PROCESS CALCULATIONS

PROCESS CALCULATIONS

SECOND EDITION

V. VENKATARAMANI Formerly ProfessorDepartment of Chemical Engineering National Institute of Technology Tiruchirappalli

N. ANANTHARAMAN ProfessorDepartment of Chemical Engineering National Institute of Technology Tiruchirappalli

K.M. MEERA SHERIFFA BEGUM Associate ProfessorDepartment of Chemical Engineering National Institute of Technology Tiruchirappalli

New Delhi-110001 2011 PROCESS CALCULATIONS, Second EditionV. Venkataramani, N. Anantharaman and K.M. Meera Sheriffa Begum

© 2011 by PHI Learning Private Limited, New Delhi. All rights reserved. No part of this book may be reproduced in any form, bymimeograph or any other means, without permission in writing from the publisher.

ISBN-978-81-203-4199-9The export rights of this book are vested solely with the publisher.Second Printing (Second Edition) … … February, 2011

Published by Asoke K. Ghosh, PHI Learning Private Limited, M-97, Connaught Circus, New Delhi-110001 and Printed by MeenakshiArt Printers, Delhi-110006. To My Parents— V. Venkataramani— K.M. Meera Sheriffa BegumTo My Mother— N. Anantharaman

Contents

Preface .............................................................................................................. xi Preface to the FirstEdition ............................................................................ xiii Acknowledgements........................................................................................... xv

1 UNITS AND DIMENSIONS 1–6

1.1 Introduction ..................................................................................... 11.2 Basic Units and Notations ............................................................... 11.3 Derived Units ................................................................................... 21.4 Definitions ....................................................................................... 3Worked Examples ..................................................................................... 3Exercises ................................................................................................... 6

2 MASS RELATIONS 7–34

2.1 Mass Relations in Chemical Reaction ............................................. 72.2 Conservation of Mass ...................................................................... 82.3 Avogadro’s Hypothesis .................................................................... 92.4 Limiting Reactant and Excess Reactant .......................................... 92.5 Conversion and Yield ....................................................................... 92.6 Composition of Mixtures and Solutions ........................................ 102.6.1 Weight Percent ................................................................... 102.6.2 Volume Percent .................................................................. 102.6.3 Mole Fraction and Mole Percent ....................................... 112.6.4 Atomic Fraction and Atomic Percent ................................ 112.6.5 Composition of Liquid Systems ........................................ 112.7 Density and Specific Gravity ......................................................... 122.7.1 Baume’ (°Be’) Gravity Scale ............................................. 122.7.2 API Scale (American Petroleum Institute) ........................ 122.7.3 Twaddell Scale ................................................................... 132.7.4 Brix Scale .......................................................................... 13Worked Examples ................................................................................... 13Exercises ................................................................................................. 32

viiviiiCONTENTS

3 IDEAL GASES 35–73

3.1 Relation between Mass and Volume for Gaseous Substances ....... 353.1.1 Standard Conditions .......................................................... 353.1.2 Ideal Gas Law .................................................................... 353.2 Gaseous Mixture ............................................................................ 363.2.1 Partial Pressure (PP) .......................................................... 363.2.2 Pure Component Volume (PCV) ....................................... 363.2.3 Dalton’s Law ..................................................................... 373.2.4 Amagat’s Law (or) Leduc’s Law ....................................... 373.3 Average Molecular weight ............................................................. 383.4 Density of Mixture ......................................................................... 38Worked Examples ................................................................................... 38Exercises ................................................................................................. 70

4 VAPOUR PRESSURE 74–86

4.1 Effect of Temperature on Vapour Pressure .................................... 744.2 Hausbrand Chart ............................................................................ 75Worked Examples ................................................................................... 75Exercises ................................................................................................. 85

5 PSYCHROMETRY 87–110

5.1 Humidity ........................................................................................ 875.2 Definitions ..................................................................................... 87Worked Examples ................................................................................... 90Exercises ............................................................................................... 106

6 CRYSTALLIZATION 111–121 Worked Examples................................................................................. 112 Exercises............................................................................................... 120

7 MASS BALANCE 122–179Worked Examples ................................................................................. 122Exercises ............................................................................................... 173

8 RECYCLE AND BYPASS 180–195

8.1 Recycle ........................................................................................ 1808.2 Bypass .......................................................................................... 1808.3 Purge............................................................................................ 180Worked Examples ................................................................................. 181Exercises ............................................................................................... 195

CONTENTS ix 9 ENERGY BALANCE 196–220

9.1 Definitions ................................................................................... 1969.1.1 Standard State .................................................................. 1969.1.2 Heat of Formation ............................................................ 1969.1.3 Heat of Combustion ......................................................... 1979.1.4 The Heat of Reaction ....................................................... 1979.1.5 Heat of Mixing ................................................................ 1979.2 Hess’s Law ................................................................................... 1979.3 Kopp’s Rule ................................................................................. 1989.4 Adiabatic Reaction Temperature ................................................. 1989.5 Theoretical Flame Temperature ................................................... 198Worked Examples ................................................................................. 198Exercises ............................................................................................... 217

10 PROBLEMS ON UNSTEADY STATE OPERATIONS 221–228 Worked Examples................................................................................. 221 Exercises............................................................................................... 227

T ables .................................................................................................... 229–234I Important Conversion Factors .............................................................229II Atomic Weights and Atomic Numbers of Elements ......................... 231

III(a) Empirical Constants for Molal Heat Capacities of Gases at Constant Pressure................................................................................. 234

III(b)Molal Heat Capacities of Hydrocarbon Gases .................................. 234Answers to Exercises........................................................................... 235–245Index...................................................................................................... 247–248

Preface

The objective of this book is to enrich a budding chemical engineer the techniques involved inanalyzing a process plant by introducing the concepts on units and conversions, mass and energybalances. This will enable him to achieve a proper design of process equipment. An attempt has beenmade to explain the principles involved through numerical examples. The problems are not onlyconfined to SI system of units but also worked out in other systems like FPS, CGS and MKS systems.We feel that our attempt will be more rewarding if students come across data presented in FPS, CGSand MKS systems while designing equipment, since different reference books give standard valuesand data in various units.

The book covers various interesting topics such as units and dimensions, mass relations, properties ofgases, vapour pressure, psychrometry, crystallization, mass balance including recycle and bypass,energy balance and unsteady state operations. The second edition is now enriched with additionalworked examples and exercises to give additional exposure and practice to students.

The text is designed for a one semester programme as a four credit course and takes care of thesyllabus on ‘Process Calculations’ of most of the universities in India.

V. Venkataramani N. Anantharaman K.M. Meera Sheriffa Begum

xi

Preface to the First Edition

Chemical engineers in process industries generally need to focus on design, operation, control andmanagement of a process plant. It is, therefore, absolutely essential for them to be conversant with themass and energy conservation techniques at every stage of the process to achieve economy in thedesign of process equipment in various units of the plant. This book aims at imparting knowledge ofthe basic chemical engineering principles and techniques used in analyzing a chemical process. Byapplying the relevant techniques, a chemical engineer is able to evaluate material and energy balancesin different units and present the information in a proper form so that the data can be used by themanagement in taking correct decisions. Keeping this in mind, an attempt has been made to give abrief theory on the principle involved and more emphasis on numerical examples.

Since data are generally obtained in different units, the worked examples are not confined to SI units,but to other systems as well, such as FPS, CGS and MKS systems of units. The examplesincorporated in the text are simple and concrete to make the book useful for self-instruction.

The text is organized into ten chapters and appends three important tables. The organization is suchthat the topics are presented in order of easy comprehension rather than following a logical sequence,e.g. the chapter on unsteady state operations has been included as the last chapter so that students canabsorb the problems easily. We strongly feel that once the student understands the topics presented inthis book, he will find other advanced courses in chemical engineering simple and easy to follow.

The topics covered in this book cater to the syllabi on ‘Process Calculations’ of most universitiesoffering courses in chemical engineering and its allied branches at the undergraduate level.

V. Venkataramani N. Anantharaman xiii

Acknowledgements

At the outset, we wish to thank the almighty for his blessings.

V. Venkataramani wishes to acknowledge his wife, Prof. (Mrs.) Booma Venkataramani, sons Mr. V.Ravi Chandar, Mr. V. Hari Sundar and his daughters-in-law Mrs. Vandana Ravi Chandar and Mrs.Ramya Hari Sundar and granddaughters Miss Vaishnavi Ravi Chandar and Miss Sadhana Hari Sundarfor their support, cooperation and patience shown during the preparation of this book.

N. Anantharaman wishes to thank his mother, wife, Dr. Usha Anantharaman, sons Master A. Srinivasand A. Varun for all their patience, cooperation and support shown during the preparation of thisbook. The encouragement received from his brothers and sisters and their family members isgratefully acknowledged. He also wishes to place on record the support received from his brothers-in-law and sisters-in-law and their family members.

K.M. Meera Sheriffa Begum wishes to acknowledge her mother, husband Mr. S. Malik Raj and babyM. Rakshana Roshan for all their encouragement, support and cooperation while preparing this book.She also wishes to place on record the support received from her parents-in-law. The supportreceived from her brothers, sisters, in-laws and their families is gratefully acknowledged.

We also thank Director, NIT, Tiruchirappalli for extending all the facilities and his words ofappreciation.We wish to acknowledge the support and encouragement received from the Head of ChemicalEngineering Department and all our colleagues during the course of preparation of this book.We also wish to place on record the suggestions received from students, especially those at NIT,Tiruchirappalli, and also the faculty from other institutions.

xv xviACKNOWLEDGEMENTS

We gratefully acknowledge all the well-wishers.Finally, we wish to thank the publishers, PHI Learning, New Delhi for encouraging us to bring out thesecond edition of the book.

V. Venkataramani N. Anantharaman K.M. Meera Sheriffa Begum

Units and Dimensions 1

1.1 INTRODUCTION

Chemical engineers are concerned with the design and development of processes which involvechanges in the bulk properties of matter. To make a quantitative estimation of these processes,chemical equations showing the quantities of reactants and products are used. Though internationallywe follow SI system of units, a chemical engineer is expected to be familiar and conversant with allthe systems so far adopted for measuring and expressing various quantities. A review of literature anddata over the years will be available in various units. These are used to express properties, processvariables and design parameters in FPS, CGS, MKS and SI systems of Units. Hence, one has to beconversant with their use and applications. This chapter deals with the basic notations and conversionof a given quantity from one system of units to another.

The quantities used in our analysis are classified as fundamental quantities and derived quantities.The fundamental quantities comprise length, mass, time and temperature. The quantities such as force,density, pressure, mass flow rate derived from the fundamental quantities are called derivedquantities. While handling these quantities, we come across different systems of units as mentionedearlier. Now let us see in detail these systems of units and their conversion from one unit to another.

1.2 BASIC UNITS AND NOTATIONSEnglish Metric Engineering System Engineering,CGS MKSInternational, FPS SI

Mass ( m) lb g kg kg Length (L)ft cmm m Time (t)ssss Temperature (T)°F °C °C K

1

Mass (m)1 kg = 2.205 lbLength (L)1 ft = 30.48 cm= 0.3048 mTime (t)1 h = 3600 sTemperature (T )

°C=°F 32(Celsius and Centigrade are same)1.8 1.3 DERIVED UNITS

Area: = length ´ breadth (L2):1 ft2 = 0.0929 m2

10.76 ft2 = 1 m2

Force: = mass ´ acceleration (mL/t2):1 dyne = 1 g cm/s2

(Force applied on a mass of 1 g, which gives an acceleration of 1 cm/s2)1 Newton (N) = 1 kg m/s2

= (1000 g) (100 cm)/s2

1 N = 105 g cm / s2 = 105 dynesWork/energy: = 1 kg m/s2

1 erg is the amount of work done on a mass of 1 g when it is displaced by 1 cm by applying a force of1 dyne.1 erg = [1 dyne] ´ [1 cm]= [1 g cm/s2] ´ [1 cm]= 1 g cm2/s2

1 Joule = (1 N) ´ (1 m)= 105 g cm/s2 ´ 100 cm= 107 g cm2/s2

1 Joule = 107 ergHeat Unit:1 Btu = 0.252 kcal = 252 cal1 cal = 4.18 J1 J/s = 1 W

1.4 DEFINITIONSSystem. This refers to a substance or group of substances under consideration, e.g. storage tank,water in a tank, hydrogen stored in cylinder, etc.

Process. Changes taking place within the system is called process, e.g. burning of fuel, or reactionbetween two substances like hydrogen and oxygen to form water.

Isolated system. Boundaries of the system are limited by a mass of material, and its energy content iscompletely detached from all other matter and energy. In an isolated system, the mass of the systemremains constant, regardless of the changes taking place within the system.

Extensive property. It is a state of system, which depends on the mass under consideration, e.g.volume.Intensive property. This state of a system is independent of mass. An example of this property istemperature.WORKED EXAMPLES1.1 The superficial mass velocity is found to be 200 lb/h.ft2. Find its equivalent in kg/s.m2

G (Mass velocity) = (200) lb/h.ft2=(200)ʈ¥¥ ¥11 1

m2

Á˜Ë¯ (3600 s) (0.0929) = 0.2712 kg/s.m2

1.2 Convert the heat transfer coefficient of value 100 Btu/h.ft2.°F into W/m2 °Ch (Heat transfer coefficient) = (100)È˘Í˙Î˚

(0.0929 m2)

(1.8 °C) [1 degree variation in Farh. scale is equivalent to 1.8 times the variation in celsius scale]

= 4.186 ¥ 10–2 kcal/s.m2 °C = 4.186 ¥ 10–2¥ 103¥ 4.18 W/m2 °C = 174.98 W/m2 °C

1.3 The rate of heat loss per unit area is given by (0.5) [(DT)1.25/(D)0.25] Btu/h ft2 for a process,where, DT is in °F and D is in ft. Convert this relation to estimate the heat flux in terms of kcal/h. m2

using DT in °C and D in m.We know that,

9 C1 + 32 = F15 9 C2 + 32 = F25 Therefore,1.8 [DC] = (DF) qT()1.25

0.5()0.25A

q , Btu/h ft2 = 0.5 '(°F)1.25A (ft)0.25

We know that,

1 Btu = 0.252 kcal1 ft2 = 0.0929 m2

1 ft = 0.3048 m

For DT °F = 1.8 DT °C ËÛ1.25

Btu/h ft2 = 0.5 'ÌÜ (ft)0.25 (a) ÌÜ ÍÝ (1.8 ' T °C) 1.25

= (0.5) (Dm/0.3048)0.25(b)

= (0.7746) '(°C)1.25

0.25(m)

Btu/h ft2 = (0.252 kcal)/h (0.0929 m2)= 2.713 kcal/h m2 (c)

From (b) and (c) we get the expression for heat flux in units of kcal/h m2 with temperature differencein Celsius and diameter in metre as:

Heat flux, kcal/h m2 =(m)0.25ËÛT °C)

1.25

'ÌÜ 2.713ÌÜ ÍÝ (2.101)( ' T °C) 1.25

= (m)0.25 (d)

Now let us check the conversion with the following data: D = 0.2 ft, i.e. D¢ = 0.06096 mDT = 18 °F, i.e. DT = 10 °CFrom Eq. (a), heat flux is = 0.5

(18)1.25

0.25

= 27.72 Btu/h ft2 = 75.2 kcal/h m2 (0.2) Also, from Eq. (d), heat flux = 2.101 (10)1.25 (0.06096)0.25

= 75.2 kcal/h m2

Both the values agree.

1.4 If Cp of SO2 is 10 cal/g mole K, what is the value in FPS units? The Cp value is the same in allunits, i.e. 10 Btu/lb mole °R.

1.5 Iron metal weighs 500 lb and occupies a volume of 29.25 litres. Find the density in kg/m3.Basis: 500 lb of Iron = 500/2.2 = 227.27 kg29.25 lit = 29.25 ´ 10–3 m3227.27´ 10–3 = 7770 kg/m3Density = 29.25

1.6 Etching operation follows the relation d = 16.2 – 16.2e–0.021t, where t is in s. and d is in microns.Convert this equation to evaluate d in mm with t in min.

d = 16.2 [1 – e–0.021t]Let d¢ be in mm and t¢ be in min. (d = d¢ ´ 103 and t = t¢ ´ 60) Then, d = d¢ × 103 = 16.2 [1 – e–0.021t

× 60]d¢ = 0.0162 [1 – e–1.26t¢]

1.7 The density of fluid is given by r = 70.5 exp (8.27 × 10–7). Convert this equation to calculate thedensity in kg/m3 with pressure in N/m2.

1000 kg/m3 = 62.43 lb/ft314.7 psi = 1.0133 × 105 N/m2

1 kg/m3 = 62.43 × 10–3 lb/ft3Let, r¢ be in kg/m3 and p¢ be in N/m2.Then, (r¢ × 62.43 × 10–3) = 70.5 × exp (8.27 × 10–7× p¢ × 14.7/1.0133 × 105) r¢ (kg/m3) = 1.129 ×103 × exp [119.97 × 10–12 × p¢ (N/m2)]

1.8 Vapour pressure of benzene in the range of 7.5 °C to 104 °C is given by log10 (p) = 6.9057 –1211/(T + 220.8), where T is in °C and p is in torr 1 torr = 133.3 N/m2. Convert it to SI units.

Let p¢ be in N/m2and T¢ be in K.Then, log ÉÙ ÊÚ pÈØ = 6.9057 – 1211

(T273) 220.8 1211log p¢ – log 133.3 = 6.90305 – T 52.2 log (p1211.¢) = 9.0305 – T52.2 EXERCISES1.1 Convert the following quantities:

(a) 42 ft2/h to cm2/s(b) 25 psig to psia(c) 100 Btu to hp-h(d) 30 N/m2 to lbf/ft2(e) 100 Btu/h ft2 °F to cal/s cm2 °C(f) 1000 kcal/h m°C to W/m K

1.2 The heat transfer coefficient for a stream to another is given by h = 16.6 Cp G0.8/D0.2

where h = Heat transfer coefficient in Btu/(h)(ft)2(°F) D = Flow diameter, inchesG = Mass velocity, lb/(s)(ft)2

Cp = Specific heat, Btu/(lb) (°F)

Convert this equation to express the heat transfer coefficient in kcal/ (h)(m)2(°C)With D = Flow diameter in m, G = Mass velocity in kg/(s)(m)2 and Cp = Specific heat, kcal/(kg) (°C)

1.3 Mass flow through a nozzle as a function of gas pressure and temperature is given by m = 0.0549

p/(T)0.5 where m is in lb/min, p is in psia and T is in °R, where T(°R) = T °F + 460. Obtain anexpression for the mass flow rate in kg/s with p in atmospheres (atm) and T in K.

1.4 The flow past a triangular notch weir can be calculated by using the following empirical formula:

q = [0.31 h2.5/g0.5] tan Fwhere q = Volumetric flow rate, ft3/sh = Weir head, ft

g = Local acceleration due to gravity, ft/s2

F = Angle of V-notch with horizontal plane

1.5 In the case of liquids, the local heat transfer coefficient, for long tubes and using bulk-temperatureproperties, is expressed by the empirical equation,

h= 0.023 G0.8 k0.67 C0.5/(D0.2m0.47

p )

where G = Mass velocity of liquids, lb/ft2.sk = Thermal conductivity, Btu/ft.h.°FCp = Specific heat, Btu/lb °FD = Diameter of tube, ftm = Viscosity of liquid, lb/ft.s

Convert the empirical equation to SI units.

Mass Relations 2

2.1 MASS RELATIONS IN CHEMICAL REACTION

In stoichiometric calculations, the mass relations between reactants and products of a chemicalreaction are considered and are based on the atomic weight of each element involved in the reaction.

For the following reactions the material balance is established as indicated below:(i) CaCO3 ® CaO + CO2 (2.1) [40 + 12 + 3 ´ 16] ® [40 + 16] + [12 + 32] 100 ® 56 + 44(ii) 3Fe + 4H2O ® Fe3O4 + 4H2 (2.2) (3 ´ 55.84) + 4(2 ´ 1 + 16) ® (55.84 ´ 3 + 4 ´ 16) + 4(2 ´ 1)167.52 + 72 ® 231.52 + 8 239.52 ® 239.52

Based on the reactions given by Eqs. (2.1) and (2.2) we conclude that when 100 parts by weight ofCaCO3 reacts, 56 parts by weight of CaO and 44 parts by weight of CO2 are formed. Similarly, when167.52 parts by weight of iron reacts with 72 parts by weight of steam (water), we get 231.52 partsby weight of magnetite and 8 parts by weight of hydrogen. Thus the total weight of reactants is always

equal to the total weight of products.

Such computations will help one to estimate the quantity of reactants needed to obtain a specifiedamount of product.

gram atom (or g atom) = Mass in grams/Atomic weight katom (or kg atom) = Mass in kg/Atomicweightgram mole (or g mole) = Mass in grams/Molecular weight kmole (or kg mole) = Mass inkg/Molecular weight

7

The conclusions based on reactions (2.1) and (2.2) on material balance can be expressed in otherforms too, as per the definitions given above:1 kmole of CaCO3 gives 1 kmole of CaO and 1 kmole of CO2Similarly, 3 kmoles of iron reacts with 4 kmoles of steam (water) to yield 1 kmole of oxide and 4kmoles of hydrogen. When such balances (on molar basis) are made, the number of moles on thereactants side need not be equal to the total numbers of moles on the product side.

One atom of oxygen weighs = 16 grams OOne atom of hydrogen weighs = 1 gram HOne molecule of oxygen weighs = 32 grams OOne molecule of hydrogen weighs = 2 grams H

In other words,

16 grams of oxygen32 pounds of oxygen2 g atoms of oxygen1 gram of hydrogen2 kg of hydrogen2 kg atoms of hydrogen = 1 kmole of hydrogen

\ g atom or lb atom = Mass in grams or pounds/Atomic weight \ g mole or lb mole = Mass in grams orpounds/Molecular weight = 1 g atom of oxygen = 1 lb mole of oxygen = 1 g mole of oxygen = 1 gatom of hydrogen = 1 kmole of hydrogen

2.2 CONSERVATION OF MASS

The law of conservation of mass states that mass can neither be created nor be destroyed. It is thebasic principle adopted in solving the material balance problems in chemical process calculations,whether a chemical reaction is involved or not. However, while applying the law of conservation ofmass, one should not apply it for the conservation of molecules. We frequently come across chemicalreactions in which the total number of moles on the reactant side is not equal to the total number ofmoles on the product side. For example:

Na2CO3 + Ca(OH)2 ® CaCO3 + 2NaOH

The total number of moles on the reactant side is 2 and on the product side 3. Here the mass balanceis ensured but not the mole balance. Now consider the reaction:

2Cr2O3 + 3CS2 ® 2Cr2S3 + 3CO2

Here the total number of moles both on the reactant side and the product side is 5. Hence both theconservation of mass and the conservation of moles are observed.

2.3 AVOGADRO’S HYPOTHESIS

1 g mole of any gaseous substance at NTP occupies 22,414 cc or22.414 litres and 1 lb mole of the same substance occupies 359 ft3at NTP.1 kmole of any gaseous substance occupies 22.414 m3 at NTP.

Note: NTP = Normal temperature (273 K) and pressure (1 atmosphere) are also referred to at timesas standard conditions (SC).2.4 LIMITING REACTANT AND EXCESS REACTANT

For most of the chemical reactions the reactants will not be used in stoichiometric proportion orquantities. One of the reactants will be present in excess and remain unreacted even when the otherreactant has completely reacted. The reactant thus present in excess is termed excess reactant and theother reactant which is present in a lesser quantity and cannot react with whole of the other reactant(excess reactant) is called limiting reactant. All calculations involved in estimating the quantity ofproduct and conversion are always based on the limiting reactant. The amount by which any reactantis present in excess to that required to combine with the limiting reactant is usually expressed aspercentage excess. The percentage excess of any reactant is defined as the percentage ratio of theexcess to that theoretically required by the stoichiometric equation for combining with the limitingreactant. A limiting reactant is the one, which will not be present in the product, whereas the excessreactant is the one, which will always be present in the product.

Let us consider that 18 kg of carbon is burnt with 32 kg of oxygen. As per stoichiometryC + O2Æ CO2 i.e. 12 kg of carbon will burn with 32 kg of oxygen to form 44 kg of CO2.

Hence, for 18 kg of carbon to react fully we should have 48 kg of oxygen. Since 32 kg of oxygenalone is available, it is called the limiting reactant and carbon is called the excess reactant. For 32 kgof oxygen to react fully, it is sufficient to have 12 kg of carbon. However 6 kg of carbon is present inexcess.

Hence % excess of carbon is =6¥ 100 = 50%.12 2.5 CONVERSION AND YIELD These terms are used for a chemical reaction where the reactantsgive out new compounds or products.

Conversion is the ratio of the amount of material actually converted to that present initially, whereasthe yield is the amount of desired product actually formed compared to that which can be formedtheoretically.

Conversion for a reaction is based on the limiting reactant whereas the yield is based on the productformed.2.6 COMPOSITION OF MIXTURES ANDSOLUTIONSDifferent methods are available for expressing composition of mixtures of gases, liquids and solids.

Conventionally the composition of solids is either expressed on weight basis or mole basis. Let usconsider a binary system comprising components A and B.

W = Total weight of the system.WA, WB = Weight of components A and B respectively. MA, MB = Molecular weight of components Aand B respectively,

if they are compounds.AA, AB = Atomic weight of components A and B respectively, if they are elements.V = Volume of the system.VA and VB= Pure component volume of components A and B respectively.2.6.1 Weight PercentThis is defined as the ratio of weight of a particular component to the total weight of the system inevery 100 part, i.e.Weight % of A =WA ¥ 100W

This method of expressing composition is generally employed in solid and liquid systems and notused in gaseous system. One major advantage of weight percent is, its independence to changes intemperature and pressure.

The composition of a solid mixture is to be always taken as weight % when nothing is mentionedabove its units.2.6.2 Volume PercentThe ratio of the volume of each component and the total volume of the system, for each 100 part of thetotal volume is called volume percent Volume % of A =VA´ 100V

This method of expressing composition is employed always for gases, rarely for liquids and seldomfor solids. The composition of a gas mixture is to be taken as volume % when nothing is mentioningabout its units.

The volume % is also equal to mole % for ideal gases but not for liquids and solids. This is based onAvogadro’s law.2.6.3 Mole Fraction and Mole Percent

These concepts are generally adopted in the case of a mixture containing molecules of differentspecies. WA

MAMole fraction of A = WA WB MA MB Mole % of A = Mole fraction of A ´ 1002.6.4 Atomic Fraction and Atomic Percent This is adopted when a mixture contains two or moreatoms. WA Atomic fraction of A=

AA WA WB AA AB

Atomic % of A = Atomic fraction of A ´ 1002.6.5 Composition of Liquid SystemsIn the case of liquids we come across more number of methods of expressing compositions of theliquid constituents.

(i) Weight ratio (ii) Mole ratio (iii) Molality(iv) Molarity= Weight of solute/weight of solvent= g moles of solute/g moles of solvent= g moles of solute/1 kg of solvent= Number of g moles of solute/1 litre of solution

(v) Normality (N) = Number of gram equivalents of solute/1 litre of solution

Hence, concentration in grams per litre = Normality ( N) ´ Equivalent weight of soluteFor very dilute aqueous solutions, molality = molarity

2.7 DENSITY AND SPECIFIC GRAVITY

Density is defined as mass per unit volume and it varies with temperature. Specific gravity is the ratioof density of a liquid to that of water. However, in the case of gases, it is defined as the ratio of itsdensity to that of air at same conditions of temperature and pressure.

Over a narrow range of temperature, the variation in density of solids is not high. However, in thecase of liquids and gases the variation in density is significant. Similarly, the densities varysignificantly with concentration also. This property of density and specific gravity varying withconcentration is very widely used both in industries and markets as an index for finding thecomposition of a system comprising a specific solute and a specific solvent.

Several scales are in use in which specific gravities are expressed in terms of a degree, which arerelated to specific gravities and densities by arbitrary mathematical definitions.

2.7.1 Baume (°Be) Gravity ScaleFor liquids lighter than water

Degrees Baume’ =140– 130G Gis the specific gravity at 60°F 15¦µ 60 15 §¶ · Thus, water will have a gravity of 10° Be’ and this degree decreases with increase in specificgravity.For liquids heavier than water 145Degrees Baume’ = 145 – G In this scale the degree increases with increase in specific gravity.2.7.2 API Scale (American Petroleum Institute) This scale is used for expressing gravities ofpetroleum products. This is similar to Baume’ scale for liquids lighter than water.Degrees API = 141.5– 131.5G 2.7.3 Twaddell ScaleThis scale is used for liquids heavier than water. Degrees Twaddell (Tw) = 200 (G – 1.0)2.7.4 Brix ScaleThis scale is used in sugar industry and 1 degree Brix is equal to 1% sugar in solution.Degree Brix = 400– 400G WORKED EXAMPLES 2.1 Convert 5000 ppm into weight %. 5000 ×100= 0.5%106 2.2 The strength of H3PO4 was found to be 35% P2O5. Find the weight % of the acid.The acid can be split into 2H PO → P O + 3H O 34 25 2(2 98)142(3 18)×

196 units of the acid contains 142 units of the pentaoxide. The weight % of pentaoxide is (142/196)which is 72.5% for pure acid. When the strength of pentoxide is 35%, the weight % of acid is �� =48.3%����

2.3 What is the volume of 25 kg of chlorine at standard condition?25 kg Cl2 = 25kmoles of Cl22 ×35.46 25 × 22.414= 7.9 m3Volume = 2 × 35.46

2.4 How many grams of liquid propane will be formed by the liquefaction of 500 litres. of the gas atNTP? Molecular weight of propane (C3H8) is 44

1 g mole of any gas occupies 22.414 litres at NTP500 litres of propane at NTP = 500 = 22.31 g moles22.414

22.31 g moles of propane weighs = 22.31 ´ 44 = 981.52 g. 2.5 Find the volume of (a) 100 kg ofhydrogen and (b) 100 lb of hydrogen at standard conditions?(a) 100 kg of H2 = 50 kmoles of hydrogenvolume occupied by 50 kmoles of hydrogen º 50 ´ 22.414

= 1120.7 m3 (b) 100 lb of H2 º 50 lb moles of hydrogenVolume occupied by 50 lb moles of hydrogen º 50 ´ 359 = 17950 ft3

2.6 A solution of naphthalene in benzene contains 25 mole % Naphthalene. Express the compositionin weight %. Basis: 100 g moles of solution

Component Molecular weight Weight, Actual g mole weight, g

Composition in weight percent

Naphthalene C10H8 128 Benzene C6H6 78 25 25 128 = 3200 75 75 78 = 5850 3200 100/9050 = 35.35 5850 100/9050 = 64.65

Total 9050 100.002.7 What is the weight of one litre of methane CH4 at standard conditions? 22.414 litres of any gas atNTP is equivalent to 1 g mole of that gas 1 = 0.0446 g mole\ 1 litre of methane º 1 ´22.414 \ Weight of one litre methane = 0.0446 ´ 16 = 0.714 g

2.8 A compound whose molecular weight is 103 analyses C : 81.5, H : 4.9 and N : 13.6 by weight.What is the formula?Basis: 100 g of substance

Element Atomic weight

Carbon 12Hydrogen 1Nitrogen 14

Weight, g Weight, Rounding of Weight of g atom atoms each element

81.5 81.5/12 = 6.8 7 844.9 4.9/1 = 4.9 5 513.6 13.6/14 = 0.9 1 14

Total 103Hence the formula obtained after rounding is correct. So the molecular formula is C7H5N

2.9 An analysis of a glass sample yields the following data. Find the mole %.

Na2O : 7.8%, MgO : 7.0%, ZnO : 9.7%, Al2O3: 2.0%, B2O3 : 8.5% and rest SiO2.Basis: 100 g of glass sample

Component Weight, g Molecular weight g mole mole %

Na2O 7.8 62.0 0.1258 7.665MgO 7.0 40.3 0.1737 10.583ZnO 9.7 81.4 0.1192 7.262Al2O3 2.0 102.0 0.0196 1.194B2O3 8.5 69.6 0.1221 7.439SiO2 65.0 60.1 1.0815 65.857

Total 100.0 — 1.6419 100.0

2.10 A gaseous mixture analyzing CH4: 10%, C2H6: 30% and rest H2 at 15 °C and 1.5 atm is flowingthrough an equipment at the rate of 2.5 m3/min. Find (a) the average molecular weight of the gasmixture, (b) weight % and (c) the mass flow rate.

Basis: 100 g moles of the gaseous mixture.Component Weight, Molecular Weight, Weight % g mole weight g

CH4 10 16 160 13.56C2H6 30 30 900 76.27H2 60 2 120 10.17

Total 100 — 1180 100The average molecular weight =1180= 11.8100 Volumetric flow rate at standard conditions = 2.5¥ ������¥ (273/288) = 3.555 m3/minMoles of the gas = 3.555 = 0.156 kmole22.414 Mass flow rate = moles ¥ average molecular weight = 0.156 ¥ 11.8 = 1.84 kg/min 2.11 In anevaporator a dilute solution of 4% NaOH is concentrated to 25% NaOH. Calculate the evaporation ofwater per kg of feed. Basis: 1 kg of feed.NaOH present is 0.04 kg, which appears as 25% in the thick liquor formedWeight of thick liquor formed = 0.04 = 0.16 kg0.25 Weight of water evaporated = (1 – 0.16) = 0.84 kgWater evaporated per kg of feed = 0.84 kg

2.12 The average molecular weight of a flue gas sample is calculated by two different engineers. Oneengineer used the correct molecular weight of N2 as 28, while the other used an incorrect value of 14.They got the average molecular weight as 30.08 and the incorrect one as 18.74. Calculate the %volume of N2 in the flue gases. If the remaining gases are CO2 and O2 calculate their compositionalso.

Basis: 100 g moles of flue gasComponent g mole I Engineer II Engineer

N2x 28x 14x CO2y 44y 44y O2z 32z 32z

Total 100 3008 1874

x + y + z = 100 (i)28x + 44y + 32z = 3008 (ii)14x + 44y + 32z = 1874 (iii) Solving Eqs. (i), (ii) and (iii), we getx = Moles of nitrogen = 81%y = Moles of carbon dioxide = 11%z = Moles of oxygen = 8%

2.13 An aqueous solution contains 40% of Na2CO3 by weight. Express the composition in molepercent.Basis: 100 g of solution

Component grams Molecular g mole Composition in weight mole % Na2CO3 40 106 40/106 = 0.3770.377 ´ 100/3.71 = 10.16 Water 60 18 60/18 = 3.333 3.333 ´ 100/3.71 = 89.84 Total 3.710 100.002.14 What is the weight of iron and water required for the production of 100 kg of hydrogen? 3Fe 4H O +→Fe O + 4H ↑234 2

(3×55.84) (4 18) (355.84) (4 16) (4 2 1)×+××××167.52 72 231.52 8

239.52 239.52

Method 1 (Based on absolute mass)167.52 kg of Fe is required for producing 8 kg of H2\ For producing 100 kg of H2 (by stoichiometry)

100 t167.52Iron (Fe) required = 8 = 2094 kgSimilarly, for getting 100 kg of H2 the amount of steam (H2O) required is= 100 t 72= 900 kg8 The total weight of reactants is2094 kg Fe and 900 kg H2O = 2994 kgThe weight of Fe 231.52´ 2094 = 2894 kg3O4 formed is = 167.52

\ The total weight of products is (100 kg H2 + 2894 kg Fe3O4) = 2994 kgThe total weight of reactants is (2094 kg Fe + 900 kg H2O) = 2994 kg

Method 2 (Based on moles)

100 kg of H24 kmoles H2 comes from

50 kmoles of H2 comes from Weight of 37.5 katoms Fe50 kmoles H2 from

Weight of 50 kmoles H2O Moles of Fe3O4 formed isWeight of Fe3O4 formed is

Total weight of reactants Total weight of products = 50 kmoles= 3 katoms of Fe (by stoichiometry)

= ¦µ50

t§¶ = 37.5 katoms Fe¨·4= (37.5 ´ 55.84) = 2094 kg of iron

= 50¦µ kmoles of water4t§¶ · = (50 ´ 18) = 900 kg H2O=37.5kmoles3 37.5 = ¦µ §¶´ 231.52 = 2894 kg¨· = 2994 kg= 2994 kg

2.15 How much super phosphate fertilizer can be made from one ton of calcium phosphate 93.5%pure?Atomic weights are: Ca : 40, P : 31, O : 16, S : 32

Ca3(PO4)2 + 2H2SO4Æ CaH4(PO4)2 + 2CaSO4310 (2 ¥ 98) 234 (2 ¥ 136) 506 506One ton of raw calcium phosphate contains 0.935 tons of pure calcium phosphateWeight of super phosphate formed is = 234 ¥0.935= 0.70577 tonne\310 2.16 SO2 is produced by the reaction between copper and sulphuric acid. How much Cu must be usedto get 10 kg of SO2? Cu +2H SO

24

→ CuSO +SO +2H O 4 2 263.54 64

64 kg of sulphur dioxide is obtained from 63.54 kg of copper. 10 kg of sulphur dioxide will beobtained from 9.93 kg of copper. 2.17 How much potassium chlorate must be taken to produce thesame amount of oxygen that will be produced by 2.3 g of mercuric oxide? 2KClO3 Æ 2KCl + 3O2(2¥ 122.46) (2¥ 74.46) (6¥ 16) 244.92 148.92 96

2HgO Æ 2Hg + O2(2¥ 216.6) (2¥ 200.6) (2¥ 16)

433.2 g HgO gives 32 g of oxygen2.3 g of HgO will give (32 ¥ 2.3/433.2) = 0.1698 g of O20.1698 g O2 is obtained from (244.92 ¥ 0.1698)/96 = 0.4332 g of KClO3.

2.18 Ammonium phosphomolybdate is made up of the radicals NH3, H2O, P2O5 and MoO3. What is %composition of the molecule with respect to these radicals?

The formula of ammonium phosphomolybdate is(NH4)3PO4◊ 12MoO3◊ 3H2OFirst let us form the final product from the radicals:3NH3 + 4.5H2O + 12MoO3 + ½P2O5Æ (NH4)3PO4 12MoO3·3H2O

(3¥17 = 51) (4.5¥18 = 81) (12¥144 = 1728) (½¥142 = 71) 1931

% of NH3=51 ¥100= 2.641931 % of H2O= 81 ¥100= 4.191931 % of MoO3 = 1728 ´100= 89.491931 % of P2O5=71100= 3.68´1931 Total = 100.002.19 How many grams of salt are required to make 2500 g of salt cake? How much Glauber’s salt canbe obtained from this?The molecular formula of Glauber’s salt is Na2SO4 × 10H2O(142 + 180 = 322)

2NaClH SO Na SO + 2HCl (2×58.46 =116.92)

+→24 24 98 142 (2×36.46)

Thus, 142 g of Na2SO4 is obtained from 116.92 g NaCl. 2500 g of salt cake is obtained from 116.92 ´2500/142 = 2058.45 g NaClHence, salt needed is 2058.45 gGlauber’s salt (Na2SO4 × 10H2O) obtained is 2500 ´ 322/142 = 5669 g

2.20 (a) How many grams of K2Cr2O7 are equivalent to 5 g KMnO4? (b) How many grams of KMnO4are equivalent to 5 g K2Cr2O7? 2KMnO4 + 8H2SO4 + 10FeSO4 ® 5Fe2(SO4)3 + K2SO4 + 2MnSO4

+ 8H2OK2Cr2O7 + 7H2SO4 + 6FeSO4 ® 3Fe2(SO4)3 + K2SO4 + (Cr2SO4)3 + 7H2O2KMnO4 gives 5Fe2(SO4)3(2 158 = 316) (5 400 = 2000)

K2Cr2O7 gives 3Fe2(SO4)3(294) (3 400 = 1200)

1200 t 316= 189.6 g KMnO4\ 294 g K2Cr2O7 is equivalent to 2000 3 t189.6= 1.935 g KMnO4\ 3 g K2Cr2O7 º294 Similarly, 5 g KMnO4

5 t 294= 7.75 g K2Cr2O7 º189.6 Alternatively, 53t= 7.75 g K2Cr2O71.935 2.21 If 45 g of iron react with H2SO4, how many litres of hydrogen are liberated at standardcondition?There are two possible reactions in this case: (a) Case I Fe H SO FeSO + H (55.85)

+→ 24 4 2 (i)(2) The weight of hydrogen formed by reaction (i) is = 45 ¥2 = 1.611 g,55.85 i.e. 1.611= 0.806 g mole2 0.806 g mole ∫ 0.806 ¥ 22.414 = 18.06 litres(b) Case II 2Fe3H SO Fe (SO ) + 3H (111.7) +→24 2 43 2 (ii)(6) The moles of hydrogen formed by reaction (ii) is 45 ¥6 = 2.418 g111.7 2.418 g H2 = 1.209 g moles of hydrogen ∫ 1.209 ¥ 22.414 ∫ 27.1 litres2.22 A natural gas has the following composition by volume CH4 : 83.5%, C2H6: 12.5%, and N2: 4%.Calculate the following:

(a) composition in mole % (b) composition in weight % (c) average molecular weight (AVMWT) (d)density at standard condition (kg/m3) Basis: 100 kmoles of gas mixture

Component Molecular mole % Weight, kg Weight % weight

CH4 16 83.5 83.5¥ 16 = 1336 1336¥ 100/1823 = 73.29C2H6 30 12.5 12.5¥30 = 375 375¥ 100/1823 = 20.57N2 28 4.0 4.0¥28 = 112 112¥100/1823 = 6.14

Total 1823 100.0

(c) Average molecular weight =1823= 18.23100 Volume at standard condition = 00 ¥ 22.414 = 2241.4 m3

(d) Density of gas at standard condition = 1823= 0.813 kg/m3 2241.4 2.23 Convert 54.75 g/litre of HCl into molarity.Molarity = g moles/litre of solution= 54.75 = 1.536.45

2.24 A solution of NaCl in water contains 230 g of NaCl per litre at 20 °C. The density of thesolution at this temperature is 1.148 g/cc. Find the composition in (a) weight % (b) volume % ofwater (c) mole % (d) atomic % (e) molality and (f) g NaCl/g water.

Basis: (a) 1 litre of solution has a weight of 1148 gComponent Molecular Weight, g Weight, %g mole mole % weightNaCl 58.5 230 20.03 230/58.5 = 3.93 3.93/54.93 = 7.15 Water 18 918 79.97 918/18 = 51.00 51/54.93 = 92.85 Total 1148 100.00 54.93100.00

(b) Volume % of water: 918 g is present in 1 litre of solution, i.e.918 cc water is present in 1000 cc. of solution (density of water is1 g/cc)Volume % = 91.8%(d)

Element g atoms Atomic %

Na 3.93 2.443Cl 3.93 2.443H 102.00 63.409O 51.00 31.705

Total 160.86 100.000(All are based on the molecular formula)

(e) Molality = g moles of solute in 1 kg of solvent (3.93 1000/918) or, 3.93 g moles of NaCl ispresent in 918 g of water (i.e.) 4.28 g moles/1000 g of solventMolality = 4.28Molarity = Moles of solute per litre of the solution = 3.93

(f) g NaCl/g water = 230 = 0.2529182.25 A benzene solution of anthracene contains 10% by weight of the solute. Find the composition interms of (a) molality (b) mole fraction. Basis: 100 g of solution

Component Molecular Weight, g Weight, mole weight g mole fraction Anthracene 178 10 (10/178)0.046 0.0562 Benzene 78 90 (90/78) 0.954 1.1538 Total 1.2100 1.00Molality = g moles of anthracene in 1000 g benzene=0.0562 1000 = 0.62490¥

2.26 Calculate the weight of NaCl that should be placed in a 1 litre volumetric flask to prepare asolution of 1.8 molality. Density of this solution is 1.06 g/ccMolality = g moles of NaCl/1000 g of water

= 1.8or, 1.8 g moles NaCl = (1.8 ¥ 58.46) = 105.228 g

Component Weight, g Weight % NaCl 105.228 9.52H2O 1000.000 90.481105.228 100.00

Density of this solution = 1.06 g/ccVolume of this solution, i.e. mass/density =1105.228= 1042.67 cc1.06 or 1042.67 cc of this solution contains 105.228 g of NaCl1000 cc of this solution will have = 1000 ¥ 105.228\1042.67= 100.92 g of NaCl. NaCl needed = 100.92 g

2.27 For the operation of a refrigeration plant it is desired to prepare a solution of 20% by weight ofNaCl solution.(a) Find the weight of salt that should be added to one gallon of

water at 30°C?(b) What is the volume of this solution?Basis: 100 lb of solutionIt will have 20 lb NaCl and 80 lb water80 lb water = 1.28 ft3 (since the density of water is 62.47 lb/ft3) We know that 1 ft3 = 7.48 gallonsTherefore, 1.28 ft3 = 9.57 gallons.

20 (a) Weight of salt per gallon of water = ¦µ §¶ = 2.09 lb.¨· (b) Specific gravity of NaCl solution at 30 °C = 1.14\ Density of solution = 1.14 ´ 62.4 = 71.14 lb/ft3 Weight of 1 gallon of water = 62.47 = 8.35 lb.7.48 Total weight of solution = weight of water + weight of salt = 8.35 + 2.09 = 10.44 lb.Hence, volume of the above solution = 10.44 = 0.147 ft3 71.14 = 1.1 gallons.2.28 (a) A solution has 100° Tw gravity. What is its specific gravity and °Be’?(b) An oil has a specific gravity of 0.79. Find °API and °Be’ (a) 100 = 200 (G – 1) \ G = 1.5°Be’ = 145 – 145 145

= 145 – ¦µ = 48.3 °Be’G §¶ · 141.5 141.5 ¦µ – 131.5 = 47.6 · – 131.5 = §¶(b) °API =¦µ

§¶ ¨· 140 °Be’= §¶ ¦µ – 130 = 47.2¨·

2.29 An aqueous solution contains 15% ethyl alcohol by volume. Express the composition in weight% and mole %. Density of ethyl alcohol and water are 790 kg/m3 and 1000 kg/m3 respectively. Basis:1 m3 of solution.

Compound Molecular Volume, Density, Weight, Number Weight mole weight m3 kg/m3 kg of moles %%

Ethanol 46 0.15 790 118.5 2.576 12.235 5.173 Water 18 0.85 1000 850 47.222 87.765 94.827 Total 1.00 968.5 49.798 100 100

2.30 The quality of urea is expressed in terms of nitrogen content. If the nitrogen content in the sampleis only 40%, estimate the purity of sample in terms of urea content.

The molecular weight of urea (NH2CONH2) is 60 and that of N2 is 28. Basis: 100 kg of sample

60 kg of urea has 28 kg of N2

100 kg of urea will have = 28 × 100 = 46.67 kg of N260

(Theoretically)The given sample has 40% N2

Hence, the % purity is = 40 × 100 = 85.71%46.67 2.31 If the nitrogen content in ammonium nitrate sample is 28%, estimate the purity of ammoniumnitrate.Molecular weight of ammonium nitrate, NH4NO3 = 80% Nitrogen in pure ammonium nitrate = 28 × 100= 35%80 The % of nitrogen in the sample is 28 28 Hence, the purity of ammonium nitrate is §¶ ¦µ × 100 = 80%¨·

2.32 Nitrobenzene is produced by reacting nitrating mixture with benzene. The nitrating mixturecontains 31.5% HNO3, 60% H2SO4 and 8.5% H2O. A charge contains 663 kg of benzene and 1700 kg

of nitrating mixture which sent into the reactor. If the reaction is 95%, then calculate the amount ofnitrobenzene and spent acid produced. The reaction is

C6H6 + HNO3 ® C6H5NO2 + H2OFeed, C6H6 : 663/78 = 8.5 kmoleHNO3 : 31.5% of 1700 kg = 535.5 kg = 8.5 kmoles H2SO4 : 60 % of 1700 kg = 1020 kg = 10.408kmoles H2O : 8.5 % of 1700 kg = 144.5 kg = 8.028 kmoles Reaction is 95% completeHence, HNO3 unreacted : 0.05 × 8.5 = 0.425 kmole C6H6 unreacted : 0.05 × 8.5 = 0.425 kmoleH2SO4 unreacted : 10.408 kmolesH2O unreacted : 8.028 kmolesH2O formed : 8.5 × 0.95 = 8.075 kmoles Nitrobenzene formed : 8.5 × 0.95 = 8.075 kmoles

Component Weight, Molecular Weight, Weight, kmole weight kg %

HNO3 0.425 63 26.775 1.133H2SO4 10.408 98 1020.000 43.165H2O (8.075 + 8.028) = 16.103 18 289.850 12.266Nitrobenzene 8.075 123 993.225 42.032C6H6 0.425 78 33.150 1.403Total 2363.00 100%

Nitrobenzene produced = 993.225 kgSpent acid = 26.775 + 1020.000 + 289.85 = 1336.63 kg 2.33 A sample of caustic soda flake contains74.6% Na2O by weight. Estimate the purity of flakes.

Reaction is as follows:2NaOH ® Na2O + H2OAmount of Na2O in pure flakes = 62 × 100/80 = 77.5% % Purity = 0.746/0.775 × 100 = 96.26%

2.34 Two kg of CaCO3 and MgCO3 was heated to a constant weight of 1.1 kg. Calculate the %amount of CaCO3 and MgCO3 in reacting mixture.

Reaction is as follows:CaCO3 ® CaO + CO2(100) (56) (44)MgCO3 ® MgO + CO2(84) (40) (44)

Let, x be the amount of CaCO3.Therefore, (2 – x) be the weight of MgCO3100 kg of CaCO3 gives 56 kg of CaOTherefore, x kg of CaCO3 gives 56x = 0.56x kg of CaO.100

Similarly, 84 kg of MgCO3 gives 40 kg of MgO

40 Therefore, (2 – x) kg of MgCO3

gives ©¸ ª¹× (2 – x) kg of MgO

ǼThe weight of product left behind is 1.1 kg, i.e. weight of MgO + CaO left behind0.56x + (0.4672)(2 Рx) = 1.10.0838x = 1.1 Р0.96524Therefore, x = 1.761 kg

Component Weight, kg Weight, %

CaCO3 1.761 88.05MgCO3 0.239 11.95Total 2.000 100.00

2.35 The composition of NPK fertilizer is expressed in terms of N2, P2O5 and K2O each of about 15weight %. Anhydrous ammonia, 100% phosphoric acid and 100% KCl are mixed to get 1 ton offertilizer. Estimate the amount of filler in the NPK fertilizer.Basis: 1000 kg of fertilizerReactions are:

2NH3 ® N2 +3H2(34) (28) (6)2H3PO4 ® P2O5 +3H2O(196) (142) (54)2KCl + H2O ® K2O + 2HCl(149) (18) (94) (73)N2, K2O and P2O5 are each equivalent to 15 weight % = 150 kg each Ammonia reacted = 34 ×150= 182.14 kg28 H150 = 207.04 kg3PO4 needed = 196 × 142 KCl needed = 149 × 150= 237.77 kg94 The amount of inert material/filler = 1000 – 626.95 = 373.05 kg

2.36 A solution whose specific gravity is 1 contains 35% A by weight and the rest is B. If the specificgravity of A is 0.7, find the specific gravity of B.Basis: 1000 kg of solution

Weight of A: 350 kgWeight of B: 650 kg

Volume of solution = 1 m3 (since density is 1000 kg/m3 due to specific gravity being unity)Mass/volume = densityAssuming ideal behaviour 350 650= 1700 SB rB = 1300 kg/m3

Therefore, specific gravity of B = 1.32.37 An aqueous solution contains 47% of A on volume basis. If the density of A is 1250 kg/m3,express the composition of A in weight %.

