Heat And Mass Transfer

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[ ANNA UNIVERSITY SYLLABUS

I

_

ME1351 : HEAT AND MASS TRANSFER

_

hanical Engineering -(R'egulation For B.E. VI Semester Mec ----

CONTENTS

:201l4f,

l. CO~OU,~TlONt_

Mechanism of Heat Transfer - Conduction BaSIC Concep!) ... f ' . d Radiation - General Differential equation 0 Heat Convection an .. d C I' . y mdncal Con ducii UCllon - FOllrl'er Law of Conduction - Cartesian an . Steady State Heat . Conduction C oor d·lilates - One Dimensional Conduction through Plane Wall, Cylinders and Spherical Systems _ Composite Systems - Conduction with Internal Heat Generation _ Extended Surfaces - Unsteady Heat Conduction - Lumped Analysis _ Use of Heislers Chart. 2. CONVECTION Basic Concepts - Convective Heat Transfer Coefficients - Boundary Layer Concept - Types of Convection - Forced Convection _ Dimensional Analysis - External Flow - Flow over Plates, Cylinders and Spheres - Internal Flow - Laminar and Turbulent Flow - Combined Laminar and Turbulent - Flow over Bank of tubes - Free Convection _ Dimensional Analysis - Flow over vertical plate, Horizontal plate, Inclined plate, Cylinders and Spheres. 3. PHASE CHANGE HEAT TRANSFER AND HEAT EXCHANGERS Nusselts theory of condensation - Pool boiling, flow boiling, correlations in boiling and condensation. Types of Heat Exchangers _ LMTD Method of Heat Exchanger Analysis -- Effectiveness _ NTU method of Heat Exchanger Analysis - Overall Heat Transfer Coefficient _ Fouling Factors. 4. RADIATION

CHAPTER 1: CONDUCTION

~H;at .

1.1.4. 1.1.5. 1.1.6. 1.1.7. 1.1.8. 1.1.9. 1.1.10. 1.1.11. 1.1.12. 1.1.13. 1.1.14. 1.1.15. 1.2.

1.3.

1.3.5.

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Data

Book

is

1.3.6.

1.4.

1.1

Modes of Heat Transfer _ I. I Fourier Law of Conduction .. .1.2 General Heat Conduction Equation in Cartesian Co-ordinates . 1.2 General Heat Conduction Equation in Cylindrical Co-ordinates 1.9 Conduction of Heat through a Slab or Plane Wall.. .1.14 Conduction of Heat through a Hollow Cylinder 1.16 Conduction of Heat through a Hollow Sphere 1.17 Newton's Law ofCooling........................ . 1.19 Heat Transfer through a Composite Plane Wall with inside and Outside Convection 1.19 Heat Transfer through Composite Pipes (or) Cylinders with Inside and Outside Convection 1.22 Solved Problems 0" Slabs 1.25 Soilled University Problems 011 Slabs 1.74 Solved Problems 011 Cylinders 1.111 Solved University Problems 011 Cylinders 1.144 SO/lied Problems 011 Hollow Sphere 1.160 Radius

of Insulation

Critical Radius of Insulation Solved Problems

Heat Conduction

1.3.1. 1.3.2. 1.3.3. 1.3.4.

Basic Concepts - Diffusion Mass Transfer - Fick's law of diffusion _ Steady State Molecular Diffusion - Convective Mass Transfer _ Momentum, Heat and Mass Transfer Analogy _ Convective Mass Transfer Correlations. Transfer

Critical I? I

1:2:2:

Basic Concepts, Laws of Radiation - Stefan Boltzman Law, Kirchoff Law - Black Body Radiation - Grey Body Radiation Shape Factor Algebra - Electrical Analogy - Radiation Shields - Introduction to Gas Radiation. 5. MASS TRANSFER

Note : . (Use of Standard Heat and Mass pernulled 117 the University Examination).

Transfer

1.1.1. 1.1.2. 1.1.3.

1.167

for a Cylinder

1.167 1.169 1.179

with Heat Generation

Plane Wall with Internal Heat Generation Cylinder with Internal Heat Generation Internal Heat Generation - Formulae Used Solved Problems 011 Plane Willi with Internal Heat Generation Solved Problems 011 Cylinder with Internal Heat Generation Solved Problems Oil Sphere with lnternul Heat Generation

1.179 1.183 1.185 1.187 1.196 1.202

T;~~·~·~t~·Fi;~~·::::::::· ..::::·.:·.·.:::·.::::::::::::::: ..::=:'::::::: ::~~~

~.i:SI. 1.4.2. Temperature Distribution and Heat Dissipation in Fin 1.4.3. Application......... .

.

1.206 1.21

r.: 1.4.'-l.

Fill Ftliciellc)'

·· .

..1.217

1.~.5. Fin rfkcriVt'ness. 1.~.6. Ftll"lllllllicUsed..... I A. 7. So/I't!d Proh/ellH" 1.4.8. SII/I'd U"itl(!f.5i~I' Prublctn« 1.4.9. Pr()hkllll/Or Practice ················ ..·

15.

Transient Heat Condul~tioll (or) Unsteady Conduction

" .. 1.217 1.2IS

.

1.219 1.245 1.263 1.264 1.264 1.266 1.269

tteot AII(I~I'jiJ ........•..........•..•......••....•...•..•.........•.•...••. Heat Flow in Semi-lnfiutie Solids SO/lied Problems - Semi-illfillite Transient Heat Flow in an Infillite

1.288

2.7. 2.8.

... 1.329

1.5.8.

1.332

I.S.9.

1.351

2.9.

1.374

2.1.1.

HEAT TRANSFER -..-..-..-..-..-..- -..-..-..-..-..-..-

Dimensions

... 2.1

2 1.2. Buckingham 1I Theorem. . .. 2.1.3. Advantages cf Dimensional Analysis

. .

2.14. Limitations of Dimensional Analysis Dimensionless Numbers and their Physical

2.2.

2.11.

2.2.1. Reynolds Number (Re) 2.2.2. Prandrl Number (Pr) 2.2.3. Nusselt Number (Nu)

2.4 2.4 2.5

2.11.3. 2.12.

~e\Vlonion and Non-Newtollioll

Fluids

2.6

EL;;;~~~:~~::;~p·:~:.::::·:::::.::::~~~::: ..:::::::::::::::::::~~::::::::::::::~:7;

2.3.

L~;~;~·:::············································";'8

2.4.

2 ~ I. Types of Boundary 2.:.~. !iydrodynalllic Boundary 2.).). r~lenn;]IUoUfldarylayer i~lIlve~~~:lt~;;:~

2

-u.

1.~··~·r:·· y .. ·

'l'~~'"rC' ..···..·..:

Types ofC~nveoc!i

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······· .. ······· .. ····2·9 ····· 2·9

,..:::::::::::::::::::::::::::::::::2: 9

onveC!rOfl 011....

··

.

2.9 2.9

for Combination Flow...

2.13 of

Formulue Used lor Flow Over Balik of Tubes Solved Problem it

Cylinder

-Internal

Flow

Formulae usedfor Flow tit rough Cylinders (lnternul flow) S;)lved Problems - Flow through Cylinders (lnteruat Flow) Solved University Problems - Internal Flow

Formulae Used/or Free Convection Solved Problems 011 Free Convection (or) Natural Convection 2.12.3. Solved University Problems - Free Convection Problems for Practice TII'o Murk Questions {lilt! Allswers

CHAPTER III: 3.1.

2.10

Free Convection 2.12.1. 2.12.2.

2.13. 2.14.

for

and Spheres

~;~~t~::t~~r~,~~::; (~~?••.•.••••••••••••••.•••••••••••.•••••••••.•.••••.•.• ;; 2.2.6.

Coefficients

Formulae Usedfer Flow Over Cylinders and Spheres Solved Problems - Flow Over Cylinders

Flow through

2.11.2.

2.4

for

Flow over 'lalli, of Tubes

2.11.1.

2.3

Significance

Coefricients

Problems 011 Flat Surfaces - Forced Convection Solved University Problems 011 Flat Surfaces Forced Convection

How over Cylinders

2.10.1. 2.10.2.

2.2 2.3

2.9 ..2.10

Laminar and Turbulent .. 2.15 Layer Thickness, Shear Stress and Skin Friction Coefficient for Turbulent Flow 2.IR Heat Transfer 1'1'0111 Flat Surfaces - Formulae Used 2.23

2.9.2.

:~~

ocificient

.

Boundary

2.9.1.

2.10.

CHAPTER II : CONVECTIVE 2.1. I)i 111(~ns iona I A IIa lysis

l leat Transfer

( .3

..

Local and Average Heat Transfer Plate - Laminar Flow Local and Average Heat. Transfer Plate-Turhulcnt Flow

2.8.1. 2.8.2.

1.30R

Solids Plate

The Flat The Flat

Free (or) Natural Convection Forced Convection

2.6.1.

1.306

Solved Problems - lufintie Slllitl,· SO/lied University Problem" - Infinite Solids 1'11'0Mark QlleJtiOlIl & AII.'"II'en

1.6.

2.5. 2.6.

State

1.5. I. l3ior Number . . 1.5.:? Lumped Heat Anal)' is . 1.5.3. Solved Problems -1_llIlIped lleat AII(I~I/JiJ ....•.........•. '.5.4. Solllcd University Prohlelll.,·-Llllllped I. -.S 1.5.6. 1.5.7.

2.4.3. 2.4.4.

( 'onteuts

2.26 2.83 2.115 2.116 2.117 2. 122 2.123 2.124 2.126 2.127 2.129 2.150 2.162 2.162 2.165 2.194 2.217 2.219

PHASE CHANGE HEAT TRANSFER AND HEAT EXCHANGERS Boiling and Condensation ~.I 3.1.1. Introduction . ).1 .... 3.1 Boiling . 3.1.2. Condensation 3.1 3.1.3. . . .3.1 Applications . 3. 1.4.

C.4

Heat and Mass Tram/a

Contents

3.1.5. 3.1.6. 3.1.7.

Boilll1g Heat Transfer Phenomena Flow Boiling... ········ .. · · Boiling Correlations

3.2 3.4

3.1.8. 3.1.9.

Solved Prohlellls Solved A11IUIUniversity

3.1 10. 3.1.11. 3.1.12. 3.1.13. 3.1.14. 3.1.15.

Condensation. . Modes of Condensation · Filmwise Condensation Dropwise Condensation ..· Nusselt's Theory for Film Condensation Correlation for Filmwise Condensation Process

J.)

3.7 3.23

Problems

3.29 3.29 3.29 3.30 3.30 3.30

3.1.16.

3.2.

Solved Problems Oil Laminar Flow, Vertical Surfaces 3.1.17. Solved Problems Oil Laminar Flow, Horizontal Surfaces 3.1.18. Solved Anna University Problems 3.1.19. Problems for Practice Heat Exchangers

3.2.1. 3.2.2. 3.2.3. 3.2.4. 3.2.5. 3.2.6. 3.2.7. 3.2.8. 3.2.9.

3.32 3.54 3.61 3.65 3.66

Introduction Type of Heat Exchangers Logarithmic Mean Temperature Difference (LMTD) Assumptions Logarithmic Mean Temperature Difference for Parallel Flow Logarithmic Mean Temperature Difference for Counter Flow Fouling Factors Effectiveness by Using Number of Transfer Units (NTU)

32I

3.73 3.77 3.81

Shell and Tube Heal Exchangers Solved Anna UI1iversity Problems Solved Problems Oil NT(! Method m1 ; b"1University Solved Problems ro ems for Practice Two M k . ..· ..·..· · ur Questions and AI1swers

Introduction Emission Properties

3.82

4.29.

Electrical Network by Using Radiosity

3.1 09 3.117 3.J24 3.138

4.30. 4.31.

Radiation of Heat Exchange Solved Problems

3.145 3.146

.. ·

·

·

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·

·

Emissive Power 4.1 Monochromatic Emissive Power 4.2 Absorption, Reflection and Transmission 4.2 Concept of Black Body 4.3 Planck's Distribution Law 4.4 Wien's Displacement Law 4.4 Stefan-Boltzmann Law · 4.5 Maximum Emissive Power 4.5 Emissivity 4.6 Gray Body 4.6 Kirchoff's Law of Radiation 4.6 Intensity of Radiation 4.6 Lambert's Cosine Law 4.7 4.16. Formulae Used 4.7 4.17. Solved Problems 4.8 4.18. Solved University Problems 4.25 4.19. Radiation Exchange Between Surfaces 4.31 4.20. Radiation Exchange Between Two Black Surfaces separated by a Non-absorbing Medium 4.31 4.21. Sha pe Factor 4.36 4.22. Shape Factor Algebra 4.36 4.23. Heat Exchange Between Two Non-Black (Gray) Parallel Planes 4.37 4.24. Heat Exchange Between Two Large Cocnentric Cylinders or Spheres 4.41 4.25. Radia tion Shield 4.45

4.26. Solved Problems 4.27. Solved Problems 011 Radiation Shield 4.28. Solved University Problems

CHAPTER IV : RADIA nON 4.1. 4.2.

