Stoichiometry and Process Calculations_K. v. Narayanan and B. Lakshmikutty

Scilab Textbook Companion for Stoichiometry And Process Calculations by K. V. Narayanan And B. Lakshmikutty1 Created by Jimit Dilip Patel FOURTH YEAR Chemical Engineering Visvesvaraya National Institute Of Technology College Teacher Dr. Sachin Mandavagane Cross-Checked by

August 10, 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

Book Description Title: Stoichiometry And Process Calculations Author: K. V. Narayanan And B. Lakshmikutty Publisher: Prentice Hall Of India, New Delhi Edition: 1 Year: 2006 ISBN: 81-203-2992-9

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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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Contents List of Scilab Codes

4

2 Units and Dimensions

11

3 Fundamental concepts of stoichiometry

16

4 Ideal Gases and Gas Mixtures

28

5 Properties of Real Gases

40

6 Vapour Pressure

47

7 Solutions and Phase Behaviour

53

8 Humidity and Humidity chart

68

9 Material Balance in Unit Operations

83

10 Material Balance with Chemical Reaction

101

11 Energy Balance Thermophysics

126

12 Energy Balance Thermochemistry

152

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List of Scilab Codes Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa

2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9 3.10 3.11 3.12 3.13 3.14 3.15 3.16 3.17 3.18 3.19

Mass flow rate . . . . . . . . . . . . . . . Poundal to Newton . . . . . . . . . . . . Conversion of N per m2 . . . . . . . . . . Thermal conductivity . . . . . . . . . . . Mass and FPS system . . . . . . . . . . . Kinetic energy calculation . . . . . . . . . Force and pressure for piston cylinder . . units conversion . . . . . . . . . . . . . . units conversion . . . . . . . . . . . . . . pounds per minute to kmol per hour . . . Number of molecules . . . . . . . . . . . . Moles of sodium sulphate . . . . . . . . . Pyrites and oxygen moles . . . . . . . . . Volume of oxygen . . . . . . . . . . . . . Reaction of iron and steel . . . . . . . . . Equivalent weight . . . . . . . . . . . . . Specific gravity calculation . . . . . . . . Specific gravity of mixture . . . . . . . . . Baume scale . . . . . . . . . . . . . . . . Density using API scale . . . . . . . . . . Drying Ammonium sulphate . . . . . . . Percentage of water removed . . . . . . . Amount and percentage water removed . NaCl solution . . . . . . . . . . . . . . . . Molal absolute humidity . . . . . . . . . . K2CO3 solution . . . . . . . . . . . . . . Molarity and Molality of Alcohol solution CO to phosgene . . . . . . . . . . . . . . 4

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11 11 12 12 13 13 13 14 15 16 16 17 17 18 18 19 19 19 20 20 21 21 22 22 23 23 24 25

Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa

3.20 3.21 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9 4.10 4.11 4.12 4.13

Exa 4.14 Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa

5.1 5.2 5.3 5.4 5.5 5.6 5.7 5.8 5.9 6.1 6.2 6.3 6.4 6.5 6.6 7.1

Exa 7.2 Exa 7.3

Extent of reaction . . . . . . . . . . . . . . . . . . . . ethylene to ethanol . . . . . . . . . . . . . . . . . . . . gas constant R . . . . . . . . . . . . . . . . . . . . . . Molar volume of air . . . . . . . . . . . . . . . . . . . Ideal gas equation . . . . . . . . . . . . . . . . . . . . Maximum allowable temperature of tyre . . . . . . . . Pressure calculation . . . . . . . . . . . . . . . . . . . weight composition and density calculation . . . . . . calculations for natural gas . . . . . . . . . . . . . . . Volume of gas from absorption columnn . . . . . . . . dehumidification . . . . . . . . . . . . . . . . . . . . . Absorption column for H2S . . . . . . . . . . . . . . . Reaction stoichiometry for preparation of ammonia . . Reaction stoichiometry for preparation of producer gas Reaction stoichiometry for preparation of Chlorine from HCl . . . . . . . . . . . . . . . . . . . . . . . . . . . . Reaction stoichiometry for dissociation of Carbon Dioxide . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Van der waals equation . . . . . . . . . . . . . . . . . Van der waals equation for CO2 gas . . . . . . . . . . Redlich Kwong equation for gaseous ammonia . . . . . Molar Volume calculation for gaseous ammonia . . . . virial equation of state . . . . . . . . . . . . . . . . . . Lyderson method for n butane . . . . . . . . . . . . . Pitzer correlation for n butane . . . . . . . . . . . . . Molar volume by different methods . . . . . . . . . . . Van der waals equation and Kays method . . . . . . . Quality of steam . . . . . . . . . . . . . . . . . . . . . Calculation of vapour pressure . . . . . . . . . . . . . Clausius Clapeyron equation for acetone . . . . . . . . Antoine equation for n heptane . . . . . . . . . . . . . Cox chart . . . . . . . . . . . . . . . . . . . . . . . . . Duhring line . . . . . . . . . . . . . . . . . . . . . . . composition calculation of Liquid and vapour at equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . Composition and total pressure calculation . . . . . . Mole fraction calculation for particular component in liquid vapour mixture . . . . . . . . . . . . . . . . . . 5

25 26 28 28 29 29 30 30 31 32 33 33 34 35 37 38 40 40 41 41 42 42 43 43 45 47 47 48 49 49 51 53 53 54

Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa

7.4 7.5.a 7.5.b 7.6 7.7 7.8 7.9 7.10 7.11 7.12 7.13 7.14 7.15 8.1 8.2 8.3 8.4 8.5 8.6 8.7 8.8 8.9 8.10 8.11 8.12 8.14 8.15 8.16 8.17 8.18 8.19 9.1 9.2 9.3 9.4 9.5 9.6

Flash vapourization of benzene toluene mixture . . . . Boiling point diagram . . . . . . . . . . . . . . . . . . Equilibrium Diagram . . . . . . . . . . . . . . . . . . Bubble point temperature and vapour composition . . Dew point temperature pressure and concentration . . Partial pressure of acetaldehyde . . . . . . . . . . . . . Raults law application . . . . . . . . . . . . . . . . . . Dew point temperature and pressure . . . . . . . . . . bubble point and dew point . . . . . . . . . . . . . . . component calculations . . . . . . . . . . . . . . . . . equilibrium temperature and composition . . . . . . . Temperature composition diagram . . . . . . . . . . . Boiling point calculation . . . . . . . . . . . . . . . . . Nitrogen and ammonia gas mixture . . . . . . . . . . . Benzene vapour air mixture . . . . . . . . . . . . . . . Evaporation of acetone using dry air . . . . . . . . . . Humidity for acetone vapour and nitrogen gas mixture Percent saturation and relative saturation . . . . . . . Analysis of Moist air . . . . . . . . . . . . . . . . . . . Heating value calculation for a fuel gas . . . . . . . . . Analysis of nitrogen benzene mixture . . . . . . . . . . Drying . . . . . . . . . . . . . . . . . . . . . . . . . . Saturation lines for hexane . . . . . . . . . . . . . . . Psychometric chart application . . . . . . . . . . . . . Humid heat calculation for a sample of air . . . . . . . wet bulb temperature and dry bulb temperature . . . Humidity calculation . . . . . . . . . . . . . . . . . . . SAturation curve and adiabatic cooling line . . . . . . Adiabatic drier . . . . . . . . . . . . . . . . . . . . . . Psychometric chart application . . . . . . . . . . . . . Psychometric chart application and given wet bulb and dry bulb temperature . . . . . . . . . . . . . . . . . . Combustion of coal . . . . . . . . . . . . . . . . . . . . Drying of wood . . . . . . . . . . . . . . . . . . . . . . Effluent discharge . . . . . . . . . . . . . . . . . . . . benzene requirement calculation . . . . . . . . . . . . Fortification of waste acid . . . . . . . . . . . . . . . . Triple effect evaporator . . . . . . . . . . . . . . . . . 6

55 56 56 58 58 59 59 61 62 63 65 65 67 68 69 69 70 71 71 72 73 74 76 76 77 78 78 79 80 81 82 83 83 84 84 85 86

Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa

9.7 9.8 9.9 9.10 9.11 9.12 9.13 9.14 9.15 9.16 9.17 9.18 9.19 9.20 9.21 9.22 10.1 10.2 10.3 10.4 10.5 10.6 10.7 10.8 10.9 10.10 10.11 10.12 10.13 10.14 10.15 10.16 10.17 10.18 10.19 10.20 10.21 10.22

Crystallization operation . . . . . . . . . . . Evaporation of Na2CO3 . . . . . . . . . . . . Crystallization . . . . . . . . . . . . . . . . . Extraction . . . . . . . . . . . . . . . . . . . Leaching operation . . . . . . . . . . . . . . . Dryer and oven . . . . . . . . . . . . . . . . . Adiabatic drier . . . . . . . . . . . . . . . . . Extraction of isopropyl alcohol . . . . . . . . Absorption of acetone . . . . . . . . . . . . . Absorption of SO3 . . . . . . . . . . . . . . . Continuous distillation column . . . . . . . . Distillation operation for methanol solution . Bypass operation . . . . . . . . . . . . . . . . Recycle operation centrifuge plus filter . . . . Recycle operation granulator and drier . . . . Blowdown operation . . . . . . . . . . . . . . Combustion of propane . . . . . . . . . . . . Combustion of hydrogen free coke . . . . . . Combustion of fuel oil . . . . . . . . . . . . . Combustion of producer gas . . . . . . . . . . Combustion of coal . . . . . . . . . . . . . . . Stoichiometric analysis of combustion of coal Orsat analysis . . . . . . . . . . . . . . . . . Burning of pyrites . . . . . . . . . . . . . . . Production of sulphuric acid . . . . . . . . . Burning of limestone mixed with coke . . . . treating limestone with aqueous H2SO4 . . . Production of TSP . . . . . . . . . . . . . . . Production of sodium phosphate . . . . . . . Production of pig iron . . . . . . . . . . . . . Production of nitric acid . . . . . . . . . . . . Material balance in nitric acid production . . Electrolysis of brine . . . . . . . . . . . . . . Preparation of Formaldehyde . . . . . . . . . Recycle operation reactor and separator . . . Conversion of sugar to glucose and fructose . Purging operation . . . . . . . . . . . . . . . Purging operation for production of methanol 7

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86 87 88 89 89 90 91 92 93 94 95 95 96 97 98 99 101 102 103 105 105 106 109 109 111 112 113 115 115 116 117 118 119 121 122 122 123 124

Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa

11.1 11.2 11.3 11.4 11.5 11.6 11.7 11.8 11.9 11.10 11.11 11.12 11.13 11.14 11.15 11.16 11.17 11.18 11.19 11.20 11.21 11.22 11.23 11.24 11.25 11.26 11.27 11.28 11.29 11.30 11.31 11.32 11.33 11.34

Power calculation . . . . . . . . . . . . . . . . . . . . Kinetic energy calculation . . . . . . . . . . . . . . . . Work done calculation for a gas confined in a cylinder Power requirement of the pump . . . . . . . . . . . . . Specific enthalpy of the fluid in the tank . . . . . . . . internal energy and enthalpy change calculation . . . . change in internal energy . . . . . . . . . . . . . . . . reaction of iron with HCl . . . . . . . . . . . . . . . . Thermic fluid . . . . . . . . . . . . . . . . . . . . . . . Heat capacity . . . . . . . . . . . . . . . . . . . . . . . Enthalpy change when chlorine gas is heated . . . . . Molal heat capacity . . . . . . . . . . . . . . . . . . . Enthalpy change of a gas . . . . . . . . . . . . . . . . Combustion of solid waste . . . . . . . . . . . . . . . . Heat capacity calculation for Na2SO4 10H2O . . . . . Heat of vaporization calculation . . . . . . . . . . . . Heat requirement . . . . . . . . . . . . . . . . . . . . . Equilibrium temperature of mixture . . . . . . . . . . Estimation of mean heat of vaporisation . . . . . . . . Heat of vaporization of methyl chloride . . . . . . . . Watson equation . . . . . . . . . . . . . . . . . . . . . Kistyakowsky equation . . . . . . . . . . . . . . . . . . Quality of steam . . . . . . . . . . . . . . . . . . . . . Heat calculation . . . . . . . . . . . . . . . . . . . . . Enthalpy balance for evaporation process . . . . . . . Mean heat capacity of ethanol water solution . . . . . Evaporation of NaOH solution . . . . . . . . . . . . . Heat transfer to air . . . . . . . . . . . . . . . . . . . change in internal energy . . . . . . . . . . . . . . . . Heat liberation in oxidation of iron fillings . . . . . . . Saturated steam and saturated water . . . . . . . . . . constant volume and constant pressure process . . . . series of operations . . . . . . . . . . . . . . . . . . . . change in internal energy and enthalpy and heat supplied and work done . . . . . . . . . . . . . . . . . . . Exa 11.35 Heat removed in condenser . . . . . . . . . . . . . . . Exa 11.36 Throttling process . . . . . . . . . . . . . . . . . . . . Exa 11.37 water pumping and energy balances . . . . . . . . . . 8

126 126 127 127 128 128 128 129 129 130 130 132 133 133 134 134 135 136 137 137 137 138 138 139 140 141 142 142 143 143 143 144 145 146 147 148 148

Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa

11.38 11.39 12.1 12.2 12.3 12.4 12.5 12.6 12.7 12.8 12.9 12.10 12.11 12.12 12.13 12.14 12.15 12.16 12.17 12.18 12.19 12.20

Energy balance on rotary drier . . . . . . . . . . . . . Energy balance on the fractionator . . . . . . . . . . . Heat liberated calculation . . . . . . . . . . . . . . . . Heat of formation of methane . . . . . . . . . . . . . . Net heating value of coal . . . . . . . . . . . . . . . . Heat of reaction for esterification of ethyl alcohol . . . Vapour phase hydration of ethylene to ethanol . . . . Standard heat of formation of acetylene . . . . . . . . Standard heat of roasting of iron pyrites . . . . . . . . Standard heat of formation of liquid methanol . . . . . Gross heating value and Net heating value calculation Standard heat of reaction calculation . . . . . . . . . . Constant pressure heat of combustion . . . . . . . . . Heat of reaction for ammonia synthesis . . . . . . . . Standard heat of reaction of methanol synthesis . . . . Combustion of CO . . . . . . . . . . . . . . . . . . . . Heat added or removed calculation . . . . . . . . . . . CO2 O2 and N2 passed through a bed of C . . . . . . Partial oxidation of natural gas . . . . . . . . . . . . . Maximum allowable conversion calculation . . . . . . . Theoretical flame temperature calculation . . . . . . . Temperature of products on burning of hydrogen gas .

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149 150 152 153 153 154 154 155 155 156 157 157 158 158 159 160 161 162 163 164 165 166

List of Figures 6.1 6.2

Cox chart . . . . . . . . . . . . . . . . . . . . . . . . . . . . Duhring line . . . . . . . . . . . . . . . . . . . . . . . . . . .

50 51

7.1 7.2 7.3

Boiling point diagram . . . . . . . . . . . . . . . . . . . . . . Equilibrium Diagram . . . . . . . . . . . . . . . . . . . . . . Temperature composition diagram . . . . . . . . . . . . . . .

55 57 66

8.1 8.2

Saturation lines for hexane . . . . . . . . . . . . . . . . . . . SAturation curve and adiabatic cooling line . . . . . . . . .

75 79

11.1 Enthalpy change when chlorine gas is heated . . . . . . . . .

131

10

Chapter 2 Units and Dimensions

Scilab code Exa 2.1 Mass flow rate 1 2 3 4 5 6 7 8 9 10

clc () funcprot (0) V1 = 15; // f t ˆ3/ min ft = 0.3048; //m min = 60; // s e c s V = V1 * ft ^3/ min ; disp ( ”mˆ3/ s ” ,V , ” V o l u m e t r i c f l o w r a t e = ” ) D = 1000; // kg /mˆ3 M = V * D; disp ( ” kg / s ” ,M , ” mass f l o w r a t e = ” )

Scilab code Exa 2.2 Poundal to Newton 1 2 3 4 5 6

clc () funcprot (0) ft = 0.3048; //m lb = 0.4536; // kg P = ft * lb ; disp ( ”N” ,P , ” 1 p o u n d a l i s 1 f t ∗ l b / s ˆ2 = ” ) 11

Scilab code Exa 2.3 Conversion of N per m2 1 2 3 4 5 6 7 8 9 10 11 12

clc () funcprot (0) kgf = 9.80665; //N cm = 10^ -2; //m P = kgf / cm ^2; disp ( ”N/mˆ2 ” ,P , ” 1 k g f /cmˆ2 = ” ) lbf = 32.174; // l b ∗ f t // s ˆ2 lb = 0.4535924; // kg ft = 0.3048; //m in = 0.0254; //m P1 = lbf * lb * ft / in ^2; disp ( ”N/mˆ2 ” ,P1 , ” 1 l b f / i n ˆ2 = ” )

Scilab code Exa 2.4 Thermal conductivity 1 2 3 4 5 6 7 8 9 10 11 12 13 14

clc () Q1 = 10000; // kJ / h r kJ = 1000; // J hr = 3600; // s Q = Q1 * kJ / hr ; // J / s disp ( ” J / s ” ,Q , ”Q = ” ) x = 0.1; //m A = 1 //mˆ2 T = 800; //K k = x * Q /( A * T ) ; disp ( ”W/ (m∗K) ” ,k , ” t h e r m a l c o n d u c t i v i t y = ” ) J = 1/4.1868; // c a l k1 = k * J * hr /1000; disp ( ” k c a l / ( h∗m∗C) ” ,k1 , ” t h e r m a l c o n d u c t i v i t y = ” ) 12

Scilab code Exa 2.5 Mass and FPS system 1 2 3 4 5 6 7 8

clc () F = 300; //N a = 9.81; //m/ s ˆ2 m = F / a ; // kg disp ( ” kg ” ,m , ” mass i n kg = ” ) lb = 4.535924/10; // kg m1 = m / lb ; disp ( ” l b ” ,m1 , ” mass i n pounds = ” )

Scilab code Exa 2.6 Kinetic energy calculation 1 2 3 4 5 6 7 8 9

clc () z = 15; //m PE = 2000; // J g = 9.8067; //m/ s ˆ2 m = PE /( z * g ) ; disp ( ” kg ” ,m , ” mass = ” ) v = 50; //m/ s KE = 1/2* m *( v ^2) /1000; disp ( ” kJ ” ,KE , ” k i n e t i c e n e r g y = ” )

Scilab code Exa 2.7 Force and pressure for piston cylinder 1 2 3 4

clc () g = 9.81; //m/ s ˆ2 m = 100 * 0.4536; // kg P = 101325; //N/mˆ2 13

5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22

D1 = 4; // i n c h D = D1 * 2.54 * 10^ -2; //m A = 3.1415 * ( D ^2) /4; //mˆ2 F1 = P * A ; //N F2 = m * g ; //N F = F1 + F2 ; disp ( ”N” ,F , ” T o t a l f o r c e a c t i n g on t h e g a s = ” ) P1 = F / A ; //N/mˆ2 P2 = P1 /100000; // b a r P3 = P1 /(6.894757 * 10^3) ; // p s i disp ( ”N/mˆ2 ” ,P1 , ” P r e s s u r e i n N/mˆ2 = ” ) disp ( ” b a r ” ,P2 , ” P r e s s u r e i n b a r = ” ) disp ( ” p s i ” ,P3 , ” P r e s s u r e i n p s i = ” ) d = 0.4; //m W = F * d; disp ( ” J ” ,W , ”Work done = ” ) PE = m * g * d ; disp ( ” J ” ,PE , ” Change i n p o t e n t i a l e n e r g y = ” )

Scilab code Exa 2.8 units conversion 1 clc () 2 //kG = ( 6 . 7 ∗ 10ˆ −4) ∗ ( ( G ∗ ( d s + d t ) / d s ) ˆ 0 . 8 ) / 3 4 5 6 7 8 9 10

( ( d s ˆ 0 . 4 ) ∗ (dG ˆ 0 . 2 ) ) //kG − l b m o l / ( h f t ˆ2 atm ) , G − l b / ( f t ˆ2 h ) , ds , dG , dt − f e e t // kG1 − kmol / (mˆ2 h atm ) , G1 − kg / (mˆ2 h ) , ds1 , dG1 , dt1 − m G = 0.2048; //G1 ∗ l b / ( f t ˆ2 h ) d = 3.2808; // d1 ∗ f t ds = d ; dt = d ; dG = d ; kG = 4.885; // kG1 ( l b m o l / ( h f t ˆ2 atm ) = 4 . 8 8 5 ∗ kmol / (mˆ2 h atm ) ) 14

11 C = (6.7 * 10^ -4) * (( G * d / ds ) ^0.8) / (( ds ^0.4) 12 13 14 15

*( dG ^0.2) ) * kG ; disp (C , ”Co− e f f i c i e n t = ” ) // t h i s i s t h e c o n s t a n t f o r t h e e q u a t i o n // t h e e q u a t i o n t h u s becomes , // kG1 = C ∗ ( ( G1 ∗ ( d s 1 + d t 1 ) / d s 1 ) ˆ 0 . 8 ) / ( ( d s 1 ˆ 0 . 4 ) ∗ ( dG1 ˆ 0 . 2 ) )

Scilab code Exa 2.9 units conversion 1 clc () 2 //Cp = 7 . 1 3 + 0 . 5 7 7 ∗ (10ˆ −3) ∗ t + 0 . 0 2 4 8 ∗ (10ˆ −6) 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

∗ t ˆ2 //Cp − Btu / l b −mol F , t − F // Cp1 − kJ / kmol K , t 1 − K a = 7.13; b = 0.577 * 10^ -3; c = 0.0248 * 10^ -6; // t = 1 . 8 ∗ t 1 − 4 5 9 . 6 7 Cp = 4.1868; // Cp1 ( Btu / l b −mol F = 4 . 1 8 6 8 ∗ ( kJ / kmol K) ) // s u b s t i t u t i n g t h e above , we g e t , // Cp1 = 2 8 . 7 6 3 + 4 . 7 6 3 ∗ (10ˆ −3) ∗ t 1 + 0 . 3 3 6 6 ∗ (10ˆ −6) ∗ t ˆ2 a1 = 28.763; b1 = 4.763 * (10^ -3) ; c1 = 0.3366 * (10^ -6) ; disp ( a1 , ” a1 = ” ) disp ( b1 , ” b1 = ” ) disp ( c1 , ” c 1 = ” ) // t h i s a r e t h e c o e f f i c e n t s f o r t h e f o l l o w i n g equation ; // Cp1 = a1 + b1 ∗ t 1 + c 1 ∗ ( t 1 ) ˆ2

15

Chapter 3 Fundamental concepts of stoichiometry

Scilab code Exa 3.1 pounds per minute to kmol per hour 1 2 3 4 5

clc () m = 1000 * 0.4536; // kg / min M = 30.24; //gm/ mol m1 = m * 60 / M ; disp ( ” kmol / h r ” ,m1 , ” m o l a r f o l w r a t e = ” )

Scilab code Exa 3.2 Number of molecules 1 2 3 4 5 6 7 8

clc () MK = 39.1; MC = 12.0; MO = 16; MK2CO3 = MK * 2 + MC + MO * 3; m = 691; N = m / MK2CO3 ; A = 6.023 * 10^23; 16

9 10

molecules = N * A ; disp ( ” m o l e c u l e s ” , molecules , ” T o t a l no . o f m o l e c u l e s = ”)

Scilab code Exa 3.3 Moles of sodium sulphate 1 2 3 4 5 6

clc () Na = 23; //gm/ mol MNa = 100; // kg N = MNa * 1000 / Na ; // g−atoms NNa2SO4 = N / 2; disp ( ” kmol ” , NNa2SO4 , ” ( a ) m o l e s o f sodium s u l p h a t e = ”) 7 MNa2SO4 = 142.06; 8 m = NNa2SO4 * MNa2SO4 /1000; 9 disp ( ” kg ” ,m , ” ( b ) k i l o g r a m s o f sodium s u l p h a t e = ” )

Scilab code Exa 3.4 Pyrites and oxygen moles 1 2 3 4 5 6 7 8 9 10 11 12

clc () MFe = 55.85; MO = 16; MS = 32; MFeS2 = MFe + MS * 2; MFe2O3 = MFe * 2 + MO * 3; MSO3 = MS + MO * 3; m1SO3 = 100; // kg N1 = m1SO3 / ( MSO3 ) ; // kmol NFeS2 = N1 / 2; mFeS2 = NFeS2 * MFeS2 ; disp ( ” kg ” , mFeS2 , ” mass o f p y r i t e s t o o b t a i n 100 kg o f SO3 =” ) 13 m2SO3 = 50; // kg 17

14 N2 = m2SO3 / ( MSO3 ) ; // kmol 15 NO2 = N2 * 15/8; 16 mO2 = NO2 * MO * 2; 17 disp ( ” kg ” ,mO2 , ” mass o f Oxygen consumed t o p r o d u c e 50

kg o f SO3 =” )

Scilab code Exa 3.5 Volume of oxygen 1 2 3 4 5 6 7 8

clc () MKClO3 = 122.55 mKClO3 = 100; // kg NKClO3 = mKClO3 / MKClO3 ; NO2 = 3 * NKClO3 / 2; V1 = 22.4143; //mˆ3/ kmol ; V = V1 * NO2 ; disp ( ”mˆ3 ” ,V , ” volume o f o x y g e n p r o d u c e d = ” )

Scilab code Exa 3.6 Reaction of iron and steel 1 2 3 4 5 6 7 8 9 10 11 12

clc () mH2 = 100; // kg NH2 = mH2 /2.016; NFe = 3 * NH2 / 4; mFe = NFe * 55.85; disp ( ” kg ” ,mFe , ” ( a ) mass o f i r o n r e q u i r e d = ” ) NH2O = NH2 ; mH2O = NH2O * 18; disp ( ” kg ” , mH2O , ” mass o f steam r e q u i r e d =” ) V1 = 22.4143; //mˆ3/ kmol ; V = V1 * NH2 ; disp ( ”mˆ3 ” ,V , ” Volume o f h y d r o g e n = ” )

18

Scilab code Exa 3.7 Equivalent weight 1 clc () 2 MCaCO3 = 100.08; 3 GE = MCaCO3 / 2; 4 disp ( ” g ” ,GE , ”Gram e q u i v a l e n t wt .

o f CaCO3 =” )

Scilab code Exa 3.8 Specific gravity calculation 1 2 3 4 5 6 7 8

clc () m1 = 1; // kg ( mass i n a i r ) m2 = 0.9; // kg ( mass i n w a t e r ) m3 = 0.82; // kg ( mass i n l i q u i d ) L1 = m2 - m1 ; // kg ( l o s s o f mass i n w a t e r ) L2 = m3 - m1 ; // kg ( l o s s o f mass i n l i q u i d ) sp . g = L2 / L1 ; disp ( sp .g , ” s p e c i f i c g r a v i t y o f l i q u i d = ” )

Scilab code Exa 3.9 Specific gravity of mixture 1 2 3 4 5 6 7 8 9

clc () m1 = 10; // kg m2 = 5; // kg sp . g1 = 1.17; sp . g2 = 0.83; Dwater = 1000; // kg /mˆ3 DA = Dwater * sp . g1 ; DB = Dwater * sp . g2 ; V1 = m1 / DA ; 19

10 V2 = m2 / DB ; 11 V = V1 + V2 ; 12 Dmix = ( m1 + m2 ) / V ; 13 sp . g3 = Dmix / Dwater ; 14 disp ( sp . g3 , ” s p e c i f i c g r a v i t y

o f m i x t u r e =” )

Scilab code Exa 3.10 Baume scale 1 2 3 4 5

clc () Tw = 100; //Tw sp . g = Tw /200 + 1; Be = 145 - 145/ sp . g ; disp ( ”Be” ,Be , ” s p e c i f i c g r a v i t y on beume s c a l e =” )

Scilab code Exa 3.11 Density using API scale 1 clc () 2 API1 = 30; // API 3 sp . g1 = 141.5/(131.5 + API1 ) ; // ( s i n c e , API = 1 4 1 . 5 /

sp . g −131.5) 4 Dwater = 999; // kg /mˆ 3 ; 5 Doil1 = sp . g1 * Dwater ; 6 V1 = 250; //mˆ3 7 m1 = V1 * Doil1 ; 8 API2 = 15; // API 9 sp . g2 = 141.5/(131.5 + API2 ) ; // ( s i n c e , API = 1 4 1 . 5 / 10 11 12 13 14 15

sp . g −131.5) Dwater = 999; // kg /mˆ 3 ; Doil2 = sp . g2 * Dwater ; V2 = 1000; //mˆ3 m2 = V2 * Doil2 ; Dmix = ( m1 + m2 ) /( V1 + V2 ) ; disp ( ” kg /mˆ3 ” , Dmix , ” d e n s i t y o f t h e m i x t u r e =” ) 20

Scilab code Exa 3.12 Drying Ammonium sulphate 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23

clc () m1 = 250; // kg mwater1 = 50; // kg mdrysolid1 = m1 - mwater1 ; wfe1 = mwater1 / m1 ; wr1 = mwater1 / mdrysolid1 ; wtpercentw1 = mwater1 * 100 / m1 ; wtpercentd1 = mwater1 * 100 / mdrysolid1 ; a = 90; //% mwater2 = mwater1 * (1 - a /100) ; m2 = mdrysolid1 + mwater2 ; wfe2 = mwater2 / m2 ; wr2 = mwater2 / mdrysolid1 ; wtpercentw2 = mwater2 * 100 / m2 ; wtpercentd2 = mwater2 * 100 / mdrysolid1 ; disp ( wfe1 , ” ( a ) w e i g h t f r a c t i o n o f w a t e r a t e n t r a n c e = ”) disp ( wfe2 , ” w e i g h t f r a c t i o n o f w a t e r a t e x i t = ” ) disp ( wr1 , ” ( b ) w e i g h t r a t i o o f w a t e r a t e n t r a n c e = ” ) disp ( wr2 , ” w e i g h t r a t i o o f w a t e r a t e x i t = ” ) disp ( wtpercentw1 , ” ( c ) w e i g h t p e r c e n t o f m o i s t u r e on wet b a s i s a t e n t r a n c e = ” ) disp ( wtpercentw2 , ” w e i g h t p e r c e n t o f m o i s t u r e on wet b a s i s at e x i t = ”) disp ( wtpercentd1 , ” ( d ) w e i g h t p e r c e n t o f m o i s t u r e on dry b a s i s at e n t r a n c e = ”) disp ( wtpercentd2 , ” w e i g h t p e r c e n t o f m o i s t u r e on d r y b a s i s at e x i t = ”)

Scilab code Exa 3.13 Percentage of water removed 21

1 clc () 2 mdrysolid = 100; // kg 3 percentin = 25; 4 mwaterin = mdrysolid * percentin / 100; 5 percentout = 2.5; 6 mwaterout = mdrysolid * percentout / 100; 7 mremoved = mwaterin - mwaterout ; 8 percentremoved = mremoved *100 / mwaterin ; 9 disp ( percentremoved , ” p e r c e n t a g e o f w a t e r removed = ”

)

Scilab code Exa 3.14 Amount and percentage water removed 1 clc () 2 m = 1; // kg 3 percent1 = 20; //% 4 mwaterin = m * percent1 / 100; 5 mdrysolid = m - mwaterin ; 6 percent2 = 2.44; //% 7 mout = mdrysolid / (1 - percent2 /100) ; 8 mwaterout = mout - mdrysolid ; 9 mremoved = mwaterin - mwaterout ; 10 percentremoved = mremoved * 100 / mwaterin ; 11 disp ( ” kg ” , mremoved , ” w e i g h t o f w a t e r removed = ” ) 12 disp ( ”%” , percentremoved , ” p e r c e n t a g e o f w a t e r removed

= ”)

Scilab code Exa 3.15 NaCl solution 1 clc () 2 mwater = 100; // kg 3 mNaCl = 35.8; // kg 4 msolu = mwater + mNaCl ;

22

5 6 7 8 9 10 11 12 13 14 15 16 17 18

mfr = mNaCl / msolu ; mpr = mfr * 100; MNaCl = 58.45; // kg / kmol NNaCl = mNaCl / MNaCl ; MH2O = 18; // kg / kmol NH2O = mwater / MH2O ; Mfr = NNaCl / ( NNaCl + NH2O ) ; Mpr = Mfr * 100; N = NNaCl *1000 / mwater ; disp ( mfr , ” ( a ) mass f r a c t i o n o f NaCl =” ) disp ( mpr , ” mass p e r c e n t o f NaCl= ” ) disp ( Mfr , ” ( b ) mole f r a c t i o n o f NaCl =” ) disp ( Mpr , ” mole p e r c e n t o f NaCl = ” ) disp (N , ” kmol NaCl p e r 1 0 0 0 kg o f w a t e r =” )

