PROCESS CALCULATION

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 ` 195.00 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 ...............................................7...2 Conservation of Mass ........................................................................... 1 1............................................................4 Atomic Fraction and Atomic Percent . 3 Exercises ................................1 Introduction ........................................................................... 10 2.......................7 Density and Specific Gravity ........................................................................................... xv 1 UNITS AND DIMENSIONS 1–6 1............................... 13 Exercises ......................................................................................................................................................................... 2 1........6..................... 10 2........................... 9 2.........................................................................................................6....................... 3 Worked Examples .............3 Mole Fraction and Mole Percent ...6............................. 11 2............ 11 2........................................................3 Avogadro’s Hypothesis ...................................... 9 2............................................... xi Preface to the First Edition ........6.............................. 10 2......................... 12 2....................... xiii Acknowledgements ... 7 2.......................................................................................................7........................... 12 2......................................................... 9 2.....................................1 Mass Relations in Chemical Reaction .........................6....7.......................................5 Conversion and Yield ................................1 Baume’ (°Be’) Gravity Scale .. 13 Worked Examples ..................... 32 vii ....................................2 API Scale (American Petroleum Institute) ................................... 6 2 MASS RELATIONS 7–34 2....................................4 Definitions ......................5 Composition of Liquid Systems ...............3 Derived Units ............ 13 2..6 Composition of Mixtures and Solutions .........4 Limiting Reactant and Excess Reactant ... 12 2....... 11 2........................................3 Twaddell Scale .... 1 1.....1 Weight Percent ..............................7....2 Volume Percent .............................................2 Basic Units and Notations .......................................Contents Preface ....... 8 2.............................4 Brix Scale ............................... ..........................................................................................1 Recycle .... 195 .................2 Ideal Gas Law ............................................ 38 Worked Examples .....................................................2... 180 8.................................1 Relation between Mass and Volume for Gaseous Substances ............................................................................ 181 Exercises ..................... 38 Exercises .............. 180 8....4 Amagat’s Law (or) Leduc’s Law ..........................................................................................viii 3 CONTENTS IDEAL GASES 35–73 3...2 Definitions ........................................2...............................2........................................................................2 Gaseous Mixture ..... 120 7 MASS BALANCE 122–179 Worked Examples .......................3 Purge .....................1................................................................1 Humidity ....... 75 Exercises .......................................... 173 8 RECYCLE AND BYPASS 180–195 8....................................................... 36 3............................ 38 3................................................................................1 Standard Conditions .................................................3 Average Molecular weight ....................................................................................................... 70 4 VAPOUR PRESSURE 74–86 4............................................1 Partial Pressure (PP) .......... 87 5................................................................... 35 3......... 87 Worked Examples ...................................................................................................................................................... 37 3.. 112 Exercises ... 35 3.............................. 37 3... 122 Exercises ....................................................................................................................................... 36 3....................... 35 3...........................................................................................1......................................................................2 Bypass ..........2................... 75 Worked Examples ...................................2 Hausbrand Chart ................ 36 3.....................3 Dalton’s Law ......... 106 6 CRYSTALLIZATION 111–121 Worked Examples ...................... 90 Exercises .............. 74 4.....................................................................................................................................2 Pure Component Volume (PCV) .......................................................................................................................1 Effect of Temperature on Vapour Pressure . 85 5 PSYCHROMETRY 87–110 5........4 Density of Mixture .................................................... 180 Worked Examples ......................... ........................2 Hess’s Law ..........................................................................5 Theoretical Flame Temperature ..................................................... 234 III(b)Molal Heat Capacities of Hydrocarbon Gases ................................................... 197 9.....................................................................................................................................1 Standard State .......4 Adiabatic Reaction Temperature ........... 231 III(a) Empirical Constants for Molal Heat Capacities of Gases at Constant Pressure ..................................................................4 The Heat of Reaction ............................... 235–245 Index ......... 197 9..................................................................................... 196 9..... 198 Worked Examples ....1 Definitions .................................................................................................... 234 Answers to Exercises ............................................. 227 Tables ............1.. 229–234 I Important Conversion Factors ............ 196 9.... 198 9.....................................................................................3 Kopp’s Rule .............................................................................................1.................2 Heat of Formation ...............CONTENTS 9 ENERGY BALANCE ix 196–220 9.......... 229 II Atomic Weights and Atomic Numbers of Elements .............................................. 196 9................. 221 Exercises .............................3 Heat of Combustion .............. 247–248 ................................................................................................................. 217 10 PROBLEMS ON UNSTEADY STATE OPERATIONS 221–228 Worked Examples .................. 197 9...........................1.......... 198 Exercises ...........1.........................5 Heat of Mixing .................... 197 9.............................................. 198 9......................................................................1..................................................................................................... Meera Sheriffa Begum xi . An attempt has been made to explain the principles involved through numerical examples. Anantharaman K. CGS and MKS systems. properties of gases. We feel that our attempt will be more rewarding if students come across data presented in FPS. Venkataramani N.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. V. vapour pressure. The second edition is now enriched with additional worked examples and exercises to give additional exposure and practice to students.M. energy balance and unsteady state operations. psychrometry. CGS and MKS systems while designing equipment. mass and energy balances. The book covers various interesting topics such as units and dimensions. This will enable him to achieve a proper design of process equipment. crystallization. since different reference books give standard values and data in various units. The problems are not only confined to SI system of units but also worked out in other systems like FPS. mass relations. 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. mass balance including recycle and bypass. an attempt has been made to give a brief theory on the principle involved and more emphasis on numerical examples. The text is organized into ten chapters and appends three important tables.g. Venkataramani N. operation. the chapter on unsteady state operations has been included as the last chapter so that students can absorb the problems easily. Anantharaman xiii . e. By applying the relevant techniques. 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. We strongly feel that once the student understands the topics presented in this book. This book aims at imparting knowledge of the basic chemical engineering principles and techniques used in analyzing a chemical process. the worked examples are not confined to SI units. Since data are generally obtained in different units. V. he will find other advanced courses in chemical engineering simple and easy to follow. 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. CGS and MKS systems of units. therefore. It is. 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. such as FPS. control and management of a process plant. but to other systems as well.Preface to the First Edition Chemical engineers in process industries generally need to focus on design. The organization is such that the topics are presented in order of easy comprehension rather than following a logical sequence. The examples incorporated in the text are simple and concrete to make the book useful for self-instruction. Keeping this in mind. Vandana Ravi Chandar and Mrs. Anantharaman wishes to thank his mother. Mr. especially those at NIT. Tiruchirappalli for extending all the facilities and his words of appreciation. We also thank Director. 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. NIT. husband Mr. Rakshana Roshan for all their encouragement.Acknowledgements At the outset.) Booma Venkataramani. xv . Meera Sheriffa Begum wishes to acknowledge her mother. support and cooperation while preparing this book. We also wish to place on record the suggestions received from students. Dr. sisters. Usha Anantharaman. wife. Prof. K. The support received from her brothers. Varun for all their patience. and also the faculty from other institutions. Srinivas and A. V. He also wishes to place on record the support received from his brothers-in-law and sisters-in-law and their family members. cooperation and patience shown during the preparation of this book. (Mrs. The encouragement received from his brothers and sisters and their family members is gratefully acknowledged. sons Mr. Ravi Chandar. V. sons Master A.M. Venkataramani wishes to acknowledge his wife. N. Malik Raj and baby M. Tiruchirappalli. S. Hari Sundar and his daughters-in-law Mrs. V. in-laws and their families is gratefully acknowledged. She also wishes to place on record the support received from her parents-in-law. we wish to thank the almighty for his blessings. Ramya Hari Sundar and granddaughters Miss Vaishnavi Ravi Chandar and Miss Sadhana Hari Sundar for their support. cooperation and support shown during the preparation of this book. Meera Sheriffa Begum .xvi ACKNOWLEDGEMENTS We gratefully acknowledge all the well-wishers. we wish to thank the publishers.M. V. Venkataramani N. New Delhi for encouraging us to bring out the second edition of the book. Anantharaman K. Finally. PHI Learning. Though internationally we follow SI system of units. we come across different systems of units as mentioned earlier. FPS Mass (m) Length (L) Time (t) Temperature (T) Metric Engineering lb ft s °F 1 CGS MKS g cm s °C kg m s °C System International.2 BASIC UNITS AND NOTATIONS English Engineering. pressure.1 INTRODUCTION Chemical engineers are concerned with the design and development of processes which involve changes in the bulk properties of matter. CGS. one has to be conversant with their use and applications. chemical equations showing the quantities of reactants and products are used. density. 1. SI kg m s K . MKS and SI systems of Units.Units and Dimensions 1 1. These are used to express properties. process variables and design parameters in FPS. mass flow rate derived from the fundamental quantities are called derived quantities. Hence. The quantities used in our analysis are classified as fundamental quantities and derived quantities. This chapter deals with the basic notations and conversion of a given quantity from one system of units to another. A review of literature and data over the years will be available in various units. The fundamental quantities comprise length. a chemical engineer is expected to be familiar and conversant with all the systems so far adopted for measuring and expressing various quantities. time and temperature. To make a quantitative estimation of these processes. mass. While handling these quantities. Now let us see in detail these systems of units and their conversion from one unit to another. The quantities such as force. 0929 m2 = 1 m2 Force: 1 dyne = mass ´ acceleration (mL/t 2): = 1 g cm/s2 (Force applied on a mass of 1 g.8 1.76 ft2 = length ´ breadth (L2): = 0.48 cm = 0.205 lb Length (L) 1 ft = 30.252 kcal = 252 cal = 4.2 PROCESS CALCULATIONS Mass (m) 1 kg = 2. 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.3048 m Time (t) 1 h = 3600 s Temperature (T ) °C = °F  32 (Celsius and Centigrade are same) 1.18 J =1W . 1 erg 1 Joule 1 Joule = = = = = = = [1 dyne] ´ [1 cm] [1 g cm/s2] ´ [1 cm] 1 g cm2/s2 (1 N) ´ (1 m) 105 g cm/s2 ´ 100 cm 107 g cm2/s2 107 erg Heat Unit: 1 Btu 1 cal 1 J/s = 0.3 DERIVED UNITS Area: 1 ft2 10. 25/(D)0.8 times the variation in celsius scale] = 4. This state of a system is independent of mass. where.g.4 DEFINITIONS System. Process. DT is in °F and D is in ft. e.ft2. Convert this relation to estimate the heat flux in terms of kcal/h.1 The superficial mass velocity is found to be 200 lb/h.2712 kg/s.5) [(DT)1.ft2 1 1 Ê 1 ˆ = (200) ¥ Á kg˜ ¥ ¥ m2 Ë 2. storage tank.UNITS AND DIMENSIONS 3 1. Intensive property.m2 °C = 4. Boundaries of the system are limited by a mass of material.0929 m ) Î (3600 s) ˚ (1. An example of this property is temperature. .3 The rate of heat loss per unit area is given by (0. Extensive property.18 W/m2 °C = 174. Isolated system. or reaction between two substances like hydrogen and oxygen to form water.8 °C) [1 degree variation in Farh.186 ¥ 10–2 ¥ 103 ¥ 4. Find its equivalent in kg/s. This refers to a substance or group of substances under consideration.°F into W/m2 °C È (0. etc. e. hydrogen stored in cylinder.98 W/m2 °C 1.m2 G (Mass velocity) = (200) lb/h. WORKED EXAMPLES 1. which depends on the mass under consideration.252 kcal) ˘ 2 h (Heat transfer coefficient) = (100) Í ˙ (0. It is a state of system.186 ¥ 10–2 kcal/s.g.ft2.205 ¯ (3600 s) (0. e. the mass of the system remains constant.25] Btu/h ft2 for a process. burning of fuel.0929) = 0. In an isolated system.2 Convert the heat transfer coefficient of value 100 Btu/h.g. volume. regardless of the changes taking place within the system. m2 using DT in °C and D in m. and its energy content is completely detached from all other matter and energy. scale is equivalent to 1. water in a tank. Changes taking place within the system is called process.m2 1. 4 PROCESS CALCULATIONS We know that, 9 C1 + 32 = F1 5 9 C2 + 32 = F2 5 Therefore, 1.8 [DC] = (DF) q A 0.5 ( 'T )1.25 ( D)0.25 ( 'T °F)1.25 q , Btu/h ft2 = 0.5 A ( D 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 Btu/h ft2 Ë [ 'T °F)1.25 Û = 0.5 Ì 0.25 Ü ÌÍ (D ft) ÜÝ = (0.5) (1.8 'T °C)1.25 ( D m/0.3048)0.25 = (0.7746) Btu/h ft2 (a) (b) ( 'T °C)1.25 (D m)0.25 = (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: Ë (0.7746)( 'T °C)1.25 Û Heat flux, kcal/h m2 = Ì Ü ´ 2.713 ( D „ m)0.25 ÌÍ ÜÝ = (2.101)( 'T °C)1.25 ( D „ m)0.25 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 (d) UNITS AND DIMENSIONS From Eq. (a), heat flux is = 0.5 (18)1.25 (0.2) 0.25 5 = 27.72 Btu/h ft2 = 75.2 kcal/h m2 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 1.6 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 m3 227.27 Density = ´ 10–3 = 7770 kg/m3 29.25 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. 1211 È p Ø Then, log É = 6.9057 – Ê 133.3 ÙÚ (T „  273)  220.8 log p¢ – log 133.3 = 6.90305 – log (p¢) = 9.0305 – 1211 . T „  52.2 1211 T „  52.2 6 PROCESS CALCULATIONS EXERCISES 1.1 Convert the following quantities: (a) (b) (c) (d) (e) (f) 1.2 42 ft2/h to cm2/s 25 psig to psia 100 Btu to hp-h 30 N/m2 to lbf/ft2 100 Btu/h ft2 °F to cal/s cm2 °C 1000 kcal/h m°C to W/m K The heat transfer coefficient for a stream to another is given by h = 16.6 Cp G 0.8/D0.2 where h D G Cp = = = = Heat transfer coefficient in Btu/(h)(ft)2(°F) Flow diameter, inches Mass velocity, lb/(s)(ft)2 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 G0.8 k0.67 Cp0.5/(D0.2 m0.47) 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 2.1 2 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: CaCO3 ® CaO + CO2 (i) (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) katom (or kg atom) gram mole (or g mole) kmole (or kg mole) = = = = Mass Mass Mass Mass 7 in in in in grams/Atomic weight kg/Atomic weight grams/Molecular weight kg/Molecular weight 8 PROCESS CALCULATIONS 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 One One One atom of oxygen weighs atom of hydrogen weighs molecule of oxygen weighs molecule of hydrogen weighs = = = = 16 grams O 1 gram H 32 grams O 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 = = = = = = \ \ = Mass in grams or pounds/Atomic weight = Mass in grams or pounds/Molecular weight g atom or lb atom g mole or lb mole 1 1 1 1 1 1 g atom of oxygen lb mole of oxygen g mole of oxygen g atom of hydrogen kmole of hydrogen 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. For 32 kg of oxygen to react fully. 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. One of the reactants will be present in excess and remain unreacted even when the other reactant has completely reacted.414 litres and 1 lb mole of the same substance occupies 359 ft3at NTP. 2.5 6 ¥ 100 = 50%. 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. Hence % excess of carbon is = 2. As per stoichiometry C + O2 Æ CO2 i. Note: NTP = Normal temperature (273 K) and pressure (1 atmosphere) are also referred to at times as standard conditions (SC). Hence.414 m3 at NTP. 12 CONVERSION AND YIELD These terms are used for a chemical reaction where the reactants give out new compounds or products. for 18 kg of carbon to react fully we should have 48 kg of oxygen.MASS RELATIONS 2. 1 kmole of any gaseous substance occupies 22. Let us consider that 18 kg of carbon is burnt with 32 kg of oxygen.414 cc or 22. . A limiting reactant is the one.e. Since 32 kg of oxygen alone is available.4 LIMITING REACTANT AND EXCESS REACTANT For most of the chemical reactions the reactants will not be used in stoichiometric proportion or quantities. whereas the excess reactant is the one. All calculations involved in estimating the quantity of product and conversion are always based on the limiting reactant. which will not be present in the product. it is called the limiting reactant and carbon is called the excess reactant. However 6 kg of carbon is present in excess. 12 kg of carbon will burn with 32 kg of oxygen to form 44 kg of CO2. it is sufficient to have 12 kg of carbon. 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.3 9 AVOGADRO’S HYPOTHESIS 1 g mole of any gaseous substance at NTP occupies 22. which will always be present in the product. 2. VA and VB = Pure component volume of components A and B respectively. WA. 2. One major advantage of weight percent is. liquids and solids. i. Conversion for a reaction is based on the limiting reactant whereas the yield is based on the product formed.2 Volume Percent The ratio of the volume of each component and the total volume of the system. MB = Molecular weight of components A and B respectively. its independence to changes in temperature and pressure. if they are compounds. for each 100 part of the total volume is called volume percent . AA. V = Volume of the system. The composition of a solid mixture is to be always taken as weight % when nothing is mentioned above its units. whereas the yield is the amount of desired product actually formed compared to that which can be formed theoretically. MA. Conventionally the composition of solids is either expressed on weight basis or mole basis.e.10 PROCESS CALCULATIONS Conversion is the ratio of the amount of material actually converted to that present initially.6 COMPOSITION OF MIXTURES AND SOLUTIONS Different methods are available for expressing composition of mixtures of gases.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.6. W = Total weight of the system. Let us consider a binary system comprising components A and B. WB = Weight of components A and B respectively. 2.6. if they are elements. AB = Atomic weight of components A and B respectively. 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. This is based on Avogadro’s law.3 Mole Fraction and Mole Percent These concepts are generally adopted in the case of a mixture containing molecules of different species. WA MA Mole fraction of A = WA W . 2.6. The volume % is also equal to mole % for ideal gases but not for liquids and solids.MASS RELATIONS Volume % of A = 11 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. WA AA Atomic fraction of A = WA WB . B M A MB Mole % of A = Mole fraction of A ´ 100 2.4 Atomic Fraction and Atomic Percent This is adopted when a mixture contains two or more atoms.6. (i) (ii) (iii) (iv) (v) Weight ratio Mole ratio Molality Molarity Normality (N) = = = = = 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 Number of gram equivalents of solute/1 litre of solution Hence. molality = molarity .5 Composition of Liquid Systems In the case of liquids we come across more number of methods of expressing compositions of the liquid constituents. AA AB Atomic % of A = Atomic fraction of A ´ 100 2. concentration in grams per litre = Normality (N) ´ Equivalent weight of solute For very dilute aqueous solutions.6. 7.12 PROCESS CALCULATIONS 2. which are related to specific gravities and densities by arbitrary mathematical definitions. water will have a gravity of 10° Be’ and this degree decreases with increase in specific gravity. However. Degrees API = 141. Specific gravity is the ratio of density of a liquid to that of water. 2. 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.2 API Scale (American Petroleum Institute) This scale is used for expressing gravities of petroleum products. the variation in density of solids is not high. Similarly. Degrees Baume’ = 145 – 2. Several scales are in use in which specific gravities are expressed in terms of a degree.7. Over a narrow range of temperature. For liquids heavier than water 145 G In this scale the degree increases with increase in specific gravity.1 Baume’ (°Be’) Gravity Scale For liquids lighter than water Degrees Baume’ = 140 – 130 G G is the specific gravity at 60 ¦ 15 µ °F § °C¶ 60 ¨ 15 · Thus. in the case of liquids and gases the variation in density is significant. However. 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. the densities vary significantly with concentration also.7 DENSITY AND SPECIFIC GRAVITY Density is defined as mass per unit volume and it varies with temperature.5 G . This is similar to Baume’ scale for liquids lighter than water.5 – 131. 1 Convert 5000 ppm into weight %. of the gas at NTP? Molecular weight of propane (C3H8) is 44 1 g mole of any gas occupies 22.7.3 What is the volume of 25 kg of chlorine at standard condition? 25 kg Cl2 = Volume = 2. Degree Brix = 400 – 400 G WORKED EXAMPLES 2.414 litres at NTP .4 25 kmoles of Cl2 2 × 35.46 25 × 22.5 ⎟ = 48.0) 2.2 = 0.3 13 Twaddell Scale This scale is used for liquids heavier than water.9 m3 2 × 35. 5000 × 100 10 6 2. The acid can be split into 2H 3 PO 4 → P2 O 5 + 3H 2 O (2 × 98) 142 (3 × 18) 196 units of the acid contains 142 units of the pentaoxide. When the strength of pentoxide is 35%.MASS RELATIONS 2.5% for pure acid. Find the weight % of the acid.5% The strength of H3PO4 was found to be 35% P2O5. Degrees Twaddell (Tw) = 200 (G – 1. the weight % of acid is 100 ⎞ ⎛ ⎜ 35 × 72.414 = 7.7.46 How many grams of liquid propane will be formed by the liquefaction of 500 litres.3% ⎝ ⎠ 2.4 Brix Scale This scale is used in sugar industry and 1 degree Brix is equal to 1% sugar in solution. The weight % of pentaoxide is (142/196) which is 72. 8 1 = 0.8 4.0446 ´ 16 = 0.9/1 = 4.65 9050 100.31 g moles 22.31 g moles of propane weighs = 22. Express the composition in weight %.0446 g mole 22.9 13. g mole 25 75 Actual weight.31 ´ 44 = 981.14 PROCESS CALCULATIONS 500 = 22.6 by weight.6 A solution of naphthalene in benzene contains 25 mole % Naphthalene.9 13.7 Composition in weight percent What is the weight of one litre of methane CH4 at standard conditions? 22.414 Weight of one litre methane = 0.9 and N : 13.6/14 = 0.414 litres of any gas at NTP is equivalent to 1 g mole of that gas \ \ 2.414 22.52 g.414 = 1120. 500 litres of propane at NTP = 2. H : 4. Basis: 100 g moles of solution Component Molecular weight Naphthalene C10H8 Benzene C6H6 128 78 Weight. g atom 81. So the molecular formula is C7H5N .5/12 = 6. g 25 ´ 128 = 3200 75 ´ 78 = 5850 3200 ´ 100/9050 = 35.7 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 2.9 Rounding of atoms Weight of each element 7 5 1 84 5 14 Total 103 Hence the formula obtained after rounding is correct.00 Total 2.35 5850 ´ 100/9050 = 64.5 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. What is the formula? Basis: 100 g of substance Element Carbon Hydrogen Nitrogen Atomic weight Weight.714 g 1 litre of methane º 1 ´ A compound whose molecular weight is 103 analyses C : 81.6 Weight.5. g 12 1 14 81.5 4. 8%.857 — 1.8 100 ⎛ 1.5 65.1258 0. MgO : 7.7 2.262 1. ZnO : 9.84 kg/min Moles of the gas = 2.555 = 0. C2H6 : 30% and rest H2 at 15 °C and 1.8 7.56 76. g Na2O MgO ZnO Al2O3 B2 O 3 SiO2 7. Calculate the evaporation of water per kg of feed.156 ¥ 11.6419 100.0815 7.8 = 1.5 ⎞ Volumetric flow rate at standard conditions = 2.0 8. B2O3 : 8.6 60.156 kmole 22.0 9.414 Mass flow rate = moles ¥ average molecular weight = 0.1221 1. Basis: 100 g of glass sample Component Weight.10 A gaseous mixture analyzing CH4 : 10%.7%.0 40. Na2O : 7.439 65.0 Molecular weight 2.555 m3/min 3.0 g mole mole % 62. Al2O3 : 2.194 7.0 Total 100.0196 0. Find the mole %.9 An analysis of a glass sample yields the following data. .0 69. g mole Molecular weight Weight. Basis: 100 g moles of the gaseous mixture.17 100 1180 = 11.3 81.4 102.1192 0.0%.5 atm is flowing through an equipment at the rate of 2.15 MASS RELATIONS 2.1 0. g CH4 C2H6 H2 10 30 60 16 30 2 160 900 120 Total 100 — 1180 The average molecular weight = Weight % 13.5% and rest SiO2.0%.27 10. (b) weight % and (c) the mass flow rate. Component Weight.11 In an evaporator a dilute solution of 4% NaOH is concentrated to 25% NaOH.583 7. Find (a) the average molecular weight of the gas mixture.1737 0.665 10.5 m3/min.5 ¥ ⎜ ⎟ ¥ (273/288) ⎝ 1 ⎠ = 3. 377 ´ 100/3.04 kg.84 kg 2. (ii) and (iii). They got the average molecular weight as 30. we get x = Moles of nitrogen = 81% y = Moles of carbon dioxide = 11% z = Moles of oxygen = 8% 2.16) = 0.74.333 3.84 kg Weight of thick liquor formed = Water evaporated per kg of feed = 0.08 and the incorrect one as 18.04 = 0.13 An aqueous solution contains 40% of Na2CO3 by weight. Express the composition in mole percent.84 3.333 ´ 100/3.00 Total Composition in mole % .16 PROCESS CALCULATIONS Basis: 1 kg of feed. (i). while the other used an incorrect value of 14.377 0. Basis: 100 g of solution Component grams Molecular weight g mole Na2CO3 40 106 40/106 = 0. Calculate the % volume of N2 in the flue gases.71 = 10.25 Weight of water evaporated = (1 – 0.710 100. which appears as 25% in the thick liquor formed 0.16 Water 60 18 60/18 = 3.16 kg 0.12 The average molecular weight of a flue gas sample is calculated by two different engineers. If the remaining gases are CO2 and O2 calculate their composition also.71 = 89. One engineer used the correct molecular weight of N2 as 28. NaOH present is 0. Basis: 100 g moles of flue gas g mole I Engineer II Engineer N2 x 28x 14x CO2 y 44y 44y O2 z 32z 32z 100 3008 1874 Component Total x + y + z = 100 (i) 28x + 44y + 32z = 3008 (ii) 14x + 44y + 32z = 1874 (iii) Solving Eqs. 84) + (4 ×16) 8 231.5 katoms Fe 50 µ = 37.5 µ § 3 ¶ ´ 231.52 → Fe 3 O 4 + 4H 2 ↑ (4 × 2 ×1) (3 × 55.52 = 2894 kg ¨ · 2994 kg ¦ § ¨ .52 ´ 2094 = 2894 kg The weight of Fe3O4 formed is = 167.84) 72 167. for getting 100 kg of H2 the amount of steam (H2O) required is 100 t 72 = 900 kg 8 The total weight of reactants is = 2094 kg Fe and 900 kg H2O = 2994 kg 231.52 kg of Fe is required for producing 8 kg of H2 \ For producing 100 kg of H2 (by stoichiometry) 100 t 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 = 50 kmoles 4 kmoles H2 comes from = 3 katoms of Fe (by stoichiometry) 50 kmoles of H2 comes from = Weight of 37.52 239.5 ´ 55.52 239.52 Method 1 (Based on absolute mass) 167.5 katoms Fe 4 ¶· = (37.MASS RELATIONS 17 2.14 What is the weight of iron and water required for the production of 100 kg of hydrogen? 3Fe + 4H 2 O (4 ×18) (3 × 55.84) = 2094 kg of iron 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 = 2994 kg ¦ §3 t ¨ 50 µ t 4 ¶ kmoles of water 4 · (50 ´ 18) = 900 kg H2O 37.52 8 = 2094 kg Iron (Fe) required = Similarly.5 kmoles 3 ¦ 37. 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. 10 kg of sulphur dioxide will be obtained from 9.46) 148.54 64 kg of sulphur dioxide is obtained from 63.3 g of mercuric oxide? 2KClO3 Æ 2HgO Æ (2 ¥ 122.935 tons of pure calcium phosphate 0. P : 31.16 SO2 is produced by the reaction between copper and sulphuric acid.46) 244.70577 tonne \ Weight of super phosphate formed is = 234 ¥ 310 2.92 2Hg (6 ¥ 16) 96 + O2 (2 ¥ 200.1698 g O2 is obtained from (244. H2O.93 kg of copper.5H2O + 12MoO3 + ½P2O5 Æ (NH4)3PO4 12MoO3 · 3H2O (3 ¥ 17 = 51) (4. O : 16.6) (2 ¥ 16) 433. 2. How much Cu must be used to get 10 kg of SO2? Cu + 2H 2 SO 4 → CuSO 4 + SO 2 + 2H 2 O 64 63.92 ¥ 0. 2.3 g of HgO will give (32 ¥ 2.4332 g of KClO3.64 % of H2O = 81 ¥ 100 1931 = 4.18 Ammonium phosphomolybdate is made up of the radicals NH3.6) 2KCl + 3O2 (2 ¥ 74.18 PROCESS CALCULATIONS 2.5 ¥ 18 = 81) (12 ¥ 144 = 1728) (½ ¥ 142 = 71) % of NH3 = 51 ¥ 100 1931 = 2.2) = 0.1698 g of O2 0.19 1931 . P2O5 and MoO3.1698)/96 = 0.3/433. S : 32 Ca3(PO4)2 + 2H2SO4 Æ CaH4(PO4)2 + 2CaSO4 310 (2 ¥ 98) (2 ¥ 136) 234 506 506 One ton of raw calcium phosphate contains 0.54 kg of copper.92 (2 ¥ 216.15 How much super phosphate fertilizer can be made from one ton of calcium phosphate 93.17 How much potassium chlorate must be taken to produce the same amount of oxygen that will be produced by 2.5% pure? Atomic weights are: Ca : 40.935 = 0.2 g HgO gives 32 g of oxygen 2. 75 g K2Cr2O7 1.46) Thus.6 g KMnO4 2000 3 t 189.45 g NaCl Hence.MASS RELATIONS % of MoO3 = 1728 ´ 100 1931 71 ´ 100 1931 Total % of P2O5 = = 89.00 2.92 g NaCl. 2500 g of salt cake is obtained from 116.49 = 3.935 g KMnO4 294 5 t 294 = 7. .21 If 45 g of iron react with H2SO4. 5 g KMnO4 º 1200 t 316 = 189.935 2.92) + H 2 SO 4 → Na 2 SO 4 + 2HCl 98 142 (2 × 36.92 ´ 2500/142 = 2058. salt needed is 2058.20 (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 ® 5Fe 2(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) \ 294 g K2Cr2O7 is equivalent to \ 3 g K2Cr2O7 º Similarly.75 g K2Cr2O7 189.68 19 = 100.45 g Glauber’s salt (Na2SO4 × 10H2O) obtained is 2500 ´ 322/142 = 5669 g 2.6 5t3 = 7. how many litres of hydrogen are liberated at standard condition? Alternatively. 142 g of Na2SO4 is obtained from 116.46 =116.6 = 1.19 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) 2NaCl (2 × 58. 209 ¥ 22.611 = 0.0 (c) Average molecular weight = 1823 100 Weight % = 18.23 Volume at standard condition = 00 ¥ 22.5%.4 m3 (d) Density of gas at standard condition = 1823 2241.5 ¥ 16 = 1336 1336 ¥ 100/1823 = 73.5 83.414 = 18.813 kg/m3 .7 2.14 Total 1823 100.4 = 0.806 g mole 2 0.1 litres 2. and N2 : 4%.5 ¥ 30 = 375 375 ¥ 100/1823 = 20.29 C 2H 6 30 12. 55.806 g mole ∫ 0.418 g H2 = 1.5%.418 g 111.85) (i) (2) The weight of hydrogen formed by reaction (i) is = 45 ¥ 2 = 1.806 ¥ 22. (b) Case II 2Fe + 3H 2 SO 4 → Fe 2 (SO 4 )3 + 3H 2 (111.57 N2 28 4.0 4.5 12.414 = 2241.06 litres i.0 ¥ 28 = 112 112 ¥ 100/1823 = 6. Calculate the following: (a) composition in mole % (b) composition in weight % (c) average molecular weight (AVMWT) (d) density at standard condition (kg/m3) Basis: 100 kmoles of gas mixture Component Molecular weight mole % Weight.e.22 A natural gas has the following composition by volume CH4 : 83.414 ∫ 27.85 1.7) (ii) (6) The moles of hydrogen formed by reaction (ii) is 45 ¥ 6 = 2. C2H6 : 12.209 g moles of hydrogen ∫ 1.20 PROCESS CALCULATIONS There are two possible reactions in this case: (a) Case I Fe + H 2 SO 4 → FeSO 4 + H 2 (55.611 g. kg CH4 16 83. 86 100.23 Convert 54.93 102.00 mole % 230/58.93/54.75 g/litre of HCl into molarity. Find the composition in (a) weight % (b) volume % of water (c) mole % (d) atomic % (e) molality and (f) g NaCl/g water.409 31.24 A solution of NaCl in water contains 230 g of NaCl per litre at 20 °C.15 (b) Volume % of water: 918 g is present in 1 litre of solution.443 63. Molarity = g moles/litre of solution 54.93 g moles of NaCl is present in 918 g of water (i.28 g moles/1000 g of solvent Molality = 4.) 4.e.97 918/18 = 51.148 g/cc.00 51/54.5 = 3. of solution (density of water is 1 g/cc) Volume % = 91.000 (All are based on the molecular formula) (e) Molality = g moles of solute in 1 kg of solvent (3.5 230 20.93 = 7. g Weight.85 Total 1148 100.93 = 92.28 Molarity = Moles of solute per litre of the solution = 3.8% (d) Element g atoms Atomic % Na Cl H O 3.93 3.00 2.21 MASS RELATIONS 2.705 Total 160. 3.00 51. The density of the solution at this temperature is 1. 918 cc water is present in 1000 cc.252 918 .45 2.00 54.03 Water 18 918 79.93 ´ 1000/918) or.e.75 = 1.5 36. = Basis: (a) 1 litre of solution has a weight of 1148 g Component Molecular weight Weight.93 (f) g NaCl/g water = 230 = 0.443 2.93 3.93 100. % g mole NaCl 58. i. Find the composition in terms of (a) molality (b) mole fraction.228 g Component Weight.228 100. 1.8 g moles NaCl = (1.26 Calculate the weight of NaCl that should be placed in a 1 litre volumetric flask to prepare a solution of 1.954 1.25 A benzene solution of anthracene contains 10% by weight of the solute.00 Density of this solution = 1.06 g/cc Molality = g moles of NaCl/1000 g of water = 1. Density of this solution is 1.2100 1.48 1105.0562 90 ¥ 1000 = 0. i.67 = 100.52 H2O 1000. 1000 cc of this solution will have = 1000 ¥ NaCl needed = 100. Basis: 100 g of solution Component Molecular weight Weight.92 g .8 or.8 ¥ 58.228 g of NaCl \ 105.06 or 1042.00 Benzene Total Molality = g moles of anthracene in 1000 g benzene = 0.67 cc 1.0562 78 90 (90/78) 0.000 90.e. g Weight % NaCl 105. mass/density = 1105.624 2. g Weight.228 = 1042.1538 1.046 0.67 cc of this solution contains 105.46) = 105.06 g/cc Volume of this solution.228 1042.8 molality. g mole mole fraction Anthracene 178 10 (10/178) 0.22 PROCESS CALCULATIONS 2.228 9.92 g of NaCl. 10.3 °Be’ – 131.28 (a) A solution has 100° Tw gravity. What is its specific gravity and °Be’? Hence.47 = 8.35 lb.MASS RELATIONS 23 2.1 gallons.5 – 131.28 ft3 = 9. volume of the above solution = (b) An oil has a specific gravity of 0.147 ft3 71. Density of ethyl alcohol and water are 790 kg/m3 and 1000 kg/m3 respectively.2 0.09 lb. 1. Express the composition in weight % and mole %.48 gallons Therefore. 9.27 For the operation of a refrigeration plant it is desired to prepare a solution of 20% by weight of NaCl solution.4 = 71. .09 = 10.14 \ Density of solution = 1.79 ¶· 2.29 An aqueous solution contains 15% ethyl alcohol by volume. Find °API and °Be’ (a) 100 = 200 (G – 1) \ °Be’ = 145 – (b) °API = °Be’ = 145 = 145 – G ¦ 141.6 140 µ – 130 = 47. 7. (a) Weight of salt per gallon of water = ¦ § ¨ 20 µ = 2. (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 = 1.48 Total weight of solution = weight of water + weight of salt Weight of 1 gallon of water = = 8.14 lb/ft 3 62.5 = ¦ 145 µ § 1.79 ¶ ¨ · = 48.5 µ § 0.57 gallons. 2.79.47 lb/ft3) We know that 1 ft3 = 7.44 lb.44 = 0.35 + 2.5 ¶ ¨ · ¦ 141.5 = 47.28 ft3 (since the density of water is 62.14 ´ 62.5 µ § G ¶ ¨ · ¦ § ¨ G = 1.14 = 1.57 ¶· (b) Specific gravity of NaCl solution at 30 °C = 1. Basis: 100 kg of sample 60 kg of urea has 28 kg of N2 100 kg of urea will have = 28 × 100 = 46.00 790 1000 Weight. weight m3 kg/m3 Ethanol Water Total 46 18 0. If the reaction is 95%. 