to be given by Ye = 2X. The solvent enters the column free of H2S and leaves containing 0.013 kmol of H2S/kmol of solvent. If the flow 2

of inert gas is 0.015 kmol/s m of column cross-section, calculate, (a) The height of the absorber necessary for the desired separation. (b) The number of transfer units. The 3 overall coefficient for absorption KG a may be assumed to be 0.04 kmol/s m .

Principles of Mass Transfer Operations − I (Vol. − I)

5.96

Gas Absorption

[Solution notes : (1) Determine driving force at top, (2) Determine driving force at bottom. (3) Determine logarithmic mean driving force, (4) Apply the mass-balance equation and compute the column height, (5) Determine the height of a transfer unit. (6) Determine the number of transfer units.] NOMENCLATURE Any consistent set of units may be used, except as noted. Symbols

Meaning

a

Specific interface surface, area/packed volume, m2/m3

A

Absorption factor, L/mg, dimensionless

AE

Effective absorption factor, dimensionless

A, B, C

Components A, B and C

C

Concentration, moles/volume, mole/m3

ds

Equivalent diameter of packing, m

D

Diffusivity, m2/s

DL

Liquid diffusivity, m2/s

e

2.7183

Eo

Overall tray efficiency, fractional

FG, J FL, J

mole m2.s mole Liquid-phase mass transfer coefficient for component J, m2.s Gas phase mass transfer coefficient for component J,

G

Overall mass transfer coefficient, mole/m2.s Total gas rate for tray towers, mole/s; for packed towers, superficial molar mass velocity, mole/m2.s

GJ

Gas molar mass velocity of component, J, mole/m2 .s

G'

Gas mass velocity, kg/ m2. s

h

Heat-transfer coefficient, w/m2.s K

h'

Heat-transfer coefficient corrected for mass transfer, w/m2·s K

k

Thermal conductivity

kG

Gas mass transfer coefficient, mole/m2.s (N/m2)

kL

Liquid mass transfer coefficient, mole/m2.s (mole/m3)

kx

Liquid mass transfer coefficient, mole/m2.s (mole fraction)

ky

Gas mass transfer coefficient, mole/m2.s (mole fraction)

K

overall mass transfer coefficient (units indicated by subscripts, as for k's)

ln

Natural logarithm

Fo

Principles of Mass Transfer Operations − I (Vol. − I)

5.97

Symbols

Gas Absorption Meaning

L

Total molar liquid rate, for tray towers, mole/s

L'

Total liquid mass rate (tray towers), kg/s; total mass rate per unit Tower cross section (packed towers), kg/m2.s

Ls

Solvent mass rate (tray towers), kg/s; total mass rate per unit tower crosssection (packed towers), kg/m2.s

m

dy* Slope of the equilibrium curve, dx ; equilibrium distribution ratio, y*/x, dimensionless

M

Molecular weight, kg/mole

NJ

Mass transfer flux of component J, mole/m2.s

NP

Number of equilibrium trays

NTU (NTU)o

Number of transfer units Number of overall transfer units

P

Vapour pressure, kN/m2

p

–

Partial pressure, kN/m2

Pt

Total pressure, kN/m2

Pr

Prandtl number, Cpµ/K, dimensionless

S

Stripping factor, dimensionless

SC

Schmidt number, µ/ρD, dimensionless

t

Temperature, K

x

Concentration in the liquid, mole fraction, mole/mole

X

Concentration in the liquid, mole/ mole solvent, mole/mole

X'

Concentration in the liquid, mole/mole entering liquid, mole/mole

y

Concentration in the gas, mole fraction, mole/mole

Y

Concentration in the gas, mole/mole carrier gas, mole/mole

Y'

Concentration in the gas, mole/mole entering gas, mole/mole

Z

Height of packing, m

Greek letters : ∆

Difference

∈

Volume fraction of voids, dry packing, m3/m3

∈ο

Volume fraction of voids, irrigated packing, m3/m3

µ

Viscosity, kg/m.s

ρ

Density, kg/m3

φt

volume liquid Fractional total liquid holdup, volume packed space

Principles of Mass Transfer Operations − I (Vol. − I) Symbols

5.98

Gas Absorption Meaning

Subscripts : av A, B, C, J

Average Component A, B, C, J

G

Gas

i

Interface

L

Liquid

M

Logarithmic mean

n

Effluent from tray n

Np

Effluent from tray Np

O

Overall

r

Reference substance

0

Liquid entering top tray

1

Bottom of a packed tower; effluent from tray 1 (tray tower)

2

Top of a packed tower, effluent from tray 2 (tray tower)

Superscript : *

In equilibrium with the other phase.

✱✱✱ REFERENCES 1.

A.S. Foust, L.A. Wenzel, C.W. Clump, L. Maus and L.B. Andersen, "Principles of Unit Operations", Second Edition, John Wiley and Sons, 1980.

2.

R.E. Treybal, "Mass-Transfer Operations", Third Edition, McGraw-Hill, 1981.

3.

W.L. McCabe, J.C. Smith and P. Harriot, "Unit Operations of Chemical Engineering", Fifth Edition, McGraw-Hill, 1993.

4.

J.D. Seader and E.J. Henley, "Separation Process Principles", John Wiley and Sons, 1998.

5.

C.J. Geankoplis, “ Mass Transport Phenomena”, Columbus, Chio, 1972

6.

R.H. Perry and D. Green, "Perry's Chemical Engineers' Handbook", Seventh Edition, McGraw-Hill, 1997.

7.

Seader and Henley, “ Separation Process Principles”, John Wiley and Sons, 1998.

8.

J.M. Coulson and J.F. Richardson, "Chemical Engineering” (Vol, 1 and 2), Third Edition, Pergamon Press, 1986.

9.

J.R. Welty, R.E. Wilson and C.E. Pikes, “ Fundamentals of Momentum, Heat and Mass Transfer”, Wiley New York, 1980.

,,,

6 CHAPTER

HUMIDIFICATION AND DEHUMIDIFICATION OPERATIONS 6.1

Introduction

6.2

General Principles of Humidification and Dehumidification

6.3

Vapour Gas Mixture

6.4

Definitions of Humidity Terms

6.5

Adiabatic Saturation Process

6.6

Theory of Wet Bulb Temperature

6.7

Humidity Chart

6.8

The Lewis Relation

6.9

Gas Liquid Contact Operations

6.10 6.11

6.12

6.13

Adiabatic Operations Humidification and Dehumidification Equipments 6.11.1

Cooling Tower Overview

6.11.2

Operating Principle of Cooling Towers

6.11.3

Cooling Tower Selection

6.11.4

Hot Water Distribution Systems

6.11.5

Air Flow Distribution Systems

Cooling towers : Design and Operational Considerations 6.12.1

Types of Cooling Towers

6.12.2

Cooling Tower Theory

6.12.3

Design Considerations

Height of Cooling Tower Solved Problem Exercise for Practice Nomenclature References

(6.1)

Principles of Mass Transfer Operations − I (Vol. − I)

6.2

Humidification and Dehumidification Operations

6.1 INTRODUCTION The operations considered in this chapter are concerned with the interface transfer of mass and of energy, which result when a gas is brought into contact with a pure liquid in which it is essentially insoluble. While the term humidification operations is used to characterize these in a general fashion, the purpose of such operations may include not only humidification of the gas but dehumidification and cooling of the gas, measurement of its vapour content, and cooling of the liquid as well. The matter transferred between phases in such cases is the substance constituting the liquid phase, which either vaporizes or condenses. As in all mass transfer problems, it is necessary for a complete understanding of the operation to be familiar with the equilibrium characteristics of the systems. But since a simultaneous transfer of heat energy will invariably accompany the mass transfer in these cases as well, some consideration must also be given to the enthalpy considerations of the systems. 6.2 GENERAL PRINCIPLES OF HUMIDIFICATION AND DEHUMIDIFICATION Humidification and dehumidification operations are used frequently in industrial and domestic applications (e.g. : air conditioning). Typically in these operations, air or some other gas is contacted with water for the purpose of changing the humidity of the gas or lowering the temperature of either the gas or the water. The major industrial applications may be classified as follows : (a)

Water-cooling : This is by far the most important operation where it is used to cool water from heat exchangers, condensers etc., for eventual reuse by contacting it with atmospheric air. The cooling effect is due mainly to the latent heat of vaporization.

(b) Gas cooling : The gas is cooled by sensible heat transfer to the water and simultaneously humidified to some extent. (c)

Gas humidification and dehumidification : The principal process here is the transfer of water from or to the gas. Simultaneous heat transfer in either direction always takes place as well. The main purpose is the control of moisture content and temperature for reasons of comfort or to meet industrial process requirements.

Most industrial humidification/dehumidification operations are carried out by direct contact of the gas with water. Large packed towers fitted with grids or other high voltage packing is the preferred contacting device for air-water systems. It is essential to appreciate the fundamental properties of air-water systems. The important concepts are those of "dry bulb" and "wet bulb" temperatures, "relative humidity" or "saturation" and "dew point". In the experiment it will be necessary to measure temperatures and air humidities needed for the operating diagram. The “dry bulb” thermometer, air humidity by the “wet bulb” thermometer, measures air temperature. Knowing air temperature and humidity, air enthalpies can be directly read from a psychometric chart. 6.3 VAPOUR GAS MIXTURES Character of the System : A system of two components, only one of which will condense in the range of operation. Simultaneous heat and mass transfer between liquid and gas phase. A : is condensable component (vapour and liquid) found in both gas and liquid phase. B : is no condensable component (gas), found in gas phase only.

Principles of Mass Transfer Operations − I (Vol. − I)

6.3

Humidification and Dehumidification Operations

Basic Assumptions : The total pressure in such systems is usually fairly low. In addition, the vapour (condensable component) exists at its pure component pressure, which is lower yet. Then, we normally assume ideal gases, so that : partial pressure = pure component pressure –

pt

pA = p A + p B → yA = p t –

–

–

pB yB = p t

… (6.1)

–

pt – pA pt

=

6.4 DEFINITIONS OF HUMIDITY TERMS (1) (a)

Absolute Humidity or Molar Absolute Humidity : Molar :

(b) Mass : (2)

–

–

pA

pA

moles A Y = moles B

yA = y B

mass A Y' = mass B

MA MA pA = Y M  = M   B   B  pt – p– A

=– = – pB pt – pA

… (6.2)

–

… (6.3)

Saturated Vapour-Gas Mixtures : If an insoluble dry gas B is brought into contact with

sufficient liquid A, the liquid will evaporate into the gas until ultimately (at equilibrium) the partial pressure of A in the vapour-gas mixture reaches its saturation value, i.e. the vapour pressure pA at the prevailing temperature. For saturation values at a given temperature T, replace –

p A in above expressions by the equilibrium vapour pressure pA. pA YS = p – p t A and

Y'S

pA = p –p t A

MA M   B

… (6.4)

(3)

Dry-bulb temperature tG : Thermometer reading in vapour-gas mixture phase.

(4)

Relative Saturation or Relative Humidity φ : mole fraction of A at T and pt in gas-vapour phase φ = mole fraction of A at T and p in saturated gas-vapour phase t –

pA φ = p A (5)

–

→ φ%

pA = 100 p A

… (6.5)

Percentage Saturation or Percentage absolute Humidity, S : –

pA pt – pA Y' Y Y' Y → S% = 100 Y = 100 Y' … (6.6) S = Y = Y' = p – S S A p –p S S t A (6)

Humid Volume vH : Volumes – always associated with 1 kg of dry gas.

Principles of Mass Transfer Operations − I (Vol. − I) Dry Volume :

RT vG = p M t B

6.4

Humidification and Dehumidification Operations 3

(=)

m of dry gas  kg of dry gas 

Humid Volume = volume of 1 kg of dry gas + volume of associated vapour RT RT RT 1 Y' m3 of mixture vH = p M + Y' p M = p M + M  (=)  kg of dry gas    t B t A t  B A vH = 8315

tG + 273 pt

 1 + Y'  MB MA

for ideal gas … (6.7)

vH = (0.00283 + 0.00456 Y') (tG + 273) for the air (B) – water (A) system at 1 std. atm. pressure RT vHS = p t

3

 1 + Y'S  (=) m of mixture  kg of dry gas  MB MA

… (6.8)

Total volume of gas-vapour mixture : V = vH (mass of dry gas ) (=) [m3] Plot vG and vHS lines against tG on psychometric chart. Any other vH can be obtained from these two by linear interpolation between the values at the same temperature according to percentage saturation. (7) Dew-Point, tDP : The temperature at which condensation begins (i.e. the vapour-gas mixture is saturated) when cooled at constant total pressure. (8)

Humid Heat, CS : The heat capacity of 1 kg of dry gas + the associated vapour. The heat

required to raise the temperature of 1 kg of dry gas and its accompanying vapour one degree at constant pressure. J for mixture  CS = CB + Y' CA (=)  kg dry gas oC

… (6.9)

(9) Enthalpy, H' : The (relative) enthalpy of a vapour-gas mixture is the sum of the (relative) enthalpies of the gas and of the vapour content. Choose the reference temperature t0, then the enthalpy of : Dry gas

HB = CB (tG – t0)

… (6.10)

Vapour

HA = CA, L (tDP – t0) + λDP + CA (tG – tDP)

… (6.11)

Gas-vapour mixture H' = HB + Y'HA = CB (tG – t0) + Y' [CA, L (tDP – t0) + λDP + CA (tG – tDP)] or with insignificant error H' = CB (tG – t0) + Y' [CA (tG – t0) + λ0] = (CB + CA Y') (tG – t0) + Y' λ0

… (6.12) … (6.13)

H' = CS (tG – t0) + Y' λ0 (constant CA, CA, L and CB are assumed) … (6.14) J for mixture H'S = (CB + CA Y'S ) (–t0) + Y'S λ0 (=)  kg dry gas  … (6.15)   Enthalpies for unsaturated mixtures on the psychometric chart can be interpolated between '

the saturated. Hs and dry gas, HB, values at the same temperature according to the percent saturation.

Principles of Mass Transfer Operations − I (Vol. − I)

6.5

Humidification and Dehumidification Operations

6.5 ADIABATIC SATURATION PROCESS

Fig. 6.1 : Adiabatic Saturation Process

where, G'S

=

superficial mass velocity of gas,

L'

=

superficial mass velocity of liquid,

H'

=

enthalpy of vapour - gas mixture,

HL

=

enthalpy of liquid,

λ

=

latent heat of vapourisation,

kg dry2 air  sm  kg liquid   s m2  J   kg dry air J   kg liquid J   kg liquid

If the air at (1) is not saturated, water will evaporate from the reservoir (water in the reservoir cools down) and energy will be transferred from the air to the water reservoir; Therefore the air temperature will decrease. Now, suppose we operate this equipment, so that the air at (2) is saturated (sufficiently large contact area) and the temperature of the make up water in the reservoir is adjusted to tG . Then, we have a steady flow process and tG is called Adiabatic 2

2

Saturation Temperature : Adiabatic Process ⇒ 1st Law ⇒ ∆EH = 0 Mass balance, L' = G'S (Y'2 – Y'1) Energy balance

… (6.16) … (6.17)

G'S HL' + L'HL = G' S H'2

… (6.18)

Substituting equation (6.17) into equation (6.18), H'1 + (Y'2 – Y'1 ) HL = H'2

… (6.19)

If equilibrium is reached at point 2, then : tG = tL = tas and equation 4 can be written as : 2

(HB + Y'1 HA ) + (YaS' – Y'1 ) HL aS= (HB aS + YaS' HA aS) 1

1

… (6.19 a)

or

(HB – HB aS) + Y'1 (HA – HA aS + HA aS – HL aS) = YaS' (HA aS – HL aS)

… (6.19 b)

and

(HB – Y'1 HA ) – (HB aS + Y'1 HA aS) = (YaS' – Y'1 ) (HA aS – HL aS)

… (6.19 c)

1 1

1

1

Recall the following relationships : (HA aS – HL aS) = λaS ; (HB + Y'1 HA ) = CS (tG – t0) + Y'1 λ0 1

and

1

1

1

(HB aS + Y'1 HA aS) = CS (taS – t0) + Y'1 λ0 1

Principles of Mass Transfer Operations − I (Vol. − I)

6.6

Humidification and Dehumidification Operations

and substitute them into equation (6.19 c), CS (tG – t0) – CS (taS – t0) = (YaS' – Y'1 ) λaS 1

1

… (6.20)

1

One can rearrange equation (6.20) to obtain the expression for the adiabatic saturation curve. CS λaS (YaS' – Y'1) 1 taS = tG – (YaS' – Y'1) C ⇒ (t – t ) = – λaS 1 S aS G 1

… (6.21)

1

[

To determine taS for given conditions of gas-vapour mixture tG ‚ Y'1‚ CS 1

1

] we must do a little

trial and error. Guess taS ⇒ look up λaS and paS ⇒ calculate YaS' and check taS However, taS is a unique temperature. This temperature is a constant for this process. 6.6 THEORY OF WET BULB TEMPERATURE, tWB : If air is not saturated, liquid will evaporate form wick and wick will cool. We then have a temperature difference between the air and the wick, so heat transfer occurs. Eventually, a steady state will be reached there the heat transfer from the air to the wick just supplies the energy needed for vapourization. The temperature of the wetted wick as indicated by the thermometer is the wet-bulb temperature, tWB.

Fig. 6.2 : Wet-bulb temperature

Rate of heat transfer to wick = hG A (tG – tWB) ' – Y') Rate of mass transfer to wick = kY, A (YWB ' – Y') Energy needed for vapourization = λWB kY, A (YWB At steady state, or

hG A (tG – tWB) = λWB kY, A (YWB ' – Y') tWB = tG –

(YWB ' – Y') λWB (hG/kY')

(YWB ' – Y') (hG/kY') ⇒ (t – t ) = – λWB WB G

… (6.22)

Equation (6.22) is very similar to equation 6.21. Note that tWB and taS would be identical if hG CS = k . 1 Y' In general,

hG kY'

Sc 0.567 = CS Pr

 

… (6.23)

Principles of Mass Transfer Operations − I (Vol. − I)

6.7

Humidification and Dehumidification Operations

6.7 HUMIDITY CHART While humidity charts for any vapour gas mixture can be prepared when circumstances warrant, the system air-water occurs so frequently that unusually complete charts for this mixture are available. Fig. 6.3 shows such a chart for SI system. It is prepared for a total pressure of 1 std. atm. It should be noted that all the quantities (absolute humidity, enthalpies, humid volumes) are plotted against temperature. For the enthalpies, gaseous air and saturated liquid water at 0°C were the reference conditions, so that the chart can be used in conjunction with the stream tables. The data for enthalpy of saturated air were then plotted with two enthalpy scales o provide for the large range of values necessary. The series of curves marked ì adiabatic saturation curves on the chart were plotted according to equation. For most purposes these can be considered as curves of constant enthalpy for the vapour-gas mixture per unit mass of gas.

Fig. 6.3 : Psychrometric chart for air-water vapour, 1 std atm abs, in SI Units

6.8 THE LEWIS RELATION We have seen that for the system air-water vapour, hGky is approximately equal to CS, or, approximately, hG/kyCS = 1. This is the so-called Lewis relation (after W.K. Lewis). Not only does it lead to near equality of the wet-bulb and adiabatic saturation temperatures (as in the ease of air water vapour) but also to other simplifications to be developed later. For air water system (hG/kY') ≈ 0.227. However, for the same air water system, CS is very close to 0.227 at moderate Y'. It is customary to just set, CS ≈ (hG/kY') and say tWB = taS … (6.24) Le is essentially unity for air-water vapour but not for other systems. 6.9 GAS LIQUID CONTACT OPERATIONS Common Use of Gas Liquid Contact : Direct contact of gas with pure liquid may have any of several purposes : Adiabatic operations Non-adiabatic operations Cooling a liquid Evapourative cooling Cooling a hot gas Dehumidification Humidification Dehumidification

Principles of Mass Transfer Operations − I (Vol. − I)

6.8

Humidification and Dehumidification Operations

Cooling a liquid

⇒

transfer of sensible heat from liquid to gas + evaporation.

Cooling a hot gas

⇒

transfer of sensible heat from gas to liquid + evaporation.

Humidification

⇒

transfer of mass from liquid to gas by vapourisation + heat transfer.

Dehumidification

⇒

transfer of mass (vapour) from gas to liquid by condensation.

6.10 ADIABATIC OPERATIONS Derivation of General Equation : Consider a section of a continuous contact-tower between z and z + dz. Arbitrary take that : tG > ti . tL (G = gas, i = interface, L = liquid). (equations will come out the same no matter what assumption we make here). Also, we will consider counter-current contact and base our calculations on 1 [m2] of area normal to the flow (actual column cross section, not free area).

Fig. 6.4 : Different section of a continuous counter current adiabatic gas liquid packed tower

Control Volume I – Mass and Energy Balance of the Gas Phase (differential volume) : δme = – G'S dY' – G'S dH' + HA G'S dY' – hG' aH (tG – ti) dz = 0 dH' CS dtG + HA dY' –G'S CS dtG = hG' aH (tG – ti) dz

… (6.25) … ( 6.26) … ( 6.27) … (6.28)

Principles of Mass Transfer Operations − I (Vol. − I)

6.9

Humidification and Dehumidification Operations

Control Volume II – Mass and Energy Balance of the Liquid Phase (differential volume) : (from equation (6.25) … (6.29) dL' = – δmi = G'S dY' d (L' HAL) + HALi δmi + hL aH (ti – tL) dz = 0

… (6.30)

L' CL dtL = (G'S CL dY' – hL aH dz) (ti – tL)

… (6.31)

Control Volume III – Overall Mass and Energy Balance (differential volume) : dL' = G'S dY' d (L' HAL) = G'S dH' L' CAL dtL = G'S CS dtG + G'S [CA (tG – t0) + λ0 – CL (tL – t0)] dY'

… (6.32) … (6.33)

Now let us apply these general relationships to several specific cases. In each case we will look at typical temperature and concentration profiles. We will also make some simplifying assumptions. As we will see, in the last specific case to be considered the simplifying assumptions do not have to be made, but, without them calculations become increasingly complicated. Evapourative Cooling of Water with Air : The purpose of this operation is to decrease tL by evapourating water. Latent heat of vapourization, λ, is very compared to the sensible heat effects. Therefore, small evapouration produces large cooling effects. Amount of water evapourated is usually very small compared to the water flow rate L', usually less than 3%. We will assume that L' = constant. Overall Energy Balance CV III :

Fig. 6.5 : Differential section of a continuous countercurrent adiabatic gas-liquid packed tower for evaporative cooling of water

d (L' HAL) = G'S dH' (remember L' = constant → dL' = 0)

… (6.34)

Principles of Mass Transfer Operations − I (Vol. − I)

6.10

Humidification and Dehumidification Operations

L' CAL dtL = G'S dH' L' CAL (tL – tL ) = G'S (H'2 – H'1) 2

… (6.36) … (6.35)

1

Equation (6.35) is an operating line; straight line in H' versus tL chart :

Fig. 6.6 : H' versus tL Chart

A mass balance in the control volume I (CV I) will yield; ' G'S dY' = kY aM (Y'i – Y') dz

… (6.36)

An energy balance on the interface (CV II) at steady state will give : ' hL aH (tL – ti) dz = hG aH (ti – tG) dz + HA kYaM (Y'i – Y') dz

… (6.37 a)

= HAL + λ0 + CA (ti – t0)

… (6.37 b)

i

HA

i

0

but λ0 >>> CA (ti – t0) and HAL0 = 0 (liquid water at reference temperature) HA

∴

i

Å λ0

… (6.37 c)

Now we can substitute equation (6.37 c) into equation (6.37 a) to obtain, hL aH (tL – ti) dz = hG aH (ti – tG) dz + λ0 k' Y aM (Y'i – Y') dz Equation (6.22) A mass balance in the control volume III will give : G'S dY' = dL' (remember L' Å constant → dL' = 0) ⇒ G'S dY' = 0 … (6.37 d) Also, from equation (6.36) we have, … (6.38) L' CAL dtL = G'S dH' Now, we can substitute equation (6.37), equation (6.36) and equation (6.37d) into equation (6.31). L' cL dtL = CL (g'S dY') (ti – tL) + hL aH (tL – ti) dz (6.31) Also, G'S dH' = hGaH (ti – tG) dz + λ0 kY' aM (Y'i – Y') dz Now, assume aH = aM = a, hG = kY' CS and CS = constant for air water system G'S dH'

= kY' a = kY' a

G'S dH'

= kY' a

[CS (ti – tG) + λ0 (Y'i – Y')] dz {[CS (ti – t0) + λ0 Y'i] – [CS (tG – t0) + λ0 Y']} [H'i – H'] dz = hL a (tL – ti) dz

Now we can integrate equation (6.39), G'S dH' = kY' a [H'i – H'] dz

dz … (6.39) … (6.39 a)

Principles of Mass Transfer Operations − I (Vol. − I) 2

⌠ ⌡

kY' a dH' H'i – H' = G'S

1

2

6.11

Humidification and Dehumidification Operations

kY' a G'S dz = C' z ⇒ z = k ' a S Y

⌠ ⌡ 1

2

dH'

⌠ H' – H' ⌡ i

… (6.40)

1

Also, from equation (6.39) we can obtain, H'i – H' hLa = – k' a ti – tL Y

… (6.41)

Equation (6.41) relates points on interface to points on operating line. L' CAL

' ' H2 – H1 L' CAL ' ' (tL – tL ) = GS (H2 – H'1) ⇒ t – t = ' 2 1 L2 L1 GS

The interface represents saturated conditions. Therefore, the values needed for integration of equation (6.40) can be obtained as shown on the diagram below :

Fig. 6.7 : H' versus tL Diagram

Number of Gas-Enthalpy Transfer Units ' H2

dH'

⌠ (H' – H') ⌡ i av

Height of Gas - Enthalpy Transfer Unit

H'2 – H'1 = (H' – H') = NtG i av

G'S HtG = k ' a Y

' H1

where, kg of vapour m2 interface ; a (=) 3 kg vapour m of tower packing m2 interfaces  kg dry air    Height of the active part of a cooling tower, Z : Z = HtG NtG Recirculating Liquid – Gas Humidification – Cooling : This is a special case where the liquid enters the equipment at the adiabatic saturation temperature of the entering gas. This can be achieved by continuously reintroducing the exit liquid to the contact tower without addition or removal of heat. In such a system, the temperature of the entire liquid will fall to and remain at, the adiabatic saturation temperature. The gas will be cooled and humidified, following along the path of the adiabatic saturation curve on the psychometric chart which passes through the entering gas conditions. G'S (=)

kg dry air ; kY' (=) s m2 of tower

If mass transfer is used as a basis for design : G'S dY' = kY' aM (Y'i – Y') dz

… (6.42)

Principles of Mass Transfer Operations − I (Vol. − I) But,

6.12

Yi = YaS' and ti = taS and aM = aH = a G'S dY' = kY' aM (YaS' – Y') dz ' Y2

⌠ ⌡

dY' YaS' – Y'

kY' a = G' S

Z

YaS' – Yi

Z

⌠ dz ⇒ ln Y ' – Y'2 = kY' a G' ⌡ aS S 0

' Y 1

(YaS' – Yi) – (YaS' – Y'2) YaS' – Y'1 ln Y ' – Y' aS 2

G'S (Y'2 – Y'1)

= kY' a Z

G'S (Y'2 – Y'1)

= kY' a Z AEYln' Y'2 – Y'1 YaS' – Y'1 = AEY ' = ln Y ' – Y' ln aS 2

NtG and

Humidification and Dehumidification Operations

G'S HtG = k ' a Y

thus Z = HtG NtG If heat transfer is used as the basis for design : – G'S CS dtG = hG aH (tG – ti) dz; assume constant CS = CB + Y' CA – G'S CS dtG = hG a (tG – taS) dz

Fig. 6.8

… (6.43)

… (6.44)

Principles of Mass Transfer Operations − I (Vol. − I)

6.13

Humidification and Dehumidification Operations

Fig. 6.9 tG 2

⌠ ⌡

dtG hG a = tG – taS G'S CS

tG 1

z

G'S CS

(tG – taS) 1

⌠ dz ⇒ Z = h a ln (t – t ) ⌡ G G2 aS 0

(tG – taS) – (tG – taS) G'S CS (tG – tG ) = hG a Z 1

1

2

1

2

tG – taS ln t

1

G2

– taS

G'S CS (tG – tG ) = hG a Z AEln and Z = HtG NtG 1

1

2

General Case : (L' – constant) Take a dehumidifier as an example. We want to develop a method which involves numerical solution.

Fig. 6.10

Note that tL – taS, so even though the process is adiabatic, we do not know tL throughout the 2

column. Also, we are not assuming L' = constant. Thus our unknowns are : L'1

Z tL

2

tG

2

Principles of Mass Transfer Operations − I (Vol. − I)

6.14

Humidification and Dehumidification Operations

Overall mass balance on column yields, dL' = G'S dY' ⇒ L'1 – L'2 = G'S (Y'1 – Y'2) (solve for L'1)

… (6.45)

Overall energy balance on column yields, we get, d (L' HAL) = G'S dH' ⇒ L'1 HAL – L'2 HAL = G'S (H'1 – H'2) 1

but,

2

… (6.46)

HAL = CL (tL – t0) H' = CS (tG – t0) + Y' λ0

[

L'1 CL (tL – t0) – L'2 CL (tL – t0) = G'S (CS (tG – t0) + Y'1 λ0) – (CS (tG – t0) + Y'2 λ0) 1

2

1

1

2

2

]

… (6.47) Equation (6.47) contains two unknowns, tL and tG . We have a trial and error solution. Note 1

2

that we do not know the path between the two ends of the column. As a matter of fact we do not really know one of the end points. Now, we can make energy balance on the interface : hG aH (tG – ti) dz + λi kY' aM (Y' – Y'i ) dz = hL aH (ti – tL) dz … (6.48) λi is an approximation for (H' – HL ); very small sensible heat term is ignored. i

Now assume, and

aH = kY' a CS =

aM hG a for air water system

Fig. 6.11

Then;

hG a hG a (tG – ti) + λi C (Y' – Y'i) = hL a (ti – tL) S dtG dz

… (6.49)

AEtG Å AEz

=

hG a (tG – ti)AVG G'S CSAVG

… (6.50)

dY' AEY' dz Å AEz

=

kY' a (Y'i – Y')AVG G'S

… (6.51)

dtL AEtL Å dz AEz

=

hL a (ti – tL)AVG G'S (t – t ) i L AVG LAVG ' LAVG ' CL

… (6.52)

Where subscript AVG indicates the average conditions over the increment AEZ dY' dL' G'S dz = dz

Principles of Mass Transfer Operations − I (Vol. − I)

6.15

Humidification and Dehumidification Operations

AEY' AEL' G'S AEZ = AEZ … (6.53) At a point where we know tG, Y' and tL, only the conditions at the interface are unknown. Thus, we could start by guessing ti and then : (a) evaluate interfacial conditions at point (1) from equation (6.49) (establish ti and Y'i ). or

1

1

(b) find (tG – ti)1 , (Y'i – Y')1, (ti – tL)1, CS , L'1 and use these values as average the AVG. value 1

in equation (6.33), (6.34) and (6.35) to obtain tG, Y' and tL at the end of the increment AEZ. Also calculate L' form equation (6.53). (c) Repeat step (a) for the values determined in (b), i.e. for the end of the increment AEZ. (d) Look at the average of the corresponding values calculated in (a) and (c) and see if they are close enough to values used in (b). If not, reiterate using a new average. The result will be a set of average values AEZ meters up the column. (e) Return to step (a) with these values at Z = Z + AEZ and do not next increment to get to Z = Z + 2 AEZ. (f) Continue until tL = known tL . Now check your values of Y'2 and tG (assumed). 2

2

If everything checks, your original assumption was correct. If not, change assumed tG

2

(calculate tL from equation 6.47 and repeat the whole procedure.) 1

6.11 HUMIDIFICATION AND DEHUMIDIFICATION EQUIPMENTS (A) Cooling Tower (B) Tray Towers (C) Spray Chambers (D) Spray Ponds Out of these equipments, Cooling Towers are most important. This is discussed in detail. 6.11.1 Cooling Tower Overview The primary task of a cooling tower is to reject heat into the atmosphere. This heat rejection is accomplished through the natural process of evapouration that takes place when air and water are brought into direct contact in the cooling tower. The evapouration is most efficient when the maximum water surface area is exposed to the maximum flow of air, for the longest possible period of time. Cooling towers are designed in two different configurations, Counterflow and Crossflow. The specific configuration indicates the direction of airflow through the tower relative to the direction of the water flow. Cooling tower water and air distribution systems are designed in concert, with each playing an equally important role in determining the efficiency and proper application of the cooling tower. 6.11.2 Operating Principle of Cooling Towers Just like air, water is one of our most important natural resources and vital to all our lives. The world’s growing water consumption combined with its increasing scarcity is making it more important than ever to use water intelligently and carefully. The re-cooling and recycling of water as a transport medium for waste heat for which there is no further intelligent use should therefore be the first rules of economy and ecology. Water-cooling systems : Water-cooling systems can be subdivided into various categories on the basis of the cooling water temperatures they require. At temperatures of below approx. + 20°C refrigeration machines are generally used. Above this temperature cooling towers are used. At temperature over 45 °C dry-type cooling units can also be employed.

Principles of Mass Transfer Operations − I (Vol. − I)

6.16

Humidification and Dehumidification Operations

Types of cooling towers : Cooling towers are distinguished according to various criteria : (1)

The forced of the air-stream : Naturally ventilated cooling towers (natural draught cooling towers). Artificially ventilated cooling towers (mechanical draught cooling tower). Fans may be of the induced or forced draught type design.

(2)

The relationship of the air-stream to the water flow : Counter-flow cooling towers, Cross-flow cooling towers, Combination of these two designs.

The design of the fill material : Open systems in which the water is cooled by direct contact with the surrounding air (wet-type cooling towers). Closed systems in which the medium to be cooled does not come into direct contact with the surrounding air. Hybrid cooling towers, a combination of open and closed systems. Wet-type cooling towers are able to achieve water temperatures below the ambient temperature, the decisive factor being the so-called wet-bulb temperature. This represents a physical value for the relationship between ambient temperature and relative air humidity. (3)

Wet-type cooling towers can reach the minimum cold water temperatures of approx. 3-4 K above the wet-bulb temperature. In India standard wet-bulb temperatures are dependent on the location. (4)

Open wet-type cooling towers : The water to be cooled is sprayed and trickled over fill material by a water distribution system, which, owing to its shape and position, guarantees high water and air contact times. At the same time, the surrounding air is fed through the tower in counter-flow, thereby evapourating a small part of the circulating water. The heat required for this evapouration process is drawn-off by the cooling water and provides the majority of the cooling capacity. The remainder of the cooling capacity is created by the convection of the warm water to the cold air. The re-cooled water collects in the collecting basin from where it is fed back to the cooling points. The saving in terms of cooling water in comparison with continuous flow cooling can be up to 97% of the circulation water, the remainder being required to compensate for the water loss due to evapouration and bleed-off.

(5)

Closed-type cooling towers and coolers : The medium to be cooled flows through a closed heat exchanger and does not come into direct contact with the surrounding air. Water is trickled over the heat exchanger by a water distribution system (secondary circuit) in order to use the evapouration heat.

(6)

Dry-type coolers : The medium to be cooled flows through a heat exchanger just as for closed cooling towers. Heat is removed by means of convection to the surrounding air thus allowing the system to attain cooling water temperatures in excess of ambient temperature.

6.11.3 Cooling Tower Selection Cooling towers are designed and manufactured in various sizes and configurations. Recognizing and understanding the different configurations and the advantages and limitations of each is essential to specifying the most cost effective solution for the end user. The purpose of this bulletin is to highlight the differences between Crossflow and Counterflow cooling towers and to describe applications where each configuration should be specified.

Principles of Mass Transfer Operations − I (Vol. − I)

6.17

Humidification and Dehumidification Operations

6.11.4 Hot Water Distribution Systems The overall efficiency of a cooling tower is directly related to the design of the tower's hot water distribution system. The primary consideration in selecting the type of hot water distribution system for a specific application is pump head. The pump head imposed by a cooling tower consists of the static lift (related to the height of the inlet) plus the pressure necessary to move the water through the distribution system and over the fill. The pump head varies according to the cooling tower configuration. Counterflow towers use a high-pressure spray nozzle hot water distribution system to achieve water coverage of the fill. The nozzle spray pattern is sensitive to changes in water flow, and consequent change in nozzle pressure. The air movement is vertically upward through the fill, counter to the downward fall of the water (Figure 6.12). Counterflow towers typically have a smaller footprint than Crossflow towers, but require additional height, static lift, and dynamic head to achieve the same cooling effect. Crossflow towers utilize a distinctly different type of water distribution system. Hot water is distributed to the fill by gravity through metering orifices in the floor of the inlet basin. There is no pressure spray distribution system. The air movement is horizontally through the fill, across the downward fall of the water (Figure 6.13). In Crossflow towers, the internal pressure component of pump head is insignificant because maximum flow is achieved by gravity. Compared to Crossflow towers, Counterflow towers may require up to five or six psig added pump head to achieve the proper spray distribution. The high Counterflow pumping head requirement (tower height plus nozzle pressure) leads to a higher may result in inadequate hot water flow, reducing tower efficiency and performance. First cost pumping system and significantly higher annual pump energy consumption and operating costs. If the system condenser pumps are not properly sized, the additional pump head required in Counterflow towers 6.11.5 Air Flow Distribution Systems Cooling tower performance is also related to the amount of air moving through the tower and coming into direct contact with the water. In Counterflow towers the air movement is vertically upward through the fill, counter to the downward fall of the water. This configuration, along with the finer water droplet size available from pressurized spray nozzles, allows Counterflow towers to make more efficient use of available air. However, the resistance to upward air travels against the falling water results in higher static pressure loss and greater fan horsepower than a Crossflow system. Crossflow towers have a fill configuration through which air flows horizontally across the downward flow of the water. Crossflow towers utilize essentially the full tower height for inlet louvers, reducing air inlet velocity and minimizing recalculation and drift loss. The air inlet louvers in Counterflow towers are restricted to the tower base, increasing inlet velocities and susceptibility to airborne trash and other debris. Crossflow : Advantages of Crossflow : Cooling towers due to their gravity flow hot water distribution system : (a) Low pumping head. (b) Lower first cost pumping systems. (c) Lower annual energy consumption and operating costs.

Principles of Mass Transfer Operations − I (Vol. − I)

6.18

Humidification and Dehumidification Operations

(d) Accepts larger variation in water flow without adverse effect on the water distribution pattern (flat plate heat exchanger operation in winter). (e) Easy maintenance access to distribution nozzles.

Fig. 6.12 : Crossflow cooling towers

Disadvantages of Crossflow : Cooling towers due to their gravity flow hot water distribution system : (a) Low-pressure head on the distribution pan may encourage orifice clogging and less water breakup at spray nozzle. (b) Exposure to air in the hot water basin may accelerate algae growth. Larger footprint. Counterflow : Advantages of Counterflow : Cooling towers due to their pressurized spray water distribution system : (a) Increased tower height accommodates longer ranges and closer approaches. (b) More efficient use of air due to finer droplet size from pressure sprays.

Fig. 6.13 : Counterflow cooling towers

Disadvantages of Counterflow : cooling towers due to their pressurized spray water distribution system : (a) Increased system pumping head requirements. (b) Increased energy consumption and operating costs. (c) Distribution nozzles difficult to inspect and clean. (d) Requires individual risers for each cell, increasing external piping costs.

Principles of Mass Transfer Operations − I (Vol. − I)

6.19

Humidification and Dehumidification Operations

Conclusions and Recommendations : The air and water distribution systems for counterflow and crossflow cooling towers have advantages and disadvantages inherent in their respective designs. It cannot be said that one is better than the other. Rather, with the proper application, both configurations are cost effective and can serve the end user well. Crossflow cooling towers should be specified when the following criteria and limitations are important : (a) To minimize pump head. (b) To minimize pumping and piping first costs. (c) To minimize operating costs. (d) When condenser water flow variance is expected. (e) When ease of maintenance is a concern. 6.12 COOLING TOWERS : DESIGN AND OPERATION CONSIDERATIONS Cooling towers are a very important part of many chemical plants. They represent a relatively inexpensive and dependable means of removing low-grade heat from cooling water.

Fig. 6.14 : Closed loop cooling tower system

The make-up water source is used to replenish water lost to evapouration. Hot water from heat exchangers is sent to the cooling tower. The water exits the cooling tower and is sent back to the exchangers or to other units for further cooling. 6.12.1

Types of Cooling Towers

Cooling towers fall into two main sub-divisions : natural draft and mechanical draft. Natural draft designs use very large concrete chimneys to introduce air through the media. Due to the tremendous size of these towers (500 ft high and 400 ft in diameter at the base) they are generally used for water Flowrates above 200,000 gal/min. Usually these types of towers are only used by utility power stations in the United States. Mechanical draft cooling towers are much more widely used. These towers utilize large fans to force air through circulated water. The water falls downward over fill surfaces, which help, increase the contact time between the water and the air. This helps maximize heat transfer between the two.

Principles of Mass Transfer Operations − I (Vol. − I)

6.20

Humidification and Dehumidification Operations

Types of Mechanical Draft Towers :

Fig. 6.15 : Mechanical draft counterflow tower

Fig. 6.16 : Mechanical draft cross flow tower

Mechanical draft towers offer control of cooling rates in their fan diameter and speed of operation. These towers often contain several areas (each with their own fan) called cells. 6.12.2 Cooling Tower Theory The transfer of sensible and latent heat transfers heat from water drops to the surrounding air.

Fig. 6.17 : Water Drop with Interfacial Film

This movement of heat can be modeled with a relation known as the Merkel Equation : KaV L

T1

=

dT

⌠ h –h ⌡ w a

… (6.54)

T2

KaV L

T1

=

dT

⌠ h –h = ⌡ w a

T1 – T2 4

 1 + 1 + 1 + 1  ∆h ∆H ∆j ∆h  2 3 4  1

T2

… (6.55)

Principles of Mass Transfer Operations − I (Vol. − I)

6.21

Humidification and Dehumidification Operations

where, ∆h1 = Value of hw – ha at T2 + 0.1 (T1 – T2) ∆h2 = Value of hw – ha at T2 + 0.4 (T1 – T2) ∆h3 = Value of hw – ha at T1 – 0.4 (T1 – T2) ∆h4 = Value of hw – ha at T1 – 0.1 (T1 – T2) where, KaV/L = Tower characteristic K = Mass transfer coefficient (lb water/h ft2) a = Contact area/tower volume V = Active cooling volume/plan area L = Water rate (lb/h ft2) T1 = Hot water temperature (oF or oC) T2 = Cold water temperature (oF or oC) T = Bulk water temperature (oF or oC) Enthalpy of air-water vapour mixture at bulk water hw = temperature (J/kg dry air or Btu/lb dry air) Enthalpy of air-water vapour mixture at wet bulb ha = temperature (J/kg dry air or Btu/lb dry air) Thermodynamics also dictate that the heat removed from water must be equal to the heat absorbed by the surrounding air : L (T1 – T2) = G (h2 – h1) So, where,

L G

h2 – h1 = T –T 1 2

… (6.56)

L/G = Liquid to gas mass flow ratio (lb/lb or kg/kg) T1 = Hot water temperature (oF or oC) T2 = Cold water temperature (oF or oC) Enthalpy of air-water vapour mixture at exhaust h2 = wet-bulb temperature (same units as above) Enthalpy of air-water vapour mixture at h1 = inlet wet-bulb temperature (same units as above)

The tower characteristic value can be calculated by solving equation 1 with the Chebyshev numerical method.

