elementary principles of chemical processes_r. m. felder and r. w. rousseau.pdf

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Book Description

Title: Elementary Principles Of Chemical Processes

Author: R. M. Felder And R. W. Rousseau

Publisher: Wiley India Pvt. Ltd., New Delhi.

Edition: 3

Year: 2010

ISBN: 978-81-265-1582-0

1



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

Exa Example (Solved example)

Eqn Equation (Particular equation of the above book)

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

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

2



Contents

List of Scilab Codes 4

2 Introduction To Engineering Calculations 13

3 Processes and Process Variables 20

4 Fundamentals Of Material Balances 30

5 Single Phase Systems 62

6 Multiphase Systems 72

7 Energy And Energy Balances 89

8 Balances On Nonreactive Processes 101

9 Balances On Reactive Processes 115

11 Balances on Transient Processes 129

3



List of Scilab Codes

Exa 2.2.1 chapter 2 example 1 . . . . . . . . . . . . . . . . . . . 13Exa 2.3.1 chapter 2 example 2 . . . . . . . . . . . . . . . . . . . 14Exa 2.4.1 chapter 2 example 3 . . . . . . . . . . . . . . . . . . . 15Exa 2.5.2 chapter 2 example 4 . . . . . . . . . . . . . . . . . . . 16Exa 2.7.1 chapter 2 example 5 . . . . . . . . . . . . . . . . . . . 17Exa 2.7.2 chapter 2 example 6 . . . . . . . . . . . . . . . . . . . 19Exa 3.1.1 chapter 3 example 1 . . . . . . . . . . . . . . . . . . . 20Exa 3.1.2 chapter 3 example 2 . . . . . . . . . . . . . . . . . . . 21Exa 3.3.1 chapter 3 example 3 . . . . . . . . . . . . . . . . . . . 22Exa 3.3.2 chapter 3 example 4 . . . . . . . . . . . . . . . . . . . 24Exa 3.3.3 chapter 3 example 5 . . . . . . . . . . . . . . . . . . . 24Exa 3.3.4 chapter 3 example 6 . . . . . . . . . . . . . . . . . . . 26Exa 3.3.5 chapter 3 example 7 . . . . . . . . . . . . . . . . . . . 26

Exa 3.4.1 chapter 3 example 8 . . . . . . . . . . . . . . . . . . . 27Exa 3.4.2 chapter 3 example 9 . . . . . . . . . . . . . . . . . . . 28Exa 3.5.2 chapter 3 example 10 . . . . . . . . . . . . . . . . . . 28Exa 4.2.1 chapter 4 example 1 . . . . . . . . . . . . . . . . . . . 30Exa 4.2.2 chapter 4 example 2 . . . . . . . . . . . . . . . . . . . 31Exa 4.2.3 chapter 4 example 3 . . . . . . . . . . . . . . . . . . . 32Exa 4.2.4 chapter 4 example 4 . . . . . . . . . . . . . . . . . . . 33Exa 4.3.1 chapter 4 example 5 . . . . . . . . . . . . . . . . . . . 34Exa 4.3.2 chapter 4 example 6 . . . . . . . . . . . . . . . . . . . 35Exa 4.3.3 chapter 4 example 7 . . . . . . . . . . . . . . . . . . . 37Exa 4.3.5 chapter 4 example 8 . . . . . . . . . . . . . . . . . . . 37

Exa 4.4.1 chapter 4 example 9 . . . . . . . . . . . . . . . . . . . 39Exa 4.4.2 chapter 4 example 10 . . . . . . . . . . . . . . . . . . 42Exa 4.5.1 chapter 4 example 11 . . . . . . . . . . . . . . . . . . 43Exa 4.5.2 chapter 4 example 12 . . . . . . . . . . . . . . . . . . 46

4



Exa 4.6.1 chapter 4 example 13 . . . . . . . . . . . . . . . . . . 47

Exa 4.6.3 chapter 4 example 14 . . . . . . . . . . . . . . . . . . 49Exa 4.7.2 chapter 4 example 15 . . . . . . . . . . . . . . . . . . 50Exa 4.7.3 chapter 4 example 16 . . . . . . . . . . . . . . . . . . 52Exa 4.8.1 chapter 4 example 17 . . . . . . . . . . . . . . . . . . 54Exa 4.8.2 chapter 4 example 18 . . . . . . . . . . . . . . . . . . 55Exa 4.8.3 chapter 4 example 19 . . . . . . . . . . . . . . . . . . 56Exa 4.8.4 chapter 4 example 20 . . . . . . . . . . . . . . . . . . 58Exa 4.9.1 chapter 4 example 21 . . . . . . . . . . . . . . . . . . 59Exa 5.1.1 chapter 5 example 1 . . . . . . . . . . . . . . . . . . . 62Exa 5.2.1 chapter 5 example 2 . . . . . . . . . . . . . . . . . . . 63Exa 5.2.2 chapter 5 example 3 . . . . . . . . . . . . . . . . . . . 64

Exa 5.2.3 chapter 5 example 4 . . . . . . . . . . . . . . . . . . . 64Exa 5.2.4 chapter 5 example 5 . . . . . . . . . . . . . . . . . . . 65Exa 5.2.5 chapter 5 example 6 . . . . . . . . . . . . . . . . . . . 66Exa 5.3.1 chapter 5 example 7 . . . . . . . . . . . . . . . . . . . 67Exa 5.3.2 chapter 5 example 8 . . . . . . . . . . . . . . . . . . . 68Exa 5.4.1 chapter 5 example 9 . . . . . . . . . . . . . . . . . . . 69Exa 5.4.2 chapter 5 example 10 . . . . . . . . . . . . . . . . . . 70Exa 5.4.3 chapter 5 example 11 . . . . . . . . . . . . . . . . . . 71Exa 6.1.1 chapter 6 example 1 . . . . . . . . . . . . . . . . . . . 72Exa 6.3.1 chapter 6 example 2 . . . . . . . . . . . . . . . . . . . 73

Exa 6.3.2 chapter 6 example 3 . . . . . . . . . . . . . . . . . . . 74Exa 6.3.3 chapter 6 example 4 . . . . . . . . . . . . . . . . . . . 75Exa 6.4.1 chapter 6 example 5 . . . . . . . . . . . . . . . . . . . 76Exa 6.4.2 chapter 6 example 6 . . . . . . . . . . . . . . . . . . . 78Exa 6.5.1 chapter 6 example 7 . . . . . . . . . . . . . . . . . . . 79Exa 6.5.2 chapter 6 example 8 . . . . . . . . . . . . . . . . . . . 80Exa 6.5.3 chapter 6 example 9 . . . . . . . . . . . . . . . . . . . 82Exa 6.5.4 chapter 6 example 10 . . . . . . . . . . . . . . . . . . 83Exa 6.6.1 chapter 6 example 11 . . . . . . . . . . . . . . . . . . 84Exa 6.6.2 chapter 6 example 12 . . . . . . . . . . . . . . . . . . 85Exa 6.7.1 chapter 6 example 13 . . . . . . . . . . . . . . . . . . 86

Exa 7.2.1 chapter 7 example 1 . . . . . . . . . . . . . . . . . . . 89Exa 7.2.2 chapter 7 example 2 . . . . . . . . . . . . . . . . . . . 90Exa 7.4.1 chapter 7 example 3 . . . . . . . . . . . . . . . . . . . 90Exa 7.4.2 chapter 7 example 4 . . . . . . . . . . . . . . . . . . . 92Exa 7.5.1 chapter 7 example 5 . . . . . . . . . . . . . . . . . . . 92

5



AP 8 9.5.1.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 134

AP 9 9.4.1.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 134AP 10 9.3.1.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 134AP 11 9.1.2.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 134AP 12 9.1.1.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 135AP 13 8.5.5.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 135AP 14 8.5.2.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 135AP 15 8.5.1.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 135AP 16 8.4.4.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 136AP 17 8.4.3.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 136AP 18 8.4.2.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 136AP 19 8.4.1.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 136

AP 20 8.3.5.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 136AP 21 8.3.4.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 137AP 22 8.3.3.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 137AP 23 8.3.2.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 137AP 24 8.3.1.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 137AP 25 7.7.3.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 137AP 26 7.7.2.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 138AP 27 771.sci . . . . . . . . . . . . . . . . . . . . . . . . . . . 138AP 28 7.6.3.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 138AP 29 7.6.2.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 138


AP 41 6.5.3.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 141AP 42 6.5.2.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 141AP 43 6.5.1.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 141AP 44 6.4.2.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 142AP 45 6.4.1.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 142

7



AP 46 6.3.3.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 142


AP 58 5.2.2.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 145AP 59 5.2.1.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 145AP 60 5.1.1.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 145AP 61 4.9.1.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 146AP 62 4.8.4.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 146AP 63 4.8.3.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 146AP 64 4.8.2.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 146AP 65 4.8.1.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 146AP 66 4.7.3.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 147AP 67 4.7.2.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 147


AP 79 4.2.3.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 150AP 80 4.2.2.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 150AP 81 4.2.1.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 150AP 82 3.5.2.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 150AP 83 3.4.2.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 150

8



AP 84 3.4.1.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 151


AP 96 2.3.1.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 153AP 97 2.2.1.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 153

9



List of Figures

2.1 chapter 2 example 1 . . . . . . . . . . . . . . . . . . . . . . 142.2 chapter 2 example 2 . . . . . . . . . . . . . . . . . . . . . . 152.3 chapter 2 example 3 . . . . . . . . . . . . . . . . . . . . . . 152.4 chapter 2 example 4 . . . . . . . . . . . . . . . . . . . . . . 162.5 chapter 2 example 5 . . . . . . . . . . . . . . . . . . . . . . 172.6 chapter 2 example 6 . . . . . . . . . . . . . . . . . . . . . . 18

3.1 chapter 3 example 1 . . . . . . . . . . . . . . . . . . . . . . 213.2 chapter 3 example 2 . . . . . . . . . . . . . . . . . . . . . . 223.3 chapter 3 example 3 . . . . . . . . . . . . . . . . . . . . . . 233.4 chapter 3 example 4 . . . . . . . . . . . . . . . . . . . . . . 243.5 chapter 3 example 5 . . . . . . . . . . . . . . . . . . . . . . 253.6 chapter 3 example 6 . . . . . . . . . . . . . . . . . . . . . . 253.7 chapter 3 example 7 . . . . . . . . . . . . . . . . . . . . . . 263.8 chapter 3 example 8 . . . . . . . . . . . . . . . . . . . . . . 273.9 chapter 3 example 9 . . . . . . . . . . . . . . . . . . . . . . 283.10 chapter 3 example 10 . . . . . . . . . . . . . . . . . . . . . . 28

4.1 chapter 4 example 1 . . . . . . . . . . . . . . . . . . . . . . 304.2 chapter 4 example 2 . . . . . . . . . . . . . . . . . . . . . . 314.3 chapter 4 example 3 . . . . . . . . . . . . . . . . . . . . . . 324.4 chapter 4 example 4 . . . . . . . . . . . . . . . . . . . . . . 334.5 chapter 4 example 5 . . . . . . . . . . . . . . . . . . . . . . 344.6 chapter 4 example 6 . . . . . . . . . . . . . . . . . . . . . . 35

4.7 chapter 4 example 7 . . . . . . . . . . . . . . . . . . . . . . 364.8 chapter 4 example 8 . . . . . . . . . . . . . . . . . . . . . . 384.9 chapter 4 example 9 . . . . . . . . . . . . . . . . . . . . . . 404.10 chapter 4 example 10 . . . . . . . . . . . . . . . . . . . . . . 414.11 chapter 4 example 11 . . . . . . . . . . . . . . . . . . . . . . 44

10



4.12 chapter 4 example 12 . . . . . . . . . . . . . . . . . . . . . . 45

4.13 chapter 4 example 13 . . . . . . . . . . . . . . . . . . . . . . 474.14 chapter 4 example 14 . . . . . . . . . . . . . . . . . . . . . . 494.15 chapter 4 example 15 . . . . . . . . . . . . . . . . . . . . . . 504.16 chapter 4 example 16 . . . . . . . . . . . . . . . . . . . . . . 524.17 chapter 4 example 17 . . . . . . . . . . . . . . . . . . . . . . 544.18 chapter 4 example 18 . . . . . . . . . . . . . . . . . . . . . . 554.19 chapter 4 example 19 . . . . . . . . . . . . . . . . . . . . . . 564.20 chapter 4 example 20 . . . . . . . . . . . . . . . . . . . . . . 584.21 chapter 4 example 21 . . . . . . . . . . . . . . . . . . . . . . 60

5.1 chapter 5 example 1 . . . . . . . . . . . . . . . . . . . . . . 62


6.1 chapter 6 example 1 . . . . . . . . . . . . . . . . . . . . . . 726.2 chapter 6 example 2 . . . . . . . . . . . . . . . . . . . . . . 736.3 chapter 6 example 3 . . . . . . . . . . . . . . . . . . . . . . 746.4 chapter 6 example 4 . . . . . . . . . . . . . . . . . . . . . . 756.5 chapter 6 example 5 . . . . . . . . . . . . . . . . . . . . . . 776.6 chapter 6 example 6 . . . . . . . . . . . . . . . . . . . . . . 786.7 chapter 6 example 7 . . . . . . . . . . . . . . . . . . . . . . 796.8 chapter 6 example 8 . . . . . . . . . . . . . . . . . . . . . . 816.9 chapter 6 example 9 . . . . . . . . . . . . . . . . . . . . . . 826.10 chapter 6 example 10 . . . . . . . . . . . . . . . . . . . . . . 836.11 chapter 6 example 11 . . . . . . . . . . . . . . . . . . . . . . 846.12 chapter 6 example 12 . . . . . . . . . . . . . . . . . . . . . . 856.13 chapter 6 example 13 . . . . . . . . . . . . . . . . . . . . . . 87

7.1 chapter 7 example 1 . . . . . . . . . . . . . . . . . . . . . . 89

11



7.2 chapter 7 example 2 . . . . . . . . . . . . . . . . . . . . . . 90


8.1 chapter 8 example 1 . . . . . . . . . . . . . . . . . . . . . . 1018.2 chapter 8 example 2 . . . . . . . . . . . . . . . . . . . . . . 1028.3 chapter 8 example 3 . . . . . . . . . . . . . . . . . . . . . . 1038.4 chapter 8 example 4 . . . . . . . . . . . . . . . . . . . . . . 1048.5 chapter 8 example 5 . . . . . . . . . . . . . . . . . . . . . . 1058.6 chapter 8 example 6 . . . . . . . . . . . . . . . . . . . . . . 1068.7 chapter 8 example 7 . . . . . . . . . . . . . . . . . . . . . . 1078.8 chapter 8 example 8 . . . . . . . . . . . . . . . . . . . . . . 1088.9 chapter 8 example 9 . . . . . . . . . . . . . . . . . . . . . . 1108.10 chapter 8 example 10 . . . . . . . . . . . . . . . . . . . . . . 1118.11 chapter 8 example 11 . . . . . . . . . . . . . . . . . . . . . . 112

8.12 chapter 8 example 12 . . . . . . . . . . . . . . . . . . . . . . 113


11.1 chapter 11 example 1 . . . . . . . . . . . . . . . . . . . . . . 12911.2 chapter 11 example 2 . . . . . . . . . . . . . . . . . . . . . . 130

12



Chapter 2

Introduction To Engineering

Calculations

check Appendix AP 97 for dependency:

