Unformatted text preview: en below is adiabatically
flashed to 150 psia before entering a stripper containing 7 equilibrium stages and a partial
reboiler. The bottoms flow rate from the stripper = B = 99.3 lbmol/h.
Find: Compositions of the vapor and liquid products from stripper.
Analysis: First, adiabatically flash the feed given below, using Chemcad with the SRK equation
of state. The results of the flash are as follows, where the calculated flash temperature = 26.3oF:
Lbmol/h:
Component
Feed Vapor
C1
59.5
37.67
C2
73.6
14.83
C3
153.2
8.90
nC 4
173.5
2.58
nC 5
58.2
0.23
nC 6
33.6
0.04
Total:
551.6
64.25 Liquid
21.83
58.77
144.30
170.92
57.97
33.56
487.35 The liquid from the adiabatic flash is sent to the reboiled stripper where it is separated into an
overhead vapor and a liquid bottoms. The overhead vapor is mixed with the vapor from the
adiabatic flash to give the final vapor product.
To solve the reboiled stripper by the group method, apply Edmister's method, rather than
Kremser's method, because we do not know the composition and flow rate of the vapor leaving
the reboiler and entering the column. For a reboiled stripper, Eq. (564) applies:
lF S B φ AX + 1
KV
=
(1) where S B = B B (2)
b
φ SX
B
For φAX and φΕΞ, use average absorption and stripping factors computed for L = entering flash
liquid = 487.35 kmol/h, V = vapor leaving the reboiler, which can be taken as L  B = 487.35 99.30 = 388.05 kmol/h, and Kvalues from Fig. 2.8 at an average temperature of say 200oF and a
pressure of 150 psia. The feed gas composition and the resulting Kvalues, values of S = KV/L
and A = 1/S, and corresponding values of φAX and φEX , computed from Eqs. (548) and (550),
φ AX =
are as follows: A −1
S −1
and φ SX = N +1
N +1
A −1
S
−1 Exercise 9.27 (continued)
Analysis: (continued)
Component
C1
C2
C3
nC 4
nC 5
nC 5
Total: lF , lbmol/h
21.83
58.77
144.30
170.92
57.97
33.56
487.35 Kvalue
22
7.1
2.9
1.23
0.52
0.23 S = KV/L
17.5
5.65
2.31
0.979
0.414
0.183 A = 1/S
0.0571
0.177
0.433
1.021
2.415
5.46 φSX
0.000
0.000
0.004
0.152
0.587
0.817 φAX
0.943
0.823
0.569
0.134
0.003
0.000 To compute values of SB from Eq. (2), the assumed value of VB is critical. We seek the value of
VB that will give values of SB , which when substituted into Eq. (1) will give values of bi that will
sum to the specified value of B = 99.3 lbmol/h. Assume the temperature at the bottom is 250oF
and obtain from Fig. 2.8 values of K at 250oF and 150 psia. To begin, assume a value of VB =
500 lbmol/h, calculate B =
bi , and then adjust the values of bi so they sum to the specified B.
The results are as follows:
Component
C1
C2
C3
nC 4
nC 5
nC 5
Total: KB
23
8.8
4.0
1.85
0.88
0.45 SB
116
44.3
20.1
9.31
4.43
2.27 lF/b
3100
14.8
1.73
1.22 lbmol/h:
b
0.00
0.00
0.05
11.55
33.51
27.44
72.55 Adj. b
0.00
0.00
0.07
25.92
45.87
27.44
99.30 The final product distribution is as follows, where it is compared to the result obtained by
using Chemcad with the Tower model to also calculate the reboiled stripper:
Edmis...
View
Full
Document
This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.
 Spring '11
 Levicky
 The Land

Click to edit the document details