Fundamentals of Engineering Heat and Mass Transfer_R. C. Sachdeva

8/20/2019 Fundamentals of Engineering Heat and Mass Transfer_R. C. Sachdeva

1/213

Scilab Textbook Companion for

Fundamentals Of Engineering Heat And Mass

Transfer

by R. C. Sachdeva1

Created byNittala Venkata Krishna

B.TECHMechanical Engineering

SASTRA UNIVERSITY

College TeacherDr. Anjan Kumar Dash

Cross-Checked byLavitha Pereira

August 31, 2013

1Funded by a grant from the National Mission on Education through ICT,http://spoken-tutorial.org/NMEICT-Intro. This Textbook Companion and Scilabcodes written in it can be downloaded from the ”Textbook Companion Project”section at the website http://scilab.in


2/213

Book Description

Title: Fundamentals Of Engineering Heat And Mass Transfer

Author: R. C. Sachdeva

Publisher: New Age Science Ltd., New Delhi

Edition: 4

Year: 2009

ISBN: 9781906574123

1


3/213

Scilab numbering policy used in this document and the relation to theabove book.

Exa Example (Solved example)

Eqn Equation (Particular equation of the above book)

AP Appendix to Example(Scilab Code that is an Appednix to a particularExample of the above book)

For example, Exa 3.51 means solved example 3.51 of this book. Sec 2.3 meansa scilab code whose theory is explained in Section 2.3 of the book.

2


4/213

Contents

List of Scilab Codes 4

1 Basic Concepts 10

3 OneDimensional Steady State Heat Conduction 16

5 Transient Heat Conduction 51

6 Fundamentals of convective heat transfer 77

7 Forced Convection Systems 82

8 Natural Convection 111

9 Thermal radiation basic relations 128

10 Radiative Heat exchange between surfaces 136

11 Boiling and Condensation 158

12 Heat Exchangers 169

13 Diffusion Mass Transfer 192

14 Convective Mass Transfer 202

3


5/213

List of Scilab Codes

Exa 1.1 Rate of heat transfer . . . . . . . . . . . . . . . . . . . 10Exa 1.2 Rate of heat transfer . . . . . . . . . . . . . . . . . . . 11Exa 1.3 Rate of radiant heat . . . . . . . . . . . . . . . . . . . 11Exa 1.5 Overall heat transfer coefficient . . . . . . . . . . . . . 12Exa 1.6 Heat loss per unit length . . . . . . . . . . . . . . . . 13Exa 1.7 Surface temperature . . . . . . . . . . . . . . . . . . . 14Exa 3.1 Rate of heat loss and interior temperature . . . . . . . 16Exa 3.2 Tempertaure and heat flow . . . . . . . . . . . . . . . 17Exa 3.3 Heat flow and teperature . . . . . . . . . . . . . . . . 18Exa 3.4 Heat loss per square meter surface area . . . . . . . . 19Exa 3.5 Rate of heat flow . . . . . . . . . . . . . . . . . . . . . 20Exa 3.6 Heat loss per unit length . . . . . . . . . . . . . . . . 20Exa 3.8 Thickness of insulation . . . . . . . . . . . . . . . . . . 21

Exa 3.9 Rate of heat leaking . . . . . . . . . . . . . . . . . . . 22Exa 3.10 Heat transfer through the composite wall . . . . . . . 23Exa 3.11 Surface temperature and convective conductance . . . 24Exa 3.12 Heat loss and thickness of insulation . . . . . . . . . . 26Exa 3.13 Heat loss from the pipe . . . . . . . . . . . . . . . . . 27Exa 3.14 Heat loss per meter length of pipe . . . . . . . . . . . 28Exa 3.15 Change in heat loss . . . . . . . . . . . . . . . . . . . 29Exa 3.16 Mass of steam condensed . . . . . . . . . . . . . . . . 30Exa 3.17 Rate of heat flow . . . . . . . . . . . . . . . . . . . . . 31Exa 3.18 Heat loss and surface temperature . . . . . . . . . . . 32Exa 3.19 Effect of insulation on the current carrying conductor 33

Exa 3.20 Surface temperature and maximum temperature in thewall . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

Exa 3.21 Surface temperature . . . . . . . . . . . . . . . . . . . 36Exa 3.22 Heat transfer coefficient and maximum temperature . 37

4


6/213

Exa 3.23 Diameter of wire and rate of current flow . . . . . . . 38

Exa 3.24 Temperature drop . . . . . . . . . . . . . . . . . . . . 39Exa 3.25 Temperature at the centre of orange . . . . . . . . . . 40Exa 3.26 Total heat dissipated . . . . . . . . . . . . . . . . . . . 40Exa 3.27 Thermal conductivity . . . . . . . . . . . . . . . . . . 41Exa 3.28 Temperature distribution and rate of heat flow . . . . 42Exa 3.29 Rate of heat transfer and temperature . . . . . . . . . 43Exa 3.31 Rate of heat flow . . . . . . . . . . . . . . . . . . . . . 44Exa 3.32 Efficiency of the plate . . . . . . . . . . . . . . . . . . 46Exa 3.33 Heat transfer coefficient . . . . . . . . . . . . . . . . . 47Exa 3.34 Insulation of fin . . . . . . . . . . . . . . . . . . . . . 48Exa 3.35 Measurement error in temperature . . . . . . . . . . . 49

Exa 5.1 Heat transfer and temperature . . . . . . . . . . . . . 51Exa 5.2 Time taken . . . . . . . . . . . . . . . . . . . . . . . 52Exa 5.3 Time taken . . . . . . . . . . . . . . . . . . . . . . . . 53Exa 5.4 Time required to cool aluminium . . . . . . . . . . . . 55Exa 5.5 Heat transfer coefficient . . . . . . . . . . . . . . . . . 56Exa 5.6 Time constant and time period . . . . . . . . . . . . . 57Exa 5.8 Temperature and total heat . . . . . . . . . . . . . . . 57Exa 5.9 Time taken to cool . . . . . . . . . . . . . . . . . . . . 59Exa 5.10 Temperature . . . . . . . . . . . . . . . . . . . . . . . 60Exa 5.11 Time required and depth . . . . . . . . . . . . . . . . 60

Exa 5.12 Temperature and total thermal energy . . . . . . . . . 61Exa 5.13 Temperature . . . . . . . . . . . . . . . . . . . . . . . 63Exa 5.14 Time required for cooling . . . . . . . . . . . . . . . . 64Exa 5.15 Temperature . . . . . . . . . . . . . . . . . . . . . . . 65Exa 5.16 Centreline temperature . . . . . . . . . . . . . . . . . 66Exa 5.17 Temperature . . . . . . . . . . . . . . . . . . . . . . . 67Exa 5.18 Surface temperature . . . . . . . . . . . . . . . . . . . 69Exa 5.19 Time elapsed . . . . . . . . . . . . . . . . . . . . . . . 70Exa 5.20 Temperature . . . . . . . . . . . . . . . . . . . . . . . 71Exa 5.21 Amplitude of temperature and time lag . . . . . . . . 72Exa 5.22 Time lag and heat flow . . . . . . . . . . . . . . . . . 73

Exa 5.23 Depth of penetration . . . . . . . . . . . . . . . . . . . 74Exa 5.24 Instantaneous heat removal rate . . . . . . . . . . . . 75Exa 6.2 Type of flow . . . . . . . . . . . . . . . . . . . . . . . 77Exa 6.6 Maximum temperature rise and heat flux . . . . . . . 78Exa 6.7 Type of flow and entry length . . . . . . . . . . . . . . 79

5


7/213

Exa 6.8 Head loss and power required . . . . . . . . . . . . . . 80

Exa 6.9 Pressure drop . . . . . . . . . . . . . . . . . . . . . . . 81Exa 7.1 Thickness of hydrodynamic boundary layer and skinfriction coefficient . . . . . . . . . . . . . . . . . . . . 82

Exa 7.2 Local heat transfer coefficient . . . . . . . . . . . . . . 83Exa 7.3 Boundary layer thickness and total drag force . . . . . 84Exa 7.4 Rate of heat to be removed . . . . . . . . . . . . . . . 85Exa 7.5 Drag force . . . . . . . . . . . . . . . . . . . . . . . . 86Exa 7.6 Thickness of boundary layer and heat transfer coefficient 88Exa 7.7 Friction coefficient and heat transfer coefficient . . . . 89Exa 7.8 Heat loss . . . . . . . . . . . . . . . . . . . . . . . . . 90Exa 7.9 Heat transfer rate and percentage of power lost . . . . 91

Exa 7.10 Heat transfer coefficient and current intensity . . . . . 92Exa 7.11 Rate of heat transfer . . . . . . . . . . . . . . . . . . . 93Exa 7.12 Heat transfer coefficient and pressure drop . . . . . . . 94Exa 7.13 Convection coefficient . . . . . . . . . . . . . . . . . . 96Exa 7.14 Convective heat transfer coefficient . . . . . . . . . . . 97Exa 7.15 Average temperature of the fluid . . . . . . . . . . . . 98Exa 7.16 Length and heat transfer coefficient . . . . . . . . . . 99Exa 7.17 Heat transfer coefficient and heat transfer rate . . . . 100Exa 7.18 Heat transfer coefficient and length of the tube . . . . 100Exa 5.19 Nusselt number . . . . . . . . . . . . . . . . . . . . . . 101

Exa 7.20 Heat transfer coefficient . . . . . . . . . . . . . . . . . 103Exa 7.21 Heat transfer coefficient and heat transfer rate . . . . 104Exa 7.22 Heat transfer coefficient . . . . . . . . . . . . . . . . . 105Exa 7.23 Heat transfer coefficient . . . . . . . . . . . . . . . . . 106Exa 7.24 Heat leakage . . . . . . . . . . . . . . . . . . . . . . . 107Exa 7.25 Heat transfer coefficient . . . . . . . . . . . . . . . . . 108Exa 7.26 Minimum length of the tube . . . . . . . . . . . . . . 109Exa 8.1 Boundary layer thickness . . . . . . . . . . . . . . . . 111Exa 8.2 Heat transfer coefficient . . . . . . . . . . . . . . . . . 112Exa 8.3 Heat transfer coefficient and rate of heat transfer . . . 113Exa 8.4 Convective heat loss . . . . . . . . . . . . . . . . . . . 114

Exa 8.5 Rate of heat input . . . . . . . . . . . . . . . . . . . . 115Exa 8.6 Heat gained by the duct . . . . . . . . . . . . . . . . . 116Exa 8.7 Coefficient of heat transfer . . . . . . . . . . . . . . . 118Exa 8.8 Rate of heat loss . . . . . . . . . . . . . . . . . . . . . 119Exa 8.9 Rate of heat loss . . . . . . . . . . . . . . . . . . . . . 120

6


8/213

Exa 8.10 Coefficient of heat transfer and current intensity . . . 121

Exa 8.11 Rate of convective heat loss . . . . . . . . . . . . . . . 122Exa 8.12 Rate of heat loss . . . . . . . . . . . . . . . . . . . . . 123Exa 8.13 Thermal conductivity and heat flow . . . . . . . . . . 123Exa 8.14 Free convection heat transfer . . . . . . . . . . . . . . 124Exa 8.15 Convective heat transfer coefficient . . . . . . . . . . . 125Exa 8.16 Heat transfer . . . . . . . . . . . . . . . . . . . . . . . 126Exa 9.1 Rate of solar radiation . . . . . . . . . . . . . . . . . . 128Exa 9.2 Fraction of thermal radiation emmitted by the surafce 129Exa 9.3 Hemispherical transmittivity . . . . . . . . . . . . . . 130Exa 9.4 Surface temperature and emmisive power . . . . . . . 130Exa 9.5 Emmissivity and wave length . . . . . . . . . . . . . . 131

