Solution Manual for Fluid Mechanics for Engineers, 2017 Edition

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SOLUTIONS MANUALFluid Mechanicsfor EngineersFirst EditionDavid A. ChinJanuary 2018

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Chapter1PropertiesofFluids1.1.(a)FindthedimensionsofCLasfollows,CL=FL12ρV2A→CL=MLT−2(ML−3)(LT−1)2(L2)=MLT−2MLT−2=1ThereforeCLisdimensionlessanddoesnotdependonthesystemofunitsused.(b)NoadjustmentfactorisnecessarywhenUSCSunitsareusedinsteadofSIunits.1.2.(a)Insertingthedimensionsofthevariablesinthegivenequationyieldsρd2zdt2+adzdt+bz=c┌ML3LT2+aLT+b[L] =c┐ ┌┐┌┐┌ML2T2+aLT+b[L] =c┐┌┐Therefore,therequireddimensionsoftheparametersa,b, andcare,a=ML3T,b=ML3T2,c=ML2T2┌┐┌┐┌┐(b)Ifρ∗,z∗, andt∗are the givenvariables in nonstandard units, then the conversionfactorsare:ρρ∗= 103,zz∗= 10−3,tt∗= 36001.3.(a)InsertingthedimensionsofthevariablesinthegivenequationyieldsQ=1nA53P23S120→L3T−1=1()(L2)53L23(−)12→L3T−1= L83−Sincethe dimensionofthe left-hadsideoftheequationisnot equaltothe dimensiononthe right-hand side of the equation, the given equation isnotdimensionallyhomogeneous.

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(b)Ifthelengthunitsarechangedfrommtoftand1m=3.281ft,theninsertingthisconversionfactorintothegivenequationsrequiresthat(3.281)−3Q'=1n(3.281)−2A'53[(3.281)−1P']23S120┌┐where theprimedquantitieshavelengthunits offt.Simplifying theaboveequationandremovingtheprimesgivesQ=3.28113nA53P23S120→Q=1.486nA53P23S120Therefore,theconversionfactortobeaddedis1.486.1.4.−−−··−QuantityDimensionTypicalSIUnitenergy2T−2FL=MLJforceF=MLT−2Nheat2T2FL = MLJmoment2T−2FL=MLN mmomentumMV =MLT−1kg m/spower2FLT−1T−3= MLWpressureT−2FL−2= ML−1PastrainLL−1=stressT−2FL−2= ML−1Pawork2T2FL = MLJ1.5.GivenWithPrefix5.63×107N56.3 MN8.27×105Pa827 kPa23.86×10−4m20.0386 km7.88×105m788 km1.6.QuantityUSCSAbbreviationInSIUnits12gallonsperminute12gpm45.4L/min55milesperhour55mph88.5km/h5feetpersecond5ft/s1.5m/s125cubicfeetperminute125cfm3.54m3/min1000gallons1000gal3785L25acres25ac10.1ha500horsepower500hp373kW

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1.7.(a)1 hp = 550ft·lbs×0.3048mft×4.448Nlb=745.7 W(b)1 lb/in2= 1lbin2×4.448Nlb×10.02542in2m2= 6.894×103Pa =6.894 kPa1.8.Fromthegivendata:ρ0=1000kg/m3,andthedensitydeviatesmostfrom1000kg/m3atT=100◦C,whereρ=958.4kg/m3.Hence,themaximumerrorinassumingadensityof1000kg/m3iserror =1000−958.4958.4×100 =4.3%1.9.Fromthegivendata:V1= 3 L,ρ1= 1030 kg/m3,V2= 5 L,andρ= 920 kg/m3.Thedensityofthemixture,ρm,isgivenbyρm=ρ1V1+ρ2V2V1+V2=(1030)(3) + (920)(5)3 + 5=961 kg/m3Note:ThevolumesdonotneedtobeconvertedfromLtom3sincetheconversionfactorwouldcancelout.1.10.(a)The specific weight,γ, is derived from the density,ρ, using the relation:γ=ρg= 9.807ρ.ObtainingρfromAppendixB.1gives:Temperatureρ(◦C)(kg/m3)(N/m3)γ0999.8980520998.29789100958.49399(b)The specific gravity,SG, is derivedfrom the density,ρ, using the relation:SG =ρ/ρ4◦C.ObtainingρfromAppendixB.1gives:TemperatureρSG(◦C)(kg/m3)(–)0999.81.00020998.20.998100958.40.958V1=400 L,T1= 15◦C,andT2= 90◦C.1.11.Fromthegivendata:Thedensitiesofwatercorresponding toT1andT2(from Appendix B.1) are:ρ1= 999.1 kg/m3andρ2= 965.3 kg/m3.(a)Theinitialmass,m1,inthetankisgivenbym1=ρ1V1= (999.1)(0.4)= 399.6 kg

