Scilab Textbook Companion for

Textbook Of Heat Transfer

by S. P. Sukhatme1

Created by
Tushar Goel
B.Tech
Chemical Engineering
Indian Institute of Technology (BHU) , Varanasi
College Teacher
Dr. Prakash Kotecha
Cross-Checked by

May 20, 2016

1 Funded by a grant from the National Mission on Education through ICT,

http://spoken-tutorial.org/NMEICT-Intro. This Textbook Companion and Scilab
codes written in it can be downloaded from the ”Textbook Companion Project”
section at the website http://scilab.in
Book Description

Title: Textbook Of Heat Transfer

Author: S. P. Sukhatme

Publisher: Universities Press

Edition: 4

Year: 2005

ISBN: 9788173715440

1
 Scilab numbering policy used in this document and the relation to the
above 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 particular

Example of the above book)

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

2
Contents

List of Scilab Codes 4

1 Introduction 5

2 Heat Conduction in Solids 9

3 Thermal Radiation 33

4 Principles of Fluid Flow 50

5 Heat Transfer by Forced Convection 58

6 Heat Transfer by Natural convection 76

7 Heat Exchangers 86

8 Condensation and boiling 96

9 Mass Transfer 107

3
List of Scilab Codes

Exa 1.1 Viscosity in SI system . . . . . . . . . . . . . . . . . . 5

Exa 1.2 Useful heat gain and thermal efficiency . . . . . . . . . 6
Exa 1.3 Exit velocity and Temperature . . . . . . . . . . . . . 7
Exa 2.1 Heat flow rate . . . . . . . . . . . . . . . . . . . . . . 9
Exa 2.2 Heat flow rate . . . . . . . . . . . . . . . . . . . . . . 10
Exa 2.3 Engineers decision . . . . . . . . . . . . . . . . . . . . 11
Exa 2.4 Thickness of insulation . . . . . . . . . . . . . . . . . . 12
Exa 2.5 Heat loss rate . . . . . . . . . . . . . . . . . . . . . . . 14
Exa 2.6 Critical radius . . . . . . . . . . . . . . . . . . . . . . 15
Exa 2.7 Maximum temperature . . . . . . . . . . . . . . . . . 16
Exa 2.8 Steady state temperature . . . . . . . . . . . . . . . . 17
Exa 2.9 Time taken by the rod to heat up . . . . . . . . . . . 18
Exa 2.10.i Heat transfer coefficient at the centre . . . . . . . . . 19
Exa 2.10.ii heat transfer coefficient at the surface . . . . . . . . . 21
Exa 2.11.a Time taken by the centre of ball . . . . . . . . . . . . 22
Exa 2.11.b time taken by the centre of ball to reach temperature . 23
Exa 2.12 Temperature at the centre of the brick . . . . . . . . . 24
Exa 2.13.a Temperature at the copper fin tip . . . . . . . . . . . 26
Exa 2.13.b Temperature at the steel fin tip . . . . . . . . . . . . . 27
Exa 2.13.c Temperature at the teflon fin tip . . . . . . . . . . . . 28
Exa 2.14 Heat loss rate . . . . . . . . . . . . . . . . . . . . . . . 29
Exa 2.15 Decrease in thermal resistance . . . . . . . . . . . . . 30
Exa 2.16 Overall heat transfer coefficient . . . . . . . . . . . . . 31
Exa 3.1 Monochromatic emissive power . . . . . . . . . . . . . 33
Exa 3.2 Heat flux . . . . . . . . . . . . . . . . . . . . . . . . . 34
Exa 3.3 Absorbed radiant flux and absorptivity and reflectivity 35
Exa 3.4.a Total intensity in normal direction . . . . . . . . . . . 36
Exa 3.4.b Ratio of radiant flux to the emissive power . . . . . . 37

4
Exa 3.5 Rate of incident radiation . . . . . . . . . . . . . . . . 38
Exa 3.6 Shape factor F12 . . . . . . . . . . . . . . . . . . . . . 38
Exa 3.7 Shape factor . . . . . . . . . . . . . . . . . . . . . . . 40
Exa 3.8 Shape factor F12 . . . . . . . . . . . . . . . . . . . . . 41
Exa 3.9 Shape factor . . . . . . . . . . . . . . . . . . . . . . . 42
Exa 3.10 Net radiative heat transfer . . . . . . . . . . . . . . . 42
Exa 3.11 steady state heat flux . . . . . . . . . . . . . . . . . . 43
Exa 3.12 Rate of heat loss . . . . . . . . . . . . . . . . . . . . . 45
Exa 3.13 Rate of nitrogen evaporation . . . . . . . . . . . . . . 45
Exa 3.14 Rate of energy loss from satellite . . . . . . . . . . . . 46
Exa 3.15 Net radiative heat transfer . . . . . . . . . . . . . . . 47
Exa 4.1 Pressure drop in smooth pipe . . . . . . . . . . . . . . 50
Exa 4.2.a Pressure drop and maximum velocity calculation . . . 51
Exa 4.2.b Pressure drop and maximum velocity calculation . . . 52
Exa 4.3 Pressure drop and power needed . . . . . . . . . . . . 54
Exa 4.4 Thickness of velocity boundary layer . . . . . . . . . . 55
Exa 4.5 Drag coefficient and drag force . . . . . . . . . . . . . 56
Exa 5.1.a Local heat transfer coefficient . . . . . . . . . . . . . . 58
Exa 5.1.b Wall temperature . . . . . . . . . . . . . . . . . . . . 59
Exa 5.2 ratio of thermal entrance length to entrance length . . 60
Exa 5.3.i Length of tube . . . . . . . . . . . . . . . . . . . . . . 61
Exa 5.3.ii Exit water temperature . . . . . . . . . . . . . . . . . 62
Exa 5.4 Length of tube over which temperature rise occurs . . 64
Exa 5.5 Rate of heat transfer to the plate . . . . . . . . . . . . 66
Exa 5.6.i Heat transfer rate . . . . . . . . . . . . . . . . . . . . 67
Exa 5.6.ii Average wall tempeature . . . . . . . . . . . . . . . . 68
Exa 5.7.i Pressure drop . . . . . . . . . . . . . . . . . . . . . . . 70
Exa 5.7.ii Exit temperature of air . . . . . . . . . . . . . . . . . 71
Exa 5.7.iii Heat transfer rate . . . . . . . . . . . . . . . . . . . . 73
Exa 6.1 Average nusselt number . . . . . . . . . . . . . . . . . 76
Exa 6.2 Reduce the equation . . . . . . . . . . . . . . . . . . . 78
Exa 6.3 Time for cooling of plate . . . . . . . . . . . . . . . . 79
Exa 6.4 True air temperature . . . . . . . . . . . . . . . . . . . 81
Exa 6.5 Rate of heat flow by natural convection . . . . . . . . 83
Exa 6.6 Average Heat transfer coeffficient . . . . . . . . . . . . 84
Exa 7.1 Heat transfer coeffficient . . . . . . . . . . . . . . . . . 86
Exa 7.2 Area of heat exchanger . . . . . . . . . . . . . . . . . 87
Exa 7.3 Mean temperature difference . . . . . . . . . . . . . . 88

5
Exa 7.4.a Area of heat exchanger . . . . . . . . . . . . . . . . . 89
Exa 7.4.b Exit temperature of hot and cold streams . . . . . . . 90
Exa 7.5 Exit Temperature . . . . . . . . . . . . . . . . . . . . 92
Exa 8.1 Average Heat Transfer Coefficient . . . . . . . . . . . 96
Exa 8.2 Average heat transfer coefficient and film Reynolds num-
ber . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97
Exa 8.3 Length of the tube . . . . . . . . . . . . . . . . . . . . 99
Exa 8.4 boiling regions . . . . . . . . . . . . . . . . . . . . . . 101
Exa 8.5 Initial heat transfer rate . . . . . . . . . . . . . . . . . 105
Exa 9.1 Composition on molar basis . . . . . . . . . . . . . . . 107
Exa 9.2 Diffusion coefficient of napthalene . . . . . . . . . . . 108
Exa 9.3.a Rate of hydrogen diffusion . . . . . . . . . . . . . . . . 108
Exa 9.3.b Rate of hydrogen diffusion . . . . . . . . . . . . . . . . 109
Exa 9.4.a Rate of loss of ammonia . . . . . . . . . . . . . . . . . 110
Exa 9.4.b Rate at which air enters the tank . . . . . . . . . . . . 111
Exa 9.5 Rate of evaporation . . . . . . . . . . . . . . . . . . . 112
Exa 9.6 Rate of evaporation . . . . . . . . . . . . . . . . . . . 113
Exa 9.7.a Mass transfer coefficient Colburn anology . . . . . . . 114
Exa 9.7.b Mass transfer coefficient Gnielinski equation . . . . . . 115
Exa 9.7.c To show mass flux of water vapour is small . . . . . . 117
Exa 9.8 Mass fraction . . . . . . . . . . . . . . . . . . . . . . . 118

6
 Chapter 1

Introduction

Scilab code Exa 1.1 Viscosity in SI system

1 clear ;
2 clc ;
3 // A Textbook on HEAT TRANSFER by S P SUKHATME
4 // C h a p t e r 1
5 // I n t r o d u c t i o n
6
7
8 // Example 1 . 1
9 // Page 5
10 // Given t h a t t h e v i s c o s i t y o f w a t e r a t 100 d e g r e e
C e l s i u s i s 2 8 . 8 ∗ 10ˆ −6 k g f s /mˆ2 i n MKS s y s t e m ,
e x p r e s s t h i s value in SI system .
11 printf ( ” Example 1 . 1 , Page 5 \n \n ” )
12
13 // S o l u t i o n :
14
15 // a t 100 d e g r e e C e l s i u s
16 v1 =28.8 * 10^ -6; // [ k g f s /mˆ 2 ]
17 v2 =28.8 * 10^ -6 * 9.8; // [ N s /mˆ 2 ]
18 printf ( ” V i s c o s i t y o f w a t e r a t 100 d e g r e e
c e l s i u s in
t h e S I s y s t e m i s %e N . s /mˆ−2 ( o r kg /m s ) ” , v2 )

7
 Scilab code Exa 1.2 Useful heat gain and thermal efficiency

1 clear ;
2 clc ;
3 // Textbook o f Heat T r a n s f e r ( 4 t h E d i t i o n ) ) , S P
Sukhatme
4 // C h a p t e r 1 − I n t r o d u c t i o n
5
6 // Example 1 . 2
7 // Page 14
8 printf ( ” Example 1 . 2 , Page 14 \n \n ” )
9 // S o l u t i o n :
10 i =950; // r a d i a t i o n f l u x [W/mˆ 2 ]
11 A =1.5; // a r e a [mˆ 2 ]
12 T_i =61; // i n l e t t e m p e r a t u r e
13 T_o =69; // o u t l e t t e m p e r a t u r e
14 mdot =1.5; // [ kg / min ] , mass f l o w r a t e
15 Mdot =1.5/60; // [ kg / s e c ]
16 Q_conductn =50; // [W]
17 t =0.95; // t r a n s m i s s i v i t y
18 a =0.97; // a b s o p t i v i t y
19 // from a p p e n d i x t a b l e A. 1 a t 65 d e g r e e C
20 C_p = 4183 ; // [ J / kg K ]
21 // U s i n g E q u a t i o n 1 . 4 . 1 5 , a s s u m i n g t h a t t h e f l o w
t h r o u g h t h e t u b e s i s s t e a d y and one d i m e n s i o n a l .
22 // i n t h i s c a s e (dW/ d t ) s h a f t = 0
23 // a s s u m i n g (dW/ d t ) s h e a r i s n e g l i g i b l e
24 // eqn ( 1 . 4 . 1 5 ) r e d u c e s t o
25 q = Mdot * C_p *( T_o - T_i ) ;
26
27 // l e t ’ n ’ be t h e r m a l efficiency
28 n = q /( i * A ) ;
29 n_percent = n *100;
30

8
31
32 // e q u a t i o n 1 . 4 . 1 3 y i e l d s dQ/ d t = 0
33 Q_re_radiated =( i * A * t * a ) - Q_conductn - q ; // [W]
34
35
36 printf ( ” U s e f u l h e a t g a i n r a t e i s %f W \n ” ,q ) ;
37 printf ( ” Thermal e f f i c i e n c y i s %e i . e . %f p e r c e n t \n
” ,n , n_percent ) ;
38 printf ( ” The r a t e a t which e n e r g y i s l o s t by r e −
r a d i a t i o n and c o n v e c t i o n i s %f W” , Q_re_radiated )

Scilab code Exa 1.3 Exit velocity and Temperature

1 clear ;
2 clc ;
3 // A Textbook on HEAT TRANSFER by S P SUKHATME
4 // C h a p t e r 1
5 // I n t r o d u c t i o n
6
7
8 // Example 1 . 3
9 // Page 16
10 printf ( ” Example 1 . 3 , Page 16\ n\n ” ) ;
11
12 // S o l u t i o n :
13 // Given
14 v_i =10; // [m/ s ]
15 q =1000; // [W]
16 d_i =0.04; // [m]
17 d_o =0.06; // [m]
18
19 // From a p p e n d i x t a b l e A. 2
20 rho1 =0.946; // [ kg /mˆ 3 ] a t 100 d e g r e e C
21 C_p =1009; // [ J / kg K ]
22

9
23 mdot = rho1 *( %pi /4) *( d_i ^2) * v_i ; // [ kg / s ]
24
25
26 // I n t h i s c a s e (dW/ d t ) s h a f t =0 and ( z o − z i ) =0
27 // From eqn 1 . 4 . 1 5 , q=mdot ∗ ( h o −h i )
28 // L e t dh = ( h o −h i )
29 dh = q / mdot ; // [ J / kg ]
30 // L e t T o be t h e o u t l e t t e m p e r a t u r e
31 T_o = dh / C_p +100;
32
33 rho2 =0.773; // [ kg /mˆ 3 ] a t T o = 1 8 3 . 4 d e g r e e C
34 // From eqn 1 . 4 . 6
35 v_o = mdot /( rho2 *( %pi /4) *( d_o ) ^2) ; // [m/ s ]
36
37 dKE_kg =( v_o ^2 - v_i ^2) /2; // [ J / kg ]
38
39
40 printf ( ” E x i t T e m p e r a t u r e i s %f d e g r e e C \n ” , T_o ) ;
41 printf ( ” E x i t v e l o c i t y i s %f m/ s \n ” , v_o ) ;
42 printf ( ” Change i n K i n e t i c Energy p e r kg = %f J / kg ” ,
dKE_kg ) ;

10
 Chapter 2

Heat Conduction in Solids

Scilab code Exa 2.1 Heat flow rate

1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 2
6 // Heat C o n d u c t i o n i n S o l i d s
7
8 // Example 2 . 1
9 // Page 27
10 printf ( ” Example 2 . 1 , Page 27 \n\n ” )
11
12 d_i =0.02; // [m] i n n e r r a d i u s
13 d_o =0.04; // [m] o u t e r r a d i u s
14 r_i = d_i /2; // [m] i n n e r r a d i u s
15 r_o = d_o /2; // [m] o u t e r r a d i u s
16 k =0.58; // [ w/m K ] t h e r m a l c o n d u c t i v i t y o f t u b e
material
17 t_i =70; // [ d e g r e e C ]
18 t_o =100; // [ d e g r e e C ]
19 l =1; // [m] p e r u n i t l e n g t h
20 // u s i n g e q u a t i o n 2 . 1 . 5

11
21 q = l *2*( %pi ) * k *( t_i - t_o ) / log ( r_o / r_i ) ;
22 printf ( ” Heat f l o w p e r u n i t l e n g t h i s %f W/m” ,q ) ;

Scilab code Exa 2.2 Heat flow rate

1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 2
6 // Heat C o n d u c t i o n i n S o l i d s
7
8 // Example 2 . 2
9 // Page 31
10 printf ( ” Example 2 . 2 , Page 31 \n\n ” )
11
12 d_i =0.02; // [m] i n n e r r a d i u s
13 d_o =0.04; // [m] o u t e r r a d i u s
14 r_i = d_i /2; // [m] i n n e r r a d i u s
15 r_o = d_o /2; // [m] o u t e r r a d i u s
16 k =0.58; // [ w/m K ] t h e r m a l c o n d u c t i v i t y o f t u b e
material
17 t_i =70; // [ d e g r e e C ]
18 t_o =100; // [ d e g r e e C ]
19 l =1; // [m] p e r u n i t l e n g t h
20
21 // t h e r m a l r e s i s t a n c e o f t u b e p e r u n i t l e n g t h
22 R_th_tube =( log ( r_o / r_i ) ) /(2* %pi * k * l ) ; // [ K/W]
23
24 // from t a b l e 1 . 3 , h e a t t r a n s f e r co− e f f i c i e n t f o r
c o n d e n s i n g steam may be t a k e n a s
25 h =5000; // [W/mˆ2 K ]
26 // t h e r m a l r e s i s t a n c e o f c o n d e n s i n g steam p e r u n i t
length
27 R_th_cond =1/( %pi * d_o * l * h ) ;

12
28
29 // s i n c e R t h t u b e i s much l e s s t h a n R t h c o n d , we
can assume o u t e r s u r f a c e t o be a t 100 d e g r e e C
30 // h e n c e h e a t f l o w r a t e p e r u n i t m e t e r i s
31 q = l *2*( %pi ) * k *( t_i -100) / log ( r_o / r_i ) ;
32
33 printf ( ” Thermal r e s i s t a n c e o f t u b e p e r u n i t l e n g t h
i s %f K/W\n ” , R_th_tube ) ;
34 printf ( ” Thermal r e s i s t a n c e o f c o n d e n s i n g steam p e r
u n i t l e n g t h i s %f K/W\n ” , R_th_cond ) ;
35 printf ( ” Heat f l o w p e r u n i t l e n g t h i s %f W/m” ,q ) ;

Scilab code Exa 2.3 Engineers decision

1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 2
6 // Heat C o n d u c t i o n i n S o l i d s
7
8 // Example 2 . 3
9 // Page 31
10 printf ( ” Example 2 . 3 , Page 31 \n\n ” )
11
12 h_w =140; // h e a t t r a n s f e r c o e f f i c i e n t on w a t e r s i d e ,
[W/mˆ2 K ]
13 h_o =150; // h e a t t r a n s f e r c o e f f i c i e n t on o i l s i d e , [
W/mˆ2 K ]
14 k =30; // t h e r m a l c o n d u c t i v i t y [W/m K ]
15 r_o =0.01; // i n n e r d i a m e t e r o f GI p i p e on i n s i d e
16 r_i =0.008; // o u t e r d i a m e t e r GI p i p e on i n s i d e
17 l =1; // [m] , p e r u n i t l e n g t h
18
19 // Thermal r e s i s t a n c e o f i n n e r GI p i p e

13
20 R_inner_GI = log (( r_o / r_i ) ) /(2* %pi * k * l ) ;
21
22
23 // Thermal r e s i s t a n c e on t h e o i l s i d e p e r u n i t
length
24 R_oilside =1/( h_o * %pi *2* r_i * l ) ;
25
26
27 // Thermal r e s i s t a n c e on c o l d w a t e r s i d e p e r u n i t
length
28 R_waterside =1/( h_w * %pi *2* r_o * l ) ;
29
30
31 // we s e e t h e r m a l r e s i s t a n c e o f i n n e r GI p i p e
c o n t r i b u t e s l e s s than 0 . 5 p e r c e n t to the t o t a l
resistance
32
33
34 printf ( ” Thermal r e s i s t a n c e o f i n n e r GI p i p e = %f K/W
\n ” , R_inner_GI ) ;
35 printf ( ” Thermal r e s i s t a n c e on t h e o i l s i d e p e r u n i t
l e n g t h = %f K/W \n ” , R_oilside ) ;
36 printf ( ” Thermal r e s i s t a n c e on c o l d w a t e r s i d e p e r
u n i t l e n g t h = %f K/W \n ” , R_waterside ) ;
37 printf ( ” So , E n g i n e e r i n −c h a r g e h a s made a bad
d e c i s i o n ”);

Scilab code Exa 2.4 Thickness of insulation

1 clear all ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 2
6 // Heat C o n d u c t i o n i n S o l i d s

14
 7
8 // Example 2 . 4
9 // Page 32
10 printf ( ” Example 2 . 4 , Page 32 \n\n ” )
11
12 Ti = 300; // I n t e r n a l temp o f h o t g a s i n
degree Celsius
13 OD = 0.1; // Outer d i a m e t e r o f l o n g m e t a l
pipe in meters
14 ID = 0.04; // I n t e r n a l d i a m t e r e o f l o n g m e t a l
pipe in meters
15 ki = 0.052; // t h e r m a l c o n d u c t i v i t y o f m i n e r a l
wood i n W/mK
16 To = 50; // Outer s u r f a c e t e m p e r a t u r e i n
degree c e l s i u s
17 hi = 29; // 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
t h e i n n e r s i d e i n W/mˆ2 K
18 ho = 12; // 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
t h e o u t e r p i p e W/mˆ2 K
19
20 // D e t e r m i n a t i o n o f t h i c k n e s s o f i n s u l a t i o n
21 function [ f ] = thickness ( r )
22 f = r *(10.344 + 271.15* log ( r *(0.05) ^ -1) ) -11.75
23 funcprot (0) ;
24 endfunction
25 r = 0.082;
26 while 1
27 rnew = r - thickness ( r ) / diff ( thickness ( r ) ) ;
28 if rnew == r then
29 r3 = rnew ;
30 break ;
31 end
32 r = rnew ;
33 end
34 t = r3 - OD /2;
35 printf ( ” \n T h i c k n e s s o f i n s u l a t i o n = %f cm” ,t *100) ;
36 // Heat l o s s p e r u n i t l e n g t h
37 q = 600*(22/7) * r3 ;

15
38 printf ( ” \n Heat l o s s p e r u n i t l e n g t h = %. 1 f W/m” ,q ) ;

