Introduction to Chemical Engineering

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Scilab Textbook Companion for

Introduction To Chemical Engineering

by S. K. Ghoshal, S. K. Sanyal And S. Datta

1 Created by

Himanshu Bhatia

Btech

Chemical Engineering

IIT Guwahati

College Teacher

Dr Prakash Kotecha

Cross-Checked by

Ganesh R

October 3, 2013

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

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Book Description

Title: Introduction To Chemical Engineering Author: S. K. Ghoshal, S. K. Sanyal And S. Datta

Publisher: Tata McGraw Hill Education Pvt. Ltd., New Delhi Edition: 1

Year: 2006

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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.

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List of Scilab Codes

Exa 1.1 Air composition . . . 9

Exa 1.2 Volume calculation . . . 10

Exa 1.3 Gas Composition . . . 10

Exa 1.4 Volume calculation . . . 12

Exa 1.5 Amount of CO2 released. . . 12

Exa 1.6 Vapor pressure . . . 13

Exa 1.7 Duhring Plot calculations . . . 14

Exa 1.8 Vapor Pressure of Mixture . . . 14

Exa 1.9 Vapor pressure . . . 15

Exa 1.10 Flow relation . . . 16

Exa 1.11 Average Velocity . . . 16

Exa 1.12 Velocity determination . . . 17

Exa 1.13 Velocity determination . . . 18

Exa 1.14 Dimensional analysis . . . 19

Exa 1.15 Dimensional analysis . . . 19

Exa 1.16 Dynamic similarity . . . 19

Exa 1.17 Dynamic similarity . . . 20

Exa 1.18 Nomographic chart . . . 21

Exa 1.19 Calculation using Nomograph . . . 22

Exa 2.1 Ideal gas system . . . 23

Exa 2.2 Mixture properties . . . 23

Exa 2.3 Equivalent metal mass . . . 24

Exa 2.4 Purity of Sodium Hydroxide. . . 25

Exa 2.5 Carbon content formulation . . . 26

Exa 2.6 Combustion of gas . . . 26

Exa 2.7 Sulphuric acid preparation . . . 27

Exa 2.8 Molarity Molality Normality Calculation . . . 28

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Exa 2.10 Precipitation of KClO3 . . . 29

Exa 2.11 Solubility of CO2 . . . 29

Exa 2.12 Vapor pressure calculation. . . 30

Exa 2.13 Boiling point calculation . . . 30

Exa 2.14 Colligative properties . . . 31

Exa 2.15 Huggins Equation . . . 32

Exa 2.16 Molecular Formula . . . 32

Exa 2.20 Metal deposition . . . 35

Exa 2.21 EMF of cell . . . 36

Exa 2.24 Silver deposition . . . 38

Exa 2.25 Electroplating time. . . 38

Exa 2.26 Water hardness . . . 39

Exa 2.29 Mixture composition . . . 41

Exa 2.30 Mixture composition . . . 42

Exa 2.31 Mixture properties . . . 43

Exa 2.32 Humidity . . . 44

Exa 3.1 Coal consumption . . . 46

Exa 3.2 Nitric acid preparation. . . 47

Exa 3.3 HCl production . . . 47

Exa 3.4 Acetylene consumption . . . 48

Exa 3.5 Screen effectiveness . . . 49

Exa 3.6 Absorption . . . 50 Exa 3.7 Extraction . . . 51 Exa 3.8 Distillation . . . 52 Exa 3.9 Distillation . . . 53 Exa 3.10 Crystallization . . . 54 Exa 3.11 crystallization . . . 55 Exa 3.12 Drying . . . 55

Exa 3.13 Conditioning of air . . . 56

Exa 3.14 Ammonia Synthesis . . . 57

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Exa 3.16 Enthalpy calculation . . . 59

Exa 3.17 Enthalpy of formation . . . 59

Exa 3.18 Combustion . . . 60

Exa 3.19 Heat of reaction . . . 61

Exa 3.20 Heat transfer . . . 61

Exa 3.21 Calorific value . . . 62

Exa 3.22 Coal combustion . . . 62

Exa 3.23 Coal combustion . . . 64

Exa 3.24 Petrol combustion . . . 66

Exa 3.25 Air supply . . . 67

Exa 3.26 CO2 cooling . . . 68

Exa 3.27 Heating area . . . 68

Exa 3.28 Distillation column . . . 69

Exa 3.29 Crystallization . . . 71

Exa 3.30 Combustion . . . 72

Exa 4.1 Water compressibility . . . 73

Exa 4.2 Isothermal Compressibility . . . 73

Exa 4.3 Viscosity . . . 74

Exa 4.4 Streamline flow . . . 74

Exa 4.5 Frictional losses . . . 74

Exa 4.6 Velocity profile . . . 75

Exa 4.7 Velocity profile . . . 75

Exa 4.8 Boundary layer . . . 75

Exa 4.9 Pipe flow . . . 76

Exa 4.10 Temperature rise . . . 77

Exa 4.11 Bernoulli equation . . . 78

Exa 4.12 Power requirements . . . 78

Exa 4.13 Hagen Poiseulle equation . . . 79

Exa 4.14 Pressure Head calculation . . . 80

Exa 4.15 Level difference calculation . . . 80

Exa 4.16 Energy cost calculation . . . 81

Exa 4.17 Pressure loss . . . 82

Exa 4.18 Pressure gradient . . . 82

Exa 4.19 Flow rate . . . 83

Exa 4.20 Pipe dimensions . . . 84

Exa 4.21 Pressure gradient . . . 84

Exa 4.22 Minimum fluidization velocity . . . 85

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Exa 5.1 Heat conduction . . . 87

Exa 5.2 Heat conduction . . . 87

Exa 5.3 Heat conduction through sphere . . . 88

Exa 5.4 Composite wall . . . 89

Exa 5.5 Composite Pipeline . . . 90

Exa 5.6 Parellel Resistance . . . 91

Exa 5.7 Heat transfer coefficient . . . 92

Exa 5.8 Heat transfer coefficient . . . 93

Exa 5.9 Earth Temperature. . . 94

Exa 5.10 Earth Temperature. . . 94

Exa 5.11 Equilibrium temperature . . . 95

Exa 5.12 Equilibrium temperature . . . 95

Exa 5.13 Temperature calculation . . . 96

Exa 5.14 Solar constant . . . 96

Exa 5.15 Evaporator . . . 97

Exa 6.1 Diffusivity . . . 100

Exa 6.2 Absorption . . . 100

Exa 6.3 Equimolar counter diffusion . . . 101

Exa 6.4 Resistane to diffusion . . . 101

Exa 6.5 Vapor diffusion . . . 102

Exa 6.6 Flux of HCl . . . 102

Exa 6.7 Vaporization . . . 103

Exa 6.8 Gas Absorption. . . 104

Exa 6.9 Equilibrium Composition . . . 105

Exa 6.10 Equilibrium Composition . . . 105

Exa 6.11 Vapor Liquid Equilibrium . . . 106

Exa 6.12 Distillation Column . . . 106

Exa 6.13 Distillation . . . 107

Exa 6.14 Steam Distillation . . . 108

Exa 6.15 Mcabe Thiele Method . . . 108

Exa 6.16 Liquid liquid extraction . . . 109

Exa 6.17 Liquid liquid extraction . . . 110

Exa 6.18 Humidity calculation . . . 110

Exa 6.19 Drying operation . . . 111

Exa 6.20 Crystallization . . . 112

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Exa 7.1 Constant volume reaction . . . 115

Exa 7.2 Rate of reaction . . . 115

Exa 7.4 Order of reaction . . . 116

Exa 7.5 Rate Expression . . . 116

Exa 7.6 Volume function . . . 117

Exa 7.7 Pressure time relation . . . 117

Exa 7.8 Entropy changes . . . 117

Exa 7.9 Hydrocarbon cracking . . . 118

Exa 7.10 Equilibrium conversion . . . 118

Exa 7.12 Concentration calculation . . . 119

Exa 7.14 Equilibrium shifts . . . 121

Exa 7.15 Rate equation . . . 121

Exa 8.1 Specific gravity . . . 124

Exa 8.4 Mixture density . . . 125

Exa 8.5 Viscosity calculation . . . 126

Exa 8.6 Solution viscosity . . . 126

Exa 8.7 Flow rate calculation . . . 127

Exa 8.8 Venturi meter. . . 128

Exa 8.9 Venturi meter. . . 129

Exa 8.10 Pitot tube . . . 130

Exa 8.11 Rotameter capacity . . . 131

Exa 8.12 Flow rate calculation . . . 131

Exa 9.1 Coiled tube pressure drop . . . 133

Exa 9.2 Heat exchanger pressure drop . . . 134

Exa 9.3 Heat exchanger area . . . 135

Exa 9.4 Batch distillation . . . 136

Exa 9.5 Gas mixture exit temperature . . . 137

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Chapter 1 Introduction

Scilab code Exa 1.1 Air composition

1 clc 2 c l e a r 3 p r i n t f( ” e x a m p l e 1 . 1 p a g e number 1 9 \ n \ n ” ) 4 // t o f i n d c o m p o s i t i o n o f a i r by w e i g h t 5 y _ o x y g e n = 0.21 // m o l e f r a c t i o n o f o x y g e n 6 y _ n i t r o g e n = 0.79 // m o l e f r a c t i o n o f n i t r o g e n 7 m o l a r _ m a s s _ o x y g e n = 32 8 m o l a r _ m a s s _ n i t r o g e n = 28 9 10 m o l a r _ m a s s _ a i r = y _ o x y g e n * m o l a r _ m a s s _ o x y g e n + y _ n i t r o g e n * m o l a r _ m a s s _ n i t r o g e n ; 11 m a s s _ f r a c t i o n _ o x y g e n = y _ o x y g e n * m o l a r _ m a s s _ o x y g e n / m o l a r _ m a s s _ a i r ; 12 m a s s _ f r a c t i o n _ n i t r o g e n = y _ n i t r o g e n * m o l a r _ m a s s _ n i t r o g e n / m o l a r _ m a s s _ a i r ; 13 14 p r i n t f( ” mass f r a c t i o n o f o x y g e n = %f \ n \ n ” , m a s s _ f r a c t i o n _ o x y g e n ) 15 p r i n t f( ” mass f r a c t i o n o f n i t r o g e n = %f \ n \ n ” , m a s s _ f r a c t i o n _ n i t r o g e n ) 16

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17 V1 = 22.4 // i n l i t e r s 18 P1 = 760 // i n mm Hg 19 P2 = 7 3 5 . 5 6 // i n mm Hg 20 T1 = 273 // i n K 21 T2 = 298 // i n K 22 23 V2 = ( P1 * T2 * V1 ) /( P2 * T1 ) ; 24 d e n s i t y = m o l a r _ m a s s _ a i r / V2 ; 25 26 p r i n t f( ” d e n s i t y = %f gm/ l ” , d e n s i t y )

Scilab code Exa 1.2 Volume calculation

1 clc 2 c l e a r 3 p r i n t f( ” e x a m p l e 1 . 2 p a g e number 2 0 \ n \ n ” ) 4 // f i n d t h e v o l u m e o c c u p i e d by p r o p a n e 5 6 m a s s _ p r o p a n e = 1 4 . 2 // i n kg 7 m o l a r _ m a s s =44 // i n kg 8 m o l e s =( m a s s _ p r o p a n e * 1 0 0 0 ) / m o l a r _ m a s s ; 9 v o l u m e = 2 2 . 4 * m o l e s ; // i n l i t e r s 10 11 p r i n t f( ” v o l u m e = %d l i t e r s \ n \ n ” , v o l u m e )

Scilab code Exa 1.3 Gas Composition

1 clc 2 c l e a r 3 p r i n t f( ” e x a m p l e 1 . 3 p a g e number 2 0 \ n \ n ” ) 4 // t o f i n d t h e a v e r a g e w e i g h t , w e i g h t c o m p o s i t i o n , g a s v o l u m e i n a b s e n c e o f SO2 5 y _ C O 2 = 0 . 2 5 ;

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y_CO = 0 . 0 0 2 ;

7 y _ S O 2 = 0 . 0 1 2 ;

8 y_N2 = 0 . 6 8 0 ;

9 y_O2 = 0 . 0 5 6 ;

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11 Mm = y _ C O 2 *44+ y_CO *28+ y _ S O 2 *64+ y_N2 *28+ y_O2 *32;

