UnitOperationsofChemicalEngineering(化工单元操作技巧)

\\1.1Whatwillbethe(a)thegaugepressureand(b)theabsolutepressureofwateratdepth12mbelowthesurface?ρwater=1000kg/m3,andPatmosphere=101kN/m2.Solution:Rearrangingtheequation1.1-4Setthepressureofatmospheretobezero,thenthegaugepressureatdepth12mbelowthesurfaceisAbsolutepressureofwateratdepth12m1.3AdifferentialmanometerasshowninFig.issometimesusedtomeasuresmallpressuredifference.Whenthereadingiszero,thelevelsintworeservoirsareequal.AssumethatfluidBismethane（甲烷）,thatliquidCinthereservoirsiskerosene(specificgravity=0.815),andthatliquidAintheUtubeiswater.TheinsidediametersofthereservoirsandUtubeare51mmand6.5mm,respectively.Ifthereadingofthemanometeris145mm.,whatisthepressuredifferenceovertheinstrumentInmetersofwater,(a)whenthechangeinthelevelinthereservoirsisneglected,(b)whenthechangeinthelevelsinthereservoirsistakenintoaccount?Whatisthepercenterrorintheanswertothepart(a)?Solution：pa=1000kg/m3pc=815kg/m3pb=0.77kg/m3D/d=8R=0.145mWhenthepressuredifferencebetweentworeservoirsisincreased,thevolumetricchangesinthereservoirsandUtubes(1)so(2)andhydrostaticequilibriumgivesfollowingrelationship(3)so(4)substitutingtheequation(2)forxintoequation(4)gives(5)（a）whenthechangeinthelevelinthereservoirsisneglected,（b）whenthechangeinthelevelsinthereservoirsistakenintoaccounterror=1.4TherearetwoU-tubemanometersfixedonthefluidbedreactor,asshowninthefigure.ThereadingsoftwoU-tubemanometersareR1=400mm，R2=50mm,respectively.Theindicatingliquidismercury.Thetopofthemanometerisfilledwiththewatertopreventfromthemercuryvapordiffusingintotheair,andtheheightR3=50mm.TrytocalculatethepressureatpointAandB.Figureforproblem1.4Solution:ThereisagaseousmixtureintheU-tubemanometermeter.Thedensitiesoffluidsaredenotedby,respectively.ThepressureatpointAisgivenbyhydrostaticequilibriumissmallandnegligibleincomparisonwithandρH2O,equationabovecanbesimplified==1000×9.81×0.05+13600×9.81×0.05=7161N/m²=7161+13600×9.81×0.4=60527N/mDdpapaHhAFigureforproblem1.51.5Waterdischargesfromthereservoirthroughthedrainpipe,whichthethroatdiameterisd.TheratioofDtodequals1.25.TheverticaldistancehbetweenthetankAandaxisofthedrainpipeis2m.WhatheightHfromthecenterlineofthedrainpipetothewaterlevelinreservoirisrequiredfordrawingthewaterfromthetankAtothethroatofthepipe?Assumethatfluidflowisapotentialflow.Thereservoir,tankAandtheexitofdrainpipeareallopentoair.Solution:Bernoulliequationiswrittenbetweenstations1-1and2-2,withstation2-2beingreferenceplane:Wherep1=0,p2=0,andu1=0,simplificationoftheequation1Therelationshipbetweenthevelocityatoutletandvelocityuoatthroatcanbederivedbythecontinuityequation:2Bernoulliequationiswrittenbetweenthethroatandthestation2-23Combiningequation1,2,and3givesSolvingforHH=1.39m1.6Aliquidwithaconstantdensityρkg/m3isflowingatanunknownvelocityV1m/sthroughahorizontalpipeofcross-sectionalareaA1m2atapressurep1N/m2,andthenitpassestoasectionofthepipeinwhichtheareaisreducedgraduallytoA2m2andthepressureisp2.Assumingnofrictionlosses,calculatethevelocitiesV1andV2ifthepressuredifference(p1-p2)ismeasured.Solution:InFig1.6,theflowdiagramisshownwithpressuretapstomeasurep1andp2.Fromthemass-balancecontinuityequation,forconstantρwhereρ1=ρ2=ρ,FortheitemsintheBernoulliequation,forahorizontalpipe,z1=z2=0ThenHYPERLINK"javascript:moveTo('ch02equ101');"Bernoulliequationbecomes,aftersubstitutingforV2,Rearranging,PerformingthesamederivationbutintermsofV2,1.7AliquidwhosecoefficientofviscosityisµflowsbelowthecriticalvelocityforlaminarflowinacircularpipeofdiameterdandwithmeanvelocityV.Showthatthepressurelossinalengthofpipeis.Oilofviscosity0.05Pasflowsthroughapipeofdiameter0.1mwithaaveragevelocityof0.6m/s.Calculatethelossofpressureinalengthof120m.Solution:TheaveragevelocityVforacrosssectionisfoundbysummingupallthevelocitiesoverthecrosssectionanddividingbythecross-sectionalarea1Fromvelocityprofileequationforlaminarflow2substitutingequation2foruintoequation1andintegrating3rearrangingequation3givesFigureforproblem1.81.8.Inaverticalpipecarryingwater,pressuregaugesareinsertedatpointsAandBwherethepipediametersare0.15mand0.075mrespectively.ThepointBis2.5mbelowAandwhentheflowratedownthepipeis0.02m3/s,thepressureatBis14715N/m2greaterthanthatatA.AssumingthelossesinthepipebetweenAandBcanbeexpressedaswhereVisthevelocityatA,findthevalueofk.IfthegaugesatAandBarereplacedbytubesfilledwithwaterandconnectedtoaU-tubecontainingmercuryofrelativedensity13.6,giveasketchshowinghowthelevelsinthetwolimbsoftheU-tubedifferandcalculatethevalueofthisdifferenceinmetres.Solution:dA=0.15m;dB=0.075mzA-zB=l=2.5mQ=0.02m3/s,pB-pA=14715N/m2Whenthefluidflowsdown,writingmechanicalbalanceequation0.295makingthestaticequilibriumFigureforproblem1.91.9．Theliquidverticallyflowsdownthroughthetubefromthestationatothestationb,thenhorizontallythroughthetubefromthestationctothestationd,asshowninfigure.Twosegmentsofthetube,bothabandcd，havethesamelength,thediameterandroughness.Find:（1）theexpressionsof,hfab,andhfcd,respectively.（2）therelationshipbetweenreadingsR1andR2intheUtube.Solution:(1)FromFanningequationandsoFluidflowsfromstationatostationb,mechanicalenergyconservationgiveshence2fromstationctostationdhence3Fromstaticequationpa-pb=R1（ρˊ-ρ）g-lρg4pc-pd=R2（ρˊ-ρ）g5Substitutingequation4inequation2，thentherefore6Substitutingequation5inequation3，then7ThusR1=R21.10Waterpassesthroughapipeofdiameterdi=0.004mwiththeaveragevelocity0.4m/s,asshowninFigure.1)Whatisthepressuredrop–PwhenwaterflowsthroughthepipelengthL=2m,inmH2Ocolumn?LrFigureforproblem1.102)Findthemaximumvelocityandpointratwhichitoccurs.3)Findthepointratwhichtheaveragevelocityequalsthelocalvelocity.4）ifkeroseneflowsthroughthispipe，howdothevariablesabovechange？（theviscosityanddensityofWaterare0.001Pasand1000kg/m3，respectively；andtheviscosityanddensityofkeroseneare0.003Pasand800kg/m3，respectively）solution:1）fromHagen-Poiseuilleequation2）maximumvelocityoccursatthecenterofpipe,fromequation1.4-19soumax=0.4×2=0.8m3）whenu=V=0.4m/sEq.1.4-174)kerosene:1.12Asshowninthefigure,thewaterlevelinthereservoirkeepsconstant.Asteeldrainpipe(withtheinsidediameterof100mm)isconnectedtothebottomofthereservoir.OnearmoftheU-tubemanometerisconnectedtothedrainpipeattheposition15mawayfromthebottomofthereservoir,andtheotherisopenedtotheair,theUtubeisfilledwithmercuryandtheleft-sidearmoftheUtubeabovethemercuryisfilledwithwater.Thedistancebetweentheupstreamtapandtheoutletofthepipelineis20m.Whenthegatevalveisclosed,R=600mm,h=1500mm;whenthegatevalveisopenedpartly,R=400mm,h=1400mm.Thefrictioncoefficientλis0.025,andthelosscoefficientoftheentranceis0.5.Calculatetheflowrateofwaterwhenthegatevalveisopenedpartly.