5 Edition - Stoichiometry_B. I. Bhatt and S. B. Thakore (2) (1)

Scilab Textbook Companion for Stoichiometry by B. I. Bhatt And S. B. Thakore1 Created by Sumit Kumar Singh B.Tech Chemical Engineering NIT Rourkela, Odhisa College Teacher Hara Mohan Jena Cross-Checked by Ganesh November 9, 2013

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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 Author: B. I. Bhatt And S. B. Thakore Publisher: Tata McGraw - Hill Education Edition: 5 Year: 2010 ISBN: 9780070681149

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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

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1 Dimensions and Units

9

2 Basic Chemical Calculations

13

3 Material Balances without Chemical Reaction

39

4 Material Balances Involving Chemical Reactions

56

5 Energy Balances

83

6 Stoichiometry and Unit Operations

138

7 Combustion

172

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List of Scilab Codes Exa 1.1 Exa 1.2 Exa 1.3 Exa 1.4 Exa 2.1 Exa 2.2 Exa 2.3 Exa 2.4 Exa 2.5 Exa 2.6 Exa 2.7 Exa 2.8 Exa 2.9 Exa 2.10 Exa 2.11 Exa 2.12 Exa 2.13 Exa 2.14 Exa 2.15 Exa 2.16 Exa 2.17 Exa 2.18 Exa 2.19 Exa 2.20 Exa 2.21 Exa 2.22 Exa 2.23 Exa 2.24

Mass flow rate . . . . . . . . . . . . steam velocity in pipeline . . . . . . conversion of TR . . . . . . . . . . . conversion of equation into SI units . gm of NH4Cl . . . . . . . . . . . . . equivalent moles of CuSO4 . . . . . moles of K2CO3 . . . . . . . . . . . no of atoms of BaCl2 . . . . . . . . equivalent mass . . . . . . . . . . . . equivalents . . . . . . . . . . . . . . composition of mixture . . . . . . . composition and molar mass . . . . actual urea content . . . . . . . . . . mass percent . . . . . . . . . . . . . no of ions . . . . . . . . . . . . . . . composition of solution . . . . . . . composition of solution . . . . . . . Na2O percentage . . . . . . . . . . . TOC and ThOD . . . . . . . . . . . conc of salts . . . . . . . . . . . . . ppm unit . . . . . . . . . . . . . . . molarity normality and molality . . molarity of solution . . . . . . . . . conc of CO2 . . . . . . . . . . . . . pH of HOCl . . . . . . . . . . . . . . Mavg and composition of air . . . . Composition and specific gravity . . percentage error . . . . . . . . . . . 4

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9 10 10 11 13 14 14 15 16 17 17 18 19 20 21 22 23 24 25 25 26 27 28 29 29 30 31 33

Exa 2.25 Exa 2.26 Exa 2.27 Exa 2.28 Exa 2.29 Exa 3.1 Exa 3.2 Exa 3.3 Exa 3.4 Exa 3.5 Exa 3.6 Exa 3.7 Exa 3.8 Exa 3.9 Exa 3.10 Exa 3.11 Exa 3.12 Exa 3.14 Exa 3.15 Exa 3.16 Exa 3.17 Exa 3.18 Exa 3.20 Exa 4.1 Exa 4.2 Exa 4.3 Exa 4.4 Exa 4.5 Exa 4.6 Exa 4.7 Exa 4.8 Exa 4.9 Exa 4.10 Exa 4.11 Exa 4.12 Exa 4.13 Exa 4.14 Exa 4.15

molar volume . . . . . . . . . . . . . . . . . . ternary mix analysis . . . . . . . . . . . . . . vapour mix composition . . . . . . . . . . . . absolute humidity . . . . . . . . . . . . . . . nozzle outlet T . . . . . . . . . . . . . . . . . Lancashire boiler . . . . . . . . . . . . . . . . Textile mill . . . . . . . . . . . . . . . . . . . recovered tannin . . . . . . . . . . . . . . . . Extraction of dry neem leaves . . . . . . . . . Extraction of mix of Acetone and Chloroform Pressure Swing Adsorption . . . . . . . . . . Required Oleum strength . . . . . . . . . . . Mixed acid formation . . . . . . . . . . . . . Actual analysis of borewell water . . . . . . . Matrix use . . . . . . . . . . . . . . . . . . . Flowrate calculation . . . . . . . . . . . . . . solving eqs with graphical plot . . . . . . . . ion exclusion process . . . . . . . . . . . . . . Air Conditioning plant . . . . . . . . . . . . . Overall efficiency of Pulp Mill . . . . . . . . . 2 stage membrane CO separation . . . . . . . 2 stage reverse osmosis . . . . . . . . . . . . . Purging by atmospheric pressure method . . Manufacture of MCA . . . . . . . . . . . . . Bechamp Process . . . . . . . . . . . . . . . . Pilot Plant Calculations . . . . . . . . . . . . Manufacturing of Acetaldehyde . . . . . . . . Lime Soda process . . . . . . . . . . . . . . . Manufacture of Ammonia by Fertilizer plant . Saponification of Tallow . . . . . . . . . . . . Sulphur Burner . . . . . . . . . . . . . . . . . Hydrogenation of Refined Soybean oil . . . . Material Balance in Formox Process . . . . . Pyrites fines roasting . . . . . . . . . . . . . . Burning of Pyrites and ZnS . . . . . . . . . . Raising pH with NaOH . . . . . . . . . . . . Solving eg 10 with Linear Model Method . . Electrochemical cell . . . . . . . . . . . . . . 5

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33 35 36 36 37 39 40 40 41 42 43 44 45 45 47 47 48 49 50 51 52 53 55 56 57 58 59 61 61 62 63 64 66 68 69 71 72 73


Hooker type Diaphragm Cell . . . . . . . . . . Naptha Reforming to Ammonia . . . . . . . . . Additional membrane separator in eg 17 . . . . Partial Demineralisation Plant . . . . . . . . . Capacity increment by Second Reactor . . . . . Blast Furnace Calculations . . . . . . . . . . . Pumping of water . . . . . . . . . . . . . . . . Heating of CH4 . . . . . . . . . . . . . . . . . . Calculation of heat added . . . . . . . . . . . . Heating of Toulene . . . . . . . . . . . . . . . . Aq caustic soda heating . . . . . . . . . . . . . Heating Chlorinated diphenyl . . . . . . . . . . Roasting of pyrites fine . . . . . . . . . . . . . Anniline and water mix subcooled . . . . . . . Vapor Pressure calculations . . . . . . . . . . . Calculations on O zylene . . . . . . . . . . . . latent heat of vaporization of ethanol . . . . . . Saturation P of steam . . . . . . . . . . . . . . Bubble and Dew pt calculations . . . . . . . . Hot air drying machine . . . . . . . . . . . . . Flow of saturated vapors of R134 . . . . . . . . Liquifaction of Cl2 . . . . . . . . . . . . . . . . Melting of Tin . . . . . . . . . . . . . . . . . . steam fluctuation calculations . . . . . . . . . . Manufacture of dry ice . . . . . . . . . . . . . . Steam produced in S burner . . . . . . . . . . . Equimoar pentane and hexane mix . . . . . . . Flashing of saturated liq mix . . . . . . . . . . H2 recovery from Refinery off gases . . . . . . . Refrigiration calculations . . . . . . . . . . . . Chlorination of benzene . . . . . . . . . . . . . Heat of formation of ethylene . . . . . . . . . . Heat of combustion of ethyl mercaptan . . . . . Std heat of formation of gaseous di ethyl ether Heat of formation of motor spirit . . . . . . . . Mean heat capacity . . . . . . . . . . . . . . . Heat of reaction . . . . . . . . . . . . . . . . . Std heat of reaction . . . . . . . . . . . . . . . 6

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74 75 76 77 78 80 83 84 85 85 86 87 88 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 106 107 108 108 109 110


Burning of SO2 . . . . . . . . . . . . . . . . . . . . . Esterification of acetic acid . . . . . . . . . . . . . . Heat transfer in intercoolers . . . . . . . . . . . . . . Enthalpy balance in the reactor . . . . . . . . . . . . Calculation of circulation rate . . . . . . . . . . . . . Loop reactor for EDC manufacture . . . . . . . . . . Calculations in adiabatic converter . . . . . . . . . . Burning of HCl . . . . . . . . . . . . . . . . . . . . . Dehydrogenation of EB . . . . . . . . . . . . . . . . Heat of crystallization . . . . . . . . . . . . . . . . . Heat of crystallization . . . . . . . . . . . . . . . . . Heat of sol of Boric acid . . . . . . . . . . . . . . . . Heat of dissolution . . . . . . . . . . . . . . . . . . . T change in dissolution . . . . . . . . . . . . . . . . Using std heat of formations . . . . . . . . . . . . . Heat effect of the solution . . . . . . . . . . . . . . . Integral heats of solution . . . . . . . . . . . . . . . Hx for H2SO4 . . . . . . . . . . . . . . . . . . . . . Using heat of formations of H2SO4 . . . . . . . . . . Heat to be removed for cooling it to 308K . . . . . . Heat changes in formation of MNB . . . . . . . . . . Final T of solution in absorption of NH3 . . . . . . . Using table 5 60 . . . . . . . . . . . . . . . . . . . . Heat removed in cooler . . . . . . . . . . . . . . . . Hx vs x1 . . . . . . . . . . . . . . . . . . . . . . . . repat of 5 57 using heat capacities . . . . . . . . . . He vs x1 of acetone and ethylacetate . . . . . . . . . Heat of dilution . . . . . . . . . . . . . . . . . . . . eg 5 60 with use of ice at 273K . . . . . . . . . . . . Overall material and energy balance . . . . . . . . . Cryogenic Separation of Nitrogen . . . . . . . . . . . Azeotropic distillation of IPA and water . . . . . . . CO2 absorption in aq MEA solution . . . . . . . . . Heat effect of Scrubbing . . . . . . . . . . . . . . . . Extraction of Acetic Acid . . . . . . . . . . . . . . . Multiple contact counter current Extractor . . . . . Recovery of Acetic Acid by Ethyl Acetate Extraction Yield of Glauber salt . . . . . . . . . . . . . . . . . . 7

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110 111 112 113 114 114 116 117 118 120 120 121 122 122 123 124 125 126 127 128 128 129 130 131 132 134 134 135 136 138 139 141 142 143 146 147 147 149

Exa 6.10 Exa 6.11 Exa 6.12 Exa 6.13 Exa 6.14 Exa 6.15 Exa 6.16 Exa 6.17 Exa 6.18 Exa 6.19 Exa 6.20 Exa 6.21 Exa 6.22 Exa 6.23 Exa 6.24 Exa 6.25 Exa 7.1 Exa 7.2 Exa 7.3 Exa 7.4 Exa 7.5 Exa 7.6 Exa 7.7 Exa 7.8 Exa 7.9 Exa 7.10 Exa 7.11 Exa 7.12 Exa 7.13

Cooling in a Crystallizer . . . . . . . . . . . . . . . . . Recovery of p DCB . . . . . . . . . . . . . . . . . . . Extractive Crystallization of o and p nitrochlorobenzenes Calculation of Dew Point . . . . . . . . . . . . . . . . Calculations on Ambient Air . . . . . . . . . . . . . . Humidification of Air in a Textile Industry . . . . . . Induced draft cooling tower . . . . . . . . . . . . . . . Waste Heat recovery unit . . . . . . . . . . . . . . . . Recovery of CS2 by adsorption . . . . . . . . . . . . . Hooker type diaphragm cell . . . . . . . . . . . . . . . Absorption of NH3 from pure gas . . . . . . . . . . . . Direct contact counter current rotary drier . . . . . . . Hot air dryer of textile mill . . . . . . . . . . . . . . . Quadruple effect forward feed evaporator . . . . . . . Triple effect evaporation system . . . . . . . . . . . . . Four compartment washing thickner . . . . . . . . . . GCV and NCV calculations . . . . . . . . . . . . . . . NCV of crude oil . . . . . . . . . . . . . . . . . . . . . Gaseous propane . . . . . . . . . . . . . . . . . . . . . GCV NCV for natural gas . . . . . . . . . . . . . . . . Coal burnt in excess air . . . . . . . . . . . . . . . . . Residue fuel oil sample . . . . . . . . . . . . . . . . . . Orsat analysis of flue gases . . . . . . . . . . . . . . . Sugar factory boiler . . . . . . . . . . . . . . . . . . . Stoker fired water tube boiler . . . . . . . . . . . . . . Atomization of fuel . . . . . . . . . . . . . . . . . . . . Water tube boiler . . . . . . . . . . . . . . . . . . . . Gassification by coal . . . . . . . . . . . . . . . . . . . Open Hearth steel furnace . . . . . . . . . . . . . . . .

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150 151 152 154 154 156 157 158 159 161 162 163 165 166 167 169 172 173 174 174 175 176 177 178 180 181 182 184 185

Chapter 1 Dimensions and Units

Scilab code Exa 1.1 Mass flow rate

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

clear ; clc ;

/ / S t o i ch i o m et r y / / C ha pt er 1 / / D i me n s io n s and U n i t s

/ / E xa mp le 1 . 1 / / P ag e 1 2 printf ( ” E xa mp le 1 . 1 , P ag e 1 2 \ n \ n ” ) ; // s o l u t i o n / / U si n g c o n v e r s i o n f a c t o r s fro m t a b l e 1 . 3 ( Pg 9 ) q 1 = 75 / / [ g a l l o n / min ] ( v o l u m e t r i c f l o w r a t e ) q 2 = 7 5 /( 6 0* . 21 9 96 9 ) // [dmˆ3/ s ] r ow = 0.8 / / [ k g /dm ˆ 3 ] q3 = q2 * row / / [ kg / s ] ( m ass f l o w r a t e ) printf ( ” mass f l o w r a t e = ” + string ( q 3 ) + ” [ kg / s ] \ n ” )

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Scilab code Exa 1.2 steam velocity in pipeline

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

clear ; clc ;


/ / E xa mp le 1 . 2 / / P ag e 1 2 printf ( ” E xa mp le 1 . 2 , P ag e 1 2 \ n \ n ” ) ; // s o l u t i o n q m = 20 00 / / [ kg / h ] ( m ass f l o w r a t e ) d 1 = 3 .0 68 // [ i n ] ( i n t e r n a l d ia o f p ip e )

/ / U si n g c o n v e r s i o n f a c t o r s fro m t a b l e 1 . 3 ( Pg 9 ) d 2 = 3 . 06 8 /. 0 39 3 70 1 // [mm] A = ( ( %p i / 4) * d 2 ^2 ) / 10 ^6 // [mˆ 2 ] ( c r o s s s e c t i o n a re a ) / / U s in g s te am t a b l e s ; A pp en di x IV . 3 v = 0 .4 61 66 // [mˆ 3/ k g ] ( s p . v o l . o f s te am a t 4 40 kPa) q v = ( q m *v ) / 36 00 // [mˆ3/ s ] v s = qv /A / / [ m/ s ] printf ( ” v e l o c i t y o f t h e stea m i n t h e p i p e l i n e i s ” + string ( v s ) + ” m / s ” )

Scilab code Exa 1.3 conversion of TR

10

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

clear ; clc ;


/ / E xa mp le 1 . 3 / / P ag e 1 3 printf ( ” E xa mp le 1 . 3 , P ag e 1 3 \ n \ n ” ) ; // s o l u t i o n m = 2000 // [ l b ] ( m ass f l o w r a t e ) t = 2 4 // [ hr ] l f = 144 // [ Btu / l b ] ( l a t e n t h ea t o f f u s i o n )

/ / U si n g c o n v e r s i o n f a c t o r s fro m t a b l e 1 . 3 ( Pg 9 ) T R = ( m * l f * . 2 5 1 99 6 * 4 .1 8 4 ) / ( 3 6 0 0 *2 4 ) printf ( ” 1 T R = ” + string ( T R ) + ” kW” )

Scilab code Exa 1.4 conversion of equation into SI units

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

clear ; clc ;


/ / E xa mp le 1 . 4 / / P ag e 1 3 printf ( ” E xa mp le 1 . 4 , P ag e 1 3 \ n \ n ” ) ;

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

// s o l u t i o n // C = 8 9 . 2 ∗ A ∗ (T/Mˆ ) . 5 [ ft ˆ3/s ] k = 89.2 // C 1 = 1 // [ f t ˆ3/ s ] / / U si n g c o n v e r s i o n f a c t o r s fro m t a b l e 1 . 3 ( Pg 9 )

C 2 = 3 5 .3 1 46 7 * C1 T1 = 1 // [ dgre e R] T 2 = 1 .8 * T1 // [ K] A 1 = 1 // [ f t ˆ 2 ] A 2 = 1 0. 76 39 1 k 2 = ( k * A 2 * ( 1 .8 ) ^ . 5 ) / 3 5 . 34 1 6 7 printf ( ” eq i n S I beco mes \ n C = ” + string ( k 2 ) + ” ∗ (T/M) ˆ.5 [mˆ 3/ s ] ” )

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Chapter 2 Basic Chemical Calculations

Scilab code Exa 2.1 gm of NH4Cl

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

clear ; clc ;

/ / S t o i ch i o m et r y / / C ha pt er 2 / / B a si c Ch em ic al C a l c u l a t i o n s

/ / E xa mp le 2 . 1 / / P ag e 1 7 printf ( ” E xa mp le 2 . 1 , P ag e 1 7 \ n \ n ” ) ; // s o l u t i o n // NH4Cl M = 1 4+ 4+ 35 .5 / / [ g ] ( m o l ar m as s o f NH4Cl ) n =5 / / [ m ol ] m = M*n // [ g ] printf ( ” 5 mo l o f NH4Cl = ” + string ( m ) + ” [ g ] ” )

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Scilab code Exa 2.2 equivalent moles of CuSO4

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

clear ; clc ;


/ / E xa mp le 2 . 2 / / P ag e 1 7 printf ( ” E xa mp le 2 . 2 , P ag e 1 7 \ n \ n ” ) ; // s o l u t i o n // CuSO4 . 5H2O M 1 = 1 59 .5 / / [ g ] ( m o l ar m as s o f CuSO4 ) M 2 = 1 5 9. 5 +5 * (2 + 16 ) / / ( m o l a r m as s o f CuSO4 . 5 H2O ) m = 499 n = m / M2 // [ mol ] printf ( ” I n t h e f o r m u l a CuSO4 . 5 H2O , t h e m o l es o f

CuSO4 i s o n e h e nc e , \ n t he e q u i v a l e n t m o l e s o f CuSO4 i n t he c r y s t a l i s ” + string ( n ) + ” . ” )

Scilab code Exa 2.3 moles of K2CO3

1 2 3 4 5

clear ; clc ;

/ / S t o i ch i o m et r y / / C ha pt er 2 14

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

/ / B a si c Ch em ic al C a l c u l a t i o n s

/ / E xa mp le 2 . 3 / / P ag e 1 7 printf ( ” E xa mp le 2 . 3 , P ag e 1 7 \ n \ n ” ) ; // s o l u t i o n / / K2CO3 m = 117 // [ kg ] ( wt o f K) M k = 39 // [ g ] ( a t wt o f K) a = m / Mk / / [ k g a to ms ]

// 1 mol o f K2CO3 c o n t a i n s 2 atoms o f K n = a /2 / / [ k m ol ] ( m o l e s o f K2CO3 ) printf ( ” ” + string ( n ) + ” km ol o f K2CO3 c o n t a i n s 1 17 k g of K. ”)

Scilab code Exa 2.4 no of atoms of BaCl2

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

clear ; clc ;


/ / E xa mp le 2 . 4 / / P ag e 1 8 printf ( ” E xa mp le 2 . 4 , P ag e 1 8 \ n \ n ” ) ; // s o l u t i o n / / B aC l2 15

16 17 18 19 20

M = 1 3 7. 3 +2 * 35 . 5 / / [ g ] ( m o la r m as s o f Ba Cl 2 ) m = 41 6. 6 / / [ g ] n = m / M / / [ m ol ] N = n * 6 . 02 2 *1 0 ^2 3 / / ( no . o f atoms ) printf ( ” Atoms p r e s e n t i n 4 1 6 . 6 g BaCl2 = ” + string ( N ) +” ”)

Scilab code Exa 2.5 equivalent mass

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

clear ; clc ;


/ / E xa mp le 2 . 5 / / P ag e 1 9 printf ( ” E xa mp le 2 . 5 , P ag e 1 9 \ n \ n ” ) ; // s o l u t i o n printf ( ” ( a ) \ n \ n ” )

/ /PO4 r a d i c a l M = 3 1+ 4* 16 // [ g ] V = 3 / / ( v a l e n c e o f PO4 )

eqm = M /V printf ( ” eq . mass o f PO4 i s ” + string ( e q m ) + ” [ g ] \ n \ n \n”) 21 printf ( ” ( b ) \ n \ n ” ) 22 //Na3PO4 23 M = 3 *2 3+ 95 // [ g ] 24 V = 3 25 eqm = M /V

16

26 printf ( ” e q . m as s o f Na3PO4 i s \n”)

” + string ( e q m ) + ” [ g ] \ n

Scilab code Exa 2.6 equivalents

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

clear ; clc ;


/ / E xa mp le 2 . 6 / / P ag e 1 9 printf ( ” E xa mp le 2 . 6 , P ag e 1 9 \ n \ n ” ) ; // s o l u t i o n / / A lC l3 v = 3 // v al en cy o f Al i on e q = 3*3 / / [ m ol ] printf ( ” no . o f e q u i v a l e n t s i n 3 kmol o f A l Cl 3 i s ” + string ( e q ) + ” k e q . ” )

Scilab code Exa 2.7 composition of mixture

1 2 3 4 5 6

clear ; clc ;

/ / S t o i ch i o m et r y / / C ha pt er 2 / / B a si c Ch em ic al C a l c u l a t i o n s 17

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

/ / E xa mp le 2 . 7 / / P ag e 2 0 printf ( ” E xa mp le 2 . 7 , P ag e 2 0 \ n \ n ” ) ; // s o l u t i o n // ( a ) printf ( ” ( a ) \ n \ n ” )

/ / m as s % m 1 = 600 / / [ k g ] ( N aC l ) m 2 = 200 // [ kg ] (KCl) m = m1 +m2 // t o t a l mass W a = ( m1 / m) * 10 0 W b = ( m2 / m) * 10 0 printf ( ” mass p e rc e n t ag e o f NaCl i s ” + string ( W a ) + ” \ nmass p e r ce n t a ge o f KCl i s ” + string ( W b ) + ” \ n \ n \ n”)

// (b) printf ( ” ( b ) \ n \ n ” )

//mol % M 1 = 2 3+ 35 .5 / / m ol ar mass o f NaCl n 1 = m1 / M1 // no . o f m ol es o f NaCl M 2 = 3 9+ 35 .5 / / m ol ar mass o f KCl n 2 = m2 / M2 // no . o f m ol es o f KCl n = n1 +n2 N 1 = ( n1 / n) * 10 0 N 2 = ( n2 / n) * 10 0 printf ( ” mol p e rc e nt a ge o f NaCl i s ” + string ( N 1 ) + ” \ nmol p e rc e n t ag e o f KCl i s ” + string ( N 2 ) + ” \ n ” )

Scilab code Exa 2.8 composition and molar mass

1 clear ;

18

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 ;


/ / E xa mp le 2 . 8 / / P ag e 2 1 printf ( ” E xa mp le 2 . 8 , P ag e 2 1 \ n \ n ” ) ; // s o l u t i o n // CH.35O.35S.14 / / m as s % C = 1 2. 01 07 // [ kg ] H = 1 . 00 7 94 * .3 5 / / [ k g ] O = 1 5 .9 9 94 * .3 5 / / [ k g ] S = 3 2. 06 5* .1 4 // [ kg ]

m = C +H +O +S m 1 = ( C /m ) *1 00 m 2 = ( H /m ) *1 00 m 3 = ( O /m ) *1 00 m 4 = ( S /m ) *1 00 printf ( ” mass p e r c e n t a g e o f C i s ” + string ( m 1 ) + ” \ nmass p e rc e n t a g e o f H i s ” + string ( m 2 ) + ” \ nmass p e r c e n t a g e o f O i s ” + string ( m 3 ) + ” \ nmass p er ce nt ag e o f S i s ” + string ( m 4 ) + ” \ n \ n ” ) 26 M = m / ( 1 +. 3 5+ . 35 + .1 4 ) 27 printf ( ” m o l ar m as s = ” + string ( M ) + ” k g / k m o l . ” )

Scilab code Exa 2.9 actual urea content

1 clear ; 2 clc ; 3

19

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


/ / E xa mp le 2 . 9 / / P ag e 2 2 printf ( ” E xa mp le 2 . 9 , P ag e 2 2 \ n \ n ” ) ; // s o l u t i o n / / b a s i s 1 0 0 kg u re a m 1 = 45 / / [ kg ] ( m a ss o f N p r e s e n t ) M u = 60 / / ( m ol a r ma ss o f u r e a ) m 2 = 14 *2 / / [ kg ] ( m ass o f N i n 1 kmol o f u r ea ) m = ( Mu / m2 ) * m1 printf ( ” The s am pl e c o n t a i n s ” + string ( m ) + ” p e rc e n t ur ea . ” )

Scilab code Exa 2.10 mass percent

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

clear ; clc ;


/ / E xa mp le 2 . 1 0 / / P ag e 2 2 printf ( ” E xa mp le 2 . 1 0 , P ag e 2 2 \ n \ n ” ) ; // s o l u t i o n

20

15 16 17 18 19

// NaOH I mp ur it y = 60 // [ ppm ] S iO 2 m = ( 6 0/ 1 00 0 00 0 ) * 10 0 printf ( ” Mass p e rc e n t o f SiO2 i s ” + string ( m ) + ” . ” )

Scilab code Exa 2.11 no of ions

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

clear ; clc ;


/ / E xa mp le 2 . 1 1 / / P ag e 2 2 printf ( ” E xa mp le 2 . 1 1 , P ag e 2 2 \ n \ n ” ) ; // s o l u t i o n C a = 4 0. 07 8 / / a t . wt o f Ca F = 1 8. 99 84 03 2 // a t wt o f F M 1 = 3 * Ca + 2 *( 3 0. 9 7 76 2 +( 4 *1 5 .9 9 94 ) ) // m ol ar mass o f

Ca3PO4 18 19 20 21 22 23

M2 M3 m1 m2 m3 n1

= = = = = =

C a + 1 2. 0 10 7 +3 * 15 . 99 9 4 / / m o l a r m as s o f CaCO3 Ca +2* F // m o la r m as s o f CaF2 800 // [ mg ] Ca3PO4 200 // [ mg ] CaCO3 5 // [mg] CaF2 ( ( 3* C a ) / M 1 ) * m 1 + ( Ca / M 2 ) * m 2 + ( Ca / M 3 ) * m 3 // [ mg ]

t o t a l Ca i o n s 24 n 2 = ( F /M3 ) * 2* 5 / / [ mg ] t o t a l F i o n s 25 printf ( ” T ot al no . o f Ca+ i o n s i s ” + string ( n 1 ) + ” and 21

\ n t o t a l no . o f F− i o n s i s ” + string ( n 2 ) + ” . ” )

Scilab code Exa 2.12 composition of solution

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

clear ; clc ;


/ / E xa mp le 2 . 1 2 / / P ag e 2 3 printf ( ” E xa mp le 2 . 1 2 , P ag e 2 3 \ n \ n ” ) ; // s o l u t i o n // ( a ) printf ( ” ( a ) \ n \ n ” )

/ / m as s % m 1 = 100 / / [ k g ] m e t ha n o l ( b a s i s ) m 2 = 64 // [ kg ] s a l i c y l i c a ci d m = m1 +m2 / / [ kg ] mass o f s o l u t i o n w 1 = m2 / m *1 00 w2 = 100 - w1 printf ( ” mass p e r c e n t o f s a l i c y l i c a c i d i s ” + string ( w 1 ) + ” and \ nmass p e r ce n t o f m et ha no l i s ” + string ( w2)+” . \n \n \n”)

//(b) printf ( ” ( b ) \ n \ n ” )

//mole % M 1 = 32 // m ol ar mass o f m et ha no l M 2 = 138 / / m o l a r m as s o f s a l i c y l i c a c i d 22

30 31 32 33 34 35

n 1 = m1 / M1 / / [ k mo l ] m e t h a n o l n 2 = m2 / M2 // [ kmol ] s a l i c y l i c a c id n = n1 +n2 N 1 = n1 / n *1 00 N 2 = n2 / n *1 00 printf ( ” Mole p e rc e n t o f m e t h a no l i s ” + string ( N 1 ) + ” and \ nMo le p e r c e n t o f s a l i c y l i c a c i d i s ” + string ( N2)+” . ”)

Scilab code Exa 2.13 composition of solution

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

clear ; clc ;



//mass %

m 1 = 1 3. 70 / / HCl m 2 = 8. 67 / / N aC l m 3 = 100 // H2O m = m1 + m2 + m3 // mass o f s o l u t i o n w 1 = m1 / m *1 00 w 2 = m2 / m *1 00 w 3 = m3 / m *1 00

23

25 printf ( ” mass p e rc e n t o f HCl i s ” + string ( w 1 ) + ” , \ nmass p e rc e n t o f NaCl i s ” + string ( w 2 ) + ”and \ nmass p e rc e n t o f H2O i s ” + string ( w 3 ) + ” . \ n \ n \ n ” ) 26 M 1 = 3 6 . 4 6 0 9 //HCl 27 M 2 = 5 8 . 4 4 2 8 //NaCl 28 M 3 = 1 8 . 0 1 5 3 //H2O 29 n 1 = m 1 / M 1 //HCl 30 n 2 = m 2 / M 2 //NaCl 31 n 3 = m 3 / M 3 //H2O 32 n = n 1 + n 2 + n 3 33 N 1 = n 1 / n * 1 0 0 34 N 2 = n 2 / n * 1 0 0 35 N 3 = n 3 / n * 1 0 0 36 printf ( ” Mole p e rc e n t o f HCl i s ” + string ( N 1 ) + ” , \ nMole p e r ce n t o f NaCl i s ” + string ( N 2 ) + ”and \ nMole p e rc e n t o f H2O i s ” + string ( N 3 ) + ” . ” )

Scilab code Exa 2.14 Na2O percentage

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

clear ; clc ;


/ / E xa mp le 2 . 1 4 / / P ag e 2 4 printf ( ” E xa mp le 2 . 1 4 , P ag e 2 4 \ n \ n ” ) ; // s o l u t i o n m = 100 / / [ k g ] L ye ( b a s i s ) m 1 = 73 // [ kg ] NaOH

24

17 18 19 20

M 1 = 40 / / NaOH M 2 = 62 // Na2O p = ( M 2 * m1 ) / (2 * M 1 ) printf ( ” p e rc e n t ag e o f Na2O i n t he s o l u t i o n string ( p ) + ” . ” )

i s ”+

Scilab code Exa 2.15 TOC and ThOD

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

clear ; clc ;



// (CH2OH) 3 M = 9 2 // m o la r m ass o f g l y c e r i n C = 600 / / [ mg/ l ] g l y c e r i n c o nc . T OC = ( 3 *1 2 /9 2 ) * 60 0 // [ mg/ l ] / / by c om bu st io n r e a c t i o n we s e e 3 . 5 O2 i s r e q u i r e d f o r 1 m ol o f (CH2OH) 3 20 T h OD = ( 3 . 5* 3 2 * 60 0 ) / 9 2 // [mg/ l ] 21 printf ( ”TOC = ” + string ( T O C ) + ” mg/ l \ nThOD = ” + string ( T h O D ) + ” mg/ l ” )

Scilab code Exa 2.16 conc of salts

25

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

clear ; clc ;


/ / E xa mp le 2 . 1 6 / / P ag e 2 5 printf ( ” E xa mp le 2 . 1 6 , P ag e 2 5 \ n \ n ” ) ; // s o l u t i o n M 1 = 100 / / CaCO3 v 1 = 2 // v a l e n c e o f CaCO3 e qm 1 = M 1/ v1 // e q u i v a l e n t m as s o f CaCO3 M 2 = 162 // Ca(HCO3) 2 v2 = 2 e qm 2 = M 2/ v2 m = 500 // [ mg/ l ] CaCO3 C 1 = ( e qm 2 / e qm 1 ) * m *. 6 / / [ mg/ l ] c o n c . o f Ca ( HCO3 ) 2 M 3 = 1 46 .3 // Mg(HCO3) 2 v3 = 2 e qm 3 = M 3/ v3 C 2 = ( e qm 3 / e qm 1 ) * m *. 4 // [mg/ l ] conc . o f Mg(HCO3) 2 printf ( ” A c t ua l c o n c e n t r a t i o n o f Ca (HCO3) 2 i n t h e s a mp l e w at er i s ” + string ( C 1 ) + ” mg/ l a nd o f Mg ( HCO3 ) 2 i s ” + string ( C 2 ) + ” mg/ l . ” )

Scilab code Exa 2.17 ppm unit

1 clear ; 2 clc ; 3

26

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



S = .68 // s u l ph u r c o n te n t by mass d = .85 / / k g / l s = ( S * d *1 0 ^6 ) / 10 0 / / [ mg/ l ] o r [ ppm ] printf ( ” S u lp hu r c o nt e nt i n LDO i s ” + string ( s ) + ” ppm . ”)

Scilab code Exa 2.18 molarity normality and molality

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

clear ; clc ;

/ / S t o i ch i o m et r y / / C ha pt er 2 / / B a si c Ch em ic al C a l c u l a t i o n s / / E xa mp le 2 . 1 8 / / P ag e 2 6 printf ( ” E xa mp le 2 . 1 8 , P ag e 2 6 \ n \ n ” ) ; // s o l u t i o n m 1 = 100 // [ kg ] s o l u t i o n m 2 = 20 // [ kg ] NaCl

( basi s )

27

17 18 19 20 21 22 23 24

d = 1. 12 7 // [ kg / l ] V = m1 /d // volume o f 10 0 kg s o l . n = ( m 2 / 58 .5 ) * 1 00 / / [ m ol ] Na Cl M = n / V // [M] v = 1 / / v a l e n ce o f NaCl so , N = M m = n /( m1 - m 2) // [ mol / kg ] printf ( ” M o l a r i t y = ” + string ( M ) + ” M \ n N o r m a l i t y = ” + string ( N ) + ” N \ n M o la l i ty = ” + string ( m ) + ” m ol / k g . ” )

Scilab code Exa 2.19 molarity of solution

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

clear ; clc ;

/ / S t o i ch i o m et r y / / C ha pt er 2 / / B a si c Ch em ic al C a l c u l a t i o n s / / E xa mp le 2 . 1 9 / / P ag e 2 7 printf ( ” E xa mp le 2 . 1 9 , P ag e 2 7 \ n \ n ” ) ; // s o l u t i o n m 1 = 100 / / [ k g ] TEA s o l u t i o n ( b a s i s ) m 2 = 50 // [ kg ] TEA M 1 = 149 / / m o l ar m as s o f TEA d = 1.05 // [ kg / l ] V = m1 /d // volume o f 10 0 kg s o l . n = ( m2 / M1 ) * 10 0 / / [ m ol ] N aC l M = n / V // [M] printf ( ” M o l a r it y o f s o l u t i o n = ” + string ( M ) + ” M. ” )

28

Scilab code Exa 2.20 conc of CO2

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

clear ; clc ;

/ / S t o i ch i o m et r y / / C ha pt er 2 / / B a si c Ch em ic al C a l c u l a t i o n s / / E xa mp le 2 . 2 0 / / P ag e 2 7 printf ( ” E xa mp le 2 . 2 0 , P ag e 2 7 \ n \ n ” ) ; // s o l u t i o n m 1 = 100 / / [ k g ] MEA s o l u t i o n ( b a s i s ) m 2 = 20 // [ kg ] MEA M 1 = 61 / / m o l ar m as s o f MEA n 1 = m2 / M1 / / [ k m ol ] C = .206 n 2 = C *n1 / / [ k mo l ] d i s s o l v e d CO2 m 3 = n2 *44 / / [ k g ] m as s o f CO2 n 3 = ( m1 - m 2 - m3 ) / 18 / / [ k m ol ] w a t e r n = ( n 2 /( n 1 + n2 + n 3 )) * 1 00 m = ( m 3 / 10 0) * 1 00 printf ( ” Mass p e r c e n t o f CO2 = ” + string ( m ) + ” a nd Mol p e r c e n t = ” + string ( n ) + ” . ” )

Scilab code Exa 2.21 pH of HOCl

29

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

clear ; clc ;

/ / S t o i ch i o m et r y / / C ha pt er 2 / / B a si c Ch em ic al C a l c u l a t i o n s / / E xa mp le 2 . 2 1 / / P ag e 2 7 printf ( ” E xa mp le 2 . 2 1 , P ag e 2 9 \ n \ n ” ) ; // s o l u t i o n //HOCl M a = .1 / / m o l a r i t y K a = 9 .6 *1 0^ - 7 C = ( Ma * Ka ) ^. 5 / / c on c . o f H+ i o n s p H = -log10 ( C ) printf ( ”pH o f t h e s o l i s ” + string ( p H ) + ” . ” )

Scilab code Exa 2.22 Mavg and composition of air

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

clear ; clc ;

/ / S t o i ch i o m et r y / / C ha pt er 2 / / B a si c Ch em ic al C a l c u l a t i o n s / / E xa mp le 2 . 2 2 / / P ag e 3 9 printf ( ” E xa mp le 2 . 2 2 , P ag e 3 9 \ n \ n ” ) ;

30

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

// s o l u t i o n

n = 100 // [ mol ] a i r ( b a s i s ) n 1 = 21 // [ mol ] O2 n 2 = 78 // [ mol ] N2 n 3 = 1 // [ mol ] Ar M 1 = 3 1. 99 88 / / O2 M 2 = 2 8. 01 34 / / N2 M 3 = 3 9. 94 8 / / Ar m 1 = n1 * M1 m 2 = n2 * M2 m 3 = n3 * M3 M a = ( m 1 + m2 + m 3 )/ n printf ( ” a v e ra ge m o l a r mass o f a i r g . ”)

i s ” + string ( M a ) + ”

Scilab code Exa 2.23 Composition and specific gravity

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

clear ; clc ;

/ / S t o i ch i o m et r y / / C ha pt er 2 / / B a si c Ch em ic al C a l c u l a t i o n s / / E xa mp le 2 . 2 3 / / P ag e 3 9 printf ( ” E xa mp le 2 . 2 3 , P ag e 3 9 \ n \ n ” ) ; // s o l u t i o n // ( a ) printf ( ” ( a ) \ n \ n ” ) n = 100 / / [ kmol ] c r a ck e d g a s ( b a s i s )

31

n 1 = 45 / / m et h an e n 2 = 10 / / e t ha n e n 3 = 25 // e t h y l e ne n 4 = 7 / / p r op a ne n 5 = 8 // p r o p y l en e n 6 = 5 // n−b u t a n e M 1 = 16 M 2 = 30 M 3 = 28 M 4 = 44 M 5 = 42 M 6 = 58 m 1 = n1 * M1 m 2 = n2 * M2 m 3 = n3 * M3 m 4 = n4 * M4 m 5 = n5 * M5 m 6 = n6 * M6 m = m 1 + m2 + m 3 + m4 + m 5 + m6 M = m/n printf ( ” A v er a g e m ol a r m ass o f g as i s ” + string ( M ) + ” g .”)

