Uploaded by:
Ebooks Chemical Engineering https://www.facebook.com/pages/Ebooks-Chemical-Engineering/238197077030
For More Books, softwares & tutorials Related to Chemical Engineering Join Us @facebook: https://www.facebook.com/pages/Ebooks-ChemicalEngineering/238197077030 @facebook: https://www.facebook.com/AllAboutChemcalEngineering @facebook: https://www.facebook.com/groups/10436265147/
ADMIN: I.W
<< If you like this Book, than support the author and BuY it >>
PROCESS CALCULATIONS
PROCESS CALCULATIONS SECOND EDITION V. VENKATARAMANI Formerly Professor Department of Chemical Engineering National Institute of Technology Tiruchirappalli N. ANANTHARAMAN Professor Department of Chemical Engineering National Institute of Technology Tiruchirappalli K.M. MEERA SHERIFFA BEGUM Associate Professor Department of Chemical Engineering National Institute of Technology Tiruchirappalli New Delhi-110001 2011 PROCESS CALCULATIONS, Second Edition V. 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, by mimeograph or any other means, without permission in writing from the publisher. ISBN-978-81-203-4199-9 The 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 Meenakshi Art Printers, Delhi-110006.
To My Parents — V. Venkataramani — K.M. Meera Sheriffa Begum To My Mother — N. Anantharaman
Contents Preface .............................................................................................................. xi Preface to the First Edition ............................................................................ xiii Acknowledgements ........................................................................................... xv 1 UNITS AND DIMENSIONS 1–6 1.1 Introduction ..................................................................................... 1 1.2 Basic Units and Notations ............................................................... 1 1.3 Derived Units ................................................................................... 2 1.4 Definitions ....................................................................................... 3 Worked Examples ..................................................................................... 3 Exercises ................................................................................................... 6
2 MASS RELATIONS 7–34 2.1 Mass Relations in Chemical Reaction ............................................. 7 2.2 Conservation of Mass ...................................................................... 8 2.3 Avogadro’s Hypothesis .................................................................... 9 2.4 Limiting Reactant and Excess Reactant .......................................... 9 2.5 Conversion and Yield ....................................................................... 9 2.6 Composition of Mixtures and Solutions ........................................ 10 2.6.1 Weight Percent ................................................................... 10 2.6.2 Volume Percent .................................................................. 10 2.6.3 Mole Fraction and Mole Percent ....................................... 11 2.6.4 Atomic Fraction and Atomic Percent ................................ 11 2.6.5 Composition of Liquid Systems ........................................ 11 2.7 Density and Specific Gravity ......................................................... 12 2.7.1 Baume’ (°Be’) Gravity Scale ............................................. 12 2.7.2 API Scale (American Petroleum Institute) ........................ 12 2.7.3 Twaddell Scale ................................................................... 13 2.7.4 Brix Scale .......................................................................... 13 Worked Examples ................................................................................... 13 Exercises ................................................................................................. 32 vii viii CONTENTS
3 IDEAL GASES 35–73 3.1 Relation between Mass and Volume for Gaseous Substances ....... 35 3.1.1 Standard Conditions .......................................................... 35 3.1.2 Ideal Gas Law .................................................................... 35 3.2 Gaseous Mixture ............................................................................ 36 3.2.1 Partial Pressure (PP) .......................................................... 36 3.2.2 Pure Component Volume (PCV) ....................................... 36 3.2.3 Dalton’s Law ..................................................................... 37 3.2.4 Amagat’s Law (or) Leduc’s Law ....................................... 37 3.3 Average Molecular weight ............................................................. 38 3.4 Density of Mixture ......................................................................... 38 Worked Examples ................................................................................... 38 Exercises ................................................................................................. 70 4 VAPOUR PRESSURE 74–86 4.1 Effect of Temperature on Vapour Pressure .................................... 74 4.2 Hausbrand Chart ............................................................................ 75 Worked Examples ................................................................................... 75 Exercises ................................................................................................. 85 5 PSYCHROMETRY 87–110
5.1 Humidity ........................................................................................ 87 5.2 Definitions ..................................................................................... 87 Worked Examples ................................................................................... 90 Exercises ............................................................................................... 106 6 CRYSTALLIZATION 111–121 Worked Examples ................................................................................. 112 Exercises ............................................................................................... 120 7 MASS BALANCE 122–179 Worked Examples ................................................................................. 122 Exercises ............................................................................................... 173 8 RECYCLE AND BYPASS 180–195 8.1 Recycle ........................................................................................ 180 8.2 Bypass .......................................................................................... 180 8.3 Purge............................................................................................ 180 Worked Examples ................................................................................. 181 Exercises ............................................................................................... 195 CONTENTS
ix
9 ENERGY BALANCE 196–220 9.1 Definitions ................................................................................... 196 9.1.1 Standard State .................................................................. 196 9.1.2 Heat of Formation ............................................................ 196 9.1.3 Heat of Combustion ......................................................... 197 9.1.4 The Heat of Reaction ....................................................... 197 9.1.5 Heat of Mixing ................................................................ 197 9.2 Hess’s Law ................................................................................... 197 9.3 Kopp’s Rule ................................................................................. 198 9.4 Adiabatic Reaction Temperature ................................................. 198 9.5 Theoretical Flame Temperature ................................................... 198 Worked Examples ................................................................................. 198 Exercises ............................................................................................... 217 10 PROBLEMS ON UNSTEADY STATE OPERATIONS 221–228 Worked Examples ................................................................................. 221 Exercises ............................................................................................... 227 T ables .................................................................................................... 229–234 I Important Conversion Factors .............................................................229 II 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 .................................. 234 Answers to Exercises........................................................................... 235–245 Index...................................................................................................... 247–248
Preface The objective of this book is to enrich a budding chemical engineer the techniques involved in analyzing a process plant by introducing the concepts on units and conversions, mass and energy balances. This will enable him to achieve a proper design of process equipment. An attempt has been made to explain the principles involved through numerical examples. The problems are not only confined 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, CGS and MKS systems while designing equipment, since different reference books give standard values and data in various units. The book covers various interesting topics such as units and dimensions, mass relations, properties of gases, vapour pressure, psychrometry, crystallization, mass balance including recycle and bypass, energy balance and unsteady state operations. The second edition is now enriched with additional worked 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 the syllabus 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 and management of a process plant. It is, therefore, absolutely essential for them to be conversant with the mass and energy conservation techniques at every stage of the process to achieve economy in the design of process equipment in various units of the plant. This book aims at imparting knowledge of the basic chemical engineering principles and techniques used in analyzing a chemical process. By applying the relevant techniques, a chemical engineer is able to evaluate material and energy balances in different units and present the information in a proper form so that the data can be used by the management in taking correct decisions. Keeping this in mind, an attempt has been made to give a brief 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 examples incorporated 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 such that 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 can absorb the problems easily. We strongly feel that once the student understands the topics presented in this 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 universities offering 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 Sundar for 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. Srinivas and A. Varun for all their patience, cooperation and support shown during the preparation of this book. The encouragement received from his brothers and sisters and their family members is gratefully acknowledged. He also wishes to place on record the support received from his brothersin-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 baby M. 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 support received 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 of appreciation. We wish to acknowledge the support and encouragement received from the Head of Chemical Engineering 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 xvi ACKNOWLEDGEMENTS
We gratefully acknowledge all the well-wishers. Finally, we wish to thank the publishers, PHI Learning, New Delhi for encouraging us to bring out the second 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 involve changes 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 internationally we follow SI system of units, a chemical engineer is expected to be familiar and conversant with all the systems so far adopted for measuring and expressing various quantities. A review of literature and data over the years will be available in various units. These are used to express properties, process variables and design parameters in FPS, CGS, MKS and SI systems of Units. Hence, one has to be conversant with their use and applications. This chapter deals with the basic notations and conversion of 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 derived quantities. While handling these quantities, we come across different systems of units as mentioned earlier. Now let us see in detail these systems of units and their conversion from one unit to another. 1.2 BASIC UNITS AND NOTATIONS English 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 lb Length (L) 1 ft = 30.48 cm = 0.3048 m Time (t) 1 h = 3600 s Temperature (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 dynes Work/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 of 1 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 erg Heat Unit: 1 Btu = 0.252 kcal = 252 cal 1 cal = 4.18 J 1 J/s = 1 W 1.4 DEFINITIONS System. 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 reaction between 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 is completely detached from all other matter and energy. In an isolated system, the mass of the system remains 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 is temperature. WORKED EXAMPLES 1.1 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 = 11 1 2 (200)ʈ¥¥ ¥ m 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 °C h (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, 9C + 1 9C + 2
32 = F15 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 kcal 1 ft2 = 0.0929 m2 1 ft = 0.3048 m For DT °F = 1.8 DT °C ËÛ Btu/h ft = 0.5 'ÌÜ (ft)0.25 (a) ÌÜ ÍÝ (1.8 ' T °C) 1.25
2
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 difference in Celsius and diameter in metre as: Heat flux, kcal/h m2 = (m)0.25 1.25 °C) 'ÌÜ´ 2.713ÌÜ ËÛT
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 m DT = 18 °F, i.e. DT = 10 °C From Eq. (a), heat flux is = 0.5 (18)1.25 0.25
= 27.72 Btu/h ft 2 = 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 all units, 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 kg 29.25 lit = 29.25 ´ 10–3 m 3 227.27´ 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 the density in kg/m3 with pressure in N/m2. 1000 kg/m3 = 62.43 lb/ft3 14.7 psi = 1.0133 × 105 N/m2 1 kg/m3 = 62.43 × 10–3 lb/ft3 Let, 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 1211 log (p .¢) = 9.0305 – T52.2 EXERCISES 1.1 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, inches G = 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 an expression 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 F where q = Volumetric flow rate, ft3/s h = 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-temperature properties, is expressed by the empirical equation, h = 0.023 G 0.8 k0.67 C0.5/(D0.2m0.47 ) p
where G = Mass velocity of liquids, lb/ft2.s k = Thermal conductivity, Btu/ft.h.°F Cp = Specific heat, Btu/lb °F D = Diameter of tube, ft m = 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 chemical reaction 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 of CaCO3 reacts, 56 parts by weight of CaO and 44 parts by weight of CO2 are formed. Similarly, when 167.52 parts by weight of iron reacts with 72 parts by weight of steam (water), we get 231.52 parts by 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 specified amount of product. gram atom (or g atom) = Mass in grams/Atomic weight katom (or kg atom) = Mass in kg/Atomic weight gram mole (or g mole) = Mass in grams/Molecular weight kmole (or kg mole) = Mass in kg/Molecular weight 7
The conclusions based on reactions (2.1) and (2.2) on material balance can be expressed in other forms too, as per the definitions given above: 1 kmole of CaCO3 gives 1 kmole of CaO and 1 kmole of CO2 Similarly, 3 kmoles of iron reacts with 4 kmoles of steam (water) to yield 1 kmole of oxide and 4 kmoles of hydrogen. When such balances (on molar basis) are made, the number of moles on the reactants side need not be equal to the total numbers of moles on the product side. One atom of oxygen weighs = 16 grams O One atom of hydrogen weighs = 1 gram H One molecule of oxygen weighs = 32 grams O One molecule of hydrogen weighs = 2 grams H In other words, 16 grams of oxygen 32 pounds of oxygen 2 g atoms of oxygen 1 gram of hydrogen 2 kg of hydrogen 2 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 or pounds/Molecular weight = 1 g atom of oxygen = 1 lb mole of oxygen = 1 g mole of oxygen = 1 g atom 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 the basic 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 of mass, one should not apply it for the conservation of molecules. We frequently come across chemical reactions in which the total number of moles on the reactant side is not equal to the total number of moles 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 balance is 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 the conservation 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 or 22.414 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 times as 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 or quantities. One of the reactants will be present in excess and remain unreacted even when the other reactant has completely reacted. The reactant thus present in excess is termed excess reactant and the other 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 of product and conversion are always based on the limiting reactant. The amount by which any reactant is present in excess to that required to combine with the limiting reactant is usually expressed as percentage excess. The percentage excess of any reactant is defined as the percentage ratio of the excess to that theoretically required by the stoichiometric equation for combining with the limiting reactant. A limiting reactant is the one, which will not be present in the product, whereas the excess reactant 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 stoichiometry C + 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 oxygen alone is available, it is called the limiting reactant and carbon is called the excess reactant. For 32 kg of oxygen to react fully, it is sufficient to have 12 kg of carbon. However 6 kg of carbon is present in excess. Hence % excess of carbon is = 6¥ 100 = 50%. 12 2.5 CONVERSION AND YIELD These terms are used for a chemical reaction where the reactants give out new compounds or products.
Conversion is the ratio of the amount of material actually converted to that present initially, whereas the yield is the amount of desired product actually formed compared to that which can be formed theoretically. Conversion for a reaction is based on the limiting reactant whereas the yield is based on the product formed. 2.6 COMPOSITION OF MIXTURES AND SOLUTIONS Different 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 us consider 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 A and 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 Percent This is defined as the ratio of weight of a particular component to the total weight of the system in every 100 part, i.e. Weight % of A = WA ¥ 100 W This method of expressing composition is generally employed in solid and liquid systems and not used in gaseous system. One major advantage of weight percent is, its independence to changes in temperature and pressure. The composition of a solid mixture is to be always taken as weight % when nothing is mentioned above its units. 2.6.2 Volume Percent The ratio of the volume of each component and the total volume of the system, for each 100 part of the total volume is called volume percent Volume % of A = VA´ 100 V This method of expressing composition is employed always for gases, rarely for liquids and seldom for solids. The composition of a gas mixture is to be taken as volume % when nothing is mentioning about its units. The volume % is also equal to mole % for ideal gases but not for liquids and solids. This is based on Avogadro’s law. 2.6.3 Mole Fraction and Mole Percent
These concepts are generally adopted in the case of a mixture containing molecules of different species. WAMAMole fraction of A = WA WB MA MB Mole % of A = Mole fraction of A ´ 100 2.6.4 Atomic Fraction and Atomic Percent This is adopted when a mixture contains two or more atoms. WA Atomic fraction of A = AA WA WB AA AB Atomic % of A = Atomic fraction of A ´ 100 2.6.5 Composition of Liquid Systems In the case of liquids we come across more number of methods of expressing compositions of the liquid 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 solute For 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 ratio of density of a liquid to that of water. However, in the case of gases, it is defined as the ratio of its density 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 the case of liquids and gases the variation in density is significant. Similarly, the densities vary significantly with concentration also. This property of density and specific gravity varying with concentration is very widely used both in industries and markets as an index for finding the composition 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 are related to specific gravities and densities by arbitrary mathematical definitions. 2.7.1 Baume (°Be) Gravity Scale For liquids lighter than water
Degrees Baume’ = 140– 130 G G is 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 specific gravity. 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 of petroleum products. This is similar to Baume’ scale for liquids lighter than water. Degrees API = 141.5– 131.5G 2.7.3 Twaddell Scale This scale is used for liquids heavier than water. Degrees Twaddell (Tw) = 200 (G – 1.0) 2.7.4 Brix Scale This 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 at NTP? Molecular weight of propane (C3H8) is 44
1 g mole of any gas occupies 22.414 litres at NTP 500 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 of hydrogen and (b) 100 lb of hydrogen at standard conditions? (a) 100 kg of H2 = 50 kmoles of hydrogen volume occupied by 50 kmoles of hydrogen º 50 ´ 22.414 = 1120.7 m3 (b) 100 lb of H2 º 50 lb moles of hydrogen Volume 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 composition in 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 100.00
2.7 What is the weight of one litre of methane CH4 at standard conditions? 22.414 litres of any gas at NTP 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 12 Hydrogen 1 Nitrogen 14 Weight, g Weight, Rounding of Weight of g atom atoms each element 81.5 81.5/12 = 6.8 7 84 4.9 4.9/1 = 4.9 5 5 13.6 13.6/14 = 0.9 1 14 Total 103 Hence 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 7.665 MgO 7.0 40.3 0.1737 10.583 ZnO 9.7 81.4 0.1192 7.262 Al2O3 2.0 102.0 0.0196 1.194 B2O3 8.5 69.6 0.1221 7.439 SiO2 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 flowing through an equipment at the rate of 2.5 m3/min. Find (a) the average molecular weight of the gas mixture, (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 13.56 C2H6 30 30 900 76.27 H2 60 2 120 10.17 Total 100 — 1180 100 The average molecular weight = 1180= 11.8 100 Volumetric flow rate at standard conditions = 2.5¥ ¥ (273/288) = 3.555 m3/min Moles 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 an evaporator a dilute solution of 4% NaOH is concentrated to 25% NaOH. Calculate the evaporation of water per kg of feed. Basis: 1 kg of feed. NaOH present is 0.04 kg, which appears as 25% in the thick liquor formed Weight of thick liquor formed = 0.04 = 0.16 kg0.25 Weight of water evaporated = (1 – 0.16) = 0.84 kg Water 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. One engineer 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 composition also.
