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Unformatted text preview: sults are obtained: Component
Propane
Isobutane
nButane
Isopentane
nPentane
nHexane
nHeptane
nOctane αi, iC5 at top
9.06
3.55
1.00
0.783
0.272
0.094
0.038 αi, iC5 at bottom αi, iC5 average
3.93
5.97
2.24
2.82
1.00
0.848
0.448
0.228
0.128 0.815
0.349
0.146
0.070 di , kmol/h
2499.996
399.63
594
15
6.12
0.001
8.2x107
1.92x1010 bi , kmol/h
4.2x103
0.37
6
85
193.88
39.999
50
40 Exercise 9.8
Subject:
Type condenser and operating pressure. Minimum equilibrium stages and
distribution of nonkey components using the Fenske equation for distillation of a light
hydrocarbon mixture.
Given: Feed composition and split of two key components in Fig. 9.24. Kvalues from Figs.
2.8 and 2.9.
Find: Type condenser.
Operating pressure.
Minimum equilibrium stages.
Distribution of nonkey components at total reflux.
Analysis: To determine the type condenser and operating pressure, follow the algorithm in Fig.
7.16. The estimated distillate and bottoms compositions from the given data are:
Component Feed, lbmol/h Methane
Ethylene
Ethane
Propylene
Propane
nButane Distillate,
lbmol/h
1000
2500
1999
5
0
0 1000
2500
2000
200
100
50 Bottoms, lbmol/h
0
0
1
195
100
50 Using Fig. 2.8, it is quickly shown that at 120oF, both the bubblepoint and the dewpoint
pressures for the estimated distillate composition are greater than 1000 psia because the Kvalue
for ethane is greater than 1.0 at 120oF and 1000 psia. Therefore, select the operating pressure as
415 psia and use a partial condenser.
Run a dew point on the distillate at 415 psia. Use Eq. (413) in the following form: i di
1000 2500 1999
5
=D=
+
+
+
= 5504
Ki
K C1
KC =
KC 2
KC =
2 (1) 3 By trial and error, using Fig. 2.9 for Kvalues, the temperature that satisfies Eq. (1) is 0oF, which
gives:
di
1000 2500 1999
5
=D=
+
+
+
= 5505
5.0
105 0.69 018
.
.
i Ki
The relative volatility of the two key components is αC2, C3= = 0.69/0.18 = 3.83 Exercise 9.8 (continued) Analysis:
(continued)
Now run a bubblepoint on the estimated bottoms composition at 415 psia. Use Eq. (412) in the
following form:
bi Ki = B = KC 2 + 195KC= + 100 KC3 + 50 KnC4 = 346
(2)
3 i By trial and error, using Fig. 2.8 for Kvalues, the temperature that satisfies Eq. (1) is 155oF,
.
.
which gives:
bi Ki = B = 2.7 + 195(115) + 100(101) + 50(0.39) = 347
i The relative volatility of the two key components is αC2, C3= =2.7/1.15 = 2.35
The geometric mean relative volatility = [(3.83)(2.35)]1/2 = 3.0
From the Fenske equation, (912),
d C2 N min = bC= d C= log bC2 3 3 log α C log
= =
2 , C3 1999 195
5
1
log 3.0 = 10. 3 To compute the distribution of nonkey components at total reflux, use for the lighter than light
key components, LLK, Eq. (915), and use Eq. (916) for the heavier than heavy key, HHK, with
C= as the reference component, r, and the above value of Nmin.
3
fi
fi
=
Thus, for the LLK, i, bi =
(3)
5
d C= N
10.3
3
min
1+
α i,C= 1 + 195 α i,C=
3
3
b
C=
3 fi
For the HHK, i, di = d C=
3 bC = α iN min
,C =
3 3 1+ d C=
3 bC= α iN min
,C =
3 5
α 10.3
=
195 i ,C3...
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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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