Volume 4 coulson and richardson s chemical engineering (solutions to the problems in chemical

CHEMICAL ENGINEERING

VOLUME 4

Solutions to the Problems in Chemical Engineering Volume 1

Coulson & Richardson’s Chemical Engineering

Chemical Engineering, Volume 1, Sixth edition

Fluid Flow, Heat Transfer and Mass Transfer

J. M. Coulson and J. F. Richardson

with J. R. Backhurst and J. H. Harker

Chemical Engineering, Volume 2, Fourth edition

Particle Technology and Separation Processes

J. M. Coulson and J. F. Richardson

with J. R. Backhurst and J. H. Harker

Chemical Engineering, Volume 3, Third edition

Chemical & Biochemical Reactors & Process Control

Edited by J. F. Richardson and D. G. Peacock

Solutions to the Problems in Volume 1, First edition

J. R. Backhurst and J. H. Harker

with J. F. Richardson

Chemical Engineering, Volume 5, Second edition

Solutions to the Problems in Volumes 2 and 3

J. R. Backhurst and J. H. Harker

Chemical Engineering, Volume 6, Third edition

Chemical Engineering Design

R. K. Sinnott

Coulson & Richardson’s

CHEMICAL ENGINEERING

J. M. COULSON and J. F. RICHARDSON

Solutions to the Problems in Chemical Engineering

Volume 1

By

J. R. BACKHURST and J. H. HARKER

University of Newcastle upon Tyne

With

J. F. RICHARDSON

University of Wales Swansea

OXFORD AUCKLAND BOSTON JOHANNESBURG MELBOURNE NEW DELHI

Butterworth-Heinemann

Linacre House, Jordan Hill, Oxford OX2 8DP

225 Wildwood Avenue, Woburn, MA 01801-2041

A division of Reed Educational and Professional Publishing Ltd

First published 2001

 J. F. Richardson, J. R. Backhurst and J. H. Harker 2001

All rights reserved. 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 W1P 9HE.

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 4950 X

Typeset by Laser Words, Madras, India

Contents

Preface iv

1. Units and dimensions 1

2. Flow of fluids — energy and momentum relationships 16

3. Flow in pipes and channels 19

4. Flow of compressible fluids 60

5. Flow of multiphase mixtures 74

6. Flow and pressure measurement 77

7. Liquid mixing 103

8. Pumping of fluids 109

9. Heat transfer 125

10. Mass transfer 217

11. The boundary layer 285

12. Momentum, heat and mass transfer 298

13. Humidification and water cooling 318

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 4,

is an attempt to satisfy this demand as far as the problems in Volume 1 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 relates to the sixth edition of Volume 1 and incor￾porates 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.

SECTION 1

Units and Dimensions

PROBLEM 1.1

98% sulphuric acid of viscosity 0.025 N s/m2 and density 1840 kg/m3 is pumped at

685 cm3/s through a 25 mm line. Calculate the value of the Reynolds number.

Solution

Cross-sectional area of line D /40.0252 D 0.00049 m2.

Mean velocity of acid, u D 685 ð 106/0.00049 D 1.398 m/s.

∴ Reynolds number, Re D du

/ D 0.025 ð 1.398 ð 1840/0.025 D 2572

PROBLEM 1.2

Compare the costs of electricity at 1 p per kWh and gas at 15 p per therm.

Solution

Each cost is calculated in p/MJ.

1 kWh D 1 kW ð 1 h D 1000 J/s3600 s D 3,600,000 J or 3.6 MJ

1 therm D 105.5 MJ

∴ cost of electricity D 1 p/3.6 MJ or 1/3.6 D 0.28 p/MJ

cost of gas D 15 p/105.5 MJ or 15/105.5 D 0.14 p/MJ

PROBLEM 1.3

A boiler plant raises 5.2 kg/s of steam at 1825 kN/m2 pressure, using coal of calorific

value 27.2 MJ/kg. If the boiler efficiency is 75%, how much coal is consumed per day?

If the steam is used to generate electricity, what is the power generation in kilowatts

assuming a 20% conversion efficiency of the turbines and generators?

1

2 CHEMICAL ENGINEERING VOLUME 1 SOLUTIONS

Solution

From the steam tables, in Appendix A2, Volume 1, total enthalpy of steam at 1825 kN/m2 D

2798 kJ/kg.

∴ enthalpy of steam D 5.2 ð 2798 D 14,550 kW

Neglecting the enthalpy of the feed water, this must be derived from the coal. With an

efficiency of 75%, the heat provided by the coal D 14,550 ð 100/75 D 19,400 kW.