18 19 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

//(b) printf ( ” ( b ) \ n \ n ” )

/ / c o m p o s it i o n

p 1 = ( m1 / m) * 10 0 p 2 = m2 * 10 0/ m p 3 = m3 * 10 0/ m p 4 = m4 * 10 0/ m p 5 = m5 * 10 0/ m p 6 = m6 * 10 0/ m printf ( ” GAS

Mass ” + string ( p 1 ) + ” ” + string ( p 2 ) + ” ” + string ( p 3 ) + ” ” + string ( p 4 ) + ” ” + string ( p 5 ) + ” ” + string ( p 6 ) + ” 32

P e r c e n t \ n M et h an e \ n E th a ne \ n E t hy l en e \ n P r op a ne \ n P r op y le n e \ n n−Butane \ n \ n \ n” )

49 50 51 52 53

// ( c ) printf ( ” ( c ) \ n \ n ” )

// s p e c i f i c gra vity g = M /2 / 2 8. 8 . 97 97 printf ( ” S p e c i f i c g r a v i t y

i s ” + string ( g ) + ” . ” )

Scilab code Exa 2.24 percentage error

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

clear ; clc ;

/ / S t o i ch c h i o m et et r y // Ch haa pt p t er er 2 / / B a si s i c C h em e m ic i c al al C a l c u l a t i o n s

/ / E xa xa mp mp le le 2 . 2 4 / / P aagg e 4 0 printf ( ” E xa xa mp mp le le 2 . 2 4 , P ag ag e 4 0 \ n \ n ” ) ; // s o l u t i o n

p = 1 0 0 // [ ba r ] T = 6 23 2 3 .1 . 1 5 // [K] R = . 08 0 8 31 3 1 45 45 ol ] m o l ar a r v ol o l um um e V = R *T * T /p / p / / [ l / m ol v = V / 18 1 8 .0 . 0 15 15 3 / / printf ( ” S p e c i f i c v o l u m e = ” + string ( v ) + ” mˆ3/ kg . ” )

Scilab code Exa 2.25 molar volume

1 clear ; 2 clc ;

33

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


/ / E xa xa mp mp le le 2 . 2 5 / / P aagg e 4 0 xa mp mp le le 2 . 2 5 , P ag ag e 4 0 \ n \ n ” ) ; printf ( ” E xa // s o l u t i o n

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

p = 4 // [ ba r ] T = 7 73 7 3 .1 . 1 5 // [K] R = . 08 0 8 31 3 1 45 45 ol ] m o l ar a r v ol o l um um e V = R *T * T /p / p / / [ l / m ol printf ( ” M o la l a r v ol o l um u m e = ” + string ( V ) + ” l / m o l . \ n \ n \ n ” )

/ / u s in i n g a pp p p en e n di di x I I I / / c a l c u l a t i n g Tc an a n d Pc Pc o f d i f f e r e n t g a s e s a cc o r di n g t o t h e i r mass f r a c t i o n s T c 1 = . 3 52 5 2 * 32 3 2 . 20 2 0 / / H2 T c 2 = . 1 48 4 8 * 19 1 9 0 .5 . 5 6 / / m et e t h an an e T c 3 = . 1 28 2 8 * 28 2 8 2 .3 .3 4 / / e t h y l e n e T c 4 = . 3 39 3 9 * 13 1 3 2 .9 . 9 1 / / CO T c 5 = . 0 15 1 5 * 30 3 0 4 .1 . 1 0 / / CO2 T c 6 = . 0 18 1 8 * 12 1 2 6 .0 . 0 9 / / N2 T c = T c 1 + T c2 c2 + T c 3 + T c 4 + T c5 c 5 + T c 6 // / / Tc o f g as as / / s i m i l a r l y f i n d i n g Pc Pc

Pc1=.352*12.97 Pc2=.148*45.99 Pc3=.128*50.41 Pc4=.339*34.99 Pc5=.015*73.75 Pc6=.018*33.94 / / Pc o f g a s P c = P c 1 + P c 2 + P c 3 + P c 4 + P c 5 + P c 6 // a = ( 27 2 7 * R ^ 2* 2* T c ^ 2) 2 ) / ( 64 6 4 * P c ) / / [ b a r / m ol ol ˆ 2 ] ol b = ( R * Tc T c ) / (8 ( 8 * P c) c ) / / l / m ol

34

39 / / s u b s t i t u t i n g

t h e s e v a l u es e s i n v an a n de d e rw r w al a l l e q and s o l v i n g b y N e w t o n R ha ha ps ps on o n m et et h od od we g e t 40 V = 1 5 .7 .7 4 / / [ l / m o l ] 41 printf ( ” by b y V a n de d e r wa w a l l e q m o la l a r v ol o l um u m e = ” + string ( V ) + ” l / m ol ol ” )

Scilab code Exa 2.26 ternary mix analysis

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

clear ; clc ;


/ / E xa xa mp mp le le 2 . 2 6 / / P aagg e 4 3 printf ( ” E xa xa mp mp le le 2 . 2 6 , P ag ag e 4 3 \ n \ n ” ) ; // s o l u t i o n m = 6 .5 . 5 06 06 5 / / [ g ] m i x t u r e ( b a s i s ) P v = 2 .1 . 1 75 7 5 / / [ k P a ] V . P . o f w a t e r o v e r KOH P a = 1 02 0 2 .5 . 5 - 2 . 17 1 7 5 / / [ k P a ] P a r t i a l P o f n−b u ta t a n e a nd nd 1

butene 17 18 19 20 21 22 23 24 25 26

V = 4 15 1 5 .1 . 1 *1 * 1 0^ 0 ^ - 3 // [ l ] R = 8 .3 . 3 14 1 4 47 47 2 T = 2 9 6. 6 . 4 // [K] / / m ol o l es e s o f b ut u t en e n e a n d b ut u t an an e n = ( Pa P a * V) V ) /R / R * T // n 1 = n *. * . 43 4 3 1 / / n−b u t a n e m 1 = n 1 *5 *58 / / [ g ] n 2 = n -n - n 1 / / 1 b ut u t en en e m 2 = n 2 *5 * 5 6 // [ g ] m 3 = m -m -m1 // [ g ] f u r f u r a l n 3 = m 3 /9 /96

35

27 printf ( ” component

mol p e r c e n t mass p e r c e n t \ nn−Butane ” + string ( n 1 / n * 1 0 0 ) + ” ” + string ( m 1 / m * 1 0 0 ) + ” \ n1 −Butene ” + string ( n 2 / n * 1 0 0 ) + ” ”+ ” + string ( n 3 / string ( m 2 / m * 1 0 0 ) + ” \ n F u r f u r a l n*100)+” ” + string ( m 1 / m * 1 0 0 ) + ” ” )

Scilab code Exa 2.27 vapour mix composition

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

clear ; clc ;


/ / E xa mp le 2 . 2 7 / / P ag e 4 4 printf ( ” E xa mp le 2 . 2 7 , P ag e 4 4 \ n \ n ” ) ; // s o l u t i o n P = 5 .7 +1 .0 1 // [ b ar ]

absolute tota l P

/ / u s i n g R ou lt ’ s l aw v p = 3 . 29 3 *. 7 73 7 // [ kPa ] vap P o f f u r f u r a l // u s in g Da lt on ’ s l aw o f p a r t i a l P n 1 = v p /( P * 1 00 ) // m ol f r a c t i o n o f f u r f u r a l n 2 = 1 -n1 // m ol f r a c t i o n o f 1 − b u t e n e printf ( ” mol f r a c t i o n o f F u r f u r a l i s ” + string ( n 1 ) + ” \ nmol f r a c t i o n o f 1− B utene i s ” + string ( n 2 ) + ” . ” )

Scilab code Exa 2.28 absolute humidity

36

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

clear ; clc ;


/ / E xa mp le 2 . 2 8 / / P ag e 4 4 printf ( ” E xa mp le 2 . 2 8 , P ag e 4 4 \ n \ n ” ) ; // s o l u t i o n P = 100 / / [ kPa ] t o t a l P P w = 2 .5 32 6 // [ kPa ] V. P> of w at er a t dew p o i nt

/ / a b s o l u t e h umi ty = m ass o f w at er v ap ou r / mass o f dry a i r 18 H = ( P w / ( P - Pw ) ) * ( 1 8 . 0 1 5 3/ 2 8 . 9 69 7 ) // a b s o l u t e humidity 19 printf ( ” a b s o l u t e h um id i ty = ” + string ( H ) + ” . ” )

Scilab code Exa 2.29 nozzle outlet T

1 2 3 4 5 6 7 8 9 10 11

clear ; clc ;


/ / E xa mp le 2 . 2 9 / / P ag e 4 5 printf ( ” E xa mp le 2 . 2 9 , P ag e 4 5 \ n \ n ” ) ; 37

12 13 14 15 16 17 18 19 20 21

// s o l u t i o n / / T i −Tf = mu ∗ ( Pi −Pf ) P i = 20 .7 // [ ba r ] P f = 8.7 / / [ b a r ] m u = 1 .6 16 // [K/ bar ] T i = 3 55 .1 5 // [K] T f = Ti - m u * ( Pi - P f ) printf ( ” O u tl et t em pe ra tu re i s ” + string ( T f ) + ” K” )

38

Chapter 3 Material Balances without Chemical Reaction

Scilab code Exa 3.1 Lancashire boiler

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

clear ; clc ;

/ / S t o i ch i o m et r y / / C ha pt er 3 / / M a t e r i a l B a l a nc e s Wi th out C he mi ca l R e a ct i o n

/ / E xa mp le 3 . 1 / / P ag e 6 0 printf ( ” E xa mp le 3 . 1 , P ag e 6 0 \ n \ n ” ) ; // s o l u t i o n

m = 1 / / [ kg ] f e e d w a te r m 1 = 12 00 // [ mg ] d i s s o l v e d s o l i d s i n 1 kg f e e d w a t e r m 2 = 35 00 // [ mg ] max d i s s o l v e d s o l i d c o nt e nt x = ( m* m1 ) / m2 / / [ k g ] b lo wn down w a t e r printf ( ” P e r ce n ta g e o f f e e d w at er t o be blown down i s

39

” + string ( x ) + ” . ” )

Scilab code Exa 3.2 Textile mill

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

clear ; clc ;


/ / E xa mp le 3 . 2 / / P ag e 6 1 printf ( ” E xa mp le 3 . 2 , P ag e 6 1 \ n \ n ” ) ; // s o l u t i o n m = 100 / / [ kg ] weak l i q u o r ( f e e d ) m 1 = 4 // [ kg ] NaOH

p = .25 x = 4/ p / / w a t e r l e f t y = 100 - 16 / / [ kg ] e v a po r a te d w at er printf ( ” Amount o f w at er t h a t e v a po r a te d i s ” + string ( y)+” kg . ”)

Scilab code Exa 3.3 recovered tannin

1 2 3 4 5 6

clear ; clc ;

/ / S t o i ch i o m et r y / / C ha pt er 3 / / M a t e r i a l B a l a nc e s Wi th out C he mi ca l R e a ct i o n 40

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


m = 100 / / [ kg ] b a bu l b a rk ( b a s i s ) m 1 = 5.8 / / [ k g ] m o i s t u r e m 2 = 12 .6 // [ kg ] Tannin m 3 = 8.3 // [ kg ] s o l u b l e non t a nn i n o r g a n i c m a t e r i a l m 4 = m - m1 - m2 - m 3 / / [ kg ] L i g n i n

// l i g n i n c o n t e nt r em ai ns u n a f f ec t e d d ur in g l e a c h i n g m 5 = 1 00 - . 92 - . 65 // [ kg l i g n i n / kg d ry r e s i d u e ] x = ( m4 * 10 0) / m5 / / [ k g ] T 1 = x * .0 09 2 / / [ kg ] Ta nn in p r e s e n t i n r e s i d u e T2 = m2 - T 1 / / [ kg ] T an ni n r e c o v e r e d T = ( T2 / m2 ) * 10 0 printf ( ” P e r ce n ta g e o f Tannin r e c o v er e d d u ri n g l e a c h i n g i s ” + string ( T ) + ” . ” )

Scilab code Exa 3.4 Extraction of dry neem leaves

1 2 3 4 5 6 7 8 9 10 11

clear ; clc ;


/ / E xa mp le 3 . 4 / / P ag e 6 2 printf ( ” E xa mp le 3 . 4 , P ag e 6 2 \ n \ n ” ) ; 41

12 13 14 15 16 17 18 19

// s o l u t i o n m = 1 / / [ kg ] d ry neem l e a v e s ( b a s i s ) m 1 = . 01 /1 00 / / [ kg ] b et a c a r t e n e c o nt e nt o f l e a v e s E x = ( m 1 * 10 0) / . 41 / / [ kg ] e x t r a c t q u a n t i t y T c1 = Ex * .1 55 // [ kg ] Al pha T o co p he r ol i n t h e e x t r a c t T c 2 = . 46 / 10 0 / / [ kg ] A lp ha T o c o p he r o l i n t h e neem

leaves 20 R = ( T c1 * 1 00 ) / T c2 // r e c o ve r y o f Alpha T oc op he ro l 21 printf ( ” ( a ) \ n \ nmass o f e x t r a ct p ha se p er kg o f d r y l e a v e s i s ” + string ( E x ) + ” kg \ n \ n \ n ( b ) \ n \ n p er ce n t r e co v e r y o f Alpha T oc op he ro l i s ” + string (R)+” . ”)

Scilab code Exa 3.5 Extraction of mix of Acetone and Chloroform

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

clear ; clc ;


/ / E xa mp le 3 . 5 / / P ag e 6 2 printf ( ” E xa mp le 3 . 5 , P ag e 6 2 \ n \ n ” ) ; // s o l u t i o n m = 1 00 // [ kg ] o r i g i n a l m ix tu re ( b a s i s ) A = 27.8 // [ kg ] B = 72.2 // [ kg ]

/ / l e t x a nd y be u pe r and l o w er l a y e r amounts 42

19 20 21 22 23 24 25 26 27 28

// t o t a l m ix tu re = ( x+y ) kg // b a l a n ci n g A a nd B X = [ .0 75 . 20 3; .0 35 . 67 3] d = [ 2 7. 8 ;7 2 .2 ] x = X\d M = X ( 1 ,1 ) +X ( 2 , 1) / / [ kg ] t o t a l m ix tu re Ms = M - m / / [ k g ] m ix ed s o l v e n t M r = Ms /m // mixed s o l v e n t / o r i g i n a l m ix tu re S 1 = x ( 1 , 1 ) * . 57 4 + x ( 2 , 1) * . 0 2 8 / / [ k g ] w a t e r b a l a n c e S 2 = x ( 1 , 1 ) * . 31 6 + x ( 2 , 1) * . 0 9 6 // [ kg ] a c e t i c a c i d

balance 29 30 31 32

Q s = S1 + S2 p S 1 = ( S 1 * 10 0) / Q s p S2 = 10 0 - pS 1 printf ( ” ( a ) \ n \ nUpper l a y e r = ” + string ( x ( 1 , 1 ) ) + ” kg and Lower l a y e r = ” + string ( x ( 2 , 1 ) ) + ” \ n \ n \ n ( b )

\ n \ nmass r a t i o o f t he mixed s o l v e n t t o t he o r i g i n a l m i x t u r e i s ” + string ( M r ) + ” \ n \ n \ n ( c ) \ n \ n wa te r mass p e r c e n t = ” + string ( p S 1 ) + ” and a c e t i c a c id mass p e rc e nt = ” + string ( p S 2 ) + ” . ” )

Scilab code Exa 3.6 Pressure Swing Adsorption

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

clear ; clc ;


/ / E xa mp le 3 . 6 / / P ag e 6 3 printf ( ” E xa mp le 3 . 6 , P ag e 6 3 \ n \ n ” ) ;

43

13 14 15 16 17 18 19 20 21 22

// s o l u t i o n m = 170 / / [ Nmˆ 3 / h ] a i r ( b a s i s ) m 1 = 5 0* .9 9 / / [ Nmˆ 3 / h ] N2 c o n t e n t o f t h e s t r e a m m 2 = 5 0* .0 1 // [Nmˆ3/ h ] N = m *. 79 - m 1 // [Nmˆ3/ h ] N2 O = m *. 21 - m 2 // [Nmˆ3/ h ] O2 V 1 = N * 10 0/ ( N+ O ) V 2 = O * 10 0/ ( N+ O ) printf ( ” Vol p e r c e n t o f N2 i s ” + string ( V 1 ) + ” and Vo l p er ce nt o f O2 i s ” + string ( V 2 ) + ” . ” )

Scilab code Exa 3.7 Required Oleum strength

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

clear ; clc ;


/ / E xa mp le 3 . 7 / / P ag e 6 4 printf ( ” E xa mp le 3 . 7 , P ag e 6 4 \ n \ n ” ) ; // s o l u t i o n m = 100 / / [ kg ] SO3 f r e e m ix ed a c i d ( b a s i s ) m 1 = 55 // [ kg ] HNO3 m 2 = 45 // [ kg ] H2SO4

// SO3 + H2O −−> H2SO4 m 3 = ( 80 /1 8) * 3 // [ kg ] SO3 e q u i v a l e n t t o 3 kg o f water 20 Q = m2 +m3 / / [ k g ] o le um t o b e m ix ed 44

21 S = ( m3 / Q) * 10 0 // s t r e n g t h o f oleum 22 R = m1 /Q 23 printf ( ” S t re n g t h o f Oleum r e q u i r e d i s ” + string ( S ) + ”

\nHNO3 a nd Oleum a r e r e q u i r e d t o b e m ix ed i n t h e p r o p o r ti o n o f ” + string ( R ) + ” : 1 . ” )

Scilab code Exa 3.8 Mixed acid formation

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

clear ; clc ;


/ / E xa mp le 3 . 8 / / P ag e 6 4 printf ( ” E xa mp le 3 . 8 , P ag e 6 4 \ n \ n ” ) ; // s o l u t i o n m = 1000 / / [ kg ] m ix ed a c i d ( b a s i s )

/ / d o in g o v e r a l l mass b a la n ce , H2SO4 b a l a n c e and HNO3 ba la nc e A = [1 1 1;.444 0 .9 8;. 113 .9 0] d = [ 1 00 0 ;6 0 0; 3 20 ] x = A\d printf ( ” q u a n t i t i e s o f a c i d s r e q u i r e d a re : \ n S p e n t = ” + string ( x ( 1 , 1 ) ) + ” kg \ n HNO3 = ” + string ( x ( 2 , 1 ) ) + ” kg \ n H2SO4 = ” + string ( x ( 3 , 1 ) ) + ” k g . ” )

Scilab code Exa 3.9 Actual analysis of borewell water

45

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

clear ; clc ;

28 29 30 31 32

m 7 = ( m 6 /5 0) * 2 25 m8 = 384 -225 // [ mg ] CaCO3 from Ca(HCO3) 2



l = 1 // [ l i t r e ] water ( ba si s ) C l = 4 75 .6 // [ mg ] m 1 = ( 5 8. 5 /3 5 .5 ) * C l / / [ mg ] NaCl p r e s e n t i n w at er S O 4 = 1 02 .9 / / [ mg ] / / SO4 m 3 = ( 14 2 /9 6 ) * SO 4 / / [ mg ] Na2SO4 p r e s e n t i n w a te r

/ / c a r b o n a t e s a r e p r e s e n t du e t o Na2CO3 / / e q m as s o f CaCO3 = 5 0 / / e q m as s o f Na2CO3 = 5 3 m 4 = ( 5 3/ 5 0) * 6 5 .9 // [ mg ] Na2CO3 p r e s e n t i n w a te r / / NaHCO3 i n w a t e r = b i c a r b o n a t e s − t e m p o r a r y hardness 25 m 5 = 3 90 .6 - 3 84 / / [ mg ] NaHCO3 p r e s e n t a s CaCO3 26 m 6 = ( 84 /5 0) * m5 // [ mg ] NaHCO3 p r e s e n t i n w a t e r 27 / / e q u i v a l e n t m as s o f Mg (HCO3 ) 2 = 7 3 . 1 5

/ / e q u i v a l e n t mass o f Ca (HCO3) 2 i s 8 1 m 9 = ( m 8 /5 0) * 1 59 / / [ mg ] Ca ( HCO3 ) 2 p r e s e n t i n w a t e r printf ( ” Component a n a l y s i s o f raw w a te r : \ n \ n \ nCompound mg/ l \ n \ nCa (HCO3) 2 ”+ ” + string ( m 7 ) + ” \ string ( m 9 ) + ” \ nMg (HCO3) 2 nNaHCO3 ” + string ( m 6 ) + ” \ nNa2CO3 ” + string ( m 4 ) + ” \ nNaCl ”+ ” + string ( m 3 ) + ” ” ) string ( m 1 ) + ” \ nNa2SO4 46

Scilab code Exa 3.10 Matrix use

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

clear ; clc ;


/ / E xa mp le 3 . 1 0 / / P ag e 6 7 printf ( ” E xa mp le 3 . 1 0 , P ag e 6 7 \ n \ n ” ) ; // s o l u t i o n / / s e e e x am pl es 3 . 5 and 3 . 8

Scilab code Exa 3.11 Flowrate calculation

1 2 3 4 5 6 7 8 9 10 11

clear ; clc ;


/ / E xa mp le 3 . 1 1 / / P ag e 6 8 printf ( ” E xa mp le 3 . 1 1 , P ag e 6 8 \ n \ n ” ) ; 47

12 13 / / s o l u t i o n 14 15 / / b a s i s : 1 0 0 0 kg /h o f f e e d 16 / / b a l a n c i n g H2SO4 , HNO3 and H2O i n

a l l the three

p r od u ct s t r ea m s 17 M = [1 0 0 1 0 0 1 0 0 ;0 1 0 0 1 0 0 1 0 ;0 0 1 0 0 1 0 0 1;1 0 0 0 0 0 0 0 0;0 1 0 0 0 0 0 0 0;0 0 1 0 0 0 0 0 0;0 0 0 1 0 0 0 0 0;0 0 0 0 1 0 0 0 0;0 0 0 0 0 1 0 0 0] 18 v = [ 4 0 0 ; 10 0 ; 5 0 0 ;4 ; 9 4 ; 6 0; 1 6 ; 6 ; 40 0 ] 19 s = M \ v 20 A = s ( 1) + s ( 2) + s ( 3) 21 B = s ( 4) + s ( 5) + s ( 6) 22 C = s ( 7) + s ( 8) + s ( 9) 23 printf ( ” F l o w ra t e s a r e : \ n A = ” + string ( A ) + ” kg / h \ n B = ” + string ( B ) + ” kg / h \ n C = ” + string ( C ) + ” k g /h ” )

Scilab code Exa 3.12 solving eqs with graphical plot

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

clear ; clc ;


/ / E xa mp le 3 . 1 2 / / P ag e 7 0 printf ( ” E xa mp le 3 . 1 2 , P ag e 7 0 \ n \ n ” ) ; // s o l u t i o n m = 100 // kg x = linspace ( 7 0 , 1 1 0 , 5 ) ;

48

16 17 18 19 20 21 22

y = linspace ( 1 0 0 , 1 1 5 , 4 ) ; y 1 = 2 7. 8/ .2 03 - . 07 5* x / .2 03 y 2 = 7 2. 2/ .6 73 - . 03 5* x / .6 73 x = linspace ( 7 0 , 1 1 0 , 5 ) ; plot ( x , y 1 , s t y l e = 4 ) plot ( x , y 2 , s t y l e = 8 )

23 24 25 26 27 28 29 30 31 32 33

x = 93 .4 ; y = 1 02 .4 ; M = x + y / / [ kg ] t o t a l m ix tu re Ms = M - m / / [ k g ] m ix ed s o l v e n t M r = Ms /m // mixed s o l v e n t / o r i g i n a l m ix tu re S 1 = x * . 57 4+ y * . 02 8 / / [ k g ] w a t e r b a l a n c e S 2 = x * . 31 6+ y * . 09 6 // [ kg ] a c e t i c a c i d b a l an c e Q s = S1 + S2 p S 1 = ( S 1 * 10 0) / Q s p S2 = 10 0 - pS 1 printf ( ” ( a ) \ n \ nUpper l a y e r = ” + string ( x ) + ” k g a n d Lower l a y e r = ” + string ( y ) + ” \ n \ n \ n ( b ) \ n \ nmass

// fro m g r a p h i t s .

c l e a r x = 9 3 . 4 kg and y = 1 0 2 . 4 kg

r a t i o o f t h e mixed s o l v e n t t o t h e o r i g i n a l m ix tu re i s ” + string ( M r ) + ” \ n \ n \ n ( c ) \ n \ n w a t e r mass p e r c e n t = ” + string ( p S 1 ) + ” and a c e t i c a c i d mass p e r ce n t = ” + string ( p S 2 ) + ” . ” )

Scilab code Exa 3.14 ion exclusion process

1 2 3 4 5 6 7 8

clear ; clc ;


49

9 10 11 12 13 14 15 16 17 18 19 20 21 22 23

/ / E xa mp le 3 . 1 4 / / P ag e 7 3 printf ( ” E xa mp le 3 . 1 4 , P ag e 7 3 \ n \ n ” ) ; // s o l u t i o n // u s in g t a b l e 2 . 7 on p a g e no 75 R g = 8 1 24 * 10 0 /9 4 48 // r e c o v e r y o f g l y c e r i n e L g = ( 1 6+ 8 3) * 1 0 0 /9 4 48 // l o s s o f g l y c e r i n e i n w a s t e R e g = 1 00 - R g - Lg // r e c y c l e o f g l y c e r i n e m 1 = 2 38 /8 12 4 // NaCl i n p r od u ct m 2 = Rg * 1 2/ 10 0 // g l y c e r i n e i n p ro du ct m 3 = m1 + m2 // t o t a l s o l u t e n = m1 * 10 0/ m 3 // NaCl p e r c e n t i n t o t a l s o l u t e printf ( ” ( a ) \ n \ n re co ve ry p er ce nt o f g l y c e r i n e i s ” + string ( R g ) + ” \ n \ n \ n ( b ) \ n \ n pe rc en t l o s s o f g l y c e r i n r i s ” + string ( L g ) + ” \ n \ n \ n ( c ) \ n \ n pr od uc t c o nt a mi n at i on w it h r e s p e c t t o s a l t NaCl i s ” + string ( n ) + ” . ” )

Scilab code Exa 3.15 Air Conditioning plant

1 2 3 4 5 6 7 8 9 10 11 12 13

clear ; clc ;


/ / E xa mp le 3 . 1 5 / / P ag e 7 6 printf ( ” E xa mp le 3 . 1 5 , P ag e 7 6 \ n \ n ” ) ; // s o l u t i o n 50

14 15 f 1 = 1. 25 // [mˆ 3/ s ]

f r e s h a mb ie nt a i r a s f e e d ( b a s i s

) 16 f 2 = 5 .8 06 / / [mˆ / s ] a i r e n t e r i n g a u d it o r iu m 17 v 1 = 8 . 31 4 *2 9 0/ 1 01 . 3 / / [mˆ 3/ k mo l ] s p . v o l . o f m o i s t 18 19 20 21 22 23 24 25 26 27 28 29

a i r a t 1 0 1. 3 kPa and 2 90 K n a 1 = f 2 * 10 00 / v 1 // [ mol / s ] m o la r f l ow r a t e o f a i r e n t e r i n g a u di t or i um n w 1 = 2 4 3 . 9 5* . 0 1 6 3/ 1 . 0 1 63 / / [ m ol / s ] n a2 = 2 43 .9 5 - nw 1 // [ mol / s ] d ry a i r f l o w n w2 = 2 4 0. 0 4* . 02 2 5 // [ mol / s ] m o is t ur e e n t e r i n a i r c o n d i t i o n i n g p l an t / / u si ng t a b l e 3 . 8 m 1 = ( nw2 - n w1 ) / / [ k g /h ] m o i s t u r e r em ov ed i n a c plant m 2 = n a2 - . 0 18 1 // [ mol / s ] m o is t ur e i n a i r l e a v i n g auditorium m 3 = ( m2 - n w 1 ) *1 8 / / [ kg / h ] m o i s t ur e a dd ed i n auditorium V m2 = 8 . 31 4 *3 0 8/ 1 01 . 3 / / [ mˆ 3 / k m o l ] n a3 = ( f 1 / 25 . 28 ) * 1 00 0 // [ mol/ s ] n 4 = 5 .4 0 - 1 .9 25 / / [ mol / s ] m o i s t ur e i n r e c y c l e s tr ea m m r = 2 4 0. 04 - 4 7 .5 2 5 / / [ mol / s ] m ol ar f l o w r a t e o f wet r e c y c l e s tr ea m

30 R = mr / na3 31 printf ( ” ( a ) \ n \ n m o is t u re rem oved i n AC p l a n t = ” + string ( m 1 ) + ” \ n \ n \ n ( b ) \ n \ n m o is t u re a dd ed i n a u d i t o r i u m = ” + string ( m 3 ) + ” \ n \ n \ n ( c ) \ n \

n r e c y c l e r a t i o o f m o l e s o f a i r r e c y c l e d p e r mole mole o f f r e s h a m b i e n t a i r i np ut = ” + string ( R ) + ” . ” )

Scilab code Exa 3.16 Overall efficiency of Pulp Mill

1 clear ;

51

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 ;


/ / E xa mp le 3 . 1 6 / / P ag e 7 8 printf ( ” Exa mpl e 3 . 1 6 Pa ge 7 8 \ n \ n ” ) ; // s o l u t i o n // // // //

s c re e n 1 f e ed = N kg O v er s iz e p a r t i c l e = NE1 k g U nd er si ze p a r t i c l e = N−NE1

//screen 2 // fe e d = NE1+X kg / / O v e r s i z e p a r t i c l e = ( NE1+X) ∗ E2 k g / / U n d e r s i z e p a r t i c l e = ( NE1+X) (1 − E2 ) k g

//screen 3 // fe ed = (NE1+X) ∗ E2 k g / / O v e r s i z e p a r t i c l e = ( NE1+X) ∗ E2 ∗ E3 k g / / U n d e r s i z e p a r t i c l e = ( NE1+X) ∗ E2 ∗ (1 − E3 ) kg printf ( ” O v e ra l l E f f i c i e n c y = ( E1 E2 E3 ) ∗ 1 0 0 / [ ( 1 − E1 ) (1 − E2)+E2 E3 ] . ” )

Scilab code Exa 3.17 2 stage membrane CO separation

1 clear ; 2 clc ;

52

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

printf ( ” ( a ) \ n \ n ” ) F = 5000 / / [ k mo l / h ] f e e d ( b a s i s ) m 1 = F *.4 7 / / [ k mo l / h ] CO i n F m 2 = F -m1 / / [ k mo l / h ] H2 i n F m 3 = m1 * .9 32 / / CO i n p r od u ct s tr ea m n 2 = m3 / .9 8 // [ kmol /h ] printf ( ” Flow r a t e o f p ro du ct s tr ea m i s ” + string ( n 2 ) + ” k mo l / h . \ n \ n \ n ( b ) \ n \ n ” ) n 2 = n2 - m3 / / [ k mo l / h ] H2 i n CO s t r e a m P r o d u c t H2 s t r e a m : \ n H2 = ” + string ( m2 printf ( ” - n 2 ) + ” k mo l / h \ n CO = ” + string ( m 1 - m 3 ) + ” k mo l / h \ n \ n \ n ( c ) \ n \ n” ) n H 2 = 2 69 7 .3 9 // [ kmol /h ] n C O = 3 00 0 - n H2 / / [ k mo l / h ] n4 = m2 + nH2 n5 = m1 + nCO n 6 = n4 + n5

C o m p o s i t i o n o f Mixed f ee d : \ n H2 = ” + printf ( ” string ( n 4 * 1 0 0 / n 6 ) + ” \ n CO = ” + string ( n 5 * 1 0 0 / n 6 ) + ””)

