流体力学与传热习地题目参考解答(英文)

流体力学与传热习地题目参照解答(英文)流体力学与传热习地题目参照解答(英文)10/10流体力学与传热习地题目参照解答(英文)合用标准文档1.Waterispumpedataconstantvelocity1m/sfromlargereservoirrestingonthefloortotheopentopofanabsorptiontower.Thepointofdischargeis4meterabovethefloor,andthefrictionlossesfromthereservoirtothetoweramountto30J/kg.Atwhatheightinthereservoirmustthewaterlevelbekeptifthepumpcandeveloponly60J/kg?w=Z2g+U22hfP2(Z1g+U12P1)22U1=01210W=60j/kghf30/kgU2=1m/s2(60300.5)/g3mZZ2Z1431m2.Thefluid(density1200kg/m3)ispumpedataconstantrate20m3/hfromthelargereservoirtotheevaporator.Thepressureabovethereservoirmaintainsatmospherepressureandthepressureoftheevaporatorkeeps200mmHg(vacuum).Thedistancebetweenthelevelofliquidinthereservoirandtheexitofevaporatoris15meterandfrictionallossinthepipeis120J/kgnotincludingtheexitofevaporator,whatisthepumpeffectiveworkandpowerifthediameterofpipeis60mm?1Z1g+U12W2Z2g+U22hf2210P2200x1.013x1054N/m27601200Kg/m3U10hf120J/kgU2V201.97m/sA3600*π2Z10Z215W415x9.812120246.88J/kg12002Watercomesoutofthepipe(Φ108x4mm),asshowninFig.Thefrictionlossofthepipelinewhichdoesnotcoverthelossattheexitofpipecanbecalculatedbythefollowingequation:优秀文案合用标准文档hf=6.5U2whereUisthevelocityinthepipe,finda.watervelocityatsectionA-A'.b.waterflowrate,inm3/h.1Z1g+U122U22hf2Z2g+2U1012Z16mZ20hf2U222U2.9m/sV=UA=2.9π/4x01.2x360082m3/hWaterpassesthroughthevariablepipe.Thevelocityinthesmallpipeis2.5m/s.TheverticalglasstubesareinsertedrespectivelyatthesectionAandBtomeasurethepressure(seefig.)Ifthefrictionlossbetweentwosectionis15J/kg,whatisthewatercolumndifferencebetweentwoglasstubes?Bytheway,drawtherelativeliquidcolumnheightoftwotubesintheFig.UaAaUbAbUb2.5*(33/47)2aZag+Ua2bZbg+Ub2Za=Zb2hf2abUb2Ua222h22fPaPbRgR=31.29x103m5.Acentrifugalpumptakesbrine(density1180kg/m3,viscosity1.2cp)fromthebottomofasupplytank优秀文案合用标准文档anddeliversitintoanothertank.Thelinebetweenthetanksis300mof25mmdiameterpipe(innerdiameter).Theflowrateis2m3/h.Inthisline,therearetwogatevalves,fourelbows(90o)andonereturnbend,whatisthefrictionlossiftheroughnessofpipeis0.025mm?hfhfsthflocallU2U24f2kd21180kg/m3v2m3/h-3kc=0.4kl=1krkel4x0.753kreuv/A2/(3600x2)Reud2.78x104fhf2/2+2/2Theorificemeter(diameteroforifice0.0001m)isinstalledformeasuringtheflowrate.TheindicatingliquidoforificeismercuryifUshapepressuregaugereadingis0.6meterandorificecoefficientcanbetakenas0.61,whatistheflowrateofwater?2gR(0)uocoV0uos022gx0.6x(136001000)0.61x/4x0.0001x10005.8x108m3/sWaterflowsthroughapipewithadiameterdi100mmasshowninfigure.a.whenthevalveisclosed,Ris600mmandhequals1500mm.Whilethevalveopenspartially,R=400mmandh=1400mm,f=0.00625(Finningfactor)andkc=0.5(contractioncoefficient),whatistheflowrateofwater,inm3/h?b.Ifthevalveopensfully,whatisthepressureofsection2-2',inN/m2?Theequivalentlengthofthevalveis1.5mandtheFanningfactorfkeepsthesame?