coulson richardsons chemical engineering volume 1 solutions

Chemical Engineering, Volume 1, Sixth edition Fluid Flow, Heat Transfer and Mass TransferJ.. PROBLEM 1.5 It is found experimentally that the terminal settling velocity u0 of a spherical

CHEMICAL ENGINEERING

Solutions to the Problems in Chemical Engineering Volume 1

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

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 RICHARDSONUniversity of Wales Swansea

OXFORD AUCKLAND BOSTON JOHANNESBURG MELBOURNE NEW DELHI

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

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ISBN 0 7506 4950 X

Typeset by Laser Words, Madras, India

2 Flow of fluids — energy and momentum relationships 16

Each of the volumes of the Chemical Engineering Series includes numerical examples toillustrate the application of the theory presented in the text In addition, at the end of eachvolume, there is a selection of problems which the reader is invited to solve in order toconsolidate 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 forsolutions 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 thereader arrives at a different answer from that in the book, it does not necessarily meanthat it is wrong

This edition of the solutions manual relates to the sixth edition of Volume 1 and porates many new problems There may therefore be some mismatch with earlier editionsand, as the volumes are being continually revised, they can easily get out-of-step witheach other

incor-None of the authors claims to be infallible, and it is inevitable that errors will occurfrom time to time These will become apparent to readers who use the book We havebeen very grateful in the past to those who have pointed out mistakes which have thenbeen corrected in later editions It is hoped that the present generation of readers willprove to be equally helpful!

J F R

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 /40.0252D0.00049 m2

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

If the power P Df , 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 3, diameter, D D L, viscosity, D M/LT, and speed of rotation, N D T1

Thus the power number is a function of the Reynolds number to the power m Infact NP is also a function of the Froude number, DN2/g The previous equation may bewritten as:

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:

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

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

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

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

L/T D kLaML/T2bM/L3cM/LTdEquating dimensions:

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

D M/LT 3, g D L/T2, and  D M/T2 There are 6 variables and 3fundamentals and hence 6  3 D 3 dimensionless groups Taking as the recurring set,and g, then:

variable are: d D L, V D L3 3, g D L/T2,  D M/T2 There are 5 variablesand g as the recurring set, then:



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 filmthickness If the flow is streamline, show that the volumetric flowrate is directly propor-tional to the density of the liquid

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

The dimensions of each variable are: Q D L2/T 3, g D L/T2, D M/LT and d D L.

Solution

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

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

h D fu, di, d0 p, k

The dimensions of each variable are: h D H/L2Tq , u D L/T, diDL, d0D 3,

D M/LT, CpDH/Mq , k D H/LT q 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

the group H/ q and in effect the fundamental dimensions are 4 (M, L, T and H/ q) andthere will be 8  4 D 4 groups For the recurring set, the variables di, , k

coef-It is found by experiment that, when the flow is turbulent, increasing the flowrate by afactor 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 remainingconstant?

Solution

For heat transfer for a fluid flowing through a circular pipe, the dimensional analysis isdetailed in Section 9.4.2 and, for forced convection, the heat transfer coefficient at thewall is given by equations 9.64 and 9.58 which may be written as:

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

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

d DM/L3 cDM/L3, and g D L/T2 There are 7 variables and hence with 3 mental dimensions, there will be 7  3 D 4 dimensionless groups The variables d, uand  will be chosen as the recurring set and hence:

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

and the function becomes: dpD ddu2 cdu2/, gd/u2

PROBLEM 1.11

Liquid flows under steady-state conditions along an open channel of fixed inclination tothe horizontal On what factors will the depth of liquid in the channel depend? Obtain arelationship between the variables using dimensional analysis

Solution

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

and the angle of inclination, ",

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,

to the square of the particle diameter when other variables are kept constant What will

be the effect of doubling the viscosity of the liquid? What does this suggest regarding thenature of the flow?

