A Textbook Of Chemical Engineering Thermodynamics_K. V. Narayanan (1).pdf

8/20/2019 A Textbook Of Chemical Engineering Thermodynamics_K. V. Narayanan (1).pdf

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Scilab Textbook Companion for

A Textbook Of Chemical Engineering

Thermodynamics

by K. V. Narayanan1

Created byApurti Marodia

B.TechChemical Engineering

NIT Tiruchirappalli

College TeacherDr. Prakash Kotecha

Cross-Checked by

August 10, 2013

1Funded by a grant from the National Mission on Education through ICT,http://spoken-tutorial.org/NMEICT-Intro. This Textbook Companion and Scilabcodes written in it can be downloaded from the ”Textbook Companion Project”section at the website http://scilab.in


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Book Description

Title: A Textbook Of Chemical Engineering Thermodynamics

Author: K. V. Narayanan

Publisher: Prentice Hall Of India, New Delhi

Edition: 15

Year: 2011

ISBN: 978-81-203-1732-1

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

Exa Example (Solved example)

Eqn Equation (Particular equation of the above book)

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

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

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Contents

List of Scilab Codes 4

1 Introduction and Basic Concepts 11

2 First Law of Thermodynamics 18

3 PVT Behaviour And Heat Effects 34

4 Second Law of Thermodynamics 51

5 Some Applications of the Laws of Thermodynamics 72

6 Thermodynamic Properties of Pure Fluids 105

7 Properties of Solutions 130

8 Phase equilibria 155

9 Chemical Reaction Equilibria 194

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List of Scilab Codes

Exa 1.1 To find mans mass and weight on earth . . . . . . . . 11Exa 1.2 To find height of manometer fluid . . . . . . . . . . . 12Exa 1.3 To find height from ground and Kinetic Energy . . . . 13Exa 1.4 To determine the power developed in man . . . . . . . 13Exa 1.5 To determine the force exerted pressure work done and

change in potential energy . . . . . . . . . . . . . . . . 14Exa 1.6 To determine work done by gas . . . . . . . . . . . . . 15Exa 1.7 To find the work done on surrounding . . . . . . . . . 16Exa 2.1 To find change in internal energy . . . . . . . . . . . . 18Exa 2.2 To find heat liberated work done and change in internal

energy . . . . . . . . . . . . . . . . . . . . . . . . . . . 19Exa 2.3 To find the heat energy dissipated by brakes . . . . . . 20Exa 2.4 To find internal energy change during each step and

work done during adiabatic process . . . . . . . . . . . 20Exa 2.5 To find change in internal energy and enthalpy . . . . 22Exa 2.6 To find internal energy of saturated liquid and internal

energy and enthalpy of saturated vapour . . . . . . . . 23Exa 2.7 To calculate molar internal energy change and molar

enthalpy change . . . . . . . . . . . . . . . . . . . . . 24Exa 2.8 To determine the theoretical horsepower developed . . 25Exa 2.9 To find temperature of water delivered to second storage

tank . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26Exa 2.10 To find change in enthalpy and maximum enthalpy change 27Exa 2.11 To determine heat transfer rates . . . . . . . . . . . . 28

Exa 2.12 To find change in internal energy enthalpy heat suppliedand work done . . . . . . . . . . . . . . . . . . . . . . 29

Exa 2.13 To determine change in internal energy and change inenthalpy . . . . . . . . . . . . . . . . . . . . . . . . . 31

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Exa 3.1 To find the molar volume of air . . . . . . . . . . . . . 34

Exa 3.2 Theoretical problem . . . . . . . . . . . . . . . . . . . 35Exa 3.3 To determine heat and work effects for each step . . . 35Exa 3.4 To calculate change in internal energy change in en-

thalpy work done and heat supplied . . . . . . . . . . 37Exa 3.5 To determine work done and amount of heat transferred 39Exa 3.6 To compare the pressures . . . . . . . . . . . . . . . . 40Exa 3.7 To calculate the volume . . . . . . . . . . . . . . . . . 41Exa 3.8 Theoretical problem . . . . . . . . . . . . . . . . . . . 43Exa 3.9 To calculate compressibility factor and molar volume . 43Exa 3.10 To calculate heat of formation of methane gas . . . . . 46Exa 3.11 To calculate heat of formation of chloroform . . . . . . 47

Exa 3.12 To calculate standard heat of reaction at 773 K . . . . 47Exa 3.13 To determine heat added or removed . . . . . . . . . . 48Exa 3.14 To calculate theoretical flame temperature . . . . . . . 49Exa 4.1 To calculate the maximum efficiency . . . . . . . . . . 51Exa 4.2 To determine minimum amount of work done and heat

given to surrounding . . . . . . . . . . . . . . . . . . . 52Exa 4.3 To determine efficiency of proposed engine . . . . . . . 53Exa 4.4 To calculate entropy of evaporation . . . . . . . . . . . 54Exa 4.5 To determine change in entropy . . . . . . . . . . . . . 54Exa 4.6 To calculate the entropy change . . . . . . . . . . . . . 55

Exa 4.7 To determine change in entropy . . . . . . . . . . . . . 56Exa 4.8 To determine the change in entropy . . . . . . . . . . 57Exa 4.9 To calculate the total entropy change . . . . . . . . . . 58Exa 4.10 To calculate entropy of 1 kmole of air . . . . . . . . . 59Exa 4.11 To determine change in entropy for the reaction . . . . 59Exa 4.12 Theoretical problem . . . . . . . . . . . . . . . . . . . 61Exa 4.13 To calculate change in entropy and check whether the

process is reversible . . . . . . . . . . . . . . . . . . . 61Exa 4.14 To determine the change in entropy of system . . . . . 62Exa 4.15 To calculate entropy change . . . . . . . . . . . . . . . 64Exa 4.16 To calculate entropy change in the process . . . . . . . 66

Exa 4.17 To calculate loss in capacity of doing work . . . . . . . 67Exa 4.18 To calculate total change in entropy and available work 68Exa 4.19 To calculate the molar entropy of metal . . . . . . . . 69Exa 4.20 To calculate the absolute entropy of water vapour . . . 70Exa 5.1 To calculate the pressure at exit . . . . . . . . . . . . 72

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Exa 5.2 To determine quality of steam flowing through the pipe 73

Exa 5.3 To determine the discharge velocity . . . . . . . . . . 74Exa 5.4 To determine thermodynamic properties at throat andcritical pressure . . . . . . . . . . . . . . . . . . . . . . 75

Exa 5.5 Theoretical problem . . . . . . . . . . . . . . . . . . . 77Exa 5.6 Theoretical problem . . . . . . . . . . . . . . . . . . . 77Exa 5.7 To calculate work required and temperature after com-

pression . . . . . . . . . . . . . . . . . . . . . . . . . . 78Exa 5.8 To calculate work required and temperature . . . . . . 79Exa 5.9 To determine the least amount of power . . . . . . . . 80Exa 5.10 To determine COP heat rejected and lowest temperature 81Exa 5.11 To determine COP at given conditions . . . . . . . . . 82

Exa 5.12 To determine power requirement and refrigeration ca-pacity in tonnes . . . . . . . . . . . . . . . . . . . . . 83

Exa 5.13 To calculate the COP and refrigeration circulation rate 84Exa 5.14 To determine the COP and air circulation rate . . . . 86Exa 5.15 To verify that given heat pump is equivalent to 30 kW

pump . . . . . . . . . . . . . . . . . . . . . . . . . . . 87Exa 5.16 To determine the amount of fuel burned . . . . . . . . 88Exa 5.17 To calculate fraction of liquid in inlet stream and tem-

perature . . . . . . . . . . . . . . . . . . . . . . . . . . 89Exa 5.18 To determine fraction of air liquified and temperature

of air . . . . . . . . . . . . . . . . . . . . . . . . . . . 90Exa 5.19 To determine ideal Rankine cycle efficiency thermal ef-ficiency and rate of steam production . . . . . . . . . 92

Exa 5.20 To determine the work output thermal efficiency andrate of steam circulation . . . . . . . . . . . . . . . . . 94

Exa 5.21 To determine fraction of steam withdrawn and thermalefficiency of cycle . . . . . . . . . . . . . . . . . . . . . 96

Exa 5.22 To determine mean effective pressure . . . . . . . . . . 98Exa 5.23 To determine work done thermal effeciency and mean

effective pressure . . . . . . . . . . . . . . . . . . . . . 100Exa 5.24 To determine temperature pressure work and thermal

effeciency . . . . . . . . . . . . . . . . . . . . . . . . . 102Exa 6.1 To determine change in entropy of system . . . . . . . 105Exa 6.2 To calculate vapour pressure of water at 363 K . . . . 106Exa 6.3 To determine the melting point of mercury at 10 bar . 107Exa 6.4 To calculate increase in entropy of solid magnesium . . 108

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Exa 6.5 Theoretical problem . . . . . . . . . . . . . . . . . . . 109

Exa 6.6 Theoretical problem . . . . . . . . . . . . . . . . . . . 109Exa 6.7 To calculate internal energy enthalpy entropy and freenergy for 1 mole of nitrogen . . . . . . . . . . . . . . 110

Exa 6.8 To calculate entropy change and mean heat capacity . 111Exa 6.9 Theoretical problem . . . . . . . . . . . . . . . . . . . 113Exa 6.10 To calculate Cv for mercury . . . . . . . . . . . . . . . 114Exa 6.11 Theoretical problem . . . . . . . . . . . . . . . . . . . 115Exa 6.12 Theoretical problem . . . . . . . . . . . . . . . . . . . 115Exa 6.13 Theoretical problem . . . . . . . . . . . . . . . . . . . 116Exa 6.14 Theoretical problem . . . . . . . . . . . . . . . . . . . 117Exa 6.15 Theoretical problem . . . . . . . . . . . . . . . . . . . 117

Exa 6.16 Theoretical problem . . . . . . . . . . . . . . . . . . . 118Exa 6.17 Theoretical problem . . . . . . . . . . . . . . . . . . . 118Exa 6.18 Theoretical problem . . . . . . . . . . . . . . . . . . . 119Exa 6.19 Theoretical problem . . . . . . . . . . . . . . . . . . . 119Exa 6.20 Theoretical problem . . . . . . . . . . . . . . . . . . . 120Exa 6.21 To estimate the fugacity of ammonia . . . . . . . . . . 121Exa 6.22 To determine the fugacity of gas . . . . . . . . . . . . 121Exa 6.23 To determine the fugacity coeffeceint at given pressure 122Exa 6.24 Theoretical problem . . . . . . . . . . . . . . . . . . . 123Exa 6.25 To determine the fugacity of pure ethylene . . . . . . . 124