Scilab code Exa 2.5 Heat loss rate

1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 2
6 // Heat C o n d u c t i o n i n S o l i d s
7
8 // Example 2 . 5
9 // Page 34
10 printf ( ” Example 2 . 5 , Page 34 \n\n ” )
11
12 Ti = 90; //Temp on i n n e r
side in degree c e l s i u s
13 To = 30; //Temp on o u t e r
side in degree c e l s i u s
14 hi = 500; // h e a t t r a n s f e r
c o e f f c i e n t i n W/mˆ2 K
15 ho = 10; // h e a t t r a n s f e r
c o e f f c i e n t i n W/mˆ2 K
16 ID = 0.016; // I n t e r n a l
diameter in meters
17 t = [0 0.5 1 2 3 4 5]; // I n s u l a t i o n
t h i c k n e s s i n cm
18 OD = 0.02; // Outer d i a m e t e r
in meters
19 r3 = OD /2 + t /100; // r a d i u s a f t e r
i n s u l a t i o n in meters
20
21 i =1;
22 printf ( ” \n I n s u l a t i o n t h i c k n e s s ( cm ) r 3 (m)
h e a t l o s s r a t e p e r m e t e r (W/m) ” ) ;

16
23 while i <=7
24 ql ( i ) = [2*( %pi ) *( ID /2) *( Ti - To ) ]/[(1/ hi )
+(0.008/0.2) * log ( r3 ( i ) /0.01) + (0.008/ r3 ( i ) )
*(1/ ho ) ];
25 printf ( ” \n %. 1 f %. 3 f
%. 1 f ” ,t ( i ) , r3 ( i ) , ql ( i ) ) ;
26 i = i +1;
27 end
28 plot (t , ql ) ;
29 xtitle ( ” ” ,” I n s u l a t i o n t h i c k n e s s ( cm ) ” ,” Heat l o s s r a t e
p e r u n i t l e n g t h ,W/m” ) ;
30 printf ( ” \n The maxima i n t h e c u r v e i s a t r 3 = 0 . 0 2
m” ) ;

Scilab code Exa 2.6 Critical radius

1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 2
6 // Heat C o n d u c t i o n i n S o l i d s
7
8 // Example 2 . 6
9 // Page 36
10 printf ( ” Example 2 . 6 , Page 34 \n\n ” )
11
12 h_natural = 10; // h e a t t r a n s f e r c o e f f i c i e n t f o r
n a t u r a l c o n v e c t i o n i n W/mˆ2 K
13 h_forced = 50; // h e a t t r a n s f e r c o e f f i c i e n t f o r
f o r c e d c o n v e c t i o n i n W/mˆ2 K
14 // f o r a s b e s t o s
15 k1 = 0.2; // t h e r m a l c o n d u c t i v i t y i n W/m K
16 // f o r m i n e r a l w o o l
17 k2 = 0.05; // t h e r m a l c o n d u c t i v i t y i n W/m K

17
18 printf ( ” \n 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 cm” ) ;
19 printf ( ” \n h = 10
h = 50 ” ) ;
20 printf ( ” \n A s b e s t o s %. 1 f
%. 1 f ” , k1 *100/ h_natural , k1
*100/ h_forced ) ;
21 printf ( ” \n M i n e r a l w o o l %. 1 f
%. 1 f ” , k2 *100/ h_natural , k2
*100/ h_forced ) ;

Scilab code Exa 2.7 Maximum temperature

1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 2
6 // Heat C o n d u c t i o n i n S o l i d s
7
8 // Example 2 . 7
9 // Page 43
10 printf ( ” Example 2 . 7 , Page 43 \n\n ” )
11
12 H = 5 ; // H e i g h t , [m]
13 L = 10 ; // Length , [m]
14 t = 1 ; // t h i c k n e s s , [m]
15 b = t /2;
16 k = 1.05 ; // [W/m K ]
17 q = 58 ; // [W/mˆ 3 ]
18 T = 35 ; // [ C ]
19 h = 11.6 ; // Heat t r a n s f e r c o e f f i c i e n t , [W/mˆ2 K ]
20 // S u b s t i t u t i n g t h e v a l u e s i n e q u a t i o n 2 . 5 . 6
21 T_max = T + q * b *( b /(2* k ) +1/ h ) ;
22 printf ( ”Maximum T e m p e r a t u r e = %f d e g r e e C” , T_max ) ;

18
 Scilab code Exa 2.8 Steady state temperature

1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 2
6 // Heat C o n d u c t i o n i n S o l i d s
7
8 // Example 2 . 8
9 // Page 47
10 printf ( ” Example 2 . 8 , Page 47 \n\n ” )
11
12 // The b a r w i l l have two d i m e n s i o n a l v a r i a t i o n i n
temperature
13 // t h e d i f f e r e n t i a l e q u a t i o n i s s u b j e c t t o boundary
conditions
14 x1 = 0; // [ cm ]
15 Tx1 = 30; // [ C ]
16 x2 = 5; // [ cm ]
17 Tx2 = 30; // [ C ]
18 y1 = 0; // [ cm ]
19 Ty1 = 30; // [ C ]
20 y2 = 10; // [ cm ]
21 Ty2 = 130; // [ C ]
22 // s u b s t i t u t i n g t h e t a = T−30 and u s i n g eqn 2 . 6 . 1 1
23 // p u t t i n g x = 2 . 5 cm and y = 5cm i n i n f i n i t e
summation s e r i e s
24
25
26 n = 1;
27 x1 = (1 - cos ( %pi * n ) ) /( sinh (2* %pi * n ) ) * sin ( n ^ %pi /2) *
sinh ( n * %pi ) ;
28

19
29 n = 3;
30 x3 = (1 - cos ( %pi * n ) ) /( sinh (2* %pi * n ) ) * sin ( n ^ %pi /2) *
sinh ( n * %pi ) ;
31
32 n = 5;
33 x5 = (1 - cos ( %pi * n ) ) /( sinh (2* %pi * n ) ) * sin ( n ^ %pi /2) *
sinh ( n * %pi ) ;
34
35 x = x1 + x3 + x5 ;
36
37 T = x *100+30;
38 printf ( ” S t e a d y s t a t e t e m p e r a t u r e = %f C” ,T ) ;

Scilab code Exa 2.9 Time taken by the rod to heat up

1 clear all ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 2
6 // Heat C o n d u c t i o n i n S o l i d s
7
8 // Example 2 . 9
9 // Page 51
10 printf ( ” Example 2 . 9 , Page 51 \n\n ” )
11
12 k = 330; // t h e r m a l c o n d u c t i v i t y i n W/m K
13 a = 95*10^( -6) ; // t h e r m a l e x p a n s i o n c o e f f i c i e n t
14 R = 0.01; // r a d i u s i n m e t e r s
15 To = 77; // t e m p e r a t u r e i n k e l v i n s
16 Tf = 273+50; // t e m p e r a t u r e i n k e l v i n s
17 theta1 = To - Tf ;
18 T = 273+10; // t e m p e r a t u r e i n k e l v i n s
19 theta = T - Tf ;
20 h = 20; // h e a t t r a n s f e r c o e f f i c i e n t in W

20
 /mˆ2 K
21 printf ( ” \n Theta1 = %d K” , theta1 ) ;
22 printf ( ” \n Theta = %d K ” , theta ) ;
23 printf ( ” \n v /A = %. 3 f m” ,R /2) ;
24 printf ( ” \n k / a = %. 4 f ∗ 1 0 ˆ ( 6 ) J /mˆ3 K” ,( k / a ) *10^( -6) )
;
25
26 time = ( k / a ) *( R /2) / h * log ( theta1 / theta ) ;
27
28 printf ( ” \n Time t a k e n by t h e r o d t o h e a t up = %. 1 f
s e c s ” , time ) ;
29 Bi = h * R / k ;
30 printf ( ” \n B i o t number Bi = %. 2 f ∗10ˆ( −4) ” , Bi *10^4) ;
31 printf ( ” \n S i n c e B i o t number i s much l e s s t h a n 0 . 1 ,
t h e r e f o r e assumption that i n t e r n a l temperature
g r a d i e n t s a r e n e g l i g i b l e i s a good one ” ) ;

Scilab code Exa 2.10.i Heat transfer coefficient at the centre

1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 2
6 // Heat C o n d u c t i o n i n S o l i d s
7
8 // Example 2 . 1 0 ( i )
9 // Page 58
10 printf ( ” Example 2 . 1 0 ( i ) , Page 58 \n\n ” )
11
12 // C e n t r e o f t h e s l a b
13 // Given d a t a
14 b = 0.005 ; // [m]
15 t = 5*60; // time , [ s e c ]
16 Th = 200 ; // [ C ]

21
17 Tw = 20 ; // [ C ]
18 h = 150 ; // [W/mˆ2 K ]
19 rho = 2200 ; // [ kg /mˆ 3 ]
20 Cp = 1050 ; // [ J / kg K ]
21 k = 0.4 ; // [W/m K ]
22 // U s i n g c h a r t s i n f i g 2 . 1 8 and 2 . 1 9 and eqn 2 . 7 . 1 9
and 2 . 7 . 2 0
23
24 theta = Th - Tw ;
25 Biot_no = h * b / k ;
26 a = k /( rho * Cp ) ; // a l p h a
27 Fourier_no = a * t / b ^2;
28
29 // From f i g 2 . 1 8 , r a t i o = t h e t a x b 0 / t h e t a o
30 ratio_b0 = 0.12;
31 // From f i g 2 . 1 8 , r a t i o = t h e t a x b 1 / t h e t a o
32 ratio_b1 = 0.48;
33
34 // T h e r e f o r e
35 theta_x_b0 = theta * ratio_b0 ; // [ C ]
36 T_x_b0 = theta_x_b0 + Tw ; // [ C ]
37 theta_x_b1 = theta * ratio_b1 ; // [ C ]
38 T_x_b1 = theta_x_b1 + Tw ; // [ C ]
39
40 // From T a b l e 2 . 2 f o r Bi = 1 . 8 7 5
41 lambda_1_b = 1.0498;
42 x = 2* sin ( lambda_1_b ) /[ lambda_1_b +( sin ( lambda_1_b ) )
*( cos ( lambda_1_b ) ) ];
43
44 // From eqn 2 . 7 . 2 0
45 theta_x_b0 = theta * x *( exp (( - lambda_1_b ^2) * Fourier_no
));
46 T_x_b0 = theta_x_b0 + Tw ;
47 printf ( ” T e m p e r a t u r e a t b=0 i s %f d e g r e e C\n ” , T_x_b0 )
;

22
 Scilab code Exa 2.10.ii heat transfer coefficient at the surface

1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 2
6 // Heat C o n d u c t i o n i n S o l i d s
7
8 // Example 2 . 1 0 ( i i )
9 // Page 58
10 printf ( ” Example 2 . 1 0 ( i i ) , Page 58 \n\n ” )
11
12 // ( i i ) S u r f a c e o f t h e s l a b
13
14 b = 0.005 ; // [m]
15 t = 5*60; // time , [ s e c ]
16 Th = 200 ; // [ C ]
17 Tw = 20 ; // [ C ]
18 h = 150 ; // [W/mˆ2 K ]
19 rho = 2200 ; // [ kg /mˆ 3 ]
20 Cp = 1050 ; // [ J / kg K ]
21 k = 0.4 ; // [W/m K ]
22 // U s i n g c h a r t s i n f i g 2 . 1 8 and 2 . 1 9 and eqn 2 . 7 . 1 9
and 2 . 7 . 2 0
23 theta = Th - Tw ;
24 Biot_no = h * b / k ;
25 a = k /( rho * Cp ) ; // a l p h a
26 Fourier_no = a * t / b ^2;
27
28 // From f i g 2 . 1 8 , r a t i o = t h e t a x b 0 / t h e t a o
29 ratio_b0 = 0.12;
30 // From f i g 2 . 1 8 , r a t i o = t h e t a x b 1 / t h e t a o
31 ratio_b1 = 0.48;

23
32
33 // T h e r e f o r e
34 theta_x_b0 = theta * ratio_b0 ; // [ C ]
35 T_x_b0 = theta_x_b0 + Tw ; // [ C ]
36 theta_x_b1 = theta * ratio_b1 ; // [ C ]
37 T_x_b1 = theta_x_b1 + Tw ; // [ C ]
38
39 // From T a b l e 2 . 2 f o r Bi = 1 . 8 7 5
40 lambda_1_b = 1.0498;
41 x = 2* sin ( lambda_1_b ) /[ lambda_1_b +( sin ( lambda_1_b ) )
*( cos ( lambda_1_b ) ) ];
42
43 // From 2 . 7 . 1 9
44 theta_x_b1 = theta_x_b0 *( cos ( lambda_1_b *1) ) ;
45 T_x_b1 = theta_x_b1 + Tw ;
46 printf ( ” T e m p e r a t u r e a t b=1 i s %f d e g r e e C\n ” , T_x_b1 )
;

Scilab code Exa 2.11.a Time taken by the centre of ball

1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 2
6 // Heat C o n d u c t i o n i n S o l i d s
7
8 // Example 2 . 1 1 ( a )
9 // Page 65
10 printf ( ” Example 2 . 1 1 ( a ) , Page 65 \n\n ” )
11
12 D = 0.05 ; // [m]
13 To = 450 ; // [ degree C]
14 Tf = 90 ; // [ degree C]
15 T = 150 ; // [ degree c ]

24
16 h = 115 ; // [W/mˆ2 K ]
17 rho = 8000 ; // [ kg /mˆ 3 ]
18 Cp = 0.42*1000 ; // [ J / kg K ]
19 k = 46 ; // [W/m K ]
20 R = D /2;
21
22 // ( a )
23 // From eqn 2 . 7 . 3 f o r a s p h e r e
24 t1 = rho * Cp * R /(3* h ) * log (( To - Tf ) /( T - Tf ) ) ; // [ s e c ]
25 t1_min = t1 /60 ; // [ min ]
26 printf ( ” Time t a k e n by t h e c e n t r e o f t h e b a l l t o
r e a c h 150 d e g r e e C i f i n t e r n a l g r a d i e n t s a r e
n e g l e c t e d i s %f s e c o n d s i . e . %f m i n u t e s \n ” ,t1 ,
t1_min ) ;

Scilab code Exa 2.11.b time taken by the centre of ball to reach temper-
ature

1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 2
6 // Heat C o n d u c t i o n i n S o l i d s
7
8 // Example 2 . 1 1 ( b )
9 // Page 65
10 printf ( ” Example 2 . 1 1 ( b ) , Page 65 \n\n ” )
11
12 D = 0.05 ; // [m]
13 To = 450 ; // [ d e g r e e C ]
14 Tf = 90 ; // [ d e g r e e C ]
15 T = 150 ; // [ d e g r e e c ]
16 h = 115 ; // [W/mˆ2 K ]
17 rho = 8000 ; // [ kg /mˆ 3 ]

25
18 Cp = 0.42*1000 ; // [ J / kg K ]
19 k = 46 ; // [W/m K ]
20 R = D /2;
21
22 // ( b )
23 // l e t r a t i o = t h e t a R 0 / t h e t a o
24 ratio = (T - Tf ) /( To - Tf ) ;
25 Bi = h * R / k ;
26 // From T a b l e 2 . 5
27 lambda_1_R = 0.430;
28 x = 2*[ sin ( lambda_1_R ) - lambda_1_R * cos ( lambda_1_R )
]/[ lambda_1_R - sin ( lambda_1_R ) * cos ( lambda_1_R ) ];
29
30 // S u b s t i t u t i n g i n e q u a t t i o n 2 . 7 . 2 9 , we have an
e q u a t i o n i n v a r i a b l e y(= a t /Rˆ 2 )
31 // S o l v i n g
32 function [ eqn ] = parameter ( y )
33 eqn = ratio - x * exp ( -( lambda_1_R ^2) *( y ) ) ;
34 funcprot (0) ;
35 endfunction
36
37 y = 5; // ( i n i t i a l
g u e s s , assumed v a l u e f o r f s o l v e
function )
38 Y = fsolve (y , parameter ) ;
39
40 a = k /( Cp * rho ) ; // a l p h a
41 t2 = Y *( R ^2) /( a ) ; // [ s e c ]
42 t2_min = t2 /60; // [ min ]
43 printf ( ” Time t a k e n by t h e c e n t r e of the b a l l to
r e a c h 150 d e g r e e C i f i n t e r n a l t e m p e r a t u r e
g r a d i e n t s a r e n o t n e g l e c t e d i s %f s e c o n d s i . e . %f
m i n u t e s ” ,t2 , t2_min ) ;

Scilab code Exa 2.12 Temperature at the centre of the brick

26
 1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 2
6 // Heat C o n d u c t i o n i n S o l i d s
7
8 // Example 2 . 1 2
9 // Page 67
10 printf ( ” Example 2 . 1 2 , Page 67 \n\n ” )
11
12 a = 0.12 ; // [m]
13
14 T = 400 ; // [ C ]
15 To = 25 ; // [ C ]
16 t = 100/60 ; // [ h o u r ]
17 h = 10 ; // [W/mˆ2 K ]
18 k = 1.0 ; // [W/m K ]
19 alpha = 3.33*10^ -3 ; // [mˆ2/ h ]
20 // u s i n g f i g 2 . 1 8 and eqn 2 . 7 . 2 0
21
22 x1 = h * a / k ;
23 x2 = k /( h * a ) ;
24 x3 = alpha * t / a ^2;
25
26 // L e t r a t i o x = t h e t a / t h e t a o f o r x d i r e c t i o n , from
f i g 2.18
27 ratio_x = 0.82 ;
28
29 // S i m i l a r l y , f o r y d i r e c t i o n
30 ratio_y = 0.41;
31
32 // S i m i l a r l y , f o r z d i r e c t i o n
33 ratio_z = 0.30;
34
35 // T h e r e f o r e
36 total_ratio = ratio_x * ratio_y * ratio_z ;
37

27
38 T_centre = To + total_ratio *( T - To ) ; // [ d e g r e e C ]
39 printf ( ” T e m p e r a t u r e a t t h e c e n t r e o f t h e b r i c k = %f
d e g r e e C \n\n ” , T_centre ) ;
40
41 // A l t e r n a t i v e l y
42 printf ( ” A l t e r n a t i v e l y , o b t a i n i n g B i o t number and
v a l u e s o f l a m b d a 1 b and u s i n g eqn 2 . 7 . 2 0 , we g e t
\n ” )
43
44 ratio_x = 1.1310* exp ( -(0.9036^2) *0.385) ;
45 ratio_y = 1.0701* exp ( -(0.6533^2) *2.220) ;
46 ratio_z = 1.0580* exp ( -(0.5932^2) *3.469) ;
47 ratio = ratio_x * ratio_y * ratio_z ;
48
49 T_centre = To + total_ratio *( T - To ) ; // [ d e g r e e C ]
50 printf ( ” T e m p e r a t u r e a t t h e c e n t r e o f t h e b r i c k = %f
d e g r e e C \n ” , T_centre ) ;

Scilab code Exa 2.13.a Temperature at the copper fin tip

1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 2
6 // Heat C o n d u c t i o n i n S o l i d s
7
8 // Example 2 . 1 3 ( a )
9 // Page 73
10 printf ( ” Example 2 . 1 3 ( a ) , Page 73 \n\n ” )
11
12 D = 0.003 ; // [m]
13 L = 0.03 ; // [m]
14 h = 10 ; // [W/mˆ 2 ]
15 Tf = 20 ; // [ C ]

28
16 T1 = 120 ; // [ C ]
17
18 // ( a ) Copper f i n
19 k = 350 ; // [W/m K ]
20
21 // For a c i r c u l a r c r o s s s e c t i o n
22 m = [4* h /( k * D ) ]^(1/2) ;
23 mL = m *0.03 ;
24 // T a t x = L
25 T = Tf + ( T1 - Tf ) / cosh ( m * L ) ;
26 printf ( ”mL = %f \n ” , mL ) ;
27 printf ( ” T e m p e r a t u r e a t t h e t i p o f f i n made o f c o p p e r
i s %f d e g r e e C \n ” ,T ) ;

Scilab code Exa 2.13.b Temperature at the steel fin tip

1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 2
6 // Heat C o n d u c t i o n i n S o l i d s
7
8 // Example 2 . 1 3 ( b )
9 // Page 73
10 printf ( ” Example 2 . 1 3 ( b ) , Page 73 \n\n ” )
11
12 D = 0.003 ; // [m]
13 L = 0.03 ; // [m]
14 h = 10 ; // [W/mˆ 2 ]
15 Tf = 20 ; // [ C ]
16 T1 = 120 ; // [ C ]
17
18
19 // ( b ) S t a i n l e s s s t e e l f i n

29
20 k = 15 ; // [W/m K ]
21
22 // For a c i r c u l a r c r o s s s e c t i o n
23 m = [4* h /( k * D ) ]^(1/2) ;
24 mL = m *0.03 ;
25 // T a t x = L
26 T = Tf + ( T1 - Tf ) / cosh ( m * L ) ;
27 printf ( ”mL = %f \n ” , mL ) ;
28 printf ( ” T e m p e r a t u r e a t t h e t i p o f f i n made o f s t e e l
i s %f d e g r e e C \n ” ,T ) ;

Scilab code Exa 2.13.c Temperature at the teflon fin tip

1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 2
6 // Heat C o n d u c t i o n i n S o l i d s
7
8 // Example 2 . 1 3 ( c )
9 // Page 73
10 printf ( ” Example 2 . 1 3 ( c ) , Page 73 \n\n ” )
11
12 D = 0.003 ; // [m]
13 L = 0.03 ; // [m]
14 h = 10 ; // [W/mˆ 2 ]
15 Tf = 20 ; // [ C ]
16 T1 = 120 ; // [ C ]
17
18 // ( c ) T e f l o n f i n
19 k = 0.35 ; // [W/m K ]
20
21 // For a c i r c u l a r c r o s s s e c t i o n
22 m = [4* h /( k * D ) ]^(1/2) ;

30
23 mL = m *0.03 ;
24 // T a t x = L
25 T = Tf + ( T1 - Tf ) / cosh ( m * L ) ;
26 printf ( ”mL = %f \n ” , mL ) ;
27 printf ( ” T e m p e r a t u r e a t t h e t i p o f f i n made o f t e f l o n
i s %f d e g r e e C \n ” ,T ) ;

