Table Of ContentCHEMICAL ENGINEERING
SolutionstotheProblemsinChemicalEngineeringVolume1
Coulson &Richardson’s ChemicalEngineering
Chemical Engineering,Volume1, Sixthedition
Fluid Flow,Heat Transfer andMassTransfer
J.M. CoulsonandJ.F. Richardson
withJ.R.BackhurstandJ.H.Harker
Chemical Engineering,Volume2, Fourthedition
ParticleTechnology andSeparationProcesses
J.M. CoulsonandJ.F. Richardson
withJ.R.BackhurstandJ.H.Harker
Chemical Engineering,Volume3, Thirdedition
Chemical &BiochemicalReactors&ProcessControl
Editedby J.F. RichardsonandD. G.Peacock
SolutionstotheProblems inVolume 1,Firstedition
J.R.BackhurstandJ.H.Harker
withJ.F.Richardson
Chemical Engineering,Volume5, Secondedition
SolutionstotheProblems inVolumes 2and3
J.R.BackhurstandJ.H.Harker
Chemical Engineering,Volume6, Thirdedition
Chemical EngineeringDesign
R.K.Sinnott
Coulson & Richardson’s
CHEMICAL ENGINEERING
J. M. COULSON and J. F. RICHARDSON
SolutionstotheProblemsinChemicalEngineering
Volume1
By
J. R. BACKHURST and J. H. HARKER
UniversityofNewcastleuponTyne
With
J. F. RICHARDSON
UniversityofWalesSwansea
OXFORD AUCKLAND BOSTON JOHANNESBURG MELBOURNE NEWDELHI
Butterworth-Heinemann
LinacreHouse,JordanHill,OxfordOX28DP
225WildwoodAvenue,Woburn,MA01801-2041
AdivisionofReedEducationalandProfessionalPublishingLtd
Firstpublished2001
J.F.Richardson,J.R.BackhurstandJ.H.Harker2001
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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
illustratetheapplicationofthetheorypresentedinthetext.Inaddition,attheendofeach
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.
Manyreaderswhodonothavereadyaccesstoassistancehaveexpressedthedesirefor
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 relatesto the sixth edition of Volume 1 andincor-
poratesmanynewproblems.Theremaythereforebesomemismatchwithearliereditions
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 lineD(cid:3)(cid:4)/4(cid:5)0.0252 D0.00049 m2.
Mean velocity of acid, uD(cid:3)685ð10(cid:3)6(cid:5)/0.00049D1.398 m/s.
∴ Reynolds number, ReDdu(cid:10)/(cid:11)D(cid:3)0.025ð1.398ð1840(cid:5)/0.025D2572
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 kWhD1 kWð1 hD(cid:3)1000 J/s(cid:5)(cid:3)3600 s(cid:5)D3,600,000 J or 3.6 MJ
1 thermD105.5 MJ
∴ cost of electricityD1 p/3.6 MJ or (cid:3)1/3.6(cid:5)D0.28 p/MJ
cost of gasD15 p/105.5 MJ or (cid:3)15/105.5(cid:5)D0.14 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 CHEMICALENGINEERINGVOLUME1SOLUTIONS
Solution
Fromthesteamtables,inAppendixA2,Volume 1,totalenthalpyofsteamat1825 kN/m2 D
2798kJ/kg.
∴ enthalpy of steamD(cid:3)5.2ð2798(cid:5)D14,550 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 coalD(cid:3)14,550ð100(cid:5)/75D19,400 kW.
For a calorific value of 27,200 kJ/kg, rate of coal consumptionD(cid:3)19,400/27,200(cid:5)
D0.713 kg/s
or: (cid:3)0.713ð3600ð24(cid:5)/1000D61.6 Mg/day
20% of the enthalpy in the steam is converted to power or:
(cid:3)14,550ð20(cid:5)/100D2910 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
IfthepowerPDf(cid:3)DN(cid:10)(cid:11)(cid:5),thenatypicalformofthefunctionisPDkDaNb(cid:10)c(cid:11)d,where
k is a constant. The dimensions of each parameter in terms of M, L, and T are: power,
PDML2/T3, density, (cid:10) DM/L3, diameter, DDL, viscosity, (cid:11)DM/LT, and speed of
rotation, NDT(cid:3)1
Equating dimensions:
M: 1 DcCd
L: 2 Da(cid:3)3c(cid:3)d
T: (cid:3)3 D(cid:3)b(cid:3)d
Solving in terms of d:aD(cid:3)5(cid:3)2d(cid:5), bD(cid:3)3(cid:3)d(cid:5), cD(cid:3)1(cid:3)d(cid:5)
(cid:1) (cid:2)
D5 N3 (cid:10)
∴ PDk (cid:11)d
D2dNd(cid:10)d
or: P/D5N3(cid:10) Dk(cid:3)D2N(cid:10)/(cid:11)(cid:5)(cid:3)d
that is: N DkRem
P
UNITSANDDIMENSIONS 3
Thus the power number is a function of the Reynolds number to the power m. In
fact N is also a function of the Froude number, DN2/g. The previous equation may be
P
written as:
P/D5N3(cid:10) Dk(cid:3)D2N(cid:10)/(cid:11)(cid:5)m
Experimentally: P/N2
From the equation, P/NmN3, that is mC3D2 and mD(cid:3)1
Thus for the same fluid, that is the same viscosity and density:
(cid:3)P /P (cid:5)(cid:3)D5N3/D5N3(cid:5)D(cid:3)D2N /D2N (cid:5)(cid:3)1 or: (cid:3)P /P (cid:5)D(cid:3)N2D3(cid:5)/(cid:3)N2D3(cid:5)
2 1 1 1 2 2 1 1 2 2 2 1 2 2 1 1
In this case, N DN and D D2D .
