Unformatted text preview: ve volatility as (1.167 + 1.159)/2 = 1.163.
From Eq. (726), the slope of the qline = q/(q  1) = 0.9/(0.9  1) = 9. Feed is 10 mol%
vaporized.
Apply the FenskeUnderwoodGilliland method to obtain an initial estimate of reflux and stage
requirements. Can use the SHOR model in CHEMCAD. The results are:
Nmin = 48,
Rmin = 9.37
R = 1.15Rmin = 10.77 N = 103 (102 + reboiler)
N of feed = stage 50 from the top
To construct the equilibrium curve for the McCabeThiele method, use Eq. (73), yP = α PO xP
1163xP
.
=
1+ xP α PO − 1 1 + 0163xP
. From Eq. (79), the slope of the rectifying section operating line is,
L/V = R/(R+1) = 10.77/(10.77 + 1) = 0.9150 (1) Exercise 7.39 (continued) Analysis: (continued)
This line passes through x = 0.98 on the 45o line. The equation for this line is given by Eq. (79),
L
1
1
(2)
x+
x D = 0.915x +
(0.98) = 0.915x + 0.08326
V
R +1
11.77 + 1
Below the feed stage, with 10 mol% vaporization of the feed, L / V = 1.0535 . From a
rearrangement of Eq. (712), the boilup ratio, VB, is 18.6916. The equation of the stripping
section operating line is given by Eq. 714),
L
1
1
y=
x−
x D = 10535x −
.
(0.04) = 10535x − 0.00214
.
(3)
V
VB
18.6916
y= The equation for the qline is given by Eq. (726),
y= q
1
0.9
1
x−
zF =
x−
(0.62) = −9 x + 6.2
q −1
q −1
0.9 − 1
0.9 − 1 (4) Based on Eqs. (1) to (4), the McCabeThiele diagram in terms of P, the more volatile component,
is drawn below for three regions: (2) x = 0.2 to 0.4, (3) x = 0.4 to 0.6, and (4) x = 0.6 to 0.8, in
order to gain accuracy. In these three regions, 28 stages are stepped off in the rectifying section
up to x = 0.8, and 34.3 stages are stepped off in the stripping section down to x = 0.2.
Let region (1) extend from x = 0.8 to 0.98 (i.e. xD). Apply the Kremser equation, Eq.
(739) to this region. Apply this equation to the heavy component, O. Obtain the Kvalue for O
from the top α of 1.167, taking the Kvalue for P = 1.00. Therefore, KO = 1/1.167 = 0.857.
Therefore, the absorption factor, A = L/KV = 0.915/0.857 = 1.067. Other quantities needed in
Eq. (739) are: xo = 1  (xD)P = 1  0.98 = 0.02
y1 = xo = 0.02
From Eq. (2), for xN = 0.8, yN+1 for P = 0.8153. Therefore, for O, yN+1 = 1  0.8153 = 0.1847. log
NR = 1
1
+ 1−
A
A
logA yN +1 − xo K
y1 − xo K log
= 1
1
+ 1−
1.067
1067
. 0.1847 − 0.02(0.857)
0.02 − 0.02(0.857) log1.067 = 23.6 Let region (5) extend from x = 0.04 (i.e. xB) to 0.20. Apply the Kremser equation, Eq.
(740) to this region. Apply this equation to the light component, P. Obtain the Kvalue for P
from the bottom α of 1.159, taking the Kvalue for O = 1.00. Therefore, KP = 1.159. Therefore,
the absorption factor in the stripping section is A = L / KV = 1.0535 / 1.159 = 0.9085 . Other
values needed in Eq. (740) are x1 = xB = 0.04 and xN+1 = 0.20. Therefore,
x −x /K
0.20 − 0.04 / 1159
.
log A + 1 − A N +1 1
log 0.9085 + 1 − 0.9085
x1 − x1 / K
0.04 − 0.04 / 1159
.
NS =
=
= 135
.
1
1
log
log
A
0.0985 Analysis: (continu...
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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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