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[ ANNA UNIVERSITY SYLLABUS I _ME1351 : HEAT AND MASS TRANSFER _

hanical Engineering -(R'egulation :201l4fFor B.E. VI Semester Mec -- --- ,l. CO~OU,~TlONt_ Mechanism of Heat Transfer - Conduction

BaSIC Concep!) - ... f '. d Radiation - General Differential equation 0 HeatConvection an .. d C I' .

ducii FOllrl'er Law of Conduction - Cartesian an y mdncalCon UCllon - .C d t s One Dimensional Steady State Heat Conduction -oor lila e - .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 - BoundaryLayer Concept - Types of Convection - Forced Convection _Dimensional Analysis - External Flow - Flow over Plates, Cylinders andSpheres - Internal Flow - Laminar and Turbulent Flow - CombinedLaminar 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

EXCHANGERSNusselts theory of condensation - Pool boiling, flow boiling,

correlations in boiling and condensation. Types of Heat Exchangers _LMTD Method of Heat Exchanger Analysis -- Effectiveness _ NTUmethod of Heat Exchanger Analysis - Overall Heat Transfer Coefficient _Fouling Factors.4. RADIATION

Basic Concepts, Laws of Radiation - Stefan Boltzman Law, KirchoffLaw - Black Body Radiation - Grey Body Radiation Shape FactorAlgebra - Electrical Analogy - Radiation Shields - Introduction to GasRadiation.5. MASS TRANSFER

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

Note : . (Use of Standard Heat and Mass Transfer Data Book ispernulled 117 the University Examination).

CONTENTS

CHAPTER 1: CONDUCTION~H;at Transfer 1.1

. 1.1.1. Modes of Heat Transfer _ I. I1.1.2. Fourier Law of Conduction .. .1.21.1.3. General Heat Conduction Equation in

Cartesian Co-ordinates . 1.21.1.4. General Heat Conduction Equation in

Cylindrical Co-ordinates 1.91.1.5. Conduction of Heat through a Slab or Plane Wall.. .1.141.1.6. Conduction of Heat through a Hollow Cylinder 1.161.1.7. Conduction of Heat through a Hollow Sphere 1.171.1.8. Newton's Law ofCooling........................ . 1.191.1.9. Heat Transfer through a Composite Plane Wall

with inside and Outside Convection 1.191.1.10. Heat Transfer through Composite Pipes (or) Cylinders

with Inside and Outside Convection 1.221.1.11. Solved Problems 0" Slabs 1.251.1.12. Soilled University Problems 011 Slabs 1.741.1.13. Solved Problems 011 Cylinders 1.1111.1.14. Solved University Problems 011 Cylinders 1.1441.1.15. SO/lied Problems 011 Hollow Sphere 1.160

1.2. Critical Radius of Insulation 1.167I ? I Critical Radius of Insulation for a Cylinder 1.1671:2:2: Solved Problems 1.169

1.3. Heat Conduction with Heat Generation 1.1791.3.1. Plane Wall with Internal Heat Generation 1.1791.3.2. Cylinder with Internal Heat Generation 1.1831.3.3. Internal Heat Generation - Formulae Used 1.1851.3.4. Solved Problems 011 Plane Willi with Internal

Heat Generation 1.1871.3.5. Solved Problems 011 Cylinder with

Internal Heat Generation 1.1961.3.6. Solved Problems Oil Sphere with

lnternul Heat Generation 1.202

1.4. ~.i:SI. T;~~~~t~Fi;~~::::::::..::::.:..:::.:::::::::::::::..::=:':::::::::~~~1.4.2. Temperature Distribution and Heat

Dissipation in Fin . 1.2061.4.3. Application......... . 1.21


r.:1.4.'-l. Fill Ftliciellc)' . ..1.2171.~.5. Fin rfkcriVt'ness. " .. 1.2171.~.6. Ftll"lllllllicUsed..... . 1.2ISIA. 7. So/I't!d Proh/ellH" 1.2191.4.8. SII/I'd U"itl(!f.5i~I' Prublctn 1.2451.4.9. Pr()hkllll/Or Practice .. 1.263

15. Transient Heat Condul~tioll (or) Unsteady StateConduction 1.2641.5. I. l3ior Number . . 1.2641.5.:? Lumped Heat Anal)' is . 1.2661.5.3. Solved Problems -1_llIlIped lleat AII(I~I/JiJ .............. 1.269'.5.4. Solllcd University Prohlelll.,-Llllllped

tteot AII(I~I'jiJ ................................................. 1.288I. -.S Heat Flow in Semi-lnfiutie Solids 1.3061.5.6. SO/lied Problems - Semi-illfillite Solids 1.30R1.5.7. Transient Heat Flow in an Infillite Plate ... 1.3291.5.8. Solved Problems - lufintie Slllitl, 1.332I.S.9. SO/lied University Problem" - Infinite Solids 1.351

1.6. 1'11'0Mark QlleJtiOlIl & AII.'"II'en 1.374

CHAPTER II : CONVECTIVE HEAT TRANSFERI)i 111(~nsiona I A IIa lysis -..-..-..-..-..-..- -..-..-..-..-..-..- :~~2.1.1. Dimensions ... 2.12 1.2. Buckingham 1I Theorem. . .. . 2.22.1.3. Advantages cf Dimensional Analysis . 2.32.14. Limitations of Dimensional Analysis 2.3Dimensionless Numbers and their Physical Significance 2.42.2.1. Reynolds Number (Re) 2.42.2.2. Prandrl Number (Pr) 2.42.2.3. Nusselt Number (Nu) 2.5

~;~~t~::t~~r~,~~::;(~~?........; ;2.2.6. ~e\Vlonion and Non-Newtollioll Fluids 2.6

EL;;;~~~:~~::;~p:~:.:::::::::.::::~~~:::..:::::::::::::::::::~~::::::::::::::~:7;2 ~ I. Types of Boundary L~;~;~:::";'82.:.~. !iydrodynalllic Boundary 1.~~r: .. .. 292.).). r~lenn;]IUoUfldarylayer y .. 29

i~lIlve~~~:lt~;;:~ 'l'~~'"rC' ......: ,..:::::::::::::::::::::::::::::::::2: 92 -u. Types ofC~nveoc!i onveC!rOfl 2.9

011.... . 2.9

2.1.

2.2.

2.3.

2.4.

( 'onteuts ( . 3

2.4.3. Free (or) Natural Convection . 2.92.4.4. Forced Convection .. ..2.10

2.5. The Local and Average Heat Transfer Coefricients forFlat Plate - Laminar Flow 2.10

2.6. The Local and Average Heat. Transfer Coefficients forFlat Plate-Turhulcnt Flow 2.132.6.1. l leat Transfer ocificient for Combination of

Laminar and Turbulent Flow... .. 2.152.7. Boundary Layer Thickness, Shear Stress and Skin Friction

Coefficient for Turbulent Flow 2.IR2.8. Heat Transfer 1'1'0111 Flat Surfaces - Formulae Used 2.23

2.8.1. Problems 011 Flat Surfaces - Forced Convection 2.262.8.2. Solved University Problems 011 Flat Surfaces -

Forced Convection 2.832.9. How over Cylinders and Spheres 2.115

2.9.1. Formulae Usedfer Flow Over Cylindersand Spheres 2.116

2.9.2. Solved Problems - Flow Over Cylinders 2.1172.10. Flow over 'lalli, of Tubes 2. 122

2.10.1. Formulue Used lor Flow Over Balik of Tubes 2.1232.10.2. Solved Problem 2.124

2.11. Flow through it Cylinder -Internal Flow 2.1262.11.1. Formulae usedfor Flow tit rough

Cylinders (lnternul flow) 2.1272.11.2. S;)lved Problems - Flow through Cylinders

(lnteruat Flow) 2.1292.11.3. Solved University Problems - Internal Flow 2.150

2.12. Free Convection 2.1622.12.1. Formulae Used/or Free Convection 2.1622.12.2. Solved Problems 011 Free Convection (or)

Natural Convection 2.1652.12.3. Solved University Problems - Free Convection 2.194

2.13. Problems for Practice 2.2172.14. TII'o Murk Questions {lilt! Allswers 2.219

CHAPTER III: PHASE CHANGE HEAT TRANSFER AND HEATEXCHANGERS

3.1. Boiling and Condensation ~.I3.1.1. Introduction . ).1

.... 3.13.1.2. Boiling .Condensation 3.1

. . .3.1Applications .3.1.3.3. 1.4.


C.4 Heat and Mass Tram/a

Boilll1g Heat Transfer Phenomena 3.2Flow Boiling... .. 3.4Boiling Correlations J.)Solved Prohlellls 3.7Solved A11IUIUniversity Problems 3.23Condensation. . 3.29Modes of Condensation 3.29Filmwise Condensation 3.29

3.1.13. Dropwise Condensation .. 3.303.1.14. Nusselt's Theory for Film Condensation 3.303.1.15. Correlation for Filmwise Condensation Process 3.303.1.16. Solved Problems Oil Laminar Flow,

Vertical Surfaces 3.323.1.17. Solved Problems Oil Laminar Flow,

Horizontal Surfaces 3.543.1.18. Solved Anna University Problems 3.613.1.19. Problems for Practice 3.65

3.2. Heat Exchangers 3.663.2.1. Introduction 3.663.2.2. Type of Heat Exchangers 3.663.2.3. Logarithmic Mean Temperature Difference (LMTD) 3.733.2.4. Assumptions 3.733.2.5. Logarithmic Mean Temperature Difference for

Parallel Flow 3.733.2.6. Logarithmic Mean Temperature Difference for

Counter Flow 3.773.2.7. Fouling Factors 3.813.2.8. Effectiveness by Using Number of

Transfer Units (NTU) 3.823.2.9. Problems on Parallel Flow and Counter

Flow Heat \:cllangers , 3.823.2.10. Problems 011 Cross Flow Heal Exchangers (or)3 2 I Shell and Tube Heal Exchangers 3.1 093'2'1~' Solved Anna UI1iversity Problems 3.1173'2'13' Solved Problems Oil NT(! Method 3.J243'2' 14' ;m1b"1University Solved Problems 3.138" . ro ems for Practice

3.2.15. Two M k . .. .. .. 3.145ur Questions and AI1swers 3.146

3.1.5.3.1.6.3.1.7.3.1.8.3.1.9.3.1 10.3.1.11.3.1.12.

CHAPTER IV : RADIA nON4.1. Introduction ..4.2. Emission Properties 4.1

4.1

Contents C.5

4.3. Emissive Power 4.14.4. Monochromatic Emissive Power 4.24.5. Absorption, Reflection and Transmission 4.24.6. Concept of Black Body 4.34.7. Planck's Distribution Law 4.44.8. Wien's Displacement Law 4.44.9. Stefan-Boltzmann Law 4.54.10. Maximum Emissive Power 4.54.11. Emissivity 4.64.12. Gray Body 4.64.13. Kirchoff's Law of Radiation 4.64.14. Intensity of Radiation 4.64.15. Lambert's Cosine Law 4.74.16. Formulae Used 4.74.17. Solved Problems 4.84.18. Solved University Problems 4.254.19. Radiation Exchange Between Surfaces 4.314.20. Radiation Exchange Between Two Black Surfaces separated

by a Non-absorbing Medium 4.314.21. Sha pe Factor 4.364.22. Shape Factor Algebra 4.364.23. Heat Exchange Between Two Non-Black (Gray)

Parallel Planes 4.374.24. Heat Exchange Between Two Large Cocnentric Cylinders or

Spheres 4.414.25. Radia tion Shield 4.454.26. Solved Problems 4.494.27. Solved Problems 011 Radiation Shield 4.604.28. Solved University Problems 4.794.29. Electrical Network Analogy for Thermal Radiation Systems

by Using Radiosity and Irradiation 4.IOO4.30. Radiation of Heat Exchange for Three Gray Surfaces 4. 1044.31. Solved Problems 4.1054.32. University Solved Problems 4.1294.33. Radiation from Gases and Vapours 4.1534.34. M ea n Bea m Length 4.1544.35. Solved Problems 4.1554.36. Problems for Practice 4.1664.37. 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. J ntroductlon 5.15.2. Modes of Mass Transfer .. .. S.15.3. Diffusion Mass Transfer .. .. S.15.4. Molecu~ar ~iffusion .. .... 5.25.5. Eddy Dlffuslon 5.25.6. Convection Mass Transfer .. 5.15.7. Cocentrations .. .... 5.25.8. Fick'~ Law of Diffusion ...... 5.35.9. Steady State Diffusion through a Plane Membrane 5.4~.I O. So/J'ed Problems Oil Concentrations 5.6~.II. Solved Problems Oil Membrane 5.175.12. Solved Univeristy Problems on Membrane 5.215.13. Steady State Equimolar Counter Diffusion 5.235.14. Solved Problems Oil Equimolar Counter Diffusion 5.265.15. Solved University Problems 011 Equimolar

Counter Diffusion 5.315.I6. Isothermal Evaporation of Water into Air 5.345.17. Solved Problems on Isothermal Evaporation

of Water into Air 5.355. J 8. Solved University Problems Oil Isothermal Evaporation

of Water into Air 5.445.19. Convective Mass Transfer S.545.20. Types of Convective Mass Transfer 5.545.21. Free Convective Mass Transfer 5.545.22. Forced Convective Mass Transfer 5.S45.23. Significance of Dimensionless Groups 5.545.24. Formulae Used for Flat Plate Problem.') 5.565.25. Solved Problems on Flat Plate 5.575.26. Anna University Solved Problems 011 Flat Plate 5.655.27. Formulue Usedfor Internal Flow Problems 5.685.28. Solved Problems on Intemal Flow 5.695.29. University Solved Problems 5.725.30. Problems for Practice 5.755.31. Two Mark Questions and Answers 5.76

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

Chapter 1: Conduction

cr Basic Concepts

CF General DifferentialEquation

0" Fourier Law of Conduction

C7 Internal Heat Generation

c:r Extended Surfaces

c- Unsteady Heat Conduction

cr Solved Problems

(7' Solved University Problems


CHAPTER-I

1.CONDUCTION

1.1 HEAT TRANSFERHeat transfer can be defined as the transmission of energy

from one region to another region due to temperature difference.

