天津大学化工流体课程下册英文版答案.docx
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天津大学化工流体课程下册英文版答案.docx
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天津大学化工流体课程下册英文版答案
Chapter6
6.1.Carbondioxideisdiffusingthroughnitrogeninonedirectionatatmosphericpressureand0°C.ThemolefractionofCO2atpointAis0.2;atpointB,3maway,inthedirectionofdiffusion,itis0.02.DiffusivityDis0.144cm2/s.Thegasphaseasawholeisstationary;thatis,nitrogenisdiffusingatthesamerateasthecarbondioxide,butintheoppositedirection.(a)WhatisthemolalfluxofCO2,inkilogrammolespersquaremeterperhour?
(b)Whatisthenetmassflux,inkilogramspersquaremeterperhour?
(c)Atwhatspeed,inmeterspersecond,wouldanobserverhavetomovefromonepointtotheothersothatthenetmassflux,relativetohimorher,wouldbezero?
(d)Atwhatspeedwouldtheobserverhavetomovesothat,relativetohimorher,thenitrogenisstationary?
(e)Whatwouldbethemolalfluxofcarbondioxiderelativetotheobserverundercondition(d)?
solution:
(a)from(6.1-8)
fromequation(6.1-19)
(b)netmassflux
forcarbondioxide(molecularweight=44)
massfluxofCO2=44×1.388×10-4kg/m2h
fornitrogen(molecularweight=28)
massfluxofN2=28×1.388×10-4kg/m2h
sothenetmassfluxinthedirectionofCO2diffusion
m=(44-28)×1.388×10-4=2.221×10-3kg/m2h
(c)HereJA=NA=NB,sincethediffusionisequimolal.Theconcentrationatanypointdependsonpositionduetotheconcentrationprofileoftheequimolaldiffusion,sodoesvelocitybasedonequations(6.1-3a)and(6.1-3b).Toselecttwopoints,yA=0.2and0.02,respectively,tocalculatethepositionsofobserver
foryA=0.2,CA=Cm×yA=
fromequation(6.1-3a)
thediffusingvelocityofA:
forB:
CB=Cm×(1-yA)=
thediffusingvelocityofB:
letuoisthevelocityoftheobservermovinginthedirectionofCO2diffusion,thennetvelocity(uA-uo)givesamasstransferratemAequaltothatinoppositedirectionmBcorrespondingto(uB+uo)
fromequations(6.1-4)and(6.1-5)
formA=mB,andrearrangingtwoequationsabovegives
ItissimilartocalculatinguoatthepointofyA=0.02
(d)
Whenthevelocityofobservermovingisequaltothatofnitrogendiffusing,thenitrogenisstationary
uo=
(e)Whenthevelocityofobservermovingisequaltothatofnitrogendiffusing,themolalfluxofcarbondioxidediffusingisindicatedby
Chapter7
7.1
solution:
Thedatafromthirdcolumninthetableareusedascalculating
Molefractionxofammoniainwater:
molarratioofammoniatowaterXinliquid
molarratioofammoniatoinertgas
theresultsofcalculationarelistinthetable
p/kPa
00.40.81.21.62.02.433.324.236.679.28
x
00.005270.010480.015630.020740.02580.030790.040630.050280.073570.09574
X
00.005300.010590.015880.021180.02650.031770.042350.052940.079400.1059
Y
00.003960.007960.011980.016000.02010.024530.033870.043530.070400.1008
ThedatainthetableareusedtoplotthemolefractionxversuspartialpressurepdiagramandmolarratioX-Ydiagram
虚线范围表示符合
Herry’slaw
7.3Vapor-pressuredataforamixtureofpentane(C5H12)andhexane(C6H14)aregivenbythetable.Calculatethevaporandliquidcompositioninequilibriumforthepentane-hexaneat13.3kpapressureonassumingthatthevaporofmixtureapproachesidealbehaviorandliquidfollowsRaoult’slow.
