CHEMICAL ENGINEERING
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

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

Solution
Each cost is calculated in p/MJ.
1kWhD 1kWð 1hD 1000 J/s3600 s D 3,600,000 J or 3.6MJ
1thermD 105.5MJ
∴ cost of electricity D 1p/3.6MJor1/3.6 D 0.28 p/MJ
cost of gas D 15 p/105.5MJor15/105.5 D 0.14 p/MJ
PROBLEM 1.3
A boiler plant raises 5.2 kg/s of steam at 1825 kN/m
2
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/m
2
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:

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

D
5
D
2d
N
3
N
d


d

d

or: P/D
5
N
3
 D kD
2
N/
d
that is: N
P
D kRe

5
1
N
3
1
/D
5
2
N
3
2
 D D
2
1
N
1
/D
2
2
N
2

1
or: P
2
/P
1
 D N
2
2

A similar solution may be obtained using the Recurring Set method as follows:
P D
fD,N,,,fP,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 Á L
N Á T
1
 Á ML
3

Thus:

L Á D
T Á N
1
M Á L
3
D D
3
First group, 
1
,isPML
2
T
3

1
Á PD

2
N
Thus: f

P
D
5
N
3
,

D
2
N

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 u
0
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 u
0
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

L: 1 D a C b  3c  d
T:1 D2b  d
Solving in terms of b:
a D1,cD b  1, and d D 1 2b
∴ u
0
D k1/dW
b

b
//
2b
 where k is a constant,
or: u
0
D k/dW/
2

b
Rearranging:
du
0
/ D kW/
2

b
where (W/
2
) is a function of a form of the Reynolds number.
For u

∴ M D L
3
D d
3
g Á L/T
2
∴ T
2
D L/g D d/g and T D d
0.5
/g
0.5
Thus, dimensionless group 1: RT/L D Rd
0.5
/dg
0.5
D R/dg
0.5
dimensionless group 2: LT/M D dd
0.5
/g
0.5
d
3
 D /g
0.5
d
1.5

dimensionless group 3: T

d
3
,

gd
2

(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,andthe
surface tension of the liquid, . In this case: d D fV, , g, . The dimensions of each
variable are: d D L, V D L
3
,  D M/L
3
, g D L/T
2
,  D M/T
2
. 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 Á L, L D d
 Á M/L
3
∴ M D L
3
D d

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, ,thefilm
thickness, d, and the acceleration due to gravity, g,
or: Q D f, g, , d, or: Q D K
a
g
b

c
d
d
where K is a constant.
The dimensions of each variable are: Q D L
2
/T,  D M/L
3
, g D L/T
2
,  D M/LT
and d D L.
Equating dimensions:
M: 0 D a C c
L: 2 D3a C b  c Cd
T:1 D2b  c
from which, c D 1  2b, a Dc D 2b  1, and d D 2 C3a  b C c
D 2 C 6b  3 b C 1  2b D 3b

/
2
, Q D Kgd
3
/
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, , , C
p
and k, respectively, and of the inside
and outside diameters of the annulus, d
i
and d
0
respectively, then:
h D fu, d
i
,d
0
,,,C
p
,k
The dimensions of each variable are: h D H/L
2
Tq, u D L/T, d

i
/d
i
 D d
2
i
/
k Á H/
q/LT,H/q D kLT D kd
i
d
2
i
/ D kd
3
i
/
Dimensionless group 1: hL
2
T/H/q D hd
2
i
d
2
i
/kd
3
i
/ D hd
i

/k D fd
i
u/, C
p
/k, d
0
/d
i
 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 fdu/, C
p
/k
or: hd/k D Kdu/
n
C
p
/k

 D 1.50, h
2
/h
1
D 1.50
0.585
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, d
p
, 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
 
c
g to take account of gravitational
forces and there may be some justification in also taking into account the viscosity of the
continuous phase.

2
D d
2
/u
2
Thus, dimensionless group 1: LT/M D dd/u/d
2
/u
2
 D u/
dimensionless group 2: 
d
L
3
/M D 
d
d
3
/d
2
/u
2
 D 
d
du
2
/
dimensionless group 3: 
c
L

du
2
/, gd/u
2

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, Â,
or: d D f,,g,Q,Â
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/L
3
,  D M/LT, g D L/T
2
, Q D L
2
/T, and, choosing Q,  and g as the recurring
set, then:
Q D L
2
/TTD L
2
/Q

/g D Q
2
/g
Thus, dimensionless group 1: d/L D dg
1/3
/Q
2/3
or d
3
g/Q
2
dimensionless group 2: LT/M D Q
2/3
/g
1/3
Q
1/3
/g
2/3
/Q
2
g D /Q
and the function becomes: d
3
g/Q
2
D f/Q, Â
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

and in (ii) 1 D 3b  b C c  2b 4c or: c D1/3
∴ d D 2/3  b
10 CHEMICAL ENGINEERING VOLUME 1 SOLUTIONS
Thus: d D K
b
Ð 
b
Ð g
1/3
Ð Q
2/3b

dg
1/3
/Q
2/3
D K/Q
b
and: d
3
g/Q
2
D K/Q
b
Â
e
as before.
PROBLEM 1.13
A glass particle settles under the action of gravity in a liquid. Upon what variables
would the terminal velocity of the particle be expected to depend? Obtain a relevant