Exa 9.6 True temperature of the body . . . . . . . . . . . . . . 132Exa 9.7 Rate of absorbption and emmission . . . . . . . . . . . 133Exa 9.8 Energy absorbed and transmitted . . . . . . . . . . . . 134Exa 10.1 Surface temperature . . . . . . . . . . . . . . . . . . . 136Exa 10.4 Net exchange of energy . . . . . . . . . . . . . . . . . 137Exa 10.5 Shape factor . . . . . . . . . . . . . . . . . . . . . . . 138Exa 10.8 Net heat exchange . . . . . . . . . . . . . . . . . . . . 138Exa 10.14 Net heat exchange and equilibrium temperature . . . . 139Exa 10.15 Radiant heat exchange . . . . . . . . . . . . . . . . . . 140Exa 10.16 Radiant heat exchange . . . . . . . . . . . . . . . . . . 141

Exa 10.17 Reduction in heat loss . . . . . . . . . . . . . . . . . . 142Exa 10.18 Loss of heat . . . . . . . . . . . . . . . . . . . . . . . . 143Exa 10.19 Radiation heat transfer . . . . . . . . . . . . . . . . . 144Exa 10.20 Net radiant heat exchange . . . . . . . . . . . . . . . . 145Exa 10.21 Loss of heat . . . . . . . . . . . . . . . . . . . . . . . . 146Exa 10.22 Net heat transfer . . . . . . . . . . . . . . . . . . . . . 147Exa 10.23 Percentage reduction in heat transfer and temperature

of the shield . . . . . . . . . . . . . . . . . . . . . . . 148Exa 10.24 Number of screens . . . . . . . . . . . . . . . . . . . . 149Exa 10.25 Loss of heat and reduction in heat . . . . . . . . . . . 150Exa 10.26 Error in temperature . . . . . . . . . . . . . . . . . . . 151

Exa 10.27 Rate of heat loss . . . . . . . . . . . . . . . . . . . . . 152Exa 10.28 Heat transfer coefficient . . . . . . . . . . . . . . . . . 153Exa 10.29 Extinction coefficient for radiation . . . . . . . . . . . 154Exa 10.30 Emmissivity of the mixture . . . . . . . . . . . . . . . 155Exa 10.31 Net radiation exchange and coefficient of heat transfer 156

7


9/213

Exa 11.1 Temperature of the surface . . . . . . . . . . . . . . . 158

Exa 11.2 Burnout heat flux . . . . . . . . . . . . . . . . . . . . 159Exa 11.3 Heat transfer coefficient and power dissipation . . . . 160Exa 11.4 Heat transfer coefficient . . . . . . . . . . . . . . . . . 161Exa 11.5 Heat transfer . . . . . . . . . . . . . . . . . . . . . . . 162Exa 11.6 Thickness of condensate film . . . . . . . . . . . . . . 163Exa 11.7 Rate of heat transfer and condensate mass flow rate . 164Exa 11.8 Heat transfer rate . . . . . . . . . . . . . . . . . . . . 165Exa 11.9 Rate of formation of condensate . . . . . . . . . . . . 166Exa 11.10 Condensation rate . . . . . . . . . . . . . . . . . . . . 167Exa 12.1 Overall heat transfer coefficient . . . . . . . . . . . . . 169Exa 12.2 Overall heat transfer coefficient . . . . . . . . . . . . . 170

Exa 12.3 Heat exchanger area . . . . . . . . . . . . . . . . . . . 171Exa 12.4 Heat exchanger area . . . . . . . . . . . . . . . . . . . 172Exa 12.5 Length of heat exchanger . . . . . . . . . . . . . . . . 174Exa 12.6 Surface area and rate of condensation of steam . . . . 175Exa 12.7 Effective log mean temperature difference . . . . . . . 176Exa 12.8 Area of heat exchanger . . . . . . . . . . . . . . . . . 177Exa 12.9 Area of heat exchanger . . . . . . . . . . . . . . . . . 178Exa 12.10 Surface area . . . . . . . . . . . . . . . . . . . . . . . 179Exa 12.11 Number of tube passes and number of tubes per pass . 181Exa 12.12 Surface area and temperature . . . . . . . . . . . . . . 182

Exa 12.13 Total heat transfer and surface temperature . . . . . . 184Exa 12.14 Parameters of heat exchanger . . . . . . . . . . . . . . 185Exa 12.15 Exit temperature . . . . . . . . . . . . . . . . . . . . . 187Exa 12.16 Outlet temperature . . . . . . . . . . . . . . . . . . . 188Exa 12.17 Outlet temperature . . . . . . . . . . . . . . . . . . . 189Exa 12.18 Diameter and length of heat exchanger . . . . . . . . . 190Exa 13.1 Average molecular weight . . . . . . . . . . . . . . . . 192Exa 13.2 Mass and molar average velocities and flux . . . . . . 193Exa 13.3 Diffusion flux . . . . . . . . . . . . . . . . . . . . . . . 194Exa 13.4 Initial rate of leakage . . . . . . . . . . . . . . . . . . 195Exa 13.5 Loss of oxygen by diffusion . . . . . . . . . . . . . . . 196

Exa 13.6 Mass transfer rate . . . . . . . . . . . . . . . . . . . . 197Exa 13.7 Diffusion rate . . . . . . . . . . . . . . . . . . . . . . 198Exa 13.8 Diffusion coefficient . . . . . . . . . . . . . . . . . . . 199Exa 13.9 Time required . . . . . . . . . . . . . . . . . . . . . . 200Exa 14.1 Convection mass transfer coefficient . . . . . . . . . . 202

8


10/213

Exa 14.2 Local Mass transfer coefficient . . . . . . . . . . . . . 203

Exa 14.3 Mass Transfer coefficient . . . . . . . . . . . . . . . . . 204Exa 14.4 Mass transfer coefficient . . . . . . . . . . . . . . . . . 205Exa 14.5 Average mass transfer coefficient . . . . . . . . . . . . 206Exa 14.6 Mass transfer coefficient . . . . . . . . . . . . . . . . . 206Exa 14.7 Steady state temperature . . . . . . . . . . . . . . . . 207Exa 14.8 True air temperature . . . . . . . . . . . . . . . . . . . 208Exa 14.9 Relative humidity . . . . . . . . . . . . . . . . . . . . 209Exa 14.10 Specific humidity . . . . . . . . . . . . . . . . . . . . . 210Exa 14.11 Rate of evaporation . . . . . . . . . . . . . . . . . . . 211

9


11/213

Chapter 1

Basic Concepts

Scilab code Exa 1.1 Rate of heat transfer

1 //C hapt e r −1 , Exa mpl e 1 . 1 , Pa ge 92 //

============================================================

3 clc

4 clear

56 //INPUT DATA7 L = 0 . 0 2 ; // T h ic ne s s o f s t a i n l e s s s t e e l p l at e i n m8 T = [ 5 5 0 , 5 0 ] ; // T em pe ra tu re s a t b ot h t he f a c e s i n

d e g r e e C9 k = 1 9 . 1 ; // Thermal C on du ct iv it y o f s t a i n l e s s s t e e l a t

3 0 0 d e g r e e C i n W/m .K10

11 //CALC9ULATIONS12 q = ( ( k * ( T ( 1 ) - T ( 2 ) ) ) / ( L * 1 0 0 0 ) ) ; // H eat t r a n s f e r e d p er

u n i a r e a i n kW/mˆ 21314 //OUTPUT15 mprintf ( ’ The h ea t t r a n s f e r e d t hr ou gh t he m a t e r i a l

p e r u n i t a r e a i s %3 . 1 f kW/mˆ 2 ’ ,q )

10


12/213

16

17 //=================================END OF PROGRAM==============================

Scilab code Exa 1.2 Rate of heat transfer

1 //C hapt e r −1 , Exa mpl e 1 . 2 , Pa ge 1 12 //

============================================================

3 clc

4 clear

5

6 //INPUT DATA7 L = 1 ; // Length o f t h e f l a t p l a t e i n m8 w = 0 . 5 ; // Width o f t he f l a t p l a te i n m9 T = 3 0 ; // A i r s tr ea m t em p er a tu re i n d e g re e C

10 h = 3 0 ; / / C o n v e c t i v e h e a t t r a n s f e r c o e f f i c i e n t i n W/mˆ 2 . K

11 T s = 3 0 0 ; // T em pe ra tu re o f t he p l a t e i n d e gr e e C

1213 //CALCULATIONS14 A = ( L * w ) ; // Area o f t he p l a t e i n mˆ215 Q = ( h * A * ( T s - T ) / ( 1 0 0 0 ) ) ; // H eat t r a n s f e r i n kW16

17 //OUTPUT18 mprintf ( ’ H eat t r a n s f e r r a t e i s %3 . 2 f kW ’ ,Q )19

20 //=================================END OF PROGRAM==============================

Scilab code Exa 1.3 Rate of radiant heat

11


13/213

1 //C hapt e r −1 , Exa mpl e 1 . 3 , Pa ge 1 1

2 // ============================================================

3 clc

4 clear

5

6 //INPUT DATA7 T = 5 5 ; // S u r f a ce t em p er at u re i n d e g re e C8

9 //CALCULATIONS10 q = ( 5 . 6 6 9 7 * 1 0 ^ - 8 * ( 2 7 3 + T ) ^ 4 ) / 1 0 0 0 ; // The r a t e a t wh ic h

t h e r a d i a t o r e mi ts r a di a nt h ea t p er u ni t a re a i f i t b e ha v es a s a b l a c k body i n kW/mˆ 2

11

12 //OUTPUT13 mprintf ( ’ The r a t e a t which t he r a d i a t o r e mi ts

r a d i a n t h e at p e r u n i t a r e a i s %3 . 2 f kW/mˆ 2 ’ ,q )14

15 //=================================END OF PROGRAM==============================

Scilab code Exa 1.5 Overall heat transfer coefficient


============================================================

3 clc

4 clear

56 //INPUT DATA7 k = 0 . 1 4 5 ; / / Therma l c o n d u c t i v i t y o f F i r e b r i c k i n W/m. K8 e = 0 . 8 5 ; / / E m i s s i v i t y9 L = 0 . 1 4 5 ; // T hi ck ne ss o f t he w a ll i n m

12


14/213

10 T g = 8 0 0 ; // Gas t e m pe r a tu r e i n d e g r e e C

11 T w g = 7 9 8 ; // Wall t em pe ra tu re i o n g as s i d e i n d e g r ee C12 h g = 4 0 ; / / F il m c o n d u ct a n c e on g a s s i d e i n W/mˆ 2 . K13 h c = 1 0 ; / / Fi lm c o nd u ct a nc e on c o o l a n t s i d e i n W/mˆ 2 .K14 F = 1 ; // R a d i at i o n Sh ap e f a c t o r b et we en w a l l and g a s15

16 //CALCULATIONS17 R 1 = ( ( ( e * 5 . 6 7 * 1 0 ^ - 8 * F * ( ( T g + 2 7 3 ) ^ 4 - ( T w g + 2 7 3 ) ^ 4 ) ) / ( T g -

T w g ) ) + ( 1 / h g ) ) ; // Thermal r e s i s t a n c e i n v e r s e18 R 2 = ( L / k ) ; // T herm al r e s i s t a n c e19 R 3 = ( 1 / h c ) ; / / Th erma l r e s i s t a n c e20 U = 1 / ( ( 1 / R 1 ) + R 2 + R 3 ) ; // O v er a l l h ea t t r a n s f e r

c o e f f i c i e n t i n W/mˆ 2 .K21

22 //OUTPUT23 mprintf ( ’ O v e r a l l h e a t t r a n s f e r c o e f f i c i e n t i s %3 . 3 f

W/mˆ2.K’ ,U )