Basis: 100 g moles of flue gas Component 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 get x = 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 mole percent. Basis: 100 g of solution Component grams Molecular g mole Composition in weight mole % Na2CO3 40 106 40/106 = 0.377 0.377 ´ 100/3.71 = 10.16 Water 60 18 60/18 = 3.333 3.333 ´ 100/3.71 = 89.84 Total 3.710 100.00 2.14 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 kg Similarly, for getting 100 kg of H2 the amount of steam (H2O) required is = 100 t 72= 900 kg8 The total weight of reactants is 2094 kg Fe and 900 kg H2O = 2994 kg The 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 kg The total weight of reactants is (2094 kg Fe + 900 kg H2O) = 2994 kg Method 2 (Based on moles)
100 kg of H2 4 kmoles H2 comes from 50 kmoles of H2 comes from Weight of 37.5 katoms Fe 50 kmoles H2 from Weight of 50 kmoles H2O Moles of Fe3O4 formed is Weight of Fe3O4 formed is Total weight of reactants Total weight of products = 50 kmoles = 3 katoms of Fe (by stoichiometry) = ¦µ50 = 37.5 katoms Fe¨·4 = (37.5 ´ 55.84) = 2094 kg of iron t§¶
= 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 + 2CaSO4 310 (2 ¥ 98) 234 (2 ¥ 136) 506 506
One ton of raw calcium phosphate contains 0.935 tons of pure calcium phosphate Weight 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 used to get 10 kg of SO2? Cu +2H SO
24
→ CuSO +SO +2H O 422 63.54 64
64 kg of sulphur dioxide is obtained from 63.54 kg of copper. 10 kg of sulphur dioxide will be obtained from 9.93 kg of copper. 2.17 How much potassium chlorate must be taken to produce the same 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 oxygen 2.3 g of HgO will give (32 ¥ 2.3/433.2) = 0.1698 g of O2 0.1698 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◊ 3H2O First 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= 2.64 1931 % of H2O= 81 ¥100= 4.191931 % of MoO3 = 1728 ´100= 89.491931 % of P2O5=71100= 3.68´1931 Total = 100.00 2.19 How many grams of salt are required to make 2500 g of salt cake? How much Glauber’s salt can be 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 NaCl Hence, salt needed is 2058.45 g Glauber’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 KMnO4 are equivalent to 5 g K2Cr2O7? 2KMnO4 + 8H2SO4 + 10FeSO4 ® 5Fe2(SO4)3 + K2SO4 + 2MnSO4
+ 8H2O K2Cr2O7 + 7H2SO4 + 6FeSO4 ® 3Fe2(SO4)3 + K2SO4 + (Cr2SO4)3 + 7H2O 2KMnO4 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 KMnO \ 3 g K Cr O º 4 2 2 7 294 Similarly, 5 g KMnO 4
5 t 294=
7.75 g K2Cr2O7 º 189.6 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 standard condition? There are two possible reactions in this case: (a) Case I Fe H SO FeSO + H (55.85)
+→ 24 (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 (ii)(6) (111.7) +→ 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 litres 2.22 A natural gas has the following composition by volume CH4 : 83.5%, C2H6: 12.5%, and N2: 4%. Calculate the following: 42
24 2 43 2
(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 = 73.29 C2H6 30 12.5 12.5¥30 = 375 375¥ 100/1823 = 20.57 N2 28 4.0 4.0¥28 = 112 112¥100/1823 = 6.14 Total 1823 100.0
(c) Average molecular weight = 1823= 18.23 100 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 the solution at this temperature is 1.148 g/cc. Find the composition in (a) weight % (b) volume % of water (c) mole % (d) atomic % (e) molality and (f) g NaCl/g water. Basis: (a) 1 litre of solution has a weight of 1148 g Component Molecular Weight, g Weight, %g mole mole % weight NaCl 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 54.93 100.00
(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 is 1 g/cc) Volume % = 91.8% (d) Element g atoms Atomic % Na 3.93 2.443 Cl 3.93 2.443 H 102.00 63.409 O 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 is present in 918 g of water (i.e.) 4.28 g moles/1000 g of solvent Molality = 4.28 Molarity = Moles of solute per litre of the solution = 3.93 (f) g NaCl/g water = 230 = 0.252918 2.25 A benzene solution of anthracene contains 10% by weight of the solute. Find the composition in terms 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 1.00 Molality = 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 a solution of 1.8 molality. Density of this solution is 1.06 g/cc Molality = g moles of NaCl/1000 g of water = 1.8 or, 1.8 g moles NaCl = (1.8 ¥ 58.46) = 105.228 g Component Weight, g Weight % NaCl 105.228 9.52 H2O 1000.000 90.48 1105.228 100.00 Density of this solution = 1.06 g/cc Volume of this solution, i.e. mass/density = 1105.228= 1042.67 cc 1.06 or 1042.67 cc of this solution contains 105.228 g of NaCl 1000 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 of NaCl 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 solution It will have 20 lb NaCl and 80 lb water 80 lb water = 1.28 ft3 (since the density of water is 62.47 lb/ft3) We know that 1 ft3 = 7.48 gallons Therefore, 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 sample is 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 ammonium nitrate. 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 mixture contains 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 of nitrobenzene and spent acid produced. The reaction is C6H6 + HNO3 ® C6H5NO2 + H2O Feed, C6H6 : 663/78 = 8.5 kmole HNO3 : 31.5% of 1700 kg = 535.5 kg = 8.5 kmoles H2SO4 : 60 % of 1700 kg = 1020 kg = 10.408 kmoles H2O : 8.5 % of 1700 kg = 144.5 kg = 8.028 kmoles Reaction is 95% complete Hence, HNO3 unreacted : 0.05 × 8.5 = 0.425 kmole C6H6 unreacted : 0.05 × 8.5 = 0.425 kmole H2SO4 unreacted : 10.408 kmoles H2O unreacted : 8.028 kmoles H2O 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 1.133 H2SO4 10.408 98 1020.000 43.165 H2O (8.075 + 8.028) = 16.103 18 289.850 12.266 Nitrobenzene 8.075 123 993.225 42.032 C6H6 0.425 78 33.150 1.403 Total 2363.00 100% Nitrobenzene produced = 993.225 kg Spent acid = 26.775 + 1020.000 + 289.85 = 1336.63 kg 2.33 A sample of caustic soda flake contains 74.6% Na2O by weight. Estimate the purity of flakes. Reaction is as follows: 2NaOH ® Na2O + H2O Amount 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 MgCO3 100 kg of CaCO3 gives 56 kg of CaO Therefore, 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 MgCO 3
gives ©¸ ª¹× (2 – x) kg of MgO «º The weight of product left behind is 1.1 kg, i.e. weight of MgO + CaO left behind 0.56x + (0.4672)(2 – x) = 1.1 0.0838x = 1.1 – 0.96524 Therefore, x = 1.761 kg Component Weight, kg Weight, % CaCO3 1.761 88.05 MgCO3 0.239 11.95 Total 2.000 100.00 2.35 The composition of NPK fertilizer is expressed in terms of N2, P2O5 and K2O each of about 15 weight %. Anhydrous ammonia, 100% phosphoric acid and 100% KCl are mixed to get 1 ton of fertilizer. Estimate the amount of filler in the NPK fertilizer. Basis: 1000 kg of fertilizer Reactions 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 kg 28 150 H = 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 specific gravity of A is 0.7, find the specific gravity of B. Basis: 1000 kg of solution
Weight of A: 350 kg Weight of B: 650 kg Volume of solution = 1 m3 (since density is 1000 kg/m3 due to specific gravity being unity) Mass/volume = density Assuming ideal behaviour 350 650= 1700 SB rB = 1300 kg/m3 Therefore, specific gravity of B = 1.3 2.37 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 solution Volume of A in solution = 1 × 0.47 = 0.47 m3 Weight of A = 0.47 × 1250 = 587.5 kg Volume 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 1.3 g/cc. Find the composition in molarity and molality. Basis: 100 g of solvent Weight of K2CO3 = 43 g Weight of solution = 143 g Density of solution = 1.3 g/cc Volume of solution = 143 = 110 cc 1.3 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.0.1 2.39 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 %, average molecular weight and density. Basis: 100 kmoles of mixture Compound mole % Molecular weight Weight, kg Weight %
C2H4 30.6 28 856.8 22.00 C6H6 24.5 78 1911.0 49.07 O2 1.3 32 41.6 1.07 CH4 15.5 16 248.0 6.37 C2H6 25 30 750.0 19.26 N2 3.1 28 86.8 2.23 Total 3894.2 100.00 Density = 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 by weight. Find the molecular formula of this compound. Basis: 100 g of compound Compound Weight, Atomic weight Number g or Molecular of moles weight Mg 9.76 24 0.410 S 13.01 32 0.410 O 26.01 16 1.615 H2O 57.22 18 2.8738 Converting to whole numbers dividing by 0.41 1 1 3.94 6.92 Therefore, molecular formula of the compound = MgSO4.7H2O 2.41 A substance on analysis gave 1.978 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 whole numbers dividing by 9.156 × 10–3 Ag 1.978 108 S 0.293 32 O2 0.587 16 Molecular formula = Ag2SO4 0.0183 2 9.156 × 10–3 1 0.0367 4.04
2.42 Two engineers are estimating the average molecular weight of gas containing oxygen and another gas. One uses the molecular weight as 32 and finds the average molecular weight as 39.8 and the other 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 of oxygen, the value is 39.8. Let x be the mole fraction of oxygen in the mixture and the molecular weight of the other gas be M 39.8 = (x) × (32) + (1 – x) × (M) 33.4 = (x) × (16) + (1 – x) × (M) Solving, we get x = 0.4 i.e the fraction of oxygen in the mixture is 0.4 2.43 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 the composition of initial mixture. Reactions involved are: 2FeO + 0.5 O2 ® Fe2O3 2Fe3O4 + 0.5 O2 ® 3 Fe2O3 Basis: 100 kg of feed mixture Let X be the weight of FeO in the mixture From 144 kg of FeO, Fe2O3 formed is 160 kg Therefore, from X kg of FeO, Fe 2
O 3
formed is 160 × X 144 From 232 kg of Fe3O4, Fe2O3 formed is 480 kg Therefore, from (100 – X ) of Fe 3
O 4
,Fe 2
O 3
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) 105 144 464 Solving, X, the weight of FeO = 20.25 kg Fe3O4 = (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 stone 100 kg of CaCO3 will have 56% CaO If 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 81 SiO2 14 H2O5 Compound Weight % Molecular weight moles mole % MgCO3 81 84 0.964 65.34 SiO2 14 60 0.233 15.82 H2O 5 18 0.278 18.83 2.47 The concentration of H3PO4 is expressed in terms of P2O5 content. If 35% P2O5 is reported, find the 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 the reaction shown below: PbS + O2 ® Pb + SO2 (1) PbS + 2O2 ® PbO2 + SO2 (2) Estimate (i) unreacted PbS, (ii) % excess oxygen 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 comes from 239.2 × 6 = 6.927 kg of PbS207.2 239.2 kg of PbO2 comes from 239.2 kg of PbS Therefore, 1 kg of PbO2 comes from 1 kg of PbS Therefore, total PbS reacted [from Eqs. (1) and (2)] = 6.927 + 1 = 7.927 kg Unreacted PbS = 10 – 7.927 = 2.073 kg O2 required for this process From 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 kg 207.2 From Reaction 2: 239.2 kg of PbO2 requires 64 kg of oxygen Therefore, 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 O 2
supplied = ©¸ ª¹× 100 = 151%«º Amount of SO2 formed If 207.2 kg of Pb is formed, SO2 formed is 64 kg If 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 kg If 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, 0.5 kmole of B and 10 wt of % C. The molecular weights of A, B and C are 40, 50 and 60 respectively and their densities are 0.75 g/cc, 0.8 g/cc and 0.9 g/cc respectively. Express the composition in weight %, mole %. Also give its average molecular weight and density assuming ideal behaviours. Let the weight of mixture be W kg
Weight of A = 11 kg Weight of C (10%) = 0.1W kg Weight of B = W – 11 – 0.1W = 0.5 kmole = 0.5 × 50 = 25 kg i.e. weight of B = W – 11 – 0.1W = 25 kg 0.9W = 36 kg W = 40 kg i.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 0.0147 B 25 62.5 50 0.500 59.38 800 0.0313 C 4 10.0 60 0.067 7.96 900 0.0044 Total 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) EXERCISES 2.1 How many g moles are equivalent to 1.0 kg of hydrogen? 2.2 How many kilograms of charcoal is required to reduce 3 kg of arsenic trioxide? As2O3 + 3C Æ 3CO + 2As 2.3 Oxygen is prepared according to the following equation: 2KClO3Æ 2KCl + 3O2. What is the yield of oxygen when 9.12 g of potassium chlorate is decomposed? How many grams of potassium chlorate must be decomposed to 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 ratio and mole percent. 2.6 A solution of caustic soda in water contains 20% NaOH by weight. The density of the solution is 1196 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 kg methanol 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 this solution 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 by microorganisms. The initial concentration of sugar determines the final alcohol content and sweetness of the wine. The general convention is to adjust the specific gravity of the starting stock to achieve a desired 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 ® 2NH3 In 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 according to 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. For a 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 the laboratory can be prepared by the reaction of sulphuric acid with zinc metal H2SO4 (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/h of hydrogen at standard conditions. Assume all the acid used reacts completely. 2.16 Sulphur dioxide may be produced by the reaction Cu + 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 shown below: Al2O3 + 3 H2SO4 ® Al2 (SO4)3 + 3 H2O Bauxite ore contains 55.4% by weight Al2O3, the reminder being impurities. The sulphuric acid contains 77.7% H2SO4, the rest being water. To produce crude aluminium sulphate containing 1798 kg 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 % and mole %. 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 the gas 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 SUBSTANCES 3.1.1 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 under standard conditions will occupy 359 cu.ft. T(K) = T °C + 273.16 T(°R) = T °F + 459.69 3.1.2 Ideal Gas Law The ideal gas law states that, PV = nRT P = Pressure of gas V = Volume of n moles of gas n = Number of moles of gas R = Gas constant T = Absolute temperature Using the ideal gas law (PV = nRT) and the above information one can always determine the weight of a gas if the volume is known and vice-versa. Parameters Pressure Molar volume Absolute temperature Gas constant
Normal Temperature and Pressure/Standard Conditions English Metric SI 1.033 kgf/cm2 1.01325 bar 359 ft 1 atm3/lb mole 22.414 m3/kmole 22.414 m3/kmole 491.69 273.16 K 273.16 K 0.73 atm ft oR 3/lb mole oR 0.085 kgf m3/kmole K 0.083 Bar m3/kmole K 35
Pressure 1 atm = 1.033 kgf/cm2 = 1.01325 bar 1 atm = 14.67 psia = 105 N/m2 = 760 mm Hg 1 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 °R nT R = 10.73 lbf ft3/in2 lb mole °R Avogadro’s Number: 6.023 ´ 1023 molecules per g mole 2.73 ´ 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, and Pa. 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 K or °R. Thus, the gas constant is a dimensional quantity. The following table gives the gas constant in different units. Temperature Pressure Volume Gas constant ‘Rg’ R psia in3 18.51 R psia ft3 10.73 R atmospheres ft3 0.73 KPa m3 8314 K atmospheres m3 0.08206 K atmospheres cm3 82.06 K cm Hg cm3 6239.79 Units of gas constant in3psia/lb mole R ft3psia/lb mole R ft3atm/lb mole R m3Pa/kmole K m3atm/kmole K cm3atm/g mole K (cm3cm Hg)g mole K Rg = 8.314 J/(g mole) (K) = 1545 ft lb/lb mole °R 3.2 GASEOUS MIXTURE 3.2.1 Partial Pressure (PP) The partial pressure of a component gas that is present in a mixture of gases is the pressure that would be exerted by that component gas if it alone were present in the same volume and at the same temperature 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 be occupied by that component gas if it alone were present at the same pressure and temperature as the mixture. 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 Law The total pressure (P) exerted by a gaseous mixture in a definite volume is equal to the sum of partial pressures. pA + pB + pC = P where pA, pB, pC, … represent partial pressure of components A, B, C, … . 3.2.4 Amagats Law (or) Leducs Law The total volume (V ) occupied by a gaseous mixture is equal to the sum of the pure component volumes VA + VB + VC = V VA, VB, VC, … stand for pure component volume of components A, B, C, … Where ideal gas law is applicable; n RT; p B
= Bn RT; pC = Cn RT(a) pA = A
VVV Adding all the partial pressures of A, B and C, we have, RT P = p A
+ p B
+ p C
= ËÛ ÌÜ ´ (nA + nB + nC)ÍÝ Dividing, pRT V nA ;A/P = V ()RT n×× ++n A BC
pressure fraction = mole fraction. (b) PVA = nART; PVB = nBRT; PVC = nCRT Adding P(VA + VB + VC) = RT (nA + nB + nC) = nRT = PV nRTP = N A VA AV = Dividing, ()AB Cn RT ++ or, VA = NA × V 3.3 AVERAGE MOLECULAR WEIGHT The weight of unit mole of the mixture is called average molecular weight, which is also equal to total 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 kmole Number of moles of nitrogen = 79 and those of oxygen = 21 Weight of a component = Number of moles ¥ respective molecular weight \ Weight of nitrogen = 79 ¥ 28 = 2212 kg Weight of oxygen = 21 ¥ 32 = 672 kg 2884 kg \ The weight of 1 kmole = 2884/100 = 28.84 kg Hence, the average molecular weight of air = 28.84 3.4 DENSITY OF MIXTURE Density is defined as the weight of a mixture per unit volume and is independent of temperature. As the volume of liquids and gases is a strong function of temperature, density also varies significantly with temperature for a specified composition. However, in the case of solids the variation of density with 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¯ ˯ 760 530 ÈØÈØ = ÉÙÉÙ ÊÚÊÚ = 33.4 ft3 3.3 Calculate the weight of 200 cu.ft. of water vapour at 15.5 mm Hg and 23 °C Basis: 200 ft3 of gas at given condition PV ÈØ ÈØT0Volume at standard condition =11 ÉÙ ÉÙ ÊÚ ÈØ15.5 273= 3.76 ft = ÉÙ296ÊÚ 3.76 Moles of water = 01
3
ÉÙ ÈØ = 0.01047 lb mole ÊÚ Weight of water = (0.01047 ´ 18) = 0.18846 lb 3.4 It is desired to compress 30 lb of CO2 to a volume of 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 given condition = 20 ´ 0.02832 = 0.5664 m3 PV ÈØT1Pressure
at the given condition “P1”= 00ÈØ
ÉÙ ÉÙ ÊÚ ÊÚ 6.9528 303 ÈØÈØ =1 ÉÙÉÙ273ÊÚÊÚ = 13.62 atm 3.5 Calculate the maximum temperature to which 10 lb of nitrogen enclosed in a 30 ft3 chamber may be heated without exceeding 100 psi pressure. Basis: 10 lb of N2 = 10/28 = 0.357 lb mole Volume 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)
Pressure Temperature Volume 720 mm Hg 760 mm Hg 373 K 273 K 11,450 cc ? PVÈØ ÈØ0Volume
at standard condition = 11
ÉÙ ÊÚ ÊÚ 720 273 = 11450ÈØÈØ ÉÙÉÙ373ÊÚÊÚ = 7939.2 cc Therefore, 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 gas Component Volume % = Molecular g mole Weight, g mole % weight 01
CO2 13.1 44 0.131 0.131 ´ 44 = 5.764 O2 7.7 32 0.077 0.077 ´ 32 = 2.464 N2 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 oxygen that contains 11.1% of hydrogen by weight. Basis: 1 lb of gas mixture Component Hydrogen Oxygen Weight % Molecular weight lb mole 0.111 2 0.111/2 = 0.0555 0.889 32 0.889/32 = 0.0278 Total 0.0833 lb moles Temperature = 30 °C = 86° F = 546° R Volume of the gas at the given condition = 0.0833 ´ 359 ´ (29.92/29) 3
´ (546/492) = 34.24 ft 3
Density = (1/34.24) = 0.0292 lb/ft 3.9 Calculate the density in g/litre at 70 °F and 741 mm Hg of air Basis: 1 g mole of air Component Volume % = mole % Molecular weight Weight, g Oxygen 0.21 32 6.72 Nitrogen 0.79 28 22.12 Total 28.84 g Volume of air = 1
´ 22.414 ´ 530ÈØÈ Ø760 ÉÙÉ Ù = 24.8 litresÊÚÊ Ú 28.84 Density of air = ÉÙ ÈØ = 1.162 g/litre ÊÚ 3.10 In 1000 ft3 of a mixture of hydrogen, nitrogen and carbon-dioxide at 250 °F, the partial pressures are 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 of H2; (e) volume fraction and volume % of H2; (f) weight of H2; (g) weight fraction and weight % of H2; (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 ´ 0.26 492 1.00 710 = 180.169 ft3 180.169 ÈØ = 0.502 lb moles = 2 ÉÙlb moles of HÊÚ (b) Total pressure = (0.26 + 0.32 + 1.31) = 1.89 atm Total 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 H 2
= 0.502 ´ 359 ´ ÈØÈ Ø = 138 ft 1.89ÉÙÉ Ù492ÊÚÊ Ú (e) Volume fraction of hydrogen = 138 = 0.1381000 3
Thus volume % = pressure % = mole % (f) Weight of hydrogen = 0.502 ´ 2 = 1.004 lb (g) Basis 100 lb moles of gas mixture Mole fraction of nitrogen = 0.32 = 0.169 189 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 0.78 Nitrogen 28 16.9 16.9 16.9 ´ 28 = 473.2 13.33 Carbon 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 = 35.5 ÊÚ (i) 1000 ft 3
of gas at given condition º 1000 ´ 492 1.89 ÈØÈ Ø ÉÙÉ Ù ÊÚÊ Ú = 1309.7 ft3 at NTP condition Number 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 is burnt 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. What is the compound? Basis: 1 mole of hydrocarbon. Let the hydrocarbon be CxH y
y y O2 ® xCO2 + ÈØ H2OÉÙ4 ÉÙCxHy + xÈØ ÊÚ ÊÚ ÈØy® xyMoles: 1 x ÉÙ4 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+7 x + ÈØ ÉÙ = 6x + 3yÊÚ (7 x –6 x )+ 7yÈØ ÉÙ = –7ÊÚ x – 5y = –74
Since the hydrocarbon has carbon atoms less than 5, we have x < 5; solving above equation assuming carbon 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; – 45 x = 3; – 5y = –10; y = 40 = 84 5 5y = –11 y = 44x = 4 – 45 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 and the gases leave at 85 °C and 740 mm Hg with the following composition N2: 48.3%, O2: 4.4% and H2O : 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 entering 473ÈØÈ 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 = 164 g 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 K PV = nRT, where R = 8.314 J/g mole K = 8.314 (Pa) (m3)/g mole K PV = 112 10 n = 1000= 33,678 g molesRT 8.314 400 3
Alternatively , P1 = 1.01325 ´ 105 N/m2 T1 = 273 K since PVPV T= T V 11
22
12
1
(at NTP) = 112 103
tt1000 273= t
754.4 m3 400 1.01325 t105 1000 g moles of gas occupies 22.414 m3 at NTP, Hence, 754.4 m 3
contains ÉÙ ÈØ1000 = 33,658 g moles of nitrogen ÊÚ (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 g moles 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 by volume. This gas is passed through an absorption system in which 98% of the HCl is removed. The gas enters the system at 48.8 °C and 743 mm Hg and leaves at 26.7 °C and 738 mm Hg. Applying the pure 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 of air = 75 litres Pure component volume of HCl = 25 litres Pure component volume of HCl absorbed = (25 ´ 0.98) = 24.5 litres Pure component volume of HCl remaining = 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 0.66 Air 75.0 99.34 Total: 75.5 100.00 Volume 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 the Deacon process enters the absorption apparatus at 740 mm Hg and 75 °F. The partial pressure of Cl2 is 59 mm Hg and the remainder being inert gas. The gas leaves at 80 °F and 743 mm Hg with Cl2 having 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 entering Partial 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) 681 299.7 Volume
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 = 100 59 273 = 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 reaction takes place. 4NH3 + 5O2 ® 6H2O + 4NO. The gases from this process are passed into towers where they are cooled and the oxidation is completed according to the reactions: 2NO + O2 ® 2NO2 3NO2 + H2O ® 2HNO3 + NO The NO liberated is re-oxidized in part and forms more nitric acid in successive repetitions of the above 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 complete oxidation of ammonia to nitric acid. The gases leave the catalyst at 700 °C and 743 mm Hg. Given the overall 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 of gases 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 tower is oxidized to acid. NH 3 Catalyzer
Air Exit gas Absorber HNO3
Basis: 1 g mole of NH3 overall reaction is NH3 + 2O2 ® HNO3 + H2O O2 required is 2 g moles But O2 supplied is 2 ´ 1.2 = 2.4 g moles i.e. air supplied is 2.4 = 11.42 g moles 0.21 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 = ÉÙ4ÊÚ O2 leaving = (2.4 – 1.06) = 1.34 g moles NO 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 12.635 mole % = volume % 71.40 10.60 1.20 6.70 10.10 100.000 (d) Volume of gases leaving the catalyzer = 12.635 973 ¥760= 1031.8 litres¥ 22.414 ¥ 273 243 This is the volume of gases leaving for 24.2 litres of ammonia entering Therefore, for 100 litres of NH3 entering = (1031.8 ¥ 100)/24.2 = 4264 litres of gas leaves (e) NO entering the tower = 0.85 g mole NO converted = 0.85 ¥ 0.9 = 0.765 g mole HNO3 produced = 0.765 g mole = (0.765 ¥ 63) = 48 g For 100 litres of NH3 weight of HNO3 produced = 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 kg C19H36O2 + H2Æ C19H38O2 296 2 298
296 kg of ester reacts with 2 kg of H2 168000 kg of ester reacts with 2 ¥168000= 1135 kg of hydrogen 296 1 kmole of any gas occupies 22.414 m3 at NTP 1 ÈØ ´ÈØH initially present in the 2 10
cylinder (10 m3) = (1) ´ÉÙ ÉÙ ÊÚ ÊÚ = 0.45 kmole (by reducing to
NTP condition) 12 H 2
after use = ÈØÈØ ÉÙÉÙ = 0.09 kmoleÊÚÊÚ \ H2 available from one cylinder = 0.36 kmole 1135 H 2
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. 1000 ft3 of gases enter the cooler per hour at 100 °C and 100 mm Hg (gauge). The partial pressure of toluene is 300 mm Hg. 740 ft3 of gases leave cooler per hour at 40 mm Hg and 50 °C. Calculate the weight of toluene condensed per hour. Vapour pressure of toluene at 50 °C = 90 mm Hg. Basis: One hour 100 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 much water is removed per 1000 ft3 of entering air? Basis: 1000 ft3 of entering air.