Principles of Mass Transfer Operations − I (Vol. − I)

6.22

Humidification and Dehumidification Operations

Fig. 6.18 : Graphical representation of tower characteristic

The following represents a key to Figure 6.18 : C'

=

Entering air enthalpy at wet-bulb temperature, Twb

BC

=

Initial enthalpy driving force

CD

=

Air operating line with slope L/G

DEF

=

Projecting the exiting air point onto the water operating line and then onto the temperature axis shows the outlet air web-bulb temperature

As shown by Equation (6.54), by finding the area between ABCD in Fig. 6.18, one can find the tower characteristic. An increase in heat load would have the following effects on the diagram in Fig. 6.19 : (1) Increase in the length of line CD, and a CD line shift to the right. (2) Increases in hot and cold-water temperatures. (3) Increases in range and approach areas. The increased heat load causes the hot water temperature to increase considerably faster than thoes the cold-water temperature. Although the area ABCD should remain constant, it actually decreases about 2% for every 10 oF increase in hot water temperature above 100 oF. To account for this decrease, an "adjusted hot water temperature" is used in cooling tower design.

Principles of Mass Transfer Operations − I (Vol. − I)

6.23

Humidification and Dehumidification Operations

Fig. 6.19 : Graph of Adjusted Hot Water Temperatures

The area ABCD is expected to change with a change in L/G, this is very key in the design of cooling towers. Cooling Tower Design : Although KaV/L can be calculated, designers typically use charts found in the Cooling Tower Institute Blue Book to estimate KaV/L for given design conditions. It is important to recall three key points in cooling tower design : (1) A change in wet bulb temperature (due to atmospheric conditions) will not change the tower characteristic (KaV/L). (2) A change in the cooling range will not change KaV/L. (3) Only a change in the L/G ratio will change KaV/L. The straight line shown in Figure 6.20 is a plot of L/G vs KaV/L at a constant airflow. The slope of this line is dependent on the tower packing, but can often be assumed to be – 0.60. Figure 6.20 represents a typical graph supplied by a manufacturer to the purchasing company. From this graph, the plant engineer can see that the proposed tower will be capable of cooling the water to a temperature that is 10 oF above the wet-bulb temperature. This is another key point in cooling tower design. Cooling towers are designed according to the highest geographic wet bulb temperatures. This temperature will dictate the minimum performance available by the tower. As the wet bulb temperature decreases, so will the available cooling water temperature. For example, in the cooling tower represented by Figure 6.16, if the wet bulb temperature dropped to 75 oF, the cooling water would still be exiting 10 oF above this temperature (85 oF) due to the tower design. Below is the summary of steps in the cooling tower design process in industry. More detail on these steps will be given later. (1) Plant engineer defines the cooling water flowrate, and the inlet and outlet water temperatures for the tower.

Principles of Mass Transfer Operations − I (Vol. − I)

6.24

Humidification and Dehumidification Operations

Fig. 6.20 : A Typical Set of Tower Characteristic Curves

(2) Manufacturer designs the tower to be able to meet these criteria on a "worst case scenario" (i.e. during the hottest months). The tower characteristic curves and the estimate is given to the plant engineer. (3) Plant engineer reviews bids and makes a selection. 6.12.3 Design Considerations Once a tower characteristic has been established between the plant engineer and the manufacturer, the manufacturer must design a tower that matches this value. The required tower size will be a function of : (1) Cooling range. (2) Approach to wet bulb temperature. (3) Mass flowrate of water. (4) Web bulb temperature. (5) Air velocity through tower or individual tower cell. (6) Tower height. In short, monographs available in literature utilize the cold-water temperature, wet bulb 2 temperature, and hot water temperature to find the water concentration in gal/min ft . The

Principles of Mass Transfer Operations − I (Vol. − I)

6.25

Humidification and Dehumidification Operations

tower area can then be calculated by dividing the water circulated by the water concentration. General rules are usually used to determine tower height depending on the necessary time of contact : Approach to Wet Bulb Cooling Range Tower Height 0 0 (ft) ( F) ( F) 15 – 20 25 – 35 15 – 20 10 – 15 25 – 35 25 – 30 5 – 10 25 – 35 35 – 40 Other design characteristics to consider are fan horsepower, pump horsepower, make-up water source, fogging abatement, and drift eliminators Operation Considerations. Water Make-up : Water losses include evaporation, drift (water entrained in discharge vapour), and blow down (water released to discard solids). Drift losses are estimated to be between 0.1 and 0.2% of water supply. … (6.57) Evapouration Loss = 0.00085 × water flowrate (T1 – T2) Blow down Loss = Evapouration Loss/(cycles-1) : Where cycles is the ratio of solids in the circulating water to the solids in the make-up water. Total Losses = Drift Losses + Evapouration Losses + Blow down Losses Cold Weather Operation : Even during cold weather months, the plant engineer should maintain the design water flowrate and heat load in each cell of the cooling tower. If less water is needed due to temperature changes (i.e. the water is colder), one or more cells should be turned off to maintain the design flow in the other cells. The water in the base of the tower should be 0 maintained between 60 and 70 F by adjusting air volume if necessary. Usual practice is to run the fans at half speed or turn them-off during colder months to maintain this temperature range. 6.13 HEIGHT OF COOLING TOWER The height of a water cooling tower can be determined by setting up a material balance on the water, an energy balance and rate equation for the transfer of heat in the liquid and gas and mass transfer in the gas phase. There is no resistance to mass transfer in the liquid phase and hence no concentration gradient in the liquid. Consider the counter current flow of water and air in a tower of height z as shown in above Fig. 6.21, the mass rate of flow of air per unit cross sectional area is G' is constant throughout the whole height of the tower and because only a small proportion of the total supply of water is normally evaporated (1 – 5%) the liquid rate per unit area L' can be taken as constant.

Fig. 6.21 : Flow in water cooling tower

Principles of Mass Transfer Operations − I (Vol. − I) Let,

6.26

Humidification and Dehumidification Operations

θ = temperature E = Enthalpy H = Humidity

Suffixes G, L, 1, 2 and f being used to denote condition in the gas and liquid at the bottom and top of the tower and of the air in contact with water. The five basic equations for an incremental height of column dz are as follows : (i)

Water Balance :

(ii) Enthalpy Balance :

dL' = G' dE

… (6.57)

G'dHG = L'dHL

… (6.58)

Since only small propositions (1 – 5%) of liquid is evapourated, Now,

Thus,

HG = S (θG – θo) + λE

… (6.59)

HL = CL (θL – θo)

… (6.60)

G' dHG = L' CL dθL

and

… (6.61)

dHG = S dθG + λdE

… (6.62)

Assuming the physical properties of the material do not changes appreciably, integration gives above equation over the whole height of the column, … (6.63) G' (HG – GG ) = L' CL (θL – θL ) 2

1

2

1

(iii) Heat Transfer from the body of the liquid to the interface : hL a dZ (θL – θf) = L' CL dθL … (6.64) where hL = Heat transfer coefficient in the liquid phase a = Interfacial area per unit volume of column Arranging equation (6.64) we get, dθL θL – θf

hL · a = L' C dZ L

(iv) Heat transfer from the interface to the bulk of gas : hG· a dZ (θf – θG) = G'S dθG where hG = Heat transfer in the gas phase. Rearranging this equation, we get, dθG θf – θG

=

hG·a GS · dZ

(v) Mass transfer from interface to the gas : hD ρa dz (Ef – E) = G' dE where, hD = Mass transfer coefficient for the gas

… (6.65) … (6.66)

… (6.67) … (6.68)

ρ = Mean density of air Rearranging this equation, we get, dE Ef – E

=

hD a ρ dZ G'

… (6.69)

These equations can't be integrated directly since the conditions at the interface are not necessarily constant. Using Lewis relation and writing hG as hD ρS from equation (6.66), G'S dθG = hD θ · a dZ (Sθf – SθG)

… (6.70)

Principles of Mass Transfer Operations − I (Vol. − I)

6.27

Humidification and Dehumidification Operations

from equation 6.68, G' λ dE = hD ρ · a dZ (λE – λE)

… (6.71)

f

Adding equation (6.70) and (6.71) we get, G' (S dθG + λdE) = hD ρ · adz [Sθf + λEf] – [Sθ4 + λE] G' dHG = hD ρ · a dz [Hf – HG] This gives,

dHG Hf – HG

=

hD · a ρ dz G'

Combining equation (6.61), (6.64) and (6.72) we get, hL HG – Hf = θL – θf hD ρ

(from equation 6.59) … (6.72)

… (6.73)

From equation (6.70) and (6.72) we have, HG dHG = θG – θf dθG

… (6.74)

From equation (6.70) and (6.68) we get, E – Ef dE = θG – θf dθG

… (6.75)

These equations will now be employed in the determination of the required height of cooling tower for a given duty. The method consists of the graphical evaluation of the relation between the enthalpy of the body of gas and enthalpy of the gas at the interface with the liquid. Consider equation (6.72), dHG hD · a · ρ = dz Hf – HG G' Integrating above equation we get, 2

Z=

⌠ ⌡

G' dZ hD a ρ

1

2

dHG

⌠ H –H ⌡ f G

(6.76)

1

assuming hD to remain approximately constant. Since (Hf – HG) is known as a function of HG, 1 (Hf – HG) can be plotted against HG and the integral is evaluated between the required limits. The height of cooling tower is then determined. SOLVED PROBLEMS (1) A humidifier is conditioning 15000 kg air per hour at 49 oC dry bulb and 32 oC wet bulb temperature by heating outside air, passing it through an adiabatic spray chamber in which it reaches 90% humidity and then reheating to the desired temperature. The outside air is at 4.5 oC and is fogy carrying 0.006 kg of liquid water per m3. To what temperature must the air be heated in the first heating operation ? What is the temperature of air as it emerges from the spray chamber before the final heating operations. Calculate the heat supplied in the first and final heating.

Principles of Mass Transfer Operations − I (Vol. − I)

6.28

Humidification and Dehumidification Operations

Sol. : Neglect the volume of liquid water and assume ideal gas law is applicable for inlet air. 0.082 × (273 + 4.5) = 22.75 m3/kg mole Thus, V = 1 1 1 ρ = V = 22.75 kg mole/m3 29 So, ρair = 22.75 = 1.275 kg/m3 1 kg wter Humidity = 0.006 × 1.275 = 0.047 kg dry air The complete process in a humidity chart is as follows : Exit Conditions of air : Dry bulb temperature = 49oC Wet bulb temperature = 32oC So, humidity = 0.023 kg water/kg dry air Inlet conditions of air : Dry bulb temperature = 4.5 oC Humidity = 0.0047 kg water/kg dry air Temperature of air after first heating = 68oC.

Fig. 6.22 : Process on a humidity chart

Temperature of air as it emerges from the spray chamber = 29 oC. 15,000 kg/hour of air at dry bulb, 49 oC and wet bulb, 32 oC. 1 Dry air = 1.023 × 15,000 = 14662.8 kg/hour Heat supplied : (i) First Heating : Humid heat = 0.24 + 0.45 × 0.0047

(ii) Final heating :

= Heat supplied = = Humid heat = = Heat supplied = =

0.242 k cal/kg dry air oC. 14662.8 × 0.242 (68 – 4.5) 22,5323 k cal/hour 0.24 + 0.45 × 0.023 0.250 k cal/kg dry air oC 14,662.8 × 0.25 (49 – 29) 73,314 kcal/hour

(Ans.)

(Ans.)

Principles of Mass Transfer Operations − I (Vol. − I)

6.29

Humidification and Dehumidification Operations

(2) Air containing 0.005 kg water vapour per kg of dry air is heated to 325 ok in a dryer and then passed to the lower shelves. It leaves these shelves at 60% relative humidity and is reheated to 325 ok and passed over another set of shelves, again leaving at 60% relative humidity. In all there are four sets of shelves, after which the air leaves the dryer. On the assumption that the material on each shelf has reached the wet-bulb temperature and that heat losses from the dryer can be neglected, determine : (a)

the temperature of the material on each set of shelves.

(b) the amount of water removed in kg/s, if 5 m3/sec moist air leaves the dryer. (c)

the temperature to which the inlet air would have to be raised to carry out the drying in a single stage.

Sol. :

Fig. 6.23 : The drying process

Assumptions : (1) Material on each shelf has attained the wet-bulb temperature. (2) Heat losses are negligible. (a)

Air leaves the dryer at 325 ok and H = 0.005 kg water/kg dry air and enters the first shelf. This corresponds to a wet-bulb temperature of 25 oC (298 oK). Moisture is removed along the constant wet bulbs line till 60% relative humidity, which is the exit condition of air from first shelf. Humidity of air leaving the 1st shelf = 0.016 kg water/kg dry air (from humidity chart).

Exit air at 25 oC W. B. and H = 0.016 is heated to 325 K (52 oC) D.B. in H2 (heater 2). Air leaves the heater at 325 K and 0.016 kg water/kg air which corresponds to wet bulb temperature of 32.5 oC. This is the temperature of material on the second shelf. This process is repeated for third and fourth shelves. The conditions of material and temperature of air are as follows : Shelf No.

Temperature of tray material oC

1 2 3 4

25.0 32.5 35.5 38.0

Air Condition (humidity) Entering Leaving 0.005 0.016 0.016 0.026 0.026 0.032 0.032 0.038 (Ans.)

Principles of Mass Transfer Operations − I (Vol. − I)

6.30

Humidification and Dehumidification Operations

The process on a humidity chart is given as under (Fig. 6.24).

Fig. 6.24 : Process path on humidity chart

(b) Final moist air conditions : Humidity = 0.038 Wet bulb temperature = 38oC Humid volume = 0.96 m3/kg dry air (Ans.) 5 Amount of dry air/sec = 0.96 = 5.21 kg. Water removed = 5.21 (0.038 – 0.005) = 0.172 kg/sec. (Ans.) (c) The path of the process for this condition is indicated by dotted line. The requisite temperature is 380 K(107 oC) from humidity chart. (3) Fresh air at 21.2 oC in which partial pressure of water vapour is 0.0118 atmosphere is blown at the rate of 214 m3/hour first through a pre-heater and then adiabatically saturated in a spray chamber to 100% saturation and again reheated. This reheated air has a humidity of 0.024 kg water vapour per kg dry air. It is assumed that the fresh air and the air leaving the reheater have the same percentage humidity. Determine : (a) the temperature of pre-heater, spray chamber and reheater. (b) heat requirements for pre-heating and reheating. 0.0118 18 Sol. : Humidity of entering air = 1 – 0.0118 × 29 = 0.0074 kg water/kg dry air From humidity chart, % humidity of fresh air = 48. So humidity of air leaving reheater = 48. (i) From humidity chart, temperature of air after pre heater = 65 oC. Preheated air is adibatically saturated till 10% saturation. Temperature of air from spray chamber = 28oC. The air is reheated back to 48% saturation. Temperature of air leaving the reheater = 41oC. (Ans.)

Principles of Mass Transfer Operations − I (Vol. − I)

6.31

Humidification and Dehumidification Operations

Fig. 6.25 : Complete process shown on a humidity chart

(ii) Heat requirements : Humid heat of air = 0.24 + 0.45 H where H = Humidity Humid heat = 0.24 + 0.45 × 0.0074 = 0.243 k cal/kg dry air oC 3 o Air rate is 214 m /hour at 21.2 C dry bulb and 0.0074 kg water vapour/kg dry air. H 1 T Humid volume = 18 + 29 × 0.82 × P   T = 21.2 oC = 273 + 21.2 = 294.2 k P = 1 atm. 1 kg dry air = 214 × 0.843 Humid volume, hour = 253.9 Heat requirements = 253.9 × 0.243 (65 – 21.2) = 2702.4 k cal/hour (Ans.) In the reheater, Humid heat = 0.24 + 0.024 × 0.45 = 0.251 Heat requirement = 253.9 × 0.251 (41 – 28) = 828.5 k cal/hour (Ans.) (4) Air at 1 atm is blown past the bulb of a mercury thermometer. The bulb is covered with a wick. The wick is immersed in an organic liquid (molecular weight = 58). The reading of the thermometer is 7.6°C. At this temperature, the vapour pressure of the liquid is 5 kPa. Find the air temperature, given that the ratio of heat transfer coefficient to the mass transfer coefficient (psychrometric ratio) is 2 kJ/kg.K and the latent heat of vapourization of the liquid is 360 kJ/kg. Assume that the air, which is blown, is free from the organic vapour. Sol. : For simultaneous mass and heat transfer, heat flux q and mass flux NA are related as q = NA λ … (1) where, λ is the latent heat of vapourization. Mass flux is given by '

NA = kY (Yw – Y') where,

kY = mass transfer coefficient; '

Yw

= mass ratio of vapour in surrounding air at satuation; and

Y' = mass ratio of vapour is surrounding air.

… (2)

Principles of Mass Transfer Operations − I (Vol. − I)

6.32

Humidification and Dehumidification Operations

q = h (T – Tw)

… (3)

Connectivity heat flux is given by,

where,

h = Heat transfer coefficient; Tw = Wet bulb temperature of air; and T = Dry bulb temperature of air

Substituting for NA and q from equation (2) and equation (3) in equation (1), '

h (T – Tw) 6.13 = ky (Yw – Y') λ '

T – Tw =

λ (Yw – Y')

… (4)

h/kY

Given : Y' = 0; λ = 360 kJ/k; h/kY = 2 kJ/kg.K; and Tw = 7.6°C. '

Yw

=

kg organic vapour at saturation kg dry air

5 58 = 101.3 – 5 29 = 0.1038 Substituting these in eqution (4) T – 7.6 =

(360) (0.1038 – 0) = 18.69 2

T = 18.69 + 7.6 = 26.29°C

(Ans.)

Temperature of air = 26.29°C (5) Air at 30°°C and 150 kPa in a closed container is compressed and cooled. It is found that the first droplet of water condenses at 200 kPa and 15°°C. Calculate the percent relative humidity of the original air. The vapour pressures of water at 15°°C and 30°°C are 1.7051 kPa and 4.246 kPa respectively. Sol. : Since the first droplet of water condenses at 15°C and 200 kPa, at this conditions the air is 100% humidity. pA (pt − ps) i.e. percentage humidity or percentage absolute humidity = 100 × p (p − p ) = 100 s

Therefore, where,

pA ps pt − pA = pt − ps pA = Partial pressure of water vapour ps = Vapour pressure of water pt = Total pressure of system

t

A

Principles of Mass Transfer Operations − I (Vol. − I)

6.33

Humidification and Dehumidification Operations

Assuming water-vapour air mixture as an ideal gas, 1.7051 Number of moles of water vapour per mole or dry air = 200 − 1.7051 = 0.0086 This ratio (moles of water vapour/mole of dry air) is not going to change for a closed system and before condensation. Therefore, partial pressure of A at 30°C and 150 kPa is found by equating this mole ratio as, pA pt − pA = 0.0086 pA 150 − pA = 0.0086 pA = 1.279 kPa Percentage relative humidity of the original air : Percentage relative humidity = 100 × (pA/ps) = 100 × (1.279/4.246) = 30.12% (6) A forced-draft counter current water-cooling tower is cool 3 kg/s water from 40°C to 20°C. The air is drawn into the tower from ambient conditions at 20°C and 1 atmosphere. The ambient air has a wet bulb temperature of 10°C. The gas film mass transfer coefficient is estimated using a correlation to give kya = 0.02 kmol/(m3.s). The liquid film heat transfer resistance is negligible, in other words, hL a is very large. The air rate is opened at 25% above the minimum. (a)

What is the enthalpy of the entering air stream per unit mass of the dry air ?

(b) What is the operating line ? Determining the slope for the possible operating line. (c)

Estimate the enthalpy of air stream leaving the tower per unit mass of the dry.

(d) What is the connecting line ? Determine the slope of the connecting line. Additional data are given as follows. Molar masses : 18.015 kg/kmol for H2O and 28.96 kg/kmol for dry air. Heat capacities : 1.005 kJ (kg.K) for dry air, 1.884 kJ/(kg.K) for water vapour and 4.141 kJ/(kg.K) for liquid water. The latent heat of water is 2501.4 kJ/kg. Density and viscosity of liquid water are approximately 997 kg/m3 and 0.89 cP; respectively. Vapor pressuer of water at 10°C, 20°C, 25°C, 30°C, 35°C and 40°C are : 1.228 kPa, 2.338 kPa, 3.169 kPa, 4.246 kPa, 5.643 kPa and 7.384 kPa, respectively. Data : Enthalpy of saturated air in reference to the zero degree Celsius dry air and liquid water.

Principles of Mass Transfer Operations − I (Vol. − I)

6.34

Humidification and Dehumidification Operations

200

Hy, kJ/kg-dry air

150

100

50

0

20

25

30 o TL, C

35

40

Graph 6.1

Sol. : 2

1

Fig. 6.26

L = 3 kg/s,

TL2

= 40°C,

TL1

= 20°C

TG1

= 20°C,

Tw 1

= 10°C

'

kya (a)

= 0.02 kmol/(m3.s), *

Tw1 ≈ Tambient air ⇒ Hy1 = Hy at 10°C PAS = 1.228 kPa PAS MA H1 = P – P × M t AS B 1.228 18.15 = 101.325 – 1.228 × 28.96 = 7.632 × 10–3 kg/kg.dry air

G = 1.25 Gmin

Principles of Mass Transfer Operations − I (Vol. − I)

6.35

Humidification and Dehumidification Operations

Cs = 1.005 + 1.884 H Hy 1

= (1.005 + 1.884 H1) (TG1 – TR) + 2501.4 × H1 kJ/kg.dry air = (1.005 + 1.884 × 7.632 × 10–3) (20 – 0) + 2501.4 × 7.632 × 10–3

Hy 1

= 39.48 kJ/kg.dry air

(Ans.)

(b) Operating line is the overall energy balance equation between the middle of the tower (any point) and one end of the tower. It is given by, Hy – Hy 1 TL – TL1

LG = G ,TL1 = 20°C, TL2 = 40°C

TL at Pinch = 33°C, Hy at Pinch = 116.8.

LG G

∴

=

max

Hy at Pinch – Hy 1 TL at Pinch – TL1

=

116.8 – 39.48 33 – 20

= 5.948 kJ/kg.K kg air Gmin =

(c)

(Ans.)

L · CL L  CL G

max

=

3 kg/s × 4.141 kJ/kg.K 5.948 kJ/k.kg dry air

= 2.0887 kg.dry air/s G = 1.25

Gmin = 2.611 kg.dry air/s LG G

LG 1 = 1.25  G   

max

kJ = 4.7584 kg.kg dry air Hy 2 – Hy 1 TL 2 – TL 1 Hy 2

=

L · CL G

LG = Hy1 + G (TL2 – TL1) = 39.48 + 4.7584 (40 – 20) = 134.648 kJ/kg-dry air.

(Ans.)

(d) Connecting line is the energy flux conditions at the interface. Energy transfer from the gas phase to the interface is the same as the energy transferred to the liquid ⇒ vertical line.

Principles of Mass Transfer Operations − I (Vol. − I)

6.36

Humidification and Dehumidification Operations

200

Hy, kJ/kg-dry air

150

100

50

0

25

20

30 o TL, C

35

40

ati

on

Graph 6.2 (7) A mixture of air and water-vapour has a dry bulb temperature of 60°C and an absolute humidity of 0.03 kg-water vapour/kg-dry air. The system pressure is at 1 atmosphere absolute. Evaluate. (a) Saturation absolute humidity (b) Relative humidity or relatie saturation (c) Dew point temperature (d) Humid volume (e) Humid heat (f) Enthalpy (g) Heat required to heat 1.2 m3 of this mixture to 120°C (h) Adiabatic saturation temperature (i) Wet-bulb temperature Sol. : (a) Saturation absolute humidity : Use the Psychometric chart.

1 00 % sa t

ur

0.15

0.03

o

60 C

Hs = 0.15 kg H2O/kg-dry air

Fig. 6.27 : Determination of saturation absolute humidity

Principles of Mass Transfer Operations − I (Vol. − I)

6.37

Humidification and Dehumidification Operations

(b) Relative humidity : HR =

PA ° × 100 % PA

PA MA H = (P – P ) M t A B HS =

° PA MA ° (Pt – PA) MB

⇒

MB H PA = M + M H · Pt A B

⇒

° PA

M B HS = M + M H · Pt A B S

M A + M B HS H HR = M + M H × H × 100 % A B S 18.02 + 28.97 × 0.15 0.03 = × × 100 % 18.02 + 28.97 × 0.03 0.15

Thus,

⇒

HR = 23.68 %

(Ans.)

sa tu

ra t

io n

(b) Dew point temperature :

10 0%

0.03

30.9

o

60 C

Tdp = 30.9°C

(Ans.)

Fig. 6.28 : Determination of dew point

1 H T υH = 8314 M + M  · P A t  B 1 0.03 60 + 273.15 = 8314 × 28.97 + 18.02 × m3-mixture/kg-dry 101325  

(d)

air (e)

(f)

υH = 0.9831 m3-mixture/kg-dry air

(Ans.)

CS = = = =

(Ans.)

Humid heat : 1.005 + 1.884 H kJ/kg-dry air.K 1.005 + 1.884 × 0.003 kJ/kg-dry air.K 1.0615 kJ/kg-dry air.K 1.0615 kJ/kg-dry air.K

Enthalpy : Hy = CS (T – T0) + Hλ0 Hy = 1.0615 × (60 – 0) + 2501.4 × 0.03 kJ/kg-dry air Hy = 138.73 kJ/kg.dry air

(g)

(Ans.)

Q = WB · CS ∆T 1.2 = 0.9891 × 1.0615 × (120 – 60) kJ = 77.27 kJ

(Ans.)

Principles of Mass Transfer Operations − I (Vol. − I)

6.38

Humidification and Dehumidification Operations

1 00 %

sa tu

ra t

io n

(h) Adiabatic saturation temperature :

adiabatic saturation curve 0.03

o

TS = 35.6°C

(Ans.)

o

35.6 C 60 C

Fig. 6.29 : Determination of adiabatic saturation temperature

(i)

Wet bulb temperature : For air water system Twb ≈ TS

⇒

Twb = 35.6°C

(Ans.)

(8) A plant requires 5000 kg/h of cooling water to flow through its overhead condenser in a distillation column. Hot water will leave the condenser at 46°C. It is planned to cool the water for reuse by contact with air in an induced draft cooling tower. Entering air is at 24°C dry bulb temperature and 20°C wet bulb temperature. Water is to be cooled within 6°C of the inlet air wet bulb temperature. An air flow rate of 1.4 times the minimum (based on the dry air basis) should be used. For the packing to be used kya = 8.25 mol/(m3.s) and hLa = 10.0 kJ/(m3.s.K). The tower diameter is 2 meters. (a) Determine the inlet air stream humidity air and enthalpy (b) Calculate the air rate to be used. (c) Calculate the packing height. Molar mases are : 18.02 for water and 28.97 for dry air. The heat capacity of liquid water is 4.181 kJ/(kg.K). Sol. : G1

HY 2 H2 TG2

L TL2

G1

HY1 H2 TG1

L TL1

Fig. 6.30 : Schematic for Ex. (7)

Principles of Mass Transfer Operations − I (Vol. − I)

6.39

Humidification and Dehumidification Operations

L = 5000 kg/hr, TL2 = 46°C, TG1 = 24°C, Tw1 = 20°C, TL1 = 26°C, G = 1.4 Gmin kya = 8.25 mol/(m3.s); hLa = 10.0 kJ/(s.m3.K) (a)

Hy 1

= Hy at Tw1 = 57.8 kJ/kg-dry air (read from graph)

Hy 1

= Cs (TG1 – T0) + λ0 H1 = (CB + H1 CA) (TG1 – T0) + λ0 H1

⇒

H1 = =

Hy1 – CB (TG1 – T0 ) λ0 + CA (TG1 – T0) 57.8 – 1.005 × (24 – 0) kg-H2O/kg-dry air 2501.4 + 1.884 × (24 – 0)

= 0.0132 kg-H2O/kg-dry air

(Ans.)

(b) Operating line (overall balance) ⇒

Hy – Hy 1 TL – TL1

=

LCL G

Plot pinch line passing (TL1 , Hy1 ) on Hys ⇒

LCL  G 

=

max

244 – 57.8 50 – 26 kJ/(kg.K)

= 7.758 kJ/(kg.K) G = 1.4 Gmin = 1.4 × 7.758–1 × 5000 × 4.181 kg/h = 3772.5 kg/hv (c)

Operating line Hy – Hy 1 TL – TL1

LCL = G

=

5000 × 4.181 kJ/(kg.K) = 5.54 kJ/kg 3772.5

TL2 = 46°C, Hy2 = 168.6 kJ/kg-dry air Hy 1 + Hy 2 2 Connecting line (slope). Hy i – Hy hL a Ti – TL = ' kG a Pt MB

= 113.2 kJ/kg-dry air ⇒ TL = 36°C

hL a = –k aM y B

= –

10.0 × 103 kJ/(kg.K) 8.25 × 28.97

= – 418.41 kJ/(kg.K) Ti =

Hy = 57.8

Hy = 113.2

Hy = 168.6

Hy i =

TL = 26°C 141.51

TL = 36°C 196.91

TL = 46°C 252.31

(Ans.)

Principles of Mass Transfer Operations − I (Vol. − I)

6.40

Humidification and Dehumidification Operations

Hy Hy i

57.8 79.0

113.2 133.0

168.6 234

1 Hy i – Hy

0.0472

0.050505

0.01529

Hy2

NG =

dHy 113.2 – 57.8 ⌠ (0.0472 + 0.050505 × 4 + 0.01528) ⌡ Hy i – Hy = 3

Hy1

= 4.885

G HG = K a M y B

=

3772.5  360  π × 22 4  8.25 × 28.97

= 1.396 m Z = HG · NG, Z = 6.28 m

(Ans.)

280 260 240 220

Hy, kJ/kg-dry air

200 180 160 (TL2, Hy2) 120 100 80 60 40

(TL1, Hy1)

20 20 22 24 26 28 30 32 34 36 38 40 42 44 46 48 50 o T, C

Graph 6.3 : Graphical analysis for Ex. (7)

Principles of Mass Transfer Operations − I (Vol. − I)

6.41

Humidification and Dehumidification Operations

EXERCISE FOR PRACTICE (1) Water is to be cooled from 43 oC to 27 oC in a forced draft cooling tower under conditions such that the height of the transfer unit is 0.5 m. Air enters the bottom of the tower at 24oC and a wet bulb temperature of 21 oC. Find the tower height if 1.33 times the minimum air is used. Neglect the heat transfer resistance of liquid phase. Enthalpy Data : Enthalpy of moist air, Ei, k cal/kg dry air air, Ti (oC) 15.6 14.7 18.3 16.7 21.1 18.9 23.9 21.5 26.7 24.3 29.4 27.5 32.2 31.1 35.0 35.2 37.8 39.9 40.6 45.2 43.3 51.3 (Ans. : Height of cooling tower = 2.012 m.) (2) Hot water at 45 oC is to be cooled in an induced draft cooling tower by contact with air to a temperature of 29 oC; calculate the following : (a) The minimum air rate (dry air) required in kg/sec. (b) The height of packed tower required if the actual air flow rate is taken as 11 kg dry air per second. (Ans. : (a) Minimum dry air rate = 6.94 kg/sec. (b) HTU = 12.2 m, NTU = 3.125, Z = Height of packed tower = 38.2 m.) (3) At a temperature of 37.8°C and at 1.0 atm if the partial pressure of water vapour is 3.59 kPa, calculate the following : (Use charts or equations) (a) Humidity (b) Saturation Humidity (c) Percentage Humidity (d) Percentage Relative Humidity (4) The air entering a dryer has a dry bulb temperature of 65.6°C and a dew point of 15.6°C. Using appropriate charts or equations, calculate the following : (a) Humidity (b) Percentage Humidity (c) Humid volume (d) Humid heat o

(5) An air stream at 87.8 C having a humidity of 0.030 kg H2O/kg dry air is contacted in an adiabatic saturator with water. It is cooled and humidified to 90% saturation. (a) Calculate the final H (enthalpy) and T (temperature). (b) For 100% saturation, what would be the values of H and T ? (6) A counter current water-cooling tower using an air flow rate that is 1.5 times the 2

o

o

minimum required is to cool 2000 lb/(hr.ft ) water from 110 F to 85 F. The entering air o

o

has a dry bulb temperature of 85 F and a wet-bulb temperature of 75 F. The product of mass transfer coefficient (overall, gas phase) and the interfacial area per unit volume (or 3

KGa) is known to be 6.90 lb mol/(hr.ft atm). (a) Calculate the minimum airflow rate. (b) Calculate the height of the tower using an air rate 1.5 times the minimum.

Principles of Mass Transfer Operations − I (Vol. − I)

6.42

Humidification and Dehumidification Operations

(7) The following data were obtained during a test run of a packed cooling tower of 0.4 m diameter and 1 m packed height, operating at atmospheric pressure. Calculate the humidity of the exit air by means of an enthalpy balance. Average temperature of entering and leaving air is 38°C and 39°C respectively. Average temperature of water entering and leaving is 46°C and 35°C respectively. Rate of entering air = 13.6 m3/min Rate of entering water = 1000 kg/hr Humidity of entering air = 0.0175 kg water vapour/kg dry air Latent heat of evaporation of water = 589 kcal/kg Specific heat of air = 0.245 kcal/kg°C Specific heat of water vapour = 0.45 kcal/kg°C (8) A wet solid material is dried from 0.7 kg water/kg dry solid to 0.08 kg water/kg dry solid in a continuous counter-current drier from which the product flows out at the rate of 500 kg/hr. The inlet air to the drier is at 54°C with an initial humidity of 0.015 kg water/kg of dry air and the exit air is at 32.2°C with 80% saturated humidity. Calculate the inlet air rate in m3/hr and the heat supplied by the preheater if the atmospheric temperature is 24°C. Data : Saturated humidity at 32.2°C = 0.025 kg water/kg dry air Specific volume of dry air at 54°C = 0.925 m3/kg Saturated volume = 1.09 m3/kg Humid heat 0.015 humidity = 0.243 cal/gm°C Humid heat at 32.2°C and at 80% saturated humidity = 0.250 cal/gm°C (9) For a drying process, air is required at 38°C with a dew point of 21°C. To make this air at 15°C and 40 percent relative humidity is first heated and passed through an adiabatic humidifier until the required conditions are obtained. Estimate the temperature to which air is to be heated first. Vapour pressure of water : 12.8 mm Hg at 15°C and 18.7 mm Hg at 21°C. Latent heat of vapourization at 38°C = 576 cal/gm. Specific heat of dry air = 0.24 and that of water vapour = 0.48. (10) A continuous counter-current drier is to be used to dry 10000 kg per hour of wet solid containing 5% water (wet basis) to water content of 0.1% (wet basis). Ambient air at 27°C and a humidity of 0.0075 will be heated to 150°C and the heated air is passed through the drier. The air leaving the drier is at 70°C with a percentage humidity of 10 percent. Calculate the air required and the heat supplied in the preheater. Saturation humidity at 70°C = 0.299 Humid heat of inlet air = 0.243 cal/gmol°K (11) Conditioned air at 24°C and 40% saturation is to be supplied to a laboratory room of size 4 × 20 × 6 meters, with no facility for recirculation. The air-conditioner takes outside air with 90% relative humidity at 38°C that it refrigerates and separates out the condensed water and then reheats it in a heat exchanger using condensing steam at 1 atm pressure. Determine the following : (a) The volume of outside air at entry conditions, and the temperature to which it must be cooled. (b) The tones refrigeration required and the kg of saturated steam required per hour. (12) A mixture of oxygen and acetone vapour at a total pressure of 1050 mm Hg at 25°C has a percentage saturation of 75%. Calculate : (a) The molal humidity (b) Absolute humidity

Principles of Mass Transfer Operations − I (Vol. − I) (c) (e)

Relative humidity Molal humid volume and

6.43

Humidification and Dehumidification Operations

(d) Volume percent acetone (f) Molal humid heat. o

The saturation vapour pressure of acetone at 25 C is 290 mm Hg and the specific heats of o

oxygen and acetone vapour are 0.25 and 0.35 kcal/kg C respectively. (13) A mechanical draft-cooling tower is to be designed to cool 75000 kg/hr of water from o

o

45 C using 62500 kg of dry air per hour. 24 C is suggested as the design air wet bulb temperature. Calculate the number of transfer units and hence the height of the packed section if the height of a transfer unit for the condition stated above is 4 m. It may be o

assumed that the liquid phase resistance to heat transfer is negligible. Temperature t C saturated enthalpy in kcal/kg dry air (H) data is as follows : o 24 29 32.5 38 43.5 t C H 20 25 28 36 46 (14) A horizontal spray chamber with recirculated water is to be used for the adiabatic humidification and cooling of air. The active part of the chamber is 1.5 m long and a cross section of 6 m with the nozzle arrangements provided and when operated with water circulation rate recommended by the nozzle manufacturer, the coefficient of heat transfer o

o

is expected to be hGa = 1600 kcal/hr.cm. C. An amount of 200 cc/min of air at 65 C, Y' = 0.017 kg water/kg dry air, is to be blown through the spray. Determine (a) What exit temperature and humidity can be expected from the air ? (b) Express the performance in terms of kya, NtOG, and HG and stage efficiency. Note : For air water system CS (humid heat) = 0.24 + 0.45 Y'. o

(15) The air supply to a drier has a dry bulb temperature of 21 C and wet bulb temperature of o

o

15 C. It is heated to 90 C by heating coils and introduced into the drier. In the drier it cools along adiabatic cooling line and leaves the drier fully saturated. (a) What is the dew point of the initial air ? (b) What is its humidity ? (c) How much water will be evaporated per 100 cubic meters of entering air ? o

(d) How much heat is needed to heat 100 cubic meters to 90 C ? (e) At what temperature does the air leave the drier ? (16) Air at 15°C having a moisture content of 0.010 kg water/kg dry air is heated to 90°C and used in a dryer operating under essentially adiabatic conditions which it leaves at 50°C. Calculate (a) The heat aded to the dry air. (b) The moisture expanded per kg dry air. (c) The specific volume of air leaving the dryer. (17) 6000 kg/hr of dry air at 10°C and 85% humidity is to be humidified to 40°C and 50% humidity by first heating the air, contacting it with water in an adiabatic humidifier from which the air exists at 85% saturation and finally reheating the air to the desired temperature. Estimate the load on each heater and the volume of the humidifier. Mass Transfer coefficient kGa = 0.8 kg m–3 s–1 (18) Air at 25°C and 90% humidity is to be conditioned to 50°C and 30% humidity by first heating the air then humidifying it in an adiabatic humidifier. Estimate the load on the heater and the volume of the humidifier. Mass Transfer coefficient, kGa = 0.9 kg m–3 s–1 Air flow at heater inlet = 1.2 m3 s–1

Principles of Mass Transfer Operations − I (Vol. − I)

6.44

Humidification and Dehumidification Operations

(19) Air saturated at 18°C is to be conditioned to 35°C dry bulb and 27°C wet bulb temperatures by first heating the air, then contacting the air in an adibatic humidifier. A tower of 1.5 m diameter is available and a maximum air velocity of 1.05 m s–1 is permitted. Calculate : (a) The volume of air (based on initial conditions) which may be humidified. (b) The height of the tower. (c) The load on the heater. Mass Transfer coefficient, kGa = 0.9 kg m–3 s–1 (20) Air saturated at 20°C is to be used to dry solids in a drier operating under essentially adiabatic conditions. Air is expected to leave the drier at 35°C and 90% saturated with water vapour. The mass velocity of the gas, G is estimated at 1.2 kg m-2 s-1 and the value of a, the contact area/m3 of drier volume at 30 m2 m–3. Calculate the temperature at which the air must enter the drier and the length of the drier. The heat transfer coefficient, h, may be estimated from h = 0.0241 G0.37/kW m–2 K–1. NOMENCLATURE Any consistent set of units may be used as noted, except as noted. Symbols Meaning a Specific interfacial surface, based on volume of packed section, m2/m3 aH Specific interfacial surface for heat transfer m2/m3 aM Specific interfacial surface of mass transfer, m2/m3 AO Outside surface of tubes, m2 c Molar density, mole/m3 C Heat capacity of a gas or vapour, unless otherwise indicated, at constant pressure. Cs Humid heat, heat capacity of a vapour-gas mixture per unit mass of dry gas content, CT Heat capacity of tube - side fluid, dav Mean diameter of a tube, m di Inside diameter of a tube, m do Outside diameter of a tube, m DAB Molecular diffusivity of vapour A in mixture with gas B, m2/sec ED Eddy diffusivity of mass, m2/s EΗ Eddy diffusivity of heat, m2/s EMG Murphree gas-phase stage efficiency, fractional F Mass transfer coefficient, mole m2/s G Molar mass velocity of dry gas, kg/m2.s G's Superficial mass velocity, mole/m2.s h

Convective heat transfer coefficient,

h'

Convective heat transfer coefficient corrected for mass transfer,

h L' , h L"

Water heat - transfer coefficients, evapourative cooler

hR

Radiation heat transfer coefficient in convection form

H

Enthalpy,

Principles of Mass Transfer Operations − I (Vol. − I) Symbols H' Ht G

Ht

OG

6.45

Humidification and Dehumidification Operations

Meaning Enthalpy of a vapour-gas mixture per unit mass of dry gas Height of a gas transfer unit, m Overall height of a gas-enthalpy transfer unit, m

JA k kG kY KY ln log L' Le m

Flux of mass for no bulk flow, mole/m2.sec Thermal conductivity Gas-phase mass transfer coefficient, Gas-phase mass transfer coefficient, Overall gas-phase mass transfer coefficient, Natural logarithm Common logarithm Superficial mass velocity of liquid Lewis number, Sc/Pr, dimensionless Slope of a chord of the saturated - enthalpy curve on a H' * – tL diagram;

M N Nt

dH'* /dtL Mol wt, M/mole Mass transfer flux, mole/m2.s Number of gas transfer units, dimensionless

Nt

Number of overall gas transfer units, dimensionless

G

OG

p –

p –

pB, M PR qs qT Q S Sc Sh t tas tDP tW t0 T Uo

Vapour pressure, kN/M2 Partial pressure, kN/M2 Mean partial pressure of non-diffusing gas, kN/M2 Prandtl number, C µ/k, dimensionless Sensible - heat - transfer flux, Total - heat- transfer flux, Evaporative - cooler heat load, gain in enthalpy, Interfacial surface, m2 Schmidt number, µ/ρ DAB, dimensionless Sherwood number, kydo/MBc DAB, dimensionless Temperature, K Adiabatic saturation temperature, K Dew-point temperature , K Wet-bulb temperature, K Reference temperature, K Absolute temperature, K Overall heat-transfer coefficient based on outside tube surface

v

Molar specific volume, m3/mole

vH

Humid volume, volume vapour - gas mixture/mass dry gas, m3 /kg Mass rate, kg/sec Mass rate of dry gas, kg/s Mass of dry gas, M

w wS WB

Principles of Mass Transfer Operations − I (Vol. − I)

6.46

Symbols x Y' zm Z α Greek Letters : ∆ λ λ' µ ρ Subscripts : 0 1, 2 as av A bp B G i L m min nbp o r s T w Superscript : *

Humidification and Dehumidification Operations Meaning

Fraction of heat-transfer surface traversed in the direction of airflow, dimensionless Absolute humidity, mass vapour/mass dry gas, kg/kg Metal thickness, m Length or height of active part of equipment, m Thermal diffusivity, m2/s Difference Latent heat of vapourization Molar latent heat of vapourization Viscosity, kg/m.s Density, kg/m3 Reference condition Positions 1, 2 Adiabatic saturation Average Substance A, the vapour Boiling point Substance B, the gas Pertaining to the gas Interface Pertaining to the liquid Metal Minimum Normal boiling point Overall; outside Reference substance Saturation Tube - side fluid Wet-bulb temperature At saturation

✱✱✱ REFERENCES 1. 2. 3. 4. 5.

R.E. Treybal, "Mass-Transfer Operations", Third Edition, McGraw-Hill, 1981. W.L. McCabe, J.C. Smith and P. Harriot, "Unit Operations of Chemical Engineering", Fifth Edition, McGraw-Hill, 1993. J.D. Seader and E.J. Henley, "Separation Process Principles", John Wiley and Sons, 1998. C.J. Geankoplis, “ Mass Transport Phenomena”, Columbus, Chio, 1972 R.H. Perry and D. Green, "Perry's Chemical Engineers' Handbook", Seventh Edition, McGraw-Hill, 1997.