Scilab code Exa 1.6 Heat loss per unit length


============================================================

3 clc

4 clear

5

6 //INPUT DATA7 D = 0 . 0 5 ; // O ut si de d i am et er o f t he p i pe i n m8 e = 0 . 8 ; / / E m m i s s i v i t y

9 T = 3 0 ; / /Room T e mp e ra t ur e i n d e g r e e C10 T s = 2 5 0 ; // S u r f a ce t em p er at u re i n d e gr e e C11 h = 1 0 ; / / C o n v e c t i v e h e a t t r a n s f e r c o e f f i c i e n t i n W/m

ˆ 2 . K12

13


15/213

13 //CALCULATIONS

14 q = ( ( h * 3 . 1 4 * D * ( T s - T ) ) + ( e * 3 . 1 4 * D * 5 . 6 7 * 1 0 ^ - 8 * ( ( T s + 4 7 3 )^ 4 - ( T + 2 7 3 ) ^ 4 ) ) ) ; // Heat l o s s p er u n i t l e n g t h o f p i p e i n W/m

15

16 //OUTPUT17 mprintf ( ’ H eat l o s s p er u n i t l e n g t h o f p ip e i s %3 . 1 f

W/m ’ ,q )18

19 //=================================END OF PROGRAM==============================

Scilab code Exa 1.7 Surface temperature


============================================================

3 clc

4 clear

56 //INPUT DATA7 A = 0 . 1 ; // S u r f ac e a r ea o f w at er h e a t er i n mˆ28 Q = 1 0 0 0 ; // Heat t r a n s f e r r a t e i n W9 T w a t e r = 4 0 ; // T em pe ra tu re o f w at er i n d e g re e C

10 h 1 = 3 0 0 ; //H e at t r a n s f e r c o e f f i c i e n t i n W/mˆ2. K11 T a i r = 4 0 ; // T em per at ure o f a i r i n d e gr e e C12 h 2 = 9 ; //H e at t r a n s f e r c o e f f i c i e n t i n W/mˆ2. K13

14 //CALCULATIONS

15 T s w = ( Q / ( h 1 * A ) ) + T w a t e r ; / / T e m pe r at u re when u s e d i nw at er i n d e g r ee C16 T s a = ( Q / ( h 2 * A ) ) + T a i r ; / / T em pe ra tu re when u s ed i n a i r

i n d eg re e C17

14


16/213

18 //OUTPUT

19 mprintf ( ’ T em pe ra tu re when u s ed i n w a te r i s %3 . 1 f d e g r e e C \n T em pe ra tu re when u se d i n a i r i s %id e g r e e C ’ , T s w , T s a )

20

21 //=================================END OF PROGRAM==============================

15


17/213

Chapter 3

OneDimensional Steady State

Heat Conduction

Scilab code Exa 3.1 Rate of heat loss and interior temperature


============================================================

3 clc

4 clear

5

6 //INPUT DATA7 l = 5 ; // L ength o f t he w a ll i n m8 h = 4 ; // H ei gh t o f t h e w al l i n m9 L = 0 . 2 5 ; // T hi ck ne ss o f t he w a ll i n m

10 T = [ 1 1 0 , 4 0 ] ; / / Te mp er at ur e on t h e i n n e r and o u t e rs u r f a c e i n d eg r e e C

11 k = 0 . 7 ; / / T he rm al c o n d u c t i v i t y i n W/m .K

12 x = 0 . 2 0 ; // D i st an c e from t he i n n er w a ll i n m1314 //CALCULATIONS15 A = l * h ; // A re a r o f t he w a ll i n mˆ216 Q = ( k * A * ( T ( 1 ) - T ( 2 ) ) ) / L ; // Heat t r a n s f e r r a te i n W

16


18/213

17 T = ( ( ( T ( 2 ) - T ( 1 ) ) * x ) / L ) + T ( 1 ) ; // T em pe ra tu re a t i n t e r i o r

p o in t o f t he w al l , 20 cm d i s t a n t from t he i n n erw al l i n d eg re e C18

19 //OUTPUT20 mprintf ( ’ a ) Heat t r a n s f e r r a t e i s %i W \n b )

T empe r atu re a t i n t e r i o r p o in t o f t he w al l , 20 cmd i s t a n t from t he i n ne r w al l i s %i d eg r e e C ’ , Q , T )

21

22 //=================================END OF PROGRAM==============================

Scilab code Exa 3.2 Tempertaure and heat flow


============================================================

3 clc

4 clear

56 D i = 0 . 0 5 ; // I nn e r d i a me te r o f h ol lo w c y l i n d e r i n m7 D o = 0 . 1 ; // O uter d ia me te r o f h ol lo w c y l i n d e r i n m8 T = [ 2 0 0 , 1 0 0 ] ; // I n n er and o u t er s u r f a c e t em p er a tu re i n

d e g r e e C9 k = 7 0 ; / / T he rm al c o n d u c t i v i t y i n W/m .K

10

11 //CALCULATIONS12 r o = ( D o / 2 ) ; // Outer r a di u s o f h ol lo w c y l i n d e r i n m13 r i = ( D i / 2 ) ; // I nn er r a di u s o f h ol lo w c y l i n d e r i n m

14 Q = ( ( 2 * 3 . 1 4 * k * ( T ( 1 ) - T ( 2 ) ) ) / ( log ( r o / r i ) ) ) ; //H e att r a n s f e r r a t e i n W15 r 1 = ( r o + r i ) / 2 ; // R ad iu s a t h al fw ay b et we en r o and r i

i n m16 T 1 = T ( 1 ) - ( ( T ( 1 ) - T ( 2 ) ) * ( log ( r 1 / r i ) / ( log ( r o / r i ) ) ) ) ; //

17


19/213

T em pe ra tu re o f t h e p o i n t h a l fw a y b et we en t h e

i n n e r and o ut er s u r f a c e i n d eg re e C1718 //OUTPUT19 mprintf ( ’ H eat t r a n s f e r r a t e i s %3 . 1 f W /m\n

T em pe ra tu re o f t h e p o i n t h a l fw a y b et we en t h ei n n er and o u te r s u r f a c e i s %3 . 1 f d e g r e e C ’ , Q , T 1 )

20

21 //=================================END OF PROGRAM==============================

Scilab code Exa 3.3 Heat flow and teperature


============================================================

3 clc

4 clear

5

6 D i = 0 . 1 ; // I n ne r d ia me te r o f h ol lo w s p he r e i n m7 D o = 0 . 3 ; // O uter d ia me te r o f h ol lo w s p he r e i n m8 k = 5 0 ; / / T he rm al c o n d u c t i v i t y i n W/m .K9 T = [ 3 0 0 , 1 0 0 ] ; // I n n er and o u t er s u r f a c e t em p er a tu re i n

d e g r e e C10

11 //CALCULATIONS12 r o = ( D o / 2 ) ; // O uter r a d i us o f h ol lo w s ph e r e i n m13 r i = ( D i / 2 ) ; // I n ne r r a d i us o f h ol lo w s ph e r e i n m14 Q = ( ( 4 * 3 . 1 4 * r o * r i * k * ( T ( 1 ) - T ( 2 ) ) ) / ( r o - r i ) ) / 1 0 0 0 ; //H e at

t r a n s f e r r a t e i n W15 r = r i + ( 0 . 2 5 * ( r o - r i ) ) ; / /The v a l u e a t one−f o u r t h way o f t e i n n e r and o ut er s u r f a c e s i n m

16 T = ( ( r o * ( r - r i ) * ( T ( 2 ) - T ( 1 ) ) ) / ( r * ( r o - r i ) ) ) + T ( 1 ) ; //T em pe ra tu re a t a p o i nt a q u a r t er o f t he way

18


20/213

bet w ee n t he i n n er and o u te r s u r f a c e s i n d e g r e e C

1718 //OUTPUT19 mprintf ( ’ H eat f l o w r a t e t hr ou gh t he s p he r e i s %3 . 2 f

kW \n Tem per atu re a t a p o i nt a q u a r t er o f t he waybet w ee n t he i n n er and o u te r s u r f a c e s i s %i d e g r ee

C ’ , Q , T )20

21 //=================================END OF PROGRAM==============================

Scilab code Exa 3.4 Heat loss per square meter surface area


============================================================

3 clc

4 clear

5

6 //INPUT DATA7 L = 0 . 4 ; // T hi ck ne ss o f t he f u rn a c e i n m8 T = [ 3 0 0 , 5 0 ] ; // S u r f ac e t e mp e ra t ur e s i n d e gr e e C9 //k = 0. 005T−5∗10ˆ−6Tˆ2

10

11 //CALCULATIONS12 q = ( ( 1 / L ) * ( ( ( 0 . 0 0 5 / 2 ) * ( T ( 1 ) ^ 2 - T ( 2 ) ^ 2 ) ) - ( ( 5* 1 0 ^ - 6 * ( T

( 1 ) ^ 3 - T ( 2 ) ^ 3 ) ) / 3 ) ) ) ; // H eat l o s s p er s q ua r e m et ers u r f a c e a r e a i n W/mˆ 2

13

14 //OUTPUT15 mprintf ( ’ H eat l o s s p er s qu ar e me t e r s u r f a c e a re a i s%3 . 0 f W/mˆ2 ’ ,q )

16

17 //=================================END OF PROGRAM

19


21/213

==============================

Scilab code Exa 3.5 Rate of heat flow


============================================================

3 clc

4 clear5

6 //INPUT DATA7 L = 0 . 2 ; // T hi ck ne ss o f t he w a ll i n m8 T = [ 1 0 0 0 , 2 0 0 ] ; // S u r f ac e t e mp e ra t ur e s i n d e gr e e C9 k o = 0 . 8 1 3 ; // V al ue o f t he rm al c o n d u c t i v i t y a t T=0 i n W

/m.K10 b = 0 . 0 0 0 7 1 5 8 ; // T e mpe r at ur e c o e f f i c i e n t of t he r mal

c o n d u ct i v i t y i n 1/K11

12 //CALCULATIONS

13 k m = k o * ( 1 + ( ( b * ( T ( 1 ) + T ( 2 ) ) ) / 2 ) ) ; / / C o n st a nt t h e r ma lc o n d u c t i v i t y i n W/m. K

14 q = ( ( k m * ( T ( 1 ) - T ( 2 ) ) ) / L ) ; / / R at e o f h e a t f l o w i n W/mˆ 215

16 //OUTPUT17 mprintf ( ’ R at e o f h e a t f l o w i s %3 . 0 f W/mˆ 2 ’ ,q )18

19 //=================================END OF PROGRAM==============================

Scilab code Exa 3.6 Heat loss per unit length


20


22/213

2 //

============================================================

3 clc

4 clear

5

6 //INPUT DATA7 r = [ 0 . 0 1 , 0 . 0 2 ] ; // I n ne r and o u te r r a d i u s o f a c op pe r

c y l i n d e r i n m8 T = [ 3 1 0 , 2 9 0 ] ; // I n n e r and Ou te r s u r f a c e t e m pe r a tu r e i n

d e g r e e C9 k o = 3 7 1 . 9 ; // V al ue o f t he rm al c o n d u c t i v i t y a t T=0 i n W

/m.K10 b = ( 9 . 2 5 * 1 0 ^ - 5 ) ; // T e mpe r at ur e c o e f f i c i e n t of t he r mal

c o n d u ct i v i t y i n 1/K11

12 //CALCULATIONS13 Tm=(( T(1) -150)+(T(2) -150))/2; / / Mean t e m p e r a t u r e i n

d e g r e e C14 k m = k o * ( 1 - ( b * T m ) ) ; // C o ns ta nt t he rm al c o n d u c t i v i t y i n

W/m.K15 q = ( ( 2 * 3 . 1 4 * k m * ( T ( 1 ) - T ( 2 ) ) ) / log ( r ( 2 ) / r ( 1 ) ) ) / 1 0 0 0 ; //