Exa 6.26 To determine fugacity and fugacity coeffecient of steam 124Exa 6.27 To estimate fugacity of ammonia . . . . . . . . . . . . 125Exa 6.28 To calculate the fugacity of liquid water . . . . . . . . 126Exa 6.29 To determine the fugacity of n butane in liquid state at

given conditions . . . . . . . . . . . . . . . . . . . . . 127Exa 6.30 To determine the activity of solid magnesium . . . . . 128Exa 7.1 Theoretical problem . . . . . . . . . . . . . . . . . . . 130Exa 7.2 To find the volume of mixture . . . . . . . . . . . . . . 130Exa 7.3 To find the required volume of methanol and water . . 132Exa 7.4 To calculate the volume of water to be added and vol-

ume of dilute alcohol solution . . . . . . . . . . . . . . 133

Exa 7.5 Theoretical problem . . . . . . . . . . . . . . . . . . . 134Exa 7.6 To determine enthalpies of pure components and at in-

finite dilution . . . . . . . . . . . . . . . . . . . . . . . 135Exa 7.7 To calculate the partial molar volume of the components 136Exa 7.8 Theoretical problem . . . . . . . . . . . . . . . . . . . 137

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Exa 7.9 Theoretical problem . . . . . . . . . . . . . . . . . . . 138

Exa 7.10 To estimate the solubility of oxygen in water at 298 K 139Exa 7.11 To confirm that mixture conforms to Raoults Law andto determine Henrys Law constant . . . . . . . . . . . 140

Exa 7.12 To calculate activity and activity coeffecient of chloro-form . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141

Exa 7.13 To determine fugacity fugacity coeffecient Henrys Lawconstant and activity coeffecient . . . . . . . . . . . . 143

Exa 7.14 Theoretical problem . . . . . . . . . . . . . . . . . . . 144Exa 7.15 Theoretical problem . . . . . . . . . . . . . . . . . . . 145Exa 7.16 Theoretical problem . . . . . . . . . . . . . . . . . . . 146Exa 7.17 To determine enthalpies at infinite dilution . . . . . . 146

Exa 7.18 Theoretical problem . . . . . . . . . . . . . . . . . . . 147Exa 7.19 To determine change in entropy for the contents of the

vessel . . . . . . . . . . . . . . . . . . . . . . . . . . . 148Exa 7.20 To determine heat of formation of LiCl in 12 moles of

water . . . . . . . . . . . . . . . . . . . . . . . . . . . 149Exa 7.21 To calculate the free energy of mixing . . . . . . . . . 150Exa 7.22 To calculate the mean heat capacity of 20 mol percent

solution . . . . . . . . . . . . . . . . . . . . . . . . . . 151Exa 7.23 To find the final temperature attained . . . . . . . . . 153Exa 7.24 Theoretical problem . . . . . . . . . . . . . . . . . . . 153

Exa 8.1 Theoretical problem . . . . . . . . . . . . . . . . . . . 155Exa 8.2 Theoretical problem . . . . . . . . . . . . . . . . . . . 155Exa 8.3 Theoretical problem . . . . . . . . . . . . . . . . . . . 156Exa 8.4 Theoretical problem . . . . . . . . . . . . . . . . . . . 157Exa 8.5 Theoretical problem . . . . . . . . . . . . . . . . . . . 157Exa 8.6 To determine composition of vapour and liquid in equi-

librium . . . . . . . . . . . . . . . . . . . . . . . . . . 158Exa 8.7 To determine pressure at the beginning and at the end

of the process . . . . . . . . . . . . . . . . . . . . . . . 159Exa 8.8 To determine temperature pressure and compositions . 160Exa 8.9 To construct boiling point and equilibrium point diagram 163

Exa 8.10 Theoretical problem . . . . . . . . . . . . . . . . . . . 165Exa 8.11 To calculate van Laar constants . . . . . . . . . . . . . 165Exa 8.12 To calculate activity coeffecients in a solution containing

10 percent alcohol . . . . . . . . . . . . . . . . . . . . 166

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Exa 8.13 To calculate equilibrium vapour composition for solu-

tion containing 20 mole percent hydrazine . . . . . . . 167Exa 8.14 Theoretical problem . . . . . . . . . . . . . . . . . . . 169Exa 8.15 To determine the total pressure . . . . . . . . . . . . . 170Exa 8.16 To construct the Pxy diagram . . . . . . . . . . . . . 171Exa 8.17 To determine the composition and total pressure of azeotrope 172Exa 8.18 Theoretical problem . . . . . . . . . . . . . . . . . . . 174Exa 8.19 To calculate equilibrium pressure and composition . . 174Exa 8.20 To determine parameters in Wilsons equation . . . . . 177Exa 8.21 To alculate bubble and dew point and the composition 179Exa 8.22 To calculate bubble and dew point temperatures . . . 182Exa 8.23 To test whetherthe given data are thermodynamically

consistent or not . . . . . . . . . . . . . . . . . . . . . 185Exa 8.24 Theoretical problem . . . . . . . . . . . . . . . . . . . 186Exa 8.25 To estimate the constants in Margules equation . . . . 187Exa 8.26 To calculate the partial pressure of water in vapour phase 188Exa 8.27 to calculate under three phase equilibrium . . . . . . . 189Exa 8.28 To prepare temperature composition diagram . . . . . 191Exa 9.1 Theoretical problem . . . . . . . . . . . . . . . . . . . 194Exa 9.2 Theoretical problem . . . . . . . . . . . . . . . . . . . 194Exa 9.3 Theoretical problem . . . . . . . . . . . . . . . . . . . 195Exa 9.4 Theoretical problem . . . . . . . . . . . . . . . . . . . 196

Exa 9.5 Theoretical problem . . . . . . . . . . . . . . . . . . . 196Exa 9.6 To calculate equilibrium constant . . . . . . . . . . . . 197Exa 9.7 To calculate equilibrium constant at 500 K . . . . . . 198Exa 9.8 To alculate standard free energy change and heat of for-

mation . . . . . . . . . . . . . . . . . . . . . . . . . . 199Exa 9.9 To estimate free energy change and equilibrium constant

at 700 K . . . . . . . . . . . . . . . . . . . . . . . . . 200Exa 9.10 to calculate equilibrium constant at 600 K . . . . . . . 201Exa 9.11 To calculate equilibrium constant at 500K . . . . . . . 202Exa 9.12 To find the value of n . . . . . . . . . . . . . . . . . . 203Exa 9.13 To determine the percent conversion . . . . . . . . . . 205

Exa 9.14 To calculate fractional dissociation of steam . . . . . . 206Exa 9.15 To determine conversion of nitrogen affected by argon 207Exa 9.16 To calculate the fractional dissociation of steam . . . . 208Exa 9.17 To calculate the fractional distillation of steam . . . . 210Exa 9.18 To evaluate the percent conversion of CO . . . . . . . 211

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Exa 9.19 To determine the composition of gases leaving the reactor 212

Exa 9.20 To evaluate the equilibrium constant . . . . . . . . . . 214Exa 9.21 To calculate the decomposition pressure and tempera-ture at 1 bar . . . . . . . . . . . . . . . . . . . . . . . 215

Exa 9.22 To evaluate wt of iron produced per 100 cubic m of gasadmitted . . . . . . . . . . . . . . . . . . . . . . . . . 216

Exa 9.23 To calculate the composition at equilibrium assumingideal behaviour . . . . . . . . . . . . . . . . . . . . . . 218

Exa 9.24 To determine the number of degrees of freedom . . . . 220

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Chapter 1

Introduction and Basic

Concepts

Scilab code Exa 1.1 To find mans mass and weight on earth

1 / /A T ex tb o ok o f C h em i c al E n g i n e e r i n g T he rm od yn am ic s2 / / C ha p te r 13 / / I n t r o d u c t i o n and B a s i c C on ce pt s4 // E xampl e 15

6 c l e ar a ll ;

7 clc ;

8

9

10 // Given :11 F = 300; // [N]12 g _ l oc a l = 4 .5 ; / / l o c a l g r a v i t a t i o n a l a c c e l e r a t i o n [m/

s ̂ 2 ]13 g _ e ar t h = 9 .8 1; // e ar th ’ s g r a v i t a t i o n a l a c c e l e r a t i o n

[m/ s ̂ 2 ]1415

16 / /To f i n d man ’ s mass and w ei g ht on e a r t h17 m = F / g_ lo ca l ; / / m a ss o f man [ k g ]

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18 w = m * g_ ea rt h ; / / w ei g h t o f man on e a r t h [ N]

19 mprintf ( ’ Mass o f man i s %f kg ’ , m ) ;20 mprintf ( ’ \nWeight o f man o n e a r th i s %f N ’ , w ) ;21

22

23 //e nd

Scilab code Exa 1.2 To find height of manometer fluid


6 c l e ar a ll ;

7 clc ;

8

9

10 // Given :11 p 1 = 1 . 15 * 10 ^ 5; / / m e a su r ed p r e s s u r e [ N/mˆ 2 ]

12 p 2 = 1 . 01 3 25 * 10 ^ 5; / / a t m o s p h e r i c p r e s s u r e [ N/mˆ 2 ]13 s g = 2 .9 5; / / s p e c i f i c g r a v i t y o f f l u i d14

15 / /To f i n d h e i g ht o f manometer f l u i d16 p = p 1 - p 2 ; / / d i f f e r e n c e i n p r e s s u r e17 // U si ng e q ua t i on 1 . 2 ( P age no . 6 )18 h = p / ( sg * ( 1 0^ 3 ) * 9 .8 0 67 ) ; / / h e i g h t o f m an om et er

f l u i d [m]19 mprintf ( ’ H ei gh t o f manometer f l u i d i s %f m’ , h ) ;20

2122 //e nd

12


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Scilab code Exa 1.3 To find height from ground and Kinetic Energy


6 c l e ar a ll ;

7 clc ;

8

9

10 //Gi v e n

11 P E = 1 . 5* 1 0^ 3 ; / / p o t e n t i a l e n e r g y [ J ]12 m = 10; // m ass i n kg13 u = 50; // v e l o c i t y i n m/ s14

15 / /To f i n d h e i g ht from g ro un d and k i n e t i c e ne rg y16 // U si ng e q ua t i on 1 . 8 ( P age no . 8 )17 h = P E /( m * 9 .8 0 67 ) ; // h e i g h t fro m g ro un d i n m18

19 // U si ng e q ua t i on 1 . 9 ( P age no . 8 )20 K E = 0 .5 * m *( u ^ 2) ; // K i ne t i c e ne rg y i n J21 mprintf (

’ H ei g ht f ro m g ro un d i s %f m ’, h ) ;

22 mprintf ( ’ \ n K i n e t i c E ne rg y o f body i s %3 . 2 e J ’ , K E ) ;23

24

25 //e nd

Scilab code Exa 1.4 To determine the power developed in man


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6 c l e ar a ll ;

