Unformatted text preview: tial reboiler and partial condenser, using subscripts
B for streams leaving the reboiler and D for streams leaving the partial condenser.
Total material balance:
F + LD = B + VB or 100 + LD = 75 + VB
Benzene material balance: FxF + LDxD = BxB + VByB
69.4 + LDxD = 75(0.625) + VByB = 46.9 + VByB
Total material balance:
VB = D + L D = 7 5 + L D
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. (7-2). 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
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)
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
69.4 + LD(0.783) = 46.9 + VB(0.806)
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)
On the McCabe-Thiele diagram below, the equilibrium curve is obtained from Eq. (7-3),
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. (7-3) gives the equilibrium relation at the still or any of the 4 plates, n,
yn = αxn
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.
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