,,,

7

CHAPTER

EQUIPMENT FOR GAS LIQUID OPERATIONS 7.1 7.2 7.3

7.4

7.5

7.6 7.7 7.8

Introduction Gas Dispersals Sparged Vessels (Bubble Columns) Mechanically Agitated Vessels 7.3.1 Mechanical Agitation of Single-Phase Liquids 7.3.2 Vortex Formation and Prevention 7.3.3 Similarity considerations in Agitated Vessels Tray Towers 7.4.1 General Characteristics 7.4.2 Bubble Cap Trays. 7.4.3 Proprietary Trays : Linde Trays, Valve Trays and Counterflow Trays 7.4.4 Tray Efficiency 7.4.5 Design Considerations Packed Towers 7.5.1 Introduction 7.5.2 Construction Details 7.5.3 Gas and Liquid Phase Coefficients 7.5.4 Design Criteria for Packed Towers 7.5.5 Comparision between Tray Towers and Packed Columns Liquid Dispersals Ventrui Scrubbers Wetted Wall Tower Spray Tower Solved Problems Nomenclature References

7.1 INTRODUCTION Gas and liquid can conveniently be contacted, with gas dispersed as bubbles, in agitated vessels whenever multistage, countercurrent effects are not required. This is particularly the case when a chemical reaction between the dissolved gas and a constituent of the liquid is required. The carbonation of a lime slurry, the chlorination of paper stock, the hydrogenation of vegetable oils, the aeration of fermentation broths, as in the production of penicillin, the production of citric acid from beet sugar by action of microorganisms, and the aeration of activated sludge for biological oxidation are all examples. It is perhaps significant that in most of them solids are suspended in the liquids. (7.1)

Principles of Mass Transfer Operations − I (Vol. − I)

7.2

Equipment for Gas Liquid Operations

7.2 SPARGED VESSELS (BUBBLE COLUMNS) A sparger is a device for introducing a stream of gas in the form of small bubbles into a liquid. If the vessel diameter is small, the sparger, located at the bottom of vessel, may simply be an open tube through which the gas issues into the liquid. For vessels of diameter greater than roughly 0.3 m, it is better to use several orifices for introducing the gas to ensure better gas distribution. In that case, the orifices may be holes, from 1.5 to 3 mm (1/16 to 1/4 in.) in diameter, drilled in a pipe distributor placed horizontally at the bottom of the vessel. Porous plates made of ceramics, plastics, or sintered metals are also used, but their fine pores are more readily plugged with solids, which may be present in the gas or the liquid. The purpose of the sparging may be contacting the sparged gas with the liquid. On the other hand, it may be simply a device for agitation. It can provide the gentlest of agitation, used for example in washing nitroglycerin with water; it can provide vigorous agitation as in the Pachuca tank. Air agitation in the extraction of radioactive liquids offers the advantages of freedom from moving parts, but it may require decontamination of the effluent air. There is no standardization of the depth of liquid; very deep tanks, 15 m (50 ft) or more, may be advantageous despite the large work of compression required for the gas. Ecknenfelder developed an expression for transfer of oxygen from air bubbles rising in a column of still air : 1+n A h0.78 KL V  = θg · QG · V … (7.1)   A The gas disperser must be designed so as to optimize the combined KL V  factor.   where, θg = A correlating constant dependent on the type of dispersor Nm3 QG = Gas flow rate, min (Nm3 = Normal Cubic Meter) n = A correlating constant dependent on the size of small orifices in the disperser. h = Depth below the liquid at which air is introduced to the aeration tank. (I) Gas Bubble Diameter : (a) Very Slow gas Flow Rate : 1

5  20 (σ · do · gc) 6  QG <    3  (g ∆ ρ)2 ρL   

… (7.2) 1

dP where,

 do · σ gc3 = σ· ·  ∆ρ g  

σ = Surface tension, N/m do = Diameter of orifice, m gc = Conversion factor = 1 kg.m/N.s2 ∆ρ = Density difference between the liquid and gas bubble = ρL – ρG, kg/m3 dP = Bubble diameter, m

Principles of Mass Transfer Operations − I (Vol. − I)

7.3

Equipment for Gas Liquid Operations

For liquids with high viscosities, µL upto 1 kg/m sec.

 

dP = 2.312 µL ·

QG ρL

 0.25 

· g

(b) Intermediate Flow Rate : 1

5   (σ · do · gc) 6  QG > 20. 3    (g· ∆ ρ)2 · ρL   

… (7.3)

but Reo < 2100 at the orifice outlet. 1 2

dP = 0.0287 (do) (Reo)

For air-water system :

1 3

where dP and do are in meters and Re

o

=

d o Vo ρ G µG

for other gases and liquids : 1

dP

 72 ρL 5 0.4 =  2  QG π g ∆ρ

(c) Larger gas rates : for orifice diameter – 0.4 – 1.6 mm for air-water : Re

o

= 10,000 – 50,000

dP = 0.0071 + 0.05 Reo (II) Terminal Velocity of Single Bubble : The steady-state rising velocity (terminal velocity) of single bubbles is Region 1 : for dP < 0.7 mm ∆ρ 2 Vt = g · dP · (18 µ ) (Stoke's law region) L Region 2 : For 0.7 mm < dP < 1.4 mm Vt is obtained from the line AB of Fig. (7.1).

Fig. 7.1 : Terminal Velocity Vs. Bubble Diameter Plot

… (7.4)

Principles of Mass Transfer Operations − I (Vol. − I)

7.4

Equipment for Gas Liquid Operations

Slip velocity (VS) is the relative velocity of the gas and liquid. VG VL – (A) For cocurrent flow : VS = φG 1 – φG (B) For countercurrent flow :

VS =

VG φG

+

… (7.5)

VL

… (7.6)

1 – φG

The gas hold up for sparged vessel is correlated through slip velocity by the following figure.

Fig. 7.2 : Slip Velocity (Sparged Vessel)

(III) Specific–Interfacial Area : It is the interfacial area of a gas-liquid mixture in unit-volume. 2

a = n × π × dP Again,

π 3 φG = n ×  6 · dP



… (7.7)



a = Interfacial area, m2 n = Number of bubbles in a gas volume φG dP = Bubble diameter, m VL For air-water system, and φG in the range 0.1 to 0.4, and = 0.15 to 15 m/s. (1 – φG)

where,

2.344  VL  dP = (1000)   (1 – φG)

… (7.8)

(IV) Mass Transfer : In all liquid-gas bubble operations, the liquid phase mass transfer resistance overwhelms gas-phase resistance. So mass transfer is calculated on the basis of liquid-phase coefficients : FL · dP ShL = C · D L where,

0.779

= 2 + bRe

G

0.546

· SC

L

1 0.116

  9 3 dP 2   D     L 

b = 0.061 for single gas bubbles = 0.0187 for swarm of bubbles

Principles of Mass Transfer Operations − I (Vol. − I) Re For single gas bubble,

G

7.5

Equipment for Gas Liquid Operations

= Gas Reynolds number =

dP × VS × ρL µL

φG = 0 and VS = Vt FL = Liquid mass-transfer coefficient, mole/m2.s DL = Diffusivity of gas in the liquid, m2/S SC

L

= Schmidt number for the liquid phase

(V) Gas Hold-up : By gas holdup φG is meant the volume fraction of the gas-liquid mixture in the vessel, which is occupied by the gas. If the superficial gas velocity, defined as the volume rate of gas flow divided by the cross-sectional areas of the vessel, is VG, then VG/ϕG can be taken as the true gas velocity relative to the vessel walls. If the liquid flows upward, co-currently with the gas, at a velocity relative to the vessel walls VL/(1 - φG), the relative velocity of gas and liquid, or slip velocity is given by : VS =

VG φG

–

VL 1 – φG

… (7.9)

Equation (7.9) will also give the slip velocity for countercurrent flow of liquid if VL for the downward liquid flow is assigned a negative sign. 7.3 MECHANICALLY AGITATED VESSELS Mechanical agitated of a liquid, usually by a rotating device, is especially suitable for dispersing solids, liquids, or gases into liquids, and it is used for many of the mass-transfer operations. Agitators can produce very high turbulence intensities (u’/u average as high as 0.35 ranging up to 0.75 in the vicinity of an agitating impeller, contrasted with about 0.04 for turbulent flow in pipes), which not only produce good mass-transfer coefficients but also are necessary for effective dispersion of liquids and gases. High liquid velocities, particularly desirable for suspending solids, are readily obtained. 7.3.1 Mechanical Agitation of Single-Phase Liquids Typical agitated vessels are vertical circular cylinder; rectangular tanks are unusual, although not uncommon in certain liquid-extraction applications. The liquids are usually maintained at a depth of one to two tank diameters. Impellers : There are literally scores of designs. The discussion here is limited to the most popular, as shown in Fig. 7.3. There are usually mounted on an axially arranged, motor driven shaft, as in fig. In the smaller sizes, particularly, the impeller and shaft may enter the vessel at an angle to the vessel axis, with the motor drive clamped to the rim of the vessel. The marine-type propeller, Fig. 7.3 (a), is characteristically operated at relatively high speed, particularly in low viscosity liquids, and is especially useful for its high liquid-circulating capacity. The ratio of impeller diameter to vessel diameter, di/T, is usually set at 1 : 5 or less. The liquid flow is axial, and the propeller is turned so that it produces downward flow toward the bottom of the vessel. In describing the propeller, pitch refers to the ratio of the distance advanced per revolution by a free propeller operating without slip to the propeller diameter; square pitch, which is most common in agitator designs, means a pitch equal to unity. Propellers are more frequently used for liquid blending operations than for mass-transfer purposes.

Principles of Mass Transfer Operations − I (Vol. − I)

7.6

Equipment for Gas Liquid Operations

Fig. 7.3 : Impellers with typical proportions (a) marine-type properleers; (b) fiat blade turbine, w = di /5, (c) disk flat-blade turbine, w = di / 5, di = 2d/3, B = di / 4; (d) curved-blade turbine, w = di / 8, (e) pitched blade turbine, w = di / 8; and (f) shrouded turbine, w = di / 8.

Fig. 7.4 : Liquid agitation in presence of a gas-liquid interface, with and without wall baffles : (a) marine impeller and (b) disk flat-blade turbines; (c) in full vessels without a gas liquid interface (continuous flow) and without baffles

Turbines, particularly the flat-blade designs of Fig. 7.3 (b) and (c), are frequently used for mass-transfer operations. The cured blade design is useful for suspension of fragile pulps, crystals, and the like and the pitched-blade turbine more frequently for blending liquids. The

Principles of Mass Transfer Operations − I (Vol. − I)

7.7

Equipment for Gas Liquid Operations

surrounded impeller of fig. 7.3 (f) has limited use in gas-liquid contracting. Flow of liquid from the impeller is radial except for the pitched-blade design, where it is axial. Although the best ratio di/T depends upon the application, a value of 1:3 is common. Turbines customarily operate with peripheral speeds of the order of 2.5 to 4.6 m/s (450 to 850 ft/min), depending the service. 7.3.2 Vortex Formation and Prevention Typical flow patterns for single-phase newtonian liquids of moderate viscosity are shown in Fig 7.4. For an axially located impeller operating at low speeds in an open vessel with gas-liquid surfaces, the liquid surface is level and the liquid circulates about the axis. As the impeller speed is increased to produce turbulent conditions, the power required to turn the impeller increases and a vortex begins to form around the shaft. At higher speeds the vortex eventually reaches the impeller, as in the left-hand sketches of Fig. 7.4 (a) and (b). Air is drawn into the liquid, the impeller operates partly in air, and the power required drops. The drawing of gas into the liquid is frequently undesirable, and, in addition, vortex formation leads to difficulties in scaling up of model experiments and pilot-plant studies, so that steps are usually taken to prevent vortices. A possible exception is when a solid that is difficult to wet is to be dissolved in a liquid, e.g., powdered boric acid in water, when it is useful to pour the solid into the vortex. Vortex formation can be prevented by the following means : (1) Open tanks with gas-liquid surfaces : (a) Operation only in the laminar range for the impeller (NRe 10 to 20). This is usually impractically slow for mass-transfer purposes. (b) Off-center location of the impeller on a shaft entering the vessel at an angle to the vessel axis. This is used relatively infrequently in permanent installations, and is used mostly in small-scale work with impeller where the agitator drive is clamped to the rim of the vessel. (c) Installation of baffles. This is by far the most common method. Standard baffling consists of 0 four flat, vertical strips arranged radially at 90 intervals around and tank wall, extending for the full liquid depth, as in the right-hand sketches of Fig. 7.4 (a) and (b). The standard baffle width is usually T/12, less frequently T/10 (“10 percent baffles”). Baffles are sometimes set at a clearance from the vessel wall of about one-sixth the baffle width to eliminate stagnant pockets in which the solids can accumulate. They may be used as supports for helical heating or cooling coils immersed in the liquid. The condition of fully baffled turbulence is considered to exist for these vessels at impeller Reynolds numbers above 10,000. The presence of baffles reduces swirl and increases the vertical liquid currents in the vessel, as shown on the right of Fig. 7.4. The power required to turn the impeller is also increased. (2) Closed tanks, operated full, with no gas-liquid surface : This is especially convenient for cases where the liquid flows continuously through the vessel, as in Fig. 7.4 (c). A circular flow pattern is superimposed upon the axial flow directed toward the center of the impeller. The arrangement is not practical, of course, for gas-liquid contact, where baffles should be used. 7.3.3 Similarity Considerations in Agitated Vessels While discussing the characteristics of any agitated vessels, it is necessary to consider the similarity and this is particularly important when it is desired to obtain simple laws describing vessels of different sizes. For example, when proper regard to similarity and by describing results in terms of dimensionless groups, it is possible to experiment with a vessel filled with air and

Principles of Mass Transfer Operations − I (Vol. − I)

7.8

Equipment for Gas Liquid Operations

expect the results for power, degree of turbulence, to be applicable to vessels filled with liquid. Three types of similarity are significant when dealing with liquid motion : (1) Geometric Similarity : It refers to linear dimensions. Two vessels of different sizes are geometrically similar if the ratios of corresponding dimensions on the two scales are the same, and this refers to tank, baffles, impellers and the liquid depths. If photographs of two vessels are completely – superimposable, they are geometrically similar. (2) Kinematic Similarity : It refers to motion and requires geometric similarity and the same ratio of velocities for corresponding positions in the vessels. This is especially important for mass-transfer studies. (3) Dynamic Similarity : It deals with forces and requires all force ratios for corresponding positions to be equal in kinematically similar vessels. 7.4 TRAY TOWERS Introduction : The purpose of the equipment used for gas-liquid operations is to provide intimate contact of the two fluids in order to permit interphase diffusion of the constituents. The rate of mass transfer is directly dependent upon the interfacial surface exposed between the phases, and the nature and degree of dispersion of one fluid in the other. The equipment can be classified according to whether its principle action is to disperse the gas or the liquid, although in many devices both phases because dispersed. Tray towers are the most important of the gas-dispersed group since they produce countercurrent, multistage contact. Other equipment is sparged and agitated vessels. Tray Towers : Tray towers are vertical cylinders in which the liquid and the gas contacted in stepwise fashion on trays or plates as shown in Fig. 7.5. The liquid enters at the top and flow downward by gravity. On the way it flows across each tray and through a downspout to the tray below. The gas passes upward through opening at one sort or another in the tray, then bubbles through the liquid to form a froth, disengages from the froth and passes on the next tray. The overall effect is a multiple counter current contact of gas and liquid. Each tray of the tower is a stage, since on the tray the fluids are brought into intimate contact, interphase diffusion occurs and fluids are separated. (i) Flooding : High pressure drop may lead directly to a condition of flooding. When a large pressure difference in the space between trays, the level of liquid leaving a tray relatively low pressure and entering one of high pressure must necessarily assume an elevated position in the downspouts as shown in Fig. 7.5. As the pressure difference is increased due to increased rate of flow of either gas or liquid, the level in the downspout will rise further to permit the liquid to enter the lower tray. Ultimately the liquid level may reach that on the tray above. Further increases in either flow rate then aggravates the conditions rapidly and the liquid will fill the entire space between the trays. The tower is then flooded, the tray efficiency falls to a low value, the flow of gas is erratic and liquid may be forced out of the exit pipe at the top of the tower. (ii) Priming : For liquid-gas combinations which tend to foam excessively, high gas velocities may lead to a condition of priming. Here the foam persists throughout the space between trays and great amount of liquid is carried by the gas from one tray to the tray above. This is an exaggerated condition of back mixing. The liquid so carried recalculates between trays and the added liquid – handling load increases the gas pressure drop sufficiently to lead to flooding.

Principles of Mass Transfer Operations − I (Vol. − I)

7.9

Equipment for Gas Liquid Operations

Fig. 7.5 : Schematic section through sieve-tray tower

(iii) Coning : If the liquids are too low, the gas rising through the openings of the tray may push the liquid away is allied as coning. In this situation, contact of the gas and liquid is poor. (iv) Weeping : If the gas rate is too low, much of the liquid may rain down through the openings of the tray is called as weeping. Thus failing to obtain the benefit of complete flow over the trays. (v) Dumping : At very low gas rates, none of the liquid reaches the downspouts, this is called as dumping. The relation between these conditions are shown in Fig. (7.6) and all types of trays are subject to these difficulties in same form. 7.4.1 General Characteristics Certain design features common to most frequently used tray designs are as follows : (1) Shell and Trays : The tower may be made of any number of materials depending upon the corrosion conditions expected. Glass, glass-lined metal, plastics and metals (frequently) are used. For metal tower the shells are usually cylindrical for reasons of cost. In order to facilitate cleaning, small-diameter tower are fitted with hand-holes & large towers with manways. The trays are usually made of sheet metals, the thickness governed by the corrosion resistivity. The trays must be stiffen or supported and must be fastened to the shell to prevent movement owing to surges of gas with allowance for thermal expansion.

Principles of Mass Transfer Operations − I (Vol. − I)

7.10

Equipment for Gas Liquid Operations

(2) Tray spacing : Tray spacing is usually chosen on the basis of expediency in construction, maintenance and cost. It should provide adequate insurance against flooding and excessive entrainment. For special cases where tower height is an important consideration, spacing of 15 cm (6 in) have been used. for all except the smallest tower diameters, 50 cm (20 in) in would seem to be more workable minimum from the point of view of cleaning the tray. (3) Tower Diameter : The tower diameter and consequently its cross-sectional area must be sufficiently large to handle the gas and liquid rates within the region of satisfactory operation of figure (7.6). The tower diameter required may be decreased by use of increased tray spacing, so well as diameter, passes through a minimum at some optimum tray spacing.

Fig. 7.6 : Operating Characteristics of Sieve Trays

(4) Downspouts : The liquid is led from one tray to the next by means of downspouts, or downcomers. These may be circular pipes or preferably portions of the tower cross-section set wide for liquid by vertical plates. Since the liquid is agitated into a froth on the tray, adequate residence time must be allowed in the downspout to permit disengaging the gas from the liquid, so that only clear liquid enter the tray below. The downspout must be brought close enough to the tray below to seal into the liquid on that tray thus preventing gas from rising up the downspout to short-circuit the tray above. (5) Weirs : The depth of liquid on the tray required for gas contacting is maintained by an overflow (outlet) weir, which may or may not be a continuation of the downspout plate. Straight weirs are mast common, multiple V-notch weirs maintain a liquid depth, which is less sensitive to variations. In liquid flow rate and consequently also from departure of the tray from levelness. Inlet weirs may results in a hydraulic jump of liquid and are not generally recommended. A weir of length of from 60 to 80 percent of the tower diameter is used. (6) Liquid-Flow : Several schemes are used for directing the liquid flow as shown in Fig. 7.7. Reverse flow can be used for relatively small towers but the most common arrangement is the single pass cross flow trays for large diameter towers, radial or split flow can be used but crossflow tray are more preferred due to their low cost. Commercial columns upto 50 ft in diameter use cascade designs of bubble cap trays while two pass trays are common for diameters of 3 to 6 m and more passes for larger diameter.

Principles of Mass Transfer Operations − I (Vol. − I)

7.11

Equipment for Gas Liquid Operations

Fig. 7.7 : Tray Arrangements Arrows Show Direction of Liquid Flow

7.4.2 Bubble – Cap Trays

Fig. 7.8 (a) : Typical Bubble-Cap Designs

Principles of Mass Transfer Operations − I (Vol. − I)

7.12

Equipment for Gas Liquid Operations

On these trays chimneys or risers lead the gas through the tray and underneath caps surmounting the risers. A series of slots is cut into the rim of each cap and the gas passes through them to contact the liquid, which flows past the caps. The liquid depth is such that the caps are covered or nearly so. Design characteristics are available in some standard books. They offer the distinct advantage of being able to handle very wide ranges of liquid and gas flow rates satisfactorily (the ratio of design rate to minimum rate is the turndown ratio) but they cost double of that of sieve, counterflow and valve trays. Sieve (Perforated) Trays :

Fig. 7.8 (b) : Typical Bubble–Cap Tray Arrnagement

These are perforated trays and now-a-days are very popular due to their low cost. The principal part of the tray is a horizontal sheet of perforated metal across which the liquid flows with the gas passing upward through the perforations. The gas such dispersed expands the liquid into a turbulent froth, characterized by a very large. Interfacial surface for mass transfer. The trays are subject to flooding because of backup of liquids in the downspouts or excessive entrainment (priming). Standard books provide design considerations of these trays. 7.4.3 Proprietary Trays Many decades the basic design of tray tower internals remained relatively static, recent years following types of trays are gaining much importance. (i) Linde Trays : These designs have involved improvements both in perforation design and the tray arrangements. Figure (7.9), shows the slotted tray, an alteration in the perforation pattern

Principles of Mass Transfer Operations − I (Vol. − I)

7.13

Equipment for Gas Liquid Operations

to influence the flow of liquid. The slots distributed throughout the tray, not only reduce the hydraulic gradient in large trays (over 10 m diameter) but are also so deployed that they influence the direction of liquid flow to eliminate stagnant areas and achieve, as nearly as possible, desirable plug flow of liquid across the tray. Fig. 7.9 shows a bubbling promoter, or inclined perforated area at the liquid entrance to the tray. This reduces excessive weeping and produces more uniform froth throughout the tray.

Fig. 7.9 : Linde Sieve Trays : (a) Slotted Sieve Tray, (b) Bubbling Promoter, Cross flow Tray

(ii) Valve Trays : These are sieve trays with largely (roughly 35 to 40 mm diameter) variable openings for gas flow. The perforations are covered with movable caps which rise as the flow rate of gas increases. At low gas rates and correspondingly small openings, the tendency to weep is reduced. At high gas rates the gas pressure drop remains low but not as low as that for sieve trays. (iii) Counterflow Trays : These tray-resembling devices differ from conventional trays in that there are no ordinary downspouts : liquid and vapour flow counter-currently through the same openings. 7.4.4 Tray Efficiency It is the fractional approach to an equilibrium stage, which is attained by a real tray since the conditions at various locations on the tray may differ, we may consider local or point efficiency of mass transfer at particular place on the tray surface. (i) Overall Tray Efficiency (Eo) : Performance of a tray tower can also be calculated as overall tray efficiency (Eo). Number of ideal trays requried Eo = Number of real trays required (ii) Point Efficiency (Eo ) : Figure (7.10) is a schematic representation of one tray of a G

multitray tower. The tray n is fed from tray n – 1 above by liquid of average composition xn – 1

Principles of Mass Transfer Operations − I (Vol. − I)

7.14

Equipment for Gas Liquid Operations

mole fraction of transferred component and it delivers liquid of average composition xn to the tray below. At the place under considerations, a pencil of gas composition yn + 1, local local rises from below, and as a result of mass transfer, leaves with a concentration, yn, local. At the place in question, it is assumed that the liquid concentration xlocal is constant in the vertical direction. The point efficiency is then defined by, EOG =

yn‚ local – yn + 1‚ local

… (7.10)

*

ylocal – yn + 1‚ local

*

where ylocal is the concentration in equilibrium with xlocal, and equation (7.10) then represents the change in gas concentration which actually occurs as a fraction of that which would occur if equilibrium were established. The subscripts G signifies the gas concentrations are used and the 0 emphasises that EOG is a measure of the overall resistance to mass-transfer for both phases. (iii) Murphree Tray Efficiency : The bulk-average concentrations of all the local pencils of gas of Fig. 7.10 are yn + 1 and yn. The Murphree efficiency of the entire tray is given by : yn – yn + 1 EMG = * … (7.11) yn – yn + 1 *

where yn is the value in equilibrium with the leaving liquid of concentration xn. Entertainment : It represents a from of back-mixing, which acts to destroy the concentration changes produced by the trays. The Murphree efficiency corrected for entertainment is given by : EMG EMGE = … (7.12) E 1 + EMG (1 – E)  

Fig. 7.10 : Tray Efficiency

7.4.5 Design Considerations The number of equilibrium stages in a column or tower is dependent only upon the difficulty of the separation to be carried out and is determined solely from material balances and equilibrium consideration. The stage or tray efficiency and therefore the no. of real trays is determined by the mechanical design used and the conditions of operation. The diameter of tower on the other hand depends upon the quantities of liquid and gas flowing through the tower per unit time. Another major problems in design of tower is to choose dimensions and arrangements which will represent the best compromise between several opposing tendencies, since it is generally found that conditions leading to high tray efficiencies will ultimately lead to operational difficulties. For high efficiency the time of contact should be long to permit the

Principles of Mass Transfer Operations − I (Vol. − I)

7.15

Equipment for Gas Liquid Operations

diffusion to occur, the interfacial surface between phases must be made large and a relatively high intensity of turbulence is required to obtain high mass transfer coefficients. There are many ways to fulfils all these conditions but at the same time they causes flooding, priming, coning weeping, dumping. 7.5 PACKED TOWERS 7.5.1 Introduction Packed columns/towers are generally used for two phases in contact with each other. Normally one of the fluids will preferentially wet the packing and will flow as a thin film over its surface while the second fluid passes through the remaining volume of the column with gas/vapour liquid systems, the liquid will normally be wetting the packing and gas/vapour will rise through the column making close contact with the down flowing fluid. In a packed column used for distillation, the more volatile component in transferred to vapour phase progressively while less volatile condenses but in liquid. Packed columns have also been used extensively for liquid-liquid extraction processes where a solute is transferred from one solvent to another. 7.5.2 Construction Details The device consists of a cylindrical column equipped with a gas inlet and distributing space at the bottom, a liquid inlet and distribution at the top, gas and liquid outlets at the top and bottom respectively and a supported mass of invert solid shapes called packings, the above description is for gas absorption equipment. In general, shell of the column may be constructed from metal, ceramics, glass or plastics material or from metal with corrosion resistant lining. The columns should be mounted truly vertically for uniform liquid distribution. Distributors : It the top of the packed column bed, a liquid distributor of suitable design provides for the uniform irrigation of the packing. Different types of distributor are as follows : (1) Simple orifice type giving very fine distribution but correct sizing is essential for a particular duty and should not be used where there is a risk of hole plugging. (2) Notched chimney type, having good range of flexibility for medium and high flow rates and is not prone to blockage. (3) Notched trough distributor type, suitable for large tower sizes and high gas rates. (4) Perforated ring type of distributor for use with absorption columns where high gas rates and relatively small liquid rates are encountered. The type is specially suitable where pressure drop must be minimised. Redistribution Plate : If the tower is high, distributing plates are necessary for uniform liquid flow. These plates are needed at intervals of about 2.5 to 3 column diameters for Rasching ring and 5 – 10 columns diameters for pall rings, but are usually not more than 6 m apart. Packing Supports : An open space at the bottom of the tower is required for ensuring good distribution of the gas into the packing. Consequently, the packing must be supported above the open space. The support must be reasonably strong to carry the weight of a reasonable height of packing and must have ample free area for the flow of liquid and gas with a minimum of restriction. The simplest support is a grid with relatively wide spaced bars which a few layers of relatively large Rasching or partition rings are stacked. The gas injection plate is designed to provide separate passages for gas and liquid so that they need not complete for passage through the same opening. This is aerated by providing the gas inlets to the bed at a point above the level at which liquid leaves the bed.

Principles of Mass Transfer Operations − I (Vol. − I)

7.16

Equipment for Gas Liquid Operations

Entrainment Eliminators : When gas velocities are very high, the gas leaving the top of the packing may carry off droplets of liquid as a mist. This can be removed by mist eliminators, through which gas must pass, installed above the liquid inlet. A layer of mesh (of wire, Teflon, polyethylene etc.). Specially knitted with 98 to 99% voids, about 100 mm thick will virtually collect all mist particles. Other types of eliminators include cyclones, venetian blind arrangements. A matter of dry random packing is also effective.

Fig. 7.11 : Packed Tower

Hold Down Plate : It is placed at the top of a packed column to minimise movement and breakage of the packing caused by the surges in flow rates. The gas inlet should be design for uniform flow over cross section of the gas and should be separate from liquid inlet.

Principles of Mass Transfer Operations − I (Vol. − I)

7.17

Equipment for Gas Liquid Operations

Packings : They are of two major types, random and regular. Random packings are slightly dumped into the tower during installation and allowed to fall at random. The materials used earlier like stone, gravel, lumps of coke etc., through inexpensive provided small surfaces and had poor fluid flow characteristics. Common types of random packings are : (1) Rasching rings made of chemical stoneware of porcelain, metals, plastics having diameter ranging from 6 to 100 mm. (2) Lessing rings. (3) Saddle shaped packigns : Berl and Intalox available in series from 6 to 35 mm, made of chemical stoneware or plastics. (4) Pall rings (Flexi rings). (5) Cascade rings. (6) Fellerates made of plastics. Generally, the random packings offer larger specific surface in the smaller sizes, but cost less per unit volume in larger sizes. During installation packings are poured into the tower to fall at random and in order to prevent breakage of ceramic or carbon packings. The tower may first be filled with water to reduce the velocity of fall. Regular packings are of great variety. They offer the advantages of low-pressure drop for the gases and grater possible fluid flow rates, usually at the expense of more costly installation than random packings. Stacked Rashing rings are economically practical only in very large sizes, wood grids or hurdle is inexpensive and frequently used where large void volumes are required like liquid carrying suspended solids. Knitted or gauge like arrangements provides a large interfacial surface of a contacted liquid and gas and a very gas pressure drop used for vacuum distillation. Random Versus Structured Packing : (1) At low liquid rates (< 20 gpm/ft2), a higher ap for structured packings relative to random packings makes them more efficient. (2) For random and structured packings of the same ap values, the latter yield lower values of the packing parameter, Fp. This indicates the latter yield a greater capacity (3) (4) (5)

(6) (7) (8)

(9) (10)

(< 20 gpm/ft2). Structured packings yield much lower pressure drop per unit length of tower height. Structured packigns do not perform well at high pressures and/or liquid flow rates : > 10 gpm/ft2 > 100 – 200 psia. Structured packings perform less well with aqueous liquid systems and those which possess high stress. The latter property indicates that larger droplets form, which indicates that poorer wetting will occur on the surface of metal structured packings. Liquid hold up : similar values for structured and random packings. Structured packings are very susceptible to corrosion. Structured packings are much less sensitive to surges and plant upsets. A packed tower operating in the loading region can be easily induced into the flooding region of operation with plant upsets. Inspection and maintenance of structured packings are much more difficult than random packings. Cost : A trade-off exists. The cost per unit mass is 3 – 10 times more expensive for structured compared to random packings; but, the former is more efficient (lower HETP) due to lower pressure drop. Also, pumping costs are less for structured packings because of the lower pressure drop and the shorter columns.

Principles of Mass Transfer Operations − I (Vol. − I)

7.18

Equipment for Gas Liquid Operations

Packing Restrainers : These are necessary when gas velocities are high and are used to guard against lifting of packing during sudden gas surge. Heavy screens or bars may be used for heavy ceramic packings; heavy bar plats resting freely on top of the packing may be used. For plastics and the light-weights packings, restrained is attained to tower shell. Contact between Liquid and Gas : Ideally, the liquid once distributed on top of the packing should flow in a thin film over the entire packing surface all the way down the tower. Actually the films tend to grow thicker in some pales and thinner in some, so that liquid collects in small rivables and flows along localised paths through the packing. At low liquid rates much of liquid surface may be deep or at best covered by a stagnant film of liquid. This is called channeling and the main reason for poor performance of large packed towers. Channeling is severe in towers filled with stacked packing, which is the main reason why they are not used much. It is less severe in dumped packings. In towers of moderate size having the tower diameters at least 8 times the packing diameter can minimize channeling. If the ratio of tower diameter to packing diameter is less than 8 : 1 liquid tends to flow out of the packing and down the walls of the column. In larger tower initial distribution is specially important. Countercurrent Flow of Liquid Gas Through Packing : For most random packings, pressure drop suffered by the gas is influenced by the gas and liquid flow rates in a manner similar to that shown in the Fig. 7.12. The slope of the line for dry packing is about 1.8 to 2.0 indicating turbulent flow for practically most gas velocities. At a fixed gas velocity, the gas pressure drop increases with increased liquid rate, principally because of the reduced free cross section available for flow of gas resulting in pressure of liquid.

Fig. 7.12 : Typical gas pressure drop for counter flow of liquid and gas in random packings

In the region below A, the liquid hold up is generally constant and with changing gas velocities although it increases with liquid rate. In the region between A and B liquid hold up increases sharply with gas rate, free area for gas flow becomes smaller and pressure drop rises more rapidly. This is called loading. As the gas flow rate is increased to B at fixed liquid rate of changes occurs : (1) A layer of liquid, through which gas bubbles may appear at the top of the packing. (2) Liquid may fill the tower, starting at the bottom or at any intermediate restriction like, packing support so that there is a change from gas-continuous liquid dispersed to liquid-continuous gas-dispersed (inversion).

Principles of Mass Transfer Operations − I (Vol. − I)

7.19

Equipment for Gas Liquid Operations

(3) Slugs of foam may rise rapidly upwards through the packing. At same time entrainment of the liquid by effluent gas increases rapidly and tower is flooded. The gas pressure drop then increases very rapidly. The change in conditions in region A to B gradual and initial loading and flooding are frequently determined by the change in slope of the pressure drop curves rather than through visible effect. It is practical to operate just below or in lower part of the loading region. Flooding conditions in random packings depend upon method of packing (wet/dry) and sorting of packing. Typically, absorbers and strippers are designed for gas-pressure drops of 200 to 400 N/m2 per meter of packed depths (0.25 to 0.5 in H2O/ft), atmospheric pressure fractionates from 400 to 600 N/m2 per meter and vacuum stills for 8 to 4 N/m2 per meter (0.01 to 0.05 in H2O/ft). Flooding velocities for regular or stacked packings will generally be considerable greater than for random packing. Pressure drop for single-phase flow : When a single fluid flows the a bed of packed solids like spheres, cylinders, gravel, sand etc. when it alone fills the voids, then the correlation for pressured drop is given by the Ergun equation. 3 ∆P (gc ∈ dP ρg) Z (1 – ∈) G'2

=

150 (1 – ∈) + 1.75 Re

… (7.13)

This is applicable equally well to flow of gases and liquids. The term on the left is a friction factor. Those on the right represent, contribution to the friction factor, the first for purely laminar flow, the second for completely turbulent flow. There is a gradual transition from one type of flow to the other as a result of diverse character of void spaces, the two terms of the equation changing their relative importances as flow rate changes and dP is the effective G' Re = dP µ

… (7.14)

diameter of a sphere of same surface/volume ratio as the packing in place. If the specific surface is dP, the surface per unit volume of the particles is aP (1 – ∈) and from the properties of sphere, (1 – ∈) dP = 6 a . This will not normally be the same as normal size of the particles for flow of gases P a greater than about 0.7 kg/m2s. The first term on R.H.S. of equation is negligible. For a specific type and size of manufactured tower packing equation can be simplified to the empirical equation. ∆P Z

= CD

G'2 ρG

… (7.15)

Pressure Drop for two phase flow : Simultaneous flow, counter current flow of liquid and gas the pressure drop data of various investigators show wide discrepancies due to differences in packing density and manufacture such as changes in wall thickness. Estimate cannot be therefore expected to be very accurate. For most purposes the correlation of Fig. 7.13 will serve this purpose.

Principles of Mass Transfer Operations − I (Vol. − I)

7.20

Equipment for Gas Liquid Operations

Fig. 7.13 : Flooding and Pressure Drop in Random-Packed Towers

7.5.3 Gas and Liquid Phase Coefficients Uses of KG and KL : (1) From physical properties of system determine KG and KL, if system is known or is assumed as all gas or liquid film controlling, then only the controlling K to be calculated. (2) Combine effective interfacial area, to calculate KGa or KLa. 1 1 1 (3) Determine KGa by K =K + 1 h1 KLa = 1 KLa H Ga Ga (H1 – Henry's constant, lb mole/ft3. atm) Height Equivalent to Theoretical Plate (HETP) : Distillation operations are expressed in terms of equilibrium relation and theoretical plates. αµ where, HETP = K1G'K2 DKa ZK3   ρ G' = Vapour mass velocity, lb/ft2·hr. Z = Packed height, feet D = Tower diameter, inches α = Average relative, velocity µ = Liquid viscosity cP, feet/sec. ρ = Liquid density, gm/cm3 and K1, K2, K3 are constants. HETP is unique to packing size and configuration. Large packing requires greater HETP than small size but pressure drop is more in small packing. HETP gradients for industrial process equipment are : (1) Never use HETP less than 12 inch, if tower is 12 inch diameter or longer. (Use HETP = 1.5 – 2 feet).

Principles of Mass Transfer Operations − I (Vol. − I)

7.21

Equipment for Gas Liquid Operations

(2) Use HETP = HOG or HOL if other data is not available. (3) Use HETP = column diameter (over 12 inch) 7.5.4 Design Criteria for Packed Towers The relationship in packed tower performance that are concerned specifically with gas and liquid, flows rates through a bed are expressed as a function of pressure drop. This may be created by poor packing arrangement i.e. tight and open sections in the bed or plugging of void spaces by solids or reaction products. Following points are to be considered :

   

L' (1) Calculate the absicca of Fig. 7.13 = G'

  ρL – ρG  ρG

(2) Select a design or operating pressure drop as shown in the curves of Fig. 7.13. Selection basis is as follows : (i) Low to medium pressure column operation, select pressure drop of 0.75 to 1 inch water/foot of packing height. (ii) Absorption and similar systems select a pressure drop of 0.5 to 0.1 H2O/feet. (iii) Atmospheric or pressure distillation, select a pressure drop of 0.5 to 1 inch H2O/ft. (iv) Vacuum distillation selects a low pressure drop in the 0.1 to 0.25 inch H2O/foot range. (v) Foaming materials should be operated at 0.1 inch to 6.25 inch water/feet. (3) Select packing and determine its packing factor from literature depending upon the packing type and its material. Packing is selected for its expected process performance, pressure drop and material of construction. (i) As packing factor, for (a/t2) becomes larger by selection of smaller size packing, gas capacity for the column is reduced and pressure drop is increased for a fixed gas flow. (ii) Not all material are manufactured in all materials of construction i.e. ceramic, plastic materials. (iii) Some packings are sized by general dimensions (inches) while some shapes are identified by number #, #2, #3 for increasing size. (iv) Packing size versus tower diameter recommendations are given as below : Tower Diameter (feet) < 1.0 1.0 – 3.0 > 3.0 Normal Packing Size (Inches) < 1.0 1 – 1.5 2–3 (4) From Fig. (7.13) read up from the absicca to pressure drop line selected and read across the ordinate. 0.1

G'2 Cf µL J ρG (ρL – ρG) gc C = Concentration in lb mole/feet3 G' = Superficial mass gas rate, lb/hr square feet f = Packing factor. Substitute f and other known values and solve for G. Then determine the required tower cross sectional area and diameter. G' = Gas rate, lb/sec. Q = Superficial gas rate, lb/feet2 feet of tower where,

Principles of Mass Transfer Operations − I (Vol. − I) Hence, but ∴

7.22

Equipment for Gas Liquid Operations

G' = aG π a = 4 · d2 π G' = 4 · d2 G 1

∴

d

Gas Rate2  = 1.1283  Q  

… (7.16)

Effect of Physical Properties : (a) For non-foaming liquids, capacity of packing is independent of surface tension. (b) Foaming conditions reduce capacity significantly. (c) For liquid of viscosity 30 cP and lower, effect on capacity is small. Number of Transfer Units : The transfer of mass between phases in a packed tower occurs either as gas film controlling or liquid film controlling. This has been used to express the ease or difficulty of transfer under the condition of operation with respect to system equilibrium, the system is evaluated as the number of transfer units, NOG or NOL required. Z … (7.17) NOG = H OG or

Z NOL = H OL

… (7.18)

Z Z or NOL = H NOL = H OL OL

… (7.19)

where, NOG = Number of transfer units based on overall gas film coefficient. NOL = Number of transfer units based on overall liquid film coefficient. HOG = Height of transfer units based on overall gas film coefficient. HOL = Height of transfer units based on overall liquid film coefficient. The transfer process is termed as gas film controlling if all the resistance of mass transfer is in the gas film, it means gas in soluble with liquid of the system. If the system is liquid film controlling, it means controlling gas is relatively insoluble in liquid and resistance to transfer is in liquid film. y1

For dilute solutions :

NOG =

⌠ ⌡ y2 x1

NOL

=

NOG =

(1 – y)M dy

⌠ (1 – y) (y – y*) ⌡ y2 x1

NOL =

dx

⌠ x* – x ⌡ x2 y1

For concentrated solutions :

dy y – y*

(1 – x)M dx

⌠ (1 – x) (x* – x) ⌡

… (7.20)

… (7.21)

… (7.22)

… (7.23)

x2

where (1 – y)M and (1 – x)M = log mean average of concentrations at the opposite ends of the diffusion process.

Principles of Mass Transfer Operations − I (Vol. − I)

7.23

Equipment for Gas Liquid Operations

7.5.5 Comparison Between Tray and Packed Columns (i) Liquid Hold Up : Packed towers will provide a substantially smaller liquid hold up. This is important where liquid deterioration occur with high temperature and short holding times are essential. It is also important in obtaining sharp separations in batch distillation. (ii) Liquid Cooling : Cooling coils are more readily built into tray tower and liquid can more readily be removed from trays, to be passed through coolers and returned, than packed towers. (iii) Liquid/Gas Ratios : Very low values of this ratio are best handled in tray towers. High values are best handled in packed towers. (iv) Gas Pressure Drop : Packed towers will ordinarily require a smaller pressure drop. (v) Side Streams : These are more readily removed from tray towers. (vi) Corrosion : Packed towers for different corrosion problems are likely to be less costly. (vii) Floor Loading : Plastic packed towers are lighter in weight, than tray towers, which in turn are lighter than ceramic or metal packed towers. In any event, the floor loading should be designed for accidental complete filling of tower with liquid. (viii) Foaming Systems : Packed towers operate with less bubbling of gas through the liquid and are more suitable. (ix) Cleaning : Frequent cleaning is easier with tray towers. (x) Large Temperature Fluctuations : Fragile packing tend to be crushed. Tray or metal packings are satisfactory. Advantages of Packed Column/Towers : (i) Packed towers will ordinarily require a smaller pressure drop, this is useful for vacuum distillation. (ii) Packed tower will produce a substantially lower liquid hold up. (iii) Packed towers can handle high values of liquid to gas ratio. (iv) Packed towers can operate with less bubbling of gas through liquid and are suitable for foaming systems. Disadvantages of Packed Column/Towers : (i) Very low values of liquid/gas ratio cannot be handled in packed columns. (ii) Liquid cooling cannot be easily achieved in packed column. (iii) Side streams are not easily removed from packed towers. (iv) Frequent cleaning is troublesome in packed towers. 7.6 VENTURI SCRUBBERS In these devices, which are similar to ejectors, the gas is drawn into the throat of a Venturi by a stream of absorbing liquid sprayed into the convergent duct section, as shown in Fig. 7.14. The device is used especially where the liquid contains as suspended solid, which would plug the otherwise more commonly used tray and packed towers, and where low gas-pressure drop is required. These applications have become increasingly important in recent years, as in the absorption of sulfur dioxide from furnace gases with slurries of limestone, lime, or magnesia. Some very large installations (10 m diameter) are in service for electric utilities. The cocurrent flow produces only a single stage, but this becomes less important when a chemical reaction occurs, as in the case of the sulfur dioxide absorbers. Multistage countercurrent effects can be obtained by using several venturies. The device is also used for removing dust particles from gases.