Heat l o s s p er u n i t l e n g th i n kW/m1617 //OUTPUT18 mprintf ( ’ H eat l o s s p er u n i t l e n g th i s %3 . 2 f kW/m ’ ,q )

Scilab code Exa 3.8 Thickness of insulation


2 // ============================================================

3 clc

4 clear

21


23/213

5

6 //INPUT DATA7 L 1 = 0 . 5 ; // T hi ck ne ss o f t he w a ll i n m8 k 1 = 1 . 4 ; / / T he rm al c o n d u c t i v i t y i n W/m .K9 k 2 = 0 . 3 5 ; // Thermal c o n d uc t i v i t y o f i n s u l a t i n g

m a t e r i a l i n W/m .K10 q = 1 4 5 0 ; // Heat l o s s p er s qu ar e me t r e i n W11 T = [ 1 2 0 0 , 1 5 ] ; // I n n er and o u t er s u r f a c e t e mp e ra t ur e s

i n d eg re e C12

13 //CALCULATIONS14 L 2 = ( ( ( T ( 1 ) - T ( 2 ) ) / q ) - ( L 1 / k 1 ) ) * k 2 ; ; / / T h i ck n e ss o f t h e

i n s u l a t i o n r e q u i r e d i n m15

16 //OUTPUT17 mprintf ( ’ T hi ck ne ss o f t h e i n s u l a t i o n r e q u i r e d i s %3

. 3 f m ’ , L 2 )18

19 //=================================END OF PROGRAM==============================

Scilab code Exa 3.9 Rate of heat leaking


============================================================

3 clc

4 clear

5

6 L 1 = 0 . 0 0 6 ; // T hi ck ne ss o f e a ch g l a s s s h e e t i n m7 L 2 = 0 . 0 0 2 ; // T hi ck ne ss o f a i r gap i n m8 T b = - 2 0 ; // T empe ra tur e o f t he a i r i n s i d e t he room i n

d e g r e e C9 T a = 3 0 ; // Ambient t em p er a tu re o f a i r i n d e g re e C

22


24/213

10 h a = 2 3 . 2 6 ; / / H ea t t r a n s f e r c o e f f i c i e n t b e tw e en g l a s s

a nd a i r i n W/mˆ 2 . K11 k g l a s s = 0 . 7 5 ; // T herm al c o n d u c t i v i t y o f g l a s s i n W/m. K12 k a i r = 0 . 0 2 ; / / Th erma l c o n d u c t i v i t y o f a i r i n W/m. K13

14 //CALCULATIONS15 q = ( ( T a - T b ) / ( ( 1 / h a ) + ( L 1 / k g l a s s ) + ( L 2 / k a i r ) + ( L 1 / k g l a s s )

+ ( 1 / h a ) ) ) ; // Rate o f h ea t l e a k i n g i n t o t he roomp er u n i t a r ea o f t he d oo r i n W/mˆ2

16

17 //OUTPUT18 mprintf ( ’ R ate o f h ea t l e a k i n g i n t o t he room p er u n it

a r e a o f t h e d o o r i s %3 . 1 f W/mˆ 2 ’ ,q )19

20 //=================================END OF PROGRAM==============================

Scilab code Exa 3.10 Heat transfer through the composite wall

1 //C hapt e r −3 , Exa mpl e 3 . 1 0 , P ag e 6 5

2 //============================================================

3 clc

4 clear

5

6 //INPUT DATA7 L A = 0 . 0 5 ; // L ength o f s e c t i o n A i n m8 L B = 0 . 1 ; // L engt h o f s e c t i o n A i n m9 L C = 0 . 1 ; // L engt h o f s e c t i o n A i n m

10 L D = 0 . 0 5 ; // L ength o f s e c t i o n A i n m11 L E = 0 . 0 5 ; // L ength o f s e c t i o n A i n m12 k A = 5 0 ; / / Therma l c o n d u c t i v i t y o f s e c t i o n A i n W/m. K13 k B = 1 0 ; / / Therma l c o n d u c t i v i t y o f s e c t i o n B i n W/m. K14 k C = 6 . 6 7 ; / / Therma l c o n d u c t i v i t y o f s e c t i o n C i n W/m. K

23


25/213

15 k D = 2 0 ; / / Therma l c o n d u c t i v i t y o f s e c t i o n D i n W/m. K

16 k E = 3 0 ; / / Therma l c o n d u c t i v i t y o f s e c t i o n E i n W/m. K17 A a = 1 ; // Area o f s e c t i o n A i n mˆ218 A b = 0 . 5 ; // Area o f s e c t i o n B i n mˆ219 A c = 0 . 5 ; // Area o f s e c t i o n C i n mˆ220 A d = 1 ; // Area o f s e c t i o n D i n mˆ221 A e = 1 ; // Area o f s e c t i o n E i n mˆ222 T = [ 8 0 0 , 1 0 0 ] ; // T em pe ra tu re a t i n l e t and o u t l e t

t e mp e ra t u re s i n d e g re e C23

24 //CALCULATIONS25 R a = ( L A / ( k A * A a ) ) ; // Thermal R e s is t a nc e o f s e c t i o n A i n

K/W26 R b = ( L B / ( k B * A b ) ) ; // Thermal R e s is t a nc e o f s e c t i o n B i n

K/W27 R c = ( L C / ( k C * A c ) ) ; // Thermal R e s is t a nc e o f s e c t i o n C i n

K/W28 R d = ( L D / ( k D * A d ) ) ; // Thermal R e s is t a nc e o f s e c t i o n D i n

K/W29 R e = ( L E / ( k E * A e ) ) ; // Thermal R e s is t a nc e o f s e c t i o n E i n

K/W30 R f = ( ( R b * R c ) / ( R b + R c ) ) ; // E q u i v a le nt r e s i s t a n c e o f

s e c t i o n B and s e c t i o n C i n K/W31 R = R a + R f + R d + R e ; // E qu i v al en t r e s i s t a n c e o f a l ls e c t i o n s i n K/W

32 Q = ( ( T ( 1 ) - T ( 2 ) ) / R ) / 1 0 0 0 ; // Heat t r a n s f e r t hr ou gh t hec o m po s i te w a l l i n kW

33

34 //OUTPUT35 mprintf ( ’ H eat t r a n s f e r t hr ou gh t he c om po si te w a ll i s

%3. 1 f kW’ ,Q )36

37 //=================================END OF PROGRAM

==============================

24


26/213

Scilab code Exa 3.11 Surface temperature and convective conductance

1 //C hapt e r −3 , Exa mpl e 3 . 1 1 , P ag e 6 62 //

============================================================

3 clc

4 clear

5

6 //INPUT DATA7 T 1 = 2 0 0 0 ; // T em per at ure o f h ot g as i n d e g re e C8 T a = 4 5 ; / /Room a i r t e mp e r at u r e i n d e g r e e C

9 Q r 1 = 2 3 . 2 6 0 ; // Heat f l o w by r a d i a t i o n from g a s es t oi n s i d e s u r f a c e o f t he w a l l i n kW/mˆ2

10 h = 1 1 . 6 3 ; / / C o n v e c t i v e h e a t t r a n s f e r c o e f f i c i e n t i n W/m ˆ 2 .

11 C = 5 8 ; / / Th er ma l c o n d u ct a n c e o f t h e w a l l i n W/mˆ 2 . K12 Q = 9 . 3 ; // Heat f l o w by r a d i a t i o n from e x t e r n a l s u r f a c e

t o t h e s u r r o u n d i n g i n kW .mˆ 213 T 2 = 1 0 0 0 ; // I n t e r i o r w a ll t em pe ra tu re i n d e g r e e C14

15 //CALCULATIONS16 q r 1 = Q r 1 ;

/ / Ha et by r a d i a t i o n i n kW/mˆ 217 q c 1 = h * ( ( T 1 - T 2 ) / 1 0 0 0 ) ; / / H e a t b y c o n d u c t i o n i n kW/mˆ 218 q = q c 1 + q r 1 ; / / T o ta l h e at e n t e r i n g t h e w a l l i n kW/mˆ 219 R = ( 1 / C ) ; / / Th er ma l r e s i s t a n c e i n mˆ 2 . K/W20 T 3 = T 2 - ( q * R * 1 0 0 0 ) ; // E x t e rn a l w a l l t e m pe r a tu r e i n

d e g r e e C21 Q l = q - Q ; / / He at l o s s d ue t o c o n v e c t i o n kW/mˆ 222 h 4 = ( Q l * 1 0 0 0 ) / ( T 3 - T a ) ; / / C o n v e c t i v e c o n d u c t a n c e i n W/m

ˆ 2 . K23

24 mprintf ( ’ The s u r f a c e t em pe ra tu re i s %i d e g r e e C \

nThe c o n v e c t i v e c o n d u ct a n c e i s %3 . 1 f W/mˆ 2 . K ’ , T 3 ,h4 )

25

26 //=================================END OF PROGRAM==============================

25


27/213

Scilab code Exa 3.12 Heat loss and thickness of insulation


============================================================

3 clc

4 clear

56 //INPUT DATA7 L 1 = 0 . 1 2 5 ; // T hi ck ne ss o f f i r e c l a y l a y e r i n m8 L 2 = 0 . 5 ; // T hi ck ne ss o f r ed b r i c k l a y e r i n m9 T = [ 1 1 0 0 , 5 0 ] ; // T e mp er at ur es a t i n s i d e and o u t s i d e t he

f u rn a c e s i n d eg re e C10 k 1 = 0 . 5 3 3 ; // Thermal c o n d u c t i v i t y o f f i r e c l a y i n W/m. K11 k 2 = 0 . 7 ; // T herm al c o n d u c t i v i t y o f r ed b r i c k i n W/m. K12

13 //CALCULATIONS14 R 1 = ( L 1 / k 1 ) ; // R e s i s t a n c e o f f i r e c l a y p e r u ni t a re a i n

K/W15 R 2 = ( L 2 / k 2 ) ; // R e s is t a nc e o f r ed b r i c k p er u n it a re a

i n K/W16 R = R 1 + R 2 ; / / T o ta l r e s i s t a n c e i n K/W17 q = ( T ( 1 ) - T ( 2 ) ) / R ; / / H ea t t r a n s f e r i n W/mˆ 218 T 2 = T ( 1 ) - ( q * R1 ) ; / / Te mp er at ur e i n d e g r e e C19 T 3 = T ( 2 ) + ( q * R 2 * 0 . 5 ) ; // T em per at ur e a t t he i n t e r f a c e

bet w ee n t he two l a y e r s i n d e g r ee C20 k m = 0 . 1 1 3 + ( 0 . 0 0 0 2 3 * ( ( T 2 + T 3 ) / 2 ) ) ; //Mean t he r mal

c o n d u c t i v i t y i n W/m. K

21 x = ( ( T 2 - T 3 ) / q ) * k m ; // T hi c kn e ss o f d i a t o mi t e i n m22

23 //OUTPUT24 mprintf ( ’ Amount o f h e a t l o s s i s %3 . 1 f W/mˆ 2 \n

T hi ck ne ss o f d i at o mi t e i s %3 . 4 f m ’ ,q , x )

26


28/213

25

26 //=================================END OF PROGRAM==============================

Scilab code Exa 3.13 Heat loss from the pipe


============================================================

3 clc

4 clear

5

6 //INPUT DATA7 D i = 0 . 1 ; // I . D o f t he p i p e i n m8 L = 0 . 0 1 ; // T hi ck ne ss o f t he w a ll i n m9 L 1 = 0 . 0 3 ; // T hi ck ne ss o f i n s u l a t i o n i n m

