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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 J.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. 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,

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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 J.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. 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,

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