Unformatted text preview: 11.19
12.58
12.60
12.50
12.60 C3
26.09
34.28
35.26
35.30
35.30 nC4
51.46
77.04
87.94
93.34
98.64 nC5
74.29
93.08
95.24
95.33
95.33 Total Gas
36.26
40.32
40.98
41.15
41.33 oil
0.05
0.05
0.05
0.05
0.05 Additional stages have little effect on the absorption of C1, C2, and C3 because their
absorption factors are less than one. Exercise 5.14 (continued) Exercise 5.15
Subject: Multistage absorption of a hydrocarbon gas with a highmolecularweight oil.
Given: Hydrocarbon feed gas and absorbent oil of Example 5.3 in Fig. 5.11. Absorber
pressure of 400 psia and average temperature of 97.5oF, with Kvalues at these
conditions.
Find: Material balance for absorbent rate of 330 lbmol/h instead of the 165 lbmol/h and
3 stages instead of the 6 in Example 5.3. Compare results with Example 5.3 and discuss
effect of trading stages for absorbent flow rate.
Analysis: With two times the absorbent rate of Example 5.3, double the absorption
factors and halve the stripping factors given in Example 5.3. Then, using the same
procedure as in that example, the following results are obtained and compared.
Component
C1
C2
C3
nC 4
nC 5
Oil
Total N = 3 and L0 = 330 lbmol/h: N = 6 and L0 = 165 lbmol/h:
υ1 , lbmol/h l3 , lbmol/h υ1 , lbmol/h l6 , lbmol/h
150.1
9.9
155.0
5.0
278.1
91.9
323.5
46.5
93.8
146.2
155.4
84.6
1.5
23.6
3.0
22.0
0.3
6.3
0.3
5.5
0.1
328.3
0.1
164.1
523.9
606.2
637.3
327.7 The results show that considerably more absorption takes place when the absorbent rate
is doubled and the number of stages is halved. Exercise 5.16
Subject: Multistage absorption of a hydrocarbon gas with a highmolecularweight oil.
Given: Hydrocarbon feed gas and absorbent oil of Example 5.3 in Fig. 5.11. Absorber
pressure of 400 psia and average temperature of 97.5oF, with Kvalues at these
conditions.
Find: Minimum absorbent rate to reduce the propane flow rate to 155.4 lbmol/h in exit
vapor.
Analysis: The minimum absorbent rate corresponds to N = ∞ . Use Eq. (548) to
obtain the absorption factor, A = L/KV, needed for propane. Because A for propane < 1
for N = 6, it will be < 1 for N = ∞ , since the absorbent rate, L, will be lower. From Eq.
(548),
Fraction of propane not absorbed = 155.4/240 = 0.648 = φΑ = (Α − 1)/(ΑΝ+1 −1)
When A < 1, ΑΝ+1 = 0, therefore, Eq. (548) reduces to:
A = 1  φΑ = 1  0.648 = 0.352
From Eq. (538), L = AKV = 0.352(0.584)(800) = 165 lbmol/h = minimum rate
This is the same L as in Example 5.3 because for φΑ > about 0.5, N above 5 has no effect
on A. Exercise 5.17
Subject: Multistage absorption of a hydrocarbon gas with a highmolecularweight oil.
Given: Hydrocarbon feed gas and absorbent oil of Example 5.3 in Fig. 5.11. Absorber
pressure of 400 psia and isothermal operation with N = 6 equilibrium stages at:
(a) 125oF.
(b) 150oF.
Assumptions: Applicability of Kremser's method.
Find: % absorption of components and % stripping of oil.
Analysis: From Fig. 2.8, the following Kvalues are obtained, where for methane, the oil
is as...
View
Full
Document
 Spring '11
 Levicky
 The Land

Click to edit the document details