化工基础学习知识原理习题集规范标准答案英文

1、ProblemsandSolutionsDistillation1、Acontinuousfractionatingcolumnisusedtoseparate4000kg/hofamixtureof30percentCS?and70percentCCl4.Bottomproductcontain5percentCS2atleast,andtherateofrecoveryofCSzintheoverheadproductis88%byweight,required.Calculate(a)themolesflowofoverheadproductperhour.(b)themolefract

2、ionsofCS2andCCl4intheoverheadproduct,respectivelySolution:FormoverallmaterialbalanceFDW(1)FXFDXDWxW(2)KnownbythejusticeoftheproblemTaketheplaceof3typesandenter2typesFxF0.88FXFWxwDxD0.88FXF(3)W.0.12FXF0.12Xw0.052880kg/hDFW400028801220kg/h0.88FxF0.8840000.3xDF0.943D1120Theunitconverts：XD0.943/760.943/

3、760.057/1540.97(molefraction)Mm0.97760.0315478.3kg/kmol1120D14.3Kmol/h78.32、Aliquidcontaining40molepercentmethanoland60molepercentwateristobeseparatedinacontinuousfractionalcolumnat1atmpressure.Calculatethevalueofqunderthethreefollowingconditions(a)thefeedingisliquidat40C(b)thefeedingissaturatedliqu

4、id.Theequilibriumdataformethanol-waterliquidat1atmpressurearegivenintheattachedtable.Ifthecolumnisfedwith100koml/h.Themolarfractionsofmethanolinoverheadproductandbottomproductare0.95and0.04,respectively.Arefluxratioatthetopofcolumnis2.5.Calculate(a)themoleflowofoverheadproductperhour(b)themoleflowof

5、liquidinrectifyingcolumn(c)themoleflowofvaporinstrippingcolumn.Assumethattheconstantmolarflowappliestothissystem.Solution:FormoverallmaterialbalanceFDWFxfDXDWxwSolvetheequtionwecanhave:Dg3100(0.40.04)39.6kmol/hXDXw0.950.04W10039.660.4kmol/hLRD2.539.699kmol/hAndVDL39.699138.6kmol/hTheseuppervaluesare

6、fixedunderthethreefeedconditionsVV(q1)FThefeedat40c,q=1.07V138.6(1.071)100145.6kmol/hFeedsaturatedliquid,q=1VV138.6kmol/h3、Thefeedingatdewpointisfedtoacontinuousfractionatingcolumn,andtheequilibriumrelationsaregivenbyequations:y=0.723x+0.263(inrectifyingcolumn);y=1.25x-0.0187(instrippingcolumn).Calc

7、ulate(a)thecompositionsoffeeding,overheadproductandbottomproduct,respectively(b)therefluxratioSolution:TheslopeoftheoperatinglineinrectifyingcolumnisR0.723R1Wecanget:R=2.61TheinterceptoftherectifyinglineonyaxisisXD屋0.263R1XD0.263(2.611)0.95TheintersectionofthestrippinglineandthediagonalisY=X=XwsoXw=

8、0.0748Formtheintersectionofthesetwooperatinglines,gives:0.723X0.2631.25X0.0187X0.535Y0.7230.53350.2630.65Becausethefeedatdewpoint,sothefeedlineishorizon,andthecompositionoffeedisXFY0.656.12Itisdesiredtoproduceanoverheadproductcontaining95.7molepercentAformaidealmixtureof44moleAand50molepercentBfeedi

9、ngintoacontinuousfractionatingcolumn.Theaveragerelativevolatilityequalsto2.5,andaminimumrefluxratiois1.63.Explaintheheatstateoffeedingandcalculatethevalueofq.Solution:Formequilibriumequation:X2.5XY1(1)X11.5XFormoperatingequation:RXn1.630.957YXDXR1R11.6311.6310.62X0.364Substitutingineq.1),gives:X=0.3

