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Unformatted text preview: tial reboiler and partial condenser, using subscripts
B for streams leaving the reboiler and D for streams leaving the partial condenser.
Partial reboiler:
Total material balance:
F + LD = B + VB or 100 + LD = 75 + VB
(3)
Benzene material balance: FxF + LDxD = BxB + VByB
(4)
or
69.4 + LDxD = 75(0.625) + VByB = 46.9 + VByB
Partial condenser:
Total material balance:
VB = D + L D = 7 5 + L D
(5)
Benzene material balance: VByB = DyD + LDxD = 75(0.9) + LDxD = 67.5 + LDxD (6)
Assume equilibrium in the partial condenser and partial reboiler. Using the given α with its
definition in Eq. (72). For the partial condenser,
α = 2.5 = yD (1  xD)/xD(1  yD) = 0.9(1  xD)/xD(1 0.9) = 9(1  xD)/xD
(7)
Solving Eq. (7), xD = 0.783
For the partial reboiler,
α = 2.5 = yB (1  xB)/xB(1  yB) = yB (1  0.625)/0.625(1  yB) = 0.6 yB/(1  yB)
(8)
Solving Eq. (8), yB = 0.806
Eqs. (3) through (6), are 4 equations in 2 unknowns, VB and LD. We only need 2 of the 4
equations. Using Eqs. (3) and (4),
100 + LD = 75 + VB
(3)
69.4 + LD(0.783) = 46.9 + VB(0.806)
(9)
Solving Eqs. (3) and (9),
LD = 98 moles /100 moles of feed and VB = 123 moles/100 moles of feed Exercise 7.24 (continued) Analysis: (continued)
Graphical Method:
On the McCabeThiele diagram below, the equilibrium curve is obtained from Eq. (73),
αx
2.5x
y=
=
1 + x (α − 1) 1 + 15x
.
The rectification section operating line is located, as shown, so that two equilibrium steps, one
for the partial condenser and one for the partial reboiler, are stepped off between xC = 0.9 (from
the total condenser) and xB = 0.625. The measured slope of the operating line = LD/VB = 0.8.
Combining this with Eq. (3),
VB = 125 moles/100 moles of feed, which is close to the analytical value. Exercise 7.25
Subject: Rectification of a mixture of benzene and chlorobenzene at total reflux. Given: Feed of 100 kmol of 20 mol% benzene and 80 mol% chlorobenzene. Column has 4
theoretical plates, a total condenser, a reflux drum, and a still to vaporize the feed. At an
operating pressure of 1 atm, relative volatility of benzene with respect to chlorobenzene = α =
4.13. Operate at total reflux with holdups only in the reflux drum and the still. Want liquid in
the still with 0.1 mol% benzene.
Assumptions: Perfect mixing to give uniform compositions in the reflux drum and the still.
Find: Moles of liquid in the still at steady state.
Analysis: This exercise can be solved analytically or graphically. Since benzene is the more
volatile component, Eq. (73) gives the equilibrium relation at the still or any of the 4 plates, n,
as,
yn = αxn
4.13xn
=
1 + xn (α − 1) 1 + 313xn
. (1) At total reflux, numbering stages up from the bottom,
xn+1 = yn (2) Analytical Method:
Start at the bottom, stage 1, with x1 = 0.001. Solve for y1 (vapor leaving the still) with
Eq. (1). Then, from Eq. (2), x2 = y1. Continue in this manner, solving alternately Eq. (1) and
then Eq. (2), until y5 (vapor leaving the top plate) is reached. Then, x in the reflux drum = y5.
The results...
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 Spring '11
 Levicky
 The Land

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