Scilab code Exa 3.16 Molal absolute humidity 1 2 3 4 5 6 7 8 9 10

clc () Y = 0.015; // kg w a t e r v a p o u r / kg d r y a i r Mair = 29; // kg / kmol Mwater = 18.016; // kg / kmol Nwater = Y / Mwater ; // kmol Nair = 1 / Mair ; // kmol Mpr = Nwater *100 / ( Nwater + Nair ) ; Mr = Nwater / Nair ; disp ( Mpr , ” ( a ) mole p e r c e n t o f w a t e r v a p o u r = ” ) disp ( ” kmol w a t e r / kmol d r y a i r ” ,Mr , ” ( b ) m o l a l a b s o l u t e h u m i d i t y =” )

Scilab code Exa 3.17 K2CO3 solution 1 clc () 2 msolu = 100; // g

23

3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23

MK2CO3 = 138.20; // g / mol percent1 = 50; //% mK2CO3 = percent1 * msolu / 100; NK2CO3 = mK2CO3 / MK2CO3 ; mwater = msolu - mK2CO3 ; Nwater = mwater / 18.06; Mpr = NK2CO3 * 100 / ( NK2CO3 + Nwater ) ; sp . gr =1.53; Vsolu = msolu / sp . gr ; //mL Vwater = mwater / 1; //mL Vpr = Vwater * 100/ Vsolu ; Molality = NK2CO3 / ( mwater * 10^ -3) ; Molarity = NK2CO3 / ( Vsolu * 10^ -3) ; Eq . wt = MK2CO3 / 2; No = mK2CO3 / Eq . wt ; N = No / ( Vsolu * 10^ -3) ; disp ( ”%” ,Mpr , ” ( a ) Mole p r c e n t o f s a l t = ” ) disp ( ”%” ,Vpr , ” ( b ) Volume p e r c e n t o f w a t e r = ” ) disp ( ” mol / kg ” , Molality , ” ( c ) M o l a l i t y = ” ) disp ( ” mol /L” , Molarity , ” ( d ) M o l a r i t y = ” ) disp ( ”N” ,N , ” ( e ) N o r m a l i t y ” )

Scilab code Exa 3.18 Molarity and Molality of Alcohol solution 1 clc () 2 msolu = 100; // kg 3 percent1 = 60; //% 4 Dwater = 998; // kg /mˆ3 5 Dalco = 798; // kg /mˆ3 6 Dsolu = 895; // kg /mˆ3 7 Vsolu = msolu / Dsolu ; 8 malco = msolu * percent1 / 100; 9 Valco = malco / Dalco ; 10 Vpr = Valco * 100 / Vsolu ; 11 Malco = 46.048; // kg / kmol

24

12 N = malco / Malco ; 13 Molarity = N /( Vsolu ) ; 14 mwater = msolu - malco ; 15 Molality = N * 1000 / mwater ; 16 disp ( ”%” ,Vpr , ” ( a ) Volume p e r c e n t

of ethanol in

s o l u t i o n = ”) 17 disp ( ” mol /L” , Molarity , ” ( b ) M o l a r i t y = ” ) 18 disp ( ” mol / ( kg o f w a t e r ) ” , Molality , ” ( c ) M o l a l i t y = ” )

Scilab code Exa 3.19 CO to phosgene 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

clc () //CO + CL2 = COCl2 Np = 12; // m o l e s NCl2 = 3; // m o l e s NCO = 8; // m o l e s N1Cl2 = NCl2 + Np ; N1CO = NCO + Np ; pr . ex = ( N1CO - N1Cl2 ) * 100/ N1Cl2 ; pr . co = ( N1Cl2 - NCl2 ) * 100/ N1Cl2 ; T = Np + NCl2 + NCO ; T1 = N1Cl2 + N1CO ; N = T / T1 ; disp ( ”%” , pr . ex , ” ( a ) p e r c e n t e x c e s s o f CO = ” ) disp ( ”%” , pr . co , ” ( b ) p e r c e n t c o n v e r s i o n = ” ) disp (N , ” ( c ) M o l e s o f t o t a l p r o d u c t s p e r mole o f t o t a l r e a c t a n t s = ”)

Scilab code Exa 3.20 Extent of reaction 1 clc () 2 Nn2 = 2; // m o l e s 3 Nh2 = 7; // m o l e s

25

4 Nnh3 = 1; // mole 5 n0 = Nn2 + Nh2 + Nnh3 ; 6 v = 2 - 1 - 3; 7 //YN2 = ( 2 − E) / ( 1 0 − 2∗E) 8 // Yh2 = (7 −3∗E) / ( 1 0 − 2∗E) 9 // Ynh3 = ( 1 + 2∗E) / ( 1 0 − 2∗E) 10 disp ( ” mole f r a c t i o n o f N2 = ( 2 − E) / ( 1 0 − 2∗E) ” ) 11 disp ( ” mole f r a c t i o n o f H2 = (7 −3∗E) / ( 1 0 − 2∗E) ” ) 12 disp ( ” mole f r a c t i o n o f NH3 = ( 1 + 2∗E) / ( 1 0 − 2∗E) ” )

Scilab code Exa 3.21 ethylene to ethanol 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23

clc () msolu = 100; // g MK2CO3 = 138.20; // g / mol percent1 = 50; //% mK2CO3 = percent1 * msolu / 100; NK2CO3 = mK2CO3 / MK2CO3 ; mwater = msolu - mK2CO3 ; Nwater = mwater / 18.06; Mpr = NK2CO3 * 100 / ( NK2CO3 + Nwater ) ; sp . gr =1.53; Vsolu = msolu / sp . gr ; //mL Vwater = mwater / 1; //mL Vpr = Vwater * 100/ Vsolu ; Molality = NK2CO3 / ( mwater * 10^ -3) ; Molarity = NK2CO3 / ( Vsolu * 10^ -3) ; Eq . wt = MK2CO3 / 2; No = mK2CO3 / Eq . wt ; N = No / ( Vsolu * 10^ -3) ; disp ( ”%” ,Mpr , ” ( a ) Mole p r c e n t o f s a l t = ” ) disp ( ”%” ,Vpr , ” ( b ) Volume p e r c e n t o f w a t e r = ” ) disp ( ” mol / kg ” , Molality , ” ( c ) M o l a l i t y = ” ) disp ( ” mol /L” , Molarity , ” ( d ) M o l a r i t y = ” ) disp ( ”N” ,N , ” ( e ) N o r m a l i t y ” ) 26

27

Chapter 4 Ideal Gases and Gas Mixtures

Scilab code Exa 4.1 gas constant R 1 2 3 4 5 6 7 8 9 10 11 12

clc () P1 = 760; //mmHg T1 = 273.15; //K V1 = 22.4143 * 10^ -3; //mˆ3/ mol R1 = P1 * V1 / T1 ; disp ( ”mˆ3 mmHg / ( molK ) ” ,R1 , ” Gas c o n s t a n t R =” ) P2 = 101325; //N/mˆ2 T2 = 273.15; //K V2 = 22.4143 * 10^ -3; //mˆ3/ mol R2 = P2 * V2 / T2 ; // J /molK R3 = R2 / 4.184; // c a l /molK disp ( ” c a l /molK” ,R3 , ” Gas c o n s t a n t R i n MKS s y s t e m =” )

Scilab code Exa 4.2 Molar volume of air 1 clc () 2 T = 350; //K 3 P = 1; // b a r

28

4 V1 = 22.4143 * 10^ -3; //mˆ3 ( s u f f i x 1 r e p r e s e n t s a t

STD) 5 P1 = 1.01325; // b a r 6 T1 = 273.15; //K 7 V = P1 * V1 * T /( T1 * P ) ; 8 disp ( ”mˆ3/ mol ” ,V , ” Molar volume =” )

Scilab code Exa 4.3 Ideal gas equation 1 2 3 4 5 6 7 8 9 10 11 12

clc () P = 10; // b a r T = 300; //K V = 150; //L P1 = 1.01325; // b a r ( \ s u f f i x 1 r e p r e s e n t s a t STD) T1 = 273.15; //K V2 = T1 * P * V /( T * P1 ) ; //mˆ3 V1 = 22.4143; //mˆ3/ mol N = V2 / V1 ; // mol MO2 = 32; m = N * MO2 /1000; disp ( ” kg ” ,m , ” Mass o f o x y g e n i n t h e c y l i n d e r = ” )

Scilab code Exa 4.4 Maximum allowable temperature of tyre 1 2 3 4 5 6

clc () P = 195; // kPa T = 273; //K P1 = 250; // kPa T1 = P1 * T / P ; disp ( ”K” ,T1 , ”Maximum t e m p e r a t u r e t o which t y r e may be h e a t e d = ” )

29

Scilab code Exa 4.5 Pressure calculation 1 2 3 4 5 6 7 8

clc () V = 250; //L T = 300; //K V1 = 1000; //L P1 = 100; // kPa T1 = 310; //K P = T * P1 * V1 /( T1 * V ) ; disp ( ” kPa ” ,P , ” O r i g i n a l p r e s s u r e i n t h e c y l i n d e r = ” )

Scilab code Exa 4.6 weight composition and density calculation 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

clc () Vper1 = 70; //% ( 1 = HCl ) Vper2 = 20; //% ( 2 = Cl2 ) Vper3 = 10; //% ( 3 = CCl4 ) M1 = 36.45; M2 = 70.90; M3 = 153.8; m1 = Vper1 * M1 ; m2 = Vper2 * M2 ; m3 = Vper3 * M3 ; mper1 = m1 * 100/( m1 + m2 + m3 ) ; mper2 = m2 * 100/( m1 + m2 + m3 ) ; mper3 = m3 * 100/( m1 + m2 + m3 ) ; disp ( mper1 , ” ( a ) w e i g h t p e r c e n t o f HCl= ” ) disp ( mper2 , ” w e i g h t p e r c e n t o f Cl2= ” ) disp ( mper3 , ” w e i g h t p e r c e n t o f CCl4= ” ) m = ( m1 + m2 + m3 ) /( Vper1 + Vper2 + Vper3 ) ; disp ( ” kg ” ,m , ” ( b ) a v e r a g e m o l e c u l a r w e i g h t = ” ) v = 22.4143; //mˆ3/ kmol 30

20 Vtotal = v * ( Vper1 + Vper2 + Vper3 ) ; 21 D = ( m1 + m2 + m3 ) / Vtotal ; 22 disp ( ” kg /mˆ3 ” ,D , ” ( c ) D e n s i t y a t s t a n d a r d c o n d i i o n s =

”)

Scilab code Exa 4.7 calculations for natural gas 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28

clc () per1 = 93; //% ( 1 = methane ) per2 = 4.5; //% ( 2 = e t h a n e ) per3 = 100 - ( per1 + per2 ) ; //% ( 3 = N2 ) ; T = 300; //K p = 400; // kPa P3 = p * per3 / 100; v = 10; //mˆ3 V2 = per2 * v / 100; M1 = 16.032; M2 = 30.048; M3 = 28; N1 = per1 ; N2 = per2 ; N3 = per3 ; m1 = M1 * N1 ; m2 = M2 * N2 ; m3 = M3 * N3 ; m = m1 + m2 + m3 ; Vstp = 100 * 22.4143 * 10^ -3; //m3 a t STP D = m /(1000 * Vstp ) ; Pstp = 101.325; // kPa T1 = 273.15; //K V = T * Pstp * Vstp / ( T1 * p ) ; D1 = m /(1000 * V ) ; Mavg = m /100; mper1 = m1 * 100 / ( m1 + m2 + m3 ) ; mper2 = m2 * 100 / ( m1 + m2 + m3 ) ; 31

29 30 31 32 33 34 35 36 37

mper3 = m3 * 100 / ( m1 + m2 + m3 ) ; disp ( ” kPa ” ,P3 , ” ( a ) P a r t i a l p r e s s u r e o f n i t r o g e n = ” ) disp ( ”mˆ3 ” ,V2 , ” ( b ) pure −component volume o f e t h a n e = ”) disp ( ” kg /mˆ3 ” ,D , ” ( c ) D e n s i t y a t s t a n d a r d c o n d i t i o n s = ”) disp ( ” kg /mˆ3 ” ,D1 , ” ( d ) D e n s i t y a t g i v e n c o n d i t i o n = ” ) disp ( Mavg , ” ( e ) A v e r a g e m o l e c u l a r w e i g h t = ” ) disp ( ”%” , mper1 , ” ( f ) w e i g h t p e r c e n t o f Methane = ” ) disp ( ”%” , mper2 , ” w e i g h t p e r c e n t o f Ethane = ” ) disp ( ”%” , mper3 , ” w e i g h t p e r c e n t o f N i t r o g e n = ” )

Scilab code Exa 4.8 Volume of gas from absorption columnn 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

clc () per1 = 20; //% ( 1 = ammonia ) Vstp = 22.4143; //mˆ3/ kmol Pstp = 101.325; // kPa Tstp = 273.15; //K V1 = 100; //mˆ3 P1 = 120; // kPa T1 = 300; //K P2 = 100; // kPa T2 = 280; //K per2 = 90; //% ( a b s o r b e d ) N = V1 * P1 * Tstp / ( Vstp * Pstp * T1 ) ; // kmol Nair = (1 - per1 / 100) * N ; N1 = per1 * N /100; Nabs = per2 * N1 / 100; N2 = N1 - Nabs ; // l e a v i n g Ntotal = Nair + N2 ; Vstp1 = Ntotal * Vstp ; //mˆ3 V2 = Vstp1 * Pstp * T2 / ( Tstp * P2 ) ; disp ( ”mˆ3 ” ,V2 , ” Volume o f g a s l e a v i n g = ” ) 32

Scilab code Exa 4.9 dehumidification 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

clc () V = 100; //mˆ3 Ptotal = 100; // kPa Pwater = 4; // kPa Pair = Ptotal - Pwater ; T = 300; //K T1 = 275; //K Vstp = 22.4143; //mˆ3/ kmol Tstp = 273.15; //K Pstp = 101.325; // kPa Pwater1 = 1.8; // kPa Pair1 = Ptotal - Pwater1 ; V1 = V * Pair * T1 / ( T * Pair1 ) ; Nwater = V * Pwater * Tstp / ( Vstp * Pstp * T ) ; Nwater1 = V1 * Pwater1 * Tstp / ( Vstp * Pstp * T1 ) ; m = ( Nwater - Nwater1 ) * 18.02; disp ( ”mˆ3 ” ,V1 , ” ( a ) volume o f a i r a f t e r d e h u m i d i f i c a t i o n = ”) 18 disp ( ” kg ” ,m , ” ( b ) Mass o f w a t e r v a p o u r removed = ” )

Scilab code Exa 4.10 Absorption column for H2S 1 2 3 4 5 6 7

clc () V = 100; //mˆ3 P = 600; // kPa T = 310; //K per1 = 20; //% ( H2S e n t e r i n g ) per2 = 2; //% ( H2S l e a v i n g ) Pstp = 101.325; // kPa 33

8 Tstp = 273.15; //K 9 Vstp = 22.414; //mˆ3/ kmol 10 Vstp1 = V * P * Tstp / ( T * Pstp ) 11 N = Vstp1 / Vstp ; 12 N1 = N * per1 / 100; 13 N2 = N - N1 ; // ( 2 = i n e r t s ) 14 Nleaving = N2 / ( 1 - per2 / 100) ; 15 N1leaving = per2 * Nleaving / 100; 16 mabsorbed = ( N1 - N1leaving ) * 34.08; // ( m o l e c u l a r 17 18 19 20 21 22 23 24 25 26 27

wt . = 3 4 . 0 8 ) mgiven = 100; // kg / h Vactual = mgiven * V / mabsorbed ; Nactual = Nleaving * Vactual / V ; // a c t u a l m o l e s leaving Vstpl = Nactual * Vstp ; // volume l e a v i n g a t STP P2 = 500; // kPa T2 = 290; //K V2 = Vstpl * Pstp * T2 / ( P2 * Tstp ) ; Precovery = ( N1 - N1leaving ) *100 / N1 ; disp ( ”mˆ3/ h ” , Vactual , ” ( a ) Volume o f g a s e n t e r i n g p e r hour ”) disp ( ”mˆ3/ h ” ,V2 , ” ( b ) Volume o f g a s l e a v i n g p e r h o u r ” ) disp ( ”%” , Precovery , ” ( c ) P e r c e n t a g e r e c o v e r y o f H2S” )

Scilab code Exa 4.11 Reaction stoichiometry for preparation of ammonia 1 clc () 2 //N2 + 3H2 = 2NH3 3 V1 = 100; //mˆ3 ( 1 = N2 ) 4 V2 = V1 * 3; // ( A c c o r d i n g t o A v a g a d r o s p r i n c i p l e ,

e q u a l volumes o f a l l g a s e s under s i m i l a r c o n d i t i o n c o n t a i n s same no . o f m o l e s ) 5 disp ( ”mˆ3 ” ,V2 , ” ( a ) Volume o f h y d r o g e n r e q u i r e d a t same c o n d i t i o n = ” ) 6 P1 = 20; // b a r 34

7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27

T1 = 350; //K P2 = 5; // b a r T2 = 290; //K V3 = 3 * V1 * P1 * T2 / ( P2 * T1 ) ; disp ( ”mˆ3 ” ,V3 , ” ( b ) Volume r e q u i r e d a t 50 b a r and 290K = ”) m = 1000; // kg ( ammonia ) N = m / 17.03; // kmol N1 = N /2; // ( n i t r o g e n ) N2 = N * 3 / 2; // ( h y d r o g e n ) P3 = 50; // b a r T3 = 600; //K Pstp = 1.01325; // b a r Tstp = 273.15; //K Vstp = 22.414; //mˆ3/ kmol V1stp = N1 * Vstp ; V4 = V1stp * Pstp * T3 / ( P3 * Tstp ) ; // ( n i t r o g e n a t 50 b a r and 600K) V5 = V4 * 2 ; // ( ammonia a t 50 b a r and 600K) V6 = V4 * 3 ; // ( h y d r o g e n a t 50 b a r and 600K) disp ( ”mˆ3 ” ,V4 , ” ( c ) Volume o f n i t r o g e n a t 50 b a r and 600K” ) disp ( ”mˆ3 ” ,V6 , ” Volume o f h y d r o g e n a t 50 b a r and 600K” ) disp ( ”mˆ3 ” ,V5 , ” Volume o f ammonia a t 50 b a r and 600K” )

Scilab code Exa 4.12 Reaction stoichiometry for preparation of producer gas 1 2 3 4 5

clc () N = 100; // kmol p r o d u c e r g a s P1 = 25; //% ( Carbon monoxide ) P2 = 4; //% ( Carbon D i o x i d e ) P3 = 3; //% ( Oxygen ) 35

6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39

P4 = 68; //% ( N i t r o g e n ) N1 = N * P1 /100; N2 = N * P2 /100; N3 = N * P3 /100; N4 = N * P4 /100; NC = N1 + N2 ; m = NC * 12; Ngas = N / m ; // m o l e s o f g a s f o r 1 kg o f Carbon Vstp = 22.4143; //mˆ3/ kmol Vstp1 = Vstp * Ngas ; P = 1; // b a r T = 290; // k Pstp = 1.01325; // b a r Tstp = 273.15; //K V = T * Vstp1 * Pstp / ( Tstp * P ) ; disp ( ”mˆ3 ” ,V , ” ( a ) Volume o f g a s a t 1 b a r and 290 K p e r kg Carbon = ” ) //CO + 1/2 ∗ O2 = CO2 Nrequired = N1 /2 - N3 ; // ( o x y g e n r e q u i r e d ) Nsupplied = Nrequired * 1.2; PO1 = 21; //% ( Oxygen p e r c e n t i n a i r ) Nair = Nsupplied * 100/ PO1 ; V1 = 100; //mˆ 3 ; Vair = V1 * Nair / N ; disp ( ”mˆ3 ” , Vair , ” ( b ) Volume o f a i r r e q u i r e d = ” ) NCO2 = N2 + N1 ; NO2 = Nsupplied - Nrequired ; NN2 = N4 + ( Vair * (1 - PO1 / 100) ) ; Ntotal = NCO2 + NO2 + NN2 ; PCO2 = NCO2 * 100 / Ntotal ; PO2 = NO2 * 100 / Ntotal ; PN2 = NN2 * 100 / Ntotal ; disp ( ”%” , PCO2 , ” P e r c e n t c o m p o s i t i o n o f Carbon D i o x i d e = ”) disp ( ”%” ,PO2 , ” P e r c e n t c o m p o s i t i o n o f Oxygen = ” ) disp ( ”%” ,PN2 , ” P e r c e n t c o m p o s i t i o n o f N i t r o g e n = ” )

36

Scilab code Exa 4.13 Reaction stoichiometry for preparation of Chlorine from HCl 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31

clc () // 4HCl + O2 = 2 Cl2 + 2H2O n = 1; // mol ( B a s i s 1 mol o f HCl ) NO2 = n / 4; NO2supp = 1.5 * NO2 ; Nair = NO2supp * 100 / 21; V = 100; //mˆ3 Vair = V * Nair / n ; disp ( ”mˆ3 ” , Vair , ” ( a ) Volume o f a i r a d m i t t e d = ” ) P1 = 80; //% ( HCl c o n v e r t e d ) Ncon = n * P1 /100; N2 = Ncon /4; // o x y g e n r e q u i r e d NH2O = Ncon / 2; NCl2 = Ncon / 2; nHCl = n - Ncon ; nO2 = NO2supp - N2 ; Nnitro = Nair - NO2supp ; Ntotal = nHCl + nO2 + NH2O + NCl2 + Nnitro ; V1 = V * Ntotal ; P1 = 1; // b a r T1 = 290; //K P2 = 1.2; // b a r T2 = 400; //K V2 = V1 * P1 * T2 / ( P2 * T1 ) ; disp ( ”mˆ3 ” ,V2 , ” ( b ) Volume o f g a s l e a v i n g = ” ) VCl2 = NCl2 * V ; Pstp = 1.01325; // b a r Tstp = 273; //K Vstp = 22.4143; //mˆ3/ kmol Vstp1 = Tstp * P1 * VCl2 / ( T1 * Pstp ) ; Nstp = Vstp1 / Vstp ; 37

32 33 34 35 36 37 38 39

m = Nstp * 70.90; disp ( ” kg ” ,m , ” ( c ) K i l o g r a m s o f C h l o r i n e p r o d u c e d = ” ) Ntotaldry = nHCl + nO2 + NCl2 + Nnitro ; // d r y b a s i s p1 = nHCl *100/ Ntotaldry ; p2 = nO2 *100/ Ntotaldry ; p3 = NCl2 *100/ Ntotaldry ; p4 = Nnitro *100/ Ntotaldry ; disp ( ”%” ,p1 , ” ( d ) P e r c e n t c o m p o s i t i o n o f HCl i n e x i t stream = ”) 40 disp ( ”%” ,p2 , ” P e r c e n t c o m p o s i t i o n o f Oxygen i n e x i t stream = ”) 41 disp ( ”%” ,p3 , ” Percent composition of Chlorine in e x i t stream = ”) 42 disp ( ”%” ,p4 , ” Percent composition of nitrogen in e x i t stream = ”)

Scilab code Exa 4.14 Reaction stoichiometry for dissociation of Carbon Dioxide 1 2 3 4 5 6 7 8 9 10 11 12 13 14

clc () // CO2 = CO + 1/2 ∗ O2 P1 = 1; // b a r T1 = 3500; //K P2 = 1; // b a r T2 = 300; //K V2 = 25; //L V1 = V2 * P2 * T1 / ( P1 * T2 ) ; disp ( ”L” ,V1 , ” ( a ) F i n a l volume o f g a s i f no d i s s o c i a t i o n occured = ”) Pstp = 1.01325; // b a r Tstp = 273; //K Vstp = 22.4143; //mˆ3 N2 = V2 * P2 * Tstp / ( Vstp * Pstp * T2 ) ; // l e t x be t h e f r a c t i o n d i s s o c i a t e d , t h e n a f t e r dissociation , 38

15 16 17 18 19 20 21 22

// CO2 = ( 1 − x ) mol , CO = xmol , O2 = ( 0 . 5 ∗ x ) mol // t o t a l m o l e s = 1 − x + x + o . 5 ∗ x = 1 + 0 . 5 ∗ x V = 350; //L N1 = V * P1 * Tstp / ( Vstp * Pstp * T1 ) ; // 1 + 0 . 5 ∗ x = N1 , t h e r e f o r e x = ( N1 - 1) / 0.5 ; p = x *100; disp ( ”%” ,p , ” ( b )CO2 c o n v e r t e d = ” )

39

Chapter 5 Properties of Real Gases

Scilab code Exa 5.1 Van der waals equation 1 2 3 4 5 6 7 8 9 10 11

clc () V = 0.6; //mˆ 3 ; T = 473; //K N = 1 * 10 ^ 3; // mol R = 8.314; // Pa ∗ mˆ3/ molK P = N * R * T / ( V * 10^5) ; disp ( ” b a r ” ,P , ” ( a ) P r e s s u r e c a l c u l a t e d u s i n g i d e a l g a s equation = ”) a = 0.4233; //N ∗ mˆ4 / mol ˆ2 b = 3.73 * 10^ -5; //mˆ3/ mol P1 = ( R * T /( V / N - b ) -a /( V / N ) ^2) /10^5; disp ( ” b a r ” ,P1 , ” ( a ) P r e s s u r e c a l c u l a t e d u s i n g van d e r waals equation = ”)

Scilab code Exa 5.2 Van der waals equation for CO2 gas 1 clc () 2 P = 10^7; // Pa ;

40

3 T = 500; //K 4 R = 8.314; // Pa ∗ L / mol K 5 V = N * R * T / ( P * 1000) ; 6 disp ( ”mˆ3 ” ,V , ” ( a ) Volume o f CO2 c a l c u l a t e d

using

i d e a l gas equation = ”)

Scilab code Exa 5.3 Redlich Kwong equation for gaseous ammonia 1 2 3 4 5 6 7 8 9 10

clc () V = 0.6 * 10^ -3; //mˆ3 T = 473; //K Tc = 405.5; //K Pc = 112.8 * 10 ^ 5 // Pa R = 8.314; a = 0.4278 * ( R ^2) * ( Tc ^ 2.5) / Pc ; b = 0.0867 * R * Tc / Pc ; P1 = ( R * T /( V - b ) - a /(( T ^0.5) * V *( V + b ) ) ) /10^5; disp ( ” b a r ” ,P1 , ” P r e s s u r e d e v e l o p e d by g a s = ” )

Scilab code Exa 5.4 Molar Volume calculation for gaseous ammonia 1 2 3 4 5 6 7 8 9 10

clc () P = 10^6; // Pa T = 373; //K Tc = 405.5; //K Pc = 112.8 * 10 ^ 5 // Pa R = 8.314; a = 0.4278 * ( R ^2) * ( Tc ^ 2.5) / Pc ; b = 0.0867 * R * Tc / Pc ; // P1 = (R∗T/ (V − b ) − a / ( ( T ˆ 0 . 5 ) ∗V∗ (V + b ) ) ) / 1 0 ˆ 5 ; // 1 0 ˆ 6 = ( ( 8 . 3 1 4 ∗ 3 7 3 ) / (V−2.59∗10ˆ −5) ) − 8 . 6 8 / ( ( 3 7 3 ˆ 0 . 5 ) ∗ V∗ (V+2.59∗10ˆ −5) 11 // s o l v i n g t h i s we g e t , 41

12 V = 3.0 * 10^ -3; //mˆ3/ mol 13 disp ( ”mˆ3/ mol ” ,V , ” m o l a r volume o f g a s = ” )

Scilab code Exa 5.5 virial equation of state 1 2 3 4 5 6 7 8

9 10 11

clc () B = -2.19 * 10^ -4; //mˆ3/ mol C = -1.73 * 10^ -8; //mˆ6/ mol ˆ2 P = 10; // b a r T = 500; //K // v i r i a l e q u a t i o n i s g i v e n as , Z = PV/RT = 1 + B/V + C/Vˆ2 //V = (RT/P) ∗ ( 1 + B/V + C/Vˆ 2 ) // now by a s s u m i n g d i f f e r e n t v a l u e s f o r V on RHS and c h e c k i n g f o r c o r r e s p o n d i n g V on LHS , we have t o assume s u c h v a l u e o f V on RHS by which we g e t t h e same v a l u e f o r LHS V // by t r i a l and e r r o r we g e t , V = 3.92 * 10^ -3; //mˆ3 disp ( ”mˆ3 ” ,V , ” Molar volume o f m e t h a n o l = ” )

Scilab code Exa 5.6 Lyderson method for n butane 1 2 3 4 5 6 7 8 9 10

clc () T = 510; //K P = 26.6; // b a r Tc = 425.2; //K Pc = 38; // b a r Zc = 0.274; R = 8.314; Pr = P / Pc ; Tr = T / Tc ; disp ( Pr , ” Pr = ” ) 42

11 12 13 14 15 16 17

disp ( Tr , ” Tr = ” ) // From f i g . 5 . 4 and 5 . 5 from t h e t e x t book Z = 0.865; D = 0.15; Z1 = Z + D * ( Zc - 0.27) ; V = R * T * Z1 / ( P * 10^5) ; disp ( ”mˆ3/ mol ” ,V , ” Molar volume o f n−b u t a n e = ” )

Scilab code Exa 5.7 Pitzer correlation for n butane 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

clc () T = 510; //K P = 26.6; // b a r Tc = 425.2; //K Pc = 38; // b a r w = 0.193; R = 8.314; Pr = P / Pc ; Tr = T / Tc ; disp ( Pr , ” Pr = ” ) disp ( Tr , ” Tr = ” ) // From f i g . 5 . 6 and 5 . 7 from t h e t e x t book Z0 = 0.855; Z1 = 0.042; Z = Z0 + w * Z1 ; disp (Z , ”Z = ” ) V = R * T * Z / ( P * 10^5) ; disp ( ”mˆ3/ mol ” ,V , ” Molar volume o f n−b u t a n e = ” )

Scilab code Exa 5.8 Molar volume by different methods 1 clc () 2 P = 6000; // kPa

43

3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36

T = 325; //K xn2 = 0.4; xethane = 0.6; an2 = 0.1365; //N mˆ4 / mol ˆ2 bn2 = 3.86 * 10^ -5; //mˆ3/ mol aethane = 0.557; //N mˆ4 / mol ˆ2 bethane = 6.51 * 10^ -5; //mˆ3/ mol Pcn2 = 3394; // kPa Tcn2 = 126.2; //K Pcethane = 4880; // kPa Tcethane = 305.4; //K R = 8.314; V = R * T / ( P *1000) ; disp ( ”mˆ3/ mol ” ,V , ” ( a ) Molar volume by i d e a l g a s e q u a t i o n =” ) a = ( xn2 * ( an2 ^0.5) + xethane * ( aethane ^0.5) ) ^2; b = ( xn2 * bn2 + xethane * bethane ) ; // s u b s t i t u t i n g t h e a b o v e v a l u e s i n van d e r w a a l s e q u a t i o n , and s o l v i n g , we g e t V1 = 3.680 * 10^ -4; //mˆ3/ mol disp ( ”mˆ3/ mol ” ,V1 , ” ( b ) Molar volume by van d e r w a a l s e q u a t i o n =” ) Prin2 = P / Pcn2 ; Trin2 = T / Tcn2 ; Priethane = P / Pcethane ; Triethane = T / Tcethane ; // u s i n g c o m p r e s s i b i l t y c h a r t , Zn2 = 1; Zethane = 0.42; Z = xn2 * Zn2 + xethane * Zethane ; V2 = Z * R * T / P ; disp ( ”mˆ3/ mol ” ,V2 , ” ( c ) Molar volume b a s e d on c o m p r e s s i b i l t y f a c t o r =” ) Pri1n2 = xn2 * P / Pcn2 ; Tri1n2 = T / Tcn2 ; Pri1ethane = xethane * P / Pcethane ; Tri1ethane = T / Tcethane ; // u s i n g c o m p r e s s i b i l t y c h a r t , 44

37 Zn21 = 1; 38 Zethane1 = 0.76; 39 Z1 = xn2 * Zn21 + xethane * Zethane1 ; 40 V3 = Z1 * R * T / P ; 41 disp ( ”mˆ3/ mol ” ,V3 , ” ( c ) Molar volume b a s e d on d a l t o n s 42 43 44 45 46

law =” ) Tc = xn2 * Tcn2 + xethane * Tcethane ; Pc = xn2 * Pcn2 + xethane * Pcethane ; Zc = 0.83; V4 = Zc * R * T / P ; disp ( ”mˆ3/ mol ” ,V4 , ” ( d ) Molar volume by k a y s method =” )

Scilab code Exa 5.9 Van der waals equation and Kays method 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

clc () P1 = 40; //% ( n i t r o g e n ) P2 = 60; //% ( e t h a n e ) T = 325; //K V = 4.5 * 10^ -4; //mˆ3/ mol a1 = 0.1365; //N∗mˆ4/ mol ˆ2 b1 = 3.86 * 10 ^ -5; //mˆ3/ mol a2 = 0.557; //N∗mˆ4/ mol ˆ2 b2 = 6.51 * 10 ^ -5; //mˆ3/ mol Pc1 = 3394; // kPa Tc1 = 126.1; //K Pc2 = 4880; // kPa Tc2 = 305.4; //K R = 8.314; Pideal = R * T / ( V * 1000) ; // kPa disp ( ” kPa ” , Pideal , ” ( a ) P r e s s u r e o f Gas by t h e i d e a l gas equation = ”) 17 y1 = P1 /100; 18 y2 = P2 /100; 19 a = ( y1 * ( a1 ^(1/2) ) + y2 * ( a2 ^(1/2) ) ) ^2; 45