Obtain the relevant dimensionless groups

For streamline flow it is found that the film thickness is proportional to the one thirdpower of the volumetric flowrate per unit width Show that the heat transfer coefficient

is expected to be inversely proportional to the one third power of viscosity

Solution

For a film of liquid flowing down a vertical surface, the variables influencing the film

3, Q D L2/T, and g D L/T2 Thus, with 5 variables and 3 fundamental dimensions, 5  3 D 2 dimensionless

Thus, dimensionless group 1: QT/L2 DQ1/3 1/3g2/3/4/3 4/3g2/3

dimensionless group 2: υL D υ2/3 2/3g1/3or, cubing D υ3 2g/2

For streamline flow, υ / Q1/3 or n D 1

As the resistance to heat transfer is attributable to the thermal resistance of the sate layer which in turn is a function of the film thickness, then: h / k/υ where k is thethermal conductivity of the film and since υ / 1/3, h / k/1/3, that is the coefficient isinversely proportional to the one third power of the liquid viscosity

conden-PROBLEM 1.15

A spherical particle settles in a liquid contained in a narrow vessel Upon what variableswould you expect the falling velocity of the particle to depend? Obtain the relevantdimensionless 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

For particles settling in a vessel of large diameter, u / 1/ But u/ / 1/2n and,when n D 1, n / 1/ In this case:

2d3/2

Thus the settling velocity is proportional to the square of the particle size

PROBLEM 1.16

A liquid is in steady state flow in an open trough of rectangular cross-section inclined at

an angle " to the horizontal On what variables would you expect the mass flow per unittime to depend? Obtain the dimensionless groups which are applicable to this problem

Force on particle from Stokes’ Law D 3du; where is the fluid viscosity, d is theparticle diameter and u is the velocity of the particle relative to the fluid

What will be the terminal falling velocity of a particle of diameter 10 µm and of density

1600 kg/m3 settling in a liquid of density 1000 kg/m3 and of viscosity 0.001 Ns/m2?

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 andliquid?

Solution

The accelerating force due to gravity D mass of particle  mass of liquid displacedg.For a particle of radius r, volume D 4r3/3, or, in terms of diameter, d, volume D4d3/23/3 D d3/6 Mass of particle D d3

s/ s is the density of the solid.Mass of liquid displaced D d3

force due to gravity D d3 s/6  d3 6g D d3/ s

At steady state, that is when the terminal velocity is attained, the accelerating force due

to gravity must equal the drag force on the particle F, or: d3/ s 3du0where u0 is the terminal velocity of the particle

It is assumed that the resistance per unit projected area of the particle, R0, is a function

D M/LT and u D L/T With 5 variables and 3 fundamental dimensions, there will be

A sphere, initially at a constant temperature, is immersed in a liquid whose temperature

is maintained constant The time t taken for the temperature of the centre of the sphere

to reach a given temperature "c is a function of the following variables:

Diameter of sphere, d

Thermal conductivity of sphere, k

Specific heat capacity of sphere, Cp

Temperature of fluid in which it is immersed, "s

Obtain relevant dimensionless groups for this problem

Solution

p, "c, "s The dimensions of each variable are: t D T, d D

L, k D ML/Tq, CpDL2/T2q, "cDq, "s Dq There are 7 variables and hence with 4

pand "c as the recurring set, then:

PROBLEM 1.19

Upon what variables would the rate of filtration of a suspension of fine solid particles beexpected to depend? Consider the flow through unit area of filter medium and express thevariables in the form of dimensionless groups

It is found that the filtration rate is doubled if the pressure difference is doubled Whatwould be the effect of raising the temperature of filtration from 293 to 313 K?

The viscosity of the liquid is given by:

D 01  0.015T  273

where is the viscosity at a temperature T K and 0 is the viscosity at 273 K

Solution

The volume flow of filtrate per unit area, u m3/m2

fluid viscosity, ; particle size, d; pressure difference across the bed, P, and the voidage

of the cake, e or: vD The dimensions of each of these variables are

u D L/T 3, D M/LT, d D L, P D M/LT2 and e D dimensionless Thereare 6 variables and 3 fundamental dimensions and hence 6  3 D 3 dimensionless

Since the filtration rate is doubled when the pressure difference is doubled, then:

Flow of Fluids — Energy and

Momentum Relationships

PROBLEM 2.1

Calculate the ideal available energy produced by the discharge to atmosphere through anozzle of air stored in a cylinder of capacity 0.1 m3 at a pressure of 5 MN/m2 The initialtemperature of the air is 290 K and the ratio of the specific heats is 1.4

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 Pand temperature T If the small cylinders are initially at a temperature T and pressure P0,what is the final temperature of the gas in the cylinders if heat losses can be neglectedand if the compression can be regarded as reversible? Assume that the ideal gas laws areapplicable

Solution

From equation 2.1, dU D υq  υW For an adiabatic operation, q D 0 and υq D 0 and

υW D Pdv or dU D Pdv The change in internal energy for any process involving an

ideal gas is given by equation 2.25:

Flow in Pipes and Channels

PROBLEM 3.1

Calculate the hydraulic mean diameter of the annular space between a 40 mm and a

50 mm tube

Solution

The hydraulic mean diameter, dm, is defined as four times the cross-sectional area divided

by the wetted perimeter Equation 3.69 gives the value dmfor an annulus of outer radius

r and inner radius ri as:

Cross-sectional area of pipe D /40.0752 D0.0044 m2

Velocity of acid in the pipe, u D 0.015/0.0044 D 3.4 m/s

Reynolds number D ud/ D 1060 ð 3.4 ð 0.07/2.5 ð 103 D1.08 ð 105

Pipe roughness e D 6 ð 105 m and e/d D 6 ð 105/0.075 D 0.0008

The pressure drop is calculated from equation 3.18 as: PfD4R/u2l/du2From Fig 3.7, when Re D 1.08 ð 105 and e/d D 0.0008, R/u2D0.0025

Substituting: PfD4 ð 0.002570/0.0751060 ð 3.42

D114,367 N/m2 or: 114.4 kN/m2

19

PROBLEM 3.3

A cylindrical tank, 5 m in diameter, discharges through a mild steel pipe 90 m long and

230 mm diameter connected to the base of the tank Find the time taken for the waterlevel in the tank to drop from 3 m to 1 m above the bottom The viscosity of water is

1 mNs/m2

Solution

If at any time the depth of water in the tank is h and levels 1 and 2 are the liquid levels

in the tank and the pipe outlet respectively, then the energy balance equation states that:

u2/2 C gz CvP2P1 C F D0

In this example, P1 DP2 Datmospheric pressure and vP2P1 D0 Also u1/u2D

0.23/52 D0.0021 so that u1 may be neglected The energy balance equation thenbecomes:

u2/2  hg C 4R/u2l/du2D0The last term is obtained from equation 3.19 and z D h

Substituting the known data:

u2/2  9.81h C 4R/u290/0.23u2 D0

[1 C 3130R/u2]

In falling from a height h to h  dh,

the quantity of water discharged D /452dh D 19.63dh m3

Volumetric flow rate D /40.232u D0.0415u D 0.184ph/

[1 C 3130R/u2], andthe time taken for the level to fall from h to h  dh is:

19.63

0.184

dhph

From equation 3.23:

R/u2 Re2DRd2/ 2Dh2gd3/4l 2

Dh ð10002ð9.81 ð 0.233/4 ð 90 ð 106 D3.315 ð 108hThus as h varies from 3 m to 1 m, R/u2Re2 varies from (9.95 ð 108) to (3.315 ð

108.)

If R/u2 is taken as 0.002, Re will vary from (7.05 ð 105) to (4.07 ð 105) FromFig 3.7 this corresponds to a range of e/d of between 0.004 and 0.005 or a roughness ofbetween 0.92 and 1.15 mm, which is too high for a commercial pipe.