221.sci

Scilab code Exa 2.2.1 chapter 2 example 1

1 clc

2 p a t h n a m e = g e t _ a b s o l u t e _ f i l e _ p a t h ( ’ 2 2 1 . s c e ’ )

3 f i l e n a m e = p a t h n a m e + f i l e s e p ( ) + ’ 2 2 1 . s c i ’4 exec ( f i l e n a m e )

5 printf ( ” A l l t h e v a l u es i n t he t ex tb oo k a reA pp ro xi ma te d h en ce t h e v a l u e s i n t h i s c od e d i f f e r

f ro m t h o s e o f Te xt bo ok ” )

6 A c c l F i n a l = A c c l I n i t i a l * ( ( 3 6 0 0 * 2 4 * 3 6 5 ) ^ 2 ) / 1 0 ^ 5 ;

7 / / t he c a l c u l a t i o n s i n v o l v ed a re t he c o n ve r si o nf a c t o r s

8 printf ( ” \n f i n a l a c c e l e r a t i o n =%E Km/ Yr ˆ2 ” , A c c l F i n a l

)

13



Figure 2.1: chapter 2 example 1


231.sci


1 clc

2 / / t h i s p ro gr am i s u se d t o c o n v e r t l b . f t / min ˆ2 t o kg .c m/sˆ2


4 f i l e n a m e = p a t h n a m e + f i l e s e p ( ) + ’ 2 3 1 . s c i ’

5 exec ( f i l e n a m e )6 F i n a l = I n i t i a l * 0 . 4 5 3 5 9 3 * 1 0 0 / ( 3 . 2 8 1 * 6 0 * 6 0 )

7 // t h e c a l c u l a t i o n s i n vo l ve d a re c o n v er s io n f a c t o r s8 disp ( ” f i n a l =” )

9 disp ( F i n a l ) ; disp ( ”k g . cm/s ˆ2” )

14




7 printf ( ” mass o f t h e w at er = vo lu me x d e n s i t y =%f lbm ”, m a s s )

8 printf ( ” \n At s e a l e v e l , g = 32 .1 74 f t / s ˆ 2 ”)

9 g = 3 2 . 1 7 4 ;10 w e i g h t = m a s s * g / 3 2 . 1 7 4 ;

11 printf ( ” \n w ei gh t a t s e a l e v e l = %f l b f \n” , w e i g h t )

12 printf ( ” \n At d en ve r , g = 32 .1 39 f t / s ˆ 2 ”)

13 g = 3 2 . 1 3 9 ;

14 w e i g h t = m a s s * g / 3 2 . 1 7 4 ;

15 printf ( ”\n w e ig h t a t d en v er= %f l b f ” , w e i g h t )

16 // t he d i v i s i o n w it h 3 2 . 17 4 i s t o c o n ve r t lbm . f t / s ˆ2t o l b f

check Appendix AP 94 for dependency:252.sci


1 clc

2 p a t h n a m e = g e t _ a b s o l u t e _ f i l e _ p a t h ( ’ 2 5 2 . s c e ’ )3 f i l e n a m e = p a t h n a m e + f i l e s e p ( ) + ’ 2 5 2 . s c i ’4 exec ( f i l e n a m e )

5 // H ere We u se d s t an d ar d l i b r a r y f u n c t i o n s mean ands t d e v i a t i o n

16




6 y b a r = mean ( y ) ;7 sy = s t _ d e v i a t i o n ( y ) ;

8 d e f a u l t v a l u e = y b a r + 3 * s y + 1 ;

9 printf ( ” t h e maximum a l l o w e d v a l u e o f y i . e . badb a tc he s i n a week i s %d \n” , d e f a ul t v a lu e )

10 disp ( ” i n c as e o f 2 s ta nd ar d d e v i a t i o n s ” ) ;

11 d e f a u l t v a l u e = y b a r + 2 * s y + 1 ;

12 printf ( ” t he l i m i t i n g v al ue o f y i . e . bad b at ch es i na week i s %d” , d e f a u l t v a l u e )


271.sci


1 clc


5 // t h i s program u se s l e a s t s q u a r e s f i t t o s o l v e f o rs l o p e and i n t e r c e p t .

17




6 / / h en ce t h e v a l u e d i f f e r s f ro m t e xt b o ok a b i t .7 sx = sum ( x ) ; s x 2 = sum ( x ^ 2 ) ; s y = sum ( y ) ; s x y = sum ( x . * y ) ; n =

length ( x ) ;

8 A = [ s x , n ; s x 2 , s x ] ; B = [ s y ; s x y ] ; p = A \ B ;

9 m = p ( 1 , 1 ) ; b = p ( 2 , 1 ) ;

10 c l f ( )11 xtitle ( ’ 2 . 7 1 . s c e ’ , ’ V dot ( L/mi n) ’ , ’R ’ , ’ box ed ’ )

12 plot2d ( x , y , s t y l e = 3 )

13 disp ( ” i n c a s e 2 , R=36 ”)

14 R = 3 6 ;

15 V = m * R + b ;

16 printf ( ” the n V=%f” , V ) ;


272.sci

18




1 clc



5 disp ( s q r t T ) ;

6 sx = sum ( s q r t T ) ; s x 2 = sum ( T ) ; s y = sum ( M ) ; s x y = sum ( s q r t T . * M )

; n = length ( T ) ;

7 A = [ s x , n ; sx 2 , s x ] ; B = [ s y ; s xy ] ; p = A \ B ;

8 a = p ( 1 , 1 ) ; b = p ( 2 , 1 ) ;

9 c l f ( )

10 xtitle ( ’ 2 . 7 . 2 . s c e ’ , ’T1/2 ’ , ’ mdot ’ , ’ box ed ’ )11 plot2d ( s q r t T , M , s t y l e = 3 ) ;

12 printf ( ” s l o p e=%f”, a ) ;

13 printf ( ”\n i n t e r c e p t =%f ”, b ) ;

19



Chapter 3

Processes and Process

Variables


311.sci


1 clc



5 d e n s i t y = 1 3 . 5 4 6 * 6 2 . 4 3

6 printf ( ” d e n s i t y o f m e rc u ry=%f lbm / f t ˆ 3 ” , d e n s i t y ) ;

7 / / t h e m u l t i p l i c a t i o n f a c t o r i s t o c on v e rt d e ns i t yf ro m gm/ c c t o lbm / f t ˆ 3 .

8 v o l um e = m a s s / ( . 45 4 * d e n s it y ) ; // f t ˆ39 / / t h e d i v i s i o n by 0 . 4 54 i s t o c on ve rt mass i n kg t o

lbm .10 printf ( ” \n The volume o f %d kg o f me rcu ry i s %f f t

ˆ3 ” , m a s s , v o l u m e )

20





312.sci


1 clc



5 disp ( ”we know t ha t V (T )=Vo[ 1+ 0 . 18 182 x 10ˆ(−3)xT

+ 0 . 0 0 7 8 x 1 0 ˆ (−6)xTxT] ” )6 V a t0 = V a t 2 0 / ( 1 + 0 .1 8 1 82 * 1 0^ ( - 3 ) * T 1 + 0 . 00 7 8 *1 0 ^ ( - 6 ) * T 1 *

T1 )

7 / / t h e f u nc t i o n i s d e f i n e d w i t h t h e v a r i a b l e a st e m p e r a t u r e

21




8 function [ v o l u m e ] = v o l u m e ( T )

9 v o lu m e = V a t0 * ( 1 +0 . 1 81 8 2 *1 0 ^ ( - 3 ) * T + 0 . 00 7 8 *1 0 ^ ( - 6 )

* T * T ) ;

10 e n d f u n c t i o n

11 printf ( ” v at 20=%f ” , v o l u m e ( T 1 ) )

12 printf ( ” \n v at 1 00=%f ” , v o l u m e ( T 2 ) )

13 c h a n g e = ( ( v o l u m e ( T 2 ) ) - ( v o l u m e ( T 1 ) ) ) * 4 / ( % p i * D * D )

14 printf ( ” \n c ha ng e i n t he h e ig h t o f mer cury l e v e l =%f f t ” , c h a n g e )

15 / / t h e a n sw e r i s a b i t d i f f e r e n t due t o r ou nd in g o f f o f v ol um e ( T2 ) i n t e x t b o o k


331.sci


22




1 clc



5 m o l e s = m a s s / M

6 printf ( ”\n no . of mol e s=%f” , m o l e s )

7 l b m o l e = m o l e s / 4 5 3 . 6

8 printf ( ”\n n o . o f l b m o l e s=%f ” , l b m o l e )

9 C m o l e s = m o l e s

10 printf ( ”\n no . o f m o l es o f c a rb o n=%f ”, C m o l e s )

11 O m o l e s = 2 * m o l e s

12 printf ( ”\n n o . o f m o l e s o f o x yg e n=%f ” , O m o l e s )

13 O 2 m o l e s = m o l e s

14 printf ( ”\n no . o f m o l es o f d i o x yg e n=%f ” , O 2 m o l e s )15 g r a m s O = O m o l e s * 1 6

16 printf ( ”\n n o . o f g ra ms o f o x yg e n=%f ” , g r a m s O )

17 g r a m s O 2 = O 2 m o l e s * 3 2

18 printf ( ”\n n o . o f g ra ms o f o x yg e n=%f ” , g r a m s O 2 )

19 m o l e c u l e s C O 2 = m o l e s * 6 . 0 2 * 1 0 ^ ( 2 3 )

20 printf ( ”\n n o . o f m o l e c u l e s o f CO2 = %E” , m o l e c u l e s C O 2

)


332.sci

23





1 clc



4 exec ( f i l e n a m e )5 m a s s A = m a s s * x A

6 printf ( ” \n Mass o f A i n %d k g o f s o l u t i o n = %f kg A” , m a s s , m a s s A )

7 f l o w r a t e A = f l o w r a t e 1 * x A

8 printf ( ” \n Mass f lo w r a t e o f A i n a s t r e a m f l ow i n gat %d l bm/h =%f l bm A/h” , f l o w r at e 1 , f l o w r a t e A )

9 f l o w r a t e B = f l o w r a t e 2 * y B

10 printf ( ” \n Molar f l o w r a t e o f B i n a s t r e a m f l ow i n gat %d mol /mi n = %f mol B/mi n” ,flowrate2 ,flowrateB)

11 T o t a l f l o w r a t e = m o l a r B / y B12 printf ( ” \n T o ta l f lo w r a t e o f a s o l u t i o n w i t h %dkmolB/s=%f”, m o l a r B , T o t a l f l o w r a t e )

13 M a s s S o l u t i o n = m a s s o f A / x A

14 printf ( ” \n Mass o f s o l u t i o n t ha t c o n t a i n s %d lbm o f A = % f ” ,massofA ,MassSolution)


333.sci


24





1 clc



5 m o l O 2 = m a s s O 2 / M O 2

6 m o l C O = m a s s C O / M C O

7 m o l C O 2 = m a s s C O 2 / M C O 2

8 m o l N 2 = m a s s N 2 / M N 2

9 T o t a l M o l = m o l O 2 + m o l C O + m o l C O 2 + m o l N 2

10 printf ( ” \n m o l e f r a c t i o n o f O2=%f ”, m o l O 2 / T o t a l M o l )

11 printf ( ” \n m o l e f r a c t i o n o f CO=%f ”, m o l C O / T o t a l M o l )

12 printf ( ” \n m o l e f r a c t i o n o f CO2=%f ” , m o l C O 2 / T o t a l M o l )

13 printf ( ” \n m o l e f r a c t i o n o f N2=%f ”, m o l N 2 / T o t a l M o l )


334.sci

25





1 clc2 p a t h n a m e = g e t _ a b s o l u t e _ f i l e _ p a t h ( ’ 3 3 4 . s c e ’ )


5 M b a r = y N 2 * M N 2 + ( 1 - y N 2 ) * M O 2

6 printf ( ” \n a v e ra ge m o le cu l ar w ei gh t o f a i r fromm o l ar c o m p o s i t i o n =%f ” , M b a r )

7 I n v M b ar = x N 2 / 2 8 + ( 1 - x N2 ) / 3 2

8 printf ( ” \n a v e ra ge m o le cu l ar w ei gh t o f a i r fromm as s c o m p o s i t i o n =%f ” , 1 / I n v M b a r )


335.sci




5 m a s s _ c o n c = c o n c * 9 8

26




6 printf ( ” mass c o n c e n t r a t i o n o f s u l f u r i c a c i d=%f kg /mˆ3 ” , m a s s _ c o n c )

7 m a s s _ f l o w r a t e = r a t e * m a s s _ c o n c / 6 0

8 printf ( ” \n Mass f l o w r a t e o f s u l f u r i c a c id=%f kg / s ”, m a s s _ f l o w r a t e )

9 m a s s f r a c t i o n = 1 / ( r a t e * D * 1 0 0 0 / 6 0 )

10 printf ( ” \n Mass f r a c t i o n o f s u l f u r i c a c id=%f” ,

ma s sf r a ct i on )


341.sci


1 clc



5 P r e s s u r e = P r e s s u r e * 1 0 0 0 / ( 1 3 6 0 0 * 9 . 8 0 7 )

6 printf ( ” Pr es su re =%E mm of Hg” , P r e s s u r e )


342.sci

27






1 clc



5 P h = P 0 + D * g * h

6 printf ( ” P r e s s u r e a t t h e b ot to m o f t h e l a k e =%E N/mˆ 2 ”, P h )



1 clc


5 // I n t h i s c o d e I u s e d a f u nc t i o n t o a c h i e v e t h ec o n v e r s i o n

28



Chapter 4

Fundamentals Of Material

Balances


421.sci


1 clc



5 printf ( ” A l l t he v a l ue s i n t he t ex tb oo k a r eA pp ro xi ma te d h en ce t h e v a l u e s i n t h i s c od e d i f f e r



30




6 a c c u m u la t i o n = input + g e n e r a t i on - o u t pu t - c o n s u m p t i o n

7 disp ( ” We know t h a t a c c u m u l a t i o n =i n p u t +g e n e r a t i o n −out put−c o n s u m p t i o n ” )

8 printf ( ” Hence , Each y e a r p o p u l a t i o n d e c r e a s e s by %dp e o p l e ” , - a c c u m u l a t i o n )


422.sci


1 clc





31




8 x = ( m 1 * x1 + m 2 * x2 ) / m9 printf ( ” \n The c o mp o s it i on o f t he m et ha no l i n t he

p ro du ct i s %f a nd w a te r i s %f ” , x , 1 - x )


424.sci


1 clc





6 n d o t = r a t e / ( 1 - x 1 )7 d el t aN = - v ol * d * 10 ^3 / M

8 t f = de l ta N / ( - 0. 1 * n do t )

9 printf ( ” \n The t im e R eq ui re d f o r t he T ot al p r o c e s s=%d min” , t f )

33




9 n 3 = n 2 / x

10 printf ( ”n3=%f mol/ min” , n 3 )

11 disp ( ” U s in g t o t a l m ol e b a la n ce , ” )

12 n 1 = ( n 3 - n 2 ) / ( 1 + x 1 )

13 printf ( ”n1=%f mol/ min” , n 1 )

14 disp ( ” U s i n g N2 b a l a n c e , ” )

15 y = 1 - x - 0 . 7 9 * n 1 / n 3

16 printf ( ”y=%f mol O2/mol ” ,y )