20 b = y1 * b1 + y2 * b2 ; 21 Pv = (( R * T / ( V - b ) ) - a / ( V ^2) ) /1000; 22 disp ( ” kPa ” ,Pv , ” ( b ) P r e s s u r e o f Gas by Van d e r w a a l s 23 24 25 26 27 28 29 30 31

equation = ”) Tc = y1 * Tc1 + y2 * Tc2 ; Pc = y1 * Pc1 + y2 * Pc2 ; Vc = R * Tc / Pc ; // Pseudo c r i t i c a l i d e a l volume Vr = V / Vc ; // Pseudo r e d u c e d i d e a l volume Tr = T / Tc ; // Pseudo r e d u c e d t e m p e r a t u r e // From f i g 5 . 3 , we g e t Pr = 1 . 2 Pr = 1.2; Pk = Pr * Pc ; disp ( ” kPa ” ,Pk , ” ( b ) P r e s s u r e o f Gas by t h e Kays method = ”)

46

Chapter 6 Vapour Pressure

Scilab code Exa 6.1 Quality of steam 1 clc () 2 P = 500; // kPa 3 SV = 0.2813; //mˆ3/ kg 4 Vsaturatedl = 1.093 * 10^ -3; //mˆ3/ kg 5 Vsaturatedv = 0.3747; //mˆ3/ kg 6 // l e t t h e f r a c t i o n o f v a p o u r be y 7 // (1−y ) ∗ V s a t u r a t e d l + y ∗ V s a t u r a t e d v = SV 8 // t h e n we g e t , (1−y ) ∗ ( 1 . 0 9 3 ∗ 1 0 ˆ − 3 ) + y ∗ ( 0 . 3 7 4 7 ) = 9 10 11 12 13

0.2813 y = ( SV - Vsaturatedl ) /( Vsaturatedv - Vsaturatedl ) ; P1 = y * 100; P2 = 100 - P1 ; disp ( ”%” ,P1 , ” P e r c e n t a g e o f Vapour = ” ) disp ( ”%” ,P2 , ” P e r c e n t a g e o f L i q u i d = ” )

Scilab code Exa 6.2 Calculation of vapour pressure 1 clc ()

47

2 3 4 5 6 7 8 9

T1 = 363; //K T2 = 373; //K P2s = 101.3; // kPa J = 2275 * 18; // kJ / kmol R = 8.314; // kJ / kmolK // l n ( P2s / P1s ) = J ∗ ( 1 / T1 − 1/T2 ) / R P1s = P2s / exp ( J * (1/ T1 - 1/ T2 ) / R ) ; disp ( ” kPa ” ,P1s , ” Vapour p r e s s u r e o f w a t e r a t 363 K = ”)

Scilab code Exa 6.3 Clausius Clapeyron equation for acetone 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

clc () P1s = 194.9; // kPa P2s = 8.52; // kPa T1 = 353; //K T2 = 273; //K T3 = 300; //K Pair = 101.3; // kPa // l o g ( P2s / P1s ) = J ∗ ( 1 / T1 − 1/T2 ) / R // l e t J / R = L L = log ( P2s / P1s ) /(1/ T1 - 1/ T2 ) ; P3s = P1s * exp ( L * (1/ T1 - 1/ T3 ) ) ; Ptotal = P3s + Pair ; // a t s a t u r a t i o n v a p o u r p r e s s u r e = partial pressure disp ( ” kPa ” , Ptotal , ” ( a ) F i n a l p r e s s u r e o f t h e m i x t u r e = ”) MP = P3s * 100 / Ptotal ; // mole p e r c e n t = m o l e s o f a c e t o n e ∗ 100 / t o t a l moles //= P a r t i a l p r e s s u r e o f a c e t o n e ∗ 100 / t o t a l Pressure disp ( ”%” ,MP , ” ( b ) Mole p e r c e n t o f a c e t o n e i n t h e f i n a l mixture = ”)

48

Scilab code Exa 6.4 Antoine equation for n heptane 1 2 3 4 5 6 7 8 9 10 11

clc () A = 13.8587; B = 2911.32; C = 56.56; T1 = 325; //K // P r e s s u r e a t n o r m a l c o n d i t i o n = 1 0 1 . 3 kPa P2 = 101.3; // kPa // A n t o i n e e q u a t i o n − lnP = A − B / (T − C) lnP = A - ( B / ( T - C ) ) ; P1 = exp ( lnP ) ; disp ( ” kPa ” ,P1 , ” ( a ) Vapour p r e s s u r e o f n−h e p t a n e a t 325K = ” ) 12 T2 = B /( A - log ( P2 ) ) + C ; 13 disp ( ”K” ,T2 , ” ( b ) Normal b o i l i n g p o i n t o f n−h e p t a n e = ”)

Scilab code Exa 6.5 Cox chart 1 2 3 4 5 6

clc () T = [273 293 313 323 333 353 373]; Ps = [0.61 2.33 7.37 12.34 19.90 47.35 101.3]; plot2d ( ’ l l ’ ,T , Ps , rect =[250 ,0.1 ,380 ,195]) ; P = get ( ” h d l ” ) ; xtitle ( ’ C o n s t r u c t i o n o f c o x c h a r t ’ , ’ Temperature , K ’ , ’ P r e s s u r e , kPa ’ ) ; 7 T1 = [273 353] 8 Ps1 = [8.52 194.9] 9 plot2d ( ’ l l ’ ,T1 , Ps1 ) ; 49

Figure 6.1: Cox chart

50

Figure 6.2: Duhring line

Scilab code Exa 6.6 Duhring line 1 clc () 2 Pswater1 = 6.08; // kPa 3 T1 = 313; //K 4 // l n P s = 1 6 . 2 6 2 0 5 − 3 7 9 9 . 8 8 7 / ( T − 4 6 . 8 5 4 ) 5 Tb1 =3799.887/(16.26205 - log ( Pswater1 ) ) + 46.854; 6 disp ( ”K” ,Tb1 , ” b o i l i n g p o i n t o f w a t e r a t 6 . 0 8 kPa 7 8 9 10 11 12 13

vapour p r e s s u r e = ”) Pswater2 = 39.33; // kPa T2 = 353; //K Tb2 =3799.887/(16.26205 - log ( Pswater2 ) ) + 46.854; disp ( ”K” ,Tb2 , ” b o i l i n g p o i n t o f w a t e r a t 3 9 . 3 3 kPa vapour p r e s s u r e = ”) Tb = [ Tb1 Tb2 ]; T = [ T1 T2 ]; plot (T , Tb ) ; 51

14

15 16 17 18 19

xtitle ( ’ Equal p r e s s u r e r e f e r e n c e p l o t f o r s u l p h u r i c a c i d ’ , ’ B o i l i n g p o i n t o f s o l u t i o n ,K ’ , ’ B o i l i n g p o i n t o f water , K ’ ) ; T3 = 333; //K // c o r r e s p o n d i n g t o T3 on x a x i s , on y we g e t Tb3 = 329; //K Pswater3 = exp (16.26205 - 3799.887/( Tb3 - 46.854) ) ; disp ( ” kPa ” , Pswater3 , ” Vapour p r e s s u r e o f s o l u t i o n a t 333K” )

52

Chapter 7 Solutions and Phase Behaviour

Scilab code Exa 7.1 composition calculation of Liquid and vapour at equilibrium 1 2 3 4 5 6 7 8 9 10 11 12

clc () Pas = 71.2; // kPa Pbs = 48.9; // kPa P = 65; // kPa //P=(Pas−Pbs ) ∗ xa+Pbs , xa=mole f r a c t i o n o f n−h e p t a n e , l i q . condition , therefore xa = ( P - Pbs ) /( Pas - Pbs ) ; // ya = Pa / P , Vapour c o n d i t i o n ya = Pas * xa / P ; P1 = xa * 100; P2 = ya * 100; disp ( ”%” ,P1 , ” P e r c e n t a g e o f h e p a t n e i n l i q u i d = ” ) disp ( ”%” ,P2 , ” P e r c e n t a g e o f h e p a t n e i n v a p o u r = ” )

Scilab code Exa 7.2 Composition and total pressure calculation 1 clc ()

53

2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

P1 = 100; // kPa ( Vapour p r e s s u r e o f l i q A ) P2 = 60; // kPa ( Vapour p r e s s u r e o f l i q B ) T = 320; //K // Pa = xa ∗ P1 = 100 ∗ xa // Pa = xb ∗ P2 = 60 ∗ xb //P = xa ∗ P1 + ( 1 − xa ) ∗ P2 // = 100 xa + ( 1 − xa ) ∗ 60 // = 60 + 40∗ xa // ya = Pa / P // 0 . 5 = 100 ∗ xa / ( 60 + 40 ∗ xa ) xa = 60 * 0.5 / (100 - 20) ; Per1 = xa * 100; disp ( ”%” , Per1 , ” ( a ) P e r c e n t a g e o f A i n l i q u i d = ” ) Ptotal = 60 + 40 * xa ; disp ( ” kPa ” , Ptotal , ” ( b ) T o t a l p r e s s u r e o f t h e v a p o u r = ”)

Scilab code Exa 7.3 Mole fraction calculation for particular component in liquid vapour mixture 1 2 3 4 5 6 7 8 9 10 11 12

clc () xa = 0.25; xb = 0.30; xc = 1 - xa - xb ; Ptotal = 200; // kPa Pcs = 50; // kPa ( Vapour p r e s s u r e o f c ) Pc = xc * Pcs ; // ( p a r t i a l p r e s s u r e o f c ) yc = Pc / Ptotal ; yb = 0.5; ya = 1 - yb - yc ; per1 = ya * 100; disp ( ”%” , per1 , ” P e r c e n t a g e o f A i n v a p o u r = ” )

54

Figure 7.1: Boiling point diagram Scilab code Exa 7.4 Flash vapourization of benzene toluene mixture 1 2 3 4 5 6 7 8 9 10 11

clc () P = 101.3; // kPa Pbs = 54.21; // kPa Pas = 136.09; // kPa xf = 0.65; xw = ( P - Pbs ) /( Pas - Pbs ) ; yd = xw * Pas / P ; // f = ( x f − xw ) / ( yd − xw ) f = ( xf - xw ) / ( yd - xw ) ; per1 = f * 100; disp ( ”%” , per1 , ” mole p e r c e n t o f t h e f e e d t h a t i s vapourised = ”)

55

Scilab code Exa 7.5.a Boiling point diagram 1 2 3 4 5 6 7 8 9 10 11 12 13 14

clc () T = [371.4 378 383 388 393 398.6] Pas = [101.3 125.3 140 160 179.9 205.3] Pbs = [44.4 55.6 64.5 74.8 86.6 101.3] Ptotal = 101.3; // kPa for i = 1:6 x ( i ) = ( Ptotal - Pbs ( i ) ) /( Pas ( i ) - Pbs ( i ) ) ; end for i = 1:6 y ( i ) = x ( i ) * Pas ( i ) / Ptotal ; end plot (x ,T , ’−o ’ ) ; plot (y ,T , ’−x ’ ) ; xtitle ( ’ B o i l i n g p o i n t d i a g r a m ’ , ’ Mole f r a c t i o n x o r y ’ , ’ Temperature K ’ )

Scilab code Exa 7.5.b Equilibrium Diagram 1 2 3 4 5 6 7 8 9 10 11 12 13

clc () T = [371.4 378 383 388 393 398.6] Pas = [101.3 125.3 140 160 179.9 205.3] Pbs = [44.4 55.6 64.5 74.8 86.6 101.3] Ptotal = 101.3; // kPa for i = 1:6 x ( i ) = ( Ptotal - Pbs ( i ) ) /( Pas ( i ) - Pbs ( i ) ) ; end for i = 1:6 y ( i ) = x ( i ) * Pas ( i ) / Ptotal ; end w = x; plot (x , w ) ; 56

Figure 7.2: Equilibrium Diagram

57

14 15

plot (x ,y , ’−o ’ ) ; xtitle ( ’ E q u i l i b r i u m c u r v e ’ , ’ x , mole f r a c t i o n i n l i q u i d ’ , ’ y , mole f r a c t i o n i n v a p o u r ’ ) ;

Scilab code Exa 7.6 Bubble point temperature and vapour composition 1 2 3 4 5 6 7 8 9 10 11 12

clc () Ps = 100; // kPa A1 = 13.8587; // ( 1 = n−h e p t a n e ) A2 = 13.8216; // ( 2 = n−h e x a n e ) B1 = 2911.32; B2 = 2697.55; C1 = 56.51; C2 = 48.78; // l n P s = A − B / ( T − C) T1 = B1 / ( - log ( Ps ) + A1 ) + C1 ; T2 = B2 / ( - log ( Ps ) + A2 ) + C2 ; x2 = 0.25;

Scilab code Exa 7.7 Dew point temperature pressure and concentration 1 clc () 2 // l n P a s = 1 4 . 5 4 6 3 − 2 9 4 0 . 4 6 / ( T − 3 5 . 9 3 ) 3 // l n P b s = 1 4 . 2 7 2 4 − 2 9 4 5 . 4 7 / (T − 4 9 . 1 5 ) 4 // xa = (P − Pbs ) / ( Pas − Pbs ) 5 //Ya = Pas ∗ (P − Pbs ) / (P ∗ ( Pas − Pbs ) ) 6 Ya = 0.4; 7 P = 65; // kPa 8 // v a r i o u s t e m p e r a t u r e v a l u e a r e assumed and t r i e d

t i l l LHS = RHS , we g e t 9 T = 334.15; //K 10 Pas = exp (14.5463 - 2940.46/( T - 35.93) ) ; 11 Pbs = exp (14.2724 - 2945.47 / ( T - 49.15) ) ;

58

12 xa = ( P - Pbs ) /( Pas - Pbs ) ; 13 disp ( ”K” ,T , ” ( a ) The Dew p o i n t t e m p e r a t u r e a t 65 kPa = 14 15 16 17 18 19 20 21

”) disp ( xa , ” C o n c e n t r a t i o n o f the f i r s t drop o f l i q u i d = ”) T1 = 327; //K Pas1 = exp (14.5463 - 2940.46/( T1 - 35.93) ) ; Pbs1 = exp (14.2724 - 2945.47 / ( T1 - 49.15) ) ; xa1 = Ya * Pbs1 / ( Pas1 - Ya *( Pas1 - Pbs1 ) ) ; P1 = xa1 * Pas1 / Ya ; disp ( ” kPa ” ,P1 , ” ( b ) The dew p o i n t p r e s s u r e a t 327 K = ”) disp ( xa1 , ” C o n c e n t r a t i o n a t 327K = ” )

Scilab code Exa 7.8 Partial pressure of acetaldehyde 1 2 3 4 5 6 7 8 9 10 11 12

clc () MW = 44.032; Mwater = 18.016; x = 2; //% Pa = 41.4; // kPa Mfr = ( x / MW ) /( x / MW + (100 - x ) / Mwater ) ; // henry ’ s law g i v e s Pa = Ha ∗ xa Ha = Pa / Mfr ; Molality = 0.1; Mfr1 = Molality / (1000/ Mwater + Molality ) ; Pa1 = Ha * Mfr1 ; disp ( ” kPa ” ,Pa1 , ” P a r t i a l P r e s s u r e = ” )

Scilab code Exa 7.9 Raults law application 1 clc () 2 // 1 − p e n t a n e , 2 − hexane , 3 − h e p t a n e

59

3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36

x1 = 0.6; x2 = 0.25; x3 = 0.15; A1 = 13.8183; A2 = 13.8216; A3 = 13.8587; B1 = 2477.07; B2 = 2697.55; B3 = 2911.32; C1 = 39.94; C2 = 48.78; C3 = 56.51; // As r a o u l t s law i s a p p l i c a b l e , Ki = y i / x i = P i s /P // y i = x i ∗ P i s /P // l n P = A− B/ (T−C) // Assuming , P = 400; // kPa T = 369.75; //K Pas1 = exp ( A1 - B1 / ( T - C1 ) ) ; Pas2 = exp ( A2 - B2 / ( T - C2 ) ) ; Pas3 = exp ( A3 - B3 / ( T - C3 ) ) ; Yi = ( x1 * Pas1 + x2 * Pas2 + x3 * Pas3 ) / P ; disp ( ”K” ,T , ” ( a ) b u b b l e p o i n t t e m p e r a t u r e o f t h e mixture = ”) y1 = x1 * Pas1 / P ; y2 = x2 * Pas2 / P ; y3 = x3 * Pas3 / P ; disp ( ”%” , y1 *100 , ” ( b ) c o m p o s i t i o n o f n−p e n t a n e i n vapour = ”) disp ( ”%” , y2 *100 , ” c o m p o s i t i o n o f n−h e x a n e i n v a p o u r = ”) disp ( ”%” , y3 *100 , ” c o m p o s i t i o n o f n−h e p t a n e i n v a p o u r = ”) T1 = 300; //K Ps1 = exp ( A1 - B1 / ( T1 - C1 ) ) ; Ps2 = exp ( A2 - B2 / ( T1 - C2 ) ) ; Ps3 = exp ( A3 - B3 / ( T1 - C3 ) ) ; P1 = x1 * Ps1 + x2 * Ps2 + x3 * Ps3 ; 60

37

disp ( ” kPa ” ,P1 , ” ( c ) Bubble p o i n t p r e s s u r e =” )

Scilab code Exa 7.10 Dew point temperature and pressure 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

clc () // 1 − p e n t a n e , 2 − hexane , 3 − h e p t a n e y1 = 0.6; y2 = 0.25; y3 = 0.15; A1 = 13.8183; A2 = 13.8216; A3 = 13.8587; B1 = 2477.07; B2 = 2697.55; B3 = 2911.32; C1 = 39.94; C2 = 48.78; C3 = 56.51; P = 400; // kPa T = 300; //K // As r a o u l t s law i s a p p l i c a b l e , Ki = y i / x i = P i s /P // x i = y i ∗P/ P i s // l n P = A− B/ (T−C) // Assuming , T1 = 385.94; //K Pas1 = exp ( A1 - B1 / ( T1 - C1 ) ) ; Pas2 = exp ( A2 - B2 / ( T1 - C2 ) ) ; Pas3 = exp ( A3 - B3 / ( T1 - C3 ) ) ; disp ( ”K” ,T , ” ( a ) Dew p o i n t t e m p e r a t u r e o f t h e m i x t u r e = ”) Ps1 = exp ( A1 - B1 / ( T - C1 ) ) ; Ps2 = exp ( A2 - B2 / ( T - C2 ) ) ; Ps3 = exp ( A3 - B3 / ( T - C3 ) ) ; P1 = 1/( y1 / Ps1 + y2 / Ps2 + y3 / Ps3 ) ; disp ( ” kPa ” ,P1 , ” ( b ) Dew p o i n t p r e s s u r e = ” ) 61

Scilab code Exa 7.11 bubble point and dew point 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28

clc () // 1 − methanol , 2 − e t h a n o l , 3 − p r o p a n o l x1 = 0.45; x2 = 0.3; x3 = 1 - ( x1 + x2 ) ; P = 101.3; // kPa // by d r a w i n g t h e t e m p e r a t u r e v s v a p o u r p r e s s u r e g r a p h and i n t e r p o l a t i o n , a s s u m i n g , T = 344.6; //K Ps1 = 137.3; Ps2 = 76.2; Ps3 = 65.4; y1 = x1 * Ps1 / P ; y2 = x2 * Ps2 / P ; y3 = x3 * Ps3 / P ; disp ( ”K” ,T , ” ( a ) Bubble p o i n t t e m p e r a t u r e = ” ) disp ( ”%” , y1 *100 , ” C o m p o s i t i o n o f m e t h a n o l i n v a p o u r = ”) disp ( ”%” , y2 *100 , ” C o m p o s i t i o n o f e t h a n o l i n v a p o u r = ”) disp ( ”%” , y3 *100 , ” C o m p o s i t i o n o f p r o p a n o l i n v a p o u r = ”) // a g a i n , f o r x i = 1 T1 = 347.5; //K P1 = 153.28; P2 = 85.25; P3 = 73.31; xa = x1 * P / P1 ; xb = x2 * P / P2 ; xc = x3 * P / P3 ; disp ( ”K” ,T1 , ” ( b ) Dew p o i n t t e m p e r a t u r e = ” ) disp ( ”%” , xa *100 , ” C o m p o s i t i o n o f m e t h a n o l i n l i q u i d = 62

”) 29 disp ( ”%” , xb *100 , ” C o m p o s i t i o n o f e t h a n o l i n l i q u i d = ”) 30 disp ( ”%” , xc *100 , ” C o m p o s i t i o n o f p r o p a n o l i n l i q u i d = ”)

Scilab code Exa 7.12 component calculations 1 2 3 4 5 6

7 8 9 10 11 12 13 14 15 16 17 18 19

clc () xp = 0.25; xnb = 0.4; xnp = 0.35; P = 1447.14; // kPa // a s s u m i n g t e m p e r a t u r e s 3 5 5 . 4 K and 3 6 6 . 5 K , c o r r e s p o n d i n g Ki v a l u e s a r e f o u n d from nomograph and t o t a l Ki v a l u e a r e 0 . 9 2 8 and 1 . 0 7 5 r e s p , t h u s b u b b l e p o i n t t e m p e r a t u r e l i e s between , u s i n g i n t e r p o l a t i o n bubble point temperature i s found t o be , Tb = 361; //K disp ( ”K” ,Tb , ” ( a ) The b u u b l e p o i n t t e m p e r a t u r e = ” ) // At 3 6 1 , Kip = 2.12; Kinb = 0.85; Kinp = 0.37; xp1 = Kip * xp ; xnb1 = Kinb * xnb ; xnp1 = Kinp * xnp ; disp ( xp1 , ” c o n c e n t r a t i o n o f p r o p a n e a t b u b b l e p o i n t = ”) disp ( xnb1 , ” c o n c e n t r a t i o n o f n−b u t a n e a t b u b b l e p o i n t = ”) disp ( xnp1 , ” c o n c e n t r a t i o n o f n−p e n t a n e a t b u b b l e point = ”) // At dew p o i n t Yi / Ki = 1 , a t 3 7 7 . 6K t h i s i s 1 . 1 5 9 8 63

20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49

and a t 3 8 8 . 8K i t i s 0 . 9 6 7 7 , by i n t e r p o l a t i o n dew p o i n t i s f o u n d t o be Td = 387; //K Kip1 = 2.85; Kinb1 = 1.25; Kinp1 = 0.59; yp1 = xp / Kip1 ; ynb1 = xnb / Kinb1 ; ynp1 = xnp / Kinp1 ; disp ( ”K” ,Td , ” ( b ) The dew p o i n t t e m p e r a t u r e = ” ) disp ( yp1 , ” c o n c e n t r a t i o n o f p r o p a n e a t dew p o i n t = ” ) disp ( ynb1 , ” c o n c e n t r a t i o n o f n−b u t a n e a t dew p o i n t = ”) disp ( ynp1 , ” c o n c e n t r a t i o n o f n−p e n t a n e a t dew p o i n t = ”) // summation z i / ( 1 + L/VKi )= 0 . 4 5 , u s i n g t r i a l and e r r o r , we f i n d T = 374.6; //K L = 0.55; V = 0.45; Kip2 = 2.5; Kinb2 = 1.08; Kinp2 = 0.48; t = ( xp /(1+ L /( V * Kip2 ) ) ) +( xnb /(1+ L /( V * Kinb2 ) ) ) + ( xnp /(1+ L /( V * Kinp2 ) ) ) ; yp2 = ( xp /(1+ L /( V * Kip2 ) ) ) / t ; ynb2 = ( xnb /(1+ L /( V * Kinb2 ) ) ) / t ; ynp2 = ( xnp /(1+ L /( V * Kinp2 ) ) ) / t ; xp2 = ( xp - V * yp2 ) / L ; xnb2 = ( xnb - V * ynb2 ) / L ; xnp2 = ( xnp - V * ynp2 ) / L ; disp ( ”K” ,T , ” ( c ) T e m p e r a t u r e o f t h e m i x t u r e = ” ) disp ( yp2 , ” v a p o u r p h a s e c o n c e n t r a t i o n o f p r o p a n e = ” ) disp ( ynb2 , ” v a p o u r p h a s e c o n c e n t r a t i o n o f n−b u t a n e = ”) disp ( ynp2 , ” v a p o u r p h a s e c o n c e n t r a t i o n o f n−p e n t a n e = ”) disp ( xp2 , ” l i q u i d p h a s e c o n c e n t r a t i o n o f p r o p a n e = ” ) 64

disp ( xnb2 , ” l i q u i d p h a s e c o n c e n t r a t i o n o f n−b u t a n e = ”) 51 disp ( xnp2 , ” l i q u i d p h a s e c o n c e n t r a t i o n o f n−p e n t a n e = ”) 50

Scilab code Exa 7.13 equilibrium temperature and composition 1 2 3 4 5 6 7 8 9 10 11 12 13

clc () P = 93.30; // kPa T1 = 353; //K T2 = 373; //K Pwater1 = 47.98; // kPa Pwater2 = 101.3; // kPa Pliq1 = 2.67; // kPa Pliq2 = 5.33; // kPa T = T1 + ( T2 - T1 ) *( P - ( Pwater1 + Pliq1 ) ) /( Pwater2 + Pliq2 - ( Pwater1 + Pliq1 ) ) ; disp ( ”K” ,T , ” ( a ) The e q u i l i b r i u m t e m p e r a t u r e = ” ) Pwater = 88.50; y = Pwater * 100 / P ; disp ( ”%” ,y , ” ( b ) Water v a p o u r i n v a p o u r m i x t u r e = ” )

Scilab code Exa 7.14 Temperature composition diagram 1 clc () 2 // t h e t h r e e p h a s e t e m p e r a t u r e

i s f i r s t f i n d out , which comes t o be 342K, t h e c o r r e s p o n d i n g Ps1 = 7 1 . 1 8 , Ps2 = 3 0 . 1 2 3 T = [342 343 348 353 363 373]; 4 Ps2 = [30.12 31.06 37.99 47.32 70.11 101.3]; 5 Ps1 = [71.18 72.91 85.31 100.5 135.42 179.14]; 65

Figure 7.3: Temperature composition diagram

66

6 7 8 9 10 11 12 13 14 15

P = 101.3; // kPa for i = 1:4 y1 ( i ) = 1 - ( Ps1 ( i ) ) / P ; end for i = 1:6 y2 ( i ) = 1 - ( Ps2 ( i ) ) / P ; end plot2d ( y2 , T ) ; plot2d (1 - y1 ,T , rect = [0 ,320 ,1 ,380]) ; xtitle ( ’ T e m p e r a t u r e − c o m p o s i t o n d i a g r a m ’ , ’ x , y ( mole f r a c t i o n o f b e n z e n e ) ’ , ’ T e m p e r a t u r e ’ )

Scilab code Exa 7.15 Boiling point calculation 1 2 3 4 5 6 7 8 9 10

clc () T = 379.2; //K P = 101.3; // kPa Ps = 70; // kPa Molality = 5; Pws = exp (16.26205 - 3799.887/( T - 46.854) ) ; k = P / Pws ; Pws1 = Ps / k ; T1 = 3799.887 / (16.26205 - log ( Pws1 ) ) + 46.854; disp ( ”K” ,T1 , ” B o i l i n g p o i n t o f t h e s o l u t i o n = ” )

67

Chapter 8 Humidity and Humidity chart

Scilab code Exa 8.1 Nitrogen and ammonia gas mixture 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

clc () T = 280; //K P = 105; // kPa Pas = 13.25; // kPa ( Vpaour p r e s s u r e o f a c e t o n e ) Pa = Pas ; // ( As g a s i s s a t u r a t e d , p a r t i a l p r e s s u r e = vapour p r e s s u r e ) Mfr = Pa / P ; // ( Mole f r a c t i o n ) Mpr = Mfr * 100; disp ( ”%” ,Mpr , ” ( a ) The mole p e r c e n t o f a c e t o n e i n t h e mixture = ”) Ma = 58.048; // ( m o l e c u l a r w e i g h t o f a c e t o n e ) Mn = 28; // ( m o l e c u l a r w e i g h t o f n i t r o g e n ) N = 1; // mole Na = Mfr * N ; Nn = N - Na ; ma = Na * Ma ; mn = Nn * Mn ; mtotal = ma + mn ; maper = ma *100 / mtotal ; mnper = mn *100/ mtotal ; disp ( ”%” , maper , ” ( b ) Weight p e r c e n t o f a c e t o n e = ” ) 68

20 disp ( ”%” , mnper , ” Weight p e r c e n t o f n i t r o g e n = ” ) 21 Vstp = 22.4; //mˆ3/ kmol 22 Pstp = 101.3; // kPa 23 Tstp = 273.15; //K 24 V = Vstp * Pstp * T / ( Tstp * P ) ; 25 C = ma / V ; 26 disp ( ” kg /mˆ3 ” ,C , ” ( c ) C o n c e n t r a t i o n o f v a p o u r = ” )

Scilab code Exa 8.2 Benzene vapour air mixture 1 2 3 4 5 6 7 8

clc () P = 101.3; // kPa Per1 = 10; //% Pa = P * Per1 / 100; // ( a − b e n z e n e ) Ps = Pa ; // ( s a t u r a t i o n ) // l n P s = 1 3 . 8 8 5 8 − 2 7 8 8 . 5 1 / ( T − 5 2 . 3 6 ) T = 2788.51 / ( 13.8858 - log ( Ps ) ) + 52.36; disp ( ”K” ,T , ” T e m p e r a t u r e a t which s a t u r a t i o n o c c u r s = ”)

Scilab code Exa 8.3 Evaporation of acetone using dry air 1 clc () 2 Pdryair = 101.3; // kPa 3 Pacetone = 16.82; // kPa 4 Nratio = Pacetone / ( Pdryair - Pacetone ) ; 5 mratio = Nratio * 58.048 / 29; // ( Macetone = 6 7 8

5 8 . 0 4 8 , Mair = 29 ) macetone = 5; // kg ( g i v e n ) mdryair = macetone / mratio ; disp ( ” kg ” , mdryair , ”Minimum a i r r e q u i r e d = ” )

69

Scilab code Exa 8.4 Humidity for acetone vapour and nitrogen gas mixture 1 clc () 2 Pa = 15; // kPa ( p a r t i a l p r e s s u r e o f a c e t o n e ) 3 Ptotal = 101.3; // kPa 4 Mfr = Pa / Ptotal ; 5 disp ( Mfr , ” ( a ) Mole f r a c t i o n o f a c e t o n e = ” ) 6 Macetone = 58.048; 7 Mnitrogen = 28; 8 mafr = Mfr * Macetone / ( Mfr * Macetone + (1 - Mfr ) *

Mnitrogen ) ; 9 disp ( mafr , ” ( b ) Weight f r a c t i o n o f a c e t o n e = ” ) 10 Y = Mfr / ( 1 - Mfr ) ; 11 disp ( ” m o l e s o f a c e t o n e / m o l e s o f n i t r o g e n ” ,Y , ” ( c ) 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26

Molal humidity = ”) Y1 = Y * Macetone / Mnitrogen ; disp ( ” kg a c e t o n e / kg n i t r o g e n ” ,Y1 , ” ( d ) A b s o l u t e humidity = ”) Pas = 26.36; // kPa ( v a p o u r p r e s s u r e ) Ys = Pas / ( Ptotal - Pas ) ; // s a t u r a t i o n h u m i d i t y disp ( ” m o l e s o f a c e t o n e / m o l e s o f n i t r o g e n ” ,Ys , ” ( e ) S a t u r a t i o n humidity = ”) Y1s = Ys * Macetone / Mnitrogen ; disp ( ” kg a c e t o n e / kg n i t r o g e n ” ,Y1s , ” ( f ) A b s o l u t e s a t u r a t i o n humidity = ”) V = 100; //mˆ3 Vstp =22.4143; //mˆ3/ kmol Pstp = 101.3; // kPa Tstp = 273.15; //K T = 295; //K N = V * Ptotal * Tstp / ( Vstp * Pstp * T ) ; Nacetone = N * Mfr ; macetone = Nacetone * Macetone ; 70

27

disp ( ” kg ” , macetone , ” ( g ) Mass o f a c e t o n e i n 100mˆ3 o f the t o t a l gas = ”)

Scilab code Exa 8.5 Percent saturation and relative saturation 1 2 3 4 5 6 7 8 9

clc () ; Pa = 15; // kPa ( P a r t i a l p r e s s u r e ) Pas = 26.36; // kPa ( Vapour p r e s s u r e ) RS = Pa * 100 / Pas ; Y = 0.1738; Ys = 0.3517; PS = Y * 100 / Ys ; disp ( ”%” ,RS , ” R e l a t i v e h u m i d i t y = ” ) disp ( ”%” ,PS , ” P e r c e n t h u m i d i t y = ” )

