Absorptionofacetoneinacountercurrentstagetower
Agasiscontaining1.0mol%acetoneinair.Itisdesiredtoabsorb90%ofthe
acetoneinacountercurrentstagetower.Thetotalinletflowtothetoweris60.0
kgmol/h.Thetotalamountofpurewateris180kgmolH20/h.Theprocessis
isothermicandoperatesatatemperatureof300Kat1atm(101.3kPa).The
equilibriumrelationfortheacetone(A)inthegas‐liquidisyA=2.53xA.
(Absorb90%acetonefromagascontain1.0mol%acetoneinair,yAN+1=0.01
TotalinletgasflowVN+1=60kgmol/h
Totalinletpurewater(xA0=0)flowL0=180kgmolH2O/h
Equilibriumrelationforacetone(A)inthegas‐liquidyA=2.53xA)
a.Giventhematerialbalanceforacetone.
L0x0+VN+1yN+1=LNxN+V1y1=MxM
Explainthesymbols
Drawafigureofthetowerindicatingtheinputandexitflowsandcompositions
Solution:
Seepage631
b.Showthatthematerialbalancecanbewritten
L’x0/(1‐x0)+V’yN+1/(1‐yN+1)=L’xN/(1‐xN)+V’y1/(1‐y1)
WhatisL’andV’?
Solution:
Seepage663
Assumethatgasphaseisnotsolubleintheliquidphaseandthattheliquiddoes
notvaporizeintothegasphase.
Then,L’=L0(1‐x0),V’=V1(1‐y1),etc.andinsert.
L’istheinertsolvent
V’istheinertair
c.UsingmaterialsbalancefindyA1andxAN
Solution:
Seepage633
Materialbalanceacetone
L0x0+VN+1yN+1=LNxN+V1y1=MxM
InputacetoneyN+1VN+1=0.01×60=0.60kgmol/h
Enteringair(1‐yN+1)VN+1=(1‐0.01)×60=59.4kgmolair/h
AcetoneleavinginLN(90%ofinput):0.90×0.60=0.54kgmol/h
AcetoneleavinginV1:(1‐0.90)×0.60=0.10×0.60=0.060kgmol/h
TotalgasstreamV1leavingV1=59.4+0.06=59.46kgmolair+acetone/h
MolfracofacetoneleavingyA1=0.60kgmol/h/59.46kgmolair+acetone/h
yA1=0.60/59.46=0.00101
TotalliquidstreamLNleavingLN=180+0.54=180.54kgmolwater+acetone/h
MolfracofacetoneleavingxAN=0.54kgmol/h/180.54kgmolwater+
acetone/h
xAN=0.54/180.54=0.00300
Alt.yN+1=0.01,90%isabsorbed=>y1=yN+1(1‐0.90)=0.001
d.Assumetheoperatinglineisstraightinthisconcentrationrange.Ifyouhave
notfoundxANandyA1usexAN=0.003andyA1=0.001.
Plottheoperatinglineandequilibriumlineinayx‐plotandfindnumberof
theoreticalstages
Solution:
Seepage633
OnlyslightvariationinLandV

operatinglineisstraightthrough(xA0,yA1)and(xAN,yAN+1)
y=0.001+3x
Equilibriumlineisgivenbytherelation
yA=2.53xA
=>numberoftheoreticalstagesis5.2
e.UsingtheKremserequationfindthetheoreticalnumberofstages.
 y  mx 0 
1  1 
log  N  1
1 

y 1  mx 0 
A  A 

N 
log A
yN  1  mx 0  1  1 
ln
1  
 y 1  mx 0  A  A 
N
ln A
Solution:
Seepage635
Thesimplifiedversionisacceptedinthispart.Thedetailedsolutionisleftasthe
bonusquestion.
A=L0/(mVN+1)=180/(2.53×60.0)=1.186
N=5.16
Bonusquestion(g)
Atstage1
V1=29.73kgmol/h
yA1=0.00101
L0=90.0kgmol/h
xA0=0
yA=2.53xA=>m=2.53
A1=L0/(mV1)=180/(2.53×59.46)=1.197
AtstageN
VN+1=60.0kgmol/h
yAN+1=0.01
LN=180.54kgmol/h
xAN=0.00300
yA=2.53xA=>m=2.53
AN=LN/(mVN+1)=180.54/(2.53×60.0)=1.189
GeometricaverageA=(A1AN)0.5=(1.197×1.189)0.5=1.193
AcetoneistransferredfromVtoL(absorption)
N=ln[(yN+1‐mx0)/(y1‐mx0)](1‐1/A)+1/A]/ln(A)
X0=0
=>N=ln[(yN+1/y1)(1‐1/A)+1/A]/ln(A)
=ln[(0.01/0.00101)(1‐1/1.193)+1/1.193]/ln(1.193)
N=5.09stages
f.Explainthedifferencebetweenadsorptionandabsorption.Whatisdesorption?
•Adsorptionisaprocessthatoccurswhenagasorliquidsoluteaccumulateson
thesurfaceofasolidoraliquid(adsorbent),formingafilmofmoleculesoratoms
(theadsorbate).
•Itisdifferentfromabsorption,inwhichasubstancediffusesintoaliquidor
solidtoformasolution.
•Thetermsorptionencompassesbothprocesses
•Desorptionisthereverseprocess.