Chemical Engineering

CHEMICAL ENGINEERING Solutions to the Problems in Chemical Engineering Volumes 2 and 3 Related Butterworth-Heinemann Titles in the Chemical Engineering Series by J. M. COULSON & J. F. RICHARDSON Chemical Engineering, Volume 1, Sixth edition Fluid Flow, Heat Transfer and Mass Transfer (with J. R. Backhurst and J. H. Harker) Chemical Engineering, Volume 3, Third edition Chemical and Biochemical Reaction Engineering, and Control (edited by J. F. Richardson and D. G. Peacock) Chemical Engineering, Volume 6, Third edition Chemical Engineering Design (R. K. Sinnott) Chemical Engineering, Solutions to Problems in Volume 1 (J. R. Backhurst, J. H. Harker and J. F. Richardson) Chemical Engineering, Solutions to Problems in Volume 2 (J. R. Backhurst, J. H. Harker and J. F. Richardson) Coulson & Richardson’s CHEMICAL ENGINEERING J. M. COULSON and J. F. RICHARDSON Solutions to the Problems in Chemical Engineering Volume 2 (5th edition) and Volume 3 (3rd edition) By J. R. BACKHURST and J. H. HARKER University of Newcastle upon Tyne With J. F. RICHARDSON University of Wales Swansea OXFORD AMSTERDAM BOSTON LONDON NEW YORK PARIS SAN DIEGO SAN FRANCISCO SINGAPORE SYDNEY TOKYO Butterworth-Heinemann An imprint of Elsevier Science Linacre House, Jordan Hill, Oxford OX2 8DP 225 Wildwood Avenue, Woburn, MA 01801-2041 First published 2002 Copyright  2002, J.F. Richardson and J.H. Harker. All rights reserved The right of J.F. Richardson and J.H. Harker to be identified as the authors of this work has been asserted in accordance with the Copyright, Designs and Patents Act 1988 No part of this publication may be reproduced in any material form (including photocopying or storing in any medium by electronic means and whether or not transiently or incidentally to some other use of this publication) without the written permission of the copyright holder except in accordance with the provisions of the Copyright, Designs and Patents Act 1988 or under the terms of a licence issued by the Copyright Licensing Agency Ltd, 90 Tottenham Court Road, London, England W1T 4LP. Applications for the copyright holder’s written permission to reproduce any part of this publication should be addressed to the publishers British Library Cataloguing in Publication Data A catalogue record for this book is available from the British Library Library of Congress Cataloguing in Publication Data A catalogue record for this book is available from the Library of Congress ISBN 0 7506 5639 5 For information on all Butterworth-Heinemann publications visit our website at www.bh.com Contents Preface Preface to the Second Edition of Volume 5 Preface to the First Edition of Volume 5 Factors for Conversion of SI units vii ix xi xiii Solutions to Problems in Volume 2 2-1 Particulate solids 2-2 Particle size reduction and enlargement 2-3 Motion of particles in a fluid 2-4 Flow of fluids through granular beds and packed columns 2-5 Sedimentation 2-6 Fluidisation 2-7 Liquid filtration 2-8 Membrane separation processes 2-9 Centrifugal separations 2-10 Leaching 2-11 Distillation 2-12 Absorption of gases 2-13 Liquid–liquid extraction 2-14 Evaporation 2-15 Crystallisation 2-16 Drying 2-17 Adsorption 2-18 Ion exchange 2-19 Chromatographic separations 1 8 14 34 39 44 59 76 79 83 98 150 171 181 216 222 231 234 235 Solutions to Problems in Volume 3 3-1 Reactor design — general principles 3-2 Flow characteristics of reactors — flow modelling 3-3 Gas–solid reactions and reactors 3-4 Gas–liquid and gas–liquid–solid reactors 237 262 265 271 v 3-5 3-7 Biochemical reaction engineering Process control 285 294 (Note: The equations quoted in Sections 2.1–2.19 appear in Volume 2 and those in Sections 3.1–3.7 appear in Volume 3. As far as possible, the nomenclature used in this volume is the same as that used in Volumes 2 and 3 to which reference may be made.) vi Preface Each of the volumes of the Chemical Engineering Series includes numerical examples to illustrate the application of the theory presented in the text. In addition, at the end of each volume, there is a selection of problems which the reader is invited to solve in order to consolidate his (or her) understanding of the principles and to gain a better appreciation of the order of magnitude of the quantities involved. Many readers who do not have ready access to assistance have expressed the desire for solutions manuals to be available. This book, which is a successor to the old Volume 5, is an attempt to satisfy this demand as far as the problems in Volumes 2 and 3 are concerned. It should be appreciated that most engineering problems do not have unique solutions, and they can also often be solved using a variety of different approaches. If therefore the reader arrives at a different answer from that in the book, it does not necessarily mean that it is wrong. This edition of the Solutions Manual which relates to the fifth edition of Volume 2 and to the third edition of Volume 3 incorporates many new problems. There may therefore be some mismatch with earlier editions and, as the volumes are being continually revised, they can easily get out-of-step with each other. None of the authors claims to be infallible, and it is inevitable that errors will occur from time to time. These will become apparent to readers who use the book. We have been very grateful in the past to those who have pointed out mistakes which have then been corrected in later editions. It is hoped that the present generation of readers will prove to be equally helpful! J. F. R. vii This Page Intentionally Left Blank Preface to the Second Edition of Volume 5 IT IS always a great joy to be invited to prepare a second edition of any book and on two counts. Firstly, it indicates that the volume is proving useful and fulfilling a need, which is always gratifying and secondly, it offers an opportunity of making whatever corrections are necessary and also adding new material where appropriate. With regard to corrections, we are, as ever, grateful in the extreme to those of our readers who have written to us pointing out, mercifully minor errors and offering, albeit a few of what may be termed ‘more elegant solutions’. It is important that a volume such as this is as accurate as possible and we are very grateful indeed for all the contributions we have received which, please be assured, have been incorporated in the preparation of this new edition. With regard to new material, this new edition is now in line with the latest edition, that is the Fourth, of Volume 2 which includes new sections, formerly in Volume 3 with, of course, the associated problems. The sections are: 17, Adsorption; 18, Ion Exchange; 19, Chromatographic Separations and 20, Membrane Separation Processes and we are more than grateful to Professor Richardson’s colleagues at Swansea, J. H. Bowen, J. R. Conder and W. R. Bowen, for an enormous amount of very hard work in preparing the solutions to these problems. A further and very substantial addition to this edition of Volume 5 is the inclusion of solutions to the problems which appear in Chemical Engineering, Volume 3 — Chemical & Biochemical Reactors & Process Control and again, we are greatly indebted to the authors as follows: 3.1 3.2 3.3 3.4 3.5 3.6 Reactor Design — J. C. Lee Flow Characteristics of Reactors — J. C. Lee Gas–Solid Reactions and Reactors — W. J. Thomas and J. C. Lee Gas–Liquid and Gas–Liquid–Solid Reactors — J. C. Lee Biological Reaction Engineering — M. G. Jones and R. L. Lovitt Process Control — A. P. Wardle and also of course, to Professor Richardson himself, who, with a drive and enthusiasm which seems to be getting ever more vigorous as the years proceed, has not only arranged for the preparation of this material and overseen our efforts with his usual meticulous efficiency, but also continues very much in master-minding this whole series. We often reflect on the time when, in preparing 150 solutions for the original edition of Volume 4, the worthy Professor pointed out that we had only 147 correct, though rather reluctantly agreed that we might still just merit first class honours! Whatever, we always have and we are sure that we always will owe him an enormous debt of gratitude. ix We must also offer thanks to our seemingly ever-changing publishers for their drive, efficiency and encouragement and especially to the present staff at Butterworth-Heinemann for not inconsiderable efforts in locating the manuscript for the present edition which was apparently lost somewhere in all the changes and chances of the past months. We offer a final thought as to the future where there has been a suggestion that the titles Volume 4 and Volume 5 may find themselves hijacked for new textural volumes, coupled with a proposal that the solutions offered here hitherto may just find a new resting place on the Internet. Whatever, we will continue with our efforts in ensuring that more and more solutions find their way into the text in Volumes 1 and 2 and, holding to the view expressed in the Preface to the First Edition of Volume 4 that ‘. . . worked examples are essential to a proper understanding of the methods of treatment given in the various texts’, that the rest of the solutions are accessible to the widest group of students and practising engineers as possible. J. R. BACKHURST J. H. HARKER Newcastle upon Tyne, 1997 (Note: Some of the chapter numbers quoted here have been amended in the later editions of the various volumes.) x Preface to the First Edition of Volume 5 IN THE preface to the first edition of Chemical Engineering, Volume 4, we quoted the following paragraph written by Coulson and Richardson in their preface to the first edition of Chemical Engineering, Volume 1: ‘We have introduced into each chapter a number of worked examples which we believe are essential to a proper understanding of the methods of treatment given in the text. It is very desirable for a student to understand a worked example before tackling fresh practical problems himself. Chemical Engineering problems require a numerical answer, and it is essential to become familiar with the different techniques so that the answer is obtained by systematic methods rather than by intuition.’ It is with these aims in mind that we have prepared Volume 5, which gives our solutions to the problems in the third edition of Chemical Engineering, Volume 2. The material is grouped in sections corresponding to the chapters in that volume and the present book is complementary in that extensive reference has been made to the equations and sources of data in Volume 2 at all stages. The book has been written concurrently with the revision of Volume 2 and SI units have been used. In many ways these problems are more taxing and certainly longer than those in Volume 4, which gives the solutions to problems in Volume 1, and yet they have considerable merit in that they are concerned with real fluids and, more importantly, with industrial equipment and conditions. For this reason we hope that our efforts will be of interest to the professional engineer in industry as well as to the student, who must surely take some delight in the number of tutorial and examination questions which are attempted here. We are again delighted to acknowledge the help we have received from Professors Coulson and Richardson in so many ways. The former has the enviable gift of providing the minimum of data on which to frame a simple key question, which illustrates the crux of the problem perfectly, whilst the latter has in a very gentle and yet thorough way corrected our mercifully few mistakes and checked the entire work. Our colleagues at the University of Newcastle upon Tyne have again helped us, in many cases unwittingly, and for this we are grateful. J. R. BACKHURST J. H. HARKER Newcastle upon Tyne, 1978 xi Factors for conversion of SI units mass 1 lb 1 ton length 1 in 1 ft 1 mile time 1 min 1 h 1 day 1 year area 1 in2 1 ft2 volume 1 in3 1 ft3 1 UK gal 1 US gal force 1 pdl 1 lb 1 dyne energy 1 ft lb 1 cal 1 erg 1 Btu power 1 h.p. 1 Btu/h 0.454 kg 1016 kg 25.4 mm 0.305 m 1.609 km 60 s 3.6 ks 86.4 ks 31.5 Ms 645.2 mm2 0.093 m2 16,387.1 mm3 0.0283 m3 4546 cm3 3786 cm3 0.138 N 4.45 N 10−5 N 1.36 J 4.187 J 10−7 J 1.055 kJ 745 W 0.293 W pressure 1 lbf/in2 1 atm 1 bar 1 ft water 1 in water 1 in Hg 1 mm Hg viscosity 1 P 1 lb/ft h 1 stoke 1 ft2 /h mass flow 1 lb/h 1 ton/h 1 lb/h ft2 thermal 1 Btu/h ft2 1 Btu/h ft2 ◦ F 1 Btu/lb 1 Btu/lb ◦ F 1 Btu/h ft ◦ F energy 1 kWh 1 therm 6.895 kN/m2 101.3 kN/m2 100 kN/m 2.99 kN/m2 2.49 N/m2 3.39 kN/m2 133 N/m2 0.1 N s/m2 0.414 mN s/m2 10−4 m2 /s 0.258 cm2 /s 0.126 g/s 0.282 kg/s 1.356 g/s m2 3.155 5.678 2.326 4.187 1.731 W/m2 W/m2 K kJ/kg kJ/kg K W/m K 3.6 MJ 106.5 MJ calorific value 1 Btu/ft3 1 Btu/lb 37.26 kJ/m3 2.326 kJ/kg density 1 lb/ft3 16.02 kg/m3 SECTION 2-1 Particulate Solids PROBLEM 1.1 The size analysis of a powdered material on a mass basis is represented by a straight line from 0 per cent at 1 µm particle size to 100 per cent by mass at 101 µm particle size. Calculate the surface mean diameter of the particles constituting the system. Solution See Volume 2, Example 1.1. PROBLEM 1.2 The equations giving the number distribution curve for a powdered material are dn/dd = d for the size range 0–10 µm, and dn/dd = 100,000/d 4 for the size range 10–100 µm where d is in µm. Sketch the number, surface and mass distribution curves and calculate the surface mean diameter for the powder. Explain briefly how the data for the construction of these curves may be obtained experimentally. Solution See Volume 2, Example 1.2. PROBLEM 1.3 The fineness characteristic of a powder on a cumulative basis is represented by a straight line from the origin to 100 per cent undersize at a particle size of 50 µm. If the powder is initially dispersed uniformly in a column of liquid, calculate the proportion by mass which remains in suspension in the time from commencement of settling to that at which a 40 µm particle falls the total height of the column. It may be assumed that Stokes’ law is applicable to the settling of the particles over the whole size range. 1 Solution For settling in the Stokes’ law region, the velocity is proportional to the diameter squared and hence the time taken for a 40 µm particle to fall a height h m is: t = h/402 k where k a constant. During this time, a particle of diameter d µm has fallen a distance equal to: kd 2 h/402 k = hd 2 /402 The proportion of particles of size d which are still in suspension is: = 1 − (d 2 /402 ) and the fraction by mass of particles which are still in suspension is:  40 = [1 − (d 2 /402 )]dw 0 Since dw/dd = 1/50, the mass fraction is:  40 = (1/50) [1 − (d 2 /402 )]dd 0 = (1/50)[d − (d 3 /4800)]40 0 = 0.533 or 53.3 per cent of the particles remain in suspension. PROBLEM 1.4 In a mixture of quartz of density 2650 kg/m3 and galena of density 7500 kg/m3 , the sizes of the particles range from 0.0052 to 0.025 mm. On separation in a hydraulic classifier under free settling conditions, three fractions are obtained, one consisting of quartz only, one a mixture of quartz and galena, and one of galena only. What are the ranges of sizes of particles of the two substances in the original mixture? Solution Use is made of equation 3.24, Stokes’ law, which may be written as: u0 = kd 2 (ρs − ρ), where k (= g/18μ) is a constant. For large galena: u0 = k(25 × 10−6 )2 (7500 − 1000) = 4.06 × 10−6 k m/s For small galena: u0 = k(5.2 × 10−6 )2 (7500 − 1000) = 0.176 × 10−6 k m/s For large quartz: u0 = k(25 × 10−6 )2 (2650 − 1000) = 1.03 × 10−6 k m/s For small quartz: u0 = k(5.2 × 10−6 )2 (2650 − 1000) = 0.046 × 10−6 k m/s 2 If the time of settling was such that particles with a velocity equal to 1.03 × 10−6 k m/s settled, then the bottom product would contain quartz. This is not so and hence the maximum size of galena particles still in suspension is given by: 1.03 × 10−6 k = kd 2 (7500 − 1000) or d = 0.0000126 m or 0.0126 mm. Similarly if the time of settling was such that particles with a velocity equal to 0.176 × 10−6 k m/s did not start to settle, then the top product would contain galena. This is not the case and hence the minimum size of quartz in suspension is given by: 0.176 × 10−6 k = kd 2 (2650 − 1000) or d = 0.0000103 m or 0.0103 mm. It may therefore be concluded that, assuming streamline conditions, the fraction of material in suspension, that is containing quartz and galena, is made up of particles of sizes in the range 0.0103–0.0126 mm PROBLEM 1.5 A mixture of quartz and galena of a size range from 0.015 mm to 0.065 mm is to be separated into two pure fractions using a hindered settling process. What is the minimum apparent density of the fluid that will give this separation? How will the viscosity of the bed affect the minimum required density? The density of galena is 7500 kg/m3 and the density of quartz is 2650 kg/m3 . Solution See Volume 2, Example 1.4. PROBLEM 1.6 The size distribution of a dust as measured by a microscope is as follows. Convert these data to obtain the distribution on a mass basis, and calculate the specific surface, assuming spherical particles of density 2650 kg/m3 . Size range (µm) 0–2 2–4 4–8 8–12 12–16 16–20 20–24 Number of particles in range (−) 2000 600 140 40 15 5 2 3 Solution From equation 1.4, the mass fraction of particles of size d1 is given by: x1 = n1 k1 d13 ρs , where k1 is a constant, n1 is the number of particles of size d1 , and ρs is the density of the particles = 2650 kg/m3 . x1 = 1 and hence the mass fraction is:  x1 = n1 k1 d13 ρs nkd 3 ρs . In this case: d n kd 3 nρs x 1 3 6 10 14 18 22 200 600 140 40 15 5 2 5,300,000k 42,930,000k 80,136,000k 106,000,000k 109,074,000k 77,274,000k 56,434,400k 0.011 0.090 0.168 0.222 0.229 0.162 0.118  = 477,148,400k  = 1.0 The surface mean diameter is given by equation 1.14:  ds = (n1 d13 ) (n1 d12 ) and hence: Thus: d n nd 2 nd 3 1 3 6 10 14 18 22 2000 600 140 40 15 5 2 2000 5400 5040 4000 2940 1620 968 2000 16,200 30,240 40,000 41,160 29,160 21,296  = 21,968  = 180,056 ds = (180,056/21,968) = 8.20 µm This is the size of a particle with the same specific surface as the mixture. The volume of a particle 8.20 µm in diameter = (π/6)8.203 = 288.7 µm3 . 4 The surface area of a particle 8.20 µm in diameter = (π × 8.202 ) = 211.2 µm2 and hence: the specific surface = (211.2/288.7) = 0.731 µm2 /µm3 or 0.731 × 106 m2 /m3 PROBLEM 1.7 The performance of a solids mixer was assessed by calculating the variance occurring in the mass fraction of a component amongst a selection of samples withdrawn from the mixture. The quality was tested at intervals of 30 s and the data obtained are: mixing time (s) sample variance (−) 30 0.025 60 0.006 90 0.015 120 0.018 150 0.019 If the component analysed represents 20 per cent of the mixture by mass and each of the samples removed contains approximately 100 particles, comment on the quality of the mixture produced and present the data in graphical form showing the variation of mixing index with time. Solution See Volume 2, Example 1.3. PROBLEM 1.8 The size distribution by mass of the dust carried in a gas, together with the efficiency of collection over each size range is as follows: Size range, (µm) Mass (per cent) Efficiency (per cent) 0–5 10 20 5–10 15 40 10–20 35 80 20–40 20 90 40–80 10 95 80–160 10 100 Calculate the overall efficiency of the collector and the percentage by mass of the emitted dust that is smaller than 20 µm in diameter. If the dust burden is 18 g/m3 at entry and the gas flow is 0.3 m3 /s, calculate the mass flow of dust emitted. Solution See Volume 2, Example 1.6. PROBLEM 1.9 The collection efficiency of a cyclone is 45 per cent over the size range 0–5 µm, 80 per cent over the size range 5–10 µm, and 96 per cent for particles exceeding 10 µm. 5 Calculate the efficiency of collection for a dust with a mass distribution of 50 per cent 0–5 µm, 30 per cent 5–10 µm and 20 per cent above 10 µm. Solution See Volume 2, Example 1.5. PROBLEM 1.10 A sample of dust from the air in a factory is collected on a glass slide. If dust on the slide was deposited from one cubic centimetre of air, estimate the mass of dust in g/m3 of air in the factory, given the number of particles in the various size ranges to be as follows: Size range (µm) Number of particles (−) 0–1 2000 1–2 1000 2–4 500 4–6 200 6–10 100 10–14 40 It may be assumed that the density of the dust is 2600 kg/m3 , and an appropriate allowance should be made for particle shape. Solution If the particles are spherical, the particle diameter is d m and the density ρ = 2600 kg/m3 , then the volume of 1 particle = (π/6)d 3 m3 , the mass of 1 particle = 2600(π/6)d 3 kg and the following table may be produced: Size (µm) 0–1 1–2 2–4 4–6 Number of particles (−) 2000 1000 500 200 Mean diameter (µm) 0.5 1.5 3.0 5.0 −6 −6 −6 1.5 × 10 3.0 × 10 5.0 × 10−6 (m) 0.5 × 10 Volume (m3 ) 6.54 × 10−20 3.38 × 10−18 1.41 × 10−17 6.54 × 10−17 Mass of one particle (kg) 1.70 × 10−16 8.78 × 10−15 3.68 × 10−14 1.70 × 10−13 Mass of one particles in size range (kg) 3.40 × 10−13 8.78 × 10−12 1.83 × 10−11 3.40 × 10−11 Size (µm) Number of particles (−) Mean diameter (µm) (m) 3 Volume (m ) Mass of one particle (kg) Mass of one particles in size range (kg) 6–10 100 8.0 8.0 × 10−6 2.68 × 10−16 6.97 × 10−13 10–14 40 12.0 12.0 × 10−6 9.05 × 10−16 2.35 × 10−12 6.97 × 10−11 9.41 × 10−11 6 Total mass of particles = 2.50 × 10−10 kg. As this mass is obtained from 1 cm3 of air, the required dust concentration is given by: (2.50 × 10−10 ) × 103 × 106 = 0.25 g/m3 PROBLEM 1.11 A cyclone separator 0.3 m in diameter and 1.2 m long, has a circular inlet 75 mm in diameter and an outlet of the same size. If the gas enters at a velocity of 1.5 m/s, at what particle size will the theoretical cut occur? The viscosity of air is 0.018 mN s/m2 , the density of air is 1.3 kg/m3 and the density of the particles is 2700 kg/m3 . Solution See Volume 2, Example 1.7. 7 SECTION 2-2 Particle Size Reduction and Enlargement PROBLEM 2.1 A material is crushed in a Blake jaw crusher such that the average size of particle is reduced from 50 mm to 10 mm, with the consumption of energy of 13.0 kW/(kg/s). What will be the consumption of energy needed to crush the same material of average size 75 mm to average size of 25 mm: (a) assuming Rittinger’s Law applies, (b) assuming Kick’s Law applies? Which of these results would be regarded as being more reliable and why? Solution See Volume 2, Example 2.1. PROBLEM 2.2 A crusher was used to crush a material with a compressive strength of 22.5 MN/m2 . The size of the feed was minus 50 mm, plus 40 mm and the power required was 13.0 kW/(kg/s). The screen analysis of the product was: Size of aperture (mm) through 6.0 on 4.0 on 2.0 on 0.75 on 0.50 on 0.25 on 0.125 through 0.125 Amount of product (per cent) all 26 18 23 8 17 3 5 What power would be required to crush 1 kg/s of a material of compressive strength 45 MN/m2 from a feed of minus 45 mm, plus 40 mm to a product of 0.50 mm average size? 8 Solution A dimension representing the mean size of the product is required. Using Bond’s method of taking the size of opening through which 80 per cent of the material will pass, a value of just over 4.00 mm is indicated by the data. Alternatively, calculations may be made as follows: Size of aperture (mm) Mean d1 (mm) n1 nd13 nd12 nd1 nd14 6.00 5.00 0.26 1.3 6.5 32.5 162.5 3.00 0.18 0.54 1.62 4.86 1.375 0.23 0.316 0.435 0.598 0.822 0.67 0.08 0.0536 0.0359 0.0241 0.0161 0.37 0.17 0.0629 0.0233 0.0086 0.00319 0.1875 0.03 0.0056 0.00105 0.00020 0.000037 0.125 0.05 0.00625 0.00078 0.000098 0.000012 2.284 8.616 4.00 14.58 2.00 0.75 0.50 0.25 0.125 Totals: 37.991 From equation 1.11, the mass mean diameter is:  dv = n1 d14 n1 d13 = (177.92/37.991) = 4.683 mm. From equation 1.14, the surface mean diameter is:  ds = n1 d13 n1 d12 = (37.991/8.616) = 4.409 mm. From equation 1.18, the length mean diameter is:  d1 = n1 d12 n1 d1 = (8.616/2.284) = 3.772 mm. From equation 1.19, the mean length diameter is:  d1′ = n1 d1 n1 = (2.284/1.0) = 2.284 mm. 9 177.92 In the present case, which is concerned with power consumption per unit mass, the mass mean diameter is probably of the greatest relevance. For the purposes of calculation a mean value of 4.0 mm will be used, which agrees with the value obtained by Bond’s method. For coarse crushing, Kick’s law may be used as follows: Case 1: mean diameter of feed = 45 mm, mean diameter of product = 4 mm, energy consumption = 13.0 kJ/kg, compressive strength = 22.5 N/m2 In equation 2.4: 13.0 = KK × 22.5 ln(45/4) and: KK = (13.0/54.4) = 0.239 kW/(kg/s) (MN/m2 ) Case 2: mean diameter of feed = 42.5 mm, mean diameter of product = 0.50 mm compressive strength = 45 MN/m2 Thus: E = 0.239 × 45 ln(42.5/0.50) = (0.239 × 199.9) = 47.8 kJ/kg or, for a feed of 1 kg/s, the energy required = 47.8 kW. PROBLEM 2.3 A crusher reducing limestone of crushing strength 70 MN/m2 from 6 mm diameter average size to 0.1 mm diameter average size, requires 9 kW. The same machine is used to crush dolomite at the same output from 6 mm diameter average size to a product consisting of 20 per cent with an average diameter of 0.25 mm, 60 per cent with an average diameter of 0.125 mm and a balance having an average diameter of 0.085 mm. Estimate the power required, assuming that the crushing strength of the dolomite is 100 MN/m2 and that crushing follows Rittinger’s Law. Solution The mass mean diameter of the crushed dolomite may be calculated thus: n1 d1 n1 d13 n1 d14 0.20 0.60 0.20 0.250 0.125 0.085 0.003125 0.001172 0.000123 0.00078 0.000146 0.000011 0.00442 0.000937 Totals: and from equation 1.11:  dv = n1 d14 n1 d13 = (0.000937/0.00442) = 0.212 mm. 10 For Case 1: E = 9.0 kW, fc = 70.0 MN/m2 , L1 = 6.0 mm, and L2 = 0.1 mm and in equation 2.3: 9.0 = KR × 70.0[(1/0.1) − (1/6.0)] or: KR = 0.013 kW mm/(MN/m2 ) For Case 2: fc = 100.0 MN/m2 , Hence: L1 = 6.0 mm and L2 = 0.212 mm E = 0.013 × 100.0[(1/0.212) − (1/6.0)] = 5.9 kW PROBLEM 2.4 If crushing rolls 1 m diameter are set so that the crushing surfaces are 12.5 mm apart and the angle of nip is 31◦ , what is the maximum size of particle which should be fed to the rolls? If the actual capacity of the machine is 12 per cent of the theoretical, calculate the throughput in kg/s when running at 2.0 Hz if the working face of the rolls is 0.4 m long and the feed density is 2500 kg/m3 . Solution See Volume 2, Example 2.2. PROBLEM 2.5 A crushing mill reduces limestone from a mean particle size of 45 mm to the following product: Size (mm) Amount of product (per cent) 12.5 0.5 7.5 7.5 5.0 45.0 2.5 19.0 1.5 16.0 0.75 8.0 0.40 3.0 0.20 1.0 It requires 21 kJ/kg of material crushed. Calculate the power required to crush the same material at the same rate, from a feed having a mean size of 25 mm to a product with a mean size of 1 mm. 11 Solution The mean size of the product may be obtained thus: n1 d1 n1 d13 n1 d14 0.5 7.5 45.0 19.0 16.0 8.0 3.0 1.0 12.5 7.5 5.0 2.5 1.5 0.75 0.40 0.20 3906 3164 5625 296.9 54.0 3.375 0.192 0.008 48,828 23,731 28,125 742.2 81.0 2.531 0.0768 0.0016 Totals: 13,049 101,510 and from equation 1.11, the mass mean diameter is:  dv = n1 d14 n1 d13 = (101,510/13,049) = 7.78 mm Kick’s law is used as the present case may be regarded as coarse crushing. Case 1: E = 21 kJ/kg, L1 = 45 mm and L2 = 7.8 mm. In equation 2.4: 21 = KK fc ln(45/7.8) KK fc = 11.98 kJ/kg and: Case 2: L1 = 25 mm and L2 = 1.0 mm. Thus: E = 11.98 ln(25/1.0) = 38.6 kJ/kg PROBLEM 2.6 A ball-mill 1.2 m in diameter is run at 0.8 Hz and it is found that the mill is not working satisfactorily. Should any modification in the condition of operation be suggested? Solution See Volume 2, Example 2.3. 12 PROBLEM 2.7 Power of 3 kW is supplied to a machine crushing material at the rate of 0.3 kg/s from 12.5 mm cubes to a product having the following sizes: 80 per cent 3.175 mm, 10 per cent 2.5 mm and 10 per cent 2.25 mm. What power should be supplied to this machine to crush 0.3 kg/s of the same material from 7.5 mm cube to 2.0 mm cube? Solution The mass mean diameter is calculated thus: n1 d1 n1 d13 n1 d14 0.8 0.1 0.1 3.175 2.5 2.25 25.605 1.563 1.139 81.295 3.906 2.563 and from equation 1.11:  dv = n1 d14 n1 d13 = (87.763/28.307) = 3.1 mm (Using Bond’s approach, the mean diameter is clearly 3.175 mm.) For the size ranges involved, the crushing may be considered as intermediate and Bond’s law will be used. Case 1: E = (3/0.3) = 10 kW/(kg/s), L1 = 12.5 mm and L2 = 3.1 mm. Thus in equation 2.5: or: √ q = (L1 /L2 ) = 4.03 and E = 2C (1/L2 )(1 − 1/q 0.5 ) √ 10 = 2C (1/3.1)(1 − 1/4.030.5 ) = (2C × 0.568 × 0.502). C = 17.54 kW mm0.5 /(kg/s) Thus: Case 2: L1 = 7.5 mm, Hence: L2 = 2.0 mm and q = (7.5/2.0) = 3.75. E = 2 × 17.54(1/2.0)(1 − 1/3.750.5 ) = (35.08 × 0.707 × 0.484) = 12.0 kJ/kg For a feed of 0.3 kg/s, the power required = (12.0 × 0.3) = 3.6 kW 13 SECTION 2-3 Motion of Particles in a Fluid PROBLEM 3.1 A finely ground mixture of galena and limestone in the proportion of 1 to 4 by mass, is subjected to elutriation by a current of water flowing upwards at 5 mm/s. Assuming that the size distribution for each material is the same, and is as follows, estimate the percentage of galena in the material carried away and in the material left behind. The absolute viscosity of water is 1 mN s/m2 and Stokes’ equation should be used. Diameter (µm) Undersize (per cent mass) 20 15 30 28 40 48 50 54 60 64 70 72 80 78 100 88 The density of galena is 7500 kg/m3 and the density of limestone is 2700 kg/m3 . Solution See Volume 2, Example 3.2. PROBLEM 3.2 Calculate the terminal velocity of a steel ball, 2 mm diameter and of density 7870 kg/m3 in an oil of density 900 kg/m3 and viscosity 50 mN s/m2 . Solution For a sphere: ′ (R0′ /ρu20 )Re02 = (2d 3 /3μ2 )ρ(ρs − ρ)g (equation 3.34) = (2 × 0.0023 /3 × 0.052 )900(7870 − 900)9.81 = 131.3 log10 131.3 = 2.118 From Table 3.4: log10 Re0′ = 0.833 or: Thus: Re0′ = 6.80 u0 = (6.80 × 0.05)/(900 × 0.002) = 0.189 m/s 14 PROBLEM 3.3 What is the terminal velocity of a spherical steel particle of 0.40 mm diameter, settling in an oil of density 820 kg/m3 and viscosity 10 mN s/m2 ? The density of steel is 7870 kg/m3 . Solution See Volume 2, Example 3.1. PROBLEM 3.4 What are the settling velocities of mica plates, 1 mm thick and ranging in area from 6 to 600 mm2 , in an oil of density 820 kg/m3 and viscosity 10 mN s/m2 ? The density of mica is 3000 kg/m3 . Solution A′ dp dp3 volume k′ Smallest particles Largest particles 6× 10−6 m2 [(4 × 6 × 10−6 )/π] = 2.76 × 10−3 m 2.103 × 10−8 m3 6 × 10−9 m3 0.285 6 × 10−4 m2 √ [(4 × 6 × 10−4 )/π] = 2.76 × 10−2 m 2.103 × 10−5 m3 6 × 10−7 m3 0.0285 ′ (R0′ /ρu2 )Re02 = (4k ′ /μ2 π)(ρs − ρ)ρdp3 g (equation 3.52) = [(4 × 0.285)/(π × 0.012 )](3000 − 820)(820 × 2.103 × 10−8 × 9.81) = 1340 for smallest particle and 134,000 for largest particle Smallest particles ′ log10 (R0′ /ρu2 )Re02 log10 Re0′ Correction from Table 3.6 Corrected log10 Re0′ Re0′ u 3.127 1.581 −0.038 1.543 34.9 0.154 m/s Largest particles 5.127 2.857 (from Table 3.4) −0.300 (estimated) 2.557 361 0.159 m/s Thus it is seen that all the mica particles settle at approximately the same velocity. PROBLEM 3.5 A material of density 2500 kg/m3 is fed to a size separation plant where the separating fluid is water which rises with a velocity of 1.2 m/s. The upward vertical component of 15 the velocity of the particles is 6 m/s. How far will an approximately spherical particle, 6 mm diameter, rise relative to the walls of the plant before it comes to rest in the fluid? Solution See Volume 2, Example 3.4. PROBLEM 3.6 A spherical glass particle is allowed to settle freely in water. If the particle starts initially from rest and if the value of the Reynolds number with respect to the particle is 0.1 when it has attained its terminal falling velocity, calculate: (a) the distance travelled before the particle reaches 90 per cent of its terminal falling velocity, (b) the time elapsed when the acceleration of the particle is one hundredth of its initial value. Solution When Re′ < 0.2, the terminal velocity is given by equation 3.24: u0 = (d 2 g/18μ)(ρs − ρ) Taking the densities of glass and water as 2750 and 1000 kg/m3 , respectively, and the viscosity of water as 0.001 Ns/m2 , then: u0 = [(9.81d 2 )/(18 × 0.001)](2750 − 1000) = 9.54 × 105 d 2 m/s The Reynolds number, Re′ = 0.1 and substituting for u0 : d(9.54 × 105 d 2 )(1000/0.001) = 0.1 or: d = 4.76 × 10−5 m a = 18μ/d 2 ρs = (18 × 0.001)/[4.76 × 10−5 )2 × 2750] (equation 3.89) b = [1 − (ρ/ρs )]g = [1 − (1000/2750)]9.81 = 6.24 m/s2 (equation 3.90) = 2889 s−1 and: In equation 3.88: y= v b b t+ − 2+ a a a   v −at b e − a2 a In this case v = 0 and differentiating gives: ẏ = (b/a)(1 − e−at ) 16 or, since (b/a) = u0 , the terminal velocity: ẏ = u0 (1 − e−at ) 0.9 = (1 − e−2889t ) When ẏ = 0.9u0 , then: 2889t = 2.303 or: and t = 8.0 × 10−4 s Thus in equation 3.88: y = (6.24 × 8.0 × 10−4 )/2889 − (6.24/28892 ) + (6.24/28892 ) exp(−2889 × 8.0 × 10−4 ) = (1.73 × 10−6 ) − (7.52 × 10−7 ) + (7.513 × 10−8 ) = 1.053 × 10−6 m or 1.05 mm From equation 3.86: ÿ = b − a ẏ At the start of the fall, ẏ = 0 and the initial acceleration, ÿ = b. When ÿ = 0.01b, then: 0.01b = b − a ẏ or: Thus: or: and: ẏ = (0.89 × 6.24)/2889 = 0.00214 m/s 0.00214 = (6.24/2889)(1 − e−2889t ) 2889t = 4.605 t = 0.0016 s PROBLEM 3.7 In a hydraulic jig, a mixture of two solids is separated into its components by subjecting an aqueous slurry of the material to a pulsating motion, and allowing the particles to settle for a series of short time intervals such that their terminal falling velocities are not attained. Materials of densities 1800 and 2500 kg/m3 whose particle size ranges from 0.3 mm to 3 mm diameter are to be separated. It may be assumed that the particles are approximately spherical and that Stokes’ Law is applicable. Calculate approximately the maximum time interval for which the particles may be allowed to settle so that no particle of the less dense material falls a greater distance than any particle of the denser material. The viscosity of water is 1 mN s/m2 . Solution For Stokes’ law to apply, Re′ < 0.2 and equation 3.88 may be used:   v b b b v −at − e y= t+ − 2+ a a a a2 a 17 or, assuming the initial velocity v = 0: y= b b b t − 2 + 2 e−at a a a and a = 18μ/d 2 ρs . (equations 3.89 and 3.90) b = [1 − (ρ/ρs )]g where: For small particles of the dense material: b = [1 − (1000/2500)]9.81 = 5.89 m/s2 a = (18 × 0.001)/[(0.3 × 10−3 )2 2500] = 80 s−1 For large particles of the light material: b = [1 − (1000/1800)]9.81 = 4.36 m/s2 a = (18 × 0.001)/[(3 × 10−3 )2 1800] = 1.11 s−1 In order that these particles should fall the same distance, from equation 3.88: (5.89/80)t − (5.89/802 )(1 − e−80t ) = (4.36/1.11)t − (4.36/1.112 )(1 − e−1.11t ) 3.8504t + 3.5316 e−1.11t − 0.00092 e−80t = 3.5307 Thus: and, solving by trial and error: t = 0.01 s PROBLEM 3.8 Two spheres of equal terminal falling velocity settle in water starting from rest at the same horizontal level. How far apart vertically will the particles be when they have both reached 99 per cent of their terminal falling velocities? It may be assumed that Stokes’ law is valid and this assumption should be checked. The diameter of one sphere is 40 µm and its density is 1500 kg/m3 and the density of the second sphere is 3000 kg/m3 . The density and viscosity of water are 1000 kg/m3 and 1 mN s/m2 respectively. Solution Assuming Stokes’ law is valid, the terminal velocity is given by equation 3.24 as: u0 = (d 2 g/18μ)(ρs − ρ) For particle 1: u0 = {[(40 × 10−6 )2 × 9.81]/(18 × 1 × 10−3 )}(1500 − 1000) = 4.36 × 10−4 m/s 18 Since particle 2 has the same terminal velocity: 4.36 × 10−4 = [(d22 × 9.81)/(18 × 1 × 10−3 )](3000 − 1000) d2 = (2 × 10−5 ) m or 20 µm From which: a = 18μ/d 2 ρs From equation 3.83: a1 = (18 × 1 × 10−3 )/((40 × 10−6 )2 × 1500) = 7.5 × 103 s−1 For particle 1: a2 = (18 × 1 × 10−3 )/((20 × 10−6 )2 × 3000) = 1.5 × 104 s−1 and for particle 2: From equation 3.90: b = (1 − ρ/ρs )g b1 = (1 − 1000/1500)9.81 = 3.27 m/s2 For particle 1: b2 = (1 − 1000/3000)9.81 = 6.54 m/s2 and for particle 2: The initial velocity of both particles, v = 0, and from equation 3.88: y= b b b t − 2 + 2 e−at a a a Differentiating: ẏ = (b/a)(1 − e−at ) or, from equation 3.24: ẏ = ut (1 − e−at ) When ẏ = u0 , the terminal velocity, it is not possible to solve for t and hence ẏ will be taken as 0.99u0 . For particle 1: (0.99 × 4.36 × 10−4 ) = (4.36 × 10−4 )[1 − exp(−7.5 × 103 t)] t = 6.14 × 10−4 s and: The distance travelled in this time is given by equation 3.88: y = (3.27/7.5 × 103 )6.14 × 10−4 − [3.27/(7.5 × 103 )2 ] [1 − exp(−7.5 × 103 × 6.14 × 10−4 )] = 2.10 × 10−7 m For particle 2: (0.99 × 4.36 × 10−4 ) = (4.36 × 10−4 )[1 − exp(−1.5 × 104 t)] and: t = 3.07 × 10−4 s Thus: y = ((6.54/1.5 × 104 )3.07 × 10−4 ) − [6.54/(1.5 × 104 )2 ] [1 − exp(−1.5 × 104 × 3.07 × 10−4 )] = 1.03 × 10−7 m Particle 2 reaches 99 per cent of its terminal velocity after 3.07 × 10−4 s and it then travels at 4.36 × 10−4 m/s for a further (6.14 × 10−4 − 3.07 × 10−4 ) = 3.07 × 10−4 s during which time it travels a further (3.07 × 10−4 × 4.36 × 10−4 ) = 1.338 × 10−7 m. 19 Thus the total distance moved by particle 1 = 2.10 × 10−7 m and the total distance moved by particle 2 = (1.03 × 10−7 + 1.338 × 10−7 ) = 2.368 × 10−7 m. The distance apart when both particles have attained their terminal velocities is: (2.368 × 10−7 − 2.10 × 10−7 ) = 2.68 × 10−8 m For Stokes’ law to be valid, Re′ must be less than 0.2. For particle 1, Re = (40 × 10−6 × 4.36 × 10−4 × 1500)/(1 × 10−3 ) = 0.026 and for particle 2, Re = (20 × 10−6 × 4.36 × 10−4 × 3000)/(1 × 10−3 ) = 0.026 and Stokes’ law applies. PROBLEM 3.9 The size distribution of a powder is measured by sedimentation in a vessel having the sampling point 180 mm below the liquid surface. If the viscosity of the liquid is 1.2 mN s/m2 , and the densities of the powder and liquid are 2650 and 1000 kg/m3 respectively, determine the time which must elapse before any sample will exclude particles larger than 20 µm. If Stokes’ law applies when the Reynolds number is less than 0.2, what is the approximate maximum size of particle to which Stokes’ Law may be applied under these conditions? Solution The problem involves determining the time taken for a 20 µm particle to fall below the sampling point, that is 180 mm. Assuming that Stokes’ law is applicable, equation 3.88 may be used, taking the initial velocity as v = 0. Thus: y = (bt/a) − (b/a 2 )(1 − e−at ) where: b = g(1 − ρ/ρs ) = 9.81[1 − (1000/2650)] = 6.108 m/s2 and: a = 18μ/d 2 ρs = (18 × 1.2 × 10−3 )/[(20 × 10−6 )2 × 2650] = 20,377 s−1 In this case: Thus: y = 180 mm or 0.180 m 0.180 = (6.108/20,377)t − (6.108/20,3772 )(1 − e−20,377t ) = 0.0003t + (1.4071 × 10−8 e−20,377t ) Ignoring the exponential term as being negligible, then: t = (0.180/0.0003) = 600 s The velocity is given by differentiating equation 3.88 giving: ẏ = (b/a)(1 − e−at ) 20 When t = 600 s: ẏ = [(6.108d 2 × 2650)/(18 × 0.0012)]{1 − exp[−(18 × 0.0012 × 600)/d 2 × 2650]} = 7.49 × 105 d 2 [1 − exp(−4.89 × 10−3 d −2 )] For Re′ = 0.2, then d(7.49 × 105 d 2 )[1 − exp(−4.89 × 10−3 d −2 )] × 2650/0.0012 = 0.2 1.65 × 1012 d 3 [1 − exp(−4.89 × 10−3 d −2 )] = 0.2 or: When d is small, the exponential term may be neglected and: d 3 = 1.212 × 10−13 d = 5.46 × 10−5 m or or: 54.6 µm PROBLEM 3.10 Calculate the distance a spherical particle of lead shot of diameter 0.1 mm settles in a glycerol/water mixture before it reaches 99 per cent of its terminal falling velocity. The density of lead is 11,400 kg/m3 and the density of liquid is 1000 kg/m3 . The viscosity of liquid is 10 mN s/m2 . It may be assumed that the resistance force may be calculated from Stokes’ Law and is equal to 3πμdu, where u is the velocity of the particle relative to the liquid. Solution The terminal velocity, when Stokes’ law applies, is given by: or: 1 3 πd (ρs − ρ)g = 3πμdu0 6 d 2 ρs d 2g (ρs − ρ) = g(1 − ρ/ρs ) = (b/a) (equations 3.24, 3.89 and 3.90) u0 = 18μ 18μ where: b = g(1 − ρ/ρs ) = 9.81[1 − (1000/11,400)] = 8.95 m/s2 and: Thus: a = 18μ/d 2 ρs = (18 × 10 × 10−3 )/[(0.1 × 10−3 )2 11,400] = 1579 s−1 u0 = (8.95/1579) = 5.67 × 10−3 m/s When 99 per cent of this velocity is attained, then: ẏ = (0.99 × 5.67 × 10−3 ) = 5.61 × 10−3 m/s Assuming that the initial velocity v is zero, then equation 3.88 may be differentiated to give: ẏ = (b/a)(1 − e−at ) Thus: (5.61 × 10−3 ) = (5.67 × 10−3 )(1 − e−1579t ) 21 and t = 0.0029 s Substituting in equation 3.88: y = (b/a)t − (b/a 2 )(1 − e−at ) = (5.67 × 10−3 × 0.0029) − (5.67 × 10−3 /1579)(1 − e−1579×0.0029 ) = (1.644 × 10−5 ) − (3.59 × 10−6 × 9.89 × 10−1 ) = 1.29 × 10−5 m or 0.013 mm PROBLEM 3.11 What is the mass of a sphere of material of density 7500 kg/m3 whose terminal velocity in a large deep tank of water is 0.6 m/s? Solution R0′ 2μg ′ Re −1 = 2 3 (ρs − ρ) ρu20 0 3ρ u0 (equation 3.41) Taking the density and viscosity of water as 1000 kg/m3 and 0.001 N s/m2 respectively, then: (R0′ /ρu20 )/Re0′ = [(2 × 0.001 × 9.81)/(3 × 10002 × 0.63 )](7500 − 1000) = 0.000197 log10 (R0′ /ρu20 )/Re0′ = 4.296 Thus: From Table 3.5, log10 Re0′ = 3.068 Re0′ = 1169.5 and: d = (1169.5 × 0.001)/(0.6 × 1000) = 0.00195 m or 1.95 mm. The mass of the sphere = πd 3 ρs /6 = (π × 0.001953 × 7500)/6 = 2.908 × 10−5 kg or 0.029 g PROBLEM 3.12 Two ores, of densities 3700 and 9800 kg/m3 are to be separated in water by a hydraulic classification method. If the particles are all of approximately the same shape and each is sufficiently large for the drag force to be proportional to the square of its velocity in the fluid, calculate the maximum ratio of sizes which can be completely separated if the particles attain their terminal falling velocities. Explain why a wider range of sizes can be 22 separated if the time of settling is so small that the particles do not reach their terminal velocities. An explicit expression should be obtained for the distance through which a particle will settle in a given time if it starts from rest and if the resistance force is proportional to the square of the velocity. The acceleration period should be taken into account. Solution If the total drag force is proportional to the square of the velocity, then when the terminal velocity u0 is attained: F = k1 u20 dm2 since the area is proportional to dp2 and the accelerating force = (ρs − ρ)gk2 dp3 where k2 is a constant depending on the shape of the particle and dp is a mean projected area. When the terminal velocity is reached, then: k1 u20 dp2 = (ρs − ρ)gk2 dp3 and: u0 = [(ρs − ρ)gk3 dp ]0.5 In order to achieve complete separation, the terminal velocity of the smallest particle (diameter d1 ) of the dense material must exceed that of the largest particle (diameter d2 ) of the light material. For equal terminal falling velocities: [(9800 − 1000)9.81k3 d1 ]0.5 = [(3700 − 1000)9.81k3 d2 ]0.5 and: (d2 /d1 ) = (8800/2700) = 3.26 which is the maximum range for which complete separation can be achieved if the particles settle at their terminal velocities. If the particles are allowed to settle in a suspension for only very short periods, they will not attain their terminal falling velocities and a better degree of separation may be obtained. All particles will have an initial acceleration g(1 − ρ/ρs ) because no fluid frictional force is exerted at zero particle velocity. Thus the initial acceleration is a function of density only, and is unaffected by both size and shape. A very small particle of the denser material will therefore always commence settling at a higher rate than a large particle of the less dense material. Theoretically, therefore, it should be possible to effect complete separation irrespective of the size range, provided that the periods of settling are sufficiently short. In practice, the required periods will often be so short that it is impossible to make use of this principle alone. As the time of settling increases some of the larger particles of the less dense material will catch up and then overtake the smaller particles of the denser material. If the total drag force is proportional to the velocity squared, that is to ẏ 2 , then the equation of motion for a particle moving downwards under the influence of gravity may be written as: mÿ = mg(1 − ρ/ρs ) − k1 ẏ 2 Thus: or: ÿ = g(1 − ρ/ρs ) − (k1 /m)ẏ 2 ÿ = b − cẏ 2 23 where b = g(1 − ρ/ρs ), c = k1 /m, and k1 is a proportionality constant. dẏ/(b − cẏ 2 ) = dt Thus: dẏ/(f 2 − ẏ 2 ) = c dt or: where f = (b/c)0.5 . Integrating: When t = 0, then: Thus: (1/2f ) ln[(f + ẏ)/(f − ẏ)] = ct + k4 ẏ = 0 and k4 = 0 (1/2f ) ln[(f + ẏ)/(f − ẏ)] = ċt (f + ẏ)/(f − ẏ) = e2f ct f − ẏ = 2f/(1 + e2f ct )  t dt/(1 − e2f ct ) y = f t − 2f 0 y = f t − (1/c) ln[e2f ct /(1 + e2f ct )] + k5 when t = 0, then: Thus: y = 0 and k5 = (1/c) ln 0.5 y = f t − (1/c) ln (0.5e2f ct )/(1 + e2f ct ) where f = (b/c)0.5 , b = g(1 − ρ/ρs ), and c = k1 /m. PROBLEM 3.13 Salt, of density 2350 kg/m3 , is charged to the top of a reactor containing a 3 m depth of aqueous liquid of density 1100 kg/m3 and of viscosity 2 mN s/m2 and the crystals must dissolve completely before reaching the bottom. If the rate of dissolution of the crystals is given by: dd − = 3 × 10−6 + 2 × 10−4 u dt where d is the size of the crystal (m) at time t (s) and u is its velocity in the fluid (m/s), calculate the maximum size of crystal which should be charged. The inertia of the particles may be neglected and the resistance force may be taken as that given by Stokes’ Law (3πμdu) where d is the equivalent spherical diameter of the particle. Solution See Volume 2, Example 3.5. PROBLEM 3.14 A balloon of mass 7 g is charged with hydrogen to a pressure of 104 kN/m2 . The balloon is released from ground level and, as it rises, hydrogen escapes in order to maintain a 24 constant differential pressure of 2.7 kN/m2 , under which condition the diameter of the balloon is 0.3 m. If conditions are assumed to remain isothermal at 273 K as the balloon rises, what is the ultimate height reached and how long does it take to rise through the first 3000 m? It may be assumed that the value of the Reynolds number with respect to the balloon exceeds 500 throughout and that the resistance coefficient is constant at 0.22. The inertia of the balloon may be neglected and at any moment, it may be assumed that it is rising at its equilibrium velocity. Solution Volume of balloon = (4/3)π(0.15)3 = 0.0142 m3 . Mass of balloon = 7 g or 0.007 kg. The upthrust = (weight of air at a pressure of P N/m2 ) − (weight of hydrogen at a pressure of (P + 2700) N/m2 ). The density of air ρa at 101,300 N/m2 and 273 K = (28.9/22.4) = 1.29 kg/m3 , where the mean molecular mass of air is taken as 28.9 kg/kmol. The net upthrust force W on the balloon is given by: W = 9.81{0.0142[(ρa P /101,300) − ρa (2/28.9)(P + 2700)/101,300] − 0.007} = 0.139[0.0000127P − 0.000000881(P + 2700)] − 0.0687 = (0.00000164P − 0.0690) N (i) The balloon will stop using when W = 0, that is when: P = (0.0690/0.00000164) = 42,092 N/m2 . From equation 2.43 in Volume 1, the variation of pressure with height is given by: g dz + v dP = 0 For isothermal conditions: v = (1/ρa )(101,300/P ) m3 Thus: dz + [101,300/(9.81 × 1.29P )] dP = 0 (z2 − z1 ) = 8005 ln(101,300/P ) and, on integration: When P = 42,092 N/m2 , (z2 − z1 ) = 8005 ln(101,300/42,092) = 7030 m The resistance force per unit projected area R on the balloon is given by: (R/ρa u2 ) = 0.22 or: R = 0.22ρa (P /101,300)(π × 0.32 /4)(dz/dt)2 N/m2 = 1.98 × 10−7 P (dz/dt)2 25 This must be equal to the net upthrust force W , given by equation (i), 0.00000164P − 0.0690 = (1.98 × 10−7 P )(dz/dt)2 or: (dz/dt)2 = (8.28 − 3.49 × 105 )/P and: z = 8005 ln(101,300/P ) But: (dz/dt)2 = 8.28 − [(3.49 × 105 ) ez/8005 ]/101,300 Therefore: (dz/dt) = 1.89(2.41 − e1.25×10 and: −4z 0.5 ) The time taken to rise 3000 m is therefore given by:  3000 −4z dz/(2.41 − e1.25×10 )0.5 t = (1/1.89) 0 I= Writing the integral as:  3000 dz/(a − ebz )0.5 0 (a − ebz ) = x 2 and putting: dz = 2x dx/[b(a − x 2 )] then: I = (−2/b) and:  dx/(a − x 2 ) √  √ 3000 a − (a − ebz ) √ = (−2/b)[1/2( a)] ln √ √ a + (a − ebz ) 0 √ √ √ √ √ [ a − (a − e3000b )][ a + (a − 1)] = [1/(b a)] ln √ √ √ √ [ a + (a − e3000b )][ a − (a − 1)] Substituting: a = 2.41 and b = 1.25 × 10−4 then: I = 5161 ln[(1.55 − 0.977)/(1.55 + 0.977)][(1.55 − 1.19)/(1.55 + 1.19)] = 2816 Thus: t = [2816(1/1.89)] = 1490 s (25 min) PROBLEM 3.15 A mixture of quartz and galena of densities 3700 and 9800 kg/m3 respectively with a size range of 0.3 to 1 mm is to be separated by a sedimentation process. If Stokes’ Law is applicable, what is the minimum density required for the liquid if the particles all settle at their terminal velocities? A separating system using water as the liquid is considered in which the particles were to be allowed to settle for a series of short time intervals so that the smallest particle of galena settled a larger distance than the largest particle of quartz. What is the approximate maximum permissible settling period? 26 According to Stokes’ Law, the resistance force F acting on a particle of diameter d, settling at a velocity u in a fluid of viscosity μ is given by: F = 3πμ du The viscosity of water is 1 mN s/m2 . Solution For particles settling in the Stokes’ law region, equation 3.32 applies: dg /dA = [(ρA − ρ)/(ρB − ρ)]0.5 For separation it is necessary that a large particle of the less dense material does not overtake a small particle of the dense material, or: (1/0.3) = [(9800 − ρ)/(3700 − ρ)]0.5 and ρ = 3097 kg/m3 Assuming Stokes’ law is valid, the distance travelled including the period of acceleration is given by equation 3.88: y = (b/a)t + (v/a) − (b/a 2 ) − [(b/a 2 ) − (v/a)]e−at When the initial velocity v = 0, then: y = (b/a)t + (b/a 2 )(e−at − 1) where: a = 18μ/d 2 ρs (equation 3.89) and: b = g(1 − ρ/ρs ) (equation 3.90) For a small particle of galena b = 9.81[1 − (1000/9800)] = 8.81 m/s2 a = (18 × 1 × 10−3 )/[(0.3 × 10−3 )2 × 9800] = 20.4 s−1 For a large particle of quartz b = 9.81[1 − (1000/3700)] = 7.15 m/s2 a = (18 × 1 × 10−3 )/[(1 × 10−3 )2 × 3700] = 4.86 s−1 In order to achieve separation, these particles must travel at least the same distance in time t. Thus: (8.81/20.4)t + (8.81/20.42 )(e−20.4t − 1) = (7.15/4.86)t + (7.15/4.862 )(e−4.86t − 1) or: (0.0212 e−20.4t − 0.303 e−4.86t ) = 1.039t − 0.282 and solving by trial and error: t = 0.05 s 27 PROBLEM 3.16 A glass sphere, of diameter 6 mm and density 2600 kg/m3 , falls through a layer of oil of density 900 kg/m3 into water. If the oil layer is sufficiently deep for the particle to have reached its free falling velocity in the oil, how far will it have penetrated into the water before its velocity is only 1 per cent above its free falling velocity in water? It may be assumed that the force on the particle is given by Newton’s law and that the particle drag coefficient R ′ /ρu2 = 0.22. Solution The settling velocity in water is given by equation 3.25, assuming Newton’s law, or: u20 = 3dg(ρs − ρ)/ρ For a solid density of 2600 kg/m3 and a particle diameter of (6/1000) = 0.006 m, then: u20 = (3 × 0.006 × 9.81)(2600 − 1000)/1000 and u0 = 0.529 m/s The Reynolds number may now be checked taking the viscosity of water as 0.001 Ns/m2 . Thus: Re′ = (0.529 × 0.006 × 1000)/0.001 = 3174 which is very much in excess of 500, which is the minimum value for Newton’s law to be applicable. The settling velocity in an oil of density 900 kg/m3 is also given by equation 3.25 as: u20 = (3 × 0.006 × 9.81)(2600 − 900)/900 and u0 = 0.577 m/s. Using the nomenclature of Chapter 3 in Volume 2, a force balance on the particle in water gives: mÿ = mg(1 − ρ/ρs ) − Aρ ẏ 2 (R ′ /ρu2 ) Substituting R ′ /ρu2 = 0.22, then: ÿ = g(1 − ρ/ρs ) − 0.22(Aρ/m)ẏ 2 = 9.81(1 − (1000/2600)) − 0.22((π/4)d 2 ρ ẏ 2 )/((π/6)d 3 ρs ) = 6.03 − (0.33ẏ 2 (1000/2600)/0.006) = 6.03 − 21.4ẏ 2 or from equation 3.97: ÿ = b − cẏ 2 Following the reasoning in Volume 2, Section 3.6.3, for downward motion, then: y = f t + (1/c) ln(1/2f )[f + v + (f − v)e−2f ct ] 28 (equation 3.101) where f = (b/c)0.5 . Thus: ẏ = f + (1/c){1/[(f + v) + (f − v)e−2f ct ][(f − v)e−2f ct (−2f c)]} = f [1 − {c/[1 + (f + v)e2f ct /(f − v)]}]   c =f 1− 1 + (f + v)e2f ct /(f − v) y = f = (b/c)0.5 When t = ∞: = (6.03/21.4)0.5 = 0.529 m/s, as before. The initial velocity, v = 0.577 m/s. (f + v)/(f − v) = (0.529 + 0.577)/(0.529 − 0.577) = −23.04 Thus: 2f c = (2 × 0.529 × 21.4) = 22.6 When (ẏ/f ) = 1.01, then: 1.01 = 1 − 2/(1 − 23.04 e22.6t ) e22.6t = 8.72 and: and t = 0.096 s In equation 3.101: y = (0.529 × 0.0958) + (1/21.4) ln(1/(2 × 0.529))(0.529 + 0.577) + (0.529 − 0.577) exp[−(22.6 × 0.0958)] and: y = 0.048 m or 48 mm PROBLEM 3.17 Two spherical particles, one of density 3000 kg/m3 and diameter 20 µm, and the other of density 2000 kg/m3 and diameter 30 µm start settling from rest at the same horizontal level in a liquid of density 900 kg/m3 and of viscosity 3 mN s/m2 . After what period of settling will the particles be again at the same horizontal level? It may be assumed that Stokes’ Law is applicable, and the effect of added mass of the liquid moved with each sphere may be ignored. Solution For motion of a sphere in the Stokes’ law region equation 3.88 is valid: y = (b/a)t + (v/a) − (b/a 2 ) + [(b/a 2 ) − (v/a)]e−at When the initial velocity, v = 0, then: y = (b/a)t − (b/a 2 )(1 − e−at ) 29 (i) From equation 3.89, a = 18μ/(d 2 ρs ) and hence, for particle 1: a1 = (18 × 3 × 10−3 )/[(20 × 10−6 )2 × 3000] = 45,000 and for particle 2: a2 = (18 × 3 × 10−3 )/[(30 × 10−6 )2 × 2000] = 30,000 Similarly: b = g(1 − (ρ/ρs )) (equation 3.30) For particle 1: b1 = 9.81[1 − (900/3000)] = 6.867 and for particle 2: b2 = 8.81[1 − (900/2000)] = 5.395 Substituting for a1 , a2 , b1 , b2 in equation (i), then: y1 = (6.867/45,000)t − (6.867/45,0002 )(1 − e−45,000t ) (ii) y2 = (5.395/30,000)t − (5.395/30,000 )(1 − e (iii) 2 −30,000t ) Putting y1 = y2 , that is equating (ii) and (iii), then: t = 0.0002203(1 − e−30000t ) − 0.0001247(1 − e−45000t ) and solving by trial and error: t = 7.81 × 10−5 s PROBLEM 3.18 A binary suspension consists of equal masses of spherical particles of the same shape and density whose free falling velocities in the liquid are 1 mm/s and 2 mm/s, respectively. The system is initially well mixed and the total volumetric concentration of solids is 0.2. As sedimentation proceeds, a sharp interface forms between the clear liquid and suspension consisting only of small particles, and a second interface separates the suspension of fines from the mixed suspension. Using a suitable model for the behaviour of the system, estimate the falling rates of the two interfaces. It may be assumed that the sedimentation velocity uc in a concentrated suspension of voidage e is related to the free falling velocity u0 of the particles by: uc /u0 = e2.3 Solution In the mixture, the relative velocities of the particles, uP are given by: for the large particles: uP L = u0L en−1 (from equation 5.108) 30 and for the small particles: uP S = u0S en−1 If the upward fluid velocity is uF m/s, then the sedimentation velocities are: for the large particles: ucL = u0L en−1 − uF and for the small particles: ucS = u0S en−1 − uF Combining these equations and noting that the concentrations of large and small particles are equal then: uF e = ucL (1 − e)/2 + ucS (1 − e)/2 = (u0L en−1 − uF )(1 − e)/2 + (u0S en−1 − uF )(1 − e)/2 uF = (en−1 (1 − e)/2)(u0L + u0S ) Thus: ucL = u0L en−1 − (en−1 (1 − e)/2)(u0L + u0S ) and: = en−1 [u0L (1 + e)/2 − ucS (1 − e)/2] (i) Similarly: uCS = en−1 [u0S (1 + e)/2 − u0L (1 − e)/2] (ii) If, in the upper zone, the settling velocity of the fine particles and the voidage are ux and ex respectively, (ux /u0S ) = exn then: (iii) The rate at which solids are entering the upper, single-size zone is (ucL − ucS )(1 − e)/2, per unit area, and the rate at which the zone is growing = (ucL − uS ) (1 − ex ) = (ucL − ucS )(1 − e)/2(ucL − ux ) Thus: (iv) In equation (i): ucL = (1 − 0.2)2.3−1 (2(1 + 1 − 0.2)/2 − 1(1 − (1 − 0.2))/2) = 0.733 mm/s and: ucS = (1 − 0.2)2.3−1 (1(1 + 1 − 0.2)/2 − 2(1 − (1 − 0.2))/2) = 0.523 mm/s In equation (iii): (ux /1) = ex2.3 (v) and in equation (iv): (1 − ex ) = (0.733 − 0.523)(1 − 0.8)/2(0.733 − ux ) = 0.021/(0.733 − ux ) (vi) By solving equations (v) and (vi) simultaneously and, by assuming values of ex in the range 0.7–0.9, it is found that ex = 0.82, at which ux = 0.634 mm/s 31 This is the settling rate of the upper interface. The settling rate of the lower interface is, as before: ucL = 0.733 mm/s PROBLEM 3.19 What will be the terminal falling velocity of a glass sphere 1 mm in diameter in water if the density of glass is 2500 kg/m3 ? Solution ′ (R0′ /ρu20 )Re02 = (2d 3 /3μ2 )ρ(ρs − ρ)g For a sphere, d = 1 mm Noting that: or (equation 3.34) 0.001 m μ = 1 mNs/m = 0.001 Ns/m2 , per water and: then: ′ 2 ρ = 1000 kg/m3 per water (R0′ /ρu20 )Re0 2 = [(2 × 0.0013 )/(3 × 0.0012 )]1000(2500 − 1000)9.81 = 9810 log10 9810 = 3.992 From Table 3.4: or: Thus: log10 Re0′ = 2.16 Re0′ = 144.5 u0 = (1445 × 0.001)/(1000 × 0.001) = 0.145 m/s PROBLEM 3.20 What is the mass of a sphere of density 7500 kg/m3 which has a terminal falling velocity of 0.7 m/s in a large tank of water? Solution For a sphere diameter d, the volume = πd 3 /6 = 0.524d 3 m3 The mass of the sphere is then: m = 0.524d 3 × 7500 = 3926d 3 kg or: d = 0.0639 m0.3 m From equation 3.34: R0′ /ρu20 = (2dg/3ρu20 )(ρs − ρ) 32 and: (R0′ /ρu20 )Re′−1 = [(2dg/3ρu20 )(ρs − ρ)](μ/du0 ρ) = [2g(ρs − ρ)μ]/(3ρ 2 u30 ) = [(2 × 9.81)(7500 − 1000) × 1 × 10−3 ]/(3 × 10002 × 0.73 ) = 1.24 × 10−4 From Figure 3.6: Re′ = 1800 d = 1800μ/(u0 ρ) and: = (1800 × 1 × 10−3 )/(0.7 × 1000) = 2.57 × 10−3 m or 2.6 mm The mass of the sphere is then: m = 3926(2.57 × 10−3 )3 = 6.6 × 10−5 kg or 0.066 g As Re is in the Newton’s law region, it is more accurate to use: R ′ /ρu20 = 0.22 or: (equation 3.18) [2dg(ρs − ρ)]/3ρu20 = 0.22 from which: d = ρu20 /[3g(ρs − ρ)] = (1000 × 0.72 )/[(3 × 9.81)(7500 − 1000)] = 2.56 × 10−3 m or 2.6 mm, as before. 