化工原理习题答案英文

ProblemsandSolutionsDistillation1、Acontinuousfractionatingcolumnisusedtoseparate4000kg/hofamixtureof30percentCSand70percentCCl.Bottomproductcontain5percentCSatleast,andtherateofrecoveryofCSintheoverheadproductis88%byweight,required.Calculate(a)themolesflowofoverheadproductperhour.(b)themolefractionsofCSandCClintheoverheadproduct,respectivelySolution:Formoverallmaterialbalance Knownbythejusticeoftheproblem Taketheplaceof3typesandenter2types Theunitconverts： （molefraction） 2、Aliquidcontaining40molepercentmethanoland60molepercentwateristobeseparatedinacontinuousfractionalcolumnat1atmpressure.Calculatethevalueofqunderthethreefollowingconditions(a)thefeedingisliquidat40C(b)thefeedingissaturatedliquid.Theequilibriumdataformethanol-waterliquidat1atmpressurearegivenintheattachedtable.Ifthecolumnisfedwith100koml/h.Themolarfractionsofmethanolinoverheadproductandbottomproductare0.95and0.04,respectively.Arefluxratioatthetopofcolumnis2.5.Calculate(a)themoleflowofoverheadproductperhour(b)themoleflowofliquidinrectifyingcolumn(c)themoleflowofvaporinstrippingcolumn.Assumethattheconstantmolarflowappliestothissystem.Solution:Formoverallmaterialbalance Solvetheequtionwecanhave： And Theseuppervaluesarefixedunderthethreefeedconditions。 Thefeedat40℃，q=1.07 Feedsaturatedliquid，q=1 3、Thefeedingatdewpointisfedtoacontinuousfractionatingcolumn,andtheequilibriumrelationsaregivenbyequations:y=0.723x+0.263(inrectifyingcolumn);y=1.25x–0.0187(instrippingcolumn).Calculate(a)thecompositionsoffeeding,overheadproductandbottomproduct,respectively(b)therefluxratioSolution:Theslopeoftheoperatinglineinrectifyingcolumnis Wecanget：R=2.61 Theinterceptoftherectifyinglineonyaxisis Theintersectionofthestrippinglineandthediagonalis Y=X=XwsoXw=0.0748 Formtheintersectionofthesetwooperatinglines,gives: Becausethefeedatdewpoint,sothefeedlineishorizon,andthecompositionoffeedis 6.12Itisdesiredtoproduceanoverheadproductcontaining95.7molepercentAformaidealmixtureof44moleAand50molepercentBfeedingintoacontinuousfractionatingcolumn.Theaveragerelativevolatilityequalsto2.5,andaminimumrefluxratiois1.63.Explaintheheatstateoffeedingandcalculatethevalueofq.Solution:：Formequilibriumequation: Formoperatingequation: Substitutingineq.1),gives: X=0.365Y=0.59 Formdefiniensoftheminimumrefluxratio，theintersectionofthetwoupperequationsisalsothatoftheequilibriumcurveandthefeedline. sothefeedisamixtureofliquidandvapor. Formthefeedequation(feedline),gives: Solveandhave:4、Aliquidofbenzeneandtolueneisfedtocontinuousdistillationinaplatecolumn.Underthetotalrefluxratiocondition,thecompositionsofliquidontheclosetogetherplatesare0.28,0.41,and0.57,respectively.CalculatetheMurphreeplateefficiencyofrelativelytwolowlayers.Theequilibriumdataforbenzene—tolueneliquidattheoperatingconditionaregivenas:x0.260.380.51y0.450.600.72Solution:Underthetotalrefluxratiocondition