12 p r i n t f ( ” \ n m o l a r mass = %d \ n ” , Mm ) 13 14 p r i n t f( ” \ n f i n d i n g w e i g h t c o m p o s i t i o n \ n ” ) 15 w _ C O 2 = y _ C O 2 * 4 4 * 1 0 0 / Mm ; 16 p r i n t f ( ” \ n w e i g h t C O 2 = %f \ n \ n ” , w _ C O 2 ) 17 w_CO = y_CO * 2 8 * 1 0 0 / Mm ; 18 p r i n t f ( ” w e i g h t C O = %f \ n \ n ” , w_CO ) 19 w _ S O 2 = y _ S O 2 * 6 4 * 1 0 0 / Mm ; 20 p r i n t f ( ” w e i g h t S O 2 = %f \ n \ n ” , w _ S O 2 ) 21 w_N2 = y_N2 * 2 8 * 1 0 0 / Mm ; 22 p r i n t f ( ” w e i g h t N 2 = %f \ n \ n ” , w_N2 ) 23 w_O2 = y_O2 * 3 2 * 1 0 0 / Mm ; 24 p r i n t f ( ” w e i g h t O 2 = %f \ n \ n ” , w_O2 ) 25 26 p r i n t f( ” i f SO2 i s r emoved \ n \ n ” ) 27 v _ C O 2 = 25; 28 v_CO = 0.2; 29 v_N2 = 6 8 . 0 ; 30 v_O2 = 5.6; 31 v = v _ C O 2 + v_CO + v_N2 + v_O2 ; 32 v 1 _ C O 2 = ( v _ C O 2 * 1 0 0 / 9 8 . 8 ) ; 33 34 p r i n t f ( ” volume CO2 = %f \ n \ n ” , v 1 _ C O 2 ) 35 v 1 _ C O = ( v_CO * 1 0 0 / 9 8 . 8 ) ; 36 p r i n t f ( ” volume CO = %f \ n \ n ” , v 1 _ C O ) 37 v 1 _ N 2 = ( v_N2 * 1 0 0 / 9 8 . 8 ) ; 38 p r i n t f ( ” v o l u m e N 2 = %f \ n \ n ” , v 1 _ N 2 ) 39 v 1 _ O 2 = ( v_O2 * 1 0 0 / 9 8 . 8 ) ; 40 p r i n t f ( ” v o l u m e O 2 = %f \ n \ n ” , v 1 _ O 2 )

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Scilab code Exa 1.4 Volume calculation 1 clc 2 c l e a r 3 p r i n t f( ” e x a m p l e 1 . 4 p a g e number 2 4 \ n \ n ” ) 4 // t o f i n d v o l u m e o f NH3 d i s s o l v a b l e i n w a t e r 5 6 p =1 // atm 7 H =2.7 // atm 8 x = p / H ; 9 10 m o l e _ r a t i o = ( x ) /(1 - x ) ; 11 m o l e s _ o f _ w a t e r = ( 1 0 0 * 1 0 0 0 ) /18; 12 m o l e s _ o f _ N H 3 = m o l e _ r a t i o * m o l e s _ o f _ w a t e r ; 13 14 p r i n t f( ” m o l e s o f NH3 d i s s o l v e d = %f \ n \ n ” , m o l e s _ o f _ N H 3 ) 15 16 v o l u m e _ N H 3 =( m o l e s _ o f _ N H 3 * 2 2 . 4 * 2 9 3 ) / 2 7 3 ; 17 p r i n t f( ” v o l u m e o f NH3 d i s s o l v e d = %f l i t e r s ” , v o l u m e _ N H 3 )

Scilab code Exa 1.5 Amount of CO2 released

1 clc 2 c l e a r 3 4 p r i n t f( ” e x a m p l e 1 . 5 p a g e number 2 4 \ n \ n ” ) 5 6 // t o c a l c u l a t e amount o f CO2 r e l e a s e d by w a t e r 7 p =746 // i n mm Hg 8 H = 1 . 0 8 * 1 0 ^ 6 // i n mm Hg , Henry ’ s c o n s t a n t

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10 x = p / H ; // m o l e f r a c t i o n o f CO2 11 X = x * ( 4 4 / 1 8 ) ; // mass r a t i o o f CO2 i n w a t e r 12 13 i n i t i a l _ C O 2 = 0 . 0 0 5 ; // kg CO2/ kg H20 14 G = 1 0 0 0 * ( i n i t i a l _ C O 2 - X ) ; 15 16 p r i n t f( ”CO2 g i v e n up by 1 c u b i c m e t e r o f w a t e r = %f kg CO2/ c u b i c m e t e r H20 ” ,G )

Scilab code Exa 1.6 Vapor pressure

1 clc 2 c l e a r 3 4 p r i n t f( ’ e x a m p l e 1 . 6 p a g e number 27 \ n \ n ’ ) 5 // t o f i n d v a p o r p r e s s r e o f e t h y l a l c h o h a l 6 7 pa1 = 2 3 . 6 ; //VP o f e t h y l a l c h o h a l a t 10 d e g r e e C 8 pa3 =760 //VP o f e t h y l a l c h o h a l a t 7 8 . 3 d e g r e e C i n mm Hg 9 pb1 = 9.2 //VP o f e t h y l w a t e r a t 10 d e g r e e C i n mm Hg 10 pb3 =332 //VP o f e t h y l w a t e r a t 7 8 . 3 d e g r e e C i n mm Hg 11 12 C =(l o g 1 0( pa1 / pa3 ) /(l o g 1 0( pb1 / pb3 ) ) ) ; 13 14 pb2 =149 //VP o f w a t e r a t 60 d e g r e e C i n mm Hg 15 16 pas =( pb3 / pb2 ) ; 17 pa = C *l o g 1 0( pas ) ; 18 pa2 = pa3 / ( 1 0 ^ pa ) ; 19

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20 p r i n t f( ” v a p o r p r e s s u r e o f e t h y l a l c h o l o h a t 60 d e g r e e C = %f mm Hg” , pa2 )

Scilab code Exa 1.7 Duhring Plot calculations

1 clc 2 c l e a r 3 4 p r i n t f( ’ e x a m p l e 1 . 7 p a g e number 28 \ n \ n ’ ) 5 6 // t o f i n d v a p o r p r e s s u r e u s i n g d u h r i n g p l o t 7 8 t1 = 41 // i n d e g r e e C 9 t2 =59 // i n d e g r e e C 10 t h e t a _ 1 =83 // i n d e g r e e C 11 t h e t a _ 2 =100 // i n d e g r e e C 12 13 K = ( t1 - t2 ) /( theta_1 - t h e t a _ 2 ) ; 14 t = 5 9 + ( K * ( 1 0 4 . 2 - 1 0 0 ) ) ; 15 16 p r i n t f ( ” b o i l i n g p o i n t o f S C l 2 a t 8 8 0 T o r r = %f d e g r e e c e l c i u s ” ,t )

Scilab code Exa 1.8 Vapor Pressure of Mixture

1 clc 2 c l e a r 3 p r i n t f( ’ e x a m p l e 1 . 8 p a g e number 2 9 \ n \ n ’ ) 4 // t o f i n d t h e amount o f s t e a m r e l e a s e d 5 6 v p _ C 6 H 6 = 520 // i n t o r r 7 v p _ H 2 O = 225 // i n t o r r 8 m a s s _ w a t e r =18

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m a s s _ b e n z e n e =78 10 11 a m o u n t _ o f _ s t e a m = ( v p _ H 2 O / v p _ C 6 H 6 ) /( m a s s _ b e n z e n e / m a s s _ w a t e r ) ; 12 13 p r i n t f( ” amount o f s t e a m = %f ” , a m o u n t _ o f _ s t e a m )

Scilab code Exa 1.9 Vapor pressure

1 clc 2 c l e a r 3 p r i n t f( ’ e x a m p l e 1 . 9 p a g e number 3 0 \ n \ n ’ ) 4 5 // t o f i n d e q u i l i b r i u m v a p o r l i q u i d c o m p o s i t i o n 6 p0b = 385 // v a p o r p r e s s u e o f b e n z e n e a t 60 d e g r e e C i n t o r r 7 p0t =140 // v a p o r p r e s s u e o f t o l u e n e a t 60 d e g r e e C i n t o r r 8 xb = 0 . 4 ; 9 xt = 0 . 6 ; 10 11 pb = p0b * xb ; 12 pt = p0t * xt ; 13 P = pb + pt ; 14 15 p r i n t f( ” t o t a l p r e s s u r e = %f t o r r \ n \ n ” ,P ) 16 17 yb = pb / P ; 18 yt = pt / P ; 19 p r i n t f( ” v a p o r c o m p o s i t i o n o f b e n z e n e = %f \ n v a p o r c o m p o s i t i o n o f t o l u e n e = %f \ n \ n ” ,yb , yt ) 20 21 // f o r l i q u i d b o i l i n g a t 90 d e g r e e C and 7 6 0 t o r r , l i q u i d p h a s e c o m p o s i t i o n 22 // x =(760 −408) / ( 1 0 1 3 − 4 0 8 ) ;

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23 ( 1 0 1 3 * x ) + ( 4 0 8 * ( 1 - x ) ) = = 7 6 0 ;

24 p r i n t f( ” m o l e f r a c t i o n o f b e n z e n e i n l i q u i d m i x t u r e = %f \ n m o l e f r a c t i o n o f t o l u e n e i n l i q u i d m i x t u r e = %f ” ,x ,1 - x )

Scilab code Exa 1.10 Flow relation

1 clc 2 c l e a r 3 4 p r i n t f( ’ e x a m p l e 1 . 1 0 p a g e number 3 3 \ n ’ ) 5 6 // t o f i n d r e l a t i o n b e t w e e n f r i c t i o n f a c t o r and r e y n o l d ’ s number 7

8 // l o g f=y , l o g Re=x , l o g a=c

9 s i g m a _ x = 2 3 . 3 9 3 ; 10 s i g m a _ y = - 1 2 . 4 3 7 ; 11 s i g m a _ x 2 = 9 1 . 4 5 6 12 s i g m a _ x y = - 4 8 . 5 5 4 ; 13 m = ( ( 6 * s i g m a _ x y ) -( s i g m a _ x * s i g m a _ y ) ) /(6* sigma_x2 -( s i g m a _ x ) ^2) ; 14 p r i n t f( ”m = %f \ n ” , m ) 15 16 c =(( s i g m a _ x 2 * s i g m a _ y ) -( s i g m a _ x y * s i g m a _ x ) ) /(6* sigma_x2 -( s i g m a _ x ) ^2) ; 17 p r i n t f( ” c = %f \ n ” , c ) 18 19 p r i n t f( ” f = 0 . 0 8 4 ∗ Re ˆ − 0 . 2 5 6 ” )

Scilab code Exa 1.11 Average Velocity

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c l e a r 3 p r i n t f( ” e x a m p l e 1 . 1 1 p a g e number 3 5 \ n \ n ” ) 4 5 // t o f i n d t h e a v e r a g e v e l o c i t y 6 7 u = [ 2 ; 1 . 9 2 ; 1 . 6 8 ; 1 . 2 8 ; 0 . 7 2 ; 0 ] ; 8 r = [ 0 ; 1 ; 2 ; 3 ; 4 ; 5 ] ; 9 10 z = u .* r ; 11 plot( r , z ) 12 t i t l e ( ” v a r i a t i o n o f u r w i t h r ” ) 13 x l a b e l ( ” r ” ) 14 y l a b e l ( ” u r ” ) 15 16 // by g r a p h i c a l i n t e g r a t i o n , we g e t 17 u _ a v g = ( 2 / 2 5 ) * 1 2 . 4 18 p r i n t f( ” a v e r a g e v e l o c i t y = %f cm/ s \ n ” , u_avg )

Scilab code Exa 1.12 Velocity determination

1 clc 2 c l e a r 3 4 p r i n t f( ’ e x a m p l e 1 . 1 2 p a g e number 3 7 \ n ’ ) 5 6 // t o f i n d t h e a v e r a g e v e l o c i t y 7 p r i n t f( ’ u s i n g t r a p e z o i d r u l e \ n ’ ) 8 9 n = 6; 10 h = (3 - 0) / n ; 11 I = ( h /2) * ( 0 + 2 * 0 . 9 7 + 2 * 1 . 7 8 + 2 * 2 . 2 5 + 2 * 2 . 2 2 + 2 * 1 . 5 2 + 0 ) ; 12 u _ a v g = ( 2 / 3 ^ 2 ) * I ; 13 14 p r i n t f( ” a v e r a g e v e l o c i t y = %f cm/ s \ n ” , u_avg ) 15

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16 disp( ’ S i m p s o n s r u l e ’ ) 17 18 n = 6; 19 h = 3/ n ; 20 I = ( h /3) * ( 0 + 4 * ( 0 . 9 7 + 2 . 2 5 + 1 . 5 2 ) + 2 * ( 1 . 7 8 + 2 . 2 2 ) +0) ; 21 u _ a v g = ( 2 / 3 ^ 2 ) * I ; 22 23 p r i n t f( ” a v e r a g e v e l o c i t y = %f cm/ s \ n ” , u_avg )

Scilab code Exa 1.13 Velocity determination

1 clc 2 c l e a r 3 4 p r i n t f( ’ e x a m p l e 1 . 1 3 p a g e number 3 8 \ n \ n ’ ) 5 6 // t o f i n d t h e s e t t l i n g v e l o c i t y a s a f u n c t i o n o f t i m e 7 z0 = 3 0 . 8 4 ; 8 z1 = 2 9 . 8 9 ; 9 z2 = 2 9 . 1 0 ; 10 h = 4; 11 12 u 1 _ t 0 = ( -3* z0 +4* z1 - z2 ) /(2* h ) ; 13 u 1 _ t 4 = ( - z0 + z2 ) /(2* h ) ; 14 u 1 _ t 8 = ( z0 -4* z1 +3* z2 ) /(2* h ) ; 15 16 // c o n s i d e r i n g d a t a s e t f o r t = 4 , 8 , 1 2 min 17 z0 = 2 9 . 8 9 ; 18 z1 = 2 9 . 1 0 ; 19 z2 = 2 8 . 3 0 ; 20 u 2 _ t 4 = ( -3* z0 +4* z1 - z2 ) /(2* h ) ; 21 u 2 _ t 8 = ( - z0 + z2 ) /(2* h ) ; 22 u 2 _ t 1 2 = ( z0 -4* z1 +3* z2 ) /(2* h ) ; 23

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24 // c o n s i d e r i n g d a t a s e t f o r t = 8 , 1 2 , 1 6 min 25 z0 = 2 9 . 1 0 ; 26 z1 = 2 8 . 3 0 ; 27 z2 = 2 7 . 5 0 ; 28 u 3 _ t 8 = ( -3* z0 +4* z1 - z2 ) /(2* h ) ; 29 u 3 _ t 1 2 = ( - z0 + z2 ) /(2* h ) ; 30 u 3 _ t 1 6 = ( z0 -4* z1 +3* z2 ) /(2* h ) ; 31 32 // t a k i n g a v e r a g e 33 u_t4 = ( u 1 _ t 4 + u 2 _ t 4 ) /2; 34 u_t8 = ( u 1 _ t 8 + u 2 _ t 8 + u 3 _ t 8 ) /3; 35 u _ t 1 2 = ( u 2 _ t 1 2 + u 3 _ t 1 2 ) /2; 36 37 p r i n t f( ” u t 0 = %f cm/ min \ n u t 4 = %f cm/ min \ n u t 8 = %f cm/ min \ n u t 1 2 = %f / n cm/ min \ n u t 1 6 =%f / n cm/ min ” , u1_t0 , u_t4 , u_t8 , u_t12 , u3_t16 )

Scilab code Exa 1.14 Dimensional analysis

1 p r i n t f( ’ e x a m p l e 1 . 1 4 p a g e number 45 ’ )

2 disp ( ” t h i s i s a t h e o r i t i c a l q u e s t i o n , book s h a l l be r e f e r r e d f o r s o l u t i o n ” )

Scilab code Exa 1.15 Dimensional analysis

1 p r i n t f( ’ e x a m p l e 1 . 1 5 p a g e number 46 ’ )

2 disp ( ” t h i s i s a t h e o r i t i c a l q u e s t i o n , book s h a l l be r e f e r r e d f o r s o l u t i o n ” )

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1 clc 2 c l e a r 3 p r i n t f( ’ e x a m p l e 1 . 1 6 p a g e number 4 9 \ n ’ ) 4 5 // t o f i n d t h e f l o w r a t e and p r e s s u r e d r o p 6 d e n s i t y _ w a t e r =988 // i n kg /m3 7 v i s c o s i t y _ w a t e r = 5 5 * 1 0 ^ - 5 // i n Ns /m2 8 d e n s i t y _ a i r = 1 . 2 1 // i n kg /m3 9 v i s c o s i t y _ a i r = 1 . 8 3 * 1 0 ^ - 5 // i n Ns /m2 10 L =1 // l e n g t h i n m 11 12 L1 =10* L // l e n g t h i n m 13 Q = 0 . 0 1 3 3 ; 14 15 Q1 =(( Q * d e n s i t y _ w a t e r * v i s c o s i t y _ a i r * L ) /( L1 * v i s c o s i t y _ w a t e r * d e n s i t y _ a i r ) ) 16 17 p r i n t f( ” f l o w r a t e = %f c u b i c m e t e r / s \ n ” , Q1 ) 18 19 // e q u a t i n g e u l e r number 20 21 p = 9 . 8 0 6 7 * 1 0 ^ 4 ; // p r e s s u r e i n p a s c a l 22 p1 =( p * d e n s i t y _ w a t e r * Q ^2* L ^4) /( d e n s i t y _ a i r * Q1 ^2* L1 ^4) ; 23 24 p r i n t f( ” p r e s s u r e d r o p c o r r e s p o n d i n g t o 1 kp / s q u a r e cm = %f kP/ s q u a r e cm” , p1 / p )

Scilab code Exa 1.17 Dynamic similarity

1 clc

2 c l e a r

3 p r i n t f( ’ e x a m p l e 1 . 1 7 p a g e number 5 0 \ n ’ )

4

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7 L =1 // l e n g t h o f p r o t o t y p e i n m 8 L1 =10* L // l e n g t h o f mo del i n m 9 d e n s i t y _ p r o t o t y p e = 2 . 6 5 //gm/ c c 10 d e n s i t y _ w a t e r =1 //gm/ c c 11 12 d e n s i t y _ m o d e l =( L ^3*( d e n s i t y _ p r o t o t y p e - d e n s i t y _ w a t e r ) ) /( L1 ^3) +1; 13 14 p r i n t f( ” s p e c i f i c g r a v i t y o f p l a s t i c = %f ” , d e n s i t y _ m o d e l )

Scilab code Exa 1.18 Nomographic chart

1 clc 2 c l e a r 3 p r i n t f( ’ e x a m p l e 1 . 1 8 p a g e number 5 3 \ n \ n ’ ) 4 5 // t o f i n d e r r o r i n a c t u a l d a t a and n o m o g r a p h i c c h a t v a l u e 6 7 // f o r my 8 ly = 8 // i n cm 9 my = ly / ( ( 1 / 0 . 2 5 ) - ( 1 / 0 . 5 ) ) ; 10 lz = 1 0 . 1 5 // i n cm 11 mz = lz / ( ( 1 / 2 . 8 5 ) - ( 1 / 6 . 7 6 ) ) ; 12 mx = ( my * mz ) /( my + mz ) ; 13 p r i n t f( ”mx = %f cm\ n ” , mx ) 14 err = ( ( 1 - 0 . 9 9 4 5 ) / 0 . 9 9 4 5 ) * 1 0 0 ; 15 p r i n t f( ” \ n e r r o r = %f \ n ” , err ) 16 17 x = 2 18 y = 0 . 5 : 0 . 5 : 2 . 5 ; 19 20 plot( x , y )

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21 t i t l e ( ” nomograph ” ) 22 x l a b e l ( ” x ” ) 23 y l a b e l ( ” y ” ) 24 25 x = 3 26 y = 0 . 4 : 0 . 2 : 2 ; 27 plot( x , y )

Scilab code Exa 1.19 Calculation using Nomograph

1 clc 2 c l e a r 3 p r i n t f( ’ e x a m p l e 1 . 1 9 p a g e number 5 4 \ n ’ ) 4 5 // t o f i n d t h e e c o n o m i c p i p e d i a m e t e r f r o m nomograph 6 // f r o m t h e nomograph , we g e t t h e v a l u e s o f w and d e n s i t y 7 8 w =450 // i n kg / h r 9 d e n s i t y = 1 0 0 0 // i n kg /m3 10 d =16 // i n mm 11 12 u =( w / d e n s i t y ) / ( 3 . 1 4 * d ^ 2 / 4 ) ; 13 Re = u * d e n s i t y * d / 0 . 0 0 1 ; 14 15 if Re > 2 1 0 0 then p r i n t f( ” f l o w i s t u r b u l e n t and d= %f mm” ,d )

16 else disp ( ” f l o w i s l a m i n a r and t h i s nomograph i s n o t v a l i d ” )

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Chapter 2 Physico Chemical Calculations

Scilab code Exa 2.1 Ideal gas system

1 clc 2 c l e a r 3 p r i n t f( ” e x a m p l e 2 . 1 p a g e number 7 1 \ n \ n ” ) 4 5 // t o f i n d t h e v o l u m e o f o x y g e n t h a t c a n be o b t a i n e d 6 7 p1 =15 // i n b a r 8 p2 = 1 . 0 1 3 // i n b a r 9 t1 =283 // i n K 10 t2 =273 // i n K 11 v1 =10 // i n l 12 13 v2 = p1 * v1 * t2 /( t1 * p2 ) ; 14 15 p r i n t f( ” v o l u m e o f o x y g e n = %f l i t e r s ” , v2 )

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1 clc 2 c l e a r 3 p r i n t f( ” e x a m p l e 2 . 2 p a g e number 7 1 \ n \ n ” ) 4 5 // t o f i n d v o l u m e t r i c c o m p o s i t i o n , p a r t i a l p r e s s u e o f e a c h g a s and t o t a l p r e s s u r e o f m i x t u r e 6 7 nCO2 = 2 / 4 4 ; // m o l e s o f CO2 8 nO2 = 4 / 3 2 ; // m o l e s o f O2 9 nCH4 = 1 . 5 / 1 6 ; // m o l e s o f CH4 10 11 t o t a l _ m o l e s = nCO2 + nO2 + nCH4 ; 12 yCO2 = nCO2 / t o t a l _ m o l e s ; 13 yO2 = nO2 / t o t a l _ m o l e s ; 14 yCH4 = nCH4 / t o t a l _ m o l e s ; 15 16 p r i n t f ( ” C o m p o s i t i o n o f m i x t u r e = \nCH4 = %f \nO2 = %f \ n CO2 = %f \ n \ n ” , yCH4 , yO2 , yCO2 )

17 18 pCO2 = nCO2 * 8 . 3 1 4 * 2 7 3 / ( 6 * 1 0 ^ - 3 ) ; 19 pO2 = nO2 * 8 . 3 1 4 * 2 7 3 / ( 6 * 1 0 ^ - 3 ) ; 20 pCH4 = nCH4 * 8 . 3 1 4 * 2 7 3 / ( 6 * 1 0 ^ - 3 ) ; 21 22 p r i n t f ( ” p r e s s u r e o f CH4 = %f kPa \ n p r e s s u r e o f O2 = %f kPa \ n p r e s s u r e o f CO2 =%f kPa \ n \ n ” , pCH4

*10^ -3 , pO2 *10^ -3 , pCO2 *10^ -3)

23

24 t o t a l _ p r e s s u r e = pCO2 + pCH4 + pO2 ;

25 p r i n t f ( ” t o t a l p r e s s u r e = %f Kpa” , t o t a l _ p r e s s u r e *10^ -3)

Scilab code Exa 2.3 Equivalent metal mass

1 clc

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3 p r i n t f( ” e x a m p l e 2 . 3 p a g e number 7 2 \ n \ n ” ) 4 5 // t o f i n d e q u i v a l e n t mass o f m e t a l 6 7 P = 1 0 4 . 3 // t o t a l p r e s s u r e i n KPa 8 pH2O =2.3 // i n KPa 9 pH2 = P - pH2O ; // i n KPa 10 11 VH2 = 2 0 9 * pH2 * 2 7 3 / ( 2 9 3 * 1 0 1 . 3 ) 12 13 p r i n t f( ” v o l u m e o f h y d r o g e n o b t a i n e d = %f ml \ n \ n ” , VH2 ) 14 15 // c a l c u l a t i n g amount o f m e t a l h a v i n g 1 1 . 2 l o f h y d r o g e n 16 17 m = 3 5 0 / 1 9 6 . 0 8 * 1 1 . 2 // mass o f m e t a l i n grams 18 p r i n t f( ” mass o f m e t a l e q u i v a l e n t t o 1 1 . 2 l i t r e / mol o f h y d r o g e n = %f gm” ,m )

Scilab code Exa 2.4 Purity of Sodium Hydroxide

1 clc 2 c l e a r 3 p r i n t f( ” e x a m p l e 2 . 4 p a g e number 7 2 \ n \ n ” ) 4 5 // t o f i n d NaCl c o n t e n t i n NaOH s o l u t i o n 6 7 w =2 // i n gm 8 m = 0 . 2 8 7 // i n gm 9 10 // p r e c i p i t a t e f r o m 5 8 . 5 gm o f NaCl = 1 4 3 . 4gm 11 12 m N a C l = 5 8 . 5 / 1 4 3 . 4 * m ; 13

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14 p r i n t f( ” mass o f NaCl = %f gm\ n ” , mNaCl )

15

16 p e r c e n t a g e _ N a C l = m N a C l / w * 1 0 0 ;

17 p r i n t f( ” amount o f NaCl = %f ” , p e r c e n t a g e _ N a C l )

Scilab code Exa 2.5 Carbon content formulation

1 clc 2 c l e a r 3 p r i n t f( ” e x a m p l e 2 . 5 p a g e number 7 2 \ n \ n ” ) 4 5 // t o f i n d t h e c a r b o n c o n t e n t i n s a m p l e 6 7 w = 4 . 7 3 // i n gm5 8 VCO2 = 5 . 3 0 // i n l i t e r s 9 10 w e i g h t _ C O 2 = 4 4 / 2 2 . 4 * VCO2 ; 11 c a r b o n _ c o n t e n t = 1 2 / 4 4 * w e i g h t _ C O 2 ; 12 13 p e r c e n t a g e _ c o n t e n t =( c a r b o n _ c o n t e n t / w ) * 1 0 0 ; 14 15 p r i n t f( ” p e r c e n t a g e amount o f c a r b o n i n s a m p l e = %f ” , p e r c e n t a g e _ c o n t e n t )

Scilab code Exa 2.6 Combustion of gas

1 clc 2 c l e a r 3 p r i n t f( ” e x a m p l e 2 . 6 p a g e number 7 3 \ n \ n ” ) 4 // t o f i n d t h e v o l u m e o f a i r 5 6 v o l u m e _ H 2 =0.5 // i n m3 7 v o l u m e _ C H 4 = 0 . 3 5 // i n m3

