Unformatted text preview: 8 mole fraction of 1 in instantaneous distillate.
Second specification of 100 kmol/h of boilup
Stop operation when mole fraction of 1 in the bottoms (residue) = 0.05
Step size = 0.02 h recorded every 5 increments with stop tolerance of 0.001
Results are as follows:
(a) 7.28 h for operation step 1.
(b) kmol of cumulative distillate (from material balance) = 200 kmol at 78oC.
(c) Initial (minimum) reflux ratio = 2.35
Final (maximum) reflux ratio = 3.7
(d) The variation of the distillate rate, computed from the reflux ratios, is as follows:
0
2
4
6 7.28
Time, h
29.9 28.6 26.9 24.1 21.3
Distillate, kmol/hr Exercise 13.16
Subject: Batch rectification with variable reflux ratio and constant distillation composition.
Given: 500 lbmol charge of 48.8 mol% A and 51.2 mol% B. αA,B = 2.0. Total condenser, 7
theoretical trays in column, and partial reboiler. Constant 95 mol% A in distillate. Boilup rate =
213.5 lbmol/h.
Assumptions: Negligible holdup on trays and in condensing system. Perfect mixing in
reboiler. No pressure drop. Constant molar overflow.
Find: When mole fraction of A in boiler drops to 0.192:
(a) Time in h.
(b) Total amount of distillate, lbmol.
Analysis: (a) Using the procedure outlined in Section 13.3, a McCabeThiele construction is
applied, as in Fig. 13.7, to obtain the relationship between the reflux ratio and the mole fraction
of A in the residue for xD = constant = 0.95 and 7 + 1 = 8 equilibrium stages. The construction
for the initial residue composition of xW = x0 = 0.488 is shown on the following page, where the
correct slope of the operating line was obtained by trial and error to be L/V = 0.706. From Eq.
(77), the corresponding reflux ratio = R = (L/V)/[1  (L/V)] = 0.706/(1  0.706) = 2.40. In the
table below, other values of increasing L/V by the McCabeThiele construction give the values
of xW listed below until the final value of 0.192 is reached. These values are used to obtain
time, t, for distillation starting from a bubblepoint charge, with Eq. (1316): t= W0 x D − x0
V dxW x0
xWt 1 − L / V ( x D − xW
L/V 0.706
0.744
0.783
0.828
0.870
0.898 xW
0.488
0.425
0.365
0.290
0.230
0.192 2 R
2.4
2.9
3.6
4.8
6.7
8.8 = 500 0.95 − 0.488
2135
.
I= dxW 0.488
0.192 1 − L / V 0.95 − xW 1
1 − L / V ( x D − xW
15.9
14.2
13.5
13.4
14.8
17.1 2 2 ∆ t,
h
1.03
0.90
1.09
0.92
0.66 The above table includes the numerical integration of Eq. (1) by the trapezoidal rule. For
example, the ∆t of 0.95 in the 3rd row is obtained from:
500 0.95 − 0.488
∆t =
Iavg (∆xW) = 1.08[(15.9 + 14.2)/2](0.488  0.425) = 1.03 h
213.5 (1) Exercise 13.16 (continued) Analysis: (continued)
The total time is the sum of ∆t values from the above table:
t = 1.03 + 0.90 + 1.09 + 0.92 + 0.66 = 4.60 h
This is close to the value of 4.75 h obtained using the batch distillation program of Chemcad. (b) The total distillate is obtained by material balance on component A, giving an equation like
(1313): D = W0 1 − xD − x0
xD − xW = 500 1 − 0.95 − 0.48...
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 Spring '11
 Levicky
 The Land

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