Scilab code Exa 8.6 Analysis of Moist air 1 2 3 4 5 6 7 8 9 10 11 12 13

clc () mwater = 0.0109; // kg V = 1; //mˆ3 T = 300; //K P = 101.3; // kPa Vstp =22.4143; //mˆ3/ kmol Pstp = 101.3; // kPa Tstp = 273.15; //K N = V * P * Tstp / ( Vstp * Pstp * T ) ; Nwater = mwater / 18.016; Nfr = Nwater / N ; Pwater = Nfr * P ; disp ( ” kPa ” , Pwater , ” ( a ) P a r t i a l p r e s s u r e o f w a t e r vapour = ”) 14 Ps = exp (16.26205 - 3799.887/( T - 46.854) ) ; 15 RS = Pwater * 100 / Ps ; 71

16 disp ( ”%” ,RS , ” ( b ) R e l a t i v e s a t u r a t i o n = ” ) 17 Y1 = Pwater *18 / (( P - Pwater ) *29) ; 18 disp ( ” kg w a t e r / kg d r y a i r ” ,Y1 , ” ( c ) A b s o l u t e 19 20 21 22 23 24 25 26 27

humidity = ”) Y1s = Ps *18 / (( P - Ps ) *29) ; PS1 = Y1 * 100 / Y1s ; disp ( ”%” ,PS1 , ” ( d ) P e r c e n t s a t u r a t i o n = ” ) PS = 10; //% Y1S = Y1 * 100/ PS ; // Y1S = Pas / (P − Pas ) ∗ 18 /29 Pas1 = 29 * P * Y1S / (18 + 29* Y1s ) ; T1 = 3799.887 / (16.26205 - log ( Pas1 ) ) + 46.854; disp ( ”K” ,T1 , ” ( e ) T e m p e r a t u r e a t which 10% s a t u r a t i o n occurs = ”)

Scilab code Exa 8.7 Heating value calculation for a fuel gas 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

clc () T = 300; //K P = 100; // kPa S = 25000 // kJ /mˆ3 T1 = 295; //K P1 = 105; // kPa RS = 50; //% Ps = 3.5; // kPa Ps1 = 2.6; // kPa Vstp = 22.4143; //mˆ3/ kmol Pstp = 101.3; // kPa Tstp = 273.15; //K V = 1; //mˆ3 N = V * P * Tstp /( Vstp * Pstp * T ) ; Nfuel = N * ( P - Ps ) / P ; Smol = S / Nfuel ; // kJ / kmol N1 = V * P1 * Tstp /( Vstp * Pstp * T1 ) ; Pwater = Ps1 * RS /100; 72

19 Nfuel1 = N1 * ( P1 - Pwater ) / P1 ; 20 S1 = Smol * Nfuel1 ; 21 disp ( ” kJ /mˆ3 ” ,S1 , ” H e a t i n g v a l u e o f g a s a t 295K and

105 kPa = ” )

Scilab code Exa 8.8 Analysis of nitrogen benzene mixture 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

clc () T = 300; //K T1 = 335; //K P = 150; // kPa // l n P s = 1 3 . 8 8 5 8 − 2 7 8 8 . 5 1 / ( T − 5 2 . 3 6 ) Ps = exp (13.8858 - 2788.51 / ( T - 52.36) ) ; Ps1 = exp (13.8858 - 2788.51 / ( T1 - 52.36) ) ; Pa = Ps ; // ( Vapour p r e s s u r e a t dew p o i n t i s e q u a l t o the p a r t i a l p r e s s u r e o f the vapour ) Y = Pa / ( P - Pa ) ; Ys = Ps1 / ( P - Ps1 ) ; PS = Y * 100 / Ys ; disp ( ”%” ,PS , ” ( a ) P e r c e n t s a t u r a t i o n = ” ) Ma = 78.048; Mb = 28; Q = Y * Ma / Mb ; disp ( ” kg b e n z e n e / kg n i t r o g e n ” ,Q , ” ( b ) Q u a n t i t y o f benzene per kilgram of n i t r o g e n = ”) V = 1; //mˆ3 ( b a s i s ) Vstp = 22.4143; //mˆ3/ kmol Pstp = 101.3; // kPa Tstp = 273.15; //K N = V * P * Tstp /( Vstp * Pstp * T1 ) ; y = Y / ( 1 + Y ); Nbenzene = N * y ; C = Nbenzene * Ma ; disp ( ” kg /mˆ3 ” ,C , ” ( c ) K i l o g r a m o f b e n z e n e p e r mˆ3 o f n i t r o g e n = ”) 73

26 27 28 29 30 31 32 33 34

P1 = 100; // kPa Pbenzene = y * P1 ; T1 = 2788.51 / ( 13.8858 - log ( Pbenzene ) ) + 52.36; disp ( ”K” ,T1 , ” ( d ) Dew p o i n t = ” ) Per1 = 60; //% Y2 = Y * (1 - Per1 /100) ; //Y2 = Pa / (P − Pa ) P = Pa / Y2 + Pa ; disp ( ” kPa ” ,P , ” ( e ) P r e s s u r e r e q u i r e d = ” )

Scilab code Exa 8.9 Drying 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

clc () T = 300; //K T1 = 285; //K Pwater = 3.56; // kPa Pwater1 = 1.4; // kPa V = 1; //mˆ3 ( B a s i s ) Vstp = 22.4143; //mˆ3/ kmol N = V / Vstp ; Pstp = 101.3; // kPa Y = Pwater / ( Pstp - Pwater ) ; Y1 = Pwater1 / ( Pstp - Pwater1 ) ; Nremoved = Y - Y1 ; Ndryair = N * 1 / (1 + Y ) ; mremoved = Ndryair * Nremoved * 18.016; disp ( ” kg ” , mremoved , ” ( a ) amount o f w a t e r removed = ” ) Nremaining = Ndryair * Y1 ; V1 = ( Ndryair + Nremaining ) * Vstp ; disp ( ”mˆ3 ” ,V1 , ” ( b ) Volume o f g a s a t s t p a f t e r d r y i n g = ”)

74

Figure 8.1: Saturation lines for hexane

75

Scilab code Exa 8.10 Saturation lines for hexane 1 2 3 4 5 6 7 8 9 10 11 12 13

clc () P = 100; // kPa T = [273 280 290 300 310 320 330 340]; for i =1:8 Ps ( i ) = exp (13.8216 - 2697.55/( T ( i ) -48.78) ) ; end disp (( Ps ) ) for j = 1:8 Ys ( j ) = Ps ( j ) * 86.11 / (( P - Ps ( j ) ) *28) ; end disp ( Ys ) plot (T , Ys , rect =[273 ,0 ,333 ,10]) ; xtitle ( ’ 100% s a t u r a t i o n l i n e f o r n i t r o g e n −h e x a n e s y s t e m ’ , ’ Temperature , K ’ , ’ Humidity , kg h e x a n e / kg n i t r o g e n ’ );

Scilab code Exa 8.11 Psychometric chart application 1 2 3 4 5 6 7 8 9

clc () Td = 328; //K ( d r y b u l b ) P = 101.3; // kPa PS = 10; //% // r e f e r i n g t o t h e p s y c h o m e t r i c c h a r t , c o r r e s p o n d i n g t o 328 K and 10% s a t u r a t i o n Y1 = 0.012; // kg w a t e r / kg d r y a i r disp ( ” kg w a t e r / kg d r y a i r ” ,Y1 , ” ( a ) A b s o l u t e humidity = ”) //Y1 = Pa ∗ 18 / ( P − Pa ) ∗ 29 Pa = Y1 * P * 29 /( 18 + Y1 * 29 ) ;

76

10 11 12 13 14 15 16 17 18 19

20 21

disp ( ” kPa ” ,Pa , ” ( b ) P a r t i a l P r e s s u r e o f w a t e r v a p o u r = ”) // u s i n g p s y c h o m e t r i c c h a r t , s a t u r a t i o n h u m i d i t y a t 328 K i s g i v e n a s Y1s = 0.115; // kg w a t e r / kg d r y a i r disp ( ” kg w a t e r / kg d r y a i r ” ,Y1s , ” ( c ) The a b s o l u t e h u m i d i t y a t 328K = ” ) // a t 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 u r p r e s s u r e Pas = Y1s * P * 29 /( 18 + Y1s * 29 ) ; disp ( ” kPa ” ,Pas , ” ( d ) Vapour P r e s s u r e o f w a t e r v a p o u r = ”) RS = Pa * 100 / Pas ; disp ( ”%” ,RS , ” ( e ) P e r c e n t r e l a t i v e s a t u r a t i o n = ” ) // u s i n g p s y c h o m e t r i c c h a r t , moving h o r i z o n t a l l y k e e p i n g h u m i d i t y c o n s t a n t t o 100% s a t u r a t i o n , we g e t dew p o i n t as , T = 290; //K disp ( ”K” ,T , ” ( f ) Dew p o i n t = ” )

Scilab code Exa 8.12 Humid heat calculation for a sample of air 1 2 3 4 5 6 7 8 9 10 11 12 13 14

clc () Ca = 1.884; // kJ /kgK Cb = 1.005; // kJ /kgK Y1 = 0.012; // Cs = Cb + Y1 ∗ Ca Cs = Cb + Y1 * Ca ; disp ( ” kJ /kgK” ,Cs , ”Humid h e a t o f t h e s a m p l e = ” ) P = 101.3; // kPa V = 100; //mˆ3 R = 8.314; T = 328; //K T1 = 373; //K N = P * V / ( R * T ); Pa = 1.921; // kPa 77

15 Ndryair = N * ( P - Pa ) / P ; 16 mdryair = Ndryair * 29; 17 Ht = mdryair * Cs * ( T1 - T ) ; 18 disp ( ” kJ ” ,Ht , ” Heat t o be s u p p l i e d = ” )

Scilab code Exa 8.14 wet bulb temperature and dry bulb temperature 1 2 3 4 5 6 7 8 9 10

clc () P = 101.3; // kPa MW = 58; T1 = 280.8; //K Ps = 5; // kPa pr = 2; // kJ /kgK ( P s y c h o m e t r i c r a t i o ) Hvap = 360; // kJ / kg Tw = T1 ; Yw1 = Ps * MW / (( P - Ps ) * 29) ; // Tw = Tg − Hvap ∗ ( Yw1 − Y1 ) / (hG / kY ) , where hG/kY i s t h e p s y c h m e t r i c r a t i o p r 11 Y1 = 0; 12 Tg = Tw + Hvap * ( Yw1 - Y1 ) / pr ; 13 disp ( ”K” ,Tg , ” The a i r t e m p e r a t u r e = ” )

Scilab code Exa 8.15 Humidity calculation 1 2 3 4 5 6 7 8 9

clc () Td = 353.2; //K Tw = 308; //K Hvap = 2418.5; // kJ / kg pr = 0.950; // kJ / kg Ps = 5.62; // kPa P = 101.3; // kPa Yw1 = ( Ps * 18) / (( P - Ps ) * 29) ; Y1 = Yw1 - pr * ( Td - Tw ) / Hvap ; 78

Figure 8.2: SAturation curve and adiabatic cooling line 10 11

disp ( ” kg w a t e r / kg d r y a i r ” ,Y1 , ” Humidity = ” ) // h u m i d i t y can a l s o be d i r e c t l y o b t a i n e d from p s y c h o m e t r i c c h a r t , which we g e t t o be 0 . 0 1 8 kg w a t e r / kg d r y a i r

Scilab code Exa 8.16 SAturation curve and adiabatic cooling line 1 2 3 4

clc () P = 101.3; // kPa T = [283 293 303 313]; for i =1:4 79

5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

Ps ( i ) = exp (13.8858 - 2788.51/( T ( i ) -52.36) ) ; end for j =1:4 Ys ( j ) = Ps ( j ) * 78.048 / (( P - Ps ( j ) ) *29) ; end disp ( Ps ) disp ( Ys ) plot (T , Ys , rect = [270 ,0 ,323 ,0.9]) ; // Tas = Tg − L ∗ ( Y1as − Y1 ) / Cs // Cs = Cb + Y1 ∗Ca = 1 . 0 0 5 + Y1 ∗ 1 . 2 , L = 435.4; // kJ /kgK // f o r d i f f e r e n t v a l u e o f Tg and Y1 t r i e d , we have the f o l l o w i n g s e t of values Tg = [283 290.4 300 310.1 320.8]; Y1 = [0.1701150 0.15 0.125 0.1 0.075]; plot ( Tg , Y1 ) ; xtitle ( ’ S a t u r a t i o n c u r v e and a d i a b a t i c c o o l i n g l i n e ’ , ’ Temperature , K ’ , ’Y, kg b e n z e n e / kg d r y a i r ’ ) ;

Scilab code Exa 8.17 Adiabatic drier 1 2 3 4 5 6 7 8 9 10 11 12

clc () Tin = 380.7; //K Pin = 101.3; // kPa Tdew = 298; //K mremoved = 2.25; // kg V = 100; //mˆ3 // u s i n g h u m i d i t y c h a r t , h u m i d i t y o f a i r a t d r y b u l b t e m p e r a t u r e o f 3 8 0 . 7K and dew p o i n t = 298K i s , Y = 0.02; // kg w a t e r / kg d r y a i r disp ( ” kg w a t e r / kg d r y a i r ” ,Y , ” ( a ) Humidity o f a i r e n t e r i n g the d r i e r = ”) Tstp = 273.15; //K Vstp = 22.4143; //mˆ3/ kmol N = V * Tstp / ( Vstp * Tin ) ; 80

13 MY = Y * 29 / 18; // m o l a l h u m i d i t y 14 Ndryair = N / ( 1 + MY ) ; 15 mdryair = Ndryair *29; 16 mwaterin = mdryair * Y ; 17 mwaterout = mwaterin + mremoved ; 18 Yout = mwaterout / mdryair ; 19 // p e r c e n t h u m i d i t y i s c a l c u l a t e d u s i n g t h e c h a r t , 20 21 22 23 24 25 26 27 28 29 30 31

and i s PY = 55; //% disp ( ” kg w a t e r / kg d r y a i r ” , Yout , ” ( b ) e x i t a i r humidity = ”) disp ( ”%” ,PY , ” P e r c e n t h u m i d i t y = ” ) // from t h e h u m i d i t y c h a r t Twet = 313.2; //K Td = 322.2; //K disp ( ”K” , Twet , ” ( c ) e x i t a i r wet b u l b t e m p e r a t u r e = ” ) disp ( ”K” ,Td , ” ( c ) e x i t a i r d r y b u l b t e m p e r a t u r e = ” ) MYout = Yout * 29 / 18; Nout = Ndryair * ( 1 + MYout ) / 1; V1 = Nout * Vstp * Td / Tstp ; disp ( ”mˆ3 ” ,V1 , ” ( d ) Volume o f e x i t a i r = ” )

Scilab code Exa 8.18 Psychometric chart application 1 2 3 4 5 6 7 8 9 10 11

clc () P = 101.3; // kPa Td = 303; //K Tw = 288; //K // u s i n g p s y c h o m e t r i c c h a r t , Y1 = 0.0045; // kg w a t e r / kg d r y a i r PY = 18; //% Theated = 356.7; //K Cb = 1.005; Ca = 1.884; Cs = Cb + Y1 * Ca ; 81

12 Q = 1 * Cs * ( Theated - Td ) ; 13 disp ( ” kg w a t e r / kg d r y a i r ” ,Y1 , ” ( a ) Humidity o f t h e

i n i t i a l a i r = ”) 14 disp ( ”%” ,PY , ” ( b ) P e r c e n t h u m i d i t y = ” ) 15 disp ( ”K” , Theated , ” ( c ) T e m p e r a t u r e t o which t h e a i r i s heated = ”) 16 disp ( ” kJ ” ,Q , ” ( d ) Heat t o be s u p p p l i e d = ” )

Scilab code Exa 8.19 Psychometric chart application and given wet bulb and dry bulb temperature 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

clc () Tw = 313; //K Td = 333; //K // U s i n g t h p s y c h o m e t r i c c h a r t , Y = 0.04; // kg w a t e r / kg d r y a i r PS = 26.5; //% VS = 1.18; //mˆ3/ kg d r y a i r ( volume o f s a t u r a t e d a i r ) VD = 0.94; //mˆ3/ kg d r y a i r ( volume o f d r y a i r ) VH = VD + PS * ( VS - VD ) /100; HS = 470; // J / kg d r y a i r ( e n t h a l p y o f s a t u r a t e d air ) HD = 60; // J / kg d r y a i r ( e n t h a l p y o f d r y a i r ) H = HD + PS * ( HS - HD ) /100; disp ( ” kg w a t e r / kg d r y a i r ” ,Y , ” ( a ) A b s o l u t e Humidity of the a i r = ”) disp ( ”%” ,PS , ” ( b ) P e r c e n t h u m i d i t y = ” ) disp ( ”mˆ3/ kg d r y a i r ” ,VH , ” ( c ) Humid volume = ” ) disp ( ” kJ / kg d r y a i r ” ,H , ” ( d ) E n t h a l p y o f wet a i r = ” )

82

Chapter 9 Material Balance in Unit Operations

Scilab code Exa 9.1 Combustion of coal 1 2 3 4 5 6 7 8 9 10 11 12

clc () PC1 = 85; //% ( P e r c e n t c a r b o n i n c o a l ) PA1 = 15; //% ( P e r c e n t a s h i n c o a l ) PA2 = 80; //% ( P e r c e n t a s h i n c i n d e r ) PC2 = 20; //% ( P e r c e n t c a r b o n i n c i n d e r ) m = 100; // kg ( w e i g h t o f c o a l ) mash = PA1 * m / 100; w = mash * 100 / PA2 ; // w e i g h t o f c i n d e r mcarbon = w - mash ; Plost = mcarbon * 100 / ( m - mash ) ; disp ( ” kg ” ,w , ” w e i g h t o f c i n d e r f o r m e d = ” ) disp ( ”%” , Plost , ” P e r c e n t f u e l l o s t = ” )

Scilab code Exa 9.2 Drying of wood 1 clc ()

83

2 m = 1; // kg ( mass o f c o m p l e t e l y d r y wood ) 3 P1 = 40; //% ( p e r c e n t a g e m o i s t u r e i n wet wood ) 4 P2 = 5; //% ( P e r c e n t a g e m o i s t u r e i n d r y wood ) 5 mwaterin = P1 * m / ( 100 - P1 ) ; 6 mwaterout = P2 * m / ( 100 - P2 ) ; 7 mevaporated = mwaterin - mwaterout ; 8 disp ( ” kg ” , mevaporated , ” mass o f w a t e r e v a p o r a t e d p e r

kg o f d r y wood = ” )

Scilab code Exa 9.3 Effluent discharge 1 2 3 4 5 6 7 8 9

clc () F1 = 6*1000; //L/ s BOD1 = 3 * 10^ -5; // g /L BOD2 = 5 * 10^ -3; // g /L V = 16 * 10^3; //mˆ3/ day v = V * 10^3 / (24 * 3600) ; //L/ s // L e t BOD o f t h e e f f l u e n t be BODeff , BODeff = ( BOD2 * ( F1 + v ) - BOD1 * F1 ) / ( v ) ; disp ( ” g /L” , BODeff , ”BOD o f t h e e f f l u e n t o f t h e p l a n t = ”)

Scilab code Exa 9.4 benzene requirement calculation 1 2 3 4 5 6 7 8 9

clc () D = 100; // kg o f o v e r h e a d p r o d u c t xfa = 0.956; xdw = 0.074; xdb = 0.741; xda = 0.185; // w a t e r b a l a n c e g i v e s F = D * xdw / (1 - xfa ) ; W = F * xfa - xda * D ; 84

10 W1 = 100; 11 B = xdb * D ; 12 Bused = B * W1 / W ; 13 disp ( ” kg ” , Bused , ” Q u a n t i t y o f b e n z e n e r e q u i r e d = ” )

Scilab code Exa 9.5 Fortification of waste acid 1 clc () 2 // l e t , W − w a s t e a c i d , S − S u l f u r i c 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

acid , N − n i t r i c

a c i d , M − mixed a c i d xsh2so4 = 0.95; xsh2o = 0.5; xwh2so4 = 0.3; xwhno3 = 0.36; xwh2o = 0.34; xmh2so4 = 0.4; xmhno3 = 0.45; xmh2o = 0.15; xnhno3 = 0.8; xnh2o = 0.2; M = 1000; // kg // t o t a l m a t e r i a l b a l a n c e , W + S + N = 1 0 0 0 ; //H2SO4 b a l a n c e , xwh2so4 ∗ W + x s h 2 s o 4 ∗ S = xmh2so4 ∗M //HNO3 b a l a n c e , xwhno3 ∗ W + xnhno3 ∗ N = xmhno3 ∗M //H2O b a l a n c e , xwh2o ∗W+xnh2o ∗N + x s h 2 o ∗S = xmh2o∗M // s o l v i n g t h e a b o v e e q u a t i o n s s i m u l t a n e o u s l y , we g e t , W = 70.22; // kg S = 398.88; // kg N = 530.9; // kg disp ( ” kg ” ,W , ” Waste a c i d = ” ) disp ( ” kg ” ,S , ” C o n c e n t r a t e d H2SO4 = ” ) disp ( ” kg ” ,N , ” C o n c e n t r a t e d HNO3 = ” )

85

Scilab code Exa 9.6 Triple effect evaporator 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

clc () F = 1000; // kg Psolute1 = 20; //% Psolute2 = 80; //% // t a k i n g s o l u t e b a l a n c e L3 = F * Psolute1 / Psolute2 ; // t a k i n g t o t a l m a t e r i a l b a l a n c e V = ( F - L3 ) / 3; // f o r f i r s t e f f e c t , t o t a l b a l a n c e g i v e s , L1 = F - V ; // s o l u t e b a l a n c e g i v e s , Psolute3 = F * Psolute1 / L1 ; // For s e c o n d e f f e c t , t o t a l b a l a n c e g i v e s , L2 = L1 - V ; // s o l u t e b a l n c e g i v e s , Psolute4 = L1 * Psolute3 / L2 ; disp ( ”%” , Psolute3 , ” s o l u t e e n t e r i n g s e c o n d e f f e c t = ” ) 18 disp ( ” kg ” ,L1 , ” Weight e n t e r i n g s e c o n d e f f e c t ” ) 19 disp ( ”%” , Psolute4 , ” s o l u t e e n t e r i n g t h i r d e f f e c t = ” ) 20 disp ( ” kg ” ,L2 , ” Weight e n t e r i n g t h i r d e f f e c t ” )

Scilab code Exa 9.7 Crystallization operation 1 2 3 4 5

clc () F = 100; // kg xf = 0.25; x2 = 7/107; P1 = 10; //%

86

6 W3 = P1 * F * (1 - xf ) /100; // (W3 − w e i g h t o f w a t e r 7

8 9 10 11 12 13 14

evaporated ) // l e t W1 and W2 be w e i g h t o f c r y s t a l and w e i g h t o f mother l i q u o r r e m a i n i n g a f t e r c r y s t a l l i z a t i o n resp . , //F = W1 + W2 + W3 // 100 = W1 + W2 + 7 . 5 // s o l u t e b a l a n c e g i v e s , F∗ x f = W1∗ x1 + W2∗ x2 // 1 0 0 ∗ 0 . 2 5 = W1∗1+W2 ∗ 0 . 0 6 5 4 W2 = ( F - W3 - F * xf ) /(1 - x2 ) ; W1 = F - W3 - W2 ; disp ( ” kg ” ,W1 , , , ” y e i l d o f t h e c r y s t a l s = ” )

Scilab code Exa 9.8 Evaporation of Na2CO3 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

clc () F = 100; // kg xf = 0.15; P1 = 80; //% ( C a r b o n a t e r e c o v e r e d ) M1 = 106; // ( M o l e c u l a r w e i g h t o f Na2CO3 ) M2 = 286; // ( M o l e c u l a r w e i g h t o f Na2CO3 . 1 0 H2O) x1 = M1 / M2 ; // ( Weight f r a c t i o n o f Na2CO3 i n crystals ) Mrecovered = P1 * F * xf / 100; Wcrystal = Mrecovered / x1 ; disp ( ” kg ” , Wcrystal , ” ( a ) q u a n t i t y o f c r y s t a l s f o r m e d = ”) //Na2CO3 b a l a n c e g i v e s , F∗ x f = W c r y s t a l ∗ x1 + W2∗ x2 //W2 w e i g h t o f mother l i q u o r r e m a i n i n g a f t e r crystallization // l e t M = W2 ∗ x2 , t h e r e f o r e M = F * xf - Mrecovered ; x2 = 0.09; W2 = M / x2 ; W3 = F - Wcrystal - W2 ; // w e i g h t o f w a t e r e v a p o r a t e d 87

18

disp ( ” kg ” ,W3 , ” ( b ) Weight o f w a t e r e v a p o r a t e d = ” )

Scilab code Exa 9.9 Crystallization 1 clc () 2 m = 100; // kg ( o f 60% s o l u t i o n ) 3 //w − w a t e r added t o t h e o r i g i n a l s o l u t i o n 4 //w1 − wt . o f Na2S2O3 . 5 H2O c r y s t a l l i z e d 5 //w2 − wt . o f mother l i q u o r o b t a i n e d 6 //w3 − s o l u t i o n c a r r i e d away by t h e c r y s t a l s 7 xf = 0.6; 8 Mna2s2o3 = 158; 9 Mna2s2o35h2o = 248; 10 mcrystals = m * xf * Mna2s2o35h2o / Mna2s2o3 ; 11 // f r e e w a t e r a v a i l a b l e = m + w − 1 − m c r y s t a l s 12 // c o n c e n t r a t i o n o f i m p u r i t y = 1 / (w+ 4 . 8 2 3 ) 13 // t o t a l b a l a n c e , 100 − 1 + w = w1 + w2 + w3 14 //w1 + w2 + w3 − w = 99 15 // Na2S2O3 b a l a n c e , 60 = ( w1 + w2 ∗ 1 . 5 / 2 . 5 + w3 ∗ 16 17 18 19 20 21 22 23 24 25 26 27 28

1 . 5 / 2 . 5 ) ∗158/248 //w1 + 0 . 6 ∗ w2 + 0 . 6 ∗ w3 = 9 4 . 1 7 7 // e a c h gram c r y s t a l s c a r r y 0 . 0 5 kg s o l u t i o n , //w3 = 0 . 0 5 ∗ w1 // i m p u r i t y % = 0 . 1 // i m p u r i t y = w3 / ( 2 . 5 ∗ (w+ 4 . 8 2 3 ) ) // s o l v i n g a b o v e e q u a t i o n s , we g e t w = 14.577; // kg w1 = 65.08; // kg w2 = 45.25; // kg w3 = 0.05 * w1 ; disp ( ” kg ” ,w , ” ( a ) amount o f w a t e r added = ” ) disp ( ” kg ” ,w1 , ” ( b ) amount o f Na2S2O3 . 5 H2O c r y s t a l s added = ” ) m1 = w1 * Mna2s2o3 / Mna2s2o35h2o + w3 * 1.5 * Mna2s2o3 / (2.5 * Mna2s2o35h2o ) ; 88

29 P = m1 *100/( m * xf ) ; 30 disp ( ”%” ,P , ” ( c ) P e r c e n t a g e r e c o v e r y

o f Na2S2O3 = ” )

Scilab code Exa 9.10 Extraction 1 clc () 2 m = 100; // kg 3 Pin1 = 40; //% ( t a n n i n ) 4 Pin2 = 5; //% ( m o i s t u r e ) 5 Pin3 = 23; //% ( s o l u b l e non t a n n i n m a t e r i a l ) 6 Pin4 = 100 - Pin1 - Pin2 - Pin3 ; //% ( i n s o l u b l e 7 8 9 10 11 12 13 14 15

lignin ) // s i n c e , l i g n i n i s i n s o l u b l e , a l l o f i t w i l l be present in the r e s i d u e Pout1 = 3; //% Pout2 = 50; //% Pout3 = 1; //% Pout4 = 100 - Pout1 - Pout2 - Pout3 ; // l e t W be t h e mass o f r e s i d u e , t h e n we g e t W = Pin4 * m / Pout4 ; Ptannin = W * Pout1 * 100 / ( m * Pin1 ) ; disp ( ”%” , Ptannin , ” P e r c e n t o f o r i g i n a l t a n n i n unextracted = ”)

Scilab code Exa 9.11 Leaching operation 1 clc () 2 F = 100; // kg 3 //F − f e e d , R − o v e r f l o w , U − u n d e r f l o w , S − s o l v e n t 4 //F + S = U + R ( T o t a l b a l a n c e ) 5 Poil1 = 49; //% ( 1 − f e e d ) 6 Ppulp1 = 40; //% 7 Psalts1 = 3; //%

89

8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27

Pwater = 100 - Poil1 - Ppulp1 - Psalts1 ; Phexane2 = 25; //%( 2 − u n d e r f l o w ) Psalts2 = 2.5; //% Poil2 = 15; //% Pwater2 = 7.5; //% Ppulp2 = 100 - Phexane2 - Poil2 - Pwater2 - Psalts2 ; Poil3 = 25; //% ( 3 − e x t r a c t ) // t a k i n g p u l p ( i n e r t ) b a l a n c e U = Ppulp1 * F / Ppulp2 ; // o i l b a l a n c e g i v e s , F ∗ P o i l 1 = U ∗ P o i l 2 + R ∗ P o i l 3 , from t h e s e , we g e t R = ( F * Poil1 - U * Poil2 ) / Poil3 ; S = U + R - F; disp ( ” kg ” ,S , ” ( a ) The amount o f s o l v e n t u s e d f o r e x t r a c t i o n = ”) Precovered = 95; //% mhexane2 = Phexane2 * U / 100; mrecovered = mhexane2 * Precovered / 100; P = mrecovered * 100 / S ; disp ( ”%” ,P , ” ( b ) P e r c e n t o f h e x a n e u s e d t h a t i s r e c o v e r e d from t h e u n d e r f l o w = ” ) Poil = Poil3 * R * 100 / ( F * Poil1 ) ; disp ( ”%” , Poil , ” ( c ) P e r c e n t r e c o v e r y o f o i l = ” )

Scilab code Exa 9.12 Dryer and oven 1 clc () 2 //F = f e e d ( wet s o l i d ) , V1 = w a t e r e v a p o r a t e d ( d r i e r ) ,

3 4 5 6

V2 = w a t e r e v a p o r a t e d ( oven ) , S1 = Dry s o l i d ( d r i e r ) , S2 = Dry s o l i d ( oven ) F = 1000; // kg xf = 0.8; x1 = 0.15; x2 = 0.02; 90

7 8 9 10 11 12 13 14 15 16 17 18 19 20

// m o i s t u r e f r e e s o l i d b a l a n c e f o r d r i e r , F ∗ ( 1 − x f ) = S1 ∗ ( 1 − x1 ) S1 = F * ( 1 - xf ) /(1 - x1 ) ; // t o t a l b a l a n c e f o r d r i e r , F = S1 + V1 V1 = F - S1 ; // For oven , S1 ∗ ( 1 − x1 ) = S2 ∗ ( 1 −x2 ) S2 = S1 * ( 1 - x1 ) /(1 - x2 ) ; // Also , S1 = S2 + V2 V2 = S1 - S2 ; disp ( ” kg ” ,S1 , ” ( a ) Weight o f p r o d u c t l e a v i n g t h e d r i e r = ”) disp ( ” kg ” ,S2 , ” Weight o f p r o d u c t l e a v i n g t h e oven = ”) P1 = V1 *100/ ( F * xf ) ; P2 = V2 *100/ ( F * xf ) ; disp ( ”%” ,P1 , ” ( b ) P e r c e n t a g e o f o r i g i n a l w a t e r removed in d r i e r = ”) disp ( ”%” ,P2 , ” P e r c e n t a g e o f o r i g i n a l w a t e r removed i n oven = ” )

Scilab code Exa 9.13 Adiabatic drier 1 clc () 2 // S s = s o l i d f l o w r a t e , 3 Pwaterin = 25; //% 4 Pwaterout = 5; //% 5 X1 = Pwaterin /(100 - Pwaterin ) ; // kg w a t e r / kg d r y a i r 6 X2 = Pwaterout /(100 - Pwaterout ) ; // kg w a t e r / kg d r y 7 8 9 10 11 12

air // form h u m i d i t y c h a r t , Y2 = 0.015; // kg w a t e r / kg d r y a i r Y1 = 0.035; // kg w a t e r / kg d r y a i r m = 1; // kg o f d r y a i r // S s ∗ X1 + Y2 = S s ∗ X2 + Y1 Ss = ( Y1 - Y2 ) / ( X1 - X2 ) ; 91

T = 87.5 + 273.15; //K P = 101.3; // kPa Tstp = 273.15; //K Pstp = 101.3; // kPa Vstp = 22.4143; //mˆ3/ mol V = 100; //mˆ3 N = V * P * Tstp / ( Vstp * Pstp * T ) ; Nr2 = Y2 * 29 / 18; // kmol o f w a t e r / kmol o f d r y a i r Ndryair = N * 1 / (1 + Nr2 ) ; mdryair = Ndryair * 29; mevaporated = mdryair * ( Y1 - Y2 ) ; disp ( ” kg ” , mevaporated , ” ( a ) t o t a l m o i s t u r e e v a p o r a t e d p e r 100mˆ3 o f a i r e n t e r i n g = ” ) 25 Ss1 = mdryair * Ss ; 26 mproduct = Ss1 * ( 1 + X2 ) ; 27 disp ( ” kg ” , mproduct , ” ( b ) mass o f f i n i s h e d p r o d u c t p e r 100mˆ3 o f a i r e n t e r i n g = ” )