7 clc ;8

9

10 //Gi v e n11 F = 600; // w e ig h t i n N12 t = 120; // t im e i n s e c13 h = 0. 18 ; / / h e i gh t o f s t a i r s i n m14

15 / /To d e t er m i ne t h e p ow er d e v el o p ed i n man16 S = 20* h; // t o t a l v e r t i c a l d i s pl ac em en t i n m17 W = F * S ; // work d on e i n J

18 P = W / t ; // p o we r d e v e l o p e d19 mprintf ( ’ P ower d e v el o p ed i s %i W’ , P ) ;20

21

22 //e nd

Scilab code Exa 1.5 To determine the force exerted pressure work doneand change in potential energy


6 c l e ar a ll ;

7 clc ;

8

9

10 // Given :11 A = ( % pi / 4 ) * (0 .1 ^ 2) ; / / a r e a i n mˆ 212 P = 1 . 01 3 25 * 10 ^ 5; / / p r e s s u r e i n N/mˆ 213 m = 50; // mass o f p i s t on and w ei gh t i n kg14 g = 9. 81 ; / / a c c e l e r a t i o n due t o g r a v i t y (N/mˆ 2)

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15

1617 / /To d et er m in e t he f o r c e e x e rt e d p r e s s u r e work do ne

and c ha ng e i n p o t e n t i a l e ne rg y18 / / ( a )19 F a = P *A ; / / f o r c e e x e rt e d by a tm os ph er e i n N20 F p = m *g ; // f o r c e e x er t ed by p i s t o n and w ei gh t i n N21 F = Fp + Fa ; // t o t a l f o r c e e x e r t e d i n N22 mprintf ( ’ T ot al f o r c e e x e rt e d by t he a tm os ph er e , t he

p i s t o n and t he w ei gh t i s %f N ’ , F ) ;23

24 / / ( b )

25 P g = F /A ; / / p r e s s u r e o f g a s i n N/mˆ226 mprintf ( ’ \ n P re s su r e o f g as i s %5 . 4 e Pa ’ , P g ) ;27

28 // ( c )29 S = 0.4; / / d i s p la c e me n t o f g as i n m30 W = F * S ; // work done by g as i n J31 mprintf ( ’ \nWork d on e by g a s i s %f J ’ , W ) ;32

33 / / ( d )34 P E = m *g *S ; // c ha n ge i n p o t e n t i a l e ne rg y i n J35 mprintf (

’ \nChange i n p o t e n t i a l e ne rg y i s %f J ’, P E ) ;

36

37 //e nd

Scilab code Exa 1.6 To determine work done by gas

1 / /A T ex tb o ok o f C h em i c al E n g i n e e r i n g T he rm od yn am ic s2 / / C ha p te r 1

3 / / I n t r o d u c t i o n and B a s i c C on ce pt s4 // E xampl e 65

6

7 c l e ar a ll ;

15


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8 clc ;

910

11 // Given :12 //P/D = c o n st an t , wh ere P i s p r e s s u r e and D i s

d i a m e t e r13 / / P = ( 2∗1 0 ˆ 5 )∗D14 D f = 2. 5; / / f i n a l d i a m et e r (m)15 D i = 0. 5; // i n i t i a l di am et er (m)16

17 / / To d e t e r m i ne wo rk d on e by g a s18 / / Work d o ne = i n t e g r a l ( PdV )

19 //W = i n t g ( ( 2 ∗1 0 ˆ 5∗D) d( pi /6) ( Dˆ3) ) . . . . t ha t i s20 W = ( % p i / 4) * 1 0 ^ 5 * (( D f ^ 4 ) - D i ^ 4) ;

21 mprintf ( ’ Work d on e by g a s i s %6 . 4 e J ’ , W ) ;22

23 //e nd

Scilab code Exa 1.7 To find the work done on surrounding


6

7 c l e ar a ll ;

8 clc ;

9

10

11 // Given :12 T = 300; / / t e m p e r a tu r e i n K13 P = 6 .5 *1 0^ 5; / / p r e s s u r e i n N/mˆ 214 P a = 1 . 01 3 25 * 10 ^ 5; / / a t m o s p h e r i c p r e s s u r e i n N/mˆ 215 R = 8 .3 14 ; / / i d e a l g as c o n st a n t

16


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16 m = 2 ; // mass o f g as ( kg )

17 M = 44; / / m o l ec u l ar w ei hg t o f g as1819 / /To f i n d t h e work d on e o n s u r r o un d i n g20 n = m / M ; // n i s number o f k mo le s21 V i = ( n * R *1 0^ 3* T ) / P ; / / i n i t i a l v o lu m e i n mˆ 322 V f = 2* Vi ; / / f i n a l vo lum e i n mˆ 323 V = V f - V i ; // c h a n ge i n v ol um e24 P s = P a + (5 0 00 * 9. 8 06 7 ) ; / / p r e s s u r e on s u r r o u n d i n g s25 W = Ps *V ; // work d on e on t h e s u r r o u n d i n g s26 mprintf ( ’ Work d on e on s u r r o u n d i n g s i s %5 . 2 e J ’ , W ) ;27

2829 //e nd

17


19/222

Chapter 2

First Law of Thermodynamics

Scilab code Exa 2.1 To find change in internal energy

1 / /A T ex tb o ok o f C h em i c al E n g i n e e r i n g T he rm od yn am ic s2 / / C ha p te r 23 / / F i r s t Law o f T he rm o dy na m ic s4 // E xampl e 15

6

7 c l e ar a ll ;8 clc ;

9

10

11 //Gi v e n12 W = - 2. 25 *7 45 . 7; // w ork d on e on s ys te m i n J / s13 Q = - 3 40 0 *( 1 0 ^3 ) / 3 6 0 0; / / h ea t t r a n s f e r r e d t o t he

s ur r o un d in g i n J / s14

15 //To f i n d t he c ha nge i n i n t e r n a l e ne rg y

16 // U si ng e q ua t i on 2 . 4 ( P age no . 2 6)17 U = Q - W; // c ha ng e i n i n t e r n a l e ne rg y i n J / s18 mprintf ( ’ I n t e r n a l e ne rg y o f sy st e m i n c r e a s e s by %f J

/ s ’ , U ) ;19

18


20/222

20 //e nd

Scilab code Exa 2.2 To find heat liberated work done and change in in-ternal energy


56

7 c l e ar a ll ;

8 clc ;

9

10

11 //Gi v e n12 T = 298; / / t e m p e r a tu r e i n K13 P = 101; / / p r e s s u r e i n kPa14 n _ ir on = 2; // m ol es o f i r o n r e a ct e d15 Q = - 83 1. 08 ; // h ea t l i b e r a t e d i n kJ

16 R = 8 .3 14 ; / / i d e a l g as c o n st a n t17

18 / /To f i n d h ea t l i b e r a t e d work done and c ha ng e i ni n t e r n a l e ne rg y

19 mprintf ( ’ H eat l i b e r a t e d d ur in g t h e r e a c t i o n i s %f k J’ , Q ) ;

20 n _ o xy g en = 1 .5 ; // m o le s o f o xyg en r e a c t ed21

22 // U si ng i d e a l g as e q ua t i on P ( Vf −Vi )=nRT and W=P( Vf −Vi )

23 W = - 1.5 * R* T ; // work d on e b y s ys te m i n J2425 // U si ng e q ua t i on 2 . 4 ( P age no . 2 6)26 U = ( Q *1 0^ 3) - W ; // c h a ng e i n i n t e r n a l e n er gy i n J27 mprintf ( ’ \nWork d on e by g a s i s %f J ’ , W ) ;

19


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28 mprintf ( ’ \nChange i n i n t e r n a l e ne rg y i s %6 . 3 e J ’ , U ) ;

2930 //e nd

Scilab code Exa 2.3 To find the heat energy dissipated by brakes

1 / /A T ex tb o ok o f C h em i c al E n g i n e e r i n g T he rm od yn am ic s2 / / C ha p te r 23 / / F i r s t Law o f T he rm o dy na m ic s

4 // E xampl e 356

7 c l e ar a ll ;

8 clc ;

9

10

11 //Gi v e n12 u = 20; / / s pe ed o f c a r i n m/ s13 z = 30; // h e i g ht v e r t i c a l l y a bo ve t he bottom o f h i l l

i n m

14 m = 14 00 ; // mass o f c ar i n kg15

16 / /To f i n d t he h ea t e ne rg y d i s s i p a t e d by b ra k e s17 // U si ng e q ua t i on 2 . 3 ( P age no . 2 6)18 K E = - 0. 5* m * ( u ^2 ) ; // c ha n ge i n k i n e t i c e ne rg y i n J19 P E = - m *9 .8 1* z ; // c h a ng e i n p o t e n t i a l e ne rg y i n J20 Q = -( KE + PE ) ; // h ea t d i s s i p a t e d by b ra k e s i n J21 mprintf ( ’ H eat d i s s i p a t e d by b ra k e s i s %3 . 2 e J ’ , Q ) ;22

23 //e nd

Scilab code Exa 2.4 To find internal energy change during each step andwork done during adiabatic process

20


22/222

1 / /A T ex tb o ok o f C h em i c al E n g i n e e r i n g T he rm od yn am ic s

2 / / C ha p te r 23 / / F i r s t Law o f T he rm o dy na m ic s4 // E xampl e 45

6

7 c l e ar a ll ;

8 clc ;

9

10

11 // Given :12 // S te p 1 : c o ns t an t p r e s s ur e p r o c e s s

13 // S t ep 2 : c o s t a n t volume p r o c e s s14 // S te p 3 : a d i b a t i c p r oc e ss15

16 / /To f i n d i n t e r n a l e ne rg y c ha nge d ur in g e a ch s t epand work done d u ri n g a d i a b a t i c p r o c e s s

17

18 / / For s t e p 119 W 1 = -50; // work r e c e i v e d i n J20 Q 1 = -25; / / h ea t gv en o ut i n J21 U 1 = Q1 - W1 ; // i n t e r n a l e ne rg y c h ang e i n J22 mprintf (

’ Change i n i n t e r n a l e ne rg y f o r c on s ta n tp r e s s u r e p r oc e ss i s %i J ’ , U 1 ) ;23

24 / / For s t e p 225 W 2 = 0; // work done f o r c o n st a n t volume p r o c e s s i s

z e r o26 Q 2 = 75; / / h ea t r e c e i v e d i n J27 U 2 = Q2 ; // i n t e r n a l e ne rg y c h an ge i n J28 mprintf ( ’ \nChange i n i n t e r n a l e ne rg y f o r c on s ta n t

volume p r o c e s s i s %i J ’ , U 2 ) ;29

30 / / For s t e p 331 Q 3 = 0; // no h ea t e xc ha ng e i n a d i a b a t i c p r o c es s32 / / S i n ce t he p r o c e s s i s c y c l i c33 // U3+U2+U1 = 0 ;34 U 3 = -( U1 + U2 ) ;