60% H2SO4 and 8.24 PROCESS CALCULATIONS Basis: 1 m3 of solution.5 kmoles HNO3 : 60 % of 1700 kg = 1020 kg = 10.5% of 1700 kg = 535.32 Nitrobenzene is produced by reacting nitrating mixture with benzene.5 % of 1700 kg = 144.576 47.71% 46. then calculate the amount of nitrobenzene and spent acid produced. The nitrating mixture contains 31.408 kmoles H2SO4 : 8.5 kmole : 31.765 94.15 0.67 2.85 1.222 49. Compound Molecular Volume. A charge contains 663 kg of benzene and 1700 kg of nitrating mixture which sent into the reactor.235 5. kg Number of moles 118. estimate the purity of ammonium nitrate.798 Weight % mole % 12.31 If the nitrogen content in ammonium nitrate sample is 28%.5% H2O. the % purity is = 40 × Molecular weight of ammonium nitrate. the purity of ammonium nitrate is ¦ § ¨ 28 µ × 100 = 80% 35 ¶· 2.5 850 968. The reaction is C6H6 + HNO3 ® C6H5NO2 + H2O Feed. C6H6 : 663/78 = 8.30 The quality of urea is expressed in terms of nitrogen content. If the nitrogen content in the sample is only 40%. The molecular weight of urea (NH2CONH2) is 60 and that of N2 is 28. NH4NO3 = 80 % Nitrogen in pure ammonium nitrate = 28 × 100 = 35% 80 The % of nitrogen in the sample is 28 Hence.173 87.5% HNO3.5 kg = 8.028 kmoles H2O . Hence. estimate the purity of sample in terms of urea content.67 kg of N2 60 (Theoretically) The given sample has 40% N2 100 = 85.5 kg = 8.5 2. Density.827 100 100 2. % HNO3 0.05 × 8.266 Nitrobenzene 8.34 Two kg of CaCO3 and MgCO3 was heated to a constant weight of 1. HNO3 unreacted C6H6 unreacted H2SO4 unreacted H2O unreacted H2O formed Nitrobenzene formed : : : : : : 0.5 × 0.225 kg Spent acid = 26.33 A sample of caustic soda flake contains 74.63 kg 2.075 123 993.028 kmoles 8.775 1.05 × 8.95 = 8.408 98 1020.075 8.028) = 16.403 Total 2363. Therefore.775 × 100 = 96.032 C6H6 0. Calculate the % amount of CaCO3 and MgCO3 in reacting mixture.000 + 289.775 + 1020.000 43.425 0.408 kmoles 8.075 + 8.225 42.5 = 0.00 100% Nitrobenzene produced = 993.26% 2.95 = 8.MASS RELATIONS 25 Reaction is 95% complete Hence.425 10. Reaction is as follows: CaCO3 ® CaO + CO2 (100) (56) (44) MgCO3 ® MgO + CO2 (84) (40) (44) Let.5 × 0.133 H2SO4 10. kmole Molecular weight Weight.85 = 1336.165 H2O (8.425 63 26.5% % Purity = 0. (2 – x) be the weight of MgCO3 100 kg of CaCO3 gives 56 kg of CaO .150 1.746/0. kg Weight. Reaction is as follows: 2NaOH ® Na2O + H2O Amount of Na2O in pure flakes = 62 × 100/80 = 77.5 = 0.425 78 33.850 12.103 18 289.1 kg. Estimate the purity of flakes. x be the amount of CaCO3.6% Na2O by weight.075 kmole kmole kmoles kmoles Component Weight. 0838x = 1.761 kg Component Weight. .35 The composition of NPK fertilizer is expressed in terms of N2.1 0. Estimate the amount of filler in the NPK fertilizer.96524 Therefore. Basis: 1000 kg of fertilizer Reactions are: ® N2 2NH3 (34) + 3H2 (28) 2H3PO4 (6) ® P2O5 + 3H2O (196) (142) (54) 2KCl + H2O ® K2O + 2HCl (149) (18) (94) (73) N2. (2 – x) kg of MgCO3 gives © 40 ¸ ª 84 ¹ « º × (2 – x) kg of MgO The weight of product left behind is 1.04 kg 142 = 196 × = 182. Anhydrous ammonia.95 = 373. If the specific gravity of A is 0. x = 1.239 2.95 100. P2O5 and K2O each of about 15 weight %.36 A solution whose specific gravity is 1 contains 35% A by weight and the rest is B.14 kg = 149 × 150 = 237.00 2.77 kg 94 The amount of inert material/filler = 1000 – 626.05 11. x kg of CaCO3 gives Therefore. weight of MgO + CaO left behind 0.26 PROCESS CALCULATIONS 56 x = 0. 100 Similarly.000 88. 100% phosphoric acid and 100% KCl are mixed to get 1 ton of fertilizer.1 – 0. find the specific gravity of B.1 kg. i.56x + (0. % CaCO3 MgCO3 Total 1.e.56x kg of CaO.4672)(2 – x) = 1. 84 kg of MgCO3 gives 40 kg of MgO Therefore.05 kg KCl needed 2.7. K2O and P2O5 are each equivalent to 15 weight % = 150 kg each Ammonia reacted = 34 × 150 28 H3PO4 needed 150 = 207. kg Weight.761 0. MASS RELATIONS 27 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 =1 . 47 = 0.37 An aqueous solution contains 47% of A on volume basis. 700 SB rB = 1300 kg/m3 Therefore.47 × 1250 = 587.53 m3 × 1000 = 530 kg Hence.57% 530 .53 m3 Therefore. specific gravity of B = 1.5 kg Volume of water = (1 – 0. Basis: 1 m3 of solution Volume of A in solution = 1 × 0.47) = 0. express the composition of A in weight %.47 m3 Weight of A = 0.5 = 52. If the density of A is 1250 kg/m3. weight % of A = 587. the weight of water = 0.3 2. 3 g/cc Volume of solution = 143 = 110 cc 1. 587. Find the composition in molarity and molality. The density of solution is 1.1 .5 2.3 g/cc.312 = 2.38 An aqueous solution contains 43 g of K2CO3 in 100 g of water.312 = 3.3 Moles of solute = Weight/Molecular weight = 43 = 312 g moles 138 Molarity = g mole/volume of solution in lit = 0. Basis: 100 g of solvent Weight of K2CO3 = 43 g Weight of solution = 143 g Density of solution = 1. 0.12 g moles/kg solvent.833M 0.11 Molality = g mole/kg of solvent = 0. 26 2.6 24.0 41. Compound Weight.00 49.94 6. O2: 1. g Atomic weight or Molecular weight Number of moles 9.293 0.00 3894.23 3894.615 2.414 = 1.587 108 32 16 Molecular formula = Ag2SO4 Number of moles 0.587 g of O2. molecular formula of the compound = MgSO4. 26.8 22. average molecular weight and density.737 kg/m3 = 1.28 PROCESS CALCULATIONS 2.156 × 10–3 0.01% O2 and 57. benzene: 24.978 g of Ag.92 Therefore.01% S.39 A gaseous mixture contains ethylene: 30. Find the molecular formula of this compound.410 0.410 1.0 750.41 1 1 3. Basis: 100 kmoles of mixture Compound mole % C2H4 C6H6 O2 CH4 C2H6 N2 30.0367 Converting to whole numbers dividing by 9. 0. g Atomic weight or Molecular weight Ag S O2 1.76% Mg.737 g/l.6 248.22% H2O by weight.5 1.04 .22 24 32 16 18 0.2 100. ethane: 25%.1% and methane: 15.156 × 10–3 2 1 4.2 × 22.40 A compound has a composition of 9. Basis: 100 g of compound Compound Mg S O H2O Weight.293 g of S and 0.07 6.76 13.07 1. 100 2. Find the molecular formula of the compound. 13. N2: 3. Estimate the composition in mole %.01 26.0 86.1 Molecular weight 28 78 32 16 30 28 Total Density = Weight/Volume = Weight.0183 9.8 1911.3%. weight %.01 57.5%.5% in volume basis.978 0.5 25 3.7H2O 2.6%.3 15.8738 Converting to whole numbers dividing by 0.37 19. kg Weight % 856.41 A substance on analysis gave 1. 44 A mixture of FeO and Fe3O4 was heated in air and is found to gain 5% in mass. the value is 39.e the fraction of oxygen in the mixture is 0. Let the mole fraction of methane be X 21. Fe2O3 formed is (100 – X) × 160 t X (480) .MASS RELATIONS 29 2. we get x = 0.6 = (Molecular weight of CH4)(X) + (Molecular weight of C2H6) (1 – X) 21. Estimate the composition of the gas mixture. Fe2O3 formed is 160 kg Therefore. from X kg of FeO.42 Two engineers are estimating the average molecular weight of gas containing oxygen and another gas.e.43 A mixture of methane and ethane has an average molecular weight of 21.4 and by using the molecular weight of oxygen. Let x be the mole fraction of oxygen in the mixture and the molecular weight of the other gas be M 39. i. One uses the molecular weight as 32 and finds the average molecular weight as 39. the weight of final product is 105 kg.4 2. Find the composition.6.4 = (x) × (16) + (1 – x) × (M) Solving.4. Fe2O3 formed is 480 kg (480) 464 Since 5% gain in mass is observed. the value is 33. we get X = 0. Reactions involved are: 2FeO + 0. Therefore.8.8 = (x) × (32) + (1 – x) × (M) 33.4 i. By using the atomic weight of oxygen as 16.8 and the other uses the atomic weight of oxygen as 16 and finds the average molecular weight as 33.6 2. Fe2O3 formed is 160 × X 144 From 232 kg of Fe3O4. Find the composition of initial mixture.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. from (100 – X) of Fe3O4.6 = 16 × X + 30 × (1 – X) Solving.5 O2 ® Fe2O3 2Fe3O4 + 0. 75 kg . the weight of FeO = 20. X. 100  X t  105 144 464 Solving.45) = 79.25 kg Fe3O4 = (100 – 83. 30 PROCESS CALCULATIONS 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 SiO2 H2O 81 14 5 Compound Weight % Molecular weight moles mole % MgCO3 SiO2 H2O 81 14 5 84 60 18 0.964 0.233 0.278 65.34 15.82 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 H3PO4 142 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 PbS + 2O2 ® PbO2 + SO2 (1) (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 (239.2) (32) (207.2) (1) (64) PbS + 2O2 ® PbO2 + SO2 (239.2) (64) (239.2) (2) (64) 207.2 kg of Pb comes from 239.2 kg of PbS [from stoichiometry Eq. (1)] MASS RELATIONS 31 6 = 6.927 kg of PbS 207.2 239.2 kg of PbO2 comes from 239.2 kg of PbS Therefore, 6 kg of Pb comes from 239.2 × 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) ¸ ¹ × 100 = 151% « 1.195 º Percentage excess O2 supplied = ª Amount of SO2 formed If 207.2 kg of Pb is formed, SO2 formed is 64 kg 6 = 1.853 kg 207.2 If 239.2 kg of PbO2 is formed, SO2 formed is 64 kg If 6 kg of Pb is formed, SO2 formed is 64 × 64 = 0.268 kg 239.2 Total SO2 formed = 1.853 + 0.268 = 2.121 kg If 1 kg of PbO2 is formed, SO2 formed is % conversion of PbS fed to Pb = Mass of PbS converted to Pb/Total mass of PbS 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. = 6.927 × 32 PROCESS CALCULATIONS 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 kg % weight A B C 11 25 4 27.5 62.5 10.0 40 50 60 Total 40 100.00 Average density = Mass/Volume = moles, kmole mole % Density, Volume, kg/m3 m3 0.275 0.500 0.067 32.66 59.38 7.96 0.842 100.00 750 800 900 0.0147 0.0313 0.0044 0.0504 40 = 793.65 kg/m3 0.0504 Average molecular weight = Weight/Total moles = 40 = 47.5 0.842 (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: 2KClO3 Æ 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. MASS RELATIONS 33 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 34 PROCESS CALCULATIONS 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. 033 kgf/cm 22. pressure or 760 mm Hg or 29.414 litres 1 lb mole of any gas under standard conditions will occupy 359 cu.1 Standard Conditions 1 atm.16 K 0.1. Normal Temperature and Pressure/Standard Conditions Parameters Pressure Molar volume Absolute temperature Gas constant English Metric 2 SI 1 atm 359 ft3/lb mole 1. 1 g mole of any gas under standard conditions will occupy 22.3 Ideal Gases 3.2 Ideal Gas Law The ideal gas law states that.083 Bar m3/kmole K 35 .1.414 m3/kmole 1.69 oR 0.16 T(°R) = T °F + 459.ft.085 kgf m3/kmole K 273.73 atm ft3/lb mole oR 273.01325 bar 22.414 m3/kmole 491.1 RELATION BETWEEN MASS AND VOLUME FOR GASEOUS SUBSTANCES 3.69 3.92 inches of Hg and 0 °C or 32 °F By Avogadro’s Hypothesis.16 K 0. PV P V n R T = = = = = = nRT Pressure of gas Volume of n moles of gas Number of moles of gas Gas constant 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. T(K) = T °C + 273. 73 lbf ft3/in2 lb mole °R Avogadro’s Number: 6. °F. Temperature R R R K K K K Pressure Volume psia psia atmospheres Pa atmospheres atmospheres cm Hg in3 ft3 ft3 m3 m3 cm3 cm3 Gas constant ‘Rg’ 18.cc/g mole K = 0.06 6239.023 ´ 1026 molecules per kg mole Different units are used to express pressure like atmosphere.67 psia = 105 N/m2 = 760 mm Hg 1 atm = 760 Torr = 29.73 0. kg/cm2.1 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. Similarly. bar. However. .023 ´ 1023 molecules per g mole 2. N/m2. m3. litre.08206 82.01325 bar 1 atm = 14. 3. The following table gives the gas constant in different units. The temperature is expressed in °C.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.2.36 PROCESS CALCULATIONS Pressure 1 atm = 1. volume is expressed in cm3.92 inches of Hg = 76 cm Hg PV R = = 82.033 kgf/cm2 = 1.2.73 ´ 1026 molecules per lb mole 6.51 10. Thus. and Pa.314 J/(g mole) (K) = 1545 ft lb/lb mole °R 3. psia.73 atm ft3/lb mole °R nT R = 10.79 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. the temperature used in the application of Ideal gas law is in terms of K or °R.2 GASEOUS MIXTURE 3.73 8314 0.06 atm. mm of Hg. the gas constant is a dimensional quantity. K and °R. ft3 and gallon. (a) n RT n RT nA RT . C.4 Amagat’s Law (or) Leduc’s Law The total volume (V ) occupied by a gaseous mixture is equal to the sum of the pure component volumes VA + V B + V C = V VA. … Where ideal gas law is applicable. VC. PVC = nCRT Adding P(VA + VB + VC) = RT (nA + nB + nC) = nRT = PV Dividing. pC = C V V V Adding all the partial pressures of A. … . Dividing.IDEAL GASES 37 Components A B C Sum of the quantities Partial pressure Number of moles Pure component volume pA nA VA pB nB VB pC nC VC P n V 3.3 Dalton’s Law The total pressure (P) exerted by a gaseous mixture in a definite volume is equal to the sum of partial pressures. V RT (n A + nB + nC ) pressure fraction = mole fraction. 3. pA = Ë RT Û P = pA + p B + p C = Ì Ü ´ (nA + nB + nC) ÍV Ý nA RT V × × . VB.2. pA /P = (b) PVA = nART.2. PVB = nBRT. B. pB = B . pA + pB + p C = P where pA. pB. or. n A RTP VA = NA = P(n A + nB + nC ) RT V VA = NA × V . B. we have. … stand for pure component volume of components A. B and C. … represent partial pressure of components A. pC. C. 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. Basis: 15 kg Cl2 = 15/71.2113 ¥ 22. the average molecular weight of air = 28. density also varies significantly with temperature for a specified composition. WORKED EXAMPLES 3.1 Calculate the volume of 15 kg of Chlorine at a pressure of 0.9 273 Calculate the volume occupied by 6 lb of chlorine at 743 mm Hg and 70 °F Its volume is 0.84 kg Hence.0 293 ¥ = 5.84 3.0845 ¥ 359) = 30. However.2 Basis: 6 lb of Cl2 ∫ 6/71 = 0. This is applicable only for gaseous mixtures and not for solid or liquid mixtures.34 ft3 Volume at given condition ÊPV ˆ ÊT ˆ = Á 0 0˜ ¥Á 1˜ Ë T0 ¯ Ë P1 ¯ .38 3.2113 kmole 1.4 DENSITY OF MIXTURE Density is defined as the weight of a mixture per unit volume and is independent of temperature.3 PROCESS CALCULATIONS AVERAGE MOLECULAR WEIGHT The weight of unit mole of the mixture is called average molecular weight.9 bar and 293 K.0845 lb mole of chlorine Volume at standard condition = (0. which is also equal to total weight of the gas mixture divided by the total number of moles in the mixture. in the case of solids the variation of density with temperature is not very significant.643 m3 0.414 ¥ 3. air contains 79% nitrogen and 21% oxygen by volume.0 = 0. For example. As the volume of liquids and gases is a strong function of temperature. Find the pressure of the gas stored (required)? 0.5 Calculate the maximum temperature to which 10 lb of nitrogen enclosed in a 30 ft3 chamber may be heated without exceeding 100 psi pressure.3102 ´ 22.5664 m3 ÈPV Ø ÈT Ø Pressure at the given condition “P1” = É 0 0 Ù – É 1 Ù Ê V1 Ú Ê T0 Ú 6.34 – Ù –É Ù Ê 743 Ú Ê 492 Ú = 33.2046 Ê 44 Ú = 0.4 °C .67 Ú = 435.76 ft3 ٖ Ê 760 Ú 296 3.9528 m3 Volume at given condition = 20 ´ 0.21Ú Ê 14.4 K = 162.5664 Ú Ê 273 Ú = 13.IDEAL GASES 39 760 Ø È 530 Ø È = É 30.9528 Ø È 303 Ø È = É1 – Ù –É Ù Ê 0.4 ft3 3.02832 = 0.6818 È 30 Ø Basis: 30 lb CO2 = É Ù = 0.01047 lb mole Ê 359 ÙÚ Weight of water = (0.21 ft3 \ Temperature ‘T1’ ÈT V Ø È P Ø = É 0 1Ù – É 1 Ù Ê V0 Ú Ê P0 Ú 30 Ø È 100 Ø = ÉÈ 273 – Ù –É Ù Ê 128.18846 lb It is desired to compress 30 lb of CO2 to a volume of 20 ft3 at 30 oC.357 lb mole Volume at standard condition = 0.5 Ø 273 È = É 200 – = 3.76 Ø = É = 0.4 Moles of water È 3.6818 lb mole = 2.ft. Basis: 10 lb of N2 = 10/28 = 0.5 mm Hg and 23 °C Basis: 200 ft3 of gas at given condition È PV Ø È T Ø Volume at standard condition = É 1 1 Ù – É 0 Ù Ê P0 Ú Ê T1 Ú 15.01047 ´ 18) = 0.3102 kmole Volume at standard condition = 0.3 Calculate the weight of 200 cu.414 = 6.62 atm 3. of water vapour at 15.357 ´ 359 = 128. 792 0.187 – x + 2x) = 0. calculate the percentage dissociation of N2O4 to NO2? N 2 O 4 → 2NO 2 (92) (2 × 46) N2O4 present initially = 17.764 0.792 ´ 28 = 22.2 = 0.2% Basis: 1 g mole of the gas Component Volume % = mole % CO2 O2 N2 13.40 3.450 cc.2/92 = 0. 414 (0.0 Molecular weight g mole 44 32 28 0. ‘2x’ g moles of NO2 is formed.167 Ø Percentage dissociation = É ´ 100 = 89.7 Calculate the average molecular weight of a gas having the following composition by volume.2 cc Therefore. Total moles after dissociation = (0.187 g mole.077 0. then.131 0.167 È 0. Assuming that the ideal gas law applies. of g moles remaining = \ \ 7939.450 cc ? È PV Ø È T Ø Volume at standard condition = É 1 1 Ù – É 0 Ù Ê P0 Ú Ê T1 Ú 720 Ø È 273 Ø È = É11450 – Ù –É Ù Ê 760 Ú Ê 373 Ú = 7939. Let ‘x’ g mole of N2O4 dissociate.354 22. CO2: 13. no.42% Ê 0.6 PROCESS CALCULATIONS When heated to 100 °C and 720 mm Hg.354 x = 0.1%. 17.7% and N2: 79. g .1 7.000 30.404 Average molecular weight is 30.7 79.2 Total 100.464 0.187 + x) = 0.131 ´ 44 = 5.187 + x Given condition (1) Parameter Standard condition (0) Pressure 720 mm Hg 760 mm Hg Temperature 373 K 273 K Volume 11.2 g of N2O4 gas occupies a volume of 11.187 ÙÚ 3. O2: 7.404.077 ´ 32 = 2.176 1. Weight. 0278 Total 0.IDEAL GASES 3. find the following: (a) lb moles of H2.8 41 Calculate the density in lb/ft3 at 29.21 0. (c) pressure fraction of H2 (d) partial volume of H2. (b) mole fraction and mole % H2.26 atm V = 1000 ft3.79 32 28 6. (i) density of gas mixture.31 atm.0555 0.92/29) ´ (546/492) = 34. 0. Basis: 1 lb of gas mixture Component Weight % Molecular weight lb mole Hydrogen Oxygen 0. (j) density at standard condition Also. (h) average molecular weight.1% of hydrogen by weight.84 g È 530 Ø È 760 Ø – = 24.889/32 = 0.414 ´ É Ê 492 ÙÚ ÉÊ 741 ÙÚ È 28. g Oxygen Nitrogen 0.111 0.12 Total 28.72 22. Assuming ‘Ideal Gas’ behaviour.0292 lb/ft3 3.8 litres Volume of air = 1 ´ 22.84 Ø Density of air = É = 1. (e) volume fraction and volume % of H2. (f) weight of H2.24 ft3 Density = (1/34. show that volume % = pressure % = mole % Basis: 1000 ft3 of gas mixture (a) Partial pressure of hydrogen = 0. T = 710 °R .162 g/litre Ê 24.24) = 0. (g) weight fraction and weight % of H2.0833 lb moles Temperature = 30 °C = 86° F = 546° R Volume of the gas at the given condition = 0.0 inches of Hg and 30 °C for a mixture of hydrogen and oxygen that contains 11. the partial pressures are 0.26.10 In 1000 ft3 of a mixture of hydrogen.8 ÙÚ 3.889 2 32 0.32 and 1.0833 ´ 359 ´ (29.9 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.111/2 = 0. nitrogen and carbon-dioxide at 250 °F. 89 atm È 1000 Ø È 1.89 Ø È 492 Ø Total moles = É = 3.00 .502 ´ 2 = 1.26 Ø = 0.8 16. lb Weight % 13.89 atm and 710 °R È 1 Ø È 710 Ø Volume of H2 = 0.648 È 0.89 0.3 ´ 44 = 3049.31 = 0.169 189 Mole fraction of carbon dioxide = Mole fraction of hydrogen = Component Molecular weight Hydrogen 2 Nitrogen 28 Carbon dioxide 44 Total mole % 13.33 85.00 710 = 180.89 ÙÚ (d) Partial volume of hydrogen is the volume occupied by 0.8 13.138 3.32 + 1.2 69.502 ´ 359 ´ É = 138 ft3 – Ê 1.648 lb moles – – Ê 359 ÚÙ ÊÉ 1.42 PROCESS CALCULATIONS 0.138 (c) Pressure fraction of hydrogen = É Ê 1.89 lb mole Weight.138 1.9 ´ 28 = 473.502 lb moles of it at 1.26 492 – 1.89 100.0 3550.3 100.0 0.502 = 0.26 + 0.3 69.00 ÚÙ ÊÉ 710 ÚÙ mole fraction of hydrogen = 0.0 1.169 Ø lb moles of H2 = É = 0.004 lb (g) Basis 100 lb moles of gas mixture Mole fraction of nitrogen = 0.9 16.138 1000 Thus volume % = pressure % = mole % (e) Volume fraction of hydrogen = (f) Weight of hydrogen = 0.169 ft3 Volume of H2 at standard condition = 1000 ´ È 180.31) = 1.502 lb moles Ê 359 ÙÚ (b) Total pressure = (0.8 ´ 2 = 27.2 100.26 = 0.9 69.32 = 0.89 ÚÙ ÊÉ 492 ÚÙ 138 = 0.693 1.6 16.78 13. 00 ÙÚ ÉÊ 710 ÙÚ \ = 1309. Density at given condition = ÈÉ 129.5 Ê 100 ÙÚ È 492 Ø È 1.55 Ø Density = É = 0.7 ft3 \ È 129.7 ÙÚ 3.55 lb.IDEAL GASES 43 È 3550 Ø (h) Average molecular weight = É = 35.12955 lb/ft3 Ê 1000 Ú ( j) Density at standard condition: È 1. The volume of reactants (all gases) is 600 ml and the volume of products (all gases) under the same condition is 700 ml.5 359 = 129.11 A certain gaseous hydrocarbon is known to contain less than 5 carbon atoms.648 lb moles º 3. Let the hydrocarbon be Cx Hy Moles: y y CxHy + ÈÉ x  ØÙ O2 ® xCO2 + ÈÉ ØÙ H2O Ê Ê 2Ú 4Ú y yØ È ® x 1 ÉÊ x  ÙÚ 2 4 È Total moles of reactants = É1  x  Ê yØ Ù 4Ú yØ È Total moles of products = É x  Ù Ê 2Ú Reactants (1 + x + y /4) 600 6 = = = Products ( x + y /2) 700 7 È 7y Ø 7 + 7x + É Ù = 6x + 3y Ê 4Ú .7 = 3.09892 lb/ft3 Ê 1309. What is the compound? Basis: 1 mole of hydrocarbon.648 ´ 35.0 ÙÚ = 1309.7 ft3 at NTP condition Number of moles = 1309. This compound is burnt with exactly the volume of oxygen required for complete combustion.55 ØÙ = 0.89 Ø È 492 Ø – Volume at standard condition = 1000 ´ É Ê 1.89 Ø – (i) 1000 ft3 of gas at given condition º 1000 ´ É Ê 710 ÙÚ ÉÊ 1. the hydrocarbon is C3H8 : (Propane) The combustion reaction is C3H8 + 5O2 ® 3CO2 + 4H2O 3.44 PROCESS CALCULATIONS È 7y Ø  3 y Ù = –7 (7x – 6x) + É Ê 4 Ú 5y = –7 4 Since the hydrocarbon has carbon atoms less than 5. solving above equation assuming carbon atoms as 1.61 = 164 g moles. 4 y= 40 =8 5 5y 44 = –11 y= 4 5 The value of y is to be an integer. – 5y = –8. – 5y = –10. O2. CO2 : 13. 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.414 ´ ÈÉ – Ê 740 ÙÚ ÉÊ 273 ÙÚ = 4950. we have x < 5. CO2) + H2O ­ Water Basis: 100 g moles gas entering È 473 Ø È 760 Ø = 3972. up to 5 we get the values of y as indicated below: x– x = 1. CO2) ® Evaporator ® (N2. – 5y = –9. (N2.3%.12 Combustion gases having the following molal composition are passed into an evaporator at 200 °C and 743 mm Hg (N2 : 79. 760 Ø È 358 Ø Volume of gas leaving = 164 ´ 22. x=4 – Hence. \ g moles of gases leaving = 100/0. O2 : 4. O2 : 7. Water added = 164 – 100 = 64 g moles = 1152 g.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 : 48.2%.4% and H2O : 39%.414 ´ É – Ê 273 ÙÚ ÉÊ 743 ÙÚ This 100 g moles of entering gas º 61% of gases leaving. 4 y= 36 5 x = 3. 4 y= 32 5 x = 2.7 litres . 2.31 litres Volume = 100 ´ 22.2%. O2. 658 g moles of nitrogen Ê 22.IDEAL GASES (a) 45 Volume of gas leaving 4950.14 In the manufacture of hydrochloric acid.31 3.16 K the g moles of nitrogen will be 33. Applying the pure component volume method.4 – Ù = 33.678 g moles RT 8.0 t 100   29 g.5 litres .677) 3.01325 t 10 1000 g moles of gas occupies 22.7 °C and 738 mm Hg.0 t 100  100 litres gas-entering 3972.6 litres (b) Volume of water added 1152.4 m3 contains É 754. a gas is obtained that contains 25% HCl and 75% air by volume.314 – 400 Alternatively. 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 ´ 0.8 °C and 743 mm Hg and leaves at 26. 100 litres gas-entering 3972. This gas is passed through an absorption system in which 98% of the HCl is removed. calculate: (a) The volume of gas leaving per 1000 litres entering the absorption column.4 m3 t 5 400 1.01325 ´ 105 N/m2 T1 = 273 K since P1V1 P2V2 = T1 T2 V1 (at NTP) = 112 t 10 3 t 1000 273 = 754. where R = 8. P1 = 1.314 J/g mole K = 8. 754.414 Ú (Error is due to the fact that T1 is taken as 273 K and not as 273.98) = 24. The gas enters the system at 48.31 = 124.314 (Pa) (m3)/g mole K n= PV 112 – 103 – 1000 = = 33.414 m3 at NTP.16 K on taking T1 as 273.13 How many g moles of nitrogen will occupy 1000 m3 at 112 ´ 103 N/m2 and 400 K PV = nRT. (b) The weight of HCl removed per 100 litres entering. 1000 Ø È Hence. 8 Litre Volume % (or) mole % HCl 0.5 litres Volume of gas leaving = 75 + 0. by applying the partial pressure method: Weight of HCl absorbed = (a) volume of gas leaving per 100 litres entering (b) weight of Cl2 absorbed.8 lit.057 g.0 99.46 PROCESS CALCULATIONS Pure component volume of HCl remaining = 0.7 = 70.5 litres ? Volume of (entering) gas at leaving condition = 75.3 litres – 760 580 20.5 297 = 92. 22.5 100. 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 ´ 743 492 = 20.5 litres (a) Parameter Entering condition Leaving condition Pressure 743 mm Hg 738 mm Hg Temperature 48.34 Total: 75. The partial pressure of Cl2 is 59 mm Hg and the remainder being inert gas.7 °C Volume 75.5 mm Hg.5 = 33.3 ´ 36.5 ´ 743 299.15 Absorbing chlorine in milk of lime produces calcium hypochlorite. The gas leaves at 80 °F and 743 mm Hg with Cl2 having a partial pressure of 0.5 litres Volume of inert gas leaving = 100 ´ .5 0.8 °C 26.5 = 75. – 738 321. Calculate. A gas produced by the Deacon process enters the absorption apparatus at 740 mm Hg and 75 °F.414 3.5 mm Hg Volume of inert gases entering = 100 litres (681 mm Hg) 681 299.7 – 742.00 Component (b) Composition: Volume of HCl absorbed at standard condition = 24.5 = 742.66 Air 75. 5 litres (743 mm Hg) Volumes of Cl2 entering and leaving are 100 litres and 92.055 litre – 760 299.14 – 0. 4NH3 + 5O2 ® 6H2O + 4NO.7 Volume at standard condition of Cl2 absorbed = 7. The gases from this process are passed into towers where they are cooled and the oxidation is completed according to the reactions: (b) Weight of Cl2 absorbed = 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.5 273 = 0. The gases leave the catalyst at 700 °C and 743 mm Hg. The following reaction takes place.14 litres – 760 297 Volume at standard condition of Cl2 leaving = 100 ´ 0.5 ´ 7.16 Nitric acid is produced by the oxidation of ammonia with air.055 = 7. In the first step of the process.45 g. 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%) . ammonia and air are mixed together and passed over a catalyst at 700 °C. 22. Given the overall reaction: 2NO + 1.414 3. The ammonia and the air enter the process at 20 °C and 755 mm Hg.5 litres Entering condition: Parameter Pressure Temperature Volume Given condition (1) Standard condition (0) 59 mm Hg 297 K 100 litres 760 mm Hg 273 K ? Volume at standard condition of Cl2 entering 59 273 = 7. 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.5O2 + H2O ® 2HNO3.285 ´ 71 = 22.IDEAL GASES 47 (a) Volume of gas leaving = 92.085 litres = 92. 40 19.2 = 2.4 litres 273 755 Volume of NH3 = 1 ´ 22.85 – Ù = 1.7 12.21 Thus. (a) Volume of air = 11.06 g moles Ê 4Ú .0 Component Total (c) Gases leaving catalyzer are nitrogen. NH3 Air Exit gas Catalyzer 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 ´ 1.15 g mole 5Ø È O2 consumed = É 0. air supplied is 2.4 – 100 = 1142.3 N2 9. ammonia.2 litres – 273 755 Volume of air/100 litres of NH3 = 276.2 (b) g mole mole % = volume % NH3 1.4 g moles i.42 g moles 0.2 litres 24.e. oxygen. assuming 90% of the nitric oxide entering the tower is oxidized to acid.02 g moles. nitric oxide and water: N2 (all that enters) = 9. N2 = 9.48 PROCESS CALCULATIONS (d) The volume of gases leaving the catalyzer for 100 litres ammonia entering (e) Weight of acid produced per 100 litres NH3.0 O2 2.02 g moles NH3 (85% conversion) = (1 – 0.00 8.414 ´ 293 760 – = 276.85) = 0.42 100.02 72.414 ´ 293 760 = 24.4 = 11.42 ´ 22. 10 100.635 ¥ 22. The company works 24 h/day.635 mole % = volume % 71. for 100 litres of NH3 entering = (1031.85 g mole NO converted = 0.414 m3 at NTP .20 6.8 litres ¥ 273 243 This is the volume of gases leaving for 24.15 0.34 g moles NO formed = 0.40 10.85 g mole Ê H2O formed = Á 0.000 kg C19H36O2 + H2 Æ 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 22.70 10.06) = 1.275 12.000 (d) Volume of gases leaving the catalyzer 973 760 = 1031.2 = 4264 litres of gas leaves (e) NO entering the tower = 0.85 ¥ Ë 6ˆ ˜ = 1.4 – 1.765 ¥ 63) = 48 g For 100 litres of NH3 weight of HNO3 produced 48 ¥ 100 = 199 g 24.IDEAL GASES 49 O2 leaving = (2.02 1. The company purchases its H2 in cylinders of 1 m3 capacity. How many cylinders are needed per week? = Basis: Ester being processed in one week = (1000 ¥ 24 ¥ 7) = 1. 7 days a week.34 0.765 g mole HNO3 produced = 0.414 ¥ Therefore.68.85 ¥ 0.) and drops to 2 kg/cm2 (abs. The pressure in cylinder is initially 10 kg/cm2 (abs.9 = 0.2 3.2 litres of ammonia entering = 12.17 1000 kg/h of an organic ester C19H36O2 is hydrogenated to C19H38O2 in a process.85 1.8 ¥ 100)/24.275 g moles 4¯ Component N2 O2 NH3 NO H2O Total g moles 9.) after use.60 1.765 g mole = (0. ) 300 Ø È 3 Toluene entering = É1000 – Ù = 348.09 kmole H2 after use = É – Ê 22.5 Ø = 1577 Cylinders required per week = É Ê 0. The temperature of entering air is 500 °F and the pressure remains constant at 760 mm Hg.19 Air is dried from a partial pressure of 50 mm of water vapour to a partial pressure of 10 mm.8 Ø È 860 Ø È 273 Ø Weight = É ´ 92 = 74 lb. 3. How much water is removed per 1000 ft3 of entering air? .36 kmole 1135 Ø = 567. rounded to next highest value \ È 567. Basis: One hour 100 mm Hg (gauge) = 860 mm Hg (abs.18 A mixture of toluene and air is passed through a cooler where some toluene is condensed.36 ÙÚ 3. 1000 ft3 of gases enter the cooler per hour at 100 °C and 100 mm Hg (gauge).8 ft Ê 860 Ú È 348.414 ÙÚ ÉÊ 1 ÙÚ \ H2 available from one cylinder = 0. air is saturated with toluene. The partial pressure of toluene is 300 mm Hg. È 90 Ø Toluene leaving = 740 ´ É = 90 ft3 Ê 740 ÙÚ È 90 Ø È 740 Ø È 273 Ø – – Weight = É ´ 92 = 19 lb. Ê 359 ÙÚ ÉÊ 760 ÙÚ ÉÊ 323 ÙÚ \ Weight of toluene condensed = 74 – 19 = 55 lb. 740 ft3 of gases leave cooler per hour at 40 mm Hg and 50 °C. Vapour pressure of toluene at 50 °C = 90 mm Hg.414 Ú = 0. – – Ê 359 ÙÚ ÉÊ 760 ÙÚ ÉÊ 373 ÙÚ At exit. Calculate the weight of toluene condensed per hour.50 PROCESS CALCULATIONS È 10 Ø È 1 Ø H2 initially present in the cylinder (10 m3) = (1) ´ É Ù ´ É Ù Ê 1 Ú Ê 22.45 kmole (by reducing to NTP condition) È 1 Ø È 2Ø = 0.5 kmoles H2 needed = 1135 kg = ÈÉ Ê 2 ÙÚ The number of cylinders required should be full number. 5 kg of gas = Weight.5%.978 0.43 – 0.0 44) 28) 32) 28) Weight % = 418.416 ÙÚ (d) Specific gravity = density of gas/density of air = 1. È 1000 Ø È 492 Ø = 1.43 – Ù = 0.5 0.187 10.416 m È 0.2 (9.4 100.2 = 2259.5 (0.01672 kmole 29. Using ideal gas law.094 lb mole Ê 760 Ú Moles of dry air = (1.37 lb 3.000 0.43 lb moles Moles of air entering = É – Ê 359 ÙÚ ÉÊ 960 ÙÚ 50 Ø È Moles of water in it = É1. kg ´ ´ ´ ´ 100.7 (9. kmole 9.0 = 5. O2 : 9. weight CO2 CO O2 N2 44 28 32 28 Total (b) 0.0178 lb mole Ê 750 Ú Water condensed = (0.7%. 760 mm Hg = 0.202 = 0.7 Weight. (c) density of the gas in kg/m3 at condition of (b) (d) specific gravity of the gas mixture.01672 ´ ÈÉ 303 ØÙ ´ 22.5 Ø (c) Density = É = 1.0178) = 0.5 kg of gas at 30 °C and 760 mm Hg.904 Volume at 30 °C.094) = 1. CO : 0.51 IDEAL GASES Basis: 1000 ft3 of entering air.2 9. calculate: (a) its weight percentage (b) volume occupied by 0.6 = 307.5 = 0.300 .3 g/cc) Basis: 100 kmoles of chimney gas (a) Component Mol.925 1.6 13.0762 lb mole = 1.414 Ê 273 Ú 3 = 0.2%.562 2990.202 kg/m3 Ê 0.20 Chimney gas has the following composition: CO2 : 9.6 80. (Density of air may be taken as 1.6 (80.094 – 0.6% and N2 : 80.336 – Ù = 0.273 75.336 lb moles 10 Ø È Moles of water leaving = É1. 28 m3 È 298 Ø È 760 Ø – Volume of air = 52. CO = 23 kmoles O2 needed = 23/2 = 11.5 – 2.4 + 23) = 27.6% and N2 : 70%. CO : 23%.92 – 100 = 295.9 ´ 0.98 ´ 0.414 ´ É = 1313.98 kmoles 0. O2 in feed = 2.6 kmoles O2 actually needed = (11.43 – 100 = 52.4%.475 kmoles Composition of gases leaving (volume % = mole %): CO2 : 19.76 m3 2479.475 ´ 22.98 ´ 0.57% and N2 : 79.98 m3 2479.92 m3 Volume of gases leaving/100 m3 of feed = 7332.225 kmoles = 111.06% È 623 Ø È 760 Ø Volume of gases leaving = 141.850 kmoles N2 leaving = (70 + 41.85) 141.414 ´ É Ê 273 ÙÚ ÉÊ 750 ÙÚ = 2479.37%.52 PROCESS CALCULATIONS 3.98 ´ 22.21 A producer gas has the following composition CO2 : 4.9 kmoles 8. 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.21 = 11.400 kmoles O2 leaving = (8.25) = 2.414 ´ É – Ê 273 ÙÚ ÉÊ 750 ÙÚ = 7332.28 .125 kmoles N2 in air = 52. O2 : 1. Basis: 100 kmoles of producer gas.5 kmole.6) = 8.85 kmoles Air supplied = È 298 Ø È 760 Ø – Volume of feed = 100 ´ 22.28 (b) (assuming complete combustion) CO2 leaving the burner = (4.79 = 41. O2 : 2.43 m3 Ê 273 ÚÙ ÊÉ 750 ÚÙ (a) Volume of air/100 m3 of feed = 1313.21 O2 supplied = 52.9 – 1.25 = 52. 22 Natural gas has the following composition: CH4 : 94.0 2.120 Total 100.38 lb moles.029 ´ 100 = 2. 100 (b) Pure component volume of N2 = 0. R = 8.32 psia. of moles of gas = 497 = 1.9%.38 ´ 0.157 Ø Density at standard condition = É = 0.260 (0.1 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 nV where. (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 80 – 2.042 ´ 30) = 1. (a) Ideal gas law P = V = 381 ´ 10–6 m3 a Ø È (b) van der Waals equation É P  2 Ù (V – b) = nRT Ê V Ú a = 0. Calculate the following.0 1.1%.9 1.029 = 0.28 ´ 10–5 rest same as above Ë RT Û È a Ø P = Ì ÜÉ 2Ù Í (V  b) Ý Ê V Ú . C2H6 : 3% and N2 : 2. lb Component mole % CH4 C2H6 N2 94.m/g mole K (J/g mole K).298 1.38 ´ 0.23157 lb/ft3 Ê 100 ÙÚ È 23.38 ´ 0.042 1.040 ´ 28) = 1.157 Ø Density of gas = ÈÉ = 0.0466 lb/ft3 Ê 497 ÙÚ 3. b = 4.298 ´ 16) = 20. This gas is piped from the well at 80°F and 80 psi. 359 lb mole Weight.03 = 0.9 ft3 (a) Partial pressure of N2 = 80 Ø È 492 Ø (c) Volume of gas at NTP = 100 ´ ÈÉ = 497 ft3 – Ê 14.157 23.67 ÙÚ ÉÊ 540 ÙÚ No.040 (1. T = 313 K.3646.9 = 2.941 = 1.380 23.777 (0.53 IDEAL GASES 3.314 N. 2 ´ 10–6 N/m2 3.24 ft3 – Ê 492 ÙÚ ÉÊ 743 ÙÚ È 6. Ê 2 ÙÚ .29 ´ 10–3 g/cc 22414 3.28 t 10 º « (381 t 10 ) º « 381 t 10 = 5.25 A tire is inflated to 35 psig at 0 °F.031 lb mole.54 PROCESS CALCULATIONS P = © ¸ © 8.3646 ¸  ª 6 5 ¹ ª 6 2 ¹  4.73 ft3 atm/lb mole °R T = 120 + 460 = 580 °R nRT 580 = 0.3 ´ 10–3 g/cc 22414 1 – 28.5 atm V 0.0156 ´ 359 ´ É = 6. n1 = n2.414 cc.27 Acetylene gas is produced according to the reaction Density of air = 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. volume remaining same? P= P1 = 35 + 14.24 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.314 t 313 0. 1 = 0. To what temperature it can be heated up to a pressure of 50 psig.26 Calculate densities of C2H6 and air at NTP.031 ´ 0.12 h. 64 CaC2 + 2H2O ® C2H2 + Ca(OH)2 Basis: 1 lb of CaC2 = 64 36 26 74 È 535 Ø È 760 Ø Volume of acetylene got = 0.84 = 1. V1 = V2. P2 = 50 + 14.67 psia. T2 T2 ( P2T1 ) P1 (64.67 600 °R Temperature = 140 °F 3.67 = 64. Density of C2H6 = 1 – 30 = 1. What is the pressure? Basis: 1 lb of O2 = 1/32 = 0.67 = 49.67 psia.5 3.73 ´ = 26. R = 0.67 – 460) 49.0156 lb mole. At NTP 1 g mole occupies 22.24 Ø Time of burning = É = 3. T1 = 460 °R. T2 = ? ( P1V1 ) T1 \ ( P2V2 ) . Br2 : 28%. Ê 40 ÙÚ 3. g Weight.414 ´ É = 32 litres – Ê 273 ÚÙ ÊÉ 740 ÚÙ È 100 Ø = 3.15 g/litre – Ê 298 ÚÙ ÊÉ 760 ÚÙ Alternatively. 740 mm Hg.7 12.0156 kmole.6 Ø Time of burning = ÈÉ = 9. for a lamp burning gas 40 litres h. 64 298 Ø È 760 Ø = 392.12 g/litre Density = É Ê 32 ÙÚ (c) Density of air = 1. Basis: 100 g of gas mixture (a) Component Molecular weight Weight.3 Total 100 1.175 (5/32) = 0.293 g/litre (at standard condition) Density of air at 25 °C and 740 mm Hg È 273 Ø È 740 Ø = 1. volume of air at 25 °C and 740 mm Hg È 760 Ø È 298 Ø = 22.8 h.293 ´ É = 1.0 13.945 (28/160) = 0.6 litres.414 ´ É = 25. 25 °C.276 100 È 298 Ø È 760 Ø (b) Volume of gases = 1.156 74.28 By electrolyzing a mixed brine a mixture of gases is obtained at the cathode having the following composition by weight: Cl2 : 67%. Basis: 1 kg CaC2 = 1 = 0.IDEAL GASES 55 A similar problem has been solved in MKS system also: Data: 1 kg CaC2.0156 ´ 22414 ´ ÈÉ – Ê 273 ÚÙ ÊÉ 740 ÚÙ 392. Calculate: (a) composition by volume (b) density at 25 °C and 740 mm Hg (c) specific gravity of the gas mixture.13 litres – Ê 740 ÚÙ ÊÉ 273 ÚÙ . g mole Volume % = mole % Cl2 Br2 O2 71 160 32 67 28 5 (67/71) = 0. O2 : 5%. Volume of C2H2 = 0.276 ´ 22. 05% NH3 by volume. 30 °C) È 730 Ø È 293 Ø Volume of air at exit condition = 94.4 Ø Volume of exit gas = ÈÉ = 92.7 mm Hg. The gases leave the tower at 725 mm Hg and 20 °C having 0.372 ft3 4.7 1.012 lb mole = 0. Volume of NH3 = 5.1 NH3 by volume.0482) = 5.207020 lb 359 3.15 g/litre 25.047 ft3 È 725 Ø È 303 Ø Volume of exit NH3 at inlet condition = 0.372 = 0.9 ft3 (730 mm Hg. Without the total pressure being changed. Basis: 100 ft3 of entering gases. through an absorption tower in which NH3 is removed.9 ´ É = 92.9995 ÙÚ Volume of NH3 in exit = 0.447 ft3/min. After that the partial pressure of water is 12.0518 ft3 È 730 Ø È 273 Ø Volume of NH3 absorbed at NTP = 5.0482 ft3 (b) Volume of NH3 absorbed = (5.56 PROCESS CALCULATIONS 28. Using partial pressure method.15 3.047 ´ É – Ê 730 ÚÙ ÊÉ 293 ÚÙ = 0.12 Specific gravity of gas mixture = = 2. the temperature is reduced to 15 °C and some water condenses.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. This gas is passed at a rate of 100 ft3/min.13 3. Calculate: Density of air = (a) the rate of flow of gases leaving the tower and (b) weight of NH3 absorbed. Basis: 1000 ft3 of moist air at 740 mm Hg and 30 °C Partial pressure of water = 22 mm Hg . (a) \ 92.84 = 1. Ê 0.4 ft3 – Ê 725 ÙÚ ÉÊ 303 ÙÚ 92.29 A mixture of NH3 and air at 730 mm Hg and 30 °C contains 5.0518 ´ É – Ê 760 ÚÙ ÊÉ 303 ÚÙ = 4.95% of exit gas.4 ft3 of air º 99. find the following: Moles of NH3 = (a) Volume of gas after cooling and (b) Weight of water condensed.1 ft3 and of air = 94.1 – 0. 2 ft3/lb. Basis: 100 lb moles of gas i. 15 °C) (12.8 ÙÚ 23 (b) O2 required for the combustion of CO = ÈÉ ØÙ = 11. carbon in the feed = 23 + 4.e. CO2 : 4.562 lb – – Ê 359 ÙÚ ÉÊ 760 ÙÚ ÉÊ 303 ÙÚ 3. (Volume of gas/lb of carbon) = É Ê 328.4 = 27.7) = 727. (d) Calculate the volume of gases leaving at 600 °F and 750 mm Hg per 100 ft3 gas burnt.31 A producer gas has the following composition by volume CO : 23%.9 ´ 1.6) = 8.6 lb moles Theoretical O2 required = (11.3 ÙÚ ÉÊ 303 ÙÚ After cooling volume of gases = Volume of water vapour + Dry air (740 mm.3 mm Hg È 718 Ø È 288 Ø = 938 ft3 (a) Volume of air after cooling = 1000 ´ É – Ê 727. È 39200 Ø = 119.200 ft3 – Ê 492 ÙÚ ÉÊ 750 ÙÚ Weight of carbon present = (27.8 lb. O2 : 2.6% and N2 : 70% (a) Calculate the ft3 of gas at 70 °F and 750 mm Hg per lb of carbon present.5 lb moles Ê 2Ú O2 available in feed = 2. (b) Calculate the volume of air required for the combustion of 100 ft3 of the gas if 20% excess air is used.5 – 2.7 mm.7 Ø È 303 Ø Volume of air leaving at inlet condition = 938 ´ É – Ê 22 ÙÚ ÉÊ 288 ÙÚ = 570 ft3 Volume of water vapour condensed = 1000 – 570 = 430 ft3 430 Ø È 22 Ø È 273 Ø (b) Water condensed = ÈÉ ´ 18 = 0. 