10 T a = 8 5 ; // T emper at ur e o f h ot l i q u i d i n d e g r e e C11 T b = 2 5 ; // T em per at ur e o f s u r ro u n d i ng s i n d e gr e e C12 k 1 = 5 8 ; / / Therma l c o n d u c t i v i t y o f s t e e l i n W/m. K

13 k 2 = 0 . 2 ; // Thermal c o n d u ct i v i t y o f i n s u l a t i n g m a t e r i a lin W/m.K

14 h a = 7 2 0 ; / / I n s i d e h e a t t r a n s f e r c o e f f i c i e n t i n W/mˆ 2 . K15 h b = 9 ; / / O u t s i d e h e a t t r a n s f e r c o e f f i c i e n t i n W/m ˆ 2 . K16 D 2 = 0 . 1 2 ; / / I n n e r d i a m et e r i n m17 r 3 = 0 . 0 9 ; / / R ad iu s i n m18

19 //CALCULATIONS20 q = ( ( 2 * 3 . 1 4 * ( T a - T b ) ) / ( ( 1 / ( h a * ( D i / 2 ) ) ) + ( 1 / ( h b * r 3 ) ) + (

log ( D 2 / D i ) / k 1 ) + ( log ( r 3 / ( D 2 / 2 ) ) / k 2 ) ) ) ; // H eat l o s s

f r o an i n s u l a t e d p i p e i n W/m2122 //OUTPUT23 mprintf ( ’ H eat l o s s f r o an i n s u l a t e d p ip e i s %3 . 2 f W/

m ’ ,q )

27


29/213

24

25 //=================================END OF PROGRAM==============================

Scilab code Exa 3.14 Heat loss per meter length of pipe


============================================================

3 clc

4 clear

5

6 //INPUT DATA7 D i = 0 . 1 ; // I . D o f t he p i p e i n m8 D o = 0 . 1 1 ; //O. D o f t he p ip e i n m9 L = 0 . 0 0 5 ; // T hi ck ne ss o f t he w a ll i n m

10 k 1 = 5 0 ; // Thermal c o n d u c t i v i t y o f s t e e l p ip e l i n e i n W/m.K

11 k 2 = 0 . 0 6 ; // Thermal c o n d u c t i v i t y o f f i r s t i n s u l a t i n g

m a t e r i a l i n W/m .K12 k 3 = 0 . 1 2 ; // Thermal c o n d u c t i v i t y o f s ec on d i n s u l a t i n g

m a t e r i a l i n W/m .K13 T = [ 2 5 0 , 5 0 ] ; // T em per at ur e a t i n s i d e t ub e s u r f a c e and

o u t s i d e s u r f a c e o f i n s u l a t i o n i n d eg re e C14 r 3 = 0 . 1 0 5 ; // R adi u s o f r 3 i n m a s shown i n f i g . 3 . 1 4 on

p a ge no . 7 115 r 4 = 0 . 1 5 5 ; // R adi u s o f r 4 i n m a s shown i n f i g . 3 . 1 4 on

p a ge no . 7 116

17 //CALCULATIONS18 r 1 = ( D i / 2 ) ; // R adi us o f t he p ip e i n m19 r 2 = ( D o / 2 ) ; // R adi us o f t he p ip e i n m20 q = ( ( 2 * 3 . 1 4 * ( T ( 1 ) - T ( 2 ) ) ) / ( ( ( log ( r 2 / r 1 ) ) / k 1 ) + ( ( log ( r 3 /

r 2 ) ) / k 2 ) + ( ( log ( r 4 / r 3 ) ) / k 3 ) ) ) ; // L os s o f h ea t p er

28


30/213

m et re l e n g t h o f p i p e i n W/m

21 T 3 = ( ( q * log ( r 4 / r 3 ) ) / ( 2 * 3 . 1 4 * k 3 ) ) + T ( 2 ) ; // I n t e r f a c et em p er a tu r e i n d e gr e e C22

23 //OUTPUT24 mprintf ( ’ L o ss o f h ea t p er m et r e l e n g t h o f p ip e i s %3

. 1 f W/m \n I n t e r f a c e t em pe ra tu re i s %3 . 1 f d e g r e eC ’ , q , T 3 )

25

26 //=================================END OF PROGRAM==============================

Scilab code Exa 3.15 Change in heat loss


============================================================

3 clc

4 clear

56 //INPUT DATA7 D i = 0 . 1 ; // I . D o f t he p i p e i n m8 D o = 0 . 1 1 ; //O. D o f t he p ip e i n m9 L = 0 . 0 0 5 ; // T hi ck ne ss o f t he w a ll i n m

10 k 1 = 5 0 ; // Thermal c o n d u c t i v i t y o f s t e e l p ip e l i n e i n W/m.K

11 k 3 = 0 . 0 6 ; // Thermal c o n d u c t i v i t y o f f i r s t i n s u l a t i n gm a t e r i a l i n W/m .K

12 k 2 = 0 . 1 2 ; // Thermal c o n d u c t i v i t y o f s ec on d i n s u l a t i n g

m a t e r i a l i n W/m .K13 T = [ 2 5 0 , 5 0 ] ; // T em per at ur e a t i n s i d e t ub e s u r f a c e ando u t s i d e s u r f a c e o f i n s u l a t i o n i n d eg re e C

14 r 3 = 0 . 1 0 5 ; // R adi u s o f r 3 i n m a s shown i n f i g . 3 . 1 4 onp a ge no . 7 1

29


31/213

15 r 4 = 0 . 1 5 5 ; // R adi u s o f r 4 i n m a s shown i n f i g . 3 . 1 4 on

p a ge no . 7 11617 //CALCULATIONS18 r 1 = ( D i / 2 ) ; // R adi us o f t he p ip e i n m19 r 2 = ( D o / 2 ) ; // R adi us o f t he p ip e i n m20 q = ( ( 2 * 3 . 1 4 * ( T ( 1 ) - T ( 2 ) ) ) / ( ( ( log ( r 2 / r 1 ) ) / k 1 ) + ( ( log ( r 3 /

r 2 ) ) / k 2 ) + ( ( log ( r 4 / r 3 ) ) / k 3 ) ) ) ; // L os s o f h ea t p erm et re l e n g t h o f p i p e i n W/m

21

22 //OUTPUT23 mprintf ( ’ L o ss o f h ea t p er m et r e l e n g t h o f p ip e i s %3

. 2 f W/m ’ ,q )24 // Comparing t he r e s u l t w it h t he p r e v i o us ex am ple Ex

. 3 . 1 4 , i t i s s e e n t h a t t h e l o s s o f h e a t i si n c r e as e d by ab ou t 1 8 .1 1%. S i nc e t he p ur po se o f i n s u l a t i o n i s t o r e d u c e t h e l o s s o f hea t , i t i sa lw a ys b e t t e r t o p ro v i de t he i n s u l a t i n g m a t e r i a lw i th low t he rm al c o n d u ct i v i t y on t he s u r f a c e o f t h e t u b e f i r s t

25

26 //=================================END OF PROGRAM

==============================

Scilab code Exa 3.16 Mass of steam condensed


============================================================

3 clc4 clear

5

6 //INPUT DATA7 D 1 = 0 . 1 ; //O. D o f t he p i p e i n m

30


32/213

8 P = 1 3 7 3 ; // P r e s su r e o f s a t u r a t ed s team i n kPa

9 D 2 = 0 . 2 ; // D i am et er o f m ag ne si a i n m10 k 1 = 0 . 0 7 ; / / Th er ma l c o n d u c t i v i t y o f m a gn e si a i n W/m. K11 k 2 = 0 . 0 8 ; / / Therma l c o n d u c t i v i t y o f a s b e s t o s i n W/m. K12 D 3 = 0 . 2 5 ; // D ia me t er o f a s b e s t o s i n m13 T 3 = 2 0 ; / / T em er at ur e u nd er t h e c an v as i n d e g r e e C14 t = 1 2 ; // Time f o r c o n de n s at i o n i n h ou rs15 l = 1 5 0 ; / / Lemgth o f p i p e i n m16 T 1 = 1 9 4 . 1 4 ; / / S a t u r a t i o n t e m pe r a tu r e o f s te am i n

d e g r e e C f r om T ab l e A . 6 ( A pp en di x A) a t 1 3 73 kPaon p ag e no . 6 43

17 h f g = 1 9 6 3 . 1 5 ; // L a te n t h e at o f s te am i n kJ / kg f ro m

T ab l e A . 6 ( A pp en di x A ) a t 1 37 3 kPa on p ag e no .643

18

19 //CALCULATIONS20 r 1 = ( D 1 / 2 ) ; // R adi us o f t he p ip e i n m21 r 2 = ( D 2 / 2 ) ; // R ad iu s o f m ag ne si a i n m22 r 3 = ( D 3 / 2 ) ; // R adi us o f a s b e s t o s i n m23 Q = ( ( ( 2 * 3 . 1 4 * l * ( T 1 - T 3 ) ) / ( ( log ( r 2 / r 1 ) / k 1 ) + ( log ( r 3 / r 2 ) /

k 2 ) ) ) * ( 3 6 0 0 / 1 0 0 0 ) ) ; // Heat t r a n s f e r r a t e i n k J/h24 m = ( Q / h f g ) ; / / Mass o f s tea m c o nd en s ed p e r h ou r25 m 1 = ( m * t ) ;

// Mass o f s te am c o nd e ns e d i n 1 2 h o ur s2627 //OUTPUT28 mprintf ( ’ Mass o f stea m c on de ns ed i n 12 h ou rs i s %3 . 2

f kg ’ , m 1 )29

30 //=================================END OF PROGRAM==============================



31


33/213

============================================================

3 clc

4 clear

5

6 //INPUT DATA7 D 1 = 0 . 1 ; // I . D o f t h e f i r s t p i p e i n m8 D 2 = 0 . 3 ; //O .D o f t h e f i r s t p i p e i n m9 k 1 = 7 0 ; / / Therma l c o n d u c t i v i t y o f f i r s t m a t e r i a l i n W/

m. K10 D 3 = 0 . 4 ; //O. D o f t he s ec on d p i p e i n m11 k 2 = 1 5 ; // Thermal c o n d u c t i v i t y o f s ec on d m a t e r i a l i n W

/m.K12 T = [ 3 0 0 , 3 0 ] ; // I n s i d e and o u t s i d e t e mp e ra t u re s i n

d e g r e e C13

14 //CALCULATIONS15 r 1 = ( D 1 / 2 ) ; / / I n n e r R ad iu s o f f i r s t p i p e i n m16 r 2 = ( D 2 / 2 ) ; / / Ou te r R ad iu s o f f i r s t p i p e i n m17 r 3 = ( D 3 / 2 ) ; // R adi us o f s ec on d p i p e i n m18 Q = ( ( 4 * 3 . 1 4 * ( T ( 1 ) - T ( 2 ) ) ) / ( ( ( r 2 - r 1 ) / ( k 1 * r 1 * r 2 ) ) + ( ( r 3 -

r 2 ) / ( k 2 * r 2 * r 3 ) ) ) ) / 1 0 0 0 ; // Rate o f h ea t f l ow

t hr ou gh t he s p h er e i n kW1920 //OUTPUT21 mprintf ( ’ R ate o f h ea t f l o w t hr ou gh t he s p he r e i s %3

. 2 f kW’ ,Q )22

23 //=================================END OF PROGRAM==============================

Scilab code Exa 3.18 Heat loss and surface temperature


32


34/213

============================================================

3 clc

4 clear

5

6 //INPUT DATA7 D i = 0 . 1 ; // I . D o f a s team p i pe i n m8 D o = 0 . 2 5 ; // I . D o f a stea m p i pe i n m9 k = 1 ; // Thermal c o n d u ct i v i t y o f i n s u l a t i n g m a t e r i a l i n