(H2O=1000kg/m3,Hg=13600kg/m3)优秀文案合用标准文档(1)thevalveopenspartially,forselection1-1’and2-2’,wehavegZ1u121gZ2u222hf1-222102g(HgRH2oh)39630N/m2u10Z2=0hf1-24flu2u2d2+kc2WecangetZ1fromthevalveclosedR=0.6mZ1HggR/H2Oh9.81x6.66u2/2239630/1000Vh3600x23/h(2)whenthevalveopensfully,forsection1-1’and3-3’,wehavegZ1u121gZ3u323hf1-322Z30Z1u1=0hf1-3(4fllekc)u2+0.5)2d2u22u3.51m/sForsection1-1’and2-2’u21u222gZ11gZ2hf1-22210Z16.66Z20u10u22hf1-2(4flkc)u2/226.2J/kgd22/2232970Nm2Therotameterisinstalledtomeasurethewaterflowrate,asshowninfigure.IfthetotallengthincludingequivalentlengthofpipelineAis10mandthereadingofrotameteris2.72m3/h,whatistheflowrateforpipelineB?(fA=0.0075,fB=0.0045)优秀文案合用标准文档ForparallelpipelinehfAhfBVtotalVA+VBhfA4fA(l+le)uA24x0.0075x10/0.053/2()2dA22hfB4fB(l+le)uB2B2dB2uBB=uBAB2600m3/h10.Aflatfurnacewallisconstructedof120mmlayerofsil-o-celbrick,withaw/(moC),backedbya150mmofcommonbrick,ofconductivity0.8w/(moC),thetemperatureofinnerfaceofthewallis1400oo,andthatoftheouterfaceis200C.a.Whatistheheatlossthroughthewallinwpersquaremeter.b.Toreducetheheatlossto600w/m2byaddingalayerofcorkwithk0.2w/(moC)ontheoutsideofcommonbrick,howmanymetersofcorkarerequied?a.Qt1400200711N/m2LR1113.Airatthenormalpressurepassesthroughthepipe(di20mm)andisheatedfrom20oCto100oC.Whatisthefilmheattransfercoefficientbetweentheairandpipewalliftheaveragevelocityofairis10m/s?Thepropertiesofairat60oCareasfollows:density1.06kg/m3,viscosity0.02cp,conductivity0.0289w/(moC),andheatcapacity1kJ/kg-KRe=1.06x1041040.02x103优秀文案合用标准文档T=T1+T22010060℃122cpPr=kPr1/31.06x1041/3hidihi2.kk14.Ahotfluidwithamassflowrate2250kg/hpassesthrougha25x2.5mmtube.Thephysicalpropertiesoffluidareasfollows:k=0.5w/(moC),Cp=4kJ/kg-K,viscosity10-3N-s/m2,density1000kg/m3Find:a.Heattransferfilmcoefficienthi,inw/(m2-K).b.Iftheflowratedecreasesto1125kg/handotherconditionsarethesame,whatistheih?c.Ifthediameteroftube(insidediameter)decreasesto10mm,andthevelocityukeepsthesameasthatofcasea,calculatehi.d.Whentheaveragetemperatureoffluidandquantityofheatflowpermeteroftubeare40oCand400w/m,respectively,whatistheaveragetemperatureofpipewallforcasea?e.Fromthisproblem,inordertoincreasetheheattransferfilmcoefficientandenhanceheattransfer,whatkindsofmethodscanyouuseandwhichisbetter,explain?Hint:forlaminarflow,Nu=1.86[RePr]1/3forturbulentflowNu=0.023RePr1/3udGd4N2250x4d3600x44(1)Re10=.1/3Nu1/3440023RePrhiNuk5500w/m2kd(2)w12w2Re2Re1/2=2x104104Nu2Re2hi2Nu1Re1hi1hi25500x0.53159w/m2k(3)Re3u3d32x104104Nu0.023RePr1/3hi=6347w/m2k(4)Q=hiAi(t-tw)=400=500x2x0.02(t-tw)t=40℃tw℃theremethods:increaseuorhiordecreasedThefirstisbetterInadoublepipeexchange(Φ23x2mm),thecoldfluid(Cp=1kJ/kg,flowrate500kg/h)passesthrough优秀文案合用标准文档thepipeandthehotfluidgoesthroughtheoutside.Theinletandoutlettemperaturesofcoldfluidare20and80o,andtheinletandoutlettemperaturesofhotfluidare150and90o,respectively.Thehi(filmcoefficientinsidepipe)is700w/(m2oC)andoverallheattransfercoefficientUo(basedontheoutsidesurfaceofpipe)is300w/(m2oC),respectively.Iftheheatlossisignoredandtheconductivityofpipewall(steel)istakenas45w/(moC),find:heattransferfilmcoefficientoutsidethepipeho?thepipelengthrequiredforcounterflow,inm?whatisthepipelengthrequirediftheheatingmediumchangestosaturatedvapor(140oC)anditcondensestosaturatedliquidandotherconditionskeepunchanged?Whentheexchangerisusedforayear,itisfoundthatitcannotmeettheneedofproduction(theoutlettemperatureofcoldfluidcannotreach80oC),explainwhy?11d0lmdo123(a)Vohidikdm300700x1945x21h01/h0=1/U0-(do/hidi+bdo/kdm)=1/300-23/700*19-0.002*23/45*21h0642.9w/2mkLMTD=70℃=t1+t2Q=UoAoTm=mcCp(Tcb-Tca)2300*2*0.023*70L=500/3600*1000*(80-20)8020(c)LMTD=℃ln1402014080L1t1L20.81L14.4mL2t2scaleisformedontheoutside,V0isdecreasedWaterflowsturbulentlyinthepipeofΦ25x2.5mmshelltubeexchanger.Whenthevelocityofwateruis1m/s,overallheattransfercoefficientUo(basedontheoutersurfaceareaofpipe)is2115w/(m2oC).Iftheu2obecomes1.5m/sandotherconditionskeepunchanged,Uois2660w/(mC).Whatisthefilmcoefficienthooutsidethepipe?(Heatresistancesofpipewallandscaleareignored)111111(2)Uohi(1)U'oh'ihoho111111(1)-(2)=2660u1Cu20.8C1CC2115111C=2859ho=8127W/(m2K)hoUohi17.Waterandoilpassparallellythroughanexchangerwhichis1mlong.Theinletandoutlettemperaturesofwaterare15and40oC,andthoseofoilare150and100oC,respectively.Iftheoutlettemperatureofoildecreasesto80oC,andtheflowratesandphysicalpropertiesandinlettemperaturesofwaterandoilmaintainthesame,whatisthepipelengthofnewexchanger?(Heatlossandpipewallresistanceareneglected)QWhChT1-T2WcCct2t1VAtmWhChT-T'WCt'1t12cc2优秀文案合用标准文档150100401550℃15080tt2215L2T1T2'tm11508092.585L1T1T2tm21.15010069.8L21.85mL1=1mm1t9m2.25t69.818.Airwhichpassesthroughthepipeinturbulentflowisheatedfrom20to80oC.Thesaturatedvaporat116.3oCcondensestosaturatedwateroutsidethepipe.Ifairflowrateincreasesto120%oftheoriginandinletandoutlettemperaturesofairstayconstant,whatkindofmethodcanyouemployinordertodothat?(Heatresistanceofpipewallandscalecanbeignored)Q1h1iAiTm1mc1Cpc(TcbTca)Qh12AiTm2('cbTca)2mc2CpcThi2Tm2mc2Tm2hi1Tm1mc1Tm1Tm1802080)8020Tm2ln(Th20)/(Th80)19.WaterflowsthroughthepipeofaΦ25x2.5mmshell-tubeexchangerfrom20to50oC.Thehotfluid(Cp1.9kJ/kgoC,flowrate1.25kg/s)goesalongtheshellandthetemperatureschangefrom80to30oC.Filmcoefficientsofwaterandhotfluidare0.85kw/(m2oC)and1.7kw/(m2oC).WhatistheoverallheattransfercoefficientUoandheattransferareaifthescaleresistancecanbeignored?(theconductivityofsteelis45w/(moC).℃Q2WhCpT1T21.25x1.9x80-30119Kwtm=3010V0=1lmd0=472w/m2kAiQ119x1032301d0V0tmlnh0hidikdm1020.Asphericalparticle(density2650kg/m3)settlesfreelyinairat20oC(densityofair1.205kg/m3,viscosity1x10-5Pa.s).CalculatethemaximumdiameterofparticleifthesettleobeystheStokes’Law?Dp2g3182Re≤1Ut18DpDpg18x10-101/3Dp-5优秀文案合用标准文档Afilterpress(A=0.1m2)isusedforfilteringslurry.Thevacuuminsidethefilteris500mmHg.Oneliterfiltratecanbegotafterfilteringof5minand0.6moreliterfiltrateisobtainedafter5moremin.How