15 °C) È 12.68 lb moles . (c) Calculate the volumetric composition of gases leaving assuming complete combustion.4 atoms 530 Ø È 760 Ø (a) Volume of gases = 100 ´ 359 ´ ÉÈ = 39.4 ´ 12) = 328.4%.IDEAL GASES 57 Partial pressure of air = (740 – 22) = 718 mm Hg Partial pressure of air after cooling = (740 – 12.9 lb moles O2 supplied = (8.2) = 10. (a) Calculate the composition of the flue gases.06 moles moles moles moles moles moles (d) Volume of gases leaving È 760 Ø È 1060 Ø = 1. C + O2 ® CO2 C : 89.68 – 8. Carbon burnt = 89.35 ´ 359 ´ É – Ê 750 ÙÚ ÉÊ 492 ÙÚ Volume of gases/100 ft3 of feed = 1.19 lb . It may be assumed that 97% of the carbon burnt is oxidized to carbon dioxide and the rest to carbon monoxide.4) = 27.200 3.9 lb.1 ´ 0.930 ft3 – Ê 492 ÙÚ ÉÊ 750 ÙÚ (70 oF.68) = 40. Ash : 10.218 – 100 = 278.85 lb moles Ê 21 ÙÚ È 530 Ø È 760 Ø Volume of air = 50.85 ´ 359 ´ É = 19.66 1.9 = 80.58 PROCESS CALCULATIONS È 100 Ø Air supplied = 10.1% and ash : 10.09.00 lb N2 entering along with feed = 110. 750 mm Hg) È 530 Ø È 760 Ø = 39. calculate the rate of flow of gases in ft3/min.28 79.6 ft3 39.40 lb = 1.9%.218 ft3 = 139. Air supplied is 30% in excess of that required for complete combustion.200 ft3 Volume of feed = 100 ´ 359 ´ É – Ê 492 ÙÚ ÉÊ 750 ÙÚ 19930 – 100 = 50.9) N2 leaving (from air) = (50.85 – 10.35 lb Volume of air/100 ft3 of feed = Composition of CO2 O2 N2 Volume % 19.85 ft3 39200 (c) CO2 leaving = (23 + 4.78 lb O2 remaining = (10.32 A furnace is to be designed to burn coke at the rate of 200 lb/h having a composition C : 89.17 lb N2 Total Total = 139. Basis: 100 lb of coke.1 lb. (b) If the flue gases leave the furnace at 550 °F and 743 mm Hg.09. The grate efficiency of the furnace is such that 90% of the carbon present in the coke charged is burnt.68 ´ É = 50.17 lb = 70. 3.65) = 36.19 ´ 0. 66% will be associated to N2O4 at 26 °C. volume of flue gases is = (34. Of the NO2 formed.96 – 9.e.48 14. oxidizes to NO2. The gases are cooled to 26 °C at 750 mm Hg before entering the absorption column.07 lb moles (a) Component lb mole mole % CO2 6.31 78.83 Total 46.4 ft3/min. 3NO2 + H2O ® 2HNO3 + NO NO liberated in this reaction will be reoxidized in cooler.07 CO 0. Volumetric flow rate of flue gases = 1157.88 Ø O2 supplied = 89.97 = 77. air is passed through a magnetically flattened electric arc.58) = 3.41 lb = 0.00 È 760 Ø È 1010 Ø (b) Volume of flue gases = 46.1 ´ É Ù ´ 1.43 O2 3. for 200 lb/h of coke charge.2 Ø O2 required = É ÙÚ = 6.48 lb moles Carbon burnt to CO = 80. which on cooling.722 ´ 2) = 69.65 lb moles È 100 Ø Air supplied = 9. 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.19 ´ 0.444 ft3/h i.2 lb mole È 32 Ø È 308.33 In the fixation of nitrogen by the arc process.06 100.65 ´ É = 45. The gases are then passed into water washed absorption towers where nitric acid is formed.78 lb = 6.3 = 308.06 ´ 359 É = 34.88 lb = É Ê 12 Ú Ê 32 ÙÚ = 9.58 lb moles Ê 2 Excess O2 leaving = (9.IDEAL GASES 59 Carbon burnt to CO2 = 80.07 6. .48  0.722 ft3 – Ê 743 ÙÚ ÉÊ 492 ÙÚ Hence.31 lb moles È 6.67 N2 36.03 = 2.96 lb moles Ê 21 ÙÚ N2 in air = (45.20 0. Some of the nitrogen is oxidized to NO.65 – 6. O2 Arc NO 2% Cooler NO2 26 °C. Col. we have. (a) Reaction in arc furnace is given as N2 + O2 ® 2NO which means ‘x’ mole of N2 react to give 2x mole of NO.66 g mole NO2 remaining = 2 – 1.68 + 0.32 \ N2O4 formed = 0.66) = 1. Total moles of gas leaving = (79 – x) + (21 – x) + 2x = 100 (by stoichiometry) N2 O2 NO when x = 1.32 = 0.68 g mole NO2 Total moles: N 2O 4 O2 N2 Total 0. % NO = 2 = N2 O2 NO ¯ ¯ ¯ 78 20 2 2x 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 ´ 0.60 PROCESS CALCULATIONS (a) Calculate the composition of hot gases leaving the furnace assuming air is at NTP. (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. (b) Calculate the partial pressure of NO2 and N2O4 in the gas entering the absorption apparatus.66 + 19 + 78 = 98. H 2O Air N2. HNO3 Basis: 100 g moles of air contain N2 : 79 g moles and O2 : 21 g moles.34 . 750 mm Chamber N 2O 4 Abs. Moles of NO2 = (40.000 ft3/h is fed to gas reforming plant where the following reaction takes place. CnH2n + 2 + nH2O ® nCO + (2n + 1)H2 CO + H2O ® CO2 + H2 (1) (2) Reaction (1) is 95% complete and Reaction (2) is 90% complete. Find (a) Average molecular weight of the gas leaving stabilizer.68 – 750 = 5.25) ´ È 2Ø \ 0.34 0.66 – 750 = 5. (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. È 492 Ø È 16 Ø Volume of gas at NTP = 70.25 g moles (0. 98.278 ´ É Ù ´ 0.000 ´ É = 68.295 Ø Moles of gas = É = 190.66 ´ 2) = 1.278 g mole 98.082)(299) ÜÝ = 40.278 kmole of NO2 yields 0.000 ft3 of gas.95 g. CH4 : 78%.68) = 0.24 lb moles Ê 359 ÙÚ . 98.295 ft3 – Ê 550 ÙÚ ÉÊ 14.32 g moles/2 g moles 3NO2 + H2O ® 2HNO3 + NO Thus. Total moles of gas in 1000 litres of entering gas = PV RT 1000 Û È 750 Ø Ë = É – Ê 760 ÙÚ ÌÍ (0.85 = 0. 3.68 g mole Partial pressure of N2O4 = in N2O4 – (0.34 (c) Combined N2: in NO2 – 0.030 mm Hg.34 The gas leaving a gasoline stabilizer has the following analysis by volume C3H8 : 8%.67 ÙÚ È 68.188 mm Hg. C2H6 : 10% and C4H10 : 4% This gas leaving at 90 oF and 16 psia at the rate of 70.IDEAL GASES Partial pressure of NO2 = 61 0.34 3 moles of NO2 yields 2 moles of HNO3.158 kmole of HNO3 Ê 3Ú (85% conversion) \ Weight of HNO3 formed = 0.158 ´ 63 = 9. 32 190.39 + (19.54 lb.22 ´ 3 ´ 18) = 821.535 lb moles Total CO formed = (262.506.9 + (4725.535 + 224.39 ´ 0.22 ´ 3) + 148.9 (a) Average molecular weight of the gas leaving stabilizer = 4055.02 ´ 2) + (7.39 ´ 1 ´ 18) = 2.02 lb Weight of water produced from C2H6 = (19.39 16 2374.24 (b) C 3 H 8 + 3H 2 O → 3CO + 7H 2 44 54 84 14 CH 4 + H 2 O → CO + 3H 2 16 18 28 6 C 2 H 6 + 2H 2 O → 2CO + 5H 2 30 36 56 10 C 4 H10 + 4H 2 O → 4CO + 9H 2 58 72 112 18 Weight of water produced from C3H8 = (15.98 lb/h.02 ´ 0.54 lb CO + H2O ® CO2 + H2 Total CO formed = (15.68 78 148.46) ´ 2 = 1808 lb/h .22 44 669.54 ´ 2)] = 13.88 lb Weight of water produced from CH4 = (148.95 ´ 7) + (148.38 100 190.62 PROCESS CALCULATIONS Components Volume % lb mole Molecular weight Weight.6 4 7.40 ´ 0.02 ´ 2 ´ 18) = 684.95 ´ 5) + (7.02 30 570.671.9 = 21. we have (15.53 ´ 0.61 ´ 0.40 lb moles Hydrogen from CO (90% conversion) = (249. lb C3H8 CH4 C2H6 C4H10 Total 8 15.24 4055.92 lb = 4. (c) Hydrogen formed: From Reaction (1) (hydrocarbons undergo 95% conversion).90) = 224.53 ´ 18) = 4725. Weight of gas fed to reformer = [4055.53 lb moles.61 58 441.24 10 19.46 lb moles \ Total hydrogen leaving reformer = (679.72 lb Weight of water produced from C4H10 = (7.22 ´ 0.95 ´ 9) = 679.61 ´ 4 ´ 18) Total = 547.61 ´ 4) = 262.95 ´ 3) + (19.725.95) = 249. H2O needed for forming CO = (262. 320 NH3 17 2 Total — 100 Average molecular weight = 2 ´ 17 = 34 2.42 100 (b) Volumetric flow rate at standard condition È 273 Ø È 2 Ø = 600 ´ É = 1081.537 Weight % 0.892 lb Component C 3H 8 CH4 C 2H 6 C4H10 CO CO2 H 2O H2 Weight. CO2 : 30% and NH3 : 2%.484 lb 118.39 ´ 0.63 IDEAL GASES (d) Gases leaving unreacted = HC.725. lb 33.35 Analysis of a sewage gas sample from municipal sewage plant is given.54 – (224.32 9.876.000 lb 33.05 ´ 44 148.260 lb (from Reaction 2) 13.1 ´ 28 249.248 0.4 ´ 0.506.712 lb 28.12 Total 1808 13.823 13.05 ´ 58 249.484 118.53 ´ 0.53 22. (a) the average molecular weight can be calculated from the following table.4 ´ 0.0 ´ 0.712 28.61 ´ 0. Gas Molecular weight Weight.442 2.069 698. kg CH4 16 68 68 ´ 16 = 1.069 lb 698.506.000 3.240 lb 236. CO2.05 ´ 16 19.24 921. 600 m3/h of this gas at 30 oC and 2 atmospheres is flowing through a pipe.163 5.530 lb 22. CO. 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.088 CO2 44 30 30 ´ 44 = 1.320 lb 9876. H2 C3 H 8 CH4 C2 H 6 C4H10 CO CO2 H2O H2 15. Basis: 100 kmoles of the sewage gas. 442 = 24. kmole Weight.385 100. H2O.212 0.2 m3/h – Ê 303 ÙÚ ÉÊ 1 ÙÚ .05 ´ 30 7.9 ´ 44 262.17 73. CH4 : 68%.46 ´ 18)] = 1808.22 ´ 0.05 ´ 18 = = = = = = = = H2O [4.879 0.892 6.277 lb (from Reaction 1) 685. 27 6.00 3.64 PROCESS CALCULATIONS È 1081.89 1.80 0.1% NH3. Calculate (a) the rate of flow of gas leaving the tower and (b) weight of NH3 absorbed in kg/h.24 kmoles 2.442 kg Therefore.8 kmole Cl2 formed = (2 ´ 0.24 = 1. HCl gas is oxidized with air as follows: (c) Density of the gas is 4HCl + O2 ® 2Cl2 + 2H2O If the air used is 30% excess and oxidation is 80% complete.178 kg/h = 2442 = 2.3 ´ ÈÉ ØÙ = 4. the weight of 48.3 kmoles N2 entering 79 = 1. Basis: 4 kmoles HCl gas.8) = 0. 1081.79 100.37 A mixture of NH3 and air at 730 mm Hg and 30 °C contains 5.24 kmoles/h Ê 22.50 4.42 62. .60 10. 100 Hence.8) = 1. find the composition of dry gases leaving.89 kmoles Ê 21 Ú O2 remaining = (1. The gas is passed through an absorption tower at the rate of 100 m3/h where NH3 is removed. 442 – 48.2 Ø Molar flow rate of gases = É = 48.6 kmoles Composition of dry gases leaving is presented in the following table Gas mole mole % HCl O2 N2 Cl2 0.54 Total 7.36 In the process of manufacturing Cl2.414 ÙÚ We know that 100 kmoles of this gas weighs = 2.2 3.5 kmole HCl un-reacted = (4 ´ 0.26 kg/m3.3 – 0. The gases leave the tower at 725 mm Hg and 20 °C having 0.05% NH3.178 kg. the mass flow rate of gas in kg/h = 1. O2 needed = 1 kmole O2 supplied = 1.2) = 0.77 20. 30 °C.05% NH3 while leaving the absorber Therefore the total moles of gases leaving the absorber is 3.051 = 0.56 ´ 0.663 kmoles of air contains 0.002 kmole.197 kmole Air entering = (3.56 kmoles (from nitrogen balance) 0.21 = 22.26 kmoles O2 needed for conversion of CO to CO2 = 3.663 = 3.17 kmoles O2 consumed = (22. NH3 leaving the tower = 0.9995 Hence.42 m3/h – Ê 725 ÙÚ ÉÊ 273 ÙÚ 3. O2 : 8.1% NH3 65 Basis: One hour of operation È 100 Ø È 730 Ø È 273 Ø Moles of gas entering per hour = É – – Ê 22. 83.665 ´ 22.38 The analysis of a flue gas. 20 °C 725 mm Hg Abs.IDEAL GASES 0.197 – 0. from a fuel gas containing no N2 has CO2 : 4.91% and N2 : 83.39%.39 = 105.86 kmoles NH3 entering = 3. Basis: 100 kmoles of dry flue gas.317 kg Volumetric flow rate of exit gas È 760 Ø È 293 Ø = 3.17 – 8.663 kmoles 3. = (b) NH3 absorbed = (0.414 ´ É = 92. 730 mm Hg 5.197) = 3.0556 (a) Air supplied = CH4 ® C 2H6 ® Air ® CO2 CO O2 N2 Reactions are: CH4 + 2O2 ® CO2 + 2H2O C2H6 + 31/2O2 ® 2CO2 + 3H2O (b) O2 supplied = 105. CO : 3.79 kmole of dry air/kmole of dry flue gas = 1.86 ´ 0.08%. 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 C2H6.002) ´ 17 = 3.665 kmoles 0.86 – 0.08 = 1.54 kmoles 2 .91) = 13.414 ÙÚ ÉÊ 760 ÙÚ ÉÊ 303 ÙÚ = 3. Tower 100 m3/h.62%.05% NH3. 3 and y = 1.6/0.62 + 3. (b) % Excess air used (c) % Of carbon charged which is lost in the ash.4) = 78. The ash pit residue after being washed with water analyze 10% carbon.46 Total 6.8% % Excess air = É Ê 14.8 Solving for x and y.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. The flue gas analysis shows CO2 : 14%.5 kmoles.80 ÙÚ (c) Let CH4 be ‘x’ kmole and C2H6 be ‘y’ kmole Making a CO2 balance.1P (1) . O2 : 6. Basis: 100 kmoles of exit gas. air supplied = 78.91 – 1.39 A furnace is fired with coke containing 90% carbon and 10% ash.3 1. 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: 0.54 18. we get x = 5.9 – 6.4 = 14. CO : 1%.5 kmoles.2 81.08) = 7. O2 in air supplied is = 20. 40% ash and rest water.00 3. kmole mole % CH4 C2 H 6 5.6 kmoles From the foregoing.2 Gas Weight.5y = 14.5 100.80 kmoles Excess oxygen = (8. x + 2y = (4.9F = (15 ´ 12) + 0.70 O2 needed (by stoichiometry) 2x + 3.37 Ø ´ 100 = 49.54) = 7. N2 = 100 – (14 + 1 + 6.79 = 99.66 PROCESS CALCULATIONS \ O2 needed for complete combustion = 14.9 kmoles Furnace Flue gas Coke (F) Ash (P) O2 reacted = 20.37 kmoles È 7. CO + ½O2 ® CO2 Carbon lost in ash: 10% = 51.6P – 0.72 kg (coke) \ P= 2C + O2 ® 2CO C + O2 ® CO2.471 kmoles È 5.9 – 15.IDEAL GASES Ash balance: 0.9 ´ 205.471Ø ´ 100 = 35. carbon in coke: 1000 × 0.43 × 0.148 kg = 15.148 kg 205.72 × 0.72 Carbon reacted = 900 – 25 = 875 kg = 72.9 = 900 kg 1000 = 25 kg 205.429 kmoles (by stoichiometry) Excess O2 = 20.1 kmoles of the gas Ê 205.5 F = 205.143 kg (for 205.429 katoms O2 needed for conversion of C to CO2 is 15.1 kmoles flue gas .) F = 4P Substituting for (1) from (2).1 = 5.72 ÙÚ Aliter For 1000 kg of coke fed.72 kg of coke fed) O2 needed theoretically: O2 supplied: 20.1P) = 180 180 = 51.916 katoms reacted º 486.e.429 = 5.72 kg of coke gives 100 kmoles of gas È 100 Ø 1000 kg of coke gives 1000 ´ É = 486.9 ´ 4P) = 180 + 0.9 kmoles O2 needed (theoretically): Total Carbon fed = 205.72) = 185.43 kg (ash) 3.1P (3.1F = 0.46 % (b) % Excess air = É Ê 15 ÙÚ (a) Total carbon fed = (0.916 katoms Carbon lost in ash = 5. we get (0.4P 67 (2) (i.143 kg × \ 15 katoms of carbon reacted º 100 kmoles of flue gas 72.9 = 185. V2 = 20 m3. Basis: 1 m3 of inlet air = 1000 litres = 106 cc È 1 Ø È 768 Ø È 273 Ø Moles of air entering = É = 0. For a total pressure of 760 mm Hg at 111.1 ´ 22.02707 (1 + 0.2 m3/ton of coke.03063 kmole Volume of exit air È 384.414 ÙÚ ÉÊ 760 ÙÚ ÉÊ 439. The hot air enters the drier at 768 mm Hg and 166. moles of water/mole of wet air = 100 = 0.454 kmole 44 Volume at standard conditions: 0.02799 kmole – – Ê 22.414 ´ É – Ê 750 ÙÚ ÉÊ 273 ÙÚ = 21.9659 m3 3. moles of water/mole of wet air 25 = 0. The partial pressure of water vapour in this air is 25 mm Hg.0326) = 0.151. È 51.1Ø (c) Carbon lost = É ´ 100 = 2.41 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.1316) = 0. 20 = 0.414 = 10.0326 768 In the outgoing stream.19 m3 Basis: 20 kg of carbon dioxide = P1 = 1 bar. calculate the volume of exit air per cubic metre of inlet gas.19 m3. At the exit partial pressure of water is 100 mm Hg.7 °C.7 ÙÚ In the incoming stream.1Ø = 0. T1 = 273 K P2 = 20 bar.78% Ê 185.03063 ´ 22.454 ´ 22.414 ´ É Ê 273 ÙÚ = 0.02707 kmole = moles of wet air leaving = 0.1316 768 moles of dry air entering = 0.02799 (1 – 0.43 – 0. T2 = ? .148 ÙÚ 3.68 PROCESS CALCULATIONS Volume of flue gas produced È 760 Ø È 523 Ø = 486. V1 = 10.1 °C in this process.40 Hot air is being used to dry a wet wallboard. 716.42 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.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.79 m3. Find the mole fraction. Vapour pressure of water is 20 mm Hg.111 kmole = 20 kg of water Amount of water = 22.39 = 736.IDEAL GASES We know that È P1V1 Ø ÉÊ T ÙÚ 1 69 È P2V2 Ø ÉÊ T ÙÚ 2 Thus T2 is found to be 10. M = 27.39 mm Hg. Find the weight of water in the gas. CO2 and H2 in the ratio of 4 : 3 : 1 is at 150 °C and 2 atm pressure.39 mm Hg. È 4Ø Partial pressure of N2 = É Ù × 2 = 1 atm Ê 8Ú . weight % of each.433. Volume at standard conditions = 1000 – 273 – 736.414 3.3 K = 10. average molecular weight.43 A gaseous mixture has the following three components X. Y. mole % of Y = 40% Weight.3 °C 3.44 One hundred m3 of mixture of N2. Find the molecular weight of Y component. and total weight of the mixture. 902. Z and the composition is as expressed below.97 3. Basis: 1000 m3 of the gas at the given conditions 155 mm of water = 11. Given pressure = 725 + 11.79 – Component mole % Weight % Molecular weight Weight X 35 – 85 35 × 85 Y – 20.39 = 902. % of Y = 20 = (40 × M) × 100 (35 – 85)  (40 – M )  (25 – 60) 89500 + 800 M = 4000 M Therefore. 293 – 760 20 725 = 1. The barometer reads 725 mm Hg. 1 A liquefied mixture of n-butane.16 kmoles ˙¥Í ˙¥Á Î 423 ˚ Î 1 ˚ Ë 22.75 Total weight = 177.125) + 28 × (0.5) + 44 × (0. n-pentane and n-hexane has the following composition in percent.76 177.25 ¥ 100 ¥ 273 ˘ Number of moles of H2 = Í ˙ = 0.72 2 1. n-C4H10 : 50 n-C5H12 : 30 n-C6H14 : 20 Calculate the weight fraction.25 atm Ë 8¯ v ˘ È T0 ˘ È 1 ˘ È Number of moles of N2 = Í pN 2 ¥ ˙ ¥ Í ˙ ¥ Í T ˚ Î P0 ˚ Î 22.414 ˙˚ Î È 0.375 H2 0.197 00.00 1.16 44 95.375) = 30.70 PROCESS CALCULATIONS Ê 3ˆ Partial pressure of CO2 = Á ˜ × 2 = 0.500 CO2 2.676 45.197 kg EXERCISES 3.75 atm Ë 8¯ Ê 1ˆ Partial pressure of H2 = Á ˜ × 2 = 0.080 53.414 ¯ È 0. kmoles Molecular weight Weight.441 0.66 0.000 Average molecular weight = 2 × (0.414 ¯ Î v ˘ È T0 ˘ È 1 ˘ È Number of moles CO2 = Í pCO2 ¥ ˙ ¥ Í ˙ ¥ Í T ˚ Î P0 ˚ Î 22.125 Total 5.88 kmoles 423 ˙˚ ÍÎ 1 ˙˚ Ë 22.414 ˚ Compound Weight.88 28 80.531 0. kg Weight % mole % N2 2. .414 ˙˚ Î È 100 ˘ È 273 ˘ Ê 1 ˆ = Í1 ¥ ¥ ¥Á ˜ = 2.81 0.72 kmole Î 473 ¥ 1 ¥ 22. mole fraction and mole percent of each component and also the average molecular weight of the mixture.75 ¥ 100 ˘ È 273 ˘ Ê 1 ˆ = Í ˜ = 2. 5 atm has the following composition by weight. Nitrogen : 65% CO2 : 15% H2O : 12% . 50% ethane and rest hydrogen by volume at a temperature of 283 K and a pressure of 5 atmosphere. 3.6 atm. 3. Calculate the volume.3 A steel container has a volume of 200 m3. Express the concentration in kmole/m3.IDEAL GASES 71 3.2 A steel tank having a capacity of 25 m3 holds carbon dioxide at 30 °C and 1. The internal pipe diameter is 50 mm. Carbon monoxide : 4%. It is desired to store 12.5 A typical flue from a chimney is found to contain the following composition by weight: Oxygen : 16%.66 g. velocity in pipeline and density of the gas mixture.4 Natural gas has the following composition in volumetric percent: CH4 : 80%. calculate the fraction of the nitrogen which leaves the container. 3.12 The flue gas of a burner at 800 °C and a pressure 2. 3. flows through a pipe line at the rate of 60 m3/h. Calculate average molecular weight and density of the gas at NTP. Estimate its density at 20 °C and 741 mm Hg pressure. Calculate the pressure inside the cylinder by assuming that the mixture obeys the ideal gas laws.2% hydrogen by weight. and (d) density at standard condition.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.7 Air contains 79% nitrogen and 21% oxygen by volume.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. Carbon dioxide : 17% and rest nitrogen. what is the formula of the gas? 3. If 369 ml of the gas at 22 °C and 748 mm Hg weighs 0.6 A mixture of gases analyzing 20% methane. 3. 3.11 A gas analyzes 60% methane and 40% ethylene by volume. (b) composition in weight % (c) Average molecular weight. 3. If the container valve is opened and the container heated to 200 °C. C2H6 : 15% and N2 : 5% Calculate (a) composition in mole %.10 A gas contains 81.8 One kilogram of benzene is stored at a temperature of 50 °C and a pressure of 600 atmospheres. 3.8% carbon and 18. Calculate the weight of the carbon dioxide in grams. 3. It is filled with nitrogen at 22 °C and at atmospheric pressure. Assuming the validity of ideal gas law. g mole/cc. find: (a) Partial pressure of nitrogen (b) Volume of nitrogen/100 m3 of gas (c) Density of mixture (d) Average molecular weight of mixture 3.16 A gas flowing at 1000 litres/s has the following composition: CH4 : 10%. and (c) % conversion of hydrogen to ammonia. . Find (a) partial pressure of N2.17 Two hundred eighty kg of nitrogen and 64. 3. mole % and the average molecular weight. 3. (b) volume of N2 per 100 m3 of gas and (c) density of the gas.5 kg of hydrogen are brought together and allowed to react at 550 °C and 300 atm. (e) mass flow rate of the gas and (f) average molecular weight.14 The following is the analysis of a mixture of gases by weight: chlorine : 65%. (d) the molar density of the mixture. 3.5 kg of carbide in air lamp burning 100 m3 of gas/hour at 25 °C and 760 mm Hg. Calculate the composition by volume %. Calculate (a) The mole fraction of each component. 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. bromine : 27% and rest oxygen. (c) partial pressure of each component. C2H6 : 3% and N2 : 3% is piped from the well at 298 K and 3 atm pressure. 5% ethane and 5% nitrogen is piped from a well at 25 °C and 1 atm pressure.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. It is found that there are 38 kmoles of gases present at equilibrium. (b) what is the % excess of excess reactant.18 A natural gas containing 90% methane.72 PROCESS CALCULATIONS O2 : 7% CO : 1% Find (a) composition by volume (b) the average density of the flue gas (c) mole fraction of the components 3.15 A natural gas having CH4 : 94%. (b) the concentration of each component. C2H6 : 30% and H2 : 60% at 303 K and 2000 mm Hg pressure. (a) What is the limiting reactant. (b) average molecular weight.2% and ammonia 9. 3. ammonia gas and air are mixed at 7 atm pressure and 650 °C and passed over a catalyst. find (a) composition in mole % and (b) density in kg/m3. 3.5%.8%. Assuming the validity of the ideal gas law. Calculate (a) the composition by mole %. and (c) density at 298 K and 740 mm Hg.IDEAL GASES 73 3.22 Calculate the volume occupied by 30 g of chlorine at a pressure of 743 mm Hg and 21. 3. bromine: 25% and rest nitrogen. oxygen 18.23 Propane is liquefied for storage in cylinders.5%. The composition of this gas mixture on weight basis is nitrogen: 70.20 The following is the analysis of a mixture of gases by weight: Chlorine: 60%. How many kilograms of it will be formed by liquefying 500 litres of the gas at standard conditions? . water vapour: 1.1 °C.21 In the manufacture of nitric acid. Vapour Pressure 4 Whenever we come across a liquid system. the above reaction reduces to Clausius–Clapeyron equation. T : absolute temperature. which is called vapour pressure. l : heat of vaporization at T. the boiling point is referred as normal boiling point (NBP).1 EFFECT OF TEMPERATURE ON VAPOUR PRESSURE Illustrated by the effect of temperature on vapour pressure is Clapeyron equation dp l = dT T (VG . VG : volume of gas and VL : volume of liquid. p represents vapour pressure. 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. When that pressure becomes equal to the atmospheric pressure. if the sum of their partial pressures equals the surrounding pressure then the system will boil. it is generally in contact with its own vapour over the liquid surface. molal latent heat of vaporization.VL ) where. When the temperature does not vary over a wide range. Whenever we have a binary system. or d(ln p) = 74 . dp l dT = p RT 2 1 -l ¥d T R where R is gas law constant and l . This vapour exerts a pressure like gases. If the volume of liquid is neglected and applicability of the ideal gas law assumed. When the liquid is at its boiling condition the vapour pressure will be equal to the surrounding pressure. 4. (ii) Pressures below 10 bar. 4. p* is the T +C vapour pressure and A.5 cal/g. Calculate vapour pressure at 20 °C and 35 °C. l = (92.VAPOUR PRESSURE 75 Ê p ˆ l Ê 1 1ˆ ln Á ˜ = Á . B.303 R ¯ ÁË T0 T ˜¯ Effect of Temperature on V. T2 = 308 K .2 HAUSBRAND CHART Steam distillation takes place at point of intersection of the curves at which p A = p – pW \ p = p A + pW Total pressure = vapour pressure of component + vapour pressure of water p – pW pA Vapour pressure Temperature WORKED EXAMPLES 4. Latent heat of vaporization is 92. log p* (mm Hg) = A – (where. C. B.99 cal/g mole K T0 = 273 K. C are constants) Limitations: (i) Applicable only in the range of applicability of A.P: B Antoine equation.˜ Ë p0 ¯ R Ë T0 T ¯ or Ê p ˆ Ê l ˆÊ 1 1ˆ log Á ˜ = Á ˜ Ë p0 ¯ Ë 2.5 ¥ 74) = 6845 cal/g mole R = 1.1 The vapour pressure of ethyl ether at 0 °C is 185 mm Hg. Molecular weight of ethyl ether = 74. T1 = 293 K. 69 °C 785 p – pW pB Vapour pressure Temperature Basis: 1 g mole of mixed vapour.31 – 0.99 ÚÙ ÉÊ 273  293 ÙÚ È p Ø log É Ê 185 ÙÚ Therefore. Temperature.1 78 18 10. mm Hg Vapour pressure Total pressure. of water. g mole of C6 H 6 g mole of water g C6 H 6 g water g water g benzene Vapour pressure of C6 H 6 Vapour pressure of water 2.303 – 1.01 520 225 2.31 . mm Hg mm Hg 60 65 68 69 390 460 510 520 150 190 215 225 70 550 235 540 650 725 745 distillation temperature. 4. Calculate the temperature at which the distillation will proceed and the weight of steam accompanying 1 g of benzene vapour.76 PROCESS CALCULATIONS At 20 °C. p = 773 mm Hg.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. °C Vapour pressure of benzene.99 ÚÙ ÉÊ 273  308 ÙÚ È p Ø log É Ê 185 ÙÚ Hence. p = 437 mm Hg. È ØÈ 1 6845 1 Ø ÉÊ 2. At 35 °C. È ØÈ 1 6845 1 Ø ÉÊ 2.303 – 1. 1840 kg of steam is needed.3 ´ 10–5 kmoles = 0.7 Ø = É = 4. (b) Calculate the weight of steam per kg of acid if 26 inches of Hg vacuum is maintained.4 kg Steam 15. Moles of the acid È 0. 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.9843 kmole = 17.3 is to be distilled at 200 °C by use of superheated steam.6 mm Hg Basis: 1 kmole of mixed vapour È 11. Calculate the temperature and the weight of steam to be used per kg of acid.3 and 4.4: When ordinary steam is used at atmospheric pressure.1 ÙÚ kg of acid Discussion on examples 4. quantity of steam needed is 0.0098 Ú 1840 kg The acid given in example 4.032 Ø : É ´ 1 = 4.0098 kg Ê 740 ÙÚ Mole of water Moles of water = 1 kmole = 18 kg Weight of steam kg of acid 4.855 kmole = 15.58 ÙÚ kg of acid (b) 26 inches of Hg.0157 kmole = 3.1 kg Ê 80 ÙÚ Weight of water (steam) = 0.70 kg Steam È 17. At 99 °C: Vapour pressure of H2O = 740 mm Hg Vapour pressure of acid = 0.5 ´ 0.95 kg Ê 3.5 mm Hg Partial pressure of acid (80% saturation) = (14.8) = 11.4) = 80 mm Hg È 11.465 kg only. Due to low pressure and hence low boiling point under vacuum.95 kg is needed.VAPOUR PRESSURE 4.3 77 It is decided to purify myristic acid (C13H27COOH) by steam distillation under 740 mm Hg pressure. vacuum = 740 – (26 ´ 25.6 Ø Weight of acid = É ´ 1 = 0. .465 kg Ê 33.58 kg Ê 740 ÙÚ Weight of water = (1 – 0.4 È 18 Ø ÉÊ Ù 0.4 Ø = ÈÉ = 0. When superheated steam at 200 °C is used. (a) Vapour pressure of acid at 200 °C = 14.032 mm Hg \ The distillation temperature can be taken as 99 °C Basis: 1 kmole of mixed vapour.0157) = 0.145 kmole = 33. 4.6 Ø Weight of acid = É ´ 1 = 0. 435 0.000 Total Weight. wt Vapour pressure. Vapour pressure of water at 108.236 0.5 mm Hg. Composition of vapours and their partial pressure Component Benzene (C6H6) Toluene (C7H8) Xylene (C8H10) Mol.000 797 1.388 ´ 560 = 217 0. Find the vapour pressure of the solution at 30 °C and the boiling point elevation.055 100 1.8 = 5. toluene (Molecular weight : 92) is 560 mm Hg and Xylene (molecular weight : 106) is 210 mm Hg. Assume that the relative vapour pressure of the solution is independent of temperature. Vapour pressure of water at 30 °C = 31.449 0.5 PROCESS CALCULATIONS Calculate the total pressure and the composition of the vapours in contact with a solution at 100 °C containing 35% Benzene. kmole Mole fraction in liquid phase Partial pressure. È from Cox chart Ø Boiling point of water at 23.211 ´ 210 = 44 0.74.5 mm Hg = 24. the vapour pressures of benzene (molecular weight : 78) is 1340 mm Hg. mm Hg Mole fraction in vapour phase Total pressure = 797 mm Hg. At 100 °C.8 °C É Ê from Steam Tables ÙÚ Boiling point elevation = (Boiling point of solution – Boiling point of solvent) = 30 – 24.2 °C .8 mm Hg (from Tables) Vapour pressure of solution at 30 °C = (31. mm Hg Weight % 78 1340 35 35/78 = 0. Basis: 100 kg of solution.673 92 560 40 40/92 = 0.12 1.7 °C = 1030 mm Hg (from Steam Tables) k Vapour pressure of solution Vapour pressure of solvent 760 1030 0.78 4.401 ´ 1340 = 536 0. Since the solution boils at 108.211 0.272 106 210 25 25/106 = 0.7 °C at 760 mm Hg. the vapour pressure of solution = 760 mm Hg.8 ´ 0. 40% Toluene and 25% Xylene by weight.388 0.7 °C.401 0. 4.74) = 23.6 An aqueous solution of NaNO3 having 10 g moles of salt/1 kg of water boils at 108. VAPOUR PRESSURE 4. \ xA = 0. \ xA = 0. mm Hg 760 780 915 1060 1225 1405 1577 1765 Heptane (B).21 ´ 1405) = 295.0945 ´ 1577) = 149. The resulting vapour contains . \ xA = 0. mm Hg 295 302 348 426 498 588 675 760 Assuming Raoult’s law is valid. \ yA = 149/760 = 0.2 °C Boiling point of heptane xA (1765 – 760) = (760 – 760).985 yA = At 75 °C xA (915 – 348) = (760 – 348). \ yA = 441/760 = 0.53 pA = (0. pA = yAP. \ xA = 0 = yA 4.8.53 ´ 1060) = 562.880 At 80 °C xA (1060 – 426) = (760 – 426).8/760 = 0. \ xA = 0. 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.73 pA = (0.39 At 95 °C xA (1577 – 675) = (760 – 675). \ yA = 668/760 = 0. p A = xA P A . \ yA = 562/760 = 0.7 79 The following table gives vapour pressure data: Temperature °C 69 70 75 80 85 90 95 99. (xA + xB) = 1 where PA.8 A solution of methanol in water containing 0.196 At 99. (yA + yB) = 1 pA + pB = P.158 mole fraction alcohol boils at 84. p B = xB P B .96 pA = (0. xB = (1 – xA ) (xAPA) + (1 – xA)PB = P xA(PA – PB) = (P – PB) xA 760 + (1 – xA)295 = 760 At 69 °C.21 pA = (0. \ xA = 1 760 =1 760 At 70 °C xA (780 – 302) = (760 – 302).740 At 85 °C xA (1225 – 498) = (760 – 498).1 °C (760 mm Hg).96 ´ 780) = 748.36 pA = (0.0945 pA = (0.58 At 90 °C xA (1405 – 588) = (760 – 588).73 ´ 915) = 668. \ yA = 748. (Boiling point of hexane) xA(760 – 295) = (760 – 295).36 ´ 1225) = 441. \ xA = 0. PB are vapour pressures of hexane (A) and heptane (B) respectively. pB = yBP. \ xA = 0. \ yA = 295/760 = 0.2 Hexane (A). 10 Bottled liquid gas is sold. 240 yA = = 0.553 mole fraction of alcohol.2 and B = 3420. È 530 Ø È 760 Ø (a) Volume of dry exit gas = 1. O2 : 0. pA = xAPA = (0.1 = 2. Vapour pressure at 84. How does the actual composition of the vapour compare with composition calculated from Raoult’s Law? Temperature. According to Raoult’s law. ( p. vapour pressure in mm Hg and T. (b) Partial pressure of O2 in exit (c) Weight of water removed. calculate (a) volume of dry exit gas at 70 °F and 750 mm Hg.2 lb mole.80 PROCESS CALCULATIONS 0.2 – 750 = 125 mm Hg 1.8% Ê 0.1 °C. (b) vapour composition. the equation ln p = A – B/(T – 43) can be used. The composition in mole % in liquid phase is given as follows: n-Butane : 50%.316 760 Ë Actual value  Calculated value Û % Error = Ì Ü ´ 100 Actual value Í Ý È 0. Propane : 45%.553 4.2 ´ 359 ´ É – Ê 492 ÙÚ ÉÊ 750 ÙÚ = 470.9 Methane burns to form CO2 and water.2 (c) Water removed = 2 lb moles = 36 lb (b) Partial pressure of O2 in exit = 4.0 lb mole.316 Ø = É ÙÚ ´ 100 = 42. mm Hg 80 1340 100 2624 To get vapour pressure at 84. temperature in Kelvin A and B are constants) Solving A = 18. °C Vapour pressure. CH4 + 2O2 ® CO2 + 2H2O O2 supplied = 2 ´ 1. If 1 lb mole is burnt with 10% excess pure O2 and the resulting gas mixture is cooled and dried. Basis: 1 lb mole of methane.0.553 . Determine (a) the pressure of the system.2 lb moles Gases leaving after drying: CO2 : 1.26 ft3 0.158 ´ 1520) = 240. and Ethane : 5% .1 ºC = 1520 mm Hg. 0987 Ø Benzene after compression = 36.0987 atm.012 = 5 × 10–4 kmole/h ¦ 760 µ ¦ 323 µ 22.741 kmole Benzene condensed = (4. Molar flow rate of N2 = 0.414 § ¶§ ¶ ¨ 840 · ¨ 273 · .6 Gas n-Butane Propane Ethane Total Mole % Vapour pressure at 30 °C (bar) Partial pressure (bar) 50 3.86 (10.1 °C after compression.67 5 46.5) 19.1ÙÚ = 40. This gas is at 21.6 2.1 °C and 750 mm Hg. Basis: 1000 m3 of original mixture È 1000 Ø È 750 Ø È 273 Ø Moles of original mixture = É – – Ê 22.4.12 N2 from a cylinder is bubbled through acetone at 840 mm Hg and 323 K at the rate of 0.087 = 36.8 ´ 0.8 and ethane: 46.087 – 0. O2 : 4% and N2 : 81%.012 m3/h.89 100 4.0987 ÙÚ = 0.4 1. (other gas is Tie element) È 0. leaves at 760 mm Hg and 308 K at 0.414 ÙÚ ÉÊ 760 ÙÚ ÉÊ 294.087 kmoles Ê 750 ÙÚ Moles of gas other than benzene = 40.8 4.4 ´ 0.11 A solvent recovery system delivers a gas saturated with benzene vapour which analyzes on a benzene free basis as follows: CO : 15%.33 (46. How many kilograms of benzene are condensed by this process per 1000 m3 of original mixture? The vapour pressure of benzene at 21.783 ´ É Ê 5  0.VAPOUR PRESSURE 81 Vapour pressure (in bar) at 30 °C are n-butane: 3. It is compressed to 5 atm and cooled to 21.45) 54.05) 26. The N2 saturated with acetone vapour.6 ´ 0. propane: 10.12 45 10. Find the vapour pressure of acetone at 308 K.87 kmoles È 75 Ø Benzene present originally = 40.023 m3/h.741) ´ 78 = 261 kg 4.21 100 Vapour composition % 8.783 kmoles Vapour pressure of benzene = 75 mm Hg = 75/760 = 0.87 ´ É = 4.87 – 4.70 (3.1 °C is 75 mm Hg. 50 0. x. mole fraction mm Hg × 100 1. x.21 for ethane.05 3.50 0.8 46.7 4.95 kmole.13 Determine the pressure of the system and equilibrium VP at 30 oC.00 Partial pressure Total pressure È 1. VP at 30 °C. in liquid phase mm Hg 0.89 ÙÚ 54.86 25.8 Partial pressure. in liquid phase mm Hg n-butane 0.5263 n-propane 0.4 10. Compound Mole fraction.00 .45 0.67 26.89 100.21 8.789 4.91 74.7 Ø = É × 100 = 19.33 19.09 6.86 2.023 = 9.4737 Total 3.67 for n-propane 26. Y.414 § ¶§ ¶ ¨ 1. When ethane is removed.45 0.09 × 10–4 kmole/h ¦ 1. in liquid phase n-butane n-propane Ethane 0. 9.12 for n-butane Ê 8.0) = 342 mm Hg 4. Compound Mole fraction.09 × 10–4y = 5 × 10–4 Solving.12 54. estimate the pressure and vapour phase composition components of the system at 30 oC. Y.05 Compound n-butane n-propane Ethane Mole fraction. mole fraction of acetone = 0. Assuming all ethane is removed.904 100.45) = (VP) (1.82 PROCESS CALCULATIONS (N2 + CH3COCH3) = 0.55 Thus. Then.50/0. we get y = 0.4 10.013 · ¨ 273 · Let y be the mole fraction of N2 in leaving stream.45 (PT) × (y) = (Partial pressure of acetone) = (VP)acetone (y)acetone (760)(0. mole fraction mm Hg × 100 1. x.6 Total y (Mole fraction) = Partial pressure.013 µ ¦ 308 µ 22.95 = 0. VP at 30 °C. total moles will be 0. 8) Similarly for toluene. Vapour pressure of benzene is 1340 mm Hg at 80 oC Vapour pressure toluene is 560 mm Hg at 80 oC At equilibrium.15 A certain quantity of an organic solvent (molecular weight 125 and density 1.2) PT × (0. xB + xT = 1 4 × (1 – xB) × (560) = 4 × PT × (0. evacuated and the contents reach equilibrium at 30 °C.e.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. P = 240 mm Hg Moles at NTP = 3990 t 240 273 1 t t = 0. partial pressure of benzene = (Mole fraction.05065 g mole 303 760 22414 . of benzene) × (1340) = (Total pressure) × (Mole fraction in vapour phase) = PT (0. xB. (xB) × (1340) = PT × (0. Vapour pressures of benzene and toluene are 1340 mm Hg and 560 mm Hg respectively at 80 oC.8) 4 (1 – xB) 560 = xB (1340) 2240 – 1340xB – 2240xB = 0 Solving. Vapour pressure of solvent at 30 °C is 240 mm Hg.VAPOUR PRESSURE 83 4. It is observed that only 10 ml of liquid solvent is present finally.2) xB × (1340) = PT × (0.8) + PT × (0. Volume of vapour space = 4000 – 10 = 3990 ml T = 303 K .2) = PT We also know that. The flask is closed. xB = 0.6 4.505 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. (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. xT × (560) = PT × (0.8) i. Total pressure = 760 mm Hg Let x and y be the liquid and vapour phase compositions respectively.505 g/cc = 15.3 respectively.331 = 21.331/21.2961 4.05065 × 125 = 6. by Raoult’s law.05 + 6.381) = 0. xB = 0. yT = PT xT 380 t 0.173 PTot 760 .16 Benzene and toluene form an ideal solution.3)(20) = 49 + 6 = 55 mm Hg Hence.3)(20)/55 = 0. 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. yA = (0.345  = 0.345 yB = PB x B 960 t 0.0 °C.05 g Total weight = 15.655 xT = 0. 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. Suffix B and T denote benzene and toluene respectively Then.331 g Mass of liquid left behind = 10 ml × 1.84 PROCESS CALCULATIONS Molecular weight = 125 Weight = 0. calculate their vapour phase composition. PTot = (xB) (PB) + (xT) (PT) 760 = (xB) (960) + (1 – xB) 380 Solving.11 4.17 Vapour pressure of pure benzene and toluene are 960 mm Hg and 380 mm Hg respectively at 86.89 yB = (0.381 g Mole fraction of vapour = (6.7)(70)/55 = 0.7 and 0.655  = 0.7)(70) + (0.827 PTot 760 In the same way. Calculate the liquid phase and vapour phase composition at that temperature. 6 mole toluene is vaporized. Assuming the vaporization efficiency to be 75%.5 314 345 378 414 452 494 538 585 635 689 747 760 The following data gives the vapour pressure data for a binary system.26 K. Estimate the mole fraction of benzene in vapour phase. The vapour pressure of ethyl acetate is 200 mm Hg abs. . Compute the VLE vapour–liquid equilibrium data at a total pressure of 760 mm Hg.VAPOUR PRESSURE 85 4.3 The following data gives the vapour pressure (VP) for benzene (A)– toluene (B) system.0 °C. estimate the weight of nitrobenzene per kg of steam in distillate.2 Prepare a Cox chart for ethyl acetate. and compare with the experimental value. 4.333)(0. ln (p*) = 15. 760 811 882 957 1037 1123 1214 1310 1412 1530 1625 1756 — mm Hg VPB. Estimate the weight of nitro benzene associated per kg of steam in distillate. Distillation takes place at 99 °C. Compute the vapour-liquid equilibrium data at a total pressure of 760 mm Hg.1 The vapour pressure of benzene can be calculated from the Antoine equation.18 Vapour pressure of benzene is 3 atm and that of toluene is 1.2/2.9008 – [2788.4 mole benzene and 0. VPA. A liquid fuel containing 0.1 °C). — mm Hg 4. Partial pressure of benzene = (3)(0. at 42 °C and 5. By using the chart estimate the boiling point of ethyl acetate at 760 mm Hg and compare with the experimental value (77. Determine the latent heat of vaporization of benzene at its normal boiling point of 353.0 atm ybenzene = 1.6 EXERCISES 4. mm Hg 433 498 588 698 760 It is desired to purify nitrobenzene by steam distillation under a pressure of 760 mm Hg.51/(–52.36 + T)] where p* is in mm Hg and T is in K.4) = 1.2 Partial pressure of toluene = (1. Vapour pressure of A.0 atm at 126.6) = 0.8 Total pressure = 2.333 atm.4 4. 4. mm Hg 760 860 1002 1160 1262 Vapour pressure of B. Vapour pressure of nitrobenzene is 20 mm Hg.0 =0. i-C4H10 : 2250. n-C4H10 : 47% and C5H12 : 4%.15 molal solution in water. n-C3H8 : 6525. n-C4H10 : 1560 and C5H12 : 430.245 kg of CH3CHO in 20 kg of water is 190 mm Hg at 93.8 Calculate the total pressure and the composition of the vapours and liquid in molar quantities at 100 °C for a solution containing 45% benzene. 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. Express the composition by (a) volume %. 