Scilab code Exa 3.18 2 stage reverse osmosis

53

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 32 33 34 35 36 37 38

clear ; clc ;


/ / E xa mp le 3 . 1 8 / / P ag e 7 9 printf ( ” E xa mp le 3 . 1 0 , P ag e 7 9 \ n \ n ” ) ; // s o l u t i o n // // // // // //

O v e ra l l b a l an c e F=R1+P2 B al an ce a c r o s s Module I F+R2 = R1+P1 ==> R1+P2+R2 = R1+P1 b a l a nc e a c r o s s module I I P1 = P2+R2 P 2 = 5 // [mˆ3/ h ] P 1 = P2 /.8 // [mˆ3/ h ] R 2 = P1 - P2 // [mˆ3/ h ] F = P1 /.66 - R2 // [mˆ3 / h ] R 1 = F -P2 // [mˆ3 / h ] / / O v er a l l b a la n c e o f DS i n w a te r x R1 = ( F * 42 00 - P 2 * 5 ) / R1 // [mg/ l ] x P1 = ( P 2 *5 ) / ( .0 1 5* P 1 ) / / [ mg / l ] x R2 = ( P 1 * xP1 - P 2 * 5 ) / R2 // [mg/ l ] m 1 = F * 4 20 0+ R 2 * x R2 / / [ g ] DS m i x e e d i n MF C 1 = m1 / (F + R2 ) / / [ mg / l ] m2 = R1 * xR1 / / [ g ] DS i n R1 r = m2 * 10 0/ m 1 // r e j e c t i o n i n module i n I m 3 = m1 - m2 / / [ g ] DS i n P1 C 2 = m3 / P1 / / [ mg / l ] R = R2 /F R 1 = P2 * 10 0/ F

54

39 printf ( ” F = ” + string ( F ) + ” m ˆ 3 / h \ n R 1 = ” + string ( R 1 ) + ” m ˆ 3 / h \ n P = ” + string ( P 1 + P 2 ) + ” m ˆ 3 / h \ n R 2 = ” + string ( R 2 ) + ” m ˆ 3 / h \ n r e c y c l e r a t i o = ” + string ( R ) + ” \ n r e j e c t i o n p e r ce n t a g e o f s a l t i n module I = ” + string ( r ) + ” ” )

Scilab code Exa 3.20 Purging by atmospheric pressure method

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

clear ; clc ;



/ / c o n c e t r a t i o n o f t he component a f t e r n t im es i n tr o d u c t i o n o f v volume o f i n e r t g a s : 16 / / Cn = Co / (1 +1 / n ) ˆ n 17 / / we know l i m n −−> i n f i n i t y (1+1/ n ) ˆ n = e 18 // t h e r e f o r e Cv = Co/ e

55

Chapter 4 Material Balances Involving Chemical Reactions

Scilab code Exa 4.1 Manufacture of MCA

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

clear ; clc ;

/ / S t o i ch i o m et r y / / C ha pt er 4 / / M a t er i a l B a la n ce s i n v o l v i n g C hem ic al R ea ct i on

/ / E xa mp le 4 . 1 / / P ag e 1 16 printf ( ” E xa mp le 4 . 1 , P ag e 1 1 6 \ n \ n ” ) ; // s o l u t i o n / / b a s i s o ne day o p e ra t i o n / / C l2 i s t he l i m i t i n g component n 1 = 4 53 6/ 71 / / [ k mo l ] C l 2 c h a r g e d / / 1 mol MCA r e q u i r e s 1 mol Cl2 , s o n 2 = 5 0 00 / 94 . 5 / / [ kmo l ] C l2 u s ed f o r MCA p r o d u c t i o n 56

20 21 22 23 24 25 26

// 1 mol DCA r e q u i r e s 2 mol o f Cl 2 n 3 = 2 6 3* 2 /1 2 9 / / [ k mo l ] C l2 u s e d f o r DCA p r o d u c t i o n n 4 = n2 + n3 // t o t a l Cl2 u s e d a = n4 * 10 0/ n 1 // c o n v e r s i o n %age b = n2 * 10 0/ n 4 / / y i e l d % o f MCA s = n2 /n3 printf ( ” P e r ce n ta g e c o n v e rs i o n = ” + string ( a ) + ” \ n \ n P e r c en t a g e y i e l d o f MCA = ” + string ( b ) + ” \ n \ n s e l e c t i v i t y o f MCA = ” + string ( s ) + ” . ” )

Scilab code Exa 4.2 Bechamp Process

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

clear ; clc ;


/ / E xa mp le 4 . 2 / / P ag e 1 17 printf ( ” E xa mp le 4 . 2 , P ag e 1 1 7 \ n \ n ” ) ; // s o l u t i o n

m = 700 / / [ kg ] ONT c h a r g ed t o r e a c t o r ( b a s i s ) m 1 = 5 05 *. 99 / / [ k g ] OT p r o d u c e d m 2 = ( 4 * 13 7 * 5 00 ) / ( 4 * 1 0 7) / / [ k g ] ONT r e q u i r e d m 3 = m *.9 8 / / [ k g ] ONT r e a c t e d n 1 = m1 * 10 0/ m 3 // y i e l d o f OT m 4 = ( 9* 5 6* m ) / ( 4* 1 37 ) / / [ kg ] t h e o r e t i c a l i r o n

reqiurement 21 m 5 = 8 00 *. 9 / / [ kg ] i r o n c h a r g ed 22 E = ( m5 - m 4 ) * 10 0/ m 4 // e x ce s s i r o n 57

23 printf ( ” ( a ) \ n \ n Y i el d o f OT = ” + string ( n 1 ) + ” \ n \ n \ n ( b ) \ n \ n Ex ce ss q u a nt i t y o f i r o n powder = ” + string ( E ) + ” . ” )

Scilab code Exa 4.3 Pilot Plant Calculations

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

clear ; clc ;


/ / E xa mp le 4 . 3 / / P ag e 1 18 printf ( ” E xa mp le 4 . 3 , P ag e 1 1 8 \ n \ n ” ) ; // s o l u t i o n printf ( ” ( a ) \ n \ n ” ) m = 100 / / kg c h l o r o b e n z e n e ( b a s i s ) m 1 = 1 0 6. 5 *. 6 55 / / kg HNO3 m 2 = 1 08 *. 93 6 // kg H2SO4 m 3 = 1 0 6. 5 *. 3 45 + 1 08 *. 0 64 / / kg w a te r M = m1 + m2 + m3 printf ( ” A n a l y s i s o f c ha rg e : \ n Component

m as s p e r c e n t \ n C h l o r o b e n z e n e ” + string ( m ” + string ( m1 * 1 0 0 / M ) + ” \ n HNO3 * 1 0 0 / M ) + ” \ n H2SO4 ” + string ( m2 ” + string ( m3 * 1 0 0 / M ) + ” \ n H2O *100/M)+” \n \n \n ( b ) \n \n” ) 22 / / ( b ) 23 / / t o t a l c ha rg e mass i s c on s t a n t 24 m 4 = 3 1 4. 5 *. 0 2 / / [ k g ] u n r e a c t e d CB i n t h e p r o d u ct 58

25 m5 = 100 - m4 // [ kg ] CB t h a t r e a c t e d 26 c = m5 * 1 00 /1 00 / / c o n ve r s i o n o f CB 27 printf ( ” P e r ce nt c o n ve r s i o n o f C h lo ro b en ze ne i s ” + string ( c ) + ” \ n \ n \ n ( c ) \ n \ n ” ) 28 / / ( c ) 29 m 6 = 6 3* c / 1 12 .5 // [ kg ] HNO3 consu med 30 m 7 = m1 - m6 / / u n r e a c t e d HNO3 31 m 7 = 1 57 . 5* c / 1 12 . 5 / / [ k g ] t o t a l NCB p r od u ce d 32 m 8 = m7 * .6 6 // [ kg ] p−NCB 33 m 9 = m7 * .3 4 // [ kg ] o−NCB 34 m 1 0 = 1 8* c / 1 12 .5 / / [ k g ] w a t e r p r o d u c e d 35 m 1 1 = m 10 + m3 // t o t a l w at er i n p ro du ct 36 m 1 2 = m 4 + m8 + m 9 + m 7 + m 2 + m 11 37 printf ( ” C om po si ti on o f p ro du ct s tr ea m : \ n

Component NCB NCB HNO3 H2SO4 H2O

mass p e r c e n t \ n CB ” + string ( m 4 * 1 0 0 / m 1 2 ) + ” \ n p− ” + string ( m 8 * 1 0 0 / m 1 2 ) + ” \ n o− ” + string ( m 9 * 1 0 0 / m 1 2 ) + ” \ n ” + string ( m 7 * 1 0 0 / m 1 2 ) + ” \ n ” + string ( m 2 * 1 0 0 / m 1 2 ) + ” \ n ” + string ( m 1 1 * 1 0 0 / m 1 2 ) + ” ” )

Scilab code Exa 4.4 Manufacturing of Acetaldehyde

1 2 3 4 5 6 7 8 9 10 11

clear ; clc ;


/ / E xa mp le 4 . 4 / / P ag e 1 19 printf ( ” E xa mp le 4 . 4 , P ag e 1 1 9 \ n \ n ” ) ; 59

12 13 14 15 16 17 18 19 20

// s o l u t i o n

n = 1 0 0 / / [ kmo l ] o u t g o i n g g a s f ro m 2 nd s c r u b b e r n 1 = . 8 5 2 * n // [ kmol ] N2 n 2 = 2 1 * n 1 / 7 9 // [ kmol ] O2 n 3 = n 2 - 2 . 1 / / [ kmo l ] r e a c t e d O2

// O2 balance / / O2 c on su me d i n r x n ( i i ) , ( i i i ) , ( v ) − O2 p r o d u c e d by r xn ( i v ) = 2 0 . 5 5 kmol 21 / / l e t a , b , c b e e t h a n o l r e a c t e d ( i i ) , ( i i i ) , ( i v ) and d be H2 r e a c t e d i n ( v ) 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37

/ / CO b a l a n c e a = 2 . 3 / 2 //kmol / / CO2 b a l a n c e b = .7/2

/ /CH4 b a l a n c e c=2.6/2

/ / O2 b a l a n c e d = 41 .1 - a - 3* b + c

/ / H2 b a l a n c e e = 7.1 + c+ d / / km ol ( t o t a l H2 p r od u ce d ) f = e -(3* b + 3* a) / / kmol ( H2 p ro du ce d i n ( i ) =

e t h a no l r e a c t ed i n ( i ) ) 38 g = f +a +b +c // t o t a l e th a n o l r e a c te d 39 h = 2 *( n 1 +n 2) // t o t a l e th a n o l e n t e r i n g 40 c 1 = g * 10 0/ h 41 printf ( ” ( a ) \ n \ n C on ve rs io n p e rc e nt o f e t ha n o l = ” + string ( c 1 ) + ” \ n \ n \ n” ) 42 y = f *1 00 /g 43 printf ( ” ( b ) \ n \ n Y i e l d o f a c et a ld e hy d e = ” + string ( y )+” . ” )

60

Scilab code Exa 4.5 Lime Soda process

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

clear ; clc ;


/ / E xa mp le 4 . 5 / / P ag e 1 21 printf ( ” E xa mp le 4 . 5 , P ag e 1 2 1 \ n \ n ” ) ; // s o l u t i o n v = 1 / / [ l ] w a te r ( b a s i s )

/ / 1 mo l ( 1 0 0mg ) CaCO3 g i v e s 1 mol ( 5 6 ) Cao / / u s e t a b l e 3 . 3 and eg 3 . 9 x = 5 6 *3 9 0. 6 /1 0 0 / / [ mg/ l ] l i m e p r o d u c e d printf ( ” Amount o f l i m e r e q u i r e d = ” + string ( x ) + ” mg/ l .”)

Scilab code Exa 4.6 Manufacture of Ammonia by Fertilizer plant

1 2 3 4 5 6 7

clear ; clc ;


61

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

/ / E xa mp le 4 . 5 / / P ag e 1 21 printf ( ” E xa mp le 4 . 5 , P ag e 1 2 1 \ n \ n ” ) ; // s o l u t i o n m = 1 0 0 / / [ k mo l ] ( b a s i s ) d r y m ix ed g a s

// // // //

x = kmol o f w at er g as y =kmol o f p r o du c er g a s o v e r a l l m a t e r i a l b al an ce : x+y = 10 0 (i)

// r 2 =.43 x +.25 y // H2 fo rm ed by s h i f t rx n / / r 2 =. 51 x +. 25 y / / H2 e n t e r i n g w it h w at er and p r o du c er g a s // r = r 1+r 2 // t a o a l H2 / / /n = .0 2x +. 63 y / / N2 e n t e r i n g //N2:H2=1:3 // ==> x − 1 . 8 0 7 y = 0 ( i i ) / / s o l v i n g ( i ) and ( i i )

A = [1 1;1 -1 .80 7] d = [ 10 0; 0] x = A\d s = . 43 * x ( 1) + . 2 5* x ( 2 ) / / s te am r e q . printf ( ” x = ” + string ( x ( 1 ) ) + ” and y = ” + string ( x ( 2 ) ) + ” \ nAmount o f s te am r e q u i r e d = ” + string ( s ) + ” k mo l ” )

Scilab code Exa 4.7 Saponification of Tallow

1 clear ; 2 clc ; 3 4 / / S t o i ch i o m et r y

62

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

/ / C ha pt er 4 / / M a t er i a l B a la n ce s i n v o l v i n g C hem ic al R ea ct i on


m = 100 / / [ k g ] T a l l o w m 1 = 3 *4 0 3* m / 8 90 / / [ k g ] m 2 = 9 2* m / 89 0 printf ( ” ( a ) \ n \ n NaOH r e q u i r e d = ” + string ( m 1 ) + ” kg

\ n \ n \ n ( b ) \ n \ n a mount o f g l y c e r i n e l i b e r a t e d = ” + string ( m 2 ) + ” k g . ” )

Scilab code Exa 4.8 Sulphur Burner

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

clear ; clc ;


/ / E xa mp le 4 . 8 / / P ag e 1 24 printf ( ” E xa mp le 4 . 8 , P ag e 1 2 4 \ n \ n ” ) ; // s o l u t i o n n = 100 // [ kmol ] SO3 f r e e g a s b a s i s n 1 = 16 .5 // [ kmol ] SO2

63

17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38

n 2 = 3 // [ kmol ] O2 n 3 = 80 .5 // [ kmol ] N2

/ / S + O2 = SO2 // S + 3/2 O2 = SO3 n 4 = ( 2 1/ 7 9) * 8 0 .5 / / [ k mo l ] O2 s u p p l i e d n 5 = n4 - n1 - n 2 / / [ k mo l ] U n a c co u n t ed O2 / / O2 u s e d i n 2 nd eq i s m5 n 6 = ( 2/ 3) * n5 / / [ k m ol ] SO3 p r o d u c e d n 7 = n1 + n6 // s u l ph u r b ur nt m 7 = n7 *32 // [ kg ] f 1 = n6 / n7 // f r a c t i o n o f SO3 b ur nt // O2 r eq . f o r c om pl et e c om bu st io n o f S = n7 n 8 = n4 - n7 / / [ k mo l ] e x c e s s O2 p 1 = n8 * 10 0/ n 7 // %age o f e x c es s a i r n 9 = n4 + n3 / / [ kmol / s ] a i r s u p p l i e d F 1 = n9 * .3 / n7 // a i r s up pl y r a te v = 2 2 . 4 1 4* ( 3 0 3 .1 5 / 2 7 3. 1 5 ) * ( 1 0 1 . 3 2 5/ 1 0 0 ) // [mˆ3/ kmol ] sp . v o l o f a i r V 1 = F1 *v / / [mˆ 3 / s ] f l o w r a t e o f f r e s h a i r n10 = n + n7 / / [ kmol ] t o t a l g a s f ro m b u rn e r n 1 1 = n 10 * . 3/ m 7 / / [ kmol / s ] g as r eq . f o r . 3 kg / s S V 2 = 2 2 0 4 14 * 1 0 73 . 1 5 * n 1 1 / 2 7 3. 1 5 // f l o w r a t e o f b ur ne r gases printf ( ” ( a ) \ n \ n The f r a c t i o n o f S b u r n t = ” + string ( f 1 ) + ” \ n \ n \ n ( b ) \ n \ n p er ce nt ag e o f e x c e s s a i r o ve r t he amount r eq . f o r S o x i d i s i n g t o SO2 = ” + string ( p 1 ) + ” \ n \ n \ n ( c ) \ n \ n v olume o f d ry a i r = ” + string ( V 1 ) + ” mˆ3/ s \ n \ n \ n ( d ) \ n \ n v ol um e o f b ur ne r g a s e s = ” + string ( V 2 ) + ” mˆ3 / s . ” )

Scilab code Exa 4.9 Hydrogenation of Refined Soybean oil

1 clear ; 2 clc ; 3

64

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


/ / E xa mp le 4 . 9 / / P ag e 1 25 printf ( ” E xa mp le 4 . 9 , P ag e 1 2 5 \ n \ n ” ) ; // s o l u t i o n m = 100 // [ kg ] s oy a f a t t y

acid ( basis )

// use t a bl e 4 .6 M 1 = m / .3 59 7 // M( avg ) o f s oy a f a t t y a c i d / / 3 m ol o f f a t t y a c i d + 1 m ol o f g l y c e r o l = 1 m ol t r i g l y c e r i d e + 3 mol o f w a t e r M 2 = M 1 * 3+ 9 2. 0 9 - 3 *1 8 .0 2 // Mavg o f s oy a be an o i l q 1 = M 2 *m / ( M1 * 3 ) // s oy ab ea n o i l p er 10 0 kg f a t t y acid / / b as ed on r e a c t i o n s o c cu r i n g q 2 = . 0 9 6 7 +. 1 8 2 2 *2 + . 0 2 41 * 3 / / kmol H2 r e q . p e r 1 00 kg s o ya f a t t y a c i d q 3 = . 51 01 / / kmol H2 r eq . p er 1 00 kg s oy ab ea n o i l q 4 = 1 1. 43 4 / / Nmˆ 3 /1 0 0 k g s o y ab e an o i l // x = l i n o l e i c a c i d c o nv e rt e d t o o l e i c a c i d / / y = o l e i c a ci d c on ve rt e d t o s t e a r i c a ci d q 5 = 2 8 2. 4 6* 6 .7 / 27 8 .4 3 // / / q6 = 2 8 2 . 4 6 ∗ x / 2 8 0 . 1 5 = 1 . 0 0 71 7 x [ kg ] o l e i c a c id by l i n o l e i c a ci d / / q7 = 2 8 4 . 4 8 ∗ y / 2 8 2 . 4 6 = 1 . 0 0 71 5 y [ kg ] s t e a r i c a c id by o l e i c a ci d / / q8 = 1 00 . 0 9 7 + . 0 0 7 1 7 x + . 00 7 1 5 y t o t a l f a t t y a c i d // s t e a r i c b al a n ce : − . 00 10 5 x + 1 . 0 0 6 1 1 y = 1 0 . 8 1 4 2 (i) // l i n o l e i c balance : 1.0019 x + .00019 y = 48.4975 (ii) // s o l v i n g ( i ) and ( i i ) we g e t x = 48.5 / / k g 65

35 36 37 38 39

y = 10.8 / / k g M 3 = 1 0 0. 5 2/ . 35 9 6 // Mavg o f f a t t y a c id H 2 r e q1 = . 5 33 4 - . 2 8 64 // p er 1 00 kg f a t t y a c i d H 2r eq = 5 2. 95 //Nmˆ3 / t I 2 s = 1 29 .5 // kg I 2 p er 10 0 kg s o ya b e a n o i l // f o r

s oy ab ea n o i l 40 I2h = 6 9. 2 // kg I 2 p e r 1 00 kg o f f a t 41 printf ( ” ( a ) \ n \ n t h e o r e t i c a l H2 r e qu i r e d = ” + string ( q 4 ) + ” Nmˆ 3 /1 0 0 kg s o ya b ea n o i l \ n \ n \ n ( b ) \ n \ n a c t u a l H2 r e q u i r e d = ” + string ( H 2 r e q ) + ” \ n \ n \ n ( c ) \ n \ n I o di n e v a l ue f o r s o y a b e a n o i l = ” + string ( I 2 s ) + ” . \ n \ n \ n ( d ) \ n \ n I o d i n e v a lu e o f h ar de ne d f a t = ” + string ( I 2 h ) + ” . ” )

Scilab code Exa 4.10 Material Balance in Formox Process

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

clear ; clc ;


/ / E xa mp le 4 . 1 0 / / P ag e 1 28 printf ( ” E xa mp le 4 . 1 0 , P ag e 1 2 8 \ n \ n ” ) ; // s o l u t i o n

F1 F2 F3 F4 n1

= = = = =

40 00 / / kg / h m et ha n ol ( b a s i s ) F1 /32 / / k m o l / h F2 / .0 84 // kmol / h g a s e o u s mix f l o w r a t e F2 - F3 // kmol /h f l ow o f wet a i r . 0 11 * 29 / 18 / / kmol / kmol d ry a i r

66

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

F 5 = F4 / (1 + n1 ) // kmol /h d ry a i r f l o w r a t e O 2 = F5 * .2 1 / / k m o l / h N 2 = F5 - O2 / / k m o l / h M r ea c te d 1 = F 2 * .9 9 / / k m o l / h M u n r e a ct e d 1 = F2 - M r e ac t e d1 / / k m o l / h

// r e a c t i o n ( i )

M r e a c te d 2 = M r e ac t e d1 * .9 / / k m o l / h H C H O p r od u c e d1 = 1 1 1 .3 7 5 O 2 c o n s um e d 1 = 1 1 1 .3 7 5 /2 H 2 O p r o du c e d 1 = 1 1 1 .3 7 5

/ / f or rxn i i to i v M c o n s um e d = M r e ac t e d1 * .1

//rxn ( i i ) C H 3 O H r e ac t e d 1 = M c o ns u m ed * . 71 O 2 c o n s um e d 2 = 8 . 7 86 * 1 .5 C O 2p r od u ce d = 8 .7 86 H 2 O p r o du c e d 2 = 8 . 7 86 * 2

// rxn ( i i i ) C H 3 O H r e ac t e d 2 = 1 2 . 37 5 * . 08 C O pr o du c ed = . 99 H 2 pr o du c ed = 2 *. 99

// rxn ( iv ) C H 3 O H r e ac t e d 3 = 1 2 . 37 5 * . 05 C H 4p r od u ce d = . 61 9 O 2 pr o du c ed = . 6 19 /2

47 48 49 50 51

//rxn (v ) C H 3 O H r e a c t e d 4 = 1 2 . 37 5 - C H 3 O H r ea c t e d 1 - C H 3 O H r e a c t e d2 CH3OHreacted3 D M E p r o du c e d = 1 . 98 / 2 H 2 O p r o du c e d 3 = 1 . 9 8/ 2 O 2 = 2 8 1. 27 - O 2 co n su m ed 1 - O 2 c o n s um e d 2 + O 2 p r o du c e d H 2O = 2 3 .7 3 + H 2 O p r o du c e d1 + H 2 O p ro d u c ed 2 + H 2 O p r o d u ce d 3 printf ( ” C om po si ti on o f e x i t g as s tr ea m : \ n \ n CH3OH = ” + string ( M u n r e a c t e d 1 ) + ” \ n HCHO = ” + string ( H C H O p r o d u c e d 1 ) + ” \ n C O 2 = ” + string ( C O 2 p r o d u c e d ) + ” \ n C O = ” + string ( C O p r o d u c e d ) + ” \ n H 2 = ” + string ( H 2 p r o d u c e d ) + ” \ n C H 4 = ” + string ( C H 4 p r o d u c e d ) + ” \ n (CH3)2O = ” + string ( D M E p r o d u c e d ) + ” \ n O2 = ” +

67

string ( O 2 ) + ” \ n N 2 = ” + string ( N 2 ) + ” \ n H 2 O = ” + string ( H 2 O ) + ” . ” )

Scilab code Exa 4.11 Pyrites fines roasting

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

clear ; clc ;


/ / E xa mp le 4 . 1 1 / / P ag e 1 32 printf ( ” E xa mp le 4 . 1 1 , P ag e 1 3 2 \ n \ n ” ) ; // s o l u t i o n m = 100 // kg p y r i t e s ( b a s i s )

//(a) printf ( ” ( a ) \ n \ n ” ) S 1 = 42 / / k g i 1 = 58 // kg i n er t s

// 8 m ol l S = 3 mol O2 i n Fe2O3 m 1 = 3 * 32 * 42 / 8* 3 2 / / kg O2 c o n v e r t e d t o Fe2O3 m 2 = i1 + m1 // mass o f SO3 f r e e c i n d e r // 2 . 3 kg S i s i n 10 0 kg c i n d e r

m 3 = 1 0 0 - ( 2. 3 *8 0 /3 2 ) m 4 = ( 10 0/ m 3 ) * m2 m 5 = m4 * .0 23 // kg S i n c i n d e r p 1 = 1 . 8* 1 00 / 42 printf ( ” p e rc e n t ag e o f c i n d e r r em ai ne d i n c i n d e r = ” + string ( p 1 ) + ” . \ n \ n \ n ( b ) \ n \ n” )

29 / / ( b )

68

30 31 32 33 34 35 36 37 38 39 40

m 6 = 100 / / kmol SO3 f r e e r o a s t e r g as ( b a s i s ) m 7 = 7. 12 / / k m ol O2 a s SO2 m 8 = 10 .6 //O2 m 9 = 100 - m8 - m 7 //N2 m 1 0 = ( 21 / 79 ) * m9 // O2 e n t e r i n g r o a s t e r a l o n g N2 m 11 = m 7 + m8 + ( 3 * 7. 1 2/ 8 ) / / a c c o un t e d O2 m 12 = m10 - m 11 / / u n a c c ou n t e d O2 m 1 3 = ( 8/ 15 ) * m 12 / / SO3 f o r m e d m 1 4 = m 13 + m7 / / S b u r n t p 2 = ( m 13 / m 14 ) * 10 0 printf ( ” p e r ce n t ag e o f S b ur nt t o fo rm SO3 = ” + string (p2)+ ” \n \n \n ( c ) \n \n”)

41 42 43 44 45 46 47 48

// ( c ) / / b a s i s 1 0 0 kg p y r i t e m 1 5 = 3 7 .8 1 /3 2 / / SO2 f o r m ed m 16 = ( m 9 + m1 0 ) * 1. 18 1 / m7 // a i r s u p p l ie d // 4 kg p y ri t e i s r oa st ed t o t al a i r s up pl ie d m 1 7 = m 16 * 4 / 10 0 // kmol / s

49 50 51 52

// (d)

53 54 55 56 57

// ( f )

v 1 = m 17 * 2 4 .9 57 printf ( ” v o l u m e t r i c f lo w r a t e o f a i r = ” + string ( v 1 ) + ” mˆ3/ s \ n \ n \ n ( d ) \ n \ n ” ) m 1 8 = ( 1 0 0. 4 5 5* m 1 7 ) / ( m 9 + m 10 ) // r o a s t e r g a s e s v 2 = m 18 * 6 6 .3 86 printf ( ” v o l u m e t r i c f lo w r a t e o f r o a s t e r g a s e s = ” + string ( v 2 ) + ” mˆ3/ s \ n \ n \ n ( f ) \ n \ n ” ) m 19 = 4 . 83 8* 1 0^ - 2 * .9 8 // SO3 a b so r be d i n a b s o r b e r

// SO3 + H2O = H2SO4 m 2 0 = ( m 1 9 * 9 8 * 2 4* 3 6 0 0) / ( . 98 * 1 0 00 ) // [ t /d ] printf ( ” Amount o f 9 8 p e r c e n t a c i d s t r e n g t h p ro du ce d = ” + string ( m 2 0 ) + ” t / d . ” )

Scilab code Exa 4.12 Burning of Pyrites and ZnS

69

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 32 33 34 35 36

clear ; clc ;


/ / E xa mp le 4 . 1 2 / / P ag e 1 36 printf ( ” E xa mp le 4 . 1 2 , P ag e 1 3 6 \ n \ n ” ) ; // s o l u t i o n // b a s i s 1 0 0 kg mixed c ha rg e = 75 kg p y r i t e + 2 5 kg ZnS // p y r i t e s F eS 2 m 1 = 7 5* .9 2 / / [ k g ] G 1 = 75 - m1 / / g a ng u e / / 4 F e S2 + 1 1 O2 = 2 F e2 O3 + 8 SO2 / / 4 F e S2 + 1 5 O2 = 2 F e2 O3 + 8 SO3 / / Zn o r e m 2 = 2 5* .6 8 / / ZnS I 1 = 25 - m2 // i n e r t s / / 2 ZnS + 3 O2 = 2 ZnO + 2 SO2 I 2 = I1 +6 // t o t a l i n e r t s / / new b a s i s : 1 00 kg c i n d e r m 3 = 3 .5 *. 7 / / S a s SO3 m4 = 3.5 - m3 // S a s FeS2 m 5 = 100 - m3 - m 4 // S f r e e c i n d e r m 6 = ( 8 1. 4 /9 7 .4 ) * 17 // ZnO // FeS2 r e a c t ed = x // ( FeS2 i n c i n d e r / S f r e e c i n d e r ) = (69 − x ) / ( 28 . 2 +. 6 6 7 x ) = 1 . 9 6 9 / 9 1 . 9 06 / / s o l v i n g t h i s we g e t x = 67 .4 3 / / k g m7 = m6 + .667* x + 14 // S f r e e c i n d e r m 8 = 69 -x // FeS2 i n c i n d e r 70

m 9 = 6 .1 2 5* m 7 / m5 // SO3 m 1 0 = . 66 7* x // Fe2O3 m 11 = m 6 + m1 0 + m8 + m 9 + I2 printf ( ” ( a ) \ n \ n T ot al c i n d e r p ro du ce d = ” + string ( m11 ) + ” kg \ n C o mp os it io n o f c i n d e r : \ n Z n O = ” + string ( m 6 ) + ” kg \ n Fe2O3 = ” + string ( m 1 0 ) + ” kg \ n S a s FeS2 = ” + string ( m 8 ) + ” kg \ n S a s SO3 = ” + string ( m 9 ) + ” kg \ n i n e r t s = ” + string ( I 2 ) + ” kg \ n \ n \ n ( b ) \ n \ n” ) 41 S 1 = ( 64 / 12 0 ) * 69 + ( 3 2/ 9 7. 4 ) * 17 // [ kg ] S c ha rg ed t o 37 38 39 40

burner 42 S 2 = . 0 35 * 79 . 63 // S i n c i n d e r 43 p = S2 * 10 0/ S 1 44 printf ( ” p e rc e n t ag e o f S l e f t i n c i n d e r = ” + string ( p ) +” ”)

Scilab code Exa 4.13 Raising pH with NaOH

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

clear ; clc ;


/ / E xa mp le 4 . 1 3 / / P ag e 1 38 printf ( ” E xa mp le 4 . 1 3 , P ag e 1 3 8 \ n \ n ” ) ; // s o l u t i o n m 1 = 1 20 0* 1. 2 // [ kg ] mass o f r e a c t a n t s p O H1 = 1 4 -6 //pOH o f r e a c t a n t s pOH = 14 -9 //pOH o f f i n a l mass

71

18 19 20 21 22 23 24 25 26 27 28

// ROWs = 1/ si gm a (Wi/ROWsi) / /Ms = m as s o f . 5 % NaOH r e q u i r e d / /ROWs = d e n s i t y o f f i n a l s o l u t i o n //ROWs = 1 / { ((m1 ∗ 1 0 ˆ 3 ∗ 1 ) / ( ( ( m1 ∗ 10ˆ3+Ms) ∗ 1 . 2 ) + ( M s / ( ( m1∗ 10ˆ3+Ms) ∗ 1 . 0 0 5 ) ) } (i) / / b a l a n c e o f OH− i o n s / / 1 2 0 0 ∗ 10ˆ − 8 +Ms ∗ 1 0 ˆ − 1 . 1 5 / ( 1 . 0 0 5 ∗ 1 0 ˆ − 5 ) = ( 1 2 0 0 ∗ 1 . 2 ∗ 10ˆ3+Ms) ∗ 10ˆ − 5/ROWs∗ 10ˆ − 5 ( ii ) / / s o l v i n g ( i ) and ( i i ) M s = 1 70 .2 1 // g R O Ws = 1 .2 01 6 // [ kg / l ] printf ( ” Mass o f 0 . 5 p e r c n t NaOH r e q u i r e d t o be a dd ed t o r a i s e t he pH = ” + string ( M s ) + ” g . ” )

Scilab code Exa 4.14 Solving eg 10 with Linear Model Method

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

clear ; clc ;


/ / E xa mp le 4 . 1 4 / / P ag e 1 40 printf ( ” E xa mp le 4 . 1 4 , P ag e 1 4 0 \ n \ n ” ) ; // s o l u t i o n / / u s in g e q u a t i o n s o f e xa m p l e 4 . 1 0 / / s o v i ng 4 . 1 0 by l i n e a r model method M = [ 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0;0 1 0 0 0 0 0 0 0

72

0 1 1 1 1 2;0 0 1 0 0 0 0 0 0 0 .5 1.5 0 -.5 0;0 0 0 1 0 0 0 0 0 0 -1 -2 0 0 -1;0 0 0 0 1 0 0 0 0 0 -1 0 0 0 0;0 0 0 0 0 1 0 0 0 0 0 -1 0 0 0;0 0 0 0 0 0 1 0 0 0 0 0 -1 0 0;0 0 0 0 0 0 0 1 0 0 0 0 -2 0 0;0 0 0 0 0 0 0 0 0 0 0 0 0 -1 0;0 0 0 0 0 0 0 0 0 1 0 0 0 0 -1;0 0 0 0 0 0 0 0 0 0 1 1 1 1 2;0 0 0 0 0 0 0 0 0 0 .5 1.5 0 -.5 0;0 0 0 0 0 0 0 0 0 0 -1 -2 0 0 -1;0 0 0 0 0 0 0 0 0 0 0 0 0 1 0;0 0 0 0 0 0 0 0 0 0 0 0 0 0 2] 19 V = [ 1 0 5 8 . 1 ; 1 2 5 ; 2 8 1 . 2 7 ; 2 3 . 7 3 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; . 9 9 * 1 2 5 ; . 2 4 3 7 *2 8 1 . 2 7 ; - 5 . 4 7 5 6 20 X = M \ V 21 disp ( X )

Scilab code Exa 4.15 Electrochemical cell

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

clear ; clc ;


/ / E xa mp le 4 . 1 5 / / P ag e 1 43 printf ( ” E xa mp le 4 . 1 5 , P ag e 1 4 3 \ n \ n ” ) ; // s o l u t i o n / / b a s i s = 1 . 1 2 M63 O2 a t NTP m 1 = 1 . 12 * 10 0 0* 3 2/ 2 2. 4 // [ g ] O2 m 2 = m1 /8 // g e q O2 / / a t c a t h o d e : Cu++ +2 e = Cu 73

19 20 21 22

/ / a t a no de : SO4−− − 2 e = S O 4 e q w tC u = 6 3. 5/ 2 d e p o s i te d C u = e q wt C u * m 2 E = ( 1 13 0 *1 8 00 0 ) / 9 64 85 // f a r a d ay s

T ot al e ne rg y

passed to c e l l 23 l i b C u = ( 1 1 3 0* 1 8 00 0 * e q w t C u ) / 9 64 8 5 // [ g ]

theoritica l

l i b e r a t i o n o f Cu 24 e f f = ( d e p o s i te d C u / l i b Cu ) * 1 0 0 // c u r r e n t e f f i c i e n c y 25 printf ( ” ( a ) \ n \ n Amount o f Cu l b e r a t e d = ” + string ( l i b C u ) + ” \ n \ n \ n ( b ) \ n \ n C ur r e n t e f f i c i e n c y o f t h e c e l l = ” + string ( e f f ) + ” p e r c e n t . ” )