1.1.1 Modes of Heat Transfer* Conduction* Convection* Radiation

Conduction

Heat conduction is a mechanism of heat transfer from a regionof high temperature to a region of low temperature within a medium(solid, liquid or gases) or between different medium in direct physicalcontact.

In conduction, energy exchange takes place by the kinematicmotion or direct impact of molecules. Pure conduction is found onlyin solids.

Convection

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

Convection is possible only in the presence offluid medium.

Radiation

The heat transfer from one body to another without anytransmitting medium is known as radiation. It is an electromagneticwave phenomenon.

'2


I.': Heat tnd HII.\.\' Transfer

1.1.2 Fourier tal" or ConductionRate of heal conduction is proponional 10 the area mea lIred

1101'111:11 to the iirection of heat OO\V and io the temperature gradientin that direction.

Q

\ here

A - Area in 111-

dT _ Temperature gradient in k/mdr

k - Thermal nducti iry in W/m"-

Thermal conducti it)' is defined a the abilit fa ub tan eto conduct heat.

[The negative sign indicates that the h at 0 w in a dire ti nalong which there is a decrease in temperature]

1.1.3 General heat conduction equation incartesian coordinates

Consider a small rectangular element fide dx, d I andd: as shown in Fig.I.I.

The energ balance of this rectangular element i btain dfrom first law of thermodynam ics.

Net heatconducted into Heat 1 Heat l

=> element from generated st red

all the coordinate \\ ithin the t = in rhel directions element j elern nt../... 1.1

Conduction 1.3

O.Element volume

O.

C.Ch)

Fig. 1.1.

Net heat conducted into element from all the coordinatedirections.

Let q x be the heat flux in a direction of face ABO andq d x be the heat flux in a direction f face EF H.

The rate f heat' fl '" . t th I . dinine e ernent In x irection throughthe face AB 0 i

I Q, dz I ... (1.2where k hermal nducti ity, W/mK

ernperature gradient

The rate heat fl \ t f tJ I .u re e ernent In x directi n thr ughthe fa e EFGH i

Q +dx Q ax ) dxaT

-k -d I d:x


/4 Heata~_ .. ~

Subtracting ( 1.2) - (1.3) .

= -k aT dydz _I-k ~.QI dydz=Ox - Q(I' + dxl x AX l. . ox

~ [kx aT Jdx dy dZ]ax ax

er of= -k. - dydz + kx -8 dydz +.t ax x

=> Q _ Q = .1_ [kx or] dx dy dz.I' (.I' + dx) ax ax ... (1.4)

Similarly

Q)' .- Q (y + (M = ~ [ky :;] dx dy dz .. (1.5)

(1.6)

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

Net heat conducted = a~ [k'l: g: Jdt dy dz +

Conduction 1.5

Net heat conducted into element from all the coordinatedirections

~[ :.[ kx :] + M ky :] + ![k, :] ] dx dy dz... (1.7)

Heat Stored in the element

We know that,

{He~t stored}

m the =element

{Mass Of} { SpeCifiC} { Rise in }

the x heat of the x temperatureelement element of element

aTm x Cp?< at

!,. aTp x dx dy dz x Cp x at

[v Mass = Density x Volume]

{Heat stored in } er

the element = p Cp at dx dy dz ... (1.8)

Heat generated within the element

Heat generated within the element is given by

Q = q dx dy dz ... (1.9)


1.6 Heal ami Mass Tran.~ler

Substituting eqllation (1.7), (1.8) and (1.9) in equation {I.I)

(\'1) ~ I;_ [k\ ~ 1 +....c [k,. c~] + a~ [k: ~ J] dx L~l' dz.. ,\ 0' 0 0_+ q dx dy dz P Cp or dx dy dzat

Considering the material is isotropic. So,

k, = ky = k, = k = constant.

&r &r -~r] or-+-+-k+q=pC-ax2 0~ c:z2 P at

Divided by k,

iJ1r in q p Cp or+-+-+- =--0,1 &2 k' k at

... (1.10)ora: at

It is a general three dimensional heat conduction equationin cartesian coordinates

where,k

a: =;: Thermal diffusivity = -- - m2/s. . .. pCp .

Thermal diffusivity is nothing but how fast heat is diffusedthrough a material during changes of temperature with time.

Conduction I.?

Case (i) : No heat sources

In the absence of internal heat generation, equation (1.10)reduces to

02r 02r 02r+-+-ax2 0,2 az2

oroc at .. (1.11 )

This equation is known as diffusion equation (or) Fourier'sequation.

Case (ii) : Steady state conditions

In steady state condition, the temperature does not change

with time. So, or = O. The heat ~onduction equation (1.10)at'

reduces to

iJ2r 02r q+-+- +-0'2 o:z2 k =0 ... (1.12)

(or)

) qV-T + - = 0

k

This equation is known as Poisson's equation.

In the absence of internal heat generation, equation (1.12)becomes :,

... (1.13)

(or)

This equation is known as Laplace equation.


I.8 Heal and Ma.H' Transfer

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

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

o2T q---- 1 z- 0ax'; I.; . (1.14)

In the absence of interns I heat generation, equation ( 1.14)becomes:

'" (1.15)

Case (iv): Two dimensional steady slate "eat conductio"

If the temperature varies only in the x and y directions, theequation (I. 10) becomes:

... (1.16)

In the absence of internal heat generation, equation (I. 16)redcues to

tYT if1-j---- =0ax2 oyl ... (I. 17)

Case (,~: Unsteady state, one dimensional, without internalheal generation :

, oin un~teady state, the temperature changes with time,i.e., -a, :t: O.So, the general conduction equation (I. J 0) reduces to

()

o: ("'" ... (1.18)

-Conduction J, 9

1.1.4 General Heat Conduction Equation in CylindricalCo-ordinates

The general heat conduction equation in cartesiancoordinates derived in the previous section is used for solids withrectangular boundaries like squares, cubes, slabs etc. But, thecartesian coordinate system is not applicable for the solids likecylinders, cones, spheres etc. For cylindrical solids, a cylindricalcoordinate system is used.

Consider a small cylindrical element of sides dr, dcj> and dzas shown in fig.I.2.

I

: drI

J~_(r,4J,z/ '

//

dz

Fig.J.2

Elemental volume

Q(r+dr)

The volume of the element dv = r d~ drdz .

I Let us assume that thermal conductivity k, Specific heat epand density p are constant.


1.10 Hea/I.Jf1J Mass Transfer

The energy balance of this cylindrical element is obtainedfrom first law of rhrmodynamiocs.

INet heat

conducted intoelement from

all the coordinatedirections

{

Heat}-' (Heat }generated I stored

+ within the = ~ in theelement L element

... (1.19)

Ne! heat conducted into element from (II/ the co-ordinatedirections

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

Q_ = -k (r d~ dr) ~ de- az

/ Heat lea ing from the element through (r, ~) plane in time de.oQ= T d: = Q= + oz (Qz) dz

Net heat conducted into the elem~nt through (r, ~) plane intime de.

= Q= - Qz + dz

. = - -#-- (Q_) dzaz -

= ! [k (rde.dr). [~J de J dz- . [OZTJ- .- k 0:;2 (dr~rd~.dz) de

l... (1.20)

Net heat couducred _ OZT .through (r,~) plane - k [oz2] (dr.rdq,.dz)de

Conduction 1.11

Heat entering in the element through (~, z) plane in time dB.

Or=-k(rd~dz) cr deorHeat leaving from the element through (~, z) plane in time de.

Q = Q + _E_(O )drr + ell' r or r

Net heat conducted into the element through (~, z) plane intime de.

= Qr- 0,. + dr

= - -#- (Q ) dror r

= - :,. [-k (rd~.dz). [ :Jde Jdrk (dr d~.dz). +L1/: aT] de

or or

Net heat conductedthrough (~, z) plane k (dr. rd~.dz) [~T + _!_ ~TJ dea,.1 r cr

I... (1.21)Heat entering in the element through (z, r) plane in time de.

Q~ = -k tdr.dz) or dera~Heat leaving from the element through (z, r) plane in time de.


I. 12 Heat and Mass Transfer

Net heat conducted into the element through (z, r) plane in

rime de.t,

Q, - Q~.,d~ = - ,.a~ (Q+) rd$

= _ __L [ -k tdr dz). or de] rdro$ ro~

a [' OT]= k - - - (dr dtk dz) de04> r D~ If

= k [l_ &TJ (dr rd~ dz) de_r2 8$2 .

/Net heat conducted _ . [' cPT .L~hrOugh (z, r) plane - k 7i WJ (dl rd$ dz) de ... (1.22)

Net heat conducted into element from all the co-ordinatedirections

k ~; (dr rd$ dz) de

+

k (dr rd~ dz) r&T + .!.._ OT] deLor2 r 0,.+

k C~~n(dr rd~ dz) de[Adding equation 1.20, 1.21 and 1.22J

= k (dr rdqldz) de r- &T + iYT +_}_ oT I a2r J-oz2 or2 ,. or +-;2 ~)~2

Conduction 1.1J

Net heat conducted into element from all the co-ordinatedirections

= k (dr rddz) de r 0.2T +.!.._ aT1_ or2 r or

I &T &T-/+-- - +-r2 ocp2 i3z2 _

... (1.23)

Heat generated within the element

Total heat generated within the element is given by

Q = ci (dr rddz) de . .. (1.24)

Heat stored in the element

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

Increase in internal energy

= Net heat stored in the element

= p (dr rd$ dz)Cp or x de00 ... (1.25)

Substituting equation (1.23). (1.24) and (1.25) in (1.19)

(1.19) ::::::> k (dr rd$ dz) de 1- &T + L or + .L &T I ~!J-_ 0,.2 ,. a,. ,.2 acp2 OZ~+ q (dr rddz) de

p (dr rd$ dz) ep L'T x dereDivided by (dr rd$ dz) de

::::::> k r&T + _!_ OT_ + .L &T + &TJ' + .L or2 ,. 8,. ,.2 0$2 oz2 q =- p. C, :-


... (1.26)

It is a general three dimensional heat conduction equat'~ IOnin c~'lindrical co-ordinates.

flT + L aT + .L &T + &T + ~ = l_ aTOIl r r ,.2 13cp2 az2 k ex ae

If the flow is steady, one dimensional and no heat generaion ,equation ( 1.26) becomes:

&T + _!_ aT = 013,.2 r a,. ... (1.27)

(or)

... (1.28)_!__ _{__ l,. dT J = 0,. dr . dr

1.1.5 Conduction of heat through a slab or plane wall

--J d~'1---Consider a slab of

uniform thermal conductivityk, thickness L, with innertemperature T I, and outertemperature T2'

Let us consider a smallelemental area of thickness 'dx'.