4.10 Ethyl acetate at 30 °C exerts a vapour pressure of 110 mm Hg. 4. 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. Toluene (C7H8) (molecular weight: 92) : 560 mm Hg. xylene (C7H8) (molecular weight: 106) : 210 mm Hg. 40% toluene and 15% xylene by weight. 4. (a) Calculate the total pressure in cylinder at 21 °C and the composition of the vapour evolved. Calculate the partial pressure of CH3CHO over 0.6 Methyl alcohol and ethyl alcohol at 100 °C have vapour pressures of 2710 mm Hg and 1635 mm Hg respectively. It is stored in cylinders for sale. (b) Find the total pressure at 21 °C if all C2H6 were removed.7 A liquefied fuel has the following analysis: C2H6 : 2%.9 The partial pressure of actetaldehyde in a solution containing 0. n-C3H8 : 40%. the vapour pressures are benzene (C6H6) (molecular weight: 78) : 1340 mm Hg.5 °C. and (b) weight %. i-C4H10 : 7%. At 100 °C. Vapour pressure data (mm Hg): C2H6 : 28500.86 PROCESS CALCULATIONS 4. 4. . under equilibrium condition.2) . nv„ ËM Û [ nv ] – Ì A Ü Í MB Ý Ë pv Û Ë M A Û Ì Ü–Ì Ü Í p  pv Ý Í M B Ý 87 mass of A mass of B (5.1) When the quantities are expressed in mass. is called the wet bulb temperature.5 Psychrometry 5. It is defined as the moles of vapour carried by unit mole of vapour free gas.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. Absolute humidity (nv): The substance that is transferred (vapour) is designated by A and the main gas phase is designated by B. Wet bulb temperature (WBT): The temperature measured by a thermometer or thermocouple with a wet wick covering the bulb. The matter transferred between phases in both the cases is the substance constituting the liquid phase.2 DEFINITIONS Dry bulb temperature (DBT): The temperature measured by a bare thermometer or thermocouple is called the dry bulb temperature. 5. which either vaporizes or condenses indicating respectively the humidification process or the dehumidification process. then it is called mass absolute humidity (n¢v ) or Grosvenor humidity. [ nv ] Ë yA Û Ì Ü Í yB Ý pA pB Ë pv Û Ì Ü Í p  pv Ý moles of A moles of B (5. 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. 3) 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.L (tDP – t0)] where (5.6) will reduce to Eq. Enthalpy: The enthalpy H¢ of a vapour-gas mixture is the sum of the enthalpies of the gas and the vapour content. The above Eq. If pv is the partial pressure under a given condition and ps is the vapour pressure at DBT of the mixture.4) 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.7) when tDP itself is taken as the reference temperature. The expression for humid volume in m3/kg is .L = Specific heat of component ‘A’ (vapour) in liquid phase. (5.5) where CA and CB are specific heats of vapour and gas respectively.7) Humid volume: The humid volume vH of a vapour-gas mixture is the volume of unit mass of dry gas and its accompanying vapour at the prevailing temperature and pressure. (5. The moment the temperature is reduced below the dew point. 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. with a humidity of n¢v . CA. the vapour will condense as a liquid dew. For a gas at a DBT of tG. the enthalpy relative to the reference state ‘t0’ is.6) lDP = Latent heat of vaporization at dew point. CS = CAn¢v + CB (5. = CB(tG – t0) + n¢v [CA(tG – tDP) + lDP + CA. then Relative saturation = yr % = pv ¥ 100 ps (5.88 PROCESS CALCULATIONS Relative humidity or relative saturation ( yr%): It is normally expressed in percentage. Percentage saturation = yP % = nv ¥ 100 ns (5. t0 H¢ = CB (tG – t0) + n¢v [CA(tG – t0) + l 0] = CS (tG – t0) + n¢v l 0 (5. H¢ = enthalpy of gas + enthalpy of vapour component. 8) where. 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. the equations given can be used to determine the properties. 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 mixture is partially saturated.013 + ¥ ¥ ¥ ¥ vH = Í Á Í ˙ ˙ Í 273 ˙ ˜ Á ˜ p ˙˚ Î ˚ ÍÎ ÎË M B ¯ Ë M A ¯ ˚ È Ê 1 ˆ Ê nv¢ ˆ ˘ È tG¢ + 273 ˘ = (8315) Í Á ˙ ˜ +Á ˜˙ ¥ Í p ˚ ÎË MB ¯ Ë M A ¯ ˚ Î (5. 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.414 1.PSYCHROMETRY 89 È Ê 1 ˆ Ê nv¢ ˆ ˘ 105 ˘ È tG¢ + 273 ˘ È 22. 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. 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. (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. p = total pressure N/m2 A typical psychrometric chart is shown at the end of the chapter. p Relative saturation % = yr = v ¥ 100 ps . 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. 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. Alternatively. Various properties of air-water system can be obtained from the chart. the gas is said to be ‘saturated’ with the vapour. nv = moles of vapour per mole of vapour free gas actually present ns = moles of vapour per mole of vapour free gas at saturation.0483 = pA p .pv ˜¯ and Ê ps ˆ ns = Á Ë p .0133 ¥ 105 N/m2 Molecular weight of air = 28. If p is the total pressure.672 ¥ 103 N/m2 where.pA Ê 105 ˆ 20. then from Dalton’s law.a = 0.pv ˆ yp = (yr) Á Ë p .03 kg w. (ii) % humidity (chart) = 35% (iii) % relative saturation = partial pressure/vapour pressure Partial pressure under the given condition is given by molal humidity (0. pA = Partial pressure of water p = Total pressure .69 = Á1.0133 ¥ –1 p A ¯˜ Ë pA = 4.pv ˜¯ WORKED EXAMPLES 5.84 (i) n¢v (chart) = 0. Ë p .1 An air (B)-water (A) sample has a dry bulb temperature of 50 °C and a wet bulb temperature of 35 °C. Ê pv ˆ nv = Á . 1 atm = 1.pA pA 1.v/kg.ps ˆ = ns ÁË ps ˜¯ ÁË p .0483) = 0.d.pv ˜¯ Ê p .0133 ¥ 105 .90 PROCESS CALCULATIONS where.a. pv = Partial pressure of vapour ps = Vapour pressure of pure liquid Percentage saturation = yp = nv ¥ 100 ns where. Estimate its properties at a total pressure of 1 atm.04833 kmole of w.ps ˜¯ nv Ê pv ˆ Ê p .v/kmole of d. 84 = 0. 5.332 31.969 m3 mixture/kg of dry air Alternatively. 5.PSYCHROMETRY 91 Vapour pressure of water (Steam Tables) at 50 °C = 92. (b) Specific volume of saturated air = 1.16 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)(0.91 + (1.03) 1. = 4. 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.2 Ether at a temperature of 20 °C exerts a vapour pressure of 442 mm Hg.055 – 0.03) (2502) = 128.062 (50 – 0) + (0.35) = 128.4 kJ/kg (vii) Humid volume vH Ë È 1 Ø È nv„ Ø Û Ë tG„  273 Û (a) vH = (8315) Ì É Ü Ù É ÙÜ – Ì Ý Í Ê M B Ú Ê M A Ú Ý Í pt Û Ë È 1 Ø È nv„ Ø Û Ë (325) = (8315) Ì É  É ÙÜ – Ì Ü Ù 5 Ê Ú Ê 18 Ú Ý Í 1. (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 .35) = 0.005 + 1.062 kJ/kg of dry air °C % R.672 ´ (iv) Dew point = (v) Humid heat = (Eq.H.7) (a) H¢ = CS (tG – t0) + n¢v l0 l0 = 2502 kJ/kg H¢ = 1.91 m3/kg of dry air By interpolation vH = 0.055 m3/kg of dry air Specific volume of dry air = 0.884 (0.0133 – 10 Ý Í 28.5°C CS = CB + CA n¢v 1.91)(0.332 ´ 103 N/m2 \ 103 ´ 103 = 37.88% 12.961 m3/kg of dry air 5.5) = = (vi) Enthalpy (For a reference temperature of 0 °C) (Eq.51 mm Hg = 12. 407 ´ 28) = 11.4 Nitrogen 28 0.9% 0.593 ´ 74) = 43.9 = 3.8 ÙÚ 0.248 Ø = 0.3 Vapour pressure of acetone at 20 °C = 184.407 ft3 Percentage of ethyl ether = (b) Basis: 1 lb mole of mixture Molecular weight lb mole Weight.148 ´ 745) = 110 mm Hg È 110 Ø ´ 100 = 59.8 mm Hg Partial pressure of acetone = (0.174 nv = ÈÉ Ê 0.457 0.7% = 0.593 (0.8 Ø Û Ì ÉÊ 745 ÙÚ Ü Ü ns = Ì Ì È 560.000 55.852 ÙÚ Ë È 184.4 20.593 ft3 = 59.3 100.9 79.593 = 1.2 Ø Ü Ì ÉÊ 745 ÙÚ Ü Í Ý yp = nv ns È 0.407 (0.407 A mixture of acetone vapour and nitrogen contains 14.4 \ 0.0 Component Total Weight % 760 Ø È 293 Ø (c) Volume of gas mixture = 1 ´ 359 ´ ÈÉ = 393 ft3 – Ê 745 ÙÚ ÉÊ 273 ÙÚ È 43.7% yr = relative saturation = É Ê 184.329 ÊÉ 0. Calculate yr and yp at 20 °C and 745 mm Hg (e) lb mole vapour/lb moles vapour free gas = 5. 745 Nitrogen = 40.148 Ø = 0. lb Ether 74 0.8% acetone by volume.92 PROCESS CALCULATIONS (a) Basis: 1ft3 of gas mixture 442 – 1 = 0.174 = 52.3 Volume %.6 1.329 Always yr > yp .112 Ê 393 ÙÚ 43.752 ÚÙ 0.9 Ø lb of vapour/ft3 = É = 0.85 (d) lb of vapour/lb of vapour free gas = 11. 1% benzene by volume.032 ´ 0. 5. 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 8.PSYCHROMETRY 5.5 ´ ns = 0.1 grains of water vapour per ft3 at 30 °C.4 = 130 mm Hg 0.52 ´ É Ê 1. Basis: 1000 ft3 of entering gas 70 °F and 14. .00 ÙÚ 19.013 ÙÚ Amount of water removed = 0. Vapour pressure of water at 70 °F = 0. Ë 0.013 mole of water/mole of dry gas È 0.518 lb.9 = 0.3  0.5 A stream of gas at 70 °F and 14.032 lb mole Amount of water entering = 2.5 ´ Ì Ü Í 14.0288 lb mole Weight of water removed = (0.36 Ý = 0.0256 Ø Partial pressure of water vapour = 760 ´ É = 19.36 Û nv = 0.15 This corresponds to a temperature of 57 °C.3 psi and 50% saturated with water vapour is passed through a drying tower. where 90% of water vapour is removed. (a) Calculate the dew point of the mixture when at a temperature of 25 °C and 750 mm Hg.3 Ø È 492 Ø lb mole of gas mixture = É – – Ê 359 ÙÚ ÉÊ 14.0288 ´ 18) = 0.1 = 1.6 A mixture of benzene vapour and air contains 10.0256 ft3 Ê 273 ÙÚ È 0. Vapour pressure of water vapour = 5.4 93 Moist air is found to contain 8.16 ´ 10–3 lb º 6.4 mm Hg Ê 1.52 lb moles n yp = 0.013 Ø = 0.42 ´ 10–5 lb moles 7000 Pure component volume of water at 30 °C Amount of water = È 303 Ø = 6.5 = v ns where nv = lb mole of water/lb mole of vapour free gas.36 psi.67 ÙÚ ÉÊ 530 ÙÚ = 2. Calculate the pound of water removed per 1000 ft3 of entering gas.3 psi È 1000 Ø È 14.42 ´ 10–5 ´ 359 ´ É = 0. 75. (b) Partial pressure of benzene remains same.7 °C. (c) Calculate the dew point when the mixture is at a temperature of 30 °C and 700 mm Hg. (b) Calculate the moles of acetone evaporated per mole of the vapour free gas passing through the evaporator. The nitrogen enters the evaporator at 30 °C containing acetone such that its dew point is 10 °C.7 mm Hg. It leaves at a temperature of 25 °C with a dew point of 20 °C.328 565 .94 PROCESS CALCULATIONS (b) Calculate the dew point when the mixture is at a temperature of 30 °C and 750 mm Hg. (c) Calculate the weight of acetone evaporated per 1000 ft3 of gases entering. i. 5. Data: Vapour pressure of acetone: At 10 °C = 116 mm Hg and at 20 °C = 185 mm Hg. The atmospheric pressure is 750 mm Hg. it is found that this pressure corresponds to a temperature of 20 °C.e. it is seen that the dew point does not depend on the temperature but does vary with total pressure.e. (a) Calculate the vapour concentration of the gases entering and leaving the evaporator expressed in moles of vapour/mole of vapour free gas.101 ´ 750 = 75.101 ´ 700 = 70.7 It is proposed to recover acetone which is used as a solvent in an extraction process by evaporation into a stream of nitrogen. 20 °C (only temperature of gas mixture has changed. i.7 mm Hg.) (c) Partial pressure of benzene = 0. Basis: 1 mole of mixture. From these results. (a) Entering gases: partial pressure of acetone = 116 mm Hg partial pressure of N2 = 634 mm Hg 116 mole of acetone/mole of N2 = = 0. the dew point. (a) Partial pressure of benzene = 0. Dew point for this partial pressure is 18. (d) Calculate the volume of gases leaving the evaporator per 1000 ft3 of gases entering.7 mm Hg From the vapour pressure data for benzene.183 634 Leaving gases: partial pressure of acetone = 185 mm Hg partial pressure N2 = 750 – 185 = 565 mm Hg 185 mole of acetone/mole of N2 = = 0. Hence dew point also remains same. 328 lb moles Acetone evaporated per 1000 ft3 = Ê 760 ˆ Ê 303 ˆ Volume of gases leaving = 1.4 ¥ 1000 = 17.145 (c) Total moles entering = (1 + 0. .) Initial humidity = (b) Final partial pressure of water = 9. Data: Vapour pressure of H2O at 20 °C =17.5 mm Hg and at 10 °C = 9.328 – 0.8 526 ¥ 1000 = 1102 ft3 477 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.183 lb moles Ê 760 ˆ Ê 303 ˆ ¥ Volume of entering gas = 1. (b) Calculate the humidity of this air if its temperature is reduced to 10 °C and its pressure increased to 35 psi.8) = 14 mm Hg 14 750 .5 ¥ 0.2 È ˘ Final humidity = Í ˙ Î 1810 .a. 20 °C and 750 mm Hg.PSYCHROMETRY 95 Basis: for (b). condensing out some of the water.7 9.183) = 0. (b) Moles of acetone evaporated = (0.2 ˚ = 0.183 ¥ 359 ¥ Á = 477 ft3 Ë 750 ˜¯ ÁË 273 ˜¯ Weight of acetone evaporated = (0. (c) and (d): 1 lb mole of N2 passing through evaporator.328 ¥ 359 ¥ Á ¥ Ë 750 ˜¯ ÁË 273 ˜¯ = 526 ft3 Volume of gases leaving per 1000 ft3 = 5.328) = 1.6 lb 477 (d) Total moles of gas leaving = (1 + 0.v.9.4 lb 8.2 mm Hg (saturated) Final total pressure = 35 ¥ 760 = 1810 mm Hg 14.019 kmole of water vapour (w.)/kmole of dry air (d.0051 kmole of water vapour/kmole of dry air 3 Basis: 1000 ft of wet air. (c) Calculate the weight of water condensed from 1000 ft3 of gas.145 ¥ 58) = 8.14 = 0. (a) Initial partial pressure of water = (17.183) = 1.2 mm Hg. (d) Calculate the final volume of wet air leaving. 1 = 2110 mm Hg 0. 5.019 ´ 2. Vapour pressure at 5 °C = 89.10 Air at a temperature of 20 oC and 750 mm Hg has a relative humidity of 80%.1 = 0. (d) Total moles finally present = (2. Calculate.1 mm Hg = 89.035 lb moles = 0.0422 = 40. \ Final pressure of gas mixture = (i) The molal humidity of the air . Calculate the final pressure in psi for acetone.0477 lb mole lb mole of water finally = (0.96 PROCESS CALCULATIONS (c) Volume at standard condition and hence weight of mixture in lb mole È 1000 Ø È 750 Ø È 273 Ø ÉÊ Ù –É Ù –É Ù = 2.8 psi.0306 = 0.7246 lb mole lb mole of acetone in final gas/lb mole of final gas mixture 0.694 + 0. It is proposed to condense 90% of the acetone by cooling to 5 °C and compressing.2/750 = 0.56 lb moles 359 Ú Ê 760 Ú Ê 293 Ú È 736 Ø lb mole of dry air = 2.0127 lb mole \ water condensed = 0.306) = 0.2 mm Hg.0422 0.2 mm Hg.63 lb.0051 ´ 2.51 + 0.1 mm Hg.0306 lb mole Final gas = 0.56 ´ É = 2.5227 lb moles È 760 Ø È 283 Ø – = 395 ft3 Volume of wet air = 2.5227 ´ 359 ´ É Ê 1810 ÙÚ ÉÊ 273 ÙÚ 5. Basis: 1 lb mole of mixture of gases.51) = 0. (equal to vapour pressure as it is at dew point) lb moles of acetone in it = 229.7246 Final pressure of acetone = 89.51) = 0.9 A mixture of dry flue gases and acetone at a pressure of 750 mm Hg has a dew point of 25 °C.306 lb mole Dry flue gases = (1 – 0.306 ´ 0. Partial pressure of acetone = 229.0306 = 0.0127) lb moles = 2. Vapour pressure at 25 °C = 229.51 lb moles Ê 750 ÙÚ lb mole of water vapour initially = (0.694 lb mole Acetone finally present = 0. 25 ¥ 18 = 10.11 A material is to be dried from 16% moisture by weight (wet basis) to 0.43 g 5.09 kg of water/kg of dry air. .9.019 g mole of water vapour/g mole of dry air (ii) Molal humidity at 10 °C. ps = pv = 9. Vapour pressure of water at 20 °C = 17. The air enters at 301 K and at atmospheric pressure.5 mm Hg Vapour pressure of water at 10 °C = 9.8 = v = v ps 17. 2000 mm Hg (when saturated.019 ˜¯ \ water condensed = (difference in humidity) ¥ (flow rate of dry air) ¥ (molecular weight of water) = (0.pv 750 . (i) Molal humidity at 20 °C. The exit humidity of air is 0.414 ˜¯ ÁË 760 ˜¯ ÁË 293 ˜¯ = 41.5% by circulation of hot air.02 Ê 1 ˆ = 40. vapour pressure = partial pressure.PSYCHROMETRY 97 (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.2 ˙˚ = 4.2 mm Hg At 20 °C and 750 mm Hg. Find the volume of fresh air required if 1000 kg/h of dried material is to be produced. p p yr = 0.6 ¥ 10–3) ¥ 40. partial pressure of H2O vapour = pv = 14 mm Hg.02 ¥ Á Ë 1.02 kg of water/kg of dry air. The fresh air contains 0.6 ¥ 10–3 g mole of water vapour/g mole of dry air (iii) Basis: 1000 litres of original wet air Ê 1000 ˆ Ê 750 ˆ Ê 273 ˆ g mole of wet air = Á ¥ ¥ Ë 22.5 where. we have.2 mm Hg) È ˘ ps 9. 750 mm Hg = pv 14 = p .25 g mole of dry air = 41.2 =Í p .ps Î 2000 .14 = 0.019 – 4. 5% Air 70% Drier 1000 lb/h.07 kg H2O \ Weight of dry air needed = 184.17 m3 Volume of wet air = 94.995 . Dry solid = 995 kg Moisture = 5 kg \ feed = 995 = 1184. Air enters at 760 mm Hg. Air is passed counter current to the flow of paint.5 – 1000 = 184.02) = 0.33 kmoles Water in entering air = \ 301 Ø = 2331.5 kg Dry solid is the Tie Element Water removed = 1184.07 2636 Ø Dry air needed = ÈÉ = 91.98 PROCESS CALCULATIONS 301 K Air 0.84 ÙÚ 2636 – 0.84 Water in feed = 189.40 kmoles Ê 28.02 = 2.5 kg 1 kg of dry air removes = (0. What flow rate of air is needed? Air 10% Dry product 0.33 ´ 22.5%. 140 °F and 10% humidity while it leaves at 750 mm Hg 95 °F and 70% humidity.5% Basis: 1000 kg/h product. 1000 lb/h of feed having 10% water is to be dried to 0. 10% moisture Basis: 1000 lb/h of feed Dry material = 900 lb Moisture = 100 lb 900 Product = = 904.12 A tunnel drier is used to dry an organic paint.414 ´ ÈÉ Ê 273 ÙÚ 5.09 – 0.02 16% wet solid 0.93 kmoles 18 Total moles of wet air = 94.5 = 2636 kg/h 0.5 lb 0.09 Drier Dried solid 0.5 kg 0. 9325 ´ 0.8754 inches of Hg. (a) The pounds of water evaporated per pound of dry air.6128 inches of Hg (lb mole H2O/lb mole dry air) in entering air = 0.77 lb mole/h È 600 Ø Volumetric flow rate of air = (543.8754 ´ 0.3865 (lb mole H2O/lb mole dry air) in outgoing air = 0. Atmospheric pressure is 29.2 = 0.13 Air at 100 oF and 20% relative humidity is forced through a cooling tower.01 lb mole H2O = 0. 5.7 = 0.5) = 95.3865 = 0.18 ÙÚ \ Total air entering = 530.3865 inches of Hg Partial pressure of H2O in exit air = 0.065 ft3/h.618 lb mole H2O evaporated/lb mole of dry air = 0.5 Ø = 530.48  0.447 lb moles – Ê 359 ÙÚ ÉÊ 560 ÙÚ . (a) Partial pressure of H2O in entering air = 1. (b) Volume of air leaving per 1000 ft3 of entering air.00795 ´ É = 5 ´ 10–3 lb = 0.38.5 lb/h From humidity chart the following details are obtained: Humidity of entering air = 0.005 lb Ê 28.035 lb mole water vapour/lb mole dry air 1 lb mole of dry air removes 0. Calculate the following.5 lb mole/h Dry air needed = É Ê 0.6128 = 0.01328 29.00795 È 18 Ø lb H2O/lb of dry air = 0.5 ´ 1.PSYCHROMETRY 99 Water removed = (1000 – 904.025 lb mole water vapour/lb mole dry air Humidity of leaving air 0.48 inches of Hg.18 lb È 95.9325 inches of Hg and at 85°F = 0.025 = 543. Vapour pressure of water at 100 °F = 1.77 ´ 359) ´ É Ê 492 ÙÚ = 2.48  0.02123 29. The air leaves at 85 °F and 70% relative humidity.84 ÙÚ È 1000 Ø È 43.2 Ø (b) 1000 ft3 of entering air = É = 2. 00527 ÙÚ Weight of water removed/h = 1989.3 750  6.00399 lb of water vapour/lb of dry air È 2000 Ø Dry air entering per hour = É = 1989. Basis: One lb mole of dry air Water in entering air = 6.447 lb moles of wet air entering = 2.52 lb Ê 1.00926 lb of water vapour/lb of dry air \ water vapour removed = 0. Recovery of the acetone is accomplished by evaporation into a stream of N2 followed by cooling and compression of the final gas mixture.00926 – 0.00847 lb mole of water vapour/lb mole of dry air = 0.466 ´ 359 ´ É Ê 492 ÙÚ 5.8 ft3 Volume of air leaving = 2.466 lb moles of wet air leaves 1.01328 lb moles of wet air entering.02123 lb moles of wet air leaves \ for 2. how many lb of water is removed per hour. The nitrogen leaves at 85 °F. In the solvent recovery unit 50 lb of acetone are removed per hour. 5. 1.00527 = 0.00527 lb of water vapour/lb of dry air 11 Water in leaving air = 750  11 = 0. If 2000 lb of wet air is forced through the drier per hour.52 ´ 0. 740 mm Hg and 85% saturation.447 ´ 1.00399 = 7. Total pressure 750 mm Hg. The dew point of entering air is 40 °F while that of leaving air is 55 °F.01328 È 535 Ø = 962.02123 = 2.100 PROCESS CALCULATIONS For 1.938 lb.01488 lb mole of water vapour/lb mole of dry air = 0.15 Acetone is used as a solvent in a certain process.14 Leather containing 100% of its own weight of water is dried by means of air.3 mm Hg and at 55 °F = 11 mm Hg. . The N2 is admitted at 100 °F and 750 mm Hg partial pressure of acetone in incoming nitrogen is 10 mm Hg. Vapour pressure of water at 40 °F = 6.3 = 0. 664 ´ 359 ´ É Ê 492 ÙÚ ÉÊ 750 ÙÚ = 689.525 lb mole of acetone 0.525 Total moles of gas entering = 1.642 lb mole 0. Acetone 100 °F 750 mm Hg partial pressure of acetone 10 mm Hg Drier N2.33 Ø lb mole of acetone/lb mole of N2 in entering stream = ÈÉ Ê 98.67 ÙÚ = 0.0135 Ë Û pv yp = 0.67 lb moles Í 750 Ý Acetone entering = 100 – 98.PSYCHROMETRY 101 (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? N2.67 = 1.5385 lb mole of acetone/lb mole of N2 in leaving stream = \ 1 lb mole of N2 removes: (0.642 ´ 1.33 lb moles 1.0135) = 0.862 lb mole 58 Ë 750  10 Û N2 in entering mixture = Ì Ü ´ 100 = 98. 740 mm Hg 85% saturation (Molecular weight of Acetone (CH3)2CO is 58.0 ft3 .664 lb moles lb mole N2 needed/h = È 560 Ø È 760 Ø – (a) Volume of gases admitted = 1. Acetone 85°F. Vapour pressure of acetone at 85 °F = 287 mm Hg) Basis: 100 lb moles of gas mixture entering Acetone removed = 50 = 0.0135 = 1.5385 – 0.85 = Ì Ü Í 740  pv Ý Ë 287 Û Ì 740  287 Ü Í Ý where yp (percentage saturation) = 85% pv (partial pressure) = 258 mm Hg 258 740  258 = 0.862 = 1. /kg of d.052 kmole Water removed: 0. 28 Final humidity = 0.16 By adsorption in silica gel you are able to remove all the weight (0.v.17 At 297 K and 1 bar.217 kg of w.93 kg = ÊÁ Ë 18 ˜¯ Total wet air = 44.217 = 0.32 × 10 3 3 × 78 = 0.052 = 44.642 ¥ 1.895 kmole of the dry air: Á 1000 ¥ 273 ¥ Ë ¯ 22. yr = Á ¥ 100 = 5. yr = pv Ê 0.114 ˆ .895 + 0.6 ˆ Ê pv = Partial pressure of water = Á 0. What was the relative humidity of the air? Basis: 1000 m3 of dry air at the given conditions. 0.3 × 10 − 7.2 × 60 × 10 3 = 7. 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.947 ¯ ps = Vapour pressure of water = 15 mm Hg = 2 kPa Q Relative humidity.414 0. the total pressure = 7.052 ¥ ˜ = 0. The same air measures 1000 m3 at 20°C and 108 kPa when dry.642 × 105 N/m2 . PT.6 kPa.0217 kg of w.v.93 ˆ = 0. Ë 2 ˜¯ ps 5.0217 = 6 × 10 3 PT − 6 × 10 3 × 78 28 Therefore. Relative humidity = 0.114 kPa Ë 44.32 × 103 N/m2 100 7.526 ¥ 359 ¥ Á ¥ Ë 492 ˜¯ ÁË 740 ˜¯ = 1031.10 × 0.a. the mixture of N2 and C6H6 has a relative humidity of 60%.a./kg of d.6 ft3 5.526 lb moles Ê 545 ˆ Ê 760 ˆ Volume of exit gases = 2.6 = Partial pressure/Vapour pressure Partial pressure = Humidity = 12.102 PROCESS CALCULATIONS (b) Total moles of exit gas = 1.7%.32 × 10 3 101.5385 = 2. It is desired to recover 90% C6H6 present by cooling to 283 K and compressing to a suitable pressure.947 kmoles 98. 108 Ê ˆ ¥ 293 ¥ 100˜ = 44.93 kg) of water from moist air at 15 °C and 98. 95 Weight of water removed = 0.PSYCHROMETRY 103 5.15 kg/s containing 60% moisture is dried in a countercurrent dryer to give a fixed product containing 5% moisture content./kmole d.09 kg/s Weight of dried wood = 0. 1.4 × 103 N/m2 Water in incoming wood = 0.99 kmoles/h Dry air rate = 38 – 3. absolute humidity.0868 kg/s Weight of final product = .36  86.15 × 0.7/825.01 kmoles/h 86. The air used is at 100 °C and is having water vapour at a partial pressure of 103 N/m2. percentage humidity and dew point. 1. The vapour pressure of water at 40 °C = 7.1 bar 1000 t 825 273 1 t t  38 kmoles/h 348 760 22.18 Humid air at 75 °C.105 = 3.a.19 A piece of wood entering furnace at 0. Humidity = 825.1 × 105 N/m2 1 atm = 760 mm Hg = 1.7 mm Hg Total pressure = 1.7% 289 /(825  289) Dew point = 49 °C 5.15 – 0.15 × 0.1 bar = 825.84 Percentage humidity = 0. The air leaves the dryer at 70% RH and at 40 °C.4 = 0.a.06 kg/s Weight of water in dried wood is 5% 0.1 bar = 1. Vapor pressure at 75 °C = 289 mm Hg Partial pressure of water vapour = Vapour pressure of water × Relative humidity = 86.36) = 0.105 Volumetric flow rate = 1000 m3/h at 75 °C and 1.99 = 34. Determine the molar flow rates of water and dry air entering the process.7 = 0.0632 = 0.06 = 0.v.36 mm Hg Mole fraction of water vapour = (86.1 bar and 30% relative humidity is fed at a rate of 1000 m3/h.0632 kg/s 0.414 Water flow rate = 38 × 0.117 × 100 = 21. 28./kg d.7 18 Mass humidity = 0.013 × 105 N/m2 Therefore. molar humidity.v.6 = 0.117 kmole w.117 × = 0.073 kg w. Estimate the air required to remove the moisture. 3 – 103  5.17) × (1.17 kg/s 0.7 = 0.02973 740 Mole fraction of water vapour in original air when cooled to 15 °C Mole fraction of water vapour in original air (30 °C) = 12.00662 = 0.02738 kg w. mole fraction of dry air = 0./kg d.63 kmoles 0.14 kmoles Í 22.0868 = = 3.v.414 – 740 – 273 Ý 22 = 0.976 = 38.1897 kg/s Dry air needed = Weight of wet air leaving = (3.v.983 .0336 – 0. Estimate the amount of air at new condition Moles of humidified air at initial condition Ë 1000 – 760 – 303 Û = Ì Ü = 39.983 = Moles of wet air at 15 °C = 37. water gets partially condensed.v.00622) = 3.17 × (1. During this process.00623 kg w.97 × 10–3 × Humidity of leaving air.18 × 103 N/m2 Humidity = 5.02738 Weight of wet air needed = 3.18 × 103 101.0336 kg w.277 kg/s 5.104 PROCESS CALCULATIONS Humidity of entering air = 103 1.20 Air at 740 mm Hg at 30 °C contains water vapour at a partial pressure of 22 mm Hg.a. = 9. Weight of water in kg to be removed Water removed per kg of dry air 0.013 – 10 5  103 18 28.017 740 Therefore.18 – 103 = 0.a.4 × 103 Partial pressure = 5./kg d.84 = 0./kg d.a. Increase in humidity = 0. Relative humidity = Partial pressure/Vapour pressure 0.7 mm Hg.0336) = 3.7 = Partial pressure/7. 1000 m3 of the air is cooled to 15 °C and the partial pressure of water vapour is brought to 12. The column has 2.01 × 0.1 kg Water adsorbed/h = 2.7 h 3600 5.1 = 0.PSYCHROMETRY 105 Volume of dry air at new condition (assuming ideal behaviour) 760 Û Ë 288 Û Ë 3 = Ì38.01 kg of water vapour per kg of dry air is passed through a bed of silica gel to get air containing 0.14) × (0. Calculate the flow rate of air both at inlet and outlet.004 kg of water vapour per kg of dry air is bubbled through water at a rate of 0. = Water adsorbed by silica gel in 5 h = 2.004 = 0.63 × 0.017) = 0. Weight of water evaporated = 0. The air leaves with a humidity of 0.1 kg of silica gel initially and after 5 h of operation. Estimate the time needed to vaporize 0.126 kg 5. the weight of silica gel is found to be 2.014 kg of water vapour per kg of dry air.1 = 0.1 m3 of water.01 kg Water removed per second by using 0.11 m /kg dry air 273 Í Ý Í Ý Water removed is (39.02973) – (38.001 kg of water vapour per kg of dry air for a specific application.02 .01 – 0.1 = 0.1 kg/s.1 kg of air = 0.2  2.22 Air containing 0.001 100000 = 27.414 – Ü – Ì 740 Ü = 938.001) + 0.1 × 1000 = 100 kg (since the density is 1000 kg/m3) Weight of water removed/kg of dry air = 0.02 kg/h 5 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.014 – 0. time needed to evaporate 100 kg of water = 100 = 100000 s 0. (G) (0.001) = 0.001 kg/s Hence.63 – 22.02 G (0.2 kg.507 kmole = 0.21 Air with humidity of 0.2 – 2.547 × 18 kg = 9.01) = (G) (0. 1 Toluene–air mixture is available such that the partial pressure of the vapour is 1. Find the volume of fresh air required if 1000 kg/h of dried material is to be produced.5 A mixture of benzene–air is available at 750 mm Hg and 60 °C. the flow rate of dry air in kg/h = 2.222 kg/h 0.222 (1 + 0. The fresh air contains 0. 5.0% by circulation of hot air. 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. humid heat and enthalpy.3 A material is to be dried from 20% moisture by weight (wet basis) to 1.33 kPa.001) = 2. 5. Air is available at a DBT of 35 °C and 27 °C.7 Air at 50% relative humidity (RH) is to be used for a specific operation. 5. % saturation and humid volume using humidity chart. 5.02 kg of water/kg of dry air.e.2 Air is available at a DBT of 50 °C and a wet bulb temperature of 30 °C. Estimate its humidity. The air enters at 300 K and at atmospheric pressure. The vapour pressure of benzene is 275 mm Hg at 50 °C. 5. (c) the weight of toluene per unit weight of vapour free gas.02 = 2. Estimate the mass humidity and molal humidity.4 A mixture of benzene–air is saturated at 1 atm and 50 °C.3 kPa and the temperature is 21 °C. dew point. Estimate the mass humidity and molal humidity. 5.106 PROCESS CALCULATIONS 0. (d) the percent saturation and (e) the percentage of toluene by volume.222[1 + 0. (b) The moles of toluene per mole of vapour free gas. The partial pressure of benzene is 150 mm Hg at 50 °C.222 kg/h G = Flow rate of wet air at inlet = 2.244 kg/h Flow rate of wet air at outlet = 2. The total pressure is 99.09 kg water vapour/kg dry air. How will you obtain this condition? Indicate the exact temperature to which it . humid volume. Estimate its humidity. adiabatic saturation temperature. The exit humidity of air is 0. Calculate (a) the relative humidity.224 kg/h EXERCISES 5.6 Air-water vapour mixture is available at a DBT of 35 °C and a relative saturation of 70%.01] {since G = Gs (1 + x)} = 2.009 i. 9 °C and is heated to 68. calculate (a) the relative humidity. Calculate (a) air consumption rate. and (d) dew point of the air leaving the drier. Air at a DBT of 41 °C and WBT of 24 °C is available. atm is contacted with water at the adiabatic saturation temperature and is thereby humidified and cooled to 70% saturation. condensing out some of the water. The air leaving the drier has a wet-bulb temperature of 37.4 °C. (c) wet-bulb temperatures (WBT) of air before and after preheating. (c) weight of water condensed from 100 m3 of the original wet air in cooling to 15 °C and compressing to 200 kPa.PSYCHROMETRY 107 should be taken before bringing it to 30 °C and 50% RH. atm pressure.24 kPa Vapour pressure of water at 15 °C = 1.03 kg water vapour per kg dry air. 5. 5. (b) molal humidity of this air if its temperature is reduced to 15 °C and its pressure increased to 200 kPa. This air for weaving room is obtained by saturating it initially and then heating it to 80% saturation. 1 std. and (d) the final volume of the wet air of part (c) Data: Vapour pressure of water at 30 °C = 4.70 kPa 5.8 Air at 80% saturation is to be maintained in a weaving room.10 Air at a temperature of 30 °C and a pressure of 100 kPa has a relative humidity of 80%. estimate the quantity of air needed at its original condition.13 A drier is used to remove 100 kg of water per hour from the material being dried.11 Air at 85 °C and absolute humidity of 0. Calculate (a) the molal humidity of air. Water is evaporated in the drier at a rate of 25 kg/h. The available air has a humidity of 0. and a temperature of 23.12 An air-water vapour sample has a dry bulb temperature (DBT) of 55 °C and an absolute humidity 0. (b) humid volume of air before and after preheating. 5.8 °C and a dry-bulb temperature of 54. Estimate the final temperature of the air and its humidity.030 kg water/kg dry air at 1 std.3 °C before entering the drier. . Using humidity chart. If it is necessary to use 5000 m3/h of air at the final condition mentioned above. What are the final temperature and humidity of the air? 5.010 kg bone dry air.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. Calculate (a) percentage humidity of air leaving the drier and (b) volume of wet air leaving the drier per hour. The air leaves at a temperature of 35 °C and a pressure of 742 mm Hg. if vapour pressure of water at 55 °C is 118 mm Hg and (b) the humid volume in m3/kg dry air. (use vapour pressure data from Steam Tables). 5. 24 °C and 8 °C are 100 mm Hg. Water at the rate of 20 kg/h is to be removed from the wet material.15 A mixture of carbon disulphide vapour and air contains 23. dew point. % saturation. Vapour pressure of CS2 at 20 °C = 300 mm Hg 5. 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. Calculate (a) percent humidity of leaving air and (b) volumetric flow rate of leaving air. The mixture is then reheated to 24 °C at 1 atm.17 Air is available at a DBT of 310 K and a WBT of 305 K. (c) The weight of water condensed from 1000 m3 of the original wet air.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.5 mm Hg respectively.3% carbon disulphide (CS2) by volume. 5. and (b) amount of water condensed? Data: Vapour pressure of water at 49 °C. Estimate humid volume and enthalpy using equations and compare the results. 5. In the drier. condensing out some of the water. so that part of water vapour is condensed. VP of water at 30 °C and 20 °C are 32 mm Hg and 17. .108 PROCESS CALCULATIONS 5. It is heated to 85 °C by heating coils and introduced into the drier.14 The air supply for a drier has a dry-bulb temperature of 23 °C and a wet-bulb temperature of 16 °C.19 Air at a temperature of 30 °C and pressure 760 mm Hg has a relative humidity of 60%. Estimate humidity. 5. 27 mm Hg and 8 mm Hg respectively. (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. humid volume and enthalpy using humidity chart. Calculate the relative saturation and percent saturation of the mixture at 20 °C and 740 mm Hg pressure. The air leaves the drier at 35 °C and 742 mm Hg. it cools along the adiabatic cooling line and leaves the drier fully saturated. What is (a) the volumetric flow rate of leaving mixture.18 One thousand m3/min of methane saturated with water vapour at 1 atm. 49 °C is cooled to 10 °C. The volume of the mixture is 5 litres and the partial pressure of water vapour in the mixture is 27. Water is removed at a rate of 2. and vapour pressure at 60 °C = 149. relative saturation and % saturation at 20 °C and 745 mm Hg. partial pressure of acetone.7 mm Hg. Compute the following: (a) Partial pressure of water vapour (b) Partial pressure of dry air (c) Total pressure.20 A mixture of air water vapour is present at 40 °C and 762 mm Hg. vapour pressure at 35 °C = 43 mm Hg.PSYCHROMETRY 109 5.8 mm Hg. 5.21 A mixture of acetone vapour and nitrogen contains 15% acetone by volume. Calculate molal humidity. 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. Vapour pressure of acetone at 20 °C = 184. Air leaves at 35 °C at 742 mm Hg. and (d) Molal humidity. The mixture is heated to 60 °C at constant volume.5 kg/h.38 mm Hg Estimate: (a) Percentage humidity of air leaving (b) Volumetric flow rate of wet air leaving . 110 PROCESS CALCULATIONS Psychrometric chart for air–water system at 760 mm Hg pressure. . A simple illustration of a crystallization process is shown below.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. xF xM xE we can find C and M by solving the Eqs. E kg/h xE Feed. xC is 1. 111 .2) Since we normally know F. C kg/h xC Feed rate of solution of concentration. xE will always be zero as it is free from salt.1) and (6. M kg/h xM Crystal. 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. E. The above principles are explained through the following problems. FxF = CxC + MxM + ExE : C kg/h (6. Water evaporated.1) (6. (6. Some times we get hydrated salt during crystallization process.0. The mixture of crystals and mother liquor is known as magma. F = C + M + E Component balance gives.6 Crystallization Crystallization is a process in which the solid particles are formed from a homogeneous phase. The solution left behind after the separation of crystals is known as mother liquor which will also be a saturated solution. xC Total mass balance gives.2). During crystallization. the crystals form when a saturated solution gets cooled. When we get anhydrous salt. Under such circumstances xC will not be 1. F kg/h xF Crystallizer Mother liquor. 7) = 0.3) kg Solubility of CaCl2 at saturated condition for 1000 kg water = 7.3) 0.12 ¥ 58.7 kg of CaCl2 and 123.3) kg Ë 1000 ˜¯ Equating (1) and (2).38 kmoles = 819.12 kg mole/1000 kg H2O Solubility of NaCl at 65 °C = 6.38 kg mole CaCl2/1000 kg H2O) = 6.62x + 126.112 PROCESS CALCULATIONS WORKED EXAMPLES 6. Calculate the weight of NaCl that can be dissolved by 100 kg of this solution if it is heated to 65 °C.3 kg of 6H2O Total CaCl2 entering the process = (0.37 ¥ 58.3) kg of water Ê 819.7) kg (1) Water entering the process = (x + 123.19918x = – 25. a solution of calcium chloride in water contains 62 kg CaCl2 per 100 kg of water.5) = 373 kg NaCl that may be dissolved further = (373 – 358) = 15 kg Ê 1000 ˆ Water in 100 kg of saturated solution at 15 °C = 100 ¥ Á Ë 1358 ˜¯ = 73.62x + 126.18 ˆ = Á (x + 123.81918 (x + 123.1 kg 1000 After a crystallization process.6 = 1. Solubility of NaCl at 15 °C = 6.37 kg mole/1000 kg H2O Basis: 1000 kg of water NaCl in saturated solution at 15 °C = (6.1 A solution of sodium chloride in water is saturated at a temperature of 15 °C.5) = 358 kg NaCl in saturated solution at 65 °C = (6.2 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 126.6 kg \ amount of additional NaCl that will be dissolved 15 ¥ 73.18 kg \ CaCl2 leaving the process along with (x + 123.7 = 0.81918x + 101 – 0.7 \ x = 129.62x + 126. we get (0. Calculate the weight of this solution necessary to dissolve 250 kg of CaCl2 ◊ 6H2O at 25 °C (solubility = 7.03 kg (2) . 03 kg of water is equivalent to 129.96 Ø % saturation of given solution = É ´ 100 = 90.96 51 È kg of NaNO3 Ø at 40 °C.4 = 1.06 ÙÚ (b) Let a kg be NaNO3 crystallized out NaNO3 balance = (1000 ´ 0.53% Ê 490 ÙÚ 6.6 È 0.49) = (1000 – a) ´ (0.39 kmole/1000 kg H2O Basis: 1000 kg of original solution K2Cr2O7 = 130 kg.3 A solution of sodium nitrate in water at a temperature of 40 °C contains 49% NaNO3 by weight.445) + a.4% by weight. Solubility of NaNO3 at 10 °C = 44.8% Ê 1. The remaining solution is cooled to 20 °C.5% by weight.4 A solution of K2Cr2O7 in water contains 13% by weight. Solubility at 20 °C = 0. (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. É Ê kg of water ÙÚ saturation 51. Water = 870 kg Water present finally = (870 – 640) = 230 kg .06 48. From 1000 kg of this solution is evaporated 640 kg of water. a = 81 kg È 81 Ø (c) % Yield = É 100 = 16. Calculate the amount and the percentage yield of K2Cr2O7 crystals formed. 6.CRYSTALLIZATION 113 100 kg of water is equivalent to 162 kg of solution 162 100 = 209 kg of solution 129. Solving.03 ´ Original solution needed = 209 kg. Basis: 1000 kg of original solution (a) kg of NaNO3 kg of water 49 = 0. Solubility of NaNO3 at 40 °C = 51. a = 700 kg.114 PROCESS CALCULATIONS 0. b = 230 kg Check (water balance): Ë È 180 Ø Û Ë È 100 Ø Û Ì 700 – ÉÊ 286 ÙÚ Ü  Ì 230 – ÉÊ 121.2 – 85 \ Ice formed = Original water – water retained along with NaNO3 = (909 – 172. Salt = 300 kg Water = 700 kg Water evaporated = 700 ´ 0. what weight of salt crystallizes out? Basis: 1000 kg of solution. 