33 SECTION 2-4 Flow of Fluids Through Granular Beds and Packed Columns PROBLEM 4.1 In a contact sulphuric acid plant the secondary converter is a tray type converter, 2.3 m in diameter with the catalyst arranged in three layers, each 0.45 m thick. The catalyst is in the form of cylindrical pellets 9.5 mm in diameter and 9.5 mm long. The void fraction is 0.35. The gas enters the converter at 675 K and leaves at 720 K. Its inlet composition is: SO3 6.6, SO2 1.7, O2 10.0, N2 81.7 mole per cent and its exit composition is: SO3 8.2, SO2 0.2, O2 9.3, N2 82.3 mole per cent The gas flowrate is 0.68 kg/m2 s. Calculate the pressure drop through the converter. The viscosity of the gas is 0.032 mN s/m2 . Solution From the Carman equation: and: e3 (−P ) 1 R = 2 S(1 − e) l ρu2c ρu1 (equation 4.15) R = 5/Re1 + 0.4/Re10.1 ρu21 (equation 4.16) G′ S(1 − e)μ (equation 4.13) Re1 = S = 6/d = 6/(9.5 × 10−3 ) = 631 m2 /m3 Hence: and: Re1 = 0.68/(631 × 0.65 × 0.032 × 10−3 ) = 51.8     R 0.4 5 = 0.366 + = ρu2 51.8 (51.8)0.1 From equation 4.15: −P = 0.366 × 631 × 0.65 × (3 × 0.45) × 0.569 × (1.20)2 /(0.35)3 = 3.87 × 103 N/m2 or 3.9 kN/m2 34 PROBLEM 4.2 Two heat-sensitive organic liquids of an average molecular mass of 155 kg/kmol are to be separated by vacuum distillation in a 100 mm diameter column packed with 6 mm stoneware Raschig rings. The number of theoretical plates required is 16 and it has been found that the HETP is 150 mm. If the product rate is 5 g/s at a reflux ratio of 8, calculate the pressure in the condenser so that the temperature in the still does not exceed 395 K (equivalent to a pressure of 8 kN/m2 ). It may be assumed that a = 800 m2 /m3 , μ = 0.02 mN s/m2 , e = 0.72 and that the temperature changes and the correction for liquid flow may be neglected. Solution See Volume 2, Example 4.1. PROBLEM 4.3 A column 0.6 m diameter and 4 m high is, packed with 25 mm ceramic Raschig rings and used in a gas absorption process carried out at 101.3 kN/m2 and 293 K. If the liquid and gas properties approximate to those of water and air respectively and their flowrates are 2.5 and 0.6 kg/m2 s, what is the pressure drop across the column? In making calculations, Carman’s method should be used. By how much may the liquid flow rate be increased before the column floods? Solution Carman’s correlation for flow through randomly packed beds is given by: where: and: R1 /ρu21 = 5/Re1 + 1.0/Re10.1     (−P ) 1 e3 R/ρu2 = S(1 − e) l ρu2c Re1 = G′ S(1 − e)μ  29 22.4  273 293  = 1.21 kg/m3 G′ = 0.6 kg/m2 s and: u = (0.6/1.21) = 0.496 m/s 35 (equation 4.15) (equation 4.13) Using the data given, then: ρair = (equation 4.19) From Table 4.3 for 25 mm Raschig rings: S = 190 m2 /m3 and e = 0.71 Re1 = 0.6/(190 × 0.29 × 0.018 × 10−3 ) = 605     R (0.71)3 (−P ) 1 = ρu2 190 × 0.29 4 1.21 × (0.496)2 Thus: = 5.90 × 10−3 (−P ) Hence: 5.90 × 10−3 (−P ) = (5/605) + (1.0/(605)0.1 ) = 0.535 and: −P = 90.7 N/m2 or 0.091 kN/m2 The pressure drop of the wet, drained packing is given by: −Pw = 90.7[1 + 3.30/25] = 102.5 N/m2 (equation 4.46) To take account fully of the liquid flow, reference 56 in Chapter 4 of Volume 2 provides a correction factor which depends on the liquid flowrate and the Raschig size. This factor acts as a multiplier for the dry pressure drop which in this example, is equal to 1.3, giving the pressure drop in this problem as: (1.3 × 90.7) = 118 N/m2 or 0.118 kN/m2 Figure 4.18 may be used to calculate the liquid flowrate which would cause the column to flood. At a value of the ordinate of 0.048, the flooding line gives: L′ G′ from which:  ρV ρL  = 2.5 L′ = 4.3 kg/m2 s PROBLEM 4.4 A packed column, 1.2 m in diameter and 9 m tall, is packed with 25 mm Raschig rings, and used for the vacuum distillation of a mixture of isomers of molecular mass 155 kg/kmol. The mean temperature is 373 K, the pressure at the top of the column is maintained at 0.13 kN/m2 and the still pressure is 1.3–3.3 kN/m2 . Obtain an expression for the pressure drop on the assumption that this is not appreciably affected by the liquid flow and may be calculated using a modified form of Carman’s equation. Show that, over the range of operating pressures used, the pressure drop is approximately directly proportional to the mass rate of flow rate of vapour, and calculate the pressure drop at a vapour rate of 0.125 kg/m2 . The specific surface of packing, S = 190 m2 /m3 , the mean voidage of bed, e = 0.71, the viscosity of vapour, μ = 0.018 mN s/m2 and the molecular volume = 22.4 m3 /kmol. 36 Solution The proof that the pressure drop is approximately proportional to the mass flow rate of vapour is given in Problem 4.5. Using the data specified in this problem: Re1 = G/S(1 − e)μ = 0.125/(190 × 0.29 × 0.018 × 10−3 ) = 126.0 The modified Carman’s equation states that: R/ρu2 = 5/Re1 + 1/Re10.1 (equation 4.19) 0.1 = (5/126.0) + (1/(126.0) ) = 0.656 As in Problem 4.2: R e3 (−dP ) 1 = 2 ρu S(1 − e) dl ρu2 e3 (−dP ) ρ S(1 − e) dl G′2  R S(1 − e) ′ 2 G l − ρ dP = ρu2 e3 (equation 4.15) = Thus: 2 2 = (0.656 × 190 × 0.29 × 9G′ )/(0.71)3 = 909G′ kg/m3 ρ/P = ρs /Ps where subscript s refers to the still.     273 Ps 155 = 5 × 10−5 Ps kg/m3 ρs = 22.4 373 101.3 × 103 ρs /Ps = 5 × 10−5 , and ρ = 5 × 10−5 P and: −  Ps Pc ρ dP = 2.5 × 10−5 (Ps2 − Pc2 ) (Ps − Pc ) = −P , and if −P ≃ Ps , then (Ps2 − Pc2 ) ≃ (−P )2  Ps 2 Thus: − ρ dP = 2.5 × 10−5 (−P )2 = 909G′ or −P ∝ G′ Pc If G′ = 0.125, −P = [909 × (0.125)2 /2.5 × 10−5 ]0.5 = 754 N/m2 or 0.754 kN/m2 PROBLEM 4.5 A packed column, 1.22 m in diameter and 9 m high, and packed with 25 mm Raschig rings, is used for the vacuum distillation of a mixture of isomers of molecular mass 155 kg/kmol. The mean temperature is 373 K, the pressure at the top of the column is maintained at 0.13 kN/m2 , and the still pressure is 1.3 kN/m2 . Obtain an expression for 37 the pressure drop on the assumption that this is not appreciably affected by the liquid flow and may be calculated using the modified form of Carman’s equation. Show that, over the range of operating pressures used, the pressure drop is approximately directly proportional to the mass rate of flow of vapour, and calculate approximately the flow of vapour. The specific surface of the packing is 190 m2 /m3 , the mean voidage of the bed is 0.71, the viscosity of the vapour is 0.018 mN s/m2 and the kilogramme molecular volume is 22.4 m3 /kmol. Solution The modified form of Carman’s equation states that: R/ρu2 = 5/Re1 + (1/Re1 )0.1 (equation 4.19) ′ Re1 = G /S(1 − e)μ where: In this case: Re1 = G′ /[190(1 − 0.71) × 0.018 × 10−3 ] = 1008G′ where G′ is in kg/m2 s ′ Thus: (R/ρu2 ) = (5/1008G′ ) + (1/1008G′ )0.1 = (0.005/G′ ) + (0.501/G 0.1 ) ′ (R/ρu2 ) = [e3 /S(1 − e)][(−dP )/dl](ρ/G 2 ) so that, as in Problem 7.4:  2 − ρdP = (R/ρu2 )[S(1 − e)/e3 ]G′ 1 = [(0.005/G′ ) + (0.501/G′ = 6.93G′ + 694G′ 1.9 0.1 )][{190 × 0.29 × 9)/0.713 ]G′ As before: − Thus:  ρdP = 2.5 × 10−5 (−P )2 (−P )2 = 2.8 × 105 G′ + 2.8 × 107 G′ Neglecting the first term: −P = 5.30 × 103 G′ 0.95 N/m2 and, when −P = (1300 − 130) = 1170 N/m2 , then: G′ = 0.018 kg/m2 s 38 1.9 2 SECTION 2-5 Sedimentation PROBLEM 5.1 A slurry containing 5 kg of water/kg of solids is to be thickened to a sludge containing 1.5 kg of water/kg of solids in a continuous operation. Laboratory tests using five different concentrations of the slurry yielded the following results: concentration Y (kg water/kg solid) rate of sedimentation uc (mm/s) 5.0 0.17 4.2 0.10 3.7 0.08 3.1 0.06 2.5 0.042 Calculate the minimum area of a thickener to effect the separation of 0.6 kg/s of solids. Solution Basis: 1 kg of solids: 1.5 kg water is carried away in underflow so that U = 1 Concentration Y (kg rate/kg solids) Water to overflow (Y − U ) Sedimentation rate uc (mm/s) (Y − U )/uc (s/mm) 5.0 4.2 3.7 3.1 2.5 3.5 2.7 2.2 1.6 1.0 0.17 0.10 0.08 0.06 0.042 20.56 27.0 27.5 26.67 23.81 The maximum value of (Y − U )/uc = 27.5 s/mm or 27,500 s/m. A= Q(Y − U ) Cρs uc ρ Cρs = 0.6 kg/s Hence: and (equation 5.54) ρ = 1000 kg/m3 A = (27,500 × 0.6)/1000 = 16.5 m2 PROBLEM 5.2 A slurry containing 5 kg of water/kg of solids is to be thickened to a sludge containing 1.5 kg of water/kg of solids in a continuous operation. 39 Laboratory tests using five different concentrations of the slurry yielded the following data: concentration (kg water/kg solid) rate of sedimentation (mm/s) 5.0 0.20 4.2 0.12 3.7 0.094 3.1 0.070 2.5 0.052 Calculate the minimum area of a thickener to effect the separation of 1.33 kg/s of solids. Solution See Volume 2, Example 5.1. PROBLEM 5.3 When a suspension of uniform coarse particles settles under the action of gravity, the relation between the sedimentation velocity uc and the fractional volumetric concentration C is given by: uc = (1 − C)n , u0 where n = 2.3 and u0 is the free falling velocity of the particles. Draw the curve of solids flux ψ against concentration and determine the value of C at which ψ is a maximum and where the curve has a point of inflexion. What is implied about the settling characteristics of such a suspension from the Kynch theory? Comment on the validity of the Kynch theory for such a suspension. Solution The given equation is: uc /u0 = (1 − C)2.3 The flux is the mass rate of sedimentation per unit area and is given by: ψ = uc C = u0 C(1 − C)2.3 m/s (from equation 5.31) A plot of ψ as a function of C is shown in Figure 5a. To find the maximum flux, this equation may be differentiated to give: dψ = u0 [(1 − C)2.3 − 2.3C(1 − C)1.3 ] dC = u0 (1 − C)1.3 (1 − 3.3C) For a maximum, dψ/dC = 0 and C = 0.30 40 At the point of inflexion: d2 ψ/dC 2 = 0 Thus: d2 ψ = u0 [−3.3(1 − C)1.3 − 1.3(1 − C)0.3 (1 − 3.3C)] dC 2 = u0 (1 − C)0.3 (7.6C − 4.6) When d2 ψ/dC 2 = 0, C = 0.61 The maximum flux and the point of inflexion are shown in Figure 5a. The Kynch theory is discussed fully in Section 5.2.3. MAXIMUM, C = 0.30 0.14 0.12 Flux, y (m/s) 0.10 INFLEXION, C = 0.61 0.08 0.06 0.04 0.02 0.30 0 0 0.2 0.4 0.6 0.8 1.0 Concentration, C Figure 5a. Flux-concentration curve for suspension when n = 2.3 PROBLEM 5.4 For the sedimentation of a suspension of uniform fine particles in a liquid, the relation between observed sedimentation velocity uc and fractional volumetric concentration C is given by: uc = (1 − C)4.6 u0 where u0 is the free falling velocity of an individual particle. Calculate the concentration at which the rate of deposition of particles per unit area is a maximum and determine 41 this maximum flux for 0.1 mm spheres of glass of density 2600 kg/m3 settling in water of density 1000 kg/m3 and viscosity 1 mN s/m2 . It may be assumed that the resistance force F on an isolated sphere is given by Stokes’ Law. Solution See Volume 2, Example 5.3. PROBLEM 5.5 Calculate the minimum area and diameter of a thickener with a circular basin to treat 0.1 m3 /s of a slurry of a solids concentration of 150 kg/m3 . The results of batch settling tests are: Solids concentration (kg/m3 ) Settling velocity (µm/s) 100 200 300 400 500 600 700 800 900 1000 1100 148 91 55.33 33.25 21.40 14.50 10.29 7.38 5.56 4.20 3.27 A value of 1290 kg/m3 for underflow concentration was selected from a retention time test. Estimate the underflow volumetric flow rate assuming total separation of all solids and that a clear overflow is obtained. Solution The settling rate of the solids, G′ kg/m2 s, is calculated as G′ = us c where uc is the settling velocity (m/s) and c the concentration of solids (kg/m3 ) and the data are plotted in Figure 5b. From the point u = 0 and c = 1290 kg/m3 , a line is drawn which is tangential to the curve. This intercepts the axis at G′ = 0.0154 kg/m2 s. The area of the thickener is then: A = (0.1 × 150)0.0154 = 974 m2 and the diameter is: d = [(4 × 974)/π]0.5 = 35.2 m 42 0.0016 0.0154 kg/m2s Settling velocity, G ′ (kg/m2s) 0.0014 0.0012 0.0010 0.008 0.006 0.004 1290 kg/m3 0.002 0 200 400 600 800 1000 1200 1400 Solids concentration, c (kg/m3) Figure 5b. Construction for Problem 5.5 The volumetric flow rate of underflow, obtained from a mass balance, is: = [(0.1 × 150)/1290] = 0.0116 m3 /s 43 SECTION 2-6 Fluidisation PROBLEM 6.1 Oil, of density 900 kg/m3 and viscosity 3 mN s/m2 , is passed vertically upwards through a bed of catalyst consisting of approximately spherical particles of diameter 0.1 mm and density 2600 kg/m3 . At approximately what mass rate of flow per unit area of bed will (a) fluidisation, and (b) transport of particles occur? Solution See Volume 2, Example 6.2. PROBLEM 6.2 Calculate the minimum velocity at which spherical particles of density 1600 kg/m3 and of diameter 1.5 mm will be fluidised by water in a tube of diameter 10 mm on the assumption that the Carman-Kozeny equation is applicable. Discuss the uncertainties in this calculation. Repeat the calculation using the Ergun equation and explain the differences in the results obtained. Solution The Carman-Kozeny equation takes the form: umf = 0.0055[e3 /(1 − e)][d 2 (ρs − ρ)g/μ] (equation 6.4) As a wall effect applies in this problem, use is made of equation 4.23 to determine the correction factor, fw where: fw = (1 + 0.5Sc /S)2 where: Sc = surface area of the container/volume of bed = (π × 0.01 × 1)/[(π/4)(0.012 × 1)] = 400 m2 /m3 S = 6/d for a spherical particle = [6/(1.5 × 10−3 )] = 4000 m2 /m3 Thus: fw = [1 + 0.5(400/4000)]2 = 1.10 44 The uncertainty in this problem lies in the chosen value of the voidage e. If e is taken as 0.45 then: umf = 0.0055[0.453 /(1 − 0.45)][(1.5 × 10−3 )2 (1600 − 1000) × 9.81]/(1 × 10−3 ) = 0.0120 m/s Allowing for the wall effect: umf = (0.0120 × 1.10) = 0.0133 m/s By definition: Galileo number, Ga = d 3 ρ(ρs − ρ)g/μ2 = (1.5 × 10−3 )3 × 1000(1600 − 1000) × 9.81)/(1 × 10−3 )2 = 1.99 × 104 Assuming a value of 0.45 for emf , equation 6.14 gives:  ′ Remf = 23.6{ [1 + (9.39 × 10−5 )(1.99 × 104 )] − 1} = 16.4 and from equation 6.15: umf = [(1 × 10−3 ) × 16.4]/(1.5 × 10−3 × 1000) = 0.00995 m/s As noted in Section 6.1.3 of Volume 2, the Carman-Kozeny equation applies only to conditions of laminar flow and hence to low values of the Reynolds number for flow in the bed. In practice, this restricts its application to fine particles. Approaches based on both the Carman-Kozeny and the Ergun equations are very sensitive to the value of the voidage and it seems likely that both equations overpredict the pressure drop for fluidised systems. PROBLEM 6.3 In a fluidised bed, iso-octane vapour is adsorbed from an air stream onto the surface of alumina microspheres. The mole fraction of iso-octane in the inlet gas is 1.442 × 10−2 and the mole fraction in the outlet gas is found to vary with time as follows: Time from start (s) 250 500 750 1000 1250 1500 1750 2000 Mole fraction in outlet gas (×102 ) 0.223 0.601 0.857 1.062 1.207 1.287 1.338 1.373 45 Show that the results may be interpreted on the assumptions that the solids are completely mixed, that the gas leaves in equilibrium with the solids and that the adsorption isotherm is linear over the range considered. If the flowrate of gas is 0.679 × 10−6 kmol/s and the mass of solids in the bed is 4.66 g, calculate the slope of the adsorption isotherm. What evidence do the results provide concerning the flow pattern of the gas? Solution See Volume 2, Example 6.4. PROBLEM 6.4 Cold particles of glass ballotini are fluidised with heated air in a bed in which a constant flow of particles is maintained in a horizontal direction. When steady conditions have been reached, the temperatures recorded by a bare thermocouple immersed in the bed are: Distance above bed support (mm) 0 0.64 1.27 1.91 2.54 3.81 Temperature (K) 339.5 337.7 335.0 333.6 333.3 333.2 Calculate the coefficient for heat transfer between the gas and the particles, and the corresponding values of the particle Reynolds and Nusselt numbers. Comment on the results and on any assumptions made. The gas flowrate is 0.2 kg/m2 s, the specific heat in air is 0.88 kJ/kg K, the viscosity of air is 0.015 mN s/m2 , the particle diameter is 0.25 mm and the thermal conductivity of air 0.03 is W/mK. Solution See Volume 2, Example 6.5. PROBLEM 6.5 The relation between bed voidage e and fluid velocity uc for particulate fluidisation of uniform particles which are small compared with the diameter of the containing vessel is given by: uc = en u0 where u0 is the free falling velocity. 46 Discuss the variation of the index n with flow conditions, indicating why this is independent of the Reynolds number Re with respect to the particle at very low and very high values of Re. When are appreciable deviations from this relation observed with liquid fluidised systems? For particles of glass ballotini with free falling velocities of 10 and 20 mm/s the index n has a value of 2.39. If a mixture of equal volumes of the two particles is fluidised, what is the relation between the voidage and fluid velocity if it is assumed that complete segregation is obtained? Solution The variation of the index n with flow conditions is fully discussed in Chapters 5 and 6 of Volume 2. The ratio uc /u0 is in general, dependent on the Reynolds number, voidage, and the ratio of particle diameter to that of the containing vessel. At low velocities, that is when Re < 0.2, the drag force is attributable entirely to skin friction, and at high velocities when Re > 500 skin friction becomes negligible and in these regions the ratio uc /u0 is independent of Re. For intermediate regions, the data given in Table 5.1 apply. Considering unit volume of each particle say 1 m3 , then: Voidage of large particles = e1 , volume of liquid = e1 /(1 − e1 ). Voidage of small particles = e2 , volume of liquid = e2 /(1 − e2 ). Total volume of solids = 2 m3 . Total volume of liquid = e1 /(1 − e1 ) + e2 /(1 − e2 ). Total volume of system = 2 + e1 /(1 − e1 ) + e2 /(1 − e2 ) m3 Thus: voidage = = That is: e= e1 /(1 − e1 ) + e2 /(1 − e2 ) 2 + e1 /(1 − e1 ) + e2 /(1 − e2 ) e1 (1 − e2 ) + e2 (1 − e1 ) 2(1 − e1 )(1 − e2 ) + e1 (1 − e2 ) + e2 (1 − e1 ) e1 + e2 − 2e1 e2 2 − e1 − e2 But, since the free falling velocities are in the ratio 1 : 2, then: e1 = Thus: at:  u u01 1/2.4 and e2 =  u u01 /2 1/2.4 e2 = e1 21/2.4 e= e= e1 + e1 × 21/2.4 − 23.4/2.4 × e12 2 − e1 − 21/2.4 × e1 (u/20)1/2.4 (1 + 21/2.4 ) − 23.4/2.4 (u/20)1/1.2 2 − (1 + 21/2.4 )(u/20)1/2.4 47 e= 3u0.42 − u0.83 (with u in mm/s) 9 − 3u0.42 9e = 3eu0.42 = 3u0.42 − u0.84 and: u0.84 − 3(1 + e)u0.42 + 9e = 0 or: u0.42 = 1.5(1 + e) + [2.24(1 + e)2 − 9e] and: This relationship is plotted in Figure 6a. 14 12 Velocity u (mm/s) 10 8 6 4 2 0 0.1 0.2 0.3 0.4 0.5 0.6 Voidage e Figure 6a. Plot of the relationship for u in Problem 6.5 0.7 0.8 0.9 1.0 PROBLEM 6.6 Obtain a relationship for the ratio of the terminal falling velocity of a particle to the minimum fluidising velocity for a bed of similar particles. It may be assumed that Stokes’ Law and the Carman-Kozeny equation are applicable. What is the value of the ratio if the bed voidage at the minimum fluidising velocity is 0.4? Solution In a fluidised bed the total frictional force must be equal to the effective weight of bed. Thus: −P = (1 − e)(ρs − ρ)lg (equation 6.1) Substituting equation 6.1 into equation 4.9, and putting K ′′ = 5, gives: umf = 0.0055 e3 d 2 (ρs − ρ)g 1−e μ 48 (equation 6.4) u0 = umf Hence:  d 2 g(ρs − ρ) 18μ × 0.0055  (1 − e) e3  μ d 2 (ρs − ρ)g  (1 − e) = 10.1(1 − e)/e3 (18 × 0.0055e3 ) = 94.7 . = If e = 0.4, then: u0 /umf The use of the Carman–Kozeny equation is discussed in Section 4.2.3 of Chapter 4. It is interesting to note that if e = 0.48, which was the value taken in Problem 6.1, then u0 /uf = 47.5, which agrees with the solution to that problem. PROBLEM 6.7 A packed bed consisting of uniform spherical particles of diameter 3 mm and density 4200 kg/m3 , is fluidised by means of a liquid of viscosity 1 mN s/m2 and density 1100 kg/m3 . Using Ergun’s equation for the pressure drop through a bed height l and voidage e as a function of superficial velocity, calculate the minimum fluidising velocity in terms of the settling velocity of the particles in the bed. State clearly any assumptions made and indicate how closely the results might be confirmed by an experiment. Ergun’s equation: −P (1 − e)2 μu (1 − e) ρu2 = 150 + 1.75 3 2 l e d e3 d Solution The pressure drop through a fluidised bed of height l is: −P / l = (1 − e)(ρs − ρ)g (equation 6.1) Writing u = umf , the minimum fluidising velocity in Ergun’s equation and substituting for −P /H gives: 1.75(1 − e)ρu2mf 150(1 − e)2 /μumf + e3 d 2 e3 d 1.75ρu2mf 150(1 − e)μumf + (ρs − ρ)g = e3 d 2 e3 d (1 − e)(ρs − ρ)g = or: (equation 6.7) If d = 3 × 10−3 m, ρs = 4200 kg/m3 , ρ = 1100 kg/m3 , μ = 1 × 10−3 Ns/m2 , and if e is taken as 0.48 as in Problem 6.1, these values may be substituted to give: 3.04 = 7.84umf + 580u2mf or: umf = 0.066 m/s neglecting the negative root. 49 If Stokes’ law applies, then: u0 = d 2 g(ρs − ρ)/18μ (equation 3.24) −3 −3 = (3 × 10 )(9.81 × 3100/18 × 10 ) = 15.21 m/s Thus: Re = (3 × 10−3 × 15.21 × 4200/10−3 ) = 1.92 × 105 which is outside the range of Stokes’ law. A Reynolds number of this order lies in the region (c) of Figure 3.4 where: u20 = 3dg(ρs − ρ)/ρ −3 Thus: u20 and: u0 = 0.5 m/s = (3 × 3 × 10 (equation 3.25) × 9.81 × 3100)/1100 A check on the value of the Reynolds number gives: Re = (3 × 10−3 × 0.5 × 4200)/10−3 = 6.3 × 103 which is within the limits of region (c). Hence: u0 /umf = (0.5/0.066) = 7.5 Empirical relationships for the minimum fluidising velocity are presented as a function of Reynolds number and this problem illustrates the importance of using the equations applicable to the particle Reynolds number in question. PROBLEM 6.8 Ballotini particles, 0.25 mm in diameter, are fluidised by hot air flowing at the rate of 0.2 kg/m2 cross-section of bed to give a bed of voidage 0.5 and a cross-flow of particles is maintained to remove the heat. Under steady state conditions, a small bare thermocouple immersed in the bed gives the following data: Distance above bed support (mm) 0 0.625 1.25 1.875 2.5 3.75 Temperature (◦ C) (K) 66.3 64.5 61.8 60.4 60.1 60.0 339.5 337.7 335.0 333.6 333.3 333.2 Assuming plug flow of the gas and complete mixing of the solids, calculate the coefficient for heat transfer between the particles and the gas. The specific heat capacity of air is 0.85 kJ/kg K. 50 A fluidised bed of total volume 0.1 m3 containing the same particles is maintained at an approximately uniform temperature of 425 K by external heating, and a dilute aqueous solution at 375 K is fed to the bed at the rate of 0.1 kg/s so that the water is completely evaporated at atmospheric pressure. If the heat transfer coefficient is the same as that previously determined, what volumetric fraction of the bed is effectively carrying out the evaporation? The latent heat of vaporisation of water is 2.6 MJ/kg. Solution See Volume 2, Example 6.6. PROBLEM 6.9 An electrically heated element of surface area 12 cm2 is completely immersed in a fluidised bed. The resistance of the element is measured as a function of the voltage applied to it giving the following data: Potential (V) Resistance (ohms) 1 15.47 2 15.63 3 15.91 4 16.32 5 16.83 6 17.48 The relation between resistance Rw and temperature Tw is: Rw = 0.004Tw − 0.092 R0 where R0 , is the resistance of the wire at 273 K is 14 ohms and Tw is in K. Estimate the bed temperature and the value of the heat transfer coefficient between the surface and the bed. Solution The heat generation rate by electrical heating = V 2 /R The rate of heat dissipation = hA(Tw − TB ) where Tw and TB are the wire and bed temperatures respectively. At equilibrium, V 2 /Rw = hA(Tw − TB ) But: Rw /R0 = 0.004Tw − 0.092 so that: Tw = 250(Rw /R0 ) + 23 Thus: V2 = 250hAR̄w Rw − hAR̄w (TB − 23) R0 where R̄w is a mean value of Rw noting that the mean cannot be used inside the bracket in the equation for Tw . 51 Thus a plot of V 2 against Rw should yield a line of slope = 250 hAR̄w /R0 . This is shown in Figure 6b from which the value of the slope is 17.4. 40 30 V 2 20 Slope = 17.4 10 0 15 16 17 18 Rw (ohm) Figure 6b. Hence: A plot of V 2 and Rw for Problem 6.9 h = (17.4 × 14)/(250 × 12 × 10−4 × 16.5) = 49.2 W/m2 K The bed temperature is found by the intercept at V 2 = 0 that is when Rw = 15.4 ohm Thus: TB = 250(15.4/14) + 23 = 298 K . PROBLEM 6.10 (a) Explain why the sedimentation velocity of uniform coarse particles in a suspension decreases as the concentration is increased. Identify and, where possible, quantify the various factors involved. (b) Discuss the similarities and differences in the hydrodynamics of a sedimenting suspension of uniform particles and of an evenly fluidised bed of the same particles in the liquid. (c) A liquid fluidised bed consists of equal volumes of spherical particles 0.5 mm and 1.0 mm in diameter. The bed is fluidised and complete segregation of the 52 two species occurs. When the liquid flow is stopped the particles settle to form a segregated two-layer bed. The liquid flow is then started again. When the velocity is such that the larger particles are at their incipient fluidisation point what will be the approximate voidage of the fluidised bed composed of the smaller particles? It may be assumed that the drag force F of the fluid on the particles under the free falling conditions is given by Stokes’ law and that the relation between the fluidisation velocity uc and voidage, e, for particles of terminal velocity, u0 , is given by: uc /u0 = e4.8 For Stokes’ law, the force F on the particles is given by F = 3πμdu0 , where d is the particle diameter and μ is the viscosity of the liquid. Solution Parts (a) and (b) of this Problem are considered in Volume 2, Chapter 6. Attention is now concentrated on part (c). A force balance gives: 3πμduc = (π/6)d 3 (ρs − ρ)g Thus: That is: uc = (d 2 g/18μ)(ρs − ρ)g uc = ki d 2 (i) For larger particles and assuming that the voidage at umf = 0.45, then: uc /u0L = 0.45n (ii) For smaller particles: uc /u0S = en uc has the same ki both large and small particles and from equation (i): u0S = u0L /4 Thus: 4uc /u0L = en (iii) Dividing equation (iii) by equation (ii) gives: 4 = (e/0 45)n = (e/0 45)4.8 Thus: and: 1.334 = e/0 45 e = 0.60 PROBLEM 6.11 The relation between the concentration of a suspension and its sedimentation velocity is of the same form as that between velocity and concentration in a fluidised bed. Explain 53 this in terms of the hydrodynamics of the two systems. A suspension of uniform spherical particles in a liquid is allowed to settle and, when the sedimentation velocity is measured as a function of concentration, the following results are obtained: Fractional volumetric concentration (C) 0.35 0.25 0.15 0.05 Sedimentation velocity (uc m/s) 1.10 2.19 3.99 6.82 Estimate the terminal falling velocity u0 of the particles at infinite dilution. On the assumption that Stokes’ law is applicable, calculate the particle diameter d. The particle density, ρs = 2600 kg/m3 , the liquid density, ρ = 1000 kg/m3 , and the liquid viscosity, μ = 0.1 Ns/m2 . What will be the minimum fluidising velocity of the system? Stokes’ law states that the force on a spherical particle = 3πμdu0 . Solution The relation between uc and C is: uc /u0 = (1 − C)n (i) u0 and n may be obtained by plotting log10 uc and log10 (1 − C). This will give a straight line of slope n and u0 is given by the intercept which corresponds to c = 0 at ln(1 − c) = 0. Alternatively on algebraic solution may be sought as follows: C 0.35 0.25 0.15 0.05 1−c 0.65 0.75 0.85 0.95 log10 (1 − C) −0.187 −0.125 −0.0706 −0.0222 uc 1.10 2.19 3.99 6.82 log10 uc 0.0414 0.8404 0.6010 0.8338 Taking logarithms of equation (i) gives: log10 uc = log10 u0 + n log10 (1 − c) (ii) Inserting values from the table into equation (ii) gives: 0.0414 = log10 uc + n(−0.187) (iii) 0.8404 = log10 uc + n(−0.125) (iv) 0.6010 = log10 uc + n(−0.0706) (v) 0.8338 = log10 uc + n(−0.0222) (vi) Any two of the equations (iii)–(vi) may be used to evaluate log10 u0 and n. Substituting (iii) from (v): 0.5396 = 0.116n 54 and n = 4.8 Substituting (iv) from (vi): 0.4934 = 0.103n n = 4.8 which is consistent. and Substituting in equation (v): log10 u0 = 0.6010 + (0.0706 × 4.8) = 0.94 Substituting in equation (vi): log10 u0 = 0.8338 + (0.0222 × 4.8) = 0.94 which is consistent and hence: u0 = 8.72 mm/s A force balance gives: 3πμdu0 = (π/6)d 3 (ρs − ρ)g and: d = [(18μu0 /(g(ρs − ρ))]0.5 = [(18 × 0.1 × 8.72 × 10−3 )/(9.81(2600 − 1000))]0.5 = 0.001 m or 1 mm Assuming that emf = 0.45, then: umf = u0 emf = 8.72 × 0.454.8 = 0.19 mm/s PROBLEM 6.12 A mixture of two sizes of glass spheres of diameters 0.75 and 1.5 mm is fluidised by a liquid and complete segregation of the two species of particles occurs, with the smaller particles constituting the upper portion of the bed and the larger particles in the lower portion. When the voidage of the lower bed is 0.6, what will be the voidage of the upper bed? The liquid velocity is increased until the smaller particles are completely transported from the bed. What is the minimum voidage of the lower bed at which this phenomenon will occur? It may be assumed that the terminal falling velocities of both particles may be calculated from Stokes’ law and that the relationship between the fluidisation velocity u and the bed voidage e is given by: (uc /u0 ) = e4.6 Solution A force balance gives: (π/6)d 3 (ρs − ρ)g = 3πμdu0 from which: u0 = (d 2 g/18μ)(ρs − ρ) 55 and hence, for constant viscosity and densities: u0 = k1 d 2 For large particles of diameter dL : u0 = k1 dL2 The voidage, 0.6, achieved at velocity u is given by: u/k1 dL2 = 0.64.6 For small particles of diameter ds , the voidage e at this velocity is given by: u/k1 ds2 = e4.6 ds2 /dL2 = (0.6/e)4.6 Dividing: Since ds /dL = 0.5, (ds /dL )2 = 0.25 and: 0.25 = (0.6/e)4.6 e = 0.81 from which: For transport of the smaller particles just to occur, the voidage of the upper bed is unity and: 0.25 = (e/1)4.6 from which, for the large particles: e = 0.74 PROBLEM 6.13 (a) Calculate the terminal falling velocities in water of glass particles of diameter 12 mm and density 2500 kg/m3 , and of metal particles of diameter 1.5 mm and density 7500 kg/m3 . It may be assumed that the particles are spherical and that, in both cases, the friction factor, R ′ /ρu2 is constant at 0.22, where R ′ is the force on the particle per unit of projected area of the particle, ρ is the fluid density and u the velocity of the particle relative to the fluid. (b) Why is the sedimentation velocity lower when the particle concentration in the suspension is high? Compare the behaviour of the concentrated suspension of particles settling under gravity in a liquid with that of a fluidised bed of the same particles. (c) At what water velocity will fluidised beds of the glass and metal particles have the same densities? The relation between the fluidisation velocity uc terminal velocity u0 and bed voidage e is given for both particles by: (uc /u0 ) = e2.30 56 Solution For spheres, a take balance gives: R ′ (π/4)d 2 = (π/6)d 3 (ρs − ρ)g or: Thus: R ′ = (2d/3)(ρs − ρ)g = (R ′ /ρu20 )ρu20 = 0.22ρu20 ≈ (2/g)ρu20 u0 = [3dg(ρs − ρ)/ρ]0.5 For the metal particles: u0 = [(3 × 1.5 × 10−3 × 6500 × 9.81)/1000]0.5 = 0.536 m/s For the glass particles: u0 = [(3 × 12 × 10−3 × 1500 × 9.81)/1000]0.5 = 0.727 m/s. For the fluidised bed: The density of the suspension = eρ + (1 − e)ρs For the metal particles: (uc /0.536) = e1 2.30 (i) For the glass particles: (uc /0.727) = e2 2.30 (ii) and from equations (i) and (ii): (e1 /e2 ) = (0.727/0.536)1/2.30 = 1.142 For equal bed densities: e1 ρ(1 − e1 )ρs1 = e2 ρ + (1 − e2 )ρs2 Thus: from which: and: (iii) (1.142e1 )1000 + (1 − 1.142e2 )7500 = 1000e2 + (1 − e2 )2500 e3 = 0.844 uc = (0.727 × 0.8442.30 ) = 0.492 m/s Substituting in equation (iii): e1 = 0.964 and: uc = (0.536 × 0.9642.30 ) = 0.493 m/s PROBLEM 6.14 Glass spheres are fluidised by water at a velocity equal to one half of their terminal falling velocities. Calculate: 57 (a) the density of the fluidised bed, (b) the pressure gradient in the bed attributable to the presence of the particles. The particles are 2 mm in diameter and have a density of 2500 kg/m3 . The density and viscosity of water are 1000 kg/m3 and 1 mNs/m2 respectively. Solution The Galileo number is given by: Ga = d 3 ρ(ρs − ρ)g/μ2 = [(2 × 10−3 )3 × 1000(2500 − 1000) × 9.81]/(1 × 10−3 )2 = 117,720 From equation 5.79: (4.8 − n)/(n − 2.4) = 0.043Ga 0.57 = 0.043 × 117,7200.57 = 33.4 n = 2.47 and: u/u0 = 0.5 = e2.47 e = 0.755 and hence: The bed density is given by: (1 − e)ρs + eρ = (1 − 0.755) × 2500 + (0.755 × 1000) = 1367 kg/m3 The pressure gradient due to the solids is given by: {[(1 − e)ρs + eρ] − ρ}g = (1 − e)(ρs − ρ)g = (1 − 0.755)(2500 − 1000)9.81 = 3605 (N/m2 )/m 58 SECTION 2-7 Liquid Filtration PROBLEM 7.1 A slurry, containing 0.2 kg of solid/kg of water, is fed to a rotary drum filter, 0.6 m in diameter and 0.6 m long. The drum rotates at one revolution in 360 s and 20 per cent of the filtering surface is in contact with the slurry at any given instant. If filtrate is produced at the rate of 0.125 kg/s and the cake has a voidage of 0.5, what thickness of cake is formed when filtering at a pressure difference of 65 kN/m2 ? The density of the solid is 3000 kg/m3 . The rotary filter breaks down and the operation has to be carried out temporarily in a plate and frame press with frames 0.3 m square. The press takes 120 s to dismantle and 120 s to reassemble, and, in addition, 120 s is required to remove the cake from each frame. If filtration is to be carried out at the same overall rate as before, with an operating pressure difference of 75 kN/m2 , what is the minimum number of frames that must be used and what is the thickness of each? It may be assumed that the cakes are incompressible and the resistance of the filter media may be neglected. Solution See Volume 2, Example 7.6. PROBLEM 7.2 A slurry containing 100 kg of whiting/m3 of water, is filtered in a plate and frame press, which takes 900 s to dismantle, clean and re-assemble. If the filter cake is incompressible and has a voidage of 0.4, what is the optimum thickness of cake for a filtration pressure of 1000 kN/m2 ? The density of the whiting is 3000 kg/m3 . If the cake is washed at 500 kN/m2 and the total volume of wash water employed is 25 per cent of that of the filtrate, how is the optimum thickness of cake affected? The resistance of the filter medium may be neglected and the viscosity of water is 1 mN s/m2 . In an experiment, a pressure of 165 kN/m2 produced a flow of water of 0.02 cm3 /s though a centimetre cube of filter cake. Solution See Volume 2, Example 7.2. 59 PROBLEM 7.3 A plate and frame press gave a total of 8 m3 of filtrate in 1800 s and 11.3 m3 in 3600 s when filtration was stopped. Estimate the washing time if 3 m3 of wash water is used. The resistance of the cloth may be neglected and a constant pressure is used throughout. Solution For constant pressure filtration with no cloth resistance: t= rμv 2A2 (−P ) V2 (equation 7.1) At t1 = 1800 s, V1 = 8 m3 , and when t2 = 3600 s, V2 = 11 m3 Thus: (3600 − 1800) = rμv (112 − 82 ) 2A2 (−P ) rμv = 316 2A2 (−P ) Since: A2 (−P ) dV = dt rμvV 0.0158 1 = = (2 × 31.6V ) V The final rate of filtration = (0.0158/11) = 1.44 × 10−3 m3 /s. For thorough washing in a plate and frame filter, the wash water has twice the thickness of cake to penetrate and half the area for flow that is available to the filtrate. Thus the flow of wash water at the same pressure will be one-quarter of the filtration rate. Hence: and: rate of washing = (1.44 × 10−3 )/4 = 3.6 × 10−4 m3 /s time of washing = 3/(3.6 × 10−4 ) = 8400 s (2.3 h) PROBLEM 7.4 In the filtration of a sludge, the initial period is effected at a constant rate with the feed pump at full capacity, until the pressure differences reaches 400 kN/m2 . The pressure is then maintained at this value for a remainder of the filtration. The constant rate operation requires 900 s and one-third of the total filtrate is obtained during this period. Neglecting the resistance of the filter medium, determine (a) the total filtration time and (b) the filtration cycle with the existing pump for a maximum daily capacity, if the time for removing the cake and reassembling the press is 1200 s. The cake is not washed. 60 Solution For a filtration carried out at a constant filtration rate for time t1 in which time a volume V1 is collected and followed by a constant pressure period such that the total filtration time is t and the total volume of filtrate is V , then: V 2 − V12 = 2A2 (−P ) (t − t1 ) rμv (equation 7.13) Assuming no cloth resistance, then: rμv V2 A2 (−P ) 1 for the constant rate period: t1 = Using the data given: t1 = 900 s, volume = V1 Thus: 900 rμv = 2 A2 (−P ) V1 (equation 7.10) (a) For the constant pressure period: V = 3V1 and (t − t1 ) = tp 2V12 tp 900 tp = 3600 s 8V12 = Thus: Thus: and: total filtration time = (900 + 3600) = 4500 s total cycle time = (4500 + 1200) = 5700 s (b) For the constant rate period: t1 = V12 rμv 2 = V 1 A2 (−P ) K For the constant pressure period: t − t1 = Total filtration time, Rate of filtration t= rμv V 2 − V12 2 2 (V − V ) = 1 2A2 (−P ) 2K 1 K V12 + V 2 − V12 2 = (V 2 + V12 ) 2K V where td is the downtime t + td 2KV = 2 V + V12 + 2Ktd = For the rate to be a maximum, d(rate) =0 dV or V12 − V 2 + 2Ktd = 0 61 Thus: But: Thus: 1 (V 2 − V12 ) = (t − t1 ) 2K td = 1200 = (t − 900) and t = 2100 s td = total cycle time = (2100 + 1200) = 3300 s PROBLEM 7.5 A rotary filter, operating at 0.03 Hz, filters at the rate of 0.0075 m3 /s. Operating under the same vacuum and neglecting the resistance of the filter cloth, at what speed must the filter be operated to give a filtration rate of 0.0160 m3 /s? Solution For constant pressure filtration in a rotary filter: V2 = 2A2 (−P )t rμv V 2 ∝ t ∝ 1/N or: (equation 7.11) where N is the speed of rotation. As V ∝ 1/N 0.5 and the rate of filtration is V /t, then: V /t ∝ (1/N 0.5 )(1/t) ∝ (N/N 0.5 ) ∝ N 0.5 Thus: (V /t)1 /(V /t)2 = N10.5 /N20.5 0.0075/0.0160 = 0.030.5 /N20.5 and: N2 = 0.136 Hz (7.2 rpm) PROBLEM 7.6 A slurry is filtered in a plate and frame press containing 12 frames, each 0.3 m square and 25 mm thick. During the first 180 s, the filtration pressure is slowly raised to the final value of 400 kN/m2 and, during this period, the rate of filtration is maintained constant. After the initial period, filtration is carried out at constant pressure and the cakes are completely formed in a further 900 s. The cakes are then washed with a pressure difference of 275 kN/m2 for 600 s, using thorough washing. What is the volume of filtrate collected per cycle and how much wash water is used? A sample of the slurry was tested, using a vacuum leaf filter of 0.05 m2 filtering surface and a vacuum equivalent to a pressure difference of 71.3 kN/m2 . The volume of filtrate collected in the first 300 s was 250 cm3 and, after a further 300 s, an additional 150 cm3 was collected. It may be assumed that cake is incompressible and the cloth resistance is the same in the leaf as in the filter press. 62 Solution See Volume 2, Example 7.1. PROBLEM 7.7 A sludge is filtered in a plate and frame press fitted with 25 mm frames. For the first 600 s the slurry pump runs at maximum capacity. During this period the pressure difference rises to 500 kN/m2 and a quarter of the total filtrate is obtained. The filtration takes a further 3600 s to a complete at constant pressure and 900 s is required for emptying and resetting the press. It is found that, if the cloths are precoated with filter aid to a depth of 1.6 mm, the cloth resistance is reduced to 25 per cent of its former value. What will be the increase in the overall throughput of the press if the precoat can be applied in 180 s? Solution See Volume 2, Example 7.7. PROBLEM 7.8 Filtration is carried out in a plate and frame filter press, with 20 frames 0.3 m square and 50 mm thick, and the rate of filtration is maintained constant for the first 300 s. During this period, the pressure is raised to 350 kN/m2 , and one-quarter of the total filtrate per cycle is obtained. At the end of the constant rate period, filtration is continued at a constant pressure of 350 kN/m2 for a further 1800 s, after which the frames are full. The total volume of filtrate per cycle is 0.7 m3 and dismantling and refitting of the press takes 500 s. It is decided to use a rotary drum filter, 1.5 m long and 2.2 m in diameter, in place of the filter press. Assuming that the resistance of the cloth is the same in the two plants and that the filter cake is incompressible, calculate the speed of rotation of the drum which will result in the same overall rate of filtration as was obtained with the filter press. The filtration in the rotary filter is carried out at a constant pressure difference of 70 kN/m2 , and the filter operates with 25 per cent of the drum submerged in the slurry at any instant. Solution Data from the plate and frame filter press are used to evaluate the cake and cloth resistance for use with the rotary drum filter. For the constant rate period: V12 + A2 (−P ) LA V1 = t1 v rμv 63 (equation 7.17) For the subsequent constant pressure period: (V 2 − V12 ) + 2A2 (−P ) 2LA (V − V1 ) = (t − t1 ) v rμv (equation 7.18) From the data given: t1 = 300 s, (−P ) = (350 − 101.3) = 248.7 kN/m2 , V1 = 0.175 m3 Thus: (0.175)2 + and A = (2 × 20 × 0.3 × 0.3) = 3.6 m2 L (3.6)2 × 248.7 × 103 × 300 × 3.6 × 0.175 = v rμv 0.0306 + 0.63(L/v) = 9.68 × 108 /rμv or: (i) For the constant pressure period: V = 0.7 m3 , V1 = 0.175 m3 , A = 3.6 m2 (t − t1 ) = 1800 s, Thus: (0.72 − 0.1752 ) + 2(L/v) × 3.6(0.7 − 0.175) = (2 × 3.6)2 × 248.7 × 103 × 1800 rμv 0.459 + 3.78(L/v) = 116.08 × 108 /rμv or: Solving equations (i) and (ii) simultaneously gives: rμv = 210.9 × 108 and L/v = 0.0243 For the rotary drum filter: D = 2.2 m, L = 1.5 m, (−P ) = 70 kN/m2 A = (2.2π × 1.5) = 10.37 m2 (−P ) = 70 × 103 N/m2 If θ is the time of one revolution, then as the time of filtration is 0.25θ : 2A2 (−P ) L V 2 + 2A V = × 0.25θ v rμv V 2 + (2 × 10.37 × 0.0243V ) = or: 2(10.37)2 × 70 × 103 × 0.25θ 210.9 × 108 V 2 + 0.504V = 1.785 × 10−4 θ The rate of filtration = V /t = 0.7/(300 + 1800 + 500) = 2.7 × 10−4 m3 /s 64 (ii) V = 2.7 × 10−4 t Thus: (2.7 × 10−4 t)2 + (0.504 × 2.7 × 10−4 )t = (1.785 × 10−4 )t and: t = 580 s from which: Hence: speed = (1/580) = 0.002 Hz (0.12 rpm) PROBLEM 7.9 It is required to filter a slurry to produce 2.25 m3 of filtrate per working day of 8 hours. The process is carried out in a plate and frame filter press with 0.45 m square frames and a working pressure difference of 348.7 kN/m2 . The pressure is built up slowly over a period of 300 s and, during this period, the rate of filtration is maintained constant. When a sample of the slurry is filtered, using a pressure difference of 66.3 kN/m2 on a single leaf filter of filtering area 0.05 m2 , 400 cm3 of filtrate is collected in the first 300 s of filtration and a further 400 cm3 is collected during the following 600 s. Assuming that the dismantling of the filter press, the removal of the cakes and the setting up again of the press takes an overall time of 300 s, plus an additional 180 s for each cake produced, what is the minimum number of frames that need be employed? The resistance of the filter cloth may be taken as the same in the laboratory tests as on the plant. Solution For constant pressure filtration on the leaf filter: 2A2 (−P )t L V 2 + 2 AV = v rμv (equation 7.18) When t = 300 s, V = 0.0004 m3 , A = 0.05 m2 , (−P ) = 66.3 kN/m2 , and: (0.0004)2 + 2(L/v) × 0.05 × 0.0004 = or: 2 × (0.05)2 × 66.3 × 300 rμv 1.6 × 10−7 + 4 × 10−5 (L/v) = 99.4/rμv When t = 900 s, V = 800 cm3 or 0.0008 m3 and substituting these values gives: (6.4 × 10−7 ) + (8 × 10−5 )(L/v) = 298.4/rμv Thus: L/v = 4 × 10−3 and rμv = 3.1 × 108 In the filter press For the constant rate period: V12 + A2 (−P )t1 LA V1 = v rμv 65 (equation 7.17) A = 2 × 0.45n = 0.9n where n is the number of frames, t1 = 300 s Thus: or: V12 + (4 × 10−3 × 0.9)nV1 = 0.81n2 × 348.7 × (300/3.1) × 108 V12 + (3.6 × 10−3 )nV1 = (2.73 × 10−4 )n2 V1 = 0.0148n and: For the constant pressure period: V 2 − V12 2 + LA A2 (−P ) (V − V1 ) = (t − t1 ) v rμv (equation 7.18) Substituting for L/v and rμv, t1 = 300 and V1 = 0.0148n gives:   2 (0.81n2 × 348.7) V − 2.2 × 10−4 n2 (tf − 300) + (V − 0.0148n)4 × 10−3 × 0.9n = 2 (3.1 × 108 ) or: 0.5V 2 + 1.1 × 10−4 n2 + 3.6 × 10−3 nV = 9.11 × 10−7 n2 t (i) The total cycle time = (tf + 300 + 180n) s. Required filtration rate = 2.25/(8 × 3600) = 7.81 × 10−5 m3 /s. Volume of filtrate = V m3 . Thus: and: V = 7.81 × 10−5 (tf + 300 + 180n) tf = 1.28 × 104 V − 300 − 180n (ii) Thus the value of tf from equation (ii) may be substituted in equation (i) to give: V 2 + V (7.2 × 10−3 n − 2.34 × 10−2 n2 ) + (7.66 × 10−4 n2 + 3.28 × 10−4 n3 ) = 0 (iii) This equation is of the form V 2 + AV + B = 0 and may thus be solved to give:  −A ± (A2 − 4B) V = 2 where A and B are the expressions in parentheses in equation (iii). In order to find the minimum number of frames. dV /dn must be found and equated to zero. From above (V − a)(V − b) = 0, where a and b are complex functions of n. Thus V = a or V = b and dV /dn can be evaluated for each root. Putting dV /dn = 0 gives, for the positive value, n = 13 PROBLEM 7.10 The relation between flow and head for a slurry pump may be represented approximately by a straight line, the maximum flow at zero head being 0.0015 m3 /s and the maximum head at zero flow 760 m of liquid. Using this pump to feed a slurry to a pressure leaf filter, 66 (a) how long will it take to produce 1 m3 of filtrate, and (b) what will be the pressure drop across the filter after this time? A sample of the slurry was filtered at a constant rate of 0.00015 m3 /s through a leaf filter covered with a similar filter cloth but of one-tenth the area of the full scale unit and after 625 s the pressure drop across the filter was 360 m of liquid. After a further 480 s the pressure drop was 600 m of liquid. Solution For constant rate filtration through the filter leaf: V2 + LA A2 (−P )t V = v rμv (equation 7.17) At a constant rate of 0.00015 m3 /s when the time = 625 s: V = 0.094 m3 , (−P ) = 3530 kN/m2 and at t = 1105 s: V = 0.166 m3 and (−P ) = 5890 kN/m2 Substituting these values into equation 7.17 gives: (0.094)2 + LA/v × 0.094 = (A2 /rμv) × 3530 × 625 or: and: or: 0.0088 + 0.094LA/v = 2.21 × 106 A2 /rμv (0.166)2 + LA/v × 0.166 = (A2 /rμv) × 5890 × 1105 0.0276 + 0.166LA/v = 6.51 × 106 A2 /rμv Equations (i) and (ii) may be solved simultaneously to give: LA/v = 0.0154 and A2 /rμv = 4.64 × 10−9 As the filtration area of the full-size plant is 10 times that of the leaf filter then: LA/v = 0.154 and A2 /rμv = 4.64 × 10−7 If the pump develops a head of 760 m of liquid or 7460 kN/m2 at zero flow and has zero head at Q = 0.0015 m3 /s, its performance may be expressed as: (−P ) = 7460 − (7460/0.0015)Q or: (−P ) = 7460 − 4.97 × 106 Q kN/m2 A2 (−P ) dV = dt rμv(V + LA/v) Substituting for (−P ) and the filtration constants gives: dV A2 (7460 − 4.97 × 106 dV /dt) = dt rμv (V + 0.154) 67 (equation 7.16) Since Q = dV /dt, then: dV 4.67 × 10−7 [7460 − 4.97 × 106 (dV /dt)] = dt (V + 0.154) (V + 0.154)dV = 3.46 × 10−3 − 2.31dV /dt The time taken to collect 1 m3 is then given by:  1  t (V + 0.154 + 2.31)dV = (3.46 × 10−3 )dt 0 0 t = 857 s and: The pressure at this time is found by substituting in equation 7.17 with V = 1 m3 and t = 857 s2 . This gives: 12 + 0.154 × 1 = 4.64 × 10−7 × 857(−P ) and: (−P ) = 2902 kN/m2 PROBLEM 7.11 A slurry containing 40 per cent by mass solid is to be filtered on a rotary drum filter 2 m diameter and 2 m long which normally operates with 40 per cent of its surface immersed in the slurry and under a pressure of 17 kN/m2 . A laboratory test on a sample of the slurry using a leaf filter of area 200 cm2 and covered with a similar cloth to that on the drum, produced 300 cm3 of filtrate in the first 60 s and 140 cm3 in the next 60 s, when the leaf was under pressure of 84 kN/m2 . The bulk density of the dry cake was 1500 kg/m3 and the density of the filtrate was 1000 kg/m3 . The minimum thickness of cake which could be readily removed from the cloth was 5 mm. At what speed should the drum rotate for maximum throughput and what is this throughput in terms of the mass of the slurry fed to the unit per unit time? Solution See Volume 2, Example 7.4. PROBLEM 7.12 A continuous rotary filter is required for an industrial process for the filtration of a suspension to produce 0.002 m3 /s of filtrate. A sample was tested on a small laboratory filter of area 0.023 m2 to which it was fed by means of a slurry pump to give filtrate at a constant rate of 0.0125 m3 /s. The pressure difference across the test filter increased from 14 kN/m2 after 300 s filtration to 28 kN/m2 after 900 s, at which time the cake thickness had reached 38 mm. What are suitable dimensions and operating conditions for the rotary filter, assuming that the resistance of the cloth used is one-half that on the test filter, 68 and that the vacuum system is capable of maintaining a constant pressure difference of 70 kN/m2 across the filter? Solution Data from the laboratory filter may be used to find the cloth and cake resistance of the rotary filter. For the laboratory filter operating under constant rate conditions: V12 + A2 (−P )t LA V1 = v rμv (equation 7.17) A = 0.023 m2 and the filtration rate = 0.0125 m3 /s At t = 300 s, then: (−P ) = 14 kN/m2 and V1 = 3.75 × 10−3 m3 When t = 900 s, then: (−P ) = 28 kN/m2 and V1 = 1.125 × 10−2 m3 Hence: (3.75 × 10−3 )2 + (L/v) × 0.023 × 3.75 × 10−3 = 1.41 × 10−5 + 8.63 × 10−5 (L/v) = 2.22/rμv or: and: 14 × (0.023)2 × 300 rμv (1.25 × 10−2 )2 + (L/v) × 0.023 × 1.125 × 10−3 = 28 × (0.023)2 × 900 rμv 1.27 × 10−4 + 2.59 × 10−4 (L/v) = 13.33/rμv from which: L/v = 0.164 m, and rμv = 7.86 × 104 kg/m3 s If the cloth resistance is halved by using the rotary filter, L/v = 0.082. As the filter operates at constant pressure, then:   2LA 2A2 (−P )t V2 (equation 7.18) = v rμv If θ is the time for 1 rev × fraction submerged and V ′ is volume of filtrate/revolution (given by equation 7.18), the speed = 0.0167 Hz (1 rpm) and 20 per cent submergence, then: θ = (60 × 0.2) = 12 s Thus: or: from which: V ′2 + 2 × 0.082AV ′ = (2A2 × 70 × 12) (7.86 × 104 ) V ′2 + 0.164AV ′ = 0.0214A2 (A/V ′ ) = 11.7 m−1 The required rate of filtration = 0.002 m3 /s. Thus: The volume/revolution, V ′ = (0.002 × 60) = 0.12 m3 A = (11.7 × 0.12) = 1.41 m2 69 If L = D, then: area of drum = πDL = πD 2 = 1.41 m2 D = L = 0.67 m and: The cake thickness on the drum should now be checked. v = AL/V and from data on the laboratory filter: v = (0.023 × 0.038)/(1.125 × 10−2 ) = 0.078 Hence, the cake thickness on the drum, vV ′ /A = (0.078/11.7) = 0.0067 m or 6.7 mm which is acceptable. PROBLEM 7.13 A rotary drum filter, 1.2 m diameter and 1.2 m long, handles 6.0 kg/s of slurry containing 10 per cent of solids when rotated at 0.005 Hz. By increasing the speed to 0.008 Hz it is found that it can then handle 7.2 kg/s. What will be the percentage change in the amount of wash water which may be applied to each kilogram of cake caused by the increased speed of rotation of the drum, and what is the theoretical maximum quantity of slurry which can be handled? Solution For constant pressure filtration: A2 (−P ) a dV = = (say) dt rμv[V + (LA/v)] V +b or: V 2 /2 + bV = at For Case 1: 1 revolution takes (1/0.005) = 200 s and the rate = V1 /200. For Case 2: 1 revolution takes (1/0.008) = 125 s and the rate = V2 /125. But: or: V1 /200 6.0 = V2 /125 7.2 V1 /V2 = 1.33 and V2 = 0.75V1 For case 1, using the filtration equation (7.16): V12 + 2bV1 = 2a × 200 and for case 2: V22 + 2bV2 = 2a × 125 70 (equation 7.16) Substituting V2 = 0.75V1 in these two equations allows the filtration constants to found as: a = 0.00375V12 and b = 0.25V1 The rate of flow of wash water will equal the final rate of filtration so that for case 1: Wash water rate = a/(V1 + b). Wash water per revolution ∝ 200a/(V1 + b). Wash water/revolution per unit solids ∝ 200a/V1 (V1 + b), or: ∝ (200 × 0.00375V12 )/V1 (V1 + 0.25V1 ) ∝ 0.6. Similarly for case 2, the wash water per revolution per unit solids is proportional to: 125a/V 2 (V2 + b) which is: ∝ (125 × 0.00375V12 )/0.75V1 (0.75V1 + 0.25V1 ) ∝ 0.625 Hence: per cent increase = [(0.625 − 0.6)/0.6] × 100 = 4.17 per cent As 0.5V 2 + bV = at, the rate of filtration V /t is given by: a/(0.5V + b) The highest rate will be achieved as V tends to zero and: (V /t)max = a/b = 0.00375V12 /0.25V1 = 0.015V1 For case 1, the rate = (V1 /200) = 0.005V1 . Hence the limiting rate is three times the original rate, that is: 18.0 kg/s PROBLEM 7.14 A rotary drum with a filter area of 3 m3 operates with an internal pressure of 71.3 kN/m2 below atmospheric and with 30 per cent of its surface submerged in the slurry. Calculate the rate of production of filtrate and the thickness of cake when it rotates at 0.0083 Hz, if the filter cake is incompressible and the filter cloth has a resistance equal to that of 1 mm of cake. It is desired to increase the rate of filtration by raising the speed of rotation of the drum. If the thinnest cake that can be removed from the drum has a thickness of 5 mm, what is the maximum rate of filtration which can be achieved and what speed of rotation of the drum is required? The voidage of the cake = 0.4, the specific resistance of cake = 2 × 1012 m−2 the density of solids = 2000 kg/m3 , the density of filtrate = 1000 kg/m3 , the viscosity of filtrate = 10−3 N s/m2 and the slurry concentration = 20 per cent by mass solids. Solution A 20 per cent slurry contains 20 kg solids/80 kg solution. Volume of cake = 20/[2000(1 − 0.4)] = 0.0167 m3 . 71 Volume of liquid in the cake = (0.167 × 0.4) = 0.0067 m3 . Volume of filtrate = (80/1000) − 0.0067 = 0.0733 m3 . v = (0.0167/0.0733) = 0.23 Thus: The rate of filtration is given by: dV A2 (−P ) = dt rμv[V + (LA/v)] (equation 7.16) In this problem: A = 3 m2 , (−P ) = 71.3 kN/m2 or (71.3 × 103 ) N/m2 , r = 2 × 1012 m−2 , μ = 1 × 10−3 Ns/m2 , v = 0.23 and L = 1 mm or 1 × 10−3 m dV (32 × 71.3 × 103 ) = dt 0.23 × 2 × 1012 × 1 × 10−3 [V + (1 × 10−3 × 3/0.23)] Thus: = From which: (1.395 × 103 ) (V + 0.013) V 2 /2 + 0.013V = (1.395 × 10−3 )t If the rotational speed = 0.0083 Hz, 1 revolution takes (1/0.0083) = 120.5 s and a given element of surface is immersed for (120.5 × 0.3) = 36.2 s. When t = 36.2 s, V may be found by substitution to be 0.303 m3 . Hence: rate of filtration = (0.303/120.5) = 0.0025 m3 /s . Volume of filtrate for 1 revolution = 0.303 m3 . Volume of cake = (0.23 × 0.303) = 0.07 m3 . Thus: cake thickness = (0.07/3) = 0.023 m or 23 mm As the thinnest cake = 5 mm, volume of cake = (3 × 0.005) = 0.015 m3 . As v = 0.23, volume of filtrate = (3 × 0.005)/0.23 = 0.065 m3 . Thus: (0.065)2 /2 + (0.013 × 0.065) = 1.395 × 10−3 t t = 2.12 s and: Thus: time for 1 revolution = (2.12/0.3) = 7.1 s speed = 0.14 Hz (8.5 r.p.m) and: Maximum filtrate rate = 0.065 m3 in 7.1 s or: (0.065/7.1) = 0.009 m3 /s 72 PROBLEM 7.15 A slurry containing 50 per cent by mass of solids of density 2600 kg/m3 is to be filtered on a rotary drum filter, 2.25 m in diameter and 2.5 m long, which operates with 35 per cent of its surface immersed in the slurry and under a vacuum of 600 mm Hg. A laboratory test on a sample of the slurry, using a leaf filter with an area of 100 cm2 and covered with a cloth similar to that used on the drum, produced 220 cm3 of filtrate in the first minute and 120 cm3 of filtrate in the next minute when the leaf was under a vacuum of 550 mm Hg. The bulk density of the wet cake was 1600 kg/m3 and the density of the filtrate was 1000 kg/m3 . On the assumption that the cake is incompressible and that 5 mm of cake is left behind on the drum, determine the theoretical maximum flowrate of filtrate obtainable. What drum speed will give a filtration rate of 80 per cent of the maximum? Solution a) For the leaf filter: A = 100 cm2 or 0.01 m2 , (−P ) = 550 mm Hg. when t = 1 min, V = 220 cm3 = 0.00022 m3 when t = 2 min, V = 340 cm3 = 0.00034 m3 These values are substituted into the constant pressure filtration equation: V2 + 2(−P )A2 t 2LAV = v rμv (equation 7.18) to give the filtration constants as: L/v = 9.4 × 10−3 m and rμv = 1.23 × 106 Ns/m4 b) Cake properties: The densities are: solids = 2600 kg/m3 , cake = 1600 kg/m3 and filtrate = 1000 kg/m3 . For the cake; with a voidage e: 1 m3 of cake contains (1 − e) m3 of solids and e m3 of liquid or: 2600(1 − e) kg of solids and 1000e kg of liquid Thus the cake density is: 1600 = 2600(1 − e) + 1000e e = 0.625 and: −4 3.8461 × 10 = (1/2600) m3 solids form: (1/2600)(1/0.375) = 1.0256 × 10−3 m3 cake and: (1/2600)(0.625/0.375) = 6.4101 × 10−4 m3 liquid. 73 Thus: (1/1000) − 6.4101 × 10−4 m3 liquid form 1.0256 × 10−3 m3 cake and: v = 0.358 Thus: and: L = (9.4 × 10−3 × 0.358) = 0.365 × 10−3 m 1/rμ = (8.16 × 10−7 × 0.358) = 2.92 × 10−7 c) For the rotary filter 5 mm of cake is left on the drum. effective L = 8.365 × 10−3 m Thus: Considering 1 revolution of the filter taking tr min, then the filtration time is 0.35tr . A = (π × 2.25 × 2.5) = 17.67 m2 and −P = 600 mm Hg. Thus, in equation 7.18: V 2 + [(2 × 17.67 × 8.365 × 10−3 )/0.358]V = (2 × 17.672 × 600 × 8.16 × 10−7 × 0.35tr ) V 2 + 0.826V = 0.107tr or: The filtration rate is: V /tr = 0.107V /(V 2 + 0.826V ) = 0.107/(V + 0.826) This is a maximum when V = 0, that is when the rate is 0.13275 m3 /s The actual rate is: (0.80 × 0.13275) = 0.1064 m3 /s Thus: 0.107/(V + 0.826) = 0.1064 V = 0.180 and: Hence: tr = (1/0.107)(0.180 + 0.826)0.180 = 1.69 min and: speed of rotation = (0.35 × 1.69) = 0.59 rpm (0.0099 Hz) PROBLEM 7.16 A rotary filter which operates at a fixed vacuum gives a desired rate of filtration of a slurry when rotating at 0.033 Hz. By suitable treatment of the filter cloth with a filter aid, its effective resistance is halved and the required filtration rate is now achieved at a rotational speed of 0.0167 Hz (1 rpm). If, by further treatment, it is possible to reduce the effective cloth resistance to a quarter of the original value, what rotational speed is required? If the filter is now operated again at its original speed of 0.033 Hz, by what factor will the filtration rate be increased? 74 Solution From equation 7.11: V 2 = 2A2 (−P )t/(rμv) V 2 ∝ (t/r) ∝ (1/N r)  V ∝ 1/ (N r)    V /t ∝ (1/ (N r))(1/t) ∝ (1/ (N r))N ∝ (N/r) or: Thus: In this way:  V /t = F = k (N/r) where k is a constant. For a constant value of F : F /k = In the second case: Thus:  √ (0.033/r) = 0.182/ r N = 0.0167 Hz and the specific resistance is now 0.5r  √ F /k = (0.0167/0.5r) = 0.182/ r which is consistent. In the third case, the specific resistance is 0.25r and the speed is N Hz  √ Thus: 0.182/ r = (N/0.25r) and: Since: For the third case: and: N = 0.0083 Hz (0.5 rpm)  F1 = k (0.033/r)  F3 = k (0.033/0.25r)   F3 /F1 = (0.033/0.25r)/ (0.033/r) = 2.0 in this way, the filtration rate will be doubled. 75 SECTION 2-8 Membrane Separation Processes PROBLEM 8.1 Obtain expressions for the optimum concentration for minimum process time in the diafiltration of a solution of protein content S in an initial volume V0 , (a) If the gel–polarisation model applies. (b) If the osmotic pressure model applies. It may be assumed that the extent of diafiltration is given by: Vd = Volume of liquid permeated Vp = Initial feed volume V0 Solution See Volume 2, Example 8.1. PROBLEM 8.2 In the ultrafiltration of a protein solution of concentration 0.01 kg/m3 , analysis of data on gel growth rate and wall concentration Cw yields the second order relationship: dl = Kr Cw2 dt where l is gel thickness, and Kr is a constant, 9.2 × 10−6 m7 /kg2 s. The water flux through the membrane may be described by: J = |P | μw Rm where |P | is pressure difference, Rm is membrane resistance and μw is the viscosity of water. This equation may be modified for protein solutions to give: J = |P |   l μp Rm + Pg where Pg is gel permeability, and μp is the viscosity of the permeate. 76 The gel permeability may be estimated from the Carman–Kozeny equation:   2  e3 d Pg = 180 (1 − e)2 where d is particle diameter and e is the porosity of the gel. Calculate the gel thickness after 30 minutes of operation. Data: Flux mm/s 0.02 0.04 0.06 Viscosity of water Viscosity of permeate Diameter of protein molecule Operating pressure Porosity of gel Mass transfer coefficient to gel, hD |P | (kN/m2 ) 20 40 60 = = = = = = 1.3 mNs/m2 1.5 mNs/m2 20 nm 10 kN/m2 0.5 1.26 × 10−5 m/s Solution The gel growth rate as a function of the wall concentration, Cw , is given by: dl/dt = Kr Cw2 where l is the gel thickness, Kr is a constant = 9.2 × 10−6 m7 /kg s and Cw , the wall concentration given by: Cw = Cf exp(u/ hD ) The permeate flux is given by: Jsoln = |P |/[μp (Rm + l/Pg )] where |P | is the pressure difference, μp the viscosity of the permeate, Rm the membrane resistance and Pg , the gel permeability, which may be estimated from the Carman–Kozeny equation:   2  e3 d Pg = 180 (1 − e)2 where d is the particle diameter and e the porosity of the gel. For water: Jw = P /μw Rm and hence: Rm = |P |/J μw = (2.0 × 103 )/(1.3 × 10−3 × 0.02 × 10−3 ) = 7.60 × 1011 1/m 77 Also: Thus: and: Pg =  d2 180  e3 (1 − e)2  =  (20 × 10−9 )2 180  = 1.11 × 10−18 m2 (0.5)3 (1 − 0.5)2  dl/dt = Kr Cf2 exp{2P /[hD μp (Rm + l/Pg )]}  l  l dt = dl/[Kr Cf2 exp{2P /[hD μp (Rm + l/Pg )]} 0 0 The function of l is plotted against l in Figure 8a and the area under the curve is then plotted in the same figure. It is found that when t = 30 min, the area under the curve is 1800 at which l = 3.25 µm. Function of l 10 × 108 5 × 108 0 0 2 4 6 8 10 4 5 l (µm) 2000 1800 3.25 µm Area under curve 3000 1000 0 0 1 2 3 l (µm) Figure 8a. Graphical integration for Problem 8.2 78 SECTION 2-9 Centrifugal Separations PROBLEM 9.1 If a centrifuge is 0.9 m diameter and rotates at 20 Hz, at what speed should a laboratory centrifuge of 150 mm diameter be run if it is to duplicate the performance of the large unit? Solution If a particle of mass m is rotating at radius x with an angular velocity ω, it is subjected to a centrifugal force mxω2 in a radial direction and a gravitational force mg in a vertical direction. The ratio of the centrifugal to gravitational forces, xω2 /g, is a measure of the separating power of the machine, and in order to duplicate conditions this must be the same in both machines. In this case: x1 = 0.45 m, Thus: 2 ω1 = (20 × 2π) = 40π rad/s, 0.45(40π) /g = ω2 = 0.075ω22 /g  x2 = 0.075 m [6(40π)2 ] = (2.45 × 40π) = 98π rad/s and the speed of rotation = (98π/2π) = 49 Hz (2940 rpm) PROBLEM 9.2 An aqueous suspension consisting of particles of density 2500 kg/m3 in the size range 1–10 µm is introduced into a centrifuge with a basket 450 mm diameter rotating at 80 Hz. If the suspension forms a layer 75 mm thick in the basket, approximately how long will it take for the smallest particle to settle out? Solution Where the motion of the fluid with respect to the particle is turbulent, the time taken for a particle to settle from h1 to distance h2 from the surface in a radial direction, is given by the application of equation 3.124 as: t= 2 [(x + h2 )0.5 − (x + h1 )0.5 ] a′ 79 √ where a ′ = [3dω2 (ρs − ρ)/ρ], d is the diameter of the smallest particle = 1 × 10−6 m, ω is the angular velocity of the basket = (80 × 2π) = 502.7 rad/s, ρs is the density of the solid = 2500 kg/m3 , ρ is the density of the fluid = 1000 kg/m3 , and x is the radius of the inner surface of the liquid = 0.150 m.  