Unformatted text preview: quilibrium curve.
From the equilibrium data, this intersection is at y = 0.50 and x = 0.293. Then, the slope of the
rectifying section operating line, (L/V)min is (0.96  0.50)/(0.96  0.293) = 0.690. From a
rearrangement of Eq. (77), Rmin = (L/V)min /[1  (L/V)min ] = 0.69/(1  0.69) = 2.23. (2) Reflux ratio = 2(2.23) = 4.46. From Eq. (77), the slope of the rectifying section
operating line = L/V = R/(1 + R) = 4.46/5.46 = 0.817. To determine the bottoms composition,
use a McCabeThiele diagram in terms of benzene, the more volatile component. The qline and
the rectifying section operating line are fixed and 4 trays are stepped off from the top, starting at
the distillate mole fraction for benzene, xD , of 0.96. Then, the stripping section operating line is positioned by trial and error so that 3 more stages plus the reboiler stage are stepped off to arrive
at the point where the assumed location of the stripping section operating line intersects the 45o
line. The result is shown below, where it is seen that xB = 0.08. Analysis: (b) (continued Exercise 7.17 (continued) (3) The products are now computed by overall material balances: F = 100 = D + B and
50 = xDD + xBB = 0.96D + 0.08B. Solving these two equations, D = 47.7 mol/100 mol feed and
B = 52.3 mol/100 mol feed. Exercise 7.17 (continued) Analysis: (continued)
(c)
(1) A saturated vapor feed is fed to the reboiler. The slope of the rectifying section
operating line, (L/V), is 0.9. To determine the bottoms composition, use a McCabeThiele
diagram in terms of benzene, the more volatile component. The qline and the rectifying section
operating line are fixed and 7 trays are stepped off from the top, starting at the distillate mole
fraction for benzene, xD , of 0.96. Then, the stripping section operating line is positioned by trial
and error so that the reboiler stage is stepped off to arrive at the point where the assumed location
of the stripping section operating line intersects the 45o line. The result is shown below, where it
is seen that xB = 0.07.
(2) The products are now computed by overall material balances: F = 100 = D + B and
50 = xDD + xBB = 0.96D + 0.07B. Solving these two equations, D = 48.3 mol/100 mol feed and
B = 51.7 mol/100 mol feed. Exercise 7.18
Subject:
Conversion of a distillation column to a reboiled stripper to obtain very pure
toluene from a mixture of benzene and toluene at 101 kPa.
Given: A column with 8 theoretical plates, a total condenser, and a partial reboiler. Feed
contains 36 mol% benzene and 64 mol% toluene. Reboiler produces 100 kmol/h of vapor. To
obtain nearly pure toluene bottoms, feed is introduced to the top plate as a saturated liquid, with
no reflux. Vaporliquid equilibrium data are in Exercise 7.13.
Assumptions: Constant molar overflow.
Find: (a) Minimum feed rate and corresponding bottoms composition.
(b) Bottoms rate and composition for a feed rate 25% above the minimum.
Analysis:
(a)
The minimum feed rate corresponds to a rate equal to the boilup rate...
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 Spring '11
 Levicky
 The Land

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