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Unformatted text preview: corresponding flow rate of exiting solvent is SD = 15,893.6  1,986.7 = 13,906.9 lb/h
The slope of the solventfree operating line at the top of the cascade =
LD/V'N = 1,644.2/1,986.7 = 0.828. However, this slope is not constant because as we move down
from the top of the cascade, all three components undergo mass transfer between the two liquid
phases in an overall nonequimolar fashion. At the bottom of the cascade, by overall solvent
material balance, the entering solvent flow rate = SB = 13,906.9 + 67.4  140 = 13,834.3 lb/h
Next, determine two Y  X points on the operating line above the feed entry by writing the
following balances around a section of stages from Stage n above the feed to Stage N at the top,
in terms of total stream mass flows, V and L, and mass fractions (including solvent), y and x:
Total mass balance:
Solute mass balance:
Solvent mass balance: Vn1 = Ln + 14,249.4
(yn1)A Vn+1  (xn)A Ln = 284.3
(yn1)C Vn+1  (xn)C Ln = 13,906.9 (5)
(6)
(7) Eqs. (5) to (7) are solved by trial and error for a selected value of Xn. To do this, combine Eq. (5)
with (6) to eliminate Vn1, and combine Eq. (5) with (7) to eliminate Vn1. We now have two
equations for Ln. The selected value of Xn fixes the mass fractions of A and C in the raffinate.
Then find a value of Yn1 (which fixes the mass fractions of A and C in the extract) that gives
identical values of Ln. The results are as follows, with these two points for Y and X, together
with Y = X = 0.83 establishing the upper operating line, as shown in the Y  X diagram on the
second page of this exercise: Xn
0.621
0.405 Yn1
0.670
0.520 Vn1, lb/h
15,790
15,400 Ln, lb/h
1,541
1,151 Exercise 8.26 (continued)
Analysis: (continued)
Next, determine two Y  X points on the operating line below the feed entry by solving the
following balances around a section of stages from Stage m below the feed to bottom Stage1:
Total mass balance:
Solute mass balance:
Solvent mass balance: Vm = Lm+1  13,109.4
(ym)A Vm  (xm+1)A Lm+1 = 65.7
(ym)C Vm  (xm+1)C Lm+1 = 13,766.9 (8)
(10) (9) . The results are as follows, with these two points for Y and X, together with Y = X = 0.10
establishing the lower operating line, as shown in the Y  X diagram on the second page of this
exercise: Xm+1
0.287
0.166 Ym
0.390
0.205 Vm, lb/h
15,182
15,064 Lm+1,lb/h
2,073
1,955 From the Y  X plot on the second page of this exercise, 10 equilibrium stages are stepped off,
with 2 above the feed stage and 7 below.
The component material balance in lb/h is as follows: Component
A
B
C
Total Feed
350.0
650.0
140.0
1140.0 Solvent in
0.0
0.0
13834.3
13834.3 Extract
284.3
58.2
0.0
342.5 Raffinate
65.7
591.8
67.4
724.9 Solvent out
0.0
0.0
13906.9
13906.9 Reflux, LD
1364.7
279.5
0.0
1644.2 Exercise 8.27
Subject: Extraction of methylcyclohexane (MCH) from nheptane (C) with aniline (S) at 25oC
in a countercurrentstage extractor.
Given: Feed of 50 wt% MCH in C. On a solventfree basis, extract contains 95 wt% MCH and
raffinate contains 5 wt% MCH. Reflux at both ends as in...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.
 Spring '11
 Levicky
 The Land

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