4.3. 4.4. 4.5. 4.6. 4.7. 4.8. 4.9. 4.10. 4.11. 4.12. 4.13. 4.14. 4.15.

3.82

Problems on Parallel Flow and Counter

Flow Heat £\:cllangers , 3.2.10. Problems 011 Cross Flow Heal Exchangers (or) 3'2'1~' 3'2'13' 3'2' 14' " . 3.2.15.

3.66 3.66 3.73 3.73

4.1 4.1

C.5

Analogy for Thermal and Irradiation for Three

4.32.

University

Solved Problems

4.33.

Radiation

from Gases and Vapours

4.34.

M ea n Bea m Length Problems

4.49 4.60 Radiation Gray Surfaces

4.79 Systems 4.IOO 4. 104 4.105 4.129 4.153 4.154

4.35.

Solved

4.36.

Problems for Practice

· 4.155 4.166

4.37.

Two Mark. Qlte.5tiOl1!iand Al1swers

4.168

~C~.6~~R~ea~t~a~n~d~~~a~s~s~r.~ra~n~sfi~e_r

-

=C-H-AP-T-E-R-V-:~M7.A~S~S~T~RA~NS~F~E~R~-----------------5.1. 5.2. 5.3. 5.4. 5.5. 5.6. 5.7. 5.8. 5.9. ~.I O. ~.II.

5.12. 5.13. 5.14. 5.15. 5.I6. 5.17. 5. J 8. 5.19. 5.20. 5.21. 5.22. 5.23. 5.24. 5.25. 5.26. 5.27. 5.28. 5.29. 5.30. 5.31.

J ntroductlon · . Modes of Mass Transfer ·..· · · ·..· Diffusion Mass Transfer ..· ·..· ·· · · · Molecu~ar ~iffusion ·..· · · ·..·..· · Eddy Dlffuslon Convection Mass Transfer ·..· · Cocentrations ·· · ·..·· ·· ·..··..·· Fick'~ Law of Diffusion ·..·..·..· Steady State Diffusion through a Plane Membrane So/J'ed Problems Oil Concentrations Solved Problems Oil Membrane Solved Univeristy Problems on Membrane Steady State Equimolar Counter Diffusion Solved Problems Oil Equimolar Counter Diffusion Solved University Problems 011 Equimolar Counter Diffusion Isothermal Evaporation of Water into Air Solved Problems on Isothermal Evaporation of Water into Air Solved University Problems Oil Isothermal Evaporation of Water into Air Convective Mass Transfer Types of Convective Mass Transfer Free Convective Mass Transfer Forced Convective Mass Transfer Significance of Dimensionless Groups Formulae Used for Flat Plate Problem.') Solved Problems on Flat Plate Anna University Solved Problems 011 Flat Plate Formulue Used for Internal Flow Problems Solved Problems on Intemal Flow University Solved Problems Problems for Practice Two Mark Questions and Answers

------

5.1 S.1 S.1 5.2 5.2 5.1 5.2 5.3 5.4 5.6 5.17 5.21 5.23 5.26

cr

Basic Concepts

CF

General Differential Equation

5.31 5.34 0"

Fourier Law of Conduction

C7

Internal Heat Generation

c:r

Extended Surfaces

c-

Unsteady Heat Conduction

cr

Solved Problems

(7'

Solved University

5.35 5.44 S.54 5.54 5.54 5.S4 5.54 5.56 5.57 5.65 5.68 5.69

5.72 5.75 5.76

ANNA UNIVERSITY SOLVED QUESTION PAPERS ........ S.1 - S.71

DO

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Chapter 1: Conduction

Problems

CHAPTER-I 1.CONDUCTION 1.1 HEAT TRANSFER Heat transfer can be defined as the transmission from one region to another region due to temperature

of energy

difference.

1.1.1 Modes of Heat Transfer

* * *

Conduction Convection Radiation

Conduction Heat conduction is a mechanism of heat transfer from a region of high temperature to a region of low temperature within a medium (solid, liquid or gases) or between different medium in direct physical contact. In conduction, energy exchange takes place by the kinematic motion or direct impact of molecules. Pure conduction is found only in solids.

Convection Convection is a process of heat transfer that will occur between a solid surface and a fluid medium when they are at different temperatures. Convection

is possible only in the presence offluid

medium.

Radiation The heat transfer from one body to another without any transmitting medium is known as radiation. It is an electromagnetic wave phenomenon.

'2

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I.': Heat tnd

HII.\.\'

1.1.2 Fourier

tal"

Transfer Conduction

or Conduction

Rate of heal conduction is proponional 10 the area mea lIred to the iirection of heat OO\V and io the temperature gradient

1.3

O.

1101'111:11

in that direction. Q O.

°C.·Ch)

Element volume

\ here A - Area in

111-

dT _ Temperature gradient in k/m dr k - Thermal nducti iry in W/m"Thermal conducti to conduct heat.

Fig. 1.1.

it)' is defined a the abilit

fa

ub tan e

[The negative sign indicates that the h at 0 w in a dire ti along which there is a decrease in temperature] 1.1.3 General heat conduction cartesian coordinates

equation

Consider a small rectangular

Net

heat

conducted

into

element

q¥

dx

be the heat flux in a direction

The rate f heat' fl '"nine. t th the face AB 0 i fide

all the

coordinate

directions.

Let q x be the heat flux in a direction n

in

element

from

f face EF

and

H.

e Iernent .In x diirection

I Q,

dx, d I and

of face ABO

dz

through

I

... (1.2

d: as shown in Fig.I.I.

where The energ balance of this rectangular from first law of thermodynam ics.

=>

Net heat conducted into element from all the coordinate directions

l

Heat generated

element

element j

t

Heat st red =

l

hermal

The rate heat fl \ the fa e EFGH i

Q

in rhe elern nt../ ...

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k

nducti

btain d ernperature

1

\\ ithin the

i

1.1

+dx

ity, W/mK

gradient

f tJre e Iernent In . x directi

ut

Q

ax

-k

-d x

) dx

aT

I

d:

n

thr

ugh

/4 Heata~_

..

Subtracting ( 1.2) - (1.3)

Ox -

= -k

Net heat conducted directions

. dydz= l. .~.QI ox

aT dydz _I-k x

Q(I' + dxl

Conduction 1.5

~

AX

~[ :.[ ~

er ax

= ·-k..t

dydz

kx :]

+

[kx aT Jdx dy dZ]

ax

ax

into element from all the coordinate

M

ky :]

+

![

k, :]

] dx dy dz ... (1.7)

of + kx -8

dydz +

x

Heat Stored in the element We know that,

=>

Q _Q .I'

(.I'

+ dx)

or] dx dy dz

= .1_ ax [kx ax

{ ... (1.4)

He~t stored} m the element

Mass Of} { SpeCifiC} the x heat of the element element

= {

m

x Cp?<

!

Similarly Q)' .- Q (y + (M

= ~

[k

y

:;]

dx

dy dz

{

Rise in } temperature of element

aT at

,. .•. (1.5)

x

aT

p x dx dy dz x Cp x

at

[v Mass ••• (1.6) Heat stored in } { the element = p Cp

= Density x Volume]

er

at dx

dy dz

... (1.8)

Adding (1.4) + (1.5) + (1.6)

Net heat conducted =

a~ [k'l:

g: Jdt

Heat generated within the element dy dz +

Heat generated within the element is given by

Q

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=

q dx

dy dz

... (1.9)

1.6 Heal ami Mass Tran.~ler Substituting (\'1) ~

eqllation

I;_ [k\

(1.7), (1.8) and (1.9) in equation

~,\ 1 + 0'.... c [k,.

..

c~] a~ [k: 0·

0_

or at

I.?

Case (i) : No heat sources dx

L~l'

In the absence

dz reduces

P Cp

+ q dx dy dz

J]

~

+

Conduction

{I.I) of internal

heat generation,

equation

(1.10)

to

02r ax2

dx dy dz

02r 02r 0,2 az2

or

+-+-

This equation equation.

oc

..•

at

is known as diffusion

equation

(1.11 )

(or) Fourier's

Case (ii) : Steady state conditions Considering

k,

the material =

ky = k, = k

is isotropic.

=

In steady

So,

with time.

constant.

reduces

&r &r -~r] -+-+-k+q=pCax2 0~ c:z2

P

or at

to

So,

state condition,

or at'

iJ1r

in

q k·'

p Cp =--

or

or

It is a general three dimensional in cartesian coordinates

through

.

02r

0'2

o:z2

Thermal diffusivity . ..

Thermal diffusivity a material during

) q V-T + k

at

k

a:

=;:

iJ2r

q

+k

does not change equation

=0

...

(1.10)

(1.12)

(or)

+-+-+0,1 &2

a:

the temperature

O. The heat ~onduction

+-+-

Divided by k,

where,

=

...

at

(1.10)

In the absence becomes :,

heat conduction

k

= --

- m2/s

pCp

0

is known of internal

as Poisson's

equation.

heat generation,

equation

.

is nothing but how fast heat is diffused changes of temperature with time.

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This equation

=

equation

... (or)

This equation

is known

as Laplace

equation.

(1.12)

(1.13)

I. 8 Heal and Ma.H' Transfer Conduction J, 9

Out! (iii): One dimensional steady state "e(lf condllcliOll---

If the temperature varies only in thex direction the e . , ,quatloll (1.10) reduces to q

o2T ----

1

ax';

z-

0

•.• (1.14)

I.;

In the absence of interns I heat generation, becomes:

equation ( 1.14)

'"

(1.15)

1.1.4 General

Heat

Conduction

Equation

in Cylindrical

Co-ordinates

The general heat conduction equation in cartesian coordinates derived in the previous section is used for solids with rectangular boundaries like squares, cubes, slabs etc. But, the cartesian coordinate system is not applicable for the solids like cylinders, cones, spheres etc. For cylindrical solids, a cylindrical coordinate system is used. Consider a small cylindrical as shown in fig.I.2.

element of sides dr, dcj> and dz

Case (iv): Two dimensional steady slate "eat conductio" If the temperature varies only in the x and y directions, equation (I. 10) becomes: ...

In the absence of internal heat generation, redcues to

the

(1.16)

equation

I

(I. 16)

: dr I

J~_(r,4J,z tYT

ax2

if1 -j----

oyl

/ /

=0

...

(I. 17)

Elemental

'

volume

/

Q(r+dr)

dz

Case (,~: Unsteady state, one dimensional, without internal heal generation :

, oin un~teady i.e.,

-a,

:t:

state, the temperature changes with time, O. So, the general conduction equation (I. J 0) reduces to Fig.J.2 ()

o:

("'"

...

(1.18)

The volume of the element

dv

=

r d~ drdz .

I Let us assume that thermal conductivity and density p are constant.

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k, Specific heat

ep

1.10 Hea/I.Jf1J Mass Transfer The energy balance of this cylindrical from first law of rhrmodynamiocs.

I

Net heat conducted into element from all the coordinate directions

Heat}-' generated + within the { element

=

(Heat } I stored ~ in the

Heat entering in the element

through

Or=-k(rd~dz)

cr or

Qr (1.19)

(II/ the co-ordinate

= Qr-

/

= -k

(r d~ dr)

~ de az

ell'

=

T

d: =

Net heat conducted time de. =

= -

:,.

r

[-k (rd~.dz).

=

k (dr d~.dz).

o

oz (Qz) dz

Qz

+

+

[ :Jde

Jdr

aT] de or

L1/:

or

-#-(Q_) az -

dz

Net heat conducted through (~, z) plane

dz

! [k (rde.dr). [~J de J dz

-- k. [OZTJ. 0:;2 (dr~rd~.dz)

Heat entering in the element through Q~ = -k tdr.dz)

de

[~T a,.1

k (dr. rd~.dz)

or

ra~

(z, r)

_

- k

OZT

[oz2]

. (dr.rdq,.dz)de

l

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...