Principles of Mass Transfer Operations − I (Vol. − I)

7.24

Equipment for Gas Liquid Operations

Fig. 7.14 : Venturi Scrubber

7.7 WETTED WALL TOWER (i) A thin film of liquid running down inside of a vertical pipe, with gas flowing either co-currently or counter currently, constitutes wetted-wall tower. (ii) Such devices have been used for theoretical studies of mass transfer as the interfacial surface between the passes is readily kept under control and measurable. (iii) Industrially, they have been used as absorbers for Hydrochloric acid, where absorption is accompanied by a very large evolution of heat. In this case the wetted all tower is surrounded with rapidly flowing cooling water. (iv) Multitube devices have also been used for distillation where the liquid film is generated at the top by partial condensation of rising vapour. (v) Gas-pressure drop in these towers is probably lower than in any other gas-liquid contacting device a given set of operating conditions. (vi) Measurement of the rate of evaporation of the liquid into the gas stream over knows surface permits calculation of mass transfer coefficient for the gas phase. Use of different gases and liquid provides variation of Schemidt number.

Fig. 7.15 : Wetted Wall Tower

Principles of Mass Transfer Operations − I (Vol. − I)

7.25

Equipment for Gas Liquid Operations

Thus, Sherwood & Gilliland covered the values of NRe from 2000 – 35,000, NSc from 0.6 to 2.5 gas pressures from 0.1 to 0.3 atm. Lintion and Sheeword Correlation : Shav = 0.023 Re

0.83

1 3 SC Re

upto 3000

… (7.24)

Over NRe 500 to 200,000, Friction factor in smooth pipe. 1 –0.2 2 f = 0.023 Re Also,

2 3 SC

Shav Re

1.03

SC

1 = 2

… (7.25)

f

… (7.26)

For highly turbulent conditions : Shav = 0.0149 Re

0.88

1 3 SC

(For Liquid, SC > 100)

… (7.27)

7.8 SPRAY TOWER Introduction : In the spray tower, gas enters at the bottom and the liquid is introduced through a series of sprays at the top. The performance of the units in general is rather poor because the droplets tend to coalesce after they have fallen through a new feet and the interfacial surface is thereby seriously reduced. Although there is considerably turbulence in the gas phase, there is little circulation of liquid within the drops and the residence of equipment liquid of equivalent liquid film tends to be high. The flow may be countercurrent as in vertical columns with liquid sprayed downward or parallel as in horizontal spray chambers.

Fig. 7.16 : Schematic Diagram of Spray Column

The devices have the advantage of pressure drop through the spray nozzle. The tendency for entrainment of the liquid by gas leaning is considerable and mist eliminator will almost always be

Principles of Mass Transfer Operations − I (Vol. − I)

7.26

Equipment for Gas Liquid Operations

the necessary. The interface can be run above the top distribution, below the bottom distribution or in the middle depending upon where the best performance is achieved. Because of severe back mixing (axial) it is difficult to achieve the equivalent of more than one or two theoretical stages or transfer upto on one side of the surface. (1) Distributor : The onfices or nozzle for introducing the dispersed phase are usually not smaller than 0.13 cm diameter in order to avoid cloggings not larger than 0.64 cm in order to avoid the formation of excessively large drops. (2) Applications : Spray towers are largely used for the removal of SO2 from edges of towers, flue gases that are exhausted from large coal forced power-generating stations. They are used for absorption of NH3 in water and also in air humidification. Spray Chambers : Spray absorbers are currently being applied on a large scale on a commercial basis in system for removing SO2 from boiler flue gases that are being generated from large coal fixed power generator station. Spray chambers are particularly advantageous when the lower pressure drop is in the incoming gas stream. As there is no packing in spray absorber liquid phase, residence time in spray absorber is very low in order of 1 to 10 seconds as the gases contact in time. Spray absorbers are limited to absorption duties and especially applicable to the system in which the rate of transfer in gas phase, the mass transfer is limited. The condition will however exist in all liquid phase. Residence can be neglected and backpressure of the solute over the liquid is small. Advantages of Spray Towers : (1) It is simple in construction than packed columns. (2) Its cost is less. (3) It can be used for scale forming or solids containing liquid. (4) It can operate over a wide range of flowrate. (5) Cooling can be installed easily. (6) Pressure drop is low as compared to packed towers. Disadvantages of Spray Towers : (1) It requires high recalculation of continuous phase, which increase pumping cost. (2) Its efficiency is lower than packed towers. (3) There is turbulence due to axial mixing resulting in poor performance. (4) Due to high recalculation, the countercurrent flow is not maintained upto 6 m, may require obtaining a theoretical stage.

SOLVED PROBLEMS (1) In a contact sulphuric acid plant the secondary conversion reactor is a tray-type converter at atmospheric pressure, 2.3 meters in diameter with the catalyst arranged in three 0.45 meter thick layer. The catalyst consists of cylindrical pellets 6.35 mm in both diameter and length. The gas enters at 675 K and leavs at 720 K. Its inlet and exit compositions are, respectively : SO2 O2 N2 SO3 Component Mol % 6.6 1.7 10.0 81.7 SO3 SO2 O2 N2 Component Mol % 8.2 0.2 9.3 82.3

Principles of Mass Transfer Operations − I (Vol. − I)

7.27

Equipment for Gas Liquid Operations

The gas flow rate is 0.68 kg/s. Calculate the pressure drop across the converter µ = 0.032 mNs/m2 Sol. : First we need a molecular weight and temperature that best describe the properties of the gas : Mean mol. weight at inlet = (0.066 × 80) + (0.017 × 64) + (0.1 × 32) + (0.817 × 28) = 32.444 kg/kmol Mean mol. weight at outlet = (0.082 × 80) + (0.002 × 64) + (0.093 × 32) + (0.823 × 28) = 32.708 kg/kmol 1 Mean mol. weight in column = 2 (32.444 + 32.708) = 32.576 kg/kmol 1 Mean temperature in column = 2 (675 + 720) = 697.5 K Assuming that the gas ideal, we can calculate its density for future use in packed column formulae : pV = nRT n p V = RT

∴ ∴

nM pM V = ρ = RT

=

101325 × 32.576 = 0.5692 kg/m3 8314 × 697.5

The superificial velocity will be a function of the specific flow rate, the density and the cross-sectional area of the circular vessel : π D2G π 2.32 × 0.68 AG = 4 = 4 0.5962 = 4.964 m/s v = ρ ρ L = 3 × 0.45 m = 1.35 m B = 4.6 × 10–8 (from literature) µ = 0.032 × 10–3 Ns/m2 Using the Darcy equation : ∆P =

vµL B

=

4.964 × 0.032 × 10–3 × 1.35 = 4661.44 Pascals 4.6 × 10–8

So, the pressure drop across the converter = 4661.144 Pascals

(Ans.)

(2) An organic liquid mixture at 395 K (average molecular weight = 155 kg/kmol) is separated by vacuum distillation in a 12.74 metre diameter tower packed with 9.65 mm ceramic Intalox saddles. The HETP (Height Equivalent of a Theoretical Plate) is 150 mm and the number of theoretical plates required is 16. If the product rate is 4.16 kg/s at a reflux ratio of 11.6, calculate the pressure in the condenser so that the pressure in the still does not exceed 8 kPa. Neglect temperature changes. Remember to use water physical properties at 293 K. ρL = 800 kg/m3, µL = 100 mNs/m2. Sol. : Assuming that the vapour follows the ideal gas law, we can calculate its density as a function of temperature, pressure and molecular weight : pV = nRT

Principles of Mass Transfer Operations − I (Vol. − I) n V

∴

7.28

Equipment for Gas Liquid Operations

p = RT

pM 8000 × 155 nM 3 V = ρG = RT = 8314 × 395 = 0.3776 kg/m ρL = 800 kg/m3 The specific flow rates will be a function of tower cross-sectional area and product rate. From knowledge of distillation, assuming constant molal overflow : Vn = L n + D

∴

Ln = RD ∴

Ln Vn D = Ln Ln + Ln

1 R+1 = 1+R = R

where, Vn, Ln are molar flows of vapour and liquid, D that of the distillate and R is the molar reflux ratio. As we assume the fluid is of uniform molecular mass : Ln L' R ∴ Vn = G' = R + 1 Although we do not need to calculate the value of L', we do need the value of G'. MVn MD (R + 1) = G' = A A As D is a MOLAR proceeding :

flow of distillate, we must convert the mass flow rate before 4.16 kg/s D = 155 kg/kmol = 0.02684 kmol/s G' =

∴

MD (R + 1) 155 × 0.2684 (11.6 + 1) = = 0.4113 kg/m2s (π/4) d2 (π/4) (12.74)2

From literature, F = 1080 From steam tables we find that : ρW = 998 kg/m2 µW = 0.001 Ns/m2 ∴

L' G'

ρG ρL

11.6 = 11.6 + 1

0.3776 800

= 0.0200

 ρL ρW (G')2 F   ρW ρL 

0.1

∴

ρG (ρL – ρG) g

From monographs from literature ∆P

=

0.1 998 0.1 (0.4113)2 1080 0.001 1800





0.3776 (800 – 0.3776) 9.807

= 0.1

= 83 mm water per metre packed height

Packed height = Number of stages × HETP = 16 × 0.15 = 2.4 metres ∴ ∆P = 2.4 × 83 = 199.2 mm water ≅ 0.1992 × 998 × 9.807 = 1949.6 Pascals If pressure drops across, packing is 1949.6 Pa and maximum pressure (at base of column) cannot exceed 8000 Pa, then condenser pressure will be (8000 – 1949.6 =) 6050.4 Pascals. (Ans.)

Principles of Mass Transfer Operations − I (Vol. − I)

7.29

Equipment for Gas Liquid Operations

(3) A column, packed with 25.4 mm Raschig rings to a bed height of 4 m, is used in a gas absorption process carried out at 293 K and atmospheric pressure. If the liquid and gas are assumed to have the same properties as water and air, and their flowrates are 2.5 and 0.6 kg/m2s respectively, what will be the pressure drop across the column ? What is the maximum liquid rate before flooding occurs ? Sol. : If gas and liquid have the same properties as air and water respectively, then, 0.1 µW ρW0.1 = 1  µL µW  = µW ρ  µW µL  W  Assuming air to be ideal : pV = nRT p n ∴ V = RT pM 101325 × 28.84 nM = 1.1996 kg/m3 ∴ V = ρ = RT = 8314 × 293 As F = 525 (from literature), ρL = 998 kg/m3 (from steam tables), L' = 2.5 kg/m2s and G' = 0.6 kg/m2s : ∴

L' G'

ρG ρL

 µL ρW0.1 (G')2 F µ   W ρL  ∴

ρG (ρL – ρG) g

2.5 = 0.6

1.1996 998 = 0.1444576

(0.6)2 525 [1] = 1.1996 (998 – 1.1996) 9.807 = 0.0161169

From monograph from = 14 mm water per metre packed height literature ∆P Packed height = 4 metres ∴ ∆P = 4 × 14 = 56 mm water ≡ 0.056 × 998 × 9.807 = 548.09 Pascals As the point falls between two graph (See Fig. 7.13) curves, it is possible for reading to be anywhere between 8 and 21 mm water per metre packing i.e. between 313 and 822 Pascals. By keeping the gas flowrate fixed, our y-axis co-ordinate stays at 0.016. However, if we are to reach flooding at this flowrate, our x-axis co-ordinate becomes 1.1 '

1.1 =

∴ ∴

'

Lflood

Lflood 0.6

1.1996 ' 998 = 0.05778 Lflood

= 19.0367 kg/m2s

So Maximum liquid rate before flooding occurs = 19.0367 kg/m2 · s.

(Ans.)

(4) A column packed with a bed of 25 mm Raschig rings 9 m high is used in the vacuum distillation of an isomer mixture of molecular weight 155 kg/kmol. The mean temperature is 373 K, the pressure at the top of the column is maintained at 0.13 kPa and the still pressure must lie between 1.3 and 3.3 kPa. Obtain an expression for the pressure drop from the Carman-Kozeny equation and gas flow rate, assuming that liquid flow does not significantly affect results. Calculate pressure drop when this flow rate is 0.125 kg/m2s. µ = 0.018 mNs/m2

Principles of Mass Transfer Operations − I (Vol. − I)

7.30

Equipment for Gas Liquid Operations

Sol. : From literature, S = 756 and e = 0.416, which are then inserted into the Carman-Kozeny equation : 5 (1 – e)2 s2 µ Lvs ∆P =  e3   5 (1 – 0.416)2 7562 (0.018 × 10–3) Q =  × 9× A 0.4163   m' = 2193.1802 ρG A G' = 2193.1802 ρG Assuming that the vapour follows the ideal gas law, we can calculate its density as a function of temperature, pressure and molecular weight : pV = nRT p n ∴ V = RT Ps × 155 nM pM ∴ = ρ = = = 4.9982 × 10–5 Ps kg/m3 G V RT 8314 × 373 G' ∴ ∆P = 4.387948 × 107 P s Ps = Pc + ∆P = ∆P + 130 ∴ ∴ ∴

∆P = 4.387948 × 107

G' ∆P + 130

∆P2 + 130 ∆P – 4.387948 × 107 G' = 0 ∆P =

– 130 ±

1302 – 4 (– 4.387948 × 107) G' 2

= – 65 ± 4225 + 4 (4.387948 × 107) G' For G' = 0.125 kg/m2s ∆P = – 65 ± 2342.874 = 2277.874 Pascals ∴ Ps = 130 + 2277.874 = 2407.874 Pascals (Ans.) (5) Calculate the interfacial area per unit volume of dispersion, in a gas-liquid contactor, for fractional hold-up of gas = 0.1 and gas bubble diameter = 0.5 mm. Sol. : Basis : Total volume = 1 m3 Volume of bubble = (hold-up fraction) (Total volume) = 0.1 × 1 = 0.1 m3 Let n be the number of gas bubbles. 4 ∴ Volume of n bubbles = n × 3 πr3 = 0.1 m3 0.5 Given : r = 2 = 0.25 mm 3 0.1 ∴ n = 4 = 1.528 × 109 π × (0.25 × 10–3) ∴ Surface area for n bubbles = n × 4πr2 = 1.528 × 109 × 4 × π × (0.25 × 10–3)2 = 1200 m2 ∴ Interfacial area per unit volume = 1200 m2/m3 (Ans.) ∴

Principles of Mass Transfer Operations − I (Vol. − I)

7.31

Equipment for Gas Liquid Operations

NOMENCLATURE Any consistent set of units may be used, except as noted. Symbols Meaning a Average specific interfacial surface for mass transfer, area/volume, m2/m3 ap specific surface of packing, area/volume, m2/m3 av specific interfacial surface for contact of a gas with a pure liquid, area/volume, m2, m3 A Projected area, m2 Aa Active area, area of perforated sheet, m2 At Tower cross-sectional area, m2 b Baffle width, m B blade Length, m c molar density of the liquid , mole/m3, solute concentration, mole/m3 C distance from impeller to tank bottom, m CD empirical constant Cf characterization factor of packing, two phase flow, empirical constant CF flooding constant for trays Co orifice coefficient, dimensionless Cp specific heat at constant pressure dd disk diameter, m di impeller diameter, m do orifice or perforation diameter, m dp average bubble diameter, m nominal diameter of tower - packing particle, m ds diameter of sphere of same surface as a single packing particle, m D diffusivity, m2/s DE eddy diffusivity of back mixing, m2/s E

fractional entertainment, entrained liquid/entrained liquid + net liquid flow), mole/mole

EMG

Murphree gas - phase stage efficiency, fractional

EMGE

Murphree gas - phase stage efficiency corrected for fractional

EO

overall tray efficiency of a tower, fractional

EOG

point gas - phase tray efficiency, fractional

f

function Fanning friction factor, dimensionless

F

mass - transfer coefficient, mole/m2.s

FD

drag force, N

Fr

impeller Froude number = diN2/ g, dimensionless

g

acceleration of gravity, m/s2

entertainment,

Principles of Mass Transfer Operations − I (Vol. − I)

7.32

Symbols

Equipment for Gas Liquid Operations Meaning

G

superficial molar gas mass velocity, mole/ m2.s

G'

superficial gas mass velocity, kg/m2.s

h

heat - transfer coefficient , w/m2 k

hD

dry-plate gas-pressure drop as head of clear liquid, m

hG

gas-pressure drop as head of clear liquid , m

hL

gas-pressure drop due to liquid holdup on tray, as head of clear liquid, m

hR

residual gas-pressure drop as head of clear liquid, L

hW

weir height, m

h1

weir crest, m

Ht

OG

jD jH

overall height of a transfer unit, m 2/3

mass-transfer group = FG ScG /G, dimensionlesss 2/3

heat -transfer group = h Pr G /CpG', dimensionless

kG

gas - phase mass-transfer coefficient, mole/sec. m2 (N/m2)

kL

liquid - phase mass - transfer coefficient, mole/sec.m2 (mol/ m3)

kth

thermal conductivity

kx

liquid-phase mass-transfer coefficient, mole/sec.m2 (mole fraction)

ky

gas-phase mass-transfer coefficient, mole/sec.m2 (mole fraction)

KG

overall gas-phase mass-transfer coefficient

Ky

overall gas-phase mass-transfer coefficient, mole/sec.m2 (mole fraction)

l

plate thickness, m

ln

natural logarithm

log

common logarithm

L L'

superficial liquid molar mass velocity, mole/m2·sec superficial liquid mass velocity, kg/m2.sec characteristic length, m

–

L m m' n Nt

G

Nt Nt

L

OG

average slope of a chord of equilibrium curve, mole fraction in gas/mole fraction in liquid , empirical constant slope of a chord of equilibrium curve, Eq. (6.53), mole fraction gas/mole fraction in liquid number of baffles, dimensionless tray number, dimensionless empirical constant number of gas-phase transfer units, dimensionless number of liquid-phase transfer units, dimensionless number of overall gas-phase transfer units, dimensionless

Principles of Mass Transfer Operations − I (Vol. − I) Symbols Nu

7.33

Equipment for Gas Liquid Operations

Meaning Nusselt number = hd3/ kth, dimensionless

p ∆p ∆pR

pressure, kN/m2 pressure difference, kN/m2 residual gas-pressure drop, F/ L2

P PG

power delivered by an impeller, no gas flow, kN/m·sec power delivered by an impeller with gas flow, kN/m·sec

PC

Peclet number for liquid flow rate, m3/sec

Po

power number = Pgc/ ρLN3 d i , dimensionless

Pr

Prandtl number = Cρµ/ kth, dimensionless

q Q QG O Ra Rc

5

volumetric liquid flow rate per orifice, m3/sec volumetric flow rate, m3/ sec volumetric gas flow rate, m3 /sec 3

Rayleigh number = dP ∆p g/ DLµL, dimensionless 2

impeller Reynolds number = d i NρL/ µL, dimensionless

RcG

gas Reynolds number = dpVS ρL/ µL , dimensionless

Rco

orifice Reynolds number = do Voρ G / µG = 4wo/ πdo µ G,, dimensionless

Sc Sh T u u' vL V

Schmidt number = µ/ρD, dimensionless Sherwood number = Fdp / cD, dimensionless tank diameter , tower diameter average velocity, m/s root- mean - square fluctuating velocity, m/s liquid volume, m3 superficial velocity based on tower cross section; for tray towers, based on An,, m/s gas velocity based on Aa, m/s flooding velocity based on An, m/s velocity thorugh an orifice, m/s minimum gas velocity through perforations below which excessive weeping occurs, m/s slip velocity, terminal settling velocity of a single bubble, m/s blade width, m mass rate of flow per orifice, kg/s weir length, m Weber number = ρu2dp/σg, dimensionless

Vg VF Vo Vm Vg Vt w wo W We Greek letters : x

concentration in liquid, mole fraction

y

concentration in gas, mole fraction

z

average flow width for liquid on a tray, m

Principles of Mass Transfer Operations − I (Vol. − I)

7.34

Equipment for Gas Liquid Operations

Symbols

Meaning

Z

liquid depth in an agitated vessel; depth of packing; length of travel on a tray, m empirical constant empirical constant, for flooding velocity for holdup in packing fractional void volume in a dry packed bed, volume voids/ volume bed, dimensionless constant, dimensionless time of residence of a liquid on a tray, sec viscosity, kg/m.sec density, kg/m3 difference in density, kg/m3 surface tension, N/M gas holdup, volume fraction dimensionless liquid holdup, fraction of packed volume, dimensionless

α β

η θL µ ρ ∆ρ σ φG φL Subscripts : av AW F G L n o s Superscript : *

average air - water at flooding gas liquid tray number orifice; operating or moving(holdup and packing void space) surface; static (holdup in packing) in equilibrium with bulk liquid

✱✱✱ REFERENCES 1.

R.E. Treybal, "Mass-Transfer Operations", Third Edition, McGraw-Hill, 1981.

2.

W.L. McCabe, J.C. Smith and P. Harriot, "Unit Operations of Chemical Engineering", Fifth Edition, McGraw-Hill, 1993.

3.

A.S. Foust, L.A. Wenzel, C.W. Clump, L. Maus and L.B. Andersen, "Principles of Unit Operations", Second Edition, John Wiley and Sons, 1980.

4.

J.M. Coulson and J.F. Richardson, "Chemical Engineering” Vol., Third Edition, Pergamon Press, 1986.

5.

R.H. Perry and D. Green, "Perry's Chemical Engineers' Handbook", Seventh Edition, McGraw-Hill, 1997.

6.

C.J. Geankoplis, "Transport Processes and Unit Operations", Fourth Edition, Prentice Hall, 2003.






8 CHAPTER

DRYING OPERATIONS 8.1 8.2 8.3 8.4 8.5 8.6 8.7

Introduction Expression of Moisture Content Equilibrium in Drying Types of Moisture Mechanism of Batch Drying Drying Tests Rate of Drying Curve

8.8

Time required for Drying

8.9

Drying Time for Droplets 8.9.1

Constant Rate Period

8.9.2

Falling Rate Period

8.9.3

Droplet Trajectory

8.10

Heat Transfer in Dryers

8.11

Mechanism of Moisture Movement within the Solid

8.12

Equipments for Drying

8.13

Selection, Sizing and Costs

8.14

Efficient Energy Utilization in Drying Solved Problems Exercise For Practice Nomenclature

8.1 INTRODUCTION The drying of materials – whether solids, liquids or slurries – to improve storage life or reduce transportation costs is one of the oldest and most commonly used unit operations. Drying of fruit, meat and various building and craft materials date back before the discovery of fire. The physical laws governing drying remain the same, even though the machinery to accomplish it has improved considerably ! Today, dryers are in operation in most manufacturing industries including chemical, pharmaceutical, process and food. Products that are dried range from organic pigments to proteins, as well as minerals to dairy products. Because of the spectrum of duties required, there are a great variety of dryers available. The correct choice depends on the properties of the feed material and the desired characteristics of the final product. (8.1)

Principles of Mass Transfer Operations − I (Vol. − I)

8.2

Drying Operations

The term drying refers generally to the removal of moisture from a substance. It is so loosely and inconsistently applied that some restriction in its meaning is necessary in the treatment to be given the subject here. For example, a wet solid such as wood, cloth, or paper can be dried by evaporation of the moisture either into a gas stream or without the benefit of the gas to carry away the vapour, but the mechanical removal of such moisture by expression or centrifuging is not ordinarily considered drying. A solution can be “dried” by spraying it in fine droplets into a hot, dry gas, which results in evaporation of the liquid, but evaporation of the solution by boiling in the absence of gas to carry away the moisture is not ordinarily considered a drying operation. A liquid such as benzene can be “dried” of any small water content by an operation, which is really distillation, but the removal of a small amount of acetone by the same process would not usually be called drying. Gases and liquids containing small amounts of water can be dried by adsorption operations. Purpose of Drying Operations : It is carried out for a number of reasons : (1) It is used for purifying a crystalline product so that the solvent adhering to the crystals is removed or the solvent can be recovered. (2) Storage of dry solid as compared to wet solid is easy. (3) The cost of transportation for dry material would be reduced. (4) Drying can provided definite desired properties to the material as in the case of pharmaceutical substances such as tablets, syrups etc. (5) Presence of traces of moisture may lead to problems of corrosion as in the case of chlorine gas. Dry chlorine gas is not corrosive but traces of moisture make it very corrosive. (6) Sometimes can be an essential part of the process, e.g. drying of paper. However, care must be taken to avoid shrinkage of material or loss of flavour in food product or cracking etc.; which may takes place during drying operations. 8.2 EXPRESSION OF MOISTURE CONTENT Moisture content can be expressed one of two ways; Dry weight or wet weight. mass of water m.c. (m) = mass of sample × 100 Where mass of sample can be made up of water and dry matter or solids. Thus mass of water m.c. (m) = mass of water + solid × 100 On a dry weight basis, moisture is calculated as mass of water M = mass of solids This can sometimes be expressed on a percentage dry weight basis, i.e. 100 multiplied by the moisture. It can be shown by eliminating the mass of solids that 100M m m = 1 + M or M = 100 (1 – m/100) Moisture content (W/W basis) is most often used in food composition tables, whereas moisture (Dry basis) is more often encountered with sorption isotherms and drying curves.

Principles of Mass Transfer Operations − I (Vol. − I)

8.3

Drying Operations

8.3 EQUILIBRIUM IN DRYING The moisture contained in a wet solid or liquid solution exerts a vapour pressure to an extent depending upon the nature of the moisture, the nature of the solid, and the temperature. If then a wet solid is exposed to a continuous supply of fresh gas containing a fixed partial pressure of the vapour p, the solid will either lose moisture by evaporation or gain moisture from the gas until the vapour pressure of the moisture of the solid equals p. The solid and the gas are then in equilibrium, and the moisture content of the solid is termed its equilibrium-moisture content at the prevailing conditions. 8.4 TYPES OF MOISTURE

Fig. 8.1 : Types of Moisture *

Equilibrium moisture X : This is the moisture content of a substance when at equilibrium with a given partial pressure of the vapour. Bound moisture : This refers to the moisture contained by a substance, which exerts an equilibrium vapour pressure less than that of the pure liquid at the same temperature. Unbound moisture : This refers to the moisture contained by a substance, which exerts an equilibrium vapour pressure equal to that of the pure liquid at the same temperature. Free moisture : Free moisture is that moisture contained by a substance in excess of the * equilibrium moisture : X – X . Only free moisture can be evaporated, and the free-moisture content of a solid depends upon the vapour concentration in the gas. These relations are shown graphically in Fig. 8.1 for a solid of moisture content X exposed to a gas of relative humidity A. 8.5 MECHANISM OF BATCH DRYING In order to set up drying schedules and to determine the size of equipment, it is necessary to know the time required to dry a substance from one moisture content to another under specified conditions. We shall also wish to estimate the influence that different drying conditions will have upon the time for drying. Our knowledge of the mechanism of drying is so incomplete that it is necessary with few exceptions to rely upon at least some experimental measurements for these purposes. Measurements of the rate of batch drying are relatively simple to make and provide much information not only for batch but also for continuous operation.

Principles of Mass Transfer Operations − I (Vol. − I)

8.4

Drying Operations

8.6 DRYING TESTS The rate of drying can be determined for a sample of a substance by suspending it in a cabinet or duct, in a stream of air, from a balance. The weight of the drying sample can then be measured as a function of time. Certain precautions must be observed if the data are to be of maximum utility. The sample should not be too small. Further, the following conditions should resemble as closely as possible those expected to prevail in the contemplated large-scale operation : (1) the sample should be similarly supported in a tray or frame; (2) it should have the same ratio of drying to non-drying surface; (3) it should be subjected to similar conditions of radiant-heat transfer; and (4) the air should have the same temperature, humidity, and velocity (both speed and direction with respect to the sample). If possible, several tests should be made on samples of different thicknesses. The dry weight of the sample should also be obtained. The exposure of the sample to air of constant temperature, humidity, and velocity constitutes drying under constant drying conditions. 8.7 RATE-OF-DRYING CURVE From the data obtained during such a test, a curve of moisture content as a function of time Fig. 8.2 can be plotted. This will be useful directly in determining the time required for drying larger batches under the same drying conditions. Much information can be obtained if the data are converted into rates (or fluxes) of drying, expressed as N mass/(area)(time), and plotted against moisture content, as in Fig. 8.3. This can be done by measuring the slopes of tangents drawn to the curve of Fig. 8.2 or by determining from the curve small changes in moisture content –Ss ∆X . ∆X for corresponding small changes in time ∆θ and calculating the rate as N = A∆θ

Fig. 8.2 : Batch drying constant drying operations

Here SS is the mass of dry solid, A is the wet surface over which the gas blows and through which evaporation takes place in the case of cross-air circulation drying. In the case of throughcirculation drying, A is the cross section of the bed measured at right angles to the direction of the gas flow. The rate-of-drying curve is sometimes plotted with the ordinate expressed as mass moisture evaporated/(mass dry solid) (time), which in the present notation is – d/dθ. There are usually two major parts to the rate curve of Fig. 8.3, a period of constant rate and one of falling rate, as marked on the figure. While different solids and different conditions of drying often give rise to

Principles of Mass Transfer Operations − I (Vol. − I)

8.5

Drying Operations

curves of very different shape in the falling-rate period, the curve shown occurs frequently. Some of the differences which may arise will be considered later, but for the present let use briefly review the reasons generally advanced for the various parts of the curve shown. If a solid is initially very wet, the surface will be covered with a thin film of liquid, which we shall assume is entirely unbound moisture. When it is exposed to relatively dry air, evaporation will take place from the surface. The rate at which moisture evaporates can be described in terms of a gas mass transfer coefficient KY and the difference in humidity of the gas at the liquid surface YS and in the main stream Y. Thus, for cross-circulation drying. NC = KY (YS – Y)

… (8.1)

The coefficient KY can be expected to remain constant as long as the speed and direction of gas flow past the surface do not change. The humidity YS is the saturated humidity at the liquidsurface temperature ts and will therefore depend upon this temperature. Since evaporation of moisture absorbs latent heat, the liquid surface will come to, and remain at, an equilibrium temperature such that the rate of heat flow from the surroundings to the surface exactly equals the rate of heat absorption. Ys therefore remains constant. The capillaries and interstices of the solid, filled with liquid, can deliver liquid to the surface as rapidly as it evaporates there. Since in addition Y remains unchanged under constant drying conditions, the rate of evaporation must remain constant at the value NC, as shown in Fig. 8.2 and 8.3 between points B and C. In the beginning, the solid and the liquid surface are usually colder than the ultimate surface temperature ts, and the evaporation rate will increase while the surface temperature rises to its ultimate value during the period AB on these curves. Alternatively the equilibrium temperature ts may be lower than the initial value, which will give rise to a curve A’B while the initial adjustment occurs. The initial period is usually so short that it is ordinarily ignored in subsequent analysis of the drying times. When the average moisture content of the solid has reached a value Xc, the critical moisture content, the surface film of moisture has been so surface; these spots occupy increasingly larger proportions of the exposed surface as drying proceeds. Since, however, the rate N is computed by means of the constant gross surface A, the value of N must fall even though the rate per unit of wet surface remains constant. This gives rise to the first part of the falling-rate period, the period of unsaturated surface drying, from points C to D (Fig. 8.2 and Fig. 8.3). Ultimately the original surface film of liquid will have entirely evaporated at an average moisture content for the solid corresponding to point D. This part of the curve may be missing entirely, or it may constitute the whole of the falling-rate period. With some textiles, other explanations for the linear falling-rate period have been necessary. 8.8 TIME REQUIRED FOR DRYING If one wishes to determine the time of drying a solid under the same conditions for which a drying curve such as Fig. 8.3 has been completely determined, one need merely read the difference in the times corresponding to the initial and final moisture contents from the curve. Within limits, it is sometimes possible to estimate the appearance of a rate-of-drying curve such as Fig. 8.3, for conditions different from those used in the experiments. In order to determine the time for drying for such a curve, we proceed as follows.

Principles of Mass Transfer Operations − I (Vol. − I)

8.6

Drying Operations

Fig. 8.3 : Rate of Batch Drying Curve

As evident from above figure, the rate of drying curve consists of two major zones : (a)

a period of constant rate drying.

(b) a period of falling rate drying. Since the mechanism of drying during the initial adjustment is not very clear, hence this period is not included in the total drying time. Rate of drying is given by :

LS dx N = –A · dθ

… (8.2)

From equation (8.2), the expression for drying time is obtained as : LS θ = A

x1

dx

⌠ N ⌡

… (8.3)

x2

If drying takes place within the constant rate period only (so that x1 > xo and x2 > xC, N = NC) LS (x1 – x2) then equation (8.3) becomes, θ = … (8.4) A NC For more general situations involving both constant and falling rate regimes, equation (8.4) can be rewritten to give the constant rate time period as, θC =

LS (x1 – xC) A NC

… (8.5)

Time for the falling rate period is determined with the help of graphical integration of 1 equation (8.5) by determining the area under the curve of N Vs x So,

LS θf = A × (area under the curve of Fig. 8.4)

Principles of Mass Transfer Operations − I (Vol. − I)

8.7

Drying Operations

If the falling rate curve over the complete range (DC and DE of Fig. 8.3) is approximated as a straight line, then equation on simplification yields time for the falling rate period is,

Fig. 8.4 : Graphical integration for falling rate period

θf =

LS (XC – X*) ANC

x1 – x* · ln x – x* 2

(when x1 < xC) …(8.6)

If the initial moisture content of the solid is more than the critical moisture, equation (8.6) is written as, θf =

LS (xC – x*) xC – x* ln x – x* ANC 2

… (8.7)

For such a case, the total time for drying with the help of equations (8.5) and (8.7) is given by, θT = θC + θf LS θT = AN C

(x – x ) + (x – x*) ln xC – x*  1 C C x2 – x*  

… (8.8)

Estimation of drying rate : The heat or mass transfer equation can be used to estimate constant drying rate. We have heat transfer equation : . hG (Ta – Ti) m NC = A = … (8.9) λi where,

. m A Ta Ti λi

= = = = =

Rate of evaporation, kg/hour Drying surface, m2 Temperature of air Temperature of interface Latent heat at Ti

For air flow parallel to surface the heat transfer coefficient can be estimated by the following dimensional equation : hG = 0.0176 G0.8 … (8.10 a) where,

hG = heat transfer coefficient, k cal/hr. m2 oC G = Mass velocity of air, kg/hr. m2

When a flow perpendicular to surface is considered, the above equation becomes, hG = 1.004 G0.37 Note that numerical constants of equation (8.10 a) and (8.10 b) are dimensional.

… (8.10 b)

Principles of Mass Transfer Operations − I (Vol. − I)

8.8

Drying Operations

8.9 DRYING TIME FOR DROPLETS The constant rate and falling rate periods must be considered separately. 8.9.1 Constant Rate Period The rate of mass transfer during the constant rate period is given by the following expression. This will be dependent on the amount of heat present in the product i.e. it can be related to the rate of heat transfer. dQ hA (Ta – Tw) dt dw = = – dt λ λ Ta = Dry bulb temperature Tw = Wet bulb temperature = Tsurface Heat transfer coefficient for a spherical particle at Re < 20 2ka h = D ka = Thermal conductivity of air; D = Droplet diameter. Droplet diameter will change during the constant rate period 2

Surface area of a sphere = πD = 4πr2. – 2ka πD2 dw = (Ta – Tw) dt D λ tr

wc

⌠ ⌠ λ  dt =  ⌡ ⌡ 2kaπD (Ta – Tw) dw 0

wo wc

–λ tc = 2k π(T – T ) a a w

⌠ 1  ⌡ D dw wo

Once we remove all the constants, you are left with an incompatible differential. We must account for the change in weight as a change in the diameter of the particle. 3

πD ρ 4πr3 ρ = 6 3 π 3 3 ∆w = ∆Vρw = 6 ρw (D – (D – ∆D) ) πρw 3 3 2 2 3 {D – D + 3D ∆D – 3D∆D + ∆D } 6 πρw ∆w 2 2 = 6 (3D – 3D∆D + ∆D ) ∆D As ∆D → 0 then w =

dw dD

2

πpw πD ρw 2 = 6 3D = 2 2

πD ρw dw = dD Rate of change in weight, related to rate of change in diameter 2

Principles of Mass Transfer Operations − I (Vol. − I)

8.9

Drying Operations

Dc

⌠ 1 πD2  ⌡ D 2 ρw dD

–λ tc = 2k π (T – T ) a a w

Do Dc

⌠  DdD ⌡

– λπρw = 4k π (T – T ) a a w

Do

Dc Do – λρw = 4k (T – T )  2 – 2   a a w  2

tc =

Thus,

(

2

2

λρw Do – Dc

2

)

8ka (Ta – Tw)

Note that the minus sign has been taken inside the upper bracket. 8.9.2 Falling Rate Period The rate of heat transfer during the falling rate period is given by the following expression : dQ q = dt = ha A(Ta – Ts), where Ts is the surface temperature of the particle. Unfortunately Ts varies from Tw at the end of the falling rate period. Use then

Ts = dQ dt

Ta + Tw 2

 Ta Tw  = haA (T – T ) = ha A Ta – 2 + 2 a w 2   

We thus need to convert this expression relating heat transfer to time to one relating mass transfer to time. This is achieved by using the latent heat multiplied by the mass of the particle. Mass in equal to the density of the dry particle multiplied by the volume of the particle w = ρd V dw dt

dQ 1 dQ 1 = dt = dt λw λρdV

dw dt

– haA (Ta – Tw) dQ 1 = dt = λρdV 2λρdV

Now area and volume can be expressed in terms of the diameter of the particle, thus A V

and

2

=

ha =

πD

3

πD 6

2ka D

6 =D

Principles of Mass Transfer Operations − I (Vol. − I)

8.10

Drying Operations

Assume in falling rate period that D is constant = Dc

Thus

dw dt

2ka 6 (Ta – Tw) = – D ·D · 2λρd c c

dw dt

=

– 6ka (Ta – Tw) 2

λρdDc

tp

wf

2 λρd Dc

– ⌠ ⌠  dt =  6ka(Ta – Tw) ⌡ dw ⌡ 0

wc 2

– λρd Dc tf = 6k (T – T ) (wf – wc) a a w 2

Thus,

λρd Dc tf = 6k (T – T ) (wc – wf) a a w

Note that the minus sign has been taken inside the upper bracket. 8.9.3 Droplet trajectory Three forces act on a particle in its trajectory. These are Buoyancy; Gravity and Drag force-Stoke's Law. Maximum distance travelled by particle before it reaches low moisture content 2

Vpo Dp ρp Vpo Smax = K = 18µ 8.10 HEAT TRANSFER IN DRYERS Drying of solids in a heat transfer process coupled with diffusional transfer of moisture. However, most dryers are designed conveniently on heat transfer conditions : Heat must be applied for : (i) heating the feed to vapourisation temperature. (ii) vapourizing the liquid. (iii) heating solid to final exit temperature. (iv) heating vapour to final exit temperature, and Thus the total rate of heat transfer is, qT = CP S (TS – TS ) + Xa CP (TV – TS ) + . b a L a mS (Xa – Xb) λ + Xb · CP (TS – TV) + L

(Xa – Xb) CP (TV – TV) V

where,

b

qT = Amount of heat transfer, k cal/hr. .

m S = Amount of bone-dry solid, kg/hr. Xa = kg moisture/kg dry solid (entry) Xb = kg moisture/kg dry solid (exit)

b

… (8.11)

Principles of Mass Transfer Operations − I (Vol. − I) CP CP CP

8.11

Drying Operations

= Specific heat of solid k cal/kg oC

S

= Specific heat of liquid, k cal/kg oC

L

= Specific heat of vapour k cal/kg oC

V

= Inlet solid temperature, oC

TS

a

TV = Vapourisation temperature, oC TV

= Final vapour (exit) temperature, oC

b

λ = Latent heat of vapour, k cal/kg The temperature distribution pattern in a continuous counter-current dryer is given below :

Fig. 8.5 : Temperature pattern in a continuous counter current adiabatic dryer

For adiabatic dryers, TV and TV of equation (8.11) become equal to wet bulb temperature and b

exit temperature of the drying medium respectively. For a continuous adiabatic dryer heat balance is given by, . qT = m g (1 + Ha) CS (Th – Th ) a

where

a

b

… (8.12)

Mass rate or drying medium‚ . m g = kg of humidity-free/hour Ha = Humidity of drying medium at entry CS

a

= Humid heat of drying medium at entry

In dryer calculations, the basic heat transfer equation used is, qT = UA ∆ T

… (8.13)

There is uncertainty in area available for heat transfer. So many dryers are designed on the basis of a volumetric heat-transfer coefficient, Ua. where 'a' is the (unknown) heat transfer 0 area per unit dryer volume. The governing equation is, qT = Ua V · ∆ T where,

Ua = volumetric heat transfer coefficient, k cal/m2 hr oC V = dryer volume, m3 ∆T = average temperature difference, oC

… (8.14)

Principles of Mass Transfer Operations − I (Vol. − I)

8.12

Drying Operations

8.11 MECHANISM OF MOVEMENT OF MOISTURE WITHIN THE SOLID When surface evaporation occurs, there must be a movement of moisture from the depths of the solid to the surface. The nature of the movement influences the drying the falling-rate periods. In order to appreciate the diverse nature of the falling-rate portions of the drying curve, which have been observed, let us review very briefly some of the theories advanced to explain moisture movement and their relation to the falling-rate curves. Liquid Diffusion : Diffusion of liquid moisture may result because of concentration gradients between the depths of the solid, where the concentration is high, and the surface, where it is low. These gradients are set up during drying from the surface. This method of moisture transport is probably limited to cases where single-phase solid solutions are formed with the moisture, as in the case of soap, glue, gelatin, and the like, and to certain cases where bound moisture is being dried, as in the drying of the last portions of water from clays, flour, textiles, paper, and wood. The general mechanism of this process is described in chapter. 4. It has been found that the moisture diffusivity usually decreases rapidly with decreased moisture content. During the constant-rate period of drying such solids, the surface-moisture concentration is reduced, but the concentration in the depths of the solid remains high. The resulting high diffusivities permit movement of the moisture to the surface as fast as it can be evaporated, and the rate remains constant. When dry spots appear because portions of the solid project into the gas film, a period of unsaturated surface evaporation results. The surface eventually dries to the equilibrium-moisture content for the prevailing gas. Further drying occurs at rates which are entirely controlled by the diffusion rates within the solid, since these are slow at low moisture contents. If the initial constant-rate drying is very rapid, the period of unsaturated surface evaporation may not appear, and the diffusion-controlled falling-rate period begins immediately after the constant rate period is completed, as in Fig. 8.6.

Fig. 8.6 : Diffusion – Controlled Falling Rate

Capillary Movement : With certain pastes dried in pans, the adhesion of the wet cake to the bottom of the pan may not permit ventilation of the subsurface passageways by gas. This can give rise to curves of the sort shown in Fig. (8.7). In this case, the usual constant-rate period prevailed during a. When the surface moisture was first depleted, liquid could not be brought to the surface by the tension in the capillaries since no air could enter to replace the liquid, the surface of moisture crumpled, admitting air to replace the liquid, whereupon capillary action brought this to the surface and the rate rose again, as at c.

Principles of Mass Transfer Operations − I (Vol. − I)

8.13

Drying Operations

Fig. 8.7 : Effect of a adhesion of drying paste to the pan

Vapour Diffusion : Especially if heat is supplied to one surface of a solid while drying proceeds from another, the moisture may evaporate beneath the surface and diffuse outward as a vapour. Moisture particles in granular solid, which have been isolated from the main portion of the moisture flowing through capillaries, may also be evaporated below the surface. 8.12 EQUIPMENT FOR DRYING (1) Through-Circulation Driers : Granular solids can be arranged in thin beds for through circulation of the gas, and, if necessary, pastes and filters cakes can be preformed into granules, pellets, or noodles, as described in the case of batch driers. In the continuous through circulation dryer of Fig. 8.8, the solid is spread to a depth of 38 to 50 mm upon a moving endless conveyor, which passes through the dryer. The conveyor is made of perforated plates or woven wire screens in hinged sections in order to avoid failure from repeated flexing of the screen. Fans blow the heated air through the solid, usually upward through the wet solid and downward after initial drying has occurred. In this way, a more uniform moisture concentration portion is discarded continuously at each fan position in the dryer. For materials which permit the flow of gas through the bed in the manner shown, drying is much more rapid than for tray-type tunnel dryers.

