Unformatted text preview: 0.8D + xRB
Solving Eqs. (14), (15), (16), and (17),
yR = 0.565,
xR = 0.342,
D = 78.3 mol/s or 78.3 mol/100 mol feed, B = 21.7 mol/s
Therefore, vapor generated = V = 1.5D = 1.5(78.3) = 117.5 mol/s
The rectifying section operating line for the yx diagram passes through the (y, x) point (0.8, 0.8)
with a slope, L/V = 1/3, the qline (feed line) is a vertical line, and the stripping section
operating line passes throught the (y, x) point (0.565, 0.342) and the intersection of the other two
lines, as shown in the diagram on the following page.
The results from the 5 procedures are summarized as follows:
Procedure D/100 moles of feed V/mole of D x of benzene in B
1
62.9
1.5
0.530
2
67.2
1.5
0.495
3
68.8
1.5
0.480
4
67.2
1.5
0.495
5
78.3
1.5
0.342 Exercise 7.11 (continued) Analysis: Procedure 5: (continued) (e) Procedure 5 is recommended because it produces by far the most distillate, which
corresponds to the highest recovery of benzene. Exercise 7.12
Subject: Distillation of a mixture of benzene (A) and toluene (B) at 101 kPa. Given: Column consisting of a partial reboiler, one theoretical plate, and a total condenser.
Produce a distillate of 75 mol% benzene from a saturated liquid feed of 50 mol% benzene.
Assumptions: Constant molar overflow. Constant relative volatility = αΑ,Β = 2.5.
Find: Number of moles of distillate per 100 moles of feed for:
(a) Feed to the reboiler and no reflux.
(b) Feed to the reboiler and a reflux ratio, L/D = 3.
(c) Feed to the plate and a reflux ratio of 3.
(d) Same as (c) except a partial condenser.
(e) Feed to the reboiler with minimum reflux.
(f) Feed to the reboiler with total reflux.
Analysis: Either a graphical or analytical method can be used. Because the relative volatility is
assumed constant, use an analytical method. For each part, the theoretical plate and the partial
reboiler are equilibrium stages. Benzene is the more volatile component, so the yx diagram is
based on benzene. Because the relative volatility = constant = 2.5, the equilibrium relationship is
given by Eq. (73),
αx
2.5x
y=
=
(1)
1 + x (α − 1) 1 + 15x
.
Take as a basis, 100 moles of feed. Therefore, the feed is 50 moles of A and 50 moles of B.
(a) With no reflux, separation occurs only in the reboiler. The vapor leaving the reboiler
is totally condensed to become the distillate with yD = xD = 0.75. Solve Eq. (1) for equilibrium x,
0.750
yD
=
= 0.545
(2)
y D + α (1 − y D ) 0.750 + 2.5(1 − 0.750)
Because the distillate and bottoms have benzene mole fractions greater than the mole fraction of
the feed (0.5), it is impossible to obtain a distillate with a benzene mole fraction of 0.75. xB = (b) From the reflux ratio, L = 3D, V = L + D = 4D. Therefore, D/V = 1/4 and L/V =3/4.
Use a subscript of D for distillate, R for reflux, B for streams leaving the reboiler, and 1 for the
theoretical plate, when used. With 1 theoretical plate, from part (a),
y1 = 0.75
xD = 0.75
x1 = 0.545
Benzene material balance around plate 1,
y BV + x D L = y1V + x1...
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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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