10、65Y=0.59Formdefiniensoftheminimumrefluxratio,theintersectionofthetwoupperequationsisalsothatoftheequilibriumcurveandthefeedline.QXXF,YXFsothefeedisamixtureofliquidandvapor.Formthefeedequation(feedline),gives:XF(1q)YqX0.44(1q)0.59q0.3652Solveandhave:q0.667一(1)Em2Y2Y3*Y2Y30.570.410.6280.410.7373%Witht

11、hereasonEm30.410.280.4750.2867%4、Aliquidofbenzeneandtolueneisfedtocontinuousdistillationinaplatecolumn.Underthetotalrefluxratiocondition,thecompositionsofliquidontheclosetogetherplatesare0.28,0.41,and0.57,respectively.CalculatetheMurphreeplateefficiencyofrelativelytwolowlayers.Theequilibriumdataforb

12、enzenetolueneliquidattheoperatingconditionaregivenas:x0.260.380.51y0.450.600.72Solution:UnderthetotalrefluxratioconditionYN1XNFormMurphreeefficiency:YnY(n1)*YYn(n1)Y3X20.41Y2X10.57Checked丫；0.628by20.41Em,v5、Acontinuousfractionatingcolumnisusedtoseparate2500kg/hofamixtureof25percentacetoneand75percen

13、twater.Overheadproductcontaining99percentacetone,required.Thesepercentagesarebyweight,and80molepercentofacetoneinfeedingisenteringtotheoverheadproduct.Thefeedingat20c,andtheoperaterefluxratiois2.7timesasmuchastheminimum.therearetwocondensersatthetopofcolumn.Partialcondenserisusedtocontroltherefluxfl

14、owtobesaturatedliquid.Thesteamnotcondensedenterthefinalcondensertobecondensedandcooled,regardedastheproducts.Iftheoverallefficiencyis60%,thencalculatethenumberofidealplates.Solution:ThecompositionsofeachflowrateKnownbythemeaningofthequestionDXD0.8FXFXF25/580.0937XD25/5875/1899/580.968MFF99/581/180.0

15、937580.90631821.8kg/kmol2500/21.8114.7Kmol/h0.8114.70.0937D8.9Kmol/h0.968WFD114.78.9105.8Kmol/hXFXFWWDXD114.70.09738.90.968D0.0202105.8CalculatethenumberofactualplatesFromtheattachedfigure1,atthepointXF0.0937,thecorrespondingbubblepointis672,sotheaverageofthefeedtemperatureandthebubblepointis1/2(67+

16、20)=43.5C。Thespecificheatsofeachcomponentat4315are:CPACPBCPm2.25kj/(kgK)4.19kj/(kgK)0.09372.25580.90634.191880.6kj/(kmolK)thelatentheatsofeachcomponentat67are:505kj/kgBrm2400kj/kg5050.09375824000.90631842000kj/kmol4200080.6(6720)10942000.,qq1.09theslopeofthefeedline=12.1q11.091Fromfig.2,theminimumre

17、fluxratiomustcomputedfromtheslopeofoperatinglineagthatistangenttotheequilibriumcurve,andmeasuretheslopeis0.475.Rmin0.475Rmin1SolveandhaveRmin0.905R2.70.9052.44theinterceptoftherectifyinglineonyaxis-XD-0.9680.281R12.441theinterceptofthestrippinglineonyaxis春令0709theoperatinglineoftheattachedfigure2isd

18、rawnaccordingtotheslope0.281,andthisoperatinglineoftheattachedfigure3isdrawaccordingtotheintercept0.709.7idealplatesareneeded(exceptreboilerandpartialcondenser)6、TheequilibriumdataforthesystemCS2-CC14atthe1atmpressurearegiveninExample1-11.Acontinuousfractionatingcolumnisusedtoseparateafeedcontains30

19、molepercentCS?and70molepercentCCl4.Overheadproductcontaining95molepercentCSandabottomproductcontaining2.5molepercentCS2,required.Thefeedissaturatedliquidandthemassflowis400kg/h.Arefluxratioequalsto1.7timesastheminimum.Thedistillationtakesplaceat6TCand1atm.thesuperficialvelocitybasedonemptytoweris0.8