Thus: a ′ = [3 × 1 × 10−6 × 502.72 (2500 − 1000)/1000] = 1.066 t = (2/1.066)[(0.150 + 0.075)0.5 − (0.150 + 0)0.5 ] and: = 1.876(0.474 − 0.387) = 0.163 s This is a very low value, equivalent to a velocity of (0.075/0.163) = 0.46 m/s. Because of the very small diameter of the particle, it is more than likely that the conditions are laminar, even at this particle velocity. For water, taking μ = 0.001 N s/m2 , then: Re = (1 × 10−6 × 0.46 × 1000)/0.001 = 0.46 and hence by applying equation 3.120: t = {18μ/[d 2 ω2 (ρs − ρ)]} ln[(x + h2 )/(x + h1 )] = {18 × 0.001/[10−12 × 502.72 (2500 − 1000)]} ln[(0.150 + 0.075)/(0.150 + 0)] = 47.5 ln(0.225/0.150) = 19.3 s PROBLEM 9.3 A centrifuge basket 600 mm long and 100 mm internal diameter has a discharge weir 25 mm diameter. What is the maximum volumetric flow of liquid through the centrifuge such that, when the basket is rotated at 200 Hz, all particles of diameter greater than 1 µm are retained on the centrifuge wall? The retarding force on a particle moving liquid may be taken as 3πμ du, where u is the particle velocity relative to the liquid μ is the liquid viscosity, and d is the particle diameter. The density of the liquid is 1000 kg/m3 , the density of the solid is 2000 kg/m3 and the viscosity of the liquid is 1.0 mN s/m2 . The inertia of the particle may be neglected. Solution With a basket radius of b m, a radius of the inner surface of liquid of x m and h m the distance radially from the surface of the liquid, the equation of motion in the radial direction of a spherical particle of diameter d m under streamline conditions is: (πd 3 /6)(ρs − ρ)(x + h)ω2 − 3πμ du − (πd 3 /6)ρs du/dt = 0 (from equation 3.108) Replacing u by dh/dt and neglecting the acceleration term gives: (dh/dt) = d 2 (ρs − ρ)ω2 (x + h)/18μ (equation 3.109) The time any element of material remains in the basket is V ′ /Q, where Q is the volumetric rate of feed to the centrifuge and V ′ is the volume of liquid retained in the basket at any 80 time. If the flow rate is adjusted so that a particle of diameter d is just retained when it has to travel through the maximum distance h = (b − x) before reaching the wall, then: h = d 2 (ρs − ρ)bω2 V ′ /(18μQ) Q = d 2 (ρs − ρ)bω2 V ′ /(18μh) or: V ′ = (π/4)(0.12 − 0.0252 ) × 0.6 = 0.0044 m3 In this case: h = (0.10 − 0.025)/2 Thus: Q= [(1 × 10−6 )2 (2000 − 1000) × 0.1 × (200 × 2π)2 × 0.0044] (18 × 0.001 × 0.0375) = 1.03 × 10−3 m3 /s (1 cm3 /s) PROBLEM 9.4 When an aqueous slurry is filtered in a plate and frame press, fitted with two 50 mm thick frames each 150 mm square at a pressure difference of 350 kN/m2 , the frames are filled in 3600 s. The liquid in the slurry has the same density as water. How long will it take to produce the same volume of filtrate as is obtained from a single cycle when using a centrifuge with a perforated basket 300 mm in diameter and 800 mm deep? The radius of the inner surface of the slurry is maintained constant at 75 mm and speed of rotation is 65 Hz (3900 rpm). It may be assumed that the filter cake is incompressible, that the resistance of the cloth is equivalent to 3 mm of cake in both cases and that the liquid in the slurry has the same density as water. Solution See Volume 2, Example 9.3. PROBLEM 9.5 A centrifuge with a phosphor bronze basket, 380 mm in diameter, is to be run at 67 Hz with a 75 mm layer of liquid of density 1200 kg/m3 in the basket. What thickness of walls are required in the basket? The density of phosphor bronze is 8900 kg/m3 and the maximum safe stress for phosphor bronze is 87.6 MN/m2 . Solution The stress at the walls is given by: S = (rb /t)(Pc − ρm trb ω2 ) 81 where: rb t ρp ω and Pc is is is is is the the the the the radius of the basket = (380/2) = 190 mm or 0.19 m wall thickness (m) density of the metal = 8900 kg/m3 rotational speed of the basket = (67 × 2π)2 = 1.77 × 105 rad/s pressure at the walls, given by: Pc = 0.5ρω2 (rb2 − x 2 ) where: ρ is the density of the fluid = 1200 kg/m3 and x is the radius at the liquid surface = [380 − (2 × 75)]/2 = 115 mm or 0.115 m Thus: ρc = (0.5 × 1200 × 1.77 × 105 )(0.192 − 0.1152 ) = 2.43 × 106 N/m2 and, the stress at the walls is: s = (0.19/t)[2.43 × 106 + (8900t × 0.19 × 1.77 × 10t 5 )] = (0.19/t)(2.43 × 106 + 2.993 × 108 t) = (4.62 × 105 /t) + 5.69 × 107 N/m2 The safe stress = 87.6 × 106 N/m2 Thus: and: 87.6 × 106 = (4.62 × 105 )/t + 5.63 × 107 t = 1.51 × 102 m or 15.1 mm 82 SECTION 2-10 Leaching PROBLEM 10.1 0.4 kg/s of dry sea-shore sand, containing 1 per cent by mass of salt, is to be washed with 0.4 kg/s of fresh water running countercurrently to the sand through two classifiers in series. It may be assumed that perfect mixing of the sand and water occurs in each classifier and that the sand discharged from each classifier contains one part of water for every two of sand by mass. If the washed sand is dried in a kiln dryer, what percentage of salt will it retain? What wash rate would be required in a single classifier in order to wash the sand to the same extent? Solution The problem involves a mass balance around the two stages. If x kg/s salt is in the underflow discharge from stage 1, then: salt in feed to stage 2 = (0.4 × 1)/100 = 0.004 kg/s. The sand passes through each stage and hence the sand in the underflow from stage 1 = 0.4 kg/s, which, assuming constant underflow, is associated with (0.4/2) = 0.2 kg/s water. Similarly, 0.2 kg/s water enters stage 1 in the underflow and 0.4 kg/s enters in the overflow. Making a water balance around stage 1, the water in the overflow discharge = 0.4 kg/s. In the underflow discharge from stage 1, x kg/s salt is associated with 0.2 kg/s water, and hence the salt associated with the 0.4 kg/s water in the overflow discharge = (x × 0.4)/0.2 = 2x kg/s. This assumes that the overflow and underflow solutions have the same concentration. In stage 2, 0.4 kg/s water enters in the overflow and 0.2 kg/s leaves in the underflow. Thus: water in overflow from stage 2 = (0.4 − 0.2) = 0.2 kg/s. The salt entering is 0.004 kg/s in the underflow and 2x in the overflow — a total of (0.004 + 2x) kg/s. The exit underflow and overflow concentrations must be the same, and hence the salt associated with 0.2 kg/s water in each stream is: (0.004 + 2x)/2 = (0.002 + x) kg/s Making an overall salt balance: 0.004 = x + (0.002 + x) 83 and x = 0.001 kg/s This is associated with 0.4 kg/s sand and hence: salt in dried sand = (0.001 × 100)/(0.4 + 0.001) = 0.249 per cent The same result may be obtained by applying equation 10.16 over the washing stage: Sn+1 /S1 = (R − 1)/(R n+1 − 1) In this case: (equation 10.16) R = (0.4/0.2) = 2, n = 1, S2 = x, S1 = (0.002 + x) and: x/(0.002 + x) = (2 − 1)/(22 − 1) = 0.33 x = (0.000667/0.667) = 0.001 kg/s and the salt in the sand = 0.249 per cent as before. Considering a single stage: If y kg/s is the overflow feed of water then, since 0.2 kg/s water leaves in the underflow, the water in the overflow discharge = (y − 0.2) kg/s. With a feed of 0.004 kg/s salt and 0.001 kg/s salt in the underflow discharge, the salt in the overflow discharge = 0.003 kg/s. The ratio (salt/solution) must be the same in both discharge streams or: (0.001)/(0.20 + 0.001) = 0.003/(0.003 + y − 0.2) and y = 0.8 kg/s PROBLEM 10.2 Caustic soda is manufactured by the lime-soda process. A solution of sodium carbonate in water containing 0.25 kg/s Na2 CO3 is treated with the theoretical requirement of lime and, after the reaction is complete, the CaCO3 sludge, containing by mass 1 part of CaCO3 per 9 parts of water is fed continuously to three thickeners in series and is washed countercurrently. Calculate the necessary rate of feed of neutral water to the thickeners, so that the calcium carbonate, on drying, contains only 1 per cent of sodium hydroxide. The solid discharged from each thickener contains one part by mass of calcium carbonate to three of water. The concentrated wash liquid is mixed with the contents of the agitated before being fed to the first thickeners. Solution See Volume 2, Example 10.2. PROBLEM 10.3 How many stages are required for a 98 per cent extraction of a material containing 18 per cent of extractable matter of density 2700 kg/m3 and which requires 200 volumes of liquid/100 volumes of solid for it to be capable of being pumped to the next stage? The strong solution is to have a concentration of 100 kg/m3 . 84 Solution Taking as a basis 100 kg solids fed to the plant, this contains 18 kg solute and 82 kg inert material. The extraction is 98 per cent and hence (0.98 × 18) = 17.64 kg solute appears in the liquid product, leaving (18 − 17.64) = 0.36 kg solute in the washed solid. The concentration of the liquid product is 100 kg/m3 and hence the volume of the liquid product = (17.64/100) = 0.1764 m3 . Volume of solute in liquid product = (17.64/2700) = 0.00653 m3 . Volume of solvent in liquid product = (0.1764 − 0.00653) = 0.1699 m3 . Mass of solvent in liquid product = 0.1699ρ kg where ρ kg/m3 is the density of solvent. In the washed solids, total solids = 82 kg or (82/2700) = 0.0304 m3 . Volume of solution in the washed solids = (0.0304 × 200)/100 = 0.0608 m3 . Volume of solute in solution = (0.36/2700) = 0.0001 m3 . Volume of solvent in washed solids = (0.0608 − 0.0001) = 0.0607 m3 . and mass of solvent in washed solids = 0.0607ρ kg Mass of solvent fed to the plant = (0.0607 + 0.1699)ρ = 0.2306ρ kg The overall balance in terms of mass is therefore; Inerts Solute Solvent 82 — 82 — 18 — 0.36 17.64 — 0.2306ρ 0.0607ρ 0.1699ρ Feed to plant Wash liquor Washed solids Liquid product Solvent discharged in the overflow , R = (0.2306ρ/0.0607ρ) = 3.80 Solvent discharged in the underflow The overflow product contains 100 kg solute/m3 solution. This concentration is the same as the underflow from the first thickener and hence the material fed to the washing thickeners contains 82 kg inerts and 0.0608 m3 solution containing (100 × 0.0608) = 6.08 kg solute. Thus, in equation 10.16: (3.80 − 1)/(3.80n+1 − 1) = (0.36/6.08) or: 3.80n+1 = 48.28 and n = 1.89, say 2 washing thickeners. Thus a total of 3 thickeners is required. 85 PROBLEM 10.4 Soda ash is mixed with lime and the liquor from the second of three thickeners and passed to the first thickener where separation is effected. The quantity of this caustic solution leaving the first thickener is such as to yield 10 Mg of caustic soda per day of 24 hours. The solution contains 95 kg of caustic soda/1000 kg of water, whilst the sludge leaving each of the thickeners consists of one part of solids to one of liquid. Determine: (a) the mass of solids in the sludge, (b) the mass of water admitted to the third thickener and (c) the percentages of caustic soda in the sludges leaving the respective thickeners. Solution Basis: 100 Mg CaCO3 in the sludge leaving each thickener. In order to produce 100 Mg CaCO3 , 106 Mg Na2 CO3 must react giving 80 Mg NaOH according to the equation: Na2 CO3 + Ca(OH)2 = 2NaOH + CaCO3 . For the purposes of calculation it is assumed that a mixture of 100 Mg CaCO3 and 80 Mg NaOH is fed to the first thickener and w Mg water is the overflow feed to the third thickener. Assuming that x1 , x2 and x3 are the ratios of caustic soda to solution by mass in each thickener then the mass balances are made as follows: Overall Underflow feed Overflow feed Underflow product Overflow product Thickener 1 Underflow feed Overflow feed Underflow product Overflow product Thickener 2 Underflow feed Overflow feed Underflow product Overflow product Thickener 3 Underflow feed Overflow feed Underflow product Overflow product CaCO3 NaOH Water 100 — 100 — 80 — 100x3 (80 − 100w3 ) — w 100(1 − x3 ) w − 100(1 − x3 ) 100 — 100 — 80 100(x1 − x3 ) 100x1 80 − 100x3 — w + 100(x3 − x1 ) 100(1 − x1 ) w − 100(1 − x3 ) 100 — 100 — 100x1 100(x2 − x3 ) 100x2 100(x1 − x3 ) 100(1 − x1 ) w + 100(x3 − x2 ) 100(1 − x2 ) w + 100(x3 − x1 ) 100 — 100 — 100x2 — 100x3 100(x2 − x3 ) 100(1 − x2 ) w 100(1 − x3 w + 100(x3 − x2 ) 86 In the overflow product, 0.095 Mg NaOH is associated with 1 Mg water. Thus: x1 = 0.095/(1 + 0.095) = 0.0868 Mg/Mg solution (i) Assuming that equilibrium is attained in each thickener, the concentration of NaOH in the overflow product is equal to the concentration of NaOH in the solution in the underflow product. Thus: x3 = [100(x2 − x3 )]/[100(x2 − x3 ) + w − 100(x2 − x3 )] = 100(x2 − x3 )/w (ii) x2 = [100(x1 − x3 )]/[100(x1 − x3 ) + w − 100(x1 − x3 )] = 100(x1 − x3 )/w (iii) x3 = (80 − 100x3 )/[80 − 100x3 + w − 100(1 − x3 )] = (80 − 100x3 )/(w − 20) (iv) Solving equations (i)–(iv) simultaneously, gives: x3 = 0.0010 Mg/Mg, x2 = 0.0093 Mg/Mg, x1 = 0.0868 Mg/Mg w = 940.5 Mg/100 Mg CaCO3 and: The overflow product = w − 100(1 − x3 ) = 840.6 Mg/100 Mg CaCO3 . The actual flow of caustic solution = (10/0.0868) = 115 Mg/day. Thus: mass of CaCO3 in sludge = (100 × 115)/840.6 = 13.7 Mg/day The mass of water fed to third thickener = 940.5 Mg/100 Mg CaCO3 = (940.5 × 13.7)/100 = 129 Mg/day or: The total mass of sludge leaving each thickener = 200 Mg/100 Mg CaCO3 . The mass of caustic soda in the sludge = 100x1 Mg/100 Mg CaCO3 and hence the concentration of caustic in sludge leaving, thickener 1 = (100 × 0.0868 × 100)/200 = 4.34 per cent thickener 2 = (100 × 0.0093 × 100)/200 = 0.47 per cent thickener 3 = (100 × 0.0010 × 100)/200 = 0.05 per cent PROBLEM 10.5 Seeds, containing 20 per cent by mass of oil, are extracted in a countercurrent plant and 90 per cent of the oil is recovered as a solution containing 50 per cent by mass of oil. If the seeds are extracted with fresh solvent and 1 kg of solution is removed in the underflow in association with every 2 kg of insoluble matter, how many ideal stages are required? 87 Solution See Volume 2, Example 10.4. PROBLEM 10.6 It is desired to recover precipitated chalk from the causticising of soda ash. After decanting the liquor from the precipitators the sludge has the composition 5 per cent CaCO3 , 0.1 per cent NaOH and the balance water. 1000 Mg/day of this sludge is fed to two thickeners where it is washed with 200 Mg/day of neutral water. The pulp removed from the bottom of the thickeners contains 4 kg of water/kg of chalk. The pulp from the last thickener is taken to a rotary filter and concentrated to 50 per cent solids and the filtrate is returned to the system as wash water. Calculate the net percentage of CaCO3 in the product after drying. Solution Basis: 1000 Mg/day sludge fed to the plant If x1 and x2 are the solute/solvent ratios in thickeners 1 and 2 respectively, then the mass balances are: Overall Underflow feed Overflow feed Underflow product Overflow product Thickener 1 Underflow feed Overflow feed Underflow product Overflow product Thickener 2 Underflow feed Overflow feed Underflow product Overflow product CaCO3 NaOH Water 50 — 50 — 1 — 200x2 (1 − 200x2 ) 949 200 200 949 50 — 50 — 1 200(x1 − x2 ) 200x1 (1 − 200x2 ) 949 200 200 949 50 — 50 — 200x1 — 200x2 200(x1 − x2 ) 200 200 200 200 Assuming that equilibrium is attained, the solute/solvent ratio will be the same in the overflow and underflow products of each thickener and: x2 = 200(x1 − x2 )/200 and: x1 = (1 − 200x2 )/949 Thus: x1 = 0.000954 and 88 or x2 = 0.5x1 x2 = 0.000477 The underflow product contains 50 Mg CaCO3 , (200 × 0.000477) = 0.0954 Mg NaOH and 200 Mg water. After concentration to 50 per cent solids, the mass of NaOH in solution = (0.0954 × 50)/200.0954 = 0.0238 Mg and the CaCO3 in dried solids = (100 × 50)/50.0238 = 99.95 per cent This approach ignores the fact that filtrate is returned to the second thickener together with wash water. Taking this into account, the calculation is modified as follows. The underflow product from the second thickener contains: 50 Mg CaCO3 , 200x2 Mg NaOH and 200 Mg water After filtration, the 50 Mg CaCO3 is associated with 50 Mg solution of the same concentration and hence this contains: 50x2 /(1 + x2 ) Mg NaOH and 50/(1 + x2 ) Mg water The remainder is returned with the overflow feed to the second thickener. The filtrate returned contains: 200x2 − 50x2 /(1 + x2 ) Mg NaOH 200 − 50/(1 + x2 ) Mg water and: The balances are now: Overall Underflow feed Overflow feed Underflow product Overflow product Thickener 1 Underflow feed Overflow feed Underflow product Overflow product Thickener 2 Underflow feed Overflow feed Underflow product Overflow product CaCO3 NaOH Water 50 — 50 — 1 200x2 − 50x2 /(1 + x2 ) 200x2 1 − 50x2 /(1 + x2 ) 949 400 − 50/(1 + x2 ) 200 1149 − 50/(1 + x2 ) 50 — 50 — 1 200x1 − 50x2 /(1 + x2 ) 200x1 1 − 50x2 /(1 + x2 ) 949 400 − 50/(1 + x2 ) 200 1149 − 50/(1 + x2a ) 50 — 50 — 200x1 200x2 − 50x2 /(1 + x2 ) 200x2 200x1 − 50x2 /(1 + x2 ) 200 400 − 50/(1 + x2 ) 200 400 − 50/(1 + x2 ) Again, assuming equilibrium is attained, then: x2 = [200x1 − 50x2 /(1 + x2 )]/[400 − 50/(1 + x2 )] and: x1 = [1 − 50x2 /(1 + x2 )]/[1149 − 50/(1 + x2 )] 89 Solving simultaneously, then: x1 = 0.000870 Mg/Mg and x2 = 0.000435 Mg/Mg The solid product leaving the filter contains 50 Mg CaCO3 and: (50 × 0.000435)/(1 + 0.000435) = 0.02175 Mg NaOH in solution. After drying, the solid product will contain: (100 × 50)/(50 + 0.02175) = 99.96 per cent CaCO3 PROBLEM 10.7 Barium carbonate is to be made by reacting sodium carbonate and barium sulphide. The quantities fed to the reaction agitators per 24 hours are 20 Mg of barium sulphide dissolved in 60 Mg of water, together with the theoretically necessary amount of sodium carbonate. Three thickeners in series are run on a countercurrent decantation system. Overflow from the second thickener goes to the agitators, and overflow from the first thickener is to contain 10 per cent sodium sulphide. Sludge from all thickeners contains two parts water to one part barium carbonate by mass. How much sodium sulphide will remain in the dried barium carbonate precipitate? Solution Basis: 1 day’s operation The reaction is: Molecular masses: BaS + Na2 CO3 = BaCO3 + Na2 S 169 106 197 78 kg/kmol. Thus 20 Mg BaS will react to produce (20 × 197)/169 = 23.3 Mg BaCO3 and: (20 × 78)/169 = 9.23 Mg Na2 S The calculation may be made on the basis of this material entering the washing thickeners together with 60 Mg water. If x1 , x2 , and x3 are the Na2 S/water ratio in the respective thickeners, then the mass balances are: Overall Underflow feed Overflow feed Underflow product Overflow product BaCO3 Na2 S Water 23.3 — 23.3 — 9.23 — 46.6x3 9.23 − 46.6x3 60 w (say) 46.6 w + 13.4 90 Thickener 1 Underflow feed Overflow feed Underflow product Overflow product Thickener 2 Underflow feed Overflow feed Underflow product Overflow product Thickener 3 Underflow feed Overflow feed Underflow product Overflow product BaCO3 Na2 S Water 23.3 — 23.3 — 9.23 46.6(x1 − x3 ) 46.6x1 9.23 − 46.6x3 60 w 46.6 w + 13.4 23.3 — 23.3 — 46.6x1 46.6(x2 − x3 ) 46.6x2 46.6(x1 − x3 ) 46.6 w 46.6 w 23.3 — 23.3 — 46.6x2 — 46.6x3 46.6(x2 − x3 ) 46.6 w 46.6 w In the overflow product leaving the first thickener: (9.23 − 46.6x3 )/(13.4 + w + 9.23 − 46.6x3 ) = 0.10 (i) Assuming equilibrium is attained in each thickener, then: and: x1 = (9.23 − 46.6x3 )/(13.4 + w), (ii) x2 = 46.6(x1 − x3 )/w, (iii) x3 = 46.6(x2 − x3 )/w (iv) Solving equations (i)–(iv) simultaneously: x1 = 0.112, x2 = 0.066, x3 = 0.030, and w = 57.1 Mg/day In the underflow product from the third thickener, the mass of Na2 S is (46.6 × 0.030) = 1.4 Mg associated with 23.3 Mg BaCO3 When this stream is dried, the barium carbonate will contain: (100 × 1.4)/(1.4 + 23.3) = 5.7 per cent sodium sulphide PROBLEM 10.8 In the production of caustic soda by the action of calcium hydroxide on sodium carbonate, 1 kg/s of sodium carbonate is treated with the theoretical quantity of lime. The sodium carbonate is made up as a 20 per cent solution. The material from the extractors is fed to a countercurrent washing system where it is treated with 2 kg/s of clean water. The washing thickeners are so arranged that the ratio of the volume of liquid discharged in 91 the liquid offtake to that discharged with the solid is the same in all the thickeners and is equal to 4.0. How many thickeners must be arranged in series so that not more than 1 per cent of the sodium hydroxide discharged with the solid from the first thickener is wasted? Solution The reaction is: Molecular masses: Na2 CO3 + Ca(OH)2 = 2NaOH + CaCO3 106 74 80 100 kg/kmol. Thus 1 kg/s Na2 CO3 forms (1 × 80)/106 = 0.755 kg/s NaOH (1 × 100)/106 = 0.943 kg/s CaCO3 and: In a 20 per cent solution, 1 kg/s Na2 CO3 is associated with (1 − 0.20)/0.2 = 4.0 kg/s water. If x kg/s NaOH leaves in the underflow product from the first thickener, then 0.01x kg/s NaOH should leave in the underflow product from the nth thickener. The amount of NaOH in the overflow from the first thickener is then given from an overall balance as = (0.755 − 0.01x) kg/s: Since the volume of the overflow product is 4x, the volume of solution in underflow product, then: (0.755 − 0.01x) = 4x and x = 0.188 kg/s and the NaOH leaving the nth thickener in the underflow = (0.01 × 0.188) = 0.00188 kg/s. Thus the fraction of solute fed to the washing system which remains associated with the washed solids, f = (0.00188/0.755) = 0.0025 kg/kg In this case R = 4.0 and in equation 10.17: n = {ln[1 + (4 − 1)/0.0025]}/(ln 4) − 1 = 4.11, say 5 washing thickeners PROBLEM 10.9 A plant produces 100 kg/s of titanium dioxide pigment which must be 99 per cent pure when dried. The pigment is produced by precipitation and the material, as prepared, is contaminated with 1 kg of salt solution containing 0.55 kg of salt/kg of pigment. The material is washed countercurrently with water in a number of thickeners arranged in series. How many thickeners will be required if water is added at the rate of 200 kg/s and the solid discharged from each thickeners removes 0.5 kg of solvent/kg of pigment? What will be the required number of thickeners if the amount of solution removed in association with the pigment varies with the concentration of the solution in the thickener as follows: kg solute/kg solution kg solution/kg pigment 0 0.30 0.1 0.32 0.2 0.34 0.3 0.36 0.4 0.38 0.5 0.40 The concentrated wash liquor is mixed with the material fed to the first thickener. 92 Solution Part I The overall balance in kg/s, is: Feed from reactor Wash liquor added Washed solid Liquid product TiO2 Salt Water 100 — 100 — 55 — 0.1 54.9 45 200 50 195 The solvent in the underflow from the final washing thickener = 50 kg/s. The solvent in the overflow will be the same as that supplied for washing = 200 kg/s. This: Solvent discharged in overflow = (200/50) = 4 for the washing thickeners. Solvent discharged in underflow The liquid product from plant contains 54.9 kg of salt in 195 kg of solvent. This ratio will be the same in the underflow from the first thickener. Thus the material fed to the washing thickeners consists of 100 kg TiO2 , 50 kg solvent and (50 × 54.9)/195 = 14 kg salt. The required number of thickeners for washing is given by equation 10.16 as: (4 − 1)/(4n+1 − 1) = (0.1/14) 4n+1 = 421 giving: or: 4 < (n + 1) < 5 Thus 4 washing thickeners or a total of 5 thickeners are required. Part 2 The same nomenclature will be used as in Volume 2, Chapter 10. By inspection of the data, it is seen that Wh+1 = 0.30 + 0.2Xh . 2 Sh+1 = Wh+1 Xh = 0.30Xh + 0.2Xh2 = 5Wh+1 − 1.5Wh+1 Thus: Considering the passage of unit quantity of TiO2 through the plant: Ln+1 = 0, wn+1 = 2, Xn+1 = 0 since 200 kg/s pure solvent is used. Sn+1 = 0.001 S1 = 0.55 and therefore and Wn+1 = 0.3007. W1 = 1.00 Thus the concentration in the first thickener is given by equation 10.23: X1 = Ln+1 + S1 − Sn+1 = (0 + 0.55 − 0.001)/(2 + 1 − 0.3007) = 0.203 Wn+1 + W1 − Wn+1 93 From equation 10.26: Xh+1 = (−0.001 + Sh+1 ) Ln+1 − Sn+1 + Sh+1 (0 − 0.001 + Sh+1 ) = = Wn+1 − Wn+1 + Wh+1 (2 − 0.3007 + Wh+1 ) (1.7 + Wh+1 ) Since X1 = 0.203, then W2 = (0.30 + 0.2 × 0.203) = 0.3406 and: S2 = (0.3406 × 0.203) = 0.0691 Thus: X2 = (0.0691 − 0.001)/(1.7 + 0.3406) = 0.0334 Since X2 = 0.0334, then W3 = (0.30 + 0.2 × 0.0334) = 0.30668 and: S2 = (0.3067 × 0.0334) = 0.01025 Thus: X3 = (0.01025 − 0.001)/(1.7 + 0.3067) = 0.00447 Since X3 = 0.00447, then W4 = 0.30089 Hence, by the same method: X4 = 0.000150 and S4 = 0.0013 Since X4 = 0.000150, then W5 = 0.30003 and S5 = 0.000045. Thus S5 is less than Sn+1 and therefore 4 thickeners are required. PROBLEM 10.10 Prepared cottonseed meats containing 35 per cent of extractable oil are fed to a continuous countercurrent extractor of the intermittent drainage type using hexane as the solvent. The extractor consists of ten sections and the section efficiency is 50 per cent. The entrainment, assumed constant, is 1 kg solution/kg solids. What will be the oil concentration in the outflowing solvent if the extractable oil content in the meats is to be reduced by 0.5 per cent by mass? Solution Basis: 100 kg inert cottonseed material Mass of oil in underflow feed = (100 × 0.35)/(1 − 0.35) = 53.8 kg. In the underflow product from the plant, mass of inerts = 100 kg and hence mass of oil = (100 × 0.005)/(1 − 0.005) = 0.503 kg. This is in 100 kg solution and hence the mass of hexane in the underflow product = (100 − 0.503) = 99.497 kg. The overall balance in terms of mass is: Underflow feed Overflow feed Underflow product Overflow product Inerts Oil Hexane 100 — 100 — 53.8 — 0.503 53.297 — h (say) 99.497 (h − 99.497) 94 Since there are ten stages, each 50 per cent efficient, the system may be considered, as a first approximation as consisting of five theoretical stages each of 100 per cent efficiency, in which equilibrium is attained in each stage. On this basis, the underflow from stage 1 contains 100 kg solution in which the oil/hexane ratio = 53.297/(h − 99.497) and hence the amount of oil in this stream is: S1 = 100[1 − (h − 99.497)/(h − 46.2)] kg Sn+1 = 0.503 kg With constant underflow, the amount of solution in the overflow from each stage is say, h kg and the solution in the underflow = 100 kg. R = (h/100) = 0.01h Thus: and in equation 10.16: 0.503/[100 − 100(h − 99.497)/(h − 46.2)] = (0.01h − 1)/[(0.01h)5 − 1] (0.503h − 23.24) = (53.30h − 5330)/[(0.01h)5 − 1] or: Solving by trial and error: h = 238 kg and in the overflow product: mass of hexane = (238 − 99.497) = 138.5 kg, mass of oil = 53.3 kg and concentration of oil = (100 × 53.3)/(53.3 + 138.5) = 27.8 per cent . PROBLEM 10.11 Seeds containing 25 per cent by mass of oil are extracted in a countercurrent plant and 90 per cent of the oil is to be recovered in a solution containing 50 per cent of oil. It has been found that the amount of solution removed in the underflow in association with every kilogram of insoluble matter, k is given by: k = 0.7 + 0.5ys + 3ys2 kg/kg where ys is the concentration of the overflow solution in terms of mass fraction of solute kg/kg. If the seeds are extracted with fresh solvent, how many ideal stages are required? Solution Basis: 100 kg underflow feed to the first stage The first step is to obtain the underflow line, that is a plot of xs against xA . The calculations are made as follows: ys 0 0.1 Ratio (kg/kg inerts) k 0.70 0.78 Mass fraction oil (kys ) solvent k(1 − ys ) underflow (k + 1) oil (xA ) solvent (xs ) 0 0.078 0.70 0.702 1.70 1.78 0 0.044 0.412 0.394 95 ys 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 Ratio (kg/kg inerts) k 0.92 1.12 1.38 1.70 2.08 2.52 3.02 3.58 4.20 Mass fraction oil (kys ) solvent k(1 − ys ) underflow (k + 1) oil (xA ) solvent (xs ) 0.184 0.336 0.552 0.850 1.248 1.764 2.416 3.222 4.20 0.736 0.784 0.828 0.850 0.832 0.756 0.604 0.358 0 1.92 2.12 2.38 2.70 3.08 3.52 4.02 4.58 5.20 0.096 0.159 0.232 0.315 0.405 0.501 0.601 0.704 0.808 0.383 0.370 0.348 0.315 0.270 0.215 0.150 0.078 0 A plot of xA against xs is shown in Figure 10a. Considering the underflow feed, the seeds contain 25 per cent oil and 75 per cent inerts, and the point xs1 = 0, xA1 = 0.25 is drawn in as x1 . In the overflow feed, pure solvent is used and hence: ys·n+1 = 1.0, yA·n+1 = 0 This point is marked as yn+1 . 100% solvent y 1.0 n +1 Mass fraction of solvent, xs , ys y4 0.8 y3 y2 0.6 y1 0.4 xn+1 x5 x4 x3 x2 0.2 x1 0.4 0.6 0.8 Mass fraction of oil, xA, yA 0 100% inerts ∆ Figure 10a. Graphical construction for Problem 10.11 96 100% oil 1.0 In the overflow product, the oil concentration is 50 per cent and ys1 = 0.50 and yA1 = 0.50. This point lies on the hypotenuse and is marked y1 . 90 per cent of the oil is recovered, leaving 25(1 − 0.90) = 2.5 kg in the underflow product associated with 75 kg inerts; that is: ratio (oil/inerts) = (2.5/75) = 0.033 = kys 0.033 = (0.7ys + 0.5ys2 + 3ys3 ) Thus: Solving by substitution gives: ys = 0.041 and hence k = (0.033/0.041) = 0.805 xA = 0.0173 and xs = 0.405 This point is drawn as xn+1 on the xs against xA curve. The pole point  is obtained where yn+1 · xn+1 and y1 · x1 extended meet and the construction described in Chapter 10 is then followed. It is then found that xn+1 lies between x4 and x5 and hence 4 ideal stages are required . PROBLEM 10.12 Halibut oil is extracted from granulated halibut livers in a countercurrent multi-batch arrangement using ether as the solvent. The solids charge contains 0.35 kg oil/kg of exhausted livers and it is desired to obtain a 90 per cent oil recovery. How many theoretical stages are required if 50 kg of ether are used/100 kg of untreated solids. The entrainment data are: Concentration of overflow (kg oil/kg solution) Entrainment (kg solution/ kg extracted livers) 0 0.1 0.2 0.3 0.4 0.5 0.6 0.67 0.28 0.34 0.40 0.47 0.55 0.66 0.80 0.96 Solution See Volume 2, Example 10.5. 97 SECTION 2-11 Distillation PROBLEM 11.1 A liquid containing four components, A, B, C and D, with 0.3 mole fraction each of A, B and C, is to be continuously fractionated to give a top product of 0.9 mole fraction A and 0.1 mole fraction B. The bottoms are to contain not more than 0.5 mole fraction A. Estimate the minimum reflux ratio required for this separation, if the relative volatility of A to B is 2.0. Solution The given data may be tabulated as follows: A B C D Feed Top Bottoms 0.3 0.3 0.3 0.1 0.9 0.1 — — 0.05 The Underwood and Fenske equations may be used to find the minimum number of plates and the minimum reflux ratio for a binary system. For a multicomponent system nm may be found by using the two key components in place of the binary system and the relative volatility between those components in equation 11.56 enables the minimum reflux ratio Rm to be found. Using the feed and top compositions of component A:    1 (1 − xd ) xd −α (equation 11.50) Rm = α−1 xf (1 − xf )    1 (1 − 0.9) 0.9 Thus: Rm = −2 = 2.71 2−1 0.3 (1 − 0.3) PROBLEM 11.2 During the batch distillation of a binary mixture in a packed column the product contained 0.60 mole fraction of the more volatile component when the concentration in the still was 0.40 mole fraction. If the reflux ratio used was 20 : 1, and the vapour composition y is related to the liquor composition x by the equation y = 1.035x over the range of 98 concentration concerned, determine the number of ideal plates represented by the column. x and y are in mole fractions. Solution It is seen in equation 11.48, the equation of the operating line, that the slope is given by R/(R + 1)(= L/V ) and the intercept on the y-axis by: xd /(R + 1) = (D/Vn ) yn = R xd xn+1 + R+1 R+1 (equation 11.41) In this problem, the equilibrium curve over the range x = 0.40 to x = 0.60 is given by y = 1.035x and it may be drawn as shown in Figure 11a. The intercept of the operating line on the y-axis is equal to xd /(R + 1) = 0.60/(20 + 1) = 0.029 and the operating line is drawn through the points (0.60, 0.60) and (0, 0.029) as shown. 0.60 0.50 Equilibrium curve y = 1.035 x 0.40 0.40 0.50 0.60 Mole fraction MVC in vapour 0.30 0.50 Operating line xD 0.40 xs 0.30 0.20 0.10 Intercept = xd / (R + 1) = 0.029 0 Figure 11a. 0.20 0.40 0.60 Mole fraction MVC in liquid Construction for Problem 11.2 In this particular example, all these lines are closely spaced and the relevant section is enlarged in the inset of Figure 11a. By stepping off the theoretical plates as in the McCabe–Thiele method, it is seen that 18 theoretical plates are represented by the column. 99 PROBLEM 11.3 A mixture of water and ethyl alcohol containing 0.16 mole fraction alcohol is continuously distilled in a plate fractionating column to give a product containing 0.77 mole fraction alcohol and a waste of 0.02 mole fraction alcohol. It is proposed to withdraw 25 per cent of the alcohol in the entering stream as a side stream containing 0.50 mole fraction of alcohol. Determine the number of theoretical plates required and the plate from which the side stream should be withdrawn if the feed is liquor at its boiling point and a reflux ratio of 2 is used. Solution Taking 100 kmol of feed to the column as a basis, 16 kmol of alcohol enter, and 25 per cent, that is 4 kmol, are to be removed in the side stream. As the side-stream composition is to be 0.5, that stream contains 8 kmol. An overall mass balance gives: F =D+W +S That is: 100 = D + W + 8 or 92 = D + W A mass balance on the alcohol gives: (100 × 0.16) = 0.77D + 0.02W + 4 or: 12 = 0.77D + 0.02W. from which: distillate, D = 13.55 kmol and bottoms, W = 78.45 kmol. In the top section between the side-stream and the top of the column: R = Ln /D = 2, and hence Ln = (2 × 13.55) = 27.10 kmol Vn = Ln + D and Vn = (27.10 + 13.55) = 40.65 kmol For the section between the feed and the side stream: Vs = Vn = 40.65, and: Ln = S + Ls Ls = (27.10 − 8) = 19.10 kmol At the bottom of the column: Lm = Ls + F = (19.10 + 100) = 119.10, if the feed is at its boiling-point. Vm = Lm − W = (119.10 − 78.45) = 40.65 kmol. The slope of the operating line is always L/V and thus the slope in each part of the column can now be calculated. The top operating line passes through the point (xd , xd ) and has a slope of (27.10/40.65) = 0.67. This is shown in Figure 11b and it applies until xs = 0.50 where the slope becomes (19.10/40.65) = 0.47. The operating line in the bottom of the column applies from xf = 0.16 and passes through the point (xw , xw ) with a slope of (119.10/40.65) = 2.92. 100 1.0 0.90 Mole fraction alcohol in vapour 0.80 0.70 2 Slope = Ln /Vn = 0.67 3 0.60 1 x = 0.77 D 4 0.50 5 6 0.40 0.30 Xs = 0.50 Slope = Ls /Vs = 0.47 7 0.20 Slope = Lm /Vm = 2.92 8 0.10 xF = 0.16 xw = 0.02 0 Figure 11b. 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 Mole fraction alcohol in liquid 1.0 Graphical construction for Problem 11.3 The steps corresponding to the theoretical plates may be drawn in as shown, and 8 plates are required with the side stream being withdrawn from the fourth plate from the top. PROBLEM 11.4 In a mixture to be fed to a continuous distillation column, the mole fraction of phenol is 0.35, o-cresol is 0.15, m-cresol is 0.30 and xylenols is 0.20. A product is required with a mole fraction of phenol of 0.952, o-cresol 0.0474 and m-cresol 0.0006. If the volatility to o-cresol of phenol is 1.26 and of m-cresol is 0.70, estimate how many theoretical plates would be required at total reflux. Solution The data may be tabulated in terms of mole fractions as follows. Component Feed Top P O M X 0.35 0.15 0.30 0.20 0.952 0.0474 0.0006 — 1.0000 101 Bottoms α 1.26 1.0 0.7 Fenske’s equation may be used to find the minimum number of plates. Thus the number of plates at total reflux is given by: n+1= log[(xA /xB )d (xB /xA )s ] log αAB (equation 11.58) For multicomponent systems, components A and B refer to the light and heavy keys respectively. In this problem, o-cresol is the light key and m-cresol is the heavy key. A mass balance may be carried out in order to determine the bottom composition. Taking as a basis, 100 kmol of feed, then: 100 = D + W For phenol: (100 × 0.35) = 0.952D + xwp W D = 36.8 W = 63.2 If xwp is zero then: For o-cresol: (100 × 0.15) = (0.0474 × 36.8) + (xwo × 63.2) and xwo = 0.21 For m-cresol: (100 × 0.30) = (0.0006 × 36.8) + (xwm × 63.2) and xwm = 0.474 By difference: xwx = 0.316 αom = (1/0.7) = 1.43 Hence, substituting into Fenske’s equation gives: n+1= and: log[(0.0474/0.0006)(0.474/0.21)] log 1.43 n = 13.5 PROBLEM 11.5 A continuous fractionating column, operating at atmospheric pressure, is to be designed to separate a mixture containing 15.67 per cent CS2 and 84.33 per cent CCl4 into an overhead product containing 91 per cent CS2 and a waste of 97.3 per cent CCl4 all by mass. A plate efficiency of 70 per cent and a reflux of 3.16 kmol/kmol of product may be assumed. Using the following data, determine the number of plates required. The feed enters at 290 K with a specific heat capacity of 1.7 kJ/kg K and a boiling point of 336 K. The latent heats of CS2 and CCl4 are 25.9 kJ/kmol. 0 8.23 15.55 26.6 33.2 49.5 63.4 CS2 in the vapour (Mole per cent) CS2 in the 0 2.36 6.15 11.06 14.35 25.85 33.0 liquid (Mole per cent) 74.7 82.9 87.8 93.2 53.18 66.30 75.75 86.04 Solution The equilibrium data are shown in Figure 11c and the problem may be solved using the method of McCabe and Thiele. All compositions are in terms of mole fractions so that: 102 1.0 1 (xd, xd) 0.90 2 0.80 3 Mole fraction CS2 in vapour 0.70 4 0.60 5 0.50 6 0.40 q -line 0.30 0.20 7 (xf , xf ) 0.23 8 0.10 9 0.196 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 0 (xw, xw) Mole fraction CS2 in liquor Figure 11c. 1.0 Equilibrium data for Problem 11.5 Top product: xd = Feed: xf = Bottom product: xw = (91/76) = 0.953 (91/76) + (9/154) (15.67/76) = 0.274 (15.67/76) + (84.33/154) (2.7/76) = 0.053 (2.7/76) + (97.3/154) In this problem, the feed is not at its boiling-point so the slope of the q-line must be determined in order to locate the intersection of the operating lines. q is defined as the heat required to vaporise 1 kmol of feed/molar latent heat of feed, or q = (λ + Hf s − Hf )/λ where λ is the molar latent heat. Hf s is the enthalpy of 1 kmol of feed at its boiling-point, and Hf is the enthalpy of 1 kmol of feed. The feed composition is 27.4 per cent CS2 and 72.6 per cent CCl4 so that the mean molecular mass of the feed is given by: (0.274 × 76) + (0.726 × 154) = 132.6 kg/kmol Taking a datum of 273 K: Hf = 1.7 × 132.6(290 − 273) = 3832 kJ/kmol Hf s = 1.7 × 132.6(336 − 273) = 14,200 kJ/kmol 103 λ = 25,900 kJ/kmol q = (25,900 + 14,200 − 3832)/25,900 = 1.4 Thus: The intercept of the q-line on the x-axis is shown from equation 11.46 to be xf /q or:     q xf yq = xq − (equation 11.46) q −1 q −1 xf /q = (0.274/1.4) = 0.196 Thus the q-line is drawn through (xf , xf ) and (0.196, 0) as shown in Figure 11c. As the reflux ratio is given as 3.16, the top operating line may be drawn through (xd , xd ) and (0, xd /4.16). The lower operating line is drawn by joining the intersection of the top operating line and the q-line with the point (xw , xw ). The theoretical plates may be stepped off as shown and 9 theoretical plates are shown. If the plate efficiency is 70 per cent, the number of actual plates = (9/0.7) = 12.85, Thus: 13 plates are required PROBLEM 11.6 A batch fractionation is carried out in a small column which has the separating power of 6 theoretical plates. The mixture consists of benzene and toluene containing 0.60 mole fraction of benzene. A distillate is required, of constant composition, of 0.98 mole fraction benzene, and the operation is discontinued when 83 per cent of the benzene charged has been removed as distillate. Estimate the reflux ratio needed at the start and finish of the distillation, if the relative volatility of benzene to toluene is 2.46. Solution The equilibrium data are calculated from the relative volatility by the equation: yA = αxA 1 + (α − 1)xA (equation 11.16) to give: xA yA 0 0 0.1 0.215 0.2 0.380 0.3 0.513 0.4 0.621 0.5 0.711 0.6 0.787 0.7 0.852 0.8 0.908 0.9 0.956 1.0 1.0 If a constant product is to be obtained from a batch still, the reflux ratio must be constantly increased. Initially S1 kmol of liquor are in the still with a composition xs1 of the MVC and a reflux ratio of R1 is required to give the desired product composition xd . When S2 kmol remain in the still of composition xs2 , the reflux ratio has increased to R2 when the amount of product is D kmol. 104 From an overall mass balance: S1 − S2 = D S1 xs1 − S2 xs2 = Dxd (xs1 − xs2 ) D = S1 (xd − xs2 ) For the MVC: from which: (equation 11.98) In this problem, xs1 = 0.6 and xd = 0.98 and there are 6 theoretical plates in the column. It remains, by using the equilibrium data, to determine values of xs2 for selected reflux ratios. This is done graphically by choosing an intercept on the y-axis, calculating R, drawing in the resulting operating line, and stepping off in the normal way 6 theoretical plates and finding the still composition xs2 . This is shown in Figure 11d for two very different reflux ratios and the procedure is repeated to give the following table. Intercept on y-axis (φ) Reflux ratio (φ = xd /R + 1) xs2 0.45 0.40 0.30 0.20 0.10 0.05 1.18 1.20 2.27 3.90 8.8 18.6 0.725 0.665 0.545 0.46 0.31 0.26 1.0 0.90 4 6 5 xs 2 3 0.80 0.70 30 25 4 0.60 20 5 0.50 0.40 15 6 xs 2 10 0.30 0.20 5 0.10 0 0 Figure 11d. 0.2 0.4 xs 2 0.6 0.8 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 Mole fraction benzene in liquid Equilibrium data for Problem 11.6 105 0 1.0 1.0 Reflux ratio R Mole fraction benezene in vapour 2 1 3 2 xd From the inset plot of xs2 against R in Figure 11d: At the start: xs2 = 0.6 and R = 1.7. At the end: xs2 is calculated using equation 11.98 as follows. If S1 = 100 kmol, kmol of benzene initially = (100 × 0.60) = 60 kmol. kmol of benzene removed = (0.83 × 60) = 49.8 kmol. D = (49.8/0.98) = 50.8 Thus: (0.6 − xs2 ) (0.98 − xs2 ) xs2 = 0.207 and R = 32 50.8 = 100 and: from which: PROBLEM 11.7 A continuous fractionating column is required to separate a mixture containing 0.695 mole fraction n-heptane (C7 H16 ) and 0.305 mole fraction n-octane (C8 H18 ) into products of 99 mole per cent purity. The column is to operate at 101.3 kN/m2 with a vapour velocity of 0.6 m/s. The feed is all liquid at its boiling-point, and this is supplied to the column at 1.25 kg/s. The boiling-point at the top of the column may be taken as 372 K, and the equilibrium data are: mole fraction of heptane in vapour mole fraction of heptane in liquid 0.96 0.91 0.83 0.74 0.65 0.50 0.37 0.24 0.92 0.82 0.69 0.57 0.46 0.32 0.22 0.13 Determine the minimum reflux ratio required. What diameter column would be required if the reflux used were twice the minimum possible? Solution The equilibrium curve is plotted in Figure 11e. As the feed is at its boiling-point, the q-line is vertical and the minimum reflux ratio may be found by joining the point (xd , xd ) with the intersection of the q-line and the equilibrium curve. This line when produced to the y-axis gives an intercept of 0.475. Thus: If 2Rm is used, then: 0.475 = xD /(Rm + 1) and Rm = 1.08 R = 2.16 and Ln /D = 2.16 Taking 100 kmol of feed, as a basis, an overall mass balance and a balance for the n-heptane give: 100 = (D + W ) and: 100 × 0.695 = 0.99D + 0.01W since 99 per cent n-octane is required. 106 1.0 xd 0.90 Mole fraction heptane in vapour 0.80 Vertical q - line for boiling feed 0.70 xf 0.60 0.50 xd Rm + 1 0.40 0.30 0.20 0.10 0 Figure 11e. 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 Mole fraction heptane in liquid 1.0 Geometrical construction for Problem 11.7 Hence: D = 69.9 and W = 30.1 and: Ln = 2.16D = 151 and Vn = Ln + D = 221 The mean molecular mass of the feed = (0.695 × 100) + (0.305 × 114) = 104.3 kg/kmol. Thus: feed rate = (1.25/104.3) = 0.0120 kmol/s The vapour flow at the top of the column = (221/100) × 0.0120 = 0.0265 kmol/s. The vapour density at the top of the column = (1/22.4)(273/372) = 0.0328 kmol/m3 . Hence the volumetric vapour flow = (0.0265/0.0328) = 0.808 m3 /s. If the vapour velocity = 0.6 m/s, the area required = (0.808/0.6) = 1.35 m2 equivalent to a column diameter of [(4 × 1.35)/π]0.5 = 1.31 m. PROBLEM 11.8 The vapour pressures of chlorobenzene and water are: Vapour pressure (kN/m2 ) (mm Hg) Temperatures, (K) Chlorobenzene Water 13.3 100 6.7 50 4.0 30 2.7 20 343.6 324.9 326.9 311.7 315.9 303.1 307.7 295.7 107 A still is operated at 18 kN/m2 and steam is blown continuously into it. Estimate the temperature of the boiling liquid and the composition of the distillate if liquid water is present in the still. Solution For steam distillation, assuming the gas laws to apply, the composition of the vapour produced may be obtained from:     PA mB mA yA PA = = = (equation 11.120) MA MB PB yB (P − PA ) where the subscript A refers to the component being recovered and B to steam, and m is the mass, M is the molecular mass, PA and PB are the partial pressures of A and B and P is the total pressure. If there is no liquid phase present, then from the phase rule there will be two degrees of freedom. Thus both the total pressure and the operating temperature can be fixed independently, and PB = P − PA (which must not exceed the vapour pressure of pure water if no liquid phase is to appear). With a liquid water phase present, there will only be one degree of freedom, and setting the temperature or pressure fixes the system and the water and the other component each exert a partial pressure equal to its vapour pressure at the boiling-point of the mixture. In this case, the distillation temperature will always be less than that of boiling water at the total pressure in question. Consequently, a high-boiling organic material may be steamdistilled at temperatures below 373 K at atmospheric pressure. By using reduced operating pressures, the distillation temperature may be reduced still further, with a consequent economy of steam. A convenient method of calculating the temperature and composition of the vapour, for the case where the liquid water phase is present, is by using Figure 11.47 in Volume 2 where the parameter (P − PB ) is plotted for total pressures of 101, 40 and 9.3 kN/m2 and the vapour pressures of a number of other materials are plotted directly against temperature. The intersection of the two appropriate curves gives the temperature of distillation and the molar ratio of water to organic material is given by (P − PA )/PA . The relevance of the method to this problem is illustrated in Figure 11f where the vapour pressure of chlorobenzene is plotted as a function of temperature. On the same graph (P − PB ) is plotted where P = 18 kN/m2 (130 mm Hg) and PB is the vapour pressure of water at the particular temperature. These curves are seen to intersect at the distillation temperature of 323 K. The composition of the distillate is found by substitution in equation 11.120 since PA = 5.5 kN/m2 (41 mmHg) at 323 K. Hence: PA yA 5.5 = = = 0.44 yB (P − PA ) (18 − 5.5) 108 Chlorobenzene ( P − PB ) P = 130 mmHg PA 100 90 Vapour pressure (mmHg) 80 70 60 50 40 30 20 323K 10 0 280 Figure 11f. 290 300 310 320 330 340 350 Temperature (K) Vapour pressure as a function of temperature, Problem 11.8 PROBLEM 11.9 The following values represent the equilibrium conditions in terms of mole fraction of benzene in benzene–toluene mixtures at their boiling-point: Liquid Vapour 0.521 0.72 0.38 0.60 0.26 0.45 0.15 0.30 If the liquid compositions on four adjacent plates in a column were 0.18, 0.28, 0.41 and 0.57 under conditions of total reflux, determine the plate efficiencies. Solution The equilibrium data are plotted in Figure 11g over the range given and a graphical representation of the plate efficiency is shown in the inset. The efficiency EMl in terms of the liquid compositions is defined by: EMl = (xn+1 − xn ) (xn+1 − xe ) 109 (equation 11.125) Equilibrium curve K Mole fraction benzene in vapour 0.70 0.60 J D 0.50 H 0.40 C E 0.30 F G y E = bd / bc B a xd c d b 0.20 A 0.10 xn xn +1 ideal plate xn +1 actual plate x 0.10 0.20 0.30 0.40 0.50 0.60 Mole fraction benzene in liquid Figure 11g. Graphical construction for Problem 11.9 0 In the inset, the line ab represents an operating line and bc is the enrichment achieved on a theoretical plate. bd is the enrichment achieved on an actual plate so that the efficiency is then the ratio ba/bc. Referring to the data given, at total reflux, the conditions on actual plates in the column are shown as points A, B, C, and D. Considering point A, if equilibrium were achieved on that plate, point E would represent the vapour composition and point F the liquid composition on the next plate. The liquid on the next plate is determined by B however so that the line AGE may be located and the efficiency is given by AG/AE = 0.59 or 59 per cent In an exactly similar way, points H, J, and K are located to give efficiencies of 66 per cent, 74 per cent, and 77 per cent. PROBLEM 11.10 A continuous rectifying column handles a mixture consisting of 40 per cent of benzene by mass and 60 per cent of toluene at the rate of 4 kg/s, and separates it into a product containing 97 per cent of benzene and a liquid containing 98 per cent toluene. The feed is liquid at its boiling-point. (a) Calculate the mass flows of distillate and waste liquor. (b) If a reflux ratio of 3.5 is employed, how many plates are required in the rectifying part of the column? 110 (c) What is the actual number of plates if the plate-efficiency is 60 per cent? Mole fraction of benzene in liquid Mole fraction of benzene in vapour 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 0.22 0.38 0.51 0.63 0.7 0.78 0.85 0.91 0.96 Solution The equilibrium data are plotted in Figure 11h. As the compositions are given as mass per cent, these must first be converted to mole fractions before the McCabe–Thiele method may be used. (40/78) = 0.440 (40/78) + (60/92) Mole fraction of benzene in feed, xf = Similarly: xd = 0.974 and xw = 0.024 As the feed is a liquid at its boiling-point, the q-line is vertical and may be drawn at xf = 0.44. 1.0 1 xD 0.90 2 Mole fraction benzene in vapour 0.80 3 0.70 4 0.60 5 0.50 6 xF 0.40 7 0.30 8 0.20 0.10 0 Figure 11h. 9 10 xw 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 Mole fraction benzene in liquid Graphical construction for Problem 11.10 111 1.0 (a) A mass balance over the column and on the more volatile component in terms of the mass flow rates gives: 4.0 = W ′ + D ′ (4 × 0.4) = 0.02W ′ + 0.97D ′ from which: and: bottoms flowrate, W ′ = 2.4 kg/s top product rate, D ′ = 1.6 kg/s (b) If R = 3.5, the intercept of the top operating line on the y-axis is given by xd /(R + 1) = (0.974/4.5) = 0.216, and thus the operating lines may be drawn as shown in Figure 11h. The plates are stepped off as shown and 10 theoretical plates are required. (c) If the efficiency is 60 per cent, the number of actual plates = (10/0.6) = 16.7 or 17 actual plates PROBLEM 11.11 A distillation column is fed with a mixture of benzene and toluene, in which the mole fraction of benzene is 0.35. The column is to yield a product in which the mole fraction of benzene is 0.95, when working with a reflux ratio of 3.2, and the waste from the column is not to exceed 0.05 mole fraction of benzene. If the plate efficiency is 60 per cent, estimate the number of plates required and the position of the feed point. The relation between the mole fraction of benzene in liquid and in vapour is given by: Mole fraction of benzene in liquid Mole fraction of benzene in vapour 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 0.20 0.38 0.51 0.63 0.71 0.78 0.85 0.91 0.96 Solution The solution to this problem is very similar to that of Problem 11.10 except that the data are presented here in terms of mole fractions. Following a similar approach, the theoretical plates are stepped off and it is seen from Figure 11i that 10 plates are required. Thus (10/0.6) = 16.7 actual plates are required and 17 would be employed. The feed tray lies between ideal trays 5 and 6, and in practice, the eighth actual tray from the top would be used. 112 1.0 0.90 1 2 Mole fraction benzene in vapour 0.80 3 0.70 0.60 4 5 0.50 6 0.40 0.30 7 xF 8 0.20 9 0.10 Figure 11i. xD 10 11 xw 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 0 Mole fraction benzene in liquid Graphical construction for Problem 11.11 1.0 PROBLEM 11.12 The relationship between the mole fraction of carbon disulphide in the liquid and in the vapour during the distillation of a carbon disulphide–carbon tetrachloride mixture is: x y 0 0 0.20 0.445 0.40 0.65 0.60 0.795 0.80 0.91 1.00 1.00 Determine graphically the theoretical number of plates required for the rectifying and stripping portions of the column. The reflux ratio = 3, the slope of the fractionating line = 1.4, the purity of product = 99 per cent, and the concentration of carbon disulphide in the waste liquors = 1 per cent. What is the minimum slope of the rectifying line in this case? Solution The equilibrium data are plotted in Figure 11j. In this problem, no data are provided on the composition or the nature of the feed so that conventional location of the q-line 113 1.0 3 0.90 1 xD = 0.99 4 0.80 Mole fraction CS2 in vapour 2 5 0.70 6 0.60 7 0.50 8 0.40 Slope = Lm /Vm = 1.4 0.30 9 0.20 10 0.10 0 Figure 11j. 11 12 xw = 0.01 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.0 Mole fraction CS2 in liquid Equilibrium data for Problem 11.12 is impossible. The rectifying line may be drawn, however, as both the top composition and the reflux ratio are known. The intercept on the y-axis is given by xd /(R + 1) = (0.99/4) = 0.248. The slope of the lower operating line is given as 1.4. Thus the line may be drawn through the point (xw , xw ) and the number of theoretical plates may be determined, as shown, as 12. The minimum slope of the rectifying line corresponds to an infinite number of theoretical stages. If the slope of the stripping line remains constant, then production of that line to the equilibrium curve enables the rectifying line to be drawn as shown dotted in Figure 11j. The slope of this line may be measured to give Ln /Vn = 0.51 PROBLEM 11.13 A fractionating column is required to distill a liquid containing 25 per cent benzene and 75 per cent toluene by mass, to give a product of 90 per cent benzene. A reflux ratio of 3.5 is to be used, and the feed will enter at its boiling point. If the plates used are 100 per cent efficient, calculate by the Lewis–Sorel method the composition of liquid on the third plate, and estimate the number of plates required using the McCabe–Thiele method. 114 Solution The equilibrium data for this problem are plotted as Figure 11k. Converting mass per cent to mole fraction gives xf = 0.282 and xd = 0.913. There are no data given on the bottom product so that usual mass balances cannot be applied. The equation of the top operating line is: D Ln yn = xn+1 + xd (equation 11.35) Vn Vn 1.0 0.90 xd = 0.913 1 Mole fraction benzene in vapour 0.80 2 0.70 0.60 3 0.50 0.40 4 5 0.30 0.20 xf = 0.282 0.10 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 Mole fraction benzene in liquid Equilibrium data for Problem 11.13 0 Figure 11k. 1.0 Ln /Vn is the slope of the top operating line which passes through the points (xd , xd ) and 0, xd /(R + 1). This line is drawn in Figure 11k and its slope measured or calculated as 0.78. The reflux ratio which is equal to Ln /D is given as 3.5, so that D/Vn may then be found since:   D 1 D × 0.78 = 0.22 = = Vn Ln 3.5 Thus: yn = 0.78n+1 + 0.22xd = 0.78n+1 + 0.20 The composition of the vapour yt leaving the top plate must be the same as the top product xd since all the vapour is condensed. The composition of the liquid on the top plate xt is found from the equilibrium curve since it is in equilibrium with vapour of composition yt = xd = 0.913. Thus: xt = 0.805 115 The composition of the vapour rising to the top plate yt−1 is found from the equation of the operating line. That is: yt−1 = (0.78 × 0.805) + 0.20 = 0.828 xt−1 is in equilibrium with yt−1 and is found to be 0.66 from the equilibrium curve. Then: yt−2 = (0.78 × 0.66) + 0.20 = 0.715 Similarly: xt−2 = 0.50 and: yt−3 = (0.78 × 0.50) + 0.20 = 0.557 and: xt−3 = 0.335 The McCabe–Thiele construction in Figure 11k shows that 5 theoretical plates are required in the rectifying section. PROBLEM 11.14 A 50 mole per cent mixture of benzene and toluene is fractionated in a batch still which has the separating power of 8 theoretical plates. It is proposed to obtain a constant quality product containing 95 mole per cent benzene, and to continue the distillation until the still has a content of 10 mole per cent benzene. What will be the range of reflux ratios used in the process? Show graphically the relation between the required reflux ratio and the amount of distillate removed. Solution If a constant product is to be obtained from a batch still, the reflux ratio must be constantly increased. Initially S1 kmol of liquor is in the still with a composition xs1 of the MVC and a reflux ratio of R1 is required to give the desired product composition xd . When S2 kmol remain in the still of composition xs2 , the amount of product is D kmol and the reflux ratio has increased to R2 . From an overall mass balance: For the MVC: from which: (S1 − S2 ) = D S1 xs1 − S2 xs2 = Dxd D = S1 (xs1 − xs2 ) (xd − xs2 ) (equation 11.98) In this problem, xs1 = 0.5 and xd = 0.95 and there are 8 theoretical plates in the column. It remains, by using the equilibrium data, to determine values of xs2 for selected reflux ratios. This is done graphically by choosing an intercept on the y-axis, calculating R, drawing in the resulting operating line, and stepping off in the usual way 8 theoretical plates and finding the still composition xs2 and hence D. The results of this process are as follows for S1 = 100 kmol. 116 φ = xd /(R + 1) R xs2 D 0.4 0.35 0.30 0.25 0.20 0.15 0.10 1.375 1.71 2.17 2.80 3.75 5.33 8.50 0.48 0.405 0.335 0.265 0.195 0.130 0.090 4.2 17.3 26.8 34.3 40.3 45.1 47.7 The initial and final values of R are most easily determined by plotting R against xs2 as shown in Figure 11l. The initial value of R corresponds to the initial still composition of 0.50 and is seen to be 1.3 and, at the end of the process when xs2 = 0.1, R = 7.0. 50 8.0 40 6.0 30 R D 4.0 20 2.0 0 10 0.10 0.20 0.30 0.40 0.50 0 xs2 Figure 11l. 2.0 4.0 6.0 8.0 10.0 R Reflux ratio data for Problem 11.14 Figure 11l includes a plot of reflux ratio against the quantity of distillate. When R = 7.0, D = 47.1 kmol/100 kmol charged initially. PROBLEM 11.15 The vapour composition on a plate of a distillation column is: mole fraction relative volatility C1 0.025 36.5 C2 0.205 7.4 i − C3 0.210 3.0 n − C3 0.465 2.7 i − C4 0.045 1.3 n − C4 0.050 1.0 What will be the composition of the liquid on the plate of it is in equilibrium with the vapour? 117 Solution In a mixture of A, B, C, D, and so on, if the mole fractions in the liquid are xA , xB , xC , xD , and so on, and in the vapour yA , yB , yC , and yD , then: xA + xB + xC + · · · = 1 xB xC 1 xA + + +··· = xB xB xB xB yA xA = xB yB αAB Thus: But: yB yC 1 yA + + + ··· = yB αAB yB αBB yB αCB xB   yB yA = αAB xB yA xB yB = αAB xA   yA xB yA = αAB αAB xA xB Thus: or: But: and substituting: xA = Thus xB = Similarly: (yA /αAB ) (yA /αAB ) (yB /αBB ) (yA /αAB ) and xC = (yC /αBC ) (yA /αAB ) These relationships may be used to solve this problem and the calculation is best carried out in tabular form as follows. Component C1 C2 i − C3 n − C3 i − C4 n − C4 yi α yi /α xi = (yi /α)/ (yi /α) 0.025 0.205 0.210 0.465 0.045 0.050 36.5 7.4 3.0 2.7 1.3 1.0 0.00068 0.0277 0.070 0.1722 0.0346 0.050 0.002 0.078 0.197 0.485 0.097 0.141 (yi /α) = 0.355 1.000 PROBLEM 11.16 A liquor of 0.30 mole fraction of benzene and the rest toluene is fed to a continuous still to give a top product of 0.90 mole fraction benzene and a bottom product of 0.95 mole fraction toluene. 