432.sci


1 clc



35






6 m u l t i p l y = F i n a l B a s i s / b a s i s

7 F e e d = 1 0 0 * m u l t i p l y

8 T o p S t r e a m = 5 0 * m u l t i p l y

9 B o t t o m S t r e a m 1 = 1 2 . 5 * m u l t i p l y10 B o t t o m S t r e a m 2 = 3 7 . 5 * m u l t i p l y

11 printf ( ” \n F i n a l B a s i s =%d l b−m o le s / h” , F e e d )

12 printf ( ” \n F i n a l Top S t re a m F ee d=%d l b−m o le s / h” ,

T o p S t r e a m )

13 printf ( ” \n F i n a l B ott om S tr ea m F ee d 1 =%d l b−m o l e sA/h” , B o t t o m S t r e a m 1 )

14 printf ( ” \n F i n a l B ott om S tr ea m F ee d 2 =%d l b−m o l e sB/h” , B o t t o m S t r e a m 2 )


433.sci

36




1 clc




f ro m t h o s e o f Te xt bo ok ” )6 disp ( ” U s i n g NaOH b a l a n c e ” )

7 m 2 = i n p u t x * b a s i s / o u t p u t x

8 printf ( ”m2=%f Kg NaOH” , m 2 )

9 disp ( ” U s in g T o ta l mass b a l a n c e ” )

10 m 1 = m 2 - b a s i s

11 printf ( ”m1=%f Kg Water ” , m 1 )

12 V 1 = m 1 / D

13 printf ( ” \n V1=%f L i t r e s ” , V 1 )

14 R a t i o 1 = V 1 / b a s i s

15 R a t i o 2 = m 2 / b a s i s

16 printf ( ” \n R at io o f l t w at er /Kg F eed = %f l t w at er /Kg F e ed” , R a t i o 1 )

17 printf ( ” \n R a t io o f Kg p r o d u ct / Kg Fe ed = %f Kgp r o d u c t / Kg F ee d ” , R a t i o 2 )


435.sci


37




1 clc



4 exec ( f i l e n a m e )5 printf ( ” A l l t h e v a l u es i n t he t ex tb oo k a re

A pp ro xi ma te d h en ce t h e v a l u e s i n t h i s c od e d i f f e rf ro m t h o s e o f Te xt bo ok ” )

6 m a s s 1 = o u t p u t B a s i s * M 1 * o u t p u t x

7 m a s s 2 = o u t p u t B a s i s * M 2 * ( 1 - o u t p u t x )

8 m a s s = m a s s 1 + m a s s 2

9 y B 2 = m a s s 1 / m a s s

10 m 1 = b a s i s * D

11 printf ( ” \n m1=%f Kg/h” , m 1 )

12 m B 3 = z * i n p u t x * m 113 printf ( ” \n mB3=%f Kg/h ” , m B 3 )

14 disp ( ” U s i n g B e nz en e b a l a n c e , ” )

15 m 2 = ( i n p u t x * m 1 - m B 3 ) / y B 2

16 printf ( ” \n m2=%f Kg/h” , m 2 )

38




40




16 x 1 = ( i n p u t M a s s 1 * i n p ut x 1 - o u t p u t M a s s 1 * o u t p u t x 1 ) / m 1

17 printf ( ”x1=%f Kg A/ kg” , x 1 )

18 disp ( ” u s i n g Mass b a la n ce on m ix in g p oi nt , ” )

19 m 2 = i n p u t M a s s 2 + m 1

20 printf ( ”m2=%d Kg/ h” , m 2 )

21 disp ( ” u s i n g A b a l an c e on m ix in g p oi nt , ” )

22 x 2 = ( i n p u t M a s s 2 * i n p u t x 2 + m 1 * x 1 ) / m 2

23 printf ( ”x2=%f Kg A/ kg” , x 2 )


442.sci

41




1 clc




f ro m t h o s e o f Te xt bo ok ” )6 disp ( ” U s i ng B a l a n c e s a ro u nd two−e x t r a x t o r s ys te m , ” )

7 disp ( ” B a l an c e on t o t a l mass , ” )

8 printf ( ”%d + %d + %d = %f + m1+m3” , m a s s i n , M 1 , M 2 ,

ma sso ut )

9 disp ( ” B a l a n c e on A” )

10 printf ( ”%d ∗ % f = % f ∗ %f + m1 %f + m3 %f ” , m a s s i n ,

inputx ,massout ,outputxA ,m1xA ,m3xA )

11 A = [ 1 1 ; m 1x A , m 3 x A ]

12 b = [ m a s s i n + M 1 + M2 - m a s so u t ; m a s si n * i n p u tx - m a s so u t *

o u t p u t x A ]

13 C = A \ b

14 m 1 = C ( 1 , 1 )

15 m 3 = C ( 2 , 1 )

16 printf ( ” \n m1=%f Kg” , m 1 )

17 printf ( ” \n m3=%f Kg” , m 3 )

18 disp ( ” B a l a n c e o n M” )

19 x M 1 = ( m a s s i n + M 2 - m a s s o u t * o u t p u tx M - m 3 * m 3 x M ) / m 1

20 printf ( ”xM1=%f kg MIBK/ kg ” , x M 1 )

21 disp ( ” B a l a n c es a ro un d E x t ra c t m ix in g p o in t , ” )

22 disp ( ” B a l a n c e on A” )

23 m A 4 = m 1 * m 1 x A + m 3 * m 3 x A24 printf ( ” \n mA4=%f Kg Ace ton e ” , m A 4 )

25 disp ( ” B a l a n c e o n M” )

26 m M 4 = m 1 * x M 1 + m 3 * m 3 x M

27 printf ( ” \n mM4=%f Kg MIBK” , m M 4 )

42



28 disp ( ” B a l a n c e o n W” )

29 m W 4 = m 1 *( m 1 xM - x M 1 ) + m 3 * m 3x W30 printf ( ” \n mW4=%f Kg Wate r ” , m W 4 )

31 disp ( ” B a l a nc es a ro un d t he F i r s t e x t r a c t o r ” )

32 disp ( ” B a la nc e on A” )

33 m A 2 = m a s s i n * i n p u t x - m 1 * m 1 x A

34 printf ( ” \n mA2=%f Kg Ace ton e ” , m A 2 )

35 disp ( ” B a l an c e on M” )

36 m M 2 = m a s s i n - x 1 * x M 1

37 printf ( ” \n xM1=%f Kg MIBK” , x M 1 )

38 disp ( ” B a l an c e o n W” )

39 m W2 = m a s s i n * i np ut x - m 1 * ( m1 xM - x M 1 )

40 printf ( ” \n mW2=%f Kg Wate r ” , m W 2 )


451.sci


1 clc





6 disp ( ” O v e r a l l d ry A ir b a la n ce , ” )

7 n 1 = x 3 * b a s i s / x 1

8 printf ( ” n 1=%f m ol F r e s h f e e d ” , n 1 )9 disp ( ” O v e r a l l m ol e b a l a n ce , ” )

10 n 3 = n 1 - b a s i s

11 printf ( ” n 3=%f m ol W at er c o n d e n s e d ” , n 3 )

12 disp ( ” Mo le b a l a n c e on m i xi n g p o i n t , ” )

43




13 disp ( ”n1+n5=n2” )

14 disp ( ” Water b a l a n c e on m i xi n g p o i n t ” )

15 printf ( ” % f n 1 + % f n 5 = % f n 2 ” , 1 - x 1 , 1 - x 3 , 1 - x 2 )

16 A = [ 1 - 1; 1 - x2 , - (1 - x 3 ) ]17 b = [ n 1 ; ( 1 - x 1 ) * n 1 ]

18 C = A \ b

19 n 2 = C ( 1 , 1 )

20 n 5 = C ( 2 , 1 )

21 printf ( ” \n n2=%f mol” , n 2 )

22 printf ( ” \n n 5=%f m ol R e c y c l e d ” , n 5 )


452.sci

44




45




1 clc





6 m 6 x = m 5 x

7 printf ( ” Give n , m4=%f (m4+m5) ” ,x )

8 disp ( ” u s i n g O v e r a l l K2CrO4 b a l a n ce , ” )

9 printf ( ”%f ∗ %f = m4+ %f m5” , f e e d x , f e e d , m 6 x )10 A = [ 1 - x , - x ; 1 , m 6 x ]

11 b = [ 0 ; f e e d x * f e e d ]

12 C = A \ b

13 m 4 = C ( 1 , 1 )

14 m 5 = C ( 2 , 1 )

15 printf ( ” \n m4=%f K2CrO4 c r y s t a l s /h” , m 4 )

16 printf ( ” \n m5=%f e n t r a i n e d s o l u t i o n / h” , m 5 )

17 disp ( ” O v e ra l l T ot al mass b a l an c e , ” )

18 m 2 = f e e d - m 4 - m 5

19 printf (”m2=%f Kg H2O ev ap or at ed /h”

, m 2 )

20 disp ( ” Mass b a l an c e a ro un d t he c r y s t a l l i z e r , ” )

21 disp ( ”m3=m4+m5+m6”)

22 disp ( ” Water b al a nc e a round t he c r y s t a l l i z e r , ” )

23 printf ( ” %f m3= %f m5 + %f m6” , 1 - m 3 x , 1 - m 5 x , 1 - m 6 x )

24 D = [ 1 - 1; 1 ( - 1 + m 6x ) / ( 1 - m 3 x ) ]

25 e = [ m 4 + m 5 ; ( 1 - m 5 x ) * m 5 / ( 1 - m 3 x ) ]

26 F = D \ e

27 m 3 = F ( 1 , 1 )

28 m 6 = F ( 2 , 1 )

29 printf ( ” \n m3=%f Kg/h fe d to th e c r y s t a l l i z e r ” , m 3 )

30 printf ( ” \n m6=%f Kg/h ” , m 6 )31 r a t i o = m 6 / f e e d

32 printf ( ” \n r a t i o =%f Kg r e c y c l e / Kg f r e s h f e e d ” ,

r a t i o )

46




33 disp ( ” m as s b a l a n c e a ro un d R e cy c le−f r e s h f e ed mi xi ngp oi nt , ” )

34 m 1 = f e e d + m 6

35 printf ( ”m1=%f kg / h f e e d t o t h e e v a p o r a t o r ” , m 1 )

36 disp ( ” With o ut r e c y c l e , ” )

37 disp ( ”m3=622 Kg/h” )

38 disp ( ”m5=2380 Kg/h” )


461.sci


1 clc



47



5 printf ( ” A l l t h e v a l u es i n t he t ex tb oo k a re


6 n P = b a s i s * x P

7 n N = b a s i s * x N

8 n O 2 = b a s i s * x A * 0 . 2 1

9 if ( n N / n P > 1 )

10 disp ( ”NH3 i s i n e x c e s s ” )

11 else

12 disp ( ” Pr ope ne i s i n e x c e s s ” )

13 end

14 if ( n O 2 / n P > 1 )

15 disp ( ”O2 i s i n e x c es s ” )16 else

17 disp ( ” p ro pe ne i s i n e x c e s s ” )

18 end

19 n O 2 r e a c t e d = n P * 1 . 5

20 n N r e a c t e d = n P * 1

21 E x c e s s A m m o n i a = ( n N - n N r e a c t e d ) * 1 0 0 / n N r e a c t e d

22 E x c e s s O 2 = ( n O 2 - n O 2 r e a c t e d ) * 1 0 0 / n O 2 r e a c t e d

23 printf ( ” \n p e r c e n t a g e e x c e s s Ammonia=%f ” ,

E x c e s s A m m o n i a )

24 printf (” \n p e r c e n t a g e e x c e s s Oxyg en=%f ”

, E x c e s s O 2 )

25 n P o u t = ( 1 - x ) * n P

26 printf ( ” \n no . o f m o le s o f P r o p yl e n e l e f t = %d mol ” ,

n P o u t )

27 E = n P - n P o u t

28 n N o u t = n N - E

29 nO2out= nO2 -1.5*E

30 n A c = E

31 n W = 3 * E

32 printf ( ” \n n o . o f m o l e s o f Ammonia l e f t = %f m ol ” ,

n N o u t )

33 printf ( ” \n no . o f m o le s o f o xy ge n l e f t = %f mol ” ,n O 2 o u t )

34 printf ( ” \n n o . o f m o l e s o f ACN f o r m e d= %d m ol ” , n A c )

35 printf ( ” \n no . o f m o le s o f w a te r f or me d= %d mol ” , n W )

48





463.sci


1 clc





6 n 1 = ( 1 - c o n v 1 ) * b a s i s * x

7 n 2 = c o n v 2 * b a s i s * x8 E 1 = n 2

9 E 2 = b a si s * x - E1 - n 1

10 n 3 = E 1 - E 2

11 n 4 = 2 * E 2

49




12 n 5 = b a s i s * ( 1 - x )

13 n t = n 1 + n 2 + n 3 + n 4 + n 5

14 s e l e c t i v i t y = n 2 / n 415 printf ( ” s e l e c t i v i t y=%f mol Ethe ne /mol methane ” ,

s e l e c t i v i t y )


472.sci


1 clc

50







6 disp ( ” O v e r a l l P ro pa ne c o n v e rs i o n , ” )

7 n 6 = ( 1 - E ) * b a s i s

8 printf ( ”n6=%d mol prop ane ” , n 6 )

9 disp ( ” O v e r a l l C b a la n ce , ” )

10 n 7 = b as is - n6

11 printf ( ”n7=%d mol pr op e ne ” , n 7 )

12 disp ( ” O v e ra l l H b a l an c e , ” )13 n 8 = ( b a s i s * 8 - n 6 * 8 - n 7 * 6 ) / 2

14 printf ( ” n8=%d mol H2” , n 8 )

15 P e r c e n t a g e P r o p a n e = n 6 * 1 0 0 / ( n 6 + n 7 + n 8 )

16 printf ( ”\n Mo le p e r c e n t a g e o f p r op a ne=%f ”,

P e r c e n t a g e P r o p a n e )

17 P e r c e n t a g e P r o p e n e = ( 1 0 0 - P e r c e n t a g e P r o p a n e ) / 2

18 printf ( ”\n M ole p e r c e n t a g e o f p r o pe n e=m ol ep e r c e n t a g e o f h y d ro g e n=%f ” , P e r c e n t a g e P r o p e n e )

19 disp ( ” U si ng g i v en r e l a t i o n s among s e p a r a t o r

v a r i a b l e s , ”)

20 n 3 = n 6 / 0 . 0 0 5 5 5

21 n 1 0 = 0 . 0 5 * n 7

22 printf ( ” \n n3=%d mol Propane ” , n 3 )

23 printf ( ” \n n 10=%f m ol P r o p en e ” , n 1 0 )

24 disp ( ” u s i n g Pro pa ne b a l an c e a bo ut s e p a r a a t i o n u n i t ” )

25 n 9 = n 3 - n 6

26 printf ( ”n9=%d mol Propane ” , n 9 )

27 n 1 = b a s i s + n 9

28 disp ( ” U s i ng P ro pa ne b a l a n c e a b ou t m i xi n g p o i n t , ” )

29 printf ( ” n1=%d mol H2” , n 1 )

30 R e c y c l e R a t i o = ( n 9 + n 1 0 ) / b a s i s31 printf ( ” \n r e c y c l e r a t i o =%d m ol r e c y c l e / mol f r e s h

f e e d ” , R e c y c l e R a t i o )

32 S i n g l e P a s s = ( n 1 - n 3 ) * 1 0 0 / n 1

33 printf ( ” \n s i n g l e −p a s s c o n v e r s i o n =%f ” , S i n g l e P a s s )