1 2 2 1
∴ (cid:3)P /P (cid:5)D8D3/D3 D8
2 1 1 1
A similar solution may be obtained using the Recurring Set method as follows:
PDf(cid:3)D,N,(cid:10),(cid:11)(cid:5),f(cid:3)P,D,N,(cid:10),(cid:11)(cid:5)D0
Using M, L and T as fundamentals, there are five variables and three fundamentals
and therefore by Buckingham’s (cid:4) theorem, there will be two dimensionless groups.
Choosing D, N and (cid:10) as the recurring set, dimensionally:
(cid:3) (cid:4)
D (cid:6)L L (cid:6)D
N (cid:6)T(cid:3)1 Thus: T (cid:6)N(cid:3)1
(cid:10) (cid:6)ML(cid:3)3 M (cid:6)(cid:10)L3 D(cid:10)D3
P
First group, (cid:4) , is P(cid:3)ML2T(cid:3)3(cid:5)(cid:3)1 (cid:6)P(cid:3)(cid:10)D3D2N3(cid:5)(cid:3)1 (cid:6)
1 (cid:10)D5N3
(cid:11)
Second group, (cid:4) , is (cid:11)(cid:3)ML(cid:3)1T(cid:3)1(cid:5)(cid:3)1 (cid:6)(cid:11)(cid:3)(cid:10)D3D(cid:3)1N(cid:5)(cid:3)1 (cid:6)
2 (cid:10)D2N
(cid:1) (cid:2)
P (cid:11)
Thus: f , D0
(cid:10)D5N3 (cid:10)D2N
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 of a spherical particle in
0
a fluid is a function of the following quantities:
particle diameter, d; buoyant weight of particle (weight of particle(cid:3)weight of displaced
fluid), W; fluid density, (cid:10), and fluid viscosity, (cid:11).
Obtain a relationship for u using dimensional analysis.
0
Stokes established, fromtheoreticalconsiderations, that for small particleswhich settle
at very low velocities, the settling velocity is independent of the density of the fluid
4 CHEMICALENGINEERINGVOLUME1SOLUTIONS
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: u DkdaWb(cid:10)c(cid:11)d, then working in dimensions of M, L and T:
0
(cid:3)L/T(cid:5)Dk(cid:3)La(cid:3)ML/T2(cid:5)b(cid:3)M/L3(cid:5)c(cid:3)M/LT(cid:5)d(cid:5)
Equating dimensions:
M: 0 DbCcCd
L: 1 DaCb(cid:3)3c(cid:3)d
T: (cid:3)1 D(cid:3)2b(cid:3)d
Solving in terms of b:
aD(cid:3)1,cD(cid:3)b(cid:3)1(cid:5), and dD(cid:3)1(cid:3)2b(cid:5)
∴ u Dk(cid:3)1/d(cid:5)(cid:3)Wb(cid:5)(cid:3)(cid:10)b/(cid:10)(cid:5)(cid:3)(cid:11)/(cid:11)2b(cid:5) where k is a constant,
0
or: u Dk(cid:3)(cid:11)/d(cid:10)(cid:5)(cid:3)W(cid:10)/(cid:11)2(cid:5)b
0
Rearranging:
(cid:3)du (cid:10)/(cid:11)(cid:5)Dk(cid:3)W(cid:10)/(cid:11)2(cid:5)b
0
where (W(cid:10)/(cid:11)2) is a function of a form of the Reynolds number.
For u to be independent of (cid:10), b must equal unity and u DkW/d(cid:11)
0 0
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, (cid:11); volume of the drop, V expressed in terms of d, the drop diameter; density of
the liquid, (cid:10); acceleration due to gravity, g and possibly, surface tension of the liquid,