, M D L
3
∴ L
3
D LT, T D L
2
/
g Á L/T
2
D 
2
L/
2
L
4
D 
2
/
2
L
3
∴ L
3
D 
2
/
2
g and L D 
2/3
/

2
/g
UNITS AND DIMENSIONS 11
Thus, dimensionless group 1: QT/L
2
D Q
1/3
/
1/3
g
2/3
/
4/3
/
4/3
g
2/3
 D Q/
dimensionless group 2: υL D υ
2/3
/
2/3
g
1/3
 or, cubing D υ
3

2
g/
2

thermal conductivity of the film and since υ / 
1/3
, h / k/
1/3
, that is the coefficient is
inversely proportional to the one third power of the liquid viscosity.
PROBLEM 1.15
A spherical particle settles in a liquid contained in a narrow vessel. Upon what variables
would you expect the falling velocity of the particle to depend? Obtain the relevant
dimensionless groups.
For particles of a given density settling in a vessel of large diameter, the settling velocity
is found to be inversely proportional to the viscosity of the liquid. How would this depend
on particle size?
Solution
This problem is very similar to Problem 1.13, although, in this case, the liquid through
which the particle settles is contained in a narrow vessel. This introduces another variable,
D, the vessel diameter and hence the settling velocity of the particle is given by: u D
fd, , , D, 
s
,g. The dimensions of each variable are: u D L/T, d D L,  D M/L
3
,
 D M/LT, D D L, 
s
D M/L
3
,andg D L/T
2
. With 7 variables and 3 fundamental
dimensions, there will be 7  3 D 4 dimensionless groups. Taking d,  and  as the

and dimensionless group 4: gT
2
/L D g
2
d
4
/
2
d D g
2
d
3
/
2
Thus: du/ D fD/d
s
/g
2
d
3
/
2

In particular, du/ D Kg
2
d
3
/
2


Thus: G D f,,h,a,g,Â. The dimensions of each variable are: G D M/T,  D M/L
3
,
 D M/LT, h D L, a D L, g D L/T
2
and neglecting  at this stage, with 6 variables
with dimensions and 3 fundamental dimensions, there will be 6 3 D 3 dimensionless
groups. Taking h,  and  as the recurring set then:
h Á L, L D h
 Á M/L
3
, M D L
3
D h
3
 Á M/LT, T D M/L D h
3
/h D h
2
/
Thus: dimensionless group 1: GT/M D Gh
2
/h
3
 D G/h
dimensionless group 2: a/L D a/h
dimensionless group 3: gT
2
/L D g
2

3
and of viscosity 0.001 Ns/m
2
?
If Stokes’ Law applies for particle Reynolds numbers up to 0.2, what is the diameter
of the largest particle whose behaviour is governed by Stokes’ Law for this solid and
liquid?
Solution
The accelerating force due to gravity D mass of particle  mass of liquid displacedg.
For a particle of radius r, volume D 4r
3
/3, or, in terms of diameter, d, volume D
4d
3
/2
3
/3 D d
3
/6. Mass of particle D d
3

s
/6, where 
s
is the density of the solid.
Mass of liquid displaced D d
3
/6, where  is the density of the liquid, and accelerating
force due to gravity D d
3

R
0
D fd, , , u. The dimensions of each variable are R
0
D M/LT
2
, d D L,  D M/L
3
,
 D M/LT and u D L/T. With 5 variables and 3 fundamental dimensions, there will be
5  3 D 2 dimensionless groups. Taking d,  and u as the recurring set, then:
d Á L, L D d
 Á M/L
3
, M D L
3
D d
3
u Á L/T, T D L/u D d/u
Thus: dimensionless group 1: R
0
LT
2
/M D R
0
dd
2
/u
2
/d

s
D 1600 kg/m
3
;
 D 1000 kg/m
3
and  D 0.001 Ns/m
2
.
Thus, in equation (i): u
0
D 10
5

2
ð 9.81/18 ð 0.0011600  1000
D 3.27 ð 10
5
m/s or 0.033 mm/s
When Re D 0.2, du/ D 0.2 or when the terminal velocity is reached:
du
0
D 0.2/ D 0.2 ð 0.001/1000 D 2 ð 10
7
or: u
0
D 2 ð 10
7
/d
14 CHEMICAL ENGINEERING VOLUME 1 SOLUTIONS

Temperature of fluid in which it is immersed, Â
s
.
Obtain relevant dimensionless groups for this problem.
Solution
In this case, t D fd, k, , C
p
,Â
c
,Â
s
. The dimensions of each variable are: t D T, d D
L, k D ML/T
q, C
p
D L
2
/T
2
q, Â
c
D q, Â
s
D q. There are 7 variables and hence with 4
fundamental dimensions, there will be 7 4 D 3 dimensionless groups. Taking d, , C
p
and Â
c
as the recurring set, then:
d Á L, L D d,