If e is taken as 0.05 mm, e/d D 0.0002, and, for Reynolds numbers near 106, R/u2D0.00175 Substituting R/u2 D0.00175 and integrating gives a time of 398 s for the level

to fall from 3 m to 1 m If R/u2 D0.00175, Re varies from (7.5 ð 105) to (4.35 ð 105),and from Fig 3.7, e/d D 0.00015, which is near enough to the assumed value Thus thetime for the level to fall is approximately 400 s

PROBLEM 3.4

Two storage tanks A and B containing a petroleum product discharge through pipes each0.3 m in diameter and 1.5 km long to a junction at D From D the product is carried by

a 0.5 m diameter pipe to a third storage tank C, 0.8 km away The surface of the liquid

in A is initially 10 m above that in C and the liquid level in B is 7 m higher than that in

A Calculate the initial rate of discharge of the liquid if the pipes are of mild steel Thedensity of the petroleum product is 870 kg/m3 and the viscosity is 0.7 mNs/m2

Solution

See Volume 1, Example 3.4

PROBLEM 3.5

Find the drop in pressure due to friction in a glazed porcelain pipe 300 m long and

150 mm diameter when water is flowing at the rate of 0.05 m3/s

Solution

For a glazed porcelain pipe, e D 0.0015 mm, e/d D 0.0015/150 D 0.00001

Cross-sectional area of pipe D /40.152D0.0176 m2

Velocity of water in pipe, u D 0.05/0.0176 D 2.83 m/s

Reynolds number D ud/ D 1000 ð 2.83 ð 0.15/103 D4.25 ð 105

From Fig 3.7, R/u2D0.0017

The pressure drop is given by equation 3.18: PfD4R/u2l/du2

or: 4 ð 0.0017300/0.151000 ð 2.832 D108,900 N/m2 or 1 MN/m2

PROBLEM 3.6

Two tanks, the bottoms of which are at the same level, are connected with one another

by a horizontal pipe 75 mm diameter and 300 m long The pipe is bell-mouthed at each

end so that losses on entry and exit are negligible One tank is 7 m diameter and containswater to a depth of 7 m The other tank is 5 m diameter and contains water to a depth of

3 m

If the tanks are connected to each other by means of the pipe, how long will it takebefore the water level in the larger tank has fallen to 6 m? Assume the pipe to be of agedmild steel

Solution

The system is shown in Fig 3a If at any time t the depth of water in the larger tank is

hand the depth in the smaller tank is H, a relationship between h and H may be found

Area of larger tank D /472 D38.48 m2,area of smaller tank D /452 D19.63 m2

The energy balance equation is: u2/2 C gz CvP1P2 D F

u2/2 may be neglected, and P1DP2 Datmospheric pressure, so that:

If R/u2 is taken as 0.002, then:

2.95h  16.72 and t D10590 sAverage volumetric flowrate D 38.487  6/10590 D 0.00364 m3/s

Cross-sectional area of pipe D 0.00442 m2

Average velocity in the pipe D 0.00364/0.00442 D 0.82 m/s

Reynolds number D 1000 ð 0.82 ð 0.75/103 D6.2 ð 104

From Fig 3.7, if e D 0.05 mm, e/d D 0.00067 and R/u2D0.0025, which is near enough

to the assumed value of 0.002 for a first estimate

Thus the time for the level to fall is approximately 10590 s (2.94 h)

PROBLEM 3.7

Two immiscible fluids A and B, of viscosities Aand B, flow under streamline conditionsbetween two horizontal parallel planes of width b, situated a distance 2a apart (where a

is much less than b), as two distinct parallel layers one above the other, each of depth a

Show that the volumetric rate of flow of A is:

RA = Shear stress at centre-plane on A

RB = Shear stress at centre-plane on B



A

CRAlwhere RAis the shear stress at the centre plane,

s C k2where R is the shear stress at the centre plane on B

2 Bl C

RAa B

2 a2l



1

A

1 ...

Re D1 .1 ð 10 4 Ddu/

∴ u D Re /d D 1. 1 ð 10 4ð0. 01/ 0.075 D 1. 47... pipe, e D 0.0 015 mm, e/d D 0.0 015 /15 0 D 0.000 01

Cross-sectional area of pipe D /40 .15 2D0. 017 6 m2

Velocity of water in pipe, u D 0.05/0. 017 6 D 2.83 m/s...

d1. 25 0.25 DK0u1. 75/d1. 25For pipe in which the velocity is u1< /small>, P D K0u1. 751< /sub> /0.31. 25