Moles of air entering = 1000ÈØÈ Ø492 = 1.43 lb moles 359ÉÙÉ Ù960ÊÚÊ Ú ÈØ50 Moles of water in it = = 0.094 lb moleÉÙ760ÊÚ Moles of dry air = (1.43 – 0.094) = 1.336 lb moles Moles of water leaving = ÈØ ÉÙ 10 = 0.0178 lb mole ÊÚ Water condensed = (0.094 – 0.0178) = 0.0762 lb mole = 1.37 lb 3.20 Chimney gas has the following composition: 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 at condition 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 13.978 CO 28 0.2 (0.2 ´ 28) = 5.6 0.187 O2 32 9.6 (9.6 ´ 32) = 307.2 10.273 N2 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 kmole 29.904 303 ÈØ´ 22.414´ Volume at 30 °C, 760 mm Hg = 0.01672 3 ÉÙ ÊÚ= 0.416 m 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 the following 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 of gas burnt. Basis: 100 kmoles of producer gas. CO = 23 kmoles O2 needed = 23/2 = 11.5 kmole; O2 in feed = 2.6 kmoles O2 actually needed = (11.5 – 2.6) = 8.9 kmoles Air supplied = 8.9 1.25= 52.98 kmoles 0.21 O2 supplied = 52.98 ´ 0.21 = 11.125 kmoles N2 in air = 52.98 ´ 0.79 = 41.85 kmoles 298ÈØÈ Ø760Volume of feed = 100 ´ 22.414 ´ ÉÙÉ Ù 273 750ÊÚÊ Ú 3 = 2479.28 m Volume of air = 52.98 ´ 22.414 ´ 298ÈØÈ Ø760 = 1313.43 m3ÉÙÉ Ù ÊÚÊ Ú (a) Volume of air/100 m 3 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 N2 leaving = (70 + 41.85) = 111.850 kmoles 141.475 kmoles Composition of gases leaving (volume % = mole %): CO2: 19.37%, O2: 1.57% and N2 : 79.06% Volume of gases leaving = 141.475 ´ 22.414 ´ 623ÈØÈ Ø760 273ÉÙÉ Ù750ÊÚÊ Ú = 7332.92 m3 Volume of gases leaving/100 m 3 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 is piped 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) Density Basis: 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) = 20.777 C2H6 3.0 1.38 ´ 0.03 = 0.042 (0.042 ´ 30) = 1.260 N2 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 lb of gas at 120 °F. What is the pressure? Basis: 1 lb of O2 = 1/32 = 0.031 lb mole; R = 0.73 ft3 atm/lb mole °R T = 120 + 460 = 580 °R P= nRT= 0.031 ´ 0.73 ´580 = 26.5 atm V 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 50 psig, 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 11 22
T ()( ) ( )460);600°R TT P2 49.6712 1
\ Temperature = 140 °F 3.26 Calculate densities of C2H6 and air at NTP. At NTP 1 g mole occupies 22,414 cc. Density of C 2H6 = 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 reaction CaC2 + 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 of gas/hour at 75 °F and 743 mm Hg. Basis: 1 lb of CaC 1 = 0.0156 lb mole. = 2 64 CaC2 + 2H2O ® C2H2 + Ca(OH)2 64 36 26 74
Volume 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 a lamp burning gas 40 litres h. Basis: 1 kg CaC2 = 1 = 0.0156 kmole.64 Volume of C 2
H 2
= 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 following composition 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 74.0 Br2 160 28 (28/160) = 0.175 13.7 O2 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
= 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 ÉÙ
3.29 A mixture of NH3 and air at 730 mm Hg and 30 °C contains 5.1 NH3 by volume. This gas is passed at a rate of 100 ft3/min. through an absorption tower in which NH3 is removed. The gases leave 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 NH 3
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 NH 3
absorbed at NTP = 5.0518 ´ 730ÈØÈ Ø273 ÉÙÉ Ù ÊÚÊ Ú = 4.372 ft3 Moles of NH3 = 4.372= 0.012 lb mole = 0.207020 lb359 3
3.30 1000 ft3 of moist air at 740 mm Hg and 30 °C contains water vapour in such proportions that its partial pressure is 22 mm Hg. Without the total pressure being changed, the temperature is reduced to 15 °C and some water condenses. After that the partial pressure of water is 12.7 mm Hg. Using partial 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 °C Partial pressure of water = 22 mm Hg Partial pressure of air = (740 – 22) = 718 mm Hg Partial 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) (12.7 mm, 15 °C) 12.7ÈØÈ Ø303Volume of air leaving at inlet condition = 938 ´ÉÙÉ Ù ÊÚÊ Ú = 570 ft3 Volume 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% and N2 : 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 is used. (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 gas i.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) O 2
required for the combustion of CO =
ÉÙ ÈØ = 11.5 lb moles ÊÚ O2 available in feed = 2.6 lb moles Theoretical O2 required = (11.5 – 2.6) = 8.9 lb moles O2 supplied = (8.9 ´ 1.2) = 10.68 lb moles 100 ÈØ = 50.85 lb moles´ Air supplied = 10.68 ÉÙ 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 ft 3 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 N2 entering along with feed N2 Total Total Composition of CO2 Volume % 19.66 = 70.00 lb moles = 110.17 lb moles = 139.35 lb moles O2 N2 1.28 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 coke charged is burnt. Air supplied is 30% in excess of that required for complete combustion. It may be assumed 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 in ft3/min. Basis: 100 lb of coke. C + O2 ® CO2 C : 89.1 lb. Ash : 10.9 lb. Carbon burnt = 89.1 ´ 0.9 = 80.19 lb Carbon 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 O 2
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 0.43 O2 3.07 6.67 N2 36.31 78.83 Total 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/h i.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 flattened electric arc. Some of the nitrogen is oxidized to NO, which on cooling, oxidizes to NO2. Of the NO2 formed, 66% will be associated to N2O4 at 26 °C. The gases are then passed into water washed absorption towers where nitric acid is formed. 3NO2 + H2O ® 2HNO3 + NO NO 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 is 2% by volume while hot. The gases are cooled to 26 °C at 750 mm Hg before entering the absorption column. (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 entering the absorption apparatus. (c) Calculate the weight of acid formed per 1000 litres of gas entering the absorption system if the combustion to HNO3 of the combined nitrogen in the furnace gases is 85% complete. H2O Air ArcNO CoolerNO2 ChamberN2O4 Abs. N ,O 2% 26 °C,Col. 750 mm 2 2 HNO3
Basis: 100 g moles of air contain N2: 79 g moles and O2: 21 g moles. (a) Reaction in arc furnace is given as
N2 + O2 ® 2NO which 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 ® NO2 2 g moles of NO gives 2 g moles of NO2 O2 remaining = (20 – 1) = 19 g moles Reaction in chamber is 2NO2 ® N2O4 66% of NO2 remains as N2O4 Moles of NO2 associated to N2O4 = (2 ´ 0.66) = 1.32 \ N2O4 formed = 0.66 g mole NO2 remaining = 2 – 1.32 = 0.68 g mole NO2 N2O4 O2 N2 Total Total moles: 0.68 + 0.66 + 19 + 78 = 98.34 Partial 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 mole in N2O4 – (0.66 ´ 2) = 1.32 g moles/2 g moles Thus, 3NO2 + H2O ® 2HNO3 + NO Total moles of gas in 1000 litres of entering gas PV= RT 750 1000 = ËÛ ÈØ ÉÙÌÜ760ÊÚÍÝ = 40.25 g moles Moles 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 NO 2
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 the following reaction takes place. CnH2n+2 + nH2O ® nCO + (2n + 1)H2 (1) CO + H2O ® CO2 + H2 (2) Reaction (1) is 95% complete and 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 of gases 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 58 Weight, lb 669.68 2374.24 570.6 441.38 4055.9 (a) Average molecular weight of the gas leaving stabilizer = 4055.9 = 21.32 190.24 (b) CH 38 + 3H O2 → 3CO + 7H2 44 54 84 14
CH +HO 42 → CO + 3H2 16 18
28 6
26 2 →CH 30 3656 10
CH
+2H O 2CO+5H2
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 from CH4 = (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 = 1808 lb/h (d) Gases leaving unreacted = HC, H2O, CO, CO2, H2 H2 = 1808.000 lb C3H8 15.22 ´ 0.05 ´ 44 = 33.484 lb CH4 148.39 ´ 0.05 ´ 16 = 118.712 lb C2H6 19.0 ´ 0.05 ´ 30 = 28.530 lb C4H10 7.61 ´ 0.05 ´ 58 = 22.069 lb CO 249.4 ´ 0.1 ´ 28 = 698.320 lb CO2 249.4 ´ 0.9 ´ 44 = 9876.240 lb H2O 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 lb Component C3H8 CH4 C2H6 C4H10 CO CO2 H2OH2Total Weight, 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 6.823 13.385 100.000
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) the average 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 = 1,088 CO2 44 30 30 ´ 44 = 1,320 NH3 17 2 2 ´ 17 = 34 Total — 100 2,442 Average molecular weight = 2, 442= 24.42 (b) Volumetric flow rate at standard condition 100 273ÈØÈØ2 = 600 = 1081.2 m3/h´303ÉÙÉÙ1ÊÚÊÚ 1081.2 Molar flow rate of gases = ÉÙ ÈØ = 48.24 kmoles/h ÊÚ 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 gases leaving. Basis: 4 kmoles HCl gas. O2 needed = 1 kmole O2 supplied = 1.3 kmoles 79 N 2
entering = 1.3 ´ ÈØ ÉÙ = 4.89 kmolesÊÚ O2 remaining = (1.3 – 0.8) = 0.5 kmole HCl un-reacted = (4 ´ 0.2) = 0.8 kmole Cl2 formed = (2 ´ 0.8) = 1.6 kmoles Composition of dry gases leaving is presented in the following table Gas mole mole %
HCl 0.80 10.27 O2 0.50 6.42 N2 4.89 62.77 Cl2 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 passed through an absorption tower at the rate of 100 m3/h where NH3 is removed. The gases leave the tower at 725 mm Hg and 20 °C having 0.05% NH3. Calculate (a) the rate of flow of gas leaving the tower and (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 Hg 5.1% NH3
Basis: One hour of operation Moles of gas entering per hour = 100730 273ÈØÈØÈØ 22.414ÉÙÉÙÉÙ303ÊÚÊÚÊÚ = 3.86 kmoles NH3 entering = 3.86 ´ 0.051 = 0.197 kmole Air entering = (3.86 – 0.197) = 3.663 kmoles 3.663 kmoles of air contains 0.05% NH3 while leaving the absorber Therefore the total moles of gases 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 kg Volumetric 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 = 1.0556 CH4 ®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 kmoles O2 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 81.54 C2H6 1.2 18.46 Total 6.5 100.00 3.39 A furnace is fired with coke containing 90% carbon and 10% ash. The ash pit residue after being washed 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 kmoles From the foregoing, air supplied = 78.6/0.79 = 99.5 kmoles; O2 in air supplied is = 20.9 kmoles Furnace Flue gas Coke (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 in flue gas = 15 katoms Carbon balance: 0.9F = (15 ´ 12) + 0.1P (1) Ash balance: 0.1F = 0.4P (2) (i.e.) F = 4P Substituting 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 ® 2CO C + O2 ® CO2; CO + ½O2 ® CO2 Carbon 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 kmoles O2 needed (theoretically): Total Carbon fed = 205.72 × 0.9 = 185.148 kg = 15.429 katoms O2 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 kg 205.72 kg of coke gives 100 kmoles of gas 100 1000 kg of coke gives 1000 ´ ÈØ
= 486.1 kmoles of the gasÊÚ Aliter For 1000 kg of coke fed, carbon in coke: 1000 × 0.9 = 900 kg Carbon lost in ash = 5.143 kg × 1000 = 25 kg205.72 ÉÙ
\ Carbon reacted = 900 – 25 = 875 kg = 72.916 katoms 15 katoms of carbon reacted º 100 kmoles of flue gas 72.916 katoms reacted º 486.1 kmoles flue gas Volume 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 and 166.7 °C. The partial pressure of water vapour in this air is 25 mm Hg. At the exit partial pressure of water is 100 mm Hg. For a total pressure of 760 mm Hg at 111.1 °C in this process, calculate the volume of exit air per cubic metre of inlet gas. Basis: 1 m3 of inlet air = 1000 litres = 106 cc Moles of air entering = 1 768ÈØÈØÈ Ø273 = 0.02799 kmole 22.414ÉÙÉÙÉ 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 kmole moles 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 °C 3.42 A telescopic gas holder contains 1000 m3 of gas saturated with water vapour 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 of water is 20 mm Hg. Basis : 1000 m3 of the gas at the given conditions 155 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 as expressed 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 mixture Let M be the molecular weight of component Y Therefore, mole % of Y = 40% 100 Weight, % of Y = 20 = (40 × M)×(35 85) (40 M) (25 60) 89500 + 800 M = 4000 M Therefore, 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 atm pressure. Find the mole fraction, weight % of each, average molecular weight, and total weight of the mixture. 4 Partial pressure of N 2
= 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 N = p TP 2 Í˙ 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 CO = p TPÍ˙ Í ˙ 2 22.414Î˚ Î ˚Î˚ = ¥¥Á˜0.75 100 273¥È˘È˘Êˆ1 = 2.16 kmolesÍ˙Í˙˯ Î˚Î˚ Number of moles of H 2
= 0.25 100 273 ¥¥Í˙= 0.72 kmole 473 1 22.414 ¥¥Î˚ Compound Weight, Molecular Weight, Weight mole kmoles weight kg % % N2 2.88 28 80.676 45.531 0.500 CO2 2.16 44 95.080 53.66 0.375 H2 0.72 2 1.441 0.81 0.125 Total 5.76 177.197 00.00 1.000 Average molecular weight = 2 × (0.125) + 28 × (0.5) + 44 × (0.375) = 30.75 Total weight = 177.197 kg EXERCISES 3.1 A liquefied mixture of n-butane, n-pentane and n-hexane has the following composition in percent. n-C4H10 : 50
n-C5H12 : 30 n-C6H14 : 20 Calculate the weight fraction, mole fraction and mole percent of each component and also the average molecular weight of the mixture. 3.2 A steel tank having a capacity of 25 m3 holds carbon dioxide at 30 °C and 1.6 atm. Calculate the weight of the carbon dioxide in grams. 3.3 A steel container has a volume 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 molecular weight and density of the gas at NTP. 3.6 A mixture of gases analyzing 20% methane, 50% ethane and rest hydrogen by volume at a temperature 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 and density of the gas mixture. 3.7 Air contains 79% nitrogen and 21% oxygen by volume. Estimate its density at 20 °C and 741 mm Hg 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 heated without 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 and 748 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 this gas 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 by weight. 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 500 kPa 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 3 atm pressure. Find (a) partial pressure of N2, (b) volume of N2 per 100 m3 of gas and (c) density of the 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) the concentration of each component, g mole/cc, (c) partial pressure of each component, (d) the molar density 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 to react 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 of hydrogen to ammonia. 3.18 A natural gas containing 90% methane, 5% ethane and 5% nitrogen is piped from a well at 25 °C and 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 reaction CaC2 + 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 burning 100 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 °C and 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 by liquefying 500 litres of the gas at standard conditions?
Vapour Pressure 4 Whenever we come across a liquid system, it is generally in contact with its own vapour over the liquid surface. This vapour exerts a pressure like gases, which is called vapour pressure. When the liquid is at its boiling condition the vapour pressure will be equal to the surrounding pressure. When that pressure becomes equal to the atmospheric pressure, the boiling point is referred as normal boiling point (NBP). Whenever we have a binary system, if the sum of their partial pressures equals the surrounding pressure then the system will boil. 4.1 EFFECT OF TEMPERATURE ON VAPOUR PRESSURE Illustrated by the effect of temperature on vapour pressure is Clapeyron equation l dT 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 the ideal gas law assumed, the above reaction reduces to Clausius–Clapeyron equation. dp=ldT p RT2 or d (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 of vaporization is constant and the above equation may be integrated between the limits p0 and p and T0 and T. 74
ln ʈ Ê ˆ =Á˜ Á ˜ or log ʈ Ê ˆ ʈ=Á˜ Á ˜ Effect of Temperature on V.P: Antoine equation, log p* (mm Hg) = A – B(where, p* is the vapour pressure and A , B , C are constants) TC + 00
00
Limitations: (i) Applicable only in the range of applicability of A, B, C. (ii) Pressures below 10 bar. 4.2 HAUSBRAND CHART Steam distillation takes place at point of intersection of the curves at which pA = p – pW \ p = pA + pW Total pressure = vapour pressure of component + vapour pressure of water p–p W pA Vapour pressure Temperature
WORKED 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 mole R = 1.99 cal/g mole K T0 = 273 K, T1 = 293 K, T2 = 308 K At 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-volatile solutes by subjecting it to distillation with saturated steam under atmospheric pressure of 745 mm Hg. Calculate the temperature at which the distillation will proceed and 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 60 390 65 460 68 510 69 520 150 540 190 650 215 725 225 745 distillation temperature, 69 °C 70 550 235 785 pp W p B Vapour pressure Temperature
Basis: 1 g mole of mixed vapour. g mole of C H66 Vapour pressure of C H66 520 2.31g mole gC H66 2.31 78 10.01gwater
of water Vapour pressure of water 225
18
g water 0.1g benzene
4.3 It is decided to purify myristic acid (C13H27COOH) by steam distillation under 740 mm Hg pressure. 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 be assumed that the relative saturation of the steam with acid vapours will be 80% (a) Find the weight of steam required per kg of acid distilled at 740 mm Hg. (b) Calculate the weight of steam per kg of acid if 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 kg kg of acid ÉÙ ÊÚ Discussion on examples 4.3 and 4.4: When ordinary steam is used at atmospheric pressure, 1840 kg of steam is needed. When superheated steam at 200 °C is used, 4.95 kg is needed. Due to low pressure 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 by weight. 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, fraction mm Hg in vapour phase Benzene (C6H6) Toluene (C7H8) Xylene (C8H10) 78 1340 35 35/78 = 0.449 0.401 0.401 ´ 1340 = 536 0.673 92 560 40 40/92 = 0.435 0.388 0.388 ´ 560 = 217 0.272 106 210 25 25/106 = 0.236 0.211 0.211 ´ 210 = 44 0.055 Total 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 is independent of temperature. Find the vapour pressure of the solution at 30 °C and the boiling point elevation. 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 1030 Vapour 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 °C 4.7 The following table gives vapour pressure data: Temperature °C 69 70 75 80 85 90 95 99.2 Hexane (A), mm Hg 760 780 915 1060 1225 1405 1577 1765 Heptane (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 the mole 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) = 1 where 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 = P xA(PA – PB) = (P – PB) xA 760 + (1 – xA)295 = 760 At 69 °C, (Boiling point of hexane) xA(760 – 295) = (760 – 295); \ xA = 1 y 760 = 1 = A 760 At 70 °C xA(780 – 302) = (760 – 302); \ xA = 0.96 pA = (0.96 ´ 780) = 748.8; \ yA = 748.8/760 = 0.985 At 75 °C xA(915 – 348) = (760 – 348); \ xA = 0.73 pA = (0.73 ´ 915) = 668; \ yA = 668/760 = 0.880 At 80 °C xA(1060 – 426) = (760 – 426); \ xA = 0.53 pA = (0.53 ´ 1060) = 562; \ yA = 562/760 = 0.740 At 85 °C xA(1225 – 498) = (760 – 498); \ xA = 0.36 pA = (0.36 ´ 1225) = 441; \ yA = 441/760 = 0.58 At 90 °C xA(1405 – 588) = (760 – 588); \ xA = 0.21 pA = (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 heptane xA(1765 – 760) = (760 – 760); \ xA = 0 = yA 4.8 A solution of methanol in water containing 0.158 mole fraction alcohol boils at 84.1 °C (760 mm Hg). The resulting vapour contains 0.553 mole fraction of alcohol. How does the actual composition of the vapour compare with composition calculated from Raoult’s Law? Temperature, °C 80 100 Vapour pressure, mm Hg 1340 2624 To get vapour pressure at 84.1 °C, the equation ln 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 = 0.316 760 % 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 resulting gas mixture is cooled and dried, calculate (a) volume of dry exit gas at 70 °F and 750 mm Hg. (b) Partial pressure of O2 in exit (c) Weight of water removed. Basis : 1 lb mole of methane. CH4 + 2O2 ® CO2 + 2H2O O2 supplied = 2 ´ 1.1 = 2.2 lb moles Gases 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 lb 4.10 Bottled liquid gas is sold. Determine (a) the pressure of the system; (b) vapour composition. The composition in mole % in liquid phase is given 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 5 100 Vapour 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 a benzene 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 atm and cooled to 21.1 °C after compression. How many kilograms of benzene are condensed by this process 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 kmole Benzene 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 0.012 m3/h. The N2 saturated with acetone vapour, leaves at 760 mm Hg and 308 K at 0.023 m3/h. Find the vapour 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 = 0.55 Thus, mole fraction of acetone = 0.45 (PT) × (y) = (Partial pressure of acetone) = (VP)acetone (y)acetone (760)(0.45) = (VP) (1.0) = 342 mm Hg 4.13 Determine the pressure of the system and equilibrium VP at 30 oC. Assuming all ethane is removed, estimate the pressure and vapour phase composition components of the system at 30 oC. Compound Mole fraction, x, in liquid phase n-butane 0.50 n-propane 0.45 Ethane 0.05 Compound n-butane n-propane Ethane Total 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 19.12 0.45 10.8 4.86 54.67
0.05 46.6 2.33 26.21 8.89 100.00
y (Mole fraction) = Partial pressure Total pressure 1.7 = ÈØ ÉÙ × 100 = 19.12 for n-butaneÊÚ 54.67 for n-propane 26.21 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 25.91 n-propane 0.4737 10.8 4.86 74.09 Total 6.904 100.00
4.14 Estimate the liquid phase composition of a mixture of benzene and toluene at 80 oC when their gas phase compositions are 80% benzene and 20% toluene. Vapour pressures of benzene and toluene are 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 oC At 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) = PT We also know that, xB + xT = 1 4 × (1 – xB) × (560) = 4 × PT× (0.2) xB × (1340) = PT × (0.8) 4 (1 – xB) 560 = xB(1340) 2240 – 1340xB – 2240xB = 0 Solving, xB = 0.6 4.15 A certain quantity of an organic solvent (molecular weight 125 and density 1.505 g/cc) is kept in an 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 is present 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 Hg Moles at NTP = 3990 240 273 1= 0.05065 g mole 303 t tt 760 22414 Molecular weight = 125 Weight = 0.05065 × 125 = 6.331 g Mass of liquid left behind = 10 ml × 1.505 g/cc = 15.05 g Total weight = 15.05 + 6.331 = 21.381 g Mole 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 that of toluene is 20 mm Hg and their mole fractions in liquid phase are 0.7 and 0.3 respectively, calculate their 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 PA PPB = (PT)(yB) = xB PB PT(yA + yB) = xA (PA) + (1 – xA) PB = PP PT = PPA + PPB PT = (0.7)(70) + (0.3)(20) = 49 + 6 = 55 mm Hg Hence, yA= (0.7)(70)/55 = 0.89 yB = (0.3)(20)/55 = 0.11 4.17 Vapour pressure of pure benzene and toluene are 960 mm Hg and 380 mm Hg respectively at 86.0 °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 benzene and toluene respectively Then, by Raoult’s law, PTot = (xB) (PB) + (xT) (PT) 760 = (xB) (960) + (1 – xB) 380 Solving,
xB = 0.655 xT = 0.345 Px yB = BB 960 t0.655= 0.827PTot760 In the same way, yP Px 380t 0.345= 0.173T = TT760Tot 4.18 Vapour pressure of benzene is 3 atm and that of toluene is 1.333 atm. A liquid fuel containing 0.4 mole benzene and 0.6 mole toluene is vaporized. Estimate the mole fraction of benzene in vapour phase. Partial pressure of benzene = (3)(0.4) = 1.2 Partial pressure of toluene = (1.333)(0.6) = 0.8 Total pressure = 2.0 atm ybenzene = 1.2/2.0 =0.6 EXERCISES 4.1 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 normal boiling 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. at 42 °C and 5.0 atm at 126.0 °C. By using the chart estimate the boiling point of ethyl acetate at 760 mm Hg 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. Compute the 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 Hg VPB, — 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 1262 Vapour 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 weight of nitro benzene associated per kg of steam in distillate. Assuming the vaporization efficiency to be 75%, 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 liquid containing 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 at 21 °C and the composition of the vapour evolved. (b) Find the total pressure at 21 °C if all C2H6 were 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 at 100 °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 of water is 190 mm Hg at 93.5 °C. Calculate the partial pressure of CH3CHO over 0.15 molal solution in water. 4.10 Ethyl acetate at 30 °C exerts a vapour pressure of 110 mm Hg. Calculate the composition of the saturated mixture of ethyl acetate and air at a temperature of 30 °C and an absolute pressure of 900 mm 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 a gas and a pure liquid are brought into intimate contact. The term humidification is used to designate a process where the liquid is transferred to gas phase and dehumidification indicates a process where the transfer is from the gas phase to the liquid phase. The matter transferred between phases in both the cases is the substance constituting the liquid phase, which either vaporizes or condenses indicating respectively the humidification process or the dehumidification process. 5.2 DEFINITIONS Dry bulb temperature (DBT): The temperature measured by a bare thermometer or thermocouple is called the dry bulb temperature. Wet bulb temperature (WBT): The temperature measured by a thermometer or thermocouple with a wet 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 main gas 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 Grosvenor humidity. pMM nn [] ËÛ vAA