Basis: 1 m3 of solutionVolume of A in solution = 1 × 0.47 = 0.47 m3

Weight of A = 0.47 × 1250 = 587.5 kgVolume of water = (1 – 0.47) = 0.53 m3

Therefore, the weight of water = 0.53 m3 × 1000 = 530 kg

Hence, weight % of A =587.5= 52.57%530 587.5

2.38 An aqueous solution contains 43 g of K2CO3 in 100 g of water. The density of solution is 1.3g/cc. Find the composition in molarity and molality.

Basis: 100 g of solventWeight of K2CO3 = 43 gWeight of solution = 143 gDensity of solution = 1.3 g/cc

Volume of solution =143 = 110 cc1.3 Moles of solute = Weight/Molecular weight = 43 = 312 g moles138 Molarity = g mole/volume of solution in lit = 0.312= 2.833M0.11

Molality = g mole/kg of solvent = 0.312= 3.12 g moles/kg solvent.0.12.39 A gaseous mixture contains ethylene: 30.6%, benzene: 24.5%, O2: 1.3%, ethane: 25%, N2: 3.1%and methane: 15.5% in volume basis. Estimate the composition in mole %, weight %, averagemolecular weight and density.Basis: 100 kmoles of mixture

Compound mole % Molecular weight Weight, kg Weight %

C2H4 30.6 28 856.8 22.00C6H6 24.5 78 1911.0 49.07O2 1.3 32 41.6 1.07CH4 15.5 16 248.0 6.37C2H6 25 30 750.0 19.26N2 3.1 28 86.8 2.23

Total 3894.2 100.00Density = Weight/Volume =3894.2× 22.414 = 1.737 kg/m3 = 1.737 g/l.100

2.40 A compound has a composition of 9.76% Mg, 13.01% S, 26.01% O2 and 57.22% H2O byweight. Find the molecular formula of this compound.

Basis: 100 g of compoundCompound Weight, Atomic weight Number g or Molecular of moles weight

Mg 9.76 24 0.410S 13.01 32 0.410O 26.01 16 1.615H2O 57.22 18 2.8738

Converting to whole numbers dividing by 0.41

113.946.92

Therefore, molecular formula of the compound = MgSO4.7H2O 2.41 A substance on analysis gave1.978 g of Ag, 0.293 g of S and 0.587 g of O2. Find the molecular formula of the compound.Compound Weight, Atomic weight g or Molecular weight Number Converting to of moles wholenumbers dividing by 9.156 × 10–3

Ag 1.978 108S 0.293 32O2 0.587 16

Molecular formula = Ag2SO4

0.0183 29.156 × 10–3 10.0367 4.04

2.42 Two engineers are estimating the average molecular weight of gas containing oxygen and anothergas. One uses the molecular weight as 32 and finds the average molecular weight as 39.8 and theother uses the atomic weight of oxygen as 16 and finds the average molecular weight as 33.4.Estimate the composition of the gas mixture.

By using the atomic weight of oxygen as 16, the value is 33.4 and by using the molecular weight ofoxygen, the value is 39.8. Let x be the mole fraction of oxygen in the mixture and the molecular weightof the other gas be M39.8 = (x) × (32) + (1 – x) × (M)33.4 = (x) × (16) + (1 – x) × (M)

Solving, we get x = 0.4i.e the fraction of oxygen in the mixture is 0.42.43 A mixture of methane and ethane has an average molecular weight of 21.6. Find the composition.Let the mole fraction of methane be X

21.6 = (Molecular weight of CH4)(X) + (Molecular weight of C2H6) (1 – X)21.6 = 16 × X + 30 × (1 – X)

Solving, we get X = 0.6

2.44 A mixture of FeO and Fe3O4 was heated in air and is found to gain 5% in mass. Find thecomposition of initial mixture.Reactions involved are:

2FeO + 0.5 O2 ® Fe2O32Fe3O4 + 0.5 O2 ® 3 Fe2O3Basis: 100 kg of feed mixtureLet X be the weight of FeO in the mixtureFrom 144 kg of FeO, Fe2O3 formed is 160 kg

Therefore, from Xkg of FeO, Fe2

O3

formed is 160 × X 144 From 232 kg of Fe3O4, Fe2O3 formed is 480 kgTherefore, from (100 – X) of Fe3

O4

,Fe2

O3

formed is (100 – X ) × (480) 464 Since 5% gain in mass is observed, the weight of final product is 105 kg, i.e. 160Xtt (480) 105144 464 Solving, X, the weight of FeO = 20.25 kgFe3O4 = (100 – 83.45) = 79.75 kg

2.45 A sample of lime stone has 54.5% CaO. Find the weight % of lime stone.Basis: 100 kg of lime stone100 kg of CaCO3 will have 56% CaOIf the CaO is 54.5%, then % of CaCO3 in the sample is

54.5 × 100/50 = 97.32%2.46 Express the composition of magnesite in mole %.

Compound Weight %MgCO3 81SiO2 14H2O5

Compound Weight % Molecular weight moles mole % MgCO3 81 84 0.964 65.34SiO2 14 60 0.233 15.82H2O 5 18 0.278 18.83

2.47 The concentration of H3PO4 is expressed in terms of P2O5 content. If 35% P2O5 is reported, findthe composition of H3PO4 by weight. P2O5 + 3H2O ® 2H3PO4(142) (54) (98)

i.e. 142 kg of P2O5 º 196 kg of H3PO4

Therefore, 35 kg of P2O5 º 35 × 196 = 48.3 H3PO4142

i.e. H3PO4 is 48.3%

2.48 Ten kg of PbS and 3 kg of oxygen react to yield 6 kg of Pb and 1 kg of PbO2 according to thereaction shown below:PbS + O2 ® Pb + SO2 (1) PbS + 2O2 ® PbO2 + SO2 (2) Estimate (i) unreacted PbS, (ii) % excessoxygen supplied, (iii) total SO2 formed, and (iv) the % conversion of PbS to Pb.

PbS + O2 ® Pb + SO2 (1)(239.2) (32) (207.2) (64)PbS + 2O2 ® PbO2 + SO2 (2)(239.2) (64) (239.2) (64)

207.2 kg of Pb comes from 239.2 kg of PbS [from stoichiometry Eq. (1)] Therefore, 6 kg of Pb comesfrom 239.2 × 6 = 6.927 kg of PbS207.2

239.2 kg of PbO2 comes from 239.2 kg of PbSTherefore, 1 kg of PbO2 comes from 1 kg of PbSTherefore, total PbS reacted [from Eqs. (1) and (2)]

= 6.927 + 1 = 7.927 kgUnreacted PbS = 10 – 7.927 = 2.073 kgO2 required for this processFrom Reaction 1:32 kg of oxygen is needed to produce 207.2 kg of Pb

Therefore, to produce 6 kg of PbO2, oxygen needed = 6 ×32= 0.927 kg207.2

From Reaction 2:239.2 kg of PbO2 requires 64 kg of oxygenTherefore, to produce 1 kg of PbO2, oxygen required is 268 kg Therefore, total oxygen used = 0.927 +0.268 = 1.195 kg

(3 1.195) Percentage excess O2

supplied = ©¸ ª¹× 100 = 151%«º Amount of SO2 formedIf 207.2 kg of Pb is formed, SO2 formed is 64 kgIf 6 kg of Pb is formed, SO6 = 1.853 kg2 formed is 64 × 207.2 If 239.2 kg of PbO2 is formed, SO2 formed is 64 kgIf 1 kg of PbO64 = 0.268 kg2 is formed, SO2 formed is 239.2 Total SO2 formed = 1.853 + 0.268 = 2.121 kg% conversion of PbS fed to Pb = Mass of PbS converted to Pb/Total mass of PbS= 6.927 × 100= 69.27%10

2.49 The composition of a liquid mixture containing A, B and C is peculiarly given as 11 kg of A, 0.5kmole of B and 10 wt of % C. The molecular weights of A, B and C are 40, 50 and 60 respectivelyand their densities are 0.75 g/cc, 0.8 g/cc and 0.9 g/cc respectively. Express the composition inweight %, mole %. Also give its average molecular weight and density assuming ideal behaviours.Let the weight of mixture be W kg

Weight of A = 11 kgWeight of C (10%) = 0.1W kgWeight of B = W – 11 – 0.1W = 0.5 kmole

= 0.5 × 50 = 25 kgi.e. weight of B = W – 11 – 0.1W = 25 kg0.9W = 36 kgW = 40 kgi.e. total weight of mixture is 40 kg.

Component Weight, Weight, Molecular moles, mole Density, Volume, kg % weight kmole % kg/m3 m3

A 11 27.5 40 0.275 32.66 750 0.0147B 25 62.5 50 0.500 59.38 800 0.0313C 4 10.0 60 0.067 7.96 900 0.0044Total 40 100.00 0.842 100.00 0.0504

40 = 793.65 kg/m3Average density = Mass/Volume = 0.0504 Average molecular weight = Weight/Total moles = 40 = 47.50.842(Check: Average molecular weight= 40 × 0.3266 + 50 × 0.5938 + 60 × 0.0796 = 47.5)EXERCISES2.1 How many g moles are equivalent to 1.0 kg of hydrogen? 2.2 How many kilograms of charcoal isrequired to reduce 3 kg of arsenic trioxide?

As2O3 + 3C Æ 3CO + 2As2.3 Oxygen is prepared according to the following equation: 2KClO3Æ 2KCl + 3O2. What is theyield of oxygen when 9.12 g

of potassium chlorate is decomposed? How many grams of potassium chlorate must be decomposedto get 5 g of oxygen?

2.4 An aqueous solution of sodium chloride contains 28 g of NaCl per 100 cc of solution at 293 K.Express the composition in (a) percentage NaCl by weight (b) mole fraction of NaCl and (c) molality.Density of solution is 1.17 g/cc.

2.5 An aqueous solution has 20% sodium carbonate by weight. Express the composition by mole ratioand mole percent.2.6 A solution of caustic soda in water contains 20% NaOH by weight. The density of the solution is1196 kg/m3. Find the molarity, normality and molality of the solution.

2.7 A saturated solution of salicylic acid in methanol contains 64 kg salicylic acid per 100 kgmethanol at 298 K. Find the composition by weight % and volume %.

2.8 A solution of sodium chloride in water contains 270 g per litre at 323 K. The density of thissolution is 1.16 g/cc. Estimate the composition by weight %, volume %, mole %, atomic %, molality

and kg of salt per kg of water.

2.9 A mixture of gases has the following composition by weight at 298 K and 740 mm Hg.

Chlorine: 60%, Bromine: 25% and Nitrogen: 15%.Express the composition by mole % and determine the average molecular weight.

2.10 Wine making involves a series of very complex reactions most of which are performed bymicroorganisms. The initial concentration of sugar determines the final alcohol content and sweetnessof the wine. The general convention is to adjust the specific gravity of the starting stock to achieve adesired quality of wine. The starting solution has a specific gravity of 1.075 and contains 12.7 weight% of sugar. If all the sugar is assumed to be C12H22O11, determine

(a) kg sugar/kg H2O(b) kg solution/m3 solution(c) g sugar/litre solution

2.11 The synthesis of ammonia proceeds according to the following reaction

N2 + 3H2 ® 2NH3In a given plant, 4202 lb of nitrogen and 1046 lb of hydrogen are fed to the synthesis reactor per hour.Production of pure ammonia from this reactor is 3060 lb/h.(a) What is the limiting reactant?(b) What is the percent excess reactant?

(c) What is the percent conversion obtained (based on the limiting reactant)?2.12 How many grams of chromic sulphide will be formed from 0.718 g of chromic oxide accordingto the following equation?2Cr2O3 + 3CS2 ® 2Cr2S3 + 3CO2

2.13 How many kilograms of silver nitrate are there in 55.0 g mole silver nitrate?2.14 Phosphoric acid is used in the manufacture of fertilizers and as a flavouring agent in drinks. Fora given 10 weight % phosphoric acid solution of specific gravity 1.10, determine:

(a) the mole fraction composition of this mixture.(b) the volume of this solution, which would contain 1 g mole H3PO4. 2.15 Hydrogen gas in thelaboratory can be prepared by the reaction of sulphuric acid with zinc metalH2SO4 (l) + Zn(s) ® ZnSO4(s) + H2 (g)

How many grams of sulphuric acid solution (97%) must act on an excess of zinc to produce 12.0 m3/hof hydrogen at standard conditions. Assume all the acid used reacts completely.

2.16 Sulphur dioxide may be produced by the reactionCu + 2 H2SO4 ® CuSO4 + 2H2O + SO2.

Find how much copper and how much 94% sulphuric acid must be used to obtain 32 kg of SO2.

2.17 Aluminium sulphate is produced by reacting crushed bauxite ore with sulphuric acid as shownbelow:Al2O3 + 3 H2SO4 ® Al2 (SO4)3 + 3 H2O

Bauxite ore contains 55.4% by weight Al2O3, the reminder being impurities. The sulphuric acidcontains 77.7% H2SO4, the rest being water. To produce crude aluminium sulphate containing 1798kg of pure Al2(SO4)3, 1080 kg of bauxite ore and 2510 kg of sulphuric acid solution are used. Find (a)the excess reactant, (b) % of excess reactant consumed, and (c) degree of completion of the reaction.

2.18 600 kg of sodium chloride is mixed with 200 kg of KCl. Find the composition in weight % andmole %.

2.19 What is the weight of iron and water required to produce 100 kg of hydrogen.2.20 Cracked gas from petroleum refinery has the following composition by volume:Methane: 42%, ethane: 13%, ethylene: 25%, propane: 6%, propylene: 9%, and rest n-butane. Find:(a) average molecular weight of mixture, (b) Composition by weight, and (c) specific gravity of thegas mixture.

2.21 A gas contains methane: 45% and carbon dioxide: 45% and rest nitrogen. Express (i) the weight%, (ii) average molecular weight, and (iii) density of the gas at NTP.

Ideal Gases 3

3.1 RELATION BETWEEN MASS AND VOLUME FOR GASEOUS SUBSTANCES3.1.1 Standard Conditions

1 atm. pressure or 760 mm Hg or 29.92 inches of Hg and 0 °C or 32 °F By Avogadro’s Hypothesis,1 g mole of any gas under standard conditions will occupy 22.414 litres 1 lb mole of any gas understandard conditions will occupy 359 cu.ft.

T(K) = T °C + 273.16T(°R) = T °F + 459.693.1.2 Ideal Gas LawThe ideal gas law states that,

PV = nRTP = Pressure of gasV = Volume of n moles of gasn = Number of moles of gasR = Gas constantT = Absolute temperature

Using the ideal gas law (PV = nRT) and the above information one can always determine the weightof a gas if the volume is known and vice-versa.Parameters

PressureMolar volume Absolute

temperature Gas constant

Normal Temperature and Pressure/Standard Conditions English Metric SI 1.033 kgf/cm2 1.01325 bar 359 ft1 atm3/lb mole 22.414 m3/kmole 22.414 m3/kmole491.69 273.16 K 273.16 K 0.73 atm ftoR 3/lb mole oR 0.085 kgf m3/kmole K 0.083 Bar m3/kmole K35

Pressure 1 atm = 1.033 kgf/cm2 = 1.01325 bar1 atm = 14.67 psia = 105 N/m2 = 760 mm Hg1 atm = 760 Torr = 29.92 inches of Hg = 76 cm Hg

R =PV = 82.06 atm.cc/g mole K = 0.73 atm ft3/lb mole °RnT R = 10.73 lbf ft3/in2 lb mole °R

Avogadro’s Number: 6.023 ´ 1023 molecules per g mole2.73 ´ 1026 molecules per lb mole

6.023 ´ 1026 molecules per kg mole

Different units are used to express pressure like atmosphere, mm of Hg, psia, kg/cm2, bar, N/m2, andPa. Similarly, volume is expressed in cm3, m3, litre, ft3 and gallon. The temperature is expressed in°C, °F, K and °R. However, the temperature used in the application of Ideal gas law is in terms of Kor °R.

Thus, the gas constant is a dimensional quantity. The following table gives the gas constant indifferent units.Temperature Pressure Volume Gas constant ‘Rg’

R psia in3 18.51R psia ft3 10.73R atmospheres ft3 0.73KPa m3 8314K atmospheres m3 0.08206K atmospheres cm3 82.06K cm Hg cm3 6239.79

Units of gasconstant

in3psia/lb mole R ft3psia/lb mole Rft3atm/lb mole Rm3Pa/kmole Km3atm/kmole Kcm3atm/g mole K (cm3cm Hg)g mole K

Rg = 8.314 J/(g mole) (K) = 1545 ft lb/lb mole °R

3.2 GASEOUS MIXTURE3.2.1 Partial Pressure (PP)

The partial pressure of a component gas that is present in a mixture of gases is the pressure that wouldbe exerted by that component gas if it alone were present in the same volume and at the sametemperature as the mixture.

3.2.2 Pure Component Volume (PCV)

The PCV of a component gas that is present in a mixture of gases is the volume that would beoccupied by that component gas if it alone were present at the same pressure and temperature as themixture.

Components A B C Sum of the quantities

Partial pressure pA pB pC P Number of moles nA nB nC n Pure component volume VA VB VC V

3.2.3 Daltons LawThe total pressure (P) exerted by a gaseous mixture in a definite volume is equal to the sum of partialpressures.pA + pB + pC = Pwhere pA, pB, pC, … represent partial pressure of components A, B, C, … .3.2.4 Amagats Law (or) Leducs LawThe total volume (V ) occupied by a gaseous mixture is equal to the sum of the pure componentvolumesVA + VB + VC = V

VA, VB, VC, … stand for pure component volume of components A, B, C, …Where ideal gas law is applicable;

n RT; pB = Bn RT; pC = Cn RT(a) pA = A

V V V Adding all the partial pressures of A, B and C, we have, RT P= pA

+ pB

+ pC

= ËÛ ÌÜ ´ (nA + nB + nC)ÍÝ Dividing, pRT V nA ;A/P = V ()RT n×× ++nA BC

pressure fraction = mole fraction.(b) PVA = nART; PVB = nBRT; PVC = nCRTAdding P(VA + VB + VC) = RT (nA + nB + nC) = nRT = PVnRTP = NAV = ADividing,VA

()AB Cn RT ++ or, VA = NA × V3.3 AVERAGE MOLECULAR WEIGHT

The weight of unit mole of the mixture is called average molecular weight, which is also equal tototal weight of the gas mixture divided by the total number of moles in the mixture. This is applicable

only for gaseous mixtures and not for solid or liquid mixtures. For example, air contains 79%nitrogen and 21% oxygen by volume.

Basis: 100 kmoleNumber of moles of nitrogen = 79and those of oxygen = 21

Weight of a component = Number of moles ¥ respective molecular weight\ Weight of nitrogen = 79 ¥ 28 = 2212 kgWeight of oxygen = 21 ¥ 32 = 672 kg2884 kg

\ The weight of 1 kmole = 2884/100 = 28.84 kgHence, the average molecular weight of air = 28.843.4 DENSITY OF MIXTURE

Density is defined as the weight of a mixture per unit volume and is independent of temperature. Asthe volume of liquids and gases is a strong function of temperature, density also varies significantlywith temperature for a specified composition. However, in the case of solids the variation of densitywith temperature is not very significant.

WORKED EXAMPLES

3.1 Calculate the volume of 15 kg of Chlorine at a pressure of 0.9 bar and 293 K.Basis: 15 kg Cl2 = 15/71.0 = 0.2113 kmole

1.0¥293 = 5.643 m3Its volume is 0.2113 ¥ 22.414 ¥0.9 273

3.2 Calculate the volume occupied by 6 lb of chlorine at 743 mm Hg and 70 °F Basis: 6 lb of Cl2∫6/71 = 0.0845 lb mole of chlorine Volume at standard condition = (0.0845 ¥ 359) = 30.34 ft3

PVʈʈ1Volume at given condition = 00

Á˜Ë¯ ˯01

760 530= ÈØÈØ ÉÙÉÙ ÊÚÊÚ = 33.4 ft33.3 Calculate the weight of 200 cu.ft. of water vapour at 15.5 mm Hg and 23 °CBasis: 200 ft3 of gas at given condition PV ÈØ ÈØT0Volume at standard condition =11 ÉÙÉÙ ÊÚ ÈØ15.5 273= 3.76 ft3= ÉÙ296ÊÚ 3.76 Moles of water=

ÉÙ ÈØ = 0.01047 lb moleÊÚ

Weight of water = (0.01047 ´ 18) = 0.18846 lb 3.4 It is desired to compress 30 lb of CO2 to a volumeof 20 ft3 at 30 oC.Find the pressure of the gas stored (required)?

30 0.6818Basis: 30 lb CO2 = ÈØ

ÉÙ = 0.6818 lb mole = 2.2046ÊÚ

= 0.3102 kmole Volume at standard condition = 0.3102 ´ 22.414 = 6.9528 m3 Volume at givencondition = 20 ´ 0.02832 = 0.5664 m3

PV ÈØT1Pressure at the given condition “P1”= 00ÈØ

ÉÙ ÉÙ ÊÚ ÊÚ 6.9528 303= 1ÈØÈØ ÉÙÉÙ273ÊÚÊÚ

= 13.62 atm3.5 Calculate the maximum temperature to which 10 lb of nitrogen enclosed in a 30 ft3 chamber maybe heated without exceeding 100 psi pressure. Basis: 10 lb of N2 = 10/28 = 0.357 lb moleVolume at standard condition = 0.357 ´ 359 = 128.21 ft3

TV ÈØÈØP1\ Temperature ‘T1’= 01ÉÙÉÙVÊÚÊÚ

= 27330ÈØÈØ100ÉÙÉÙ ÊÚÊÚ= 435.4 K = 162.4 °C

3.6 When heated to 100 °C and 720 mm Hg, 17.2 g of N2O4 gas occupies a volume of 11,450 cc.Assuming that the ideal gas law applies, calculate the percentage dissociation of N2O4 to NO2?

NO 24 → 2NO2 (92) (2 × 46)

N2O4 present initially = 17.2/92 = 0.187 g mole;Let ‘x’ g mole of N2O4 dissociate,then, ‘2x’ g moles of NO2 is formed.Total moles after dissociation = (0.187 – x + 2x) = 0.187 + x

Parameter Given condition (1) Standard condition (0)

PressureTemperature Volume720 mm Hg 760 mm Hg 373 K 273 K 11,450 cc ?

PVÈØ ÈØ0Volume at standard condition = 11

ÉÙ ÊÚ ÊÚ01

720 273= 11450ÈØÈØ ÉÙÉÙ373ÊÚÊÚ = 7939.2 ccTherefore, no. of g moles remaining = 7939.2 = 0.35422,414\ (0.187 + x) = 0.354\ x = 0.167 0.167 Percentage dissociation = ÉÙ ÈØ´ 100 = 89.42%ÊÚ 3.7 Calculate the average molecular weight of a gas having the following composition by volume.CO2: 13.1%, O2: 7.7% and N2: 79.2% Basis: 1 g mole of the gasComponent Volume % = Molecular g mole Weight, g mole % weight

CO2 13.1 44 0.131 0.131 ´ 44 = 5.764O2 7.7 32 0.077 0.077 ´ 32 = 2.464N2 79.2 28 0.792 0.792 ´ 28 = 22.176

Total 100.0 1.000 30.404 Average molecular weight is 30.404.

3.8 Calculate the density in lb/ft3 at 29.0 inches of Hg and 30 °C for a mixture of hydrogen and oxygenthat contains 11.1% of hydrogen by weight.

Basis: 1 lb of gas mixtureComponent Hydrogen OxygenWeight % Molecular weight lb mole 0.111 2 0.111/2 = 0.0555 0.889 32 0.889/32 = 0.0278 Total0.0833 lb moles Temperature = 30 °C = 86° F = 546° RVolume of the gas at the given condition = 0.0833 ´ 359 ´ (29.92/29)3

´(546/492) = 34.24 ft3

Density = (1/34.24) = 0.0292 lb/ft3.9 Calculate the density in g/litre at 70 °F and 741 mm Hg of airBasis: 1 g mole of air

Component Volume % = mole % Molecular weight Weight, g Oxygen 0.21 32 6.72 Nitrogen 0.79 2822.12 Total 28.84 gVolume of air = 1

´22.414 ´ 530ÈØÈ Ø760 ÉÙÉ Ù = 24.8 litresÊÚÊ Ú 28.84 Density of air = ÉÙ ÈØ = 1.162 g/litreÊÚ

3.10 In 1000 ft3 of a mixture of hydrogen, nitrogen and carbon-dioxide at 250 °F, the partial pressuresare 0.26, 0.32 and 1.31 atm. Assuming ‘Ideal Gas’ behaviour, find the following:

(a) lb moles of H2; (b) mole fraction and mole % H2; (c) pressure fraction of H2 (d) partial volume ofH2; (e) volume fraction and volume % of H2; (f) weight of H2; (g) weight fraction and weight % ofH2; (h) average molecular weight; (i) density of gas mixture; (j) density at standard condition

Also, show that volume % = pressure % = mole %Basis: 1000 ft3 of gas mixture(a) Partial pressure of hydrogen = 0.26 atm V = 1000 ft3, T = 710 °R Volume of H

2

at standard condition = 1000 ´

0.26 4921.00 710 = 180.169 ft3

180.169 ÈØ = 0.502 lb moles2 = ÉÙlb moles of HÊÚ (b) Total pressure = (0.26 + 0.32 + 1.31) = 1.89 atmTotal moles = ÉÙÉ ÙÉ Ù 1000 1.89ÈØÈ ØÈ Ø492 = 3.648 lb molesÊÚÊ ÚÊ Ú mole fraction of hydrogen = 0.502 = 0.1383.648 0.26 ÈØ = 0.138(c) Pressure fraction of hydrogen = ÉÙ ÊÚ (d) Partial volume of hydrogen is the volume occupied by 0.502 lb moles of it at 1.89 atm and 710 °R 1710 Volume of H2

= 0.502 ´359 ´ ÈØÈ Ø = 138 ft3 1.89ÉÙÉ Ù492ÊÚÊ Ú (e) Volume fraction of hydrogen = 138 = 0.1381000

Thus volume % = pressure % = mole % (f) Weight of hydrogen = 0.502 ´ 2 = 1.004 lb (g) Basis 100lb moles of gas mixture

Mole fraction of nitrogen =0.32 = 0.169189 Mole fraction of carbon dioxide = 1.31 = 0.6931.89 Mole fraction of hydrogen = 0.26 = 0.1381.89 Component Molecular mole % lb mole Weight, lb Weight % weight

Hydrogen 2 13.8 13.8 13.8 ´ 2 = 27.6 0.78Nitrogen 28 16.9 16.9 16.9 ´ 28 = 473.2 13.33Carbon dioxide 44 69.3 69.3 69.3 ´ 44 = 3049.2 85.89

Total 100.0 100.0 3550.0 100.00(h) Average molecular weight =ÉÙ 3550 ÈØ = 35.5ÊÚ (i) 1000 ft3

of gas at given condition º1000 ´ 492 1.89 ÈØÈ Ø ÉÙÉ Ù ÊÚÊ Ú = 1309.7 ft3 at NTP conditionNumber of moles = 1309.7= 3.648 lb moles º 3.648 ´ 35.5359 = 129.55 lb. 129.55 \Density at given condition = ÈØ = 0.12955 lb/ft3 ÉÙ ÊÚ

( j) Density at standard condition:Volume at standard condition = 1000 ´ 1.89ÈØÈ Ø492 ÉÙÉ Ù ÊÚÊ Ú = 1309.7 ft3 129.55 ÈØ = 0.09892 lb/ft3\ Density = ÉÙ ÊÚ

3.11 A certain gaseous hydrocarbon is known to contain less than 5 carbon atoms. This compound isburnt with exactly the volume of oxygen required for complete combustion. The volume of reactants(all gases) is 600 ml and the volume of products (all gases) under the same condition is 700 ml. Whatis the compound?Basis: 1 mole of hydrocarbon.

Let the hydrocarbon be CxHyyy O2 ® xCO2 + ÈØ H2OÉÙ4 ÉÙCxHy + xÈØ

ÊÚ ÊÚ ÈØy® xyMoles: 1 xÉÙ4 2ÊÚ Total moles of reactants = 1xÈØyÉÙ4ÊÚ ÈØyTotal moles of products = xÉÙ ÊÚ Reactants (1xy/4) 600 6 Products ++=== (xy/2) 700 7 7 y 7 + 7x+ ÈØ ÉÙ = 6x + 3yÊÚ (7x– 6x) + 7yÈØ ÉÙ = –7ÊÚ x – 5y = –74

Since the hydrocarbon has carbon atoms less than 5, we have x < 5; solving above equation assumingcarbon atoms as 1, 2, up to 5 we get the values of y as indicated below:

5y = –8; y = 32x = 1; –4 5 5y = –9; y = 36x = 2; –4 5 x = 3; – 5y = –10; y = 40 = 84 5 5y = –11 y = 44x = 4 –4 5

The value of y is to be an integer.Hence, the hydrocarbon is C3H8 : (Propane)The combustion reaction is C3H8 + 5O2 ® 3CO2 + 4H2O

3.12 Combustion gases having the following molal composition are passed into an evaporator at 200°C and 743 mm Hg (N2: 79.2%, O2 : 7.2%, CO2: 13.6 %) Water is added to the stream as vapour andthe gases leave at 85 °C and 740 mm Hg with the following composition N2: 48.3%, O2: 4.4% andH2O : 39%. Calculate (a) volume of gases leaving evaporator per 100 litres of gas entering and (b)weight of water added per 100 litres of gas entering.

(N2, O2, CO2) ® Evaporator ® (N2, O2, CO2) + H2O Water

Basis: 100 g moles gas entering473ÈØÈ Ø760 = 3972.31 litres´ 22.414 ´ÉÙÉ ÙVolume = 100 ÊÚÊ Ú

This 100 g moles of entering gas º 61% of gases leaving. \ g moles of gases leaving = 100/0.61 = 164g moles. Water added = 164 – 100 = 64 g moles = 1152 g.

760ÈØÈ Ø358Volume of gas leaving = 164 ´ 22.414 ´740ÉÙÉ Ù273ÊÚÊ Ú (a) Volume of gas leaving 4950.0t100 100 litres gas-entering 3972.31 = 124.6 litres Volume of water added 1152.0 (b) 100 29 g.100 litres gas-enteringt

3972.31 3.13 How many g moles of nitrogen will occupy 1000 m3 at 112 ´ 103 N/m2 and 400 KPV = nRT, where R = 8.314 J/g mole K = 8.314 (Pa) (m3)/g mole K PV = 112 103

n = 1000= 33,678 g molesRT 8.314 400

Alternatively ,P1 = 1.01325 ´ 105 N/m2

T1 = 273 K

since 11PVPV2 2 T= T12

V1

(at NTP) = 112 103tt1000

t273= 754.4 m3 400 1.01325 t105

1000 g moles of gas occupies 22.414 m3 at NTP,Hence, 754.4 m3

contains ÉÙ ÈØ1000 = 33,658 g moles of nitrogenÊÚ (Error is due to the fact that T1 is taken as 273 K and not as 273.16 K on taking T1 as 273.16 K the gmoles of nitrogen will be 33,677)

3.14 In the manufacture of hydrochloric acid, a gas is obtained that contains 25% HCl and 75% air byvolume. This gas is passed through an absorption system in which 98% of the HCl is removed. Thegas enters the system at 48.8 °C and 743 mm Hg and leaves at 26.7 °C and 738 mm Hg. Applying thepure component volume method, calculate:

(a) The volume of gas leaving per 1000 litres entering the absorption column.

(b) The weight of HCl removed per 100 litres entering. Basis: In 100 litres of entering gas volume ofair = 75 litres Pure component volume of HCl = 25 litresPure component volume of HCl absorbed = (25 ´ 0.98) = 24.5 litres Pure component volume of HClremaining = 0.5 litres Volume of gas leaving = 75 + 0.5 = 75.5 litres

(a) Parameter Entering condition Leaving condition Pressure 743 mm Hg 738 mm Hg

Temperature 48.8 °C 26.7 °C Volume 75.5 litres ?Volume of (entering) gas at leaving condition= 75.5 ´743 299.7= 70.8 lit.738 321.8

Component Litre Volume % (or) mole % (b) Composition: HCl 0.5 0.66Air 75.0 99.34Total: 75.5 100.00Volume of HCl absorbed at standard condition

= 24.5 ´743 492= 20.3 litres760 580 Weight of HCl absorbed =20.3 ´ 36.5 = 33.057 g.22.414

3.15 Absorbing chlorine in milk of lime produces calcium hypochlorite. A gas produced by theDeacon process enters the absorption apparatus at 740 mm Hg and 75 °F. The partial pressure of Cl2is 59 mm Hg and the remainder being inert gas. The gas leaves at 80 °F and 743 mm Hg with Cl2having a partial pressure of 0.5 mm Hg. Calculate, by applying the partial pressure method:

(a) volume of gas leaving per 100 litres entering(b) weight of Cl2 absorbed.Basis: 100 litres of gases enteringPartial pressure of inert gas entering = 740 – 59 = 681 mm Hg Partial pressure of inert gas leaving =743 – 0.5 = 742.5 mm Hg Volume of inert gases entering = 100 litres (681 mm Hg)

681 299.7Volume of inert gas leaving = 100 ´742.5 297(a) Volume of gas leaving = 92.5 litres (743 mm Hg)Volumes of Cl2 entering and leaving are 100 litres and 92.5 litres Entering condition:

Parameter Given condition (1) Standard condition (0)

Pressure 59 mm Hg 760 mm Hg Temperature 297 K 273 K Volume 100 litres ?

Volume at standard condition of Cl2 entering= 10059 273= 7.14 litres´760 297 Volume at standard condition of Cl2 leaving= 92.5 0.5 273= 0.055 litre´760 299.7 Volume at standard condition of Cl2 absorbed= 7.14 – 0.055 = 7.085 litres(b) Weight of Cl 7.285 ´ 71 = 22.45 g.2 absorbed = 22.414

3.16 Nitric acid is produced by the oxidation of ammonia with air. In the first step of the process,ammonia and air are mixed together and passed over a catalyst at 700 °C. The following reactiontakes place. 4NH3 + 5O2 ® 6H2O + 4NO. The gases from this process are passed into towers wherethey are cooled and the oxidation is completed according to the reactions:

2NO + O2 ® 2NO23NO2 + H2O ® 2HNO3 + NO

The NO liberated is re-oxidized in part and forms more nitric acid in successive repetitions of theabove reactions. The ammonia and the air enter the process at 20 °C and 755 mm Hg. The air is

present in such proportion that the oxygen will be 20% in excess of that required for completeoxidation of ammonia to nitric acid. The gases leave the catalyst at 700 °C and 743 mm Hg. Given theoverall reaction: 2NO + 1.5O2 + H2O ® 2HNO3, calculate the following:

(a) The volume of air to be used per 100 litres of NH3 entering the process (b) The composition ofgases entering the catalyzer

(c) The composition of gases leaving (assuming the reaction in catalyzer is 85%)

(d) The volume of gases leaving the catalyzer for 100 litres ammonia entering(e) Weight of acid produced per 100 litres NH3, assuming 90% of the nitric oxide entering the toweris oxidized to acid.

NH3 CatalyzerAirExit gas AbsorberHNO3

Basis: 1 g mole of NH3 overall reaction is NH3 + 2O2 ® HNO3 + H2O O2 required is 2 g molesBut O2 supplied is 2 ´ 1.2 = 2.4 g moles

i.e. air supplied is2.4 = 11.42 g moles0.21 Thus, N2 = 9.02 g moles.(a) Volume of air = 11.42 ´ 22.414 293 760= 276.4 litres273 755 Volume of NH3 = 1 293 760= 24.2 litres´ 22.414 ´273 755 Volume of air/100 litres of NH3 = 276.4 100= 1142.2 litres24.2 (b)

Component g mole mole % = volume % NH3 1.00 8.0

O2 2.40 19.3 N2 9.02 72.7 Total 12.42 100.0

(c) Gases leaving catalyzer are nitrogen, oxygen, ammonia, nitric oxide and water:N2 (all that enters) = 9.02 g moles

NH3 (85% conversion) = (1 – 0.85) = 0.15 g mole

OÈØ5 = 1.06 g moles2 consumed = ÉÙ4ÊÚO2 leaving = (2.4 – 1.06) = 1.34 g molesNO formed = 0.85 g mole

Hʈ6 = 1.275 g moles2O formed = ¥Á˜Ë¯

Component N2 O2 NH3 NO H2O Total g moles 9.02 1.34 0.15 0.85 1.275 12.635mole % = volume % 71.40 10.60 1.20 6.70 10.10 100.000

(d) Volume of gases leaving the catalyzer= 12.635973 ¥760= 1031.8 litres¥ 22.414 ¥273 243 This is the volume of gases leaving for 24.2 litres of ammonia enteringTherefore, for 100 litres of NH3 entering= (1031.8 ¥ 100)/24.2 = 4264 litres of gas leaves

(e) NO entering the tower = 0.85 g moleNO converted = 0.85 ¥ 0.9 = 0.765 g moleHNO3 produced = 0.765 g mole = (0.765 ¥ 63) = 48 g For 100 litres of NH3 weight of HNO3produced

=48 ¥100= 199 g24.2

3.17 1000 kg/h of an organic ester C19H36O2 is hydrogenated to C19H38O2

in a process. The company purchases its H2 in cylinders of 1 m3

capacity. The pressure in cylinder is initially 10 kg/cm2 (abs.) and drops to 2 kg/cm2 (abs.) after use.The company works 24 h/day, 7 days a week. How many cylinders are needed per week?

Basis: Ester being processed in one week = (1000 ¥ 24 ¥ 7) = 1,68,000 kgC19H36O2 + H2Æ C19H38O2296 2 298296 kg of ester reacts with 2 kg of H2168000 kg of ester reacts with2 ¥168000= 1135 kg of hydrogen296 1 kmole of any gas occupies 22.414 m3 at NTP 10 1ÈØ ´ÈØH2 initially present in the cylinder (10 m3) = (1) ´ÉÙ ÉÙ ÊÚ ÊÚ = 0.45 kmole (by reducing toNTP condition) 12 H2

after use = ÈØÈØ ÉÙÉÙ = 0.09 kmoleÊÚÊÚ \ H2 available from one cylinder = 0.36 kmole 1135 H2

needed = 1135 kg = ÈØ ÉÙ = 567.5 kmolesÊÚ The number of cylinders required should be full number, rounded to next highest value 567.5 ÈØ = 1577\ Cylinders required per week = ÉÙ ÊÚ

3.18 A mixture of toluene and air is passed through a cooler where some toluene is condensed. 1000ft3 of gases enter the cooler per hour at 100 °C and 100 mm Hg (gauge). The partial pressure oftoluene is 300 mm Hg. 740 ft3 of gases leave cooler per hour at 40 mm Hg and 50 °C. Calculate theweight of toluene condensed per hour. Vapour pressure of toluene at 50 °C = 90 mm Hg.

Basis: One hour100 mm Hg (gauge) = 860 mm Hg (abs.) 300 = 348.8 ft3Toluene entering = ÈØ

ÉÙ ÊÚ Weight = 348.8 860ÈØÈØÈØ ÉÙÉÙÉÙ 273 92 = 74 lb.ÊÚÊÚÊÚ At exit, air is saturated with toluene. 90 Toluene leaving = 740 ´ ÈØ = 90 ft3ÉÙ

ÊÚ Weight = 90 740ÈØÈ ØÈØ ÉÙÉ ÙÉÙ 273 92 = 19 lb.ÊÚÊ ÚÊÚ \ Weight of toluene condensed = 74 – 19 = 55 lb.

3.19 Air is dried from a partial pressure of 50 mm of water vapour to a partial pressure of 10 mm.The temperature of entering air is 500 °F and the pressure remains constant at 760 mm Hg. How muchwater is removed per 1000 ft3 of entering air?Basis: 1000 ft3 of entering air.

Moles of air entering =1000ÈØÈ Ø492 = 1.43 lb moles359ÉÙÉ Ù960ÊÚÊ Ú Moles of water in it = ÈØ50 = 0.094 lb mole ÉÙ760ÊÚ Moles of dry air = (1.43 – 0.094) = 1.336 lb molesMoles of water leaving = ÈØ ÉÙ 10 = 0.0178 lb moleÊÚ

Water condensed = (0.094 – 0.0178) = 0.0762 lb mole = 1.37 lb 3.20 Chimney gas has the followingcomposition:

CO2: 9.5%, CO : 0.2%, O2: 9.6% and N2: 80.7%. Using ideal gas law, calculate:

(a) its weight percentage(b) volume occupied by 0.5 kg of gas at 30 °C and 760 mm Hg. (c) density of the gas in kg/m3 atcondition of (b)(d) specific gravity of the gas mixture.(Density of air may be taken as 1.3 g/cc)Basis: 100 kmoles of chimney gas(a)

Component Mol. weight Weight, kmole Weight, kg Weight %

CO2 44 9.5 (9.5 ´ 44) = 418.0 13.978CO 28 0.2 (0.2 ´ 28) = 5.6 0.187O2 32 9.6 (9.6 ´ 32) = 307.2 10.273N2 28 80.7 (80.7 ´ 28) = 2259.6 75.562

Total 100.0 2990.4 100.000(b) 0.5 kg of gas =0.5 = 0.01672 kmole29.904 303 ÈØ´ 22.414´ÉÙVolume at 30 °C, 760 mm Hg = 0.01672 ÊÚ= 0.416 m3

0.5 ÈØ = 1.202 kg/m3(c) Density = ÉÙ ÊÚ

(d) Specific gravity = density of gas/density of air = 1.202 = 0.9251.3003.21 A producer gas has thefollowing composition CO2: 4.4%, CO : 23%, O2: 2.6% and N2: 70%. Calculate the following:(a) volume of air at 25 °C and 750 mm Hg required for the

combustion of 100 m3 of gas at the same condition if 25% excess air is used

(b) the composition and volume of gases leaving the burner at 350 °C and 750 mm Hg per 100 m3 ofgas burnt.Basis: 100 kmoles of producer gas.

CO = 23 kmolesO2 needed = 23/2 = 11.5 kmole; O2 in feed = 2.6 kmoles O2 actually needed = (11.5 – 2.6) = 8.9kmoles

Air supplied =8.9 1.25= 52.98 kmoles0.21 O2 supplied = 52.98 ´ 0.21 = 11.125 kmolesN2 in air = 52.98 ´ 0.79 = 41.85 kmoles 298ÈØÈ Ø760Volume of feed = 100 ´ 22.414 ´273ÉÙÉ Ù750ÊÚÊ Ú = 2479.28 m3

Volume of air = 52.98´22.414´ 298ÈØÈ Ø760 = 1313.43 m3ÉÙÉ Ù

ÊÚÊ Ú (a) Volume of air/100 m3 of feed = 1313.43100= 52.98 m3 2479.28 (b) (assuming complete combustion)

CO2 leaving the burner = (4.4 + 23) = 27.400 kmoles O2 leaving = (8.9 ´ 0.25) = 2.225 kmoles N2leaving = (70 + 41.85) = 111.850 kmoles

141.475 kmolesComposition of gases leaving (volume % = mole %): CO2: 19.37%, O2: 1.57% and N2 : 79.06%Volume of gases leaving = 141.475´22.414´

623ÈØÈ Ø760273ÉÙÉ Ù750ÊÚÊ Ú = 7332.92 m3

Volume of gases leaving/100 m3 of feed = 7332.92 100= 295.76 m3 2479.28

3.22 Natural gas has the following composition: CH4: 94.1%, C2H6:3% and N2: 2.9%. This gas ispiped from the well at 80°F and 80 psi. Calculate the following.

(a) Partial pressure of N2

(b) Pure component volume of N2 per 100 ft3 of the gas (c) DensityBasis: 100 ft3 of the natural gas

(a) Partial pressure of N2 =80 2.9= 2.32 psia.100 (b) Pure component volume of N2 = 0.029 ´ 100 = 2.9 ft3

(c) Volume of gas at NTP = 100 ´ 80ÈØÈ Ø492 = 497 ft3 14.67ÉÙÉ Ù540ÊÚÊ Ú No. of moles of gas = 497 = 1.38 lb moles.359 Component mole % lb mole Weight, lb

CH4 94.1 1.38 ´ 0.941 = 1.298 (1.298 ´ 16) = 20.777C2H6 3.0 1.38 ´ 0.03 = 0.042 (0.042 ´ 30) = 1.260N2 2.9 1.38 ´ 0.029 = 0.040 (0.040 ´ 28) = 1.120

Total 100.0 1.380 23.157 23.157 ÈØ = 0.23157 lb/ft3Density of gas = ÉÙ ÊÚ 23.157 ÈØ = 0.0466 lb/ft3Density at standard condition = ÉÙ ÊÚ 3.23 Compare pressures given by the ideal gas and van der Waals equation for 1 mole of CO2

occupying a volume of (381 ´ 10–6) m3 at 40 °C RT = 6.831 ´ 106 N/m2(a) Ideal gas law P = nV where, R = 8.314 N.m/g mole K (J/g mole K); T = 313 K; V = 381 ´ 10–6 m3

(b) van der Waals equation P ÈØa ÉÙ (V – b) = nRTÊÚ a = 0.3646; b = 4.28 ´ 10–5 rest same as above RTa P= ËÛÈØ ÌÜÊÚÉÙ

ÍÝV

t©¸©¸P = 65 62ª¹ª¹381«º«º

= 5.2 ´ 10–6 N/m2

3.24 It is desired to market O2 in small cylinders having volumes of 0.5 ft3 and exactly containing 1 lbof gas at 120 °F. What is the pressure? Basis: 1 lb of O2 = 1/32 = 0.031 lb mole; R = 0.73 ft3 atm/lbmole °R

T = 120 + 460 = 580 °RP =nRT= 0.031 ´ 0.73 ´580 = 26.5 atmV 0.5 3.25 A tire is inflated to 35 psig at 0 °F. To what temperature it can be heated up to a pressure of 50psig, volume remaining same? P1 = 35 + 14.67 = 49.67 psia; P2 = 50 + 14.67 = 64.67 psia. T1 = 460°R; V1 = V2; n1 = n2; T2 = ? PV P V 112 2

T ()( ) ( )460);600°RTT P2 49.6712 1

\ Temperature = 140 °F3.26 Calculate densities of C2H6 and air at NTP.At NTP 1 g mole occupies 22,414 cc.

Density of C2H6 =

130= 1.3 ´ 10–3 g/cc22414 Density of air = 128.84= 1.29 ´ 10–3 g/cc22414

3.27 Acetylene gas is produced according to the reactionCaC2 + 2H2O ® C2H2 + Ca(OH)2

Calculate the number of hours of service that can be got from 1 lb of carbide in lamp burning 2 ft3 ofgas/hour at 75 °F and 743 mm Hg.Basis: 1 lb of CaC1 = 0.0156 lb mole.2 = 64 CaC2 + 2H2O ® C2H2 + Ca(OH)264 36 26 74Volume of acetylene got = 0.0156 ´359 ´ 535ÈØÈ Ø760 = 6.24 ft3 492ÉÙÉ Ù743ÊÚÊ Ú 6.24 Time of burning =

ÈØ ÉÙ = 3.12 h.ÊÚ A similar problem has been solved in MKS system also: Data: 1 kg CaC2; 25 °C; 740 mm Hg; for alamp burning gas 40 litres h.Basis: 1 kg CaC2 = 1 = 0.0156 kmole.64 Volume of C2

H2

= 0.0156´22414´ 298ÈØÈ Ø760 ÉÙÉ Ù = 392.6 litres.ÊÚÊ Ú 392.6 Time of burning = ÉÙ ÈØ = 9.8 h.ÊÚ

3.28 By electrolyzing a mixed brine a mixture of gases is obtained at the cathode having the followingcomposition by weight: Cl2 : 67%, Br2: 28%, O2: 5%. Calculate:

(a) composition by volume(b) density at 25 °C and 740 mm Hg(c) specific gravity of the gas mixture.Basis: 100 g of gas mixture(a)

Component Molecular Weight, g Weight, Volume % weight g mole = mole %

Cl2 71 67 (67/71) = 0.945 74.0Br2 160 28 (28/160) = 0.175 13.7O2 32 5 (5/32) = 0.156 12.3

Total 100 1.276 100(b) Volume of gases = 1.276 ´22.414 ´ 298ÈØÈ Ø760 ÉÙÉ Ù = 32 litresÊÚÊ Ú 100 Density = ÈØ

ÉÙ = 3.12 g/litreÊÚ (c) Density of air = 1.293 g/litre (at standard condition) Density of air at 25 °C and 740 mm Hg 273ÈØÈ Ø740 = 1.15 g/litre´ÉÙÉ Ù= 1.293 ÊÚÊ Ú Alternatively, volume of air at 25 °C and 740 mm Hg 760ÈØÈ Ø298 = 25.13 litres´ÉÙÉ Ù= 22.414 ÊÚÊ Ú Density of air = 28.84 = 1.15 g/litre25.13 Specific gravity of gas mixture = 3.12 = 2.71.15

3.29 A mixture of NH3 and air at 730 mm Hg and 30 °C contains 5.1 NH3 by volume. This gas ispassed at a rate of 100 ft3/min. through an absorption tower in which NH3 is removed. The gasesleave the tower at 725 mm Hg and 20 °C having 0.05% NH3 by volume. Calculate:

(a) the rate of flow of gases leaving the tower and(b) weight of NH3 absorbed.Basis: 100 ft3 of entering gases.Volume of NH3 = 5.1 ft3 and of air = 94.9 ft3 (730 mm Hg, 30 °C)

730ÈØÈ Ø293 = 92.4 ft3Volume of air at exit condition = 94.9 ´725ÉÙÉ Ù303ÊÚÊ Ú 92.4 ft3 of air º 99.95% of exit gas. 92.4 (a) \Volume of exit gas = ÈØ = 92.447 ft3 ÉÙ /min.ÊÚ Volume of NH3 in exit = 0.047 ft3

Volume of exit NH3

at inlet condition = 0.047´ 725ÈØÈ Ø303 ÉÙÉ Ù ÊÚÊ Ú = 0.0482 ft3(b) Volume of NH3 absorbed = (5.1 – 0.0482) = 5.0518 ft3

Volume of NH3

absorbed at NTP = 5.0518´ 730ÈØÈ Ø273 ÉÙÉ Ù ÊÚÊ Ú = 4.372 ft3Moles of NH3 = 4.372= 0.012 lb mole = 0.207020 lb359

3.30 1000 ft3 of moist air at 740 mm Hg and 30 °C contains water vapour in such proportions that its

3.30 1000 ft3 of moist air at 740 mm Hg and 30 °C contains water vapour in such proportions that itspartial pressure is 22 mm Hg. Without the total pressure being changed, the temperature is reduced to15 °C and some water condenses. After that the partial pressure of water is 12.7 mm Hg. Usingpartial pressure method, find the following: (a) Volume of gas after cooling and

(b) Weight of water condensed.Basis: 1000 ft3 of moist air at 740 mm Hg and 30 °CPartial pressure of water = 22 mm HgPartial pressure of air = (740 – 22) = 718 mm HgPartial pressure of air after cooling = (740 – 12.7) = 727.3 mm Hg

718ÈØÈØ288 = 938 ft3(a) Volume of air after cooling = 1000 ´ÉÙÉÙ

ÊÚÊÚ After cooling volume of gases = Volume of water vapour + Dry air (740 mm, 15 °C) (12.7mm, 15 °C) 12.7ÈØÈ Ø303Volume of air leaving at inlet condition = 938 ´ÉÙÉ Ù ÊÚÊ Ú = 570 ft3Volume of water vapour condensed = 1000 – 570 = 430 ft3 43022273 (b) Water condensed = ÉÙÉ ÙÉ Ù ÈØÈ ØÈ Ø 18 = 0.562 lbÊÚÊ ÚÊ Ú

3.31 A producer gas has the following composition by volume CO : 23%, CO2: 4.4%, O2 : 2.6% andN2 : 70%(a) Calculate the ft3 of gas at 70 °F and 750 mm Hg per lb of carbon

present.(b) Calculate the volume of air required for the combustion of 100 ft3 of the gas if 20% excess air isused.(c) Calculate the volumetric composition of gases leaving assuming complete combustion.(d) Calculate the volume of gases leaving at 600 °F and 750 mm Hg per 100 ft3 gas burnt.Basis: 100 lb moles of gasi.e. carbon in the feed = 23 + 4.4 = 27.4 atoms 530ÈØÈ Ø760 = 39,200 ft3(a) Volume of gases = 100 ´ 359 ´ÉÙÉ Ù ÊÚÊ Ú Weight of carbon present = (27.4 ´ 12) = 328.8 lb. 39200 (Volume of gas/lb of carbon) = ÈØ = 119.2 ft3 ÉÙ /lb.ÊÚ 23 (b) O2

required for the combustion of CO =

ÉÙ ÈØ = 11.5 lb molesÊÚ

O2 available in feed = 2.6 lb molesTheoretical O2 required = (11.5 – 2.6) = 8.9 lb molesO2 supplied = (8.9 ´ 1.2) = 10.68 lb moles

100 ÈØ = 50.85 lb moles´ÉÙAir supplied = 10.68 ÊÚ Volume of air = 50.85 ´359 ´ 530ÈØÈ Ø760 = 19,930 ft3 ÉÙÉ Ù ÊÚÊ Ú (70 oF, 750 mm Hg)Volume of feed = 100 ´359 ´ 530ÈØÈ Ø760 = 39,200 ft3ÉÙÉ Ù

ÊÚÊ Ú Volume of air/100 ft3 of feed = 19930 100 39200(c) CO2 leaving = (23 + 4.4)O2 remaining = (10.68 – 8.9)= 50.85 ft3

= 27.40 lb moles = 1.78 lb moles N2 leaving (from air) = (50.85 – 10.68) = 40.17 lb moles N2entering along with feed

N2 TotalTotalComposition of CO2 Volume % 19.66 = 70.00 lb moles = 110.17 lb moles = 139.35 lb moles

O2 N21.28 79.06(d) Volume of gases leaving= 139.35 ´359 ´

760 1060 ÈØÈ Ø = 1,09,218 ft3 ÉÙÉ Ù ÊÚÊ Ú 1,09,218100= 278.6 ft3Volume of gases/100 ft3 of feed = 39,200

3.32 A furnace is to be designed to burn coke at the rate of 200 lb/h having a composition C : 89.1%and ash : 10.9%. The grate efficiency of the furnace is such that 90% of the carbon present in the cokecharged is burnt. Air supplied is 30% in excess of that required for complete combustion. It may beassumed that 97% of the carbon burnt is oxidized to carbon dioxide and the rest to carbon monoxide.