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8 v o l u m e _ C O = 0 . 0 8 // i n m3 9 v o l u m e _ C 2 H 4 = 0 . 0 2 // i n m3 10 v o l u m e _ o x y g e n = 0 . 2 1 // i n m3 i n a i r 11 12 // r e q u i r e d o x y g e n f o r v a r i o u s g a s e s 13 H2 = 0 . 5 * v o l u m e _ H 2 ; 14 CH4 =2* v o l u m e _ C H 4 ; 15 CO = 0 . 5 * v o l u m e _ C O ; 16 C2H4 =3* v o l u m e _ C 2 H 4 ; 17 18 t o t a l _ O 2 = H2 + CH4 + CO + C2H4 ; 19 o x y g e n _ r e q u i r e d = t o t a l _ O 2 / v o l u m e _ o x y g e n ; 20 21 p r i n t f( ” amount o f o x y g e n r e q u i r e d = %f c u b i c m e t e r ” , o x y g e n _ r e q u i r e d )

Scilab code Exa 2.7 Sulphuric acid preparation

1 clc 2 c l e a r 3 p r i n t f( ” e x a m p l e 2 . 7 p a g e number 7 3 \ n \ n ” ) 4 5 // t o f i n d t h e v o l u m e o f s u l p h u r i c a c i d and mass o f w a t e r consumed 6 7 d e n s i t y _ H 2 S O 4 = 1.10 // i n g / ml 8 m a s s _ 1 = 100* d e n s i t y _ H 2 S O 4 ; // mass o f 1 0 0 ml o f 15% s o l u t i o n 9 m a s s _ H 2 S O 4 = 0 . 1 5 * m a s s _ 1 ; 10 d e n s i t y _ s t d = 1.84 // d e n s i t y o f 96% s u l p h u r i c a c i d 11 m a s s _ s t d = 0 . 9 6 * d e n s i t y _ s t d ; // mass o f H2SO4 i n 1 ml 96% H2SO4 12 13 v o l u m e _ s t d = m a s s _ H 2 S O 4 / m a s s _ s t d ; // v o l u m e o f 96 %H2SO4

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14 m a s s _ w a t e r = m a s s _ 1 - m a s s _ H 2 S O 4 ;

15

16 p r i n t f( ” v o l u m e o f 0 . 9 6 H2SO4 r e q u i r e d = %f ml ” , v o l u m e _ s t d )

17 p r i n t f( ” \ nmass o f w a t e r r e q u i r e d = %f g ” , m a s s _ w a t e r )

Scilab code Exa 2.8 Molarity Molality Normality Calculation

1 clc 2 c l e a r 3 p r i n t f( ” e x a m p l e 2 . 8 p a g e number 7 3 \ n \ n ” ) 4 5 // t o f i n d m o l a r i t y , m o l a l i t y and n o r m a l i t y 6 7 w _ H 2 S O 4 = 0 . 1 5 // i n gm/1gm s o l u t i o n 8 d e n s i t y = 1 . 1 0 // i n gm/ ml 9 m = d e n s i t y * 1 0 0 0 ; // mass p e r l i t e r 10 w e i g h t = m * w _ H 2 S O 4 ; // H2SO4 p e r l i t e r s o l u t i o n 11 m o l a r _ m a s s =98; 12 13 M o l a r i t y = w e i g h t / m o l a r _ m a s s ; 14 p r i n t f( ” M o l a r i t y = %f mol / l \ n \ n ” , M o l a r i t y ) 15 16 e q u i v a l e n t _ m a s s =49; 17 n o r m a l i t y = w e i g h t / e q u i v a l e n t _ m a s s ; 18 p r i n t f( ” N o r m a l i t y = %f N\ n \ n ” , n o r m a l i t y ) 19 20 m o l a l i t y = 1 7 6 . 5 / m o l a r _ m a s s ; 21 p r i n t f( ” M o l a l i t y = %f ” , m o l a l i t y )

Scilab code Exa 2.9 Normality calculation

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c l e a r 3 p r i n t f( ” e x a m p l e 2 . 9 p a g e number 7 4 \ n \ n ” ) 4 5 m o l a r _ m a s s _ B a C l 2 = 2 0 8 . 3 ; // i n gm 6 e q u i v a l e n t _ H 2 S O 4 = 0 . 1 4 4 ; 7 n o r m a l i t y = e q u i v a l e n t _ H 2 S O 4 * 1 0 0 0 / 2 8 . 8 ; 8 9 p r i n t f( ” N o r m a l i t y = %f N” , n o r m a l i t y )

Scilab code Exa 2.10 Precipitation of KClO3

1 clc 2 c l e a r 3 p r i n t f( ” e x a m p l e 2 . 1 0 p a g e number 7 4 \ n \ n ” ) 4 5 // t o f i n d amount o f KClO3 p r e c i p i t a t e d 6 7 s o l u b i l i t y _ 7 0 = 3 0 . 2 // i n gm/ 1 0 0gm 8 w _ s o l u t e = s o l u b i l i t y _ 7 0 * 3 5 0 / 1 3 0 . 2 ; // i n gm 9 10 w _ w a t e r =350 - w _ s o l u t e ; 11 s o l u b i l i t y _ 3 0 = 1 0 . 1 // i n gm/ 1 0 0gm 12 p r e c i p i t a t e =( s o l u b i l i t y _ 7 0 - s o l u b i l i t y _ 3 0 ) * w _ w a t e r /100 13 14 p r i n t f( ” amount p r e c i p i t a t e d = %f gm” , p r e c i p i t a t e )

Scilab code Exa 2.11 Solubility of CO2

1 clc

2 c l e a r

3 p r i n t f( ” e x a m p l e 2 . 1 1 p a g e number 7 4 \ n \ n ” )

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5 // t o f i n d t h e p r e s s u r e f o r s o l u b i l i t y o f CO2 6 7 a b s o r b t i o n _ c o e f f i c i e n t = 1 . 7 1 // i n l i t e r s 8 m o l a r _ m a s s =44; 9 s o l u b i l i t y = a b s o r b t i o n _ c o e f f i c i e n t * m o l a r _ m a s s / 2 2 . 4 ; // i n gm 10 p r e s s u r e =8/ s o l u b i l i t y * 1 0 1 . 3 ; 11 12 p r i n t f( ” p r e s s u r e r e q u i r e d = %f kPa ” , pr es su re )

Scilab code Exa 2.12 Vapor pressure calculation

1 clc 2 c l e a r 3 p r i n t f( ” e x a m p l e 2 . 1 2 p a g e number 7 4 \ n \ n ” ) 4 5 // t o f i n d t h e v a p o r p r e s s u r e o f w a t e r 6 7 w _ w a t e r =540 // i n gm 8 w _ g l u c o s e =36 // i n gm 9 m _ w a t e r =18; // m o l a r mass o f w a t e r 10 m _ g l u c o s e = 1 8 0 ; // m o l a r mass o f g l u c o s e 11 12 x =( w _ w a t e r / m _ w a t e r ) /( w _ w a t e r / m _ w a t e r + w _ g l u c o s e / m _ g l u c o s e ) ; 13 p = 8 . 2 * x ; 14 d e p r e s s i o n =8.2 - p ; 15 16 p r i n t f( ” d e p r e s s i o n i n v a p o r p r e s s u r e = %f Pa ” , d e p r e s s i o n * 1 0 0 0 )

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clc 2 c l e a r 3 p r i n t f( ” e x a m p l e 2 . 1 3 p a g e number 7 5 \ n \ n ” ) 4 5 // t o f i n d t h e b o i l i n g p o i n t o f s o l u t i o n 6 7 w _ g l u c o s e =9 // i n gm 8 w _ w a t e r =100 // i n gm 9 E = 0 . 5 2 ; 10 m = 9 0 / 1 8 0 ; // m o l e s / 1 0 0 0gm w a t e r 11 12 d e l t a _ t = E * m ; 13 b o i l i n g _ p o i n t = 1 0 0 + d e l t a _ t ; 14 15 p r i n t f( ” b o i l i n g p o i n t o f w a t e r = %f d e g r e e C ” , b o i l i n g _ p o i n t )

Scilab code Exa 2.14 Colligative properties

1 clc 2 c l e a r 3 p r i n t f( ” e x a m p l e 2 . 1 4 p a g e number 7 5 \ n \ n ” ) 4 5 // t o f i n d t h e m o l a r mass and o s m o t i c p r e s s u r e 6 7 K = 1 . 8 6 ; 8 c =15 // c o n c e n t r a t i o n o f a l c o h o l 9 d e l t a _ t = 1 0 . 2 6 ; 10 11 m = d e l t a _ t / K ; // m o l a l i t y 12 M = c /( m *85) ; // m o l a r mass 13 p r i n t f( ” m o l a r mass = %f gm\ n \ n ” , M * 1 0 0 0 ) 14 15 d e n s i t y = 0 . 9 7 // g / ml 16 cm = c * d e n s i t y /( M * 1 0 0 ) ;

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17 p r i n t f( ” m o l a r c o n c e n t r a t i o n o f a l c o h o l = %f m o l e s / l \ n \ n ” , cm )

18

19 p = cm * 8 . 3 1 4 * 2 9 3 // o s m o t i c p r e s s u r e

20 p r i n t f( ” o s m o t i c p r e s s u r e = %f Mpa\ n \ n ” ,p /1000)

Scilab code Exa 2.15 Huggins Equation

1 clc 2 c l e a r 3 p r i n t f( ” e x a m p l e 2 . 1 5 p a g e number 7 5 \ n \ n ” ) 4 5 // t o f i n d u i n , M v , k ’ 6 7 u_in = 0 . 5 7 5 // f r o m t h e g r a p h 8 u_s = 0 . 2 9 5 // i n mPa−s 9 10 M_v = ( u_in / ( 5 . 8 0 * 1 0 ^ - 5 ) ) ^ ( 1 / 0 . 7 2 ) ; 11 u _ r e d = 0 . 6 2 8 ; // i n d l / g 12 13 c = 0.40 // i n g / d l

14 k = ( u_red - u_in ) /(( u_in ^2) * c ) ;

15

16 p r i n t f( ” k = %f \nMv = %f \ n u i n = %f d l /gm” ,k , M_v , u_in )

Scilab code Exa 2.16 Molecular Formula

1 clc

2 c l e a r

4

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7 C = 5 4 . 5 //% o f c a r b o n 8 H2 =9.1 //% o f h y d r o g e n 9 O2 = 3 6 . 4 //% o f o x y g e n 10 x = C /12; // number o f c a r b o n m o l e c u l e s 11 y = O2 /16; // number o f o x y g e n m o l e c u l e s 12 z = H2 /2 // number o f h y d r o g e n m o l e c u l e s 13 m o l a r _ m a s s =88; 14 d e n s i t y =44; 15 16 r a t i o = m o l a r _ m a s s / d e n s i t y ; 17 x = r a t i o *2; 18 y = r a t i o *1; 19 z = r a t i o *4; 20 21 p r i n t f( ” x = %f , y = %f , z = %f ” ,x ,y , z ) 22 p r i n t f( ” \ n \ n f o r m u l a o f b u t y r i c a c i d i s = C4H8O2” )

1 clc 2 c l e a r 3 p r i n t f( ” e x a m p l e 2 . 1 7 p a g e number 7 7 \ n \ n ” ) 4 5 // t o f i n d m o l e c u l a r f o e m u l a 6 C = 9 3 . 7 5 //% o f c a r b o n 7 H2 = 6 . 2 5 //% o f h y d r o g e n 8 x = C /12 // number o f c a r b o n atoms 9 y = H2 /2 // number o f h y d r o g e n atoms 10 m o l a r _ m a s s =64 11 d e n s i t y = 4 . 4 1 * 2 9 ; 12 13 r a t i o = d e n s i t y / m o l a r _ m a s s ; 14 15 x = r a t i o *5;

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16 y = r a t i o *4;

17 18

19 p r i n t f( ” x = %f , y = %f ” ,x , y )

20 p r i n t f( ” \ n \ n f o r m u l a o f b u t y r i c a c i d i s = C10H8 ” )

1 clc 2 c l e a r 3 p r i n t f( ” e x a m p l e 2 . 1 8 p a g e number 7 7 \ n \ n ” ) 4 5 // t o f i n d m o l e c u l a r f o r m u l a 6 C = 5 0 . 6 9 //% o f c a r b o n 7 H2 = 4 . 2 3 //% o f h y d r o g e n 8 O2 = 4 5 . 0 8 //% o f o x y g e n 9 a = C /12; // number o f c a r b o n m o l e c u l e s 10 c = O2 /16; // number o f o x y g e n m o l e c u l e s 11 b = H2 /2; // number o f h y d r o g e n m o l e c u l e s 12 m o l a r _ m a s s =71; 13 14 f u n c t i o n M = f ( m ) 15 M = ( 2 . 0 9 * 1 0 0 0 ) / ( 6 0 * m ) ; 16 17 e n d f u n c t i o n 18 19 M = f ( ( 1 . 2 5 / 5 . 1 ) ) ; 20 21 p r i n t f( ” a c t u a l m o l e c u l a r mass = %f \ n \ n ” ,M ) 22 23 r a t i o = M / m o l a r _ m a s s ; 24 a = r a t i o *3; 25 b = r a t i o *3; 26 c = r a t i o *2; 27

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29 p r i n t f( ” a = %f , b = %f , c = %f ” ,a ,b , c )

30 p r i n t f( ” \ n \ n f o r m u l a o f b u t y r i c a c i d i s = C6H6O4” )