13 14 15 16 17 18 19 20 21 22 23 24

Scilab code Exa 9.14 Extraction of isopropyl alcohol 1 clc () 2 //F = f e e d , E = e x t r a c t , S = s o l v e n t , R = R a f f i n a t e 3 xwaterF = 0.7; // Feed 4 xalcoholF = 0.3; 5 xwaterR = 0.71; // r a f f i n a t e 6 xalcoholR = 0.281; 7 xethyR = 0.009; 8 xwaterE = 0.008; // E x t r a c t 9 xalcoholE = 0.052; 10 xethyE = 0.94; 11 // T o t a l b a l a n c e , R + E = F + S 12 F = 100; // kg 13 //R + E = 100 + S (1) 14 // I s o p r o p y l b a l a n c e , x a l c o h o l R ∗ R + x a l c o h o l E ∗E =

xalcoholF ∗ F 92

15 16 17 18 19 20 21 22 23 24 25 26 27

// 0 . 2 8 1 ∗R + 0 . 0 5 2 ∗ E = 30 (2) // E t h y l e n e t e t r a c h l o r i d e b a l a n c e , xethyR ∗ R + xethyE ∗ E = S // 0 . 0 0 9 ∗R + 0 . 9 4 ∗ E = S (3) // S o l v i n g e q u a t i o n 1 , 2 and 3 s i m u l t a n e o u s l y , we g e t , S = 45.1; E = 47.04; R = 98.06; disp ( ” kg ” ,S , ” ( a ) Amount o f s o l v e n t u s e d = ” ) disp ( ” kg ” ,E , ” ( b ) Amount o f e x t r a c t = ” ) disp ( ” kg ” ,R , ” Amount o f r a f f i n a t e = ” ) mextracted = E * xalcoholE ; P1 = mextracted * 100 / ( F * xalcoholF ) ; disp ( ”%” ,P1 , ” ( c ) P e r c e n t o f i s o p r o p y l a l c o h o l e x t r a c t e d = ”)

Scilab code Exa 9.15 Absorption of acetone 1 clc () 2 G1 = 100; // kmol 3 //G1 and G2 be t h e m o l a r f l o w

4 5 6 7 8 9 10 11

rate of the gas at the i n l e t and t h e e x i t o f t h e a b s o r b e r r e s p . , y1 and y2 mole f r a c t i o n a t e n t r a n c e and e x i t r e s p . , y1 = 0.25; //% y2 = 0.05; //% // a i r b a l a n c e g i v e s , G1 ∗ ( 1−y1 ) = G2 ∗ ( 1−y2 ) G2 = G1 * ( 1 - y1 ) / (1 - y2 ) ; maleaving = G2 * y2 ; maentering = G1 * y1 ; Pabsorbed = ( maentering - maleaving ) * 100 / ( maentering ) ; disp ( ”%” , Pabsorbed , ” P e r c e n t a g e o f a c e t o n e a b s o r b e d = ”)

93

Scilab code Exa 9.16 Absorption of SO3 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23

24 25 26

clc () F = 5000; // kg / h P1 = 50; //% ( H2O4 i n ) MH2SO4 = 98.016; P1gas = 65; // ( n i t r o g e n i n g a s e n t e r i n g ) P2gas = 35; // ( SO3 ) MN2 = 28; MSO3 = 80; Mavg = ( MN2 * P1gas + MSO3 * P2gas ) /100; // avg m o l e c u l a r wt . o f e n t e r i n g g a s G = 4500; // kg / h Ng = G / Mavg ; NN2 = Ng * P1gas / 100; NSO3 = Ng - NN2 ; P2 = 75; //% ( H2O4 o u t ) //W be t h e mass o f 75% H2SO4 , x and y be t h e m o l e s o f SO3 and w a t e r v a p o u r l e a v i n g r e s p . , Pwater = 25; // kPa Ptotal = 101.3; // kPa // Pwater / P t o t a l = y / ( NN2 + x + y ) // we g e t , y = 0 . 3 2 7 6 5 ∗ x + 2 . 7 4 4 (1) // T o t a l b a l a n c e Feed + G = W + (NN2 ∗ 28 + x ∗ 80 + y ∗ 18.016) // we g e t , W + 80∗ x + 1 8 . 0 1 6 ∗ y = 7 7 2 7 . 3 2 (2) // from 1 and 2 , 8 4 . 9 1 7 4 ∗ x + W = 7 3 5 2 . 6 8 (3) // SO3 b a l a n c e , So3 e n e t e r i n w i t h 50% H2SO4 + SO3 i n f e e d g a s = SO l e a v i n g w i t h 75%H2SO4 + SO3 l e a v i n g in e x i t gas // 5 0 0 0 ∗ 0 . 5 ∗ 8 0 / 9 8 . 0 1 6 + 3 4 . 0 9 ∗ 8 0 = 80∗ x + 0 . 7 5 ∗W ∗ 80/98.016 (4) // from 3 and 4 , x = 9.74; 94

27 28 29

Nabsorbed = NSO3 - x ; Pabsorbed = Nabsorbed * 100 / NSO3 ; disp ( ”%” , Pabsorbed , ” P e r c e n t a g e o f SO3 a b s o r b e d = ” )

Scilab code Exa 9.17 Continuous distillation column 1 clc () 2 F = 200; // kmol / h 3 //F , D and W be t h e f l o w

4 5 6 7 8 9 10 11 12 13 14 15 16

r a t e s of the feed , the d i s t i l l a t e and r e s i d u e r e s p . , x f , xd and xw be t h e mole f r a c t i o n o f e t h a n o l i n t h e f e e , d i s t i l l a t e and t h e r e s i d u e r e s p . xf = 0.10; xd = 0.89; xw = 0.003; // t o t a l b a l a n c e g i v e s , F = D + W //D + W = 200 (1) // A l c o h o l b a l a n c e g i v e s , F∗ x f = D∗ xd + W∗xw // 0 . 8 9 ∗D+ 0 . 0 0 3 ∗W = 20 (2) // s o l v i n g 1 and 2 D = 21.87; // kmol / h W = 178.13; // kmol / h Nawasted = W * xw ; mmakeup = Nawasted * 46*24; disp ( ” kg ” , mmakeup , ” The make up a l c o h o l r e q u i r e d p e r day = ” )

Scilab code Exa 9.18 Distillation operation for methanol solution 1 clc () 2 F = 100; // kg 3 //F , D and W be t h e f l o w

r a t e s of the feed , the d i s t i l l a t e and bottom p r o d u c t r e s p . , x f , xd and 95

4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

xw be t h e mole f r a c t i o n o f m e t h a n o l i n t h e f e e , d i s t i l l a t e and t h e bottom p r o d u c t r e s p . xf = 0.20; xd = 0.97; xw = 0.02; // u s i n g , F = D + W and F∗ x f + D∗ xd + W∗xw , we g e t D = 18.95; // kg / h W = 81.05; // kg / h R = 3.5; //R = L / D // f o r d i s t i l l a t e = 1 kg D1 = 1; // kg L = R * D1 ; // Taking b a l a n c e a r o u n d t h e c o n d e n s e r , G = L + D1 ; mcondensed = G * D / F ; disp ( ” kg ” ,D , ” ( a ) Amount o f d i s t i l l a t e = ” ) disp ( ” kg ” ,W , ” Amount o f Bottom P r o d u c t = ” ) disp ( ” kg ” ,G , ” ( b ) Amount o f v a p o u r c o n d e n s e d p e r kg o f d i s t i l l a t e = ”) disp ( ” kg ” , mcondensed , ” ( c ) Amount o f v a p o u r c o n d e n s e d p e r kg o f f e e d = ” )

Scilab code Exa 9.19 Bypass operation 1 clc () 2 mdryair = 1; // kg 3 Pwater1 = 1.4; // kPa ( P a r t i a l p r e s s u r e a t 285K ) 4 Pwater2 = 10.6; // kPa ( P a r t i a l p r e s s u r e a t 320K ) 5 P = 101.3; // ( T o t a l ) 6 Ys1 = Pwater2 * 18 / (( P - Pwater2 ) *29) ; // (

s a t u r a t i o n h u m i d i t y a t 320K ) 7 Ys2 = Pwater1 * 18 / (( P - Pwater1 ) *29) ; // ( s a t u r a t i o n h u m i d i t y a t 285K ) 8 Ys = 0.03; // kg w a t e r / kg d r y a i r . ( f i n a l h u m i d i t y ) 96

9

10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

26 27 28 29 30 31 32

// h u m i d i t y o f a i r l e a v i n g d e h u m i d i f i e r i s Ys2 and h u m i d i t y o f b y p a s s e d a i r i s Ys1 . t h e s e 2 s t r e a m s combine t o g i v e h u m i d i t y o f 0 . 0 3 kg w a t e r / kg d r y air . // t h e r e f o r e , t a k i n g b a l a n c e we g e t , 1∗ Ys2 + x ∗ Ys1 = ( 1 + x ) ∗Ys x = (1* Ys2 - 1* Ys ) /( Ys - Ys1 ) ; disp ( ” kg d r y a i r ” ,x , ” ( a ) Mass o f d r y a i r b y p a s s e d p e r kg o f d r y a i r s e n t t h r o u g h t h e d e h u m i d i f i e r = ” ) mcondensed = Ys1 - Ys2 ; mwetair = mdryair + Ys1 ; Nwetair = mdryair /29 + Ys1 /18.016; Vstp = 22.4143; //mˆ3/ kmol Vstp1 = Nwetair * Vstp ; T = 320; //K P = 101.3; // kPa Tstp = 273.15; //K Pstp = 101.325; // kPa V = Vstp1 * Pstp * T / ( P * Tstp ) ; Vgiven = 100; //mˆ3 mcondensed1 = mcondensed * Vgiven / V ; disp ( ” kg ” , mcondensed1 , ” ( b ) mass o f w a t e r v a p o u r c o n d e n s e d i n t h e d e h u m i d i f i e r p e r 100mˆ3 o f a i r sent through i t = ”) mfinal = mdryair + x ; mfinalair = mfinal * Vgiven / V ; N = mfinalair / 29; Ysn = Ys * 29/18; // kmol w a t e r / kmol d r y a i r Ntotal = N * ( Ysn + 1) ; Vfinal = Ntotal * Vstp * Pstp * T / ( Tstp * P ) ; disp ( ”mˆ3 ” , Vfinal , ” ( c ) Volume o f f i n a l a i r o b t a i n e d p e r 100 c u b i c m e t r e s f a i r p a s s e d t h r o u g h d e h u m i d i f i e r = ”)

Scilab code Exa 9.20 Recycle operation centrifuge plus filter 97

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

clc () F = 100; // kg / h xf = 0.2; xp = 0.93; xr = 0.5/1.5; xx = 0.65; //R − r e c y c l e s t r e a m , P − P r o d u c t s t r e a m , W − w a t e r s e p a r e t e d and removed // component A b a l a n c e , F ∗ x f = P ∗ xp , t h a t i s , P = F * xf / xp ; // T o t a l b a l a n c e , F = P + W, t h e r e f o r e W = F - P; // x be t h e f l o w r a t e o f s t r e a e n t e r i n g t h e f i l t e r // t o t a l b a l a n c e , x = P + R (1) // component A b a l a n c e , 0 . 6 5 ∗ x = 0 . 5 ∗R/ 1 . 5 + 0 . 9 3 P (2) // S o l v i n g 1 and 2 , we g e t , R = ( xx * P - xp * P ) /( xr - xx ) ; disp ( ” kg / h ” ,R , ” Flow r a t e o f t h e r e c y c l e s t r e a m = ” )

Scilab code Exa 9.21 Recycle operation granulator and drier 1 2 3 4 5 6 7 8 9

clc () F = 1000; // kg / h xfwater = 0.7; xpwater = 0.2; xrwater = 0.20; xswater = 0.5; y1 = 0.0025; y2 = 0.05; //R − r e c y c l e , S − s t r e a m e n t e r i n g g r a n u l a t o r , P − Product , G1 − a i r e n t e r i n g t h e d r i e r , G2 − a i r l e a v i n g the drier , 10 // t a k i n o v e r a l l , m o i s t u r e f r e e b a l a n c e , F ∗ x f = P ∗ 98

11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

xp P = F * ( 1 - xfwater ) /(1 - xpwater ) ; // t a k i n g m a t e r i a l b a l a n c e a t p o i n t where r e c y c l e s t r e a j o i n s the feed , // F = R + S // w a t e r b a l a n c e , F∗ x f w a t e r = R∗ x r w a t e r + S∗ x s w a t e r , s o l v i n g t h i s we g e t , R = ( - F * xfwater + F * xswater ) /( xrwater - xswater ) ; S = F + R; mleaving = P + R ; // s o l i d l e a v i n g t h e d r i e r // d r y a i r e n t e r i n g w i l l t h e r e be i n a i r l e a v i n g , therefore //G1 ∗ ( 1 − y1 ) = G2 ∗ ( 1 − y2 ) // w a t e r b a l a n c e o v e r t h e d r i e r g i v e s , S∗ x s w a t e r+G1∗ y1=G2∗ y2 +(P+R) ∗ x p w a t e r // from a b o v e 2 e q u a t i o n s , we g e t G1 = (( mleaving * xpwater - S * xswater ) /( y1 - y2 *(1 - y1 ) /(1 - y2 ) ) ) ; disp ( ” kg /h ” ,R , ” ( a ) Amount o f s o l i d r e c y c l e d = ” ) mdryair = G1 * (1 - y1 ) ; disp ( ” kg /h ” , mdryair , ” ( b ) c i r c u l a t i o n r a t e o f a i r i n t h e d r i e r on d r y b a s i s = ” )

Scilab code Exa 9.22 Blowdown operation 1 2 3 4 5

clc () xf = 500 * 10^ -6; xp = 50 * 10^ -6; xb = 1600 * 10^ -6; //F − Feed w a t e r r a t e , B − blow down r a t e , S − h i g h p r e s s u r e steam , P − p r o c e s s s t r e a m r a t e 6 // t o t a l b a l a n c e , F = P + B 7 // S o l i d b a l a n c e , F ∗ x f + P ∗ xp = B ∗ xb 8 // e l i m i n a t i n g P , we g e t , F ∗ x f + ( F − B) ∗ xp = B ∗ xb 99

9 // l e t F/B be X 10 X = ( xb + xp ) /( xf + xp ) ; 11 disp (X , ” t h e r a t i o o f f e e d w a t e r t o t h e blowdown

water = ”)

100

Chapter 10 Material Balance with Chemical Reaction

Scilab code Exa 10.1 Combustion of propane 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

clc () mair = 500; // kg mCO2 = 55; // kg mCO = 15; // kg //C3H8 + 5O2 = 3CO2 + 4H20 MCO2 = 44; MCO = 28; NCO2 = mCO2 / MCO2 ; NCO = mCO / MCO ; Mair = 29; Nair = mair / Mair ; // c a r b o n b a l a n c e g i v e s , F = ( NCO2 + NCO ) /3; MC3H8 = 44.064; mC3H8 = MC3H8 * F ; disp ( ” kg ” , mC3H8 , ” ( a ) mass o f p r o p a n e b u r n t = ” ) // one mole o f p r o p a n e r e q u r e s 5 m o l e s o f o x y g e n f o r combustion 18 NO2 = F * 5; 101

19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38

Nairt = NO2 * 100 /21; // t h e o r e t i c a l a i r r e q u i r e d Pexcess = ( Nair - Nairt ) * 100 / Nairt ; disp ( ”%” , Pexcess , ” ( b ) The p e r c e n t e x c e s s a i r = ” ) //C3H8 + 7/2 ∗ O2 = 3CO + 4H2O NH2O = F * 4; // Taking o x y g e n b a l a n c e , unburned o x y g e n i s calculated , //O2 s u p p l i e d = O2 p r e s e n t i n form o f CO2 , CO and H2O + unburned O2 Nunburnt = Nair * 21 / 100 - NCO2 - NCO /2 - NH2O /2; NN2 = Nair * 79 / 100; Ntotal = NCO2 + NCO + NH2O + NN2 + Nunburnt ; PCO2 = NCO2 * 100 / Ntotal ; PCO = NCO *100/ Ntotal ; PH2O = NH2O *100/ Ntotal ; PN2 = NN2 *100/ Ntotal ; PO2 = Nunburnt *100 / Ntotal ; disp ( ”%” , PCO2 , ” ( c ) P e r c e n t c o m p o s i t i o n o f CO2 = ” ) disp ( ”%” ,PCO , ” P e r c e n t c o m p o s i t i o n o f CO = ” ) disp ( ”%” , PH2O , ” P e r c e n t c o m p o s i t i o n o f H2O = ” ) disp ( ”%” ,PN2 , ” P e r c e n t c o m p o s i t i o n o f N2 = ” ) disp ( ”%” ,PO2 , ” P e r c e n t c o m p o s i t i o n o f O2 = ” )

Scilab code Exa 10.2 Combustion of hydrogen free coke 1 2 3 4 5 6 7 8 9

clc () Nflue = 100; // kmol NCO2 = 14.84; NCO = 1.65; NO2 = 5.16; NN2 = 78.35; PCF = 85; //PERCENT CARBON IN FEED PIF = 15; //PERCENT INERT IN FEED //F − amount o f c o k e c h a r g e d , W − mass o f c o k e l e f t , W = 0.05F 102

10 NCflue = NCO2 + NCO ; 11 MC = 12; 12 mC = MC * NCflue ; 13 // c a r b o n b a l a n c e g i v e s , F ∗ PCF / 100 = W ∗ PCF + mC 14 F = mC / ( PCF / 100 - 0.05* PCF / 100) ; 15 // l e t A kmol a i r s u p p l i e d , t a k i n g N2 b a l a n c e , 16 Nair = NN2 * 100/79; 17 NO2supplied = Nair - NN2 ; 18 Ntheoretical = F * PCF / (100 * MC ) ; 19 Pexcess = ( NO2supplied - Ntheoretical ) * 100 / ( 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35

Ntheoretical ) ; disp ( ”%” , Pexcess , ” ( a ) P e r c e n t a g e e x c e s s a i r = ” ) mair = Nair * 29; m = mair / F ; // a i r s u p p l i e d p e r kg o f c o k e c h a r g e d disp ( ” kg ” ,m , ” ( b ) a i r s u p p l i e d p e r kg o f c o k e c h a r g e d = ”) P = 100; // kPa T = 500; //K V = Nflue *22.4143*101.325 * T / ( F * P * 273.15) ; disp ( ”mˆ3 ” ,V , ” ( c ) volume o f f l u e g a s p e r kg o f c o k e = ”) W = 0.05* F ; mCr = W * PCF /100; // c a r b o n i n r e f u s e mir = F * (1 - PCF /100) ; // i n e r t i n r e f u s e mr = mCr + mir ; C = mCr * 100 / mr ; I = mir *100/ mr ; disp ( ”%” ,C , ” ( d ) Carbon = ” ) disp ( ”%” ,I , ” I n e r t = ” )

Scilab code Exa 10.3 Combustion of fuel oil 1 clc () 2 Nflue = 100; // kmol 3 NCO2 = 9;

103

4 5 6 7 8 9 10 11 12 13

14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34

NCO = 2; NO2 = 3; NN2 = 86; NCflue = NCO2 + NCO ; MC = 12; mC = MC * NCflue ; // l e t A kmol a i r s u p p l i e d , t a k i n g N2 b a l a n c e , Nair = NN2 * 100/79; NO2supplied = Nair - NN2 ; // i f CO i n t h e f l u e g a s was t o be c o m p l e t e l y c o n v e r t e d t o CO2 , then , t h e m o l e s o f o x y g e n p r e s e n t i n t h e f l u e g a s would be 3−1 =2kmol Noexcess = NO2 - NCO /2; Pexcess = Noexcess * 100 / ( NO2supplied - Noexcess ); disp ( ”%” , Pexcess , ” ( a ) P e r c e n t a g e e x c e s s a i r = ” ) NwaterO = NO2supplied - NCO2 - NCO /2 - NO2 ; NH2 = NwaterO *2; mH2 = NH2 * 2; xCF = 0.7 R = mC / mH2 ; disp (R , ” ( b ) R a t i o o f c a r b o n t o h y d r o g e n i n t h e f u e l = ”) // l e t x be t h e amount o f m o i s t u r e i n t h e f e e d , n i t i s g i v e n t h a t 70% i s c a r b o n , t h e r e f o r e , // 0 . 7 = 3 . 3 2 / ( 1 + 3 . 3 2 + x ) x = R / xCF - 1 - R ; mH = x * 2.016 / 18.016; mHtotal = mH + mH2 ; Rtotal = mC / mHtotal ; disp ( Rtotal , ” ( c ) R a t i o o f c a r b o n t o t o t a l h y d r o g e n i n the f u e l = ”) ntotal = R + 1 + x ; PH2 = 1*100/ ntotal ; PH2O = x * 100 / ntotal ; disp ( ”%” ,PH2 , ” ( d ) p e r c e n t a g e o f c o m b u s t i b l e h y d r o g e n in the f u e l = ”) disp ( ”%” , PH2O , ” p e r c e n t a g e o f m o i s t u r e i n t h e f u e l = 104

”) 35 nH2Ototal = ( PH2O + PH2 * 18.016 / 2.016) /100; 36 disp ( ” kg ” , nH2Ototal , ” ( e ) The mass o f m o i s t u r e i n t h e f l u e g a s p e r kg o f f u e l b u r n e d = ” )

Scilab code Exa 10.4 Combustion of producer gas 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23

clc () Nflue = 100; // k m o l e s NCO2 = 9.05; NCO = 1.34; NO2 = 9.98; NN2 = 79.63; PCO2F = 9.2; //% ( Feed ) PCOF = 21.3; //% PH2F = 18; //% PCH4F = 2.5; //% PN2F = 49; //% // Taking c a r b o n b a l a n c e , F = ( NCO2 + NCO ) / ( ( PCO2F + PCOF + PCH4F ) /100) ; // N i t r o g e n b a l a n c e g i v e s , Nair = ( NN2 - F * PN2F /(100) ) * 100 / 79; R = Nair / F ; disp (R , ” ( a ) m o l a r R a t i o o f a i r t o f u e l = ” ) Oexcess = NO2 - NCO / 2; Pexcess = Oexcess *100/ ( Nair *21/100 - Oexcess ) ; disp ( ”%” , Pexcess , ” ( b ) P e r c e n t e x c e s s o f a i r = ” ) NN2F = F * PN2F / 100; PN2F = NN2F *100/ NN2 ; disp ( ”%” , PN2F , ” ( c ) P e r c e n t o f n i t r o g e n i n t h e f l u e g a s t h a t came from f u e l = ” )

Scilab code Exa 10.5 Combustion of coal 105

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26

clc () Nflue = 100; // kmole NCO2 = 16.4; NCO = 0.4; NO2 = 2.3; NN2 = 80.9; PCF = 80.5; //% ( Feed ) PO = 5.0; //% PHF = 4.6; //% PN = 1.1; //% Pash = 8.8; //% // Taking Carbon b a l a n c e , W = ( NCO2 + NCO ) *12 / ( PCF / 100) ; mCO2 = NCO2 * 44; mCO = NCO * 32; mO2 = NO2 * 28; mN2 = NN2 * 28.014; mtotal = mCO2 + mCO + mO2 + mN2 ; Mdryflue = mtotal * 100/ W ; disp ( ” kg ” , Mdryflue , ” ( a ) The w e i g h t o f d r y g a s e o u s p r o d u c t s f o r m e d p e r 100 kg o f c o a l f i r e d = ” ) // t a k i n g n i t r o g e n b a l a n c e , x = ( mN2 - W * PN /100) /28.014; Noxygen = x * 21 / 79; Nrequired = W * ( PCF /12 + PHF /(2*2.016) - PO /32) /100; Pexcess = ( Noxygen - Nrequired ) *100/ Nrequired ; disp ( ”%” , Pexcess , ” ( b ) P e r c e n t e x c e s s a i r s u p p l i e d f o r combustion = ”)

Scilab code Exa 10.6 Stoichiometric analysis of combustion of coal 1 clc () 2 mcoal = 100; // kg 3 mC = 63; // kg

106

4 mH = 12; // kg 5 mO = 16; // kg 6 mash =9; // kg 7 mfixC = 39; // kg 8 mH2O = 10; // kg 9 mCvolatile = mC - mfixC ; 10 mHH2O = mH2O *2.016/18.016; // ( mass o f h y d r o g e n i n 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36

moisture ) mHvolatile = mH - mHH2O ; mOH2O = mH2O - mHH2O ; mOvolatile = mO - mOH2O ; mtvolatile = mCvolatile + mHvolatile + mOvolatile ; PC = mCvolatile * 100 / mtvolatile ; PH = mHvolatile * 100 / mtvolatile ; PO = mOvolatile * 100 / mtvolatile ; disp ( ”%” ,PC , ” ( a ) p e r c e n t c a r b o n i n v o l a t i l e m a t t e r = ”) disp ( ”%” ,PH , ” p e r c e n t hydrogen in v o l a t i l e matter = ”) disp ( ”%” ,PO , ” p e r c e n t oxygen i n v o l a t i l e matter = ”) PCflue = 10.8; //% Pvflue = 9.0; //% Pashflue = 80.2; //% // t a k i n g a s h b a l a n c e , Wis t h e w e i g h t o f t h e r e f u s e , W = mash *100 / Pashflue ; mvflue = Pvflue * W /100; mCflue = W * PCflue / 100; Ctflue = mCflue + mvflue * PC / 100; // t o t a l c a r b o n in f l u e Htflue = mvflue * PH / 100; Otflue = mvflue * PO / 100; PCflue = Ctflue *100/ W ; PHflue = Htflue *100/ W ; POflue = Otflue *100/ W ; disp ( ”%” , PCflue , ” ( b ) p e r c e n t Carbon i n r e f u s e = ” ) disp ( ”%” , PHflue , ” p e r c e n t Hydrogen i n r e f u s e = ” ) disp ( ”%” , POflue , ” p e r c e n t Oxygen i n r e f u s e = ” ) 107

37 disp ( ”%” , Pashflue , ” p e r c e n t Ash i n r e f u s e = ” ) 38 Coalburnt = mcoal - W ; 39 NCburnt = ( mC - Ctflue ) /12; 40 NHburnt = ( mH - Htflue ) /2.016; 41 NOburnt = ( mO - Otflue ) /32; 42 PCO2 = 80; // P e r c e n t a g e o f c a r b o n b u r n t 43 NCO2 = PCO2 * NCburnt / 100; 44 NCO = ( 1 - PCO2 /100 ) * NCburnt ; 45 Vair = 1000; //mˆ3 46 Nair = Vair / 22.4143; 47 NN2 = Nair * 79 / 100; 48 NO2 = Nair * 21 / 100; 49 Ocompounds = NCO2 + NCO /2 + NHburnt /2; // Oxygen 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68

p r e s e n t i n CO2 ,CO and H2O // Oxygen b a l a n c e g i v e s f r e e o x y g e n as , Ofree = NO2 + mO /32 - Otflue /32 - Ocompounds ; Ntotal = NN2 + Ofree + NCO2 + NCO ; // d r y b a s i s PCO21 = NCO2 *100/ Ntotal ; PCO1 = NCO * 100/ Ntotal ; PO21 = Ofree * 100/ Ntotal ; PN21 = NN2 * 100/ Ntotal ; disp ( ”%” , PCO21 , ” ( c ) p e r c e n t CO2 i n f l u e = ” ) disp ( ”%” , PCO1 , ” p e r c e n t CO i n f l u e = ” ) disp ( ”%” , PO21 , ” p e r c e n t O2 i n f l u e = ” ) disp ( ”%” , PN21 , ” p e r c e n t N2 i n f l u e = ” ) NOrequired = mC /12 + mH /(2.016*2) - mO /32; Oexcess = NO2 - NOrequired ; Pexcess = Oexcess * 100 / NOrequired ; disp ( ”%” , Pexcess , ” ( d ) P e r c e n t e x c e s s a i r s u p p l i e d = ” ) NH2Oflue = NHburnt ; mH2O = NH2Oflue * 18.016; m = mH2O * 100/ Ntotal ; disp ( ” g w a t e r v a p o u r / 100 kmol d r y f l u e g a s ” ,m , ” ( e ) mass o f w a t e r v a p o u r p e r 100 m o l e s o f d r y f l u e gas = ”)

108

Scilab code Exa 10.7 Orsat analysis 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

clc () Pexcess = 20; //% PSO3 = 5; //% ( P e r c e n t o f s u l p h u r b u r n t t o SO3 ) // S + O2 = SO2 N = 1; // kmol s u l p h u r Orequired = N ; // kmol Osupplied = Orequired * ( 1 + Pexcess /100) ; Nsupplied = Osupplied * 79/21; NSO2 = (1 - PSO3 /100) * N ; NSO3 = PSO3 * N /100; Oconsumed = NSO2 + 3/2 * PSO3 /100; Oremaining = Osupplied - Oconsumed ; Ntotal = NSO2 + NSO3 + Oremaining + Nsupplied ; PSO2 = NSO2 * 100 / Ntotal ; PSO3 = NSO3 * 100 / Ntotal ; PO2 = Oremaining * 100 / Ntotal ; PN2 = Nsupplied * 100 / Ntotal ; disp ( ”%” , PSO2 , ” P e r c e n t SO2 i n b u r n e r g a s = ” ) disp ( ”%” , PSO3 , ” P e r c e n t SO3 i n b u r n e r g a s = ” ) disp ( ”%” ,PO2 , ” P e r c e n t O2 i n b u r n e r g a s = ” ) disp ( ”%” ,PN2 , ” P e r c e n t N2 i n b u r n e r g a s = ” )

Scilab code Exa 10.8 Burning of pyrites 1 clc () 2 Nburner = 100; // kmol 3 NSO2b = 9.5; // kmol 4 NO2b = 7; // kmol 5 NN2 = Nburner - NSO2b - NO2b ; 6 NOsupplied = NN2 * 21 / 79; // Oxygen s u p p l i e d

109

7 // 4 FeS2 + 11O2 = 2 Fe2O3 + 8SO2 8 // 4 FeS2 + 15O2 = 2 Fe2O3 + 8SO3 9 NOtotal = NO2b + NSO2b + NSO2b * 3 / 8; 10 NOunaccounted = NOsupplied - NOtotal ; 11 NSO31 = NOunaccounted * 8 /15; 12 NStotal = NSO2b + NSO31 ; 13 mS = NStotal * 32.064; 14 Pburnt = 50; //% ( p e r c e n t a g e o f p y r i t e s b u r n t ) 15 mFeS2 = mS * 100/ Pburnt ; 16 disp ( ” kg ” , mFeS2 , ” ( a ) T o t a l p y r i t e s b u r n t = ” ) 17 NFeS2 = NStotal / 2; 18 MFeS2 = 119.975; 19 mFeS21 = MFeS2 * NFeS2 ; 20 mgangue = mFeS2 - mFeS21 ; 21 NFe2O3 = NFeS2 * Pburnt / 100; 22 MFe2O3 = 159.694; 23 mFe2O3 = MFe2O3 * NFe2O3 ; 24 PSO3c = 2.5; //% ( p e r c e n t a g e s u l p h u r a s SO3 i n 25 26 27 28 29 30 31 32 33 34 35 36 37 38

cinder ) mc = 100; // kg ( b a s i s ) NSO3 = PSO3c / 32.064; mSO3 = NSO3 * 80.064; mremaining = mc - mSO3 ; // ( Fe2O3 + g a n g u e ) // x be t h e w e i g h t o f t h e c i n d e r x = ( mFe2O3 + mgangue ) *100/ mremaining ; disp ( ” kg ” ,x , ” ( b ) w e i g h t o f c i n d e r p r o d u c e d = ” ) Slost = x * NSO3 / 100; PSlost = Slost *100/ NStotal ; disp ( ”%” , PSlost , ” ( c ) P e r c e n t o f t o t a l S l o s t i n t h e c i n d e r = ”) mSO3c = mSO3 * x / 100; NSO3b = NSO31 - Slost ; P = NSO3b * 100 / NStotal ; disp ( ”%” ,P , ” ( d ) P e r c e n t a g e o f S c h a r g e d t h a t i s p r e s e n t a s SO3 i n t h e b u r n e r g a s = ” )