118 If the reflux ratio is 5.0, how many plates are required: (a) if the feed is saturated vapour? (b) if the feed is liquid at 283 K? Solution In this problem, the q-lines have two widely differing slopes and the effect of the feed condition is to alter the number of theoretical stages as shown in Figure 11m. 1.0 0.90 xD = 0.9 Mole fraction benzene in vapour 0.80 0.70 0.60 0.50 0.40 q -line for cold feed 0.30 xF = 0.3 q -line for saturated vapour 0.20 0.10 xw = 0.05 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 Mole fraction benzene in liquid Figure 11m. Equilibrium data for Problem 11.16 0 q= 1.0 λ + Hf s − Hf λ where λ is the molar latent heat of vaporisation, Hf s is the molar enthalpy of the feed at its boiling-point, and Hf is the molar enthalpy of the feed. For benzene and toluene: and: λ = 30 MJ/kmol specific heat capacity = 1.84 kJ/kg K. The boiling-points of benzene and toluene are 353.3 and 383.8 K respectively. 119 (a) If the feed is a saturated vapour, q = 0. (b) If the feed is a cold liquid at 283 K, the mean molecular mass is: (0.3 × 78) + (0.7 × 92) = 87.8 kg/kmol and the mean boiling-point = (0.3 × 353.3) × (0.7 × 383.8) = 374.7 K. Using a datum of 273 K: Hf s = 1.84 × 87.8(374.7 − 273) = 16,425 kJ/kmol Hf = 1.84 × 87.8(283 − 273) = 1615 kJ/kmol or or 16.43 MJ/kmol 1.615 MJ/kmol q = (30 + 16.43 − 1.615)/30 = 1.49. Thus: From equation 11.46, the slope of the q-line is q/(q − 1). Hence the slope = (1.49/0.49) = 3.05. Thus for (a) and (b) the slope of the q-line is zero and 3.05 respectively, and in Figure 11m these lines are drawn through the point (xf , xf ). By stepping off the ideal stages, the following results are obtained: Theoretical plates Feed Saturated vapour Cold liquid Stripping section Rectifying section Total 4 4 5 3 9 7 Thus a cold feed requires fewer plates than a vapour feed although the capital cost saving is offset by the increased heat load on the reboiler. PROBLEM 11.17 A mixture of alcohol and water containing 0.45 mole fraction of alcohol is to be continuously distilled in a column to give a top product of 0.825 mole fraction alcohol and a liquor at the bottom containing 0.05 mole fraction alcohol. How many theoretical plates are required if the reflux ratio used is 3? Indicate on a diagram what is meant by the Murphree plate efficiency. Solution This example is solved by a simple application of the McCabe–Thiele method and is illustrated in Figure 11n, where it is seen that 10 theoretical plates are required. The Murphree plate efficiency is discussed in the solution to Problem 11.9. 120 1.0 0.90 Mole fraction alcohol in vapour 0.80 0.70 6 5 4 3 2 1x = 0.825 D 7 0.60 8 0.50 0.40 xF = 0.45 9 0.30 0.20 0.10 10 xw = 0.05 0 Figure 11n. 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 Mole fraction alcohol in liquid 1.0 Graphical construction for Problem 11.17 PROBLEM 11.18 It is desired to separate 1 kg/s of an ammonia solution containing 30 per cent NH3 by mass into 99.5 per cent liquid NH3 and a residual weak solution containing 10 per cent NH3 . Assuming the feed to be at its boiling point, a column pressure of 1013 kN/m2 , a plate efficiency of 60 per cent and that an 8 per cent excess over the minimum reflux requirements is used, how many plates must be used in the column and how much heat is removed in the condenser and added in the boiler? Solution Taking a material balance for the whole throughput and for the ammonia gives: D ′ + W ′ = 1.0 and: Thus: (0.995D ′ + 0.1W ′ ) = (1.0 × 0.3) D ′ = 0.22 kg/s and W ′ = 0.78 kg/s The enthalpy-composition chart for this system is shown in Figure 11o. It is assumed that the feed F and the bottom product W are liquids at their boiling-points. Location of the poles N and M Nm for minimum reflux is found by drawing a tie line through F, representing the feed, to cut the line x = 0.995 at Nm . 121 y6 Vapour 2200 y5 2000 1800 (Switch) N Nm y4 y3 y2 1600 y1 A Enthalpy (kJ/kg) 1400 1200 1000 800 600 x6 W x5 400 x4 x3 F 200 L x2 x1 Liquid 0 −200 0 Figure 11o. M Distillate Feed Bottoms 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.0 Mass fraction NH3 (kg/kg) Graphical construction for Problem 11.18 The minimum reflux ratio is given by: Rm = (1952 − 1547) length Nm A = = 0.323 length AL (1547 − 295) Since the actual reflux is 8 per cent above the minimum, NA = 1.08Nm A = (1.08 × 405) = 437 Point N therefore has an ordinate (437 + 1547) = 1984 and abscissa 0.995. Point M is found by drawing NF to cut the line x = 0.10, through W, at M. The number of theoretical plates is found, as on the diagram, to be 5+. The number of plates to be provided = (5/0.6) = 8.33, say 9. The feed is introduced just below the third ideal plate from the top, or just below the fifth actual plate. The heat input at the boiler per unit mass of bottom product is given by: QB /W = 582 − (−209) = 791 kJ/kg (from equation 11.88). The heat input to boiler = (791 × 0.78) = 617 kW. 122 The condenser duty = length NL × D = (1984 − 296) × 0.22 = 372 kW . PROBLEM 11.19 A mixture of 60 mole per cent benzene, 30 per cent of toluene and 10 per cent xylene is handled in a batch still. If the top product is to be 99 per cent benzene, determine: (a) (b) (c) (d) the liquid composition on each plate at total reflux, the composition on the 2nd and 4th plates for R = 1.5, as for (b) but R = 5, as for (c) but R = 8 and for the condition when the mol per cent benzene in the still is 10, (e) as for (d) but with R = 5. The relative volatility of benzene to toluene may be taken as 2.4, and the relative volatility of xylene to toluene as 0.43. Solution Although this problem is one of multicomponent batch distillation, the product remains of constant composition so that normal methods can be used for plate-to-plate calculations at a point value of the varying reflux ratio. (a) At total reflux, for each component the operating line is: yn = xn+1  y = αx Also: (αx) The solution is given in tabular form as: ys = x1 αx1 y1 = x2 αx2 y2 = x3 Similarly x4 x5 B 2.4 0.60 1.44 T 1.0 0.30 0.30 X 0.43 0.10 0.043 0.808 0.168 0.024 1.164 0.168 0.010 0.867 0.125 0.008 2.081 0.125 0.003 0.942 0.057 0.001 0.975 0.025 — 0.989 0.011 — 1.783 1.000 1.342 1.000 2.209 1.000 1.000 1.000 α xs αxs (b) The operating line for the rectifying section is: Ln D xn+1 + xd Vn Vn R = Ln /D and V = Ln + D yn = Thus: yn = xd R xn+1 + R+1 R+1 123 If R = 1.5: for benzene, ynb = 0.6xn+1 + 0.396, toluene, ynt = 0.6xn+1 + 0.004, and xylene, ynx = 0.6xn+1 The liquid composition on each plate is then found from these operating lines. ys x1 αx1 y1 x2 αx2 y2 x3 αx3 y3 x4 B 0.808 0.687 1.649 0.850 0.757 1.817 0.848 0.754 1.810 0.899 0.838 T 0.168 0.273 0.273 0.141 0.228 0.228 0.107 0.171 0.171 0.085 0.135 X 0.024 0.040 0.017 0.009 0.015 0.098 0.045 0.075 0.032 0.016 0.027 1.000 1.000 1.939 1.000 1.000 2.143 1.000 1.000 2.013 1.000 1.000 (c) If R = 5, the operating line equations become: ynb = 0.833xn+1 + 0.165 ynt = 0.833xn+1 + 0.0017 ynx = 0.833xn+1 ys x1 αx1 y1 x2 αx2 y2 x3 αx3 y3 x4 B 0.808 0.772 1.853 0.897 0.879 2.110 0.947 0.939 2.254 0.974 0.971 T 0.168 0.200 0.200 0.097 0.114 0.114 0.051 0.059 0.059 0.025 0.028 X 0.024 0.028 0.012 0.006 0.007 0.003 0.002 0.002 0.001 0.001 0.001 1.000 1.000 2.065 1.000 1.000 2.227 1.000 1.000 2.314 1.000 1.000 (d) When the benzene content is 10 per cent in the still, a mass balance gives the kmol of distillate removed, assuming 100 kmol initially, as: D = 100(0.6 − 0.1)/(0.99 − 0.1) = 56.2 kmol Thus 43.8 kmol remain of which 4.38 are benzene, xb = 0.10 29.42 are toluene, xt = 0.67 10.00 are xylene, xx = 0.23 and: If R = 8, the operating lines become: ynb = 0.889xn+1 + 0.11, xs αxs ys x1 αx1 ynt = 0.889xn+1 + 0.001 and ynx = 0.889xn+1 y1 x2 αx2 y2 x3 αx3 y3 x4 B 0.10 0.24 0.24 0.146 0.350 0.307 0.222 0.533 0.415 0.343 0.823 0.560 0.506 T 0.67 0.67 0.66 0.741 0.741 0.650 0.730 0.730 0.569 0.639 0.639 0.435 0.488 X 0.23 0.10 0.10 0.113 0.049 0.043 0.048 0.021 0.016 0.018 0.008 0.005 0.006 1.01 1.00 1.000 1.140 1.000 1.000 1.284 1.000 1.000 1.470 1.000 1.000 124 (e) Exactly the same procedure is repeated for this part of the question, when the operating lines become: ynb = 0.833xn+1 + 0.165, ynt = 0.833xn+1 + 0.0017 and ynx = 0.833xn+1 PROBLEM 11.20 A continuous still is fed with a mixture of 0.5 mole fraction of the more volatile component, and gives a top product of 0.9 mole fraction of the more volatile component and a bottom product containing 0.10 mole fraction. If the still operates with an Ln /D ratio of 3.5 : 1, calculate by Sorel’s method the composition of the liquid on the third theoretical plate from the top: (a) for benzene–toluene, and (b) for n-heptane–toluene. Solution A series of mass balances as described in other problems enables the flows within the column to be calculated as follows. For a basis of 100 kmol of feed and a reflux ratio of 3.5: D = 50, Ln = 175, Vn = 225 kmol The top operating line equation is then: yn = 0.778xn+1 + 0.20 (a) Use is made of the equilibrium data from other examples involving benzene and toluene. The vapour leaving the top plate has the same composition as the top product, or yt = 0.9. From the equilibrium data, xt = 0.78. Thus: yt−1 = (0.778 × 0.78) + 0.20 = 0.807 and xt−1 , from equilibrium data = 0.640 Similarly: yt−2 = (0.778 × 0.640) + 0.20 = 0.698 and xt−2 = 0.49 yt−3 = (0.778 × 0.490) + 0.20 = 0.581 and xt−3 = 0.36 (b) Vapour pressure data from Perry1 for n-heptane and toluene are plotted in Figure 11p. These data may be used to calculate an mean value of the relative volatility α from: α = PH0 /PT0 1 PERRY, R. H., GREEN, D. W. and MALONEY, J. O.: Perry’s Chemical Engineers’ Handbook, 6th edn. (McGrawHill, New York, 1987). 125 1000 Vapour pressure (mm Hg) n -heptane 100 toluene 10 1.0 2.6 2.8 3.0 3.2 3.4 3.6 3.8 4.0 1/ T × 103 (K−1) Figure 11p. Vapour pressure data for Problem 11.20 T 103 /T PH0 PT0 α 370 333 303 278 2.7 3.0 3.3 3.6 730 200 55 15 510 135 35 9.4 1.43 1.48 1.57 1.60 Mean α = 1.52 As an alternative to drawing the equilibrium curve for the system, point values may be calculated from: y (equation 11.16) x= α − (α − 1)y Thus: yt = 0.9 and xt = 0.9/(1.52 − 0.52 × 0.9) = 0.856 From the same operating line equation: yt−1 = (0.778 × 0.856) + 0.20 = 0.865 Similarly: xt−1 = 0.808, yt−3 = 0.792 yt−2 = 0.829, and xt−2 = 0.761, xt−3 = 0.715 126 PROBLEM 11.21 A mixture of 40 mole per cent benzene with toluene is distilled in a column to give a product of 95 mole per cent benzene and a waste of 5 mole per cent benzene, using a reflux ratio of 4. (a) Calculate by Sorel’s method the composition on the second plate from the top. (b) Using the McCabe and Thiele method, determine the number of plates required and the position of the feed if supplied to the column as liquid at the boiling-point. (c) Find the minimum reflux ratio possible. (d) Find the minimum number of plates. (e) If the feed is passed in at 288 K find the number of plates required using the same reflux ratio. Solution The equilibrium data for benzene and toluene are plotted in Figure 11q. 1.0 0.90 1 1′ xD = 0.95 2 Mole fraction benzene in vapour 0.80 2′ 0.70 3 0.60 4 3′ 0.50 0.40 5 5 xD Rm+1 0.30 6 0.20 0.10 0 Figure 11q. 5′ 7 6 4′ xF = 0.40 7 8 8 6′ xw = 0.05 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 Mole fraction benzene in liquid 1.0 Equilibrium data for Problem 11.21 xf = 0.40, xd = 0.95, xw = 0.05, and R = 4.0 and mass balances may be carried out to calculate the operating line equations. Taking a basis of 100 kmol, then: 100 = D + W 127 (100 × 0.4) = 0.95D + 0.05W and: D = 38.9 and W = 61.1 kmol from which: Ln /D = 4 so that Ln = 155.6 kmol Vn = Ln + D = 194.5 kmol (a) The top operating line equation is: Ln D xn+1 + xd Vn Vn yn = (155.6/194.5)xn+1 + (38.9/194.5) × 0.95 yn = or: (equation 11.35) yn = 0.8xn+1 + 0.19 Vapour yt leaving the top tray has a composition, xd = 0.95 so that xt , the liquid composition on the top tray, is found from the equilibrium curve to be equal to 0.88. yt−1 is found from the operating line equation as: yt−1 = (0.8 × 0.88) + 0.19 = 0.894 xt−1 = 0.775, from the equilibrium curve. yt−2 = (0.8 × 0.775) + 0.19 = 0.810 Thus: xt−2 = 0.645 , from the equilibrium curve. (b) The steps in the McCabe–Thiele determination are shown in Figure 11q where 8 theoretical plates are required with a boiling liquid feed. (c) The minimum reflux ratio corresponds to an infinite number of plates. This condition occurs when the top operating line passes through the intersection of the q-line and the equilibrium curve. This line is seen to give an intersection on the y-axis equal to 0.375. 0.375 = xd /(RM + 1) and Thus: RM = 1.53 (d) The minimum number of plates occurs at total reflux and may be determined by stepping between the equilibrium curve and the diagonal y = x to give 6 theoretical plates as shown. Alternatively, Fenske’s equation may be used. Thus: n + 1 = log = [(xA /xB )d (xB /xA )s ] log αAB log(0.95/0.05)(0.95/0.05) log 2.4 (equation 11.58) and n = 5.7 or 6 plates (e) If a cold feed is introduced, the q-line is no longer vertical. The slope of the line may be calculated as shown in Problem 11.16. In this problem, q is found to be 1.45 and the q-line has a slope of 3.22. This line is shown in Figure 11q and the number of theoretical plates is found to be unchanged at 8. 128 PROBLEM 11.22 Determine the minimum reflux ratio using Fenske’s equation and Colburn’s rigorous method for the following three systems: (a) 0.60 mole fraction C6 , 0.30 mole fraction C7 , and 0.10 mole fraction C8 to give a product of 0.99 mole fraction C6 . (b) Components (c) Components A B C A B C D Mole fraction 0.3 0.3 0.4 0.25 0.25 0.25 0.25 Relative volatility, α 2 1 0.5 2 1 0.5 0.25 xd 1.0 — — 1.0 — — — Solution (a) Under conditions where the relative volatility remains constant, Underwood developed the following equations from which Rm may be calculated: and: αB xf B αC xf C αA xf A + + + ··· = 1 − q αA − θ αB − θ αC − θ αB xdB αC xdC αA xdA + + + · · · = Rm + 1 αA − θ αB − θ αC − θ (equation 11.114) (equation 11.115) where xf A , xf B , xdA , xdB , and so on, are the mole fractions of components A and B, and so on, in the feed and distillate with A the light key and B the heavy key. αA , αB , αC , etc., are the volatilities with respect to the least volatile component. θ is the root of equation 11.114 and lies between the values of αA and αB . Thus θ may be calculated from equation 11.114 and substituted into 11.115 to give Rm . (b) Colburn’s method allows the value of Rm to be calculated from approximate values of the pinch compositions of the key components. This value may then be checked again empirical relationships as shown in Example 11.15 in Volume 2. The method is long and tedious and only the first approximation will be worked here.     xdB xdA 1 Rm = − αAB (equation 11.108) αAB − 1 xnA xnB where xdA and xnA are the top and pinch compositions of the light key component, xdB and xnB are the top and pinch compositions of the heavy key component, and αAB is the volatility of the light key relative to the heavy key component. The difficulty in using this equation is that the values of xnA and xnB are known only for special cases where the pinch coincides with the feed composition. Colburn has suggested that an approximate value for xnA is given by: 129 and: rf (1 + rf )(1 + αxf h ) xnA (approx.) = rf xnA (approx.) = (equation 11.109) xnB (equation 11.110) where rf is the estimated ratio of the key components on the feed plate. For an all liquid feed at its boiling-point, rf is equal to the ratio of the key components in the feed. Otherwise rf is the ratio of the key components in the liquid part of the feed, xf h is the mole fraction of each component in the liquid portion of feed heavier than the heavy key, and α is the volatility of the component relative to the heavy key. (a) Relative volatility data are required and it will be assumed that α6,8 = 5, α6,7 = 2.5, and also that xd7 = 0.01 and that q = 1. Substituting in Underwood’s equation gives:       5 × 0.60 2.5 × 0.30 0.1 × 1 + + =1−q =0 5−θ 2.5 − θ 1−θ from which by trial and error, θ = 3.1.     2.5 × 0.01 5 × 0.99 + = Rm + 1 Then: 5 − 3.1 2.5 − 3.1 and Rm = 1.57 In Colburn’s equation, using C6 and C7 as the light and heavy keys respectively: rf = (0.6/0.3) = 2.0 αxf h = (1/2.5) × 0.1 = 0.04, xnA = 2/(3 × 1.04) = 0.641 αAB = (5.0/2.5) = 2.0, xnB = (0.641/2) = 0.32     1 0.99 2 × 0.01 Rm = − and Rm = 1.48 1 0.641 0.32 (b) The light key = A, the heavy key = B, and αA = 4, αB = 2, αC = 1, and if q is assumed to equal 1, then substitution into Underwood’s equations gives θ = 2.8 and Rm = 2.33. In Colburn’s method, xdB = 0 and rf = (0.3/0.3) = 1.0. αnf h = (1/2) × 0.4 = 0.2 xnA = 1/(2 × 1.2) = 0.417 αAB = (4/2) = 2.0 and Rm = 2.40 (c) In this case, αA = 8, αB = 4, αC = 2, αD = 1. If q = 1, then θ is found from Underwood’s equation to be equal to 5.6 and Rm = 2.33. In Colburn’s method, xdB = 0 and rf = 1.0, the light key = A, the heavy key = B. αxf h = (0.5 × 0.25) + (0.25 × 0.25) = 0.188 130 xnA = 1/(2 × 1.188) = 0.421 αAB = 2 and Rm = 2.38 In all cases, good agreement is shown between the two methods. PROBLEM 11.23 A liquor consisting of phenol and cresols with some xylenols is fractionated to give a top product of 95.3 mole per cent phenol. The compositions of the top product and of the phenol in the bottoms are: phenol o-cresol m-cresol xylenols Compositions (mole per cent) Feed Top Bottom 35 95.3 5.24 15 4.55 — 30 0.15 — 20 — — 100 100 If a reflux ratio of 10 is used, (a) (b) (c) (d) Complete the material balance over the still for a feed of 100 kmol. Calculate the composition on the second plate from the top. Calculate the composition on the second plate from the bottom. Calculate the minimum reflux ratio by Underwood’s equation and by Colburn’s approximation. The heavy key is m-cresol and the light key is phenol. Solution (a) An overall mass balance and a phenol balance gives, on a basis of 100 kmol: 100 = D + W (100 × 0.35) = 0.953D + 0.0524W and: from which: D = 33.0 kmol and W = 67.0 kmol. Balances on the remaining components give the required bottom product composition as: o-cresol: (100 × 0.15) = (0.0455 × 33) + 67xwo and xwo = 0.2017 m-cresol: (100 × 0.30) = (0.0015 × 33) + 67xwm and xwm = 0.4472 xylenols: (100 × 0.20) = 0 + 67xwx , 131 and xwx = 0.2987 Ln /D = 10 (b) and Ln = 330 kmol Vn = Ln + D and Vn = 363 kmol The equation of the top operating line is: yn = Ln D xn+1 + xd = (330/363)xn+1 + (33/330)xd = 0.91xn+1 + 0.091xd Vn Vn The operating lines for each component then become: phenol: ynp = 0.91xn+1 + 0.0867 o-cresol: yno = 0.91xn+1 + 0.0414 m-cresol: ynm = 0.91xn+1 + 0.00014 ynx = 0.91xn+1 xylenols: Mean α-values are taken from the data given in Volume 2, Table 11.2 as: αPO = 1.25, αOO = 1.0, αMO = 0.63, αXO = 0.37 The solution may be set out as a table as follows, using the operating line equations and the equation: y/α x= (y/α) phenol o-cresol m-cresol xylenols y = xd yt /α xt yt−1 yt−1 /α xt−1 0.953 0.0455 0.0015 — 0.762 0.0455 0.0024 — 0.941 0.056 0.003 — 0.943 0.054 0.003 — 0.754 0.054 0.005 — 0.928 0.066 0.006 — (yt /α) = 0.8099 1.000 1.000 0.813 1.000 (c) In the bottom of the column: Lm = Ln + F = 430 kmol and: Vm = Lm − W = 363 kmol W Lm ym = xn+1 − xw = 1.185xm+1 − 0.185xw Vm Vm Hence for each component: phenol: ymp = 1.185xm+1 − 0.0097 o-cresol: ymo = 1.185xm+1 − 0.0373 m-cresol: ymm = 1.185xm+1 − 0.0827 xylenols: ymx = 1.185xm+1 − 0.0553 132 Using these operating lines to calculate x and also y = αx/αx gives the following data: phenol o-cresol m-cresol xylenols xs αxs ys x1 αx1 y1 x2 0.0524 0.2017 0.4472 0.2987 0.066 0.202 0.282 0.111 0.100 0.305 0.427 0.168 0.093 0.289 0.430 0.188 0.116 0.289 0.271 0.070 0.156 0.387 0.363 0.094 0.140 0.358 0.376 0.126 1.000  = 0.661 1.000 1.000  = 0.746 1.000 1.000 (d) Underwood’s equations defined in, Problem 11.22, are used with αp = 3.4, αo = 2.7, αm = 1.7, αx = 1.0 to give:         2.7 × 0.15 1.7 × 0.30 1.0 × 0.20 3.4 × 0.35 + + + = (1 − q) = 0 3.4 − θ 2.7 − θ 1.7 − θ 1−θ 3.4 > θ > 1.7 and θ is found by trial and error to be 2.06.       3.4 × 0.953 2.7 × 0.0455 1.7 × 0.0015 Then: + + = Rm+1 3.4 − 2.06 2.7 − 2.06 1.7 − 2.06 Rm+1 = 1.60 and: Colburn’s equation states that: Rm = xnA = 1 αAB − 1  xdA xnA  − αAB rf   (1 + rf ) 1 + xf h xnB = xnA /rf  xdB xnB  (equation 11.108) (equation 11.109) (equation 11.110) where A and B are the light and heavy keys, that is phenol and m-cresol. rf = (0.35/0.30) = 1.17 αxf h = (0.37/0.63) × 0.20 = 0.117 xnA = 1.17/(2.17 × 1.117) = 0.482 xnB = (0.482/1.17) = 0.413 Thus: αAB = (1.25/0.63) = 1.98     1 0.953 1.98 × 0.0015 Rm = − = 1.95 0.98 0.482 0.413 This is the first approximation by Colburn’s method and provides a good estimate of Rm . 133 PROBLEM 11.24 A continuous fractionating column is to be designed to separate 2.5 kg/s of a mixture of 60 per cent toluene and 40 per cent benzene, so as to give an overhead of 97 per cent benzene and a bottom product containing 98 per cent toluene by mass. A reflux ratio of 3.5 kmol of reflux/kmol of product is to be used and the molar latent heat of benzene and toluene may be taken as 30 MJ/kmol. Calculate: (a) The mass flow of top and bottom products. (b) The number of theoretical plates and position of the feed if the feed is liquid at 295 K, of specific heat capacity 1.84 kJ/kg K. (c) How much steam at 240 kN/m2 is required in the still. (d) What will be the required diameter of the column if it operates at atmospheric pressure and a vapour velocity of 1 m/s. (e) If the vapour velocity is to be 0.75 m/s, based on free area of column, determine the necessary diameter of the column. (f) The minimum possible reflux ratio, and the minimum number of plates for a feed entering at its boiling-point. Solution (a) An overall mass balance and a benzene balance permit the mass of product and waste to be calculated directly: 2.5 = D ′ + W ′ and: from which: 2.5 × 0.4 = 0.97D ′ + 0.02W ′ W ′ = 1.5 kg/s and D ′ = 1.0 kg/s (b) This part of the problem is solved by the McCabe–Thiele method. If the given compositions are converted to mole fractions, then: xf = 0.44, xw = 0.024, xd = 0.974 and a mass balance gives for 100 kmol of feed: 100 = D + W (100 × 0.44) = 0.974D + 0.024W from which D = 43.8 and W = 56.2 kmol/100 kmol of feed If R = 3.5, then: Ln /D = 3.5 and Ln = 153.3 kmol Vn = Ln + D and Vn = 197.1 kmol The intercept on the y-axis = xd /(R + 1) = 0.216. As the feed is a cold liquid, the slope of the q-line must be found. Using the given data and employing the method used in earlier problems, q is found to be 1.41 and the slope = q/(q + 1) = 3.44. This enables the diagram to be completed as shown in 134 Figure 11r from which it is seen that 10 theoretical plates are required with the feed tray as the fifth from the top. 1.0 1 0.90 2 Mole fraction benzene in vapour 0.80 3 0.70 4 0.60 5 slope = 3.44 0.50 0.40 6 xD Rm +1 xF = 0.3 7 0.30 0.20 0.10 0 Figure 11r. xD = 0.974 8 9 10 xw = 0.05 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 Mole fraction benzene in liquid 1.0 Equilibrium data for Problem 11.24 (c) The boil-up rate at the bottom of the column = Vm . Vm = 238.1 kmol/100 kmol feed feed rate = 2.5/(mean mol mass) = (2.5/86.4) = 0.0289 kmol/s Thus: vapour rate = (238.1/100) × 0.0289 = 0.069 kmol/s The heat load = (0.069 × 30) = 2.07 MW or 2070 kW The latent heat of steam at 240 kN/m2 = 2186 kJ/kg (from the Appendix in Volume 2) Thus: steam required = (2070/2186) = 0.95 kg/s (d) At the top of the column the temperature is the boiling-point of essentially pure benzene, that is 353.3 K. Thus: C = (1/22.4)(273/353.3) = 0.034 kmol/m3 and: Vn = 197.1 kmol/100 kmol of feed. Vapour flow = (197.1/100) × 0.0289 = 0.057 kmol/s Thus: volumetric flowrate = (0.057/0.034) = 1.68 m3 /s 135 If the vapour velocity is 1.0 m/s, then: the area = 1.68 m2 and the diameter = 1.46 m If the diameter is calculated from the velocity at the bottom of the column, the result is a diameter of 1.67 m so that, if the velocity is not to exceed 1 m/s in any part of the column, its diameter must be 1.67 m. (e) The velocity based on the free area (tower area − downcomer area) must not exceed 0.75 m/s. The vapour rate in the bottom of the column is 2.17 m3 /s and, for a single-pass crossflow tray, the free area is approximately 88 per cent of the tower area. Thus: At = 2.17/(0.75 × 0.88) = 3.28 m2 and: Dt = 2.05 m PROBLEM 11.25 For a system that obeys Raoult’s law, show that the relative volatility αAB is PA0 /PB0 , where PA0 and PB0 are the vapour pressures of the components A and B at the given temperature. From vapour pressure curves of benzene, toluene, ethyl benzene and of o-, m- and p-xylenes, obtain a plot of the volatilities of each of the materials relative to m-xylene in the range 340–430 K. Solution The volatility of A = PA /xA and the volatility of B = PB /xB and the relative volatility αAB is the ratio of these volatilities, that is: αAB = PA xB xA PB For a system that obeys Raoult’s law, P = xP 0 . Thus: αAB = (xA PA0 )xB P0 = A0 0 xA (xB PB ) PB The vapour pressures of the compounds given in the problem are plotted in Figure 11s and are taken from Perry.1 For convenience, the vapour pressures are plotted on a logarithmic scale against the reciprocal of the temperature (1/K) given a straight line. The relative volatilities may then be calculated in tabular form as follows: 1 PERRY, R. H., GREEN, D. W., and MALONEY, J. O.: Perry’s Chemical Engineer’s Handbook, 6th edn. (McGrawHill, New York, 1987). 136 10 000 B T Vapour pressure (mm Hg) E 1000 O 100 B M+P E O 10 2.40 T 2.60 2.80 3.00 3.20 3.40 3.60 I /T × 103 (K−1) Figure 11s. Vapour pressure data for Problem 11.25 Temperature (K) 103 /T (1/K) (m − x) P 0 (mmHg)/ (p − x) (o − x) E B T 340 360 380 400 415 430 2.94 2.78 2.63 2.50 2.41 2.32 65 141 292 554 855 1315 64 140 290 550 850 1300 56 118 240 450 700 1050 72 155 310 570 860 1350 490 940 1700 2900 4200 6000 180 360 700 1250 1900 2800 Temperature (K) αpm αom αEm αBm αT m 340 360 380 400 415 430 0.98 0.99 0.99 0.99 0.99 0.99 0.86 0.84 0.82 0.81 0.82 0.80 1.11 1.10 1.06 1.03 1.01 1.03 7.54 6.67 5.82 5.23 4.91 4.56 2.77 2.55 2.40 2.26 2.22 2.13 137 These data are plotted in Figure 11t. 10.0 9.0 8.0 Relative volatility, a 7.0 6.0 5.0 aBM 4.0 3.0 aTM 2.0 aEM aPM aOM 1.0 0 340 Figure 11t. 360 380 400 Temperature (K) 420 440 Relative volatility data for Problem 11.25 PROBLEM 11.26 A still contains a liquor composition of o-xylene 10 per cent, m-xylene 65 per cent, pxylene 17 per cent, benzene 4 per cent and ethyl benzene 4 per cent. How many plates are required at total reflux to give a product of 80 per cent m-xylene, and 14 per cent p-xylene? The data are given as mass per cent. Solution Fenske’s equation may be used to find the number of plates at total reflux. log[(xA /xB )d (xB /xA )s ] (equation 11.58) Thus: n+1= log αAB In multicomponent distillation, A and B are the light and heavy key components respectively. In this problem, the only data given for both top and bottom products are for mand p-xylene and these will be used with the mean relative volatility calculated in the previous problem. Thus: xA = 0.8, xB = 0.14, xBs = 0.17, xAs = 0.65 αAB = (1/0.99) = 1.0101 Thus: n + 1 = log[(0.8/0.14)(0.17/0.65)]/log 1.0101 and 138 n = 39 plates PROBLEM 11.27 The vapour pressures of n-pentane and of n-hexane are: Pressure (kN/m2 ) (mm Hg) Temperature (K) C5 H12 C6 H14 1.3 10 2.6 20 5.3 40 8.0 60 13.3 100 26.6 200 53.2 400 101.3 760 223.1 248.2 233.0 259.1 244.0 270.9 257.0 278.6 260.6 289.0 275.1 304.8 291.7 322.8 309.3 341.9 The equilibrium data at atmospheric pressure are: x y 0.1 0.21 0.2 0.41 0.3 0.54 0.4 0.66 0.5 0.745 0.6 0.82 0.7 0.875 0.8 0.925 0.9 0.975 (a) Determine the relative volatility of pentane to hexane at 273, 293 and 313 K. (b) A mixture containing 0.52 mole fraction pentane is to be distilled continuously to give a top product of 0.95 mole fraction pentane and a bottom of 0.1 mole fraction pentane. Determine the minimum number of plates, that is the number of plates at total reflux, by the graphical McCabe–Thiele method, and analytically by using the relative volatility method. (c) Using the conditions in (b), determine the liquid composition on the second plate from the top by Lewis’s method, if a reflux ratio of 2 is used. (d) Using the conditions in (b), determine by the McCabe–Thiele method the total number of plates required, and the position of the feed. It may be assumed that the feed is all liquid at its boiling-point. Solution The vapour pressure data and the equilibrium data are plotted in Figures 11u and 11v. αP H = Pp0 /PH0 (a) Using: The following data are obtained: Temperature (K) Pp0 PH0 αP H 273 293 313 24 55 116 6.0 16.0 36.5 4.0 3.44 3.18 Mean = 3.54 (b) The McCabe–Thiele construction is shown in Figure 11v where it is seen that 4 theoretical plates are required at total reflux. Using Fenske’s equation at total reflux: n + 1 = log[(0.95/0.05)(0.90/0.10)]/log 3.54 and 139 n = 3.07 pentane 100.0 Vapour pressure (mm Hg) n -hexane 10.0 1.0 210 Figure 11u. 230 250 270 290 310 Temperature (K) 330 350 Vapour pressure data for Problem 11.27 1.0 0.90 1 1 Mole fraction pentane in vapour 0.80 xD 2 0.70 3 2 0.60 4 0.50 xF 0.40 5 3 0.30 0.20 6 4 0.10 xw 0 Figure 11v. 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 Mole fraction pentane in liquid Equilibrium data for Problem 11.27 140 1.0 The discrepancy here is caused by using a mean value of α although α does in fact vary considerably. (c) From a mass balance it is found that for 100 kmol of feed and R = 2: Then: and: D = 49.4, W = 50.6, D Ln yn = xn+1 + xd Vn Vn yn = 0.67xn+1 + 0.317 Ln = 98.8, Vn = 148.2 (equation 11.34) The vapour leaving the top plate has the composition of the distillate, that is: yt = xd = 0.95. The liquid on the top plate is in equilibrium with this vapour and from the equilibrium curve has a composition xt = 0.845. The vapour rising to the top tray yt−1 is found from the operating line: yt−1 = 0.67 × 0.845 + 0.317 = 0.883 xt−1 = from the equilibrium curve = 0.707 yt−2 = (0.67 × 0.707) + 0.317 = 0.790 and: xt−2 = 0.56 (d) From Figure 11v, 6 theoretical plates are required and the feed tray is the third from the top of the column. PROBLEM 11.28 The vapour pressures of n-pentane and n-hexane are as given in Problem 11.27. Assuming that both Raoult’s and Dalton’s Laws are obeyed, (a) Plot the equilibrium curve for a total pressure of 13.3 kN/m2 . (b) Determine the relative volatility of pentane to hexane as a function of liquid composition for a total pressure of 13.3 kN/m2 . (c) Would the error caused by assuming the relative volatility constant at its mean value be considerable? (d) Would it be more advantageous to distil this mixture at a higher pressure? Solution (a) The following equations are used where A is n-pentane and B is n-hexane: xA = P − PB0 PA0 − PB0 yA = PA0 xA /P 141 (equation 11.5) (equation 11.4) At P = 13.3 kN/m2 : Temperature (K) PA0 PB0 xA yA α = PA0 /PB0 260.6 265 270 275 280 285 289 13.3 16.5 21.0 26.0 32.5 40.0 47.0 2.85 3.6 5.0 6.7 8.9 11.0 13.3 1.0 0.752 0.519 0.342 0.186 0.079 0 1.0 0.933 0.819 0.669 0.455 0.238 0 4.67 4.58 4.20 3.88 3.65 3.64 3.53 Mean α = 4.02 These figures are plotted in Figure 11w. (b) The relative volatility is plotted as a function of liquid composition in Figure 11w. 1.0 0.9 0.7 5.0 0.6 0.5 4.0 0.4 0.3 Values at a = 4.02 Equilibrium curve at 13.3 kN/m2 Equilibrium curve at 100 kN/m2 0.2 Relative volatility, a Mole fraction n -pentane in vapour 0.8 3.0 0.1 0 Figure 11w. 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Mole fraction n -pentane in liquid 0.9 1.0 Equilibrium data for Problem 11.28 (c) If α is taken as 4.02, yA may be calculated from: αxA yA = 1 + (α − 1)xA (equation 11.15) Using equation 11.15, a new equilibrium curve may be calculated as follows: xA yA 0 0 0.05 0.174 0.10 0.308 0.20 0.500 142 0.40 0.727 0.60 0.857 0.80 0.941 1.0 1.0 These points are shown in Figure 11w where it may be seen that little error is introduced by the use of this mean value. (d) If a higher pressure is used, the method used in (a) may be repeated. If P = 100 kN/m2 , the temperature range increases to 309–341 K and the new curve is drawn in Figure 11w. Clearly, the higher pressure demands a larger number of plates for the same separation and is not desirable. PROBLEM 11.29 It is desired to separate a binary mixture by simple distillation. If the feed mixture has a composition of 0.5 mole fraction, calculate the fraction which it is necessary to vaporise in order to obtain: (a) a product of composition 0.75 mole fraction, when using a continuous process, and (b) a product whose composition is not less than 0.75 mole fraction at any instant, when using a batch process. If the product of batch distillation is all collected in a single receiver, what is its mean composition? It may be assumed that the equilibrium curve is given by: y = 1.2x + 0.3 for liquid compositions in the range 0.3–0.8. Solution (a) If F = number of kmol of feed of composition xf , L = kmol remaining in still with compositionx, and V = kmol of vapour formed with compositiony, then: F =V +L and F xf = V y + Lx For 1 kmol of feed: xf = V y + Lx and y = L xf − x V V This equation is a straight line of slope −L/V which passes through the point (xf , xf ), so that, if y is known, L/V may be found. This is illustrated in Figure 11x where: −L/V = −5.0 As: F = 1, 1=V +L and: V = 0.167 kmol/kmol of feed or 16.7 per cent is vaporised (b) For a batch process it may be shown that:   m−1  S y−x = y0 − x0 S0 143 (equation 11.29) 1.0 0.9 Equilibrium curve y = 1.2 x + 0.3 0.8 y = 0.75 Mole fraction in vapour 0.7 Slope = −5.0 = −L /V y=x 0.6 0.5 xf = 0.5 0.4 0.3 0.2 0.1 0 Figure 11x. 0.1 0.2 0.3 0.4 0.5 0.6 0.7 Mole fraction in liquid 0.8 0.9 1.0 Graphical construction for Problem 11.29 where S is the number of kmol charged initially = 100 kmol (say), S0 is the number of kmol remaining, x is the initial still composition = 0.5, y is the initial vapour composition = (1.2 × 0.5) + 0.3 = 0.90, y0 is the final vapour composition = 0.75 and x0 is the final liquid composition, is found from: 0.75 = 1.2x0 + 0.3 or x0 = 0.375 and m is the slope of equilibrium curve = 1.2.   (0.90 − 0.50) 100 0.2 Thus: = 1.07 = (0.75 − 0.375) S0 and: Therefore: S0 = 71.3 kmol/100 kmol feed   100 − 71.3 × 100 = 28.7 per cent amount vaporised = 100 The distillate composition may be found from a mass balance as follows: Charge Distillate Residue Total kmol kmol A kmol B 100 28.7 71.3 50 (50 − 26.7) = 23.3 (0.374 × 71.3) = 26.7 50 (50 − 44.6) = 5.4 (71.3 − 26.7) = 44.6 Distillate composition = (23.3/28.7) × 100 = 81.2 per cent. 144 PROBLEM 11.30 A liquor, consisting of phenol and cresols with some xylenol, is separated in a plate column. Given the following data complete the material balance: Component Mole per cent Feed C6 H5 OH o-C7 H7 OH m-C7 H7 OH C8 H9 OH Total 35 15 30 20 100 Top Bottom 95.3 4.55 0.15 — 5.24 — — — 100 — Calculate: (a) the composition on the second plate from the top, (b) the composition on the second plate from the bottom. A reflux ratio of 4 is used. Solution The mass balance is completed as in Problem 11.24, where it was shown that: xwo = 0.2017, xwm = 0.4472, xwx = 0.2987 (a) For 100 kmol feed, from mass balances with R = 4.0, the following values are obtained: Ln = 132, Vn = 165, Ln = 232, Vn = 165 The equations for the operating lines of each component are obtained from: D Ln xn+1 + xd Vn Vn = 0.8xn+1 + 0.191 yn = as: phenols: ynp o-cresol: yno = 0.8xn+1 + 0.009 (equation 11.35) m-cresol: ynm = 0.8xn+1 + 0.0003 xylenols: ynx = 0.8xn+1 The compositions on each plate may then be found by calculating y from the operating y/α line equations and x from x = to give the following results: (y/α) 145 phenols o-cresol m-cresol xylenols α yt = xd yt /α xt = (yt /α)/0.81 yt−1 yt−1 /α xt−1 1.25 1.0 0.63 0.37 0.953 0.0455 0.0015 — 0.762 0.046 0.002 — 0.940 0.057 0.003 — 0.943 0.055 0.002 — 0.762 0.055 0.087 — 0.843 0.061 0.096 — 1.000  = 0.81 1.000 1.000 0.904 1.000 (b) In a similar way, the following operating lines may be derived for the bottom of the column: phenols: ymp = 1.406xn+1 − 0.0212 o-cresol: ymo = 1.406xn+1 − 0.0819 m-cresol: ymm = 1.406xn+1 − 0.1816 xylenols: ymx = 1.406xn+1 − 0.1213 Thus: phenols o-cresol m-cresol xylenols α xs αxs ys = αxs /0.661 x1 αx1 y1 x2 1.25 1 0.63 0.37 0.0524 0.2017 0.4472 0.2987 0.066 0.202 0.282 0.111 0.100 0.305 0.427 0.168 0.086 0.275 0.433 0.206 0.108 0.275 0.273 0.076 0.148 0.375 0.373 0.104 0.121 0.324 0.394 0.161  = 0.661 1.000 1.000 0.732 1.000 1.000 PROBLEM 11.31 A mixture of 60, 30, and 10 mole per cent benzene, toluene, and xylene respectively is separated by a plate-column to give a top product containing at least 90 mole per cent benzene and a negligible amount of xylene, and a waste containing not more than 60 mole per cent toluene. Using a reflux ratio of 4, and assuming that the feed is boiling liquid, determine the number of plates required in the column, and the approximate position of the feed. The relative volatility of benzene to toluene is 2.4 and of xylene to toluene is 0.45, and it may be assumed that these values are constant throughout the column. Solution Assuming 100 kmol of feed, the mass balance may be completed to give: D = 60, and: xdt = 0.10, W = 40 kmol xwb = 0.15, 146 xwx = 0.25 If R = 4 and the feed is at its boiling-point then: Ln = 240, Vn = 300, Lm = 340, Vm = 300 and the top and bottom operating lines are: yn = 0.8xn+1 + 0.2xd ym = 1.13xn+1 − 0.133xw and: A plate-to-plate calculation may be carried out as follows: In the bottom of the column. α xs αxs ys x1 αx1 y1 x2 αx2 y2 x3 2.4 benzene 0.15 0.360 0.336 0.314 0.754 0.549 0.503 1.207 0.724 0.65 1.0 toluene 0.60 0.600 0.559 0.564 0.564 0.411 0.432 0.432 0.258 0.29 0.45 xylene 0.25 0.113 0.105 0.122 0.055 0.040 0.065 0.029 0.018 0.04 1.073 1.000 1.000 1.373 1.000 1.000 1.668 1.000 1.00 The composition on the third plate from the bottom corresponds most closely to the feed and above this tray the rectifying equations will be used. benzene toluene xylene x3 αx3 y3 x4 αx4 y4 0.657 0.298 0.045 1.577 0.298 0.020 0.832 0.157 0.011 0.815 0.171 0.014 1.956 0.171 0.006 0.917 0.080 0.003 1.000 1.895 1.000 1.000 2.133 1.000 As the vapour leaving the top plate will be totally condensed to give the product, 4 theoretical plates will be required to meet the given specification. PROBLEM 11.32 It is desired to concentrate a mixture of ethyl alcohol and water from 40 mole per cent to 70 mole per cent alcohol. A continuous fractionating column, 1.2 m in diameter with 10 plates is available. It is known that the optimum superficial vapour velocity in the column at atmosphere pressure is 1 m/s, giving an overall plate efficiency of 50 per cent. Assuming that the mixture is fed to the column as a boiling liquid and using a reflux ratio of twice the minimum value possible, determine the location of the feed plate and the rate at which the mixture can be separated. Equilibria data: Mole fraction alcohol in liquid Mole fraction alcohol in vapour 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.89 0.43 0.526 0.577 0.615 0.655 0.70 0.754 0.82 0.89 147 Solution The equilibrium data are plotted in Figure 11y, where the operating line corresponding to the minimum reflux ratio is drawn from the point (xd , xd ) through the intersection of the vertical q-line and the equilibrium curve to give an intercept of 0.505. 1.0 0.90 Mole fraction alcohol in vapour 0.80 0.70 xD 1 0.60 2 3 0.50 0.40 xF 0.30 0.20 0.10 0 Figure 11y. Thus: 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 Mole fraction alcohol in liquid 1.0 Equilibrium data for Problem 11.32 xd /(Rm + 1) = 0.505 and Rm = 0.386 The actual value of R is then (2 × 0.386) = 0.772 so that the top operating line may be constructed as shown. This column contains the equivalent of (10 × 0.5), that is 5 theoretical plates, so that these may be stepped off from the point (xd , xd ) to give the feed plate as the third from the top. The problem as stated gives no bottom-product composition, so that whilst all flowrates in the top of the column may be calculated, no information about the lower half can be derived. In the absence of these data, the feed rate cannot be determined, though the rate of distillate removal may be calculated as follows. Mean molecular mass of top product = (46 × 0.7 + 18 × 0.3) = 37.6 kg/kmol If the top temperature is assumed to be 353 K, then: C = (1/22.4)(273/353) = 0.0345 kmol/s3 148 If the vapour velocity = 1 m/s, the volumetric vapour flow at the top of the column is: (π/4)(1.2)2 × 1 = 1.13 m3 /s. Vn = (1.13 × 0.0345) = 0.039 kmol/s Hence: From the slope of the operating line: Ln /Vn = 0.436 Ln = (0.436 × 0.039) = 0.017 kmol/s Thus: As R = 0.772, then: D = (Ln /0.772) = 0.022 kmol/s 149 or 0.828 kg/s distillate SECTION 2-12 Absorption of Gases PROBLEM 12.1 Tests are made on the absorption of carbon dioxide from a carbon dioxide–air mixture in a solution containing 100 kg/m3 of caustic soda, using a 250 mm diameter tower packed to a height of 3 m with 19 mm Raschig rings. The results obtained at atmospheric pressure were: Gas rate, G′ = 0.34 kg/m2 s. Liquid rate, L′ = 3.94 kg/m2 s. The carbon dioxide in the inlet gas was 315 parts per million and the carbon dioxide in the exit gas was 31 parts per million. What is the value of the overall gas transfer coefficient KG a? Solution At the bottom of the tower: y1 = 315 × 10−6 , and: G′m = (0.34/29) = 0.0117 kmol/m2 s At the top of the tower: y2 = 31 × 10−6 , and: G′ = 0.34 kg/m2 s L′ = 3.94 kg/m2 s x2 = 0 The NaOH, solution contains 100 kg/m3 NaOH. The mean molecular mass of liquid is: (100 × 40) + (900 × 18) = 20.2 kg/kmol 1000 Thus: L′m = (3.94/20.2) = 0.195 kmol/m2 s For dilute gases, y = Y and a mass balance over the tower gives: G′m (y1 − y2 )A = KG aP (y − ye )lm ZA It may be assumed that as the solution of NaOH is fairly concentrated, there will be a negligible vapour pressure of CO2 over the solution, that is all the resistance to transfer lies in the gas phase. 150 Therefore the driving force at the top of the tower = (y2 − 0) = 31 × 10−6 and: at the bottom of the tower = (y1 − 0) = 315 × 10−6 The log mean driving force, (y − ye )lm = Therefore: (315 − 31) × 106 = 122.5 × 10−6 ln(315/31) 0.0117(315 − 31)10−6 = KG a(101.3 × 122.5 × 10−6 × 3) KG a = 8.93 × 10−5 kmol/m3 s (kN/m2 ) from which: PROBLEM 12.2 An acetone–air mixture containing 0.015 mole fraction of acetone has the mole fraction reduced to 1 per cent of this value by countercurrent absorption with water in a packed tower. The gas flowrate G′ is 1 kg/m2 s of air and the water flowrate entering is 1.6 kg/m2 s. For this system, Henry’s law holds and ye = 1.75x, where ye is the mole fraction of acetone in the vapour in equilibrium with a mole fraction x in the liquid. How many overall transfer units are required? Solution See Volume 2, Example 12.3. PROBLEM 12.3 An oil containing 2.55 mole per cent of a hydrocarbon is stripped by running the oil down a column up which live steam is passed, so that 4 kmol of steam are used/100 kmol of oil stripped. Determine the number of theoretical plates required to reduce the hydrocarbon content to 0.05 mole per cent, assuming that the oil is non-volatile. The vapour–liquid relation of the hydrocarbon in the oil is given by ye = 33x, where ye is the mole fraction in the vapour and x the mole fraction in the liquid. The temperature is maintained constant by internal heating, so that steam does not condense in the tower. Solution If the steam does not condense, Lm /Gm = (100/4) = 25. Inlet oil concentration = 2.55 mole per cent, x2 = 0.0255 and X2 = x2 /(1 − x2 ) = 0.0262 Exit oil concentration = 0.05 mol per cent and x1 = 0.0005 151 A mass balance between a plane in the tower, where the concentrations are X and Y , and the bottom of the tower gives: Lm (X − X1 ) = Gm (Y − Y1 ) Y1 = 0 Y = 25X − 25x1 = 25X − 0.0125 Therefore: This is the equation of the operating line and as the equilibrium data are ye = 33x, then: 33X Y = 1+Y 1+X or Y = 33X 1 − 32X Using these data, the equilibrium and lines may be drawn as shown in Figure 12a. The number of theoretical plates is then found from a stepping-off procedure as employed for distillation as 8 plates. 0.9 Equlibrium line Y = 33X 1−32X 0.8 0.7 0.6 Y X2 = 0.0262 Operating line Y = 25X − 0.0125 0.5 0.4 0.3 0.2 0.1 0 0.005 0.010 0.015 0.020 0.025 X Figure 12a. Equilibrium data for Problem 12.3 PROBLEM 12.4 Gas, from a petroleum distillation column, has its concentration of H2 S reduced from 0.03 kmol H2 S/kmol of inert hydrocarbon gas to 1 per cent of this value, by scrubbing with a triethanolamine-water solvent in a countercurrent tower, operating at 300 K and at atmospheric pressure. H2 S is soluble in such a solution and the equilibrium relation may be taken as Y = 2X, where Y is kmol of H2 S kmol inert gas and X is kmol of H2 S/kmol of solvent. 152 The solvent enters the tower free of H2 S and leaves containing 0.013 kmol of H2 S/kmol of solvent. If the flow of inert hydrocarbon gas is 0.015 kmol/m2 s of tower cross-section and the gas-phase resistance controls the process, calculate: (a) the height of the absorber necessary, and (b) the number of transfer units required. The overall coefficient for absorption KG′′ a may be taken as 0.04 kmol/s m3 of tower volume (unit driving force in Y ). Solution The driving force at the top of column, (Y2 − Y2e ) = 0.0003. The driving force at bottom of column, (Y1 − Y1e ) = (0.03 − 0.026) = 0.004. The logarithmic mean driving force = (0.004 − 0.0003)/ ln(0.004/0.0003) = 0.00143. From equation 12.70: Gm (Y1 − Y2 )A = KG aP (Y − Ye )lm AZ G′m (Y1 − Y2 ) = KG′′ a(Y − Ye )lm Z That is: Thus: 0.015(0.03 − 0.0003) = 0.04 × 0.00143Z and: Z = (0.000446/0.0000572) = 7.79 m or 7.8 m The height of transfer unit, HOG = G′m /KG′′ a = (0.015/0.04) = 0.375 m. The number of transfer units, NOG = (7.79/0.375) = 20.8 or 21. PROBLEM 12.5 It is known that the overall liquid transfer coefficient KL a for absorption of SO2 in water in a column is 0.003 kmol/s m3 (kmol/m3 ). Obtain an expression for the overall liquid-film coefficient KL a for absorption of NH3 in water in the same equipment using the same water and gas rates. The diffusivities of SO2 and NH3 in air at 273 K are 0.103 and 0.170 cm2 /s. SO2 dissolves in water, and Henry’s constant H is equal to 50 (kN/m2 )/(kmol/m3 ). All the data are expressed for the same temperature. Solution See Volume 2, Example 12.1. PROBLEM 12.6 A packed tower is used for absorbing sulphur dioxide from air by means of a caustic soda solution containing 20 kg/m3 NaOH. At an air flow of 2 kg/m2 s, corresponding to a Reynolds number of 5160, the friction factor R/ρu2 is found to be 0.020. 153 Calculate the mass transfer coefficient in kg SO2 /s m2 (kN/m2 ) under these conditions if the tower is at atmospheric pressure. At the temperature of absorption, the diffusion coefficient SO2 is 1.16 × 10−5 m2 /s, the viscosity of the gas is 0.018 mN s/m2 and the density of the gas stream is 1.154 kg/m3 . Solution See Volume 2, Example 12.2. PROBLEM 12.7 In an absorption tower, ammonia is absorbed from air at atmospheric pressure by acetic acid. The flowrate of 2 kg/m2 s in a test corresponds to a Reynolds number of 5100 and hence a friction factor R/ρu2 of 0.020. At the temperature of absorption the viscosity of the gas stream is 0.018 mN s/m2 , the density is 1.154 kg/m3 and the diffusion coefficient of ammonia in air is 1.96 × 10−5 m2 /s. Determine the mass transfer coefficient through the gas film in kg/m2 s (kN/m2 ). Solution  From equation 12.25: hd u  PBm P  μ ρD 0.56 = jd jd ≃ R/ρu2 and: Substituting the given data gives:  Therefore: kG =  (μ/ρD)0.56 = 0.88   PBm hd = (0.0199/0.88) = 0.0226 u P hd RT u = G′ /ρ = 1.733 m/s  PBm P  =  0.0226 × 1.733 8.314 × 298  = 1.58 × 10−5 kmol/m2 s (kN/m2 ) and: kG = (1.58 × 10−5 × 17) = 2.70 × 10−4 kg/m2 s (kN/m2 ) PROBLEM 12.8 Acetone is to be recovered from a 5 per cent acetone–air mixture by scrubbing with water in a packed tower using countercurrent flow. The liquid rate is 0.85 kg/m2 s and the gas rate is 0.5 kg/m2 s. 154 The overall absorption coefficient KG a may be taken as 1.5 × 10−4 kmol/[m3 s (kN/m2 ) partial pressure difference] and the gas film resistance controls the process. What height of tower is required tower to remove 98 per cent of the acetone? The equilibrium data for the mixture are: Mole fraction of acetone in gas Mole fraction of acetone in liquid 0.0099 0.0076 0.0196 0.0156 0.0361 0.0306 0.0400 0.0333 Solution At the bottom of the tower: y1 = 0.05 G′ = (0.95 × 0.5) kg/m2 s and G′m = 0.0164 kmol/m2 s At the top of the tower: y2 = (0.02 × 0.05) = 0.001, L′ = 0.85 kg/m2 s and Lm = 0.0472 kmol/m2 s The height and number of overall transfer units are defined as HOG and NOG by:  y2 dy HOG = Gm /KG aP and NOG = y1 y − ye (equations 12.80 and 12.77) Thus: −4 HOG = 0.0164/(1.5 × 10 × 101.3) = 1.08 m The equilibrium data given are represented by a straight line of slope m = 1.20. As shown in Problem 12.12, the equation for NOG may be integrated directly when the equilibrium line is given by ye = mx to give:    1 mG′ y1 mG′ + ′m NOG = ln 1 − ′ m (1 − mGm /Lm ) Lm y2 Lm m(G′m /L′m ) = 1.20(0.0164/0.0472) = 0.417 y1 /y2 = (0.05/0.001) = 50 Thus: NOG =   1 ln[(1 − 0.417)50 + 0.417] = 5.80 1 − 0.417 The packed height = NOG × HOG = (5.80 × 1.08) = 6.27 m PROBLEM 12.9 Ammonia is to be removed from a 10 per cent ammonia–air mixture by countercurrent scrubbing with water in a packed tower at 293 K so that 99 per cent of the ammonia is removed when working at a total pressure of 101.3 kN/m2 . If the gas rate is 0.95 kg/m2 s 155 of tower cross-section and the liquid rate is 0.65 kg/m2 s, what is the necessary height of the tower if the absorption coefficient KG a = 0.001 kmol/m3 s (kN/m2 ) partial pressure difference. The equilibrium data are: Concentration (kmol NH3 /kmol water) Partial pressure NH3 (kN/m2 ) 0.021 0.031 0.042 0.053 0.079 0.106 1.6 2.4 3.3 4.2 6.7 9.3 0.150 15.2 Solution See Volume 2, Example 12.5. PROBLEM 12.10 Sulphur dioxide is recovered from a smelter gas containing 3.5 per cent by volume of SO2 , by scrubbing it with water in a countercurrent absorption tower. The gas is fed into the bottom of the tower, and in the exit gas from the top the SO2 exerts a partial pressure of 1.14 kN/m2 . The water fed to the top of the tower is free from SO2 , and the exit liquor from the base contains 0.001145 kmol SO2 /kmol water. The process takes place at 293 K, at which the vapour pressure of water is 2.3 kN/m2 . The water flow rate is 0.43 kmol/s. If the area of the tower is 1.85 m2 and the overall coefficient of absorption for these conditions KL′′ a is 0.19 kmol SO2 /s m3 (kmol of SO2 /kmol H2 O), what is the height of the column required? The equilibrium data for SO2 and water at 293 K are: kmol SO2 /1000 kmol H2 O kmol SO2 /1000 kmol inert gas 0.056 0.14 0.28 0.42 0.7 1.6 4.3 7.9 0.56 11.6 Solution At the top of the column: PSO2 = 1.14 kN/m2 That is: 1.14 = 101.3y2 and y2 = 0.0113 ≃ Y2 At the bottom of the column:y1 = 0.035, that is Y1 = 0.036 X1 = 0.001145 Lm = 0.43 kmol/s 156 0.84 19.4 1.405 35.3 The quantity of SO2 absorbed = 0.43(0.001145 − 0) NA = 4.94 × 10−4 kmol SO2 /s That is: NA = KL′′ a(Xe − X)lm The log mean driving force in terms of the liquid phase must now be calculated. Values of Xe corresponding to the gas composition Y may be found from the equilibrium data given (but are not plotted here) as: Y2 = 0.0113, When: Y1 = 0.036, Thus: Xe2 = 0.54 × 10−3 Xe1 = 1.41 × 10−3 (Xe1 − X1 ) = (1.41 − 1.145)10−3 = 0.265 × 10−3 kmol SO2 /kmol H2 O (Xe2 − X2 ) = 0.5 × 10−3 kmol SO2 /kmol H2 O Thus: (Xe − X)lm = (0.54 − 0.265)10−3 = 3.86 × 10−4 kmol SO2 /kmol H2 O ln(0.54/0.265) 4.94 × 10−4 = 0.19V × 3.86 × 10−4 , from which the packed volume, V = 6.74 Thus: packed height = (6.74/1.35) = 5.0 m PROBLEM 12.11 Ammonia is removed from a 10 per cent ammonia–air mixture by scrubbing with water in a packed tower, so that 99.9 per cent of the ammonia is removed. What is the required height of tower? The gas enters at 1.2 kg/m2 s, the water rate is 0.94 kg/m2 s and KG a is 0.0008 kmol/s m3 (kN/m2 ). Solution The molecular masses of ammonia and air are 17 and 29 kg/kmol respectively. The data in mass per cent must be converted to mole ratios as the inlet gas concentration is high. Thus: 17y1 and y1 = 0.159 17y1 + 29(1 − y1 )   0.159 = 0.189 Y1 = 1 − 0.159 0.10 = Y2 ≃ y2 = 0.000159 The rates of entering gases are: total = 1.2 kg/m2 s, ammonia = 0.12 kg/m2 s, and air = 1.08 kg/m2 s. Thus: G′m = 0.0372 kmol/m2 s, L′m = (0.94/18) = 0.0522 kmol/m2 s and: X2 = 0 that is ammonia free 157 The equation of the operating line is found from a mass balance between a plane where the compositions are X and Y and the top of the tower as: 0.0372(Y − 0.000159) = 0.0522X Y = (1.4X + 0.000159) or: This equilibrium line is plotted on Figure 12b. 0.20 kmol NH3/kmol air 0.15 Operating line (Problem 12.11) Equilibrium curve 0.10 Operating line (Problem 12.9) 0.05 0 Figure 12b. 0.05 0.10 kmol NH3 /kmol H2O 0.15 Operating lines, Problem 12.11 The integral in the following equation may be obtained graphically from Figure 12c as 40.55 using the following data. G′m Z= kG aP Y 0.20 0.19 0.15 0.10 Yi 0.152 0.138 0.102 0.063  Y1 Y2 (1 + Y )(1 + Yi )dY (Y − Yi ) (Y − Yi ) (1 + Y )(1 + Yi ) 0.048 0.052 0.048 0.037 3.18 1.35 1.27 1.17 158 (1 + Y )(1 + Yi ) (Y − Yi ) 66.3 26.0 26.4 31.6 Y Yi (Y − Yi ) (1 + Y )(1 + Yi ) 0.05 0.04 0.03 0.02 0.01 0.00015 0.028 0.022 0.016 0.011 0.005 0.000 0.022 0.018 0.014 0.009 0.005 0.00015 1.08 1.06 1.05 1.03 1.015 1.00015 (1 + Y )(1 + Yi ) (Y − Yi ) 49.1 58.8 74.7 114.6 203.0 6670.0 To 6670 at Y = 0.00015 (1+Y )(1+Yi ) / (Y−Yi ) 200 150 100 50 Area under curve = 40.55 Y1 = 0.189 Y2 = 0.00015 0 Figure 12c. 0.05 0.10 Y 0.15 0.20 Evaluation of integral, Problem 12.11 KG a is approximately equal to kG a for a very soluble gas so that: Z= (0.0372 × 40.55) = 18.6 m (0.0008 × 101.3) It is interesting to note that if Y = 0.01 rather than 0.00015, the integral has a value of 8.25 and Z is equal to 3.8 m. Thus 14.8 m of packing is required to remove the last traces of ammonia. 159 PROBLEM 12.12 A soluble gas is absorbed from a dilute gas–air mixture by countercurrent scrubbing with a solvent in a packed tower. If the liquid fed to the top of the tower contains no solute, show that the number of transfer units required is given by:    mG′m y1 mGm 1  ln 1 − + N=  mG′m Lm y2 Lm 1− Lm where G′m and Lm are the flowrates of the gas and liquid in kmol/s m2 tower area, and y1 and y2 the mole fractions of the gas at the inlet and outlet of the column. The equilibrium relation between the gas and liquid is represented by a straight line with the equation ye = mx, where ye is the mole fraction in the gas in equilibrium with mole fraction x in the liquid. In a given process, it is desired to recover 90 per cent of the solute by using 50 per cent more liquid than the minimum necessary. If the HTU of the proposed tower is 0.6 m, what height of packing will be required? Solution By definition: NOG =  y1 y2 dy y − ye (equation 12.77) A mass balance between the top and some plane in the tower where the mole fractions are x and y gives: Gm (y − y2 ) = Lm (x − x2 ) If the inlet liquid is solute free, then: x2 = 0 and x = G′m (y − y2 ) L′m If the equilibrium data are represented by: ye = mx then substituting for ye = m(G′m /L′m )(y − y2 ) gives:  y1 dy NOG = ′ mG y2 y − ′ m (y − y2 ) Lm  y1 dy  = ′  mG mG′ y2 y 1 − ′ m + ′ m y2 Lm Lm      mG′m y1 mG′m −1 mG′m ln 1 − ′ = 1− ′ + ′ Lm Lm y2 Lm 160 Operating line y Equilibrium line ye = mx y1 y2 Slope = (L′/G ′)min x2 = O Top of column Figure 12d. x1 x Bottom of column Graphical construction for Problem 12.12 Referring to Figure 12d:  ′       L y1 − y2 y1 − y2 y2 = = =m 1− G′ min x1 y1 /m y1   0.1y1 = 0.9 m =m 1− y1 If 1.5 (L′ /G′ )min is actually employed, L′m /G′m = (1.5 × 0.9) m = 1.35 m Thus: mGm m = = 0.74 Lm 1.35 m y1 /y2 = 10 Therefore: NOG = 1 ln[(0.26 × 10) + 0.74] = 4.64 0.26 HOG = 0.6 m and the height of packing = (0.6 × 4.64) = 2.78 m PROBLEM 12.13 A paraffin hydrocarbon of molecular mass 114 kg/kmol at 373 K, is to be separated from a mixture with a non-volatile organic compound of molecular mass 135 kg/kmol by stripping with steam. The liquor contains 8 per cent of the paraffin by mass and this is to be reduced to 0.08 per cent using an upward flow of steam saturated at 373 K. If three times the minimum amount of steam is used, how many theoretical stages will be required? The vapour pressure of the paraffin at 373 K is 53 kN/m2 and the process takes place at atmospheric pressure. It may be assumed that the system obeys Raoult’s law. 161 Solution If Raoult’s law applies, the partial pressure = x (vapour pressure) PA = xPA0 That is: y = PA /P and hence ye = x(PA0 /P ) = (53/101.3)x = 0.523x 0.523X Ye = 1 + Ye 1+X Thus the equilibrium curve may be obtained as follows. In terms of mole ratios: X X/(1 + X) Ye /(1 + Ye ) Ye 0 0.02 0.04 0.06 0.08 0.10 0.12 0 0.0196 0.0385 0.0566 0.0741 0.0909 0.107 0 0.0103 0.020 0.0296 0.0387 0.0475 0.0560 0 0.0104 0.0204 0.0305 0.0403 0.0499 0.059 This curve is plotted in Figure 12e. As the inlet gas contains 8 per cent by mass of paraffin, then: X2 = (8/114)/(92/135) = 0.103 X1 = 0.00103 and Y1 = 0 and: 0.006 Low concentrations Equilibrium line Y 0.004 0.06 0.002 Operating line 0.05 0.04 0 0.005 X 0.01 Y 0.03 0.02 Operating line 0.01 0 Figure 12e. 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10 0.11 0.12 X Equilibrium data for Problem 12.13 162 The minimum amount is required occurs when the exit streams are in equilibrium, that is X2 = 0.103, when: Ye2 = 0.0513 From an overall mass balance: Lmin (0.103 − 0.00103) = Gmin (0.0513 − 0) and: (L/G)min = 0.503 and (L/G)actual = 0.167 Thus the operating line, passing through the point (0.00103, 0) with a slope = 0.167, may be drawn in Figure 12e and the number of theoretical stages is found to be 4. This problem may also be solved analytically by the use of the absorption factor method. This is illustrated in Problem 12.16. PROBLEM 12.14 Benzene is to be absorbed from coal gas by means of a wash-oil. The inlet gas contains 3 per cent by volume of benzene, and the exit gas should not contain more than 0.02 per cent benzene by volume. The suggested oil circulation rate is 480 kg oil/100 m3 of inlet gas measured at 273 K and 101.3 kN/m2 . The wash oil enters the tower solute-free. If the overall height of a transfer unit based on the gas phase is 1.4 m, determine the minimum height of the tower which is required to carry out the absorption. The equilibrium data are: Benzene in oil (per cent by mass) Equilibrium partial pressure of benzene in gas (kN/m2 ) 0.05 0.01 0.50 1.0 2.0 3.0 0.013 0.033 0.20 0.53 1.33 3.33 Solution At the top and bottom of the tower respectively: y2 = 0.0002, and: y1 = 0.03, x2 = 0 x1 = exit oil composition Taking 100 m3 of inlet gas at 273 K and 101.3 kN/m2 as the basis of calculation, Volume of benzene at inlet = (0.03 × 100) = 3.0 m3 in 97.0 m3 of gas. Volume of benzene at exit = (0.0002 × 97) = 0.0194 m3 . At 273 K and 101.3 kN/m2 , the kilogramme molecular volume = 22.4 m3 . Thus: kmol of gas = (97/22.4) = 4.33 kmol 163 Volume of benzene absorbed = (3.0 − 0.0194) = 2.9806 m3 . Density of benzene at 273 K and 101.3 kN/m2 = (78/22.4) = 3.482 kg/m3 . mass of benzene absorbed = (2.9806 × 3.482) = 10.38 kg Thus: As the oil rate = 490 kg/100 m3 of gas, the mass per cent of benzene at the exit = (10.38 × 100)/490 = 2.12 per cent Y1 = 0.03/(1 − 0.03) = 0.031 Y2 ≃ y2 = 0.0002 Thus the operating line may be plotted as shown in Figure 12f. The equilibrium data converted to the appropriate units, are as follows and are plotted in Figure 12f. Mass per cent benzene Mass fraction Equilibrium partial pressure (kN/m2 ) Ye = P 0 /P 0 0.05 0.10 0.50 1.0 2.0 3.0 0 0.0005 0.001 0.005 0.01 0.02 0.03 0 0.013 0.033 0.20 0.53 1.33 3.33 0 0.00013 0.00033 0.00197 0.00523 0.01313 0.3287 0.03 Bottom of tower Y1 = 0.031 0.02 Y Equilibrium curve 0.01 0 0.005 0.01 0.015 Mass fraction benzene in oil Figure 12f. Operating line for Problem 12.14 164 0.02 X 0.025 NOG =  Y2 Y1 dY Y − Ye The value of this integral may be evaluated from the operating and equilibrium line by graphical means for values of Y between 0.0002 and 0.031. Y Ye (Y − Ye ) 1/(Y − Ye ) 0.0002 0.0015 0.003 0.005 0.0075 0.010 0.015 0.020 0.025 0.030 0 0.00033 0.007 0.0012 0.0021 0.0031 0.0055 0.008 0.0106 0.0137 0.002 0.00117 0.0023 0.0038 0.0054 0.0069 0.0095 0.012 0.0144 0.0163 5000 855 435 263 185 145 105 83 69 61 From Figure 12g, the area under the curve, NOG = 8.27 Thus: height of column = NOG × HOG = (8.27 × 1.4) = 11.6 m 900 1 = 5000 at Y = 0.0002 Y − Ye 800 700 1 Y − Ye 600 Area under curve = NOG = 8.27 500 400 300 200 100 0 0.01 0.02 Y Figure 12g. Evaluation of integral in Problem 12.14 165 0.03 PROBLEM 12.15 Ammonia is to be recovered from a 5 per cent by volume ammonia–air mixture by scrubbing with water in a packed tower. The gas rate is 1.25 m3 /m2 s measured at 273 K and 101.3 kN/m2 and the liquid rate is 1.95 kg/m2 s. The temperature of the inlet gas is 298 K and the temperature of the inlet water 293 K. The mass transfer coefficient is KG a = 0.113 kmol/m3 s (mole ratio difference) and the total pressure is 101.3 kN/m2 . What is the required height of the tower to remove 95 per cent of the ammonia. The equilibrium data and the heats of solutions are: Mole fraction in liquid Integral heat of solution (kJ/kmol of solution) Equilibrium partial pressures (kN/m2 ) at 293 K at 298 K at 303 K 0.005 181 0.01 363 0.4 0.48 0.61 0.015 544 0.77 0.97 1.28 0.02 723 1.16 1.43 1.83 0.03 1084 1.55 1.92 2.47 2.33 2.93 3.86 Adiabatic conditions may be assumed and heat transfer between phases neglected. Solution The data provided are presented in Figure 12h. The entering gas rate = 12.5 m3 /m2 s at 273 K and 101.3 kN/m2 . Density at 273 K and 101.3 kN/m2 = (1/22.4) = 0.0446 kmol/m3 . At the bottom of the tower, y1 = 0.05. Thus: Y1 = (0.05/0.95) = 0.0526 G′m = (0.95 × 1.25 × 0.0446) = 0.053 kmol/m2 s Y2 = (0.05 × 0.0526) = 0.00263 L′m = (1.95/18) = 0.108 kmol/m2 s and X2 = 0 An overall mass balance gives: 0.108(X1 − 0) = 0.053(0.0526 − 0.00263) and Xi = 0.0245 In this problem, the temperature varies throughout the column and the tower will be divided into increments so that by heat and mass balances the terminal conditions over each section may be found. Knowing the compositions and the temperature, the data given may be used in conjunction with the mass coefficient to calculate the height of the chosen increment. Adiabatic conditions will be assumed and, as the sensible heat change of the gas is small, the heat of solution will be used only to raise the temperature of the liquid. The gas temperature will therefore remain constant at 295 K. 166 303 K 3.0 298 K 2.5 1200 1100 2.0 1000 900 800 1.5 700 600 1.0 500 400 300 0.5 Heat of solution (kJ/kmol of solution) Equilibrium partial pressure (kN/m2) 293 K 200 100 0 0 Figure 12h. 0.01 0.02 0.03 Mole fraction in liquid x Equilibrium data, Problem 12.15 Considering conditions at the top of the tower: X2 = 0, T1 = 293 K, Tg = 295 K, Y2 = 0.00263 Choosing an increment such that the exit liquor stream and inlet gas streams have compositions X = 0.005 and Y , a mass balance taking 1 m2 as a basis gives: Lm (X − 0) = Gm (Y − Y2 ) That is: (0.108 × 0.005) = 0.053(Y − 0.00263) and Y = 0.0128 NH3 absorbed in the section = 0.108 × 0.005 = 5.4 × 10−4 kmol/s. Heat of solution when X = 0.005 = 181 kJ/kmol of solution. Thus: and: heat liberated = (181 × 0.005) = 19.55 kW temperature rise = 19.55/(0.108 × 18 × 4.18) = 2.4 deg K and the liquid exit temperature = 295.4 K At 295.4 K when X = 0.005, Pe = 0.44 kN/m2 , and Ye = 0.00434 167 At the top of the section, Y − Ye = (0.00263 − 0) = 0.00263. At the bottom of the section, Y − Ye = (0.0128 − 0.00434) = 0.00846. Thus: and: (Y − Ye )lm = (0.00846 − 0.00263)/ ln(0.00846/0.00263) = 0.00499 5.4 × 10−4 = 0.113 × 1 × H × 0.00499 (since A = 1 m2 ) and H = 0.958 m. In a similar way, further increments may be taken and the heights of each found. A summary of these calculations is as follows. Increment 1 2 3 4 5 Increment 1 2 3 4 5 Inlet Outlet X Y X Y 0 0.005 0.010 0.015 0.020 0.00263 0.0128 0.023 0.0332 0.0434 0.005 0.010 0.015 0.020 0.0245 0.0128 0.023 0.0332 0.0434 0.0526 Outlet liquid temperature (K) Outlet Pe (kN/m2 ) (Y − Ye ) top 295.4 297.8 300.2 302.6 304.8 0.44 0.95 1.62 2.40 3.30 0.00263 0.00846 0.0136 0.0172 0.0197 (Y − Ye ) bottom (Y − Ye )lm Quantity absorbed (kmol/s) Height of section (m) 0.00846 0.0136 0.0172 0.0197 0.020 0.00499 0.01083 0.0153 0.0185 0.0198 5.4 × 10−4 5.4 × 10−4 5.4 × 10−4 5.4 × 10−4 5.4 × 10−4 0.958 0.441 0.328 0.258 0.241 = 2.23 m Thus the required height of packing = 2.23 m. PROBLEM 12.16 A thirty-plate bubble-cap column is to be used to remove n-pentane from a solvent oil by means of steam stripping. The inlet oil contains 6 kmol of n-pentane/100 kmol of pure oil and it is desired to reduce the solute content of 0.1 kmol/100 kmol of solvent. Assuming isothermal operation and an overall plate efficiency of 30 per cent, what is the specific steam consumption, that is kmol of steam required/kmol of solvent oil treated, and the ratio of the specific and minimum steam consumptions. How many plates would be required if this ratio is 2.0? The equilibrium relation for the system may be taken as Ye = 3.0X, where Ye and X are expressed in mole ratios of pentane in the gas and liquid phases respectively. 168 Solution See Volume 2, Example 12.6. PROBLEM 12.17 A mixture of ammonia and air is scrubbed in a plate column with fresh water. If the ammonia concentration is reduced from 5 per cent to 0.01 per cent, and the water and air rates are 0.65 and 0.40 kg/m2 s, respectively, how many theoretical plates are required? The equilibrium relationship may be written as Y = X, where X is the mole ratio in the liquid phase. Solution Assuming that the compositions are given as volume per cent, then: 0.05 = 0.0526. At the bottom of the tower: y1 = 0.05 and Y1 = (1 − 0.05) Operating line Y1 = 0.0523 0.05 Equilibrium line Y=X 0.04 0.03 0.002 Y 0.02 Y=X 1 Y 0.001 4 0.01 5 2 Y2 = 0.0001 4 0 3 6 0.001 X 0.01 0.02 0.03 X Figure 12i. 0 Graphical construction for Problem 12.17 169 0.04 0.05 At the top of the tower: y2 = 0.0001 = Y2 . L′m = (0.65/18) = 0.036 kmol/m2 s G′m = (0.40/29) = 0.0138 kmol/m2 s A mass balance gives the equation of the operating line as: 0.0138(Y − 0.0001) = 0.036(X − 0) or: Y = 2.61 + 0.0001 The operating line and equilibrium line are then drawn in and from Figure 12i, 6 theoretical stages are required. 170 SECTION 2-13 Liquid–Liquid Extraction PROBLEM 13.1 Tests are made on the extraction of acetic acid from a dilute aqueous solution by means of a ketone in a small spray tower of diameter 46 mm and effective height of 1090 mm with the aqueous phase run into the top of the tower. The ketone enters free from acid at the rate of 0.0014 m3 /s m2 , and leaves with an acid concentration of 0.38 kmol/m3 . The concentration in the aqueous phase falls from 1.19 to 0.82 kmol/m3 . Calculate the overall extraction coefficient based on the concentrations in the ketone phase, and the height of the corresponding overall transfer unit. The equilibrium conditions are expressed by: (Concentration of acid in ketone phase) = 0.548 (Concentration of acid in aqueous phase). Solution The solvent flowrate, L′E = 0.0014 m3 /s m2 and the increase in concentration of the extract stream is given by: (CE2 − CE1 ) = (0.38 − 0) = 0.38 kmol/m3 At the bottom of the column, CR1 = 0.82 kmol/m3 and the equilibrium concentration is then: CE∗ 1 = (0.548 × 0.82) = 0.449 kmol/m3 , Thus: CE1 = 0 C1 = (CE∗ 1 − CE1 ) = 0.449 kmol/m3 At the top of the column: CR2 = 1.19 kmol/m3 and hence CE∗ 2 = (0.548 × 1.19) = 0.652 kmol/m3 CE2 = 0.38 kmol/m3 Therefore: C2 = (CE∗ 2 − CE2 ) = (0.652 − 0.38) = 0.272 kmol/m3 The logarithmic mean driving force, Clm = (0.449 − 0.272)/ ln(0.449/0.272) = 0.353 kmol/m3 171 The effective height, Z = 1.09 m and in equation 13.30: 0.0014(0.38 − 0) = KE a(0.353 × 1.09) KE a = 0.00138 s−1 from which: The height of an overall transfer unit based on concentrations in the extract phase is given by equation 13.26: HOE = L′E /KE a = (0.0014/0.00138) = 1.02 m PROBLEM 13.2 A laboratory test is carried out into the extraction of acetic acid from dilute aqueous solution, by means of methyl iso-butyl ketone, using a spray tower of 47 mm diameter and 1080 mm high. The aqueous liquor is run into the top of the tower and the ketone enters at the bottom. The ketone enters at the rate of 0.0022 m3 /s m2 of tower cross-section. It contains no acetic acid, and leaves with a concentration of 0.21 kmol/m3 . The aqueous phase flows at the rate of 0.0013 m3 /s m2 of tower cross-section, and enters containing 0.68 kmol acid/m3 . Calculate the overall extraction coefficient based on the driving force in the ketone phase. What is the corresponding value of the overall HTU, based on the ketone phase? Using units of kmol/m3 , the equilibrium relationship under these conditions may be taken as: (Concentration of acid in the ketone phase) = 0.548 (Concentration in the aqueous phase.) Solution The increase in concentration of the extract phase, (CE2 − CE1 ) = 0.21 kmol/m3 and the amount of acid transferred to the ketone phase is given by: L′E (CE2 − CE1 ) = (0.0022 × 0.21) = 0.000462 kmol/m2 s Making a mass balance by means of equation 13.21 gives: 0.000462 = 0.0013(0.68 − CR1 ) and CR1 = 0.325 kmol/m3 At the top of the column: CR2 = 0.68 kmol/m3 and hence CE∗ 2 = (0.548 × 0.68) = 0.373 kmol/m3 CE2 = 0.21 kmol/m3 and hence C2 = (0.373 − 0.21) = 0.163 kmol/m3 172 At the bottom of the column: CR1 = 0.325 kmol/m3 and hence CE∗ 1 = (0.548 × 0.325) = 0.178 kmol/m3 CE1 = 0 and hence C1 = (0.178 − 0) = 0.178 kmol/m3 The logarithmic mean driving force is then: Clm = (0.178 − 0.163)/ ln(0.178/0.163) = 0.170 kmol/m3 The height Z = 1.08 m, and in equation 13.30: 0.0022(0.21 − 0) = KE a(0.170 × 1.08) and KE a = 0.0025/s In equation 13.26: HOE = (0.0022/0.0025) = 0.88 m PROBLEM 13.3 Propionic acid is extracted with water from a dilute solution in benzene, by bubbling the benzene phase into the bottom of a tower to which water is fed at the top. The tower is 1.2 m high and 0.14 m2 in area, the drop volume is 0.12 cm3 , and the velocity of rise is 12 cm/s. From laboratory tests the value of Kw during the formation of drops is 7.6 × 10−5 kmol/s m2 (kmol/m3 ) and for rising drops Kw = 4.2 × 10−5 kmol/s m2 (kmol/m3 ). What is the value of Kw a for the tower in kmol/sm3 (kmol/m3 )? Solution Considering drop formation KW = 6 × 10−5 kmol/s m2 (kmol/m3 ) Droplet volume = 0.12 cm3 or 1.2 × 10−7 m3 . Radius of a drop = [(3 × 1.2 × 10−7 )/4π]0.33 = 3.08 × 10−3 m. Mean area during formation, as in Problem 13.6 = 12πr 2 /5 = 12π(3.08 × 10−3 )2 /5 = 7.14 × 10−5 m2 Mean time of exposure = (3tf /5) s, where tf , the time of formation, may be taken as tf = (volume of one drop)/(volumetric throughput) = (1.2 × 10−7 /Q), where Q m3 /s is the volumetric throughput of the benzene phase. Thus: mean time of exposure = (3 × 1.2 × 10−7 )/5Q = (7.2 × 10−8 )/Q s and mass transferred = (6 × 10−5 × 7.14 × 10−5 × 7.2 × 10−8 )/Q = (3.24 × 10−16 )/Q kmol/(kmol/m3 ) 173 Considering drop rise KW = 4.2 × 10−5 kmol/s m2 (kmol/m3 ) Residence time = (1.2/0.12) = 10 s Thus: and: volume in suspension = 10Q m3 number of drops rising = 10Q/(1.2 × 10−7 ) = (8.3 × 107 )Q Area of one drop = 4π(3.08 × 10−3 )2 = (1.19 × 10−4 ) m2 and interfacial area available = (8.3 × 107 Q × 1.19 × 10−4 ) = 9.88Q × 103 m2 Thus: mass transferred = (4.2 × 10−5 × 10 × 9.88Q × 103 ) = 4.15Q kmol/(kmol/m3 ) Total mass transferred = 4.15Q + (3.24 × 10−16 )/Q kmol/(kmol/m3 ). Total residence time = 10 + (1.2 × 10−7 /Q) s. Volume of column = (1.2 × 0.14) = 0.168 m3 . Therefore: KW a = (4.15Q + 3.24 × 10−16 /Q)/[0.168(10 + 1.2 × 10−7 /Q)] which is approximately equal to 2.47Q kmol/s m3 (kmol/m3 ) Further solution is not possible without information on the volumetric throughput of the benzene phase. PROBLEM 13.4 A 50 per cent solution of solute C in solvent A is extracted with a second solvent B in a countercurrent multiple contact extraction unit. The mass of B is 25 per cent that of the feed solution, and the equilibrium data are: 100% B 100% C 100% A Determine the number of ideal stages required, and the mass and concentration of the first extract if the final raffinate contains 15 per cent of solute C. 174 Solution See Volume 2, Example 13.2. PROBLEM 13.5 A solution of 5 per cent acetaldehyde in toluene is to be extracted with water in a five stage co-current unit. If 25 kg water/100 kg feed is used, what is the mass of acetaldehyde extracted and the final concentration? The equilibrium relation is given by: (kg acetaldehyde/kg water) = 2.20 (kg acetaldehyde/kg toluene) Solution For co-current contact with immiscible solvents where the equilibrium curve is a straight line as in the present case: Xn = [A/(A + Sm)]n Xf (equation 13.6) On the basis of 100 kg feed: mass of solvent in the feed, A = 95 kg, mass of solvent added, S = 25 kg, slope of the equilibrium line, m = 2.20 kg/kg, number of stages, n = 5, and mass ratio of solute in the feed Xf = (5/95) = 0.0527 kg/kg. Thus: mass ratio of solute in raffinate, Xn = [95/(95 + 2.2 × 25)]5 × 0.0527 = 0.00538 kg/kg solvent Thus, the final solution consists of 0.00538 kg acetaldehyde in 1.00538 kg solution and the concentration = (100 × 0.00538)/1.00538 = 0.536 per cent With 95 kg toluene in the raffinate, the mass of acetaldehyde = (0.00538 × 95) = 0.511 kg and the mass of acetaldehyde extracted = (5.0 − 0.511) = 4.489 kg/100 kg feed PROBLEM 13.6 If a drop is formed in an immiscible liquid, show that the average surface available during formation of the drop is 12πr 2 /5, where r is the final radius of the drop, and that the average time of exposure of the interface is 3tf /5, where tf is the time of formation of the drop. 175 Solution If it is assumed that the volumetric input to the drop is constant or, (4πr 3 /3t) = k, r 2 = (3k/4π)2/3 t 2/3 then The surface area at radius r is given by as = 4πr 2 or substituting for r 2 : as = 4π(3k/4π)2/3 t 2/3 The mean area exposed between time 0 and t is then:  t as dt ās = (1/t) 0 = (1/t)4π(3k/4π)2/3 2/3 = 4π(3k/4π) (3t 2/3  t t 2/3 dt 0 /5) = 12πr 2 /5 The area under the curve of as as a function of t is:  t as dt or (12πr 2 t/5) 0 t and the mean time of exposure, t¯ = (1/as ) 0 as dt = (12πr 2 t/5as ) When t = tf , the time of formation, as = 4πr 2 t¯ = (12πr 2 tf /20πr 2 ) = 3tf /5 and: PROBLEM 13.7 In the extraction of acetic acid from an aqueous solution with benzene a packed column of height 1.4 m and cross-sectional area 0.0045 m2 , the concentrations measured at the inlet and the outlet of the column are: acid concentration in the inlet water phase, CW2 = 0.69 kmol/m3 . acid concentration in the outlet water phase, CW1 = 0.684 kmol/m3 . flowrate of benzene phase = 5.6 × 10−6 m3 /s = 1.24 × 10−3 m3 /m2 s. inlet benzene phase concentration, CB1 = 0.0040 kmol/m3 . outlet benzene phase concentration, CB2 = 0.0115 kmol/m3 . Determine the overall transfer coefficient and the height of the transfer unit. Solution The acid transferred to the benzene phase is: 5.6 × 10−6 (0.0115 − 0.0040) = 4.2 × 10−8 kmol/s The equilibrium relationship for this system is: CB∗ = 0.0247CW or CB∗ 1 = (0.0247 × 0.684) = 0.0169 kmol/m3 176 and: CB∗ 2 = (0.0247 × 0.690) = 0.0171 kmol/m3 Thus: driving force at bottom, C1 = (0.0169 − 0.0040) = 0.0129 kmol/m3 and: driving force at top, C2 = (0.0171 − 0.0115) = 0.0056 kmol/m3 Logarithmic mean driving force, Clm = 0.0087 kmol/m3 . Therefore: KB a = (kmol transferred)/(volume of packing × Clm ) = (4.2 × 10−8 )/(1.4 × 0.0045 × 0.0087) = 7.66 × 10−4 kmol/s m3 (kmol/m3 ) and: HOB = (1.24 × 10−3 )/(7.66 × 10−4 ) = 1.618 m PROBLEM 13.8 It is required to design a spray tower for the extraction of benzoic acid from its solution in benzene. Tests have been carried out on the rate of extraction of benzoic acid from a dilute solution in benzene to water, in which the benzene phase was bubbled into the base of a 25 mm diameter column and the water fed to the top of the column. The rate of mass transfer was measured during the formation of the bubbles in the water phase and during the rise of the bubbles up the column. For conditions where the drop volume was 0.12 cm3 and the velocity of rise 12.5 cm/s, the value of Kw for the period of drop formation was 0.000075 kmol/s m2 (kmol/m3 ), and for the period of rise 0.000046 kmol/s m2 (kmol/s m3 ). If these conditions of drop formation and rise are reproduced in a spray tower of 1.8 m in height and 0.04 m2 cross-sectional area, what is the transfer coefficient, Kw a, kmol/s m3 (kmol/m3 ), where a represents the interfacial area in m2 /unit volume of the column? The benzene phase enters at the flowrate of 38 cm3 /s. Solution See Volume 2, Example 13.5. PROBLEM 13.9 It is proposed to reduce the concentration of acetaldehyde in aqueous solution from 50 per cent to 5 per cent by mass, by extraction with solvent S at 293 K. If a countercurrent multiple-contact process is adopted and 0.025 kg/s of the solution is treated with an equal quantity of solvent, determine the number of theoretical stages required and the mass flowrate and concentration of the extract from the first stage. The equilibrium relationship for this system at 293 K is: 177 100% Acetaldehyde 50 50 100% Water Figure 13a. 50 100% Solvent S Equilibrium data for Problem 13.9 Solution The data are replotted in Figure 13b and the point F, representing the feed, is drawn in at 50 per cent acetaldehyde, 50 per cent water. Similarly, Rn , the raffinate from stage n located on the curve corresponding to 5 per cent acetaldehyde. (This solution then contains 2 per cent S and 93 per cent water.) FS is joined and point M located such that FM = MS, since the ratio of feed solution to solvent is unity. Rn M is projected to meet the equilibrium curve at E1 and FE1 and Rn S are projected to meet at P. The tie-line E1 R1 is drawn in and the line R1 P then meets the curve at E2 . The working is continued in this way and it is found that R4 is below Rn and hence four theoretical stages are required. 100% acetaldehyde 50 F 50 E1 R1 M E2 R2 E3 E4 R3 R4 Rn 100% water Figure 13b. 50 100% S Graphical construction for Problem 13.9 178 P From Figure 13b, the composition of the extract from stage 1, E1 , is: 3 per cent water, 32 per cent acetaldehyde, and 65 per cent S Making an overall balance, F + S = Rn + E1 (0.025 + 0.025) = (Rn + E1 ) = 0.050 kg/s Thus: Making an acetaldehyde balance: (0.50F + 0) = (0.05Rn + 0.32E1 ) 0.05(0.050 − E1 ) + 0.32E1 = (0.50 × 0.025) and E1 = 0.037 kg/s PROBLEM 13.10 160 cm3 /s of a solvent S is used to treat 400 cm3 /s of a 10 per cent by mass solution of A in B, in a three-stage countercurrent multiple contact liquid–liquid extraction plant. What is the composition of the final raffinate? Using the same total amount of solvent, evenly distributed between the three stages, what would be the composition of the final raffinate, if the equipment were used in a simple multiple-contact arrangement? Equilibrium data: kg A/kg B kg A/kg S Densities (kg/m3 ) 0.05 0.069 A = 1200, 0.10 0.159 B = 1000, 0.15 0.258 S = 800 Solution See Volume 2, Example 13.1 PROBLEM 13.11 In order to extract acetic acid from a dilute aqueous solution with isopropyl ether, the two immiscible phases are passed countercurrently through a packed column 3 m in length and 75 mm in diameter. It is found that if 0.5 kg/m2 of the pure ether is used to extract 0.25 kg/m2 s of 4.0 per cent acid by mass, then the ether phase leaves the column with a concentration of 1.0 per cent acid by mass. Calculate: (a) the number of overall transfer units, based on the raffinate phase, and (b) the overall extraction coefficient, based on the raffinate phase. The equilibrium relationship is given by: (kg acid/kg isopropyl ether) = 0.3 (kg acid/kg water). 179 Solution See Volume 2, Example 13.5 PROBLEM 13.12 It is proposed to recover material A from an aqueous effluent by washing it with a solvent S and separating the resulting two phases. The light product phase will contain A and the solvent S and the heavy phase will contain A and water. Show that the most economical solvent rate, W (kg/s) is given by: W = [(F 2 ax0 )/mb)]0.5 − F /m where the feedrate is F kg/s water containing x0 kg A/kg water, the value of A in the solvent product phase = £a/kg A, the cost of solvent S = £b/kg S and the equilibrium data are given by: (kg A/kg S)product phase = m(kg A/kg water)water phase where a, b and m are constants Solution The feed consists of F kg/s water containing F x0 kg/s A, and this is mixed with W kg/s of solvent S. The product consists of a heavy phase containing F kg/s water and, say, F x kg/s of A, and a heavy phase containing W kg/s of S and Wy kg/s A, where y is the concentration of A in S, that is kg A/kg S. The equilibrium relation is of the form: y = mx Making a balance in terms of the solute A gives: F x0 = Wy + F x or: F x0 = (W mx + F x) = x(mW + F ) and x = F x0 /(mW + F ) The amount received of A recovered = F (x0 − x) kg/s which has a value of F a(x0 − x)£/s. Substituting for x, the value of A recovered = F ax0 − F 2 ax0 /(mW + F )£/s The cost involved is that of the solvent used, W b £/s. Taking the profit P as the value of A recovered less the cost of solvent, all other costs being equal, then: P = F ax0 [1 − F /(mW + F )] − W b £/s Differentiating: dP /dW = F 2 ax0 m/(mW + F )2 − b Putting the differential equal to zero for maximum profit, then: F 2 ax0 m = b(mW + F )2 and: W = (F 2 ax0 /mb)0.5 − F /m 180 SECTION 2-14 Evaporation PROBLEM 14.1 A single-effect evaporator is used to concentrate 7 kg/s of a solution from 10 to 50 per cent of solids. Steam is available at 205 kN/m2 and evaporation takes place at 13.5 kN/m2 . If the overall heat transfer coefficient is 3 kW/m2 K, calculate the heating surface required and the amount of steam used if the feed to the evaporator is at 294 K and the condensate leaves the heating space at 352.7 K. The specific heat capacity of a 10 per cent solution is 3.76 kJ/kg K, the specific heat capacity of a 50 per cent solution is 3.14 kJ/kg K. Solution From the Steam Tables in Volume 2, assuming the steam is dry and saturated at 205 kN/m2 , the steam temperature = 394 K and the enthalpy = 2530 kJ/kg. At 13.5 kN/m2 water boils at 325 K and in the absence of data as to the boiling-point rise, this will be taken as the temperature of evaporation, assuming an aqueous solution. The total enthalpy of steam at 325 K is 2594 kJ/kg. The feed containing 10 per cent solids has to be heated therefore from 294 to 325 K at which the evaporation takes place. In the feed, the mass of dry solids = (7 × 10)/100 = 0.7 kg/s and for x kg/s of water in the product: (0.7 × 100)/(0.7 + x) = 50 and x = 0.7 kg/s Thus: water to be evaporated = (7.0 − 0.7) − 0.7 = 5.6 kg/s Summarising: Feed Product Solids (kg/s) Liquid (kg/s) Total (kg/s) 0.7 0.7 6.3 0.7 7.0 1.4 5.6 5.6 Evaporation Using a datum of 273 K: Heat entering with feed = (7.0 × 3.76)(294 − 273) = 552.7 kW. Heat leaving with product = (1.4 × 3.14)(325 − 273) = 228.6 kW. 181 Heat leaving with evaporated water = (5.6 × 2594) = 14,526 kW. Thus: heat transferred from steam = (14,526 + 228.6) − 552.7 = 14,202 kW. The condensed steam leaves at 352.7 K at which the enthalpy is: = 4.18(352.7 − 273) = 333.2 kJ/kg Thus the heat transferred from 1 kg steam = (2530 − 333.2) = 2196.8 kJ/kg and the steam required = (14,202/2196.8) = 6.47 kg/s. The temperature driving force will be taken as the difference between the temperature of the condensing steam and that of the evaporating water as the preheating of the solution and subcooling of the condensate represent but a small proportion of the total heat load, that is T = (394 − 325) = 69 deg K. Thus from equation 14.1: A = Q/U T = 14,202/(3 × 69) = 68.6 m2 PROBLEM 14.2 A solution containing 10 per cent of caustic soda is to be concentrated to a 35 per cent solution at the rate of 180,000 kg/day during a year of 300 working days. A suitable single-effect evaporator for this purpose, neglecting the condensing plant, costs £1600 and for a multiple-effect evaporator the cost may be taken as £1600N , where N is the number of effects. Boiler steam may be purchased at £0.2/1000 kg and the vapour produced may be assumed to be 0.85N kg/kg of boiler steam. Assuming that interest on capital, depreciation, and other fixed charges amount to 45 per cent of the capital involved per annum, and that the cost of labour is constant and independent of the number of effects employed, determine the number of effects which, based on the data given, will give the maximum economy. Solution The first step is to prepare a mass balance. Mass of caustic soda in 180,000 kg/day of feed = (180,000 × 10)/100 = 18,000 kg/day and mass of water in feed = (180,000 − 18,000) = 162,000 kg/day. For x kg water in product/day: (18,000 × 100)/(18,000 + x) = 35 and water in product x = 33,430 kg/day. Therefore: evaporation = (162,000 − 33,430) = 128,570 kg/day 182 Summarising: Feed Product Solids (kg/day) Liquid (kg/day) Total (kg/day) 18,000 18,000 162,000 33,430 180,000 51,430 128,570 128,570 Evaporation The evaporation in one year is (128,570 × 300) = 3.86 × 107 kg The boiler steam required = (1/0.85N ) kg/kg of vapour produced (3.86 × 107 )/0.85N = 4.54 × 107 /N kg/year or: Thus the annual cost of steam = (0.2 × 4.54 × 107 /N )/1000 = 9076/N £/year The capital cost of the installation = £1600N and the annual capital charges = (1600N × 45)/100 = £702N/year. The labour cost is independent of the number of effects and hence the total annual cost C is made up of the capital charges plus the cost of steam, or: C = (720N + 9076/N ) £/year Thus: dC/dN = (720 − 9076/N 2 ) In order to minimise the costs: dC/dN = 0 or (9076/N 2 ) = 720 from which N = 3.55 Thus N must be either 3 or 4 effects. When N = 3, C = (720 × 3) + (9076/3) = 5185 £/year When N = 4, C = (720 × 4) + (9076/4) = 5149 £/year and hence 4 effects would be specified. PROBLEM 14.3 Saturated steam leaves an evaporator at atmospheric pressure and is compressed by means of saturated steam at 1135 kN/m2 in a steam jet to a pressure of 135 kN/m2 . If 1 kg of the high pressure steam compresses 1.6 kg of the evaporated atmospheric steam, what is the efficiency of the compressor? Solution See Volume 2, Example 14.3. 183 PROBLEM 14.4 A single effect evaporator operates at 13 kN/m2 . What will be the heating surface necessary to concentrate 1.25 kg/s of 10 per cent caustic soda to 41 per cent, assuming a value of U of 1.25 kW/m2 K, using steam at 390 K? The heating surface is 1.2 m below the liquid level. The boiling-point rise of the solution is 30 deg K, the feed temperature is 291 K, the specific heat capacity of the feed is 4.0 kJ/kg deg K, the specific heat capacity of the product is 3.26 kJ/kg deg K and the density of the boiling liquid is 1390 kg/m3 . Solution Making a mass balance: In 1.25 kg/s feed, mass of caustic soda = (1.25 × 10)/100 = 0.125 kg/s For x kg/s water in the product: (100 × 0.125)/(x + 0.125) = 41.0 from which x = 0.180 kg/s. Thus: Feed Product Solids (kg/s) Liquid (kg/s) Total (kg/s) 0.125 0.125 1.125 0.180 1.250 0.305 0.945 0.945 Evaporation At a pressure of 13 kN/m2 , from the Steam Tables in Volume 2, water boils at 324 K. Thus at the surface of the liquid the temperature will be (324 + 30) = 354 K. The pressure due to the hydrostatic head of liquid = (1.2 × 1390 × 9.81)/1000 = 16.4 kN/m2 and hence the pressure at the heating surface = (16.4 + 13) = 29.4 kN/m2 at which pressure the temperature of saturated steam = 341 K. Thus the temperature at which the liquid boils at the heating surface is 371 K, at which the enthalpy of steam = 2672 kJ/kg. Thus: temperature difference, T = (390 − 371) = 19 deg K The heat load Q is the heat in the vapour plus the enthalpy of the product minus the enthalpy of the feed, or: Q = (0.945 × 2672) + [0.305 × 3.26(371 − 273)] − [1.250 × 4.0(291 − 273)] assuming the product is withdrawn at 371 K. Therefore: Q = (2525 + 97.4 − 90) = 2532.4 kW Thus in equation 14.1: A = 2532.4/(1.25 × 19) = 106.6 m2 184 PROBLEM 14.5 Distilled water is produced from sea water by evaporation in a single-effect evaporator working on the vapour compression system. The vapour produced is compressed by a mechanical compressor of 50 per cent efficiency, and then returned to the calandria of the evaporator. Extra steam, dry and saturated at 650 kN/m2 , is bled into the steam space through a throttling valve. The distilled water is withdrawn as condensate from the steam space. 50 per cent of the sea water is evaporated in the plant. The energy supplied in addition to that necessary to compress the vapour may be assumed to appear as superheat in the vapour. Calculate the quantity of extra steam required in kg/s. The production rate of distillate is 0.125 kg/s, the pressure in the vapour space is 101.3 kN/m2 , the temperature difference from steam to liquor is 8 deg K, the boiling-point rise of sea water is 1.1 deg K and the specific heat capacity of sea water is 4.18 kJ/kg K. The sea water enters the evaporator at 344 K from an external heater. Solution See Volume 2, Example 14.4. PROBLEM 14.6 It is claimed that a jet booster requires 0.06 kg/s of dry and saturated steam at 700 kN/m2 to compress 0.125 kg/s of dry and saturated vapour from 3.5 kN/m to 14.0 kN/m2 . Is this claim reasonable? Solution With the nomenclature used in Volume 2, Example 14.3, H1 = 2765 kJ/kg and, assuming isentropic expansion to 3.5 kN/m2 , from the entropy — enthalpy chart: H2 = 2015 kJ/kg Making an enthalpy balance across the system, noting that the enthalpy of saturated steam at 3.5 kN/m2 is 2540 kJ/kg then: (0.06 × 2765) + (0.125 × 2540) = (0.06 + 0.125)H4 and: H4 = 2612 kJ/kg Assuming isentropic compression from 3.5 kN/m2 to 14.0 kN/m2 , then H3 = 2420 kJ/kg (again using the chart). Using the equation for efficiency given in Example 14.3: η = (0.06 + 0.125)(2612 − 2420)/[0.06(2765 − 2015)] = 0.79 As stated in the solution to Example 14.3, with good design, overall efficiencies may approach 0.75–0.80 and the claim here is therefore reasonable. 185 In Example 14.3 and Problem 14.6 use has been made of entropy — enthalpy diagrams. The change in enthalpy due to isentropic compression or expansion may also be calculated, however, using equations 8.30 and 8.32 in Volume 1. PROBLEM 14.7 A forward-feed double-effect evaporator, having 10 m2 of heating surface in each effect, is used to concentrate 0.4 kg/s of caustic soda solution from 10 to 50 per cent by mass. During a particular run, when the feed is at 328 K, the pressures in the two calandrias are 375 and 180 kN/m2 respectively, and the condenser operates at 15 kN/m2 . For these conditions, calculate: (a) the load on the condenser; (b) the steam economy and (c) the overall heat transfer coefficient in each effect. Would there be any advantages in using backward feed in this case? Heat losses to the surroundings are negligible. Physical properties of caustic soda solutions: Solids (per cent by mass) Boiling-point rise (deg K) Specific heat capacity (kJ/kg K) Heat of dilution (kJ/kg) 10 20 30 50 1.6 6.1 15.0 41.6 3.85 3.72 3.64 3.22 0 2.3 9.3 220 Solution A mass balance may be made as follows: Feed Product Evaporation Solids (kg/s) Liquor (kg/s) Total (kg/s) 0.04 0.04 0.36 0.04 0.40 0.08 — 0.32 0.32 (D1 + D2 ) = 0.32 kg/s Thus: From the given data: at 375 kN/m2 , at 180 kN/m2 , at 15 kN/m2 , T0 = 414.5 K, λ0 = 2141 kJ/kg T1 = 390 K, λ1 = 2211 kJ/kg T2 = 328 K, λ2 = 2370 kJ/kg 186 Making a heat balance around stage 1: D0 λ0 = W Cp1 (T1′ − Tf ) + W hd1 + D1 λ1 (i) where Cp1 is the mean specific heat capacity between T1′ and Tf ; T1′ = T1 + the boilingpoint rise in stage 1; and hd1 is the difference in the heat of dilution between the concentration in the first effect and the feed concentration. Similarly, around stage 2: D1 λ1 + (W − D1 )Cp2 (T1′ − T2′ ) = D2 λ2 + (W − D1 )hd2 (ii) Values of D1 and D2 are now selected such that a balance is obtained in equation (ii) Trying D1 = 0.17 kg/s, D2 = 0.15 kg/s. Thus: concentration of solids in the first effect = 0.04/[0.04 + (0.36 − 0.17)] = 0.174 kg/kg solution at which, the boiling-point rise = 5.0 deg K the specific heat capacity = 3.75 kJ/kg K and the heat of dilution = 1.6 kJ/kg T1′ = (390 + 5.0) = 395 K Thus: T2′ = (328 + 41.6) = 369.6 K Cp2 = (3.75 + 3.22)/2 = 3.49 kJ/kg K The heat of dilution to be provided in the second effect = (220 − 1.6) = 218.4 kJ/kg Thus in equation (ii): (2211 × 0.17) + (0.4 − 0.17)3.49(395 − 369.6) = (0.15 × 2370) + (0.4 − 0.17)218. 396.3 = 405.7 or: which is close enough for the purposes of this calculation, and hence the load on the condenser, D2 = 0.15 kg/s. For the first effect: Cp1 = (3.85 + 3.75)/2 = 3.80 kJ/kg K h1 = (1.6 − 0) = 1.6 kJ/kg and in equation (i): 2141D0 = (0.4 × 3.80)(395 − 328) + (0.4 × 1.6) + (0.17 × 2211) and D0 = 0.23 kg/s and the economy is (0.36/0.23) = 1.57 The overall heat transfer coefficient in the first effect is given by: U1 = D0 λ0 /A1 T1 = (0.23 × 2141)/[10(414.5 − 395)] = 2.53 kW/m2 K 187 and for the second effect: U2 = (0.17 × 2211)/10(390 − 369.6) = 1.84 kW/m2 K With a backward feed arrangement, the concentration of solids in the first effect would be 50 per cent, which gives a boiling-point rise of 41.6 deg K. The vapour passing to the second effect must condense at 390 K in the calandria of the second effect to give a pressure there of 180 kN/m2 . Thus the temperature of the liquor in the first effect would be (390 + 41.6) = 431.6 K, which is higher than the feed steam temperature, 414.5 K, and thus there is no temperature gradient. There is therefore no advantage in using backward feed. PROBLEM 14.8 A 12 per cent glycerol — water mixture produced as a secondary product in a continuous process plant flows from the reactor at 4.5 MN/m2 and at 525 K. Suggest, with preliminary calculations, a method of concentration to 75 per cent glycerol in a plant where no lowpressure steam is available. Solution Making a mass balance on the basis of 1 kg feed: Feed Product Evaporation Glycerol (kg) Water (kg) Total (kg) 0.12 0.12 0.88 0.04 1.00 0.16 — 0.84 0.84 The total evaporation is 0.84 kg water/kg feed. The first possibility is to take account of the fact that the feed is at high pressure which could be reduced to, say, atmospheric pressure and the water removed by flash evaporation. For this to be possible, the heat content of the feed must be at least equal to the latent heat of the water evaporated. Assuming evaporation at 101.3 kN/m2 , that is at 373 K, and a specific heat capacity for a 12 per cent glycerol solution of 4.0 kJ/kg K, then: heat in feed = (1.0 × 4.0)(525 − 375) = 608 kJ heat in water evaporated = (0.84 × 2256) = 1895 kJ and hence only (608/2256) = 0.27 kg water could be evaporated by this means, giving a solution containing (100 × 0.12)/[0.12 + (0.88 − 0.27)] = 16.5 per cent glycerol. It is therefore necessary to provide an additional source of heat. Although low-pressure steam is not available, presumably a high-pressure supply (say 1135 kN/m2 ) exists, and vapour recompression using a steam-jet ejector could be considered. 188 Assuming that the discharge pressure from the ejector, that is in the steam-chest, is 170 kN/m2 at which the latent heat, λ0 = 2216 kN/m2 , then a heat balance across the unit gives: D0 λ0 = W Cp (T1 − Tf ) + D1 λ1 Thus: 2216D0 = (1.0 × 4.0)(373 − 525) + (0.84 × 2256) and D0 = 0.58 kg. Using Figure 14.12b in Volume 2, with a pressure of entrained vapour of 101.3 kN/m2 , a live-steam pressure of 1135 kN/m2 and a discharge pressure of 170 kN/m2 , then 1.5 kg live steam is required/kg of entrained vapour. Thus, if x kg is the amount of entrained vapour, then: (1 + 1.5x) = 0.58 and x = 0.23 kg. The proposal is therefore to condense (0.84 − 0.23) = 0.61 kg of the water evaporated and to entrain 0.23 kg with (1.5 × 0.23) = 0.35 kg of steam at 1135 kN/m2 in an ejector to provide 0.58 kg of steam at 170 kN/m2 which is then fed to the calandria. These values represent only one solution to the problem and variation of the calandria and live-steam pressures may result in even lower requirements of high-pressure steam. PROBLEM 14.9 A forward-feed double-effect standard vertical evaporator with equal heating areas in each effect is fed with 5 kg/s of a liquor of specific heat capacity of 4.18 kJ/kg K, and with no boiling-point rise, so that 50 per cent of the feed liquor is evaporated. The overall heat transfer coefficient in the second effect is 75 per cent of that in the first effect. Steam is fed at 395 K and the boiling-point in the second effect is 373 K. The feed is heated to its boiling point by an external heater in the first effect. It is decided to bleed off 0.25 kg/s of vapour from the vapour line to the second effect for use in another process. If the feed is still heated to the boiling-point of the first effect by external means, what will be the change in the steam consumption of the evaporator unit? For the purposes of calculation, the latent heat of the vapours and of the live steam may be taken as 2230 kJ/kg. Solution The total evaporation, (D1 + D2 ) = (5.0/2) = 2.5 kg/s. In equation 14.7: U1 A1 T1 = U2 A2 T2 Since: Therefore: Also: A1 = A2 and U2 = 0.75U1 T1 = 0.75T2 T = T1 + T2 = (395 − 373) = 22 deg K 189 and solving between these two equations gives: T1 = 9.5 deg K and T2 = 12.5 deg K For steam to the first effect, T0 = 395 K For steam to the second effect, T1 = (395 − 9.5) = 385.5 K and T2 = 373 K. The latent heat, λ is 2230 kJ/kg in each case. Making a heat balance across the first effect then: D0 λ0 = W Cp (T1 − Tf ) + D1 λ1 or: 2230D0 = (5.0 × 4.18)(385.5 − 385.5) + 2230D1 and D0 = D1 . Making a heat balance across the second effect then: D1 λ1 + (W − D1 )Cp (T1 − T2 ) = D2 λ2 or: 2230D1 + (5.0 − D1 )4.18(385.5 − 273) = 2230D2 and: 2178D1 = 2230D2 − 261.3 But: D2 = (2.5 − D1 ) Therefore: D1 = 1.21 kg/s and the steam consumption, D0 = 1.21 kg/s Considering the case where 0.25 kg/s is bled from the steam line to the calandria of the second effect, a heat balance across the first effect gives: D0 = D1 as before. Making a heat balance across the second effect, gives: (D1 − 0.25)λ1 + (W − D1 )Cp (T1 − T2 ) = D2 λ2 Thus: 2230(D1 − 0.25) + (5.0 − D1 )4.18(385.5 − 373) = 2230D2 and: 177.7D1 = 2230D2 + 296.2 Substituting (2.5 − D1 ) for D2 , then: D1 = 1.33 kg/s and the steam consumption, D0 = 1.33 kg/s The change in steam consumption is therefore an increase of 0.12 kg/s. PROBLEM 14.10 A liquor containing 15 per cent solids is concentrated to 55 per cent solids in a doubleeffect evaporator operating at a pressure of 18 kN/m2 in the second effect. No crystals are formed. The feedrate is 2.5 kg/s at a temperature of 375 K with a specific heat capacity of 3.75 kJ/kg K. The boiling-point rise of the concentrated liquor is 6 deg K and the pressure of the steam fed to the first effect is 240 kN/m2 . The overall heat transfer coefficients in 190 the first and second effects are 1.8 and 0.63 kW/m2 K, respectively. If the heat transfer area is to be the same in each effect, what areas should be specified? Solution Making a mass balance based on a flow of feed of 2.5 kg/s, gives: Feed Product Solids (kg/s) Liquor (kg/s) Total (kg/s) 0.375 0.375 2.125 0.307 2.50 0.682 — 1.818 1.818 Evaporation (D1 + D2 ) = 1.818 kg/s Thus: (i) At 18 kN/m2 pressure, T2 = 331 K and T2′ , the temperature of the liquor in the second effect, allowing for the boiling-point rise, = (331 + 6) = 337 K. At 240 kN/m2 pressure, T0 = 399 K, then: T1 + T2 = (399 − 337) = 62 deg K (ii) U1 T1 = U2 T2 From equation 14.8: Substituting U1 = 1.8 and U2 = 0.63 kW/m2 K, and combining with equation (ii), gives: T1 = 16 deg K and T2 = 46 deg K Thus T1 = (399 − 16) = 383 K and hence the feed enters slightly cooler than the temperature of the liquor in the first effect. Hence T1 will be slightly greater and the following values will be assumed: T1 = 17 deg K, T2 = 45 deg K Thus, for steam to 1: T0 = 399 K, λ0 = 2185 kJ/kg For steam to 2: T1 = 382 K, λ1 = 2232 kJ/kg and: T2 = 331 K, λ2 = 2363 kJ/kg Making a heat balance around each effect: (1) D0 λ0 = W Cp (T1 − Tf ) + D1 λ1 or (2) 2185D0 = (2.5 × 3.75)(382 − 375) + 2232D1 D1 λ1 + (W − D1 )Cp (T1 − or T2′ ) (iii) = D2 λ2 2232D1 + (2.5 − D1 )3.75(382 − 337) = 2363D2 191 (iv) Solving equations (i), (iii), and (iv) simultaneously: D0 = 0.924 kg/s, D1 = 0.875 kg/s, and D2 = 0.943 kg/s. The areas are given by: A1 = D0 λ0 /U1 T1 = (0.924 × 2185)/(1.8 × 17) = 66.1 m2 A2 = D1 λ1 /U2 T2 = (0.875 × 2232)/(0.63 × 45) = 68.8 m2 which are sufficiently close to justify the assumed values of T1 and T2 . A total area of 134.9 m2 is required and hence an area of, say, 67.5 m2 would be specified for each effect. PROBLEM 14.11 Liquor containing 5 per cent solids is fed at 340 K to a four-effect evaporator. Forward feed is used to give a product containing 28.5 per cent solids. Do the following figures indicate normal operation? If not, why not? Effect Solids entering (per cent) Temperature in steam chest (K) Temperature of boiling solution (K) 1 5.0 382 369.5 2 6.6 374 364.5 3 9.1 367 359.6 4 13.1 357.5 336.6 Solution Examination of the data indicates one obvious point in that the temperatures in the steam chests in effects 2 and 3 are higher than the temperatures of the boiling solution in the previous effects. The explanation for this is not clear although a steam leak in the previous effect is a possibility. Further calculations may be made as follows, starting with a mass balance on the basis of 1 kg feed. Feed Product Product Product Product from from from from 1 2 3 4 Solid (kg) Liquor (kg) Total (kg) 0.05 0.05 0.05 0.05 0.05 0.950 0.708 0.500 0.332 0.125 1.00 0.758 0.550 0.382 0.175 D1 D2 D3 D4 = (0.950 − 0.708) = 0.242 = (0.708 − 0.500) = 0.208 = (0.500 − 0.332) = 0.168 = (0.332 − 0.125) = 0.207 and the total evaporation = (0.242 + 0.208 + 0.168 + 0.207) = 0.825 kg. The steam fed to the plant is obtained by a heat balance across stage 1, given: D0 λ0 = W Cp (T1 − Tf ) + D1 λ1 192 kg kg kg kg Taking Cp as 4.18 kJ/kg K and λ0 and λ1 as 2300 kJ/kg in all effects, 2300D0 = (1.0 × 4.18)(369.5 − 340) + (2300 × 0.242) and D0 = 0.296 kg The overall coefficient in each effect assuming equal areas, A m2 , is: U1 = D0 λ0 /AT1 = (0.296 × 2300)/(12.5A) = (54.5/A) kW/m2 K U2 = D1 λ1 /AT2 = (0.242 × 2300)/(9.5A) = (58.6/A) kW/m2 K U3 = D2 λ2 /AT3 = (0.208 × 2300)/(6.6A) = (72.5/A) kW/m2 K U4 = D3 λ3 /AT4 = (0.168 × 2300)/(20.9A) = (18.5/A) kW/m2 K These results are surprising in that a reduction in U is normally obtained with a decrease in boiling temperature. On this basis U3 is high, which may indicate a change in boiling mechanism although T3 is reasonable. Even more important is the very low value of U in effect 4. This must surely indicate that part of the area is inoperative, possibly due to the deposition of crystals from the highly concentrated liquor. PROBLEM 14.12 1.25 kg/s of a solution is concentrated from 10 to 50 per cent solids in a triple-effect evaporator using steam at 393 K, and a vacuum such that the boiling point in the last effect is 325 K. If the feed is initially at 297 K and backward feed is used, what is the steam consumption, the temperature distribution in the system and the heat transfer area in each effect, each effect being identical? For the purpose of calculation, it may be assumed that the specific heat capacity is 4.18 kJ/kg K, that there is no boiling point rise, and that the latent heat of vaporisation is constant at 2330 kJ/kg over the temperature range in the system. The overall heat transfer coefficients may be taken as 2.5, 2.0 and 1.6 kW/m2 K in the first, second and third effects, respectively. Solution Making a mass balance: Feed Product Evaporation Solid (kg/s) Liquor (kg/s) Total (kg/s) 0.125 0.125 1.125 0.125 1.250 0.250 1.0 1.0 — D1 + D2 + D3 = 1.0 kg/s Thus:  From equation 14.8: T = (T0 − T3 ) = (393 − 325) = 68 deg K 2.5T1 = 2.0T2 = 1.6T3 193 (i) (ii) (iii) and from equations (ii) and (iii): T1 = 18 deg K, T2 = 22 deg K, and T3 = 28 deg K Modifying the figures to take account of the effect of the feed temperature, it will be assumed that: T1 = 19 deg K, T2 = 24 deg K, and T3 = 25 deg K The temperatures in each effect and the corresponding latent heats are then: T0 = 393 K, λ0 = 2202 kJ/kg T1 = 374 K, λ1 = 2254 kJ/kg T2 = 350 K, λ2 = 2315 kJ/kg T3 = 325 K, λ3 = 2376 kJ/kg Making a heat balance for each effect: (1) D0 λ0 = (W − D3 − D2 )Cp (T1 − T2 ) + D1 λ1 or (2) 2202D0 = (1.25 − D2 − D3 )4.18(374 − 350) + 2254D1 D1 λ1 = (W − D3 )Cp (T2 − T3 ) + D2 λ2 2254D1 = (1.25 − D3 )4.18(350 − 325) + 2315D2 or (3) (iv) (v) D2 λ2 = W Cp (T3 − Tf ) + D3 λ3 or 2315D2 = (1.25 × 4.18)(325 − 297) + 2376D3 (vi) Solving equations (i), (iv), (v), and (vi) simultaneously: D0 = 0.432 kg/s, D1 = 0.393 kg/s, D2 = 0.339 kg/s, and D3 = 0.268 kg/s. The areas of transfer surface are: A1 = D0 λ0 /U1 T1 = (0.432 × 2202)/(2.5 × 19) = 20.0 m2 A2 = D1 λ1 /U2 T2 = (0.393 × 2254)/(2.0 × 24) = 18.5 m2 A3 = D2 λ2 /U3 T3 = (0.268 × 2315)/(1.6 × 25) = 15.5 m2 which are probably sufficiently close for design purposes; the mean area being 18.0 m2 . The steam consumption is therefore, D0 = 0.432 kg/s The temperatures in each effect are: (1) 374 K, (2) 350 K, and (3) 325 K. PROBLEM 14.13 A liquid with no appreciable elevation of boiling-point is concentrated in a triple-effect evaporator. If the temperature of the steam to the first effect is 395 K and vacuum is 194 applied to the third effect so that the boiling-point is 325 K, what are the approximate boiling-points in the three effects? The overall transfer coefficients may be taken as 3.1, 2.3, and 1.1 kW/m2 K in the three effects respectively. Solution For equal thermal loads in each effect, that is Q1 = Q2 = Q3 , then: U1 A1 T1 = U2 A2 T2 = U3 A3 T3 (equation 14.7) or, for equal areas in each effect: U1 T1 = U2 T2 = U3 T3 (equation 14.8) In this case: 3.1T1 = 2.3T2 = 1.1T3 T1 = 0.742T2 Thus: and T3 = 2.091T2 T = T1 + T2 + T3 = (395 − 325) = 70 deg K 0.742T2 + T2 + 2.091T2 = 70 deg K and T2 = 18.3 deg K Thus: T1 = 13.5 deg K, and: T3 = 38.2 deg K The temperatures in each effect are therefore: T1 = (395 − 13.5) = 381.5 K T2 = (381.5 − 18.3) = 363.2 K, and T3 = (363.2 − 38.2) = 325 K PROBLEM 14.14 A three-stage evaporator is fed with 1.25 kg/s of a liquor which is concentrated from 10 to 40 per cent solids. The heat transfer coefficients may be taken as 3.1, 2.5, and 1.7 kW/m2 K in each effect respectively. Calculate the required steam flowrate at 170 kN/m2 and the temperature distribution in the three effects, if: (a) if the feed is at 294 K, and (b) if the feed is at 355 K. Forward feed is used in each case, and the values of U are the same for the two systems. The boiling-point in the third effect is 325 K, and the liquor has no boiling-point rise. Solution (a) In the absence of any data to the contrary, the specific heat capacity will be taken as 4.18 kJ/kg K. 195 Making a mass balance: Feed Product Evaporation Solids (kg/s) Liquor (kg/s) Total (kg/s) 0.125 0.125 1.125 0.188 1.250 0.313 — 0.937 0.937 (D1 + D2 + D3 ) = 0.937 kg/s Thus: 2 T0 = 388 K For steam at 170 kN/m : T = (388 − 325) = 63 deg K Therefore: From equation 14.8: 3.1T1 = 2.5T2 = 1.7T3 and hence: T1 = 15.5 deg K, T2 = 19 deg K, and T3 = 28.5 deg K Allowing for the cold feed (294 K), it will be assumed that: T1 = 20 deg K, and hence: T2 = 17 deg K, and T3 = 26 deg K T0 = 388 K, λ0 = 2216 kJ/kg T1 = 368 K, λ1 = 2270 kJ/kg T2 = 351 K, λ2 = 2312 kJ/kg T3 = 325 K, λ3 = 2376 kJ/kg Making a heat balance across each effect: (1) 2216D0 = (1.25 × 4.18)(368 − 294) + 2270D1 (2) 2270D1 + (1.25 − D1 )4.18(368 − 351) = 2312D2 (3) 2312D2 + (1.25 − D1 − D2 )4.18(351 − 325) = 2376D3 or D0 = (0.175 + 1.024D1 ) or D2 = (0.951D1 + 0.038) D3 = (0.927D2 − 0.046D1 + 0.057) or: Noting that D1 + D2 + D3 = 0.937 kg/s, these equations may be solved to give: D0 = 0.472 kg/s, D1 = 0.290 kg/s, D2 = 0.313 kg/s and D3 = 0.334 kg/s The area of each effect is then: A1 = D0 λ0 /U1 T1 = (0.472 × 2216)/(3.1 × 20) = 16.9 m2 A2 = D1 λ1 /U2 T2 = (0.290 × 2270)/(2.5 × 17) = 15.5 m2 A3 = D2 λ2 /U3 T3 = (0.334 × 2312)/(1.7 × 26) = 17.4 m2 These are probably sufficiently close for a first approximation, and hence the steam consumption, D0 = 0.472 kg/s 196 and the temperatures in each effect are: (1) 368 K, (2) 351 K, (3) 325 K (b) In this case the feed is much hotter and hence less modification to the estimated values of T will be required. It is assumed that: T1 = 17 deg K, and hence: T2 = 18 deg K, and T3 = 28 deg K T0 = 388 K, λ0 = 2216 kJ/kg T1 = 371 K, λ1 = 2262 kJ/kg T2 = 353 K, λ2 = 2308 kJ/kg T3 = 325 K, λ3 = 2376 kJ/kg The heat balances are now: (1) 2216D0 = (1.25 × 4.18)(371 − 355) + 2262D1 (2) 2262D1 + (1.25 − D1 )4.18(371 − 353) = 2308D2 (3) 2308D2 + (1.25 − D1 − D2 )4.18(353 − 325) = 2376D3 or or D2 = (0.948D1 + 0.041) D3 = (0.922D2 − 0.049D1 + 0.062) Again: and: D0 = 0.331 kg/s, Thus: D0 = (0.038 + 1.021D1 ) or (D1 + D2 + D3 ) = 0.937 kg/s D1 = 0.287 kg/s, D2 = 0.313 kg/s, and D3 = 0.337 kg/s A1 = (0.331 × 2216)/(3.1 × 17) = 13.9 m2 A2 = (0.287 × 2262)/(2.5 × 18) = 14.4 m2 A3 = (0.313 × 2308)/(1.7 × 28) = 15.1 m2 These are close enough for design purposes and hence: the steam consumption D0 = 0.331 kg/s and the temperatures in each effect are: (1) 371 K, (2) 353 K, (3) 325 K PROBLEM 14.15 An evaporator operating on the thermo-recompression principle employs a steam ejector to maintain atmospheric pressure over the boiling liquid. The ejector uses 0.14 kg/s of steam at 650 kN/m2 , and is superheated by 100 K and the pressure in the steam chest is 205 kN/m2 . A condenser removes surplus vapour from the atmospheric pressure line. What is the capacity and economy of the system and how could the economy be improved? The feed enters the evaporator at 295 K and the concentrated liquor is withdrawn at the rate of 0.025 kg/s. The concentrated liquor exhibits a boiling-point rise of 10 deg K. Heat losses to the surroundings are negligible. The nozzle efficiency is 0.95, the efficiency of momentum transfer is 0.80 and the efficiency of compression is 0.90. 197 Solution See Volume 2, Example 14.5. PROBLEM 14.16 A single-effect evaporator is used to concentrate 0.075 kg/s of a 10 per cent caustic soda liquor to 30 per cent. The unit employs forced circulation in which the liquor is pumped through the vertical tubes of the calandria which are 32 mm o.d. by 28 mm i.d. and 1.2 m long. Steam is supplied at 394 K, dry and saturated, and the boiling-point rise of the 30 per cent solution is 15 deg K. If the overall heat transfer coefficient is 1.75 kW/m2 K, how many tubes should be used, and what material of construction would be specified for the evaporator? The latent heat of vaporisation under these conditions is 2270 kJ/kg. Solution Making a mass balance: Feed Product Evaporation Solids (kg/s) Liquor (kg/s) Total (kg/s) 0.0075 0.0075 0.0675 0.0175 0.0750 0.0250 — 0.0500 0.0500 The temperature of boiling liquor in the tubes, assuming atmospheric pressure, T1′ = (373 + 15) = 388 K. In the absence of any other data it will be assumed that the solution enters at 373 K and the specific heat capacity will be taken as 4.18 kJ/kg K. A heat balance then gives: D0 λ0 = W Cp (T1′ − Tf ) + D1 λ1 = (0.0750 × 4.18)(388 − 373) + (0.050 × 2270) = 118.2 kW T1 = (394 − 388) = 6 deg K and the area: A1 = D0 λ0 /U1 T1 = 118.2/(1.75 × 6) = 11.25 m2 . The tube o.d. is 0.032 m and the outside area per unit length of tubing is given by: (π × 0.032) = 0.101 m2 /m Thus: total length of tubing required = (11.25/0.101) = 112 m and number of tubes required = (112/1.2) = 93 Mild steel does not cope with caustic soda solutions, and stainless steel has limitations at higher temperatures. Aluminium bronze, copper, nickel, and nickel — copper alloys may 198 be used, together with neoprene and butyl rubber, though from the cost viewpoint and the need for a good conductivity, graphite tubes would probably be specified. PROBLEM 14.17 A steam-jet booster compresses 0.1 kg/s of dry and saturated vapour from 3.4 kN/m2 to 13.4 kN/m2 . The high-pressure steam consumption is 0.05 kg/s at 690 kN/m2 . (a) What must be the condition of the high pressure steam for the booster discharge to be superheated by 20 deg K? (b) What is the overall efficiency of the booster if the compression efficiency is 100 per cent? Solution (a) Considering the outlet stream at 13.4 kN/m2 and 20 deg K superheat, this has an enthalpy, H4 = 2638 kJ/kg. The enthalpy of the entrained vapours, H3′ = 2540 kJ/kg assuming that these are dry and saturated. If H1 is the enthalpy of the high pressure steam, then an enthalpy balance gives: 0.05H1 + (0.1 × 2540) = (0.15 × 2638) and H1 = 2834 kJ/kg At 690 kN/m2 , this corresponds to a temperature of 453 K. At 690 kN/m2 , steam is saturated at 438 K, and the high pressure steam must be superheated by 15 deg K. (b) Assuming the high pressure steam is expanded isentropically from 690 kN/m2 (and 453 K) to 3.4 kN/m2 , its enthalpy H2 (assuming 100 per cent efficiency) = 2045 kJ/kg. If the outlet stream is expanded isentropically from 3.4 kN/m2 to 13.4 kN/m2 then H3 = 2435 kJ/kg and the efficiency is given by: (0.1/0.05) = [(2834 − 2045)/(2638 − 2435)]η − 1 from which: η = 0.77 PROBLEM 14.18 A triple-effect backward-feed evaporator concentrates 5 kg/s of liquor from 10 per cent to 50 per cent solids. Steam is available at 375 kN/m2 and the condenser operates at 13.5 kN/m2 . What is the area required in each effect, assumed identical, and the economy of the unit? The specific heat capacity is 4.18 kJ/kg K at all concentrations and that there is no boiling-point rise. The overall heat transfer coefficients are 2.3, 2.0 and 1.7 kW/m2 K respectively in the three effects, and the feed enters the third effect at 300 K. 199 Solution As a variation, this problem will be solved using Storrow’s Method (Volume 2 page 786). A mass balance gives the total evaporation as follows: Feed Product Evaporation For steam at 375 kN/m2 , 2 For steam at 13.5 kN/m , Solids (kg/s) Liquor (kg/s) Total (kg/s) 0.50 0.50 4.50 0.50 5.0 1.0 — 4.0 4.0 T0 = 414 K T3 = 325 K T = (414 − 325) = 89 deg K Thus: For equal heat transfer rates in each effect: 2.3T1 = 2.0T2 = 1.7T3 and hence: T1 = 26 deg K, T2 = 29 deg K, (equation 14.8) T3 = 34 deg K Modifying the values to take account of the feed temperature, it will be assumed that: T1 = 27 deg K, and hence: T2 = 30 deg K, and T1 = 387 K, T3 = 32 deg K T2 = 357 K, and T3 = 325 K As a first approximation, equal evaporation in each effect will be assumed, or: D1 = D2 = D3 = 1.33 kg/s With backward feed, the liquor has to be heated to its boiling-point as it enters each effect. Heat required to raise the feed to the second effect to its boiling-point is given by: (4.0 − 1.33)4.18(357 − 325) = 357.1 kW Heat required to raise the feed to the first effect to its boiling-point is [4.0 − (2 × 1.33)]4.18(387 − 357) = 168.1 kW At T0 = 414 K: λ0 = 2140 kJ/kg At T3 = 325 K: λ3 = 2376 kJ/kg A mean value of 2258 kJ/kg which will be taken as the value of the latent heat in all three effects. 200 The relation between the heat transferred in each effect and in the condenser is given by: (Q1 − 168.1) = Q2 = (Q3 + 357.1) = (Qc + 357.1) + [5.0 × 4.18(325 − 300)] = (Qc + 879.6) The total evaporation is: (Q2 + Q3 + Qc )/2258 = 4.0 or: Q2 + (Q2 − 357.1) + (Q2 − 879.6) = 9032 Therefore: Q2 = 3423 = (AT2 × 2.0) Q3 = 3066 = (AT3 × 1.7) Q1 = 3591 = (AT1 × 2.3) Thus: AT1 = 1561 m2 K, AT2 = 1712 m2 K, and AT3 = 1804 m2 K (T1 + T2 + T3 ) = 89 deg K Values of T1 , T2 , and T3 deg K are now chosen by trial and error to give equal value of A m2 in each effect as follows: T1 A1 T2 A2 T3 A3 27 27.5 27.25 57.8 56.8 57.3 30 30.5 30.25 57.1 56.1 56.6 32 31 31.5 56.4 58.1 57.3 These areas are approximately equal and the assumed values of T are acceptable The economy = 4.0/(3591/2258) = 2.52 and the area to be specified for each effect = 57 m2 PROBLEM 14.19 A double-effect climbing-film evaporator is connected so that the feed passes through two preheaters, one heated by vapour from the first effect and the other by vapour from the second effect. The condensate from the first effect is passed into the steam space of the second effect. The temperature of the feed is initially 289 K, 348 K after the first heater and 383 K after the second heater. The vapour temperature in the first effect is 398 K and in the second effect 373 K. The feed flowrate is 0.25 kg/s and the steam is dry and saturated at 413 K. What is the economy of the unit if the evaporation rate is 0.125 kg/s? Solution A heat balance across the first effect gives: 201 D0 λ0 = W Cp (T1 − Tf ) + D1 λ where Tf is the temperature of the feed leaving the second preheater, 383 K. When T0 = 413 K, λ0 = 2140 kJ/kg When T1 = 398 K, λ1 = 2190 kJ/kg Taking Cp = 4.18 kJ/kg K throughout, then: 2140D0 = (0.25 × 4.18)(398 − 383) + 2190D1 D0 = (0.0073 + 0.023D1 ) kg/s Thus: Assuming that the first preheater is heated by steam from the first effect, then the heat transferred in this unit is: (0.25 × 4.18)(348 − 289) = 61.7 kW and hence the steam condensed in the preheater = (61.7/2190) = 0.028 kg/s. Therefore the flowrate of vapour from the first effect which is condensed in the steam space of the second effect = (D1 − 0.028) kg/s. A heat balance on the second effect is thus: (D1 − 0.028)λ1 + (W − D1 )Cp (T1 − T2 ) = D2 λ2 Since the total evaporation = 0.125 kg/s, D2 = (0.125 − D1 ) and: (D1 − 0.028)2190 + (0.25 − D1 )4.18(398 − 373) = (0.125 − D1 )2256 Therefore: Thus: and the economy is: D1 = 0.073 kg/s D0 = 0.0073 + (1.023 × 0.073) = 0.082 kg/s (0.125/0.082) = 1.5 kg/kg PROBLEM 14.20 A triple-effect evaporator is fed with 5 kg/s of a liquor containing 15 per cent solids. The concentration in the last effect, which operates at 13.5 kN/m2 , is 60 per cent solids. If the overall heat transfer coefficients in the three effects are 2.5, 2.0, and 1.1 kW/m2 K, respectively, and the steam is fed at 388 K to the first effect, determine the temperature distribution and the area of heating surface required in each effect? The calandrias are identical. What is the economy and what is the heat load on the condenser? The feed temperature is 294 K and the specific heat capacity of all liquors is 4.18 kJ/kg K. If the unit is run as a backward-feed system, the coefficients are 2.3, 2.0, and 1.6 kW/m2 K respectively. Determine the new temperatures, the heat economy, and the heating surface required under these conditions. 