(1.20)

+

_!_ r

~T J cr

de

I...

(1.21)

plane in time de.

de

Heat leaving from the element through (z, Net heat couducred through (r,~) plane

plane in

0,. + dr

-#- (Q ) dr or

z)

into the elem~nt through (r, ~) plane in

Q= -

.=-

Q= +

dB.

de

into the element through (~,

Heat lea ing from the element through (r, ~) plane in time de.

Q=

plane in time

Q r + _E_(O )dr or r

= -

Heat entering in the element through (r, ~) plane in time de -

+

Net heat conducted time de.

directions

Q_

(~, z)

Heat leaving from the element through (~, z) plane in time de.

L element

... Ne! heat conducted into element from

Conduction 1.11

element is obtained

r)

plane in time de.

I. 12 Heat and Mass Transfer Net heat conducted rime

into the element

de.

=_

=

=

t,

Net heat conducted into element from all the co-ordinate

__L [ ro$

-k

directions

de]

tdr dz). or

ro~

k -04>a ['-r D~ -OT] (dr k

Conduction 1.1J

(z, r) plane in

,.a~ (Q+) rd$

= -

Q, - Q~ ., d~

through

=

rd¢

If

de

Q

.

=

r&T

... (1.22)

=

OT] 0,.

Net heat stored in the element

de

C~~n

Substituting

(dr rd$ dz)Cp

equation

+

k

(1.19)

::::::>

k (dr rd$ dz) de

(dr rd~ dz) de

or

x

de

...

00

1- &T _ 0,.2

+L

or

,. a,.

1.20, 1.21 and 1.22J p (dr rd$ dz)

=

k (dr rdql dz) de

r- &T + iYT +_}_ oT oz2

or2

,. or

I a2r J+-;2 ~)~2

Divided ::::::>

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k

r &T

L or2

by (dr rd$ dz)

(1.25)

(1.23). (1.24) and (1.25) in (1.19) +

.L acp2 &T

,.2

+ q (dr rddz) de [Adding equation

is equal to

Increase in internal energy

de

r

by

. .. (1.24)

The increase in internal energy of the element the net heat stored in the element.

from all the co-ordinate

+ .!.._

within the element is given

ci (dr rddz) de

= p

Lor2

(1.23)

Heat stored in the element

+ k (dr rd~ dz)

i3z _

Heat generated within the element Total heat generated

W J (dl rd$ dz) de

Net heat conducted into element directions k ~; (dr rd$ dz)

r2 ocp2

... dtk dz)

_r2

cPT

I -&T +&T-/ +-2

aT

or

1_

[l_ 8$2 &TJ (dr rd~ dz) de .

/Net heat conducted _ . [' L~hrOugh (z, r) plane - k 7i

r 0.2T +.!.._ or2 r

k (dr rddz) de

ep

L'T x de

re

de

+ _!_ OT_ + .L &T + &TJ' + . ,. 8,. ,.2 0$2 oz2 q =- p. C, :-

I

~! J-

OZ~

Conduction 1./5 . From Fourier law of conduction, dT dr

Q=-kA ... It is a ~general three dimensional in c~'lindrical co-ordinates. flT OIl

+ L aT + r r

heat conduction

.L

&T + &T + ~

,.2

13cp2

az2

=

k

(1.26) =>

equat' IOn

ae

Q.dr

=

-k A dT

. the above equation

Integrating

l_ aT ex

L

=>

r

a,.

= 0

...

1.1.5 Conduction

Q

(1.27)

=>

T2

f dr = - kA f dT

of heat through

...

0

TI

o

a slab or plane wall

Q [L - 0]

=

-k A [T2 - Tj]

Q

[T

1 -

=

--J d~'1---

kA L

T 2]

~T overall R

where ~T L

--.J

Fig 1.3

R

=

=

T1-

(1.29)

...

(1.30)

L kA

Q =

Let us consider a small elemental area of thickness 'dx'.

...

TI- T2

Q

T 2'

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TI

(1.28)

Consider a slab of uniform thermal conductivity k, thickness L, with inner temperature T I, and outer temperature

T2

Q [x] =-k A [T]

=> =

1

k A dT

L

(or)

l,. . drdT J

tl e limits of 0 to L

TI'