ÌÜ ÌÜ ËÛ mass of A(5.2) ÌÜ mass of B ÍÝM pp M BvBÍÝ ÍÝ vv
87
Relative humidity or relative saturation ( yr%): It is normally expressed in percentage. If pv is the partial 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 the ratio 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 at constant total pressure out of contact with a liquid. The moment the temperature is reduced below the dew 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 and its 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 the vapour content. For a gas at a DBT of tG, with a humidity of nv¢, the enthalpy relative to the reference state ‘t0’ is, H¢ = enthalpy of gas + enthalpy of vapour component. = CB(tG – t0) + nv¢[CA(tG – tDP) + lDP + CA,L (tDP – t0)] (5.6) where lDP = 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 to Eq. (5.7) when tDPitself is taken as the reference temperature, t0 H¢ = 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 gas and its accompanying vapour at the prevailing temperature and pressure. The expression for humid volume in m3/kg is È˘¢¢È˘2731105 v MMÍ˙ Í˙273 p Î˚
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 will acquire vapour from the liquid until the partial pressure of the vapour in the gas mixture equals the vapour 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 than the 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 ratio of 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 its temperature as well as of the nature of the vapour. Relative saturation also represents the following ratios: (a) The ratio of the percentage of the vapour by volume to the percentage by volume that would be present if the gas were saturated at the existing temperature and total pressure, (b) the ratio of the weight of vapour per unit volume of mixture to the weight 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 unit weight of vapour free gas to the weight of vapour that would exist per unit weight of vapour free gas if 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 of vapour that would be present per mole of vapour free gas if the mixture were saturated at the existing temperature and pressure. Relative saturation % = yr = pv ¥ 100 where, ps 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 of vapour free gas at saturation. If p is the total pressure, then from Dalton’s law, ʈv
, and ns = ʈsnv = Á˜ Á˜ -˯ -˯ np ppÊˆÊ ˆ - =Á˜Á ˜ Ë¯Ë ¯ y v
s
vv s
ss v
p
=( y r
) ʈv Á˜ -˯ WORKED EXAMPLES v
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 pressure Partial pressure under the given condition is given by pAmolal
humidity (0.0483) = pp
-A
pA0.0483
=1.0133 105
¥p ʈ
A
5
20.69 =
¥Á˜Ë¯
–1
where, pA = 4.672 ¥ 103 N/m2 pA = Partial pressure of water p = Total pressure Vapour 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/kg H¢ = 1.062 (50 – 0) + (0.03) (2502) = 128.16 kJ/kg da (b) Enthalpy of saturated air = 274 kJ/kg Enthalpy 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 air Alternatively, (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 the composition of a saturated mixture of nitrogen and ethyl ether vapour at 20 °C and 745 mm Hg and express 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 3 Nitrogen = 40.7% = 0.407 ft (b) Basis: 1 lb mole of mixture Component Molecular lb mole Weight, lb Weight % weight Ether 74 0.593 (0.593 ´ 74) = 43.9 79.4 Nitrogen 28 0.407 (0.407 ´ 28) = 11.4 20.6 Total 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/ft = ÈØ ÉÙ = 0.112ÊÚ (d) lb of vapour/lb of vapour free gas = 43.9 = 3.8511.4 3
(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 yp at 20 °C and 745 mm Hg Vapour pressure of acetone at 20 °C = 184.8 mm Hg Partial pressure of acetone = (0.148 ´ 745) = 110 mm Hg 110 y r
= relative saturation = ÈØ ÉÙ ´ 100 = 59.7%ÊÚ 0.148 ÈØ = 0.174 = v ÉÙnÊÚ 184.8 ËÛ ÉÙÌÜ 0.248= 0.329 = ÊÚÈØÌÜÉÙn s 560.2 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 temperature to 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 mixture Amount of water = 8.1 = 1.16 ´ 10–3 lb º 6.42 ´ 10–5 lb moles 7000 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 a drying tower, where 90% of water vapour is removed. Calculate the pound of water removed per 1000 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 moles y p
= 0.5 = nv ns where nv = lb mole of water/lb mole of vapour free gas. 0.36 n v
= 0.5 ´ n s
= 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 dew point 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 for benzene, 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 by evaporation into a stream of nitrogen. The nitrogen enters the evaporator at 30 °C containing acetone such that its dew point is 10 °C. It leaves at a temperature of 25 °C with a dew point of 20 °C. The atmospheric 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 the evaporator. (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 Hg mole of acetone/mole of N2 = 116 = 0.183 634 Leaving gases: partial pressure of acetone = 185 mm Hg partial pressure N2 = 750 – 185 = 565 mm Hg mole of acetone/mole of N2 = 185 = 0.328565 Basis: for (b), (c) and (d): 1 lb mole of N2 passing through evaporator. (b) Moles of acetone evaporated = (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 lb Acetone 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 ft3 Volume of gases leaving per 1000 ft 3 = 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 increased to 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 of wet 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 = 9.2 mm 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 moles 359ÉÙÉ ÙÉ Ù293ÊÚÊ ÚÊ Ú 736 ÈØ = 2.51 lb moles´ lb mole of dry air = 2.56 ÉÙ ÊÚ lb mole of water vapour initially = (0.019 ´ 2.51) = 0.0477 lb mole lb 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. It is proposed to condense 90% of the acetone by cooling to 5 °C and compressing. Calculate the final pressure 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 mole Acetone finally present = 0.306 ´ 0.1 = 0.0306 lb mole Final gas = 0.694 + 0.0306 = 0.7246 lb mole lb 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 to 2000 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 at 20 °C = 17.5 mm Hg Vapour pressure of water at 10 °C = 9.2 mm Hg At 20 °C and 750 mm Hg, pp y r
= 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) p s
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˜ ˯˯˯ = 41.02 g 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 by weight (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 volume of fresh air required if 1000 kg/h of dried material is to be produced. The exit humidity of air is 0.09 kg of water/kg of dry air. The air enters at 301 K and at atmospheric pressure. 301 K Air 0.02 0.09 16% wet solidDrier Dried solid 0.5%
Basis: 1000 kg/h product. Dry solid = 995 kg Moisture = 5 kg 995 =
1184.5 kg\ feed = 0.84
Water in feed = 189.5 kg Dry solid is the Tie Element Water removed = 1184.5 – 1000 = 184.5 kg 1 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 ÊÚ 5.12 A tunnel drier is used to dry an organic paint. 1000 lb/h of feed having 10% water is to be dried to 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 Hg 95 °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 lb Moisture = 100 lb Product = 900 = 904.5 lb0.995 Water removed = (1000 – 904.5) = 95.5 lb/h From humidity chart the following details are obtained: Humidity of entering air = 0.025 lb mole water vapour/lb mole dry air Humidity of leaving air 0.035 lb mole water vapour/lb mole dry air 1 lb mole 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 of entering 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 H 2
O/lb of dry air = 0.00795 ´ ÈØ –3 ÉÙ = 5 ´ 10 lb = 0.005 lbÊÚ (b) 1000 ft 3
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
= 2.447 1.02123 = 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 of entering air is 40 °F while that of leaving air is 55 °F. If 2000 lb of wet air is forced through the drier per 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 Hg and at 55 °F = 11 mm Hg. Basis: One lb mole of dry air Water in entering air = 6.3 750 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 = 11 750 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 a certain process. Recovery of the acetone is accomplished by evaporation into a stream of N2 followed by cooling and compression of the final gas mixture. In the solvent recovery unit 50 lb of acetone are removed per hour. The N2 is admitted at 100 °F and 750 mm Hg partial pressure of acetone in incoming nitrogen is 10 mm Hg. The nitrogen 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 per hour?
N2, Acetone 100 °F 750 mm Hg N2, Acetone Drier partial 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 mole 58 N 2
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 N in entering stream = 2 ÉÙ ÊÚ = 0.0135 p vËÛËÛ287yp = 0.85 = 740 pÌÜÌÜ287ÍÝÍÝ
where yp (percentage saturation) = 85% pv (partial pressure) = 258 mm Hg lb mole of acetone/lb mole of N 2
in leaving stream = 258 740 258 = 0.5385 \ 1 lb mole of N2 removes: (0.5385 – 0.0135) = 0.525 lb mole of acetone lb 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 ft3 5.16 By adsorption in silica gel you are able to remove all the weight (0.93 kg) of water from moist air at 15 °C and 98.6 kPa. The same air measures 1000 m3 at 20°C and 108 kPa when dry. What was the 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 kmoles pʈ98.6 = 0.114 kPav = Partial pressure of water = ¥Á˜Ë¯ ps = Vapour pressure of water = 15 mm Hg = 2 kPa pv, y = 0.114¥ 100 = 5.7%, Relative humidity, y = r Q r 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 to recover 90% C6H6 present by cooling to 283 K and compressing to a suitable pressure. What is the pressure to which the mixture should be pressurised? Vapour pressures of benzene at 283 K and 297 K are 6 kN/m2 and 12.2 kN/m2 respectively. Relative humidity = 0.6 = Partial pressure/Vapour pressure 12.2 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 air entering 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 Hg Mole fraction of water vapour = (86.7/825.36) = 0.105 Volumetric flow rate = 1000 m3/h at 75 °C and 1.1 bar 1000 825 273 138 kmoles/h348 t tt 760 22.414
Water flow rate = 38 × 0.105 = 3.99 kmoles/h Dry 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 a countercurrent dryer to give a fixed product containing 5% moisture content. The air used is at 100 °C and is having water vapour at a partial pressure of 103 N/m2. The air leaves the dryer at 70% RH and at 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/s Weight of water in dried wood is 5% Weight of final product = 0.06 = 0.0632 kg/s 0.95 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 be removedDry 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 of the air is cooled to 15 °C and the partial pressure of water vapour is brought to 12.7 mm Hg. During this 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 kmolesÍÝ Mole fraction of water vapour in original air (30 °C) = 22 = 0.02973740Mole fraction of water vapour in original air when cooled to 15 °C = 12.7 = 0.017740 Therefore, mole fraction of dry air = 0.983 Moles 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 a rate 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/s Hence, 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 to get air containing 0.001 kg of water vapour per kg of dry air for a specific application. The column has 2.1 kg of silica gel initially and after 5 h of operation, the weight of silica gel is found to be 2.2 kg. 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 kg Water 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 in kg/h. (G) (0.01) = (G) (0.001) + 0.02 G (0.01 – 0.001) = 0.02 G = 0.02 = 2.222 kg/h0.009 i.e. the flow rate of dry air in kg/h = 2.222 kg/h Flow 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/h EXERCISES 5.1 Toluene–air mixture is available such that the partial pressure of the vapour is 1.33 kPa. The total pressure is 99.3 kPa and the temperature is 21 °C. Calculate (a) the relative humidity, (b) The moles of 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 hot air. The fresh air contains 0.02 kg of water/kg of dry air. Find the volume of fresh air required if 1000 kg/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 275 mm 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 is 150 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 and enthalpy. 5.7 Air at 50% relative humidity (RH) is to be used for a specific operation. Air is available at a DBT of 35 °C and 27 °C. What should be the final temperature to which it should be heated? It is found that a DBT of 30 °C will be quite comfortable to the labourers. How will you obtain this condition? Indicate the exact temperature to which it should be taken before bringing it to 30 °C and 50% RH. If it is necessary to use 5000 m3/h of air at the final condition mentioned above, estimate the quantity 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 of 24 °C is available. This air for weaving room is obtained by saturating it initially and then heating it to 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 to a 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 a temperature of 35 °C and a pressure of 742 mm Hg. Calculate (a) percentage humidity of air leaving the drier and (b) volume of wet air leaving the drier per hour. (use vapour pressure data from Steam Tables). 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 to 15 °C and its pressure increased to 200 kPa, condensing out some of the water, (c) weight of water condensed 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 kPa Vapour 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 is contacted with water at the adiabatic saturation temperature and is thereby humidified and cooled to 70% 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 absolute humidity 0.030 kg water/kg dry air at 1 std. atm pressure. Using humidity chart, calculate (a) the relative humidity, if vapour pressure of water at 55 °C is 118 mm Hg and (b) the humid volume in m3/kg dry air. 5.13 A drier is used to remove 100 kg of water per hour from the material being dried. The available air has a humidity of 0.010 kg bone dry air, and a temperature of 23.9 °C and is heated to 68.3 °C before entering the drier. The air leaving the drier has a wet-bulb temperature of 37.8 °C and a drybulb temperature of 54.4 °C. Calculate (a) air consumption rate, (b) humid volume of air before and after preheating, (c) wet-bulb temperatures (WBT) of air before and after preheating, and (d) dew point 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 the adiabatic 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 the air leave the drier?
5.15 A mixture of carbon disulphide vapour and air contains 23.3% carbon disulphide (CS2) by volume. Calculate the relative saturation and percent saturation of the mixture at 20 °C and 740 mm Hg 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 rate of 1100 m3/h. Water at the rate of 20 kg/h is to be removed from the wet material. The air leaves the drier at 35 °C and 742 mm Hg. Calculate (a) percent humidity of leaving air and (b) volumetric flow rate of leaving air. 5.17 Air is available at a DBT of 310 K and a WBT of 305 K. Estimate humidity, % saturation, dew point, humid volume and enthalpy using humidity chart. Estimate humid volume and enthalpy using equations 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 Hg respectively. 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 27.7 mm 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 molal humidity, 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 1500 m3/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 left behind after the separation of crystals is known as mother liquor which will also be a saturated solution. 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/h xE Feed, F kg/h xF Crystallizer Mother liquor, M kg/h xM Crystal, C kg/h xC
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 concentration xM : 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 by molecular weight of anhydrous salt divided by the molecular weight of hydrated salt. The above principles are explained through the following problems. 111
WORKED EXAMPLES 6.1 A solution of sodium chloride in water is saturated at a temperature of 15 °C. Calculate the weight 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 water NaCl 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 per 100 kg of water. Calculate the weight of this solution necessary to dissolve 250 kg of CaCl2◊ 6H2O at 25 °C (solubility = 7.38 kg mole CaCl2/1000 kg H2O) Basis: x kg of water is present in original solution CaCl2◊ 6H2O has the molecular weight = 111 + 108 = 219 250 kg of solution will have 126.7 kg of CaCl2 and 123.3 kg of 6H2O Total CaCl2 entering the process = (0.62x + 126.7) kg (1) Water entering the process = (x + 123.3) kg Solubility 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 kg 100 kg of water is equivalent to 162 kg of solution
162129.03
kg of water is equivalent to 129.03 ´100 = 209 kg of solution Original 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 be crystallized from 1000 kg of solution by reducing the temperature to 10 °C (c) Calculate the percentage yield of the process. 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 = 1.06ÊÚ saturation 48.6 0.96 % saturation of given solution = ÉÙ ÈØ´ 100 = 90.8% ÊÚ (b) Let a kg be NaNO3 crystallized out NaNO3 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 is evaporated 640 kg of water. The remaining solution is cooled to 20 °C. Calculate the amount and the percentage yield of K2Cr2O7 crystals formed. Solubility at 20 °C = 0.39 kmole/1000 kg H2O Basis: 1000 kg of original solution K2Cr2O7 = 130 kg; Water = 870 kg Water present finally = (870 – 640) = 230 kg
0.39 ÈØ´ 294 = 26.4 kg´ K Cr O present then = 230 ÉÙ 2 2 7 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 the amount of ice formed in cooling 1000 g of this solution to –15 °C (solubility at –15 °C = 6.2 g mole/1000 g H2O) Basis: 1000 g of original solution 100 ÈØ
= 91 g´ÉÙNaNO3 present initially = 1000 ÊÚWater present initially = 909 g Water 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 anhydrous Na2CO3 at 20 °C is 21.5 kg/100 kg of water, what weight of salt crystallizes out? Basis: 1000 kg of solution. Salt = 300 kg Water = 700 kg Water evaporated = 700 ´ 0.1 = 70 kg Let 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 kg Check (water balance): 180 ËÛË Û 700 ÈØ È Ø ÌÜÌ Ü 100 = 630 kgÊÚ
ÊÚ
ÍÝÍ Ý 6.7 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, how many 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 saturated solution at 50 °C. Solubility at 50 °C = (4.48 ´ 106) kg Na2CO3/1000 kg H2O i.e. 474.88 kg Na2CO3/1474.88 kg solution. 100 474.88In 100
kg of solution, Na2CO3 present = 1474.88
= 32.2 kg water present = (100 – 32.2) = 67.8 kg Molecular 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 is 32.2 180 = 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 an evaporator. Saturated solution is produced and some salt crystallizes out. The hot crystals with some adhering solution are centrifuged to remove some of the solution. Then the crystals are dried to remove the rest of water. During an hour test, 837.6 kg of concentrated solution was removed in evaporator, 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) the solubility of the salt; (b) water removed in evaporator; (c) water removed in drier; and (d) water leaving the evaporator. Discussion: The solution leaving the evaporator and centrifuge maintains the same concentration and is saturated. From evaporator, the various streams leaving are water vapour, saturated solutions and crystals with adhering solution. Basis: One hour. 60% adhering solution (removed in centrifuge) = 198.7 kg \ Total adhering solution entering centrifuge (along with crystal) = 198.7= 331.2 kg 0.6
Water vapour 10,000 kg/hrCentrifuge6% solution Evaporator AdheringWater solution + Dry crystal crystalDrier 361.7 kg 837.6 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 kg water and 0.23 kg of salt) (d) Weight of material entering drier = Final dry salt + Water evaporated = (361.7 + 102) = 463.7 kg Weight 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 kg Check: Water balance = 8,500 + 102 + (837.6 + 198.7) ´ (0.77) = 8,500 + 102 + 798 = 9,400 kg 6.9 In a solution of naphthalene in benzene, mole fraction of naphthalene is 0.12. Calculate the weight of this solution necessary to dissolve 100 kg of naphthalene at 40 °C (solubility of naphthalene is 57% by weight) 100 kg X kg solution (100 + X) kg 0.817 wt fr. of benzeneTank 0.183 wt fr. of naphthalene0.57 wt. fr. of naphthalene
Basis: 1 mole of feed solution Component mole Molecular Weight, kg Weight weight fraction Naphthalene (C10H8) 0.12 128 15.4 0.183 Benzene (C6H6) 0.88 78 68.6 0.817 Total — — 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% by weight 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 water Molecular weight of Glauber’s salt (Na2SO4×10H2O) = 142 + 180 = 322 Basis: 7,500 kg of aqueous solution. Na2SO4 present = 7500 ´ 0.296 = 2220 kg H2O present = (7500 – 2220) = 5280 kg H2O lost in evaporation = 5280 ´ 0.05 = 264 kg H2O 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 ØH O balance: 5016 = 180 2 ÉÙÉ Ù ÊÚÊ Ú Solving, a = 3749.5 kg and b = 3486.5 kg Weight of Glauber salt = 3749.5 kg Weight of solution = 3486.5 kg Check: (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 in Weight %). Calculate the feed rate kg/h. Water 20% NaCl Evaporator
Feed F crystallizer
13,000 kg dry salt/h
Basis: 1 h of operation 13,000 kg of salt with 20% brine Weight of feed brine = 13,000 ´ 0.2 = 2,600 NaCl in this solution = 2,600 ´ 0.27 = 702 NaCl 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 an open 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 of crystals obtained. (b) Calculate the yield of crystals neglecting loss of water by evaporation; KCl crystallizes without any water of crystals.
3% Water 5000 kg KCl Crystallizer 20 °C 80 °C Saturated solution Saturated 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 water or, solubility of KCl = 55 kg/155 kg solution = 0.355 kg/kg solution 5000 kg KCl is equivalent to 5000/0.355 = 140,90.91 kg of solution Water present = 140,90.91 – 5000 = 9090.91 kg Water 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 = 1818.18 100 = 36.37%´ 5000 6.13 A crystallizer is charged with 6,400 kg of an aqueous solution containing 29.6% of anhydrous sodium 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% water 6,400 kg 18.3% Na2SO4Crystallizer Mother liquor M 29.6% Na2SO4× 10H2O (142 + 180) = 322
Basis: 6,400 kg solution. Weight of Na2SO4 = 6,400 ´ 0.296 = 1,894.4 kg Water 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 Na 2
SO
4
gives, 1894.4 = CÈØ ÉÙ + (0.183 ´ M)ÊÚ322 C + M = (6400 – 450.56) = 5949.44 kg Solving, M = mother liquor = 2826.7 kg C = Crystal = 3122.7 kg Check: (water balance) 450.56 + 2826.7 81.7ÈØÈ Ø180 ÉÙÉ Ù
322ÊÚÊ Ú = 4505.6 kg 6.14 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 is 19.4 g per 100 g of water. Feed 32% Crystallizer Mother Liquor, Y Crystals, X
Basis: 1,000 g of the feed solution. Na2SO4.10H2O (142 + 180 = 322) Na2SO4 present in the feed = 320 g Water present in the feed = 680 g Overall balance is X + Y = 1000 g 142 XY Na 2
SO 4
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 g Yield 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 sodium sulphate by weight at 104 °C. The solution is cooled to 20 °C to crystallize out the desired Glauber’s salt. The mother liquor is found to contain 16.1% anhydrous sodium sulphate. Estimate the weight of mother liquor. 6.2 A vacuum crystallizer is to produce 10,000 kg of FeSO4.5H2O crystals per hour. The feed solution 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 water lost 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 anhydrous Na2CO3 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. During cooling 15% water originally present evaporates. The crystal formed is Na2CO3.10H2O. If the solubility of anhydrous Na2CO3 at 20 °C is 21.5 kg/100 kg of water, what is the weight of Na2CO3.10H2O crystals formed? 6.5 1500 kg of a solution containing 20% of Na2SO4 and 80% water by weight is subjected to evaporative cooling and 20% of original water evaporates. Calculate the yield of Na2SO4.10H2O crystals formed from solution, if the temperature of solution is reduced to 20 °C. The solubility of anhydrous 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 concentrated liquor is sent to a crystallizer where crystals of KNO3 are formed and separated. The mother liquor from the crystallizer is recycled and mixed with the evaporator feed. The recycle stream is a saturated solution containing 0.6 kg KNO3/kg water. The crystals carry 4% water. Compute water evaporated and crystals obtained. 6.7 A crystallizer is charged with 9000 kg of an aqueous solution of 20% Na2SO4 and it is subjected to evaporative cooling and 10% of original water evaporates. Glauber’s salt crystallizes. The mother liquor 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 the weight of crystals formed and the percentage yield obtained by cooling 100 kg of this solution to a temperature of 5 °C. Solubility Data: At 40 °C : 32.6% Na2SO4 At 5 °C : 5.75% Na2SO4
6.9 A solution of K2Cr2O7 in water contains 16% K2Cr2O7 by weight. 1000 kg of this solution are evaporated to remove some water. The resulting solution is cooled to 20 °C. If the yield of K2Cr2O7 crystals 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 and feed, if 1500 kg of MgSO4.7H2O crystallizes. Solubility data: At 80 °C : 64.28 g MgSO4 per 100 g water 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% Na2S2O3 solution is cooled to 293 K from 333 K. The solubility at 293 K is 70 parts anhydrous salt per 100 parts of water. Estimate the amount of feed needed.
