Unformatted text preview: ol/h of nC10 absorbent at
90oF. Absorber operating at 4 atm. Kvalues in Fig.2.8.
Find: Percent absorption for each component for (a) 4 stages, (b) 10 stages, and (c) 30 stages.
Analysis: : Apply Kremser's method, with absorption factors, A = L/KV, computed for L =
entering absorbent rate = 500 kmol/h, V = entering gas rate = 1,000 kmol/h, and Kvalues from
Fig. 2.8 at the average of the two entering temperatures, (90 + 70)/2 = 80oF and a pressure of 4
atm = 59 psia. The feed gas composition and the resulting Kvalues and values of A are as
follows:
Component
C1
C2
C3
nC 4
nC 5
Total: f, kmol/h
250
150
250
200
150
1000 Kvalue
35
7.7
2.2
0.65
0.20 A = L/KV
0.0142
0.065
0.227
0.769
2.50 The fraction of a gas component not absorbed, φA, is given by the Kremser Eq. (548), and by
material balance the amount absorbed follows:
A −1
φ A = N +1
and Percent absorbed = 100 1 φ A
A −1
The results for all three cases are as follows: Component
C1
C2
C3
nC 4
nC 5 N = 4 stages:
φA
0.986
0.935
0.773
0.316
0.0155 %
absorbed
1.4
6.5
22.7
68.4
98.4 N = 10 stages:
N = 30 stages:
%
%
φA
φA
absorbed
absorbed
0.986
1.4
0.986
1.4
0.935
6.5
0.935
6.5
0.773
22.7
0.773
22.7
0.245
75.5
0.231
76.9
0.0001
99.99 0.0000
100.0 The above results show that as N increases, the percent absorption of the heavier components
increases, but the absorption of the lighter components does not increase. However, beyond 10
stages, little change occurs. Exercise 9.24
Subject: Flow rate of stripping steam by the Kremser method
Given: Feed of composition below enters a flash drum operating at 150oF and 2 atm.
Equilibrium liquid is sent to a 5equilibriumstage stripper operating at 2 atm to give 0.5 kmol/h
of nC5 in stripper bottoms.
Find: Flow of steam needed in the stripper.
Analysis: Use Chemcad with the SRK equation of state to flash the feed to determine the liquid
feed to the stripper. The result is:
Component
C1
C2
C3
nC 4
nC 5
nC12
Total: Feed, kmol/h Distillate, kmol/h
13.7
13.6
101.3
98.2
146.9
134.9
23.9
19.1
5.7
3.4
196.7
0.4
488.2
269.6 Bottoms, kmol/h
0.1
3.1
12.0
4.8
2.3
196.3
218.6 The stripping steam enters the stripper at 2 atm and 300oF. For the stripper, the average
temperature of the two feeds = (150 + 300) = 225oF. From Fig. 2.8, the Kvalue of nC5 at 2 atm
and 225oF = 3.0. The average stripping factor for npentane can be taken as:
S = KV/L = 3V/218.6 = 0.0137 V
The fraction of nC5 not stripped = φS = 0.5/2.3 = 0.217
Use Kremser Eq. (550) to compute the value of S needed for N = 5 theoretical stages.
By trial and error, the result from:
φS =
is S = 0.89. Therefore, S −1
= 0.217
S N +1 − 1 Flow rate of stripping steam = V = S/0.0137 = 0.89/0.0137 = 65 kmol/h Exercise 9.25
Subject: Stripping of a hydrocarbon liquid with superheated steam.
Given: 1000 kmol/h of feed liquid at 250oF with composition below. 100 kmol/h of
superheated steam at 300oF and 50 psia. Stripper operates at 50 psia and has 3 equilibrium
stages.
Assumptions: Negligible stripping of nC10 and negligible condensation of steam
Find: Flow ra...
View
Full Document
 Spring '11
 Levicky
 The Land

Click to edit the document details