o

&T + _!_ aT

_!__ _{__ ,. dr

f

= -

L

and no heat generaion ,

( 1.26) becomes: 13,.2

etween

T2

f Q dr o

equation

b

and TI to T2· =>

If the flow is steady, one dimensional

we knoW that,

T2

C. - Thermal

resistance

of slab.

~~~~~~~~--------~----------/.16 Heal and

-

l.L

Tram/er

MLI.\"J

6 C' nduction of Heat Through Hollow Cylinder

Conduction 1.17

0

onsidcr a hollow cylinder ofillllcr radiu rl' outer radius r2, inner tcrnperatllr~ T I' outer temperature T2 and thermal

Q==

Q

cOllducti, it ". Let II c 11 idcr a small elemental area of thickness "dz" From Fourier law c nduction, we know that,

T2

~

of

Q ==

In(;n

...

(1.31)

...

(1.32)

TI- T2 1 ('2rl ) --In2nLk

.1Toverall R

where

dT -

Q=-kA

27tLk [T, - T2]

dr

Fig 1.4

1

R = 2nLk

(r21 Thermal Inrtr

resistance of the hollow cylinder.

Area of a cylinder is 27trL A

=

1.1.7 Conduction of Heat Through Hollow Sphere

27trL

s, Q = -k27trL Q

d,.

x -

,.

=

Consider a hollow sphere of inner radius rl, outer radius r2, inner .ernperature T I, outer temperature T2 and thermal conductivity k.

dT dr

-k27tL dT from rl to "2 and TI to T2·

Integrating the above equation ~, dr

Q

J

r

T2 = -

k27tL

f dT TI

rl

Let us consider a small elemental area of thickness 'dr'. From Fourier law of heat conduction, we know that

Q = -kA dT dr Area of sphere is 4m-2

=

Q

[111'2

Q

/11 [:~ 1 = 27tLk [T, - T2J

= -

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k27tL [T2 - TI1

Fig 1.5

A = 47tr2 Q

= - k 4m2

ciT e1r

.,.

(1.33)

p'n

eep •

------------------

Conduction 1.19 1.1.8 Newton's

Law of Cooling

Heat transfer by convection lntegrating

...

12

r

Q

dr:;;;

:::>

_

,.1

rl

j

47tk dT

'2

I

Q

:::>

A - Area exposed

(-

d~ == - 41tk . dT

. r '1

1

'2

Q

:::>

\=1-1 r

Q

:::>

:::>

Q

:::>

Q ==

I

h - Heat transfer

1

12

== - 41tk [T]

'2

41tk[T1-T2]

(r2-rl)== rl r2

...

(1.34)

r2 - '1

in W Im2K

Ts

-

Temperature

of the surface in K

T a:

-

Temperature

of the fluid in K.

T1-

Q

to conduction.

T2

r2 - rl

41tk (rl r2)

Q=

:::::>

ilT overall R

...

(1.35) Convection

where

_

r2 - '1.

R - 4 k( 1t

.

- Iherrnal resistance of hollow sphere.

A

'1 '2)

Fig 1.6

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Plane Wall with

Consider a composite wall of thickness L1, L2 and L3 having thermal conductivity kl> k2 and k3 respectively. It is assumed that the interior and exterior surface of the system are subjected to convection at mean temperatures T and T b with heat transfer coefficient hQ and hb respectively. Within the composite wall, the slabs are subjected

r1 r2

Q==

co-efficient

in m2

1.1.9 Heat Transfer Through a Composite Inside and Outside Convection

== - 41tk[T2 - Td

41tk [T, - T2]

:::::>

to heat transfer

Tl

r1

lL- l1 '1

(1.36)

where

11 T)

I

is given by Newtons law of cooling

on both sides

~

il

Conduction /.21

1.20 Heal and Mass Transfer From Newton'S law of cooling, we know that,

Adding both sides of the above eq ua tiIons

Heat transfer by convection at side A is

Q = ha A [Ta - T,

J

[From equn. (1.36)]

...

(1.37)

...

(1.38)

...

(1.39)

=> Ta - Tb = Q·hA[_1_ + a

_!j_ + -+ L2 k( A k2 A

L3 + I hb A

k3 A

1

Heat transfer by condl1ction at slab (I) is

Q = k, A [T, - T.,]- -[From equn. (1.29)]

=> Q=

__s_

_I + L2 L3 +-+-+ [ ha A k( A k2 A k3 A

L(

I] hb A

Heat transfer by conduction at slab (2) is Q=

k2A[T2-T3] L2

=> Q

=

~Toverall R

...

(1.42)

where

Similarly at slab (3) is Q=

k3A[T3-T4J

...

(lAO) Thermal resistance , R

L3

= R a + R I + R 2 + R 3+ R b

Heat transfer by convection at side B is ...

We know that,

We know that,

To-T,

=Qx_1

[From equn. (1.37)] haA

T,-T2

=Qx~ . k( A

T2 - T~)

=

Q

(1.41)

x~

L2 k2 A

T3 - T4 =Q x ~

[From equn. (1.38)] [From equn. (1.39)]

k3 A

[From equn. (1.40)]

hb

[From equn. (1.41)]

R=_l_ UA

Ta-Tb .:::... => Q= __ _I_ UA

=> Q

A

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U A [T a - T b ]/

...

where '(0"

T4-Tb=Qx_'_

=

IS t he overall heat transfer co-efficient

(W /m2K).

(1.43)

~I

r

1.24 Heal and Mass Transfer Q

-~

= .1Toverall •"

R

Conduction I. 25

(1.48)

t,1.1J Solved Problems

where

fZJ

On Slabs

Determine the heat transfer through the plane of length 6 111, heigh' 4 m and thickness 0.30 m. The temperature of inner and outer surfaces are 100 C and 40 C. Thermal conductivity of wall is 0.55 WlmK. 0

0

Give" : Inner surface Temperature, T I

we know that,

=

100° C + 273

= 373

K

Outer surface Temperature, T 2 = 40° C + 273 = 313 K I R= VA

Thickness, L = 0.30 m Area, A

Ta-Tb

~

Q=

~

Q = VA

x

4

=

24 m2

Thermal conductivity, k

_I_ VA [To - Tb J

=6

...

=

0.55 W/mK

(1.49)

where U = Overall heat transfer co-efficient, W/m2K

Tofilld:

I. Heat transfer (Q) Solution : We know that, heat transfer through plane wall is

Q = .1Toverall R HMT DOlO book (C P Kothandaraman)

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[From Equn. 110. {I. 30) or page no. 43 (Sixth editiont]

1.• 6

Heat and Mass Transfer

where

Conduction 1.27

Tofi"d: Thickness

of insulation (L2)

SOIUlion: Let the thickness of insulation be L2 We know that,

373-313

= 2640 watts

0.30 0.55

Q

=

x

Q =

[From £qun no. 1.42 (or) HMT Data book page no. 43 & 44 (Sixth edition)]

where

24

2640 watts

AT=Ta-Tb

I

Result: Heat transfer, Q

AToverall R

R =

(or) T)-T3

L I +_)_+ ha A k) A

__~ + __L3 +_ J k2A k3A hb A

= 2640 W

In A

wall of 0.6 m thickness having thermal conductivity oJ 1.1 WlmK. The wall is to he insulated with a material having an average thermal conductivity of 0.3 WlmK.lnllerandouter surface temperatures are 1000 C and 10 C respectively. If heat transfer rate is 1400 Wlm1 calculate the thickness oj insulation. Wall Insulation I-I-I Given: 0

Thickness of wall, LJ

=

0.6

0

::::> Q =

I haA+

Heat transfer

Heat transfer per unit area, 0/A = 1400 W/m2

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ha' hb and thickness L3 are not

[T)- T31 L)

L2

k) A

k2 A

--+-[T)-T31 L)

L2

k)

k2

-+-

of

Outer surface Temperature, T) = 10° C + 273 = 283 K

co-efficients

::::>Q=

III

Inner surface Temperature, TJ = 1000° C + 273 = 1273 K

~ L3 I k2A + k3A + hbA

given. So, neglect that terms.

Thermal conductivity of wall, kJ = 1.2 W/mK Thermal conductivity insulation, k2 = 0.3 W/mK

L) k)A+

~nl

1273-283 ::::> 1400 =

[L2

Result :

Thickness

0.6 +

.!:1.

1.2

OJ

= 0.0621 .'

~

of I11sulatJOll, Lz

= 00621 .

m

ZFF

1.28 Heal and Mass Transfer

III The wall of

{I cold room is composed of three layer. Til layer is brick 30 em thick. The middle layer is cork e OilIer thick, the inside layer is cement J 5 em thick. The temp 20 c", of the outside air is 25° C and 011 the inside air is _20~~/II'es film co-efficient for outside air and brick is 55.4 Wlm2/( . ~he co-efficient for inside air and cement is J 7 Wlm2 K. Fin~ ~i/", ~wro~ ~

Conduction

1.29

Tofintl: Heat flow rate (Q/A) solution: . . b Heat flow through composite wall IS given y

Take ~Toverall

k for brick = 2.5 WImK

Q ==

k for cork = 0.05 WlmK

R

[From Equn no. 1.42 or HMT DolO book page No. 43 and 44]

where

k for cement = 0.28 WlmK Given: Thickness of brick, L3

= 30 em = 0.3 m

Thickness of cork, L2 = 20 em = 0.2 m Thickness of cement, L) = IS em Inside air temperature.T a Outside air temperature, Film co-efficient Film co-efficient kbrick kcork

=

k3

=

k2

=

=

k)

Tb

0.15 m

-20 C + 273

=

253 K

=> Q

=

= 2S C + 273 = 298 K 0

for inner side, ha = 17 W/m2K for outside, hb = 5S.4 W/m2K

=> Q/A

2.S W/mK

=

kcement

=

=

0

1 L( L2 L3 -+-+-+-+ha k( k2 k3

O.OSW/mK =

1 hb

253 - 298 => Q/A ==

0.28 W/mK

1 + 0.l5 +_Q1__+..Ql_+_l_ 0.28 0.05 2.5 55.4

17 Inside

Cement

Cork

Brick

k(

k2

k)

Outside

I Q/A == -9.S W/m2! The negative sign indicates that the heat flows from the outside into the cold room. Result: Heat flow rate, Q/A == -9.5 W/m2

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1.30 Heat and Mass Transfer

(!]

A wat! of II cold room is composed of three layer; Tile (JUte, layer is brick 20 em thick, t"e middle layer is Cork 10 c", thick, the inside layer is cement 5 em thick. The temperature of the outside air is 25° C and tltat on the inside air is -20 C TI,efilm co-efficient for outside air and brick is 45.4 WI",2 K and for inside air 011(1 cement is 17W/m2 K. 0

Conduction 1.31 Film co-efficient

for outside air and brick, hb = 45.4 W/m2K

Film co-efficient

for inside air and cement, ha = 17 W/m2K

K) = 3.45 W/mK

K2

= 0.043

W/mK

K(

= 0.294

W/mK

Find i) Thermal resistance ii) The heat flow rate. Tofind:

Take

I. Heat flow rate

k for brick = 3.45WlmK Ii for cork

2. Thermal resistance of the wall

= 0.043 WlmK

sotutio» :

k for cement = 0.294 WlmK

Heat flow through composite wall is given by Given : Q= In ide

Outside ement

kJ

Cork

Brick

k2

kJ

~Toverall R

[From Equn (1.42) (or)

HMT Data book page No.43 &44J

where ~T=T{/-Tb

I L( L2 L) R =--+--+--+--+-ha A kJ A k2 A k) A

I hb A

=>Q

j

kne

f brick LJ

= 20 em

=

0.2 m => O/A

f ernent LJ

= 5 em

= 0.05 m

=> QIA UI

ide air temp ~rature, Tb = 250 C

III ide air len perature, To = -200

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273 = 298 K 273 = 253 K

253 - 298 _1_ + 0.05 + __Q:l_ + 0.2 + _1_ 17 0.294 0.043 3.45 45.4

lOlA =-17.081 W/m21

/. 32 Ileal and Mass Transfer TIle negative sign indicates that the heat flows fro~ into the cold room.

COnd/l(;liml I. JJ

e OUt. s~

(i;"C'II .'

(PT2

outer layer

Inner layer middle laye

For Unit Area ~

I

k, LI

L2

L)

R =-+-+-+-+ha kl k2 k)

= 2.634 KJW

kJ

k2

I hb

_1_ +0.05 + _QL + _Q1_ +_1_ 17 0.294 0.043 3.45 45.4

~ IR

Thermal conductivity

of inner layer, k I = 8.5 W/mK

Thermal conductivity

of middle layer, k2 = 0.25 W/mK

Thermal conductivity

of outer layer, kJ = 0.08 W/mK

Inner thickness, LI

I

= 2S em = 0.25

Midddle layer thickness, L2 = S ern Rault:

Outer layer thickness,

= -17.081

I) Heat flow rate, Q/A 2) Thermal resistance,

R

= 2.634

W/m2

LJ

Inside wall temperature, OUI

K/W

ide wall temperature,

=

3 em

(I] A furnace wall is made up 0/ three layers, imide layer

T4

I. Equivalent electrical circuit MI~~

2. Heat flow per m2

thermal conductivity 8.5 WlmK, the middle layer MIll' conductivity 0.15 WlmK, the outer layer with cO/l(luClivi~

