Unformatted text preview: nalysis: First, verify the separation for a feed rate of 100 kmol/h. This is shown in the
McCabeThiele plot below in terms of mole fractions of A, the more volatile component. The
equilibrium curve is computed from Eq. (73),
3x
αx
y=
=
1 + x (α − 1) 1 + 2 x
The given mole fractions are:
xF = 0.50
xD = 0.90
xB = 0.20
The rectification section operating line has a slope of 0.75 and passes through point {0.9, 0.9}.
The qline is vertical at x = 0.5. The plot shows almost perfect agreement with the desired
separation for a feed rate of 100 kmol/h. Analysis: (continued) Exercise 7.20 (continued) For the base case of F = 100 kmol/h, the material balance equations are
F = D + B and xFF = 50 = xDD + xBB = 0.90D + 0.20B. Solving, these equations, along with
V = L + D, V/L = 0.75, V = V , and L = L + F , gives the following results: Stream
Feed
Distillate
Bottoms
Reflux, L
Overhead vapor, V
Liquid to reboiler, L
Vapor from reboiler, V Flow rate, kmol/h
100.00
42.86
57.14
128.58
171.44
228.58
171.44 Mol% A
50
90
20 Mol% B
50
10
80 When the feed rate is reduced to 25 kmol/h, the reflux rate, L, is maintained at 128.58
kmol/h and the boilup, V , is maintained at 171.44 kmol/h. Therefore by material balances,
L = L + F = 128.58 + 25 = 153.58 and B = L  V = 153.58  171.44 = 17.86 kmol/h. This is
impossible. Therefore, the column can not be operated with the same boilup rate. That rate
would have to be reduced to achieve a desired bottoms rate, e. g. the 57.14 % of the feed, as in
the base case or 0.5714(25) = 14.29 kmol/h. If this were done, we would now have, V = L  B =
153.58  14.29 = 139.29 kmol/h = V. Thus, in the rectifying section, L/V = 128.58/139.29 =
0.923 and in the stripping section, L / V = 153.58/139.29 = 1.103. The resulting distillate and
bottoms compositions are determined by positioning the operating lines so that 3 stages + a
reboiler can be stepped off. The result is shown below, where the mole fractions of A are 0.93 in
the distillate and reflux, 0.18 in the bottoms, and 0.38 in the reboiler vapor. Analysis: (continued) Exercise 7.20 (continued) Exercise 7.21
Subject:
Distillation of a saturated vapor of maleic anhydride (A) and benzoic acid (B)
under vacuum at 13.3 kPa.
Given: Feed contains 90 mol% A and 10 mol% B. Distillate to contain 99.5 mol% anhydride
and bottoms to contain 0.5 mol% acid. Vapor pressure data.
Assumptions: Raoult's law to compute Kvalues from vapor pressure data. Constant molar
overflow.
Find: Number of theoretical plates needed if a reflux ratio, L/D = 1.6 times minimum.
Analysis: First compute an equilibrium y, x curve using Raoult's law with the vapor pressure
data. Eq. (244 ) applies, as well as the sum of the mole fractions in the phases in equilibrium.
Thus,
Ps T
Ps T
y
y
KA = A = A
,
KB = B = B
(1, 2)
xA
P
xB
P
yA + y B = 1
,
xA + x B = 1
(3, 4)
Equations (1) to (4) can be reduced to the following equations for the mole fractions of
maleic anhydride (A) in terms of the Kvalues:
1 − KB
(5, 6)
,
y A...
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 Spring '11
 Levicky
 The Land

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