21


23/222

35 W 3 = - U3 ; // work d on e i n J

36 mprintf ( ’ \nWork d one d u ri n g a d i a b a t i c p r o c e s s i s %iJ ’ , W 3 ) ;37

38 //e nd

Scilab code Exa 2.5 To find change in internal energy and enthalpy



6

7 c l e ar a ll ;

8 clc ;

9

10

11 // Given :12 n _ w at e r = 1 0^ 3; // m o le s o f w at er

13 T = 373; // t e m pe a r t u r e (K)14 P = 1 01 .3 ; / / p r e s s u r e ( k Pa )15 s v _ l i qu i d = 0 . 0 01 0 4 ; / / s p e c i f i c v o lu m e o f l i q u i d (m

ˆ 3 / k m o l )16 s v _v a po u r = 1 .6 75 ; // s p e c i f i c volume o f vapou r (mˆ3/

k mol )17 Q = 1 .0 3* 10 ^3 ; / / h e a t a dd ed i n kJ18

19 / /To f i n d c h an ge i n i n t e r n a l e ne rg y and e nt ha l p y20 W = P * n _ w a te r * ( s v _ va p ou r - s v _ l i qu i d ) * 1 0^ - 3 ; //

e x p an s i o n work d on e i n kJ21 U = Q - W; // c ha ng e i n i n t e r n a l e ne rg y i n kJ22

23 / / For c o ns t an t p r e s su r e p r o c es s24 H = Q ; / / e n t ha l p y c ha ng e i n kJ

22


24/222

25 mprintf ( ’ Change i n i n t e r n a l e ne rg y i s %f kJ ’ , U ) ;

26 mprintf ( ’ \nChange i n e n t ha l p y i s %3 . 2 e kJ ’ , H ) ;2728 //e nd

Scilab code Exa 2.6 To find internal energy of saturated liquid and inter-nal energy and enthalpy of saturated vapour



6

7 c l e ar a ll ;

8 clc ;

9

10

11 // Given :12 T = 233; / / t e m p e r a tu r e i n K

13 V P = 1 . 00 5 *1 0 ^3 ; // v a po ur p r e s s u r e o f CO2 i n kPa14 s v _ l i qu i d = 0 . 9* 1 0^ - 3 ; / / s p e c i f i c v ol um e o f l i q u i d

CO2 i n mˆ3/k g15 s v _ v a po u r = 3 8 .2 * 10 ^ - 3 ; // s p e c i c i f i c volume o f CO2

v a po u r i n mˆ 3 / k g16 L = 3 20 .5 ; // l a t e n t h ea t o f v a p o r i s a t i o n o f CO2 i n

k J /k g17 / / Assuming a t t h e s e c o n d i t i o n s CO2 i s s a t u r a t ed

l i q u i d s o18 H 1 = 0; // e nt ha lp y i n l i q u i d s t a t e

1920 / /To f i n d i n t e r n a l e ne rg y o f s a tu r a t ed l i q u i d andi n t e r n a l e ne rg y and e nt ha l p y o f s a t ur a t ed v a pou r

21 // For s a t ur a t e d l i q u i d22 U 1 = H1 - ( V P * s v_ l iq u id ) ; // i n t e r n a l e n e r g y i n l i q u i d

23


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s t a t e i n kJ /kg

23 / / For s a t u r a t e d v ap ou r24 H v = H1 +L ; / / e n th a lp y o f s a t u r a t ed v ap ou r i n kJ / kg25 U v = Hv - ( V P * s v_ v ap o ur ) ; // i n t e r n a l e ne rg y i n v ap ou r

s t a t e i n kJ /kg26 mprintf ( ’ I n t e r n a l Energy o f s a tu r at e d l i q u i d i s %f

kJ/kg ’ , U 1 ) ;27 mprintf ( ’ \ n En th al py o f v ap ou r s t a t e i s %f kJ / kg ’ , H v )

;

28 mprintf ( ’ \ n I n t e rn a l Energy o f v apo ur s t a t e i s %f k J/kg ’ , U v ) ;

29

30 //e nd

Scilab code Exa 2.7 To calculate molar internal energy change and molarenthalpy change


4 // E xampl e 75

6

7 c l e ar a ll ;

8 clc ;

9

10

11 // Given :12 I = 0.5; / / c u r r e n t i n A mp er es13 V = 12; / / v o l t a g e i n v o l t s

14 t = 5* 60 ; // t im e i n s e c15 m = 0 .7 98 ; // mass o f w at er v a p or i se d i n g16 M = 18; / / m o l ec u l ar mass o f w at er i n g17

18 //To c a l c u l a t e m ol ar i n t e r n a l e ne rg y c ha ng e and

24


26/222

m ol a r e n t h a l p y c ha n ge

19 Q = ( I * V* t / 10 0 0) ; / / e l e c t r i c e n e r g y s u p p l i e d i n k J20 // R e f e r r i n g e q ua t i on 2 . 1 0 ( P age no . 2 9)21 H = ( Q* M) /m ; // m o la r e n t ha l p y c ha ng e i n kJ / mo le22

23 / /BY i d e a l g a s e q u a t i o n PV=RT24 / / R e f e r r i n g e qu a ti o n 2 . 9 f o r c on s ta n t p r e s s ur e

p r o c e s s ( P age no . 2 9)25 U = H - ( 8 .3 1 4* 1 0^ - 3 * 3 73 ) ; // m ol ar i n t e r n a l e ne rg y

c ha n ge i n kJ / m ol e26 mprintf ( ’ Mo lar E nt ha lp y c ha ng e d u ri n g t he p r o c e s s i s

%i k J / m o l e ’ , H ) ;

27 mprintf ( ’ \n Mo la r I n t e r a n l E ne rg y c h a ng e d u r in g t h ep r o c e s s i s %f kJ / mole ’ , U ) ;

28

29 //e nd

Scilab code Exa 2.8 To determine the theoretical horsepower developed



6

7 c l e ar a ll ;

8 clc ;

9

10

11 // Given :

12 m = 16 50 ; // mass o f s tea m u se d i n kg / hr13 H 1 = 3 20 0; / / e n t ha l p y a t 1 36 8 kPa and 6 45 K i n kJ / kg14 H 2 = 2 69 0; / / e n t ha l p y a t 1 37 kPa and 6 45 K i n kJ / kg15

16 / /To d et er m in e t he t h e o r e t i c a l h or se po we r d e ve l op e d

25


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17 / / U s in g e q u a t i o n 2 . 1 3 ( P age no . 3 2 )

18 Q = 0 ; // s i n c e t he p r oc e ss i s a d i a b a t i c19 z = 0 ; // a ss umi ng t ha t i n l e t and d i s c ha r g e o f t u r b i n e a re a t s ame l e v e l

20 u = 0 ; / / f e e d a nd d i s c h a r g e v e l o c i t i e s b e i n g e q u a l21 W s = -( H2 - H 1) ;

22 W j = W s * 10 ^3 * m / 3 60 0; // work done by t u r b i n e i n J23 W = Wj / 7 45 .7 ; // work done by t u r b i n e i n hp24 mprintf ( ’ Work d on e by t u r b i n e i s %f hp ’ , W ) ;25

26 //e nd

Scilab code Exa 2.9 To find temperature of water delivered to second stor-age tank



8

9

10 // Given :11 m = 2 5* 10 ^3 ; // mass f l o w r a t e o f w a te r i n kg /h12 P = 2 ; // p ow er s u p p l i e d by m oto r i n hp13 q = 4 20 00 ; / / h e at g i v e n i n kJ / min14 z = 20; // e l e v a t i o n i n m15 T = 368; / / t e m p e r a tu r e i n K

16 T o = 27 3; / / s t a nd a r d t e m pe r a tu r e i n K17 C p = 4. 2; / / s p e c i f i c h e a t o f w a te r i n kJ / kg K18

19 / /To f i n d t em pe ra tu re o f w a te r d e l i v e r e d t o s ec on ds t o r a g e t an k

26


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20 W = ( P * 7 45 .7 * 10 ^ - 3 * 36 0 0) / m ; // work d on e p e r kg o f

w a t e r pumped i n k J / kg21 Q = q *6 0/ m ; // h ea t g i v en o ut p er kg o f f l u i d22 P E = 0 9. 8 1* z * 1 0^ - 3 ; // c ha ng e i n p o t e n t i a l e ne rg y i n

k J /k g23

24 // U si ng e q ua t i on 2 . 1 3 ( P age no . 3 2)25 H = - Q+ W -PE ;

26 //H = H2−H127 H 1 = Cp * (T - To ) ;

28 H 2 = H1 +H ;

29 / / L et T1 b e t he t em p er a tu re a t s ec on d s t o r a g e t an k

30 T 1 = T o +( H 2 / Cp ) ;31 mprintf ( ’ T em pe ra tu re o f w at er a t s e co n d s t o r a g e t an k

i s %i K ’ , T 1 ) ;32

33 //e nd

Scilab code Exa 2.10 To find change in enthalpy and maximum enthalpychange


6

7 c l e ar a ll ;

8 clc ;

9

1011 // Given :12 D 1 = 25; // i n t e r n a l d i am et e r o f p i pe i n mm13 u 1 = 10; // u ps tr ea m v e l o c i t y i n m/ s14 D 2 = 50; // d o wn st re am d i a m e t e r o f p i p e i n mm

27


29/222

15 / / S i n ce t h e r e i s no e x t e r n al d e v i c e f o r a d di n g o r

r em o vi n g e n e r g y a s wo rk16 / /Q = 0 , Ws = 017

18 / / To f i n d c h an g e i n e n t h a l p y and maximum e n t h a l p yc h a n g e

19

20 / / ( a )21 / / L e t A1 n ad A2 b e u p s tr e am a nd d ow ns tr ea m

c r o s s s e c t i o n a l a r e as o f p ip e22 u 2 = ( ( D1 / D 2 ) ^2 ) * u1 ; // d own st re am v e l o c i t y i n m/ s23 H = 0 .5 *( u 1 ^2 - u 2 ^ 2) ; // c ha ng e i n e n th a lp y i n J /kg

24 mprintf ( ’ C hange i n e n t ha l p y i s %f J / kg ’ , H ) ;25

26 / / ( b )27 / / F o r maximum e n t h a l p y c h a n g e28 u 2 = 0;

29 H m ax = 0 .5 * u1 ^ 2 ; / / ( J / k g )30 mprintf ( ’ \nMaximum e n t h a l p y c h n ag e f o r a s u dd e n

e nl a rg e me n t i n p i pe i s %f J /kg ’ , H m a x ) ;31

32 //e nd

Scilab code Exa 2.11 To determine heat transfer rates


67 c l e ar a ll ;