20、m/sandthedistancebetweentwoboardsis0.4m.Theoverallefficiencyis50%.Calculate(a)Thenumberofactualplatesneeded.(b)themassflowoftwoproductsc)thecross-sectionaldiameteroftower(d)theeffectiveheightoftower.Solution:(a)calculatethenumberofactualplatesPaintedx-ypicturefortheequilibriumdatabythequestion,asfig

21、urehence,thenumberofactualplaten70.612showsSincethefeedissaturatedliquid,XqXF0.3,fromfig.X-YYq0.54故Rmin迎上0.950.541.71YqXq0.540.3R=1.51.71=2.56Theinterceptofoperatinglineinrectifyingcolumn=含黑。须CalculatethenumberofidealplateswithThegraphicmethod,andthestepsisomitted.Theresultindicates,besidesreboiler,

22、11idealplatesareneededandfeedshouldbeintroducedontheseventhplatefromthetop1)ThemassflowrateofproductTheaveragemolecularweightoffeedisMm=0.316+0.1154=130.6kg/kmolF400030.6kmol/h130.6FormoverallmaterialbalanceD+W=30.60.95D+0.025W=0.330.6Fromthetwoupperequations,gives:W=21.5mol/handMW0.025760.975154152

23、W21.51523270kg/hD40003270730kg/h2)Thecross-sectionaldiameteroftowerActualplatesN1110.520Sincethefeedissaturatedliquid,hence:VV(R1)DAssumethattherisingvaporisidealgas,henceV273t1Vs22.43600273ThevalidityheightoftowerisHhNP0.4208.0mAbsorption7-1Thevaporofmethanolmixedwithairisabsorbedinwater,thetempera

24、tureis270candthepressureis101.3kpa.Themolardensityofmethanolinbulkofliquidandgasphaseareveryweak.Henrysawappliestothissystem,H=1.995kmol/m3*kpa,KG1.55*105kmol/(m2*s*kpa),kl2.08*105kmol/(m2*s*kmol/m3)Calculate(1)KG(2)ThepercentofgasresistanceinthewholeAmongthem：FW30.621.59.1kmol/h(2.561)9.132.4kmol/h

25、32.427361122.436000.247m/s2731D,/VS0.627m40.2470.83)resistanceWhereH=solubilitycoefficient冗individualmass-transfercoefficientforliquidphasekG=lndividualmass-transfercoefficientforgasphaseKG=Overallmass-transfercoefficientbasedongasphaseSolution:1)KGkGHkL7-2AcountercurrentflowtowerisusedtoabsorbH2sfr

26、omtheair-H2ssteamfedtoit,usingpurewaterastheabsorbingliquid,(PurewaterisusedinthecountercurrentflowtowertoabsorbH2Sfromtheair-H2Smixedgas),thetowerisoperatingat25cand101.3kpa.ThedensityofH2Sischanged_(reduced)from2%into1%(involume).Henrysawappliestothissystem,andHenryconstantE=5.52*10kpa.iftheamount

27、oftheabsorbentis1.2timesasmuchasthesmallestamountaccordingtotheory,(1)calculatetheliquid-gasratioL/Vand(outletliquid)concentrationsX1(2)Repeatcalculatetheliquid-gasratioL/Vand(outletliquid)concentrations,iftheoperatingpressureis101.3kpa(1.55105m1.122105m2)1skmkmoliZ2skm1“kmolc”“5kmol、1(1.95522.08103

28、)2)YkGYKGY“51.5510100%Y572.3%Solution:1)FromEq(7-6),theslopeofequilibriumlinemis:545atm5451atmAnd:丫V11V10.020.020410.02Y2X2V21V200.0010.00110.001FormEq(7-54),calculatethelimitingliquid-gasratioLY1Y20.02040.001(V)minY一二0.0204c5181X20Sooperatingliquid-gasratioisL1.2(L)min1.2518622VVtheterminalconcentr

29、ationX1VXIX2L(YIY2)=0+工(0.02040.001)6223.12105kmol(H2S)/kmol(H2O)E5455452)TheslopeofequilibriumlinemP1054.5TheterminalconcentrationX1vXiX2-(YiY2)=0+(0.02040.001)62.2)3.12104kmol(H2S)/kmol(H2O)7-3Thebutanemixedwiththeairisabsorbedinasieve-platetowercontainingeightidealplates.Theabsorbingliquidisanon-

