Unformatted text preview: o of which are nonlinear, in 6 unknowns gives the following
results: Component
A
B
C zF
0.4
0.2
0.4 xD
0.662
0.307
0.031 xB
0.007
0.039
0.954 (b) Assuming that component A does not distribute, the Underwood Eq. (928) for Class 2
applies: α i , D zi , F
i αi ,D − θ Analysis: (continued) = 1− q = 0 = 5(0.40) 3(0.20) 1(0.40)
+
+
5− θ
3− θ
1− θ (1) Exercise 9.19 (continued) There are two roots to Eq. (1). The one of interest is the one between 3 and 1.0, which from a
spreadsheet calculation is θ = 1.425. If component A does not distribute, the composition of the
distillate by material balances is as follows, if the total distillate rate of 60 kmol/h is maintained:
kmol/h:
xD
xB
Component
Feed Distillate Bottoms
A
40
40
0.0 0.6667 0.0000
B
20
18.3
1.7 0.3050 0.0425
C
40
1.7
38.3 0.0283 0.9575
Total:
100
60
40 1.0000 1.0000
Apply Underwood Eq. (929):
α i , D xi , D
i αi ,D − θ = 5(0.6667) 3(0.3050) 1(0.0283)
+
+
= 1 + Rmin
5 − 1425
.
3 − 1.425 1 − 1.425 = 0.9324 + 0.5810 − 0.0666 = 1447 = 1 + Rmin
.
Therefore, Rmin = 1.447  1.000 = 0.447. Assume this is also the external reflux ratio.
Assume constant molar overflow. Then, the reflux rate in the rectifying section = 0.447(60) =
26.8 kmol/h. The reflux rate in the stripping section = 100 + 26.8 = 126.8 kmol/h. The boilup
rate in the stripping section = 126.8  40 = 86.8 kmol/h. The boilup ratio = 86.8/40 = 2.17.
(c) Operating reflux ratio = 1.2(0.0.447) = 0.536. In the Gilliland equation, (934),
X = (R  Rmin)/(R + 1) = (0.536  0.447)/0.536 + 1) = 0.058. Using Eq. (9.34), Y = 0.60 =
(N  Nmin)/(N + 1).
Solving, N = 14 stages.
For feed stage location, apply the Kirkbride equation (936):
From above, using the Fenske distribution with a material balance, D/B = 1.5.
NR
=
NS zC , F xB, B zB , F xC , D 2 0.206 B
D = 0.40
0.20 0.039
0.031 2 1
15
. 0.206 = 1166
. Therefore, of 14 equilibrium stages, (1.166/2.166)(14) = 7.5 stages are in the rectifying section.
Assuming a total condenser, the feed stage is stage 7 or 8 from the top. Exercise 9.20
Subject:
Comparison of the Kirkbride, Fenske, and McCabeThiele methods for
determining feedstage location.
Given: A feed mixture of 25 mol% acetone (A) and 75 mol% water (W). Distillate to contain
95 mol% A, and bottoms to contain 2 mol% A. Infinite dilution activity coefficients from
Exercise 9.6.
Find: Ratio of rectifying to stripping stages by:
(a) Fenske equation.
(b) Kirkbride equation.
(c) McCabeThiele method.
Analysis: (a) Fenske equation:
From Exercise 9.6:
zF
xD
xB αi,W, D αi,W, B
Component
Acetone
0.25 0.95 0.02
1.48
27.7
Water
0.75 0.05 0.98
1.00
1.0
The relative volatility at feed conditions is also needed. Assume a bubblepoint liquid feed.
Also from vaporliquid, yx table of Exercise 9.6, αι,W = [(0.789/0.250)]/[(0.211/0.750)] = 11.2
From Eq. (935),
1/ 2
N R log ( 0.95 / 0.25 ) ( 0.75 / 0.05 ) log[(27.7)(11.2)]
1.756(1.246)
=
=
= 2. 96
1/ 2
1.213(0.6097)
N S log ( 0.25 / 0.02 )( 0.98 / 0.75 ) log[(1.48)(11.2)]
(b) Kirkb...
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 Spring '11
 Levicky
 The Land

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