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Thevolumeofwaterafterheatingto90◦C is givenbyV2=m1ρ2=399.6965.3= 0.4140 m3= 414.0 LTherefore,thespilledvolume,ΔV,isgivenbyΔV=V2V1= 414400 =14 L−−(b)Thespilledmass,ΔmisgivenbyΔm=ρ2ΔV= (965.3)(1410−3) = 13.5 kg×Thepercentchangeinthemass(=percentchangeinweight)is13.5/399.6×100=3.4%1.12.Fromthegivendata:γ= 12 kN/m3=12000N/m3.Forwaterat4◦C:ρw=1000kg/m3.Accordingtothedefinitionsofdensityandspecificgravity,ρ=γg= (12000)(9.807)=1224 kg/m3SG =ρρw=12241000= 1.2241.13.Fromthegivendata:SG=1.5.Forwaterat4C:◦ρw=1000kg/m3.Accordingtothedefinitionsofdensityandspecificweight,ρ= SG·ρw= (1.5)(1000)=1500 kg/m3γ=ρg= (1500)(9.807)= 14710 N/m314.7 kN m3≃·1.14.Forany givenvolume,V,containinga mixture, letCm= mass ratio,ρf= density of the purefluid,ρm=densityofthemixture,mf=massofpurefluid,mm= massofmixture,ms=mass of solids in the mixture,SGf= specific gravityof pure fluid, andSGm= specific gravityofmixture.Therefore,mf=ρfV,mm=ρmV,ms=mmmf= (ρmρf)V−−Usingtheserelationshipsyields,Cm=msmm=(ρm−ρf)//Vρm//V= 1−ρfρm→Cm= 1−SGfSGm1.35.Fromthegivendata:ρ=800kg/m3.Forwaterat4◦C:ρw=1000kg/m3.AccordingtothedefinitionsgiveninEquations1.9and1.10,γ=ρg= (800)(9.807)= 7846 N/m37.85 kN/m3≃SG =ρρw=8001000= 0.80

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1.16.Fromthegivendata:M=200kg,andSG=1.5.At4Cthedensityofwateris◦ρw=1000kg/m3.Thevolume,V,ofthereservoirisgivenbyV=MρwSG=200(1000)(1.5)= 0.133 m3·1.17.Fromthegivendata:Wc= 10 N,andWt=50 N.Forkeroseneat20◦C,ρ= 808 kg/m3andγ= 7924 N/m3(AppendixB.4).Usingthesedatagivesthefollowing,weightofkerosene,Wk=WtWc= 5010 = 40 N−−volumeofkerosene,Vk=Wkγ=407924= 5.048×10−3m3= 5.05 Lmassofkerosene =Wkg=409.807= 4.08 kg1.18.Thebulkmodulus,Ev,isdefinedbyEquation1.12asEv=−dpdV /V(1)wherethedensityofthefluid,ρ,isdefinedbyρ=MV(2)whereMisthe(constant)massoffluidandVisthevolumeoffluidthatiscompressedbythefluidpressure.DifferentiatingEquation2withrespecttoVgivesdρdV=−MV2(3)CombiningEquations2and3toeliminateMyieldsdρdV=−ρVordVV=−dρρ(4)Finally,combiningEquations1and4givesEv=dpdρ/ρ1.19.Equation1.13canbeapproximatedbyΔρρ=ΔpEv(1)andat20◦C,Ev= 2.18106kPa(TableB.1).For×Δρ/ρ=0.01,Equation1becomes0.01 =Δp2.18106→Δp= 2.18×104kPa =21.8 MPa×