1 clc 2 c l e a r 3 p r i n t f( ” e x a m p l e 2 . 1 9 p a g e number 7 8 \ n \ n ” ) 4 5 // t o f i n d t h e m o l e c u l a r f o r m u l a 6 C = 6 4 . 6 //% o f c a r b o n 7 H2 =5.2 //% o f h y d r o g e n 8 O2 = 1 2 . 6 //% o f o x y g e n 9 N2 =8.8 //% o f n i t r o g e n 10 Fe =8.8 //% o f i r o n 11 12 a = C /12; // number o f c a r b o n m o l e c u l e s 13 c = 8 . 8 / 1 4 ; // number o f n i t r o g e n m o l e c u l e s 14 b = H2 /2; // number o f h y d r o g e n m o l e c u l e s 15 d = O2 /16; // number o f o x y g e n m o l e c u l e s 16 e = Fe /56 // number o f i r o n atoms 17 18 cm = 2 4 3 . 4 / ( 8 . 3 1 * 2 9 3 ) // c o n c e n t r a t i o n 19 20 m o l a r _ m a s s = 6 3 . 3 / cm ; 21 22 p r i n t f( ” a = %f , b = %f , c = %f , d = %f , e = %f ” ,a *6.5 , b *6.5 , c *6.5 , d *6.5 , e * 6 . 5 ) 23 p r i n t f( ” \ n \ n f o r m u l a o f b u t y r i c a c i d i s = C34H33N4O5Fe ” )

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1 clc 2 c l e a r 3 p r i n t f( ” e x a m p l e 2 . 2 0 p a g e number 7 8 \ n \ n ” ) 4 5 // t o f i n d s e q u e n c e o f d e p o s i t i o n 6 E1 = -0.25; 7 E2 = 0 . 8 0 ; 8 E3 = 0 . 3 4 ; 9 10 a =[ E1 ; E2 ; E3 ]; 11 b =g s o r t( a ) ; 12 13 p r i n t f( ” s o r t e d p o t e n t i a l i n v o l t s =” ) 14 disp ( b ) 15 disp ( ” E2>E3>E1 ” ) 16 disp ( ” s i l v e r >c o p p e r > n i c k e l ” )

Scilab code Exa 2.21 EMF of cell

1 clc 2 c l e a r 3 p r i n t f( ” e x a m p l e 2 . 2 1 p a g e number 7 9 \ n \ n ” ) 4 5 // t o f i n d t h e emf o f c e l l 6 7 E 0 _ Z n = -0.76; 8 E 0 _ P b = -0.13; 9 c_Zn = 0 . 1 ; 10 c_Pb = 0 . 0 2 ; 11 12 E_Zn = E 0 _ Z n + ( 0 . 0 5 9 / 2 ) *l o g 1 0( c_Zn ) ; 13 E_Pb = E 0 _ P b + ( 0 . 0 5 9 / 2 ) *l o g 1 0( c_Pb ) ; 14 E = E_Pb - E_Zn ; 15 16 p r i n t f( ” emf o f c e l l = %f V” ,E )

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17 p r i n t f( ” \ n \ n S i n c e p o t e n t i a l o f l e a d i s g r e a t e r t h a n t h a t o f z i n c t h u s r e d u c t i o n w i l l o c c u r a t l e a d e l e c t r o d e and o x i d a t i o n w i l l o c c u r a t z i n c e l e c t r o d e ” )

1 clc 2 c l e a r 3 p r i n t f( ” e x a m p l e 2 . 2 2 p a g e number 7 9 \ n \ n ” ) 4 5 // t o f i n d t h e emf o f c e l l 6 E 0 _ A g = 0 . 8 0 ; 7 E 0 _ A g N O 3 = 0 . 8 0 ; 8 c_Ag = 0 . 0 0 1 ; 9 c _ A g N O 3 = 0 . 1 ; 10 11 E_Ag = E 0 _ A g + ( 0 . 0 5 9 ) *l o g 1 0( c_Ag ) ; 12 E _ A g N O 3 = E 0 _ A g N O 3 + ( 0 . 0 5 9 ) *l o g 1 0( c _ A g N O 3 ) ; 13 E = E_AgNO3 - E_Ag ; 14 15 p r i n t f( ” emf o f c e l l = %f V” ,E ) 16 p r i n t f( ” \ n \ n s i n c e E i s p o s i t i v e , t h e l e f t hand e l e c t r o d e w i l l be a n o d e and t h e e l e c t r o n w i l l t r a v e l i n t h e e x t e r n a l c i r c u i t f r o m t h e l e f t hand t o t h e r i g h t hand e l e c t r o d e ” )

1 clc

2 c l e a r

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5 // t o f i n d emf o f c e l l 6 pH =12; //pH o f s o l u t i o n 7 E_H2 =0; 8 E2 = -0.059* pH ; 9 E = E_H2 - E2 ; 10 p r i n t f( ”EMF o f c e l l = %f V” ,E )

Scilab code Exa 2.24 Silver deposition

1 clc 2 c l e a r 3 p r i n t f( ” e x a m p l e 2 . 2 4 p a g e number 8 0 \ n \ n ” ) 4 5 // t o f i n d amount o f s i l v e r d e p o s i t e d 6 I =3 // i n Ampere 7 t =900 // i n s 8 m_eq = 1 0 7 . 9 // i n gm/ mol 9 F = 9 6 5 0 0 ; 10 11 m =( I * t * m_eq ) / F ; 12 p r i n t f( ” mass = %f gm” , m )

Scilab code Exa 2.25 Electroplating time

1 clc 2 c l e a r 3 p r i n t f( ” e x a m p l e 2 . 2 5 p a g e number 8 0 \ n \ n ” ) 4 5 // t o f i n d t h e t i m e f o r e l e c t r o p l a t i n g 6 v o l u m e = 1 0 * 1 0 * 0 . 0 0 5 ; // i n cm3 7 mass = v o l u m e * 8 . 9 ; 8 F = 9 6 5 0 0 ; 9 a t o m i c _ m a s s = 5 8 . 7 // i n amu

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10 c u r r e n t =2.5 // i n Ampere 11 12 c h a r g e = ( 8 . 9 * F *2) / a t o m i c _ m a s s ; 13 y i e l d = 0 . 9 5 ; 14 a c t u a l _ c h a r g e = c h a r g e /( y i e l d * 3 6 0 0 ) ; 15 t = a c t u a l _ c h a r g e / c u r r e n t ; 16 17 p r i n t f( ” t i m e r e q u i r e d = %f h o u r s ” ,t )

Scilab code Exa 2.26 Water hardness

1 clc 2 c l e a r 3 p r i n t f( ” e x a m p l e 2 . 2 6 p a g e number 8 0 \ n \ n ” ) 4 5 // t o f i n d h a r d n e s s o f w a t e r 6 m _ M g S O 4 =90 // i n ppm 7 M g S O 4 _ p a r t s = 1 2 0 ; 8 C a C O 3 _ p a r t s = 1 0 0 ; 9 10 h a r d n e s s =( C a C O 3 _ p a r t s / M g S O 4 _ p a r t s ) * m _ M g S O 4 ; 11 12 p r i n t f( ” h a r d n e s s o f w a t e r = %f mg/ l ” , ha rd ne ss )

Scilab code Exa 2.27 Water hardness

1 clc 2 c l e a r 3 p r i n t f( ” e x a m p l e 2 . 2 6 p a g e number 8 0 \ n \ n ” ) 4 5 m1 = 162 // mass o f c a l c i u m b i c a r b o n a t e i n mg 6 m2 = 73 // mass o f magnesium b i c a r b o n a t e i n mg 7 m3 = 136 // mass o f c a l s i u m s u l f a t e i n mg

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8 m4 = 95 // mass o f magnesium c l o r i d e 9 m5 = 500 // mass o f s o d i u m c l o r i d e i n mg 10 m6 = 50 // mass o f p o t a s s i u m c l o r i d e i n mg 11 12 c o n t e n t _ 1 = m1 * 1 0 0 / m1 ; // c o n t e n t o f c a l c i u m b i c a r b o n a t e i n mg 13 c o n t e n t _ 2 = m2 * 1 0 0 / ( 2 * m2 ) ; // c o n t e n t o f magnesium b i c a r b o n a t e i n mg 14 c o n t e n t _ 3 = m3 * 1 0 0 / m3 ; // c o n t e n t o f c a l s i u m s u f a t e i n mg 15 c o n t e n t _ 4 = m4 * 1 0 0 / m4 ; // c o n t e n t o f magnesium c l o r i d e 16 17 // p a r t 1 18 19 t e m p _ h a r d n e s s = c o n t e n t _ 1 + c o n t e n t _ 2 ; // d e p e n d s on b i c a r b o n a t e o n l y 20 t o t a l _ h a r d n e s s = c o n t e n t _ 1 + c o n t e n t _ 2 + c o n t e n t _ 3 + c o n t e n t _ 4 ; 21 p r i n t f( ” t o t a l h a r d n e s s = %f \ n t e m p o r a r y h a r d n e s s = %f \ n ” , temp_hardness , t o t a l _ h a r d n e s s ) 22 23 // p a r t 2 24 w t _ l i m e = ( 7 4 / 1 0 0 ) *( c o n t e n t _ 1 +2* c o n t e n t _ 2 + c o n t e n t _ 4 ) ; 25 a c t u a l _ l i m e = w t _ l i m e / 0 . 8 5 ; 26 p r i n t f( ” amount o f l i m e r e q u i r e d = %f \ n ” , a c t u a l _ l i m e ) 27 28 s o d a _ r e q u i r e d = ( 1 0 6 / 1 0 0 ) *( c o n t e n t _ 1 + c o n t e n t _ 4 ) ; 29 a c t u a l _ s o d a = s o d a _ r e q u i r e d / 0 . 9 8 ; 30 p r i n t f( ” amount o f s o d a r e q u i r e d = %f \ n ” , a c t u a l _ s o d a )

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clc 2 c l e a r 3 p r i n t f( ” e x a m p l e 2 . 2 8 p a g e number 8 2 \ n \ n ” ) 4 5 // t o f i n d h a r d n e s s o f w a t e r 6 7 v o l u m e _ N a C l =50 // i n l 8 c _ N a C l = 5 0 0 0 // i n mg/ l 9 10 m = v o l u m e _ N a C l * c _ N a C l ; 11 e q u i v a l e n t _ N a C l = 5 0 / 5 8 . 5 ; 12 13 h a r d n e s s = e q u i v a l e n t _ N a C l * m ; 14 15 p r i n t f( ” h a r d n e s s o f w a t e r = %f mg/ l ” , ha rd ne ss /1000)

Scilab code Exa 2.29 Mixture composition

1 clc 2 c l e a r 3 p r i n t f( ” e x a m p l e 2 . 2 9 p a g e number 8 2 \ n \ n ” ) 4 5 // t o f i n d t h e t o t a l v a p o r p r e s s u r e and m o l a r c o m p o s i t i o n s 6 7 m _ b e n z e n e = 55 // i n kg 8 m _ t o l u e n e = 28 // i n kg 9 m _ x y l e n e = 17 // i n kg 10 11 m o l e _ b e n z e n e = m _ b e n z e n e /78; 12 m o l e _ t o l u e n e = m _ t o l u e n e /92; 13 m o l e _ x y l e n e = m _ x y l e n e / 1 0 6 ; 14 15 m o l e _ t o t a l = m o l e _ b e n z e n e + m o l e _ t o l u e n e + m o l e _ x y l e n e ; 16 x _ b e n z e n e = m o l e _ b e n z e n e / m o l e _ t o t a l ;

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17 x _ t o l u e n e = m o l e _ t o l u e n e / m o l e _ t o t a l ; 18 x _ x y l e n e = m o l e _ x y l e n e / m o l e _ t o t a l ; 19 20 P = x _ b e n z e n e * 1 7 8 . 6 + x _ t o l u e n e * 7 4 . 6 + x _ x y l e n e *28; 21 p r i n t f( ” t o t a l p r e s s u r e = %f kPa \ n ” ,P ) 22 23 b e n z e n e = ( x _ b e n z e n e * 1 7 8 . 6 * 1 0 0 ) / P ; 24 t o l u e n e = ( x _ t o l u e n e * 7 4 . 6 * 1 0 0 ) / P ; 25 x y l e n e = ( x _ x y l e n e * 2 8 * 1 0 0 ) / P ; 26 27 p r i n t f( ” x y l e n e = %f \ n t o l u e n e = %f \ n b e n z e n e = %f ” , xylene , toluene , b e n z e n e )

Scilab code Exa 2.30 Mixture composition

1 clc 2 c l e a r 3 p r i n t f( ” e x a m p l e 2 . 3 0 p a g e number 8 3 \ n \ n ” ) 4 5 // t o f i n d t h e m i x t u r e c o m p o s i t i o n 6 7 v a p o r _ p r e s s u r e =8 // i n kPa 8 p r e s s u r e =100 // i n kPa 9 10 // p a r t 1 11 v o l u m e =1 // i n m3 12 v o l u m e _ e t h a n o l = v o l u m e *( v a p o r _ p r e s s u r e / p r e s s u r e ) ; 13 v o l u m e _ a i r =1 - v o l u m e _ e t h a n o l ; 14 p r i n t f( ” v o l u m e t r i c c o m p o s i t i o n : − \ n a i r c o m p o s i t i o n = %f \ n e t h a n o l c o m p o s t i o n = %f ” , v o l u m e _ a i r *100 , v o l u m e _ e t h a n o l * 1 0 0 ) 15 16 // p a r t 2 17 m o l a r _ m a s s _ e t h a n o l =46; 18 m o l a r _ m a s s _ a i r = 2 8 . 9 ;