t,K
260.6
265
270
275
280
285
289
PA,kPa
13.3
17.3
21.9
26.5
34.5
42.5
48.9
PB,kPa
2.83
3.5
4.26
5.0
8.53
11.2
13.3
Solution:
FromRaoult’slow(equations(7.1-4)and(7.1-5))
Andthetotalpressureofsystemisequaltosumofthepartialpressuresoftwosubstances
rearrangingequationabove
1
and
rearrangingequationabovegives
2
substitutingthedataforthetableintotheequations1and2givestheresultsinthefollowingtable
t,K
260.6
265
270
275
280
285
289
PA,kPa
13.3
17.3
21.9
26.5
34.5
42.5
48.9
PB,kPa,
2.83
3.5
4.26
5.0
8.53
11.2
13.3
x
1.0
0.710
0.513
0.386
0.184
0.067
0
y
1.0
0.924
0.845
0.769
0.477
0.214
0
7.4Usingtheconditionsintheproblem7.3forthepentane-hexanemixture,doasfollows:
(a)Calculatetherelativevolatility.
(b)Usetherelativevolatilitytocalculatethevaporandliquidcompositioninequilibriumandcomparewiththeresultgivenbytheproblem7.3.
Solution:
(a)calculatethevalueof
by
Theresultsaregiveninthetable:
t,K
260.6
265
270
275
280
285
289
PA,kPa
13.3
17.3
21.9
26.5
34.5
42.5
48.9
PB,kPa,
2.83
3.5
4.26
5.0
8.53
11.2
13.3
α
4.70
4.94
5.14
5.3
4.04
3.79
3.68
Averagerelativevolatility
andequilibriumrelationwithrelativevolatilityforamixtureofpentane(C5H12)andhexane(C6H14)isexpressedby
(b)Usingtheequationabovetocalculatethevaporandliquidcompositioninequilibriumandcomparewiththeresultgivenbytheproblem7.3.
T,K
α
x
y
Theresultsofproblem7.3
Thecalculatingresultbyusing
260.6
265
270
275
280
285
289
4.51
1.0
0.710
0.513
0.386
0.184
0.067
0
1.0
0.924
0.845
0.769
0.477
0.214
0
1.0
0.917
0.826
0.739
0.504
0.245
0
*7.5BoilingPointandRaoult'sLaw.Forthesystembenzene–toluene,doasfollows,usingthedatafromTable7.1-1:
①At378.2K,calculateyAandxAusingRaoult'slawattotalpressureof101.325kPa.
②IfamixturehasacompositionofxA=0.40andisat358.2Kand101.32kPapressure,willitboil?
Ifnot,atwhattemperaturewillitboilandwhatwillbethecompositionofthevaporfirstcomingoff?
solution:
①At378.2KfromTable7.1-1forbenzene,vaporpressurePA=204.2kPa,fortoluene,vaporpressurePB=86.0kPa,substitutingthesedataintoEq(7.1-11)andsolvingforxA(molefractionofbenzene)
substitutingintoEq(7.1-12)
②Fromfigure7.1-2,ifamixturehasacompositionofxA=0.40andisat358.2Kand101.32kPapressure,itwillnotboil,andwillboilattemperatureof95oC(368.2K);
At95oC(368.2K),fromTable7.1-1forbenzene,vaporpressurePA=155.7kPa,PB=63.3kPa,andsubstitutingintoEq(7.1-12)
与原料组成基本一致,达到泡点
7.7WhataretheeffectsontheconcentrationsoftheexitgasandliquidstreamsofthefollowingchangesintheoperatingconditionsofthecolumnofExample7.5?