W/m.K10 T = [ 2 0 0 , 2 0 ] ; / / S tea m t e m p e r a t u r e a nd a m bi e n t

t e mp e ra t u re s i n d e g re e C

11 h = 8 ; / / C o n v e c t i v e h e a t t r a n s f e r c o e f f i c i e n t b e tw e ent he i n s u l a t i o n s u r f a c e and a i r i n W/mˆ 2 .K

12

13 //CALCULATIONS14 r i = ( D i / 2 ) ; // I n n er R ad iu s o f s team p i pe i n m15 r o = ( D o / 2 ) ; // O uter R ad iu s o f s team p i pe i n m16 r c = ( k / h ) * 1 0 0 ; // C r i t i c a l r a d i u s o f i n s u l a t i o n i n cm17 q = ( ( T ( 1 ) - T ( 2 ) ) / ( ( log ( r o / r i ) / ( 2 * 3 . 1 4 * k ) ) + ( 1 / ( 2 * 3 . 1 4 *

r o * h ) ) ) ) ; // Heat l o s s p e r me t r e o f p ip e a tc r i t i c a l ra di u s in W/m

18 R o = ( q / ( 2 * 3 . 1 4 * r o * h ) ) + T ( 2 ) ;/ / Ou te r s u r f a c et em p er a tu r e i n d e gr e e C

19

20 //OUTPUT21 mprintf ( ’ H eat l o s s p e r m et re o f p i p e a t c r i t i c a l

r a d i u s i s %i W/m \n Outer s u r f a c e t em pe ra tu re i s%3 . 2 f d e g r e e C ’ , q , R o )

22

23 //=================================END OF PROGRAM==============================

Scilab code Exa 3.19 Effect of insulation on the current carrying conduc-tor

33


35/213


2 // ============================================================

3 clc

4 clear

5

6 //INPUT DATA7 D i = 0 . 0 0 1 ; // D i am et er o f c op pe r w i re i n m8 t = 0 . 0 0 1 ; // T hi ck ne ss o f i n s u l a t i o n i n m;9 T o = 2 0 ; // T em per at ur e o f s u r r o n d i ng s i n d e g re e C

10 T i = 8 0 ; / /Maximum t e m p e ra t u re o f t h e p l a s t i c i n d e g r e e

C11 k c o p p e r = 4 0 0 ; / / Th er ma l c o n d u c t i v i t y o f c o p p er i n W/m.

K12 k p l a s t i c = 0 . 5 ; // Thermal c o n d u c t i v i t y o f p l a s t i c i n W/

m. K13 h = 8 ; //H e at t r a n s f e r c o e f f i c i e n t i n W/mˆ2. K14 p = ( 3 * 1 0 ^ - 8 ) ; // S p e c i f i c e l e c t r i c r e s i s t a n c e o f c o pp e r

i n Ohm.m15

16 //CAlCULATIONS17 r = ( D i / 2 ) ;

// R ad iu s o f c op pe r t ub e i n m18 r o = ( r + t ) ; // R a di us i n m19 R = ( p / ( 3 . 1 4 * r * r * 0 . 0 1 ) ) ; // E l e c t r i c a l r e s i s t a n c e p er

m et er l e n g t h i n ohm/m20 R t h = ( ( 1 / ( 2 * 3 . 1 4 * r o * h ) ) + ( log ( r o / r ) / ( 2 * 3 . 1 4 * k p l a s t i c ) )

) ; // Thermal r e s i s t a n c e o f c o nv e ct i on f i l mi n s u l a t i o n p er me t r e l e ng t h

21 Q = ( ( T i - T o ) / R t h ) ; // Heat t r a n s f e r i n W22 I = sqrt ( Q / R ) ; // Maximum s a f e c u r r e n t l i m i t i n A23 r c = ( ( k p l a s t i c * 1 0 0 ) / h ) ; // C r i t i c a l r a di u s i n cm24

25 //OUTPUT26 mprintf ( ’ The maximum s a f e c u r r e n t l i m i t i s %3 . 3 f A \

n As t he c r i t i c a l r a d i u s o f i n s u l a t i o n i s muchg r e a t e r t ha n t h at p r ov i de d i n t he pro blem , t hec u r r e nt c a r r y i n g c a pa c i ty o f t he c on du ct or can be

34


36/213

r a i s e d c o n s i d e r a b l y i n i n c r e a s i n g t h e r a d i u s o f

p l a s t i c c o v e r i ng u pt o %3 . 1 f cm .\ n T h i s mayhowe ve r l e ad t o t he pr obl e m o f h av in g t oo h ig h at em pe ra tu re a t t he c a b le c e n tr e i f t het em pe r at ur e i n s i d e t h e p l a s t i c c o at i ng ha s t o b ek ep t w i t hi n t he g i v e n l i m i t s ’ , I , r c )

27

28 //=================================END OF PROGRAM==============================

Scilab code Exa 3.20 Surface temperature and maximum temperature inthe wall


============================================================

3 clc

4 clear

5

6 //INPUT DATA7 L = 0 . 1 ; // T hi ck ne ss o f t he w a ll i n m8 Q = ( 4 * 1 0 ^ 4 ) ; / / Hea t t a n s f e r r a t e i n W/mˆ 39 h = 5 0 ; / / C o n v e c t i v e h e a t t r a n s f e r c o e f f i c i e n t i n W/m

ˆ 2 . K10 T = 2 0 ; / / Ambient a i r t e mp e ra t u re i n d e g r e e C11 k = 1 5 ; / / Th erma l c o n d u c t i v i t y o f t h e m a t e r i a l i n W/m. K12

13 //CALCULATIONS14 T w = ( T + ( ( Q * L ) / ( 2 * h ) ) ) ; / / S u r f a c e t e m pe r a tu r e i n d e g r e e

C15 T m a x = ( T w + ( ( Q * L * L ) / ( 8 * k ) ) ) ; //Maximum t e m pe r at u r e i nt h e w al l i n d eg re e C

16

17 //OUTPUT

35


37/213

18 mprintf ( ’ S u r f a ce t em p er a tu r e i s %i d e g re e C \n

Maximum t e mp e r at u r e i n t h e w a l l i s %3 . 3 f d e g r e e C’ , T w , T m a x )19

20 //=================================END OF PROGRAM==============================

Scilab code Exa 3.21 Surface temperature


============================================================

3 clc

4 clear

5

6 //INPUT DATA7 D o = 0 . 0 0 6 ; // O uter d ia me te r o f h al l o w c y l i n d e r i n m8 D i = 0 . 0 0 4 ; // I nn e r d ia me te r o f h al l o w c y l i n d e r i n m9 I = 1 0 0 0 ; // C u rr en t i n A

10 T = 3 0 ; // T em pe ra tu re o f w at er i n d e g re e C11 h = 3 5 0 0 0 ; //H e at t r a n s f e r c o e f f i c i e n t i n W/mˆ2. K12 k = 1 8 ; / / Th erma l c o n d u c t i v i t y o f t h e m a t e r i a l i n W/m. K13 R = 0 . 1 ; / / E l e c t r i c a l r e i s i t i v i t y o f t h e m a t e r i a l i n

ohm .mmˆ2 /m14

15 //CALCULATIONS16 r o = ( D o / 2 ) ; // Outer r a di u s o f h al lo w c y l i n d e r i n m17 r i = ( D i / 2 ) ; // I nn er r a di u s o f h al lo w c y l i n d e r i n m18 V = ( ( 3 . 1 4 * ( r o ^ 2 - r i ^ 2 ) ) ) ; // V ol . o f w i re i n mˆ2

19 R t h = ( R / ( 3 . 1 4 * ( r o ^ 2 - r i ^ 2 ) * 1 0 ^ 6 ) ) ; // R e s i s t i v i t y i n ohm/mmˆ220 q = ( ( I * I * R t h ) / V ) ; / / Hea t t r a n s f e r r a t e i n W/mˆ 321 T o = T + ( ( ( q * r i * r i ) / ( 4 * k ) ) * ( ( ( ( 2 * k ) / ( h * r i ) ) - 1 ) * ( ( r o / r i )

^ 2 - 1 ) + ( 2 * ( r o / r i ) ^ 2 * log ( r o / r i ) ) ) ) ; / / T e m pe r at u re a t

36


38/213

t h e o ut er s u r f a c e i n d eg re e C

2223 //OUTPUT24 mprintf ( ’ T em pe ra tu re a t t he o u t er s u r f a c e i s %3 . 2 f

d e g r e e C ’ , T o )25

26 //=================================END OF PROGRAM==============================

Scilab code Exa 3.22 Heat transfer coefficient and maximum temperature


============================================================

3 clc

4 clear

5

6 //INPUT DATA7 D = 0 . 0 2 5 ; // D ia me te r o f a n ne al e d c op pe r w i re i n m

8 I = 2 0 0 ; // C u rr en t i n A9 R = ( 0 . 4 * 1 0 ^ - 4 ) ; / / R e s i s t a n c e i n ohm/ cm

10 T = [ 2 0 0 , 1 0 ] ; / / S u r f a c e t e m p e r a tu r e and a mb i en tt em p er a tu r e i n d e gr e e C

11 k = 1 6 0 ; / / T he rm al c o n d u c t i v i t y i n W/m .K12

13 //CALCULATIONS14 r = ( D / 2 ) ; // R ad ius o f a nn ea le d c op pe r w ir e i n m15 Q = ( I * I * R * 1 0 0 ) ; // H eat t r a n s f e r r a t e i n W/m16 V = ( 3 . 1 4 * r * r ) ; // V ol . o f w i re i n mˆ2

17 q = ( Q / V ) ; / / Hea t l o s s i n c o n du c to r i n W/mˆ218 T c = T ( 1 ) + ( ( q * r * r ) / ( 4 * k ) ) ; / / Maximum t e m p e r a t u r e i n t h ew ir e i n d eg r e e C

19 h = ( ( r * q ) / ( 2 * ( T ( 1 ) - T ( 2 ) ) ) ) ; // H eat t r a n s f e rc o e f f i c i e n t i n W/mˆ 2 .K

37


39/213

20

21 //OUTPUT22 mprintf ( ’ Maximum t e m p e r at u r e i n t h e w i r e i s %3 . 2 f d e g r e e C \n H ea t t r a n s f e r c o e f f i c i e n t i s %3 . 2 f W/

m ˆ 2 . K ’ , T c , h )23

24 //=================================END OF PROGRAM==============================

Scilab code Exa 3.23 Diameter of wire and rate of current flow


============================================================

3 clc

4 clear

5

6 //INPUT DATA7 p = 1 0 0 ; // R e s i s t i v i t y o f ni c h r o m e i n ohm−cm

8 Q = 1 0 0 0 0 ; // Heat i np ut o f a h e at e r i n W9 T = 1 2 2 0 ; // S u r f a ce t em p er at u re o f n ic hr om e i n d e gr e e C

10 T a = 2 0 ; // T em per at ur e o f s u rr o u nd i n g a i r i n d e gr e e C11 h = 1 1 5 0 ; // O u ts i de s u r f a c e c o e f f i e n t i n W/mˆ 2 . K12 k = 1 7 ; / / Th er ma l c o n d u c t i v i t y o f n i ch r om e i n W/m. K13 L = 1 ; // L ength o f h e at e r i n m14

15 //CALCULATIONS16 d = ( Q / ( ( T - T a ) * 3 . 1 4 * h ) ) * 1 0 0 0 ; / / D i am et er o f n i ch r om e

w i r e i n mm

17 A = ( 3 . 1 4 * d * d ) / 4 ; // Area o f t he w i re i n mˆ218 R = ( ( p * 1 0 ^ - 8 * L ) / A ) ; // R e s i s t a n ce o f t he w i re i n ohm19 I = sqrt ( Q / R ) / 1 0 0 0 ; // Rate o f c u r r e n t f lo w i n A20