FormMurphreeefficiency:Checkedby5、Acontinuousfractionatingcolumnisusedtoseparate2500kg/hofamixtureof25percentacetoneand75percentwater.Overheadproductcontaining99percentacetone,required.Thesepercentagesarebyweight,and80molepercentofacetoneinfeedingisenteringtotheoverheadproduct.Thefeedingat20c,andtheoperaterefluxratiois2.7timesasmuchastheminimum.therearetwocondensersatthetopofcolumn.Partialcondenserisusedtocontroltherefluxflowtobesaturatedliquid.Thesteamnotcondensedenterthefinalcondensertobecondensedandcooled,regardedastheproducts.Iftheoverallefficiencyis60%,thencalculatethenumberofidealplates.Solution:Thecompositionsofeachflowrate Knownbythemeaningofthequestion：CalculatethenumberofactualplatesFromtheattachedfigure1，atthepoint，thecorrespondingbubblepointis67℃，sotheaverageofthefeedtemperatureandthebubblepointis1/2（67+20）=43.5℃。Thespecificheatsofeachcomponentat43.5℃are：thelatentheatsofeachcomponentat67℃are:theslopeofthefeedline=Fromfig.2,theminimumrefluxratiomustcomputedfromtheslopeofoperatinglineagthatistangenttotheequilibriumcurve,andmeasuretheslopeis0.475.Solveandhave： theinterceptoftherectifyinglineonyaxis theinterceptofthestrippinglineonyaxis theoperatinglineoftheattachedfigure2isdrawnaccordingtotheslope0.281，andthisoperatinglineoftheattachedfigure3isdrawaccordingtotheintercept0.709. 7idealplatesareneeded(exceptreboilerandpartialcondenser)hence,thenumberofactualplates6、TheequilibriumdataforthesystemCS-CClatthe1atmpressurearegiveninExample1-11.Acontinuousfractionatingcolumnisusedtoseparateafeedcontains30molepercentCSand70molepercentCCl.Overheadproductcontaining95molepercentCSandabottomproductcontaining2.5molepercentCS,required.Thefeedissaturatedliquidandthemassflowis400kg/h.Arefluxratioequalsto1.7timesastheminimum.Thedistillationtakesplaceat61Cand1atm.thesuperficialvelocitybasedonemptytoweris0.8m/sandthedistancebetweentwoboardsis0.4m.Theoverallefficiencyis50%.Calculate(a)Thenumberofactualplatesneeded.(b)themassflowoftwoproductsc)thecross-sectionaldiameteroftower(d)theeffectiveheightoftower.Solution:(a)calculatethenumberofactualplates Paintedx-ypicturefortheequilibriumdatabythequestion,asfigureshows。 Sincethefeedissaturatedliquid, ，fromfig.X-Y Theinterceptofoperatinglineinrectifyingcolumn= CalculatethenumberofidealplateswithThegraphicmethod,andthestepsisomitted.Theresultindicates,besidesreboiler,11idealplatesareneededandfeedshouldbeintroducedontheseventhplatefromthetop Actualplates1）ThemassflowrateofproductTheaveragemolecularweightoffeedisM=0.316+0.1154=130.6kg/kmolFormoverallmaterialbalanceD+W=30.60.95D+0.025W=0.330.6 Fromthetwoupperequations,gives: W=21.5mol/hand 2）Thecross-sectionaldiameteroftowerSincethefeedissaturatedliquid,hence:AmongthemAssumethattherisingvaporisidealgas，hence3）ThevalidityheightoftowerisAbsorption7-1Thevaporofmethanolmixedwithairisabsorbedinwater,thetemperatureis27candthepressureis101.3kpa.Themolardensityofmethanolinbulkofliquidandgasphaseareveryweak.Henry’slawappliestothissystem,H=1.995kmol/m*kpa,,Calculate(1)(2)ThepercentofgasresistanceinthewholeresistanceWhereH=solubilitycoefficient=Individualmass-transfercoefficientforliquidphase=Individualmass-transfercoefficientforgasphase=Overallmass-transfercoefficientbasedongasphaseSolution:1) 2) 