Scilab code Exa 2.14 Heat loss rate

1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 2
6 // Heat C o n d u c t i o n i n S o l i d s
7
8 // Example 2 . 1 4
9 // Page 74
10 printf ( ” Example 2 . 1 4 , Page 74 \n\n ” )
11
12 L = 0.02 ; // [m]
13 t = 0.002 ; // [m]
14 b = 0.2 ; // [m]
15 theta1 = 200 ; // [ C ]
16 h = 15 ; // [W/mˆ2 K ]
17 k = 45 ; // [W/m K ]
18
19 Bi = h *( t /2) / k ;
20
21 // We have
22 P = 2*( b + t ) ; // [m]
23 A = b * t ; // [mˆ 2 ]
24 // T h e r e f o r e
25 mL = ([( h * P ) /( A * k ) ]^(1/2) ) * L ;
26

31
27 // From e q u a t i o n 2 . 8 . 6 , f i n e f f e c t i v e n e s s n
28 n = tanh ( mL ) / mL ;
29 printf ( ” Fin E f f e c t i v e n e s s = %f \n ” ,n ) ;
30
31 q_loss = n * h *40.4*2*10^ -4*200; // [W]
32 printf ( ” Heat l o s s r a t e from f i n s u r f a c e = %f W” ,
q_loss ) ;

Scilab code Exa 2.15 Decrease in thermal resistance

1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 2
6 // Heat C o n d u c t i o n i n S o l i d s
7
8 // Example 2 . 1
9 // Page 74
10 printf ( ” Example 2 . 1 5 , Page 74 \n\n ” )
11
12
13 // Find D e c r e a s e i n t h e r m a l R e s i s t a n c e
14 // Find I n c r e a s e i n h e a t t r a n s f e r r a t e
15
16 h = 15 ; // [W/mˆ 2 .K ]
17 k = 300; // [W/m. K ]
18 T = 200; // [ C ]
19 Tsurr = 30; // [ C ]
20 d = .01; // [m]
21 L = .1; // [m]
22 A = .5*.5 // [mˆ 2 ]
23 n = 100 // Number o f P i n s
24
25 Bi = h * d /2/ k ; // B i o t Number

32
26 // V a l u e o f B i o t Number i s much l e s s t h a n . 1
27 // Thus u s i n g e q u a t i o n 2 . 8 . 6
28 mL = ( h *4/ k / d ) ^.5* L ;
29 zi = tanh ( mL ) / mL ;
30 Res1 = 1/ h / A ; // Thermal r e s i s t a n c e w i t h o u t
f i n s , [ K/W]
31 Res2 = 1/( h *( A - n * %pi /4* d ^2 + zi *( n * %pi * d * L ) ) ) ; //
Thermal r e s i s t a n c e w i t h f i n s , [ K/W]
32
33 delRes = Res1 - Res2 ; // [ K/W]
34 // I n c r e a s e i n h e a t t r a n s f e r r a t e
35 q = (T - Tsurr ) / Res2 - (T - Tsurr ) / Res1 ; // [W]
36
37 printf ( ” \n\n D e c r e a s e i n t h e r m a l r e s i s t a n eat
s u r f a c e %. 4 f K/W. \ n I n c r e a s e i n h e a t t r a n s f e r
r a t e %. 1 f W” , delRes , q )
38 //END

Scilab code Exa 2.16 Overall heat transfer coefficient

1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 2
6 // Heat C o n d u c t i o n i n S o l i d s
7
8 // Example 2 . 1 6
9 // Page 75
10 printf ( ” Example 2 . 1 6 , Page 75 \n\n ” )
11
12 // T h e o r e t i c a l Problem
13
14 printf ( ’ \n\n T h i s i s a T h e o r e t i c a l Problem , d o e s n o t
i n v o l v e any m a t h e m a t i c a l c o m p u t a t i o n . ’ ) ;

33
15 //END

34
 Chapter 3

Thermal Radiation

Scilab code Exa 3.1 Monochromatic emissive power

1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 3
6 // Thermal R a d i a t i o n
7
8 // Example 3 . 1
9 // Page 114
10 printf ( ” Example 3 . 1 , Page 114 \n\n ” ) ;
11
12 T = 5779 ; // [ Temperature , i n K e l v i n ]
13 // From Wein ’ s law , eqn 3 . 2 . 8
14 lambda_m = 0.00290/ T ; // [m]
15 // S u b s t i t u t i n g t h i s v a l u e i n p l a n k ’ s law , we g e t
16 e = 2*( %pi ) *0.596*(10^ -16) /(((0.5018*10^ -6) ^5) *( exp
(0.014387/0.00290) -1) ) ; // [W/mˆ2 m]
17
18 e_bl_max = e / 10^6 ;
19
20 printf ( ” V a l u e o f e m i s s i v i t y on sun s u r f a c e i s %f W/m

35
 ˆ2 um \n ” , e_bl_max ) ; // [W/mˆ2 um ]
21
22 e_earth = e_bl_max *((0.695*10^6) /(1.496*10^8) ) ^2 ;
23
24 printf ( ” The v a l u e o f e m m i s s i v i t y on e a r t h s s u r f a c e
i s %f W/mˆ2 um” , e_earth )

Scilab code Exa 3.2 Heat flux

1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 3
6 // Thermal R a d i a t i o n
7
8 // Example 3 . 2
9 // Page 115
10 printf ( ” Example 3 . 2 , Page 115 \n\n ” )
11
12 // Heat e m i s s i o n
13 Stefan_constt = 5.67*10^( -8) ; // (W/mˆ 2 .Kˆ 4 )
14 T = 1500; // t e m p e r a t u r e is
in kelvins
15 eb = ( Stefan_constt ) *( T ^(4) ) ; // e n e r g y
r a d i a t e d by b l a c k b o d y
16 // e m i s s i o n i n 0 . 3 um t o 1um
17 e = 0.9; // e m i s s i v i t y
18 lamda1 = 1; // w a v e l e n g t h i s i n um
19 lamda2 = 0.3; // w a v e l e n g t h i s i n um
20 D0_1 =0.5*(0.01972+0.00779) ; // From t a b l e 3 . 1
page− 114
21 D0_2 =0; // From t a b l e 3 . 1 page
− 114
22 q = e *( D0_1 - D0_2 ) * Stefan_constt * T ^(4) ; // i n W/mˆ2

36
23 printf ( ” \n w a v e l e n g t h ∗ temp = %d um K” ,1*1500) ;
24 printf ( ” \n w a v e l e n g t h ∗ temp a t 0 . 3 um = %d um K”
,0.3*1500) ;
25 printf ( ” \n\n R e q u i r e d h e a t f l u x , q = %d W/mˆ2 ” ,q ) ;

Scilab code Exa 3.3 Absorbed radiant flux and absorptivity and reflectiv-
ity

1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 3
6 // Thermal R a d i a t i o n
7
8 // Example 3 . 3
9 // Page 119
10 printf ( ” Example 3 . 3 , Page 119 \n\n ” )
11
12
13 a0_2 =1; // a b s o r p t i v i t y
14 a2_4 =1; // a b s o r p t i v i t y
15 a4_6 =0.5; // a b s o r p t i v i t y
16 a6_8 =0.5; // a b s o r p t i v i t y
17 a8_ =0; // a b s o r p t i v i t y
18 H0_2 =0; // I r r a d i a t i o n i n W/mˆ2 um
19 H2_4 =750; // I r r a d i a t i o n i n W/mˆ2 um
20 H4_6 =750; // I r r a d i a t i o n i n W/mˆ2 um
21 H6_8 =750; // I r r a d i a t i o n i n W/mˆ2 um
22 H8_ =750; // I r r a d i a t i o n i n W/mˆ2 um
23 Absorbed_radiant_flux =1*0*(2 -0) +1*750*(4 -2)
+0.5*750*(8 -4) +0;
24 H = 750*(8 -2) ; // I n c i d e n t f l u x
25 a = Absorbed_radiant_flux / H ;
26 p = 1 - a ; // S i n c e t h e s u r f a c e i s opaque

37
27 printf ( ” \n Absorbed r a d i a n t f l u x = %d W/mˆ2 ” ,
Absorbed_radiant_flux ) ;
28 printf ( ” \n I n c i d e n t f l u x = %d W/mˆ2 ” ,H ) ;
29 printf ( ” \n A b s o r p t i v i t y = %. 3 f ” ,a ) ;
30 printf ( ” \n S i n c e t h e s u r f a c e i s opaque r e f l e c t i v i t y
= %. 3 f ” ,p ) ;

Scilab code Exa 3.4.a Total intensity in normal direction

1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 3
6 // Thermal R a d i a t i o n
7
8 // Example 3 . 4 ( a )
9 // Page 123
10 printf ( ” Example 3 . 4 ( a ) , Page 123 \n\n ” )
11
12
13 e = 0.08; // e m i s s i v i t y
14 T = 800; // t e m p e r a t u r e , [ K ]
15
16 Stefan_constt = 5.67*10^( -8) ; // [W/mˆ 2 .Kˆ 4 ]
17 // From S t e f a n Boltzmann law , e q u a t i o n 3 . 2 . 1 0
18 q = e * Stefan_constt * T ^4; // [W/mˆ 2 ]
19 printf ( ” \n Energy e m i t t e d = %. 1 f W/mˆ2 ” ,q ) ;
20
21 // ( a )
22 // T h e r e f o r e
23 in = ( q /( %pi ) ) ;
24 printf ( ” \n Energy e m i t t e d n o r m a l t o t h e s u r f a c e = %
. 1 f W/mˆ2 s r ” , in ) ;

38
 Scilab code Exa 3.4.b Ratio of radiant flux to the emissive power

1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 3
6 // Thermal R a d i a t i o n
7
8 // Example 3 . 4 ( b )
9 // Page 123
10 printf ( ” Example 3 . 4 ( b ) , Page 123 \n\n ” )
11
12
13 e = 0.08; // e m i s s i v i t y
14 T = 800; // t e m p e r a t u r e , [ K ]
15
16 Stefan_constt = 5.67*10^( -8) ; // [W/mˆ 2 .Kˆ 4 ]
17 // From S t e f a n Boltzmann law , e q u a t i o n 3 . 2 . 1 0
18 q = e * Stefan_constt * T ^4; // [W/mˆ 2 ]
19 in = ( q /( %pi ) ) ;
20
21 // ( b )
22 // R a d i a n t f l u x e m i t t e d i n t h e c o n e 0 <= p z i <= 50
d e g r e e , 0 <= t h e t a <= 2∗ p i
23 q_cone =2*( %pi ) * in *( - cos (100*( %pi /180) ) + cos (0) ) /4;
24
25 printf ( ” \n R a d i a n t f l u x e m i t t e d i n t h e c o n e =%. 1 f W/
mˆ2 ” , q_cone ) ;
26
27 Ratio = q_cone / q ;
28 printf ( ” \n R a t i o = %. 3 f ” , Ratio ) ;

39
 Scilab code Exa 3.5 Rate of incident radiation

1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 3
6 // Thermal R a d i a t i o n
7
8 // Example 3 . 5
9 // Page 124
10 printf ( ” Example 3 . 5 , Page 124 \n\n ” )
11
12 l1 = 0.5 ; // w a v e l e n g t h , [ um ]
13 l2 = 1.5 ; // w a v e l e n g t h , [ um ]
14 l3 = 2.5 ; // w a v e l e n g t h , [ um ]
15 l4 = 3.5 ; // w a v e l e n g t h , [ um ]
16 H1 = 2500 ; // [W/mˆ2 um ]
17 H2 = 4000 ; // [W/mˆ2 um ]
18 H3 = 2500 ; // [W/mˆ2 um ]
19
20 // S i n c e t h e i r r i d i a t i o n i s d i f f u s e , t h e s p e c t r a l
i n t e n s i t y i s g i v e n by eqn 3 . 4 . 1 4 and 3 . 4 . 8
21 // I n t e g r a t i n g i l a m b d a o v e r t h e d i r e c t i o n s o f t h e
s p e c i f i e d s o l i d a n g l e and u s i n g f i g 3 . 1 2
22
23
24 flux = 3/4*[ H1 *( l2 - l1 ) + H2 *( l3 - l2 ) + H3 *( l4 - l3 ) ];
25 printf ( ” Rate a t which r a d i a t i o n i s i n c i d e n t on t h e
s u r f a c e = %f W/mˆ2 ” , flux ) ;

Scilab code Exa 3.6 Shape factor F12

40
 1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 3
6 // Thermal R a d i a t i o n
7
8 // Example 3 . 6
9 // Page 132
10 printf ( ” Example 3 . 6 , Page 132 \n\n ” )
11
12 // T h i s i s a t h e o r e t i c a l p r o b l e m w i t h no n u m e r i c a l
data
13 printf ( ” T h i s i s a t h e o r e t i c a l p r o b l e m w i t h no
n u m e r i c a l d a t a \n ” ) ;
14
15 // C o n s i d e r i n g an e l e m e n t a r y r i n g dA2 o f w i d t h d r a t
an a r b i t a r y r a d i u s r , we have
16 // r = h∗ tanB1
17 // dA2 = 2∗ %pi ∗ r ∗ d r
18 // dA2 = 2∗ %pi ∗ ( h ˆ 2 ) ∗ t a n ( B1 ) ∗ s e c ˆ 2 ( B1 ) ∗dB1
19 // B2 = B1 , s i n c e s u r f a c e s a t e p a r a l l e l , and
20 // L = h / c o s ( B1 )
21 // S u b s t i t u t i n g i n eqn 3 . 6 . 7
22 // F12 = s i n ˆ 2 ( a )
23
24
25 printf ( ” C o n s i d e r i n g an e l e m e n t a r y r i n g dA2 o f w i d t h
d r a t an a r b i t a r y r a d i u s r , we have \n ” ) ;
26 printf ( ” r = h∗ tanB1 \n ” ) ;
27 printf ( ”dA2 = 2∗ p i ∗ r ∗ d r \n ” ) ;
28 printf ( ”dA2 = 2∗ p i ∗ ( h ˆ 2 ) ∗ t a n ( B1 ) ∗ s e c ˆ 2 ( B1 ) ∗dB1 \n ” ) ;
29 printf ( ”B2 = B1 , s i n c e s u r f a c e s a t e p a r a l l e l , and \n
”);
30 printf ( ”L = h / c o s ( B1 ) \n ” ) ;
31 printf ( ” S u b s t i t u t i n g i n eqn 3 . 6 . 7 \n ” ) ;
32 printf ( ” F12 = s i n ˆ 2 ( a ) \n ” ) ;

41
 Scilab code Exa 3.7 Shape factor

1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 3
6 // Thermal R a d i a t i o n
7
8 // Example 3 . 7
9 // Page 134
10 printf ( ” Example 3 . 7 , Page 134 \n\n ” )
11
12 // T h i s i s a t h e o r e t i c a l p r o b l e m w i t h no n u m e r i c a l
data
13 printf ( ” T h i s i s a t h e o r e t i c a l p r o b l e m w i t h no
n u m e r i c a l d a t a \n ” ) ;
14
15
16 // C o n s i d e r i n g an e l e m e n t a r y c i r c u l a r r i n g on t h e
s u r f a c e o f t h e s p h e r e ’ s s u r f a c e a t any a r b i t a r y
anglr B,
17 // we have B1 = B , B2 = 0 , L = R and dA 2 = 2∗ %pi ∗ (R
ˆ 2 ) ∗ ( s i n (B) ) dB
18 // T h e r e f o r e , from e q u a t i o n 3 . 6 . 7
19 // F12 = s i n ˆ 2 ( a )
20
21 printf ( ” C o n s i d e r i n g an e l e m e n t a r y c i r c u l a r r i n g on
t h e s u r f a c e o f t h e s p h e r e s u r f a c e a t any a r b i t a r y
a n g l r B \n ” ) ;
22 printf ( ”we have B1 = B , B2 = 0 , L = R and dA 2 = 2∗
p i ∗ (Rˆ 2 ) ∗ ( s i n (B) ) dB \n ” ) ;
23 printf ( ” T h e r e f o r e , from e q u a t i o n 3 . 6 . 7 \n ” ) ;
24 printf ( ” F12 = s i n ˆ 2 ( a ) ” ) ;

42
 Scilab code Exa 3.8 Shape factor F12

1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 3
6 // Thermal R a d i a t i o n
7
8 // Example 3 . 8
9 // Page 135
10 printf ( ” Example 3 . 8 , Page 135 \n\n ” )
11
12 // From eqn 3 . 7 . 5 o r f i g 3 . 1 9
13 F65 = 0.22;
14 F64 = 0.16;
15 F35 = 0.32;
16 F34 = 0.27;
17 A1 = 3; // [mˆ 2 ]
18 A3 = 3; // [mˆ 2 ]
19 A6 = 6; // [mˆ 2 ]
20
21 // U s i n g a d d i t i v e and r e c i p r o c a l r e l a t i o n s
22 // We have F12 = F16 − F13
23
24 F61 = F65 - F64 ;
25 F31 = F35 - F34 ;
26
27 F16 = A6 / A1 * F61 ;
28 F13 = A3 / A1 * F31 ;
29
30 F12 = F16 - F13 ;
31
32 printf ( ” F 1 −2 = %f ” , F12 ) ;

43
 Scilab code Exa 3.9 Shape factor

1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 3
6 // Thermal R a d i a t i o n
7
8 // Example 3 . 9
9 // Page 136
10 printf ( ” Example 3 . 9 , Page 136 \n\n ” )
11
12 // T h i s i s a t h e o r e t i c a l problem , d o e s n o t i n v o l v e
any n u m e r i c a l c o m p u t a t i o n
13 printf ( ” T h i s i s a t h e o r e t i c a l problem , d o e s n o t
i n v o l v e any n u m e r i c a l c o m p u t a t i o n \n ” ) ;
14 // D e n o t i n g a r e a o f c o n i c a l s u r f a c e by A1
15 // C o n s i d e r i n g an i m a g i n a r y f l a t s u r f a c e A2 c l o s i n g
the c o n i c a l cavity
16
17 F22 = 0 ; // F l a t surface
18
19 // from eqn 3 . 7 . 2 , we have F11 + F12 = 1 and F22 +
F21 = 1
20 F21 = 1 - F22 ;
21
22 // F12 = A2/A1∗ F21 ;
23 // F11 = 1 − F12 ;
24 // F11 = 1 − s i n ( a )

Scilab code Exa 3.10 Net radiative heat transfer

44
 1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 3
6 // Thermal R a d i a t i o n
7
8 // Example 3 . 1 0
9 // Page 138
10 printf ( ” Example 3 . 1 0 , Page 138 \n\n ” )
11
12 sigma = 5.670*10^ -8 ;
13 T1 = 473 ; // [ K ]
14 T2 = 373 ; // [ K ]
15 A1 = 1*2 ; // a r e a , [mˆ 2 ]
16 X = 0.25;
17 Y = 0.5 ;
18 // From eqn 3 . 7 . 4
19 F12 = (2/( %pi * X * Y ) ) *[ log ((((1+ X ^2) *(1+ Y ^2) ) /(1+ X ^2+ Y
^2) ) ^(1/2) ) + Y *((1+ X ^2) ^(1/2) ) * atan ( Y /((1+ X ^2)
^(1/2) ) ) + X *((1+ Y ^2) ^(1/2) ) * atan ( X /((1+ Y ^2)
^(1/2) ) ) - Y * atan ( Y ) - X * atan ( X ) ];
20
21
22 q1 = sigma * A1 *( T1 ^4 - T2 ^4) *[(1 - F12 ^2) /(2 -2* F12 ) ];
23
24 printf ( ” Net r a d i a t i v e h e a t t r a n s f e r from t h e s u r f a c e
= %f W \n ” , q1 ) ;

Scilab code Exa 3.11 steady state heat flux

1 clear all ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME

45
 5 // C h a p t e r 3
6 // Thermal R a d i a t i o n
7
8 // Example 3 . 1 1
9 // Page 141
10 printf ( ” Example 3 . 1 1 , Page 141 \n\n ” )
11
12 // A l l modes o f h e a t t r a n s f e r a r e i n v o l v e d
13 // l e t s t e a d y s t a t e h e a t f l u x f l o w i n g t h r o u g h t h e
c o m p o s i t e s l a b be ( q / a )
14 h1 = 20; // [W/mˆ2 K ]
15 w1 = 0.2; // [m]
16 k1 = 1; // [W/m K ]
17 e1 = 0.5; // e m m i s i v i t y a t s u r f c e 1
18 e2 = 0.4; // e m m i s i v i t y a t s u r f c e 2
19 w2 = 0.3; // [m]
20 k2 = 0.5; // [W/m K ]
21 h2 = 10; // [W/mˆ2 K ]
22 T1 = 473; // [ K e l v i n ]
23 T2 = 273+40; // [ K e l v i n ]
24 stefan_cnst = 5.67 e -08; // [W/mˆ2 Kˆ 4 ]
25
26 // For r e s i s t a n c e s 1 and 2
27 function [ f ]= temperature ( T )
28 f (1) = ( T1 - T (1) ) /(1/ h1 + w1 / k1 ) - ( T (2) - T2 ) /(
w2 / k2 + 1/ h2 ) ;
29 f (2) = stefan_cnst *( T (1) ^4 - T (2) ^4) /(1/ e1 + 1/
e2 -1) - ( T (2) - T2 ) /( w2 / k2 + 1/ h2 ) ;
30 funcprot (0) ;
31 endfunction
32
33 T = [10 10]; // assumed i n i t i a l values for fsolve
function
34 y = fsolve (T , temperature ) ;
35
36 printf ( ” \n S t e a d y s t a t e h e a t f l u x q /A = %. 1 f W/mˆ2 ”
,( T1 - y (1) ) /(1/ h1 + w1 / k1 ) ) ;

