SAMPLE PROBLEM #9:SAMPLE PROBLEM #9:

One thousand Kilogram per hour of a 45 One thousand Kilogram per hour of a 45 % acetone – in – water solution is to be % acetone – in – water solution is to be extracted at 25 extracted at 25 ooC in a continuous counter C in a continuous counter current system with pure 1,1,2- current system with pure 1,1,2- trichloroethane to obtain a raffinate trichloroethane to obtain a raffinate containing 10 wt % actone. Using the containing 10 wt % actone. Using the following equilibrium data determine with following equilibrium data determine with an equilateral triangle diagram:an equilateral triangle diagram:

a) minimum flow rate of solventa) minimum flow rate of solventb) the number of stages required for a b) the number of stages required for a solvent rate equal to 1.5 times the solvent rate equal to 1.5 times the minimum.minimum.

AcetoneAcetonewt fractionwt fraction

WaterWaterwt fractionwt fraction

TrichloroeTrichloroethanethane(wt (wt fraction)fraction)

RaffinateRaffinate 0.550.55 0.350.35 0.100.10

0.500.50 0.430.43 0.070.07

0.400.40 0.570.57 0.030.03

0.300.30 0.680.68 0.020.02

0.200.20 0.790.79 0.010.01

0.100.10 0.8950.895 0.0050.005

AcetoneAcetonewt fractionwt fraction

WaterWaterwt fractionwt fraction

TrichloroeTrichloroethanethane(wt (wt fraction)fraction)

ExtractExtract 0.60.6 0.130.13 0.270.27

0.50.5 0.040.04 0.460.46

0.40.4 0.030.03 0.570.57

0.30.3 0.020.02 0.680.68

0.20.2 0.0150.015 0.7850.785

0.10.1 0.010.01 0.890.89

RaffinateRaffinate

Wt Fraction Wt Fraction AcetoneAcetone

Extract Extract

Wt Fraction Wt Fraction AcetoneAcetone

0.440.44 0.560.56

0.290.29 0.400.40

0.120.12 0.180.18

The Tie Line DataThe Tie Line Data

REQUIREDREQUIREDSSn+1 minn+1 min

XXmm

SOLUTIONSOLUTION (a) S(a) Sn+1n+1,min detn ,min detn SS11 is detd by trial, so that tie line is detd by trial, so that tie line

from Sfrom S11 passes thru passes thru by inverse-lever arm:by inverse-lever arm:

SSn+1,min n+1,min = 239.67= 239.67

M

M0

0

min,1n

x

xx

L

S

363.0

363.045.0

1000

S min,1n

A

L0

Sn+1

M

S1

T W

DXM=0.363

(b) Detn of n(b) Detn of n Sn+1 = 1.5 (Sn+1, min) = Sn+1 = 1.5 (Sn+1, min) =

1.5(239.67) = 359.51.5(239.67) = 359.5 to calc act xto calc act xMM

xxMM = 0.331 = 0.331

M

M0

0

1n

x

xx

L

S

M

M

x

x45.0

1000

5.359