流体力学与传热习题参考解答(英文)

....下载可编辑..1.Waterispumpedataconstantvelocity1m/sfromlargereservoirrestingonthefloortotheopentopofanabsorptiontower.Thepointofdischargeis4meterabovethefloor,andthefrictionlossesfromthereservoirtothetoweramountto30J/kg.Atwhatheightinthereservoirmustthewaterlevelbekeptifthepumpcandeveloponly60J/kg?2222112f1UPUPw=Zg+h(Zg+)22U1=01210W=60j/kgfh30/kg2U=1m/s2(60300.5)/g3m21ZZZ431m2.Thefluid(density1200kg/m3)ispumpedataconstantrate20m3/hfromthelargereservoirtotheevaporator.Thepressureabovethereservoirmaintainsatmospherepressureandthepressureoftheevaporatorkeeps200mmHg(vacuum).Thedistancebetweenthelevelofliquidinthereservoirandtheexitofevaporatoris15meterandfrictionallossinthepipeis120J/kgnotincludingtheexitofevaporator,whatisthepumpeffectiveworkandpowerifthediameterofpipeis60mm?22112212fUUZg+WZg+h22105422200Px1.013x102.67x10N/m76031200Kg/m1U0fh120J/kg22V20U1.97m/sA3600*4006π/*.1Z02Z15422.67x101.97W15x9.81120246.88J/kg12002NWQ246.88x1200x20/3600=1646W....下载可编辑..3.Watercomesoutofthepipe(Φ108x4mm),asshowninFig.Thefrictionlossofthepipelinewhichdoesnotcoverthelossattheexitofpipecanbecalculatedbythefollowingequation:hf=6.5U2whereUisthevelocityinthepipe,finda.watervelocityatsectionA-A'.b.waterflowrate,inm3/h.22112212fUUZg+Zg+h221U0121Z6m2Z02fh6.5U22U6x9.816.5U2U2.9m/s23V=UA=2.94x01x360082m/hπ/.4.Waterpassesthroughthevariablepipe.Thevelocityinthesmallpipeis2.5m/s.TheverticalglasstubesareinsertedrespectivelyatthesectionAandBtomeasurethepressure(seefig.)Ifthefrictionlossbetweentwosectionis15J/kg,whatisthewatercolumndifferencebetweentwoglasstubes?Bytheway,drawtherelativeliquidcolumnheightoftwotubesintheFig.aabbUAUA2bU2.5*(33/47)1.23m/s22aabbabfUUZg+Zg+h22abZ=Z....下载可编辑..22abbafUUh22221.23/22.5/21512.63abPPRg3312.63R=1.29x10m9.8x105.Acentrifugalpumptakesbrine(density1180kg/m3,viscosity1.2cp)fromthebottomofasupplytankanddeliversitintoanothertank.Thelinebetweenthetanksis300mof25mmdiameterpipe(innerdiameter).Theflowrateis2m3/h.Inthisline,therearetwogatevalves,fourelbows(90o)andonereturnbend,whatisthefrictionlossiftheroughnessofpipeis0.025mm?22ffstflocalUUhhh4fkd22l31180kg/m300m,d=0.025ml3-3v2m/h=1.2cp=1.2x10Pa.sk=0.025mmk/d=0.025/25=0.001clrk=0.4k=1k=2x0.07=0.14elrek4x0.753k1.5-2.22uv/A2/(3600x/4x0.025)1.13m/s4udRe2.78x10f0.0632f2h4x0.0063x300/0.025x1.13/2+(0.4+1+2x0.07+4x0.7+1.5)x1.13/2=197.86J/kg6.Theorificemeter(diameteroforifice0.0001m)isinstalledformeasuringtheflowrate.TheindicatingliquidoforificeismercuryifUshapepressuregaugereadingis0.6meterandorificecoefficientcanbetakenas0.61,whatistheflowrateofwater?0oo2gR()uc20o02gx0.6x(136001000)Vus0.61x/4x0.0001x1000835.8x10m/s7.Waterflowsthroughapipewithadiameterdi100mmasshowninfigure.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,....下载可编辑..Hg=13600kg/m3)(1)thevalveopenspartially,forselection1-1’and2-2’,wehave22112212f1-2uugZgZh222212HgHo0g(Rh)39630N/m2212f1-2cuuu0Z=0h4f+k2.13ud22lWecangetZ1fromthevalveclosed21HgHOh=1.5mR=0.6mZgR/h6.66m229.81x6.66u/22.13u39630/100023hu=3.13m/sV3600x/4x0.1x3.1388.5m/h(2)whenthevalveopensfully,forsection1-1’and3-3’,wehave22331113f1-3uugZgZh22311Z0Z6.66mu=022ef1-3cu3.1.5h(4fk)(4x0.00625x+0.5)4.81ud20.01ll229.81x6.66u/24.81uu3.51m/sForsection1-1’and2-2’22112212f1-2uugZgZh22112120Z6.66Z0u0u3.5122f1-2cluh(4fk)(4x0.00625x15/0.10.5)3.51/226.2J/kgd222229.81x6.663.15/226.2N32970m8.Therotameterisinstalledtomeasurethewaterflowrate,asshowninfigure.IfthetotallengthincludingequivalentlengthofpipelineAis10mandthereadingofrotameteris2.72m3/h,whatistheflowrateforpipelineB?(fA=0.0075,fB=0.0045)....下载可编辑..ForparallelpipelinefAfBtotalAB22AfAA2AhhVV+Vu(l+le)2.72h4f4x0.0075x10/0.053/2()d23600x/4x0.0530.333J/kg22BfBBBB23BBBBu(l+le)h4f4x0.0045x2/0.3/2xu0.333d2u2.36m/sV=uA2.36x/4x0.23600m/h10.Aflatfurnacewallisconstructedof120mmlayerofsil-o-celbrick,withathermalconductivity0.08w/(moC),backedbya150mmofcommonbrick,ofconductivity0.8w/(moC),thetemperatureofinnerfaceofthewallis1400o,andthatoftheouterfaceis200oC.a.Whatistheheatlossthroughthewallinwpersquaremeter.b.Toreducetheheatlossto600w/m2byaddingalayerofcorkwithk0.2w/(moC)ontheoutsideofcommonbrick,howmanymetersofcorkarerequied?a.2Qt1400200711N/m11LR0.080.80.120.15b.600=(1400-200)/(0.12/0.08+0.15/0.8+x/0.2)x=0.0625m13.Airatthenormalpressurepassesthroughthepipe(di20mm)andisheatedfrom20oCto100oC.Whatisthefilmheattransfercoefficientbetweentheairandpipewalliftheaveragevelocityofairis10m/s?Thepropertiesofairat60oCareasfollows:density1.06kg/m3,viscosity0.02cp,conductivity0.0289w/(moC),andheatcapacity1....下载可编辑