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composition of the cumulative distillate would be 60.5 mol% ethanol. Exercise 13.12 (continued)
Analysis: Distillation with 2 stages plus a reboiler (continued) Exercise 13.13
Subject: Batch distillation of a mixture of acetone (A) and ethanol (E) at 101 kPa in a column
with equilibrium stages and a reboiler.
Given: Feed of 50 mol% A and 50 mol% E. Vapor-liquid equilibrium data at 101 kPa.
Assumptions: Perfect mixing in the still. No holdup on the stages or in the condenser.
Find: (a) Number of equilibrium stages if the instantaneous distillate composition is 90 mol%
acetone when the residue is 10 mol% acetone for a reflux ratio, L/D = 1.5 times the minimum
(b) Plot of reflux ratio versus the still pot composition and amount of residue left if the
column has 8 stages (9 including the reboiler) and the reflux rate is varied to maintain the
distillate composition constant at 90 mol% acetone.
(c) Total vapor generated for distillation with 8 stages (9 including the reboiler) at
constant reflux ratio to obtain a residue of 10 mol% acetone and a cumulative distillate
composition of 90 mol% acetone
Which method of operation requires more energy.
A more efficient operating policy to produce a distillate of 90 mol% acetone.
(a) First compute the minimum L/V. The slope of the operating line for this
condition is given by: L
min ( y at 0.9) - ( y in equilibrium with x at 0.10) 0.9 − 0.25
( x at 0.9) - ( x at 0.10)
0.9 − 01
. From Eq. (7-27), Rmin = 0.8125/(1 - 0.8125) = 4.33. Therefore, R = 1.5Rmin = 1.5(4.33) = 6.5.
Operating L/V = R/(R+1) = 6.5/7.5 = 0.867. On the next page, the number of equilibrium stages
are stepped off by the McCabe-Thiele method, where the given vapor-liquid equilibrium data are
plotted, operating line has a slope of 0.867, and the steps start at x = 0.9 and end at x = 0.1.
Almost 9 equilibrium stages are stepped off. Analysis: (continued) Exercise 13.13 (continued) Analysis: (continued) Exercise 13.13 (continued) (b) For constant acetone mole fraction in the distillate = 0.90, and 8 equilibrium stages plus a
reboiler, the McCabe-Thiele diagram can be used to determine the residue composition and
amount of residue as a function of reflux ratio. For example, as shown at the bottom of the
following page, when L/V = 0.90, R = 0.9/(1-0.9) = 9, and xW = 0.065. From Eq. (13-6),
x − (W / W0 ) xW
For xW = 0.065, W / W0 = 0.48
( yD )avg = 0.90 = 0
1 − (W / W0 )
W0 0.9 − xW
For other reflux ratios, the following results are obtained and plotted on the next page.
0.48 0.35 0.22 0.13 0.065 0.035 0.006
0.95 0.73 0.59 0.52
(c) The case of constant reflux ratio is a tedious trial and error problem. A guess of the reflux
ratio is made and a series of McCabe-Thiele constructions is carried out as in Fig. 13.4 or 13.5
with 9 total stages, such that the initial xW = 0.5 and the final xW = 0.1. This gives a series of
values of instantaneous yD and xW. Using these values, Eq. (13-2) is integrated to obtain W when
the final xW = 0.1is reached. Then, Eq. (13-6) is solved for y D avg . If the value is not 0.9, the
procedure is repeated with different values of the reflux ratio until the va...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.
- Spring '11
- The Land