3. Thermal resistance

0.08 WlmJ(. T,.'lerespective thickness of the inner, n~i{ldlea~ outer layers are 15 em, 5 em , and 3 em respectively- T~

4. Interface temperatures

inside {/~d outside wall temperatures lire 6000 C and ~O·~ respectively. Draw the equivalent electrical circuit f~ conduction 0/ Ileal through the wall and find t"ernl ~ and interface temperatures4

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=

III

=

0.05111

0.03

III

T t = 6000 C + 273

Tofintl .'

resistance, heat flewlml

(

= 50° C

= 873 K

273 = 323 K

1.34 Heal and Mass Transfer

.---=--------------..._

Solution:

Conduction 1.35

electrical circuit for COIl(llICtioll

1. Equivalent

873 - 323

9.25 + 0.05 + ~QL. 8.5 0.25 0.08

~IQ/A

W/I11~

= 909.97

3) Thermal Resistance

2. Heat flow through composite wall is given by L\ToveraJl

Q=

[From Equn.

R

110.

(1.42) (or)

(Neglecting ha' hb terms)

HMT Data book page No. 43 & 44J

where L\T=Ta-Tb

=T,-T4 For unit area

,

R

=

L, L2 L3 I ha A + k, A + k2 A + k3 A + hb A

~ Q =

0.25 -I-

0.05

8.5

0.25

+

0.03.

0.08

4) Interface temperatures We know that,

[Convective heat transfer co-efficients ha and hb are not given. So, neglect rh{lt terms] = - ... -.-.--~.-.----

r, =:)

()

--

=-">

Q

=

'14

-,-{---

T,-T2

R,

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T, - 1'2

T:2 - T 3

= -R-,-= --~

=

T 3 -- T4

R3

... (

1)

I: 36 Hear and Mass Transfer

Conduction

1.37

Resllll:

= 909.97

I. Heat flow per m2, Q/A

klA

2. Thermal Resistance, R

0.604 K/W

TI -T2

Q/A

T 2 = 846.23 K

3. Interface temperatures,

LI kl

909.97 =

T)

873 - T2 0.25 8.5

I T2 = 846.23 Kj

@

= 664.23

A mild steel tank of wall thickness

20

K

111111

contains

water fll

100" C. Estimate lite loss of heat per square metre area of lite tank surface, if lite 11IIIk is exposed 10 OIl atmosphere til

15° C. Thermal conductivity of steel is 50 WlmK, while hem transfer co-efficient for lite outside wid inside II,£, tank are 10 WI1112K and 28.50 WI",2K respectively. WIItII will he lite

Silllilarly (1)

=

W/m2

o

=:>

= T2-T3

,

where,

lemperalllN'

R2

Oil

lite outside o.f lite tank wall.

Given :

L2 R2=-k2A

Thickness of steel wall

Q =

Inner water temperature

LI

T2-T3

L2 k2A

Ta

0.05

-

0.25 664.23 K

= 373 K

air temperature

Thermal conductivity steel, k I = 50 W /mK

= 846.23 - T3

I T3 =

100° C + 273

Atmospheric

~ k2 909.97

=

Inside

In

Th= WC+273=288K

T2 - T3

Q/A

= 20111111 = 0.02

I

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of

Inside heat transfer co2 efficient, ITa = 2850 Whn K Outside heat transfer coefficient, hb = I 0 W/I11~K

Outside

1.36 Heal and Mass Transfer where Conduction

Q

=

TI - T2 LI

1.37

Reslllt: I. Heat flow per m2, Q/A

=

909.97 W/m2

klA 2. Thermal Resistance, R

TI - T2

Q/A

=-L-

=

0.604 K/W

-

I

3. Interface temperatures, T 2

=

846.23 K

~ T 3 = 664.23 K

909.97 =

873 - T2 ----=0.25 8.5

I T2 = 846.23 Kj

@]

A mild steel tank of wall til ick ness 20 111mcontains water (It / 00° C Estimate tile loss of heat per square metre area of tile tank surface, if tile tank is exposed to an {Itmo."plwre (It /50 C. Thermal conductivity of steel i...50 WlmK. while heat transfer co-efficient for tile out s ide and in ...ide tile ttlnk are JOWl",] K am/ 2850 Wlm2 K respectively. What will he the

Similarly

Q = T2-T3

(1) ~

temperature

R2

Q

=

T2 - T3

T2 - T3 k2 846.23 - T3

0.05 0.25 / T3 = 664.23 K

Outside

Inside

Ta=100°C+273=373K

To

Atmospheric air temperature Th= 15°C+273=288K

ha

Thermal conductivity steel, k I = 50 W ImK

~

=

tank wall.

Inner water temperature

k2A

909.97

01 the

Thickness of steel wall L, = 20mm = 0.02 m

L2

Q/A

tile outside

Given:

L2 R2=-k2A

where,

Oil

I

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of

Inside heat transfer coefficient, "a = 2850 W 1m2 K Outside heat transfer coefficient, lIb = 10 W/m2K

1\ -----\

...-

I. 38

!...~:!.!~ and

Mass Transfer -------

To .Ii" tI : i)

" II)

..

_--- ----------_ " /

Ileal loss per square metre area of the tank Surface ( T ,Hl1\"doutsi c temperature, T 2 , Q 1\)

Conduction 1.39 We know that, TG -Tl

T1I -T,

~

Q =

-R-

=> Q

= T -T,

=R;--

Solutio" : Heal loss,

_G __

AToverall

Q =

---R---

where

Ra [From Equn, I/o. (I.42j~

I-IMT Data book page NO.4J

& 44J

1

= -, A

where, Ra

1£1

I-H=Ta-Tb R

I

L, k, A

L2 k2 A

LJ +__ I k3 A hb A

= --+--+--+-

ha A

_ Ta-T, I/h

=> Q/A -

a

=>

843.66

=

373 - T, 1/2850

(Neglect L2, L3 terms)

=>

IT,

=

372.7

Similarly

=> f)/A I LJ I -+--+-h{/ kJ fib

3 73 - 288

'_.) Q/;\

,

where, R, --'-

----- --------

:- (V/\

- _. ---.

-I-

2S50

_O_:_Q£ 50

+

.L '()

T,-

--------.-

-, S43.6(j

1

kJA => Q/A

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T2

L,

Willi ~1

-_- .-._------ "_-

Ll

= k,A

KI

_~

T,-T2 __

R,

=

T2-Tb_

Rh

1.40 Heat and Mass Transfer =>

843.66

=

372.7 - T2 0.02

Conduction

50

I

T]

=

Hot gas temperature.

372.4 K

I

Result: =

2. Outside surface temperature, A steam boiler furnace

843.66 \Vlm2

Heal now b) radiation

from gases

T2

=

Convective

372.4 K

is made of fire clay. rite

temnerature imide the boiler furnace r

is 2100

transfer co-efficient at tlte interior surface is 12.2

Illleril r wall

to

""(11

8.2 kWln'; and interior wall sill/ace temperature is 1080"('. Calculatefor external surface I. Sur/ace temperature 2. Convective conductance Given:

Scanned by CamScanner

inside surface

"0

=

=

of the

from external

J

2.2 W/m2K

58 W/mK surface

to surroundinc

10·) WlrnK

+ 273

1'1 = 1080°(,

urface temperature,

= 1353

is

K

Tafind :

1¥111I'/(,

slirrouluJi"K

323 K

10

i) External

thermal conductance of the wall is 58 WII1IK, heat flow ~v susface

10

=

103 W/m]

x

of the wall

QI{2 = 8.2 kW/11l2 = 8.2

t, roo-.., (II,. caSel"

at interior,

conductance

Ileal now by radiation

hOI CQJ

inside surface of the wall is 25.2 k Wlm], cOllveclio"

extemal

transfer

Thermal

temperature is 50 'r, heat flow by radiation from

radiation from

T b = 50 C + 273

= 25.2

I..J I

C

Room air temperature.

wall. Ol{ 1 = 25.2 kW/m2

I. Heat loss per m2, (Q/A)

o

Ta = 2100

surface

ii) E.\lerll,il

C

temperature,

'1']

nvc rive c nductancc.

hh

Solution : We kn: \ Total

hca:

e n Ic r i n ~

the

\I

all

Ihill

} ()

=

Heat nvc

Inn fer

ti

h~

n (II interior

l le at radiation

Irat I . fer

h)

at interior

r ,I

1.42 Heal

"

Mass

(11/(/

TrLlm!er ('(lnductiun

We kn \\ !haL

'1 - T, .-E-- _ R

o

R

-s:

Resull:

T, - T2 _ T2 - '1'"

T-l"_.-!!----R

I

-

1.43

I. [.x!<.:rfl
0

,,/;

"'2 = 703.4 K

(I

2. _xlemal convective

co-efficient,

hh = 77.3 W/m~K.

T, - TZ ~Q==~

@ A wall

is constructed

of several layers. The first layer o/I/Ia.\'OII(/ry brick 2() cm thick (lJ tlt er m al conductivity fl.66 WlntlC, lite second layer consists of 3 em thick 1/101'1111' of tltermal conductivity 0.6 WlmK, lite third layer consists oJ 8 em thick lime MOlle of thermal conductivity 0.58 WlmK and the outer layer consists of 1.2 em thick plaster 0/ thermal conductivity 0.6 WlmK. The heat transfer coefficient Oil the interior 1I11(1 exterior of the wall lire 5.6 WIIII1K and II Wlm2K respectively. Interior room temperature is 2rc and outside air temperature is -5" C. consists

WJ- T2 RI

LA,I--.V't"" -

J

1353 - T2 I -8

l''-

Re5istance =

External surface temperature,

T 2 = 703.9 K

3 .644

==

Heal loss due

d

con uctancc

Calculate (I) Overall heat transfer co-efficient

radiation at exterior

to

J

b) Overall thermal resistance c] The rate of heat transfer Hear 10>5by

vection at exterior

= T! -I hea entering - Heat

loss by radiation

at exterio

d) The temperature {II the junction between the mortar and lite limestone Given :

=

37.644 - 8.2

Y

103 k2

--..'Qc = _9.!

rr

~Tl

---- lib A

II

T

,=

29, 4

!

/ A / IT! - Th) -= 29,441

!

hb / I / [ O~.9 - 323 j :: 29,444 F.\lCrnal

'omccl; e

CO-Cf!lCI'CIlt

__

I , 'b::

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) 77.3 W/III-K

I

eDT,

2

I

asouary

Mortar

I .

. Limestone

Plaster'

lib

I

-1 ----'

/I('a1 and Mm.' Transfer ,=

Thcnna I (I

IH.lIII.. '1ivity

1 hidlll::-'l

~I

ondu

f Plasicr.

I..n~ss

OJ1dll 111;31

Exterior

.~

rJ1

Conduction

-,

_.

uvity,

z:

_'_'1_ ~ "1 k A kl A

I

L

~

k A

k" A

== 0,08 III -

2C)S - 268

O/A I 5.6

= I.~ em == 0.012111

"4

0.08 + 0.012 + .L 0.58 0.6 II

0.20 + 0.03 0.66 0.6

0.6 W/mK

hctlllrallsfcrco-cfficient,

hh

Tb = -

Ileat trallsfer

h" = 5.6 W/m2K

trnnsfcr, co-efficient

Ollh';dc air Itl1lpcratIJre

0 h,A

"3 == 0.58 W/I11K

L4

=

o

V. e knov

II W/m2K 273

per L1l1itarea, Q/A

34.56

W/m2

that.

Q=

Heat transfer,

= 295 K

=

UA (Til _ T b) [From equation

no.I.4J

where. U - overall heat transfer co-efficient

C - 27" = 26R K.

Tolind: . ) (h

c.:rall lint transfer

11) (h'~rjJII !lrcnlltll . Ih·allrall·fl:r/lll-.

co-efficient,

reo istancc,

U

(I<)

34.56 U == _.::.._:_.:.;:_::295 - 268

(()fA)

d) '111(' temperature at the junction

Ilw limes!

between

ihc Mortar

and

Overall heat transfer co-efficient,

U

=

1.28 W/m2K

ne. (Tl) We know (hat,

Solution :

Overall Thermal

.II "11 Il( \\ C)

'\ I

thrnugll )\I'J'

COIJIP(

sire wall

I IAtr I (I

R

by

/ h()1JI Equn.

"'1.'

"

given

is

For

11

H \111

1.../5

T,,- T"

IIIIK

r room ll:rnpcratIJrc, To = 22" C

lntcr:

0.20

r-,

"2:;- 0.6 W/IIlK

= ~ ern

ue. L,

ThLTm. I 'Olldllclivily,

Interior

0.66 \\

tivuy Oflllol1ar.

of limcst

~1l~:\S

Thermal

.!-'

0 em

~_

f me rtar, L_ = 3 em == 0.0" rn

henna!

Thi

=

, (1I1;1:-ollary. 1.,

Thickncss

. hi

_~ ~

,~._,_c_

__=

/)11/0

hook

j}(lJ,{('

I/O.

= --

(R)

LI L2 LJ L4 --+--+--+--+-kl A k2 A k3 A k4 A

I

hb A

unit Area

(In) or

No. -I3cC-I-Ij

I

hI A

resistance

R

=

I L, L) LJ L4 -+ -+-- +-+-+h(l k1 "2 kJ "<j

1 hb

_I + 0.20 + 0.03 + O.OS + 0.012 +_1 56 0.66 0.6 0.58 0.6 II

I;,

L,

I.,

.1

1< \

/\

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0.78

K!WJ

I

I 46 Hid

__:_____ eo a17 Mass Tran.~fer

Interface te mperature between mortar and ------..' the Interface temperatures relation

Q _

"1-T2

To--TI

--~=-R-I-=~

"2-T3

T -..........

linrel'IOIIe,

_ T3

.

-"4

-~-==~R

..>

Conduction 1.47

3

T4-l

5 4

TS-Tb

=>

278_3 - T3

Q=

=~

~

T,,-' TI

Q=

k2A

Ra

278.3 - T3

Q/A

295 - TI