7-2AcountercurrentflowtowerisusedtoabsorbHSfromtheair-HSsteamfedtoit,usingpurewaterastheabsorbingliquid,(PurewaterisusedinthecountercurrentflowtowertoabsorbH2Sfromtheair-H2Smixedgas),thetowerisoperatingat25cand101.3kpa.ThedensityofHSischanged(reduced)from2%into1%(involume).Henry’slawappliestothissystem,andHenryconstantE=5.52*10kpa.iftheamountoftheabsorbentis1.2timesasmuchasthesmallestamountaccordingtotheory,(1)calculatetheliquid-gasratioL/Vand(outletliquid)concentrationsX1(2)Repeatcalculatetheliquid-gasratioL/Vand(outletliquid)concentrations,iftheoperatingpressureis101.3kpaSolution:1)FromEq(7-6),theslopeofequilibriumlinemis: And: FormEq(7-54),calculatethelimitingliquid—gasratio Sooperatingliquid-gasratiois theterminalconcentrationX1 2)TheslopeofequilibriumlineSo TheterminalconcentrationX1 7-3Thebutanemixedwiththeairisabsorbedinasieve-platetowercontainingeightidealplates.Theabsorbingliquidisanon-volatilizationoilhavingamolecularweightof250andadensityof900kg/m.Theabsorptiontakesplaceat101.3kpaand15c.5percentofbutaneintheenteringgas.Thebutaneistoberecoveredtotheextentof95%,thevaporpressureofbutaneat15Cis194.5kpa,andliquidbutanehasadensityof580kg/m.AssumethatRaoult’sandDalton’slawsapply.(1)calculatethecubicmetersoffreshabsorbingoilpercubicmeterofbutanerecovered.(Caculatetheamoutofabsorbingoilisrequired(involume)whenpercubicmeterofbutaneisrecovered)(2)Repeatcaculate(1),ontheassumptionthattheoperatingpressureis3034.4kpaandthatallotherfactorsremainconstant.(conditionsreainthesame)Solution:1）calculatethemassflowrateofwater 2）Calculateidealplates(a)SchematicallyadPlottheoperatinglineBEandtheequilibriumlineOE(Y=26.7X)intheX-Yright-anglecoordinate.DrawtherectangularstepsbetweentheoperatinglineandtheequilibriumlinefromthepointB,andwecangetNSo:idealplates(b)UsingKremser.A.mapTheabsorptioncoefficientis95%，andX2=0，sotheabsorptioncoefficientofrelativity，underthelimitingamountofwateraccordingtotheorycondition，。Lookingintofig7-21accordingtothesedata,wecanget:TheabsorptionfactorinthestateofoperationLookingintofig7-21（orcalculateaccordingtoequation7-77）,wecanknowwhenA=1.43，,theidealplates。Sotheresultisthesameasthatofschematicallyad3)ThisquestionshouldbeestimatebyKremser.A.mapabsorptioncoefficient，idealplatesFormfig7-21，readingSo: 7-4Anabsorptiontoweristorecover99percentofammoniainanairstream,usingpurewaterastheabsorbingliquid.(Purewaterisusedinanabsorptiontowertoabsorbammoniainanairstream,theabsorptivityis99percent),theheightofthepackedsectionis3m,theabsorptiontakesplaceat101.3kpaand20c.ThemassflowrateofgasVis580kg/(m.h),and6percentofammoniainthegasinvolume.ThemassflowrateofwaterLis770kg/(m.h).Thetoweriscountercurrentoperatedunderisothermaltemperature,theequilibriumequationY=0.9X,ka,buthasnothingtodowithL,trytomakeouthowtheheightofthepackedsectiontochangeinordertokeeptheabsorptioncoefficientunchangedwhentheconditionsofoperationhavebechangedasfollowing(1)theoperatingpressureis2timesasmuchastheoriginal.