L --.JFig 1.3

Conduction 1./5

. we knoW that,From Fourier law of conduction,

dTQ=-kA -dr

=> Q.dr = -k A dT

. b tl e limits of 0 to LIntegrating the above equation etween 1

and TI to T2L T2

=> f Q dr = - f k A dTo TI'

L T2

=> Q f dr = - kA f dTo TI

L T2=> Q [x] =-k A [T]

o TI

=> Q [L - 0] = -k A [T2 - Tj]

Q = kA [T 1 - T2]L

TI- T2

... (1.29)

Q LkA

~T overallR

... (1.30)Q =

where

~T = T1- T2

R = C. - Thermal resistance of slab.


/.16 Heal and MLI.\"J Tram/er~~~~~~~~--------~------------ 6 C' nduction of Heat Through Hollow Cylinderl.L 0

onsidcr a hollow cylinderofillllcr radiu rl' outer radius r2,inner tcrnperatllr~ T I' outertemperature T2 and thermalcOllducti, it ".

Let II c 11 idcr a smallelemental area of thickness "dz"

From Fourier law ofc nduction, we know that,

dTQ=-kA -dr

Area of a cylinder is 27trL

A = 27trL

s ,dTQ = -k27trL -dr

Fig 1.4

d,.Q x - = -k27tL dT,.Integrating the above equation from rl to "2 and TI to T2

~, dr T2Q J r = - k27tL f dT

rl TI

Q [111'2

p'neep

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

lntegrating on both sides1r 2

:::>Q dr:;;; _ j 47tk dT

rl ,.1 11T)

'2 (-

:::>Q I d~ == - 41tk . dT

. r 1'1 1

'2 12

:::>Q \=1-1 == - 41tk [T]

r Tlr 1

:::> Q lL- l1== - 41tk[T2 - Td'1 '2:::> Q (r2-rl)== 41tk[T1-T2]

I I rl r2

Q ==41tk [T, - T2]

:::> r2 - '1r1 r2

Q==T1- T2

:::::> r2 - rl

41tk (rl r2)

Q=ilT overall

:::::>R

where

... (1.34)

... (1.35)

_ r2 - '1. .R - 4 k ( - Iherrnal resistance of hollow sphere.

1t '1 '2)

Conduction 1.191.1.8 Newton's Law of Cooling

Heat transfer by convection is given by Newtons law of cooling

... (1.36)

where

A - Area exposed to heat transfer in m2

h - Heat transfer co-efficient in W Im2K

T s - Temperature of the surface in KT a: - Temperature of the fluid in K.

1.1.9 Heat Transfer Through a Composite Plane Wall withInside and Outside Convection

Consider a composite wall of thickness L1, L2 and L3 havingthermal conductivity kl> k2 and k3 respectively. It is assumed thatthe interior and exterior surface of the system are subjected toconvection at mean temperatures T Q and T b with heat transfer co-efficient hQand hb respectively. Within the composite wall, the slabsare subjected to conduction.

Convection

A

Fig 1.6


~

il1.20 Heal and Mass Transfer

From Newton'S law of cooling, we know that,

Heat transfer by convection at side A is

Q = ha A [Ta - T, J [From equn. (1.36)]

Heat transfer by condl1ction at slab (I) is

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

Heat transfer by conduction at slab (2) is

Q= k2A[T2-T3]

L2

Similarly at slab (3) is

Q= k3A[T3-T4JL3

Heat transfer by convection at side B is

We know that,

To-T, =Qx_1haA

T,-T2 =Qx~. k( A

T2 - T~ = Q L2) x~k2 A

T3 - T4 =Q x ~k3 A

T4-Tb=Qx_'_hb A

[From equn. (1.37)]

[From equn. (1.38)]

[From equn. (1.39)]

[From equn. (1.40)]

[From equn. (1.41)]

... (1.37)

... (1.38)

... (1.39)

... (lAO)

... (1.41)

Conduction /.21

Adding both sides of the above eq tiua Ions

=> Ta - Tb = Q [_1_ + _!j_ + L2 L3 I 1hA -+ +a k ( A k2 A k3 A hb A

=> Q=

[_I +__s_ L2 L3 I]h A +-+-+

a k( A k2 A k3 A hb A

~ToverallR ... (1.42)=> Q =where

Thermal resistance R = R + R + R + R R, a I 2 3+ b

We know that,

R=_l_UA

Ta-Tb=> Q= __ .:::..._I_UA

=> Q = UA [T a - Tb ]/

where

... (1.43)

'(0" hIS t e overall heat transfer co-efficient (W /m2K).


~I

r1.24 Heal and Mass Transfer

Q = .1ToverallR

-~ " (1.48)

where

we know that,

IR= VA

Ta-Tb~ Q=

_I_VA

~ Q = VA [To - Tb J ... (1.49)

where

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

Conduction I.25

t,1.1J Solved Problems On Slabs

fZJ Determine the heat transfer through the plane of length6 111, heigh' 4 m and thickness 0.30 m. The temperature ofinner and outer surfaces are 1000 C and 400 C. Thermalconductivity of wall is 0.55 WlmK.

Give" :Inner surface Temperature, T I = 100 C + 273 = 373 KOuter surface Temperature, T2 = 40 C + 273 = 313 KThickness, L = 0.30 m

Area, A = 6 x 4 = 24 m2Thermal conductivity, k = 0.55 W/mK

Tofilld:

I. Heat transfer (Q)Solution :

We know that, heat transfer through plane wall is

Q = .1ToverallR

[From Equn. 110. {I. 30) orHMT DOlO book (C P Kothandaraman) page no. 43 (Sixth editiont]


1. 6 Heat and Mass Transfer

where

373-313= 2640 watts

0.300.55 x 24

Q = 2640 watts IResult:

Heat transfer, Q = 2640 W

In A wall of 0.6 m thickness having thermal conductivity oJ1.1 WlmK. The wall is to he insulated with a material havingan average thermal conductivity of 0.3 WlmK.lnllerandoutersurface temperatures are 10000 C and 100 C respectively. Ifheat transfer rate is 1400 Wlm1 calculate the thickness ojinsulation. Wall Insulation

I- -I- -IGiven:

Thickness of wall, LJ = 0.6 III

Thermal conductivity of wall,kJ = 1.2 W/mKThermal conductivity ofinsulation, k2 = 0.3 W/mK ~nlInner surface Temperature,TJ = 1000 C + 273 = 1273 KOuter surface Temperature,T) = 10 C + 273 = 283 KHeat transfer per unit area,0/A = 1400 W/m2

Conduction 1.27

Tofi"d:Thickness of insulation (L2)

SOIUlion:Let the thickness of insulation be L2

We know that,AToverall

Q = R [From qun no. 1.42 (or) HMT Data bookpage no.43 & 44 (Sixth edition)]where

AT=Ta-Tb (or) T)-T3

L ~ L3 JI +_)_+ __ + __ +_R = ha A k) A k2A k3A hb A

::::> Q =I L) ~ L3 I

haA+ k)A+ k2A + k3A + hbA

Heat transfer co-efficients ha' hb and thickness L3 are not

given. So, neglect that terms.

[T)- T31::::>Q=

L) L2--+--k) A k2 A

[T)-T31L) L2-+-k) k2

1273-283::::> 1400 = 0.6 + .!:1.

1.2 OJ

[L2 = 0.0621 ~

Result : .' = 00621 mThickness of I11sulatJOll, Lz .


1.28 Heal and Mass Transfer

III The wall of {I cold room is composed of three layer. Tillayer is brick 30 em thick. The middle layer is cork e OilIerthick, 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'esfilm co-efficient for outside air and brick is 55.4 Wlm2/( . ~heco-efficient for inside air and cement is J 7 Wlm2 K. Fin~~i/",~wro~ ~

Takek for brick = 2.5 WImK

k for cork = 0.05 WlmK

k for cement = 0.28 WlmKGiven:

Thickness of brick, L3 = 30 em = 0.3 mThickness of cork, L2 = 20 em = 0.2 m

Thickness of cement, L) = IS em = 0.15 m

Inside air temperature.T a = -200 C + 273 = 253 KOutside air temperature, Tb = 2S0 C + 273 = 298 KFilm co-efficient for inner side, ha = 17 W/m2KFilm co-efficient for outside, hb = 5S.4 W/m2K

kbrick = k3 = 2.S W/mKkcork = k2 = O.OSW/mKkcement = k) = 0.28 W/mK

Inside Cement Cork Brick

k( k2 k)

Outside

ZFF

Conduction 1.29

Tofintl:Heat flow rate (Q/A)

solution: . . bHeat flow through composite wall IS given y

~ToverallQ == R

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

where

=> Q =

=> Q/A1 L( L2 L3 1-+-+-+-+-

ha k( k2 k3 hb

253 - 298=> Q/A ==

1 + 0.l5 +_Q1__+..Ql_+_l_17 0.28 0.05 2.5 55.4

I Q/A == -9.S W/m2!

The negative sign indicates that the heat flows from the outsideinto the cold room.

Result:Heat flow rate, Q/A == -9.5 W/m2


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 temperatureof the outside air is 25 C and tltat on the inside air is -200 CTI,efilm co-efficient for outside air and brick is 45.4 WI",2Kand for inside air 011(1 cement is 17W/m2 K.

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

Takekfor brick = 3.45WlmK

Ii for cork = 0.043 WlmKk for cement = 0.294 WlmK

Given :

In ide

ement Cork Brick

k k2 kJJ

Outside

j kne f brick LJ = 20 em = 0.2 m

f ernent LJ = 5 em = 0.05 m

UI ide air temp ~rature, Tb = 250 C 273 = 298 K

III ide air len perature, To = -200 273 = 253 K

Conduction 1.31Film co-efficient for outside air and brick, hb = 45.4 W/m2KFilm co-efficient for inside air and cement, ha = 17 W/m2KK) = 3.45 W/mK

K2 = 0.043 W/mKK( = 0.294 W/mK

Tofind:I. Heat flow rate

2. Thermal resistance of the wallsotutio :

Heat flow through composite wall is given by

Q= ~ToverallR

[From Equn (1.42) (or)

HMT Data book page No.43 &44Jwhere

~T=T{/-Tb

I L( L2 L) IR =--+--+--+--+--

ha A kJ A k2 A k) A hb A

=>Q

=> O/A

=> QIA253 - 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~ e OUt.into the cold room. s~ (i;"C'II .'

COnd/l(;liml I.JJ

For Unit Area

I LI L2 L) I~ R =-+-+-+-+-ha kl k2 k) hb

_1_ +0.05 + _QL + _Q1_ +_1_17 0.294 0.043 3.45 45.4

~ IR = 2.634 KJW IRault:

I) Heat flow rate, Q/A = -17.081 W/m2

2) Thermal resistance, R = 2.634 K/W

(I] A furnace wall is made up 0/ three layers, imide layer MI~~thermal conductivity 8.5 WlmK, the middle layer MIll'conductivity 0.15 WlmK, the outer layer with cO/l(luClivi~0.08 WlmJ(. T,.'lerespective thickness of the inner, n~i{ldlea~outer layers are 15 em, 5 em , and 3 em respectively- T~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 ~resistance, heat flewlml and interface temperatures-

1.34 Heal and Mass TransferSolution: .---=--------------..._

1. Equivalent electrical circuit for COIl(llICtioll

2. Heat flow through composite wall is given by

Q=L\ToveraJl

R[From Equn. 110. (1.42) (or)

HMT Data book page No. 43 & 44Jwhere

L\T=Ta-Tb =T,-T4

, L, L2 L3 IR = ha A + k, A + k2 A + k3 A + hb A

~ Q =

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

= - ...-.-.--~.-.----

Conduction 1.35

873 - 323

9.25 + 0.05 + ~QL.8.5 0.25 0.08

~IQ/A = 909.97 W/I11~

3) Thermal Resistance

(Neglecting ha' hb terms)

For unit area

0.25 -I- 0.05 + 0.03.8.5 0.25 0.08

4) Interface temperatures

We know that,

r, - '14 T, - 1'2 T:2 - T 3 T 3 -- T4 1)=:) () -- -,-{--- = -R-,-= --~ = R3 ... (

=-"> Q = T,-T2R,


I:36 Hear and Mass Transfer

klA

TI -T2LIkl

909.97 = 873 - T20.258.5

Q/A

IT2 = 846.23 KjSilllilarly

(1) =:> o = T2-T3, R2L2R2=--k2A

Q = T2-T3L2

k2A

where,

T2 - T3

~k2

909.97 = 846.23 - T30.05 -0.25

Q/A

I T3 = 664.23 K I

Conduction 1.37

Resllll:

I. Heat flow per m2, Q/A = 909.97 W/m2

2. Thermal Resistance, R = 0.604 K/W

3. Interface temperatures, T 2 = 846.23 K

T) = 664.23 K

@ A mild steel tank of wall thickness 20 111111 contains water fll100" C. Estimate lite loss of heat per square metre area of litetank surface, if lite 11IIIk is exposed 10 OIl atmosphere til15 C. Thermal conductivity of steel is 50 WlmK, while hemtransfer co-efficient for lite outside wid inside II,, tank are10 WI1112Kand 28.50 WI",2K respectively. WIItII will he litelemperalllN' Oil lite outside o.f lite tank wall.

