Unformatted text preview: sumed to have a MW of 300. The Kvalues for the absorbent oil are rough guesses.
The values are compared to those in Example 5.3.
Component
C1
C2
C3
nC 4
nC 5
Oil Kvalues at 400 psia:
T = 97.5F T = 125oF
6.65
8.0
1.64
2.0
0.584
0.73
0.195
0.26
0.0713
0.098
0.0001
0.0002 T = 150oF
8.8
2.4
0.90
0.34
0.135
0.0003 Absorption and stripping factors are obtained with Eqs. (538) and (551), respectively,
using the above Kvalues with the entering vapor and liquid flow rates.
Using the same procedure as in Example 5.3 and letting:
The % absorption of a component in the entering gas = l N / l0 + υ N +1 × 100%
The % stripping of the oil in the absorbent = υ 1 oil / l0 oil × 100% The following results are obtained and compared with those of Example 5.3.
% absorption for N = 6 stages at 400 psia:
Component
T = 97.5F T = 125oF T = 150oF
C1
3.10
2.58
2.34
C2
12.60
10.31
8.59
C3
35.26
28.24
22.91
nC 4
87.94
74.09
59.32
nC 5
95.24
93.08
88.93
Oil
0.05
0.10
0.15
These results show that increasing temperature decreases the % absorption of gaseous
components and increases the % stripping of the absorbent. Exercise 5.18
Subject: Multistage absorption of a lighthydrocarbon gas mixture with nheptane. Given: VN+1 = 1,000,000 lbmol/day of feed gas containing in mole %, C1 94.9, C2 4.2,
C3 0.7, nC4 0.1, nC5 0.1. Absorption at 550 psia and 30oF with N = 10 equilibrium
stages and Kvalues below.
Assumptions: Applicability of Kremser's method.
Find: Absorbent flow rate, L0 , and component distribution for 50% absorption of C2.
Analysis: From Eq. (548), φ A C2 = fraction of C 2 not absorbed = 0.50 =
Solving and using Eq. (538), AC 2 = 0.50 = AC2 − 1
N
AC2+1 − 1 L0
L0
=
KC 2VN +1 (0.36)(1,000,000) Therefore, L0 = 0.50(0.36)(1,000,000) = 180,000 lbmol/day
To compute the distribution of the other components,i, between products V1 and L10 ,
compute Ai = L0/KiVN+1 and φA for i from Eq. (548). Then Eq. (554) is used to compute
the component flow rates in the exit vapor. Then an overall material balance is used to
compute the component flow rates in the exit liquid:
l N = l0 + υ N +1 − υ 1
The results are as follows, using a spreadsheet:
Flow rates, lbmol/day:
Component
Kvalue
A
l10
φA υn+1 = υ11
υ1
C1
2.85 0.063
0.937
949,000 889,200
59,800
C2
0.36
0.50
0.500
42,000
21,000
21,000
C3
0.066
2.73 0.0000276
7,000
0.2 6,999.8
nC 4
0.017 10.59
0
1,000
0
1,000
nC 5
0.004
45.0
0
1,000
0
1,000 Exercise 5.19
Subject: Stripping of a hydrocarbon liquid with steam at 50 psia and 300oF. Given: In Fig. 5.8(b), LN+1 = 1,000 kmol/h of 0.03% C1, 0.22% C2, 1.82% C3, 4.47%
nC4, 8.59 nC5, and 84.87% nC10 , in mol%. V0 = 1,000 kmol/h of superheated steam.
Stripper with N = 3 equilibrium stages. Kvalue for nC10 = 0.20. Other Kvalues from
Fig. 2.8.
Assumptions: Applicability of Kremser's equation. Negligible absorption of steam.
Find: Compositions and flow rates of exiting stripped liquid and rich gas. Dew point.
Analysis: For a stripper, Eq. (555) applies, which for no hydro...
View
Full Document
 Spring '11
 Levicky
 The Land

Click to edit the document details