Coulson and Richardson's Chemical Engineering Volume 4 - Solutions to the Problems in Chemical Engineering from Volume 1

290 Pages·2001·10.07 MB·English

Coulson & Richardson's
CHEMICAL ENGINEERING
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
1
E I N E M A N N
OXFORD AUCKLAND BOSTON JOHANNESBURG MELBOURNE NEW DELHI
Butterworth-Heinemann
Linacre House, Jordan Hill, Oxford OX2 8DP
225 Wildwood Avenue, Wobum, MA 01801-2041
A division of Reed Educational and Professional Publishing Ltd
-@A member of the Reed Elsevier PIC group
First published 2001
Transferred to digital printing 2004
0 J. F. Richardson, J. R. Backhurst and J. H. Harker 2001
All rights resewed. 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 WlP 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
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.
Contents
Preface iv
1. Units and dimensions 1
2. Flow of fluids -e nergy 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
1 1. The boundary layer 285
12. Momentum, heat and mass transfer 298
13. Humidification and water cooling 318
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 = (n/4)0.025* = 0.00049 m2.
Mean velocity of acid, u = (685 x 10-6)/0.00049 = 1.398 d s .
:. Reynolds number, Re = dup/p = (0.025 x 1.398 x 1840)/0.025 = -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 = 1 kW x 1 h = (loo0 J/s)(3600 s) = 3,600,000 J or 3.6 MJ
1 therm = 105.5 MJ
:.
cost of electricity = 1 p/3.6 MJ or (1/3.6) = 0.28 p/MJ
cost of gas = 15 p/105.5 MJ or (15/105.5) = 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 MJkg. 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 =
2798 Hkg.
.. enthalpy of steam = (5.2 x 2798) = 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 = (14,550 x 100)/75 = 19,400 kW.
For a calorific value of 27,200 kJkg, rate of coal consumption = (19,400/27,200)
= 0.713 kg/s
or: (0.713 x 3600 x 24)/1000 = 61.6 Mg/day
20% of the enthalpy in the steam is converted to power or:
(14,550 x 20)/100 = 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 = 4(DNpp),t hen a typical form of the function is P = kDnNbpCpd,w here
k is a constant. The dimensions of each parameter in terms of M, L, and T are: power,
P = ML2/T3, density, p = M/L3, diameter, D = L, viscosity, p = M/LT, and speed of
rotation, N = T-'
Equating dimensions:
M: I=c+d
L: 2=a-3c-d
T:-3=-b-d
Solving in terms of d : a = (5 - 2d), b = (3 - d), c = (1 - d)
or: PI@ N3p = k( D2N p l ~ ) - ~
that is: Np = 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/DSN3p= k(D2Np/p)"'
Experimentally: P cx N2
+
From the equation, P cx NmN3,t hat is m 3 = 2 and m = -1
Thus for the same fluid, that is the same viscosity and density:
(P~/PI)(D:N:/D~=N (D~):N l/D;N2)-' or: (P2/P1)= (N;Di)/(N:Di)
In this case, NI = N2 and D2 = 201.
.. (P2/Pl)= 8D:/D: =
A similar solution may be obtained using the Recurring Set method as follows:
P = 4W, N, P, p), f (P. D, N, P,p ) = 0
Using M, L and T as fundamentals, there are five variables and three fundamentals
and therefore by Buckingham's 17 theorem, there will be two dimensionless groups.
Choosing D, N and p as the recurring set, dimensionally:
] [
D r L L = D
N = T-' Thus: T ~ N - I
p E ML-3 M G pL3 = pD3
P
First group, XI, is P(ML2T-3)-1= P(pD3D2N3)-' E -
pDSN3
Second group, 7r2, is p(ML-'T-I)-l = p(pD3D-lN)-' E -P
pD2N
Thus:
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 uo of a spherical particle in
a fluid is a function of the following quantities:
particle diameter, d; buoyant weight of particle (weight of particlc - weight of displaced
fluid), W;fl uid density, p, and fluid viscosity, p.