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1.20.Fromthegivendata:T= 20C,D= 3 m,R=D/2 = 1.5 m,and◦Δp= 9 MPa.Forwaterat20C,◦ρ0= 998.2 kg/m3andEv= 2.18106kPa×(fromAppendixB.1).Usingthesedata,thevolume,V,ofthetank,andtheinitialmass,m0isthetankarecalculatedasV=43πR3=43π(1.5)3= 14.14 m3,m0=ρ0V= (998.2)(14.14)= 1.411104kg×Fromthedefinitionofthebulkmodulus,Ev,Ev≈ΔpΔρ/ρ0→Δmm0=ΔpEv→Δm=m0ΔpEvSubstitutingthegivenandderiveddataintothisrelationshipyields,Δm= (1.411×104)9×1032.18106= 58.3 kg×1.21.Fromthegivendata:p1=100kPa,p2=20000kPa,V1=1.700m3,andV2=1.650m3.UsingthedefinitionofthebulkmodulusgivenbyEquation1.12,Ev≈ −ΔpΔV /V=−20000−1001.650−1.7001.700= 6.766×105kPa =677 MPa1.22.From the given data:V1= 10 m3,andΔp= 10 MPa.For benzene,Ev= 1030 MPa(AppendixB.4).UsingthedefinitionofthebulkmodulusgivenbyEquation1.12,Ev≈ −ΔpΔV /V1→1030≈ −10ΔV10→ΔV≈0.0971 m31.23.Fromthegivendata:T1= 10◦C,andT2=100◦C.Theaveragecoefficientofvolumeexpansion,β,betweenT1andT2isderivedfromAppendixB.1as¯β= 0.41810−3K−1.×(a)ApplyingEquation1.19(withΔp= 0 )givesΔρρ≈ −¯βΔT=−0.418×10−3(100−10) =−0.0376=−3.76%(b)LetAbethesurfaceareaofthewaterinthepot(assumedtobeconstant),h1isthedepthofwateratT1,andh1+ ΔhbethedepthofwateratT2.Therefore,sincethemassofwaterisconstant,ρ1//Ah1=ρ2//A(h1+ Δh)→ρ1h1=ρ2h1+ρ2Δh→Δhh1=ρ1−ρ2ρ2(1)ThedensityofwateratT1= 10◦CandT2= 100◦CareobtainedfromAppendixB.1asρ1= 999.7 kg/m3andρ2= 958.4 kg/m3,respectively.UsingEquation1givesΔhh1=999.7−958.4958.4= 0.0431 =4.31%

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1.24.fromthegivendata:β= 5.710−4K−1,×T1= 10◦C,andT2= 90◦C.ApplyingEquation1.19(withΔp= 0)givesΔρρ≈ −βΔT=−5.7×10−4(90−10) =−0.0432 =−4.32%1.25.Fromthegivendata:T1= 15◦C,ΔV /V1= 0.01,andβ= 9.510−4K−1×.Usingthegivendata,V2−V1V= 0.01→V2V−1 = 0.01→V2V= 1.01(1)111Δρρ1=//m/V2−//m/V1//m/V1→Δρρ1=V1V2−1(2)Δρρ1=−5.7×10−4ΔT(3)whereΔTisthemaximumallowabletemperaturerise.CombiningEquations1to3yieldsΔT= 10.4◦C.1.26.Fromthegivendata:T= 20◦C=293K.Notingthatthespeedofsound,c, isgivenbyc=√vE/ρ,thecalculationofthespeedofsoundinwaterandmercuryaresummarizedinthefollowingtable:MediumEv(×106Pa)ρ(kg/m3)c(m/s)Water21719981475Mercury26200135501390Therefore,thespeedofsoundinwaterat20◦Cis1475m/s,andthespeedofsoundinmercuryat20◦Cis1390m/s.1.27.Fromthegivendata:c=1700m/s,andSG=1.8.Forwaterat4◦C,ρw=1000kg/m3.UsingEquation1.15,thebulkmodulus,Ev,iscalculatedasfollows:c=EvSGρw→1700 =Ev(1.8)(1000)→Ev= 3.06×106Pa =3.06 MPa/·/1.28.Fromthegivendata:n1= 1010moleculespermm3,andT= 15◦C=288 K.Foranidealgas,Ru= 8.314 J/(kg K).·(a)Usingthegivendatawiththeideal-gaslaw,n=1010×6.0231023= 1.660×10−14molesV= 1 mm3= 10−9m3p=nVRuT=1.660×10−1410−9(8.314)(288)=0.0397 Pa