202 Solution (a) Forward feed A mass balance gives: Feed Product Solids (kg/s) Liquor (kg/s) Total (kg/s) 0.75 0.75 4.25 0.50 5.0 1.25 — 3.75 3.75 Evaporation For steam saturated at 13.5 kN/m2 : T3 = 325 K and λ3 = 2375 kJ/kg T0 = 388 K and λ0 = 2216 kJ/kg T = (388 − 325) = 63 deg K Thus: For equal heat transfer rates in each effect: U1 T1 = U2 T2 = U3 T3 2.5T1 = 2.0T2 = 1.1T3 Thus: and: (equation 14.8) T1 = 14 deg K, T2 = 17.5 deg K, and T3 = 31.5 deg K Allowing for the cold feed it will be assumed that: T1 = 18 deg K, and hence: T2 = 16 deg K, and T3 = 29 deg K T0 = 388 K, λ0 = 2216 kJ/kg T1 = 370 K, λ1 = 2266 kJ/kg T2 = 354 K, λ2 = 2305 kJ/kg T3 = 325 K, λ3 = 2375 kJ/kg Making heat balances over each effect: (1) D0 λ0 = W Cp (T1 − Tf ) + D1 λ1 or 2216D0 = (5.0 × 4.18)(370 − 294) + 2266D1 (2) D1 λ1 + (W − D1 )Cp (T1 − T2 ) = D2 λ2 or 2266D1 + (5.0 − D1 )4.18(370 − 354) = 2305D2 (3) D2 λ2 + (W − D1 − D2 )Cp (T2 − T3 ) = D3 λ3 or 2305D2 + (5.0 − D1 − D2 )4.18(354 − 325) = 2375D3 (D1 + D2 + D3 ) = 3.75 kg/s Also: Solving simultaneously: D0 = 1.90 kg/s, D1 = 1.16 kg/s, D2 = 1.25 kg/s, 203 and D3 = 1.35 kg/s The areas are now given by: A1 = D0 λ0 /U1 T1 = (1.90 × 2216)/(2.5 × 18) = 93.6 m2 A2 = D1 λ1 /U2 T2 = (1.16 × 2266)/(2.0 × 16) = 82.1 m2 A3 = D2 λ2 /U3 T3 = (1.25 × 2305)/(1.1 × 29) = 90.3 m2 It is apparent that the modification to the temperature difference made to take account of the effect of the cold feed has been incorrect and it is now assumed that: T1 = 19 deg K, T2 = 15 deg K, and T3 = 29 deg K T0 = 388 K and λ0 = 2216 kJ/kg Thus: T1 = 369 K and λ1 = 2267 kJ/kg T2 = 354 K and λ2 = 2305 kJ/kg T3 = 325 K and λ3 = 2375 kJ/kg The heat balances now become: (1) 2216D0 = (5.0 × 4.18)(369 − 294) + 2267D1 (2) 2267D1 + (5.0 − D1 )4.18(369 − 354) = 2305D2 (3) 2305D2 + (5.0 − D1 − D2 )4.18(354 − 325) = 2375D3 and solving: D0 = 1.89 kg/s, Hence: D1 = 1.16 kg/s, D2 = 1.25 kg/s, and D3 = 1.34 kg/s A1 = (1.89 × 2216)/(2.5 × 19) = 88.2 m2 A2 = (1.16 × 2267)/(2.0 × 15) = 87.8 m2 A3 = (1.25 × 2305)/(1.1 × 29) = 90.3 m2 giving much closer values for the three areas. The temperature distribution is now: (1) 369 K, (2) 354 K, (3) 325 K The area in each effect should be about 89 m2 . The economy is (3.75/1.89) = 2.0 and the heat load on the condenser is: D3 λ3 = (1.34 × 2375) = 31.8 kW (b) Backward feed (D1 + D2 + D3 ) = 3.75 kg/s As before: and: In this case: and:  T1 = 17.5 deg K, T = (388 − 325) = 63 deg k 2.3T1 = 2.0T2 = 1.6T3 T2 = 20.0 deg K, 204 and T3 = 25.5 deg K Modifying for the effect of the cold feed, it is assumed that: T1 = 18.5 deg K, T2 = 20.5 deg K, T0 = 388 K, Thus: and T3 = 24 deg K λ0 = 2216 kJ/kg T1 = 369.5 K, λ1 = 2266 kJ/kg T2 = 349 K, λ2 = 2318 kJ/kg T3 = 325 K, λ3 = 2375 kJ/kg The heat balances become: (1) D0 λ0 = (W − D3 − D2 )Cp (T1 − T2 ) + D1 λ1 or 2216D0 = (5.0 − D3 − D2 )4.18(369.5 − 349) + 2266D1 (2) D1 λ1 = (W − D3 )Cp (T2 − T3 ) + D2 λ2 or 2266D1 = (5.0 − D3 )4.18(349 − 325) + 2318D2 (3) D2 λ2 = W Cp (T3 − Tf ) + D3 λ3 or 2318D2 = (5.0 × 4.18)(325 − 294) + 2375D3 Solving: D0 = 1.62 kg/s, D1 = 1.49 kg/s, D2 = 1.28 kg/s, and D3 = 0.98 kg/s The areas are therefore: A1 = (1.62 × 2216)/(2.3 × 18.5) = 84.4 m2 A2 = (1.49 × 2266)/(2.0 × 20.5) = 82.4 m2 A3 = (1.28 × 2318)/(1.6 × 24) = 77.3 m2 which are probably close enough for design purposes. Thus, the temperatures in each effect are now: (1) 369.5 K, (2) 349 K, and (3) 325 K The economy is (3.75/1.62) = 2.3, and the area required in each effect is approximately 81 m2 . PROBLEM 14.21 A double-effect forward-feed evaporator is required to give a product which contains 50 per cent by mass of solids. Each effect has 10 m2 of heating surface and the heat transfer coefficients are 2.8 and 1.7 kW/m2 K in the first and second effects respectively. Dry and saturated steam is available at 375 kN/m2 and the condenser operates at 13.5 kN/m2 . The concentrated solution exhibits a boiling-point rise of 3 deg K. What is the maximum permissible feed rate if the feed contains 10 per cent solids and is at 310 K? The latent heat is 2330 kJ/kg and the specific heat capacity is 4.18 kJ/kg under all the above conditions. 205 Solution Making a mass balance on the basis of W kg/s feed: Feed Product Evaporation Solids (kg/s) Liquor (kg/s) Total (kg/s) 0.10W 0.10W 0.90W 0.10W W 0.20W — 0.80W 0.80W (D1 + D2 ) = 0.8W kg/s Thus: At 375 kN/m2 : T0 = 413 K At 13.5 kN/m2 : T2 = 325 K and the temperature of the boiling liquor in the second effect, T2′ = (325 + 3) = 328 K. T = (413 − 328) = 85 deg K Therefore: At this stage, values may be assumed for T1 and T2 and heat balances made until equal areas are obtained. There are sufficient data, however, to enable an exact solution to be obtained as follows. Making heat balances: 1st effect D0 λ0 = W Cp (T1 − Tf ) + D1 λ1 U1 A(T0 − T1 ) = W (T1 − Tf ) + U2 A(T1 − T2′ ) or: Thus: (2.8 × 10)(413 − T1 ) = (W × 4.18)(T1 − 310) + (1.7 × 10)(T1 − 328) W = (17,140 − 45T1 )/(4.18T1 − 1296) and: 2nd effect D1 λ1 = (W − D1 )Cp (T1 − T2′ ) + D2 λ2 But: D1 = (1.7 × 10)(T1 − 328)/2330 = (0.0073T1 − 2.393) and: D2 = (0.8W − D1 ) = (0.8W − 0.0073T1 − 2.393) Thus: 2330(0.0073T1 − 2.393) = (W − 0.0073T1 + 2.393)4.18(T1 − 328) + (0.8W − 0.0073T1 − 2.393)2330 and: W = (14T1 − 7872 + 0.0305T12 )/(4.18T1 + 493) Equations (i) and (ii) are plotted in Figure 14a and the two curves coincide when T1 = 375 K and W = 0.83 kg/s. 206 8 7 6 equation (i) 5 W (kg/s) 4 3 2 1 W = 0.83 kg/s equation (ii) 375 K 0 −1 320 340 360 T1(K) 400 −2 −3 Figure 14a. Graphical construction for Problem 14.21 PROBLEM 14.22 For the concentration of fruit juice by evaporation, it is proposed to use a falling-film evaporator and to incorporate a heat-pump cycle with ammonia as the medium. The ammonia in vapour form enters the evaporator at 312 K and the water is evaporated from the juices at 287 K. The ammonia in the vapour — liquid mixture enters the condenser at 278 K and the vapour then passes to the compressor. It is estimated that the work required to compress the ammonia is 150 kJ/kg of ammonia and that 2.28 kg of ammonia is cycled/kg water evaporated. The following proposals are made for driving the compressor: (a) To use a diesel engine drive taking 0.4 kg of fuel/MJ. The calorific value of the fuel is 42 MJ/kg, and the cost £0.02/kg. (b) To pass steam, costing £0.01/10 kg, through a turbine which operates at 70 per cent isentropic efficiency, between 700 and 101.3 kN/m2 . Explain by means of a diagram how this plant will work, illustrating all necessary major items of equipment. Which method for driving the compressor is to be preferred? 207 Solution See Volume 2, Example 14.6. PROBLEM 14.23 A double-effect forward-feed evaporator is required to give a product consisting of 30 per cent crystals and a mother liquor containing 40 per cent by mass of dissolved solids. Heat transfer coefficients are 2.8 and 1.7 kW/m2 K in the first and second effects respectively. Dry saturated steam is supplied at 375 kN/m2 and the condenser operates at 13.5 kN/m2 . (a) What area of heating surface is required in each effect, assuming they are both identical, if the feed rate is 0.6 kg/s of liquor, containing 20 per cent by mass of dissolved solids, and the feed temperature is 313 K? (b) What is the pressure above the boiling liquid in the first effect? The specific heat capacity may be taken as constant at 4.18 kJ/kg K, and the effects of boiling point rise and of hydrostatic head may be neglected. Solution The final product contains 30 per cent crystals and hence 70 per cent solution containing 40 per cent dissolved solids. The total percentage of dissolved and undissolved solids in the final product is 0.30 + (0.40 × 0.70) = 0.58 or 58 per cent, and the mass balance becomes: Feed Product Evaporation At 375 kN/m2 : Thus: Solids (kg/s) Liquor (kg/s) Total (kg/s) 0.12 0.12 0.48 0.087 0.60 0.207 — (D1 + D2 ) = 0.393 0.393 T0 = 413 K and at 13.5 kN/m2 , T2 = 325 K  T = (413 − 325) = 88 deg K For equal heat transfer areas in each effect: 2.8T1 = 1.7T2 and: T1 = 33 deg K and T2 = 55 deg K Since the feed is cold, it will be assumed that: T1 = 40 deg K and hence: and T2 = 48 deg K T0 = 413 K, λ0 = 2140 kJ/kg T1 = 373 K, λ1 = 2257 kJ/kg T2 = 325 K, λ1 = 2375 kJ/kg 208 (equation 14.8) Taking the specific heat capacity of the liquor as constant at 4.18 kJ/kg K at all times, then the heat balance over each effect becomes: (1) D0 λ0 = W Cp (T1 − Tf ) + D1 λ1 or 2140D0 = (0.6 × 4.18)(373 − 313) + 2257D1 (2) D1 λ1 + (W − D1 )Cp (T1 − T2 ) = D2 λ2 or Solving: D0 = 0.263 kg/s, 2257D1 + (0.6 − D1 )4.18(373 − 325) = 2375D2 D1 = 0.184 kg/s, and D2 = 0.209 kg/s The areas are now given by: A1 = D0 λ0 /U1 T1 = (0.263 × 2140)/(2.8 × 40) = 5.04 m2 A2 = D1 λ1 /U2 T2 = (0.184 × 2257)/(1.7 × 48) = 5.08 m2 which show close agreement. Thus: the area of heating to be specified = 5.06 m2 or approximately 5 m2 in each effect. The temperature of liquor boiling in the first effect, assuming no boiling-point rise, is 373 K at which temperature steam is saturated at 101.3 kN/m2 . The pressure in the first effect is, therefore, atmospheric. PROBLEM 14.24 1.9 kg/s of a liquid containing 10 per cent by mass of dissolved solids is fed at 338 K to a forward-feed double-effect evaporator. The product consists of 25 per cent by mass of solids and a mother liquor containing 25 per cent by mass of dissolved solids. The steam fed to the first effect is dry and saturated at 240 kN/m2 and the pressure in the second effect is 20 kN/m2 . The specific heat capacity of the solid may be taken as 2.5 kJ/kg K, both in solid form and in solution, and the heat of solution may be neglected. The mother liquor exhibits a boiling point rise of 6 deg K. If the two effects are identical, what area is required if the heat transfer coefficients in the first and second effects are 1.7 and 1.1 kW/m2 K respectively? Solution The percentage by mass of dissolved and undissolved solids in the final product = (0.25 × 75) + 25 = 43.8 per cent and hence the mass balance becomes: Feed Product Evaporation Solids (kg/s) Liquor (kg/s) Total (kg/s) 0.19 0.19 1.71 0.244 1.90 0.434 — (D1 + D2 ) = 1.466 1.466 209 At 240 kN/m2 : T0 = 399 K At 20 kN/m2 , T2′ Thus: T2 = 333 K = (333 + 6) = 339 K and, allowing for the boiling-point rise in the first effect: T1 + T2 = (399 − 339) − 6 = 54 deg K From equation 14.8: 1.7T1 = 1.1T2 and: T1 = 21 deg K and T2 = 33 deg K Modifying these values to allow for the cold feed, it will be assumed that: T1 = 23 deg K and T2 = 31 deg K Assuming that the liquor exhibits a 6 deg K boiling-point rise at all concentrations, then, with T1′ as the temperature of boiling liquor in the first effect and T2′ that in the second effect: T0 = 399 K at which λ0 = 2185 kJ/kg T1′ = (399 − 23) = 376 K T1 = (376 − 6) = 370 K at which λ1 = 2266 kJ/kg T2′ = 339 K T2 = 333 K at which λ2 = 2258 K Making a heat balance over each effect: (1) D0 λ0 = W Cp (T1′ − Tf ) + D1 λ1 or 2185D0 = (1.90 × 2.5)(376 − 338) + 2266D1 (2) D1 λ1 + (W − D1 )Cp (T1′ − T2′ ) = D2 λ2 or: Solving: 2358D2 = (1.90 − D1 )2.5(376 − 339) + 2266D1 D0 = 0.833 kg/s, D1 = 0.724 kg/s, and D2 = 0.742 kg/s The areas are then given by: A1 = D0 λ0 /U1 (T0 − T1′ ) = 0.833 × 2185/[1.7(399 − 376)] = 46.7 m2 A2 = D1 λ1 /U2 (T1 − T2′ ) = 0.724 × 2266/[1.1(370 − 339)] = 48.0 m2 which are close enough for design purposes. The area to be specified for each effect is approximately 47.5 m2 . PROBLEM 14.25 2.5 kg/s of a solution at 288 K containing 10 per cent of dissolved solids is fed to a forward-feed double-effect evaporator, operating at 14 kN/m2 in the last effect. If the 210 product is to consist of a liquid containing 50 per cent by mass of dissolved solids and dry saturated steam is fed to the steam coils, what should be the pressure of the steam? The surface in each effect is 50 m2 and the coefficients for heat transfer in the first and second effects are 2.8 and 1.7 kW/m2 K respectively. It may be assumed that the concentrated solution exhibits a boiling-point rise of 5 deg K, that the latent heat has a constant value of 2260 kJ/kg and that the specific heat capacity of the liquid stream is constant at 3.75 kJ/kg K. Solution A mass balance gives: Feed Product Evaporation Solids (kg/s) Liquor (kg/s) Total (kg/s) 0.25 0.25 2.25 0.25 2.50 0.50 — 2.0 2.0 (D1 + D2 ) = 2.0 kg/s Thus: 2 T2 = 326 K At 14.0 kN/m : and allowing for the boiling-point rise in the second effect, T2′ = (326 + 5) = 331 K. Writing a heat balance for each effect, then: (1) D0 λ0 = W (T1 − Tf ) + D1 λ1 (2) D1 λ1 + (W − D1 )(T1 − or: T2′ ) or 2260D0 = (2.5 × 3.75)(T1 − 288) + 2260D1 (i) = D2 λ2 2260D1 + (2.5 − D1 )3.75(T1 − 331) = 2260D2 since D2 = (2.0 − D1 ), Thus: But: Therefore: 4520(1 − D1 ) = (9.375 − 3.75D1 )(T1 − 331) D1 λ1 = U2 A2 (T1 − 2260D1 = (1.7 × 50)(T1 − 331) T2′ ) or T1 = (26.6D1 + 331) Substituting in equation (ii) for T1 : (D12 − 48.26D1 + 45.31) = 0 and: D1 = 47.30 kg/s (which is clearly impossible) or 0.96 kg/s Thus: T1 = (26.6 × 0.96) + 331 = 356.5 K 211 (ii) In equation (i): 2260D0 = (2.5 × 3.75)(356.5 − 288) + (2260 × 0.96) and D0 = 1.24 D0 λ0 = U1 A1 (T0 − T1 ) But: (1.24 × 2260) = (2.8 × 50)(T0 − 356.5) or: and T0 = 376.5 K Steam is dry and saturated at 376.5 K at a pressure of 115 kN/m2 . PROBLEM 14.26 A salt solution at 293 K is fed at the rate of 6.3 kg/s to a forward-feed triple-effect evaporator and is concentrated from 2 per cent to 10 per cent of solids. Saturated steam at 170 kN/m2 is introduced into the calandria of the first effect and a pressure of 34 kN/m2 is maintained in the last effect. If the heat transfer coefficients in the three effects are 1.7, 1.4 and 1.1 kW/m2 K respectively and the specific heat capacity of the liquid is approximately 4 kJ/kg K, what area is required if each effect is identical? Condensate may be assumed to leave at the vapour temperature at each stage, and the effects of boiling point rise may be neglected. The latent heat of vaporisation may be taken as constant throughout. Solution The mass balance is as follows: Feed Product Evaporation Solids (kg/s) Liquor (kg/s) Total (kg/s) 0.126 0.126 6.174 1.134 6.30 1.26 — 5.04 5.04 (D1 + D2 + D3 ) = 5.04 kg/s Thus: At 170 kN/m2 : 2 At 34 kN/m : T0 = 388 K and λ0 = 2216 kJ/kg T3 = 345 K and λ3 = 2328 kJ/kg Thus the latent heat will be taken as 2270 kJ/kg throughout and: T = (388 − 345) = 43 deg K From equation 14.8: 1.7T1 = 1.4T2 = 1.1T3 and hence: T1 = 11.5 deg K, T2 = 14 deg K, 212 and T3 = 17.5 deg K Modifying these values for the cold feed, it will be assumed that: T1 = 15 deg K, T0 = 388 K, Thus: T2 = 12 deg K, T1 = 373 K, and T3 = 16 deg K T2 = 361 K, and T3 = 345 K λ0 = λ1 = λ2 = λ3 = 2270 kJ/kg The heat balance over each effect is now: (1) D0 λ0 = W Cp (T1 − Tf ) + D1 λ1 or 2270D0 = (6.3 × 4)(373 − 293) + 2270D1 (2) D1 λ1 + (W − D1 )Cp (T1 − T2 ) = D2 λ2 2270D1 + (6.3 − D1 )4(373 − 361) = 2270D2 or: (3) D2 λ2 + (W − D1 − D2 )Cp (T1 − T2 ) = D3 λ3 2270D2 + (6.3 − D1 − D2 )4(361 − 345) = 2270D3 or: Solving: D0 = 2.48 kg/s, D1 = 1.59 kg/s, D2 = 1.69 kg/s, D3 = 1.78 kg/s The areas are given by: A1 = D0 λ0 /U1 T1 = (2270 × 2.48)/(1.7 × 15) = 220.8 m2 A2 = D1 λ1 /U2 T2 = (2270 × 1.59)/(1.4 × 12) = 214.8 m2 A3 = D2 λ2 /U3 T3 = (2270 × 1.69)/(1.1 × 16) = 217.9 m2 which are close enough for design purposes. The area specified for each stage is therefore 218 m2 . PROBLEM 14.27 A single-effect evaporator with a heating surface area of 10 m2 is used to concentrate a NaOH solution flowing at 0.38 kg/s from 10 per cent to 33.3 per cent. The feed enters at 338 K and its specific heat capacity is 3.2 kJ/kg K. The pressure in the vapour space is 13.5 kN/m2 and 0.3 kg/s of steam is used from a supply at 375 K. Calculate: (a) The apparent overall heat transfer coefficient. (b) The coefficient corrected for boiling point rise of dissolved solids. (c) The corrected coefficient if the depth of liquid is 1.5 m. Solution Mass of solids in feed = (0.38 × 10/100) = 0.038 kg/s. Mass flow of product = (0.038 × 100/33.3) = 0.114 kg/s. Thus: evaporation, D1 = (0.38 − 0.114) = 0.266 kg/s 213 At a pressure of 13.5 kN/m2 , steam is saturated at 325 K and λ1 = 2376 kJ/kg. At 375 K, steam is saturated at 413 K and λ0 = 2140 kJ/kg. (a) Ignoring any boiling-point rise, it may be assumed that the temperature of the boiling liquor, T1 = 325 K. Thus: T1 = (375 − 325) = 50 deg K U1 = D0 λ0 /A1 T1 = (0.3 × 2140)/(10 × 50) = 1.28 kW/m2 K (b) Allowing for a boiling-point rise, the temperature of the boiling liquor in the effect, T1′ may be calculated from a heat balance: D0 λ0 = W Cp (T1′ − Tf ) + D1 λ1 or: and: (0.3 × 2140) = (0.38 × 3.2)(T1′ − 338) + (0.266 × 2376) T1′ = 346 K T1 = (375 − 346) = 29 deg K and: U1 = (0.3 × 2140)/(10 × 29) = 2.21 kW/m2 K (c) Taking the density of the fluid as 1000 kg/m3 , the pressure due to a height of liquid of 1.5 m = (1.5 × 1000 × 9.81) = 14,715 N/m2 or 14.7 kN/m2 . The pressure outside the tubes is therefore (13.5 + 14.7) = 28.2 kN/m2 at which pressure, water boils at 341 K. Thus: and: T1 = (375 − 341) = 34 deg K U1 = (0.3 × 2140)/(10 × 34) = 1.89 kW/m2 K A value of the boiling liquor temperature T1′ = 346 K obtained in (b) by heat balance must take into account the effects of hydrostatic head and of boiling-point rise. The true boiling-point rise is (346 − 341) = 5 deg K. Thus: T1′ = (325 + 5) = 330 K T1′ = (375 − 330) = 45 K and: U1 = (0.3 × 2140)/(10 × 45) = 1.43 kN/m2 K . PROBLEM 14.28 An evaporator, working at atmospheric pressure, is used to concentrate a solution from 5 per cent to 20 per cent solids at the rate of 1.25 kg/s. The solution, which has a specific heat capacity of 4.18 kJ/kg K, is fed to the evaporator at 295 K and boils at 380 K. Dry saturated steam at 240 kN/m2 is fed to the calandria, and the condensate leaves at the temperature of the condensing stream. If the heat transfer coefficient is 2.3 kW/m2 K, 214 what is the required area of heat transfer surface and how much steam is required? The latent heat of vaporisation of the solution may be taken as being the same as that of water. Solution A material balance gives: Feed Product Solids (kg/s) Liquor (kg/s) 0.0625 0.0625 1.1875 0.2500 1.2500 0.3125 — 0.9375 0.9375 = D1 Evaporation At 240 kN/m2 , Total (kg/s) T0 = 399 K and λ0 = 2185 kJ/kg At a pressure of 101.3 kN/m2 , λ1 = 2257 kJ/kg Making a heat balance across the unit: D1 λ1 + W Cp (T1 − Tf ) = D0 λ0 or: (2257 × 0.9375) + (1.25 × 4.18)(380 − 295) = 2185D0 and The heat transfer area is given by: A = D0 λ0 /U T1 = (1.17 × 2185)/[2.3(399 − 380)] = 58.5 m2 215 D0 = 1.17 kg/s SECTION 2-15 Crystallisation PROBLEM 15.1 A saturated solution containing 1500 kg of potassium chloride at 360 K is cooled in an open tank to 290 K. If the density of the solution is 1200 kg/m3 , the solubility of potassium chloride/100 parts of water by mass is 53.55 at 360 K and 34.5 at 290 K calculate: (a) the capacity of the tank required, and (b) the mass of crystals obtained, neglecting any loss of water by evaporation. Solution At 360 K, 1500 kg KCl will be dissolved in (1500 × 100)/53.55 = 2801 kg water. The total mass of the solution = (1500 + 2801) = 4301 kg. The density of the solution = (1.2 × 1000) = 1200 kg/m3 and hence the capacity of the tank = (4301/1200) = 3.58 m3 . At 290 K, the mass of KCl dissolved in 2801 kg water = (2801 × 34.5)/100 = 966 kg Thus: mass of crystals which has come out of solution = (1500 − 966) = 534 kg PROBLEM 15.2 Explain how fractional crystallisation may be applied to a mixture of sodium chloride and sodium nitrate, given the following data. At 290 K, the solubility of sodium chloride is 36 kg/100 kg water and of sodium nitrate 88 kg/100 kg water. Whilst at this temperature, a saturated solution comprising both salts will contain 25 kg sodium chloride and 59 kg sodium nitrate/100 parts of water. At 357 K these values, again per 100 kg of water, are 40 and 176, and 17 and 160 kg respectively. Solution See Volume 2, Example 15.9. 216 PROBLEM 15.3 10 Mg of a solution containing 0.3 kg Na2 CO3 /kg solution is cooled slowly to 293 K to form crystals of Na2 CO3 .10H2 O. What is the yield of crystals if the solubility of Na2 CO3 at 293 K is 21.5 kg/100 kg water and during cooling 3 per cent of the original solution is lost by evaporation? Solution The initial concentration of the solution = 0.3 kg/kg solution c1 = 0.3/(1 − 0.3) = 0.428 kg/kg water. or: The final concentration of the solution, c2 = (21.5/100) = 0.215 kg/kg water. The feed of 10 Mg of solution contains (10 × 0.3) = 3 Mg of anhydrous salt and (10 − 3) = 7 Mg of water. Thus: the initial mass of solvent in the liquid, w1 = (7 × 1000) = 7000 kg. 3 per cent of the original solution or (10 × 1000 × 3)/100 = 300 kg is evaporated. Thus the mass of solvent evaporated/mass of solvent in initial solution is given by: E = [300/(10 × 1000)] = 0.03 kg/kg solution. The final mass of solvent in the liquid, w2 = (7000 − 300) = 6700 kg. The molecular mass of Na2 CO3 = 106 kg/kmol and the molecular mass of Na2 CO3 .10H2 O = 286.2 kg/kmol and hence the molecular mass of hydrate/molecular mass of anhydrous salt is given by: R = (286.2/106) = 2.7 Therefore from equation 15.22: y = Rw1 [c1 − c2 (1 − E)]/[1 − c2 (R − 1)] and by substituting, the yield is: y = (2.7 × 7000)[0.428 − 0.215(1 − 0.03)]/[1 − 0.215(2.7 − 1.0)] = 6536 kg . PROBLEM 15.4 The heat required when 1 kmol of MgSO4 .7H2 O is dissolved isothermally at 291 K in a large mass of water is 13.3 MJ. What is the heat of crystallisation per unit mass of the salt? Solution The molecular mass of MgSO4 .7H2 O = 246.5 kg/kmol Thus: heat of crystallisation = (13.3 × 1000)/246.5 = 53.9 kJ/kg 217 PROBLEM 15.5 A solution of 500 kg of Na2 SO4 in 2500 kg water is cooled from 333 K to 283 K in an agitated mild steel vessel of mass 750 kg. At 283 K, the solubility of the anhydrous salt is 8.9 kg/100 kg water and the stable crystalline phase is Na2 SO4 .10H2 O. At 291 K, the heat of solution is −78.5 MJ/kmol and the specific heat capacities of the solution and mild steel are 3.6 and 0.5 kJ/kg deg K respectively. If, during cooling, 2 per cent of the water initially present is lost by evaporation, estimate the heat which must be removed. Solution It is assumed that the heat of crystallisation = −(the heat of solution) = 78.5 MJ/kmol or: (78.5 × 1000)/322 = 244 kJ/kg where 322 kg/kmol is the molecular mass of the hydrate. Molecular mass of Na2 SO4 = 142 kg/kmol, and R = (322/142) = 2.27 The latent heat of vaporisation of water will be taken as 2395 kJ/kmol. The initial concentration of the solution, c1 = (500/2500) = 0.2 kg/kg water and the final concentration of the solution, c2 = (8.9/100) = 0.089 kg/kg water at 283 K. The initial mass of water, w1 = 2500 kg and the evaporation is 2 per cent of the initial water, or E = 0.02 kg/kg water. Thus: In equation 15.22, the yield is: y = (2.27 × 2500)[0.2 − 0.089(1 − 0.02)]/[1 − 0.089(2.27 − 1)] = 723 kg. Therefore: The heat of crystallisation = (723 × 244) = 176,412 kJ. Heat removed from the solution, assuming crystallisation takes place at 283 K, is then: = (500 + 2500)3.6(333 − 283) = 540,000 kJ Heat removed from the mild steel vessel = (750 × 0.5)(333 − 283) = 18,750 kJ and total heat to be removed = (176,412 + 540,000 + 18,750) = 735,162 kJ But: [(2500 × 2)/100]2395 = 119,750 kJ is lost by evaporation. Thus: net heat to be removed = (735,162 − 119,750) ≈ 615,000 kJ PROBLEM 15.6 A batch of 1500 kg of saturated potassium chloride solution is cooled from 360 K to 290 K in an unagitated tank. If the solubilities of KCl are 53 and 34 kg/100 kg water at 360 K and 290 K respectively and water losses due to evaporation may be neglected, what is the yield of crystals? 218 Solution The initial concentration of the solution is (53/100) = 0.53 kg/kg water c1 = 0.53/(1 + 0.53) = 0.346 kg/kg solution. or: Mass of potassium chloride in the original batch = (1500 × 0.346) = 520 kg and hence, mass of water in the original batch, w1 = (1500 − 520) = 980 kg. The concentration of potassium chloride in the final solution is: c2 = (34/100) = 0.34 kg/kg water. With no hydrate formation and negligible evaporation, R = 1.0 and E = 0 respectively and in equation 15.22, the yield is: y = (1.0 × 980)[0.53 − 0.34(1 − 0)]/[1 − 0.34(1 − 1)] = 186 kg PROBLEM 15.7 Glauber’s salt, Na2 SO4 .10H2 O, is to be produced in a Swenson–Walker crystalliser by cooling to 290 K a solution of anhydrous Na2 SO4 which saturates between 300 K and 290 K. If cooling water enters and leaves the unit at 280 K and 290 K respectively and evaporation is negligible, how many sections of crystalliser, each 3 m long, will be required to process 0.25 kg/s of the product? The solubilities of anhydrous Na2 SO4 in water are 40 and 14 kg/100 kg water at 300 K and 290 K respectively, the mean heat capacity of the liquor is 3.8 kJ/kg K and the heat of crystallisation is 230 kJ/kg. For the crystalliser, the available heat transfer area is 3 m2 /m length, the overall coefficient of heat transfer is 0.15 kW/m2 K and the molecular masses are Na2 SO4 .10H2 O = 322 kg/kmol and Na2 SO4 = 142 kg/kmol. Solution The ratio of the molecular mass of the hydrate to that of the anhydrous salt, R = (322/142) = 2.27 and the evaporation, E may be neglected. The concentration of salt in the feed = (40/100) = 0.40 kg/kg water and: c1 = 0.40/(1 − 0.40) = 0.286 kg/kg solution. Similarly: c2 = (14/100) = 0.140 kg/kg solution. In 1 kg of feed, the mass of salt = 0.286 kg and the water present, w1 = (1 − 0.286) = 0.714 kg/kg feed. In equation 15.22, the yield is: y = (2.27 × 0.714)[0.40 = 0.14(1 − 0)]/[1 − 0.14(2.27 − 1)] = 0.573 kg/kg feed. 219 In order to produce 0.25 kg/s of crystals, a feed rate of: (1 × 0.25)/0.573 = 0.487 kg/s is required. The heat to be removed from the solution is given by: (0.487 × 3.8)(300 − 290) = 18.5 kW (assuming crystals are formed at 290 K) The heat of crystallisation is given by: (0.25 × 330) = 57.5 kW and the total heat to be removed = (18.5 + 57.5) = 76.0 kW. The logarithmic mean temperature difference, assuming counter current flow is: Tm = [(300 − 290) − (290 − 280)]/ ln[(300 − 290)/(290 − 280)] = 10 deg K and with an overall coefficient of heat transfer of 0.15 kW/m2 deg K, the area required is: A = Q/U Tm = 76.0/(0.15 × 10) = 50.67 m2 Area per unit length of crystallise section = 3 m2 /m Hence: Total length of unit required = (50.67/3) = 16.9 m and 6 sections each of 3 m length would be specified. PROBLEM 15.8 What is the evaporation rate and yield of the sodium acetate hydrate CH3 COONa.3H2 O from a continuous evaporative crystalliser operating at 1 kN/m2 when it is fed with 1 kg/s of a 50 per cent by mass aqueous solution of sodium acetate hydrate at 350 K? The boiling point elevation of the solution is 10 deg K and the heat of crystallisation is 150 kJ/kg. The mean heat capacity of the solution is 3.5 kJ/kg K and, at 1 kN/m2 , water boils at 280 K at which temperature the latent heat of vaporisation is 2.482 MJ/kg. Over the range 270–305 K, the solubility of sodium acetate hydrate in water s at T (K) is given approximately by: s = 0.61T − 132.4 kg/100 kg water Molecular masses: CH3 COONa.3H2 O = 136 kg/kmol, CH3 COONa = 82 kg/kmol. Solution Allowing for the boiling point elevation, the temperature of the solution at equilibrium is: (280 + 10) = 290 K The concentration of the feed solution is (50/100) = 0.5 kg/kg solution and the initial concentration of solution is: c1 = 0.5/(1 − 0.5) = 1.0 kg/kg water. 220 Using the given relationship for the solubility, at 290 K, the final concentration of solution is: c2 = [(0.61 × 290) − 132.4]/100 = 0.445 kg/kg water. The heat of crystallisation, qc = 150 kJ/kg and the ratio of the molecular masses, R = (136/82) = 1.66. The evaporation rate is given by equation 15.23 as: E = [qc R(c1 − c2 ) + Cp (T1 − T2 )(1 + c1 )(1 − c2 (R − 1))] /[L(1 − c2 (R − 1)) − qc Rc2 ] or: E = [150 × 1.66(1.0 − 0.445) + 3.5(350 − 290)(1 + 1.0)(1 − 0.445(1.66 − 1))] /[2482(1 − 0.445(1.66 − 1)) − (150 × 1.66 × 0.445)] = 0.265 kg/kg The actual feed is 1 kg/s of solution containing 0.5 kg/s of salt and 0.5 kg/s of water = w1 Therefore: Actual evaporation rate = (0.265 × 0.5) = 0.132 kg/s The yield is then given by equation 15.22: y = Rw1 [c1 − c2 (1 − E)]/[1 − c2 (R − 1)] = (1.66 × 0.5)(1.0 − 0.445(1 − 0.265))/(1 − 0.445(1.66 − 1)) = 0.791 kg/s 221 SECTION 2-16 Drying PROBLEM 16.1 A wet solid is dried from 35 to 10 per cent moisture under constant drying conditions in 18 ks (5 h). If the equilibrium moisture content is 4 per cent and the critical moisture content is 14 per cent, how long will it take to dry to 6 per cent moisture under the same conditions? Solution See Volume 2, Example 16.1. PROBLEM 16.2 Strips of a material 10 mm thick are dried under constant drying conditions from 28 per cent to 13 per cent moisture in 25 ks. If the equilibrium moisture content is 7 per cent, what is the time taken to dry 60 mm planks from 22 to 10 per cent moisture under the same conditions, assuming no loss from the edges? All moisture contents are expressed on the wet basis. The relation between E, the ratio of the average free moisture content at time t to the initial free moisture content, and the parameter f is given by: E f 1 0 0.64 0.1 0.49 0.2 0.38 0.3 0.295 0.5 0.22 0.6 0.14 0.7 where f = kt/ l 2 , k is a constant, t is the time in ks and 2l is the thickness of the sheet of material in mm. Solution See Volume 2, Example 16.2. PROBLEM 16.3 A granular material containing 40 per cent moisture is fed to a countercurrent rotary dryer at 295 K and is withdrawn at 305 K containing 5 per cent moisture. The air supplied, which contains 0.006 kg water vapour/kg of dry air, enters at 385 K and leaves at 310 K. The dryer handles 0.125 kg/s wet stock. 222 Assuming that radiation losses amount to 20 kJ/kg of dry air used, determine the mass flow of dry air supplied to the dryer and the humidity of the outlet air. The latent heat of water vapour at 295 K = 2449 kJ/kg, the specific heat capacity of dried material = 0.88 kJ/kg K, the specific heat capacity of dry air = 1.00 kJ/kg K, and the specific heat capacity of water vapour = 2.01 kJ/kg K. Solution See Volume 2, Example 16.3. PROBLEM 16.4 1 Mg of dry mass of a non-porous solid is dried under constant drying conditions in an air stream flowing at 0.75 m/s. The area of surface drying is 55 m2 . If the initial rate of drying is 0.3 g/m2 s, how long will it take to dry the material from 0.15 to 0.025 kg water/kg dry solid? The critical moisture content of the material may be taken as 0.125 kg water/kg dry solid. If the air velocity were increased to 4.0 m/s, what would be the anticipated saving in time if the process were surface-evaporation controlled? Solution During the constant rate period, that is whilst the moisture content falls from 0.15 to 0.125 kg/kg, the rate of drying is: (dw/dt)/A = (0.3/1000) = 0.0003 kg/m2 s At the start of the falling rate period, w = wc = 0.125 kg/kg and: or: and: (dw/dt)/A = m(wc − we ) 0.0003 = m(0.125 − 0.025) m = 0.003 kg/m2 s kg dry solid = (0.003/1000) = 3.0 × 10−6 kg/m2 s Mg dry solid The total drying time is given by equation 16.14: t = (1/mA)[ln(fc /f ) + (f1 − fc )/fc ] where: f = (0.025 − 0) = 0.025 kg/kg (taking we as zero) fc = (0.125 − 0) = 0.125 kg/kg f1 = (0.15 − 0) = 0.150 kg/kg Thus: t = [1/(3.0 × 10−6 × 55)][ln(0.125/0.025) + (0.150 − 0.125)/0.125] = 10,960 s or 10.96 ks (3 h) 223 As a first approximation it may be assumed that the rate of evaporation is proportional to the air velocity raised to the power of 0.8. For the second case, m may then be calculated as: m = (3.0 × 10−6 )(4.0/0.75)0.8 = (1.15 × 10−5 ) kg water/m2 s Mg dry solid The time of drying is then: t = [1/(1.15 × 10−5 × 55)](1.609 + 0.20) = 2860 s or 2.86 ks and the time saved is therefore: (10.96 − 2.86) = 8.10 ks (2.25 h) PROBLEM 16.5 A 100 kg batch of granular solids containing 30 per cent of moisture is to be dried in a tray dryer to 15.5 per cent moisture by passing a current of air at 350 K tangentially across its surface at the velocity of 1.8 m/s. If the constant rate of drying under these conditions is 0.7 g/s m2 and the critical moisture content is 15 per cent, calculate the approximate drying time. It may be assumed that the drying surface is 0.03 m2 /kg dry mass. Solution See Volume 2, Example 16.4. PROBLEM 16.6 A flow of 0.35 kg/s of a solid is to be dried from 15 per cent to 0.5 per cent moisture on a dry basis. The mean specific heat capacity of the solids is 2.2 kJ/kg deg K. It is proposed that a co-current adiabatic dryer should be used with the solids entering at 300 K and, because of the heat sensitive nature of the solids, leaving at 325 K. Hot air is available at 400 K with a humidity of 0.01 kg/kg dry air and the maximum allowable mass velocity of the air is 0.95 kg/m2 s. What diameter and length should be specified for the proposed dryer? Solution See Volume 2, Example 16.5. PROBLEM 16.7 0.126 kg/s of a solid product containing 4 per cent water is produced in a dryer from a wet feed containing 42 per cent water on a wet basis. Ambient air at 294 K and of 40 per cent relative humidity is heated to 366 K in a preheater before entering the dryer from which it leaves at 60 per cent relative humidity. Assuming that the dryer operates 224 adiabatically, what must be the flowrate of air to the preheater and how much heat must be added to the preheater? How will these values be affected if the air enters the dryer at 340 K and sufficient heat is supplied within the dryer so that the air again leaves at 340 K with a relative humidity of 60 per cent? Solution From the humidity chart, Figure 13.4 in Volume 1, air at 294 K and of 40 per cent relative humidity has a humidity of 0.006 kg/kg. This remains unchanged on heating to 366 K. At the dryer inlet, the wet bulb temperature of the air is 306 K. In the dryer, the cooling takes place along the adiabatic cooling line until 60 per cent relative humidity is reached. At this point: the humidity = 0.028 kg/kg and the dry bulb temperature = 312 K. The water picked up by 1 kg of dry air = (0.028 − 0.006) = 0.22 kg water/kg dry air. The wet feed contains: 42 kg water/100 kg feed or 42/(100 − 42) = 0.725 kg water/kg dry solids The product contains: 4 kg water/100 kg product or 4/(100 − 4) = 0.0417 kg water/kg dry solids. Thus: water evaporated = (0.725 − 0.0417) = 0.6833 kg/kg dry solids The throughput of dry solids is: 0.126(100 − 4)/100 = 0.121 kg/s and the water evaporated is: (0.121 × 0.6833) = 0.0825 kg/s The required air throughput is then: 0.0825/(0.078 − 0.006) = 3.76 kg/s At 294 K, from Figure 13.4 in Volume 1: specific volume of the air = 0.84 m3 /kg and the volume of air required is: (0.84 × 3.76) = 3.16 m3 /s 225 At a humidity of 0.006 kg water/kg dry air, from Figure 13.4, the humid heat is 1.02 kJ/kg deg K and hence the heat required in the preheater is: 3.76 × 1.02(366 − 294) = 276 kW In the second case, air both enters and leaves at 340 K, the outlet humidity is 0.111 kg water/kg dry air and the water picked up by the air is: (0.111 − 0.006) = 0.105 kg/kg dry air The air requirements are then: (0.0825/0.105) = 0.786 kg/s or (0.786 × 0.84) = 0.66 m3 /s The heat to be added in the preheater is: 0.66 × 1.02(340 − 294) = 31.0 kW The heat to be supplied within the dryer is that required to heat the water to 340 K plus its latent heat at 340 K. 0.0825[4.18(340 − 290) + 2345] = 211 kW That is: Thus: Total heat to be supplied = (81 + 211) = 242 kW . PROBLEM 16.8 A wet solid is dried from 40 to 8 per cent moisture in 20 ks. If the critical and the equilibrium moisture contents are 15 and 4 per cent respectively, how long will it take to dry the solid to 5 per cent moisture under the some drying conditions? All moisture contents are on a dry basis. Solution For the first drying operation: w1 = 0.40 kg/kg, Thus: w = 0.08 kg/kg, wc = 0.15 kg/kg and we = 0.04 kg/kg f1 = (w1 − we ) = (0.40 − 0.04) = 0.36 kg/kg fc = (wc − we ) = (0.15 − 0.04) = 0.11 kg/kg f = (w − we ) = (0.08 − 0.04) = 0.04 kg/kg. From equation 16.14, the total drying time is: t = (1/mA)[(f1 − fc )/fc + ln(fc /f )] or: and: 20 = (1/mA)[(0.36 − 0.11)/0.11 + ln(0.11/0.04) mA = 0.05(2.27 + 1.012) = 0.164 kg/ks 226 For the second drying operation: w1 = 0.40 kg/kg, w = 0.05 kg/kg, wc = 0.15 kg/kg, and we = 0.04 kg/kg f1 = (w1 − we ) = (0.40 − 0.04) = 0.36 kg/kg Thus: fe = (wc − we ) = (0.15 − 0.04) = 0.11 kg/kg f = (w − we ) = (0.05 − 0.04) = 0.01 kg/kg The total drying time is then: t = (1/0.164)[(0.36 − 0.11)/0.11 + ln(0.11/0.01)] = 6.098(2.273 + 2.398) = 28.48 ks (7.9 h) PROBLEM 16.9 A solid is to be dried from 1 kg water/kg dry solids to 0.01 kg water/kg dry solids in a tray dryer consisting of a single tier of 50 trays, each 0.02 m deep and 0.7 m square completely filled with wet material. The mean air temperature is 350 K and the relative humidity across the trays may be taken as constant at 10 per cent. The mean air velocity is 2.0 m/s and the convective coefficient of heat transfer is given by: hc = 14.3G′ 0.8 W/m2 deg K where G′ is the mass velocity of the air in kg/m2 s. The critical and equilibrium moisture contents of the solid are 0.3 and 0 kg water/kg dry solids respectively and the bulk density of the dry solid is 6000 kg/m3 . Assuming that the drying is by convection from the top surface of the trays only, what is the drying time? Solution From Figure 13.4 in Volume 1, the specific volume of moist air at 350 K and 10 per cent relative humidity is 1.06 m3 /kg. Thus: mass velocity, G′ = (2.0/1.06) = 1.88 kg/m2 s and: convective heat transfer coefficient, hc = (14.3 × 1.880.8 ) = 23.8 W/m2 K or 0.0238 kW/m2 K. If it may be assumed that the temperature of the surface is equal to the wet bulb temperature of the air which, from Figure 13.4, is 317 K, then: mean temperature driving force, T = (350 − 317) = 33 deg K. 227 The area of the top surface of the trays is: A = (50 × 0.72 ) = 24.5 m2 From steam tables in the Appendix of Volume 2, the latent heat of vaporisation of water at 317 K is λ = 2395 kJ/kg. The drying rate during the constant drying period is then given by: W = hc AT /λ (equation 16.9) W = (0.0238 × 24.5 × 33)/2359 or: = 0.00816 kg/s The total volume of dry material is: (50 × 0.72 × 0.02) = 0.49 m3 and the mass of dry material is: (0.49 × 6000) = 2940 kg The rate of drying during the constant rate period is then: Rc = 0.00816/(2940 × 24.5) = 1.133 × 10−7 kg water/m2 s kg dry solid In this problem: w1 = 1.0 kg/kg, w = 0.01 kg/kg, wc = 0.3 kg/kg and we = 0 f1 = (w1 − we ) = (1.0 − 0) = 1.0 kg/kg Thus: fc = (wc − we ) = (0.3 − 0) = 0.3 kg/kg f = (w − we ) = (0.01 − 0) = 0.01 kg/kg and: From equation 16.13: Rc = mfc or: 1.133 × 10−7 = m × 0.3 m = 3.78 × 10−7 and: Thus, in equation 16.14, the total drying time is: t = [1/(3.78 × 10−7 × 24.5)][(1.0 − 0.3)/0.3 + ln(0.3/0.01)] = 1.081 × 105 (2.33 + 3.40) = 6.19 × 105 s or 619 ks (172 h) PROBLEM 16.10 Skeins of a synthetic fibre are dried from 46 per cent to 8.5 per cent moisture on a wet basis in a 10 m long tunnel dryer by a countercurrent flow of hot air. The air mass 228 velocity, G′ is 1.36 kg/m2 s and the inlet conditions are 355 K and a humidity of 0.03 kg moisture/kg dry air. The air temperature is maintained at 355 K throughout the dryer by internal heating and, at the outlet, the humidity of the air is 0.08 kg moisture/kg dry air. The equilibrium moisture content is given by: we = 0.25 (per cent relative humidity) and the drying rate by: R = 1.34 × 10−4 G′ 1.47 (w − we )(Hw − H ) kg/s kg dry fibres where H is the humidity of dry air and Hw the saturation humidity at the wet bulb temperature. Data relating w, H and Hw are as follows: w (kg/kg dry fibre) 0.852 0.80 0.60 0.40 0.20 0.093 H (kg/kg dry air) 0.080 0.0767 0.0635 0.0503 0.0371 0.030 Hw (kg/kg dry air) 0.095 0.092 0.079 0.068 0.055 0.049 relative humidity (per cent) 22.4 21.5 18.2 14.6 11.1 9.0 At what velocity should the skeins be passed through the dryer? Solution For G′ = 1.36 kg/m2 s, the rate of drying is given by: R = 1.34 × 10−4 × 1.361.47 (w − we )(Hw − H ) = −2.11 × 10−4 (we − w)(Hw − H ) kg/s kg dry fibre The working is now laid out in tabular form, noting that an inlet moisture content of 46.0 per cent on a wet basis is equivalent to: 46.0/(100 − 46.0) = 0.852 kg/kg dry fibre. and an outlet moisture content of 8.5 per cent on a wet basis is equivalent to: 8.5/(100 − 8.5) = 0.093 kg/kg dry fibre. w (kg/kg dry fibre) relative humidity (per cent) we (kg/kg dry fibre) (= 0.25 RH/100) (w − we ) (kg/kg dry fibre) 0.852 22.4 0.80 21.5 0.60 18.2 0.40 14.6 0.20 11.1 0.093 9.0 0.056 0.054 0.046 0.037 0.028 0.023 0.794 0.746 0.537 0.363 0.172 0.070 229 Hw (kg/kg dry air) 0.095 0.092 H (kg/kg dry air) 0.0800 0.0767 (Hw − H ) (kg/kg dry air) 0.0150 0.0153 R (kg/s kg dry fibre) −0.0112 −0.0103 1/R (kg dry fibre s/kg) −89.0 −97.0 0.079 0.0635 0.068 0.0503 0.055 0.0371 0.049 0.0300 0.0155 0.0177 0.0179 0.0190 −0.0076 −0.0043 −0.0020 −0.0008 −132 −233 −500 −1250 −1/R (s kg dry solid/kg water) Because R/w is equal to the drying time in s, the area under a plot of 1/R and w is the drying time. Such a plot is shown in Figure 16a from which the area between w = 0.852 and w = 0.093 kg/kg is equivalent to 203 s. Area under curve = 202.8 s 800 600 400 200 0 Figure 16a. w = 0.093 0 0.1 w = 0.852 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Moisture content (w kg water/kg dry solids) Drying curve for Problem 16.10 Thus: the velocity at which material should be passed through the dryers = (10/203) = 0.0494 or 230 0.05 m/s 0.9 1.0 SECTION 2-17 Adsorption PROBLEM 17.1 Spherical particles 15 nm in diameter and of density 2290 kg/m3 are pressed together to form a pellet. The following equilibrium data were obtained for the sorption of nitrogen at 77 K. Obtain estimates of the surface area of the pellet from the adsorption isotherm and compare the estimates with the geometric surface. The density of liquid nitrogen at 77 K is 808 kg/m3 . P /P 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 m3 liquid N2 × 106 /kg solid 66.7 75.2 83.9 93.4 108.4 130.0 150.2 202.0 348.0 where P is the pressure of the sorbate and P 0 is its vapour pressure at 77 K. Solution See Volume 2, Example 17.1. PROBLEM 17.2 In a volume of 1 m3 of a mixture of air and acetone vapour, the temperature is 303 K and the total pressure is 100 kN/m2 . If the relative saturation of the air by acetone is 40 per cent, what mass of activated carbon must be added to the space so that at equilibrium the value is reduced to 5 per cent at 303 K? If 1.6 kg carbon is added, what is relative saturation of the equilibrium mixture assuming the temperature to be unchanged? The vapour pressure of acetone at 303 K is 37.9 kN/m2 and the adsorption equilibrium data for acetone on carbon at 303 K are: Partial pressure acetone × 10−2 (N/m2 ) xr (kg acetone/kg carbon) 0 0 5 10 30 50 90 0.14 0.19 0.27 0.31 0.35 Solution The data are plotted in Figure 17a. The final partial pressure of acetone = (0.05 × 37.9 × 1000) = 1895 N/m2 231 From the isotherm, the acetone in the carbon at equilibrium = 0.23 kg/kg carbon. The mass of acetone in the air initially = (1/22.4)(273/303)(0.4×37,900 × 58/105 ) kg where the molecular mass of acetone = 58 kg/kmol. Thus the acetone removed from the air is: = (1/22.4)(273/303)(37,900 × 58/105 )(0.4 − 0.05) = 0.3095 kg. Thus the mass of carbon to be added = (0.3095/0.23) = 1.35 kg If the mass of carbon added is 1.6 kg, then: 1.6x = (1/22.4)(273/303)(37,900 × 58/105 )(0.4 − y) x = 0.221 − 0.553y = 0.221 − 1.458 × 10−3 P . and: From Figure 17a, the two curves intersect at x = 0.203 kg/kg 0.40 xr (kg acetone /kg carbon) 0.30 0.203 kg/kg 0.20 1220 N/m2 0.10 0 2000 4000 6000 Partial pressure of acetone (N/m2) Figure 17a. Adsorption data for Problem 17.2 232 8000 10000 and: P = 1220 N/m2 = 100(1220/37,900) = 3.2 per cent relative saturation. PROBLEM 17.3 A solvent contaminated with 0.03 kmol/m3 of a fatty acid is to be purified by passing it through a fixed bed of activated carbon which adsorbs the acid but not the solvent. If the operation is essentially isothermal and equilibrium is maintained between the liquid and the solid, calculate the length of a bed of 0.15 m in diameter which will allow operation for one hour when the fluid is fed at 1 × 10−4 m3 /s. The bed is free of adsorbate initially and the intergranular voidage is 0.4. Use an equilibrium, fixed-bed theory to obtain the length for three types of isotherm: (a) Cs = 10 C. (b) Cs = 3.0 C 0.3 (use the mean slope). (c) Cs = 104 C 2 (the breakthrough concentration is 0.003 kmol/m3 ). C and Cs refer to concentrations in kmol/m3 in the gas phase and the absorbent, respectively. Solution See Volume 2, Example 17.3. 233 SECTION 2-18 Ion Exchange PROBLEM 18.1 A single pellet of resin is exposed to a flow of solution and the temperature is maintained constant. The take-up of exchanged ion is followed automatically and the following results are obtained: t (min) xr (kg/kg) 2 0.091 4 0.097 10 0.105 20 0.113 40 0.125 60 120 0.128 0.132 On the assumption that the resistance to mass transfer in the external film is negligible, predict the values of xr , the mass of sorbed phase per unit mass of resin as a function of time t, for a pellet of twice the radius. Solution See Volume 2, Example 18.1. 234 SECTION 2-19 Chromatographic Separations PROBLEM 19.1 Describe the principle of separation involved in elution chromatography and derive the retention equation:   (1 − ε) tR = tM 1 + K ε Solution The principles of separation by elution chromatography are considered in Volume 2, Sections 19.1 and 19.2. The derivation of the retention equation is given by equations 19.1–19.8 in Section 19.2.2 of Volume 2. PROBLEM 19.2 In chemical analysis, chromatography may permit the separation of more than a hundred components from a mixture in a single run. Explain why chromatography can provide such a large separating power. In production chromatography, the complete separation of a mixture containing more than a few components is likely to involve two or three columns for optimum economic performance. Why is this? Solution In chemical analysis by the elution technique, only very small samples are injected into the column. The resolution is then given by equation 19.14 and can be very high for most components if the stationary phase is suitably chosen. The equation governs the resolution regardless of the relative amounts of adjacent eluted components. In production chromatography using the elution method, the injected feed bands are much larger in both height (solute concentration) and width (duration of the injection). Equation 19.14 then no longer applies. Components present in high concentrations will give asymmetrical, broader bands than adjacent components of smaller concentration, which may then appear only as unresolved shoulders on the broader bands of their larger neighbours. To optimise the overall separation performance and achieve high throughout and product purity (Section 19.5.4), it may then be best to use one column for the main separation and further columns to complete the resolution of the closest adjacent pairs. The alternative, 235 of resolving all components on one column, might require a much longer column which would not be optimum for other component pairs already resolved on a short column. In the selective adsorption of biological macromolecules more than one type of chromatographic column is often used for a different reason, namely to take advantage of the different types of chromatography such as ion-exchange, affinity and size exclusion, as described in Section 19.6.2. PROBLEM 19.3 By using the chromatogram shown in Figure 19.3, of Volume 2, show that k1′ = 3.65, k2′ = 4.83, α = 1.32, Rs = 1.26 and N = 500. Show also that, if ε = 0.8 and L = 1.0 m, then K1 = 14.6, K2 = 19.3 and H = 2.0 mm. Calculate the ratio of plate height to particle diameter to confirm that the column is inefficient, as might be anticipated from the wide bands shown in Figure 19.3. It may be assumed that the particle size is that of a typical GC column, as given in Table 19.3 in Volume 2. Solution See Volume 2, Example 19.1 PROBLEM 19.4 Suggest one or more types of chromatography to separate each of the following mixtures: (a) (b) (c) (d) α- and β-pinenes blood serum proteins hexane isomers purification of cefonicid, a synthetic β-lactam antibiotic. Solution Examination of Table 19.2 in Volume 2 suggests that the following types of chromatograph might be suitable for separating the various mixtures: (a) (b) (c) (d) Gas — liquid (GLC) or reverse phase (RP-BPC) chromatography Ion — exchange (IEC) or affinity (AC) chromatography Gas chromatography (GLC or GSC) Reverse phase (RP-BPC) chromatography It may be noted that these techniques are not exclusive and reference may be made to Section 19.6.2. 236 SECTION 3-1 Reactor Design — General Principles PROBLEM 1.1 A preliminary assessment of a process for the hydrodealkylation of toluene is to be made. The reaction involved is: − ⇀ C6 H5 · CH3 + H2 − ↽ −− − − C6 H6 + CH4 The feed to the reactor will consist of hydrogen and toluene in the ratio 2H2 : 1C6 H5 · CH3 . (a) Show that with this feed and an outlet temperature of 900 K, the maximum conversion attainable, that is the equilibrium conversion, is 0.996 based on toluene. The equilibrium constant of the reaction at 900 K is Kp = 227. (b) Calculate the temperature rise which would occur with this feed if the reactor were operated adiabatically and the products were withdrawn at equilibrium. For reaction at 900 K, −H = 50,000 kJ/kmol. (c) If the maximum permissible temperature rise for this process is 100 deg K suggest a suitable design for the reactor. Specific heat capacities at 900 K (kJ/kmol K): C6 H6 = 198, C6 H5 CH3 = 240, CH4 = 67, H2 = 30. Solution (a) If, on the basis of 1 kmol toluene fed to the reactor, α kmol react, then a material balance is: C6 H5 · CH3 H2 C 6 H6 CH4 Total In (kmol) (kmol) Out (mole fraction) (partial pressure) 1 2 0 0 3 (1 − α) (2 − α) α α 3 (1 − α)/3 (2 − α)/3 α/3 α/3 1 (1 − α)P /3 (2 − α)P /3 αP /3 αP /3 P 237 If equilibrium is reached at 900 K at which Kp = 227, then: Kp = (PC6 H6 × PCH4 )/(PC6 H5 ·CH3 × PH2 ) = α 2 /(1 − α)(2 − α) = 227 Thus: (1 − α)(2 − α) = α 2 /227 = 0.00441α 2 0.996α 2 − 3α + 2 = 0 or: Because α cannot exceed 1, only one root of this quadratic equation is acceptable and: α = 0.996 It may be noted that, if a feed ratio of 1C6 H5 · CH3 : 1H2 had been used, the conversion at equilibrium would have been only 0.938, illustrating the advantage of using an excess of hydrogen, the less expensive reactant, rather than the stoichiometric proportion. (b) For the adiabatic reaction, a thermodynamically equivalent path may be considered such that the reactants enter at a temperature Tin and the heat produced by the reaction is used to heat the product mixture to 900 K. On the basis of 1 kmol toluene fed to the reactor and neglecting any changes in the heat capacity and enthalpy of reaction with temperature, then: Heat released in the reaction = increase in sensible heat of the gas or: α(−H ) = ni (Cpi (900 − Tin ) Thus: (0.996 × 50,000) = [(0.996 × 198) + (0.996 × 67) + ((1 − 0.996) × 240) + ((1 − 0.996) × 30)](900 − Tin ) 49,800 = 295(900 − Tin ) and: (900 − Tin ) = 169 deg K . This is the temperature rise which would occur if the reactor were designed and operated as a single adiabatic stage. (c) If the maximum permissible temperature rise in the process is 100 deg K, then a suitable design for the reactor would be a two-stage arrangement with inter-stage cooling, where: 1st stage: inlet = 800 K outlet = (800 + 100) = 900 K. 2nd stage: inlet = 900 − (169 − 100) = 831 K outlet = 900 K. although the inlet temperatures would be adjusted in order to optimise the process. 238 PROBLEM 1.2 In a process for the production of hydrogen required for the manufacture of ammonia, natural gas is to be reformed with steam according to the reactions: − ⇀ CH4 + H2 O − ↽ −− − − CO + 3H2 , − ⇀ CO + H2 O − ↽ −− − − CO2 + H2 , Kp (at 1173 K) = 1.43 × 1013 N2 /m4 Kp (at 1173 K) = 0.784 The natural gas is mixed with steam in the mole ratio 1CH4 : 5H2 O and passed into a catalytic reactor which operates at a pressure of 3 MN/m2 (30 bar). The gases leave the reactor virtually at equilibrium at 1173 K. (a) Show that for every 1 mole of CH4 entering the reactor, 0.950 mole reacts, and 0.44 mole of CO2 is formed. (b) Explain why other reactions such as: − ⇀ CH4 + 2H2 O − ↽ −− − − CO2 + 4H2 need not be considered. (c) By considering the reaction: − ⇀ 2CO − ↽ −− − − CO2 + C 2 for which Kp = PCO2 /PCO = 2.76 × 10−7 m2 /N at 1173 K, show that carbon deposition on the catalyst is unlikely to occur under the operating conditions. (d) What will be the effect on the composition of the exit gas of increasing the total pressure in the reformer? Why, for ammonia manufacture, is the reformer operated at 3 MN/m2 (30 bar) instead of at a considerably lower pressure? The reforming step is followed by a shift conversion: − ⇀ CO + H2 O − ↽ −− − − CO2 + H2 , absorption of the CO2 , and ammonia synthesis according to the reaction: − ⇀ N2 + 3H2 − ↽ −− − − 2NH3 Solution (a) If, for 1 kmol of CH4 fed to the reactor, α kmol is converted and β kmol of CO2 is formed, then at a total pressure, P , a material balance is: In (kmol) (kmol) CH4 H2 O CO H2 CO2 1 5 0 0 0 Total 6 Out (mole fraction) (partial pressure) (1 − α) (5 − α − β) (α − β) (3α + β) β (1 − α)/(6 + 2α) (5 − α − β)/(6 + 2α) (α − β)/(6 + 2α) (3α + β)/(6 + 2α) β/(6 + 2α) (1 − α)P /(6 + 2α) (5 − α − β)P /(6 + 2α) (α − β)P /(6 + 2α) (3α + β)P /(6 + 2α) βP /(6 + 2α) 6 + 2α 1 P 239 based on the stoichiometry of the reactions: − ⇀ CH4 + H2 O − ↽ −− − − CO2 + 3H2 − ⇀ CO + H2 O − ↽ −− − − CO2 + H2 and: (I) (II) For reaction I: KpI = (PCO × PH32 )/(PCH4 × PH2 O ) = (α − β)(3α + β)3 P 2 (1 − α)(5 − α − β)(6 + 2α)2 If the units of P are bar and KpI = 1430 then: (α − β)(3α + β)3 = (1430/302 ) = 1.59 (1 − α)(5 − α − β)(6 + 2α)2 Substituting α = 0.950 and β = 0.44, then the left-hand side of this equation = 1.61 and hence the equation is satisfied. For reaction II: KpII = (PCO2 × PH2 )/(PCO × PH2 O ) = β(3α + β) = 0.784 (α − β)(5 − α − β) Again, substituting α = 0.950 and β = 0.44, the left-hand side of this equation = 0.786 and hence the equation is satisfied. (b) The reaction: − ⇀ CH4 + 2H2 O − ↽ −− − − CO2 + 4H2 (III) need not be considered because it is not a further independent reaction. It can in fact be obtained by the addition of reactions I and II and correspondingly: KpI × KpII = KpIII . (c) If solid and pure carbon is to be in equilibrium with the product gas mixture: − ⇀ 2CO − ↽ −− − − CO2 + C (IV) then the equilibrium constant for this reaction, which can apply only if solid C is present, 2 is: KpIV = PCO2 /PCO . With P in bar, KpIV = 0.0276, and from the material balance:       1 1 β(6 + 2α) 0.44(6 + 2 × 0.950) PCO2 = = = 0.45 2 2 2 (α − β) P (0.950 − 0.44) 30 PCO Thus it may be seen that, in the reaction mixture the partial pressure of CO2 , PCO2 , relative to PCO , greatly exceeds the value that would be required to satisfy KPIV and therefore solid C cannot co-exist at equilibrium. The high value of PCO2 may be thought of as driving reaction IV completely to the left. (d) It may be noted that reaction I involves an increase in the number of moles: CH4 + H2 O − − ⇀ ↽ −− − − CO + 3H2 (2 moles) (4 moles) Qualitatively, Le Chatelier’s principle indicates that increasing the pressure will tend to decrease the fractional conversion of CH4 at equilibrium. At first it may therefore seem surprising that the reformer should be operated at a pressure as high as 30 bar. 240 The reason for this lies in the relatively high cost of gas compression, which depends on the ratio of the inlet and outlet pressures. The methane feed to the reformer will probably be available at a pressure much above 1 bar and similarly with the steam. Le Chatelier’s principle indicates also that excess steam, as used in practice, will favour a higher conversion of methane compared with the stoichiometric proportions of the reactants. The same conclusion follows quantitatively from KpI for which the equation involves the total pressure P . At a pressure of 30 bar and with excess steam the fractional conversion of methane in the reformer is reasonably satisfactory. The high pressure of 30 bar will favour the removal of carbon dioxide, following the shift reaction: CO + H2 O ⇀ ↽ CO2 + H2 , and reduce the cost of compressing the purified hydrogen to a value, typically in the range 50–200 bar, required for ammonia synthesis. PROBLEM 1.3 An aromatic hydrocarbon feedstock consisting mainly of m-xylene is to be isomerised catalytically in a process for the production of p-xylene. The product from the reactor consists of a mixture of p-xylene, m-xylene, o-xylene and ethylbenzene. As part of a preliminary assessment of the process, calculate the composition of this mixture if equilibrium were established over the catalyst at 730 K. Equilibrium constants at 730 K are: − ⇀ m-xylene − ↽ −− − − p-xylene, − ⇀ m-xylene − ↽ −− − − o-xylene, Kp = 0.45 Kp = 0.48 − ⇀ m-xylene − ↽ −− − − ethylbenzene, Kp = 0.19 Why is it unnecessary to consider reactions such as: kf − ⇀ o-xylene − ↽ −− − − p-xylene? kr Solution The following equilibria at 730 K and total pressure P may be considered, noting that the positions of the equilibria are not dependent on the value of P . − ⇀ m-xylene − ↽ −− − − ethylbenzene: KE = (PE /PM ) = (yE P )/(yM P ) = (yE /yM ) (i) − ⇀ m-xylene − ↽ −− − − o-xylene: KO = (PO /PM ) = (yO P )/(yM P ) = (yO /yM ) (ii) − ⇀ m-xylene − ↽ −− − − p-xylene: KP = (Pp /PM ) = (yP P )/(yM P ) = (yP /yM ) 241 (iii) where PM , PE , PO and PP are the partial pressures of the various components and yM , yE , yO and yP are the mole fractions. Thus: (yE /yM ) = 0.19, (yO /yM ) = 0.48 and (yP /yM ) = 0.45 Noting that: yE + yO + yP + yM = 1 then: yM = 0.473, yE = 0.30, yO = 0.225 and yP = 0.212 . It is unnecessary to consider other equations for the set of equilibrium such as: − ⇀ o-xylene − ↽ −− − − p-xylene − ⇀ o-xylene − ↽ −− − − ethylbenzene − ⇀ p-xylene − ↽ −− − − ethylbenzene (i) (ii) (iii) because these are not independent equations, since they can be derived from combinations of the equations already considered. For example, subtracting equation (ii) from equation (iii) gives equation (iv) and, correspondingly, dividing the equilibrium constant KP by KO gives:    PP PM PP = = KP′ the equlibrium constant for equlibrium I. (iv) KP /KO = PM PO PC Thus the complete set of equilibrium may be depicted as: Ethylbenzene m -xylene o -xylene p -xylene Any three of these equations may be taken as independent relations provided that the ones chosen involve all the species present. PROBLEM 1.4 The alkylation of toluene with acetylene in the presence of sulphuric acid is carried out in a batch reactor. 6000 kg of toluene is charged in each batch, together with the required amount of sulphuric acid and the acetylene is fed continuously to the reactor under pressure. Under circumstances of intense agitation, it may be assumed that the liquid is always saturated with acetylene, and that the toluene is consumed in a simple pseudo first-order reaction with a rate constant of 0.0011 s−1 . 242 If the reactor is shut down for a period of 900 s (15 min) between batches, determine the optimum reaction time for the maximum rate of production of alkylate, and calculate this maximum rate in terms of mass P toluene consumed per unit time. Solution The toluene is consumed in a simple pseudo first-order reaction with a rate constant of 0.0011 s−1 It may be assumed that the volume of the liquid phase does not change appreciably as the reaction proceeds, although, in practice there will be some departure from this assumption. If the reaction is considered complete at the stage when 1 kmol C2 H2 (molecular mass = 26 kg/kmol) has been added to 1 kmol C7 H8 (molecular mass = 92 kg/kmol), then per 1 kmol of toluene, the total mass in the reactor will have increased from 92 kg initially to 118 kg of product having a mass density similar to that of the original toluene. For a first-order reaction, the integrated form of the rate equation for a constant volume batch reactor, from Table 1.1 in Volume 3, is: t= 1 C0 ln k1 (C0 − x) Writing this in terms of the fractional conversion α = x/C0 , then: t= 1 1 ln k1 (1 − α) To find the reaction time corresponding to the maximum production rate the method outlined in Section 1.6.3 of Volume 3 is adopted as follows. From the relation:   1 1 ln t= 0.0011 (1 − α) the following data are obtained: Fractional conversion, α Time (s) 0.3 324 0.4 464 0.5 630 0.6 833 0.65 954 0.70 1095 0.75 1260 0.80 1463 These data are plotted in Figure 1a. A tangent to the curve is then drawn from the point on the time axis at −(15 × 60) = −900 s. The reaction time at the tangent point is 1050 s, at which the fractional conversion is 0.68. Thus toluene is consumed at the rate of: (6000 × 0.68) = 2.09 kg/s. (1050 + 900) and the maximum production rate in terms of toluene consumed = 2.09 kg/s (7530 kg/h) 243 1.0 0.9 0.8 0.68 0.7 0.6 Fractional Conversion 0.5 a 1050 0.4 0.3 0.2 0.1 −900 0 1000 2000 Time (s) Figure 1a. Graphical construction for Problem 1.4 PROBLEM 1.5 Methyl acetate is hydrolysed by water according to the reaction: kf − ⇀ CH3 · COOCH3 + H2 O − ↽ −− − − CH3 · COOH + CH3 OH kr A rate equation is required for this reaction taking place in dilute solution. It is expected that reaction will be pseudo first-order in the forward direction and second-order in reverse. The reaction is studied in a laboratory batch reactor starting with a solution of methyl acetate and with no products present. In one test, the initial concentration of methyl acetate was 0.05 kmol/m3 and the fraction hydrolysed at various times subsequently was: Time (s) Fractional conversion 0 0 1350 0.21 3060 0.43 5340 0.60 7740 0.73 ∞ 0.90 (a) Write down the rate equation for the reaction and develop its integrated form applicable to a batch reactor. (b) Plot the data in the manner suggested by the integrated rate equation, confirm the order of reaction and evaluate the forward and reverse rate constants, kf and kr . 