(inm³/h)Whenthegatevalveiswidelyopen,calculatethestaticpressureatthetap(ingaugepressure,N/m²).le/d≈15whenthegatevalveiswidelyopen,andthefrictioncoefficientλisstill0.025.Figureforproblem1.12Solution：(1)Whenthegatevalveisopenedpartially,thewaterdischargeisSetupBernoulliequationbetweenthesurfaceofreservoir1—1’andthesectionofpressurepoint2—2’，andtakethecenterofsection2—2’asthereferringplane,then（a）Intheequation(thegaugepressure)Whenthegatevalveisfullyclosed,theheightofwaterlevelinthereservoircanberelatedtoh(thedistancebetweenthecenterofpipeandthemeniscusofleftarmofUtube).（b）whereh=1.5mR=0.6mSubstitutetheknownvariablesintoequationbSubstitutetheknownvariablesequationa9.81×6.66=thevelocityisV=3.13m/stheflowrateofwateris2）thepressureofthepointwherepressureismeasuredwhenthegatevalveiswide-open.Writemechanicalenergybalanceequationbetweenthestations1—1’and3-3´，then（c）sinceinputtheabovedataintoequationc，9.81thevelocityis:V=3.51m/sWritemechanicalenergybalanceequationbetweenthestations1—1’and2——2’,forthesamesituationofwaterlevel（d）sinceinputtheabovedataintoequationd，9.81×6.66=thepressureis:1.14Waterat20℃passesthroughasteelpipewithaninsidediameterof300mmand2mlong.Thereisaattached-pipe(Φ603.5mm)whichisparallelwiththemainpipe.Thetotallengthincludingtheequivalentlengthofallformlossesoftheattached-pipeis10m.Arotameterisinstalledinthebranchpipe.Whenthereadingoftherotameteris2.72m3/h,trytocalculatetheflowrateinthemainpipeandthetotalflowrate,respectively.Thefrictionalcoefficientofthemainpipeandtheattached-pipeis0.018and0.03,respectively.Solution：Thevariablesofmainpipearedenotedbyasubscript1,andbranchpipebysubscript2.ThefrictionlossforparallelpipelinesisTheenergylossinthebranchpipeisIntheequationinputthedataintoequationcTheenergylossinthemainpipeisSoThewaterdischargeofmainpipeisTotalwaterdischargeis1.16AVenturimeterisusedformeasuringflowofwateralongapipe.ThediameteroftheVenturithroatistwofifthsthediameterofthepipe.TheinletandthroatareconnectedbywaterfilledtubestoamercuryU-tubemanometer.Thevelocityofflowalongthepipeisfoundtobem/s,whereRisthemanometerreadinginmetresofmercury.DeterminethelossofheadbetweeninletandthroatoftheVenturiwhenRis0.49m.(Relativedensityofmercuryis13.6).Figureforproblem1.16Solution:Writingmechanicalenergybalanceequationbetweentheinlet1andthroatoforVenturimeter1rearrangingtheequationabove,andset(z2-z1)=x2fromcontinuityequation3substitutingequation3forVointoequation2gives4fromthehydrostaticequilibriumformanometer5substitutingequation5forpressuredifferenceintoequation4obtains6rearrangingequation61.17.Sulphuricacidofspecificgravity1.3isflowingthroughapipeof50mminternaldiameter.Athin-lippedorifice,10mm,isfittedinthepipeandthedifferentialpressureshownbyamercurymanometeris10cm.Assumingthattheleadstothemanometerarefilledwiththeacid,calculate(a)theweightofacidflowingpersecond,and(b)theapproximatefrictionlossinpressurecausedbytheorifice.Thecoefficientoftheorificemaybetakenas0.61,thespecificgravityofmercuryas13.6,andthedensityofwateras1000kg/m3Solution:a)approximatepressuredrop12066.3Papressuredifferenceduetoincreaseofvelocityinpassingthroughtheorificepressuredropcausedbyfrictionloss2.1Waterisusedtotestfortheperformancesofpump.Thegaugepressureatthedischargeconnectionis152kPaandthereadingofvacuumgaugeatthesuctionconnectionofthepumpis24.7kPaastheflowrateis26m3/h.Theshaftpoweris2.45kwwhilethecentrifugalpumpoperatesatthespeedof2900r/min.Iftheverticaldistancebetweenthesuctionconnectionanddischargeconnectionis0.4m,thediametersofboththesuctionanddischargelinearethesame.Calculatethemechanicalefficiencyofpumpandlisttheperformanceofthepumpunderthisoperatingcondition.Solution:WritethemechanicalenergybalanceequationbetweenthesuctionconnectionanddischargeconnectionwheretotalheadsofpumpisefficiencyofpumpissinceN=2.45kWThenmechanicalefficiencyTheperformanceofpumpisFlowrate，m³/h26Totalheads，m18.41Shaftpower，kW2.45Efficiency，%53.12.2Wateristransportedbyapumpfromreactor,whichhas200mmHgvacuum,tothetank,inwhichthegaugepressureis0.5kgf/cm2,asshowninFig.Thetotalequivalentlengthofpipeis200mincludingalllocalfrictionalloss.Thepipelineis57×3.5mm,theorificecoefficientofCoandorificediameterdoare0.62and25mm,respectively.Frictionalcoefficientis0.025.Calculate:DevelopedheadHofpump,inm(thereadingRofUpressuregaugeinorificemeteris168mmHg)Solution:Equation(1.6-9)Massflowrate2)Fluidflowthroughthepipefromthereactortotank,theBernoulliequationisasfollowsforV1=V2z=10mp/g=7.7mTherelationbetweentheholevelocityandvelocityofpipeFrictionlosssoH=7.7+10+5.1=22.8mHg2.3.Acentrifugalpumpistobeusedtoextractwaterfromacondenserinwhichthevacuumis640mmofmercury,asshowninfigure.Attherateddischarge,thenetpositivesuctionheadmustbeatleast3mabovethecavitationvaporpressureof710mmmercuryvacuum.Iflossesinthesuctionpipeaccountedforaheadof1.5m.Whatmustbetheleastheightoftheliquidlevelinthecondenserabovethepumpinlet?Solution:Fromanenergybalance,WherePo=760-640=120mmHgPv=760-710=50mmHgUseoftheequationwillgivetheminimumheightHgas2.4Sulphuricacidispumpedat3kg/sthrougha60mlengthofsmooth25mmpipe.Calculatethedropinpressure.Ifthepressuredropfallsbyonehalf,whatwillthenewflowratebe?Densityofacid1840kg/m3Viscosityofacid25×10-3PasSolution:Velocityofacidinthepipe:Reynoldsnumber:fromFig.1.22forasmoothpipewhenRe=6109,f=0.0085pressuredropiscalculatedfromequation1.4-9orfrictionfactoriscalculatedfromequation1.4-25ifthepressuredropfallsto783.84/2=391.92kPasonewmassflowrate=0.785d2uρ=0.785×0.0252×2.27×1840=2.05kg/s2.4Sulphuricacidispumpedat3kg/sthrougha60mlengthofsmooth25mmpipe.Calculatethedropinpressure.Ifthepressuredropfallsbyonehalfonassumptionthatthechangeoffrictionfactorisnegligible,whatwillthenewflowratebe?Densityofacid1840kg/m3Viscosityofacid25×10-3PaFrictionfactorforhydraulicallysmoothpipeSolution:Writeenergybalanceequation:Δp=46.92×1840×9.81=847.0kpa2.6ThefluidispumpedthroughthehorizontalpipefromsectionAtoBwiththeφ382.5mmdiameterandlengthof30meters,shownasfigure.Theorificemeterof16.4mmdiameterisusedtomeasuretheflowrate.OrificecoefficientCo＝0.63.thepermanentlossinpressureis3.5×104N/m2,thefrictioncoefficientλ=0.024.find:(1)WhatisthepressuredropalongthepipeAB?(2)WhatistheratioofpowerobliteratedinpipeABtototalpowersuppliedtothefluidwhentheshaftworkis500W,60%efficiency?(Thedensityoffluidis870kg/m3)solution：∴u=(16.4/33)2×8.5=2.1m/s∴(2)sotheratioofpowerobliteratedinfrictionlossesinABtototalpowersuppliedtothefluid3.2Asphericalquartzoseparticle（颗粒）withadensityof2650kg/m³settles（沉淀）freelyinthe20℃air,trytocalculatethemaximumdiameterobeyingStocks’lawandtheminimumdiameterobeyingNewton’slaw.Solution:ThegravitysettlingisfollowedStocks’law,somaximumdiameterofparticlesettledcanbecalculatedfromRethatissetto1,thenequation3.2-16fortheterminalvelocitysolvingforcriticaldiameterCheckuptheappendixThedensityof20℃airρ=1.205kg/m³andviscosityµ=1.81×10-3N·s/m2whenReynoldsnumber≥1000,theflowpatternfollowsNewton’slawandterminalvelocitycanbecalculatedbyequation3.2-191criticalReynoldsnumberis，2rearrangingtheequation2gives3combinationofequation1withequation3solvingforcriticaldiameter3.3Itisdesiredtoremovedustparticles50micronsindiameterfrom226.5m3/minofair,usingasettlingchamberforthepurpose.Thetemperatureandpressureare21oCand1atm.Theparticledensityis2403kg/m3.Whatminimumdimensionsofthechamberareconsistentwiththeseconditions?