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 106 Ø È 21.7 A batch of saturated Na2CO3 solution of 100 kg is to be prepared at 50 °C.5 Ú Solving.5 ÙÚ Ü = 630 kg Í Ý Í Ý 6. If the solubility of anhydrous Na2CO3 at 20 °C is 21.5 Ø È Na2CO3 balance = É a – Ù  Éb – Ù = 300 kg Ê 286 Ú Ê 121.4 g 6.7% 130 A solution of sodium nitrate in water contains 100 g of salt per 1000 g of water. 10% water originally present evaporates.6 1000 kg of a 30% aqueous solution of Na2CO3 is slowly cooled to 20 °C. The solubility is 4.6) = 736.6 = 79. .4 = 103.1 = 70 kg Let a kg of salt crystallize out.5 Basis: 1000 g of original solution È 100 Ø = 91 g NaNO3 present initially = 1000 ´ É Ê 1100 ÙÚ Water present initially = 909 g 1000 – 91 = 172.2 g mole/1000 g H2O) % Yield = 6.6 kg 103.6 g Water retained along with NaNO3 = 6.4 kg Ê 1000 ÙÚ \ K2Cr2O7 crystallized = 130 – 26. The crystal is Na2CO3× 10H2O. During cooling.5 kg/100 kg of water.39 Ø K2Cr2O7 present then = 230 ´ ÈÉ ´ 294 = 26.48 g mole/1000 g H2O at 50 °C. Calculate the amount of ice formed in cooling 1000 g of this solution to –15 °C (solubility at –15 °C = 6. Solubility at 50 °C = (4. The hot crystals with some adhering solution are centrifuged to remove some of the solution. water available in the salt is (ii) Weight of decahydrate needed = 32.8 10.88 = 32. and (d) water leaving the evaporator.6 kg of concentrated solution was removed in evaporator.000 kg/hr of 6% solution of salt in water is fed to an evaporator.2 ´ 180 = 54.2 ´ 6.e. Saturated solution is produced and some salt crystallizes out.7 kg of solution was separated in the centrifuge and 361. thus making the total water to 13. Previous test on the centrifuge show that it removed 60% of adhering solution.88 kg 106 Discussion: In case (ii).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 Ë 32.7 kg of dried crystals got.88 1474.2) = 67.2 – 18 Û = (67.33 kg 286 = 86. Then the crystals are dried to remove the rest of water.68 kg 106 The additional water needed hence is 13. Discussion: The solution leaving the evaporator and centrifuge maintains the same concentration and is saturated.2 kg In 100 kg of solution. (b) water removed in evaporator. saturated solutions and crystals with adhering solution.8) – Ì Ü Í 106 Ý = 62. . During an hour test. From evaporator.88 kg solution. Na2CO3 present = water present = (100 – 32. Calculate (a) the solubility of the salt. 837. the various streams leaving are water vapour. 100 – 474. 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.12 kg. (c) water removed in drier.88 kg Na2CO3/1474. 198.8 kg (which corresponds to solubility data) 32.68 = 67.CRYSTALLIZATION 115 (i) If the monohydrate were available.12 + 54. 474.48 ´ 106) kg Na2CO3/1000 kg H2O i. 5 ´ 0.500 kg Check: Water balance = 8.000 ´ 0.7) ´ (0.23 kg salt/0.400 kg 6.7)s + 361.7 kg Let s be the solubility of salt.7 + 198.4) = 8.77) = 8.183 wt fr.7) = 132. 1 kg of solution contains 0.000 kg/hr Evaporator Centrifuge 837.7 + 102) = 463. 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 0.9 In a solution of naphthalene in benzene. mole fraction of naphthalene is 0.6 + 198. of naphthalene Tank (100 + X) kg 0.7 kg 6% solution Adhering solution + crystal Water Drier Dry crystal 361.5 kg (c) Water in it (removed in drier) = 132.23 kg of salt) (d) Weight of material entering drier = Final dry salt + Water evaporated = (361. which is given by (s kg of salt/kg of solution) Making salt balance.2 – 198. of benzene 0.116 PROCESS CALCULATIONS Basis: One hour.6 Water vapour 10.7 kg \ Total adhering solution entering centrifuge (along with crystal) 198.06) = (837.7 = = 331.7) = 662.500 + 102 + 798 = 9.77 = 102 kg (since.2 kg 0.23 kg of salt/kg of solution. fr.7 kg Weight of material entering centrifuge = (463.817 wt fr.6 + 198. 1 kg solution = 0.77 kg H2O = 0. (10.6 kg solution 198.000 – (837.6 + 662. 60% adhering solution (removed in centrifuge) = 198. of naphthalene .77 kg water and 0.7 Solving. (a) s = 0.500 + 102 + (837.57 wt.3 kg salt/kg H2O (b) Weight of adhering solution entering drier = (331.4 kg Weight of water leaving the evaporator = 10.12. 10 A crystallizer is charged with 7.296 = 2220 kg H2O present = (7500 – 2220) = 5280 kg H2O lost in evaporation = 5280 ´ 0. Naphthalene balance in tank gives. Solubility at 20 °C = 194 g Na2SO4/100 g water Molecular weight of Glauber’s salt (Na2SO4 ×10H2O) = 142 + 180 = 322 Basis: 7.183 Benzene (C6H6) 0. Glauber salt crystallizes out.57 X = 111 kg of original feed solution.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.4 0. kg Weight fraction Naphthalene (C10H8) 0. 0.4 Ú a Ø È b Ø È H2O balance: 5016 = É180 – Ù  É100 – Ù Ê 322 Ú Ê 119.12 128 15.0 1.5 kg Check: (total balance) 3749.4 Ú Solving.CRYSTALLIZATION 117 Basis: 1 mole of feed solution Component mole Molecular weight Weight.88 78 68. 6. The salt removed carries 20% of its weight of brine containing 27% NaCl (All Composition in Weight %). a Ø È b Ø È Na2SO4 balance: 2220 = É142 – ÙÚ  ÉÊ19. The solution is then cooled to 20 °C. Calculate the feed rate kg/h.5 kg and b = 3486. Na2SO4 present = 7500 ´ 0.000 Total Let us assume that X kg of feed solution enters the tank.500 kg of aqueous solution at 104 °C containing 29. During this operation 5% of water is lost by evaporation.4 – Ù Ê 322 119.183X + 100 = (100 + X)0.000 kg dry salt/h from a feed solution having 20% NaCl.11 An evaporator-crystallizer is to produce 13. .5 kg Weight of solution = 3486. a = 3749.5 kg Weight of Glauber salt = 3749.6% by weight of anhydrous Na2SO4.5 + 3486.500 kg of aqueous solution.5 + 264 = 7500 kg 6.6 0. Find the yield of crystals.817 — — 84. 91 – 422.355 = 140.90.000 kg of KCl salt in a solution at 80 oC (a) Solubility at 80 °C = 55 kg salt/100 kg water or.2F = (13.000 kg dry salt/h Basis: 1 h of operation 13.090.118 PROCESS CALCULATIONS Water 20% NaCl Evaporatorcrystallizer Feed ‘F’ 13.14 kg .000 ´ 0. The solution is cooled to 20 °C in an open tank.91 ´ 0. (a) Assuming water equal to 3% by weight of solution is lost by evaporation.18 kg È 35 Ø KCl in leaving solution at 20°C = 8668.90.12 5000 kg of KCl are present in a saturated solution at 80 °C. solubility of KCl = 55 kg/155 kg solution = 0.600 NaCl in this solution = 2.91 – 5000 = 9090.18 ´ É = 3033.73 = 8668. 0.91 kg Water evaporated = 3% weight if solution = 140.91 kg of solution Water present = 140.702/0. 6.27 = 702 NaCl balance gives.510 kg/h.03 = 422. (b) Calculate the yield of crystals neglecting loss of water by evaporation.73 kg Water remaining = 9.2 = 68.2 = 2.000 + 702) \ Feed = 13. KCl crystallizes without any water of crystals. 3% Water 5000 kg KCl 80 °C Saturated solution Crystallizer 20 °C Saturated solution Crystal Basis: 5. The solubilities of KCl at 80 °C and 20 °C are 55 and 35 parts per 100 parts of water.000 kg of salt with 20% brine Weight of feed brine = 13.355 kg/kg solution 5000 kg KCl is equivalent to 5000/0. calculate the weight of crystals obtained.600 ´ 0.90.86 kg Ê 100 ÙÚ \ KCl crystallized out = 5000 – 3033.86 = 1966. The solubility of Na2SO4 in water at 20 °C is 19. calculate the weight of the mother liquor.6 kg = 450.10H2O crystallizes out. Water lost by evaporation = 10% of 4505.56 kg 142 Ø È Overall balance of Na2SO4 gives.505.44 kg Solving. If the mother liquor (after crystallization) is found to contain 18. . Na2SO4.3% Na2SO4.37% 5000 6.6 kg Let M be the weight of mother liquor and C the weight of crystal formed.7 – ÙÚ  ÉÊ 3122.56) = 5949.6 kg 6.400 kg of an aqueous solution containing 29.7 Ø È 180 Ø È Check: (water balance) 450.6% 18. 1894. M = mother liquor = 2826.4 = É C – Ù + (0.183 ´ M) Ê 322 Ú C + M = (6400 – 450. % Yield of crystals = 1818.CRYSTALLIZATION 119 (b) KCl in solution at 20 °C = 9090.91 ´ 0.14 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.4 kg Water present in feed = 4.7 kg C = Crystal = 3122.18 ´ 10% water 6.894.13 A crystallizer is charged with 6.7 kg 81.400 ´ 0.4 g per 100 g of water. Weight of Na2SO4 = 6.35 = 3181.82 kg \ KCl crystallized out = 1818. The solution is cooled and 10% of the initial water is lost by evaporation.7 – Ù Ê 100 322 Ú = 4505.400 kg Crystallizer 29.296 = 1.18 kg 100 = 36.400 kg solution.3% Na2SO4 Mother liquor ‘M’ Na2SO4 × 10H2O (142 + 180) = 322 Basis: 6.6% of anhydrous sodium sulphate.56 + É 2826. 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 X ˆ Ê 19.3 10.1% anhydrous sodium sulphate.5H2O crystals per hour.3 parts FeSO4/100 parts water. The mother liquor is found to contain 16.10H2O formed? . Determine the quantity of feed and water lost as vapour.2 A vacuum crystallizer is to produce 10.5 kg/100 kg of water. X Basis: 1.000 kg of FeSO4 .6% anhydrous sodium sulphate by weight at 104 °C.420 kg of aqueous solution containing 29. 6. 3% water originally present evaporates.000 kg of a 30% aqueous solution of Na2CO3 is slowly cooled to 20 °C. The crystal is Na2CO3 .4 ˜¯ Ê 180 X ˆ Ê 100Y ˆ Water balance = Á = 680 + Ë 322 ˜¯ ÁË 119. Na2SO4. we get X = Weight of crystals = 566.1 Feed to a crystallizer contains 6.4 ˜¯ Solving.120 PROCESS CALCULATIONS Feed 32% Crystallizer Mother Liquor. The solubility at 27 °C is 30.761 g and Y = Weight of mother liquor = 433. Y Crystals. what is the weight of Na2CO3.000 g of the feed solution. The feed enters at 70 °C and is cooled to 27 °C. If the solubility of anhydrous Na2CO3 at 20 °C is 21.761 = 56. During cooling.10H2O.4Y ˆ Na2SO4 balance = ÊÁ = 320 + Ë 322 ˜¯ ÁË 119. 6.239 g Yield of the crystals = 566.68% 1000 EXERCISES 6. The solution is cooled to 20 °C to crystallize out the desired Glauber’s salt. Estimate the weight of mother liquor.9 parts FeSO4/100 parts water. The feed solution contains 38. mother liquor leaving and feed. The solubility of anhydrous Na2SO4 in water at 20 °C is 19. The mother liquor from the crystallizer is recycled and mixed with the evaporator feed. The concentrated liquor is sent to a crystallizer where crystals of KNO3 are formed and separated.28 g MgSO4 per 100 g water At 25 °C : 40.10H2O. The resulting solution is cooled to 20 °C.4 1000 kg of a solution containing 25% of Na2CO3 by weight is slowly cooled to 20 °C.8 A solution of sodium sulphate in water is saturated at a temperature of 40 °C. If the solubility of anhydrous Na2CO3 at 20 °C is 21. Compute water evaporated and crystals obtained.11 Hypo crystals Na2S2O3 × 5H2O are to be produced at the rate of 2000 kg/h. Find the weight of water lost. Estimate the amount of feed needed. Calculate the weights of (a) water loss (b) the mother liquor leaving and (c) crystals formed. 1000 kg of this solution are evaporated to remove some water. Glauber’s salt crystallizes. Solubility data: At 80 °C : 64. 6. 6. During this process 5% of the original water is lost. The recycle stream is a saturated solution containing 0.4 gm per 100 gm of water.9 A solution of K2Cr2O7 in water contains 16% K2Cr2O7 by weight. what is the weight of Na2CO3. Calculate the weight of crystals formed and the percentage yield obtained by cooling 100 kg of this solution to a temperature of 5 °C.5 kg/100 kg of water. Solubility of K2Cr2O7 at 20 °C = 0. if the temperature of solution is reduced to 20 °C. The crystal formed is Na2CO3. . calculate the amount of water evaporated. Calculate the yield of Na2SO4.6 An evaporator concentrates 10.10H2O crystals formed from solution. Solubility Data: At 40 °C : 32. The mother liquor leaves with 16% salt. During cooling 15% water originally present evaporates.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.39 k mole/1000 kg H2O 6.88 g MgSO4 per 100 g water 6.7H2O crystallizes.000 kg/h of 20% KNO3 solution to 50% KNO3.6% Na2SO4 At 5 °C : 5.75% Na2SO4 6.10 A saturated solution of MgSO4 at 80 °C is cooled to 25 °C. If the yield of K2Cr2O7 crystals is 80%. if 1500 kg of MgSO4 . 6. The crystals carry 4% water. A 60% Na2S2O3 solution is cooled to 293 K from 333 K.10H2O crystals formed? 6.6 kg KNO3/kg water. The solubility at 293 K is 70 parts anhydrous salt per 100 parts of water.CRYSTALLIZATION 121 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. WORKED EXAMPLES 7.6% % oil extracted = Á Ë 18.2 A multiple effect evaporator handles 100 tonnes/day of pure cane sugar.7% of cake \ cake = 78. Basis: 100 kg of seeds Solids free from oil is the tie element Weight of solids in feed = 69 kg = 87. 69% solids and 12. The application of mass balance in unit operations and processes is illustrated through worked examples and exercise problems. While the concentrate is leaving with 75% solids concentration.6% oil.971 kg Ê 17.68 ¥ 0. 87. The principles given in Chapter 2 still hold good. At the end of the extraction process.4% moisture. calculate the amount of water evaporated per day.7% solids and 11.6 ˜¯ 7. The seeds contain 18.1 Soyabean seeds are extracted with hexane in batch extractors.008) = 17.68 kg Oil extracted = (18.971ˆ ¥ 100 = 96.6 – 78.8% oil. The analysis of cake reveals 0. ? 30% solids Evaporator Basis: 100 tonnes of feed = 105 kg Solids in the feed = 30.000 kg 122 75% solids .7 Mass Balance This chapter deals with the mass balance both with and without chemical reactions. the residual cake is separated from hexane. Find the % recovery of oil. The feed to the evaporator contains 30% solids.5% moisture. 6%.56 ÙÚ Water evaporated = (325 – 200) = 125 kg 7.000 – 40.91) = 182 kg 182 Ø Wet fabric entering = ÈÉ = 325 kg Ê 0.5%. Calculate the weight of water removed per 200 kg of dried fabric.3 Concentrated liquid leaving = A water soaked fabric is dried from 44% moisture to a final moisture of 9%. From the flue gas.000 kg \ 7.0% and O2 : 8. CO : 89% and N2 : 3%. 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 furnace Air Burner CO2. O2. CO : 22. This gas is burnt with air under conditions selectively to oxidize phosphorus. ? 44% moisture Drier 9% moisture 200 kg Basis: 200 kg of product.9%.000 = 40. the oxides of phosphorus precipitate on cooling and is separated from the remaining gas. Assume oxidation of phosphorus is complete and that it exists in the flue gas partly as P4O6 and partly as P4O10.00.000) = 60. Dry fabric = (200 ´ 0.75 Water evaporated/day = (1.MASS BALANCE 123 30. N2 : 68. 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 .4 The exit gas from a phosphate reduction furnace analyzes P4 : 8%. CO. Analysis of the latter shows CO2 : 0.000 kg 0. 712 g moles O2 leaving = 380.7654 g moles \ P4 converted to P4O10 = (4.7% H2SO4 are added to the tank and the final acid is 18.46) = 552 kg Water 54% moisture Drier 9% moisture . 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.342 ´ 0. If 200 kg of 77.1243x + (200 ´ 0.234 (a) % CO oxidized to CO2 = (380.243 g moles Making a balance for oxygen.953 – 32.68) = 258.905. (Total carbon in CO and CO2) 89 = 380.6 A cotton mill dries a water soaked fabric in a drier from 54% to 9% moisture.777) = 0.342 ´ 0.953 g moles 79 O2 for formation of P4O10 = 5x g mole O2 from air = 255.63%.4% of final gas.5 \ x = 4.852 g moles \ Total final gas moles = (b) N2 in final gas = (380. Basis: 1200 kg of feed. x = 1.632 ´ O2 for formation of P4O6 = 3(8 – x) g mole O2 for formation of CO2 = 380.105.1863y Solving.632 – 3) = 255.009 ´ 0.009/89) ´ 100 = 3.632 g moles N2 from air = (258.243 7.71 g moles O2 reacted = (67. How many kilogram of water are removed by drying operation per 1200 kg of feed.342 g moles 0.71) = 35.086 = 32.342 ´ 0.43% acid.632 g moles 21 = 67. Overall balance is x + 200 = y Acid balance is given as 0.5 = 1.5 kg 7.342 ´ 0. Dry fabric = (1200 ´ 0.7654 ´ 100/8) = 60% A tank of weak H2SO4 contains 12.124 PROCESS CALCULATIONS 89 g atoms of C º 23.5 kg and y = 2.712) = 35. (5x + 24 – 3x + 1. 8 kg MgCO3 = 0.MASS BALANCE 125 552 = 606.6 kg Water removed = 648 – 54.16% and rest insoluble.4 kg 7. MgCO3 : 4. 25. The concentration of NH3 in the down stream is 6%.4 kg CaCO3 ® CaO + CO2 100 56 44 MgCO3 ® MgO + CO2 84.6 kg 0.06 Solving. To measure the flow rate an ammonia rich gas containing 96% NH3. Find the flow rate of gas at inlet.3 CaO got = 40.8 – 56 = 2. gas Pipe NH3 rich gas (96%) 100 cc/min. x cc/min. (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 = 0.7 A gas containing 2% NH3.6 ´ 0.91 Water removed = 1200 – 606. Making a balance for ammonia: 0.52%. 7.02x + 96 = (100 + x) 0.6 kg = 593.8 Exit (100 + x) 6% NH3 A lime stone analysis shows CaCO3 : 94.9452 ´ 4000 = 3.3 44 3780.6 = 593. 3% H2 and 1% N2 is sent at a rate of 100 cc/min.117.09 = 54.3 kg 100 .780.4 kg Weight of product = Alternative method: Water in feed = 1200 ´ 0.54 = 648 kg Water in product = 606.4% N2 and the rest H2 is flowing in a pipe. x = 2250 cc/min. Basis: Let x cc/min of feed gas enter the pipe Ammonia rich gas entering = 100 cc/min \ Exit gas = (100 + x) cc/min.0416 ´ 4000 = 166. 126 PROCESS CALCULATIONS (b) Basis: 1 kg of limestone CaCO3 = 0. 81. The .5203 g moles Exit stream = SO2 converted to SO3 = SO2 entering – SO2 leaving the converter = (9.34 g moles 100 % SO2 converted = 9.86%. 4.54%.4) 98 142 (2 × 36.6/0.00605 = 0.9 The gases from a sulphur burner have the following analysis: SO2 : 9. The HCl gas is absorbed in water.5% O2.4) È 2 ¥ 36. 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.4 ˚ È 142 ˘ (b) Na2SO4 formed = 1960 ¥ Í ˙ = 2382.10 Hydrochloric acid is commercially prepared in a Mannheim furnace by the reaction between NaCl and H2SO4.4376 kg (by stoichiometry) 7. Calculate (a) the weight of HCl formed when 2 tonnes of 98% salt reacts.94895 SO2 leaving = 86 ¥ 0.9 kg Î 2 ¥ 58.98 = 1960 kg 2NaCl + H 2 SO 4 → Na 2 SO 4 + 2HCl (2 × 58.9452 kg. N2 : 81.86 – 0.7% 9. After passage of gases through a catalytic converter.4 ˘ (a) HCl formed = 1960 ¥ Í ˙ = 1221.11 Dolomite chiefly a mixture of calcium and magnesium carbonates is to be leached with concentrated H2SO4 to recover Mg as MgSO4.0416 ¥ ˜ Ë 100 84.34 ¥ = 94.0416 kg 44 ˆ Ê 44 ˆ Ê CO2 produced = Á 0. the analysis is 0. (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 ¥ 0.895% N2.52) = 9.4 ˚ 40% acid formed = 1221.6 = 86 g moles 0.605% SO2.4 = 3054 kg 7.6 kg Î 2 ¥ 58.3 ¯ = 0.86 7. 94.6%. MgCO3 = 0.9452 ¥ ˜¯ + ÁË 0. O2 : 8. 25 g Mg in sludge cake = 250 ´ 100 Mg extracted = (10 – 1.13 Phosphate rock at ` 80/tonne is being added to a cheap fertilizer to enrich it. After reaction the soluble material is filtered off and the solid are washed and discarded. Analysis of the sludge cake shows 50% moisture. 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 \ Weight of dried soap = 500 = 625 kg 0. 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.5 = 1.875 10 7. x = 0.23 = cost of production 1.12 A certain soap contains 50% moisture on the wet basis when raw. 2% SiO2.25) = 8.5% 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 5 = 250 g 0.23 Actual cost (30% profit) = Solving.75 = 0.MASS BALANCE 127 rock analysis indicates 20% Ca.02 0.3 ÙÚ (The number of cakes has to be an integer) 7. Blended fertilizer = (1 + x) tonne Final cost of blended fertilizer = ` 64/tonne 64 = ` 49.3 Cost balance: 32 + 80x = (1 + x) 49. If the cheap fertilizer costs ` 32/tonne and the final blended product including 30% profit is to be marketed at ` 64/tonne. The moisture is reduced to 20% before the soap is pressed into cakes. of cakes pressed = ÈÉ = 2083 Ê 0.75 g \ weight of sludge cake = Fraction Mg extracted = 8.56 tonne (Phosphate rock) . 10% Mg and 5% SiO2.8 625 Ø No. 0. 0. The fractionating system consists of two towers No. The bottom product from I is feed to II resulting in an overhead product of y kg/h of 3% benzene. The overhead product from I is x kg/h of 95% benzene. Lime is made up calcining the carbonates.741.000 Individual balance yields: For benzene. 30% toluene and 20% xylene.000 Solving.3 44 Overall balance: x + y = 100 (since pure) CO2 balance = 0.04z = 9. The feed enters tower I.01z = 15. 3% toluene and 2% xylene. 0.8 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.e.3 kg/h z = 5.2 kg/h y = 8. x = 90 g (CaCO3) and y = 10 g (MgCO3) 7. 0.000 kg/h of 50% benzene.0 kg/h .95y + 0. i.000 For xylene.02y + 0. 95% toluene and 2% xylene while the bottom from II tower is z kg/h of 1% benzene.95z = 6.5 kg/h Total 30. I and No.000 For toluene.128 PROCESS CALCULATIONS 7. When pure limestone is calcined.03y + 0. x = 15. 44.44x + 0.000. 4% toluene and 95% xylene. driving away all CO2 gas. y and z.8 Solving.95x + 0. II.15 The feed to a fractionating system is 30.52y = 44.03x + 0. Find x.806. x Feed I y II z Overall balance yields x + y + z = 30.14 Limestone is a mixture of Ca and Mg carbonates.02x + 0.3 40.452. 07  Ù Ê 2 Ú = 0.4763 kmole = 0.2667 kmole (79%) N2 total = 0. H2 : 15. Basis: 1 kg of fuel C : 0.07 = 0.071 kmole (21%) N2 = 0.5143 kmole 0.07 kmole H2O formed = 0.076 kmole.4063 kmole N2 entering (along with stoichiometrically needed O2) = 0.814 – (0.6% CO2 by volume.MASS BALANCE 129 7.21 Exhaust gas contains CO2. Estimate the amount of air supplied per kg of fuel and find the products of combustion (air contains 23.5143 + 0.008 kg C + O2 12 32 n CO .076 kmole Total kmole of exhaust stream = 0.16 A sample of fuel has the following composition by mass: C : 84%.814 – 0. The fuel was completely burnt with excess air in an internal combustion engine. H2 : 0. N2 and O2 Air needed (stoichiometric) = Excess air in exhaust = Total moles in exhaust stream (dry) – (N2 entering as per stoichiometry + CO2 formed) = 0.07 kmole.84 kg = 0. Noncombustibles – 0. 2 44 H 2 + ½O 2 2 16 nH O 2 18 CO2 formed º 0.5143 ÙÚ .4063 + 0.6% % excess air = É Ê 0.814 (dry) 0.4063 + 0.108 kmole 79 21 = 0.8520 kmole È 0.108 ´ 0.3377 kmole We know what air contains O2 = 0.673 kmole Total air supplied = (0.3377 Ø ´ 100 = 65. The dry exhaust gas has 8.3377) = 0.1% oxygen by weight).076 Ø È O2 needed for forming CO2 and H2O = É 0.086 0.152 kg = 0.2% and rest comprising noncombustibles.108 = 0.2667 = 0.07) = 0. 18 A mixture of 5% ethylene and 95% air is passed through a suitable catalyst in a reactor.12 (b) Freight charge = 4.7 82. If one mole of water is sprayed per 100 mole of gas mixture. It is also agreed that the price can be suitably adjusted if the moisture content differs from the specified value.607 kg and weight of moisture = 393 kg (a) Cost of paper = 4. Some of the ethylene does not react.0 7. some form oxide.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.8140 100.055% and N2 : 81. Basis: 100 kg of 5% moist paper.000 kg paper with 7. (b) If it were agreed that the freight was to be borne by the company and adjusted according to the moisture content.852 ´ 28. calculate the composition of ethylene glycol–water product formed.515% on dry basis.6 8.4 mm Hg while total pressure is 745 mm Hg.743.000 kg of paper was found to contain 7.607 ´ 7. The gas leaving the absorber analyzes C2H4 : 1.57 kg Components Weight. . what would the company pay to the manufacturer if the freight rate is ` 40/1000 kg of dry paper.0700 0. kmole Composition of dry exhaust gas CO2 O2 N2 0.16 95 In the case of 5.16 = ` 14.607 ´ 3.085%. The partial pressure (pp) of H2O in this gas is 15. (a) Find the price to be paid to manufacturer as per contract. Dry paper 95 kg.12 40 = ` 185 1000 Total cost including freight = ` 14.86% moisture.558. A shipment of 5. The oxide is converted to glycol. The entire gas mixture enters an absorption tower where water is sprayed. CO2 : 4.345%. moisture 5 kg 3 = ` 3.7 Total 0.86% moisture Cost of dry paper = 100 ´ Weight of dry paper = 4.130 PROCESS CALCULATIONS Weight of air supplied/kg of fuel = (0. O2 : 13. some turn to CO2 and water.84) = 24.0710 0.6730 8. MASS BALANCE 5% C2H4 Reactor Gases 131 Ethylene. O 2.00 kmoles . O2 : (95 ´ 0.4 mm Hg Absorber 95% air Ethylene glycol–water Reactions: 1.H2O 15.4 C2H4 reacted : 5 – 1 = 4 kmoles H 2O = 4 = 2 kmoles 2 C2H4 converted to C2H4O : 4 – 2 = 2 kmoles C2H4 converted to CO2 : 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. C 2 H 4 O + H 2 O → (CH 2 OH) 2 Ethylene glycol Basis: 100 kmoles of feed gas.4 = 1.04345 = 4 kmoles O2 : 92 ´ 0.81515 C2H4 : 92 ´ 0.98) = 4. C2H4O : 2. C2H4 + 3O2 ® 2CO2 + 2H2O 3.98 kmoles Water sprayed at the top of absorber = 98 ´ Water in exit gas = 1. 2C 2 H 4 + O 2 → 2C 2 H 4 O Ethylene Ethylene oxide 2. 1 = 0. N 2 pp. N2 : 75 Total number of moles = 98. CO2 : 4. O2 : 12. H2O : 4.94 kmoles 745  15.01085 = 1 kmole CO2 : 92 ´ 0. C2H4 : 5 kmoles. N2 : 75 kmoles N2 balance: Moles leaving the absorber = 75 = 92 kmoles 0.21) = 20 kmoles. CO2.13055 = 12 kmoles 92 – 15.98 100 Total moles of water entering absorber = (4 + 0.94 kmoles Water reacted in absorption column = 2. 132 PROCESS CALCULATIONS 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 1.04 18 18.72 13.12 142.72 100.00 Ethylene glycol (H2O) Water Total 3.04 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. 21 = 21.8 kmoles (using nitrogen balance) 79 O2 leaving = 7.8 kmoles; O2 used = 14 kmoles (a) O2 fed = 82 ´ 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 y 0.1 = 0.02 5 (b) Total FeS2 reacted = 5.1 moles O2 needed by Reaction (2) = 5.1 ´ 5.5 = 14.025 moles 2 MASS BALANCE 133 Excess O2 supplied = 21.8 – 14.025 = 7.775 moles È 7.775 Ø ´ 100 = 55.4% % excess air = É Ê 14.025 ÙÚ 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 C2 H 4 CO H2 CH4 N2 9.2 0.4 20.9 15.6 1.9 52.0 9.2 0.8 20.9 — 1.9 — — 0.4 ´ 3 = 1.2 20.9 ´ 0.5 = 10.45 15.6 ´ 0.5 = 7.8 1.9 ´ 2 = 3.8 — Total 100.0 — 23.25 Oxygen needed for combustion Reactions C2H4 + 3O2 CO + 0.5O2 H2 + 0.5O2 CH4 + 2O2 ® ® ® ® 2CO2 + 2H2O CO2 H2 O CO2 + 2H2O Carbon is the tie element C entering = 32.8 g atoms º 11.2% exit. 32.8 \ Total moles of exit gas = = 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 21 O2 from air = 181 ´ = 48; Total air = 229 g moles m3 of air/m3 79 229 feed = g mole of air/g mole of feed = = 2.29 100 134 PROCESS CALCULATIONS (b) O2 supplied = 48 g moles needed = 23.25 g moles Excess O2 supplied = (48 – 23.25) = 24.75 g moles 100 = 106.45% 23.25 (c) Total N2 leaving = 233 g moles % excess air = 24.75 ´ 100 Û Ë % N2 from air = Ì 52 – = 22.3% 233 ÜÝ Í 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 ´ (b) % lost in second reaction = 0.6 ´ (c) Air fed (using nitrogen balance) = Moles of feed/moles of air = 100 = 30% 12.3 100 = 4.9% 12.3 68 = 86 g moles; O2 = 18 g moles 0.79 12.3 = 0.143 86.0 O2 needed for converting CH3OH to HCHO = 12.3 = 6.15 2 MASS BALANCE 135 Excess O2 supplied = (18 – 6.15) = 11.85 g moles 100 = 192.7% 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. % excess air = 11.85 ´ 2CuS + 2.5O 2 → Cu 2 O + 2SO 2 2 × 95 2.5 × 32 (1) 2 × 64 142 2FeS2 + 5.5O2 ® Fe2O3 + 4SO2 (2) Cu 2 O + C → 2Cu + CO (3) 142 12 126 28 Basis: 1 tonne of ore = 1000 kg Ore Crushing and Flotation Inert (2/3) P1 Roaster P2 Cu, CO CuS : 100 : 16.67% FeS2 : 300 : 50.00% Inert : 200 : 33.33% 600 2 = 400 kg 3 Inert remaining = 200 kg = 0.333 P1 \ P1 = 600 kg Inert in ore = 600 kg. Removed = 600 ´ CuS in P1 = 600 ´ 0.166 = 100 kg; FeS2 = 600 ´ 0.5 = 300 kg 142 = 74.74 kg 190 (b) Cu2O reacted = 74.74 ´ 0.95 = 71.0 kg (a) Cu2O got from CuS = 100 ´ 126 = 63 kg 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. (c) Cu got from Cu2O = 71 ´ (a) Excess air (b) Volume of air supplied (c) Volume of burner gas leaving. 136 PROCESS CALCULATIONS Basis: 100 kg of pyrites weight of FeS2 = Total weight – weight of inert = 100 – 25 = 75 kg 4FeS2 + 11O 2 → 2Fe 2 O3 + 8SO 2 480 FeS2 = 75 kg = 352 512 320 75 = 0.625 kmole 120 8 = 1.25 kmoles 4 SO2 = 1.25 kmoles º 9% of exit gas. SO2 produced = 0.625 ´ 1.25 = 13.89 kmoles 0.09 O2 in exit gas = 13.89 ´ 0.086 = 1.19 kmoles \ Total exit gas in mole = N2 in exit gas = 13.89 ´ 0.824 = 11.44 kmoles O2 consumed = 0.625 ´ 11 = 1.72 kmoles 4 (a) % excess air = 1.19 ´ 100 = 69.2% 1.72 1.19  1.72 = 13.86 kmoles 0.21 Volume of air supplied = 13.86 ´ 22.414 = 310.6 m3 at NTP. (b) Air supplied = 773 = 881.5 m3 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. (c) Volume of exit gas = 13.89 ´ 22.414 ´ 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 H2O 1 0.06 Net oxygen needed – 0.03 10.25 87 g moles 0. Basis: 1 g atom of S = 32 g S + O 2 → SO 2 S + 1.73 g moles 100 = 16.8 g moles O2 reacted = 0.8 ´ 79/21 = 6.3 ´ 1.MASS BALANCE 137 Carbon in exit stream = (9.70 Total 6. Calculate the molar ratio of SO3 to SO2 in the exit gas.15) = 0.98 – 10.3 + x O2 N2 7.7 + (0.25 =1.15 g moles Exit gas stream: Component SO2 SO3 Weight.5% 7 = 60.04 Ø Air supplied = ÈÉ = 57.5O 2 → SO 3 32 32 32 O2 needed 64 48 80 = 1.8 – 1.04 g moles \ g moles of exit gas = 45.0 . S to SO3 is the desired reaction Hence % excess air = 1.2 = 1.5 ´ 1.65 Mole % 8.5 g mole O2 supplied = 1.9% 10.87 ´ 0.74) = 45.8 8.02 g moles Ê 0.8 g moles N2 supplied = 1.7 – x SO3 = 0.98 g moles and excess O2 = 11.73 ´ (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 4.6) = 11.3 (1.3% SO2.25 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 SO2.45 80.7 0.28 3.25 7.47 100.5 O2 ® SO3 Exit gas stream consists of the following: SO2 = 0.5) = 1. g mole 0.115 N2 in exit = (60.9 + 1.55 Let x moles of SO2 react to form SO3 SO2 + 0.79 ÙÚ \ O2 supplied = 11. 6% and that of the gas leaving is SO2 : 2.5 O 2 → SO3 32 48 80 1 katom of sulphur requires 1 kmole of oxygen to form SO2 Therefore oxygen supplied = 15.6 \ Total moles in exit = 0.5x N2 = 6.78 2.05 = 0. Find the exit gas analysis.00 7.9 16.75 – 16.015) = 2.5x SO2 = 0.625 ´ 0.34 100.625 katoms 32 S + O 2 → SO 2 32 32 64 S + 1.735 70.845 0. O2 : 13.7  x º 0.27 The composition of the gas entering a converter is SO2 : 7.54 88.70 0.625 ´ 0.015 kmoles O2 remaining = (18.08 79.26 500 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. 500 Basis: 500 kg sulphur = = 15.2%.625 ´ 1.5 x Solving.34 È SO3 Ø ÊÉ SO ÚÙ 2 exit È 0. we get x = 0. Determine the % of SO2 oxidized to SO3 Basis: 100 g moles entering 79. kmole mole % SO2 SO3 O2 N2 Total 14. O2 : 11.75 ´ S to SO3 = 15.64 Ø = 1.845 kmoles N2 entering = 18.1 g moles (making nitrogen balance) .45 – 0.845 + (1.78 ÊÉ 0.735 kmoles Exit gas Weight.855 = 93.138 PROCESS CALCULATIONS O2 = 0.45  0.043 8.95 = 14.5 ´ 0.8 Total = 8.2 = 18.65 – 0.8%.36 ÚÙ 7.2% and N2 : 79.88 3.75 kmoles 79 = 70.78)] = 16.5%.780 kmole Total O2 consumed = [14.7% and N2 : 85.54 kmoles 21 S to SO2 = 15. 4. 1.59 ´ (a) Ratio of lime produced to coal burnt (b) Stack gas produced/tonne of lime produced.75% 7.2 (0.028 = 2.595. we get z = 187 kg y = 2.4% CO2.7% H2.28 Limestone containing (on dry basis) 42.047z – Ù = (4.2 7.61 g moles SO2 converted to SO3 = moles of SO2 entering – moles of SO2 in exit stream = 7.6% O2 and 7.5% N2.046z + 0.8% N2.5% CO2 is burnt with coal having an analysis of 81% C.3 32 2 × 44 Stack gas Limestone z coal Burner CaO Lime y Air Basis: 100 kmoles of stack gas Let z kg of coal and y kg of air enter the burner (air contains 23% O2. N2 balance gives.MASS BALANCE 139 \ SO2 in exit stream = 93.018z + 0.3 12 40. 4. 0.1% O2 and 71. 77% N2 by weight %) C + O2 ® CO2 H2 + 0. 4.81z – Ê 12 2Ú Solving the above equations.61 = 4.2 – 2.6 kg .23y) – É 0.5 ´ 28 = 2002 O2 balance gives. 32 16 Ø È  0.59 g moles 100 = 63.9% ash. Reactions: CaCO3 + C + O 2 → CaO + 2CO 2 100 12 56 32 88 MgCO 3 + C + O 2 → MgO + 2CO 2 84.5O2 ® H2O On weight basis. Compute % SO2 oxidized = 4. The stack gas analyzes 24.1 ´ 32) = 131.1 ´ 0.77y = 71. 002 + 131.140 PROCESS CALCULATIONS CO2 balance gives: 24.5 = 3.2 + 1. Calculate the stack gas composition.2 = 1.1 ´ 2 = 16 g moles .81 12 = 555. 80% ethane goes to CO2.3 kg 0.206.5 kg lime produces 3.5O2 ® 2CO2 + 3H2O C2H6 + 2.219.6 – 555.934 g moles CO2 formed = 80 ´ 0.8 kg stack gas Hence.4 kmoles CO2 = 1. for 1 tonne or 1000 kg lime.073.4) = 518.073.5 kg 7.2 kg 518.8 kg \ 659.81 kg C gives 44 – 187 – 0.862. Basis: 100 g moles of gas we have.5 187 N2 O2 CO2 (b) Weight of stack gas 2.4 kg of CO2 \ CO2 obtained from limestone = total CO2 – CO2 from coal = (1.29 A gas containing 80% ethane and 20% oxygen is burnt with 200% excess air.5O2 ® 2CO + 3H2O O2 needed = 80 ´ 3.8 ´ 2 = 128 g moles CO formed = 80 ´ 0.206. C2H6 : 80 g moles O2 : 20 g moles C2H6 + 3.6 kg 12 kg C gives 44 kg CO2 Carbon in coal = 187 ´ 0. and 10% remains unburnt.073.5 = 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) = 2.5 kg 44 \ Limestone needed = (a) Lime produced/coal burnt = 659.6 = 3. 10% to CO.2 ´ = 659. the stack gas produced is 4.425 56 Lime obtained from limestone = 518. 1 ´ 3) = 216 g moles C2H6 remaining = 80 ´ 0.56%.05%. H2O : 84. The acid used contains 12% H2SO4 by weight.3 kg of CaCO3 136 84.000 7. Basis: 100 kg of residue Inert : 0.56 ´ 5.207 Total 3. Inert : 0.415 14.12%.5) + (80 ´ 0.8 ´ 3.412 76.53%.1 ´ 2.67 kg of MgCO3 120. The residue from the process had the following composition: CaSO4 : 8.30 Pure CO2 may be prepared by treating limestone with sulphuric acid.23 ´ 18 100 = 6.318 0. CO2 : 0.23 kg CaCO3 + H2SO4 ® CaSO4 + CO2 + H2O 100 98 136 44 18 MgCO3 + H2SO4 ® MgSO4 + CO2 + H2O 84.MASS BALANCE 141 O2 used = (80 ´ 0. 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.53 kg. H2SO4 : 1.8 ´ 3) + (80 ´ 0.51%.1 = 8 g moles Composition of stack gases: Gases g moles mole % CO2 CO O2 N2 H2O C2H6 128 16 556 2. During the process.23 kg of MgSO4 obtained from 5.23%. MgCO3 and inert compounds.56 kg of CaSO4 obtained from 8. MgSO4 : 5.3 = 3.934 216 8 3.56 kg.598 0. MgSO4 : 5.050 5. CaSO4 : 8.3 44 8. The limestone used in the process contains CaCO3.3 98 120.5) = 244 g moles O2 remaining = Oxygen available – oxygen used = (800 – 244) = 556 g moles H2O formed = (80 ´ 0.858 100.3 . 444 + 1.689 – 0.694 kg and CO2 in gas phase = (4.286 + 1.3 Ú H2O in acid = (95.569 kg 27 % (weight %) 73 % (weight %) 47.78 kg Ê 0.263 kg (c) \ 1000 kg limestone.44 mole % Total 6.78) = 106.689 kg Ê 100 84.12) = 4.142 PROCESS CALCULATIONS (a) Limestone analysis Weight.3 Ú Total acid needed = 6.78 – 11.28 kg Residue + vapours = (100 + 6.918) = 86.95 0.67 – Ù = 1.569 kg vapours: H2O CO2 1.05 kg 100 Ø È % Excess acid used = É1.12 kg \ Amount of H2O vapours = (86.263 ´ 100 = 596.67 – Ù kg of H2SO4 = 4.263 kg For 10.05 – Ù = 10. vapours formed = 6.3 Ú 18 Ø È 18 Ø È H2O formed = É 6.204 – 84.174 kg Ê 100 Ú 98 Ø È 3.3 60 3.694 kg 4.3 – Ù kg of H2SO4 = 6.5 kg 10.51) = 1.3 – Ù  É 3.263) = 106. kg Weight % CaCO3 MgCO3 Inert Total 6.5 + 95.) 12% acid supplied = É = 95.3 – ÙÚ  ÉÊ 3.05) = 11.27 kg Ê 84.e.50 100.5 kg limestone.05% Ê 10.67 – Ù = 4.494) = 84.12 ÙÚ 44 Ø È 44 Ø È CO2 formed = É 6.5 .918 kg Ê 100 Ú Ê 84.56 mole % 52.494 Ø (i.67 34.53 5.444 kg Excess acid remaining = 1.3 kg of CaCO3 requires É 6.174 + 4.0 98 Ø È (b) 6. vapours formed = 6.204 kg H2O in residue = 84.263 kg 100 100 Check Limestone + acid = (10.286 kg Total water = water from acid + water formed from reaction = (84.27 = 10.494 kg È 11.05 10.67 kg of MgCO3 requires É 3.51 kg CO2 in residue = 0.444 Ú Total acid used = (10. If 40% excess air is used. O2 : 8.2%.88) = 1.07 ´ N2 leaving = (17.32 A furnace using hydrocarbon fuel oil has a dry stack gas analysis as follows: CO2 : 10.63kg = 3.3%.07 – ØÙ = 2.5 ´ 21 = 21.232 kg of air/kg of coal = 17.38 kg 0.035 kmole 32 16 (a) O2 needed = ÈÉ 0.032) = 13.MASS BALANCE 143 7.5 kmoles Air supplied = 81.79 O2 entering = 81.152 kg = 6.88 kg [by stoichiometry] Ê 12 Ú Ê 2Ú O2 supplied = (2.00 7.16 kmoles 0.19kg = 17.38 – 4.5 = 103.032 = 17. 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 = 81.07 kg = 0.0725 katom H2 = 0.032 kg 4.87 kg = 0.44 wt % 2 O2 remaining = (4.032 – 2.4) = 4.29 wt % H2O produced = 0.41 wt % 12 18 = 0.5%.348 kg = 72. (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 = 0.31 A coal containing 87% C and 7% unoxidized hydrogen is burnt in air.88 ´ 1. O2 and N2 CO2 produced = 0.32 kg 100. N2 : 81.87 – ØÙ  ÈÉ 0.87 ´ 44 = 3.38 Air supplied = (b) Gases leaving are: CO2. H2O.86 wt % Total 18.66 kmoles 79 . 36 kmoles 8. kg Weight % C H 10. HNO3 = 192 kg Total = 1. H2SO4 = 390 kg z = con. 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.3 kmoles O2 consumed = (21.000 kg .20 12.60 = 600 Balance for HNO3 gives.00 Total (c) Air supplied = 103.04 Ú 7.13% (a) % excess air = ÈÉ Ê 13.414 Ø È m3 of air/kg of fuel = É103.64 90.144 PROCESS CALCULATIONS O2 leaving = 8.16 kmoles 22.2 katoms Composition of fuel: Element katom Weight. Calculate the weight of acids needed to obtain 1000 kg of desired acid.23x + 0.36 kmoles (O2 reacted) \ y = 12.36 135.64 122. x = waste acid = 418 kg y = con. 0.2 kmoles (CO2).33 The waste acid from a nitrating process contains 23% HNO3. 57% H2SO4. 0.16 – Ù = 17.57x + 0. Basis: 1000 kg of desired acid Let x kg be the weight of waste acid.27 = 270 Solving.66 – 8.9z = 1000 ´ 0. This acid is to be concentrated to 27% HNO3.64 katoms and x = 10.93y = 1000 ´ 0.3) = 13.40 12.04 100.3 Ø ´ 100 = 62.36 ÙÚ (b) Let the fuel be Cx Hy CxHy + (x + y/4)O2 ® xCO2 + y/2H2O x = 10. 20% water.12 at NTP Ê 135. x + y/4 = 13. 60% H2SO4 by addition of 93% H2SO4 and 90% HNO3.64 9. 2 ´ 418) + (0.7 ´ 233 = 45. The remainder is barium carbonate and infusible matter.3% barium sulphate. IM = (40.68 kg 142 197 = 38.07 ´ 390) + (0.03 kg BaCO3 formed = 27.75 + 41.34 In order to obtain barium in a form that may be put into solution. 20.MASS BALANCE 145 Check (using water balance): (0.13) = 130 7. Fused BaSO4 + IM + Na2CO3 (IM = infusible matter) BaCO3 + Na2SO4 + IM Analysis of product: 11. Basis: 100 kg of fusion mass. the natural sulphate (barites) containing only pure barium sulphate and infusible matter is fused with an excess of pure anhydrous soda ash.35) = 40. 27.3% BaSO4.7 ´ È 45.43 kg 142 BaSO4 initially present = (45.3) = 56. 