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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. Vaporliquid 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
value.
(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.
Analysis:
(a) First compute the minimum L/V. The slope of the operating line for this
condition is given by: L
V =
min ( y at 0.9)  ( y in equilibrium with x at 0.10) 0.9 − 0.25
=
= 0.8125
( x at 0.9)  ( x at 0.10)
0.9 − 01
. From Eq. (727), 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 McCabeThiele method, where the given vaporliquid 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 McCabeThiele 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/(10.9) = 9, and xW = 0.065. From Eq. (136),
x − (W / W0 ) xW
W
0.4
Solving,
=
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.
R=L/D
2.33
3.0
4.0
5.67
9.0
19
∞
xW
0.48 0.35 0.22 0.13 0.065 0.035 0.006
W/W0
0.95 0.73 0.59 0.52
0.48
0.46
0.45
(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 McCabeThiele 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. (132) is integrated to obtain W when
the final xW = 0.1is reached. Then, Eq. (136) 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
 Levicky
 The Land

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