(a) Calculate the composition of the flue gases.(b) If the flue gases leave the furnace at 550 °F and 743 mm Hg, calculate the rate of flow of gases inft3/min.

Basis: 100 lb of coke.C + O2 ® CO2C : 89.1 lb. Ash : 10.9 lb.Carbon burnt = 89.1 ´ 0.9 = 80.19 lbCarbon burnt to CO2 = 80.19 ´ 0.97 = 77.78 lb = 6.48 lb moles Carbon burnt to CO = 80.19 ´ 0.03 =2.41 lb = 0.2 lb mole

32 308.88

O2 supplied = 89.1 ´ÈØ´ 1.3 = 308.88 lb = ÈØÉÙ ÉÙÊÚ ÊÚ = 9.65 lb moles 100 Air supplied = 9.65 ´ ÈØ ÉÙ = 45.96 lb molesÊÚ N2 in air = (45.96 – 9.65) = 36.31 lb moles 6.48 0.2 O2

required = ÈØ ÉÙ = 6.58 lb molesÊÚ Excess O2 leaving = (9.65 – 6.58) = 3.07 lb moles(a)

Component lb mole mole % CO2 6.48 14.07

CO 0.20 0.43O2 3.07 6.67N2 36.31 78.83Total 46.06 100.00

(b) Volume of flue gases = 46.06 ´359 760 1010 ÈØÈ Ø = 34,722 ft3 ÉÙÉ Ù ÊÚÊ Ú

Hence, for 200 lb/h of coke charge, volume of flue gases is = (34,722 ´ 2) = 69,444 ft3/hi.e. Volumetric flow rate of flue gases = 1157.4 ft3/min.

3.33 In the fixation of nitrogen by the arc process, air is passed through a magnetically flattenedelectric arc. Some of the nitrogen is oxidized to NO, which on cooling, oxidizes to NO2. Of the NO2formed, 66% will be associated to N2O4 at 26 °C. The gases are then passed into water washedabsorption towers where nitric acid is formed.

3NO2 + H2O ® 2HNO3 + NONO liberated in this reaction will be reoxidized in cooler.

In the operation of such a plant it is possible to produce gases from the arc furnace in which the NO is2% by volume while hot. The gases are cooled to 26 °C at 750 mm Hg before entering the absorptioncolumn.(a) Calculate the composition of hot gases leaving the furnace

assuming air is at NTP.(b) Calculate the partial pressure of NO2 and N2O4 in the gas enteringthe absorption apparatus.

(c) Calculate the weight of acid formed per 1000 litres of gas entering the absorption system if thecombustion to HNO3 of the combined nitrogen in the furnace gases is 85% complete.

H2OAir ArcNO CoolerNO2 ChamberN2O4 Abs. N2,O2 2% 26 °C,Col. 750 mmHNO3Basis: 100 g moles of air contain N2: 79 g moles and O2: 21 g moles. (a) Reaction in arc furnace isgiven as

N2 + O2 ® 2NOwhich means ‘x’ mole of N2 react to give 2x mole of NO. Total moles of gas leaving = (79 – x) + (21– x) + 2x = 100 (by stoichiometry) N2 O2 NO when x = 1, we have, N2 O2 NO

¯¯¯78 20 2 2x% NO = 2 = 100

\ x = 1(b) Reaction in cooler NO + ½O2 ® NO22 g moles of NO gives 2 g moles of NO2O2 remaining = (20 – 1) = 19 g molesReaction in chamber is 2NO2 ® N2O466% of NO2 remains as N2O4Moles of NO2 associated to N2O4 = (2 ´ 0.66) = 1.32 \ N2O4 formed = 0.66 g moleNO2 remaining = 2 – 1.32 = 0.68 g moleNO2 N2O4 O2 N2 TotalTotal moles: 0.68 + 0.66 + 19 + 78 = 98.34Partial pressure of NO2 = 0.68 750= 5.188 mm Hg.98.34

Partial pressure of N 0.66 750= 5.030 mm Hg.2O4 = 98.34

(c) Combined N2: in NO2 – 0.68 g molein N2O4 – (0.66 ´ 2) = 1.32 g moles/2 g molesThus, 3NO2 + H2O ® 2HNO3 + NOTotal moles of gas in 1000 litres of entering gasPV= RT

750 1000 = ËÛ ÈØ ÉÙÌÜ760ÊÚÍÝ = 40.25 g molesMoles of NO2 = (40.25) (0.68) = 0.278 g mole´98.34 3 moles of NO2 yields 2 moles of HNO3. 2 \0.278 kmole of NO2

yields 0.278

´ ÈØ ÉÙ ´ 0.85 = 0.158 kmole of HNO3ÊÚ (85% conversion)\ Weight of HNO3 formed = 0.158 ´ 63 = 9.95 g.

3.34 The gas leaving a gasoline stabilizer has the following analysis by volume C3H8: 8%, CH4:78%, C2H6: 10% and C4H10:4%

This gas leaving at 90 oF and 16 psia at the rate of 70,000 ft3/h is fed to gas reforming plant where thefollowing reaction takes place.

CnH2n+2 + nH2O ® nCO + (2n + 1)H2 (1) CO + H2O ® CO2 + H2 (2) Reaction (1) is 95% completeand Reaction (2) is 90% complete. Find (a) Average molecular weight of the gas leaving stabilizer;(b) weight of gas fed to reforming plant (lb/h) (c) weight of H2 leaving (lb/h) and (d) composition ofgases leaving (weight %)

Basis: One hour = 70,000 ft3 of gas. 492ÈØÈ Ø16 = 68,295 ft3Volume of gas at NTP = 70,000 ´550ÉÙÉ Ù ÊÚÊ Ú 68,295 Moles of gas = ÈØ ÉÙ = 190.24 lb molesÊÚ Components C3H8 CH4 C2H6 C4H10Total

Volume % 8 78 10 4 100 lb mole 15.22 148.39 19.02 7.61 190.24 Molecular weight 44 16 30 58Weight, lb 669.68 2374.24 570.6 441.38 4055.9

(a) Average molecular weight of the gas leaving stabilizer=4055.9 = 21.32190.24 (b) C H 38 + 3H O2 → 3CO + 7H2 44 54 84 14

CH + H O 42 → CO + 3H2 16 1828 6

26 2 →CH +2H O 2CO+5H2 30 3656 10

C H

410 + 4H O2 →4CO + 9H2 58 72 112 18

Weight of water produced from C3H8 = (15.22 ´ 3 ´18) = 821.88 lb Weight of water produced fromCH4 = (148.39 ´ 1 ´18) = 2,671.02 lb Weight of water produced from C2H6 = (19.02 ´ 2 ´18) =684.72 lb Weight of water produced from C4H10 = (7.61 ´ 4 ´18) = 547.92 lb

Total = 4,725.54 lb CO + H2O ® CO2 + H2

Total CO formed = (15.22 ´ 3) + 148.39 + (19.02 ´ 2) + (7.61 ´ 4) = 262.53 lb moles.H2O needed for forming CO = (262.53 ´ 18) = 4725.54 lb. Weight of gas fed to reformer = [4055.9 +(4725.54 ´ 2)] = 13,506.98 lb/h.(c) Hydrogen formed:

From Reaction (1) (hydrocarbons undergo 95% conversion), we have(15.22 ´ 0.95 ´ 7) + (148.39 ´ 0.95 ´ 3) + (19.02 ´ 0.95 ´ 5) + (7.61 ´ 0.95 ´ 9) = 679.535 lb moles

Total CO formed = (262.53 ´ 0.95) = 249.40 lb moles Hydrogen from CO (90% conversion) =(249.40 ´ 0.90) = 224.46 lb moles \ Total hydrogen leaving reformer = (679.535 + 224.46) ´ 2 = 1808lb/h(d) Gases leaving unreacted = HC, H2O, CO, CO2, H2

H2 = 1808.000 lbC3H8 15.22 ´ 0.05 ´ 44 = 33.484 lbCH4 148.39 ´ 0.05 ´ 16 = 118.712 lbC2H6 19.0 ´ 0.05 ´ 30 = 28.530 lbC4H10 7.61 ´ 0.05 ´ 58 = 22.069 lbCO 249.4 ´ 0.1 ´ 28 = 698.320 lbCO2 249.4 ´ 0.9 ´ 44 = 9876.240 lbH2O 262.53 ´ 0.05 ´ 18 = 236.277 lb

(from Reaction 1) H2O [4,725.54 – (224.46 ´18)] = 685.260 lb(from Reaction 2) 13,506.892 lbComponent C3H8 CH4 C2H6 C4H10 CO CO2 H2OH2TotalWeight, lb 33.484 118.712 28.53 22.069 698.32 9,876.24 921.537 1808 13,506.892 Weight % 0.248 0.879 0.212 0.163 5.17 73.12 6.82313.385 100.000

3.35 Analysis of a sewage gas sample from municipal sewage plant is given. CH4 : 68%, CO2: 30%and NH3: 2%. 600 m3/h of this gas at 30 oC and 2 atmospheres is flowing through a pipe. Find (a) theaverage molecular weight of the sewage gas and (b) the mass rate of flow of gas in kg/h and (c)density of the gas.

Basis: 100 kmoles of the sewage gas.(a) the average molecular weight can be calculated from the following table.

Gas Molecular weight Weight, kmole Weight, kg

CH4 16 68 68 ´ 16 = 1,088CO2 44 30 30 ´ 44 = 1,320NH3 17 2 2 ´ 17 = 34

Total — 100 2,442Average molecular weight =2, 442= 24.42100(b) Volumetric flow rate at standard condition= 600 273ÈØÈØ2 = 1081.2 m3/h´303ÉÙÉÙ1ÊÚÊÚ 1081.2 Molar flow rate of gases = ÉÙ ÈØ = 48.24 kmoles/hÊÚ We know that 100 kmoles of this gas weighs = 2,442 kg Therefore, the weight of 48.24 kmoles 2,44248.24= 1,178 kg.= 100 Hence, the mass flow rate of gas in kg/h = 1,178 kg/h(c) Density of the gas is 2442 = 2.26 kg/m3.1081.2 3.36 In the process of manufacturing Cl2, HCl gas is oxidized with air as follows:4HCl + O2 ® 2Cl2 + 2H2O

If the air used is 30% excess and oxidation is 80% complete, find the composition of dry gasesleaving.Basis: 4 kmoles HCl gas.

O2 needed = 1 kmoleO2 supplied = 1.3 kmoles 79 N2

entering= 1.3 ´ ÈØ ÉÙ = 4.89 kmolesÊÚ

O2 remaining = (1.3 – 0.8) = 0.5 kmoleHCl un-reacted = (4 ´ 0.2) = 0.8 kmoleCl2 formed = (2 ´ 0.8) = 1.6 kmolesComposition of dry gases leaving is presented in the following table

Gas mole mole %

HCl 0.80 10.27O2 0.50 6.42N2 4.89 62.77Cl2 1.60 20.54

Total 7.79 100.00

3.37 A mixture of NH3 and air at 730 mm Hg and 30 °C contains 5.1% NH3. The gas is passedthrough an absorption tower at the rate of 100 m3/h where NH3 is removed. The gases leave the towerat 725 mm Hg and 20 °C having 0.05% NH3. Calculate (a) the rate of flow of gas leaving the towerand (b) weight of NH3 absorbed in kg/h.

0.05% NH3,

Abs. 20 °C 100 m3/h, Tower 725 mm Hg 30 °C,

730 mm Hg5.1% NH3

Basis: One hour of operationMoles of gas entering per hour = 100730 273ÈØÈØÈØ 22.414ÉÙÉÙÉÙ303ÊÚÊÚÊÚ

= 3.86 kmolesNH3 entering = 3.86 ´ 0.051 = 0.197 kmoleAir entering = (3.86 – 0.197) = 3.663 kmoles3.663 kmoles of air contains 0.05% NH3 while leaving the absorber Therefore the total moles ofgases leaving the absorber is

=3.663 = 3.665 kmoles0.9995

Hence, NH3 leaving the tower = 0.002 kmole.(b) NH3 absorbed = (0.197 – 0.002) ´ 17 = 3.317 kgVolumetric flow rate of exit gas

= 3.665 ´22.414 ´ 760ÈØÈ Ø293 ÉÙÉ Ù = 92.42 m3/hÊÚÊ Ú

3.38 The analysis of a flue gas, from a fuel gas containing no N2 has CO2: 4.62%, CO : 3.08%, O2:8.91% and N2: 83.39%. Calculate (a) kmole of dry air supplied per kmole of dry flue gas (b) %excess air (c) analysis of the fuel gas which is a mixture of CH4 and C2H6.

Basis: 100 kmoles of dry flue gas.(a) Air supplied =83.39 = 105.56 kmoles (from nitrogen balance)0.79 kmole of dry air/kmole of dry flue gas = 1.0556CH4 ®CO2

C2H6 ® CO Reactions are:Air®O2 CH4 + 2O2 ® CO2 + 2H2O N2 C2H6 + 31/2O2 ® 2CO2 + 3H2O

(b) O2 supplied = 105.56 ´ 0.21 = 22.17 kmolesO2 consumed = (22.17 – 8.91) = 13.26 kmoles

O2 needed for conversion of CO to CO2 = 3.08= 1.54 kmoles2\ O2 needed for complete combustion = 14.80 kmoles Excess oxygen = (8.91 – 1.54) = 7.37 kmoles

7.37 % Excess air = ÈØ ÉÙ´ 100 = 49.8%ÊÚ

(c) Let CH4 be ‘x’ kmole and C2H6 be ‘y’ kmole Making a CO2 balance, x + 2y = (4.62 + 3.08) =7.70 O2 needed (by stoichiometry) 2x + 3.5y = 14.8 Solving for x and y, we get x = 5.3 and y = 1.2

Gas Weight, kmole mole %CH4 5.3 81.54C2H6 1.2 18.46Total 6.5 100.00

3.39 A furnace is fired with coke containing 90% carbon and 10% ash. The ash pit residue after beingwashed with water analyze 10% carbon; 40% ash and rest water. The flue gas analysis shows CO2 :14%, CO : 1%, O2: 6.4% and the rest N2

Calculate the following:(a) Volume of flue gas produced at 750 mm Hg and 250 °C per tonne of coke charged.

(b) % Excess air used(c) % Of carbon charged which is lost in the ash.Basis: 100 kmoles of exit gas.

N2 = 100 – (14 + 1 + 6.4) = 78.6 kmolesFrom the foregoing, air supplied = 78.6/0.79 = 99.5 kmoles; O2 in air supplied is = 20.9 kmoles

Furnace Flue gasCoke (F)Ash (P)

O2 reacted = 20.9 – 6.4 = 14.5 kmoles.Let F be the coke supplied and P be the ash in the pit (in kg) Total carbon reacted = Total carbon influe gas = 15 katoms Carbon balance: 0.9F = (15 ´ 12) + 0.1P (1)

Ash balance: 0.1F = 0.4P (2) (i.e.) F = 4PSubstituting for (1) from (2), we get(0.9 ´ 4P) = 180 + 0.1P(3.6P – 0.1P) = 180

180 = 51.43 kg (ash)\ P = 3.5

F = 205.72 kg (coke)2C + O2 ® 2COC + O2 ® CO2;

CO + ½O2 ® CO2Carbon lost in ash: 10% = 51.43 × 0.1 = 5.143 kg (for 205.72 kg of coke fed)O2 needed theoretically:O2 supplied: 20.9 kmolesO2 needed (theoretically):Total Carbon fed = 205.72 × 0.9

= 185.148 kg= 15.429 katomsO2 needed for conversion of C to CO2 is 15.429 kmoles (by stoichiometry) Excess O2 = 20.9 –15.429 = 5.471 kmoles

5.471 (b) % Excess air = ÈØ ÉÙ´ 100 = 35.46 %ÊÚ (a) Total carbon fed = (0.9 ´ 205.72) = 185.148 kg205.72 kg of coke gives 100 kmoles of gas 100 1000 kg of coke gives 1000 ´ ÈØ

ÉÙ = 486.1 kmoles of the gasÊÚ AliterFor 1000 kg of coke fed, carbon in coke: 1000 × 0.9 = 900 kgCarbon lost in ash = 5.143 kg × 1000 = 25 kg205.72

\ Carbon reacted = 900 – 25 = 875 kg = 72.916 katoms15 katoms of carbon reacted º 100 kmoles of flue gas72.916 katoms reacted º 486.1 kmoles flue gasVolume of flue gas produced

760ÈØÈ Ø523= 486.1 ´ 22.414 ´750ÉÙÉ Ù273ÊÚÊ Ú = 21,151.2 m3/ton of coke. 51.43 (c) Carbon lost = ÈØ ÉÙ 0.1´ 100 = 2.78%ÊÚ

3.40 Hot air is being used to dry a wet wallboard. The hot air enters the drier at 768 mm Hg and166.7 °C. The partial pressure of water vapour in this air is 25 mm Hg. At the exit partial pressure ofwater is 100 mm Hg. For a total pressure of 760 mm Hg at 111.1 °C in this process, calculate thevolume of exit air per cubic metre of inlet gas.

Basis: 1 m3 of inlet air = 1000 litres = 106 ccMoles of air entering =1 768ÈØÈØÈ Ø273 = 0.02799 kmole22.414ÉÙÉÙÉ Ù439.7ÊÚÊÚÊ Ú In the incoming stream, moles of water/mole of wet air=25 = 0.0326768 In the outgoing stream, moles of water/mole of wet air = 100 = 0.1316768

moles of dry air entering = 0.02799 (1 – 0.0326)= 0.02707 kmolemoles of wet air leaving = 0.02707 (1 + 0.1316)= 0.03063 kmole

384.1 ÈØVolume of exit air = 0.03063 ´ 22.414 ´ÉÙ ÊÚ= 0.9659 m3

3.41 Applying Ideal gas law find out the maximum temperature which 20 kg of CO2 enclosed in 20 m3

chamber may be heated to with pressure not exceeding 20 bar.

Basis: 20 kg of carbon dioxide =

20 = 0.454 kmole44

Volume at standard conditions: 0.454 ´ 22.414 = 10.19 m3P1 = 1 bar, V1 = 10.19 m3, T1 = 273 K P2 =20 bar, V2 = 20 m3, T2 = ?

PVPV ÈØÈ Ø2 2We know that 11

ÉÙÉ Ù 12

Thus T2 is found to be 10,716.3 K = 10,433.3 °C3.42 A telescopic gas holder contains 1000 m3 of gas saturated with watervapour at 20 °C and a pressure of water 155 mm Hg above

atmosphere. The barometer reads 725 mm Hg. Find the weight of water in the gas. Vapour pressure ofwater is 20 mm Hg.

Basis : 1000 m3 of the gas at the given conditions155 mm of water = 11.39 mm Hg.Given pressure = 725 + 11.39 = 736.39 mm Hg.Volume at standard conditions

= 1000 273 736.39= 902.79 m3.293760 20902.79 Amount of water = 725= 1.111 kmole = 20 kg of water22.414

3.43 A gaseous mixture has the following three components X, Y, Z and the composition is asexpressed below. Find the molecular weight of Y component.

Component mole % Weight % Molecular weight Weight X 35 – 85 35×85 Y – 20.0 ? 40 × M Z 25 –60 25×60

Total 4475 + 40M Basis: 100 moles of mixtureLet M be the molecular weight of component YTherefore, mole % of Y = 40%

100 Weight, % of Y = 20 = (40 × M)×(35 85) (40 M) (25 60) 89500 + 800 M = 4000 MTherefore, M = 27.97

3.44 One hundred m3 of mixture of N2, CO2 and H2 in the ratio of 4 : 3 : 1 is at 150 °C and 2 atmpressure. Find the mole fraction, weight % of each, average molecular weight, and total weight of themixture.

4 Partial pressure of N2

= ÈØ ÉÙ × 2 = 1 atmÊÚ Partial pressure of CO2 = 3 × 2 = 0.75 atmʈÁ˜Ë¯ Partial pressure of H2 = 1 × 2 = 0.25 atmʈÁ˜Ë¯ 8 T v È˘È˘ È ˘1Number of moles of N2 = p TPÍ Í ˙22.414Î˚ Î ˚Î˚ = 1 ¥¥ ¥Á˜ 100 273È˘È˘Ê ˆ1 = 2.88 kmolesÍ˙Í˙˯ Î˚Î˚ v È˘È˘ È ˘1Number of moles CO2 = p TPÍ˙ Í ˙22.414Î˚ Î ˚Î˚ = ¥¥Á˜0.75 100 273¥È˘È˘Êˆ1 = 2.16 kmolesÍ˙Í˙˯ Î˚Î˚ Number of moles of H2

= 0.25 100 273 ¥¥Í˙= 0.72 kmole473 1 22.414 ¥¥Î˚ Compound Weight, Molecular Weight, Weight mole kmoles weight kg % %

N2 2.88 28 80.676 45.531 0.500CO2 2.16 44 95.080 53.66 0.375H2 0.72 2 1.441 0.81 0.125Total 5.76 177.197 00.00 1.000Average molecular weight = 2 × (0.125) + 28 × (0.5) + 44 × (0.375) = 30.75Total weight = 177.197 kg

EXERCISES3.1 A liquefied mixture of n-butane, n-pentane and n-hexane has the following composition in percent.

n-C4H10 : 50

n-C5H12 : 30n-C6H14 : 20Calculate the weight fraction, mole fraction and mole percent of each component and also the averagemolecular weight of the mixture. 3.2 A steel tank having a capacity of 25 m3 holds carbon dioxide at30 °C and 1.6 atm. Calculate the weight of the carbon dioxide in grams. 3.3 A steel container has avolume of 200 m3. It is filled with nitrogen at

22 °C and at atmospheric pressure. If the container valve is opened and the container heated to 200°C, calculate the fraction of the nitrogen which leaves the container.

3.4 Natural gas has the following composition in volumetric percent: CH4 : 80%, C2H6 : 15% and N2: 5%Calculate (a) composition in mole %, (b) composition in weight % (c) Average molecular weight,and (d) density at standard condition.

3.5 A typical flue from a chimney is found to contain the following composition by weight: Oxygen :16%, Carbon monoxide : 4%, Carbon dioxide : 17% and rest nitrogen. Calculate average molecularweight and density of the gas at NTP.

3.6 A mixture of gases analyzing 20% methane, 50% ethane and rest hydrogen by volume at atemperature of 283 K and a pressure of 5 atmosphere, flows through a pipe line at the rate of 60 m3/h.The internal pipe diameter is 50 mm. Express the concentration in kmole/m3, velocity in pipeline anddensity of the gas mixture.

3.7 Air contains 79% nitrogen and 21% oxygen by volume. Estimate its density at 20 °C and 741 mmHg pressure.3.8 One kilogram of benzene is stored at a temperature of 50 °C and a pressure of 600 atmospheres.Calculate the volume.

3.9 Find the maximum temperature to which 20 kg of CO2 enclosed in 20 m3 chamber may be heatedwithout exceeding a pressure of 20 bars.

3.10 A gas contains 81.8% carbon and 18.2% hydrogen by weight. If 369 ml of the gas at 22 °C and748 mm Hg weighs 0.66 g, what is the formula of the gas?

3.11 A gas analyzes 60% methane and 40% ethylene by volume. It is desired to store 12.3 kg of thisgas mixture in a cylinder having a capacity of 5·14 ´ 10–2 m3 at a maximum temperature of 45 °C.Calculate the pressure inside the cylinder by assuming that the mixture obeys the ideal gas laws.

3.12 The flue gas of a burner at 800 °C and a pressure 2.5 atm has the following composition byweight.

Nitrogen : 65%CO2 : 15%

H2O : 12%O2 :7%CO : 1%Find (a) composition by volume

(b) the average density of the flue gas(c) mole fraction of the components

3.13 How many kilogram of liquid propane will be formed by liquefaction of 6 m3 of the gas at 500kPa and 300 K?3.14 The following is the analysis of a mixture of gases by weight: chlorine : 65%, bromine : 27%and rest oxygen. Calculate the composition by volume %, mole % and the average molecular weight.

3.15 A natural gas having CH4 : 94%, C2H6 : 3% and N2: 3% is piped from the well at 298 K and 3atm pressure. Find (a) partial pressure of N2, (b) volume of N2 per 100 m3 of gas and (c) density ofthe gas.

3.16 A gas flowing at 1000 litres/s has the following composition: CH4: 10%, C2H6: 30% and H2:60% at 303 K and 2000 mm Hg pressure. Calculate (a) The mole fraction of each component, (b) theconcentration of each component, g mole/cc, (c) partial pressure of each component, (d) the molardensity of the mixture, (e) mass flow rate of the gas and (f) average molecular weight.

3.17 Two hundred eighty kg of nitrogen and 64.5 kg of hydrogen are brought together and allowed toreact at 550 °C and 300 atm. It is found that there are 38 kmoles of gases present at equilibrium. (a)What is the limiting reactant, (b) what is the % excess of excess reactant, and (c) % conversion ofhydrogen to ammonia.

3.18 A natural gas containing 90% methane, 5% ethane and 5% nitrogen is piped from a well at 25 °Cand 1 atm pressure. Assuming the validity of ideal gas law, find:(a) Partial pressure of nitrogen

(b) Volume of nitrogen/100 m3 of gas(c) Density of mixture(d) Average molecular weight of mixture

3.19 Acetylene gas is produced according to the reactionCaC2 + 2 H2O ® C2H2 + Ca (OH)2

Calculate the number of hours of service that can be got from 2.5 kg of carbide in air lamp burning100 m3 of gas/hour at 25 °C and 760 mm Hg.3.20 The following is the analysis of a mixture of gases by weight:

Chlorine: 60%, bromine: 25% and rest nitrogen.Calculate (a) the composition by mole %, (b) average molecular weight, and (c) density at 298 K and

740 mm Hg.

3.21 In the manufacture of nitric acid, ammonia gas and air are mixed at 7 atm pressure and 650 °Cand passed over a catalyst. The composition of this gas mixture on weight basis is nitrogen: 70.5%,oxygen 18.8%, water vapour: 1.2% and ammonia 9.5%. Assuming the validity of the ideal gas law,find (a) composition in mole % and (b) density in kg/m3.

3.22 Calculate the volume occupied by 30 g of chlorine at a pressure of 743 mm Hg and 21.1 °C.3.23 Propane is liquefied for storage in cylinders. How many kilograms of it will be formed byliquefying 500 litres of the gas at standard conditions?

Vapour Pressure4

Whenever we come across a liquid system, it is generally in contact with its own vapour over theliquid surface. This vapour exerts a pressure like gases, which is called vapour pressure. When theliquid is at its boiling condition the vapour pressure will be equal to the surrounding pressure. Whenthat pressure becomes equal to the atmospheric pressure, the boiling point is referred as normalboiling point (NBP). Whenever we have a binary system, if the sum of their partial pressures equalsthe surrounding pressure then the system will boil.

4.1 EFFECT OF TEMPERATURE ON VAPOUR PRESSUREIllustrated by the effect of temperature on vapour pressure is Clapeyron equation ldT dp =T V() GL

where, p represents vapour pressure, T : absolute temperature, l: heat of vaporization at T, VG :volume of gas and VL : volume of liquid. If the volume of liquid is neglected and applicability of theideal gas law assumed, the above reaction reduces to Clausius–Clapeyron equation. dp=ldTp RT2

ord(ln p) = -l d 1 R ¥T where R is gas law constant and l, molal latent heat of vaporization.

When the temperature does not vary over a wide range, it may be assumed that the molal latent heat ofvaporization is constant and the above equation may be integrated between the limits p0 and p and T0and T.

74

ln ʈ Ê ˆ =Á˜ Á ˜ 00

or log ʈ Ê ˆ ʈ=Á˜ Á ˜ 00

Effect of Temperature on V.P:Antoine equation, log p* (mm Hg) = A – B(where, p* is the vapour pressure and A, B, Care constants) TC +

Limitations:(i) Applicable only in the range of applicability of A, B, C. (ii) Pressures below 10 bar.

4.2 HAUSBRAND CHARTSteam distillation takes place at point of intersection of the curves at whichpA = p – pW

\ p = pA + pWTotal pressure = vapour pressure of component + vapour pressure of water

p–pWpAVapour pressureTemperatureWORKED EXAMPLES

4.1 The vapour pressure of ethyl ether at 0 °C is 185 mm Hg. Latent heat of vaporization is 92.5 cal/g.Calculate vapour pressure at 20 °C and 35 °C.Molecular weight of ethyl ether = 74.

l = (92.5 ¥ 74) = 6845 cal/g moleR = 1.99 cal/g mole KT0 = 273 K, T1 = 293 K, T2 = 308 KAt 20 °C,

p ÈØ6845 1È Ø1ÈØÉÙ 2.303ÉÙ 273É Ùlog 185ÊÚÊ Ú Ê Ú Therefore, p = 437 mm Hg.

At 35 °C,

p ÈØÈØ6845 1È Ø1log 185ÉÙ ÉÙ É Ù ÊÚ 2.303Ê Ú Ê Ú

Hence, p = 773 mm Hg.4.2 It is proposed to purify benzene from small amount of non-volatilesolutes by subjecting it to distillation with saturated steam under

atmospheric pressure of 745 mm Hg. Calculate the temperature at which the distillation will proceedand the weight of steam accompanying 1 g of benzene vapour.

Temperature, °C Vapour pressure of Vapour pressure Total pressure, benzene, mm Hg of water,mm Hg mm Hg

60 39065 46068 51069 520150 540 190 650 215 725 225 745 distillation temperature, 69 °C70 550 235 785

ppWpB VapourpressureTemperatureBasis: 1 g mole of mixed vapour. g mole of C H66 Vapour pressure of C H66 520 2.31g mole of water Vapour pressure of water 225 gC H66 2.31 78 10.01gwater 18 g water 0.1g benzene

4.3 It is decided to purify myristic acid (C13H27COOH) by steam distillation under 740 mm Hgpressure. Calculate the temperature and the weight of steam to be used per kg of acid.At 99 °C: Vapour pressure of H2O = 740 mm Hg

Vapour pressure of acid = 0.032 mm Hg\ The distillation temperature can be taken as 99 °C Basis: 1 kmole of mixed vapour.Moles of the acid

0.032 : ÈØ ÉÙ ´ 1 = 4.3 ´ 10–5 kmoles = 0.0098 kgMole of water ÊÚ

Moles of water = 1 kmole = 18 kg Weight of steam 18 ÈØ 1840 kgkg of acidÉÙ

ÊÚ

4.4 The acid given in example 4.3 is to be distilled at 200 °C by use of superheated steam. It may beassumed that the relative saturation of the steam with acid vapours will be 80% (a) Find the weight ofsteam required per kg of acid distilled at 740 mm Hg. (b) Calculate the weight of steam per kg of acidif 26 inches of Hg vacuum is maintained.

(a) Vapour pressure of acid at 200 °C = 14.5 mm Hg Partial pressure of acid (80% saturation) =(14.5 ´ 0.8) = 11.6 mm Hg Basis: 1 kmole of mixed vapour 11.6 Weight of acid = ÈØ ÉÙ´ 1 = 0.0157 kmole = 3.58 kgÊÚ Weight of water = (1 – 0.0157) = 0.9843 kmole = 17.70 kg Steam 17.7 = ÈØ ÉÙ = 4.95 kgkg of acid ÊÚ (b) 26 inches of Hg. vacuum = 740 – (26 ´ 25.4) = 80 mm Hg 11.6 Weight of acid = ÈØ ÉÙ´ 1 = 0.145 kmole = 33.1 kgÊÚ Weight of water (steam) = 0.855 kmole = 15.4 kg Steam 15.4 = ÈØ = 0.465 kgkg of acid ÉÙ ÊÚ

Discussion on examples 4.3 and 4.4: When ordinary steam is used at atmospheric pressure, 1840 kgof steam is needed. When superheated steam at 200 °C is used, 4.95 kg is needed. Due to lowpressure and hence low boiling point under vacuum, quantity of steam needed is 0.465 kg only.4.5 Calculate the total pressure and the composition of the vapours

in contact with a solution at 100 °C containing 35% Benzene, 40% Toluene and 25% Xylene byweight. At 100 °C, the vapour pressures of benzene (molecular weight : 78) is 1340 mm Hg, toluene(Molecular weight : 92) is 560 mm Hg and Xylene (molecular weight : 106) is 210 mm Hg.Basis: 100 kg of solution.

Composition of vapours and their partial pressure

Component Mol. Vapour Weight Weight, Mole

wt pressure, % kmole fraction mm Hg in liquid phase

Partial Mole pressure, fractionmm Hg in vapour phase

Benzene (C6H6)Toluene (C7H8)Xylene (C8H10)

78 1340 35 35/78 = 0.449 0.401 0.401 1340 = 536 0.67392 560 40 40/92 = 0.435 0.388 0.388 560 = 217 0.272106 210 25 25/106 = 0.236 0.211 0.211 210 = 44 0.055Total 100 1.12 1.000 797 1.000

Total pressure = 797 mm Hg. 4.6 An aqueous solution of NaNO3 having 10 g moles of salt/1 kg of

water boils at 108.7 °C at 760 mm Hg. Assume that the relative vapour pressure of the solution isindependent of temperature. Find the vapour pressure of the solution at 30 °C and the boiling pointelevation.

Since the solution boils at 108.7 °C, the vapour pressure of solution = 760 mm Hg.

Vapour pressure of water at 108.7 °C = 1030 mm Hg (from Steam Tables)

k Vapour pressure of solution 760 0.74.Vapour pressure of solvent 1030Vapour pressure of water at 30 °C = 31.8 mm Hg (from Tables) Vapour pressure of solution at 30 °C= (31.8 ´ 0.74) = 23.5 mm Hg,

from Cox chart Boiling point of water at 23.5 mm Hg = 24.8 °C ÈØ ÉÙ

ÊÚ Boiling point elevation= (Boiling point of solution – Boiling point of solvent) = 30 – 24.8 = 5.2 °C4.7 The following table gives vapour pressure data:

Temperature °C 69 70 75 80 85 90 95 99.2Hexane (A), mm Hg 760 780 915 1060 1225 1405 1577 1765Heptane (B), mm Hg 295 302 348 426 498 588 675 760

Assuming Raoult’s law is valid, use the above data to calculate for each of the above temperature themole percent ‘x’ of hexane in the liquid and the mole percent ‘y’ of hexane in vapour at 760 mm Hg.

pA = xA PA; pB = xBPB; (xA + xB) = 1where PA, PB are vapour pressures of hexane (A) and heptane (B) respectively.pA = yAP; pB = yBP; (yA + yB) = 1

pA + pB = P; xB = (1 – xA)(xAPA) + (1 – xA)PB = PxA(PA – PB) = (P – PB)xA 760 + (1 – xA)295 = 760At 69 °C, (Boiling point of hexane)

xA(760 – 295) = (760 – 295); \ xA = 1y760 = 1A = 760

At 70 °C xA(780 – 302) = (760 – 302); \ xA = 0.96pA = (0.96 ´ 780) = 748.8; \ yA = 748.8/760 = 0.985 At 75 °C xA(915 – 348) = (760 – 348); \ xA =0.73pA = (0.73 ´ 915) = 668; \ yA = 668/760 = 0.880 At 80 °C xA(1060 – 426) = (760 – 426); \ xA = 0.53pA = (0.53 ´ 1060) = 562; \ yA = 562/760 = 0.740 At 85 °C xA(1225 – 498) = (760 – 498); \ xA = 0.36pA = (0.36 ´ 1225) = 441; \ yA = 441/760 = 0.58 At 90 °C xA(1405 – 588) = (760 – 588); \ xA = 0.21pA = (0.21 ´ 1405) = 295; \ yA = 295/760 = 0.39 At 95 °C xA(1577 – 675) = (760 – 675); \ xA =0.0945 pA = (0.0945 ´ 1577) = 149; \ yA = 149/760 = 0.196 At 99.2 °C Boiling point of heptanexA(1765 – 760) = (760 – 760); \ xA = 0 = yA 4.8 A solution of methanol in water containing 0.158mole fraction alcohol boils at 84.1 °C (760 mm Hg). The resulting vapour contains 0.553 molefraction of alcohol. How does the actual composition of the vapour compare with compositioncalculated from Raoult’s Law? Temperature, °C 80 100Vapour pressure, mm Hg 1340 2624To get vapour pressure at 84.1 °C, the equationln p = A – B/(T – 43)can be used, ( p, vapour pressure in mm Hg and T, temperature in Kelvin A and B are constants)Solving A = 18.2 and B = 3420;Vapour pressure at 84.1 ºC = 1520 mm Hg.According to Raoult’s law; pA = xAPA = (0.158 ´ 1520) = 240;

yA =240 = 0.316760 % Error = Actual value Calculated value ËÛ ÌÜ ´ 100ÍÝ 0.553 - 0.316 = ÈØ ÉÙ ´ 100 = 42.8%ÊÚ

4.9 Methane burns to form CO2 and water. If 1 lb mole is burnt with 10% excess pure O2 and the

4.9 Methane burns to form CO2 and water. If 1 lb mole is burnt with 10% excess pure O2 and theresulting gas mixture is cooled and dried, calculate (a) volume of dry exit gas at 70 °F and 750 mmHg. (b) Partial pressure of O2 in exit (c) Weight of water removed.

Basis : 1 lb mole of methane.CH4 + 2O2 ® CO2 + 2H2OO2 supplied = 2 ´ 1.1 = 2.2 lb molesGases leaving after drying: CO2 : 1.0 lb mole; O2 : 0.2 lb mole.

530 ÈØÈ Ø760(a) Volume of dry exit gas = 1.2 ´ 359 ´492ÉÙÉ Ù750ÊÚÊ Ú

= 470.26 ft3(b) Partial pressure of O2 in exit = 0.2 750= 125 mm Hg1.2

(c) Water removed = 2 lb moles = 36 lb4.10 Bottled liquid gas is sold. Determine (a) the pressure of the system;(b) vapour composition. The composition in mole % in liquid phase isgiven as follows:n-Butane : 50%, Propane : 45%, and Ethane : 5%Vapour pressure (in bar) at 30 °C are n-butane: 3.4, propane: 10.8 and ethane: 46.6

Gas n-Butane Propane Ethane Total

Mole % 50 45 5 100Vapour pressure at 30 °C (bar) 3.4 10.8 46.6 Partial pressure (bar) 1.70 4.86 2.33 8.89(3.4 0.5) (10.8 0.45) (46.6 0.05) Vapour composition % 19.12 54.67 26.21 100

4.11 A solvent recovery system delivers a gas saturated with benzene vapour which analyzes on abenzene free basis as follows:

CO : 15%, O2 : 4% and N2 : 81%. This gas is at 21.1 °C and 750 mm Hg. It is compressed to 5 atmand cooled to 21.1 °C after compression. How many kilograms of benzene are condensed by thisprocess per 1000 m3 of original mixture? The vapour pressure of benzene at 21.1 °C is 75 mm Hg.

Basis: 1000 m3 of original mixture 1000 750ÈØÈØÈ Ø273Moles of original mixture = 22.414ÉÙÉÙÉ Ù294.1ÊÚÊÚÊ Ú = 40.87 kmoles 75 Benzene present originally = 40.87 ´ ÈØ ÉÙ = 4.087 kmolesÊÚ

Moles of gas other than benzene = 40.87 – 4.087 = 36.783 kmoles Vapour pressure of benzene = 75

mm Hg = 75/760 = 0.0987 atm; (other gas is Tie element)

0.0987 ÈØBenzene after compression = 36.783 ´ÉÙ Ê50.0987Ú = 0.741 kmoleBenzene condensed = (4.087 – 0.741) ´ 78 = 261 kg

4.12 N2 from a cylinder is bubbled through acetone at 840 mm Hg and 323 K at the rate of 0.012m3/h. The N2 saturated with acetone vapour, leaves at 760 mm Hg and 308 K at 0.023 m3/h. Find thevapour pressure of acetone at 308 K.

0.012= 5 × 10–4 kmole/h 22.414 760 323Molar flow rate of N2 = ¦µ¦ µ 840 §¶§ ¶ ·¨ · (N2 + CH3COCH3) = 0.023= 9.09 × 10–4 kmole/h1.013 308 22.414 1.013 ¦µ¦µ §¶§¶

¨·¨·Let y be the mole fraction of N2 in leaving stream.Then, 9.09 × 10–4y = 5 × 10–4

Solving, we get y = 0.55Thus, mole fraction of acetone = 0.45(PT) × (y) = (Partial pressure of acetone) = (VP)acetone (y)acetone (760)(0.45) = (VP) (1.0) = 342 mmHg

4.13 Determine the pressure of the system and equilibrium VP at 30 oC. Assuming all ethane isremoved, estimate the pressure and vapour phase composition components of the system at 30 oC.

Compound Mole fraction, x, in liquid phase n-butane 0.50n-propane 0.45Ethane 0.05

Compound

n-butane n-propane EthaneTotal

Mole fraction, VP at 30 °C, Partial pressure, Y, mole fraction x, in liquid phase mm Hg mm Hg × 100

0.50 3.4 1.7 19.120.45 10.8 4.86 54.67

0.05 46.6 2.33 26.218.89 100.00

y(Mole fraction) = Partial pressure Total pressure 1.7 = ÈØ ÉÙ × 100 = 19.12 for n-butaneÊÚ

54.67 for n-propane26.21 for ethane. When ethane is removed, total moles will be 0.95 kmole.

Compound Mole fraction, VP at 30 °C, Partial pressure, Y, mole fraction x, in liquid phase mm Hg mm Hg × 100

n-butane 0.50/0.95 = 0.5263 3.4 1.789 25.91n-propane 0.4737 10.8 4.86 74.09Total 6.904 100.00

4.14 Estimate the liquid phase composition of a mixture of benzene and toluene at 80 oC when theirgas phase compositions are 80% benzene and 20% toluene. Vapour pressures of benzene and tolueneare 1340 mm Hg and 560 mm Hg respectively at 80 oC.Vapour pressure of benzene is 1340 mm Hg at 80 oC

Vapour pressure toluene is 560 mm Hg at 80 oCAt equilibrium, partial pressure of benzene= (Mole fraction, xB, of benzene) × (1340)= (Total pressure) × (Mole fraction in vapour phase)= PT (0.8)i.e. (xB) × (1340) = PT × (0.8)Similarly for toluene,xT × (560) = PT × (0.2)PT × (0.8) + PT × (0.2) = PTWe also know that, xB + xT = 14 × (1 – xB) × (560) = 4 × PT× (0.2)xB × (1340) = PT × (0.8)4 (1 – xB) 560 = xB(1340)2240 – 1340xB – 2240xB = 0Solving, xB = 0.6

4.15 A certain quantity of an organic solvent (molecular weight 125 and density 1.505 g/cc) is kept inan open flask and boiled long enough so that the vapour fills the vapour space completely by

displacing all the air. The flask is closed, evacuated and the contents reach equilibrium at 30 °C.Vapour pressure of solvent at 30 °C is 240 mm Hg. It is observed that only 10 ml of liquid solvent ispresent finally.

(i) What is the pressure in the flask at equilibrium?(ii) What is the total mass in grams of organic liquid in the flask?

(iii) What fraction of the organic liquid present in the flask is in vapour phase at equilibrium?Equilibrium vapour pressure = 240 mm Hg at 30 °C and the pressure in the system is 240 mm Hg.Volume of vapour space = 4000 – 10 = 3990 ml

T = 303 K; P = 240 mm HgMoles at NTP =3990 240 273 1= 0.05065 g mole303 t tt

760 22414

Molecular weight = 125Weight = 0.05065 × 125 = 6.331 gMass of liquid left behind = 10 ml × 1.505 g/cc = 15.05 g Total weight = 15.05 + 6.331 = 21.381 gMole fraction of vapour = (6.331/21.381) = 0.2961

4.16 Benzene and toluene form an ideal solution. If vapour pressure of benzene is 70 mm Hg and thatof toluene is 20 mm Hg and their mole fractions in liquid phase are 0.7 and 0.3 respectively, calculatetheir vapour phase composition.Let A be benzene and B be toluene

Let PP and PT be partial pressure and total pressure respectively PPA = (PT) (yA) = xA PAPPB = (PT)(yB) = xB PBPT(yA + yB) = xA (PA) + (1 – xA) PB = PPPT = PPA + PPBPT = (0.7)(70) + (0.3)(20)= 49 + 6 = 55 mm HgHence, yA= (0.7)(70)/55 = 0.89yB = (0.3)(20)/55 = 0.11

4.17 Vapour pressure of pure benzene and toluene are 960 mm Hg and 380 mm Hg respectively at86.0 °C. Calculate the liquid phase and vapour phase composition at that temperature.Total pressure = 760 mm Hg

Let x and y be the liquid and vapour phase compositions respectively. Suffix B and T denote benzeneand toluene respectively Then, by Raoult’s law,PTot = (xB) (PB) + (xT) (PT)760 = (xB) (960) + (1 – xB) 380Solving,

xB = 0.655xT = 0.345

Px yB = BB 960 t0.655= 0.827PTot760 In the same way, yP Px 380t 0.345= 0.173T = TT

760Tot

4.18 Vapour pressure of benzene is 3 atm and that of toluene is 1.333 atm. A liquid fuel containing0.4 mole benzene and 0.6 mole toluene is vaporized. Estimate the mole fraction of benzene in vapourphase.

Partial pressure of benzene = (3)(0.4) = 1.2Partial pressure of toluene = (1.333)(0.6) = 0.8Total pressure = 2.0 atmybenzene = 1.2/2.0 =0.6

EXERCISES4.1 The vapour pressure of benzene can be calculated from the Antoine equation.

ln ( p*) = 15.9008 – [2788.51/(–52.36 + T)]where p* is in mm Hg and T is in K. Determine the latent heat of vaporization of benzene at its normalboiling point of 353.26 K, and compare with the experimental value.

4.2 Prepare a Cox chart for ethyl acetate. The vapour pressure of ethyl acetate is 200 mm Hg abs. at42 °C and 5.0 atm at 126.0 °C. By using the chart estimate the boiling point of ethyl acetate at 760 mmHg and compare with the experimental value (77.1 °C).

4.3 The following data gives the vapour pressure (VP) for benzene (A)– toluene (B) system. Computethe vapour-liquid equilibrium data at a total pressure of 760 mm Hg.

VPA, 760 811 882 957 1037 1123 1214 1310 1412 1530 1625 1756 — mm HgVPB, — 314 345 378 414 452 494 538 585 635 689 747 760 mm Hg

4.4 The following data gives the vapour pressure data for a binary system. Compute the VLE vapour–liquid equilibrium data at a total pressure of 760 mm Hg.

Vapour pressure of A, mm Hg 760 860 1002 1160 1262Vapour pressure of B, mm Hg 433 498 588 698 760

4.5 It is desired to purify nitrobenzene by steam distillation under a pressure of 760 mm Hg.Distillation takes place at 99 °C. Vapour pressure of nitrobenzene is 20 mm Hg. Estimate the weightof nitro benzene associated per kg of steam in distillate. Assuming the vaporization efficiency to be75%, estimate the weight of nitrobenzene per kg of steam in distillate.

4.6 Methyl alcohol and ethyl alcohol at 100 °C have vapour pressures of 2710 mm Hg and 1635 mm

Hg respectively. Calculate the total pressure and composition of the vapour in contact with a liquidcontaining 30 weight % of methyl alcohol and 70 weight % of ethyl alcohol at 100 °C.

4.7 A liquefied fuel has the following analysis: C2H6 : 2%, n-C3H8 : 40%, i-C4H10 : 7%, n-C4H10 :47% and C5H12 : 4%. It is stored in cylinders for sale. (a) Calculate the total pressure in cylinder at21 °C and the composition of the vapour evolved. (b) Find the total pressure at 21 °C if all C2H6were removed.Vapour pressure data (mm Hg):C2H6 : 28500, n-C3H8 : 6525, i-C4H10 : 2250, n-C4H10 : 1560 and C5H12 : 430.

4.8 Calculate the total pressure and the composition of the vapours and liquid in molar quantities at100 °C for a solution containing 45% benzene, 40% toluene and 15% xylene by weight. At 100 °C,the vapour pressures are benzene (C6H6) (molecular weight: 78) : 1340 mm Hg; Toluene (C7H8)(molecular weight: 92) : 560 mm Hg, xylene (C7H8) (molecular weight: 106) : 210 mm Hg.