27.75.43) = 2. (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.35% Na2CO3 Therefore.75 ÙÚ (b) Composition of original barites BaSO4 = 56.75 kg Na2CO3 supplied = (20.3 + 27.45 + 11.7% sodium sulphate and 20.03] = 2. IM = 100 – [56.65% BaSO 4 + Na 2 CO 3 → BaCO 3 + Na 2 SO 4 233 106 197 142 Na2SO4 = 27.22 kg .7 kg Weight of BaSO4 reacted = 27.1 ´ 192) = (1000 ´ 0.68 + 20. Calculate the following: (a) % conversion of barium sulphate to the carbonate. Upon analysis of the fusion mass it is found to contain 11.35% sodium carbonate.45 kg 142 106 = 20.45 Ø ´ 100 = 80% (a) % conversion of BaSO4 to BaCO3 = É Ê 56.7% Na2SO4.22 kg Also.35) = 41. (BaCO3 + IM) = 100 – (11.65 – 38.7 ´ Na2CO3 reacted = 27. total mass of products = total mass of reactants = 100 kg Therefore.7 + 20. 95 2 O2 needed (theoretically) = (7.9825) = 0.36 kmoles .146 PROCESS CALCULATIONS Therefore original barites = 56.3 kmoles Oxygen used for this reaction is = O2 supplied = 10.22 Ø Hence %IM = É ´ 100 = 3.9825 katoms Oxygen used for this reaction = 6.5O2 ® H2O 95% C is converted to CO2 = 7.5O2 ® CO H2 + 0.9% Ê 125.97 kg È 2.75 + 2.8% H2 is burnt with 20% excess air.2 11.35 – 6. Determine the orsat analysis of the flue gas (dry flue gas).2 = 12.75 Ø %BaSO4 = É ´ 100 = 96.82 ÙÚ 7.82) = 15. Basis: 100 kg of fuel oil Component Weight.2% C and 11.95 = 6.35 katoms 5.21 Ø % excess = É ´ 100 = 58.75 ´ È 15.3675 katoms 0.8 7.24% Ê 58.97 ÙÚ (c) Excess Na2CO3: 106 = 25. 95% of carbon is burnt to CO2 and the rest to CO.21 Na2CO3 needed for all BaSO4 = 56.3 ´ 1.3675 = 0.9825 kmole 5% C is converted to CO = (7.35 + 2.95) = 10. kg kmole or katom C H2 88.9 = 2.76% Ê 58.35 ´ 0.90 kmoles Total 100.35 A fuel oil containing 88.22 = 58.18375 kmoles 2 Conversion of H2 to H2O = 5.9 kmoles Oxygen used for this reaction is = 5.82 kg 233 Excess Na2CO3 = (41.0 — C + O2 ® CO2 C + 0. All the Hydrogen is converted to water.97 ÙÚ È 56.03 – 25. 5.9825 12.36 ´ 21 Component mole mole % CO2 6.9 a = 5b By nitrogen balances oxygen supplied = 87.5 + 1.5%.6600 4.9% and rest is N2. CO : 1. Total = 100.95)] = 2. bx = 1.MASS BALANCE 147 O2 remaining = [12.9) = 11.4%.0937 0.9%.9 Ø È by Ø Oxygen reacted.45 CO O2 N2 0.50 56. 9.36 A furnace uses a natural gas which consists entirely of hydrocarbons.24375 46.2%. CO : 1.0% Carbon balance = a + b = (9.78 Ê 4Ú Ê 2Ú Ê 4Ú È ay Ø È 1.3675 2.89 100.9 ax 9.78 Ê 4Ú Ê 2 Ú Ê 4Ú .78 kmoles È ay Ø È bx Ø È by Ø Oxygen reacted = ax + É Ù  É Ù  É Ù = 21.5%.4%.00000 82.5 =5 bx 1.18375 + 2.4.9825 + 0.21 = 23.79 Oxygen left = 1.24375 kmoles 79 = 46. The flue gas analysis is: CO2 : 9. O2 : 1. ax = 9.5 kmoles N2 entering = 12.18 kmoles 0. N2 : 87.0000 Total 7.2 ´ 0. 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 Cx Hy Basis: 100 mole of the flue gas Let us assume the hydrocarbon (HC) to be Cx Hy 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(Cx Hy) + a(x + y/4)O2 ® axCO2 + ay/2H2O b(Cx Hy) + b(x/2 + y/4)O2 ® bxCO + by/2H2O The composition in percentage of the gases are: CO2 : 9. O2 : 1.5 + É Ù  É Ù  É Ù = 21.36 – (6.40 kmoles Oxygen reacted = 21. let us consider the following: (a) Atomic ratio of (c) CO + 0.45% O2 N2 Converter SO 2 + 0.78 kmoles È 0. a = 9.) É Ù (a + b) = 11.45 kmole Theoretical O2 required = 11.975 C 1 (b) For % of excess air.98% \ Excess air = É Ê 22.32.975 = CH4 (methane) 7.975 a + b = 11.45 Ø ´ 100 = 1. O2 : 13% and N2 : 83%.5 x È 0.4) = 45. x = 1 H 3.37 The analysis of gas entering the converter in a contact H2SO4 plant is SO2 : 4%.33 Ê 4Ú \ y(11. Basis: 100 g mole of gas entering the converter Let x g mole SO2 get converted to SO3 SO2 : 4 O2 : 13 N2 : 83 SO2 : 0.5.5x) 4x 100  1.4 – 0.e.4 (from CO and CO2 values) + 11.148 PROCESS CALCULATIONS È yØ (i.9.5O2 ® CO2 One kmole of CO requires 0.95 mole of O2 will be used for converting CO to CO2 Excess O2 remaining = 1.5O 2 → SO 3 x gases leaving (SO3 free basis) x /2 x SO2 ® (4 – x) O2 ® (13 – x/2) N2 ® 83 Total % SO2 in exit = (100 – 1. Calculate the % of SO2 entering the converter getting converted to SO3.5 kmole of oxygen Therefore.33 (for H2O formation) = 22. \ y = 3.45% SO2 on SO3 free basis.4 = 6b.78 ÙÚ Hence.45 Ø ÉÊ Ù of exit gas. 100 Ú .95 = 0. The gas leaving the converter contains 0. Hydrocarbon = C1H3. 0.975 = 3.5 since ax = 9. \ b = 1. 44 + 3.858 Component Total we have CO + 0.904 100. O2 : 0.5) = 30. we get x = 3. If the combustion is 98% complete.5% and N2 : 68%. find the weight and volumetric composition of gases produced per 100 kg of gas burnt.2 + 0.5 ´ 1.5%. CO2 : 3.25% 4 7.57 100 = 89.5 44 3.56 kmole Nitrogen balance 79 = 60.02) = 0. This gas is burnt with 20% excess of the net O2 needed for combustion.MASS BALANCE 149 Solving the above equations.5 ´ 32 = 16 N2 68. kg .7 kmoles O2 remaining = (16.0 28 28 ´ 28 = 784 CO2 3.98) = 27.5 kmoles Net O2 needed = 13.0 — 2.9) = 128.98 = 13.9 kmoles 21 N2 in exit = (68 + 60.7) = 3 kmoles Carbon balance CO2 formed = (28 ´ 0.2 ´ Weight.57 ´ Basis: 100 kmoles of fuel gas mole Molecular weight CO 28.2 kmoles O2 consumed = 14 ´ 0.5 = 14 kmoles O2 available in feed = 0.38 A producer gas made from coke has the composition CO : 28%.5O2 ® CO2 Oxygen balance O2 needed = 28 ´ 0.5 ´ 44 = 154 O2 0.9 kmoles N2 from air = 16.44 kmoles CO2 total = (27. % SO2 converted to SO3 = 3.5 – 13.2 = 16.0 28 68 ´ 28 = 1.94 kmoles CO unreacted = (28 ´ 0.5 kmoles O2 supplied = 13.5 32 0. 04) = 395.7 0.107.40 A laundry can purchase soap containing 30% water at the rate of ` 7 per 100 kg f.89 3. (b) Calculate the weight of Na2CO3 extracted per tonne of black ash.o. The same manufacturer offers another soap having 5% water.84 96.6 per 100 kg of soap what is the maximum price that the laundry should pay the manufacturer for the soap with 5% water? (f.5%.: freight on board) . other water soluble material : 6% and insoluble 52%.359. insoluble : 85% and moisture : 10.00 5.0 1. If the freight rate is ` 0.00 100 kmoles of fuel gas = 2. The solid residue left behind has the composition Na2CO3 : 4%.7 kg 2.39 In the manufacture of soda ash by Le Blanc process.7 ´ 100 = 178.b.34 15.94 44 18.52 ´ 1000) = 0. Insoluble: (0.93 1. The resulting black ash has the following composition: Na2CO3 : 42%.0 71.o.858 kg \ Weight of product/100 kg of feed = 5. kg Weight % CO2 30.88 N2 128.56 28 0.107.00 32 1.21 Total 163.0 26. other water soluble material : 0.40 100.5% Na CO 2 3 Solution y Basis: 1 tonne (1000 kg) of black ash Let x be the weight of residue and y be the weight of solution. Feed I Unit Black ash Water Residue x II Unit Na2CO3: 4% Water solubles: 0.858 7.5 kg 7.31 O2 3.60 CO 0.5% Insoluble: 85% Moisture: 10. sodium sulphate is heated with charcoal and calcium carbonate. The black ash is treated with water to extract the sodium carbonate.637.5% (a) Calculate the weight of residue remaining from the treatment of 1 tonne black ash.90 28 78.150 PROCESS CALCULATIONS Exit gas kmole Molecular weight mole % = volume % Weight.42 ´ 1000) – (612 ´ 0.7 100.b. the factory.85x \ x = 612 kg Na2CO3 extracted = (0. to form CO.0 kmoles Feed Ca3(PO4)2 SiO2 C Total Weight.0 11.25 = cost of 105.72 105.3 33. The amount of silica used is 10% in excess. kg Composition in weight % 1. sand and charcoal.88 – 0.6 100 = ` 10. It may be assumed that in a charge the following conditions exist.25 ´ (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.wt or At.3 .3 7. Charcoal is 40% in excess of that required to combine with P2O5 liberated. I Ca 3 (PO 4 ) 2 + 3SiO 2 → 3CaSiO 3 + P2 O 5 (3 × 60) 310 II 5C (5 × 12) (3 × 116) 142 + P2 O 5 → 2P + 5CO 142 (2 × 31) (5 × 28) (a) Basis: 1 kmole of calcium phosphate = 310 kg Silica charged = 3 ´ 1.6 = ` 0.63) = ` 10.85 » 43 kg 70 Case I: (30% water) water present = 100 ´ Total weight = (100 + 43) = 143 kg 7.1 = 3.63 100 Cost of 100 kg dry soap = (10.25 ´ 100 = ` 9.25 kg 0.25 kg of wet soap Freight charge alone = 105. kmole or katom Mol.25 kg 95 Case II: (5% water) water present = 100 ´ Total weight = 105.0 3.wt Weight. \ Cost of 100 kg wet soap = 10.5 14.4 = 7.0 310 60 12 310 198 84 52.25 7.41 Phosphorus is prepared by heating in an electric furnace a thoroughly mixed mass of calcium phosphate.2 592 100.3 kmoles Carbon charged = 5 ´ 1.88 Cost of 100 kg dry soap including freight at laundry = 143 ´ 5 = 5.MASS BALANCE 151 Basis: 100 kg of dry soap 30 = 42. 152 PROCESS CALCULATIONS (b) Basis: 100 kg of charge: calcium phosphate present = 52.5 = 1.6 kg 310 62 = 6.281) = 0.281  ÈÉ 0.6 ´ 0.3 2.22 ´ 0. which promotes oxidation of the acid.281 0.7) ´ (a) The weight of air supplied per kg of acid.9) = 47. a dry mixture of HCl gas and air is passed over a hot catalyst.0 (c) HCl consumed = 0.6 kg HCl remaining = 0.22 kg of air 146 (b) O2 entering = 1.220 100.5) =146 O2 (2 ×16) = 32 → 2Cl 2 (2 × 71) =142 + 2H 2 O (2 ×18) = 36 Basis: 4 kmoles of HCl º 146 kg HCl O2 Converter N2 Cl2 HCl O2 N2 H 2O (a) Oxygen supplied = 1 ´ 1.3 kg Calcium phosphate consumed = (52.000 0.23 = 0.939 kg Weight of air/weight of acid = Gases Weight.22 – 0. kg Weight % HCl O2 N2 1. Phosphorus produced = (21. Calculate the following.7 42.1495 kg Ê 146 Í Ý . (b) The composition (weight %) of gases entering (c) The composition (weight %) of gases leaving.6 – ÙÚ Ü = 0.07 ´ 142 = 21.3 ´ É Ê 21 ÙÚ 178.3 kmoles È 100 Ø ´ 28.6 kg 142 7.07 kg P2O5 formed = 47.84 = 178. 30% excess air is used.939 45.42 In the Deacon process for the manufacture of chlorine.4 kg Ë 32 Ø Û O2 remaining = Ì 0.0 12.5 kg Weight of air supplied = 1.3 = 1. assuming that 60% of the acid is oxidized in the process 4HCl + (4 × 36.281 kg N2 entering = (1.3 ´ 0. 5 × 32) 160 (2) 4SO 2 + 2O 2 → 4SO 3 (4 × 64) (2 × 32) (4 × 80) (4 × 64) Unit II (Converter) Gases .6 ´ Gases Weight.585 kg 146 36 = 0. Of the pyrites charged 15% is lost in grate and not burnt. Air supplied is 40% in excess of the sulphur actually burnt to SO3. kg Weight % HCl Cl2 O2 N2 H2O 0.2215 100.148 kg H2O formed = 0. N 2 Fe2O 3 FeS2 Basis: 100 kg of iron pyrites Pyrites burnt = 85 kg lost = 15 kg (1) 2FeS 2 + 5.939 0. iron pyrites are burnt in dry air. 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.MASS BALANCE 153 142 = 0.1480 18. (c) The catalyst converts 96% of SO2 to SO3.5 O 2 → Fe 2 O 3 + 4SO 2 (2 × 120) (5.43 In the manufacture of H2SO4 by contact process.5850 0.6 ´ 146 Cl2 formed = 0.00 7.27 6. (d) Calculate the composition (by weight %) of gases leaving the II unit.4000 0. SO2 formed is further oxidized to SO3 by passing the gases mixed with air over hot catalyst. SO2 O 2. Calculate the composition (by weight %) of gases leaving I unit.66 2. Iron is oxidized to Fe2O3.1495 0. Calculate the weight of SO3 formed.0 26. (e) Calculate the overall degree of completion of sulphur to SO3 FeS2 Unit I (Burner) Air SO3.33 6.73 42. 4 kg 240 32 Ø È O2 used for forming SO3 = É 85 – 0. kg Weight % SO2 54.6) ´ 256 = 54.4 – 0.04) = 2.4 = 517.5 × 32) (4 × 80) 160 In Reaction (3) equal weight of oxygen is needed for FeS2.30 O2 47. N2 = 398.72 N2 398.4 kg O2 remaining (entering Unit II) = (119 – 71.0 kg Ê 240 Ú 32 Ø È O2 used for forming SO2 = É 85 – 0.96 – 4 – 64 ÜÝ Í O2 remaining = (47.7 100.3 8.6 – 13.4) = 47.3 kg 240 SO2 produced (Unit I) = (85 ´ 0.97 SO3 45.6 – 5.4) ´ 320 = 45.5O 2 → Fe 2 O 3 + 4SO 3 [Reaction (1) + Reaction (2)] (2 × 120) (7.4 kg Ê 240 Ú Total oxygen used = 71.4 ´ 0.01 Total 545. 85 – 1.176 kg .4 kg (a) Air supplied = (b) SO3 produced (Unit I) = (85 ´ 0.4 – 7.3 + 65.154 PROCESS CALCULATIONS (3) 2FeS2 + 7.5 – Ù = 37. (40% excess oxygen) 0.4 73.6 8.00 320 Ø È (c) SO3 produced (II Unit) = É 54.58 kg Ë 2 – 32 Û = 13.4 9.05 kg (d) O2 consumed = Ì 54.55 kg SO2 remaining = (54.96 – Ù = 65.4 kg.05) = 34.28) = 110.4 – 0.6 kg Gases Weight.23 O2 = 119 kg.5 – Ù = 34.28 kg Ê 256 Ú \ Total SO3 produced = (45. 65 kg 170 HNO3 remaining in cake = 74 ´ 0. (c) Calculate the weight of HNO3 and water vapour distilled from the niter cake. In order that the resulting ‘niter cake’ may be fluid. Na2SO4 formed = 100 ´ HNO3 formed = 100 ´ 142 = 83.32 ´ (a) Calculate the composition of niter cake by weight %. kg Weight % SO2 2.48 kg (H2SO4 + H2O)% in cake = (34 + 1. formed per 100 kg of sodium nitrate charged.3 kg 120 Sulphur in SO3 = 110. 2% of HNO3 formed will remain in the cake.5% H2SO4 required = 100 ´ .400 73. 2NaNO 3 + H 2 SO 4 → Na 2 SO 4 + 2HNO 3 (2 × 85) (98) (142) (2 × 63) Basis: 100 kg of NaNO3 charged.3 7.0 170 kg kg 98 = 57.004 Total 545.000 Gases (e) Sulphur in FeS2 = 100 ´ 155 64 = 53.550 6.706 100.5% water and that the reaction will go to completion.MASS BALANCE Weight.264 O2 34.44 In the common process for the production of nitric acid sodium nitrate is treated with 95% H2SO4.400 SO3 110.580 20.58 ´ 32 = 44.4% 53. % conversion of S to SO3 = 44.5 170 126 = 74.332 N2 398.5) = 35.02 = 1.32 kg 80 100 = 83. It may be assumed that the cake will contain 1. (b) Calculate the weight of sulphuric acid to be used. so that final cake contains 34% sulphuric acid. it is desirable to use excess acid.176 0. 5 + 1.97 1. In such a process.5% cake. BaS is first prepared by heating the barites the natural sulphate with carbon.50 Total 76.00 Components 7. H2SO4.65 kg Total acid to be supplied = 102.50 131.8 kg H2SO4 for the reaction = 57.64 3.38 34. The BaS is extracted from this mass with water and the solution treated with Na2CO3 to precipitate BaCO3. In its manufacture.97 kg Water distilled off = (5. kg Weight % HNO3 H2O 72.45 kg 102.06 100.45 Barium carbonate is a commercially important chemical.97) = 3.12 63.39 – 1. kg Weight % HNO3 Na2SO4 H2SO4 H2O 1.34 = 44.5% (83.64 kg Composition of vapours removed: Weight.45) = 5.00 1. it is found that the solution of BaS formed also contains some CaS originating from impurities in barites.50 44.48 kg HNO3 formed = 126 ´ HNO3 distilled off = 72.5 = 64.39 kg 95% acid needed = Water remaining in cake = 1.12 kg 98 HNO3 in cake = 1.42 95.156 PROCESS CALCULATIONS Niter cake has Na2SO4. HNO3 and H2O \ (Na2SO4 + HNO3) % cake = 100 – 35.75 ´ 0.95 (c) Water in the acid = (107.75 100.98/0.84 – 102.48) = 84.00 Total (b) H2SO4 in the cake = 131.50 4.75 kg (a) Composition of niter cake: Components Weight. The solution .98 kg º 64.84 kg 0. \ Weight of niter cake = 84.45 = 107.645 = 131.80 1.65 = 74.42 kg 57.48 83. 85 – 5.45 kg of dry precipitate is removed from each 100 kg of filtrate collected.28 kg 78 CaCO3 impurity in soda ash = (1.85 kg 197.28) = 0. Na2CO3 H 2O Barites (X) Carbon (CaS impurity) Reactions in (X): Cake Filter Filtrate BaS + Na2CO3 ® BaCO3 + Na2S CaS + Na2CO3 ® CaCO3 + Na2S Molecular weights Na2CO3 : 106.85) = 1.25 kg (2.9%.63 kg (a) Na2CO3 required for forming 14.45 – 14.4.82 ´ 106 = 7.35 kg CaCO3 formed along with Na2S = 1 ´ 106 = 1. Na2S : 78. The analysis of the precipitate is BaCO3 90.1% and CaCO3 9.28 ´ .28 kg CaCO3 = 1. CaCO3 : 100.4.25% and H2O : 90. It is found that 16.82) = 1.00 kg Na2S formed along with BaCO3 = 14. (a) Determine the percentage excess Na2CO3 used above than that required for BaS and CaS.45 ´ 0. BaCO3 : 197. Na2CO3 : 2.9%. Basis: 100 kg of filtrate Precipitate 16.901 = 14.25% in filtrate) Na2CO3 required for 1.96 kg 197. BaCO3 is filtered off.63 – 1.45 kg Amount of BaCO3 in precipitate = 16. (b) Calculate the composition of the original solution of BaS and CaS (c) Calculate the composition of dry soda ash used.4 \ Na2S formed along with CaCO3 = (6.MASS BALANCE 157 is treated with Na2CO3 and the precipitated mass of CaCO3.357 kg 100 Na2CO3 present in filtrate = 2.82 kg Amount of CaCO3 in precipitate = (16.85%. The Na2CO3 for the precipitation used contained CaCO3 as impurity.82 ´ 100 = 1. The analysis of filtrate is reported to be Na2S : 6. CaS : 72.4 78 = 5.82 kg BaCO3 = 14. BaS : 169. 158 PROCESS CALCULATIONS Total Na2CO3 needed = (7. In a beater containing 5000 gallons of pulp it is found that there is lime equivalent to 0. (c) Find the weight of acid.567 kg We have CaCO3 as impurity in Na2CO3 = 0.00 7.28 ´ Weight.357) = 9. .567 0.4 = 12.317 + 2.917 100.1% 9.25 ´ (b) BaS formed = 14.82 ´ 100 = 24.06 2.72 kg 197.35 kg Composition of dry Na2CO3 is shown as follows: Components Weight.5416 100. (a) Calculate the kmole of lime in the beater.46 In the manufacture of straw pulp for the production of cheap straw board paper.9216 kg 100 Components of original solution CaS formed = 1.4 72 = 0.17 0.94 Total 11.9000 12.96 + 1.25 = 11. kg Weight % BaS CaS Water 12. kg Weight % Na2CO3 CaCO3 11. (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.317 kg \ Excess Na2CO3 = 2.01 Total 104.317 169. (d) Find CaSO4 formed in kg (1 gallon = 4.82 87.5 g of CaO per litre.00 Components Total Na2CO3 = Na2CO3 used in reaction + Na2CO3 in filtrate = 9. It is proposed to neutralize this lime with acid of 67% H2SO4.4 litres). a certain amount of lime is carried into the beater.350 97.7200 0.9216 90. 714 g = 26.7 kmoles 100 = 4. (d) CaSO4 formed = 11. CS2 = 1.84 The product gases are found to contain CCl4 33.4% and Cl2 32%.3 kmoles CS2 remaining = 1.4) = 22.47 The available nitrogen content in a urea sample is 45%.000 g = 196.4 ´ (based on the kmole left in the product) .4666 7.000 ´ CO(NH2)2 = Molecular weight 60 Therefore.48 Carbon tetrachloride is made as follows: Purity of sample = 0.67 136 = 26.3%.5 = 11. Basis: 100 kmoles of product gas (a) CS2 reacted = 33.3 CS2 is the limiting reactant and Cl2 is the excess reactant.01) = 198. N2 in urea is = 28 = 0.MASS BALANCE 159 Basis: 5000 gallons = (5000 ´ 4.76 153.4432 = 29. Calculate the following: (a) the percentage of the excess reactants used.000 litres CaO + H 2 SO 4 → CaSO 4 + H 2 O 56 98 136 18 (a) Lime in this solution = 22.43% 0.02 kg 0. S2Cl2 33.2% 33.4 ´ 98 = 19. Find the actual urea content in the sample. (b) the percentage conversion.000 ´ 0.4 kmoles CS2 taken = 34.000 g 11.4 g moles = 198.45 ´ CS2 + 3Cl 2 → CCl 4 + S2 Cl 2 212.43 g moles (b) Acid needed to neutralize = 196.714 kg 56 7. % excess reactant.3%. 1% excess = (196.4432 kg (c) 67% acid needed = 19. CS2 1.43 ´ 1.4666 60 100 = 96. (c) kg of CCl4 produced per 100 kg Cl2 converted.43 g moles Acid used is. 44 0.1 (b) % Conversion: CS2 = 33.0 ´ 100 = 30.44x + 0. The calcination is complete.7 100 = 75.73% 104.84 ´ 7087.0 =131.44 gives. (2) – Eq.9 (c) Cl2 reacted = 99. Analysis of coke C : 76%.87 ´ (a) Compute the analysis of limestone.160 PROCESS CALCULATIONS Cl2 required (theoretical) = 34.0 kmoles \ Cl2 taken = 99. If the limestone contains 10% inert.9 + 32.84 kg CCl4 \7084. ash : 21% and moisture : 3% Basis: 100 kg of limestone containing 10 kg inert.3 ´ 100 = 95.08y = \ y = 55 kg MgCO3. When 100 kg of limestone is calcined 44 kg of CO2 is obtained.87 kg CCl4 212.9 kmoles But Cl2 unreacted = 32.76 kg Cl2 gives 153.91 7.4 .3 CO2 balance gives.7 ´ 3 = 104.6 (3) Eq.91 kg Cl2 gives 153. (b) Calculate the composition of gases leaving the kiln.1 kmoles (100% conversion) Cl2 reacted = 99.52y = 44.91 kg \ Cl2 = 99.0 (2) Eq.3 40. (1) ´ 0.44y = 39.31 kg 7084. calculate the following: CCl4 produced/100 kg Cl2 reacted = 5122.74% 131. 0. 10 kg of above limestone is mixed with 2 kg of coke and is burnt with 100% excess air.97% 34.49 Limestone is a mixture of calcium and magnesium carbonates and inert.9 ´ 212. (a) Let the limestone contain x kg of CaCO3 and y kg of MgCO3 \ x + y = 90 (1) CaCO 3 → CaO + CO 2 100 56 44 MgCO3 → MgO + CO 2 84. x = 35 kg CaCO3 4.44x + 0.76 100 = 72.9 kmoles % excess reactant = 32. (3) 0.91 = 5122. Lime made by calcining the limestone by heating until the CO2 is driven off.9 kmoles = 7084. 1266 0.5 ´ 88 = 5.3 12 32 40. a substantial fraction of the ethylene converted to ethylene oxide. some unconverted.23 Total 1.00333 kmole 18 Total CO2 in gases leaving H2O (3%) = 0.3266 kmole Composition of gases leaving: kmole mole % CO2 O2 (Excess) N2 H2O 0.4090 100.1266 kmole = 0.2532 ´ 79 = 0.3 8. 3.06 = 0. Gases leaving kiln CO2.50 A chemical manufacturer produces ethylene oxide (EO) by burning ethylene gas with air in the presence of catalyst.08 kg 100 From MgCO3 = 5.52 kg Carbon.0033 23.82 = 0. O2. If the conditions are carefully controlled.5 kg MgCO3.9525 0. some completely oxidized to form .82 kg = O2 supplied (100% excess) = (0. N2 and H2O CO2 From CaCO3 = 3.1266 ´ 2) = 0.2532 kmole N2 entering = 0.74 kg 84.3266 0.06 kg moisture.1266 k atom Total O2 CO2 = 8.MASS BALANCE 161 (b) CaCO 3 + C + O 2 → CaO + 2CO 2 100 12 32 56 88 MgCO 3 + C + O 2 → MgO + 2CO 2 84.2 kmole + 0.5 ´ 88 = 3.3 88 10 kg of limestone = 5.00 Gas 7.99 67.18 8.9525 kmole 21 0. 0.5 kg CaCO3.60 0.2 kmole 44 Carbon present = 1.52 kg = 0.06 kg = = CO2 from limestone + CO2 from Carbon (Coke) = 0. and 1 kg inert 2 kg Coke = 1. O2 consumed = oxygen supplied – oxygen remaining = (21.26 g moles C2H4 supplied = C2H4 remaining + C2H4 consumed by Reaction (1) + C2H4 consumed by Reaction (2) = 6.3% salt. they are passed through an absorber in which the ethylene oxide is removed. It is concentrated at a rate of 5.26 + 4. Find: C2H4 converted to C2H4O = 8.53 – 3) = 18.8 = 19.26 ´ (a) the amount of salt crystallized out in the salt box of the evaporator and (b) the evaporation taken place in the system. O2 : 3%. Assume that 4.4 + 8. Of the ethylene entering the reactor.4 g moles Therefore. After the gases leave. Basis: One hour 9. C2H4.46 7. Formation of CO2 is negligible.53 g moles O2 consumed by Reaction (2) = (4.4 = 4. O2.6%.4 g moles of C2H4 unreacted.000 kg/h in a double effect evaporator until the solution contains 80% glycerol and 6% salt. H 2O Reactor Exit Gases CO2. C2H4 : 6. 9.000 ´ 0.13 g moles \ C2H4 converted to C2H4O = 8.4% 19.51 A spent dye sample obtained from a soap-making unit contains 9. O2.46 g moles 100 = 42. O2 consumed by Reaction (1) = 18.005 kg .162 PROCESS CALCULATIONS CO2 and H2O. A typical orsat analysis of the gases leaving the absorber is CO2 : 9.6 = 480 kg 100 Salt = 5.103 = 515 kg Glycerol = 5. what percent is converted to oxide? C 2H 4 Air Reactor CO2.4% and N2 : 81%.8 ´ 3) = 14. N2 EO. N2 Reaction (1): 2C2H4 + O2 ® 2C2H4O Reaction (2): C2H4 + 3O2 ® 2CO2 + 2H2O Basis: 100 g moles of exit gas 6.53 g moles of O2 supplied.6% glycerol and 10.5% glycerol is lost by entrainment.53 – 14.000 ´ Water = 4.6 g moles of CO2 º 4.8 mole of C2H4 converted to CO2 81 g moles of N2 º (81 ´ 21/79) = 21. 6 kg Glycerol remaining = (480 – 21.8 1.06 = 34.25 = 9.875 g moles N2 from air: Exit gases CO2 O2 N2 Total g mole mole % 7.38 kg Check: Vapour + solution + salt = (3.6 = 3.5 g atoms C + O2 ® CO2 O2 needed for the above reaction is 7.375 g moles (25% excess) 9.5 g moles O2 supplied: 7.0 44.4 kg 458.000 kg = Feed 7.000 ´ 0.375 – 7.5 = 1.875 4.2 35.22 kg Salt crystallized = (515 – 34.62) = 5.38) = 80.000 kg of wet material.0 7.645 100.35 = 350 kg Dry material is 1.5 ´ 1.38 + 573 + 480.000 kg of wet substance from 35% to 5%.000 – 350 = 650 kg .375 – 79 = 35.52 Coal with 90% purity and rest ash is burnt with 25% excess air.000 kg/hr Evaporator Solution Salt Final solution: Loss of glycerol = 480 ´ 0.27 g moles 21 O2 remaining 9.22) + 21.4 = 573 kg 0. Basis: 1. Basis: 100 g of coal: carbon present is 90 g = 7.27 79.005 – 80.38 kg \ Solution leaving = Water in this leaving solution = (573 – 458.946.045 = 21.53 Determine the weight of water removed while drying 1.8 Salt in it = 573 ´ 0. Moisture present is 1.4 – 34.62 kg Vapours leaving = Water in vapour + Glycerol in vapour = (4.MASS BALANCE 163 Entrained Vapour 5.6) = 458. Find the analysis of the flue gases.946. Dry material is the ‘Tie element’.38) = 480.5 16. 86R Solving the above.21 kg 0.14R Water balance: 0. Find: (a) CO2 produced (b) acid required per tonne of the ore and (c) the composition of the solid left behind.475 kg of acid and 0.95 So water removed during the drying is 1.000 – 684. Ê 0.493 kg and R = 0. MgCO3 : 300 kg and SiO2 : 20 kg The reactions taking place are: Reaction (i): CaCO 3 + H 2 SO 4 → CaSO 4 + CO 2 + H 2 O 100 98 136 44 18 Reaction (ii): MgCO 3 + H 2 SO 4 → MgSO 4 + CO 2 + H 2 O 84. The ore analyzes CaCO3 : 68%.14 – Ù = 14.94%. 650 = 684.3 kg Let E kg and R kg be the weight of extract and raffinate Acid balance: 0.3 98 120.55 A plant makes liquid carbon-dioxide by treating Dolomite with commercial sulphuric acid.3 44 18 .5% of acetic acid is being separated by extraction in a counter current multistage unit.525 = 0. the final extract composition on a solvent free basis is found to be 82% of acid.164 PROCESS CALCULATIONS Dry material appears as 95% in the exit.21 = 315.507 kg 0.18E + 0. Acid used is 94% pure. The raffinate is found to contain 14% of acid on solvent free basis. Using the solvent in the ratio of 1. Basis: 1 tonne of the ore Weight of CaCO3 : 680 kg. The operating temperature is 24 °C and the solvent used is iso-propyl ether. we have.507 Ø È \ Acid unextracted: É 0.525 kg of water Solvent used = 1.82E + 0. E = 0.475 Ú 7.475 = 0. Find the percentage of acid unextracted? Acetic acid Isopropyl ether Solvent Extraction unit Extract ‘E’ Raffinate ‘R’ Basis: 1 kg of feed contains 0.54 A mixture containing 47.79 kg Therefore the total weight of material leaving is 7. MgCO3 : 30% and rest silica.3 kg/kg of feed. 080 kg Commercial acid required = É Ê 0. 10% of the hydrocarbon remains unburnt. 20% ethane and 10% oxygen.2 Ø = 1.6 kg 84.46.5O2 ® CO + 2H2O C2H6 + 3.2 kg 100 300 – 44 = 156.36%.5O2 ® 2CO2 + 3H2O C2H6 + 2.8 kg Reaction (ii): (b) H2SO4 required during Reaction (i): 680 – 98 = 666.94 ÙÚ (c) The solid residue contains CaSO4.3 Total weight of CO2 produced = 299. Calculate the composition of the flue gas on dry and wet basis.1 kg Ê 84. Basis: 100 g moles of the fuel gas The reactions taking place are: CH4 + 2O2 ® CO2 + 2H2O CH4 + 1.015.8 kg 84.2 + 156. Of the total carbon burnt 90% forms CO2 and the rest forms CO.5) – 10 = 200 g moles .56 A fuel gas contains 70% methane.8 kg Ê 100 Ú 120.3 Total weight of H2SO4 required = 1.3 Ú Silica remaining from the ore = 20 kg Composition of the solid left behind CaSO4 : 67.18% and SiO2 : 1. MgSO4 : 31.015. MgSO4 and SiO2 136 Ø È CaSO4 formed: É 680 – Ù = 924.6 = 455.3 Ø È MgSO4 formed: É 300 – Ù = 428.5O2 ® 2CO + 3H2O Oxygen required for complete combustion = (70 ´ 2) + (20 ´ 3.MASS BALANCE 165 (a) CO2 produced by Reaction (i): 680 – 44 = 299.2 kg Reaction (ii): È 1.% 7.4 kg 100 300 – 98 = 348. The fuel-air mixture contains 200% excess O2 before combustion. 334 kmoles and it appears as 13.1 g moles CO formed: (63 ´ 0.9 180. .257. unburnt = 2 g moles CO2 formed: (63 ´ 0.067 3.9 ´ 3) + (18 ´ 0.257 g moles Methane burnt: (70 ´ 0.5) + (18 ´ 0. 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.072 3.252 0.166 PROCESS CALCULATIONS Oxygen supplied: 200 ´ 3 = 600 mole (200% excess) Nitrogen entering from air = (600 ´ 79/21) = 2.1 ´ 1.9) = 18 mole.0 89.959 81.4% of exit gases.1 ´ 2) = 9.236 0.1 ´ 2) + (18 ´ 0. O2 : 3.334 katoms H2 : 12 kg = 6 kmoles The reactions are: C + O2 ® CO2 H2 + 0.1 ´ 3) = 180 g moles Component gases Weight.00 — — 7.9 g moles O2 used: (63 ´ 0.010 0.5) + (18 ´ 0.9 100.356 14.9 ´ 3.9 2.5) = 184. CO2 : 13.157 100.0 2.57 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.0 2.9 ´ 2) + (63 ´ 0.334 14.5O2 ® H2O CO2 produced will be 7.9) + (18 ´ 0.960.0 0.1) + (18 ´ 0.1 ´ 2.049 76.4%. C : 88 kg = 7.204 0.224 — 6.9) = 63 mole.080 Total (wet) 2.9 ´ 2) + (63 ´ 0. unburnt = 7 g moles Ethane burnt: (20 ´ 0.000 0.1 9.6% and N2 : 83%.9 415.9 ´ 2) = 89. g mole Composition % Wet basis Dry basis CH4 C2 H 6 CO2 CO O2 N2 Total (dry) H2O 7.780.1 moles H2O formed: (63 ´ 0. 7 ´ 8) = 5.4 kmoles Oxygen reacted: (1 ´ 0.5) = 11.73 kmoles Ê 0.7 kmoles CO2 CO O2 N2 Total mole 5.59 Isothermal and isobaric absorption of SO2 is carried out in a packed tower containing Raschig rings.43 82.52 5.3 ´ 8.334 Ø Thus the exit gas moles is É = 54. C8H18 + 12.6 kmoles CO formed by the reaction is (1 ´ 0. What is the analysis of the exit gases (on dry basis)? Basis: 1 kmole of the isooctane.000 litres per minute. Calculate the percentage of SO2 in the outlet water and express it in weight %.MASS BALANCE 167 È 7.5O2 ® 8CO2 + 9H2O.5) + (1 ´ 0. Water is distributed at the top of the column at the rate of 1.334 kmoles È 3.7 56.6 2.334 ÙÚ 7.3 kmoles Oxygen remaining is (15 – 11.425 m3/h.5O2 ® 8CO + 9H2O.8% of SO2.2) = 15 kmoles 79 Ø È Nitrogen coming from air É15 – Ù = 56. The total volume of gas handled at 1 atm and 30 °C is 1.4 3.7 ´ 12.334 + 3) = 10.43 68. The gases leaving the tower are found to contain 1% of SO2.83 100.3 ´ 8) = 2.43 kmoles Ê 21 Ú CO2 formed by the reaction is (1 ´ 0.58 Aviation gasoline is isooctane C8H18.5 kmoles Oxygen supplied is (12. Oxygen required for complete combustion is 12. .134 ÙÚ 12 = 6 kmoles 2 Dry flue gas leaving is = 54. The reactions that are taking place are: C8H18 + 8.3) = 3.84% Ê 10.73 kmoles Water present in the exit gases = Oxygen reacted: (7.5 ´ 1.22 3.0 Gases 7. The gases enter the bottom of the tower with 14.13 mole % 8. It is burned with 20% excess air and 30% of the carbon forms CO and rest goes to carbon dioxide.6 Ø Percentage excess air: É ´ 100 = 34. 4) = 80.281 – 8.4776 – 0.5 kmole of O2 for conversion to CO2 \ 21 Ø È O2 supplied = É 80.9846 kmoles = 511 kg Amount of water flowing = 1.281 ´ 0.148) = 8. N2 in flue gas = 100 – (10. Volume of the gases at standard conditions: 273 Ø È 3 1425 – Ù = 1283.4776) = 48. Basis: 100 kmoles of the flue gases.8034 kmoles È 48.414 ÙÚ SO2 entering absorber: (57.37 kmoles Ê 79 Ú .493 kmole If 99% is equal to 48.9 m ÊÉ Ú 303 È 1283.8034 Ø = 0.60 A pure gaseous hydrocarbon is burnt with excess air.511ÙÚ 7.511 kg 100 Ø È = 0.000 ´ 60 = 60.2%. \ 1 kmole of CO requires 0.e.4 kmoles Let the hydrocarbon be CxHy.000 kg/h Weight of the total liquid leaving the tower is 60.493 = 7. CO : 1%. Weight % of SO2 = É 511 – Ê 60.84%.4% and rest nitrogen. The Orsat analysis of the flue gas: CO2 : 10. 2C + O2 ® 2CO i.4 – Ù = 21.2 + 1 + 8.168 PROCESS CALCULATIONS Basis: One hour. We also know that. What is the atomic ratio of H to C in the fuel? Find the % excess oxygen supplied.9 Ø Amount of gases = É = 57. CxHy + (x/2 + y/4)O2 ® xCO + y/2H2O. The reactions are: CxHy + (x + y/4)O2 ® xCO2 + y/2H2O.4776 kmoles Inert gases are the tie element Inert gases: (57.8034 kmoles then 1% is É Ê 99 ÙÚ SO2 absorbed in the tower: 8. Let a kmole of it get oxidized to CO2 and b kmole of it get oxidized to CO. 2 kmoles of CO requires 1 kmole of O2.281 kmoles Ê 22. O2 : 8. 8 Hence.37 – 8. acetylene.9 t 100  58. F.47 kmoles 7. The hydrocarbon will be C2H2.Xf = D. After thorough mixing and allowing the system to reach equilibrium.4 – 0. 7. F = 100 kmoles/h.37 – 7.65 13.95) + W(0.47 Carbon balance = a + b = 11.1) (4) Solving (3) and (4).1 100 = D + W (3) 1000(0. O2 required = 21.97 Ê 4Ú Ê 2Ú Ê 4Ú Solving. The composition of both the layers have been analyzed and given below: . we get D = 353 kmoles/h W = 647 kmoles/h 7.2 % excess air = CO2 formed = ax = 10.XD + W.MASS BALANCE 169 O2 reacted = (21. Xf = 0. Making a total material balance.62 One hundred kilograms of liquid mixture containing 30% A and 70% B is extracted with a solvent mixture containing C and D. y is approximated to a full number “1” Molar ratio.5 kmoles = 7. Estimate the flow rate of distillate and residue.4) = D(0.2 ay bx by Oxygen consumed.9 = 13. H/C = y/x = 1. we get x = 1 and y = 0.Xw (2) Substituting.5 kmoles \ O2 excess = 8.4) = 12.9 kmoles i. two separate layers are observed. XD = 0.2.4.61 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. CO formed: bx = 1 ax/bx = 10.97 kmoles for forming CO and CO2 For converting CO to CO2 additional O2 required is 0. ax + ÉÈ ÙØ  ÉÈ ÙØ  ÉÈ ÙØ = 12.95 and Xw = 0. F=D+W (1) Making a component balance.e. 7) = x(0.05) + y(0.5 = 53.6) + 110(0. and (iii) composition of C and D in solvent. we get 100(0.5 kg . making component balance for B.6y Solving. and the weight of bottom layer be y from the extractor. 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) 70 = (0. we get x = 80 kg.05) = 48 + 5. (ii) weight of solvent.170 PROCESS CALCULATIONS Layer Component A B C D Top 10 05 60 25 Bottom 20 60 05 15 Estimate (i) the weight of each layer.2) 30 = 0. we get 100(0.1) + y(0.05)x + 0.3) = x(0.2y Similarly. Basis: 100 kg of feed Solvent Feed x Extractor y Let the weight of top layer be x.1x + 0. 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. xC = 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(0.2) + x(0.MASS BALANCE 171 Weight of D = 80(0. 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) (0.5 = 36. it is to be taken as mole %.25) + 110(0. Basis: 1 kmole of incoming gas mixture Since the composition of gas is given in volume %.39y Solving. y = 0.63 One hundred kg of a mixture containing 80% alcohol and 20% water is mixed with another mixture containing 40% alcohol and 60% water.5 kg Since.4) 0.15) = 20 + 16.5 = 0. x = 300 kg Check: by making water balance 100(0.128 kmole Gas to air ratio is 1/0.05 = 0.1x = 30 Solving. 80 + 0.5945 90 36. Estimate the gas/air ratio to be maintained to achieve the composition as 40% in the final air. x = 300 kg 7. Air is assumed to contain 79% N2 and 21% O2 by mole %. 0.4x = 50 + 0.6) = (100 + x)(0.5 = 0. It is desired to have the nitrogen composition as 40% in the final mixture by mixing it with fresh air.8) + x(0. If it is desired to produce a mixture containing 50% alcohol and 50% water.4055 xD = 90 7.35) + y(0. CO2 : 50%.e. Also. estimate the composition of leaving air. the total weight of solvent is 90 kg 53.5) i.5x Solving. estimate the quantity of 40% alcohol and 60% water mixture needed.79) = (1 + y) (0.64 A gas mixture has CO : 10%.5) i.e.128 = 7. O2 : 5% and rest nitrogen by volume.4) = (100 + x)(0.813 . 21 = 0.35 + 0.4432) × (40) + (0. 10% O2.5 0.1 0.82 40. Estimate the fractional conversion of SO2 to SO3 and also the composition of gases leaving catalytic chamber.5912 7.07688 0.005 = 100 – .86 44.128 100.0886) × (28) + (0.0682) × (32) + (0. kmoles mole % CO CO2 O2 N2 0. and rest 85% N2 enters a catalytic chamber where the leaving gas contains only 0.05 + 0. unreacted SO2 = 5 – x Then.128 × 0.5% SO2.4) × (28) = 33. SO3 formed is x and the corresponding O2 reacted is x/2 moles (by stoichiometry) xÛ Ë Balance O2 = Ì10  Ü 2Ý Í xÛ Ë Total amount of gas leaving = 85 + Ì10  Ü + x + [5 – x] 2 Í Ý x 2 Mole fraction of SO2 in the leaving stream = 0.128 × 0.65 A gas containing 5% SO2.32 6.45112 8.172 PROCESS CALCULATIONS y kmole O2 : 21% N2 : 79% 1 kmole CO = 10% CO2 = 50% O2 = 5% N2 = 35% Mixer (1 + y) kmole N2 = 40% Leaving stream: Component Weight. Basis: 100 kmoles of gas entering the chamber The reaction is SO2 + ½ O2 ® SO3 Let x be the kmoles of SO2 reacted So.00 1.00 Total Average molecular weight: (0.79 = 0. 000 0.500 4. H2: 5%. 7.3 The flue gas from an industrial furnace has the following composition by volume.17%. Calculate the percentage excess air used.81% N2: 81. 7.923 86.511/2 = 7. 7. and rest N2.9%. Calculate: (a) percentage excess air. CO2: 11. Assuming complete combustion.744 100.000 kg of desired acid. C2H4: 7%.511 100 - Ê 4.2%. (b) C:H ratio in the fuel and (c) kg of air supplied per kg of fuel burnt. H2O: 9% S: 1% and ash: 5%.489 4. O2: 7.615 7. CO: 0. H2: 0.4 A fuel gas contains CO2: 2%.962 97. estimate the composition of leaving gases. Calculate the weight of acids needed to obtain 10.2 The gas obtained from a furnace fired with a hydrocarbon fuel oil analyses CO2: 10. H2: 41%. producing a gas of 28% CO2.005 x 2 Solving.000 EXERCISES 7.73%.511 = 0. if the loss of carbon in the clinker and the ash is 1% of the fuel used.x = 0. 50% H2SO4 and 25% water. .MASS BALANCE i.2%. N2: 81.744 85. It is burnt with 25% excess air. 173 5.1 The waste acid from a nitrating process contains 25% HNO3.3% H2SO4 and 90% HNO3. 60% H2SO4 by addition of 95. O2: 5%.09% .9%. CH4: 11% and rest N2.511 10 – 4.O2: 6.511ˆ × 100 = 90. Calculate the amount of lime produced per 100 kg of coke burnt and the amount of excess air. 7. This acid is to be concentrated to 30% HNO3. The fuel gas has the following composition by weight: C: 74%. O2: 1%.5 Limestone is burnt with coke having 85% carbon. 5% O2. kmoles mole % 5 – 4. CO: 34%.e.22% Fractional conversion of SO2 to SO3 = Á Ë 5 ˜¯ Gas analysis: Component SO2 SO3 O2 N2 Total Weight. we get x = 4. N2: 1%. 0. is to be separated by fractional distillation into a distillate containing 95 mole % butane.2% sulphur.10 Butane is burnt with 80% of the theoretical air. 3. 4 mole % pentane and rest hexane and a bottom product. A certain limestone analyzes: CaCO3 : 96.9% carbon. 7.70%. ammonia is reacted with air at 650 °C and 7 bar. 7. Estimate the composition of gases leaving.6%.12 A synthetic fuel oil is known to contain only H and C.15 Gypsum (plaster of Paris: CaSO4 × 2H2O) is produced by the reaction of calcium carbonate and sulphuric acid. No sulphur is left in the cinder. 7.41% and inerts : 1. Calculate the C:H ratio in the fuel. The distillate is expected to contain 90% of the butane in the feed.8%.4% oxygen and 0.1% ash by weight is burnt with 25% excess air. oxygen: 18.6 Pure S is burnt in a furnace with 65% excess air. Find the average molecular weight. O2: 2. What is the percentage of excess air? 7. The composition of the mixed vapour is nitrogen: 70%.9% N2. For 5 metric tonne of limestone reacted completely. 7. Calculate the composition of the bottom product. 7. Calculate the analysis of the gases leaving assuming that all H2 present is converted to water.8% and N2: 80. composition of leaving gases in weight % and the density of gases.4% hydrogen. During combustion 90% of S is burnt to SO2 and rest to SO3. Fresh air is supplied at 30 °C and 755 mm Hg.11 Determine the combustion gas analysis when a medium fuel oil with 84.000 kg/h. Gases leave the burner at 800 °C and 760 mm Hg pressure.7% O2 and 83.174 PROCESS CALCULATIONS 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. determine: . MgCO3 : 1. gives on combustion an Orsat analysis of CO2: 2%.89%. Assume complete combustion. 7.5% SO2 and 3% O2 and rest nitrogen on SO3 free basis.9 Pure sulphur is burnt in a burner at a rate of 1. 