Unformatted text preview: be favored
over the Gilliland correlation, which emphasizes rectification, rather than stripping. Subject:
mixture. Exercise 9.16
Use of the FUG method for the distillation of a normal paraffin hydrocarbon Given: Bubblepoint liquid feed at 300 psia, with composition in mole fractions:
C3
nC 4
nC 5
nC 6
nC 7
Component
C2
Mole fraction
0.08 0.15 0.20 0.27 0.20 0.10
Find: (a) For a sharp separation between nC4 and nC5, determine column pressure and type
condenser for condenser outlet temperature of 120oF.
(b) At total reflux, determine the separation for 8 equilibrium stages, with 0.01 mole
fraction of nC4 in the bottoms.
(c) Minimum reflux ratio for separation in Part (b).
(d) Number of equilibrium stages for R/Rmin = 1.5 using Gilliland correlation.
Analysis: (a) For a sharp split, the distillate composition is as follows:
Component
C2
C3
nC 4
Total: xd
0.186
0.349
0.465
1.000 K at 120oF,
250 psia
2.8
1.13
0.34 K xd
0.520
0.394
0.158
1.072 In this table, a distillate bubblepoint pressure of 250 psia is assumed and the bubblepoint
equation (412) is applied, using Kvalues in Fig. 2.8. The summation is 1.072. In order for the
sum to be 1.0, an even higher pressure is needed, since the Kvalue is almost inversely
proportional to pressure. But, in Fig. 7.16, we are already > 215 psia, therefore use a partial
condenser. To determine the pressure, run a dew point using Eq. (413). Component
C2
C3
nC 4
Total: yD
0.186
0.349
0.465
1.000 K at 120oF,
140 psia
4.7
1.58
0.53 yD/K
0.040
0.221
0.877
1.138 K at 120oF,
120 psia
5.4
1.80
0.60 yD/K
0.030
0.194
0.775
0.999 On the second trial in the above table, the dewpoint pressure is found to be 120 psia. Set the
column pressure at this value with a partial condenser. Exercise 9.16 (continued)
Analysis: (continued)
(b) Run a bubble point on an assumed bottoms composition, using Eq. (412):
xB K at 320oF,
xB K K at 290oF,
xB K
Component
120 psia
120 psia
nC 4
0.010
3.3 0.033
2.75
0.028
nC 5
0.469
1.7 0.797
1.38
0.647
nC 6
0.347
0.92 0.319
0.72
0.250
nC 7
0.174
0.48 0.084
0.36
0.063
Total:
1.000
1.233
0.988
o
A third trial gives a bottoms temperature of 295 F
The Kvalues and relative volatilities referred to the nC5, the heavy key, at distillate and bottoms
conditions are as follows for a pressure of 120 psia:
Component
K at 120oF, αI, nC5 at 120oF, K at 295oF, αI, nC5 at 295oF, Geometric
120 psia
120 psia
120 psia
120 psia average α
C2
5.40
25.7
12.0
8.28
14.6
C3
1.80
8.6
5.8
4.0
5.9
nC 4
0.60
2.86
2.8
1.93
2.35
nC 5
0.21
1.00
1.45
1.00
1.00
nC 6
0.074
0.352
0.74
0.51
0.424
nC 7
0.028
0.133
0.38
0.262
0.187
log
From the Fenske Eq. (912), N min = 8 = d nC4 bnC5
d nC5 bnC4 log α nC4 ,nC5 avg log
= d nC 4 bnC5
d nC5 bnC4
log 2.35 (1) Take as a basis, 1 mol of feed. Assume that the distribution of the nonkeys is negligible. That is,
only the LK and the HK are found in both the distillate and bottoms. Assume the total bottoms,
B, is 0.57 mol. Then bnC4 = 0.01(0.57) = 0.0057. Therefore, dnC4 = 0.20  0.0057 = 0.1943 mol.
Solving Eq. (1), bnC5/dnC5 = 27.3. Since bnC5 + dnC5 = fnC5 = 0...
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 Spring '11
 Levicky
 The Land

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