4.9 The partial pressure of actetaldehyde in a solution containing 0.245 kg of CH3CHO in 20 kg ofwater is 190 mm Hg at 93.5 °C. Calculate the partial pressure of CH3CHO over 0.15 molal solutionin water.

4.10 Ethyl acetate at 30 °C exerts a vapour pressure of 110 mm Hg. Calculate the composition of thesaturated mixture of ethyl acetate and air at a temperature of 30 °C and an absolute pressure of 900mm Hg pressure. Express the composition by (a) volume %, and (b) weight %.

Psychrometry 5

5.1 HUMIDITY

Humidification operation is a classical example of an interphase transfer of mass and energy, when agas and a pure liquid are brought into intimate contact. The term humidification is used to designate aprocess where the liquid is transferred to gas phase and dehumidification indicates a process wherethe transfer is from the gas phase to the liquid phase. The matter transferred between phases in boththe cases is the substance constituting the liquid phase, which either vaporizes or condensesindicating respectively the humidification process or the dehumidification process.

5.2 DEFINITIONSDry bulb temperature (DBT): The temperature measured by a bare thermometer or thermocouple iscalled the dry bulb temperature.

Wet bulb temperature (WBT): The temperature measured by a thermometer or thermocouple with awet wick covering the bulb, under equilibrium condition, is called the wet bulb temperature.

Absolute humidity (nv): The substance that is transferred (vapour) is designated by A and the maingas phase is designated by B.It is defined as the moles of vapour carried by unit mole of vapour free gas. yp ËÛp moles of A (5.1)

vËÛ ÌÜ[] ÌÜy ppp moles of BÍÝ ÍÝ

When the quantities are expressed in mass, then it is called mass absolute humidity (n¢) or Grosvenorhumidity. pMMnn [] ËÛvAA

ÌÜ vv ÌÜ ËÛ mass of A(5.2)

ÍÝM pp M ÌÜ mass of BBvBÍÝ ÍÝ 87

Relative humidity or relative saturation ( yr%): It is normally expressed in percentage. If pv is thepartial pressure under a given condition and ps is the vapour pressure at DBT of the mixture, then

Relative saturation = yr % =pv ¥ 100 (5.3)ps

Percentage saturation or percentage absolute humidity ( yP): It is defined as the percentage of theratio of humidity under given condition to the humidity under the saturated condition.

Percentage saturation = yP % =nv ¥ 100 (5.4)ns

Dew point: This is the temperature at which a vapour-gas mixture becomes saturated when cooled atconstant total pressure out of contact with a liquid. The moment the temperature is reduced below thedew point, the vapour will condense as a liquid dew.

Humid heat: The humid heat CS is the heat required to raise the temperature of unit mass of gas andits accompanying vapour by one degree centigrade at constant pressure.

CS = CAnv¢+ CB (5.5) where CA and CB are specific heats of vapour and gas respectively.

Enthalpy: The enthalpy H¢ of a vapour-gas mixture is the sum of the enthalpies of the gas and thevapour content. For a gas at a DBT of tG, with a humidity of nv¢, the enthalpy relative to the referencestate ‘t0’ is,

H¢ = enthalpy of gas + enthalpy of vapour component.= CB(tG – t0) + nv¢[CA(tG – tDP) + lDP + CA,L (tDP – t0)] (5.6) wherelDP = Latent heat of vaporization at dew point.CA,L = Specific heat of component ‘A’ (vapour) in liquid phase. The above Eq. (5.6) will reduce toEq. (5.7) when tDPitself is taken as the reference temperature, t0H¢ = CB (tG – t0) + nv¢[CA(tG – t0) + l0]= CS (tG – t0) + nv¢ l0 (5.7)

Humid volume: The humid volume vHof a vapour-gas mixture is the volume of unit mass of dry gasand its accompanying vapour at the prevailing temperature and pressure. The expression for humidvolume in m3/kg is

È˘¢¢È˘2731105 v MMÍ Í˙273 pÎ˚ Î˚ È˘ ÊˆÊ ˆ =+¥Í˙Í˙ nt¢¢È˘(5.8) Ë¯Ë ¯Î˚ Î˚

where, p= total pressure N/m2

A typical psychrometric chart is shown at the end of the chapter.Various properties of air-water system can be obtained from the chart.Alternatively, the equations given can be used to determine the properties.

Saturated vapour: When a gas or gaseous mixture remains in contact with a liquid surface it willacquire vapour from the liquid until the partial pressure of the vapour in the gas mixture equals thevapour pressure of the liquid at its existing temperature when the vapour concentration reaches this

equilibrium concentration, the gas is said to be ‘saturated’ with the vapour.

Partial saturation: If a gas contains a vapour in such proportions that its partial pressure is less thanthe vapour pressure of the liquid at the existing temperature, the mixture is partially saturated.

Relative saturation: The relative saturation of such a mixture may be defined as the percentage ratioof the partial pressure of the vapour to the vapour pressure of the liquid at the existing temperature.The relative saturation is therefore a function of both the composition of the mixture and itstemperature as well as of the nature of the vapour.

Relative saturation also represents the following ratios: (a) The ratio of the percentage of the vapourby volume to the percentage by volume that would be present if the gas were saturated at the existingtemperature and total pressure, (b) the ratio of the weight of vapour per unit volume of mixture to theweight per unit volume present at saturation at the existing temperature and total pressure.

Percentage saturation: It is defined as the percentage ratio of the existing weight of vapour per unitweight of vapour free gas to the weight of vapour that would exist per unit weight of vapour free gasif the mixture were saturated at the existing temperature and pressure.

This represents the ratio of the existing moles of vapour per mole of vapour free gas to the moles ofvapour that would be present per mole of vapour free gas if the mixture were saturated at the existingtemperature and pressure.

Relative saturation % = yr =pv ¥ 100pswhere,pv = Partial pressure of vapour ps = Vapour pressure of pure liquid Percentage saturation = yp = nv ¥100ns

where,nv = moles of vapour per mole of vapour free gas actually present ns = moles of vapour per mole ofvapour free gas at saturation. If p is the total pressure, then from Dalton’s law,

ʈv , and ns = ʈsnv = Á˜ Á˜ -˯v -˯s

np ppÊˆÊ ˆ vv s- =Á˜Á ˜ ss vË¯Ë ¯ yp

= (yr

) ʈv Á˜ -˯v

WORKED EXAMPLES

5.1 An air (B)-water (A) sample has a dry bulb temperature of 50 °C and a wet bulb temperature of

35 °C. Estimate its properties at a total pressure of 1 atm.1 atm = 1.0133 ¥ 105 N/m2

Molecular weight of air = 28.84(i) nv¢(chart) = 0.03 kg w.v/kg.d.a = 0.04833 kmole of w.v/kmole of d.a.

(ii) % humidity (chart) = 35%(iii) % relative saturation = partial pressure/vapour pressurePartial pressure under the given condition is given by

pAmolal humidity (0.0483) = pp-A

pA0.0483 =1.0133 105 ¥pA

ʈ 5 20.69 = ¥Á˜Ë¯– 1

where,pA = 4.672 ¥ 103 N/m2

pA = Partial pressure of water p = Total pressureVapour pressure of water (Steam Tables) at 50 °C = 92.51 mm Hg = 12.332 ´ 103 N/m2

\% R.H. = 4.672 ´ 103

12.332

´ 103 = 37.88%(iv) Dew point = 31.5°C(v) Humid heat = CS = CB + CA n¢ (Eq. 5.5) = 1.005 + 1.884 (0.03)= 1.062 kJ/kg of dry air °C(vi) Enthalpy

(For a reference temperature of 0 °C)(a) H¢ = CS (tG – t0) + n¢ l0 (Eq. 5.7)l0 = 2502 kJ/kgH¢ = 1.062 (50 – 0) + (0.03) (2502) = 128.16 kJ/kg da (b) Enthalpy of saturated air = 274 kJ/kgEnthalpy of dry air = 50 kJ/kg\ Enthalpy of wet air = 50 + (274 – 50)(0.35) = 128.4 kJ/kg (vii) Humid volume vH

ËÛËÛ(a) vH = (8315) ÉÙÉ Ù ÌÜÌÜÍÝÊÚÊ ÚÍÝ 1(325) n ËÛ= (8315)ËÛÈØ

ÉÙ ÉÙ ÈØÌÜÊÚ ÊÚÍÝÍÝ

= 0.969 m3 mixture/kg of dry airAlternatively,(b) Specific volume of saturated air = 1.055 m3/kg of dry air

Specific volume of dry air = 0.91 m3/kg of dry air By interpolation vH = 0.91 + (1.055 – 0.91)(0.35)= 0.961 m3/kg of dry air

5.2 Ether at a temperature of 20 °C exerts a vapour pressure of 442 mm Hg. Calculate thecomposition of a saturated mixture of nitrogen and ethyl ether vapour at 20 °C and 745 mm Hg andexpress the same in the following forms.

(a) Percentage composition by volume(b) Composition by weight(c) lb of vapour/ft3 of mixture(d) lb of vapour/lb of vapour free gas(e) lb moles of vapour/lb mole of vapour free gas(a) Basis: 1ft3 of gas mixture

Percentage of ethyl ether =442 1= 0.593 ft3 = 59.3 Volume %;745 Nitrogen = 40.7% = 0.407 ft3(b) Basis: 1 lb mole of mixture Component Molecular lb mole Weight, lb Weight % weight

Ether 74 0.593 (0.593 ´ 74) = 43.9 79.4Nitrogen 28 0.407 (0.407 ´ 28) = 11.4 20.6Total 1.000 55.3 100.0

(c) Volume of gas mixture = 1´359´ 760ÈØÈ Ø293 = 393 ft3 ÉÙÉ Ù ÊÚÊ Ú 43.9 \lb of vapour/ft3 = ÈØ ÉÙ = 0.112ÊÚ (d) lb of vapour/lb of vapour free gas = 43.9 = 3.8511.4

(e) lb mole vapour/lb moles vapour free gas = 0.593 = 1.4570.407 5.3 A mixture of acetone vapour and nitrogen contains 14.8% acetone by volume. Calculate yr and ypat 20 °C and 745 mm HgVapour pressure of acetone at 20 °C = 184.8 mm HgPartial pressure of acetone = (0.148 ´ 745) = 110 mm Hg 110 yr

= relative saturation = ÈØ ÉÙ ´ 100 = 59.7%ÊÚ 0.148 ÈØ = 0.174v = ÉÙnÊÚ 184.8 ËÛ ÉÙÌÜ 0.248= 0.329s = ÊÚÈØÌÜÉÙn560.2 0.752 ÈØÌÜ ÊÚÉÙÌÜ ÊÚÍÝ yp = nv 0.174= 52.9%ns 0.329

Always yr > yp

5.4 Moist air is found to contain 8.1 grains of water vapour per ft3 at 30 °C. Calculate the temperatureto which it must be heated in order that its relative saturation will be 15% (7000 grains = 1 lb)

Basis: 1 ft3 of water vapour–air mixtureAmount of water =8.1 = 1.16 ´ 10–3 lb º 6.42 ´ 10–5 lb moles7000 Pure component volume of water at 30 °C 303 = 6.42 ´10–5 ´ 359 ´ÈØ = 0.0256 ft3 ÉÙ ÊÚ 0.0256 ÈØ = 19.4 mm Hg´ÉÙPartial pressure of water vapour = 760 ÊÚ Vapour pressure of water vapour = 19.4 = 130 mm Hg0.15 This corresponds to a temperature of 57 °C.

5.5 A stream of gas at 70 °F and 14.3 psi and 50% saturated with water vapour is passed through adrying tower, where 90% of water vapour is removed. Calculate the pound of water removed per1000 ft3 of entering gas. Vapour pressure of water at 70 °F = 0.36 psi.

Basis: 1000 ft3 of entering gas 70 °F and 14.3 psi 1000 14.3ÈØÈ ØÈ Ø492lb mole of gas mixture = ÉÙÉ ÙÉ Ù ÊÚÊ ÚÊ Ú = 2.52 lb molesyp

= 0.5 = nvns where nv = lb mole of water/lb mole of vapour free gas. 0.36 nv

= 0.5 ´ns

= 0.5 ´ ËÛ 14.3 ÌÜ ÍÝ = 0.013 mole of water/mole of dry gas 0.013 Amount of water entering = 2.52 ´ ÈØ ÉÙ = 0.032 lb moleÊÚ Amount of water removed = 0.032 ´ 0.9 = 0.0288 lb mole Weight of water removed = (0.0288 ´ 18) =0.518 lb.5.6 A mixture of benzene vapour and air contains 10.1% benzene by volume. (a) Calculate the dewpoint of the mixture when at a temperature of 25 °C and 750 mm Hg.

(b) Calculate the dew point when the mixture is at a temperature of 30 °C and 750 mm Hg.(c) Calculate the dew point when the mixture is at a temperature of 30 °C and 700 mm Hg.

Basis: 1 mole of mixture.(a) Partial pressure of benzene = 0.101 ´ 750 = 75.7 mm Hg From the vapour pressure data forbenzene, it is found that this pressure corresponds to a temperature of 20 °C, the dew point.

(b) Partial pressure of benzene remains same, i.e. 75.7 mm Hg. Hence dew point also remains same,i.e. 20 °C (only temperature of gas mixture has changed.)

(c) Partial pressure of benzene = 0.101 ´ 700 = 70.7 mm Hg. Dew point for this partial pressure is

18.7 °C. From these results,

it is seen that the dew point does not depend on the temperature but does vary with total pressure.

5.7 It is proposed to recover acetone which is used as a solvent in an extraction process byevaporation into a stream of nitrogen. The nitrogen enters the evaporator at 30 °C containing acetonesuch that its dew point is 10 °C. It leaves at a temperature of 25 °C with a dew point of 20 °C. Theatmospheric pressure is 750 mm Hg. (a) Calculate the vapour concentration of the gases entering and

leaving the evaporator expressed in moles of vapour/mole of vapour free gas.(b) Calculate the moles of acetone evaporated per mole of the vapour free gas passing through theevaporator.

(c) Calculate the weight of acetone evaporated per 1000 ft3 of gases entering.(d) Calculate the volume of gases leaving the evaporator per 1000 ft3 of gases entering.Data: Vapour pressure of acetone:

At 10 °C = 116 mm Hg and at 20 °C = 185 mm Hg.(a) Entering gases: partial pressure of acetone = 116 mm Hg partial pressure of N2 = 634 mm Hgmole of acetone/mole of N2 =116 = 0.183634 Leaving gases: partial pressure of acetone = 185 mm Hg partial pressure N2 = 750 – 185 = 565 mmHg

mole of acetone/mole of N2 = 185 = 0.328565Basis: for (b), (c) and (d): 1 lb mole of N2 passing through evaporator. (b) Moles of acetoneevaporated = (0.328 – 0.183) = 0.145 (c) Total moles entering = (1 + 0.183) = 1.183 lb moles

760ÊˆÊ ˆ303 = 477 ft3Volume of entering gas = 1.183 ¥ 359 ¥Á˜Á ˜ 750 273 Weight of acetone evaporated = (0.145 ¥ 58) = 8.4 lbAcetone evaporated per 1000 ft3 = 8.4¥1000= 17.6 lb477 (d) Total moles of gas leaving = (1 + 0.328) = 1.328 lb moles 760ÊˆÊ ˆ303Volume of gases leaving = 1.328 ¥ 359 ¥Á˜Á ˜ Ë¯Ë ¯ = 526 ft3Volume of gases leaving per 1000 ft3 = 526¥1000= 1102 ft3 477 5.8 Air at a temperature of 20 °C and pressure 750 mm Hg has a relative humidity of 80%(a) Calculate the molal humidity of air.(b) Calculate the humidity of this air if its temperature is reduced to 10 °C and its pressure increasedto 35 psi, condensing out some of the water.(c) Calculate the weight of water condensed from 1000 ft3 of gas. (d) Calculate the final volume ofwet air leaving.Data: Vapour pressure of H2O at 20 °C =17.5 mm Hg and at 10 °C = 9.2 mm Hg.

(a) Initial partial pressure of water = (17.5 ¥ 0.8) = 14 mm Hg 14Initial humidity = 750 -14 = 0.019 kmole of water vapour (w.v.)/kmole of dry air (d.a.) (b) Final partial pressure of water = 9.2mm Hg (saturated)Final total pressure = 35 ¥760= 1810 mm Hg14.7 9.2Final humidity = È˘Í˙ -Î˚ = 0.0051 kmole of water vapour/kmole of dry air Basis: 1000 ft3 of wet air, 20 °C and 750 mm Hg.(c) Volume at standard condition and hence weight of mixture in lb mole 1000 750ÈØÈ ØÈ Ø 273 = 2.56 lb moles359ÉÙÉ ÙÉ Ù293ÊÚÊ ÚÊ Ú 736 ÈØ = 2.51 lb moles´ÉÙlb mole of dry air = 2.56 ÊÚ

lb mole of water vapour initially = (0.019 ´ 2.51)= 0.0477 lb molelb mole of water finally = (0.0051 ´ 2.51) = 0.0127 lb mole\ water condensed = 0.035 lb moles = 0.63 lb.(d) Total moles finally present = (2.51 + 0.0127) lb moles = 2.5227 lb moles

760ÈØÈ Ø283 = 395 ft3Volume of wet air = 2.5227 ´ 359 ´ÉÙÉ Ù

ÊÚÊ Ú

5.9 A mixture of dry flue gases and acetone at a pressure of 750 mm Hg has a dew point of 25 °C. Itis proposed to condense 90% of the acetone by cooling to 5 °C and compressing. Calculate the finalpressure in psi for acetone.

Vapour pressure at 25 °C = 229.2 mm Hg.Vapour pressure at 5 °C = 89.1 mm Hg.Basis: 1 lb mole of mixture of gases.

Partial pressure of acetone = 229.2 mm Hg; (equal to vapour pressure as it is at dew point)lb moles of acetone in it = 229.2/750 = 0.306 lb mole

Dry flue gases = (1 – 0.306) = 0.694 lb moleAcetone finally present = 0.306 ´ 0.1 = 0.0306 lb mole Final gas = 0.694 + 0.0306 = 0.7246 lb molelb mole of acetone in final gas/lb mole of final gas mixture

0.0306 = 0.0422= 0.7246 Final pressure of acetone = 89.1 mm Hg 89.1 = 2110 mm Hg\ Final pressure of gas mixture = 0.0422 = 40.8 psi.5.10 Air at a temperature of 20 oC and 750 mm Hg has a relative humidity of 80%. Calculate,

(i) The molal humidity of the air(ii) The molal humidity of this air if its temperature is reduced to 10 °C and pressure increased to2000 mm Hg condensing out some of the water and

(iii) Weight of water condensed from 1000 litre of the original wet air. Vapour pressure of water at20 °C = 17.5 mm HgVapour pressure of water at 10 °C = 9.2 mm HgAt 20 °C and 750 mm Hg,

pp yr

= 0.8 = vv

p=17.5s where, partial pressure of H2O vapour = pv = 14 mm Hg. pv 14(i) Molal humidity at 20 °C, 750 mm Hg =ppv =-14 = 0.019 g mole of water vapour/g mole of dry air (ii) Molal humidity at 10 °C, 2000 mm Hg(when saturated, vapour pressure = partial pressure, we have, ps = pv = 9.2 mm Hg) ps

È˘ =Í˙ pp2000 9.2--Î˚ = 4.6 ¥ 10–3 g mole of water vapour/g mole of dry air (iii) Basis: 1000 litres of original wet air 1000 750ʈʈʈ273g mole of wet air = ¥¥

Á˜Á˜Á˜ ˯˯˯ = 41.02g mole of dry air = 41.02 ¥1 = 40.25ʈÁ˜Ë¯ \ water condensed = (difference in humidity) ¥ (flow rate of dry air) ¥ (molecular weight of water) =(0.019 – 4.6 ¥ 10–3) ¥ 40.25 ¥ 18 = 10.43 g 5.11 A material is to be dried from 16% moisture byweight (wet basis) to

0.5% by circulation of hot air. The fresh air contains 0.02 kg of water/kg of dry air. Find the volumeof fresh air required if 1000 kg/h of dried material is to be produced. The exit humidity of air is 0.09kg of water/kg of dry air. The air enters at 301 K and at atmospheric pressure.

301 K Air 0.02 0.0916% wet solidDrier Dried solid 0.5%

Basis: 1000 kg/h product.Dry solid = 995 kgMoisture = 5 kg

995 = 1184.5 kg\ feed = 0.84

Water in feed = 189.5 kgDry solid is the Tie ElementWater removed = 1184.5 – 1000 = 184.5 kg1 kg of dry air removes = (0.09 – 0.02) = 0.07 kg H2O

184.5 = 2636 kg/h\ Weight of dry air needed = 0.07 2636 Dry air needed = ÈØ ÉÙ = 91.40 kmolesÊÚ Water in entering air = 2636 0.02= 2.93 kmoles18 \ Total moles of wet air = 94.33 kmoles 301 Volume of wet air = 94.33 ´22.414 ´ ÈØ = 2331.17 m3

ÉÙ

ÊÚ5.12 A tunnel drier is used to dry an organic paint. 1000 lb/h of feed having 10% water is to be driedto 0.5%. Air is passed counter current

to the flow of paint. Air enters at 760 mm Hg. 140 °F and 10% humidity while it leaves at 750 mm Hg95 °F and 70% humidity. What flow rate of air is needed?

Air 10% Air 70%Dry product 0.5%Drier 1000 lb/h; 10% moisture

Basis: 1000 lb/h of feed Dry material = 900 lbMoisture = 100 lb

Product = 900 = 904.5 lb0.995Water removed = (1000 – 904.5) = 95.5 lb/hFrom humidity chart the following details are obtained: Humidity of entering air = 0.025 lb molewater vapour/lb mole dry air Humidity of leaving air 0.035 lb mole water vapour/lb mole dry air 1 lbmole of dry air removes 0.01 lb mole H2O = 0.18 lb

95.5 Dry air needed = ÈØ ÉÙ = 530.5 lb mole/hÊÚ \ Total air entering = 530.5 ´ 1.025 = 543.77 lb mole/h

600 Volumetric flow rate of air = (543.77 ´359) ´ ÈØ ÉÙ ÊÚ= 2,38,065 ft3/h.

5.13 Air at 100 oF and 20% relative humidity is forced through a cooling tower. The air leaves at 85°F and 70% relative humidity. Atmospheric pressure is 29.48 inches of Hg. Calculate the following.

(a) The pounds of water evaporated per pound of dry air. (b) Volume of air leaving per 1000 ft3 ofentering air.Vapour pressure of water at 100 °F = 1.9325 inches of Hg and at 85°F = 0.8754 inches of Hg. (a)Partial pressure of H2O in entering air = 1.9325 ´ 0.2 = 0.3865 inches of Hg

Partial pressure of H2O in exit air = 0.8754 ´ 0.7= 0.6128 inches of Hg(lb mole H2O/lb mole dry air) in entering air

=0.3865= 0.0132829.48 0.3865 (lb mole H2O/lb mole dry air) in outgoing air= 0.6128= 0.0212329.48 0.618 lb mole H2O evaporated/lb mole of dry air = 0.00795 18 lb H2

O/lb of dry air = 0.00795 ´ ÈØ ÉÙ = 5 ´ 10–3lb = 0.005 lbÊÚ (b) 1000 ft3

of entering air = 1000 43.2 ÈØÈ Ø ÉÙÉ Ù = 2.447 lb molesÊÚÊ Ú

For 1.01328 lb moles of wet air entering, 1.02123 lb moles of wet air leaves\ for 2.447 lb moles of wet air entering

= 2.4471.02123 = 2.466 lb moles of wet air leaves´1.01328 535 ÈØ = 962.8 ft3Volume of air leaving = 2.466 ´ 359 ´ÉÙ ÊÚ

5.14 Leather containing 100% of its own weight of water is dried by means of air. The dew point ofentering air is 40 °F while that of leaving air is 55 °F. If 2000 lb of wet air is forced through the drierper hour, how many lb of water is removed per hour. Total pressure 750 mm Hg.

Vapour pressure of water at 40 °F = 6.3 mm Hgand at 55 °F = 11 mm Hg.Basis: One lb mole of dry airWater in entering air

=

6.3750 6.3= 0.00847 lb mole of water vapour/lb mole of dry air = 0.00527 lb of water vapour/lb of dry air

Water in leaving air =

11750 11= 0.01488 lb mole of water vapour/lb mole of dry air = 0.00926 lb of water vapour/lb of dry air

\ water vapour removed = 0.00926 – 0.00527= 0.00399 lb of water vapour/lb of dry air 2000 Dry air entering per hour = ÈØ ÉÙ = 1989.52 lbÊÚ

Weight of water removed/h = 1989.52 ´ 0.00399 = 7.938 lb. 5.15 Acetone is used as a solvent in acertain process. Recovery of the

acetone is accomplished by evaporation into a stream of N2 followed by cooling and compression ofthe final gas mixture. In the solvent recovery unit 50 lb of acetone are removed per hour. The N2 isadmitted at 100 °F and 750 mm Hg partial pressure of acetone in incoming nitrogen is 10 mm Hg. Thenitrogen leaves at 85 °F, 740 mm Hg and 85% saturation.(a) How many ft3 of incoming gas must be admitted per hour to

obtain the required rate of evaporation of the acetone? (b) How many ft3 of gases leave the unit perhour?

N2, Acetone 100 °F 750 mm Hg N2, AcetoneDrierpartial pressure of acetone 10 mm Hg 85°F, 740 mm Hg 85% saturation

(Molecular weight of Acetone (CH3)2CO is 58. Vapour pressure of acetone at 85 °F = 287 mm Hg)Basis: 100 lb moles of gas mixture entering

Acetone removed =50 = 0.862 lb mole58 N2

in entering mixture = 750 10 ËÛ ÌÜ ´ 100 = 98.67 lb moles ÍÝ Acetone entering = 100 – 98.67 = 1.33 lb moles 1.33 ÈØlb mole of acetone/lb mole of N2 in entering stream = ÉÙ ÊÚ = 0.0135

p vËÛËÛ287yp = 0.85 = 740 pÌÜÌÜ287ÍÝÍÝwhereyp (percentage saturation) = 85%pv (partial pressure) = 258 mm Hg

lb mole of acetone/lb mole of N2

in leaving stream = 258 740 258 = 0.5385 \ 1 lb mole of N2 removes: (0.5385 – 0.0135)= 0.525 lb mole of acetonelb mole N2 needed/h = 0.862 = 1.642 lb mole0.525 Total moles of gas entering = 1.642 ´ 1.0135 = 1.664 lb moles 560ÈØÈ Ø760(a) Volume of gases admitted = 1.664 ´ 359 ´ÉÙÉ Ù ÊÚÊ Ú = 689.0 ft3(b) Total moles of exit gas = 1.642 ¥ 1.5385 = 2.526 lb moles 545ÊˆÊ ˆ760Volume of exit gases = 2.526 ¥ 359 ¥Á˜Á ˜ Ë¯Ë ¯

= 1031.6 ft35.16 By adsorption in silica gel you are able to remove all the weight (0.93 kg) of water from moistair at 15 °C and 98.6 kPa. The same air measures 1000 m3 at 20°C and 108 kPa when dry. What wasthe relative humidity of the air?Basis: 1000 m3 of dry air at the given conditions.

108 ʈ kmole of the dry air: ¥¥ ¥¥Á˜Ë¯ = 44.895

Water removed: 0.93 kg = 0.93 = 0.052 kmoleʈÁ˜Ë¯ Total wet air = 44.895 + 0.052 = 44.947 kmolespʈ98.6 = 0.114 kPav = Partial pressure of water = ¥Á˜Ë¯ ps = Vapour pressure of water = 15 mm Hg = 2 kPa pv, yr = 0.114¥ 100 = 5.7%,Q Relative humidity, yr =ps ʈÁ˜Ë¯

5.17 At 297 K and 1 bar, the mixture of N2 and C6H6 has a relative humidity of 60%. It is desired torecover 90% C6H6 present by cooling to 283 K and compressing to a suitable pressure. What is thepressure to which the mixture should be pressurised? Vapour pressures of benzene at 283 K and 297K are 6 kN/m2 and 12.2 kN/m2 respectively.

Relative humidity = 0.6 = Partial pressure/Vapour pressure12.2 60 103

Partial pressure = ××= 7.32 × 103 N/m2

100 7.32 × 10 3 Humidity = 78= 0.217 kg of w.v./kg of d.a.101.3 1033 ×28×− ×10 Final humidity = 0.10 × 0.217 = 0.0217 kg of w.v./kg of d.a. 3 780.0217 610 ×=×3 28PT −×610

Therefore, PT, the total pressure = 7.642 × 105 N/m25.18 Humid air at 75 °C, 1.1 bar and 30%relative humidity is fed at a rate of 1000 m3/h. Determine the molar flow rates of water and dry airentering the process, molar humidity, absolute humidity, percentage humidity and dew point.Vapor pressure at 75 °C = 289 mm Hg

Partial pressure of water vapour= Vapour pressure of water × Relative humidity = 86.7 mm Hg

Total pressure = 1.1 bar = 1.1 × 105 N/m2

1 atm = 760 mm Hg = 1.013 × 105 N/m2

Therefore, 1.1 bar = 825.36 mm HgMole fraction of water vapour = (86.7/825.36) = 0.105 Volumetric flow rate = 1000 m3/h at 75 °Cand 1.1 bar 1000 825 273 138 kmoles/h348 t tt

760 22.414Water flow rate = 38 × 0.105 = 3.99 kmoles/hDry air rate = 38 – 3.99 = 34.01 kmoles/h

Humidity = 86.7= 0.117 kmole w.v./kmole d.a.825.36 86.7 Mass humidity = 0.117 × 18 = 0.073 kg w.v./kg d.a.28.84 Percentage humidity = 0.117 × 100= 21.7%289 /(825289) Dew point = 49 °C

5.19 A piece of wood entering furnace at 0.15 kg/s containing 60% moisture is dried in acountercurrent dryer to give a fixed product containing 5% moisture content. The air used is at 100 °Cand is having water vapour at a partial pressure of 103 N/m2. The air leaves the dryer at 70% RH andat 40 °C. Estimate the air required to remove the moisture. The vapour pressure of water at 40 °C =7.4 × 103 N/m2

Water in incoming wood = 0.15 × 0.6 = 0.09 kg/s Weight of dried wood = 0.15 × 0.4 = 0.06 kg/sWeight of water in dried wood is 5%

Weight of final product =0.06 = 0.0632 kg/s0.95 Weight of water removed = 0.15 – 0.0632 = 0.0868 kg/s 10 3 Humidity of entering air = 1.013 10 5310 = 9.97 × 10–3 × 18

28.84 = 0.00623 kg w.v./kg d.a. Humidity of leaving air,Relative humidity = Partial pressure/Vapour pressure 0.7 = Partial pressure/7.4 × 103

Partial pressure = 5.18 × 103 N/m2

10 3 Humidity = 5.18 × 101.3 1033= 0.0336 kg w.v./kg d.a. 10 Increase in humidity = 0.0336 – 0.00662 = 0.02738 kg w.v./kg d.a. Weight of water in kg to beremovedDry air needed = Water removed per kg of dry air =0.0868 = 3.17 kg/s0.02738 Weight of wet air needed = 3.17 × (1.00622) = 3.1897 kg/s Weight of wet air leaving = (3.17) ×(1.0336) = 3.277 kg/s

5.20 Air at 740 mm Hg at 30 °C contains water vapour at a partial pressure of 22 mm Hg. 1000 m3 ofthe air is cooled to 15 °C and the partial pressure of water vapour is brought to 12.7 mm Hg. Duringthis process, water gets partially condensed. Estimate the amount of air at new condition

Moles of humidified air at initial condition 1000

760 303 = ÌÜ 22.414 740 273= 39.14 kmoles ÍÝ Mole fraction of water vapour in original air (30 °C) = 22 = 0.02973740Mole fraction of water vapourin original air when cooled to 15 °C= 12.7 = 0.017740 Therefore, mole fraction of dry air = 0.983Moles of wet air at 15 °C = 37.976 = 38.63 kmoles0.983 Volume of dry air at new condition (assuming ideal behaviour)= 38.63 ËÛËÛ ÌÜÌÜ 760 288= 938.11 m3/kg dry air ÍÝÍÝ

Water removed is (39.14) × (0.02973) – (38.63 × 0.017) = 0.507 kmole= 0.547 × 18 kg= 9.126 kg

5.21 Air with humidity of 0.004 kg of water vapour per kg of dry air is bubbled through water at arate of 0.1 kg/s. The air leaves with a humidity of 0.014 kg of water vapour per kg of dry air.Estimate the time needed to vaporize 0.1 m3 of water.

Weight of water evaporated = 0.1 × 1000 = 100 kg (since the density is 1000 kg/m3)Weight of water removed/kg of dry air = 0.014 – 0.004 = 0.01 kg

Water removed per second by using 0.1 kg of air= 0.01 × 0.1 = 0.001 kg/sHence, time needed to evaporate 100 kg of water

=100 = 100000 s0.001 = 100000= 27.7 h3600

5.22 Air containing 0.01 kg of water vapour per kg of dry air is passed through a bed of silica gel toget air containing 0.001 kg of water vapour per kg of dry air for a specific application. The columnhas 2.1 kg of silica gel initially and after 5 h of operation, the weight of silica gel is found to be 2.2kg. Calculate the flow rate of air both at inlet and outlet.

Water adsorbed by silica gel in 5 h = 2.2 – 2.1 = 0.1 kgWater adsorbed/h =

2.2 2.1= 0.02 kg/h5

Making a balance for water:Water in incoming air = Water in leaving air + Water adsorbed Let G be the flow rate of dry air inkg/h.(G) (0.01) = (G) (0.001) + 0.02G (0.01 – 0.001) = 0.02G = 0.02 = 2.222 kg/h0.009

i.e. the flow rate of dry air in kg/h = 2.222 kg/hFlow rate of wet air at inlet = 2.222[1 + 0.01] {since G = Gs (1 + x)} = 2.244 kg/h

Flow rate of wet air at outlet = 2.222 (1 + 0.001)= 2.224 kg/hEXERCISES

5.1 Toluene–air mixture is available such that the partial pressure of the vapour is 1.33 kPa. The totalpressure is 99.3 kPa and the temperature is 21 °C. Calculate (a) the relative humidity, (b) The molesof toluene per mole of vapour free gas, (c) the weight of toluene per unit weight of vapour free gas,(d) the percent saturation and (e) the percentage of toluene by volume.

5.2 Air is available at a DBT of 50 °C and a wet bulb temperature of 30 °C. Estimate its humidity, %saturation and humid volume using humidity chart.

5.3 A material is to be dried from 20% moisture by weight (wet basis) to 1.0% by circulation of hotair. The fresh air contains 0.02 kg of water/kg of dry air. Find the volume of fresh air required if 1000kg/h of dried material is to be produced. The exit humidity of air is 0.09 kg water vapour/kg dry air.The air enters at 300 K and at atmospheric pressure.

5.4 A mixture of benzene–air is saturated at 1 atm and 50 °C. The vapour pressure of benzene is 275mm Hg at 50 °C. Estimate the mass humidity and molal humidity.

5.5 A mixture of benzene–air is available at 750 mm Hg and 60 °C. The partial pressure of benzene is150 mm Hg at 50 °C. Estimate the mass humidity and molal humidity.

5.6 Air-water vapour mixture is available at a DBT of 35 °C and a relative saturation of 70%.Estimate its humidity, dew point, humid volume, adiabatic saturation temperature, humid heat andenthalpy.

5.7 Air at 50% relative humidity (RH) is to be used for a specific operation. Air is available at aDBT of 35 °C and 27 °C. What should be the final temperature to which it should be heated? It isfound that a DBT of 30 °C will be quite comfortable to the labourers. How will you obtain thiscondition? Indicate the exact temperature to which it should be taken before bringing it to 30 °C and50% RH. If it is necessary to use 5000 m3/h of air at the final condition mentioned above, estimate thequantity of air needed at its original condition.

5.8 Air at 80% saturation is to be maintained in a weaving room. Air at a DBT of 41 °C and WBT of24 °C is available. This air for weaving room is obtained by saturating it initially and then heating itto 80% saturation. Estimate the final temperature of the air and its humidity.

5.9 Air at a temperature of 60 °C and a pressure of 745 mm Hg and a % humidity of 10 is supplied toa drier at a rate of 120 m3/h. Water is evaporated in the drier at a rate of 25 kg/h. The air leaves at atemperature of 35 °C and a pressure of 742 mm Hg. Calculate (a) percentage humidity of air leavingthe drier and (b) volume of wet air leaving the drier per hour. (use vapour pressure data from SteamTables).

5.10 Air at a temperature of 30 °C and a pressure of 100 kPa has a relative humidity of 80%.Calculate (a) the molal humidity of air, (b) molal humidity of this air if its temperature is reduced to15 °C and its pressure increased to 200 kPa, condensing out some of the water, (c) weight of watercondensed from 100 m3 of the original wet air in cooling to 15 °C and compressing to 200 kPa, and(d) the final volume of the wet air of part (c)

Data: Vapour pressure of water at 30 °C = 4.24 kPaVapour pressure of water at 15 °C = 1.70 kPa

5.11 Air at 85 °C and absolute humidity of 0.03 kg water vapour per kg dry air, 1 std. atm iscontacted with water at the adiabatic saturation temperature and is thereby humidified and cooled to70% saturation. What are the final temperature and humidity of the air?

5.12 An air-water vapour sample has a dry bulb temperature (DBT) of 55 °C and an absolutehumidity 0.030 kg water/kg dry air at 1 std. atm pressure. Using humidity chart, calculate (a) therelative humidity, if vapour pressure of water at 55 °C is 118 mm Hg and (b) the humid volume inm3/kg dry air.

5.13 A drier is used to remove 100 kg of water per hour from the material being dried. The availableair has a humidity of 0.010 kg bone dry air, and a temperature of 23.9 °C and is heated to 68.3 °Cbefore entering the drier. The air leaving the drier has a wet-bulb temperature of 37.8 °C and a dry-bulb temperature of 54.4 °C. Calculate (a) air consumption rate, (b) humid volume of air before andafter preheating, (c) wet-bulb temperatures (WBT) of air before and after preheating, and (d) dewpoint of the air leaving the drier.

5.14 The air supply for a drier has a dry-bulb temperature of 23 °C and a wet-bulb temperature of 16°C. It is heated to 85 °C by heating coils and introduced into the drier. In the drier, it cools along theadiabatic cooling line and leaves the drier fully saturated.(a) What is its humidity?

(b) What is the dew point of the initial air?(c) How much water will be evaporated per 100 cubic metre of entering air?(d) How much heat is needed to heat 100 cubic metre air to 85 °C? (e) At what temperature does theair leave the drier?

5.15 A mixture of carbon disulphide vapour and air contains 23.3% carbon disulphide (CS2) byvolume. Calculate the relative saturation and percent saturation of the mixture at 20 °C and 740 mmHg pressure. Vapour pressure of CS2 at 20 °C = 300 mm Hg

5.16 Air at 60 °C and 745 mm Hg having a percent humidity of 10% is supplied to a drier at the rateof 1100 m3/h. Water at the rate of 20 kg/h is to be removed from the wet material. The air leaves thedrier at 35 °C and 742 mm Hg. Calculate (a) percent humidity of leaving air and (b) volumetric flowrate of leaving air.

5.17 Air is available at a DBT of 310 K and a WBT of 305 K. Estimate humidity, % saturation, dewpoint, humid volume and enthalpy using humidity chart. Estimate humid volume and enthalpy usingequations and compare the results.

5.18 One thousand m3/min of methane saturated with water vapour at 1 atm, 49 °C is cooled to 10 °C,so that part of water vapour is condensed. The mixture is then reheated to 24 °C at 1 atm. What is (a)the volumetric flow rate of leaving mixture, and (b) amount of water condensed?Data: Vapour pressure of water at 49 °C, 24 °C and 8 °C are 100 mm Hg, 27 mm Hg and 8 mm Hgrespectively.

5.19 Air at a temperature of 30 °C and pressure 760 mm Hg has a relative humidity of 60%,calculate:(a) The absolute humidity of air(b) The humidity of this air if its temperature is reduced to 20 °C and

the pressure increased to 2600 mm Hg, condensing out some of the water.

(c) The weight of water condensed from 1000 m3 of the original wet air. VP of water at 30 °C and 20°C are 32 mm Hg and 17.5 mm Hg respectively.5.20 A mixture of air water vapour is present at 40 °C and 762 mm Hg.

The volume of the mixture is 5 litres and the partial pressure of water vapour in the mixture is 27.7mm Hg. The mixture is heated to 60 °C at constant volume. Compute the following:(a) Partial pressure of water vapour(b) Partial pressure of dry air(c) Total pressure, and(d) Molal humidity.

5.21 A mixture of acetone vapour and nitrogen contains 15% acetone by volume. Calculate molalhumidity, partial pressure of acetone, relative saturation and % saturation at 20 °C and 745 mm Hg.Vapour pressure of acetone at 20 °C = 184.8 mm Hg.

5.22 Air at 60 °C and pressure of 745 mm Hg and a % humidity of 20 is supplied to a dryer at 1500m3/h. Water is removed at a rate of 2.5 kg/h. Air leaves at 35 °C at 742 mm Hg, vapour pressure at 35°C = 43 mm Hg, and vapour pressure at 60 °C = 149.38 mm Hg Estimate:(a) Percentage humidity of air leaving

(b) Volumetric flow rate of wet air leaving

Psychrometric chart for air–water system at 760 mm Hg pressure.

Crystallization 6

Crystallization is a process in which the solid particles are formed from a homogeneous phase.During crystallization, the crystals form when a saturated solution gets cooled. The solution leftbehind after the separation of crystals is known as mother liquor which will also be a saturatedsolution. The mixture of crystals and mother liquor is known as magma.

A simple illustration of a crystallization process is shown below.Water evaporated, E kg/hxE

Feed, F kg/hxF Crystallizer

Mother liquor, M kg/h xM

Crystal, C kg/hxC

Feed rate of solution of concentration, xF : F kg/h Amount of water evaporated(xE = 0 as salt in water evaporated will be zero) : E kg/h Amount of mother liquor of concentrationxM : M kg/h (Mother liquor will always be a saturated solution)Weight of crystal formed of concentration, xC : C kg/h Total mass balance gives, F = C + M + E (6.1)Component balance gives, FxF = CxC + MxM + ExE (6.2)

Since we normally know F, E, xF xM xEwe can find C and M by solving the Eqs. (6.1) and (6.2). xE

will always be zero as it is free from salt.When we get anhydrous salt, xCis 1.0. Some times we get hydrated salt during crystallization process.Under such circumstances xC will not be 1.0 but will be less than unity and this can be obtained bymolecular weight of anhydrous salt divided by the molecular weight of hydrated salt. The aboveprinciples are explained through the following problems.

111WORKED EXAMPLES

6.1 A solution of sodium chloride in water is saturated at a temperature of 15 °C. Calculate theweight of NaCl that can be dissolved by 100 kg of this solution if it is heated to 65 °C.Solubility of NaCl at 15 °C = 6.12 kg mole/1000 kg H2O

Solubility of NaCl at 65 °C = 6.37 kg mole/1000 kg H2O Basis: 1000 kg of waterNaCl in saturated solution at 15 °C = (6.12 ¥ 58.5) = 358 kg NaCl in saturated solution at 65 °C =(6.37 ¥ 58.5) = 373 kg NaCl that may be dissolved further = (373 – 358) = 15 kg

1000Water in 100 kg of saturated solution at 15 °C = 100 ¥ÊˆÁ˜Ë¯ 1358 = 73.6 kg \ amount of additional NaCl that will be dissolved= 15 ¥73.6= 1.1 kg1000

6.2 After a crystallization process, a solution of calcium chloride in water contains 62 kg CaCl2 per100 kg of water. Calculate the weight of this solution necessary to dissolve 250 kg of CaCl2◊ 6H2O at25 °C (solubility = 7.38 kg mole CaCl2/1000 kg H2O)

Basis: x kg of water is present in original solutionCaCl2◊ 6H2O has the molecular weight = 111 + 108 = 219 250 kg of solution will have 126.7 kg ofCaCl2 and 123.3 kg of 6H2O Total CaCl2 entering the process = (0.62x + 126.7) kg (1) Waterentering the process = (x + 123.3) kgSolubility of CaCl2 at saturated condition for 1000 kg water

= 7.38 kmoles = 819.18 kg\ CaCl2 leaving the process along with (x + 123.3) kg of water=819.18 (x + 123.3) kg (2)ʈÁ˜Ë¯ 1000

Equating (1) and (2), we get(0.62x + 126.7) = 0.81918 (x + 123.3)0.62x + 126.7 = 0.81918x + 101 – 0.19918x = – 25.7\ x = 129.03 kg100 kg of water is equivalent to 162 kg of solution

162129.03 kg of water is equivalent to 129.03 ´100= 209 kg of solutionOriginal solution needed = 209 kg.

6.3 A solution of sodium nitrate in water at a temperature of 40 °C contains 49% NaNO3 by weight.(a) Calculate the percentage saturation of this solution (b) Calculate the weight of NaNO3 that may becrystallized from

1000 kg of solution by reducing the temperature to 10 °C (c) Calculate the percentage yield of theprocess.Solubility of NaNO3 at 40 °C = 51.4% by weight.Solubility of NaNO3 at 10 °C = 44.5% by weight.Basis: 1000 kg of original solution(a) kg of NaNO3 49 = 0.96kg of water 51

kg of NaNO at 40 °C, ÈØ3 ÉÙ 51.4 = 1.06ÊÚsaturation 48.6 0.96 % saturation of given solution = ÉÙ ÈØ´ 100 = 90.8%ÊÚ

(b) Let a kg be NaNO3 crystallized outNaNO3 balance = (1000 ´ 0.49) = (1000 – a) ´ (0.445) + a; Solving, a = 81 kg

81 (c) % Yield = ÈØ ÉÙ 100 = 16.53%ÊÚ

6.4 A solution of K2Cr2O7 in water contains 13% by weight. From 1000 kg of this solution isevaporated 640 kg of water. The remaining solution is cooled to 20 °C. Calculate the amount and thepercentage yield of K2Cr2O7 crystals formed.Solubility at 20 °C = 0.39 kmole/1000 kg H2O

Basis: 1000 kg of original solutionK2Cr2O7 = 130 kg;Water = 870 kgWater present finally = (870 – 640) = 230 kg

0.39 ÈØ´ 294 = 26.4 kg´ÉÙK2Cr2O7 present then = 230 ÊÚ \ K2Cr2O7 crystallized = 130 – 26.4 = 103.6 kg% Yield = 103.6= 79.7%130

6.5 A solution of sodium nitrate in water contains 100 g of salt per 1000 g of water. Calculate theamount of ice formed in cooling 1000 g of this solution to –15 °C (solubility at –15 °C = 6.2 gmole/1000 g H2O)

Basis: 1000 g of original solution 100

ÈØ = 91 g´ÉÙNaNO3 present initially = 1000 ÊÚWater present initially = 909 gWater retained along with NaNO3 = 1000 91= 172.6 g6.2 85

\ Ice formed = Original water – water retained along with NaNO3 = (909 – 172.6) = 736.4 g

6.6 1000 kg of a 30% aqueous solution of Na2CO3 is slowly cooled to 20 °C. During cooling, 10%water originally present evaporates. The crystal is Na2CO3× 10H2O. If the solubility of anhydrousNa2CO3 at 20 °C is 21.5 kg/100 kg of water, what weight of salt crystallizes out?

Basis: 1000 kg of solution.Salt = 300 kgWater = 700 kgWater evaporated = 700 ´ 0.1 = 70 kgLet a kg of salt crystallize out. Let the weight of mother liquor be b kg Molecular weight of Na2CO3=106 and that of Na2CO3 × 10H2O = 286 a + b = (1000 – 70) = 930 kg

106ÈØÈ Ø21.5 = 300 kg2CO3 balance = abNaÉÙÉ Ù ÊÚÊ Ú Solving, a = 700 kg; b = 230 kgCheck (water balance): 180 ËÛË Û 700 ÈØ È Ø ÌÜÌ Ü 100 = 630 kgÊÚ Ê Ú

ÍÝÍ Ý6.7 A batch of saturated Na2CO3 solution of 100 kg is to be prepared at

50 °C. The solubility is 4.48 g mole/1000 g H2O at 50 °C. (i) If the monohydrate were available, howmany kg of water would be required to form the solution?(ii) If the decahydrate is available how many kg of salt will be required? Basis: 100 kg of saturatedsolution at 50 °C.Solubility at 50 °C = (4.48 ´ 106) kg Na2CO3/1000 kg H2O i.e. 474.88 kg Na2CO3/1474.88 kgsolution.

100 474.88In 100 kg of solution, Na2CO3 present = 1474.88

= 32.2 kgwater present = (100 – 32.2) = 67.8 kgMolecular weight of Na2CO3 × H2O = 106 + 18 = 124 kg Molecular weight of Na2CO3 × 10H2O =106 + 180 = 286 kg (i) Water needed for monohydrate = Actually needed-water from

hydrated salt= (67.8) – 32.2 18 ËÛ ÌÜ ÍÝ = 62.33 kg(ii) Weight of decahydrate needed = 32.2 ´286 = 86.88 kg106 Discussion: In case (ii), water available in the salt is32.2 180 = 54.68 kg´106

The additional water needed hence is 13.12 kg, thus making the total water to 13.12 + 54.68 = 67.8 kg(which corresponds to solubility data) 6.8 10,000 kg/hr of 6% solution of salt in water is fed to anevaporator. Saturated solution is produced and some salt crystallizes out. The hot crystals with someadhering solution are centrifuged to remove some of the solution. Then the crystals are dried toremove the rest of water. During an hour test, 837.6 kg of concentrated solution was removed inevaporator, 198.7 kg of solution was separated in the centrifuge and 361.7 kg of dried crystals got.Previous test on the centrifuge show that it removed 60% of adhering solution. Calculate (a) thesolubility of the salt; (b) water removed in evaporator; (c) water removed in drier; and (d) waterleaving the evaporator.

Discussion: The solution leaving the evaporator and centrifuge maintains the same concentration andis saturated. From evaporator, the various streams leaving are water vapour, saturated solutions andcrystals with adhering solution.Basis: One hour.

60% adhering solution (removed in centrifuge) = 198.7 kg \ Total adhering solution enteringcentrifuge (along with crystal) =198.7= 331.2 kg0.6

Water vapour10,000 kg/hrCentrifuge6% solution EvaporatorAdheringWatersolution+ Dry crystal crystalDrier 361.7 kg837.6 kg solution 198.7 kg

Let s be the solubility of salt, which is given by (s kg of salt/kg of solution) Making salt balance,(10,000 ´ 0.06) = (837.6 + 198.7)s + 361.7

Solving, (a)s = 0.23 kg of salt/kg of solution.1 kg solution = 0.23 kg salt/0.77 kg H2O = 0.3 kg salt/kg H2O

(b) Weight of adhering solution entering drier = (331.2 – 198.7) = 132.5 kg(c) Water in it (removed in drier) = 132.5 ´ 0.77 = 102 kg (since, 1 kg of solution contains 0.77 kgwater and 0.23 kg of salt)(d) Weight of material entering drier = Final dry salt + Water evaporated

= (361.7 + 102) = 463.7 kgWeight of material entering centrifuge = (463.7 + 198.7) = 662.4 kg

Weight of water leaving the evaporator = 10,000 – (837.6 + 662.4) = 8,500 kgCheck:Water balance = 8,500 + 102 + (837.6 + 198.7) ´ (0.77)

= 8,500 + 102 + 798 = 9,400 kg6.9 In a solution of naphthalene in benzene, mole fraction of naphthalene is 0.12. Calculate the weightof this solution necessary to dissolve 100 kg of naphthalene at 40 °C (solubility of naphthalene is57% by weight)

100 kgX kg solution (100 + X) kg0.817 wt fr. of benzeneTank

0.183 wt fr. of naphthalene0.57 wt. fr. of naphthalene

Basis: 1 mole of feed solutionComponent mole Molecular Weight, kg Weight weight fraction

Naphthalene (C10H8) 0.12 128 15.4 0.183Benzene (C6H6) 0.88 78 68.6 0.817Total — — 84.0 1.000

Let us assume that X kg of feed solution enters the tank. Naphthalene balance in tank gives, 0.183X +100 = (100 + X)0.57 X = 111 kg of original feed solution.