51





473.sci


1 clc


52




4 exec ( f i l e n a m e )5 printf ( ” A l l t h e v a l u es i n t he t ex tb oo k a reA pp ro xi ma te d h en ce t h e v a l u e s i n t h i s c od e d i f f e r


6 disp ( ” R ea ct or a n a l y s i s , ” )

7 n 2 = ( 1 - s i n g l e _ p a s s ) * b a s i s * i n p u t x H 2

8 printf ( ” n2=%d mol H2” , n 2 )

9 disp ( ”H2 b a l a n c e ” )

10 consH2= basis* inputxH2 -n2

11 printf ( ”H2 mol es consumed=%d mol H2” , c o n s H 2 )

12 disp ( ”CO2 b a l a n c e ” )

13 n 1 = b a s i s * i n p u t x C O 2 - c o n s H 2 / 314 printf ( ” n1=%d mol CO2” , n 1 )

15 disp ( ” M et ha no l b a l a n c e ” )

16 n 3 = c o n s H 2 / 3

17 printf ( ”n3=%d mol Methanol ” , n 3 )

18 disp ( ”H2O b a l a n c e ” )

19 n 4 = c o n s H 2 / 3

20 printf ( ” n4=%d mol H2O” , n 4 )

21 disp ( ” c o n d en s er a n a l y s i s ” )

22 disp ( ” T o ta l mol e b a la n ce ” )

23 n 5 = n 1 + n 2 + m o l I

24 printf ( ” n5=%d mol ” , n 5 )

25 disp ( ”CO2 b a l a n c e ” )

26 x 5 C = n 1 / n 5

27 printf ( ”x5C=%d mol CO2/ mol ” , x 5 C )

28 disp ( ”H2 b a l a n c e ” )

29 x 5 H = n 2 / n 5

30 printf ( ”x5H=%d mol CO2/ mol ” , x 5 H )

31 x 1 = 1 - x 5 C - x 5 H

32 printf ( ”x1=%d mol I /mol” , x 1 )

33 disp ( ” F r e s h F ee d−R e cy cl e m ix in g p o i nt a n a l y s i s ” )

34 disp ( ” T o ta l mol e b a la n ce ” )35 printf ( ”n0+nr=%d” , b a s i s )

36 disp ( ” I b al a nc e ” )

37 printf ( ” n 0 %f + n r %f = %d” , I x , x 1 , m o l I )

38 A = [1 1 ; Ix x 1]

53




39 b = [ b a s i s ; m o l I ]

40 C = A \ b // H ere We s o l v e two l i n e a r e q u a ti o n ss i m u l t a n e o u s l y

41 n 0 = C ( 1 , 1 )

42 n r = C ( 2 , 1 )

43 printf ( ” \n n0=%f mol f r e s h f e e d ” , n 0 )

44 printf ( ” \n n r= %f mol r e c y c l e ” , n r )45 x 0 C = ( b a s i s * i n p u t x C O 2 - n r * x 5 C ) / n 0

46 printf ( ” \n x0C=%f mol CO2/mol ” , x 0 C )

47 x 0 H = 1 - x 0 C - I x

48 printf ( ” \n x0h=%f mol H2/mol” , x 0 H )

49 disp ( ” R e c y c l e−P urge s p l i t t i n g A n al y si s ” )

50 disp ( ” T o ta l mol e b a la n ce ” )

51 n p = n 5 - n r

52 printf ( ”np=%f mol pur ge ” , n p )

53 disp ( ” Flow c h a rt s c a l i n g ” )

54 F a c t o r = f i n a l / n 3

55 printf ( ” F a c t o r f o r s c a l i n g =%f Kmol /h / mo l” , F a c t o r )


481.sci


1 clc



54




4 exec ( f i l e n a m e )



6 disp ( ” p a rt 1 ”)

7 w e t = x N 2 w e t + x C O 2 w e t + x O 2 w e t

8 x N 2 d r y = x N 2 w e t / w e t

9 x C O 2 d r y = x C O 2 w e t / w e t10 x O 2 d r y = x O 2 w e t / w e t

11 printf ( ” \n xN2 d ry = % f mo l N2 /mo l d ry g a s ” , x N 2 d r y )

12 printf ( ” \n xO2 d ry = %f mol O2/ mo l d ry g a s ” , x O 2 d r y )

13 printf ( ” \n xCO2 d r y = % f m ol CO2/ m ol d r y g a s ” ,

x C O 2 d r y )


482.sci


1 clc





6 n O 2 T h e o r e t i c a l = b a s i s B u t a n e * 6 . 5

7 n A i r T h e o r e t i c a l = n O 2 T h e o r e t i c a l * 4 . 7 6

55



7 n O 2 t h e o r e t i c a l = b a s i s * 3 . 5

8 n 0 = n O 2 t h e o r e t i c a l * ( 1 + e x c e s s ) / 0 . 2 19 printf ( ” n 0=%d mo l a i r f e d ” , n 0 )

10 disp ( ” 9 0% e t h a n e c o n v e r s i o n ” )

11 n 1 = ( 1 - E 1 ) * b a s i s

12 printf ( ”No . o f m ol es o f e t ha n e u n r e ac t e d= %d” , n 1 )

13 disp ( ” 2 5% c o n v e r s i o n t o CO” )

14 n 4 = E 2 * ( b a s i s - n 1 ) * 2

15 printf ( ” n4= %d mol CO” , n 4 )

16 disp ( ” n i t r o g e n b a l an c e ” )

17 n 3 = 0 . 7 9 * n 0

18 printf ( ”n3= %d mol N2” , n 3 )

19 disp ( ” A to mi c c a rb o n b a l a n c e ” )20 n5=2* basis -2*n1- n4

21 printf ( ” n5= %d mol CO2” , n 5 )

22 disp ( ” A to mi c h y dr o ge n b a l a n c e ” )

23 n 6 = ( b a s i s * 6 - n 1 * 6 ) / 2

24 printf ( ” n6= %d mol H2O”, n 6 )

25 disp ( ” A to mi c o xy ge n b a l a n c e ” )

26 n 2 = ( n O 2 t h e o r e t i c a l * 1 . 5 * 2 - n 4 - n 5 * 2 - n 6 ) / 2

27 printf ( ”n2= %d mol O2” , n 2 )

28 d r y = n 1 + n 2 + n 3 + n 4 + n 5

29 w e t = d r y + n 6

30 y 1 = n 1 / d r y

31 printf ( ”\n y 1= %f mol C2H6/mol ” , y 1 )

32 y 2 = n 2 / d r y

33 printf ( ”\n y 2= % f m ol O2 / mo l ” , y 2 )

34 y 3 = n 3 / d r y

35 printf ( ”\n y 3= % f m ol N2 / mo l ” , y 3 )

36 y 4 = n 4 / d r y

37 printf ( ”\n y 4= %f mol CO/mol ” , y 4 )

38 y 5 = n 5 / d r y

39 printf ( ”\n y 5= %f mol CO2/mol ” , y 5 )

40 r a t i o = n 6 / d r y41 printf ( ” \n r a t i o =%f mol H2O/ mol d ry s t a c k g a s ” ,

r a t i o )


57




484.sci


1 clc





6 disp ( ”N2 b a l a n c e ” )

7 n a = b a s i s * x N 2 / 0 . 7 9

8 printf ( ”na=%f mol a i r ” , n a )

9 disp ( ” A t o m i c C b a l a n c e ” )10 n c = b a s is * x C O + b a si s * x C O 2

11 printf ( ”nc=%f mol C” , n c )

12 disp ( ” A t o m i c O b a l a n c e ” )

13 n w = 0. 21 * n a *2 - b a si s * ( xC O + x CO 2 * 2 + x O2 * 2 )

58



14 printf ( ”nw=%f mol oxygen ” , n w )

15 disp ( ” A to mi c H2 b a l a n c e ” )16 n h = n w * 2

17 printf ( ”nh=%f mol H2” , n h )

18 r a t i o = n h / n c

19 printf ( ”\n C/H r a t i o i n f u e l =%f mol H/ mol C” , r a t i o )

20 disp ( ” p e r ce n t e x c e s s a i r ” )

21 n O 2 t h e or e t i c al = n c + n h / 4

22 printf ( ” nO2 t h e o r e t i c a l =%f mol O2” , n O 2 t h e o r e t i c a l )

23 n O 2 f e d = 0 . 2 1 * n a

24 printf ( ” \n nO2fed=%f mol O2” , n O 2 f e d )

25 p e r c e n t = ( n O 2 f e d - n O 2 t h e o r e t i c a l ) * 1 0 0 / n O 2 t h e o r e t i c a l

26 printf ( ”\n p e rc e nt a ge e x c e s s a i r =%f e x c es s a i r ” ,p e r c e n t )


491.sci


1 clc





6 disp ( ” D e s i g n ” )

7 M E K i n 1 = f e e d * x8 M E K ou t 1 = T o p F l ow 1 * o u t p u t x1 + B o t t om F l o w1

9 c l o s u r e 1 = M E K o u t 1 * 1 0 0 / M E K i n 1

10 printf ( ” c l o s u r e 1 =%d p e r c e n t ” , c l o s u r e 1 )

11 disp ( ” E x p e r i m e n t ” )

59



12 M E K i n 2 = f e e d * x

13 M E K ou t 2 = T o p F l ow 2 * o u t p u t x2 + B o t t om F l o w214 c l o s u r e 2 = M E K o u t 2 * 1 0 0 / M E K i n 2

15 printf ( ” c l o s u r e 2 =%d p e r c e n t ” , c l o s u r e 2 )

61



Chapter 5

Single Phase Systems


511.sci




5 i n v Pb a r = w t p e rc t / D w a t er + ( 1 - w t p er c t ) / D s u l fu r i c

6 printf ( ” D e n si t y c a l c u l a t e d u s i ng volume a d d i t v i t y =%f ” , 1 / i n v P b a r )

7 P b ar = w t p e r c t * D w a te r + ( 1 - w t p er c t ) * D s u l fu r i c

8 printf ( ” \n D en si ty c a l c u l a t e d u si ng mass a d d i t i v i t y=%f” , P b a r )


62





1 clc



5 n d o t = V d o t / M //k mol /h6 v d o t = n d o t * 2 2 . 4 * T / ( 2 7 3 * P )

7 printf ( ” The v o l u m et r i c f l o w r a t e o f t he s tr ea m=%f mˆ3/h” , v d o t )



1 clc


5 V 2 = V 1 * P 1 * T 2 / ( P 2 * T 1 )

6 printf ( ” Volume i n f i n a l s t a t e =%f f t ̂ 3 ”, V 2 )

64






524.sci


1 clc



4 exec ( f i l e n a m e )5 //SCFH means f t ˆ3 (STP) /h6 n d o t = 3 . 9 5 * 1 0 ^ 5 / 3 5 9

7 printf ( ” M ol ar f l o w r a t e =%E l b−m o l e s / h r ” , n d o t )

8 V 2 d o t = V 1 d o t * T 2 * P 1 / ( T 1 * P 2 )

65




9 printf ( ” \n T ru e v o l u m e t r i c f l o w r a t e =%E f t ˆ 3/ h ”,

V 2 d o t )


525.sci


1 clc




f ro m t h o s e o f Te xt bo ok ” )6 n 2 c a p = f l o w i n A * D a c e t o n e / M a c e t o n e

7 printf ( ” \n M ol ar f l o w r a t e o f A ce to n e=%f m ol A ce to n e/min” , n 2 c a p )

8 P = P f i na l * 7 60 + 7 63

66




9 y 4 = P a c e t o n e / P

10 printf ( ” \n Mole f r a c t i o n o f A c e t o ne i n t he f i n a lf l o w = %f m ol A c et o n e / m ol ” , y 4 )

11 printf ( ” \n Mole f r a c t i o n o f N it ro ge n i n t h e f i n a lf l o w= %f mol N i t r o g e n / m ol ” ,1-y4)

12 n 3 c a p = f l o w i n N / 0 . 0 2 2 413 n 4 c a p = n 2 c a p / y 4

14 disp ( ”By u s i n g O v e r a l l Mo la r b a la n ce , ” )

15 n 1 c a p = n 4 c a p - n 2 c a p - n 3 c a p

16 V 1 c a p = n 1 c a p * 0 . 0 2 2 4 * T 1 * 7 6 0 / ( 1 * 2 7 3 * P 1 )

17 printf ( ” V o l u m et r ic F l ow r at e o f N i t ro g e n = %f N i t r o g e n / m i n ” , V 1 c a p )


531.sci


1 clc



4 exec ( f i l e n a m e )5 printf ( ” A l l t he v a l ue s i n t he t ex tb oo k a r e


6 P i d e a l = 0 . 0 8 2 0 6 * T / V c a p

67




7 printf ( ” \n The v al ue o f p r e s s u r e a s p er I d e a l g ase q u a t i o n = % f a t m ” , P i d e a l )

8 T r = T / T c

9 B 0 = 0 . 0 83 - ( 0 . 42 2 / ( T r ) ^ 1 .6 )

10 B 1 = 0 . 1 39 - ( 0 . 17 2 / ( T r ) ^ 4 .2 )

11 B = 0 . 0 8 2 0 6 * T c * ( B 0 + w * B 1 ) / P c

12 P v i ri a l = 0 . 0 8 20 6 * T * ( 1 + B / V c ap ) / V c a p13 printf ( ”\n The v a lu e o f p r e s s u r e a s p e r V i r i a l g a s

e q u a t i o n = % f a t m ” , P v i r i a l )

14 e = ( P i d e a l - P v i r i a l ) * 1 0 0 / P v i r i a l

15 printf ( ” \n P er ce nt ag e e r r o r due t o I d e a l g asE q u a t i o n = % f ” ,e )


532.sci


1 clc



5 printf ( ” A l l t he v a l ue s i n t he t ex tb oo k a r eA pp ro xi ma te d h en ce t h e v a l u e s i n t h i s c od e d i f f e rf ro m t h o s e o f Te xt bo ok ” )

6 V c a p = V / n

7 a = 0 . 42 7 4 7 *( R * T c ) ^ 2 / P c

68




8 b = 0 . 0 8 6 6 4 * R * T c / P c

9 m = 0 . 4 8 5 08 + 1 . 5 17 1 * w - 0 .1 5 61 * w * w

10 T r = T / T c

11 a l p h a = ( 1+ m * (1 - sqrt ( T r ) ) ) ^ 2

12 P = ( R * T / ( V c a p - b ) ) - ( a l p h a * a / ( V c a p * ( V c a p + b ) ) )

13 printf ( ” \n P r es s ur e o f g as c a l c u l a t e d u s in g SRK

e q u a t i o n = % f a t m ” ,P )


541.sci


1 clc





6 z = 0 . 9 3 4

7 printf ( ” \n From t h e T ab le , z= %f ” ,z )

8 n c a p = P * V c a p * 1 0 1 . 3 2 5 / ( z * R * T * 1 . 0 1 3 2 5 )9 m c a p = n c a p * M

10 printf ( ” \n Mass f l o w r a t e o f Methane = %f Kg/ h r ” ,

mcap )

69





542.sci


1 clc





6 T r = T / T c7 P r = P / P c

8 V r i d e a l = V * P c / ( n * R * T c )

9 printf ( ” \n Tr= %f” , T r )

10 printf ( ” \n P r= %f ” , P r )

