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Unformatted text preview: ride equation:
By material balances on A and W, B/D = 0.753/0.247 = 3.05
NR
=
NS zW, F xA, B zA, F xW, D 2 0.206 B
D = 0.75
0.25 0.02
0.05 0.206 2 3.05 = 1.08 (c) McCabeThiele method:
From the McCabeThiele graph of Exercise 9.6, shown on the next page, where the total number
of minimum stages = 5.
N R 4.2
=
= 5.25
N S 0.8
The McCabeThiele method is the most accurate. The Fenske equation is better than the
Kirkbride equation because the Kirkbride equation does not take into account the drastic change
in relative volatility from the distillate to the bottoms. Exercise 9.20 (continued)
Analysis: McCabeThiele method (continued) Exercise 9.21
Subject: Derivation of Eq. (937) for the effective stripping factor Given: Equations (546) and (547)
Analysis: By analogy to Eqs. (546) and (547), the fraction of species in the entering feed
liquid that is not stripped is given for just 2 stages by:
φS = 1
1
=2
S N S1 + S N + 1 Se + Se + 1 Therefore,
Se2 + Se + 1 = S N S1 + S N + 1 : Solving for Se ,
Se2 + Se − S N S1 + 1 = 0
Se = −1 + 1 + 4 S N S1 + 1 Which is Eq. (937). 2 = 0.25 + S N S1 + 1 − 0.5 Exercise 9.22
Subject: Effect of number of stages and pressure on multicomponent absorption
Given: Feed gas of 2,000 lbmol/h of light paraffin hydrocarbons at 60oF, and 500 lbmol/h of
nC10 absorbent at 90oF. Kvalues in Fig. 2.8.
Find: Separation at following conditions:
(a) 6 stages and 75 psia.
(b) 3 stages and 150 psia.
(c) 6 stages and 150 psia
Analysis: Apply Kremser's method, with absorption factors, A = L/KV, computed for L =
entering absorbent rate = 500 lbmol/h, V = entering gas rate = 2,000 lbmol/h, and Kvalues from
Fig. 2.8 at the average of the two entering temperatures, (90 + 60)/2 = 75oF and each of the two
pressures. The feed gas composition and the resulting Kvalues and values of A are as follows:
Component
C1
C2
C3
nC 4
nC 5
Total: Feed, lbmol/h
1660
168
96
52
24
2000 K at 75 psia
29
5.8
1.68
0.48
0.145 A at 75 psia
0.00862
0.0431
0.149
0.521
1.724 K at 150 psia
15
3.1
0.93
0.27
0.08 A at 150 psia
0.0167
0.0806
0.269
0.926
3.13 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 l N = 1 − φ A f
A −1
The results for all three cases are as follows: Component
C1
C2
C3
nC 4
nC 5
Total: Case (a):
Case (b):
φA lN, lbmol/h
φA lN, lbmol/h
0.991
14.9
0.983
28.2
0.957
7.2
0.919
13.6
0.851
14.3
0.731
25.8
0.484
26.8
0.280
37.4
0.0164
23.6 0.0224
23.5
86.8
128.5 Case (c):
φA lN, lbmol/h
0.983
28.2
0.919
13.6
0.731
25.8
0.178
42.7
0.00072
23.98
134.28 The results show that doubling the pressure is much more effective than doubling the number of
equilibrium stages. Even Case (c) results in little additional absorption over Case (b) when the
stages are doubled for the same pressure. Exercise 9.23
Subject: Effect of the number of stages on the absorption of a light normal paraffin
hydrocarbon gas by the Kremser method.
Given: 1000 kmol/h of gas at 70oF of composition below, and 500 km...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.
 Spring '11
 Levicky
 The Land

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