6.10 A crystallizer is charged with 7,500 kg of aqueous solution at 104 °C containing 29.6% byweight of anhydrous Na2SO4. The solution is then cooled to 20 °C. During this operation 5% of water

is lost by evaporation. Glauber salt crystallizes out. Find the yield of crystals.

Solubility at 20 °C = 194 g Na2SO4/100 g waterMolecular weight of Glauber’s salt (Na2SO4×10H2O) = 142 + 180 = 322 Basis: 7,500 kg of aqueoussolution.Na2SO4 present = 7500 ´ 0.296 = 2220 kgH2O present = (7500 – 2220) = 5280 kgH2O lost in evaporation = 5280 ´ 0.05 = 264 kgH2O remaining = (5280 – 264) = 5016 kg

Let a kg of Glauber salt crystallize and b kg be the weight of the solution. ab ÈØÈ ØNa2SO4 balance: 2220 = 142ÉÙÉ Ù ÊÚÊ Ú ab ÈØÈ ØH2O balance: 5016 = 180

ÉÙÉ Ù

ÊÚÊ ÚSolving, a = 3749.5 kg and b = 3486.5 kgWeight of Glauber salt = 3749.5 kgWeight of solution = 3486.5 kgCheck: (total balance) 3749.5 + 3486.5 + 264 = 7500 kg

6.11 An evaporator-crystallizer is to produce 13,000 kg dry salt/h from a feed solution having 20%NaCl. The salt removed carries 20% of its weight of brine containing 27% NaCl (All Composition inWeight %). Calculate the feed rate kg/h.

Water20% NaCl EvaporatorFeed F crystallizer13,000 kg dry salt/h

Basis: 1 h of operation13,000 kg of salt with 20% brineWeight of feed brine = 13,000 ´ 0.2 = 2,600NaCl in this solution = 2,600 ´ 0.27 = 702NaCl balance gives, 0.2F = (13,000 + 702)\ Feed = 13,702/0.2 = 68,510 kg/h.

6.12 5000 kg of KCl are present in a saturated solution at 80 °C. The solution is cooled to 20 °C in anopen tank. The solubilities of KCl at 80 °C and 20 °C are 55 and 35 parts per 100 parts of water.

(a) Assuming water equal to 3% by weight of solution is lost by evaporation, calculate the weight ofcrystals obtained. (b) Calculate the yield of crystals neglecting loss of water by evaporation; KClcrystallizes without any water of crystals.

3% Water5000 kg KCl Crystallizer 20 °C

80 °CSaturated solutionSaturated solution

Crystal

Basis: 5,000 kg of KCl salt in a solution at 80 oC(a) Solubility at 80 °C = 55 kg salt/100 kg wateror, solubility of KCl = 55 kg/155 kg solution = 0.355 kg/kg solution 5000 kg KCl is equivalent to5000/0.355 = 140,90.91 kg of solution Water present = 140,90.91 – 5000 = 9090.91 kgWater evaporated = 3% weight if solution = 140,90.91 ´0.03 = 422.73 kg Water remaining =9,090.91 – 422.73 = 8668.18 kg

35 KCl in leaving solution at 20°C = 8668.18 ´ ÈØ ÉÙ = 3033.86 kgÊÚ

\ KCl crystallized out = 5000 – 3033.86 = 1966.14 kg (b) KCl in solution at 20 °C = 9090.91 ´ 0.35 =3181.82 kg \ KCl crystallized out = 1818.18 kg

% Yield of crystals = 1818.18100 = 36.37%´5000

6.13 A crystallizer is charged with 6,400 kg of an aqueous solution containing 29.6% of anhydroussodium sulphate. The solution is cooled and 10% of the initial water is lost by evaporation.Na2SO4.10H2O crystallizes out. If the mother liquor (after crystallization) is found to contain 18.3%Na2SO4, calculate the weight of the mother liquor.

10% water6,400 kg 18.3% Na2SO4Crystallizer Mother liquor M29.6%Na2SO4× 10H2O (142 + 180) = 322

Basis: 6,400 kg solution.Weight of Na2SO4 = 6,400 ´ 0.296 = 1,894.4 kgWater present in feed = 4,505.6 kg

Let M be the weight of mother liquor and C the weight of crystal formed.Water lost by evaporation = 10% of 4505.6 kg = 450.56 kg

142 Overall balance of Na2

SO

4

gives, 1894.4 = CÈØ ÉÙ + (0.183 ´ M)ÊÚ322

C + M = (6400 – 450.56) = 5949.44 kgSolving, M = mother liquor = 2826.7 kgC = Crystal = 3122.7 kg

Check:(water balance) 450.56 + 2826.7 81.7ÈØÈ Ø180

ÉÙÉ Ù 322ÊÚÊ Ú = 4505.6 kg6.14 What will be the yield of Glauber salt if pure 32% solution is cooled

to 20 °C from hot condition? There is no loss of water. The solubility of Na2SO4 in water at 20 °C is19.4 g per 100 g of water.Feed 32%CrystallizerMother Liquor, YCrystals, X

Basis: 1,000 g of the feed solution.Na2SO4.10H2O (142 + 180 = 322)Na2SO4 present in the feed = 320 gWater present in the feed = 680 gOverall balance is X + Y = 1000 g

142 XY Na2

SO4

balance = ÊˆÊ ˆ Á˜Á ˜ = 320Ë¯Ë ¯ 180 XY Water balance = ÊˆÊ ˆ = 680Á˜Á ˜ 322 119.4 Solving, we get X = Weight of crystals = 566.761 g and Y = Weight of mother liquor = 433.239 gYield of the crystals = 566.761= 56.68%1000 EXERCISES

6.1 Feed to a crystallizer contains 6,420 kg of aqueous solution containing 29.6% anhydrous sodiumsulphate by weight at 104 °C. The solution is cooled to 20 °C to crystallize out the desired Glauber’ssalt. The mother liquor is found to contain 16.1% anhydrous sodium sulphate. Estimate the weight ofmother liquor.

6.2 A vacuum crystallizer is to produce 10,000 kg of FeSO4.5H2O crystals per hour. The feedsolution contains 38.9 parts FeSO4/100 parts water. The feed enters at 70 °C and is cooled to 27 °C.The solubility at 27 °C is 30.3 parts FeSO4/100 parts water. Determine the quantity of feed and waterlost as vapour.

6.3 10,000 kg of a 30% aqueous solution of Na2CO3 is slowly cooled to 20 °C. During cooling, 3%water originally present evaporates. The crystal is Na2CO3.10H2O. If the solubility of anhydrousNa2CO3 at 20 °C is 21.5 kg/100 kg of water, what is the weight of Na2CO3.10H2O formed?

6.4 1000 kg of a solution containing 25% of Na2CO3 by weight is slowly cooled to 20 °C. Duringcooling 15% water originally present evaporates. The crystal formed is Na2CO3.10H2O. If thesolubility of anhydrous Na2CO3 at 20 °C is 21.5 kg/100 kg of water, what is the weight ofNa2CO3.10H2O crystals formed?

6.5 1500 kg of a solution containing 20% of Na2SO4 and 80% water by weight is subjected toevaporative cooling and 20% of original water evaporates. Calculate the yield of Na2SO4.10H2Ocrystals formed from solution, if the temperature of solution is reduced to 20 °C. The solubility ofanhydrous Na2SO4 in water at 20 °C is 19.4 gm per 100 gm of water.

6.6 An evaporator concentrates 10,000 kg/h of 20% KNO3 solution to 50% KNO3. The concentratedliquor is sent to a crystallizer where crystals of KNO3 are formed and separated. The mother liquorfrom the crystallizer is recycled and mixed with the evaporator feed. The recycle stream is asaturated solution containing 0.6 kg KNO3/kg water. The crystals carry 4% water. Compute waterevaporated and crystals obtained.

6.7 A crystallizer is charged with 9000 kg of an aqueous solution of 20% Na2SO4 and it is subjectedto evaporative cooling and 10% of original water evaporates. Glauber’s salt crystallizes. The motherliquor leaves with 16% salt. Calculate the weights of (a) water loss (b) the mother liquor leaving and(c) crystals formed.

6.8 A solution of sodium sulphate in water is saturated at a temperature of 40 °C. Calculate theweight of crystals formed and the percentage yield obtained by cooling 100 kg of this solution to atemperature of 5 °C.

Solubility Data: At 40 °C : 32.6% Na2SO4At 5 °C : 5.75% Na2SO4

6.9 A solution of K2Cr2O7 in water contains 16% K2Cr2O7 by weight. 1000 kg of this solution areevaporated to remove some water. The resulting solution is cooled to 20 °C. If the yield of K2Cr2O7crystals is 80%, calculate the amount of water evaporated.

Solubility of K2Cr2O7 at 20 °C = 0.39 k mole/1000 kg H2O 6.10 A saturated solution of MgSO4 at 80°C is cooled to 25 °C. During this

process 5% of the original water is lost. Find the weight of water lost, mother liquor leaving andfeed, if 1500 kg of MgSO4.7H2O crystallizes. Solubility data: At 80 °C : 64.28 g MgSO4 per 100 gwater At 25 °C : 40.88 g MgSO4 per 100 g water

6.11 Hypo crystals Na2S2O3 × 5H2O are to be produced at the rate of 2000 kg/h. A 60% Na2S2O3solution is cooled to 293 K from 333 K. The solubility at 293 K is 70 parts anhydrous salt per 100parts of water. Estimate the amount of feed needed.

Mass Balance7

This chapter deals with the mass balance both with and without chemical reactions. The principlesgiven in Chapter 2 still hold good. The application of mass balance in unit operations and processesis illustrated through worked examples and exercise problems.

WORKED EXAMPLES

7.1 Soyabean seeds are extracted with hexane in batch extractors. The seeds contain 18.6% oil, 69%solids and 12.4% moisture. At the end of the extraction process, the residual cake is separated fromhexane. The analysis of cake reveals 0.8% oil, 87.7% solids and 11.5% moisture. Find the %recovery of oil.

Basis: 100 kg of seedsSolids free from oil is the tie elementWeight of solids in feed = 69 kg = 87.7% of cake \ cake = 78.68 kgOil extracted = (18.6 – 78.68 ¥ 0.008) = 17.971 kg

% oil extracted =17.971¥ 100 = 96.6%ʈÁ˜Ë¯ 18.6

7.2 A multiple effect evaporator handles 100 tonnes/day of pure cane sugar. The feed to theevaporator contains 30% solids. While the concentrate is leaving with 75% solids concentration,calculate the amount of water evaporated per day.

?30% solids 75% solidsEvaporator

Basis: 100 tonnes of feed = 105 kg Solids in the feed = 30,000 kg122\ Concentrated liquid leaving =30,000= 40,000 kg0.75

Water evaporated/day = (1,00,000 – 40,000) = 60,000 kg 7.3 A water soaked fabric is dried from44% moisture to a final moistureof 9%. Calculate the weight of water removed per 200 kg of driedfabric.

?44% moisture 9% moisture Drier200 kgBasis: 200 kg of product.Dry fabric = (200 ´ 0.91) = 182 kg 182

Wet fabric entering = ÈØ ÉÙ = 325 kgÊÚ Water evaporated = (325 – 200) = 125 kg

7.4 The exit gas from a phosphate reduction furnace analyzes P4 : 8%, CO : 89% and N2: 3%. Thisgas is burnt with air under conditions selectively to oxidize phosphorus. From the flue gas, the oxidesof phosphorus precipitate on cooling and is separated from the remaining gas. Analysis of the lattershows CO2: 0.9%, CO : 22.5%, N2 : 68.0% and O2: 8.6%. Assume oxidation of phosphorus iscomplete and that it exists in the flue gas partly as P4O6 and partly as P4O10. Calculate:

(a) % of CO entering the furnace that is oxidized to CO2 and (b) % of P4 that is oxidized to P4O10P4 + 3O2 ® P4O6P4 + 5O2 ® P4O10CO + ½O2 ® CO2

Exit gasfrom phosphatereduction furnaceBurner CO2, O2, Air CO, N2Oxides of P

Basis: 100 g moles of exit gas from phosphate reduction furnace P4 is 8 g moles.Let x g mole of P4 get converted to P4O10\ (8 – x) g mole of P4 gets converted to P4O6 ‘Carbon’ is the tie element89 g atoms of C º 23.4% of final gas. (Total carbon in CO and CO2)

89 = 380.342 g moles\ Total final gas moles = 0.234 (a) % CO oxidized to CO2 = (380.342 ´ 0.009/89) ´ 100 = 3.852 g moles(b) N2 in final gas = (380.342 ´ 0.68) = 258.632 g moles N2 from air = (258.632 – 3) = 255.632 gmolesO21 = 67.953 g moles2 from air = 255.632 ´79

O2 for formation of P4O10 = 5x g moleO2 for formation of P4O6 = 3(8 – x) g moleO2 for formation of CO2 = 380.342 ´ 0.009 ´ 0.5 = 1.712 g moles O2 leaving = 380.342 ´ 0.086 =32.71 g molesO2 reacted = (67.953 – 32.71) = 35.243 g molesMaking a balance for oxygen,

(5 x + 24 – 3x + 1.712) = 35.243\ x = 4.7654 g moles\ P4 converted to P4O10 = (4.7654 ´ 100/8) = 60%

7.5 A tank of weak H2SO4 contains 12.43% acid. If 200 kg of 77.7% H2SO4 are added to the tank andthe final acid is 18.63%, how many kg of weak acid is used?

Let the weight of weak acid be x kg and the weight of final acid be y kg.Overall balance is x + 200 = y

Acid balance is given as 0.1243x + (200 ´ 0.777) = 0.1863y Solving, x = 1,905.5 kg and y = 2,105.5kg

7.6 A cotton mill dries a water soaked fabric in a drier from 54% to 9% moisture. How manykilogram of water are removed by drying operation per 1200 kg of feed.

Basis: 1200 kg of feed.Dry fabric = (1200 ´ 0.46) = 552 kg Water54% moisture 9% moisture Drier Weight of product = 552 = 606.6 kg0.91

Water removed = 1200 – 606.6 = 593.4 kgAlternative method:Water in feed = 1200 ´ 0.54 = 648 kgWater in product = 606.6 ´ 0.09 = 54.6 kgWater removed = 648 – 54.6 kg

= 593.4 kg7.7 A gas containing 2% NH3, 25.4% N2 and the rest H2 is flowing in a pipe. To measure the flowrate an ammonia rich gas containing 96%

NH3, 3% H2 and 1% N2 is sent at a rate of 100 cc/min. The concentration of NH3 in the down streamis 6%. Find the flow rate of gas at inlet.

Basis: Let x cc/min of feed gas enter the pipeAmmonia rich gas entering = 100 cc/min\ Exit gas = (100 + x) cc/min.Making a balance for ammonia: 0.02x + 96 = (100 + x) 0.06 Solving, x = 2250 cc/min.

x cc/min. gas Exit (100 + x) Pipe6% NH3NH3rich gas(96%) 100 cc/min.

7.8 A lime stone analysis shows CaCO3: 94.52%, MgCO3: 4.16% and rest insoluble. (a) How manykg of CaO could be obtained from 4 tonnes of limestone? (b) How many kg of CO2 are given off perkg of limestone?(a) Basis: 4 tonnes of limestone = 4,000 kg

CaCO3 = 0.9452 ´ 4000 = 3,780.8 kgMgCO3 = 0.0416 ´ 4000 = 166.4 kg

CaCO3 ® CaO + CO2100 56 44MgCO3 ® MgO + CO284.3 40.3 44

CaO got =3780.8 56= 2,117.3 kg100 (b) Basis: 1 kg of limestoneCaCO3 = 0.9452 kg, MgCO3 = 0.0416 kg 44ÊˆÊ ˆ44CO2 produced = 0.9452 ¥+ ¥

Á˜Á ˜ Ë¯Ë ¯ = 0.4376 kg (by stoichiometry)

7.9 The gases from a sulphur burner have the following analysis: SO2: 9.86%, O2: 8.54%, N2: 81.6%.After passage of gases through a catalytic converter, the analysis is 0.605% SO2; 4.5% O2; 94.895%N2. What percentage of SO2 entering the converter has been oxidized to SO3?

Basis: 100 g moles of gas entering the converterN2 is the tie element.Exit stream =81.6 = 86 g moles0.94895

SO2 leaving = 86 ¥ 0.00605 = 0.5203 g molesSO2 converted to SO3= SO2 entering – SO2 leaving the converter= (9.86 – 0.52) = 9.34 g moles

% SO2 converted = 9.34 ¥100 = 94.7%9.86

7.10 Hydrochloric acid is commercially prepared in a Mannheim furnace by the reaction betweenNaCl and H2SO4. The HCl gas is absorbed in water. Calculate (a) the weight of HCl formed when 2tonnes of 98% salt reacts; (b) How much sodium sulphate is produced and how many kg of 40% acidwill be produced?

Basis: 2 tonnes of salt = 2000 kgNaCl = 2000 ¥ 0.98 = 1960 kg 2NaCl + H SO →Na SO + 2HCl24 24

(2 58.4) 98××142 (2 36.4)

2 36.4 (a) HCl formed = 1960 ¥ ¥Í˙ 2 58.4= 1221.6 kg¥Î˚ (b) Na2SO4 formed = 1960 142= 2382.9 kg¥2È˘Í˙ ¥Î˚ 40% acid formed = 1221.6/0.4 = 3054 kg

7.11 Dolomite chiefly a mixture of calcium and magnesium carbonates is to be leached withconcentrated H2SO4 to recover Mg as MgSO4. The rock analysis indicates 20% Ca, 10% Mg and 5%SiO2. After reaction the soluble material is filtered off and the solid are washed and discarded.Analysis of the sludge cake shows 50% moisture, 2% SiO2; 0.5% Mg and 1% Al. What fraction ofMg was extracted? Basis: 100 g of dolomite

SiO2 is the tie element5 g SiO2 º 2% sludge cake\ weight of sludge cake =5 = 250 g0.02 Mg in sludge cake = 250 0.5 = 1.25 g´100 Mg extracted = (10 – 1.25) = 8.75 gFraction Mg extracted = 8.75= 0.87510

7.12 A certain soap contains 50% moisture on the wet basis when raw. The moisture is reduced to20% before the soap is pressed into cakes. How many 300 g cakes can be pressed from 1 tonne ofraw soap?

Basis: 1 tonne of raw soap = 1000 kgDry soap = 500 kg º 80% after drying 500 = 625 kg\ Weight of dried soap = 0.8 625 No. of cakes pressed = ÈØ ÉÙ = 2083ÊÚ (The number of cakes has to be an integer)

7.13 Phosphate rock at ` 80/tonne is being added to a cheap fertilizer to enrich it. If the cheapfertilizer costs ` 32/tonne and the final blended product including 30% profit is to be marketed at `64/tonne, how much phosphate rock should be added to each tonne of cheap fertilizer?

Basis: 1 tonne of cheap fertilizer and x tonne of phosphate rock. Blended fertilizer = (1 + x) tonne

Final cost of blended fertilizer = ` 64/tonne

Actual cost (30% profit) =64 = ` 49.23 = cost of production1.3 Cost balance: 32 + 80x = (1 + x) 49.23Solving, x = 0.56 tonne (Phosphate rock)

7.14 Limestone is a mixture of Ca and Mg carbonates. Lime is made up calcining the carbonates, i.e.driving away all CO2 gas. When pure limestone is calcined, 44.8 g of CO2 is got per 100 g oflimestone. What is the composition of limestone?Basis: 100 g of limestone (pure)

Let x g be the weight of CaCO3 and y g be the weight of MgCO3CaCO3 ® CaO + CO2

100 56 44MgCO3 ® MgO + CO284.3 40.3 44Overall balance: x + y = 100 (since pure)CO2 balance = 0.44x + 0.52y = 44.8Solving, x = 90 g (CaCO3) and y = 10 g (MgCO3) 7.15 The feed to a fractionating system is 30,000kg/h of 50% benzene, 30% toluene and 20% xylene. The fractionating system consists of two towersNo. I and No. II. The feed enters tower I. The overhead product from I is x kg/h of 95% benzene, 3%toluene and 2% xylene. The bottom product from I is feed to II resulting in an overhead product of ykg/h of 3% benzene, 95% toluene and 2% xylene while the bottom from II tower is z kg/h of 1%benzene, 4% toluene and 95% xylene. Find x, y and z.

x yFeed I IIz

Overall balance yieldsx + y + z = 30,000Individual balance yields:For benzene, 0.95x + 0.03y + 0.01z = 15,000 For toluene, 0.03x + 0.95y + 0.04z = 9,000 For xylene,0.02x + 0.02y + 0.95z = 6,000 Solving, x = 15,452.2 kg/h

y = 8,741.3 kg/hz = 5,806.5 kg/hTotal 30,000.0 kg/h

7.16 A sample of fuel has the following composition by mass: C : 84%, H2: 15.2% and restcomprising noncombustibles. The fuel was completely burnt with excess air in an internal combustionengine. The dry exhaust gas has 8.6% CO2 by volume. Estimate the amount of air supplied per kg offuel and find the products of combustion (air contains 23.1% oxygen by weight).

Basis: 1 kg of fuelC : 0.84 kg = 0.07 kmole; H2: 0.152 kg = 0.076 kmole; Noncombustibles – 0.008 kgC+O

CO ; H +½O nn H O222 2 212 32 44 2 16 18

CO2 formed º 0.07 kmoleH2O formed = 0.076 kmoleTotal kmole of exhaust stream = 0.07 = 0.814 (dry)0.086 O2

needed for forming CO2

and H2

O = ÈØ0.076 ÉÙ ÊÚ = 0.108 kmoleN2

entering (along with stoichiometrically needed O2

) = 0.108 ´

7921= 0.4063 kmole

Air needed (stoichiometric) = 0.108 = 0.5143 kmole0.21 Exhaust gas contains CO2, N2 and O2Excess air in exhaust = Total moles in exhaust stream (dry) –

(N2 entering as per stoichiometry + CO2 formed) = 0.814 – (0.4063 + 0.07)= 0.814 – 0.4763 kmole= 0.3377 kmole

We know what air containsO2 = 0.071 kmole (21%) N2 = 0.2667 kmole (79%)N2 total = 0.4063 + 0.2667 = 0.673 kmoleTotal air supplied = (0.5143 + 0.3377) = 0.8520 kmole

0.3377 % excess air = ÈØ ÉÙ ´ 100 = 65.6%ÊÚ Weight of air supplied/kg of fuel = (0.852 ´ 28.84) = 24.57 kg

Components Weight, kmole Composition of dry exhaust gas

CO2 0.0700 8.6O2 0.0710 8.7N2 0.6730 82.7

Total 0.8140 100.0

7.17 A company buys its paper from the manufacturer at the contract price of ` 3/kg on a specificationthat the paper should not have more than 5% moisture. It is also agreed that the price can be suitablyadjusted if the moisture content differs from the specified value.

A shipment of 5,000 kg of paper was found to contain 7.86% moisture. (a) Find the price to be paidto manufacturer as per contract. (b) If it were agreed that the freight was to be borne by the companyand adjusted according to the moisture content, what would the company pay to the manufacturer if thefreight rate is ` 40/1000 kg of dry paper.

Basis: 100 kg of 5% moist paper.Dry paper 95 kg; moisture 5 kgCost of dry paper = 1003 = ` 3.16´95

In the case of 5,000 kg paper with 7.86% moisture Weight of dry paper = 4,607 kgand weight of moisture = 393 kg

(a) Cost of paper = 4,607 ´ 3.16 = ` 14,558.12(b) Freight charge = 4,60740 = ` 185´1000 Total cost including freight = ` 14,743.12

7.18 A mixture of 5% ethylene and 95% air is passed through a suitable catalyst in a reactor. Some ofthe ethylene does not react, some form oxide, some turn to CO2 and water. The entire gas mixtureenters an absorption tower where water is sprayed. The oxide is converted to glycol. The gas leavingthe absorber analyzes C2H4 : 1.085%, CO2: 4.345%, O2: 13.055% and N2: 81.515% on dry basis.The partial pressure (pp) of H2O in this gas is 15.4 mm Hg while total pressure is 745 mm Hg. If onemole of water is sprayed per 100 mole of gas mixture, calculate the composition of ethylene glycol–water product formed.

5% C2H4Reactor

95% air Gases Ethylene, CO2, Absorber O2, N2

pp.H2O 15.4 mm HgEthylene glycolwater

Reactions:1.2C H + O → 24 2 Ethy lene

2C H O 24

Ethy lene oxide

2. C2H4 + 3O2 ® 2CO2 + 2H2O3. 24 2 →C H O+HO (CHOH)2 2 Ethy lene gly col

Basis: 100 kmoles of feed gas.C2H4: 5 kmoles; O2 : (95 ´ 0.21) = 20 kmoles; N2: 75 kmolesN75 = 92 kmoles2 balance: Moles leaving the absorber = 0.81515

C2H4 : 92 ´ 0.01085 = 1 kmoleCO2 : 92 ´ 0.04345 = 4 kmolesO2 : 92 ´ 0.13055 = 12 kmoles

H2O =92 15.4= 1.94 kmoles74515.4 C2H4 reacted : 5 – 1 = 4 kmolesC2H4 converted to CO2 : 4 = 2 kmoles2

C2H4 converted to C2H4O : 4 – 2 = 2 kmolesWater formed by reaction 2 = 4 mole º CO2 kmole [oxygen sent] – [used up oxygen] = 20 – [1 + 6] =13 kmoles

Moles of gases entering absorber: C2H4 : 1; CO2 : 4; C2H4O : 2; H2O:4; O2 : 12; N2:75Total number of moles = 98.Water sprayed at the top of absorber = 98 ´1 = 0.98100

Total moles of water entering absorber = (4 + 0.98) = 4.98 kmoles Water in exit gas = 1.94 kmolesWater reacted in absorption column = 2.00 kmolesEthylene glycol formed = 2 kmoles

Water leaving along with ethylene glycol = (4.98 – 2 – 1.94) = 1.04 kmolesComposition of liquid stream:Component moles Molecular weight Weight Weight % (CH2OH)2 2.00 62 124.00 86.88Ethylene glycol(H2O) 1.04 18 18.72 13.12WaterTotal 3.04 142.72 100.007.19 The first step in the manufacture of H2SO4 from pyrites consists of burning pyrites in air. Thereactions taking place are:

FeS2 + 2.5O2 ® FeO + 2SO2 (1) 2FeS2 + 5.5O2 ® Fe2O3 + 4SO2 (2)

The flue gas analysis shows SO2: 10.2%; O2: 7.8%; N2: 82% on dry basis at 600 °C and 780 mm Hg.(a) In what ratio does the two reactions take place?

(b) How much excess air was fed if the reaction (2) is desired?

Basis: 100 kmoles of flue gasSO2: 10.2%; O2: 7.8%; N2 : 82%Let x kmole of FeS2 by reaction (1) andy kmole of FeS2 by reaction (2) be reacted.

(a) O21 = 21.8 kmoles (using nitrogen balance)2 fed = 82 ´79

O2 leaving = 7.8 kmoles; O2 used = 14 kmolesSO2 balance = 2x + 2y = 10.2; \ x + y = 5.1 (a) O2 used = 2.5x + 2.75y = 14.0 (b) Solving (a) and (b),we get

x = 0.1 and y = 5; ratio ofx 0.1= 0.02y 5 (b) Total FeS2 reacted = 5.1 molesO2 needed by Reaction (2) = 5.1 ´5.5= 14.025 moles2 Excess O2 supplied = 21.8 – 14.025 = 7.775 moles 7.775 % excess air = ÈØ ÉÙ ´ 100 = 55.4%ÊÚ

7.20 A producer gas contains CO2: 9.2%, C2H4: 0.4%, CO : 20.9%, H2: 15.6%, CH4: 1.9% and N2:52%, when it is burnt, the products of combustion are found to contain CO2: 10.8%, CO : 0.4, O2 :9.2%, N2: 79.6%. Calculate the following:

(a) m3 of air used per m3 of producer gas (b) % excess air(c) % N2 that has come from producer gas. Basis: 100 g mole of producer gas

Component Weight, g mole ‘C’ atom

CO2 9.2 9.2C2H4 0.4 0.8CO 20.9 20.9H2 15.6 — CH4 1.9 1.9

N2 52.0 —

Total 100.0 — Reactions

C2H4 + 3O2 ® 2CO2 + 2H2O CO + 0.5O2 ® CO2H2 + 0.5O2 ® H2OCH4 + 2O2 ® CO2 + 2H2O

Carbon is the tie elementC entering = 32.8 g atoms º 11.2% exit.Oxygen needed for combustion—0.4 ´ 3 = 1.2

20.9 ´ 0.5 = 10.45 15.6 ´ 0.5 = 7.8 1.9 ´ 2 = 3.8 —

23.25\ Total moles of exit gas =32.8 = 293 g moles0.112

N2 from air: N2 in exit = 293 ´ 0.796 = 233 g molesN2 in feed = 52 g molesN2 from air = 181 g moles

O2

from air = 181 ´ 21 = 48; Total air = 229 g moles m3 of air/m3 79

feed = g mole of air/g mole of feed = 229 = 2.29100(b) O2 supplied = 48 g molesneeded = 23.25 g molesExcess O2 supplied = (48 – 23.25) = 24.75 g moles

% excess air = 24.75 100 = 106.45%´23.25 (c) Total N2 leaving = 233 g moles% NËÛ100= 22.3%2 from air = ÌÜ233ÍÝ 7.21 Formaldehyde is manufactured by the catalytic oxidation of methanol using an excess of air. Asecondary reaction also takes place:

CH3OH + 0.5O2 ® HCHO + H2O (1) HCHO + 0.5O2 ® HCOOH (2)

The product gases have the following composition. CH3OH : 8.6%; HCHO : 3.1%; HCOOH : 0.6%;

H2O : 3.7%; O2: 16%; N2: 68%. Find the following:(a) the percentage conversion of CH3OH to HCHO

(b) % methanol lost in second reaction and(c) molar ratio of feed to air and the % excess air used. Basis: 100 g moles of product gases.

[From the product gas analysis, total quantity of HCHO formed is equivalent to the sum of free HCHOavailable + HCOOH formed by the secondary reaction (2)]

HCHO formed = HCHO formed by reaction (1) + HCOOH formed from HCHO by reaction (2) = (3.1+ 0.6) = 3.7 g moles CH3OH supplied = reacted by reaction (1) + unconverted at the end = (3.7 +8.6) = 12.3 g moles(a) % conversion of CH3OH to HCHO = 3.7 ´100 = 30%12.3 (b) % lost in second reaction = 0.6 100 = 4.9%´12.3 (c) Air fed (using nitrogen balance) =68= 86 g moles; O2 = 18 g moles0.79 Moles of feed/moles of air = 12.3 = 0.14386.0 O2 needed for converting CH3OH to HCHO = 12.3= 6.152Excess O2 supplied = (18 – 6.15) = 11.85 gmoles 100 = 192.7%% excess air = 11.85 ´6.15

7.22 A rich copper ore analysis gives the following constituent percent: CuS : 10%; FeS2: 30% andinert : 60%. By crushing and floating, 2/3 inert is eliminated. The ore is roasted with carbon. Inert areunchanged. In the reduction of the Cu2O to Cu, there is 5% loss. Find the weight of copper obtainablefrom 1 tonne of ore.

2CuS + 2.5O →Cu O + 2SO

295 22 2 (1)×××64

2FeS2 + 5.5O2 ® Fe2O3 + 4SO2 (2) 2Cu O + C → 2Cu + CO(3)142 12 126 28 Basis: 1 tonne of ore = 1000 kgOre Crushing andP

1

Roaster P2 Cu, COFlotation

Inert (2/3) CuS : 100 : 16.67% FeS2 : 300 : 50.00% Inert : 200 : 33.33%

600

Inert in ore = 600 kg. Removed = 600 ´2 = 400 kg3 Inert remaining = 200 kg = 0.333 P1 \ P1 = 600 kgCuS in P1 = 600 ´ 0.166 = 100 kg; FeS2 = 600 ´ 0.5 = 300 kg(a) Cu2O got from CuS = 100 142 = 74.74 kg´190

(b) Cu2O reacted = 74.74 ´ 0.95 = 71.0 kg(c) Cu got from Cu 126 = 63 kg2O = 71 ´142

7.23 Pyrites is roasted in producing SO2. The gases leaving the roaster have a temperature of 500 °Cwith composition SO2: 9%, O2: 8.6% and N2: 82.4%. Composition of pyrites by weight is Fe : 35%,S : 40% and gangue 25%. Calculate for 100 kg of pyrites roasted.

(a) Excess air(b) Volume of air supplied(c) Volume of burner gas leaving.Basis: 100 kg of pyritesweight of FeS2 = Total weight – weight of inert = 100 – 25 = 75 kg

4FeS

+ 11O→ 2Fe O + 8SO22 23 2480 352 320 512

FeS 75 = 0.625 kmole2 = 75 kg = 120 SO2 produced = 0.625 ´8 = 1.25 kmoles4 SO2 = 1.25 kmoles º 9% of exit gas. 1.25 = 13.89 kmoles\ Total exit gas in mole = 0.09 O2 in exit gas = 13.89 ´ 0.086 = 1.19 kmoles N2 in exit gas = 13.89 ´ 0.824 = 11.44 kmolesO 11 = 1.72 kmoles2 consumed = 0.625 ´4 (a) % excess air = 1.19 ´100 = 69.2%1.72 (b) Air supplied = 1.19 1.72= 13.86 kmoles0.21 Volume of air supplied = 13.86 ´ 22.414 = 310.6 m3 at NTP. 773 = 881.5 m3(c) Volume of exit gas = 13.89 ´ 22.414 ´273

7.24 A fuel oil with the following composition C : 84%, H2: 13%, O2 : 1%, S : 1% and H2O : 1%; isburnt and the flue gas obtained gives the following analysis: CO2 : 9.9%, CO : 1.6%, H2O : 10.75%,SO2 : 0.05%, O2 : 3.7% and N2 : 74%. Calculate the % excess air used.

Basis: 100 g of oilg g atoms O2 needed ReactionC 84 7.00 7.00 C + O2 ® CO2H2 13 6.50 3.25 H2 + ½O2 ® H2OS 1 0.03 0.03 S + O2 ® SO2 10.28O2 1 0.03 –0.03H2O 1 0.06Net oxygen needed 10.25

Carbon in exit stream = (9.9 + 1.6) = 11.5%

7 = 60.87 g moles\ g moles of exit gas = 0.115 N2 in exit = (60.87 ´ 0.74) = 45.04 g moles 45.04 Air supplied = ÈØ ÉÙ = 57.02 g molesÊÚ \ O2 supplied = 11.98 g moles andexcess O2 = 11.98 – 10.25 =1.73 g molesHence % excess air = 1.73 100 = 16.9%´10.25

7.25 Pure sulphur is burnt in air. Even when 20% excess dry air is passed only 30% of the S burns toSO3 and the remaining goes to SO2. S to SO3 is the desired reaction

(a) What is the analysis of resulting gases?(b) The gases from the burner are passed through a converter where

SO2 is converted to SO3 (without addition of any more air) if the gases leaving the converter has4.3% SO2. Calculate the molar ratio of SO3to SO2 in the exit gas.

Basis: 1 g atom of S = 32 g

S+O SO S + 1.5O→SO→22 23

32 32 64 32 48 80

O2 needed = 1.5 g moleO2 supplied = 1.5 ´ 1.2 = 1.8 g molesN2 supplied = 1.8 ´ 79/21 = 6.8 g molesO2 reacted = 0.7 + (0.3 ´ 1.5) = 1.15 g molesExit gas stream:

Component SO2 SO3 O2 N2Total Weight, g mole 0.7 0.3 (1.8 – 1.15) = 0.65 6.8 8.45Mole % 8.28 3.55 7.70 80.47 100.0Let x moles of SO2 react to form SO3

SO2 + 0.5 O2 ® SO3Exit gas stream consists of the following:SO2 = 0.7 – xSO3 = 0.3 + xO2 = 0.65 – 0.5xN2 = 6.8

Total = 8.45 – 0.5x

SO0.7 xº 0.0432 = 8.45 0.5x Solving, we get x = 0.34 SOÈØ ÈØ0.64 SO = 1.78

ÊÚexitÉÙ ÊÚ

7.26 500 kg/h of pure sulphur is burnt with 20% excess air (based on S to SO2) 5% S is oxidized toSO3 and rest to SO2. Find the exit gas analysis.

Basis: 500 kg sulphur =500 = 15.625 katoms32 S+O →

SO2232 32 64

S+1.5O →

SO 23 32 48 80

1 katom of sulphur requires 1 kmole of oxygen to form SO2 Therefore oxygen supplied = 15.625 ´ 1.2= 18.75 kmolesN79 = 70.54 kmoles2 entering = 18.75 ´21

S to SO2 = 15.625 ´ 0.95 = 14.845 kmolesS to SO3 = 15.625 ´ 0.05 = 0.780 kmoleTotal O2 consumed = [14.845 + (1.5 ´ 0.78)] = 16.015 kmoles O2 remaining = (18.75 – 16.015) =2.735 kmoles

Exit gas SO2 SO3 O2 N2Total Weight, kmole 14.845 0.78 2.735 70.54 88.9 mole % 16.70 0.88 3.0879.34 100.00

7.27 The composition of the gas entering a converter is SO2 : 7.2%, O2: 13.2% and N2: 79.6% andthat of the gas leaving is SO2 : 2.8%, O2: 11.7% and N2: 85.5%. Determine the % of SO2 oxidized toSO3Basis: 100 g moles entering

79.6\ Total moles in exit = 0.855 = 93.1 g moles (making nitrogen balance) \ SO2 in exit stream = 93.1 ´ 0.028 = 2.61 g moles

SO2 converted to SO3= moles of SO2 entering – moles of SO2 in exit stream = 7.2 – 2.61 = 4.59 g moles

% SO100 = 63.75%2 oxidized = 4.59 ´7.2

7.28 Limestone containing (on dry basis) 42.5% CO2 is burnt with coal having an analysis of 81% C;4.7% H2; 1.8% N2; 4.6% O2 and 7.9% ash. The stack gas analyzes 24.4% CO2, 4.1% O2 and 71.5%N2. Compute

(a) Ratio of lime produced to coal burnt(b) Stack gas produced/tonne of lime produced.Reactions: CaCO +C +O→ CaO +2CO 32 2

100 12 32 56 88

MgCO + C + O →

MgO + 2CO 32 2 84.3 12 32 40.3 2×44

Stack gasLimestonez coal Burnery Air CaO Lime

Basis: 100 kmoles of stack gasLet z kg of coal and y kg of air enter the burner (air contains 23% O2; 77% N2 by weight %)

C + O2 ® CO2H2 + 0.5O2 ® H2OOn weight basis,N2 balance gives, 0.018z + 0.77y = 71.5 ´ 28 = 2002 O2 balance gives,

3216

(0.046z+ 0.23y) – 0.81 zzÈØ

ÉÙ = (4.1 32) = 131.2ÊÚ

Solving the above equations, we getz = 187 kgy = 2,595.6 kgCO2 balance gives: 24.4 kmoles CO2 = 1,073.6 kg 12 kg C gives 44 kg CO2

Carbon in coal = 187 ´0.81 kg C gives 44187 0.81 12 = 555.4 kg of CO2\ CO2 obtained from limestone = total CO2 – CO2 from coal = (1,073.6 – 555.4) = 518.2 kg 518.2 = 1,219.3 kg\ Limestone needed = 0.425 Lime obtained from limestone = 518.2 ´56 = 659.5 kg44 (a) Lime produced/coal burnt = 659.5= 3.5187 N2 O2 CO2

(b) Weight of stack gas 2,002 + 131.2 + 1,073.6 = 3,206.8 kg \ 659.5 kg lime produces 3,206.8 kgstack gasHence, for 1 tonne or 1000 kg lime, the stack gas produced is 4,862.5 kg

7.29 A gas containing 80% ethane and 20% oxygen is burnt with 200% excess air. 80% ethane goesto CO2; 10% to CO; and 10% remains unburnt. Calculate the stack gas composition.

Basis: 100 g moles of gaswe have,C2H6: 80 g molesO2: 20 g molesC2H6 + 3.5O2 ® 2CO2 + 3H2OC2H6 + 2.5O2 ® 2CO + 3H2OO2 needed = 80 ´ 3.5 = 280 g molesO2 available in gas = 20 g molesO2 theoretically needed = 260 g moles200% excess oxygen is supplied\ O2 supplied = (260 ´ 3) = 780 g molesTotal O2 available = 20 + 780 = 800 g molesN2 entering = (780 ´ 79/21) = 2,934 g molesCO2 formed = 80 ´ 0.8 ´ 2 = 128 g molesCO formed = 80 ´ 0.1 ´ 2 = 16 g molesO2 used = (80 ´ 0.8 ´ 3.5) + (80 ´ 0.1 ´ 2.5) = 244 g moles

O2 remaining = Oxygen available – oxygen used = (800 – 244) = 556 g moles

H2O formed = (80 ´ 0.8 ´ 3) + (80 ´ 0.1 ´ 3) = 216 g moles C2H6 remaining = 80 ´ 0.1 = 8 g moles

Composition of stack gases:

Gases g moles mole %

CO2 128 3.318CO 16 0.415O2 556 14.412N2 2,934 76.050H2O 216 5.598C2H6 8 0.207

Total 3,858 100.000

7.30 Pure CO2 may be prepared by treating limestone with sulphuric acid. The limestone used in theprocess contains CaCO3, MgCO3 and inert compounds. The acid used contains 12% H2SO4 byweight. The residue from the process had the following composition: CaSO4 : 8.56%, MgSO4 :5.23%, H2SO4 : 1.05%, Inert : 0.53%, CO2 : 0.12%, H2O : 84.51%. During the process, the masswas warmed where CO2 and H2O got removed. Calculate the following:(a) The analysis of limestone

(b) The % excess acid used(c) The weight and analysis of the material distilled from the reaction mass per 1000 kg of limestonetreated.Basis: 100 kg of residueInert : 0.53 kg, CaSO4: 8.56 kg, MgSO4: 5.23 kgCaCO3 + H2SO4 ® CaSO4 + CO2 + H2O100 98 136 44 18

MgCO3 + H2SO4 ® MgSO4 + CO2 + H2O84.3 98 120.3 44 18

8.56 kg of CaSO4 obtained from 8.56100 = 6.3 kg of CaCO3´136

5.23 kg of MgSO4 obtained from 5.23 84.3 = 3.67 kg of MgCO3´120.3(a) Limestone analysis CaCO3 MgCO3 Inert Total Weight, kg 6.3 3.67 0.53 10.50 Weight % 6034.95 5.05 100.0

98 kg of H2SO4 = 6.174 kg3 requires ÈØ(b) 6.3 kg of CaCOÉÙ ÊÚ 3.67 kg of MgCO3

requires ÉÙ ÈØ98 kg of H2SO4 = 4.27 kgÊÚ84.3 Total acid needed = 6.174 + 4.27 = 10.444 kg

Excess acid remaining = 1.05 kg% Excess acid used = ÈØ ÉÙ 100 = 10.05%ÊÚ Total acid used = (10.444 + 1.05) = 11.494 kg 11.494 (i.e.) 12% acid supplied = ÈØ ÉÙ = 95.78 kgÊÚ 44ÈØÈ Ø44 = 4.689 kg2 formed = 6.3 COÉÙÉ Ù ÊÚÊ Ú H2

O formed = 6.3 18ÈØÈ Ø ÉÙÉ Ù

18 = 1.918 kgÊÚÊ Ú84.3H2O in acid = (95.78 – 11.494) = 84.286 kgTotal water = water from acid + water formed from reaction

= (84.286 + 1.918) = 86.204 kgH2O in residue = 84.51 kgCO2 in residue = 0.12 kg\ Amount of H2O vapours = (86.204 – 84.51) = 1.694 kg

and CO2 in gas phase = (4.689 – 0.12) = 4.569 kg vapours: H2O 1.694 kg 27 % (weight %) 47.56mole % CO2 4.569 kg 73 % (weight %) 52.44 mole % Total 6.263 kg 100 100

Check Limestone + acid = (10.5 + 95.78) = 106.28 kgResidue + vapours = (100 + 6.263) = 106.263 kgFor 10.5 kg limestone, vapours formed = 6.263 kg

(c) \ 1000 kg limestone, vapours formed = 6.263 100 = 596.5 kg10.57.31 A coal containing 87% Cand 7% unoxidized hydrogen is burnt in air. If 40% excess air is used;(a) Calculate kg of air per kg of coal burnt(b) Assuming complete combustion calculate the composition of gases

by weight %

Basis: 1 kg of coal

Carbon = 0.87 kg = 0.0725 katomH2 = 0.07 kg = 0.035 kmole

32ÈØÈ Ø16= 2.88 kg [by stoichiometry]2 needed = 0.87 (a) OÉÙÉ Ù ÊÚÊ Ú O2 supplied = (2.88 ´ 1.4) = 4.032 kgAir supplied = 4.032 = 17.38 kg0.232 kg of air/kg of coal = 17.38(b) Gases leaving are: CO2, H2O, O2 and N2

CO44 = 3.19kg = 17.41 wt %2 produced = 0.87 ´12 H2O produced = 0.07 18 = 0.63kg = 3.44 wt %´2 O2 remaining = (4.032 – 2.88) = 1.152 kg = 6.29 wt % N2 leaving = (17.38 – 4.032) = 13.348 kg =72.86 wt % Total 18.32 kg 100.00 7.32 A furnace using hydrocarbon fuel oil has a dry stack gasanalysis as follows:

CO2: 10.2%, O2: 8.3%, N2: 81.5%; Find the following: (a) % excess air used(b) the composition of fuel oil in weight %(c) m3 of air supplied/kg of fuel.Basis: 100 kmoles dry stock gas

N2 in stack gas = 81.5 kmolesAir supplied =81.5 = 103.16 kmoles0.79

O2 entering = 81.5 ´21 = 21.66 kmoles79O2 leaving = 8.3 kmolesO2 consumed = (21.66 – 8.3) = 13.36 kmoles

8.3 (a) % excess air = ÈØ ÉÙ´ 100 = 62.13%ÊÚ

(b) Let the fuel be CxHyCxHy + (x + y/4)O2 ® xCO2 + y/2H2O x = 10.2 kmoles (CO2);x + y/4 = 13.36 kmoles (O2 reacted)\ y = 12.64 katomsand x = 10.2 katoms

Composition of fuel:Element katom Weight, kg Weight %

C 10.20 122.40 90.64 H 12.64 12.64 9.36 Total 135.04 100.00(c) Air supplied = 103.16 kmolesm3 of air/kg of fuel =ÈØ22.414ÉÙ = 17.12 at NTPÊÚ

7.33 The waste acid from a nitrating process contains 23% HNO3; 57% H2SO4; 20% water. Thisacid is to be concentrated to 27% HNO3, 60% H2SO4 by addition of 93% H2SO4 and 90% HNO3.Calculate the weight of acids needed to obtain 1000 kg of desired acid. Basis: 1000 kg of desiredacid

Let x kg be the weight of waste acid;y kg be the weight of H2SO4 andz kg be the weight of HNO3

Overall balance: x + y + z = 1000Balance for H2SO4 gives, 0.57x + 0.93y = 1000 ´ 0.60 = 600 Balance for HNO3 gives, 0.23x + 0.9z =1000 ´ 0.27 = 270 Solving, x = waste acid = 418 kg

y = con. H2SO4 = 390 kgz = con. HNO3 = 192 kg

Total = 1,000 kgCheck (using water balance):(0.2 ´ 418) + (0.07 ´ 390) + (0.1 ´ 192) = (1000 ´ 0.13) = 130

7.34 In order to obtain barium in a form that may be put into solution, the natural sulphate (barites)containing only pure barium sulphate and infusible matter is fused with an excess of pure anhydroussoda ash. Upon analysis of the fusion mass it is found to contain 11.3% barium sulphate, 27.7%sodium sulphate and 20.35% sodium carbonate. The remainder is barium carbonate and infusiblematter. Calculate the following:(a) % conversion of barium sulphate to the carbonate.

(b) composition of the original barites(c) % excess of the sodium carbonate used in excess of the theoretically needed amount for all thebarium sulphate. Basis: 100 kg of fusion mass.Fused

BaSO4 + IM + Na2CO3 BaCO3 + Na2SO4 + IM (IM = infusible matter)Analysis of product: 11.3% BaSO4; 27.7% Na2SO4; 20.35% Na2CO3 Therefore, (BaCO3 + IM) =100 – (11.3 + 27.7 + 20.35) = 40.65%BaSO + Na CO→ BaCO + Na SO 423 3 24233 106 197 142

Na2SO4 = 27.7 kgWeight ofBaSO4 reacted = 27.7 ´233 = 45.45 kg142 Na2CO3 reacted = 27.7 106 = 20.68 kg´142 BaCO 197 = 38.43 kg3 formed = 27.7 ´142 BaSO4 initially present = (45.45 + 11.3) = 56.75 kgNa2CO3 supplied = (20.68 + 20.35) = 41.03 kg 45.45 (a) % conversion of BaSO4

to BaCO3

= ÈØ ÉÙ´ 100 = 80%ÊÚ

(b) Composition of original baritesBaSO4 = 56.75, IM = (40.65 – 38.43) = 2.22 kgAlso, total mass of products = total mass of reactants = 100 kg Therefore, IM = 100 – [56.75 + 41.03]= 2.22 kgTherefore original barites = 56.75 + 2.22 = 58.97 kg

2.22 Hence %IM = ÈØ ÉÙ´ 100 = 3.76%ÊÚ 56.75 %BaSO4

= ÈØ ÉÙ´ 100 = 96.24%ÊÚ (c) Excess Na2CO3:Na2CO3 needed for all BaSO4 = 56.75 106 = 25.82 kg233 Excess Na2CO3 = (41.03 – 25.82) = 15.21 15.21 % excess = ÈØ ÉÙ´ 100 = 58.9%ÊÚ

7.35 A fuel oil containing 88.2% C and 11.8% H2 is burnt with 20% excess air. 95% of carbon isburnt to CO2 and the rest to CO. All the Hydrogen is converted to water. Determine the orsat analysisof the flue gas (dry flue gas).