11 printf ( ”\n V r i d e a l =%f ” , V r i d e a l )

12 z = 1 . 7 7

13 printf ( ”\n From t h e g r a p h s , z=%f ” ,z )

14 P = z * R * T * n / V

15 printf ( ” \n P r es s ur e i n t he c y l i n d e r = %f atm” ,P )


543.sci

70





1 clc





6 disp ( ” A p pl yi ng n ewt on c o r r e c t i o n s f o r Hydrogen , ” )

7 T c a H 2 = T c H 2 + 8

8 P c a H 2 = P c H 2 + 8

9 T c b a r = y H2 * T c a H 2 + y N 2 * T cN 2

10 P c b a r = y H2 * P c a H 2 + y N 2 * P cN 211 T r b a r = T / T c b a r

12 P r b a r = P / P c b a r

13 printf ( ” \n Trbar=%f” , T r b a r )

14 printf ( ” \n Pcbar=%f” , P c b a r )

15 Z m = 1 . 8 6

16 printf ( ” \n From the graph , Zm=%f”, Z m )

17 V c a p = Z m * R * T / P

18 printf ( ” \n S p e c i f i c Volume o f M ix tu re= %f L / mol ” ,

V c a p )

71



Chapter 6

Multiphase Systems


611.sci






72





6 disp ( ” l e t d e l ta H v /R = S ” )

7 S = - ( T1 * T2 * log ( P 2 / P 1 ) ) / ( T 1 - T 2 )

8 d e l t a h v = S * R

9 printf ( ” \n L a t e nt H ea t o f V a p o r i z a t i o n =%d” , d e l t a h v )

10 B = log ( P1 ) + S /T1

11 printf ( ”\n B=%f” ,B )

12 P = exp ( - S / T + B )

13 printf ( ”\n P∗ a t %f K = %f ”, T , P )


631.sci


1 clc





6 y = P s t a r / P7 printf ( ” \n Molar c o mp o s it i o n o f Water i s %f and A ir

i s %f” , y , 1 - y )


73




632.sci


1 clc




6 P = y * P T

7 if ( P < 7 6 0 )

8 disp ( ” The V apo ur i s S u pe r h e a t ed ” )

9 elseif ( P = 7 6 0 )

10 disp ( ” The v a po u r i s At Dew p o i n t ” )11 else

12 disp ( ” The v a po ur i s n ot S up er h e at e d ”)

13 end

14 disp ( ” From t a b l e s Tdp=90 C ”)

74




5 printf ( ” A l l t h e v a l u es i n t he t ex tb oo k a reA pp ro xi ma te d h en ce t h e v a l u e s i n t h i s c od e d i f f e rf ro m t h o s e o f Te xt bo ok ” )

6 P = h r * P 7 5

7 y = P / P o r i g

8 n d o t = P o r i g B a r * V d o t / ( R * T )

9 n d o t W a t e r = n d o t * y

10 printf ( ” \n M ol ar f l o w r a t e o f W ater=%f K mol / h” ,

n d o t W a t e r )

11 n d o t B D A = n d o t * ( 1 - y )

12 printf ( ” \n M ol ar f l o w r a t e o f Dry A i r=%f K mol /h ” ,

n d o t B D A )13 n d o t O 2 = n d o t B D A * 0 . 2 1

14 printf ( ” \n M ol ar f l o w r a t e o f Oxyg en=%f Kmol / h” ,

n d o t O 2 )

15 h m = P / ( P o r i g - P )

16 h a = h m * 1 8 / 2 9

17 h m d o t = P 7 5 / ( P o r i g - P 7 5 )

18 h p = 1 0 0 * h m / h m d o t

19 printf ( ” \n M ol al H umi di t y=%f mol w at e r /mol BDA” , h m )

20 printf ( ” \n A b s o l u t e H u m id i t y=%f k g w a t e r / k g BDA” , ha

)

21 printf ( ” \n P e r c e n t a g e H u m id i t y=%f ” , h p )


641.sci


1 clc



76







6 y H 2 O = P H 2 O / P

7 y S O 2 = P S O 2 / P

8 y A i r = 1 - y H 2 O - y S O 2

9 disp ( ” U si ng A ir b al an ce , ” )

10 n G 2 = ( 1 - x ) * b a s i s / y A i r11 printf ( ”nG2=%f lbm/ h” , n G 2 )

12 x S O 2 = y / 1 0 2

13 x H 2 O = 1 - x S O 2

14 disp ( ” U s i ng SO2 b a l a nc e , ” )

15 n L 2 = ( b a s i s * x - n G 2 * y S O 2 ) * M 1 / ( x S O 2 )

16 printf ( ”nL2=%d lbm/ h” , n L 2 )

17 disp ( ” U s i n g H2O b a l a n c e , ” )

18 n L 1 = n G2 * y H 2 O * M 2 + n L2 * x H 2 O

19 printf ( ” nL1=%d lbm H2O/h ” , n L 1 )

20 S O 2 A b s o r b e d = n L 2 * x S O 221 S O 2 F e d = b a s i s * x * M 1

22 F r a c t i o n = S O 2 A b s o r b e d / S O 2 F e d

23 printf ( ” \n F r a c t i o n SO2 a b s o r b e d= %f lbm SO2a b s o r b e d / l bm SO2 f e d ” , F r a c t i o n )

77





642.sci


1 clc





6 x = y * P / H

7 P B s ta r = 1 0 ^ (6 . 90 6 - 1 2 1 1/ ( T + 2 2 0 . 8) )

8 P T s ta r = 1 0 ^ (6 . 95 3 3 - 1 3 4 3. 9 / ( T + 2 1 9. 3 8 ) )

9 P B = x B * P B s t a r

10 P T = ( 1 - x B ) * P T s t a r

11 P t o t a l = P B + P T

12 y B = P B / P t o t a l13 y T = P T / P t o t a l

14 printf ( ” \n T o t a l s y st em p r e s s u r e =%f mm o f Hg” ,

P t o t a l )

15 printf ( ” \n C o m p s o it i o n o f b e n z e ne=%f ” , y B )

78




16 printf ( ” \n C o mp s o it i on o f t o l u e n e =%f ” , y T )


651.sci


1 clc





6 disp ( ” C Om po si ti on o f F i l t e r c ak e , ” )

79



7 disp ( ”m2=4m3”)

8 disp ( ” Water b a la n ce a ro un d t he c r y s t a l l i z e r , ” )9 printf ( ”\n %f k g H2O = % f m1 + %f m1” , b a s i s * i n p u t x ,

outputx ,outpu tx)

10 disp ( ” Mass b a la nc e ar ound c r y s t a l l i z e r , ” )

11 printf ( ” \n %d=m1+m2+m3” , b a s i s )

12 A = [0 1 -4; o ut pu tx 0 o ut pu tx ; 1 1 1 ]

13 b = [ 0 ; b a s i s * i n p u t x ; b a s i s ]

14 C = A \ b

15 / / Here we s o l ve d two l i n e a r e q ua t i on s s i m ul t a ne o us l y16 m 1 = C ( 1 , 1 )

17 printf ( ” \n m1=%f Kg” , m 1 )

18 m 2 = C ( 2 , 1 )19 printf ( ” \n m2=%f Kg” , m 2 )

20 m 3 = C ( 3 , 1 )

21 printf ( ” \n m3=%f Kg” , m 3 )

22 disp ( ” O v e r al l AgNO3 bal anc e , ” )

23 m 5 = ( 1 - i n p ut x ) * b a s is - ( 1 - o u t pu t x ) * m 1

24 printf ( ”m5=%f k g AgNO3 c r y s t a l s r e c o v e r e d ” , m 5 )

25 p e r c e n t a g e = m 5 * 1 0 0 / ( b a s i s * ( 1 - i n p u t x ) )

26 printf ( ” \n P e r c e n t a g e r e c o v e r y =%f ” , p e r c e n t a g e )

27 disp ( ” O v e r a l l mass b a l a n c e ” )

28 m 4 = b a s i s - m 1 - m 5

29 printf ( ”m4=%d Kg w a t e r r e mo v ed i n t h e D r y e r ” , m 4 )


652.sci


1 clc



80







6 o u t p u t x = S / ( S + 1 0 0 )

7 printf ( ” x=%f Kg KNO3/Kg” , o u t p u t x )

8 disp ( ” Water b a l a n c e ” )9 m 1 = b a s i s * ( 1 - i n p u t x ) / ( 1 - o u t p u t x )

10 printf ( ” \n m1=%f Kg” , m 1 )

11 disp ( ” Mass b a l a n c e ” )

12 m 2 = b a s i s - m 1

13 printf ( ” \n m2=%f kg ” , m 2 )

14 p e r c e n t a g e = m 2 * 1 0 0 / ( b a s i s * i n p u t x )

15 printf ( ” \n P e rc e nt a ge o f KNO3 i n t he f e e d t h atc r y s t a l l i z e s i s %f” , p e r c e n t a g e )


653.sci

81





1 clc





6 disp ( ” T o t a l mass b a l a n c e ” )

7 disp ( ”m1=1+m2” )

8 disp ( ”MgSO4 ba l a nc e ” )

9 printf ( ” \n %f m1 = %d∗ % f / %f + m2 %f ” , i n p u t x ,

b a s i s , M , M 1 , o u t p u t x )

10 A = [ 1 - 1; i n p u tx - o u t pu t x ]11 b = [ 1 ; b a s i s * M / M 1 ]

12 C = A \ b

13 / / Here we s o l ve d two l i n e a r e q ua t i on s s i m ul t a ne o us l y14 m 1 = C ( 1 , 1 )

82




15 m 2 = C ( 2 , 1 )

16 printf ( ” \n m1=%f Tonne/h ” , m 1 )

17 printf ( ” \n m2=%f Tonne/h ” , m 2 )


654.sci


1 clc




6 x = ( T f - 1 0 0 ) * 4 0 6 5 6 / ( R * 3 7 3 . 1 6 ^ 2 )

7 y = m 2 / 1 8 . 0 1 6

8 M s = m 1 * ( 1 - x ) / ( y * x )

9 printf ( ” \n Ms=%f ” , M s )

10 d e l ta T m = R * ( 2 7 3 .1 6 ) ^ 2 * x / 6 0 0 9. 5

11 T m s = 0 - d e l t a T m12 printf ( ” \n Tms=%f ” , T m s )

13 P s t a r = ( 1 - x ) * 2 3 . 7 5 6

14 printf ( ” \n S o l v e n t V ap ou r p r e s s u r e =%f mm Hg ”, P s t a r )

83





661.sci


1 clc





6 D b a r = D A * DW / ( x * D W + ( 1 - x ) * DA )

7 m a s s 1 = V 1 * D b a r

8 m a s s 2 = V 2 * D C

9 disp ( ” C b a l a n c e ” )10 m 4 = m a s s 2

11 printf ( ” \n m4=%f ” , m 4 )

12 disp ( ”W b a l a n c e ” )

13 m 2 = ( 1 - x ) * m a s s 1

84




14 printf ( ” \n m2=%f ” , m 2 )

15 disp ( ” A b a l a n c e ” )

16 printf ( ”m1+m3=%f ∗ %f ” , x , m a s s 1 )

17 disp ( ” D i s t r i b u t i o n C o o e f f i c i e n t ,K=m3∗(m1+m2) /m1∗ (m3+m4) ” )

18 disp ( ”On s o l v i n g , ” )

19 m 1 = 2 . 7

20 m 3 = 1 6 . 8

21 p e r c e n t a g e = m 3 * 1 0 0 / ( x * m a s s 1 )22 printf ( ” \n m1=%f ” , m 1 )

23 printf ( ” \n m3=%f ” , m 3 )

24 printf ( ” \n p e r c e nt ag e o f a ce to ne t r a n s f e r r e d t oc h l o r o f o r m = % f ”, p e r c e n t a g e )


662.sci


85




1 clc





6 n = P * V / ( R * T )

7 printf ( ” \n No . o f m o l e s=%f m ol ” ,n )

8 y 0 = y * P s t a r / P m m

9 printf ( ” \n Y0=%f mol CCl4/mol” , y 0 )

10 P f i n a l = x F * P m m

11 b = 0 . 0 9 6 * P f i n a l

12 X s t a r = 0 . 7 9 4 * b / ( 1 + b )

13 printf ( ” \n Mass o f CCl4 a d so r be d t o Ca rbon a t

e q u i l i b r i u m =%f g CCl4 a d s / g C” , X s t a r )14 M a ds = ( y 0 * n - x F * n ) * 15 4

15 printf ( ” \n Ma ss o f CCl4 a d s o r b e d=%f g ” , M a d s )

16 M c = M a d s / X s t a r

17 printf ( ” \n Mass o f c a rb o n R e qu i r ed=%f g ” , M c )

87



88



Chapter 7

Energy And Energy Balances


721.sci


1 clc





6 u = V d o t * 1 0 0 ^2 / ( % pi * ( I D / 2 ) ^ 2 * 3 6 00 )

7 m d ot = V d o t * 1 0 ^3 / 3 60 0


89




8 E k = m d ot * u ^ 2 / 2

9 printf ( ” \n Ek=%f J / s ” , E k )


722.sci


1 clc





6 P o w e r = m d o t * g * ( z 2 - z 1 )

7 printf ( ” \n Power=%d J/ s ” , P o w e r )


741.sci


90





1 clc





6 H c a p = U + P * V c a p * 1 0 1 . 3

7 H = n d o t * H c a p * 1 0 ^ 3

8 printf ( ” \n S p e c i f i c E n th a lp y=%d J / m ol ” , H c a p )

9 printf ( ”\n E n t h a l p y o f H e li u m=%E J / h ” ,H )


742.sci

91





1 clc





6 E k = m d o t * 1 0 ^ ( - 3 ) * ( u 2 ^ 2 - u 1 ^ 2 ) / 2

7 E p = m d o t * g * d e l t a Z / 1 0 ^ 38 Q d o t = Q d o t / ( 0 . 2 3 9 * 3 6 0 0 )

9 H d o t = Q d o t - W s - E k - E p

10 printf ( ” \n De lt aH=%f KW” , H d o t )

11 H c a p = H d o t / m d o t

12 printf ( ”\n S p e c i f i c E n th a lp y=%f Kj /Kg” , H c a p )


751.sci


92




1 clc





6 d e l t a H = H 0 - H 5 0

7 d e l t a U = d e l t a H + ( ( P f i n a l * V f i n a l - P i n i t i a l * V i n i t i a l )

* 1 . 9 8 7 / 1 0 . 7 3 )

8 printf ( ” \n c ha n ge i n S p e c i f i c E nt h al p y=%f B tu / lbm ” ,

d e l t a H )

9 printf ( ” \n c ha ng e i n S p e c i f i c I n t e r n a l En erg y=%f Btu/lbm” , d e l t a U )


753.sci


1 clc





93




6 disp ( ” From Stea m t a b l e s , ” )

7 H i n = 3 2 0 1 //Kj/Kg8 H o u t = 2 6 7 5 //Kj/Kg9 W s = - m d ot * ( H o ut - H i n ) / 3 60 0

10 printf ( ” Work d e l i v e r e d by T u rb i ne t o s u r r o u n d i n g s =%dKw” , W s )


761.sci


1 clc





94




6 d e l t a H = m 3 * H 3 - m 1 * H 1 - m 2 * H 2

7 disp ( ” From t a b l e s , V do t = 0 . 1 1 6 6 mˆ 3 / k g ” )

8 V d o t = 0 . 1 1 6 6

9 A = % p i * ( I D / 2) ^ 2 / 1 0^ 4

10 u = m 3 * V d o t / ( A * 6 0 )

11 E k = m 3 * u ^ 2 / ( 2 *1 0 ^ 3)

12 Q d o t = d e l t a H + E k

13 printf ( ”H e at r e q u i r e d=%E Kj /min” , Q d o t )


762.sci


1 clc





95




6 Q d o t = b a s i s * ( x * H o u t 1 + ( 1 - x ) * H o u t 2 - x * H i n 1 - ( 1 - x ) * H i n 2 )

7 printf ( ” \n H ea t r e q u i r e d =%f KJ /Kg ”, Q d o t / b a s i s )


763.sci


1 clc





6 disp ( ” M as s b a l a n c e o n W ate r , ” )

7 disp ( ”m3+m1=m2” )

8 disp ( ” E n e r g y b a l a n c e , ” )

96




9 disp ( ”m3∗H3+m1∗H1=m2∗H2” )

10 A = [ 1 , - 1 ; H 2 , - H 1 ]

11 b = [ m 3 ; m 3 * H 3 ]

12 C = A \ b

13 // h er e we s o l ve d two l i n e a r e q ua t i on s s i m ul t a ne o us l y.