110

Scilab code Exa 10.9 Production of sulphuric acid 1 clc () 2 Ncgas = 100; // kmol ( b a s i s − SO3 f r e e

converter gas

) 3 NSO2 = 4.5; // kmol 4 NO2 = 7.5; // kmol 5 NN2 = 88.0; // kmol 6 NOsupplied = NN2 * 21/ 79; 7 NOconverter = NO2 + NSO2 ; 8 NOconsumed = NOsupplied - NOconverter ; // ( Oxygen 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28

consumed f o r SO3 ) NSO3c = NOconsumed / 1.5; NStotal = NSO3c + NSO2 ; Nbgas = 100; // kmol ( b a s i s − SO3 f r e e b u r n e r g a s ) NSO21 = 15; //% NO21 = 5; //% NN21 = 80; //% NOburner = NO21 + NSO21 ; NOsupplied1 = NN21 * 21 / 79; NOconsumed1 = NOsupplied1 - NOburner ; // ( Oxygen consumed f o r SO3 ) NSO3b = NOconsumed1 / 1.5; NStotal1 = NSO3b + NSO21 ; mS = 100; // kg ( b a s i s − s u l p h u r c h a r g e d ) Pburned = 95; //% mburned = mS * Pburned / 100; Nburned = mburned / 32.064; // l e t x be t h e SO3 f r e e b u r n e r g a s p r o d u c e d , t h e n sulphur balance gives , x = Nburned * Nbgas / NStotal1 ; NSO2b = NSO21 * x / 100; NO2b = NO21 * x / 100; NN2b = NN21 * x / 100; 111

29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50

Ntotalb = NSO2b + NO2b + NN2b ; NSO3b1 = NSO3b * x / 100; // l e t y be t h e no . o f c o n v e r t e r g a s p r o d u c e d y = Nburned * Ncgas / NStotal ; NSO2c = NSO2 * y / 100; NO2c = NO2 * y / 100; NN2c = NN2 * y / 100; Ntotalc = NSO2c + NO2c + NN2c ; NSO3c1 = NSO3c * y / 100; Nairsec = ( NN2c - NN2b ) * 100 / 79; P = 100; // kPa T = 300; //K V = Nairsec * 22.414 * 101.3 * T / ( P * 273.15) ; disp ( ”mˆ3/ h ” ,V , ” ( a ) The volume o f s e c o n d a r y a i r a t 100 kPa and 300K = ” ) NSabsorbed = 95; //% mSO3abs = NSabsorbed * NSO3c1 * 80.064 / 100; // l e t z be t h e amount o f 98% H2SO4 , t h e r e f o r e , 100% H2SO4 p r o d u c e d = z + mSO3abs // t a k i n g SO3 b a l a n c e z = ( mSO3abs - mSO3abs * 80.064 / 98.08) / ( 80.064 / 98.08 - 0.98 * 80.064/98.08) ; disp ( ” kg ” ,z , ” ( b ) 98% H2SO4 r e q u i r e d p e r h o u r = ” ) w = z + mSO3abs ; disp ( ” kg ” ,w , ” ( c ) 100% H2SO4 p r o d u c e d p e r h o u r = ” )

Scilab code Exa 10.10 Burning of limestone mixed with coke 1 clc () 2 mlime = 3 mcoke = 4 PCaCO3l 5 PMgCO3l 6 NCaCO3l 7 NMgCO3l

5; // kg 1; // kg = 84.5; //% = 11.5; //% = PCaCO3l * mlime / (100.09*100) ; = PMgCO3l * mlime / (84.312*100) ; 112

8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27

mInertsl = mlime * ( 100 - PCaCO3l - PMgCO3l ) / 100; PCc = 76; //% Pashc = 21; //% Pwaterc = 3; //% NCc = mcoke * PCc /(100*12) ; Nwaterc = mcoke * Pwaterc / ( 100 * 18.016 ) ; mash = Pashc * mcoke / 100; //CaCO3 + C + O2 = CaO + 2CO2 //MgCO3 + C + O2 = MgO + 2CO2 PCaCO3conv = 95; // ( P e r c e n t c a l c i n a t i o n o f CaCO3 ) PMgCO3conv = 90; // ( P e r c e n t c a l c i n a t i o n o f MgCO3) NCaO = PCaCO3conv * NCaCO3l / 100; mCaO = NCaO * 56.08; NMgO = PMgCO3conv * NMgCO3l / 100; mMgO = NMgO * 40.312; mCaCO3 = ( NCaCO3l * (1 - PCaCO3conv /100) *100.09) ; mMgCO3 = ( NMgCO3l * (1 - PMgCO3conv /100) *84.312) ; mtotal = mCaO + mMgO + mCaCO3 + mMgCO3 + mInertsl + mash ; PCaO = mCaO * 100 / mtotal ; disp ( ”%” , PCaO , ” The w e i g h t p e r c e n t o f CaO i n t h e product l e a v i n g the k i l n = ”)

Scilab code Exa 10.11 treating limestone with aqueous H2SO4 1 clc () 2 R = 100; // kg ( b a s i s − r e s i d u e ) 3 MCaSO4 = 136.144; 4 MMgSO4 = 120.376; 5 mCaSO4r = 9; // kg 6 mMgSO4r = 5; // kg 7 mH2SO4r = 1.2; // kg 8 minertr = 0.5; // kg 9 mCO2r = 0.2; // kg

113

10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39

mH2O = 84.10; // kg NCaSO4 = mCaSO4r / MCaSO4 ; NMgSO4 = mMgSO4r / MMgSO4 ; //CaCO3 + H2SO4 = CaSO4 + H2O + CO2 //MgSO4 + H2SO4 = MgSO4 + H2O + CO2 mCaCO3 = NCaSO4 * 100.08; mMgCO3 = NMgSO4 * 84.312; mtotallime = minertr + mCaCO3 + mMgCO3 ; PCaCO3 = mCaCO3 * 100/ mtotallime ; PMgCO3 = mMgCO3 *100/ mtotallime ; Pinerts = minertr *100/ mtotallime ; disp ( ”%” , PCaCO3 , ” ( a ) P e r c e n t a g e o f CaCO3 i n l i m e s t o n e = ”) disp ( ”%” , PMgCO3 , ” P e r c e n t a g e o f MgCO3 i n l i m e s t o n e = ”) disp ( ”%” , Pinerts , ” Percentage of i n e r t s in l i m e s t o n e = ”) NH2SO4 = NCaSO4 + NMgSO4 ; mH2SO4 = NH2SO4 * 98.08; Pexcess = mH2SO4r * 100 / ( mH2SO4 ) ; disp ( ”%” , Pexcess , ” ( b ) The p e r c e n t a g e e x c e s s o f a c i d used = ”) macidt = mH2SO4 + mH2SO4r ; Pacidic = 12; //% mwaterin = macidt * (100 - Pacidic ) / Pacidic ; mwaterr = ( NCaSO4 + NMgSO4 ) *18.016; mwatert = mwaterin + mwaterr ; mvaporized = mwatert - mH2O ; m = mvaporized * 100/ mtotallime ; // w a t e r v a p o r i z e d p e r 100 kg o f l i m e s t o n e disp ( ” kg ” ,m , ” ( c ) t h e mass o f w a t e r v a p o r i z e d p e r 100 kg o f l i m e s t o n e = ” ) mCO2pr = ( NCaSO4 + NMgSO4 ) *44; mCO2rel = mCO2pr - mCO2r ; m1 = mCO2rel * 100 / mtotallime ; //CO2 p e r 100 kg o f limestone disp ( ” kg ” ,m1 , ” ( d ) t h e mass o f CO2 p e r 100 kg o f l i m e s t o n e = ”) 114

Scilab code Exa 10.12 Production of TSP 1 clc () 2 macid = 1000; // kg ( b a s i s − d i l u t e p h o s p h o r i c a c i d ) 3 Mphacid = 97.998; 4 P = 1.25; //% ( d i l u t e % ) 5 mphacid = macid * P /100; 6 Nphacid = mphacid / Mphacid ; 7 // 1 mole o f p h o s p h o r i c a c i d − 1 mole o f t r i s o d i u m 8 9 10 11 12 13 14 15 16 17 18 19 20

phosphate NTSP = Nphacid ; MTSP = 380.166; mTSP = NTSP * MTSP ; disp ( ” kg ” , mTSP , ” ( a ) Maximum w e i g h t o f TSP o b t a i n e d = ”) NCO2 = NTSP ; Pwater = 6.27 // kPa // s i n c e g a s i s s a t u r a t e d w i t h w a t e r vapour , v a p o u r pressure = partial pressure Nwater = NCO2 * Pwater / ( 100 - Pwater ) ; Ntotal = Nwater + NCO2 ; P = 100; // kPa T = 310; //K V = Ntotal * 101.3 * T *22.4143 / ( P * 273.15 ) ; disp ( ”mˆ3 ” ,V , ” ( b ) Volume o f CO2 = ” )

Scilab code Exa 10.13 Production of sodium phosphate 1 clc () 2 mTSPd = 1000; // kg ( b a s i s − 20% d i l u t e TSP ) 3 P = 20; //%

115

4 mTSP = mTSPd * P / 100; 5 NTSP = mTSP / 163.974; 6 msodaashd = NTSP * 106; 7 mphacidd = NTSP * 97.998; 8 mNaOHd = NTSP * 40.008; 9 Pphacid = 85; //% ( 8 5% s o l u t i o n p h o s p h o r i c a c i d ) 10 PNaOH = 50; //% ( 5 0% s o l u t i o n NaOH) 11 // l e t x be t h e w a t e r i n s o d a ash , 12 // t a k i n g w a t e r b a l a n c e , 13 x = ( mTSPd - mTSP ) - mNaOHd * PNaOH /(100 - PNaOH ) 14 15 16 17 18 19

mphacidd * (100 - Pphacid ) / Pphacid ; msodaash = msodaashd + x ; C = msodaashd *100 / msodaash ; disp ( ”%” ,C , ” ( a ) C o n c e n t r a t i o n o f s o d a a s h s o l u t i o n = ”) mphacid = mphacidd * 100 / Pphacid ; R = msodaash / mphacid ; disp (R , ” ( b ) Weight r a t i o i n which s o d a a s h and c o m m e r c i a l p h o s p h o r i c a c i d a r e mixed = ” )

Scilab code Exa 10.14 Production of pig iron 1 clc () 2 m = 1000; // kg ( b a s i s − p i g i r o n p r o d u c e d ) 3 // l e t x be t h e i r o n o r e c h a r g e d and y be t h e amount

4 5 6 7 8 9 10 11

o f f l u x added and z be t h e w e i g h t o f s l a g produced PFepg = 95; //% ( Fe% i n p r o d u c t ) PCpg = 4; //% PSipg = 1; //% PFech = 85; //% ( Fe% i n f e e d ) mcoke = 1000; // kg PCcoke = 90; //% PSicoke = 10; //% PSislag = 60; //% 116

12 PSiflux = 5; //% 13 PCaCO3fx = 90; //% 14 PMgCO3fx = 5; //% 15 PCMslag = 40; //% 16 // i r o n b a l a n c e g i v e s , 17 x = PFepg * m *159.694 / ( PFech * 111.694) ; 18 // s i l i c o n b a l a n c e g i v e s , 19 // x ∗ ( 1 0 0 − PFech ) ∗ 2 8 . 0 8 6 / ( 1 0 0 ∗ 6 0 . 0 8 6 )+mcoke ∗ P s i c o k e

20 21 22 23 24

∗ 2 8 . 0 8 6 / ( 1 0 0 ∗ 6 0 . 0 8 6 )+y ∗ P S i f l u x ∗ 2 8 . 0 8 6 / ( 1 0 0 ∗ 6 0 . 0 8 6 ) = 10 + z ∗ P s i s l a g ∗ 2 8 . 0 8 6 / ( 100∗60.086 ) // t a k i n g (CaO + MgO) b a l a n c e // y ∗ ( ( PCaCO3fx ) ∗ 5 6 . 8 8 / ( 1 0 0 ∗ 1 0 0 . 8 8 ) +(PMgCO3fx ∗ 4 0 . 3 1 2 / ( 1 0 0 ∗ 8 4 . 3 1 2 ) )=z ∗ PCMslag / 1 0 0 // s o l v i n g a b o v e 2 e q u a t i o n s , we g e t y = 403.31; disp ( ” kg ” ,y , ” t h e amount o f f l u x r e q u i r e d t o p r o d u c e 1 0 0 0 kg o f p i g i r o n = ” )

Scilab code Exa 10.15 Production of nitric acid 1 clc () 2 N = 100; // mol ( b a s i s − s c r u b b e r ) 3 NNOs = 2.4; // mol 4 NN2s = 92; // mol 5 NO2s = 5.6; // mol 6 PNOs = 20; //% ( P e r c e n t a g e NO l e a v i n g s c r u b b e r ) 7 NNOreac = NNOs * 100 / PNOs ; 8 // l e t x mol o f n i t r o g e be p r o d u c e d i n t h e r e a c t i o n ,

t h e n t h e amount x mol − ( 1 ) 9 // 4NH3 + 5O2 = 4NO 10 // 4NH3 + 3O2 = 2N2 11 // 4 m o l e s o f NO − 5 m o l e s o f O2

o f N2 p r e s e n t i n t h e a i r = NN2s − + 6H2O + 6H2O m o l e s o f O2 , 2 m o l e s o f N2 − 3

117

12 13 14 15 16 17 18 19 20 21 22 23 24

// T o t a l o x y g e n u s e d up , O = NNOreac ∗ 5/4 + x ∗ 3/2 // t o t a l o x y g e n s u p p l i e d , N O t o t a l= (O) + NO2s // N i t r o g e n a s s o c i a t e d w i t h O2 s u p p l i e d NN2 = N O t o t a l ∗79/21 − ( 2 ) // c o m p a r i n g 1 and 2 , x = 2.1835; // 12 m o l e s NO r e q u i r e s 12 m o l e s ammonia , 1 mole N2 r e q u i r e s 2 mole ammonia Nammonia = x *2 + NNOreac ; Oreq = Nammonia * 5 / 4; Osupp = NNOreac * 5/4 + x *3/2 + NO2s ; Pexcess = ( Osupp - Oreq ) *100/ Oreq ; disp ( ”%” , Pexcess , ” ( a ) P e r c e n t a g e e x c e s s o x y g e n = ” ) fr = x * 2 / Nammonia ; disp ( fr , ” F r a c t i o n o f ammonia t a k i n g p a r t i n s i d e r e a c t i o n = ”)

Scilab code Exa 10.16 Material balance in nitric acid production 1 clc () 2 m = 100; // kg ( b a s i s sodium n i t r a t e r e a c t e d ) 3 NNaNO3 = m /85; 4 // 2NaNO3 + H2SO4 = 2HNO3 + Na2SO4 5 mh2so4 = NNaNO3 * 98.08/2; 6 mhno3 = NNaNO3 *63.008; 7 mna2so4 = NNaNO3 * 142.064 /2; 8 Phno3 = 2; //%( p e r c e n t n i t r i c a c i d r e m a i n i n g i n t h e 9 10 11 12 13 14

cake ) mhno3cake = mhno3 * Phno3 / 100; Ph2so4 = 35; //% Pwater = 1.5; //% mtotal = ( mna2so4 + mhno3cake ) *100/(100 - Ph2so4 Pwater ) ; mwater = Pwater * mtotal / 100; mh2so4c = Ph2so4 * mtotal / 100; 118

15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33

Pna2so4 = mna2so4 *100/ mtotal ; Phno3c = mhno3cake * 100 / mtotal ; disp ( ” kg ” , mna2so4 , ” ( a ) Mass o f Na2SO4 i n t h e c a k e = ” ) disp ( ” kg ” , mhno3 , ” Mass o f HNO3 i n t h e c a k e = ” ) disp ( ” kg ” , mwater , ” Mass o f w a t e r i n t h e c a k e = ” ) disp ( ” kg ” , mh2so4c , ” Mass o f H2SO4 i n t h e c a k e = ” ) disp ( ”%” , Pna2so4 , ” P e r c e n t a g e o f Na2SO4 i n t h e c a k e = ”) disp ( ”%” , Phno3c , ” P e r c e n t a g e o f HNO3 i n t h e c a k e = ” ) disp ( ”%” , Pwater , ” P e r c e n t a g e o f w a t e r i n t h e c a k e = ” ) disp ( ”%” , Ph2so4 , ” P e r c e n t a g e o f H2SO4 i n t h e c a k e = ” ) mh2so4req = mh2so4 + mh2so4c ; P = 95; //% ( 9 5% d i l u t e s u l p h u r i c a c i d ) w = mh2so4req * 100 / P ; disp ( mh2so4 ) disp ( ” kg ” ,w , ” ( b ) Weight o f 95% s u l p h u r i c a c i d r e q u i r e d = ”) mnitric = mhno3 - mhno3cake ; disp ( ” kg ” , mnitric , ” ( c ) w e i g h t o f n i t r i c a c i d p r o d u c t obtained = ”) mwaterd = w *(1 - P /100) - mwater ; disp ( ” kg ” , mwaterd , ” ( d ) t h e w a t e r v a p o u r t h a t i s d i s t i l l e d from t h e n i t r e c a k e = ” )

Scilab code Exa 10.17 Electrolysis of brine 1 clc () 2 m = 50; // kg ( b a s i s − mass o f b r i n e c h a r g e d ) 3 // l e t x be t h e amount o f NaCl i n t h e b r i n e 4 Pelect = 50; //% ( e l e c t r o l y z e d ) 5 // 2 NaCl + 2H2O = 2NaOH + Cl2 + H2 6 // amount o f NaCl r e a c t e d =x ∗ P e l e c t / ( 1 0 0 ∗ 5 8 . 4 5 ) kmol=x

119

∗ P e l e c t / 1 0 0 kg ( 1 ) 7 // amount o f w a t e r r e a c t e d = x ∗ P e l e c t ∗ 1 8 . 0 1 6 / ( 100 ∗ 5 8 . 4 5 ) kg ( 2 ) 8 // G a s e s p r o d u c e d , Cl2 = x ∗ P e l e c t / ( 1 0 0 ∗ 5 8 . 4 5 ∗ 2 ) kmol = x ∗ P e l e c t ∗ 71/ ( 1 0 0 ∗ 5 8 . 4 5 ∗ 2 ) kg ( 3 ) 9 //H2 = x ∗ P e l e c t / ( 1 0 0 ∗ 5 8 . 4 5 ∗ 2 ) kmol = x ∗ P e l e c t ∗ 2 . 0 1 6 / ( 1 0 0 ∗ 5 8 . 4 5 ∗ 2 ) kg ( 4 ) 10 Nwater = 0.03; // mol w a t e r v a p o u r / mol o f g a s 11 // w a t e r v a p o u r p r e s e n t = Nwater ∗ 2 ∗ ( Cl2 + H2 ) kmol = Nwater ∗ 2 ∗ ( Cl2 + H2 ) ∗ 1 8 . 0 1 6 kg ( 5 ) 12 //NaoH = x ∗ P e l e c t ∗ 4 0 . 0 0 8 / ( 1 0 0 ∗ 5 8 . 4 5 ) kg ( 6 ) 13 // w a t e r = w a t e r i n b r i n e − w a t e r r e a c t e d − w a t e r present in gas ( 7 ) 14 //= (m − P e l e c t / 1 0 0 ) − w a t e r r e a c t e d ( 2 ) − w a t e r present in the gas ( 5 ) 15 // T o t a l w e i g h t o f s o l u t i o n = NaCl ( 1 ) + NaOH ( 6 ) + Water ( 7 ) 16 // s i n c e NaOH i s 10 p e r c e n t o f t h e t o t a l w e i g h t , we have NaOH = 0 . 1 ∗ t o t a l w e i g h t , from t h e s e we g e t , 17 x = 0.1 * 50 / (0.1* 0.3165 + 0.3422 ) ; 18 NaOH = x * Pelect * 40.008/ (100 * 58.45 ) ; 19 NaCl = x * Pelect / 100; 20 water = 34.5032; // kg 21 Pevap = 50; //NaOh p e r c e n t a g e i n s o l u t i o n l e a v i n g evaporator 22 // t a k i n g NaOH b a l a n c e 23 mevap = NaOH * 100 / Pevap ; 24 disp ( ” kg ” , mevap , ” ( a ) amount o f 50% NaOH s o l u t i o n produced = ”) 25 Cl2 = x * Pelect *71/ (100 * 58.45 * 2 ) ; // kg 26 H2 = x * Pelect *2.016/ (100 * 58.45 * 2 ) ; // kg 120

disp ( ” kg ” ,Cl2 , ” ( b ) C h l o r i n e p r o d u c e d = ” ) disp ( ” kg ” ,H2 , ” Hydrogen p r o d u c e d = ” ) Pleav = 1.5; //% NaCl l e a v i n g t h e e v a p o r a t o r NaClleav = mevap * Pleav / 100; mcrystal = NaCl - NaClleav ; disp ( ” kg /h ” , mcrystal , ” ( c ) Amount o f NaCl c r y s t a l l i z e d = ”) 33 mwaterleav = mevap - NaOH - NaClleav ; 34 Mwaterevap = water - mwaterleav ; 35 disp ( ” kg ” , Mwaterevap , ” ( d ) Weight o f w a t e r e v a p o r a t e d = ”) 27 28 29 30 31 32

Scilab code Exa 10.18 Preparation of Formaldehyde 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

clc () m = 100; // mol ( b a s i s r e a c t o r e e x i t g a s ) //CH3OH + O2 = HCOOH + H2O //CH3OH + O2 / 2 = HCHO + H2O Nn2 = 64.49; // mol No2 = 13.88; // mol Nh2o = 5.31; // mol Nch3oh = 11.02; // mol Nhcho = 4.08; // mol Nhcooh = 1.22; // mol // x be t h e m o l e s o f m e t h a n o l r e a c t e d , t a k i n g C b a l a n c e , we g e t , x = Nch3oh + Nhcho + Nhcooh ; Pconv = Nhcho * 100 / x ; disp ( ”%” , Pconv , ” ( a ) P e r c e n t c o n v e r s i o n o f formaldehyde = ”) Nair = Nn2 * 100 / 79; R = Nair / x ; disp (R , ” ( b ) R a t i o o f a i r t o m e t h a n o l i n t h e f e e d = ” )

121

Scilab code Exa 10.19 Recycle operation reactor and separator 1 clc () 2 NA = 100; // mol ( b a s i − 100 mol A i n t h e

fresh feed

) 3 Pconv = 95; //% 4 NApro = NA * (100 - Pconv ) /100; 5 //A = 2B + C 6 NB = NA * Pconv * 2 / 100; 7 NC = NA * Pconv /100; 8 PAent = 0.5; //% 9 NAent = NApro * 100 / PAent ; 10 PBrec = 1; //% 11 NBent = NB * 100 / (100 - PBrec ) ; 12 m = ( NAent - NApro + NA ) ; 13 conv = (( NAent - NApro + NA ) - NAent ) *100/( NAent -

NApro + NA ) ; 14 disp ( ”%” , conv , ” ( a ) s i n g l e p a s s c o n v e r i o n = ” ) 15 Nrecycled = ( NAent - NApro ) + ( NBent - NB ) ; 16 R = Nrecycled / NA ; 17 disp (R , ” ( b ) r e c y c l e r a t i o = ” )

Scilab code Exa 10.20 Conversion of sugar to glucose and fructose 1 clc () 2 m = 100; // kg ( b a s i s − s u c r o s e 3

feed ) //R − r e c y c l e r e a c t o r e x i t , f r a c t i o n o f s u c r o s e and y of i n v e r s i o n sugar in the combined s t r e a m fraction 0.04 122

s o l u t i o n as f r e s h

l e t x be t h e w e i g h t be t h e w e i g h t f r a c t o n r e c y c l e stream , f o r of Glucose + f r u c t o s e =

4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22

// z be t h e w e i g h t f r a c t i o n o f s u c r o s e i n t h e combined s t r e a m e n t e r i n g t h e r e a c t o r Psfeed = 25; //% p e r c e n t s u c r o s e i n f r e s h f e e d // s u c r o s e b a l a n c e g i v e s , 25 + R∗ x = (100+R) ∗ z (A) // G l u c o s e + f r u c t o s e b a l a n c e , R ∗ y = ( 1 0 0 + R ) ∗0.04 (B) Sucrosecon = 71.7; //% s u c r o s e consumed // s u c r o s e b a l a n c e a r o u n d t h e r e a c t o r , ( 1 0 0 +R) z =0 . 71 7 ∗ ( 10 0+R) z +(100+R) x (C) // From (C) , x = 0 . 2 8 3 ∗ z (D) // Amount c o n v e r t e d t o G l u c o s e + f r u c t o s e = 0 . 7 1 7 ( 100 + R ) ∗ z // = 0 . 7 1 7 ( 100 + R ) ∗ z ∗ 3 6 0 . 1 9 2 / 3 4 2 . 1 7 6 kg // G l u c o s e and f r u c t o s e b a l a n c e a r o u n d t h e r e a c t o r , // (100+R) ∗ 0 . 0 4 + 0 . 7 1 7 ( 1 0 0 +R) ∗ z ∗ 3 6 0 . 1 9 2 / 3 4 2 . 1 7 6 = (100+R) ∗ y (E) // S o l v i n g (E) , y = 0 . 0 4 + 0 . 7 5 4 8 ∗ z (F) // S o l v i n g , (A) , (B) , (C) and ( F ) x = 0.06; y = 0.2; z = 0.212; R = 25; disp ( ” kg ” ,R , ” ( a ) R e c y c l e f l o w = ” ) disp ( ”%” ,y *100 , ” ( b ) Combined c o n c e n t r a t i o n o f G l u c o s e and F r u c t o s e i n t h e r e c y c l e s t r e a m = ” )

Scilab code Exa 10.21 Purging operation 1 clc () 2 N = 1; // mol ( b a s i s − combined f e e d ) 3 //F − m o l e s o f f r e s h f e e d 4 Pinert = 0.5; //%

123

5 Pconv = 60; //% 6 P1inert = 2; //% 7 NA1 = N * ( 1 - P1inert /100 ) ; 8 NA2 = NA1 * ( 1 - Pconv / 100 ) ; 9 NB2 = NA1 - NA2 ; 10 N1inert = N * P1inert / 100; 11 N2inert = N1inert ; 12 // L e t R be t h e m o l e s r e c y c l e d and P be t h e m o l e s

purged 13 //W = R + P 14 W = NA2 + N2inert ; // 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29

(A) PWinert = N2inert * 100/ ( NA2 + N2inert ) ; // component A b a l a n c e , A f r e s h f e e d = A p u r g e s t r e a m + A r e c y c l e stream //F ∗ 0 . 9 = P ∗ 0 . 9 5 1 5 + 0 . 5 8 8 (B) // i n e r t b a l a n c e a t t h e p o i n t where f r e s h f e e d i s mixed w i t h t h e r e c y c l e , //F ∗ 0 . 0 0 5 + R∗ 0 . 0 4 8 5 = 1∗ 0 . 0 2 (C) // S o l v i n g (A) , ( B) and (C) F = 0.6552; // mol P = 0.0671; // mol R = 0.3448; // mol disp ( ” mol ” ,R , ” ( a ) m o l e s o f r e c y c l e s t r e a m = ” ) disp ( ” mol ” ,P , ” ( b ) m o l e s o f p u r g e s t r e a m = ” ) NAconv = NA1 - NA2 ; NAf = F * (1 - Pinert / 100) ; Conv = NAconv *100/ NAf ; disp ( ”%” , Conv , ” ( c ) O v e r a l l c o n v e r s i o n = ” )

Scilab code Exa 10.22 Purging operation for production of methanol 1 clc ()

124

2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26

N = 100; // m o l e s ( B a s i s − F r e s h f e e d ) Pconv = 20; //% xco = 0.33; xh2 = 0.665; xch4 = 0.005; //R − r e c y c l e s t r e a m , P − p u r g e s t r e a m // x − mole f r a c t i o n o f CO i n r e c y c l e s t r e a m , xch4r = 0.03; //CO = x , H2 = 1 − x c h 4 r − CO = 0.97 − x ; // methane b a l a n c e o v e r t h e e n t i r e system , P = xch4 * N / xch4r ; // t a k i n g c a r o n b a l a n c e , 3 3 . 5 = M + P ( 0 . 0 3 + x ) // Hydrogen b a l a n c e , 6 6 . 5 + 2 ∗ 0 . 5 = 2M + P ( 2 ∗ 0 . 0 3 + 0.97 − x) // s u b s t i t u t i n g P , M + 1 6 . 6 7 x = 3 3 . 0 and 2M − 1 6 . 6 7 x = 50.33 M = (33.0 + 50.33) /3; x = (( xco + xch4 ) * N - M ) / P - xch4r ; // m e t h a n o l b a l a n c e , ( x c o ∗N+Rx ) ∗ Poncv / 1 0 0 = M R = ( M *100 / Pconv - ( xco * N ) ) / x ; disp ( ” mol ” ,R , ” ( a ) m o l e s o f r e c y c l e s t r e a m = ” ) disp ( ” mol ” ,P , ” ( b ) m o l e s o f p u r g e s t r e a m = ” ) H2 = 1 - xch4r - x ; disp ( ”%” , xch4r *100 , ” ( c )CH4 i n p u r g e s t r e a m = ” ) disp ( ”%” ,x *100 , ”CO i n p u r g e s t r e a m = ” ) disp ( ”%” , H2 *100 , ” h y d r o g e n i n p u r g e s t r e a m = ” ) disp ( ” mol ” ,M , ” ( d ) Methanol p r o d u c e d = ” )

125

Chapter 11 Energy Balance Thermophysics

Scilab code Exa 11.1 Power calculation 1 2 3 4 5 6 7 8 9 10

clc () m = 75; // kg g = 9.81 //mˆ2/ s d = 10; //m t = 2.5*60; // s f = m*g; w = f * d; P = w / t; disp ( ”Nm” ,w , ” The work done = ” ) disp ( ”W” ,P , ” Power r e q u i r e d = ” )

Scilab code Exa 11.2 Kinetic energy calculation 1 2 3 4 5

clc () PE = 1.5*10^3; // J m = 10; // kg g = 9.81; //m/ s ˆ2 v = 50; //m/ s 126

6 //PE = mgz 7 z = PE / ( m * g ) ; 8 KE = m * ( v ^2) / 2; 9 disp ( ”m” ,z , ” H e i g h t o f t h e body from t h e g r o u n d = ” ) 10 disp ( ” kJ ” , KE /1000 , ” K i n e t i c e n e r g y o f t h e body = ” )

Scilab code Exa 11.3 Work done calculation for a gas confined in a cylinder 1 2 3 4 5 6 7 8 9 10 11 12 13

clc () d = 100 /1000; //m m = 50; // kg P = 1.01325*10^5; // Pa A = %pi * ( d ^2) /4; Fatm = P * A ; Fwt = m * g ; Ftotal = Fatm + Fwt ; P = Ftotal / A ; disp ( ” b a r ” ,P /10^5 , ” ( a ) P r e s s u r e o f t h e g a s = ” ) z = 500/1000; //m w = Ftotal * z ; disp ( ” J ” ,w , ” ( b ) Work done by t h e g a s = ” )

Scilab code Exa 11.4 Power requirement of the pump 1 2 3 4 5 6 7 8

clc () Sgr = 0.879; F = 5; //mˆ3/ h D = Sgr * 1000; m = F * D /3600; // kg / s P = 3500; // kPa W = P * m * 1000/ D ; disp ( ”W” ,W , ” Power r e q u i r e m e n t f o r t h e pump = ” ) 127

Scilab code Exa 11.5 Specific enthalpy of the fluid in the tank 1 2 3 4 5 6 7 8 9

clc () d = 3; //m m = 12500; // kg P = 7000; // kPa U = 5.3*10^6; // kJ Vtank = 4* %pi *(( d /2) ^3) / 3; Vliq = Vtank / 2; H = U + P * Vliq ; disp ( ” kJ / kg ” ,H /m , ” S p e c i f i c e n t h a l p y o f t h e f l u i d i n the tank = ”)

Scilab code Exa 11.6 internal energy and enthalpy change calculation 1 2 3 4 5 6 7 8 9 10

clc () P = 101.3; // kPa SVl = 1.04 * 10^ -3; //mˆ3/ kmol SVg = 1.675; //mˆ3/ kmol Q = 1030; // kJ W = P * 10^3 * ( SVg - SVl ) /1000; U = Q - W; H = U + P * 10^3 * ( SVg - SVl ) /1000; disp ( ” kJ / kmol ” ,U , ” Change i n i n t e r n a l e n e r g y = ” ) disp ( ” kJ / kmol ” ,H , ” Change i n e n t h a l p y = ” )

Scilab code Exa 11.7 change in internal energy 1 clc ()

128

2 // work i s done on t h e system , hence , W i s n e g a t i v e 3 W = - 2 * 745.7; // J / s 4 // h e a t i s t r a n s f e r r e s t o t h e s u r r o u n d i n g , hence ,

heat t r a n s f e r r e d i s negative , 5 Q = -3000; // kJ /h 6 U = Q *1000/3600 - W ; 7 disp ( ” J / s ” ,U , ” Change i n

i n t e r n a l energy = ”)

Scilab code Exa 11.8 reaction of iron with HCl 1 2 3 4 5 6 7 8 9 10 11 12

clc () // Fe ( s ) + 2HCl ( aq ) = FeCl2 ( aq ) + H2 ( g ) MFe = 55.847; m = 1; // kg Nfe = m * 10^3/ MFe ; Nh2 = Nfe ; // ( s i n c e 1 mole o f Fe p r o d u c e s 1 mole o f H2 ) T = 300; //K R = 8.314; // t h e c h a n g e i n volume i s e q u a l t o t h e volume o c c u p i e d by h y d r o g e n p r o d u c e d PV = Nh2 * R * T ; W = PV ; disp ( ” kJ ” ,W , ”Work done = ” )