Given :

Thickness of steel wall

LI = 20111111= 0.02 InInner water temperature

T a = 100 C + 273 = 373 KAtmospheric air temperatureTh= WC+273=288K

Thermal conductivity ofsteel, k I = 50 W /mKInside heat transfer co-efficient, ITa = 2850 Whn2KOutside heat transfer co-efficient, hb = I0 W/I11~K

Inside Outside



where

Q = TI - T2LI

klA

Q/A TI - T2=-L-I-

~

873 - T2909.97 = ----=-

0.258.5

IT2 = 846.23 KjSimilarly

(1) ~ Q = T2-T3R2

L2R2=--k2A

Q = T2 - T3L2

k2A

T2 - T3

~k2

where,

Q/A

909.97 = 846.23 - T30.050.25

/ T3 = 664.23 K I

Reslllt:

I. Heat flow per m2, Q/A = 909.97 W/m2

2. Thermal Resistance, R = 0.604 K/W

3. Interface temperatures, T 2 = 846.23 K

T 3 = 664.23 K

Conduction 1.37

@] 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 tiletank surface, if tile tank is exposed to an {Itmo."plwre (It/50 C. Thermal conductivity of steel i...50 WlmK. while heattransfer co-efficient for tile out s ide and in ...ide tile ttlnk areJOWl",] K am/ 2850 Wlm2 K respectively. What will he thetemperature Oil tile outside 01 the tank wall.

Given:

Thickness of steel wallL, = 20mm = 0.02 m InsideInner water temperatureTa=100C+273=373K ToAtmospheric air temperature haTh= 15C+273=288KThermal conductivity ofsteel, k I= 50 W ImKInside heat transfer co-efficient, "a = 2850 W 1m2 KOutside heat transfer co-efficient, lIb = 10 W/m2K

Outside

1\...- -----\


I. 38 !...~:!.!~and Mass TransferTo .Ii" tI :

------- .._--- ----------_" Conduction 1.39

i) Ileal loss per square metre area of the tank Surface ( /" T "d T , Q 1\)II) ,Hl1\ outsi c temperature, 2

We know that,

T -Tl T -T, T,-T2 T2-TbG ~ 1I _~ __ = _-R- =R;-- R, RhQ =

Solutio" :

Heal loss, AToverallQ = ---R--- [From Equn, I/o. (I.42j~I-IMT Data book page NO.4J & 44J

T -T,=> Q = _G __Ra

where1

where, Ra = -, A11

I-H=Ta-Tb

I L, L2 LJ IR = --+--+--+- +__ha A k, A k2 A k3 A hb A

_ Ta-T,=> Q/A - I/h

a

(Neglect L2, L3 terms)

373 - T,

1/2850

=> IT, = 372.7 KI=> 843.66 =

=> f)/A SimilarlyI LJ I-+--+--

h{/ kJ fib

'_.) Q/;\ 3 73 - 288

--'- -I- _O_:_Q + .L2S50 50 '()

Llwhere, R, = k,A

T,- T2L,

kJA,------------- --------.-

:- (V/\ -, S43.6(j Willi ~1- _. ---. -_- .-._------ "_- 1

=> Q/A



372.7 - T2=> 843.66 = 0.02

50

I T] = 372.4 K IResult:

I. Heat loss per m2, (Q/A) = 843.66 \Vlm2

2. Outside surface temperature, T2 = 372.4 Ko A steam boiler furnace is made of fire clay. rite hOI CQJtemnerature imide the boiler furnace is 2100 t, roo- .

r .., (II,temperature is 50 'r, heat flow by radiation from caSel" 10inside surface of the wall is 25.2 k Wlm], cOllveclio" ""(11transfer co-efficient at tlte interior surface is 12.2 1111I'/(,thermal conductance of the wall is 58 WII1IK, heat flow ~v

radiation from extemal susface to slirrouluJi"K is8.2 kWln'; and interior wall sill/ace temperature is 1080"('.Calculatefor external surface

I. Sur/ace temperature2. Convective conductance

Given:

Conduction I..J IHot gas temperature. Ta = 2100 C

Room air temperature. T b = 50 C + 273 = 323 K

Heal now b) radiation from gases 10 inside surface of the

wall. Ol{ 1 = 25.2 kW/m2 = 25.2 x 103 W/m]

Convective transfer at interior, "0 = J 2.2 W/m2KThermal conductance of the wall = 58 W/mK

Ileal now by radiation from external surface to surroundinc

QI{2 = 8.2 kW/11l2 = 8.2 10) WlrnK

Illleril r wall urface temperature, 1'1 = 1080(, + 273= 1353 K

Tafind :i) External surface temperature, '1']

ii) E.\lerll,il C nvc rive c nductancc. hh

Solution :

We kn: \ Ihill

Total hca: }e n I c r i n ~ () =the \I all

Heat Inn fer h~nvc ti n (II interior

l le at Irat I .fer h)radiation at interior


r,I " TrLlm!er

1.42 Heal (11/(/ Mass

We kn \\ !haL'1 - T, T, - T2 _ T2 - '1'"

T-l"_.-!!----R - 0.-E-- _ R I ,,/;o -s: R (IT, - TZ

~Q==~

WJ- T2LA,I---J .V't"" - R I

1353 - T23 .644 == I

-8

l''- Re5istance = d Jcon uctancc

External surface temperature, T2 = 703.9 K

Heal loss due to radiation at exterior

Hear 10>5by vection at exterior

= T! -I hea entering - Heat loss by radiation at exterio

= 37.644 - 8.2 Y 103

--..'Qc = _9.!

---- lib A T ,= 29, 4

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

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

F.\lCrnal 'omccl; e CO-Cf!lCI'CIlt I )__ , 'b:: 77.3 W/III-K

('(lnductiun 1.43

Resull:I. [.x!

/I('a1 and Mm.' Transfer _~ ~ ~_,= __= ,~._,_c_ .~

Thickncss , (1I1;1:-ollary. 1., = 0 em r-, 0.20 rJ1 -,Thcnna I (I IH.lIII..'1ivity ~I .!-' 0.66 \\ IIIIK

1 hidlll::-'l f me rtar, L_ = 3 em == 0.0" rnhenna! ondu tivuy Oflllol1ar. "2:;- 0.6 W/IIlK

. hi ~1l~:\S of limcst ue. L, = ~ ern == 0,08 IIIThLTm. I 'Olldllclivily, "3 == 0.58 W/I11K

Thi I..n~ss f Plasicr. L4 = I.~ em == 0.012111Thermal OJ1dll uvity, "4 z: 0.6 W/mKInterior 111;31 trnnsfcr, co-efficient h" = 5.6 W/m2K

Exterior hctlllrallsfcrco-cfficient, hh = II W/m2K

lntcr: r room ll:rnpcratIJrc, To = 22" C 273 = 295 KOllh';dc air Itl1lpcratIJre Tb = - o C - 27" = 26R K.

Tolind:. ) (h c.:rall lint transfer co-efficient, U

11) (h'~rjJII !lrcnlltll reo istancc, (I

I

I

I 46 Hid__:_____ eo a17 Mass Tran.~ferInterface te -------..' mperature between mortar and the linrel'IOIIe, T -..........Interface temperatures relation . 3

Q _ To--TI "1-T2 "2-T3 _ T3-"4 T4-l--~=-R-I-=~ -~-==~R 5..> 4

TS-Tb=~

=> Q=T,,-' TI

Ra

Q295 - TI

)/ha A

Q/A295 - TI

11170

34.56295 - T,

1/5.6

IT, = 288.8 K)T,- T2

=> Q=---R,

Q288.8 - T2

Q/A

L,k,A

288.8 -, T2L,

",__ _3!8.8 - ~~

34,56 - 0.20

0,66

IT? = 27S,3E]

Conduction 1.47

278_3 - T3

~k2A

278.3 - T3

~k2

278.3 -- T334.56 =

0.Q30.6

Q=

Q/A

Temperature between Mortar and limestone (T]) is 276.5 K

Result:

I) Overall heat transfer co-efficient, U = 1.28 W/Il12K2) Overall thermal resistance, R = 0.78 K/W3) Heal transfer, Q/A = 34.56 \\ /m2,1) Temperature between mortar and lilllcstone,(T3) = 276,5 K

[!J The wall of (/ refrigerators is made lip oftwo mild steel plates2.5 111111 thick with (I 6 em tltick glass wool ill between theplates. The interior temperature is' -20" C, while tile outsideof 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 A{/III01'Oj University /l.'OI'-I.}-I)

Ll = LJ

= 2,5 111111 =-= 0.0025 III

L'l = () ern = 0.06 III


I.48 Heal and Mass Transfer

I -Glass woolMild steel Mild steel

T k, k2 k3a Tb

I I~. L, I- L2 L3-/- ..,

Ta = -200 C; Tb = 400 C

k, = k3 = 23 W/mK; k2 = 0.015 W/mK

Tofind :

i) Heat flow, (0)

Solution:

Heat flow through composite slab is given by

,1To vera IIR

Q [From Equn. no. (1.42)]

where

o

Conduction 1.49

Convective heat transfer coefficient is 1I0t given.

So, neglect ha' lib terms

~QIA.

Q/A -20 - 40- -.---.0.0025 + 0.06 + 0.0025

23 .015 23

I Q == -14.99 W/m2!The '-ve' sign indicated that the heal flows from the outside

into the refrigerator.

Result :i) Heat flow, Q == -14.99 W/m2.

IZ!fl The inside temperature of the refrigerator is -1 tl" Cand outsidesurface tempera/lire is 30t) C and area is 301111. Thisrefrigerator consists of 2.2 ntnt of steel at the inner surface,15 111mplywood at the outer surface and J() em ofglass woolill between steel (lilt!plywood. Calculate the heat Ion and thecapacity 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 Temperature, T, =_IOL'C f'273 =-=263



k k2 k)I

Steel Glass wool Plywood

(DTI ~DT2 ~ T)~~

Outside temperature, T4 == 30 C + 273 == 303 K

Area, A ==30 m2

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

Thickness of plywood, L3== 15 mm == 0.015 m

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

Thermal conductive of steel, k) == 20 W ImK

Thermal conductivity of plywood, k3 == 0.05 W/mK

Thermal conductivity of glass wool, k2 == 0.06 W/mK

Toflnd :

i) Heat loss, Q

2) Capacity of the refrigerator

Solution:

Hear flow throu?h composite wall is given by

t1To vera lJQ=

R[From Equn. no.(/.42) ~

HMT Data book page No.43 & 441

Conduction /.51

where

(Convective heat transfer co-effficients ha and hb are notgiven. So, neglect that terms)

T)-T4

Q 263 - 3030.002 + 0.10 + 0.015

20 x 30 0.06 x 30 0.05 x 30

I Q =-610.1 W==-0.610KWIThe -ve sign indicates that the heat flows from the outside

into the refrigerator.

We know that

3.5 kW ==I ton

:=)O.610kW== 0.610 ton3.5

==0.174 ton

:=) Capacity of the refrigerator ==0.174 ton


'{

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

Heat transfer, Q ::::610 WCapacity of the refrigerator:::: 0,174 ton

f1D A s'tetlm to liquid heut exclulIlger area of 25cOllslrllelell ",illi 11.5 em nickel and 0.1 en, / .'"' ;, I 7'1 .. ' p (II'"ncopper Oil lite steam S/{Ies. I lie re.\'ISIIV/~V of (I .Vfller t> OJdeposil 011 lite steam side is {}.0015Kiw. The steun, -sCa/!(I1lti/itsurftlce C{JIIt/uel'IIIee lire 5400 WI", 2K and 56() W. 14~reJpeClive~y.Tile Itealeds'lell", is filII Ou C (flllillefllet[ ~",2kis' al 70 C. "ql/id

CalculateJ) Overall steam IO/iquidllelll transfer co-efficient2) Tempemllire drop IICrOS,\'lite settle deposit

Takek(copper) = 35(1 W'ImK filii/ k(Nickel) = 55 WlmK.