Mass Balance 7 This chapter deals with the mass balance both with and without chemical reactions. The principles given in Chapter 2 still hold good. The application of mass balance in unit operations and processes is 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 from hexane. The analysis of cake reveals 0.8% oil, 87.7% solids and 11.5% moisture. Find the % recovery of oil. Basis: 100 kg of seeds Solids free from oil is the tie element Weight of solids in feed = 69 kg = 87.7% of cake \ cake = 78.68 kg Oil 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 the evaporator contains 30% solids. While the concentrate is leaving with 75% solids concentration, calculate the amount of water evaporated per day. ? 30% solids 75% solids Evaporator
Basis: 100 tonnes of feed = 105 kg Solids in the feed = 30,000 kg 122
\ Concentrated liquid leaving = 30,000= 40,000 kg 0.75 Water evaporated/day = (1,00,000 – 40,000) = 60,000 kg 7.3 A water soaked fabric is dried from 44% moisture to a final moisture of 9%. Calculate the weight of water removed per 200 kg of dried fabric. ? 44% moisture 9% moisture Drier 200 kg
Basis: 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%. This gas is burnt with air under conditions selectively to oxidize phosphorus. From the flue gas, the oxides of phosphorus precipitate on cooling and is separated from the remaining gas. Analysis of the latter shows CO2: 0.9%, CO : 22.5%, N2 : 68.0% and O2: 8.6%. Assume oxidation of phosphorus is complete 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 P4O10 P4 + 3O2 ® P4O6 P4 + 5O2 ® P4O10 CO + ½O2 ® CO2 Exit gas from phosphate reduction furnaceBurner CO2, O2, Air CO, N2 Oxides 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 element 89 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 g moles O21 = 67.953 g moles2 from air = 255.632 ´79 O2 for formation of P4O10 = 5x g mole O2 for formation of P4O6 = 3(8 – x) g mole O2 for formation of CO2 = 380.342 ´ 0.009 ´ 0.5 = 1.712 g moles O2 leaving = 380.342 ´ 0.086 = 32.71 g moles O2 reacted = (67.953 – 32.71) = 35.243 g moles Making 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 and the 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 = 2,105.5 kg 7.6 A cotton mill dries a water soaked fabric in a drier from 54% to 9% moisture. How many kilogram of water are removed by drying operation per 1200 kg of feed. Basis: 1200 kg of feed. Dry fabric = (1200 ´ 0.46) = 552 kg Water 54% moisture 9% moisture Drier Weight of product = 552 =
606.6 kg0.91
Water removed = 1200 – 606.6 = 593.4 kg Alternative method: Water in feed = 1200 ´ 0.54 = 648 kg Water in product = 606.6 ´ 0.09 = 54.6 kg Water removed = 648 – 54.6 kg = 593.4 kg 7.7 A gas containing 2% NH3, 25.4% N2 and the rest H2 is flowing in a pipe. To measure the flow rate 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 stream is 6%. Find the flow rate of gas at inlet. Basis: Let x cc/min of feed gas enter the pipe Ammonia 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) Pipe 6% NH3NH rich gas 3 (96%) 100 cc/min.
7.8 A lime stone analysis shows CaCO3: 94.52%, MgCO3: 4.16% and rest insoluble. (a) How many kg of CaO could be obtained from 4 tonnes of limestone? (b) How many kg of CO2 are given off per kg of limestone? (a) Basis: 4 tonnes of limestone = 4,000 kg
CaCO3 = 0.9452 ´ 4000 = 3,780.8 kg MgCO3 = 0.0416 ´ 4000 = 166.4 kg CaCO3 ® CaO + CO2 100 56 44
MgCO3 ® MgO + CO2 84.3 40.3 44
CaO got = 3780.8 56= 2,117.3 kg 100 (b) Basis: 1 kg of limestone CaCO3 = 0.9452 kg, MgCO3 = 0.0416 kg 44ÊˆÊ ˆ44CO produced = 0.9452 ¥+ ¥ 2 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 converter N2 is the tie element. Exit stream = 81.6 = 86 g moles 0.94895 SO2 leaving = 86 ¥ 0.00605 = 0.5203 g moles SO2 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 between NaCl and H2SO4. The HCl gas is absorbed in water. Calculate (a) the weight of HCl formed when 2 tonnes of 98% salt reacts; (b) How much sodium sulphate is produced and how many kg of 40% acid will be produced? Basis: 2 tonnes of salt = 2000 kg NaCl = 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 with concentrated 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 of Mg was extracted? Basis: 100 g of dolomite SiO2 is the tie element 5 g SiO2 º 2% sludge cake \ weight of sludge cake = 5 = 250 g 0.02 Mg in sludge cake = 250 0.5 = 1.25 g´100 Mg extracted = (10 – 1.25) = 8.75 g Fraction Mg extracted = 8.75= 0.87510 7.12 A certain soap contains 50% moisture on the wet basis when raw. The moisture is reduced to 20% before the soap is pressed into cakes. How many 300 g cakes can be pressed from 1 tonne of raw soap? Basis: 1 tonne of raw soap = 1000 kg Dry 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 cheap fertilizer 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 production 1.3 Cost balance: 32 + 80x = (1 + x) 49.23 Solving, 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 of limestone. 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 MgCO3 CaCO3 ® CaO + CO2 100 56 44
MgCO3 ® MgO + CO2 84.3 40.3 44
Overall balance: x + y = 100 (since pure) CO2 balance = 0.44x + 0.52y = 44.8 Solving, x = 90 g (CaCO3) and y = 10 g (MgCO3) 7.15 The feed to a fractionating system is 30,000 kg/h of 50% benzene, 30% toluene and 20% xylene. The fractionating system consists of two towers No. 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 y kg/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. xy Feed I II z
Overall balance yields x + y + z = 30,000 Individual 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/h z = 5,806.5 kg/h Total 30,000.0 kg/h 7.16 A sample of fuel has the following composition by mass: C : 84%, H2: 15.2% and rest comprising noncombustibles. The fuel was completely burnt with excess air in an internal combustion engine. The dry exhaust gas has 8.6% CO2 by volume. Estimate the amount of air supplied per kg of fuel and find the products of combustion (air contains 23.1% oxygen by weight).
Basis: 1 kg of fuel C : 0.84 kg = 0.07 kmole; H2: 0.152 kg = 0.076 kmole; Noncombustibles – 0.008 kg C+O CO ; H +½O nn H O222 2 2 12 32 44 2 16 18
CO2 formed º 0.07 kmole H2O formed = 0.076 kmole Total kmole of exhaust stream = 0.07 = 0.814 (dry)0.086 O 2
needed for forming CO 2
and H 2
O= ÈØ0.076 ÉÙ ÊÚ = 0.108 kmole N 2
entering (along with stoichiometrically needed O 2
) = 0.108 ´ 79 21 = 0.4063 kmole Air needed (stoichiometric) = 0.108 = 0.5143 kmole0.21 Exhaust gas contains CO2, N2 and O2 Excess 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 contains O2 = 0.071 kmole (21%) N2 = 0.2667 kmole (79%) N2 total = 0.4063 + 0.2667 = 0.673 kmole Total 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 8.6 O2 0.0710 8.7 N2 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 specification that the paper should not have more than 5% moisture. It is also agreed that the price can be suitably adjusted 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 paid to manufacturer as per contract. (b) If it were agreed that the freight was to be borne by the company and adjusted according to the moisture content, what would the company pay to the manufacturer if the freight rate is ` 40/1000 kg of dry paper. Basis: 100 kg of 5% moist paper. Dry paper 95 kg; moisture 5 kg Cost of dry paper = 100 3 = ` 3.16´ 95 In the case of 5,000 kg paper with 7.86% moisture Weight of dry paper = 4,607 kg and weight of moisture = 393 kg (a) Cost of paper = 4,607 ´ 3.16 = ` 14,558.12 (b) Freight charge = 4,607 40 = ` 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 of the ethylene does not react, some form oxide, some turn to CO2 and water. The entire gas mixture enters an absorption tower where water is sprayed. The oxide is converted to glycol. The gas leaving the 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 one mole of water is sprayed per 100 mole of gas mixture, calculate the composition of ethylene glycol– water product formed. 5% C2H4 Reactor 95% air Gases Ethylene, CO2, Absorber O2, N2 pp.H2O 15.4 mm Hg Ethylene glycolwater
Reactions: 1. 2C H + O → 2C H O
24 2 Ethy lene
24
Ethy lene oxide
2. C2H4 + 3O2 ® 2CO2 + 2H2O 3. →C H O+HO (CHOH) Basis: 100 kmoles of feed gas. C2H4: 5 kmoles; O2 : (95 ´ 0.21) = 20 kmoles; N2: 75 kmoles N75 = 92 kmoles2 balance: Moles leaving the absorber = 0.81515 24 2
2 2 Ethy lene gly col
C2H4 : 92 ´ 0.01085 = 1 kmole CO2 : 92 ´ 0.04345 = 4 kmoles O2 : 92 ´ 0.13055 = 12 kmoles H2O = 92 15.4= 1.94 kmoles 74515.4 C2H4 reacted : 5 – 1 = 4 kmoles C2H4 converted to CO2 : 4 = 2 kmoles2 C2H4 converted to C2H4O : 4 – 2 = 2 kmoles Water 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; N2:75 Total number of moles = 98. Water sprayed at the top of absorber = 98 ´ 1 = 0.98 100 Total moles of water entering absorber = (4 + 0.98) = 4.98 kmoles Water in exit gas = 1.94 kmoles Water reacted in absorption column = 2.00 kmoles Ethylene glycol formed = 2 kmoles Water leaving along with ethylene glycol = (4.98 – 2 – 1.94) = 1.04 kmoles Composition of liquid stream: Component moles Molecular weight Weight Weight % (CH2OH)2 2.00 62 124.00 86.88 Ethylene glycol
(H2O) 1.04 18 18.72 13.12 Water
Total 3.04 142.72 100.00 7.19 The first step in the manufacture of H2SO4 from pyrites consists of burning pyrites in air. The reactions 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 gas SO2: 10.2%; O2: 7.8%; N2 : 82% Let x kmole of FeS2 by reaction (1) and y kmole of FeS2 by reaction (2) be reacted. (a) O 21 = 21.8 kmoles (using nitrogen balance) fed = 82 ´ 2 79 O2 leaving = 7.8 kmoles; O2 used = 14 kmoles SO2 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 of x 0.1= 0.02 y5 (b) Total FeS2 reacted = 5.1 moles O2 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 9.2 C2H4 0.4 0.8 CO 20.9 20.9 H2 15.6 — CH4 1.9 1.9
N2 52.0 — Total 100.0 — Reactions C2H4 + 3O2 ® 2CO2 + 2H2O CO + 0.5O2 ® CO2 H2 + 0.5O2 ® H2O CH4 + 2O2 ® CO2 + 2H2O Carbon is the tie element C 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 moles 0.112 N2 from air: N2 in exit = 293 ´ 0.796 = 233 g moles N2 in feed = 52 g moles N2 from air = 181 g moles O 2
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 moles needed = 23.25 g moles Excess 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. A secondary 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 HCHO available + 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 g moles 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% and inert : 60%. By crushing and floating, 2/3 inert is eliminated. The ore is roasted with carbon. Inert are unchanged. In the reduction of the Cu2O to Cu, there is 5% loss. Find the weight of copper obtainable from 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 kg Ore Crushing andP 1 Roaster P 2 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 kg 3 Inert remaining = 200 kg = 0.333 P1 \ P1 = 600 kg CuS 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 °C with 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 pyrites weight of FeS2 = Total weight – weight of inert = 100 – 25 = 75 kg 4FeS + 11O→ 2Fe O + 8SO 22 23 2 480 352 320 512
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 kmoles O 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%; is burnt 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 oil g g atoms O2 needed Reaction C 84 7.00 7.00 C + O2 ® CO2 H2 13 6.50 3.25 H2 + ½O2 ® H2O S 1 0.03 0.03 S + O2 ® SO2 10.28 O2 1 0.03 –0.03 H2O 1 0.06 Net 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 and excess O2 = 11.98 – 10.25 =1.73 g moles Hence % 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 to SO3 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 has 4.3% 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 mole O2 supplied = 1.5 ´ 1.2 = 1.8 g moles N2 supplied = 1.8 ´ 79/21 = 6.8 g moles O2 reacted = 0.7 + (0.3 ´ 1.5) = 1.15 g moles Exit gas stream: Component SO2 SO3 O2 N2Total Weight, g mole 0.7 0.3 (1.8 – 1.15) = 0.65 6.8 8.45 Mole % 8.28 3.55 7.70 80.47 100.0 Let x moles of SO2 react to form SO3 SO2 + 0.5 O2 ® SO3 Exit gas stream consists of the following: SO2 = 0.7 – x SO3 = 0.3 + x O2 = 0.65 – 0.5x N2 = 6.8
Total = 8.45 – 0.5x SO 0.7 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 to SO3 and rest to SO2. Find the exit gas analysis. Basis: 500 kg sulphur = 500 = 15.625 katoms 32 S+O → SO 22 32 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 kmoles N79 = 70.54 kmoles2 entering = 18.75 ´21 S to SO2 = 15.625 ´ 0.95 = 14.845 kmoles S to SO3 = 15.625 ´ 0.05 = 0.780 kmole Total 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 3.08 79.34 100.00 7.27 The composition of the gas entering a converter is SO2 : 7.2%, O2: 13.2% and N2: 79.6% and that of the gas leaving is SO2 : 2.8%, O2: 11.7% and N2: 85.5%. Determine the % of SO2 oxidized to SO3Basis: 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 % SO 100 = 63.75% oxidized = 4.59 ´ 2 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 gas Limestone z coal Burner y Air CaO Lime
Basis: 100 kmoles of stack gas Let z kg of coal and y kg of air enter the burner (air contains 23% O2; 77% N2 by weight %) C + O2 ® CO2 H2 + 0.5O2 ® H2O On weight basis, N2 balance gives, 0.018z + 0.77y = 71.5 ´ 28 = 2002 O2 balance gives, 3216 (0.046 z + 0.23 y )– 0.81 zzÈØ
ÉÙ = (4.1 ´ 32) = 131.2ÊÚ
Solving the above equations, we get z = 187 kg y = 2,595.6 kg CO2 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 kg stack gas Hence, 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 goes to CO2; 10% to CO; and 10% remains unburnt. Calculate the stack gas composition. Basis: 100 g moles of gas we have, C2H6: 80 g moles O2: 20 g moles C2H6 + 3.5O2 ® 2CO2 + 3H2O C2H6 + 2.5O2 ® 2CO + 3H2O O2 needed = 80 ´ 3.5 = 280 g moles O2 available in gas = 20 g moles O2 theoretically needed = 260 g moles 200% excess oxygen is supplied \ O2 supplied = (260 ´ 3) = 780 g moles Total O2 available = 20 + 780 = 800 g moles N2 entering = (780 ´ 79/21) = 2,934 g moles CO2 formed = 80 ´ 0.8 ´ 2 = 128 g moles CO formed = 80 ´ 0.1 ´ 2 = 16 g moles O2 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 3.318 CO 16 0.415 O2 556 14.412 N2 2,934 76.050 H2O 216 5.598 C2H6 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 the process contains CaCO3, MgCO3 and inert compounds. The acid used contains 12% H2SO4 by weight. 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 mass was 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 limestone treated. Basis: 100 kg of residue Inert : 0.53 kg, CaSO4: 8.56 kg, MgSO4: 5.23 kg CaCO3 + H2SO4 ® CaSO4 + CO2 + H2O 100 98 136 44 18
MgCO3 + H2SO4 ® MgSO4 + CO2 + H2O 84.3 98 120.3 44 18
8.56 kg of CaSO4 obtained from 8.56 100 = 6.3 kg of CaCO ´ 3 136 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 % 60 34.95 5.05 100.0 98 kg of H SO = 2 4
6.174 kg3 requires ÈØ(b) 6.3 kg of CaCOÉÙ
ÊÚ 3.67 kg of MgCO 3
requires ÉÙ ÈØ98 kg of H SO = 4.27 kg 2 4 ÊÚ84.3 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 kg formed = 6.3 CO 2 ÉÙÉ Ù ÊÚÊ Ú H 2
O formed = 6.3 18ÈØÈ Ø ÉÙÉ Ù 18 = 1.918 kgÊÚÊ
Ú84.3 H2O in acid = (95.78 – 11.494) = 84.286 kg Total water = water from acid + water formed from reaction = (84.286 + 1.918) = 86.204 kg H2O in residue = 84.51 kg CO2 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 %) 47.56 mole % CO2 4.569 kg 73 % (weight %) 52.44 mole % Total 6.263 kg 100 100 Check Limestone + acid = (10.5 + 95.78) = 106.28 kg Residue + vapours = (100 + 6.263) = 106.263 kg For 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% C and 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 katom H2 = 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 kg Air 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 gas analysis 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 kmoles Air supplied = 81.5 = 103.16 kmoles 0.79 O2 entering = 81.5 ´21 = 21.66 kmoles79 O2 leaving = 8.3 kmoles O2 consumed = (21.66 – 8.3) = 13.36 kmoles 8.3 (a) % excess air = ÈØ ÉÙ´ 100 = 62.13%ÊÚ (b) Let the fuel be CxHy CxHy + (x + y/4)O2 ® xCO2 + y/2H2O x = 10.2 kmoles (CO2); x + y/4 = 13.36 kmoles (O2 reacted) \ y = 12.64 katoms and 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 kmoles m 3 of air/kg of fuel = ÈØ22.414 ÉÙ = 17.12 at NTP ÊÚ
7.33 The waste acid from a nitrating process contains 23% HNO3; 57% H2SO4; 20% water. This acid 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 desired acid Let x kg be the weight of waste acid; y kg be the weight of H2SO4 and z kg be the weight of HNO3 Overall balance: x + y + z = 1000 Balance 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 kg z = con. HNO3 = 192 kg Total = 1,000 kg Check (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 anhydrous soda 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 infusible matter. 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 the barium 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 24 233 106 197 142
Na2SO4 = 27.7 kg Weight of BaSO4 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 kg Na2CO3 supplied = (20.68 + 20.35) = 41.03 kg 45.45 (a) % conversion of BaSO 4
to BaCO 3
= ÈØ ÉÙ´ 100 = 80%ÊÚ (b) Composition of original barites BaSO4 = 56.75, IM = (40.65 – 38.43) = 2.22 kg Also, total mass of products = total mass of reactants = 100 kg Therefore, IM = 100 – [56.75 + 41.03] = 2.22 kg Therefore original barites = 56.75 + 2.22 = 58.97 kg 2.22 Hence %IM = ÈØ ÉÙ´ 100 = 3.76%ÊÚ 56.75 %BaSO 4
= 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 is burnt to CO2 and the rest to CO. All the Hydrogen is converted to water. Determine the orsat analysis of the flue gas (dry flue gas).