46
 Scilab code Exa 3.12 Rate of heat loss

1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 3
6 // Thermal R a d i a t i o n
7
8 // Example 3 . 1 2
9 // Page 145
10 printf ( ” Example 3 . 1 2 , Page 145 \n\n ” )
11
12 D = 0.02 ; // [m]
13 T1 = 1000+273 ; // [ K ]
14 T2 = 27+273 ; // [ K ]
15 s = 5.670*10^ -8 ; // s t e f a n s c o n s t a n t
16 // Assuming t h e o p e n i n g i s c l o s e d by an i m a g i n a r y
s u r f a c e a t t e m p e r a t u r e T1
17 // U s i n g e q u a t i o n 3 . 1 0 . 3 , we g e t
18 q = s *1* %pi *(( D /2) ^2) *( T1 ^4 - T2 ^4) ; // [W]
19
20 printf ( ” Rate a t which h e a t i s l o s t by r a d i a t i o n = %f
W” ,q ) ;

Scilab code Exa 3.13 Rate of nitrogen evaporation

1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 3

47
 6 // Thermal R a d i a t i o n
7
8 // Example 3 . 1 3
9 // Page 146
10 printf ( ” Example 3 . 1 3 , Page 146 \n\n ” )
11
12 D = 0.32 ; // [m]
13 D_s = 0.36 ; // [m]
14 e = 0.02 ; // e m i s s i v i t y
15 l = 201 ; // [ kJ / kg ]
16 rho = 800 ; // [ kg /mˆ 3 ]
17 s = 5.670*10^ -8 ;
18
19 T2 = 303 ; // [ K ]
20 T1 = 77 ; // [ K ]
21
22 // From e q u a t i o n 3 . 1 0 . 1
23 q1 = s *4* %pi *(( D /2) ^2) *( T1 ^4 - T2 ^4) /[1/ e +(( D / D_s ) ^2)
*(1/ e -1) ]; // [W]
24
25 evap = abs ( q1 ) *3600*24/( l *1000) ; // [ kg / day ]
26 mass = 4/3* %pi *(( D /2) ^3) * rho ;
27 boiloff = evap / mass *100 ; // p e r c e n t
28
29 T_drop = ( abs ( q1 ) ) /(4* %pi *(( D /2) ^2) ) *(1/100) ; // [ C ]
30
31 printf ( ” Rate a t which n i t r o g e n e v a p o r a t e s = %f kg /
day \n ” , evap )
32 printf ( ” B o i l − o f f r a t e = %f p e r c e n t \n ” , boiloff ) ;
33 printf ( ” T e m p e r a t u r e d r o p b e t w e e n l i q u i d N i t r o g e n and
i n n e r s u r f a c e = %f C” , T_drop ) ;

Scilab code Exa 3.14 Rate of energy loss from satellite

1 clear ;

48
 2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 3
6 // Thermal R a d i a t i o n
7
8 // Example 3 . 1 4
9 // Page 147
10 printf ( ” Example 3 . 1 4 , Page 147 \n\n ” )
11
12 D = 1 ; // [m]
13 r = 6250 ; // [ km ]
14 D_surf = 300 ; // [ km ]
15 s = 5.670*10^ -8;
16 e = 0.3 ;
17 Tc = -18+273 ; // [ K ]
18 T_surf = 27+273 ; // [ K ]
19
20 // Rate o f e m i s s i n o o f r a d i a n t e n e r g y from t h e two
faces of s a t e l l i t e disc
21 r_emission = 2* e * %pi *(( D /2) ^2) * s * Tc ^4; // [W]
22
23 // A2∗ F21 = A1∗ F12
24 sina = ( r /( r + D_surf ) ) ;
25 F12 = sina ^2;
26
27 // Rate a t which t h e s a t e l l i t er e c e i v e s and a b s o r b s
e n e r g y coming from e a r t h
28 r_receive = e * s *( %pi *(( D /2) ^2) ) * F12 * T_surf ^4; // [W]
29
30 r_loss = r_emission - r_receive ; // [W]
31
32 printf ( ” Net Rate a t which e n e r g y i s l e a v i n g t h e
s a t e l l i t e = %f W” , r_loss ) ;

49
 Scilab code Exa 3.15 Net radiative heat transfer

1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 3
6 // Thermal R a d i a t i o n
7
8 // Example 3 . 1 5
9 // Page 151
10 printf ( ” Example 3 . 1 5 , Page 151 \n\n ” )
11
12 // From e x a m p l e 3 . 1 0
13 F12 = 0.0363;
14 F11 = 0;
15 F13 = 1 - F11 - F12 ;
16 // S i m i l a r l y
17 F21 = 0.0363;
18 F22 = 0;
19 F23 = 0.9637;
20
21 // Now , F31 = A1/A3∗ F13
22 F31 = 2/24* F13 ;
23 // T h e r e f o r e
24 F32 = F31 ;
25 F33 = 1 - F31 - F32 ;
26
27 // S u b s t i t u t i n g i n t o e q u a t i o n 3 . 1 1 . 6 , 3 . 1 1 . 7 ,
3 . 1 1 . 8 , we have f ( 1 ) , f ( 2 ) , f ( 3 )
28
29 function [ f ]= flux ( B )
30 f (1) = B (1) - 0.4*0.0363* B (2) - 0.4*0.9637* B (3) -
0.6*(473^4) *(5.670*10^ -8) ;
31 f (2) = -0.4*0.0363* B (1) + B (2) - 0.4*0.9637* B (3)
- 0.6*(5.670*10^ -8) *(373^4) ;
32 f (3) = 0.0803* B (1) + 0.0803* B (2) - 0.1606* B (3) ;
33 funcprot (0) ;

50
34 endfunction
35
36 B = [0 0 0];
37 y = fsolve (B , flux ) ;
38 printf ( ” \n B1 = %. 1 f W/mˆ2 ” ,y (1) ) ;
39 printf ( ” \n B2 = %. 1 f W/mˆ2 ” ,y (2) ) ;
40 printf ( ” \n B3 = %. 1 f W/mˆ2 \n ” ,y (3) ) ;
41
42 // Therefore
43 H1 = 0.0363* y (2) + 0.9637* y (3) ; // [W/mˆ 2 ]
44 // and
45 q1 = 2*( y (1) - H1 ) ; // [W]
46
47 printf ( ” Net r a d i a t i v e h e a t t r a n s f e r = %f W” , q1 ) ;

51
 Chapter 4

Principles of Fluid Flow

Scilab code Exa 4.1 Pressure drop in smooth pipe

1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 4
6 // P r i n c i p l e s o f F l u i d Flow
7
8 // Example 4 . 1
9 // Page 172
10 printf ( ” Example 4 . 1 , Page 172 \n\n ” ) ;
11
12 L = 3 ; // Length , [m]
13 D = 0.01 ; // ID , [m]
14 V = 0.2 ; // A v e r a g e V e l o c i t y , [m/ s ]
15
16 // From T a b l e A. 1 a t 10 d e g r e e C
17 rho =999.7 ; // [ kg /mˆ 3 ]
18 v =1.306 * 10^ -6 ; // [mˆ2/ s ]
19
20 Re_D =0.2*0.01/(1.306*10^ -6) ;
21

52
22 // t h i s v a l u e i s l e s s t h a n t h e t r a n s i t i o n R e y n o l d s
number 2 3 0 0 .
23 // Hence f l o w i s l a m i n a r . From eqn 4 . 4 . 1 9
24 f = 16/ Re_D ;
25
26 // from eqn 4 . 4 . 1 7
27 delta_p = 4* f *( L / D ) *( rho * V ^2) /2;
28
29 // s i n c e f l o w i s l a m i n a r
30 V_max = 2* V ;
31
32 printf ( ” P r e s s u r e d r o p i s %f Pa \n ” , delta_p ) ;
33 printf ( ” Maximum v e l o c i t y i s %f m/ s ” , V_max ) ;

Scilab code Exa 4.2.a Pressure drop and maximum velocity calculation

1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 4
6 // P r i n c i p l e s o f F l u i d Flow
7
8 // Example 4 . 2 ( a )
9 // Page 180
10 printf ( ” Example 4 . 2 ( a ) , Page 180 \n\n ” )
11
12 L = 3 ; // [m]
13 D = 0.01 ; // [m]
14 V = 0.2 ; // [m/ s ]
15
16 // ( a )
17 printf ( ” ( a ) I f t h e t e m p e r a t u r e o f w a t e r i s i n c r e a s e d
t o 80 d e g r e e C \n ” ) ;
18

53
19
20 // P r o p e r t i e s o f w a t e r a t 80 d e g r e e C
21 rho = 971.8 ; // [ kg /mˆ 3 ]
22 v = 0.365 * 10^ -6 ; // [mˆ2/ s ]
23
24 Re_D = D * V / v ;
25
26 // f l o w i s t u r b i l e n t , s o from eqn 4 . 6 . 4 a
27
28 f =0.079*( Re_D ) ^( -0.25) ;
29 delta_p = (4* f * L * rho * V ^2) /( D *2) ; // [ Pa ]
30 printf ( ” P r e s s u r e d r o p i s %f Pa \n ” , delta_p ) ;
31
32 // from eqn 4 . 4 . 1 6
33
34 // x = ( T w/ p ) ˆ 0 . 5 = ( ( f / 2 ) ˆ 0 . 5 ) ∗V ;
35 x = (( f /2) ^0.5) * V ;
36 y_plus = 0.005* x /(0.365*10^ -6) ;
37
38 // from eqn 4 . 6 . 1 c & 4 . 6 . 2
39
40 V_max = x *(2.5* log ( y_plus ) + 5.5) ; // [m/ s ]
41 ratio = V_max / V ;
42 printf ( ”V max = %f m/ s \n ” , V_max ) ;
43 printf ( ”V max/ V bar = %f \n\n ” , ratio ) ;

Scilab code Exa 4.2.b Pressure drop and maximum velocity calculation

1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 4
6 // P r i n c i p l e s o f F l u i d Flow
7

54
 8 // Example 4 . 2 ( b )
9 // Page 180
10 printf ( ” Example 4 . 2 ( b ) , Page 180 \n\n ” )
11
12 L = 3 ; // [m]
13 D = 0.01 ; // [m]
14 V = 0.2 ; // [m/ s ]
15
16 // ( b )
17
18 V1 =0.7;
19 v1 = 1.306 * 10^ -6 ; // [mˆ2/ s ]
20
21 printf ( ” ( b ) I f t h e v e l o c i t y i s i n c r e a s e d t o 0 . 7 \n ” )
;
22 // i f v e l o c i t y o f w a t e r i s 0 . 7 m/ s
23 V1 =0.7; // [m/ s ]
24 Re_D1 = V1 * D /(1.306*10^ -6) ;
25 printf ( ” R e y n o l d s no i s %f \n ” , Re_D1 ) ;
26
27 // f l o w i s a g a i n t u r b u l e n t
28 f1 = 0.079*( Re_D1 ) ^( -0.25) ;
29
30 delta_p1 = (4* f1 * L *999.7*0.7^2) /(0.01*2) ; // [ Pa ]
31 printf ( ” P r e s s u r e d r o p i s %f Pa \n ” , delta_p1 ) ;
32
33 // x1 = ( T w/ p ) ˆ 0 . 5 = ( ( f 1 / 2 ) ˆ 0 . 5 ) ∗V ;
34 x1 = (( f1 /2) ^0.5) * V1 ;
35
36 y1_plus = 0.005* x1 /( v1 ) ;
37 printf ( ” y+ a t c e n t r e l i n e = %f \n ” , y1_plus ) ;
38
39 V_max1 = x1 *(2.5* log ( y1_plus ) + 5.5) ; // [m/ s ]
40 printf ( ”V max i s %f m/ s \n ” , V_max1 ) ;
41
42 ratio1 = V_max1 / V1 ;
43 printf ( ”Vmax/ Vbar = %f ” , ratio1 ) ;

55
 Scilab code Exa 4.3 Pressure drop and power needed

1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 4
6 // P r i n c i p l e s o f F l u i d Flow
7
8 // Example 4 . 3
9 // Page 181
10 printf ( ” Example 4 . 3 , Page 181 \n\n ” )
11 P = 80 * 10^3 ; // [ Pa ]
12 L = 10 ; // [m]
13 V_bar = 1.9 ; // [m/ s ]
14 l = 0.25 ; // [m]
15 b = 0.15 ; // [m]
16
17 // F u l l y d e v e l o p e d f l o w
18
19 // From T a b l e A . 2 , f o r a i r a t ! atm p r e s s u r e and 25
degree C
20 rho = 1.185 ; // [ kg /mˆ 3 ]
21 mew = 18.35 * 10^ -6 ; // [ kg /m s ]
22
23 // At 80 kPa and 25 d e g r e e C
24 rho1 = rho *(80/101.3) ; // [ kg /mˆ 3 ]
25
26 // For g i v e n d u c t r =(b/ a )
27 r = b / l ;
28
29 D_e = (4* l /2* b /2) /( l /2 + b /2) ; // [m]
30
31 // From eqn 4 . 6 . 7

56
32
33 D_l = [2/3 + 11/24*0.6*(2 -0.6) ]* D_e ; // [m]
34
35 // R e y n o l d s no b a s e d on D l
36
37 Re = rho1 * D_l * V_bar / mew ;
38 printf ( ” R e y n o l d s no = %f \n ” , Re ) ;
39
40 f = 0.079*( Re ^ -0.25) ;
41 printf ( ” f = %f \n ” ,f ) ;
42
43 // From eqn 4 . 4 . 1 7
44
45 delta_P = 4* f *( L / D_l ) *( rho1 *( V_bar ^2) /2) ;
46 printf ( ” P r e s s u r e d r o p = %f Pa \n ” , delta_P ) ;
47
48 power = delta_P *( V_bar * l * b )
49 printf ( ” Power r e q u i r e d = %f W” , power ) ;

Scilab code Exa 4.4 Thickness of velocity boundary layer

1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 4
6 // P r i n c i p l e s o f F l u i d Flow
7
8 // Example 4 . 4
9 // Page 189
10 printf ( ” Example 4 . 4 , Page 189 \n\n ” )
11
12 l = 2 ; // [m]
13 b = 1 ; // [m]
14 V = 1 ; // [m/ s ]

57
15
16 // From t a b l e A. 2
17
18 rho = 1.060 ; // [ kg /mˆ 3 ]
19 v = 18.97 * 10^ -6 ; // [mˆ2/ s ]
20
21 // At x = 1 . 5m
22 x = 1.5 ; // [m]
23 Re = V * x / v ; // R e y n o l d s number
24
25 // From eqn 4 . 8 . 1 2
26
27 d = 5* x /( Re ^(1/2) ) *1000 ; // [mm]
28 printf ( ” T h i c k n e s s o f Boundary l a y e r a t x = 1 . 5 i s %f
mm \n ” ,d )
29
30 Re_l = V * l / v ;
31
32 // From eqn 4 . 8 . 1 9 and 4 . 8 . 1 6
33
34 c_f = 1.328* Re_l ^ -(1/2) ; // d r a g c o e f f i c i e n t
35 printf ( ” Drag C o e f f i c i e n t c f = %f \n ” , c_f ) ;
36
37 F_d = 0.00409*(1/2) * rho *(2* l * b ) *1^2;
38 printf ( ” Drag F o r c e F D = %f N” , F_d ) ;

Scilab code Exa 4.5 Drag coefficient and drag force

1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 4
6 // P r i n c i p l e s o f F l u i d Flow
7

58
 8 // Example 4 . 5
9 // Page 195
10 printf ( ” Example 4 . 5 , Page 195 \n\n ” ) ;
11
12 l = 2 ; // [m]
13 v = 4 ; // [m/ s ]
14
15 // From T a b l e A. 2
16
17 mew = 18.1*10^ -6; // [ N s /mˆ 2 ]
18 rho = 1.205*1.5; // [ kg /mˆ 3 ]
19
20 Re_l = rho * v * l / mew ;
21 // Boundary l a y e r i s p a r t l y l a m i n a r and p a r t l y
t u r b u l e n t , we s h a l l u s e eqn 4 . 1 0 . 4
22 Cf = 0.074*(7.989*10^5) ^( -0.2) - 1050/ Re_l ;
23 printf ( ” Drag c o e f f i c i e e n t i s %f \n ” , Cf )
24
25 D_f = Cf *1/2* rho * l * v ^2;
26 printf ( ” Drag f o r c e p e r m e t e r w i d t h = %f N \n ” , D_f ) ;
27
28 // from eqn 4 . 1 0 . 1
29
30 x = 3*10^5 * (18.1*10^ -6) /(1.808*4) ;
31 printf ( ” V a l u e o f x c i s %f m” ,x ) ;

59
 Chapter 5

Heat Transfer by Forced

Convection

Scilab code Exa 5.1.a Local heat transfer coefficient

1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 5
6 // Heat T r a n s f e r by F o r c e d C o n v e c t i o n
7
8
9 // Example 5 . 1 ( a )
10 // Page 209
11 printf ( ” Example 5 . 1 ( a ) \n\n ” )
12
13 D = 0.015 ; // [m]
14 Q = 0.05 ; // [mˆ3/ h ]
15 H = 1000 ; // [W/mˆ 2 ]
16 T_b = 40 ; // [ d e g r e e C ]
17
18 // From t a b l e A . 1 , p r o p e r t i e s a t 40 d e g r e e C
19 k = 0.634 ; // [W/m K ]

60
20 v = 0.659*10^ -6 ; // [mˆ2/ s ]
21
22 V_bar = 4* Q /(( %pi ) * D ^2) ;
23
24 Re_D = V_bar * D / v ;
25
26 // T h e r e f o r e , Laminar Flow , from eqn 5 . 2 . 8
27
28 h = 4.364* k / D ; // [W/mˆ2 K ]
29
30 printf ( ” ( a ) L o c a 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 %f W/
mˆ2 K \n ” ,h ) ;

Scilab code Exa 5.1.b Wall temperature

1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 5
6 // Heat T r a n s f e r by F o r c e d C o n v e c t i o n
7
8
9 // Example 5 . 1 ( b )
10 // Page 209
11 printf ( ” Example 5 . 1 ( b ) \n\n ” )
12
13 D = 0.015 ; // [m]
14 Q = 0.05 ; // [mˆ3/ h ]
15 H = 1000 ; // [W/mˆ 2 ]
16 T_b = 40 ; // [ d e g r e e C ]
17
18 // From t a b l e A . 1 , p r o p e r t i e s a t 40 d e g r e e C
19 k = 0.634 ; // [W/m K ]
20 v = 0.659*10^ -6 ; // [mˆ2/ s ]

61
21
22 V_bar = 4* Q /(( %pi ) * D ^2) ;
23
24 Re_D = V_bar * D / v ;
25
26 // T h e r e f o r e , Laminar Flow , from eqn 5 . 2 . 8
27
28 h = 4.364* k / D ;
29
30 // From t h e d e f i n i t i o n o f h i n eqn 5 . 2 . 3 , t h e local
wal t o b u l k mean t e m p e r a t u r e d i f f e r e n c e i s given
by
31
32 T_w = H / h + T_b ;
33
34 printf ( ” ( b ) Wall T e m p e r a t u r e Tw = %f d e g r e e C” , T_w ) ;

Scilab code Exa 5.2 ratio of thermal entrance length to entrance length

1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 5
6 // Heat T r a n s f e r by F o r c e d C o n v e c t i o n
7
8
9 // Example 5 . 2
10 // Page 213
11 printf ( ” Example 5 . 2 , Page 213 \n\n ” )
12
13 // From eqn 5 . 2 . 1 2 and 4 . 4 . 2 0
14 // L e t r = Lth / Le
15 // r = 0 . 0 4 3 0 5 ∗ Pr / 0 . 0 5 7 5 ;
16

62
17 function [ T ]= r ( Pr )
18 T = 0.04305* Pr /0.0575
19 endfunction
20
21 // For Pr = 0 . 0 1
22 r1 = r (0.01) ;
23 // For Pr = 0 . 1
24 r2 = r (1) ;
25 // For Pr = 100
26 r3 = r (100) ;
27
28 printf ( ” Lth / Le a t Pr = 0 . 0 1 i s %f \n ” , r1 ) ;
29 printf ( ” Lth / Le a t Pr = 1 i s %f \n ” , r2 ) ;
30 printf ( ” Lth / Le a t Pr = 100 i s %f ” , r3 ) ;

Scilab code Exa 5.3.i Length of tube

1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 5
6 // Heat T r a n s f e r by F o r c e d C o n v e c t i o n
7
8
9 // Example 5 . 3 ( i )
10 // Page 215
11 printf ( ” Example 5 . 3 ( i ) , Page 215 \n\n ” )
12
13 D = 0.015 ; // [m]
14 V = 1 ; // [m/ s ]
15 Tw = 90 ; // [ d e g r e e C ]
16 Tmi = 50 ; // [ d e g r e e C ]
17 Tmo = 65 ; // [ d e g r e e C ]
18

63
19 // ( i )
20
21 // From T a b l e A. 1
22 k = 0.656 ; // [W/m K ]
23 rho = 984.4 ; // [ kg /mˆ 3 ]
24 v = 0.497 * 10^ -6 ; // [mˆ2/ s ]
25 Cp = 4178 ; // [ J / kg K ]
26 Pr = 3.12 ;
27 rho_in = 988.1 ; // [ kg /mˆ 3 ]
28
29 m_dot = %pi *( D ^2) * rho_in * V /4 ; // [ kg / s ]
30
31 Re = 4* m_dot /( %pi * D * rho * v ) ;
32
33 // U s i n g eqn 5 . 3 . 2 and 4 . 6 . 4 a
34 f = 0.079*( Re ) ^ -0.25 ;
35
36 Nu = ( f /2) *( Re -1000) * Pr /[1+12.7*( f /2) ^(1/2) *(( Pr
^(2/3) ) -1) ];
37 h = Nu * k / D ; // [W/mˆ2 K ]
38
39 // From t h e e n e r g y e q u a t i o n , e x t r a c t i n g t h e v a l u e o f
L
40 L = m_dot * Cp *( Tmo - Tmi ) *[ log (( Tw - Tmi ) /( Tw - Tmo ) ) ]/[((
Tw - Tmi ) -( Tw - Tmo ) ) * h * D * %pi ]; // [m]
41
42 printf ( ” The l e n g t h o f t u b e i f t h e e x i t w a t e r
t e m p e r a t u r e i s 65 d e g r e e C = %f m\n ” ,L ) ;