8 clc ;

9

10

28


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11 // Given :

12 // At i n l e t :13 T 1 = 29 3; // Tempera ture (K)14 P 1 = 3 0 0+ 1 36 . 8; / / P r e s s u r e ( k Pa )15

16 / / At e x i t :17 T 2 = 45 3; // Tempera ture (K)18 P 2 = 1 36 .8 ; / / P r e s s u r e ( k Pa )19 C p = 2 9. 4; / / s p e c i f i c h e a t c a p a c i t y a t c o n s t a n t

p r e s s u r e i n kJ / kmol20 m = 10 00 ; // m ass o f h yd ro ge n i n kg21 M = 2. 02 ; / / m o l e c u l a r m as s o f h y dr o ge n

2223 / /To d et er mi ne h ea t t r a n s f e r r a t e s24 // N e gl e c ti n g t he k i n e t i c nd p o t e n t i a l e ne rg y c ha ng es25 / / Assuming t he p r o c e s s t o be o c c u r in g t hr ou gh a

number o f s t e p s26

27 / / St e p 1 be i s ot h er m al and s t e p 2 be i s o b a r i c28 H 1 = 0; // c han ge i n e nt ha lp y f o r s t ep 129 H 2 = ( m / M ) * Cp * ( T 2 - T 1 ) / 1 00 0 ; // c ha ng e i n e n th a lp y f o r

s t e p 2 i n k J30 H = H2 + H1 ;

31 Q = H ; // h e a t t r a n s f e r r e d i n c o i l s i n kJ32 mprintf ( ’ Heat t r a n s f e r r e d i n c o i l s i s %f k J ’ , Q ) ;33

34 //e nd

Scilab code Exa 2.12 To find change in internal energy enthalpy heat sup-plied and work done


29


31/222

5

67 c l e ar a ll ;

8 clc ;

9

10

11 // Given :12 m = 10; // mass o f a i r i n kg13 P 1 = 10 0; // i n i t i a l pr e s su r e ( k Pa )14 T 1 = 30 0; // i n i t i a l te mp er at ur e (K)15 T 2 = 60 0; / / f i n a l t e m p e r a tu r e (K)16 R = 8 .3 14 ; / / i d e a l g a s c o n s t a n t ( k J / km ol K )

17 C p = 2 9. 09 9; / / s p e c i f i c h e a t c a p a c i t y a t c o n s t a n tp r e s s u r e ( k J /kmol K)

18 C v = 2 0. 78 5; / / s p e c i f i c h e a t c a p a c i t y a t c o n s t s a n tv o lu m e ( k J / k mo l K)

19 M = 29; / / m o l ec u l ar w ei gh t o f a i r20

21 / /To d et er m in e c ha ng e i n i n t e r n a l e ne rg y e n th a lp yh e at s u p p l i e d and work d on e

22 n = m / M ; / / number o f m o le s o f g a s ( k mo l )23 V 1 = ( n * R* T 1 )/ P 1 ; / / i n i t i a l v ol um e o f a i r (mˆ 3 )24

25 / / ( a )26 / / C o ns t an t v ol um e p r o c e s s27 V 2 = V1 ; // f i n a l v ol ume28 // Change i n i n t e r n a l e ne rg y U = n∗ in t g (CvdT) . . . so29 U = n * Cv * ( T2 - T 1 ) ; // c ha ng e i n i n t e r n a l e ne rg y ( kJ )30 Q = U ; / / h e a t s u p p l i e d ( k J )31 W = 0 ; / / w or k d o ne32 H = U + ( n* R *( T2 - T 1 ) ) ; / / c h an ge i n e n t ha l p y ( k J )33 disp ( ’ F or c o n s t a n t v olu me p r o c e s s ’ ) ;34 mprintf ( ’ \nChange i n i n t e r n a l e ne rg y i s %i kJ ’ , U ) ;

35 mprintf ( ’ \nHeat s u p p l i e d i s %i kJ ’ , Q ) ;36 mprintf ( ’ \nWork d on e i s %i kJ ’ , W ) ;37 mprintf ( ’ \nChange i n e n t ha l p y i s %i kJ ’ , H ) ;38

39 / / ( b )

30


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40 / / Co ns ta nt p r e s s u r e p r o c e s s

41 / / Change i n e n t ha l p y H = n∗ in t g (CpdT) . . . so42 H = n * Cp * ( T2 - T 1 ) ; / / c h an ge i n e n t h al p y ( k J )43 Q = H ; / / h e a t s u p p l i e d ( k J )44 U = H - ( n *R * ( T2 - T 1 ) ) ;// c ha ng e i n i n t e r n a l e ne rg y ( kJ )45 W = Q - U; / / w or k d o ne ( k J )46 mprintf ( ’ \n\nFor c o n st a n t p r e s s u r e p r o c e s s ’ ) ;47 mprintf ( ’ \n\nChange i n i n t e r n a l e ne rg y i s %i kJ ’ , U ) ;48 mprintf ( ’ \nHeat s u p p l i e d i s %i kJ ’ , Q ) ;49 mprintf ( ’ \nWork d on e i s %i kJ ’ , W ) ;50 mprintf ( ’ \nChange i n e n t ha l p y i s %i kJ ’ , H ) ;51

52 //e nd

Scilab code Exa 2.13 To determine change in internal energy and changein enthalpy


4 // E xampl e 135

6

7 c l e ar a ll ;

8 clc ;

9

10

11 // Given :12 R = 8 .3 14 ; / / i d e a l g a s c o n s t a n t ( k J / km ol K )13 C v = 2 0. 8; / / s p e c i f i c h e a t c a p a c i t y a t c o n s t a n t

volume (kJ/kmol K)14 C p = 2 9. 1; / / s p e c i f i c h e a t c a p a c i t y a t c o n s t a n tp r e s s u r e ( k J / k mo l K)

15 P 1 = 10; // i n i t i a l p r e s s u r e ( bar )16 T 1 = 28 0; / / i n i t i a l t e m p e r a t u r e i n K

31


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17 P 2 = 1; / / f i n a l p r e s s u r e ( b a r )

18 T 2 = 34 0; / / f i n a l t e m p e r a tu r e (K)1920 / /To d et er m in e t he c ha ng e i n i n t e r n a l e ne rg y and

c ha ng e i n e n t ha l p y21 / / S o l u t i o n22 n = 1 ; // b a s i s : 1 kmol o f i d e a l g as23 V 1 = ( n * R* T 1 ) /( P 1 * 10 0) ; // i n i t i a l volume in mˆ324 V 2 = ( n * R* T 2 ) /( P 2 * 10 0) ; / / f i n a l v ol ume i n mˆ 325

26 / / Assuming t he c h ang e i n s t a t e i s o c c u r i ng a l on g t hef o l l o w i n g two s t ep p r o c es s

27 / / 1 . A c o n st a n t volume p r o c e s s i n whi ch t he p r e s s u r ei s r ed uc e d t o t he f i n a l v al ue P2 and t he

t em p er a tu r e g e t s r ed uc ed t o T228 / / L e t Po a nd V od en ot e t h e p r e s s u r e and v ol um e o f

s y s t e m a f t e r t h i s s t e p29 P o = P2 ;

30 V o = V1 ;

31 T o = ( P o * 10 0* V o ) /( n * R );

32 U 1 = C v *( To - T 1 ) ;

33 H 1 = U 1 +( V 1 * 10 0* ( P2 - P 1 ) ) ;

34 W 1 = 0;

35 Q 1 = U1 ;

36

37 // 2 . A c on st an t p r e s s u r e p r o c e s s i n which t he g as i sh ea te d t o t he f i n a l t em pe ra tu re T2 and t he f i n a lv ol ume V2

38 H 2 = C p *( T2 - T o ) ;

39 U 2 = H2 - 10 0* ( V2 - V 1 ) ;

40 Q 2 = H2 ;

41 W 2 = Q2 - U2 ;

42

43 / / For a c t ua l p r o c e s s44 U = U1 + U2 ; // c ha ng e i n i n t e r n a l e ne rg y ( kJ )45 H = H1 + H2 ; / / c h an ge i n e n t ha l p y ( k J )46 mprintf ( ’ Change i n i n t e r n a l e ne rg y i s %f kJ ’ , U ) ;47 mprintf ( ’ \nChange i n e n t ha l p y i s %f kJ ’ , H ) ;

32


34/222

48

49 //e nd

33


35/222

Chapter 3

PVT Behaviour And Heat

Effects

Scilab code Exa 3.1 To find the molar volume of air

1 / /A T ex tb o ok o f C h em i c al E n g i n e e r i n g T he rm od yn am ic s2 / / C ha p te r 33 //P−V−T B e ha v io u r and Hea t E f f e c t s4 // E xampl e 15

6

7 c l e ar a ll ;

8 clc ;

9

10

11 // Given :12 T = 350; / / t e m p e r a tu r e i n K13 P = 10 ^5 ; / / p r e s s u r e i n N/mˆ 214 R = 8 .3 14 ; / / i d e a l g as c o n st a n t

1516 / /To f i n d t he m ol ar volume o f a i r17

18 V = ( R* T) /P ; / / m o l ar v ol um e i n mˆ 319 mprintf ( ’ Mo la r v ol um e o f a i r i s %3 . 2 e c u b i c m/ mol ’ ,V

34


36/222

) ;

2021 //e nd

Scilab code Exa 3.2 Theoretical problem

1 / /A T ex tb o ok o f C h em i c al E n g i n e e r i n g T he rm od yn am ic s2 / / C ha p te r 33 //P−V−T B e ha v io u r and Hea t E f f e c t s

4 // E xampl e 256

7 c l e ar a ll ;

8 clc ;

9

10

11 // Given :12

13

14 / /The g i v e n pr obl e m i s t h e o r e t i c a l and d oe s no t

i n v o l v e any n u m e ri c a l c o mp ut a ti on15

16 //e nd

Scilab code Exa 3.3 To determine heat and work effects for each step


3 //P−V−T B e ha v io u r and Hea t E f f e c t s4 // E xampl e 35

6

7 c l e ar a ll ;

35


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8 clc ;

910

11 // Given :12 C p = 2 9. 3; / / s p e c i f i c h e a t a t c o n s t a n t p r e s s u r e ( k J /

kmol K)13 R = 8 .3 14 ; / / i d e a l g as c o n st a n t14

15 / /To d e t er m i ne h e at and work e f f e c t s f o r e ac h s t e p16

17 // S t ep 1 : Gas i s h ea te d a t c o n st a n t volume18 T 1 = 30 0; / / t e m p e r at u r e i n K