Â
c
or: T D d/C
0.5
p
Â
0.5
c
Thus: dimensionless group 1: t/T D tC
0.5
p
Â
0.5
c
/d
dimensionless group 2: kT
q/ML D kd/C
0.5
p
Â
0.5
c
Â
c
/d
3
d D kÂ
0.5
c
/C

fluid viscosity, ; particle size, d; pressure difference across the bed, P, and the voidage
of the cake, e or:
v D f, , d, P, e. The dimensions of each of these variables are
u D L/T,  D M/L
3
,  D M/LT, d D L, P D M/LT
2
and e D dimensionless. There
are 6 variables and 3 fundamental dimensions and hence 6 3 D 3 dimensionless
groups. Taking, d,  and  as the recurring set, then:
d Á L, L D d
 Á M/L
3
, M D L
3
D d
3
 Á M/LT, T D M/L D d
3
/d D d
2
/
Thus: dimensionless group 1: uT/L D ud
2
/d D du/
dimensionless group 2: PLT
2
/M D Pdd
2
/

0
1  0.015293  273/
0
1  0.015313 273 D 0.7/0.4 D 1.75
∴ v
313
/v
293
 D 
293
/
313
 D 1.75
and the filtration rate will increase by 75%.
SECTION 2
Flow of Fluids — Energy and
Momentum Relationships
PROBLEM 2.1
Calculate the ideal available energy produced by the discharge to atmosphere through a
nozzle of air stored in a cylinder of capacity 0.1 m
3
at a pressure of 5 MN/m
2
. The initial
temperature of the air is 290 K and the ratio of the specific heats is 1.4.
Solution
From equation 2.1: dU D υq  υW. For an adiabatic process: υq D 0anddU DυW,
and for an isentropic process: dU D C
v
dT DυW from equation 2.25.

and hence: W D P
1
v
1
 P
2
v
2
/  1
P
1
v

1
D P
2
v

2
and substituting for v
2
gives:
W D [P
1
v
1
/  1]1  P
2
/P
1

3
/kg.
∴ W D [5 ð 10
6
ð 0.0166/0.4][0.1013/5
0.4/1.4
 1] D0.139 ð10
6
J/kg
Mass of gas D 0.1/0.0166 D 6.02 kg
∴ U D0.139 ð 10
6
ð 6.20 D0.84 ð 10
6
Jor  840 kJ
PROBLEM 2.2
Obtain expressions for the variation of: (a) internal energy with change of volume,
(b) internal energy with change of pressure, and (c) enthalpy with change of pressure, all
at constant temperature, for a gas whose equation of state is given by van der Waals’ law.
16
FLOW OF FLUIDS — ENERGY AND MOMENTUM RELATIONSHIPS 17
Solution
See Volume 1, Example 2.2.
PROBLEM 2.3
Calculate the energy stored in 1000 cm
3
of gas at 80 MN/m
2
at 290 K using STP as the
datum.

2
/P
2
  nRT
1
/P
1
]
Equating these expressions for W gives: C
v
T
2
 T
1
 D P
2

RT
2
P
2

RT
1
P
1

In this example:
P
1

v
C R (from equation 2.27) or: C
v
D R/ 1
Substituting: T
2
D 174.15 K.
PV D nRT and n D 80000 ð 10
3
/8.314 ð 290 D 0.033 kmol
∴ U DW D C
v
nT
2
 T
1
 D 1.5 ð 8.314 ð0.033174.15  290 D47.7kJ
PROBLEM 2.4
Compressed gas is distributed from a works in cylinders which are filled to a pressure
P by connecting them to a large reservoir of gas which remains at a steady pressure P
and temperature T. If the small cylinders are initially at a temperature T and pressure P
0
,
what is the final temperature of the gas in the cylinders if heat losses can be neglected
and if the compression can be regarded as reversible? Assume that the ideal gas laws are
applicable.
Solution
From equation 2.1, dU D υq  υW. For an adiabatic operation, q D 0andυq D 0and
υW D Pd
v or dU DPdv. The change in internal energy for any process involving an

 or T
2
/T
1
 D v
2
/v
1

1
P
1
v
1
/T
1
D P
2
v
2
/T
2
and hence v
1
/v
2
D P
2
/P
1

by the wetted perimeter. Equation 3.69 gives the value d
m
for an annulus of outer radius
r and inner radius r
i
as:
d
m
D 4r
2
 r
2
i
/2r C r
i
 D 2r r
i
 D d d
i

If r D 25 mm and r
i
D 20 mm, then:
d
m
D 225  20 D 10 mm
PROBLEM 3.2
0.015 m
3
/s of acetic acid is pumped through a 75 mm diameter horizontal pipe 70 m

2

From Fig. 3.7, when Re D 1.08 ð10
5
and e/d D 0.0008, R/u
2
D 0.0025.
Substituting: P
f
D 4 ð 0.002570/0.0751060 ð 3.4
2

D 114,367 N/m
2
or: 114.4kN/m
2
19