14 m 2 = C ( 1 , 1 )

15 m 1 = C ( 2 , 1 )

16 printf ( ” I n p u t f l o w r a t e , m1=%f Kg /h ” , m 1 )

17 printf ( ” \n O ut pu t f l o w r a t e , m2=%f Kg /h ” , m 2 )

18 disp ( ”From t a bl e s , V dot = 3. 11 mˆ3/k g ” )

19 V d o t = 3 . 1 1

20 printf ( ” V o l u m e tr i c i n p u t f l o w r a t e =%f mˆ 3/ h ”, m 1 * V d o t

)


771.sci


97




1 clc





6 u 1 = V d o t * 1 0^ 4 / ( 1 0^ 3 * 6 0* % p i * ( I D 1 / 2) ^ 2 )

7 u 2 = V d o t * 1 0^ 4 / ( 1 0^ 3 * 6 0* % p i * ( I D 2 / 2) ^ 2 )8 d el t aP = - ( ( u2 ^ 2 - u 1 ^2 ) /2 + g * d el ta Z ) * 10 ^3

9 P 1 = P 2 - d e l t a P

10 printf ( ” \n P1=%E Pa” , P 1 )


772.sci


1 clc

98








6 u2 = sqrt ( 2 * 3 2 . 1 7 4 * ( - F - g * d e l t a Z / 3 2 . 1 7 4 ) )

7 V d ot = u 2 * % p i * ( I D / 2) ^ 2 / 1 44

8 t = V * 0 . 1 3 3 7 / ( V d o t * 6 0 )

9 printf ( ” T o t a l t i m e t a k e n=%f min ” ,t )


773.sci


1 clc



99





6 m do t = - Ws / ( d el ta P / D + g * d el ta Z )

7 printf ( ” \n Water f l o w r a t e =%f kg / s ” , m d o t )

100



Chapter 8

Balances On Nonreactive

Processes


831.sci


1 clc




101





f ro m t h o s e o f Te xt bo ok \n ” )

6 function [ C v ] = f u n ( T )

7 C v = 0 . 85 5 + T * 9 . 4 2* 1 0 ^( - 4 )

8 e n d f u n c t i o n9 [ U c a p , e r r ] = intg ( T i , T f , f u n ) // i n t g i s an i n b u l t

f u n c t i o n u se d f o r d e f i n i t e i n t e g r a t i o n10 Q = m a s s * U c a p

11 printf ( ”H e at R e q ui r e d=%f KJ ”,Q )


832.sci


1 clc




A pp ro xi ma te d h en ce t h e v a l u e s i n t h i s c od e d i f f e rf ro m t h o s e o f Te xt bo ok \n ” )

6 disp ( ” p a rt 1 ”)

7 function [ C p ] = f u n 1 ( T )

102




8 C p = 0. 0 29 0 0+ T * 0. 2 19 9 *1 0 ^( - 5 ) + T ^ 2 * 0. 57 23

* 10 ^( - 8 ) - T ^3 * 2 .8 71 * 1 0^ ( -1 2)


10 [ d e l t a H , e r r ] = intg ( T 1 , T 2 , f u n 1 ) / / i n t g i s an i n b u l tf u n c t i o n u se d f o r d e f i n i t e i n t e g r a t i o n

11 Q d o t = n d o t * d e l t a H

12 printf ( ” H e a t T r a n s f e r r e d =%f KJ/ m in ” , Q d o t )13 disp ( ” p a r t 2 ”)

14 function [ C V ] = f u n 2 ( T )

15 C V = f u n1 ( T ) - 8. 14 * 1 0 ^( - 3 )


17 [ d e l t a U , e r r ] = intg ( T 3 , T 4 , f u n 2 ) / / i n t g i s an i n b u l tf u n c t i o n u se d f o r d e f i n i t e i n t e g r a t i o n

18 n = P * V / ( R * T )

19 Q = n * d e l t a U

20 printf ( ” H ea t t r a n s f e r r e d =%f KJ” ,Q )


833.sci




103






6 function [ C p ] = f u n ( T )

7 Cp = 0 .0 28 94 + T * 0. 41 47 * 1 0^ ( -5 ) + T ^2 * 0 .3 19 1 *

1 0^( - 8) - T ^3 * 1 .9 65 * 1 0^ ( -1 2)

8 e n d f u n c t i o n9 d e l t a H = intg ( T 1 , T 2 , f u n ) // i n t g i s an i n b ul t f u nc t i o n

us e d f o r d e f i n i t e i n t e g r a t i o n10 Q d ot = n d o t * d e l t aH * 1 0 ^ 3 / 60

11 printf ( ” \n R at e o f h e a t r e m o va l= %f KW” , Q d o t )


834.sci


1 clc




A pp ro xi ma te d h en ce t h e v a l u e s i n t h i s c od e d i f f e rf ro m t h o s e o f Te xt bo ok \n ” )

6 function [ C p 1 ] = f u n 1 ( T )

7 C p1 = 0 . 04 9 37 + T * 1 3. 92 * 10 ^ ( - 5) - T ^ 2

* 5. 8 16 * 10 ^( - 8) + T ^ 3 * 7. 2 80 * 1 0^ ( - 12 )

104





9 function [ C p 2 ] = f u n 2 ( T )

10 C p2 = 0 . 06 8 03 + T * 2 2. 5 9* 1 0^ ( - 5 ) - T ^ 2

* 13 . 11 * 10 ^( - 8) + T ^ 3 * 31 . 71 * 1 0^ ( - 12 )



13 C p = x * f un 1 ( T ) + ( 1 - x ) * f un 2 ( T )


15 d e l t a H = intg ( T 1 , T 2 , f u n ) // i n t g i s an i n b u l t f u nc t i o n

u s e d f o r d e f i n i t e i n t e g r a t i o n16 printf ( ” \n H ea t c a p a c i t y o f M i x tu r e=%f KJ/ m ol ” ,

d e l t a H )

17 Q d o t = n d o t * d e l t a H

18 printf ( ” \n H e at R e q ui r e d=%f KJ /h”, Q d o t )


835.sci


105




1 clc





6 n d o t = V d o t / 2 2 . 4


8 C p = 0. 0 34 3 1 + T * 5 .4 6 9* 1 0^ ( - 5 ) + T ^ 2

* 0 . 36 6 1 *1 0 ^ ( - 8 ) + T ^ 3 * 1 1 *1 0 ^( - 1 2)


10 H1 = intg ( T 1 , T 2 , f u n ) // i n t g i s an i n b u l t f u nc t i o nu s e d f o r d e f i n i t e i n t e g r a t i o n

11 printf ( ”H1=%f Kj/ mol ” , H 1 )

12 disp ( ”From T abl e s H2= −0.15 Kj / mol , H3=8.17 Kj /

mol” )13 H 2 = - 0. 15

14 H 3 = 8 . 1 7

15 Q d ot = ( n d o t * x * H 1 + n d ot * ( 1 - x ) * ( H3 - H 2 ) ) / 6 0

16 printf ( ” Heat In pu t=%f KW” , Q d o t )


841.sci


106




1 clc





6 Q d o t = m d o t * d e l t a H v / ( M * 6 0 )

7 printf ( ” R a te o f H ea t t r a n s f e r =%f KW” , Q d o t )


842.sci

107





1 clc





6 disp ( ”We h a ve d e lt a H v a t 6 9C h e n c e We f o l l o w p at hADG” )

7 d e l ta H A = C p * ( T2 - T 1 ) + V * ( 1 .0 1 3 - P ) * M / ( D * 1 0 ^4 )

8 printf ( ” \n deltaHA=%f” , d e l t a H A )

9 d e l t a H D = d e l t a H v

10 printf ( ” \n deltaHD=%f” , d e l t a H D )

11 function [ C ] = f u n 1 ( T )

12 C = 0 .1 37 4 4 + T * 4 0. 8 5* 10 ^ ( - 5 ) - T ^ 2 * 23 . 92 * 10 ^( - 8)

+ T ^ 3 * 57 . 66 * 10 ^ ( - 1 2)13 e n d f u n c t i o n

14 d e l t a H G = intg ( T 2 , T 3 , f u n 1 ) // i n t g i s an i n b ul tf u n c t i o n u se d f o r d e f i n i t e i n t e g r a t i o n

15 printf ( ” \n deltaHG=%f Kj/mol” , d e l t a H G )

16 Q d o t = n d o t * ( d e l t a H A + d e l t a H D + d e l t a H G ) / 3 6 0 0

17 printf ( ” \n r a t e o f Hea t s u p pl y= %f K j/ mol ” , Q d o t )

18 printf ( ” \n I n t h i s pr obl em we n e g l ec t e d V∗ d e lt a P a si t i s n e g l i g i b l e ” )


108




1 clc




f ro m t h o s e o f Te xt bo ok ” )6 disp ( ” u s i n g T ro ut on r u l e , ” )

7 d e l t a H v T 1 = 0 . 1 0 9 * T 1

8 disp ( ” In t h i s c a s e , t ro ut on r u l e g i v e s a b e t t e re s t i m a t e ” )

9 disp ( ” u s i n g Watson c o r r e c t i o n ” )

10 d e l t a H v T 2 = 3 6 . 8 * ( ( T c - T 2 ) / ( T c - T 1 ) ) ^ ( 0 . 3 8 )

11 printf ( ” E s t i m at e d v a l u e u s i n g T ro ut on r u l e =%f K j/ mol” , d e l t a H v T 1 )

12 printf ( ” \n E st im at ed v a l ue u s i ng wa ts on c o r r e c t i o n =%f K j / mo l ” , d e l t a H v T 2 )


844.sci


1 clc



109






6 A = [1 1; x y ]

7 b = [ b a s i s ; b a s i s / 2 ]

8 C = A \ b

9 / / Here We s o l v e d two l i n e a r e q u a ti o n s s i m u l t a n eo u s l y

10 n V = C ( 1 , 1 )11 n L = C ( 2 , 1 )

12 H 1 = 5 . 3 3 2

13 H 2 = 6 . 3 4 0

14 H 3 = 3 7 . 5 2

15 H 4 = 4 2 . 9 3

16 Q = n V * x* H 1 + n V *( 1 - x )* H 2 + n L *y * H3 + n L *( 1 - y )* H 4

17 printf ( ” \n Heat t r a n s f e r r e d = %f Kj ”,Q )

18 disp ( ” The a ns we r f o r t h i s p ro bl em i n Text i s wrong ” )


851.sci

110





1 clc





6 n H C l = m d ot * x * 1 0 ^ 3 / M 1

7 n H 2O = m d o t * (1 - x ) * 1 0 ^ 3 / M 2

8 function [ C p 1 ] = f u n ( T )

9 C p1 = 2 9 . 1 3 *1 0 ^ ( - 3 ) - T * 0 . 1 34 1 * 10 ^ ( - 5 ) + T ^ 2

* 0 . 97 1 5 *1 0 ^ ( - 8 ) - T ^ 3 * 4 . 33 5 * 10 ^ ( - 1 2 )


11 H1 = intg ( T 1 , T 2 , f u n ) // i n t g i s an i n b u l t f u nc t i o nu s e d f o r d e f i n i t e i n t e g r a t i o n

12 disp ( H 1 )

13 r = n H 2 O / n H C l

14 disp ( ” From t a b l e B . 1 1 , d e l ta H a= −67. 4 Kj /mol HCl ” )

111




15 d e l t a Ha = - 67 .4

16 y = n H C l / ( n H C l + n H 2 O )

17 C p = 0 . 7 3 * m d o t * 4 . 1 8 4 / n H C l

18 d e l t a H b = C p * ( T 3 - T 1 )

19 H 2 = d e l t a H a + d e l t a H b

20 Q d o t = n H C l * ( H 2 - H 1 )

21 printf ( ”Heat =%E kj /h” , Q d o t )


852.sci




112






6 disp ( ” s u l p h u r i c a c id b a la nc e ” )

7 m 2 = x * m d o t / y8 disp ( ” T o t a l Mass b a l a n ce ” )

9 m 1 = m d o t - m 2

10 Q d o t = m 1 * H v + m 2 * H l - m d o t * H f

11 printf ( ” Rate o f Heat t r a n s f e r = %f Btu /h ”, Q d o t )


855.sci


113



1 clc




6 disp ( ” From f i g u r e 8 .5 −2 , ” )

7 x L = 0 . 1 8 5

8 x V = 0 . 8 9

9 m L = b a s i s * ( ( x V - x F ) / ( x V - x L ) )

10 m V = b a s i s - m L

11 Q do t =m V* Hv + mL * Hl - b as is * HF12 printf ( ” R ate o f h e at t r a n s f e r =%f B tu /h ”, Q d o t )

114



Chapter 9

Balances On Reactive Processes


911.sci


1 clc





6 disp ( ” P ar t 1 ”)

7 E 1 = n do t /4

8 d e l t a H 1 = E 1 * H r 1

9 printf ( ” e n t h a l p y c h a n g e=%E K j / s ” , d e lt a H1 )10 disp ( ” p a r t 2 ”)

11 H r 2 = 2 * H r 1

12 E 2 = n d o t / 8

13 d e l t a H 2 = E 2 * H r 2

115





14 printf ( ” E n t h a l p y c h a n g e=%E k j / s ” , d e l t a H 2 )15 disp ( ” p a rt 3 ”)

16 H r 3 = H r 1 + 5 * H v W a t e r + H v B u t a n e

17 d e l t a H 3 = E 1 * H r 3

18 printf ( ” E n t h a l p y c h a n g e=%E k j / s ” , d e l t a H 3 )


912.sci


116




1 clc





6 disp ( ”From r e a c t io n , o n ly g a s eo u s a r e c ou nt ed ” )

7 l e f t = 1 + 2

8 r i g h t = 1 + 1

9 d e l t a U r = d e l t a Hr - R * T * ( r i g h t - l e f t ) / 1 0 ^ 3

10 printf ( ” d e l t a U r =%f K j / mo l ” , d e l t a U r )


931.sci


1 clc





6 H r = 5 * H C O 2 + 6 * H H 2 O - H C 5 H 1 2

7 printf ( ” \n H ea t o f t h e r xn= %f KJ/ mol ” , H r )