Fig. 8.8 : Continuous through-circulation (single conveyor) dryer with roller extruder

Principles of Mass Transfer Operations − I (Vol. − I)

8.14

Drying Operations

(2) Rotary Dryers : (Refer Fig. 8.9, 8.10, 8.11, 8.12 and 8.13) : This is a most important group of dryers, suitable for handling free-flowing granular materials, which can be tumbled about without concern over breakage. shows one form of such a dryer, a direct countercurrent hot-air dryer. The solid to be dried is continuously introduced into one end of a rotating cylinder, as shown, while heated air flows into the other. The cylinder is installed at a small angle to the horizontal, and the solid consequently moves slowly through the device. Inside the dryer, lifting flights extending from the cylinder wall for the full length of the dryer lift the solid and shower it down in a moving curtain through the air, thus exposing it thoroughly to the drying action of the gas. This lifting action also assists in the forward motion of the solid. Fig. 8.9 : Turbo-type dryer

At the feed end of the dryer, a few short spiral flights assist in imparting the initial forward motion to the solid before the principal flights are reached. The solid must clearly be one, which is neither sticky nor gummy, which might stick to the sides of the dryer or tend to ball up. In such cases, recycling of a portion of the dried product may nevertheless permit used of a rotary dryer.

Fig. 8.10 : Ruggles – Coles XW hot-air dryer

The dryer may be fed with hot flue gas rather than air, and if the gas leaves the drier at a high enough temperature, discharging it through a stack may provide adequate natural draft to provide sufficient gas for drying. Ordinarily, however an exhaust fan is used to pull the gas through the dryer, since this provides more complete control of the gas flow. A dust collector, of the cyclone, blower may also be provided at the gas entrance, thus maintaining a pressure close to atmospheric in the dryer; this prevents leakage of cool air in it at the end housings of the drier, and if the pressure is well balanced, outward leakage will also be minimized.

Principles of Mass Transfer Operations − I (Vol. − I)

8.15

(a) Indirect rotory dryer

(b) Indirect steam - tube rotory dryer

(c) Indirect - direct rotory dryer Fig. 8.11 : Some rotary dryers (schematic)

Fig. 8.12 : Continuous through-circulation rotary dryer (Roto-Lourvre)

Drying Operations

Principles of Mass Transfer Operations − I (Vol. − I)

8.16

Drying Operations

Fig. 8.13 : Dip-feed single drum dryer

The dryer may be fed with hot flue gas rather than air and if the gas leaves the dryer at a high enough temperature, discharging it through a stack may provide adequate natural draft to provide sufficient gas for drying. Ordinarily, however an exhaust fan is used to pull the gas through the direr, since this provides more complete control of the gas flow. A dust collector, of the cyclone, blower may also be provided at the gas entrance, thus maintaining a pressure close to atmospheric in the dryer; this prevents leakage of cool air in at the end housings of the dryer and if the pressure is well balanced, outward leakage will also be minimized. Rotary dryers are made for a variety of operations. The following classification includes the major types. (1) Direct heat, countercurrent flow. For materials, which can be heated to high temperature, such as minerals, sand, limestone, clays, etc. hot flue gas can be used as the drying gas. (2) Direct heat, co-current flows. Solids, which can be dried with flue gas without fear of damage, such as gypsum, iron pyrites, and organic material such as peat and alfalfa, should be dried in a co-current flow dryer. (3) Indirect heat, countercurrent flow. For solids such as white pigments, and the like, which can be heated to high temperatures but which must remain out of contact with flue gas. (3) Spray Dryers : Solutions, slurries, and pastes can be dried by spraying them as fine droplets into a stream of hot gas in a spray dryer. One such device is shown in Fig. 8.14. The liquid to be dried is atomized and introduced into the large drying chamber, where the droplets are dispersed into a steam of heated air. The particles of liquid evaporate rapidly and dry before they can be carried to the sides of the chamber, and the bulk of the dried powder which results falls to the conical bottom of the chamber to be removed by a stream of air to the dust collector. The principal portion of the exit gas is also led to a dust collector, as shown, before being discharged. Many other arrangements are possible, involving both parallel and counter flow of gas and spray. Installations may be very large, as much as 12 m in diameter and 30 m high (40 by 100 ft). Arrangements and detailed designs vary considerably, depending upon the manufacturer. Spray dryers are used for a wide variety of products, including such diverse materials as organic and inorganic chemicals, pharmaceuticals, food products such as milk, eggs, and soluble coffee, as well as soap and detergent products. Spray drying offers the advantage of extremely rapid drying for heat-sensitive products, a product particles size and density which are controllable within limits, and relatively low operating costs, especially in large-capacity dryers.

Principles of Mass Transfer Operations − I (Vol. − I)

8.17

Drying Operations

Fig. 8.14 : Spray dryer

(4) Fluidized and Spouted Beds : Granular solids, fluidized by a drying medium such as hot air, can be dried and cooled in a similar fluidized bed, shown schematically in Fig. 8.15. The principal characteristics of such beds include cross flow of solid and drying gas, a solids residence time controllable from seconds to hours, and suitability for any gas temperature. It is necessary that the solids be free flowing, of a size range 0.1 to 36 mm. Since the mass flow rate of gas for thermal requirements is substantially less than that required for fluidization. Multistage, crossflow operation (fresh air for each stage) is a possibility, as is a two-stage countercurrent arrangement, as in Fig. 8.15. A tentative design procedure has been proposed.

Fig. 8.15 : Fluidisied Bed Dryer

Principles of Mass Transfer Operations − I (Vol. − I)

8.18

Drying Operations

Coarse solids too large for ready fluidization can be handled in a spouted bed. Here the fluid is introduced into the cone-shaped bottom of the container for the solids instead of uniformly over the cross section. It flows upward through the center of the bed in a column, causing a fountainlike spout of solids at the top. The solids circulate downward around the fluid column. Such a bed has found particular use in drying wheat, peas, flax, and the like. 8.13 SELECTION, SIZING, COSTS Throughout the food, dairy, chemical and process industries, there are various requirements for thermal drying. Some involve the removal of water or other volatiles from pasty materials such as pigments, clays, synthetic rubbers and fine chemicals. Other involve the drying of solutions or liquid suspensions such as whey, milk and coffee. To assist manufacturers in arriving at a reasonably accurate first assessment of the type, size and cost of equipment for a particular duty, the articles describes the most widely used types of both batch and continuous dryers. Three basic methods of heat transfer are used in industrial dryers in various combinations. These are convection, conduction and radiation. In the processing industries the majority of dryers employ forced convection and continuous operation. With the exception of the indirectly heated rotary dryer and the film drum dryer, units in which heat is transferred by conduction are suitable only for batch use. This limitation effectively restricts them to applications involving somewhat modest production runs. Radiant or so-called "infra-red" heating is rarely used in drying materials such as fine chemicals, pigments, clays or synthetic rubbers. Its main application is in operations such as the drying of surface coatings on large plane surfaces, since efficient utilisation generally requires a line of sight between the material being irradiated and the heat source or emitter. Feed Type Solution

Thixotrop

Fine

Powder Type

Freeflow Dustless

Dilatent Cohesive

Friable

Spray or SBD Band Spray Bed

Agglom Coated Lump

Powder

Flash

Spin Flash

Spray

Granular Wettable

Granules

Spin Flash + Fluid Bed Agglom.

Tray

Band

Fluid Bed Granulation

Fig. 8.16 : A guide to dryer selection

8.14 EFFICIENT ENERGY UTILISATION IN DRYING It is generally necessary to employ thermal methods in order to achieve a product that is termed commercially dry. Thermal dryers, therefore, are an important unit operation in many industries. Products such as pigments, baby formula, kaolin and instant coffee usually have to be processed in dryers to obtain the final product. The commercial drying process can take place in a number of different types of dryers as described in the previous section. However, the thermodynamics of drying is the same regardless of the actual device. Water or another liquid

Principles of Mass Transfer Operations − I (Vol. − I)

8.19

Drying Operations

such as a solvent has to be evaporated from a solid by the application of heat. Generally, the drying takes place in an air atmosphere, although some emphasised dryers may use nitrogen or even superheated steam as the atmosphere. The most important parameter that governs the pretreatment and the dryer design is the "cost per unit weight of dried product." While drying is an extremely energy intensive operation, there are techniques that can be used to minimize the energy costs per unit output of product, including : •

Minimizing the water content of the feed prior to feeding to the dryer.

•

Maximizing the temperature drop of the drying gas. The implies maximum inlet and minimum outlet temperature.

•

Employing the maximum possible recirculation of the drying gas.

•

Considering the possibility of two stage counter flow drying.

•

Utilizing the heat in the discharge air to preheat incoming air.

•

Utilizing direct heat wherever possible.

•

Reducing radiation and convection heat loss by means of efficient thermal insulation.

SOLVED PROBLEMS (1) Slabs of paper pulp 1 m × 1 m × 0.015 m is to be dried under constant drying conditions from 66.7% to 30% moisture. The value of equilibrium moisture for the material is 0.5%. If the critical moisture content is 60% and the rate of drying of critical point is 1.5 kg/hr.m2; calculate the drying time. The dry weight of each slab is 2.5 kg. Assume all moisture contents are on wet basis. Sol. : Consider drying takes place from the two big faces. So, Area for drying = (1 × 1) × 2 = 2 m2 LS = 2.5 kg 66.7 X1 = 100 – 66.7 = 2.0 30 X2 = 100 – 30 = 0.429 0.5 X* = 100 – 0.5 = 0.005 60 XC = 100 – 60 = 1.5 NC = 1.5 kg/hour m2 We have equation for total time required for drying, θT = θC + θf XC – X*   (X1 – XC) + (XC – X*) ln  X2 – X*  

=

LS ANC

=

1.5 – 0.005  2.5  (2 – 1.5) + (1.5 – 0.005) · ln  0.429 – 0.005 = 2 × 1.5 

∴ Total drying time = 1.987 hours

1.987 (Ans.)

Principles of Mass Transfer Operations − I (Vol. − I)

8.20

Drying Operations

(2) A certain material was dried under constant drying conditions and it was found that 2 hours are required to reduce the free moisture concentration from 20% to 10%. How much longer would be required to reduce the free moisture to 4% ? Assume that no constant rate period is encountered. Sol. : Assumptions : (i) Equilibrium moisture content to zero. (ii) Falling rate period is linear. Let. θf = Falling rate period for the first case 1

θf

2

= Falling rate period for the second case

20 X1 = 100 – 20 = 0.250 10 X2 = 100 – 10 = 0.111 4 ' X2 = 100 – 4 = 0.042 LS X1 θf = AN XC · ln X 1 C 2 and

θf

LS X1 = AN XC · ln ' C X2

from equation (1),

θf

LS 0.250 = AN XC ln 0.111 = 2 C

Therefore, and from equation (2),

2

1

LS XC ANC

(... X* = 0) … (1)

= 2.46

θf

LS X1 0.250 = AN XC ln X = 2.46 ln 0.042 C 2

θf

= 4.39 hours

2 2

… (2)

… (3)

∴ Further time required = 4.39 – 2.00 = 2.39 hours (Ans.) (3) Celotax sheets are to be dried by flowing air at 60 oC, 10% relative humidity and a velocity of 10 m/sec. The critical moisture content is 0.35 kg free water/kg dry solid. The rate of drying in the falling rate period can be considered linear. The celotax must be dried from 55% to 12% moisture (Wet basis). Equilibrium moisture content at the conditions of the dryer is 5% (wet basis). The celotax sheets are kept in layer of 6 cm thick in insulated trays. The dry density of solid is 1.38 gm/cc. Calculate the time of drying the sheets. Data : The rate of drying in gm/cm2 . hour in the constant rate period is given by : NC = 0.004 V0.8 (pi – pg) where V = Air velocity, m/sec pi and pg are dew point temperature and vapour pressure of water in mm of Hg in air at the wet bulb temperature respectively. kg water Sol. : At 60o C and 10% humidity, H = 0.014 kg dry air 0.014 18 kg mole water = = 0.023 kg. mole dry air 1 29

Principles of Mass Transfer Operations − I (Vol. − I)

8.21

Drying Operations

If pg = partial pressure of water vapour, pg PT – pg = 0.023, where PT = 760 mm Hg ∴ pg = 17.09 mm Hg Corresponding to dew point of the above air, kg mole water kg water Humidity = 0.028 kg dry air = 0.046 kg mole dry air pt ∴ 760 – pi = 0.046 ∴

pi = 33.42 mm Hg at dew point temperature

Partial pressure of water = vapour pressure Hence, NC = 0.004 V0.8 (pi – pg) = 0.004 (10)0.8 (33.42 – 17.09) = 0.412 gm/cm2 . hr. where NC = constant rate drying 55 X1 = 100 – 55 = 1.222, XC = 0.35, 12 X2 = 100 – 12 = 0.136, 5 X* = 100 – 5 = 0.053 Dry density = 1.38 gm/cc Let us consider a celotax sheet of 1 m × 1 m ∴ Volume of this sheet = 100 cm × 100 cm × 6 cm = 60,000 cm3 LS = Dry weight = 60,000 × 1.38 = 8.28 × 104 gm Area = 100 × 100 = 104 cm2 (Since the trays are insulated) Therefore, time required for drying, θT = θC + θf LS  XC – X*  = A·N (X1 – XC) + (XC – X*) ln X – X*  C  2   8.28 × 104  0.35 – 0.053  (1.222 – 0.35) + (0.35 – 0.053) ln  0.136 – 0.053  104 × 0.412  θT = 25.14 hours (Ans.) (4) It is necessary to dry a batch of 160 kg of a wet material from 30% to 5% moisture content, under constant rate and falling rate period. The falling rate is assumed to be linear. Calculate the total drying time considering an available drying surface of 1 m2/40 kg of dry solid. A constant drying flux of 3 × 10–4 kg/m2 is given : 0.2 kg moisutre XC = Critical moisture content = kg solid =

X* = Equilibrium moisture content = 0.05

Principles of Mass Transfer Operations − I (Vol. − I) LS A

Sol. :

8.22

Drying Operations

160 × 0.7 112 = 40 = 2.8 m2 40

=

30 X1 = 100 – 30 = 0.4285 kg moisture/kg solid 5 X2 = 100 – 5

LS (X1 – XC) 2.8 (0.4285 – 0.20) = ANC 3 × 10–4

θC =

Constant rate period :

= 0.0526 kg moisture/kg solid

θC = 2132.66 sec. LS (XC – X*) XC – X* ln  X – X*  ANC  2 

θf =

Falling rate period :

2.8 (0.2 – 0.05) 0.2 – 0.05 ln 0.0526 – 0.05 –4   3 × 10

=

= 5677.17 sec. So, the total time required for drying θT = θC + θf

=

7809.83 sec.

= 2.17 hours

(Ans.)

(5) A batch of solid for which the following table of data applies is to be dried from 25% to 6% moisture under conditions identical to those for which the data were tabulated. The initial weight of the wet solid is 300 kg and the drying surface is 1 m2/8 kg dry weight. Determine the time for drying. X

0.35

0.25

0.20

0.18

0.16

0.14

0.12

0.10

0.09

0.08

0.064

N

0.3

0.3

0.3

0.266

0.239

0.208

0.180

0.150

0.097

0.07

0.025

where, X

kg moisture = kg dry solid kg moisture evaporated hr m2 25 = 75 = 0.333 6 = 94 = 0.064

N = X1

Sol. :

X2

LS A = 8 The rate of drying curve is plotted in Fig. 8.17 (a) the values of XC and NC are obtained kg from it. NC = 0.3 hr. m2 kg moisture XC = 0.20 kg dry solid The drying from X1 = 0.333 to X2 = 0.064 covers both the constant and the falling rate periods.

Principles of Mass Transfer Operations − I (Vol. − I) Constant rate drying period :

8.23

Drying Operations

X1 = 0.333 XC = 0.20 LS θC = AN (X1 – XC) C 8 = 0.3 (0.333 – 0.20) = 3.55 hours

Falling rate drying period :

θf

LS = A

X2

⌠ ⌡

dX N

XC

(a)

(b) Fig. 8.17 : Graphical solution for example (5)

Since the plot of X vs. N (refer to Fig. 8.17 a) for this period is not a straight line hence the dX value of ⌠ N is evaluated by the method of graphical integration (Fig. 8.17 b). ⌡ 1 The table used to plot X vs. N is as under : X

N

0.20

0.300

1 N 3.33

0.18

0.266

3.76

0.16

0.239

4.18

0.14

0.208

4.81

0.12

0.180

5.56

0.10

0.150

6.67

0.09

0.097

10.31

0.08

0.070

14.29

0.064

0.025

40.00

Principles of Mass Transfer Operations − I (Vol. − I)

8.24

Drying Operations

The area under curve (Fig. 8.17 (b)) is 1.063, 0.064

⌠ ⌡

∴

dX N

= 1.063

0.20

So,

0.064

LS = A

θf

dX

⌠ N ⌡

= 8 × 1.065 = 8.5 hours

0.20

Total drying time = θC + θf

=

3.35 + 8.5 = 12.05 hours

(Ans.)

(6) A rotary counter current dryer is fed with ammonium nitrate containing 6% moisture at the rate of 100 kg/min and discharges the nitrate with 0.2% moisture. The air enters at 135oC and leaves at 80oC. The humidity of entering air being 0.007 kg H2O per kg dry air. The nitrate enter at 21oC and leaves at 65oC. Neglecting radiation losses, calculate the kg of dry air passing through the dryer and the humidity of the air leaving the dryer. Special heat of ammonium nitrate = 0.45 Special heat of dry air = 0.238 Special heat of water vapour = 0.48 Sol. : Basis : 100 kg/min of wet ammonium nitrate. Let, M = Wt. of bone-dry product So, 100 – M = Wt. of moisture = 6.0

Fig. 8.18 : Rotary counter current dryer

M = 94 kg W2 = Wt. of water

∴ At the exit, let

W2 94 + W2

So,

0.2 = 100

W2 = 0.19 kg Therefore water removed in the dryer = 6 – 0.19 = 5.81 kg. Enthalpy balance : Let

H2 = moisture consent of exit air, kg/kg dry air

Enthalpy of inlet air = EG

1

= (Cg + H1Cv) (t – to) + λH1 With

t0 = 0 o C λ = 597.7 kcal/kg Cg = 0.238 and Cv = 0.48

Principles of Mass Transfer Operations − I (Vol. − I)

8.25

Drying Operations

EG

= (0.238 + 0.007 × 0.48) (135) + 597.7 × 0.007

EG

= 36.76 k cal/kg dry air = (0.238 + 0.48 H2) (80) + 597.7 H2

1

2

= (19.04 + 636.1 H2) k cal/kg dry air Enthalpy of solid at the inlet, Enthalpy of solid at the exit, Let, By heat balance, or,

GS EG

1

= = = = GS = + 945 =

100 × 0.45 × 21 945 k cal 94.19 × 0.45 × 65 2755 k cal. weight of dry air/min. GSEG + 2755 2

36.76 GS + 945 = GS (19.04 + 636.1 H2) + 2755

By moisture balance on air, GS (H2 – 0.007) = 5.81 Simultaneous solution of the above two equations yield, GS = 414.9 kg

(Ans.)

kg water H2 = 0.021 kg dry air So rate of dry air through the dryer = 414.9 kg and the humidity of exit air is kg H2O = 0.021 kg dry air

(Ans.)

(7) In a drying experiment, a tray dryer containing a single tray of 1 sq. metre area is used to dry crystalline solids. The following data has been collected : Sr. No.

Time, Hour

Weight of wet material, kg

1

0

5.314

2

0.4

5.238

3

0.8

5.162

4

1.0

5.124

5

1.4

5.048

6

1.8

4.972

7

2.2

4.875

8

2.6

4.819

9

3.0

4.743

10

3.4

4.667

11

4.2

4.524

12

4.6

4.468

13

5.0

4.426

14

6.0

4.340

15

infinite

4.120

Principles of Mass Transfer Operations − I (Vol. − I)

8.26

Drying Operations

(a) Calculate and plot drying rates. Find the critical moisture content. (b) If dry air is available at 40 oC with an absolute humidity of 0.01 kg H2O per kg dry air and the dryer is maintained at 90 oC, calculated the amount of air required in first 2 hours. Assume that the air is heated upto 90 oC and the dry air leaves the dryer at o 90 C with 5% saturation. (c) Test the consistency of the falling rate period (choose critical moisture content and any one point in the falling rate period). Sol. : Assumptions : (1) (2) Weight of material at

X* = 0 t∞ = LS = 4.12 kg A = 1 m2 LS 4.12 A = 1

∴

= 4.12 Amount of moisture = W – LS X = (a)

The experimental values are tabulated as under : W – LS

∆W ∆θA ∆θ 0.0 5.314 1.194 0 0.290 0 0.4 5.238 1.118 0.076 0.271 0.19 0.8 5.162 1.042 0.152 0.253 0.19 1.0 5.124 1.004 0.190 0.244 0.19 1.4 5.048 0.928 0.266 0.225 0.19 1.8 4.972 0.852 0.342 0.207 0.19 2.2 4.895 0.775 0.419 0.188 0.19 2.6 4.819 0.699 0.495 0.170 0.19 3.0 4.743 0.623 0.571 0.151 0.19 3.4 4.667 0.547 0.647 0.133 0.19 4.2 4.524 4.404 0.790 0.098 0.19 4.6 4.468 0.348 0.846 0.084 0.184 5.0 4.426 0.306 0.888 0.074 0.178 6.0 4.340 0.220 0.974 0.053 0.162 4.120 0 1.194 0 0 α θ vs. ∆W is plotted in Fig. 8.19 (a) and the drying curve is plotted (X vs. N) in Fig. 8.19 (b). kg H2O From Fig. 8.19 (b), XC = 0.133 kg dry solid (Ans.) (b) From Fig. 8.19 (a), water removed in first two hours = 1.194 – 0.39 = 0.804 kg kg H2O Humidity of inlet air = 0.01 kg dry air

S. No. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15.

W – LS kg H2O LS kg dry solid

θ, hr.

W

∆W

X

N=

Principles of Mass Transfer Operations − I (Vol. − I)

8.27

Drying Operations

Humidity of exit air (90o C and 5% saturation) kg H2O = 0.068 kg dry air Water removed = 0.068 – 0.01 kg H2O = 0.058 kg dry air 0.804 Dry air required in first two hours = 0.058 = 6.62 kg

(a)

(Ans.)

(b) Fig. 8.19 : Graphical solution for example (7)

Amount of inlet air (40 oC, H = 0.01) required = 13.86 × 0.01 = 0.13862 kg. (c) Let us assume a value of X in the falling rate period from the plot and check for the drying time. X = 0.074 (corresponding to θ = 5 hours) For drying of the material from X = 0.29 to X = 0.074, let the time required = θΤ. θT = θc + θf LS θc = AN (X1 – Xc) c 4.12 = 0.19 (0.29 – 0.133) = 3.40 hr. LS Xc – X* θf = AN (Xc– X*) ln X – X* c 2 4.12 0.133 = 0.19 × 0.133 ln 0.074 (... X* = 0)

Principles of Mass Transfer Operations − I (Vol. − I)

8.28

Drying Operations

= 1.67 hour θT = 3.40 + 1.67 = 5.07 hr. Time of drying from experimental data = 5 hour. Since both the timings are nearly equal, the falling rate period is consistent. (Ans.) (8) It is desired to dry a certain type of fibre board in sheets 0.131 metre by 0.162 metre by 0.071 metre from 58% to 5% moisture (wet basis) content. Initially from laboratory test data with this fibre board, the rate of drying at constant rate period was found to be 8.9 kg/m2 hour. The critical moisture content was 24.9% and the equilibrium moisture content was 1%. The fibre board is to be dried from one side only and has a bone-dry density of 210 kg/m3. Determine the time required for drying. The falling rate may be assumed linear. Sol. : All the moisture content are on wet basis : 58 X1 = 42

= 1.3809

5 X2 = 95

= 0.0526

24.9 XC = 75.1

= 0.3316

1 X* = 99

= 0.0101

Sheet volume = = Bone dry density = Weight of bone-dry sheet = LS =

0.131 × 0.162 × 0.071 0.001507 m3 210 kg/m3 0.001507 × 210 0.316 kg

Drying area = 0.131 × 0.162 = 0.021 m2 Drying time = θΤ = θC + θf NC = 8.9 kg/m2 hr. LS  xC – x* θT = AN (x1 – xC) + (xC – x*) ln x – x* C  2 =

  

0.316 (1.3809 – 0.3316) + (0.3316 – 0.0101) ln 0.3316 – 0.0101 0.526 – 0.0101  0.0221 × 8.9 

= 2.874 hours

(Ans.)

(9) 1400 kg (bone dry) of granular solid is to be dried under constant drying conditions from a moisture content of 0.2 kg/kg of dry solid to a final moisture content of 0.02 kg/kg dry solid. The material has an effective area of 0.0615 m2/kg. Under the same conditions the following rates were previously known. Calculate the time required for drying. Moisture content, X, kg/kg dry solid 0.3 0.20 0.14 0.096 0.056 1.71 1.71 1.71 1.46 1.29 Rate, N, kg/hr m2 Moisture content, X, kg/kg dry solid 0.046 0.026 0.016 Rate, N, kg/hr.m2 0.88 0.54 0.376

Principles of Mass Transfer Operations − I (Vol. − I)

8.29

Drying Operations

Sol. : X vs. N is plotted in Fig. 8.20. It is evident that the falling rate period is not a straight line. Hence the time for it is to be calculated by graphical integration. LS = 1400 kg X1 = 0.2 kg/kg dry solid X2 = 0.02 kg/kg dry solid Area = 0.0615 m2/kg Total drying area = 1400 × 0.0615 = 86.1 m2 Drying time : Constant rate period : LS θC = AN (X1 – XC) C =

Falling rate period :

θf

1400 (0.2 – 0.14) = 0.57 hour 86.1 × 1.71

LS = A

X2

dx

⌠ N ⌡ XC

1400 = 86.1

0.02

dx

⌠ N ⌡ 0.14

X

N

0.140 0.096 0.056 0.042 0.020

1.71 1.46 1.29 0.88 0.50

1 N 0.585 0.0685 0.775 1.136 2.000

A graph is plotted with X and 1/N, as shown in Fig. 8.20.

Fig. 8.20 : Graphical solution for example (9)

Principles of Mass Transfer Operations − I (Vol. − I)

8.30

Drying Operations

Area under the curve of x vs. 1/N = 0.104 θf

LS = A

X2

dx

⌠ N ⌡ XC

1400 = 86.1 × 0.104 = 1.69 hour Total drying time = 0.57 + 1.69 = 2.26 hour (Ans.) (10) A slab with wet weight of 5 kg originally contains 50% moisture (wet basis). The slab is 600 by 900 by 75 mm thick. The equilibrium moisture content is 5% of the total weight when in contact with air of 20 oC and 20% humidity. The drying rate is given below for contact with air of the above quality at a definite velocity. Drying is from one face only. How long will it take to dry the slab to 15% moisture content (wet basis) ? Wet slab kg 9.1 7.2 5.3 4.2 3.3 2.8 2.5 kg 4.9 4.9 4.4 3.9 3.4 2.0 1.0 Drying rate hr. m2 Sol. : Let x = kg of moisture in the wet solid, So, dry solid = 5–x 50 x 5 = 100 So, x = 2.5 Weight of dry solid = 5 – x = 2.5 kg kg water N W, kg weight of W, kg weight of X kg solid (dry) Rate kg hr.m2

Wet solid

Water

4.9

9.1

6.6

2.64

4.9

7.2

4.7

1.88

4.4

5.3

2.8

1.12

3.9

4.2

1.7

0.68

3.4

3.3

0.8

0.32

2.0

2.8

0.3

0.12

1.0

2.5

0

0

In Fig. 8.21, X is plotted against 'N' Drying area = 0.6 × 0.9 = 0.54 m2 50 X1 = 100 – 50 = 1.0 15 X2 = 100 – 15 = 0.176 Drying is in the falling rate period only. Further, the falling rate is not linear. Hence the drying time is to be evaluated by graphical integration.

Principles of Mass Transfer Operations − I (Vol. − I)

8.31

X N 1.000 4.30 0.800 4.10 0.680 3.90 0.320 3.40 0.176 2.35 Values of N has been taken from Fig. 8.21 (b). 1.0

dx

⌠ N ⌡

Drying Operations 1/N 0.232 0.244 0.256 0.294 0.426

1 = Area under curve × vs N (see Fig. 8.21 a)

0.176

θf

= 0.22 LS dx = A ⌠ N ⌡ 2.5 = 0.54 × 0.22 = 1.02 hours

(Ans.)

Required time = 1.02 hours

Fig. 8.21 (a) : Graphical solution for example (10)

Fig. 8.21 (b) : Graphical solution for example (10)

(11) A rotary dryer using counter-current flow is to be used to dry 12000 kg/hr of wet salt containing 5% water (wet basis) to 0.10% water (wet basis). Heated air at 147 oC with 50 oC wet-bulb temperature is available. The specific heat of the salt is 0.21. The outlet

Principles of Mass Transfer Operations − I (Vol. − I)

8.32

Drying Operations

temperatures of air and salt are 72oC and 93oC respectively. Calculate the length and diameter of the dryer required. Sol. : Assumption wet salt enters at ambient temperature say 30oC. . m s = bone.dry solid, kg = 12000 × 0.95 = 11400 kg W2 = final water content of salt

Let,

W2 W2 + 11400

So,

0.1 = 100

W2 = 11.41 kg

or

12000 × 0.5 = 0.0526 11400 11.41 = 11400 = 0.001

Xa = Xb

Tha = 147 oC, Twa = 50 oC Ha = 0.05 kg water/kg dry air Tha – Twa 147 – 50 Ni = T – T = 72 – 50 = 4.41 hb wb λ 50o C = 569 k cal/kg CPS = 0.21 k cal/kg oC CPV = 0.45 k cal/kg oC CPL = 1.00 k cal/kg oC . Rate of mass transfer = m p . = m s (Xa – Xb) = 11400 (0.0526 – 0.001) = 588.24 kg/hr. Heat duty is found from equation (8.11) qr . = 0.21 (93 – 30) + 0.0526 × 1 (50 – 30) + (0.0526 – 0.001) 569 + 0.001 × 1 (93 – 50) ms + (0.0526 – 0.001) × 0.45 (72 – 50) = 44.19 k cal/kg. So, qT = 44.19 × 11400 – 503766 k cal/hour Flow rate of air entering is found from heat balance and humid heat, Csa = Humid heat = 0.24 + 0.05 × 0.45 = 0.263 k cal/kg oC Heat balance for air is,

. qT = mg (1 + Ha) Csa (Tha – Thb) . 503766 = mg (1 + Ha) 0.263 (147 – 72)

Principles of Mass Transfer Operations − I (Vol. − I)

8.33

Drying Operations

. Amount of wet air = mg (1 + Ha)

. mg Outlet humidity of air

503766 = 0.263 (147 – 72) = 25539.4 kg/hr. 25539.4 = 1.05 = 24323 kg dry air/hr. 588.24 = H1 + 24323 = 0.073 kg water/kg dry air

∆T is calculated from equation (116), ∆T =

(147 – 50) – (72 – 50) 147 – 50 ln 72 – 50

= 50.55 o C A value of mass velocity is assumed within the range. Let the value be 5000 kg/hr m2. 25539.4 Cross-sectional area = 5000 = 5.108 m2 So, diameter of dryer = D = 2.55 m From equation (120), qT = 0.21 π DL G0.07 ∆T So,

L =

qT

0.21 π DG0.07 ∆T 503766 = 0.21 π (2.55) (5000)0.67 50.55 = 19.69 m. Hence the diameter and length of the dryer are 2.55 m and 19.69 m respectively.

(Ans.)

(12) In order to test the feasibility of drying a certain foodstuff, drying data were obtained in a tray dryer with airflow over the top of the exposed surface having an area of 0.186 m2. 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 H2O plus 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 drying test. Time 0 0.4 0.8 1.4 2.2 3.0 4.2 5.0 7.0 9.0 12.0 (hr.) Weight 4.944 4.885 4.808 4.699 4.554 4.404 4.241 4.150 4.019 3.978 3.955 (kg) (a) Calculate the free moisture content X kg H2O/kg dry solid for each data point and plot W versus time (Hint : for 0 hrs, 4.944 – 0.190 – 3.765 = 0.989 kg free moisture is 3.765 kg dry solid. Hence X = 0.989 ÷ 3.765) (b) Measure gradients along the drying curve, calculate the drying rate R, and plot R versus X. (c)

Using this drying-rate curve, predict the total time to dry the sample from X = 0.20 to X = 0.04. State the drying rate RC in the constant rate period, and critical free moisture content XC.

Principles of Mass Transfer Operations − I (Vol. − I)

8.34

Drying Operations

Sol. : (a)

Free moisture content (kg/kg dry solid)

Time (hours) 0 0.4 0.8 1.4 2.2 3.0 4.2 5.0 7.0 9.0 12.0

Free moisture content (kg/kg) (4.944 – 0.190 – 3.765) ÷ 3.765 (4.885 – 0.190 – 3.765) ÷ 3.765 (4.808 – 0.190 – 3.765) ÷ 3.765 (4.699 – 0.190 – 3.765) ÷ 3.765 (4.554 – 0.190 – 3.765) ÷ 3.765 (4.404 – 0.190 – 3.765) ÷ 3.765 (4.241 – 0.190 – 3.765) ÷ 3.765 (4.150 – 0.190 – 3.765) ÷ 3.765 (4.019 – 0.190 – 3.765) ÷ 3.765 (3.978 – 0.190 – 3.765) ÷ 3.765 (3.955 – 0.190 – 3.765) ÷ 3.765

= = = = = = = = = = =

0.263 0.247 0.227 0.198 0.159 0.119 0.076 0.052 0.017 0.006 0.000

0.3 0.25 0.2 0.15 0.1 0.05 0

0

2

4 6 Time (hours)

8

10

12

Fig. 8.22 : Graphical solution for example (12)

LS dX – 3.765 ∆X (b) R = – A dt = 0.186 ∆t From graph, the curve is differentiated by taking its gradient at regular intervals : X (kg/kg dry solid)

Time "t" (hours)

Change in X (kg/kg)

Change in t (hours)

∆X/∆ ∆t

R

(kg/kg.hr)

(kg/hr.m2)

0

– 0.265

5.5

– 0.0481818

0.97529326

0.265

3

– 0.265

5.5

– 0.0481818

0.97529326

0.12

4.5

– 0.074

2.6

– 0.0284615

0.57611663

0.066

6

– 0.046

2.8

– 0.0164286

0.33254608

0.032

7.5

– 0.023

3.4

– 0.0067647

0.136933074

0.013

9

– 0.016

4.3

– 0.0037209

0.07531883

0.006

12

– 0.005

4.4

– 0.0011364

0.02300220

0

i.e.

RC = 0.975 kg/hr.m2

Principles of Mass Transfer Operations − I (Vol. − I)

8.35

Drying Operations

1 0.9 2

Drying rate (kg/hr.m )

0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 –0.05

0

0.05

0.1

0.15

0.2

0.25

0.3

Free moisture content (kg/kg dry solid) Fig. 8.23 : Rate of drying curve in example (12)

(c)

From graph, XC = 0.12 kg/kg dry solid and XL = – 0.014 kg/kg dry solid. constant rate dry time

t1 =

LS (X1 – XC) 3.765 (0.20 – 0.12) = = 1.660 hours ARC 0.186 × 0.975

constant rate dry time

t1 =

LS (XC – XL) XC – XL ln  X2 – XL  ARC  

=

3.765 (0.012 + 0.014) 0 .12 + 0.014 ln  0.04 + 0.014  0.186 × 0.975  

= 2.528 hours Total drying time = t1 + t2 = 4.1880 hours

(Ans.)

(13) A batch of wet solid was dried on a tray dryer using constant drying conditions and a thickness of material on the tray of 25.4 mm. Only the top surface was exposed. The drying rate during the constant-rate period was R = 2.05 kg/kg.hr.m2. The ratio used was 24.4 dry solid/m2 exposed surface. The initial free moisture content was W = 0.55 and the critical moisture content WC = 0.22 kg moisture/kg dry solid. Calculate the time to dry a batch of this material from W1 = 0.45 to W2 = 0.30 using the same drying conditions but a thickness of 50.8 mm, with drying from the top and bottom surfaces. (Hint : First calculate surface-to-mass ratio for this new case). Solution : Assumptions are as follows : (i)

Both start and finish moisture well in excess entirely within constant rate period.

of the critical content drying is

(ii)

Wet solid is of uniform density, i.e. double thickness over an area will have twice the mass.

(iii)

Drying rate has same value for both top and bottom surfaces (i.e. no gravity effect).

Principles of Mass Transfer Operations − I (Vol. − I) (iv)

8.36

Drying Operations

The new sample has twice the surface area, but twice the mass also. Therefore it will dry just as quickly. LS X1 – XC constant rate drying time t = A RC = 24.4 ×

0.45 – 0.30 2.05

= 1.785 hours

(Ans.)

(14) The wet feed material to a continuous dryer contains 40% w/w moisture on a wet basis and is dried to 25% by counter current airflow. The dried product leaves at a rate of o

0.213 kg/s. Fresh air to the system has a dry bulb temperature of 25 C and wet bulb of o

o

15 C. The humid air leaving the dryer has a dry bulb temperature of 37 C and wet o

bulb of 28 C and part of it is recirculated and mixed with the fresh air before entering o

a heater. The heated mixed air enters the dryer with a dry bulb temperature of 60 C o

o

and wet bulb of 27 C. The solid both enters and leaves at 26.7 C. Calculate the fresh airflow, the percent air that is recycled. Sol. : Drawing a schematic diagram of the dryer to help in writing a mass balance : Humid air recycled W4 W2 Fresh air in W1

W3 Dryer

heater

Humid air out W5 Moist solid in S1

Dried solid out S2

Fig. 8.24 : Schematic of counter current dryer

Make all W and S flowrates on a DRY basis. Assume steady state (no accumulation) : Mass balance of dry air over heater : W1 + W4 = W2 Mass balance of water vapour across dryer W 1 H 1 + W 4 H 4 = W 2H 2

W4 We want to know percentage recycle, i.e. 100 × W  . Eliminate W1 from balance. 

2



Rewrite dry air balance to give : W2 – W4 = W1 Substitute into water vapour balance : W2H2 = (W2 – W4) H1 + W4H4 W 2H 2

= W 2H 1 – W 4H 1 + W 4H 4

W2(H2 – H1) = W4(H4 – H1) ∴

W4 H2 – H1 100 W = 100 H – H 2 4 1

Principles of Mass Transfer Operations − I (Vol. − I)

8.37

Drying Operations

To find the recycle rate, we need the humidities of the three air streams on a kg water per kg dry air basis : inlet air H1 = 0.0065 kg/kg (from psychometric chart) outlet air H3 = 0.0202 kg/kg (from psychometric chart) mixed air H2 = 0.0101 kg/kg (from psychometric chart) 0.0101 – 0.0065 recycle = 100 × 0.0202 – 0.0065

∴

= 26.27737% As ever, we cannot be any more accurate than our original data, so 26%. Mass balance on water in dryer : S1X1 + W2H2 = S2X2 + W3H3 No accumulation, so dry solid in = dry solid out (= S) and dry air in = dry air out (= W2) so S(X1 – X2 ) = W2(H3 – H2). ∴

W2 =

S (X1 – X2) H3 – H2

We now want to find the unknown W1 W1 = W2 – W4 W4  W4  W1 = W2 – W2 W = W2 1 – W  2



2

∴

W1 =

S(X1 – X2)  H2 – H1  H3 – H2 1 – H4 – H1

∴

W1 =

S (X1 – X2) H4 – H1 H2 – H1 H3 – H2 H4 – H1 – H4 – H1

∴

W1 =

S (X1 – X2) H4 – H2  H3 – H2 H4 – H1

H3 and H4 are equal, so : ∴

S (X1 – X2) H3 – H2 H3 – H2 H3 – H1 X1 – X2 = SH –H 3 1

W1 =

To find air inlet flow, we need the X of the two solid streams and the flow rate S on a dry basis : inlet solid X1 = 0.4 ÷ (1 – 0.4) = 0.667 kg/kg outlet solid X2 = 0.25 ÷ (1 – 0.25) = 0.333 kg/kg dry solid feed S = 0.213 (1 – 0.25) = 0.15975 kg/s 0.667 – 0.333 ∴ W1 = 0.15975 0.0202 – 0.0065 = 3.8868621 kg/s As ever, we cannot be any more accurate than our original data, so 3.89 kg/s.

(Ans.)

Principles of Mass Transfer Operations − I (Vol. − I)

8.38

Drying Operations

(15) A cabinet dryer is use to dry a food product from 68% moisture content (w/w basis) to o

5.5% moisture content (w/w basis). The drying air enters the system at 54 C and 10% o

o

RH and leaves at 30 C and 70% RH. The product temperature is 25 C throughout drying. Calculate the quantity of air required for drying on the basis of 1 kg of product solids. Use of moisture balance equation · · · · maW2 + m pw1 = maW1 + mpw2 this will be rearranged · · · · ma/mp w2 + w1 = m/mp w1 + w2

(

)

(

)

mp = Mass fraction of product w1 = Moisture content in (w/w basis) w2 = Moisture content out (w/w basis)

and Sol. :

From the psychometric chart o

W1 (30 C and 70% RH) = 0.0186 kg water kg/kg dry air o

W2 (54 C and 10% RH) = 0.0094 kg water kg/kg dry air Using the above equation · · ma ma (0.0094) + 2.125 = · · (0.0186) + 0.0582 mp mp · ma 0.00094 · mp · ma · mp

= 2.067

= 219.88 kg dry air / kg solids

Then if drying 10 kg of product will need X kg for dry air. This can be calculated as follows : Need 10 × 0.32 kg solids = 3.2 kg solids Mass of dry air = 3.2 × 219.88 = 703.6 kg dry air in order to complete the drying processes (Ans.) (16) A fluidized bed dryer is being used to dry diced carrots. The product enters the dryer o

with 60% moisture content (w.w.b.) at 25 C. The air used for drying enters the dryer at o

o

120 C after being heated from ambient air with 60% RH at 25 C. Estimate the production rate when air is entering the dryer at 700 kg dry air hr–1 and product leaving the dryer is at 10% moisture content (w.w.b). Assume product leaves the dryer at the wet bulb temperature of air and the specific heat of product solid is o

–1

o

2.0 kj kg–1 C . Air leaves the dryer 10 C above the product temperature. The product rate will be determined using the material balance, and the energy balance, alongwith parameters obtained using the psychometric chart.