Scilab code Exa 11.9 Thermic fluid 1 2 3 4 5 6

clc () //Cp = 1 . 4 3 6 + 2 . 1 8 ∗ 1 0 ˆ − 3 ∗T ; m = 1000/3600; // kg / s T1 = 380; //K T2 = 550; //K x = integrate ( ’ 1 . 4 3 6 + 2 . 1 8 ∗ 1 0 ˆ − 3 ∗T ’ , ’T ’ ,T1 , T2 ) ; 129

7 Q = m*x; 8 disp ( ”kW” ,Q , ” Heat l o a d on t h e h e a t e r = ” )

Scilab code Exa 11.10 Heat capacity 1 clc () 2 //Cp = 2 6 . 5 4 + 4 2 . 4 5 4 ∗ 1 0 ˆ − 3 ∗ T − 1 4 . 2 9 8 ∗ 10ˆ −6 ∗ T 3 4 5 6 7 8 9 10 11 12 13 14 15

ˆ2; T1 = 300; //K T2 = 1000; //K m = 1; // kg N = m /44; // kmol x = integrate ( ’ 2 6 . 5 4 + 4 2 . 4 5 4 ∗ 1 0 ˆ − 3 ∗ T − 1 4 . 2 9 8 ∗ 10ˆ −6 ∗ Tˆ2 ’ , ’T ’ ,T1 , T2 ) ; Q = N*x; disp ( ” kJ ” ,Q , ” ( a ) Heat r e q u i r e d = ” ) // f o r t e m p e r a t u r e i n t d e g r e e c e l s i u s //Cp = 2 6 . 5 4 + 4 2 . 4 5 4 ∗ 1 0 ˆ − 3 ∗ ( t + 2 7 3 . 1 5 ) − 1 4 . 2 9 8 ∗ 10ˆ −6 ∗ ( t + 2 7 3 . 1 5 ) ˆ2 //Cp = 3 7 . 0 6 8 + 3 4 . 6 4 3 ∗ 10ˆ −3∗ t − 1 4 . 2 9 8 ∗ 10ˆ −6 ∗ t ˆ2 ( kJ / kmolC ) //Cp = 8 . 8 5 4 + 8 . 2 7 4 ∗ 1 0 ˆ − 3 ∗ t −3.415∗10ˆ −6∗ t ˆ2 ( K c a l / kmolC ) // For d e g r e e F e h r e n e i t s c a l e , r e p l a c e t by ( t 1 − 3 2 ) / 1 8 , we g e t //Cp = 8 . 7 0 5 8 + 4 . 6 6 4 2 ∗ 10ˆ −3 ∗ t 1 − 1 . 0 5 4 0 ∗ 10ˆ −6 ∗ t 1 ˆ2 ( Btu / lbmolF )

Scilab code Exa 11.11 Enthalpy change when chlorine gas is heated 1 clc ()

130

Figure 11.1: Enthalpy change when chlorine gas is heated

131

2 T = [273 373 473 573 673 773 873 973 1073 1173

1273]; 3 Cp = [33.6 35.1 36 36.6 37 37.3 37.5 37.6 37.7 37.8 4 5 6 7 8 9

37.9]; plot2d (T , Cp , rect =[273 ,33 ,1273 ,38]) xtitle ( ” T v s Cp” ,” T e m p e r a t u r e K” ,” Heat C a p a c i t y , kJ / kmolK ” ) // a t c o n s t a n t P r e s s u r e , H = i n t e g r a t i o n ( Cp , T , T1 , T2 ) // Area u n d e r t h e c u r v e form t h e g r a p h , i s o b t a i n e d a s Area = 3 6 8 2 8 H = 36828; // kJ / kmol disp ( ” kJ / kmol ” ,H , ” E n t h a l p y c h a n g e = ” )

Scilab code Exa 11.12 Molal heat capacity 1 clc () 2 //Cp = 2 6 . 5 8 6 + 7 . 5 8 2 ∗ 10 ˆ−3 ∗ T − 1 . 1 2 ∗ 10ˆ −6 ∗ 3 4 5 6 7 8 9 10 11 12 13 14 15 16

Tˆ2 T1 = 500; //K T2 = 1000; //K x = integrate ( ’ 2 6 . 5 8 6 + 7 . 5 8 2 ∗ 10ˆ −3 ∗ T − 1 . 1 2 ∗ 10ˆ −6 ∗ Tˆ2 ’ , ’T ’ ,T1 , T2 ) ; Cpm = 1 * x / ( T2 - T1 ) ; disp ( ” kJ / kmolK ” ,Cpm , ” ( a ) Mean m o l a l h e a t c a p a c i t y = ” ) V = 500; //mˆ 3 ; N = V / 22.4143; Q = N * Cpm * ( T2 - T1 ) ; disp ( ” kJ / h” ,Q , ” ( b ) Heat t o be s u p p l i e d = ” ) T3 = 1500; //K Q1 = Cpm * ( T3 - T1 ) ; y = integrate ( ’ 2 6 . 5 8 6 + 7 . 5 8 2 ∗ 10 ˆ−3 ∗ T − 1 . 1 2 ∗ 10ˆ −6 ∗ Tˆ2 ’ , ’T ’ ,T1 , T3 ) ; Q2 = y ; disp ( Q2 ) 132

17 18

Perror = ( Q2 - Q1 ) * 100 / Q2 ; disp ( ”%” , Perror , ” ( c ) P e r c e n t e r r o r = ” )

Scilab code Exa 11.13 Enthalpy change of a gas 1 2 3 4 5 6 7 8

clc () T1 = 1500; //K Tr = 273; //K T2 = 400; //K Cpm1 = 50; // kJ / kmol Cpm2 = 35; // kJ / mol H = Cpm1 * ( T1 - Tr ) - Cpm2 * ( T2 - Tr ) ; disp ( ” kJ / kmol ” ,H , ” E n t h a l p y c h a n g e = ” )

Scilab code Exa 11.14 Combustion of solid waste 1 clc () 2 //CO, 2 6 . 5 8 6 + 7 . 5 8 2 ∗ 1 0 ˆ − 3 ∗T − 1 . 1 2 ∗ 1 0 ˆ − 6 ∗Tˆ2 3 //CO2 , 2 6 . 5 4 0 + 4 2 . 4 5 4 ∗ 1 0 ˆ − 3 ∗T − 1 4 . 2 9 8 ∗ 1 0 ˆ − 6 ∗Tˆ2 4 //O2 , 2 5 . 7 4 + 1 2 . 9 8 7 ∗ 1 0 ˆ − 3 ∗T − 3 . 8 6 4 ∗ 1 0 ˆ − 6 ∗Tˆ2 5 //N2 , 2 7 . 0 3 + 5 . 8 1 5 ∗ 1 0 ˆ − 3 ∗T − 0 . 2 8 9 ∗ 1 0 ˆ − 6 ∗Tˆ2 6 // Cpmix = summation ( y i ∗ Cpi ) = summation ( y i ∗ a i + 7 8 9 10 11 12 13 14

y i ∗ b i ∗T + y i ∗ c i ∗Tˆ 2 ) xco2 = 0.09; xco = 0.02; xo2 = 0.07; xn2 = 0.82; T1 = 600; //K T2 = 375; //K sumai = xco * 26.586 + xco2 * 26.540 + xo2 * 25.74 + xn2 *27.03; sumbi = xco * 7.582*10^ -3 + xco2 *2.454*10^ -3+ xo2 *12.987*10^ -3 + xn2 *5.815*10^ -3; 133

15

sumci = -( xco * 1.12*10^ -6 + xco2 *14.298*10^ -6+ xo2 *3.864*10^ -6+ xn2 *0.289*10^ -6) ; 16 H = integrate ( ’ sumai+sumbi ∗T+s u m c i ∗Tˆ2 ’ , ’T ’ ,T1 , T2 ) ; 17 disp ( ” kJ / kmol ” ,H , ” E n t h a l p y c h a n g e = ” )

Scilab code Exa 11.15 Heat capacity calculation for Na2SO4 10H2O 1 2 3 4 5 6 7 8 9

clc () Hna = 26.04; // J / g−atomK Hs = 22.6; // J / g−atomK Ho = 16.8; // J / g−atomK Hh = 9.6; // J / g−atomK Hna2so410h2o = 2* Hna + Hs + 14* Ho + 20* Hh ; Hexp = 592.2; // J /molK Deviation = ( Hexp - Hna2so410h2o ) *100/ Hexp ; disp ( ”%” , Deviation , ” D e v i a t i o n i n h e a t c a p a c i t y = ” )

Scilab code Exa 11.16 Heat of vaporization calculation 1 clc () 2 P1 = 75; // kPa 3 T1 = 573; //K 4 Tvap = 365; //K 5 Tbasis = 273; //K 6 // S i n c e , t h e b o i l i n g 7 8 9 10 11

p o i n t o f w a t e r a t 75 kPa i s 375K , t h e v a p o u r a t 573K i s s u p e r h e a t e d ; H1 = 3075; // kJ / kg Cliq = 4.2; // kJ /kgK Cvap = 1.97; // kJ / kg /K m = 1; // kg // l e t assume c o n v e r t i n g l i q . w a t e r i n t o s u p e r h e a t e d stream occurs in 3 steps , 134

12

13 14

15

16 17 18 19

// s t e p 1 − w a t e r i s h e a t e d from 273K t o 365 K a t c o n s t a n t p r e s s u r e , enthalpy change i s the heat r e q u i r e d to change the temperature , Hc1 = m * Cliq * ( Tvap - Tbasis ) ; // s t e p 2 − t h e l i q i s v a p u r i z e d a t c o n s t a n t p r e s s u r e and c o n s t a n t t e m p e r a t u r e , e n t h a l p y c h a n g e i s e q u a l t o t h e h e a t o f v a p o u r i s a t i o n , s a y Hc2 // s t e p 3 − t h e s a t u r a t e d v a p o u r a t 365K i s h e a t e d t o 573K a t c o n s t a n t p r e s s u r e , t h e e n t h a l p y c h a n g e i s the heat r e q u i r e d to r a i s e the temperature Hc3 = m * Cvap *( T1 - Tvap ) ; // t o t a l e n t h a l p y = 3 0 7 5 = Hc1 + Hc2 + Hc3 , t h e r e f o r e Hc2 = H1 - Hc1 - Hc3 ; disp ( ” kJ / kg ” ,Hc2 , ” Heat o f v a p o u r i s a t i o n = ” )

Scilab code Exa 11.17 Heat requirement 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

clc () T1 = 250; //K T = 273.15; //K T2 = 400; //K Cice = 2.037; // kJ /kgK T3 = 373.15; //K Cliq = 75.726; // kJ / kmolK //Cp = 3 0 . 4 7 5 + 9 . 6 5 2 ∗ 1 0 ˆ − 3 ∗T + 1 . 1 8 9 ∗ 1 0 ˆ − 6 ∗Tˆ2 Hfusion = 6012; // kJ / kmol Hvap = 40608; // kJ / kmol // 1 − Heat f o r r a i s i n g t h e t e m p e r a t u r e o f i c e , H1 H1 = Cice * ( T - T1 ) ; // 2 − L a t e n t h e a t o f f u s i o n o f i c e , Hf Hf = Hfusion / 18.016; // kJ // 3 − S e n s i b l e h e a t o f r a i s i n g t h e t e m p e r a t u r e o f water , H2 16 H2 = Cliq * ( T3 - T ) /18.016; 17 // 4 − L a t e n t h e a t o f v a p o r i z a t i o n o f water , Hv 135

18 Hv = Hvap / 18.016; 19 // 5 − S e n s i b l e h e a t o f

r a i s i n g the temperature of w a t e r vapou , H3 20 H3 = ( integrate ( ’ 3 0 . 4 7 5 + 9 . 6 5 2 ∗ 1 0 ˆ − 3 ∗T + 1 . 1 8 9 ∗ 1 0 ˆ − 6 ∗Tˆ2 ’ , ’T ’ ,T3 , T2 ) ) /18.016; 21 Q = H1 + H2 + H3 + Hf + Hv ; 22 disp ( ” kJ ” ,Q , ” Heat r e q u i r e d = ” )

Scilab code Exa 11.18 Equilibrium temperature of mixture 1 2 3 4 5 6 7 8

9 10 11 12

13 14 15

clc () //Cp = 0 . 1 6 + 4 . 7 8 ∗ (10ˆ −3) ∗ T ( o r g a n i c l i q u i d ) //Cp = 0 . 7 9 3 5 + 1 . 2 9 8 ∗ (10ˆ −4) ∗ T ( CCL4 ) Tb = 349.9; //K Hv = 195; // kJ / kg Cp = 0.4693; // kJ /kgK // L e t T be t h e f i n a l t e m p e r a t u r e // i n t e g r a t i o n (T − 6 5 0 ) ( 0 . 1 6 + 4 . 7 8 ∗ (10ˆ −3) ∗ T) d t = i n t e g r a t i o n ( 2 9 5 − T) ( 0 . 7 9 3 5 + 1 . 2 9 8 ∗ (10ˆ −4) ∗ T) d t // t h e a b o v e e q u a t i o n y i e l d s , 2 . 4 5 4 9 ∗ ( 1 0 ˆ − 3 ) ∗Tˆ2 + 0 . 9 5 3 5 ∗T − 1 3 5 3 . 5 1 = 0 , from t h i s we g e t T = 573.3; //K // s i n c e t h i s t e m p e r a t u r e i s a b o v e b o i l i n g p o i n t o f CCl4 , // h e a t b a l a n c e i s , i n t e g r a t i o n (T − 6 5 0 ) ( 0 . 1 6 + 4 . 7 8 ∗ (10ˆ −3) ∗ T) d t = i n t e g r a t i o n ( 2 9 5 − 3 4 9 . 9 ) ( 0 . 7 9 3 5 + 1 . 2 9 8 ∗ (10ˆ −4) ∗ T) d t + Hv + i n t e g r a t i o n ( 3 4 9 . 9 − T) ∗ 0 . 4 6 9 3 ∗ dT // s o l v i n g a b o v e e q u a t i o n , we g e t , T1 = 540.1; //K disp ( ”K” ,T1 , ” e q u i l i b r i u m t e m p e r a t u r e o f t h e m i x t u r e = ”)

136

Scilab code Exa 11.19 Estimation of mean heat of vaporisation 1 2 3 4 5 6 7 8 9 10

clc () T1 = 363; //K T2 = 373; //K P1s = 70.11; // kPa P2s = 101.3; // kPa R = 8.314; // kJ / kmolK // l n ( P2s / P1s ) = Hv / R ∗ ( 1 / T1 − 1/T2 ) ; Hv = ( log ( P2s / P1s ) * R ) /(1/ T1 - 1/ T2 ) ; Hv1 = Hv / (18) ; disp ( ” kJ / kg ” ,Hv1 , ”Mean h e a t o f v a p o r i z a t i o n = ” )

Scilab code Exa 11.20 Heat of vaporization of methyl chloride 1 clc () 2 T = 273.15 - 30; //K 3 // l n P s = 1 4 . 2 4 1 0 − 2 1 3 7 . 7 2 / (T− 2 6 . 7 2 ) 4 // d l n P s /dT = Hv / RT2 5 Hv = 2137.72 * R * T ^2 / ( T - 26.72 ) ^2; 6 disp ( ” kJ / kmol ” ,Hv , ” Heat o f v a p o r i z a t i o n = ” )

Scilab code Exa 11.21 Watson equation 1 2 3 4 5

clc () Hv1 = 2256; // kJ / kg T1 = 373; //K T2 = 473; //K Tc = 647; //K 137

6 Tr1 = T1 / Tc ; 7 Tr2 = T2 / Tc ; 8 // Hv2 / Hv1 = ((1 − Tr2 ) /(1 − Tr1 ) ) ˆ 0 . 3 8 9 Hv2 = Hv1 *(((1 - Tr2 ) /(1 - Tr1 ) ) ^0.38) ; 10 disp ( ” kJ / kg ” ,Hv2 , ” L a t e n t h e a t o f v a p o r i z a t i o n

of

w a t e r a t 473K = ” )

Scilab code Exa 11.22 Kistyakowsky equation 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

clc () //Cp = a + b∗T T1 = 293.15; //K Cp1 = 131.05; // J /molK T2 = 323; //K Cp2 = 138.04; // J /molK // a + 293 ∗ b = 1 3 1 . 0 5 // a + 323 ∗ b = 1 3 8 . 0 4 b = ( Cp1 - Cp2 ) /( T1 - T2 ) ; a = Cp1 - b * T1 ; //Cp = 6 2 . 7 8 1 + 0 . 2 3 3 ∗T // Hvb / Tb = 3 6 . 6 3 + 8 . 3 1 lnTb Tb = 273.15 + 80.1; //K Hvb = (36.63 + 8.31* log ( Tb ) ) * Tb ; m = 100; // kg H = m *(10^3) * ( integrate ( ’ 6 2 . 7 8 1 + 0 . 2 3 3 ∗T ’ , ’T ’ ,T1 , Tb ) ) /78.048 + m *(10^3) * Hvb /78.048; 17 disp ( ” J ” ,H , ” Heat r e q u i r e d = ” )

Scilab code Exa 11.23 Quality of steam 1 clc () 2 P = 10; // kPa 3 T1 = 323.15; //K

138

4 5 6 7 8 9 10 11 12 13 14 15 16 17

T2 = 373.15; //K T = 358.15; //K H1 = 2592.6; // kJ / kg H2 = 2687.5; // kJ / kg //H by i n t e r p o l a t i o n , H = H1 + (( H2 - H1 ) /( T2 - T1 ) ) *( T - T1 ) ; Hl = 697.061; // kJ / kg Hg = 2762; // kJ / kg //H = x ∗ Hl + ( 1 − x ) ∗ Hg x = ( H - Hg ) /( Hl - Hg ) ; Pmois = x *100; Psteam = ( 1 - x ) *100; disp ( ”%” , Pmois , ” P e r c e n t a g e o f m o i s t u r e = ” ) disp ( ”%” , Psteam , ” P e r c e n t a g e o f d r y s a t u r a t e d steam = ”)

Scilab code Exa 11.24 Heat calculation 1 2 3 4 5 6 7 8 9 10 11 12 13 14

clc () P = 3500; // kPa T = 673.15; //K SV = 0.08453; //mˆ3/ kg Vcondensed = 1/2; m = 100; // kg V = m * SV / ( m /2) ; //m∗ ( Vl+Vg ) ∗ Vcondensed = m ∗ SV // But Vl i s n e g l i g i b l e , Vg = m * SV / ( m * Vcondensed ) ; // u s i n g steam t a b l e T1 = 459.5; //K P1 = 1158; // kPa // i n t e r n a l e n e r g y o f s u p e r h e a t e d steam from steam table 15 I = 2928.4; // kJ / kg 16 U1 = m * I ; 139

17 18 19 20 21

Ul = 790; // kJ / kg Ug = 2585.9; // kJ / kg U2 = m * Vcondensed * Ul + m *(1 - Vcondensed ) * Ug ; Q = U2 - U1 ; disp ( ” kJ ” ,Q , ” The amount o f h e a t removed f r o m t he system = ”)

Scilab code Exa 11.25 Enthalpy balance for evaporation process 1 clc () 2 m = 1000; // kg / h ( b a s i s mass o f 10% NaOH s o l u t i o n ) 3 Pfeed = 10; //% 4 Ppro = 50; // ( P e r c e n t a g e NaOH i n p r o d u c t ) 5 // Taking NaOH b a l a n c e , P b e i n g t h e w e i g h t o f t h e 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

product P = Pfeed * m / Ppro ; //W be t h e w e i g h t o f w a t e r e v a p o r i z e d W = m - P; // s t e p 1 − c o o l i n g 1 0 0 0 kg / h o f 10% s o l u t i o n from 305K t o 298K T1 = 305; //K T2 = 298; //K Cliq = 3.67; // kJ /kgK H1 = m * Cliq * ( T2 - T1 ) ; // s t e p 2 − s e p a r a t i o n i n t o p u r e c o m p o n e n ts Hsolution = -42.85; // kJ / mol H2 = - Pfeed * m *1000 * Hsolution / (40*100) ; // s t e p 3 − W kg w a t e r i s c o n v e r t e d t o w a t e r v a p o u r Hvap = 2442.5; // kJ / kg H3 = W * Hvap ; // s t e p 4 − w a t e r v a p o u r a t 298K i s h e a t e d t o 3 7 3 . 1 5K Cvap = 1.884; // kJ /kgK T3 = 373.15; //K H4 = W * Cvap * ( T3 - T2 ) ; // s t e p 5 − f o r m a t i o n o f 200 kg o f 50% NaOH s o l u t i o n a t 140

25 26 27 28 29 30 31 32

298K Hsolu = -25.89; // kJ / mol H5 = Pfeed * m *1000 * Hsolu / (40*100) ; // s t e p 6 − H e a t i n g t h e s o l u t i o n from 298K t o 380K Csolu = 3.34; // kJ / kg T4 = 380; //K H6 = P * Csolu * ( T4 - T2 ) ; Htotal = H1 + H2 + H3 + H4 + H5 + H6 ; disp ( ” kJ ” , Htotal , ” The e n t h a l p y c h a n g e a c c o m p a n y i n g the complete p r o c e s s = ”)

Scilab code Exa 11.26 Mean heat capacity of ethanol water solution 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

clc () Nwater = 0.8; // m o l e s Nethanol = 0.2; // m o l e s T = 323; //K Cwater = 4.18*10^3; // J /kgK Cethanol = 2.58*10^3; // J /kgK Hmixing1 = -758; // J / mol ( a t 298K ) Hmixing2 = -415; // J / mol ( a t 323K ) T1 = 298; //K T2 = 523; //K // s t e p 1 − 0 . 8 mol o f w a t e r i s c o o l e d from 323 K t o 298K H1 = Nwater * 18 * Cwater * ( T1 - T ) / 1000; // s t e p 2 − 0 . 2 mol e t h a n o l c o o l e d from 323K t o 298K H2 = Nethanol * 46 * Cethanol * ( T1 - T ) /1000; // s t e p 3 − 0 . 8 mol w a t e r and 0 . 2 mol e t h a n o l a r e mixed t o g e t h e r , H3 = Hmixing1 ; // s t e p 4 s o l u t i o n i s h e a t e d t o 323K, H4 = Cpm ∗ (T − T1 ) // Hmixing2 = H1 + H2 + H3 + H4 H4 = Hmixing2 - H1 - H2 - H3 ; 141

20 Cpm = H4 / ( T - T1 ) ; 21 disp ( ” J /molK” ,Cpm , ” The mean h e a t c a p a c i t y

o f a 20

percent s o l u t i o n = ”)

Scilab code Exa 11.27 Evaporation of NaOH solution 1 clc () 2 F = 1000; // kg / h 3 H1 = 116.3; // kJ / kg ( e n t h a l p y o f 4 5 6 7 8 9 10 11

f e e d s o l u t i o n − 10%

NaOH, 305 K ) H2 = 560.57; // kJ / kg ( e n t h a l p y o f t h i c k l i q u o r − 50% NaOH, 380 K ) Hsteam = 2676; // kJ / kg ( 1 atm , 3 7 3 . 1 5K ) // by d o i n g m a t e r i a l b a l a n c e s , P = 200; // kg / h mvap = 800; // kg / h // E n t h a l p y b a l a n c e g i v e s , F∗H1 + Q = mvap∗ Hsteam + P ∗H2 Q = ( mvap * Hsteam + P * H2 ) -F * H1 ; disp ( ” kJ / h ” ,Q , ” Heat t o be s u p p l i e d = ” )

Scilab code Exa 11.28 Heat transfer to air 1 clc () 2 U2 = 0.35*10^3; // kJ 3 U1 = 0.25*10^3; // kJ 4 // s i n c e t h e t a n k i s r i g i d

t h e volume d o e s n o t c h a n g e d u r i n g h e a t i n g , Under c o n s t a n t volume , t h e change in the i n t e r n a l energy i s equal to the heat supplied 5 Q = U2 - U1 ; 6 disp ( ” kJ ” ,Q , ” Heat t r a n s f e r r e d t o t h e a i r = ” ) 142

Scilab code Exa 11.29 change in internal energy 1 clc () 2 W = -2.25*745.7; //W ( work done on t h e s y s t e m and 1

hp = 7 4 5 . 7W) 3 Q = -3400; // kJ /h ( Heat t r a n s f e r r e d

to the

surrounding ) 4 U = Q *1000/3600 - W ; 5 disp ( ” J / s ” ,U , ” R i s e i n t h e I n t e r n a l

energy of the

system = ”)

Scilab code Exa 11.30 Heat liberation in oxidation of iron fillings 1 2 3 4 5 6 7 8 9 10 11 12 13 14

clc () // 2 Fe + 3/2O2 = Fe2O3 Hliberated = 831.08; // kJ Q = - Hliberated *1000; disp ( ” J ” ,Q , ”Q = ” ) //P(V) = ( n )RT //W = P(V) = ( n )RT n = -1.5; R = 8.314; T = 298; //K W = (n) * R * T; disp ( ” J ” ,W , ”W = ” ) U = Q - W; disp ( ” J ” ,U , ”U = ” )

Scilab code Exa 11.31 Saturated steam and saturated water 143

1 2 3 4 5 6 7 8 9 10 11 12

13 14 15 16 17

clc () Vgas = 0.09; //mˆ3 Vliq = 0.01; //mˆ3 SVliq = 1.061*10^ -3; //mˆ3/ kg SVvap = 0.8857; //mˆ3/ kg mvap = Vgas / SVvap ; mliq = Vliq / SVliq ; Ul = 504.5; // kJ / kg Ug = 2529.5; // kJ / kg U1 = Ul * mliq + Ug * mvap ; SVtotal = ( Vgas + Vliq ) /( mvap + mliq ) ; // u s i n g steam t a b l e , t h e s e v a l u e o f s p e c i f i c volume c o r r e s p o n d s t o p r e s s u r e o f 1 4 8 . 6 b a r and i n t e r n a l e n e r g y o f 2 4 6 4 . 6 kJ / kg U = 2464; // kJ / kg Utotal = U * ( mvap + mliq ) ; // U t o t a l − U1 = Q − W, but W = o , hence , Q = Utotal - U1 ; disp ( ” kJ ” ,Q , ” Heat t o be added = ” )

Scilab code Exa 11.32 constant volume and constant pressure process 1 2 3 4 5 6 7 8 9 10 11 12 13

clc () m = 10; // kg ( a i r ) N = m / 29; // kmol P1 = 100; // kPa T1 = 300; //K R = 8.314; V1 = N * R * T1 / P1 ; V2 = V1 ; T2 = 600; //K Cv = 20.785; // kJ / kmolK Cp = 29.099; // kJ / kmolK U = N * Cv * ( T2 - T1 ) ; Q = U; 144

14 W = Q - U ; 15 H = U + N * R * ( T2 - T1 ) ; 16 disp ( ” kJ ” ,U , ” ( a ) Change i n i n t e r n a l 17 18 19 20 21 22 23 24 25 26 27 28

energy at

c o n s t a n t volume = ” ) disp ( ” kJ ” ,Q , ” h e a t s u p p l i e d a t c o n s t a n t volume = ” ) disp ( ” kJ ” ,W , ”Work done a t c o n s t a n t volume = ” ) disp ( ” kJ ” ,H , ” Change i n E n t h a l p y a t c o n s t a n t volume = ”) P2 = P1 ; H2 = N * Cp * ( T2 - T1 ) ; Q2 = H2 ; U2 = H2 - N * R * ( T2 - T1 ) ; W2 = Q2 - U2 ; disp ( ” kJ ” ,U2 , ” ( b ) Change i n i n t e r n a l e n e r g y a t constant Pressure = ”) disp ( ” kJ ” ,Q2 , ” h e a t s u p p l i e d a t c o n s t a n t P r e s s u r e = ” ) disp ( ” kJ ” ,W2 , ”Work done a t c o n s t a n t P r e s s u r e = ” ) disp ( ” kJ ” ,H2 , ” Change i n E n t h a l p y a t c o n s t a n t Pressure = ”)

Scilab code Exa 11.33 series of operations 1 2 3 4 5 6 7 8

clc () Cp = 29.3; // kJ / kmol R = 8.314; Cv = Cp - R ; T1 = 300; //K P1 = 1; // b a r P2 = 2; // b a r // s t e p 1 − Volume r e m a i n s c o n s t a n t , t h e r e f o r e t h e work done i s z e r o and h e a t s u p p l i e d i s Cv , A l s o T2/T1 = P2/P1 9 T2 = P2 * T1 / P1 ; 10 Q1 = Cv * ( T2 - T1 ) ; 145

11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27

W1 = 0; disp ( ” kJ ” ,W1 , ”Work done a t c o n s t a n t volume = ” ) disp ( ” kJ ” ,Q1 , ” Heat s u p p l i e d a t c o n s t a n t volume = ” ) // s t e p 2 − P r o c e s s i s a b d i a b a t i c Q2 = 0; r = 1.4; T3 = T2 * (( P1 / P2 ) ^(( r - 1) / r ) ) ; W2 = Cv * ( T2 - T3 ) ; disp ( T3 ) disp ( ” kJ ” ,W2 , ”Work done i n a d i a b a t i c p r o c e s s = ” ) disp ( ” kJ ” ,Q2 , ” Heat s u p p l i e d i n a d i a b a t i c p r o c e s s = ” ) // s t e p 3 − p r o c e s s i s i s o b a r i c Q3 = Cp * ( T1 - T3 ) ; U3 = Cv * ( T1 - T3 ) ; W3 = Q3 - U3 ; disp ( ” kJ ” ,W3 , ”Work done a t c o n s t a n t p r e s s u r e = ” ) disp ( ” kJ ” ,Q3 , ” Heat s u p p l i e d a t c o n s t a n t p r e s s u r e = ” )

Scilab code Exa 11.34 change in internal energy and enthalpy and heat supplied and work done 1 2 3 4 5 6 7 8 9 10 11 12

clc () P1 = 5; // b a r P2 = 4; // b a r T1 = 600; //K V = 0.1; //mˆ3 T2 = 400; //K T = 298; //K Cp = 30; // J /molK // s t e p 1 − i s o t h e r m a l c o n d i t i o n U1 = 0; H1 = 0; P = 1; // b a r 146

13 R = 8.314; 14 W1 = R * T1 * log ( P1 / P2 ) ; 15 Q1 = W1 ; 16 disp ( ” kJ / kmol ” ,U1 , ” ( a ) Change i n t h e 17 18 19 20 21 22 23 24 25 26 27 28 29

i n t e r n a l energy in i s o t h e r m a l c o n d i t i o n = ”) disp ( ” kJ / kmol ” ,H1 , ” Change i n t h e e n t h a l p y e n e r g y i n i s o t h e r m a l c o n d i t i o n = ”) disp ( ” kJ / kmol ” ,W1 , ”Work done i n i s o t h e r m a l c o n d i t i o n = ”) disp ( ” kJ / kmol ” ,Q1 , ” Heat s u p p l i e d i n i s o t h e r m a l c o n d i t i o n = ”) N = P * (1.01325 * 10^5) * V / ( R * T ) ; Cv = Cp - R ; U2 = Cv * ( T2 - T ) * N ; H2 = Cp * ( T2 - T ) * N ; W2 = 0; Q2 = U2 + W2 ; disp ( ” kJ / kmol ” ,U2 , ” ( b ) Change i n t h e i n t e r n a l e n e r g y a t c o n s t a n t volume c o n d i t i o n = ” ) disp ( ” kJ / kmol ” ,H2 , ” Change i n t h e e n t h a l p y e n e r g y a t c o n s t a n t volume c o n d i t i o n = ” ) disp ( ” kJ / kmol ” ,W2 , ”Work done a t c o n s t a n t volume c o n d i t i o n = ”) disp ( ” kJ / kmol ” ,Q2 , ” Heat s u p p l i e d a t c o n s t a n t volume c o n d i t i o n = ”)

Scilab code Exa 11.35 Heat removed in condenser 1 2 3 4 5 6 7

clc () m = 1; // kg u2 = 0.5; //m/ s u1 = 60; //m/ s H = -3000; // kJ / kg //KE = ( u ˆ 2 ) /2 KE = (( u2 ^ 2) - ( u1 ^2) ) /2000; 147

8 9 10 11 12 13 14 15

g = 9.81; //m/ s ˆ2 Z1 = 7.5; //m Z2 = 2; //m //PE = g ∗ ( Z ) PE = g * ( Z2 - Z1 ) /1000; W = 800; // kJ / kg Q = H + PE + KE + W ; disp ( ” kJ / kg ” ,Q , ” Heat removed from t h e f l u i d = ” )

Scilab code Exa 11.36 Throttling process 1 2 3 4 5 6 7 8 9 10

clc () PE = 0; W = 0; Q = 0; // (H) + ( u ˆ 2 ) /2 = 0 // a c c o r d i n g t o t h e r e a l t i o n u1 ∗ v1 = u2 ∗ v2 // ( u ˆ 2 ) /2 i s n e g l i g i b l e , c h a n g e i n e n t h a l p y i s 0 T1 = 623; //K P1 = 6000; // kPa H1 = 3045.8; // kJ / kg ( E n t h a l p y o f t h e steam u s i n g steam t a b l e ) 11 P2 = 1000; // kPa 12 T2 = 570; //K ( v a l u e o f t e m p e r a t u r e c o r r e s p o n d i n g t o t h e e n t h a l p y and p r e s s u r e u s i n g t h e steam t a b l e ) 13 disp ( ”K” ,T2 , ” T e m p e r a t u r e o f s u p e r h e a t e d steam = ” )