Basis: 100 kg of fuel oilComponent Weight, kg kmole or katom C 88.2 7.35 katoms H2 11.8 5.90 kmoles Total 100.0 —C + O2 ® CO2

C + 0.5O2 ® COH2 + 0.5O2 ® H2O95% C is converted to CO2 = 7.35 ´ 0.95 = 6.9825 katoms Oxygen used for this reaction = 6.9825kmole5% C is converted to CO = (7.35 – 6.9825) = 0.3675 katoms

Oxygen used for this reaction is =0.3675= 0.18375 kmoles2 Conversion of H2 to H2O = 5.9 kmolesOxygen used for this reaction is = 5.9= 2.952 O2 needed (theoretically) = (7.35 + 2.95) = 10.3 kmoles O2 supplied = 10.3 ´ 1.2 = 12.36 kmolesO2 remaining = [12.36 – (6.9825 + 0.18375 + 2.95)] = 2.24375 kmolesN79 = 46.5 kmoles2 entering = 12.36 ´21

Component CO2 CO O2 N2Total mole 6.9825 0.3675 2.24375 46.50 56.0937mole % 12.45 0.6600 4.00000 82.89 100.0000

7.36 A furnace uses a natural gas which consists entirely of hydrocarbons. The flue gas analysis is:CO2: 9.5%, O2: 1.4%, CO : 1.9% and rest is N2. Calculate the following:

(a) the atomic ratio of hydrogen to carbon in the fuel(b) % excess air(c) the composition of the fuel gas in the form CxHyBasis: 100 mole of the flue gasLet us assume the hydrocarbon (HC) to be CxHyLet a g mole of HC get oxidized to CO2and b g mole of HC get oxidized to COThe reactions to take place are

a (CxHy) + a(x + y/4)O2 ® axCO2 + ay/2H2Ob(CxHy) + b(x/2 + y/4)O2 ® bxCO + by/2H2OThe composition in percentage of the gases are:CO2: 9.5%, O2: 1.4%, CO : 1.9%, N2: 87.2%, Total = 100.0%Carbon balance = a + b = (9.5 + 1.9) = 11.4; ax = 9.5; bx = 1.9ax 9.5= 5bx 1.9

a = 5bBy nitrogen balances oxygen supplied = 87.2 ´ 0.21 = 23.18 kmoles0.79

Oxygen left = 1.40 kmoles Oxygen reacted = 21.78 kmolesOxygen reacted = ax + aybx by ÈØ È Ø È Ø = 21.78ÊÚ Ê Ú Ê Ú 424 Oxygen reacted, 9.5 + ayby ÈØ È Ø È Ø = 21.78ÊÚ Ê Ú Ê Ú y (i.e.) ÉÙ ÈØ (a + b) = 11.33ÊÚ

\ y(11.4) = 45.32; \ y = 3.975a + b = 11.4 = 6b; \ b = 1.9; a = 9.5since ax = 9.5; x = 1

(a) Atomic ratio ofH 3.975= 3.975C1

(b) For % of excess air, let us consider the following:(c) CO + 0.5O2 ® CO2One kmole of CO requires 0.5 kmole of oxygenTherefore, 0.95 mole of O2 will be used for converting CO to CO2 Excess O2 remaining = 1.4 – 0.95= 0.45 kmoleTheoretical O2 required = 11.4 (from CO and CO2 values) + 11.33 (for H2O formation) = 22.78kmoles

0.45 \Excess air = ÈØ ÉÙ ´ 100 = 1.98%ÊÚ Hence, Hydrocarbon = C1H3.975 = CH4 (methane)

7.37 The analysis of gas entering the converter in a contact H2SO4 plant is SO2: 4%, O2: 13% and N2:83%. The gas leaving the converter contains 0.45% SO2 on SO3 free basis. Calculate the % of SO2entering the converter getting converted to SO3.

Basis: 100 g mole of gas entering the converterLet x g mole SO2 get converted to SO3SO2 : 4 SO2: 0.45% O2 : 13 Converter O2N2 : 83 N2

SO + 0.5O→ SO 22 3

xx/2 x

gases leaving SO2 ® (4 – x) (SO3 free basis) O2 ® (13 – x/2) N2 ® 83Total (100 – 1.5x)% SO2

in exit = 40.45 ÈØ ÉÙ of exit gas.1001.5x ÊÚ Solving the above equations, we get x = 3.57% SO 100= 89.25%2 converted to SO3 = 3.57 ´4

7.38 A producer gas made from coke has the composition CO : 28%, CO2: 3.5%, O2: 0.5% and N2:68%. This gas is burnt with 20% excess of the net O2 needed for combustion. If the combustion is98% complete, find the weight and volumetric composition of gases produced per 100 kg of gasburnt.

Basis: 100 kmoles of fuel gas

Component mole Molecular weight Weight, kg CO 28.0 28 28 ´ 28 = 784

CO2 3.5 44 3.5 ´ 44 = 154O2 0.5 32 0.5 ´ 32 = 16N2 68.0 28 68 ´ 28 = 1,904Total 100.0 — 2,858

we have CO + 0.5O2 ® CO2Oxygen balanceO2 needed = 28 ´ 0.5 = 14 kmolesO2 available in feed = 0.5 kmolesNet O2 needed = 13.5 kmolesO2 supplied = 13.5 ´ 1.2 = 16.2 kmolesO2 consumed = 14 ´ 0.98 = 13.7 kmolesO2 remaining = (16.2 + 0.5 – 13.7) = 3 kmoles

Carbon balanceCO2 formed = (28 ´ 0.98) = 27.44 kmoles CO2 total = (27.44 + 3.5) = 30.94 kmoles CO unreacted =(28 ´ 0.02) = 0.56 kmole Nitrogen balance

N2 from air = 16.2 ´79 = 60.9 kmoles21

N2 in exit = (68 + 60.9) = 128.9 kmolesExit gas kmole Molecular mole % = Weight, Weight % weight volume % kg

CO2 30.94 44 18.93 1,359.0 26.60CO 0.56 28 0.34 15.7 0.31O2 3.00 32 1.84 96.0 1.88N2 128.90 28 78.89 3,637.0 71.21

Total 163.40 100.00 5,107.7 100.00 100 kmoles of fuel gas = 2,858 kg\ Weight of product/100 kg of feed = 5,107.7 ´100 = 178.7 kg2,858

7.39 In the manufacture of soda ash by Le Blanc process, sodium sulphate is heated with charcoal andcalcium carbonate. The resulting black ash has the following composition: Na2CO3: 42%; other watersoluble material : 6% and insoluble 52%. The black ash is treated with water to extract the sodiumcarbonate. The solid residue left behind has the composition Na2CO3: 4%; other water solublematerial : 0.5%; insoluble : 85% and moisture : 10.5%(a) Calculate the weight of residue remaining from the treatment of

1 tonne black ash.(b) Calculate the weight of Na2CO3 extracted per tonne of black ash.Black Waterash Residue x

Feed I Unit II Unit Na2CO3: 4% Water solubles: 0.5% Insoluble: 85%

Na2CO3 Moisture: 10.5% Solution y

Basis: 1 tonne (1000 kg) of black ashLet x be the weight of residue and y be the weight of solution. Insoluble: (0.52 ´ 1000) = 0.85x \ x =612 kgNa2CO3 extracted = (0.42 ´ 1000) – (612 ´ 0.04) = 395.5 kg

7.40 A laundry can purchase soap containing 30% water at the rate of ` 7 per 100 kg f.o.b. the factory.The same manufacturer offers another soap having 5% water. If the freight rate is ` 0.6 per 100 kg ofsoap what is the maximum price that the laundry should pay the manufacturer for the soap with 5%water? (f.o.b.: freight on board) Basis: 100 kg of dry soap

Case I: (30% water) water present = 100 ´30 = 42.85 » 43 kg70 Total weight = (100 + 43) = 143 kgCost of 100 kg dry soap including freight at laundry = 143 ´

7.6100 = ` 10.88

Case II: (5% water) water present = 100 ´5 = 5.25 kg95 Total weight = 105.25 kgFreight charge alone = 105.25 ´0.6 = ` 0.63100 Cost of 100 kg dry soap = (10.88 – 0.63) = ` 10.25 = cost of 105.25 kg of wet soap\ Cost of 100 kg wet soap = 10.25 ´100 = ` 9.72105.25

7.41 Phosphorus is prepared by heating in an electric furnace a thoroughly mixed mass of calciumphosphate, sand and charcoal. It may be assumed that in a charge the following conditions exist. Theamount of silica used is 10% in excess. Charcoal is 40% in excess of that required to combine withP2O5 liberated, to form CO.(a) Calculate the weight % of feed

(b) Calculate the weight of phosphorus obtained per 100 kg of charge when the reaction I is 90%complete and reaction II is 70% complete. I

Ca (PO ) + 3SiO→ 3CaSiO + P O 342 2 3 25 310 (3 × 60) (3 ×116) 142

II 5C + P O25 → 2P + 5CO (5 12)142 (2 31) (5 28)×××

(a) Basis: 1 kmole of calcium phosphate = 310 kg Silica charged = 3 ´ 1.1 = 3.3 kmoles Carboncharged = 5 ´ 1.4 = 7.0 kmoles

Feed Weight, Mol.wt Weight, kmole or katom or At.wt kg

Ca3(PO4)2 1.0 310 310SiO2 3.3 60 198C 7.0 12 84

Total 11.3 592Composition in weight %

52.333.514.2

100.0(b) Basis: 100 kg of charge: calcium phosphate present = 52.3 kg Calcium phosphate consumed =(52.3 ´ 0.9) = 47.07 kgP142 = 21.6 kg2O5 formed = 47.07 ´310 Phosphorus produced = (21.6 ´ 0.7) ´62 = 6.6 kg142

7.42 In the Deacon process for the manufacture of chlorine, a dry mixture of HCl gas and air is passed

over a hot catalyst, which promotes oxidation of the acid. 30% excess air is used. Calculate thefollowing.

(a) The weight of air supplied per kg of acid.(b) The composition (weight %) of gases entering(c) The composition (weight %) of gases leaving, assuming that 60%

of the acid is oxidized in the process 4HCl + O → 2Cl + 2H O 22 2

(4 36.5)=146 (2 16)=32 (271)=142 (2 18)=36×× × ×

Basis: 4 kmoles of HCl º 146 kg

HClCl2 HCl

O2 Converter O2N2N2 H2O

(a) Oxygen supplied = 1 ´ 1.3 = 1.3 kmoles 100 Weight of air supplied = 1.3 ´ ÈØ ÉÙ´ 28.84 = 178.5 kgÊÚ Weight of air/weight of acid = 178.5= 1.22 kg of air146 (b) O2 entering = 1.22 ´ 0.23 = 0.281 kgN2 entering = (1.22 – 0.281) = 0.939 kgGases Weight, kg Weight %

HCl 1.000 45.0O2 0.281 12.7N2 0.939 42.3

2.220 100.0(c) HCl consumed = 0.6 kg HCl remaining = 0.4 kg 32= 0.1495 kg2 remaining = ËÛÈØOÉÙÌÜ ÊÚÍÝ Cl2 formed = 0.6 142 = 0.585 kg146

H36 = 0.148 kg2O formed = 0.6 ´146 Gases Weight, kg Weight %

HCl 0.4000 18.0 Cl2 0.5850 26.33 O2 0.1495 6.73 N2 0.939 42.27 H2O 0.1480 6.66

2.2215 100.00

7.43 In the manufacture of H2SO4 by contact process, iron pyrites are burnt in dry air. Iron is oxidizedto Fe2O3. SO2 formed is further oxidized to SO3 by passing the gases mixed with air over hot catalyst.Air supplied is 40% in excess of the sulphur actually burnt to SO3. Of the pyrites charged 15% is lostin grate and not burnt. Calculate the following:

(a) Weight of air to be used/100 kg of pyrite charged(b) In the burner 40% of S burnt is converted to SO3. Calculate the composition (by weight %) ofgases leaving I unit. (c) The catalyst converts 96% of SO2 to SO3. Calculate the weight of

SO3 formed.(d) Calculate the composition (by weight %) of gases leaving the IIunit.(e) Calculate the overall degree of completion of sulphur to SO3

FeS2 Unit I SO3,SO2 Unit II GasesAir (Burner)O2,N2 (Converter)

FeS2 Fe2O3

Basis: 100 kg of iron pyritesPyrites burnt = 85 kglost = 15 kg

(1)

2FeS + 5.5 O→ Fe O + 4SO 22 23 2

(2 120) (5.5 32)160 (4 64)×× ×

(2)

4SO + 2O→ 4SO22 3(4 64)××(2 32) (4×80)

(3)2FeS + 7.5O Fe O 4SO [Reaction (1) + Reaction (2)] 22 23 3→+ (2 120)××(7.5 32) 160 (4×80)

In Reaction (3) equal weight of oxygen is needed for FeS2.(a) Air supplied = 85 1.4= 517.4 kg; (40% excess oxygen)0.23 O2 = 119 kg; N2 = 398.4 kg(b) SO3 produced (Unit I) = (85 ´ 0.4) ´320 = 45.3 kg240

SO256 = 54.4 kg2 produced (Unit I) = (85 ´ 0.6) ´240 O2

used for forming SO3

= ÈØ32

ÉÙ = 34.0 kgÊÚ O2

used for forming SO2

= ÈØ32

ÉÙ = 37.4 kgÊÚTotal oxygen used = 71.4 kgO2 remaining (entering Unit II) = (119 – 71.4) = 47.6 kg

Gases Weight, kg Weight % SO2 54.4 9.97

SO3 45.3 8.30O2 47.6 8.72N2 398.4 73.01

Total 545.7 100.00(c) SO3

produced (II Unit) = ÈØ320 ÉÙ = 65.28 kgÊÚ256 \ Total SO3 produced = (45.3 + 65.28) = 110.58 kg 232 (d) O2

consumed = ËÛ ÌÜ = 13.05 kgÍÝ O2 remaining = (47.6 – 13.05) = 34.55 kgSO2 remaining = (54.4 ´ 0.04) = 2.176 kg

Gases Weight, kg Weight % SO2 2.176 0.400

SO3 110.580 20.264O2 34.550 6.332N2 398.400 73.004

Total 545.706 100.000(e) Sulphur in FeS2 = 100 ´64 = 53.3 kg120 Sulphur in SO32 = 44.32 kg3 = 110.58 ´80 % conversion of S to SO3 = 44.32 ´100 = 83.4%53.3

7.44 In the common process for the production of nitric acid sodium nitrate is treated with 95%H2SO4. In order that the resulting ‘niter cake’ may be fluid, it is desirable to use excess acid, so thatfinal cake contains 34% sulphuric acid. It may be assumed that the cake will contain 1.5% water andthat the reaction will go to completion. 2% of HNO3 formed will remain in the cake.

(a) Calculate the composition of niter cake by weight %, formed per 100 kg of sodium nitratecharged.(b) Calculate the weight of sulphuric acid to be used.(c) Calculate the weight of HNO3 and water vapour distilled from the niter cake. 2NaNO + H SO Na SO →+ 2HNO 324 24 3

(2 85) (98)(142) (2 63)××

Basis: 100 kg of NaNO3 charged.Na 142 = 83.5 kg2SO4 formed = 100 ´170 HNO 126 = 74.0 kg3 formed = 100 ´170 H2SO4 required = 100 98 = 57.65 kg´170

HNO3 remaining in cake = 74 ´ 0.02 = 1.48 kg(H2SO4 + H2O)% in cake = (34 + 1.5) = 35.5%Niter cake has Na2SO4, H2SO4, HNO3 and H2O \ (Na2SO4 + HNO3) % cake = 100 – 35.5 = 64.5%

(83.5 + 1.48) = 84.98 kg º 64.5% cake. \ Weight of niter cake = 84.98/0.645 = 131.75 kg (a)Composition of niter cake:

Components Weight, kg Weight %

HNO3 1.48 1.12Na2SO4 83.50 63.38H2SO4 44.80 34.00H2O 1.97 1.50

Total 131.75 100.00

(b) H2SO4 in the cake = 131.75 ´ 0.34 = 44.8 kg H2SO4 for the reaction = 57.65 kgTotal acid to be supplied = 102.45 kg

95% acid needed =102.45= 107.84 kg0.95

(c) Water in the acid = (107.84 – 102.45) = 5.39 kg Water remaining in cake = 1.97 kgWater distilled off = (5.39 – 1.97) = 3.42 kg

HNO57.65= 74.12 kg3 formed = 126 ´98 HNO3 in cake = 1.48 kgHNO3 distilled off = 72.64 kg

Composition of vapours removed:Components Weight, kg Weight %

HNO3 72.64 95.50 H2O 3.42 4.50 Total 76.06 100.00

7.45 Barium carbonate is a commercially important chemical. In its manufacture, BaS is firstprepared by heating the barites the natural sulphate with carbon. The BaS is extracted from this masswith water and the solution treated with Na2CO3 to precipitate BaCO3.

In such a process, it is found that the solution of BaS formed also contains some CaS originating fromimpurities in barites. The solution is treated with Na2CO3 and the precipitated mass of CaCO3,BaCO3 is filtered off. It is found that 16.45 kg of dry precipitate is removed from each 100 kg offiltrate collected. The analysis of the precipitate is BaCO3 90.1% and CaCO3 9.9%. The analysis offiltrate is reported to be Na2S : 6.85%, Na2CO3: 2.25% and H2O : 90.9%. The Na2CO3 for theprecipitation used contained CaCO3 as impurity.

(a) Determine the percentage excess Na2CO3 used above than that required for BaS and CaS.(b) Calculate the composition of the original solution of BaS and CaS (c) Calculate the compositionof dry soda ash used.

BaritesH2O Na2CO3 (X)

Carbon(CaS

impurity) Cake FilterFiltrate

Reactions in (X): BaS + Na2CO3 ® BaCO3 + Na2S CaS + Na2CO3 ® CaCO3 + Na2S

Molecular weights Na2CO3: 106, BaCO3: 197.4, BaS : 169.4, CaCO3: 100, CaS : 72, Na2S : 78.

Basis: 100 kg of filtrate

Precipitate 16.45 kgAmount of BaCO3 in precipitate = 16.45 ´ 0.901 = 14.82 kg Amount of CaCO3 in precipitate = (16.45– 14.82) = 1.63 kg (a) Na2CO3 required for forming 14.82 kg BaCO3

= 14.82106 = 7.96 kg´197.4 Na2S formed along with BaCO3 = 14.82 78 = 5.85 kg´197.4 \ Na2S formed along with CaCO3 = (6.85 – 5.85) = 1.00 kgCaCO 100 = 1.28 kg3 formed along with Na2S = 1 ´78 CaCO3 impurity in soda ash = (1.63 – 1.28) = 0.35 kgNa2CO3 required for 1.28 kg CaCO3 = 1.28 106 = 1.357 kg´100 Na2CO3 present in filtrate = 2.25 kg (2.25% in filtrate) Total Na2CO3 needed = (7.96 + 1.357) =9.317 kg 100 = 24.1%\ Excess Na2CO3 = 2.25 ´9.317 (b) BaS formed = 14.82 169.4 = 12.72 kg´197.4 CaS formed = 1.28 72 = 0.9216 kg´100 Components of original solutionComponents Weight, kg Weight %

BaS 12.7200 12.17CaS 0.9216 0.82Water 90.9000 87.01

Total 104.5416 100.00

Total Na2CO3 = Na2CO3 used in reaction + Na2CO3 in filtrate = 9.317 + 2.25 = 11.567 kgWe have CaCO3 as impurity in Na2CO3 = 0.35 kgComposition of dry Na2CO3 is shown as follows:

Components Weight, kg Weight % Na2CO3 11.567 97.06 CaCO3 0.350 2.94 Total 11.917 100.00

7.46 In the manufacture of straw pulp for the production of cheap straw board paper, a certain amountof lime is carried into the beater. It is proposed to neutralize this lime with acid of 67% H2SO4. In abeater containing 5000 gallons of pulp it is found that there is lime equivalent to 0.5 g of CaO perlitre.

(a) Calculate the kmole of lime in the beater.

(b) Calculate kmole and kg of H2SO4 that must be added to beater in order to provide an excess of1% above needed to neutralize the lime.

(c) Find the weight of acid.

(d) Find CaSO4 formed in kg (1 gallon = 4.4 litres).Basis: 5000 gallons = (5000 ´ 4.4) = 22,000 litresCaO + H SO

CaSO 24 →+4 2H O 56 98 136 18

(a) Lime in this solution = 22,000 ´ 0.5 = 11,000 g 11,000 g = 196.43 g moles(b) Acid needed to neutralize = 196.43 g molesAcid used is, 1% excess = (196.43 ´ 1.01) = 198.4 g moles = 198.4 ´ 98 = 19.4432 kg

(c) 67% acid needed =19.4432= 29.02 kg0.67 (d) CaSO4 formed = 11,000 136 = 26,714 g = 26.714 kg56

7.47 The available nitrogen content in a urea sample is 45%. Find the actual urea content in thesample.CO(NH2)2 = Molecular weight 60

Therefore, N28 = 0.46662 in urea is = 60 Purity of sample = 0.45 100 = 96.43%´0.4666 7.48 Carbon tetrachloride is made as follows: CS + 3Cl→ CCl + S Cl 22 422212.76 153.84

The product gases are found to contain CCl4 33.3%; S2Cl2 33.3%; CS2 1.4% and Cl2 32%. Calculatethe following:(a) the percentage of the excess reactants used.

(b) the percentage conversion.(c) kg of CCl4 produced per 100 kg Cl2 converted.Basis: 100 kmoles of product gas(a) CS2 reacted = 33.3 kmoles

CS2 remaining = 1.4 kmolesCS2 taken = 34.7 kmoles% excess reactant, CS2 = 1.4 ´100 = 4.2%33.3

CS2 is the limiting reactant and Cl2 is the excess reactant. (based on the kmole left in the product)

Cl2 required (theoretical) = 34.7 ´ 3 = 104.1 kmoles (100% conversion) Cl2 reacted = 99.9 kmolesBut Cl2 unreacted = 32.0 kmoles\ Cl2 taken = 99.9 + 32.0 =131.9 kmoles

% excess reactant = 32.0100 = 30.73%´104.1 (b) % Conversion:CS2 = 33.3 100 = 95.97%´34.7 100 = 75.74%\ Cl2 = 99.9 ´131.9 (c) Cl2 reacted = 99.9 kmoles = 7084.91 kg212.76 kg Cl2 gives 153.84 kg CCl4 7087.91 = 5122.87 kg CCl4\7084.91 kg Cl2 gives 153.84 ´212.76 CCl4 produced/100 kg Cl2 reacted = 5122.87 ´100 = 72.31 kg7084.91

7.49 Limestone is a mixture of calcium and magnesium carbonates and inert. Lime made by calciningthe limestone by heating until the CO2 is driven off. When 100 kg of limestone is calcined 44 kg ofCO2 is obtained. If the limestone contains 10% inert, calculate the following:

(a) Compute the analysis of limestone. 10 kg of above limestone is mixed with 2 kg of coke and isburnt with 100% excess air. The calcination is complete.

(b) Calculate the composition of gases leaving the kiln. Analysis of coke C : 76%, ash : 21% andmoisture : 3%

Basis: 100 kg of limestone containing 10 kg inert.(a) Let the limestone contain x kg of CaCO3 and y kg of MgCO3 \ x + y = 90 (1)

CaCO →

CaO + CO32100 56 44

MgCO →

MgO + CO3284.3 40.3 44

CO2 balance gives, 0.44x + 0.52y = 44.0 (2) Eq. (1) ´ 0.44 gives, 0.44x + 0.44y = 39.6 (3) Eq. (2) –Eq. (3) 0.08y = 4.4\ y = 55 kg MgCO3; x = 35 kg CaCO3(b)

CaCO + C + O→ CaO + 2CO 32 2

100 12 32 56 88

MgCO + C + O → MgO + 2CO 32 2 84.3 12 32 40.3 88

10 kg of limestone = 5.5 kg MgCO3; 3.5 kg CaCO3; and 1 kg inert2 kg Coke = 1.52 kg Carbon; 0.06 kg moisture.Gases leaving kiln CO2, O2, N2 and H2O

CO2 From CaCO3 = 3.588 = 3.08 kg´100 From MgCO88 = 5.74 kg3 = 5.5 ´84.3 Total CO2 = 8.82 kg = 8.82= 0.2 kmole44 O2 Carbon present = 1.52 kg = 0.1266 k atomO2 supplied (100% excess) = (0.1266 ´ 2) = 0.2532 kmoleN79 = 0.9525 kmole2 entering = 0.2532 ´21 H2O (3%) = 0.06 kg = 0.06= 0.00333 kmole18

Total CO2 in gases leaving= CO2 from limestone + CO2 from Carbon (Coke) = 0.2 kmole + 0.1266 kmole= 0.3266 kmole

Composition of gases leaving:Gas kmole mole %

CO2 0.3266 23.18O2 (Excess) 0.1266 8.99N2 0.9525 67.60H2O 0.0033 0.23

Total 1.4090 100.00

7.50 A chemical manufacturer produces ethylene oxide (EO) by burning ethylene gas with air in thepresence of catalyst. If the conditions are carefully controlled, a substantial fraction of the ethyleneconverted to ethylene oxide, some unconverted, some completely oxidized to form CO2 and H2O.Formation of CO2 is negligible. After the gases leave, they are passed through an absorber in whichthe ethylene oxide is removed. A typical orsat analysis of the gases leaving the absorber is CO2:9.6%, O2: 3%, C2H4: 6.4% and N2: 81%. Of the ethylene entering the reactor, what percent isconverted to oxide?

CO2,O2,C2

H4

ReactorN2

Exit GasesReactorAir EO, CO2,O2, C2H4,N2H2O

Reaction (1): 2C2H4 + O2 ® 2C2H4OReaction (2): C2H4 + 3O2 ® 2CO2 + 2H2OBasis: 100 g moles of exit gas6.4 g moles of C2H4 unreacted.9.6 g moles of CO2 º 4.8 mole of C2H4 converted to CO2 81 g moles of N2 º (81 ´ 21/79) = 21.53 gmoles of O2 supplied. O2 consumed = oxygen supplied – oxygen remaining

= (21.53 – 3) = 18.53 g molesO2 consumed by Reaction (2) = (4.8 ´ 3) = 14.4 g moles Therefore, O2 consumed by Reaction (1) =18.53 – 14.4 = 4.13 g moles \ C2H4 converted to C2H4O = 8.26 g molesC2H4 supplied = C2H4 remaining + C2H4 consumed by Reaction (1)

+ C2H4 consumed by Reaction (2) = 6.4 + 8.26 + 4.8 = 19.46 g molesC2H4 converted to C2H4O = 8.26100 = 42.4%´19.46

7.51 A spent dye sample obtained from a soap-making unit contains 9.6% glycerol and 10.3% salt. Itis concentrated at a rate of 5,000 kg/h in a double effect evaporator until the solution contains 80%glycerol and 6% salt. Assume that 4.5% glycerol is lost by entrainment. Find:

(a) the amount of salt crystallized out in the salt box of the evaporator and(b) the evaporation taken place in the system.Basis: One hourGlycerol = 5,000 ´9.6 = 480 kg100 Salt = 5,000 ´ 0.103 = 515 kg Water = 4,005 kgEntrained Vapour5,000 kg/hr Evaporator SolutionSaltFinal solution: Loss of glycerol = 480 ´ 0.045 = 21.6 kg Glycerol remaining = (480 – 21.6) = 458.4kg 458.4= 573 kg\ Solution leaving = 0.8

Salt in it = 573 ´ 0.06 = 34.38 kgWater in this leaving solution = (573 – 458.4 – 34.38) = 80.22 kg Salt crystallized = (515 – 34.38) =

480.62 kgVapours leaving = Water in vapour + Glycerol in vapour

= (4,005 – 80.22) + 21.6 = 3,946.38 kg Check: Vapour + solution + salt = (3,946.38 + 573 + 480.62)= 5,000 kg = Feed

7.52 Coal with 90% purity and rest ash is burnt with 25% excess air. Find the analysis of the fluegases.Basis: 100 g of coal:carbon present is 90 g = 7.5 g atoms

C + O2 ® CO2O2 needed for the above reaction is 7.5 g molesO2 supplied: 7.5 ´ 1.25 = 9.375 g moles (25% excess)

N2 from air:9.375 79= 35.27 g moles21 O2 remaining 9.375 – 7.5 = 1.875 g molesExit gases CO2 O2 N2Total g mole 7.5 1.875 35.27 44.645 mole % 16.8 4.2 79.0 100.07.53 Determine the weight of water removed while drying 1,000 kg of wet substance from 35% to5%.

Basis: 1,000 kg of wet material.Dry material is the ‘Tie element’.Moisture present is 1,000 ´ 0.35 = 350 kgDry material is 1,000 – 350 = 650 kgDry material appears as 95% in the exit.

Therefore the total weight of material leaving is650 = 684.21 kg0.95 So water removed during the drying is 1,000 – 684.21 = 315.79 kg

7.54 A mixture containing 47.5% of acetic acid is being separated by extraction in a counter currentmultistage unit. The operating temperature is 24 °C and the solvent used is iso-propyl ether. Using thesolvent in the ratio of 1.3 kg/kg of feed, the final extract composition on a solvent free basis is foundto be 82% of acid. The raffinate is found to contain 14% of acid on solvent free basis. Find thepercentage of acid unextracted?

Acetic acidSolvent Extract E Isopropyl etherExtraction unit Raffinate R

Basis: 1 kg of feed contains 0.475 kg of acid and 0.525 kg of water Solvent used = 1.3 kgLet E kg and R kg be the weight of extract and raffinate Acid balance: 0.475 = 0.82E + 0.14RWater balance: 0.525 = 0.18E + 0.86RSolving the above, we have, E = 0.493 kg and R = 0.507 kg

ÈØ0.507 = 14.94%.\ Acid unextracted: ÉÙ0.475ÊÚ

7.55 A plant makes liquid carbon-dioxide by treating Dolomite with commercial sulphuric acid. Theore analyzes CaCO3 : 68%, MgCO3: 30% and rest silica. Acid used is 94% pure. Find: (a) CO2produced

(b) acid required per tonne of the ore and(c) the composition of the solid left behind.Basis: 1 tonne of the ore

Weight of CaCO3: 680 kg, MgCO3: 300 kg and SiO2: 20 kg The reactions taking place are:Reaction (i):CaCO + H SO→ CaSO + CO + H O 324 4 2 2 100 98 136 44 18

Reaction (ii): MgCO + H SO→ MgSO + CO + H O 324 4 2 2 84.3 98 120.3 44 18

(a) CO2 produced byReaction (i): 680 44= 299.2 kg100 Reaction (ii): 300 44= 156.6 kg84.3 Total weight of CO2 produced = 299.2 + 156.6 = 455.8 kg (b) H2SO4 required duringReaction (i): 680 98= 666.4 kg100 Reaction (ii): 300 98= 348.8 kg84.3 Total weight of H2SO4 required = 1,015.2 kg 1,015.2 Commercial acid required = ÈØ ÉÙ = 1,080 kgÊÚ (c) The solid residue contains CaSO4, MgSO4 and SiO2CaSO4

formed: ÈØ136 ÉÙ = 924.8 kgÊÚ MgSO4

formed: ÈØ120.3

ÉÙ = 428.1 kgÊÚSilica remaining from the ore = 20 kgComposition of the solid left behindCaSO4: 67.36%, MgSO4: 31.18% and SiO2 : 1.46.%

7.56 A fuel gas contains 70% methane, 20% ethane and 10% oxygen. The fuel-air mixture contains

200% excess O2 before combustion. 10% of the hydrocarbon remains unburnt. Of the total carbonburnt 90% forms CO2 and the rest forms CO. Calculate the composition of the flue gas on dry and wetbasis.

Basis: 100 g moles of the fuel gasThe reactions taking place are:CH4 + 2O2 ® CO2 + 2H2OCH4 + 1.5O2 ® CO + 2H2OC2H6 + 3.5O2 ® 2CO2 + 3H2OC2H6 + 2.5O2 ® 2CO + 3H2O

Oxygen required for complete combustion = (70 ´ 2) + (20 ´ 3.5) – 10 = 200 g moles

Oxygen supplied: 200 ´ 3 = 600 mole (200% excess) Nitrogen entering from air = (600 ´ 79/21) =2,257 g moles Methane burnt: (70 ´ 0.9) = 63 mole, unburnt = 7 g moles Ethane burnt: (20 ´ 0.9) = 18mole, unburnt = 2 g moles CO2 formed: (63 ´ 0.9) + (18 ´ 0.9 ´ 2) = 89.1 g moles CO formed: (63 ´0.1) + (18 ´ 0.1 ´ 2) = 9.9 g moles O2 used: (63 ´ 0.9 ´ 2) + (63 ´ 0.1 ´ 1.5) + (18 ´ 0.9 ´ 3.5)

+ (18 ´ 0.1 ´ 2.5) = 184.1 moles H2O formed: (63 ´ 0.9 ´ 2) + (63 ´ 0.1 ´ 2) + (18 ´ 0.9 ´ 3) + (18 ´ 0.1´ 3) = 180 g molesComponent Weight, g mole gases

CH4 7.0C2H6 2.0CO2 89.1CO 9.9O2 415.9N2 2,257.0Total (dry) 2,780.9H2O 180.0

Total (wet) 2,960.9Composition %Wet basis Dry basis

0.236 0.2520.067 0.0723.010 3.2040.334 0.35614.049 14.95976.224 81.157— 100.006.080 —

100.000 —

7.57 In a catalytic incinerator a liquid having a composition of 88% carbon and 12% hydrogen isvaporized and burnt with dry air to a flue gas of the following composition on a dry basis. CO2:13.4%, O2: 3.6% and N2: 83%. Find:

(a) How many kmole of dry flue gas are produced per 100 kg of the liquid feed and(b) What was the % excess air?Basis: 100 kg of feed. C : 88 kg = 7.334 katoms

H2: 12 kg = 6 kmolesThe reactions are:C + O2 ® CO2H2 + 0.5O2 ® H2O

CO2 produced will be 7.334 kmoles and it appears as 13.4% of exit gases. 7.334 Thus the exit gas moles is ÉÙ ÈØ = 54.73 kmolesÊÚ Water present in the exit gases = 12 = 6 kmoles2 Dry flue gas leaving is = 54.73 kmolesOxygen reacted: (7.334 + 3) = 10.334 kmoles 3.6 Percentage excess air: ÈØ ÉÙ´ 100 = 34.84%ÊÚ

7.58 Aviation gasoline is isooctane C8H18. It is burned with 20% excess air and 30% of the carbonforms CO and rest goes to carbon dioxide. What is the analysis of the exit gases (on dry basis)?

Basis: 1 kmole of the isooctane. The reactions that are taking place are:

C8H18 + 8.5O2 ® 8CO + 9H2O;C8H18 + 12.5O2 ® 8CO2 + 9H2O.Oxygen required for complete combustion is 12.5 kmoles Oxygen supplied is (12.5 ´ 1.2) = 15kmoles

Nitrogen coming from air ÈØ ÉÙ 79 = 56.43 kmolesÊÚ

CO2 formed by the reaction is (1 ´ 0.7 ´ 8) = 5.6 kmoles CO formed by the reaction is (1 ´ 0.3 ´ 8) =2.4 kmoles Oxygen reacted: (1 ´ 0.7 ´ 12.5) + (1 ´ 0.3 ´ 8.5) = 11.3 kmoles Oxygen remaining is (15 –11.3) = 3.7 kmoles

Gases CO2 CO O2 N2Total mole 5.6 2.4 3.7 56.43 68.13mole % 8.22 3.52 5.43 82.83 100.0

7.59 Isothermal and isobaric absorption of SO2 is carried out in a packed tower containing Raschigrings. The gases enter the bottom of the tower with 14.8% of SO2. Water is distributed at the top ofthe column at the rate of 1,000 litres per minute. The total volume of gas handled at 1 atm and 30 °Cis 1,425 m3/h. The gases leaving the tower are found to contain 1% of SO2. Calculate the percentageof SO2 in the outlet water and express it in weight %. Basis: One hour.

Volume of the gases at standard conditions: ÈØ273 = 1283.9 m3ÉÙ ÊÚ 1283.9 Amount of gases = ÉÙ ÈØ = 57.281 kmolesÊÚ

SO2 entering absorber: (57.281 ´ 0.148) = 8.4776 kmoles Inert gases are the tie elementInert gases: (57.281 – 8.4776) = 48.8034 kmoles

If 99% is equal to 48.8034 kmoles then 1% isÉÙ 48.8034 ÈØ = 0.493 kmoleÊÚ SO2 absorbed in the tower: 8.4776 – 0.493 = 7.9846 kmoles = 511 kgAmount of water flowing = 1,000 ´ 60 = 60,000 kg/hWeight of the total liquid leaving the tower is 60,511 kgWeight % of SO2

= ÈØ ÉÙ 100 = 0.84%.ÊÚ

7.60 A pure gaseous hydrocarbon is burnt with excess air. The Orsat analysis of the flue gas: CO2:10.2%, CO : 1%, O2: 8.4% and rest nitrogen. What is the atomic ratio of H to C in the fuel? Find the% excess oxygen supplied.

Basis: 100 kmoles of the flue gases.N2 in flue gas = 100 – (10.2 + 1 + 8.4) = 80.4 kmoles Let the hydrocarbon be CxHy.Let a kmole of it get oxidized to CO2 and b kmole of it get oxidized to CO.The reactions are:CxHy + (x + y/4)O2 ® xCO2 + y/2H2O;CxHy + (x/2 + y/4)O2 ® xCO + y/2H2O.We also know that,2C + O2 ® 2COi.e. 2 kmoles of CO requires 1 kmole of O2.\ 1 kmole of CO requires 0.5 kmole of O2 for conversion to CO2

ÈØ 21 = 21.37 kmoles\ O2 supplied = ÉÙ79ÊÚO2 reacted = (21.37 – 8.4) = 12.97 kmoles for forming CO and CO2 For converting CO to CO2additional O2 required is 0.5 kmoles \ O2 excess = 8.4 – 0.5 kmoles

= 7.9 kmolesi.e. O2 required = 21.37 – 7.9= 13.47 kmoles

7.9 t100 58.65% excess air = 13.47

Carbon balance = a + b = 11.2CO2 formed = ax = 10.2; CO formed: bx = 1ax/bx = 10.2

Oxygen consumed; ax + aybx by ÈØ È Ø È Ø = 12.97ÉÙ É Ù É Ù 42 4

Solving, we get x = 1 and y = 0.8Hence, y is approximated to a full number “1”Molar ratio, H/C = y/x = 1;The hydrocarbon will be C2H2, acetylene.

7.61 A feed of 100 kmoles/h containing 40 mole % A is to be distilled to yield a product containing95 mole % A and a residue containing 90 mole % B. Estimate the flow rate of distillate and residue.

Making a total material balance,F = D + W (1) Making a component balance,F.Xf = D.XD + W.Xw (2) Substituting,F = 100 kmoles/h, Xf = 0.4, XD = 0.95 and Xw = 0.1

100 = D + W (3) 1000(0.4) = D(0.95) + W(0.1) (4) Solving (3) and (4), we getD = 353 kmoles/hW = 647 kmoles/h

7.62 One hundred kilograms of liquid mixture containing 30% A and 70% B is extracted with asolvent mixture containing C and D. After thorough mixing and allowing the system to reachequilibrium, two separate layers are observed. The composition of both the layers have beenanalyzed and given below:

Layer ComponentA B C D Top 10 05 60 25Bottom 20 60 05 15

Estimate (i) the weight of each layer, (ii) weight of solvent, and (iii) composition of C and D insolvent.Basis: 100 kg of feed

Solvent

Feed Extractorx

y Let the weight of top layer be x, and the weight of bottom layer be y from the extractor.

Let the weight of solvent used be sTotal material balance gives100 + s = x + y(or)s = [x + y] – 100Making component balance for A, we get100(0.3) = x(0.1) + y(0.2)30 = 0.1x + 0.2ySimilarly, making component balance for B, we get100(0.7) = x(0.05) + y(0.6)70 = (0.05)x + 0.6ySolving, we getx = 80 kg, andy = 110 kgSolvent used iss = [x + y] – 100

= 110 + 80 – 100 = 90 kgMaking a component balance for C and DWeight of C = 80(0.6) + 110(0.05) = 48 + 5.5 = 53.5 kg Weight of D = 80(0.25) + 110(0.15) = 20 +16.5 = 36.5 kg Since, the total weight of solvent is 90 kg

xC =

53.5= 0.594590 xD = 36.5= 0.405590

7.63 One hundred kg of a mixture containing 80% alcohol and 20% water is mixed with anothermixture containing 40% alcohol and 60% water. If it is desired to produce a mixture containing 50%alcohol and 50% water, estimate the quantity of 40% alcohol and 60% water mixture needed.

Basis: 100 kg of feed mixtureLet the weight of 40% alcohol + 60% water mixture to be mixed be x kg Making a material balancefor alcohol:100(0.8) + x(0.4) = (100 + x)(0.5)i.e. 80 + 0.4x = 50 + 0.5xSolving, x = 300 kgCheck: by making water balance100(0.2) + x(0.6) = (100 + x)(0.5)i.e. 0.1x = 30Solving, x = 300 kg

7.64 A gas mixture has CO : 10%, CO2: 50%, O2: 5% and rest nitrogen by volume. It is desired tohave the nitrogen composition as 40% in the final mixture by mixing it with fresh air. Estimate thegas/air ratio to be maintained to achieve the composition as 40% in the final air. Also, estimate thecomposition of leaving air.

Basis: 1 kmole of incoming gas mixtureSince the composition of gas is given in volume %, it is to be taken as mole %.

Air is assumed to contain 79% N2 and 21% O2 by mole %. Let 1 kmole of gas be mixed with y kmoleof air to give final product containing 40 mole % N2 in (1 + y) kmole of leaving air Making a materialbalance for nitrogen(1) (0.35) + y(0.79) = (1 + y) (0.4)0.05 = 0.39ySolving, y = 0.128 kmoleGas to air ratio is 1/0.128 = 7.813

y kmole O2 : 21% N2 : 79%

1 kmoleCO = 10% (1 + y) kmoleCO2= 50% MixerN2 = 40%O2 = 5%N2 = 35%

Leaving stream:

Component Weight, kmoles mole % CO 0.1 8.86CO2 0.5 44.32O2 0.05 + 0.128 × 0.21 = 0.07688 6.82

N2 0.35 + 0.128 × 0.79 = 0.45112 40.00Total 1.128 100.00Average molecular weight:(0.0886) × (28) + (0.4432) × (40) + (0.0682) × (32) + (0.4) × (28) = 33.5912

7.65 A gas containing 5% SO2, 10% O2, and rest 85% N2 enters a catalytic chamber where theleaving gas contains only 0.5% SO2. Estimate the fractional conversion of SO2 to SO3 and also thecomposition of gases leaving catalytic chamber.

Basis: 100 kmoles of gas entering the chamberThe reaction isSO2 + ½ O2 ® SO3Let x be the kmoles of SO2 reactedSo, unreacted SO2 = 5 – xThen, SO3 formed is x and the corresponding O2 reacted is x/2 moles (by stoichiometry)

Balance O2

= 10ÌÜxËÛ ÍÝ2 Total amount of gas leaving = 85 + 10ÌÜxËÛ + x + [5 – x]ÍÝ2 x= 100 – 2 Mole fraction of SO2 in the leaving stream = 0.005 i.e. 5 -x= 0.005 100 -

x2Solving, we getx = 4.511

Fractional conversion of SO2 to SO3 = 4.511 × 100 = 90.22%ʈÁ˜Ë¯ Gas analysis:

Component Weight, kmoles mole % SO2 5 – 4.511 = 0.489 0.500SO3 4.511 4.615O2 10 – 4.511/2 = 7.744 7.923N2 85.000 86.962Total 97.744 100.000

EXERCISES

7.1 The waste acid from a nitrating process contains 25% HNO3, 50% H2SO4 and 25% water. This

acid is to be concentrated to 30% HNO3, 60% H2SO4 by addition of 95.3% H2SO4 and 90% HNO3.Calculate the weight of acids needed to obtain 10,000 kg of desired acid.

7.2 The gas obtained from a furnace fired with a hydrocarbon fuel oil analyses

CO2: 10.2%, O2: 7.9%, N2: 81.9%.Calculate: (a) percentage excess air, (b) C:H ratio in the fuel and (c) kg of air supplied per kg of fuelburnt.

7.3 The flue gas from an industrial furnace has the following composition by volume. CO2: 11.73%,CO: 0.2%, H2: 0.09% ,O2: 6.81% N2: 81.17%. Calculate the percentage excess air used, if the loss ofcarbon in the clinker and the ash is 1% of the fuel used. The fuel gas has the following composition byweight:C: 74%, H2: 5%, O2: 5%, N2: 1%, H2O: 9% S: 1% and ash: 5%.

7.4 A fuel gas contains CO2: 2%, CO: 34%, H2: 41%, O2: 1%, C2H4: 7%, CH4: 11% and rest N2. Itis burnt with 25% excess air. Assuming complete combustion, estimate the composition of leavinggases.

7.5 Limestone is burnt with coke having 85% carbon, producing a gas of 28% CO2, 5% O2, and restN2. Calculate the amount of lime produced per 100 kg of coke burnt and the amount of excess air.

7.6 Pure S is burnt in a furnace with 65% excess air. During combustion 90% of S is burnt to SO2 andrest to SO3. Estimate the composition of gases leaving.

7.7 In the manufacture of nitric acid, ammonia is reacted with air at 650 °C and 7 bar. Thecomposition of the mixed vapour is nitrogen: 70%, oxygen: 18.8%, ammonia: 10% and rest water.Find the average molecular weight, composition of leaving gases in weight % and the density ofgases.

7.8 Thirty kilograms of coal analyzing 80% carbon and 20% hydrogen are burnt with 600 kg of airyielding a gas having an orsat analysis in which the ratio of CO2 to CO is 3 : 2. What is thepercentage of excess air?

7.9 Pure sulphur is burnt in a burner at a rate of 1,000 kg/h. Fresh air is supplied at 30 °C and 755mm Hg. Gases from the burner contain 16.5% SO2 and 3% O2 and rest nitrogen on SO3 free basis.Gases leave the burner at 800 °C and 760 mm Hg pressure. Calculate

(a) fraction of sulphur burnt to SO3(b) % excess air over the amount required to oxidize sulphur to SO2. 7.10 Butane is burnt with 80%of the theoretical air. Calculate the analysis of the gases leaving assuming that all H2 present isconverted to water.

7.11 Determine the combustion gas analysis when a medium fuel oil with 84.9% carbon, 11.4%hydrogen, 3.2% sulphur, 0.4% oxygen and 0.1% ash by weight is burnt with 25% excess air. Assumecomplete combustion.

7.12 A synthetic fuel oil is known to contain only H and C, gives on combustion an Orsat analysis ofCO2: 2%, O2: 2.8% and N2: 80.6%. Calculate the C:H ratio in the fuel.

7.13 A low grade pyrites containing 32% sulphur is mixed with 10 kg of pure sulphur per 100 kg ofpyrites so that the mixture will burn readily forming a burner gas that analyzes 13.4% SO2, 2.7% O2and 83.9% N2. No sulphur is left in the cinder. Calculate the % of sulphur fired that burnt to SO3.

7.14 A mixture containing 20 mole % butane, 35 mole % pentane and rest hexane, is to be separatedby fractional distillation into a distillate containing 95 mole % butane, 4 mole % pentane and resthexane and a bottom product. The distillate is expected to contain 90% of the butane in the feed.Calculate the composition of the bottom product.

7.15 Gypsum (plaster of Paris: CaSO4 × 2H2O) is produced by the reaction of calcium carbonate andsulphuric acid. A certain limestone analyzes: CaCO3: 96.89%, MgCO3: 1.41% and inerts : 1.70%.For 5 metric tonne of limestone reacted completely, determine:(a) kg of anhydrous gypsum (CaSO4) produced.(b) kg of sulphuric acid solution (98 weight %) required. (c) kg of carbon dioxide produced.

7.16 The synthesis of ammonia proceeds according to the following reaction N2 + 3H2 ® 2NH3In a given plant, 4,202 kg of nitrogen and 1,046 kg of hydrogen are fed to the synthesis reactor perhour. Production of pure ammonia

from this reactor is 3,060 kg per hour.(a) What is the limiting reactant?(b) What is the percent excess reactant?(c) What is the percent conversion obtained (based on the limiting

reactant)?

7.17 A triple effect evaporator is designed to reduce water from an incoming brine stream from 25weight % to 3 weight %. If the evaporator unit is to produce 14,670 kg/h of NaCl (along with 3weight % H2O), determine:

(a) the feed rate of brine in lb/h.(b) the water removed from the brine in each evaporator.

7.18 A natural gas analyzes CH4: 80.0% and N2: 20.0%. It is burnt under a boiler and most of theCO2 is scrubbed out of the flue gas for the production of dry ice. The exit gas from the scrubberanalyzes CO2: 1.2%, O2: 4.9% and N2 : 93.9%.

Calculate:(a) percentage of the CO2 absorbed.(b) percent excess air used.

7.19 A synthetic gas generated from coal has the following composition: CO2: 7.2%, CO : 24.3%,H2: 14.1%, CH4: 3.5% and N2 : 50.9%. (a) Calculate the cubic metre of air necessary for completecombustion per cubic metre of synthetic gas at the same conditions.

(b) If 38% excess air were used for combustion, what volume of flue gas at 400 °C and 738 mm Hgwould be produced per cubic foot of synthetic gas at standard conditions?

(c) Calculate the flue gas analysis for (a) and (b).7.20 The gas obtained by burning pure carbon in excess oxygen analyzes 75% CO2, 14% CO, and restO2 in mole %.(a) What is the percentage of excess oxygen used?(b) What is the yield of CO2 in kg per kg of carbon burnt?

7.21 A producer gas has the following composition by volume: CO: 23%, CO2: 4.3%, oxygen: 2.7%,and nitrogen: 70% (a) Calculate the average molecular weight.

(b) Calculate the gas (in m3) at 30 °C and 760 mm Hg pressure formed per kg of carbon burnt.

(c) Calculate the volume of air at 30 °C and 760 mm Hg pressure per 100 m3 of the gas at the sameconditions if the total oxygen present be 20% in excess of that theoretically required, and

(d) Calculate the composition of gases leaving for part (c) assuming complete combustion.

7.22 A sample of coke contains 80% C, 5.8% hydrogen, 8% oxygen, and 1.4% nitrogen. Rest is ash.It is gasified and the gas produced has 5% CO2, 32% CO, 12% H2, and 51% N2.(a) Estimate the volume at 25 °C and 750 mm Hg of the gas formed

per 100 kg of coke gasified, and(b) Calculate the volume of air used per unit volume of gas produced,both measured under same conditions.7.23 A coal containing 87.5% total carbon and 7% unoxidised hydrogen is burnt in air(a) If 40% excess air is used than that of theoretically needed,calculate the kg of air used per kg of coal burned. (b) Calculate the composition by weight of gasesleaving the furnaceassuming complete combustion.7.24 In a lime manufacturing process, pure limestone is burnt with coke having 87% carbon,producing a gas of 27% CO2, 3% O2, and rest N2. Calculate (a) the amount of lime produced to cokeburnt.(b) the percentage of excess air, and(c) the amount of stack gas obtained per tonne of lime produced. 7.25 A liquid hydrocarbon feed is

passed into a flash vaporizer where it is heated and separated into vapour and liquid streams. Theanalysis of various streams in weight % is as follows:

Stream component Feed Vapour Liquid C4H10 20 71.2 8.6 C5H12 30 23.8 — C6H14 50 4.8 —

Estimate the flow rates of liquid and vapour stream for a feed rate of 100 kg/h. Also, evaluate thecomposition of other two components in liquid phase.7.26 The petrol used for petrol engine contains 84% carbon and 16%

hydrogen. The air supplied is 80% of that required theoretically for complete combustion. Assumingthat all the hydrogen is burnt and that carbon is partly burnt to CO and to CO2 without any free carbonremaining, find the volumetric analysis of the dry exhaust gas.