Q

)/ha

~ k2

A

295 - TI Q/A

34.56

=

11170

278.3 -- T3 0.Q3 0.6

295 - T, 34.56

Temperature

1/5.6

IT,

between Mortar and limestone (T]) is 276.5 K

Result:

=

288.8 K)

I) Overall heat transfer co-efficient, 2) Overall thermal resistance,

T,-

T2

=> Q=--R,

3) Heal transfer, Q/A ,1) Temperature

288.8 - T2

Q

L, k,A 288.8 -, T2

Q/A

L,

", 34,56

__ _3!8.8 - ~~ 0.20 0,66

I

IT?

=

27S,3E]

I

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R

= 34.56

U = 1.28

= 0.78

W/Il12K

K/W

\\ /m2

between mortar and lilllcstone,(T3)

=

276,5 K

[!J The wall of (/ refrigerators is made

lip oftwo mild steel plates 2.5 111111 thick with (I 6 em tltick glass wool ill between the plates. The interior temperature is' -20" C, while tile outside of the refrigerator is exposed /0 40" C. Estinuue tile heat flow: Thermal conductivity of steel alit! glass wool are 23 WllltI( and 0.015 WlmK respectively.

(Madurai

L

Ll

=

L'l

= () ern = 0.06

J

= 2,5 111111

=-=

III

0.0025

III

A{/III01'Oj

University

/l.'OI'-I.}-I)

I. 48 Heal and Mass Transfer

-

I

Glass wool

Mild steel

Conduction 1.49 Convective

Mild steel

heat transfer coefficient is 1I0t given.

So, neglect ha' lib terms

Ta

k,

k2

I

.

L,

~

Ta

k3

Tb

I

L2

• I-

-/-

L3

~QIA.

..,

= -20 C; Tb = 40 C 0

0

k, = k3 = 23 W/mK;

Q/A

- -.---.

0.0025 + 0.06 23 .015

k2 = 0.015 W/mK

I

Tofind : i) Heat flow, (0)

Heat flow through composite slab is given by vera II R

23

Q == -14.99 W/m2!

Result : i) Heat flow,

,1 To

+ 0.0025

The '-ve' sign indicated that the heal flows from the outside into the refrigerator.

Solution:

Q

-20 - 40

[From Equn. no. (1.42)]

where

o

W/m2.

IZ!fl

The inside temperature of the refrigerator is -1tl" Cand outside surface tempera/lire is 30t) C and area is 301111. This refrigerator consists of 2.2 ntnt of steel at the inner surface, 15 111m plywood at the outer surface and J() em ofglass wool ill between steel (lilt!plywood. Calculate the heat Ion and the capacity of the refrigerator ill tons of refrigeration. Assume k(sleelj == 20 WlmK. k(p()'",oot!) == 0.05 WlmK. k(g/as!J'H'oo/)== 0.06 I-Vlm/(.

Given : Inside

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Q == -·14.99

Temperature,

T, =_IOL'C

f'273

=-=263

· 1.50 Heat and Mass Transfer Conduction /.51 kI

k2

k)

Steel

Glass wool

Plywood

T4 == 30° C + 273 == 303 K

Outside temperature, Area, A ==30 m

~~

~ T)

~DT2

(DTI

where

2

Thickness

of steel, L) == 2.2 mm == 0.0022 m

Thickness

of plywood,

Thickness

of glass wool, L2 == 10 cm == 0.10 m

Thermal conductive

L3==

co-effficients ha and hb are not

15 mm == 0.015 m

of steel, k) == 20 W ImK of plywood,

Thermal conductivity

of glass wool, k2 == 0.06 W/mK

Toflnd :

263 - 303

Q

0.002 + 0.10 + 0.015 20 x 30 0.06 x 30 0.05 x 30

k3 == 0.05 W/mK

Thermal conductivity

IQ

=-610.1

W==-0.610KWI

The -ve sign indicates that the heat flows from the outside into the refrigerator.

i) Heat loss, Q 2) Capacity of the refrigerator

We know that

Solution:

3.5 kW ==I ton

Hear flow throu?h composite Q=

(Convective heat transfer given. So, neglect that terms) T)-T4

wall is given by

0.610 3.5

ton

==0.174 ton

t1To vera lJ R

:=)O.610kW==

[From Equn. no.(/.42) ~ HMT Data book page No.43 & 441

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:=) Capacity

of the refrigerator

==0.174 ton

'{

r 52 HealandMassTrc.!'!!f!!_.-------"-~ Result:

I' id .iqut surface conductance,

Heat transfer, Q ::::610 W Capacity of the refrigerator::::

f1D

A s'tetlm to liquid

Steam surface' con d uctance, hh

0,174 ton

heut exclulIlger

area

copper Oil lite steam S/{Ies.

I lie

re.\'ISIIV/~V of

en, (I

/

.'"' ;,

k2(copper)

=

OJ

k) (Nickel)

= 55

p (II'"n .Vfller

=

I 53

.

5400 W/Ill2K

Steam " , , temperature, Tb = ) ) 00 C· _,+ 27"., -- .,83 K Liquid temperature: T(I = 70" C + 273 = 343 K

of 25

11.5 •em nickel and• •0.1 I 7'1 .. '

cOllslrllelell ",illi

Conduction

h(/ - 560 W/m2K

t>

is {}.0015Kiw. The steun, (I1lti/it-sCa/! surftlce C{JIIt/uel'IIIee lire 5400 WI", 2K and 56() W. 14~ reJpeClive~y.Tile Itealeds'lell", is filII Ou C (flllillefllet[ ~",2k

350 W/IllK W/IllK

deposil 011 lite steam side

is' al 70° C.

"ql/id

Toflnd : i) Overall

2) Temperature Solutio

Calculate transfer co-efficient

J) Overall steam IO/iquidllelll

2) Tempemllire drop IICrOS,\'lite settle deposit

heat transfer co-efficient, (U) drop across the scale deposit. (T, -- T4)

II :

Heat transfer through composite wall is given

Q

=

Take

r

~Toverall

Front Equn. no, (I. 42) or lIMT Data hook page No.43 & :J:J}

R k(copper) = 35(1

W'ImK

filii/

k(Nickel)

= 55 WlmK.

by

where

Given : Inside Liquid side

[

Outside

R = _I_+_L_I_ ... L2 ,L) I ha A k A . kA kA + _-

"b

T

,

1\2

Steam

side

2

3

A

Ra + R, + R2 + RJ + Rb

T2

Til 11(/

G:~~.:

R,~ value is given,

R" = k~~ = 0.00) 5 K/W

~

Copper R

I

---t--

lin A

_-'-

__

560 x25.2

=

Thickness

of Nickel

,

L I -- 0 ).- em

Thickness

of copper t'

'

L'2-- 0 , I cm=O.1

Resistivity

of scale,

R_l = 0,0015

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0,5

K/W

x x

10-2 10--111

L,

"I

A

+ 0,5 x 10-2 + 0, ) x 10-2 350 x 25,2 55 x 25,2 I

+

111

5400

~ ~IR

J.~58~x~IO~-_J~K~/W~1

x

25,2

+ 0,0015

/.5.1

Conduction/.55 fllrll(~c~ is made up of 13 em thict: of fire day,of thermal condllctlv/~" fJ.6 WlmK alU160 em thick of red brick of conductivity 0.8 . WlmK. Tire inner ~nd outer surface . . I. temperature of wall are 1"000 C and 75 C Determine

fllJ A wall of

(I

0

0

1.

Tile amount of heat loss per square metre of lire furnace wall.

2.

It is desired 10 reduce lite thickness of lite red brick layer ill litis furnace to half by filling in the space between lite two layers by diatomite whose k = 0.11J + 0.00015 T. 'Calculate lite thickness of the material.

Give" :

Furnace

Overall heat transfer co-efficient,

k,

k2

Fire clay

Red Brick

25 W/m2K

U

Temperature drop (T~.) - T4 ) across the scale is given by L1T Q=-Rsca/e ,

25.2

x

W

[.:

L,=13cm= k,

L1T

= 0.0015

=

Result: Overall heat transfer co-efficient, (U) = 25 W/m2K

T41 = 37.8

0

C

k2

=

T,

=

T3

=

= 0.6

111

0.8 W/mK 1000° C + 273 75° C + 273

=

=

1273 K

348 K

Tofind: I) Heat loss per square metre .2) Thickness

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0.13m

0.6 W/mK

L2 = 60 em

~ltlT=37.8°CI

Temperature drop across the scale, (T3 -

I·

I:H = T3 - T4]

0

f the furnace wall.

of the material. whose k=OIII+O.OOOIST .

__ ._-" - ---

.

Sol"tim, : CondUCI;OJ1 1.57

2.

1. I teat transfer through composite

wall is given by Diatomite

... Q = Il To\'<:.~~~I

[From EC/III1. no. (1.42)'», JlMT Data book page No.43 & 44]

R where

Red brick

13

Furnace

K. .'

We know that,

TI -T3 ~Q

=

·-----;---·i,_ h~-A+

.~+~+-1-

k, A +

Neglect unknown terms ~Q

348 = ...-1273·-_. __.._ --

"2 A

Q kJ A

=

T, - T.f R

=

T I - T2 _ T2 - T3 -R-,- R2

= Tr

T4 R3

••• ( I)

hb A

=> Q =

T,- T2

R) 1273-T2

Q =

L, k, A

=> Q/A

=> 956.8

=

=

1273 - T2

1273 - T2 I.,

k, 1273 - 1"2 -0.13 0.6

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,.: A

= 1m2 J

---------------------

/. 58 Heal and Mass Trumjer

Conduction J .59

T) - T4 Q=

L3 .956.8 .. '

kJ A

='

,

358.8 L2

k;

. T) - T4 QIA =

__!2__ kJ

k

=

0.111 + 0.00015 T

::::::>k2=0.111 +0.00015T

T4 - Outer surface temperature of red brick' k) - Thermal conductivity L) _ Halfofthe

k2 = 0.111 + 0.00015 [T2: T31

of red brick

2

thickness of the red brick == 0 6

=

= 0.111 + 0.00015 [1065.6 + 2

O.3m

I k2 = 0.243 WimK I

T) - 348 Q/A ==

0.3

D.8 ~

Substitute ~

=

956.8

706!KJ

Q=

T

T2-

1.

R2 Q=

1065.6-706.8

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(

~.

0.243

~.

L2 = 0.091 m

Thickness'ofthe (I)~

,

956.8 == 358.8

T) - 348

IT =

706.81

k2 value in Equation (2)

0.3

~ IT)

is

Given thermal conductivity for diatomite

where

~

.... '(2)

diatomite, L2 = 0.091 m

Remit: Heat loss, Q = 956.8 W 1m2 Til ickness of the, d iatOl;nite, L2.=' 0.091 m

),

'"

.

I

(10

Heat and

Muss r,'(/l1s[er

-'[ijl A [urnace wull hi made of ~

inside silica hrie~ • • 'J ther contlllctivill' 1.7 W/mK, 12 em thick and outside m ",~ • •• ~ ~ IV (Ig'l~f' brick of thermal conductivity .1••1 ,,'/mK, 22 em thic ",, temperature Oil the inside of the wall of the silica bk: rh, . magnesite. bri 92(1UC (111(1olltsult! nc«t. sur/ace tem'Pe r'rk'~ rmUre' 120" C. Calclliate the heatflow tit rough tit is compos I'te IVaI(~ lf the ('011 tact resistance between the two wall is 0.003Ktlt find tile temperlllllre of the surfaces at the illter/ace.

~;:::-::--------where

~C~o'~ld1!_''!'£Cli(}~!J_._61

Sf

=

R = --I_+~+~+~ ha A kI A

I

Given:

k2

Neglect

(DT2

(~TI

Magnetic

brick

brick

k3 A

unknown

L, +_J_+

k) A

1 h A b

I fib A

terms (11(1, hb and L))

TI -T3 =:>Q=------__.:__-

~~T3

Silica

...

k)- A

I + __LI + __L2 ___ h(/ A kl A k2 A

r----.---r------~

k1

TI -lJ

LI

L)

--+--kl A k2 A TI - T3

Q=

Thermal conductivity

of silica brick, kl

Thickness of silica, LI = J 2cm Thermal conductivity

of magnesite,

Thickness of magnesite,

~

Inner surface Temperature, Outside surface temperature, Contact

O. J 2

=

1'1 - T3

=

[where, Rc is contact resistance between walls]

III

RI + R2 + Rc

k2

=

5.5 W/mK

= 22 ern = 0.22 TJ

RI + R2

J.7 W/mK

=

m

920" C + 273

= J J 93

T 3 = 120 C + 273 0

resistance between t, .•vo wall, Rc

=

K

= 393 K

Q=

1193 - 393

0.003 K/W

Tn find:

Temreralurc of thee

1193 - 393 surf lace SUI

. at the Interface,

Q/A =

(T 2)

0.12 + 0.22 + 0.003 1.7 5.5

So/utioll :

HCallraJl!.,fer .

thr

" .... oUb" composite

7042.9 W /m2

wall is given by

Q :;

{From £qlll1. no. (J.42)fI)

R

II MT 0(1£0 hook page No.

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43 c(

4J

I

1.62 Heal and Mass Tra:.:'.:.:ls~ife=-r

~ Conduction 1.63

We know that, Given:

Brick

Inner

Insulation

Timber

Outer

Cold

Hot

T2

1193 Q=

Ll kl A 1193 -

Q/A

T2

0.12

Diameter

of the aluminium

rivet, d

=

4 em

=

0.04 m

1.7 Thermal

conductivity

of the aluminium

1193 - T 2 7042.9

=

0.12

Area of the surface,

1.7

A

=

I

Thermal

conductivity

Thickness

Result:

Thermal

conductivity

insulatine wall has three layers of material /JeI~