For a calorific value of 27,200 kJ/kg, rate of coal consumption D 19,400/27,200

D 0.713 kg/s

or: 0.713 ð 3600 ð 24/1000 D 61.6 Mg/day

20% of the enthalpy in the steam is converted to power or:

14,550 ð 20/100 D 2910 kW or 2.91 MW say 3 MW

PROBLEM 1.4

The power required by an agitator in a tank is a function of the following four variables:

(a) diameter of impeller,

(b) number of rotations of the impeller per unit time,

(c) viscosity of liquid,

(d) density of liquid.

From a dimensional analysis, obtain a relation between the power and the four variables.

The power consumption is found, experimentally, to be proportional to the square of

the speed of rotation. By what factor would the power be expected to increase if the

impeller diameter were doubled?

Solution

If the power P D fDN

, then a typical form of the function is P D kDaNb

cd, where

k is a constant. The dimensions of each parameter in terms of M, L, and T are: power,

P D ML2

/T3, density,

D M/L3

, diameter, D D L, viscosity,  D M/LT, and speed of

rotation, N D T1

Equating dimensions:

M : 1 D c C d

L : 2 D a 3c d

T : 3 D b d

Solving in terms of d : a D 5 2d, b D 3 d, c D 1 d

∴ P D k

D5

D2d

N3

Nd

dd

or: P/D5N3

D kD2N

/d

that is: NP D k Rem

UNITS AND DIMENSIONS 3

Thus the power number is a function of the Reynolds number to the power m. In

fact NP is also a function of the Froude number, DN2/g. The previous equation may be

written as:

P/D5N3

D kD2N

/m

Experimentally: P / N2

From the equation, P / NmN3, that is m C 3 D 2 and m D 1

Thus for the same fluid, that is the same viscosity and density:

P2/P1D5

1N3

1/D5

2N3

2 D D2

1N1/D2

2N2

1 or: P2/P1 D N2

2D3

2/N2

1D3

1

In this case, N1 D N2 and D2 D 2D1.

∴ P2/P1 D 8D3

1/D3

1 D 8

A similar solution may be obtained using the Recurring Set method as follows:

P D fD, N,

, , fP, D, N,

,  D 0

Using M, L and T as fundamentals, there are five variables and three fundamentals

and therefore by Buckingham’s  theorem, there will be two dimensionless groups.

Choosing D, N and

as the recurring set, dimensionally:

D  L

N  T1

 ML3

Thus:  L  D

T  N1

M 

L3 D

D3

First group, 1, is PML2T31  P

D3D2N31  P

D5N3

Second group, 2, is ML1

T11  

D3D1N1  

D2N

Thus: f
P

D5N3 , 

D2N

D 0

Although there is little to be gained by using this method for simple problems, there is

considerable advantage when a large number of groups is involved.

PROBLEM 1.5

It is found experimentally that the terminal settling velocity u0 of a spherical particle in

a fluid is a function of the following quantities:

particle diameter, d; buoyant weight of particle (weight of particle weight of displaced

fluid), W; fluid density,

, and fluid viscosity, .

Obtain a relationship for u0 using dimensional analysis.

Stokes established, from theoretical considerations, that for small particles which settle

at very low velocities, the settling velocity is independent of the density of the fluid

4 CHEMICAL ENGINEERING VOLUME 1 SOLUTIONS

except in so far as this affects the buoyancy. Show that the settling velocity must then be

inversely proportional to the viscosity of the fluid.

Solution

If: u0 D kdaWb

cd, then working in dimensions of M, L and T:

L/T D kLaML/T2

bM/L3

cM/LTd

Equating dimensions:

M : 0 D b C c C d

L : 1 D a C b 3c d

T : 1 D 2b d

Solving in terms of b:

a D 1, c D b 1, and d D 1 2b

∴ u0 D k1/dWb

b

/

/2b

 where k is a constant,

or: u0 D k/d

W

/2



b

Rearranging:

du0

/ D kW

/2



b

where (W

/2) is a function of a form of the Reynolds number.

For u0 to be independent of

, b must equal unity and u0 D kW/d

Thus, for constant diameter and hence buoyant weight, the settling velocity is inversely

proportional to the fluid viscosity.

PROBLEM 1.6

A drop of liquid spreads over a horizontal surface. What are the factors which will

influence:

(a) the rate at which the liquid spreads, and

(b) the final shape of the drop?

Obtain dimensionless groups involving the physical variables in the two cases.