117





941.sci


1 clc





6 Hr= Hethane -Hethene -H hydrogen

7 printf ( ” \n Hea t o f t h e r xn= %f Kj / mol ” , H r )


951.sci


118




121



6 function [ C p ] = f u n 1 ( T )

7 C p = 34 . 31 * 10 ^( - 3) + T * 5. 4 69 * 10 ^( - 5) + T ^ 2 * 0. 36 61*1 0^ ( -8 ) - T ^3 * 11 * 1 0^( - 12 )


9 H o u t M e t h a n e = intg ( T 1 , T 2 , f u n 1 )

10 H 4 = - 74 .8 5 + H o ut M et h an e

11 function [ C ] = f u n 2 ( T )

12 C = 3 4. 2 8* 1 0^ ( - 3 ) + T * 4 .2 6 8* 10 ^ ( - 5 ) - T ^ 3 * 8. 69 4 *

1 0 ^ ( - 1 2 )


14 H o u t F o r m a l = intg ( T 1 , T 2 , f u n 2 ) / / i n t g i s an i n b u i l tf u n c t i o n which can c a l c u l a t e d e f i n i t e i n t e g r a l s

15 H 7 = - 1 15 . 90 + H o u tF o r m al16 d e l t a H = N o u t W a t e r * H 9 + N o u t C a r b o n * H 8 + N o u t F o r m a l * H 7 +

N o u t N i t r o g e n * H 6 + N o u t O x y g e n * H 5 + N o u t M e t h a n e * H 4 -

N i n N i t r o g e n * H 3 - N i n O x y g e n * H 2 - N i n M e t h a n e * H 1

17 Q = d e l t a H

18 printf ( ” \n Q=%f Kj” ,Q )


954.sci


1 clc





6 disp ( ” C ar bo n B a l a n c e ” )

122




123




7 printf ( ”%d ∗ %f ∗2 + %d ∗ %f ∗2=2 n1+2 n2” , b a s i s , x ,

basis ,1- x)

8 disp ( ” H y dr og en B a l a n c e ” )

9 printf ( ”%d ∗ %f ∗6 + %d ∗ %f ∗4 = 6 n 1+4 n 2+2 n 3 ”,

basis,x ,basis ,1-x)10 disp ( ” E n er gy B a l a nc e ” )

11 printf ( ”%d = %f n1 %f n2 + %f n3 −%d ∗ %f −%d∗ %f ” ,

Q, HoutEthanol ,HoutEtha none ,HoutHy drogen ,

NinEthanol ,HinEthanol ,NinEthanone ,H inEthanon e)

12 A = [ 1 1 0 ;3 2 1 ;2 16 .8 1 1 50 .9 - 6. 59 5]

13 b = [ 1 5 0 ; 4 3 5 ; 2 8 4 1 2 ]

14 C = A \ b

15 n 1 = C ( 1 , 1 )

16 printf ( ” \n n 1=%f m ol E t h a n o l / s ” , n 1 )

17 n 2 = C ( 2 , 1 )

18 printf ( ” \n n 2=%f m ol E t ha n o ne / s ” , n 2 )19 n 3 = C ( 3 , 1 )

20 printf ( ” \n n 3=%f m ol H y dr o ge n / s ” , n 3 )

21 disp ( ” The s o l u t i o n s i n t he Text a r e Wrong ”)

22 f r a c t i o n = ( N i n E t h a n ol - n 1 ) / N i n E t h a n o l

23 printf ( ” F r a c t i o n a l c o n v e r s i o n o f E th a no l=%f ”,

f r a c t i o n )


955.sci

124





1 clc




f ro m t h o s e o f Te xt bo ok ” )6 disp ( ” p a rt 1 ”)

7 Hr= HfSalt +3*HfWater -HfAcid -3*HfBase

8 printf ( ” Hr o f t h e r x n=%f Kj / m ol ” , H r )

9 disp ( ” p a rt 2 ”)

10 d e l t a H = H r * 5 / 3

11 printf ( ” de l t aH =%f Kj ” , d e l t a H )


956.sci

125




1 clc





6 disp ( ” U si ng S b al an ce , ” )

7 m 2 = b a s i s * x * M S * M S a l t / ( M A c i d * M S )

8 printf ( ” \n m2=%f g Na2SO4” , m 2 )

9 disp ( ” U s in g Na b a la n ce , ” )10 m 1 = 2 * M N a * m 2 * M B a s e / ( y * M N a * M S a l t )

11 printf ( ” \n m1=%f g NaOH” , m 1 )

12 disp ( ” T o t a l mass b a la n ce , ” )

13 m 3 = b a s i s + m 1 - m 2

14 printf ( ” \n m3=%f g H2O”, m 3 )

15 printf ( ” \n Mass o f p ro du ct s o l u t i o n =%f” , m 2 + m 3 )

16 m = m 2 + m 3

17 W a t e r = m 2 * 2 / M S a l t

18 printf ( ” \n Wa ter Formed i n t h e r e a c t i o n =%f m ol H2O”, W a t e r )

19 disp ( ”H2SO4( aq ) : ” )

20 a 1 = b a s i s * ( 1 - x ) / M W a t e r

21 b 1 = b a s i s * x / M A c i d

22 r A c i d = a 1 / b 1

23 printf ( ” \n r A c i d =%f m ol W at er / m o l A c i d ” , r A c i d )

24 disp ( ”NaOH( aq ) : ” )

25 a 2 = m 1 * ( 1 - y ) / M W a t e r

26 b 2 = m 1 * y / M B a s e

27 r B a s e = a 2 / b 2

28 printf ( ” \n r B a s e=%f m ol W at er / m o l B a s e ” , r B a s e )

29 disp ( ”Na2SO4( aq ) : ”)30 a 3 = m 3 / M W a t e r

31 b 3 = m 2 / M S a l t

32 r S a l t = a 3 / b 3

33 printf ( ” \n r S a l t =%f m ol W ater / m ol S a l t ” , r S a l t )

126




34 E = b 1

35 printf ( ” \n E xt en t o f r e a c t i o n =%f m ol ” ,E )

36 n H A c i d = b a s i s * 3 . 8 5 * ( T 3 - T 1 ) / 1 0 0 0

37 n H S a l t = m * 4 . 1 8 4 * ( T 2 - T 1 ) / 1 0 0 038 n H B a s e = 0

39 H f S a lt = - 13 84

40 H f A c i d = - 88 4 .6

41 H f B a s e = - 46 8 .1

42 H f W a t e r = - 2 85 . 84

43 d e lt a Hr = H f S al t + 2 * H fW a te r - H f Ac id - 2 * H fB as e

44 printf ( ” \n E n ta h lp y c ha n ge i n t h e r xn=%f Kj / mol ” ,

d e l t a H r )

45 Q = E * d e l t aH r + ( n H Sa lt - n H Ac i d - n H Ba s e )

46 printf ( ” \n Q o f t h e r xn=%f Kj ” ,Q )

47 disp ( ” The a n sw er i n t h e T ex t i s w ro ng . ” )


961.sci


1 clc



127





6 y 1 = x * 1 6

7 y 2 = ( 1 - x ) * 3 0

8 x C H 4 = y 1 / ( y 1 + y 2 )

9 H H V Me t h an e = ( H 1 + 2 * 4 4. 0 1 3) / M 1

10 H H V Et h a ne = ( H 2 + 3 * 4 4. 0 1 3) / M 2

11 H H V = x C H 4 * H H V M et h a ne + ( 1 - x C H4 ) * H H V E t h an e

12 printf ( ” \n HHV of Fue l=%f KJ/g ” , H H V )

128




3 f i l e n a m e = p a t h n a m e + f i l e s e p ( ) + ’ 1 1 1 2 . s c i ’4 exec ( f i l e n a m e )



6 function [ d v ] = f u n ( t )

7 d v = ( 1 0 ^ 6 ) * exp ( - t / 1 0 0) - 10 ^7


9 [ v o l , e r r ] = intg ( 0 , 6 0 , f u n ) / / i n t g i s an i n b u i l tf u n c t i o n which can c a l c u l a t e d e f i n i t e i n t e g r a l s

10 printf ( ” \n Vo lume a t t h e e nd o f 6 0 d a y s=%E” , v o l + v 0 )


1131.sci


1 clc

130



Appendix

Scilab code AP 1 11.3.1.sci

1 m 1 = 1 . 5 0 //k g2 m 2 = 3 //k g3 C v 1 = 0 . 9 / / c a l / g C4 C v 2 = 0 . 1 2 / / c a l / g C5 Q d o t = 5 0 0 //W6 T 1 = 2 5 0 //C7 T 2 = 2 5 //C

Scilab code AP 2 1112.sci

1 v 0 = 1 0 ^ 9


1 x = 0 . 8 5

2 H 1 = 8 02 //Kj /mol3 H 2 = 1 42 8 // k j /mol4 M 1 = 1 6

5 M 2 = 3 0


1 x = 0 . 1

2 y = 0 . 2

3 M A c i d = 9 8 . 1

4 M S = 3 2

132



5 M S a l t = 1 4 2

6 M B a s e = 4 07 M W a t e r = 1 8

8 M N a = 4 6

9 b a s i s = 1 0 0 0 // g10 T 2 = 3 5

11 T 1 = 2 5

12 T 3 = 4 0


1 disp ( ” f ro m t a b l e s , ” )

2 H f A c id = - 12 94 //Kj/mol3 H f B a s e = - 46 9 .1 //Kj /mol4 H f S a lt = - 19 74 //Kj/mol5 H f W a t er = - 28 5. 8 //Kj /mol


1 b a s i s = 1 5 0 / / m o l / s2 x = 0 . 9

3 H i n Et h a no l = - 2 12 .1 9

4 H i n Et h a n on e = - 1 47 . 075 H o u tE t h a no l = - 2 16 . 81

6 H o u t Et h a n on e = - 15 0. 9

7 H o u t H y d r o g e n = 6 . 5 9 5

8 N i n E t h a n o l = 1 3 5

9 N i n E t h a n o n e = 1 5

10 Q = 2 4 4 0 //KW


1 N i n M e t h a n e = 1 0 0 //mol2 N i n O x y g e n = 1 0 0 //mol3 N i n N i t r o g e n = 3 7 6 //mol4 N o u t M e t h a n e = 6 0 //mol5 N o u t O x y g e n = 5 0 //mol

133



6 N o u t N i t r o g e n = 3 7 6 //mol

7 N o u t F o r m a l = 3 0 //mol8 N o u t C a r b o n = 1 0 //mol9 N o u t W a t e r = 5 0 //mol10 H 1 = - 74 .8 5 //Kj /mol11 H 2 = 2 .2 35 //Kj /mol12 H 3 = 2 .1 87 //Kj /mol13 H 5 = 3 . 7 5 8 //Kj /mol14 H 6 = 3 . 6 5 5 //Kj/mol15 H 8 = - 3 93 . 5+ 4 .7 5 //Kj /mol16 H 9 = - 2 41 . 83 + 4 .2 7 //Kj/mol17 T 1 = 2 5 //C

18 T 2 = 1 5 0 //C


1 n N H 3 = 1 0 0 / / m o l / s2 n O 2 i n = 2 0 0 / / m o l / s3 H 1 = 8 . 4 7 0 //Kj/mol4 H 3 = 9 . 5 7 0 //Kj/mol5 T 1 = 2 5

6 T 2 = 3 0 0

7 H r = - 90 4. 7 //Kj /mol


1 H e t h a n e = - 1 55 9 .9 //Kj/mol2 H e t h e ne = - 14 11 //Kj /mol3 H h y dr o g en = - 2 85 . 84 //Kj/mol


1 H C O 2 = - 3 93 .5 //Kj /mol2 H H 2 O = - 2 85 . 84 //Kj/mol3 H C 5 H 12 = - 17 3 //Kj /mol


134




1 b a s i s = 1 / / mol f e e d2 x = 0 . 6 8 4 // mole f r a c t i o n Of B3 y = 0 . 4


1 T 1 = 3 3 7 . 9 //K2 T 2 = 4 7 3 //K3 T c = 5 1 3 . 2 //K


1 d e l t a H v = 2 8 . 8 5 / /Kj / mol a t 69 C2 T 1 = 2 5 //C3 T 2 = 6 9 //C4 C p = 0 . 2 1 6 3 / / K j / m o l C5 V =1 //L6 P =7 / / b a r7 D = 0 . 6 5 9 //KG/L

8 M = 8 6 . 1 7 //Kg9 n d o t = 1 0 0 //mol /h10 T 3 = 3 0 0 //C


1 m d o t = 1 5 0 0 //g/mi n2 M = 3 2 //g/mol3 d e l t a H v = 3 5 . 3 //Kj /mol


1 x = 0 . 1 //CH42 T 1 = 2 0

3 T 2 = 3 0 0

4 V d o t = 2 0 0 0 //L/min

136




1 x = 0 . 6

2 T 1 = 0

3 T 2 = 4 0 0

4 n d o t = 1 5 0 //mol /h


1 T 1 = 4 3 0 //C

2 T 2 = 1 0 0 //C3 n d o t = 1 5 //Kmol/min


1 T 1 = 2 0 //C2 T 2 = 1 0 0 //C3 T 3 = 9 0 //C4 T 4 = 3 0 //C5 P =3 / / b a r6 V =5 //L

7 R = 0 . 0 8 3 1 4 //L. bar /mol . K8 T = 3 6 3 //K9 n d o t = 1 0 0 //mol/min


1 m a s s = 2 0 0 //k g2 T i = 2 0 //C3 T f = 1 5 0 //C


1 W s = 1 0 ^ 6 //N.m/s2 d e l t a P = - 8 3* 1 0^ 3 //N/mˆ23 g = 9 . 8 1 //m/sˆ2

137



4 d e l t aZ = - 10 3 //m

5 D = 1 0 ^ 3 //kg/mˆ3


1 d e l t aZ = - 2. 5 // f t2 u 1 = 0

3 D = 5 0 //lbm/ f t ̂ 34 F = 0 . 8 0 // f t . l b f /l bm5 V =5 // ga l6 g = 3 2 . 1 7 4 / / f t / s ˆ 27 I D = 0 . 2 5 // in

Scilab code AP 27 771.sci

1 V d o t = 2 0 //L/min2 P 2 = 1 . 0 1 3 2 5 * 1 0 ^ 5 //atm3 I D 1 = 0 . 5 //cm4 I D 2 = 1 //cm5 g = 9 . 8 1 //m/sˆ26 d e l t a Z = 5 0 //m


1 m 3 = 1 1 5 0 //Kg/h2 H 3 = 2 6 7 6 //KJ/Kg3 H 2 = 3 0 7 4 //KJ/Kg4 H 1 = 3 2 7 8 //KJ/Kg


1 b a s i s = 1 //Kg/s2 x = 0 . 6

// e t ha ne3 T 1 = 1 5 0 //K4 T 2 = 2 5 0 //K5 P =5 / / b a r6 H o u t 1 = 9 7 3 . 3 //KJ/Kg

138



7 H o u t 2 = 2 3 7 . 0 //KJ/Kg

8 H i n 1 = 3 1 4 . 3 //KJ/Kg9 H i n 2 = 3 0 //KJ/Kg


1 m 1 = 1 2 0 //k g2 m 2 = 1 7 5 //k g3 m 3 = 2 9 5 //k g4 I D = 6 //cm5 P = 1 7 / / b a r6 H 1 = 1 2 5 . 7 //Kj/Kg