Obtain a relationship for uo 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 musr then be
inversely proportional to the viscosity of the fluid.
Solution
If: uo = kdaWbpcpd, then working in dimensions of M, L and T:
(L/T) = k(L"( ML/T2)b( M/L3)'
Equating dimensions:
M: O = b + ~ + d
L: 1 =a+b-3c-d
T: -1 =-2b-d
Solving in terms of b:
a=-l,c=(b-l), andd=(l -2b)
.. uo = k( l/d)( Wb)(pb/p)(p/p2bw)h ere k is a constant,
or: uo = k(P/dP)(WP/LL2)b
Rearranging:
(dUOP/P) = k(WP/P2)b
where (Wp/p2)i s a function of a form of the Reynolds number.
For uo to be independent of p, b must equal unity and uo = kW/dp
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, p; volume of the drop, V expressed in terms of d, the drop diameter; density of
the liquid, p; acceleration due to gravity, g and possibly, surface tension of the liquid,
UNITS AND DIMENSIONS 5
a. In this event: R = f(p, d, p, g, a). The dimensions of each variable are: R = L/T,
p = M/LT, d = L, p = M/L3, g = L/T2, and a = M/T2. There are 6 variables and 3
fundamentals and hence (6 - 3) = 3 dimensionless groups. Taking as the recurring set,
d, p and g, then:
d = L, L = d
p = M/L3 :. M = pL3 = pd3
g = L/T2 :. T2 = L/g = d/g and T = d0.5/g0.5
Thus, dimensionless group 1: RT/L = Rdo.5/dgo.5= R/(dg)o.5
dimensionless group 2: pLT/M = ,~d(dO.~)/(g'.~p=d ~~)/ (gO.~pd'.~)
dimensionless group 3: oT2/M = ad/(gpd3) = a/(gpd2)
or:
(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 = 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, p; acceleration due to gravity, g, and the
surface tension of the liquid, a. In this case: d = f(V, p, g, a). The dimensions of each
variable are: d = L, V = L3, p = M/L3, g = L/T2, a = M/T2. There are 5 variables
and 3 fundamentals and hence (5 - 3) = 2 dimensionless groups. Taking, as before, d, p
and g as the recurring set, then:
d L, L=d
E
p M/L3 :. M = pL3 = pd3
g E L/T2 :. T2 = L/g = d/g and T = d0.5/g0.5
Dimensionless group 1: V/L3 = V/d3
Dimensionless group 2: oT2/M = ad/(gpd3) = a/(gpd2)
(z)
and hence: (d3/V) = f
nPd2
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, p, and viscosity, p, the film
thickness, d, and the acceleration due to gravity, g,
or: Q = f (p, g, p, d), or: Q = Kpagbp'dd where K is a constant.
The dimensions of each variable are: Q = L2/T, p = M/L3, g = L/T2, p = M/LT
and d = L.
Equating dimensions:
M: O=a+c
L: 2=-3a+b-~+d
T: -1 =-2b-c
from which, c = 1 - 2b,a = -c = 2b- 1, and d = 2+ 3a - b+c
= 2 + 6b - 3 - b + 1 - 2b = 3b
.. Q=K(P 26-1 g6 p1 -26 d 36 )
QP
or: - = K ( ~ ~ g d ~and/ Qp a~ ) ~
LL
For streamline flow, Q a p-'
and: -1 = 1 - 2b and b = 1
.. QP/P = K(p2gd3/p2)Q, = K(pgd3/p)
and: Q is directly proportional to the density, p
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, p, p, C, and k, respectively, and of the inside
and outside diameters of the annulus, di and do respectively, then:
h = f(u, di, do, P, ~9 Cp, k)
The dimensions of each variable are: h = H/L2T0, u = L/T, di = L, do = L, p = M/L3,
p = M/LT, C, = H/M8, k = H/LTB. There are 8 variables and 5 fundamental dimen-
sions and hence there will be (8 - 5) = 3 groups. H and 8 always appear however as