together by 4cm dian:eter aluminium (k :: 200WlmK) riV~ per O.J m2 of surface. The layers of materials consist 0 12 em thick brick (k= 0.90 WlmK) with hot surface 01

Thermal

material,

conductivity

=

22 em

Cold surface temperature.

T1

=

T4

0.90 W/mK

=

L2

=

of the timber, k)

temperature,

200 W ImK

0.12 m

of the Insulation,

of the timber, L3

Hot surface

=

of the brick, kl

of the Insulating

Thickness

IE] A composite

=

O.I m2

Th ickness of the brick, L I = 12 ern

IT2 = 695.8 K

rivet,

kriv<:l

=

I.S em

k2

=

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Percentage

of increase

0.015 111

0.170 W/mK

0.22111 =

0.11 W/mK

2200 C + 273

=

493 K

ISO C + 273

=

288 K

=

220°C, 22 em thick timber (k = O.J 10 WlmK) with colt!sl'rf{lC~Tofind: at 150 C. These two layers are interposed by a third layer0, .. I . J 70 W/"'~' IIlSU ttttng material 1.5 em thick of conductivity O. CI I . .~r~ a ell ate the percentage of increase ill "eat trans)' due to rivets.

=

in heat transfer rate.

1.6./

Heal Gild

Mas ,. Transfer _ ...

----------

Solution:

(',,"dUCli(J11

I.fij

Lrivel

Heat transfer (without

Rrivd

Q

rivet)

:::

krivCI x Arivet

R

where

0,12 + 0.015 + 0.22

~T

==

[Front fI!IIT d

T I - lt~

R

17(/ A

1/(1

book '0I')

20a x 1.25 x 10-.1 I~e

/IV.';]

& Rriv":l == 1.42 K/W

1.1

L..,

L"

kl A

k] A

k,A

We know that

h,

__

"1"1- T-t

L__

_I _ +- _'-

==

Rc:quiv:llenl

=> Q II

Ila !\

~--

L

L,

k1 A

k, !\

--T--

kl A

.'

Rwall

-

I{riwt

I x ~1.·~_~i.!!IEivct_. Arca \\ ithour rivet

/ill !\

-I-

R

-_1I{rivel

L .. , (I) where

=

<)

R==

__ I ",,/\

+ __LI klA

l.~ +_k!A

I.}

I

f----

f--.--.

kjA

"I,A

{From 1I.\.It dat« hook page no. Neglect

O.(H)

O. I

O. I 70

- ----

~-----

idcriug

0.1

0.11

0.1

area

LI

L,

LJ

klA

kzA

kjA

R==--+---+--

rivet I{

Rivet

"b terms

+-. _Q·~L_ R == _.Q.JL_ + 0.015 O.QO x 0.1 0.170 x 0.1 0.11 . 0.1

'----------_ ow con

A

11(/.

= 22.2

K/W

~'rrf.l J2 Substituting

- It/t! " (O.04)~

[!\rca~~~~).-~:·.I>S_-· 10-3

Scanned by CamScanner

JlI21

R value ill Equn (I)

_1.---R.;qul\'al.:nl

==_L 22.2

,(i~'! .. 1.~5,( 10-·;) 0.1

n/

r

is 23') C.

fC!"'flt!rllfllre

iusid« resiSfllll

c!

",I'glcclillf:

thr th ermal

of.5 nun mortar joint between marble lI11dbrick.

I ttiefotlowing.

fill

1. Overall

transmitance

•. He II loss through

:. Temperature of

T I - T~

-I. Thermal

for the wall

II,e 14'(/11

lire brick - pine interfa

conductivity

e

0/ tire mortar.

[Assume ,\lorlar brick interface temperature i 21" C, Given :

heat loss is reduced

. :e='

, n.age (}

ing thickne s of mortar

v

Q p~

by J 0%/

• T,

TI

rivet.

I

e

Q,

lnvidc

0

T

l{l"

4.13-9.22

=

15 .13

hb

ha

/ 100

'% ] ., ickncvs of marble, L = 75 mm = J.O 5 m ~

A lurJ(e composite

wall ill made

up (Jf 75 mm tnJlrhltl

thermal conductivity 1.25WlmK, 7fJmm brick fJ.fi2WlmK,

2fJ5mm pine

I 75mm Imide ptaster

0/

O,J2WIK

(J.25WlmK. 1he (Julsj . ..Ie uJ I,,, 33Wlm2X and m'ilu

I,,, 12Wlm2K. Outside

tempet"

'l hcn (II conducti Lhickncw

brick

i y of marble, k4 I.}

IUft j

T hcrmal c. nducii

70 mm

=

',25

ImK

O,07() m

'I hickncss

it)' of brick, k3 = 0,62 W/mK

of pine, 1'2 = 205 rnrn - 0,205 m

Thermal c nductivity of pine k2 = 0,12 W/mK Thicknc:

s f plaster, LI = 175 mrn = 0,17

'1hcrmal conductivity

Scanned by CamScanner

=

=

aI

conductivity

unit surface conductance ,'iur/tu:e .onductance 2fJ0(' and

of conductivity

0/ conducfit1

of plaster, kl

=

m

0,25 W/mK

.e

!_._(i8 .!"I
--------_

Outside

...

.__._-----

surl;lce cOllductance ,

Inside surtilce

conducrauce

"

hl == 3'J WI I11K 2. -----~

---._---

('OIlC/llc1ioll

h == I'"- U'/ 'K '0 no 111-

ICllIpt:ratlln..:, Tb::: 20" C + 273 :.::293 K

Outside

Insidl.:· kll1pl:ralllrc,

To ==

2]0

C + 273

==

= .5

111111

== 0.005

111

l'vlortar joinllhid:lh.::-;s To/illll: I. Overall

-------.

_L

2% K

12

__

oo

2.69

~-----

Q/A __rate, -.--

;'.11

==

- -_.

W/m2/

of brick-pine

interface, (TJ)

I lcat transfer,

Q

= IJ x

ACTa" Tb)

IFrom cqun,

of the mortar

[Mortar hrick interface lelllperu/ilre i.v] It: heal/ass

is reduced

1.11 = lJ

by 10%/

y.

(296 - 293)

SOIIlI;oll :

Heat transfer through composite wall is given by ~=

We know that, Interface temperatures relation

,\Tovl.!rall

R

where 1'4 - 1 ~ _ 15 - ""

Sf:; Ttl ·-lb

= ~-

1

R

=

7;-;;11.-

LI

-I-

_

We know that,

-L Thermal conductivity

()

••

+ 0.175 + _Q)J!~~ + O.O.~, + _QJ~~ + ,_!_ 0.25 0.12 0.62 1.25 33

r.lcallransfer

(U)

2. Heal luss. (Q) 3. Tempcnuurc

---

3 'C)I/\

trunsuuuance,

-.

I.

"'1-"

L2 + "'2"

LJ

-

-Ti;,--

. , . (,)

L,J_'

+ kJ ,,- + k,J" + lib A

(1)

::::>

Q= /.,'

Scanned by CamScanner

I{"

"0. (1.43)1

( 'r,IIIItIl:I!lIn

}) .12

I 'II

,'f !

II. () (1.1 ~ 2() •.

'f

1

I

II. 115 0.12

Il, c=

~I) -:-

C

I)

.I~

'1',

L"

K\

293.22 K

TClllpcralUJ'c

T

or brick

lical loss is reduced

RI

I - pille interface

2() .22 K

I

by 10%

L\ KIA

CmlsitierillJ;

tl,iclme.\·s of tire mortar,

295.9 - T Q/A

L\

TI

kl

295.9-T2 1.11

\T2

Inside

0.175 0.25 295.12

KI 0

Mortar brick interface temper~tllre is 21 C 1

-'

=>T4=21°C

Q

T4

=

210 + 273

295.12-T3 ~ k2/\

££IIIi1t1M'6~'_

Scanned by CamScanner

I. 72 Heal and Mass li'an~k/' _--_._----.---_. - ..__ .-.. _---_----_-

Mortar thickness,

1'4 =

5

Illlll =

0.005

III

0.99:: We know that,

T5 -- 293.03 0.075 1.25

IT5::

293.08 K]

294 - 293.08 Q ::----~

L4

T(,- Tb

(2) => Q =

k4 A

---_._Rb Q/A =

Q= T6-2~~.

0.92 0.005 k4

1

hhA Q/A=

T6 - 293

_--

.i. Thermal conductivity

T6 - 293

0.99

= --.'

of the Mortar, k4 = 538

x

3

10- W/I11K

..-~-

j_

Result :

33

I. Overall transmittance,

293.03 ~"]

U

=

2

0.37 W/m K

1. Heat loss, Q

=

3. Temperature

ofhric~ - pine interface

1.11 W/1112 =

293.12 K

T< - T(J =:~

L~ ~~ ;\

Q/A

15 - 293.03 0.075 1.25

Scanned by CamScanner

4. Thermal

COlldllctj\,jl~

of the Mortar

3

:=

5.38;.10-

\\'/IllK

/. 74 Heal and Mass Trans er

1.1.12 SOLVED UNIVERSITY

f1]

PROBLEMS

ON SLAIl

.r Inner surface temperature

A [urnace wall consists of three layers. Th~ litem thickness is made offire

brick (k= 1.04 Wlt,,~Yer~

intermediate layer of 25 em thickness brick (k = IJ.69 WlmK) followed by

(I

is made if ). 1~ o nzlll'o 5cm/hick co ' ~~

(k = 1.37 WlmK). When the furnace is in continuou' tire inner surface of tirefurnace

trcrele II' ,\ °perllJ'

~

is at 800 C while t'

I~

concrete surface is at 50"C. Calculate the rate oifl,

~

per unit area r., '

0/ tire wall, the

0

Ire 0llt

.

temperature

rellllo~

at the illl"" erj

•

tirefirebrick (lilt/ masonry brick am/the lel1l1Jerlilll I Ire interface 0/ tire masonry brick (111(/ COil crete.

lice q 11/ IN

'

Outer surface temperature,

T

800 C 0

Conduction

2

1.75

1+ 73=1073K T 4 = 50 C + 273 = 323 K 0

Tofilltl :

I) Rate of heat loss per unit area of the wall, 2) Temperature

(QI A)

brick, T2

at the interface of the fire brick and masonry .

3) Temperature

at the interface of the masonry brick and

concrete, T3. Solution : (i) Heat loss per square metre (QIA)

[Anna Uuiv -June'06j

Heat transfer Q

,

Give" :

= ~ Toverall

R

where Fire

Inner side

Masonry brick

brick

( T,

Concrete wall

[From HMT data book

page

no.43 & 44 (Sixth editionj]

Outer

side

~DTJ

=> Q

=

I L, LJ L3 --+--+---+--+ha A

k, A

k2 A

k3 A

I hb A

[Convective heat transfer co-efficients ha, hb are not given. So, neglect that terms] Thickness

of firebrick,

Thermal conductivity Thid;ness

L) ==. I Ocm == 0, 10m

of

fire brick,

of masonry brick; L2

==

kl

== 1,04

25cm = 0.~5

W/Il1K III

Thermal conductivity of masonry brick, k? = 0,69 W/IllK Th ickness of '-"0 ncre t e wa II,L.1 = 5cI11 == 0,05- m Thermal conductivity

of concrete

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wall, k3

==

1.3 7 W/n1K

Q/A

_~--:---:-

~C~'o_o'lI(jIIlCliO" I. 7

Similarly Q= TrTJ R2

(I)~

where.

LJ

RJ

= _-_

-

k2 A

ii) Iff1nftK,~ tepualU,es (T] and Tj) We

T2 - T3

~ Q=

that,

J rdL-rf

see temperat ures re lat iem ~

Q/A

1515.24 I

= 927.30 - T)

0.25 0.69

ere,

~78.30KJ

'r ,- T2

0"---'

~-

Result : I. Q/A = 1515.24 W/m2

kl A

'1',-

T2

2. T2=927.30K

3. T3=378.30K

Q/A

III All 1073··· "'2 1515.24 = _-' .,.0.10 1.04

[li~_~~~. 3·~i.·.Kl

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external wall of II house is made up of 10 em commo« brick (k = O.7 WlmK) followed by II 4 em tayer (II gypsum plaster (k = 0.48 WI",K). Whu: thickness of Iml.,·(v plIL'ked

insulatlon (k

= 0.065 WlmK) slwulll

be lidded to reduce

the I,eat loss II"ougl, the wal! by 80%. [May 200-1 _ AIIIW Univ. Ocl-99 & Oct 20tJ/-

,\4.

Uj

1.78 Heat {llId Mass Tran.lfer

Give" :

Conduction

Thermal conductivity

= O.lm of brick, kl = 0.7 W/IllK

Thickness

L2

Thi(;kness of brick, LI

=

10 em

L2 kL) -,.1] R. = _l_A [-1-'ha ,_.!::Lk,. +-+-+ k -r

2

of gypsum,

=

4 em

=

0.04

of gypsum, k2

Thermal

of insulation

conductivity

=

k-

'J'

/7'1' ,

0.48 W/mK R

= 0 065 WI IllK

=

Ib

and hb are not eziven . S0, neg Iect t Iiat terms)

_l_[.!::L A

Considering

Gypsum

Brick

,3

III

[The term's

Thermal conductivity

I. 79

. where

J:L + _s_ 1

+ k,

k2

k)

two slabs, i.e., neglect L3 term

sr

[''- A

Q =----

=

I 1112)

_s_+~ k,

k2

kI

k2

6T 100 ",,' - .....__::::...:_~ _QJ_ -I- 0.04 0.7 0.41\

Ilril:~

(iYI

~IIIII

II IS

()

'III

luss is red" .cd h)' IilJ%

Thi

.(d')

I I ·!.:II('. S 01'

wall hy 80%,

msulnrion

to reduce tile

(L1)·

insulllt ion. So, hcnr trnnsfcr

()

o .

e111l:10

r Assume heat CO) == 100 WI

W,

" I

1il [lnd :

transfer

I leA

t I

. thrOl1uh

oss

10.1 0.7

I ().()il I

OJIS

1'1 (l.()(,.

I

II

'

r~

0.0. ~Ii 111/

.

Result :

Solution : J

[eat flow rate, Q

=

,1T overall

Thickness of insulauon

R

'[From I-1M?'data hook page

Scanned by CamScanner

,/1

4J I~ /11)

I.J

0.0 XX III

__j

I.SO Heat atld Mass Tratlsfer

-II)

A composite

brick, "

wt,lI consists of 10cIII Ihick lay~

= fl. 7 WlmK

Conduction /.8/

and 3('''' II,ick pla.'iter,1c == 0.5 W/~

ilI.mltlti"g nUlterialof"

= fI.ORWIIIIK is to he tltldedlo (~

the I,et,tlrtlmft!r 111T0III:" II,e wull by 411%. Fintl its th' rr~ .

{Dec- 200-1

.11I1I1I

& Dec-lO(lj

Uuiv

Id~

Considering two slabs, i.e., neglect LJ term

:ll/l/u Viii!

Give" : Thickness of brick, L I = 10 em Thermal conductivity