Scilab code Exa 5.3.ii Exit water temperature

1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME

64
 5 // C h a p t e r 5
6 // Heat T r a n s f e r by F o r c e d C o n v e c t i o n
7
8
9 // Example 5 . 3 ( i )
10 // Page 215
11 printf ( ” Example 5 . 3 ( i i ) , Page 215 \n\n ” )
12
13 D = 0.015 ; // [m]
14 V = 1 ; // [m/ s ]
15 Tw = 90 ; // [ d e g r e e C ]
16 Tmi = 50 ; // [ d e g r e e C ]
17 Tmo = 65 ; // [ d e g r e e C ]
18
19 // From T a b l e A. 1
20 k = 0.656 ; // [W/m K ]
21 rho = 984.4 ; // [ kg /mˆ 3 ]
22 v = 0.497 * 10^ -6 ; // [mˆ2/ s ]
23 Cp = 4178 ; // [ J / kg K ]
24 Pr = 3.12 ;
25 rho_in = 988.1 ; // [ kg /mˆ 3 ]
26
27 m_dot = %pi *( D ^2) * rho_in * V /4 ; // [ kg / s ]
28
29 Re = 4* m_dot /( %pi * D * rho * v ) ;
30
31 // U s i n g eqn 5 . 3 . 2 and 4 . 6 . 4 a
32 f = 0.079*( Re ) ^ -0.25 ;
33
34 Nu = ( f /2) *( Re -1000) * Pr /[1+12.7*( f /2) ^(1/2) *(( Pr
^(2/3) ) -1) ];
35 h = Nu * k / D ; // [W/mˆ2 K ]
36
37 // From t h e e n e r g y e q u a t i o n , e x t r a c t i n g t h e v a l u e o f
L
38 L = m_dot * Cp *( Tmo - Tmi ) *[ log (( Tw - Tmi ) /( Tw - Tmo ) ) ]/[((
Tw - Tmi ) -( Tw - Tmo ) ) * h * D * %pi ]; // [m]
39

65
40 // ( i i )
41 printf ( ” \ n T r i a l and e r r o r method \n ” ) ;
42
43 // T r i a l 1
44 printf ( ” T r i a l 1\ n ” ) ;
45 printf ( ” Assumed v a l u e o f Tmo = 70 d e g r e e C\n ” ) ;
46 T_mo = 70 ; // [ d e g r e e C ]
47 T_b = 60 ; // [ d e g r e e C ]
48
49 k1 = 0.659 ; // [W/m K ]
50 rho1 = 983.2 ; // [ kg /mˆ 3 ]
51 v1 = 0.478 * 10^ -6 ; // [mˆ2/ s ]
52 Cp1 = 4179 ; // [ J / kg K ]
53 Pr1 = 2.98 ;
54
55 Re1 = 4* m_dot /( %pi * D * rho1 * v1 ) ;
56
57 // From B l a s i u s eqn ( 4 . 6 . 4 a ) , we g e t
58 f1 = 0.005928;
59
60 // S u b s t i t u t i n g t h i s v a l u e i n t o t h e G n i e l i n s k i Eqn
61 Nu_d = 154.97;
62 h = Nu_d * k1 / D ; // [W/mˆ2 K ]
63
64 // from eqn 5 . 3 . 3 , we g e t
65 Tmo1 = 73.4 ; // [ d e g r e e C ]
66 printf ( ” V a l u e o f Tmo o b t a i n e d = 7 3 . 4 d e g r e e C\n ” ) ;
67
68 // T r i a l 2
69 printf ( ” T r i a l 2\ n ” ) ;
70 printf ( ” Assume Tmo = 7 3 . 4 d e g r e e C\n ” ) ;
71 printf ( ” V a l u e o f Tmo o b t a i n e d = 7 3 . 6 d e g r e e C which
i s i n r e a s o n a b l y c l o s e a g r e e m e n t w i t h assumed
v a l u e . \ n”)

66
 Scilab code Exa 5.4 Length of tube over which temperature rise occurs

1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 5
6 // Heat T r a n s f e r by F o r c e d C o n v e c t i o n
7
8
9 // Example 5 . 4
10 // Page 219
11 printf ( ” Example 5 . 4 , Page 219 \n\n ” )
12
13 D_i = 0.05 ; // [m]
14 m = 300 ; // [ kg / min ]
15 m1 = m /60 ; // [ kg / s e c ]
16 rho = 846.7 ; // [ kg /mˆ 3 ]
17 k = 68.34 ; // [W/m K ]
18 c = 1274; // [ J / kg K ]
19 v = 0.2937*10^ -6 ; // [mˆ2/ s ]
20 Pr = 0.00468 ;
21
22 Re_D = 4* m1 /( %pi * D_i * rho * v ) ;
23
24 // Assuming b o t h t e m p e r a t u r e and v e l o c i t y p r o f i l e
are f u l l y developed over the l e n g t h o f tube
25 // u s i n g eqn 5 . 3 . 6
26 Nu_D = 6.3 + 0.0167*( Re_D ^0.85) *( Pr ^0.93) ;
27
28 h = Nu_D * k / D_i ;
29
30 // E q u a t i n g t h e h e a t
t r a n s f e r r e d through the wall of
t h e t u b e t o t h e c h a n g e o f e n t h a l p y p f sodium
31 L = 300/60*1274*(500 -400) /( h * %pi * D_i *30)
32
33 printf ( ” Length o f t u b e o v e r which t h e t e m p e r a t u r e
r i s e o c c u r s = %f m” ,L )

67
 Scilab code Exa 5.5 Rate of heat transfer to the plate

1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 5
6 // Heat T r a n s f e r by F o r c e d C o n v e c t i o n
7
8
9 // Example 5 . 5
10 // Page 231
11 printf ( ” Example 5 . 5 , Page 231 \n ” )
12
13 V = 15 ; // [m/ s ]
14 s =0.2 ; // [m]
15 T_m = (20+60) /2; // [ d e g r e e C ]
16 // P r o p e r t i e s a t mean temp = 40 d e g r e e C
17 v = 16.96*10^ -6; // [mˆ2/ s ]
18 rho = 1.128 ; // [ kg /mˆ 3 ]
19 k = 0.0276; // [W/m K ]
20 Pr = 0.699;
21 A = s ^2;
22 Re_L = V *0.2/ v ;
23 // T h i s i s l e s s t h a n 3 ∗ 1 0 ˆ 5 , h e n c e t h e boundary
l a y e r may be assumed t o be l a m i n a r o v e r t h e
entire length .
24 // from eqn 4 . 8 . 1 9
25
26 Cf = 1.328/( Re_L ) ^0.5
27 Fd = 2* Cf *1/2* rho * A * V ^2;
28
29 // From eqn 5 . 5 . 1 0
30 Nu_l = 0.664*( Pr ^(1/3) ) *( Re_L ^(1/2) ) ;

68
31
32 h = Nu_l * k / s ;
33 // T h e r e f o r e r a t e o f h e a t t r a n s f e r q i s
34 q = 2* A * h *(60 -20) ; // [W]
35
36 // With a t u r b u l e n t boundary l a y e r from l e a d i n g edge
, t h e d r a g c o e f f i c i e n t i s g i v e n by eqn 4 . 1 0 . 4
37 Cf1 = 0.074*( Re_L ) ^( -0.2) ;
38 Fd1 = 2* Cf1 *1/2* rho * A * V ^2; // [ N ]
39
40 // from eqn 5 . 8 . 3 w i t h C1 = 0
41 Nu_l1 = 0.0366*(0.699^(1/3) ) *( Re_L ^(0.8) ) ;
42
43 h1 = Nu_l1 * k / s ; // [W/mˆ2 K ]
44 q1 = 2* A * h1 *(60 -20) ;
45
46 printf ( ” For Laminar Boundary L a y e r \n ” ) ;
47 printf ( ” Rate o f Heat t r a n s f e r = %f W\n ” ,q ) ;
48 printf ( ” Drag f o r c e = %f N \n \n ” , Fd )
49
50 printf ( ” For T u r b u l e n t Boundary L a y e r from t h e
l e a d i n g e d g e \n ” ) ;
51 printf ( ” Rate o f Heat t r a n s f e r = %f W\n ” , q1 ) ;
52 printf ( ” Drag f o r c e = %f N\n ” , Fd1 )

Scilab code Exa 5.6.i Heat transfer rate

1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 5
6 // Heat T r a n s f e r by F o r c e d C o n v e c t i o n
7
8

69
 9 // Example 5 . 6 ( i )
10 // Page 235
11 printf ( ” Example 5 . 6 ( i ) , Page 235 \n\n ” )
12
13 D = 0.075 ; // [m]
14 V = 1.2 ; // [m/ s ]
15 T_air = 20 ; // [ d e g r e e C ]
16 T_surface = 100 ; // [ d e g r e e C ]
17 T_m = ( T_air + T_surface ) /2;
18
19 v = 18.97*10^ -6 ; // [mˆ2/ s ]
20 k = 0.0290 ; // [W/m K ]
21 Pr = 0.696 ;
22
23 Re_D = V * D / v ;
24
25 Nu = 0.3 + [(0.62*( Re_D ^(1/2) ) *( Pr ^(1/3) ) )
/[(1+((0.4/ Pr ) ^(2/3) ) ) ^(1/4) ]]*([1+(( Re_D /282000)
^(5/8) ) ]^(4/5) ) ;
26
27 h = Nu * k / D ; // [W/mˆ2 K ]
28
29 flux = h *( T_surface - T_air ) ; // [W/mˆ 2 ]
30 q = flux * %pi * D *1; // [W/m]
31
32 printf ( ” Heat t r a n s f e r r a t e p e r u n i t l e n g t h = %f W/m\
n ” ,q ) ;

Scilab code Exa 5.6.ii Average wall tempeature

1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 5

70
 6 // Heat T r a n s f e r by F o r c e d C o n v e c t i o n
7
8
9 // Example 5 . 6 ( i i )
10 // Page 235
11 printf ( ” Example 5 . 6 ( i i ) , Page 235 \n\n ” )
12
13 D = 0.075 ; // [m]
14 V = 1.2 ; // [m/ s ]
15 T_air = 20 ; // [ d e g r e e C ]
16 T_surface = 100 ; // [ d e g r e e C ]
17 T_m = ( T_air + T_surface ) /2;
18
19 v = 18.97*10^ -6 ; // [mˆ2/ s ]
20 k = 0.0290 ; // [W/m K ]
21 Pr = 0.696 ;
22
23 Re_D = V * D / v ;
24 Nu = 0.3 + [(0.62*( Re_D ^0.5) *( Pr ^(1/3) ) ) /[(1+((0.4/
Pr ) ^(2/3) ) ) ^(1/4) ]]*[1+( Re_D /282000) ^(5/8) ]^(5/8)
;
25 h = Nu * k / D ; // [W/mˆ2 K ]
26 flux = h *( T_surface - T_air ) ; // [W/mˆ 2 ]
27
28 // ( i i ) U s i n g T r i a l and e r r o r method
29 T_avg = 1500/ flux *( T_surface - T_air ) ;
30
31 T_assumd = 130 ; // [ d e g r e e C ]
32 Tm = 75 ; // [ d e g r e e C ]
33
34 v1 = 20.56*10^ -6 ; // [mˆ2/ s ]
35 k1 = 0.0301 ; // [W/m K ]
36 Pr1 = 0.693 ;
37
38 Re_D1 = V * D / v1 ;
39
40
41 // U s i n g eqn 5 . 9 . 8

71
42 Nu1 = 33.99;
43 h = Nu1 * k1 / D ;
44 // T h e r e f o r e
45 T_diff = 1500/ h ; // [ d e g r e e C ]
46 T_avg_calc = 129.9 ; // [ d e g r e e C ]
47 printf ( ” Assumed a v e r a g e w a l l t e m p e r a t u r e = %f d e g r e e
C\n ” , T_assumd ) ;
48 printf ( ” C a l c u l a t e d a v e r a g e w a l l T e m p e r a t u r e = %f
d e g r e e C\n ” , T_avg_calc ) ;
49 printf ( ” Hence , A v e r a g e w a l l T e m p e r a t u r e = %f d e g r e e C
” , T_avg_calc ) ;

Scilab code Exa 5.7.i Pressure drop

1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 5
6 // Heat T r a n s f e r by F o r c e d C o n v e c t i o n
7
8
9 // Example 5 . 7 ( i )
10 // Page 241
11 printf ( ” Example 5 . 7 ( i ) , Page 241 \n \n ” ) ;
12
13 // Given d a t a
14 D = 0.0125 ; // [m]
15 ST = 1.5* D ;
16 SL = 1.5* D ;
17 V_inf = 2 ; // [m/ s ]
18
19 N = 5;
20 Tw = 70; // [ d e g r e e C ]
21 Tmi = 30; // [ d e g r e e C ]

72
22 L = 1; // [m]
23 // P r o p e r t i e s o f a i r a t 30 d e g r e e C
24 rho = 1.165 ; // [ kg /mˆ 3 ]
25 v = 16.00 *10^ -6 ; // [mˆ2/ s ]
26 Cp = 1.005 ; // [ kJ / kg K ]
27 k = 0.0267 ; // [W/m K ]
28 Pr = 0.701;
29
30 // From eqn 5 . 1 0 . 2
31 Vmax = ST /( SL - D ) * V_inf ; // [m/ s ]
32 Re = Vmax * D / v ;
33
34 // From f i g 5 . 1 5
35 f = 0.37/4;
36 // Also , t u b e a r r a n g e m e n t i s s q u a r e
37 X = 1;
38 // From eqn 5 . 1 0 . 6
39 delta_P = 4* f * N * X *( rho * Vmax ^2) /2 ; // [ N/mˆ 2 ]
40
41 printf ( ” ( i ) P r e s s u r e d r o p o f a i r a c r o s s t h e bank i s
%f N/mˆ2 \n ” , delta_P ) ;

Scilab code Exa 5.7.ii Exit temperature of air

1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 5
6 // Heat T r a n s f e r by F o r c e d C o n v e c t i o n
7
8
9 // Example 5 . 7 ( i i )
10 // Page 241
11 printf ( ” Example 5 . 7 ( i i ) , Page 241 \n \n ” ) ;

73
12
13 D = 0.0125 ; // [m]
14 ST = 1.5* D ;
15 SL = 1.5* D ;
16 V_inf = 2 ; // [m/ s ]
17 N = 5;
18 Tw = 70; // [ d e g r e e C ]
19 Tmi = 30; // [ d e g r e e C ]
20 L = 1; // [m]
21
22 rho = 1.165 ; // [ kg /mˆ 3 ]
23 v = 16.00 *10^ -6 ; // [mˆ2/ s ]
24 Cp = 1.005*1000 ; // [ J / kg K ]
25 k = 0.0267 ; // [W/m K ]
26 Pr = 0.701;
27
28 // From eqn 5 . 1 0 . 2
29 Vmax = ST /( SL - D ) * V_inf ; // [m/ s ]
30 Re = Vmax * D / v ;
31
32 // From f i g 5 . 1 5
33 f = 0.37/4;
34 // Also , t u b e a r r a n g e m e n t i s s q u a r e
35 X = 1;
36 // From eqn 5 . 1 0 . 6
37 delta_P = 4* f * N * X *( rho * Vmax ^2) /2 ; // [ N/mˆ 2 ]
38
39 // At 70 d e g r e e C
40 Pr1 = 0.694 ;
41 // From t a b l e 5 . 4 and 5 . 5
42
43 C1 = 0.27;
44 m = 0.63;
45 C2 = 0.93;
46
47 // S u b s t i t u t i n g i n Eqn 5 . 1 0 . 5
48 Nu = C1 * C2 *( Re ^ m ) *( Pr ^0.36) *( Pr / Pr1 ) ^(1/4) ;
49 h = Nu * k / D ; // [W/mˆ2 K ]

74
50
51 // For 1 m l o n g t u b e
52 m_dot = rho *(10*1.5* D *1) *2; // [ kg / s ]
53
54 // S u b s t i t u t i n g m dot i n 5 . 3 . 4 and s o l v i n g , we g e t
55 function [ f ]= temp ( Tmo )
56 f (1) = h *( %pi * D * L ) *50*[( Tw - Tmi ) -( Tw - Tmo (1) ) ]/[
log (( Tw - Tmi ) /( Tw - Tmo (1) ) ) ] - m_dot * Cp *( Tmo (1) -
Tmi ) ;
57 // h ∗ ( %pi ∗D∗L ) ∗N∗ ( ( Tw−Tmi ) −(Tw−Tmo) ) / l o g [ ( Tw−Tmi
) / (Tw−Tmo) ] − m dot ∗Cp ∗ (Tmo − Tmi ) ;
58 funcprot (0) ;
59 endfunction
60
61 Tmo = 40; // I n i t i a l assumed v a l u e f o r f s o l v e
function
62 y = fsolve ( Tmo , temp ) ;
63 printf ( ”Tmo = %f \n ” ,y ) ;
64
65 printf ( ” ( i i ) E x i t t e m p e r a t u r e o f a i r = %f d e g r e e C \
n ” ,y ) ;

Scilab code Exa 5.7.iii Heat transfer rate

1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 5
6 // Heat T r a n s f e r by F o r c e d C o n v e c t i o n
7
8
9 // Example 5 . 7 ( i i i )
10 // Page 241
11 printf ( ” Example 5 . 7 ( i i i ) , Page 241 \n \n ” ) ;

75
12
13 D = 0.0125 ; // [m]
14 ST = 1.5* D ;
15 SL = 1.5* D ;
16 V_inf = 2 ; // [m/ s ]
17 N = 5;
18 Tw = 70; // [ d e g r e e C ]
19 Tmi = 30; // [ d e g r e e C ]
20 L = 1; // [m]
21
22 rho = 1.165 ; // [ kg /mˆ 3 ]
23 v = 16.00 *10^ -6 ; // [mˆ2/ s ]
24 Cp = 1.005*1000 ; // [ J / kg K ]
25 k = 0.0267 ; // [W/m K ]
26 Pr = 0.701;
27
28 // From eqn 5 . 1 0 . 2
29 Vmax = ST /( SL - D ) * V_inf ; // [m/ s ]
30 Re = Vmax * D / v ;
31
32 // From f i g 5 . 1 5
33 f = 0.37/4;
34 // Also , t u b e a r r a n g e m e n t i s s q u a r e
35 X = 1;
36 // From eqn 5 . 1 0 . 6
37 delta_P = 4* f * N * X *( rho * Vmax ^2) /2 ; // [ N/mˆ 2 ]
38
39 // At 70 d e g r e e C
40 Pr1 = 0.694 ;
41 // From t a b l e 5 . 4 and 5 . 5
42
43 C1 = 0.27;
44 m = 0.63;
45 C2 = 0.93;
46
47 // S u b s t i t u t i n g i n Eqn 5 . 1 0 . 5
48 Nu = C1 * C2 *( Re ^ m ) *( Pr ^0.36) *( Pr / Pr1 ) ^(1/4) ;
49 h = Nu * k / D ; // [W/mˆ2 K ]

76
50
51 // For 1 m l o n g t u b e
52 m_dot = rho *(10*1.5* D *1) *2; // [ kg / s ]
53
54 // S u b s t i t u t i n g m dot i n 5 . 3 . 4 and s o l v i n g , we g e t
55 function [ f ]= temp ( Tmo )
56 f (1) = h *( %pi * D * L ) *50*[( Tw - Tmi ) -( Tw - Tmo (1) ) ]/[
log (( Tw - Tmi ) /( Tw - Tmo (1) ) ) ] - m_dot * Cp *( Tmo (1) -
Tmi ) ;
57 // h ∗ ( %pi ∗D∗L ) ∗N∗ ( ( Tw−Tmi ) −(Tw−Tmo) ) / l o g [ ( Tw−Tmi
) / (Tw−Tmo) ] − m dot ∗Cp ∗ (Tmo − Tmi ) ;
58 funcprot (0) ;
59 endfunction
60
61 Tmo = 40; // I n i t i a l assumed v a l u e f o r f s o l v e
function
62 y = fsolve ( Tmo , temp ) ;
63
64 // Heat t r a n s f e r r a t e q
65 q = h *( %pi * D * L ) *50*(( Tw - Tmi ) -( Tw - y ) ) /( log (( Tw - Tmi ) /(
Tw - y ) ) ) ;
66
67 printf ( ” ( i i i ) Heat t r a n s f e r r a t e p e r u n i t l e n g t h t o
a i r = %f W” ,q ) ;

77
 Chapter 6

Heat Transfer by Natural

convection

Scilab code Exa 6.1 Average nusselt number

1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 6
6 // Heat T r a n s f e r by N a t u r a l C o n v e c t i o n
7
8
9 // Example 6 . 1
10 // Page 258
11 printf ( ” Example 6 . 1 , Page 258 \n \n ” ) ;
12
13 H = 0.5 ; // [m]
14 T_h = 100; // [ d e g r e e C ]
15 T_l = 40; // [ d e g r e e C ]
16
17 v = 20.02*10^ -6 ; // [m/ s ]
18 Pr = 0.694;
19 k = 0.0297; // [W/m K ]

78
20
21 T = ( T_h + T_l ) /2 + 273 ; // [ K ]
22 printf ( ”Mean f i l m t e m p e r a t u r e = %f K \n ” ,T ) ;
23 B = 1/ T ;
24
25 Gr = 9.81* B *(( T_h - T_l ) * H ^3) /( v ^2) ;
26 Ra = Gr * Pr ;
27
28 // ( a )
29 // Exact a n a l y s i s − E q u a t i o n 6 . 2 . 1 7
30 disp ( ” ( a ) ” ) ;
31 printf ( ” Exact a n a l y s i s \n ” ) ;
32 Nu_a = 0.64*( Gr ^(1/4) ) *( Pr ^0.5) *((0.861+ Pr ) ^( -1/4) ) ;
33 printf ( ” Nu L = %f \n ” , Nu_a ) ;
34
35 // ( b )
36 // I n t e g r a l method − E q u a t i o n 6 . 2 . 2 9
37 disp ( ” ( b ) ” ) ;
38 printf ( ” I n t e g r a l method \n ” ) ;
39 Nu_b = 0.68*( Gr ^(1/4) ) *( Pr ^0.5) *((0.952+ Pr ) ^( -1/4) ) ;
40 printf ( ” Nu L = %f \n ” , Nu_b ) ;
41
42 // ( c )
43 // McAdams c o r r e l a t i o n − E q u a t i o n 6 . 2 . 3 0
44 disp ( ” ( c ) ” ) ;
45 printf ( ”McAdams c o r r e l a t i o n \n ” ) ;
46 Nu_c = 0.59*( Ra ) ^(1/4) ;
47 printf ( ” Nu L = %f \n ” , Nu_c ) ;
48
49 // ( d )
50 // C h u r c h i l l and Chu c o r r e l a t i o n − E q u a t i o n 6 . 2 . 3 1
51 disp ( ” ( d ) ” )
52 printf ( ” C h u r c h i l l and Chu c o r r e l a t i o n \n ” ) ;
53 Nu_d = 0.68 + 0.670*( Ra ^(1/4) ) /[1+(0.492/ Pr ) ^(9/16)
]^(4/9) ;
54 printf ( ” Nu L = %f \n ” , Nu_d ) ;