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(b)Thetermusedtodescribeagasinwhichthecontinuumapproximationisnotvalidisararefiedgas.1.29.Fromthegivendata:p=101 kPa,andT= 25◦C=298 K.ForHe,RHe=8314/4.003=2077 J/(kg K),andforair,·Rair=8314/28.96=287.1 J/(kg K)(fromAppendixB.5).Using·theidealgaslaw,ρHe=pRHeT=101×103(2077)(298)= 0.1632 kg/m3ρair=pRairT=101×103(287.1)(298)= 1.181 kg/m3ThespecificvolumeisdefinedbyEquation1.11asthevolumeperunitmass,hencevHe=1ρHe=10.1632= 6.13m3/kgvair=1ρair=11.181= 0.847m3/kg1.30.Forairatstandardatmosphericpressure,p= 101 kPaandR= 287.1 J/(kg K).Taking·ρ1=densityfromAppendixB.2,andρ2=densityfromidealgaslaw,gives:T(◦C)ρ1(kg/m3)T( K)ρ2(kg/m3)Δ(%)−401.5142331.5098−0.27−201.3942531.3905−0.2501.2922731.2886−0.2651.2692781.2654−0.28101.2462831.2431−0.23151.2252881.2215−0.29201.2042931.2007−0.28251.1842981.1805−0.29301.1643031.1610−0.25401.1273131.1239−0.27501.0923231.0891−0.26601.0593331.0564−0.24701.0283431.0256−0.23800.99943530.9966−0.28900.97183630.9691−0.271000.94583730.9431−0.282000.74594730.7438−0.293000.61585730.6140−0.304000.52436730.5227−0.305000.45657730.4551−0.3110000.277212730.27640.31−

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Based on these results, the ideal gas law gives quite accurate estimates witherrorslessthan0.31%.1.31.Fromthegivendata:ρ=5kg/m3,andp=450kPa.PropertiesofO2fromAppendixB.5:cp=909J/kg K,cv=649J/kg K,andR=··cpcv−=909649=260J/kg K.Theideal−·gaslaw,Equation1.24,givesρ=pRT→5 =450×103(260)T→T= 346 K=73◦C1.32.Fromthegivendata:V= 2m3,T=15◦C=288K,andp=500kPa.Themolarmassofhelium is 4.003g/mol,and hence the gas constant for helium can be takenasR= 8314/4.003=2077J/kg K.Thedensity,mass,andweightofheliuminthetankaregivenby·ρ=pRT=500×103(2077)(288)= 0.08610 kg/m3M=ρV= (0.08610)(2)=0.1722 kgW=M g= (0.1722)(9.807)=1.689 N1.33.Fromthegivendata:m= 10 kg,T= 15C=288 K,p= 10 MPa,andL= 3D.Forpure◦oxygen,R=Ru/M= 8314/32= 259.8 J/(kg·K).Usingthegivendataandtheidealgaslaw,V=π4D2L=π4D2(3D) =3π4D3,m=pVRT→V=mRTpCombiningtheseequationsgives34πD3=mRTp→D=4mRT3πp13=4(10)(259.8)(288)3π(10106)13= 0.317 m┌┐┌×┐Sincethelengthmustbethreetimesthediameter,L= 3(0.317)=0.950 m.Therequireddimensionsofthetankareadiameterof317 mmandalengthof0.950 m.1.34.Fromthegivendata:M= 10kg,T= 60◦C=333K,andp=200kPa.Forair,R=287.1J/kg·K.Thevolume,V,canbederivedfromtheidealgaslaw,Equation1.24,asfollowsMV=pRT→10V=200×103(287.1)(333)→V= 4.78 m31.35.Fromthegivendata:V=200 L,m= 3 kg,andT= 15◦C=288 K.Thisassumesthatthetemperatureoftheairinthetankisthesameasintheroom.Forstandardair,R=287.1 J/(kg·K).Usingtheideal-gaslawgivesρ=pRT→p=mVRT=30.2(287.1)(288)= 1.24×106Pa =1.24 MPa