19 P 1 = 1; / / i n i t i a l p r e s s u r e i n b a r20 P 2 = 2; // f i n a l p r e s s u r e i n b a r21 T 2 = ( P2 / P1 ) * T1 ; // f i n a l t em p er at u re i n K22 C v = Cp - R; / / s p e c i f i c h e a t a t c o n s t a n t v ol um e23 W 1 = 0; // work done i s z e ro a s volume r em ai ns

c o n s t a n t24 Q 1 = C v *( T2 - T 1 ) ; / / h e at s u p p l i e d i n kJ / kmol25 mprintf ( ’ F or s t e p 1 ’ ) ;26 mprintf ( ’ \nWork done i n s t e p 1 i s %i ’ , W 1 ) ;27 mprintf ( ’ \nHeat s u p p l i e d i n s t e p 1 i s %f k J/ kmol ’ , Q1

) ;

28

29 / / S te p 2 : The p r oc e ss i s a d i a b a t i c30 Q 2 = 0; / / t he p r oc e ss i s a d i a b a t i c31 P 3 = 1; // p r e s s u r e a f t e r s t e p 2 i n b a r32 g am a = ( Cp / Cv ) ;

33 T 3 = ( ( P 3 / P2 ) ^ ( ( g am a - 1 ) / g am a ) ) * T 2 ; / / t e m p e r a t u r ea f t e r s te p 2

34 W 2 = ( C v *( T2 - T 3 ) ) ; // w or k d on e by s y s t em35 mprintf ( ’ \n\nFor s t e p 2 ’ ) ;36 mprintf ( ’ \nHeat s u pp l ie d i n s t e p 2 i s %i ’ , Q2 );

37 mprintf ( ’ \nWork done by s ys te m i n s t e p 2 i s %f kJ /kmol ’ , W 2 ) ;

38

39 // S te p 3 : The p r oc e ss i s i s o b a r i c40 T 4 = 30 0; / / t em p er at u re a f t e r s t e p 3 (K)

36


38/222

41 Q 3 = C p *( T4 - T 3 ) ; // h ea t s u p p l i e d d u ri n g s t e p 3 ( kJ /

k mol )42 U = ( C v *( T4 - T 3 ) ) ; // c ha ng e i n i n t e r n a l e ne rg y d ur in gs t e p 3 ( k J / km ol )

43 W 3 = Q3 - U; / / U s i n g f i r s t l a w o f t h er m od y n am i c s44 mprintf ( ’ \n\nFor s t e p 3 ’ ) ;45 mprintf ( ’ \nHeat g i v e n o ut by t he sy st e m i n s t ep 3 i s

%f k J /k mol ’ , Q 3 ) ;46 mprintf ( ’ \nWork d one on t he s ys te m i n s t e p 3 i s %f

kJ/kmol ’ , W 3 ) ;47

48 //e nd

Scilab code Exa 3.4 To calculate change in internal energy change in en-thalpy work done and heat supplied



6

7 c l e ar a ll ;

8 clc ;

9

10

11 // Given :12 R = 8 .3 14 ; / / i d e a l g as c o n st a n t

13 C p = 30; / / s p e c i f i c h e a t a t c o n s t a n t p r e s s u r e ( J / m olK)14

15 //To c a l c u l a t e c h an ge i n i n t e r n a l e ne rg y c ha nge i ne n t ha l p y work d on e a nd h e at s u p p l i e d

37


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16

17 // ( a ) : Gas i s exp an ded i s o t h e r m a l l y18 T = 600; / / t e m p e r a tu r e i n K19 P 1 = 5; / / i n i t i a l p r e s s u r e i n b a r20 P 2 = 4; // f i n a l p r e s s u r e i n b a r21 U 1 = 0; // s i n c e t h e p r o c e s s i s i s ot h er m a l22 H 1 = 0; // s i n c e t h e p r o c e s s i s i s ot h er m a l23 W 1 = ( R* T* log ( P 1 / P 2 ) ) ; / / work d on e d u r in g t h e

p r o c e s s24 Q 1 = W1 ; // h ea t s u p p l i e d d u ri n g t he p r o c e s s25 mprintf ( ’ When g a s i s e xp an ded i s o t h e r m a l l y ’ ) ;26 mprintf ( ’ \nChange i n i n t e r n a l e ne rg y i n i s o t he r m al

p r o c es s i s %i ’ , U 1 ) ;27 mprintf ( ’ \nChange i n e n th a lp y i n i s o t h e r m a l p r o c e s s

i s %i ’ , H 1 ) ;28 mprintf ( ”\nWork done d u ri n g t he p r o c e s s i s %f k J/

kmol” , W 1 ) ;29 mprintf ( ’ \nHeat s u p pl i e d d ur in g t he p r o c es s i s %f kJ

/kmol ’ , Q 1 ) ;30

31 / / ( b ) : Gas i s h ea te d a t c o n st a n t volume32 V = 0.1; / / v o lu m e (mˆ 3 )33 P 1 = 1;

// i n i t i a l p r es s u re ( bar )34 T 1 = 29 8; // i n i t i a l te mp er at ur e (K)35 T 2 = 40 0; / / f i n a l t e m p e r a tu r e (K)36 n = ( ( P1 * V * 10 ^ 5) / ( R * T1 ) ) ; // number o f m ol es o f g a s37 C v = Cp - R; / / s p e c i f i c h e a t a t c o n s t a n t v o lu m e ( J / m ol

K)38 U 2 = n * Cv * ( T2 - T 1 ) ; // c ha ng e i n i n t e r n a l e ne rg y ( J )39 H 2 = n * Cp * ( T2 - T 1 ) ; / / c h an ge i n e n t ha l p y ( J )40 W 2 = 0; / / i s o c h o r i c p r o c es s41 Q 2 = U2 + W2 ; / / h e a t s u p p l i e d ( J )42 mprintf ( ’ \n\nWhen g a s i s h e at e d a t c o n s t a n t v ol um e ’ )

;43 mprintf ( ’ \nChange i n i n t e r n a l e ne rg y i s %f J ’ , U 2 ) ;44 mprintf ( ’ \nChange i n e n t ha l p y i s %f J ’ , H 2 ) ;45 mprintf ( ’ \nWork d one d u ri n g t he p r o c e s s i s %i ’ , W 2 )

;

38


40/222

46 mprintf ( ’ \nHeat s u p pl i e d d ur in g t he p r o c es s i s %f J ’

, Q 2 ) ;47

48 //e nd

Scilab code Exa 3.5 To determine work done and amount of heat trans-ferred


6

7 c l e ar a ll ;

8 clc ;

9

10

11 // Given :12 function [ y] = Cv (T );

13 y = 2 7 . 45 2 8 +( 6 . 1 83 9 * (1 0 ^ - 3 ) * T ) - ( 8. 9 93 2 *( 1 0^ - 7 ) * (

T ^ 2 ) ) - R ;

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

15

16 m = 20; / / mass o f a i r ( k g )17 n = 1. 25 ; // p o l y t r o p i c c o n st a n t18 P 1 = 1; // i n i t i a l p r es s u re ( bar )19 P 2 = 5; / / f i n a l p r e s s u r e ( b a r )

20 T 1 = 30 0; // t e m pe r at u r e ( K)21 R = 8 .3 14 ; / / i d e a l g as c o n st a n t22 M = 29; / / m o l ec u l ar wt o f a i r23

24 / / To d e t e r m i ne wo rk d on e and amount o f h e a t

39


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t r a n s f e r r e d

2526 / / ( a ) : Work d on e b y t h e c o mp r es s o r p e r c y c l e27 n _m ol e = m /M ; // m o l es o f a i r ( k mol )28 V 1 = ( ( n _ mo l e * 1 0 ^ 3* R * T 1 ) / ( P 1 * 1 0 ^5 ) ) ; // i n i t i a l

vol ume (mˆ3 )29 V 2 = ( V 1 *( ( P1 / P 2 ) ^( 1/ n ) ) ) ; / / f i n a l v o lu m e (mˆ 3 )30

31 / / S i n c e t h e p r o c e s s i s p o l y t r o p i c P (Vˆ n )=c ( s ayc o n s t a n t )

32 c = P 1 * 10 ^ 5* ( V 1 ^n ) ;

33 / / f u n c t i o n [ z ] = f ( V) ;

34 // z = c / (V ˆ 1 . 25 ) ;35 / / e n d f u n c t i o n36 //W1 = i n t g ( V1 , V2 , f ) ; so37 W = ( c / (1 - n ) ) * ( ( V 2 ^ ( - n + 1) ) - ( V1 ^ ( - n + 1 ) ) ) / 1 00 0 ;

38 mprintf ( ’ Work d on e by c o m p r e ss o r i s %4 . 3 e J ’ , W * 1 0 0 0 );

39

40 / / ( b ) : Amount o f h ea t t r a n s f e r r e d t o s u rr o un d i ng41 T 2 = ( ( T1 * V 2 * P2 ) / ( V1 * P 1 )) ; / / f i n a l temp i n K42 U1 = intg ( T 1 , T 2 , C v ) ;

43 U = U1 * n _m ol e ; // c ha ng e i n i n t e r n a l e ne rg y ( kJ )44 Q = U + W ; // h e at s u p p l i e d

45 mprintf ( ’ \nChnage i n i n t e r n a l e ne rg y i s %f k J ’ , U ) ;46 mprintf ( ’ \nHeat s u p p l i e d i s %f kJ ’ , Q ) ;47

48 //e nd

Scilab code Exa 3.6 To compare the pressures


40


42/222

5

67 c l e ar a ll ;

8 clc ;

9

10

11 // Given :12 V = 0 .3 82 1* 1 0^ - 3 // mol ar v ol ume (mˆ3/mol )13 T = 313; / / t e m p e r a t u r e (K)14 R = 8 .3 14 ; / / i d e a l g as c o n st a n t15 a = 0 .3 65 ; b = 4 .2 8* 10 ^ -5 ; / / V an de r W aa ls c o n s t a n t16

17 / /To co mp ar e t h e p r e s s u r e s18

19 / / ( a ) : I d e a l g as e qu a ti o n20 P = ( ( R* T ) /( V * 1 0^ 5) ) ; / / p r e s s u r e i n b ar21 mprintf ( ’ P r es s ur e o bt ai ne d by i d e a l g as e q u at i on i s

%f b a r ’ , P ) ;22

23 / / ( b ) : Van d e r Wa al s e q u a t i o n24 P = ( ( (( R * T ) / ( V - b ) ) - (a / ( V ^ 2 ) ) ) / ( 10 ^ 5 ) ) ;

25 mprintf ( ’ \ n P r e s su r e o b t a i ne d by Van d e r Waals

e q ua t i on i s %f b ar ’, P ) ;

26

27 //e nd

Scilab code Exa 3.7 To calculate the volume



6

7 c l e ar a ll ;

41


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8 clc ;

910 / / To f i n d A pp ro x V a lu e11 function [ A ] = a p p r o x ( V , n )

