Distillation ...
Transcript of Distillation ...
Distillation
Theseparationofliquidmixturesintotheirvariouscomponentisoneofthemajor
operationsintheprocessindustriesanddistillation,themostwidelyusedmethodof
achievingthis.Inprocessing,thedemandforpurerproducts,coupledwiththeneedfor
greaterefficiencyhaspromotedcontinuousresearchintothetechniquesofdistillation.
Amasstransferoperationinvolvingtheuseofenergyandenergyseparatingagentsto
createavaporphasewhichcomesintocontactwiththeliquidphaseresultinginthe
transferofmassfrom theliquidphasetothevaporphase.
- Massseparatingagentinvolvesputtingasubstanceintoamixturetocause
separatione.g.solventextraction,extractivedistillation,evaporation,dryingetc.
- Phaseseparatingagentinvolvesosmosismembrane,technique,filtrationetc.
- Beforemasstransferoccurs,thesubstanceintheliquidandvaporphasesmustbe
incontact.
Vapor-LiquidEquilibrium
Thecompositionofthevaporinequilibrium withaliquidofgivencompositionis
determinedexperimentallyusinganequilibrium still.Whenequilibrium isestablished,
concentrationinthevaporphase(y)isinequilibrium withtheconcentrationoftheliquid
phase(x).Thedeviationfrom theequilibrium causestransferofmassesbetweenthe
states(liquid&vapor).Itmaybeimpossibletohave100%separationbydistillation.
Maximum &Minimum boilingazeotropes
Bubblepointanddewpointcurves
Maximum
boiling
Minimum
boiling
0 x,y1
T
Dewpointcurve
Bubblepoint
curve
0 x,y1
T
Vapor&liquidphasecurves
Saturatedvapor
PartialpressuresandDalton’s,Raoult’sandHenry’sLaws.
ThepartialpressurePAofacomponentAinamixtureofvaporsisthepressurethat
wouldbeexertedbycomponentAatthesametemperature,ifpresentinthesame
volumetricconcentrationasinthemixture.
ByDalton’slawofpartialpressuresP= PA,thatisthetotalpressureisequaltotheΣ
summationofthepartialpressures.Since,inanidealgasorvapor,thepartialpressure
Vaporphase
Liquidphase
0 x,y1
T
0 x,y1
T0
Saturatedliquid
Vapor
1
0 1Liquid
(x)
Vapor–Liquid curves(X1Y)
A0 A PB
OXB
isproportionaltothemolefractionoftheconstituent,then
PA=YAP…………………........................(1)
Foranidealmixture,thepartialpressureisrelatedtotheconcentrationintheliquid
phasebyRaoult'slawwhichmaybewrittenas
PA=PAoXA………………………………..(2)
WherePAo=VaporpressureofpureAatthesametemperature
PA=PartialpressureofsubstanceA
XA=MolefractionofsubstanceAintheliquidphase.
Henry’slaw:PA= XA………………….(3)H
Where =Henry’sConstantH
PA=PartialpressureofsubstanceA
XA=MolefractionofsubstanceAintheliquidphase.
Bycombiningequation(1)&equation(2)together,wehave
PA=PAoXA
PA=YAP
Hence,PAoXA¬=YAP
YA= ,YB=∴ P X
P
ButYA+YB=1& XA+XB=1
PA0XA+PB
oXB=1
P P
PA0XA+PB
o(1-XA)=1
P P
Therefore, XA= P–PB0 ……….(4)
PA0–PB
0
Example1:Thevaporpressuresofn-heptaneandtolueneat373K(1000C)are106and
P
73.7KN/m3respectively.Whatarethefractionsofn-heptaneinthevaporandinthe
liquidphaseof373Kifthetotalpressureis101.3kN/m2?
Solution
Recall,XA= P-PB0 =(101.3-73.7)=0.856
PA0–PB
0 (106-73.7)
andXA=PA0XA=106x0.856=0.896
P 101.3
Example2:Whatisthedewpointofanequimolarmixtureofbenzeneandtolueneat
101.3KN/m2?
Solution:
From Raoult’slaw,PB=XBPB0=YBP
And PT=XTPT0=YTP.