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19 m a s s _ e t h a n o l = 0 . 0 8 * m o l a r _ m a s s _ e t h a n o l ; // i n kg 20 m a s s _ a i r = 0 . 9 2 * m o l a r _ m a s s _ a i r ; // i n kg 21 f r a c t i o n _ e t h a n o l =( m a s s _ e t h a n o l * 1 0 0 ) /( m a s s _ a i r + m a s s _ e t h a n o l ) ; 22 f r a c t i o n _ a i r =( m a s s _ a i r * 1 0 0 ) /( m a s s _ a i r + m a s s _ e t h a n o l ) ; 23 p r i n t f( ” \ n \ n c o m p o s i t i o n by w e i g h t : −\ n A i r = %f E t h a n o l v a p o r = %f ” , fraction_air , f r a c t i o n _ e t h a n o l ) 24 25 // p a r t 3 26 m i x t u r e _ v o l u m e = 2 2 . 3 * ( 1 0 1 . 3 / 1 0 0 ) * ( 2 9 9 / 2 7 3 ) ; // i n m3 27 w e i g h t _ e t h a n o l = m a s s _ e t h a n o l / m i x t u r e _ v o l u m e ; 28 p r i n t f( ” \ n \ n w e i g h t o f e t h a n o l / c u b i c m e t e r = %f Kg” , w e i g h t _ e t h a n o l ) 29 30 // p a r t 4 31 w _ e t h a n o l = m a s s _ e t h a n o l / m a s s _ a i r ; 32 p r i n t f( ” \ n \ n w e i g h t o f e t h a n o l / kg v a p o r f r e e a i r = %f Kg” , w _ e t h a n o l ) 33 34 // p a r t 5 35 m o l e s _ e t h a n o l = 0 . 0 8 / 0 . 9 2 ; 36 p r i n t f( ” \ n \ nkmol o f e t h a n o l p e r kmol o f v a p o r f r e e a i r = %f ” , m o l e s _ e t h a n o l )

Scilab code Exa 2.31 Mixture properties

1 clc 2 c l e a r 3 p r i n t f( ” e x a m p l e 2 . 3 1 p a g e number 8 4 \ n \ n ” ) 4 5 // t o f i n d r e l a t i v e s a t u r a t i o n and dew p o i n t 6 7 v a p o r _ p r e s s u r e =8 // i n kPa 8 v o l u m e _ e t h a n o l = 0 . 0 5 ;

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9 10 // b a s i s 1 kmol o f m i x t u r e 11 12 p a r t i a l _ p r e s s u r e = v o l u m e _ e t h a n o l * 1 0 0 ; 13 r e l a t i v e _ s a t u r a t i o n = p a r t i a l _ p r e s s u r e / v a p o r _ p r e s s u r e ; 14 m o l e _ r a t i o = v o l u m e _ e t h a n o l /(1 - v o l u m e _ e t h a n o l ) ; 15 p r i n t f( ” m o l e r a t i o = %f \ n r e l a t i v e s a t u r a t i o n = %f ” , m o l e _ r a t i o , r e l a t i v e _ s a t u r a t i o n * 1 0 0 ) 16 17 // b a s i s 1 kmol s a t u r a t e d g a s m i x t u r e a t 1 0 0 kPa 18 v o l u m e _ v a p o r = ( 8 / 1 0 0 ) * 1 0 0 ; 19 e t h a n o l _ v a p o r = v o l u m e _ v a p o r / 1 0 0 ; 20 a i r _ v a p o r =1 - e t h a n o l _ v a p o r ; 21 s a t u r a t i o n _ r a t i o = e t h a n o l _ v a p o r / a i r _ v a p o r ; 22 p e r c e n t a g e _ s a t u r a t i o n = m o l e _ r a t i o / s a t u r a t i o n _ r a t i o ; 23 24 p r i n t f( ” \ n \ n p e r c e n t a g e s a t u r a t i o n = %f ” , p e r c e n t a g e _ s a t u r a t i o n ) 25 26 // dew p o i n t 27 p r i n t f( ” \ n \ n c o r r e s p o n d i n g t o p a r t i a l p r e s s u r e o f 5 kPa we g e t a dew p o i n t o f 1 7 . 3 d e g r e e c e l c i u s ” )

Scilab code Exa 2.32 Humidity

1 clc 2 c l e a r 3 p r i n t f( ” e x a m p l e 2 . 3 2 p a g e number 8 4 \ n \ n ” ) 4 5 // t o f i n d t h e p r o p e r t i e s o f humid a i r 6 7 p = 4.24 // i n kPa 8 H _ r e l = 0.8; 9 p _ p a r t i a l = p * H _ r e l ; 10 m o l a l _ H = p _ p a r t i a l /(100 - p _ p a r t i a l ) ;

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11 p r i n t f( ” i n i t i a l m o l a l h u m i d i t y = %f \ n \ n ” , molal_H ) 12 13 // p a r t 2 14 P = 200 // i n kPa 15 p _ p a r t i a l = 1.70 // i n kPa 16 f i n a l _ H = p _ p a r t i a l /( P - p _ p a r t i a l ) ; 17 p r i n t f( ” f i n a l m o l a l h u m i d i t y = %f \ n \ n ” , final_H ) 18 19 // p a r t 3 20 p _ d r y a i r = 100 - 3 . 3 9 ; 21 v = 1 0 0 * ( p _ d r y a i r / 1 0 1 . 3 ) * ( 2 7 3 / 3 0 3 ) ; 22 m o l e s _ d r y a i r = v / 2 2 . 4 ; 23 v a p o r _ i n i t i a l = m o l a l _ H * m o l e s _ d r y a i r ; 24 v a p o r _ f i n a l = f i n a l _ H * m o l e s _ d r y a i r ; 25 w a t e r _ c o n d e n s e d = ( v a p o r _ i n i t i a l - v a p o r _ f i n a l ) *18; 26 p r i n t f( ” amount o f w a t e r c o n d e n s e d = %f \ n \ n ” , w a t e r _ c o n d e n s e d ) 27 28 // p a r t 4 29 t o t a l _ a i r = m o l e s _ d r y a i r + v a p o r _ f i n a l ; 30 f i n a l _ v = 2 2 . 4 * ( 1 0 1 . 3 / 2 0 0 ) * ( 2 8 8 / 2 7 3 ) * t o t a l _ a i r ; 31 p r i n t f( ” f i n a l v o l u m e o f wety a i r = %f \ n \ n ” , final_v )

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Chapter 3 Material and Energy Balances

Scilab code Exa 3.1 Coal consumption

1 clc 2 c l e a r 3 p r i n t f( ” e x a m p l e 3 . 1 p a g e number 9 0 \ n \ n ” ) 4 5 // t o f i n d t h e c o a l c o n s u m p t i o n 6 w_C = 0.6; // amount o f c a r b o n i n c o a l 7 N 2 _ c o n t e n t = 40 // i n m3 p e r 1 0 0m3 a i r 8 9 a i r _ c o n s u m e d = N 2 _ c o n t e n t / 0 . 7 9 ; 10 w e i g h t _ a i r = a i r _ c o n s u m e d * ( 2 8 . 8 / 2 2 . 4 ) ; 11 O 2 _ c o n t e n t = a i r _ c o n s u m e d * 3 2 * ( 0 . 2 1 / 2 2 . 4 ) ; // i n kg 12 13 H 2 _ c o n t e n t = 20 // i n m3 14 15 s t e a m _ c o n s u m e d = H 2 _ c o n t e n t * ( 1 8 / 2 2 . 4 ) ; 16 17 C _ c o n s u m p t i o n 1 = ( 1 2 / 1 8 ) * s t e a m _ c o n s u m e d ; // i n r e a c t i o n 1 18 C _ c o n s u m p t i o n 2 = ( 2 4 / 3 2 ) * O 2 _ c o n t e n t ; // i n r e a c t i o n 2 19

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t o t a l _ c o n s u m p t i o n = C _ c o n s u m p t i o n 1 + C _ c o n s u m p t i o n 2 ;

21 c o a l _ c o n s u m p t i o n = t o t a l _ c o n s u m p t i o n / w_C ;

22

23 p r i n t f( ” c o a l c o n s u m p t i o n = %f kg ” , c o a l _ c o n s u m p t i o n )

Scilab code Exa 3.2 Nitric acid preparation

1 clc

2 c l e a r

3 p r i n t f( ” e x a m p l e 3 . 2 p a g e number 9 1 \ n \ n ” )

4

5 // t o f i n d amount o f ammonia and a i r consumed

6 7 N H 3 _ r e q u i r e d = ( 1 7 / 6 3 ) * 1 0 0 0 ; //NH3 r e q u i r e d f o r 1 t o n o f n i t r i c a c i d 8 N O _ c o n s u m p t i o n = 0 . 9 6 ; 9 H N O 3 _ c o n s u m p t i o n = 0 . 9 2 ; 10 N H 3 _ c o n s u m e d = N H 3 _ r e q u i r e d /( N O _ c o n s u m p t i o n * H N O 3 _ c o n s u m p t i o n ) ; 11 v o l u m e _ N H 3 = N H 3 _ c o n s u m e d * ( 2 2 . 4 / 1 7 ) ; 12 p r i n t f( ” v o l u m e o f ammonia consumed= %f c u b i c m e t r e / h ” , v o l u m e _ N H 3 ) 13 14 N H 3 _ c o n t e n t = 11 //% by v o l u m e 15 a i r _ c o n s u m p t i o n = v o l u m e _ N H 3 * ( ( 1 0 0 - 1 1 ) /11) ; 16 p r i n t f( ” \ n \ n v o l u m e o f a i r consumed = %f c u b i c m e t r e / h ” , a i r _ c o n s u m p t i o n )

Scilab code Exa 3.3 HCl production

1 clc

2 c l e a r

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4

5 // t o f i n d t h e c o n s u m p t i o n o f NaCl and H2SO4 i n HCl c o n s u m p t i o n 6 7 H C l _ p r o d u c t i o n = 500 // r e q u i r e d t o be p r o d u c e d i n kg 8 N a C l _ r e q u i r e d = ( 1 1 7 / 7 3 ) * H C l _ p r o d u c t i o n ; 9 y i e l d = 0 . 9 2 ; 10 p u r i t y _ N a C l = 0 . 9 6 ; 11 12 a c t u a l _ N a C l = N a C l _ r e q u i r e d /( p u r i t y _ N a C l * y i e l d ) ; 13 p r i n t f( ” amount o f NaCl r e q u i r e d = %f kg ” , a c t u a l _ N a C l ) 14 15 p u r i t y _ H 2 S O 4 = 0 . 9 3 ; 16 H 2 S O 4 _ c o n s u m p t i o n = ( 9 8 / 7 3 ) *( H C l _ p r o d u c t i o n /( y i e l d * p u r i t y _ H 2 S O 4 ) ) ;

17 p r i n t f( ” \ n \ namount o f H2SO4 consumed = %f kg ” , H 2 S O 4 _ c o n s u m p t i o n )

18

19 N a 2 S O 4 _ p r o d u c e d = ( 1 4 2 / 7 3 ) * H C l _ p r o d u c t i o n ;

20 p r i n t f( ” \ n \ namount o f Na2SO4 p r o d u c e d = %f kg ” , N a 2 S O 4 _ p r o d u c e d )

Scilab code Exa 3.4 Acetylene consumption

1 clc 2 c l e a r 3 p r i n t f( ” e x a m p l e 3 . 4 p a g e number 9 2 \ n \ n ” ) 4 5 // t o f i n d t h e p e r i o d o f s e r v i c e 6 7 C 2 H 2 _ p r o d u c e d = ( 1 / 6 4 ) * 0 . 8 6 ; // i n kmol 8 v o l u m e _ C 2 H 2 = C 2 H 2 _ p r o d u c e d * 2 2 . 4 * 1 0 0 0 ; // i n l 9

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10 // a s s u m i n g i d e a l b e h a v i o u r ,

11 v o l u m e = ( 1 0 0 / 1 0 1 . 3 ) * ( 2 7 3 / ( 2 7 3 + 3 0 ) ) ;

12 time = ( v o l u m e _ C 2 H 2 / v o l u m e ) * ( 1 / 6 0 ) ;

13 p r i n t f( ” t i m e o f s e r v i c e = %f h r ” , time )

Scilab code Exa 3.5 Screen effectiveness

1 clc 2 c l e a r 3 p r i n t f( ” e x a m p l e 3 . 5 p a g e number 9 2 \ n \ n ” ) 4 5 // t o f i n d t h e s c r e e n e f f e c t i v e n e s s 6 7 xv = 0 . 8 8 ; 8 xf = 0 . 4 6 ; 9 xl = 0 . 3 2 ; 10 F = 100 // i n kg 11 12 L = ( F *( xf - xv ) ) /( xl - xv ) ; 13 V = F - L ; 14 p r i n t f( ”L = %f Kg \nV = %f Kg” ,L , V ) 15 Eo = ( V * xv ) /( F * xf ) ; 16 17 p r i n t f( ” \ n \ n e f f e c t i v e n e s s b a s e d on o v e r s i z e d p a r t i c e s = %f \ n \ n ” , Eo ) 18 Eu = ( L *(1 - xl ) ) /( F *(1 - xf ) ) ; 19 20 p r i n t f( ” e f f e c t i v e n e s s b a s e d on u n d e r s i z e d p a r t i c e s = %f ” , Eu ) 21 E = Eu * Eo ; 22 23 p r i n t f( ” \ n \ n o v e r a l l e f f e c t i v e n e s s = %f ” ,E )