79
 Scilab code Exa 6.2 Reduce the equation

1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 6
6 // Heat T r a n s f e r by N a t u r a l C o n v e c t i o n
7
8
9 // Example 6 . 2
10 // Page 259
11 printf ( ” Example 6 . 2 , Page 259 \n \n ” ) ;
12
13 Tm = 150 ; // [ d e g r e e C ]
14 // From t a b l e A. 2
15 v = 28.95*10^ -6 ; // [mˆ2/ s ]
16 Pr = 0.683;
17 k = 0.0357 ; // [W/m K ]
18
19 B = 1/(273+ Tm ) ; // [ Kˆ −1]
20
21 // from eqn 6 . 2 . 3 0
22 printf ( ” E q u a t i o n 6 . 2 . 3 0 \n h = k /L ∗ 0 . 5 9 ∗ [ 9 . 8 1 ∗ B∗ (Tw−
T i n f ) ∗ ( L ˆ 3 ) ∗ 0 . 6 8 3 / ( v ˆ 2 ) ] ˆ ( 1 / 4 ) \n ” )
23 // h = k /L ∗ 0 . 5 9 ∗ [ 9 . 8 1 ∗ B∗ (Tw−T i n f ) ∗ ( L ˆ 3 ) ∗ 0 . 6 8 3 / ( v ˆ 2 )
]ˆ(1/4) ;
24 // s i m p l i f y i n g we g e t
25 // h = 1 . 3 8 ∗ [ ( Tw−T i n f ) /L ] ˆ ( 1 / 4 )
26 printf ( ” R e d u c e s t o h = 1 . 3 8 ∗ [ ( Tw−T i n f ) /L ] ˆ ( 1 / 4 ) \n ” )
27
28
29 // From eqn 6 . 2 . 3 3
30 // h∗L/ k = 0 . 1 0 ∗ [ 9 . 8 1 ∗ B∗ (Tw−T i n f ) ∗ ( L ˆ 3 ) ∗ 0 . 6 8 3 / ( v ˆ 2 )

80
 ]ˆ(1/3) ;
31 printf ( ” E q u a t i o n 6 . 2 . 3 3 \n h∗L/ k = 0 . 1 0 ∗ [ 9 . 8 1 ∗ B∗ (Tw−
T i n f ) ∗ ( L ˆ 3 ) ∗ 0 . 6 8 3 / ( v ˆ 2 ) ] ˆ ( 1 / 3 ) \n ” ) ;
32 // s i m p l i f y i n g
33 // h = 0 . 9 5 ∗ ( Tw−T i n f ) ˆ 1 / 3
34 printf ( ” R e d u c e s t o h = 0 . 9 5 ∗ ( Tw−T i n f ) ˆ 1 / 3 \n ” ) ;
35
36 printf ( ” where h i s e x p r e s s e d i n W/mˆ2 K, (Tw−T i n f )
i n C and L i n m e t r e s \n ” ) ;

Scilab code Exa 6.3 Time for cooling of plate

1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 6
6 // Heat T r a n s f e r by N a t u r a l C o n v e c t i o n
7
8
9 // Example 6 . 3
10 // Page 260
11 printf ( ” Example 6 . 3 , Page 260 \n \n ” ) ;
12
13 s = 0.2 ; // [m]
14 d = 0.005 ; // [m]
15 rho = 7900 ; // [ kg /mˆ 3 ]
16 Cp = 460 ; // [ J / kg K ]
17
18 T_air = 20 ; // [ C ]
19 // For 430 C t o 330 C
20 T_avg = 380 ; // [ C ]
21 Tm = ( T_avg + T_air ) /2 ; // [ C ]
22
23

81
24 v = 34.85*10^ -6 ; // [mˆ2/ s ]
25 Pr = 0.680 ;
26 k = 0.0393 ; // [W/m K ]
27
28 Re = 9.81*1/(273+ Tm ) *( T_avg - T_air ) *( s ^3) /( v ^2) * Pr ;
29
30 // From eqn 6 . 2 . 3 1
31 Nu = 0.68 + 0.670*( Re ^(1/4) ) /[1+(0.492/ Pr ) ^(4/9)
]^(4/9) ;
32
33 h = Nu * k / s ; // [W/mˆ2 K ]
34 t1 = rho * s * s * d * Cp /(( s ^2) *2* h ) * log ((430 - T_air ) /(330 -
T_air ) ) ; // [ s ]
35 printf ( ” Time r e q u i r e d f o r t h e p l a t e t o c o o l from 430
C t o 330 C i s %f s \n ” , t1 ) ;
36
37 // f o r 330 t o 230
38 h2 = 7.348 ; // [W/mˆ2 K ]
39 t2 = rho * s * s * d * Cp /(( s ^2) *2* h2 ) * log ((330 - T_air ) /(230 -
T_air ) ) ; // [ s ]
40 printf ( ” Time r e q u i r e d f o r t h e p l a t e t o c o o l from 330
C t o 230 C i s %f s \n ” , t2 ) ;
41
42 // f o r 230 t o 130
43 h3 = 6.780; // [W/mˆ2 K ]
44 t3 = rho * s * s * d * Cp /(( s ^2) *2* h3 ) * log ((230 - T_air ) /(130 -
T_air ) ) ; // [ s ]
45 printf ( ” Time r e q u i r e d f o r t h e p l a t e t o c o o l from 230
C t o 130 C i s %f s \n ” , t3 ) ;
46
47 // T o t a l t i m e
48
49 time = t1 + t2 + t3 ;
50 minute = time /60;
51 printf ( ” Hence , t i m e r e q u i r e d f o r t h e p l a t e t o c o o l
from 430 C t o 130 C \n = %f s \n = %f min ” , time ,
minute ) ;

82
 Scilab code Exa 6.4 True air temperature

1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 6
6 // Heat T r a n s f e r by N a t u r a l C o n v e c t i o n
7
8
9 // Example 6 . 4
10 // Page 264
11 printf ( ” Example 6 . 4 , Page 264 \n \n ” ) ;
12
13 D = 0.006 ; // [m]
14 e = 0.1 ;
15 Ti = 800 ; // [ C ]
16 Ta = 1000 ; // [ C ]
17 // Rate a t which h e a t g a i n e d = n e t r a d i a n t h e a t ,
g i v e s h ∗ ( Ta−800) = 1 3 0 6 . 0 ; // [W/mˆ 2 ]
18
19 // U s i n g t r i a l and e r r o r method
20 // T r i a l 1
21 printf ( ” T r i a l 1 \n ” ) ;
22 // L e t Ta = 1 0 0 0 d e g r e e C
23 printf ( ” L e t Ta = 1 0 0 0 0 C \n ” ) ;
24
25 Tm = ( Ta + Ti ) /2;
26 // From t a b l e A. 2
27 v = 155.1*10^ -6 ; // [mˆ2/ s ]
28 k = 0.0763 ; // [W/m K ]
29 Pr = 0.717 ;
30
31 Gr = 9.81*1/1173*(200* D ^3) /( v ^2) ;

83
32 Ra = Gr * Pr ;
33
34 // From eqn 6 . 3 . 2
35 Nu = 0.36 + 0.518*( Ra ^(1/4) ) /[1+(0.559/ Pr ) ^(9/16)
]^(4/9) ;
36 h = Nu * k / D ;
37 x = h *( Ta - Ti ) ; // [W/mˆ 2 ]
38 printf ( ” V a l u e o f h ( Ta−800) = %f W/mˆ 2 , which i s much
l a r g e r t h a n t h e r e q u i r e d v a l u e o f 1 3 0 6 W/mˆ2 \n ”
,x ) ;
39
40 // T r i a l 2
41 printf ( ” \ n T r i a l 2 \n ” ) ;
42 // L e t Ta = 900
43 printf ( ” L e t Ta = 900 C \n ” ) ;
44 Ra2 = 6.42 ;
45 Nu2 = 0.9841 ;
46 h2 = 12.15 ;
47 x2 = h2 *(900 -800) ;
48 printf ( ” V a l u e o f h ( Ta−800) = %f W/mˆ 2 , which i s a
l i t t l e l e s s t h a n t h e r e q u i r e d v a l u e o f 1 3 0 6 W/mˆ2
\n ” , x2 ) ;
49
50 // T r i a l 3
51 printf ( ” \ n T r i a l 3 \n ” ) ;
52 // L e t Ta = 910
53 printf ( ” L e t Ta = 910 C \n ” ) ;
54 Ra3 = 6.93 ;
55 Nu3 = 0.9963 ;
56 h3 = 12.33 ;
57 x3 = h3 *(910 -800) ;
58 printf ( ” V a l u e o f h ( Ta−800) = %f W/mˆ2 \ n T h i s v a l u e
i s l i t t l e more t h a n t h e r e q u i r e d v a l u e o f 1 3 0 6 W/
mˆ2 \n ” , x3 ) ;
59 // I n t e r p o l a t i o n
60 T = 900 + (910 -900) *(1306 - x2 ) /( x3 - x2 ) ;
61 printf ( ” \ nThe c o r r e c t v a l u e o f Ta o b t a i n e d by
i n t e r p o l a t i o n i s %f C” ,T ) ;

84
 Scilab code Exa 6.5 Rate of heat flow by natural convection

1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 6
6 // Heat T r a n s f e r by N a t u r a l C o n v e c t i o n
7
8
9 // Example 6 . 5
10 // Page 269
11 printf ( ” Example 6 . 5 , Page 269 \n \n ” ) ;
12
13 T_p = 75 ; // T e m p e r a t u r e o f a b s o r b e r p l a t e , d e g r e e
C
14 T_c = 55 ; // T e m p e r a t u r e o f g l a s s c o v e r , d e g r e e C
15 L = 0.025 ; // [m]
16
17 H = 2 ; // [m]
18 Y = 70 ; // d e g r e e
19
20 a = 19/180* %pi ; // [ R a d i a n s ]
21
22 r = H/L ;
23
24 T_avg = ( T_p + T_c ) /2+273 ; // [ K ]
25 // P r o p e r t i e s a t 65 d e g r e e C
26 k = 0.0294 ; // [W/m K ]
27 v = 19.50*10^ -6 ; // [mˆ2/ s ]
28 Pr = 0.695 ;
29
30 Ra = 9.81*(1/ T_avg ) *( T_p - T_c ) *( L ^3) /( v ^2) * Pr * cos ( a ) ;
31

85
32 // From eqn 6 . 4 . 3
33 Nu = 0.229*( Ra ) ^0.252;
34
35 h = Nu * k / L ; // [W/mˆ2 K ]
36
37 Rate = h *2*1*( T_p - T_c ) ; // [W]
38
39 printf ( ” Heat t r a n s f e r r a t e = %f W” , Rate ) ;

Scilab code Exa 6.6 Average Heat transfer coeffficient

1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 6
6 // Heat T r a n s f e r by N a t u r a l C o n v e c t i o n
7
8
9 // Example 6 . 6
10 // Page 270
11 printf ( ” Example 6 . 6 , Page 270 \n \n ” ) ;
12
13 T_air = 30 ; // [ C ]
14 D = 0.04 ; // [m]
15 T_s = 70 ; // s u r f a c e t e m p e r a t u r e , [ C ]
16 V = 0.3 ; // [m/ s ]
17
18 Tm = ( T_air + T_s ) /2 ; // [ C ]
19 // P r o p e r t i e s a t Tm
20 v = 17.95*10^ -6 ; // [mˆ2/ s ]
21 Pr = 0.698 ;
22 k = 0.0283 ; // [W/m K ]
23
24 Gr = 9.81*1/323*( T_s - T_air ) *( D ^3) / v ^2;

86
25 Re = V * D / v ;
26 X = Gr / Re ^2 ;
27 printf ( ” S i n c e Gr/Re ˆ2 = %f i s > 0 . 2 , we have a
combined c o n v e c t i o n s i t u a t i o n . \n\n ” ,X ) ;
28
29 // From Eqn 5 . 9 . 8
30 Nu_forced = 0.3 + 0.62*( Re ^0.5) *( Pr ^(1/3) ) /[[1+(0.4/
Pr ) ^(2/3) ]^(1/4) ]*[1+( Re /282000) ^(5/8) ]^(4/5) ;
31
32 // S u b s t i t u t i n g i n Eqn 6 . 5 . 1
33 Nu = Nu_forced *[1+6.275*( X ) ^(7/4) ]^(1/7) ;
34 h = Nu *( k / D ) ;
35 printf ( ” The A v e r a g e h e a t t r a n s f e r c o e f f i c i e n t = %f W
/mˆ2 K” ,h ) ;

87
 Chapter 7

Heat Exchangers

Scilab code Exa 7.1 Heat transfer coeffficient

1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 7
6 // Heat E x c h a n g e r s
7
8
9 // Example 7 . 1
10 // Page 285
11 printf ( ” Example 7 . 1 , Page 285 \n \n ” ) ;
12
13 h = 2000 ; // [W/mˆ2 K ]
14 // From T a b l e 7 . 1
15 U_f = 0.0001 ; // f o u l i n g f a c t o r , mˆ2K/W
16 h_f = 1/[1/ h + U_f ];
17 printf ( ” Heat t r a n s f e r c o e f f i c i e n t i n c l u d i n g t h e
e f f e c t o f f o u l u n g = %f W/mˆ2 K \n ” , h_f ) ;
18
19 p = (h - h_f ) / h *100;
20 printf ( ” P e r c e n t a g e r e d u c t i o n = %f \n ” ,p ) ;

88
 Scilab code Exa 7.2 Area of heat exchanger

1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 7
6 // Heat E x c h a n g e r s
7
8
9 // Example 7 . 2
10 // Page 294
11 printf ( ” Example 7 . 2 , Page 294 \n \n ” ) ;
12
13 m = 1000 ; // [ kg / h ]
14 Thi = 50 ; // [C]
15 The = 40 ; // [C]
16 Tci = 35 ; // [C]
17 Tce = 40 ; // [C]
18 U = 1000 ; // OHTC, W/mˆ2 K
19
20 // U s i n g Eqn 7 . 5 . 2 5
21 q = m /3600*4174*( Thi - The ) ; // [W]
22
23 delta_T = (( Thi - Tce ) -( The - Tci ) ) / log (( Thi - Tce ) /( The -
Tci ) ) ; // [ C ]
24 printf ( ” d e l t a T = %f \n\n ” , delta_T ) ;
25
26 // T1 = Th and T2 = Tc
27 R = ( Thi - The ) /( Tce - Tci ) ;
28 S = ( Tce - Tci ) /( Thi - Tci ) ;
29 // From f i g 7 . 1 5 ,
30 F =0.91 ;
31

89
32 printf ( ” Taking T1 = Th and T2 = Tc \n ” )
33 printf ( ”R = %f , S = %f \n ” ,R , S ) ;
34 printf ( ” Hence , F = %f \n \n ” ,F ) ;
35
36 // A l t e r n a t i v e l y , t a k i n g T1 = Tc and T2 = Th
37 R = ( Tci - Tce ) /( The - Thi ) ;
38 S = ( The - Thi ) /( Tci - Thi ) ;
39
40 // Again from f i g 7 . 1 5 ,
41 F =0.91 ;
42
43 printf ( ” Taking T1 = Tc and T2 = Th \n ” )
44 printf ( ”R = %f , S = %f \n ” ,R , S ) ;
45 printf ( ” Hence , F = %f \n ” ,F ) ;
46
47 A = q /( U * F * delta_T ) ;
48 printf ( ” \ nArea = %f mˆ2 ” ,A ) ;

Scilab code Exa 7.3 Mean temperature difference

1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 7
6 // Heat E x c h a n g e r s
7
8
9 // Example 7 . 3
10 // Page 295
11 printf ( ” Example 7 . 3 , Page 295 \n \n ” ) ;
12
13 // B e c a u s e o f c h a n g e o f p h a s e , Thi = The
14 Thi = 100 ; // [ C ] , S a t u r a t e d steam
15 The = 100 ; // [ C ] , Condensed steam

90
16 Tci = 30 ; // [ C ] , C o o l i n g w a t e r i n l e t
17 Tce = 70 ; // [ C ] , c o o l i n g w a t e r o u t l e t
18
19 R = ( Thi - The ) /( Tce - Tci ) ;
20 S = ( Tce - Tci ) /( Thi - Tci ) ;
21
22 // From f i g 7 . 1 6
23 F = 1;
24
25 // For c o u n t e r f l o w a r r a n g e m e n t
26 Tm_counter = (( Thi - Tce ) -( The - Tci ) ) / log (( Thi - Tce ) /(
The - Tci ) ) ; // [ C ]
27 // T h e r e f o r e
28 Tm = F * Tm_counter ;
29 printf ( ”Mean T e m p e r a a t u r e D i f f e r e n c e = %f C” , Tm )

Scilab code Exa 7.4.a Area of heat exchanger

1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 7
6 // Heat E x c h a n g e r s
7
8
9 // Example 7 . 4 ( a )
10 // Page 302
11 printf ( ” Example 7 . 4 ( a ) , Page 302 \n \n ” ) ;
12
13 // ( a )
14 printf ( ” ( a ) \n ” ) ;
15 // U s i n g Mean T e m p e r a t u r e D i f f e r e n c e a p p r o a c h
16 m_hot = 10 ; // [ kg / min ]
17 m_cold = 25 ; // [ kg / min ]

91
18 hh = 1600 ; // [W/mˆ2 K ] , Heat t r a n s f e r coefficient
on h o t s i d e
19 hc = 1600 ; // [W/mˆ2 K ] , Heat t r a n s f e r coefficient
on c o l d s i d e
20
21 Thi = 70 ; // [ C ]
22 Tci = 25 ; // [ C ]
23 The = 50 ; // [ C ]
24
25 // Heat T r a n s f e r Rate , q
26 q = m_hot /60*4179*( Thi - The ) ; // [W]
27
28 // Heat g a i n e d by c o l d w a t e r = h e a t l o s t by t h e h o t
water
29 Tce = 25 + q *1/( m_cold /60*4174) ; // [ C ]
30
31 // U s i n g e q u a t i o n 7 . 5 . 1 3
32 Tm = (( Thi - Tci ) -( The - Tce ) ) / log (( Thi - Tci ) /( The - Tce ) ) ;
// [ C ]
33 printf ( ”Mean T e m p e r a t u r e D i f f e r e n c e = %f C \n ” , Tm ) ;
34
35 U = 1/(1/ hh + 1/ hc ) ; // [W/mˆ2 K ]
36 A = q /( U * Tm ) ; // Area , [mˆ 2 ]
37 printf ( ” Area o f Heat E x c h a n g e r = %f mˆ2 \n ” ,A ) ;

Scilab code Exa 7.4.b Exit temperature of hot and cold streams

1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 7
6 // Heat E x c h a n g e r s
7
8

92
 9 // Example 7 . 4 ( b )
10 // Page 302
11 printf ( ” Example 7 . 4 ( b ) , Page 302 \n \n ” ) ;
12
13 // U s i n g Mean T e m p e r a t u r e D i f f e r e n c e a p p r o a c h
14 m_hot = 10 ; // [ kg / min ]
15 m_cold = 25 ; // [ kg / min ]
16 hh = 1600 ; // [W/mˆ2 K ] , Heat t r a n s f e r c o e f f i c i e n t
on h o t s i d e
17 hc = 1600 ; // [W/mˆ2 K ] , Heat t r a n s f e r coefficient
on c o l d s i d e
18
19 Thi = 70 ; // [ C ]
20 Tci = 25 ; // [ C ]
21 The = 50 ; // [ C ]
22
23 // Heat T r a n s f e r Rate , q
24 q = m_hot /60*4179*( Thi - The ) ; // [W]
25
26 // Heat g a i n e d by c o l d w a t e r = h e a t l o s t by t h e h o t
water
27 Tce = 25 + q *1/( m_cold /60*4174) ; // [ C ]
28
29 // U s i n g e q u a t i o n 7 . 5 . 1 3
30 Tm = (( Thi - Tci ) -( The - Tce ) ) / log (( Thi - Tci ) /( The - Tce ) ) ;
// [ C ]
31 U = 1/(1/ hh + 1/ hc ) ; // [W/mˆ2 K ]
32 A = q /( U * Tm ) ; // Area , [mˆ 2 ]
33
34 m_hot = 20 ; // [ kg / min ]
35 // Flow r a t e on h o t s i d e i . e . ’ hh ’ i s d o u b l e d
36 hh = 1600*2^0.8 ; // [W/mˆ2 K ]
37 U = 1/(1/ hh + 1/ hc ) ; // [W/mˆ2 K ]
38 m_hC_ph = m_hot /60*4179 ; // [W/K ]
39 m_cC_pc = m_cold /60*4174 ; // [W/K ]
40 // T h e r e f o r e
41 C = m_hC_ph / m_cC_pc ;
42 NTU = U * A / m_hC_ph ;