7. Calculate the % of sulphur fired that burnt to SO3. ammonia: 10% and rest water. Gases from the burner contain 16. Calculate (a) fraction of sulphur burnt to SO3 (b) % excess air over the amount required to oxidize sulphur to SO2. 2. 7.7 In the manufacture of nitric acid.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. 35 mole % pentane and rest hexane.4% SO2.14 A mixture containing 20 mole % butane. 11. MASS BALANCE 175 (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 incoming brine evaporator unit weight % H2O), evaporator is designed to reduce water from an stream from 25 weight % to 3 weight %. If the is to produce 14,670 kg/h of NaCl (along with 3 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? 176 PROCESS CALCULATIONS 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. MASS BALANCE 177 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% 178 PROCESS CALCULATIONS 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 Air dried % 8 20 28.5 43.5 Ultimate analysis Carbon Hydrogen Nitrogen Sulphur Air dried % 81.0 4.6 1.8 0.6 Balance is oxygen 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: MASS BALANCE 179 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 H2 85% 15% Calculate the ratio of air to petrol consumption by weight if the volumetric analysis of the dry exhaust gas is: Composition CO2 O2 CO N2 Volume % 11.5 0.9 1.2 86 Also find the % excess air. 1. 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. A stage may reach when the concentration of these components may cross permissible levels. This is called bypassing operation. This also improves the performance of an equipment as in the case of absorption of sulphur trioxide using sulphuric acid rather than water. The above (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. 8. 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. and is often used in industries to have a closer control in operation. 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 is quite common in the synthesis of ammonia and electrolytic refining of copper. 180 .2 BYPASS Bypassing of a fluid stream is dividing it into two streams. as the solubility is low in pure water.1 8 RECYCLE In industries. this problem can be overcome. 8. and 8. By bleeding off a fraction of the recycle stream. 8.1.Recycle and Bypass 8. This operation is known as purging.2.3) definitions have been shown in Figure 8. 000 kg/h of a 50% benzene and 50% toluene.2 Overall balance F =D+W or.RECYCLE AND BYPASS Bypass Fresh feed Feed Mixing unit 181 Gross product Separator Process Net product Recycle Purge Figure 8.95D + 0.000 kg/h ‘F’ R Distillation column Benzene 95% B 50% B. 5000 = 0. The product recovered from the top contains 95% benzene while the bottom product contains 96% toluene.950 kg/h D . 50% T W 96% Toluene Figure 8. WORKED EXAMPLES 8.000 = D + W Benzene balance gives. Find the ratio of the amount refluxed to the product taken out. 8. bypass and purge.1 A distillation column separates 10. 10.04W Solving. A portion of the product is returned to the column as reflux and the remaining is withdrawn as top product.050 kg/h W = 4.000 kg/h. The stream entering the condenser from the top of the column is 8. D = 5.1 A scheme indicating recycle.000 kg/h Condenser V 10. 080 kg/h \ W = 7.375R Thus R = 7. 8.2 What is the flow rate in recycle stream in Figure 8.680 kg/h Metallic silver may be obtained from sulphide ores by roasting to sulphates and leaching with water and subsequently precipitating silver with copper.000 kg/h \ Crystal leaving crystallizer.e.000 kg/h 20% KNO3 0.080 kg/h F =C+W F = 10.000 = 5.375 KNO3 solution) R 100°F C Crystal with 4% H2O Figure 8. KNO3 entering = 2.5M = 0.6/1.950 kg/h Refluxed quantity R = 0. what is the recycle stream in kg/tonne of product? .4 shown below. C = Overall balance since 8.6 = 0.080 + R) KNO3 Balance gives 0. In the Figure 8. M = C + R = (2.182 PROCESS CALCULATIONS Balance around condenser gives. What percentage excess copper was used? If the reaction goes to 75% completion based on the limiting agent Ag2SO4.3 ¦ 2000 µ § ¶ ¨ 0.96 · = 2. V =D+R or.3 Basis: One hour.000 kg/h C = 2. 0. the material leaving the second separator was found to contain 90% silver and 10% copper.050 + R \ R = 2.920 kg/h Crystallizer balance gives.3 shown below? Water W 300°F Evaporator 50% KNO3 M Crystallizer F Feed 10.6 kg KNO3/kg water (i.96C + 0.584  Actual product D Reflux ratio = 8. e. 10% Cu Ag2SO4 Recycle “R” Ag2SO4 Figure 8. 25% of the limiting reactant (Ag2SO4) was unconverted which goes in the recycle steam.9) 159.300) = R \ R = 433 kg 8. D H = 0. what fraction of dry air leaving is recycled? ‘x’ g dry air A H1 = 0.0152 g of wv/g C B 0.25 (R + 1.099 g H2O/g dry solid 52.5 = 665 kg 312 1.300 t 159.733. Let x g of dry air be recycled.300 or.) 900 kg silver.RECYCLE AND BYPASS 183 CuSO4 Cu S1 Reactor F S2 90% Ag.3 kg 0.5 g dry air (e) Drier Figure 8. 0.5 Basis: 1 tonne of product = 1000 kg product (i.7% 265 We know the reaction is 75% complete.300 t 63.5 Basis: 1 g of dry solid.9 CuSO4 formed = 1. 100 kg of copper Ag2SO4 needed = 900 t 312 = 1300 kg 2 t 107. F = 1. % Excess copper used = 100 ´ Ag2SO4 balance: F = R + 1.562 g H2O/g dry solid .4 In the diagram shown in Figure 8.5 2Ag + CuSO 4 (2 × 107.4 Ag 2 SO 4 + Cu → 312 63.0525 g 2 of wv/g DA 1.5.5 = 265 kg (for forming CuSO4) 312 Total copper supplied = 265 + 100 = 365 kg Cu needed = 100 = 37.25F = R or. 4 = Aw + 33.6 Recycle 33.5 What is recycle.0525 – 0.28 = 0.19 units 8.05P Overall balance: 100 = Aw + P Solving: Aw = 15.0373 Dry air passing through drier (e) = 52. which are recycled.463 g Water removed/g of dry air = 0.3 units.184 PROCESS CALCULATIONS Water removed by drying = (1. The fresh feed enters at 35 °C and 300 atm. and (b) The recycle ratio.463 = 39. The feed to the reactor contains 2 kmoles of H2 for every kmoles of CO.5 Basis: 100 units of feed. x = 33.22 = 13. Material balance for A at  = 20 + x = (100 + x)0. . feed and waste for the system shown in Figure 8.3 + 0. The methanol product is condensed and separated from the unreacted gases.81 units Product.3 = 0.0152 = 0.28 g x 13.562 – 0.0373 g 1. Feed  80% B 40% A Product P A 5% and B 95% ‚ Recycle A only x Figure 8. To produce 6.5 g (between B and C) \ Dry air needed = \ Dry air recycled (x) = 52. P = 84.22 g 0.4 Solving.6? \ Fraction recycled = 8. ‘Aw’ waste 20% A.600 kg/h of methanol calculate: (a) Volume of fresh feed gas. Only 15% of the CO entering the reactor is converted to methanol.6 Methanol is produced by the reaction of CO with H2 according to the equation CO + 2H2 ® CH3OH.333  Feed 100 Material balance for A at ‚ = (100 + 33.5 – 39.099) = 1.253  e 52.3)0. 400 kg .750 ´ 0.675 kg Recycle stream of CO and H2 = 32.600 kg/h Unreacted CO.375 – 206.25) = 3.25 kmoles 32 CO + 2H2 ® CH3OH 15% conversion of CO entering gives 206.50 kmoles \ Total moles of feed unconverted = 3.75 kmoles (a) Volume of feed gas = 618.50 kmoles \ Total moles of fresh feed = 618.506.600 = 206.712.712.5 · = 0.944 (mole ratio) By weight 1.25 0.337.725 kg.506.506.125 kmoles Total amount of CO and H2 CO unconverted (goes into recycle) = CO entering – CO converted = (1.750 kmoles = 4.337.25 µ § ¶ ¨ 3.25) = 1.168.75 kmoles of CO = 32. 2H2 185 CH3OH 6.75 ´ 22.25 kmoles Fresh H2 needed = 412. 2.RECYCLE AND BYPASS 35 °C 300 atm Reactor CO.75 kmoles H2 unconverted (goes into recycle) = (2.5 kmoles of H2 = 4.725 + 4675 = 37.85) = 2. H2 Figure 8.25 kmoles \ Recycle ratio = ¦ 3.375 kmoles = 2.15 and H2 entering the reactor = 1.375 ´ 2 \ CO entering the reactor = = 1.506.16 m3 (b) Amount of gas leaving reactor = (3.7 Basis: One hour of operation Methanol produced = 6.25 + 206.25 kmoles of methanol 206.414 ´ ¦ 308 µ ¦ 1 µ § ¶ t§ ¶ ¨ 273 · ¨ 300 · = 52.50 kmoles Amount of gas recycled = 3.168.25 kmoles Fresh CO needed = 206. 400 µ § ¶ ¨ 44.600) = 44.9 X = F + R (overall balance) (a) 0.65% CO2) R. 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. kmole of kiln gas (8. Limestone K Kiln P X Air F Burner Coke R Figure 8.186 PROCESS CALCULATIONS Products leaving reactor = (37.000 kg \ Recycle ratio = 8. (i. 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. Heat for the reaction is supplied from a furnace burning coke. kmole of gas entering the kiln (7% CO2) X F R Figure 8.85 (weight ratio) Limestone containing 95% of CaCO3 and 5% SiO2 is being calcined.400 + 6. kmole of product gas (8.0865 R (CO2 balance) (b) CaCO 3 → CaO + CO 2 100 56 44 .7 ¦ 37.65% CO2) P.65% CO2) X. kmole of gas recycled (8.8 Basis: 12 kg of coke represent F kmole of flue gas (5% CO2) K. The kiln gas contains 8.e.) moles of gas recycled per mole of gas leaving the kiln. The hot flue gases analyze 5% CO2.000 · = 0.07X = 0.05F + 0.65% CO2. 2 kmoles.05F = 1 kmole CO2 \ F = 20 kmoles (Q 1 kmole of CO2 comes from 12 kg of Carbon) C + O2 Æ CO2 Solving (a) and (b).2 kmoles Let m kmole of CO2 be added in kiln from the calcinations of limestone \ K = 44.8 12 (a) CaO formed/kg of coke burnt = = 3.5 kg R = 5. Recycle 1.0 kmoles CaO formed = (0.07 X) + m = K ¥ 0. Waste 5.0865 or. 1.05% salt Figure 8.10D stream has 500 ppm salt = 0.25% .8 = 45.2 + m)(0. D = 413.000 = B + D Salt balance = (1. B = 586.2 + 0.1 salt % sea water 4% “B”.2 = 0.10 Basis: One hour Overall balance.RECYCLE AND BYPASS 187 Assuming complete combustion 0.8 kmole \ K = 44.8 Find (a) rate of B (b) rate of D (c) recycle R “R”. 0. we have R = 24.5 kg.0865) \ m = 0. 44.000 kg/h A 3.2 ¥ 0.73 R 24.25B + 0.538 = K 45 Sea water is desalinated by reverse osmosis using the scheme shown in Figure 8.2 + m Making CO2 balance.050 Solving.8 kg 44.05% (b) recycle ratio (mole) = 8.000 ¥ 3.8 ¥ 56) = 44. X = 44.1) = 5. (0.25% salt Reverse osmosis cell Desalinated water “D”.07 + m = (44. 000 ´ 0.9 In the feed preparation section of an ammonia plant.000 + R = A Salt balance. and H2 = 300 kmoles . NH3 formed = x ´ 0. 1. H2 x.720 kg and R = 720 kg 8.031 + R ´ 0. 1. A = 1. Enough air is used in partial oxidation to give a 3:1 H2-N2 molar ratio in the feed to the ammonia unit. Fresh Reactor feed NH3 Figure 8. N2 Reactor NH3.188 PROCESS CALCULATIONS Overall balance (At entry to cell).2 ´ 2 = 0. and 3x = 375 kmoles of H2 Unreacted N2 = 100 kmoles. The unreacted H2-N2 mixture is recycled and mixed with fresh feed.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 NH3 Reactor balance gives: 3x. The products from the reactor are cooled and NH3 is removed by condensation.04 Solving. hydrogen is produced by a combination of steam-reforming/partial oxidation process. N2 and H2 Let 3x kmole of H2 and x kmole of N2 enter the reactor Since 20% conversion takes place.0525 = A ´ 0.4x = 50 kmoles \ x = 125 kmoles of N2. On the basis of 100 kmoles per hour of fresh feed determine the NH3 produced and recycle rate. 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. 000 + (0.964.4A = 1000 (grease balance) (2) Solving Eqs.15 ´ 6.464. S = 1. (4) and (5). R and B from Figure 8.464.6435 = 1.29 kg Check B = A + R.: NH3 produced = 50 kmoles Recycle = 400 kmoles 8.12 Grease removed/day = 1.RECYCLE AND BYPASS 189 Ans. A.12 shown below.500 + 1.044. G = 1.01 ´ 4.15B = 0 + 1. 6.29 = 2500 + 4.29 (0. (1) and (2) A = 2. R = 4.29) = 1.29 kg.464.01R (grease balance) Solving.000 kg (i) Overall balance gives: G S 100% 100% Process A 40% Grease 60% Solvent S + G = A (overall) (1) G = 0. B = 6.000 + R (overall) 0.500 kg.000/100 = 1.00. ‘S’ (100% free from grease) Degreasing B (15%) R-1% (recycle) (ii) B = S + G + R B = 1.044. Greased metal Solvent S 0% grease Cleaned metal G Degreasing B 15% Grease solvent Separator A 40% grease with solvent 1% grease R Solvent Recycle Figure 8. if 1 kg of grease/ 100 m2 area is present and the degreased surface per day is 105 m2.6429 (3) (4) (5) .000 kg.10 Find S.500 kg.964.29) 1.964.000 + 0. 03 = 552 kg/h NaCl in feed = 18.87 = 16.190 PROCESS CALCULATIONS 8.400 ¥ 0.400 ¥ 0.400 kg/h. KCl : 21.8%. Calculate the flow rates in kg/h and compute the composition of feed to the evaporator (F) R Fresh feed W F Evaporator P Crystallizer NaCl only KCl only Figure 8.13 Basis: One hour Water in feed = 18.840 kg/h Overall balance: water vapour Feed.168P = 0. 3% KCl and water is fed to the process shown in Figure 8.6% and water. The compositions of the streams are as follows: Evaporator product P—NaCl : 16.13 at the rate of 18.400 ¥ 0. Recycle product R—NaCl : 18.9% and water.11 A solution containing 10% NaCl. F Process KCl NaCl Balance around crystallizer: Overall balance: P = R + 552 NaCl balance: 0.1 = 1.008 kg/h KCl in feed = 18.189R P Crystallizer KCl R . 400 ´ 0. 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%. F. . R = 4.416 kg and P = 4.42) 19.816 Solving. The oxygen for the reaction is supplied from air.72% Similarly.589.816 ´ (100 – 11. m (Conc. 18.72 – 2.6 kg = 19.87 + 4.816 kg/h Recycle. we have 18.008 + 1.589.416 ´ 0. of NaCl) n (Conc. Ethylene oxide is removed from the products completely and the unreacted ethylene is recycled.816 or.12 Ethylene oxide is produced by catalytic oxidation of ethylene and oxygen. m = 11. we have 18.1 + 0.416 = 22.03 = n ´ 22.811 = 22.42% Check: Making water balance. n = 2.968 kg Balance around evaporator: W Evaporator F P NaCl F = W + NaCl + P F = 16. of KCl) Making a balance for NaCl. for KCl.400 ´ 0.6 kg 8. R Fresh feed Feed to evaporator.RECYCLE AND BYPASS 191 Solving.840 + 4.816 kg/h Feed to evaporator = recycle + fresh feed = 18.400 ´ 0.968 = 22. Calculate: (a) inlet and outlet composition of the streams and (b) moles of fresh oxygen required for recycle gases.189 ´ 4.400 + 4.416 = m ´ 22. 62 2.23 = 0.0159 48.230 9.000 Inlet gases Ethylene recycled = (1 – 0.115 kmole 2 Oxygen remaining = 10 – 0.0 10.23 kmole Nitrogen entering: 10 ´ C2H4O formed: 0.115 = 9.62 oxygen required = 0.77 = 0.056 20.560 77.000 . Separator Ethylene Air Reactor Gases Figure 8.885 37.115 mole (c) Outlet gases mole mole % C2H4O O2 N2 0.48 20.81 Total 47.0 37.192 PROCESS CALCULATIONS Reaction: C2H4 + 0.14 Basis: 1 kmole of ethylene.23 = 0.62 100.62 kmoles 21 C2H4 reacted: 1 ´ 0.735 100.77 mole (b) Moles recycled/mole of feed = 0.5O2 ® C2H4O.23) = 0.620 0. Oxygen supplied: 10 kmoles 79 = 37.71 78.384 Total 48.23 kmole 0.885 kmoles Oxygen reacted = (a) mole mole % C2H4 O2 N2 1. 15 Let us assume that the compositions given are in weight %.053 + A. Basis: 100 kg of feed P E Feed.084 ´ 43.4.04 ´ 21. . 40% C4.9006) + (0.3% acid make up acid.0464 kg Since we know that F = E + A. P. we find P = 35.4% C3.9006) + (0. So. gas containing at the rate of by 97.01P + 0. 4% C3.01 ´ 21. A 8% SO3 (rest inert) enters a SO3 absorption tower 28 kmoles/h 98. 1% C4. E = 95% C2. C4 balance: (0.15 find E.0084B C4 balance: 40 = (0. A and B.RECYCLE AND BYPASS 193 8.916B Solving the above.95 = 21.053) + 0. The compositions are: F = 20% C2.053 kg C3 balance: 40 = (0.0464) = 39. 8.6. and observe.99 ´ 35. Also. Solving A is found to be 78.99P + 0.5% of SO3 is absorbed in this tower introduced at the top and 95. find the composition of A. 100 = 21. F A I Unit II Unit B Figure 8. 40% C3.01 ´ 35.7895 Weight % = 50.14 A contact sulphuric acid plant produces 98% acid. we substitute the values of F and E.916 ´ 43.6% C4. 91.95E. Overall balance: feed = 100 = E + P + B.0464) = 39. 1% C4.947 kg Composition of A: C3 balance: (0.1575 Weight % = 49.13 In the diagram shown in Figure 8. the value of E = 20/0.9006 kg and B = 43. C2 balance: 20 = 0.9% acid is used as the Compute tonne/day of (a) make up acid required (b) acid fed at the top of the tower and (c) acid produced.053) + 0. B = 8. P = 99% C3. 236.3 ´ 98 = 5. 0.02 ´ z) or.189.194 PROCESS CALCULATIONS Basis: In one day.52.661.52.3% SO3 Exit r y 95.003.5 kg.48. we get x = 53.7) 4. . (0.08 = 53. Overall balance gives: z = (r + 4.301 kg SO 3 + H 2 O → H 2 SO 4 80 18 98 SO3 absorbed = 4.48.985 = 4.98x + (0.48.271.02 ´ 1.98z = (0.959y = 0.3) Acid balance gives. y.9% Absorption tower z (Acid) SO3 in 28 kmoles/h x 98% Figure 8.973 ´ 1. z + y = x + r Acid balance in this stream gives: 0.507.050.7) + 0.5) Solving the above.236.17 kg 80 97.301 ´ 0.5) – 953. (0.52. y = 49.5 kg. we get r = 1.17 = (0. 0.17 = (0.5) Solving the above. z = 1.16.98x + 0.425.17 = 3.271. z and r as shown.5 kg 80 Water reacted = 4.973r or. gas entering = 28 ´ 24 = 672 kmoles SO3 entering = 672 ´ 0.33 – 953.3 ´ 18 = 953.236.973r + 5.507.98z + 0.236.189.7 kg (overall) Another balance of stream gives.027 ´ 1.16 Let us label x.021r – 953.3 kg Acid formed = 4.959y = 0.76 kmoles = 53.05 kg Check: H2O balance.98 ´ 1.271.76 ´ 80 = 4.507. 4 kg water/kg of dry polymer is dried to 0.2 In a particular drier.25 kg of water per kg of dry polymer per hour.0045 kg of water vapour per kg of dry air and fresh air supplied at a humidity of 0.1 NO is produced by burning gaseous NH3 with 20% excess O2: 4NH3 + 5O2 Æ 4NO + 6H2O The reaction is 70 percent complete.RECYCLE AND BYPASS 195 EXERCISES 8. 100 kg of a wet polymer containing 1. and (b) moles of NH3 recycled per mole of NO formed.011 kg of water vapour per kg of dry air.000 kg of dry air is passed into the drier. 5. and the latter recycled. Calculate the mass rate of fresh air supplied and fraction of air recycled per hour. The air leaving the drier is having a humidity of 0. The NO is separated from the unreacted NH3. . Compute (a) moles of NO formed per 100 moles of NH3 fed. 8. while the DH of a reaction where all reactants and products are at unit activity is represented by DH°.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 enthalpies of substances in standard states are designated by the symbol H°. The activity may be looked upon as a thermodynamically corrected pressure or concentration. In the case of dissolved substances the standard state is the concentration in each instance at which the activity is unity. The standard heat of formation is the heat of formation when all the substances involved in the reaction are each at unit activity. the following terms need to be discussed first.1 Standard State A substance at any temperature is said to be in its standard state when its activity is equal to one. liquids and gases the standard state corresponds to the substances at one atmosphere pressure.1. 196 .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. 9. In order to understand the basic principles pertaining to the generation. the pressure in the standard state is not 1 atmosphere but the difference from unity is not large. The enthalpies of all elements in their standard states at 298 K are zero.1. 9. For pure solids. For real gases.Energy Balance 9 9. transformation and uses of energy. 5 Heat of Mixing When two solutions are mixed.1. The standard heat of reaction for the reaction.e. The standard heat of combustion is that resulting from the combustion of a substance. it is more convenient to calculate the standard heat of reaction directly from the standard heats of combustion instead of standard heats of reaction.2 HESS’S LAW If a reaction proceeds in several steps. D] – [a DH°r. B] where DH°r. (reactants) – SDHc.1. 9. the heat evolved or absorbed during the mixing process is known as heat of mixing. By convention. and this sum in turn will be identical with the heat.3 Heat of Combustion It is the heat liberated per mole of substance burned. 9. the heat of the overall reaction will be the algebraic sum of the heats of the various stages. C + d DH°r.4 The Heat of Reaction Heat of reaction from enthalpy data It is defined as the enthalpy of products minus the enthalpy of reactants. i.e. [DHreaction = SDHc. products]25°C 9. the reaction would evolve or absorb if it were to proceed in a single step.ENERGY BALANCE 197 9. i. 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. A + b DH°r. negative sign indicates that heat is given out and denotes exothermic reaction positive sign indicates that heat is absorbed. with the combustion beginning and ending at 298 K.1. endothermic reaction Heat of reaction from heats of combustion data For reactions involving organic compounds. in the state that is normal at 298 K and atmospheric pressure. i is the standard heat of formation of i th component. . aA + bB Æ cC + dD is given by DH°r = [c DH°r. Silicon: 3.1 kJ 9.3. the following atomic heat capacities are assigned to the elements at 20 °C: Carbon: 1.2 Find the enthalpy of formation of liquid ethanol from the following data: –DH.8 – 393. 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.1 Calculate the enthalpy of sublimation of Iodine from the following reactions and data (a) H2 (g) + I2 (s) Æ 2HI(g) DH = 57. WORKED EXAMPLES 9. Heats of reaction.6 kJ .8.8. 9.4 and all others: 6.8 (3) H2(g) + ½O2(g) Æ H2O(l) Solution: [2 ¥ (2) – (1) + (3 ¥ (3))] = 2C + 3H2 + ½O2 Æ C2H5OH.2.7.4 ADIABATIC REACTION TEMPERATURE Adiabatic reaction temperature is the temperature attained by reaction products. The enthalpy of formation of ethanol = –276. Boron: 2.5 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. Since the heat capacities of solids increase with temperature.0. Oxygen: 4. it is clear that these values do not apply over a wide range of temperature.3 PROCESS CALCULATIONS KOPP’S RULE The heat capacity of a solid compound is approximately equal to the sum of the heat capacities of the constituent elements.2 kJ The desired reaction is I2(s) Æ I2 (g) Solution: (a) – (b) = DH = 67. Fluorine: 5. Hydrogen: 2. Phosphorus: 5.5 (2) C (graphite) + O2(g) Æ CO2(g) – 285.198 9.. kJ (1) C2H5OH (l) + 3O2(g) Æ 2CO2 (g) + 3H2O(l) – 1367. 9.9 kJ (b) H2 (g) + I2 (g) Æ 2HI(g) DH = –9. As per Kopp’s rule. (The melting point is 320. at steady state.ENERGY BALANCE 9. Cp = (6 + 0.78 ª(6 t 320.005T) kcal/kmole °C and T in °C. The heat supply is from a system.3 199 200 kg of Cadmium at 27 °C is to be melted. Latent heat of fusion = 2050 kcal/kmole Basis: 200 kg of Cadmium º 1.9) . Atomic weight of Cadmium = 112.4. Find the quantity of heat to be supplied by the system.9 °C). which supplies 210 kcal/ kg.78 katoms © ¦ ª « ¨ Sensible heat = 1. 005 t 320.190.04. Quantity of steam to be supplied = 9.04.92 µ ¸ ¹ = 3885.5 Calculate the standard heat of reaction: CaC2 + 2H2O ® Ca(OH)2 + C2H2 DHf cal/mole –15. Heat balance: Heat input by steam + heat in.000 kg/h Vapour formed is 5000 kg/h Thick liquor is 5000 kg/h Enthalpy of feed = 10.000 = 5. 9.8 ´ 5.000) \ Ms (540) = 33.20. steam is 540 kcal/kg and that of the vapour is 644 kcal/kg.5 kcal 7534. Thus the weight of steam required.35.5 kcal 2 ¶· ¹º Latent heat of fusion = 1.194 .000 ´ 38. Ms = 6.(540) + 38. Enthalpy of the vapour = 644 ´ 5. Feed = 10.000 kg/h of a solution having 1% solids.1 = 38.4 Basis: One hour. It is to be concentrated to 2% solids.4 –2.75 kg/hr. in vapour + Heat out. thick liquor or.000 kcal. Enthalpies of feed are 38.20. [Ms.8 kcal/kg. by feed = Heat out. Steam at 108 °C is used.1 kcal/kg. Find the weight of vapour formed and the weight of steam used.000 = 32.1 ´ 104] = (32. § 0.43.1 ´ 104 kcal Enthalpy of the thick liquor = 100.88 kg 210 An evaporator is to be fed with 10.317. product solution is 100.000 kcal.000 + 5.0 kcal Total heat to be supplied = 7534.5 = 35.78 ´ 2050 = 3649. Heat supplied by steam = Msls = Ms ´ 540 kcal.000.800 54.000 –68. The feed is at 38 °C. 9.76 ´ 87) – (3.200 PROCESS CALCULATIONS Solution: DHrxn = (–2.21 ´ 10–6/3)(373. 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) = 7.000) – (2 ´ 68.317 (iv) H2 (g) to H2O (g) = 59.59. Q = nòCpSO3 dt At 273.25 kJ/kmole.194) – (–15.35.75 kmole 80 At 373.6 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.33 + 42.5O2 ® 6CO2 (g) + 3H2O (l) (b) C6H6 (l) ® C6H6 (g) .16 – 273.520 (iii) H2 (g) to H2O (l) = 68.7 How many Joules are needed to heat 60 kg of sulphur trioxide from 273.162 – 273.76 kJ/kg K Solution: Q = (m.800) + (54.162)} + {(–13.16 K Q = [{34. Data: Specific heat at 7 °C = 3.86 ´ 10–3/2)(373.t)1 – (mCpt)2 = 1 [(3.86 ´ 10–3T – 13.798 (v) Graphite to CO2 (g) = 94.16 K to 373.16)} + {(42.2 kJ 9.4) The standard heat of reaction is –29.16 K? CpSO3 = 34.16 K.2 cal/mole 9.980 (ii) C6H6 (g) to CO2 (g) and H2O (g) = 7.971.56 ´ 7)] = 302.Cp.317.163 – 273.56 and at 87 °C = 3.21 ´ 10–6T2 J/mole K Solution: Number of moles of the trioxide = 60 = 0.052 Desired reactions: (a) C6H6 (l) + 7.80.33 ´ (373.8 Using the following data of heats of combustion in cal/g mole.509.163)}] Q = 3. 9) = 3. Steps for equation. 9.88 (ii) 2ZnS + 3O2 ® 2ZnO + 2SO2 (iii) 2SO2 + O2 ® 2SO3 –46.5O2 ® 6CO2 (g) + 3H2O (g) (iii) H2 (g) + ½O2 ® H2O (l) (iv) H2 (g) + ½O2 ® H2O (g) (v) C + O2 ® CO2 (g) 9. the materials are at 212 °F.1 (iv) ZnSO4 ® ZnO + SO3 Desired equation: Zn + S + 2O2 ® ZnSO4 kcal/mole Steps: ½ [(ii) + (iii) – 2(i) – 2(iv)] = –233.66.80.10 Steam that is used to heat a batch reaction vessel enters the steam chest.88 55.25.9 Btu/lb we have.2 lb/h 498.78 (212 – 70) = 36.980 cal/g mole.000 \ Q = total heat = 3. Heat loss from the steam chest to the surroundings is 5000 Btu/h.9 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. DHc = – 7.66.ENERGY BALANCE 201 (a) Equation (i) itself gives value. Basis: One hour: Datum 70 °F Btu Reaction absorbs heat = 1000 ´ 325 = 3. At the end of the reaction.000 Btu \ ms = 498. at 250 °C. The reaction absorbs 1000 Btu/lb of charge in the reactor. The charge remains for an hour in the vessel. which is segregated from the reactants.000 Heat in products: 325 ´ 0. how many lb of steam are needed per lb of charge.e. Q = msls = (ms) (734.000 Heat loss to surroundings = 5. is saturated and completely condensed. If the charge contains 325 lb of material and the products and reactants have an average Cp of 0. (b) = (i) + 3(iv) – (ii) – 3(iii) l = 8. The reactants are placed in the vessel at 70 °F.097 cal/g mole Find the heat of formation of ZnSO4 from its elements and from these data: kcal/mole (i) ZnS ® Zn + S 44 –221.2 lb of steam/lb of charge = = 1.000 From steam tables at 482 °F(250 °C) ls = 734.533 325 .48 kcal/mole. (ii) C6H6 (g) + 7.78 Btu/1b °F. ft/s) Ws = (219 – 210.12 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. ¦ § ¨ 'v 2 µ 2 gc ¶·  (200)2 = 0. DH = [2.0446 ∫ C dT = 0.2656 ´ 10–6/3) (1733 – 7733)] Q = –191. Hence.5) = 8.202 PROCESS CALCULATIONS 9.386 ´ 600 + (1. (Ws = shaft work.5 Btu/lb.601 ´ 10–3T – 87.386 + 1.26 ´ 10–7/3}{T23 – T13}] Heat added = 2. Calculate the heat added per kmole Cp = 2. heat is given off 9.83 + 28.886.0446 kmole 22.8 Btu/lb (2 t 32.5 + 0. Btu/lb v: velocity.2656 ´ 10–6 T2.345 kcal.0446 [6. lbf.73 HP 2545 [1 Btu = 778 ft.2 t 778) 200 = 0.601 ´ 10–3/2) {T22 – T12} – {87.26 ´ 10 –7T2 where Cp is in kcal/kmole K and T in K T2 DH = n ∫C p dT T1 = 303 K. Cp air = 6.11 kcal 9. What is the horse power required for the compressor if the load is 200 lb of air/hour? Basis: One hour.762 ´ 10–3 T – 0.5 Btu/lb) to 10 atm and 500 °R (enthalpy 219 Btu/lb). T2 = 523 K T1 n = 1 kmole Or. where Cp is in kcal/kmole K and T in K. 1 HP = 2545 Btu/h] Horse power needed = (8. The exit velocity of air is 200 ft/s.11 Pure ethylene is heated from 30 °C to 250 °C at a constant pressure.83 {T2 – T1} + (28.762 ´ 10 –3 p 773 /2)(1732 – 7732) – (0.13 Air being compressed from 2 atm and 460 °C (enthalpy 210. 1 m3 = 1 = 0.8) ´ .414 173 Q = 0. 85 ´ 900 + (8.15 Calculate the heat input to raise the temperature of 132 kg of CO2 from 100°C to 1000°C.29] = 32.533 ´ 10–3/2)(12732 – 3732) T1 – (2. C 2 H 6   84.577. C 2 H 4  52. kcal/kmole K Basis: 132 kg of CO2 º 132/44 = 3 kmoles T2 (a) DH = n ∫C p dT = 3 [6.000 – 90.7)}(1200 – 298)] = 1.000) – [1 ´ 100 ´ (1200 – 298)] + [{(1 ´ 78. 298 K: 52. C 2H 6 ® C2H 4 + H 2 ' H f.6952 ´ (1273 – 373) = 31.85 + 8.14 Find the heat of reaction at 1200 K.04 kcal 9.576.776.826.37.37.16 SO2 gas is oxidized in 100% excess air with 70% conversion to SO3.280 – (–84.280 kJ/kmole DH°rxn.ENERGY BALANCE 203 9.200 + 97. (a) by integrating the expression for Cp and (b) by using mean heat capacity value Cp in kcal/kmole K.6884 kcal/kmole K Cpav = 11.475 ´ 10–6 /3)(12733 – 3733)] DH = 3[10.7) + (1 ´ 29.702 kcal/kmole K and Cp at 373 K = 9.8 = 1.37. Perform the calculation in the following ways.44.87 kcal Cp values at 1273 K and 373 K are: Cp at 1273 K = 13.533 ´ 10–3 T – 2.8 kJ/kmole (Heat to be supplied) 9. SO2 + ½O2 ® SO3 .475 ´ 10–6 T2. 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. Cp = 6.478.000 kJ DHrxn = DH°rxn – nCpReactants(1200 – 298) + nCpProducts (1200 – 298) = (1.720) = 1.6952 kcal/kmole K DH = òm Cpav dT = 3 ´ 11.720 kJ/kmole ' H f. gases leaving kmoles Cp mean.882.85 + (2.41 — DHrxn = –23.9 ´ 298) – 7.7 kmole SO2 remaining = 0.96) = –21.65 7.20 kcal 450 = 4.013.76 + (6.703 + (0.06 ´ 10–4T) + (13 ´ 10–8T2) Cp H2 = 6.204 PROCESS CALCULATIONS O2 sent = 0. cal/g mole °C SO3 formed = 0. Reaction: DHf /kmole N2 + 3H2 ® 2NH3 ­ ­ ­ 0 0 –10.5 ´ 2 = 1.76 0.76 ´ 7.3 0.3 kmole SO3 SO2 O2 N2 Total 0.75 kcal 450 = 12.45 kcal –16.5 kcal 9.7 ´ ´ ´ ´ 11 7.000 kcal O2 O2 = 1 N2 = 3. Find DCp = 2NH3 – (N2 + 3H2) Da = (2 ´ 6.76 + (3 ´ 6. N2 in air = 1 ´ 79/21 = 3.17 From the following data compute the enthalpy change of formation for NH3 at 480 °C.944.7 = –16.1 ´ 400 = 10.400 kcal SO2 ´ 7.676 kcal N2 SO3 Total = = = = 0.443 kcal Datum: 0°C Heat in Heat out SO2 = 1 ´ 11 ´ 400 = 4.3 11.076 kcal Total DHrxn ´ ´ ´ ´ 450 = 1.490 ´ 0.06 ´ 10–4 + (3 ´ 2.5 0.8 ´ 10–5)] = 0. 490 cal/g mole (given) DHrxn = –23.0119 t 2982 – 3 2 .0 0. DHf at 25°C for NH3 = –10.92 – (–13.1 5.0063T) where T is in K.85)] = –13.76 kmole.76 7.92 kcal.96 Basis: 1 mole of N2 (Feed at 273 K) DHrxn 298 K: (2 ´ –10.519.00 kcal 450 = 2.96 kcal/kmole Cp N2 = 6.574.0119 Dg = (2 ´ 0) – (13 ´ 10–8 + 3 ´ 22 ´ 10–8) = –7.443 kcal \ Heat in –34.0063) – [6.9 t 10 7 t 2983 0.00 kcal Hence. heat absorbed in heat exchanger = –13.65 3.50 kcal = + 20.9 ´ 10–7 ¦ 'C µ 2 ¦ ' H µ 3 DHo= DHrxn – Da T – § ¶ T –§ ¶T ¨ 2 · ¨ 3 · DHo= –21.7 15.193.1 15.5 3.485.8 ´ 10–5T) + (22 ´ 10–8T2) Cp NH3 = 6.9 Db = (2 ´ 0.5 = –18.5 ´ 400= 3.5 7.703) – [7. 5% CO2 and 62. (a) C + O2 ® CO2.9 t 10 7 µ § ¶ 3 ¨ · ´ (753)3 = –3. calculate the theoretical flame temperature for the combustion of this gas assuming theoretical amount of air is used. DHrxn: –26 kcal Also. N2 = ¦ § 62.0119 µ § ¶ ¨ 2 · ´ (753)2 ¦ 7. the combustion reaction is complete and reactants enter at 25 °C.87 kcal 480 °C = 753 K DHrxn. Cp = a + bT + cT2.56 kcal/kmole 9.5 g moles O2 needed 12. b and c are all dimensional constants and available in literature Gas a b ´ 103 c ´ 105 CO2 N2 10.02 –2.5 g moles.04 — C I Combustion Air CO CO2 O2 N2 II Combustion CO2 N2 Basis: 100 g moles of inlet gas CO entering 25 g moles.9 ´ 753) + + \ DHrxn 480°C 'H µ 3 ¶T 3 · ¦ 0.ENERGY BALANCE 205 DHo= 3.55 6. 480°C = DHo + Da T + ¦ § ¨ 'C µ 2 ¶ T + 2 · ¦ § ¨ = 3. DHrxn: –94 kcal (b) C + ½O2 ® CO. cal/kmole K where a.598.87 + (–13.66 2.18 Calculate the calorific value of a blast furnace gas analyzing 25% CO.57% N2.606. CO2 exit = 25 + 12.5 = 37.5 .598.16 1. 12. 52 g moles 21 · CO + ½O2 ® CO2 Reaction (a) – (b) gives DHrxn = –94 + 26 = –68 kcal/kmole Calorific value: Heat given out = 68 ´ 25 = 1. .5 t ¨ 79 µ ¶ = 109.700 kcal Exit gases carry this heat away. 12. 047 t 10 3 ¨ ¦ – § 1.430 – 491 + 15.052] = –39.14 – (4 ´ 1.48 ´ 10–6T2 Da = [3.206 PROCESS CALCULATIONS This gas temperature is called Theoretical flame temperature which is calculated as follows: –17 ´ 105 cal = 37.19 An inventor thinks he has developed a new catalyst which can make the gas phase reaction CO2 + 4H2 ® CH4 + 2H2O proceed to 100% conversion.085 K  2721 K = 2448 °C.483 ´ 10–6T2 CH4 = 3.889 –57.479 kcal Next we find DHrxn at 500 °C (or) 773 K: DH773 = DHo + Da T + ¦ § ¨ 'C µ 2 ¶T + 2 · ¦ § ¨ 'H 3 µ ¶ · T3 .047 ´ 10–3T – 1. DHf CO2 CH4 H2O kcal/kmole –94.415 – (4 ´ 0.078 ´ 10–6T2 H2O = 6.464 ´ 10–3T – 0.889 – (2 ´ 57.41 ´ 10–3T – 4. 9.16 = –35.433 + 4.798)] – [–94.424)] = –14.052 –17.14 ´ 10–3T – 3.891 Db = [18.204 + (2 ´ 6.5 [{10.039) = 11.339 – (4 ´ 6.719 ´ 10–6 T2 We find DHo using data at 298 K DHo = DHrxn – DaT – ¦ § ¨ 'C µ 2 ¶T – 2 · ¦ § ¨ 'H µ 3 ¶T 3 · ¦ = –39.039 ´ 10–3T – 0.415 ´ 10–6T2 H2 = 6.433 – (–14.464) – 10.97) – 6.716 t 10 6 ¨ t t 2982 µ 2 ¶· 2983 µ 3 ¶· DHo = –39.047 ´ 10–3 Dg = [–4.16 ´ 103(T2 – 2982)/2 – 2.433 kcal/kmole of CO2 Cp for CO2 = 6.424 + 1.97 + 3.891 + 11.52[6.719 ´ 10–6 DCp = –14.204 + 18.891 ´ 298) – § 11.483) – {–3.04 ´ 10–5(T3 – 2983)/3] + 109.66 (T – 298) + 1. we have to calculate the heat of reaction at 500 °C).02 ´ 10–3(T2 – 2982)/2] Solving the above equation we have T = 2721.078)}] = –1.41 + (2 ´ 3.339 + 10.48 – (2 ´ 0.798 at 298 K \ DHrxn = [–17. Estimate the heat that must be provided or removed if the reactants enter and products leave at 500 °C (in effect.55 (T – 298)} + 2. 479 + (–14.446 Btu/lb mole DH reactants: (DH1000°F – DH77°F)air + (DH50°F – DH77°F)CO = 3.2 — — — — 77 313.719 t 10 6 t 7733 3 = – 43.2 – 313.1 312. 9.75 lb mole (50% excess) Air supplied = 3.047 t 10 3 t 7732 2 1. °F CO Air O2 N2 CO2 50 125.026 – 392.7) + 1(125.628 = –1.943 kcal/kmole \ 43.612 Btu/lb mole Q = –1.927 Btu/lb mole Heat evolved by combustion = 1.21.21. Calculate the heat evolved from the combustion chamber in terms of Btu/lb of CO entering.354.1) = 23.026 1000 — 6. which is at 1000 °F.ENERGY BALANCE = –35.3 312.745 Btu/lb mole Q = DHrxn + DHproducts – DHreactants O2 remaining = 0.690 5.82(5.2 800 — — 5.21.25(5. Basis: 1 lb mole of CO = 28 lb.2) + 2. N2 : 2.25 lb mole CO2 : 1 lb mole.2 392.891 ´ 773) + + 207 11.82 lb moles DHrxn = –1.3) = 23.984 — — — DHproducts = DH800°F – DH77°F = 1(8.5 lb mole CO + ½O2 ® CO2 O2 supplied = 0.7 315.5 Btu/lb of CO .57 (6.927/28 = 4.745 + 23. The products of combustion leave the combustion chamber at 800 °F.690 – 315.443 – 312. O2 needed = 0.20 CO at 50 °F is completely burnt at 2 atm pressures with 50% excess air.5 ´ 1.943 kcal of heat must be removed.446 – 23.21.984 – 312.57 lb mole and N2 = 2.5 = 0.443 8.2) + 0.82 lb moles Datum: 32°F Data: DH (Btu/lb mole) Temperature. 2000 °F and 3000 °F.64 — 43.63 N2 3.942 Btu DH = – 4.15 9.32 — — 39.44 4.373 –65.35.39 7.18 ´ 123) = – 4.55 28.246 83.15 26.85 10.830 –88.04 — Cp — nCp 12. N2 = 3.895 + 0.76 lb moles Exit: CO2 : 1 lb mole.95 7.05 12.10 6.5 7. if the product temperatures are 200 °F.51 1500 °F 2000 °F — Cp — nCp 11.95 6.21.76 lb moles Assuming a base temperature of 25 oC.3 4.745 Btu/lb mole Gas n Cp nC p CO O2 N2 1 1 3.9 7.13 7.15 9.75 8.810 –21.9 9.7851 ´ 10–3 T – 0.2.76 6.5 11.13 Total 40.75 12. 500 °F.5.3.857 38.911 –1.88 7.229 +9.942 –1.88 29.15 3.0 4.42 — — 46.123 9.8 3.758 1.15 7.22 Coal is burnt to a gas of the following composition: CO2 : 9. O2 : 7.09. CO : 1.5 7.10 26. N2 : 3.21.76 6.0 9.0 26.32 44. O2 : 0.18 \ SnCppr (77 – 200) = –(40.7 ´ 10–7 T2 Cp of O2 = 7. DH = Hp – HR.9 O2 0. Basis: 1 lb mole of CO CO + ½O2 ® CO2 O2 supplied = 1 lb mole. N2 : 82%.0 3000 °F — Cp — nCp 12. 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. (77 oF) and using mean heat capacities.745 + SnCpR (t – 77°) 200 °F 500 °F 1000 °F — Cp — nCp — Cp — nCp CO2 1.5528 ´ 10–7 T2 .85 n SnCp — R SnCp (t–77) Q = DH 38. The reactants are originally at 200 °F.83 — Cp — nCp 10.776 16. Q = DH DH = SnCppr (77 – 200) + DHrxn77 °F + SnCpR (t – 77) Reactants: DHrxn = –1.314 61.95 7.88 — 41.7624 ´ 10–3 T – 0.10 3.5 lb mole.35 27. 1500 °F.95 26.441 –41.85 7.104 + 0. Determine the heat added or removed.05 8.55 7.25 3.208 PROCESS CALCULATIONS 9.21 Pure CO is mixed with 100% excess air and completely burnt at constant pressure. 1000 °F. 76 (all in moles) Gases leaving: CO2–1. CO + ½O2 ® CO2 Basis: 1 g mole CO Temperature of reactants: 200 °F = 93.993 2.736 ´ 10–7 T2 Basis: 1 lb mole of CO2: Multiplying these equations by the respective mole fractions of each component and adding them together.465 – 160.76 (all in moles) \ DHrxn 25 °C = – 67.049 + 1.015 ´ CpCO = 7.ENERGY BALANCE 209 Cp of CO2 = 8. O2–1.23 Calculate the theoretical flame temperature for CO burnt at constant pressure with 100% excess air? The reactants enter at 200 °F. we have for N2 = 0.757 ´ 10–3 T – 21. DH = –2.073 ´ CpO2 for CO2 = 0.3 – 25) 6.2815 t ¨ 10 10 µ (2004 – 5504) 4 ¶· or.612 Btu 9.8024 ´ 10–3 T – 0.382 cal . Gas mole DT Cpm DH = nCpmDT CO 1.6164 t ¨ 10 3 µ (2002 – 5502) 2 ¶· 10 7 µ (2003 – 5503) + 3 ¶· ¦ § 0.2815 ´ 10–10 T3 Cpnet 200 \ DH = ∫C p net dT 550 ¦ = 7.76 (93. O2–0.633 = –2.636) = –70.0 (93.448 + 5.746 + 67.2243 ´ 10–3 T – 2.270 Total 2.3 – 25) 6.092 ´ CpCO2 for CO = 0. N2–3.8 – 0.865 + 0.049 (200 – 550) + § 1. N2–3.3 °C Gases entering: CO–1.746 cal Next we have DHproduct = – (DHreactants – DHrxn) = – (2.2243 t ¨ ¦ – § 2.82 ´ CpN2 for O2 = 0.981 476 Air 4.6 + 13.59 ´ 10–7 T2 + 3 ´ 10–10 T3 Cp of CO = 6.636 cal.5.6164 ´ 10–7 T2 + 0. K 67.460 Ý = 1500 + 39 = 1539 °C º 2798 °F 9. Let the Theoretical flame temperature be 1500 °C.300  68.0 0. since air and gas are at 25 °C. then DT = (1800 – 25) = 1775 °C Gas mole DT Cpm DH CO2 O2 N2 1.400 52.75 ´ 7.25 g moles (150% excess) N2 in feed È 80 Ø = 1 É Ù = 4 g moles Ê 20 Ú 79 Ø È = É1.382  68. Both air and gas being at 25 °C.5 ´ 8.25 – Ù = 4.1.1 ´ (T – 298)] + [8.382 cal.76 1775 1775 1775 12. Data: Heat of formation of CO2 = – 94.5 3.88 ´ 1475) = 68. O2 : 0.24 Calculate the theoretical flame temperature of a gas having 20% CO and 80% N2 when burnt with 150% excess air.640 cal/g mole.460 cal Making linear interpolation for the theoretical flame temperature.5 ´ 2.052 cal/g mole.94 8.92 23.900 This total of –83. Cpm: CO2 : 12.000 7.412 cal/g mole at 25 °C.9. the value calculated.7 g moles Ê 21 Ú Exit gas: CO2 : 1 g mole.31 ´ 1475) + (3. CO + 0. Theoretical flame temperature Ë 70.55 cal/g mole K (from literature) Basis: 1 g mole CO.35 7.9 ´ (T – 298)].55 ´ (T – 298)] + [0. DHrxn = DHCO2 – DHCO = –94.300 cal is not matching with –70.460 Û = 1500 + Ì Ü ´ (1800 – 1500) Í 83. Let the “Theoretical Flame Temperature” be T.7 g moles Q = SHproducts + SHrxn – SHreactants.210 PROCESS CALCULATIONS Let us assume exit temperature as 1800 °C.052 – (–26.7 ´ 1475) + (0. we have. then DT = 1475 °C DH = (1 ´ 12. N2 : 7. O2 : 7. N2 in air . CO = –26.640 = [1 ´ 12.412) = –67. (Datum 298 K) SHreactants is zero.76 ´ 7. N2 : 8.5O2 ® CO2 O2 supplied = (0.7 ´ 7.75 g mole.5) = 1. 