(themaximumpermissiblevelocityoftheairis3m/s)solution:tocalculateterminalvelocityfromtheequation3.2-16Thedensityof21℃airρ=1.205kg/m³andviscosityµ=1.81×10-5N·s/m2so1fromequation3.3-4themaximumpermissiblevelocityoftheairis3m/s2setBtobe3m,thenfromequation1L=7mAndH=0.42m3.4Astandardcycloneistobeusedtoseparatethedustofdensityof2300kg/m³fromthegas.Theflowrateofgasis1000m³/h,theviscosityofthegasis3.610-5N·s/m²,andthedensityis0.674kg/m3.Ifthediameterofcycloneis400mm,attempttoestimatethecriticaldiameter.Solution:D=0.4mB=D/4=0.1mh=D/2=0.2mAccordingtotheequation3.3-12forN=5：3.6Afilterpressof0.1m2filteringareaisusedforfilteringasampleoftheslurry.Thefiltrationiscarriedoutatconstantpressurewithavacuum500mmHg.Thevolumeoffiltratecollectedinthefirst5minwasoneliterand,afterafurther5min,anadditional0.6literwascollected.Howmuchfiltratewillbeobtainedwhenthefiltrationhasbeencarriedoutfor15minonassumingthecaketobeincompressible?Solution:Theequationfortheconstant-pressurefiltration5min10minsolvingtheequationsaboveforVmandKandK=48ForSolvingforV=2.073lThefollowingdataareobtainedforafilterpressof0.0093m2filteringareainthetestPressuredifference/kgf/cm2filteringtime/sfiltrate/m31.05502.27×10-36609.10×10-33.5017.12.27×10-32339.10×10-3Calculate:(1)filtrationconstantK,Vmatthepressuredifferenceof1.05(2)iftheframeofthefilterisfilledwiththecakeat660s,whatisthefinalrateoffiltration(3)andwhatisthecompressibleconstantofcaken?solution:①fromequation3.4-19aForpressuredifference㎏/㎝212solvingtheequations1and2gives②=Forpressuredifference㎏/㎝234solvingtheequations3and4givesthensolvingforn=0.1423.8Aslurryiffilteredbyafilterpressof0.1m2filteringareaatconstantpressure,theequationforaconstantpressurefiltrationisasfollowswhereq=filtratevolumeperunitfilteringarea,inl/m2,t=filteringtime,inmincalculate:(1)howmuchfiltratewillbegottenafter249.6min?(2)Ifthepressuredifferenceisdoubleandboththeresistancesofthefiltrationmediumandcakeareconstant,howmuchfiltratewillbeobtainedafter249.6min?solution:(1)q=240l/m2V=24l(2)thepressuredifferenceisdoublesoK´=500V=34.25l3.9Filtrationiscarriedoutinaplateandframefilterpress,with20frames0.3msquareand50mmthick.Ataconstantpressuredifferenceof248.7kN/m2,one-quarterofthetotalfiltratepercycleisobtainedforthefirst300s.Filtrationiscontinuedataconstantpressureforafurther1800s,afterwhichtheframesarefull.Thetotalvolumeoffiltratepercycleis0.7m3anddismantlingandrefittingofthepresstakes500sItisdecidedtousearotarydrumfilter,1.5mlongand2.2mindiameter,inplaceofthefilterpress.Assumingthattheresistanceoftheclothisthesameandthatthefiltercakeisincompressible,calculatethespeedofrotationofthedrumwhichwillresultinthesameoverallrateoffiltrationaswasobtainedwiththefilterpress.Thefiltrationintherotaryfilteriscarriedoutataconstantpressuredifferenceof70kN/m2andwith25%ofthedrumsubmergedSolution:Areaoffiltration:A=2×0.32×20=3.6m2Δp=248.7kN/m2t1=300s,V1=1/4×0.7=0.175m3t2-t1=1800s,ΔV=0.7-0.175=0.525m3