244 Solution At the stage where χ mole of methyl acetate has been converted per unit volume of reaction mixture, the concentrations of the various species are: kf − ⇀ CH3 · COOCH3 + H2 O − ↽ −− − − CH3 · COOH + CH3 OH kr χ χ (C0 − χ) If the forward reaction is pseudo first order and the reverse reaction second order, then, as discussed in Sections 1.4.4 and 1.4.5 in Volume 3, the rate equation may be written as: R= dχ = kf (C0 − χ) − kr χ 2 dt (i) At equilibrium the value of χ is χe and the net rate of advancement of the reaction is dχ = 0. zero, or dt 0 = kf (C0 − χe ) − kr χe2 Thus: or: Substituting χe2 kf = Kc , the equilibrium constant. = (C0 − χe ) kr (ii) kf for kr in the rate equation, then: Kc dχ kf 2 = kf (C0 − χ) − χ dt Kc and hence, the integrated form applicable to a batch reactor is given by:  1 χ dχ t= kf 0 (C0 − χ) − χ 2 /Kc (iii) It is convenient at this stage to introduce the numerical values. The experimental data indicate that, after a long period of time such that equilibrium is established, the fractional conversion is given by: αe = 0.90 and χe = αe C0 = (0.90 × 0.05) = 0.045 kmol/m3 and, from equation (ii): Kc = 0.0452 = 0.405 (0.05 − 0.045) For the experimental data given, the integral to be evaluated is:   1 χ 1 χ dχ dχ t= = kf 0 kf 0 (0.05 − χ − 2.47χ 2 ) χ2 (0.05 − χ) − 0.405 245 The factors for the denominator are:  1 χ dχ t= kf 0 2.47(0.45 + χ)(0.045 − χ) Expressing this in partial fractions, then:   χ 1 1 1 · dχ t= + 1.222kf 0 (0.45 + χ) (0.045 − χ) and: t= 1 χ [ln(0.45 + χ) − ln(0.045 − χ)]0 1.222kf t= (0.45 + χ) 1 + constant. ln 1.222kf (0.045 − χ) (iv) The experimental data are plotted in the form indicated by equation (iv), noting that χ = αC0 = 0.05α, or: ln (0.45 + χ) is plotted against t, (0.045 − χ) This should give a straight line of slope 1.222 kf . The data are processed as follows and plotted in Figure 1b. Time (s) Fractional conversion 0 0 1350 0.21 3060 0.43 5340 0.60 7740 0.73 ∞ 0.90 χ(kmol/m3 ) (0.45 + χ) ln (0.045 − χ) 0 0.0105 0.0215 0.0300 0.0365 0.0450 2.303 2.591 3.00 3.466 4.047 ∞ The slope of the line = 0.224 × 10−3 s−1 Thus: kf = (0.224 × 10−3 /1.222) = 0.183 × 10−3 s−1 From equation (ii), Kc = kf /kr and: kr = kf /Kc = (0.183 × 10−3 /0.405) = 0.452 × 10−3 m3 /kmol s. As an alternative to the introduction of the numerical values after equation (iii), it is possible to proceed with the integration of equation (iii) algebraically as follows:  dχ 1 χ t= kf 0 χ2 (C0 − χ) − Kc  χ Kc dχ =    0.5 kf 0 Kc + Kc (1 + 4C0 /Kc ) −Kc + Kc (1 + 4C0 /Kc )0.5 +χ −χ 2 2 246 ln [(0.45 + c)/(0.045 − c)] 4.0 3.0 slope = 0.224 × 10−3 s−1 2.0 0 1000 2000 3000 4000 5000 6000 Time, t (s) Figure 1b. = Graphical construction for Problem 1.5 1 kf (1 + 4C0 /Kc )0.5  χ 0 ⎡ ⎢ 1 ⎢ × ⎢  ⎣ Kc + Kc (1 + 4C0 /Kc )0.5 +χ 2 ⎤ ⎥ 1 ⎥ ⎥  ⎦ −Kc + Kc (1 + 4C0 /Kc )0.5 −χ 2 ⎡   ⎤χ Kc + Kc (1 + 4C0 /Kc )0.5 +χ ⎥ ⎢ 1 2 ⎢ ⎥ = ⎢ln  ⎥ 0.5 ⎦ kf (1 + 4C0 /Kc )0.5 ⎣ −Kc + Kc (1 + 4C0 /Kc ) −χ 2 0 ⎧   0.5 Kc + Kc (1 + 4C0 /Kc ) ⎪ ⎪ ⎪ +χ ⎨ 1 2 = ln   kf (1 + 4C0 /Kc )0.5 ⎪ −Kc + Kc (1 + 4C0 /Kc )0.5 ⎪ ⎪ −χ ⎩ 2 ⎫ ⎪ ⎪ ⎪ [1 + (1 + 4C0 /Kc )0.5 ] ⎬ − ln [−1 + (1 + 4C0 /Kc )0.5 ] ⎪ ⎪ ⎪ ⎭ +  247 7000 8000 0.5 Introducing the numerical values and noting that (1 + 4C0 /Kc )   4 × 0.05 0.5 = 1+ 0.405 = 1.222, then:  1 χ dχ kf 0 χ2 (0.05 − χ) − 0.405  ⎧  ⎫ 0.405 + 0.405 × 1.222 ⎪ ⎪ ⎪ ⎪ +χ ⎨ 1 [1 + 1.222] ⎬ 2  − ln ln  = −0.405 + 0.405 × 1.222 1.222 kf ⎪ [−1 + 1.222] ⎪ ⎪ ⎪ ⎩ ⎭ −χ 2   0.45 + χ 1 − ln 10 ln = 1.222 kf 0.045 − χ This approach is rather more complicated, although it has the advantage that it permits exploration of the effect of the change in numerical values. PROBLEM 1.6 Styrene is to be produced by the catalytic dehydrogenation of ethylbenzene according to the reaction: − ⇀ C6 H5 · CH2 · CH3 − ↽ −− − − C6 H5 · CH: CH2 + H2 The rate equation for this reaction takes the form:   1 R = k PEt − PSt PH Kp where PEt , PSt and PH are partial pressures of ethylbenzene, styrene and hydrogen respectively. The reactor will consist of a number of tubes each of 80 mm diameter packed with catalyst with a bulk density of 1440 kg/m3 . The ethylbenzene will be diluted with steam, the feed rates are ethylbenzene 1.6 × 10−3 kmol/m2 s and steam 29 × 10−3 kmol/m2 s. The reactor will be operated at a mean pressure of 120 kN/m2 (1.2 bar) and the temperature will be maintained at 833 K (560◦ C) throughout. If the fractional conversion of ethylbenzene is to be 0.45, estimate the length and number of tubes required to produce 0.231 kg/s (20 tonne styrene/day). At 833 K, k = 6.6 × 10−11 kmol/(N/m2 ) s kg catalyst (= 6.6 × 10−6 kmol (kg catalyst bar s) and Kp = 1.0 × 104 N/m2 . Solution The fractional conversion of ethylbenzene is 0.45 or 1 kmol of ethylbenzene feed produces 0.45 kmol styrene. Taking the molecular mass of styrene as 104 kg/kmol, then: 248 1 kmol ethylbenzene feed produces (0.45 × 104) = 46.8 kg styrene. Hence, the feed rate of ethylbenzene required = 1 × (0.231/46.8) = 0.00495 kmol/s. The feed rate of ethylbenzene per unit cross-sectional area = 1.6 × 10−3 kmol/m2 s. Thus: cross-sectional area of all the tubes in the reactor = (0.00495/0.0016) = 3.09 m2 cross-sectional area of one tube = (π × 0.0802 )/4 = 0.00503 m2 number of tubes required = (3.09/0.00503) = 615 and: In estimating the length of the tubes, the mass of catalyst, W , is calculated from the design equation for a tubular reactor as: W = FA  αf dα Rm A 0 (equation 1.35, Volume 3) although, as a first stage, the rate of reaction, Rm A , must be expressed in terms of the fractional conversion, α. Basis: 1 kmol ethylbenzene entering the reactor at a total pressure of P bar. In (kmol) (kmol) ethylbenzene 1.0 1−α styrene hydrogen steam — — 18.1 α α 18.1 TOTAL 19.1 19.1 + α Thus: Noting that Rm = k  Out (mole fraction) (partial pressure) (1 − α) (19.1 + α) α/(19.1 + α) α/(19.1 + α) 18.1/19.1 + α) (1 − α)P (19.1 + α) αP /(19.1 + α) αP /(19.1 + α) 18.1P /(19.1 + α) 1 P (1 − α)P P2 α2 − 2 (19.1 + α) (19.1 + α) KP P = 1.2 bar  and Kp = 0.1 bar, then: Rm = (1.2k /(19.1 + α)2 )(19.1 − 18.1α − 13α 2 ) In equation 1.35: 1 W = FA 1.2k  0 αf (19.1 + α)2 dα (19.1 − 18.1α − 13α 2 ) where αf = 0.45. 249 Thus: W 1 = FA 1.2k  1 = 1.2k  αf 0 αf 0  −0.0769 +  −0.0769 +  28.18 + 2.832α · dα 1.469 − 1.392α − α 2  10.79 7.96 · dα + (0.702 − α) (2.094 + α) 1 = [−0.0769α − 10.79 ln(0.702 − α) + 7.96 ln(2.094 + α)]00.45 1.2k = 12.57/1.2k Noting that FA = 0.00495 kmol/s and k = 6.6 × 10−6 kmol/kg catalyst bar s, then: (W/0.00495) = 12.57/(1.2 × 6.6 × 10−6 ) and: W = 7856 kg Total volume of catalyst required = (7856/1440) = 5.46 m3 Volume of catalyst in each tube = (5.46/615) = 0.00887 m3 Thus: Length of each tube = 0.00887/[(π × 0.0802 )/4] = 1.76 m PROBLEM 1.7 Ethyl formate is to be produced from ethanol and formic acid in a continuous flow tubular reactor operated at a constant temperature of 303 K (30◦ C). The reactants will be fed to the reactor in the proportions 1 mole HCOOH: 5 moles C2 H5 OH at a combined flowrate of 0.0002 m3 /s (0.72 m3 /h). The reaction will be catalysed by a small amount of sulphuric acid. At the temperature, mole ratio, and catalyst concentration to be used, the rate equation determined from small-scale batch experiments has been found to be: R = k CF2 where: R is formic acid reacting/(kmol/m3 s) CF is concentration of formic acid kmol/m3 , and k = 2.8 × 10−4 m3 /kmol s. The density of the mixture is 820 kg/m3 and this may be assumed constant throughout. Estimate the volume of the reactor required to convert 70 per cent of the formic acid to the ester. If the reactor consists of a pipe of 50 mm i.d. what will be the total length required? Determine also whether the flow will be laminar or turbulent and comment on the significance of this in relation to the estimate of reactor volume. The viscosity of the solution is 1.4 × 10−3 N s/m2 . Solution Although the ethanol fed to the reactor is partly consumed in the reaction, the rate equation indicates that the rate of reaction depends only on the concentration of the formic acid. 250 R = k CF2 Thus: In this liquid phase reaction, it may be assumed that the mass density of the liquid is unaffected by the reaction, allowing the material balance for the tubular reactor to be applied on a volume basis (Section 1.7.1, Volume 3) with plug flow.  χf Vt dχ = Thus: (equation 1.37) v R 0 R = k CF2 = k (C0 − χ)2 , In this case: where C0 is the concentration of formic acid in the feed.    χf 1 Vt 1 dχ 1 Thus: = = − v k (C0 − χ)2 k (C0 − χf ) C0 0 In terms of fractional conversion χf = αf C0 , then:     1 Vt 1 1 αf = −1 = v k C0 (1 − αf ) k C0 1 − αf For HCOOH, the molecular mass = 46 kg/kmol. For C2 H5 OH, the molecular mass = 46 kg/kmol. Thus: 1 kmol HCOOH is present in (1 × 46) + (5 × 46) = 276 kg feed mixture. (276/820) = 0.337 m3 feed mixture. or: C0 = (1/0.337) = 2.97 kmol/m3 Thus: The volume of the reactor required to convert a fraction of the feed of 0.7 is given by: Vt = (0.72/3600)(1/(2.8 × 10−4 × 2.97))(0.7/(1 − 0.7)) = 0.561 m3 The equivalent length of a 50 mm ID pipe is then: 0.561/[(π/4)0.0502 ] = 286 m say 29 lengths, each of 10 m length, connected by U-bends. The mean velocity in the tube, u = (0.72/3600)/(π/4)0.0502 = 0.102 m/s and, the Reynolds Number is then: Re = uρd/μ = (0.102 × 820 × 0.050)/(1.4 × 10−3 ) = 2980 confirming turbulent flow and the validity of the assumed plug flow. 251 PROBLEM 1.8 Two stirred tanks are available, one 100 m3 in volume, the other 30 m3 in volume. It is suggested that these tanks be used as a two-stage CSTR for carrying out a liquid phase reaction A + B → product. The two reactants will be present in the feed stream in equimolar proportions, the concentration of each being 1.5 kmol/m3 . The volumetric flowrate of the feed stream will be 0.3 × 10−3 m3 /s. The reaction is irreversible and is of first order with respect to each of the reactants A and B, i.e. second order overall, with a rate constant 1.8 × 10−4 m3 /kmol s. (a) Which tank should be used as the first stage of the reactor system, the aim being to effect as high an overall conversion as possible? (b) With this configuration, calculate the conversion obtained in the product stream leaving the second tank after steady conditions have been reached. If there is any doubt regarding which tank should be used as the first stage, the conversions for both configurations should be calculated and compared. Accurate calculations may be required in order to distinguish between the two. Solution (a) From the treatment of reactor output given in Volume 3, Section 1.9, it may be seen that placing the smaller tank first would be an advantage. Bringing the feed into the smaller tank will give a relatively high concentration of reactants in this tank. If the overall order of the reaction is greater than 1 (the overall order being 2 in this example), high concentrations at some stage in the system will lead to higher rates of reaction and larger final fractional conversion compared with any alternative configuration. The case for preferring the smaller tank first is difficult to sustain however by strictly logical argument. The most convincing way is to calculate the fractional conversion for both configurations as the problem suggests. For a flow of feed of v m3 /s with a concentration of either A or B — the concentrations of A and B are equal throughout — of C0 kmol/m3 , with the flow passing through tanks of volumes V1 and V2 , then a steady-state mass balances gives: Tank 1 : vC0 − v(C0 − χ1 ) − V1 k (C0 − χ1 )2 = 0 (in) (out) (reaction) (accumulation) or: vχ1 = V1 k (C0 − χ1 )2 If χ1 = α1 C0 , where α is the fractional conversion, then: vα1 C0 = V1 k C02 (1 − α1 )2 or: Tank 2 : or: or: α1 = (V1 /v)k C0 (1 − α1 )2 v(C0 − χ1 ) − v(C0 − χ2 ) = V2 k (C0 − χ2 )2 = 0 v(χ2 − χ1 ) = V2 k (C0 − χ2 )2 (α2 − α1 ) = (V2 /v)k C0 (1 − α2 )2 252 In both cases, v = 0.3 × 10−3 m3 /s, k = 1.8 × 10−4 m3 /kmol s and C0 = 1.5 kmol/m3 . For configuration 1: V1 = 30 m3 , V2 = 100 m3 α1 = (30/0.3 × 10−3 )1.8 × 10−4 × 1.5(1 − α1 )2 Tank 1 : α1 = 27(1 − α1 )2 or: α12 − 2.03704α1 + 1 = 0 Thus: Noting that α1 < 1, then: α1 = 0.8252 Tank 2 : or: (α2 − α1 ) = (100/0.3 × 10−3 ) × 1.8 × 10−4 × 1.5(1 − α2 )2 α2 − 0.8252 = 90(1 − α2 )2 α22 − 2.01111α2 + 1.009169 = 0 Thus: Again, noting that α2 < 1 then: α2 = 0.961 For configuration 2: V1 = 100 m3 , V2 = 30 m3 . Tank 1 : and: α1 = (100/0.3 × 10−3 )1.8 × 10−4 × 1.5(1 − α1 )2 α12 − 2.01111α1 + 1 = 0 Noting that α1 < 1, then: α1 = 0.9000 Tank 2 : (α2 − α1 ) = (30/0.3 × 10−3 )1.8 × 10−4 × 1.5(1 − α2 )2 α2 − 0.9 = 27(1 − α2 )2 and: α22 − 2.03704α2 + 1.03333 = 0 Thus: α2 = 0.955 It may be noted that: (a) There is seemingly very little difference between the final conversions calculated for the two cases and it may be concluded that, although the final conversion will depend on the total volume of the tanks, it is not very sensitive to how that volume is distributed. (b) In operating a reactor, however, often it is the fraction which is unreacted which is important for downstream separation. In this problem, these values are: configuration 1 = 0.039 kmol/m3 configuration 2 = 0.045 kmol/m3 . 253 PROBLEM 1.9 The kinetics of a liquid-phase chemical reaction are investigated in a laboratory-scale continuous stirred-tank reactor. The stoichiometric equation for the reaction is A → 2P and it is irreversible. The reactor is a single vessel which contains 3.25 × 10−3 m3 of liquid when it is filled just to the level of the outflow. In operation, the contents of the reactor are well stirred and uniform in composition. The concentration of the reactant A in the feed stream is 0.5 kmol/m3 . Results of three steady-state runs are: Feed rate (m3 /s × 105 ) 0.100 0.800 0.800 Concentration of P in outflow (kmol/m3 ) Temperature (K) (◦ C) 298 298 333 25 25 60 0.880 0.698 0.905 Determine the constants in the rate equation: p RA = A exp(−E/RT )CA . Solution The rate equation: p RA = A exp(−E/RT )CA involves three parameters: p — the order of reaction, A — the frequency factor and E — the energy of activation. Substituting the three sets of data to RA , CA and T in this equation provides three simultaneous equations in which there are three unknowns, A, E and p, and hence, in principle, a solution may be found. In practice more than three sets of experimental data is desirable in order to validate the experimental technique and the consistency of the measurements. For a tank volume, V m3 , and a flow of feed of v m3 /s of concentration of A of C0 , a mass balance over the tank at steady state for material A gives: vC0 − vCA − V RA = 0 (in) (out) (reaction) or: RA = (v/V )(C0 − CA ) It may be noted that the experimental data give the concentration of product P but not that of reactant A, although these may be linked as follows. For the reaction: A −−−→ 2P then, for 1 m3 of feed solution, CP kmol of product is formed from (CP /2) kmol of A leaving (C0 − CP /2) kmol of unreacted A or: CA = (C0 − CP /2) 254 Using this equation and the mass balance, the following data may be obtained: Run Feed rate (m3 /s × 105 ) Temperature (K) Concentration, CA (kmol/m3 ) Rate of reaction, RA (kmol/m3 s × 103 ) 1 2 3 0.100 0.800 0.800 298 298 333 0.06 0.151 0.048 0.135 0.859 1.114 In the tests, Runs 1 and 2 were carried out at the same temperature and hence: (A exp(−E/RT ))1 = (A exp(−E/RT ))2 and: Thus: or: (RA2 /RA1 ) = (CA2 /CA1 )p p = ln(RA2 /RA1 )/ ln(CA2 /CA1 ) p = ln((0.859 × 10−3 )/(0.135 × 10−3 ))/ ln(0.151/0.06) = 2.005 Taking the order of reaction to be exactly 2, although there is no absolute necessity for the order to be an integer, the rate constants for each of the runs, 1 and 2, may now be calculated. Run 1: Run 2: 2 k1 = RA1 /CA1 = (0.135 × 10−3 )/0.062 = 0.0375 m3 /kmol s. 2 k2 = RA2 /CA2 = (0.859 × 10−3 )/0.1512 = 0.0377 m3 /kmol s. giving a mean value at 298 K = 0.0376 m3 /kmol s. Similarly, for Run 3 at 333 K with an order of reaction of p = 2: 2 k3 = RA3 /CA3 = (1.114 × 10−3 )/0.0482 = 0.495 m3 /kmol s At 298 K: k12 = A exp(−E/298R) and ln k12 = ln A − E/298R k3 = A exp(−E/333R) and ln k3 = ln A − E/333R At 333 K: Subtracting: k3 E = ln k12 R Thus:  1 1 − 298 333  where R = 8.314 kJ/kmol K E = 8.314 ln(0.495/0.0376)/((1/298) − (1/333)) = 60,700 kJ/kmol 255 Using the data from Run 3: k = A exp(−E/RT ) 0.495 = A exp(−60,700/(8.314 × 333)) or: A = 1.66 × 109 m3 /kmol s or: The complete rate equation is therefore: RA = 1.66 × 109 exp(−60.7 × 106 /RT )CA2 PROBLEM 1.10 A reaction A + B → P, which is first-order with respect to each of the reactants, with a rate constant of 1.5 × 10−5 m3 /kmol s, is carried out in a single continuous flow stirredtank reactor. This reaction is accompanied by a side reaction 2B → Q, where Q is a waste product, the side reaction being second-order with respect to B, with a rate constant of 11 × 10−5 m3 /kmol s. An excess of A is used for the reaction, the feed rates to the tank being 0.014 kmol/s of A and 0.0014 kmol/s of B. Ultimately reactant A is recycled whereas B is not. Under these circumstances the overflow from the tank is at the rate of 1.1 × 10−3 m3 /s, while the capacity of the tank is 10 m3 . Calculate (a) the fraction of B converted into the desired product P, and (b) the fraction of B converted into Q. If a second tank of equal capacity becomes available, suggest with reasons in what manner it might be incorporated (a) if A but not B is recycled as above, and (b) if both A and B are recycled. Solution The reactions are: A + B −−−→ P, Rate, with respect to A or B = kP CA CB 2B −−−→ Q, Rate with respect to B but not Q = kQ CB2 For flows of a0 kmol/s and b0 kmol/s of A and B respectively into a vessel of volume V with an outflow of v kmol/s, then a mass balance at steady-state gives: Component A: a0 − vCA − (in) (out) V kP CA CB (reaction) = Component B: b0 − vCB − V (kP CA CB + kQ CB2 ) = 0 (i) (accumulation) 0 (ii) It is necessary to solve equations (i) and (ii) simultaneously in order to obtain CA and CB , although, if CA in terms of CB from equation (i) is substituted in equation (ii), a cubic equation results. Noting that component A is in excess, the cubic equation may 256 be avoided by, as a first approximation, neglecting the reaction term in equation (i) as compared with the flow term, or: CA ≈ a0 /v = (0.014/1.1 × 10−3 ) = 12.73 kmol/m3 If this value is substituted into equation (ii), then: 0.0014 − 1.1 × 10−3 CB − 10[(1.5 × 10−5 × 12.73CB ) + (11 × 10−5 CB2 )] = 0 CB2 + 2.736CB − 1.273 = 0 and: Disregarding the negative root, then: CB = 0.405 kmol/m3 This value for CB may now be substituted into equation (i), in order to obtain an ‘improved’ value of CA . Thus: 0.014 − (1.1 × 10−3 CA′ ) − (10 × 1.5 × 10−5 × 0.405CA′ ) = 0 CA′ = 12.06 kmol/m3 or: Equation (ii) may now be used to give a more accurate value for CB . Thus: 0.0014 − (1.1 × 10−3 CB′ ) − 10[(1.5 × 10−5 × 12.06CB′ ) + (11 × 10−5 CB2 )] = 0 or: and: ′ CB2 + 2.645CB − 1.273 = 0 CB′ = 0.416 kmol/m3 These values of CA and CB are used for the rest of the calculations. It may be noted that the method of successive substitution used here is not the best or the most reliable method of iteration. In this particular example however, the convergence is rapid leading to final values of: CA = 12.0438 kmol/m3 and CB = 0.41613 kmol/m3 . If there were no reaction, that is no reaction term in equation (ii), then the concentration of B in the outflow would be given by: CB0 = (0.0014/1.1 × 10−3 ) = 1.273 kmol/m3 Thus the total amount of B converted by both reactions is: (1.273 − 0.416) = 0.857 kmol/m3 Considering the two reaction terms in equation (ii), then, in unit time: (material converted to P)/(total material converted) = kP CA CB /(kP CA CB + kQ CB2 ) −5 = (1.5 × 10 (iii) −5 × 12.06 × 0.416)/[(1.5 × 10 + (11 × 10−5 × 0.4162 )] = 0.798 257 × 12.06 × 0.416) Summarising: For 1 m3 solution leaving the reactor, 0.416 kmol B remains unreacted. (0.857 × 0.798) = 0.684 kmol B reacts to produce P. (0.857(1 − 0.798)) = 0.173 kmol B reacts to produce Q. and: Total amount of B fed to the system = (0.416 + 0.684 + 0.173) = 1.273 kmol. (a) The fraction of B which is converted into the desired product P = (0.684/1.273) = 0.537 (b) The fraction of B which is converted into the waste product Q = (0.173/1.273) = 0.136 By definition, the relative yield is given by equation (iii), where: relative yield = 0.798 By definition, the operational yield is the fraction of B which is converted to the desired product, P, or: operational yield = 0.537 The question of using a second tank and whether it should be connected in series or parallel is now considered. (a) If B is not recycled, that is not recovered, then unreacted B is lost. The aim should be therefore to achieve the highest fractional conversion of B to P, that is the highest operational yield, even though a substantial amount of B may go to form Q. High concentrations of B will favour a high degree of conversion even though the reaction to Q, being of a higher order with respect to B than the reaction to P, is favoured by high concentrations of B. Such high concentration of B will obtained if a second tank is installed in series with the first tank. (b) If B is recycled, that is recovered, then unreacted B is not wasted. The aim should be, of course, to maximise the relative yield of B and, because the unwanted reaction to Q is suppressed at low concentrations of B, the tanks should be connected in parallel so that the feed stream is further diluted compared to the system which utilises one tank only. The argument for the case where B is not recycled presupposes that, at a high concentration of B, the advantage of increasing the rate of reaction to P outweighs the disadvantage of also increasing the amount which reacts to form Q. This may not be the situation in all circumstances, since it depends on the magnitudes of kP and kQ . If the argument put forward is not found to be convincing, then the way forward is to carry out calculations for both configurations as in the solution to Problem 1.8. 258 PROBLEM 1.11 A substance A reacts with a second substance B to give a desired product P, but B also undergoes a simultaneous side reaction to give an unwanted product Q as follows: A + B −−−→ P; 2B −−−→ Q; rate = kP CA CB rate = kQ CB2 where CA and CB are the concentrations of A and B respectively. A single continuous stirred tank reactor is used for these reactions. A and B are mixed in equimolar proportions such that each has the concentration C0 in the combined stream fed at a volumetric flowrate v to the reactor. If the rate constants above are kP = kQ = k and the total conversion of B is 0.95, that is the concentration of B in the outflow is 0.05C0 , show that the volume of the reactor will be 69 v/k C0 and that the relative yield of P will be 0.82, as for case α in Figure 1.24, Volume 3. Would a simple tubular reactor give a larger or smaller yield of P than the C.S.T.R.? What is the essential requirement for a high yield of P? Suggest any alternative modes of contacting the reactants A and B which would give better yields than either a single C.S.T.R. or a simple tubular reactor. Solution A material balance over the tank for material A at steady state gives: vCA0 − vCA − kP CA CB V (in) (out) (reaction) = 0 (accumulation) The fractional conversion of B, β = 0.95. If the fractional conversion of A is α, noting that α < β since A is consumed in the first reaction only, whereas B is consumed in both, then: CA = CA0 (1 − α) CB = CB0 (1 − β) CA0 = CB0 = C0 and hence the material balance is: vC0 − vC0 (1 − α) − kP V C02 (1 − α)(1 − β) = 0 or: α = (kP C0 V /v)(1 − α)(1 − β) (i) Similarly, a material balance for material B is: vCB0 − vCB − V (kP CA CB + kQ CB2 ) = (in) (out) (reaction) or: Thus: 0 (accumulation) vC0 − vC0 (1 − β) − V C02 [kP (1 − α)(1 − β) + kQ (1 − β)2 ] = 0 β = (C0 V /v)(1 − β)[kP (1 − α) + kQ (1 − β)] 259 (ii) For the single stirred tank, the relative yield of the desired product, φ, is given by the ratio of the rates of reaction, or: φ = RP /(RP + RQ ) = (kP CA CB )/(kP CA CB + kQ CB2 ) = (kP CA )/(kP CA + kQ CB ) (iii) When, as in the problem, the reactant proportions are chosen such that CA0 = CB0 = C0 , then: φ= kP (1 − α) kP C0 (1 − α) = kP C0 (1 − α) + kQ C0 (1 − β) kP (1 − α) + kQ (1 − β) (iv) For the special case of kP = kQ = k , the equations become: k C0 V (1 − α)(1 − β) v k C0 V (1 − β)[1 − α + 1 − β] β= v 1−α φ= [1 − α + 1 − β] α= Since the fractional conversion of B, β = 0.95, then:   k C0 V (1 − α) × 0.05 from equation (v): α = v   k C0 V (1 − α) or: 20α = v   k C0 V From equation (vi): 0.95 = × 0.05[1 − α + 1 − 0.95] v   k C0 V (1.05 − α). or: 19 = v   k C0 V Dividing to eliminate the term then: v (20α/19) = (1 − α)/(1.05 − α) α 2 − 2α + 0.95 = 0 or: Solving, noting that α < 0: α = 0.776 From equation (ix): (k C0 V /v) = (19/(1.05 − 0.776)) = 69.4 Thus, Volume of the reactor, V = (69v/k C0 ) From equation (vii): Relative yield of P, φ = (1 − 0.776) = 0.818 (1 − 0.776 + 1 − 0.95) 260 (v) (vi) (vii) (viii) (ix) It may be noted that these calculations confirm the values shown in Figure 1.24, in Volume 3. The essential requirement for a high relative yield of P can be deduced from the orders of the two competing reactions. With respect to B the undesired reaction to Q has an order of two whilst the desired reaction to P has an order of one and hence the order to Q is greater than the order to P. Thus the reaction to Q will tend to be suppressed at low concentration of B and the reaction to P will be favoured by high concentration of A. In a simple tubular reactor, the concentration of both reactants will be high at the inlet and the question is whether the disadvantage of a high concentration of B outweighs the advantage of a high concentration of A. This is nearly always the case because the relative kP CA RP yield, φ = = which decreases monotonically with increasing RP + RQ kP CA + kQ CB CB irrespective of the particular values of CA , kP , kQ . In the special case where kP = kQ , a simple tubular reactor gives a poor relative yield of the desired product P, as indicated in Figure 1.24 in Volume 3. Alternative modes of contacting which give better yields are also indicated in Figure 1.24 in Volume 3. The principle is to use multiple injection points for B along the reactor so that, although concentrations of A are high in the initial stages of the reaction, the concentrations of B are comparatively low. 261 SECTION 3-2 Flow Characteristics of Reactors — Flow Modelling PROBLEM 2.1 A batch reactor and a single continuous stirred-tank reactor are compared in relation to their performance in carrying out the simple liquid phase reaction A + B → products. The reaction is first order with respect to each of the reactants, that is second order overall. If the initial concentrations of the reactants are equal, show that the volume of the continuous reactor must be 1/(1 − α) times the volume of the batch reactor for the same rate of production from each, where α is the fractional conversion. Assume that there is no change in density associated with the reaction and neglect the shutdown period between batches for the batch reactor. In qualitative terms, what is the advantage of using a series of continuous stirred tanks for such a reaction? Solution For the batch reactor of volume Vb The rate equation is: R = k a2 dx/dt = k (a − x)2 or: After time t, when x = αa0 : kt =  dx/(a0 − x)2 = 1/(a0 − x) + constant. When t = 0, x = 0 and the constant = −1/a0 . Thus: k t = 1/(a0 − x) − 1/a0 = (1/a0 )((a0 − a0 + x)/(a0 − x)) = (1/a0 )(x/(a0 − x)) = α/(a0 (1 − α)) Thus: rate of output, Vb x/t = Vb αa0 /t = Vb k a02 (1 − α) 262 For the continuous stirred-tank reactor of volume Vc A steady-state balance on reactant A gives: va0 − v(a0 − x) − Vc k (a0 − x)2 = 0 (in) (out) (reacted) Thus: vx − Vc k (a0 − x)2 = 0 The rate of output = vx = Vc k (a0 − x)2 = Vc k a02 (1 − α)2 Thus, for the same rate of output: Vb = Vc (1 − α) Vc /Vb = 1/(1 − α) or: Qualitatively, the advantage of using a series of equal-sized tanks for the process is that the total volume of the series will be less than the volume of a single tank assuming the same conversion. This is considered in detail in Section 1.10, Volume 3. PROBLEM 2.2 Explain carefully the dispersed plug-flow model for representing departure from ideal plug flow. What are the requirements and limitations of the tracer response technique for determining Dispersion Number from measurements of tracer concentration at only one location in the system? Discuss the advantages of using two locations for tracer concentration measurements. The residence time characteristics of a reaction vessel are investigated under steadyflow conditions. A pulse of tracer is injected upstream and samples are taken at both the inlet and outlet of the vessel with the following results: Inlet sample point: Time (s) Tracer concentration (kmol/m3 ) × 103 30 40 50 60 70 80 90 100 110 120 <0.05 0.1 3.1 10.0 16.5 12.8 3.7 0.6 0.2 <0.05 Outlet sample point: Time (s) 110 120 130 140 150 160 170 180 190 200 210 220 Tracer concentration <0.05 0.1 0.9 3.6 7.5 9.9 8.6 5.3 2.5 0.8 0.2 <0.05 (kmol/m3 ) × 103 Calculate (a) the mean residence time of tracer in the vessel and (b) the dispersion number. If the reaction vessel is 0.8 m in diameter and 12 m long, calculate also the volume flowrate through the vessel and the dispersion coefficient. 263 Solution For equally-spaced sampling intervals, the following approximate approach may be used. Mean residence time, t¯ = Ci ti /Ci Variance, σ 2 = (Ci ti2 /Ci ) − t¯2 If, for convenience, all times are divided by 100, then the following data are obtained. Inlet sampling point ti Ci Ci ti Ci ti2 0.3 — — — 0.4 0.1 0.04 0.016 0.5 3.1 1.55 0.775 0.6 10.0 6.0 3.6 0.7 16.5 11.55 8.085 0.8 12.8 10.24 8.192 t¯1 = (33.53/47.0) = 0.7134 or 0.9 3.7 3.33 2.997 1.0 0.6 0.6 0.6 1.1 0.2 0.22 0.242  = 47.0  = 33.53  = 24.507 71.34 s σ12 = (24.507/47.0) − 0.71342 = 0.01248 or 124.8 s2 Outlet sampling point ti Ci Ci ti Ci ti2 1.2 0.1 0.12 0.144 1.3 0.9 1.17 1.521 1.4 1.5 1.6 1.7 1.8 3.6 7.5 9.9 8.6 5.3 5.04 11.25 15.84 14.62 9.54 7.056 16.875 25.344 24.854 17.172 t¯2 = (64.35/39.4) = 1.633 or 1.9 2.5 4.75 9.025 2.1 0.2  = 39.4 0.42  = 64.35 0.882  = 106.073 163.3 s σ22 = (106.073/39.4) − 1.6332 = 0.02471 or For two sample points: 2.0 0.8 1.6 3.2 247.1 s2 (σ 2 )/(t¯)2 = 2(D/uL) (equation 2.21) Thus the Dispersion Number, (D/uL) = (247.1 − 124.8)/[2(163.3 − 71.34)2 ] = 0.0072 The mean residence time, (t¯2 − t¯1 ) = (163.3 − 71.34) = 92 s The volume of the vessel, V = 12(π0.82 /4) = 6.03 m3 and hence: volume flowrate, v = (6.03/92) = 0.0656 m3 /s The velocity in the pipe, u = (L/t) = (12/92) = 0.13 m/s and hence: Dispersion coefficient, D = (0.0072 × 0.13 × 12) = 0.0113 m2 /s 264 SECTION 3-3 Gas–Solid Reactions and Reactors PROBLEM 3.1 An approximate design procedure for packed tubular reactors entails the assumption of plug flow conditions through the reactor. Discuss critically those effects which would: (a) invalidate plug flow assumptions, and (b) enhance plug flow. Solution The general question of whether or not plug flow can be attained is discussed in Volume 3, Section 1.7. (Tubular Reactors) and the special case of Plug-Flow (Fermenters) is considered in Chapter 5, Section 5.11.3. A more detailed consideration of dispersion in packed bed reactors and those effects which enhance and invalidate plug flow is given in Chapter 3, Section 3.6.1. PROBLEM 3.2 A first-order chemical reaction occurs isothermally in a reactor packed with spherical catalyst pellets of radius R. If there is a resistance to mass transfer from the main fluid stream to the surface of the particle in addition to a resistance within the particle, show that the effectiveness factor for the pellet is given by:   coth λR − 1/λR 3 η= λR 1 + (2λR/Sh′ )(coth λR − 1/λR) hD dp , De k is the first-order rate constant per unit volume of particle, De is the effective diffusivity, and hD is the external mass transfer coefficient. where: λ = (k /De )1/2 and Sh′ = Discuss the limiting cases pertaining to this effectiveness factor. Solution A shell of radii r and r + δr within the solid pellet particle of radius R may be considered. Making a material balance for the reactant, then: 265 Rate of transfer in at radius r + δr = rate of transfer out at radius r + reaction rate. or: 4π(r + δr)2 De (dC/dr)r+δr = 4πr 2 De dC/dr + 4πr 2 δrk C where k is the chemical rate constant and C the total molar concentration. Expanding the left-hand term in this equation and neglecting (δr)2 terms and higher then: d2 C/dr 2 + (2/r)(dC/dr) − (k /De )C = 0 (i) There are two boundary conditions: A: When r = 0, B: When r = R, dC/dr = 0 and C remains finite De dC/dr = hD (C0 − Ci ) where C0 is the concentration in the bulk gas and Ci the concentration at the interface, r = R. It is convenient to define λ = (k /De )0.5 as in equation 3.13 and the Sherwood Number as Sh = hD dp /De = 2hD R/De . At r = R: (2R/Sh)dC/dr + Ci = C0 Solving equation (i): C = (1/r)(Aeλr + Be−λr ) From boundary condition A: A = −B C = A/r sinh λr and: Substituting in boundary condition B:   2 (λR cosh λR − sinh λR) + sinh λR A = RC0 Sh   2r Thus: C = RC0 sinh λR (λR cosh λR − sinh λR) + sinh λR Sh From Example 3.2 in Volume 3, for a sphere, the effectiveness factor is given by: η= 4πR 2 De (dC/dr)R (4/3)πR 3 k C0 = (3/λ2 R)(dC/dr)R /C0 From the relationship between C and r given by equation (ii), then: η = (3/λR)(coth λR − 1/λR)/((2λR)/Sh)(coth λR − 1/λR) + 1) 266 (ii) PROBLEM 3.3 Two consecutive first-order reactions: k1 k2 A−−−→B−−−→C occur under isothermal conditions in porous catalyst pellets. Show that the rate of formation of B with respect to A at the exterior surface of the pellet is:  1/2 CB k2 (k1 /k2 )1/2 − 1 + (k1 /k2 )1/2 k1 CA when the pellet size is large, and: 1− k2 CB k1 CA when the pellet size is small. CA and CB represent the concentrations of A and B respectively at the exterior surface of the pellet, and k1 and k2 are the specific rate constants of the two reactions. Comparing these results, what general conclusions can be deduced concerning the selective formation of B on large and small catalyst pellets? Solution If the flat plate model for the catalyst pellet as shown in Figure 3.2, Volume 3, is assumed, then a material balance gives: For component A: For component B: De (d2 CA /dx 2 ) − k1 CA = 0 2 (equation 3.10) 2 De (d CA /dx ) − k1 CA − k2 CB = 0 There are two boundary conditions: A: When x = L, CA = CA∞ and CB = CB∞ B: When x = 0, dCA /dx = dCB /dx = 0 Solving these two equations: and: where: √ CA /CA∞ = cosh λ1 x/ cosh λ1 L, where: λ1 = (k1 /De )   cosh λ1 x CB∞ cosh λ2 x cosh λ2 x − + CB /CA∞ = [k1 /(k1 − k2 )] cosh λ2 L cosh λ1 L CA∞ cosh λ2 L  (equation 3.46) λ2 = (k2 /De ) If the selectivity is defined as the rate of formation of B at the catalyst exterior surface compared to the rate of formation of A at the surface, then: 267 −De (dCB /dx)x=L = −De (dCA /dx)x=L  k1 k1 − k2   φ2 tanh φ2 CB∞ φ2 tanh φ2 1− − φ1 tanh φ1 CA∞ φ1 tanh φ1 (equation 3.47) where φ1 = Lλ1 and φ2 = Lλ2 . For large particles, where both φ1 and φ2 are large, the selectivity becomes: √ (k1 /k2 ) CB √ √ − (k2 /k1 ) 1 + (k1 /k2 ) CA which is the limiting form of equation 3.47. For small particles, the other asymptote in equation 3.47 gives the selectivity as:   k2 CB 1− k1 CA PROBLEM 3.4 A packed tubular reactor is used to produce a substance D at a total pressure of 100 kN/m2 (1 bar) utilising the exothermic equilibrium reaction: − ⇀ A+B− ↽ −− − −C+D 0.01 kmol/s (36 kmol/h) of an equimolar mixture of A and B is fed to the reactor and plug flow conditions within the reactor may be assumed. What are the optimal isothermal temperature for operation and the corresponding reactor volume for a final fractional conversion zf of 0.68. Is this the best way of operating the reactor? The forward and reverse kinetics are second order with the velocity constant k1 = 4.4 × 1013 exp(−105 × 106 /RT ) and k2 = 7.4 × 1014 exp(−125 × 106 /RT ) respectively. k1 and k2 are expressed in m3 /kmol s and R in J/kmol K. Solution For the reaction: − ⇀ A+B− ↽ −− − −C+D The mole fractions at any cross-section, starting with an equimolar mixture of A and B are: (1 − z)/2 moles A, (1 − z)/2 moles B, (z/2) moles C and (z/2) moles D. where z is the fractional conversion and the concentrations are: (1 − z) P CA = CB = 2 RT P . CC = CD = (z/2) RT 268 The reaction rate is then: R=  P 2RT 2 (k1 (1 − z)2 − (k2 z2 )) For plug flow, the reactor volume, V = F  dz/R where F is the rate of feed to the reactor. Substituting for R and integrating with respect to z at constant temperature then: V =F  =F  2RT P 2  2RT P 2 zf dz k1 (1 − z)2 − k2 z2 0 K k1  zf 0 dz K(1 − z)2 − z2 1 1 = √ √ 2 2 K(1 − z) − z ( K(1 − z) + z)( K(1 − z) − z) 1 = √ √ √ √ ( K − ( K + 1)z)( K − ( K − 1)z) √ √ √ If α = K, β = K + 1 and γ = K − 1, then: γ β 1 1 = − = 2 2 K(1 − z) − z (α − βz)(α − γ z) α(γ − β)(α − γ z) α(γ − β)(α − βz) The integral then becomes:  zf   zf 1 γ dz βdz − α(γ − β) 0 (α − γ z) (α − βz) 0 √ √ Noting that (γ − β) = ( K − 1) − ( K + 1) = −2, the integral is:   1 α − βzf ln α(γ − β) α − γ zf   1 α − βzf ln Thus: = 2α α − γ zf Reactor volume, V = (F /2)(2RT /P )2 (α/k1 )[ln(α − γ zf )/(α − βzf )] Substituting R = 8314 J/kmol K, P = 1 × 105 N/m2 and F = 0.010 kmol/s then: V = 0.000138T 2 (α/k1 )[ln(α − γ zf )/(α − βzf )] 6 k1 = 4.4 × 1013 e−(105×10 and: /8314T ) 6 K = (4.4/7.4) × 10−1 e(20×10 269 /8314T ) . Making calculations for T = 540, 545 and 550 K, the following data are obtained: T (K) 540 545 550 k1 (s−1 ) K α β γ ln(α − γ zf )/(α − βzf ) V (m3 ) 3064 5.116 2.262 3.262 1.262 3.466 0.103 3798 4.911 2.216 3.216 1.216 3.346 0.080 4688 4.718 2.172 3.172 1.172 4.516 0.087 from which the optimum isothermal temperature = 545 K and the reactor volume = 0.080 m3 270 SECTION 3-4 Gas–Liquid and Gas–Liquid–Solid Reactors PROBLEM 4.1 What is the significance of the parameter β = (k2 CBL DA )0.5 /kL in the choice and the mechanism of operation of a reactor for carrying out a second-order reaction, rate constant k2 , between a gas A and a second reactant B of concentration CBL in a liquid? In this expression, DA is the diffusivity of A in the liquid and kL is the liquid-film mass transfer coefficient. What is the ‘reaction factor’ and how is it related to β? Carbon dioxide is to be removed from an air stream by reaction with a solution containing 0.4 kmol/m3 of NaOH at 258 K (25◦ C) at a total pressure of 110 kN/m2 (1.1 bar). A column packed with 25 mm Raschig rings is available for this purpose. The column is 0.8 m internal diameter and the height of the packing is 4 m. Air will enter the column at a rate of 0.015 kmol/s (total) and will contain 0.008 mole fraction CO2 . If the NaOH solution is supplied to the column at such a rate that its concentration is not substantially changed in passing through the column, calculate the mole fraction of CO2 in the air leaving the column. Is a packed column the most suitable reactor for this operation? Data: Effective interfacial area for 25 mm packing = 280 m2 /m3 Mass transfer film coefficients: Liquid, kL = 1.3 × 10−4 m/s Gas, kG a = 0.052 kmol/m3 s bar For a 0.4 kmol/m3 concentration of NaOH at 298 K: Solubility of CO2 : PA = H CA where PA is partial pressure of CO2 , CA is the equilibrium liquid-phase concentration and H = 32 bar m3 /kmol. Diffusivity of CO2 , D = 0.19 × 10−8 m2 /s The second-order rate constant for the reaction CO2 + OH− = HCO3 − , k2 = 1.35 × 104 m3 /kmol s. Under the conditions stated, the reaction may be assumed pseudo first-order with respect to CO2 . Solution The reaction factor is such that: NA = kL CAi fi 271 For a fast, first-order reaction in the film: fi = β/ tanh β (equation 4.13) 0.5 β = (k2 CBL DA ) /kL where: In the present case: β = (1.35 × 104 × 0.4 × 0.19 × 10−8 )0.5 /(1.3 × 10−4 ) = 24.6 This value of β confirms the regime of a fast reaction occurring mainly in the film for which a packed column is a suitable reactor. For large values of β, tanh β = 1 and: NA = kL CAi β = kL′ CAi where kL′ is the effective liquid film mass transfer coefficient enhanced by the chemical reaction; that is: kL′ = kL β In the present case: kL′ = (1.3 × 10−4 × 24.6) = 32 × 10−4 m/s Combining the liquid- and gas-film resistances and replacing kL by kL′ since the mass transfer is enhanced by the reaction, then: 1/KG a = 1/kG a + H /kL′ a or: and: 1/KG a = (1/0.052) + (32/(32 + 10−4 × 280)) KG a = 0.0182 kmol/m3 s bar If G kmol/m2 s is the carrier gas flowrate per unit area and y is the mole fraction of CO2 , equal approximately to the mole ratio, then a balance across an element of the column δh high, gives: −Gδy = KG a(P − P ∗ )δh where P ∗ is the partial pressure that would be in equilibrium with the bulk liquid. Noting that, in this case, P ∗ = 0 and P = yPT , where PT is the total pressure, then: −Gδy = KG ayPT δh Thus: h = (G/KG aPT )  yin yout dy y = (G/KG aPT ) ln(yin /yout ) In this case: G = 0.015/(π0.82 /4) = 0.0298 kmol/m2 s G/(KG aPT ) = 0.0298/(0.0182 × 1.1) = 1.49 m Thus: ln(yin /yout ) = (4/1.49) = 2.68 (yin /yout ) = 14.6 and: yout = (0.008/14.6) = 0.00055. 272 PROBLEM 4.2 A pilot-scale reactor for the oxidation of o-xylene by air according to the following reaction has been constructed and its performance is being tested. CH3 CH3 CH3 COOH + 1.5 O2 = + H2O The reactor, an agitated tank, operates under a pressure of 1.5 kN/m2 (15 bar) and at 433 K (160◦ C). It is charged with a batch of 0.06 m3 of o-xylene and air introduced at the rate of 0.0015 m3 /s (5.4 m3 /h) measured at reactor conditions. The air is dispersed into small bubbles whose mean diameter is estimated from a photograph to be 0.8 mm, and from level sensors in the reactor, the volume of the dispersion produced is found to be 0.088 m3 . Soon after the start of the reaction (before any appreciable conversion of the o-xylene) the gas leaving the reactor is analysed (after removal of condensibles) and found to consist of 0.045 mole fraction O2 , 0.955 mole fraction N2 . Assuming that under these conditions, the rate of the above reaction is virtually independent of the o-xylene concentration and is thus pseudo first-order with respect to the concentration of dissolved O2 in the liquid, calculate the value of the pseudo first-rate constant. Data: Estimated liquid-phase mass transfer coefficient, kL = 4.0 × 10−4 m/s. Equilibrium data: PA = H CA where PA is the partial pressure and CA the equilibrium concentration in liquid, H = 127 m3 bar/kmol Diffusivity of O2 in liquid o-xylene, D = 1.4 × 10−9 m2 /s. Gas constant, R = 8314 J/kmol K Composition of air (molar): O2 = 20.9 per cent, N2 = 79.1 per cent State clearly any further assumptions made and discuss their validity. Is an agitated tank the most suitable type of reactor for this process. Solution Noting that, for an ideal gas, n = (P V /RT ), then: feed rate of air to the reactor = (15 × 105 × 5.4)/(8314 × 433 × 3600) = 0.625 × 10−3 kmol/s Air is 20.9 per cent oxygen and hence: oxygen in the feed = (0.625 × 10−3 × 20.9/100) = 0.1306 × 10−3 kmol/s nitrogen in the feed = nitrogen in the exit gas = 0.791 kmol/kmol feed Thus: oxygen leaving the reactor = (0.791 × 0.045/0.955) = 0.0372 kmol/kmol feed 273 and: Flow of oxygen from the reactor = 0.625 × 10−3 (0.791 × 0.045/0.955) = 0.0233 × 10−3 kmol/s oxygen reacting = (0.1306 × 10−3 ) − (0.0233 × 10−3 ) Thus: = 0.1073 × 10−3 kmol/s From which: fraction of oxygen fed which reacts = (0.1073 × 10−3 )/(0.1306 × 10−3 ) = 0.822 and: rate of oxygen per unit volume dispersed = (0.1073 × 10−3 )/(88 × 10−3 ) = 0.00123 kmol/m3 s. The volume fraction of gas in dispersion, εG = (88 − 60)/88 = 0.318 The mean bubble diameter is: db = 6εG /a (equation 4.30) −3 2 3 a = 6εG /db = (6 × 0.318)/(0.8 × 10 ) = 2390 m /m Thus: For a pseudo first-order reaction, the rate with respect to oxygen = k C (per unit volume of liquid). If Ci is the concentration at the bubble interface and Cb the concentration in the bulk, then assuming that: (a) Cb is significant, that is not equal to zero, and (b) a pseudo steady-state exists: Rate of mass transfer of oxygen from the interface to the bulk ≡ Rate of reaction in the bulk or, per unit volume of dispersion, or: that is: or: kL a(Ci − Cb ) = k Cb (1 − εG ) (Ci − Cb ) = (k (1 − εG )/kL a)Cb Ci = Cb (1 + (k (1 − εG )/kL a)) If it is further assumed that: (c) the gas phase in the bubbles in dispersion is well-mixed and has the same composition as the outlet gas, and (d) the partial pressure of the o-xylene in the gas phase may be neglected in that the total pressure of 1.5 kN/m2 (1.5 bar) is relatively high even at 433 K (160◦ C), then, using Henry’s law: Ci = y02 P /H = (0.045 × 15)/127 = 0.00532 kmol/m3 . Cb = Ci /[1 + k (1 − εG )/kL a] 274 and the rate reaction per unit volume of dispersion Rd is given by: Rd = k Ci (1 − εG )/[1 + k (1 − εG )/kL a] = Ci /[1/k (1 − εG ) + 1/kL a] (Ci /Rd ) = 1/[k (1 − εG )] + (1/kL a) Thus: or: Thus: (0.00532/0.00123) = 1/[k (1 − εG )] + 1/(40 × 10−4 × 2390) 1/[k (1 − εG )] = 3.28 k = 1/[3.28(1 − 0.318)] = 0.447 s−1 and: It may be noted that: Cb = Ci /[1 + (0.447 × 0.682)/(4 × 10−4 × 2390)] = 0.76 Ci which is in line with assumption (a). PROBLEM 4.3 It is proposed to manufacture oxamide by reacting cyanogen with water using a strong solution of hydrogen chloride which acts as a catalyst according to the reaction: (CN)2 + 2H2 O −−−→ (CONH2 )2 The reaction is pseudo first-order with respect to dissolved cyanogen, the rate constant at the operating temperature of 300 K being 0.19 × 10−3 s−1 . An agitated tank will be used containing 15 m3 of liquid with a continuous flow of a cyanogen–air mixture at 300 kN/m2 (3 bar) total pressure, composition 0.20 mole fraction cyanogen, and a continuous feed and outflow of the hydrogen chloride solution; the gas feed flowrate will be 0.01 m3 /s total and the liquid flowrate 0.0018 m3 /s. At the chosen conditions of agitation the following estimates have been made: kL = 1.9 × 10−5 m/s a = 47 m2 /m3 Liquid-phase mass transfer coefficient Gas–liquid interfacial area per unit volume of dispersion Gas volume fraction in dispersion Diffusivity of cyanogen in solution Henry law coefficient where PA is partial pressure and CA is concentration in liquid at equilibrium Gas constant εg = 0.031 D = 0.6 × 10−9 m2 /s PA /CA = H = 1.3 bar m3 /kmol R = 8314 J/kmol K Assuming ideal mixing for both gas and liquid phases, calculate: (a) the concentration of oxamide in the liquid outflow (the inflow contains no dissolved oxamide); (b) the concentration of dissolved but unreacted cyanogen in the liquid outflow, and 275 (c) the fraction of cyanogen removed from the gas stream. What further treatment would you suggest for the liquid leaving the tank before separation of the oxamide? Calculate the value of β for this system and suggest another type of reactor that might be considered for this process. Solution (a) A material balance on cyanogen transferred from gas to liquid gives: vG0 y0 P /RT − vG yP /RT − Vd kL a(Ci − CL ) = 0 (in) (out) (transferred) or: (P /RT )(vG0 y0 − vG y) − Vd kL a[(yP /H ) − CL ] = 0 (i) (b) A material balance on the cyanogen in the liquid gives: 0 − vL CL + Vd kL a(Ci − CL ) − k V CL = 0 (in) (out) (transferred) (reaction) Vd kL a[(yP /H ) − CL ] − (vL + k V )CL = 0 or: (ii) It may be noted that Vd is the volume of dispersion and: V = (1 − εG )Vd Vd = V /(1 − εG ) = V /0.969 or: (c) A material balance for the oxamide gives: 0 − vL CM + k V CL = 0 (in) (out) (reaction) (iii) (d) A material balance for the air, which is required because the volume flow out, vG , is less than the flowrate in, vG0 , gives: vG0 (1 − y0 ) = vG (1 − y) (iv) vG = vG0 (1 − y0 )/(1 − y) or: = 0.01(1 − 0.20)/(1 − y) = 0.008/(1 − y). From equation (i): [3 × 105 /(8314 × 300)]{(0.01 × 0.2) − [0.008y/(1 − y)]} −[(15 × 1.9 × 10−5 × 47)/0.969][(3y/1.3) − CL ] = 0 or: 0.12[0.002 − 0.008y/(1 − y)] − 0.0138(2.31y − CL ) = 0 (v) From equation (ii): 0.0138(2.31y − CL ) − [0.0018 + (0.19 × 10−3 × 15)]CL = 0 or: CL = (0.0319y/0.0185) = 1.72y 276 (vi) Substituting from equation (vi) into equation (v): 0.002 − [(0.008y)/(1 − y)] − (0.0138/0.12)(2.31y − 1.72y) = 0 or: 0.002 − 0.078y + 0.068y 2 = 0 Solving the quadratic equation and noting that y < 1, then: y = 0.0265 CL = (1.72 × 0.0265) = 0.046 kmol/m3 and: From equation (iv): cyanogen absorbed from the gas stream, vG = 0.01(1 − 0.20)/(1 − 0.0265) = 0.0082 m3 /s. The fraction absorbed = (vG0 y0 − vG y)/vG0 y0 = 1 − (vG y/vG0 y0 ) = 1 − (0.0082 × 0.0265)/(0.01 × 0.20) = 0.89 From equation (iii): oxamide in the outflow CM = (k V /vL )CL = (0.19 × 10−3 × 15) × 0.046/0.0018 = 0.073 kmol/m3 β = (k D)0.5 /kL = (0.19 × 10−3 × 0.6 × 10−9 )0.5 /(1.9 × 10−5 ) = 0.018 and hence a bubble column is the preferred design. PROBLEM 4.4 (a) Consider a gas-liquid-solid hydrogenation such as that described in (b) in which the reaction takes place within a porous catalyst particle in a trickle bed reactor. Assume that the liquid containing the compound to be hydrogenated reaches a steady state with respect to dissolved hydrogen immediately it enters the reactor and that the liquid is involatile. Show that the rate of reaction per unit volume of reactor space R [(kmol H2 converted)/m3 s] for a reaction which is pseudo first-order with respect to hydrogen is given by:  −1 1 PA Vp 1 R= + + H kL a ks Sx (1 − e) k η(1 − e) where: PA H kL a ks Vp Sx e = = = = = = = Pressure of hydrogen (bar) Henry Law coefficient (bar m3 /kmol) Gas–liquid volumetric mass transfer coefficient (s−1 ) Liquid–solid mass transfer coefficient (m/s) Volume of single particle (m3 ) External surface area of a single particle (m2 ) Voidage of the bed (−) 277 k = First-order rate constant based on volume of catalyst [m3 /(m3 catalyst) s = s−1 ] η = Effectiveness factor (−) (b) Crotonaldehyde is to be selectively hydrogenated to n-butyraldehyde in a process using a palladium catalyst deposited on a porous alumina support in a trickle bed reactor. The particles will be spheres of 5 mm diameter packed into the reactor with a voidage e of 0.4. Estimated values of the parameters listed in (a) are as follows: kL a = 0.02 s−1 , ks = 2.1 × 10−4 m/s k = 2.8 m3 /(m3 cat)s, H = 357 bar m3 /kmol Also for spheres the effectiveness factor is given by:   1 1 coth 3φ − where the Thiele modulus, η= φ 3φ φ= Vp Sx  k De 1/2 (equation 3.19) For the catalyst, the effective diffusivity, De = 1.9 × 10−9 m2 /s. If the pressure of hydrogen in the reactor is 1 bar, calculate R, the rate of reaction per unit volume of reactor, and comment on the relative values of the transfer/reaction resistances involved in the process. (c) Discuss whether the trickle bed reactor and the conditions described in (b) are the best choices for this process. What alternatives might be considered? Solution (a) Hydrogenation takes place in a sequence of steps: 1. mass transfer takes place from the gas to the liquid, 2. mass transfer takes place from the liquid to the external surface of the particles, and 3. diffusion and reaction take place within the particles. At steady-state, the rates of all these are the same and equal to the overall rate of reaction, R. On the basis of unit volume of reactor space that is the volume of gas, liquid and solid, each of these steps is considered in turn. 1. If kL a is the volume mass transfer coefficient, that is referring to the whole reactor space, then: R = kL a(Ci − CL ). Ci = P /H and hence: R = kL a[(P /H ) − CL ] (i) 2. The rate of mass transfer from the liquid to the solid for one particle = ks Sx (CL − Cs ) Number of particles per unit volume of reactor space = (1 − e)/Vp Thus: Rate of mass transfer per unit volume of reactor space, R = (ks Sx /Vp )(1 − e)(CL − Cs ) 278 (ii) 3. The rate of diffusion and reaction per unit volume of particles = k Cs η. Since the particles occupy only a fraction (1 − e) of the reactor space, then: Rate of diffusion and reaction per unit volume of reactor space, R = kCs η(1 − e) (iii) From equations (i), (ii) and (iii): R/kL a = (P /R) − CL R/[(ks Sx /Vp )(1 − e)] = CL − Cs R/[k η(1 − e)] = Cs and: Adding: and: R[(1/kL a) + (Vp /(ks Sx (1 − e))) + 1/(k η(1 − e))] = P /H     −1 P Vp 1 1 R= + + H kL a ks Sx (1 − e) k η(1 − e) (b) For the crotonaldehyde hydrogenation: R = (1/357)[(1/0.02) + Vp /(2.1 × 10−4 Sx (1 − 0.4)) + 1/(2.8η(1 − 0.4))]−1 = (1/357)[(1/0.02) + (Vp /Sx )/1.26 × 10−4 + 1/(1.68η)]−1 (iv) For spherical particles: (Vp /Sx ) = (πdp3 /6)/(πdp2 ) = (dp /6) = (0.005/6) = 0.000833 The effectiveness factor, η = (1/φ)(coth 3φ − 1/3φ) where: φ = (Vp /Sx )(k /De )0.5 = 0.000833(2.8/(1.9 × 10−9 ))0.5 = 32, which is large. Thus: η = (1/32)[coth(3 × 32) − 1/(3 × 32)] = 0.031, which is low. Substituting these values in equation (iv), then: R = (1/357) = 0.0028 [(1/0.02) (50 (gas–liquid) + + (1/0.151) 6.6 (liquid–solid) = 3.7 × 10−5 kmol/m3 s. + + (1/0.052)]−1 19.2)−1 (diffusion + reaction) PROBLEM 4.5 Aniline present as an impurity in a hydrocarbon stream is to be hydrogenated to cyclohexylamine in a trickle bed catalytic reactor operating at 403 K (130◦ C). C6 H5 · NH2 + 3H2 −−−→ C6 H11 · NH2 279 The reactor, in which the gas phase will be virtually pure hydrogen, will operate under a pressure of 2 MN/m2 (20 bar). The catalyst will consist of porous spherical particles 3 mm in diameter, and the voidage, that is the fraction of bed occupied by gas plus liquid, will be 0.4. The diameter of the bed will be such that the superficial liquid velocity will be 0.002 m/s. The concentration of the aniline in the liquid feed will be 0.055 kmol/m3 . (a) From the following data, calculate what fraction of the aniline will be hydrogenated in a bed of depth 2 m. Assume that a steady state between the rates of mass transfer and reaction is established immediately the feed enters the reactor. (b) Describe, stating the basic equations of a more complete model, how the validity of the steady-state assumption would be examined further. (c) Is a trickle bed the most suitable type of reactor for this process? If not, suggest with reasons a possibly better alternative. Data: The rate of the reaction has been found to be first-order with respect to hydrogen but independent of the concentration of aniline. The first-order rate constant k of the reaction on a basis of kmol hydrogen reacting per m3 of catalyst particles at 403 K (130◦ C) is 90 s−1 . Effective diffusivity of hydrogen in the catalyst particles with liquid-filled pores, De = 0.84 × 10−9 m2 /s. Effectiveness factor η for spherical particles of diameter dp : η=   1 1 coth 3φ − φ 3φ where: φ = dp 6  k De 1/2 External surface area of particles per unit volume of reactor = 1200 m2 /m3 . Mass transfer coefficient, liquid to particles = 0.10 × 10−3 m/s Volume mass transfer coefficient, gas to liquid (basis unit volume of reactor), (kL a) = 0.02 s−1 Henry’s law coefficient H = PA /CA for hydrogen dissolved in feed liquid = 2240 bar/(kmol/m3 ). Solution (a) The effectiveness factor for the catalyst is given by: η = (1/φ)(coth 3φ − 1/3φ) where: φ = (dp /6)(k /De )0.5 In this case: φ = (0.003/6)(90/(0.84 × 10−9 ))0.5 = 164 and: η = 1/φ = (1/164) = 0.0061 (which is low). Noting that the rate of reaction is independent of the aniline concentration, then assuming that steady-state is established immediately on entry: 280 Rate of transfer of H2 from gas to liquid step (1) = rate of transfer of H2 from liquid to the catalyst surface step (2) = rate of reaction within the catalyst step (3). Converting the rate of reaction per unit volume of particles to rate of reaction per unit volume of reactor, then for step (3): Rate of reaction per unit volume of reactor, R = k Cs η(1 − e) Equating steps (1) and (2): kL a(Ci − CL ) = ks as (CL − Cs ) (i) Equating steps (2) and (3): ks as (CL − Cs ) = k Cs η(1 − e) (ii) Ci = P /H = (20/2240) = 0.0089 and hence, in equation (i): 0.02(0.0089 − CL ) + (0.1 × 10−3 × 1200)(CL − Cs ) (iii) (0.1 × 10−3 × 1200)(CL − Cs ) = (90 × Cs × 0.0061)(1 − 0.4) (iv) In equation (ii): From equation (iii): (CL − Cs ) = 2.75Cs and CL = 3.75Cs . Substituting in equation (iv) gives: 0.02(0.0089 − CL ) = 0.12(CL − CL /3.75) CL = 1.65 × 10−3 kmol/m3 . and: Cs = (0.00165/3.75) = 0.44 × 10−3 kmol/m3 and: From step (3): Rate of reaction, R = (90 × 0.44 × 10−3 × 0.0061)(1 − 0.4) = 0.145 × 10−3 kmol H2 reacting/m3 bed s. From the stoichiometry: Rate of reaction of aniline = (0.145 × 10−3 )/3 = 0.048 × 10−3 kmol/m3 bed s. The superficial velocity of the liquid in the bed = 0.002 m/s or 0.002 m3 /m2 s Thus: Aniline feed rate = (0.002 × 0.055) = 0.11 × 10−3 kmol/m2 s. Noting that the reaction rate of aniline is independent of the position in the bed, for bed, 2 m deep, 1 m2 in area, then: aniline reacting = (0.048 × 10−3 × 2) = 0.096 × 10−3 kmol/m2 s and: fraction of aniline reacted = (0.096 × 10−3 /0.11 × 10−3 ) = 0.87 281 (b) For a more complete model, a section of bed of depth δz and unit cross-sectional are considered. A material balance for hydrogen in the liquid phase gives: uL CL − uL (CL + (dCL /dz)δz) + kL a(Ci − CL )δz − ks as (CL − Cs )δz = 0 (in) (out) (gas to liquid) (liquid to solid) or: − uL (dCL /dz) + kL a(Ci − CL ) − ks as (CL − Cs ) = 0 Since the rate of hydrogen transfer from the liquid to the solid is equal to the rate of reaction in the solid, then: ks as (CL − Cs )δz = k Cs η(1 − ε)δz Finally, a material balance on the aniline in the liquid gives: uL CLA − uL (CLA + (dCLA /dz)δz) − k Cs η(1 − e)δz/3 = 0 or: − uL (dCLA /dz) = k Cs η(1 − e)/3. (c) The trickle bed reactor is probably not the most suitable because of the very low value of the effectiveness factor and a suspended-bed catalyst system with a smaller particle size would be a much better option. PROBLEM 4.6 Describe the various mass transfer and reaction steps involved in a three-phase gas–liquid–solid reactor. Derive an expression for the overall rate of a catalytic hydrogenation process where the reaction is pseudo first-order with respect to the hydrogen with a rate constant k (based on unit volume of catalyst particles). Aniline is to be hydrogenated to cyclohexylamine in a suspended-particle agitated-tank reactor at 403 K (130◦ C) at which temperature the value of k is 90 s−1 . The diameter dp of the supported nickel catalyst particles will be 0.1 mm and the effective diffusivity De for hydrogen when the pores of the particle are filled with aniline is 1.9 × 10−9 m2 /s. For spherical particles the effectiveness factor is given by:     1 dp k 1/2 1 coth 3φ − where φ = η= φ 3φ 6 De The proposed catalyst loading, that is the ratio by volume of catalyst to aniline, is to be 0.03. Under the conditions of agitation to be used, it is estimated that the gas volume fraction in the three-phase system will be 0.15 and that the volumetric gas–liquid mass transfer coefficient (also with respect to unit volume of the whole threephase system) kL a, 0.20 s−1 . The liquid–solid mass transfer coefficient is estimated to be 2.2 × 10−3 m/s and the Henry’s law coefficient H = PA /CA for hydrogen in aniline at 403 K (130◦ C) = 2240 bar m3 /kmol where PA is the partial pressure in the gas phase and CA is the equilibrium concentration in the liquid. (a) If the reactor is operated with a partial pressure of hydrogen equal to 1 MN/m2 (10 bar) calculate the rate at which the hydrogenation will proceed per unit volume of the three-phase system. 282 (b) Consider this overall rate in relation to the operating conditions and the individual transfer resistances. Discuss the question of whether any improvements might be made to the conditions specified for the reactor. Solution For a first-order reaction, the reaction rate, R = k Cs η where Cs is the concentration at the particle surface. On the basis of unit volume of the three-phase dispersion, the reaction rate becomes Rt kmol/m3 s. For mass transfer from the gas to the liquid: Rt = kL a(Ci − CL ) (i) where Ci is the concentration of the gas–liquid interface and CL in the bulk liquid. For mass transfer from the liquid to the particle surface: Rt = ks ap εp (CL − Cs ) (ii) where ap is the external surface per unit volume of particles and εp the volume fraction of the solid particles. For diffusion and reaction within the particles: Rt = k Cs ηεp (iii) For spherical particles: ap = πdp2 /(πdp3 /6) = 6/dp Rearranging and adding equations (i), (ii) and (iii) gives:   1 1 1 + = Ci + Rt kL a ks (6/dp )εp k ηεp (gas–liquid) (liquid–solid) (diffusion + reaction) Writing Rt = Kv Ci , then: (1/Kv ) = (1/kL a) + 1/(ks εp (6/dp )) + 1/k ηεp For transfer from the gas to the liquid: 1/kL a = (1/0.20) = 5.0 s The ratio: volume of solid/volume of liquid = 0.03 εp /εL = 0.03 or: εG = 0.15 and hence: (εp + εL ) = (1 − 0.15) = 0.85 εp (1 + 1/0.03) = 0.85 and: εp = 0.0248 283 Thus: 1/(ks (6/dp )εp ) = 1/[2.2 × 10−3 (6/0.1 × 10−3 )0.0248] = 0.305 s The Thiele Modulus, φ = (dp /6)(k /De )0.5 = (0.1 × 10−3 /6)(90/1.9 × 10−9 )0.5 = 3.63 and hence: Thus: and: η = (1/3.63)(coth(3 × 3.63) − 1/(3 × 3.63)) = 0.251 1/k ηεp = 1/(90 × 0.251 × 0.0248) = 1.78 s 1/Kv = (5.0 + 0.305 + 1.78) = 7.09 s Ci = P /H = (10/2240) = 4.46 × 10−3 kmol/m3 and hence: Rate of hydrogenation, Rt = Kv Ci = (4.46 × 10−3 /7.09) = 0.63 × 10−3 kmol/m3 s Considering the three resistances, that of the gas to liquid transfer is the greatest. kL a might be improved, although, since kL a and εG are already fairly high, there is probably little scope for this. Solids loading might be increased and the overall rate of reaction could be increased by increasing the overall pressure. 