Itnowremainstoestimatethesaturationvaporpressureasafunctionoftemperature,
usingtheAntioneequationandthendeterminebyprocessoftrialanderrorwhen(XB+XT)
=1.0
Thedata,withpressureinkN/m2are:
T(K) PB0 XB PT
0 XT XB+XT
373.2 180.006 0.2813 74.152 0.6831
0.9644
371.2 170.451 0.2872 69.760 0.7261
1.0233
371.7 172.803 0.2931 70.838 0.7150
1.0081
371.9 173.751 0.2915 71.273 0.7107
1.0021
372.0 174.225 0.2907 71.491 0.7085
0.9992
As0.0995isnearenoughto1.000,thenearpointmaybetakenas372.0K
Oneofthemostwidelyusedcorrelationsofsaturatedvaporpressureisproposedby
Antoine
InP0=K1-(K2/T+K3)………………..(5)
Forconstantequation
Benzene Toluene
K1=6.90565 K1=6.95334
K2=1211.033 K2=1343.943
K3=220.79 K3=219.377
WhereP0isinmmHgandTisin0C.
Example3:Determinethevaporphasecompositionofamixtureinequilibrium witha
liquidmixtureof0.5molefractionoftolueneat338K.Willtheliquidvaporizeata
pressureof101.3KN/m2?.
Solution:Saturationvaporpressureofbenzeneat338K=650Cisgivenby
Benzene:log10PB0=6.90565-1211.033 =2.668157
65+220.79
Recall,1mmHg=133.32N/m2
PB0=465.75mmHgor62.10KN/m2
Toluene:log10PT0=6.95334-1343.943 =2.22742
65+219.377
PT0=168.82mmHgor22.5KN/m2
Partialpressuresinthemixtureare:
PB=(0.50x62.10)=31.05KN/m2andPT=(0.5x22.51)
Totalpressure=(PB+PT)=42.305kN/m2
:-Hencecompositionofvaporphaseis
Dalton’slaw YB=PB=31.05=0.734
P 42.305
YT=PT=11.255=0.266
P 42.305
Sincethetotalpressureisonly42.305KN/m2,thenwithatotalpressureof101.3KN/m2,
theliquidwillnotvaporizeunlessthepressureisdecrease.
ConstantBoilingMixtures:Mixturesinwhichtheirboilingpointremainsconstanton
movementoftemperatureduetothefacttheirconcentrationremainsconstant.
\Theconcentrationoftheliquidandvaporphasewillbethesame
Azeotropicmixture(2types)
0 x 1
(i) Maximum Boilingmixture
(ii) Minimum
i. Maximum BoilingMixture:Theboilingpointofthemixtureishigherthanthe
boilingpointofthepurecomponent
ii. Minimum BoilingMixture:Theboilingpointofthemixtureislowerthanthe
boilingpointofthepurecomponent
MethodsofDistillation–TwoComponentMixtures
Forabinarymixturewithanormalcurvex-ycurve,thevaporisalwaysricherinthemore
volatilecomponentthantheliquidfrom whichitisformed.Therearethree(3)methods
usedindistillationpracticewhichallrelyonthisbasicfact.Theseare
1.DifferentialDistillation
2.FlashorEquilibrium Distillation
3.Rectification
1.DifferentialDistillation
Thesimplestexampleofbatchdistillationisasinglestage,differentialdistillation,
startingwithastillpot,initiallyfull,heatedataconstantrate.
Inthisprocess,thevaporformedonboilingtheliquidisremovedatoncefrom the
system.Sincethisvaporisricherinthemorevolatilecomponentthantheliquid,it
followsthattheliquidremainingbecomessteadilyweakerinthiscomponent,with
theresultthatthecompositionoftheproductprogressivelyalters.
TheanalysisofthisprocesswasfirstproposedbyRAYLEIGH.
IfSisthenumberofmolesofmaterialinthestill,xisthemolefractionof
componentAandanamountdS,containingamolefractionyofAisvaporized,then
amaterialbalanceoncomponentAgives
ydS=d(Sx)
ydS=Sdx+xdS
=∫sS0
ds
s∫
xxo(dx
y-x)In = ……………………………..1S
So∫
uxo(du
y-u)Theintegralonthenighthandsideofequation(1)maybesolvedgraphicallyifthe
equilibrium relationshipbetweenyandxisavailable.Ifovertherangeconcernedthe
equilibrium relationshipisastraightlineoftheform y=mx+c,then
InS= 1 In (m-1)x+c
So m-1 (m-1)xo+c
y–x = S m-1
y0–x0 S0 …………………………..2
2.FlashEquilibrium Distribution
Flashequilibrium,frequentlycarriedoutasacontinuousprocess;consistsof
vaporizingadefinitefractionoftheliquidfeedinsuchawaythatthevaporevolved
isinequilibrium withtheresidualliquid.Thefeedisusuallypumpedthroughafired
heaterandentersthestillthroughavalvewherethepressureisreduced.Thestillis
essentiallyaseparatorinwhichtheliquidandvaporproducedbythereductionin
pressurehavesufficienttimetoreachequilibrium.Thevaporisremovedfrom the
topoftheseparatorandisthenusuallycondensed,whiletheliquidleavesfrom the
bottom.