21 //OUTPUT

38


40/213

22 mprintf ( ’ D i am et er o f n i ch r om e w i r e i s %3 . 4 f mm \n

Rate o f c u r r e n t f lo w i s %i A ’ , d , I )2324 //=================================END OF PROGRAM

==============================

Scilab code Exa 3.24 Temperature drop


2 // ============================================================

3 clc

4 clear

5

6 //INPUT DATA7 D o = 0 . 0 2 5 ; //O. D o f t he r o d i n m8 k = 2 0 ; / / T he rm al c o n d u c t i v i t y i n W/m .K9 Q = ( 2 . 5 * 1 0 ^ 6 ) ; / / R a te o f h e a t r e m o va l i n W/mˆ 2

10

11 //CALCULATIONS12 r o = ( D o / 2 ) ; // Outer r a di u s o f t h e r o d i n m13 q = ( ( 4 * Q ) / ( r o ) ) ; / / Hea t t r a n s f e r r a t e i n W/mˆ314 T = ( ( - 3 * q * r o ^ 2 ) / ( 1 6 * k ) ) ; / / T em p er at ur e d ro p f ro m t h e

c e n t r e l i n e t o t h e s u r f a c e i n d eg re e C15

16 //OUTPUT17 mprintf ( ’ T em pe ra tu re d ro p fro m t he c e n t r e l i n e t o

t h e s u r f a c e i s %3 . 3 f d eg r e e C ’ ,T )18

19 //=================================END OF PROGRAM==============================

39


41/213

Scilab code Exa 3.25 Temperature at the centre of orange


============================================================

3 clc

4 clear

5

6 //INPUT DATA7 Q = 3 0 0 ; / / He at p r od u ce d by t h e o r a n g e s i n W/mˆ 28 s = 0 . 0 8 ; // S i z e o f t he o ra ng e i n m

9 k = 0 . 1 5 ; // T herm al c o n d u c t i v i t y o f t h e s p h e r e i n W/m. K10

11 //CALCULATIONS12 q = ( 3 * Q ) / ( s / 2 ) ; / / H e at f l u x i n W/mˆ 213 T c = 1 0 + ( ( q * ( s / 2 ) ^ 2 ) / ( 6 * k ) ) ; / / T em pe ra tu re a t t h e

c e n t r e o f t he s ph er e i n d eg r e e C14

15 //OUTPUT16 mprintf ( ’ T em pe ra tu re a t t he c e n t r e o f t he o ra ng e i s

%i d e gr e e C ’ , T c )17

18 //=================================END OF PROGRAM==============================

Scilab code Exa 3.26 Total heat dissipated


============================================================

3 clc

4 clear

5

40


42/213

6 //INPUT DATA

7 T o = 1 4 0 ; // T em pe ra tu re a t t he j u n c t i o n i n d e g re e C8 T i = 1 5 ; // T em per at ur e o f a i r i n t he room i n d e gr e e C9 D = 0 . 0 0 3 ; // D ia me t er o f t he r o d i n m

10 h = 3 0 0 ; //H e at t r a n s f e r c o e f f i c i e n t i n W/mˆ2. K11 k = 1 5 0 ; / / T he rm al c o n d u c t i v i t y i n W/m .K12

13 //CALCULATIONS14 P = ( 3 . 1 4 * D ) ; // P er i me t er o f t he r od i n m15 A = ( 3 . 1 4 * D ^ 2 ) / 4 ; // Area o f t he r od i n mˆ216 Q = sqrt ( h * P * k * A ) * ( T o - T i ) ; // T ot al h ea t d i s s i p a t e d by

t h e r od i n W

1718 //OUTPUT19 mprintf ( ’ T ot al h ea t d i s s i p a t e d by t he r od i s %3 . 3 f W

’ ,Q )20

21 //=================================END OF PROGRAM==============================

Scilab code Exa 3.27 Thermal conductivity


============================================================

3 clc

4 clear

5

6 //INPUT DATA

7 D = 0 . 0 2 5 // D iam et e r o f t he r o d i n m8 T i = 2 2 ; // T em per at ur e o f a i r i n t he room i n d e gr e e C9 x = 0 . 1 ; // D i st a n ce b et wee n t he p o i n t s i n m

10 T = [ 1 1 0 , 8 5 ] ; // T em per at ur e s a t two p o i n t s i n d e g re e C11 h = 2 8 . 4 ; //H e at t r a n s f e r c o e f f i c i e n t i n W/mˆ2. K

41


43/213

12

13 //CALCULATIONS14 m = - log ( ( T ( 2 ) - T i ) / ( T ( 1 ) - T i ) ) / x ; // C a l c ul a t i on o f m f o ro b t ai n i ng k

15 P = ( 3 . 1 4 * D ) ; // P er i me t er o f t he r od i n m16 A = ( 3 . 1 4 * D ^ 2 ) / 4 ; // Area o f t he r od i n mˆ217 k = ( ( h * P ) / ( ( m ) ^ 2 * A ) ) ; // Thermal c o n d u c t i v i t y o f t he

r o d m a t e r i a l i n W/m. K18

19 //OUTPUT20 mprintf ( ’ T hermal c o n d u c t i v i ty o f t he r o d m a t e r i a l i s

%3 . 1 f W/m.K ’ ,k )

2122 //=================================END OF PROGRAM

==============================

Scilab code Exa 3.28 Temperature distribution and rate of heat flow


============================================================

3 clc

4 clear

5

6 //INPUT DATA7 L = 0 . 0 6 ; // L engt h o f t he t u r bi n e b la de i n m8 A = ( 4 . 6 5 * 1 0 ^ - 4 ) ; // C ro ss s e c t i o n a l a r ea i n mˆ29 P = 0 . 1 2 ; // P e r im e t er i n m

10 k = 2 3 . 3 ; // Thermal c o n d u c t i v i t y o f s t a i n l e s s s t e e l i n

W/m.K11 T o = 5 0 0 ; // T em pe ra tu re a t t he r o o t i n d e g re e C12 T i = 8 7 0 ; // T em pe ra tu re o f t he h ot g as i n d e gr e e C13 h = 4 4 2 ; //H e at t r a n s f e r c o e f f i c i e n t i n W/mˆ2. K14

42


44/213

15 //CALCULATIONS

16 m = sqrt ( ( h * P ) / ( k * A ) ) ; // C a l c ul a t i on o f m f o rc a l c u l a t i n g h e a t t r a n s f e r r a t e17 X = ( T o - T i ) / cosh ( m * L ) ; //X f o r c a l c u l a t i n g t em pe ta ru re

d i s t r i b u t i o n18 Q = sqrt ( h * P * k * A ) * ( T o - T i ) * tanh ( m * L ) ; // H eat t r a n s f e r

r a t e i n W19

20 //OUTPUT21 mprintf ( ’ T empe ra tur e d i s t r i b u t i o n i s g i ve n by : T−Ti

= %i c osh [ %3. 2 f ( %3. 2 f −x ) ] \n

−−−−−−−−−−−−−−−−−−−−\n

co sh [ %3. 2 f (%3. 2 f ) ] \n Heat t r a n s f e r r a te i s %3 . 1 f W’ ,( T o - T i ) , m , L , m , L , Q )

22

23 //=================================END OF PROGRAM==============================

Scilab code Exa 3.29 Rate of heat transfer and temperature


============================================================

3 clc

4 clear

5

6 //INPUT DATA

7 W = 1 ; // Length o f t h e c y l i n d e r i n m8 D = 0 . 0 5 ; // D iam et e r o f t he c y l i n d e r i n m9 T a = 4 5 ; / / Ambient t e m pe r a tu r e i n d e g r e e C

10 n = 1 0 ; / /Number o f f i n s11 k = 1 2 0 ; // Thermal c o n d u ct i v i t y o f t he f i n m a t e r i a l i n

43


45/213

W/m.K

12 t = 0 . 0 0 0 7 6 ; // T hi ck ne ss o f f i n i n m13 L = 0 . 0 1 2 7 ; // H ei g h t o f f i n i n m14 h = 1 7 ; //H e at t r a n s f e r c o e f f i c i e n t i n W/mˆ2. K15 T s = 1 5 0 ; // S u r f a ce t em pe ra tu re o f c y l i n d e r i n m16

17 //CALCULATIONS18 P = ( 2 * W ) ; // P er i m et er o f c y l i n d e r i n m19 A = ( W * t ) ; // S u r f ac e a r ea o f c y i n de r i n mˆ220 m = sqrt ( ( h * P ) / ( k * A ) ) ; // C a l c ul a t i on o f m f o r

d e t er mi n in g h ea t t r a n s f e r r a t e21 Qfin=( sqrt ( h * P * k * A ) * ( T s - T a ) * ( ( tanh ( m * L ) + ( h / ( m * k ) ) )

/ ( 1 + ( ( h / ( m * k ) ) * tanh ( m * L ) ) ) ) ) ; // H eat t r a n s f e rt hr ou gh t he f i n i n kW

22 Q b = h * ( ( 3 . 1 4 * D ) - ( n * t ) ) * W * ( T s - T a ) ; / / He at f ro m u n f i n n e d( b as e ) s u r f a c e i n W

23 Q = ( ( Q f i n * 1 0 ) + Q b ) ; // T ot al h ea t t r a n s f e r i n W24 T i = ( ( T s - T a ) / ( cosh ( m * L ) + ( ( h * sinh ( m * L ) ) / ( m * k ) ) ) ) ; / / T i

t o c a l c u l a t e t em pe ra tu re a t t h e e nd o f t h e f i n i nd e g r e e C

25 T = ( T i + T a ) ; // T empe r atu re a t t he end o f t he f i n i nd e g r e e C

26

27 //OUTPUT28 mprintf ( ’ R ate o f h ea t t r a n s f e r i s %3 . 2 f W\

n Te mp er atu re a t t he end o f t he f i n i s %3 . 2 f d e g r e e C ’ ,Q , T )

29

30 //=================================END OF PROGRAM==============================



44


46/213

============================================================

3 clc

4 clear

5

6 //INPUT DATA7 t = 0 . 0 2 5 ; // T hi ck ne ss o f f i n i n m8 L = 0 . 1 ; // Length o f f i n i n m9 k = 1 7 . 7 ; // Thermal c o n d u ct i v i t y o f t he f i n m a t e r i a l i n

W/m.K10 p = 7 8 5 0 ; / / D e n s i t y i n k g /mˆ 311 T w = 6 0 0 ; // T em pe ra tu re o f t he w a l l i n d e g re e C

12 T a = 4 0 ; // T emper at ur e o f t he a i r i n d e gr e e C13 h = 2 0 ; //H e at t r a n s f e r c o e f f i c i e n t i n W/mˆ2. K14 I 0 ( 1 . 9 ) = 2 . 1 7 8 2 ; // I o v al ue t ak en from t a b l e 3 . 2 on

p a g e n o . 1 0 815 I 1 ( 1 . 9 ) = 1 . 4 8 8 7 1 ; // I 1 v al ue t ak en from t a b l e 3 . 2 on

p ag e no . 1 0816

17 //CALCULATIONS18 B = sqrt ( ( 2 * L * h ) / ( k * t ) ) ; // C a l c u l a t i on o f B f o r

d e te r mi n i ng t em pe ra t ur e d i s t r i b u t i o n19 X = ( ( T w - T a ) / I 0 ( 2 * B * sqrt ( 0 . 1 ) ) ) ;

// C a l c ul a t i on o f X f o rd e te r mi n i ng t em p er at u re d i s t r i b u t i o n20 Y = ( 2 * B ) ; // C a l c ul a t i on o f Y f o r d e t e r mi ni n g

t em pe ra tu re d i s t r i b u t i o n21 Q =( sqrt ( 2 * h * k * t ) * ( T w - T a ) * ( ( I 1 ( 2 * B * sqrt ( 0 . 1 ) ) ) / ( I 0 ( 2 *

