Distillation ...

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Distillation Theseparationofliquidmixturesintotheirvariouscomponentisoneofthemajor operationsintheprocessindustriesanddistillation,themostwidelyusedmethodof achievingthis.Inprocessing,thedemandforpurerproducts,coupledwiththeneedfor greaterefficiencyhaspromotedcontinuousresearchintothetechniquesofdistillation. Amasstransferoperationinvolvingtheuseofenergyandenergyseparatingagentsto createavaporphasewhichcomesintocontactwiththeliquidphaseresultinginthe transferofmassfromtheliquidphasetothevaporphase. -Massseparatingagentinvolvesputtingasubstanceintoamixturetocause separatione.g.solventextraction,extractivedistillation,evaporation,dryingetc. -Phaseseparatingagentinvolvesosmosismembrane,technique,filtrationetc. -Beforemasstransferoccurs,thesubstanceintheliquidandvaporphasesmustbe incontact. Vapor-LiquidEquilibrium Thecompositionofthevaporinequilibriumwithaliquidofgivencompositionis determinedexperimentallyusinganequilibriumstill.Whenequilibriumisestablished, concentrationinthevaporphase(y)isinequilibriumwiththeconcentrationoftheliquid phase(x).Thedeviationfromtheequilibriumcausestransferofmassesbetweenthe states(liquid&vapor).Itmaybeimpossibletohave100%separationbydistillation. Maximum&Minimumboilingazeotropes Bubblepointanddewpointcurves Maximum boiling Minimum boiling 0 x,y 1 T Dewpointcurve Bubblepoint curve 0 x,y 1 T

Transcript of Distillation ...

Page 1: 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

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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)

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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

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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

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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

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(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

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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%

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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

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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

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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

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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

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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

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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

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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,

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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

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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

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Thetotalnooftheoreticalplates=7

ʅo=Theoretical/idealplate

Actualplate

0.75= 7

Actual

Actualplate=7/0.75=9.33plates=9plates