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Unformatted text preview: the ratio of d to b for any component is
given by a rearrangement of Eq. (566),
d1
=
b AF
From Eq. (565),
From Eq. (559),
From just below Eq. (564), S B φ AX + 1 / φ SX (1) AC φ SE + 1 / φ AE AF = LF / K FVF (2) AC = LC / KCVC = LC / KC D (3) S B = K BVB / LB = K BVB / B (4) where values of K are assumed to be the same as for Example 5.4 because of the
assumption of no change in stage temperatures. Thus, AF , AC , and SB change only by
changes in the reflux rate, which changes stagewise vapor and liquid rates as given
below.
From Eqs. (548) and (550), A −1
A N +1 − 1
S −1
φ S = N +1
S
−1
φA = where A = L/KV and S = KV/L. (5)
(6) Analysis: (continued) Exercise 5.21 (continued) φAX and φSX pertain to the stripping (exhauster)section,
φAE and φSE pertain to the rectifying (enricher) section,
Because the temperatures change over the stripping and rectifying sections, average
absorption and stripping factors are used. In Eq. (1), For all three cases, the new flow rates needed in the absorption and stripping factors are
determined from the assumption of constant molar overflow, with the following results: Section
Condenser
Enricher
Feed
Exhauster
Reboiler Flow rates, lbmol/h:
L0 = 1,000
(a) L0 = 1,500
Vapor Liquid Vapor Liquid
530
1,000
530
1,500
1,530
1,000
2,030
1,500
1,530
1,000
2,030
1,500
730
1,000
1,230
1,500
730
270
1,230
270 (b) L0 = 2,000
(c) L0 = 2,500
Vapor Liquid Vapor Liquid
530
2,000
530
2,500
2,530
2,000
3,030
2,500
2,530
2,000
3,030
2,500
1,730
2,000
2,230
2,500
1,730
270
2,230
270 Assume that as with Example 5.4 and at higher reflux rates, components C1, C2,
and nC5 will not distribute to any extent. Therefore, make distribution calculations only
for the component of interest, propane. Using the above flow rates for Cases (a), (b), and
(c) , the absorption and stripping factors for C3 with L0 = 1,000 lbmol/h become as
follows, where, for example, AC for Case (a) is given from Eq. (3) by: AC Factor
AC
AE
SE = 1/AE
AF
AX
SX = 1/AX
SB Case (a) = L0
=
DKC L0 = 1,000
10.48
1.72
0.581
1.108
1.73
0.577
2.78 L0
D L0
AC D = Example 5.4 (a) L0 = 1,500
15.72
1.94
0.515
1.250
1.54
0.649
4.68 1,500
= 15.72
1,000
530
10.48(530) (b) L0 = 2,000
20.96
2.08
0.481
1.340
1.46
0.685
6.59 (c) L0 = 2,500
26.20
2.17
0.461
1.397
1.416
0.706
8.49 Analysis: (continued) Exercise 5.21 (continued) Using Eqs. (5) and (6), the following φ factors are determined for the enricher and
exhauster from the above values of AE , SE , AX , and SX :
Factor
φAE
φAX
φSE
φSX L0 = 1,000
0.0289
0.028
0.435
0.439 (a) L0 = 1,500
0.0180
0.0438
0.494
0.379 (b) L0 = 2,000
0.0135
0.0530
0.526
0.351 (c) L0 = 2,500
0.0113
0.0589
0.544
0.336 Using Eq. (1) with the above factors, the following d/b ratios for C3 are obtained: Case
Example5.4
(a)
(b)
(c) d/b for C3
0.0114
0.00522
0.00322
0.00237 As the reflux rate increases, the separation is enhanced. Exercise 5.22
Subject:
Effect of numbers of rectification and stripping equilibrium stages on keycomponent distribution in multicomponent distillation by the group method.
Given: Result...
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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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