Unformatted text preview: ,D 6 5 75
0.01
=
= 2.86
log(3.4 / 0.15)
log For component distribution, use the following rearrangements of the Fenske equation: xD
xB
xD
xB =
C2 H 4 HCl xD
xB x
=D
xB N
α C2min4 ,C2 H5Cl = 0.01
H
C 2 H 5Cl α
C 2 H 5Cl N min
HCl,C 2 H 5Cl 51
.
015
. 3.8
= 0.01
015
. 2 .86 = 240 2 .86 = 103 Because the separation between the two key components is quite sharp, assume that the two
nonkeys do not distribute at minimum reflux. The relative volatilities with respect to the heavy
key are: 34 for A, 25.3 for B, and 22.7 for C. Now, use the Class 2 Underwood equation, Eq. (928),which for a bubblepoint feed is: Exercise 9.18 (continued) Analysis: (continued)
α i , D zi , F
i αi ,D − θ = 1− q = 0 = 34(0.05) 25.3(0.05) 22.7(0.10) 1.00(0.80)
+
+
+
(1)
.
34 − θ
25.3 − θ
22.7 − θ
100 − θ There are three roots to Eq. (1). The one of interest is the one between 22.7 and 1.0, which from
a spreadsheet calculation is θ = 4.318. The composition of the distillate by material balances is:
Component
distillate,
mole fraction
C2H4 , A
0.25405
HCl , B
0.25405
C2H6 , C
0.48195
C2H5Cl , D
0.00995
Apply Underwood Eq. (929):
α i , D xi , D 34(0.25405) 25.3(0.25405) 22.7(0.48195) 100(0.00995)
.
=
+
+
+
= 1 + Rmin
34 − 4.318
25.3 − 4.318
22.7 − 4.318
1.00 − 4.318
i αi ,D − θ
= 0.291 + 0.306 + 0.595 − 0.003 = 1189 = 1 + Rmin
.
Therefore, Rmin = 0.189. Assume this is equal to the external minimum reflux ratio.
Operating reflux ratio = 1.5(0.189) = 0.284. In the Gilliland equation, (934),
X = (R  Rmin)/(R + 1) = (0.284  0.189)/0.284 + 1) = 0.074. Using Eq. (9.34), Y = 0.581 =
(N  Nmin)/(N + 1).
Solving, N = 8.2 stages.
For feed stage location, apply the Kirkbride equation (936):
From above, using the Fenske distribution with a material balance, D/B = 0.245.
Also, mole fractions in the bottoms product are: xC = 0.0064 and xD = 0.9936
NR
=
NS zD , F xC , B zC , F xD , D 2 0.206 B
D = 0.80
0.10 0.0064
0.00995 2 1
0.245 0.206 = 171
. Therefore, of 8.2 equilibrium stages, (1.71/2.71)(8.2) = 5.2 stages are in the rectifying section.
With a partial condenser acting an equilibrium stage, the feed is about stage 4 from the top stage
in the column. Exercise 9.19
Subject: Use of the FUG method for the distillation of a ficticious ternary mixture. Given: 100 kmol/h of a bubblepoint feed of:
Component
A
B
C Feed rate, kmol/h
40
20
40 αi,C
5
3
1 Find: (a) Product distribution for a distillate rate of 60 kmol/h and 5 minimum theoretical
stages.
(b) Minimum reflux and boilup ratios for separation of part (a).
(c) Number of equilibrium stages and feedstage location for R = 1.2Rmin.
Analysis: (a) With a distillate rate of 60 kmol/h, the LK is B and the HK is C. From the
rearranged form of the Fenske equation, given by Eq. (914), using component C as the
reference,
dA
d
d
= C 55 = 3125 C
bA
bC
bC
dB
d
d
= C 35 = 243 C
bB
bC
bC
The following 4 material balances also apply:
dA + bA = 40
dB + bB = 20
dC + bC = 40
dA + dB + dC = 60
Solving these 6 equations, tw...
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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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