Scilab code Exa 4.16 Hooker type Diaphragm Cell

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

clear ; clc ;


/ / E xa mp le 4 . 1 6 / / P ag e 1 44 printf ( ” E xa mp le 4 . 1 6 , P ag e 1 4 4 \ n \ n ” ) ; // s o l u t i o n / / b a s i s : 1 day o p er a ti o n / / N aC l = N a+ + C l − //H2O = H+ + OH− //Na+ + OH− = NaOH / /H+ + e = ( 1 / 2 ) H 2 / / C l − − e = ( 1 / 2 ) C l2 E = ( 1 5 0 00 * 3 6 00 * 2 4 ) / 9 6 48 5 // f ar a d ay / day

74

T o ta l

energy passed thrrough c e l l 22 N a OH = ( 1 5 0 00 * 3 6 0 0* 2 4 * 4 0) / ( 9 6 48 5 * 1 00 0 ) // [ kg / day ]

t h e o r e t i c a l NaOH 23 24 25 26 27 28 29 30 31 32 33 34

e ff = ( 51 4 .1 / N a OH ) * 1 00 // c u r r e n t C l2 = ( 3 5. 5 /4 0 ) * 51 4 .1 H 2 = ( 4 56 . 3* 2 ) / ( 35 . 5* 2 )

efficiency

/ / 4 0 g NaOH = 5 8 . 5 g N aC l c o n s N aC l = ( 5 8 .5 / 4 0) * 5 1 4. 1 / / N aC l c o ns u m ed tota l cel l liqu or T l iq u or = 5 1 4. 1 /. 1 1 // [ kg / day ] r e mN a Cl = 5 1 4. 1 *1 . 4 t o t a l N aC l = c o n sN a C l + r e m Na C l F b r i ne = t o t al N a Cl / . 26 6 // f e ed r a t e o f b r i ne c o n s H 2O = ( 1 8 /4 0 ) * 5 1 4 .1 l o s s H 2 O = F b r in e - T l i q u or - c o n s H 2 O printf ( ” ( a ) \ n \ n C u r r e n t e f f i c i e n c y o f t he c e l l = ” + string ( e f f ) + ” p e r c e n t . \ n \ n \ n ( b ) \ n \ n C l 2 p r o d u c e d = ” + string ( C l 2 ) + ” kg / d ay \ n H2 p r o d u ce d = ” + string ( H 2 ) + ” kg / d ay \ n \ n \ n ( c ) \ n \ n l o s s o f w a t e r = ” + string ( l o s s H 2 O ) + ” k g / da y ” )

Scilab code Exa 4.17 Naptha Reforming to Ammonia

1 2 3 4 5 6 7 8 9 10 11 12 13

clear ; clc ;


/ / E xa mp le 4 . 1 7 / / P ag e 1 46 printf ( ” E xa mp le 4 . 1 7 , P ag e 1 4 6 \ n \ n ” ) ; // s o l u t i o n 75

14 15 //M = mix f e ed r at e , F = f r e s h 16 17 18 19 20 21 22 23 24 25 26 27 28

f e ed r a t e , R =

r e c y c l e s tr ea m / / u si ng f i g 4 .3 / / N2 b a l a n c e / / a = 2 4 . 7 5M/ ( . 2 5M+7 .5M) (i) // P = ( 4 . 1 5M + 1 7. 75 a ) /M ( ii ) / / . 5 8 5M − 1 . 77 5 a + ( 4 . 1 5M+ 1 7 . 75 a ) /M = 1 0 0 ( i i i ) // so lvi ng ( i ,) ( i i ) , ( i i i ) M = 4 38 .5 89 // [ kmol/ s ] a = ( 2 4 .7 5 * M ) / ( ( . 25 * M ) + 7 . 5) / / k m o l / s P = ( 4 . 1 5 *4 3 8 . 5 8 9+ 1 7 . 7 5* 9 2 . 6 6 2) / M / / k m o l / s R = M -100 / / km ol / s r = R / 1 0 0 // r e c y c l e r a t i o N H 3 = ( . 58 5 * M - 2 . 2 75 * a ) * 1 7 . 0 30 5 / / k g / s printf ( ” ( a ) \ n \ n r e c y c l e f ee d r a t e = ” + string ( R ) + ” k m o l / s \ n \ n \ n ( b ) \ n \ n p u r g e g as r a t e = ” + string ( P ) + ” kmo l / s \ n \ n \ n ( c ) \ n \ n m ass r a t e o f NH3 = ” + string ( N H 3 ) + ” k g / s ” )

Scilab code Exa 4.18 Additional membrane separator in eg 17

1 2 3 4 5 6 7 8 9 10 11 12 13

clear ; clc ;



14 15 / / g i ve n 16 17 / / ( . 1 ∗M∗ R1 ) / ( . 4 1 5M+ 1 .7 7 5 a ) + ( . 1 1 2 5 a ∗ P ) / ( . 4 1 5 M +

1 . 77 5 a ) + 1 = . 1M 18 / / R 1 ∗ ( . 3 1 5 M− 1 .2 25 a ) / ( . 4 1 5M + 1 . 7 7 5 a ) = . 9M−4a 19 / / M = 1 00 + R1 + ( 2 . 2 5 a ∗ p ) / ( . 4 1 5M + 1 . 7 7 5 a ) 20 / / . 1M∗ P / ( . 4 1 5M + 1 . 7 7 5 a ) − ( . 1 1 2 5 a ∗ P ) / ( . 4 1 5 M1 . 7 7 5 a ) 21 22 23 24 25 26 27

/ / s o l v i n g them M = 4 57 .0 11 / / km ol / s R 1 = 3 50 .7 71 / / k mo l / s P = 1 0. 36 8 / / kmo l / s a = 9 6. 60 8 / / kmo l / s

R 2 = 2 . 2 5 * 9 6. 6 0 8 * 1 0. 3 6 9 / ( .4 1 5 * 4 5 7. 0 1 1 + 1 . 7 7 5 * 9 6 . 6 0 8 ) / / kmo l / s 28 F = M - R1 - R 2 29 printf ( ” Mixed f e e d r a t e = ” + string ( M ) + ” km ol / s \ n R ec y cl e s tr ea m = ” + string ( R 1 ) + ” km ol / s \ n R ec o ve r ed H2 s t re a m = ” + string ( R 2 ) + ” km ol / s \ n Fr e s h f e ed r a t e = ” + string ( F ) + ” km ol / s \ n R e c y c l e r a t i o = ” + string ( ( R 1 + R 2 ) / F ) + ” kmol / kmol o f f r e s h fe ed . ” )

Scilab code Exa 4.19 Partial Demineralisation Plant

1 2 3 4 5 6 7 8 9

clear ; clc ;


/ / E xa mp le 4 . 1 9 77

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

/ / P ag e 1 53 printf ( ” E xa mp le 4 . 1 9 , P ag e 1 5 3 \ n \ n ” ) ; // s o l u t i o n m 1 = ( 5 0/ 3 5. 5 ) * 31 2 / / [ mg/ l ]

Cl 2 e x p r e s s e d a s

e q u i v a l e n t CaCO3 m 2 = ( 5 0/ 4 8) * 4 3 .2 / / [ mg/ l ] S u l ph a t es a s e q u i v a l e n t CaCO3 A = m1 +m2 // [ mg/ l a s CaCO3 ] EMA i n ra w w a te r M 1 = 550 // a l k a l i n i t y o f raw w a te r M 2 = 50 // a l k a l i n i t y o f b l e n d w a t e r / / l e t 1 00 l o f raw w at er e n t e r s b o t h i o n e xc h a n g e r s // b al an ci ng n e u t r i l a s i o n x = 1 00 *( M 1 - M2 ) / ( A+ M 1 ) // raw w a te r i n l e t t o H2 i o n echanger printf ( ” ” + string ( x ) + ” p er ce nt o f t o t a l raw w a t e r i s p a s se d t hr ou g h t h e H i o n e x ch a n ge r . ” )

Scilab code Exa 4.20 Capacity increment by Second Reactor

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

clear ; clc ;



78

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 40 41 42 43 44 45 46 47 48 49 50 51

m1 m2 m3 m4 m5 m6 m7

= = = = = = =

1 48 8. 1 // kmol /h g as mix t o r e a c t o r 1 ( b a s i s ) m1 * .0 62 5 / / CH3OH m1 - m2 // a mb i e n t a i r f l o w m 3 / 1. 0 17 7 2 // d r y a i r f l ow r a t e m3 - m4 // m o i s t ur e m2 * .9 9 / / CH3OH c o n v e r s i o n i n R1 m2 - m6 / / u n r e a c t e d CH3OH

// r xn i m 8 = m7 *.9 // CH3OH re ac te d = HCHO prod uced = H2O

produced m 9 = m8 /2 / / O2 c o ns u m ed m 10 = m6 - m 8 / / CH3OH r e a c t e d i n r xn i i t o v / / rx n i i m 11 = m 10 * .7 1 / / CH3OH r e a c t e d = CO2 p r o d u c e d m 12 = m 11 * 1. 5 / / O2 c o ns u m ed m 13 = 2* m 11 / / H2O p r o d u c e d / / r xn i i i m 14 = m 10 * .0 8 / / CH3OH r e a c t e d = CO p r o d u c e d m 15 = 2* m 14 / / H2 p r o d uc e d / / rxn i v m 16 = m 10 * .0 5 / / Ch3OH r e a c t e d = CH4 p r o d u c e d m 17 = m 16 / 2 / / O2 p r o d uc e d / / r xn v m 18 = m 10 - m 16 - m 14 - m 11 // CH3OH re a ct ed m 19 = m 18 / 2 / / ( CH3 ) 2O = H2O p r o d u c e d m 20 = 2 87 .8 7 - m 9 - m 12 + m 17 // O2 i n R1 e x i t s tr ea m m 21 = m 5 + m8 + m 13 + m 19 / / H2O i n R1 m = m 7 + m 8 + m1 1 + m 1 4 + m 15 + m 1 6 + m 1 9 + m 20 + 1 0 8 2 . 93 + m 2 1

// R2 / / x k mo l / h CH3OH i s a dd ed b /w r e a c t o r s / / ( m7+x ) / (m+x ) = . 0 8 4 s o l v i n g i t x = 1 40 .5 48 // [ kmol /h ] m22 = x + m7 / / CH3OH e n t e r i n g R2 m 23 = m 22 * .9 9 //CH3OH r e a c t e d m 24 = m22 - m 23 // CH3OH un re ac te d // r xn i m 2 5 = m 23 * .9 // CH3OH re ac te d = HCHO pro duce d = H2O 79

produced 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76

m 26 = m 25 / 2 / / O2 c o ns u m ed m 27 = m23 - m25 / / CH3OH r e a c t e d i n r xn

/ / rx n i i m 28 = m 27 * .7 1 / / CH3OH r e a c t e d = CO2 p r o d u c e d m 29 = m 28 * 1. 5 / / O2 c o ns u m ed m 30 = m 28 * 2 / / H2O p r o d u c e d

/ / r xn i i i m 31 = m 27 * .0 8 / / CH3OH r e a c t e d = CO p r o d u c e d m 32 = m 31 * 2 / / H2 p r o d uc e d

/ / rxn i v m 33 = m 27 * .0 5 / / Ch3OH r e a c t e d = CH4 p r o d u c e d m 34 = m 33 / 2 / / O2 p r o d uc e d

/ / r xn v m 35 = m 27 - m 28 - m 31 - m 33 // CH3OH re a ct ed m 36 = m 35 / 2 / / ( CH3 ) 2O = H2O p r o d u c e d m37 m38 m39 m 40

= = = =

m 20 - m26 - m 29 + m3 4 // O2 i n R2 e x i t s tr ea m m 21 + m 25 + m 36 / / H2O i n R2 9 2. 07 + m 25 // HCHO i n R2 m 24 + m 3 9 + m 2 8 + m 31 + m 3 2 + m 3 3 + m 36 + m 3 7 + m 3 8 + 1 0 82 . 9 3

m 4 1 = m 39 * 30 / / k g / h HCHO p r o d uc e d m 42 = m 41 / .3 7 // bottom s o l f l o e r a t e c = ( m 42 - 9 0 30 . 4 ) * 1 0 0 / 90 3 0 . 4 // i n c r e a s e i n c a pa c it y printf ( ” I n c r e a s e i n c a p ac i t y = ” + string ( c ) + ” p e rc e n t .”)

Scilab code Exa 4.21 Blast Furnace Calculations

1 2 3 4 5

i i to v

clear ; clc ;


6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29

/ / M a t er i a l B a la n ce s i n v o l v i n g C hem ic al R ea ct i on

/ / E xa mp le 4 . 2 1 / / P ag e 1 59 printf ( ” E xa mp le 4 . 2 1 , P ag e 1 5 9 \ n \ n ” ) ; // s o l u t i o n / / b a s i s 1 t on n e o f p ig i r o n c ok e = 1 00 0 / / k g f l ux = 4 00 / / k g F e 1 = 1 0 00 * .9 5 // Fe i n p ig i r o n F e 2 = ( 1 12 / 16 0 ) * .8 // F e a v a i l a b l e p e r kg o f o re o re = F e1 / Fe 2 / / kg S i = . 0 14 * 10 0 0 // kg // S i i n p ig i r o n s i 1 = ( 60 / 28 ) * 14 // s i l i c a p r e s e nt i n p i g i r o n s i2 = o re * .1 2 / / s i l i c a i n o r e s i3 = . 1* c ok e / / s i l i c a i n c o ke s i 4 = s i2 + s i3 - s i1 / / s i l i c a i n s l a g a l um i na = o re * . 08 / / Al2O3 i n o r e = Al2O3 i n s l a g

C aO = f lu x * ( 56 / 10 0 ) s la g = s i4 + a l u mi na + C a O printf ( ” ( a ) \ n \ n Mass o f s l a g made = ” + string ( s l a g ) + ” k g . \ n \ n \ n ( b ) \ n \ n M ass o f o re r e q u i r e d = ” + string ( o r e ) + ” k g . \ n \ n \ n ( c ) \ n \ n C o m po s i ti o n o f s l a g : \ n S i O 2 = ” + string ( s i 4 ) + ” kg \ n A l2 O3 = ” + string ( a l u m i n a ) + ” kg \ n C a O = ” + string ( C a O ) + ” kg . \ n \ n \ n ( d ) \ n \ n ” ) 30 C = . 9* c o k e + ( 1 2 / 10 0 ) * f lu x - 3 6 // t o t a l C a v a i l a b l e 31 / / CO : CO2 = 2 : 1 32 C1 = C /3 / / C c o n v e rt e d t o CO2 33 C 2 = 2* C /3 / / C c o n v e rt e d t o CO 34 O 2 1 = C 1 * ( 3 2/ 1 2 ) + C 2 * ( 1 6/ 1 2 ) / / O2 r e q u i r e d f o r CO

a nd CO2 f o r m a t i o n 35 O 2 2 = ( 32 / 28 ) * Si / / O2 f ro m S iO 2 36 O 23 = o re * ( . 8 *4 8 /1 6 0) / / O2 f r o m Fe2 O3 37 O 24 = f lu x * ( 32 / 10 0 ) // O2 from CaCO3 81

38 39 40 41 42

/ / kg k g O2 O2 t o b e s u p p l i e d O 25 2 5 = O 21 21 - O 22 2 2 - O 23 2 3 - O 24 2 4 // O 2 6 = O 25 25 / 32 3 2 //kmol a ir ir = O 26 26 / .2 . 2 1 //kmol V = a ir ir * 2 2. 2 . 41 4 1 4 //mˆ3 printf ( ” V o l u m e o f a i r t o b e s u p p l i e d = ” + string ( V ) + ” mˆ3. ”)

82

Chapter 5 Energy Balances

Scilab code Exa 5.1 Pumping of water

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

clear ; clc ;

/ / S t o i ch c h i o m et et r y // Ch haa pt p t er er 5 / / E ne n e rrgg y B a l a n c e s

/ / E xa xa mp mp le le 5 . 1 / / P aagg e 1 86 86 xa mp mp le le 5 . 1 , P ag ag e 1 8 6 \ n \ n ” ) ; printf ( ” E xa // s o l u t i o n / / b a s i s p u m p i n g o f 1 l / s o f w at a t er er H ad a d d = 5 2 // kW kW kW H lo l o st s t = 2 1 // kW f i = H a dd dd - H lo l o st st // kW kW p 1 = 1 01 0 1 32 3 2 5 / / Pa p 2 = p1 Z 1 = -50

// m 83

22 23 24 25 26 27 28 29 30

Z 2 = 10 // m g = 9 .8 . 8 06 0 6 65 6 5 / / m/ s s q m/ ( N . s s q ) g c = 1 / / k g . m/ row = 1 // kg/ l kW W = 1 .5 . 5 *. * . 55 5 5 // kW

/ / e n er e r g y b a l a n c e b /w /w A an and B / / d E = E 2−E1 = W + Q + ( Z 1−Z2 ) ∗ ( g / g c ) ∗qm d E = 3 1. 1 . 23 2 3 7 / / kW n e rg r g y b e t w e e n t he he printf ( ” I n c r e a s e i n i n t e r n a l e ne s t o r a g e t a nk n k a n d t he h e b o t t o m o f t he he w e l l = ” + string ( d E ) + ” kW. ” )

Scilab code Exa 5.2 Heating of CH4

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

clear ; clc ;


/ / E xa xa mp mp le le 5 . 2 / / P aagg e 1 97 97 xa mp mp le le 5 . 2 , P ag ag e 1 9 7 \ n \ n ” ) ; printf ( ” E xa // s o l u t i o n / / u si s i ng ng t a b l e 5 .1 / / b a s i s 1 kmol o f metha ne T 1 = 3 03 0 3 .1 .1 5 / / K T 2 = 5 23 2 3 .1 .1 5 / / K / / u si ng eq 5 .1 7 H = 1 9 .2 . 2 4 94 9 4 * ( T2 T2 - T 1 ) + 5 2 .1 . 1 1 35 3 5 * 10 1 0 ^ - 3 * ( T2 T 2 ^ 2 - T1 T 1 ^ 2) 2) / 2 + 1 1 . 97 9 7 3 * 10 10 ^ - 6 * ( T 2 ^ 3 - T 1 ^ 3 ) /3 / 3 - 1 1 . 31 3 1 7 3 *( *( T 2 ^ 4 - T 1 ^ 4 )

84

* 1 0 ^ - 9 / 4 // kJ 21 printf ( ” Hea t a dded = ” + string ( H ) + ” k J / k mo l m et ha n e . ”)

Scilab code Exa 5.3 Calculation of heat added

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

clear ; clc ;

/ / S t o i ch i o m et r y / / C ha pt er 5 / / E ne rg y B a l a n c e s

/ / E xa mp le 5 . 3 / / P ag e 1 98 printf ( ” E xa mp le 5 . 3 , P ag e 1 9 8 \ n \ n ” ) ; // s o l u t i o n / / b a s i s 1 kmol methane a t 25 b ar P c = 4 6. 04 // b ar T c = 1 90 .5 // K P r = 25 / Pc

// H−Ho = i n t g r ( f ro m3 03 . 1 5 t o 5 2 3 . 1 5 ) {CmpR dT } // s o l v i n g i t by s im ps o n ’ s r u l e H E = 2 55 .2 / / kJ / k m ol H = 9 17 5. 1+ H E printf ( ” Hea t a dded = ” + string ( H ) + ” kJ / kmol o f metha ne . ” )

Scilab code Exa 5.4 Heating of Toulene

85

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

clear ; clc ;


/ / E xa mp le 5 . 4 / / P ag e 2 06 printf ( ” E xa mp le 5 . 4 , P ag e 2 0 6 \ n \ n ” ) ; // s o l u t i o n / / u si ng t a b l e 5 . 3 / / . 2 5 kg / s t o u le n e h ea te d fro m 2 9 0. 1 5K t o 3 5 0. 1 5K q m = . 25 /9 2 / / kmo l / s // r e f er e n ce 7

f i = 2 . 7 1 7* 1 0 ^ - 3 * [ 1 . 80 8 3 * (3 5 0 .1 5 - 2 9 0 .1 5 ) + 8 1 2 .2 2 3 * 10 ^ - 3 * ( 35 0 . 1 5^ 2 - 2 9 0 .1 5 ^ 2 ) / 2 1 5 1 2. 6 7 * 10 ^ - 6 * ( 35 0 . 1 5^ 3 - 2 9 0 .1 5 ^ 3 ) / 3 + 1630.01*10^-9*(350.15^4-290.15^4)/4] 20 printf ( ” Heat r e q u i r e d t o be added t o t o u l e n e = ” + string ( f i ) + ” kW. ” )

Scilab code Exa 5.5 Aq caustic soda heating

1 2 3 4 5 6 7 8

clear ; clc ;


86

9 10 11 12 13 14 15 16 17 18 19 20 21

/ / E xa mp le 5 . 5 / / P ag e 2 06 printf ( ” E xa mp le 5 . 5 , P ag e 2 0 6 \ n \ n ” ) ; // s o l u t i o n / / b a s i s 1 kg o f 2 0% NaOH s o l // r e f e r r i n g t o f i g 5 . 4 a t 2 8 0 . 1 5K C11 = 3 .5 6 // kJ /kg . K C12 = 3 .7 1 // kJ /kg . K a t 3 6 0 . 1 5K C 1 m = ( C 11 + C 12 ) / 2 H = 1 * C1 m * ( 36 0 .1 5 - 2 8 0. 1 5) // kJ printf ( ” Heat r e q u i r e d t o be added = ” + string ( H ) + ” kJ .”)

Scilab code Exa 5.6 Heating Chlorinated diphenyl

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

clear ; clc ;


/ / E xa mp le 5 . 6 / / P ag e 2 07 printf ( ” E xa mp le 5 . 6 , P ag e 2 0 7 \ n \ n ” ) ; // s o l u t i o n / / b a s i s 1 kg D ip hy l A−30

Q = . 7 51 1 *( 5 53 . 15 - 3 13 . 15 ) + / / kJ / kg 1.465*10^-3*(553.15^2-313.15^2)/2 17 fi = Q *4 00 0 // kJ / h f o r mass f l o w r a t e 40 0 0 kg /h

87

18 19 20 21

C l m = ( 1 . 1 80 7 + 1 . 51 9 8 ) / 2 f i 1 = C lm * ( 5 5 3 . 15 - 3 1 3. 1 5 ) * 4 0 0 0 /3 6 0 0 / / kJ / h e r r = ( fi 1 - Q ) * 10 0/ Q printf ( ” Heat t o be s u p p l i e d = ” + string ( f i 1 ) + ” kW \ n P er ce nt e r r o r = ” + string ( e r r ) + ” . ” )

Scilab code Exa 5.7 Roasting of pyrites fine

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

clear ; clc ;


/ / E xa mp le 5 . 7 / / P ag e 2 08 printf ( ” E xa mp le 5 . 7 , P ag e 2 0 8 \ n \ n ” ) ; // s o l u t i o n T 1 = 2 98 .1 5 // K T 2 = 7 75 .1 5 //K

/ / u s i n g eq 5 . 1 7

Q = 2 8 . 83 9 * ( T2 - T 1 ) + 2 . 0 3 9 5* 1 0 ^ - 3 * ( T 2 ^ 2 - T1 ^ 2 ) / 2 + 6 . 9 90 7 * 10 ^ - 6 * ( T 2 ^ 3 - T 1 ^ 3 ) /3 - 3 . 2 30 4 * 10 ^ - 9 * ( T 2 ^ 4 T 1 ^ 4 ) / 4 / / k J / k mo l 19 printf ( ” Heat c o n t e n t o f 1 kmol o f g as m ix tu re a t 2 9 8 K = ” + string ( Q ) + ” kJ/ kmol . ” )

Scilab code Exa 5.8 Anniline and water mix subcooled

88

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

clear ; clc ;


/ / E xa mp le 5 . 8 / / P ag e 2 10 printf ( ” E xa mp le 5 . 8 , P ag e 2 1 0 \ n \ n ” ) ; // s o l u t i o n / / b a s i s 8 0 0 0 kg /h m i xt ur e i s t o be c o o l e d / / k g /h q n1 m = . 1 18 * 80 0 0 q n1 = q n1 m / 9 3. 12 4 2 / / km ol / h / / kg / h q n2 m = 8 00 0 - q n 1m q n2 = q n2 m /1 8 / / k mo l / h //K T 1 = 3 73 .1 5 T 2 = 3 13 .1 5 //K f i = q n 1 * [ 2 06 . 2 7 *( T 1 - T 2 ) - 2 1 1. 5 06 5 *1 0 ^ - 3 * ( T 1 ^2 - T 2 ^ 2 ) / 2 + 5 64 . 2 90 2 * 10 ^ - 6 * ( T 1 ^3 - T 2 ^ 3 ) / 3 ] + q n2 * [ 5 0. 8 4 5* ( T1-T2)+213.08*10^-3*(T1^2-T2^2)/2-631.398*10^-6*( T 1 ^ 3 - T 2 ^ 3 ) / 3 + 6 4 8 . 7 4 6 * 1 0 ^ - 9 * ( T 1 ^ 4 - T 2 ^ 4 ) / 4 ] / / kJ /

h 23 printf ( ” H eat r em ov a l r a t e o f s u b c o o l i n g z o ne o f t he c o n d e n s e r = ” + string ( f i ) + ” k J / h . ” )

Scilab code Exa 5.9 Vapor Pressure calculations


89

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

/ / C ha pt er 5 / / E ne rg y B a l a n c e s

/ / E xa mp le 5 . 9 / / P ag e 2 20 printf ( ” E xa mp le 5 . 9 , P ag e 2 2 0 \ n \ n ” ) ; // s o l u t i o n // ( a ) T = 3 05 .1 5 //K P v 1 = 1 0 ^ ( 4. 0 0 2 6 - ( 1 1 7 1. 5 3 0 /( 3 0 5 .1 5 - 4 8 . 78 4 ) ) )

// (b) T = 3 95 .1 5 // b ar P v 2 = 1 0 ^ ( 3. 5 5 9 - ( 6 4 3 . 74 8 / (3 9 5 . 15 - 1 9 8. 0 4 3 ) ) ) printf ( ” ( a ) \ n \ n V . P. o f n−h ex an e a t 3 0 5 . 1 5K = ” + string ( P v 1 ) + ” b a r . \ n \ n \ n ( b ) \ n \ n V . P . o f w at er a t 3 9 5 .1 5K = ” + string ( P v 2 ) + ” b a r . ” )

Scilab code Exa 5.10 Calculations on O zylene

1 2 3 4 5 6 7 8 9 10 11 12 13

// b ar

clear ; clc ;



14 15 16 17 18 19 20 21 22 23 24

// ( a )

P c = 37 32 / / kPa T c = 6 30 .3 // K T b = 4 17 .6 //K T Br = Tb / Tc l a m b d a v = 8 . 3 1 4 47 2 * 4 1 7. 6 * ( 1 . 09 2 * (log (3732) -5.6182) /(.930-.6625))

// (b) T 1 = 2 98 .1 5 //K l a m bd a v 1 = 3 6 2 4 0* [ ( 6 30 . 3 - 2 9 8 . 1 5 ) / ( 6 30 . 3 - 4 1 7 .6 ) ] ^ . 3 8 printf ( ” ( a ) \ n \ n L at e n t h ea t o f v a p o r i z a t i o n a t Tb u si ng R ie de l eq i s ” + string ( l a m b d a v ) + ” k J / k mo l .

\ n \ n \ n ( b ) \ n \ n L at en t h ea t o f v a p o r i z a a t i o n a t 2 9 8. 1 5 K u s in g Watson eq i s ” + string ( l a m b d a v 1 ) +” kJ/kmol . ”)

Scilab code Exa 5.11 latent heat of vaporization of ethanol

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

clear ; clc ;


/ / E xa mp le 5 . 1 1 / / P ag e 2 25 printf ( ” E xa mp le 5 . 1 1 , P ag e 2 2 5 \ n \ n ” ) ; // s o l u t i o n // ( a ) P c = 6 1. 37 // b ar

91

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

T c = 514 //K T b = 3 51 .4 P = 1 / / atm T Br = Tb / Tc

/ / R i ed e l eq l a m b d a v1 = 8 . 3 14 4 7 2* T b * 1 . 0 9 2 *(log (6137) -5.6182) /(.930-TBr)

/ / NIST e q l a mb da v 2 = 5 04 30 *exp ( - ( - . 4 4 7 5 * T B r ) ) * ( 1 - T B r ) ^ . 4 9 8 9

// (b) T 1 = 2 98 .1 5 T Br 1 = T 1/ Tc

/ / W at so n e q l a m bd a v 21 = 3 8 5 6 3* [ ( 51 4 - 2 9 8. 1 5 ) / ( 5 14 - 3 51 . 4 ) ] ^ .3 8

/ / NIST e q l a mb d av 2 2 = 5 04 30 *exp ( - ( - . 4 4 7 5 * T B r 1 ) ) * ( 1 - T B r 1 ) ^ . 4 9 6 9 printf ( ” ( a ) \ n \ n L at e n t h ea t o f v a p o r i z a t i o n a t Tb u s i n g \ n R i e d e l eq i s ” + string ( l a m b d a v 1 ) + ” k J / kmol \ n NIST eq i s ” + string ( l a m b d a v 2 ) + ” k J / km ol \

n \ n \ n ( b ) \ n \ n L at e nt h ea t o f v a p o r i za t i o n a t 2 9 8 . 1 5 K u s i n g \ n Watson eq i s ” + string ( l a m b d a v 2 1 ) + ” k J / k mo l \ n NIST eq i s ” + string ( l a m b d a v 2 2 ) + ” kJ/kmol”)

Scilab code Exa 5.12 Saturation P of steam

1 2 3 4 5 6 7 8 9

clear ; clc ;


/ / E xa mp le 5 . 1 2 92

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

/ / P ag e 2 27 printf ( ” E xa mp le 5 . 1 2 , P ag e 2 2 7 \ n \ n ” ) ; // s o l u t i o n / / u s i n g A pp en di x IV . 2

P s1 = 75 P s2 = 80 T 1 = 5 63 .6 5 T 2 = 5 68 .1 2 T = 5 65 .1 5 P s = 75* exp ( ( T 2 * ( T - T 1 ) * log ( 8 0 / 7 5 ) / ( T * ( T 2 - T 1 ) ) ) ) printf ( ” S a t ur a t i o n P r es s u r e o f stea m a t 5 6 5. 1 5K i s ” + string ( P s ) + ” b a r . ” )

Scilab code Exa 5.13 Bubble and Dew pt calculations

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

clear ; clc ;


/ / E xa mp le 5 . 1 3 / / P ag e 2 36 printf ( ” E xa mp le 5 . 1 3 , P ag e 2 3 6 \ n \ n ” ) ; // s o l u t i o n // b a s i s

1 kmol e qu im o l ar m ix n pe nt = .5 / / k mo l n hex = .5 / / k mo l P = 1 01 .3 25 / / kPa 93

19 20 21 22 23 24 25 26 27

x 1 = .5 x 2 = x1 T s 1 = 3 09 .2 // K T s 2 = 3 41 .9 // K T 1 = ( T s1 + T s2 ) / 2

/ / u s in g t h e s e data , we g e t t a b l e 5 . 1 0 and 5 . 11 T b b = 3 21 .6 //K T d p = 3 29 .9 //K printf ( ” B ub bl e p o i n t = ” + string ( T b b ) + ” K and \ n Dew p o in t = ” + string ( T d p ) + ” K . ” )

Scilab code Exa 5.14 Hot air drying machine

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

clear ; clc ;


/ / E xa mp le 5 . 1 4 / / P ag e 2 37 printf ( ” E xa mp le 5 . 1 4 , P ag e 2 3 7 \ n \ n ” ) ; // s o l u t i o n // b a s i s 1 0 0 0 kg /h o f c o n de ns a te a t t he s a t u r a t i o n t em pe ra tu re c o r r e sp o n di n g t o 8 b ar a / / u s i n g A pp en di x IV . 2 H = 7 20 .9 4 / / kJ / kg H m = 4 19 .0 6 / / kJ / kg x = poly (0 , ’ x ’ ) c o nd e ns a te = 1 00 0 - x H c on d en s at e 1 = 1 00 0* H

94

22 23 24 25

H c o n d e n sa t e 2 = c o n de n s at e * 4 1 9 . 0 6 Ht = x *2 67 6 p = H c o n de n s a te 2 + H t - H c o n d e n sa t e 1 printf ( ” The q u qn t i ty o f f l a s h s team p ro du ce d = ” + string ( roots ( p ) ) + ” k g / h . ” )

Scilab code Exa 5.15 Flow of saturated vapors of R134

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

clear ; clc ;


/ / E xa mp le 5 . 1 5 / / P ag e 2 38 printf ( ” E xa mp le 5 . 1 5 , P ag e 2 3 8 \ n \ n ” ) ; // s o l u t i o n q v1 = 50 // l / s q m = q v1 * 1. 08 // kg / s f i = q m * 3 . 0 8* ( 2 63 . 1 5 - 2 5 8 . 15 ) / / kW l v = 3 84 . 19 - 1 6 8. 7 / / kJ / k g q m2 = fi / lv H = 2 56 .3 5 / / kJ / kg x = poly (0 , ’ x ’ ) p = H * ( qm 2 + x) - 1 68 .7 * qm 2 - x * 3 84 . 19 a = qm2 +roots ( p ) printf ( ” Flo w o f v a po r f ro m h e c h i l l e r = ” + string ( a ) +” kg/s . ”)

95

Scilab code Exa 5.16 Liquifaction of Cl2

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

clear ; clc ;


/ / E xa mp le 5 . 1 6 / / P ag e 2 38 printf ( ” E xa mp le 5 . 1 6 , P ag e 2 3 8 \ n \ n ” ) ; // s o l u t i o n / / b a s i s l i q u i f a c t i o n c a p a c i t y = 0 . 1 16 kg / s p 1 = 101 / / kPa T s1 = 2 39 .1 5 l v1 = 2 88 .1 3 / / kJ / kg p 2 = 530 / / kPa T s2 = 2 90 .7 5 // K l v2 = 2 52 .9 3 / / kJ / kg

/ / r e f e r r i n g t o t a b l e 5 . 3 and u si ng eq 5 . 2 1

24 25

26 27 28

H 1 = - 3 9. 2 46 * ( Ts 2 - T s 1 ) + 1 4 0 1 .2 2 3 *1 0 ^ - 3 * ( T s 2 ^ 2 - T s 1 ^ 2) /2-6047.226*10^-6*(Ts2^3-Ts1^3)/3+8591.4*10^-9*( T s 2 ^ 4 - T s 1 ^ 4 ) / 4 / / k J / k mo l T 3 = 3 13 .1 5 H 2 = [ 2 8 .5 4 6 3* ( T 3 - T s 1 ) + 2 3 . 87 9 5 *1 0 ^ - 3 * ( T 3 ^ 2 - T s 1 ^ 2) /2-21.3631*10^-6*(T3^3-Ts1^3)/3+6.4726*10^-9*(T3 ^ 4 - T s 1 ^ 4 ) / 4 ] / 7 0 . 9 0 3 / / kJ / kg f i2 = . 11 6* H 2 C l 2 ev p = f i2 / l v1 / / kg / s C l 2 r e cy = C l 2 ev p / ( 1 - . 1 8 5)

96

29 30 31 32 33 34 35 36 37 38 39 40

R = C l2 r ec y / . 11 6 // kg / kg f r e s h

f e ed / / T4 / T1 = ( p 2 / p 1 ) ˆ [ ( gamma− 1)/gamma]

g m = 1 .3 55 p 2 2 = 3 26 .3 p 21 = 101 T 4 = T s 1 *( p 2 / p 1 ) ^ [( g m - 1 ) / g m ] T 5 = 3 13 .1 5 f i 3 = 1 . 8 8 *1 0 ^ - 3 * ( 3 4 3 .1 + 9 1 .6 - 2 6 .2 + 2 .5 ) F w a t e r1 = f i 3 / ( 8 * 4. 1 8 6 8) // kg / s

// kW

// s i m i l a r l y

T 6 = 3 79 .9 f i4 = 1 . 8 8 *1 0 ^ - 3 * [ 2 8 . 54 6 3 * ( T6 - T 5 ) + 2 3 . 8 7 95 * 1 0^ - 3 * ( T 6 ^2-T5^2)/2-21.3631*10^-6*(T6^3-T5^3) / 3 + 6 . 4 7 2 6 * 1 0 ^ - 9 * ( T 6 ^ 4 - T 5 ^ 4 ) / 4 ] // kW 41 F w a t e r2 = f i 4 / ( 8 * 4. 1 8 6 8) // kg / s 42 W r e q = F w a te r 1 + F w a t er 2 43 f i5 = 1 . 8 8 *1 0 ^ - 3 * [ 2 8 . 54 6 3 * ( T5 - T s 2 ) + 2 3 . 8 7 95 * 1 0^ - 3 * ( T 5 ^2-Ts2^2)/2-21.3631*10^-6*(T5^3-Ts2^3) / 3 + 6. 4 7 26 * 1 0^ - 9 * ( T 5 ^4 - T s 2 ^ 4 ) / 4 ] + . 1 3 33 * 2 5 2. 9 3 //

kW 44 printf ( ” ( a ) \ n \ n R ec yc le r a t i o = ” + string ( R ) + ” kg

Cl 2 / kg f r e s h f e ed \ n \ n \ n ( b ) \ n \ n C o o li n g w at er r e q u i r e d a t \ n i n t e r f a c e = ” + string ( F w a t e r 1 ) + ” k g / s \ n a f t e r c o o l e r = ” + string ( W r e q ) + ” k g / s \ n \ n \ n ( c ) \ n \ n R e f r i g i r a t i o n l o ad o f c h i l l e r = ” + string ( f i 5 ) + ” kW. ” )

Scilab code Exa 5.17 Melting of Tin

1 2 3 4 5 6

clear ; clc ;

/ / S t o i ch i o m et r y / / C ha pt er 5 / / E ne rg y B a l a n c e s 97

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

/ / E xa mp le 5 . 1 7 / / P ag e 2 42 printf ( ” E xa mp le 5 . 1 7 , P ag e 2 4 2 \ n \ n ” ) ; // s o l u t i o n // b a si s 100 kg o f t in T 1 = 3 03 .1 5 T 2 = 5 05 .1 5 n = 1 00 /1 18 .7

/ / km ol / / Q1 = n ∗ [ i n t g r f ro m T1 t o T2 ( Cms dT ) ] Q 1 = 4 97 3. 3 / / kJ

l f = 72 01 Q 2 = n *lf / / kJ Q = Q1 +Q2 l v = 278 / / kJ / k g v p = Q /lv / / k g printf ( ” Q ua nt i ty o f e u t e c t i c m ix tu re c on de ns ed = ” + string ( v p ) + ” k g p e r 10 0 kg o f t i n m e l t e d a t i t s m e l t i n g p o i n t . ”)

Scilab code Exa 5.18 steam fluctuation calculations

1 2 3 4 5 6 7 8 9 10

clear ; clc ;


/ / E xa mp le 5 . 1 8 / / P ag e 2 43 98

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

printf ( ” E xa mp le 5 . 1 8 , P ag e 2 4 3 \ n \ n ” ) ;

// s o l u t i o n

T s1 = ( 4 38 . 2+ 4 36 ) / 2 T a = 300 f i 1 = . 0 45 * ( T s1 - T a ) * 3 60 0 t he t a1 = 3 07 2 93 / f i1 // h T s 2 = ( 4 36 + 43 4 ) /2 f i 2 = . 0 45 * ( T s2 - T a ) * 3 60 0 t he t a2 = 3 02 4 15 / f i2 T s3 = ( 4 34 + 43 2 .1 ) / 2 f i 3 = . 0 45 * ( T s3 - T a ) * 3 60 0 t he t a3 = 3 13 8 59 / f i3 t h e t a = t h e ta 1 + t h e t a2 + t h e t a 3 printf ( ” t o t a l t i m e r e q u i r e d = ” + string ( t h e t a ) + ” h r s .”)