0.1

=

111

~IOO

of brick, k I = 0.7 WImK

Thickness of plaster, L2 = Jcm

=

0.03 m

0.7

= 0.5 W/mK

of plaster, k}

Thermal conductivity

of insulation, k 3 =: 0.08 W/mK

~:---~ast~

Insulation

k,

k,

~ I IH Q

~T

R

=

~T

Q=

I

60

Ttl find :

Heat now rate, Q -= where R

=

J__ A

(~+~+~l

1 A

-

----_._._ ----_. -_.__ ! f---. L I - --ic- - I. !---..j

Solulio" :

1

k3

[_QJ_ 0.7

+ 0.03 + _!::L_ 0.5

0.08

~1

:::) 60 r _QJ_ + 0.03 + 0.7 0.5 0.08

l

R

+

:::) 0.1 0.7

.!:l_ _I 1 kj

+ 0.03 +..!::L 0.5

L3

lib

:::) 0.08 =0.135

The terms I/{/ and "" are not given. So, neglect that ten11S. 7

Scanned by CamScanner

k2

20.28

_!_

sr overall +

kl

=

to reduce the heat loss through !Ii

(_L11(/ + .!:Lkl + .!:L k2

I

20.28 K

=

transfer (Q) = 100 WI

0.5

Heat loss is reduced by 40% due to insulation. So, heat transfer is60 W.

I

Thickness of insulation wall by 40%, (LJ).

[Assume heat

_QJ_ + 0.03

Thermal conductivity

I

~T __

= -~

0.08

=

1

20.28

= 0.338

I.S2 Heal 0/1(/ Mass nOl15jer

IL

= 0.0108

In

( 'ouduct inn 1.83

I 'If/jiuff

: i) II

RemIt: L) = 0.0 I08

Thicknc s of insulation,

o A sw{ace

10. s per quar

;It

ii) Interface

111

is made lip of 3 layers one of fire brick 0 insulating brick and one of red brick. The inner (III 'd Oil) nIl HlIIII

surface temperatures are 900 C lind so: C rejfJective~ll. n respective co-efficient of thermal conductivity of thelal'~ are 1.2, 0.14 and 0.9 WlmK lind tile thickness of 20 em,8 and L/ em. Assuming close bonding of the layer.\·at interfaces. Find the heal/on' per square meter and illlerf~ temperatures. [/11. U OCI-9 J 0

SO/lIlioll

(i)

metre, (011\)

(T2 and T )

temperature:

:

/JI'(I1/0.H

per square metre (QIA)

I kat Iran fer. Q

/Frolll

Tavera II

==

IIMT data hook

[lag(' 170 . .J3 Gild ./4

R

J

where

L2

1.1

--j--

I- I

k2 A

Given : lnsulatins

Fire

Inner

side

(bT ~

(

h"A

heat

[Convective

(

T3

.I-

So" ucglcct

k3

k2

kl

,

-....0

tha:

kIA

f-- LI --tc- L T,

Outer temperature,

T4

900 C + 273 0

= =

30° C + 273

1173 K

= =

k,

Thermal

of

brick '- k?

Thennal COlldllCti Thi

knes

Thic~ne Thicklle

s

it

iusulatins

of red brick, LI

=

f Iire brick, f' III ulatiug

brick, L2

r re d

L3

brick,

o

=

20

CI11 =

II em

Scanned by CamScanner

k)

=

=

= 0.14

\\/111

CIll

=

0.11

= 0.08 III

111

1 rill

J

L_ k2 A

"4 --'1'1'-- ---LI

I.,

Q/A

L

1

k, A

hb A

-'-~-

co-efliciellts

transfer

L.

~

1173 - J03

III

8

k2A

~

= 0.9 W/rnK

0.2

KIA

~'~

1.2 W/rnK

of fire brick,

conductivity

01

303 K

Thermal conductivity

L2 --

=

LI --j--

Inner temperature,

LI

--j--

brick

brick

side

-~0

Outer

Red brick

0._

0.08

ill

1.2'

0.14

0.9

h(/, hb

arc nut

given.

/. 84 (ii)

Heal and Mass rransfer

Interface teperatures (T) am/ Tj) ~ We know that, Interface temperatures relation

Condur, I.85 ---------------____:_._-;0/1

Q/A TI - T"

Q

R 1004.457 - T3 ----.::.... 0.08 0.14

1011.2546= where LI

R1=---

kl A

Result: (i) Heat loss per square meter (Q/A)

TI - T2 LI kl A

Q/A = 1011.2546 W/m2 (i i) Interface

temperatures

(T 2 and T 3)

TI - T2

Q/A

T2 = 1004.457 K

L,

1;1011.2546=

IT

T3 = 426.597 K

1l7J-T2 0.2 1.2

2 = I 004.4 57 K

[I)

I

Similarly

The wall of a furnace is made lip of 2.f0 mm fire clay of thermal conductivity 1.05 WlmK, 120 mm thick of ins Illation brick of conductivh, O.15 WlmK anti 200 mm thick red brick of conductivity

0.85 WlmK. The inner and outer surface

temperatllre oj wall are 850 C tlntl 65- C respectively. Co/cilIate the temper(l/ures at the contact surfaces. G

[Bharathida an

niversity

ov- 95}

Given: Thickne

of fire cia

Thermal conducti

Scanned by CamScanner

L1=2S0mm

=0.

5m

ity, kl = 1.05 W/mK

Thi knes of insulation

bri

Thermal

2 = 0.15 W/mK

ndu tivity,

., ~ = 120

m = 0.1

In

/. 86 Heal and

Thickness

Mass Transfer of red brick, L3

Thermal conductivity,

k3

==

==

200 nun

==

---

0.2 rn

__ Conduction 1.87 [Heat transfer co-efficients ha and hb are not given. So, neglect that terms]

0.85 W/mK

Inner surface temperature,

T1

Outer surface temperature,

T 4 = 65 + 273

==

850 + 273

== ==

1123 K 338 K

~

Fire

Insulation

clay

brick

• T,

k,

k2

T,-T4

=

L,

L2

Redbrick QfA

_s_+ ~ k,

Q/A

kJ

T

slab is gi en by

1123-338

0.25 + 1.05

= 616.46

Q/A

T, - T2

Q=--

1~

[From H \17 data book page no. 43 & 4.{

'erall

where

~ Q=

fA

Scanned by CamScanner

k3

= ------

Souaion : ....

+S_

k2

We know that,

ro eft composite

k3 A

TI-T4

=

I

Heat

~

k, A + k2 A

°nj

(~T2

Q

=

RJ

0.12 + 0.2 0.15 0.85

W/m21

> 1.88 Heal and Mass Transfer Conduction 1.89

1123 - T2 => 616.466 = 0.25 1.05

.' Similarly

wall made up of 7.5 cm offire plate anti O.65 em of mild steel plate. Inside surface exposed /0 hot glls lit 650" C anti outside air temperature 27" C. The convective Ilea/transfer co-efficient for inner side is 60 WIm1K. The cOnl'ective heattmnsfer co-efficient for outer side is 8 Wlm1K. Calculate the heat lost per square meter area of the furnace

wal! and also find outside sutface temperature. {M U. April-98]

T2 - T3

Q=~-

(I)=>

@] A furnace

Given:

where

Fire

Mild steel

plate

plate Outside

Jnside T(I'

=> Q=

n2

n,

hb

f.c-- L,

976.22 - T3 Q/A

k2

==

-----fo--- L2

--l

L2

Thickness of fire plate, LJ == 7.5 em == 0.075

k2

Thickness of mild steel, L2 == 0.65 cm == 0.0065 m

976.22 - T3 => 616.46

T/J'

Ita k,

=>

( TJ

0.12 0.15

Inside hot gas temperature, Outside air temperature, Convective

III

T a = 650 C + 273 0

= 923 0

0

T b == 27 C + 273 == 300 K

heat transfer co-efficient

for

inner side, ha == 60W/m2K Convective

heat transfer co-efficient

for

outer side, hb= 8 W/m2K.

Result: Tofi IIlI : (i) 1'2 == 976.22 K

(ii) T3;: 483.05 K

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(i) Heal lost per square meter area, (QI A) (ii) Outside surface temperature,

(T3)

K

1.90 Heat and Mass Transfer

Solution :

Conducnnn I.91

(i) Heat lost per square meter area, (QIA) Thermal

for fire _ plate (Refract ory clay) k, = 1.0035 W/IIIK.

{From H.UT data book page no. 9 (I- iflh edition •

Thermal

.

conductivity

or page

•

.

I) JlO.

-

.

Sitt"

r.~

(ii} Olltside surface temperatllre,

conductivity

We know that, Interface temperatures

retarion

edlt,

for mild steel plate

k2 = 53.6 W/I11K

... ( I)

[From HMT data book page liD. where Heat flow,

Q

Toverall R

where TrTb

~

Q= J

LI ~ --+---_ kJ A k2 A

fib A TJ

QIA

- Tb J

fib

~Q

[The term LJ is not given. So, neglect that term

,

I

8

I

Ta- Tb

~Q

T) - 300

=

2907.79

T3

=

663.473 K

I

Result : (i) Heal lost per square meter area, (Q/A)

Q/A

.. Q/A

Q/A =

_, +

I

60 QI1\

=

923 - 300 0.071+ 0.0065. 1.035 53.6

(i i) Outside

I

t-"8

= 2907.79

W/m2

surface temperature,

.. T]

=

(T 3)

663.473 K.

2907.79 W/m2!

(b!

Scanned by CamScanner

( 'flliilliNil/il

1\'1

-

[]

I.)

"j

I

h(l II t I'

I A t. IJ,

''fill ..

Fh

Imllilll(inA

bl'k~

brick

.:Iori/.

Il'hul

I

I 'J 1

I

kl A

,,~ A

II", 1..1 II lid "b nn

1'1Ilt!

Lj I

IIO(

"I

A

hi)

/I

II,lv·,1. So. nogl

tthar rerrns]

6 0

Q/A

0.23 0.115 --+-OL 0.27

kI

872 W/m2!

I---

Ll--~--

Thi kness

f fire bri k. L]

r

f insulating

L =

j

ern

0.23

=

Result:

III

Rate of heat lost per square meter, (QI A)

i kness

brick, L_

I!. - ern

=

=

0.115m Q/A

Thermal c ndu tiviry of fire brick e al conductivity

of insulating

perature difference,

6T

=

kI

=

brick, k2

=

0.2 \\/rr

[!]

TI,e 60 em

650 K

872 W/m2

inner x

dimension

of a freezer

cabinates

are

60 em. The cabinates wall consists of /HIo 2 mm

thick steel wall (k = 40 WlmK) seperated by a 4 em layer of

Tofind:

fiber

10

=

0.72 W/rnK

per square meter, Q/A

SOlUlvm:

glass

insulation

(k = 0.049 WlmK). D

TI,e inside

C and the outside

temperature

is 10 be maintained at _I5

temperature

on a hot summer day ;J 45° C. Calculate the

maximum amount of heat transfer, assuming a heat transfer

To /erall R [From IIMr data book page

co-efficient no.

4J'

of 10 Wlm] K both on inside and outside of tile

cabinate a/JO calculate outer surface temperature

of tile

cabinate. [M. U. Oct-2002}

Scanned by CamScanner

JUt" /. 94 Heat and Mass Transfer Give" "

Conduction 1.95 where Fiber

Steel

L\T=Ta-Tb

Steel

glass (DT2

R

(

k3

k2

Thickness of fibre glass, L2 Thermal conductivity Inside Temperature,

=

=

4 ern

k3

=

=

0.04

of fibre glass, k2

T b = 450 C + 273

Heat transfer co-efficient,

110 = lib

==

l:J

)

ha A

k A

k2 A

kJ A

hb A

J

258 - 318 _'--

III

+ _0.002_ +

) OxO.36

=

/Q

III

=

(ii) Outer surface temperature

+_,_ )Ox0.36

10 W/1112K.

Q

T -Tb

= __:::_[/-"-

R

inside]

(T~

TI - T2 _ Tr

R, T 3 - T4

R3

(T4)

(i) Maximum (111101111' O/hNI"Tnm/er

-21.25=

Heal flow, Q ~ .1Tovera"

0.36 x 10

R [From /I MT data book page 110.43 &

10

We know that,

318 K

"

Scanned by CamScanner

0.002 40x0.36

=-21.25W]

(i) Maximum amount of heat transfer, (Q)

Solutio"

+

0.04 0.049xO.36

[The negative sign indicates that heat flowsfrom outside

0.049 W/IllK

Tofind :

(ii) Outside surface temperature,

40x0.36

40 W/mK

To = _150 C + 273 = 258 K

Outside Temperature,

I~

J

Q/A

Thickness of steel, L) = L3 = 2 mill = 0.002 of steel, ",

L

~ Q =

Area, A = 60 ern x 60 ern = 0.36 m2

Thermal conductivity

,

--+--+--+--+-

=

4t

-

T3

R2

= T4 - Tb Rb

.•. (I)

-1. 96 Heal and Mass Transfer Conduction I. 97

Result: (i) Maximum amount of heat transfer,Q

T4

(ii) Outer surface temperature,

=

:::-2 I .25 W

Tojiml: (i) Rate

312.09 K

l!1 A mild

steel tank of wall thickness 10 mm Contllins IIIUle'l 90 C. Calculate tile rate of heat loss per ml Of tank surfll/J. area when the atmospheric temperature is 15 C. rite tile,.". conductivity of mild steel is 50 WlmK ami tile hea: trUIU!, co-efficient for inside ami outside tile tank are 2800 "II WlmlK respectively. Calculate also tile temper(lturt~ tile outside surface of tile tank. [M U. Apr-2000]

0f

heal loss per m2 of tank surface area (QI A)

(ii) Tank outside surface temperature (T2) Soilltio" :

0

Heat loss, Q

=

.1Toverall R

0

where .1T

= T{/- Tb

__ 1_+_S_+~ R - haA k, A k2A

+_!1_+_I_ kJA hbA

[LJ' ~ not given.So, neglect that terms]

Give" : Inside

k

T(I> ha

~P2

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