Solution

(a) The rate at which a drop spreads, say R m/s, will be influenced by: viscosity of the

liquid, ; volume of the drop, V expressed in terms of d, the drop diameter; density of

the liquid,

; acceleration due to gravity, g and possibly, surface tension of the liquid,

UNITS AND DIMENSIONS 5

. In this event: R D f, d,

, g, . The dimensions of each variable are: R D L/T,

 D M/LT, d D L,

D M/L3

, g D L/T2

, and  D M/T2

. There are 6 variables and 3

fundamentals and hence 6 3 D 3 dimensionless groups. Taking as the recurring set,

d,

and g, then:

d  L, L D d

 M/L3 ∴ M D

L3 D

d3

g  L/T2 ∴ T2 D L/g D d/g and T D d0.5/g0.5

Thus, dimensionless group 1: RT/L D Rd0.5/dg0.5 D R/dg0.5

dimensionless group 2: LT/M D dd0.5/g0.5

d3 D /g0.5

d1.5

dimensionless group 3: T2/M D d/g

d3 D /g

d2

∴ R/dg0.5 D f



g0.5

d1.5 , 

g

d2

or:

R2

dg D f

2

g

2d3 , 

g

d2

(b) The final shape of the drop as indicated by its diameter, d, may be obtained by

using the argument in (a) and putting R D 0. An alternative approach is to assume the

final shape of the drop, that is the final diameter attained when the force due to surface

tension is equal to that attributable to gravitational force. The variables involved here will

be: volume of the drop, V; density of the liquid,

; acceleration due to gravity, g, and the

surface tension of the liquid, . In this case: d D fV,

, g, . The dimensions of each

variable are: d D L, V D L3,

D M/L3

, g D L/T2

,  D M/T2

. There are 5 variables

and 3 fundamentals and hence 5 3 D 2 dimensionless groups. Taking, as before, d,

and g as the recurring set, then:

d  L, L D d

 M/L3 ∴ M D

L3 D

d3

g  L/T2 ∴ T2 D L/g D d/g and T D d0.5/g0.5

Dimensionless group 1: V/L3 D V/d3

Dimensionless group 2: T2/M D d/g

d3 D /g

d2

and hence: d3/V D f



g

d2

PROBLEM 1.7

Liquid is flowing at a volumetric flowrate of Q per unit width down a vertical surface.

Obtain from dimensional analysis the form of the relationship between flowrate and film

thickness. If the flow is streamline, show that the volumetric flowrate is directly propor￾tional to the density of the liquid.

6 CHEMICAL ENGINEERING VOLUME 1 SOLUTIONS

Solution

The flowrate, Q, will be a function of the fluid density,

, and viscosity, , the film

thickness, d, and the acceleration due to gravity, g,

or: Q D f

, g, , d, or: Q D K

agbcdd where K is a constant.

The dimensions of each variable are: Q D L2/T,

D M/L3, g D L/T2,  D M/LT

and d D L.

Equating dimensions:

M : 0 D a C c

L : 2 D 3a C b c C d

T : 1 D 2b c

from which, c D 1 2b, a D c D 2b 1, and d D 2 C 3a b C c

D 2 C 6b 3 b C 1 2b D 3b

∴ Q D K

2b1

gb12b

d3b



or:

Q

 D K

2

gd3

/2



b and Q / 12b

.

For streamline flow, Q / 1

and: 1 D 1 2b and b D 1

∴ Q

/ D K

2

gd3

/2

, Q D K

gd3

/

and: Q is directly proportional to the density,

PROBLEM 1.8

Obtain, by dimensional analysis, a functional relationship for the heat transfer coefficient

for forced convection at the inner wall of an annulus through which a cooling liquid is

flowing.

Solution

Taking the heat transfer coefficient, h, as a function of the fluid velocity, density, viscosity,

specific heat and thermal conductivity, u,

, , Cp and k, respectively, and of the inside

and outside diameters of the annulus, di and d0 respectively, then:

h D fu, di, d0,

, , Cp, k

The dimensions of each variable are: h D H/L2Tq, u D L/T, di D L, d0 D L,

D M/L3,

 D M/LT, Cp D H/Mq, k D H/LTq. There are 8 variables and 5 fundamental dimen￾sions and hence there will be 8 5 D 3 groups. H and q always appear however as

UNITS AND DIMENSIONS 7

the group H/q and in effect the fundamental dimensions are 4 (M, L, T and H/q) and

there will be 8 4 D 4 groups. For the recurring set, the variables di, , k and

will

be chosen. Thus:

di  L, L D di

 M/L3 M D

L3 D

d3

i

  M/LT, T D M/L D

d3

i /di D

d2

i /

k  H/q/LT, H/q D kLT D kdi

d2

i / D k

d3

i /

Dimensionless group 1: hL2T/H/q D hd2

i

d2

i /k

d3

i / D hdi/k

Dimensionless group 2: uT/L D u

d2

i /di D diu

/

Dimensionless group 3: d0/L D d0/di

Dimensionless group 4: CpM/H/q D Cp

d3

i /k

d3

i / D Cp/k

∴ hdi/k D fdiu

/, Cp/k, d0/di which is a form of equation 9.94.

PROBLEM 1.9

Obtain by dimensional analysis a functional relationship for the wall heat transfer coef￾ficient for a fluid flowing through a straight pipe of circular cross-section. Assume that

the effects of natural convection may be neglected in comparison with those of forced

convection.