7.27 Producer gases are produced by burning coke with a restricted supply of air so that more CO isproduced than CO2. The producer is producing gas having CO:CO2 mole ratio as 5:1 from a cokecontaining 80% carbon and 20% ash. The solid residue after combustion carries with it 2% unburntcarbon. Calculate: (a) moles of gas produced per 100 kg of coke burnt,(b) moles of air supplied per 100 kg of coke burnt, and (c) percentage of carbon lost in the ash

7.28 A furnace uses coke containing 80% carbon and 0.5% hydrogen and the rest ash. The furnaceoperates with 50% excess air. The ash contains 2% unburnt carbon. Of the carbon burnt 5% goes toform CO. Calculate,(a) the composition of the flue gas,

(b) the ash produced, and(c) the carbon lost per 100 kg of coke burnt.

7.29 A petroleum refinery burns a gaseous mixture containing C5H12:7%C4H10 : 10%C3H8 : 16%C2H6:9%CH4 : 55%N2:3%

at rate of 200 m3/h measured at 4.5 bars and 30 °C. Air flow rate is so adjusted that 15% excess air isused and under these conditions the ratio of moles of CO2: moles of CO in the flue gas is 20:1.Calculate

(a) m3/h of air being introduced at 1 atm and 30 °C, and (b) the composition of the flue gas on drybasis.

7.30 The off gas from a phosphate reduction furnace analyses P4:8%CO : 89%N2:3%

and is burnt with air under the conditions such that phosphorus is selectively oxidized. From the fluegas analysis, the oxides of phosphorus precipitate on cooling and are separated from the remaininggas. Analysis of the latter shows that:CO2 : 0.9%

CO : 22.5%N2 : 68%O2 : 8.6%

It may be assumed that oxidation of phosphorus is complete and phosphorus exists in the flue gaspartly as P4O6 and partly as P4O10. Calculate what % of CO entering the burner is oxidized to CO2,and what % of P4 is oxidized to P4O10?

7.31 Determine the flue gas analysis, air–fuel ratio by weight, and the volume of the combustionproducts at 250 °C, when the coal refuse of the following composition burns with 50% excess air:

Proximate analysis MoistureAshVolatile matterFixed carbon

Ultimate analysis CarbonHydrogenNitrogenSulphurBalance is oxygen

Air dried % 82028.543.5

Air dried % 81.04.61.80.6

If the rate of burning of coal is 3 tonnes/h, what is the capacity of the air blower used? Assumecomplete combustion.

7.32 Determine the flue gas analysis and the air–fuel ratio by weight when a medium viscosity offuel–oil with 84.9% C, 11.4% H2, 3.2% S, 0.4% O2 and 0.1% ash is burnt with 20% excess air.Assume complete combustion.

7.33 A furnace burns producer gas with 10% excess air at a rate of 7200 Nm3/h and discharges flue

gases at 400 °C and 760 mm Hg. Calculate the flue gas analysis, air requirement, and the volume offlue gases per hour. The gas is supplied from the gas holder and its orsat analysis is as follows:CO2:4%CO : 29%N2 : 52.4%H2 : 12%CH4 : 2.6%

Normal temperature = 30 °C. Assume complete combustion. 7.34 The following is the ultimateanalysis of a sample of petrol by weight:C 85%H2 15%

Calculate the ratio of air to petrol consumption by weight if the volumetric analysis of the dry exhaustgas is:

Composition Volume %CO2 11.5O2 0.9CO 1.2N2 86Also find the % excess air.

Recycle and Bypass 8

8.1 RECYCLE

In industries, sometimes a part of the main product stream or the intermediate product streamcomprising both reactants and products or the intermediate product is sent back along with feed to thesystem or somewhere in the middle of the system. Such a stream is called Recycle stream. This isdone to improve the conversion whenever the conversion is low and to have energy economy inoperations. This also improves the performance of an equipment as in the case of absorption ofsulphur trioxide using sulphuric acid rather than water, as the solubility is low in pure water.

8.2 BYPASS

Bypassing of a fluid stream is dividing it into two streams, and is often used in industries to have acloser control in operation. This is done if there is a sudden change in the property of a fluid streamlike excessive heating (or cooling) as it passes through a preheater (cooler) before entering anotherunit. In such cases this conditioned stream is mixed with a portion stream at its original condition andthen used in the process. This is called bypassing operation.

8.3 PURGE

One of the major problems encountered during recycling is the gradual increase in the concentrationof inert or impurities in the system. A stage may reach when the concentration of these componentsmay cross permissible levels. By bleeding off a fraction of the recycle stream, this problem can beovercome. This operation is known as purging. This is quite common in the synthesis of ammonia andelectrolytic refining of copper. The above (8.1, 8.2, and 8.3) definitions have been shown in Figure8.1.

180

BypassFresh

feed Mixing Process FeedunitGross productSepaNet productratorRecycle Purge

Figure 8.1 A scheme indicating recycle, bypass and purge. WORKED EXAMPLES

8.1 A distillation column separates 10,000 kg/h of a 50% benzene and 50% toluene. The productrecovered from the top contains 95% benzene while the bottom product contains 96% toluene. Thestream entering the condenser from the top of the column is 8,000 kg/h. A portion of the product isreturned to the column as reflux and the remaining is withdrawn as top product. Find the ratio of theamount refluxed to the product taken out.

8,000 kg/h V CondenserDistillation

RD Benzene 95% B 10,000 kg/h ‘F’column50% B, 50% TW 96% TolueneFigure 8.2 Overall balance

F = D + Wor, 10,000 = D + WBenzene balance gives,

5000 = 0.95D + 0.04W Solving, D = 5,050 kg/h W = 4,950 kg/h Balance around condenser gives,

V = D + Ror, 8,000 = 5,050 + R\ R = 2,950 kg/h

Reflux ratio =Refluxed quantityR = 0.584Actual product D 8.2 What is the flow rate in recycle stream in Figure 8.3 shown below?Water W300°FF Feed 10,000 kg/hEvaporator 20% KNO350% KNO3 0.6 kg KNO3/kg water (i.e. 0.6/1.6 = 0.375 KNO3 solution) MRCrystallizer100°FCCrystal with 4% H2O

Figure 8.3Basis: One hour. KNO3 entering = 2,000 kg/h 2000 \Crystal leaving crystallizer,C= ¦µ §¶ = 2,080 kg/h¨· Overall balance F = C + Wsince F = 10,000 kg/h

C = 2,080 kg/h\ W = 7,920 kg/hCrystallizer balance gives, M = C + R = (2,080 + R)

KNO3 Balance gives 0.5M = 0.96C + 0.375RThus R = 7,680 kg/h

8.3 Metallic silver may be obtained from sulphide ores by roasting to sulphates and leaching withwater and subsequently precipitating silver with copper. In the Figure 8.4 shown below, the materialleaving the second separator was found to contain 90% silver and 10% copper. What percentageexcess copper was used? If the reaction goes to 75% completion based on the limiting agent Ag2SO4,what is the recycle stream in kg/tonne of product?

Cu CuSO4FReactorS1

S

2

90% Ag; 10% Cu

Ag2SO4Recycle R Ag2SO4Figure 8.4 Ag SO 24 + Cu → 2Ag + CuSO4 312 63.5 (2× 107.9) 159.5

Basis: 1 tonne of product = 1000 kg product(i.e.) 900 kg silver; 100 kg of copperAg 900 t312= 1300 kg2SO4 needed = t2107.9 CuSO4 formed = 1,300 t159.5= 665 kg312 Cu needed = 1, 300 t63.5= 265 kg (for forming CuSO4)312 Total copper supplied = 265 + 100 = 365 kg% Excess copper used = 100 100 = 37.7%´265 We know the reaction is 75% complete. 25% of the limiting reactant (Ag2SO4) was unconvertedwhich goes in the recycle steam.

Ag2SO4 balance: F = R + 1,300 or, F = 1,733.3 kg0.25F = Ror, 0.25 (R + 1,300) = R \ R = 433 kg

8.4 In the diagram shown in Figure 8.5, what fraction of dry air leaving is recycled?x g dry air AH1= 0.0152 g B of wv/g0.099 g H2O/gdry solidCD H2= 0.0525 g 52.5 g dry air (e)of wv/g DA Drier 1.562 g H2O/g dry solid

Figure 8.5

Basis: 1 g of dry solid. Let x g of dry air be recycled. Water removed by drying = (1.562 – 0.099) =1.463 g Water removed/g of dry air = 0.0525 – 0.0152 = 0.0373 g

Dry air needed =1.463 = 39.22 g\0.0373 Dry air passing through drier (e) = 52.5 g (between B and C) \ Dry air recycled (x) = 52.5 – 39.22 =13.28 g x 13.28= 0.253\ Fraction recycled = e 52.5

8.5 What is recycle, feed and waste for the system shown in Figure 8.6? Basis: 100 units of feed.Material balance for A at = 20 + x = (100 + x)0.4Solving, x = 33.3 units.

Aw waste40% A20% A; Feed Product P 80% BA 5% and B 95% Recycle A onlyxFigure 8.6 Recycle 33.3= 0.333Feed 100

Material balance for A at = (100 + 33.3)0.4 = Aw + 33.3 + 0.05P Overall balance: 100 = Aw + PSolving: Aw= 15.81 units Product, P = 84.19 units

8.6 Methanol is produced by the reaction of CO with H2 according to the equation CO + 2H2 ®CH3OH. Only 15% of the CO entering the reactor is converted to methanol. The methanol product iscondensed and separated from the unreacted gases, which are recycled. The feed to the reactorcontains 2 kmoles of H2 for every kmoles of CO. The fresh feed enters at 35 °C and 300 atm. Toproduce 6,600 kg/h of methanol calculate:

(a) Volume of fresh feed gas, and(b) The recycle ratio.35 °C 300 atmReactor CH3OH CO, 2H2 6,600 kg/hUnreacted CO, H2Figure 8.7 Basis: One hour of operationMethanol produced =6,600= 206.25 kmoles32 CO + 2H2 ® CH3OH15% conversion of CO entering gives 206.25 kmoles of methanol

206.25= 1,375 kmoles\ CO entering the reactor = 0.15and H2 entering the reactor = 1,375 ´ 2 = 2,750 kmoles Total amount of CO and H2 = 4,125 kmoles

CO unconverted (goes into recycle)= CO entering – CO converted = (1,375 – 206.25) = 1,168.75 kmoles

H2 unconverted (goes into recycle) = (2,750 ´ 0.85) = 2,337.50 kmoles \ Total moles of feedunconverted = 3,506.25 kmolesFresh CO needed = 206.25 kmolesFresh H2 needed = 412.50 kmolesTotal moles of fresh feed = 618.75 kmoles\(a) Volume of feed gas = 618.75 ´22.414 ´

308 ¦µ¦ µ1273§¶§ ¶300¨·¨ ·= 52.16 m3

(b) Amount of gas leaving reactor = (3,506.25 + 206.25) = 3,712.50 kmoles

Amount of gas recycled = 3,506.25 kmoles 3,506.25 \Recycle ratio = ¦µ §¶ = 0.944 (mole ratio)¨·

By weight 1,168.75 kmoles of CO = 32,725 kg;2,337.5 kmoles of H2 = 4,675 kgRecycle stream of CO and H2 = 32,725 + 4675 = 37,400 kg Products leaving reactor = (37,400 +6,600) = 44,000 kg

37,400 \Recycle ratio = ¦µ §¶ = 0.85 (weight ratio)¨·

8.7 Limestone containing 95% of CaCO3 and 5% SiO2 is being calcined. Heat for the reaction issupplied from a furnace burning coke. The hot flue gases analyze 5% CO2. The kiln gas contains8.65% CO2. In order to conserve some of the sensible heat a portion of the kiln gas is continuouslyrecycled and mixed with fresh hot flue gas. After mixing the gas entering the kiln analyzes 7% CO2

(a) Find the kg of CaO produced/kg of coke burnt(b) Find the recycle ratio, (i.e.) moles of gas recycled per mole of gas leaving the kiln.LimestoneK KilnPX RAirF BurnerCoke

Figure 8.8

Basis: 12 kg of cokerepresent F kmole of flue gas (5% CO2) K, kmole of kiln gas (8.65% CO2)R, kmole of gas recycled (8.65% CO2) P, kmole of product gas (8.65% CO2) X, kmole of gas enteringthe kiln (7% CO2)

XR F

Figure 8.9

X = F + R (overall balance) (a) 0.07X = 0.05F + 0.0865 R (CO2 balance) (b)

CaCO → CaO + CO 32100 56 44

Assuming complete combustion0.05F = 1 kmole CO2\ F = 20 kmoles(Q 1 kmole of CO2 comes from 12 kg of Carbon)C + O2Æ CO2Solving (a) and (b), we have R = 24.2 kmoles; X = 44.2 kmoles Let m kmole of CO2 be added in kilnfrom the calcinations of limestone \ K = 44.2 + mMaking CO2 balance, (0.07 X) + m = K¥ 0.0865or, 44.2 ¥ 0.07 + m = (44.2 + m)(0.0865)\ m = 0.8 kmole\ K = 44.2 + 0.8 = 45.0 kmolesCaO formed = (0.8 ¥ 56) = 44.8 kg

(a) CaO formed/kg of coke burnt =44.8= 3.7312 (b) recycle ratio (mole) = R =24.2= 0.538K 45

8.8 Sea water is desalinated by reverse osmosis using the scheme shown in Figure 8.10D stream has500 ppm salt = 0.05%Find(a) rate of B(b) rate of D(c) recycle R

“R”, Recycle

1,000 kg/h A “B”, Waste 5.25% salt Reverse osmosis cell3.1 salt % 4%sea water

Desalinated water “D”, 0.05% salt

Figure 8.10

Basis: One hourOverall balance, 1,000 = B + DSalt balance = (1,000 ¥ 3.1) = 5.25B + 0.050 Solving, B = 586.5 kg; D = 413.5 kg

R = 5.25%Overall balance (At entry to cell), 1,000 + R = ASalt balance, 1,000 ´ 0.031 + R ´ 0.0525 = A ´ 0.04Solving, A = 1,720 kg and R = 720 kg

8.9 In the feed preparation section of an ammonia plant, hydrogen is produced by a combination ofsteam-reforming/partial oxidation process. Enough air is used in partial oxidation to give a 3:1 H2-N2molar ratio in the feed to the ammonia unit. The H2-N2 mixture is heated to reaction temperature andfed to a fixed bed reactor where 20% conversion of reactants to NH3 is obtained per pass. Theproducts from the reactor are cooled and NH3 is removed by condensation. The unreacted H2-N2mixture is recycled and mixed with fresh feed. On the basis of 100 kmoles per hour of fresh feeddetermine the NH3 produced and recycle rate.

Freshfeed Reactor NH3

Figure 8.11 Basis: 100 kmoles of feed

Given that H2 : N2 = 3 : 1hydrogen: 75 kmoles and nitrogen: 25 kmoles N2 + 3H2 ® 2NH3Overall balance gives:

Feed NH3

NH3 formed = 50 kmolesOne mole of N2 gives 2 moles of NH3Reactor balance gives:

3x, H2 Reactor NH3, N2and H2x, N2

Let 3 x kmole of H2 and x kmole of N2 enter the reactor Since 20% conversion takes place,NH3 formed = x ´ 0.2 ´ 2 = 0.4x = 50 kmoles \ x = 125 kmoles of N2; and 3x = 375 kmoles of H2Unreacted N2 = 100 kmoles, and H2 = 300 kmoles Ans.: NH3 produced = 50 kmolesRecycle = 400 kmoles

8.10 Find S, A, R and B from Figure 8.12 shown below, if 1 kg of grease/ 100 m2 area is present andthe degreased surface per day is 105 m2.

Greased metal Solvent S0% greaseGCleanedmetalB15% Grease solvent Separator ADegreasing

40% grease 1% grease Rwith

Solvent Recyclesolvent

Figure 8.12Grease removed/day = 1,00,000/100 = 1,000 kg (i) Overall balance gives:G 100%40% Grease S Process A100% 60% Solvent S + G = A (overall) (1) G = 0.4A = 1000 (grease balance) (2)Solving Eqs. (1) and (2) A = 2,500 kg; S = 1,500 kg; G = 1,000 kg.DegreasingB (15%) S (100%free fromR-1% (recycle)grease)(ii) B = S + G + R (3)

B = 1,500 + 1,000 + R (overall) (4) 0.15B = 0 + 1,000 + 0.01R (grease balance) (5) Solving, (4) and(5), B = 6,964.29 kg; R = 4,464.29 kg Check B = A + R; 6,964.29 = 2500 + 4,464.29

(0.15 ´ 6,964.29) = 1,000 + (0.01 ´ 4,464.29)1,044.6435 = 1,044.6429

8.11 A solution containing 10% NaCl, 3% KCl and water is fed to the process shown in Figure 8.13at the rate of 18,400 kg/h. The compositions of the streams are as follows: Evaporator product P—NaCl : 16.8%, KCl : 21.6% and water. Recycle product R—NaCl : 18.9% and water. Calculate theflow rates in kg/h and compute the composition of feed to the evaporator (F)

RFreshfeedF WEvaporator Crystallizer PNaCl only KCl onlyFigure 8.13

Basis: One hour Water in feed = 18,400 ¥ 0.87 = 16,008 kg/h KCl in feed = 18,400 ¥ 0.03 = 552 kg/hNaCl in feed = 18,400 ¥ 0.1 = 1,840 kg/h Overall balance: water vapour

Feed, FProcess KClNaCl

Balance around crystallizer: Overall balance: P = R + 552 NaCl balance: 0.168P = 0.189R

P Crystallizer RKCl

Solving, R = 4,416 kg and P = 4,968 kg Balance around evaporator:

WF Evaporator PNaClF = W + NaCl + PF = 16,008 + 1,840 + 4,968= 22,816 kg/hFeed to evaporator = recycle + fresh feed = 18,400 + 4,416 = 22,816 kg/hRecycle, RFresh feed Feed to evaporator, F, m (Conc. of NaCl) n (Conc. of KCl)

Making a balance for NaCl, we have18,400 ´ 0.1 + 0.189 ´ 4,416 = m ´ 22,816Solving, m = 11.72%Similarly, for KCl, we have 18,400 ´ 0.03 = n ´ 22,816 or, n = 2.42%Check:Making water balance, 18,400 ´ 0.87 + 4,416 ´ 0.811

= 22,816 ´ (100 – 11.72 – 2.42) 19,589.6 kg = 19,589.6 kg

8.12 Ethylene oxide is produced by catalytic oxidation of ethylene and oxygen. The total feed to thecatalytic bed of the reactor contains 10:1 volume ratio of oxygen to ethylene and the conversion perpass is 23%. Ethylene oxide is removed from the products completely and the unreacted ethylene isrecycled. The oxygen for the reaction is supplied from air. Calculate: (a) inlet and outlet compositionof the streams and (b) moles of fresh oxygen required for recycle gases.Reaction: C2H4 + 0.5O2 ® C2H4O.

SeparatorEthylene ReactorGasesAirFigure 8.14Basis: 1 kmole of ethylene.Oxygen supplied: 10 kmolesNitrogen entering: 1079 = 37.62 kmoles´21 C2H4 reacted: 1 ´ 0.23 = 0.23 kmoleC2H4O formed: 0.23 kmoleOxygen reacted = 0.23= 0.115 kmole2 Oxygen remaining = 10 – 0.115 = 9.885 kmoles (a)

Inlet gases mole mole % C2H4 1.0 2.056O2 10.0 20.560N2 37.62 77.384Total 48.62 100.000

Ethylene recycled = (1 – 0.23) = 0.77 mole(b) Moles recycled/mole of feed =0.77 = 0.015948.62 oxygen required = 0.115 mole(c)

Outlet gases mole mole % C2H4O 0.230 0.48O2 9.885 20.71N2 37.620 78.81Total 47.735 100.000

8.13 In the diagram shown in Figure 8.15 find E, P, A and B. Also, find the composition of A.The compositions are: F = 20% C2, 40% C3, 40% C4,

E = 95% C2, 4% C3, 1% C4,P = 99% C3, 1% C4,B = 8.4% C3, 91.6% C4.

Basis: 100 kg of feedEPA Feed, FI Unit II UnitBFigure 8.15

Let us assume that the compositions given are in weight %. Overall balance: feed = 100 = E + P + B.C2 balance: 20 = 0.95E. So, the value of E = 20/0.95 = 21.053 kg C3 balance: 40 = (0.04 ´ 21.053) +0.99P + 0.0084BC4 balance: 40 = (0.01 ´ 21.053) + 0.01P + 0.916BSolving the above, we find P = 35.9006 kg and B = 43.0464 kg

Since we know that F = E + A, we substitute the values of F and E, and observe, 100 = 21.053 + A.Solving A is found to be 78.947 kg Composition of A:C3 balance: (0.99 ´ 35.9006) + (0.084 ´ 43.0464) = 39.1575 Weight % = 49.6. C4 balance: (0.01 ´35.9006) + (0.916 ´ 43.0464) = 39.7895 Weight % = 50.4.

8.14 A contact sulphuric acid plant produces 98% acid. A gas containing 8% SO3 (rest inert) enters aSO3 absorption tower at the rate of 28 kmoles/h 98.5% of SO3 is absorbed in this tower by 97.3%acid introduced at the top and 95.9% acid is used as the make up acid. Compute tonne/day of

(a) make up acid required(b) acid fed at the top of the tower and(c) acid produced.Basis: In one day, gas entering = 28 ´ 24 = 672 kmoles SO3 entering = 672 ´ 0.08 = 53.76 kmoles

= 53.76 ´ 80 = 4,301 kgSO H O +→H SO 32 2480 18 98

SO3 absorbed = 4,301 ´ 0.985 = 4,236.3 kgAcid formed = 4,236.3 98 = 5,189.5 kg´80 Water reacted = 4,236.3 18 = 953.17 kg´80 97.3% SO3Exit r

Absorptiony 95.9% towerSO3

in 28 kmoles/hz x 98% (Acid)

Figure 8.16

Let us label x, y, z and r as shown.Overall balance gives: z = (r + 4,236.3)Acid balance gives, 0.98z = (0.973r + 5,189.5)Solving the above, we get r = 1,48,271.5 kg;

z = 1,52,507.7 kg (overall)Another balance of stream gives, z + y = x + rAcid balance in this stream gives: 0.98z + 0.959y = 0.98x + 0.973r or, (0.98 ´ 1,52,507.7) + 0.959y =0.98x + (0.973 ´ 1,48,271.5) Solving the above, we get x = 53,661.5 kg; y = 49,425.05 kg Check:H2O balance, 0.021r – 953.17 = (0.02 ´ z)or, (0.027 ´ 1,48,271.5) – 953.17 = (0.02 ´ 1,52,507.7) 4,003.33 – 953.17 = 3,050.16.

EXERCISES8.1 NO is produced by burning gaseous NH3 with 20% excess O2: 4NH3 + 5O2Æ 4NO + 6H2O

The reaction is 70 percent complete. The NO is separated from the unreacted NH3, and the latterrecycled. Compute (a) moles of NO formed per 100 moles of NH3 fed, and (b) moles of NH3recycled per mole of NO formed.

8.2 In a particular drier, 100 kg of a wet polymer containing 1.4 kg water/kg of dry polymer is driedto 0.25 kg of water per kg of dry polymer per hour. 5,000 kg of dry air is passed into the drier. Theair leaving the drier is having a humidity of 0.0045 kg of water vapour per kg of dry air and fresh airsupplied at a humidity of 0.011 kg of water vapour per kg of dry air. Calculate the mass rate of freshair supplied and fraction of air recycled per hour.

Energy Balance 9

9.1 DEFINITIONS

The following definitions are frequently used since the study of energy balance concerns conversionof our resources into energy effectively and utilize the same properly. In order to understand the basicprinciples pertaining to the generation, transformation and uses of energy, the following terms need tobe discussed first.

9.1.1 Standard State

A substance at any temperature is said to be in its standard state when its activity is equal to one. Theactivity may be looked upon as a thermodynamically corrected pressure or concentration. For puresolids, liquids and gases the standard state corresponds to the substances at one atmosphere pressure.For real gases, the pressure in the standard state is not 1 atmosphere but the difference from unity isnot large. In the case of dissolved substances the standard state is the concentration in each instance atwhich the activity is unity. The enthalpies of substances in standard states are designated by thesymbol H°, while the DH of a reaction where all reactants and products are at unit activity isrepresented by DH°.

9.1.2 Heat of Formation

The thermal change involved in the formation of 1 mole of a substance from the elements is called theheat of formation of a substance. The standard heat of formation is the heat of formation when all thesubstances involved in the reaction are each at unit activity. The enthalpies of all elements in theirstandard states at 298 K are zero.

1969.1.3 Heat of Combustion

It is the heat liberated per mole of substance burned. The standard heat of combustion is thatresulting from the combustion of a substance, in the state that is normal at 298 K and atmosphericpressure, with the combustion beginning and ending at 298 K.

9.1.4 The Heat of ReactionHeat of reaction from enthalpy dataIt is defined as the enthalpy of products minus the enthalpy of reactants. The standard heat of reactionfor the reaction,aA + bBÆ cC + dDis given byDH° = [cDH° + dDH° ] – [aDH°, A + bDH° ]r r, C r, Dr r, B

where DH°, i is the standard heat of formation of ith component. By convention,negative sign indicates that heat is given out and denotes exothermic reactionpositive sign indicates that heat is absorbed, i.e. endothermic reaction

Heat of reaction from heats of combustion data

For reactions involving organic compounds, it is more convenient to calculate the standard heat ofreaction directly from the standard heats of combustion instead of standard heats of reaction. Thestandard heat of reaction under such circumstances is the standard heat of combustion of the reactantsminus the standard heat of combustion of products.

i.e. [DHreaction = SDHc, (reactants) – SDHc, products]25°C

9.1.5 Heat of MixingWhen two solutions are mixed, the heat evolved or absorbed during the mixing process is known asheat of mixing.9.2 HESS’S LAW

If a reaction proceeds in several steps, the heat of the overall reaction will be the algebraic sum ofthe heats of the various stages, and this sum in turn will be identical with the heat, the reaction wouldevolve or absorb if it were to proceed in a single step.

9.3 KOPP’S RULE

The heat capacity of a solid compound is approximately equal to the sum of the heat capacities of theconstituent elements. As per Kopp’s rule, the following atomic heat capacities are assigned to theelements at 20 °C:

Carbon: 1.8; Hydrogen: 2.3; Boron: 2.7; Silicon: 3.8; Oxygen: 4.0; Fluorine: 5.; Phosphorus: 5.4 andall others: 6.2. Since the heat capacities of solids increase with temperature, it is clear that thesevalues do not apply over a wide range of temperature.9.4 ADIABATIC REACTION TEMPERATURE

Adiabatic reaction temperature is the temperature attained by reaction products, if the reactionproceeds without loss or gain of heat and if all the products of the reaction remain together in a singlemass or stream of materials.

9.5 THEORETICAL FLAME TEMPERATURE The temperature attained when a fuel is burnt inair or oxygen without loss or gain of heat is called the Theoretical flame temperature.WORKED EXAMPLES9.1 Calculate the enthalpy of sublimation of Iodine from the following reactions and data

(a) H2 (g) + I2 (s) Æ 2HI(g) DH = 57.9 kJ(b) H2 (g) + I2 (g) Æ 2HI(g) DH = –9.2 kJThe desired reaction is I2(s) Æ I2 (g)Solution: (a) – (b) = DH = 67.1 kJ

9.2 Find the enthalpy of formation of liquid ethanol from the following

data: –DH, Heats of reaction, kJ (1) C2H5OH (l) + 3O2(g) Æ 2CO2 (g) + 3H2O(l) – 1367.8 (2) C(graphite) + O2(g) Æ CO2(g) – 393.5 (3) H2(g) + ½O2(g) Æ H2O(l) – 285.8 Solution: [2 ¥ (2) – (1)+ (3 ¥ (3))] = 2C + 3H2 + ½O2Æ C2H5OH. The enthalpy of formation of ethanol = –276.6 kJ

9.3 200 kg of Cadmium at 27 °C is to be melted. (The melting point is 320.9 °C). The heat supply isfrom a system, which supplies 210 kcal/ kg, at steady state. Find the quantity of heat to be supplied bythe system.Atomic weight of Cadmium = 112.4, Cp = (6 + 0.005T) kcal/kmole °C and T in °C.

Latent heat of fusion = 2050 kcal/kmoleBasis: 200 kg of Cadmium º 1.78 katoms ©¸ ¦µ 2 Sensible heat = 1.78t tª¹ = 3885.5 kcal ª¹

«ºLatent heat of fusion = 1.78 ´ 2050 = 3649.0 kcalTotal heat to be supplied = 7534.5 kcal

Quantity of steam to be supplied = 7534.5= 35.88 kg210

9.4 An evaporator is to be fed with 10,000 kg/h of a solution having 1% solids. The feed is at 38 °C.It is to be concentrated to 2% solids. Steam at 108 °C is used. Find the weight of vapour formed andthe weight of steam used. Enthalpies of feed are 38.1 kcal/kg, product solution is 100.8 kcal/kg, steamis 540 kcal/kg and that of the vapour is 644 kcal/kg.

Basis: One hour.Feed = 10,000 kg/hVapour formed is 5000 kg/hThick liquor is 5000 kg/hEnthalpy of feed = 10,000 ´ 38.1 = 38.1 ´ 104 kcalEnthalpy of the thick liquor = 100.8 ´ 5,000 = 5,04,000 kcal. Enthalpy of the vapour = 644 ´ 5,000 =32,20,000 kcal. Heat supplied by steam = Msls = Ms ´ 540 kcal.Heat balance:Heat input by steam + heat in, by feed

= Heat out, in vapour + Heat out, thick liquor or, [Ms.(540) + 38.1 ´ 104] = (32,20,000 + 5,04,000)\ Ms (540) = 33,43,000.Thus the weight of steam required, Ms = 6,190.75 kg/hr.

9.5 Calculate the standard heat of reaction:CaC2 + 2H2O ® Ca(OH)2 + C2H2DHf cal/mole –15,000 –68,317.4 –2,35,800 54,194

Solution:DHrxn = (–2,35,800) + (54,194) – (–15,000) – (2 ´ 68,317.4) The standard heat of reaction is –

29,971.2 cal/mole

9.6 How much heat must be added to raise the temperature of 1 kg of a 20% caustic solution from 7°C to 87 °C? Take datum temperature as 0 °C.

Data:Specific heat at 7 °C = 3.56 and at 87 °C = 3.76 kJ/kg K

Solution: Q = (m.Cp.t)1 – (mCpt)2 = 1 [(3.76 ´ 87) – (3.56 ´ 7)] = 302.2 kJ

9.7 How many Joules are needed to heat 60 kg of sulphur trioxide from 273.16 K to 373.16 K?CpSO3 = 34.33 + 42.86 ´ 10–3T – 13.21 ´ 10–6T2 J/mole K Solution:

Number of moles of the trioxide = 60 = 0.75 kmole80At 373.16 K,

Q = nòCpSO3 dtAt 273.16 KQ = [{34.33 ´ (373.16 – 273.16)} + {(42.86 ´ 10–3/2)(373.162 – 273.162)}

+ {(–13.21 ´ 10–6/3)(373.163 – 273.163)}] Q = 3,509.25 kJ/kmole.9.8 Using the following data of heats of combustion in cal/g mole, calculate the following:(a) Heats of combustion of benzene to water(b) Heat of vaporization of benzene – cal/g mole

(i) C6H6 (l) to CO2 (g) and H2O (l) = 7,80,980(ii) C6H6 (g) to CO2 (g) and H2O (g) = 7,59,520(iii) H2 (g) to H2O (l) = 68,317(iv) H2 (g) to H2O (g) = 59,798(v) Graphite to CO2 (g) = 94,052Desired reactions:(a) C6H6 (l) + 7.5O2 ® 6CO2 (g) + 3H2O (l)(b) C6H6 (l) ® C6H6 (g)(a) Equation (i) itself gives value, DHc = – 7,80,980 cal/g mole. (ii) C6H6 (g) + 7.5O2 ® 6CO2 (g) +3H2O (g)(iii) H2 (g) + ½O2 ® H2O (l)(iv) H2 (g) + ½O2 ® H2O (g)(v) C + O2 ® CO2 (g)We can obtain the reaction (b) from the reaction (i) to (v) using

suitable multiplication factor for each step and adding or subtracting the equations as shown below:i.e. Steps for equation. (b) = (i) + 3(iv) – (ii) – 3(iii) l = 8,097 cal/g mole

9.9 Find the heat of formation of ZnSO4 from its elements and from these

data: kcal/mole(i) ZnS ® Zn + S 44(ii) 2ZnS + 3O2 ® 2ZnO + 2SO2 –221.88

(iii) 2SO2+ O2 ® 2SO3 –46.88(iv) ZnSO4 ® ZnO + SO3 55.1Desired equation: Zn + S + 2O2 ® ZnSO4 kcal/moleSteps: ½ [(ii) + (iii) – 2(i) – 2(iv)] = –233.48 kcal/mole.

9.10 Steam that is used to heat a batch reaction vessel enters the steam chest, which is segregatedfrom the reactants, at 250 °C, is saturated and completely condensed. The reaction absorbs 1000Btu/lb of charge in the reactor. Heat loss from the steam chest to the surroundings is 5000 Btu/h. Thereactants are placed in the vessel at 70 °F. At the end of the reaction, the materials are at 212 °F. Ifthe charge contains 325 lb of material and the products and reactants have an average Cp of 0.78Btu/1b °F, how many lb of steam are needed per lb of charge. The charge remains for an hour in thevessel.

Basis: One hour: Datum 70 °F BtuReaction absorbs heat = 1000 ´ 325 = 3,25,000Heat loss to surroundings = 5,000Heat in products: 325 ´ 0.78 (212 – 70) = 36,000\ Q = total heat = 3,66,000From steam tables at 482 °F(250 °C) ls = 734.9 Btu/lb we have, Q = msls = (ms) (734.9)= 3,66,000 Btu\ ms = 498.2 lb/h

lb of steam/lb of charge = 498.2= 1.5333259.11 Pure ethylene is heated from 30 °C to 250 °C at a constant pressure. Calculate the heat added perkmoleCp = 2.83 + 28.601 ´ 10–3T – 87.26 ´ 10–7T2 where

Cp is in kcal/kmole K and T in K T2

DH = n CdTp T1 = 303 K, T2 = 523 K∫T1

n = 1 kmoleOr, DH = [2.83 {T2 – T1} + (28.601 ´ 10–3/2) {T2 – T2}2 1– {87.26 ´10

–7/3}{T3 – T32 1}] Heat added = 2,886.11 kcal

9.12 Calculate the amount of heat given off when 1 m3 of air at standard conditions cools from 500 °Cto –100 °C at constant pressure. Cp air = 6.386 + 1.762 ´ 10–3 T – 0.2656 ´ 10–6 T2, where Cp is inkcal/kmole K and T in K. 1 m3 =1 = 0.0446 kmole22.414 173

Q = 0.0446 ∫C dT= 0.0446 [6.386 ´ 600 + (1.762 ´ 10–3/2)(1732 – 7732)p 773

– (0.2656 –6/3) (1733 – 7733)]´ 10Q = –191.345 kcal,Hence, heat is given off

9.13 Air being compressed from 2 atm and 460 °C (enthalpy 210.5 Btu/lb) to 10 atm and 500 °R(enthalpy 219 Btu/lb). The exit velocity of air is 200 ft/s. What is the horse power required for thecompressor if the load is 200 lb of air/hour?

Basis: One hour.(Ws = shaft work, Btu/lb v: velocity, ft/s)Ws = (219 – 210.5) = 8.5 Btu/lb.

22'v¦µ (200) = 0.8 Btu/lb¨· §¶ tt 778) Horse power needed = (8.5 + 0.8) ´200 = 0.73 HP2545 [1 Btu = 778 ft. lbf; 1 HP = 2545 Btu/h]9.14 Find the heat of reaction at 1200 K. C2H6 ® C2H4 + H2 'Hf,C H26 84,720 kJ/kmole 'Hf,C H24 52,280 kJ/kmoleDH°rxn, 298 K: 52,280 – (–84,720) = 1,37,000 kJDHrxn = DH°rxn – nCpReactants(1200 – 298) + nCpProducts (1200 – 298)

= (1,37,000) – [1 ´ 100 ´ (1200 – 298)] + [{(1 ´ 78.7) + (1 ´ 29.7)}(1200 – 298)]= 1,37,000 – 90,200 + 97,776.8

= 1,44,576.8 kJ/kmole (Heat to be supplied)

9.15 Calculate the heat input to raise the temperature of 132 kg of CO2 from 100°C to 1000°C.Perform the calculation in the following ways. (a) by integrating the expression for Cp and (b) byusing mean heat capacity value Cp in kcal/kmole K.

Cp = 6.85 + 8.533 ´ 10–3 T – 2.475 ´ 10–6 T2, kcal/kmole K Basis: 132 kg of CO2 º 132/44 = 3kmoles T2

(a) DH = n CdTp = 3 [6.85 ´ 900 + (8.533 ´ 10–3/2)(12732 – 3732)∫

T1

– (2.475 –6 /3)(12733 – 3733)]´ 10DH = 3[10,826.29] = 32,478.87 kcalCp values at 1273 K and 373 K are:

Cp at 1273 K = 13.702 kcal/kmole Kand Cp at 373 K = 9.6884 kcal/kmole KCpav = 11.6952 kcal/kmole KDH = òm Cpav dT= 3 ´ 11.6952 ´ (1273 – 373)= 31,577.04 kcal

9.16 SO2 gas is oxidized in 100% excess air with 70% conversion to SO3. The gases enter theconverter at 400 °C and leave at 450 °C. How many kcals are absorbed in the heat exchanger of theconverter per kmole of SO2 sent?

Basis: 1 kmole SO2.SO2 + ½O2 ® SO3 O2 sent = 0.5 ´ 2 = 1. SO3 formed = 0.7 kmole N2 in air = 1 ´ 79/21 = 3.76 kmole.SO2 remaining = 0.3 kmole

SO3 SO2 O2 N2 Total gases leaving kmoles 0.7 0.3 0.65 3.76 5.41 Cp mean, cal/g mole °C 15.5 11.07.5 7.1 —

Hrxn = –23, 490 cal/g mole (given)DDHrxn = –23,490 ´ 0.7 = –16,443 kcalDatum: 0°C

Heat in Heat out

SO2 = 1 11 400 = 4,400 kcal SO2 = 0.3 11 450 = 1,485.00 kcal O2 = 1 7.5 400= 3,000 kcal O2 = 0.65 7.5 450 = 2,193.75 kcalN2 = 3.76 7.1 400 = 10,676 kcal N2 = 3.76 7.1 450 = 12,013.20 kcal

SO3 = 0.7 15.5 450 = 4,882.50 kcal Total = –18,076 kcal Total = + 20,574.45 kcal DHrxn –16,443 kcal \ Heat in –34,519.00 kcal

Hence, heat absorbed in heat exchanger = –13,944.5 kcal

9.17 From the following data compute the enthalpy change of formation for NH3 at 480 °C.

D Hf at 25°C for NH3 = –10.96 kcal/kmoleCp N2 = 6.76 + (6.06 ´ 10–4T) + (13 ´ 10–8T2)Cp H2 = 6.85 + (2.8 ´ 10–5T) + (22 ´ 10–8T2)Cp NH3 = 6.703 + (0.0063T) where T is in K,Reaction: N2 +3H2 ® 2NH3

D Hf/kmole 0 0 –10.96Basis: 1 mole of N2 (Feed at 273 K)DHrxn 298 K: (2 ´ –10.96) = –21.92 kcal.Find DCp = 2NH3 – (N2 + 3H2)Da = (2 ´ 6.703) – [7.76 + (3 ´ 6.85)] = –13.9Db = (2 ´ 0.0063) – [6.06 ´ 10–4 + (3 ´ 2.8 ´ 10–5)] = 0.0119 Dg = (2 ´ 0) – (13 ´ 10–8 + 3 ´ 22 ´ 10–8)= –7.9 ´ 10–7

DHo

= DHrxn

– DaT– '¦µ T2 –'¦µ T3§¶ §¶ · ¨· 0.0119 t 298 27.9 10 73

DHo

= –21.92 – (–13.9 ´298) – 2 – t t298 3 DHo= 3,598.87 kcal480 °C = 753 KHrxn, 480°C = DHo + DaT + '¦µ T2 + '¦µ T3D§¶ §¶¨· ¨· 0.0119 = 3,598.87 + (–13.9 ´753) + ¦µ (753)2§¶ · 7.9 107¦µ (753)3+ t§¶

·\DHrxn 480°C = –3,606.56 kcal/kmole

9.18 Calculate the calorific value of a blast furnace gas analyzing 25% CO, 12.5% CO2 and 62.57%N2.(a) C + O2 ® CO2; DHrxn: –94 kcal(b) C + ½O2 ® CO; DHrxn: –26 kcal

Also, calculate the theoretical flame temperature for the combustion of this gas assuming theoreticalamount of air is used, the combustion reaction is complete and reactants enter at 25 °C.

Cp = a + bT + cT2, cal/kmole Kwhere a, b and c are all dimensional constants and available in literature 3c ´ 105Gas a b ´ 10CO2 10.55 2.16 –2.04 N2 6.66 1.02 —C COCO2

CO II Combustion2I CombustionO2 N2Air N2Basis: 100 g moles of inlet gasCO entering 25 g moles, CO2 exit = 25 + 12.5 = 37.5 g moles ¦µ79 = 109.52 g moles2 needed 12.5 g moles, N2 = t§¶O¨·

CO + ½O2 ® CO2Reaction (a) – (b) givesDHrxn = –94 + 26 = –68 kcal/kmole

Calorific value: Heat given out = 68 ´ 25 = 1,700 kcal Exit gases carry this heat away.

This gas temperature is called Theoretical flame temperature which is calculated as follows:–17 ´ 105 cal = 37.5 [{10.55 (T – 298)} + 2.16 ´ 103(T2 – 2982)/2

– 2.04 ´ 10–5(T3 – 2983)/3] + 109.52[6.66 (T – 298) + 1.02 ´ 10–3(T2 – 2982)/2] Solving the aboveequation we have T = 2721.085K 2721 K = 2448 °C.

9.19 An inventor thinks he has developed a new catalyst which can make the gas phase reaction CO2+ 4H2 ® CH4 + 2H2O proceed to 100% conversion. Estimate the heat that must be provided orremoved if the reactants enter and products leave at 500 °C (in effect, we have to calculate the heat ofreaction at 500 °C).

DHf CO2 CH4 H2O kcal/kmole –94,052 –17,889 –57,798 at 298 K

\D Hrxn = [–17,889 – (2 ´ 57,798)] – [–94,052]= –39,433 kcal/kmole of CO2

Cp for CO2 = 6.339 + 10.14 ´ 10–3T – 3.415 ´ 10–6T2 H2 = 6.424 + 1.039 ´ 10–3T – 0.078 ´ 10–6T2

H2O = 6.97 + 3.464 ´ 10–3T – 0.483 ´ 10–6T2 CH4 = 3.204 + 18.41 ´ 10–3T – 4.48 ´ 10–6T2

Da = [3.204 + (2 ´ 6.97) – 6.339 – (4 ´ 6.424)] = –14.891Db = [18.41 + (2 ´ 3.464) – 10.14 – (4 ´ 1.039) = 11.047 ´ 10–3

= [–4.48 – (2 ´ 0.483) – {–3.415 – (4 ´ 0.078)}] = –1.719 ´ 10–6DgDCp = –14.891 + 11.047 ´ 10–3T – 1.719 ´ 10–6 T2

We find DHo using data at 298 K

'¦µ T2 – '¦µ T3DHo = DHrxn – DaT – §¶ §¶ · ¨· = –39,433 – (–14.891 ´298) – 11.047 10 3¦µ2 tt§¶2¨· ¦µ 3 – 1.716 10 6

tt§¶ 3¨· DHo= –39,433 + 4,430 – 491 + 15.16 = –35,479 kcal Next we find DHrxn at 500 °C (or) 773K:

'¦µ T3DH773 = DHo + DaT + '¦µ T2 + §¶§¶ ¨·¨· = –35,479 + (–14.891 ´773) + 11.047 10 32 tt 773 2 1.719 10 63

+

t t773 3= – 43,943 kcal/kmole\ 43,943 kcal of heat must be removed.

9.20 CO at 50 °F is completely burnt at 2 atm pressures with 50% excess air, which is at 1000 °F.The products of combustion leave the combustion chamber at 800 °F. Calculate the heat evolved from

the combustion chamber in terms of Btu/lb of CO entering.

Basis: 1 lb mole of CO = 28 lb, O2 needed = 0.5 lb mole CO + ½O2 ® CO2O2 supplied = 0.5 ´ 1.5 = 0.75 lb mole (50% excess)Air supplied = 3.57 lb mole and N2 = 2.82 lb molesDHrxn = –1,21,745 Btu/lb moleQ = DHrxn + DHproducts – DHreactantsO2 remaining = 0.25 lb moleCO2: 1 lb mole, N2: 2.82 lb molesDatum: 32°FData: DH (Btu/lb mole)

Temperature , °F CO Air O2 N2 CO250 125.2 — — — —

77 313.3 312.7 315.1 312.2 392.2 800 — — 5,690 5,443 8,026 1000 — 6,984 — — —

DHproducts = DH800°F – DH77°F

= 1(8,026 – 392.2) + 2.82(5,443 – 312.2) + 0.25(5,690 – 315.1) = 23,446 Btu/lb moleDH reactants: (DH1000°F – DH77°F)air + (DH50°F – DH77°F)CO = 3.57 (6,984 – 312.7) + 1(125.2 –313.3)= 23,612 Btu/lb moleQ = –1,21,745 + 23,446 – 23,628 = –1,21,927 Btu/lb mole Heat evolved by combustion =1,21,927/28

= 4,354.5 Btu/lb of CO 9.21 Pure CO is mixed with 100% excess air and completely burnt at constantpressure. The reactants are originally at 200 °F. Determine the heat added or removed, if the producttemperatures are 200 °F, 500 °F, 1000 °F, 1500 °F, 2000 °F and 3000 °F.Basis: 1 lb mole of CO

CO + ½O2 ® CO2O2 supplied = 1 lb mole, N2 = 3.76 lb molesExit: CO2 : 1 lb mole, O2 : 0.5 lb mole, N2 : 3.76 lb moles

Assuming a base temperature of 25 oC, (77 oF) and using mean heat capacities,

D H = Hp – HR; Q = DHDH = SnCppr(77 – 200) + DHrxn77 °F + SnCpR (t – 77)Reactants: DHrxn = –1,21,745 Btu/lb mole

Gas n Cp nCp

CO 1 6.95 6.95

CO 1 6.95 6.95O2 1 7.10 7.10N2 3.76 6.95 26.13

Total 40.18\SnCppr(77 – 200) = –(40.18 ´ 123) = – 4,942 Btu DH = – 4,942 –1,21,745 + SnCpR (t – 77°)200 °F 500 °F 1000 °F 1500 °F 2000 °F 3000 °F n —— — —— — — — — — — — Cp nCp Cp nCp Cp nCp Cp nCp Cp nCp CpnCp

CO2 1.0 9.15 9.15 9.9 9.9 10.85 10.85 11.5 11.5 12.05 12.05 12.75 12.75O2 0.5 7.10 3.55 7.25 3.63 7.15 3.88 7.8 3.9 8.0 4.0 8.3 4.15N2 3.76 6.95 26.13 7.0 26.32 7.15 26.88 7.35 27.64 7.55 28.39 7.88 29.42

SnCpR — — 38.83 — 39.85 — 41.51 — 43.04 — 44.44 — 46.32SnCp(t–77) 4,776 16,857 38,314 61,246 83,758 1,35,810Q = DH –21,911 –1,09,830 –88,373 –65,441 –41,229 +9,123

9.22 Coal is burnt to a gas of the following composition: CO2 : 9.2, CO : 1.5, O2: 7.3, N2: 82%. Whatis the enthalpy difference for this gas between the bottom and the top of the stack if the temperature atthe bottom is 550 °F and at the top is 200 °F?

Cp of N2 = 6.895 + 0.7624 ´ 10–3 T – 0.7 ´ 10–7 T2Cp of O2 = 7.104 + 0.7851 ´ 10–3 T – 0.5528 ´ 10–7

T2Cp of CO2 = 8.448 + 5.757 ´ 10–3 T – 21.59 ´ 10–7 T2 + 3 ´ 10–10 T3Cp of CO = 6.865 + 0.8024 ´10–3 T – 0.736 ´ 10–7 T2Basis: 1 lb mole of CO2:

Multiplying these equations by the respective mole fractions of each component and adding themtogether, we have

for N2 = 0.82 ´ CpN2for O2 = 0.073 ´ CpO2for CO2 = 0.092 ´ CpCO2for CO = 0.015 ´ CpCO

Cpnet = 7.049 + 1.2243 ´ 10–3 T – 2.6164 ´ 10–7 T2 + 0.2815 ´ 10–10 T3

200

∫Cpnet

dT\DH = 550

= 7.049 (200 – 550) + 1.2243 103

t§¶¦µ(2002 – 5502) ¨·2

– 2.6164 107¦µ(2003 – 5503) + 0.28151010

t§¶ 3

t§¶ ¦µ 4 (2004 – 5504) ¨· ¨·

or, DH = –2,465 – 160.6 + 13.8 – 0.633 = –2,612 Btu 9.23 Calculate the theoretical flametemperature for CO burnt at constantpressure with 100% excess air? The reactants enter at 200 °F. CO + ½O2 ® CO2Basis: 1 g mole COTemperature of reactants: 200 °F = 93.3 °CGases entering: CO–1, O2–1, N2–3.76 (all in moles)Gases leaving: CO2–1, O2–0.5, N2–3.76 (all in moles)\DHrxn 25 °C = – 67,636 cal.

Gas mole DTCpm DH = nCpmDT CO 1.0 (93.3 – 25) 6.981 476Air 4.76 (93.3 – 25) 6.993 2,270Total 2,746 cal

Next we have DHproduct = – (DHreactants – DHrxn)= – (2,746 + 67,636) = –70,382 cal Let us assume exit temperature as 1800 °C, then DT = (1800 –25) = 1775 °C Gas mole DTCpm DH

CO2 1.0 1775 12.94 23,000O2 0.5 1775 8.35 7,400N2 3.76 1775 7.92 52,900

This total of –83,300 cal is not matching with –70,382 cal, the value calculated.Let the Theoretical flame temperature be 1500 °C, then DT = 1475 °C DH = (1 ´ 12.7 ´ 1475) + (0.5´ 8.31 ´ 1475) + (3.76 ´ 7.88 ´ 1475)

= 68,460 calMaking linear interpolation for the theoretical flame temperature, we have,Theoretical flame temperature

= 1500 + 70,382 68,460 ËÛ´ (1800 – 1500)83,300ÌÜ ÍÝ = 1500 + 39 = 1539 °C º 2798 °F

9.24 Calculate the theoretical flame temperature of a gas having 20% CO and 80% N2 when burntwith 150% excess air. Both air and gas being at 25 °C.