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Scilab code Exa 3.6 Absorption 1 clc 2 c l e a r 3 p r i n t f( ” e x a m p l e 3 . 6 p a g e number 9 4 \ n \ n ” ) 4 5 // t o f i n d t h e f l o w r a t e and c o n c e n t r a t i o n 6 7 G1 = 3600 // i n m3/ h 8 P = 1 0 6 . 6 // i n kPa 9 T = 40 // i n d e g r e e C 10 q = G1 *( P / 1 0 1 . 3 ) * ( 2 7 3 / ( ( 2 7 3 + T ) ) ) ; // i n m3/ s 11 m = q / 2 2 . 4 ; // i n kmol / h 12 y1 = 0 . 0 2 ; 13 Y1 = y1 /(1 - y1 ) ; 14 15 p r i n t f( ” m o l e r a t i o o f b e n z e n e = %f kmol b e n z e n e / kmol d r y g a s ” , Y1 ) 16 17 Gs = m *(1 - y1 ) ; 18 p r i n t f( ” \ n \ n m o l e s o f b e n z e n e f r e e g a s = %f kmol d r y g a s / h ” , Gs ) 19 20 // f o r 95% r e m o v a l 21 Y2 = Y1 * ( 1 - 0 . 9 5 ) ; 22 p r i n t f( ” \ n \ n f i n a l m o l e r a t i o o f b e n z e n e = %f kmol b e n z e n e / kmol d r y g a s ” , Y2 ) 23 24 x2 = 0 . 0 0 2 25 X2 = 0 . 0 0 2 / ( 1 - 0 . 0 0 2 ) ; 26 27 // a t e q u i l i b r i u m y ∗ = 0 . 2 4 0 6X 28 // p a r t 1 29 // f o r o i l r a t e t o be minimum t h e wash o i l l e a v i n g t h e a b s o r b e r must be i n e q u i l i b r i u m w i t h t h e e n t e r i n g g a s 30 31 y1 = 0 . 0 2 ;

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x1 = y1 / ( 0 . 2 4 0 6 ) ; 33 X1 = x1 /(1 - x1 ) ; 34 m i n _ L s = Gs *(( Y1 - Y2 ) /( X1 - X2 ) ) ; 35 p r i n t f( ” \ n \ nminimum Ls r e q u i r e d = %f kg / h ” , min_Ls * 2 6 0 ) 36 37 // f o r 1 . 5 t i m e s o f t h e minimum 38 Ls = 1.5* m i n _ L s ; 39 p r i n t f( ” \ n \ n f l o w r a t e o f wash o i l = %f kg / h ” , Ls *260) 40 X1 = X2 + ( Gs *(( Y1 - Y2 ) / Ls ) ) ; 41 p r i n t f( ” \ n \ n c o n c e n t r a t i o n o f b e n z e n e i n wash o i l = %f kmol b e n z e n e / kmol wash o i l ” , X1 )

Scilab code Exa 3.7 Extraction

1 clc 2 c l e a r 3 p r i n t f( ” e x a m p l e 3 . 7 p a g e number 9 5 \ n \ n ” ) 4 5 // t o f i n d t h e e x t r a c t i o n o f n i c o t i n e 6 xf = 0.01 7 Xf = xf /(1 - xf ) ; 8 Feed = 100 // f e e d i n kg 9 c _ n i c o t i n e = Feed * Xf ; // n i c o t i n e c o n c i n f e e d 10 c _ w a t e r = Feed *(1 - Xf ) // w a t e r c o n c i n f e e d 11 12 // p a r t 1 13 f u n c t i o n[ f ] = F1 ( x ) 14 f u n c p r o t(0) 15 f = ( x / 1 5 0 ) -0.9*((1 - x ) /99) ; 16 e n d f u n c t i o n 17 18 // i n i t i a l g u e s s 19 x = 10; 20 y = f s o l v e( x , F1 ) ;

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21 p r i n t f( ” amount o f n i c o t i n e remov ed N = %f kg ” ,y ) 22 // p a r t 2 23 f u n c t i o n[ f ] = F1 ( x ) 24 f = ( x /50) -0.9*((1 - x ) /99) ; 25 e n d f u n c t i o n 26 27 // i n i t i a l g u e s s 28 x = 10; 29 N1 = f s o l v e( x , F1 ) ;

30 p r i n t f( ” \ n \ namount o f n i c o t i n e rem oved i n s t a g e 1 , N1 = %f kg ” , N1 ) 31 f u n c t i o n[ f ] = F1 ( x , N1 ) 32 f = ( x /50) -0.9*((1 - x - N1 ) /99) ; 33 e n d f u n c t i o n 34 35 // i n i t i a l g u e s s 36 x = 10; 37 N2 = f s o l v e( x , F1 ) ;

38 p r i n t f( ” \ n \ namount o f n i c o t i n e rem oved i n s t a g e 2 , N2 = %f kg ” , N2 ) 39 f u n c t i o n[ f ] = F1 ( x , N1 , N2 ) 40 f = ( x /50) -0.9*((1 - x - N2 - N1 ) /99) ; 41 e n d f u n c t i o n 42 43 // i n i t i a l g u e s s 44 x = 10; 45 N3 = f s o l v e( x , F1 ) ; 46 47 p r i n t f( ” \ n \ namount o f n i c o t i n e re moved i n s t a g e 3 , N3 = %f kg ” , N3 ) 48 N = N1 + N2 + N3 ; 49 p r i n t f( ” \ n \ n t o t a l amount o f n i c o t i n e removed = %f kg ” ,N )

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clc 2 c l e a r 3 p r i n t f( ” e x a m p l e 3 . 8 p a g e number 9 6 \ n \ n ” ) 4 5 // t o f i n d t h e amount o f w a t e r i n r e s i d u e 6 7 v p _ w a t e r = 3 1 . 0 6 // i n kPa 8 v p _ b e n z e n e = 7 2 . 9 2 // i n kPa 9 10 P = v p _ w a t e r + v p _ b e n z e n e ; 11 x _ b e n z e n e = v p _ b e n z e n e / P ; 12 x _ w a t e r = v p _ w a t e r / P ; 13 14 i n i t i a l _ w a t e r = 5 0 / 1 8 ; // i n kmol o f w a t e r 15 i n i t i a l _ b e n z e n e = 5 0 / 7 8 // i n kmol o f b e n z e n e 16 w a t e r _ e v a p o r a t e d = i n i t i a l _ b e n z e n e *( x _ w a t e r / x _ b e n z e n e ) ; 17 w a t e r _ l e f t = ( i n i t i a l _ w a t e r - w a t e r _ e v a p o r a t e d ) ; 18 19 p r i n t f( ” amount o f w a t e r l e f t i n r e s i d u e = %f kg ” , w a t e r _ l e f t *18)

Scilab code Exa 3.9 Distillation

1 clc 2 c l e a r 3 p r i n t f( ” e x a m p l e 3 . 9 p a g e number 9 7 \ n \ n ” ) 4 5 // t o f i n d t h e v a p o r c o n t e n t o f d i m e t h y l a n a l i n e 6 po_D = 4.93 // i n kPa 7 po_W = 96.3 // i n kPa 8 n = 0.75 // v a p o r i z a t i o n e f f i c i e n c y 9 10 P = n * po_D + po_W ; 11 p r i n t f( ”P = %f kPa ” , P )

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12 13 x _ w a t e r = 9 6 . 3 / 1 0 0 ; 14 x _ d i m e t h y l a n a l i n e = 1 - x _ w a t e r ; 15 w t _ d i m e t h y l a n a l i n e = ( x _ d i m e t h y l a n a l i n e * 1 2 1 ) /( x _ d i m e t h y l a n a l i n e * 1 2 1 + x _ w a t e r *18) ; 16 p r i n t f( ” \ n \ n w e i g h t o f d i m e t h y l a n a l i n e i n w a t e r = %f ” , w t _ d i m e t h y l a n a l i n e * 1 0 0 ) 17 18 // p a r t 1 19 n = 0.8; 20 po_D = 32 // i n kPa 21 a c t u a l _ v p = n * po_D ; 22 p _ w a t e r = 100 - a c t u a l _ v p ; 23 s t e a m _ r e q u i r e d = ( p _ w a t e r *18) /( a c t u a l _ v p * 1 2 1 ) ; 24 p r i n t f( ” \ n \ namount o f s t e a m r e q u i r e d = %f kg s t e a m / kg d i m e t h y l a n a l i n e ” , s t e a m _ r e q u i r e d ) 25 26 // p a r t 2 27 x _ w a t e r = p _ w a t e r / 1 0 0 ; 28 w t _ w a t e r = x _ w a t e r * 1 8 / ( x _ w a t e r *18+(1 - x _ w a t e r ) * 1 2 1 ) ; 29 p r i n t f( ” \ n \ n w e i g h t o f w a t e r v a p o r = %f \ n w e i g h t o f d i m e t h y l a n a l i n e =%f ” , wt _w at er *100 ,100*(1 - wt _w at er ) )

Scilab code Exa 3.10 Crystallization

1 clc 2 c l e a r 3 p r i n t f( ” e x a m p l e 3 . 1 0 p a g e number 9 8 \ n \ n ” ) 4 5 // t o f i n d t h e amount o f w a t e r e v a p o r a t e d 6 xf = 0 . 1 5 ; 7 xl = ( 1 1 4 . 7 ) / ( 1 1 4 . 7 + 1 0 0 0 ) ; 8 xc = 1; 9

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10 K 2 C r 2 O 7 _ f e e d = 1 0 0 0 * 0 . 1 5 ; // i n kg 11 12 n = 0.8; 13 C = n * K 2 C r 2 O 7 _ f e e d ; 14 V = ( K 2 C r 2 O 7 _ f e e d -120 - 8 8 0 * 0 . 1 0 3 ) / ( - 0 . 1 0 3 ) ; 15 16 p r i n t f( ” amount o f w a t e r e v a p o r a t e d = %f kg ” ,V )

Scilab code Exa 3.11 crystallization

1 clc 2 c l e a r 3 p r i n t f( ” e x a m p l e 3 . 1 0 p a g e number 9 8 \ n \ n ” ) 4 5 // t o f i n d t h e y i e l d o f c r y s t a l s 6 7 xc = 1 0 6 / 2 8 6 ; 8 xf = 0 . 2 5 ; 9 xl = 2 7 . 5 / 1 2 7 . 5 ; 10 11 w a t e r _ p r e s e n t = 100*(1 - xf ) ; // i n kg 12 V = 0 . 1 5 * 7 5 ; // i n kg 13 C = ( 1 0 0 * xf - 8 8 . 7 * xl ) /( xc - xl ) ; 14 N a 2 C O 3 _ f e e d = 25/ xc ; 15 16 y i e l d = ( C / N a 2 C O 3 _ f e e d ) * 1 0 0 ; 17 18 p r i n t f( ” y i e l d = %f ” , y i e l d )

Scilab code Exa 3.12 Drying

1 clc

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3 p r i n t f( ” e x a m p l e 3 . 1 2 p a g e number 9 9 \ n \ n ” ) 4 5 // t o f i n d t h e f r a c t i o n o f a i r r e c i r c u l a t e d 6 7 r = 50 // w e i g h t o f d r y a i r p a s s i n g t h r o u g h d r i e r 8 w1 = 1.60 // i n kg p e r kg d r y s o l i d 9 w2 = 0.1 // i n kg / kg d r y s o l i d 10 H0 = 0 . 0 1 6 // i n kg w a t e r v a p o r / kg d r y a i r 11 H2 = 0 . 0 5 5 // i n kg w a t e r v a p o r / kg d r y a i r 12 13 y = 1 - ( w1 - w2 ) /( r *( H2 - H0 ) ) ; 14 p r i n t f( ” f r a c t i o n o f a i r r e c i r c u l a t e d = %f ” ,y ) 15 16 H1 = H2 - ( w1 - w2 ) / r ; 17 p r i n t f( ” \ n \ n h u m i d i t y o f a i r e n t e r i n g t h e d r i e r = %f kg w a t e r v a p o r / kg kg d r y a i r ” , H1 ) 18 19 // c h e c k 20 H11 = H2 * y + H0 *(1 - y ) ; 21 if H1 == H11 then p r i n t f( ” \ n \ n f r a c t i o n o f a i r r e c i r c u l a t e d = %f \ n v e r i f i e d ” ,y ) 22 end

Scilab code Exa 3.13 Conditioning of air

1 clc 2 c l e a r 3 p r i n t f( ” e x a m p l e 3 . 1 3 p a g e number 1 0 0 \ n \ n ” ) 4 5 // t o f i n d t h e v o l u m e t r i c f l o w r a t e and f r a c t i o n o f a i r p a s s i n g t h r o u g h t h e c o o l e r 6 7 // b a s i s 60m3/ h o f c o n d i t i o n e d a i r a t 25 d e g r e e C and 60% RH 8