ort2=2100s,V2=0.7m3Vm=0.2627m3AndcapacityfortherotarydrumfilterAreaoffiltration:A=πdL=3.14×2.2×1.5=10.362m2Operatingpressure:Δp=70kN/m2Andthedrumsubmerged:φ=25%Forrotarydrumfilterthefiltrationcoefficient:keepVmunchanged,KchangeswithchanginginpressuredifferencethecapacityofrotarydrumfilterisequaltothatofpressfilterSolvingfornfromtheequationabovebytrialanderrorn=0.0088=0.048rpm(转/min)3.10Apressfilterof0.093m2filteringsurfacewasusedtoseparatethesuspensioncontainingcalciumcarbonate,operatinginconstantpressure.Thevolumeoffiltratecollectedwas2.27×10-3m3duringthe50s,volumeoffiltratecollectedwas3.35×10-3m3duringthe100s.Howmuchwillthefiltratebeobtainedasfilteringfor200s?solution：filteringareaA=0.093m2;filtrateV=2.27×10-3m3fort=50sandV=3.35×10-3m3fort=100sforconstantpressurefiltrationsubstitutingvariablesgivenaboveintotheequationgives1and2combiningequations1and2Vm=3.85×10-4m3andK=1.567×10-5m2/sSubstitutionofthefiltering-constantsVmandKintotheequationforconstant-pressureequationgivessolvingforvolumeoffiltrategainedfor200sV=4.835×10-3m33.11Aslurrywithincompressiblecakeisfilteredbya1m2filterpressatconstantpressure.IfthefilteringconstantKis10m2/min,operatingpressuredifferencepis2atmandtheseptumresistancecanbeignored.Findthefiltratevolume,inm3whenthefilteringtimeis10min.iftheseptumresistancemustbeconsideredandVmis1m3,whatisthefiltratevolumeV,inm3whenthefilteringtimeis10min?(3)whatisthefiltrationratedV/dtwhenthefilteringtimeis10minforcase(1)?Solution:Equationfortheconstant-pressurefiltrationfiltermediumresistancesettobenegligible②Thefilteringmediumresistanceistakenintoaccount③theequationforconstant-pressurefilteringistakenderivativeasfilteringmediumresistancedoesn’tbetakenintoaccount4.2.Aflatfurnacewallisconstructedofa114-mmlayerofSil-o-celbrick,withathermalconductivityof0.138W/(m·°C)backedbya229-mmlayerofcommonbrick,ofconductivity1.38W/(m·°C).Thetemperatureoftheinnerfaceofthewallis760°C,andthatoftheouterfaceis76.6°C.(a)Whatistheheatlossthroughthewall?(b)Whatisthetemperatureoftheinterfacebetweentherefractorybrickandthecommonbrick?(c)Supposingthatthecontactbetweenthetwobricklayersispoorandthata“contactresistance”of0.088°C·m2/Wispresent,whatwouldbetheheatloss?Solution:Fromequation4.2-11(a)heatloss:(b)thetemperatureatinterface:t=191℃(c)contactresistanceis0.088°C·m2/W,heatloss:4.4.Thevaporpipe(do=426mm)iscoveredbya426-mminsulatinglayer(k=0.615W/moC).Ifthetemperatureofoutersurfaceofpipeis177oCandthetemperatureoutsidetheinsulatinglayeris38oC,whataretheheatlosspermeterpipeandthetemperatureprofilewithintheinsulatinglayer?Solution：Similartocompoundresistancesinseriesthroughtheflatwall,thetotalresistanceacrossinsulatinglayerisasfollowsHeatlosspermetertubethroughthewallisandtemperaturedistributionis4.5.Theouterdiameterofasteeltubeis150mm.Thetubewallisbackedbytwoinsulatinglayerstoreducetheheatloss.Theratioofthermalconductivityoftwoinsulatingmaterialsisandbothinsulatingmaterialshavethesamethickness30mm.Ifthetemperaturedifferencebetweenpipewallandoutersurfaceofinsulatingmaterialisconstant,whichinsulatingmaterialshouldbepackedinsidetoenablelessheatloss?solution：k1（lowerthermalconductivity）theinsulatingmaterialwithalowerthermalconductivityplacedoutside,rateofheatlossand,thicknessoftwolayersareequalb1=b2=b,andcontrarily,placethematerialwithlowerthermalconductivityinside,heatlossforsowithlessheatlossobtainedast