284 SECTION 3-5 Biochemical Reaction Engineering PROBLEM 5.1 The residence time, based on fresh feed, in an activated-sludge waste-water treatment unit is 21.2 Ms (5.9 h). The fresh feed has a BOD of 275 mg/l and the settler produces a recycle stream containing 6000 mg/l. Using a sludge of age 6 days, calculate the recycle ratio and the final effluent BOD, assuming that it contains no biomass, given that the yield coefficient Y is 0.54 and that the specific growth rate of the sludge is given by: μ= μm S − kd Ks + S (equation 5.70) where S is the substrate concentration (BOD), μm = 0.47 h−1 , Ks = 89 mg (BOD)/l and the endogenous respiration coefficient kd = 0.009 h−1 . Solution The flow diagram is given in Figure 5a. Sludge age, θc = Biomass in reactor/net rate of biomass generation. = XV /[(μm SXV /(Ks + S)) − kd XV ] = (Ks + S)/(μm S − kd (Ks + S)) From which: S = Ks (1 + kd θc )/[θc (μm − kd ) − 1] Thus: 89(1 + 0.009 × 6 × 24) 6 × 24(0.47 − 0.009) − 1 = 3.12 mg (BOD)/l Final concentration, S = From equation 5.151, Volume 3 a material balance for the substrate across the aeration tank gives: F0 S0 + FR S − (F0 + FR )S − (1/Y )μm SXV /(Ks + S) = V dS/dt At steady state, dS/dt = 0. 285 Fresh feed F0, S0, X0 = 0 Aeration tank Thickener-settler Final effluent Fe, S, Xe Volume, V Biomass, X Substrate, S Recycled sludge Wasted sludge Fw, S, XR FR, S, XR Figure 5a. Flow diagram for Problem 5.1 Dividing throughout by F0 , the recycle ratio, R = FR /F0 and the hydraulic residence time, θ = V /F0 , then: S0 + RS − S − RS − [μm SXθ/Y (Ks + S)] = 0 X = [(S0 − S)Y (Ks + S)]/μm Sθ and: Hence: X = (275 − 3.12)0.54(89 + 3.12)/(0.47 × 3.12 × 5.9) = 1569 mg/l Thus, the concentration factor β for the thickener–settler is: β = XR /X = (6000/1569) = 3.82 From Equation 5.152, a material balance for biomass over the aeration tank gives: F0 X0 + FR XR − (F0 + FR )X + (μm SXV /(Ks + S)) − kd XV = V (dX/dt) Since X0 is very much smaller than XR , it may be assumed that X0 = 0. At steady state, dX/dt = 0 and hence: Rβ − 1 − R + (μm S/(Ks + S))θ − kd θ = 0 Thus: R = [1 − θ ((μm S/(Ks + S)) − kd )]/(β − 1) Hence:    0.47 × 3.12 Recycle ratio, R = 1 − 5.9 − 0.009 89 + 3.12 = 0.34 286 (3.82 − 1) PROBLEM 5.2 A continuous fermenter is operated at a series of dilution rates though at constant, sterile, feed concentration, pH, aeration rate and temperature. The following data were obtained when the limiting substrate concentration was 1200 mg/l and the working volume of the fermenter was 9.8 l. Estimate the kinetic constants Km , μm and kd as used in the modified Monod equation: μm S − kd μ= Ks + S and also the growth yield coefficient Y . Feed flowrate (l/h) 0.79 1.03 1.31 1.78 2.39 2.68 Exit substrate concentration (mg/l) 36.9 49.1 64.4 93.4 138.8 164.2 Dry weight cell density (mg/l) 487 490 489 482 466 465 Solution The flow diagram is as Figure 5.56 in Volume 3, where the inlet and outlet streams are defined as F0 , X0 , S0 and F0 , X, S respectively. The accumulation = input − output + rate of formation, which for the biomass gives: V (dX/dt) = F X0 − F X + V (μm SX)/(Ks + S) − kd XV (equation 5.126) and for the substrate: V (dS/dt) = F S0 − F S − V μm SX/Y (Ks + S) (equation 5.127) At steady state, dS/dt = 0. Taking the dilution rate, D = F /V , then the balance for the substrate becomes: S0 − S − {μm SX/[DY (Ks + S)]} = 0 or: X/D(S0 − S) = (Ks Y /μm )/S + Y /μm Similarly, for the biomass: X0 − X + {μm SX/[D(Ks + S)]} − kd X/D = 0 Since the feed is sterile, X0 = 0 and: DX = [μm SX/(Ks + S)] − kd X. 287 (i) From the material balance for substrate: (μm SX)/(Ks + S) = DY (S0 − S) and substitution gives: DX = DY (S0 − S) − kd X (S0 − S)/X = kd /DY − 1/Y or: (ii) From equation (i), it is seen that a plot of X/D(S0 − S) and 1/S will produce a straight line of slope Ks Y /μm and intercept Y /μm . From equation (ii), it is seen that a plot of (S0 − S)/X against 1/D will produce a straight line of slope kd /Y and intercept 1/Y . The data are calculated as follows: Feed flowrate (F l/h) Exit substrate (S mg/l) 1/S (l/mg) X/D(S0 − S) (h) 1/D (h) (S0 − S)/X (−) 0.79 1.03 1.31 1.78 2.39 2.68 36.9 49.1 64.4 93.4 138.8 164.2 0.0271 0.0204 0.0155 0.0107 0.0072 0.0061 5.19 4.05 3.22 2.40 1.80 1.64 12.41 9.51 7.48 5.51 4.10 3.66 2.388 2.349 2.322 2.296 2.277 2.228 The data are then plotted in Figure 5b from which: Ks Y /μm = 170, Y /μm = 0.59, kd /Y = 0.0133 and 1/Y = 2.222. 2.4 5 4 slope, KsY / µm = 170 3 2 1 Y / µm = 0.59 0 (S0 − S ) / X (−) X /D (S0 − S ) (h) 6 slope, kd /Y = 0.0133/ h 2.3 X 1/Y = 2.222 2.2 0 0.01 0.02 0.03 0 2 1/S (l/mg) Figure 5b. 4 6 8 10 12 1/D (h) Graphical work for Problem 5.2 Thus: yield coefficient, Y = (1/2.222) = 0.45 endogenous respiration coefficient, kd = (0.45 × 0.0133) = 0.006 h−1 μm = (0.45/0.59) = 0.76 h−1 and: Ks = (170 × 0.76/0.45) = 300 mg/l 288 PROBLEM 5.3 When a pilot-scale fermenter is run in continuous mode with a fresh feed flowrate of 65 l/h, the effluent from the fermenter contains 12 mg/l of the original substrate. The same fermenter is then connected to a settler–thickener which has the ability to concentrate the biomass in the effluent from the tank by a factor of 3.2, and from this a recycle stream of concentrated biomass is set up. The flowrate of this stream is 40 l/h and the fresh feed flowrate is at the same time increased to 100 l/h. Assuming that the microbial system follows Monod kinetics, calculate the concentration of the final clarified liquid effluent from the system. μm = 0.15 h−1 and Ks = 95 mg/l. Solution A flow diagram is given in Figure 5c. Case 1 − no recycle Feed flowrate, F0 Biomass concentration, X0 Substrate concentration, S0 Case 2 − with recycle stream F0 X S =12 mg/ l F0′ F0 − Fw volume,V Fw , XR FR, XR, S Figure 5c. Flow diagrams for Problem 5.3 Case 1: No recycle A material balance for biomass over the aeration tank gives: biomass in feed − biomass in effluent + biomass formed by growth = accumulation of biomass or: F0 X0 − F0 X + μXV = V (dX/dt) (equation 5.126) Taking the dilution rate, D = F0 /V , noting that X0 = 0 and that at steady state, dX/dt = 0, then: D=μ and assuming Monod kinetics apply: D = μm S/(Ks + S) Thus: (equation 5.132) D = (0.15 × 12)/(95 + 12) = 0.0168 h −1 The volume of the aeration tank, V = F0 /D or: V = (6.5/0.0168) = 3864 l 289 Case 2: With recycle The material balance for biomass over the aeration tank becomes: biomass in feed + biomass in recycle − biomass leaving the settler − biomass formed = accumulation of biomass or: F0 X0 + FR XR − (F0 + FR )X + μXV = V (dX/dt) For sterile feed, X0 = 0 and, at steady state:   FR μV FR XR − 1− + =0 F0 X F0 F0 (equation 5.152) (equation 5.153) If β, the concentrating effect of the thickener–settler is (XR /X), then: Specific growth rate, μ = (F0 /V )(1 + (FR /F0 )(1 − β)) μ = [1 + (40/100)(1 − 3.2)](100/3864) or: From equation 5.133: = 0.0031 h−1 S = Ks μ/(μm − μ) = (95 × 0.0031)/(0.15 − 0.0031) = 2.0 mg/l PROBLEM 5.4 When a continuous culture is fed with substrate of concentration 1.00 g/l, the critical dilution rate for washout is 0.2857 h−1 . This changes to 0.0983 h−1 if the same organism is used but the feed concentration is 3.00 g/l. Calculate the effluent substrate concentration when, in each case, the fermenter is operated at its maximum productivity. Solution At incipient washout, the critical dilution rate, Dcrit , is related to the Monod constants by: Dcrit = μm S0 /(Ks + S0 ) (equation 5.148) where S0 is the concentration of substrate in the feed. Rearranging: μm = Dcrit (Ks + S0 )/S0 and for the initial conditions: μm = 0.2857(Ks + 1.0)/1.0 − 0.2857Ks + 0.2857 290 (i) For the increased feed rate: μm = 0.09833Ks + 0.295 (ii) From equations (i) and (ii): Ks = 0.0496 g/l and μm = 0.30 h−1 . The maximum cell productivity occurs at an optimum dilution rate, Dopt , given by: " #  Ks (equation 5.140) (iii) Dopt = μm 1 − Ks + S0 and the substrate concentration for any dilution rate below the critical value is given by: S = DKs /(μm − D) (iv) Thus, for the initial conditions: $ %  Dopt = 0.30 1 − [0.0496/(0.0496 + 1.00)] = 0.235 h−1 and: S1 = (0.235 × 0.0496)/(0.30 − 0.235) = 0.18 g/l For the increased flowrate: and: $ %  Dopt = 0.30 1 − [0.0496/(0.0496 + 3.00)] = 0.262 h−1 S2 = (0.262 × 0.0496)/(0.30 − 0.262) = 0.34 g/l PROBLEM 5.5 Two continuous stirred-tank fermenters are arranged in series such that the effluent of one forms the feed stream of the other. The first fermenter has a working volume of 100 l and the other has a working volume of 50 l. The volumetric flowrate through the fermenters is 18 h−1 and the substrate concentration in the fresh feed is 5 g/l. If the microbial growth follows Monod kinetics with μm = 0.25 h−1 , Ks = 0.12 g/l, and the yield coefficient is 0.42, calculate the substrate and biomass concentrations in the effluent from the second vessel. What would happen if the flow were from the 50 l fermenter to the 100 l fermenter? Solution The flow diagram for this operation is shown in Figure 5.62. Together with the relevant nomenclature. A material balance for biomass over the first fermenter, as discussed in section 5.11.3, leads to the equation: D1 = μ1 (equation 5.131) 291 where D1 is the dilution rate in the first vessel and μ1 is the specific growth rate for that vessel. Assuming Monod kinetics to apply: D1 = μm S1 /(Ks + S1 ) (equation 5.132) where S1 is the steady-state concentration of substrate in the first vessel, where: S1 = D1 Ks /(μm − D1 ) (equation 5.133) If D1 = F /V1 = (18/100) = 0.18 h−1 , then: S1 = (0.18 × 0.12)/(0.25 − 0.18) = 0.309 g/l Since the feed to this fermenter is sterile, X0 = 0 and from equation 5.134, Volume 3 the steady-state concentration of biomass in the first vessel is given by: X1 = Y (S0 − S1 ) (equation 5.134) = 0.42(5 − 0.309) = 1.97 g/l In a similar way, a mass-balance over the second vessel gives: D2 = μ2 X2 /(X2 − X1 ) (equation 5.167) where D2 is the dilution rate in the second vessel, μ2 is the specific growth rate in that vessel and X2 the steady-state concentration of biomass. The yield coefficient to the second vessel is then: Y = (X2 − X1 )/(S1 − S2 ) where S2 is the steady-state concentration of substrate in that vessel. Thus: X2 = X1 + Y (S1 − S2 ) = 1.97 + 0.42(0.309 − S2 ) = 2.1 + 0.42S2 Substituting this equation for X2 into equation 5.167, together with values for D2 , μ2 and X1 leads to a quadratic equation in S2 : 0.128S22 + 1.379S2 − 0.01555 = 0 from which: S2 = 0.0113 g/l and: X2 = 2.1 + (0.42 × 0.0113) = 2.1 g/l 292 When the tanks are reversed, that is with fresh feed entering the 50 litre vessel, then the dilution rate for this vessel will be as before, 0.36 h−1 , although the critical dilution rate will now be: Dcrit = μm S0 /(Ks + S0 ) (equation 5.148) = (0.25 × 5)/(0.12 + 5) = 0.244 h−1 This is lower than the dilution rate imposed and washout of the smaller vessel would take place. The concentrations of substrate and biomass in the final effluent would eventually be those attained if only the 100 litre vessel existed, that is: concentration of biomass = 1.97 g/l concentration of substrate = 0.309 g/l. 293 SECTION 3-7 Process Control PROBLEM 7.1 After being in use for some time, a pneumatic three-term controller as shown in Figure 7.118, Volume 3, develops a significant leak in the partition between the integral bellows and the proportional bellows. It is known that the rate of change of pressure in the integral bellows due to the leak is half that due to air flow through the integral restrictor. Show that the leak does not affect the form of the output response of the controller and that the ratio of the gain of the controller with the leak to that of the same controller before the leak developed is given by: 3τ2 + τ1 2τ2 + τ1 where τ1 and τ2 are the time constants of the integral and derivative restrictors respectively. Solution The relevant diagram is included as Figure 7a. The change in separation of the flapper from the nozzle at X is due to the net movement of B and C and the relative lengths of l1 and l2 . Movement of B = k2 (p1 − p2 ) Movement of C = −k1 ε where k1 and k2 are constants and ε, the error, is the difference in movement between E, the set-point, and F , the measured value. Hence: net movement of flapper at X = −k1 ε  l2 l1 + l2  + k2 (p1 − p2 ) The change in output pressure is proportional to this, or:      l1 l2 + k2 (p1 − p2 ) P = C −k1 ε l1 + l2 l1 + l2 where C is the amplification factor. 294  l1 l1 + l2  (i) Leak t3 B A p1 Output p2 l2 t1 t2 Supply X E l1 Set point D Px Nozzle C F Figure 7a. Measured value Diagram for problem 7.1 If C is large, then from equation (i): p1 − p2 =  k1 k2   l2 ε = Kε l1 (ii) where K is a constant for a given controller mechanism. If p1 > p2 , then air will flow from the proportional bellows through the integral restrictor and through the leak. As the rate of change of pressure in the bellows is proportional to this difference is pressure, then: dp2 1 1 1 (p1 − p2 ) = (p1 − p2 ) + (p1 − p2 ) = dt τ1 τ3 τA (iii) where τ1 and τ3 are the time constants for the integral restrictor and the leak respectively. For the derivative restrictor and proportional bellows: 1 dp1 = (p − p1 ) − dt τ2 1 = (p − p1 ) − τ2 1 1 (p1 − p2 ) − (p1 − p2 ) τ1 τ3 1 (p1 − p2 ) τA (iv) where P is the output pressure for t > 0. From equations (ii) and (iii): Thus: dp2 K ε = dt τA  K t ε dt p2 − Px = τA 0 (v) (p2 = Px 295 at t = 0) (vi) Substituting equations (v) and (vi) into equation (iii):    K K t 1 p1 − Px − ε= ε dt τA τA τA 0  K t Thus: p1 = Px + Kε + ε dt τA 0 (vii) and, with constant Px : dp1 dε K =K + ε dt dt τA (viii) Substituting equations (iii), (vii), and (viii) into equation (iv):    K K t K 1 dε + ε= ε dt − ε p − Px − Kε − K dt τA τ2 τA 0 τA     t 1 dε 2τ2 +1 ε+ ε dt + τ2 p − Px = K τA τA 0 dt This is the change in the controller output at time t and is in standard PID form. Hence the gain of controller with the leak is:   2τ2 + τA (Gain)1 = K τA Following the same procedure without the leak (that is τ3 = ∞) gives τA = τ1 , from equation (iii). Thus, for no leak:   2τ2 + τ1 (Gain)2 = K τ1 Hence: τ1 (2τ2 + τA ) (Gain)1 = (Gain)2 τA (2τ2 + τ1 ) (ix) The rate of change of pressure in the integral bellows due to the leak is, however, half that due to flow through the integral restrictor, or: and: τ3 = 2τ1 1 1 3 1 = + = τA τ1 2τ1 2τ1 and τA = From equation (ix): τ1 (2τ2 + 32 τ1 ) (Gain)1 = 2 (Gain)2 3 τ1 (2τ2 + τ1 ) = (3τ2 + τ1 ) (2τ2 + τ1 ) 296 2 τ1 3 PROBLEM 7.2 A mercury thermometer having first-order dynamics with a time constant of 60 s is placed in a bath at 308 K (35◦ C). After the thermometer reaches a steady state it is suddenly placed in a bath at 313 K (40◦ C) at t = 0 and left there for 60 s, after which it is immediately returned to the bath at 308 K (35◦ C). (a) Draw a sketch showing the variation of the thermometer reading with time. (b) Calculate the thermometer reading at t = 30 s and at t = 120 s. (c) What would be the reading at t = 6 s if the thermometer had only been immersed in the 313 K bath for less than 1 s before being returned to the 308 K bath? Solution (a) At t = 0 the thermometer is subjected to a step change of 5 deg K. Thus: 1 ϑ t1 = 1 + 60 s ϑb 5 ϑb = at t = 0 s    5 1 ϑ t1 = s 1 + 60 s A B + s 1 + 60 s A = 5, B = −300 5 300 5 5 ϑ t1 = − = − s 1 + 60 s s 1/60 + s = Thus: ϑ t1 = 5 − 5e−t/60 = 5(1 − e−t/60 ) After 60 s, ϑ t1 = 5(1 − e−1 ) = 3.16 deg K Hence after 60 s, the thermometer will read (308 + 3.16) = 311.16 K. At t = 60 s, a further negative step change is imposed of 3.16 deg K. The thermometer will respond immediately to this as it has only first order dynamics. Thus: ϑt2 = 3.16(1 − e−t/60 ) (b) At t = 30 s, the thermometer reading is: ϑt1 = 5(1 − e−30/60 ) = 1.97 deg K and: thermometer reading = 309.97 K 297 At t = 120 s, since a step decrease is applied at t = 60 s, then: ϑt2 = 3.16(1 − e−60/60 ) = 2 deg K Thus: Thermometer reading at 120 s = (311.16 − 2) = 309.16 K (c) At t = 0, the thermometer is immersed for less than 1 s in the 313 K bath, hence assuming that an impulse applied is: 1 ϑ t1 = 1 + 60 s ϑb From Section 7.8.5: d {F(t)}step dt d 1 −t/60 Thus: = {5(1 − e−t/60 )} = e dt 12 1 −0.1 e = 0.075 deg K Thus at t = 6 s ϑt1 = 12 and the thermometer will read 308.075 K F(t)Impulse = 40 313 39 312 38 311 37 310 36 309 Temperature (K) Temperature Θt (°C) The variation of temperature reading with time is shown in Figure 7b. 35 0 Figure 7b. 30 60 Time t (s) 90 120 Thermometer reading as a function of time, Problem 7.2 PROBLEM 7.3 A tank having a cross-sectional area of 0.2 m2 is operating at a steady state with an inlet flowrate of 10−3 m3 /s. Between the liquid heads of 0.3 m and 0.09 m the flow-head characteristics are given by: Q2 = 0.002Z + 0.0006 298 where Q2 is the outlet flowrate and Z is the liquid level. Determine the transfer functions relating (a) inflow and liquid level, (b) inflow and outflow. If the inflow increases from 10−3 to 1.1 × 10−3 m3 /s according to a step change, calculate the liquid level 200 s after the change has occurred. Solution (a) A mass balance gives: Q1 ρ − Q2 ρ = d d (ρV ) = (ρAZ) dt dt where Q1 is the inlet flow, ρ the fluid density, V the volume of fluid in the tank and A the area of the base (ρ and A are constants). Thus: Q1 − Q2 = A At steady-state: Q′1 − Q′2 = 0 dZ dt (i) (ii) Subtracting equation (ii) from equation (i), then: dZ dZ =A dt dt dZ Q1 − Q2 = A dt Q1 = Q1 − Q′1 , Q2 = Q2 − Q′2 , (Q1 − Q′1 ) − (Q2 − Q′2 ) = A where: But: Q2 = 0.002Z + 0.0006 Thus: Q2 = 0.002Z Q1 − 0.002Z = A dZ dt Transforming: Q 1 − 0.002Z = A dZ dt GA = Z/Q 1 = (b) From equations (iv) and (v): GB = Q 2 /Q 1 = 500 1 + 100 s 1 1 + 100 s For a step change in Q1 of 10−4 m3 /s: Q1 = 299 Z = Z − Z′ (iv) From equations (iii) and (iv): Hence, the transfer function, (iii) 10−4 s (v) (vi) 10−4 500 5 × 10−4 · = s 1 + 100 s s(0.01 + s)   1 1 − = 0.05 s s + 0.01 Thus, from equation (v): Z= Thus: Z = 0.05(1 − e−0.01t ) At t = 200 s: At the original steady-state: Thus: and: Z = 0.05(1 − e−2 ) = 0.043 m. Q′2 = 0.002Z ′ + 0.006 = Q′1 0.002Z ′ + 0.0006 = 0.001 original steady-state level Z1′ = 0.2 m Hence, level at t = 200 s is (0.2 + 0.043) = 0.243 m PROBLEM 7.4 A continuous stirred tank reactor is fed at a constant rate F m3 /s. The reaction is: A −−−→ B which proceeds at a rate of: R = k C0 where: R = kmoles A reacting/m3 mixture in tank (s), k = first order reaction velocity constant (s−1 ), and C0 = concentration of A in reactor (kmol/m3 ). If the density ρ and volume V of the reaction mixture in the tank are assumed to remain constant, derive the transfer function relating the concentration of A in the reactor at any instant to that in the feed stream Ci . Sketch the response of C0 to an impulse in Ci . Solution If the volume V is constant then outlet flow = inlet flow = F m3 /s. Assuming the fluid is well-mixed, then outlet concentration = C0 kmol/m3 . A mass balance on component A gives: d (V C0 ) dt dC0 F Ci − (F + k V )C0 = V dt ′ ′ F Ci − (F + k V )C0 = 0 F Ci − F C0 − RV = At steady-state: Subtracting equation (ii) from equation (i): F Ci − (F + k V )C0 = V where Ci , C0 are deviation variables. 300 (i) (ii) dC0 dt F C i − (F + k V )C 0 = V s C Transforming: 0 F (F + k V ) + V s F /(F + k V ) = 1 + [V /(F + k V )]s The transfer function = G(s) = C 0 /C i = For impulse of area (magnitude) AI : C i = AI Thus: Thus: C 0 = C0 = (equation 7.78) AI K 1 + τs where A = F /(F + kV ) AI K −t/τ e τ and τ = V /(F + kV ) The response of C0 will be as in Figure 7.30, Volume 3, with τ C0 /AI K as the abscissa and t/τ as the ordinate. PROBLEM 7.5 Liquid flows into a tank at the rate of Q m3 /s. The tank has three vertical walls and one sloping outwards at an angle β to the vertical. The base of the tank is a square with sides of length x m and the average operating level of liquid in the tank is Z0 m. If the relationship between liquid level and flow out of the tank at any instant is linear, develop an expression for the time constant of the system. Solution For constant density throughout, a mass balance gives: Q − Qout = dV /dt where: (i) V = x 2 Z + 12 (xZ 2 tan β) and Z is the liquid level at any instant. This is a non-linear relationship. Linearising using Taylor’s series, as given in equation 7.24, Volume 3:   dV V = V0 + (Z − Z0 ) + · · · = k1 + k2 Z (ii) dZ Z0 where: 1 V0 = x 2 Z0 + xZ02 tan β = k1 2   dV = x 2 + xZ0 tan β = k2 and Z = Z − Z0 dZ Z0 301 d dZ (k1 + k2 Z) = k2 dt dt =0 From equations (i) and (ii): Q − Qout = (iii) At steady-state: Q′ − Q′out (iv) Q − Qout = k2 Subtracting equation (iv) from equation (iii): dZ dt There is a linear relation between Qout and Z and Qout = k3 Z dZ From equations (v) and (vi): Q − k3 Z = k 2 dt (v) (vi) Q − k3Z = k2 s Z Transforming: Hence, the transfer function G(s) = Z/Q = = 1 k 3 + k2 s 1/k3 1 + (k2 /k3 )s and the process time constant is k2 /k3 . PROBLEM 7.6 Write the transfer function for a mercury manometer consisting of a glass U-tube 0.012 m internal diameter, with a total mercury-column length of 0.54 m, assuming that the actual frictional damping forces are four times greater than would be estimated from Poiseuille’s equation. Sketch the response of this instrument when it is subjected to a step change in an air pressure differential of 14,000 N/m2 if the original steady differential was 5000 N/m2 . Draw the frequency-response characteristics of this system on a Bode diagram. Solution From Section 7.5.4, Volume 3, for a manometer in which fractional damping is four times that predicted by Poiseuille’s equation: KMT τ 2 s 2 + 2ζ τ s + 1     l 0.54 τ= = ≈ 0.17 s 2g 2 × 9.81 G(s) = where: and: At 298 K: ζ =4× 8μ d 2ρ  2l g  μHg = 1.6 × 10−3 Ns/m2 and ρHg = 13,530 kg/m3 302 (equation 7.52) ζ = Thus: KMT =  32 × 1.6 × 10−3 0.0122 × 13,530   2 × 0.54 9.81  = 8.7 × 10−3 1 1 = = 3.8 × 10−6 m4 /kg s2 2ρg (2 × 13,530 × 9.81) Clearly ζ < 1 and the step response may be calculated from equation 7.82, Volume 3, where M = 14,000 N/m2 . ζ ≈ 0 however and thus the step response will approach a continuous oscillation with constant amplitude as shown in Figure 7.28, Volume 3. The frequency response may be determined from equations 7.94 and 7.95, Volume 3. PROBLEM 7.7 The response of an underdamped second-order system to a unit step change may be shown to be:    1 2 t ζ sin (1 − ζ ) exp(−ζ t/τ ) Y(t) = 1 − √ τ (1 − ζ 2 )     t + (1 − ζ 2 ) cos (1 − ζ 2 ) τ Prove that the overshoot for such a response is given by:  exp{−πζ / (1 − ζ 2 )} and that the decay ratio is equal to the (overshoot)2 . A forcing function, whose transform is a constant K is applied to an under-damped second-order system having a time constant of 0.5 min and a damping coefficient of 0.5. Show that the decay ratio for the resulting response is the same as that due to the application of a unit step function to the same system. Solution The overshoot is represented by the maximum of the first peak of the system response, as shown in Figure 7.57. This maximum is given by: or: dY (t) =0 dt  " √     ζ (1 − ζ 2 ) 1 ζt 2 t (1 − ζ ) − exp cos − √ τ τ τ (1 − ζ 2 )        t t (1 − ζ 2 ) + ζ sin (1 − ζ 2 ) (1 − ζ 2 ) − sin τ τ τ       1 ζt ζ 2 2 t ·√ =0 + (1 − ζ ) cos exp − (1 − ζ ) τ τ τ (1 − ζ 2 ) 303 For finite t and α = Thus: √ (1 − ζ 2 ) then: τ   (1 − ζ 2 + ζ 2 ) sin αt + [−ζ (1 − ζ 2 ) + ζ (1 − ζ 2 )] cos αt = 0 sin αt = 0 nπ t= α n = 0, 1, 2 . . . For the first peak, n = 1 and t = π/α Hence: Y(t)max|n=1    πζ 1 − √ [ζ sin π + (1 − ζ 2 ) cos π] =1− √ exp (1 − ζ 2 ) (1 − ζ 2 )   πζ = 1 + exp − √ (i) (1 − ζ 2 ) The final value of Y(t) = Y(∞) = 1   πζ Thus, the overshoot of first peak from equation 1 is exp − √ (1 − ζ 2 ) For the second peak, n = 3 and t = Thus: (ii) 3π α    1 3πζ − √ [ζ sin 3π + (1 − ζ 2 ) cos 3π] exp 2 2 (1 − ζ ) (1 − ζ )   3πζ = 1 + exp − √ (1 − ζ 2 )     3πζ πζ overshoot peak 2 exp − Decay ratio = = exp − √ √ overshoot peak 1 (1 − ζ 2 ) (1 − ζ 2 ) 2   πζ = exp − √ (1 − ζ 2 ) Y(t)max |n=3 = 1 − √ = {overshoot (peak 1)}2  (1 − ζ 2 ) = 0.866 (iii) τ = 0.5 min, ζ = 0.5 Thus: Y(t) = 1 − 1.15e−t (0.5 sin 1.73t + 0.866 cos 1.73t) α = 1.73 The forcing function of transform K is an impulse of magnitude K. The response to dY(t) an impulse given by equation 7.99, Volume 3 is: × K = X(t) dt = −1.15Ke−t (0.866 cos 1.73t − 1.5 sin 1.73t) + (0.5 sin 1.73t + 0.866 cos 1.73t)(1.15Ke−t ) X(t) = e−t (2 sin 1.73t) 1.15K 304 (iv) The overshoot for an impulse response is given by: That is: d dt  X(t) 1.15 K  =0 −2 sin 1.73t + 3.46 cos 1.73t = 0 tan 1.73t = 1.73 Thus: 1.73t = π/3 radians for peak 1 and (2π + π/3) radians for peak 2 Thus: tpeak 1 = 0.193π min. tpeak 2 = 1.35π min. The decay ratio for the impulse response from equation (iv) = " πζ Decay ratio for equivalent step response = exp −  (1 − ζ 2 ) e−1.35π = (e−0.193π )6 e−0.193π #2 = (e−0.577π )2 = (e−0.193π )6 and thus the two decay ratios are the same. PROBLEM 7.8 Air containing ammonia is contacted with fresh water in a two-stage countercurrent bubble-plate absorber. Ln and Vn are the molar flowrates of liquid and gas respectively leaving the nth plate. xn and yn are the mole fractions of NH3 in liquid and gas respectively leaving the nth plate. Hn is the molar holdup of liquid on the nth plate. Plates are numbered up the column. A. Assuming (a) temperature and total pressure throughout the column to be constant, (b) no change in molar flowrates due to gas absorption, (c) plate efficiencies to be 100 per cent, (d) the equilibrium relation to be given by yn = mxn∗ + b, (e) the holdup of liquid on each plate to be constant and equal to H , and (f) the holdup of gas between plates to be negligible, show that the variations of the liquid compositions on each plate are given by: mV 1 dx1 = (L2 x2 − L1 x1 ) + (x0 − x1 ) dt H H dx2 1 mV = (x1 − x2 ) − L2 x2 dt H H where V = V1 = V2 . B. If the inlet liquid flowrate remains constant, prove that the open-loop transfer function for the response of y2 to a change in inlet gas composition is given by: y2 c2 /(a 2 − bc) = 2 y0 {1/(a − bd)}s 2 + {2a/(a 2 − bc)}s + 1 305 where y 2 , y 0 are the transforms of the appropriate deviation variables and: L = L1 = L2 , a= mV L + , H H b= L , H c= mV H Discuss the problems involved in determining the relationship between y and changes in inlet liquid flowrate. Solution The definitions of the various symbols are shown in Figure 7c. Fresh water V 2 y2 Stage 1 L3 x3 V1y1 Stage 2 L2x2 V0y0 Air−ammonia mixture L1 x1 Figure 7c. Nomenclature for Problem 7.8 A. From assumptions (b) and (f): V1 = V 2 = V . An unsteady-state mass balance for ammonia around stage 1 gives: L2 x2 + V0 y0 − L1 x1 − V1 y1 = d(H x1 )/dt = H (dx1 /dt) From assumption (e), H , the total moles on each plate, is constant. 306 (i) A mass balance for ammonia around stage 2 gives: L3 x3 + V1 y1 − L2 x2 − V2 y2 = H dx2 dt (ii) x3 = 0. where: from assumptions (d) and (c): y0 = mx0∗ + c = mx0 + c y1 = mx1∗ + c = mx1 + c y2 = mx2∗ + c = mx2 + c Substituting in equation (i) for y and putting V = V1 = V2 , then: 1 dx1 1 (L2 x2 − L1 x1 ) + (V mx0 − V mx1 ) = H H dt Hence: mV dx1 1 = (L2 x2 − L1 x1 ) + (x0 − x1 ) dt H H (iii) Substituting in equation (ii) for y and putting V = V1 = V2 , then: mV 1 dx2 = (x1 − x2 ) − L2 x2 dt H H (iv) B. Substituting L1 = L2 = L in equations (iii) and (iv) gives: from equation (iii): mV dx1 L = (x2 − x1 ) + (x0 − x1 ) dt H H   mV mV L x0 x1 + = x2 − L/H + H H H = βx2 − αx1 + γ x0 From equation (iv): where: (v) dx2 = −αx2 + γ x1 dt (vi) α = L/H + mV /H, β = L/H and γ = mV . H At steady-state, from equations (v) and (vi): 0 = βx2′ − αx1′ + γ x0′ and: 0 = −αx2′ + γ x1′ where x0′ , x1′ and x2′ are steady-state concentrations. Subtracting steady-state relationships from equations (v) and (vi) respectively: and: dx1 = β x2 − α x1 + γ x0 dt dx2 = −α x2 + γ x1 dt (vii) (viii) 307 x0 = x0 − x0′ , where: and: x1 = x1 − x1′ d dx1 = (x1 − x1′ ) = dt dt dx2 d = (x2 − x2′ ) = dt dt dx1 dt dx2 dt and x2 = x2 − x2′ Transforming, using equations (vii) and (viii), gives: From equation (x): s x1 (s) = β x2 (s) − α x (s) + γ x0 (s) (ix) s x2 (s) = −α x2 (s) + γ x1 (s). s+α x1 (s) = x2 (s) γ (x) Substituting in equation (ix): α(s + α) s(s + α) x2 (s) = β x2 (c) − x2 (s) + γ x0 (s) γ γ Thus: γ2 x2 (s) = x0 (s) s(s + α) − βγ + α(s + α) = But: Thus: and: γ2 s 2 + 2αs + (α 2 − βγ ) yn = mxn∗ + c = mxn + c yn − yn′ = m(xn − xn′ ) or y n = mx n and yn (s)2 mx my2 (s) γ 2 /(α 2 − βγ ) x2 (s) = = 2 x0 (s) my0 (s) [1/(α − βγ )]s 2 + [2α/(α 2 − βγ )]s + 1 PROBLEM 7.9 A proportional controller is used to control a process which may be represented as two non-interacting first-order lags each having a time constant of 600 s (10 min). The only other lag in the closed loop is the measuring unit which can be approximated by a distance/velocity lag equal to 60 s (1 min). Show that, when the gain of a proportional controller is set such that the loop is on the limit of stability, the frequency of the oscillation is given by: −20ω tan ω = 1 − 100ω2 Solution A block diagram is given in Figure 7d. 308 U Controller + R Valve 1 − + + K1 1 + 10s K2 1 + 10s C B e−s Figure 7d. Block Diagram for Problem 7.9 As discussed in Section 7.10.4, Volume 3, the Bode stability criterion states that the total open loop phase shift is −π radians at the limit of stability of the closed loop system. Phase shift of each first-order lag = tan−1 (−10ω) radians Phase shift of DV lag = −ω radians Thus: (equation 7.92) (equation 7.97) Total phase shift = tan−1 (−10ω) + tan−1 (−10ω) − ω = −π tan−1 (−10ω) = 1/2(ω − π) −10ω = tan[1/2(ω − π)]   1 − cos(ω − π) =± 1 + cos(ω − π) Thus: 100ω2 = 1 + cos ω 1 − cos ω 100ω2 − 1 100ω2 + 1 20ω tan ω = 100ω2 − 1 20ω =− 1 − 100ω2 cos ω = and: PROBLEM 7.10 A control loop consists of a proportional controller, a first-order control valve of time constant τv and gain Kv and a first-order process of time constant τ1 and gain K1 . Show that, when the system is critically damped, the controller gain is given by: Kc = (E − 1)2 τv where E = 4EKv K1 τ1 If the desired value is suddenly changed by, an amount R when the controller is set to give critical damping, show that the error ǫ will be given by:        4E 1+E t t (1 − E)2 (1 − E)2 ε = exp − + + R (1 + E)2 2E(1 + E) τ1 (1 + E)2 2E τ1 309 Solution A block diagram for this problem is shown as Figure 7e. U=0 R + Kc Gc E G1 − B Figure 7e. + + Kv 1+ τv s G2 C K1 1+ τ1s Block diagram for Problem 7.10 The closed loop transfer function is:   Kv K1 C 1 + τv s 1+τ s  1   = K1 K v R (1 + Kc ) 1 + τv s 1 + τ1 s (Kc )  (i) The characteristic equation is: 2 Kc Kv K1 + (1 + τv s)(1 + τ1 s) = 0 (equation 7.119) τr τ1 s + (τv + τ1 )s + (1 + Kc Kv K1 ) = 0 For critical damping, the roots of the characteristic equation are equal, hence:  τv τ1 2 (τv + τ1 )2 = 4τv τ1 (1 + Kc Kv K1 )   τv τv + 1 = 4 Kc Kv K1 −2 τ1 τ1 E = τv /τ1 If: then: 2 E − 2E + 1 = 4EKc Kv K1 Kc = and: From the block diagrams: and: From equation (iv): where, from equation (iii): (ii) (E − 1)2 4EKv K1 (iii) ε =R−B =R−C ε R C R =1− = C (iv) R K , K + (1 + τv s)(1 + τ1 s) K = Kc Kv K1 = (E − 1)2 /4E 310 For a step of magnitude R in R, then:    K R R = R/s and C = s τv τ1 s 2 + (τv + τ1 )s + (1 + K) ⎡ ⎤   ⎥ (K/τv τ1 ) 1 ⎢ C ⎢   ⎥  Thus: = 1+K ⎦ τv + τ1 R s ⎣ 2 s+ s + τv τ1 τv τ1 As the system is critically damped, the roots of the denominator must be equal, that is it factorises to give (s + a)2 , a= where: (τv + τ1 ) 2τv τ1 K C = R τv τ1 Thus:     1 1 s (s + a)2   a K 1 1 − = − τv τ1 a 2 s (s + a)2 s+a K C [1 − (1 + at)e−at ] (t) = R τv τ1 a 2 Inverting: From equation (iii): K = Kc Kv K1 = From equation (ii): τv τ1 a 2 = Hence: where: and: (E − 1)2 4E (τv + τ1 )2 = (1 + K) 4τv τ1 K C (t) = [1 − (1 + at)e−at ] R 1+K (E − 1)2 /4E (E − 1)2 K = = 1+K 1 + (E − 1)2 /4E (E + 1)2       E+1 t t 1 + 1/E = at = 2 τ1 2E τ1 Hence, from equation (iv):       C 1 K ε =1− = + (1 + at)e−at R R 1+K 1+K 4E (E − 1)2 = + (E + 1)2 (E + 1)2          E+1 E+1 t t 1+ exp − 2E τ1 2E τ1 311 PROBLEM 7.11 A temperature-controlled polymerisation process is estimated to have a transfer function of: K G(s) = (s − 40)(s + 80)(s + 100) Show by means of the Routh–Hurwitz criterion that two conditions of controller parameters define upper and lower bounds on the stability of the feedback system incorporating this process. Solution A block diagram for this problem is given in Figure 7f. U R + + B − Gc G1 + G C H Figure 7f. Block diagram for Problem 7.11 Assuming that G1 and H are constant and written as K1 and K2 respectively, that is the time constants of the final control element and measuring element are negligible in comparison with those of the process, and that the proportional controller has a gain KC , Gc = Kc then: C and: R = Gc G1 G 1 + Gc G1 GH The characteristic equation is: 1 + Gc G1 GH = 0 or: Thus: and: Kc K1 K2 K =0 (s − 40)(s + 80)(s + 100) (s − 40)(s + 80)(s + 100) + Kx = 0 where Kx = Kc K1 K2 K. 1+ s 3 + 140s 2 + 800s + (Kx − 320,000) = 0 From Section 7.10.2 in Volume 3, for a stable system Kx ≥ 320,000 312 This is confirmed by the Routh array: 1 800 140 (K − 320,000) x   432,000 − Kx 0 140 (Kx − 320,000) 0 0 0 0 0 Also, from the Routh array: Kx ≤ 432,000 PROBLEM 7.12 A process is controlled by an industrial PI controller having the transfer function: Gc = τI s 2 + (Kc + 2τI )s + 2Kc 2τI2 s The measuring and final control elements in the control loop are described by transfer functions which can be approximated by constants of unit gain, and the process has the transfer function: 1/τ2 G2 = 2 τI τ2 s − (2τI − τ2 )s − 2 If τ1 = 21 , show that the characteristic equation of the system is given by: (s + 2){τ22 s 2 + (1/τI − 2τ2 )s + Kc /τI2 } = 0 Show also that the condition under which this control loop will be stable:  2 1 (a) when: Kc ≥ − τI 2τ2 2  1 − τI (b) when: Kc < 2τ2 is that 1/τI > 2τ2 , provided that Kc and τ2 > 0. Solution A block diagram for this problem is given in Figure 7g. G1 = H = 1 Hence, the characteristic equation is: 1 + Gc G2 = 0    τI s 2 + (Kc + 2τI )s + 2Kc 1/τ2 1+ · =0 τ1 τ2 s 2 − (2τ1 − τ2 )s − 2 2τI2 s  313 U R + + B Gc − + G1 G2 C H Figure 7g. Block Diagram for Problem 7.12 Putting τ1 = 1/2, then: τ22 τI2 s 3 − 2τI2 τ2 (1 − τ2 )s 2 + τI s 2 + (Kc + 2τI − 4τI2 τ2 )s + 2Kc = 0 and: (s + 2)[τ22 s 2 + (1/τI − 2τ2 )s + Kc /τI2 ] = 0   Kc 2Kc 2 2 3 2 2 2 2 − 4τ2 + 2 s + 2 = 0 τ2 s + 2τ2 s + 1/τI s − 2τ2 s + τI τI τI (i) (ii) Equations (i) and (ii) are the identical. Hence, the roots of the characteristic equation are: & #  ' " 2  ' 1 1 − − 2τ2 ± ( − 2τ2 − 4τ22 Kc /τI2 τI τI s = −2 or s = 2τ22 For roots to have negative real parts, that is for the system to be stable, then: 1 > 2τ2 τI Hence, for a stable system: or: Kx ≤ 432,000 and Kx ≥ 320,000     320,000 432,000 ≤ Kc ≤ K1 K2 K K1 K2 K Thus there are two value of Kc defining the upper and lower bounds on the stability of the feedback system. PROBLEM 7.13 The following data are known for a control loop of the type shown in Figure 7.3a, Volume 3: (a) The transfer function of the process is given by: G2 = 1 (0.1s 2 + 0.3s + 0.2) 314 (b) The steady-state gains of valve and measuring elements are 0.4 and 0.6 units respectively; (c) The time constants of both valves and measuring element may be considered negligible. It is proposed to use one of two types of controller in this control loop, either (a) a PD controller whose action approximates to: J = J0 + Kc ε + KD dε dt where J is the output at time t. J0 is the output t = 0, ε is the error, and KC (proportional gain) = 2 units and KD = 4 units; or (b) an inverse rate controller which has the action: J = J0 + Kc ε − KD dε dt where J, J0 , ε, Kc and KD have the same meaning and values as given previously. Based only on the amount of offset obtained when a step change in load is made, which of these two controllers would be recommended? Would this change if the system stability was taken into account? Having regard to the form of the equation describing the control action in each case, under what general circumstances would inverse rate control be better than normal PD control? Solution A block diagram for this problem is included as Figure 7h. G1 = 0.4, G2 = (0.1s 2 H = 0.6 10 1 = + 0.3s + 0.2) (s + 2)(s + 1) U R + + − Gc G1 B H Figure 7h. Block diagram for Problem 7.13 315 + G2 C (a) Gc = Kc {1 + τD s} = Kc + KD s = 2 + 4s = 2(1 + 2s) Closed-loop transfer function for load change, 10 C G2 (s + 2)(s + 1)   = = 10 1 + G G G H U c 1 2 × 0.6 1 + 2(1 + 2s) × 0.4 (s + 2)(s + 1) = 10 10 = 2 (s 2 + 3s + 2 + 4.8 + 9.6s) (s + 12.6s + 6.8) For unit step in load, U = 1/s C = s(s 2 10 + 12.6s + 6.8) Offset = R(∞) − C (∞) (equation 7.113) R(∞) = 0, that is no change in set point.     1 10 C (∞) = lim C (t) = lim s t→∞ s→0 s s 2 + 12.6s + 6.8 Section 7.8.) (final value theorem, given in = (10/6.8) ≈ 1.5 Offset = (0 − 1.5) = −1.5 Thus: (b) Gc = 2(1 − 2s) Thus: C U = 10 10 = 2 (s 2 + 3s + 2 + 4.8 − 9.6s) (s − 6.6s + 6.8) Clearly the offset will be the same as for (a) and no recommendation can be made. In case (a), the characteristic equation is: Thus: s 2 + 12.6s + 6.8 = 0 √ −12.6 ± 131.6 −12.6 ± 11.5 s= = 2 2 = −1.1 or − 24.1 Both roots have negative real parts (as expected) and the system is stable. In case (b), the characteristic equation is: s 2 − 6.6s + 6.8 = 0 Which clearly indicates an unstable system. (Section 7.10.2). Inverse rate control is used in extremely fast processes to introduce lag so that the output is more compatible in terms of time with the final control element. 316 PROBLEM 7.14 A proportional plus integral controller is used to control the level in the reflux accumulator of a distillation column by regulating the top product flowrate. At time t = 0, the desired value of the flow controller which is controlling the reflux is increased by 3 × 10−4 m3 /s. If the integral action time of the level controller is half the value which would give a critically damped response and the proportional band is 50 per cent, obtain an expression for the resulting change in level. The range of head covered by the level controller is 0.3 m, the range of top product flowrate is 10−3 m3 /s and the cross-sectional area of the accumulator is 0.4 m2 . It may be assumed that the response of the flow controller is instantaneous and that all other conditions remain the same. If there had been no integral action, what would have been the offset in the level in the accumulator? Solution Flow and block diagrams for this problem are shown in Figure 7i. The proportional band (= 50 per cent) = 100 × (error required to move the valve from fully open to fully shut)/(full range of level) Error required = (50 × 0.3)/100 = 0.15 m Thus: Variation in controller output corresponding to this error = 0.15Kc . Reflux accumulator LC h FRC Q2 Reflux Q1 Overhead product Q2 U + + R − B Gc G1 H Figure 7i. Flow and block diagrams for Problem 7.14 317 Q1 + G2 h C If the valve is included, then the total variation or range of overhead product flowrate = 0.15Kc , which is equal to the range of the reflux flowrate, or: 10−3 = 0.15Kc Kc = 6.67 × 10−3   1 For a PI Controller: Gc = Kc 1 + τI s For the valve, assuming negligible dynamics: G1 = Kv and: For the measuring element, again assuming negligible dynamics: H = Km Assuming a linear relationship between h and Q1 , then, from Section 7.5.2: G1 = K= where: K , 1 + τs steady-state change in level steady-state change in flowrate = 0.3/10−3 = 300 m s/m3 τ = AK = (0.4 × 300) = 120 s. Thus: Assuming Kv = Km = 1, then the closed loop transfer function for a load change is given by: C U = 300/(1 + 120s)   1 1 + Kc 1 + Kv Km K/(1 + τ s) τI s 300τI s , τI τ s 2 + τI (1 + KA )s + KA 300τI s = 120τI s 2 + 3τI s + 2 = where KA = KKv Km Kc = 2 For a critically-damped system, the roots of characteristic equation are equal, or: 9τI2 = 4 × 2 × 120τI τI = 106.7 s and: Thus: the integral time of the level controller is (106.7/2) = 53 s For a step change in load of 3 × 10−4 m3 /s, then: U = 3 × 10−4 s−1 Thus: C = (3 × 10−4 s−1 ) =  300τI s 120τI s 2 + 3τI s + 2 2.39s + 79.5s + 1) s(3180s 2 318  If there is no integral action, then: C U = K τ s + (KA + 1) From Section 7.9.3: the offset = lim R(t) − lim C (t) t→∞ t→∞     K 3 × 10−4 = −0.03 = 0 − lim s s→0 s τ s + (KA + 1) PROBLEM 7.15 Draw the Bode diagrams of the following transfer functions: 1 s(1 + 6s) (1 + 3s) exp(−2s) (b) G(s) = s(1 + 2s)(1 + 6s) 5(1 + 3s) (c) G(s) = 2 s(s + 0.4s + 1) (a) G(s) = Comment on the stability of the closed-loop systems having these transfer functions. Solution (a) 1 s(1 + 6s) Amplitude Ratio (AR) = AR1 × AR2 G(s) = where: AR1 is the AR of G1 (s) = and: AR2 is the AR of G2 (s) = (equation 7.104) 1 s 1 1 + 6s Phase shift = ψ1 + ψ2 where: ψ1 = phase shift of G1 (s) = 1 s and: ψ2 = phase shift of G2 (s) = AR1 (and ψ1 ) 1 1 + 6s On the Bode AR diagram, 319 (equation 7.105) In G1 (s) = Thus: 1 put s = iω (Substitution Rule — Section 7.8.4) s   1 1 G1 (iω) = =0−i iω ω AR1 = and: 1 ω ◦ and ψ1 = tan−1 (−∞) = −90 log AR1 = − log ω which is a straight line of slope −1 on the log AR/ log ω diagram passing through (1,1) AR2 (and ψ2 ) Thus: and: ψ is −90◦ for all values of ω G2 = 1 which is a first order transfer function 1 + 6s 1 AR2 =  1 + 36ω2 (equation 7.91) ψ2 = tan−1 (−6ω) (equation 7.92) The Bode diagram for a first order system is given in Figure 7.45. The Bode diagram (Figure 7j) shows plots for G1 (s), G2 (s) and G(s) as amplitude ratio against frequency. Only the asymptotes (Section 7.10.4, Volume 3) are plotted. For the phase shift plot ψ against ω for G2 is required. ω ψ ◦ = tan−1 (−6ω) 0.02 0.05 0.1 0.167 0.2 0.3 0.6 1.0 −7 −17 −31 −45 −50 −61 −74 −81 As the phase shift does not cut the −180◦ line, then the equivalent closed-loop system is stable as discussed in Section 7.10.4. (b) (1 + 3s) exp(−2s) s(1 + 2s)(1 + 6s)   1 + 9ω2 1 AR = ω (1 + 4ω2 )(1 + 36ω2 ) G(s) = π ψ = −2ω + tan−1 (3ω) + tan−1 (−2ω) + tan−1 (−6ω) − (in radians) 2 320 1000.00 Amplitude ratio 100.00 10.00 1.00 G1 0.10 G2 G 0.01 0.01 0.10 1.00 10.00 100.00 Frequency (rads/min) 0 −45 Phase shift (degrees) G2 −90 G1 G − 135 − 180 0.01 Figure 7j. 0.10 1.00 Frequency (rads/min) Bode diagram for Problem 7.15(a) 321 10.00 100.00 1000.00 Amplitude ratio 100.00 10.00 1.00 0.10 0.01 0.01 0.10 1.00 Frequency (rads/min) 10.00 100.00 Phase shift (degrees) 0 −90 −180 −270 0.01 Figure 7k. 0.10 1.00 Frequency (rads/min) Bode diagram for Problem 7.15(b) 322 10.00 100.00 1000.00 Amplitude ratio 100.00 10.00 1.00 0.10 0.01 0.01 0.10 1.00 Frequency (rads/min) 10.00 100.00 Phase shift (degrees) 0 −90 −180 −270 0.01 0.10 1.00 Frequency (rads/min) Figure 7l. Bode diagram for Problem 7.15(c) 323 10.00 100.00 This is plotted as in Section 7.10.4, Volume 3, AR > 1 at frequency at which ψ = −180◦ . Thus the system is unstable by Bode criterion. (c) s(1 + 3s) s(s 2 + 0.4s + 1)   1 + 9ω2 5 AR = ω (1 − ω2 )2 + (0.4ω)2   π −0.4ω ψ − = tan−1 (3ω) + tan−1 2 (radians) 1−ω 2 G(s) = This is plotted as in Section 7.10.4. The phase shift plot does not cut the −180◦ line, and hence the system is stable. PROBLEM 7.16 The transfer function of a process and measuring element connected in series is given by: (2s + 1)−2 exp(−0.4s) (a) Sketch the open-loop Bode diagram of a control loop involving this process and measurement lag (but without the controller). (b) Specify the maximum gain of a proportional controller to be used in this control system without instability occurring. Solution G(s) = 0.9e−0.4s /(2s + 1)2 This may be split into a DV lag, e−0.4s = G1 , and two first-order systems, each G2 = 1/(1 + 2s). Amplitude Ratio Plot G2 : For the plot of AR and ω for the first-order systems, the LFA is AR = 1. The HFA is a straight line of slope −1 passing through AR = 1, ωc = 1/τ where τ is the time constant. τ = 2 and hence ωc = 0.5. 1/(1 + 2s)2 = [1/(1 + 2s)][1/(1 + 2s)] constitutes two first-order systems in series and hence an overall AR plot may be produced for both together which will be the same as for G2 other than that the HFA will have a gradient of −2. G1 : The AR plot for a DV lag is AR = 1, which does not affect the overall AR plot. Thus, the plot for e−0.4s /(2s + 1)2 is the same as the AR plot for 1/(2s + 1)2 . The steady-state gain of 0.9 can be included in two ways: either start the overall AR plot at AR = 0.9 instead of AR = 1, or plot AR/0.9 on the vertical scale and start at AR = 1. In the latter case, the gain of the proportional controller calculated for the given 324 gain margin must be divided by the steady-state gain. The latter procedure will be used here as shown in Figure 7m. Phase-Shift Plot G2 : On the basis of two identical first order systems in series, the following is obtained: ψ for 1/(2s + 1)2 ω 0◦ −23◦ −62◦ −90◦ −127◦ −169◦ −180◦ LFA 0.1 0.3 0.5 1.0 5.0 HFA as ψ = 2 tan−1 (−ωτ ) = 2 tan−1 (−2ω). G1: DV lag: ψ = −ωτ radians ω 0.1 0.5 1.0 2.0 ψ −2◦ −11◦ −22◦ −44◦ For ωco , ω is found at ψ = −180◦ ωco = 1.6 radians/unit time. Thus: From the AR diagram at ωco : AR = 0.1 0.9 Hence the maximum allowable gain = (1/0.1) = 10 which is for (AR/0.9). Thus the maximum gain for the proportional controller if the system is to remain stable = (10/0.9) = 11.1. Thus, the value of Kc to give a gain margin of 1.8 = (11.1/1.8) = 6.2 For a phase margin of 35◦ , the allowable ψ on the phase-shift diagram = −180◦ − (−35◦ ) = −145◦ For this value of ψ, the ωco is 0.93 radians/unit time. AR = 0.3 0.9 From the AR plot: Thus: 1 (0.3 × 0.9) = 3.7 ◦ Kc for 35 phase margin = In practice the lower value of Kc = 3.7 would be used for safety. 325 LFA 1.0 AR /0.9 0.1 0.01 A HF 0.001 0.1 1 wco 10 100 w (radians/unit time) −20 0 −20 −40 −60 −80 −0.4s e y (degrees) −100 1 −140 (1 + 2s)2 −180 e−0.4s −220 (1 + 2s)2 −260 −300 10 0.1 wco w (radians/unit time) Figure 7m. Bode plot for Problem 7.16 326 100 PROBLEM 7.17 A control system consists of a process having a transfer function Gp , a measuring element H and a controller GC . If Gp = (3s + 1)−1 exp(−0.5s) and H = 4.8(1.5s + 1)−1 , determine, using the method of Ziegler and Nichols, the controller settings for P, PI, PID controllers. Solution The relevant block-diagram is shown in Figure 7n. U + R B + − Gc + G1 Gp C H Figure 7n. Block diagram for Problem 7.17 To use the Ziegler-Nichols rules, it is necessary to plot the open-loop Bode diagram without the controller. All other transfer functions are assumed to be unity. Gp : This consists of a DV lag of e−0.5s and a first order system, 1/(3s + 1) With the AR plot, only the first-order system contributes: LFA: AR = 1 HFA: The slope = −1, passing through AR = 1, ωc = 1/τ = 0.33 radians/minute. With the ψ plot, for the first order part: ψ = tan−1 (−ωτ ) = tan−1 (−3ω) which gives: ω ψ 0 0.1 0.2 0.33 0.8 2.0 0◦ −17◦ −31◦ −48◦ −67◦ −81◦ With the DV lag plot: ψDV = −ωτDV = −0.5ω radians. 327 which gives: ω ψ 0.5 1.0 2.0 3.0 5.0 8.0 −14◦ −29◦ −57◦ −86◦ −143◦ −229◦ H (measuring element): This is first order with a time constant of 1.5 min. and a steady-state gain of 4.8. With the AR plot, LFA: AR = 1 HFA: The slope is −1 and ωc = (1/1.5) = 0.67 With the ψ plot, ψ = tan−1 (−ωτ ) = tan−1 (−1.5ω) which gives: ω ψ 0 0.2 0.5 0.67 1.0 2.0 ∞ 0◦ −17◦ −37◦ −45◦ −56◦ −72◦ −90◦ The Bode diagrams are plotted for these in Figure 7o. The asymptotes on the AR plots are summed and the sums on the ψ plots are obtained by linear measurement. From the overall ψ plots, at ψ = −180◦ , ωco = 1.3 radians/min and the corresponding value of AR = 0.123. By the Ziegler–Nichols procedure, this means that: Ku = 1 = 8.13 (0.123) Taking into account the steady-state gain of the measuring elements, Ku will be reduced by a factor of 4.8 as both together will give the total gain of the loop.   8.13 Hence: Ku = ≈ 1.7 4.8   2π 2π = and: Tu = ≈ 4.8 min. ωco 1.3 328 10 Amplitude Ratio 1 H 0.1 Gp GpH 0.001 0.1 1 wCO 10 100 Phase Shift (degrees) Frequency (rads/min) 20 0 −20 −40 −60 −80 −100 H Gp(1) −140 Gp(2) −180 GpH −220 −260 −300 wco 0.1 1 10 Frequency (rads/min) Figure 7o. Bode diagram for Problem 7.17 329 100 From the Ziegler–Nichols settings: for P controller Kc = 0.5Ku = 0.85 PI Kc = 0.77 and τI = 4 min. PID Kc = 1.0, τI = 2.4 min, τs = 0.6 min PROBLEM 7.18 Determine the open-loop response of the output of the measuring element in Problem 7.17 to a unit step change in input to the process. Hence determine the controller settings for the control loop by the Cohen-Coon and ITAE methods for P, PI and PID control actions. Compare the settings obtained with those in Problem 7.17. Solution A block diagram for this problem is shown in Figure 7p. U + B P GC R G1 + + e−0.5s 3s + 1 GP C − H 4.8 1.5s + 1 Figure 7p. Block diagram for Problem 7.18 From Problem 7.17, the open-loop transfer function for generation of the process reaction curve is given by: GOL (S) = B P = 4.8e−0.5s (1.5s + 1)(3s + 1) For unit step change in P , then: B = 4.8e−0.5s 1 . s (1.5s + 1)(3s + 1) 330 In order to determine B without the dead-time, then: B = 4.8 1.07 = s(1.5s + 1)(3s + 1) s(s + 0.67)(s + 0.33) B 1 1 2 = + − 4.9 s s + 0.67 s + 0.33 Thus: B = 4.9{1 + e−0.67t − 2e−0.33t } and: (i) Including the dead-time, B will start to respond according to equation 1, 0.5 minutes after the unit step change in P . (a) Using the Cohen–Coon method, then from the process reaction curve shown in Figure 7q: τad = 1.13 min, τa = 6.0 min, Kr = 4.9 5.00 Response 4.00 3.00 Kr 2.00 1.00 0.00 0.0 tad Figure 7q. 5.0 10.0 ta 15.0 Time (min) Process reaction curve for Problem 7.18 From Table 7.4 in Volume 3: For P action: Kc =  1 Kr  τa τad   τad 1+ = 1.2 3τa For PI action: Kc = 0.99 and τI = 2.7 min. and for PID action, the results are obtained in the same way. 331 20.0 (b) Using the Integral Criteria (ITAE) method, then from Table 7.5 and equations 7.139– 7.142 in Volume 3: For P action:     1.13 −1.084 τad σ2 = 0.49 ϒ = σ1 = 2.99 = Kr Kc τa 6 Kc = 0.61 Thus: For PI action: and τI = 2.9 min (174 s) Kc = 0.90 and for PID action, the results are obtained in the same way. PROBLEM 7.19 A continuous process consists of two sections A and B. Feed of composition X1 enters section A where it is extracted with a solvent which is pumped at a rate L1 to A. The raffinate is removed from A at a rate L2 and the extract is pumped to a cracking section B. Hydrogen is added at the cracking stage at a rate L3 whilst heat is supplied at a rate Q. Two products are formed having compositions X3 and X4 . The feed rate to A and L1 and L2 can easily be kept constant, but it is known that fluctuations in X1 can occur. Consequently a feed-forward control system is proposed to keep X3 and X4 constant for variations in X1 using L3 and Q as controlling variables. Experimental frequency response analysis gave the following transfer functions: x2 x1 x3 q x4 q x4 x2 1 , s+1 1 = , 2s + 1 1 = , s+2 1 = 2 s + 2s + 1 = x3 2 = s +1 l3 x4 s+2 = 2 s + 2s + 1 l3 2s + 1 x3 = 2 x2 s + 2s + 1 where x 1 , x 2 , l 3 , q, etc., represent the transforms of small time-dependent perturbations in X1 , X2 , L3 , Q, etc. Determine the transfer functions of the feed-forward control scheme assuming linear operation and negligible distance–velocity lag throughout the process. Comment on the stability of the feed-forward controllers you design. Solution A diagram of this process is shown in Figure 7r. G11 = x 3 /x 1 , G21 = x 4 /x 1 , G12 = x 3 /S 2 , G22 = x 4 /S 2 , 332 G13 = x 3 /Q, and G23 = x 4 /Q. S2 x2 x3 B S1 x4 Q A R x1 Figure 7r. S2/x1 GB G A Q / x1 Process diagram for Problem 7.19 If x1 , s2 and Q all vary at the same time, then by the principle of superposition: x 3 = G11 x 1 + G12 S 2 + G13 Q x 4 = G21 x 1 + G22 S 2 + G23 Q. and: The control criterion is that x3 and x4 do not vary. That is x 3 = x 4 = 0 if these are deviation variables. Thus: G11 x 1 + G12 S 2 + G13 Q = 0 and: G21 x 1 + G22 S 2 + G23 Q = 0 Hence, eliminating S 2 : GA = Q/x 1 = G22 G11 − G21 G12 G12 G23 − G22 G13 From the given data: s+2 s+2 = s 2 + 2s + 1 (s + 1)2 2s + 1 = x 3 /x 1 = x 3 /x 2 · x 2 /x 1 = (s + 1)3 1 = x 4 /x 1 = x 4 /x 2 · x 2 /x 1 = (1 + s)3 2 = x 3 /S 2 = 1+s 1 = x 4 /Q = 2+s 1 = x 3 /Q = 1 + 2s G22 = x 4 /S 2 = G11 G21 G12 G23 G13 333 Thus:    s+2 2s + 1 2 1 · − (s + 1)2 (s + 1)3 (1 + s)3 1+s      and so on. GA = Q/x 1 =  2 1 s+2 1 − 1+s 2+s (s + 1)2 1 + 2s Similarly, eliminating Q: GB = S 2 /x 1 = G23 G11 − G2 G13 and so on. G13 G22 − G23 G12 PROBLEM 7.20 The temperature of a gas leaving an electric furnace is measured at X by means of a thermocouple. The output of the thermocouple is sent, via a transmitter, to a two-level solenoid switch which controls the power input to the furnace. When the outlet temperature of the gas falls below 673 K (400◦ C) the solenoid switch closes and the power input to the furnace is raised to 20 kW. When the temperature of the gas falls below 673 K (400◦ C) the switch opens and only 16 kW is supplied to the furnace. It is known that the power input to the furnace is related to the gas temperature at X by the transfer function: G(s) = 8 (1 + s)(1 + s/2)(1 + s/3) The transmitter and thermocouple have a combined steady-state gain of 0.5 units and negligible time constants. Assuming the solenoid switch to act as a standard on–off element determine the limit of the disturbance in output gas temperature that the system can tolerate. Solution Apart from the non-linear element N: the open-loop transfer function = 400°C set point 0.5 × 8 (1 + s)(1 + 1/2s)(1 + 1/3s) Solenoid switch N Power input + _ Measured temperature Thermocouple etc. 0.5 Figure 7s. Block diagram for Problem 7.20 334 G(s) Output gas temperature In order to determine the maximum allowable value of N, it is necessary to determine the real part of the open-loop transfer function when the imaginary part is zero. Thus, the open-loop transfer function: G1 (s) = Hence: 24 (1 + iω)(2 + iω)(3 + iω) 24 = 6(1 − ω2 ) + i(11ω − ω3 ) G1 (iω) = = Thus: 4 (1 + s)(1 + 1/2s)(1 + 1/3s) Re {G1 (iω)} = 24{6(1 − ω2 ) − i(11ω − ω3 )} 36(1 − ω2 )2 + (11ω − ω3 )2 24 × 6(1 − ω2 ) 36(1 − ω2 )2 + ω2 (11 − ω2 )2 −24(11ω − ω3 ) 36(1 − ω2 )2 + ω2 (11 − ω2 )2 √ ω3 = 11ω and ω = 0 or ω ≈ 11   24 × 6 × 10 Re = − 36 × 100 Tm {G1 (iω)} = For Tm = 0 : √ For ω = 11, = −0.4. The maximum N × Re = −1.0 for stability by the Nyquist criterion, Thus: N ≤ 2.5 for stability. But, for an on–off element: N= Thus: 4Y0 4×2 = πx0 πx0 x0 = (Y0 = ±2 deg K) 8 ≈ 1 deg K 2.5π PROBLEM 7.21 A gas-phase exothermic reaction takes place in a tubular fixed-bed catalytic reactor which is cooled by passing water through a coil placed in the bed. The composition of the gaseous product stream is regulated by adjustment of the flow of cooling water and, hence, the temperature in the reactor. The exit gases are sampled at a point X downstream from the reactor and the sample passed to a chromatograph for analysis. The chromatograph produces a measured value signal 600 s (10 min) after the reactor outlet stream has been sampled at which time a further sample is taken. The measured value signal from the chromatograph is fed via a zero-order hold element to a proportional controller having a proportional gain KC . The steady-state gain between the product sampling point at X and 335 the output from the hold element is 0.2 units. The output from the controller J is used to adjust the flow Q of cooling water to the reactor. It is known that the time constant of the control valve is negligible and that the steady-state gain between J and Q is 2 units. It has been determined by experimental testing that Q and the gas composition at X are related by the transfer function: Gxq (s) = 0.5 s(10s + 1) (where the time constant is in minutes) What is the maximum value of KC that you would recommend for this control system? Solution Block diagrams for this problem are given in Figure 7t. 1 − e−T s s   Kc Hence, total open loop G(s) = 0.2(1 − e−T s ) 2 s (10s) Transform for zero-order hold element = = G1 (s)G2 (s) G1 (s) = 1 − e−T s where: and G2 (s) = 0.2Kc s 2 (10s + 1) U Process Valve + _ R KC P W + 2 B Zero order hold + 0.5 s (10s + 1) C Chromatograph 0.2 (a) Original system R + _ B 0.2KC s (10s + 1) Hold element (b) Equivalent unity feedback system for stability Figure 7t. Block diagrams for Problem 7.21 336 C z−1 z   C A B G2 (s) = 0.2Kc 2 + + s s 10s + 1 Corresponding z-transform G1 (z) = For G2 (z): A(10s + 1) + Bs(10s + 1) + Cs 2 = 1 where: s = 0, putting Coefficients of s: Coefficients of s 2 : Thus: A=1 10A + B = 0, B = −10 10B + C = 0, C = 100 G2 (s) = 0.2Kc  10 10 1 − + 2 s s s + 1/10  From the Table of z-transforms in Volume 3:   10z Tz 10z G2 (z) = 0.2Kc + − (z − 1)2 z − 1 z − e−0.1T z−1 G(z) = G1 (z)G2 (z) = · G2 (z) z   T 10(z − 1) = 0.2Kc − 10 + z−1 z − e−0.1T The sampling time, T = 10 min (600 s) Thus: G(z) = 0.2Kc =  10 10(z − 1) − 10 + z−1 z − e−1  2Kc (1 − 2e−1 + ze−1 ) (z − 1)(z − e−1 ) The characteristic equation is 1 + G(z) = 0 or: (z − 1)(z − e−1 ) + 2Kc (1 − 2e−1 + ze−1 ) = 0 In order to apply Routh’s criterion, Thus: λ+1 z= λ−1       λ+1 λ+1 λ+1 =0 −1 − 0.368 + 2Kc 0.264 + 0.368 λ−1 λ−1 λ−1 0.632λ + 1.368 + 0.632 Kc λ2 − 0.528 Kc λ − 0.104 Kc2 = 0 0.632 Kc λ2 + (0.632 − 0.528 Kc )λ + (1.368 − 0.104 Kc ) = 0 337 For the system to be stable, the coefficients must be positive. Hence, the system will be unstable if: 0.104Kc ≥ 1.368 Kc ≥ 13.2 That is: 0.528Kc ≥ 0.632 or: Kc ≥ 1.2 Hence Kc < 1.2 for a stable system. The recommended value of Kc is 0.8 in order to allow a reasonable gain margin. PROBLEM 7.22 A unity feedback control loop consists of a non-linear element N and a number of linear elements in series which together approximate to the transfer function: G(s) = 3 s(s + 1)(s + 2) Determine the range of values of the amplitude x0 of an input disturbance for which the system is stable where (a) N represents a dead-zone element for which the gradient k of the linear part is 4; (b) N is a saturating element for which k = 5; and (c) N is an on–off device for which the total change in signal level is 20 units. Solution A block diagram for this problem is shown in Figure 7u. + R _ G(s) N C B Figure 7u. Block diagram for Problem 7.22 The closed loop transfer function for a set-point change is given by: C R = NG(s) , 1 + NG(s) where G(s) = 3 s(s + 1)(s + 2) (a) When N represents a dead-zone element, then: N= k (π − 2β − sin 2β) π where β = sin−1 δ/x0 and the dead-zone is 2δ as shown in Figure 7.82. 338 (equation 7.201) Using the substitution rule: G(iω) = = = 3 iω(iω + 1)(iω + 2) −3 −3[3ω2 − i(ω3 − 2ω)] = 3ω2 + i(ω3 − 2ω) 9ω4 + (ω3 − 2ω)2 ω3 − 2ω −9ω2 + i · 9ω4 + (ω3 − 2ω)2 9ω4 + (ω3 − 2ω)2 From equation 7.204 in Section 7.16.2, Volume 3, the conditionally stable conditions are given when: G(iω) = − − 1 N as shown in Figure 7.87 in Volume 3 1 lies on the real axis, that is where Im [G(iω)] = 0 N ω3 = 2ω or where: ω=0 and: √ ω= G(iω) = Re [G(iω)] = At this point: When ω = or √ 2. −9ω2 9ω4 + (ω3 − 2ω)2 2, then: −18 √ = −0.5 36 + 2(2 − 2)2 1 − = −0.5 and N = 2 N G(iω) = That is: When k = 4, N/k = 0.5 and, from Figure 7.83 in Volume 3: x0 /δ = 2.5 Hence, for k = 4, the system will be unstable if: x0 > 2.5δ (b) For a saturating non-linear element with k = 5, N/k = 2/5 = 0.4 (as shown in Figure 7.84.) From Figure 7.84(b), the corresponding value of x0 /δ is approximately 3.1. Hence, in this case, the system will be unstable if: x0 > 3.1δ 339 (c) For an on–off device having a total change in signal level of 20 units, Y0 = 10 units as shown in Figure 7.80, Volume 3. Thus: N = 4Y0 /πx0 Hence: x0 = 4Y0 /Nπ = 40/2π = 6.4 (equation 7.198) N is as in part (a) of the solution. Thus, in this case, the system will be unstable if: x0 > 6.4 units 340