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1.36.Fromthegivendata:V= 0.1m3,T= 20C=293K,andp=400kPa.Thegasconstant◦foraircanbetakenasR=287.1J/kg·K.ThedensityofairinthetankcanbecalculatedusingEquation1.24,whichgivesρ=pRT=400×103(287.1)(293)= 0.4898 kg/m3Hencetheweightofairinthetank,W, is givenbyW=ρV g= (0.4898)(0.1)(9.807)= 0.4803 N0.48 N≃Theweight(0.48N)hasbeenroundedtotwosignificantdigitstobeconsistentwiththeaccuracyofthegivendata.1.37.From the givendata:V= 10 m12 m4 m = 480 m3,××p= 101.3 kPa,T1= 20◦C = 293.15 K,andT= 10C = 283.15 K.2◦Forair,R= 287.1 J/(kg K).·(a)Usingthegivendataatatemperatureof20◦C givesρ1=pRT1=101.3×103(287.1)(293.15)= 1.204 kg/m3,m1=ρ1V= (1.204)(480)=577.7 kgW1=m1g= (577.7)(9.807)= 5.666103N =1274 lb×(b)Whenthetemperatureintheroomisreducedto10◦C,thenρ2=pRT2=101.3×103(287.1)(283.15)= 1.246 kg/m3,m2=ρ2V= (1.246)(480)= 598.1 kgchange =m2−m1m1×100 =598.1−577.7577.7×100 =3.53%1.xxFromthegivendata:V=600 mL=610−4m3,×p1=101.3 kPa,T= 30C=303.15 K,1◦T2= 1.5C=274.65 K.◦Applyingtheidealgaslaw,assumingthatthenumberofmolesofairabovethewaterremainsconstant:pV=nRT→n=pVRT→p1T1=p2T2→p2=p1T2T1= (101.3)274.65303.15= 91.78 kPaTherefore,thepressuredifferentialis101.391.78 =9.5 kPa−1.38.Fromthegivendata:p1=600 kPa,T= 20C=293 K,1◦andT2= 30◦C=303 K.TakingρandRasconstants,theidealgaslawgivesp1T1=p2T2→600293=p2303→p2= 620 kPaThereforethechangeinpressureis620 kPa−600 kPa=20 kPa.Thisresultwouldbethesameforanygasthatobeystheidealgaslaw.

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1.39.Fromthegivendata:p1=130 kPa,p2=210 kPa,V=15 L=0.015 m3,andT=30◦C.Requiredconstants:R= 8.314 kJ/kmol K,molarmassofair,M=28.97 kg/kmol,·patm= 101 kPa.Usingthesedatawiththeidealgaslaw,n1=p1V1RT1=(101 + 130)(0.015)(8.314)(273.15 + 30)= 0.00137 kmoln2=p2V2RT=(101 + 210)(0.015)(8.314)(273.15 + 30)= 0.00185 kmol2massadded = (n2−n1)M= (0.00185−0.00137)28.97= 0.0139 kg=13.9 g1.40.Considerthreestatesofthetire:State1istheinitialstate,State2istheheated-upstate,andState3isthecooled-downstate.Fromthegivendata:patm=101.3 kPa,V= 15 L,p1(gauge)=207kPa,p1(absolute)=207kPa+101.3kPa=308.3kPa,T1= 20◦C=293 K,T2= 60◦C = 333 K,andT3=T1= 293 K.(a)Betweentheinitialstateandtheheated-upstate,theairdensityinthetireremainsconstant,sotheideal-gaslawgivesρR=pT= constant→p1T1=p2T2→p2=T2T1p1=333293(207) = 350 kPaHence,theresultinggaugepressureis350kPa101.3kPa=249kPa=−36 lb/in2.(b)Betweentheheated-upstateandthecooled-downstate,theairdensityinthetireremainsconstantandtheairpressureattheheated-upstateisequalto308.3kPa.Theideal-gaslawgivesρR=pT= constant→p3T3=p2T2→p3=T3T2p2=293333(308.3)= 271 kPaHence,theresultinggaugepressureis271kPa−101.3kPa=170kPa=25 lb/in2.1.41.Fromthegivendata:T1= 20◦C=293 K,T2= 10◦C=283 K,p1g= 12.5 lb/in2,andp1=p1g+patm= 12.5 + 14.7 = 27.2 lb/in2.(a)Inthiscasetheexpectedpressure,p2in a football is givenbyp2=T2T1p1=283293(27.2) = 26.3 lb/in2→p2g=p2−patm= 26.2−14.7 =11.6 lb/in2(b)Inthiscasetherequiredtemperature,T2onthefieldisgivenbyp2=T2T1p1→(10.5 + 14.7) =T2293(27.2)= 272 K =−2◦C(c)Theory:The New England Patriots did not inflate the footballs to the required pressure.1.42.Fromthegivendata:p1= 207 kPaandp2= 241 kPa.