(a)Adropintheoperatingtemperaturethatchangestheequilibriumrelationshiptoy=0.6x.Unchangedfromtheoriginaldesign:
N,L/V,yb,andxa.(b)AreductionintheL/Vratiofrom1.5to1.25.Unchangedfromoriginaldesign:
temperature,N,yb,andxa.(c)Anincreaseinthenumberofidealstagesfrom5.02to8.Unchangedfromoriginaldesign:
temperature,L/V,yb,andxa.
solution:
(a)Foradilutesolutionandadilutegas,LandVareassumedconstant,andthestrippingfactoris
=0.6×1.5=0.9
Allconcentrationscanbeexpressedintermsofxa,themolefractionofNH3intheenteringsolution:
sinceyb=0
Fromanammoniabalance,VΔy=Vya=L(xa-xb)=Lηxa.Hence
also
FromEq.(7.2-28),
Theseparationcorrespondsto5.02idealstages,sothestageefficiencyis5.02/7=72percent.
η=0.787
Vya=L(xa-xb)=Lηxa.decreasingin
=1.5×0.787×xawithdecreasingintherecoveryη
(b)
=0.8×1.25=1
1
2
so
3
substitutingequations
(2)and(3)intoequation
(1)gives
=5.02
solvingforη=0.834
(c)
Fromanammoniabalance,Vya=LΔx=Lηxa.Hence
also
FromEq.(7.2-28),
=1.456
solvingforη=0.95
7.10.Atoxichydrocarbonisstrippedfromwaterwithairinacolumnwitheightidealstages.(a)Whatstrippingfactorisneededfor98percentremoval?
(b)Whatpercentageremovalcouldbeachievedwithastrippingfactorof2.0?
solution:
(a)fromequation(7.2-28)
1
Fromanammoniabalance,VΔy=Vya=L(xa-xb)=Lηxa.Hence
also
2
substitutingequation
(2)to
(1)gives
themethodoftrial-and-errorisusedtosolvingforstrippingfactorS
(b)
forastrippingfactorof2.0,percentageremovalcouldbecalculatedby
solvingforη=0.998
Chapter8
8.1Thegasstreamfromachemicalreactorcontains25mol%ammoniaandtherestinertgases.Thetotalflowis181.4kmol/htoanabsorptiontowerat303Kand1.013×105Papressure,wherewatercontaining0.005molfracammoniaisthescrubbingliquid.Theoutletgasconcentrationistobe2.0mol%ammonia.Whatistheminimumflow
?
Using1.5timestheminimum,plottheequilibriumandoperatinglines.
Solution:
提示:
给出E
fromFig9-2in《ChemicalEngineering》inChinese:
E=1.85×105Pa,
massbalanceoncomponentammonia
kmol/h
kmol/h=100.14mol/s
equilibriumline
operatingline
8.2Agasstreamcontains4.0mol%NH3anditsammoniacontentisreducedto0.5mol%inapackedabsorptiontowerat293Kand1.013×105Pa.Theinletpurewaterflowis68.0(改为136)kmol/handthetotalinletgasflowis57.8kmol/h.Thetowerdiameteris0.747m.Thefilmmass-transfercoefficientsareky’a=0.0739kmol/s·m3molfracandkx’a=0.169kmol/s·m3molfrac.Usingthedesignmethodsfordilutegasmixtures,calculatethetowerheight
solution:
原题中液气比小于最小液气比,无法解。
V'=57.8kmolair/h,yb=0.04molfracNH3,ya=0.005,L'=68.0kmolwater/h,xa=0.ky’a=0.0739kmol/s·m3molfracandkx’a=0.169kmol/s·m3molfrac.
equilibriumrelation
Y=1.83X
Massbalanceonammonia
overallmasstransfercoefficient
heightoftransferunitis
numberoftransferunit
theheightofpacking
Z=Hoy×Noy=0.8573×4.147=3.555m
8.3Inatower0.254mindiameterabsorbingacetonefromairat293Kand101.32kPausingpurewater,thefollowingexperimentaldatawereobtained.Heightof25.4-mmRaschigrings=4.88m,V'=3.30kmolair/h,yb=0.01053molfracacetone,ya=0.00072,L'=9.03kmolwater/h,xb=0.00363molfr
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