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Unformatted text preview: 8 − θ
100 − θ
.
.
.
.
i α i ,OX − θ
There are two roots to Eq. (2), which from a spreadsheet calculation are:
θ = 1.0280 and 1.2027
Eq. (929) is applied in the following form for each of the two values of θ, in terms of the two
unknowns, dEB and Lmin. The form used is obtained by multiplying Eq. (929) by the total
distillate rate, D, which converts distillate mole fractions to component distillate rates, and the
reflux ratio, R to L. The distillate rate of PX is not an unknown because it has the same relative
volatility as MX. Thus, it must have the same d/b ratio as MX, making its distillate rate as 99
kmol/h. D + Lmin = 12217 (d EB )
.
1.1448 (297)
1.00 (1)
+
+
12217 − 10280 11448 − 1.0280 1.00 − 10280
.
.
.
. = 6.307(d EB ) + 2875.3
D + Lmin = 12217(d EB )
.
1.1448(297)
1.00 (1)
+
+
12217 − 12027 11448 − 1.2027 1.00 − 1.2027
.
.
. = 64.30(d EB ) − 5877.2 Exercise 9.28 (continued)
Analysis: (a) (continued)
Because the total distillate rate, D, is also an unknown, the following equation is also needed:
D = dEB + 99 + 198 + 1 = dEB + 298
Solving these 3 linear equations, the following results are obtained:
D =448.9 kmol/h, Lmin = 3378 kmol/h, and dEB = 150.9 kmol/h. Thus, the assumption that EB
distributes is incorrect because only 100 kmol/h of EB is in the feed. Therefore discard the root
of θ = 1.2027. Also, D now equals 100 + 99 + 198 + 1 = 398 kmol/h. From above for the θ root
of 1.0280, D + Lmin = 6.307 dEB + 2875.3.
Therefore, Lmin = 6.307(100) + 2875.3  398 = 3108 kmol/h or Rmin = Lmin/D = 3108/398 = 7.81
The operating reflux ratio = 1.1(7.81) = 8.59. In the Gilliland equation, (934),
X = (R  Rmin)/(R + 1) = (8.59  7.81)/(8.59 + 1) = 0.0813. Using Eq. (9.34), Y = 0.573 =
(N  Nmin)/(N + 1). Solving, for Nmin = 68.3, N = 161.3 equilibrium stages.
The Kirkbride Eq. (936) is used to determine the feedstage location.
NR
=
NS zOX , F x MX , B zMX , F xOX , D 2 0.206 B
D = 0.20
0.40 0.0196
0.00251 2 102.01
397.99 0.206 = 153
. Therefore, the feed stage is located at (1.53/2.53) 161.3 = stage 97.5 from the top.
(b) The Edmister group method is applied by using Eq. (566) to estimate the b/d ratio for each
component: b
= AF
d AC φ SE + 1
φ AE
S B φ AX + 1
φ SX (3) For a total condenser, AC = the reflux ratio = 8.59 for all components. To compute values of φ
for each component, assume the following values of flow rates and stages for the enriching
(rectifying) section and exhausting (stripping) section, based on constant molar overflow and
noting that the feed and reboiler stages are not included in these two sections. Variable
Number of stages
Vapor rate, kmol/h
Liquid rate, kmol/h Enriching
Section
97
3817
3419 Exhaustin
g
Section
63
3817
3919 Exercise 9.28 (continued)
Analysis: (b) (continued)
Because in the above table the Kvalues do not vary much, use arithmetic mean values for
each section. These values and those for the feed stage and partial reboiler are as follows along
with the corresponding values of the absorption and stripping factors, as defined by equations
(538), (551), below (564), and (565), where VB/B = 3817/102 = 37.4
Variable
KE
AE
SE = 1/AE
KF
AF
KX
AX
SX = 1/AX
KB
SB EB
1.066
0.840
1.190
1.080
0.951
1.143
0.898
1.114
1.206
45.1 PX
0.999
0.897
1.115
1.012
1.015
1.072
0.958
1.044
1.131
42.3 MX
0.997
0.898
1.114
1.012
1.015
1.074
0.956
1.046
1.136
42.5 OX
0.870
1.030
0.971
0.884
1.161
0.940
1.092
0.918
0.996
37.3 Values of φA and φS are obtained from Eqs. (548) and (5...
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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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