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(a)Fromtheideal-gaslaw,ρ=pRT→p1T1=p2T2→T2T1=p2p1=241207= 1.164Usingthisrelationship,thepercentagechangeintemperatureiscalculatedasfollows,%changeinT=T2−T1T1×100 =T2T1−1×100 = [1.164−1]×100 =16.4%┌┐Notethatthispercentagechangeappliestoatemperaturegivenindegreeskelvin.(b)Fromthegivendata:T1= 25◦C=298 K.1.164(298)=347 K=74◦C.Hence,fromtheresultinPart(a),T2=Notethatthhepercentagechangein◦Cismuchhigherthan the percentage change in K.1.43.(a)Fromthegivendata:D= 6 m,T= 20◦C = 293 K,p= 200 kPa,Ru= 8312 J/(kmol K),·andV=πD3/6 = 113.1 m3.Foranidealgas:pV=nRT→(200×103)(113.1)=n(8312)(293)→n= 9.29 kmolSince the relative molecular mass of He is 4.003, the mass of 9.29 kmol of He is (9.29)(4.003)kg=37.2kg.(b)Fromthegivendata:T1= 25◦C = 298 K,p1= 210 kPa + 101 kPa= 311 kPa(assumingatmosphericpressureis101 kPa),V= 0.025 m3,andT2= 50◦C=323 K.Fromtheidealgaslaw,V=nRT1p1=nRT2p2→p2=p1T2T1= 311323298= 337 kPa()()Thiscorrespondstoapressureincreaseof337 kPa311 kPa =26 kPa.−1.44.Fromgivendata:Wrat=1.5N,Wbal=0.5N,ρair=1.17kg/m3,p=100kPa,T= 25◦C =298K. TakeRu=8.314J/mol K·andmolarmass of He,mHe,is4.003g/mol.Tolifttherat:Wrat+Wbal+ρHegVbal''''Weightlifted=ρairgVbal''''buoyantforce→Wrat+Wbal= (ρair−ρHe)gVbal(1)Forheliumintheballoon,RHe=RumHe=8.3144.003= 2.077J/g·K = 2077J/kg·KρHe=pRHeT=100×103(2077)(298)= 0.162 kg/m3Vbal=nRuTpSubstitutingintoEquation (1)gives1.5 + 0.5 = (1.17−0.162)(9.81)n(8.314)(298)100103→n= 8.16mol┌×┐Hence,themassofheliumrequiredis8.16×4.003=32.7g

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Solution Manual for Fluid Mechanics for Engineers, 2017 Edition - Page 15 preview image