Scilab code Exa 5.19 Manufacture of dry ice

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

clear ; clc ;


/ / E xa mp le 5 . 1 9 / / P ag e 2 45 printf ( ” E xa mp le 5 . 1 9 , P ag e 2 4 5 \ n \ n ” ) ; // s o l u t i o n H 1 = 4 82 .9

/ / kJ / kg 99

16 17 18 19 20

H 2 = 2 73 .4 f i 1 = 1 00 *( H 1 - H2 ) / / kJ / h T 1 = 3 13 .1 5 T 2 = 4 03 .1 5 f i 11 = 2 1 . 36 5 5 *( T 2 - T 1 ) + 6 4 . 2 84 1 * 10 ^ - 3 * ( T 2 ^ 2 - T 1 ^ 2 ) /2-41.0506*10^-6*(T2^3-T1^3)/3+9.7999*10^-9*(T2 ^ 4 - T 1 ^ 4 ) / 4 / / kJ / h

21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39

/ / a t 2 0 MPa h 1 = 2 11 .1 T s = 2 77 .6 H 1 1 = 4 27 .8 x = poly (0 , ’ x ’ ) p = x * h 1 + (1 00 - x ) * H 11 - 1 00 * H 2 a = roots ( p ) f i2 = ( 10 0 - a ) *( H 11 - h 1 ) / / kJ / h h 2 = - 14 8. 39 H 3 = 4 22 .6 1 y = poly (0 , ’ y ’ ) p 1 = 1 0 0* 1 76 . 18 - ( 1 0 0 - y ) * H 3 + h 2 * y b = roots ( p 1 ) f i3 = 1 00 *( h 1 - 1 76 . 8) H = f i3 + 2 40 21 H 4 = H / ( 10 0 - 4 3. 16 )

/ / f ro m r e f 23 T = 2 62 .1 5 printf ( ” ( a ) \ n \ n Y i e l d o f d ry i c e = ” + string ( b ) + ” kg . \ n \ n \ n ( b ) \ n \ n P e r c e n t l i q u i f a c t i o n = ” + string ( a ) + ” . \ n \ n \ n ( c ) \ n \ n Temp o f v e n t e d g as = ” + string ( T ) + ” K . ” )

Scilab code Exa 5.20 Steam produced in S burner

1 clear ; 2 clc ; 3

100

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


/ / E xa mp le 5 . 2 0 / / P ag e 2 47 printf ( ” E xa mp le 5 . 2 0 , P ag e 2 4 7 \ n \ n ” ) ; // s o l u t i o n // b a s i s 2 00 kg /h o f S ul ph ur f i r i n g F = 2 00 /3 2 / / km ol / h

O 2r eq = 6 .2 5 *1 .1 a ir in = O 2r eq / . 21 N 2i n = a ir in - O 2 re q T 1 = 1 14 4. 15 T 2 = 4 63 .1 5 f i = 7 88 85 2. 2 / / kJ / h H = 1 5 *4 . 18 6 8+ 1 94 5 .2 q m = f i * .9 / 20 0 8 / / kg / h printf ( ” Amount o f s te am p r od u ce d = ” + string ( q m ) + ” kg /h . ” )

Scilab code Exa 5.21 Equimoar pentane and hexane mix

1 2 3 4 5 6 7 8 9

clear ; clc ;


/ / E xa mp le 5 . 2 1 101

10 11 12 13 14 15 16 17 18

/ / P ag e 2 48 printf ( ” E xa mp le 5 . 2 1 , P ag e 2 4 8 \ n \ n ” ) ; // s o l u t i o n / / e n t ha l p y a t Tbb

T b b = 3 21 .6 T 1 = 2 98 .1 5 H 1 = 6 5 . 49 6 1 *( T b b - T 1 ) + 6 2 8 .6 2 8 *1 0 ^ - 3 * ( T b b ^ 2 - T 1 ^ 2) /2-1898.8*10^-6*(Tbb^3-T1^3)/3+3186.51*10^-9*(Tbb ^ 4 - T 1 ^ 4 ) / 4 / / kJ / k m ol 19 H 2 = 3 1 . 42 1 * ( T bb - T 1 ) + 9 7 6. 0 5 8* 1 0 ^ - 3 * ( T b b ^2 - T 1 ^ 2 ) /2-2353.68*10^-6*(Tbb^3-T1^3)/3+3092.73*10^-9*( T b b ^ 4 - T 1 ^ 4 ) / 4 / / k J / km ol 20 H s ol = ( H 1 + H2 ) / 2 / / k J / k mo l 21 / / e n t ha l p y a t Tdp 22 l v1 = 2 5 7 90 * ( ( 46 9 . 7 - 3 2 9 . 9) / ( 4 69 . 7 - 3 0 9 .2 ) ) ^ . 3 8 23 l v2 = 2 8 8 50 * ( ( 50 7 . 6 - 3 2 9 . 9) / ( 5 07 . 6 - 3 4 1 .9 ) ) ^ . 3 8 24 T d p = 3 29 .9 25 H 2 1i g = 6 5 . 49 6 1 *( T d p - T 1 ) + 6 2 8 .6 2 8 *1 0 ^ - 3 * ( T d p ^ 2 - T 1 ^ 2) /2-1898.8*10^-6*(Tdp^3-T1^3)/3+3186.51*10^-9*(Tdp ^4 - T 1 ^ 4) / 4 + l v1 / / k J / k mo l 26 H 2 2i g = 3 1 . 42 1 * ( T dp - T 1 ) + 9 7 6. 0 5 8* 1 0 ^ - 3 * ( T d p ^ 2 - T1 ^ 2 ) /2-2353.68*10^-6*(Tdp^3-T1^3)/3+3092.73*10^-9*( T dp ^ 4 - T 1 ^ 4 ) / 4 + l v 2 / / k J / k mo l 27 H m i x ig = ( H 2 1 ig + H 2 2 i g ) / 2 28 printf ( ” ( a ) \ n \ n H = ” + string ( H s o l ) + ” k J / km ol \ n \ n \ n ( b ) \ n \ n H = ” + string ( H m i x i g ) + ” k J / km ol ” )

Scilab code Exa 5.22 Flashing of saturated liq mix


102

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

/ / C ha pt er 5 / / E ne rg y B a l a n c e s


H 1 = 2 35 49 / / k J / k m o l H 2 = 1 63 25 H 3 = 2 83 32 H 4 = . 4* H 2 + .6 * H3 printf ( ” E n t ha l py o f v ap or − l i q u i d m ix t u re a f t e r f l a s h i n g = ” + string ( H 4 ) + ” k J / m o l . ” )

Scilab code Exa 5.23 H2 recovery from Refinery off gases

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

clear ; clc ;


/ / E xa mp le 5 . 2 3 / / P ag e 2 53 printf ( ” E xa mp le 5 . 2 3 , P ag e 2 5 3 \ n \ n ” ) ; // s o l u t i o n / / b a s i s f e e d g a s = 12 00 0 Nmˆ 3 = 5 3 5 . 4 kmol / h T 1 = 1 47 .6 5 // K 103

17 18 19 20 21 22

n 1 = 5 3 5. 4 *. 3 15 6 / / k mo l /h HP t a i l g a s s t re a m T = 11 8. 5 // K LP t a i l s tr ea m n 2 = ( 53 5. 4 - n 1 ) * .0 6 02 // kmol /h n 3 = 5 35 .4 - n 2 - n 1 // kmol / h p r od u ct H2 s t r ea m p = 3 1 5. 3 5* 1 00 / n 3 printf ( ” P u ri t y o f p ro du ct H2 s tr ea m = ” + string ( p ) + ” per cen t . ”)

Scilab code Exa 5.24 Refrigiration calculations

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

clear ; clc ;



// f i 1 = i n t e g r ( from 3 0 4 . 1 5 t o 3 1 3 . 1 5) { 1 1 8 3 1 . 6 + 2 4 9 9 7 . 4 ∗ 10Tˆ − 3 − 5979.8 ∗ 10ˆ − 6T ˆ2 − 31.7 ∗ 10ˆ − 9 T3 } d t 16 f i 1 = 1 7 07 8 7. 7 / / kJ / h 17 f i 2 = 5 35 .4 *1 20 86 [ 3 4 4 . 3 6 * 8 7 4 3 . 2 + 1 6 8 . 9 7 * 1 8 0 3 6 + 2 2 . 0 7 * 1 5 8 9 2 ] / / kJ / h 18 printf ( ” ( a ) \ n \ n R e f r i g i r a t i o n r eq ui re me nt = ” + string ( f i 1 ) + ” k J / h \ n \ n \ n ( b ) \ n \ n

R e f r i g i r a t i o n r eq u i r e me n t b as ed on r e a l e n t h a l p i e s = ” + string ( f i 2 ) + ” k J / h . ” )

104

Scilab code Exa 5.25 Chlorination of benzene

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

clear ; clc ;


/ / E xa mp le 5 . 2 5 / / P ag e 2 57 printf ( ” E xa mp le 5 . 2 5 , P ag e 2 5 7 \ n \ n ” ) ; // s o l u t i o n / / b a s i s 1 0 0 kmol /h o f b en ze n e f e ed r a t e C l2 = . 4* 10 0 H Clp = 40 B e n z en e co n = 3 7 M CB p = 1 0 0* . 37 * .9 1 89 D C Bp = B e nz e ne c on - M C B p u n r e a c t B en z e n e = 1 00 - B e n ze n e c on Nt = HC lp + MC Bp + D CB p + u nr ea ct Be nz en e

// u s i n g eq xi = 1

x i = Ni /( L(1 − K1 )+NtK i )

a nd s i g m a

24 L = 8 9. 66 9 / / km ol / h 25 V = N t - L 26 printf ( ” L i qu i d p ro du ct s tr ea m = ” + string ( L ) + ” k mo l / h \ n Vapor p r od u ct s tr ea m = ” + string ( V ) + ” k mo l / h” )

105

Scilab code Exa 5.26 Heat of formation of ethylene

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

clear ; clc ;


/ / E xa mp le 5 . 2 6 / / P ag e 2 60 printf ( ” E xa mp le 5 . 2 6 , P ag e 2 6 0 \ n \ n ” ) ; // s o l u t i o n // // // // //

2C + 2O2 = 2CO2 2H2 + O2 = 2H2O C2H4 + 3O2 = 2CO2 + 2H2O A+B−C g i v e s 2C( g ) + 2H2 = C2H4 ( g )

A B C

D H = - 2 * 39 3 . 51 - 2 * 24 1 . 82 + 1 32 3 . 1 / / kJ / m ol printf ( ” H eat o f f or ma ti o n o f E th yl en e i s ” + string ( H )+” kJ/mol . ”)

Scilab code Exa 5.27 Heat of combustion of ethyl mercaptan

1 2 3 4 5 6 7 8

clear ; clc ;


106

9 10 11 12 13 14 15

/ / E xa mp le 5 . 2 7 / / P ag e 2 60 printf ( ” E xa mp le 5 . 2 7 , P ag e 2 6 0 \ n \ n ” ) ; // s o l u t i o n H c = 2 * ( - 3 9 3 . 5 1 ) - 8 8 7 . 8 11 + 2 * ( - 2 8 5 . 8 3 ) - ( - 7 3 . 6 +0 ) / / k J

/mol 16 printf ( ” Heat o f c om bu st io n o f e t h y l m er ca pt an = ” + string ( H c ) + ” k J / m o l . ” )

Scilab code Exa 5.28 Std heat of formation of gaseous di ethyl ether

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

clear ; clc ;


/ / E xa mp le 5 . 2 8 / / P ag e 2 61 printf ( ” E xa mp le 5 . 2 8 , P ag e 2 6 1 \ n \ n ” ) ; // s o l u t i o n l v 1 = 2 66 94 / / k j / k mo l T c = 4 66 .7 4 l v2 = l v1 * ( ( T c - 2 9 8 .1 5 ) / ( Tc - 3 0 7. 7 ) ) ^ . 3 8 / 10 0 0

/ / kJ /

mol 18 H f = -252 / / k J / mo l 19 H f 1 = Hf - l v2 / / kJ / k m ol 20 printf ( ” Heat o f f or ma ti on o f l i q u i d d i e t h y l ” + string ( H f 1 ) + ” k J / m o l . ” )

107

e th er =

Scilab code Exa 5.29 Heat of formation of motor spirit

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

clear ; clc ;


/ / E xa mp le 5 . 2 9 / / P ag e 2 61 printf ( ” E xa mp le 5 . 2 9 , P ag e 2 6 1 \ n \ n ” ) ; // s o l u t i o n / / b a s i s 1 kg m ot or s p i r i t G = 1 4 1. 5 /( 1 31 . 5+ 6 4)

/ / r = C/H r = ( 74 + 15 * G ) / (2 6 - 15 * G ) C = r /6 .6 05 / / C c o n t e n t o f m ot or s p i r i t H 2 = 1- C O 2r eq = C + H2 H f = 4 40 50 - 2 7 82 9 - 1 83 06 / / kJ / kg printf ( ” Hea t o f f o r m a ti o n o f m oto r s p i r i t = ” + string ( H f ) + ” k J / k g . ” )

Scilab code Exa 5.30 Mean heat capacity

1 clear ; 2 clc ;

108

3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18


/ / E xa mp le 5 . 3 0 / / P ag e 2 67 printf ( ” E xa mp le 5 . 3 0 , P ag e 2 6 7 \ n \ n ” ) ; // s o l u t i o n / / b a s i s 1 k mol o f s t yr e n e d H = 2 4 17 49 - 1 8 93 9 8 / / kJ / m ol C mp n = d H / (6 0 0 - 29 8 .1 5) / / k J / k mo l K printf ( ” Mean h e at c a p a c i t y b et we en 6 00K and 2 9 8 . 1 5 K i s ” + string ( C m p n ) + ” k J / k mo l K . ” )

Scilab code Exa 5.31 Heat of reaction

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

clear ; clc ;


/ / E xa mp le 5 . 3 1 / / P ag e 2 69 printf ( ” E xa mp le 5 . 3 1 , P ag e 2 6 9 \ n \ n ” ) ; // s o l u t i o n / / b a s i s 1 mol o f SiO2 r e a c t e d 109

16 H f = [ - 2 87 8 7 9 + 3* 3 * ( - 2 9 6 .8 . 8 1 ) + 3 * 0 /2 / 2 ] - [ 3 * ( - 1 4 3 2. 2. 7 ) + 1 * ( - 9 0 3 . 5 ) ] / / k J / mo m o l S iO iO 2 17 printf ( ” H e a t o f r e a c t i o n = ” + string ( H f ) + ” k J / mo mo l SiO 2 . ” )

Scilab code Exa 5.32 Std heat of reaction

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

clear ; clc ;


/ / E xa xa mp mp le le 5 . 3 2 / / P aagg e 2 69 69 xa mp mp le le 5 . 3 2 , P ag ag e 2 6 9 \ n \ n ” ) ; printf ( ” E xa // s o l u t i o n / / b a s i s 1 00 0 0 k g o f 2% a m m o n i a s o l u t i o n N H 3 = 2 // // kg // kg H 2O 2 O = 9 8 // H r = - 3 61 61 . 2 - ( - 4 5 . 94 94 - 2 8 5 .8 . 8 3 ) / / k J / mo mo l NH NH3 d i s s o l v e d 10 0 kg s o l . H d = - ( Hr Hr * 2 * 1 0 0 0 / 17 1 7 . 0 3 0 5) 5 ) / / k J / 10 10 0 kg printf ( ” h e a t o f r e a c t i o n = ” + string ( H d ) + ” k J / 10 s o l u t i o n . ”)

Scilab code Exa 5.33 Burning of SO2

1 clear ; 2 clc ;

110

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


/ / E xa xa mp mp le le 5 . 3 3 / / P aagg e 2 72 72 xa mp mp le le 5 . 3 3 , P ag ag e 2 7 2 \ n \ n ” ) ; printf ( ” E xa // s o l u t i o n / / b a s i s 1 k m o l o f SO2 r e a ct ct e d

a = 2 2 . 03 0 3 6 - 2 4 . 7 7 1 - . 5 * (2 ( 2 6 . 02 02 6 ) b = ( 1 2 1. 1 . 6 2 4 - 6 2 . 94 9 4 8 - . 5 * 1 1 . 75 75 5 ) c = ( - 9 1 .8 . 8 7 6 +4 + 4 4 . 25 25 8 - . 5 * ( - 2 . 3 43 43 ) ) d = ( 2 4. 4 . 3 6 9 - 1 1 .1 . 1 2 2 - . 5 * ( - . 5 62 62 ) ) k J / k m ol ol H r = - 3 95 9 5 7 20 2 0 +2 + 2 9 68 6 8 1 0 / / kJ H r o = H r - a * 2 98 9 8 . 15 1 5 - b * 1 0 ^ - 3 * 2 98 9 8 . 15 1 5 ^ 2/ 2/ 2 - c *10^-6*298.15^3/3-d*10^-9*298.15^4/4 22 T = 7 78 7 8 .1 .1 5 23 H rt r t = - Hr H r o - 1 5 .7 . 7 4 8 * T + 2 6 .4 . 4 * 1 0^ 0 ^ - 3 * T ^ 2 - 1 5 . 48 4 8 * 10 10 ^ - 6 * T ^3+3.382*10^-9*T^4 24 printf ( ” H e a t o f r e a c t i o n a t 7 7 5K i s ” + string ( H r t ) + ” kJ/ kmol . ” )

Scilab code Exa 5.34 Esterification of acetic acid

1 2 3 4 5 6 7

clear ; clc ;


111

8 9 10 11 12 13 14 15 16

/ / E xa xa mp mp le le 5 . 3 4 / / P aagg e 2 72 72 printf ( ” E xa xa mp mp le le 5 . 3 4 , P ag ag e 2 7 2 \ n \ n ” ) ; // s o l u t i o n H r = - 4 8 0 - 2 8 5 .8 . 8 3 + 27 2 7 7 . 2+ 2 + 4 8 4. 4 . 2 / / kJ k J / m ol ol H r t 1 = H r * 1 00 0 0 0 + [ 1 4 6. 6 . 8 9 +7 + 7 5 . 7 6 - 1 1 9 . 55 5 5 - 1 2 9 .7 . 7 0 ] * 75 75

//

kJ/kmol a = 4 . 2 90 9 0 5 + 50 5 0 . 8 45 4 5 - 1 0 0. 0 . 9 2 - 1 5 5 . 48 48 b = 9 3 4 . 3 7 8+ 8 + 2 1 3 . 08 0 8 + 1 1 1 . 83 8 3 8 6 + 3 2 6. 6. 5 9 5 1 c = - 2 6 40 4 0 - 6 3 1 . 39 3 9 8 - 4 9 8 .5 . 5 4 - 7 4 4 .1 .1 9 9 d = 3 3 42 4 2 . 58 5 8 + 64 6 4 8 .7 .7 4 6 H ro r o = H r * 1 0 00 0 0 + a * ( - 2 98 9 8 . 15 1 5 ) + b * 1 0 ^ - 3 * ( - 2 9 8 .1 . 1 5 ^2 ^2 ) / 2 + c *10^-6*(-298.15^3)/3+d*10^-9*(-298.15^4)/4 7 3 .1 .1 5 22 T = 3 73 23 H rt r t = H ro r o + a * T + 7 9 2 .9 . 9 4 9 *1 * 1 0 ^ - 3 * T ^ 2 - 1 50 5 0 4 .7 . 7 1 2 *1 *1 0 ^ - 6 * T ^3+997.832*10^-9*T^4 24 printf ( ” H e a t o f r e a c t i o n a t 3 7 3 K i s ” + string ( H r t ) + ” k J / km k m ol ol r e a c t a n t . ” ) 17 18 19 20 21

Scilab code Exa 5.35 Heat transfer in intercoolers

1 2 3 4 5 6 7 8 9 10 11

clear ; clc ;


/ / E xa xa mp mp le le 5 . 3 5 / / P aagg e 2 73 73 printf ( ” E xa xa mp mp le le 5 . 3 5 , P ag ag e 2 7 3 \ n \ n ” ) ; 112

12 13 14 15 16 17

18 19

20 21 22 23

// s o l u t i o n T 2 = 800 T 1 = 2 98 .1 5 f i1 = 3 6 1 4. 5 7 7 *( T 2 - T 1 ) + 3 0 5 . 56 1 * 10 ^ - 3 * ( T 2 ^ 2 - T 2 ^ 2 ) /2+836.881*10^-6*(T2^3-T1^3)/3-393.707*10^-9*(T2 ^ 4 - T 1 ^ 4 ) / 4 // kW T 3 = 875 f i2 = 3 4 8 0. 7 3 7 *( T 3 - T 1 ) + 7 5 4 . 34 7 * 10 ^ - 3 * ( T 3 ^ 2 - T 2 ^ 2 ) /2+442.159*10^-6*(T3^3-T1^3)/3-278.735*10^-9*(T3 ^ 4 - T 1 ^ 4 ) / 4 // kW H r = - 98 91 0 // kJ / kmol SO2 r e a c t e d by eg 5 . 3 3 f i 3 = ( 8 .8 5 1 1 - . 3 51 ) * H r / 3 6 00 // kW d H = f i 2 / 3 60 0 + f i3 - f i 1 / 3 60 0 printf ( ” Net e n th a lp y c ha ng e = ” + string ( d H ) + ” kW. ” )

Scilab code Exa 5.36 Enthalpy balance in the reactor

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

clear ; clc ;


/ / E xa mp le 5 . 3 6 / / P ag e 2 75 printf ( ” E xa mp le 5 . 3 6 , P ag e 2 7 5 \ n \ n ” ) ; // s o l u t i o n / / b a s i s 1 00 kmol o u tg o in g g as m ix tu re from s c r ub b e r m o i s t in = 3 1 2 7 .7 * . 0 15 / 1 8 / / km ol 113

17 18 19 20 21

/ / km ol / / u s i n g t a b l e s 5 . 2 9 and 5 . 3 0 w a t e r in = 4 0 .2 + m o i s t in

H r = - 2 70 0 26 5 8 - ( - 2 6 8 5 33 5 9) H r 1 = H r / 24 6 .4 4 93 // k J / kmol t o t a l r e a c t a n t s printf ( ” Heat o f r e a c t i o n = ” + string ( H r 1 ) + ” k J / k mo l tota l reactants . ”)

Scilab code Exa 5.37 Calculation of circulation rate

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

clear ; clc ;


/ / E xa mp le 5 . 3 7 / / P ag e 2 76 printf ( ” E xa mp le 5 . 3 7 , P ag e 2 7 6 \ n \ n ” ) ; // s o l u t i o n f i 3 = 1 5 50 5 40 7 / / kJ / h l v = 2 96 .2 // f ro m t a b l e 5 . 6 H t = 1 71 31 55 1 / / kJ / h r = Ht /lv / / kg / h printf ( ” Downtherm c i r c u l a t i o n kg /h . ” )

r a t e = ” + string ( r ) + ”

Scilab code Exa 5.38 Loop reactor for EDC manufacture

1 clear ;

114

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 32 33 34 35 36 37 38 39

clc ;


/ / E xa mp le 5 . 3 8 / / P ag e 2 79 printf ( ” E xa mp le 5 . 3 8 , P ag e 2 7 9 \ n \ n ” ) ; // s o l u t i o n f e ed r a t e o f e t hy l e ne F = 100 / / kmol /h E co n = . 99 * F E co n1 = E co n * . 99 8 E co n2 = E co n - E c on 1 C l 2 c on = E c on 1 + 2 * E c o n2 C l2 in = F * 1. 1 C l2 s3 = C l2 in - C l 2c on H Cl s3 = E co n2 T CE p = E co n2 E DC p = E co n1 nC2H4 = 1 T = 3 28 .1 5 p v1 = exp ( 4 . 5 8 5 1 8 - 1 5 2 1 . 7 8 9 / ( T - 2 4 . 6 7 ) ) // b ar p v2 = exp ( 4 . 0 6 9 7 4 - 1 3 1 0 . 2 9 7 / ( T - 6 4 . 4 1 ) ) // b ar x E DC = E c on 1 / ( E c o n1 + E c o n 2 ) x TE C = 1 - xE DC p ED C = 3 7. 2* x E DC p TE C = 1 2. 64 * x T EC p C l 2 H C l C2 H 4 = 1 . 6* 1 00 - p ED C - p T EC y E DC = p ED C / 16 0 y T EC = p TE C / 16 0 n t = ( C l 2 s3 + E c o n 2 + 1 ) * 1 60 / p C l 2 H C lC 2 H 4 n E DC = y ED C * nt n T EC = y TE C * nt printf ( ” C om po si ti on s o f g as s tr ea ms : \ n \ n 115

Component

Strea m 3 Strea m 5 Stream 4 S tream 6 \ n C l 2 ” + string ( C l 2 s 3 ) + ” ”+ string ( C l 2 s 3 ) + ” \ n HCl ” + string ( ” + string ( H C l s 3 ) + ” \ n C2H4 HCls3)+” ” + string ( n C 2 H 4 ) + ” ” + string ( n C 2 H 4 ) + ” \ n EDC ” + string (nEDC)+” 0.2355 3.3947 9 8 . 5 6 6 5 \ n TEC ”+ string ( n T E C ) + ” Ni l ” + string ” + string ( T C E p ) + ” \ n \ n ” ) (nTEC)+” 40 f i1 = ( 1 0 . 8 0 2 * 3 3 . 9 + . 1 9 8 * 2 9 . 1 + 1 * 4 3 . 6 + 3 . 6 3 0 2 * 1 7 . 4 + . 0 0 2 5 * 8 5 .3 ) *(328.15-273.15) 41 f i2 = 3 5 . 0 5 3 *1 0 0 0 * 3 .3 9 4 7 + 3 9. 5 8 * 1 0 00 * . 0 0 2 5 42 f i 3 = ( 3 . 3 9 47 * 1 2 9 .4 + . 0 0 2 5* 1 4 4 . 4) * 5 5/ 2 43 f i = fi 1+ fi 2+ fi 3 / / kJ / h 44 printf ( ” Heavy d ut y o f Ov er he ad c o n d en s e r = ” + string (fi)+ ” k J / h . \n \n ”) 45 f i 5 = ( 1 0 0 * 43 . 6 + 1 10 * 3 3 . 9) * ( 3 28 . 1 5 - 2 7 3 .1 5 ) 46 f i6 = 3 . 6 3 0 2* 1 0 0 0 * 3 3. 6 + . 0 0 25 * 1 0 0 0* 3 8 . 1 6 6 47 f i7 = ( 9 8 . 5 66 5 * 1 2 9. 4 + . 1 9 88 * 1 4 4 . 4) * ( 3 28 . 1 5 - 2 7 3 .1 5 ) 48 f i 8 = 2 1 6 8 4 5. 5 * 9 8 . 80 2 + 3 9 2 39 4 . 5 * .1 9 8 49 f i c o l = f i 5 + fi 8 - f i1 - f i6 - f i 7 50 printf ( ” Heavy du ty o f e x t e r n a l c o o l e r = ” + string ( ficol)+” kJ/h . ” )

Scilab code Exa 5.39 Calculations in adiabatic converter

1 2 3 4 5 6

clear ; clc ;


7 8 9 10 11 12 13 14 15 16 17

/ / E xa mp le 5 . 3 9 / / P ag e 2 84 printf ( ” E xa mp le 5 . 3 9 , P ag e 2 8 4 \ n \ n ” ) ; // s o l u t i o n T o = 2 98 .1 5 T 1 = 4 83 .1 5

// f i 1 = i n t g r ( fro m To t o T1 ) { 1 2 1 9 9 . 5 + 2 2 4 1 . 4 ∗ 1 0 ˆ − 3 ∗ T +1557.7 ∗ 10ˆ − 6 ∗ Tˆ 2 − 671.3 ∗ 10ˆ − 9 ∗Tˆ 3 } dT 18 f i 1 = 2 4 55 8 74 . 6 / / kJ / h 19 d H r = 2 *( - 4 5 .9 4) / / k J / mol N2 r e a c t e d 20 f i 2 = 9 1 . 8 8 *1 0 0 0 *2 3 . 1 6 8 21 f i3 = f i1 + fi 2 22 // f i 3 = i n t g r ( fro m To t o T2 ) { 1 0 7 1 3 . 9 + 3 8 4 1∗ 1 0 ˆ − 3 ∗ T

+1278.8 ∗ 10ˆ − 6 ∗ Tˆ 2 − 752.6 ∗ 10ˆ − 9 ∗Tˆ 3 } dT 23 / / s o l v i n g i t 24 T 2 = 6 57 .4 1 // K 25 printf ( ” T em per at ur e o f t he g as m ix tu re l e a v i n g t he r e a c t o r = ” + string ( T 2 ) + ” K . ” )

Scilab code Exa 5.40 Burning of HCl

1 2 3 4 5 6 7 8 9 10

clear ; clc ;


/ / E xa mp le 5 . 4 0 / / P ag e 2 92 117

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

printf ( ” E xa mp le 5 . 4 0 , P ag e 2 9 2 \ n \ n ” ) ;

// s o l u t i o n / / b a s i s 4 kmol o f HCl g as O2req = 1 / / k mo l

O 2 sp p ly = 1 .3 5 *1 N 2 = 1 . 35 * 79 / 21 a i r = O 2s p pl y + N2 H C l br n t = . 8* 4 H C l = 4 - H C lb r nt O 2 = O 2 sp pl y - .8 Cl2 = . 8* 2 H2O = . 8* 2 printf ( ” ( a ) \ n \ n C om po si ti on o f d ry p ro d uc t g as

s tr ea m : \ n Component Dry p r o d u c t g a s stre am , kmol \ n HCl ” + string ( H C l ) + ” \ n O2 ” + string ( O 2 ) + ” \ n C l 2 ” + string ( C l 2 ) + ” \ n H2O ” + string ( H 2 O ) + ” \ n N2 ” + string ( N 2 ) + ” \ n \ n \ n ( b ) \ n \ n ”) 26 H 2 = 1 1 4. 4 *1 0 00 * .8 27 / / H2 = i n t g r ( fr om 2 9 8 . 1 5 t o T)

{ 2 8 6 . 5 5 4 + 1 2 . 5 9 6 ∗ 1 0 ˆ − 3 ∗ T+63.246 ∗ 10ˆ − 6 ∗ T ˆ2 − 25.933 ∗ 10ˆ − 9 ∗ Tˆ 3 } dT 28 / / s o l v i n g i t 29 T = 59 9. 5 // K 30 printf ( ” A d ia b a t ic r e a c t i o n t em pe ra tu re o f p ro du ct g as s tr ea m = ” + string ( T ) + ” K . ” )

Scilab code Exa 5.41 Dehydrogenation of EB

1 clear ; 2 clc ;