30、volatilizationoilhavingamolecularweightof250andadensityof900kg/m3.Theabsorptiontakesplaceat101.3kpaand150c.5percentofbutaneintheenteringgas.Thebutaneistoberecoveredtotheextentof95%,thevaporpressureofbutaneat150Cis(-),minSoVY丫2Y-X2m0.02040.001-c-51.80.020454.5(v)1.251.862.2194.5kpa,andliquidbutanehas

31、adensityof580kg/m3.AssumethatRaoultandDaltonlawsapply.(1)calculatethecubicmetersoffreshabsorbingoilpercubicmeterofbutanerecovered.(Caculatetheamoutofabsorbingoilisrequired(involume)whenpercubicmeterofbutaneisrecovered)(2)Repeatcaculate(1),ontheassumptionthattheoperatingpressureis3034.4kpaandthatallo

32、therfactorsremainconstant.(conditionsreainthesame)Solution:1)calculatethemassflowrateofwater0.10Y1丫2Lmin0.111110.100.1111(10.95)0.00565000(10.10)201kmol/h22.4V(丫Y2)201(0.11110.00556)X1X20.1111-05100kmol/h26.7L1.5Lmin1.551007650kmol(水)/h=1.377105kg(水)/hthesedata,wecanget:Theabsorptionfactorinthestate

33、ofoperationA76501.43mV26.7201Lookingintofig7-21(orcalculateaccordingtoequation7-77),wecanknowwhenA=1.43,0.95,theidealplatesNT5.5oSotheresultisthesameasthatofschematicallyad3)ThisquestionshouldbeestimatebyKremser.A.mapabsorptioncoefficient0.98,idealplatesNT5.5Formfig7-21,readingA1.75So:L1.75mV1.7526.

34、72019390kmol(水)/h7-4Anabsorptiontoweristorecover99percentofammoniainanairstream,usingpurewaterastheabsorbingliquid.(Purewaterisusedinanabsorptiontowertoabsorbammoniainanairstream,theabsorptivityis99percent),theheightofthepackedsectionis3m,thecoefficientofrelativity丫1丫2Y1mX20.95,underthelimitingamoun

35、tofwateraccordingtotheorycondition,NTOLookingintofig7-21accordingtoAninLmin0.950.95mVL1.5Lmin1.50.95mV1.50.9526.72017650kmol(水)/habsorptiontakesplaceat101.3kpaand20oc.ThemassflowrateofgasVis580kg/(m2.h),and6percentofammoniainthegasinvolume.ThemassflowrateofwaterLis770kg/(m2.h).Thetoweriscountercurre

36、ntoperatedunderisothermaltemperature,theequilibriumequationY*=0.9X,kGaV0.8,buthasnothingtodowithL,trytomakeouthowtheheightofthepackedsectiontochangeinordertokeeptheabsorptioncoefficientunchangedwhentheconditionsofoperationhavebechangedasfollowing(1)theoperatingpressureis2timesasmuchastheoriginal.(2)

37、themassflowrateofwaterisonetimemorethantheoriginal3)themassflowrateofgasistwotimesasmuchastheoriginalSolution:Z3m,p1atm,T293KY2Y1(10.99)0.000638TheaveragemolecularweightofthemixedgasM=29X0.94+170.06=28.28V5802(10.06)19.28kmol/(mh)28.28LZZ01842.Z8kmol/(m2h)mV0.919.2842.780.4056Y10.0610.060.063810,063

38、80ln(-)(10.4056)0,405610,40560.0006386.884Z3HOG0.4358mNOG6.884P2pFormEq(7-6),9cc1c0.90.452ln(100)(10.2028)0,2028FormEq(7-62)andEq(7-43a)SO：HGischangedwiththeoperatingpressureHOG_HOG1HOGHOG-0.43580.21792SoZNOGHOG5.4960.21791.198mSotheheightofthepackedsectionreduce1.802magainsttheNOGFVln(1Y1mX2Y2mX2mX

39、mV)(1TPoriginal2)L2L1)mpm-mV0.4519.280.2028SoNOG42.781丫mX27ln(-2)(11mVY2mX2mXmVT)10.20285.496HOGVVKraKGaPwhenthemassflowrateofgasincreasingahasnotremarkableeffectHOGHOG0.4358mZNOGHOG5.4960.43582.395mtheheightofthepackedsectionreduce0.605magainsttheoriginal3)V2V叱 m2(mV)20.40560.8116LLL1NOGln(100)(10.