Given :

Inside [Liquid side 1\2

GT2

Til 11(/ :~~.:CopperOutside

Steam side

Thickness of Nickel L - 0 .-, I - ) em = 0,5 x 10-2 111Thickness of copper L - 0 It' ' '2- , cm=O.1 x 10--111

Resistivity of scale, R_l = 0,0015 K/W

I' id Conduction I 53.iqut surface conductance, h(/ - 560 W/m2K .

Steam surface' dcon uctance, hh = 5400 W/Ill2KSteam temperature, Tb = ) )00 C 27" - ", , _,+ ., - .,83 KLiquid temperature: T(I = 70" C + 273 = 343 Kk2(copper) = 350 W/IllKk) (Nickel) = 55 W/IllK

Toflnd :i) Overall heat transfer co-efficient, (U)2) Temperature drop across the scale deposit. (T, --T4)

Solutio II :Heat transfer through composite wall is given by

Q = ~ToverallR

rFront Equn. no, (I. 42) orlIMT Data hook page No.43 & :J:J}

where

R = _I_+_L_I_ ... L2 ,L) Iha A k A . kA T kA + _-

, 2 3 "b ARa + R, + R2 + RJ + Rb

R,~ value is given, R" = k~~ = 0.00) 5 K/W~

RI L,

---t--lin A "I A

_-'- __ + 0,5 x 10-2560 x25.2 55 x 25,2

+ 0, ) x 10-2350 x 25,2 + 0,0015

+ I5400 x 25,2

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


/.5.1

U 25 W/m2KOverall heat transfer co-efficient,

(T~ - T ) across the scale is given byTemperature drop .) 4

L1T [.: I:H = T3 - T4]Q=--Rsca/e

, L1T25.2 x W = 0.0015

~ltlT=37.8CI

Result:

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

Temperature drop across the scale, (T3 - T41 = 37.80 C

Conduction/.55fllJ A wall of (I fllrll(~c~ is made up of 13 em thict: of fire day,of

thermal condllctlv/~" fJ.6 WlmK alU160 em thick of red brickof conductivity 0.8 WlmK. Tire inner ~nd outer surface. . I. .temperature of wall are 1"0000C and 750 C Determine

1. Tile amount of heat loss per square metre of lirefurnace wall.

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

Give" :

Furnacek, k2

Fire clay Red Brick

IL,=13cm= 0.13m

k, = 0.6 W/mK

L2 = 60 em = 0.6 111k2 = 0.8 W/mK

T, = 1000 C + 273 = 1273 K

T3 = 75 C + 273 = 348 K

Tofind: f the furnace wall.I) Heat loss per square metre 0. k=OIII+O.OOOIST.2) Thickness of the material whose .


.__ ._-" - ---Sol"tim, :

1. I teat transfer through composite wall is given by

Q = Il To\' Q = R)

Q =1273-T2

L,k, A

1273 - T2=> Q/A =

=> 956.8 = 1273- T2I.,

k,

1273 - 1"2-0.13

0.6

,.: A = 1m2 J


---/. 58 Heal and Mass Trumjer-------------------

T) - T4L3

kJ A

.T) - T4QIA = __!2__

kJ

Q=

whereT4 - Outer surface temperature of red brick'

k) - Thermal conductivity of red brick

L) _ Halfofthe thickness of the red brick == 026 = O.3mT) - 348

~ Q/A ==0.3D.8

~ 956.8 = T) - 3480.3IT

~ IT) = 706!KJ

(I)~ Q= T2-T1.R2

1065.6-706.8Q=

.... '(2)

Conduction J .59

.956.8 =' 358.8.. ' , L2

k;Given thermal conductivity for diatomite is

k = 0.111 + 0.00015 T

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

k2 = 0.111 + 0.00015 [T2: T31

= 0.111 + 0.00015 [1065.62+ 706.81

I k2 = 0.243 WimK I, (

Substitute k2 value in Equation (2)

~ 956.8 == 358.8~.0.243

~. L2 = 0.091 m

Thickness'ofthe diatomite, L2 = 0.091 m), '" .

Remit:

Heat loss, Q = 956.8 W1m2Til 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 thercontlllctivill' 1.7 W/mK, 12 em thick and outside m ",~

~ ~ IV (Ig'l~f'brick of thermal conductivity .11 ,,'/mK, 22 em thic ",,temperature Oil the inside of the wall of the silica bk: rh,

. . bri t. r'rk'92(1UC (111(1olltsult! magnesite nc sur/ace tem'Pe ~rmUre'120"C. Calclliate the heatflow tit rough tit is compos 't ~

I e IVaI(lf the ('011 tact resistance between the two wall is 0.003Ktltfind tile temperlllllre of the surfaces at the illter/ace. I

Given: r----.---r------~k2k 1

(~TI (DT2 ~~T3Silica Magnetic

brick brick

Thermal conductivity of silica brick, kl = J.7 W/mK

Thickness of silica, LI = J 2cm = O. J 2 III

Thermal conductivity of magnesite, k2 = 5.5 W/mK

Thickness of magnesite, ~ = 22 ern = 0.22 mInner surface Temperature, T J = 920" C + 273 = J J 93 KOutside surface temperature, T 3 = 1200 C + 273 = 393 KContact resistance between t, .vo wall, Rc = 0.003 K/W

Tn find:

Temreralurc of the surf .e SUI lace at the Interface, (T 2)So/utioll :

HCallraJl!.,fer thr " ..... oUb" composite wall is given by

Q :; {From qlll1. no. (J.42)fI)R 43 c( 4JII MT 0(10 hook page No.

~;:::-::--------- ~C~o'~ld1!_''!'Cli(}~!J_._61where

Sf = TI -lJ

R = --I_+~+~+~ ... 1ha A k I A k) A k A h A- 3 b

I LI L2 L, I___ + __ + __ +_J_+h(/ A kl A k2 A k) A fib A

Neglect unknown terms (11(1, hb and L))

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

LI L)--+---kl A k2 A

TI - T3

RI + R2Q=

1'1 - T3

RI + R2 + Rc

[where, Rc is contact resistancebetween walls]

1193 - 393Q=

1193 - 393Q/A =

0.12 + 0.22 + 0.0031.7 5.5

7042.9 W/m2 I


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

We know that,

1193 - T2Q=

Llkl A

Q/A1193 - T2

0.121.7

7042.9 =1193 - T 2

0.121.7

IT2 = 695.8 K IResult:

IE] A composite 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 012 em thick brick (k= 0.90 WlmK) with hot surface 01220C, 22 em thick timber (k = O.J10 WlmK) with colt!sl'rf{lC~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.

Conduction 1.63

Given:

Inner TimberBrick Insulation Outer

Hot Cold

Diameter of the aluminium rivet, d = 4 em = 0.04 m

Thermal conductivity of the aluminium rivet,

kriv

1.6./ Heal Gild Mas ,. Transfer_ ...----------

Solution:

Heat transfer (without rivet) QR

where

~T == T I - lt~ [Front fI!IIT d 1/(1 book I')'0 I~e /IV.';] &

R 1.1L.., L"

17(/ A kl A k] A k , A h,

"1"1- T-t=> Q

II L L,Ila !\

~-- --T--kl A k1 A k, !\ /ill !\.'

r

iusid fC!"'flt!rllfllre is 23') C. ",I'glcclillf: thr th ermalresiSfllll c! of.5 nun mortar joint between marble lI11dbrick.fill I ttiefotlowing.

1. Overall transmitance for the wall

. He II loss through II,e 14'(/11

:. Temperature of lire brick - pine interfa e

T I - T~ -I. Thermal conductivity 0/ tire mortar.

Q v

p~ TI T,

, n.age (} I rivet.e .e

Q, lnvidc T0l{l" hb

4.13-9.22 ha/ 100= 15 .13

'% ]

~ A lurJ(e composite wall ill made up (Jf 75 mm tnJlrhltl

thermal conductivity 1.25WlmK, 7fJmm brick 0/ conducfit1fJ.fi2WlmK, 2fJ5mm pine of conductivity O,J2WIK aII 75mm Imide ptaster 0/ conductivity (J.25WlmK. 1he (Julsj

. ..Ie uJunit surface conductance I,,, 33Wlm2X and m'iluIUft j,'iur/tu:e .onductance I,,, 12Wlm2K. Outside tempet"

2fJ0(' and

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

Given : heat loss is reduced by J 0%/

. :e=' ing thickne s of mortar

., ickncvs of marble, L = 75 mm = J.O 5 m

'l hcn (II conducti i y of marble, k4 = ',25 ImKLhickncw brick I.} = 70 mm = O,07() mT hcrmal c. nducii it)' of brick, k3 = 0,62 W/mK

'I hickncss 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 m

'1hcrmal conductivity of plaster, kl = 0,25 W/mK


!_._(i8.!"I

Outside surl;lce cOllductance hl == 3' WI 2. -----~, " J I11K

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

Outside ICllIpt:ratlln..:, Tb::: 20" C + 273 :.::293 KInsidl.: kll1pl:ralllrc, To == 2]0 C + 273 == 2% Kl'vlortar joinllhid:lh.::-;s = .5 111111 == 0.005 111

To/illll:

I. Overall trunsuuuance, (U)

2. Heal luss. (Q)

3. Tempcnuurc of brick-pine interface, (TJ)

-L Thermal conductivity of the mortar

[Mortar hrick interface lelllperu/ilre i.v] It:heal/ass is reduced by 10%/

SOIIlI;oll :

Heat transfer through composite wall is given by

() ~=,\Tovl.!rall

Rwhere

Sf:; Ttl -lb

1 LI L2 LJ L,J_'R = 7;-;;11.- -I- "'1-" + "'2" + kJ ,,- + k,J" + lib A

('OIlC/llc1ioll I. (/1)---._----------. -. --- __ oo _

'C)I/\

_L + 0.175 + _Q)J!~~+ O.O.~, + _QJ~~ + ,_!_12 0.25 0.12 0.62 1.25 33

32.69

r.lcallransfer rate, Q/A == ;'.11 W/m2/~----- __ -.-- - -_.

We know that,

I lcat transfer, Q = IJ x ACTa" Tb) IFrom cqun, "0. (1.43)1

1.11 = lJ y. (296 - 293)

We know that, Interface temperatures relation

1'4 - 1 ~ _ 15 - "" . , . ( , )= ~- - -Ti;,--

(1) ::::> Q =

/.,' I{"


( 'r,IIIItIl:I!lIn I 'II

}) .12 ,'f !II. ()

(1.1 ~

2() . 1 'f I

II. 1150.12

Il, I) .I~ K \'1', T

~I) -:- c=RI

C

L" 293.22 K ITClllpcralUJ'c or brick - pille interface 2() .22 K I

lical loss is reduced by 10%

L\KIA

295.9 - TCmlsitierillJ; tl,iclme.\s of tire mortar,

1.11

L\kl

295.9-T2

0.1750.25

Inside

T IQ/A

\T2 295.12 K IMortar brick interface temper~tllre is 21

0

C

1 -' Q =>T4=21CT4 = 210 + 273

295.12-T3~

k2/\

IIIi1t1M'6~'_


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

Mortar thickness, 1'4 = 5 Illlll = 0.005 III

We know that,

T(,- Tb---_._-Rb

Q= T6-2~~.1

(2) => Q =

hhA

T6 - 293Q/A= _--

.i.