118

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



/ / 1 kmol o f EB v ap or s e n t e r i n g t he r e a c t o r a t 811.15 K 16 / / ( f ro m 8 1 1 . 1 5 t o T1 ) i n t g r { − 36.72+671.12 ∗ 10 ˆ − 3 ∗ T − 422.02 ∗ 10ˆ − 6 ∗ Tˆ2+101.15 ∗ 10 ˆ − 9 ∗ Tˆ 3 } dT = ( f r o m T1 t o 9 7 8 . 1 5 ) i n t g r { 4 8 7 . 3 8 + 1 . 1 9∗ 1 0 ˆ − 3 ∗ T+198.16 ∗ 10ˆ − 6 ∗ Tˆ 2 − 68.21 ∗ 10ˆ − 9 ∗ Tˆ 3 } dT 17 // we g e t 18 T 1 = 9 29 .7 2 // K 19 20 21 22 23 24 25 26 27 28 29 30 31

T o = 2 98 .1 5 H 1 = 4 93 40 5 / / kJ E Br = .35 S t y r en e p = E Br * . 9 B e nz en e b = E Br * . 03 E t h y l en e b = B e n ze n e b C b = E Br * .0 1 T o ul en e d = E Br * . 06 H r1 = 1 47 .3 6 - 2 9. 9 2 / / k J / m ol EB H r 2 = 8 2 . 93 + 5 2. 5 - 2 9 . 9 2 H r 3 = - 29 .9 2 H r 4 = 5 0 .1 7 - 7 4 . 52 - 1 4 7. 3 6 / / kJ / mol s t y r e n e d Hr = 1 0 00 * ( H r 1 * ( S t yr e n ep + T o u le n e d ) + H r2 * B e n z e n eb + H r 3 *Cb+Hr4*Toulened) H2 = H1 - dHr

32 33 / / H2 = ( f ro m To t 0 T2 ) i n t g r { Comp2dT 34 // we g e t 35 T 2 = 7 98 .7 9 // K

119

36 printf ( ” A di ab at ic r e a c t i o n T a t t h e o u t l e t o f t h e r e a c t o r i s ” + string ( T 2 ) + ” K . ” )

Scilab code Exa 5.42 Heat of crystallization

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

clear ; clc ;


/ / E xa mp le 5 . 4 2 / / P ag e 2 97 printf ( ” E xa mp le 5 . 4 2 , P ag e 2 9 7 \ n \ n ” ) ; // s o l u t i o n H so l = 6 2. 86 // kJ / mol s o l u t e M c ry st a l = 2 8 6. 1 41 4 H c r y = H s ol * 1 00 0 / M c r y st a l // kJ / kg s o l u t e printf ( ” H eat o f c r y s t a l l i z a t i o n o f 1 k g c r y s t a l ” + string ( H c r y ) + ” k J . ” )

Scilab code Exa 5.43 Heat of crystallization

1 2 3 4 5 6

clear ; clc ;


is

7 8 9 10 11 12 13 14 15 16 17

/ / E xa mp le 5 . 4 3 / / P ag e 2 97 printf ( ” E xa mp le 5 . 4 3 , P ag e 2 9 7 \ n \ n ” ) ; // s o l u t i o n o f H2O H f = - 28 5. 82 // kJ / mol H c r y s t = - 4 32 7 .2 6 - ( - 1 38 7 .0 8 +1 0 * H f ) printf ( ” Heat o f c r y s t a l l i z a t i o n = ” + string ( H c r y s t ) + ” kJ/mol . ”)

Scilab code Exa 5.44 Heat of sol of Boric acid

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

clear ; clc ;



H f s = - 10 94 .3 3 H f ao = - 10 72 .3 2 H s ol = H fa o - H fs printf ( ” Heat o f s o l u t i o n H s o l ) +” k J / m o l . ” )

o f B or ic a c i d = ” + string (

121

Scilab code Exa 5.45 Heat of dissolution

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

clear ; clc ;


/ / E xa mp le 5 . 4 5 / / P ag e 2 97 printf ( ” E xa mp le 5 . 4 5 , P ag e 2 9 7 \ n \ n ” ) ; // s o l u t i o n // ( a ) H f = - 98 2. 8 H f c r ys t = - 1 05 3 .9 0 4 H d is = H f cr y st - H f

// (b) H f cr = - 30 77 .7 5 H s o l = H f c r y s t + 7 * ( - 2 8 5 . 8 3) - ( - 3 0 7 7 . 75 ) printf ( ” ( a ) \ n \ n H d i s s o l u l i t i o n = ” + string ( H d i s ) + ” kJ/mol ZnSO4 . \ n \ n \ n ( b ) \ n \ n H s ol u ti o n = ” + string ( H s o l ) + ” k J / k m o l . ” )

Scilab code Exa 5.46 T change in dissolution

1 clear ; 2 clc ; 3

122

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


/ / E xa mp le 5 . 4 6 / / P ag e 3 00 printf ( ” E xa mp le 5 . 4 6 , P ag e 3 0 0 \ n \ n ” ) ; // s o l u t i o n / / u s i n g c ha r t 5 . 1 6 we g et T = 32 9. 5 // K printf ( ” T = ” + string ( T ) + ” K . ” )

Scilab code Exa 5.47 Using std heat of formations

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

clear ; clc ;


/ / E xa mp le 5 . 4 7 / / P ag e 3 00 printf ( ” E xa mp le 5 . 4 7 , P ag e 3 0 0 \ n \ n ” ) ; // s o l u t i o n / / b a s i s 1 00 (m1 ) kg 46% s o l N aOH = 46 / / kg H 2O = 54 / / kg m 2 = N aO H /. 25 N aO Ho = 25 // kg

123

19 20 21 22 23 24 25 26 27 28 29 30 31

H 2Oo = 75 / / kg H f 1 = - 45 3. 13 8 / / kJ / m ol H f 2 = - 46 7. 67 8 / / kJ / m ol H s = Hf2 - H f1 H g = - Hs * 1 0 0 0* 1 .5 0 1

/ / u s i n g A pp en di x IV . 1 H w1 = 1 46 .6 5 H w 2 = 1 04 .9 H ad d = 8 4* ( Hw 1 - H w2 ) H = Hg + Ha dd C 1 = 3. 55 T 2 = 2 9 8. 1 5+ H / ( 1 84 * C 1 ) // K printf ( ” F i na l s o l T = ” + string ( T 2 ) + ” K . ” )

Scilab code Exa 5.48 Heat effect of the solution

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

clear ; clc ;


/ / E xa mp le 5 . 4 8 / / P ag e 3 01 printf ( ” E xa mp le 5 . 4 8 , P ag e 3 0 1 \ n \ n ” ) ; // s o l u t i o n / / b a s i s 1 0 0 kg o f s o l w i t h 32% N M N H4 N O3 = 8 0 .0 4 34 M N H2 C ON O 2 = 6 0 .0 5 53 M N 2 = 2 8. 0 13 4 n a = 3 2 /( 6 0. 9 51 6 ) U r e a d is = 1 . 1 75 8 * n a * M N H 2 CO N O 2

124

/ / kg

20 21 22 23 24 25

w a t e r = 1 00 - ( n a * M N H 4 NO 3 + U r e a d is ) n d is = 5 25 m = n di s / wa te r H E1 = 4 0 . 30 4 4 - 2 . 5 9 62 * m + . 1 5 8 2* m ^ 2 - 3 .4 7 82 * 10 ^ - 3 * m ^ 3 H E = H E1 * n di s printf ( ” Heat e f f e c t o f t h e s o l = ” + string ( H E ) + ” kJ . ” )

Scilab code Exa 5.49 Integral heats of solution

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

clear ; clc ;


/ / E xa mp le 5 . 4 9 / / P ag e 3 02 printf ( ” E xa mp le 5 . 4 9 , P ag e 3 0 2 \ n \ n ” ) ; // s o l u t i o n H m ix = 8 96 M 1 = 88 // m ol ar mass o f n−amyl a l c o h o l M 2 = 78 // m ol ar mass o f b en z en e B = .4 73 * M2 A = .5 27 * M1 H a = Hm ix / A H b = Hm ix / B printf ( ” I n t e g r a l h e a t o f s o l o f n−amyl a l c o h o l = ” + string ( H a ) + ” k J / k g n−amyl a l c o h o l and o f b en ze ne = ” + string ( H b ) + ” k J / kg b e n z e n e . ” )

125

Scilab code Exa 5.50 Hx for H2SO4

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

clear ; clc ;


/ / E xa mp le 5 . 5 0 / / P ag e 3 02 printf ( ” E xa mp le 5 . 5 0 , P ag e 3 0 2 \ n \ n ” ) ; // s o l u t i o n // f ro m f i g 5 . 1 8 T a = 3 79 .5 // K d H = - 27 4 - ( - 10 6. 5) // kJ / kg s o l C m = 2. 05 / / k J/ kg K d Hc = C m *( Ta - 2 98 .1 5 )

// b a s i s 1 0 0 kg o f 93 % a ci d / / a c id b a l a nc e x = poly (0 , ’ x ’ ) p = . 9 3 *1 0 0 + x * . 15 - ( 1 0 0 + x ) * .7 7 y = roots ( p )

// f rom f i g y 1 = 25 .3 printf ( ” ( a ) \ n \ n R es u l ta n t T o f 77 p e r c e n t s o l = ” + string ( T a ) + ” K . \ n \ n \ n ( b ) \ n \ n Heat t o be removed t o c o o l i t t o 2 9 8 .1 5 K = ” + string ( d H ) + ”

kJ / kg s o l \ n \ n \ n ( c ) \ n \ n By mean h e a t c a p a c i ty method : ” + string ( d H c ) + ” kJ / kg s o l \ n \ n \ n ( d ) \ n \ n Q u a n t i t y o f 15 p e r ce n t a c i d t o be m i x e d = ” + string ( y ) + ” k g . \ n \ n \ n ( e ) \ n \ n f ro m 126

f i g : ” + string ( y 1 ) + ” k g . ” )

Scilab code Exa 5.51 Using heat of formations of H2SO4

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

clear ; clc ;


/ / E xa mp le 5 . 5 1 / / P ag e 3 04 printf ( ” E xa mp le 5 . 5 1 , P ag e 3 0 4 \ n \ n ” ) ; // s o l u t i o n / / b a s i s 1 0 0 kg o f 93% a ci d and 2 5 . 8 kg o f 15% a ci d H fp = -8 14 H f1 = -8 30 H E1 = Hf1 - H fp H f 2 = - 88 6. 2 H E2 = Hf2 - H fp H f3 = -8 51 H E3 = Hf3 - H fp H s o l = . 9 8 7 6 * 1 0 0 0* ( - 3 7 ) - [ . 9 4 8 2* 1 0 0 0 * ( - 1 6 ) +.0394*1000*(-72.2)] H ev = 1 00 *( 30 - 2 5 ) * 1. 6 H co n = 2 5 .8 * 25 * 3. 7 n e t H ev = - H so l - H c on + H e v T = 2 9 8 .1 5 + n e t H ev / ( 1 2 5. 8 * 2 .0 5 ) printf ( ” Temp o f s o l = ” + string ( T ) + ” K . ” )

127

Scilab code Exa 5.52 Heat to be removed for cooling it to 308K

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

clear ; clc ;


/ / E xa mp le 5 . 5 2 / / P ag e 3 06 printf ( ” E xa mp le 5 . 5 2 , P ag e 3 0 6 \ n \ n ” ) ; // s o l u t i o n / / b a s i s 1 0 0 0 kg o f mixed a c id

C11 = 2 .4 5 H 1 = - 29 6 .7 + C 1 1 * ( 3 0 8. 1 5 - 2 7 3 . 15 ) C 12 = 2.2 H 2 = - 87 . 8+ C 1 2 * ( 3 0 8 .1 5 - 2 7 3 . 1 5) C13 = 1 .4 5 H 3 = - 35 . 5+ C 1 3 * ( 3 0 8 .1 5 - 2 7 3 . 1 5) C 14 = 1.8 H 4 = - 14 8 .9 + C 1 4 * ( 3 0 8. 1 5 - 2 7 3 . 15 ) H m ix = 1 0 00 * H 4 - [ 7 6 .3 * H 1 + 3 4 5 . 9* H 2 + 5 7 7 . 7 * H 3 ] printf ( ” Heat o f m ix in g = ” + string ( H m i x ) + ” kJ . ” )

Scilab code Exa 5.53 Heat changes in formation of MNB

1 2 3 4 5

clear ; clc ;


6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29

/ / E ne rg y B a l a n c e s

/ / E xa mp le 5 . 5 3 / / P ag e 3 08 printf ( ” E xa mp le 5 . 5 3 , P ag e 3 0 8 \ n \ n ” ) ; // s o l u t i o n F = 1135 B e nz en e f = 4 0 0* . 99 3 H N O 3 c on = B e n ze n e f * 6 3 /7 8 H 1 = - 18 6. 5 C11 = 1 .8 8 H 1 1 = H 1 + C 11 * ( 2 9 8 . 15 - 2 7 3. 1 5 ) H 2 = - 28 8. 9 C12 = 1 .9 6 H 2 2 = H 2 + C 12 * ( 2 9 8 . 15 - 2 7 3. 1 5 ) H3 = 0 C13 = 1 .9 8 H 3 3 = C 13 * ( 2 9 8 . 15 - 2 7 3. 1 5 ) H r = - 2 8 5. 8 3 +1 2 . 5 - ( - 1 7 4 . 1 +4 9 . 08 ) B e n z e ne r = B e n ze n e f / 7 8 . 11 1 8 f i = 9 0 3 .8 4 * H 2 2 + H N O 3c o n * H 33 - F * H 1 1 + B e nz e n er * H r * 1 0 0 0

/ / kJ / h 30 printf ( ” T ot al h ea t e xc ha ng ed = ” + string ( f i ) + ” k J / h . ”)

Scilab code Exa 5.54 Final T of solution in absorption of NH3

1 2 3 4 5

clear ; clc ;


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

/ / E ne rg y B a l a n c e s

/ / E xa mp le 5 . 5 4 / / P ag e 3 11 printf ( ” E xa mp le 5 . 5 4 , P ag e 3 1 1 \ n \ n ” ) ; // s o l u t i o n / / f ro m r e f 24

H = 1 60 0. 83 T o = 2 73 .1 5 h = 200 H f1 = - 79 .3 // t a b l e 5 . 5 9 H f 2 = - 46 .1 1 H s ol = Hf 1 - H f2 H g = H s ol * 1 0 0 0* 1 4 0 / 17 . 0 3 05 R a q = 1 40 / .1 5 / / kg / h d T = H g / (4 . 14 5 * R aq ) T = - dT + 303 printf ( ” Temp o f r e s u l t a n t s o l = ” + string ( T ) + ” K . ” )

Scilab code Exa 5.55 Using table 5 60

1 2 3 4 5 6 7 8 9 10 11

clear ; clc ;


/ / E xa mp le 5 . 5 5 / / P ag e 3 11 printf ( ” E xa mp le 5 . 5 5 , P ag e 3 1 1 \ n \ n ” ) ; 130

12 13 14 15 16 17 18 19

// s o l u t i o n

H f1 = - 80 .1 4 H f2 = - 46 .1 1 H s ol = Hf 1 - H f2 H g = H s ol * 1 0 0 0* 2 / 1 7. 0 3 0 5 printf ( ” Heat g e n er a t ed f o r making 2 p e r ce n t s o l u t i o n = ” + string ( H g ) + ” kJ / 10 0 kg s o l . ” )

Scilab code Exa 5.56 Heat removed in cooler

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

clear ; clc ;


/ / E xa mp le 5 . 5 6 / / P ag e 3 12 printf ( ” E xa mp le 5 . 5 6 , P ag e 3 1 2 \ n \ n ” ) ; // s o l u t i o n f i 3 = 1 5 50 5 40 7 f i 4 = 1 1 39 5 05 6 f i5 = fi3 - f i4 / / kJ / h f i 6 = 1 1 1 . 3 75 * 6 2 . 75 * 1 0 00 f i 7 = 1 06 3 37 9 f i 8 = 5 5 3 2 .1 5 * 4 . 18 6 8 * (3 0 3 . 15 - 2 9 8 .1 5 ) f i 9 = 9 0 3 0 .4 * 3 . 45 * ( 3 23 . 1 5 - 2 9 8 . 1 5 ) f i = f i5 + f i6 + f i8 - f i7 - f i9 printf ( ” Heat r em ov a l i n t he c o o l e r = ” + string ( f i ) + ”

131

kJ /h . ” )

Scilab code Exa 5.57 Hx vs x1

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

clear ; clc ;



T o = 2 73 .1 5 T 1 = 3 08 .1 5 H 1 = 1 2 4. 8 *( T 1 - To ) / / k J / k mo l H 2 = 1 3 4. 9 *( T 1 - To ) / / k J / k mo l H E 1 = . 1 * . 9 * [ 5 4 2 . 4 + 5 5 . 4 * ( . 9 - . 1 ) - 1 3 2 .8 * ( . 9 - . 1 ) ^ 2 - 1 6 8 . 9 * ( . 9 - . 1 ) ^ 3 ] / / kJ / kmo l o f mix H a = H E1 + H 1 * .1 + H 2 *. 9 H E 2 = . 2 * . 8 * [ 5 4 2 . 4 + 5 5 . 4 * ( . 8 - . 2 ) - 1 3 2 .8 * ( . 8 - . 2 ) ^ 2 - 1 6 8 . 9 * ( . 8 - . 2 ) ^ 3 ] / / kJ / kmo l o f mix H b = H E2 + H 1 * .2 + H 2 *. 8 H E 3 = . 3 * . 7 * [ 5 4 2 . 4 + 5 5 . 4 * ( . 7 - . 3 ) - 1 3 2 .8 * ( . 7 - . 3 ) ^ 2 - 1 6 8 . 9 * ( . 7 - . 3 ) ^ 3 ] / / kJ / kmo l o f mix H c = H E3 + H 1 * .3 + H 2 *. 7 H E 4 = . 4 * . 6 * [ 5 4 2 . 4 + 5 5 . 4 * ( . 6 - . 4 ) - 1 3 2 .8 * ( . 6 - . 4 ) ^ 2 - 1 6 8 . 9 * ( . 6 - . 4 ) ^ 3 ] / / kJ / kmo l o f mix H d = H E4 + H 1 * .4 + H 2 *. 6 H E 5 = . 5 * . 5 * [ 5 4 2 . 4 + 5 5 . 4 * ( . 5 - . 5 ) - 1 3 2 .8 * ( . 5 - . 5 )

132

28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46

^ 2 - 1 6 8 . 9 * ( . 5 - . 5 ) ^ 3 ] / / kJ / kmo l o f mix H e = H E5 + H 1 * .5 + H 2 *. 5 H E 6 = . 6 * . 4 * [ 5 4 2 . 4 + 5 5 . 4 * ( . 4 - . 6 ) - 1 3 2 .8 * ( . 4 - . 6 ) ^ 2 - 1 6 8 . 9 * ( . 4 - . 6 ) ^ 3 ] / / kJ / kmo l o f mix H f = H E6 + H 1 * .6 + H 2 *. 4 H E 7 = . 7 * . 3 * [ 5 4 2 . 4 + 5 5 . 4 * ( . 3 - . 7 ) - 1 3 2 .8 * ( . 3 - . 7 ) ^ 2 - 1 6 8 . 9 * ( . 3 - . 7 ) ^ 3 ] / / kJ / kmo l o f mix H g = H E7 + H 1 * .7 + H 2 *. 3 H E 8 = . 8 * . 2 * [ 5 4 2 . 4 + 5 5 . 4 * ( . 2 - . 8 ) - 1 3 2 .8 * ( . 2 - . 8 ) ^ 2 - 1 6 8 . 9 * ( . 2 - . 8 ) ^ 3 ] / / kJ / kmo l o f mix H h = H E8 + H 1 * .8 + H 2 *. 2 H E 9 = . 9 * . 1 * [ 5 4 2 . 4 + 5 5 . 4 * ( . 1 - . 9 ) - 1 3 2 .8 * ( . 1 - . 9 ) ^ 2 - 1 6 8 . 9 * ( . 1 - . 9 ) ^ 3 ] / / kJ / kmo l o f mix H i = H E9 + H 1 * .9 + H 2 *. 1 H E 1 0 = . 0 * 1 . * [ 5 4 2 . 4 + 5 5 . 4 * ( .0 - 1 . ) - 1 3 2 .8 * ( . 0 - 1 . ) ^ 2 - 1 6 8 . 9 * ( . 0 - 1 . ) ^ 3 ] / / kJ / kmo l o f mix H j = H E1 0 + H1 + H 2 *0 x = linspace ( 0 , 1 , 1 0 0 ) y = linspace ( 4 3 0 0 , 5 0 0 0 , 1 0 0 ) y = 4 7 21 . 5 - 5 7 . 4* x + 1 1 3 7 . 7* x ^ 2 - 3 99 3 .6 * x ^ 3 + 3 9 0 9. 2 * x ^4-1351.2*x^5 plot ( x , y ) t i t l e ( ”H v s x1 ” ) xlabel(”x1”) y l a b e l ( ”H ( k J / kg s o l . ) ” ) printf ( ” E nt ha lp y , kJ / kmol

mi x \ n

x1

HE

H \n 0 0 ” + string ( H 2 ) + ” \ n 0.1 ” + string ( H E 1 ) + ” ” + string ( Ha )+” \n 0.2 ” + string ( H E 2 ) + ” ”+ 0.3 ” + string ( H E 3 ) + ” string ( H b ) + ” \ n ” + string ( H c ) + ” \ n 0 . 4 ”+ ” + string ( H d ) + ” \ n 0 . 5 string ( H E 4 ) + ” ” + string ( H E 5 ) + ” ” + string ( He 0.6 ” + string ( H E 6 ) + ” ”+ )+” \n 0.7 ” + string ( H E 7 ) + ” string ( H f ) + ” \ n ” + string ( H g ) + ” \ n 0 . 8 ” + string ” + string ( H h ) + ” \ n 0 . 9 (HE8)+” 133

” + string ( H E 9 ) + ” ” + string ( H i ) + ” \ n 1 . 0 ” + string ( H E 1 0 ) + ” ”+ string ( H j ) + ” ” )

Scilab code Exa 5.58 repat of 5 57 using heat capacities

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

clear ; clc ;


/ / E xa mp le 5 . 5 8 / / P ag e 3 16 printf ( ” E xa mp le 5 . 5 8 , P ag e 3 1 6 \ n \ n ” ) ; // s o l u t i o n / / s ee e g 5 . 5 7 printf ( ” r e f e r t o eg 5 . 5 7 ” )

Scilab code Exa 5.59 He vs x1 of acetone and ethylacetate

1 2 3 4 5 6 7 8

clear ; clc ;


134

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

/ / E xa mp le 5 . 5 9 / / P ag e 3 18 printf ( ” E xa mp le 5 . 5 9 , P ag e 3 1 8 \ n \ n ” ) ; // s o l u t i o n / / fro m g ra ph drawn i n 5 . 5 7 we ca n s e e

H 1 E1 = 3 00 H 1E2 = 63 H 2E1 = 30 H 2 E2 = 2 14 printf ( ” H1 a t x1 =0 .3 i s ” + string ( H 1 E 1 ) + ” kJ /kg s o l \ n H2 a t x1 =0.3 i s ” + string ( H 2 E 1 ) + ” kJ / kg s o l \ n H1 a t x1 =0.6 i s ” + string ( H 1 E 2 ) + ” kJ /kg s o l \ n H2 a t x 1 =0 .6 i s ” + string ( H 2 E 2 ) + ” kJ / kg s o l . ” )

Scilab code Exa 5.60 Heat of dilution

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

clear ; clc ;


/ / E xa mp le 5 . 6 0 / / P ag e 3 20 printf ( ” E xa mp le 5 . 6 0 , P ag e 3 2 0 \ n \ n ” ) ; // s o l u t i o n / / b a s i s 1 00 kg 9 6 . 1% H2SO4 / / fro m t a b l e 5 . 6 4 m 1S O3 = 7 8. 4 // kg 135

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

m 1H 2O = 2 1. 6 n 1S O3 = m 1S O3 / 8 0 .0 63 n 1H 2O = m 1H 2O / 1 8 .0 15

/ / r e s u l t a n t s o l h as 2 3 . 2% H2SO4

m 2S O3 = 19 m 2H 2O = 81 M r s o l = m 1 SO 3 * 1 0 0 / m 2 SO 3 M w = M rs ol - 1 00 w = M rs ol - m 1 SO 3 / 1 8. 0 15 / / km ol H E o so l = n 1 SO 3 * ( - 5 69 4 0) + n 1 H 2 O * ( - 3 2 6 57 ) / / kJ H E r s o l = n 1 SO 3 * ( - 1 56 1 68 ) + w - ( - 3 3 5) H E = H Er so l - H E o so l // kJ /kg o r i g i n a l a c i d C = 3.43 / / kJ / kg K d T = - HE / ( M rs ol * C ) T = 2 91 .1 5+ d T // K printf ( ” Heat o f d i l u t i o n = ” + string ( H E ) + ” kJ / kg

o r i g i n a l s o l u t i o n \ n \ n F in al T o f r e s u l t a n t s o l u t i o n = ” + string ( T ) + ” K . ” )

Scilab code Exa 5.61 eg 5 60 with use of ice at 273K

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

clear ; clc ;



136

15 16 17 18 19

/ / b a s i s 1 0 0 k g o f o r i g i n a l a ci d l v = 3 33 .7 / / kJ / kg H = - lv - 1 8* 4 .1 8 68 H E = ( - 6 42 77 - H * 3 1 2. 6 3 ) / 1 00 / / kJ / kg printf ( ” Heat o f d i l u t i o n = ” + string ( H E ) + ” k J / k g . ” )

137

Chapter 6 Stoichiometry and Unit Operations

Scilab code Exa 6.1 Overall material and energy balance

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

clear ; clc ;

/ / S t o i ch i o m et r y / / C ha pt er 6 / / S t o i c h i o m e t r y and U ni t O p e ra t i o ns

/ / E xa mp le 6 . 1 / / P ag e 3 46 printf ( ” E xa mp le 6 . 1 , P ag e 3 4 6 \ n \ n ” ) ; // s o l u t i o n / / b a s i s = 1 00 kmol o f f e ed B e nz e ne = 1 0 0* . 72 / / k mo l T o ul e ne = 1 00 - B e n ze n e //kmol // u s e f i g 6 . 1 // D = d i s t i l l a t e , B = b ottom 138

20 21 22 23 24 25 26 27 28

// F = B + D

(i)

o v e r a l l m at e r i a l b a l a n c e

x d = .9 95 x b = .03 x f = .72

/ / xd ∗D + x b ∗B = F ∗ x f ( ii ) benzene bala nce // s o l v i n g ( i ) and ( i i ) D = 71.5 //kmol B = 28.5 //kmol printf ( ” ( a ) \ n \ n p er fo rm in g o v e r a l l m a t e r i a l b a l a nc e f o r 10 0 kmol o f f e ed we g et ” + string ( D ) + ” k m ol a s d i s t i l l a t e a nd ” + string ( B ) + ” km l a s b ot to m pr o du c t . \n \n \n ( b ) \n \n ”) 29 / / e n th a lp y b a l an c e 30 // u s e f i g 6 . 2 31 32 33 34 35 36 37 38 39 40 41 42 43

R = 1.95 v = D *( 1+ R ) // kmol T o = 2 73 .1 5 //K

t o t a l ov er h ea d v a po ur s

/ / u si ng f i g 6 .2 E v = 4 21 70 // kJ / kmol e nt ha l p y o f v ap ou rs ov er he ad e n th al py o f l i q u i d E l = 1 13 70 // kJ / kmol E 1 = Ev - El // e n t ha l p y r em ov ed i n c o n d en s e r H c = E1 *v // h ea t l o a d o f c o n d e ns er

H d = El * 71 .5 H b = 1 8 78 0 *2 8 .5 H f = 4 4 50 0 *1 0 0 H n = H d + Hc + Hb - H f // kJ heat lo a d o f r e b o i l e r printf ( ” p e r fo r m i ng o v e r a l l e nt ha l p y b a l a nc e we g et Heat l o a d o f c o n d e ns e r = ” + string ( H c ) + ” k J / k m o l and Heat l oa d o f r e b o i l e r = ” + string ( H n ) + ” k J / k m o l .”)

Scilab code Exa 6.2 Cryogenic Separation of Nitrogen

1 clear ; 2 clc ;

139

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


/ / E xa mp le 6 . 2 / / P ag e 3 49 printf ( ” E xa mp le 6 . 2 , P ag e 3 4 9 \ n \ n ” ) ; // s o l u t i o n / / b a s i s = 2 0 0 0 kg /h l i q u i d f e e d r a t e F = 2 00 0/ 28 .8 4 / / k m o l / h //D = d i s t i l l a t e , W = r e s i d u e f lo w r a t e / / N2 b a l a n c e // F ∗ . 7 9 = . 9 9 9D + . 4 2 2W (i) / / 5 4 . 8 4 0 = D + . 4 2 2 4W ( ii ) // s ol v i ng i t W = 2 5. 11 8 / / k m o l / h D = 4 4. 23 0 / / k m o l / h // u s i n g f i g 6 . 4 and 6 . 5 // t r i a l method i s u se d f o r f l a s h c a l c u l a t i o n s // T ri al I x = .75

/ / fro m f i g 6 . 4 y = .8 83 3

/ / fro m f i g 6 . 5 H l = 1 08 3. 65 H v = 6 07 1. 7 H f = . 3* H v + Hv * . 7

// c a l c u l a t i n g we g et Emix i s n o t c l o s e t o 2 5 9 2 . 2 kJ / kmol 35 // T r i a l I I 36 37 38 39

x = .71 y = .859 H l = 1 08 5. 6 H v = 6 11 8. 6

140

40 H f = . 3* 3 * H v + .7 . 7 * Hl Hl / / k J / k m o l 41 / / w h hii ch c h i s a pr p r oo o o x e q u a l t o 2 5 9 5 . 2 k J / km km o l , s o

f l a s h i n g w i l l o cc cc u r 42 printf ( ” c o m p o si s i t i o n o f v a p ou ou r l i q u i d mi x : \ n m o l f r a c t i o n N2 = ” + string ( x ) + ” i n l i q u i d p h a s e a n d ” + string ( y ) + ” i n v ap a p o ur u r p h as as e . ” )

Scilab code Exa 6.3 Azeotropic distillation of IPA and water

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

clear ; clc ;

/ / S t o i ch c h i o m et et r y // Ch haa pt p t er er 6 / / S t o i c h i o m e t r y a n d U ni n i t O p e ra r a t i o ns ns

/ / E xa xa mp mp le le 6 . 3 / / P aagg e 3 53 53 xa mp mp le le 6 . 3 , P ag ag e 3 5 3 \ n \ n ” ) ; printf ( ” E xa // s o l u t i o n / / m a t e r i a l b a l a nc nc e // V V22 v ap a p ou o u r m i x i s a t e rn r n a r y a z eo eo t ro p e i n which a l l c y c lo l o h ex e x a n e o f D1 D1 i s r e c y c l e d / / V2 V 2 s t re re a m / / C y c l o he h e x an an e b a l a n c e / / D1 = ( . 4 8 8 / . 0 2 4 ) ∗ V2 / / IPA i n V2 = . 2 0 6 V2 V2 / / w at a t er e r i n V2 = ( 1 − . 4 8 8 − . 2 0 6 ) ∗ V2 / / W2 W2 s t r e a m / / IPA i n W2 = ( . 2 3 D1 D 1 − .206V2) / / w at a t er e r i n W2 = ( 1 − . 0 2 4 − . 2 3 ) ∗ D1 − .306V2 / / W2 W2 s t re r e a m = 4 . 4 7 1 V2 V 2 + 1 4 . 8 6 2 V2 V2 141

26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47

// // // // // //

D33 i s a n a z e o tr D t r o p e c o n t a i n i n g 6 7 . 5 mol% IP IP A w at a t er e r i n W3 W3 s t re r e am a m = ( 1 − . 6 7 5 ) F b a s i s = 10 1 0 0 k m o l /h / h f r e s h f e ed ed W1+W3 = 1 0 0 (i) . 9 9 8 W1 + . 0 0 1W 1W3 = 6 7 . 5 ( ii ) s ol vi ng i t W 1 = 6 7. 7 . 60 60 3 / / k m o l / h W 3 = 3 2. 2 . 39 39 7 / / k m o l / h I P A1 A 1 = W 3 * .0 . 0 01 0 1 / / IPA i n W3 / / I P A 2 = 4 . 4 7 1 ∗ V2 − . 0 3 2 IPA in i n D3 / /C−1 = F+D3 = F1 / / w at a t er e r i n D3 = 6 . 6 2 4 V2 V 2 − . 0 4 7 − 4 . 4 7 1 V 2 + . 0 3 2 / / w a te t e r i n W3 = 1 4 . 8 6 2 V2 V 2 − 2.153V2+.015 / / s o l v i n g them V 2 = 2 .6 . 6 24 24 / / k m o l / h D 3 = 2 .1 . 1 5 3* 3 * V 2 - . 01 01 5 D 1 = 2 0. 0 . 33 3 3 3* 3* V 2 F 1 = 6 .6 . 6 2 4* 4 * V 2 + 9 9. 9 . 9 53 53 R = 1 . 75 7 5 * D1 D1 // = V1 V1+ +V2−D1 V 1 = 1 44 4 4 .1 .1 r = D 3 / 10 1 0 0 // // r e c y c l e r a t i o e r p e r fo f o r m i ng n g o v e r a l l m a t e r i a l b a l a nc n c e we we printf ( ” A f t er g e t R ef e f lu l u x , R = ” + string ( R ) + ” k m ol ol / h a nd nd \ n r e c y c l e r a t i o = ” + string ( r ) + ” k m o l / km km ol f r e s h fe ed . ”)

Scilab code Exa 6.4 CO2 absorption in aq MEA solution

1 2 3 4 5 6 7

clear ; clc ;

/ / S t o i ch c h i o m et et r y // Ch haa pt p t er er 6 / / S t o i c h i o m e t r y a n d U ni n i t O p e ra r a t i o ns ns

142

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

/ / E xa xa mp mp le le 6 . 4 / / P aagg e 3 55 55 printf ( ” E xa xa mp mp le le 6 . 4 , P ag ag e 3 5 5 \ n \ n ” ) ; // s o l u t i o n

/ / b a s i s 0 . 62 6 2 5 l / s o f ME MEA s o l u t i o n / /M c o n c o f ME MEA c = 3 . 2 // M = 6 1 / / m o l ar a r m as a s s o f MEA // g / l c on o n c o f ME MEA i n s o l C = M * c // mo l / h M E A i n = c * . 6 2 5 * 3 60 6 0 0 / 1 00 0 0 0 / / k mo C O 2d 2 d i ss s s = . 16 1 6 6 *7 * 7 . 2 / / km k m o l /h /h CO2 d i s s o l v e d i n l e a n MEA 6 . 10 1 0 7 // 21 v = 2 6. / /mˆ 3 s p . v o l o f g as a s a t 3 1 8K an an d 1 0 1. 1. 3 k Pa ( t a b l e 7 . 8 ) 0 0 0/ 0/ v / / k m o l / h 22 q v = 1 00 23 C O 2 in / / m o l e s o f CO CO2 i n i n l e t g a s i n = q v * .1 . 1 04 0 4 // 24 25 26 27 28 29 30

C O2 O 2 fr f r ee e e ga g a s = q v - C O2 O 2 in in

/ / o u t g o i n g h a s 4 . 5 % CO2 G A S o u t = C O 2 f re re e g a s / ( 1 - . 0 45 45 5 ) / / k m o l / h C O2 O 2 a bs b s = q v - G A S ou ou t C O 2 = C O2 O 2 di d i ss s s + C O2 O 2 ab ab s C O 2c 2 c o nc n c = C O2 O 2 / M E Ai A i n //kmol/kmol MEA e n t ra r a t io i o n o f d i s s o l v e d CO2 i n t he he printf ( ” C o n c en s o l u t i o n l e a v i n g t h e t o w e r = ” + string ( C O 2 c o n c ) + ” kmol/ kmol o f MEA. ” )

