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Unformatted text preview: .27, bnC5 = 0.2605 mol and dnC5 =
0.0095. Therefore, B = 0.0057 + 0.2605 + 0.20 + 0.10 = 0.5662. Repeating the calculations with
this value of B, the result is bnC4 = 0.00566, dnC4 = 0.19434, bnC5 = 0.2604, and dnC5 = 0.0096.
Using Eq. (1) for each of the nonkey components shows that they do not distribute to any extent.
Therefore the distillate and bottoms compositions at total reflux are:
xD
xB
Component
f, mol
d, mol
b, mol
C2
0.08 0.0800 0.184 0.0000 0.000
C3
0.15 0.1500 0.346 0.0000 0.000
nC 4
0.20 0.1943 0.448 0.0057 0.010
nC 5
0.27 0.0096 0.022 0.2604 0.460
nC 6
0.20 0.0000 0.000 0.2000 0.353
nC 7
0.10 0.0000 0.000 0.1000 0.177
Total
1.00 0.4339 1.000 0.5661 1.000 Exercise 9.16 (continued)
Analysis: (c)
Because the split of the two key components is quite sharp, assume a Class 2 separation
for the purpose of minimum reflux. Thus, only components methane to npentane will distribute
to the distillate. To determine the phase condition of the feed, perform an adiabatic flash with
bubblepoint conditions at 300 psia upstream of the feed valve to the column, and 120 psia
downstream of the valve, using Chemcad. The result is a V/F ratio for the feed of 0.334. From
Eq. (719), q = 1  (V/F) = 1  0.334 = 0.666 or 1  q = 0.334. From Underwood Eq. (928), α i ,nC5 zi , F
i α i ,nC5 − θ = 1 − q = 0.334 = 14.6(0.08) 5.9(015) 2.35(0.20) 1.00(0.27) 0.424(0.20) 0.187(010)
.
.
+
+
+
+
+
14.6 − θ
015 − θ
.
2.35 − θ
100 − θ
.
0.424 − θ
0187 − θ
. Solving this equation for the root between 2.35 and 1.00 gives θ = 1.5962
From Underwood Eq. ((929),
α i ,nC5 xi , D
14.6(0184)
.
5.9(0.346)
2.35(0.448)
1.00(0.022)
=
+
+
+
= 2.04 = 1 + Rmin
14.6 − 15962 5.9 − 15962 2.35 − 15962 1.00 − 15962
.
.
.
.
i α i , nC5 − θ
Therefore, the internal minimum reflux ratio = 2.04  1.00 = 1.04
Assume the external minimum reflux ratio = 1.04
(d) For the number of theoretical stages, use the Gilliland equation, (934), where for this
exercise, R = 1.5R = 1.5(1.04) = 1.56. X = (R  Rmin)/(R + 1) = (1.56  1.04)/(1.56 + 1) = 0.203.
Using Eq. (9.34), Y = 0.458 = (N  Nmin)/(N + 1).
Solving, N = 15.6 equilibrium stages. Exercise 9.17
Subject:
mixture. Use of the FUG method for the distillation of a light paraffin hydrocarbon Given: Feed mixture of:
Component
C 3 i C 4 nC 4 i C 5 nC 5
lbmol/h
5
15
25
20
35
Column pressure = 120 psia. Liquid distillate contains 92.5% of the nC4 in the feed. Bottoms
contains 82 mol% of iC5 in the feed. Use Chemcad with SRK instead of Figs. 2.8 and 2.9 for Kvalues.
Find: (a) Minimum number of equilibrium stages.
(b) Distribution of nonkey components.
(c) Minimum reflux ratio for a bubblepoint liquid feed.
(d) Number of equilibrium stages for R = 1.2 Rmin.
(e) Feedstage location.
Analysis: (a) To obtain the average relative volatilities, run bubble points on the following
assumed distillate and bottoms compositions, using the given keycomponent recoveries:
lbmol/h:
Component Feed Distillate
C3
5
5.000
iC 4
15
15.000
nC 4
25
23.125
iC...
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 Spring '11
 Levicky
 The Land

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