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Hf = 0 . 0 1 2 ; 10 Hi = 0 . 0 3 3 ; 11 H1 = 0 . 0 0 7 5 ; 12 13 w a t e r _ v a p o r = Hf /18; // i n kmol o f w a t e r v a p o r 14 d r y _ a i r = 1 / 2 8 . 9 ; // i n kmol 15 t o t a l _ m a s s = w a t e r _ v a p o r + d r y _ a i r ; 16 17 v o l u m e = 2 2 . 4 * ( 2 9 8 / 2 7 3 ) * t o t a l _ m a s s ; 18 w e i g h t = 60/ v o l u m e ; 19 p r i n t f( ” w e i g h t o f d r y a i r h a n d l e d p e r h r = %f kg ” , w e i g h t ) 20 21 // p a r t 1 22 i n l e t _ w a t e r v a p o r = 0 . 0 3 3 / 1 8 ; // i n kmol o f w a t e r v a p o r 23 v o l u m e _ i n l e t = 2 2 . 4 * ( 3 0 8 / 2 7 3 ) *( i n l e t _ w a t e r v a p o r + d r y _ a i r ) ; 24 p r i n t f( ” \ n \ n v o l u m e t r i c f l o w r a t e o f i n l e t a i r = %f c u b i c m e t e r ” , v o l u m e _ i n l e t * weight ) 25 26 // p a r t 2 27 y = ( Hf - Hi ) /( H1 - Hi ) ; 28 p r i n t f( ” \ n \ n f r a c t i o n o f i n l e t a i r p a s s i n g t h r o u g h c o o l e r = %f ” ,y )

Scilab code Exa 3.14 Ammonia Synthesis

1 clc 2 c l e a r 3 p r i n t f( ” e x a m p l e 3 . 1 4 p a g e number 1 0 2 \ n \ n ” ) 4 5 // t o f i n d t h e f r a c t i o n o f p u r g e d r e c y c l e and t o t a l y i e l d 6

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7 // x− m o l e s o f N2 and H2 r e c y c l e d ; y − m o l e s o f N2 H2 p u r g e d 8 9 A r _ f r e s h f e e d = 0.2; 10 // a r g o n i n f r e s h f e e d i s e q u a l t o a r g o n i n p u r g e 11 12 y = 0 . 2 / 0 . 0 6 3 3 ; // a r g o n i n p u r g e = 0 . 0 6 3 3 y 13 x = ( 0 . 7 9 * 1 0 0 - y ) / ( 1 - 0 . 7 9 ) ; 14 p r i n t f( ” y = %f kmol \ nx = %f kmol ” ,y , x ) 15 16 // p a r t 1 17 f r a c t i o n = y / x ; 18 p r i n t f( ” \ n \ n f r a t i o n o f r e c y c l e t h a t i s p u r g e d = %f ” , f r a c t i o n ) 19 20 // p a r t 2 21 y i e l d = 0 . 1 0 5 * ( 1 0 0 + x ) ; 22 p r i n t f( ” \ n \ n o v e r a l l y i e l d o f ammonia = %f kmol ” , y i e l d )

Scilab code Exa 3.15 Enthalpy calculation

1 clc 2 c l e a r 3 p r i n t f( ” e x a m p l e 3 . 1 5 p a g e number 1 0 7 \ n \ n ” ) 4 5 // t o f i n d c h a n g e i n e n t h a l p y 6 H 0 _ C H 4 = -74.9 // i n kJ 7 H 0 _ C O 2 = -393.5 // i n kJ 8 H 0 _ H 2 O = -241.8 // i n kJ 9 10 d e l t a _ H 0 = H 0 _ C O 2 +2* H0_H2O - H 0 _ C H 4 ; 11 p r i n t f( ” c h a n g e i n e n t h a l p y = %f kJ ” , de lt a_ H0 )

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Scilab code Exa 3.16 Enthalpy calculation 1 clc 2 c l e a r 3 p r i n t f( ” e x a m p l e 3 . 1 6 p a g e number 1 0 7 \ n \ n ” ) 4 5 // t o c o m p a r e t h e e n t h a l p y c h a n g e i n two r e a c t i o n s 6 7 H 0 _ g l u c o s e = -1273 // i n kJ 8 H 0 _ e t h a n o l = -277.6 // i n kJ 9 H 0 _ C O 2 = -393.5 // i n kJ 10 H 0 _ H 2 O = -285.8 // i n kJ 11 12 // f o r r e a c t i o n 1 13 d e l t a _ H 1 = 2* H 0 _ e t h a n o l +2* H0_CO2 - H 0 _ g l u c o s e ; 14 p r i n t f( ” e n t h a l p y c h a n g e i n r e a c t i o n 1 = %f KJ” , d e l t a _ H 1 ) 15 16 // f o r r e a c t i o n 2 17 d e l t a _ H 2 = 6* H 0 _ H 2 O +6* H0_CO2 - H 0 _ g l u c o s e ; 18 p r i n t f( ” \ n \ n e n t h a l p y c h a n g e i n r e a c t i o n 2 = %f kJ ” , d e l t a _ H 2 ) 19

20 if delta_H1 > d e l t a _ H 2 then disp ( ” r e a c t i o n 2 s u p p l i e s more e n e r g y ” )

21 else disp ( ” r e a c t i o n 1 s u p p l i e s more e n e r g y ” )

22 end

Scilab code Exa 3.17 Enthalpy of formation

1 clc

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3 p r i n t f( ” e x a m p l e 3 . 1 7 p a g e number 1 0 8 \ n \ n ” ) 4 5 // t o f i n d e n t h a l p y o f f o r m a t i o n o f CuSO4 . 5 H2O 6 7 d e l t a _ H 2 = 11.7 // i n kJ / mol 8 m _ C u S O 4 = 16 // i n gm 9 m _ H 2 O = 384 // i n gm 10 11 d e l t a _ H 3 = -(( m _ C u S O 4 + m _ H 2 O ) * 4 . 1 8 * 3 . 9 5 * 1 5 9 . 6 ) / ( 1 6 * 1 0 ^ 3 ) 12 d e l t a _ H 1 = d e l t a _ H 3 - d e l t a _ H 2 ; 13 14 p r i n t f( ” e n t h a l p y o f f o r m a t i o n = %f kJ / mol ” , de lt a_ H1 )

Scilab code Exa 3.18 Combustion

1 clc 2 c l e a r 3 p r i n t f( ” e x a m p l e 3 . 1 8 p a g e number 1 0 8 \ n \ n ” ) 4 5 // t o f i n d t h e t e m p e r a t u r e o f c o m b u s t i o n 6 7 H _ c o m b u s t i o n = 1 5 6 0 0 0 0 // i n kJ / kmol 8 H 0 _ C O 2 = 5 4 . 5 6 // i n kJ / kmol 9 H 0 _ O 2 = 35.2 // i n kJ / kmol 10 H 0 _ s t e a m = 4 3 . 3 8 // i n kJ / kmol 11 H 0 _ N 2 = 3 3 . 3 2 // i n kJ / kmol 12 13 t = H _ c o m b u s t i o n /(2* H 0 _ C O 2 +3* H 0 _ s t e a m + 0 . 8 7 5 * H 0 _ O 2 + 1 6 . 4 6 * H 0 _ N 2 ) ; 14 15 p r i n t f( ” t h e o r i t i c a l t e m p e r a t u r e o f c o m b u s t i o n = %f d e g r e e C” ,t )

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Scilab code Exa 3.19 Heat of reaction 1 clc 2 c l e a r 3 p r i n t f( ” e x a m p l e 3 . 1 9 p a g e number 1 0 9 \ n \ n ” ) 4 5 // t o f i n d t h e h e a t o f r e a c t i o n and c o n s u m p t i o n o f c o k e 6 7 H _ N a C l = 4 1 0 . 9 // i n MJ/ kmol 8 H _ H 2 S O 4 = 8 1 1 . 3 // i n MJ/ kmol 9 H _ N a 2 S O 4 = 1384 // i n MJ/ kmol 10 H _ H C l = 92.3 // i n MJ/ kmol 11 12 Q = H _ N a 2 S O 4 + 2* H _ H C l -2* H_NaCl - H _ H 2 S O 4 ; 13 p r i n t f( ” h e a t o f r e a c t i o n = %f MJ\ n \ n ” ,Q ) 14 15 h e a t _ r e q u i r e d = 6 4 . 5 * ( 5 0 0 / 7 3 ) ; 16 c o k e _ c o n s u m p t i o n = h e a t _ r e q u i r e d /19 17 p r i n t f( ” amount o f c o k e o v e n g a s consumed = %f c u b i c m e t e r ” , c o k e _ c o n s u m p t i o n )

Scilab code Exa 3.20 Heat transfer

1 clc 2 c l e a r 3 p r i n t f( ” e x a m p l e 3 . 2 0 p a g e number 1 0 9 \ n \ n ” ) 4 5 // t o f i n d t h e r a t e o f h e a t f l o w 6 7 c p _ w a t e r = 1 4 6 . 5 // i n k j / kg 8 c p _ s t e a m = 3040 // i n kJ / kg

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9 d = 0 . 1 0 2 // i n m 10 u = 1.5 // i n m/ s 11 d e n s i t y = 1000 // i n kg /m3 12 13 m = ( 3 . 1 4 / 4 ) * d ^2* u * d e n s i t y ; 14 Q = m *( cp_steam - c p _ w a t e r ) ; 15 16 p r i n t f( ” r a t e o f h e a t f l o w = %f kW” ,Q )

Scilab code Exa 3.21 Calorific value

1 clc

2 //EXAMPLE 3 . 2 1

3 // To f i n d t h e c a l o r i f i c v a l u e o f c o a l

4 disp( ’ t h i s i s a t h e o r i t i c a l p r o b l e m . R e f e r t h e book f o r s o l u t i o n ’ )

Scilab code Exa 3.22 Coal combustion

1 clc 2 c l e a r 3 p r i n t f( ” e x a m p l e 3 . 2 2 p a g e number 1 1 0 \ n \ n ” ) 4 5 // t o f i n d t h e amount o f a i r r e q u i r e d f o r c o m b u s t i o n and c o m p o s i t i o n o f f l u e g a s 6 wt_C = 0.75 // i n kg 7 w t _ H 2 = 0.05 // i n kg 8 w t _ O 2 = 0.12 // i n kg 9 w t _ N 2 = 0.03 // i n kg 10 wt_S = 0.01 // i n kg 11 w t _ a s h = 0.04 // i n kg 12 13 O2_C = wt_C * ( 3 2 / 1 2 ) ; // i n kg

(64)

14 O 2 _ H 2 = w t _ H 2 * ( 1 6 / 2 ) ; // i n kg 15 O2_S = wt_S * ( 3 2 / 3 2 ) ; // i n kg 16 O 2 _ r e q u i r e d = O2_C + O 2 _ H 2 + O2_S ; 17 18 o x y g e n _ s u p p l i e d = O 2 _ r e q u i r e d - w t _ O 2 ; 19 a i r _ n e e d e d = o x y g e n _ s u p p l i e d / 0 . 2 3 ; 20 p r i n t f( ” amount o f a i r r e q u i r e d = %f kg ” , a i r _ n e e d e d ) 21 22 v o l u m e = ( 2 2 . 4 / 2 8 . 8 ) * a i r _ n e e d e d ; 23 p r i n t f( ” \ n \ n v o l u m e o f a i r n e e d e d = %f c u b i c m e t e r ” , v o l u m e ) 24 25 a i r _ s u p p l i e d = 1 . 2 0 * a i r _ n e e d e d ; 26 N 2 _ s u p p l i e d = a i r _ s u p p l i e d * 0 . 7 7 ; 27 t o t a l _ N 2 = N 2 _ s u p p l i e d + w t _ N 2 ; 28 29 O 2 _ f l u e g a s = a i r _ s u p p l i e d * 0 . 2 3 - o x y g e n _ s u p p l i e d ; 30 31 w t _ C O 2 = wt_C + O2_C ; 32 w t _ S O 2 = wt_S + O2_S ; 33 34 m o l e s _ C O 2 = w t _ C O 2 /44; 35 m o l e s _ S O 2 = w t _ S O 2 /64; 36 m o l e s _ N 2 = t o t a l _ N 2 /28; 37 m o l e s _ O 2 = O 2 _ f l u e g a s /32; 38 t o t a l _ m o l e s = m o l e s _ C O 2 + m o l e s _ S O 2 + m o l e s _ N 2 + m o l e s _ O 2 ; 39 40 x _ C O 2 = m o l e s _ C O 2 / t o t a l _ m o l e s ; 41 x _ S O 2 = m o l e s _ S O 2 / t o t a l _ m o l e s ; 42 x_N2 = m o l e s _ N 2 / t o t a l _ m o l e s ; 43 x_O2 = m o l e s _ O 2 / t o t a l _ m o l e s ; 44 45 p r i n t f( ” \ n \nCO2 = %f ” , x _ C O 2 * 1 0 0 ) 46 p r i n t f( ” \ n \ nSO2 = %f ” , x _ S O 2 * 1 0 0 ) 47 p r i n t f( ” \ n \nN2 = %f ” , x_N2 * 1 0 0 ) 48 p r i n t f( ” \ n \nO2 = %f ” , x_O2 * 1 0 0 )

Introduction to Chemical Engineering

Scilab Textbook Companion for