93
43 printf ( ”NTU = %f \n ” , NTU ) ;
44
45 // From e q u a t i o n 7 . 6 . 8
46 e = [1 - exp ( -(1+ C ) * NTU ) ]/(1+ C ) ;
47
48 // T h e r e f o r e ( Thi − The ) / ( Thi − T c i ) = e , we g e t
49 The = Thi - e *( Thi - Tci ) ; // [ C ]
50
51 // E q u a t i n g t h e h e a t l o s t by w a t e r t o h e a t g a i n e d by
c o l d w a t e r , we g e t
52 Tce = Tci + [ m_hC_ph *( Thi - The ) ]/ m_cC_pc ;
53 printf ( ” E x i t t e m p e r a t u r e o f c o l d and h o t s t r e a m a r e
%f C and %f C r e s p e c t i v e l y . ” ,Tce , The ) ;

Scilab code Exa 7.5 Exit Temperature

1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 7
6 // Heat E x c h a n g e r s
7
8
9 // Example 7 . 5
10 // Page 304
11 printf ( ” Example 7 . 5 , Page 304 \n \n ” ) ;
12
13 mc = 2000 ; // [ kg / h ]
14 Tce = 40 ; // [ C ]
15 Tci = 15 ; // [ C ]
16 Thi = 80 ; // [ C ]
17 U = 50 ; // OHTC, [W/mˆ2 K ]
18 A = 10 ; // Area , [mˆ 2 ]
19

94
20 // U s i n g e f f e c t i v e NTU method
21 // Assuming m c ∗ C pc = (m∗ C p ) s
22 NTU = U * A /( mc *1005/3600) ;
23 e = ( Tce - Tci ) /( Thi - Tci ) ;
24 // From f i g 7 . 2 3 , no v a l u e o f C i s found
c o r r e s p o n d i n g to the above v al u es , hence
a s s u m p t i o n was wrong .
25 // So , m h∗ C ph must be e q u a l t o (m∗ C p ) s ,
p r o c e e d i n g by t r a i l and e r r o r method
26
27
28 printf ( ”m h ( kg / h NTU C e
T he (C) T he (C) ( Heat B a l a n c e ) ” ) ;
29
30 mh = rand (1:5) ;
31 NTU = rand (1:5) ;
32 The = rand (1:5) ;
33 The2 = rand (1:5) ;
34
35 mh (1) = 200
36 NTU (1) = U * A /( mh (1) *1.161) ;
37 // C o r r e s p o n d i n g V a l u e s o f C and e from f i g 7 . 2 3
38 C = .416;
39 e = .78;
40 // From E q u a t i o n 7 . 6 . 2 Page 297
41 The (1) = Thi - e *( Thi - Tci )
42 // From Heat B a l a n c e
43 The2 (1) = Thi - mc *1005/3600*( Tce - Tci ) /( mh (1)
*1.161) ;
44 printf ( ” \n\n %i %. 3 f %. 3 f %. 3 f
%. 2 f %. 2 f ” , mh (1) , NTU (1) ,C ,e , The (1) , The2 (1) )
;
45
46 mh (2) = 250
47 NTU (2) = U * A /( mh (2) *1.161) ;
48 // C o r r e s p o n d i n g V a l u e s o f C and e from f i g 7 . 2 3
49 C = .520;
50 e = .69;

95
51 // From E q u a t i o n 7 . 6 . 2 Page 297
52 The (2) = Thi - e *( Thi - Tci )
53 // From Heat B a l a n c e
54 The2 (2) = Thi - mc *1005/3600*( Tce - Tci ) /( mh (2)
*1.161) ;
55 printf ( ” \n\n %i %. 3 f %. 3 f %. 3 f
%. 2 f %. 2 f ” , mh (2) , NTU (2) ,C ,e , The (2) , The2 (2) )
;
56
57 mh (3) = 300
58 NTU (3) = U * A /( mh (3) *1.161) ;
59 // C o r r e s p o n d i n g V a l u e s o f C and e from f i g 7 . 2 3
60 C = .624;
61 e = .625;
62 // From E q u a t i o n 7 . 6 . 2 Page 297
63 The (3) = Thi - e *( Thi - Tci )
64 // From Heat B a l a n c e
65 The2 (3) = Thi - mc *1005/3600*( Tce - Tci ) /( mh (3)
*1.161) ;
66 printf ( ” \n\n %i %. 3 f %. 3 f %. 3 f
%. 2 f %. 2 f ” , mh (3) , NTU (3) ,C ,e , The (3) , The2 (3) )
;
67
68 mh (4) = 350
69 NTU (4) = U * A /( mh (4) *1.161) ;
70 // C o r r e s p o n d i n g V a l u e s o f C and e from f i g 7 . 2 3
71 C = .728;
72 e = .57;
73 // From E q u a t i o n 7 . 6 . 2 Page 297
74 The (4) = Thi - e *( Thi - Tci )
75 // From Heat B a l a n c e
76 The2 (4) = Thi - mc *1005/3600*( Tce - Tci ) /( mh (4)
*1.161) ;
77 printf ( ” \n\n %i %. 3 f %. 3 f %. 3 f
%. 2 f %. 2 f ” , mh (4) , NTU (4) ,C ,e , The (4) , The2 (4) )
;
78
79 mh (5) = 400

96
80 NTU (5) = U * A /( mh (5) *1.161) ;
81 // C o r r e s p o n d i n g V a l u e s o f C and e from f i g 7 . 2 3
82 C = .832;
83 e = .51;
84 // From E q u a t i o n 7 . 6 . 2 Page 297
85 The (5) = Thi - e *( Thi - Tci )
86 // From Heat B a l a n c e
87 The2 (5) = Thi - mc *1005/3600*( Tce - Tci ) /( mh (5)
*1.161) ;
88 printf ( ” \n\n %i %. 3 f %. 3 f %. 3 f
%. 2 f %. 2 f ” , mh (5) , NTU (5) ,C ,e , The (5) , The2 (5) )
;
89
90 clf () ;
91 plot ( mh , The , mh , The2 ,[295 295 200] ,[0 39.2 39.2])
92 xtitle ( ’ The v s mh ’ , ’mh ( kg / h r ) ’ , ’ The (C) ’ ) ;
93 printf ( ” \n\n From t h e p l o t , v a l u e o f mh = 295 kg / h r
and c o r r e s p o n d i n g l y The = 3 9 . 2 C” )

97
 Chapter 8

Condensation and boiling

Scilab code Exa 8.1 Average Heat Transfer Coefficient

1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 8
6 // C o n d e n s a t i o n and B o i l i n g
7
8
9 // Example 8 . 1
10 // Page 318
11 printf ( ” Example 8 . 1 , Page 318 \n \n ” ) ;
12 Ts = 80 ; // [ C ]
13 Tw = 70 ; // [ C ]
14 L = 1 ; // [m]
15 g = 9.8 ; // [m/ s ˆ 2 ]
16
17 // Assuming c o n d e n s a t e f i l m i s l a m i n a r and Re < 30
18 Tm = ( Ts + Tw ) /2 ;
19 // From t a b l e A. 1
20 rho = 978.8 ; // [ kg /mˆ 3 ]
21 k = 0.672 ; // [W/m K ]

98
22 u = 381 *10^ -6 ; // [ kg /m s ]
23 v = u / rho ;
24 // At 80 C ,
25 lambda = 2309 ; // [ kJ / kg ]
26 // S u b s t i t u t i n g i n eqn 8 . 3 . 9 , we g e t
27 h = 0.943*[( lambda *1000*( rho ^2) * g *( k ^3) ) /(( Ts - Tw ) * u *
L ) ]^(1/4) ; // [W/mˆ2 K ]
28
29 rate = h * L *( Ts - Tw ) /( lambda *1000) ; // [ kg /m s ]
30 Re = 4* rate / u ;
31 printf ( ” Assuming c o n d e n s a t e f i l m i s l a m i n a r and Re <
30 \n ” ) ;
32 printf ( ” h = %f W/mˆ2 K\n ” ,h ) ;
33 printf ( ” Re L = %f \n ” , Re ) ;
34 printf ( ” I n i t i a l a s s u m p t i o n was wrong , Now
c o n s i d e r i n g t h e e f f e c t o f r i p p l e s , we g e t \n ” ) ;
35
36 // S u b s t i t u t i n g h = Re ∗ ( lambda ∗ 1 0 0 0 ) ∗u / ( 4 ∗ L ∗ ( Ts−Tw) )
, i n eqn 8 . 3 . 1 2
37 Re = [[[4* L *( Ts - Tw ) * k /( lambda *1000* u ) *( g /( v ^2) )
^(1/3) ]+5.2]/1.08]^(1/1.22) ;
38 // From eqn 8 . 3 . 1 2
39 h = [ Re /(1.08*( Re ^1.22) -5.2) ]* k *(( g / v ^2) ^(1/3) ) ; //
[W/mˆ2 K ]
40 m = h * L *10/( lambda *1000) ; // r a t e o f c o n d e n s a t i o n ,
[ kg /m s ]
41
42 printf ( ”Re = %f \n ” , Re ) ;
43 printf ( ” Heat T r a n s f e r C o f f i c i e n t = %f W/mˆ2 K \n ” ,h )
;
44 printf ( ” Rate o f c o n d e n s a t i o n = %f kg /m s ” ,m ) ;

Scilab code Exa 8.2 Average heat transfer coefficient and film Reynolds
number

99
 1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 8
6 // C o n d e n s a t i o n and B o i l i n g
7
8
9 // Example 8 . 2
10 // Page 321
11 printf ( ” Example 8 . 2 , Page 321 \n \n ” ) ;
12
13 Ts = 262 ; // [ K ]
14 D = 0.022 ; // [m]
15 Tw = 258 ; // [ K ]
16
17 Tm = ( Ts + Tw ) /2;
18 // P r o p e r t i e s a t Tm
19 rho = 1324 ; // [ kg /mˆ 3 ]
20 k = 0.1008 ; // [W/m K ]
21 v = 1.90*10^ -7 // [mˆ2/ s ] ;
22 lambda = 215.1*10^3 ; // [ J / kg ]
23 g = 9.81 ; // [m/ s ˆ 2 ]
24 u = v * rho ; // V i s c o s i t y
25
26 // From eqn 8 . 4 . 1
27 h = 0.725*[ lambda *( rho ^2) * g *( k ^3) /(( Ts - Tw ) * u * D )
]^(1/4) ;
28
29 rate = h * %pi * D *( Ts - Tw ) / lambda ; // [ kg / s m]
30 Re = 4* rate / u ;
31
32 printf ( ” Heat t r a n s f e r c o e f f i c i e n t = %f W/mˆ2 K\n ” ,h )
;
33 printf ( ” C o n d e n s a t i o n r a t e p e r u n i t l e n g t h = %f kg / s
m \n ” , rate ) ;
34 printf ( ” Film R e y n o l d s number = %f \n ” , Re ) ;

100
 Scilab code Exa 8.3 Length of the tube

1 clear ;
2 clc ;
3
4 // A TeTwtbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 8
6 // C o n d e n s a t i o n and B o i l i n g
7
8
9 // ETwample 8 . 3
10 // Page 322
11 printf ( ” Example 8 . 3 , Page 322 \n \n ” ) ;
12
13 m = 25/60 ; // [ kg / s e c ]
14 ID = 0.025 ; // [m]
15 OD = 0.029 ; // [m]
16 Tci = 30 ; // [ C ]
17 Tce = 70 ; // [ C ]
18 g = 9.8 ; // [m/ s ˆ 2 ]
19
20 Ts = 100 ; // [ C ]
21 // Assuming 5 . 3 . 2 i s v a l i d , p r o p e r t i e s a t 50 C
22 // P r o p e r t i e s a t Tm
23 rho = 988.1 ; // [ kg /mˆ 3 ]
24 k = 0.648 ; // [W/m K ]
25 v = 0.556*10^ -6 // [mˆ2/ s ] ;
26 Pr = 3.54 ;
27 Re = 4* m /( %pi * ID * rho * v ) ;
28 // From eqn 4 . 6 . 4 a
29 f = 0.005635;
30 // From eqn 5 . 3 . 2
31 Nu = 198.39 ;
32 h = Nu * k / ID ;

101
33
34 // Assuming a v e r a g e w a l l t e m p e r a t u r e = 90 C
35 Tw = 90 ; // [ C ]
36 Tm = ( Tw + Ts ) /2;
37 // P r o p e r t i e s a t Tm
38 // P r o p e r t i e s a t Tm
39 rho = 961.9 ; // [ kg /mˆ 3 ]
40 k = 0.682 ; // [W/m K ]
41 u = 298.6*10^ -6 ; // [ kg /m s ]
42 lambda = 2257*10^3 ; // [ J / kg ]
43
44 h = 0.725*[ lambda *( rho ^2) * g *( k ^3) /(( Ts - Tw ) * u * OD )
]^(1/4) ;
45 // E q u a t i n g t h e h e a t f l o w from t h e c o n d e n s i n g steam
t o t h e t u b e w a l l , t o t h e h e a t f l o w from t h e t u b e
wall to the f l o w i n g water .
46 // S o l v i n g t h e s i m p l i f i e d e q u a t i o n
47 function [ f ] = temp ( Tw )
48 f =(100 - Tw ) ^(3/4) -8.3096/[ log (( Tw - Tci ) /( Tw - Tce ) )
];
49 funcprot (0) ;
50 endfunction
51
52 T = fsolve ( Tw , temp ) ;
53 printf ( ” T e m p e r a t u r e o b t a i n e d from t r i a l and e r r o r =
%f C \n ” ,T ) ;
54
55 // T h e r e f o r e
56 hc = 21338.77/(100 - T ) ^(1/4) ; // [W/mˆ2 K ]
57 printf ( ” h c = %f W/mˆ2 K \n ” , hc ) ;
58
59 // Now , e q u a t i n g t h e h e a t f l o w i n g from t h e
c o n d e n s i n g steam t o t h e t u b e w a l l t o t h e h e a t
g a i n e d by t h e water , we have
60 function [ g ] = lngth ( l )
61 g = hc *( %pi * OD * l ) *(100 - T ) -m *4174*( Tce - Tci ) ;
62 funcprot (0) ;
63 endfunction

102
64
65 l = 0; // ( i n i t i a l
g u e s s , assumed v a l u e f o r f s o l v e
function )
66 L = fsolve (l , lngth ) ;
67 printf ( ” \ nLength o f t h e t u b e = %f m \n ” ,L ) ;

Scilab code Exa 8.4 boiling regions

1 clear ;
2 clc ;
3
4 // P r o p e r t i e s a t (Tw+Ts ) /2 = 1 0 0 . 5 d e g r e e c e l s i u s
5 deltaT1 = 1; // i n d e g r e e c e l s i u s
6 p1 = 7.55 e -4; // [ Kˆ( −1) p1 i s c o e f f i c i e n t
of cubical expansion
7 v1 = 0.294 e -6; // [mˆ2/ s e c ] viscosity
at 100.5 degree c e l s i u s
8 k1 = 0.683; // [W/m−k ] t h e r m a l
conductivity
9 Pr1 = 1.74; // P r a n d t l number
10 g = 9.81; // a c c e l e r a t i o n due t o
gravity
11 L = 0.14 e -2; // d i a m e t e r i n m e t e r s
12 // P r o p e r t i e s a t (Tw+Ts ) /2 = 1 0 2 . 5
13 deltaT2 = 5; // i n d e g r e e c e l s i u s
14 p2 = 7.66 e -4; // [ Kˆ( −1) p1 i s c o e f f i c i e n t
of cubical expansion
15 v2 = 0.289 e -6; // [mˆ2/ s e c ] v i s c o s i t y at
102.5 degree c e l s i u s
16 k2 = 0.684; // [W/m−k ] t h e r m a l
conductivity
17 Pr2 = 1.71; // P r a n d t l number
18 // P r o p e r t i e s a t (Tw+Ts ) /2 =105
19 deltaT3 = 10; // i n d e g r e e c e l s i u s
20 p3 = 7.80 e -4; // [ Kˆ( −1) p1 i s c o e f f i c i e n t

103
 of cubical expansion
21 v3 = 0.284 e -6; // [mˆ2/ s e c ] v i s c o s i t y at
105 d e g r e e c e l s i u s
22 k3 = 0.684; // [W/m−k ] t h e r m a l
conductivity
23 Pr3 = 1.68; // P r a n d t l number
24
25 function [ Ra ]= Rayleigh_no (p , deltaT ,v , Pr )
26 Ra = [( p * g * deltaT * L ^3) /( v ^2) ]* Pr ;
27 funcprot (0) ;
28 endfunction
29
30 function [ q ] = flux (k , deltaT , Rai , v )
31 q =( k / L ) *( deltaT ) *{0.36+(0.518* Rai ^(1/4) )
/[1+(0.559/ v ) ^(9/16) ]^(4/9) };
32 funcprot (0) ;
33 endfunction
34
35 Ra = Rayleigh_no ( p1 , deltaT1 , v1 , Pr1 ) ;
36 q1 = flux ( k1 , deltaT1 , Ra , Pr1 ) ;
37 printf ( ” \n q /A = %. 1 f W/mˆ2 a t (Tw−Ts ) =1” , q1 ) ;
38 Ra = Rayleigh_no ( p2 , deltaT2 , v2 , Pr2 ) ;
39 q2 = flux ( k2 , deltaT2 , Ra , Pr2 ) ;
40 printf ( ” \n q /A = %. 1 f W/mˆ2 a t (Tw−Ts ) =5” , q2 ) ;
41 Ra = Rayleigh_no ( p3 , deltaT3 , v3 , Pr3 ) ;
42 q3 = flux ( k3 , deltaT3 , Ra , Pr3 ) ;
43 printf ( ” \n q /A = %. 1 f W/mˆ2 a t (Tw−Ts ) =10” , q3 ) ;
44
45 // At 100 d e g r e e c e l s i u s
46 Cpl = 4.220; // [ kJ / kg ]
47 lamda = 2257; // [ kJ / kg ]
48 ul = 282.4 e -6; // v i s c o s i t y i s i n kg /m−s e c
49 sigma = 589 e -4; // S u r f a c e t e n s i o n i s i n N/m
50 pl = 958.4; // d e n s i t y i n kg /mˆ3
51 pv =0.598; // d e n s i t y o f v a p o u r i n kg /mˆ3
52 deltap = pl - pv ;
53 Prl = 1.75; // P r a n d t l no . o f l i q u i d
54 Ksf = 0.013;

104
55 function [ q1 ]= heat_flux ( deltaT )
56 q1 =141.32* deltaT ^3;
57 funcprot (0) ;
58 endfunction
59
60 printf ( ” \n q /A a t d e l t a T = 5 d e g r e e c e l s i u s = %. 1 f W
/mˆ2 ” , heat_flux (5) ) ;
61 printf ( ” \ nq /A a t d e l t a T = 10 d e g r e e c e l s i u s = %. 1 f W
/mˆ2 ” , heat_flux (10) ) ;
62 printf ( ” \n q /A a t d e l t a T =20 d e g r e e c e l s i u s = %. 1 f W
/mˆ2 ” , heat_flux (20) ) ;
63 // q i = [ h e a t f l u x ( 5 ) , h e a t f l u x ( 1 0 ) , h e a t f l u x ( 2 0 ) ] ;
64 q = [ q1 q2 q3 ];
65 i =1;
66 while i <=10
67 T(i)=i;
68 ql ( i ) = heat_flux ( i ) ;
69 i = i +1;
70 end
71 plot2d ([1 5 10] , q ) ;
72 plot2d (T , ql ) ;
73 xtitle ( ” B o i l i n g c u r v e ” ,” (Tw − Ts ) d e g r e e c e l s i u s ” ,”
Heat f l u x , ( q /A)W/mˆ2 ” ) ;
74 L1 = ( L /2) *[ g *( pl - pv ) / sigma ]^(1/2) ;
75 printf ( ” \n Peak h e a t f l u x L = %. 3 f ” , L1 ) ;
76 f_L = 0.89+2.27* exp ( -3.44* L1 ^(0.5) ) ;
77 printf ( ” \n f ( l ) = %. 4 f ” , f_L ) ;
78 q2 = f_L *{( %pi /24) * lamda *10^(3) * pv ^(0.5) *[ sigma * g *(
pl - pv ) ]^(0.25) };
79 printf ( ” \n q /A = %. 3 e W/mˆ2 ” , q2 ) ;
80
81 Tn = poly ([0] , ’ Tn ’ ) ;
82 Tn1 = roots (141.32* Tn ^3 - q2 ) ;
83 printf ( ” \n Tw−Ts = %. 1 f d e g r e e c e l s i u s ” , Tn1 (3) ) ;
84
85
86
87 printf ( ” \n\n Minimum h e a t f l u x ” ) ;

105
 88 q3 = 0.09* lamda *10^3* pv *[ sigma * g *( pl - pv ) /( pl + pv ) ^(2)
]^(0.25) ;
89 printf ( ” \n q /A = %d W/mˆ2 ” , q3 ) ;
90 printf ( ” \n\n S t a b l e f i l m b o i l i n g ” ) ;
91 Ts1 = 140; // s u r f a c e t e m p e r a t u r e i n d e g r e e
celsius
92 Ts2 = 200; // s u r f a c e t e m p e r a t u r e i n d e g r e e
celsius
93 Ts3 = 600; // s u r f a c e t e m p e r a t u r e i n d e g r e e
celsius
94 Twm1 = (140+100) /2; // Mean f i l m t e m p e r a t u r e
95 // p r o p e r t i e s o f steam a t 120 d e g r e e c e l s i u s and
1 . 0 1 3 bar
96 kv = 0.02558; // t h e r m a l c o n d u c t i v i t y i n W/mK
97 pv1 = 0.5654; // v a p o r d e n s i t y i n kg /mˆ3
98 uv =13.185*10^( -6) ; // v i s c o s i t y o f v a p o u r i n kg /m
sec
99 lamda1 = (2716.1 -419.1) *10^(3) ; // L a t e n t h e a t o f
f u s i o n i n J / kg
100 hc = 0.62*[( kv ^3) * pv *( pl - pv ) * g * lamda1 /( L * uv
*(140 -100) ) ]^(0.25) ;
101 printf ( ” \n hc = %. 2 f W/mˆ2 ” , hc ) ;
102 qrad = 5.67*10^( -8) *(413^4 - 373^4) /[(1/0.9) +1 -1];
103 printf ( ” \n q /A due t o r a d i a t i o n = %. 2 f W/mˆ2 ” , qrad ) ;
104 hr = qrad /(413 -373) ;
105 printf ( ” \n h r = %. 2 f W/mˆ2 K ” , hr ) ;
106
107 printf ( ” \n S i n c e hr<hc ” ) ;
108 printf ( ” \n The t o t a l h e a t t r a n s f e r c o e f f i c i e n t ” ) ;
109 h = hc + 0.75* hr ;
110 printf ( ” h = %. 2 f W/mˆ2 K” ,h ) ;
111 printf ( ” \n T o t a l h e a t f l u x = %. 3 f W/mˆ2 K” ,h
*(140 -100) ) ;
112
113 hc_200 = 0.62*[( kv ^3) * pv *( pl - pv ) * g * lamda1 /( L * uv
*(200 -100) ) ]^(0.25) ;
114 qrad1 = 5.67*10^( -8) *(473^4 - 373^4) /[(1/0.9) +1 -1];
115 hr_200 = qrad1 /(200 -100) ;