Scilab code Exa 6.5 Heat effect of Scrubbing

1 2 3 4 5

clear ; clc ;

/ / S t o i ch c h i o m et et r y // Ch haa pt p t er er 6 143

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

/ / S t o i c h i o m e t r y and U ni t O p e ra t i o ns

/ / E xa mp le 6 . 5 / / P ag e 3 56 printf ( ” E xa mp le 6 . 5 , P ag e 3 5 6 \ n \ n ” ) ; // s o l u t i o n //(a) printf ( ” ( a ) \ n \ n ” )

// b a s i s 5 00 00 mˆ 3/ h o f g as mix a t 2 9 5 . 5K 10 0kPa v = 24 .5 7 //mˆ 3/ k mol s p v o l o f g a s a t 2 9 5 . 5K and 1 00 kPa 19 n 1 = 5 00 00 / v // kmol /h f lo w o f i n c o m i n g g a s 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42

N O2 in = n 1 * .0 5 46 N 2O 4 in = n 1 * .0 21 4 N 2i n = n1 - N O 2i n - N 2 O 4i n

/ /N2 i s u n a f f e c t e d o ut go i n g g as f l o w n 2 = 1 8 80 . 34 / .9 5 // kmol /h / / u s i n g t a b l e s 6 . 3 and 6 . 4 on p a g e 3 5 7 N O 2 re m = N O2 in - ( n 2 * .0 39 3 ) N 2 O 4r e m = N 2O 4 in - ( n 2 * .0 08 2 )

/ / r xn ( i i ) N a O H r ea c 2 N a N O 2 pr o 2 N a N O 3 pr o 2 H 2 Op r o2 =

= 2 * 40 * N 2 O 4 r em = 6 9* N 2 O 4 r e m = 8 5* N 2 O 4 r e m 1 8* N 2 O 4r em

// rxn ( i i i )

N O 2r ea c 3 = 3 * n2 * . 0 02 5 N a O H r ea c 3 = 2 * 4 .9 5 * 40 N a N O 3 pr o 3 = 2 * 4 .9 5 * 85 H 2 Op r o3 = 4 . 95 * 18 N O 2 a b s2 = 3 3 .3 3 - N O 2 re a c 3 N a O H r ea c 1 = 1 8 . 48 * 4 0 N a N O 2 pr o 1 = 6 9* N O 2 a b s 2 / 2 N a N O 3 pr o 1 = 8 5* N O 2 a b s 2 / 2 H 2 Op r o1 = 1 8* N O 2 ab s2 / 2

144

43 44 45 46 47 48 49 50 51 52

N a N O2 t = N a NO 2 pr o 2 + N a NO 2 pr o 1 N a N O 3 t = N a N O3 p r o2 + N a N O3 p r o3 H 2 O t = H 2 O pr o 1 + H 2 O p ro 2 + H 2 O p r o3 N a OH t = N a O Hr e a c1 + N a O Hr e a c2 + N a O Hr e a c3 l i q = 3 75 00 / / k g / h N aO H in = l iq * . 23 6 N a O H o ut = N a OH in - N a O Ht m oi st = n 2 * .0 4 5* 1 8 w a t e r = l iq - N aO H in - H 2O t - m o is t / / k g / h printf ( ” C o m p o si t i o n o f f i n a l l i q u o r : \ n Component mi ( k g /h ) \ n NaOH ”+ ”+ string ( N a O H o u t ) + ” \ n NaNO2 string ( N a N O 2 t ) + ” \ n NaNO3 ”+ ”+ string ( N a N O 3 t ) + ” \ n H2O string ( w a t e r ) + ” \ n \ n \ n ( b ) ” )

//(b) / / h ea t e f f e c t o f s c r u bb i ng / / u s in g t a b l e s 6 . 6 and 6 . 7 // f i 1 = i n t e g { 5 9 8 6 5 . 7 + 4 5 4 5 . 8 + 1 0 ˆ − 3 ∗T + 1 5 2 6 6 . 3 ∗ 1 0 ˆ − 6 ∗ Tˆ 2 − 705.11 ∗ 10ˆ − 9 ∗ Tˆ 3 } 57 f i 1 = - 1 5 59 4 1 .3 / 3 60 0 //kW 58 // s i m i l a r l y 59 f i2 = 7 5. 77 8 //kW 53 54 55 56

60 d H1 = ( - 3 4 6 . 30 3 - 4 5 0 . 1 - 2 8 5. 8 3 - ( 2 * ( - 4 6 8. 2 57 ) + 2 * 3 3 . 18 ) ) /2 // kJ/mol NO2 61 d H2 = - 3 4 6. 3 03 - 4 50 . 1 - 2 8 5. 8 3 - ( 2 *( - 4 68 . 2 57 ) + 9 . 1 6) / / k J

/ mol N2O4 62 d H 3 = ( 2 * ( - 4 5 0 . 1 ) - 2 8 5 . 8 3 +9 0 . 2 5 - ( 2 * ( - 4 6 8 . 2 5 7) + 3 * 3 3 . 1 8 ) ) / 3 / / k J / m o l NO2 63 d H d i l = - 4 69 . 83 7 - ( - 4 68 . 25 7 ) // kJ/ mol NaOH 64 f i3 = ( d H 1 * 1 0 0 0 *1 8 . 4 8+ d H 2 * 1 0 0 0 * 2 7 .3 2 + d H 3 * 1 0 0 0 * 14 . 8 5 + d H d i l * 1 0 0 0 * 1 3 8 . 2 3 ) / 3 6 0 0 //kW 65 f i 4 = - f i1 + f i2 + f i3 66 printf ( ” Heat e f e e t o f s c r u bb i ng s ys te m = ” + string ( f i 4 ) + ” kW. ” )

145

Scilab code Exa 6.6 Extraction of Acetic Acid

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

clear ; clc ;


/ / E xa mp le 6 . 6 / / P ag e 3 61 printf ( ” E xa mp le 6 . 6 , P ag e 3 6 1 \ n \ n ” ) ; // s o l u t i o n //(a) / / b a s i s 1 0 0 kg f e e d mix // F = E +R = 10 0

(i)

x f = .4 75 x e = .82 x r = .14

// a c e t i c a c id b a la n ce / / x f ∗ F = xe ∗ E + x r ∗R ( ii ) // so lvi ng ( i ) & ( i i ) E = 49.2 / / k g R = 50.8 / / k g a = R * xr // kg a c et i c a ci d l e f to v e r b = ( a / ( x f * 1 00 ) ) * 1 0 0 printf ( ” A c e t ic a c id t ha t r e ma in ed u n ex t r a ct ed = ” + string ( b ) + ” p e r c e n t . ” )

146

Scilab code Exa 6.7 Multiple contact counter current Extractor

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

clear ; clc ;


/ / E xa mp le 6 . 7 / / P ag e 3 61 printf ( ” E xa mp le 6 . 7 , P ag e 3 6 1 \ n \ n ” ) ; // s o l u t i o n // r e f e r r i n g t o f i g 6 . 9 // b a s i s 1 0 0 0 kg /h h a l i b u t l i v e r s F = 1000 / / k g / h O I L in = F * . 25 7 S in = F - O IL in // s o l i d i n t h e c h ar g e U = . 23 * Sin O IL u = U * .1 28 Eu = U - OI Lu // e t h e r i n u nd er fl o w r e c o ve r y o f o i l R = OI Li n - O IL u // kg / h p = R * 10 0/ O I Li n O = R /.7 E o = O -R E t = Eu + Eo printf ( ” Flow r a t e o f e t h e r t o t h e s y s t e m = ” + string ( E t ) + ” kg / h \ n a nd p e r ce n t a g e o f r e c o v e r y o i l = ” + string ( p ) + ” . ” )

Scilab code Exa 6.8 Recovery of Acetic Acid by Ethyl Acetate Extraction

147

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 32 33 34 35 36 37 38

clear ; clc ;


/ / E xa mp le 6 . 8 / / P ag e 3 62 printf ( ” E xa mp le 6 . 8 , P ag e 3 6 2 \ n \ n ” ) ; // s o l u t i o n F = 1000 // kg / h

B a s i s f e e d r at e

// // // //

u si ng f i g 6 . 1 1 W/A = 1 5 . 7 7 / 5 . 8 7 A+F+W = 1 0 0 0 s ol vi ng i t W = 1 5. 77 * F / 2 1. 64 / / k g / h A = F -W / / k g / h / / m a t e r i a l b a l a nc e a c r o s s C3 // R+R1 = D+W / / W/D = 1 9 . 3 1 / 1 . 8 1 // s ol v i ng i t D = 1 .8 1* W / 1 9 .3 1 / / k g / h M 1 = D+ W

/ / R1/R = 4 . 6 3 / 6 . 5 7 R 1 = 4 . 63 * 79 3 /1 1 .2 R = M1 - R1

/ / m a t e r i a l b a l a nc e a c r o s s C2 m = .89 // = E1/R1 // E = A+E1+R1 = A+M11 / / M11/A = 1 5 . 6 / 3 . 9 7 M 1 1 = 1 5. 6* A / 3 .9 7 E = M11 + A E1 = M11 - R1

/ / m a t e r i a l b a l a nc e a c r o s s C1 148

39 40 41 42 43 44

// F+S = M = E+R M = E+R S = D + E1 A A l o s s = W * . 4 * 1 0 0 /( 1 0 0 *. 3 ) A Ar ec = 1 00 - A A l os s printf ( ” Summary : \ n S tr ea m

Flow r a t e ( k g /h ) \ n F ee d

D1 from D1 from C3

” + string ( F ) + ” \ n S o l ve n t ” + string ( S ) + ” \ n E x tr a ct ” + string ( E ) + ” \ n R a f f i n a t e ” + string ( R ) + ” \ n A ce t i c a c i d ” + string ( A ) + ” \ n Top l a y e r fro m ” + string ( E 1 ) + ” \ n Bottom l a y e r ” + string ( R 1 ) + ” \ n Feed t o C3 ” + string ( M 1 ) + ” \ n O v er h ea d ” + string ( D ) + ” \ n W ater w a s t e ” + string ( W ) + ” \ n S tr ea m ” + string ( M ) + ” ” )

Scilab code Exa 6.9 Yield of Glauber salt

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

clear ; clc ;



149

15 16 17 18 19 20

/ / b a s i s 1 0 0 kg f r e e w a t e r N a2 SO 4i n = 32 / / k g W in = 68 / / k g W 1 = ( 1 80 / 14 2 ) * 32 / / kg w at er w i th Na2SO4 W f r ee 1 = W in - W 1 G S 1 = ( ( N a 2 SO 4 i n + W 1 ) * 1 00 ) / W f r e e1 // kg

glauber

s a l t p r e s e n t i n 1 00 k g f r e e w a te r 21 W 2 = ( 1 80 * 19 . 4) / 1 42 // w at er a s s o c i a t e d w it h Na2SO4 i n f i n a l mo ther l i q u o r 22 23 24 25 26

W f r ee 2 = 1 00 - W 2 G S 2 = ( ( 1 9. 4 + W 2 ) / W f re e 2 ) * 1 00 Y = GS1 - G S2 / / k g p = Y * 10 0/ G S1 printf ( ” P e r c e n t y i e l d o f g l a u b er ” . ”)

s a l t = ” + string ( p ) +

Scilab code Exa 6.10 Cooling in a Crystallizer

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

clear ; clc ;


/ / E xa mp le 6 . 1 0 / / P ag e 3 68 printf ( ” E xa mp le 6 . 1 0 , P ag e 3 6 8 \ n \ n ” ) ; // s o l u t i o n // b a s i s 1 0 0 kg f r e e w a t e r i n o r i g i n a l s o l / / i n i t i a l T = 3 5 3K W 1 = ( 1 26 / 12 0 .3 ) * 6 4. 2 / / k g 150

18 W f r ee 1 = 1 00 - W 1 19 M S 1 = ( ( 6 4. 2 0 + W 1 ) * 1 00 ) / 3 2 . 7 6 / / MgSO4 . 7 H2O i n 1 0 0 k g

f r e e w at er 20 / / 4% o f o r i g i n a l s o l e va p o r a t e s 21 22 23 24 25 26

E = ( MS 1 + 1 00 ) *. 04 W f r ee 2 = 1 00 - E // f r e e w a t e r i n mo th er l i q u o r

/ / a t 3 03 .1 5 K W 2 = ( 1 26 / 12 0 .3 ) * 4 0. 8 W f r ee 3 = 1 00 - W 2 M S 2 = ( W 2 + 4 0 . 80 ) * W f r e e 2 / W f re e 3

// c r y s t a l s o f MgSO4

.7H2O 27 y = MS1 - M S2 / / k g 28 q = 5 0 1. 2 *1 0 00 / 28 4 .6 // q u an ti ty o f o r i g i n a l

s o l to

be f e d 29 printf ( ” Q u a n t i t y i f o r i g i n a l s o l u t i o n t o be f ed t o t h e c r y s t a l l i z e r p e r 1 0 0 0 k g c r y s t a l s o f MgSO4 . 7 H2O = ” + string ( q ) + ” kg . ” )

Scilab code Exa 6.11 Recovery of p DCB

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

clear ; clc ;


/ / E xa mp le 6 . 1 1 / / P ag e 3 70 printf ( ” E xa mp le 6 . 1 1 , P ag e 3 7 0 \ n \ n ” ) ; // s o l u t i o n // ( a ) 151

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

printf ( ” ( a ) \ n \ n ” )

// u si ng f i g 6 . 1 2 / / p ef or mi n g m a t e r i a l b a la n c e a t 2 90K a 1 = 5. 76 b 1 = 4. 91 D CB s = b 1 * 10 0/ ( a 1 + b1 ) // % o f s o l i d s ep a r a t ed p−DCB D CB r1 = D CB s * 1 00 / 70 // r e c o v e r y o f p −DCB printf ( ” P e rc en ta g e r e co v e r y o f p−DCB = ” + string ( DCBr1)+” . \n \n \n ( b ) \n \n ”)

//(b) / / a t 2 55K a 2 = 5. 76 b 2 = 1 0. 22 D CB s = b 2 * 10 0/ ( a 2 + b2 ) D CB r2 = ( D C Bs * 1 0 0) / 7 0 A r = D CB r2 - D C Br 1 printf ( ” A d di t i o n a l r e co v e r y o f p−DCB = ” + string ( A r ) + ” . ”)

Scilab code Exa 6.12 Extractive Crystallization of o and p nitrochloroben-

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

clear ; clc ;


/ / E xa mp le 6 . 1 2 / / P ag e 3 71 printf ( ” E xa mp le 6 . 1 2 , P ag e 3 7 1 \ n \ n ” ) ;

152

13 / / s o l u t i o n 14 F = 5000 // kg /h 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

s o l ve n t f r e e mix f e d t o s i m pl e c r y s t a l l i z a t i o n u n it B 1 = 4 0 00 / 15 7 .5 // kmol / h p−NCB i n f e e d o−NCB i n f e e d A 1 = 1 0 00 / 15 7 .5 // kmol / h / / a f t e r c r s t a l l i z a t i o n m o th er l i q u o r h a s 3 3 . 1 m ol % B, A does n ’ t c r s t a l l i z e s m = A1 / (1 - .3 31 ) // mo ther l i q u o r e n t e r i n g e x t r a c t i v e c r y t a l l i z a t i o n u ni t

B 2 = m -A1

// o pt im iz in g s o l i d f l ux / / d Ct / dR = 1 − 2 / R ˆ 3 = 0 R = 2 ^( 1/ 3)

// r e f e r r i n g f i g 6 .1 4 / / o v e r a l l m a te r ia l b al a n ce // p− i s o m er (B ) // . 9 8D + xT = 4 0 0 0 (i) // o− i s o m e r (A) / / . 0 2 D + (1 − .05 − x )T = 1 0 0 0 ( ii ) / / m a t e r i a l b a l a nc e a ro u n d s o l v e n t r e co v er y u n it // B // 2 . 2 6 Tx = . 1 9 8G = xH ( iii ) // A / / 2 . 2 6 T( . 95 − x ) = . 5 3 1G ( iv ) / / s o l v i n g a b o v e eq T = 1 33 7. 6 / / kg / h D = 3729 / / kg / h G = 3939 / / kg / h x = .258

// p u tt i ng t h es e v a l ue s we g e t c o m p o s it i o n o f v a r i o u s streams 40 printf ( ” C om po si ti on o f v a r i o u s s tr ea m s : \ n Component T kg / h D kg / h \ n A 925.6 7 4 . 6 \n B 345.1 3 6 5 4 . 9 \n C 66.9 n i l \n \ n”) 41 printf ( ” P ur i t y o f t o p p ro du ct = 6 9 .2 p e rc e n t A \ n 153

P u ri t y o f bottom p ro du ct = 9 8 . 0 p e r ce n t \ n Make− up s o l v e n t = 6 6 . 9 kg / h . ” )

Scilab code Exa 6.13 Calculation of Dew Point

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

clear ; clc ;


/ / E xa mp le 6 . 1 3 / / P ag e 3 83 printf ( ” E xa mp le 6 . 1 3 , P ag e 3 8 3 \ n \ n ” ) ; // s o l u t i o n P w 1 = 1 2. 84 / / Pa

v . p . o f i c e a t 2 33 .1 5K

( t ab l e

6.12) 16 17 18 19 20 21

P 1 = 1 01 32 5 //Pa H m = ( P w1 / ( P1 - P w 1 ) ) // kmol / kmol d ry a i r P 2 = 8 01 32 5 //Pa P w 2 = P 2 * . 0 0 0 1 26 7 / ( 1 +. 0 0 0 1 26 7 ) from t a b l e 6 . 1 2 d p = - 20 .1 8 + 2 73 .1 5 //K printf ( ”Dew P o i n t = ” + string ( d p ) + ”K. ” )

Scilab code Exa 6.14 Calculations on Ambient Air

1 clear ; 2 clc ; 3

154

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


/ / E xa mp le 6 . 1 4 / / P ag e 3 84 printf ( ” E xa mp le 6 . 1 4 , P ag e 3 8 3 \ n \ n ” ) ; // s o l u t i o n / / Pa = v . p . a t DP P w = 2 .0 62 4 //kPa P = 100 //kPa H m = Pw / (P - Pw ) // kmol w at er v ap ou r / k mol d ry a i r H = .6 22 * Hm // kg m o i s tu r e / kg d r y a i r

/ / a t s a t u r a t i o n , DB = WB = DP P s = 4 .0 04 //kPa R H = Pw * 10 0/ P s H s = ( P s /( P - Ps ) ) * .6 22 s = H * 10 0/ H s C h = 1 . 00 6 +1 . 84 * H // kJ / kg d ry a i r K V h = ( . 0 0 07 3 + . 03 4 4 8 ) * 2 2 . 4 1 4 * 1. 1 0 6 2 *1 . 0 1 3 3 //mˆ3/kg

dry a i r // u si ng f i g 6 . 1 5 W B = 2 94 .5 5 //K ias = 6 2. 3 // kJ / kg d ry a i r d = -.28 // kJ / kg d ry a i r

27 28 29 30 31 i a = ias + d 32 printf ( ” The a b s o l u t e

m ol ar h um id i ty = ” + string ( H m ) + ” kmol w at er v ap ou r / kg d ry a i r \ n A b so l u t e h u m id i ty = ” + string ( H ) + ” kg m o is tu r e / kg d r y a i r \ n p e r c e n t RH = ” + string ( R H ) + ” \ n p e r c en t s a t u r a t i o n = ” + string ( s ) + ” \ nHumid h e a t = ” + string ( C h ) + ” kJ / kg d r y a i r K\ n Humi d v o l u me = ” + string ( V h ) + ” m ˆ 3 / k g d ry a i r . ” )

155

Scilab code Exa 6.15 Humidification of Air in a Textile Industry

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

clear ; clc ;


/ / E xa mp le 6 . 1 5 / / P ag e 3 85 printf ( ” E xa mp le 6 . 1 5 , P ag e 3 8 5 \ n \ n ” ) ; // s o l u t i o n / / b a s i s 1 kg o f d ry a i r e n t e r i n g t he a i r w a s h e r / / from f i g 6 . 1 5 H 1 = 11 .8 // g / kg d ry a i r H 2 = 1 7. 76 // g / kg d ry a i r H = H2 - H1 // m o is t ur e a dded d u ri n g s a t u r a t i o n D B = 3 00 .9 5 //K W B = 2 98 .1 5 //K D P = 2 97 .1 5 //K C h = 1 . 00 6 +1 . 84 * .0 1 77 6 // kJ / kg d ry a i r K d T = DB - DP H s = Ch * 3. 8 A = 25 00 0 //mˆ 3/ h

a c t u a l a i r a t 41 and 24 d eg re e

celcius / / a ga in fr om f i g 6 . 1 5 V h = . 90 67 / /mˆ 3/ k g d r y a i r q m = A /Vh / / kg d ry a i r / h f i = qm * Hs / / k J / h P = 300 //kPa 156

32 l a m d a = 2 1 63 . 2 // kJ / kg 33 S C = fi / l am da / / kg /h

by a p p e n d i x IV . 2 s team c o n s u m p t i o n a t t h e

heater 34 printf ( ” t h e m oi st ur e added t o t h e a i r = ” + string ( H ) + ” g /kg d ry a i r \ n DB temp o f f i n a l a i r = ” + string ( D B ) + ”K \ n WB temp o f f i n a l a i r = ” + string ( W B ) + ”K \ n The h e a t i n g l oa d o f t he s tea m c o i l p er kg dr y a i r = ” + string ( f i ) + ” kJ /h \ n S te am c o n s u m p t i o n = ” + string ( S C ) + ” k g / h . ” )

Scilab code Exa 6.16 Induced draft cooling tower

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

clear ; clc ;


/ / E xa mp le 6 . 1 6 / / P ag e 3 87 printf ( ” E xa mp le 6 . 1 6 , P ag e 3 8 7 \ n \ n ” ) ; // s o l u t i o n // M = E+B+W T a v = ( 45 + 32 ) / 2 + 27 3 .1 5 //K

avg c o o l i n g w a t e r T / / u s i n g s te am t a b l e s ( Ap pen dix A IV . 1 ) l a m da = 2 4 10 .5 / / k J / k g E = 5 30 / l am da / / k g / s

Cl Ti To fi

= = = =

4 .1 86 8 4 5 +2 7 3. 1 5 //K 3 2 +2 7 3. 1 5 //K 530 / / = m c ∗ Cl ∗ ( Ti −To )

157

24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40

m c = 5 30 /( C l * ( Ti - T o ) ) W = . 3* m c /1 00 / / k g / s

//kg/s

// // // //

d i s s o l v e d s o l i d b a l a n ce M∗xm = (B+W) ∗ xc 5 00 ∗ 10ˆ − 6 ∗M = ( B + . 0 2 9 2 ) ∗ 2000 ∗ 10ˆ − 6 s o l v i n g a b o v e e qs B = .0 44 1 / / k g / s M = .2 93 2 / / k g / s // e ne rg y b a l an c e on c o o l i n g t ow er / / f i = ma ∗ ( i 2 − i 1 ) / / i 2 − i 1 = 1 1 . 04 2 kJ / kg d r y a i r / / m o is t ur e b a la n ce //E = ma(H2−H1 )

H 2 = . 21 99 /4 8 + . 01 96 i ws = 2 54 6. 2 / / A pp en di x IV C h 1 = 1 . 0 0 6 +1 . 8 4 *. 0 1 9 6 i 1 = 1 . 0 06 * ( 2 97 . 4 5 - 2 7 3 . 1 5) + . 0 19 6 * i w s + 1 . 0 4 2 * ( 3 0 8 . 1 5 - 2 9 7 . 5 ) // kJ / kg d ry a i r 41 i2 = i1 + 11.04 42 T db = ( ( i2 - 1 . 00 6 *( 3 01 . 25 - 2 73 . 15 ) - i ws * H 2 ) /1 .0 5 ) +301.25 // K 43 printf ( ” A i r l e a v e s t h i nd uc ed d r a f t f an a t ” + string ( Tdb)+” K. ”)

Scilab code Exa 6.17 Waste Heat recovery unit

1 2 3 4 5 6 7 8 9

clear ; clc ;


/ / E xa mp le 6 . 1 7 158

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

/ / P ag e 3 89 printf ( ” E xa mp le 6 . 1 7 , P ag e 3 8 9 \ n \ n ” ) ; // s o l u t i o n / / b a s i s 1 k g d r y a i r f ed t o t o w e r / / fro m f i g 6 . 1 6 we g et // at WB=330 K and DB=393 K H 1 = . 09 72 // kg / kg d r y a i r D P = 3 25 .1 5 //K // a t 3 13 K H 2 = . 04 92 // kg / kg d r y a i r H = H1 - H2 // m o is t ur e c on de ns ed i n t ow er C h1 = 1 .0 06 + 1 .8 4* H 1 // kJ /kg d ry a i r C h2 = 1 .0 06 + 1 .8 4* H 2 i a 1 = 1 . 00 6* ( 32 5 - 2 73 ) + H 1 * 25 96 + 1 . 18 5 *( 3 93 - 3 2 5 ) //

e nt ha lp y o f e n t e r i n g a i r 26 i a 2 = 1 . 00 6* ( 31 3 - 2 73 ) + H 2 * 25 7 4. 4 // e nt ha l p y o f

o u tg o in g a i r 27 28 29 30

i = ia1 - i a2 q m = 2 00 0 /( 1 + H1 ) f i1 = qm * i // h e a t l o s s r a t e f i 2 = 1 . 1 6 7* 3 6 0 0* 4 . 1 86 8 * ( 32 3 - 3 0 5 ) // h ea t g a in e d by

water 31 r = f i2 * 1 00 / fi 1 32 printf ( ” ( a ) \ n \ n The h ea t l o s s r a t e r a t e fro m t he h o t a i r i n t h e bed = ” + string ( f i 1 ) + ” kW \ n \ n \ n (

b ) \ n \ n The p e rc e n t ag e h ea t r e co v e r y i n h ot w a t e r = ” + string ( r ) + ” p e r c e n t . ” )

Scilab code Exa 6.18 Recovery of CS2 by adsorption

1 clear ; 2 clc ; 3

159

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


/ / E xa mp le 6 . 1 8 / / P ag e 3 90 printf ( ” E xa mp le 6 . 1 8 , P ag e 3 9 0 \ n \ n ” ) ; // s o l u t i o n / / b a s i s 8 0 0 kmol o f i n l e t CS2−H2 mix P i = 1 06 .7 // kPa T ot al P r e s su r e P cs 2i = 1 6. 93 / / kPa n = 800 / / k mo l n cs 2i = P cs 2i * n / Pi / / k mo l n h 2i = n - n c s2 i P o = 1 01 .3 25 / / kPa P cs 2o = 6 .1 9 / / kPa n h2 o = 6 73 .1 / / k mo l n c s 2 o = P c s2 o * n h 2 o / ( Po - P c s2 o ) n cs 2a = n cs 2i - n c s2 o m cs 2a = n cs 2a * 7 6 .1 4 07 / / k g r = 600 // kg /h d e s i g n a d so r p t i o n r a t e M i = n * r/ m cs 2a / / k mo l / h V i = Mi * 2 2. 84 3 / / m ˆ 3 / h m cs 2 ac = . 32 - . 04 // kg CS2 a bs or be d p er kg BD

a c t i v a t e d c ar bo n 31 32 33 34 35

q m = r * 1 .0 4/ m c s 2a c / / kg / h C = n cs 2o / n h2 o / / km ol CS2 / km ol H2 = P cs 2 / ( P−Pcs2 ) P c s2 = 2 4. 76 3 / / kPa T = 28 1. 5 //K by eq 5 . 2 4 printf ( ” ( a ) \ n \ n V o lu me tr ic f l o w r a t e o f e n t e r i n g m i x t u r e = ” + string ( V i ) + ” m ˆ 3 / h \ n \ n \ n ( b ) \ n \ n Mass f l o w r a t e o f a c t i v a t e d c ar bo n = ” + string ( q m ) +

” kg / h \ n \ n \ n ( c ) \ n \ n O r i g i n a l m ix tu re must be c o o l e t o ” + string ( T ) + ” K a t 4 0 5 kPa f o r a c h i e v i n g same c o n c e n t r a t i o n o f t he o u t l e t 160

m i xt u re w it h a d s o r p t i o n . ” )

Scilab code Exa 6.19 Hooker type diaphragm cell

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

clear ; clc ;


/ / E xa mp le 6 . 1 9 / / P ag e 3 91 printf ( ” E xa mp le 6 . 1 9 , P ag e 3 9 1 \ n \ n ” ) ; // s o l u t i o n / / b a s i s 4 00 0 kg / h o f NaOH p r o du c ed C l 2 p = 3 5 . 5 *2 * 4 0 00 / 8 0 / / kg / h M c l2 = C l2 p / 71 / / k mo l / h P = 1 01 .3 25 / / kPa P w = 2 .0 62 4 / / kPa m o is t = ( P w / ( P - Pw ) ) * ( 1 8 . 0 1 5 4/ 7 0 . 90 6 ) // T mo i st = C l2 p * m oi st / / kg / h

/ / f o r 90% o nc o f a c id n = ( 1 0 / 18 . 0 1 53 ) / ( 9 0 /9 8 . 0 77 6 ) / / / k mo l H2O/ k m ol a c i d Q = 1 3 4 4 77 / ( 1 8* ( n + 1 . 7 9 8 3) ^ 2 ) / / kJ / kg H2O by eq ( i i

) 25 26 27 28

l a m bd a v = 2 45 9 / / k J / kg ( A pp en di x IV ) h e at lo a d = Q + l a mb d av f i = h ea t lo a d * 1 8. 74 / / k J / h printf ( ” The h e at l i b e r a t i o n r a t e i n t he t o w e r = ” + string ( f i ) + ” k J / h . ” )

161

Scilab code Exa 6.20 Absorption of NH3 from pure gas

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

clear ; clc ;


/ / E xa mp le 6 . 2 0 / / P ag e 3 93 printf ( ” E xa mp le 6 . 2 0 , P ag e 3 9 3 \ n \ n ” ) ; // s o l u t i o n / / b a s i s 1 0 0 kmol o f f e e d g as / / u si ng t a b l e 5 . 1 S n ia i = 2 0 . 6 * 2 9 . 5 9 0 9 + 6 2 * 2 8 . 6 1 0 5 + 4 . 1 * 2 0 . 7 7 2 3 + 1 1 . 1 * 1 9 . 2 4 9 4 + 2 .2 * 2 5 . 6 5 0 3

18 S n i b i = [ 2 0. 6 * ( - 5 . 14 1 ) +62*1.0194+11.1*52.1135+2.2*33.4806]/1000 19 S n i c i = [ 2 0 . 6* 1 3 . 18 2 9 + 62 * ( - . 1 4 7 6 ) +11.1*11.973+2.2*.3518]/10^6 20 S n id i = [ 2 0. 6 * ( - 4 . 96 8 ) + 6 2 * . 7 6 9+ 1 1 . 1* ( - 1 1 .3 1 7 3 ) +2.2*(-3.0832)]/10^9 21 H ga s = S ni ai * ( 2 83 - 2 6 3) + S ni bi * ( 2 83 ^ 2 - 2 63 ^ 2) / 2 + S n ic i * ( 2 8 3 ^3 - 2 6 3^ 3 ) / 3 + S n id i * ( 2 8 3 ^4 - 2 63 ^ 4 ) / 4

kJ 22 H n h3 = 1 53 3. 8 / / k J 23 S n i C m pi = ( H ga s - H n h 3 ) / 2 0 / / kJ / (K 9 7 . 8 kmol g a s )

NH3 f r e e g a s 24 G o = 9 7 .8 / .9 9 99 5 //kmol

162

//

25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40

N H3 a = ( 2. 2 - . 00 5) * 1 7 // kg F 1 = N H3 a /. 04 // f l o w r a t e o f 4% NH3 s o l u t i o n W a t er = F1 - N H 3a / / k g d T 1 = H g as / ( W a t e r * 4 . 1 86 8 ) // K T w a te r = 3 07 - d T 1 //K W v p = 2 .1 16 //kPa P = 5 10 1. 32 5 //kPa m oi st = G o * Wv p /( P - W vp ) // kg W = Water + moist // t o t a l d e m i ne r a l i s e d w at er d T a c t ua l = H g as / ( W * 4 . 1 8 68 ) //K

/ / fro m t a b l e 5 . 5 9 d H f1 = - 80 .0 93 / / kJm ol NH3 o f 4% NH3 s o l d H f2 = - 46 .1 1 //kJ/molNH3 H = dH f1 - d H f2 / / h e at o f 4% NH3 s o l H e v l = - ( H* N H 3 a * 1 0 00 ) / 1 7 // t o t a l h ea t e v ol ve d / / i n a b so r be r g as i s f u r t h e r h ea te d from 28 3K to 291.4K 41 H s ol = H ev l - ( 2 8 54 . 1 *( 2 9 1 .4 - 2 8 3. 1 5 ) ) / / kJ 42 // c 0 f 4% NH3 s o l = c o f w at er = 4 . 1 86 8 kJ / kg K 43 d T2 = H so l / ( F1 * 4 . 18 68 ) 44 T o = 2 91 .4 + dT 2 45 printf ( ” ( a ) \ n \ n Temp o f f e e d w at er t o a b s o rb e r = ” + string ( T w a t e r ) + ”K . \ n \ n \ n ( b ) \ n \ n Temp o f a q NH3 s o l l e a v i n g t he a b so r be r = ” + string ( T o ) + ”K. ” )

Scilab code Exa 6.21 Direct contact counter current rotary drier

1 2 3 4 5 6 7 8

clear ; clc ;


163

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

/ / E xa mp le 6 . 2 1 / / P ag e 3 96 printf ( ” E xa mp le 6 . 2 1 , P ag e 3 9 6 \ n \ n ” ) ; // s o l u t i o n // b a s i s : p ro du ct r a t e o f 10 0 k g/ h H 1 = .0 36 // kg m oi s t / k g d r y s o l i d X 1 = . 25 /. 75 // kg / kg d ry s o l i d X 2 = . 02 /. 98 // kg / kg d r y s o l i d / / m oi st b a l an c e / / ms ∗ ( X1−X2) = ma ∗ ( H2−H1 ) T o = 2 73 .1 5 //K i s 1 = 1 . 43 * (3 0 - 0 ) + X 1 * 4 . 1 8 6 8* 3 0 i s 2 = 1 . 4 3 * 80 + . 0 2 04 * 4 . 1 86 8 * 8 0 T db = 3 93 .1 5 //K T d p1 = 3 08 .1 5 //K i w b1 = 2 56 5. 4 / / k J / k g C h1 = 1 . 00 6 +1 . 84 * .0 3 6 i a1 = 1 . 0 06 * ( T dp 1 - 2 7 3. 1 5 ) + H 1 * i wb 1 + C h 1 * ( Td b - T d p 1 ) H 2 = .0 56 T d p2 = 3 15 .5 5 i w b2 = 2 57 8. 7 i a2 = 1 . 0 06 * ( T dp 2 - 2 7 3. 1 5 ) + H 2 * i wb 2 + ( 1 . 0 0 6+ 1 . 8 4* H 2 ) *(323.15-Tdp2) m a = . 08 5 /( . 05 6 - . 0 36 ) i aa = 1 . 0 06 * ( T dp 1 - 2 7 3. 1 5 ) + H 1 * i wb 1 f i = 4 . 25 *( 2 18 .6 8 - i aa ) //kW l a m bd a = 2 13 3. 0 s te am = f i / la mb da / / kg / h printf ( ” ( a ) \ n \ n F lo wr at e o f i nc om in g a i r on d r y b a s i s = ” + string ( m a ) + ” k g / s \ n \ n \ n ( b ) \ n \ n Humi di ty o f a i r l e a vi n g t h e d r i e r = ” + string ( H 2 ) +

” kg / kg d ry a i r . \ n \ n \ n ( c ) \ n \ n S te am c on su mp ti on i n t he h e a t er = ” + string ( s t e a m ) + ” k g / h.”)