Basis: 100 kg of fuel oil Component 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 ® CO H2 + 0.5O2 ® H2O 95% C is converted to CO2 = 7.35 ´ 0.95 = 6.9825 katoms Oxygen used for this reaction = 6.9825 kmole 5% C is converted to CO = (7.35 – 6.9825) = 0.3675 katoms Oxygen used for this reaction is = 0.3675= 0.18375 kmoles 2 Conversion of H2 to H2O = 5.9 kmoles Oxygen 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 kmoles O2 remaining = [12.36 – (6.9825 + 0.18375 + 2.95)] = 2.24375 kmoles N79 = 46.5 kmoles2 entering = 12.36 ´21 Component CO2 CO O2 N2Total mole 6.9825 0.3675 2.24375 46.50 56.0937 mole % 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 CxHy Basis: 100 mole of the flue gas Let us assume the hydrocarbon (HC) to be CxHy Let a g mole of HC get oxidized to CO2 and b g mole of HC get oxidized to CO The reactions to take place are a (CxHy) + a(x + y/4)O2 ® axCO2 + ay/2H2O b(CxHy) + b(x/2 + y/4)O2 ® bxCO + by/2H2O The 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 = 1.9 ax 9.5= 5bx 1.9 a = 5b By nitrogen balances oxygen supplied = 87.2 ´ 0.21 = 23.18 kmoles0.79
Oxygen left = 1.40 kmoles Oxygen reacted = 21.78 kmoles Oxygen 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) = 11.33 ÊÚ \ y(11.4) = 45.32; \ y = 3.975 a + b = 11.4 = 6b; \ b = 1.9; a = 9.5 since ax = 9.5; x = 1 (a) Atomic ratio of H 3.975= 3.975 C1 (b) For % of excess air, let us consider the following: (c) CO + 0.5O2 ® CO2 One kmole of CO requires 0.5 kmole of oxygen Therefore, 0.95 mole of O2 will be used for converting CO to CO2 Excess O2 remaining = 1.4 – 0.95 = 0.45 kmole Theoretical O2 required = 11.4 (from CO and CO2 values) + 11.33 (for H2O formation) = 22.78 kmoles 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 SO2 entering the converter getting converted to SO3. Basis: 100 g mole of gas entering the converter Let x g mole SO2 get converted to SO3 SO2 : 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 ® 83 Total (100 – 1.5x) % SO 2
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 is 98% complete, find the weight and volumetric composition of gases produced per 100 kg of gas burnt. 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 = 154 O2 0.5 32 0.5 ´ 32 = 16 N2 68.0 28 68 ´ 28 = 1,904 Total 100.0 — 2,858 we have CO + 0.5O2 ® CO2 Oxygen balance O2 needed = 28 ´ 0.5 = 14 kmoles O2 available in feed = 0.5 kmoles Net O2 needed = 13.5 kmoles O2 supplied = 13.5 ´ 1.2 = 16.2 kmoles O2 consumed = 14 ´ 0.98 = 13.7 kmoles O2 remaining = (16.2 + 0.5 – 13.7) = 3 kmoles Carbon balance CO2 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 kmoles 21
N2 in exit = (68 + 60.9) = 128.9 kmoles Exit gas kmole Molecular mole % = Weight, Weight % weight volume % kg CO2 30.94 44 18.93 1,359.0 26.60 CO 0.56 28 0.34 15.7 0.31 O2 3.00 32 1.84 96.0 1.88 N2 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 kg 2,858 7.39 In the manufacture of soda ash by Le Blanc process, sodium sulphate is heated with charcoal and calcium carbonate. The resulting black ash has the following composition: Na2CO3: 42%; other water soluble material : 6% and insoluble 52%. The black ash is treated with water to extract the sodium carbonate. The solid residue left behind has the composition Na2CO3: 4%; other water soluble material : 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 Water ash 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 ash Let x be the weight of residue and y be the weight of solution. Insoluble: (0.52 ´ 1000) = 0.85x \ x = 612 kg Na2CO3 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 of soap 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 kg 70 Total weight = (100 + 43) = 143 kg Cost of 100 kg dry soap including freight at laundry = 143 ´ 7.6 100 = ` 10.88
Case II: (5% water) water present = 100 ´5 = 5.25 kg95 Total weight = 105.25 kg Freight 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 calcium phosphate, sand and charcoal. It may be assumed that in a charge the following conditions exist. The amount of silica used is 10% in excess. Charcoal is 40% in excess of that required to combine with P2O5 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 Carbon charged = 5 ´ 1.4 = 7.0 kmoles Feed Weight, Mol.wt Weight, kmole or katom or At.wt kg Ca3(PO4)2 1.0 310 310 SiO2 3.3 60 198 C 7.0 12 84 Total 11.3 592 Composition in weight % 52.3 33.5 14.2 100.0 (b) Basis: 100 kg of charge: calcium phosphate present = 52.3 kg Calcium phosphate consumed = (52.3 ´ 0.9) = 47.07 kg P 142 = 21.6 kg O formed = 47.07 ´ 2 5 310 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 the following. (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 (2
71)=142 (2 18)=36×× × ×
Basis: 4 kmoles of HCl º 146 kg HCl Cl2 HCl O2 Converter O2 N2N2 H O 2
(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 kg N2 entering = (1.22 – 0.281) = 0.939 kg Gases Weight, kg Weight % HCl 1.000 45.0 O2 0.281 12.7 N2 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 oxidized to 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 lost in 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 %) of gases 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 II unit. (e) Calculate the overall degree of completion of sulphur to SO3 FeS2 Unit I SO3,SO2 Unit II Gases
Air (Burner)O2,N2 (Converter)
FeS2 Fe2O3
Basis: 100 kg of iron pyrites Pyrites burnt = 85 kg lost = 15 kg (1) 2FeS + 5.5 O→ Fe O + 4SO (2 120) (5.5 32)
22 23 2
160 (4 64)×× ×
(2) 4SO + 2O→ 4SO 22 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 O 2
used for forming SO 3
= ÈØ32 ÉÙ = 34.0 kgÊÚ
O 2
used for forming SO 2
= ÈØ32 ÉÙ = 37.4 kgÊÚ
Total oxygen used = 71.4 kg O2 remaining (entering Unit II) = (119 – 71.4) = 47.6 kg Gases Weight, kg Weight % SO2 54.4 9.97 SO3 45.3 8.30 O2 47.6 8.72 N2 398.4 73.01 Total 545.7 100.00 (c) SO 3
produced (II Unit) = ÈØ320 ÉÙ = 65.28 kgÊÚ256 \ Total SO3 produced = (45.3 + 65.28) = 110.58 kg 232 (d) O 2
consumed = ËÛ ÌÜ = 13.05 kgÍÝ O2 remaining = (47.6 – 13.05) = 34.55 kg SO2 remaining = (54.4 ´ 0.04) = 2.176 kg Gases Weight, kg Weight % SO2 2.176 0.400 SO3 110.580 20.264 O2 34.550 6.332 N2 398.400 73.004
Total 545.706 100.000 (e) Sulphur in FeS2 = 100 ´ 64 = 53.3 kg 120 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 that final cake contains 34% sulphuric acid. It may be assumed that the cake will contain 1.5% water and that 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 nitrate charged. (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 1.12 Na2SO4 83.50 63.38 H2SO4 44.80 34.00 H2O 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 kg Total acid to be supplied = 102.45 kg 95% acid needed = 102.45= 107.84 kg 0.95 (c) Water in the acid = (107.84 – 102.45) = 5.39 kg Water remaining in cake = 1.97 kg Water distilled off = (5.39 – 1.97) = 3.42 kg HNO 57.65= 74.12 kg formed = 126 ´ 3 98 HNO3 in cake = 1.48 kg HNO3 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 first prepared by heating the barites the natural sulphate with carbon. The BaS is extracted from this mass with 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 from impurities 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 of filtrate collected. The analysis of the precipitate is BaCO3 90.1% and CaCO3 9.9%. The analysis of filtrate is reported to be Na2S : 6.85%, Na2CO3: 2.25% and H2O : 90.9%. The Na2CO3 for the precipitation 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 composition of 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 kg Amount 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 = 14.82 106 = 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 kg CaCO 100 = 1.28 kg3 formed along with Na2S = 1 ´78 CaCO3 impurity in soda ash = (1.63 – 1.28) = 0.35 kg Na2CO3 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 Na CO = 2.25 ´ 2 3 9.317 169.4 (b) BaS formed = 14.82 = 12.72 kg´197.4 CaS formed = 1.28 72 = 0.9216 kg´100 Components of original solution Components Weight, kg Weight % BaS 12.7200 12.17 CaS 0.9216 0.82 Water 90.9000 87.01 Total 104.5416 100.00 Total Na2CO3 = Na2CO3 used in reaction + Na2CO3 in filtrate = 9.317 + 2.25 = 11.567 kg We have CaCO3 as impurity in Na2CO3 = 0.35 kg Composition 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 amount of lime is carried into the beater. It is proposed to neutralize this lime with acid of 67% H2SO4. In a beater containing 5000 gallons of pulp it is found that there is lime equivalent to 0.5 g of CaO per litre. (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 of 1% 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 litres CaO + 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 moles Acid 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 kg 0.67 (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 the sample. CO(NH2)2 = Molecular weight 60 Therefore, N 28 = 0.4666 in urea is = 2 60 100 Purity of sample = 0.45 = 96.43%´0.4666 7.48 Carbon tetrachloride is made as follows: CS + 3Cl→ CCl + S Cl 22 422 212.76 153.84
The product gases are found to contain CCl4 33.3%; S2Cl2 33.3%; CS2 1.4% and Cl2 32%. Calculate the 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 kmoles CS2 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 kmoles But Cl2 unreacted = 32.0 kmoles \ Cl2 taken = 99.9 + 32.0 =131.9 kmoles % excess reactant = 32.0 100 = 30.73%´ 104.1 (b) % Conversion: CS2 = 33.3 100 = 95.97%´34.7 100 = 75.74%\ Cl = 99.9 ´ 2 131.9 (c) Cl2 reacted = 99.9 kmoles = 7084.91 kg 212.76 kg Cl2 gives 153.84 kg CCl4 7087.91 = 5122.87 kg CCl \7084.91 kg Cl gives 153.84 ´ 4 2 212.76 100 CCl4 produced/100 kg Cl2 reacted = 5122.87 ´ = 72.31 kg7084.91 7.49 Limestone is a mixture of calcium and magnesium carbonates and inert. Lime made by calcining the limestone by heating until the CO2 is driven off. When 100 kg of limestone is calcined 44 kg of CO2 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 is burnt with 100% excess air. The calcination is complete. (b) Calculate the composition of gases leaving the kiln. Analysis of coke C : 76%, ash : 21% and moisture : 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 + CO 32 100 56 44
MgCO → MgO + CO 32 84.3 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 inert 2 kg Coke = 1.52 kg Carbon; 0.06 kg moisture. Gases leaving kiln CO2, O2, N2 and H2O CO2 From CaCO3 = 3.5 88 = 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 atom O2 supplied (100% excess) = (0.1266 ´ 2) = 0.2532 kmole N79 = 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 23.18 O2 (Excess) 0.1266 8.99 N2 0.9525 67.60 H2O 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 the presence of catalyst. If the conditions are carefully controlled, a substantial fraction of the ethylene converted 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 which the 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 is converted to oxide?
CO2,O2, C 2
H 4
Reactor N 2
Exit Gases ReactorAir EO, CO ,O , C H ,N 2 2 2 4 2H2O
Reaction (1): 2C2H4 + O2 ® 2C2H4O Reaction (2): C2H4 + 3O2 ® 2CO2 + 2H2O Basis: 100 g moles of exit gas 6.4 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 g moles of O2 supplied. O2 consumed = oxygen supplied – oxygen remaining = (21.53 – 3) = 18.53 g moles O2 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 moles C2H4 supplied = C2H4 remaining + C2H4 consumed by Reaction (1) + C2H4 consumed by Reaction (2) = 6.4 + 8.26 + 4.8 = 19.46 g moles C2H4 converted to C2H4O = 8.26 100 = 42.4%´ 19.46 7.51 A spent dye sample obtained from a soap-making unit contains 9.6% glycerol and 10.3% salt. It is 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 hour Glycerol = 5,000 ´ 9.6 = 480 kg 100 Salt = 5,000 ´ 0.103 = 515 kg Water = 4,005 kg Entrained Vapour 5,000 kg/hr Evaporator Solution Salt
Final solution: Loss of glycerol = 480 ´ 0.045 = 21.6 kg Glycerol remaining = (480 – 21.6) = 458.4 kg 458.4= 573 kg\ Solution leaving = 0.8 Salt in it = 573 ´ 0.06 = 34.38 kg Water in this leaving solution = (573 – 458.4 – 34.38) = 80.22 kg Salt crystallized = (515 – 34.38) =
480.62 kg Vapours 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 flue gases. Basis: 100 g of coal: carbon present is 90 g = 7.5 g atoms C + O2 ® CO2 O2 needed for the above reaction is 7.5 g moles O2 supplied: 7.5 ´ 1.25 = 9.375 g moles (25% excess) N2 from air: 9.375 79= 35.27 g moles 21 O2 remaining 9.375 – 7.5 = 1.875 g moles Exit gases CO2 O2 N2Total g mole 7.5 1.875 35.27 44.645 mole % 16.8 4.2 79.0 100.0 7.53 Determine the weight of water removed while drying 1,000 kg of wet substance from 35% to 5%. Basis: 1,000 kg of wet material. Dry material is the ‘Tie element’. Moisture present is 1,000 ´ 0.35 = 350 kg Dry material is 1,000 – 350 = 650 kg Dry material appears as 95% in the exit. Therefore the total weight of material leaving is 650 = 684.21 kg 0.95 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 current multistage unit. The operating temperature is 24 °C and the solvent used is iso-propyl ether. Using the solvent in the ratio of 1.3 kg/kg of feed, the final extract composition on a solvent free basis is found to be 82% of acid. The raffinate is found to contain 14% of acid on solvent free basis. Find the percentage 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 kg Let E kg and R kg be the weight of extract and raffinate Acid balance: 0.475 = 0.82E + 0.14R Water balance: 0.525 = 0.18E + 0.86R Solving 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. The ore analyzes CaCO3 : 68%, MgCO3: 30% and rest silica. Acid used is 94% pure. Find: (a) CO2 produced (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 (a) CO2 produced by Reaction (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 during Reaction (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 SiO2 CaSO 324 4 2 2 84.3 98 120.3 44 18
4
formed: ÈØ136 ÉÙ = 924.8 kgÊÚ MgSO 4
formed: ÈØ120.3 ÉÙ = 428.1 kgÊÚ
Silica remaining from the ore = 20 kg Composition of the solid left behind CaSO4: 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 carbon burnt 90% forms CO2 and the rest forms CO. Calculate the composition of the flue gas on dry and wet basis. Basis: 100 g moles of the fuel gas The reactions taking place are: CH4 + 2O2 ® CO2 + 2H2O CH4 + 1.5O2 ® CO + 2H2O C2H6 + 3.5O2 ® 2CO2 + 3H2O C2H6 + 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) = 18 mole, 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 moles Component Weight, g mole gases CH4 7.0 C2H6 2.0 CO2 89.1 CO 9.9 O2 415.9 N2 2,257.0 Total (dry) 2,780.9 H2O 180.0 Total (wet) 2,960.9 Composition % Wet basis Dry basis 0.236 0.252 0.067 0.072 3.010 3.204 0.334 0.356 14.049 14.959 76.224 81.157 — 100.00 6.080 —
100.000 — 7.57 In a catalytic incinerator a liquid having a composition of 88% carbon and 12% hydrogen is vaporized 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 kmoles The reactions are: C + O2 ® CO2 H2 + 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 kmoles ÊÚ Water present in the exit gases = 12 = 6 kmoles2 Dry flue gas leaving is = 54.73 kmoles Oxygen 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 carbon forms 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) = 15 kmoles Nitrogen coming from air ÈØ ÉÙ 79 = 56.43 kmoles ÊÚ
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 68.13 mole % 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 Raschig rings. The gases enter the bottom of the tower with 14.8% of SO2. Water is distributed at the top of the column at the rate of 1,000 litres per minute. The total volume of gas handled at 1 atm and 30 °C is 1,425 m3/h. The gases leaving the tower are found to contain 1% of SO2. Calculate the percentage of 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 kmoles ÊÚ SO2 entering absorber: (57.281 ´ 0.148) = 8.4776 kmoles Inert gases are the tie element Inert gases: (57.281 – 8.4776) = 48.8034 kmoles If 99% is equal to 48.8034 kmoles then 1% is ÉÙ
48.8034 ÈØ = 0.493 kmole ÊÚ SO2 absorbed in the tower: 8.4776 – 0.493 = 7.9846 kmoles = 511 kg Amount of water flowing = 1,000 ´ 60 = 60,000 kg/h Weight of the total liquid leaving the tower is 60,511 kg Weight % of SO 2
= 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 ® 2CO i.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 = ÉÙ79ÊÚ O2 reacted = (21.37 – 8.4) = 12.97 kmoles for forming CO and CO2 For converting CO to CO2 additional O2 required is 0.5 kmoles \ O2 excess = 8.4 – 0.5 kmoles = 7.9 kmoles i.e. O2 required = 21.37 – 7.9 = 13.47 kmoles 7.9 t100 58.65%
excess air = 13.47
Carbon balance = a + b = 11.2 CO2 formed = ax = 10.2; CO formed: bx = 1 ax/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 = 0.8 Hence, 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 containing 95 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 get D = 353 kmoles/h W = 647 kmoles/h 7.62 One hundred kilograms of liquid mixture containing 30% A and 70% B is extracted with a solvent mixture containing C and D. After thorough mixing and allowing the system to reach equilibrium, two separate layers are observed. The composition of both the layers have been analyzed and given below: Layer Component ABCD Top 10 05 60 25 Bottom 20 60 05 15 Estimate (i) the weight of each layer, (ii) weight of solvent, and (iii) composition of C and D in solvent. Basis: 100 kg of feed Solvent Feed Extractor y
x
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 s Total material balance gives 100 + s = x + y (or) s = [x + y] – 100 Making component balance for A, we get 100(0.3) = x(0.1) + y(0.2) 30 = 0.1x + 0.2y Similarly, making component balance for B, we get 100(0.7) = x(0.05) + y(0.6) 70 = (0.05)x + 0.6y Solving, we get x = 80 kg, and y = 110 kg Solvent used is s = [x + y] – 100 = 110 + 80 – 100 = 90 kg Making a component balance for C and D Weight 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 another mixture 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 mixture Let the weight of 40% alcohol + 60% water mixture to be mixed be x kg Making a material balance for alcohol: 100(0.8) + x(0.4) = (100 + x)(0.5) i.e. 80 + 0.4x = 50 + 0.5x Solving, x = 300 kg Check: by making water balance 100(0.2) + x(0.6) = (100 + x)(0.5) i.e. 0.1x = 30 Solving, x = 300 kg 7.64 A gas mixture has CO : 10%, CO2: 50%, O2: 5% and rest nitrogen by volume. It is desired to have the nitrogen composition as 40% in the final mixture by mixing it with fresh air. Estimate the gas/air ratio to be maintained to achieve the composition as 40% in the final air. Also, estimate the composition of leaving air. Basis: 1 kmole of incoming gas mixture Since 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 kmole of air to give final product containing 40 mole % N2 in (1 + y) kmole of leaving air Making a material balance for nitrogen (1) (0.35) + y(0.79) = (1 + y) (0.4) 0.05 = 0.39y Solving, y = 0.128 kmole Gas to air ratio is 1/0.128 = 7.813 y kmole O2 : 21% N2 : 79% 1 kmole CO = 10% (1 + y) kmoleCO = 50% Mixer 2 N2 = 40%O2 = 5% N2 = 35%
Leaving stream: Component Weight, kmoles mole % CO 0.1 8.86 CO2 0.5 44.32 O2 0.05 + 0.128 × 0.21 = 0.07688 6.82
N2 0.35 + 0.128 × 0.79 = 0.45112 40.00 Total 1.128 100.00 Average 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 the leaving gas contains only 0.5% SO2. Estimate the fractional conversion of SO2 to SO3 and also the composition of gases leaving catalytic chamber. Basis: 100 kmoles of gas entering the chamber The reaction is SO2 + ½ O2 ® SO3 Let x be the kmoles of SO2 reacted So, unreacted SO2 = 5 – x Then, SO3 formed is x and the corresponding O2 reacted is x/2 moles (by stoichiometry) Balance O 2
= 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 x 2 Solving, we get x = 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 0.500 SO3 4.511 4.615 O2 10 – 4.511/2 = 7.744 7.923 N2 85.000 86.962 Total 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 fuel burnt. 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 of carbon in the clinker and the ash is 1% of the fuel used. The fuel gas has the following composition by weight: 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. It is burnt with 25% excess air. Assuming complete combustion, estimate the composition of leaving gases. 7.5 Limestone is burnt with coke having 85% carbon, producing a gas of 28% CO2, 5% O2, and rest N2. 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 and rest 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. The composition 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 of gases. 7.8 Thirty kilograms of coal analyzing 80% carbon and 20% hydrogen are burnt with 600 kg of air yielding a gas having an orsat analysis in which the ratio of CO2 to CO is 3 : 2. What is the percentage 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 755 mm 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 is converted 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. Assume complete combustion. 7.12 A synthetic fuel oil is known to contain only H and C, gives on combustion an Orsat analysis of CO2: 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 of pyrites so that the mixture will burn readily forming a burner gas that analyzes 13.4% SO2, 2.7% O2 and 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 separated by fractional distillation into a distillate containing 95 mole % butane, 4 mole % pentane and rest hexane 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 and sulphuric 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 ® 2NH3 In a given plant, 4,202 kg of nitrogen and 1,046 kg of hydrogen are fed to the synthesis reactor per hour. 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 25 weight % to 3 weight %. If the evaporator unit is to produce 14,670 kg/h of NaCl (along with 3 weight % 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 the CO2 is scrubbed out of the flue gas for the production of dry ice. The exit gas from the scrubber analyzes 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 complete combustion 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 Hg would 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 rest O2 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 same conditions 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 gases leaving the furnace assuming 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 coke burnt. (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. The analysis 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 the composition 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. Assuming that all the hydrogen is burnt and that carbon is partly burnt to CO and to CO2 without any free carbon remaining, 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 is produced than CO2. The producer is producing gas having CO:CO2 mole ratio as 5:1 from a coke containing 80% carbon and 20% ash. The solid residue after combustion carries with it 2% unburnt carbon. 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 furnace operates with 50% excess air. The ash contains 2% unburnt carbon. Of the carbon burnt 5% goes to form 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 is used 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 dry basis. 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 flue gas analysis, the oxides of phosphorus precipitate on cooling and are separated from the remaining gas. 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 gas partly 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 combustion products at 250 °C, when the coal refuse of the following composition burns with 50% excess air: Proximate analysis Moisture Ash Volatile matter Fixed carbon Ultimate analysis Carbon Hydrogen Nitrogen Sulphur Balance is oxygen Air dried % 8 20 28.5 43.5 Air dried % 81.0 4.6 1.8 0.6 If the rate of burning of coal is 3 tonnes/h, what is the capacity of the air blower used? Assume complete combustion. 7.32 Determine the flue gas analysis and the air–fuel ratio by weight when a medium viscosity of fuel–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 of flue 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 ultimate analysis 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 exhaust gas is: Composition Volume % CO2 11.5 O2 0.9 CO 1.2 N2 86 Also 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 stream comprising both reactants and products or the intermediate product is sent back along with feed to the system or somewhere in the middle of the system. Such a stream is called Recycle stream. This is done to improve the conversion whenever the conversion is low and to have energy economy in operations. This also improves the performance of an equipment as in the case of absorption of sulphur 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 a closer control in operation. This is done if there is a sudden change in the property of a fluid stream like excessive heating (or cooling) as it passes through a preheater (cooler) before entering another unit. In such cases this conditioned stream is mixed with a portion stream at its original condition and then 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 concentration of inert or impurities in the system. A stage may reach when the concentration of these components may cross permissible levels. By bleeding off a fraction of the recycle stream, this problem can be overcome. This operation is known as purging. This is quite common in the synthesis of ammonia and electrolytic refining of copper. The above (8.1, 8.2, and 8.3) definitions have been shown in Figure 8.1. 180 Bypass Fresh feed Mixing Process Feedunit Gross product SepaNet product rator Recycle 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 product recovered from the top contains 95% benzene while the bottom product contains 96% toluene. The stream entering the condenser from the top of the column is 8,000 kg/h. A portion of the product is returned to the column as reflux and the remaining is withdrawn as top product. Find the ratio of the amount refluxed to the product taken out. 8,000 kg/h V Condenser Distillation
RD Benzene 95% B 10,000 kg/h ‘F’column 50% B, 50% T W 96% Toluene
Figure 8.2 Overall balance F=D+W or, 10,000 = D + W Benzene balance gives, 5000 = 0.95D + 0.04W Solving, D = 5,050 kg/h W = 4,950 kg/h Balance around condenser gives, V=D+R or, 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 W 300°F F Feed 10,000 kg/h Evaporator 20% KNO3 50% KNO3 0.6 kg KNO /kg water (i.e. 0.6/1.6 = 0.375 KNO solution) M 3 3 R Crystallizer100°F C Crystal with 4% H2O
Figure 8.3 Basis: One hour. KNO3 entering = 2,000 kg/h 2000 \ Crystal leaving crystallizer, C = ¦µ §¶ = 2,080 kg/h¨· Overall balance F = C + W since F = 10,000 kg/h C = 2,080 kg/h \ W = 7,920 kg/h Crystallizer balance gives, M = C + R = (2,080 + R) KNO3 Balance gives 0.5M = 0.96C + 0.375R Thus R = 7,680 kg/h
8.3 Metallic silver may be obtained from sulphide ores by roasting to sulphates and leaching with water and subsequently precipitating silver with copper. In the Figure 8.4 shown below, the material leaving the second separator was found to contain 90% silver and 10% copper. What percentage excess 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 CuSO4 F Reactor S 1
S 2
90% Ag; 10% Cu Ag2SO4 Recycle R Ag2SO4
Figure 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 copper Ag 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 unconverted which goes in the recycle steam. Ag2SO4 balance: F = R + 1,300 or, F = 1,733.3 kg 0.25F = R or, 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 A H1= 0.0152 g B of wv/g 0.099 g H2O/g dry solid CD H = 0.0525 g 52.5 g dry air (e)of wv/g DA Drier 1.562 g H O/g dry solid 2 2
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)0.4 Solving, x = 33.3 units. Aw waste 40% A 20% A; Feed Product P 80% B A 5% and B 95% Recycle A only x
Figure 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 + P Solving: 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 is condensed and separated from the unreacted gases, which are recycled. The feed to the reactor contains 2 kmoles of H2 for every kmoles of CO. The fresh feed enters at 35 °C and 300 atm. To produce 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/h Unreacted CO, H2
Figure 8.7 Basis: One hour of operation Methanol produced = 6,600= 206.25 kmoles 32 CO + 2H2 ® CH3OH 15% conversion of CO entering gives 206.25 kmoles of methanol 206.25=
1,375 kmoles\ CO entering the reactor = 0.15 and 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 feed unconverted = 3,506.25 kmoles Fresh CO needed = 206.25 kmoles Fresh H2 needed = 412.50 kmoles Total moles of fresh feed = 618.75 kmoles\ (a) Volume of feed gas = 618.75 ´ 22.414 ´ 308 ¦µ¦ µ1 273§¶§ ¶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 kg Recycle 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 is supplied from a furnace burning coke. The hot flue gases analyze 5% CO2. The kiln gas contains 8.65% CO2. In order to conserve some of the sensible heat a portion of the kiln gas is continuously recycled 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. Limestone K KilnP X RAir F Burner Coke
Figure 8.8 Basis: 12 kg of coke represent 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 entering the 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 32 100 56 44
Assuming complete combustion 0.05F = 1 kmole CO2\ F = 20 kmoles (Q 1 kmole of CO2 comes from 12 kg of Carbon) C + O2Æ CO2 Solving (a) and (b), we have R = 24.2 kmoles; X = 44.2 kmoles Let m kmole of CO2 be added in kiln from the calcinations of limestone \ K = 44.2 + m Making CO2 balance, (0.07 X) + m = K¥ 0.0865 or, 44.2 ¥ 0.07 + m = (44.2 + m)(0.0865) \ m = 0.8 kmole \ K = 44.2 + 0.8 = 45.0 kmoles CaO formed = (0.8 ¥ 56) = 44.8 kg (a) CaO formed/kg of coke burnt = 44.8= 3.73 12 (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 has 500 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 cell 3.1 salt % 4% sea water
Desalinated water “D”, 0.05% salt
Figure 8.10 Basis: One hour Overall balance, 1,000 = B + D Salt 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 = A Salt balance, 1,000 ´ 0.031 + R ´ 0.0525 = A ´ 0.04 Solving, 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 of steam-reforming/partial oxidation process. Enough air is used in partial oxidation to give a 3:1 H2-N2 molar ratio in the feed to the ammonia unit. The H2-N2 mixture is heated to reaction temperature and fed to a fixed bed reactor where 20% conversion of reactants to NH3 is obtained per pass. The products from the reactor are cooled and NH3 is removed by condensation. The unreacted H2-N2 mixture is recycled and mixed with fresh feed. On the basis of 100 kmoles per hour of fresh feed determine the NH3 produced and recycle rate. Fresh feed Reactor NH3
Figure 8.11 Basis: 100 kmoles of feed Given that H2 : N2 = 3 : 1 hydrogen: 75 kmoles and nitrogen: 25 kmoles N2 + 3H2 ® 2NH3 Overall balance gives: Feed NH3
NH3 formed = 50 kmoles One mole of N2 gives 2 moles of NH3Reactor balance gives: 3x, H2 Reactor NH , N and H 3 2 2x, 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 H2 Unreacted N2 = 100 kmoles, and H2 = 300 kmoles Ans.: NH3 produced = 50 kmoles Recycle = 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 and the degreased surface per day is 105 m2.
