Volume 5 coulson and richardson s chemical engineering (solutions to the problems in volume 2 and 3)

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 EINEMANN

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

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and Patents Act 1988

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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

Preface to the Second Edition of Volume 5

Preface to the First Edition of Volume 5

Factors for Conversion of SI units

Solutions to Problems in Volume 2

2-1 Particulate solids

2-2

2-3

2-4

2-5 Sedimentation

2-6 Fluidisation

2-7 Liquid filtration

2-8 Membrane separation processes

2-9 Centrifugal separations

2-10 Leaching

2-1 1 Distillation

2- 12 Absorption of gases

2-13 Liquid-liquid extraction

2- 14 Evaporation

2- 15 Cry stallisation

2-16 Drying

2- 17 Adsorption

2-18 Ion exchange

2- 19 Chromatographic separations

Solutions to Problems in Volume 3

3-1 Reactor.design - general principles

3-2

3-3 Gas-solid reactions and reactors

3-4 Gas-liquid and gas-liquid-solid reactors

Particle size reduction and enlargement

Motion of particles in a fluid

Flow of fluids through granular beds and packed columns

Flow characteristics of reactors- flow modelling

vii

ix

xi

Xiii

1

8

14

34

39

44

59

76

79

83

98

150

171

181

216

222

23 1

234

235

237

262

265

27 1

V

3-5 Biochemical reaction engineering

3-7 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

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:

= (1/50) l [ 1 - (d2/402)1dd

= (1/50)[d - (d3/4800)]p

= 0.533 or 53.3 per cent of the particles remain in suspension.

40

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

3600

6 140

14 15

18 5

22 2

10 PO

5,300,000k 0.01 1

42,930,000k 0.090

80,136,000k 0.168

106,000,000k 0.222

109,074,000k 0.229

77,274,000k 0.162

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

1

3

6

10

14

18

22

n nd2 nd3

2000 2000 2000

600 5400 16,200

140 5040 30,240

40 4Ooo 40,000

15 2940 41,160

5 1620 29,160

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

Calculate the efficiency of collection for a dust with a mass distribution of 50 per cent

0-5 pm, 30 per cent 5-10 Fm and 20 per cent above 10 Fm.

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 (pm) 0- 1 1-2 2-4 4-6 6-10 10-14

Number of particles (-) 2000 lo00 500 200 100 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 p = 2600 kg/m3,

then the volume of 1 particle = (n/6)d3 m3, the mass of 1 particle = 2600(x/6)d3 kg

and the following table may be produced

Size (km) 0- 1 1-2 2-4 4-6

Number of particles (-) 2000 1000 500 200

Mean diameter (Km) 0.5 1.5 3.0 5.0

(m) 0.5 x 1.5 x 3.0 x 5.0 x

Volume (m3) 6.54 x 3.38 x 1.41 x 6.54 x lo-’’

Mass of one particle (kg) 1.70 x 8.78 x lo-’’ 3.68 x 1.70 x

Mass of one particles in

size range (kg) 3.40 x 8.78 x 1O-l2 1.83 x lo-” 3.40 x lo-”

Size (pm) 6- 10 10-14

Number of particles (-) 100 40

(m) 8.0 x 12.0 x

Volume (m3) 2.68 x 9.05 x

Mass of one particle (kg) 6.97 x 2.35 x

Mass of one particles in

Mean diameter (pm) 8.0 12.0

size range (kg) 6.97 x lo-” 9.41 x lo-”

6

Total mass of particles = 2.50 x lo-'' kg.

As this mass is obtained from 1 cm3 of air, the required dust concentration is given by:

(2.50 x lo-'') x lo3 x lo6 = 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 ds, at what

particle size will the theoretical cut occur?

The viscosity of air is 0.018 mNs/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.

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 mrn, 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) Amount of product (per cent)

through

on

on

on

on

on

on

through

6.0

4.0

2.0

0.75

0.50

0.25

0.125

0.125

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, pius 40 mm to a product of 0.50 mm aver￾age 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:

Sizeof Mean dl nl ndl nd; nd: ndf

aperture (mm)

(mm)

6.00

4.00

2.00

0.75

0.50

0.25

0.125

5.00 0.26 1.3 6.5 32.5 162.5

3.00 0.18 0.54 1.62 4.86 14.58

1.375 0.23 0.316 0.435 0.598 0.822

0.67 0.08 0.0536 0.0359 0.024 1 0.0161

0.37 0.17 0.0629 0.0233 0.0086 0.003 19

0.1875 0.03 0.0056 0.00 105 0.00020 O.ooOo37

0.125 0.05 0.00625 0.00078 0.000098 o.ooo012

Totals: 2.284 8.616 37.991 177.92

From equation 1.1 1. the mass mean diameter is:

dv = Cnldf/Cnld:

= (177.92/37,991) = 4.683 mm.

From equation 1.14, the surface mean diameter is:

d, = Enld:/Y,nld:

= (37.991/8.616) = 4.409 IIWI.

From equation 1.18, the length mean diameter is:

dl = Xnld:/Cnldl

= (8.616/2.284) = 3.772 mm.

From equation 1.19, the mean length diameter is:

d; = Cnldl/Cnl

= (2.284/1.0) = 2.284 mm.

9

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 111111,

energy consumption = 13.0 kJAcg, compressive strength = 22.5 N/m2

In equation 2.4:

and:

13.0 = KK x 22.5ln(45/4)

KK = (13.0p4.4) = 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 x 45 ln(42.5/0.50) = (0.239 x 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:

0.20 0.250 0.003125 0.00078

0.60 0.125 0.001172 0.000146

0.20 0.085 0.000123 O.oooO11

Totals: 0.00442 0.000937

and from equation 1.1 1 :

d, = Cnldf/Znld: = (0.000937/0.00442) = 0.212 mm.

10