12 A = round ( V * 1 0 ^ n ) / 1 0 ^ n ; //V−V al ue n−To what p l a c e13 funcprot (0)


15

16

17 // Given :18 T = 300; // t e m pe r at u r e (K)19 P = 100; // pr e s su r e ( bar )

20 R = 8 .3 14 ; / / i d e a l g as c o n st a n t21 a = 0 .1 37 8; b = 3 .1 8* 10 ^ - 5; // Van d e r w a al s c o n s t a n t22

23 / /To c a l c u l a t e t he vo lume24

25 / / ( a ) : I d e a l g as e qu a ti o n26 V _ i d e a l = a p p ro x ( ( ( R * T ) /( P * 1 0 ^ 5) ) , 6) ;

27 mprintf ( ’ Volume c a l c u l a t e d by i d e a l g as e q u a ti on i s%4 . 2 e c u b i c m ’ , V _ i d e a l ) ;

28

29 / / ( b ) : Van d e r Wa al s e q u a t i o n30 function [ y ] = f ( V ) ;

31 y = ( ( P * 1 0 ^ 5 ) + ( a / ( V ^ 2 ) ) ) * ( V - b ) - ( R * T ) ; / / f u n c t i o nt o c a l c u l a t e d i f f e r e n c e be tw e en c a l c u l a t e da nd a s su m ed v o lu m e


33

34 V _ re al = 0;

35 for i = 0 . 20 : 0. 0 1: 0 .3 0 / /Van d e r w a a l s v ol um e s h o u l dbe n e a rl y e qu a l t o I d e a l g as valoume

36 r es = a p p ro x ( f ( i * 1 0^ - 3 ) , 0) ;

37 for j = -5:538 if ( j = = r e s ) // f o r v er y s ma l l d i f f e r e n c e i may

be t ak en a s e x ac t volume39 V _r ea l = i * 10 ^ - 3;

40 end

42


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41 end

42 end43 mprintf ( ’ \nVolume c a l c u l a t e d by V an d e r Wa al se q ua t i on i s %3 . 2 e c u b ic m ’ , V _ r e a l ) ;

44

45 //e nd

Scilab code Exa 3.8 Theoretical problem


6

7 c l e ar a ll ;

8 clc ;

9

10 // Given :11

12 / /The g i v e n pr obl e m i s t h e o r e t i c a l and d oe s no ti n v o l v e any n u m e ri c a l c o mp ut a ti on

13

14 //e nd

Scilab code Exa 3.9 To calculate compressibility factor and molar volume


2 / / C ha p te r 33 //P−V−T B e ha v io u r and Hea t E f f e c t s4 // E xampl e 95

6

43


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7 c l e ar a ll ;

8 clc ;9

10 / / To f i n d A pp ro x V a lu e11 function [ A ] = a p p r o x ( V , n )

12 A = round ( V * 1 0 ^ n ) / 1 0 ^ n ; //V−V al ue n−To what p l a c e13 funcprot (0)


15

16

17 // Given :18 T = 500; / / t e m p e r a t u r e (K)

19 P = 10; // pr e s su r e ( bar )20 R = 8 .3 14 ; / / i d e a l g as c o n st a n t21 B = - 2. 19 *1 0^ - 4 ; C = - 1 . 73 *1 0^ - 8 ; // V i r i a l

c o e f f e c i e n t s22 T c = 5 12 .6 ; // c r i t i c a l te mp er at ur e23 P c = 81; // c r i t i c a l pr e s su r e24

25 //To c a l c u l a t e c o m p r e s s i b i l i t y f a c t o r and m ol arvolume

26

27 / / ( a ) : T ru n ca t ed f or m o f v i r i a l e q u a t i o n28 V _ i d e a l = a p p ro x ( ( ( R * T ) /( P * 1 0 ^ 5) ) , 7) ; / / i d e a l g as

volume29 function [ z] = f1 (V )

30 z = ( ( ( R * T ) /( P * 1 0 ^ 5) ) * ( 1 + ( B / V ) + ( C /( V ^ 2 ) ) ) ) ; //f u n c t i o n f o r o b t a i n i n g v ol um e by v i r i a le q u a t i o n


32

33 / / l o o p f o r h i t a nd t r i a l metho d34 f lag = 1;

35 while ( f l a g = = 1 )36 V _ v ir i a l = a p p ro x ( f 1 ( V _ i d ea l ) , 7) ;

37 if (approx(V_ideal ,5)==approx(V_virial ,5))

38 fl ag = 0;

39 break ;

44


46/222

40 else

41 V _i d ea l = V _ vi r ia l ;42 end

43 end

44 mprintf ( ’ V olume o b t a i n e d vy v i r i a l e q u a t i o n i s %4 . 3 ec u b ic m ’ , V _ v i r i a l ) ;

45 Z = a p pr o x ( ( ( P * 1 0^ 5 * V _ v i r ia l ) / ( T * R ) ) , 3) ; //c o m p r e s s i b i l i t y f a c t o r

46 mprintf ( ’ \ n C o mp r e ss i b i lt y f a c t o r f o r v i r i a l e q ua t io ni s %f ’ , Z ) ;

47

48 / / ( b ) : R e d l i c h Kwong E q u a t io n

49 / / C o n s ta n t s i n R e d l i c h Kwong e q u a t i o n50 a = a p pr o x ( ( ( 0 . 4 27 8 * ( R ^ 2 ) * ( Tc ^ 2 . 5 ) ) / ( Pc * 1 0 ^ 5) ) , 4) ;

51 b = a p pr o x ( ( ( 0 . 08 6 7 * R * T c ) /( P c * 1 0 ^ 5) ) , 9) ;

52

53 V _ i d e a l = a p p ro x ( ( ( R * T ) /( P * 1 0 ^ 5) ) , 7) ; / / i d e a l g asvolume

54

55 / / F u n ct i on t o f i n d vo lum e b y R e d l ic h Kwong e q u a t i o n56 function [ x] = f2 (V )

57 x = ( ( R * T ) /( P * 1 0 ^ 5 ) ) + b - (( a * ( V - b ) ) / (( T ^ 0 . 5 ) * ( P

* 1 0 ^ 5 ) * V * ( V + b ) ) ) ;


59

60 / / l o o p f o r h i t a nd t r i a l metho d61 f lag = 1;

62 while ( f l a g = = 1 )

63 V _ r ed l i ch = a p pr o x ( f 2 ( V _ i de a l ) , 7 ) ;

64 if (approx(V_ideal ,5)==approx(V_redlich ,5))

65 fl ag = 0;

66 break ;

67 else

68 V _i d ea l = V _ re d li c h ;69 end

70 end

71 mprintf ( ’ \n\nVol ume o b t a i n e d by R e d l i c h KwongE q ua t io n i s %4 . 3 e c u b i c m/ mol ’ , V _ r e d l i c h ) ;

45


47/222

72 Z = a p pr o x ( ( ( P * 1 0^ 5 * V _ r e d li c h ) / ( T * R ) ) , 3) ; //

c o m p r e s s i b i l i t y f a c t o r73 mprintf ( ’ \ n C o m p r e s s b i l i ty f a c t o r by R e d l ic h Kwonge q ua t i on i s %f ’ , Z ) ;

74

75 //e nd

Scilab code Exa 3.10 To calculate heat of formation of methane gas


6

7 c l e ar a ll ;

8 clc ;

9

10

11 // Given :

12 H a = - 89 0. 94 ; // s ta nd ar d h ea t f o r r e a c t i o n a ( kJ )13 H b = - 39 3. 78 ; // s ta nd ar d h ea t f o r r e a c t i o n b ( kJ )14 H c = - 28 6. 03 ; // s ta nd ar d h ea t f o r r e a c t i o n c ( kJ )15

16 / /To c a l c u l a t e h ea t o f f or ma t i o n o f methane g as17 // c ∗2 + b − a g i v e s t he f o rm a t io n o f methane fro m

e l e m e n t s18 H f = ( 2* H c ) +Hb - H a ;

19 mprintf ( ’ H ea t o f f o r m a ti o n o f meth an e i s %f kJ / mol ’ ,H f ) ;

2021 //e nd

46


48/222

Scilab code Exa 3.11 To calculate heat of formation of chloroform


6

7 c l e ar a ll ;

8 clc ;

9

10

11 // Given :12 H a = - 50 9. 93 ; // h ea t o f c om bu st i on o f r e a c t i o n a ( kJ

)13 H b = - 29 6. 03 ; / / h ea t o f c om bu st i on o f r e a c t i o n b ( kJ

)14 H c = - 39 3. 78 ; // h ea t o f c om bu st i on o f r e a c t i o n c ( kJ

)15 H d = - 16 7. 57 ; / / h ea t o f c om bu st i on o f r e a c t i o n d ( kJ

)16

17 //To c a l c u l a t e h ea t o f f or ma t i o n o f c hl o ro f o r m18 // c + ( 3∗d ) −a −b g i v e s c hl o ro f or m from i t s e l e me nt s

19 H f = H c +( 3* H d ) - Ha - H b ;

20 mprintf ( ’ H eat o f f or ma t i o n o f c h lo r of o rm i s %f k J/mol ’ , H f ) ;

21

22 //e nd

Scilab code Exa 3.12 To calculate standard heat of reaction at 773 K

1 / /A T ex tb o ok o f C h em i c al E n g i n e e r i n g T he rm od yn am ic s2 / / C ha p te r 33 //P−V−T B e ha v io u r and Hea t E f f e c t s

47


49/222

4 // E xampl e 12

56

7 c l e ar a ll ;

8 clc ;

9

10

11 // Given :12 H o = - 16 49 87 ; // s ta nd ar d h ea t o f r e a c t i o n a t 298 K

i n J13 T 1 = 29 8;

14 T 2 = 77 3; // t e m pe r at u r e ( K)

1516 / /To c a l c u l a t e s ta nd ar d h ea t o f r e a c t i o n a t 773 K17 a l p h a = ( 2 * 29 . 1 6) + 1 3 .4 1 - 2 6 . 7 5 - ( 4 * 26 . 8 8) ;

18 b e tt a = ( ( 2 *1 4 . 4 9) + 7 7. 0 3 - 4 2 . 26 - ( 4* 4 . 35 ) ) * 1 0 ^ - 3;

19 g a m a = ( ( 2 * - 2 . 0 2 ) - 1 8 . 7 4 + 1 4 . 2 5 + (4 * 0 . 3 3 ) ) * 1 0 ^ - 6 ;

20

21 // U si ng e q ua t i on 3 . 5 4 ( P age no . 6 7)22 H 1 = H o - ( a l ph a * T 1 ) - ( b e tt a * ( T 1 ^ 2) / 2 ) - ( g a ma * ( T 1 ^ 3 ) / 3 ) ;