7 H 2 = 2 7 1 . 9 //Kj/Kg8 H 3 = 2 7 9 3 //Kj /k g


1 m d o t = 2 0 0 0 //Kg/h2 P = 1 0 / / b a r


1 H 0 = 1 9 6 . 2 3 //Btu/lbm2 H 5 0 = 2 0 2 . 2 8 //Btu/lbm

3 P f i n a l = 5 1 . 9 9 / / p s i a4 P i n i t i a l = 1 8 . 9 0 / / p s i a5 V f i n a l = 1 . 9 2 0 // f t ˆ3/ lbm6 V i n i t i a l = 4 . 9 6 9 // f t ˆ3/ lbm


1 m d o t = 5 0 0 / 3 6 0 0 //Kg/s2 u 1 = 6 0 //m/s

3 u 2 = 3 6 0 //m/s4 d e l t a Z = - 5 //m5 g = 9 . 8 1 //m/sˆ26 Q d o t = - 10 ^4 / / K c a l / h7 W s = 7 0 //KW

139




1 U = 3 8 0 0 //J /mol2 P =1 //atm3 V c a p = 2 4 . 6 3 //L/mol4 n d o t = 2 5 0 //Kmol/h


1 g = 9 . 8 1 //m/sˆ2

2 m d o t = 1 5 //Kg/s3 z 2 = 2 0 //m4 z 1 = - 2 2 0 //m


1 I D = 2 //cm2 Vdot=2 //mˆ3/h


1 V = 5 0 //L2 P =1 //atm3 T = 3 4 + 2 7 3 . 2 //K4 y = 0 . 3

5 x F = 0 . 0 0 1

6 R = 0 . 0 8 2 0 6

7 P s t a r = 1 6 9 //mm o f Hg8 P m m = 7 6 0 //mm o f Hg


1 b a s i s = 1 0 0 0 //Kg o f s o l u t i o n2 i n p u t x A = 0 . 3 //Wt . f r a c t i o n o f a c e to n e3 o u t p u t x A 1 = 0 . 0 5

4 o u t p u t x M 1 = 0 . 0 2

140



5 o u t p u t x A 2 = 0 . 1

6 o u t p u t x M 2 = 0 . 8 7


1 V 1 = 2 0 0 //CC Aceto ne2 x = 0 . 1 / /Wt a c e t o n e3 V 2 = 4 0 0 / / C C c h l o r o f o r m4 D A = 0 . 7 9 2 / / g / c c5 D C = 1 . 4 8 9 / / g / c c6 D W = 1 / / g / c c


1 m 1 = 5 // g o f s o l u t e2 m 2 = 1 0 0 // g o f Water3 P =1 //atm4 T f = 1 0 0 . 4 2 1 //C5 T i = 2 5 //C6 R = 8 . 3 1 4 //J /mol .K


1 b a s i s = 1 / / Tonn e Epsom s a l t p r o d u ce d / h2 i n p u t x = 0 . 3 0 1 / / Tonne MgSO4/ to nn e3 o u t p u t x = 0 . 2 3 2 // Tonne MgSO4/ to nn e4 M = 1 2 0 . 4

5 M 1 = 2 4 6 . 4


1 i n p u t x = 0 . 6

2 b a s i s = 1 0 0 / / k g F ee d3 S = 6 3 //Kg KNO3/1 00 Kg H2O


141



1 b a s i s = 1 5 0 // kg f e e d

2 S 1 0 0 = 0 . 9 0 5 // g AgNO3/ g3 S 2 0 = 0 . 6 8 9 // g AgNO3/g4 i n p u t x = 0 . 0 9 5 / / kg w a te r / k g5 o u t p u t x = 0 . 3 1 1 / / kg w a te r / k g


1 y = 0 . 0 1

2 T = 3 0 //C3 P = 2 0 //atm4 H = 2 . 6 3 * 1 0 ^ 4

5 x B = 0 . 5


1 b a s i s = 1 0 0 / / l b−mol e /h2 x = 0 . 4 5

3 P H 2 O = 3 1 . 6 //mm o f Hg4 P S O 2 = 1 7 6 //mm o f Hg5 P = 7 6 0 //mm o f Hg6 y =2

7 M 1 = 6 48 M 2 = 1 8


1 T = 7 5 + 27 3 //K2 P 7 5 = 2 8 9 //mm o f Hg3 h r = 0 . 3

4 P o r i g = 8 2 5 //mm o f Hg5 P o r i g B a r = 1 . 1 / / b a r6 V d o t = 1 0 0 0 //Mˆ3/h7 R = 0 . 0 8 3 1


142



1 T = 1 0 0 + 2 7 3 . 2 //K

2 P T = 5 2 6 0 //mm o f Hg3 y = 0 . 1 / / b y v o lu m e4 b a s i s = 1 00 // mol o f f e ed g as


1 P = 7 6 0 //mm o f Hg2 P s t a r = 2 8 9 //mm o f Hg


1 T 2 = 1 5 . 4 + 2 7 3 . 2 //K2 T 1 = 7 . 6 + 2 7 3 . 2 //K3 P 1 = 4 0 //mm o f Hg4 P 2 = 6 0 //mm o f Hg5 T = 4 2 . 2 + 2 7 3 . 2 //K6 R = 8 . 3 1 4 // J/ mol . k


1 y N 2 = 0 . 7 5

2 y H 2 = 1 - y N 2

3 P = 8 0 0 //atm4 T = - 7 0+ 2 73 . 2 //K5 T c H 2 = 3 3 //K6 T c N 2 = 1 2 6 . 2 //K7 P c H 2 = 1 2 . 8 //atm8 P c N 2 = 3 3 . 5 //atm


1 n = 1 0 0 //gm−m o l e s

2 V =5 / / l t r3 T = -2 0. 6 + 2 73 .2 //K4 T c = 1 2 6 . 2 //K5 P c = 3 3 . 5 //atm6 R = 0 . 0 8 2 0 6

143




1 V c a p = 5 0 //Mˆ3/hr2 P = 4 0 / / b a r3 T = 3 0 0 //K4 R = 8 . 3 1 4

5 M = 1 6 . 0 4 //k g/k mol


1 V = 2 . 5 //mˆ32 n = 1 . 0 0 //Kmol3 T = 3 00 //K4 T c = 3 0 4 . 2 //K5 P c = 7 2 . 9 //atm6 w = 0 . 2 2 5

7 R = 0 . 0 8 2 0 6


1 T = - 15 0. 8 + 2 73 .2 // k

2 Vc ap = 3 /2 //L/mol3 T c = 1 2 6 . 2 // k4 P c = 3 3 . 5 //atm5 w = 0 . 0 4 0


1 f l o w i n A = 4 0 0 //L/min2 f l o w i n N = 4 1 9 // mˆ 3 STP / min3 P f i n a l = 6 . 3 / / g a u g e4 T f i n a l = 3 2 5 // C

5 P a c e t o n e = 5 0 1 //mm o f Hg6 D a c e t o n e = 7 9 1 / / g / L7 M a c e t o n e = 5 8 . 0 8 // g8 T 1 = 3 0 0 // k9 P 1 = 1 2 3 8 / /mm Hg o r i g i n a l

144




1 // a l l t h e c a l c u l a t i o n s a re done i n R s c a l e2 T 2 = 2 8 5 + 4 6 0 //R3 T 1 = 3 2 + 4 6 0 //R4 P 2 = 1 . 3 0 //atm5 P 1 = 1 //atm6 V 1 d o t = 3 . 9 5 * 1 0 ^ 5 // f t ˆ3/h


1 V 1 = 1 0 // f t ˆ32 T 1 = 7 0 + 4 6 0 //R3 P 1 = 1 //atm4 P 2 = 2 . 5 //atm5 T 2 = 6 1 0 + 4 6 0 //R


1 T = 3 6 0 + 2 7 3 // Ke l v i n2 P =3 //atm

3 V d o t = 1 1 0 0 / / k g / h4 M = 5 8 . 1


1 T = 2 3 + 2 7 3 / / k e l v i n2 P = 3 + 1 4 . 7 / / p s i3 / / c o n v er s io n o f p r e s s u r e from p s i g t o p s i r e q u i r e s

a d di t i o n o f 1 4. 67 whi ch i s 1 atm4 R = 0 . 0 8 2 0 6 // l t −atm

5 M N 2 = 2 8 / / m o l e c u l a r wt .6 w e i g h t = 1 0 0 //gr ams


145



1 w t p e r c t = 0 . 5

2 D w a t e r = 0 . 9 9 8 //g/cmˆ33 D s u l f u r i c = 1 . 8 3 4 //g/cmˆ3


1 f e e d = 5 9 . 6 / / m o l / s2 x = 0 . 2

3 T o p F l o w 1 = 4 8 . 7 / / m o l / s4 o u t p u t x 1 = 0 . 0 2 1

5 B o t t o m F l o w 1 = 1 0 . 9

6 T o p F l o w 2 = 4 8 . 3

7 o u t p u t x 2 = 0 . 0 6 38 B o t t o m F l o w 2 = 6 . 4


1 b a s i s = 1 0 0 // mol p f p ro du ct g as2 x C O = 0 . 0 1 5

3 x C O 2 = 0 . 0 6 0

4 x O 2 = 0 . 0 8 2

5 x N 2 = 0 . 8 4 3


1 b a s i s = 1 0 0 // mol e t ha n e f e e d2 E 1 = 0 . 9

3 E 2 = 0 . 2 5

4 e x c e s s = 0 . 5


1 b a s i s B u t a n e = 1 0 0 / / m ol / h b u t an e2 b a s i s A i r = 5 0 0 0 //mol /h


146



1 x N 2 w e t = 0 . 6

2 x C O 2 w e t = 0 . 1 53 x O 2 w e t = 0 . 1

4 x H 2 O = 0 . 1 5

5 b a s i s = 1 0 0 / / m o l Wet g a s


1 x 0 = 0 . 9 9 6

2 b a s i s = 1 0 0 // mol co mbi ned f e e d t o t he r e a c t o r3 i n p u t x H 2 = 0 . 7

4 s i n g l e _ p a s s = 0 . 6

5 i n p u t x C O 2 = 0 . 2 86 molI=2

7 I x = 0 . 0 0 4

8 f i n a l = 1 5 5 //k mol /h


1 E = 0 . 9 5

2 b a s i s = 1 0 0 //mol


1 b a s i s = 1 0 0 //mol2 x = 0 . 8 5 0

3 c o n v 1 = 0 . 5 0 1

4 c o n v 2 = 0 . 4 7 1


1 b a s i s = 1 0 0 //mol2 x P = 0 . 1

3 x N = 0 . 1 2

4 x A = 0 . 7 8

5 x = 0 . 3

147




1 f e e d = 4 5 0 0 / / k g / h2 f e e d x = 0 . 3 3 3

3 m 3 x = 0 . 4 9 4

4 m 5 x = 0 . 3 6 4

5 x = 0 . 9 5


1 x 1 = 0 . 9 6 0

2 x 2 = 0 . 9 7 7

3 x 3 = 0 . 9 8 34 b a s i s = 1 0 0 //mol


1 m a s s i n = 1 0 0 / / k g2 M 1 = 1 0 0 //k g3 M 2 = 7 5 //Kg4 m a s s o u t = 4 3 . 1 //k g5 i n p u t x = 0 . 5

6 o u t p u t x A = 0 . 0 5 37 o u t p u t x M = 0 . 0 1 6

8 m 1 x A = 0 . 2 7 5

9 m 1 x M = 0 . 7 2 5

10 m 3 x W = 0 . 0 3

11 m 3 x A = 0 . 0 9

12 m 3 x M = 0 . 8 8


1 i n p u t M a s s 1 = 1 0 0 / / k g / h2 i n p u t M a s s 2 = 3 0 //Kg/h3 o u t p u t M a s s 1 = 4 0 //Kg/h4 o u t p u t M a s s 2 = 3 0 //Kg/h5 i n p u t x 1 = 0 . 5

148



6 i n p u t x 2 = 0 . 3

7 o u t p u t x 1 = 0 . 98 o u t p u t x 2 = 0 . 6


1 i n p u t x = 0 . 4 5

2 o u t p u t x = 0 . 9 5

3 b a s i s = 2 0 0 0 //L/h4 o u t p u t B a s i s = 1 0 0 //Kmol5 M 1 = 7 8 . 1 1

6 M 2 = 9 2 . 1 3

7 D = 0 . 8 7 28 z = 0 . 0 8


1 b a s i s = 1 0 0 //k g2 i n p u t x = 0 . 2

3 o u t p u t x = 0 . 0 8

4 D =1 //k g/L


1 b a s i s = 1 0 0 //mol2 F i n a l B a s i s = 1 2 5 0 / / l b−m o l e s / h


1 V d o t = 2 0 //CC/min2 x = 0 . 0 1 5

3 M H 2 O = 1 8 . 0 2 // g4 DH2O=1

//g/CC5 x 1 = 0 . 2


149



1 r a t e = 0 . 1 //kmol/min

2 x 1 = 0 . 1 // m ol e f r a c t i o n h ex an e v ap ou r3 v o l = 1 0 //mˆ34 d = 0 . 6 5 9 //k g/L5 M = 8 6 . 2 //Kg/kmol


1 m 1 = 2 0 0 // g2 m 2 = 1 5 0 // g3 x 1 = 0 . 4 / / m e t h a n o l / g4 x 2 = 0 . 7 / / m e t h a n o l / g


1 i n p u t B e n z e n e = 5 0 0 / / k g / h2 i n p u t T o l u e n e = 5 0 0 / / k g / h3 U p S t r e a m B e n z e n e = 4 5 0 / / k g / h4 D o w n S t r e a m T o l u e n e = 4 7 5 / / k g / h


1 input =50000 / / p p l / y r2 g e n e r a t i o n = 2 2 0 0 0 / / p p l / y r

3 c o n s u m p t i o n = 1 9 0 0 0 / / p p l / y r4 o u t p u t = 7 5 0 0 0 / / p p l / y r


1 T 1 = 2 0 //F2 T 2 = 8 0 //F


1 P 0 = 1 0 . 4 * 1. 0 1 3 *1 0 ^ 5 / 1 0. 3 3 //m H2O2 D = 1 0 0 0 //kg/mˆ33 g = 9 . 8 0 7 //m/sˆ24 h = 3 0 //m

150



5 f l o w r a t e 2 = 1 0 0 0 //mol/min

6 m a s s o f A = 3 0 0 //lbm7 m o l a r B = 2 8 //kmolB/s


1 m a s s = 1 0 0 / / g o f CO22 M = 4 4 . 0 1 / / m o l e c u l a r w e i g ht


1 T 1 = 2 0 // C2 T 2 = 1 0 0 // C3 V a t 2 0 = 0 . 5 6 0 // f t ̂ 34 D = 0 . 0 2 0 8 3 3 3 // f t


1 m a s s = 2 1 5 //k g


1 T = [1 0 20 40 80]2 M = [1 4 .7 6 2 0. 14 2 7. 73 3 8. 47 ]

3 s q r t T = sqrt ( T ) ;


1 x = [1 0 30 50 70 90 ]

2 y = [ 20 5 2. 1 8 4. 6 1 18 .3 1 51 .0 ]


1 / / t he b ad ba tc he s p er week a r e t ak en a s e l em e nt s o f av e c t o r y

2 y =[17 27 18 18 23 19 18 21 20 19 21 18]

152

elementary principles of chemical processes_r. m. felder and r. w. rousseau.pdf

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