1.45.Formthegivendata:z= 10 m,D1=5.0mm,T1= 20◦C,andpatm=101.3kPa.Atthebubblereleaselocation,p1=patm+γz= 101.3 + 9.79(10)= 199.2 kPa,T1= 20◦CAtthesurfaceofthelake,p2=101.3kPaandforanisothermalprocess,p1V1=p2V2→V2=p1p2V1→43πr32=p1p243πr31→r2r1=D2D1=p1p213→D2=(p1p2)13D1=(199.2101.3)13(5) = 6.3 mm()()()Therefore,thediameterofthebubblewhenitreachesthesurfaceofthelakeis6.3mm.1.46.Fromthegivendata:V1= 1.0m3,V2= 0.4m3,andp1=101kPa.Forair,theratioofspecificheatsisgivenbyk=1.40.(a)Underisentropicconditions,thepressureofthecompressedvolume,p2,is gi ven byEquations1.33and1.34asp1vk1=p2vk2→(101)(1.0)1.40=p2(0.4)1.40→p2=364 kPa(b)Underisothermalconditions,thepressureofthecompressedvolume,p2,isg ivenbyEquations1.33and1.34asp1v1=p2v2→(101)(1.0)=p2(0.4)→p2=253 kPa1.xxFromthegivendata:V0= 1 L = 10−3m3,P0=101.3 kPa,pvap= 2.337 kPa,T0= 20◦C =293 K,andR= 8.3142 J/(mol K).Since,underinitialconditionsthegasis50%CO2, the·totalnumberofmoles,nT,ofCO2osgivenbynT=(0.5P0)V0RT0=(0.5×101.3×103)(10−3)(8.3142)(293)= 0.02079 mol(1)Whenfilledwith0.75 Lofwater,thenumberofmolesofCO2inthewater,nw,isgivenbynw=CCO2·Vw=PCO2H·Vw=PCO22.937106·(0.75)→nw= 2.554×10−7PCO2(2)×Inthespaceabovethewater,theidealgaslawisobeyedsuchthatthenumberofmolesofCO2inthegaseousphaseisgivenbyng=PCO2VgRT0=PCO2(0.25×10−3)(8.3142)(293)→ng= 1.026×10−7PCO2(3)SincethetotalamountofCO2doesnotchange,thenEquations1to3canbecombinedtogivenT=nw+ng→0.02079= (2.554+1.026)×10−7PCO2→PCO2= 58078 Pa= 58.1 kPa

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Solution Manual for Fluid Mechanics for Engineers, 2017 Edition - Page 16 preview image

The(partial)pressure,PO,oftheothergasinthespaceabovethewatersurfaceisgivenbyPO= (0.5P0)V0Vg= (0.5×101.3)10.25= 202.6 kPaTherefore,thetotalpressure,P,inthespaceabovetheliquidsurfaceisequaltothesumofthepartialpressuresofthetwogassesplusthevaporpressureofwater(2.337 kPa),whichgivesP=PCO2+PO+pvap= 58.1 + 202.6 + 2.337 =263.0 kPa1.47.Fromthegivendata:T= 20◦C=293 K,p=101.3 Pa,f1= 0.20,andf2= 0.8.ForO2,R1=259.8 J/(kg K),·andforN2,R2=296.7 J/(kg K).·Underthegivenconditions,thedensitiesofO2andN2,representedbyρ1andρ2,aregivenbyρ1=pR1T=101.3×103(259.8)(293)= 1.331 kg/m3,ρ2=pR2T=101.3×103(296.7)(293)= 1.166 kg/m3(a)For each1 m3of air there isf1m3ofO2andf2m3ofN2.Therefore, the partial pressuresofO2andN2,denotedbyp1andp2,aregivenbyp1=f1p= (0.2)(293)=20.3 kPa,p2=f2p= (0.8)(293)=81.0 kPa(b)Thedensity,ρmofthemixtureisgivenbyρm=f1ρ1+f2ρ2= (0.2)(1.331) + (0.8)(1.166)=1.199 kg/m31.48.Fromthegivendata:V1= 2.0m3,V2= 4.0m3,T= 20C=293K,1◦andp1=100kPa.Foroxygen,cp=909J/kg K,·cv=649J/kg K,·andk=cp/cv=909/649=1.40.Thegasconstantforoxygen,R, isgivenbyR=cpcv= 909649 = 260J/kg K−−·(a)Underisentropicconditions,thepressure oftheexpandedvolume,p2,isgivenbyEqua-tions1.33and1.34asp1vk1=p2vk2→(100)(2.0)1.40=p2(4.0)1.40→p2= 37.9 kPaTheinitialdensity,ρ1isgivenbytheidealgaslawasfollows,ρ1=p1RT1=100×103(260)(293)= 1.131 kg/m3TheratioofthefinaldensitytotheinitialdensityisgivenbyEquation1.37asfollows:ρ2ρ1=(p2p1)1k→ρ21.131=(37.9100)11.40→ρ2= 0.566 kg/m3

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