23

24 / / At 7 73 K25 H r = H 1 + ( a lp h a * T 2 ) +( b e t t a * ( T 2 ^ 2) / 2 ) + ( g a ma * ( T 2 ^ 3 ) / 3 ) ;

26 mprintf ( ’ H eat o f r e a c t i o n a t 773 K i s %f kJ ’ , Hr/ 1 0 0 0 ) ;

27

28 //e nd

Scilab code Exa 3.13 To determine heat added or removed


48


50/222

6


9

10

11 // Given :12 T o = 29 8; / / s t a n d a r d t e m p e r a t u r e (K )13 T 1 = 40 0; / / t e m p e r at u r e o f r e a c t a n t s (K)14 T 2 = 60 0; / / t e mp e r at u r e o f p r o d u ct s (K)15 H o = - 28 3. 02 8; / / s t a nd a r d h e at o f r e a c t i o n ( k J/ mol )16

17 / / To d e t e r m i ne h e a t a dd ed o r r em ov ed

18 // Ba si s :19 n _CO = 1; / / m o le s o f CO r e a c t e d20 n _O2 = 1; // m o le s o f o xy gen s u p p l i e d21 n _ N2 = 1 * 79 / 21 ; // m ol es o f n i t r o g e n22 n 1_ O2 = 0 .5 ; // m o le s o f o xyg en r e q u i r e d23 n _C O2 = 1; // m o le s o f c ar bo n d i o x i de f or me d24

25 H 1 = ( ( n _ O2 * 2 9 .7 0 ) + ( n _ N2 * 2 9 . 1 0 ) + ( n _C O * 2 9 . 1 0) ) * ( T o - T 1

) / 1 0 0 0 ; // e nt ha l p y o f c o o l i n g o f r e a c t an t s26 H 2 = ( ( n 1 _O 2 * 2 9 . 7 0) + ( n _ N 2 * 2 9 . 10 ) + ( n _ C O2 * 4 1 .4 5 ) ) * ( T2 -

T o ) / 1 0 0 0 ; / / e n th a lp y o f h e a ti n g t he p r od u ct s27 H r = H1 + Ho + H2 ;

28 mprintf ( ’ H eat s u p pl i e d i s %f kJ ’ , H r ) ;29

30 //e nd

Scilab code Exa 3.14 To calculate theoretical flame temperature


49


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6


9

10

11 // Given :12 T o = 29 8; / / s t a n d a r d t e m p e r a tu r e (K)13 T 1 = 37 3; / / t e mp e r at u r e o f r e a c t a n t s (K)14 H o = 2 83 17 8; / / s t a n d a r d h e a t o f c o mb u s ti o n ( J / m ol )15

16 //To c a l c u l a t e t h e o r e t i c a l f la me t em pe ra tu re17 // Ba si s :

18 n _CO = 1; / / m o l es o f CO19 n _O2 = 1; // m o le s o f o xy gen s u p p l i e d20 n 1_ O2 = 0 .5 ; // m o le s o f o xyg en r e a c t ed21 n _C O2 = 1; // m o le s o f c ar bo n d i o x i de f or me d22 n _ N2 = 7 9/ 21 ; // m ol es o f n i t r o g e n23

24 H 1 = ( ( n _ O2 * 3 4 .8 3 ) + ( n _ N2 * 3 3 . 0 3 ) + ( n _C O * 2 9 . 2 3) ) * ( T o - T 1

) ; // e nt ha l p y o f c o o l i n g o f r e a c t an t s25 // U si ng e q ua t i on 3 . 5 5 ( P age no . 6 9)26 H 2 = Ho - H1 ;

27 T f = H 2 / (( n 1 _ O 2 * 3 4 . 83 ) + ( n _ N 2 * 3 3 .0 3 ) + ( n _ C O2 * 5 3 .5 9 ) )

+298; / / f l a m e t e m p e r a t u r e28 mprintf ( ’ T h e o r e t i c a l f la m e t em p er at u re i s %f K ’ , T f ) ;29

30 //e nd

50


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Chapter 4

Second Law of

Thermodynamics

Scilab code Exa 4.1 To calculate the maximum efficiency

1 / /A T ex tb o ok o f C h em i c al E n g i n e e r i n g T he rm od yn am ic s2 / / C ha p te r 43 / / S e c o n d Law o f T he rm o dy na m ic s4 // E xampl e 15

6

7 c l e ar a ll ;

8 clc ;

9

10

11 // Given :12 T 1 = 70 0; / / t e m p e r at u r e o f h e a t s o u r c e (K)13 T 2 = 30 0; / / t e m p e r at u r e o f h e a t s i n k (K)14

15 //To c a l c u l a t e t he maximum e f f i c i e n c y16 e f f = ( ( T 1 - T 2 ) / T 1 ) ; // e f f i c i e n c y o f a h ea t e ng i n e17 mprintf ( ’ Maximum e f f i c i e n c y o f h ea t e n g in e i s %f ’ ,

e f f ) ;

18

51


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19

20 //e nd

Scilab code Exa 4.2 To determine minimum amount of work done andheat given to surrounding

1 / /A T ex tb o ok o f C h em i c al E n g i n e e r i n g T he rm od yn am ic s2 / / C ha p te r 43 / / S e c o n d Law o f T he rm o dy na m ic s

4 // E xampl e 25

6

7 c l e ar a ll ;

8 clc ;

9

10 // Given :11 m = 1 ; / / m as s o f w a te r ( k g )12 T 1 = 30 0; / / t e m p e r at u r e o f s u r r o u n d i n g (K)13 T 2 = 27 3; / / t e m p e r a t u r e o f w a t e r (K)14 H f = 3 34 .1 1; // l a t e n t h ea t o f f u s i o n o f i c e ( kJ /kg )

1516

17 / / To d e t e r m i n e minimum a mo unt o f w or k a nd h e a t g i v e nu pt o s u r r o u nd i n g

18

19 / / ( a )20 Q 2 = m *Hf ; / / h e a t a b so b ed a t t e m p e ra t u r e T221 W = ( ( Q2 * ( T1 - T 2 ) ) / T2 ) ; //minimumm amount o f work

r e q u i r e d22 mprintf ( ’ Minimum amount o f wo rk r e q u i r e d i s %f kJ ’ ,W

) ;23

24 / / ( b )25 / /Q1 i s t he h ea t g i ve n up t he s ur r o u nd i n g26 Q 1 = W +Q2 ;

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27 mprintf ( ’ \nHeat g i ve n u pt o s u rr o un d i ng i s %f kJ ’ , Q 1 )

;28

29

30 //e nd

Scilab code Exa 4.3 To determine efficiency of proposed engine


2 / / C ha p te r 43 / / S e c o n d Law o f T he rm o dy na m ic s4 // E xampl e 35

6

7 c l e ar a ll ;

8 clc ;

9

10

11 // Given :12 P _o ut = 4 .5 ; / / o u t p u t p o we r ( h p )

13 P _i n = 6 .2 5; // in pu t power (kW)14 T 1 = 1 00 0; / / s o u r c e t e m p e r a t u r e (K)15 T 2 = 50 0; / / s i n k t e m p e r a t u r e (K)16

17 / /To d e t e rm in e e f f i c i e n c y o f p ro po se d e ng i ne18 e p = ( ( P _ ou t * 7 4 5 . 7) / ( P _ i n * 1 0 0 0) ) ; // pr o pos e d

e f f i c i e n c y19 mprintf ( ’ E f f i c i e n c y o f p ro po se d e ng i ne i s %f ’ , e p ) ;20

21 e m = ( ( T1 - T 2 ) / T1 ) ; // maximum e f f i c i e n c y

22 mprintf ( ’ \nThe maximum e f f i c i e n y i s %f ’ , e m ) ;23 mprintf ( ’ \nHence t he c la im o f t he p ro po se d e ng i ne i si m p o s s i b l e ’ ) ;

24

25

53


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26 //e nd

Scilab code Exa 4.4 To calculate entropy of evaporation


67 c l e ar a ll ;

8 clc ;

9

10

11 // Given :12 P = 500; / / p r e s s u r e o f d ry s a t u r a t e d s te am ( kPa )13

14 / / From s te am t a b l e s15 H v = 2 10 6; / / l a t e n t h ea t o f v a p o r i s a t i o n ( kJ / kg )16 T = 425; / / s a t u r a t i o n t e m p e r at u r e (K)

1718 / /To c a l c u l a t e t he e nt ro py o f e v ap o r a t io n19 / / By e q u a t i on 4 . 2 5 ( P age no . 9 3 )20 S v = ( Hv / T) ; / / e n t r o p y c h a ng e a c co m pa n yi n g

v a p o r i s a t i o n21 mprintf ( ’ En tr opy o f e v a p or a t i on i s %f kJ / kg K ’ , S v ) ;22

23 //e nd

Scilab code Exa 4.5 To determine change in entropy


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3 / / S e c o n d Law o f T he rm o dy na m ic s

4 // E xampl e 556

7 c l e ar a ll ;

8 clc ;

9

10

11 // Given :12 m = 2 ; / / m ass o f g a s ( k g )13 T 1 = 27 7; // i n i t i a l te mp er at ur e (K)14 T 2 = 36 8; / / f i n a l t e m p e r a tu r e (K)

15 C v = 1 .4 2; / / s p e c i f i c g e a t a t c o n s t a n t v ol um e ( k J / k gK)

16

17 / /To d e t er m i ne c ha ng e i n e n tr o py18 // U si ng e q ua t i on 4 . 3 1 ( P age no . 9 4)19 S = ( m * C v * log ( T 2 / T 1 ) ) ; / / c h a n g e i n e n t r o p y ( k J /K)20 mprintf ( ’ C hange i n e n tr o p y i s %f kJ /K ’ , S ) ;21

22

23 //e nd

Scilab code Exa 4.6 To calculate the entropy change


67 c l e ar a ll ;

8 clc ;

9

10

55


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11 // Given :

12 T = 300; / / t e m p e r a tu r e i n K13 P 1 = 10; // i n i t i a l p r e s s u r e ( bar )14 P 2 = 1; / / f i n a l p r e s s u r e ( b a r )15 R = 8 .3 14 ; / / i e a l g as c o ns t an t16

17 / /To c a l c u l a t e t he e nt ro p y c ha ng e18 / / U si ng e q u a t i o n 4 . 3 3 ( Page no . 9 4 )19 S = ( R * log ( P 1 / P 2 ) ) ; / / ( k J / k m o l K )20 mprintf ( ’ E n to py c ha n ge i s %f kJ / km ol K ’ , S ) ;21

22

23 //e nd

Scilab code Exa 4.7 To determine change in entropy


56

7 c l e ar a ll ;

8 clc ;

A Textbook Of Chemical Engineering Thermodynamics_K. V. Narayanan (1).pdf

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