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0.0385
0.1282
0.4098
1.3160 S 0.0044 φA
0.9914
0.9615
0.8718
0.5916
0.0541 φS 0.9956 υ1 l6
lbmol/h
14.0
6.5
12.3
21.2
22.7
497.8
574.5 υ1 l3
lbmol/h
30.0
12.0
22.6
33.2
22.9
498.9
619.6 lbmol/h
1,646.0
161.5
83.7
30.8
1.3
2.2
1,925.5 N = 3 and P = 150 psia: Component Kvalue A C1
C2
C3
nC 4
nC 5
nC10
Total 14.0
3.5
1.05
0.33
0.105
0.00055 0.0179
0.0714
0.2381
0.7576
2.3810 S 0.0022 φA
0.9821
0.9284
0.7644
0.3615
0.0443 φS 0.9978 lbmol/h
1,630.0
156.0
73.4
18.8
1.1
1.1
1,880.4 Exercise 6.13 (continued)
Analysis: (continued)
(c) N = 6 and P = 150 psia: Component
C1
C2
C3
nC 4
nC 5
nC10
Total Kvalue
14.0
3.5
1.05
0.33
0.105
0.00055 A
0.0179
0.0714
0.2381
0.7576
2.3810 S 0.0022 φA
0.9821
0.9284
0.7619
0.2829
0.0032 φS 0.9978 υ1 lbmol/h
1,630.0
156.0
73.1
14.7
0.1
1.1
1,875.0 l6
lbmol/h
30.0
12.0
22.9
37.3
23.9
498.9
625.0 Exercise 6.14
Subject: Absorption of a light hydrocarbon gas by nC10. Given: 1,000 kmol/h of rich gas at 70oF with, in kmol/h, 250 C1, 150 C2, 200 C3, 200 nC4, and
150 nC5. 500 kmol/h of nC10 at 90oF. Absorber operates at 4 atm. Kvalue of nC10 at 80oF and
4 atm = 0.0014.
Assumptions: Using the Kremser method, use V = 1,000 kmol/h and L = 500 kmol/h, with an
average temperature of 80oF for Kvalues from Fig. 2.8 or other source.
Find: Percent absorption of each component for 4, 10, and 30 equilibrium stages.
Analysis: The Kvalues and absorption factors, from Eq. (615), A = L/KV, are:
Component
C1
C2
C3
nC4
nC5 Kvalue
38
7.6
2.25
0.64
0.195 A
0.0132
0.0658
0.2222
0.7813
2.564 For each component in the feed gas, the percent absorption is given by Eq. (613) x 100%,
A N +1 − A
Percent absorbed = N +1
× 100%
A −1 (1) Using Eq. (1) with the values of A from the above table, the results are:
Component
C1
C2
C3
nC 4
nC 5 Percent absorbed:
for N = 4
for
1.32
6.58
22.03
65.14
96.30 N = 10
1.32
6.58
22.22
76.58
99.995 for N = 30
1.32
6.58
22.22
78.12
100.00 The number of stages affects only the heavier hydrocarbons that are absorbed to the greatest
extent. Exercise 6.15
Subject: Back calculation of tray efficiency from performance data on absorption of propane.
Given: Large commercial absorber with Na = 16 actual plates. From Example 6.3, average
total liquid rate, L, is 446.7 lbmol/h, average total gas rate, V, is 866.5 lbmol/h. Propane in gas
feed = 213.8 lbmol/h. Propane in exit liquid = 43 lbmol/h. P = 93.7 psia. Average T = 111oF.
Find: Tray efficiency from performance data. Estimated tray efficiency from DrickamerBradford and O'Connell correlations.
Analysis: Fraction of propane absorbed = 43/213.8 = 0.20
From Fig. 2.8, for propane, Kvalue = 2.0
From Eq. (615), A = L/KV = 446.7/(2.0)(866.5) = 0.258
From Eq. (613),
A N +1 − A 0.258 N +1 − 0.258
Fraction absorbed = 0.20 = N +1
=
A −1
0.258 N +1 − 1 (1) Solving Eq. (1), N = Nt = 0.93
From Eq. (61), Eo = Nt /Na = 0.93/16 = 0.058 or 5.8%
From DrickamerBradford correlation, Eq. (622), with µ L = 1.4 cP from Example 6.3,
Eo , % = 19.2  57.8 log µL = 19.2  57.8 log(1.4) = 10.8 %
Fr...
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 Spring '11
 Levicky
 The Land

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