0.99T6 - 293= --.' ..-~-

j_33

293.03 ~"]

T< - T(J=:~

L~~~ ;\

Q/A15 - 293.03

0.0751.25

0.99:: T5 -- 293.030.0751.25

IT5:: 293.08 K]

294 - 293.08Q ::----~L4

k4 A

0.92Q/A = 0.005

k4

Thermal conductivity of the Mortar, k4 = 538 x 10-3 W/I11K

Result :

I. Overall transmittance, U = 0.37 W/m2K

1. Heat loss, Q = 1.11 W/1112

3. Temperature ofhric~ - pine interface = 293.12 K

4. Thermal COlldllctj\,jl~ of the Mortar := 5.38;.10-3

\\'/IllK


/. 74 Heal and Mass Trans er

1.1.12 SOLVED UNIVERSITY PROBLEMS ON SLAIl

f1] 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 is made if ). 1~

o nzlll'obrick (k = IJ.69 WlmK) followed by (I 5cm/hick co ' ~~

trcrele II'(k = 1.37 WlmK). When the furnace is in continuou' ~

,\ perllJ'tire inner surface of tirefurnace is at 8000 C while t' I~

Ire 0lltconcrete surface is at 50"C. Calculate the rate oifl, ~

. rellllo~per unit area 0/ tire wall, the temperature at the illl""

r., ' erj lice qtirefirebrick (lilt/ masonry brick am/the lel1l1Jerlilll

I Ire 11/ INinterface 0/ tire masonry brick (111(/ COilcrete.

[Anna Uuiv -June'06j

Give" :

Inner Fire Masonry Concrete Outerside brick brick wall side

( T,

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

We that,

J rdL-rf see temperat ures re lat iem

I ere,

'r ,- T20"---'

~-kl A

'1',- T2Q/A

1073 "'21515.24 = _-' .,.-

0.101.04

[li~_~~~.3~i..Kl

_~--:---:- ~C~'o_o'lI(jIIlCliO"I. 7

Similarly

(I)~ Q= TrTJR2

where. LJRJ = _-_- k2 A

T2 - T3~ Q=

~ Q/A

1515.24 = 927.30 - T)0.250.69

~78.30KJ

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

2. T2=927.30K

3. T3=378.30K

III All external wall of II house is made up of 10 em commobrick (k = O.7 WlmK) followed by II 4 em tayer (II gypsumplaster (k = 0.48 WI",K). Whu: thickness of Iml.,(v plIL'kedinsulatlon (k = 0.065 WlmK) slwulll be lidded to reducethe 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" :Thi(;kness of brick, LI = 10 em = O.lmThermal conductivity of brick, kl = 0.7 W/IllKThickness of gypsum, L2 = 4 em = 0.04 III

Thermal conductivity of gypsum, k2 = 0.48 W/mKThermal conductivity of insulation k- = 0 065 WI'J' IllK

Ilril:~

Brick

k 2

Gypsum

k I

(iYI ~IIIII

" I

1il [lnd :. I t I . thrOl1uh II

Thi !.:II('. S 01' msulnrion to reduce tile leA oss 'wall hy 80%, (L1) .

Solution :J [eat flow rate, Q =

,1T overallR ,/14J I~

'[From I-1M?'data hook page /11)

Conduction I. 79. where

R = _l_ [-1-' ,_.!::L L2 L) .1]A -r +-+-+. ha k,. k k -,2 ,3 Ib[The term's /7'1' and hb are not ziven S I I, e . 0, neg ect t iat terms)

R = _l_[.!::L + J:L + _s_ 1A k, k2 k)

Considering two slabs, i.e., neglect L3 term

srQ =----

_s_+~k, k2

[''- A = I 1112)

6T100 ",,' - .....__::::...:_~_QJ_ -I- 0.040.7 0.41\

r Assume heattransfer CO) == 100 WI

II 'III luss is red" .cd h)' IilJ% e111l:10 insulllt ion. So, hcnr trnnsfcrIS () W,

()

o .(d')I ().()il I 1'1 I

OJIS (l.()(,.I 10.1I 0.7

r ~ 0.0. ~Ii 111/

Result :

Thickness of insulauon I.J 0.0 XX IIII

__j


I.SO Heat atld Mass Tratlsfer-II) A composite wt,lI consists of 10cIII Ihick lay~

brick, " = fl. 7 WlmK 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~. Id~

{Dec- 200-1 .11I1I1I Uuiv & Dec-lO(lj :ll/l/u Viii!

Give" :

Thickness of brick, L I = 10 em = 0.1 111

Thermal conductivity of brick, k I = 0.7 WImK

Thickness of plaster, L2 = Jcm = 0.03 mThermal conductivity of plaster, k} = 0.5 W/mKThermal conductivity of insulation, k3 =: 0.08 W/mK

~:---~ast~

I k, k, II

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

Ttl find :

Thickness of insulation to reduce the heat loss through !Iiwall by 40%, (LJ).

Insulation

Solulio" :

Heat now rate, Q sroverall-=R

where

R = J__ (_L + .!:L + .!:L + .!:l_ +_I 1A 11(/ kl k2 kj lib

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

Conduction /.8/

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

~T~IOO = -~ ___QJ_ + 0.030.7 0.5

[Assume heattransfer (Q) = 100 WI

~ I IH = 20.28 K IHeat loss is reduced by 40% due to insulation. So, heat transfer

is60 W.~T

Q = -R

~TQ=

1 (~+~+~l-A kl k2 k360 =

20.28

_!_ [_QJ_ + 0.03 + _!::L_11 0.7 0.5 0.08

:::) 60 r _QJ_ + 0.03 +~1= 20.28l0.7 0.5 0.08:::) 0.1 + 0.03 +..!::L = 0.338

0.7 0.5 0.08

L3:::) 0.08 =0.135

7


I.S2 Heal 0/1(/ Mass nOl15jer

I L = 0.0108 In IRemIt:

Thicknc s of insulation, L) = 0.0 I08 111

o A sw{ace HlIIII is made lip of 3 layers one of fire brick 0insulating brick and one of red brick. The inner ( 'd nIl

III Oil)

surface temperatures are 9000 C lind so: C rejfJective~ll.nrespective co-efficient of thermal conductivity of thelal'~are 1.2, 0.14 and 0.9 WlmK lind tile thickness of 20em,8and L/ em. Assuming close bonding of the layer.\atinterfaces. Find the heal/on' per square meter and illlerf~temperatures. [/11. U OCI-9 J

Given :

Inner

side

Fire lnsulatins Red brickbrick brick

/. 84 Heal and Mass rransfer(ii) Interface teperatures (T) am/ Tj) ~

We know that, Interface temperatures relation

Q TI - T"R

where

LIR1=---kl A

TI - T2

LIkl A

TI - T2

L,1;-

1011.2546= 1l7J-T2

Q/A

0.21.2

IT 2 = I 004.4 57 K ISimilarly

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

1004.457 - T31011.2546= ----.::....

0.080.14

Result:

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

Q/A = 1011.2546 W/m2

(i i) Interface temperatures (T 2 and T 3)

T2 = 1004.457 K

T3 = 426.597 K

[I) The wall of a furnace is made lip of 2.f0 mm fire clay ofthermal conductivity 1.05WlmK, 120 mm thick of ins Illationbrick of conductivh, O.15 WlmK anti 200 mm thick red brickof conductivity 0.85 WlmK. The inner and outer surfacetemperatllre oj wall are 850G C tlntl 65- C respectively.Co/cilIate the temper(l/ures at the contact surfaces.

[Bharathida an niversity ov- 95}Given:

Thickne of fire cia L1=2S0mm =0. 5m

Thermal conducti ity, kl = 1.05 W/mK

Thi knes of insulation bri ., ~ = 120 m = 0.1 In

Thermal ndu tivity, 2 = 0.15 W/mK


/. 86 Heal and Mass Transfer ---__Thickness of red brick, L3 == 200 nun == 0.2 rnThermal conductivity, k3 == 0.85 W/mK

Inner surface temperature, T 1 == 850 + 273 == 1123 K

Outer surface temperature, T4 = 65 + 273 == 338 K

Fire InsulationRedbrickclay brick

T, (~T2 nj

>

1123 - T2=> 616.466 = 0.25

1.05

Conduction 1.891.88 Heal and Mass Transfer

.'

@] A furnace wall made up of 7.5cm offire plate anti O.65 em ofmild steel plate. Inside surface exposed /0 hot glls lit650" C anti outside air temperature 27" C. The convectiveIlea/transfer co-efficient for inner side is 60 WIm1K. ThecOnl'ective heattmnsfer co-efficient for outer side is 8Wlm1K.Calculate the heat lost per square meter area of the furnacewal! and also find outside sutface temperature.

{M U. April-98]

SimilarlyT2 - T3

(I)=> Q=~-

whereGiven:

=> Q=

Fire Mild steel

plate plate

JnsideOutside

T(I' n, n2 ( TJ T/J'Ita hb

k, k2

f.c-- L, -----fo--- L2 --l

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

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

Inside hot gas temperature, Ta = 6500 C + 273 = 923 KOutside air temperature, Tb == 270 C + 273 == 300

0

K

Convective heat transfer co-efficient forinner side, ha == 60W/m2K

Convective heat transfer co-efficient forouter side, hb= 8 W/m2K.

976.22 - T3

L2k2

976.22 - T3=> 616.46 ==

0.120.15

=> Q/A

Result:

(ii) T3;: 483.05 K

TofiIIlI :(i) Heal lost per square meter area, (QI A)

(ii) Outside surface temperature, (T3)

(i) 1'2 == 976.22 K



Solution :(i) Heat lost per square meter area, (QIA)

Thermal conductivity for fire plate (Refract_ ory clay)k, = 1.0035 W/IIIK.

{From H.UT data book page no. 9 (I- iflh edition or page I) . . . JlO. - Sitt" edlt,

Thermal conductivity for mild steel platek2 = 53.6 W/I11K

[From HMT data book page liD.

ToverallR

Heat flow, Q

where

LI ~--+---_kJ A k2 A

~Q

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

Ta- Tb~Q

Q/A

Q/A = 923 - 300_, + 0.071+ 0.0065. I60 1.035 53.6 t-"8

IQI1\ = 2907.79 W/m2!

Conducnnn I.91

(ii} Olltside surface temperatllre, r.~We know that, Interface temperatures retarion

... ( I)

where

TrTb~ Q=

J

fib A

TJ - Tb

J

fib

QIA

T) - 3002907.79 = ,

8

IT3 = 663.473 K IResult :

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

.. Q/A = 2907.79 W/m2

(i i) Outside surface temperature, (T 3)

.. T] = 663.473 K.

(b!


t I' ''fill .. I A t. IJ, .:Iori/. I

Imllilll(inAbrick

Fhbl'k~

k I

I--- Ll--~-- LThi kness f fire bri k. L] = j ern = 0.23 III

r i kness f insulating brick, L_ = I!. - ern = 0.115mThermal c ndu tiviry of fire brick k I = 0.72 W/rnK

e al conductivity of insulating brick, k2 = 0.2 \\/rr

perature difference, 6 T = 650 KTofind:

10 per square meter, Q/A

SOlUlvm:

To /erallR

4J'[From IIMr data book page no.

( 'flliilliNil/il I 'J 1

1\'1[] -

"jI

I.) Ljh(l II I Ikl A ,,~ A "I A hi) /I

Il'hul 1'1Ilt! II", 1..1 II lid "b nn IIO( II,lv,1. So. nogl tthar rerrns]

Q/A6 0

0.23 0.115--+--OL 0.27

872 W/m2!

Result:

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

Q/A = 872 W/m2

[!] TI,e inner dimension of a freezer cabinates are60 em x 60 em. The cabinates wall consists of /HIo2 mmthick steel wall (k = 40 WlmK) seperated by a 4 em layer offiber glass insulation (k = 0.049 WlmK). TI,e insidetemperature is 10 be maintained at _I5D C and the outsidetemperature on a hot summer day ;J 45 C. Calculate themaximum amount of heat transfer, assuming a heat transferco-efficient of 10 Wlm] K both on inside and outside of tilecabinate a/JO calculate outer surface temperature of tile

cabinate.

[M. U. Oct-2002}


Give" "

/. 94 Heat and Mass Transfer

Steel Fiber Steelglass


Result:(i) Maximum amount of heat transfer,Q ::: -2 I .25 W

(ii) Outer surface temperature, T4 = 312.09 K

l!1 A mild steel tank of wall thickness 10 mm Contllins IIIUle'l900 C. Calculate tile rate of heat loss per ml Of tank surfll/J.area when the atmospheric temperature is 150 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]

Give" :

Inside

T(I>

ha

k


We know that,

Q

363 - T)Q=

1haA

363 - T)1QIA

363 - T)819.9 = ----!....