It is found by experiment that, when the flow is turbulent, increasing the flowrate by a

factor of 2 always results in a 50% increase in the coefficient. How would a 50% increase

in density of the fluid be expected to affect the coefficient, all other variables remaining

constant?

Solution

For heat transfer for a fluid flowing through a circular pipe, the dimensional analysis is

detailed in Section 9.4.2 and, for forced convection, the heat transfer coefficient at the

wall is given by equations 9.64 and 9.58 which may be written as:

hd/k D fdu

/, Cp/k

or: hd/k D Kdu

/nCp/km

∴ h2/h1 D u2/u1

n.

Increasing the flowrate by a factor of 2 results in a 50% increase in the coefficient, or:

1.5 D 2.0n and n D ln 1.5/ ln 2.0 D 0.585.

Also: h2/h1 D

2/

10.585

When

2/

1 D 1.50, h2/h1 D 1.500.585 D 1.27 and the coefficient is increased by 27%

8 CHEMICAL ENGINEERING VOLUME 1 SOLUTIONS

PROBLEM 1.10

A stream of droplets of liquid is formed rapidly at an orifice submerged in a second,

immiscible liquid. What physical properties would be expected to influence the mean size

of droplet formed? Using dimensional analysis obtain a functional relation between the

variables.

Solution

The mean droplet size, dp, will be influenced by: diameter of the orifice, d; velocity of

the liquid, u; interfacial tension, ; viscosity of the dispersed phase, ; density of the

dispersed phase,

d; density of the continuous phase,

c, and acceleration due to gravity,

g. It would also be acceptable to use the term

d

cg to take account of gravitational

forces and there may be some justification in also taking into account the viscosity of the

continuous phase.

On this basis: dp D fd, u, , ,

d,

c, g

The dimensions of each variable are: dp D L, d D L, u D L/T,  D M/T2,  D M/LT,

d D M/L3,

c D M/L3, and g D L/T2. There are 7 variables and hence with 3 funda￾mental dimensions, there will be 7 3 D 4 dimensionless groups. The variables d, u

and  will be chosen as the recurring set and hence:

d  L, L D d

u  L/T, T D L/u D d/u

  M/T2, M D T2 D d2/u2

Thus, dimensionless group 1: LT/M D dd/u/d2/u2 D u/

dimensionless group 2:

dL3/M D

dd3/d2/u2 D

ddu2/

dimensionless group 3:

cL3/M D

cd3/d2/u2 D

cdu2/

dimensionless group 4: gT2/L D gd2/u2/d D gd/u2

and the function becomes: dp D fu/,

ddu2/,

cdu2/, gd/u2

PROBLEM 1.11

Liquid flows under steady-state conditions along an open channel of fixed inclination to

the horizontal. On what factors will the depth of liquid in the channel depend? Obtain a

relationship between the variables using dimensional analysis.

Solution

The depth of liquid, d, will probably depend on: density and viscosity of the liquid,

and ; acceleration due to gravity, g; volumetric flowrate per unit width of channel, Q,

UNITS AND DIMENSIONS 9

and the angle of inclination, ",

or: d D f

, , g, Q, "

Excluding " at this stage, there are 5 variables and with 3 fundamental dimensions there

will be 5 3 D 2 dimensionless groups. The dimensions of each variable are: d D L,

D M/L3,  D M/LT, g D L/T2, Q D L2/T, and, choosing Q,

and g as the recurring

set, then:

Q D L2/T T D L2/Q

g D L/T2 L D gT2 D gL4/Q2, L3 D Q2/g,L D Q2/3/g1/3

and T D Q4/3/Qg2/3 D Q1/3/g2/3

D M/L3 M D L3

D Q2/g

D Q2

/g

Thus, dimensionless group 1: d/L D dg1/3/Q2/3 or d3g/Q2

dimensionless group 2: LT/M D Q2/3/g1/3Q1/3/g2/3/Q2

g D /Q

and the function becomes: d3g/Q2 D f/Q

, "

PROBLEM 1.12

Liquid flows down an inclined surface as a film. On what variables will the thickness of

the liquid film depend? Obtain the relevant dimensionless groups. It may be assumed that

the surface is sufficiently wide for edge effects to be negligible.

Solution

This is essentially the same as Problem 1.11, though here the approach used is that of

equating indices.

If, as before: d D K

a, b

, gc

, Qd, "e



then, excluding " at this stage, the dimensions of each variable are: d D L,

D M/L3,

 D M/LT, g D L/T2, Q D L2/T.

Equating dimensions:

M : 0 D a C b i

L : 1 D 3a b C c C 2d ii

T : 0 D b 2c d iii

Solving in terms of b and c then:

from (i) a D b

from (iii) d D b 2c

and in (ii) 1 D 3b b C c 2b 4c or: c D 1/3 ∴ d D 2/3 b