B * sqrt ( 0 . 1 ) ) ) ) ) ;

22 m = ( ( p * t * L ) / 2 ) ; // Mass o f t he f i n p er m et e r o f w i dt hi n k g /m

23 q = ( Q / m ) ; // R ate o f h ea t f l o w p er u n i t mass i n W/ kg24

25 //OUTPUT

26 mprintf ( ’ T e m p er a t u re d i s t r i b u t i o n i s T=% i+%3 . 1 f ( %3 . 4f x ) \nRate o f h ea t f l o w p er u n i t mass o f t he

f i n i s %3 . 2 f W/ kg ’ , T a , X , Y , q )27

28 //=================================END OF PROGRAM

45


47/213

==============================

Scilab code Exa 3.32 Efficiency of the plate


============================================================

3 clc

4 clear5

6 //INPUT DATA7 t = 0 . 0 0 2 ; // T hi ck ne ss o f f i n i n m8 L = 0 . 0 1 5 ; // Length o f f i n i n m9 k 1 = 2 1 0 ; / / Th er ma l c o n d u c t i v i t y o f a lu mi ni um i n W/m. K

10 h 1 = 2 8 5 ; / / H ea t t r a n s f e r c o e f f i c i e n t o f a l um i ni u m i n W/mˆ2.K

11 k 2 = 4 0 ; / / Therma l c o n d u c t i v i t y o f s t e e l i n W/m. K12 h 2 = 5 1 0 ; / / He at t r a n s f e r c o e f f i c i e n t o f s t e e l i n W/m

ˆ 2 . K

1314 //CALCULATIONS15 L c = ( L + ( t / 2 ) ) ; // C or re ct ed l e ng t h o f f i n i n m16 m L c 1 = L c * sqrt ( ( 2 * h 1 ) / ( k 1 * t ) ) ; // C a l c u l a t i o n o f mLc f o r

e f f i c i e n c y17 n1 = tanh ( m L c 1 ) / m L c 1 ; // E f f i c i e n c y o f f i n when

a lu mi ni um i s u se d18 m L c 2 = L c * sqrt ( ( 2 * h 2 ) / ( k 2 * t ) ) ; // C a l c u l a t i o n o f mLc f o r

e f f i c i e n c y19 n2 = tanh ( m L c 2 ) / m L c 2 ; // E f f i c i e n c y o f f i n when s t e e l i s

use d2021 //OUTPUT22 mprintf ( ’ E f f i c i e n c y o f f i n when al umini um i s u se d i s

%3. 4 f \ n E f f i c i e n c y o f f i n when s t e e l i s u s e d i s

46


48/213

%3. 3 f ’ , n 1 , n 2 )

2324 //=================================END OF PROGRAM

==============================

Scilab code Exa 3.33 Heat transfer coefficient


============================================================

3 clc

4 clear

5

6 //INPUT DATA7 k = 2 0 0 ; / / Th er ma l c o n d u c t i v i t y o f a lu mi ni um i n W/m. K8 t = 0 . 0 0 1 ; // T hi ck ne ss o f f i n i n m9 L = 0 . 0 1 5 ; // Width o f f i n i n m

10 D = 0 . 0 2 5 ; // D ia me te r o f t he t ub e i n m11 T b = 1 7 0 ; // F in b as e t em p er a tu re i n d e g re e C

12 T a = 2 5 ; // Ambient f l u i d t em p er a tu r e i n d e gr e e C13 h = 1 3 0 ; //H e at t r a n s f e r c o e f f i c i e n t i n W/mˆ2. K14

15 //CALCULATIONS16 L c = ( L + ( t / 2 ) ) ; // C or re ct ed l e ng t h o f f i n i n m17 r 1 = ( D / 2 ) ; // R adi u s o f t u be i n m18 r 2 c = ( r 1 + L c ) ; // C o rr e ct e d r a d i u s i n m19 A m = t * ( r 2 c - r 1 ) ; / / C o r r e c t e d a r e a i n mˆ 220 x = L c ^ ( 3 / 2 ) * sqrt ( h / ( k * A m ) ) ; // x f o r c a l c u l a t i n g

e f f i c i e n c y

21 n = 0 . 8 2 ; // From f i g . 3 . 18 on pag e no . 112 e f f i c i e n c yi s 0 .8 222 q m a x = ( 2 * 3 . 1 4 * ( r 2 c ^ 2 - r 1 ^ 2 ) * h * ( T b - T a ) ) ; //Maximum he at

t r a n s f e r i n W23 q a c t u a l = ( n * q m a x ) ; // A ct u al h ea t t r a n s f e r i n W

47


49/213

24

25 //OUTPUT26 mprintf ( ’ H eat l o s s p er f i n i s %3 . 2 f W’ , q a c t u a l )27

28 //=================================END OF PROGRAM==============================

Scilab code Exa 3.34 Insulation of fin


============================================================

3 clc

4 clear

5

6 //INPUT DATA7 k = 1 6 ; / / Th erma l c o n d u c t i v i t y o f f i n i n W/m. K8 L = 0 . 1 ; // Length o f f i n i n m9 D = 0 . 0 1 ; // D iam et e r o f f i n i n m

10 h = 5 0 0 0 ; //H e at t r a n s f e r c o e f f i c i e n t i n W/mˆ2. K11

12 //CALCULATIONS13 P = ( 3 . 1 4 * D ) ; // P er i m et er o f f i n i n m14 A = ( 3 . 1 4 * D ^ 2 ) / 4 ; // Area o f f i n i n mˆ215 m = sqrt ( ( h * P ) / ( k * A ) ) ; // C a l c ul a t i on o f m f o r

d e t er mi n in g h ea t t r a n s f e r r a t e16 n = tanh ( m * L ) / sqrt ( ( h * A ) / ( k * P ) ) ; // C a l c ul a t i on o f n f o r

c he ck in g w he t he r i n s t a l l a t i o n o f f i n i sd e s i r a b l e o r n o t

17 x = ( n - 1 ) * 1 0 0 ; // C o n ve r si o n i n t o p e r c e n t a g e1819 //OUTPUT20 mprintf ( ’ T h i s l a r g e f i n o nl y p ro du ce s an i n c r e a s e o f

%i p e r ce nt i n h ea t d i s s i p a t i o n , s o n a t u r a l l y

48


50/213

t h i s c o n f i g u r a t i o n i s u n de s i r a b l e ’ ,x )

2122 //=================================END OF PROGRAM

==============================

Scilab code Exa 3.35 Measurement error in temperature


============================================================

3 clc

4 clear

5

6 //INPUT DATA7 k = 5 5 . 8 ; // T herm al c o n d u c t i v i t y o f s t e e l i n W/m. K8 t = 0 . 0 0 1 5 ; // T hi ck ne ss o f s t e e l t ub e i n m9 L = 0 . 1 2 ; // L engt h o f s t e e l t u b e i n m

10 h = 2 3 . 3 ; //H e at t r a n s f e r c o e f f i c i e n t i n W/mˆ2. K11 T l = 8 4 ; / / T em pe ra tu re r e c o r d e d by t h e t h er mo m et er i n

d e g r e e C12 T b = 4 0 ; // T emper at ur e a t t he b as e o f t he w e l l i n

d e g r e e C13

14 //CALCULATIONS15 m = sqrt ( h / ( k * t ) ) ; // C a l c u l a t i o n o f m f o r d e te r mi n i ng

t he t em pe ra tu re d i s t r i b u t i o n16 x = 1 / cosh ( m * L ) ; // C a l c u l a t i o n o f x f o r d e t er mi n in g t he

t em pe ra tu re d i s t r i b u t i o n17 T i = ( ( T l - ( x * T b ) ) / ( 1 - x ) ) ; // T em per at ure d i s t r i b u t i o n i n

d e g r e e C18 T = ( T i - T l ) ; / / Mea su rem ent e r r o r i n d e g r e e C19

20 //OUTPUT21 mprintf ( ’ M ea su rem ent e r r o r i s %3 . 0 f d e g r e e C ’ ,T )

49


51/213

22

23 //=================================END OF PROGRAM==============================

50


52/213

Chapter 5

Transient Heat Conduction

Scilab code Exa 5.1 Heat transfer and temperature


============================================================

3 clc

4 clear

56 //INPUT DATA7 t = 0 . 5 ; // T hi ck ne ss o f s l a b i n m8 A = 5 ; // A rea o f s l a b i n mˆ29 k = 1 . 2 ; / / T he rm al c o n d u c t i v i t y i n W/m .K

10 a = 0 . 0 0 1 7 7 ; // Thermal d i f f u s i v i t y in mˆ2/ h11 / / R em pe ra ru re d i s t r i b u t i o n a s T=60−50x+12xˆ2+20x

ˆ3−15xˆ412

13 //CALCULATIONS

14 // P a r t i a l d e r i v a t i v e o f T w. r . t x i s T’=−50+24x+60xˆ2−60xˆ315 // P a r t i a l d e r i v a t i v e o f T ’ w . r . t x i s T’ ’ =24 +1 20 x

+180xˆ216 // P a r t i a l d e r i v a t i v e o f T ’ w . r . t x i s T’ ’ ’ =120 −3 6 0 x

51


53/213

17 x = 0 ;

18 y = - 5 0 + (2 4 * x ) + ( 6 0 * x ^ 2 ) - ( 60 * x ^ 3 ) ; / / T e m p e r a t ur e when x=019 Q o = ( - k * A * y ) ; // Heat e n t e r i n g t he s l a b i n W20 x = 0 . 5 ;

21 y = - 5 0 + (2 4 * x ) + ( 6 0 * x ^ 2 ) - ( 60 * x ^ 3 ) ; / / T e m p e r a t ur e when x=0.5

22 Q L = ( - k * A * y ) ; // Heat l e a v i n g t he s l ab i n W23 R = ( Q o - Q L ) ; // Rate o f h ea t s t o r a ge i n W24 x = 0 ;

25 z 1 = 2 4 + ( 1 2 0 * x ) - ( 1 8 0 * x ^ 2 ) ; //T ’ w he n x=026 p 1 = ( a * z 1 ) ; // Rate o f t em p er at u re c ha ng e a t on e s i d e

o f s l ab i n d eg r e e C/h27 x = 0 . 5 ;

28 z 2 = 2 4 + ( 1 2 0 * x ) - ( 1 8 0 * x ^ 2 ) ; / /T ’ when x = 0 . 529 p 2 = ( a * z 2 ) ; // Rate o f t em p er at u re c ha ng e a t on e s i d e

o f s l ab i n d eg r e e C/h30 / / For t he r a t e o f h e a ti n g o r c o o l i n g t o be maximum ,

T’ ’ ’= 031 x = ( 1 2 0 / 3 6 0 ) ;

32

33 //OUTPUT34 mprintf (

’ a ) i ) H eat e n t e r i n g t he s l a b i s %i W\n i i )Heat l e a v i n g t he s l ab i s %i W\nb ) R at e o f h e a ts t o r a g e i s %i W\n c ) i ) R at e o f t e m p e r a tu r e c h an g ea t one s i d e o f s l ab i s %3 . 4 f d eg r e e C/h\n i i ) R at e

o f t em pe ra tu re c h a ng e a t o t h e r s i d e o f s l ab i s%3 . 4 f d e g r e e C/ h\nd ) For t h e r a t e o f h e a t i n g o rc o o l i n g t o b e maximum x= %3 . 2 f ’ , Q o , Q L , R , p 1 , p 2 , x )

35

36 //=================================END OF PROGRAM==============================

Fundamentals of Engineering Heat and Mass Transfer_R. C. Sachdeva

Documents

Transcript of Fundamentals of Engineering Heat and Mass Transfer_R. C. Sachdeva