Principles of Mass Transfer Operations − I (Vol. − I)

8.39

Drying Operations

0.6 Initial product moisture content w1 = 0.4 = 1.5 kg water/kg solids o

Initial air condition = 20 C and 60% RH (W2 = 0.009 kg water / kg dry air) o

Air temperature entering dryer = 120 C · Air flow rate m a = 700 kg dry air hr–1 0.1 Final product moisture content (w2) = 0.9 = 0.111 kg water / kg solids –1

Specific heat of product solids (Cp) 2.0 kj kg–1 K Final product temperature = wet bulb temperature of air o

Final air temperature (Ta1 ) = Tp2 + 10 C. Sol. : Moisture balance Mass water in air in + Mass water in product in = Mass water in air out + Mass water in · product out (700 kg dry air/hr) (0.009 kg water/kg dry air) + m p (1.5 kg water/kg solids) · = (700 kg dry air/hr) W1 + m p (0.111 kg water/kg solids) Heat Balance : Heat air in + Heat product out = Heat air out + Heat product out For the air : Hair = Cs2 (120 – 0) + 0.009 HL2 where Cs2 = 1.005 + 1.88 (0.009) = 1.0219 kj/kg dry air K o

HL2 = 2202.59 kj/kg water at 120 C (from steam tables) Therefore, Hair = (1.0219 × 120) + (0.009 × 2202.59) = 142.45 kj/kg dry air o

o

Wet bulb temperature = 38 C, thus air is leaving the dryer at 48 C Then Haout = Cs1 (Ta1 – 0) + W1 (HL1) where Cs1 = 1.005 + 1.88 W1 and o

HL1 = 2387.56 kj/kg water at 48 C Therefore, Haout = (1.005 + 1.88 W1) + W1 (2387.56) For the product Hpin = (2.0 kj/kg solids K) (25 – 0) + (1.5 kg wat./kg solids) (4.178 kj/kg K) (25 – 0) Hpin = 206.75 kj/kg solids Hpout = (2.0) (Tp2 – 0) + (0.111 kg water/kg solids) (4.175 kj/kg K) (Tp2 – 0)

Principles of Mass Transfer Operations − I (Vol. − I)

8.40

Drying Operations

Then, · (700 kg dry air/hr) (142.45 kj/kg dry air/kg dry air) + mp (206.75 kj/kg solids) =

o

(700 kg dry air/hr) [(1.005 + 1.88 W1) (48 C) + W1 (2387.56 kj/kg water)] o · + mp [(2.0 kj/kg solids K) (38 C)

+ (0.111 kg water/kg solids) (4.175 kj/kg K) (38) + 0 where q = 0, indicating negligible heat loss from surface of dryer. The material balance and energy balance equations can be solved simultaneously : · · 700 (0.009) + 1.5 mp = 700 W1 + 0.111 mp

(a)

· 700 (142.45) + m p (206.75) = 700 [(1.005 + 1.88 W1) (48) + 2387.56 W1]

(b)

· + mp [(2.0) (38) + (0.111) (4.175) (38)] · · 6.3 + 1.5 mp = 700 W1 + 0.111 mp

(a) (b)

· · 99715 + 206.75 mp = 700 (48.24 + 2477.8 W1) + 93.61 mp

(a)

· (1.389 mp + 6.3) = 700

W1

(b)

· 65947 + 113.14 mp = 1734460 W1

Then

· 1734460 (1.389 mp + 6.3) · 65947 + 113.14 mp = 700 · mp = 15.12 kg solids/hr

Thus, to calculate the absolute humidity of air leaving the dryer (1.389 × 15.12 + 6.3) W1 = 700 = 0.039 kg water/kg dry air o

indicating that air leaving the dryer will be 48 C and 55% RH. (17) Calculate the drying time of a particle in a spray dryer, given : –3

o

Feed 20% solids (density 1075 kg ) To = 20 C o

Dried Product = 4% (density = 375 kg–3) T = 55 C Critical Moisture content = 50% Rotary atomiser, diameter = 0.2 m, 10,000 r.p.m. Feed rate 1000 kg hr–1 Droplets 40 - 75 microns in diameter. o

o

Ambient air temperature 20 C, 70% R.H.; is heated to 110 C Thermal conductivity of air = 0.032 W m–1 K

–1

Latent Heat = 2502 kj kg–1 Also calculate the width and height of the dryer, given that the air velocity is 1.927 ms–1.

Principles of Mass Transfer Operations − I (Vol. − I)

8.41

Drying Operations

Sol.: Drying time : tc =

Constant time :

(

2

2

λρw Do – Dc

)

8ka (Ta – Tc)

We must first calculate the diameter at the end of the constant rate period. 4πr3ρ Initially one drop weighs 3 Must account for the largest droplets. Hence weight : 3

Weight =

4π(37.5 × 10–6) (1075) 3

= 2.37 × 10

–10

kg

Weight of solids in droplets = 2.37 × 10

–10

× 0.20

= 4.75 × 10

–11

Weight of product droplet = 4.75 × 10 = 4.94 × 10 mass Volume of dry droplet = density

=

4.94 × 10 375

= 1.32 × 10

–11

× 1.04

–11 –11

–13

m3

1/3

Radius =

3V  4π 

–6

Therefore radius = 31.55 × 10 m; Diameter = 63.1 × 10 Thus, drying time can be calculated using : tc =

(

2

2

λρw Do – Dc

–6

m

)

8ka (Ta – Tc)

2.502 × 10 × 1075 × ((75 × 10 ) – (63.1 × 10 ) tc = 8 × 0.032 × (110 – 36.9) tc = 0.236 s 6

−6 2

–6 2

Falling Rate : 2

λρdDc tf = 6k (T – T ) (mc – mf) a a w mc = 0.5 w.w.b. = 1.0 d.w.b. mf = 0.04 w.w.b. = 0.0417 d.w.b. 6

tf = Hence

–6 2

2.502 × 10 × 375 × (63.1 × 10 ) (1.0 – 0.0417) 6 × 0.032 × (110 – 36.9)

tf = 0.256 s

Total drying time : t = 0.236 + 0.256 = 0.491 s

)

Principles of Mass Transfer Operations − I (Vol. − I)

8.42

Drying Operations

Dryer Width : 2

VpoDp ρp Smax = 18µ but both ρp and Dp vary. Thus, take the average thus : 2

2 ρp Dp 2

ρp Dp (average)

2

ρoDo + ρdDc = 2 =

(1075 ×

75 × 10

–6

) + (375 + 63.1 × 10–6) 2

= 3.77 × 10

–6

Vpo = Peripheral velocity of the disc = rω

hence

ω = 10,000 r.p.m. : 1 revolution = 2 π radians, 10‚000 ω =  60  × 2π  

= 1047 rad s−1 Therefore, because radius = 0.1 m then Vpo = 1047 × 0.1 = 104.7 m s–1 2

Vpo Dp ρp smax = 18 µ –6

=

(104.7) (3.77 × 1 0 ) 18 × 1.6 × 10

–5

= 1.37 m Diameter = 2 Smax = 2.74 Multiple by a Factor of Safety (1.1) Diameter of vessel = 3.02 m Dryer Height : Assume droplets fall at same speed as air Height = V × drying time = 1.927 m s−1 × 0.502 s = 0.967 m × 1.1 = 1.042 meters (18) A co-current spray dryer is to be used to dry 1,500 kg hr–1 of milk powder from 25% o

solids to 4% moisture content (dry basis). Ambient air, initially at 25 C and 70% o

relative humidity, will be heated to 120 C before entering the dryer. The product, o

o

which is initially by 25 C, will leave the dryer in equilibrium with the air at 50 C. If the dryer diameter has been calculated to be 3 m and the droplet drying time to be 15 s, determine the height of the dryer. Data :

Specific heat milk solids

1.675 kJ kg–1 K

–1

Specific heat water

4.182 kJ kg–1 K

–1

Specific heat ware vapour

1.875 kJ kg–1 K

–1

Specific heat dry air

1.005 kJ kg–1 K

–1

Latent heat of vapourization

2502 kJ kg–1 K

–1

Principles of Mass Transfer Operations − I (Vol. − I)

8.43

Drying Operations

Sol. : To calculate the air velocity, we need to calculate the volume of air passing through the dryer. This is achieved by using the mass and energy balances, as shown below : (a) Mass balance Solids in = Solids out 1500 × 0.25 = 0.96 mp mpout = 390.625 kg hr–1 Total = 1500 kg hr–1 = mpout + mwout mwout = 1109.375 kg hr–1 (b) Energy Balance Energy in = Energy out mpin hpin + main hain = mpout hpout + maout haout + mwv hwv main = maout = ma mpin = 1500 kg hr–1; mpout = 390.625 kg hr–1 ; mwout = 1109.375 kg hr–1 Let X = solids fraction hp = T (XCp + (1 – X) Cpwater) hpin = 25 (0.25 (1.675) + 0.75 (4.182)) = 88.8125 kj kg–1 hpout = 50 (0.96(1.675) + 0.04 (4.182)) = 88.764 kj kg–1 hair = ha + Ha hwv = CT + Ha (CwvT + hgs) Ha = 0.0135 from psychometric chart hairin = 1.005 (120) + 0.012 (1.857 (120) + 2502) = 157.41 kj kg–1 hairout = 1.005 (50) + 0.012 (1.857) (50) + 2502) = 85.29 kj kg–1 hwv = CwvT + hgs = 1.875 (50) + 2502 = 2595.75 kj kg–1 Hence, 1500 (88.88) + ma (157.41) = 390.625 (88.764) + ma (85.29) + 1109.375 (2595.75) ma = 38559.44 kg air hr–1 o

Specific volume at 120 C = 1.13 (38559.44 × 1.13) Air flow rate = 3600 = 12.10 m3 s–1 Air velocity = 1.71 m s–1

(Ans.)

Principles of Mass Transfer Operations − I (Vol. − I)

8.44

Drying Operations

(19) Design the width and height of a co-current spray dryer given : o

Feed 20% solids (density = 1075 kg m–3) To = 20 C o

Dried product = 4% w/w basis. (Density = 300 kgm–3) T = 55 C Critical Moisture content = 50% w/w basis Rotary atomizer, diameter = 0.2 m, 10,000 r.p.m. Feed rate 1000 kg hr–1 Droplets 40 – 70 microns from atomiser. o

o

Ambient temperature of air = 30 C and R.H = 70%, heated to 110 C Latent heat of evaporation = 2502 kj kg–1 Specific heat dry solid = 1.6575 kj kg–1 K Specific heat water = 4.182 kj kg–1 K

–1

–1

Specific heat water vapour = 1.875 kj kg–1 K Specific heat dry air = 1.005 kj kg–1 K

–1

–1

Thermal conductivity air = 0.032 W m–1 K

–1

–5

Viscosity of air = 1.6 × 10 kg m–1 s–1 Calculate : (1) Dryer width, (2) Drying time, (3) Air velocity and (4) Dryer height. Sol. : 1. Dryer Width : 2

VpoDpρp Smax = 18µ But both ρp and Dp vary. Thus take the average thus : 2

2 ρp Dp

2

ρoDo + ρdDc = 2

Assume diameter is constant during the falling rate period Initially one drop weighs

4πr3ρ 3

Must account for the largest droplets. Hence weight :

(

= 1.9306 × 10 Weight of solids in droplets = 1.9306 × 10 Weight of dry droplet = 3.8613 × 10

–10

)

–6 3

4π 35 × 10 weight = 3

–10

(1075)

kg

× 0.2 = 3.8613 × 10

–11

× 1.04 = 4.0157 × 10 mass Volume of dry droplet = density 4.0157 × 10 300

–11

= 1.3386 × 10

–13

=

Therefore

–11

m3

Radius = 31.73 × 10

–6

m;

Diameter = 63.47 × 10

–6

m

–11

Principles of Mass Transfer Operations − I (Vol. − I) 2 ρpDp(average)

8.45

Drying Operations –6

–6

(1075 × 70 × 10 ) + (300 × 63.67 × 10 ) = 2 –6

= 3.238 × 10 Vpo = Peripheral velocity the disc = rω r = 0.1 m, ω = 10000 r.p.m. : 1 revolution = 2π radians 10000 Hence, V =  60  × 2π   = 1047 rad s–1 Therefore, because radius = 0.1 m then Vpo = 1047 × 0.1 = 104.7 ms–1 2

VpoDpρp smax = 18µ –6

=

(104.7) (3.23 × 10 )

18 × 1.6 × 10 = 1.177 m Diameter = 2 Smax = 2.354

–5

(Ans.)

Multiply by a Factor of Safety (1.1) Diameter of vessel = 2.6 m 2. Drying Time : t = tc + tf Constant Rate : tc =

(

2

2

λρw Do – Dc

8ka (Ta – Tc) 6

tc =

) (

2

2.502 × 10 × 1075 × (70 × 10–6) – (63.47 × 10–6)

2

)

8 × 0.032 × (110 – 40)

tc = 0.131 s Falling Rate : 2

λρdDc tf = 6k (T – T ) (mc – mf) a a w mc = 0.5 w.w.b. = 1.0 d.w.b. mf = 0.04 w.w.b. = 0.0417 d.w.b. 6

2

2.502 × 10 × 300 × (63.47 × 10–6) (1.0 – 0.0417) 6 × 0.032 × (110 – 40) Hence tf = 0.215 s Total Drying Time : t = 0.131 + 0.215 = 0.346 s 3. Air Velocity : (a) Mass balance : Solids in = Solids out 1000 × 0.2 = 0.96 mp tf =

(Ans.)

Principles of Mass Transfer Operations − I (Vol. − I)

8.46

Drying Operations

mp = 208.3 kg hr–1 Total = 1000 kg hr–1 = mpout + mwout mwout = 791.7 kg hr–1

(Ans.)

(b) Energy Balance : Energy in = Energy out mpin hpin + main hain = mpout hpout + maout haout + mwv hwv main = maout = ma mpin = 1000 kg hr–1 ; mpout = 208.3 kg hr–1; mwout = 791.7 kg hr–1 Let X = solids fraction hp = T (XCp + (1 – X)Cpwater) hpin = 20 (0.2 (1.675) + 0.8 (4.182)) = 73.612 kj kg–1 hpout = 55(0.96(1.675) + 0.04(4.182)) = 97.64 kj kg-1 hair = ha +Ha hwv = CT + Ha(CwvT + hgs) Ha = 0.0191 from psychometric chart hairin = 1.005 (110) + 0.0191 (1.857(110) + 2502) = 162.28 kj kg–1 hairout = 1.005(55) + 0.0191 (1.857(55) + 2502) = 105.03 kj kg–1 hwv = CwvT + hgs = 1.875(55) + 2502 = 2605 kj kg–1 Hence 1000(73.612) + ma (162.28) = 208.3 (97.64) + ma (105.03) + 791.7 (2605) ma = 35095 kg air hr–1 o

Specific volume at 110 C = 1.12 (35095 × 1.05) = 10.91 m3 s–1 Air flow rate = 3600 Air velocity = 2.07 m s–1

(Ans.)

4. Dryer Height Assume droplets fall at same speed as air Height = V × drying time = 2.07 m s–1 × 0.346 s = 0.71 m × 1.1 = 0.79 metres

(Ans.)

(20) Drying of a food product is carried out in an insulated tray. The drying air has a partial o

pressure of water equal to 2360 Pa and a wet bulb temperature of 30 C. The product has a drying surface of 0.05 m2/kg dry solid. The material has a critical moisture content of 0.12 (dry basis) and negligible equilibrium moisture content. The drying rate in the falling rate period is proportional to the moisture content and the mass transfer coefficient is 5.34 × 10

–4

kg/m2.hr.Pa.

Principles of Mass Transfer Operations − I (Vol. − I)

8.47

Drying Operations

Calculate : (a)

the drying rate in the constant rate period in kg/m2 hr.

(b) the time required to dry the material from a moisture of 0.22 to 0.06 (both on dry basis). o

Vapour pressure of water at 30 C = 4232 Pa. Solution : Rate of drying at constant rate period Nc :

(–*

–

–

)

Nc = kG∆p A = kG pA – pA where,

–*

pA

… (1)

= Partial pressure of water vapour at equilibrium, or vapour pressure of water (4232 Pa)

–

p A = Partial pressure of water vapour in drying air (2360 Pa) kG = Mass transfer coefficient (5.34 × 10

–4

kg/m2.hr.Pa)

Therefore, from equation (1) Nc = 5.34 × 10

–4

(4232 – 2360) = 1 kg/hr. m2

1

N 2 (kg/hr.m )

0

0.06 0.12

0.22

X Fig. 8.25 : Graphical analysis in example (20)

Falling rate period : Given : drying rate is proportional to moisture content. Therefore, 1 Slope of falling rate period curve = 0.12 = 8.333 Rate of drying in falling rate period Nf = 8.333 X From the definition of drying rate – Ss dX N = A dθ where, Ss = Mass of dry solid; A = Drying surface area; and θ = Time of drying

… (2)

Principles of Mass Transfer Operations − I (Vol. − I) Given :

A Ss

8.48

Drying Operations

= 0.05 m2/kg dry solid

For constant period equation (2) becomes 1 – Ss dθ dX = Nc A =

1  – 1  = – 20 1 0.05

Integrating between X1 = 0.22 and X2 = 0.12, θc = – 20 (0.12 – 0.22) = 2 hr. Similarly for falling rate period – Ss 1 dθ dX = A Nf 1 = – 20 8.333 X





Integrating between X2 = 0.06 and X1 = 0.12, 1 0.06 θf = – 20 8.333 ln 0.12 = 1.664 hr. Total drying time θ = θc + θf = 2 + 1.664 = 3.664 hr. (Ans.) (21) In a laboratory drying test with a solid material the following relation for the falling rate period was obtained, dX = – 0.8 (X – 0.05) dθ θ where X is the moisture content on dry basis of θ is the time in hours. The critical moisture content is 1.4 kg moisture per kg of dry material. Calculate : (a) the time required for drying the material from X1 = 4.0 to X2 = 0.1 (b) the equilibrium moisture content. Solution : Given : Relation for falling rate period : dX = – 0.8 (X – 0.05) dθ = – 0.8 X + 0.04 … (1) Relation for constant rate period Substituting for X = 1.4 from critical moisture content data dX = – 0.8 × 1.4 + 0.04 dθ = – 1.08

… (2)

Principles of Mass Transfer Operations − I (Vol. − I)

8.49

Drying Operations

1.08

–dx dq

0

0.1

1.4

X

4

X Fig. 8.26 : Graphical analysis in example (21)

Time required for drying in constant rate period θc : Integrating Equation (2) between X1 = 4.0 and X2 = Xc = 1.4, 1.4 – 4.0 θc = – 1.08 = 2.407 hr. Time required for drying in falling rate period θf : Integrating Equation (1) between X1 = Xc = 1.4 and X2 = 0.1 1 0.1 θf = – 0.8 [ln (– 0.8 X + 0.04)]0.4= 4.12 hr. Total time required for drying θ = θc + θf = 2.407 + 4.12 = 6.527 hr. *

Equilibrium moisture content X is obtained by equating

dX in equation (1) to zero. dθ

*

i.e. – 0.8 X + 0.04 = 0 Therefore, X = 0.05

(Ans.)

(22) 160 kg of wet solid is to be dried from an initial moisture content of 25% to a final –4

value of 6%. Drying test shows that the rate of drying is constant at 3 × 10 kg H2O/m2.s in the region 0.2 – 0.44 kg H2O/kg solid. The drying rate falls linearly in the range 0.01 – 0.2 kg H2O/kg solid. If the equilibrium moisture content is 0.01 kg H2O/kg solid, calculate the time of drying. The drying surface is 1 m2/30 kg dry weight. Sol. : Wet solid = 160 kg Dry solid = 160 × (1 – 0.25) = 120 kg 0.25 Initial moisture content = 25% = 1 – 0.25 = 33.3% on dry basis 0.06 Final moisture content = 6% = 1 – 0.06 = 6.4% of dry basis

Principles of Mass Transfer Operations − I (Vol. − I)

8.50

Drying Operations

Drying rate (N) in falling rate period : –4

3 × 10 N = 0.2 – 0.01 X + C Given : at X = 0.01, N = 0. Therefore, –4

3 × 10 0 = 0.2 – 0.01 × 0.01 + C Solving, C = – 1.579 × 10 N =

–5

kg H2O/m2.sec. Therefore,

3 × 10 0.19

–4

X – 1.579 × 10

= 1.579 × 10

–3

–5

X – 1.579 × 10

–5

By formula : – Ss A – 30 = 1

dX dθ dX dθ dX = – 30 dθ Time of drying in falling rate period : N =

dX dθ dX –3 –5 1.579 × 10 X – 1.579 × 10 = – 30 dθ dθ – dX 30 = 1.579 × 10–3X – 1.579 × 10–5 N = – 30

0.064

Integrating,

θf 30

=

θf =

⌠ dX  X – 0.01 ⌡

–1 1.579 × 10

–3

0.2

30 1.579 × 10

0.2

–3

[ln (X – 0.01)]0.064

θf = 23902 sec = 6.64 hr. Time of drying in constant rate period : Ss X1 – Xc θ = A Nc 0.333 – 0.2 = 30 ×  –4   3 × 10  0.333 – 0.2 = 30 ×  –4   3 × 10  = 13300 sec = 3.69 hr. Total drying time = 6.64 + 3.69 = 10.33 hr = 10 hours and 20 minutes

(Ans.)

Principles of Mass Transfer Operations − I (Vol. − I)

8.51

Drying Operations

(23) A material having a critical moisture content of 24% is dried under constant drying conditions in slabs of dimensions 20 cm × 20 cm2 × 2 cm from a moisture content of 40% to 25% in 8 hours. Drying takes place from the two larger surfaces only. The same material is dried in slabs of 40 cm × 40 cm × 4 cm under similar drying conditions from 50% to 25% moisture content. All moisture content are on wet basis. If the critical moisture content and other physical remain unchanged, how long will the drying take ? Sol. : For drying in constant rate period, there is no change in drying rate per unit area, with the thickness of the material. For the First Case : Drying area (from 2 sides) A1 = 20 × 20 × 2 = 800 cm2 Weight of dry solid Ss1 = V1ρ(1 – 0.4) = 2 × 20 × 20 × 0.6 ρ = 480 ρ – Ss1 dX Drying rate N1 = A 1 dθ X1 – X2 480 ρ = 800 × θ On dry basis,

Therefore,

0.4 X1 = 1 – 0.4 = 0.667; 0.25 X2 = 1 – 0.25 = 0.333 480ρ 0.667 – 0.333 N1 = 800 × 8 = 0.02505 ρ = constant

For the Second Case : Drying area (from 2 sides) A2 = 40 × 40 × 2 = 3200 cm2 Weight of dry solid Ss2 = V2 ρ (1 – 0.5) = 4 × 40 × 40 × 0.5ρ = 3200ρ – Ss2 dX Drying rate N2 = A 2 dθ As drying rate per unit area does not change with thickness of material, N2 = N1 = 0.02505 ρ – Ss2 dX Also N2 = A 2 dθ 3200ρ (X1– X2) = 3200 θ

Principles of Mass Transfer Operations − I (Vol. − I) On dry basis,

Therefore, Solving, θ = = 26.63 hour.

8.52

Drying Operations

0.5 X1 = 1 – 0.5 = 1; 0.25 X2 = 1 0.25 = 0.333 3200ρ (1 – 0.333) 0.02505ρ = 3200 θ 26.63 hours. Therefore the total drying time for the second case is (Ans.) EXERCISE FOR PRACTICE

(1) A wet solid is dried from 35% to 10% moisture under constant drying conditions in 5 hours. If the equilibrium moisture content is 4% and the critical moisture content is 14%, how long will take to dry 6% moisture under the same conditions ? (Ans. : 7.1 hours) (2) A batch of solid is dried from 28% to 6% moisture, wet basis. The initial weight of solid is 380 kg and the drying surface is 0.15 m2/40 kg dry weight. The critical moisture content is 28% dry basis and the constant drying rate is 0.32 kg/hr. m2. For the falling rate period, the following data are available : Moisture content % dry basis Rate of drying kg/hr. m2 25 0.30 21.9 0.27 19 0.24 16 0.21 13.6 0.18 11 0.15 8.2 0.07 7.5 0.044 6.4 0.025 Determine the time of drying.

(Ans. : 414.2 hours)

(3) A counter-current rotary dryer is to be designed for drying 20,000 kg/hr. of wet salt containing 7.5% (wet basis) water to 0.28% (wet basis) water. The wet salt enters the dryer at 30 oC and has a specific heat of 0.25 k cal/kg oC. Heated air at 150 oC with a wet bulb temperature of 40 oC is available for drying the salt. If the outlet temperature of air and salts are 65 oC and 80 oC respectively. Calculate the length and diameter of the dryer. (4) A wet solid is dried from 40% to 10% moisture under constant drying conditions in 4 hour. The equilibrium moisture content is 4% and the critical moisture content is 14%. How long it will take to dry 6% moisture under the same conditions ? (Ans. : 5.4 hours) o

o

(5) The air supply to a rotary dryer has a dry bulb at 32.5 C and a wet bulb of 25 C. Steam coils to 95 oC heat the air before it enters the dryer. In passing through the dryer, the air cools along adiabatic cooling line and finally leaves the dryer at 90% saturation.

Principles of Mass Transfer Operations − I (Vol. − I) (a)

8.53

Drying Operations

Estimate the temperature of exit air

(b) Calculate the diameter of the dryer if it is to handle 10 tones per hour of a salt with 10% moisture on wet basis. Air mass velocity is given as 5000 kg per hour per sq.ft. and the salt is to be dried to 2% moisture content. (6) An insoluble wet granular material is dried in a pan 0.457 × 0.457 m and 25.4 mm dep. The material is 25.4 mm deep in the pan and the sides and bottom can be considered to be insulated. Heat transfer is by convection from an air stream flowing parallel to the kg H2O surface at a velocity of 6.1 m/s. The air is at 65.6 oC and has humidity of 0.01 kg dry air. Estimate the rate of drying for the constant-rate period. (Ans. : NC = 0.708 kg H2O/hr.) (7) 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 hr. was required to dry the material to a free moisture content of 0.08 kg free moisture/kg dry solid. The critical moisture content is 0.14. Assuming a drying rate in falling-rate region where the rate is a straight line from the critical point, predict the time to dry a sample from a free moisture content of 0.33 to 0.04 kg free moisture/kg dry solid. (Hint : First use the analytical equations for the constant rate and falling linear periods with known total time of 6 hour, then use the same equations for new conditions.) (8) The initial moisture content of a food product is 77% (w/w basis) and the critical moisture content is 30% (w/w basis). If the constant drying rate is 0.1 kg water m–2 s–1, calculate the time required for the product to begin the falling rate-drying period. The product has a cube shape with 5 cm sides, and the initial product density is 950 kg m–3. (Ans. : t = 53.2 seconds) (9) A tunnel dryer is being designed for the dehydration of apple slices from initial moisture content of 70% (w/w basis) to a final moisture content of 5% (w/w basis). A experimental drying curve for the product indicates that the critical moisture content is 25% (w/w basis) and the time for constant rate drying is 5 minutes. Based on the following information estimate the total drying time for the product. Assume 1 kg of product. Given : 0.7 Initial product moisture content wo = 0.3 = 2.33 kg water/kg solids. 0.25 Critical moisture content wc = 0.75 = 0.333 kg water/kgs solids. 0.05 Final moisture content w = 0.95 = 0.0526 of kg water/kg solids. Time for constant rate drying = 5 minutes. (Ans. : t = 6.54 minutes) (10) The initial moisture content of a food product is 80% (w/w basis) and the critical moisture content is 45% (w/w basis). It is dried to a final moisture content of 10% (w/w basis). If the constant drying rate is 0.01 kg water m–2 s–1, calculate the total drying time including the constant and falling rate drying periods. The product has a cube shape with 5 cm sides, and the initial product density is 1050 kg m–3. (Ans. : t = 842.7 sec.)

Principles of Mass Transfer Operations − I (Vol. − I)

8.54

Drying Operations

NOMENCLATURE Any consistent set of units can be used, except as noted Symbol a A

'

Meaning specific interfacial surface of the solid, m2/m 3 cross-sectional area perpendicular to the direction of flow for throughcirculation drying, m2 drying surface for cross-circulation drying, m2 average cross-sectional area of drying solid, m2 const heat capacity of moisture as a liquid heat capacity at constant pressure, heat capacity of dry solid, equivalent diameter, (cross-sectional area)/perimeter, m particle diameter, diffusivity, m2/s mass velocity of gas, kg/m2.s mass velocity of dry gas, kg/m2.s heat transfer coefficient for convection, heat transfer coefficient of radiation, integral heat of wetting (or of adsorption, hydration, or solution) referred to pure liquid and solid, per unit mass of dry solid, enthalpy of moist gas per unit mass of dry gas,

'

enthalpy of wet solid per unit mass of dry solid,

Am b CA Cp Cs dc dp D G GS hc hR ∆HA HG ΗS Ηt

ΟG

length of an overall gas transfer unit, m

jD

krSc2/3/Gs , dimensionless

jH

hcP r2/3/Cp G, dimensionless

kr K m

gas-phase mass transfer (humidity difference) constant constant

N

flux of drying, mass of moisture evaporated/(area) (time), kg/m2s;

coefficient,

mass evaporated/(area) (time)

rate of revolution S–1 Nc

constant flux of drying, kg/m2.s

Nt

number of gas transfer units, dimensionless

Nt

G

OG

number of overall gas transfer units, dimensionless

p

saturation vapour pressure, kN/m2

p

–

partial pressure, kN/m2

pt

total pressure, kN/m2

Principles of Mass Transfer Operations − I (Vol. − I)

8.55

Symbol

Drying Operations Meaning

Pr

Prandtl number = Cρµ/ k, dimensionless

q

flux of heat received at the drying surface, batch drying, flux of heat received by the solid/area of dryer cross section, continuous drying,

qc

heat flux for convection

qG

flux of heat transferred from the gas per unit drier cross-section,

qk

heat flux for conduction,

qR

heat flux for radiation,

Q

net flux of heat loss, per unit dryer cross-section,

Rec

Reynolds number for duct, dcG/µ, dimensionless

SS

mass of dry solid in a batch, batch drying, kg mass velocity of dry solid, continuous drying, kg/m2.sec

Sc

Schmidt number, µ/ρ D, dimensionless

tG

dry-bulb temperature of a gas, K

tR

temperature of radiating surface, K

ts

surface temperature, K

tS

solid temperature,

to

reference temperature, K

TD

diameter of drier, m

TG

absolute temperature of a gas, K

TR

absolute temperature of radiating surface, K

X

moisture content of a solid, mass moisture/mass dry solid, kg/kg

X*

equilibrium moisture content of a solid, mass moisture/mass dry solid, kg/kg

Greek Letters : ∆

difference

∈

emissivity of drying surface, dimensionless

θ

time,

λs

latent heat of vaporization at ts,

µ

viscosity, kg/m.s

π

3.1416

ρ

density, kg/m3

ρS

apparent density of solid, mass dry solid/wet volume, kg/m3

Principles of Mass Transfer Operations − I (Vol. − I)

8.56

Symbol

Drying Operations Meaning

Subscripts : as

adiabatic saturation

c

at critical moisture content

G

gas

m

average

max

maximum

s

surface

S

dry gas, dry solid

1

at beginning, batch drying; at solids - entering end, continuous dryer

2

at end, batch drying; at solids - leaving end, continuous dryer

I, II, III

zones I, II and III in a continuous dryer.

✱✱✱ REFERENCES 1.

W.L. McCabe, J.C. Smith and P. Harriot, "Unit Operations of Chemical Engineering", Fifth Edition, McGraw-Hill, 1993.

2.

A.S. Foust, L.A. Wenzel, C.W. Clump, L. Maus and L.B. Andersen, "Principles of Unit Operations", Second Edition, John Wiley and Sons, 1980.

3.

P. Chattopadhya, “Unit Operations “(Vol-I), Khanna Publishers, New Delhi, 1996.

4.

G.K.Roy, “ Fundamentals of Heat and Mass Transfer”, Second Edition, Khanna Publishers, New Delhi, 1990.

5.

R.H. Perry and D. Green, "Perry's Chemical Engineers' Handbook", Seventh Edition, McGraw-Hill, 1997.

6.

S.E. Charm, “ The Fundamentals of Food Processing”, Second Edition, Avi Publishing Co., Inc., West Port, Conn., 1971.

7.

C.J. Geankoplis, "Transport Processes and Unit Operations", Fourth Edition, Prentice Hall, 2003.

,,,

APPENDIX – A SI SYSTEMS, FUNDAMENTAL CONSTANTS AND CONVERSION UNITS 1.

SI Base Units : Quantity

Name

Symbol

Length

metre

m

Mass

kilogram

kg

Time

second

s

Electric Current

ampere

A

Thermodynamic Temperature

kelvin

K

Amount of Substance

mole

mol

Luminous Intensity

candela

2.

cd

SI Derived Units : Quantity

Area

Name square metre

Volume

cubic metre

Speed/Velocity Acceleration

metre per second metre per second squared

Wave Number

reciprocal metre

Density/Mass Density

kilogram per cubic metre

kg/m3

Specific Volume

cubic metre per kilogram

m3/kg

Current Density

ampere per square metre

Magnetic Field Strength

ampere per metre

A/m2 A/m

Concentration

mole per cubic metre

Luminous

candela per square metre

3.

Symbol m2 m3 m/s m/s2 m–1

mol/m3 cd/m2

SI Derived Units With Special Names : Quantity

Name

Symbol

Expression in terms of other units

Expression in terms of SI base units

Frequency

hertz

Hz

s–1

Force

newton

N

m kg s–2

Pressure/Stress

pascal

Pa

N/m2

m–1 kg s–2

Energy/Work/Quantity of Heat

joule

J

Nm

m2 kg s–2

Power/Radiant Flux

watt

W

J/s

m2 kg s–3

Electric Charge/Quantity of Electricity

coulomb

C

(A.1)

sA

Principles of Mass Transfer Operations − I (Vol. − I) Quantity

A.2

Name

Appendix – A : SI Systems, Fund. …… Symbol

Expression in terms of other units

Expression in terms of SI base units

Electric Potential/Potential Difference/Electromotive Force

volt

V

W/A

m2 kg s–3 A–1

Capacitance

farad

F

C/V

m–2kg–1s4A2

Electric Resistance

ohm

Ω

V/A

m2kg s–2A–1

Electric Conductance

siemens

S

A/V

m–2 kg–1 s3 A2

Magnetic Flux

weber

Wb

Vs

m2 kg s–2 A–1

Magnetic Flux Density

tesla

T

kg s–2 A–1

Inductance

henry

H

Wb/m2 Wb/A

degree Celsius lumen lux

o

Celsius Temperature Luminous Flux Illuminance

C

lm lx

M2 kg s–2 A–2 K cd sr

2

lm/m

Activity of a Radionuclide bequerel Bq Absorbed Dose/Specific Energy gray Gy J/kg Imparted/Kerma/Absorbed Dose Index Dose Equivalent/Dose Equivalent sievert Sv J/kg Index 4. SI Derived Units Expressed by Means of Special Names : Quantity Name Symbol

–2

m cd sr s–1 m2 s–2

m2 s–2

Expression in terms of SI base units

W/m2 J/K J/kg J/(kg K) J/kg W/(m K) J/m3

m–1 kg s–1 M2 kg s–2 kg s–2 kg s–3 2 M kg s–2 K–1 M2 S–2 m2 s–2 K–1 M2 s–2 m kg s–3 K–1 m–1 kg s–2

volt per metre coulomb per cubic metre

V/m C/m3

m kg s–2 A–1 m–3 s A

Electric Flux Density

coulomb per square metre

m–2 s A

Permittivity Permeability Molar Energy Molar Entropy / Molar Heat Capacity Exposure (x and y rays)

farad per metre henry per metre joule per metre joule per mole kelvin coulomb per kilogram

C/m2 F/m H/m J/mol J/(mol K) C/kg

m–3 kg–1 s4 A2 m kg s–2 A–2 m2 kg s–2 mol–1 2 m kg s–2 K–1 mol–1 kg–1 s A

Absorbed Dose Rate

gray per second

Gy/s

M2 s–3

Dynamic Viscosity Moment of Force Surface Tension Heat Flux Density/Irradiance Heat Capacity / Entropy Specific Heat Capacity Specific Entropy Specific Energy Thermal Conductivity Energy Density

pascal second newton meter newton per metre watt per square metre joule per kelvin joule per kilogram kelvin joule per kilogram watt per metre kelvin joule per cubic metre

Electric Field Strength Electric Charge Density

Pa S Nm N/m

Principles of Mass Transfer Operations − I (Vol. − I) 5.

A.3

Appendix – A : SI Systems, Fund. ……

SI Supplementary Units : Quantity

Name

Plane Angle

radian

Solid Angle

steradian

6.

Symbol

Expression in terms of SI base units

rad

m m–1 = 1

sr

m2 m–2 = 1

SI Derived Units Formed Using Supplementary Units : Quantity

Name

Symbol

Angular Velocity

radian per second

rad/s

Angular Acceleration

radian per second squared

rad/s2

Radiant Intensity

watt per steradian

W/sr

Radiance

watt per square metre steradian

W/(m2 sr)

Conversion factors gc for the common unit systems : System Fundamental Quantity

SI

English Engineering

Metric Engineering +

CGS

Mass M

Kilogram, kg

Pound mass, lb

Gram, g

Kilogram mass, kg

Length L

Meter, m

Foot, ft

Centimeter, cm

Meter, m

Time T

Second, s

Second, s, or hour, h

Second, s

Second, s

Force F

Newton, N

Pound force, lbf

Dyne, dyn

Kilogram force, kgf

and

1 kg · m/N · s2

32.714 lb · ft/lbf · s2 or

1 g · cm/dyn · s2

9.80665 kg · m/kgf · s2

4.1698 × 108 lb · ft/lbf · h2 Note that throughout this book kg alone is used as the abbreviation for kilogram mass and kgf is used for kilogram force. 7.

Basic SI Units :

Force

=

newton, N

Length

=

meter, m

Mass

=

kilogram, kg

Mole

=

kilogram mole, k mol

Temperature

=

kelvin = K

Time

=

second, s

Pressure

=

newton/meter2, N/m2 = pascal, Pa

Energy

=

newton-meter, N · m = joule, J

Power

=

newton-meter/second, N· m/s = watt, W

Frequency

=

1/second, s–1 = hertz, Hz

Principles of Mass Transfer Operations − I (Vol. − I) 8.

A.4

Appendix – A : SI Systems, Fund. ……

Prefixes for SI Units : Amount 1 000 000 1 000 100 10 0.1

Multiple 106 103 102 10 10–1

Prefix mga kilo hecto deka deci

Symbol M k h da d

0.01

10–2

centi

c

0.001

10

–3

milli

m

10

–6

micro

µ

10

–9

nano

n

0.000 001 0.000 000 001

Fundamental Constants and Conversion Factors : A. 1 – 1 Gas Law Constant R : Numerical Value 1.9872 1.9872 82.057 8314.34 82.057 × 10–3 8314.34 10.731

Units g cal/g mol · K btu/lb mol · oR cm3·atm/g mol · K J/kg mol·K m3·atm/kg mol · K kg·m2/s2·kg mol·K ft3·lbf/in2·lb mol · oR

0.7302 1545.3

ft3·atm/lb mol·oR ft·lbf/lb mol·oR

8314.34

m3·Pa/kg mol · K

A.1 – 2 Volume and Density : 1 kg mol ideal gas at 0o C, 760 mm Hg = 22.4140 litres = 22414 cm3 1 lb mol ideal gas at 0o C, 760 mm Hg = 359.05 ft3 1 kg mol ideal gas at 0o C, 760 mm Hg = 22.414 m3 Density of dry air at 0o C, 760 mm Hg = 1.2929 g/litre = 0.080711 lbm/ft3 Molecular weight of air = 28.97 lbm/lb mol = 28.97 g/g mol 1 g/cm3 = 62.43 lbm/ft3 = 1000 kg/m3 1 g/cm3 = 8.345 lbm/U.S. gal 1 lbm/ft3 = 16.0185 kg/m3 A.1 – 3 Length : 1 in. = 2.540 cm 100 cm = 1 m (meter) 1 micron = 10–6 m = 10–4 cm = 10–3 mm = 1 µm (micrometer) 1 Å (angstrom) = 10–10 m = 10–4 µm 1 mile = 5280 ft 1 m = 3.2808 ft = 39.37 in.