40、8116)0.811615.8110.8116whenmassflowrateofgasincreasingacorrespondinggrow,accordingtoPoblem:KGaV0.8KGa(/产小_02_20.43580.501mZNOGHOG15.810.501m7.92mSotheheightofthepackedsectionincrease4.92magainsttheoriginalmVLNOGmV2L5.4961mV10.40560.2028HOGVUP2V20.8KGaPG20.2HOGDrying11-1Thetotalpressureandhumidityoft

41、hehumidairare50kPaand60C,respectively,andtherelativehumidityis40percent.Calculate(a)thepartialpressureofaqueousvaporinthehumidair(b)humidity(c)thedensityofhumidair.Solution:a)ThepartialpressureofaqueousvaporinthehumidairFromaqueousvaporappendixtable,thevaporpressureofaqueousvaporat60cis149.4mmHg。pps

42、0.4149.459.76mmHgb)Humidityc)Thehumidityvolumeis:=2.24(m3iumidair)/(kgdryair)0.622Pp0.62259.7638059.760.116(kg.water)/(kgdry-air)0.7721.244Ht760273P(0.7721.2440.116)333273760380SothedensityofhumidairH1HVH10.1162.2430.498kg/m11-2Adryerisusedtodrymaterialfrom5percentmoistureto0.5percentmoisture(wetbas

43、is).Theproductioncapacityofdryeris1.5kgdrymaterial/s.Hotairentersat127cand0.007kgwater/kgdryairandleavesat82c.Thetemperatureofthematerialintheinletandoutletare21and66c,respectively.Thespecificheatofwetmaterialis1.8KJ/(kg)。Iftheheatlossofthedyercanbeignored,thencalculatea)theconsumptionofthewetair.b)

44、thetemperatureoftheairleavingthedryer.Solution:Themoisturecontentofthematerial(drybasis)is5X10.0526950.5X20.0050399.5WGC(XIX2)1.5(0.05260.0503)0.0714kg水/sFromthematerialbalance,wehave:L(H20.007)0.0714Fromtheenthalpybalance,wehaveI2)GC(IIWhere:I1(CSCWX1)Q1(1.84.1870.0526)2142.49I2(1.84.1870.00503)661

45、20.2I1(1.011.88Hl)t12490H1(1.011.880.007)12724900.007147.4I2(1.011.88)822490H282.82644H2(2)L(147.4-I2)=1.5(120.2-42.43)=116.7kgwater/kgdryair(3)Fromthese,theH20.0178,L=6.61kgwetair/s11-3Ahumidmaterialisdriedatadryer.55hoursarerequiredtoreducemoisturecontentfrom35percentto10percent.Thecriticalmoistur

46、econtentwasfoundtobe15percentandtheequilibriummoisture4percent.Ifunderthesamedryingconditions,itisrequiredthatthemoisturecontentofthematerialdropsto0.05from0.35.Assumingthattherateofdryingisproportionaltothefree-moisturecontent(XHX*),trytocalculatehowlongwillittaketodrythehumidmaterial.Solution:Thed

47、ryingtimeatconstantratestageis17C;(X1XO)USThedryingtimeatfallingratestageisGc(XX)1nXXX2X6督X、2aSX0XGcr/0.350.15、:-()aS0.150.042.424GcaS一一G一thatis2.269aSSotherequireddryingtimeunderthenewconditionis0.350.150.150.042.269()In9.57h0.150.040.050.0411-4Aatmosphereparallel-counterflowdryeristobedesignedtodryahumidmaterialfrom1.0kgmoisture/kgdrymaterialto0.1kgmoistu