_1_2800

(I) ~

Q=T)-T2

. LI

kl A

T,-T2

___!j_k,

Q/A

819.9362.7-T2

0.0150

_ T2-T- ---..Q_Rb ". (I)

where, Ra= -haA

LIwhere, R, = k A

I

Conduction 1.99

==> I T 2 = 362.5 K IResult:

(i) Heat loss per m2 surface area, Q/A = 8)9.9 W/m2(ii) Outside surface temperature, T2 = 362.5 K

I2!l Consil/ering the heating surface of a steam boiler to beplanewall oftllickness 1.2 em and having k = 50 WlmK. Determinethe rate of heat flow and surface temperatures for tilefollowing data.

Flue gas temperature 1000C

Boiling water temperature 200 C

Heat transfer co-efficient on gas side 100 Wlm1 K

Heat transfer co-efficient on steam side 500 Wlm2K

[Manonmanium Sundaranar University April- 97]

Given:

Thickness

L, = 1.2 ern = 0.012 mkl

Ta Tb

Thermal conductivity, ha hb

k, = 50 W/mK ( TI (DT2

Flue gas temperature,

T = 10000 C + 1273 Ka LlBoiling water temperature, I-

..j

Tb = 2000 C + 273 = 473 Kh 100 W/m2K

Heat transfer co-efficient on gas side, a = , 2Kid h =500W/mHeat transfer co-efficient on steam 51 e, b

. ' ild steel k = 50 W/mKThermal conductIvIty of rru ,


1.100 Heal and Mass Transfer._-----Tojintl:

(i) Heat transfer rate. Q/A

(ii) t'1I1' act' teperutures, (T, [lllll Tz)

- -~---_._. COlldl/(:/ioll J. J 0JT -T

( I) ::':> Q;-: _!.I_I-R(I

Solution :lIeat II uisfer. Q

T - '1'C.J' __ (I_I ['.: R =_1 _I(I 'I A(I'

R= -.1 A

L,~k ~.'\ },,,A

,r_I' '1'_1(.)/A =1

II"

\T=T - l(I

[lZ L, values an:' II t uiveu. oo, negle~.t(h,l!(

1_7. -1'1(l).~59 ::: - _ ___:_

_1.100

R=1,A..

_-A

=> Q/A =800

(I Q==T1- T2

R,

Q=TI - T_

LI

kl A

QfAT1 - T2

=~

k I

65,3 59 ==619 - T2

0.012SO

=> IT2 = 603.3 KJ

=:> Q = -_-"'--_=---_-1 L, I_-~--+--

h A ", A I A

I100

0.012 +_150 500

IQ/A 65.)59 W/m2 1Interface temperatures relation

1 - Tbo = .is.-R~ =


1.102 Heal andMass Transfer

Result: ~

(i) Heat transfer, Q/A = 65,359 W/m2

(i i) Surface temperatures, T 1 = 619KT2=603.3K

[ll) A composite slab is made ofthree layers 15 em, 10 c",12 em thickness respectively. Tirefirst layer is made 0'm ~

'J a'e~with k = 1.45 WlmK, for 60% of tire area and lirere

S"material with k = 2.5 WlmK. The second layer is madt;material with k = 12.5 WlmK for 50% of area and res,;material with k =18.5 WlmK. The third layer is madeols;",material of k = 0.76 WlmK. The composite slab is expOilon one side to warm at 260 C and cold air at -20 C ninside heat transfer co-efficient is 15 Wlm2 K. The outsideh,transfer co-efficient is 20 WI",2 K determine heat flow rt

and interface temperatures.

[MU Nov-~

A,a = 60% A2a = 50%A) = 100%

k1a k2a

(DT, (DT2 ( T)(

A,b = 40% A2b = 50%

k,b k2b k) -

Conduction 1./031,,:;:: 15em=0.15m

L2 = 10 em = 0.1 mL) = 12 em = 0.12 m

k'a = 1.45 W/mK,

k'b==2.5 W/mK,

k2a == 12.5 W/mK,

k2b == 18.5 W/mK,

k) = 0.76 W/rnK

T a == 260 C + 273 = 299 K

Tb == -200 C + 273 = 253 Kha == 15 W/rn2K

hb == 20 W/m2K

Ala = .60

Alb=40

A2a = .50

A2b = .50

Tofind :

(i) Heat flow rate, (Q)

(ii) Interface temperatures, (T, , T2, T3 and T4)

Solution:

Heat flow, .1ToverallQ==R

[From HMT data book. page

no.43 & 44}

where

L L II LI 2 +_3_+_- --+--+-- A k Abh- A h A k A2kJ 3 '3 ba a I I -



.. (I)

where

I Ra = 0.066 K/W I

... (2)

LR - I10 - k x AIn 10

0.15--- == o. I 724 KlW1.45 x 0.6

I Ria = 0.1724 KiwiRib = __ L_;_I_ _ 0.15 = 0.15 K/W

klb x Alb 2.5 x 0.4

IRIb = 0.15 KIW ISubstitute R In and Rib value in (2)

(2) => R 1 = o. I724 x O.) 50.1724 + 0.15

I RI =0.08 K/W I

COl/duc/ion 1.105

Similarly

... (3)

0.112.5 x 0.5 = 0.016 K/W

[R20 = 0.016 K/W]

0.118.5 x 0.5 = 0.0108 K/W

IR2b = 0.0108 Kiwi

(3) ~ R2 == 0.016 x 0.01080.0161-0.0108

I R2 = 0.0064 ~R~ = ~;_c__QJl_

.) :\3k3 ) xO.76

IR3 = O.)578~Rb = _1_ ..= _I -

Ab hb I x 20

~0.05 K/\\]

(I) =:> Q = ---=..:29~9--..::..2~53:-----0.066 -1- 0.08 + 0.0064 + 0.15789 + 0.05



(ii) Interface temperatures (Tl' T2, T3 and T.f)

We know that,

T -T](4)~ Q==T

a

299- T]0.066

127.67 == 299 - T]0.066

IT] == 290.57 K I(4):;" Q== T]-T2

R]

127:67 == 290.57 - T20.08

IT2 ==280.35 K!.

)i." 1: T(4) ~ Q ==--1:_l

"J R2

127.67 == 280.35 - Ii.0.0064 .

..

[!! == 279.532 KJ

Conduction 1.107TrT4

(4):::> Q==~

279.532 - T4127.67= 0.15789

[T4 = 259.374 K IResult:

(i) Heat now rate, Q = 127.67 W(ii) Interface temperatures, (TJ, T2, TJ and T4)

T] = 290.57 K

T2 == 280.35 K

TJ = 279.532 K

T4 = 259.374 K.

; I. I

:iI

! I

. ~ Afurnace wall consists of steel plate of20 mm thick, thermalconductivity 16.2 WlmK lined on inside with silica bricksISO mm thick with conductivity 2.2 WlmK and on the outsidewith magnesia brick 200 mm thick, of conductivity5.1 WlmK. TIre inside and outside surfaces of the walt aremaintained at 650D C and 150D C respectively. Calculate theheat loss from the wall per unit area. If the heat loss isreduced to 2850 Wlm2 byproviding an air gap between steeland silica bricks, find the necessary width of air gap if thethermal conductivity of air may be taken as 0.030 WlmJ(.

[Madurai Kamaroj University April 97J


1.J08 Heat and Mass Transfer

Give" :

T 2

k I k 2 k3

MagnesiaSteel Silica

Steel plate thickness, L, = 20 mm = 0.02 m

Thermal conductivity of steel, kl = 16.2 W/mK

Thickness of the silica, L2 = 150 mm = 0.150 m

Thermal conductivity of silica, k2 = 2.2 W/mK

Thickness of the magnesia, '-3 = 200 mm = 0.2 IIIThermal conductivity of magnesia, k3 = 5.1 W/mK

Inner surface temperature, T I = 6500 C + 273 = 923Outer surface temperature, 1'4 = 1500 C + 273 = 423 K

Heat loss reduced due to air gap is 2850 W/m2

Thermal conductivity of the air gap kair = 0.030 W/mK

Tojind:

(i) Heat loss [without considering air gap]

(ii) Thickness of the air gap

SOIIlI;oll :

Heat transfer through composite wall is given by Iwith~considering air gap]

SfQ=-

R

________ ----------------------~C~o~,,~d~uc=Il~o~n~/~./~09

where6.T= TI-T4

I LI Ll L3 IR =--+--+--+--+-ha A kl A k2 A k) A hb A

TI - T4

Neglecting unknown terms (ha and hb)

TI-T4Q=-------

LI ~ L3--+--+--kl A k2A k3 A

923 -423Q = ---~-----0,~.0:..::..2_0.150 0.2- +--+--

16.2xl 2.2xl 5.lxl

500Q = 0.1086

IQ = 4602.6 W/m2Heat loss is reduced to 2850 W1m2 due to air gap. So, the new

thermal resistance is

Q=~T

Rnew


/./10 Heal and Mass Transfer

Rnew =923 - 423

2850

Rnew = 0.1754 K/W IThermal resistance of air gap

Rair = Rnew - R

0.1754 - 0.1086

I Rair == 0.066 K/W IWe know that,

LairRair == k Aair )(

0.066 ==Lair

[.: A = 1m

0.030)( 1

I == 1.98)( 10-3 rn 1::::> Lair _Thickness of the air gap == 1.98 x 10-

3In

Result:. I . )- 4602 W/m2(i) Heat loss (Wit rout air gap -

. _ 98 x 10-3 rn(ii) Thickness of the air gap, Lair - I.

Conduction 1.111

1.1.13 Solved Problems On Cylinders

o A Itollow cylinder 5 em inner radius and 10 em outer radiushas inner surface temperature of 2000 C anti outer sur/acetemperllture of 1000 C. If the thermal conductivity is70 WlmK,jind heat transfer per unit tength.

Given:Inner radius, "1 = 5 ern = 0.05 m

Outer radius, r: = 10 cm = 0.1 mInner surface temperature,

T 1 = 200 + 273 = 473 K

Outer surface temperature,T2=100+273=373K

Thermal conductivity, k = 70 W/mK

Ttl find :Heat now per unit length

Solution :

Heat transfer through hollow cylinder is given by

6TovcrallQ= __::.:...::.:..:::ocR

[From equn. 110.1.32 orHMT data book page 110.43 & 44J

where

I [r2]R=--ln-2n:Lk rl

::::> Q =I [r2 ]--/11 -

2n:Lk rl

\------...&.~


/. 1/2 Heal and Mass Transfer

=> Q2itkL (1', - T2)

/11 [;:n-----~

=> Q/L

2rrx 70(473-373)=> Q/L = -----___:_

IIl[O~O~ 1I Q/L = 63453.04 W/m = 63.453 kW/m./

Result:

Heat transfer per unit length, Q/L = 63.453 kW/m.

III Determine thermal conductivity of asbestos powder pllckedubetween two concentric copper pipes 25 111m and 36 mdiameter length. The inner pipe housint; has (I heating Coi/Iwhich 120 HIpower is supplied. The average telllpef(/Iu't~inner (111(1outer pipes are 42.,r C (11/(127. 9 C re.\pectively

Give" ..

Inner diameter, D, = 25 mrn

Inner radius, r, = 12.5 mm=0.0125111 T2

Outer diameter, D2 = 36 mmOuter radius, r2 = 18 mm = 0.018 III

Inner temperature, T, = 42.4 C + 273

=3J5.4K

Conduction I.J 13

outer temperature, T2 = 27.9 C + 273 = 300.9 K

Heat transfer, Q = 120 W

Toft"d:Thermal conductivity, k

SolutiOJl :Heat transfer through hollow cylinder is given by

L\ TaverallQ=---R [From equn. 110.1.32 or

HMT data book page 110.43 & 44]where

R= _I_ 111 [r2]211Lk rl

~ QI [r? ]__ 111 -=-

211Lk rl

315.4 - 300.9~ 120 = --------I 111 [_0,_0'_8]

211 x I x k 0.0125

[':L=lmj

/ k 0.48 W/IllK IResult:

Thermal conductivity, k == 0.48 W/mK.

9


J.1J4 Heal and Mass Transfer

III A hotlow cylinder 5 em inner diameter an~'diamel~r has inner surface temperature of 2000 c", Olllrtsurface temperature of 1000 C Determine Iteatj1C llnd 0111(1

, Ow tilrthe eylmder per metre length. Also determine tit QlIg. e temperof the point half wa) between 'he inner and out 11/41rer Sur I:Take k = ] WlmK. 10ft!,

Gil'm:

dl = 5 em = 0.05 m

'1 = 0.025 m

d = 10 em = 0.1 m

r = 0.0- m

T 1 = _00t> C = 4 3 K

T2 = 100e C = 373 K

k = I W/mK.