(2)themassflowrateofwaterisonetimemorethantheoriginal3)themassflowrateofgasistwotimesasmuchastheoriginalSolution: TheaveragemolecularweightofthemixedgasM=29×0.94+17×0.06=28.28 1） FormEq(7-6),So FormEq(7-62)andEq(7-43a)So:ischangedwiththeoperatingpressureSo Sotheheightofthepackedsectionreduce1.802magainsttheoriginal2）whenthemassflowrateofgasincreasing,hasnotremarkableeffecttheheightofthepackedsectionreduce0.605magainsttheoriginal3） whenmassflowrateofgasincreasing,correspondinggrow，accordingtoPoblem: Sotheheightofthepackedsectionincrease4.92magainsttheoriginalDrying11-1Thetotalpressureandhumidityofthehumidairare50kPaand60C,respectively,andtherelativehumidityis40percent.Calculate(a)thepartialpressureofaqueousvaporinthehumidair(b)humidity(c)thedensityofhumidair.Solution:a）Thepartialpressureofaqueousvaporinthehumidair Fromaqueousvaporappendixtable,thevaporpressureofaqueousvaporat60℃is149.4mmHg。 b）Humidity c）Thehumidityvolumeis： =2.24(m³humidair)/(kgdryair)Sothedensityofhumidair11-2Adryerisusedtodrymaterialfrom5percentmoistureto0.5percentmoisture(wetbasis).Theproductioncapacityofdryeris1.5kgdrymaterial/s.Hotairentersat127℃and0.007kgwater/kgdryairandleavesat82℃.Thetemperatureofthematerialintheinletandoutletare21℃and66℃,respectively.Thespecificheatofwetmaterialis1.8KJ/（kg℃）。Iftheheatlossofthedyercanbeignored,thencalculatea)theconsumptionofthewetair.b)thetemperatureoftheairleavingthedryer.Solution：Themoisturecontentofthematerial(drybasis)isFromthematerialbalance,wehave:Fromtheenthalpybalance,wehave:Where： kgwater/kgdryair （3）Fromthese,the，L=6.61kgwetair/s11-3Ahumidmaterialisdriedatadryer.55hoursarerequiredtoreducemoisturecontentfrom35percentto10percent.Thecriticalmoisturecontentwasfoundtobe15percentandtheequilibriummoisture4percent.Ifunderthesamedryingconditions,itisrequiredthatthemoisturecontentofthematerialdropsto0.05from0.35.Assumingthattherateofdryingisproportionaltothefree-moisturecontent(X—X*),trytocalculatehowlongwillittaketodrythehumidmaterial.Solution:Thedryingtimeatconstantratestageis Thedryingtimeatfallingratestageis Sotherequireddryingtimeunderthenewconditionis 11-4Aatmosphereparallel-counterflowdryeristobedesignedtodryahumidmaterialfrom1.0kgmoisture/kgdrymaterialto0.1kgmoisture/kgdrymaterialusinghotair.Theairentersat135℃and0.01kgwater/kgdryair,andleavesat60℃.Assumingtheairexperiencedequalenthalpydryingprocess.Accordingtoexperiment,therateofdryingundertheconstant-rateperiodcanberepresentedbyequation-dx/dτ=30ΔHkgwater/(kgdrymaterial.H);therateofdryingcanberepresentedbyequation-dx/dτ=1.2xkgwater/(kgdrymaterial*h).Trytocalculatehowlongshouldittakestodrythismaterial?Solution：Becausetheairexperiencedequalenthalpydryingprocess,thevaluesoft1，H1，andt2canbeusedtofindthefollowingvaluesfromthehumiditychart: kgwater/kgdryair kgwater/kgdryair Frommaterialbalance. Becausetheratesofdryinginthemeetingpointbetweentwostagesarethesame,so: AssumethatthehumidityofairisHocorrespondingtothecriticalmoisturecontentofmaterialXo.Hence: （1） （2）FromEqs(1)and(2):Xo=0.74kgwater/kgdrymaterialHo=0.0184kgwater/kgdryair1）Thetimeofdrying,inconstantratestageSeparatevariablesandintegrate: 2Thetimeofdryinginfallingratestage