Greased metal Solvent S 0% grease G Cleaned metalB 15% Grease solvent Separator ADegreasing 40% grease 1% grease Rwith Solvent Recyclesolvent
Figure 8.12 Grease removed/day = 1,00,000/100 = 1,000 kg (i) Overall balance gives: G 100% 40% Grease S Process A 100% 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 8.13 at 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 the flow rates in kg/h and compute the composition of feed to the evaporator (F) R Fresh feedF W Evaporator Crystallizer P NaCl only KCl only
Figure 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/h NaCl in feed = 18,400 ¥ 0.1 = 1,840 kg/h Overall balance: water vapour Feed, F Process KCl NaCl
Balance around crystallizer: Overall balance: P = R + 552 NaCl balance: 0.168P = 0.189R P Crystallizer R KCl
Solving, R = 4,416 kg and P = 4,968 kg Balance around evaporator: W F Evaporator P NaCl
F = W + NaCl + P F = 16,008 + 1,840 + 4,968 = 22,816 kg/h Feed to evaporator = recycle + fresh feed = 18,400 + 4,416 = 22,816 kg/h Recycle, R Fresh feed Feed to evaporator, F, m (Conc. of NaCl) n (Conc. of KCl)
Making a balance for NaCl, we have 18,400 ´ 0.1 + 0.189 ´ 4,416 = m ´ 22,816 Solving, 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 the catalytic bed of the reactor contains 10:1 volume ratio of oxygen to ethylene and the conversion per pass is 23%. Ethylene oxide is removed from the products completely and the unreacted ethylene is recycled. The oxygen for the reaction is supplied from air. Calculate: (a) inlet and outlet composition of the streams and (b) moles of fresh oxygen required for recycle gases. Reaction: C2H4 + 0.5O2 ® C2H4O. Separator Ethylene ReactorGasesAir
Figure 8.14 Basis: 1 kmole of ethylene. Oxygen supplied: 10 kmoles Nitrogen entering: 10 79 = 37.62 kmoles´ 21 C2H4 reacted: 1 ´ 0.23 = 0.23 kmole C2H4O formed: 0.23 kmole Oxygen reacted = 0.23= 0.115 kmole2 Oxygen remaining = 10 – 0.115 = 9.885 kmoles (a) Inlet gases mole mole % C2H4 1.0 2.056 O2 10.0 20.560 N2 37.62 77.384 Total 48.62 100.000
Ethylene recycled = (1 – 0.23) = 0.77 mole (b) Moles recycled/mole of feed = 0.77 = 0.0159 48.62 oxygen required = 0.115 mole (c) Outlet gases mole mole % C2H4O 0.230 0.48 O2 9.885 20.71 N2 37.620 78.81 Total 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 feed EP A Feed, FI Unit II Unit B
Figure 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 + 0.0084B C4 balance: 40 = (0.01 ´ 21.053) + 0.01P + 0.916B Solving 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 a SO3 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 kg SO HO +→H SO 32 24 80 18 98
SO3 absorbed = 4,301 ´ 0.985 = 4,236.3 kg Acid 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% tower SO 3
in 28 kmoles/h z 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 + r Acid 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. EXERCISES 8.1 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 latter recycled. Compute (a) moles of NO formed per 100 moles of NH3 fed, and (b) moles of NH3 recycled 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 dried to 0.25 kg of water per kg of dry polymer per hour. 5,000 kg of dry air is passed into the drier. The air leaving the drier is having a humidity of 0.0045 kg of water vapour per kg of dry air and fresh air supplied at a humidity of 0.011 kg of water vapour per kg of dry air. Calculate the mass rate of fresh air 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 conversion of our resources into energy effectively and utilize the same properly. In order to understand the basic principles pertaining to the generation, transformation and uses of energy, the following terms need to be 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. The activity may be looked upon as a thermodynamically corrected pressure or concentration. For pure solids, 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 is not large. In the case of dissolved substances the standard state is the concentration in each instance at which the activity is unity. The enthalpies of substances in standard states are designated by the symbol H°, while the DH of a reaction where all reactants and products are at unit activity is represented 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 the heat of formation of a substance. The standard heat of formation is the heat of formation when all the substances involved in the reaction are each at unit activity. The enthalpies of all elements in their standard states at 298 K are zero. 196
9.1.3 Heat of Combustion It is the heat liberated per mole of substance burned. The standard heat of combustion is that resulting from the combustion of a substance, in the state that is normal at 298 K and atmospheric pressure, with the combustion beginning and ending at 298 K. 9.1.4 The Heat of Reaction Heat of reaction from enthalpy data It is defined as the enthalpy of products minus the enthalpy of reactants. The standard heat of reaction for the reaction, aA + bBÆ cC + dD is given by D H° = [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 reaction positive 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 of reaction directly from the standard heats of combustion instead of standard heats of reaction. The standard heat of reaction under such circumstances is the standard heat of combustion of the reactants minus the standard heat of combustion of products. i.e. [DH = SDH , – SDH , ] 9.1.5 Heat of Mixing When two solutions are mixed, the heat evolved or absorbed during the mixing process is known as heat of mixing. 9.2 HESS’S LAW reaction
c (reactants)
c products 25°C
If a reaction proceeds in several steps, the heat of the overall reaction will be the algebraic sum of the heats of the various stages, and this sum in turn will be identical with the heat, the reaction would evolve 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 the constituent elements. As per Kopp’s rule, the following atomic heat capacities are assigned to the elements at 20 °C: Carbon: 1.8; Hydrogen: 2.3; Boron: 2.7; Silicon: 3.8; Oxygen: 4.0; Fluorine: 5.; Phosphorus: 5.4 and all others: 6.2. Since the heat capacities of solids increase with temperature, it is clear that these values 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 reaction proceeds without loss or gain of heat and if all the products of the reaction remain together in a single mass or stream of materials. 9.5 THEORETICAL FLAME TEMPERATURE The temperature attained when a fuel is burnt in air or oxygen without loss or gain of heat is called the Theoretical flame temperature. WORKED EXAMPLES 9.1 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 kJ The 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 is from a system, which supplies 210 kcal/ kg, at steady state. Find the quantity of heat to be supplied by the system. Atomic weight of Cadmium = 112.4, Cp = (6 + 0.005T) kcal/kmole °C and T in °C. Latent heat of fusion = 2050 kcal/kmole Basis: 200 kg of Cadmium º 1.78 katoms ©¸ ¦µ 2
Sensible heat = 1.78t tª¹ = 3885.5 kcal ª¹ «º Latent heat of fusion = 1.78 ´ 2050 = 3649.0 kcal Total 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 and the weight of steam used. Enthalpies of feed are 38.1 kcal/kg, product solution is 100.8 kcal/kg, steam is 540 kcal/kg and that of the vapour is 644 kcal/kg. Basis: One hour. Feed = 10,000 kg/h Vapour formed is 5000 kg/h Thick liquor is 5000 kg/h Enthalpy of feed = 10,000 ´ 38.1 = 38.1 ´ 104 kcal Enthalpy 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 + C2H2 DHf 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 kmole80 At 373.16 K, Q = nòCpSO3 dt At 273.16 K Q = [{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) = 94,052 Desired 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 55.1 Desired equation: Zn + S + 2O2 ® ZnSO4 kcal/mole Steps: 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 segregated from the reactants, at 250 °C, is saturated and completely condensed. The reaction absorbs 1000 Btu/lb of charge in the reactor. Heat loss from the steam chest to the surroundings is 5000 Btu/h. The reactants are placed in the vessel at 70 °F. At the end of the reaction, the materials are at 212 °F. If the charge contains 325 lb of material and the products and reactants have an average Cp of 0.78 Btu/1b °F, how many lb of steam are needed per lb of charge. The charge remains for an hour in the vessel. Basis: One hour: Datum 70 °F Btu Reaction absorbs heat = 1000 ´ 325 = 3,25,000 Heat loss to surroundings = 5,000 Heat in products: 325 ´ 0.78 (212 – 70) = 36,000 \ Q = total heat = 3,66,000 From 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= 1.533325 9.11 Pure ethylene is heated from 30 °C to 250 °C at a constant pressure. Calculate the heat added per kmole Cp = 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 kmole Or, 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 °C to –100 °C at constant pressure. Cp air = 6.386 + 1.762 ´ 10–3 T – 0.2656 ´ 10–6 T2, where Cp is in kcal/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)]´ 10 Q = –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 the compressor 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. 'v¦µ (200) = 0.8 Btu/lb¨· §¶ tt 778)
22
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/kmole DH°rxn, 298 K: 52,280 – (–84,720) = 1,37,000 kJ DHrxn = 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) by using 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 = 3 kmoles 2
T
(a) DH = n CdTp = 3 [6.85 ´ 900 + (8.533 ´ 10–3/2)(12732 – 3732)∫
T1
– (2.475 –6 /3)(12733 – 3733)]´ 10 DH = 3[10,826.29] = 32,478.87 kcal Cp values at 1273 K and 373 K are: Cp at 1273 K = 13.702 kcal/kmole K and Cp at 373 K = 9.6884 kcal/kmole K Cpav = 11.6952 kcal/kmole K DH = 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 the converter at 400 °C and leave at 450 °C. How many kcals are absorbed in the heat exchanger of the converter 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 11.0 7.5 7.1 — Hrxn = –23, 490 cal/g mole (given)D DHrxn = –23,490 ´ 0.7 = –16,443 kcal Datum: 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 kcal N2 = 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/kmole Cp 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 –10.96 Basis: 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)] = –13.9 Db = (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 D H o
= D H rxn
– Da T – '¦µ T2 –'¦µ T3§¶ §¶ ¨· ¨· 0.0119 t 298 7.9 10 73
2
D H o
= –21.92 – (–13.9 ´ 298) – 2 – t t298 3 DHo= 3,598.87 kcal 480 °C = 753 K Hrxn, 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 theoretical amount of air is used, the combustion reaction is complete and reactants enter at 25 °C. Cp = a + bT + cT2, cal/kmole K where a, b and c are all dimensional constants and available in literature 3c ´ 105Gas a b ´ 10 CO2 10.55 2.16 –2.04 N2 6.66 1.02 — C CO CO 2
CO II Combustion2I CombustionO2 N2Air N2
Basis: 100 g moles of inlet gas CO 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 ® CO2 Reaction (a) – (b) gives DHrxn = –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 above equation 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 or removed if the reactants enter and products leave at 500 °C (in effect, we have to calculate the heat of reaction 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)] = –14.891 Db = [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 ´ 10–6Dg DCp = –14.891 + 11.047 ´ 10–3T – 1.719 ´ 10–6 T2 We find DHo using data at 298 K '¦µ T2 – '¦µ T3DH = o
DHrxn – DaT – §¶ §¶
¨· ¨· = –39,433 – (–14.891 ´ 298) – 11.047 10 2 tt§¶2¨· ¦µ 3¦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) 773
K: '¦µ T3DH 773 =
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 ® CO2 O2 supplied = 0.5 ´ 1.5 = 0.75 lb mole (50% excess) Air supplied = 3.57 lb mole and N2 = 2.82 lb moles DHrxn = –1,21,745 Btu/lb mole Q = DHrxn + DHproducts – DHreactants O2 remaining = 0.25 lb mole CO2: 1 lb mole, N2: 2.82 lb moles Datum: 32°F Data: DH (Btu/lb mole) Temperature , °F CO Air O2 N2 CO2 50 125.2 — — — — 77 313.3 312.7 315.1 312.2 392.2 800 — — 5,690 5,443 8,026 1000 — 6,984 — — — DH
products
= DH
800°F
– DH
77°F
= 1(8,026 – 392.2) + 2.82(5,443 – 312.2) + 0.25(5,690 – 315.1) = 23,446 Btu/lb mole DH 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 mole Q = –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 constant pressure. The reactants are originally at 200 °F. Determine the heat added or removed, if the product temperatures are 200 °F, 500 °F, 1000 °F, 1500 °F, 2000 °F and 3000 °F. Basis: 1 lb mole of CO CO + ½O2 ® CO2 O2 supplied = 1 lb mole, N2 = 3.76 lb moles Exit: 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 = DH DH = 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 6.95 O2 1 7.10 7.10 N2 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 Cp nCp 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 12.75 O2 0.5 7.10 3.55 7.25 3.63 7.15 3.88 7.8 3.9 8.0 4.0 8.3 4.15 N2 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 — 46.32 SnCp (t–77) 4,776 16,857 38,314 61,246 83,758 1,35,810 Q = 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%. What is the enthalpy difference for this gas between the bottom and the top of the stack if the temperature at the 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 them together, we have for N2 = 0.82 ´ CpN2 for O2 = 0.073 ´ CpO2 for CO2 = 0.092 ´ CpCO2 for CO = 0.015 ´ CpCO Cpnet = 7.049 + 1.2243 ´ 10–3 T – 2.6164 ´ 10–7 T2 + 0.2815 ´ 10–10 T3 200
∫Cp
dT\DH
net
=
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 flame temperature for CO burnt at constant pressure with 100% excess air? The reactants enter at 200 °F. CO + ½O2 ® CO2 Basis: 1 g mole CO Temperature of reactants: 200 °F = 93.3 °C Gases 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) 6.981 476 Air 4.76 (93.3 – 25) 6.993 2,270 Total 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 23,000 O2 0.5 1775 8.35 7,400 N2 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 cal Making 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 burnt with 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 N 2
in feed =1 ÈØ ÉÙ = 4 g molesÊÚ N 2
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. (Datum 298 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, K 67,640 = [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 °C 1106 K ∫ 833 °C. 9.25 Find the theoretical flame temperature of a gas containing 30% CO and 70% N2 when burnt with 100% 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/kmole K at different temperatures is given below: Temperature K CO2 O2 N2 800 45.4 31.6 30.3 1000 47.6 32.3 30.6 1200 49.4 33.0 31.2 1400 50.8 33.6 31.8 1600 52.0 34.0 32.3 1800 53.2 34.4 32.7 Basis: 1 kmole of CO, Datum: 298 K
N2 in feed = 70/30 = 2.34 kmoles O2 supplied = 0.5 ¥ 2 = 1 kmole N2 from air = 3.76 kmoles Exit gas consists of: CO2: 1, O2: 0.5, N2: (3.76 + 2.34) = 6.1 kmoles Let us consider the equation Q = SH
products
+ DH –SH rxn
reactants
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 K DHpr = (1 ¥ 50.8 ¥ 1102) + (0.5 ¥ 33.6 ¥ 1102) + (6.1 ¥ 31.8 ¥ 1102) = 2,88,261 kJ/kmole π 2,83,100 kJ/kmole Let the theoretical flame temperature be 1200 K \DT (1200 – 298) = 902 K DHpr = (1 ¥ 49.4 ¥ 902) + (0.5 ¥ 33 ¥ 902) + (6.1 ¥ 31.2 ¥ 902) = 2,31,110 kJ/kmole π 2,83,100 kJ/kmole So 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 to 425 °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 15,000 22.414
= 669.2 g moles we can then write the amount of each component SO2 : 669.2 ´ 0.1 = 66.92 g moles O2 : 669.2 ´ 0.12 = 80.30 g moles N2 : 669.2 ´ 0.78 = 521.98 g moles 669.20 g moles Reference 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 flame temperature of a gas containing 20% CO and 80% N2 when burnt with 100% excess air. Both air and gas 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 kcal Basis: 1 kmole of CO; N2 = 4 kmoles Air supplied: O2: 1 kmole, N2 from air: 3.76 kmoles Exit gas: CO2: 1, O2: 0.5, N2: 7.76 kmoles Datum 25 °C = 298 K DHrxn = –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 T 298 2 33 T 298– 3.415 ´ 10–6 T 298 3+ 6.117 ´2 –3
T
22 33
+ 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 ´ 10 298 2T –3
33
– 7.76 ´ 0.069 ´ 10 298 3 Solving for theoretical flame temperature = T = 1216 K = 943 °C –6
9.28 Dry methane and dry air at 298 K and 1 bar pressure are burnt with 100% excess air. The standard heat of reaction is –802 kJ/g mole of methane. Determine the final temperature attained by the 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 moles N2 entering = 4 79 = 15.05 g moles´ 21 Gases leaving are: CO2: 1, H2O : 2, O2: 2, and N2: 15.05 g moles. Heat given out = 802 kJ Heat 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 get SO2. The reaction is given. All the gangue plus Fe2O3 end up in the solid waste produced which analyzes 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 solid waste, 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 kg Solid waste = (0.15) + (0.029) + [0.666 ´ (0.85 – 0.029)] = 0.725786 kg A summary of the composition of the solid waste is given: Solid waste kg Weight % Gangue 0.150 20.66 FeS2 0.029 3.99 Fe2O3 0.547 75.35 Total 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 kmole SO 2
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 kJ 9.30 For the following reaction, estimate the heat of reaction at 298 K. A + B ® C + D Compound DH°f (kcal/g mole) A –269.8 B –195.2 C –337.3 D –29.05 DH° at 25 °C = SDH°f, Products – SDH°f, reactants
= [–337.3 – 29.05] – [–269.8 – 195.2] = 98.65 kcal 9.31 Estimate the standard heat of reaction DH°298 for the reaction. A + B ® C Standard heats of combustion are: DHc, 298 for A = –328000 cal/g DHc, 298 for B = –212000 cal/g DHc, 298 for C = –542000 cal/g DHc, 298 = SDH°c, reactants – SDH°c, products = [–328000 – 212000] – [–542000] = 2000 cal 9.32 Calculate the heat of formation of CHCl3 from the following data: CHCl3 + 1 O + H O ® CO + 3HCl, DH = –509.93 kJ (1) 2 2 2 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 kJ Eq. (4) × 3 + Eq. (3)/2, gives 3H + 2
Cl2 + 1 C + 1 O2 ® 3HCl + 1 CO22 2 2 2 9.33 Standard heat of reaction accompanying any chemical change is equal to the algebraic sum of the standard heat of formation of the products minus the algebraic sum. Calculate the standard heat of reaction, DH°f, 298 for the reaction 2FeS2 + 1.5O2 ® Fe2O3 + 4SO2 The standard heats of formation are: FeS2 = –42,520 cal/g mole Fe2O3 = –1,96,500 cal/g mole SO2 = –70,960 cal/g mole From the reaction, DH°r, 298 = DHFe2O3 + 4 [DHSO2] – 2[DHFeS2] = –19,6500 + 4[–70,960] – 2[42,520] = –39,5300 cal 9.34 The heat of reaction at 300 K and 1 atm pressure for the reaction A + 3B ® C is 30,000 cal/mole A converted.
