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 J E I N E M A N N OXFORD AMSTERDAM BOSTON LONDON NEW YORK PARIS SAN DlEGO 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 Transferred to digital printing 2004 Copyright Q 2002, J.F. Richardson and J.H. Harker. All rights reserved The right of J.F. Richardson and I.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 I Ininformationon; Burnorth-Heinemann publications visit our website at www.bh.com 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 Contents Preface vii Preface to the Second Edition of Volume 5 ix Preface to the First Edition of Volume 5 xi Factors for Conversion of SI units Xiii Solutions to Problems in Volume 2 2-1 Particulate solids 1 2-2 Particle size reduction and enlargement 8 2-3 Motion of particles in a fluid 14 2-4 Flow of fluids through granular beds and packed columns 34 2-5 Sedimentation 39 2-6 Fluidisation 44 2-7 Liquid filtration 59 2-8 Membrane separation processes 76 2-9 Centrifugal separations 79 2-10 Leaching 83 2-1 1 Distillation 98 2- 12 Absorption of gases 150 2-13 Liquid-liquid extraction 171 2- 14 Evaporation 181 2- 15 Cry stallisation 216 2-16 Drying 222 2- 17 Adsorption 23 1 2-18 Ion exchange 234 2- 19 Chromatographic separations 235 Solutions to Problems in Volume 3 3-1 Reactor.design- g eneral principles 237 3-2 Flow characteristics of reactors- flow modelling 262 3-3 Gas-solid reactions and reactors 265 3-4 Gas-liquid and gas-liquid-solid reactors 27 1 V 3-5 Biochemical reaction engineering 285 3-7 Process control 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 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 hm particle size to 100 per cent by mass at 101 pm 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 pm, and dn/dd = lo0,000/d4 for the size range 10-100 pm where d is in pm. 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 tine from the origin to 100 per cent undersize at a particle size of 50 pm. 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 pm 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 Fm particle to fall a height h m is: t = h/402k where k a constant. During this time, a particle of diameter d wrn has fallen a distance equal to: kd2h/402k= hd2/402 The proportion of particles of size d which are still in suspension is: = 1 - (d2/402) and the fraction by mass of particles which are still in suspension is: = lm[l - (d2/402)]dw Since dw/dd = 1/50, the mass fraction is: l 40 = (1/50) [ 1 - (d2/402)1dd = (1/50)[d - (d3/4800)]p = 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: - uo = kd2(ps PI. where k (= g/18p) is a constant. For large galena: uo = k(25 x 10-6)2(7500 - l0o0) = 4.06 x 10-6k m/s - For small galena: uo = k(5.2 x 10-6)2(7500 1OOO) = 0.176 x 10-6k m/s For large quartz: uo = k(25 x 10-6)2(26J0 - lO00) = 1.03 x 10% m/s For small quartz: uo = k(5.2 x 10-6)2(2650 - 1OOO) = 0.046 x 10-6k m/s 2 If the time of settling was such that particles with a velocity equal to 1.03 x lo-% 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 x 10% = kd2(7500 - 1OOO) or d = O.oooO126 m or 0.0126 mm. Similarly if the time of settling was such that particles with a velocity equal to 0.176 x 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 x 10-6k = kd2(2650 - 1OOO) or d = O.oooO103 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 (Fm) Number of particles in range (-) 0-2 2000 2-4 600 4-8 1 40 8- 12 40 12-16 15 16-20 5 20-24 2 3 Solution From equation 1.4, the mass fraction of particles of size dl is given by: 3 XI = nIkld1Psr where kl is a constant, n1 is the number of particles of size dl, and p, is the density of the particles = 2650 kg/m3. EX,= 1 and hence the mass fraction is: x1 = nlkld:ps/Xnkd3p,. In this case: d n kd3np, X 1 200 5,300,000k 0.01 1 3 6 0 0 42,930,000k 0.090 6 140 80,136,000k 0.168 10 PO 106,000,000k 0.222 14 15 109,074,000k 0.229 18 5 77,274,000k 0.162 22 2 56,434,400k 0.118 C = 477,148,400k X = 1.0 The surface mean diameter is given by equation 1.14: d, = Wd:)/Wld:) and hence: d n nd2 nd3 1 2000 2000 2000 3 600 5400 16,200 6 140 5040 30,240 10 40 4Ooo 40,000 14 15 2940 41,160 18 5 1620 29,160 22 2 968 2 1,296 C = 21,968 C = 180,056 Thus: d, = (180,056/21,968) = 8.20 Vrn This is the size of a particle with the same specific surface as the mixture. The volume of a particle 8.20 bm in diameter = (n/6)8.203 = 288.7 wm3. 4 The surface area of a particle 8.20 pm in diameter = (n x 8.202) = 21 1.2 pm2 and hence: the specific surface = (21 1.2/288.7) = 0.731 pm2/pm3 or 0.731 x lo6 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) 30 60 90 120 150 sample variance (-) 0.025 0.006 0.015 0.018 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, (pm) 0-5 5-10 10-20 20-40 40-80 80-160 Mass (per cent) 10 15 35 20 10 10 Efficiency (per cent) 20 40 80 90 95 100 Calculate the overall efficiency of the collector and the percentage by mass of the emitted dust that is smaller than 20 pm 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 pm, 80 per cent over the size range 5-10 pm, and 96 per cent for particles exceeding 10 pm. 5