83.2 31.03 K ∫ 833.34) = 6.76 + 2.5 ¥ 33 ¥ 902) + (6.700 kJ/kmole.8 32.0 33.100 kJ/mole By iteration method: Let the theoretical flame temperature be 1400 K: DT = (1400 – 298) = 1102 K DHpr = (1 ¥ 50.10.83.110 kJ/kmole π 2.4 30.5 ¥ 2 = 1 kmole N2 from air = 3. DHf CO = –1.1 ¥ 31. 9.6 31.4 50.0 53. N2 : (3.640 = 83.5.6 49.1 ¥ 31.945.03 °C  1106 K ∫ 833 °C.10.31.34 kmoles O2 supplied = 0.2 31.261 kJ/kmole π 2.71 T – 24.4 47.6 ¥ 1102) + (6.88.93.3 33.8 52.600 kJ/kmole Mean molar specific heat.5 ¥ 33. Datum: 298 K N2 in feed = 70/30 = 2.3 32.76 kmoles Exit gas consists of: CO2 : 1.6 32.4 ¥ 902) + (0.700) – (–1. O2 : 0.0 34.8 ¥ 1102) + (0.83.3 30.93.600) = –2.8 ¥ 1102) = 2.ENERGY BALANCE 211 67.6 \ T = 1106.25 Find the theoretical flame temperature of a gas containing 30% CO and 70% N2 when burnt with 100% excess air.2 ¥ 902) = 2.100 kJ/kmole Let the theoretical flame temperature be 1200 K \ DT (1200 – 298) = 902 K DHpr = (1 ¥ 49. DHf CO2 = –3. kJ/kmole K at different temperatures is given below: Temperature K CO2 O2 N2 800 1000 1200 1400 1600 1800 45.6 34.100 kJ/kmole . The reactants enter at 298 K.1 kmoles Let us consider the equation Q = SHproducts + DHrxn – SHreactants where SHreactants = Zero at 298 K (Q Datum is 298 K) \ DHrxn = (–3.7 Basis: 1 kmole of CO. 78 = 521.12 ´ 425) – (6.32 ´ 425) – (6.3 [(7.2 ´ 0.32 7.92 [(11 ´ 425) – (10 ´ 30)] + 80.849.31.83.12 The amount of gas mixture = 15.261  2.110 Ý So.000 22.117 + 3.20 g moles Reference temperature: 0 °C \ Q = 66. 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 N2 Cpm 30 °C 10 6.000 lit of a gas mixture at standard condition is as follows: SO2 : 10%.8 ´ 30)] = 19.414 = 669.14 ´ 10–3 T – 3.22 cal 9.1 = 66. Both air and gas are initially at 25 °C.98.98 [(7.2 g moles we can then write the amount of each component SO2 : 669.100  2.005 ´ 10–6 T2 Cp N2 = 6.27 Estimate the theoretical flame temperature of a gas containing 20% CO and 80% N2 when burnt with 100% excess air.2 ´ 0.636 kcal .96 ´ 30)] + 521.12 = 80.96 6.457 + 1.389 ´ 10–3 T – 0.80 Cpm 425 °C 11 7. 9.167 ´ 10–3 T – 1.26 The analysis of 15.31.000 litres º 15.98 g moles 669.212 PROCESS CALCULATIONS So theoretical flame temperature lies in between these two values (by interpolation ) \ Theoretical flame temperature Ë 2.339 + 10.92 g moles O2 : 669. Cp CO2 = 6. the temperature of the exit gases is 1382 K = 1109 °C. O2 : 12% and N2 : 78%.30 g moles N2 : 669.069 ´ 10–6 T2 The values of Cp are in kcal/kmole K and temperature is in K DHrxn 25 °C = –67.110 Û = 1.88.415 ´ 10–6 T2 Cp O2 = 6.2 ´ 0.200 + Ì Ü ´ (1400 – 1200) Í 2. 76 kmoles Exit gas: CO2 : 1. H2O : 40.167 ´ 10–3 T 3  2983 T 2  2982 – 1.28 Dry methane and dry air at 298 K and 1 bar pressure are burnt with 100% excess air.9.117 ´ 2 3 + 3.19.005 ´ 10–6 6 4 + 7. CO2 : 51.069 ´ 10–6 9.76 ´ 0. Datum: 298 K CH4 + 2O2 ® CO2 + 2H2O \ Oxygen supplied = 2 ´ 2 = 4 g moles 79 = 15.636 = [1 ´ òCpCO2 ´ (T – 298)] + [0. N2 : 7. O2 : 2.457 ´ (T – 298) + 7. Data: Cpm values (J/g mole K) for the components are: O2 : 31. N2 entering = 4 ´ Heat given out = 802 kJ Heat loss = (802 ´ 0.415 ´ 10–6 T  298 T 3  2983 + 6.76 ´ 6.2) = 160.5 ´ òCpO2 ´ (T – 298)] + [7.14 ´ 10–3 T 2  2982 2 – 3.79.6 kJ .76 ´ 1.05 g moles.05 g moles 21 Gases leaving are: CO2 : 1. O2 : 0.4 kJ \ Q = Heat in exit gases = (802 – 160. Basis: 1 g mole of methane.339(T – 298) + 10. N2 = 4 kmoles Air supplied: O2 : 1 kmole.76 kmoles Datum 25 °C = 298 K DHrxn = –DH products + 67. Determine the final temperature attained by the gaseous products if combustion is adiabatic and 20% of heat produced is lost to the surroundings. and N2 : 15.ENERGY BALANCE 213 Basis: 1 kmole of CO. The standard heat of reaction is –802 kJ/g mole of methane. H2O : 2.76 ´ òCpN2 ´ (T – 298)] 67.389 ´ 10–3 T 2  2982 2 T 3  2983 3 Solving for theoretical flame temperature = T = 1216 K = 943 °C – 7.4) = 641. N2 from air : 3.636 = 6.15. N2 : 32.5. 00 \ FeS2 reacted = (0. The reaction is given.35 Total 0.029) + [0.29 An iron pyrite ore contains 85% FeS2 and 15% gangue.9)] = –6.99 75.821 kg = 6.821 – 479.9) + (3. Solving the above.9 Basis: 1 kg of ore containing FeS2 = 0.15 kg Let x kg of FeS2 be in the solid waste.85 ´ 10–3 kmole Fe2O3 formed = 0.15) + 0.424 ´ 10–3 kmole 512 Û Ë = 1. It is roasted with 200% excess air to get SO2.029 0.15) + (0.029 kg Solid waste = (0.85 kg and gangue = 0. \ T = 1242 K = 969 °C.029) = 0. 319.85 ´ 10–3 ´ 177.666(0.9 ´ (T – 298)] + [15.04{(0.85 – 0.66 3.19 ´ (T – 298)] + [2 ´ 31.55 kJ .068 kg = 0.85 – x) kg. we find x = 0.85 – x) + x}.9 0 –88.726 100.016688 kmole SO2 formed = Ì 0.214 PROCESS CALCULATIONS Q = [1 ´ 51.4 ÜÝ Í = 0.79 ´ (T – 298)] + [2 ´ 40.424 ´ 10–3 ´ 822.4 ÜÝ Í Heat of reaction = –[(0.2 –296.547 kg = 3.85 – 0.05 ´ 32.8275 ´ (T – 298)] = 641.4 Û Ë then.4 352 319.85  x ) – 479.6 ´ 103 J.725786 kg A summary of the composition of the solid waste is given: Solid waste kg Weight % Gangue FeS2 Fe2O3 0.4 512 –177. FeS2 reacted: (0.3) –(6.547 20. Heat of formation data is in kJ/g mole.016688 ´ 296.666(0.029)] = 0.150 0. Determine the standard heat of reaction in kJ/kg of ore roasted and the analysis of the solid waste. 4FeS2 Weights Heats of formation (kJ/g mole) + ® 11O2 2Fe2O3 + 8SO2 479.15 ´ (T – 298)] = [679.85 – x) x = 0. Fe2O3 formed Ì(0.666 ´ (0. All the gangue plus Fe2O3 end up in the solid waste produced which analyzes 4% FeS2. 9. 05 DH° at 25 °C = SDH°f. DH = –393.31 Estimate the standard heat of reaction DH°298 for the reaction. 298 for C = –542000 cal/g DHc. (3)/2.5 kJ (4) 2 2 1 CO2 + 3HCl ® CHCl3 + O2 + H2O.ENERGY BALANCE 215 9. reactants = [–337. Products – SDH°f.8 –195.2 –337.30 For the following reaction.93 kJ 2 1 H2 + O2 ® H2O. DH = –509. DH = –296 kJ 2 C + O2CO2. 298 = SDH°c.93 kJ (1) 2 1 H2 + O2 ® H2O. DH = –296 kJ (2) 2 (3) C + O2 ® CO2.05] – [–269.8 – 195.3 –29. reactants – SDH°c. 298 for B = –212000 cal/g DHc. gives 1 1 1 3 H2 + Cl2 + C + O2 ® 3HCl + CO2 2 2 2 2 .78 kJ Eq.3 – 29. DH = –393. DH = –509. (4) × 3 + Eq. estimate the heat of reaction at 298 K.32 Calculate the heat of formation of CHCl3 from the following data: 1 CHCl3 + O2 + H2O ® CO2 + 3HCl.65 kcal 9. A+B®C Standard heats of combustion are: DHc. 298 for A = –328000 cal/g DHc. products = [–328000 – 212000] – [–542000] = 2000 cal 9.2] = 98. DH = –167. A+B®C+D Compound DH°f (kcal/g mole) A B C D –269.78 kJ 1 1 H2 + Cl2 ® HCl. 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. DH°f.500 cal/g mole SO2 = –70.4}  0.33 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.1 T2 2 = –30. 298 = DHFe2O3 + 4 [DHSO2] – 2[DHFeS2] = –19.000 + 0.6500 + 4[–70. DH°r. 298 for the reaction 2FeS2 + 1. .520] = –39.520 cal/g mole Fe2O3 = –1.05 × (6002 – 3002)] = –43.5300 cal 9.1 T ] dT 300 = –30.34 The heat of reaction at 300 K and 1 atm pressure for the reaction A + 3B ® C is 30.4T – 0. Cp data is as follows: A = –0.4 + 0.dT 300 600 = –30.960] – 2[42.000 + 0.960 cal/g mole From the reaction.5O2 ® Fe2O3 + 4SO2 The standard heats of formation are: FeS2 = –42.4[600 – 300] – [0.216 PROCESS CALCULATIONS 9.96.1 T (T in K) B = 10 C = 30 Calculate the heat of reaction at 600 K and 1 atm A + 3B ® C 600 DH600 = DH300 + ± ' C p .000 + ± [{30  3 × 10 + 0.000 cal/mole A converted. Calculate the standard heat of reaction.380 cal/g mole of A 9. 05 + 80 = 174.137 kJ/g mole CO2 = 50. {20 [50.ENERGY BALANCE 217 DH°f Cpm CO2 = –393.05 [31.835 × 20 = 5654.402 kJ/g mole O2 = 33.02] + 174.16] + 15 [33.05 g moles Gases leaving CO2 = 20 g moles O2 = 25 – 10 = 15 g moles N2 = 94.56]} (T – 25) = 5654700 Solving. –DHr = 1.760 cal/g mole .835 kJ/g mole Heat of reactants is zero (Reference temperature) Heat produced when 20 g moles of CO is burnt = 282. The reaction is complete.402 + 1 × 0] 2 = –282.91.16 kJ/kg K CO = –110. Air and methane enter at 298 K and a pressure of 1 atm.02 kJ/ kmole H2 = 31.700 kJ Heat in outgoing gas.1 Determine the theoretical flame temperature that can be attained by the combustion of methane with 20% excess air.05 g moles Atmospheric temperature = 25 °C Heat in reactants + DHR = Heat in products Standard heat of reaction.137 – [110. DHf°products – DHf°reactants = –393.56 kJ/kmole K Basis: 100 g moles of feed CO = 20 g moles 1 CO + O2 Æ CO2 2 1 O2 needed = × 20 = 10 g moles 2 O2 supplied = 2.5 × 10 = 25 g moles N2 supplied = 25 × 79/21 = 94. T = 834 °C EXERCISES 9. R is Gas constant. Cp = R (A + BT + CT2) cal/g mole where.97. B and C are constants and Cp is the heat capacity at constant pressure.081 19. 30 mole % O2 and 3 mole % N2. Component CH4 C2 H 6 9.218 PROCESS CALCULATIONS Mean heat capacities (cal/g mole K) Component 9. If the conversion of HCl is 75% and the process is isothermal.1 12.0 7.164 –5.000 kJ/kmole SO3 : –3.131 9. .4 10.000 kJ/kmole 9.95 H2O 10.2 Temperature 2000 °C 1800 °C CO2 13.4 8. A.98 kJ/kmole K Standard heat of formation for SO2 : –2.67 kJ/kmole K SO3 : 30.3 Calculate the enthalpy change in J/kmole that takes place in raising the temperature of 1 kmole of the gas mixture of 80 mole %.25 O2 8.561 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.225 –2. Methane and rest ethane from 323 K to 873 K Heat capacity equation.702 1.9 Determine the heat of reaction at 720 K and 1 atm for the reaction SO2 + 0.4 Value of constant in Cp equation A B ´ 103 C ´ 106 1.5O2 ® SO3 Mean molar specific heats of SO2 : 51.5 kJ/kmole K O2 : 45.3 N2 8. how much heat is transferred per mole of entering gas mixture.95. T is temperature in K. The lime stone is charged at 25 °C. Data: Standard heat of formation at 25 °C and 1 atm.16 9.818 J/g mole (b) Mean heat capacities: (cal/g mole K) HCl(g) : 7.52 Cl2(g) : 8.88. J/g mole HCl(g) : –92.000 cal/g mole 1.7 kcal/g mole.41.66.050 cal/g mole Mean molal heat capacity. CO + ½O2 ® CO2 . cal/g mole CaCO3 MgCO3 CaO MgO CO2 : : : : : – – – – – 2.307 J/g mole H2O(g) : –2.0 In the reaction 4FeS2(s) + 11O2 (g) ® 2Fe2O3(s) + 8SO2(g) the conversion from FeS2 to Fe2O3 is only 80% complete.2 10.7 Calculate the theoretical flame temperature for CO burnt at constant pressure with 20% excess air.450 cal/g mole 2.900 cal/g mole 1. 9.0 10.43.51.06 O2(g) : 8. If the standard heat of formation for the above is calculated to be –197.5 25. The lime is withdrawn at 900 °C and the gases leave at 200 °C.6 : : : : : : 8. The reactants enter at 366 K. 11% MgCO3 and 9% water.61 N2(g) : 7.ENERGY BALANCE 219 Data: (a) Standard heat of formation at 25 °C. what –DH°reaction should be used in energy balance per kg of FeS2 fed.840 cal/g mole 94. cal/g mole K H2O CO2 CaCO3 MgCO3 CaO MgO 9.0 14.5 Calculate the number of joules required to calcine completely 100 kg of limestone containing 80% CaCO3.0 23.54 H2O : 7. 813 9. 9.762 ´ 10–3 T – 0. . The Cp values in kcal/kmole K.13 kJ/g mole. 7.92 for CO2. 54.48 ´ 10–6 T2 for ammonia Cp = 6.017 for CO.93. 29.5 for O2 and 33.520 Carbon dioxide (g) = –3.41. assuming complete combustion.11 CO is burnt under atmospheric pressure with dry air at 773 K with 20% excess air.9 – 0. Air and methane enter at 298 K and a pressure of 1 atm. 9. Standard heats of formation at 298 K in kJ/kmole are: Methane (g) = –74. 7. 11.2656 ´ 10–6 T2 Cp is in kcal/kmole K and T is in K.28 for air.507 for N2 in kcal/kmole K.225 for air.81 ´ 10–3 T – 1.23 for CO.18 for CO2. Calculate the heat removed. 34.500 Water vapour (g) = –2.941 for O2 and 7.5 ´ 10–6 T2 9. The reaction is complete.386 + 1.8 A gaseous mixture of 1000 m3 containing 60% hydrogen and 40% ammonia is cooled from 773 K to 313 K at 1 atm pressure.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 = 6.220 PROCESS CALCULATIONS The heat capacities are 29. Standard heat of reaction: –283.636 kcal Mean specific heats are: 7.2 ´ 10–3 T + 0. and T in K are: for hydrogen Cp = 6.1 for N2 in J/g mole K. The products leave at 1223 K. Data: – DH298 K = – 67.9 Determine the theoretical flame temperature that can be obtained by the combustion of methane with 25% excess air.08 + 8. Calculate the heat involved in the reaction chamber in kcal/kmole of CO burnt. The mixture is stirred well and the product leaves the tank at a rate of 500 kg/min.1 A storage tank contains 10.000 kg of a solution containing 5% acetic acid by weight. 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. X F o. At what instant of time the acid concentration in the tank will drop to 1% acetic acid by weight? After one hour of operation. Although unsteady-state processes are difficult to formulate.10 Problems on Unsteady State Operations The term unsteady state refers to chemical processes in which the operating conditions generally fluctuate with time. A fresh feed of 500 kg/min of pure water is entering the tank and dilutes the solution in the tank. what will be the concentration in the tank? F iX i M. X o = X 221 . WORKED EXAMPLES 10. 000 É Ù = 500 ´ 0.222 PROCESS CALCULATIONS Here inlet flow rate.05X dt (i.e. (1) Let X be the concentration of acid and M the mass of solution of acid at any time. 10. at any time in the tank = Initial mass + (Inflow rate – Outflow rate)(time) = 10.000 kg We know that. Fi = 500 kg/min Outlet flow rate.000 kg The total mass M. Fo = 500 kg/min Initial mass = 10. rate of generation is zero.05 ± dt È XØ ln É Ù = –0. rate of accumulation = rate of input – rate of output or. (i. rate of input + rate of generation = rate of output + rate of accumulation Here.0 – 500X Ê dt Ú (2) (3) Here. Hence.) dX = – 0.) dX = – 0.05t (6) .e.000 + (500 – 500)t = 10. d ( MX ) = Fi X i – F o X dt M dX dM X = Fi X i – F o X dt dt dM = 0 (Since inlet and outlet flow rates are the same) dt dX Therefore.05dt X (5) Integrating. we get X t Xo t0 dX ± X   0.05(t – 0) Ê Xo Ú = –0. M = FiXi – Fo X(4) dt È dX Ø Now substituting values. (1) Let X be the concentration of acid and M the mass of solution of acid at any time. rate of accumulation = rate of input – rate of output d ( MX ) = F i Xi – Fo X i. X = e–0.2 A tank contains 10 kg of a salt solution at a concentration of 2% by weight. Fo = 1. M = 10 + 0.249% i. Fresh solution enters the tank at a rate of 2 kg/min at a salt concentration of 3% by weight. (7).e.05 (60) = 0. dt dM dX +X = F i Xi – Fo X or.5% by weight? Here inlet flow rate. t = 32.05 ÙÚ i. (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. rate of generation is zero. the concentration in the tank will be 0.05t ln ÈÉ Ê 0. we get X = 0.5t Differentiating. at any time in the tank = Initial mass + (Inflow rate – Outflow rate)(time) = 10 + (2 – 1. Hence. Fi = 2 kg/min Outlet flow rate.5 kg/min.5 dt We know that rate of input + rate of generation = rate of output + rate of accumulation Here. M dt dt (2) (3) .5 kg/min Initial mass = 10 kg The total mass M. 0.05 e–0.00249 = 0.PROBLEMS ON UNSTEADY STATE OPERATIONS Therefore. The contents are stirred well and the mixture leaves the tank at a rate of 1.e.05t Xo 223 (7) Time taken to reach a concentration of 1% is given by.5)t or.e.01 Ø = –0. after one hour of operation. we get dM = 0.249% 10.19 minutes (b) Substituting for t as 60 minutes in Eq. 06 – 2X (10 + 0. (4) (5) Integrating. dX + 0.) or.5t (10 + 0.06  2 X 10  0. by substituting values.02 ¸  2 X ¹º dt ± 10 .5t) dt dX dt 0.03 – 1.5 dt Therefore.06  « ± Xo X also.5X dt dX = 0.5t) (i.224 PROCESS CALCULATIONS dM = 0.  0.5X = 2 ´ 0. we get X dX © ª 0. we get Here.e.5 ± X o  0. 5t t0 t dX dt 2 ± 20 . 0. 01 Ì Ü Í 20  t Ý \ (6) (7) 4 (8) Time taken to reach a concentration of 2.03 Û ln Ì Ü Í 0.5t ÜÝ Ë 0. t = 3.03 Ý X  0. which is (10 + 0.5t) or.03  X Hence. (20 + t)4 = 204 ´ 0. Hence.01 0.02  0.5t Ü Í Ý (9) . (4).5X dt Û dX Ë 2 X Ì dt Í 10  0.5% is given by substituting X = 0. t X  0.03 – 1.03 0.784 minutes Aliter We shall go back to Eq. t t0 Ë (20  t ) Û  4 ln Ì Ü Í 20 Ý Ë 20 Û Ì 20  t Ü Í Ý 4 Ë 20 Û X = 0.03 – 0.5X = 2 ´ 0.01 or.025 in Eq. (7).06 Û Ì 10  0. dX + 0. we have.03 Ë X  0. (11).12 Ô 20  t dt dt + constant e 20  t (12) X ´ 4 ´ exp[ln(20 + t)] = X(20 + t)4 = Ë 0. P and Q are either functions of x or constants This equation is of the form 225 (10) The solution for this differential equation is ye òPdx = òQe òPdx dx + constant (11) Using the same analogy we can solve Eq. (9) in the following manner P= 2 10  0.12 Û Ô ÌÍ 20  t ÜÝ [20 + t]4dt + constant or.06 10  0. we get 4 X e 4 Ô 20  t dt Ô 0.5t 0.12 Û Ô ÌÍ 20  t ÜÝ ´ 4 ´ exp[ln(20 + t)] dt + constant Ë 0.01 ´ (20)4 = –1600 Equation (14) thus becomes. (14).5t 4 20  t Q 0. we get. Constant = – 0.02 Substituting in Eq.12 [20  t ]4 + constant 4 X = 0. X = 0. X = 0. X(20 + t)4 = Ô [0.12][20 + t]3dt + constant X(20 + t)4 = 0.03 + constant (13) (14) (20  t )4 Initial conditions are: t = 0.PROBLEMS ON UNSTEADY STATE OPERATIONS dy + Py = Q dx Where.03 – 1600 (20  t )4 and (15) .12 20  t and Substituting for P and Q in Eq. Hence.03 – 0. Inlet flow rate.5t Differentiating. The contents are stirred well and the mixture leaves the tank at a rate of 1. we get dV = 0. V Here. Fi = 1. The total volume V. rate of accumulation = rate of input – rate of output d (VC ) = Fi C i – F o C dt i. at any time = Initial volume + (Inflow rate – Out flow rate) (time) = 10 + (1. dV = 0.226 PROCESS CALCULATIONS Ë 20 Û X = 0. Let C be the concentration of salt and V the volume of solution at any time.5 litres/min at a salt concentration of 1 g/litre.01 Ì Ü Í (20  t ) Ý 4 (16) Comparing Eq. Fo = 1.3 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. rate of input + rate of generation = rate of output + rate of accumulation (1) Here rate of generation is zero.5 dt dC dV +C = Fi C i – F o C dt dt (2) (3) .0 litre/min.5 litres/min at a salt concentration of 1 g/litres Outlet flow rate. Estimate (a) the time at which the concentration in the tank will be 1. (16) with Eq.e.784 minutes 10. V = 10 + 0.5% is 3.5 dt We know that.0)t or.6 g/litre and (b) the contents in the tank will be 18 litres Here.0 litre/min Initial volume = 10 litres.5 – 1. (8) we find both are same and hence the time taken to reach a concentration of 2. 3 ln Í ln Í ˙ ˙ Î 2 -1˚ Î 20 ˚ C .71 minutes.6 g/litres is given by substituting C = 1.5(1 . we get C 227 (4) (5) t dt ⎛ 1 ⎞ ⎛ dC ⎞ ⎜ 1.6 = + Í ˙ Î 20 + t ˚ solving.5 ¥ 1.1 È 20 ˘ =Í ˙ 1 Î 20 + t ˚ (6) 3 (7) È 20 ˘ \ C=1+ Í ˙ Î 20 + t ˚ 3 (8) Time taken to reach a concentration of 1.1˘ È 20 + t ˘ = . 3 (b) Final volume = Initial volume + (volume flowing in – volume flowing out)(time) or.0)t Therefore. 1. (8): È 20 ˘ 0. t = 3. 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 + 0.5t ) ⎝ ⎠C ⎝ ⎠ t=0 ∫ ∫ o ⎛ 1 ⎞ ⎜ − 1.5 ⎟ ⎜ 1 − C ⎟ = + (10 0. .5t Integrating.5(1 – C) i.0C dt dC = 1. time taken for the water in the tank to reach 18 litres is = 16 minutes.C ) 10 + 10.6 in Eq.5 – 1.5t) dt dC dt = or.1 A tank contains 500 kg of a 10% salt solution.5C = 1.5C = 1.0 – 1. 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.5 ⎟ ⎝ ⎠C C t dt ⎛ dC ⎞ ⎜C −1⎟ = 2 (20 + t) ⎝ ⎠ t=0 =2 ∫ o ∫ È C . dC (10 + 0. 18 = 10 + (1. by substituting values.e.5t) + 0. EXERCISES 10.PROBLEMS ON UNSTEADY STATE OPERATIONS Therefore.5 – 1. 7 A cylindrical tank of cross-sectional area A is filled with liquid up to a height Ho.5 litres/min.5 A storage tank contains 5000 kg of a 1% sugar solution by weight. Assuming uniform mixing. A hole of diameter d at the bottom of the tank. A fresh feed of 400 kg/min of pure water is entering the tank and dilutes the solution in the tank. (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 3.2 A tank contains 1000 kg of a 10% salt solution.75% by weight? 10.3 A tank contains 50 litres of a salt solution at a concentration of 2.5 kg/min at a salt concentration of 3% by weight. at a salt concentration of 1 g/litre. Set up an unsteady state balance equation to calculate the time for the liquid level to fall to a new height H1. what will be the concentration in the tank? 10. into which water is added continuously and diluted.228 PROCESS CALCULATIONS 10. what is the concentration of Na2SO4 in the mixer at the end of 10 minutes.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 outlet concentration of the solution is Cao. Discuss about the solution of the equation.4 A tank contains 20 kg of a salt solution at a concentration of 4% by weight.0 litres/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. The mixture is stirred well and the product leaves the tank at a rate of 400 kg/min. Another salt solution enters the tank at a rate of 2. At what instant of time the sugar concentration in the tank will drop to 1% sugar by weight? After one hour of operation. 10.5 g/litre. The exit solution leaves at the rate of 9 kg/min.25 g/litre and (b) the contents in the tank will be 20 litres. The contents are stirred well and the mixture leaves the tank at a rate of 2. Fresh solution enters the tank at a rate of 2. Estimate (a) the time at which the concentration in the tank will be 1. The contents are stirred well and the mixture leaves the tank at a rate of 2.8 A solution is having solute A at a concentration CAf is fed continuously into a mixing vessel of constant volume V. which was plugged initially is opened to let the liquid drain through it. 10. Write the unsteady state solute balance equation. Volume change during mixing can be neglected.0 kg/min. 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. 10. . What will be the time at which the salt concentration in the tank will be 15%? 10. 285 ´ 10–3 Pound (lb) kg 0.4536 Gram (g.0 3 kg/m 1000 lbf N 4.452 ´ 10–4 ft2 m2 0.546 ´ 10–3 Gallons (US) m3 3.448 kgf N 9.0254 ft (¢) m 0.7 dyne N 10–5 g/litre g/cm Force 2 3 cm Mass Multiply by to kg/m 3 (Contd.0929 cm Volume Density 2 m ft3 10–4 m3 3 0.02832 10–6 litre m3 10–3 Gallons (UK) m3 4.) 229 .807 Pa N 980.3048 cm m 0.019 3 1. gm) kg 10–3 lb/ft3 kg/m3 m 16.Tables Important Conversion Factors TABLE I Quantity To convert from Length Area in (¢¢) m 0.01 Angstrom (Å) m 10–10 microns (m) m 10–6 in2 m2 6. h kmole/m2s 1.88 2 N/m = Pa in water Heat or energy Multiply by to 3 .h kg/m2s 1.m 1055 erg J = N.356 ´ 10–3 g mole/cm2s kmole g/m2s 10 Btu/lb J/kg = N.m 4187 kW.°C N.m/kg K = J/kg K 4187 ft3/s 3 Mass flow rate (Mass flux) 47.02832 ft /h.m 4.187 kcal J = N.356 ´ 10–3 g/cm2s kg/m2s 10 lb mole/ft2.1 2 133. °F N.6 ´ 106 2 m3/s 0.230 TABLES Important Conversion Factors (contd.777 ´ 10–7 lb/ft2.0133 ´ 105 torr N/m2 = Pa 133.m/kg K = J/kg K 4187 cal/g.) TABLE I Quantity To convert from Pressure lbf/ft2 N/m2 = Pa 2 lbf/in (psi) in Hg mm Hg Volumetric flow rate Molar flow rate (Molar flux) Enthalpy Heat capacity (Holds good for molal heat capacity also) 6895 2 3386 2 249.867 ´ 10–6 cm3/s m3/s 10–6 lit/h m3/s 2.m/kg 2326 cal/g = kcal/kg J/kg = N.m/kg 4187 Btu/lb.m 10–7 cal J = N.m 3. ft /h m3/s 7.807 ´ 104 Btu J = N.3 bar N/m = Pa 105 kgf/cm2 N/m2 = Pa 9.3 2 N/m = Pa N/m = Pa N/m = Pa atm N/m = Pa 1.h J = N. 94 Columbium Nb 41 92.00 Fluorine F 9 19.90 Gallium Ga 31 69.00 Beryllium Be 4 9.82 Bromine Br 35 79.00 Dysprosium Dy 66 162.36 Berkelium Bk 97 245.20 Europium Eu 63 152.91 Astatine At 85 210.76 Argon A 18 39.94 Arsenic As 33 74.54 Curium Cm 96 243.00 Carbon C 6 12.00 Boron B 5 10.00 Aluminium Al 13 26.01 Cerium Ce 58 140.00 Francium Fr 87 223.91 Chlorine Cl 17 35.91 Copper Cu 29 63.01 Cobalt Co 27 58.41 Calcium Ca 20 40.46 Chromium Cr 24 52.00 Barium Ba 56 137.60 (Contd.00 Gadolinium Gd 64 156.08 Californium Cf 98 246.13 Cesium Cs 55 132.46 Erbium Er 68 167.TABLES TABLE II 231 Atomic Weights and Atomic Numbers of Elements Element Symbol Atomic Number Atomic weight Actinium Ac 89 217.01 Bismuth Bi 83 209.) .72 Germanium Ge 32 72.92 Cadmium Cd 48 112.98 Americium Am 95 243.00 Antimony Sb 51 121. 94 Lutetium Lu 71 174.85 Krypton Kr 36 83.61 Molybdenum Mo 42 95.92 Lead Pb 82 207.95 Neodymium Nd 60 144.93 Mercury Hg 80 200.91 Nitrogen N 7 14.10 Praseodymium Pr 59 140.91 Iridium Ir 77 193.32 Manganese Mn 25 54.) Symbol Atomic Number Atomic weight Gold Au 79 197.60 Helium He 2 4.20 Oxygen O 8 16.21 Lithium Li 3 6.00 Polonium Po 84 210.92 (Contd.27 Neptunium Np 93 237.00 Indium In 49 114.70 Phosphorus P 15 30.69 Niobium Nb 41 92.98 Platinum Pt 78 195.00 Holmium Ho 67 164.80 Lanthanum La 57 138.76 Iodine I 53 126.) .18 Nickel Ni 28 58.232 TABLES TABLE II Element Atomic Weights and Atomic Numbers of Elements (contd.94 Hydrogen H 1 1.00 Neon Ne 10 20.00 Potassium K 19 39.20 Hafnium Hf 72 178.99 Magnesium Mg 12 24.23 Plutonium Pu 94 242.01 Osmium Os 76 190.00 Palladium Pd 46 106.10 Iron Fe 26 55. 92 Uranium U 92 238.88 Sodium Na 11 23.05 Radon Rn 86 222.39 Thorium Th 90 232.38 Zirconium Zr 40 91.12 Thulium Tm 69 169.63 Sulphur S 16 32.70 Titanium Ti 22 47.61 Terbium Tb 65 159.07 Vanadium V 23 50.31 Rhodium Rh 45 102.90 Tungsten W 74 183.40 Tin Sn 50 118.22 .48 Ruthenium Ru 44 101.92 Zinc Zn 30 65.96 Silicon Si 14 28.43 Scandium Sc 21 44.00 Tellurium Te 52 127.07 Tantalum Ta 73 180.70 Samarium Sm 62 150.91 Rubidium Rb 37 85.04 Yttrium Y 39 88.95 Xenon Xe 54 131.09 Silver Ag 47 107.20 Thallium Tl 81 204.00 Protactinium Pa 91 231.96 Selenium Se 34 78.00 Rhenium Re 75 186.88 Technetium Tc 43 99.00 Radium Ra 88 226.TABLES TABLE II Element 233 Atomic Weights and Atomic Numbers of Elements (contd.00 Strontium Sr 38 87.) Symbol Atomic Number Atomic weight Promethium Pm 61 145.30 Ytterbium Yb 70 173. 38 1.117 6.762 8.93 ´ 10–2 0.19 1.97 8.00 3.42 2.98 m 6.30 3.454 6.339 6.25 1.350 6.85 1.640 10.89 10.764 * Gas TABLE III(b) Molal Heat Capacities of Hydrocarbon Gases For 10 to 760 oC.040 18.628 +0.410 28.40 45.25 6.9 ´ 10–2 n 4.069 2.811 2.234 TABLES TABLE III(a) Empirical Constants for Molal Heat Capacities of Gases at Constant Pressure Cp = a + bT + cT 2.415 –3.140 10.4 ´ 10–12 8.210 2.25 27.010 19.019 7.00 .537 –0.57 ´ 10–3 3.130 0.440 7.069 –1.920 c ´ 106 – 0.85 1.386 5.3613 –10.20 ´ 10–8 2.79 1. 95 + mT n T in Rankine (oF + 460) Compound a b ´ 103 –c ´ 106 Methane Ethylene Ethane Propylene Propane n-Butane i-Butane Pentane 3.97 0. g-cal/(g-mole) (K) – Temperature range 300 to 1500 K Gas b ´ 103 a H2 N2 O2 CO NO H2O CO2 SO2 SO3 HCl C2H6 CH4 C2H4 Cl2 Air NH3(*) 6.970 –4.734 2.322 3.83 8.2656 –1.389 3.14 9.97 ´ 10–3 0.0459 –3.69 35.480 –8.91 16.167 1.4206 –0.20 23.71 1. Cp = a + bT + cT 2 For –180 to 95 oC. Cp = 7.2675 –0.25 2.457 6.41 2.4757 –0.431 38.8733 –0.653 6.005 –0. where T is in Kelvin.946 6.221 1.80 4.27 5.13 ´ 10–11 6.93 ´ 10–2 3.136 6.78 55.945 7.28 2.196 1.963 0.26 1.204 3.95 45.19 1.794 –6. 5/g0.09453) [h2.627 lbf/ft2 (e) 0.98 g moles/kg of solution Molarity: 5.572 g of O2 (b) 12.2 10.0425 Mole %: Na2CO3: 4.5] 1.3 4.1 (a) 10.1% Water: 95.378 g moles/kg of water 2.55 * 10 –3 [P/T 0.5 Mole ratio: 0.01355 cal/s cm2°C (f) 1163 W/m K 1.03929 hp-hr (d) 0.5455 kg of carbon 2.4 (a) 2.67 psia (c) 0.98 g moles/litre Volume of solution: 0.8% (b) 0.088 (c) 5.Answers to Exercises CHAPTER 1 1. CHAPTER 2 2.2] 1.6 Molality: 5.77 g of KClO3 2.5] tan F 1.2 0.8/D 0.1 500 g moles 2.93 [Cp G 0.3 (a) 3.0836 litre Normality: 5.5 As long as consistent units are used.98 235 .84 cm2/s (b) 39. the equation remains the same.4 (0.9% 2. 236 2.20 (a) 26.625 2.26 kg of 94% H2SO4 2.38% Molality: 5.00 Atomic %: Na: 2.56 g sugar/litre 2.19 2094 kg iron and 900 kg water 2.16 31.18 NaCl:75%.46 Total 100.12 0. KCl 20.93%.13 9.98 (b) 890.94 .25% (c) 89.3 76.00 100 100.10 (a) 0.98% Bromine: 10.97% 2.25%. H2: 62.8 Compound Weight % Volume % mole % NaCl H2O 23.75 kg Cu.17% and (c) 59.76%.185 g moles/kg of solution 2.17 (a) H2SO4 (b) 79.54 91.14 (a) Mole fraction of H3PO4: 0.090 g 2. 104. KCl 25% and NaCl: 79.15% Nitrogen: 34.945 g of Cr2S3 2.7 ANSWERS TO EXERCISES Weight %: 39.7 11 89 8.02 Chlorine: 54.3445 kg of AgNO3 2.75% 2. and O2: 31.11 (a) Nitrogen (b) 16. Cl2: 2.02 Mole fraction of Water: 0.8 cc of solution 2.9 AVMWT: 65.15 54.1455 kg sugar/kg water (b) 1075 kg/m3 solution (c) 136.87% 2.02% Volume %: 26% 2.93%. 3 0.03% n-butane 10.01 27.6 Methane: 0.1 Compound Weight fraction mole fraction mole % Butane Pentane Hexane 0.01338 kmole/m3 Velocity: 30.79%.00405 g/cc 3.44% and nitrogen 9.2 0.5.5701 0.00 Nitrogen Total (c) AVMWT: 18.98%. propane: 9.ANSWERS TO EXERCISES 237 (b) Methane: 24.2 70.567 litre .8 0.0008 g/cc 3.4 (a) and (b) Component mole % Weight % Methane 80 68.02231 kmole/m3 Hydrogen: 0.94%.558 m/h Density: 0.748 g 3.21 (a) Methane: 24.7 (d) Density: 0.0 1. propylene: 14.47%.58 15.33 kg/m3 CHAPTER 3 3.36 g/litre 3.41 Total 1.3 37.45 100 100.2758 0.1541 57.138 3.93 2.63% 3.8 (c) 1.008892 kmole/m3 Ethane: 0.45 Ethane 15 24. ethylene: 25.16% and carbon dioxide: 66.7 1.10 5 7.76% (c) 0.169 g/litre 3.00 AVMWT: 66.4% (b) 29. Density of gas: 1.0000 100. ethane: 14.5 0.5 AVMWT: 30. 12 (a) and (c) Component Volume % N2 mole fraction 64.8 . (b) and (c) Component mole fraction g mole/cc Concentration.719.37 atm 3.6 1.06 g/litre 3.0100 100.74 Total 100.00 3.1862 O2 6.923 kg 3.62 0.67 18.00 0.7 K 3.0952 H2O 18. mm Hg Partial pressure CH4 C2 H 6 H2 0.09 atm (b) 3 m3 (c) 2.14 Component Volume % = mole % Chlorine Bromine Oxygen 68.173 ´ 10–5 6.10 C3H8 3.00 1.200 (d) 1.16 (a).11 0.238 3.15 (a) Partial pressure: 0.3 0.6475 CO2 9.0000 Total (b) Average density: 0.9 ANSWERS TO EXERCISES 10.59 12.792 kg/m3 3.1 0.058 ´ 10–5 3.058 ´ 10–4 g mole/cc (e) 1.52 0.13 52.11 300.75 0.248 g/s (f) AVMWT: 11.0611 CO 1.347 ´ 10–5 200 600 1. 897 0.140 0.5 0. ammonia: 14.369 0.076 0.20 (a) Chlorine: 54.819 0.45 litre 3.162 0.0 0.555 0. 4.724 0. bromine: 10.758 0678 0.17 (a) Nitrogen.74% water: 1.19 34.4 s 3.459 0.5052 Mole fraction of ethanol in vapour phase: 0.773 0.1847 kg/kg of steam.030 0 x.212 0.1 31.415 0. 0. y: Mole fraction of ‘A’ in liquid and vapour phase respectively.59 g/litre 3.5%.982 kg CHAPTER 4 4.79 mm Hg Mole fraction of methanol in vapour phase: 0. oxygen: 15. y: Mole fraction of benzene in liquid and vapour phase respectively.134 0.3 3.23 0.941 J/g mole 4.21 (a) Nitrogen: 67.496 0.831 0. 4.0 0.18 (a) 0.3 x 1.958 0.013 0 y 1.79%.2 77.548 0.6 Total pressure: 2.76% 3.05 atm (b) 5 m3 (c) 0.393 0.98%.05 (c) Density: 2.4 xA 1.22 10.25 °C 4.4948 . (b) 7.590 0.0 yA 1.288 0.0 0.0 x.16% nitrogen: 34.660 0.1385 kg/kg of steam 4.ANSWERS TO EXERCISES 239 3.0 0.707 g/litre (d) 17.205 0.897 0.5% hydrogen (c) 19.044.85% (b) 65.48 kg/m3 3.281 0.97% (b) 2. 0521 kJ/kg dry air Enthalpy: 0.93% (e) Mole % = volume % = 1. y Benzene 0.907 m3/kg dry air Adiabatic saturation temperature: 30 °C Humid heat: 1.233 Xylene 0.377 0.567 mole/mole 1.240 4.2% (By weight) CHAPTER 5 5.78% (By volume) (b) Ethyl acetate: 29. n-C3H8 : 0.019 kg/kg % saturation: 21% Humid volume: 0.22 %.0433 kg toluene/kg of air (d) % saturation: 23.123 0. x Vapour phase composition.94% (b) 0.6 mm Hg 4.028 Partial pressure: 102.25 mole/mole 0.8 %.5 °C Humid volume: 0.179 and C5H12 : 0.10 (a) Ethyl acetate: 12.9 mm Hg Component Liquid phase composition.6 Humidity: 0.139.534 kg/kg 5.8 4. air: 70.951 kJ/kg dry air . air: 87.6 m3/h 5.5 0.9394 m3/kg dry air 5.676 kg/kg 5.1 (a) RH% = 24. i-C4H10 : 0.01358 mole of toluene/mole of vapour free gas (c) 0.7 ANSWERS TO EXERCISES (a) Total pressure: 4087.025 kg/kg Dew point: 28.004 (b) 3589.039.34% 5.4 0.739 Toluene 0.87 mm Hg 4.2 Humidity: 0.639.9 mm Hg C2H6 : 0.9 Total pressure: 906. n-C4H10 : 0.500 0.3 2961. 12 (a) % RH: 29.974 m3/kg dry air 5.027 kg/kg % saturation: 70% 5.03 m3/h 5.59% 5.9 (a) 96.48 kg of water 5.4 m3/min (b) 74.24 mm 810.11 46.7 kg dry air/h 0.08 m3/h 5.0377 mole/mole 241 .5%.81 kg of water 7.13 (a) (b) (c) (d) 3703.04 kg 5.17 Humid volume: 0.7 Cool to 18.20 (a) (b) (c) (d) 29.00423 kg/kg (c) 13. (b) 115.0259 kmole/kmole.ANSWERS TO EXERCISES 5.01616 kg/kg (b) 0.5 °C respectively 35°C 5.008 kg/kg 12 °C 2.8586 and 0.45 781.15 % Relative saturation: 57.19 (a) 0.2 kJ 32. 0.405.6% (b) Humid volume: 0.5 °C and then reheat it to 30 °C Air needed initially 5114. 0.9739 m3/kg dry air respectively 18 and 28.5 °C.6 kJ/kg dry air 5.019 kg/kg 5.10 (a) (b) (c) (d) 0.18 (a) 809.8 27.5 °C and 0.00857 kmole/kmole 1.858 m3 5.5% Percentage saturation: 44.16 Humidity: 0.69 Hg 0.0475 kg/kg dry air 5.0677 kmole/kmole 0.14 (a) (b) (c) (d) (e) 0.830 kg of water 49.0351 kmole/kmole 0.5 °C 5.9205 m3/kg dry air Enthalpy: 106. 6 kg/h CHAPTER 7 7.956%.811.158%.3 50.29% 7.010% 7.7% 5.35 kg.96%.5 m3/h CHAPTER 6 6. (c) 21.078. (b) 4.49% 7.2% and (c) 60% 7.2 kg/hr and (b) 1.1 H2SO4: 5.33 kg 6.1 Mother liquor: 3.165 kg 7.4% 6.5 Crystals: 342 kg and Mother liquor: 918 kg 6.25 m3 (b) 6.64 kg 6.1% H2O: 17.25% and Hexane: 55.17 (a) 18.4 kg/h in each evaporator 7.86 kg.4 CO2: 11.19 (a) 1.34 kg 6.06%.41 m3 . Pentane: 42.67 kg and crystals: 2.6 Water evaporated: 7.3 Crystals: 6.600 kg. O2: 3.81 kg 7.4% O2: 3.11 CO2: 11. 7. CO: 9. H2O: 14.9 Water evaporated: 561 kg 6. 471. HNO3: 2.66% 7.877.14 Butane: 2. SO2: 0.10 Water evaporated: 82.916.242 ANSWERS TO EXERCISES 5.920 kg and (c) 2.324. (b) 1396.22 (a) 86%.9%.14%.693%.122.16 (a) Nitrogen.2 (a) 56.51% and N2: 68.53% and N2:70.5 Excess air: 39.6 O2: 12.10 CO2: 4.87 kg and Feed: 1.85% 7.15 (a) 6.8 54.7 Water evaporated: 720 kg Mother liquor: 6589 kg and crystals: 1.21 5.434.11 2393.2 Feed: 83.67% and SO3: 0. N2: 75.4 Crystals: 479.973.8 87.34%. N2: 79.183%.310.41%.46%.636 kg 6. Mother liquor: 1.06 kg of lime/100 kg coke burnt 7.62% 7. and H2O: 9.691 kg 6.2 kg 6. SO2: 7.27 kg 7. (b) 16.85 kg and Feed: 2.083.2 kg 6.22% Excess air (b) 8.705.338. 06% 7.8% CO = 10.ANSWERS TO EXERCISES 243 (c) Component Condition (a) Condition (b) CO2 H2O N2 O2 0.642 7.3 kg.3%.08 7.08 0. moles of CO2: 1.7 m3 (b) 0. H2O = 4.0933 kg. (b) Ash produced = 21. C5H12 : 31. and N2: 72.285 m3/100 m3 CO2: 19. vapour : 18. basis) N2 = 82.089 kmoles (b) Moles of air supplied: 18. N2: 79.28%.23% (c) 5081 kg 7.5%.5%.388 kmoles (c) Percentage of carbon lost in the ash: 7.04 Total 1. N2 = 1091.63 — 0.14 0.23 (a) 18 (b) CO2: 17. C6H14 : 60.66%.25 Liquid : 81. basis): CO2 = 273.06 12.1467 kg.5 kg.965 (b) 19.1 kg (c) Carbon lost per 100 kg of coke burnt = 7.00 1.5901 kg.74 0.24 kg .8 7.9% CO2 = 6.26 Air supplied = kg of fuel combustion = 15.00 7.24 (a) 2.58% .407% 7.7% 7. CO = 9.21 (a) (b) (c) (d) 28. H2O: 3.789 kg/hr.211 kg/hr.5% 7.23 0. O2 = 118.28 (a) Composition of the flue gas (wt. volumetric analysis = (mol.20 (a) 4.3911 kg.27 (a) Moles of CO: 5.14 0.22 (a) 446.38. O2: 1.55 m3/kg carbon burnt 50.444 kmoles. air required theoretically for complete 7.49% (b) 3. O2: 6. 433 kmoles.3326 Composition CO2 Amount of air supplied = 37.4726 4.7167 0.51058.74%.92 H2O 2. basis): CO2: 10.5086 76.78434 ton 7.7382 11.846% (b) 70.32 Air–fuel ratio by mass = 16.7155 kg/kg of fuel 7.29 (a) Volume of air being introduced = 21729.54 m3/h (b) Composition of flue gas on dry basis (mol.63 O2 1. CO: 0.30 (a) 3.31 Air–fuel ratio =12.08 SO2 0.113 O2 0. O2: 3.167% .032 N2 74. N2: 85.2 H2O 8. % excess air = 16.56 Nm3/h Flue gas analysis: Composition CO2 Weight (kg) 16.59478.14 7.7165 Flue gas analysis: Composition Weight (kg) CO2 3.064 Total = 17.244 ANSWERS TO EXERCISES 7. volume = 15416.34 Air–fuel ratio = 17.9749 O2 7.2267% 7. flue gas analysis: Weight % Volume % 17.026 SO2 0.6408 N2 12.2088 6.33 Air requirement = 432.487%.8717 H2O 1.3% 7.5373%.69 N2 72. ANSWERS TO EXERCISES CHAPTER 8 8.000 + 400t + t2]/(100 + t) 10.1 minutes (b) 60 minutes 10.25.3 (a) t = 100 – 100 [(C – 1)/1.8 kJ/kmole 9.1 [100/(100 + t)]2 + 0.7179 CHAPTER 9 9.4286 mole of Ammonia/mole of NO formed 8.2 C = [10.2675 kg Na2SO4/kg of solution 245 . (b) Recycle fraction: 0.874 °C 9.1 1760 °C 9.425.2 10.2 –1.1 (a) 100 moles (b) 0.5 28.01 [40/(t + 40)]5 + 0.6 0.10 51.44 kcal CHAPTER 10 10.295.1 C = –0.5]0.7 1.369 minutes 10.78 minutes 10.4 (a) C = 0.03 (b) 2.2 (a) 1.16.8 kg. 36. 13 Bypass. 38 Avogadro’s hypothesis. 88 Humidity. 35. 196 Enthalpy. 88 Hydrated salt. 229–230 Crystal. 198 Antoine equation. 87 Law Amagat’s. 74 Composition of liquid systems. 9 Conversion factors. 37 Law of conservation of mass. 196 mixing. 111 Baume’ gravity scale. 231–233 Average molecular weight. 75 Heat. 12. 196 Heat capacity of gases (empirical constants). 8 Conversion. 37 Dalton’s. 197 Humid heat. 87 absolute. 11 Atomic weights of elements. 180 Clausius–Clapeyron equation. 88 Dry bulb temperature. 87 percentage absolute humidity. 197 formation (also standard). 12 Atomic numbers of elements. 75 API scale. 197 from enthalpy data. 197 Leduc’s. 111 Crystallization. 10 solutions. 88 relative. 37 Hess’s. 10 Conservation of mass. 35 Kopp’s rule. 8 Energy balance. 38 Dew point.Index Gas constant. 111 Ideal gas law. 198 Density. 36 Hausbrand chart. 231–233 Atomic percent. 88 Humid volume. 88 247 . 12 Brix scale. 197 from combustion data. 234 Heat of combustion. 197 reaction. 36 Adiabatic reaction temperature. 9 Avogadro’s number. 11 mixtures. 87 Pure component volume. 234 Theoretical flame temperature. 89 Specific gravity. 231–233 atomic weights. 221 generation. 197 Recycle. 3 isolated. 3 intensive. 89 Saturation. 36 Percentage saturation. 35 state. 9 Reaction endothermic. 75 System. 9 limiting. 10 Weight percent. 3 Psychrometry. 9 excess. 89 relative. 89 percentage. 13 Units and notations. 2 Unsteady state operations. 87 Yield. 197 exothermic. 221 input. 74 Normal temperature and pressure (NTP). 221 output. 111 Mass balance. 10 Wet bulb temperature. 1 derived units. 111 Normal boiling point (NBP). 7 Mole fraction. 9 . 89 partial. 221 Vapour pressure. 221 Reactant. 9 Partial pressure. 3 Property. 74 Volume percent. 196 Steam distillation. 88 Process. 229–230 molal heat capacities. 231–233 conversion factors. 36 Purge. 180 Rate of accumulation. 11 Mole percent. 198 Twaddell scale. 3 Table of atomic numbers.248 INDEX Magma. 3 extensive. 122 Mass relations. 11 Mother liquor. 180 Saturated vapour. 12 Standard condition.