2

CHEMICAL ENGINEERING

338 Pages · 2003 · 1.73 MB · English

  • CHEMICAL ENGINEERING

    CHEMICAL 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


    uf6d9J.F.Richardson,J.R.BackhurstandJ.H.Harker2001


    Allrightsreserved.Nopartofthispublication


    maybereproducedinanymaterialform(including


    photocopyingorstoringinanymediumbyelectronic


    meansandwhetherornottransientlyorincidentally


    tosomeotheruseofthispublication)withoutthe


    writtenpermissionofthecopyrightholderexcept


    inaccordancewiththeprovisionsoftheCopyright,


    DesignsandPatentsAct1988orunderthetermsofa


    licenceissuedbytheCopyrightLicensingAgencyLtd,


    90TottenhamCourtRoad,London,EnglandW1P9HE.


    Applicationsforthecopyrightholder’swrittenpermission


    toreproduceanypartofthispublicationshouldbeaddressed


    tothepublishers


    BritishLibraryCataloguinginPublicationData


    AcataloguerecordforthisbookisavailablefromtheBritishLibrary


    LibraryofCongressCataloguinginPublicationData


    AcataloguerecordforthisbookisavailablefromtheLibraryofCongress


    ISBN075064950X


    TypesetbyLaserWords,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


    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.


    null 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,


    Please note: To fully download this free PDF,EBook files you need know All free.
    Found by internet command,site not saved pdf file
You May Also Like

Related PPT Template in the same category.