Tofind:

(i) Heat flow per meter, (Q/L)

(ii) Temperature between inner and outer surfaces, (T).

Solution:

(i) Heat flow per meter (Q/L)

~ToverallQ= R[From HMT(kll(/~(X

page /10.43 & ~j

where

I l I" 1R = --/11 _1_2nLk 1"]

473 - 373

Condllclion 1.11 5

oI ln [0.05 ]

2n L x I 0.025

::::> [OIL = 906.47 W/m I

(ii) Temperature between inner and outer surfaces, (T)

Put T2 = T and 1"2 = r in heat transfer equation

::::> 0J [ r 1--In -

2rcLk '1

T]-T::::> Q/L = -------

1 / [0.0375]__ /I _-2rc x 1 0.025

473 - T::::> 906.4 7 = -..-:...:.-=----=---

_I_ / [.QJ>J 75 ]2rc /I 0.025

IT = 414.5 K I

rl + r:.: r= -2-

0.025 + 0.05::::> r=

2

::::> r = 0.0375 III

Result:

(i) Heat flow per meter, Q/L = 906.4 7 W /m(ii) Temperature between inner and outer surfaces,

T=414.5 K.



r:tl An insulated steel pipe carr) ing a hot liquid. Illller d'~ of tile pipe is 25 em, wall thickness hi 2 em ti,' 1.lallieler

. ". ' lellilessinsulation IS 5 em, temperature of hot liquid is 10 0 oftemperature of surrounding is 20 C, inside heat tr 0 (',

w../ 2K alls/erco-efficient is 730 /m an d outside Ileal tr. 2 (IIIS/erco-efficient is 12Wlm K. Calculate tile IIeat loss per

'"elrelength of the pipe.

Take kstee/:::: 55 WlmK, killslliatillg lIIateria/:::: 0.22 W/"'KGiven,'

Inner diameter, d( ::::25 em

Inner radius, 1'( :::: 12.5 ern

h =0.125 mlradius, 1'2 = /'( + thickness of wall

0.125 + 0.02

[1'2 0.14Sml

radius, 1'3== r-, + thickness of insulation

== 0.145 0.05

~3 == 0 195 mJ

Condllction 1.1/7

Temperature of hot liquid, Ta == 100 C + 273

Ta == 373 K

Temperature of surrounding, Tb == 20 C + 273

Tb == 293 K

Inside heat transfer co-efficient, ha == 730 WIm2K

Outside heat transfer co-efficient, hb == 12 W/m2K

ksteel == 55 W/mK

kinsulation == 0.22 W/mK

Tofind,'

Heat loss per metre length

SO/lItiOI1:

Heat flow through composite cylinder is given by

~ToverallQ =

R[From eqllll. no. 1.48 or HMT data book

page no. 43 & 45 (Sixth edition})where

~T=Ta-Tb

2nL [h:" +R

Ta- Tb=:> Q =

III [~~ 1

r ' +III [~~ 1 , j+ +--2rrL harl k( k2 hbr3


1.118 Heat and Mass Transfer373-293 ~

--------111 [~] 1+ +~

0.22 12x.19j

Q-;:::>--

L

[

I [.145]

I I 11 m- +21t 730)(.125 55

@/L == 281.178 W/m]

Result:Heat transfer per metre length, Q/L == 281.178 W/m.

III Hot air at 40 C flowing through a steel pipe of 10 f.diameter. The pipe is covered with two layer of ;nsulali"lmaterial of thicklless 4 em an d 3 em and Ihtucorresponding t"ermal cOlldllctivities are 0.1 a.10.32 WlmK. The ills ide and outside convective heat tram/eco-efficient are 50 WlmlK and 15 WlmlK. Tile OUIDtemperature is 10 C. Find the neat toss per meier Itngrlof steam pipe.

Given:

","78"7Conduction 1.119----Hot air temperature, Ta == 40 C + 273 == 3 13 K

Inner diameter, d, == 10 cm == O. I m

Inner radius, r, == 5 cm == 0.05 mIntermediate radius, r2 == r, + 4 em = 5 + 4 = 9 cm = 0.09 m

Outer radius. rJ = 1'2 + 3 ern = 9 + J== 12 ern = 0.12 m

k,=o.IW/mK

k2 == 0.32 W/mK

ha == 50 W/m2K

hb == 15 W/m2K

Outer temperature of air, Tb == 10 + 273 = 283 K

Tofind:Heat lost per metre length of steam pipe

Solution:

f, I

"

[From equn. no. 1.48 orHMT data book page 110.43 & 4jj

I


llToveraJlQ =-.=.:..::.:.=.:.R

where

llT= Ta-Tb

R 2.L [,,~, +

I2nL


J. J lU Heal and Mass na"'::,jPI'

Q=>-=

L

Q/L 24.37 W/m

Result,'

Heat tran fer p r unit len III / = _4. \ 1m

[1] Air at 90 C flows ill a copper tube or 5 ell, . .'J tnner d,u,"

with thermal con d uctiviry 380 Wlm/( and wirt, O. elflrwall which is healed from tit ' (/111 ide by water II( 1]0'1A scale of O.4 em thick i Iep ositcd 0" lite outer surfa I,tile tube whose th ermul on du uivitv is I.S2 W/mK. nl.(111(1water side unit urfa e ontluctance are 220 1I1"r(In,l3650 W/m? K resp tctivelv. alc ulate

l. Overall Harer 10 air tran imittance

2. Water 10 air h eat ex h ang e

3. Temperature drop a ross II,e scale deposit.

Give" :\ ter

Inner air temperature. T a '"

I T a = 363 K IInner diameter f the copper, d1 '" 5 em

radius, 1'1 '" 2.5 cm

II'I '" 0.025 III IThermal ndu tivitv. k, = 80 \\/mK

L11 r r diu' I the PI cr, 1'_ '" Inner radiu thicknes ofwall

1'2 0.025 0.007 III

~_ 0.03}3

radiu , r '" '2 + thickness of .cale

= 0.032 0.004

I r, = 0.0361nlL11 id icrnp r lure f water, Tb = LO

= "9" K27

Thermal ndu livily k = 1.82 W/I1lK

. urfa e d fair, ha = 2 ..0 W/m-Knul, n .e

urfa e ndu I n e f water, h = "650 Whn-K

To filld :

vera ll he I Iran er .eff ieru

2 v arer I air heal Iran f r, Q) Temperature dr p ihe

, T T)ale dep II, ( , - 2


1.122 Heat and Mass 1'ransfer

Solution:

Heat flow through composite cyl inlier is .given by

ToverallRQ [Fro", C(i7L1n

HM'{ I . 110. J 4(.uta book pag . e 110.43 & Iwhere

~T Ta- r,

R = _I l-I + In l:~\+ In [:~ 1 +-L2nL harl k( k2 hbr3Ta-T

~ Q=

lh:rl.L In l:~I In (~~ 1 + I 12nL k( k2 hb'32n:L 220~.02S

I [.036)II JD21.82

1-3650~. :

'j

fr m out ide to innerS'ha hea f1

Conduction I. 123--We know that,Heat transfer, Q =U A T

where

U - overall heat transfer co-efficient

A - Area = 21t rJ L

6T= Ta-Tb

Q = U x 21t r3 L x (Ta - Tb)

Q/L = U x 21t r3 x (T a - T b)

~ -739.79 = U x 2 x It x 0.036 x (363 - 393)

IU = 109.01 W/m2K IOverall heat transfer co-efficient, U = 10901 W/m2K.

Interface temperatures

T Ta - Tb = Ta - T IQ=-=R R Ra

T3 - Tb

Rb... (1)

\ here

1R)=-- 2nL


/ 1)1/ 1/('(1' (/lid M(I,VV 'l'''(lmjl'I'

'f) 'f I~()

/'''1;'11711 .() '1 r

~.L

11'

-739.79

-7.6 K

7.6 K II I depo J'I T, - T2 == 7.6 KTemperature across tne sea e 'J

Result:109.0 I \Vln/I) Overall heat transfer co-efficient, ;;::

2) Heat exchange Q/L = - 739.79 W/lll 'd L. j1 from 0/11 /[Negative sign indica I S 1/7(11 h II OW

inner side]

J) Temperature dr par the al dep it,

('(111{/," I 1m I 12

117' A 1'1('/'/ "lpI' o] 12(1 """ 1,,111'( "'"m('ll'( I"(j"f-J t" , .. "1m lillie,/111111('11" 11,/,11 1111"111,,1 ('''''(/11('111111 H W./nIK ,

( . , "/1 "",t!HI/'" two /11)11"1II/I"tl/IIIIIIIII ('11('" hlllllll)( II /"",,,,,U II/.H 11,,11. Tlu: th rr nut] ('111/(/11('1",11 II/ Ihe /Iffl Inwlll//flll"",11,1,,/ ts 1I.1I,f W/",K 11111/ III", (lj th ,rl'l'/iII" I,11.// W/",K. till' trmprrutur 1'./ tI", /111/111' tub I'llf/II('I' I"24(r C "lid IIIII'I~/I"(' IIIII,I"'C ,I/Ifjll"e 1'./11,,' IIIfIIllIlll/1l IfMJ" C. (',,/('11/"'(' 1/11' /11,1,1 11/ ""III/lI'f metre h'IIl1lh 1// pipe(//11/1"(' IlIlcrjIlC(' tempcrutur helHiCCII thr twn IlIyef.f ofill.I'II/'" irm,

Gille" :

Inner diameter, d I == 120 mm

1', == 60 111111

II', == 0.060 III 1Outer diameter, d2 == 140111111

1'2 == 70 mill

[ '2 == 0.070 Ill]


~("fr'beMMMM'."W:r::

/.1]6 Heal and Mass Transfer

radius, r3 = r2 + thickness of insulation= 0.070 + 0.055

[1"3 = 0.125 In Iradius, r4 = rJ + thickness of insulation

= 0.125 + 0.055[r4 = 0.18 m I

Thermal conductivity, k, = 55 W/mK

k2 = 0.05 W/mK

k3 = 0.11 W/mK

Inner surface temperature, T, = 2400 C + 273 :: 513 k

Outer surface temperature, T4 = 60 C + 273 :: 333k

Tofind :

i) Heat loss per metre length of pipe (Q/L)

ii) Interface temperature between two layer;insulation (TJ)

Sohaion :


6ToverallQ = _;::..;..:;.:_=R

IFrom equn. 1I0lil,HMT data hook page 1I0.m.

where

Conduction 1.127

----;:ieat transfer co-efficients ha and lib are not given. So, neglect

th,t rerms [III [;~ j + _::j;; IJ111 r2jI r,~ R=- +27tL k, k2 k3

TI- T4z Q=

I [ III [;: 1 111 [;~ j + In [;; I ]+21tL k, k2 k3Q

~ L513 - 333

[

Ill [ 0.070] In [ 0.125 ] III [...Q.:..!!_] ]I 0.060 0.070 0.125

- + +----=-21t 55 0.05 0.11

~IQ/L = 75.83 W/m IWe know that,

Interface temperatures relation

T, - T4 T, - T2Q = ---'---'-R R,

T, - T2

... (I)

(I) ~ Q =

where [ III [;:2 ] JI> - I ,"1---

21tL k I


I. 128 Heal and Mass Transfer

QCOlldll(,irm I. 1]1,1-----

_I [/11 [~Jl21tL k

I

TZ .- TJQ/L _--=---;:..___

2. [~t;IJ512.7-TJ

75.83 =

II" [.QJ.]jI 0.070

21t 0.05

=> 75.83

Resut:

=> [_rJ__ 3 7_2_.7_K_-]

I) Heat loss per metre length of pipe, Q/L '" 75.83 WIlli

2) Interface temperature between two layers of insulntion512.7 KTJ = 372.7 K.

(I) => Q III A steel pipe of /70 "'''' inner ,dame/er tllldl90 """ outerdiameter with thermal conductivity 55 WlmK is coveredwith two layers of insutation. rile thickness IIf 'lie firstlayer is 25 mm (k = 0./ WlmK) III1tI the second layerthickness is 40""". (k = 0./8 WlmK). rile temperature ofthe steam and inner surface of tile steam pipe is J20 Callll outer surface of the insulation is Sf)" C. Ambient airtemperature is 25~C. rile surface co-efficient for inside(IIId outside surfaces tire }JO Wlm]K alltl

heat and mass transfer

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