Thevaporandliquidstreamsmaycontainmanycomponentsinsuchanapplication,
althoughtheprocessmaybeanalyzedsimplyforabinarymixtureofAandBas
follows:
IfF=molesperunittimeoffeedofmolefractionxfofA
V=molesperunittimeofvaporformedwithythemolefractionofAandS=mole
perunittimeofliquidwithxthemolefractionofA.
Thenanoverallmassbalancegives
F=V+S
Fxf=Vy+Sx
ThenV= xf-x
F y-x
Ory=Fxf–x F– 1 ………….3
V V
Equation3representsatrianglelineofslope:
_ F-V =-S………………………..4
V V
Passingthroughthepoints(xf,yf),thevaluesofxandyrequiredmustsatisfynot
onlytheequationbutalsotheappropriateequilibrium data.
Example1:Anequilibrium mixtureofbenzeneandtolueneissubjectedtoflash
distillationat100kN/m2intheseparator.Usingtheequilibrium datagiven,determine
thecompositionoftheliquidandvaporleavingtheseparatorwhenthefeedis25%
vaporized.Forthistheboilingpointdiagram belowmaybeusedtodeterminethe
temperatureoftheexitliquidstream.
Molefractionofbenzene(x)
Solution:Thefractionalvaporization=V/F=f(assumption)
Slope=-F-V =-1-F –from equation4
V f
Assumef=0.25,trialanderror=Slope=- 1-0.25
0.25
Slope=-0.75=-3corresponds
0.25
Andtheconstructionismadetogivex=0.42&y=0.63.So,from theboilingpoint
diagram,theliquidtemperaturewhenx=0.42isseentogive366.5k
3.Rectification
Inthetwoprocessesconsidered,thevaporleavingthestillatanytimeisinequilibrium
withtheliquidremainingandnormallytherewillbeonlyasmallincreasein
concentrationofthemorevolatilecomponent.
Rectificationmeanspurificationofasubstancethroughrepeatedorcontinuous
distillation
Theessentialmerittofrectificationisthatitenablesavaportobeobtainedthatis
substantiallyricherinthemorevolatilecomponentthantheliquidleftinthestill.
Distillationequipmenthas3majorcomponents
Colum/still Boiler/reboiler Condenser
0.2 0.4 0.6 0.81.0
39
0
38
0
37
0
36
T=366.5k
x=0.42
Boiler/ReboilerV
ThefeedgoesintotheColum F,thecolumnisdividedinto2sections,andthepointof
exitisthepointofintroductionofthefeedintothecolumn.Theupperpartofthefeedis
therectifyingsectionandthelowerpartisthestrippingsection.Thefeedthatgoesinto
columnF,thecompositionofthefeedisxf.Thefeedcanbeinanystate.Whenthefeed
getsintothecolumn,thevaporgoesupandentersintothecondenserwhichremoves
theenergyinthevaporandtranslatesittothesaturatedvapororliquid.Theproductof
thecondensergoesthroughtherefluxdivider;separatesomeportionasproduct
(distillateD)andthecompositionofdistillate(yD).Theremainingproductofthe
condenserisreturnedtothecolumnasthereflux.Theliquidthatisbelowthefeedplate
Lgoestothereboilerandreturnstothecolumnandsomeportionisgottenasbottom
product.
Interactionbetweenvaporandliquidgivesrisetomasstransfer.Interactionbetween
vaporandliquidoccursatthesurface.Itcouldbeplatesurfaceorpackingdevicese.g.
sieveplate,bubblecapplate,valveplatesetc.packingdevicese.g.pallring,raschigring,
saddlepackingetc.
Consideringthetopsectionofthecolumn
Liquid
Rx0
–rectifying
Rectifyingsection
xf=compositionoffeed
V=vapor
Strippingsection
L
V -Vapor
L0
Feed
F,xf
n-1
Ln-1Xn-1
Ln1Xn
n+1
Vn yn
F
Materialbalance:Vn+1+Ln-1=Vn+Ln………………………………..1
:Vn+1Yn+1+Ln-1Xn-1=VnYn+LnXn……………..2
Refluxratio(R)=L(from thecondenser)
D
V1=D+R,V1=D+L0i.e.R=L0
Assumingconstantmolaroverflow,massflowratesareconstant.