164

Scilab code Exa 6.22 Hot air dryer of textile mill

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

clear ; clc ;


/ / E xa mp le 6 . 2 2 / / P ag e 3 98 printf ( ” E xa mp le 6 . 2 2 , P ag e 3 9 8 \ n \ n ” ) ; // s o l u t i o n // b a s i s c l o t h s pe e d = 1 .1 5 m/ s prod = 1 . 1 5 *1 . 2 * 3 60 0 * . 0 95 m oi st i = . 90 // kg / kg bo ne d ry c l o t h m oi st o = . 06 e v p = 4 7 1 .9 6 * ( m o is t i - m o is t o )

/ / u si ng f i g 6 . 1 5 and 6 . 1 6

H 1 = . 01 80 5 H 2 = . 08 32 d H = H2 - H1 q m 1 = e vp / dH / / kg d ry a i r / h V h = . 88 37 / /mˆ 3/ k g d ry a i r qv = qm 1* Vh D P 1 = 2 96 .5 //K D P 2 = 3 22 .5 //K l a mb da V 2 = 2 38 4. 1 / / k J / k g T o = 2 73 .1 5 //K f i1 = prod * 1 . 2 5 6 * ( 3 6 8 - 3 0 3 ) + prod * . 0 6 * ( 3 6 8 - 3 0 3 ) * 4 . 1 8 6 8

/ / kJ / h 31 f i 2 = e vp * ( 3 2 2 . 5 - 3 0 3 .1 5 ) + e v p * l a m bd a V 2 / / k J / h

165

32 i a1 = 1 . 0 0 6* ( 3 0 3. 1 5 - 2 7 3 . 1 5 ) + 2 5 5 6 . 4* . 0 1 80 5 / / k J / k g

dry a i r 33 i a 2 = 1 . 0 06 * ( 3 22 . 8 - 2 7 3 . 1 5) +2591.5*.0832+(1.006+1.84*.0832)*(393-328.8) 34 f i2 = ia2 - i a1 35 h l o st = f i2 - f i 1 / / kJ / h 36 / / u s i n g A pp en di x I V 37 h = 7 20 .9 4 / / k J / k g 38 l a mb d av = 2 04 6 .5 / / kJ / kg 39 s t e a mi = ( h + l a m bd a v ) * 8 85 / / kJ / h 40 f i 4 = h * 88 5 / / k J / h 41 q m2 = 8 85 / ev p 42 printf ( ” ( a ) \ n \ n B one d ry p r o d uc t i o n o f t he d r y er = ” + string ( prod ) + ” k g / h . \ n \ n \ n ( b ) \ n \ n The e v a p o r a t i o n t a k i n g p l ac e i n t he d ry er = ” + string ( e v p ) + ” k g / h . \ n \ n \ n ( c ) \ n \ n The a i r c i r c u l a t i o n r a t e = ” + string ( q v ) + ” mˆ3/ h . ” )

Scilab code Exa 6.23 Quadruple effect forward feed evaporator

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

clear ; clc ;


/ / E xa mp le 6 . 2 3 / / P ag e 4 01 printf ( ” E xa mp le 6 . 2 3 , P ag e 4 0 1 \ n \ n ” ) ; // s o l u t i o n / / b a s i s : weak l i q u o r f l o w r a t e = 1 0 6 0 kg /h 166

16 s 1 = 1 06 0* .0 4 // kg / h 17 l iq r = s 1 /. 25 // k g /h

s o l i d s i n weak l i q u o r conc l i q uo r l ea vi n g 4 th

effect 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42

e v p = 1 06 0 - l iq r / / kg / h l a mb d as = 2 04 6 .3 / / kJ / kg W f = 10 60 / / k g /h C1f = 4 .0 4 T 1 = 4 22 .6 T f = 303 l a mb da v 1 = 2 11 4. 4

// e nt ha lp y b a l a n ce o f 1 s t e f f e c t // Ws∗ l a m b d a s = W f ∗ C1f ∗ ( T1−T f ) + ( W f− W1) ∗ 2 1 1 4 . 4 // p u t ti n g v a l u e s we g e t / / Ws = 1 3 4 5 . 5 7 − 1 . 0 3 3 ∗ W1 / / 2 nd e f f e c t / / W1 = 5 3 1 . 3 8 + . 5 1 0 ∗ W2 // 3 rd e f f e c t // W1 − 1 . 9 9 0 ∗ W2 = − 1.027 ∗W3 // 4 th e f f e c t // W2 − 1 . 9 8 3 ∗ W3 = − 176.84 / / s o l v i n g a bo ve e q s W 1 = 862 / / kg / h W 2 = 6 48 .2 / / kg / h W 3 = 4 16 .7 / / kg / h W s = 4 55 .2 / / kg / h e c o = e vp / Ws / / kg e v a p o r a t i o n / kg s te am s pc on = 1/ e co // kg s tea m / kg e v a p o r a t i o n printf ( ” S p e c i f i c h ea t c o ns um pt i on o f t he s ys te m i s ” + string ( s p c o n ) + ” kg s te am / k g e v a p o r a t i o n . ” )

Scilab code Exa 6.24 Triple effect evaporation system

1 clear ; 2 clc ; 3

167

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



F sp d1 = 4 30 0 / / kg / h B c r t n = F s pd 1 * 6 0 0 *1 0 ^ - 6 / / kg / h F sp d2 = B cr tn / . 0 06 45 / / kg / h e vp 1 = F sp d1 - F s pd 2 F sp d3 = B cr tn / . 0 57 e vp 2 = F sp d2 - F s pd 3 C 3 = B cr tn / .4 e v p3 = F sp d3 - C 3 f i1 = F s pd 1 * 2 . 5 6 * (4 6 8 . 15 - 3 7 3 .1 5 ) + 3 9 0 0 * 45 0 / / kJ / h f i2 = F s pd 2 * 2 . 5 6 * (4 6 3 . 15 - 4 6 8 .1 5 ) + 3 5 4 . 7 3 7* 4 5 0 / / kJ / h f i3 = F s pd 3 * 2 . 5 6 * (4 5 3 . 15 - 4 6 3 .1 5 ) + 3 8 . 8 1 3 *4 5 0 / / kJ / h f i = f i1 + f i2 + f i3 m t = f i / ( 2 . 95 * ( 50 3 . 1 5 - 4 7 8 . 15 ) ) / / kg / h q t = mt / .7 1 / / l / h m c c w 1 = 1 7 5 5 0 00 / ( 8 *4 . 1 8 6 8) / / kg / h m c c w2 = m cc w1 * . 9 d T 2 = 1 5 9 63 2 / ( m c c w2 * 4 . 18 6 8 ) m cc w3 = m cc w1 - m c cw 2 d T 3 = 1 7 4 66 / ( m c c w3 * 4 . 18 6 8 ) d T = ( 1 7 5 5 00 0 + 1 5 96 3 2 + 1 74 6 6 ) / ( m c c w1 * 4 . 18 6 8 ) F w = 1 9 32 0 98 / (8 * 4. 1 86 8 ) / / kg / h printf ( ”By m ass b al an ce , r e q u i r e d c o o l i n g w at er f l o w i n e x t e r n a l c o o l e r = ” + string ( F w ) + ” k g / h . \ n \ nBy

e n th a lp y b al an ce , o v e r a l l = ” + string ( d T ) + ” K . ” )

168

r i s e i n CCW t em p er at u re

Scilab code Exa 6.25 Four compartment washing thickner

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 32

clear ; clc ;


/ / E xa mp le 6 . 2 5 / / P ag e 4 05 printf ( ” E xa mp le 6 . 2 5 , P ag e 4 0 5 \ n \ n ” ) ; // s o l u t i o n / / s t r e a m M2 V c a c o3 M 2 = . 3 4 9/ 2 . 7 11 V l iq r M2 = . 6 51 / 1. 1 67 V s l r y M2 = V c a co 3 M 2 + V l i qr M 2 s pg M2 = 1 / V sl ry M 2 F s M 2 = 2 . 8 45 * 3 6 00 * s p g M 2 s M 2 = F sM 2 * . 34 9 / / kg / h l iq r M2 = F sM 2 * . 65 1 N a 2 O M 2 = l i qr M 2 * . 1 3 4 2 /1 . 1 6 7

/ / s t r e a m O2 F s O 2 = 1 4 . 1 93 * 3 6 0 0* 1 . 0 37 / / k g /h s O 2 = F sO 2 * . 00 03 l iq r O2 = F sO 2 - s O2 N a 2 O O 2 = l i qr O 2 * . 0 2 7 2 /1 . 0 3 7

/ / s t r e a m M1 V M 1 = . 1 94 / 2. 7 11 + . 8 06 / 1. 0 37 // l s pg M1 = 1/ V M1 F sM 1 = 5 2 06 . 9/ . 19 4

169

33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54

l iq rM 1 = F sM 1 - 5 20 6. 9 N a 2 O M 1 = l i qr M 1 * . 0 2 5 2 /1 . 0 3 4

/ / s t re a m O1 F sO 1 = F sO 2 + Fs M1 - F s M2 s O 1 = F sO 1 * . 00 02 l iq rO 1 = F sO 1 - sO1 N a 2 O O 1 = l i qr O 1 * . 0 0 9 6 /1 . 0 1 4

/ / s tr ea m W V W = . 03 7/ 2. 71 1 + . 96 3 s p gW = 1 / VW F sW = 1 4 .9 7 7* 3 60 0 * s pg W s W = F sW * .0 37 l iq rW = FsW - s W N a2 OW = l iq rW * . 0 02 4

/ / s t r e a m Mo V M o = . 4 02 / 2. 7 11 + . 5 98 / 1. 0 22 s pg Mo = 1/ V Mo F s M o = 3 . 6 27 * 3 6 00 * s p g M o s M o = F sM o * . 40 2 l iq rM o = F sM o - sMo N a 2 O M o = l i qr M o * . 0 1 6 2 /1 . 0 2 2 printf ( ” M a te r i a l b a l a nc e t h i c k e n e r \ n \ n ITEM

STREAM, kg / h \ n M2 O2 M1 O1 W Mo\ n S l u r r y ” + string ( ” + string ( F s O 2 ) + ” ” + string ( F s M 1 ) F s M 2 ) +” +” ” + string ( F s O 1 ) + ” ” + string ( F s W ) + ” ” + string ( F s M o ) + ” \ n S u s p e n d ed s o l i d s ”+ string ( s M 2 ) + ” ” + string ( s O 2 ) + ” ” + string ( ” + string ( s O 1 ) + ” ” + string ( s W ) + ” sM2)+” ” + string ( s M o ) + ” \ n L i q u o r ”+ ” + string ( l i q r O 2 ) + ” ”+ string ( l i q r M 2 ) + ” string ( l i q r M 1 ) + ” ” + string ( l i q r O 1 ) + ” ” ” + string ( l i q r M o ) + ” \ n Na2O + string ( l i q r W ) + ” ” + string ( N a 2 O M 2 ) + ” ” + string ( Na2OO2)+” ” + string ( N a 2 O M 1 ) + ” ” + string ( ” + string ( N a 2 O W ) + ” ” + string ( Na2OO1)+” 170

Na2OMo)+” ” )

171

Chapter 7 Combustion

Scilab code Exa 7.1 GCV and NCV calculations

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

clear ; clc ;

/ / S t o i ch i o m et r y / / C ha pt er 7 / / C o mb u st i on


/ / b a s i s 1 0 0 kg a s r e c e i v e d c o a l O 2 = 1 8. 04 / / k g n H 2 = 2 .7 9 - ( O2 / 8) / / k g printf ( ” ( a ) \ n \ n Net H2 i n c o a l = ” + string ( n H 2 ) + ” kg . \ n \ n \ n ( b ) \ n \ n ” ) 19 c b W = 1 . 12 8 *1 8 / / k g 20 printf ( ” Combined w at er i n t h e c o a l = ” + string ( c b W ) + ” 172

kg . \ n \ n \ n ( c ) \ n \ n ” ) 21 / / D ul on g ’ s f o r m u l a 22 G CV 1 = 3 3 95 0 *( 5 0. 2 2/ 1 00 ) + 1 4 42 0 0* n H 2 / 10 0 + 9 4 0 0 * . 3 7 / 1 0 0 / / kJ / kg 23 printf ( ”GCV b y D u l o n g s f o r m u l a = ” + string ( G C V 1 ) + ” kJ /kg . \ n \ n \ n ( d ) \ n \ n ” ) 24 t H 2 = 1 .3 95 / / km ol 25 wp = tH2 *18 + 7 26 H v = 2 4 42 . 5* w p / 10 0 // kJ / kg f u e l 27 G C V 2 = 2 3 39 2 *( 1 - . 2 1 - . 0 7 ) // a s o f r e c e i v e d c o a l 28 N CV = GC V2 - H v 29 printf ( ”NCV o f t h e c o a l = ” + string ( N C V ) + ” k J / kg . \ n \n \n ( e ) \n \n ” ) 30 / / C a ld e rw o od e q 31 // T ot al C = 5 . 88 + . 0 0 5 1 2 (B− 4 0 . 5 S ) +−

. 0 0 5 3 [ 8 0 − 1 0 0 ∗ (VM/FC) ]ˆ1.55 32 C = 5 .8 8 + . 0 05 1 2* ( 72 4 0. 8 - 4 0 .5 * .3 7 ) +.0053*[80-56.52]^1.55 33 printf ( ” T o t a l Ca rb on by C al de rw oo d e q = ” + string ( C ) + ” . ”)

Scilab code Exa 7.2 NCV of crude oil

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

clear ; clc ;


/ / E xa mp le 7 . 2 / / P ag e 4 36 printf ( ” E xa mp le 7 . 2 , P ag e 4 3 6 \ n \ n ” ) ;

173

13 14 15 16 17 18 19 20 21

// s o l u t i o n // b a si s 1 kg crude o i l H 2 = .1 25 // kg burnt

H 2O = H2 * 18 /2 L h = H 2O * 2 4 42 .5 / / k J G C V = 4 50 71 N C V = GCV - L h // kJ / kg o i l printf ( ”NCV = ” + string ( N C V ) + ” k J / k g . ” )

Scilab code Exa 7.3 Gaseous propane

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

clear ; clc ;


/ / E xa mp le 7 . 3 / / P ag e 4 44 printf ( ” E xa mp le 7 . 3 , P ag e 4 4 4 \ n \ n ” ) ; // s o l u t i o n / / b a s i s 1 m ol o f g a s e o u s p ro pa ne H 2 O = 4 * 18 . 01 5 3 // g N H V = 2 2 19 . 17 - ( H 2 O * 2 4 4 2. 5 / 1 00 0 ) printf ( ”NHV = ” + string ( N H V ) + ” k J / m o l . ” )

Scilab code Exa 7.4 GCV NCV for natural gas

174

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

clear ; clc ;


/ / E xa mp le 7 . 4 / / P ag e 4 44 printf ( ” E xa mp le 7 . 4 , P ag e 4 4 4 \ n \ n ” ) ; // s o l u t i o n / / b a s i s 1 m ol o f n a t u r a l g a s / / u si ng t a b l e 7 . 7 H 2 O = [ 2 * . 8 94 + 3 * . 0 5+ . 0 1 9 + 5* ( . 0 0 4 +. 0 0 6 ) ] * 1 8 // g H v = H 2O * 2 4 4 2. 5 /1 0 00 N C V1 = 9 45 .1 6 - H v G C V = 9 4 5 . 1 6* 1 0 0 0 /1 8 . 1 32 N CV = N CV 1 * 1 0 00 / 18 . 13 2 printf ( ” GCV = ” + string ( G C V ) + ” k J / k g . \ n NCV = ” + string ( N C V ) + ” k J / k g . ” )

Scilab code Exa 7.5 Coal burnt in excess air

1 2 3 4 5 6 7 8 9

clear ; clc ;


/ / E xa mp le 7 . 5 175

10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28

/ / P ag e 4 51 printf ( ” E xa mp le 7 . 5 , P ag e 4 5 1 \ n \ n ” ) ; // s o l u t i o n / / b a s i s 1 0 0 kg f u e l O 2r eq = 4 .3 3 1* 32 / / kg r O2 r eq = O 2r eq / 1 00 N 2i n = ( 7 9/ 21 ) * 4 .3 31 / / km ol A I R r eq = O 2 re q + N 2 i n * 28 / / k g r A IR r eq = A IR r eq / 1 00 R = A IR re q / 10 0 A IR sp ld = R *2 // kg / kg c o a l O 2s p ld = 4 .3 3 1* 2 / / km ol N 2 s pl d = N 2i n * 2 N 2c o al = 2 .0 5 /2 8 / / km ol t N2 = N 2s pl d + N 2 co al m oi st = 1 . 39 5 +( 7 /1 8 ) / / km ol printf ( ” ( a ) \ n \ n T h e o r a t i c a l O2 r e qu i re m en t p er u n i t mass o f c o a l = ” + string ( r O 2 r e q ) + ” k g . \ n \ n \ n ( b ) \ n \ n T h e o r a t i c a l d r y a i r r eq u i r e me n t = ” + string ( r A I R r e q ) + ” kg / kg c o a l . ” )

Scilab code Exa 7.6 Residue fuel oil sample

1 2 3 4 5 6 7 8 9 10

clear ; clc ;


/ / E xa mp le 7 . 6 / / P ag e 4 52 176

11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31

printf ( ” E xa mp le 7 . 6 , P ag e 4 5 2 \ n \ n ” ) ;

// s o l u t i o n / / b a s i s 1 00 kg o f RFO O 2r eq = 9 .7 86 //kmol N 2r eq = ( 79 / 21 ) * O 2r eq //kmol A IR r eq = O 2r eq + N 2 re q //kmol r A I R r eq = A I R re q * 2 9 / 1 00 A I Rs p ld = A IR r eq * 1 . 25 r A I R sp l d = A I R sp l d / 1 0 0

/ / u s i n g t a b l e 7 . 1 1 and 7 . 1 2 x S O 2 = . 0 7 / (5 5 . 9 2 5+ 5 . 6 9 5) / / k mo l SO2 / km ol w et g a s v SO 2 = x SO 2 * 10 ^6 // ppm m S O 2 = 4 . 4 8 * 10 ^ 6 / ( 1 69 6 . 1 4 +1 0 2 . 5 1 )

/ / a t 5 2 3 .1 5 K and 1 0 0 .7 kPa V = [ ( 5 5 .9 2 5 + 5. 6 9 5 ) * 8 . 3 1 4 * 52 3 . 1 5 ]/ 1 0 0 . 7 c SO 2 = ( 4 .4 8 *1 0 ^6 ) / V // mg/mˆ3

// mˆ3

// from f i g 7 . 3 d p = 4 24 .4 //K printf ( ” ( a ) \ n \ n T h e o r e t i c a l a i r r e q u i r e d = ” + string ( r A I R r e q ) + ” kg / kg f u e l . \ n \ n \ n ( b ) \ n \ n A c t ua l d r y a i r s u p p l i e d = ” + string ( r A I R s p l d ) + ” kg / kg f u e l . \ n \ n \ n ( c ) \ n \ n C o n c en t r a ti o n o f SO2 = ” + string ( m S O 2 ) + ” mg/kg . \ n \ n \ n ( d ) \ n \ n C o n c en t r a t io n o f SO2 = ” + string ( v S O 2 ) + ” ppm v o l / v o l . \ n \ n \ n ( e ) \ n \ n C o n c en t r a ti o n o f SO2 i f g a s e s a r e d i s c h ar g e d a t 5 2 3 .1 5K a nd1 00 . 7 kPa = ” + string ( c S O 2 ) + ” mg/mˆ3 . \ n \ n \ n ( f ) \ n \ n Dew P o i n t o f f l u e g a s = ” + string ( d p ) + ” K . ” )

Scilab code Exa 7.7 Orsat analysis of flue gases

1 clear ; 2 clc ;

177

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


/ / E xa mp le 7 . 7 / / P ag e 4 54 printf ( ” E xa mp le 7 . 7 , P ag e 4 5 4 \ n \ n ” ) ; // s o l u t i o n / / b a s i s 1 0 0 k mol o f d ry f l u e g as O 2 ac n td = 1 1. 4 +4 . 2 / / km ol O 2 a v l bl = ( 2 1 /7 9 ) * 8 4 .4 / / k mo l O 2e xc s = 4 .2 //kmol O 2 u na c t d = O 2 av l bl - O 2 a c nt d H 2b r nt = O 2 un a ct d * 2 O 2r eq = 1 1. 4+ O 2 u na c td p e x c s AI R = O 2 ex c s * 1 0 0/ O 2 re q m H 2b r nt = H 2b r nt * 2 // kg m Cb r nt = 1 1. 4 *1 2 r = m Cb rn t / m H 2b rn t printf ( ” ( a ) \ n \ n P er ce nt e x ce s s a i r = ” + string ( pexcsAIR)+” . \n \n \n ( b ) \n \n I n f u e l C : H = ”+ string ( r ) + ” . ” )

Scilab code Exa 7.8 Sugar factory boiler

1 2 3 4 5 6

clear ; clc ;

/ / S t o i ch i o m et r y / / C ha pt er 7 / / C o mb u st i on 178

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

/ / E xa mp le 7 . 8 / / P ag e 4 59 printf ( ” E xa mp le 7 . 8 , P ag e 4 5 9 \ n \ n ” ) ; // s o l u t i o n / / b a s i s 1 0 0 k g o f b a ga s s e f i r e d i n t h b o i l e r //(a) O 2r eq = 2 .0 2 / / k mo l N 2i n = ( 7 9/ 21 ) * O 2 re q / / k mo l A I R r eq = ( O 2 r eq + N 2 i n ) * 2 9 / / k g r AI R = A IR re q / 1 00 printf ( ” ( a ) \ n \ n T h e o r e t i c a l a i r r e q u i r e d = ” + string ( r A I R ) + ” kg d ry a i r / kg f u e l . \ n \ n \ n ( b ) \ n \n ”)

22 23 24 25 26 27

// (b)

28 29 30 31 32

// ( c )

33 34 35 36

t f lu g as = 1 . 95 / .1 5 65 / / / k m o l x cs O2 N2 = t fl ug as - 1 .9 5 x = ( x c sO 2N 2 - 7 .6 ) / 4 . 76 / / km ol p x cs A IR = x * 1 00 / O 2 re q printf ( ” P er ce nt e x ce s s a i r = ” + string ( p x c s A I R ) + ” . \ n \n ( c ) \n \n ”) p W = 1 0 0* . 26 7 7 // kPa

p a r t i a l p o f w a t e r vap

// f ro m f i g 6 . 1 3 d p = 3 39 .8 5 //K printf ( ”Dew P oi nt o f f l u e g as = ” + string ( d p ) + ”K . \ n \n \n ( d ) \n \n ” ) // (d) / / f ro m a p p en d i x I V e n t h a l p y o f f ee d w a t e r a t h fw = 2 92 .9 7 // kJ / kg 343.15 K e nt ha lp y o f s u pe r he a te d H s s = 3 18 0 .1 5 // kJ /kg stea m a t 2 . 1 5 b ar and 6 4 3 .1 5K

37 H ga in = H ss - hfw 38 H 6 = H ga in * 2 . 6* 1 00

// kJ

h e a t ga i n e d by w a t e r 179

39 H 1 = 1 0 0* 1 03 0 00 0 / / kJ 40 G C V = H 6 * 10 0/ H 1 41 printf ( ” Thermal e f f i c i e n c y GCV)+” . ”)

o f t he b o i l e r = ” + string (

Scilab code Exa 7.9 Stoker fired water tube boiler

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

clear ; clc ;


/ / E xa mp le 7 . 9 / / P ag e 4 65 printf ( ” E xa mp le 7 . 9 , P ag e 4 6 5 \ n \ n ” ) ; // s o l u t i o n / / u s i ng mean h ea t c a p a c i ty d at a T ab le 7 . 2 1 / / b a s i s 1 0 0 k mol o f d ry f l u e g as H 7 = 1 . 0 8 75 * 1 0 0 *3 0 . 3 1* ( 4 2 3. 1 5 - 2 9 8 . 1 5 ) H 7 1 = 3 6 3 3 .6 5 4 * (4 2 3 . 15 - 2 9 8 .1 5 ) f i 7 = H 71 * 3 9 0 0* . 7 6 7 1/ 1 6 2 . 2 / / kJ / h / / kJ / h f i1 = 3 . 9* 1 00 0 *2 6 17 0

/ / p e rf o rm i ng h ea t b a l an c e H s t e a mg e n = 2 3 5 46 . 0 7 e ff = H s te a mg e n * 10 0/ f i 1 // o v e r a l l e f f i c i e n c y r a t e printf ( ” O ve ra l l e f f i c i e n c y r a t e = ” + string ( e f f ) + ” per cen t . ”)

180

Scilab code Exa 7.10 Atomization of fuel

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

clear ; clc ;


/ / E xa mp le 7 . 1 0 / / P ag e 4 68 printf ( ” E xa mp le 7 . 1 0 , P ag e 4 6 8 \ n \ n ” ) ; // s o l u t i o n // b as is 100 kg o f f u e l o i l O2req = 9.364 / / k mo l

N 2i n = ( 7 9/ 21 ) * O 2 re q t N 2 = N 2i n + . 03 6 A IR re q = O 2r eq * 32 + t N2 * 28 / / k g / kg r AI R = A IR re q / 1 00 w p = 4.5 / / km ol H l o s s = 2 4 42 . 8 * w p * 1 8 /1 0 0 // kJ / kg f u e l N C V = 4 35 40 - H lo ss printf ( ” ( a ) \ n \ n NCV = ” + string ( N C V ) + ” k J / kg . \ n \ n \ n ( b ) \ n \ n T h e o r e t i c a l a i r r e q u i r e d = ” + string ( r A I R ) + ” kg / kg f u e l . \ n \ n \ n ( c ) \ n \ n ” ) H 1 = 1 0 0* 4 15 6 1. 3 3 // kJ

25 26 / / f ro m t a b l e 5 . 1 27 H 7 1 = 1 3 4 9 .7 2 6 * (1 5 0 0 - 2 9 8 . 15 ) + 2 52 . 9 24 * 10 ^ - 3 * ((1500^2-298.15^2)/2) +257.436*10^-6*((1500^3-298.15^3)/3) - 1 3 7 . 5 3 2 * 1 0 ^ - 9 * ( ( 1 5 0 0 ^ 4 - 2 9 8 . 1 5 ^ 4 ) / 4 ) / / u pt o

1500 K 28 H 71 1 = H1 - H 71 / / a b o ve 1 5 0 0K 29 / / F ( T ) = { 1 5 00 t o T} i n t e g r [ 1 4 7 7 . 3 0 1 + 3 7 5 . 2 7 1 0∗ 1 0 ˆ − 3

T− 91.2760 ∗ 10ˆ − 6Tˆ2+8.146 ∗ 10ˆ − 9 T ˆ 3 ] d T − 2147118 181

(i) 30 / / s o l v i n g i t f o r T = 20 0 0 31 A F T = 2 61 2 .7 1 // K 32 printf ( ”When f l u i d i s b ur n t w i th t h e o r e t i c a l a i r AFT = ” + string ( A F T ) + ” K. \ n \ n \ n ( d ) \ n \ n ” ) 33 / / w i th 30% e x c e s s a i r 34 35 36 37 38 39 40 41 42 43 44

O 2s p ld = 9 . 36 4 *1 . 3 x cs O2 = O 2s pl d - O 2 r eq N 2 i n 1 = ( 7 9/ 2 1 ) * O 2 s pl d t N2 1 = N 2i n1 + . 0 36

// now , u s in g t a b l e 7 . 2 6 , t a b l e 7 . 2 7 and eq ( i ) we get A F T1 = 2 1 78 .6 6 // K / / fro m f i g 7 . 3 d p = 429 // K / / s i m i l a r l y f o r i n c o mp l et e c o mb us ti on we f i n d A F T2 = 2 5 61 .4 2 //K printf ( ”When 3 0 p e r ce n t e x c e s s a i r i s s u p p l i e d AFT = ” + string ( A F T 1 ) + ” K . \ n \ n \ n ( d ) \ n \ n Dew P o i n t = ” + string ( d p ) + ” K . \ n \ n \ n ( e ) \ n \ n F o r i n c o m p l e t e c o m b u st i o n AFT = ” + string ( A F T 2 ) + ” K . ” )

Scilab code Exa 7.11 Water tube boiler

1 2 3 4 5 6 7 8 9 10 11

clear ; clc ;


/ / E xa mp le 7 . 1 1 / / P ag e 4 73 printf ( ” E xa mp le 7 . 1 1 , P ag e 4 7 3 \ n \ n ” ) ; 182

12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28

// s o l u t i o n // b a s i s 100 k g o f f u e l / / m a t e r i a l b a l a nc e o f c ar bo n C O 2 = 7 . 09 2 +. 0 47 // kmol in f l u e gases N 2 = 1 1 .9 4 *7 . 13 9 /7 . 01 O 2 = 1 1 .9 4 *7 . 13 9 /7 . 01 f l ue = C O2 + N 2 + O2

/ / m a t e r i a l b a l a nc e o f O2 O 2 a ir a ir in t O2 in O 2 xc s

= = = =

2 1* N 2 / 79 N 2 + O2 ai r O 2a ir + . 0 78 // O2 i n b ur ne r t O2 in - 9 . 86 4

/ / m a t e r i a l b a l a nc e o f w a te r v a po u r fro m co mb u s t i o n o f H2 m o is t fr m d = 5 .4 5 // kmol H = .0 33 1 // kmol / kmol o f d r y a i r humidity at 1 0 0 . 7 kPa 29 m o is ta i r = H * 1 04 . 48 2 //kmol 30 31 32 33 34 35

t m o i s t = m o i st f r md + m o i st a i r p x c s a ir = O 2 xc s * 1 0 0 / 9 .7 8 6

/ / now u s in g t a b l e 7 . 3 2 H 7 = 3 3 9 1 .2 0 3 * (5 6 3 . 15 - 2 9 8 .1 5 ) / / k J F f = 400 / / kg /h f u e l f i r i n g r at e t o t a l h e a t s u p p l i ed t H = 2 7 91 .7 - 1 7 9. 9 9 // kJ / kg

in b o i le r 36 37 38 39 40 41 42 43 44 45 46 47

f i5 = tH * 43 65 / / kJ / h f i 8 = 5 . 4 5* 1 8 * F f * 2 4 0 3. 5 / 1 00 / / kJ / h G CV f = 4 22 60 / / k J / k g f i1 = Ff * G CV f F d r y a ir = 1 0 4 .4 8 2 * 29 * F f / 1 0 0 C h a = 1 . 0 0 6 +1 . 8 4 *. 0 2 0 5 // k J/ kg d ry a i r K f i 3 = F d r ya i r * C h a * ( 3 08 . 1 5 - 2 9 8 .1 5 ) f i 2 = F f * 1 . 7 58 * ( 3 53 . 1 5 - 2 9 8 . 1 5) B O I L E Re f f 1 = f i 5 * 1 00 / f i 1 N C V f = G C Vf - ( 1 8 . 0 1 5 3 / 2 . 0 1 6 ) * . 1 0 9 * 2 4 4 2 . 8 / / kJ / kg B O I L E Re f f 2 = f i 5 * 1 00 / ( F f * N C Vf ) r = 43 65 / Ff / / s t ea m : f u e l

183

48 B O I L E R c ap a c i ty = f i5 / 2 2 5 6 . 9 49 printf ( ” A f te r p e rf o rm i ng m a t e r i a l and t he rm al

b a l a nc e o p e r a t i o n s we g e t \ n \ n O v e ra l l t he rm a l e f f i c i e n c y o f t h e b o i l e r b a se d on GCV o f t h e f u e l = ” + string ( B O I L E R e f f 1 ) + ” p e r c e n t . \ n \ n O v er a l l e f f i c i e n c y o f t h e b o i l e r b a se d on NCV o f t h e f u e l = ” + string ( B O I L E R e f f 2 ) + ” p e r c e n t . \ n \ n Stea m t o f u e l r a t i o = ” + string ( r ) + ” a t 16 b a r . \ n \ n E qu i va l e n t b o i l e r c a p ac i t y = ” + string ( B O I L E R c a p a c i t y ) +” k g / h . ” )

Scilab code Exa 7.12 Gassification by coal

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

clear ; clc ;


/ / E xa mp le 7 . 1 2 / / P ag e 4 78 printf ( ” E xa mp le 7 . 1 2 , P ag e 4 7 8 \ n \ n ” ) ; // s o l u t i o n / / b a s i s 1 0 0 kmol o f d ry p ro du ce r g as C = 33 *1 2 / / k g O 2 = 1 8. 5* 32 / / k g H 2 = 20 *2 / / k g O 2a ir = 2 1* 5 1/ 79 / / k mo l C O A L g a ss i f i ed = 3 9 6 /. 6 7 2 / / kg O 2 c o a l = C O A L ga s s i fi e d * . 0 6 1 / 32 / / km ol tO2 = O 2c oa l + O 2a ir

184

23 O 2 st e am = 1 8. 5 - t O2 / / km ol 24 H 2 st e am = 2 * O 2s te a m / / km ol 25 H 2f u el = 2 0 - H 2s te a m 26 d r y p ro d u c er g a s = 1 0 0 *2 2 . 41 / C O A L ga s s i fi e d

// Nmˆ3/ kg

coal 27 P w = 2 .6 42 / / kPa 28 H a = P w /( 10 0. 7 - P w ) / / k mo l / km ol d ry g a s 29 w a t er = H a *1 00 30 m o i s tp r o d u ce r g a s = ( 1 00 + w a t e r ) * 2 2 .4 1 / C O A L g a ss i f i ed

/ / Nmˆ 3 / k g c o a l 31 d r y ai r = ( 5 1 *2 8 + O 2 a ir * 3 2 ) / C O A L g a s si f i e d

/ / kg / k g

coal t s t e a m s u p p l i e d = H 2 s t e a m + w a t er - ( C O A L g a s s i f i e d * . 0 2 6 / 1 8 ) / / k mo l 33 s t ea m = t s t e am s u p pl i e d * 1 8 / C O A L g as s i f ie d 34 printf ( ” ( a ) \ n \ n M oi s tp ro d uc er g as o b t ai n ed = ” + string ( m o i s t p r o d u c e r g a s ) + ” Nmˆ 3 / k g c o a l . \ n \ n \ n ( b ) \ n \ n A i r s u p p l i e d = ” + string ( d r y a i r ) +” k g / 32

kg c o a l g a s s i f i e d . \ n \ n \ n ( c ) \ n \ n S te am s u p p li e d = ” + string ( s t e a m ) + ” kg / kg c o a l . ” )

Scilab code Exa 7.13 Open Hearth steel furnace

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

clear ; clc ;


/ / E xa mp le 7 . 1 3 / / P ag e 4 79 printf ( ” E xa mp le 7 . 1 3 , P ag e 4 7 9 \ n \ n ” ) ;

185

5 Edition - Stoichiometry_B. I. Bhatt and S. B. Thakore (2) (1)

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