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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
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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
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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.
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
1 //C hapt e r −1 , Exa mpl e 1 . 5 , Pa ge 2 02 //
============================================================
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
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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
1 //C hapt e r −1 , Exa mpl e 1 . 6 , Pa ge 2 12 //
============================================================
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
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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
1 //C hapt e r −1 , Exa mpl e 1 . 7 , Pa ge 2 12 //
============================================================
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
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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==============================
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Chapter 3
OneDimensional Steady State
Heat Conduction
Scilab code Exa 3.1 Rate of heat loss and interior temperature
1 //C hapt e r −3 , Exa mpl e 3 . 1 , Pa ge 4 52 //
============================================================
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
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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
1 //C hapt e r −3 , Exa mpl e 3 . 2 , Pa ge 4 82 //
============================================================
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
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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
1 //C hapt e r −3 , Exa mpl e 3 . 3 , Pa ge 5 12 //
============================================================
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
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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
1 //C hapt e r −3 , Exa mpl e 3 . 4 , Pa ge 5 52 //
============================================================
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
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==============================
Scilab code Exa 3.5 Rate of heat flow
1 //C hapt e r −3 , Exa mpl e 3 . 5 , Pa ge 5 52 //
============================================================
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
1 //C hapt e r −3 , Exa mpl e 3 . 6 , Pa ge 5 8
20
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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
1 //C hapt e r −3 , Exa mpl e 3 . 8 , Pa ge 6 3
2 // ============================================================
3 clc
4 clear
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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
1 //C hapt e r −3 , Exa mpl e 3 . 9 , Pa ge 6 42 //
============================================================
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
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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
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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
==============================
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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==============================
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Scilab code Exa 3.12 Heat loss and thickness of insulation
1 //C hapt e r −3 , Exa mpl e 3 . 1 2 , P ag e 6 72 //
============================================================
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 )
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25
26 //=================================END OF PROGRAM==============================
Scilab code Exa 3.13 Heat loss from the pipe
1 //C hapt e r −3 , Exa mpl e 3 . 1 3 , P ag e 7 02 //
============================================================
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 )
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24
25 //=================================END OF PROGRAM==============================
Scilab code Exa 3.14 Heat loss per meter length of pipe
1 //C hapt e r −3 , Exa mpl e 3 . 1 4 , P ag e 7 12 //
============================================================
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
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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
1 //C hapt e r −3 , Exa mpl e 3 . 1 5 , P ag e 7 22 //
============================================================
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
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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
1 //C hapt e r −3 , Exa mpl e 3 . 1 6 , P ag e 7 32 //
============================================================
3 clc4 clear
5
6 //INPUT DATA7 D 1 = 0 . 1 ; //O. D o f t he p i p e i n m
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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==============================
Scilab code Exa 3.17 Rate of heat flow
1 //C hapt e r −3 , Exa mpl e 3 . 1 7 , P ag e 7 42 //
31
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============================================================
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
1 //C hapt e r −3 , Exa mpl e 3 . 1 8 , P ag e 7 72 //
32
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============================================================
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
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1 //C hapt e r −3 , Exa mpl e 3 . 1 9 , P ag e 7 8
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
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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
1 //C hapt e r −3 , Exa mpl e 3 . 2 0 , P ag e 8 32 //
============================================================
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
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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
1 //C hapt e r −3 , Exa mpl e 3 . 2 1 , P ag e 8 52 //
============================================================
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
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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
1 //C hapt e r −3 , Exa mpl e 3 . 2 2 , P ag e 8 82 //
============================================================
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
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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
1 //C hapt e r −3 , Exa mpl e 3 . 2 3 , P ag e 8 92 //
============================================================
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
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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
1 //C hapt e r −3 , Exa mpl e 3 . 2 4 , P ag e 9 3
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==============================
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Scilab code Exa 3.25 Temperature at the centre of orange
1 //C hapt e r −3 , Exa mpl e 3 . 2 5 , P ag e 9 52 //
============================================================
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
1 //C hapt e r −3 , Exa mpl e 3 . 2 6 , P ag e 1 022 //
============================================================
3 clc
4 clear
5
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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
1 //C hapt e r −3 , Exa mpl e 3 . 2 7 , P ag e 1 032 //
============================================================
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
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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
1 //C hapt e r −3 , Exa mpl e 3 . 2 8 , P ag e 1 032 //
============================================================
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
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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
1 //C hapt e r −3 , Exa mpl e 3 . 2 9 , P ag e 1 042 //
============================================================
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
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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==============================
Scilab code Exa 3.31 Rate of heat flow
1 //C hapt e r −3 , Exa mpl e 3 . 3 1 , P ag e 1 092 //
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============================================================
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
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==============================
Scilab code Exa 3.32 Efficiency of the plate
1 //C hapt e r −3 , Exa mpl e 3 . 3 2 , P ag e 1 162 //
============================================================
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
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%3. 3 f ’ , n 1 , n 2 )
2324 //=================================END OF PROGRAM
==============================
Scilab code Exa 3.33 Heat transfer coefficient
1 //C hapt e r −3 , Exa mpl e 3 . 3 3 , P ag e 1 172 //
============================================================
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
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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
1 //C hapt e r −3 , Exa mpl e 3 . 3 4 , P ag e 1 172 //
============================================================
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
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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
1 //C hapt e r −3 , Exa mpl e 3 . 3 5 , P ag e 1 192 //
============================================================
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 )
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22
23 //=================================END OF PROGRAM==============================
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Chapter 5
Transient Heat Conduction
Scilab code Exa 5.1 Heat transfer and temperature
1 //C hapt e r −5 , Exa mpl e 5 . 1 , Pa ge 1 592 //
============================================================
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
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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==============================