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Unformatted text preview: n from Eqs. (3), (4), and (5). The qline is vertical, passing
through x = 0.4. Note that the upper and middle section operating lines both pass through the
point {0.95, 0.95}. The theoretical stages are stepped off starting from the top, switching to the
middle section operating line after stage 2, and switching to the stripping section so as to locate
the feed stage optimally. The result is just slightly less than 5 equilibrium stages or, say, 4
stages plus a partial reboiler. Analysis: (continued) Exercise 7.31 (continued) Exercise 7.32
Subject:
Distillation of a mixture of ethyl alcohol and water at 1 atm using open steam
instead of a reboiler.
Given: 100 kmol/h of a saturated liquid feed containing 12 mol% ethyl alcohol in water.
Distillate to contain 85 mol% alcohol with a recovery of 90%. Reflux ratio, L/D = 3 with
saturated liquid reflux. Feed stage located optimally. Vaporliquid equilibrium data in Exercise
7.29.
Assumptions: Constant molar overflow. Total condenser.
Find: (a)
(b)
(c)
(d) Open steam requirement, kmol/h
Number of equilibrium stages
Optimal feed stage location.
Minimum reflux ratio. Analysis: First compute material balance. Because the ethanol mole fraction in the distillate is
less than that of the azeotrope (89.43 mol% in Exercise 7.29), the ethanol is always the more
volatile component. The feed contains 12 kmol/h of ethanol and 88 kmol/h of water. For a 90%
recovery, the distillate contains 0.9(12) = 10.8 kmol/h of ethanol. Since the distillate is 85 mol%
ethanol, the total distillate rate = D = 10.8/0.85 = 12.7 kmol/h. The bottoms contains 12  10.8 =
1.2 kmol/h of ethanol. The distillate contains 12.7  10.8 = 1.9 kmol/h of water. The bottoms
contains 88  1.9 + open steam = 89.9 + open steam in kmol/h.
(a) For a reflux ratio of 3, L = 3D = 3(12.7) = 38.1 kmol/h. Overhead vapor rate = V = L
+ D = 38.1 + 12.7 = 50.8 kmol/h. Below the feed stage, L' = L + F = 38.1 + 100 = 138.1
kmol/h. Boilup rate = V' = V = 50.8 kmol/h = flow rate of open steam.
(b) The bottoms rate = 138.1 kmol/h. The bottoms consists of 1.2 kmol/h of ethanol and
138.1  1.2 = 136.9 kmol/h of water. The mole fraction of ethanol in the bottoms = 1.2/138.1 =
0.0087. The McCabeThiele diagram is given on the next page, where the equilibrium curve is
obtained from Exercise 7.29 and the qline is vertical at x = 0.12. The rectifying section
operating line passes through the point {0.85, 0.85}and has a slope, L/V = 38.1/50.8 = 0.75. The
stripping section operating line has a slope, L'/V' = 138.1/50.8 = 2.72 and, as shown in Fig.
7.27(c), passes through the point x = xB = 0.0087 at y = 0. Because the stages are so crowded at
the high mole fraction end, a second McCabeThiele diagram is shown for the region above y = x
= 0.7. As shown, with the use of the two diagrams, just less than 20 equilibrium stages are
needed.
(c) From the first McCabeThiele plot, the optimal feed stage is Stage 18 from the top.
(d) From the third McCabeThiele diagram on the next page, the minimu...
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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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