Data: Heat of formation of CO2 = – 94,052 cal/g mole, CO = –26,412 cal/g mole at 25 °C.Cpm: CO2: 12.1, O2: 7.9, N2: 7.55 cal/g mole K (from literature) Basis: 1 g mole CO, CO + 0.5O2 ®CO2

O2 supplied = (0.5 ´ 2.5) = 1.25 g moles (150% excess) 80 N2

in feed= 1 ÈØ ÉÙ = 4 g molesÊÚ N2

in air= ÈØ79 ÉÙ = 4.7 g molesÊÚ21

Exit gas: CO2: 1 g mole, O2: 0.75 g mole, N2: 8.7 g moles Q = SHproducts + SHrxn – SHreactants. (Datum298 K)SHreactants is zero, since air and gas are at 25 °C.DHrxn = DHCO2 – DHCO= –94,052 – (–26,412) = –67,640 cal/g mole.Let the “Theoretical Flame Temperature” be T, K67,640 = [1 ´ 12.1 ´ (T – 298)] + [8.7 ´ 7.55 ´ (T – 298)] + [0.75 ´ 7.9 ´ (T – 298)], 67,640 = 83.71 T– 24,945.6\ T = 1106.03 K ∫ 833.03 °C1106 K ∫ 833 °C.9.25 Find the theoretical flame temperature of a gas containing 30% CO and 70% N2 when burnt with100% excess air. The reactants enter at 298 K.DHf CO2 = –3,93,700 kJ/kmole; DHf CO = –1,10,600 kJ/kmole Mean molar specific heat, kJ/kmoleK at different temperatures is given below:

Temperature K CO2 O2 N2

800 45.4 31.6 30.31000 47.6 32.3 30.61200 49.4 33.0 31.21400 50.8 33.6 31.81600 52.0 34.0 32.31800 53.2 34.4 32.7

Basis: 1 kmole of CO, Datum: 298 K

N2 in feed = 70/30 = 2.34 kmolesO2 supplied = 0.5 ¥ 2 = 1 kmoleN2 from air = 3.76 kmolesExit gas consists of:CO2: 1, O2: 0.5, N2: (3.76 + 2.34) = 6.1 kmolesLet us consider the equation

Q = SHproducts + DHrxn –SHreactants

where SHreactants = Zero at 298 K (Q Datum is 298 K)\DHrxn = (–3,93,700) – (–1,10,600) = –2,83,100 kJ/mole By iteration method:Let the theoretical flame temperature be 1400 K:DT = (1400 – 298) = 1102 KDHpr = (1 ¥ 50.8 ¥ 1102) + (0.5 ¥ 33.6 ¥ 1102) + (6.1 ¥ 31.8 ¥ 1102)

= 2,88,261 kJ/kmole π 2,83,100 kJ/kmoleLet the theoretical flame temperature be 1200 K\DT (1200 – 298) = 902 KDHpr = (1 ¥ 49.4 ¥ 902) + (0.5 ¥ 33 ¥ 902) + (6.1 ¥ 31.2 ¥ 902)

= 2,31,110 kJ/kmole π 2,83,100 kJ/kmoleSo theoretical flame temperature lies in between these two values (by interpolation )\ Theoretical flame temperature

2,83,100 2,31,110 = 1,200 + ÌÜ 2,88,261 2,31,110 ´ (1400 – 1200)ÍÝ So, the temperature of the exit gases is 1382 K = 1109 °C.

9.26 The analysis of 15,000 lit of a gas mixture at standard condition is as follows: SO2: 10%, O2:12% and N2: 78%. How much heat must be added to this gas to change its temperature from 30 °C to425 °C?

The Cpm values are in cal/g mole °C

Gas SO2 O2 N2Cpm 30 °C 10 6.96 6.80 Cpm 425 °C 11 7.32 7.12

The amount of gas mixture = 15,000 litres º

15,00022.414

= 669.2 g moleswe can then write the amount of each componentSO2 : 669.2 ´ 0.1 = 66.92 g molesO2 : 669.2 ´ 0.12 = 80.30 g molesN2 : 669.2 ´ 0.78 = 521.98 g moles

669.20 g molesReference temperature: 0 °C\ Q = 66.92 [(11 ´ 425) – (10 ´ 30)] + 80.3 [(7.32 ´ 425) – (6.96 ´ 30)]

+ 521.98 [(7.12 ´ 425) – (6.8 ´ 30)] = 19,98,849.22 cal 9.27 Estimate the theoretical flametemperature of a gas containing 20% CO and 80% N2 when burnt with 100% excess air. Both air andgas are initially at 25 °C.

Cp CO2 = 6.339 + 10.14 ´ 10–3 T – 3.415 ´ 10–6 T2

Cp O2 = 6.117 + 3.167 ´ 10–3 T – 1.005 ´ 10–6 T2

Cp N2 = 6.457 + 1.389 ´ 10–3 T – 0.069 ´ 10–6 T2

The values of Cp are in kcal/kmole K and temperature is in K DHrxn 25 °C = –67,636 kcalBasis: 1 kmole of CO; N2 = 4 kmolesAir supplied: O2: 1 kmole, N2 from air: 3.76 kmolesExit gas: CO2: 1, O2: 0.5, N2: 7.76 kmolesDatum 25 °C = 298 KDHrxn = –DH products + 67,636 = [1 ´ òCpCO2 ´ (T – 298)] + [0.5 ´ òCpO2 ´ (T – 298)]+ [7.76 ´ òCpN2 ´ (T – 298)]

22

67,636 = 6.339(T – 298) + 10.14 ´10–3 T 298 2 33 T 298– 3.415 ´ 10–6 T 298

3+ 6.117 ´2 22T33

+ 3.167 ´10 –3 T 298– 1.005 ´ 10–6 298 4 6 T 22

+ 7.76´6.457

´(T– 298) + 7.76´1.389´10–3 298 2 T 33

– 7.76 ´0.069 ´10–6 298 3 Solving for theoretical flame temperature = T = 1216 K = 943 °C

9.28 Dry methane and dry air at 298 K and 1 bar pressure are burnt with 100% excess air. Thestandard heat of reaction is –802 kJ/g mole of methane. Determine the final temperature attained bythe gaseous products if combustion is adiabatic and 20% of heat produced is lost to the surroundings.

Data: Cpm values (J/g mole K) for the components are: O2: 31.9, N2: 32.15, H2O : 40.19, CO2 :51.79.Basis: 1 g mole of methane.Datum: 298 K

CH4 + 2O2 ® CO2 + 2H2O\ Oxygen supplied = 2 ´ 2 = 4 g molesN2 entering = 479 = 15.05 g moles´21

Gases leaving are: CO2: 1, H2O : 2, O2: 2, and N2: 15.05 g moles. Heat given out = 802 kJHeat loss = (802 ´ 0.2) = 160.4 kJ\ Q = Heat in exit gases = (802 – 160.4) = 641.6 kJ

Q = [1 ´ 51.79 ´ (T – 298)] + [2 ´ 40.19 ´ (T – 298)] + [2 ´ 31.9 ´ (T – 298)] + [15.05 ´ 32.15 ´ (T –298)] = [679.8275 ´ (T – 298)] = 641.6 ´ 103 J.\ T = 1242 K = 969 °C.

9.29 An iron pyrite ore contains 85% FeS2 and 15% gangue. It is roasted with 200% excess air to getSO2. The reaction is given. All the gangue plus Fe2O3 end up in the solid waste produced whichanalyzes 4% FeS2. Determine the standard heat of reaction in kJ/kg of ore roasted and the analysis of

the solid waste. Heat of formation data is in kJ/g mole.

4FeS2 + 11O2 ® 2Fe2O3+8SO2 Weights 479.4 352 319.4 512

Heats of formation –177.9 0 –88.2 –296.9 (kJ/g mole)Basis: 1 kg of ore containing FeS2 = 0.85 kg and gangue = 0.15 kg Let x kg of FeS2 be in the solidwaste, 319.4

ËÛthen, FeS2 reacted: (0.85 – x) kg; Fe2O3 formed (0.85 x)ÌÜ ÍÝ= 0.666(0.85 – x)x = 0.04{(0.15) + 0.666(0.85 – x) + x};

Solving the above, we find x = 0.029 kgSolid waste = (0.15) + (0.029) + [0.666 ´ (0.85 – 0.029)] = 0.725786 kg A summary of thecomposition of the solid waste is given:

Solid waste kg Weight %Gangue 0.150 20.66FeS2 0.029 3.99Fe2O3 0.547 75.35Total 0.726 100.00

FeS2 reacted = (0.85 – 0.029) = 0.821 kg = 6.85 ´ 10–3 kmole\Fe2O3 formed = 0.547 kg = 3.424 ´ 10–3 kmoleSO2

formed = ÌÜ ËÛ512= 1.068 kg = 0.016688 kmole ÍÝ479.4

Heat of reaction = –[(0.016688 ´ 296.9) + (3.424 ´ 10–3 ´ 822.3)–(6.85 ´ 10–3 ´ 177.9)]

= –6.55 kJ9.30 For the following reaction, estimate the heat of reaction at 298 K. A + B ® C + DCompound DH°f(kcal/g mole)

A –269.8B –195.2C –337.3D –29.05DH° at 25 °C = SDH°f, Products – SDH°f, reactants

= [–337.3 – 29.05] – [–269.8 – 195.2]= 98.65 kcal9.31 Estimate the standard heat of reaction DH°298 for the reaction. A + B ® CStandard heats of combustion are:DHc, 298 for A = –328000 cal/gDHc, 298 for B = –212000 cal/gDHc, 298 for C = –542000 cal/gDHc, 298 = SDH°c, reactants – SDH°c, products= [–328000 – 212000] – [–542000]= 2000 cal9.32 Calculate the heat of formation of CHCl3 from the following data:

CHCl3 +1 O2 + H2O ® CO2 + 3HCl, DH = –509.93 kJ (1)2

H2 + 1 O2 ® H2O; DH = –296 kJ (2)2

C + O2 ® CO2; DH = –393.78 kJ (3) 1 H2 + 1 Cl2 ® HCl; DH = –167.5 kJ (4)2 2

CO2 + 3HCl 1 O2 + H2O; DH = –509.93 kJ® CHCl3 + 2 H2 + 1 O2 ® H2O; DH = –296 kJ2 C + O2CO2; DH = –393.78 kJEq. (4) × 3 + Eq. (3)/2, gives

3 H2 + Cl2 + 1 C + 1 O2 ® 3HCl + 1 CO22 2 2 29.33 Standard heat of reaction accompanying any chemical change is equal to the algebraic sum of thestandard heat of formation of the products minus the algebraic sum. Calculate the standard heat ofreaction, DH°f, 298 for the reaction2FeS2 + 1.5O2 ® Fe2O3 + 4SO2

The standard heats of formation are:FeS2 = –42,520 cal/g moleFe2O3 = –1,96,500 cal/g moleSO2 = –70,960 cal/g moleFrom the reaction,DH°r, 298 = DHFe2O3 + 4 [DHSO2] – 2[DHFeS2]

= –19,6500 + 4[–70,960] – 2[42,520]= –39,5300 cal9.34 The heat of reaction at 300 K and 1 atm pressure for the reaction A + 3B ® C is 30,000 cal/moleA converted.

Cp data is as follows:A = –0.4 + 0.1 T (T in K)B = 10C = 30Calculate the heat of reaction at 600 K and 1 atmA + 3B ® C

600

H600 = DH300 + ±'Cp.dTD 300600

= –30,000 + ±[{303 × 10 + 0.4} 0.1T] dT 300

= –30,000 + 0.4T– 0.1

T2

2= –30,000 + 0.4[600 – 300] – [0.05 × (6002 – 3002)] = –43,380 cal/g mole of A

9.35 Calculate the theoretical flame temperature of a gas containing 20% CO and 80% N2 when burntwith 150% excess air, with both air and gas being at 25 °C.

D H°f CpmCO2= –393.137 kJ/g mole CO2 = 50.16 kJ/kg K CO = –110.402 kJ/g mole O2 = 33.02 kJ/ kmole H2 =31.56 kJ/kmole KBasis: 100 g moles of feedCO = 20 g moles

CO +1 O2Æ CO22 O2 needed = 1 × 20 = 10 g moles2

O2 supplied = 2.5 × 10 = 25 g molesN2 supplied = 25 × 79/21 = 94.05 g moles Gases leavingCO2= 20 g molesO2 = 25 – 10 = 15 g molesN2 = 94.05 + 80 = 174.05 g molesAtmospheric temperature = 25 °CHeat in reactants + DHR = Heat in products Standard heat of reaction,

D Hf°products – DHf°reactants = –393.137 – [110.402 + 1 × 0]2= –282.835 kJ/g mole

Heat of reactants is zero (Reference temperature)

Heat produced when 20 g moles of CO is burnt= 282.835 × 20 = 5654.700 kJHeat in outgoing gas,{20 [50.16] + 15 [33.02] + 174.05 [31.56]} (T – 25) = 5654700 Solving, T = 834 °C

EXERCISES

9.1 Determine the theoretical flame temperature that can be attained by the combustion of methanewith 20% excess air. Air and methane enter at 298 K and a pressure of 1 atm. The reaction iscomplete.–DHr = 1,91,760 cal/g moleMean heat capacities (cal/g mole K)

Component Temperature 2000 °C 1800 °C CO2 13.1 12.95 H2O 10.4 10.25 O2 8.4 8.3 N2 8.0 7.9

9.2 Determine the heat of reaction at 720 K and 1 atm for the reaction

SO2 + 0.5O2 ® SO3Mean molar specific heats ofSO2 : 51.5 kJ/kmole KO2 : 45.67 kJ/kmole KSO3 : 30.98 kJ/kmole KStandard heat of formation forSO2 : –2,97,000 kJ/kmoleSO3 : –3,95,000 kJ/kmole

9.3 Calculate the enthalpy change in J/kmole that takes place in raising the temperature of 1 kmole ofthe gas mixture of 80 mole %. Methane and rest ethane from 323 K to 873 KHeat capacity equation, Cp = R (A + BT + CT2) cal/g mole where, R is Gas constant, T is temperaturein K. A, B and C are constants and Cp is the heat capacity at constant pressure.

Component Value of constant in Cp equation AB ´ 103C ´ 106 CH4 1.702 9.081 –2.164 C2H6 1.13119.225 –5.561

9.4 Chlorine is produced by the reaction4HCl(g) + O2(g) ® 2H2O(g) + 2Cl2(g)The feed stream to the reactor consists of 67 mole % HCl, 30 mole %

O2 and 3 mole % N2. If the conversion of HCl is 75% and the process is isothermal, how much heat istransferred per mole of entering gas mixture.Data:

(a) Standard heat of formation at 25 °C, J/g moleHCl(g) : –92,307 J/g mole

H2O(g) : –2,41,818 J/g mole

(b) Mean heat capacities: (cal/g mole K)HCl(g) : 7.06O2(g) : 8.54H2O : 7.52Cl2(g) : 8.61N2(g) : 7.169.5 Calculate the number of joules required to calcine completely 100 kg of limestone containing80% CaCO3, 11% MgCO3 and 9% water. The lime is withdrawn at 900 °C and the gases leave at200 °C. The lime stone is charged at 25 °C.Data: Standard heat of formation at 25 °C and 1 atm, cal/g mole CaCO3 : – 2,88,450 cal/g moleMgCO3 : – 2,66,000 cal/g moleCaO : – 1,51,900 cal/g moleMgO : – 1,43,840 cal/g moleCO2 : – 94,050 cal/g moleMean molal heat capacity, cal/g mole KH2O : 8.2CO2 : 10.5CaCO3 : 25.0MgCO3 : 23.0CaO : 14.0MgO : 10.09.6 In the reaction4FeS2(s) + 11O2 (g) ® 2Fe2O3(s) + 8SO2(g) the conversion from FeS2 to Fe2O3 is only 80%complete. If the standard heat of formation for the above is calculated to be–197.7 kcal/g mole, what –DH°reaction should be used in energy balance per kg of FeS2 fed.9.7 Calculate the theoretical flame temperature for CO burnt at constant pressure with 20% excessair. The reactants enter at 366 K. CO + ½O2 ® CO2The heat capacities are 29.23 for CO, 29.28 for air, 54.18 for CO2, 34.5 for O2 and 33.1 for N2 in J/gmole K. Standard heat of reaction:–283.13 kJ/g mole.9.8 A gaseous mixture of 1000 m3 containing 60% hydrogen and 40% ammonia is cooled from 773 Kto 313 K at 1 atm pressure. Calculate the heat removed. The Cp values in kcal/kmole K, and T in Kare: for hydrogen Cp = 6.9 – 0.2 ´ 10–3 T + 0.48 ´ 10–6 T2 for ammonia Cp = 6.08 + 8.81 ´ 10–3 T – 1.5´ 10–6 T29.9 Determine the theoretical flame temperature that can be obtained by the combustion ofmethane with 25% excess air. Air and methane enter at 298 K and a pressure of 1 atm. The reaction iscomplete. Standard heats of formation at 298 K in kJ/kmole are: Methane (g) = –74,520Carbon dioxide (g) = –3,93,500Water vapour (g) = –2,41,8139.10 Calculate the amount of heat given off when 1 m3 of air at standard condition cools from 600 °C

to 100 °C at constant pressure Cp air = 6.386 + 1.762 ´ 10–3 T – 0.2656 ´ 10–6 T2

Cp is in kcal/kmole K and T is in K.9.11 CO is burnt under atmospheric pressure with dry air at 773 K with 20% excess air. The productsleave at 1223 K. Calculate the heat involved in the reaction chamber in kcal/kmole of CO burnt,assuming complete combustion.Data: –DH298 K = – 67,636 kcalMean specific heats are: 7.017 for CO, 7.225 for air, 11.92 for CO2, 7.941 for O2 and 7.507 for N2 inkcal/kmole K.

Problems on Unsteady10State Operations

The term unsteady state refers to chemical processes in which the operating conditions generallyfluctuate with time. Although unsteady-state processes are difficult to formulate, the general formulaused to represent the total amount of material and energy in the process is given as

Rate of input + Rate of generation = Rate of output + Rate of accumulation This is the guidingprinciple in solving problems on the unsteady state operations.WORKED EXAMPLES

10.1 A storage tank contains 10,000 kg of a solution containing 5% acetic acid by weight. A freshfeed of 500 kg/min of pure water is entering the tank and dilutes the solution in the tank. The mixtureis stirred well and the product leaves the tank at a rate of 500 kg/min. At what instant of time the acidconcentration in the tank will drop to 1% acetic acid by weight? After one hour of operation, whatwill be the concentration in the tank?

FiXiM, XFo, Xo = X221

Here inlet flow rate, Fi = 500 kg/minOutlet flow rate, Fo= 500 kg/minInitial mass = 10,000 kgThe total mass M, at any time in the tank

= Initial mass + (Inflow rate – Outflow rate)(time) = 10,000 + (500 – 500)t

= 10,000 kgWe know that,

rate of input + rate of generation = rate of output+ rate of accumulation (1)Here, rate of generation is zero.

Let X be the concentration of acid and M the mass of solution of acid at any time.

Hence, rate of accumulation = rate of input – rate of output or, d MX()= FiXi – Fo X (2)dtdXdM MX= FiXi – Fo X (3)dtdt Here, dM= 0 (Since inlet and outlet flow rates are the same)dt Therefore, M dX = FiXi – Fo X(4)dtdX Now substituting values, 10,000 ÉÙ ÈØ = 500 ´ 0.0 – 500X (5)ÊÚ (i.e.)dX = – 0.05Xdt (i.e.)dX= – 0.05dtX Integrating, we get Xt

dX 0.05dtX ±±

oXt 0

X ln ÉÙ ÈØ = –0.05(t – 0)ÊÚ = –0.05t (6) Therefore, X= e–0.05t (7)Xo Time taken to reach a concentration of 1% is given by, 0.01 ln ÈØ ÉÙ = –0.05tÊÚ i.e. t = 32.19 minutes(b) Substituting for t as 60 minutes in Eq. (7), we get

X = 0.05 e–0.05 (60) = 0.00249 = 0.249%i.e. after one hour of operation, the concentration in the tank will be 0.249%

10.2 A tank contains 10 kg of a salt solution at a concentration of 2% by weight. Fresh solution entersthe tank at a rate of 2 kg/min at a salt concentration of 3% by weight. The contents are stirred welland the mixture leaves the tank at a rate of 1.5 kg/min.

(a) Express the salt concentration as a function of time and (b) At what instant of time the saltconcentration in the tank will reach 2.5% by weight?

Here inlet flow rate, Fi = 2 kg/min

Outlet flow rate, Fo= 1.5 kg/minInitial mass = 10 kg

The total mass M, at any time in the tank= Initial mass + (Inflow rate – Outflow rate)(time) = 10 + (2 – 1.5)t

or, M = 10 + 0.5tDifferentiating, we getdM= 0.5dt We know thatrate of input + rate of generation

= rate of output + rate of accumulation (1) Here, rate of generation is zero.Let X be the concentration of acid and M the mass of solution of acid at any time.Hence, rate of accumulation = rate of input – rate of output

i.e.d MX()= FiXi – FoX (2)dt

or, M dX + X dM = FiXi – FoX (3)dt dt

Here,dM= 0.5dtTherefore, by substituting values, we get

(10 + 0.5t) dX + 0.5X = 2 ´ 0.03 – 1.5X (4)dt (i.e.) (10 + 0.5t) dX = 0.06 – 2Xdt or, dXdt (5)0.06 2Xt Integrating, we get

Xtdt©¸ ±±Xto

ª¹ Xt 0Xt

also, 0.5 dXdt

±±

Xt oXt0 ln

X 0.03 ËÛ ËÛ(20 t)(6)0.02 0.03ÌÜ ÌÜ20ÍÝ ÍÝ 4 or, X 0.03 20 ËÛ(7)0.01 ÌÜtÍÝ204

\ X = 0.03 – 0.01ËÛ(8)20ÌÜtÍÝ Time taken to reach a concentration of 2.5% is given by substituting X = 0.025 in Eq. (7).Hence, we have,

(20 + t)4 = 204 ´0.01

0.03 X Hence, t = 3.784 minutesAliterWe shall go back to Eq. (4), which is(10 + 0.5t) dX + 0.5X = 2 ´ 0.03 – 1.5Xdt or, ÌÜÌÜ 20.06ËÛËÛ (9)dtt t ÍÝÍÝ This equation is of the form dy + Py = Q (10)dx Where, P and Q are either functions of x or constants

The solution for this differential equation isyeòPdx = òQeòPdx dx + constant (11)Using the same analogy we can solve Eq. (9) in the following manner 24P = 10 0.5 20tt

and

0.06 0.12 Q tt

Substituting for P and Q in Eq. (11), we get 44

dtdt

X t t

0.12eedt + constant (12)Ô20 t X ´ 4 ´ exp[ln(20 + t)]X(20 + t) 0.12 = ËÛ´ 4 ´ exp[ln(20 + t)] dt + constantÔ20ÌÜtÍÝ 0.12 4 = ÔËÛ[20 + t]4dt + constant20ÌÜtÍÝ or, X(20 + t)4 = Ô[0.12][20 + t]3dt + constantX

(20 + t)4 = 0.12 [20 t]4

4+ constant (13)X = 0.03 + constant(14)(20 t)4

Initial conditions are:t = 0, X = 0.02Substituting in Eq. (14), we get, Constant = – 0.01 ´ (20)4 = –1600 Equation (14) thus becomes,

X = 0.03 –1600and (15)(20 t)4204

X = 0.03 – 0.01ËÛ(16)(20ÌÜt)ÍÝ Comparing Eq. (16) with Eq. (8) we find both are same and hence the time taken to reach aconcentration of 2.5% is 3.784 minutes

10.3 A tank contains 10 litre of a salt solution at a concentration of 2 g/litre Another salt solutionenters the tank at a rate of 1.5 litres/min at a salt concentration of 1 g/litre. The contents are stirredwell and the mixture leaves the tank at a rate of 1.0 litre/min.

Estimate (a) the time at which the concentration in the tank will be 1.6 g/litre and (b) the contents inthe tank will be 18 litres

Here, Inlet flow rate, Fi = 1.5 litres/min at a salt concentration of 1 g/litres Outlet flow rate, Fo = 1.0litre/minInitial volume = 10 litres.The total volume V, at any time

= Initial volume + (Inflow rate – Out flow rate) (time) = 10 + (1.5 – 1.0)tor, V = 10 + 0.5tDifferentiating, we get

dV = 0.5dt We know that,rate of input + rate of generation

= rate of output + rate of accumulation (1) Here rate of generation is zero.

Let C be the concentration of salt and V the volume of solution at any time.Hence, rate of accumulation = rate of input – rate of outputi.e.d VC()= FiCi – Fo C (2)dt

V dC + C dV = FiCi – Fo C (3)dt dt Here, dV = 0.5dt Therefore, by substituting values,(10 + 0.5t) dC + 0.5C = 1.5 ¥ 1.0 – 1.0C (4)dt i.e. (10 + 0.5t) dC = 1.5 – 1.5C = 1.5(1 – C)dtdC or, dt (5)

1.5(1=-+Ct Integrating, we get Ct

��� �∫∫dt

1.5��� �Ct0.5 )��� �=0oCtCt

�� � �∫∫dt−=Ct)�� � �Ct 20

o ==

ln Ct È˘ È ˘(6)Í Í ˙ -Î˚ Î ˚ 3C - =Í˙(7)120+Î˚ È˘Í˙ 3 \ C = 1 + +Î˚(8)Time taken to reach a concentration of 1.6 g/litres is given by substituting C = 1.6 in Eq. (8):0.6 = + È˘Í˙3 +Î˚ solving, t = 3.71 minutes.(b) Final volume

= Initial volume + (volume flowing in – volume flowing out)(time) or, 18 = 10 + (1.5 – 1.0)t

Therefore, time taken for the water in the tank to reach 18 litres is = 16 minutes.

EXERCISES

10.1 A tank contains 500 kg of a 10% salt solution. A stream containing salt at 20% concentrationenters the tank at 10 kg/h and the mixture leaves the tank after thorough mixing at a rate of 5 kg/h.Obtain an expression for the salt concentration in the tank as a function of time and the saltconcentration in the tank after 3 hours.

10.2 A tank contains 1000 kg of a 10% salt solution. A stream containing salt at 20% concentrationenters the tank at 20 kg/min and the mixture leaves the tank after complete mixing at a rate of 10kg/min. Obtain an expression for the salt concentration in the outlet as a function of time and the saltconcentration in the tank after 1 hour. What will be the time at which the salt concentration in the tankwill be 15%?

10.3 A tank contains 50 litres of a salt solution at a concentration of 2.5 g/litre. Another salt solutionenters the tank at a rate of 2.0 litres/min. at a salt concentration of 1 g/litre. The contents are stirredwell and the mixture leaves the tank at a rate of 2.5 litres/min.

Estimate (a) the time at which the concentration in the tank will be 1.25 g/litre and (b) the contents inthe tank will be 20 litres.

10.4 A tank contains 20 kg of a salt solution at a concentration of 4% by weight. Fresh solution entersthe tank at a rate of 2.5 kg/min at a salt concentration of 3% by weight. The contents are stirred welland the mixture leaves the tank at a rate of 2.0 kg/min. (a) Express the salt concentration as a functionof time and (b) At what instant of time the salt concentration in the tank will reach 3.75% by weight?

10.5 A storage tank contains 5000 kg of a 1% sugar solution by weight. A fresh feed of 400 kg/min ofpure water is entering the tank and dilutes the solution in the tank. The mixture is stirred well and theproduct leaves the tank at a rate of 400 kg/min. At what instant of time the sugar concentration in thetank will drop to 1% sugar by weight? After one hour of operation, what will be the concentration inthe tank?

10.6 A 15% Na2SO4 solution is fed at the rate of 12 kg/min into a mixer that initially holds 100 kg ofa 50 : 50 mixture of Na2SO4 and water. The exit solution leaves at the rate of 9 kg/min. Assuminguniform mixing, what is the concentration of Na2SO4 in the mixer at the end of 10 minutes. Volumechange during mixing can be neglected.

10.7 A cylindrical tank of cross-sectional area A is filled with liquid up to a height Ho. A hole ofdiameter d at the bottom of the tank, which was plugged initially is opened to let the liquid drainthrough it. Set up an unsteady state balance equation to calculate the time for the liquid level to fall toa new height H1.

10.8 A solution is having solute A at a concentration CAfis fed continuously into a mixing vessel ofconstant volume V, into which water is added continuously and diluted. The outlet concentration of

the solution is Cao. Write the unsteady state solute balance equation. Discuss about the solution of theequation.

Tables

TABLE I Important Conversion FactorsQuantityLengthTo convert Multiply by from to

in ( ¢¢) m 0.0254 ft (¢) m 0.3048 cm m 0.01 Angstrom (Å) m 10–10

–6microns (m)m 10

Area in2 m2 6.452 ´ 10–4 ft2 m2 0.0929 cm2 m2 10–4

Volume ft3 m3 0.02832

cm3 m3 10–6 litre m3 10–3 Gallons (UK) m3 4.546 ´ 10–3 Gallons (US) m3 3.285 ´ 10–3

Mass Pound (lb) kg 0.4536 Gram (g, gm) kg 10–3 Density

Force lb/ft3 kg/m3 16.019 g/litre kg/m3 1.0 g/cm3 kg/m3 1000

lbf N 4.448kgf N 9.807Pa N 980.7dyne N 10–5TABLE I Important Conversion Factors (contd.)

(Contd.) 229 Quantity To convert Multiply by from toPressure lbf/ft2

lbf/in2 (psi) in Hgin watermm Hgatmtorrbarkgf/cm2

N/m2 = Pa 47.88 N/m2 = Pa 6895N/m2 = Pa 3386N/m2 = Pa 249.1 N/m2 = Pa 133.3 N/m2 = Pa 1.0133 ´ 105 N/m2 = Pa 133.3 N/m2 = Pa 105

N/m2 = Pa 9.807 ´ 104

Heat or energy

Volumetric flow rate BtuergcalkcalkW.h

ft3/sft3/h, ft3/h cm3/slit/hJ = N.m 1055 J = N.m 10–7 J = N.m 4.187 J = N.m 4187 J = N.m 3.6 ´ 106

m3/s 0.02832 m3/s 7.867 ´ 10–6 m3/s 10–6

3/s 2.777 ´ 10–7m

Mass flow rate (Mass flux)lb/ft2.h

g/cm2s kg/m2s 1.356 ´ 10–3 kg/m2s 10

Molar flow rate (Molar flux)

Enthalpy lb mole/ft2.h g mole/cm2s

Btu/lbcal/g = kcal/kg kmole/m2s 1.356 ´ 10–3 kmole g/m2s10

J/kg = N.m/kg 2326 J/kg = N.m/kg 4187

Heat capacity (Holds good for molal heat

capacity also)Btu/lb. °F

cal/g.°C N.m/kg K = J/kg K 4187N.m/kg K = J/kg K 4187TABLE II Atomic Weights and Atomic Numbers of Elements

Element Symbol Atomic Number Atomic weight

Actinium Ac Aluminium Al Americium Am Antimony Sb Argon A Arsenic As Astatine At Barium BaBerkelium Bk Beryllium Be Bismuth Bi Boron B Bromine Br Cadmium Cd Calcium Ca CaliforniumCf Carbon C Cerium Ce Cesium Cs Chlorine Cl Chromium Cr Cobalt Co Columbium Nb Copper CuCurium Cm Dysprosium Dy Erbium Er Europium Eu Fluorine F Francium Fr Gadolinium Gd GalliumGa Germanium Ge 89 217.00 13 26.98 95 243.00 51 121.76 18 39.94 33 74.91 85 210.00 56 137.3697 245.00

4 9.0183 209.005 10.8235 79.9248 112.4120 40.0898 246.006 12.0158 140.1355 132.9117 35.4624 52.0127 58.9441 92.9129 63.5496 243.0066 162.4668 167.2063 152.009 19.0087 223.0064 156.9031 69.7232 72.60

Element Symbol Atomic Number Atomic weight

Gold Au Hafnium Hf Helium He Holmium Ho Hydrogen H Indium In Iodine I Iridium Ir Iron FeKrypton Kr Lanthanum La Lead Pb Lithium Li Lutetium Lu Magnesium Mg Manganese Mn MercuryHg Molybdenum Mo Neodymium Nd Neptunium Np Neon Ne Nickel Ni Niobium Nb Nitrogen NOsmium Os Oxygen O Palladium Pd Phosphorus P Platinum Pt Plutonium Pu Polonium Po PotassiumK Praseodymium Pr 79 197.20 72 178.60

2 4.0067 164.941 1.0049 114.7653 126.9177 193.1026 55.8536 83.8057 138.9282 207.213 6.94

71 174.9912 24.3225 54.9380 200.6142 95.9560 144.2793 237.0010 20.1828 58.6941 92.917 14.0176 190.208 16.0046 106.7015 30.9878 195.2394 242.0084 210.0019 39.1059 140.92

Element Symbol Atomic Number Atomic weight

Promethium Pm Protactinium Pa Radium Ra Radon Rn Rhenium Re Rhodium Rh Rubidium RbRuthenium Ru Samarium Sm Scandium Sc Selenium Se Silicon Si Silver Ag Sodium Na Strontium SrSulphur S Tantalum Ta Technetium Tc Tellurium Te Terbium Tb Thallium Tl Thorium Th ThuliumTm Tin Sn Titanium Ti Tungsten W Uranium U Vanadium V Xenon Xe Ytterbium Yb Yttrium Y ZincZn Zirconium Zr 61 145.00 91 231.00 88 226.05 86 222.00 75 186.31 45 102.91 37 85.48 44 101.7062 150.43 21 44.96 34 78.96 14 28.09 47 107.88 11 23.00 38 87.63 16 32.07 73 180.88 43 99.00 52127.61 65 159.20 81 204.39 90 232.12 69 169.40 50 118.70 22 47.90 74 183.92 92 238.07 23 50.9554 131.30 70 173.04 39 88.92 30 65.38 40 91.22 TABLE III(a) Empirical Constants for Molal HeatCapacities of Gases at

Constant PressureCp = a + bT + cT2, where T is in Kelvin; g-cal/(g-mole) (K) – Temperature range 300 to 1500 KGas a b ´ 103c ´ 106

H2 6.946 – 0.196 0.4757N2 6.457 1.389 –0.069O2 6.117 3.167 –1.005CO 6.350 1.811 –0.2675NO 6.440 2.069 –0.4206H2O 7.136 2.640 –0.0459CO2 6.339 10.140 –3.415

SO2 6.945 10.010 –3.794SO3 7.454 19.130 –6.628HCl 6.734 0.431 +0.3613C2H6 2.322 38.040 –10.970CH4 3.204 18.410 –4.480C2H4 3.019 28.210 –8.537Cl2 7.653 2.221 –0.8733Air 6.386 1.762 –0.2656NH3(*) 5.920 8.963 –1.764

* Gas

TABLE III(b) Molal Heat Capacities of Hydrocarbon Gases For 10 to 760 oC, Cp = a + bT + cT2

oC, Cp = 7. 95 + mTnFor –180 to 95 T in Rankine (oF + 460)

Compound a b3 –c ´ 106m n ´ 10

Methane 3.42 9.91 1.28 6.4 ´ 10–12 4.00Ethylene 2.71 16.20 2.80 8.13 ´ 10–11 3.85Ethane 1.38 23.25 4.27 6.20 ´ 10–8 1.79Propylene 1.97 27.69 5.25 2.57 ´ 10–3 1.26Propane 0.41 35.95 6.97 3.97 ´ 10–3 1.25n-Butane 2.25 45.40 8.83 0.93 ´ 10–2 1.19i-Butane 2.30 45.78 8.89 0.93 ´ 10–2 1.19Pentane 3.14 55.85 10.98 3.9 ´ 10–2 1.00

Answers to Exercises

CHAPTER 11.1 (a) 10.84 cm2/s(b) 39.67 psia(c) 0.03929 hp-hr(d) 0.627 lbf/ft2

(e) 0.01355 cal/s cm2°C(f) 1163 W/m K1.2 10.93 [Cp G0.8/D0.2]1.3 4.55 * 10–3 [P/T0.5]1.4 (0.09453) [h2.5/g0.5] tan F1.5 As long as consistent units are used, the equation remains the same.CHAPTER 2

2.1 500 g moles2.2 0.5455 kg of carbon2.3 (a) 3.572 g of O2(b) 12.77 g of KClO32.4 (a) 2.8%(b) 0.088(c) 5.378 g moles/kg of water2.5 Mole ratio: 0.0425Mole %: Na2CO3: 4.1%Water: 95.9%2.6 Molality: 5.98 g moles/kg of solutionMolarity: 5.98 g moles/litreVolume of solution: 0.0836 litreNormality: 5.98

2352.7 Weight %: 39.02% Volume %: 26% 2.8Compound Weight % Volume % mole % NaCl 23.3 11 8.54 H2O 76.7 89 91.46 Total 100.00 100100.00

Atomic %:Na: 2.93%; Cl2: 2.93%; H2: 62.76%; and O2: 31.38% Molality: 5.185 g moles/kg of solution

2.9 AVMWT: 65.02Chlorine: 54.98%Bromine: 10.15%Nitrogen: 34.87%

2.10 (a) 0.1455 kg sugar/kg water(b) 1075 kg/m3 solution(c) 136.56 g sugar/litre

2.11 (a) Nitrogen(b) 16.17% and(c) 59.97%

2.12 0.945 g of Cr2S32.13 9.3445 kg of AgNO32.14 (a) Mole fraction of H3PO4: 0.02Mole fraction of Water: 0.98(b) 890.8 cc of solution2.15 54,090 g2.16 31.75 kg Cu, 104.26 kg of 94% H2SO4

2.17 (a) H2SO4(b) 79.25%(c) 89.6252.18 NaCl:75%, KCl 25% and NaCl: 79.25%, KCl 20.75%2.19 2094 kg iron and 900 kg water2.20 (a) 26.94(b) Methane: 24.94%, ethane: 14.47%, ethylene: 25.98%, propane: 9.79%, propylene: 14.03% n-butane 10.76%(c) 0.932.21 (a) Methane: 24.16% and carbon dioxide: 66.44% and nitrogen 9.4% (b) 29.8(c) 1.33 kg/m3

CHAPTER 3 3.1Compound Weight fraction mole fraction mole %

Butane 0.5 0.5701 57.01Pentane 0.3 0.2758 27.58Hexane 0.2 0.1541 15.41

Total 1.0 1.0000 100.00

AVMWT: 66.1383.2 70,748 g3.3 37.63%3.4 (a) and (b)

Component mole % Weight % Methane 80 68.45

Ethane 15 24.10Nitrogen 5 7.45 Total 100 100.00

(c) AVMWT: 18.7(d) Density: 0.0008 g/cc3.5 AVMWT: 30.5; Density of gas: 1.36 g/litre3.6 Methane: 0.008892 kmole/m3

Ethane: 0.02231 kmole/m3

Hydrogen: 0.01338 kmole/m3

Velocity: 30,558 m/hDensity: 0.00405 g/cc3.7 1.169 g/litre3.8 0.567 litre3.9 10,719.7 K 3.10 C3H83.11 300.37 atm 3.12 (a) and (c)

Component Volume % mole fraction N2 64.75 0.6475CO2 9.52 0.0952

H2O 18.62 0.1862O2 6.11 0.0611CO 1.00 0.0100

Total 100.00 1.0000(b) Average density: 0.792 kg/m33.13 52.923 kg 3.14 Component Volume % = mole %

Chlorine 68.59Bromine 12.67Oxygen 18.74

Total 100.00

3.15 (a) Partial pressure: 0.09 atm(b) 3 m3

(c) 2.06 g/litre

3.16 (a), (b) and (c)Component mole fraction Concentration, Partial pressure g mole/cc mm Hg

CH4 0.1 1.058 ´ 10–5 200C2H6 0.3 3.173 ´ 10–5 600H2 0.6 6.347 ´ 10–5 1,200

(d) 1.058 ´ 10–4 g mole/cc(e) 1,248 g/s(f) AVMWT: 11.83.17 (a) Nitrogen,

(b) 7.5% hydrogen(c) 19.76%

3.18 (a) 0.05 atm(b) 5 m3

(c) 0.707 g/litre(d) 17.3

3.19 34.4 s3.20 (a) Chlorine: 54.98%, bromine: 10.16% nitrogen: 34.85% (b) 65.05(c) Density: 2.59 g/litre

3.21 (a) Nitrogen: 67.5%, oxygen: 15.74% water: 1.79%, ammonia: 14.97%

(b) 2.48 kg/m3

3.22 10.45 litre3.23 0.982 kg

CHAPTER 44.1 31,941 J/g mole4.2 77.25 °C4.3

x 1.0 0.897 0.773 0.660 0.555 0.459 0.369 0.288 0.212 0.140 0.076 0.013 0y 1.0 0.958 0.897 0.831 0.758 0678 0.590 0.496 0.393 0.281 0.162 0.030 0x, y: Mole fraction of benzene in liquid and vapour phase respectively. 4.4 xA 1.0 0.724 0.415 0.1340.0

yA 1.0 0.819 0.548 0.205 0.0 x, y: Mole fraction of ‘A’ in liquid and vapour phase respectively. 4.50.1847 kg/kg of steam, 0.1385 kg/kg of steam4.6 Total pressure: 2,044.79 mm Hg

Mole fraction of methanol in vapour phase: 0.5052 Mole fraction of ethanol in vapour phase: 0.49484.7 (a) Total pressure: 4087.9 mm HgC2H6 : 0.139, n-C3H8 : 0.639, i-C4H10 : 0.039, n-C4H10 : 0.179 and C5H12 : 0.004(b) 3589.87 mm Hg4.8 Total pressure: 906.9 mm HgComponent

Benzene Toluene Xylene

Liquid phase Vapour phase composition, x composition, y

0.500 0.7390.377 0.2330.123 0.028

4.9 Partial pressure: 102.6 mm Hg4.10 (a) Ethyl acetate: 12.22 %, air: 87.78% (By volume) (b) Ethyl acetate: 29.8 %, air: 70.2% (Byweight)

CHAPTER 55.1 (a) RH% = 24.94%(b) 0.01358 mole of toluene/mole of vapour free gas (c) 0.0433 kg toluene/kg of air(d) % saturation: 23.93%(e) Mole % = volume % = 1.34%5.2 Humidity: 0.019 kg/kg

% saturation: 21%Humid volume: 0.9394 m3/kg dry air5.3 2961.6 m3/h5.4 0.567 mole/mole1.534 kg/kg5.5 0.25 mole/mole0.676 kg/kg5.6 Humidity: 0.025 kg/kgDew point: 28.5 °CHumid volume: 0.907 m3/kg dry airAdiabatic saturation temperature: 30 °CHumid heat: 1.0521 kJ/kg dry airEnthalpy: 0.951 kJ/kg dry air

5.7 Cool to 18.5 °C and then reheat it to 30 °C Air needed initially 5114.03 m3/h5.8 27.5 °C and 0.019 kg/kg5.9 (a) 96.5%, (b) 115.08 m3/h5.10 (a) 0.0351 kmole/kmole(b) 0.00857 kmole/kmole(c) 1.830 kg of water(d) 49.858 m3

5.11 46.5 °C; 0.0475 kg/kg dry air5.12 (a) % RH: 29.6%(b) Humid volume: 0.974 m3/kg dry air5.13 (a) 3703.7 kg dry air/h(b) 0.8586 and 0.9739 m3/kg dry air respectively (c) 18 and 28.5 °C respectively(d) 35°C5.14 (a) 0.008 kg/kg(b) 12 °C(c) 2.81 kg of water(d) 7,405.2 kJ(e) 32.5 °C5.15 % Relative saturation: 57.5%Percentage saturation: 44.59%5.16 Humidity: 0.027 kg/kg% saturation: 70%5.17 Humid volume: 0.9205 m3/kg dry airEnthalpy: 106.6 kJ/kg dry air5.18 (a) 809.4 m3/min(b) 74.04 kg5.19 (a) 0.0259 kmole/kmole, 0.01616 kg/kg (b) 0.0677 kmole/kmole 0.00423 kg/kg (c) 13.48 kg ofwater5.20 (a) 29.45

(b) 781.24 mm(c) 810.69 Hg(d) 0.0377 mole/mole5.21 5.7%5.22 (a) 86%, (b) 1396.5 m3/h

CHAPTER 66.1 Mother liquor: 3,324.64 kg6.2 Feed: 83,078.34 kg6.3 Crystals: 6,636 kg6.4 Crystals: 479.2 kg6.5 Crystals: 342 kg and Mother liquor: 918 kg6.6 Water evaporated: 7,916.67 kg and crystals: 2,083.33 kg

6.7 Water evaporated: 720 kg Mother liquor: 6589 kg andcrystals: 1,691 kg6.8 87.4%6.9 Water evaporated: 561 kg6.10 Water evaporated: 82.35 kg; Mother liquor: 1,122.85 kg and Feed: 2,705.2 kg6.11 2393.6 kg/h

CHAPTER 77.1 H2SO4: 5,310.86 kg, HNO3: 2,811.87 kg and Feed: 1,877.27 kg7.2 (a) 56.96%; (b) 8.338; (c) 21.81 kg7.3 50.62%7.4 CO2: 11.41%, H2O: 14.4% O2: 3.53% and N2:70.66%7.5 Excess air: 39.06%, 471.06 kg of lime/100 kg coke burnt7.6 O2: 12.14%, N2: 79.34%, SO2: 7.67% and SO3: 0.85%7.8 54.22% Excess air7.10 CO2: 4.9%, CO: 9.1% H2O: 17.51% and N2: 68.49%

7.11 CO2: 11.183%, O2: 3.956%, SO2: 0.158%, N2: 75.693%, and H2O: 9.010%7.14 Butane: 2.46%, Pentane: 42.25% and Hexane: 55.29%

7.15 (a) 6,600 kg, (b) 4,920 kg and (c) 2,165 kg7.16 (a) Nitrogen; (b) 16.2% and (c) 60%7.17 (a) 18,973.2 kg/hr and (b) 1,434.4 kg/h in each evaporator7.19 (a) 1.25 m3 (b) 6.41 m3

(c) Component Condition (a) Condition (b)

CO2 0.23 0.14H2O 0.14 0.08N2 0.63 0.74

O2 —0.04

Total 1.00 1.00

7.20 (a) 4.49%(b) 3.087.21 (a) 28.8(b) 7.55 m3/kg carbon burnt(c) 50.285 m3/100 m3

(d) CO2: 19.66%, O2: 1.28%, N2: 79.06%7.22 (a) 446.7 m3

(b) 0.6427.23 (a) 18(b) CO2: 17.5%, O2: 6.3%, H2O: 3.5%, and N2: 72.7%7.24 (a) 2.965(b) 19.23%(c) 5081 kg7.25 Liquid : 81.789 kg/hr, vapour : 18.211 kg/hr, C5H12 : 31.38, C6H14 : 60.067.26 Air supplied = 12.24 kg , air required theoretically for completekg of fuel

combustion = 15.3 kg, volumetric analysis = (mol. basis) N2 = 82.8%CO = 10.9%CO2 = 6.5%

7.27 (a) Moles of CO: 5.444 kmoles; moles of CO2: 1.089 kmoles (b) Moles of air supplied: 18.388kmoles(c) Percentage of carbon lost in the ash: 7.407%

7.28 (a) Composition of the flue gas (wt. basis): CO2 = 273.0933 kg, CO = 9.1467 kg, N2 =1091.3911 kg, O2 = 118.5901 kg, H2O = 4.5 kg,

(b) Ash produced = 21.1 kg(c) Carbon lost per 100 kg of coke burnt = 7.58%7.29 (a) Volume of air being introduced = 21729.54 m3/h(b) Composition of flue gas on dry basis (mol. basis): CO2: 10.74%, CO: 0.5373%, N2: 85.487%,O2: 3.2267%7.30 (a) 3.846% (b) 70.3%7.31 Air–fuel ratio =12.59478, flue gas analysis:

Composition Weight % Volume % CO2 17.7382 11.9749

O2 7.2088 6.69

N2 72.5086 76.92H2O 2.4726 4.08SO2 0.7167 0.3326

Amount of air supplied = 37.78434 ton 7.32 Air–fuel ratio by mass = 16.7165Flue gas analysis:

Composition Weight (kg) CO2 3.113O2 0.6408N2 12.8717H2O 1.026SO2 0.064

Total = 17.7155 kg/kg of fuel7.33 Air requirement = 432.433 kmoles, volume = 15416.56 Nm3/hFlue gas analysis:

Composition Weight (kg) CO2 16.63 O2 1.2 H2O 8.032 N2 74.14

7.34 Air–fuel ratio = 17.51058, % excess air = 16.167%

CHAPTER 88.1 (a) 100 moles (b) 0.4286 mole of Ammonia/mole of NO formed8.2 (a) 1,425.8 kg, (b) Recycle fraction: 0.7179CHAPTER 99.1 1760 °C9.2 –1,16,295.8 kJ/kmole9.7 1,874 °C9.10 51.44 kcalCHAPTER 1010.1 C = –0.1 [100/(100 + t)]2 + 0.210.2 C = [10,000 + 400t + t2]/(100 + t)10.3 (a) t = 100 – 100 [(C – 1)/1.5]0.25, 36.1 minutes(b) 60 minutes10.4 (a) C = 0.01 [40/(t + 40)]5 + 0.03(b) 2.369 minutes10.5 28.78 minutes10.6 0.2675 kg Na2SO4/kg of solution

Index

Adiabatic reaction temperature, 198 Gas constant, 35, 36 Antoine equation, 75API scale, 12

Atomic numbers of elements, 231–233 Hausbrand chart, 75Atomic percent, 11 Heat, 196Atomic weights of elements, 231–233 Heat capacity of gases (empirical Average molecular weight,38 constants), 234 Avogadro’s hypothesis, 9 Heat ofAvogadro’s number, 36 combustion, 197

formation (also standard), 196 mixing, 197Baume’ gravity scale, 12 reaction, 197Brix scale, 13 from combustion data, 197Bypass, 180 from enthalpy data, 197 Humid heat, 88Humid volume, 88Clausius–Clapeyron equation, 74 Humidity, 87Composition of absolute, 87liquid systems, 11 percentage absolute humidity, 88 mixtures, 10 relative, 88solutions, 10 Hydrated salt, 111Conservation of mass, 8Conversion, 9Conversion factors, 229–230 Ideal gas law, 35Crystal, 111Crystallization, 111Kopp’s rule, 198

Density, 12, 38Dew point, 88 LawDry bulb temperature, 87 Amagat’s, 37

Dalton’s, 37Hess’s, 197Energy balance, 196 Leduc’s, 37Enthalpy, 88 Law of conservation of mass, 8

247 248INDEX

Magma, 111Mass balance, 122 Mass relations, 7 Mole fraction, 11 Mole percent, 11 Mother liquor, 111

Normal boiling point (NBP), 74 Normal temperature and pressure

(NTP), 9Saturated vapour, 89 Saturation, 89

partial, 89percentage, 89 relative, 89

Specific gravity, 12 Standardcondition, 35

state, 196Steam distillation, 75System, 3isolated, 3

Partial pressure, 36Percentage saturation, 88 Process, 3Property, 3

extensive, 3intensive, 3Psychrometry, 87

Pure component volume, 36 Purge, 180

Table ofatomic numbers, 231–233 atomic weights, 231–233 conversion factors, 229–230 molal heatcapacities, 234

Theoretical flame temperature, 198 Twaddell scale, 13

Rate ofaccumulation, 221 generation, 221 input, 221output, 221

Reactant, 9excess, 9limiting, 9

Reactionendothermic, 197 exothermic, 197

Recycle, 180Units and notations, 1derived units, 2

Unsteady state operations, 221Vapour pressure, 74 Volume percent, 10Weight percent, 10Wet bulb temperature, 87Yield, 9