Unformatted text preview: gram on a following page. A line from RN through M to
an intersection with the equilibrium curve determines the extract point E1, as illustrated in Fig.
8.14. The operating point P is now determined, as in Fig. 8.17, by finding the intersection of
lines drawn through S and RN, and through F and E1. The number of equilibrium stages is now
stepped off, as in Fig. 8.17, giving a value of just slightly greater than Nt = 5. Exercise 8.12 (continued)
Analysis: (continued)
(c) First compute the two product flow rates, using the following product compositions
read from the plot:
Component
Feed
Solvent
Raffinate
Extract Mass fractions:
Acetone Water
0.450
0.550
0.000
0.000
0.100
0.895
0.509
0.040 1,1,2 TCE
0.000
1.000
0.005
0.451 1)
By overall total material balance, F + S = 1,000 + 350 = 1,350 = RN + E1
By overall acetone balance, 0.45F = 450 = 0.100RN + 0.509E1
(2)
Solving Eqs. (1) and (2) simultaneously, E1 = 770 kg/h and RN = 580 kg/h
The composition of each stream leaving each stage can be read from the righttriangle plot
prepared in part (b). Let x be mass fraction in the raffinate and y be mass fraction in the extract.
The stages are arbitrarily numbered from the feed end. The flow rate of each stream leaving each
stage is best obtained by total material balances around groups of stages and the inverse lever
arm rule, using the operating lines. A total balance for the first n1 stages is:
F + En= 1,000 + En = Rn1 + E1 = Rn1 + 770
Rn1  En = 1,000  770 = 230 or (3) By the inverse lever arm rule with the righttriangle diagram,
Rn −1
EP
=n
En
Rn −1 P (4) where the lines are measured from the righttriangle diagram.
Eqs. (3) and (4) are solved simultaneously for Rn1 and En for n = 2 to 6. The following results
are obtained:
Stage
1
2
3
4
5 Raffinate:
xA
xC
0.390 0.581
0.325 0.651
0.260 0.724
0.190 0.800
0.100 0.895 xS
0.029
0.024
0.016
0.010
0.005 Extract:
yA
0.509
0.440
0.360
0.270
0.165 yC
0.040
0.033
0.024
0.017
0.011 yS
0.451
0.537
0.616
0.713
0.824 kg/h:
R
935
822
738
656
580 E
770
706
592
508
426 Exercise 8.13
Subject: Extraction of isopropanol (A) from diisopropylether (C) by water (S) at 25oC.
Given: Feed containing 45 wt% A, 50 wt% C, and 5 wt% S. Raffinate to contain 2.5 wt% A.
Extract to contain 20 wt% A. Liquidliquid equilibrium data.
Find: Using VarteressianFenske method with a McCabeThiele diagram, find the number of
theoretical stages. Can an extract of 25 wt% A be obtained?
Analysis: Because the wt% of the solute, A, in both the raffinate and extract are given, the
solvent rate must be determined. Do this first by constructing a righttriangle diagram from the
equilibrium tie line and phase boundary data. As shown on the next page, for a diagram that
does not show the tie lines, points on the diagram are placed for: the feed, F, (45 wt% A, 50 wt%
C, and 5 wt% S); the solvent, S, (100% S); the raffinate, R, (2.5 wt% A on the phase boundary
line); and the extract, E, (20 wt% A on the phase boundary line). The mixing point, M, is
determined by the intersection of lines drawn from: (1) F to S and (2) R to E. The ratio of
solvent to feed i...
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 Spring '11
 Levicky
 The Land

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