Principles of Mass Transfer Operations − I (Vol. − I)

A.5

Appendix – A : SI Systems, Fund. ……

A.1 – 4 Mass : 1 lbm = 453.59 g = 0.45359 kg 1 lbm = 16 oz = 7000 grains 1 kg = 1000 g = 2.2046 lbm 1 ton (short) = 2000 lbm 1 ton (long) = 2240 lbm 1 ton (metric) = 1000 kg A.1 – 5 Standard Acceleration of Gravity : g = 9.80665 m/s2 g = 980.665 cm/s2 g = 32.174 ft/s2 gc (gravitational conversion factor) = 32.1740 lbm · ft/lbf·s2 = 980.665 gm · cm/gf·s2 A.1 – 6 Volume : = 1000 L (litre) 1 L (litre) = 1 m3 1000 cm3 3 3 = 1 U.S. gal = 54 qt 1 in. 16.387 cm 3 = 28.317 L (litre) 1 U.S. gal = 3.7854 L (litre) 1 ft 3 = 0.028317 L (litre) 1 U.S. gal = 1 ft 3785.4 cm3 3 = 7.481 U.S. gal 1 British gal = 1.20094 U.S. gal 1 ft 3 3 = 264.17 U.S. gal = 1m 1m 35.313 ft3 A.1 – 7 Force : 1 g·cm/s2 (dyn) = 10–5 kg·m/s2 = 10–5 N (newton) 1 g·cm/s2 = 7.2330 × 10–5 lbm·ft/s2 (poundal) 1 kg·m/s2 = 1 N (newton) 1 lbf = 4.4482 N 1 g·cm/s2 = 2.2481 × 10–6 lbf A. 1 – 8 Pressure : 1 bar

=

1 × 105 (pascal) = 1 × 105 N/m2

1 psia

=

1 lbf/in2

1 psia

=

2.0360 in. Hg at 0 oC

1 psia

=

2.311 ft H2O at 70 oF

1 psia

=

51.715 mm Hg at 0 oC (ρHg = 13.5955 g/m3)

1 atm

=

14.696 psia = 1.01325 × 105 N/m2 = 1.01325 bar

1 atm 1 atm 1 atm

= = =

760 mm Hg at 0 oC = 1.01325 × 105 Pa 29.921 in. Hg at 0 oC 33.90 ft H2O at 4 oC

1 psia 1 psia 1 dyn/cm2

= = =

6.89476 × 104 g/cm·s2 6.89476 × 104 dyn/cm2 2.0886 × 10–3 lbf/ft2

1 psia 1 lbf/ft2

= =

6.89476 × 103 N/m2 = 6.89476 × 103 Pa 4.7880 × 102 dyn/cm2 = 47.880 N/m2

1 mm Hg (0 oC)

=

1.333224 × 102 N/m2 = 0.1333224 kPa

Principles of Mass Transfer Operations − I (Vol. − I)

A.6

Appendix – A : SI Systems, Fund. ……

A. 1 – 9 Power : 1 hp

=

0.74570 kW

1 watt (W)

=

14.340 cal/min

1 hp

=

550 ft·lbf/s

1 btu/h

=

0.29307 W (watt)

=

1W

1 hp = 0.7068 btu/s 1 J/s (joule/s) A. 1 – 10 Heat, Energy, Work : 1 J = 1 N·m = 1 kg·m2/s2 1 kg·m2/s2 = 1 J (joule) = 107 g·cm2/s2 (erg) 1 btu = 1055.06 J = 1.00506 kJ 1 btu = 252.16 cal (thermochemical) 1 kcal (thermochemical) = 1000 cal = 4.1840 kJ 1 cal (thermochemical) = 4.1840 J 1 cal (IT) = 4.1868 J 1 btu = 251.996 cal (IT) 1 btu = 778.17 ft·lbf 1 hp·h = 0.7457 kW·h 1 hp·h = 2544·5 btu 1f ft·lbf = 1.35582 J 1 ft ·lbf/lbm = 2.9890 J/kg A.1 – 11 Thermal Conductivity : 1 btu/h·ft·oF = 4.1365 × 10–3 cal/s·cm· oC 1 btu/h·ft· oF = 1.73073 W/m·K A.1 – 12 Heat-Transfer Coefficient : 1 btu/h·ft2· oF = 1.3571 × 10–4 cal/s·cm2·oC 1 btu/h·ft2·oF = 5.6783 × 10–4 W/cm2·oC 1 btu/h·ft2·oF = 5.6783 W/m2·K 1 kcal/h·m2·oF = 0.2048 btu/h·ft2·oF A. 1 – 13 Viscosity : 1 cp = 10–2 g/cm·s (poise) 1 cp = 2.4191 lbm/ft·h 1 cp = 6.7197 × 10–4 lbm/ft·s 1 cmp = 10–3 Pa·s = 10–3 kg/m·s = 10–3 N·s/m2 1 cp = 2.0886 × 10u–5 lbf·s/ft2 1 Pa·s = 1 N·s/m2 = 1 kg/m·s = 1000 cp = 0.67197 lbm/ft·s A.1 – 14 Diffusivity : 1 cm2/s = 3.875 ft2/h 1 cm2/s = 10–4 m2/s 1 m2/h = 10.764 ft2/h 1 m2/s = 3.875 × 104 ft2/h 1 centistoke = 10–2 cm2/s

Principles of Mass Transfer Operations − I (Vol. − I)

A.7

Appendix – A : SI Systems, Fund. ……

A. 1 – 15 Mass Flux and Molar Flux : 1 g/s·cm2 = 7.3734 × 103 lbm/h·ft2 1 g mol/s·cm2 = 7.3734 × 103 lb mol/h·ft2 1 g mol/s°·cm2 = 10 kg mol/s·m2 = 1 × 104 g mol/s·m2 1 lb mol/h·ft2 = 1.3562 × 10–3 kg mol/s·m2 A. 1 – 16 Heat Flux and Heat Flow : 1 btu/h·ft2 = 3.1546 W/m2 1 btu/h = 0.29307 W 1 cal/h = 1.1622 × 10–3 W A.1 – 17 Heat Capacity and Enthalpy : 1 btu/lbm · oF = 4.1868 kJ/kg·K 1 btu/lbm ·oF = 1.000 cal/g·oC 1 btu/lbm = 2326.0J/kg 1 ft.lbf/lbm = 2.9890 J/kg 1 cal (IT)/g·oC = 4.1868 kJ/kg·K 1 kcal/g mol = 4.1840 × 103 kJ/kg mol A. 1 – 18 Mass-Transfer Coefficient : 1 kc cm/s = 10–2 m/s 1 kc ft/h = 8.4668 × 10–5 m/s 1 kx g mol/s·cm2·mol frac = 10 kg mol/s·m2·mol frac 1 kx g mol/s·cm2·mol frac = 1 × 104 g mol/s·m2·mol frac 1 kx lb mol/h·ft2·mol frac = 1.3562 × 10–3 kg·mol/s·m2·mol frac 1 kx a lb mol/h·ft3·mol frac = 4.449 × 10–3 kg mol/s·m3·mol frac 1 kG kg mol/s·m2·atm = 0.98692 × 10–5 kg mol/s·m2·Pa 1 kG a kg mol/s·m3·atm = 0.98692 × 10–5 kg mol/s·m·Pa

,,,

APPENDIX – B A FEW USEFUL OBSERVATIONS 1. Be careful when using the Ideal Gas constant, R has been given in a number of different units in Treybal and other books. If you use one that uses Rankines for the temperature o

o

scale, you must convert temperature to Rankines (for example 273 Kelvins is 0 C, 32 F and 492 Rankines). 2. Do not convert from english to metric (or vice versa) unless you have to. 3. When calculating NTU, remember that, y2

dy

⌠ y – y* ≠ ln (y2 – y*) – ln (y1 – y*) ⌡ y1

4. Remember that the operating line and the equilibrium line should not cross or touch. You calculate the minimum values for the liquid or gas rate by having them touch. 5. If mass transfer is from gas to liquid, operating line has a positive slope, a positive intercept for liquid to gas, slope is still positive, intercept is negative. 6. Remember that mole fractions are always less than 1.0 7. HTU has units of length, NTU has no units. 8. NTU is NOT equal to the number of stages (in general that is, under certain very special conditions, NTU and number of stages will be equal, in general they are not equal). 9. y = 1.9 x does not mean that L / G = 1.9 or L / G = 1/1.9. 10. You cannot simply assume Molecular Weights, Densities, Viscosities etc. Look at the book first. 11. In calculating the number of transfer units (for concentrated gas absorption or water cooling tower) make sure the limits of integration are correct (for gas absorption you know values for yin and youts for water cooling tower you know the temperatures of water in enthalpy of air in, temperature of water out and some idea of air flow rate as a function of the minimum required). 12. If the equilibrium data or enthalpy data do not have data points at precisely the beginning and end points of the interval, interpolate to calculate the values (never extrapolate unless you have no other choice). Remember that the equilibrium line always goes through (0, 0). 13. Calculation of density of the gas. If you look up the densities of the individual components of the gas and try to average them, be careful. Before averaging, convert the mole fractions to weight fractions. The easy method is to use the ideal gas relationship and calculate the density that way. For low pressures, this is an OK assumption. (B.1)

Principles of Mass Transfer Operations − I (Vol. − I)

B.2

Appendix – B : A Few Useful Observations

14. Use the flooding curve, calculate G' and then HALVE this value to calculate the required G' (or any other percentage that is specified). Do not HALVE the value of the y intercept that you read after calculating the value on the X-axis. 15. You must NOT make mathematical errors, I am not talking, about punching in the wrong numbers in your calculator. 16. Keep your grams and gram-moles different; do not confuse them. Remember gm/gmmole is the same as a lb/lb-mole and kg/kg-mole and so on. Know how to calculate moles (and mole fractions), when you are given the weight and vice-versa. 17. State your basis clearly and as you solve problems go back and make sure that you are being consistent with your original choice of basis. 18. If you have chosen a basis, remove that variable as an unknown. 19. Material balances can be written as mass, moles or mass of atomic species. Remember mass or atoms cannot be generated or lost, moles can be generated or consumed. 20. If there is no chemical reaction, you can write a total mass, mole and atomic balance; moles are not conserved when there is a reaction. The chemical equation tells you how the total moles might change. 21. In solution of material balance problems, clearly identify the system over which you are writing the balance for; in words or as a diagram. Also identify what kind of balance you are writing; mass or mole or atomic. 22. Make sure that the balances are all independent. If you write balances on the different components, the total balance equation will be a sum of the component balances. 23. Do not forget implicit constraints such as : all mole fractions add up to 1.0. So given three components and two mole fractions, you can calculate the other by subtraction. 24. In selecting, a basis, read the problem carefully. You may be able to use something already suggested in the question. If that is inconvenient as you solve the problem, go choose another basis. But remember to avoid conflicts ! Remember to scale the final answers if necessary. 25. If gas mixtures behave ideally; the sum of partial pressures is equal to the total pressure. (also for partial volumes). Partial pressure is mole fraction times total pressure. 26. Do not confuse between partial pressure and vapour pressure !

,,,

APPENDIX – C HUMIDIFICATION AND WATER COOLING IN A PACKED COLUMN Introduction : In this project a packed column, which humidifies air and cools water by counter-current contact, is evaluated. This technique is extensively used in the chemical and petroleum industries to cool water for eventual reuse (or discharge at a lower temperature) by bringing it into contact with air. Cooling occurs both due to sensible heat changes and due to evaporation. Heat transfer and mass transfer occur simultaneously in this process, as in distillation. The main objectives of the project are to : (A) Evaluate an enthalpy balance around the column.

Fig. D.1 : Experimental set up of humidification and water cooing in a packed column

(B)

Evaluate the number of transfer units (NTU), the height of a transfer unit (HTU) and the product of a mass transfer coefficient and packing factor, kya, for each run.

(C) Evaluate and quantify the effects of process variables on the enthalpy imbalance (∆), HTU, NTU and kya estimates and interpret/explain any observed trends. The apparatus is shown in figure D.1 : Enthalpy Balance : An enthalpy balance is possible if one assumes that the instruments are reading accurately and precisely. One approach is to define a rate of enthalpy imbalance, ∆, as the sum of the rate of output enthalpies minus the sum of the rate of input enthalpies : ∆ = LCL(Tw – Tw ) + G(H2 – H1) 1

2

(C.1)

… (1)

Principles of Mass Transfer Operations − I (Vol. − I)

C.2

Appendix – D : Humd. & Water Cooling …

∆ can also be thought of as a rate of enthalpy gain by the system. It is assumed that the rate of vaporization is small enough that the mass flow rate of water, L, is essentially constant throughout the column. Specific enthalpies for air, on a mass of dry air basis, may be determined from the formulae listed below which are readily incorporated into a spreadsheet. The saturation water vapour pressure, Psat at a given temperature of air, Ta, is : … (2) Psat = 6.1078 exp[17.08085Ta/(234.175 + Ta)] Given a percent relative humidity, RH, and Psat, the water vapour partial pressure, Pw, is : Pw = RH(Psat/100) … (3) The absolute humidity, Y. as kg water/kg air is : Y = 0.622Pw/(P-Pw) … (4) where P is the total pressure in mbar (1013 mbar is equivalent to one atmosphere). The enthalpy of the water vapour/air mixture as kJ/kg of dry air is : H = 1.006Ta + 1.86YTa + 2500Y … (5) The volumetric flow-rate of air plus water vapour at the outlet is determined by multiplying the gas velocity by the cross-sectional area of the gas velocity sensor. The mass flow-rate of air plus water vapour, Gw, is determined by multiplying the volumetric flow-rate by the gas density (assume the outlet stream is an ideal gas, and use an appropriate molecular weight). The mass flow-rate of dry air is then : … (6) G = Gw/(l + Y) Calculate the ratio of the rate of enthalpy imbalance to entering enthalpy for each run. The rate of enthalpy input to the column is LCL(Tw – Tref) + GH1. The reference temperature for the 2

air enthalpy is 0° C. Use your results to evaluate the validity of the assumption that the column operates adiabatically. NTU/HTU Evaluation : The design and analysis of packed humidifiers and/or water coolers are based on the same concepts used in other packed column operations such as gas-absorption/desorption and distillation in packed towers. The theory used in the HTU-NTU approach is given by Foust et al. (1990), Treybal (1980), Wankat (1988) and Geankoplis (1993) (although the form of the various derivations and the nomenclature they use may differ somewhat from the approach presented here). Only the key steps in developing the design equations are summarized below. To begin with, a differential enthalpy balance is applied to a control volume with an infinitesimally small packed bed thickness dZ at an arbitrary tower height, Z. The packing is assumed to be covered with a thin layer of liquid. The enthalpy change in air is equated to the sum of the enthalpy changes due to sensible heat transfer between the air and water and due to the transfer of latent heat from the vaporization of a small amount of water at the interface: GdH = haS (T* - Ta) dZ + kyaSλ (Y* – Y) dZ … (7) The column is assumed to be adiabatic. The driving force for water evaporation is the difference between the absolute humidity of air near the water film and the absolute humidity in the bulk air stream. The gas enthalpy of the moist air, H is given by : H = CG (Ta - Tref) + λ Y … (8a) and the gas enthalpy of the air at the interface, H*, is : H* = CG (T*- Tref) + λ Y* … (8b) Equations (7), (8a) and (8b) are awkward to use because of the many variables involved. For the case of an air-water system, a simplification can be obtained by using thefollowing

Principles of Mass Transfer Operations − I (Vol. − I)

C.3

Appendix – D : Humd. & Water Cooling …

experimentally observed relation, called the "Lewis relation", between heat and mass transfer coefficients : h/(CGky) ≅ 1 … (9) Using equation (9), we are now able to combine equations (7), (8a) and (8b) into a single expression involving air enthalpy only : GdH = kyaS (H* – H)dZ … (10) In this equation, enthalpy appears to be the driving force for mass transfer, rather than humidity differences. However, this is not actually the case - it is the simplifying assumptions (the Lewis relation, adiabatic operation, etc.) that give the equation this appearance. As pointed out by Foust et al. (1990), enthalpy is an extensive thermodynamic property, and therefore cannot be a driving force for any transfer operation. If you must, think of the (H* – H) term as a "pseudo driving force". Refer to Foust et al. (1990) for further discussion of this point, if you wish. Rearranging equation (10), we obtain : dZ = [G/(kyaS)] [dH/(H* – H)] After integration, this may be expressed as : Z = HTU × NTU … (11) where HTU and NTU are defined by : … (12) HTU = G/(kyaS) H2

dH

⌠ ⌡ H* – H

NTU =

… (13)

H1

The variable H* is the enthalpy of the saturated air at the interface temperature, T*, at a specific location in the column. We assume that the resistance to heat transfer in the liquid phase is negligible so that T* = Tw at all points in the column. Since H* can only be expressed as a function of temperature rather than enthalpy, it is necessary to translate the integral in equation (13) into a function of water temperature. We can assume the air enthalpy changes linearly with water temperature within the column between its values at the inlet and outlet, and therefore, H = H1 + [(H2 – H1) /(Tw – Tw )] (Tw – Tw ) 2

1

1

… (14)

Differentiation leads to dH = [(H2 – H1)/(Tw – Tw )] dTw 2

1

… (15)

Substitution of equations (14) and (15) into equation (13) and changing the integration limits leads us to an equation for calculating NTU : Tw 2

H2 – H1 NTU = T – T w w 2

1

⌠ dTw  ⌡ f(Tw)

… (16)

Tw 1

where, H2 – H1 f (Tw ) = H* (Tw, RH = 100%) – H1 + T – T





w 2

w 1

 T –T   [ w w1]  

… (17)

Note that H* is the enthalpy of the saturated air (saturated means at 100% RH) at the bulk temperature of the liquid, Tw.

Principles of Mass Transfer Operations − I (Vol. − I)

C.4

Appendix – D : Humd. & Water Cooling …

The integration required in calculating NTU according to equation (16) may be performed numerically in a spreadsheet using Simpson's Rule : xo ⌠ f (x) dx ≈ ∆x [ f(xo) + 4f(x1) + 2f(x2 ) + 4f (x3) + … + 2f(x n–2) + 4f (xn –1) + f(xn) ] ⌡ 3 xo where n is an even number and : xn – xo ∆x =  n    More information about Simpson's Rule and how to use it can be found in the book by Stewart (your first year calculus textbook). Once the NTU is calculated for a run, then HTU can be calculated from equation (11) and kya can be calculated from equation (12). We would like to measure kya and a individually, but this is not possible in our equipment, so we must resort to looking at the* product. Bear in mind that although we are calculating HTU from NTU via equation (11), HTU and NTU are, in fact, independent parameters. NTU depends on the driving force for mass transfer and HTU depends on the fundamental physical features of the mass transfer via kya and a. To DESIGN a packed column to achieve a given cooling or humidification goal, the NTU and the HTU are calculated independently and equation (11) is used to estimate the required packing height in the column, Z. In this project, the column is ALREADY CONSTRUCTED and operating, and the product of HTU and NTU is fixed by equation (11) since the packing height Z is fixed. In this case, the calculated values of the parameters are coupled a decrease in NTU must bring about an increase in HTU and vice versa. It is difficult to identify the causality in this situation and you will have to think carefully about why NTU or HTU or kya is increasing or decreasing. kya is of more fundamental interest than HTU or NTU, and is the main quantity that we should be focusing on in this project. Read the references for guidance. Generate plots of and kya, HTU and NTU versus L, G and L/G (or G/L or other combinations of variables). You may choose to only present the most important plots in your report (but explain why you have omitted some of the plots, if you choose to do so). If there appears to be a systematic variation in your plots, attempt to quantify the relationship between the variables using the "trendline" regression feature in Excel. Explore various model equation forms (called "trend/regression type" in Excel) to obtain a good fit (how can you evaluate the goodness of fit ?). See if you can explain any systematic relationships you find on physical grounds (i.e. WHY does A increase with increasing B ?). Based on your results, recommend optimal values of L and G for column operation (and state the criteria you used to evaluate the optimality of the settings). Explain how your results could be used in the design of a packed column for water cooling and/or humidification. Further information about the meaning (interpretation) of HTU, NTU and kya as well as a great deal of background information on humidification in packed columns can be found in the books by Foust et al. (1990), Treybal (1980), McCabe et al. (1985), Wankat (1988) and Geankoplis (1993), as well as other sources. Experimental : A suggested approach with this apparatus is to set the air flow rate to a given value and do a family of runs with different water flow rates including zero (why ?). Then perform other families of runs, each with a different air flow rate. Allow sufficient time for the column to reach steady state after setting the water and air flow rates before taking measurements. You will need to carry out as many runs as time allows. The maximum permissable water flow rate is about 700 ml/min in order to prevent water droplets from splashing onto the humidity sensor or permitting air to by-pass the column. Do all your runs with unheated air and water in order to minimize heat losses from the column.

Principles of Mass Transfer Operations − I (Vol. − I)

C.5

Appendix – D : Humd. & Water Cooling …

Nomenclature and Units : a = Water/air contact area per unit bulk volume of a packing material, m–1 o

CL

=

Heat capacity of water, kJ/(kg C)

CG Fa Fw G Gw H H*

= = = = = = =

HTU =

Heat capacity of moist air, kJ/(kg C)) velocity of moist air stream, m/s volumetric flow rate of water, ml/min Air flow rate on dry basis, kg of dry air/hr Air plus water vapour flow-rate, kg/hr Actual enthalpy of air in the column on dry basis, kJ/kg of dry air Enthalpy of the saturated air at a water temperature T* in the column, kJ/kg of dry air Height of a transfer unit defined by equation (12), m

h ky

Heat transfer coefficient between water and air in the column, kJ/(hr m C) Local mass transfer coefficient based on the absolute humidity difference, kg of

= =

o

2o

2

water vapour/(hr m ) L = Water flow rate, kg of water/hr NTU = Number of transfer units defined by equation (13) Psat = Saturated water vapour pressure, mbar Water vapour partial pressure, mbar Pw = P = Total pressure mbar RH = Relative humidity of water/air mixture, % 2

S

=

Cross-sectional area of the column, m

T*

=

Temperature at the gas-water interface at a specific height in the column, C

Tw

=

Water temperature at a specific height in the column, C

Ta Y Y*

= = =

Air temperature at a specific height in the column, C Absolute humidity at a point in the column, kg of water vapour/k; of dry air Absolute humidity of saturated air at water temperature T in the column, kg of water vapour/kg of dry air Packed bed height in the column, m Ηeat of vapourization of water, kJ/kg Rate of enthalpy imbalance, kJ/hr

o

o

o

Z = λ = ∆ = Subscripts : 1. Bottom of the packed bed 2. Top of the packed bed References : 1. 2. 3. 4. 5. 6.

Foust, A.S., Wenzel, L.A., Clump, C.W., Maus, L. and Andersen, L.B., Principles of Unit Operations, 2nd. ed., Krieger Publishing, 1990, (Chapters 16 and 17). McCabe, W.L., Smith, J.C. and Harriott, P., Unit Operations of Chemical Engineering, 4th ed., McGraw-Hill N.Y., 1985, (Chapter 23). Treybal, R., Mass Transfer Operations, 3rd ed: McGraw-Hill, 1980, (Chapter 7). Wankat, P.C., Equilibrium Staged Separations, Prentice Hall, 1988, (Chapter 19). Geankoplis, C.J., Transport Processes and Unit Operations, 3rd ed., Prentice Hall, 1993, pp. 602610, 632. Stewart, J.S., Calculus : Early Transcendentals, 3rd ed., 1994.

,,,

APPENDIX – D DIMENSIONLESS GROUPS AND CORRELATIONS IN HEAT AND MASS TRANSFER Heat Transfer Biot Number : Ratio of conductive to convective resistance Bi = hL/ksolid

Mass Transfer M.T. Biot (Damkohler) Number heat Ratio of diffusive to convective M.T. resistance Bi = hDL/Dsolid

Fourier Number : Ratio of current time to time to reach steady-state Dimensionless time in temperature curves, used in explicit finite difference stability criterion 2

2

Fo = α t /L Fo = Dt /L Reynolds Number : Ratio of convective to viscous momentum transport Pe = ρUL/η = UL/ν Prndtl Number : M.T. Prandtl (Schmidt) Number Ratio of momentum to species diffusivity Ratio of momentum to thermal diffusivity Pr = ν/α = ηcp /k Pr = ν /D = η /ρD M.T. Nusselt (Sherwood) Number Nusselt Number : Ratio of lengthscale to thermal BL thickness Ratio of lengthscale to diffusion BL thickness Used to calculate heat transfer coefficient h Used to calculate mass transfer coefficient hD Bi = hL /kfluid Bi = hDL/Dfluid Grash of Number : Ratio of natural convection buoyancy force to viscous force Control the ratio of lengthscale to natural convection boundary layer thickness 3

Gr = gβ∆TL /ν

2

3

Gr = gβc∆TL /ν

2

Peclet Number : Ratio of convective to diffusive heat/mass transport in fluid. Pe = RePr = UL / D

Pe = RePr = UL/α Rayleigh Number :

Ratio of natural convective to diffusive heat / mass transport Determines the transition to turbulence. 3

3

Ra = GrPr = gβ ∆ TL /να

Ra = GrPr = gβc ∆ TL / νD

Note : It is not necessary to know the different names of the mass transfer dimensionless numbers, just call them, e.g. “mass transfer Biot number”, as many people do. Those names are given here because some people use them, and you'll probably hear them at some point in your chemical engineering studies. (D.1)

Principles of Mass Transfer Operations − I (Vol. − I)

D.2

Appendix – D : Dim. Groups & Correlations in Heat …

Relations between mass transfer coefficients, GASES Defining equations dCA NA = yA N + JA JA = – DAB db JA = kc (CAi – CA)

NA = kc' (CAi – CA)

JA = ky (yAi – yA)

NA = ky' (yAi – yA)

JA = kg (pAi – pA)

NA = kg' (pAi – pA)

pA P cA = yA ρM = yA RT = RT P ky = kc ρM = RT = kgP

Unit of mass transfer coefficient

 mol  s cm2 mol cm3  mol  s cm2 (mol fraction)  mol  s cm2 (atm)

=

cm s

=

mol s cm2 =

cm3 atm R = gas constant = 82.05 mol K ρM = molar density,

mol cm3

Relations between mass transfer coefficients, LIQUIDS Defining equations Unit of mass transfer coefficient JA = kc (CAi – CA) NA = kc' (CAi – CA)  mol 

JA = kx (xAi – xA)

JA = kL (CAi – CA)

NA = kx' (xAi – xA)

NA= kL' (CAi – CA)

ρx cA = xA ρM = xA M ρx ρx kx = kc ρM = M = kL ρM kL = M

s cm2 mol3 cm   mol2 s cm 

(mol fraction)

 mol  s cm2 mol cm3

=

cm s

=

mol s cm2

=

mol cm3 g ρx = liquid density, cm3

ρM = molar density,

Overall mass transfer coefficients, defining equations Based on liquid side

Based on gas side

NA = Kx (xA* – xA)

NA = Ky (yA* – yA)

NA = KL (CA* – CA)

NA = Kg (pA* – pA)

KL =

Kx ρm

Ky Kg = P

Principles of Mass Transfer Operations − I (Vol. − I)

D.3

Appendix – D : Dim. Groups & Correlations in Heat …

Relationship to individual mass transfer coefficients 1 Kx

1 1 = k + mk x y

1 Ky

m 1 = k +k x y

1 KL

ρM 1 = k +Hk L g

1 Kg

=

H 1 +k ρM kL g

Equilibrium relationships pA* pAi = x H = x Ai A

yAi yA* m = x = x Ai A

Height and number of mass transfer units ZT = HTU × NTU GM V/S Hy = k a = k aP y g

Ny =

∫ ydy – yi

Nx =

∫ xidx– x

GM V/S Hoy = K a = K aP y g

NOy =

∫ y dy – y*

=

Gx/ρx L/S HOx = K a = K a x L

NOx =

∫ x*dx– x

=

L/S Hx = k a x

(y

b

=

∆yL

=

*

Gx/ρx kLa

) (

*

– yb – ya – ya

)

∆xL

*

ln

yb – yb

=

yb – ya ∆yL xb – xa ∆xL

(x* – x ) – (x* – x ) b

b

ln

*

ya – ya

a

*

xb – xb *

xa – xa

GM HOy = Hy + m L Hx M

LM HOx = Hx + mG

M

Hy

Va ya – La xa L y = V x + V V

:

gas flow rate, mol/h.

L

:

liquid flow rate, mol/h.

S

:

cross sectional area of column, m2.

a

:

packing surface area per unit volume, m2/m3.

Gx

:

Molal velocity, mol/h.m2 (= Gx/M or Gy/M but used only for gas).

LM :

Molal velocity for liquid, mol/h.m2 (= Gx/M).

a

Principles of Mass Transfer Operations − I (Vol. − I)

D.4

Appendix – D : Dim. Groups & Correlations in Heat …

Alternative Mass Transfer Coefficient Groupings Driving Force

Height of a Transfer Unit, HTU Symbol Equimolar Unimolecular Diffusion or Diffusion Dilute Unimolecular Diffusion

Number of Transfer Units, NTU Unimolecular Diffusion

Symbol

Equimolar Diffusion1 or Dilute Unimolecular Diffusion

1. (y – y*)

HOG

G KyaS

G ' Kya (1 – y)LM S

NOG

∫(y dy– y*) ∫(1 – y) (y – y*)

2. (p – p*)

HOG

G KGaPS

G ' KGa(1 – y)LMPS

NOG

∫(p dp– p*) ∫(1P– p)(p – p*)

3. (Y – Y*)

HOG

G' KYaS

G KYaS

NOG

4. (y – y1)

HG

G kyaS

G ' kya (1 – y)LMS

NG

∫(YdY– Y*) ∫(YdY– Y*) (1 – y) dy ∫(y dy– y ) ∫(1 – y) (y – y )

(1 – y)LM dy

(P – p)LM dp

LM

1

1

5. (p – p1)

HG

G kgaS

G ' kga (P – p)LM S

NG

∫

6. (x* – x)

HOL

L KxaS

L ' kxa (1 – x)LM S

NOL

∫(x*dx– x) ∫(1 – x) (x* – x)

7. (c* – c)

HOL

L ' KLa (ρ/M – c)LM S

NOL

∫(c*dc– c) ∫(ρ/M – c)(c* – c)

8. (X* – X)

HOL

L' KXaS

L' kXaS

NOL

9. (x1 – x)

HL

L kxaS

L ' kxa (1 – x)LM S

∫(X*dX– X) ∫(X*dX– X) (1 – x) dx ∫(x dx– x) ∫(1 – x) (x – x)

10. (c1 – c)

HL

L KLaS

L kLaS

L ' kLa (ρ/M – c)LM S

NL

dp (p – p1)

∫

(1 – x)LM dx

(ρ/M – c)LM dx

LM

1

NL

(P – p)LM dp (P – p) (p – p1)

1

∫(c dC– c) ∫ 1

(ρ/M – c)LM dC (ρ/M – c)(c1 – c)

,,,

1The substitution K = K' y y y BLM or its equivalent can be made.

APPENDIX – E PHYSICAL PROPERTIES Physical Properties of Water : o

E.1 : Latent Heat of Water at 273.15 K (0 C) : Latent heat of fusion = 1436.3 cal/g mol = 79.724 cal/g = 2585.3 btu/lb mol = 6013.4 kJ/kg mol o

Latent heat of vapourization at 298.15 K (25 C) Pressure (mm Hg)

Latent Heat

23.75

44 020 kJ/kg mol, 10.514 Kcal/g mol, 18925 btu/lb mol

760

44 045 kJ/kg mol, 10.520 Kcal/g mol, 18936 btu/lb mol

E.2 : Vapour Pressure of Water : Temperature

Vapour Pressure

o

Temperature

C

kPa

mm Hg

K

273.15

0

0.611

4.58

283.15

10

1.228

293.15

20

298.15

o

Vapour Pressure

C

kPa

mm Hg

323.15

50

12.333

92.51

9.21

333.15

60

19.92

149.4

2.338

17.54

343.15

70

31.16

233.7

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

K

E.3 : Density of Liquid Water : Temperature K

o

Density 3

kg/cm

Temperature 3

C

g/cm

273.15

0

0.99987

999.87

277.15

4

1.00000

283.15

10

293.15

K

o

Density 3

g/cm

323.15

50

0.98807

988.07

1000.00

333.15

60

0.98324

983.24

0.99973

999.73

343.15

70

0.97781

977.81

20

0.99823

998.23

353.15

80

0.97l83

971.83

298.15

25

0.99708

997.08

363.15

90

0.96534

965.34

303.15

30

0.99565

995.68

373.15

100

0.95838

958.38

313.15

40

0.99225

992.25 (E.1)

kg/cm

3

C

Principles of Mass Transfer Operations − I (Vol. − I)

E.2

E.4 : Viscosity of Liquid Water : Temperature Viscosity o C

K

Appendix – E : Physical Properties Temperature

3

[ (Pa – S)10 ,

K

o C

3

Viscosity 3

[ (Pa – S)10 , 3

(kg/m . s) 10 , or cp]

(kg/m . s) 10 , or cp]

273.15

0

1.7921

323.15

50

0.5494

275.15

2

1.6728

325.15

52

0.5315

277.15

4

1.5674

327.15

54

0.5146

279.15

6

1.4728

329.15

56

0.4985

281.15

8

1.3860

331.15

58

0.4832

283.15

10

1.3077

333.15

60

0.4688

285.15

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

0.3095

313.15

40

0.6560

367.15

94

0.3027

315.15

42

0.6321

369.15

96

0.2962

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

E.5 : Heat Capacity of Liquid Water at 101.325 kPa (1 Atm) : Temperature K C

o

0 10

273.15 283.15

Heat Capacity cp o

cal/g C 1.0080 1.0019

kJ/kg.K 4.220 4.195

Temperature K C

o

50 60

323.15 333.15

Heat Capacity cp o

cal/g C 0.9992 0.9992

kJ/kg.K 4.183 4.183 (cntd.)

Principles of Mass Transfer Operations − I (Vol. − I) Heat Capacity cp

Temperature o

E.3

o

Appendix – E : Physical Properties Heat Capacity cp

Temperature

C

K

cal/g C

kJ/kg.K

0

273.15

1.0080

10

283.15

20

o

o

C

K

cal/g C

kJ/kg.K

4.220

50

323.15

0.9992

4.183

1.0019

4.195

60

333.15

1.0001

4.187

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

1.0076

4.219

E.6 : Thermal Conductivity of Liquid Water : Temperature o

o

Thermal Conductivity o

F

K

but/h.ft. F

W/m.K

0

32

273.15

0.329

0.569

37.8

100

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

C

E.7 : Vapour Pressure of Saturated Ice-Water Vapour and Heat : Temperature

Vapour Pressure

Heat of Sublimation

K

o

o

F

C

kPa

273.2

32

0

6.107 × 10

266.5

20

- 6.7

3.478 × 10

261.0

10

-12.2

2.128 × 10

255.4

0

-17.8

1.275 × 10

249.9

– 10

- 23.3

7.411 × 10

244.3

– 20

- 28.9

3.820 × 10

235.8

– 30

- 34.4

2.372 × 10

233.2

– 40

- 40.0

1.283 × 10

psia –1 –1 –1 –1 –2 –2 –2 –2

8.858 × 10 5.045 × 10 3.087 × 10 1.849 × 10 1.082 × 10 6.181 × 10 3.440 × 10 1.861 × 10

–2 –2 –2 –2 –2 –3 –3 –3

mm Hg

btu/bm

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

Principles of Mass Transfer Operations − I (Vol. − I)

E.4

Appendix – E : Physical Properties

E.8 : Heat Capacity of Ice : Cp

Temperature o

o

Cp

Temperature

F

K

btu/bm. F

kJ/kg.K

32

273.15

0.500

20

266.45

0.490

10

260.95

0

255.35

o

o

F

K

btu/bm. F

kJ/kg.K

2.093

– 10

249.85

0.461

1.930

2.052

– 20

244.25

0.452

1.892

0.481

2.014

– 30

238.75

0.442

1.850

0.472

1.976

– 40

233.15

0.433

1.813

E.9 : Properties of Saturated Steam and Water (Steam Table) : SI Units TempeVapour Specific Volume Enthalpy 3 rature Pressure (kJ/kg) (m /kg) o (kPa) Liquid Sat'd Liquid Sat'd ( C) Vapour Vapour 0.01 0.6113 0.0010002 206.136 0.00 2501.4

Entropy (kJ/kg. K) Liquid 0.0000

Sat'd Vapour 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

1.3059

7.3549

Principles of Mass Transfer Operations − I (Vol. − I) E.10 : SI Units Continued : TempeVapour Specific Volume 3 rature Pressure (m /kg) o (kPa) Liquid Sat'd ( C) Vapour

E.5

Appendix – E : Physical Properties Enthalpy (kJ/kg)

Entropy (kJ/kg. K)

Liquid

Sat'd Vapour

Liquid

Sat'd Vapour

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

125

232.1

0.0010649

0.7706

524.99

2713.5

1.5813

7.0775

130

270.1

0.0010697

0.6685

546.31

2720.5

1.6344

7.0269

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

0.00l l080

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

E.11 : Prperties of Superheated Steam (Steam Table), SI units 3

(ν ν Specic volume, m / kg. enthalpy, kJ/kg; s, entropy, kJ/kg.K) : Absolute pressure kPa (Sat.

o

o Temp., C)

100

150

200

10 (45.81)

17.196 2687.5 8.4479 3.418

19.512 2783.0 8.6882 3.889

21.825 2879.5 8.9038 4.356

ν H s ν

Temperature ( C) 250 300 24.136 2977.3 9.1002 4.820

26.445 3076.5 9.2813 5.284

360

420

500

29.216 3197.6 9.4821 5.839

31.986 3320.9 9.6682 6.394

35.679 3489.1 9.8978 7.134 (Contd.)

Principles of Mass Transfer Operations − I (Vol. − I) Absolute pressure kPa (Sat.

E.6

Appendix – E : Physical Properties o

o Temp., C)

Temperature ( C) 250 300

100

150

200

360

420

500

2682.5

2780.1

2877.7

2976.0

3075.5

3196.8

3320.4

3488.7

H

7.6947

7.9401

8.1580

8.3556

8.5373

8.7385

8.9249

9.1546

s

2.270

2.587

2.900

3.211

3.520

3.891

4.262

4.755

75

ν

2679.4

2778.2

2876.5

2975.2

3074.9

3196.4

3320.0

3488.4

(91.78)

H

7.5009

7.7496

7.9690

8.1673

8.3493

8.5508

8.7374

8.9672

s

1.6958

1.9364

2.172

2.406

2.639

2.917

3.195

3.565

100

ν

2672.2

2776.4

2875.3

2974.3

3074.3

3195.9

3319.6

3488.1

(99.63)

H

7.3614

7.6134

7.8343

8.0333

8.2158

8.4175

8.6042

8.8342

s

1.2853

1.4443

1.6012

1.7570

1.9432

2.129

2.376

150

ν

2772.6

2872.9

2972.7

3073.1

3195.0

3318.9

3487.6

(111.37)

H

7.4193

7.6433

7.8438

8.0720

8.2293

8.4163

8.6466

s

0.4708

0.5342

0.5951

0.6548

0.7257

0.7960

0.8893

400

ν

2752.8

2860.5

2964.2

3066.8

3190.3

3315.3

3484.9

(143.63)

H

6.9299

7.1706

7.3789

7.5662

7.7712

7.9598

8.1913

s

0.2999

0.3363

0.3714

0.4126

0.4533

0.5070

700

ν

2844.8

2953.6

3059.1

3184.7

3310.9

3481.7

(164.97)

H

6.8865

7.1053

7.2979

7.5063

7.6968

7.9299

s

0.2060

0.2327

0.2579

0.2873

0.3162

0.3541

1000

ν

2827.9

2942.6

3051.2

3178.9

3306.5

3478.5

(179.9l)

H

6.6940

6.9247

7.1229

7.3349

7.5275

7.7622

s

0.13248

0.15195

0.16966

0.18988

0.2095

0.2352

1500

ν

2796.8

2923.3

3037.6

3.1692

3299.1

3473.1

(198.32)

H

6.4546

6.7090

6.9179

7.1363

7.3323

7.5698

s

0.11144

0.12547

0.14113

0.15616

0.17568

2000

ν

2902.5

3023.5

3159.3

3291.6

3467.6

(212.42)

H

6.5453

6.7664

6.9917

7.1915

7.4317

s

0.08700

0.09890

0.11186

0.12414

0.13998

2500

ν

2880.1

3008.8

3149 1

3284.0

3462.1

(223.99)

H

6.4085

6.6438

6.8767

7.0803

7.3234

s

0.07058

0.08114

0.09233

0.10279

0.11619

3000

ν

2855.8

2993.5

3138.7

3276.3

3456.5

(233.90)

H

6.2872

6.5390

6.7801

6.9878

7.2338

50 (81.33)

s

Principles of Mass Transfer Operations − I (Vol. − I)

E.7

Appendix – E : Physical Properties

E.12 : Physical Properties of Air at 101.325 kPa (a Atm Abs), SI Units) 3 T k NPr T p Cp µ × 10 o 3 (K) ( C) (kg/m ) (kJ/kg . (Pa . s, or kg/m - s) (w/m . K) K)

–17.8 0 10.0 37.8 65.6 93.3 121.1 148.9 176.7 204.4 232.2 260.0

255.4 273.2 283.2 311.0 338.8 366.5 394.3 422.1 449.9 477.6 505.4 533.2

1.379 1.293 1.246 1.137 1.043 0.964 0.895 0.838 0.785 0.740 0.700 0.662

1.0048 1.0048 1.0048 1.0048 1.0090 1.0090 1.0132 1.0174 1.0216 1.0258 1.0300 1.0341

1.62 1.72 1.78 1.90 2.03 2.15 2.27 2.37 2.50 2.60 2.71 2.80

0.02250 0.02423 0.02492 0.02700 0.02925 0.03115 0.03323 0.03531 0.03721 0.03894 0.04084 0.04258

0.720 0.715 0.713 0.705 0.702 0.694 0.692 0.689 0.687 0.686 0.684 0.680

β × 10 (K) 3.92 3.65 3.53 3.22 2.95 2.74 2.54 2.38 2.21 2.09 1.98 1.87

3

2

2

gβρ βρ /µ µ 3 (l/k.m ) 2.79 × 10 2.04 × 10 1.72 × 10 1.12 × 10

8 8 8 8

0.775 × 10 0.534 × 10 0.386 × 10 0.289 × 10 0.214 × 10 0.168 × 10 0.130 × 10 0.104 × 10

E.13 :

Viscosity of Gases at 101.325 kPa (1 Atm Abs) 3

[Viscosity in (Pa . s)10 , (kg/,.s) 103 or cp ] Temperature o o K O2 H2 F C 255.4 0 – 17.8 0.00800 0.0181 273.2 32 0 0.00840 0.0192 283.2 50 10.0 0.00862 0.0197 311.0 100 37.8 0.00915 0.0213 338.8 150 65.6 0.00960 0.0228 366.5 200 93.3 0.0101 0.0241 394.3 250 121.1 0.0106 0.0256 422.1 300 148.9 0.0111 0.0267 449.9 350 176.7 0.0115 0.0282 477.6 400 204.4 0.0119 0.0293 505.4 450 232.2 0.0124 0.0307 533.2 500 260.0 0.0128 0.0315 E.14 : Pradtl Number of Gases at 101.325 kPa (1 Atm Abs) : Temperature o o O2 K H2 C F –17.8 0 255.4 0.720 0.720 0 32 273.2 0.715 0.711 10.0 50 283.2 0.710 0.710 37.8 100 311.0 0.700 0.707 37.8 150 338.8 0.700 0.706 93.3 200 366.5 0.694 0.703 121.1 250 394.3 0.688 0.703 148.9 300 422.1 0.683 0.703 176.6 350 449.9 0.677 0.704 204.4 400 477.6 0.670 0.706 232.2 450 505.4 0.668 0.702 260.0 500 533.2 0.666 0.700

N2

CO

CO2

0.0158 0.0166 0.0171 0.0183 0.0196 0.0208 0.0220 0.0230 0.0240 0.0250 0.0260 0.0273

0.0156 0.0165 0.0169 0.0183 0.0195 0.0208 0.0220 0.0231 0.0242 0.0251 0.0264 0.0276

0.0128 0.0137 0.0141 0.0154 0.0167 0.0179 0.0191 0.0203 0.0215 0.0225 0.0236 0.0247

CO 0.740 0.738 0.735 0.731 0.727 0.724 0.720 0.720 0.720 0.720 0.720 0.720

CO2 0.775 0.770 0.769 0.764 0.755 0.752 0.746 0.738 0.734 0.725 0.716 0.702

N2 0.720 0.720 0.717 0.710 0.700 0.700 0.696 0.690 0.689 0.688 0.688 0.688

8 8 8 8 8 8 8 8

Principles of Mass Transfer Operations − I (Vol. − I)

E.8

Fig. E.1 : Viscosities of gases at 101.325 kPa

Appendix – E : Physical Properties

Principles of Mass Transfer Operations − I (Vol. − I)

E.9

Appendix – E : Physical Properties

E.15 : Viscosities of Gases (Co-ordinates for Use With Fig. E.1 ) No.

Gas

X

Y

No.