V.Yn+1+LXn-1=VYn+LXn
VYn+1–VYn=LXn–LXn-1
Yn+1=L/VXn–L/VXn-1+Yn
D,xD
Vn+1=Ln+D………………………………………………...1
Vn+1Yn+1=LnXn+DXD…...................................................2
V.f Yn+1=Ln/Vn+1Xn+D/Vn+1XD………………...3a
Equation3aistheoperatinglineequationforthetopsectionofthedistillationcolumn.
Vn+1 Yn+1
Liquid
Rx0
–rectifying
Rectifyingsection
xf=compositionoffeed
V=vapor
Strippingsection
L
V -Vapor
L0
Feed
F,xf
L,f Yn+1=Ln/Ln+DXn+D/Ln+DXD………..3b
Refluxratio(R)=L/D
Yn+1=R/R+1Xn+XD/R+1……….…………………….…3c
Equation3cistheoperatinglineequationfortherectifyingsection
Vm+1Ym+1
B,XB
Whereequations6aand6bareoperatinglineequationsofthestrippingsection
Equations3and6giveanequationofastraightline
Ym +1=R/R+1Xn+XD/R+1
Lm Xm
mm
m +1
Lm =Vm+1+B……………….……………………4
LmXm =Vm+1Ym+1+BXB…………………....5
Ym+1=Lm/Vm+1Xm –B/Xm+1XB……….....6a
ButVm+1=Lm –B
Hence,Ym+1=Lm/Lm-BXm –B/Lm-B…..6b
Equation3
Equation6
0 xf
1
Feedpoint
Thepointofinteractionofequations3and6isthefeedpointofthecolumn.
Feedconditions:Thestateofthefeedwhenitentersthecolumnwhichisfrom vaporto
liquidphaseandviceversaaffectstheoperatingline.
Forinstanceifthefeedisenteringthecolumnasvaporphase,it’sgoingtoaddvaporto
thevaporgoingupthecolumnbutifitisinliquidphase,itwouldaddtotheliquidgoing
downthecolumn,ifitisvapor-liquidphase,theliquidwilladdtotheliquidgoingdown
thecolumnandthevaporwilladdtotheonegoingup.
Theconditionofthefeedisdefinedby(q),heatrequiredtovaporize1moleofthefeed
enteringcolumntothemolarlatentheatofvaporizationofthefixed.
Itisgivenintermsofenthalpyas:
q=HV–HF
HV–HL
q=1+HV–HF
HV–HL
HV=Enthalpyofthefeedatdewpoint
HL=Enthalpyofthefeedatboilingpoint
HF=Enthalpyofthefeedatitsentrance
Forcoldliquid,q=1+CpL(TB-TF)
HV-HL
Feedisatbubblepointsaturatedliquid,q=1satiratedliquid,q=1ColdLiquid(q>1)
Vapor+liquid(o<q<1)
Saturated
Liquidq<0
0
Cpl=specificheatcapacityofliquidfeed
Forsuperheatedvapor,q=0+Cpv(TD–TF)
HV–HL
TF=Temperatureofthefeed
TB=bubblepointofthefeedtemperature
TD=dewpointofthefeedtemperature
From thedefinitionofq,wecanhaveanexpressionfortheq-line,andit’scalledq-line
equation
Y= q x– xf
q-1 q-1
Itisimportanttonotethatrefluxratio(R)=L/D=∞ sinceD=O
1.Totalreflux:Everythingisreturnedtothecolumn.Nothingisremovedfrom the
condenseri.e.attotalreflux,D=O.Attotalreflux,thenumberoftheoreticalstages
requiredtoachievespecifiedseparationisminimum.
Attotalreflux,boththetopandbottom operatinglinesfallonthediagonalline(450
lines)
XB XD
ThenumberofstagesisrepresentedbythenumberofstepsbetweenXBandXD.In
reality,thereisnoproductatallandthediameterofthetowerwouldbeinfinite.
2.Minimum RefluxRatio:Atminimum refluxratio,thenumberoftheoreticalstage
requiredtoachievethespecifiedseparationisinfinite.Thetopandbottom operating
linemeetattheequilibrium line.
Theredseparationoccursbetweentotalandminimum.Iftherefluxratiodecreases,
theslopeofthetopoperatinglinealsodecreasesanditwouldmovefrom the
diagonallinetotheequilibrium line.
450lines
Operatinglines
1
0XB XD1
y
o XB XD1
R1>R2>R3>R4
Minimum Refluxratio
diagram
Theslopeoftheoperatingline=R/R+1
- Optimum RefluxRatioattheoperatingline:Toachievethis,therefluxratiois
between1:2-10ofRm i.e.youmustknowhowtoobtainRm (minimum refluxratio).