Cp data is as follows: A = –0.4 + 0.1 T (T in K) B = 10 C = 30 Calculate the heat of reaction at 600 K and 1 atm A + 3B ® C 600
H600 = DH300 + ±'Cp.dTD 300 600
= –30,000 + ±[{303 × 10 + 0.4} 0.1T] dT 300
= –30,000 + 0.4 T – 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 burnt with 150% excess air, with both air and gas being at 25 °C. D H°f Cpm CO2= –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 K Basis: 100 g moles of feed CO = 20 g moles CO + 1 O Æ CO 2 22 O2 needed = 1 × 20 = 10 g moles2 O2 supplied = 2.5 × 10 = 25 g moles N2 supplied = 25 × 79/21 = 94.05 g moles Gases leaving CO2= 20 g moles O2 = 25 – 10 = 15 g moles N2 = 94.05 + 80 = 174.05 g moles Atmospheric temperature = 25 °C Heat 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 kJ Heat 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 methane with 20% excess air. Air and methane enter at 298 K and a pressure of 1 atm. The reaction is complete. –DHr = 1,91,760 cal/g mole Mean 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 ® SO3 Mean molar specific heats of SO2 : 51.5 kJ/kmole K O2 : 45.67 kJ/kmole K SO3 : 30.98 kJ/kmole K Standard heat of formation for SO2 : –2,97,000 kJ/kmole SO3 : –3,95,000 kJ/kmole 9.3 Calculate the enthalpy change in J/kmole that takes place in raising the temperature of 1 kmole of the gas mixture of 80 mole %. Methane and rest ethane from 323 K to 873 K Heat capacity equation, Cp = R (A + BT + CT2) cal/g mole where, R is Gas constant, T is temperature in 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 1.131 19.225 –5.561 9.4 Chlorine is produced by the reaction 4HCl(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 is transferred per mole of entering gas mixture. Data: (a) Standard heat of formation at 25 °C, J/g mole HCl(g) : –92,307 J/g mole
H2O(g) : –2,41,818 J/g mole (b) Mean heat capacities: (cal/g mole K) HCl(g) : 7.06 O2(g) : 8.54 H2O : 7.52 Cl2(g) : 8.61 N2(g) : 7.16 9.5 Calculate the number of joules required to calcine completely 100 kg of limestone containing 80% CaCO3, 11% MgCO3 and 9% water. The lime is withdrawn at 900 °C and the gases leave at 200 °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 mole MgCO3 : – 2,66,000 cal/g mole CaO : – 1,51,900 cal/g mole MgO : – 1,43,840 cal/g mole CO2 : – 94,050 cal/g mole Mean molal heat capacity, cal/g mole K H2O : 8.2 CO2 : 10.5 CaCO3 : 25.0 MgCO3 : 23.0 CaO : 14.0 MgO : 10.0 9.6 In the reaction 4FeS2(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% excess air. The reactants enter at 366 K. CO + ½O2 ® CO2 The 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/g mole 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 K to 313 K at 1 atm pressure. Calculate the heat removed. The Cp values in kcal/kmole K, and T in K are: 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 of methane with 25% excess air. Air and methane enter at 298 K and a pressure of 1 atm. The reaction is complete. Standard heats of formation at 298 K in kJ/kmole are: Methane (g) = –74,520 Carbon dioxide (g) = –3,93,500 Water vapour (g) = –2,41,813 9.10 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 products leave 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 kcal Mean specific heats are: 7.017 for CO, 7.225 for air, 11.92 for CO2, 7.941 for O2 and 7.507 for N2 in kcal/kmole K.
Problems on Unsteady10State Operations The term unsteady state refers to chemical processes in which the operating conditions generally fluctuate with time. Although unsteady-state processes are difficult to formulate, the general formula used 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 guiding principle 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 fresh feed of 500 kg/min of pure water is entering the tank and dilutes the solution in the tank. The mixture is stirred well and the product leaves the tank at a rate of 500 kg/min. At what instant of time the acid concentration in the tank will drop to 1% acetic acid by weight? After one hour of operation, what will be the concentration in the tank? FiXi M, X Fo , Xo = X 221
Here inlet flow rate, Fi = 500 kg/min Outlet flow rate, Fo= 500 kg/min Initial mass = 10,000 kg The total mass M, at any time in the tank = Initial mass + (Inflow rate – Outflow rate)(time) = 10,000 + (500 – 500)t = 10,000 kg We 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)dt
dXdM 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)dt dX Now substituting values, 10,000 ÉÙ ÈØ = 500 ´ 0.0 – 500X (5) ÊÚ dX (i.e.) = – 0.05Xdt (i.e.)dX= – 0.05dtX Integrating, we get Xt
dX 0.05dtX ±± o Xt 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 enters the tank at a rate of 2 kg/min at a salt concentration of 3% by weight. The contents are stirred well and 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 salt concentration in the tank will reach 2.5% by weight? Here inlet flow rate, Fi = 2 kg/min
Outlet flow rate, Fo= 1.5 kg/min Initial 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 + 0.5t Differentiating, we get dM= 0.5 dt We 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 i.e. d MX()=
FiXi – FoX (2)dt
or, M dX + X dM = FiXi – FoX (3)dt dt Here,dM= 0.5dt Therefore, 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©¸ ±±Xt o
ª¹ Xt 0 Xt
also, 0.5 dXdt
±± Xt Xt0
o
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 ÌÜtÍÝ 204
\ 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 minutes Aliter We 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 is yeòPdx = 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 tt
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 + constant 20ÌÜtÍÝ or, X(20 + t)4 = Ô[0.12][20 + t]3dt + constant X (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 = 0.02 Substituting in Eq. (14), we get, Constant = – 0.01 ´ (20)4 = –1600 Equation (14) thus becomes, X = 0.03 – 1600and (15) (20 t)4 204
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 a concentration 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 solution enters the tank at a rate of 1.5 litres/min at a salt concentration of 1 g/litre. The contents are stirred well 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 in the 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 = 1.0 litre/min Initial volume = 10 litres. The total volume V, at any time = Initial volume + (Inflow rate – Out flow rate) (time) = 10 + (1.5 – 1.0)t or, V = 10 + 0.5t Differentiating, 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 output i.e. d VC()= F C – F C (2) i i o 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)dt dC or, dt (5) 1.5(1=-+Ct Integrating, we get Ct
∫∫dt 1.5
Ct0.5 )
=0oCt
Ct
∫∫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% concentration enters 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 salt concentration in the tank after 3 hours. 10.2 A tank contains 1000 kg of a 10% salt solution. A stream containing salt at 20% concentration enters the tank at 20 kg/min and the mixture leaves the tank after complete mixing at a rate of 10 kg/min. Obtain an expression for the salt concentration in the outlet as a function of time and the salt concentration in the tank after 1 hour. What will be the time at which the salt concentration in the tank will be 15%? 10.3 A tank contains 50 litres of a salt solution at a concentration of 2.5 g/litre. Another salt solution enters the tank at a rate of 2.0 litres/min. at a salt concentration of 1 g/litre. The contents are stirred well 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 in the 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 enters the tank at a rate of 2.5 kg/min at a salt concentration of 3% by weight. The contents are stirred well and the mixture leaves the tank at a rate of 2.0 kg/min. (a) Express the salt concentration as a function of 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 of pure water is entering the tank and dilutes the solution in the tank. The mixture is stirred well and the product leaves the tank at a rate of 400 kg/min. At what instant of time the sugar concentration in the tank will drop to 1% sugar by weight? After one hour of operation, what will be the concentration in the tank? 10.6 A 15% Na2SO4 solution is fed at the rate of 12 kg/min into a mixer that initially holds 100 kg of a 50 : 50 mixture of Na2SO4 and water. The exit solution leaves at the rate of 9 kg/min. Assuming uniform mixing, what is the concentration of Na2SO4 in the mixer at the end of 10 minutes. Volume change 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 of diameter d at the bottom of the tank, which was plugged initially is opened to let the liquid drain through it. Set up an unsteady state balance equation to calculate the time for the liquid level to fall to a new height H1. 10.8 A solution is having solute A at a concentration CAf is fed continuously into a mixing vessel of constant 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 the equation.
Tables TABLE I Important Conversion Factors Quantity Length To 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 4.448 kgf N 9.807 Pa N 980.7 dyne N 10–5TABLE I Important Conversion Factors (contd.) (Contd.) 229
Quantity To convert Multiply by from to Pressure lbf/ft2 lbf/in2 (psi) in Hg in water mm Hg atm torr bar kgf/cm2 N/m2 = Pa 47.88 N/m2 = Pa 6895 N/m2 = Pa 3386 N/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 Btu erg cal kcal kW.h ft3/s ft3/h, ft3/h cm3/s lit/h J = 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/lb cal/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 4187 N.m/kg K = J/kg K 4187 TABLE 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 Ba Berkelium Bk Beryllium Be Bismuth Bi Boron B Bromine Br Cadmium Cd Calcium Ca Californium Cf Carbon C Cerium Ce Cesium Cs Chlorine Cl Chromium Cr Cobalt Co Columbium Nb Copper Cu Curium Cm Dysprosium Dy Erbium Er Europium Eu Fluorine F Francium Fr Gadolinium Gd Gallium Ga Germanium Ge 89 217.00 13 26.98 95 243.00 51 121.76 18 39.94 33 74.91 85 210.00 56 137.36 97 245.00
4 9.01 83 209.00 5 10.82 35 79.92 48 112.41 20 40.08 98 246.00 6 12.01 58 140.13 55 132.91 17 35.46 24 52.01 27 58.94 41 92.91 29 63.54 96 243.00 66 162.46 68 167.20 63 152.00 9 19.00 87 223.00 64 156.90 31 69.72 32 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 Fe Krypton Kr Lanthanum La Lead Pb Lithium Li Lutetium Lu Magnesium Mg Manganese Mn Mercury Hg Molybdenum Mo Neodymium Nd Neptunium Np Neon Ne Nickel Ni Niobium Nb Nitrogen N Osmium Os Oxygen O Palladium Pd Phosphorus P Platinum Pt Plutonium Pu Polonium Po Potassium K Praseodymium Pr 79 197.20 72 178.60 2 4.00 67 164.94 1 1.00 49 114.76 53 126.91 77 193.10 26 55.85 36 83.80 57 138.92 82 207.21 3 6.94
71 174.99 12 24.32 25 54.93 80 200.61 42 95.95 60 144.27 93 237.00 10 20.18 28 58.69 41 92.91 7 14.01 76 190.20 8 16.00 46 106.70 15 30.98 78 195.23 94 242.00 84 210.00 19 39.10 59 140.92 Element Symbol Atomic Number Atomic weight Promethium Pm Protactinium Pa Radium Ra Radon Rn Rhenium Re Rhodium Rh Rubidium Rb Ruthenium Ru Samarium Sm Scandium Sc Selenium Se Silicon Si Silver Ag Sodium Na Strontium Sr Sulphur S Tantalum Ta Technetium Tc Tellurium Te Terbium Tb Thallium Tl Thorium Th Thulium Tm Tin Sn Titanium Ti Tungsten W Uranium U Vanadium V Xenon Xe Ytterbium Yb Yttrium Y Zinc Zn Zirconium Zr 61 145.00 91 231.00 88 226.05 86 222.00 75 186.31 45 102.91 37 85.48 44 101.70 62 150.43 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 52 127.61 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 50.95 54 131.30 70 173.04 39 88.92 30 65.38 40 91.22 TABLE III(a) Empirical Constants for Molal Heat Capacities of Gases at Constant Pressure Cp = a + bT + cT2, where T is in Kelvin; g-cal/(g-mole) (K) – Temperature range 300 to 1500 K Gas a b ´ 103c ´ 106 H2 6.946 – 0.196 0.4757 N2 6.457 1.389 –0.069 O2 6.117 3.167 –1.005 CO 6.350 1.811 –0.2675 NO 6.440 2.069 –0.4206 H2O 7.136 2.640 –0.0459 CO2 6.339 10.140 –3.415
SO2 6.945 10.010 –3.794 SO3 7.454 19.130 –6.628 HCl 6.734 0.431 +0.3613 C2H6 2.322 38.040 –10.970 CH4 3.204 18.410 –4.480 C2H4 3.019 28.210 –8.537 Cl2 7.653 2.221 –0.8733 Air 6.386 1.762 –0.2656 NH3(*) 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, C = 7. 95 + mTn For –180 to 95 p T in Rankine (oF + 460) Compound a b3 –c ´ 106m n ´ 10 Methane 3.42 9.91 1.28 6.4 ´ 10–12 4.00 Ethylene 2.71 16.20 2.80 8.13 ´ 10–11 3.85 Ethane 1.38 23.25 4.27 6.20 ´ 10–8 1.79 Propylene 1.97 27.69 5.25 2.57 ´ 10–3 1.26 Propane 0.41 35.95 6.97 3.97 ´ 10–3 1.25 n-Butane 2.25 45.40 8.83 0.93 ´ 10–2 1.19 i-Butane 2.30 45.78 8.89 0.93 ´ 10–2 1.19 Pentane 3.14 55.85 10.98 3.9 ´ 10–2 1.00
Answers to Exercises CHAPTER 1 1.1 (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 K 1.2 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 F 1.5 As long as consistent units are used, the equation remains the same. CHAPTER 2
2.1 500 g moles 2.2 0.5455 kg of carbon 2.3 (a) 3.572 g of O2 (b) 12.77 g of KClO3 2.4 (a) 2.8% (b) 0.088 (c) 5.378 g moles/kg of water 2.5 Mole ratio: 0.0425 Mole %: Na2CO3: 4.1% Water: 95.9% 2.6 Molality: 5.98 g moles/kg of solution Molarity: 5.98 g moles/litre Volume of solution: 0.0836 litre Normality: 5.98 235
2.7 Weight %: 39.02% Volume %: 26% 2.8 Compound Weight % Volume % mole % NaCl 23.3 11 8.54 H2O 76.7 89 91.46 Total 100.00 100 100.00 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: 65.02 Chlorine: 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 Cr2S3 2.13 9.3445 kg of AgNO3 2.14 (a) Mole fraction of H3PO4: 0.02 Mole fraction of Water: 0.98 (b) 890.8 cc of solution 2.15 54,090 g 2.16 31.75 kg Cu, 104.26 kg of 94% H2SO4
2.17 (a) H2SO4 (b) 79.25% (c) 89.625 2.18 NaCl:75%, KCl 25% and NaCl: 79.25%, KCl 20.75% 2.19 2094 kg iron and 900 kg water 2.20 (a) 26.94 (b) Methane: 24.94%, ethane: 14.47%, ethylene: 25.98%, propane: 9.79%, propylene: 14.03% nbutane 10.76% (c) 0.93 2.21 (a) Methane: 24.16% and carbon dioxide: 66.44% and nitrogen 9.4% (b) 29.8 (c) 1.33 kg/m3 CHAPTER 3 3.1 Compound Weight fraction mole fraction mole % Butane 0.5 0.5701 57.01 Pentane 0.3 0.2758 27.58 Hexane 0.2 0.1541 15.41 Total 1.0 1.0000 100.00 AVMWT: 66.138 3.2 70,748 g 3.3 37.63% 3.4 (a) and (b) Component mole % Weight % Methane 80 68.45 Ethane 15 24.10 Nitrogen 5 7.45 Total 100 100.00 (c) AVMWT: 18.7 (d) Density: 0.0008 g/cc 3.5 AVMWT: 30.5; Density of gas: 1.36 g/litre 3.6 Methane: 0.008892 kmole/m3 Ethane: 0.02231 kmole/m3 Hydrogen: 0.01338 kmole/m3 Velocity: 30,558 m/h Density: 0.00405 g/cc 3.7 1.169 g/litre 3.8 0.567 litre 3.9 10,719.7 K 3.10 C3H8 3.11 300.37 atm 3.12 (a) and (c)
Component Volume % mole fraction N2 64.75 0.6475 CO2 9.52 0.0952 H2O 18.62 0.1862 O2 6.11 0.0611 CO 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 68.59 Bromine 12.67 Oxygen 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 ´ 10–5 200 C2H6 0.3 3.173 ´ 10–5 600 H2 0.6 6.347 ´ 10–5 1,200 (d) 1.058 ´ 10–4 g mole/cc (e) 1,248 g/s (f) AVMWT: 11.8 3.17 (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 s 3.20 (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 litre 3.23 0.982 kg CHAPTER 4 4.1 31,941 J/g mole 4.2 77.25 °C 4.3 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 0 y 1.0 0.958 0.897 0.831 0.758 0678 0.590 0.496 0.393 0.281 0.162 0.030 0
x, y: Mole fraction of benzene in liquid and vapour phase respectively. 4.4 xA 1.0 0.724 0.415 0.134 0.0 yA 1.0 0.819 0.548 0.205 0.0 x, y: Mole fraction of ‘A’ in liquid and vapour phase respectively. 4.5 0.1847 kg/kg of steam, 0.1385 kg/kg of steam 4.6 Total pressure: 2,044.79 mm Hg Mole fraction of methanol in vapour phase: 0.5052 Mole fraction of ethanol in vapour phase: 0.4948 4.7 (a) Total pressure: 4087.9 mm Hg C2H6 : 0.139, n-C3H8 : 0.639, i-C4H10 : 0.039, n-C4H10 : 0.179 and C5H12 : 0.004 (b) 3589.87 mm Hg 4.8 Total pressure: 906.9 mm Hg Component Benzene Toluene Xylene Liquid phase Vapour phase composition, x composition, y 0.500 0.739 0.377 0.233 0.123 0.028 4.9 Partial pressure: 102.6 mm Hg 4.10 (a) Ethyl acetate: 12.22 %, air: 87.78% (By volume) (b) Ethyl acetate: 29.8 %, air: 70.2% (By weight) CHAPTER 5 5.1 (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 air 5.3 2961.6 m3/h 5.4 0.567 mole/mole 1.534 kg/kg 5.5 0.25 mole/mole 0.676 kg/kg 5.6 Humidity: 0.025 kg/kg Dew point: 28.5 °C Humid volume: 0.907 m3/kg dry air Adiabatic saturation temperature: 30 °C Humid heat: 1.0521 kJ/kg dry air Enthalpy: 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/h 5.8 27.5 °C and 0.019 kg/kg 5.9 (a) 96.5%, (b) 115.08 m3/h 5.10 (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 air 5.12 (a) % RH: 29.6% (b) Humid volume: 0.974 m3/kg dry air 5.13 (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°C 5.14 (a) 0.008 kg/kg (b) 12 °C (c) 2.81 kg of water (d) 7,405.2 kJ (e) 32.5 °C 5.15 % 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 air Enthalpy: 106.6 kJ/kg dry air 5.18 (a) 809.4 m3/min (b) 74.04 kg 5.19 (a) 0.0259 kmole/kmole, 0.01616 kg/kg (b) 0.0677 kmole/kmole 0.00423 kg/kg (c) 13.48 kg of water 5.20 (a) 29.45
(b) 781.24 mm (c) 810.69 Hg (d) 0.0377 mole/mole 5.21 5.7% 5.22 (a) 86%, (b) 1396.5 m3/h CHAPTER 6 6.1 Mother liquor: 3,324.64 kg 6.2 Feed: 83,078.34 kg 6.3 Crystals: 6,636 kg 6.4 Crystals: 479.2 kg 6.5 Crystals: 342 kg and Mother liquor: 918 kg 6.6 Water evaporated: 7,916.67 kg and crystals: 2,083.33 kg 6.7 Water evaporated: 720 kg Mother liquor: 6589 kg and crystals: 1,691 kg 6.8 87.4% 6.9 Water evaporated: 561 kg 6.10 Water evaporated: 82.35 kg; Mother liquor: 1,122.85 kg and Feed: 2,705.2 kg 6.11 2393.6 kg/h CHAPTER 7 7.1 H2SO4: 5,310.86 kg, HNO3: 2,811.87 kg and Feed: 1,877.27 kg 7.2 (a) 56.96%; (b) 8.338; (c) 21.81 kg 7.3 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 burnt 7.6 O2: 12.14%, N2: 79.34%, SO2: 7.67% and SO3: 0.85% 7.8 54.22% Excess air 7.10 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 kg 7.16 (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 evaporator 7.19 (a) 1.25 m3 (b) 6.41 m3 (c) Component Condition (a) Condition (b) CO2 0.23 0.14 H2O 0.14 0.08 N2 0.63 0.74
O2 —0.04 Total 1.00 1.00 7.20 (a) 4.49% (b) 3.08 7.21 (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) 0.642 7.23 (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 kg 7.25 Liquid : 81.789 kg/hr, vapour : 18.211 kg/hr, C5H12 : 31.38, C6H14 : 60.06 7.26 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: 18.388 kmoles (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 76.92 H2O 2.4726 4.08 SO2 0.7167 0.3326 Amount of air supplied = 37.78434 ton 7.32 Air–fuel ratio by mass = 16.7165 Flue gas analysis: Composition Weight (kg) CO2 3.113 O2 0.6408 N2 12.8717 H2O 1.026 SO2 0.064 Total = 17.7155 kg/kg of fuel 7.33 Air requirement = 432.433 kmoles, volume = 15416.56 Nm3/h Flue 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 8 8.1 (a) 100 moles (b) 0.4286 mole of Ammonia/mole of NO formed 8.2 (a) 1,425.8 kg, (b) Recycle fraction: 0.7179 CHAPTER 9 9.1 1760 °C 9.2 –1,16,295.8 kJ/kmole 9.7 1,874 °C 9.10 51.44 kcal CHAPTER 10 10.1 C = –0.1 [100/(100 + t)]2 + 0.2 10.2 C = [10,000 + 400t + t2]/(100 + t) 10.3 (a) t = 100 – 100 [(C – 1)/1.5]0.25, 36.1 minutes (b) 60 minutes 10.4 (a) C = 0.01 [40/(t + 40)]5 + 0.03 (b) 2.369 minutes 10.5 28.78 minutes 10.6 0.2675 kg Na2SO4/kg of solution
Index Adiabatic reaction temperature, 198 Gas constant, 35, 36 Antoine equation, 75 API scale, 12
Atomic numbers of elements, 231–233 Hausbrand chart, 75 Atomic percent, 11 Heat, 196 Atomic weights of elements, 231–233 Heat capacity of gases (empirical Average molecular weight, 38 constants), 234 Avogadro’s hypothesis, 9 Heat of Avogadro’s number, 36 combustion, 197 formation (also standard), 196 mixing, 197 Baume’ gravity scale, 12 reaction, 197 Brix scale, 13 from combustion data, 197 Bypass, 180 from enthalpy data, 197 Humid heat, 88 Humid volume, 88 Clausius–Clapeyron equation, 74 Humidity, 87 Composition of absolute, 87 liquid systems, 11 percentage absolute humidity, 88 mixtures, 10 relative, 88 solutions, 10 Hydrated salt, 111 Conservation of mass, 8 Conversion, 9 Conversion factors, 229–230 Ideal gas law, 35 Crystal, 111 Crystallization, 111 Kopp’s rule, 198 Density, 12, 38 Dew point, 88 Law Dry bulb temperature, 87 Amagat’s, 37 Dalton’s, 37 Hess’s, 197 Energy balance, 196 Leduc’s, 37 Enthalpy, 88 Law of conservation of mass, 8 247 248 INDEX
Magma, 111 Mass balance, 122 Mass relations, 7 Mole fraction, 11 Mole percent, 11 Mother liquor, 111 Normal boiling point (NBP), 74 Normal temperature and pressure (NTP), 9 Saturated vapour, 89 Saturation, 89 partial, 89 percentage, 89 relative, 89 Specific gravity, 12 Standard condition, 35
state, 196 Steam distillation, 75 System, 3 isolated, 3 Partial pressure, 36 Percentage saturation, 88 Process, 3 Property, 3 extensive, 3 intensive, 3 Psychrometry, 87 Pure component volume, 36 Purge, 180 Table of atomic numbers, 231–233 atomic weights, 231–233 conversion factors, 229–230 molal heat capacities, 234 Theoretical flame temperature, 198 Twaddell scale, 13 Rate of accumulation, 221 generation, 221 input, 221 output, 221 Reactant, 9 excess, 9 limiting, 9 Reaction endothermic, 197 exothermic, 197 Recycle, 180 Units and notations, 1 derived units, 2 Unsteady state operations, 221 Vapour pressure, 74 Volume percent, 10 Weight percent, 10 Wet bulb temperature, 87 Yield, 9