Gas

X

Y

1

Acetic acid

7.7

14.3

29

Freon- 113

11.3

14.0

2

Acetone

8.9

13.0

30

Helium

10.9

20.5

3

Acetylene

9.8

14.9

31

Hexane

8.6

11.8

4

Air

11.0

20.0

32

Hydrogen

11.2

12.4

5

Ammonia

8.4

16.0

33

3H2 + 1N2

11.2

17.2

6

Argon

10.5

22.4

34

Hydrogen bromide

8.8

20.9

7

Benzene

8.5

13.2

35

Hydrogen chloride

8.8

18.7

8

Bromine

8.9

19.2

36

Hydrogen cyanide

9.8

14.9

9

Butene

9.2

13.7

37

Hydrogen iodide

9.0

21.3

10

Butylene

8.9

13.0

38

Hydrogen sulfide

8.6

18.0

11

Carbon dioxide

9.5

18.7

39

Iodine

9.0

18.4

12

Carbon disulfide

8.0

16.0

40

Mercury

5.3

22.9

13

Carbon monoxide

11.0

20.0

41

Methane

9.9

15.5

14

Chlorine

9.0

18.4

42

Methyl alcohol

8. 5

15.6

15

Chloroform

8.9

15.7

43

Nitric oxide

10.9

20.5

16

Cyanogen

9.2

15.2

44

Nitrogen

10.6

20.0

17

Cyclohexane

9.2

12.0

45

Nitrosyl chloride

8.0

17.6

18

Ethyne

9.1

14.5

46

Nitrous oxide

8.8

19.0

19

Ethyl acetate

8.5

13.2

47

Oxygen

11.0

21.3

20

Ethyl alcohol

9.2

14.2

48

Pentane

7.0

12.8

21

Ethyl chloride

8.5

15.6

49

Propane

9.7

12.9

22

Ethyl ether

8.9

13.0

50

Propyl alcohol

8.4

13.4

23

Ethylene

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-l l

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

Water

8.0

16.0

28

Freon-22

10.1

17.0

56

Xenon

9.3

23.0

Principles of Mass Transfer Operations − I (Vol. − I)

E.10

Fig. E.2 : Viscosity of Liquids

Appendix – E : Physical Properties

Principles of Mass Transfer Operations − I (Vol. − I)

E.11

Appendix – E : Physical Properties

E.16 : Viscosities of Liquid (Co-ordinates for Use With Fig. E.2) Liquid X Y Liquid

X

Y

Acetaldehyde

15.2

4.8

Cyclohexanol

2.9

24.3

Acetic acid, 100%

12.1

14.2

Cyclohexane

9.8

12.9

Acetic acid, 70%

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

Acetone, 35%

7.9

15.0

Diethyl ketone

13.5

9.2

Acetonitrile

14.4

7.4

Diethyl oxalate

11.0

16.4

Acrylic acid

12.3

13.9

Diethylene glycol

5.0

24.7

Allyl alcohol

10.2

14.3

Diphenyl

12.0

18.3

Allyl bromide

14.4

9.6

Dipropyl ether

13.2

8.6

Allyl iodide

14.0

11.7

Dipropyl oxalate

10.3

17.7

Ammonia, 100%

12.6

2.0

Ethyl acetate

13.7

9.1

Ammonia, 26%

10.1

13.9

Ethyl acrylate

12.7

10.4

Amyl acetate

11.8

12.5

Ethyl alcohol 100%

10.5

13.8

Amyl alcohol

7.5

18.4

Ethyl alcohol 95%

9.8

14.3

Anilinc

8.1

18.7

Ethyl alcohol, 40%

6.5

16.6

Anisole

12.3

13.5

Ethyl benzene

13.2

11.5

Arsenic trichloride

13.9

14.5

Ethyl bromide

14.5

8.1

Benzene Brine, CaCL2, 25%

12.5

10.9

2-Ethyl butyl crylate

11.2

14.0

6.6

15.9

Ethyl chlorite

14.8

6.0

Brine, NaCl, 25%

10.2

16.6

Ethyl ether

14.5

5.3

Bromine

14.2

13.2

Ethyl formate

14.2

8.4

Bromotoluene

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

Butyl alcohol

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

11.6

0.3

Ethylene bromide

11.9

15.7

Carbon disulfide

16.1

7.5

Ethylene chloride

12.7

12.2

Carbon tetrachloride

12.7

13.1

Ethylene glycol

6.0

23.6

Chlorobenzene

12.3

12.4

Ethylidene chloride

14.1

8.7

Chloroform

14.4

10.2

Fluorobenzene

13.7

10.4

Chlorosulfonic acid

11.2

18.1

Formic acid

10.7

15.8

Chlorotoluene, ortho

13.0

13.3

Freon-11

14.4

9.0

Chlorotoluene, meta

13.3

12.5

Freon-12

16.8

15.6

Chlorotoluene, para

13.3

12.5

Freon-21

15.7

7.5

Creol, meta

2.5

20.8

Freon-22

17.2

4.7

Principles of Mass Transfer Operations − I (Vol. − I)

E.12

Appendix – E : Physical Properties

E.17 : Viscosities of Liquids, (Continued) Liquid Freon 113 Glycerol 100% Glycerols 50% Heptane Hexane Hydrocloric acid 31.5% Iodobenzene Isobutyl alcohol Isobutyric acid Isopropyl alcohol Isopropyl bromide Isopropyl chloride Isopropyl iodide Kerosene Linseed oil, raw Mercury Methanol, 100% Methanol, 90% Methanol, 40%. Methyl acetate Methyl acrylate Methyl i-butyrate Methyl n - butyrate Methyl chloride Methyl ethyl ketone Methyl formate Methyl iodide Methyl propionate Methyl propyl ketone Methyl sulfide Naphthalene Nitric acid, 95% Nitric acid, 60% Nitobenzene Nitrogen dioxide Nitrotoluene Octane

X 12.5 2.0 6.9 14.1 14.7 13.0 12.8 7.1 12.2 8.2 14.1 13.9 13.7 10.2 7.5 18.4 12.4 12.3 7.8 14.2 13.0 12.3 13.2 15.0 13.9 14.2 14.3 13.5 14.3 15.3 7.9 12.8 10.8 10.6 12.9 11.0 13.7

Y 11.4 30.0 19.6 8.4 7.0 16.6 15.9 18.0 14.4 16.0 9.7 7.1 11.2 16.9 27.2 16.4 10.5 11.8 15.5 8.2 9.5 9.7 10.3 3.8 8.6 7.5 9.3 9.0 9.5 6.4 18.1 13.8 17.0 16.2 8.6 17.0 10.0

Liquid Octyl alcohol Pentachloroethane Pentane Phenol Phosphorus tribromide Phosphorus trichloride Propionic acid Propyl acetate Propyl alcohol Propyl bromide Propyl chloride Propyl formate Propyl iodide Sodium Sodium hydroxide, 50% Stannic chloride Succinonitrile Sulfur dioxide Sulfuric add, 110% Sulfuric acid, 100% Sulfuric acid, 98% Sulfuric acid, 60% Sulfuryl chloride Tetrachloroethane Thiophene Titanium tetrachloride Toluene Trichloroethylene Triethylene glycol Turpentine Vinyl acetate Vinyl toluene Water Xylene, ortho Xylene, meta Xylene, para

X 6.6 10.9 14.9 6.9 13.8 16.2 12.8 13.1 9.l 14.5 14.4 13.1 14.1 16.4 3.2 13.5 10.1 15.2 7.2 8.0 7.0 10.2 15.2 11.9 13.2 14.4 13.7 14.8 4.7 11.5 14.0 13.4 10.2 13.5 13.9 13.9

Y 21.1 17.3 5.2 20.8 16.7 10.9 13.8 10.3 16.5 9.6 7.5 9.7 11.6 13.9 25.8 12.8 20.8 7.1 27.4 25.1 24.8 21.3 12.4 15.7 11.0 12.3 10.4 10.5 24.8 14.9 8.8 12.0 13.0 12.1 10.6 10.9

,,,

APPENDIX – F DETERMINING MASS TRANSFER COEFFICIENTS It is not reasonable to expect mass transfer coefficients to be readily available for any and all systems. The "best" solution is to experimentally measure coefficients on a bench scale (using a wettedwall column, etc.) and then use the results to design a full-scale separation column. When this is not feasible, more approximate arrangements must be made. Correlations : Dimensional analysis of mass transfer suggests correlations of the form : kcL LG µ  = f µ , … (1) D ρD   Sh = f (Re, Sc) A number of correlations matching this form are presented in many textbooks of mass transfer. Treybal (1987) suggest the following correlations for use with beds packed with Raschig rings or Berl saddles : 2/3

kg pA Scg GGM

 ds GG  = 1.195   µg (1 – εLO)

0.45 kc ds ds GL 0.5 = 25.1 Scl   Dl  µ1  subject to the following definitions

… (2) … (3)

mole area × time × (partial pressure) mole mass transfer coefficient = area × time × (concentration) partial pressure of A in vopour phase Schmidt number viscosity diffusivity (in liquid phase) sphere equivalent diameter of packing superficial mass velocity (mass flux) in vapour phase superficial mass velocity (mass flux) in liquid phase superficial molar flux in gas phase operating void space in packing ε–ø

kg = mass transfer coefficient = kc = pA Sc µ DL ds GG GL GGM SLO

= = = = = = = = = =

LT

= void fraction –

total liquid holdup volume packed volume

… (4)

Analogies : Since the basic mechanisms of heat, mass, and momentum transport are essentially the same, it is sometimes possible to directly relate heat transfer coefficients, mass transfer coefficients, and friction factors by means of analogies.

(F.1)

Principles of Mass Transfer Operations − I (Vol. − I)

F.2

Appendix – F : Determining Mass Transfer Coefficients

Analogies involving momentum transfer are only valid if there is no form drag, hence they are pretty much limited to flow over flat plates and inside (but not across) conduits. Also recognize that if there is much heat or mass transfer, it may change fluid and flow characteristics enough to make analogy worthless; in some cases, a viscosity correction may be used to compensate. A simple, crude analogy recognizes that turbulent eddies transport heat and mass as well as momentum, thus one can argue that the eddie diffusivities are the same for all modes of transport, that is : ET = EH = EM. These values are seldom at hand, though. Another analogy -- probably the oldest -- is the "Reynolds Analogy", which relates the Fanning friction factor for fluid flow to heat transport : 1 h = NSt … (5) 2 fF = – ρuzcp where the right hand side is the "Stanton Number". The Stanton number is a dimensionless group made up of other, more familiar groups. It can be defined for heat transfer or for mass transfer. Nu hd µ k h StH = Re • Pr =  k  dG c µ = c G (for Heat transfer) …(6)     p  p Sh kcL µ  ρD = kc (for Mass transfer) … (7) StM = Re•Sc = D  u   Luρ  µ  The Reynolds analogy gives reasonable values for gases where the Prandtl number is roughly one. Note that one-half the friction factor is the ratio of the overall momentum transported to the wall to the inertial effects in the mainstream. The Stanton number represents the ratio of the overall heat transport to the wall to the convective effects in the mainstream. The Reynolds analogy says that these ratios are equal for mass and momentum transport. The Reynolds analogy postulates direct interaction between the turbulent core of the flow and the walls. If a laminar sublayer is included between these, the Prandtl-Taylor analogy applies : 1 2 fF NSt = … (8) uxδ 1+ (NPr – 1) uxm This form includes the ratio of the mean velocities in the sublayer and core as well as the Prandtl number for heat transfer. Note that when the Prandtl number is equal to one, this equation reduces to the Reynolds analogy. Probably the most widely used is the Colburn (or Colburn-Chilton) analogy. It is based on correlations and data rather than on assumptions about transport mechanisms. The Colburn "jfactor" for heat transfer and the Colburn-Chilton j-factor for mass transfer are : jH = StH Pr

2/3

2/3 h cpµ = c G k p   

2/3

kc µ 2/3 = u   ρD

jM = StM Sc

The heat transfer factor may be modified with the Seider-Tate viscosity correction 2/3 1 jH = StH Pr øv although this does not seem to be universally done.

…(9) … (10)

… (11)

Principles of Mass Transfer Operations − I (Vol. − I)

F.3

Appendix – F : Determining Mass Transfer Coefficients

When the j-factors are used, the fluid properties in the Stanton number are evaluated at the mean bulk average temperature and those for the Prandtl number at the film temperature (this means two heat capacities). The Colburn-Chilton analogy is simply fsmooth = jH = jM … (12) 2 valid for turbulent flow in conduits with NRe > 10000, 0.7 < NPr < 160, and tubes where L/d > 60 (the same constraints as Seider-Tate). A wider range of data is correlated by the Friend-Metzner analogy : 0.14 f  µb  NRe‚ b NPr‚ b 2 µ   w NNu, b = f –1/3 120 + 11.8 2 (NPr‚ b – 1) (NPr‚ b) NSh NRe NSc

f 2

=

f –1/3 120 + 11.8 2 (NSc – 1) (NSc) which is valid when NRe > 10000, 0.5 < NPr < 600, 0.5 < NSc < 3000.

… (13)

… (14)

Coefficients from Reference Conditions : Another possibility is to estimate mass transfer coefficients by comparison with measured values for reference systems. For instance, the overall mass transfer coefficients for the oxygen-water system has been measured and can be used to predict overall coefficients for other systems using Hx =

n 1 Gx α µ

Sc

… (15)

Literature suggest a typical value of n = 0.3, so new values can be obtained using

 µref 0.3 Hxℑnew = Hxℑref µ  

new

Scnew Scref

… (16)

For gas-film coefficients, literature provide data for ammonia-water, and recommend estimation using SC Hyℑnew = Hyℑref

new

SC

… (17)

ref

References : 1. Brodkey, R. S. and H. C. Hershey, Transport Phenomena : A Unified Approach, McGrawHill, 1988, pp. 516-20. 2.

McCabe, W. L., J. C. Smith, and P. Harriott, Unit Operations of Chemical Engineering (5th Edition), McGraw-Hill, 1993, pp. 348-52.

3.

McCabe, W. L., J. C. Smith, and P. Harriott, Unit Operations of Chemical Engineering (6th Edition), McGraw-Hill, 2001, pp. 532-40, 580-88.

,,,

APPENDIX – G ALGEBRAIC SOLUTION OF EQUILIBRIUM STAGE PROBLEMS : THE KREMSER EQUATION Solving problems involving equilibrium stage separations requires simultaneous solution of the equilibrium and operating (component balance) expressions. Choice of a solution technique -algebraic, graphical, or numerical -- depends on the form of the expressions. The Kremser Equation, an "absorption factor method", provides an algebraic solution for analyzing equilibrium cascades. It cannot be used for every problem, but is convenient for several cases, notably : • Dilute gas absorption (when set up on "solvent free" basis) • Distillation (use for the extreme ends of a high purity separation where the curvature of the equilibrium curve is not significant)

Fig. G.1

Modeling : The equations will be developed for a countercurrent cascade of N stages. Begin by writing the steady state component balance over n-1 stages : … (1) 0 = L0x0 – Lx–1 xx–1 + Vnyn – V1y1 Vnyn = Ln–1 xn–1 + (V1y1 – L0x0 ) … (2) The equilibrium expression will be written in terms of a "K-value" (McCabe and Smith develop these equations starting with a linear equilibrium expression with slope m) … (3) yn = Kn xn Vnyn = Vn(Knxn) … (4) The absorption factor will then be defined. It is the ratio of the local slope of the operating curve to that of the equilibrium curve. Similar expressions can be defined to serve as "stripping factors", or "extraction factors", or "wash factors", etc.  Ln  Vn Ln   An = K = V K … (5) n n n Ln VnKn = A … (6) n The absorption factor thus varies from stage to stage. These three expressions (component balance, equilibrium, absorption factor) are then combined and rearranged L nx n Vnyn = VnKnxn = A = Ln–1 xn–1 + (V1y1 – L0x0) …(7) n Lnxn = An (Ln–1 xn–1) + An (V1y1 – L0x0) If the same steps were taken for a balance over n-2 and n-3 stages, the results would be : Ln–1 xn–1 = An–1 (Ln–2 xn–2) + An–1 (V1y1 – L0x0 ) …(8) (G.1)

Principles of Mass Transfer Operations − I (Vol. − I)

G.2

Appendix – G : The Kremser Equation

Ln–2 xn–2 = An–2 (Ln–3 xn–3) + An–3 (V1y1 – L0x0 ) These expressions are then "nested" into the first to obtain Lnxn = An (An–1 (Ln–2 xn–2) + An–1 (V1y1 – L0x0)) + An (V1y1 – L0x0)

… (9) … (10)

= An An–1 (Ln–2 xn–2) + (An + AnAn–1) (V1y1 – L0x0) … (11) = An An–1 (An–2 (Ln–3 xn–3) + An–2 (V1y1 – L0x0)) + (An + AnAn–1) (V1y1 – L0x0) = An An–1 An–2 (Ln–3 xn–3) + (An + AnAn–1 + AnAn–1An–2) (V1y1 – L0x0)

… (12) … (13)

This process is repeated, until the balance over 1 stage is incorporated L1x1 = A1(L0x0) + A1(V1y1 – L0x0) = A1(V1y1)

(14) … (15)

Lnxn = (An An–1 An–2 … A2A1) (V1y1) + (An + AnAn–1 + AnAn–1An–2 + … + … (16) AnAn–1An–2 … A2) (V1y1 – L0x0) The balance will be written one more time, over n stages Lnxn = L0x0 + Vn+1 yn+1 – V1y1 = Vn+1yn+1 – (V1y1 – L0x0) Then the last two equations are set equal and rearranged : Vn+1yn+1 – (V1y1 – L0x0)= (An An–1 An–2 … A1) (V1y1)

… (17) … (18) … (19)

+ (An + AnAn–1 + AnAn–1An–2 + … + AnAn–1An–2 – A2) (V1y1 – L0x0) = (An + AnAn–1 + AnAn–1An–2 + … + AnAn–1An–2 – A1) (V1y1) – (An + AnAn–1 + AnAn–1An–2 + … + AnAn–1An–2 – A2) (L0x0) Vn+1yn+1 = (An + AnAn–1 + AnAn–1An–2 + … + AnAn–1An–2 – A1 + 1) (V1y1) – (An + AnAn–1 + AnAn–1An–2 + … + AnAn–1An–2 …… A2 + 1) (L0x0) If the absorbent fed is pure, x0 = 0, and the second term vanishes. It is then convenient to define the "fraction NOT absorbed", the ratio of solute leaving to solute fed Vn+1 yn+1 1 = φA = A + A A … (20) V1y1 + A A A n n n–1 n n–1 n–2 + … + AnAn–1An–2 … A1 + 1 which can sometimes be used to compact the notation. This equation allows calculation of the recovery; but it is unlikely that anyone would have all the required absorption factors. Group Method Approximation : The absorption factor A varies from stage to stage as the liquid and vapor flows and equilibrium shift. The "group method" approximation says that we can assume an average, "effective" value of the absorption factor that is defined to be the same for all stages. Note, though, that if both the equilibrium curve and operating curve are straight lines, no approximation is involved. This allows algebraic simplication of the recovery fraction a1 …(21) φA = 3 3 n Ae + Ae + Ae + … + Ae + 1 Provided one remembers the rules for geometric series from calculus 1 – rn+1 sn = 1 + r + r2 + … + rn = 1 – r …(22)

Principles of Mass Transfer Operations − I (Vol. − I) φA =

1 – Ae 1–

n+1 Ae

G.3 =

Appendix – G : The Kremser Equation

Ae – 1 n+1

Ae

–1

… (23)

A similar simplification can be done on the L0x0 term, noting that the order is one less. The full equation is thus Vn+1 yn+1

An+1 Ane – 1 – 1 e = V1y1  A – 1  – L0x0  A – 1   e   e  n Ae – 1 V1y1 = L 0 x0  A – 1  φA  e 

… (24) … (25)

The coefficient on L0x0 represents the consequences of both impure absorbent and the fact that vapor flow may do some stripping of the enriched absorbent. It thus makes sense to express this quantity in terms of the stripping factor : Se – 1 1 Se = A φs = n+1 … (26) e Se – 1 Beginning by setting up a common denominator, the L0x0 coefficient can be rewritten to obtain

Ane – 1 φA An – 1     = 1  e  Ae – 1  φA φA Ae – 1 

 Ane – 1   Ae – 1    An+1 – 1 = 1  n+1   e  φA Ae – 1

1

 Sn – 1   e = 1 1 φ  n+1 – 1 Se 

S n+1   n+1   e – Se = 1 Se – 1 + 1 – Se  Sn+1 – 1  φA  Sn+1 – 1  e  e   

=

1 φA

=

1 – Se  1 1  1 + n+1 = (1 – φs)   φA φA S –1

A



e

… (27)



So the the overall equation is Vn+1 yn+1 =

V1y1 L0x0 (1 – φs) – φA φA

… (28)

This equation is useful in solving some problems. Operating Equation Forms : From here on we will assume that the flow rates L and V and the equilibrium K-value are constants. This means that both the equilibrium and operating curves will be straight lines and that the absorption and stripping factors are constants. Vn+1yn+1 yn+1

An+1 Ane – 1 – 1 e = V1y1  A – 1  – L0x0  A – 1   e   e  n+1 n V1  Ae – 1 L0  Ae – 1   = V   A – 1  y1 – V   A – 1  x0  n+1  e  n+1  e  

… (29)

Principles of Mass Transfer Operations − I (Vol. − I)

G.4

Appendix – G : The Kremser Equation

n+1

n

–1 A L A – 1 yn+1 = A – 1 y1 – V  A – 1  x0   Define the hypothetical equilibrium vapor composition, substitute, and rearrange. *

y0

= Kx0 *

y0 x0 = K n

yn+1 =

An+1 – 1 L A – 1 *   A – 1 y1 – VK  A – 1  y0 n

=

An+1 – 1 A – 1 *   A – 1 y1 – A  A – 1  y0

yn+1 (A – 1) = (A

n+1

*

n

– 1) y1 – A(A – 1) y0 *

*

Ayn+1 – yn+1 = A

n+1

y1 – y1 – A

y1 – yn+1 = A

n+1

(y1 – y*0) + A(y*0 – yn+1)

n+1

y0 + Ay0

Next we need a rearranged version of the balance over n stages Vn+1 yn+1 = V1y1 + Lnxn – L0x0 L L yn+1 = y1 + V xn – V x0 *

L y0 yn+1 = y1 + (AK) xn – V K *

*

yn+1 = y1 + Ayn – Ay0

(*

*

yn+1 – y1 = A yn – y0

)

(Note that y*n= yn) . This can be used to calculate A from known endpoint compositions. A =

yn+1 – y1 *

yn – y0 We might also combine the last two equations to get

( * * ) = An+1 (y1 – y*0 ) + A (y*0 – yn+1) n+1 A (y1 – y*0) * n A (y1 – y0 )

y1 – yn + 1 = A y0 – yn

(*

*

*

) (yn+1 – y*n)

A y0 – yn – y0 + yn+1

= =

Principles of Mass Transfer Operations − I (Vol. − I)

A

n

=

G.5

Appendix – G : The Kremser Equation

(yn+1 – y*n) (y1 – y*0)

yn+1 – y*n  nln(A) = ln   y1 – y*  0  

n =

yn+1 – y*n  ln   y1 – y*  0   ln(A)

yn+1 – y*n  ln   y1 – y*  0   = y – y  1 n+1 ln  * *   y0 – yn 

… (30)

Which can be used to determine the number of stages needed to make a separation. References : 1.

McCabe, W.L., J. C. Smith and P. Harriott, Unit Operations of Chemical Engineering (6th Edition), McGraw-Hill, 2001, pp. 632-38.

2.

Seader, J.D. and E.J. Henley, Separation Process Principles, John Wiley, 1998, pp. 242-46.

,,,

STUDY QUESTIONS To gain the best understanding of the material presented in this book, refer these study questions for each chapter and write in your answers as you read.

Chapter 1 : Overview of Chemical Engineering Profession, Separation Processes and Introduction to Mass Transfer Operations 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30.

What are the two key process operations in chemical engineering ? What are the main auxiliary process operations in chemical engineering ? What is the difference between a block flow diagram and a process flow diagram ? Why do almost all industrial chemical processes contain separation operations ? Is the separation of a mixture a spontaneous, natural process ? What are the five general separation techniques and what do they all have in common ? Why is mass transfer a major factor in separation processes ? What limits the extent to which the separation of a mixture can be achieved ? What are the two agents that can be used to create a second phase in the separation of a chemical mixture ? What is the most common method used to separate two fluid phases ? List five separation operations that use an Energy Separation Agent (ESA) and five that use a Mass Separation Agent (MSA). Give three disadvantages of using an MSA. What is the most widely used industrial separation operation ? What is the difference between absorption and stripping ? What is liquid-liquid extraction, when should it be considered, and how does it differ from supercritical-fluid extraction and leaching ? What is the basic requirement for removing liquid from a wet solid by evaporation ? Why is crystallization important in the semiconductor industry ? Why can't osmosis be used to separate a liquid mixture ? How do reverse osmosis and dialysis differs ? What do they have in common ? In the separation of a mixture, how does the action of a membrane differ from that of a solid adsorbent ? What is the difference between adsorption and absorption ? What is the difference between dialysis and electrodialysis ? The degree of separation in a separation operation is often specified in terms of component recoveries and/or product purities. How do these two differ ? For a separation operation, is it possible to specify the recovery of a feed component and the mole (or mass) fraction of that same component in one of the products ? What factors of a separation operation influence the degree of separation ? When selecting a separation method, what factors are important ? What five separation operations are the most technologically mature ? What are the three uses of thermodynamics in separation operations ? Is energy conserved in a separation process ? In chemical engineering, does mass transfer refer to the relative movement of chemical species in a mixture or the flow rate of the entire mixture ? What are the job responsibilities of Chemical Engineer in Chemical Process Industry (CPI) ? ,,, (S.1)

Principles of Mass Transfer Operations − I (Vol. −I)

S.2

Appendix – S : Study Questions

Chapter 2 : Diffusion Mass Transfer 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

11. 12. 13. 14. 15. 16. 17. 18. 19. 20.

21. 22. 23. 24.

What is meant by diffusion ? Explain the term "FLUX" What are the two basic mechanisms of mass transfer ? When separating chemicals in commercial equipment, which mechanism is preferred ? Why ? Molecular diffusion occurs by any of what four driving forces or potentials ? Which one is the most common ? Can molecular diffusion occur in solids, liquids, and gases ? What is the bulk flow effect in mass transfer ? What is a random-walk process ? Is the mass-transfer rate proportional to the area normal to the direction of mass transfer or to the volume of the mixture ? For a binary mixture, what factors are involved in the rate of diffusion of a component, according to Fick's law ? How does Fick's law of diffusion compare to Fourier's law of heat conduction ? What is the difference between a species diffusion velocity relative to stationary coordinates and relative to the molar average velocity of the mixture that contains the species ? What is meant by equimolar counter diffusion (EMD) ? In what separation operation is it closely approached ? For a binary mixture, under what conditions is the diffusivity of A in B equal to the diffusivity of B in A, and independent of composition ? What is meant by unimolecular diffusion (UMD) ? In what separation operation is it closely approached ? For UMD, are the mole fraction profiles linear or nonlinear with distance ? Does EMD or UMD include the bulk-flow effect ? How does this effect modify Fick's law ? When measuring molecular diffusivities, why must a correction be made for bulk flow ? Distinguish between molecular and eddy diffusion. What is the difference between a mutual diffusion coefficient and a self-diffusion coefficient ? Derive from fundamentals the expression for steady state equimolal counter diffusion of gas A through another gas B Who used the kinetic theory of gases to develop a theoretical equation for estimating molecular binary diffusivities in a gas mixture at low pressures ? Why is the empirical equation of Fuller et al.preferred ? At low pressures, what is the effect of temperature and pressure on the molecular diffusivity of a species in a binary gas mixture ? Draw a graph showing concentration gradient for equimolar diffusion. What is the effect of pressure and temperature on the diffusion coefficient in gases ? Give the relationship between mass transfer coefficient and diffusivity 2

25. What is the order of magnitude of the molecular diffusivity in cm /s for a species in a binary gas mixture at low pressure ? 26. Above about what pressure is a correction to the pressure effect on binary gas diffusivity necessary ? What type of theory is used to make that correction ? 27. In general, for a binary liquid mixture, the diffusivity of A in B and the diffusivity of B in A both depend on composition. Why is this so ?

Principles of Mass Transfer Operations − I (Vol. −I)

S.3

Appendix – S : Study Questions

28. Which equation is useful for estimating diffusivities of solutes in aqueous solutions ? On what theory is that equation based ? 29. What is the order of magnitude of the molecular diffusivity in cm2/s for a species in a liquid mixture ? By how many orders of magnitude is diffusion in a liquid slower or faster than in a gas ? 30. How is liquid diffusivity related to viscosity ? 31. By what mechanisms does diffusion occur in solids ? Can Fick's law be used for diffusion in solids ? 32. What is the order of magnitude of the molecular diffusivity in cm2/s for a species in a solid ? By how many orders of magnitude is diffusion in a solid slower or faster than in a gas ? 33. Derive an expression for finding the mass flux of diffusion of A through non-diffusing B, A and B are liquids. 34. By what mechanisms does diffusion occur in porous solids ? 35. What is the effective diffusivity ? 36. Why is diffusion in crystalline solids much slower than in amorphous solids ? 37. Why is diffusion of very light gases in metals of major importance ? 38. Why is diffusion of light gases in ceramics important ? 39. Why is diffusion of gases and liquids in dense, nonporous polymers important ? 40. For diffusion across the radius of a hollow cylinder, what is the proper average area for mass transfer when applying an integrated form of Fick's law ? 41. For diffusion across the radius of a hollow sphere, what is the proper average area for mass transfer when applying an integrated form of Fick's law ? 42. What is Fick's second law ? How does it compare to Fourier's second law of heat conduction ? 43. When comparing diffusion to heat conduction, is thermal conductivity or thermal diffusivity analogous to molecular diffusivity ? 44. Is the rate of diffusion into a semi-infinite medium inversely proportional to time or inversely proportional to the square root of time ? 45. Is the rate of diffusion into a semi-infinite medium proportional to diffusivity or proportional to the square root of diffusivity ? 46. Molecular diffusion in gases, liquids, and solids ranges from slow to extremely slow. What is the best way to increase the rate of mass transfer in fluids ? What is the best way to increase the rate of mass transfer in solids ? 47. How does the Fourier number for mass transfer differ from the Fourier number for heat transfer ? 48. In laminar flow, does mass transfer in a direction normal to the direction of flow occur by molecular diffusion, eddy diffusion, or both ? 49. Describe a method to estimate the diffusivity of a volatile solvent into air. 50. Give the Wilke-Chang equation and explain the terms involved in it. ,,,

Chapter 3 : Mass Transfer Coefficients 1. 2. 3. 4.

Are the Prandtl number and Schmidt number analogous ? How ? What is the defining equation for a mass-transfer coefficient ? How does it differ from Fick's law ? How is it analogous to Newton's law of cooling ? What is the Sherwood number ? How is it analogous to the Nusselt number ? What does the entry length for flow in a circular tube mean ?

Principles of Mass Transfer Operations − I (Vol. −I) 5. 6. 7. 8. 9. 10. 11.

12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35.

S.4

Appendix – S : Study Questions

For mass transfer to or from the wall of a tube through which a fluid flows, is the appropriate driving force an arithmetic or log-mean average ? Explain what do you understand by jD and jH factors. What is the eddy mixing length of Prandtl ? What is the difference between Reynolds’s analogy and the Chilton- Colburn analogy ? Which is more useful ? What is the basic difference between the development of the Chilton- Colburn analogy and the Prandtl analogy ? What important result is given by the penetration theory ? For mass transfer across a phase interface, what is the difference between the film, penetration, and surface-renewal theories, particularly with respect to the dependence on diffusivity ? What is the two-film theory of Whitman ? Is equilibrium assumed to exist at the interface of two phases ? Why are expressions for overall mass-transfer coefficients more complex than expressions for overall heat-transfer coefficients ? Explain the two-film theory of interphase mass transfer. Discuss the reasons for modifying this theory later, by Higle and Dauckwerts. State Higbie's theory for mass transfer and the assumptions made in deriving the expression for mass transfer coefficient. Discuss the Reynolds analogy of heat and momentum transport. How Prandtl and Von Karman rectify them in their analogies ? Discuss the concept of the transfer units used in mass transfer problems. How is overall height of a transfer unit related to the individual heights of transfer unit ? In laminar flow, does mass transfer in a direction normal to the direction of flow occur by molecular diffusion, eddy diffusion, or both ? Are the Prandtl number and Schmidt number analogous ? How ? What is the Peclet number for mass transfer ? What is the defining equation for a mass-transfer coefficient ? How does it differ from Fick's law ? How is it analogous to Newton's law of cooling ? What is the Sherwood number ? How is it analogous to the Nusselt number ? Do velocity, temperature, and concentration boundary layers on a flat plate all build up at the same rate ? If not, why not ? What does the entry length for flow in a circular tube mean ? For mass transfer to or from the wall of a tube through which a fluid flows, is the appropriate driving force an arithmetic or log-mean average ? For laminar flow, can expressions for the mass-transfer coefficient can be determined from theory using Fick's law ? What is the significance of the j-factors for momentum, heat, and mass transfer ? For mass transfer across a phase interface, what is the difference between the film, penetration, and surface-renewal theories, particularly with respect to the dependence on diffusivity ? Is equilibrium assumed to exist at the interface of two phases ? Discuss the operation principles of Wetted Wall tower and Ventury Scrubber. What is momentum, thermal and mass diffusivity ? Are they analogues to each other ? Define Prandtl number, Schmidt number and Lewis number. What is Buckingham Pi Theorem ? How it is useful for dimensional analysis ? What are equations for Taylor-Prandtl analogy for heat and mass transfer ? ,,,

Principles of Mass Transfer Operations − I (Vol. −I)

S.5

Appendix – S : Study Questions

Chapter 4 : Interphase Mass Transfer 1. 2. 3.

4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15.

Discuss the need for defining the interphase mass transfer resistance in terms of overall coefficients. How they are related to individual coefficients ? Discuss the importance of equilibrium in interphase mass transfer. Write the material balance equations for following operations: (a) Steady-state co-current operations (b) Steady-state countercurrent operations (c) Cascades Define J factors for heat and mass transfer. State its applications and limitations. How will define Murphree stage efficiency ? What do you understand by 'liquid film controlling' in gas absorption ? Suggest methods of increasing the rate of transfer for this condition. Derive Kremser-Brown-Souders equation for the calculation of number of theoretical stages for absorption in a stage wise contact tower. When are the concepts of HETP and HTU used ? What is the difference between the two ? Drawing an equilibrium diagram, indicate clearly the two concepts. Illustrate the significance of operating line and equilibrium line for a steady state counter-current process. Discuss the use of Murphree efficiency for the design of an absorber. Discuss briefly how the minimum solvent requirement in counter-current gas absorption may be determined. Explain the term stage and stage efficiency. What is a separation cascade ? What is a hybrid system ? What is the difference between a countercurrent and a crosscurrent cascade ? What is the limitation of a single-section cascade ? Does a two-section cascade overcome this limitation ? ,,,

Chapter 5 : Gas Absorption 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16.

In absorption, what is a solute ? Can absorption be accompanied by a temperature change ? Why ? What is the difference between physical absorption and chemical (reactive) absorption ? What is the difference between an equilibrium-based and a rate-based calculation procedure ? What is a trayed tower ? What is a packed column ? What is the difference between entrainment and occlusion ? What are the three most common types of openings in trays for the passage of vapor ? Which of the three is rarely specified for new installations ? In a trayed tower, what are weirs and downcomers ? In a trayed tower, what do flooding and weeping mean ? Why are liquid distributors and redistributors necessary in a packed column, but not in a trayed tower ? What is the difference between random and structured packings ? What are the four generations of column packings ? What are the advantages of the more-expensive structured packings over the random packings ? For what conditions is a packed column favored over a trayed tower ? Why do most retrofits involve the replacement of trays with packing ? What factors must be considered when designing an absorber or a stripper ?

Principles of Mass Transfer Operations − I (Vol. −I)

S.6

Appendix – S : Study Questions

17. What are the characteristics for an ideal absorbent ? What are the characteristics of an ideal stripping agent ? 18. In general, why should the operating pressure be high and the operating temperature is low for an absorber, and the opposite for a stripper ? 19. For a given recovery of a key component in an absorber or stripper, does a minimum absorbent or stripping agent flow rate exist for a tower or column with an infinite number of equilibrium stages ? 20. When using a graphical method for determining the number of equilibrium stages required for an absorber or stripper, when is it advisable to express solute concentrations in mole ratios, rather than mole fractions ? 21. What is an operating-line equation ? 22. What is the difference between an operating line and an equilibrium curve ? 23. On a Y-X plot for an absorber, is the operating line above or below the equilibrium curve ? Explain. 24. On a Y-X plot for a stripper, is the operating line above or below the equilibrium curve ? Explain. 25. What is a reasonable value for the optimal absorption factor when designing an absorber ? Does that same value apply to the optimal stripping factor when designing a stripper ? 26. When stepping off stages on a Y-X plot for an absorber or a stripper, does the process start and stop with the operating line or the equilibrium curve ? 27. When should the use of an algebraic method, rather than a graphical method, be considered for determining stage requirements for an absorber or stripper ? 28. At near ambient pressure, what four methods should be considered for estimating the Kvalue of a solute ? 29. Is stage efficiency usually based on heat transfer, mass transfer, or a combination of the two ? 30. What is the definition of Lewis for the overall stage efficiency ? 31. What are the main factors that influence the overall stage efficiency ? 32. What are the four methods for predicting stage efficiency ? 33. In general, do absorbers or strippers have the higher stage efficiency ? Why ? 34. Why do longer liquid-flow paths give higher stage efficiencies ? 35. Why do large diameter towers often use multiple-pass trays ? 36. What are the assumptions in the Murphree vapor (tray) efficiency ? 37. What is the difference between the Murphree tray and point efficiencies ? 38. Under what conditions is the Murphree vapor tray efficiency equal to the Lewis overall stage efficiency ? 39. What kind of a laboratory column is used to obtain data suitable for scale-up to commercial columns ? 40. What are the four conditions for ideal stable operation of a trayed column ? 41. What is the difference between downcomer flooding and entrainment flooding ? Which is the more common type of flooding ? 42. Why does the flooding velocity depend on the tray spacing ? 43. Why does foaming tendency affect the flooding velocity ? 44. Is surface tension a major factor in determining flooding velocity ? 45. What are the major factors that determine tower diameter ? 46. What is meant by turndown ratio ? What type of tray has the best turndown ratio ? Which, the worst ? 47. How do the new high-capacity trays differ from conventional trays ?

Principles of Mass Transfer Operations − I (Vol. −I)

S.7

Appendix – S : Study Questions

48. What are the three contributing factors to the vapor pressure drop across a tray ? 49. Does vapor velocity affect both vapor-phase and liquid-phase mass transfer coefficients ? 50. Can the Chan-Fair method be used to predict both vapor-phase and liquid-phase masstransfer coefficients ? 51. Why does the vapor-phase mass-transfer coefficient decrease as the flooding velocity is approached ? 52. For what type of tray is weeping potential the most serious ? 53. Can the rate of liquid entrainment be estimated ? If so, how ? 54. Why is the clear liquid head in the downcomer greater than the vapor pressure drop per tray expressed as a liquid head ? 55. What is the HETP ? Does it have a theoretical basis ? If not, why is it so widely used ? 56. What are typical values of HETP for random packings and for structured packings ? 57. Under what conditions is the HETP equal to the HTU, and the number of theoretical stages equal to the NTU ? 58. Why are there so many different kinds of mass-transfer coefficients ? How can they be distinguished ? 59. What is the loading point in a packed column ? 60. What is the flooding point in a packed column ? 61. At low superficial gas velocities, does the holdup depend on gas velocity or just liquid rate ? 62. For operation below the loading point, what is the relationship between the pressure drop for vapor flow, but without liquid flow, and that for vapor flow with liquid flow ? 63. What is the loading region ? Is it best to operate a packed column in that region or in the pre-loading region ? 64. Can a packed column be operated efficiently without achieving complete wetting of the packing with liquid ? 65. What is an approximate value of the pressure drop in a packed column when flooding is closely approached ? 66. Are liquid density and viscosity important in determining flooding velocity in a packed column ? 67. What is the most widely used correlation for estimating dry-bed pressure drop ? 68. How does the HETP vary with the F-factor ? 69. Which has the greater influence : (a) the gas velocity on the liquid-phase volumetric mass-transfer coefficient, or (b) the liquid velocity on the gas phase volumetric masstransfer coefficient ? 70. When the solute concentration is moderate to high, instead of dilute, why are calculations for packed columns much more difficult ? 71. What is the Kremser method ? To what type of separation operations are it applicable ? What are the major assumptions of the method ? 72. What is an absorption factor ? What is a stripping factor ? 73. Why is a stripper frequently coupled with an absorber ? 74. When solving a separation problem, are the number and kind of specifications obvious ? If not, how can the required number of specifications be determined ? 75. What is the number of degrees of freedom for an N-stage countercurrent cascade of nonadiabatic equilibrium stages ? Can the degrees of freedom be determined for a hybrid system ? If so, what is the easiest way to do it ? ,,,

Principles of Mass Transfer Operations − I (Vol. −I)

S.8

Appendix – S : Study Questions

Chapter 6 : Humidification and Dehumidification Operations 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25.

What do you mean by humidification and dehumidification operations ? What is adiabatic saturation process ? What is a adiabatic saturation temperature ? Can you derive the final expression for the same ? What are common uses of gas liquid contact in industry ? Derive an expression to determine height of cooling tower. What are the assumptions in this derivation ? What is the blow down losses in cooling tower ? Discuss the significance and application of psychometric charts in simultaneous heat/mass transfer operations. Discuss the theory of wet-bulb thermometry. Under what conditions the wet-bulb temperature and adiabatic saturation temperature will be identical. Discuss in detail the design of a cooling tower based on enthalpy transfer unit concept. State the important assumptions made and the limitations in application. Describe the various types of cooling towers employed in cooling hot water from process plants. State and derive Lewis relationship. What are its applications and limitations ? What is psychometric ratio ? Indicate the methods by which the mass transfer coefficient and heat transfer coefficient are obtained when Lewis relationship fails. Define humidity What is meant by percentage humidity ? How the cooling effect in a cooling tower can be increased ? Describe the methods available for estimating humidity of a sample of air. What are the different types of cooling towers used in industries ? Briefly explain them. Explain the theory of humidification. Define all the humidity terms you know. What is Lewis relation ? What is value of Lewis number for air-water vapor system ? What is meant by wet bulb temperature approach ? How you calculate total losses in cooling towers ? What are the design considerations for cooling towers ? What is the principle of 'recirculating liquid gas humidifier' ? Define Dew point Explain the theory of adiabatic saturation curves and wet bulb temperature theory. ,,,

Chapter 7 : Equipments for Gas-Liquid Operations 1.

What are the examples of gas-liquid operations that are carried out in the chemical industry ? 2. What is sparger ? What are the typical industrial applications of the same ? 3. What is gas hold up ? How will calculate slip velocity for countercurrent flow of liquid ? 4. What are the types of impellers ? 5. What is vortex formation ? How will you prevent ? 6. What are baffles ? What is the purpose ? 7. What are the similarity considerations in agitated vessels ? 8. What is geometric similarity ? 9. What is kinematic similarity ? 10. What is dynamic similarity ? 11. What are priming, coning, weeping, and dumping ?

Principles of Mass Transfer Operations − I (Vol. −I) 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30.

S.9

Appendix – S : Study Questions

What is operating characterstics of sieve trays ? What are different schemes for directing the liquid flow ? What is turndown ratio ? What are the types of proprietary trays ? What are valve and counterflow trays ? What is entrainment ? Is it advantages ? What is purpose of Redistributor plate ? What is purpose of entrainment eliminators ? What are the types of packing ? What are the desirable properties of good packings ? Why packing restrainers are used in packed towers ? What is a design criterion for packed towers ? Compare tray towers and packed towers What is Ventury scrubber ? What are the applications ? What is wetted wall tower ? State applications What is spray tower ? How it works ? Give merits and demerits of spray towers. What are spray chambers ? What are the applications ? What are the advantages and disadvantages of packed towers ? Give a typical arrangement for Bubble cap tray. ,,,

Chapter 8 : Drying Operations 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22.

What are some of the industrial applications of drying ? Discuss the purposes of drying operations. How will express moisture content on dry basis and dry basis ? What is basic difference between these two moisture content ? What are types of moisture ? What are the equipments available for drying operations ? What are the merits and demerits ? How will select a particular dryer for given process ? What are the general principles for efficient energy utilization in drying operations ? What are the most commonly employed modes of heat transfer for drying ? Does the temperature of the solid during drying depend on the mode ? Why is there such a large variety of drying equipment ? When is a batch dryer usually preferred over a continuous dryer ? What is the difference between a direct-heat dryer and an indirect-heat dryer ? For a tray dryer, under what conditions can hot air be passed through the solids ? When must the hot air be passed over the solids in the tray ? When are agitated batch dryers used ? What types of continuous dryers can be used with wet, solid particles ? What is the difference between a direct-heat and an indirect-heat rotary dryer ? For what types of wet solids can fluidized-bed, spouted-bed, and pneumatic-conveyor dryers be used ? What types of dryers can be used with slurries ? How do infrared, dielectric, and microwave drying differ from direct heat drying ? What is freeze-drying and when is it a good choice ? What is psychrometry ? What are the differences among absolute humidity, relative humidity, and percentage humidity ? What is the wet-bulb temperature ? How is it measured ? How does it differ from the dry-bulb temperature ?

Principles of Mass Transfer Operations − I (Vol. −I)

S.10

Appendix – S : Study Questions

23. What is the adiabatic-saturation temperature ? Why is it almost identical to the wet-bulb temperature for the air-water system ? 24. For systems other than air-water, why is the wet-bulb temperature always higher than the adiabatic-saturation temperature ? 25. What is the psychrometric ratio ? How is it related to the Lewis number ? What is the Lewis number ? 26. Under what drying conditions is the moisture evaporation temperature equal to the wetbulb temperature ? 27. Distinguish between: (a) Total-moisture content (b) Free-moisture content (c) Equilibrium-moisture content (d) Unbound moisture (e) Bound moisture 28. How does the equilibrium-moisture content depend on temperature and relative humidity ? 29. How does hysteresis affect equilibrium-moisture content ? 30. What are the different periods that may occur during a drying operation ? 31. Under what conditions is a constant-rate-drying period observed ? 32. Under what conditions is a falling-rate-drying period observed ? 33. What is the critical-moisture content ? 34. What are the two most applied theories to the falling-rate-drying period ? 35. In the dryer models for a belt dryer with through-circulation and a direct-heat rotary dryer, is the rate of drying based on heat transfer or mass transfer ? 36. What is achieved by using a multiple-zone, through-circulation belt dryer with hot air flow reversal between zones ? 37. What is final equation for total time required for drying operation ? 38. What are the final expressions for drying time for droplets ? Can you derive these equations ? What are the assumptions ? 39. What are the regimes of fluidization of a bed of particles by a gas ? 40. What regime of operation is preferred for drying ? 41. Find an expression for obtaining the total drying time for the drying of a wet solid material under constant drying conditions of to final moisture content well below the critical moisture content. 42. List and explain the different ways in which the drying rate of a given substance in the constant rate period may be increased. 43. Giving reasons, indicate the type of dryer that can be used for the following purpose : (a) Removal of the last 6 percent moisture from salt (b) Drying of cakes of soap (c) Drying of heat sensitive materials like pharmaceuticals (d) Drying of paddy 44. Which drier is suitable for handling fragile crystals ? 45. Define the term Equilibrium moisture and free moisture content of solid 46. What is freeze-drying ? 47. What is meant by holdup in a rotary dryer ? 48. Which type of drier is used in the manufacture of (a) tablets (b) Paraffin wax ? 49. Define the term "Bound moisture" 50. Describe the method of obtaining the length and diameter of a rotary dryer. ,,,