- EfficiencyofDistillationColumn.
PlateEfficiency:Thisisusedtotranslatetheidealplateintoactualplate.Itis
applicabletobothabsorptionanddistillationoperations.
TypesofEfficiency
1.OverallEfficiency,ʅo= Idealplates…………….1
ActualPlate
Itconsiderstheentirecolumn.
2.Murphree: ʅm =Yn–Yn+1…………………….2
Yn*–Yn+1
Yn=Actualconcentrationofvaporleavingtheplate
Yn+1=Actualconcentrationofvaporenteringtheplate
Yn*=concentrationofvaporinequilibrium withconcentrationoftheliquid.
3.LocalorPointEfficiency:IthassomedefinitionasMurphreeefficiency.
ʅ́=Yn*–Yn+1́ …………………………………..3
Yn*́–Yn+1́
Thedifferencebetweenthetwoisthatpointefficiencydealswithpointonthetraywhile
Murphreeefficiencydealswiththeaverage.
Relationshipbetweentheefficiencies
1.Local/PointandMurphreeefficiency:Thereisuniform concentrationacrossthe
plates.
2.Murphreeandoverallefficiency
= In1+ʅm (mV/L-1)
In(mV/L)
m =slopeoftheequilibrium line
V=vaporflowrate
L=Liquidflowrate
WhenmV/L=1.0,ʅm =ʅo
Orwhenʅm =1.0,thenʅm =ʅo
Factorsinfluencingplateefficiency
1.Adequateandintimatecontactbetweenliquidandvaporwouldenhanceefficiency
2.Excessivefoamingorentrainment,poordistributionofliquid,shortcircuiting,
weepingordumpingofliquidlowersplateefficiency.
Entrainment:Whenaliquidiscarriedovervaportoaplateaboveit.
Shortcircuiting:Whereavapororliquidissupposedtopassanditdoesnotfollow
suchpartbutfollowsanotherpath.
Weeping:Whentheliquidisdroppingthroughthepreparationoftheplate.
Dumpingofliquid:Itoccursasaresultofoverloading
Example:Aliquidmixtureofbenzene-toluenedistillatedinafractionatingtowerwith
apressureof101.32kPa,thefeedis100kmol/hr.Liquidcontains45%benzene
entersat327.6K.Adistillatecontains95%molesbenzeneandthebottom contains
10%molesbenzene.Refluxratiois4:1.Theaverageheatcapacityofthefeedis
159kJ/kmolKandaveragelatentheatis32099KJ/kmol.Theequilibrium dataforthis
system isgivenintermsoftemperatureandvaporpressureasbelow.Calculatethe
distillatesandbottom inkmol/hrandthenumberoftheoreticalplates.Whatisthe
actualnumberoftraysifoverallefficiency(ʅo)is75%?
T(K) PB0(kPa) PT
0(kPa) XB YB
353.3 101.32 - 1.0 1.0
358.2 116.90 46 0.7800.900
363.2 135.50 54 0.5810.777
368.2 155.76 63.3 0.4410.632
373.2 179.26 74.3 0.2580.456
378.2 224.26 86.0 0.1110.246
383.8 224.06 101.32 0.0000.000
Solution
b.p=366.7K
X=0.42
0.42
0 xf=(0.5,0.5)trace
1
y
1
0.6
3
T(k)
0,00.20.40.60.81.0
390
380
370
360
350
F=D+B
100=D+B…………………………………………..1
100xf=DxD+BxB
100(0.45)=D(0.95)+B(0.1)
D=100–B
45=(100–B)(0.95)+0.1B
B=58.8kmol/hr
D=100–58.8=41.2kmol/hr
Yn+1=R/R+1Xn+XD/R+1
Yn+1=4/4+1Xn+0.95/4+1
Yn+1=4/5Xn+0.19
Y=q/q-1X–Xf/q-1
q=1+CpL(TB–TF)
HV–HL
HL–HF=CpL(TB–TF)
q=1+159(366.7–327.6)
32099
q=1.195
y=q/q-1x–x/q-1
y=1.195x– 0.45
1.95-1 1.195-1
y=6.128x–5.128
Molefractionofbenzeneliquid(x)B
12
34
5
67
(0,0)xB=0.1 0.951
Thetotalnooftheoreticalplates=7
ʅo=Theoretical/idealplate
Actualplate
0.75= 7
Actual
Actualplate=7/0.75=9.33plates=9plates