106
116 printf ( ” \n\n hc = %. 2 f W/mˆ2 ” , hc_200 ) ;
117 printf ( ” \n h r = %. 2 f W/mˆ2 K” , hr_200 ) ;
118 printf ( ” \n q /A due t o r a d i a t i o n = %. 2 f W/mˆ2 ” , qrad1 )
;
119 h_200 = hc_200 +0.75* hr_200 ;
120 printf ( ” \n T o t a l h e a t f l u x = %d W/mˆ2 ” , h_200 *100) ;
121 hc_600 = 0.62*[( kv ^3) * pv *( pl - pv ) * g * lamda1 /( L * uv
*(600 -100) ) ]^(0.25) ;
122 qrad2 = 5.67*10^( -8) *(873^4 - 373^4) /[(1/0.9) +1 -1];
123 hr_600 = qrad1 /(600 -100)
124 printf ( ” \n\n hc = %. 2 f W/mˆ2 ” , hc_600 ) ;
125 printf ( ” \n h r = %. 2 f W/mˆ2 K” , hr_600 ) ;
126 printf ( ” \n q /A due t o r a d i a t i o n = %. 2 f W/mˆ2 ” , qrad2 )
;

Scilab code Exa 8.5 Initial heat transfer rate

1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 8
6 // C o n d e n s a t i o n and B o i l i n g
7
8
9 // Example 8 . 5
10 // Page 337
11 printf ( ” Example 8 . 5 , Page 337 \n \n ” ) ;
12
13 D = 0.02 ; // [m]
14 l = 0.15 ; // [m]
15 T = 500+273 ; // [ K ]
16 Tc = -196+273 ; // [ K ]
17 e = 0.4;
18 s = 5.670*10^ -8;

107
19 // Film b o i l i n g w i l l o c c u r , h e n c e eqn 8 . 7 . 9 i s
applicable
20 Tm = ( T + Tc ) /2;
21
22 // P r o p e r t i e s
23 k = 0.0349 ; // [W/m K ]
24 rho = 0.80 ; // [ kg /mˆ 3 ]
25 u = 23*10^ -6 ; // [ kg /m s ]
26
27 Cp_avg = 1.048 ; // [ kJ / kg J ]
28 rho_liq = 800 ; // [ kg /mˆ 3 ]
29 latent = 201*10^3 ; // [ J / kg ]
30
31 lambda = [ latent + Cp_avg *( Tm - Tc ) *1000]; // [ J / kg ]
32 h_c = 0.62*[(( k ^3) * rho *799.2*9.81* lambda ) /( D * u *( T - Tc
) ) ]^(1/4) ; // [W/mˆ2 K ]
33
34 // Taking t h e e m i s s i v i t y o f l i q u i d s u r f a c e t o be
u n i t y and u s i n g e q u a t i o n 3 . 9 . 1 , t h e e x c h a n g e o f
radiant heat f l u x
35 flux = s *( T ^4 - Tc ^4) /(1/ e +1/1 -1) ; // [W/mˆ 2 ]
36 h_r = flux /( T - Tc ) ;
37
38 // S i n c e h r < h c , t o t a l h e a t t r a n s f e r coefficient
i s d e t e r m i n e d from eqn 8 . 7 . 1 1
39 h = h_c +3/4* h_r ; // [W/mˆ2 K ]
40
41 flux_i = h *( T - Tc ) ;
42 Rate = flux_i * %pi * D * l ; // [W]
43
44 printf ( ” I n i t i a l h e a t f l u x = %f W/mˆ2 \n ” , flux_i ) ;
45 printf ( ” I n i t i a l h e a t t r a n s f e r r a t e = %f W” , Rate ) ;

108
 Chapter 9

Mass Transfer

Scilab code Exa 9.1 Composition on molar basis

1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 9
6 // Mass T r a n s f e r
7
8
9 // Example 9 . 1
10 // Page 349
11 printf ( ” Example 9 . 1 , Page 349 \n \n ” ) ;
12
13 w_a = 0.76 ;
14 w_b = 0.24 ;
15 m_a = 28 ; // [ kg / kg mole ]
16 m_b = 32 ; // [ kg / kg mole ]
17
18 x_a = ( w_a / m_a ) /( w_a / m_a + w_b / m_b ) ;
19 x_b = ( w_b / m_b ) /( w_a / m_a + w_b / m_b ) ;
20 printf ( ” The m o l a r f r a c t i o n s a r e g i v e n by \n ” ) ;
21 printf ( ” x a = %f\n ” , x_a ) ;

109
22 printf ( ” x b = %f ” , x_b ) ;

Scilab code Exa 9.2 Diffusion coefficient of napthalene

1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 9
6 // Mass T r a n s f e r
7
8
9 // Example 9 . 2
10 // Page 350
11 printf ( ” Example 9 . 2 , Page 350 \n \n ” ) ;
12
13 // From T a b l e 9 . 1 a t 1 atm and 25 C
14 Dab = 0.62*10^ -5 ; // [mˆ2/ s ]
15 // T h e r e f o r e a t 2 atm and 50 C
16 Dab2 = Dab *(1/2) *(323/298) ^1.5 ;
17 printf ( ”Dab a t 2 atm & 50 C = %e mˆ2/ s ” , Dab2 ) ;

Scilab code Exa 9.3.a Rate of hydrogen diffusion

1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 9
6 // Mass T r a n s f e r
7
8
9 // Example 9 . 3 ( a )

110
10 // Page 352
11 printf ( ” Example 9 . 3 ( a ) , Page 352 \n \n ” ) ;
12
13 t = 0.04 ; // [m]
14 A = 2 ; // [mˆ 2 ]
15 rho1 = 0.10 ;
16 rho2 = 0.01 ;
17 D_400 = 1.6*10^ -11 ; // a t 400K [mˆ2/ s ]
18
19 // Mass D i f f u s i o n i n s o l i d s o l u t i o n , a s s u m i n g F i c k s
law i s v a l i d & s t e a d y s t a t e and one d i m e n s i o n a l
diffusion
20
21 // S u b t i t u t i n g t h e v a l u e s i n eqn 9 . 3 . 3 , At 400 K
22
23 m_400 = A * D_400 *( rho1 - rho2 ) / t ; // [ kg / s ]
24 printf ( ” Rate o f d i f f u s i o n o f Hydrogen a t 400 K = %e
kg / s \n ” , m_400 ) ;

Scilab code Exa 9.3.b Rate of hydrogen diffusion

1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 9
6 // Mass T r a n s f e r
7
8
9 // Example 9 . 3 ( b )
10 // Page 352
11 printf ( ” Example 9 . 3 ( b ) , Page 352 \n \n ” ) ;
12
13 t = 0.04 ; // [m]
14 A = 2 ; // [mˆ 2 ]

111
15 rho1 = 0.10 ;
16 rho2 = 0.01 ;
17 D_1200 = 3.5*10^ -8 ; // a t 1 2 0 0 k [mˆ2/ s ]
18
19 // Mass D i f f u s i o n i n s o l i d s o l u t i o n , a s s u m i n g F i c k s
law i s v a l i d & s t e a d y s t a t e and one d i m e n s i o n a l
diffusion
20
21 // At 1 2 0 0 K
22 // From eqn 9 . 3 . 3
23
24 m_1200 = A * D_1200 *( rho1 - rho2 ) / t ;
25 printf ( ” ( b ) Rate o f d i f f u s i o n o f Hydrogen a t 1 2 0 0 K
= %e kg / s \n ” , m_1200 ) ;

Scilab code Exa 9.4.a Rate of loss of ammonia

1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 9
6 // Mass T r a n s f e r
7
8
9 // Example 9 . 4 ( a )
10 // Page 356
11 printf ( ” Example 9 . 4 ( a ) , Page 356 \n \n ” ) ;
12
13 L = 1 ; // [m]
14 D = 0.005 ; // [m]
15 Pa1 = 1 ; // [ atm ]
16 Pa2 = 0 ;
17 R = 8314 ;
18 T = 298 ; // [ K ]

112
19
20 // Assuming E q u i m o l a l c o u n t e r d i f f u s i o n
21 // From T a b l e 9 . 1
22 Dab = 2.80*10^ -5 ; // [mˆ2/ s ]
23 // S u b s t i t u i n g i n eqn 9 . 4 . 1 2
24 Na = -[ Dab /( R * T ) *( Pa2 - Pa1 ) *(1.014*10^5) / L ]*( %pi *( D
/2) ^2) ;
25 R_NH3 = Na *17 ; // [ kg / s ]
26
27 printf ( ”Na = −Nb = %e ( kg mole ) /mˆ2 s \n ” , Na ) ;
28 printf ( ” Rate a t which ammonia i s l o s t t h r o u g h t h e
t u b e = %e kg / s \n ” , R_NH3 ) ;

Scilab code Exa 9.4.b Rate at which air enters the tank

1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 9
6 // Mass T r a n s f e r
7
8
9 // Example 9 . 4 ( b )
10 // Page 356
11 printf ( ” Example 9 . 4 ( b ) , Page 356 \n \n ” ) ;
12
13 L = 1 ; // [m]
14 D = 0.005 ; // [m]
15 Pa1 = 1 ; // [ atm ]
16 Pa2 = 0 ;
17 R = 8314 ;
18 T = 298 ; // [ K ]
19
20 // S i n c e t h e t a n k i s l a r g e and t h e p r e s s u r e and

113
 t e m p e r a t u r e a t t h e two e n d s o f t h e same t u b e a r e
same , we a r e a s s u m i n g E q u i m o l a l c o u n t e r d i f f u s i o n
21 // From T a b l e 9 . 1
22 Dab = 2.80*10^ -5 ; // [mˆ2/ s ]
23 // S u b s t i t u i n g i n eqn 9 . 4 . 1 2
24 Na = -[ Dab /( R * T ) *( Pa2 - Pa1 ) *(1.014*10^5) / L ]*( %pi *( D
/2) ^2) ;
25
26 // S i n c e e q u i m o l a l c o u n t e r d i f f u s i o n i s t a k i n g p l a c e
27 Nb = - Na ;
28 // t h e r e f o r e r a t e a t which a i r e n t e r s t h e t a n k
29 R_air = abs ( Nb ) *29 ; // [ kg / s ]
30
31 printf ( ” Rate a t which a i r e n t e r s t h e t a n k = %e kg / s ”
, R_air ) ;

Scilab code Exa 9.5 Rate of evaporation

1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 9
6 // Mass T r a n s f e r
7
8
9 // Example 9 . 5
10 // Page 359
11 printf ( ” Example 9 . 5 , Page 359 \n \n ” ) ;
12
13 // E v a p o r a t i o n o f water , one d i m e n s i o n a l
14 T_w = 20+273 ; // [ K ]
15 D = 0.04 ; // [m]
16 h = 0.20 ; // [m]
17 h_w = 0.03 ; // [m]

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18
19 P = 1.014*10^5; // [ Pa ]
20 R = 8314 ; // [ J / kg mole K ]
21 P_sat = 0.02339 ; // [ b a r ]
22 x_a1 = P_sat /1.014 ; // mole f r a c t i o n a t l i q −vap
interface
23 x_a2 = 0 ; // mole f r a c t i o n a t open t o p
24 c = P /( R * T_w ) ;
25 // From T a b l e 9 . 2
26 Dab = 2.422*10^ -5 ; // [mˆ2/ s ]
27
28 // S u b s t i t u t i n g a b o v e v a l u e s i n eqn 9 . 4 . 1 8
29 flux = 0.041626* Dab /0.17* log ((1 -0) /(1 - x_a1 ) ) ; // [ kg
mole /mˆ2 s ]
30 rate = flux *18*( %pi /4) *( D ^2) ;
31
32 printf ( ” Rate o f e v a p o r a t i o n o f w a t e r = %e kg / s ” , rate
);

Scilab code Exa 9.6 Rate of evaporation

1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 9
6 // Mass T r a n s f e r
7
8
9 // Example 9 . 6
10 // Page 364
11 printf ( ” Example 9 . 6 , Page 364 \n \n ” ) ;
12
13 l = 1; // l e n g t h , [m]
14 w = 0.25; // width , [m]

115
15 T = 293 ; // Temperature , [K ]
16 rho_infinity = 0; // [ kg /mˆ 3 ]
17 R = 8314; // [ J / kg K ]
18
19 // From T a b l e A. 2
20 v = 15.06*10^ -6; // [mˆ2/ s ]
21 // From T a b l e 9 . 2
22 Dab = 2.4224*10^ -5; // [mˆ2/ s ]
23 Re = 2.5/ v ;
24 Sc = v / Dab ;
25 // S i n c e Re > 3 ∗ 1 0 ˆ 5 , we may assume l a m i n a r boundary
layer
26 Sh = 0.664* Sc ^(1/3) * Re ^(1/2) ; // Sherwood number
27 h = Sh * Dab ;
28
29 p_aw = 2339; // S a t u r a t i o n p r e s s u r e o f w a t e r a t 20
d e g r e e C . [ N/mˆ 2 ]
30 rho_aw = p_aw /( R /18* T ) ; // [ kg /mˆ 3 ]
31 rho_a_inf = 0 ; // s i n c e a i r i n t h e f r e e s t r e a m i s
dry
32 m_h = h *(2* l * w ) *( rho_aw - rho_infinity ) ;
33 printf ( ” Rate o f e v a p o r a t i o n from p l a t e = %e kg / s ” ,
m_h ) ;

Scilab code Exa 9.7.a Mass transfer coefficient Colburn anology

1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 9
6 // Mass T r a n s f e r
7
8
9 // Example 9 . 7 ( a )

116
10 // Page 366
11 printf ( ” Example 9 . 7 ( a ) , Page 366 \n \n ” ) ;
12
13 D = 0.04 ; // [m]
14 V = 1.9 ; // [m/ s ]
15
16 // ( a ) C o l b u r n a n o l o g y and G n i e l i n s k i e q u a t i o n
17 // P r o p e r t i e s o f a i r a t 27 d e g r e e C
18 v = 15.718*10^ -6 ; // [mˆ2/ s ]
19 rho = 1.177 ; // [ kg /mˆ 3 ]
20 Pr = 0.7015 ;
21 Cp = 1005 ; // [ J / kg K ]
22 k = 0.02646 ; // [W/m K ]
23 // From T a b l e 9 . 2
24 Dab = 2.54 * 10^ -5 ; // [mˆ2/ s ]
25 Sc = v / Dab ;
26 Re = V * D / v ;
27 // The f l o w i s t u r b u l e n t and eqn 9 . 6 . 5 may be
applied
28 // l e t r = h /h m
29 r = rho * Cp *(( Sc / Pr ) ^(2/3) ) ;
30 // From B l a s i u s e q u a t i o n 4 . 6 . 4 a
31 f = 0.079* Re ^( -0.25) ;
32 // S u b s t i t u t i n g t h i s v a l u e i n t o G n i e l i n s k i e q u a t i o n
5.3.2
33 Nu = [( f /2) *( Re -1000) * Pr ]/[1+12.7*(( f /2) ^(1/2) ) *(( Pr
^(2/3) ) -1) ];
34 h = Nu * k / D ;
35 h_m = h / r ; // [m/ s ]
36
37 printf ( ”h m u s i n g C o l b u r n a n o l o g y and G n i e l i n s k i
e q u a t i o n = %f \n ” , h_m ) ;

Scilab code Exa 9.7.b Mass transfer coefficient Gnielinski equation

117
 1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 9
6 // Mass T r a n s f e r
7
8
9 // Example 9 . 7 ( b )
10 // Page 366
11 printf ( ” Example 9 . 7 ( b ) , Page 366 \n \n ” ) ;
12
13 D = 0.04 ; // [m]
14 V = 1.9 ; // [m/ s ]
15
16 // ( b ) mess t r a n s f e r c o r r e l a t i o n e q u i v a l e n t t o t h e
G l e i l i n s k i equation
17
18 // P r o p e r t i e s o f a i r a t 27 d e g r e e C
19 v = 15.718*10^ -6 ; // [mˆ2/ s ]
20 rho = 1.177 ; // [ kg /mˆ 3 ]
21 Pr = 0.7015 ;
22 Cp = 1005 ; // [ J / kg K ]
23 k = 0.02646 ; // [W/m K ]
24 // From T a b l e 9 . 2
25 Dab = 2.54 * 10^ -5 ; // [mˆ2/ s ]
26 Sc = v / Dab ;
27 Re = V * D / v ;
28
29 // From B l a s i u s e q u a t i o n 4 . 6 . 4 a
30 f = 0.079* Re ^( -0.25) ;
31
32 // S u b s t i t u t i n g i n eqn 9 . 6 . 7
33 Sh_D = [( f /2) *( Re -1000) * Sc ]/[1+12.7*(( f /2) ) *(( Sc
^(2/3) ) -1) ];
34 h_m1 = Sh_D * Dab / D ;
35
36 printf ( ” ( b ) h m = %f \n ” , h_m1 ) ;

118
 Scilab code Exa 9.7.c To show mass flux of water vapour is small

1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 9
6 // Mass T r a n s f e r
7
8
9 // Example 9 . 7 ( c )
10 // Page 366
11 printf ( ” Example 9 . 7 ( c ) , Page 366 \n \n ” ) ;
12
13 D = 0.04 ; // [m]
14 V = 1.9 ; // [m/ s ]
15
16 // ( c ) To show t h a t mass f l u x o f w a t e r i s v e r y s m a l l
compared t o t h e mass f l u x o f a i r f l o w i n g i n t h e
pipe
17 // P r o p e r t i e s o f a i r a t 27 d e g r e e C
18 v = 15.718*10^ -6 ; // [mˆ2/ s ]
19 rho = 1.177 ; // [ kg /mˆ 3 ]
20 Pr = 0.7015 ;
21 Cp = 1005 ; // [ J / kg K ]
22 k = 0.02646 ; // [W/m K ]
23 // From T a b l e 9 . 2
24 Dab = 2.54 * 10^ -5 ; // [mˆ2/ s ]
25 Sc = v / Dab ;
26 Re = V * D / v ;
27 // The f l o w i s t u r b u l e n t and eqn 9 . 6 . 5 may be
applied
28 // l e t r = h /h m
29 r = rho * Cp *(( Sc / Pr ) ^(2/3) ) ;

119
30 // From B l a s i u s e q u a t i o n 4 . 6 . 4 a
31 f = 0.079* Re ^( -0.25) ;
32
33 // From steam t a b l e
34 rho_aw = 1/38.77 ; // [ kg /mˆ 3 ]
35 // l e t X = ( m a /A) max
36 X = f * rho_aw ; // [ kg /mˆ2 s ]
37
38 // l e t Y = mass f l u x o f a i r i n p i p e = (m/A)
39 Y = rho * V ; // [ kg /mˆ2 s ]
40 ratio = X / Y ;
41 percent = ratio *100;
42
43 printf ( ” ( c ) ( m a /A) max / ( m a /A) = %f p e r c e n t Thus ,
mass f l u x o f w a t e r i s v e r y s m a l l compared t o t h e
mass f l u x o f a i r f l o w i n g i n t h e p i p e . ” , percent )
;

Scilab code Exa 9.8 Mass fraction

1 clear ;
2 clc ;
3
4 // A Textbook on HEAT TRANSFER by S P SUKHATME
5 // C h a p t e r 9
6 // Mass T r a n s f e r
7
8
9 // Example 9 . 8
10 // Page 369
11 printf ( ” Example 9 . 8 , Page 369 \n \n ” ) ;
12
13 V = 0.5 ; // [m/ s ]
14 T_h = 30 ; // [ C ]
15 T_c = 26 ; // [ C ]

120
16 Tm = ( T_h + T_c ) /2;
17 // From t a b l e A. 2
18 rho = 1.173 ; // [ kg /mˆ 3 ]
19 Cp = 1005 ; // [ J / kg K ]
20 k = 0.02654 ; // [W/m K ]
21
22 alpha = k /( rho * Cp ) ; // [mˆ2/ s ]
23
24 // From T a b l e 9 . 2 a t 301 K
25 Dab = 2.5584*10^ -5 ; // [mˆ2/ s ]
26 lambda = 2439.2*10^3 ; // [ J / kg ]
27
28 // S u b s t i t u t i n g i n e q u a t i o n 9 . 7 . 5
29 // l e t d i f f e r e n c e = rho aw −r h o a i n f i n i t y
30 difference = rho * Cp *(( alpha / Dab ) ^(2/3) ) *( T_h - T_c ) /
lambda ;
31
32 // From steam t a b l e
33 Psat = 3363;
34 rho_aw = Psat /(8314/18*299) ;
35 rho_inf = rho_aw - difference ;
36 x = rho_inf / rho ; // mole f r a c t i o n o f water vapour i n
a i r stream
37
38 PP = rho_inf *8314/18*303; // P a r t i a l pressure of
water vapour i n a i r stream
39 // From steam t a b l e p a r t i a l p r e s s u r e o f w a t e r v a p o u r
a t 30 C
40 PP_30 = 4246 ; // [ N/mˆ 2 ]
41
42 rel_H = PP / PP_30 ;
43 percent = rel_H *100;
44
45 printf ( ” R e l a t i v e h u m i d i t y = %f i . e . %f p e r c e n t ” ,
rel_H , percent ) ;

121