Scilab code Exa 11.37 water pumping and energy balances 1 clc () 2 g = 9.81; //m/ s ˆ2 3 z = 55;

148

4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

PE = g * z ; KE = 0; T2 = 288; //K f = 1.5*10^ -2; //mˆ3/ min D = 1000; // kg /mˆ3 m = f * D; Qsupp = 500; // kJ / min Qlost = 400; // kJ / min Qnet = ( Qsupp - Qlost ) * D / m ; W = 2*745.7; //W Ws = -W * 0.6 / ( m /60) ; H = Qnet - Ws - PE - KE ; Cp = 4200; T1 = H / Cp ; T = T1 + T2 ; disp ( ”K” ,T , ” The t e m p e r a t u r e o f e x i t w a t e r = ” )

Scilab code Exa 11.38 Energy balance on rotary drier 1 clc () 2 m = 1000; // kg / h ( d r i e d p r o d u c t ) 3 // S be t h e amount o f d r y s o l i d i n t h e p r o d u c t 4 5 6 7 8 9 10 11 12 13 14 15

stream Pmoisture1 = 4; //% Pmoisture2 = 0.2; //% S = m *(1 - P /1000) ; X1 = Pmoisture1 /(100 - Pmoisture1 ) ; X2 = Pmoisture2 /(100 - Pmoisture2 ) ; // l e t G be t h e w e i g h t o f d r y a i r i n t h e a i r s t r e a m Y1 = 0.01; // kg w a t e r / kg d r y s o l i d Cp = 1.507; Cw = 4.2; T1 = 298; //K T = 273; //K T2 = 333; //K 149

16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37

Tg1 = 363; //K Tg2 = 305; //K Hs1 = ( Cp + X1 * Cw ) * ( T1 - T ) ; Hs2 = ( Cp + X2 * Cw ) * ( T2 - T ) ; //Hg = Cs ( Tg − To ) + Y∗L // Cs = 1 . 0 0 5 + 1 . 8 8 4 ∗Y L = 2502.3; // kJ / kg d r y a i r Hg1 = (1.005 + 1.884 * Y1 ) *( Tg1 - T ) + Y1 * L ; Q = -40000; // kJ /h // C a l c u l a t i n g f o r T2 , Hg2 = 3 2 . 1 6 + 2 5 6 2 . 5 9 ∗Y // c h a n g e i n e n t h a l p y = Q //H1 = S ∗ Hs1 + G ∗ HG1 = 3 7 8 1 4 . 2 2 + 1 1 7 . 1 7G //H2 = 1 0 0 7 2 8 . 1 4 + G∗ ( 3 2 . 1 6 + 2 5 6 1 . 5 9 ∗Y) // c h a n g e i n e n t h a l p y = Q // 6 2 9 1 3 . 9 2 + G ∗ ( − 8 5 . 0 1 + 2 5 6 1 . 5 9 ∗Y) + 4 0 0 0 0 = 0 // 1 0 2 9 1 3 . 9 2 + G ∗ ( − 8 5 . 0 1 + 2 5 6 1 . 5 9 ∗Y) = 0 (1) // m o i s t u r e b a l a n c e , S∗X1 + G∗Y1 = S∗X2 + G∗Y2 //G∗ (Y− 0 . 0 1 ) = 3 9 . 6 2 (2) // s o l v i n g s i m u l t a n e o u s l y ( 1 ) and ( 2 ) , Gdry = 3443; // kg / h G = Gdry *(1 + Y1 ) ; disp ( ” kg /h ” ,G , ” A i r r e q u i r e m e n t = ” )

Scilab code Exa 11.39 Energy balance on the fractionator 1 clc () 2 m = 1000; // kg / h ( f e e d s o l u t i o n ) 3 //F − mass o f f e e d d i s t i l l e d , W − mass o f t h e bottom

p r o d u c t , D − mass o f t h e d i s t i l l a t e , x f , xd and xw − w e i g h t f r a c t i o n o f a c t o n e i n f e e d , d i s t i l l a t e and r e s i d u e r e s p . 4 // t o t a l b a l a n c e , F = D + W 5 // A c e t o n e b a l a n c e , F∗ x f = D∗ xd + w∗xw 150

6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39

F = 1000; xf = 0.10; xd = 0.9; xw = 0.01; // s u b s t i t u t i n g i n a b o v e e q u a t i o n s , D = F * ( xf - xw ) / ( xd - xw ) ; W = F - D; R = 8; L = R * D; // m a t e r i a l b a l a n c e a r o u n d t h e c o n d e n s e r ,G v a p o u r reaching the condenser G = L + D; Td = 332; //K T2 = 300; //K Tw = 370; //K Tf = 340; //K Lacetone1 = 620; // kJ / kg Lwater1 = 2500; // kJ / kg Ld = xd * Lacetone1 + (1 - xd ) * Lwater1 ; Cpacetone = 2.2; // kJ /kgK Cpwater = 4.2; // kJ /kgK Cp = xd * Cpacetone + (1 - xd ) * Cpwater ; H = Ld + Cp * ( Td - T2 ) ; Cpc = 4.2; // kJ / kg Tc = 30; //K ( c h a n g e i n t e m p e r a t u r e a l l o w a b l e f o r c o o l i n g water ) m = G * H / ( Cpc * Tc ) ; disp ( ” kg /h ” ,m , ” ( a ) The c i r c u l a t i o n r a t e o f c o o l i n g water = ”) Qc = G * H ; Hd = 0; Hw = ( xw * Cpacetone + (1 - xw ) * Cpwater ) *( Tw - T2 ) ; Hf = ( xf * Cpacetone + (1 - xf ) * Cpwater ) *( Tf - T2 ) ; Qb = D * Hd + W * Hw + Qc - F * Hf ; Hcondensation = 2730; // kJ / kg msteam = Qb / Hcondensation ; disp ( ” kg /h ” , msteam , ” ( b ) Amount o f steam s u p p l i e d = ” )

151

Chapter 12 Energy Balance Thermochemistry

Scilab code Exa 12.1 Heat liberated calculation 1 clc () 2 N = 100; // mol g a s m i x t u r e b u r n e d 3 //CO( g ) + 1/2 O2 ( g ) = CO2 − 4 5 6 7 8 9 10 11

12 13 14 15

Hr1 =

− 2 8 2 . 9 1 kJ / mol //H2 ( g ) + 1/2 O2 ( g ) = H2O − Hr2 = − 2 4 1 . 8 3 kJ / mol Hr1 = - 282.91; // kJ / mol Hr2 = - 241.83; // kJ / mol Nco1 = 20; Nh21 = 30; Nn21 = 50; Htotal = Nco1 * Hr1 + Nh21 * Hr2 ; disp ( ” kJ ” ,- Htotal , ” t h e amount o f h e a t l i b e r a t e d on t h e c o m p l e t e c o m b u s t i o n o f 100 mol o f t h e g a s mixture = ”) Ncoreac = Nco1 * 0.9; Nh2reac = Nh21 * 0.8; Htotal1 = Ncoreac * Hr1 + Nh2reac * Hr2 ; disp ( ” kJ ” ,- Htotal1 , ” t h e amount o f h e a t l i b e r a t e d i f 152

o n l y 90% o f CO and 80% o f H2 r e a c t o f 100 mol o f the gas mixture = ”)

Scilab code Exa 12.2 Heat of formation of methane 1 clc () 2 //C( s ) + 2H2 ( g ) = CH4( g ) Hf = ? 3 Hc = -393.51; // kJ / mol 4 Hh2 = -285.84; // kJ / mol 5 Hch4 = - 890.4; // kJ / mol 6 // h e a t o f r e a c t i o n can be c a l c u l a t e d from t h e h e a t

o f combustion data using f o l l o w i n g equation , the h e a t o f r e a c t i o n i s t h e sum o f t h e h e a t o f combustion of a l l the r e a c t a n t s in the d e s i r e d r e a c t i o n minus t h e sum o f t h e h e a t o f c o m b u s t i o n o f a l l t h e p r o d u c t s o f t h e d e s i r e d r e a c t i o n . Here t h e r e a c t a n t s a r e one mole o f Carbon and two m o l e s hydrogen , and t h e p r o d u c t i s one mole o f methane , t h e r e h e a t o f r e a c t i o n i s 7 Hf = 1 * Hc + 2 * Hh2 - 1 * Hch4 ; 8 disp ( ” kJ ” ,Hf , ” Heat o f f o r m a t i o n o f methane = ” )

Scilab code Exa 12.3 Net heating value of coal 1 2 3 4 5 6 7 8 9

clc () m = 1; // kg o f c o a l b u r n e d xc = 0.7; xh2 = 0.055; xn2 = 0.015; xs = 0.03; xo = 0.13; xash = 0.07; Hvap = 2370; // kJ / kg 153

10 C = 29000; // kJ / kg 11 Nh2 = xh2 * m / 2.016; 12 Nwater = Nh2 ; // ( amount o f w a t e r f o r m e d ) 13 mwater = Nwater * 18.016; 14 Hreq = mwater * Hvap ; 15 Hnet = C - Hreq ; 16 disp ( ” kJ / kg ” , Hnet , ” Net h e a t i n g v a l u e o f c o a l = ” )

Scilab code Exa 12.4 Heat of reaction for esterification of ethyl alcohol 1 clc () 2 //C2H5OH( l ) + CH3COOH( l ) = C2H5COOCH3( l ) + H2O( l ) H 3 4 5 6 7 8 9

= ? Hc2h5oh = -1366.91; // kJ / mol Hch3cooh = -871.69; // kJ / mol Hc2h5cooch3 = -2274.48; // kJ / mol // t o c a l c u l a t e h e a t o f r e a c t i o n from t h e h e a t o f combustion data , // H r e a c = H r e a c − Hprod Hreac = Hc2h5oh + Hch3cooh - Hc2h5cooch3 ; disp ( ” kJ ” , Hreac , ” Heat o f r e a c t i o n f o r t h e e s t e r i f i c a t i o n o f e t h y l a l c o h o l with a c e t i c acid = ”)

Scilab code Exa 12.5 Vapour phase hydration of ethylene to ethanol 1 clc () 2 //C2H4 ( g ) + H2O( g ) = C2H5OH( g ) 3 // 2CO2( g ) + 3H2O( l ) = C2H5OH( l ) + 3O2 ( g ) 4 5 6

1 3 6 6 . 9 1 kJ (A) Hc2h4 = -1410.99; // kJ / mol Hvap = 44.04; // kJ / mol Hc2h5oh = 42.37; // kJ / mol 154

H =

7 8 9 10 11 12 13 14 15 16

//C2H4 ( g ) + 3H2O( l ) = C2H5OH( l ) + 3O2 ( g ) H −1410.99 kJ (B) //H2O( l ) = H2O( g ) H 4 4 . 0 4 kJ (C) //C2H5OH( l ) = C2H5OH( g ) H 4 2 . 3 7 kJ (D) //A + B + D − C g i v e s t h e r e q u i r e d r e a c t i o n Ha = 1366.91; // kJ Hb = -1410.99; // kJ Hc = 44.04; // kJ Hd = 42.37; // kJ Hreac = Ha + Hb + Hd - Hc ; disp ( ” kJ ” , Hreac , ” The s t a n d a r d h e a t o f r e a c t i o n

= = =

= ”)

Scilab code Exa 12.6 Standard heat of formation of acetylene 1 clc () 2 //C2H5 ( g ) + 5/2O2 ( g ) = 2CO2( g ) + H2O( l ) 3 4 5 6 7 8 9 10

H1 = −1299.6 kJ (A) //C( s ) + O2 ( g ) = CO2( g ) H2 = −393.51 kJ (B) //H2 ( g ) + 1/2O2 ( g ) = H2O( l ) H3 = −285.84 kJ (C) // 2C( s ) + H2 ( g ) = C2H2 ( g ) H = ? H1 = -1299.6; // kJ H2 = -393.51; // kJ H3 = -285.84; // kJ Hreac = 2 * H2 + H3 - H1 ; disp ( ” kJ ” , Hreac , ” Heat o f f o r m a t i o n o f a c e t y l e n e = ” )

Scilab code Exa 12.7 Standard heat of roasting of iron pyrites 155

1 clc () 2 m = 100; // kg o f p y r i t e s c h a r g e d 3 xfes2in = 0.8; 4 xganguein = 0.2; 5 xfes2out = 0.05; 6 // l e t x be t h e FeS2 i n t h e f e e d , then , Fe2O3 = ( 8 0 −

7 8 9 10 11 12 13 14 15 16 17

x ) ∗ 1 5 9 . 6 9 / ( 1 1 9 . 9 8 ∗ 2 ) and g a n g u e = 2 0 , t o t a l = 7 3 . 2 4 + 0 . 3 3 4 5 , be FeS2 i s o n l y 5 % i n t h e product , hence x = 0.05 * 73.24 / (1 - 0.05*0.3345) ; mfes2reacted = m * xfes2in - x ; // 4 FeS2 + 11O2 = 2 Fe2O3 + 8SO2 Hfes2 = -178.02; // kJ / mol Hfe2o3 = -822.71; // kJ / mol Hso2 = -296.9; // kJ / mol Hreac = 2 * Hfe2o3 + 8 * Hso2 - 4 * Hfes2 ; N = mfes2reacted *1000/ 119.98; H = Hreac * N / 4; H1 = H / m ; // ( h e a t o f r e a c t i o n p e r kg o f c o a l b u r n t ) disp ( ” kJ ” ,H1 , ” Heat o f r e a c t i o n p e r 1 kg o f c o a l burned = ”)

Scilab code Exa 12.8 Standard heat of formation of liquid methanol 1 clc () 2 //CH3OH( l ) + 3/2O2 ( g ) = CO2( g ) + 2H2O( l ) 3 4 5 6 7

−726.55 kJ H1 = -726.55; // kJ Hco2 = -393.51; // kJ / mol Hh2o = -285.84; // kJ / mol Hch3oh = Hco2 + 2 * Hh2o - H1 ; disp ( ” kJ ” , Hch3oh , ” Heat o f f o r m a t i o n o f l i q u i d methanol = ”)

156

H =

Scilab code Exa 12.9 Gross heating value and Net heating value calculation 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

18 19 20 21

clc () N = 100; // mol f u e l g a s Nco = 21; Nh2 = 15.6; Nco2 = 9.0; Nch4 = 2; Nc2h4 = 0.4; Nn2 = 52; Hco = 282.99; // kJ / mol ( h e a t o f c o m b u s t i o n ) Hh2 = 285.84; // kJ / mol ( h e a t o f c o m b u s t i o n ) Hch4 = 890.4; // kJ / mol ( h e a t o f c o m b u s t i o n ) Hc2h4 = 1410.99; // kJ / mol ( h e a t o f c o m b u s t i o n ) Hvap = 44.04; // kJ / mol H = Nco * Hco + Nh2 * Hh2 + Nch4 * Hch4 + Nc2h4 * Hc2h4 ; // kJ V = N * 22.4143/1000; H1 = H / V ; // kJ /mˆ3 // on c o m b u s t i o n , 1 mol h y d r o g e n g i v e s 1 mol o f water , 1 mol o f methane g i v e s 2 mol o f w a t e r and 1 mol o f e t h y l e n e g i v e s 2 moles of water Nwater = Nh2 + 2 * Nch4 + 2 * Nc2h4 ; Hvap1 = Hvap * Nwater ; Hnet = H1 - Hvap1 ; disp ( ” kJ ” , Hnet , ” Net h e a t i n g v a l u e o f t h e f u e l = ” )

Scilab code Exa 12.10 Standard heat of reaction calculation 1 clc () 2 // C5H12 ( g ) + 8O2 ( g ) = 5CO2( g ) + 6H20 ( l )

157

3 Hfco2 = -393.51; // kJ 4 Hfh2o = - 241.826; // kJ 5 Hfc5h12 = -146.4; // kJ 6 Hvap = 43.967; // kJ / mol 7 H1 = 6* Hfh2o +5* Hfco2 - Hfc5h12 ; 8 H2 = 6 * ( - Hvap ) ; 9 Hreac = H1 + H2 ; 10 disp ( ” kJ ” , Hreac , ” S t a n d a r d h e a t o f

r e a c t i o n = ”)

Scilab code Exa 12.11 Constant pressure heat of combustion 1 2 3 4 5 6 7 8

9 10 11 12 13 14

clc () m = 1; // kg o f o i l b u r n e d mc = 0.9; // kg mh2 = 0.1; // kg Mc = mc / 12; // kmol //C( s ) + O2 ( g ) = CO2( g ) Nh2 = mh2 / 2.016; // kmol // c h a n g e i n t h e no . o f g a s e o u s c o m po n e n t s a c c o m p a n y i n g t h e c o m b u s t i o n o f 1 mole o f h y d r o g e n i n l i q u i d s t a t e i s −1/2 mol , t h e r e f o r e f o r Nh2 mol R = 8.314; T = 298; //K x = Nh2 * R * T / ( -2) ; Qv = -43000; // kJ / kg Qp = Qv + x ; disp ( ” kJ / kg ” ,Qp , ” t h e c o n s t a n t p r e s s u r e h e a t o f combustion = ”)

Scilab code Exa 12.12 Heat of reaction for ammonia synthesis 1 clc ()

158

2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

// 1 − N2 , 2 − H2 , 3 − NH3 a1 = 27.31; a2 = 29.09; a3 = 25.48; b1 = 5.2335*10^ -3; b2 = -8.374*10^ -4; b3 = 36.89 * 10^ -3; c1 = -4.1868 * 10^ -9; c2 = 2.0139*10^ -6; c3 = -6.305*10^ -6; H1 = -46191; // J T1 = 298; //K // 1/2 N2 + 3/2 H2 = NH3 H = −46.191 kJ // Ht = H + a ∗T + b∗Tˆ2 / 2+ c ∗Tˆ3 / 3 // a t 2 9 8 , a = a3 - a1 / 2 - 3 * a2 / 2; b = b3 - b1 / 2 - 3 * b2 / 2; c = c3 - c1 / 2 - 3 * c2 / 2; H = H1 -a * T1 - b * ( T1 ^2) / 2 - c * ( T1 ^3) / 3; T2 = 700; //K H2 = H + a * T2 + b * ( T2 ^2) / 2 + c * ( T2 ^3) / 3; disp ( H ) ; disp ( ” kJ ” ,H2 , ” Heat o f r e a c t i o n a t 700K = ” )

Scilab code Exa 12.13 Standard heat of reaction of methanol synthesis 1 clc () 2 //CO( g ) + 2H2 ( g ) = CH3OH( g ) 3 T1 = 298; //K 4 T2 = 1073; //K 5 //Cp (CH3OH) = 1 8 . 3 8 2 + 1 0 1 . 5 6 4 ∗ 10ˆ −3 ∗ T − 2 8 . 6 8 3

∗ 10ˆ −6 ∗ Tˆ2 6 //Cp (CO) = 2 8 . 0 6 8 + 4 . 6 3 1 ∗ 10ˆ −3 ∗ T − 2 . 5 7 7 3 ∗ 1 0 ˆ 4 ∗ Tˆ−2 7 //Cp ( H2 ) = 2 7 . 0 1 2 + 3 . 5 0 9 ∗ 10ˆ −3 ∗ T + 6 . 9 0 0 6 ∗ 159

8 9

10 11 12 13 14 15 16 17 18 19 20 21 22

1 0 ˆ 4 ∗ Tˆ−2 // f o r r e a c t a n t s , H1 = integrate ( ’ 2 8 . 0 6 8 + 4 . 6 3 1 ∗ 10ˆ −3 ∗ T − 2 . 5 7 7 3 ∗ 1 0 ˆ 4 ∗ Tˆ−2 ’ , ’T ’ ,T2 , T1 ) + 2 * integrate ( ’ 2 7 . 0 1 2 + 3 . 5 0 9 ∗ 10ˆ −3 ∗ T + 6 . 9 0 0 6 ∗ 1 0 ˆ 4 ∗ Tˆ−2 ’ , ’T ’ , T2 , T1 ) ; // f o r p r o d u c t , H2 = integrate ( ’ 1 8 . 3 8 2 + 1 0 1 . 5 6 4 ∗ 10ˆ −3 ∗ T − 2 8 . 6 8 3 ∗ 10ˆ −6 ∗ Tˆ2 ’ , ’T ’ ,T1 , T2 ) ; // H298 = H p r o d u c t s − H r e a c t a n t s ; //CO + 2H2 = CH3OH Ha1 = −238.64 kJ Ha1 = -238.64; // kJ //CH3OH( l ) = CH3OH( g ) Hvap = 3 7 . 9 8 kJ Hvap = 37.98; // kJ //CO( g ) + 2H2 ( g ) = CH3OH( g ) Ha2 = −200.66 kJ Ha2 = Ha1 + Hvap ; // kJ Hco = -110.6; // kJ / mol H298 = Ha2 - ( Hco ) ; Htotal = H1 /1000 + H298 + H2 /1000; disp ( ” kJ / mol ” , Htotal , ” The h e a t o f r e a c t i o n a t 773K = ”)

Scilab code Exa 12.14 Combustion of CO 1 clc () 2 Nco = 1; // mol CO r e a c t e d 3 //CO + 1/2 O2 = CO2 4 No2 = Nco / 2; 5 Pexcess = 100; 6 Nosupp = No2 * ( 1 + Pexcess / 100 ) ; // o x y g e n

supplied 7 Nn2 = Nosupp * 79 / 21; 8 Nco2 = Nco ; 9 Noremain = Nosupp - No2 ; 10 T1 = 298; //K

160

T2 = 400; //K Hr1 = -282.99; // kJ T3 = 600; //K SHco = 29.1; // J /molK SHo2 = 29.7; // J /molK SHn2 = 29.10; // J /molK SHco2 = 41.45; // J /molK H1 = ( Nosupp * SHo2 + Nn2 * SHn2 + Nco * SHco ) * ( T1 - T2 ) ; // e n t h a l p y o f c o o l i n g o f r e a c t a n t s from 298 t o 400 K 19 H2 = ( Nco2 * SHco2 + Nn2 * SHn2 + Noremain * SHo2 ) * ( T3 - T1 ) ; // e n t h a l p y o f h e a t i n g t h e p r o d u c t s from 298K t o 600K 20 H = H1 /1000 + Hr1 + H2 /1000; 21 disp ( ” kJ ” ,H , ” Heat c h a n g e a t 600K = ” )

11 12 13 14 15 16 17 18

Scilab code Exa 12.15 Heat added or removed calculation 1 clc () 2 //CO( g ) + H2O( g ) = CO2( g ) + H2 ( g ) 3 4 5 6 7 8 9 10 11 12 13 14 15 16

−41.190 T1 = 298; //K Pconv = 75; //% T2 = 800; //K H298 = -41.190; // kJ Hco = 30.35; // J /molK Hco2 = 45.64; // J /molK Hwater = 36; // J /molK Hh2 = 29.3; // J /molK Nco = 1; // mol Nh2o = 1; // mol Ncofinal = Nco * (1 - Pconv /100) ; Nwaterf = Ncofinal ; Nco2final = Nco - Ncofinal ; Nh2final = Nco2final ; 161

H298 =

17 H2 = ( Nco2final * Hco2 + Nh2final * Hh2 + Ncofinal * 18 19 20 21 22 23

Hco + Nwaterf * Hwater ) * ( T2 - T1 ) ; Hr1 = H298 * ( Nco - Ncofinal ) ; Hr2 = Hr1 * 1000 + H2 ; mh2 = Nh2final * 2.016 * 10^ -3; // kg // t h e r f o r e f o r 1 0 0 0 k H2 , Hr = Hr2 * 1000 / ( mh2 * 1000) ; // kJ disp ( ” kJ ” ,Hr , ”Amount o f h e a t c h a n g e f o r 1 0 0 0 kg o f hydrogen produced = ”)

Scilab code Exa 12.16 CO2 O2 and N2 passed through a bed of C 1 clc () 2 //CO2( g ) + C( s ) = 2CO( g ) 3 4 5 6 7 8 9 10

11

12 13 14 15 16

H1298 = 170

kJ / mol //O2 ( g ) + 2C( s ) = 2CO( g ) H2298 = −221.2 kJ / mol T2 = 1298; //K T1 = 298; //K Hc = 0.02; // kJ /molK Ho = 0.03; // kJ /molK Hco = 0.03; // kJ /molK Hco2 = 0.05; // kJ /molK // l e t t h e f l u e g a s c o n t a i n x mol CO2 p e r mole o f oxygen , p r o d u c t c o n t a i n s 2(1+ x ) mol CO. N i t r o g e n i n r e a c t a n t and p r o d u c t r e m a i n t h e same // e n t h a l p y o f c o o l i n g xmol CO2 , 1 mol O2 and 2 + xmol c a r b o n from 1 2 9 8 t o 298K i s g i v e n as , H1 = ( Hco2 ∗ x + Ho ∗ 1 + Hc ∗ ( 2 + x ) ) ∗ ( 2 9 8 − 1 2 9 8 ) //H1 = ( −70 x − 7 0 ) kJ // e n t h a l p y o f h e a t i n g t h e p r o d u c t , H2 = 2 ∗ ( 1 + x ) ∗ Hco ∗ ( 1 2 9 8 − 2 9 8 ) //H2 = 60 + 60 x kJ // Hr = 170 x − 2 2 1 . 2 // H t o t a l = 0 = H1 + H2 + Hr 162

17 x = (221.2 + 70 - 60) /(170 + 60 - 70) ; 18 disp ( ” mol ” ,x , ” m o l e s o f CO2 p r e s e n t p e r mol o f o x y g e n

in f e e d stream = ”)

Scilab code Exa 12.17 Partial oxidation of natural gas 1 clc () 2 N = 100; // mol f l u e g a s 3 // Carbon b a l a n c e , 4 // x i s t h e f e e d o f methane , w i s 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26

w a t e r i n f l u e ga , y

i s the oxygen s u p p l i e d xco2 = 0.019; xch2o = 0.117; xo2 = 0.038; xch4 = 0.826; xc = xco2 + xch2o + xch4 ; Nc = xc * N ; Nch4i = Nc ; // Hydrogen b a l a n c e , xh2 = xch2o + xch4 *2; w = 2 * ( Nch4i ) - xh2 * N ; // o x y g e n b a l a n c e No2s = ( xco2 + xch2o /2 + xo2 ) * N + w /2; y = No2s ; T1 = 298; //K T2 = 573; //K T3 = 673; //K // o x y g e n c o o l e d from 573K and methane from 673 t o 298K Ho573 = 30.5; // J /molK Hch4673 = 45.9; // J /molK H1 = y * Ho573 * ( T1 - T2 ) + Nch4i * Hch4673 * ( T1 T3 ) ; //CH4 + O2 = CH2O + H2O Hr1 = −282.926 kJ //CH4 + 2O2 = CO2 + 2H2O Hr2 = −802.372 kJ 163

27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42

Hr1 = -282.926; // kJ Hr2 = -802.372; // kJ H2 = xch2o * N * Hr1 + xco2 * N * Hr2 ; T4 = 873; //K Ho = 31.9 Hch4 = 51.4; Hco2 = 46.3; Hch2o = 47.1; Hh2o = 36.3; H3 = (( xco2 * Hco2 + xo2 * Ho + xch4 * Hch4 + Hch2o * xch2o ) * N + w * Hh2o ) *( T4 - T1 ) ; Htotal = H1 /1000 + H2 + H3 /1000; Nch2o = xch2o * N ; mch2o = Nch2o * 30.016/1000; // kg // f o r 1 0 0 0 kg o f f o r m a l d e h y d e p r o d u c e d , H = Htotal * 1000 / mch2o ; disp ( ” kJ ” ,H , ” The amount o f h e a t t o be removed p e r 1 0 0 0 kg o f f o r m a l d e h y d e p r o d u c e d = ” )

Scilab code Exa 12.18 Maximum allowable conversion calculation 1 clc () 2 Nn2 = 1; // kmol / s ( b a s i s − f e e d

o f N2

c o n i s t i n g o f 1 kmol

and 3 kmol o f H2 )

3 Nh2 = 3; // kmol / s 4 // l e t x be t h e f r a c t i o n c o n v e r t e d 5 T1 = 700; //K 6 Hr1 = -94.2; // kJ / mol 7 // h e a t l i b e r a t e d = Hr1 ∗ x 8 // P r o d u c t c o n s i s t s o f 2 x kmol NH3 , (1−x ) kmol N2 , and

3(1 − x ) kmol Hydrogen 9 T2 = 800; //K 10 Hn2 = 0.03; // kJ /molK 11 Hh2 = 0.0289; // kJ /molK 12 Hnh3 = 0.0492; // kJ /molK

164

13 14 15 16 17 18 19

//H2 = (1−x ) ∗ 0 . 0 3 ∗ 1 0 ˆ 3 ∗ 100 + 3∗(1 − x ) ∗ 0 . 0 2 8 9 ∗ 1 0 0 0 ∗ 1 0 0 + 2∗ x ∗ 0 . 0 4 9 2 ∗ 1 0 0 0 ∗ 1 0 0 //H2 = 1 1 . 6 7 ∗ 1 0 0 0 − 1 . 8 3 ∗ 1 0 ˆ 3 ∗ x kJ // r e a c t i o n i s a d i a b a t i c , hence , H1 = H2 // s o l v i n g t h i s we g e t , x = 0.1215; Convmax = x * 100; disp ( ”%” , Convmax , ” The maximum c o n v e r s i o n f o r n i t r o g e n s h o u l d be ” )

Scilab code Exa 12.19 Theoretical flame temperature calculation 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

clc () Nco = 1; // mol CO // CO + 1/2 O2 = CO2 O2r = 1; // mol N2r = 3.76; // mol COr = 1; // mol O2p = 0.5; // mol N2p = 3.76; // mol CO2p = 1; // mol Hco = 29.23; // J /molK Ho2 = 34.83; // J /molK Hn2 = 33.03; // J /molK Hco2 = 53.59; // J /molK Hcomb1 = -282.99; // kJ / mol T1 = 298; //K T2 = 373; //K H1 = ( O2r * Ho2 + N2r * Hn2 + COr * Hco ) * ( T1 - T2 ) ; 18 // For p r o d u c t a t temp T , H2 = ( O2p ∗ Ho2 + N2p ∗ Hn2 + CO2p ∗ Hco2 ) ∗ (T − T1 ) 19 // For a d i a b a t i c c o n d i t i o n , −(H1 + Hcomb1 ) = H2 20 T = -( H1 + Hcomb1 * 1000) / ( O2p * Ho2 + N2p * Hn2 + CO2p * Hco2 ) + T1 ; 165

21

disp ( ”K” ,T , ” T h e o r e t i c a l f l a m e t e m p e r a t u r e = ” )

Scilab code Exa 12.20 Temperature of products on burning of hydrogen gas 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

clc () N = 1; // kmol h y d r o g e n b u r n e d No = N /2; Nosupplied = 2 * No ; Nair = Nosupplied * 100 / 21; Nn2 = Nair - Nosupplied ; // R e a c t a n t s , H2 = 1 kmol , A i r = 4 . 7 6 2 kmol // Product , Water v a p o u r = 1 kmol , Oxygen = 0 . 5 kmol , N2 = 3 . 7 6 2 kmol //Cp ( w a t e r ) = 3 0 . 4 7 5 + ( 9 . 6 5 2 ∗ 1 0 ˆ − 3 ) ∗T + 1 . 1 8 9 ∗ 10ˆ −6 ∗ Tˆ2 //Cp ( n i t r o g e n ) = 2 7 . 0 3 4 + 5 . 8 1 5 ∗ 10ˆ −3 ∗T − 0 . 2 8 8 9 ∗ 10ˆ −6 ∗ Tˆ2 //Cp ( o x y g e n ) = 2 5 . 6 1 1 + 1 3 . 2 6 0 ∗ 10ˆ −3 ∗ T − 4 . 2 0 7 7 ∗ 10ˆ −6 ∗ Tˆ2 //H2 = i n t e g r a t i o n ( 2 9 8 t o T o f ( 1 ∗ Cp ( w a t e r ) + 0 . 5 ∗ Cp ( o x y g e n ) + 3 . 7 6 2 ∗ Cp ( n i t r o g e n ) ) ) // t h e r e f o r e , H2 = 1 4 0 . 3 4 ∗ T + 3 1 . 2 2 2 ∗ 10ˆ −3 ∗ Tˆ2 − 4 . 9 2 8 ∗ 10ˆ −6 ∗ Tˆ2 − 4 4 4 6 3 . 5 4 kJ H298 = -241.826 * 10^3; // kJ //H2 = −H1 − H298 //H1 = 0 // t h e r e f o r e u s i n g e q u a t i o n H2 , t h e v a l u e o f T i s o b t a i n e d t o be T = 1609.8; //K disp ( ”K” ,T , ” T e m p e r a t u r e o f t h e r e a c t i o n p r o d u c t s = ” )

166