Tài liệu Coulson & richardson_ solutions to the problems in chemical engineering volume 2 & 3 volume 5

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 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. Newcastle upon Tyne, 1997 J. R. BACKHURST J. H. HARKER (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. Newcastle upon Tyne, 1978 J. R. BACKHURST J. H. HARKER 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 − (d 2 /402 )]dd = (1/50) 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