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Unformatted text preview: 0.828)/(0.98-0.75) =
0.661. From Eq. (7-27), R = L/D = (L/V)/[1 - (L/V)] = 0.661/(1 - 0.661) = 1.95. Therefore,
L = 1.95D = 1.95(123.66) = 241.1 kmol/h and V = L + D = 241.1 + 123.66 = 364.8 kmol/h.
In the middle section, between the two feeds, L' = L + F1 = 241.1 + 100 = 341.1 kmol/h and
V' = V = 364.8 kmol/h. Therefore, the slope of the middle section operating line = L'/V' =
341.1/364.8 = 0.935. As seen in the diagram below, this line does not cause a pinched region at
Feed 2. Therefore, the assumption is correct and Rmin = 1.95.
For an operating reflux ratio of 1.2 times minimum, L = 1.2(241.1) = 289.3 kmol/h.
The vapor rate in the upper section = V = L + D = 289.3 + 123.66 = 413 kmol/h.
Therefore the upper section operating line has a slope, L/V = 289.3/413 = 0.700 and passes
through the point y = x = 0.98. It intersects the vertical q-line at x = 0.75 and, for the slope of
0.700 = (0.98 - y)/(0.98 - 0.75), at y = 0.819.
For the middle section, L' = L + F1 = 289.3 +100 = 389.3 kmol/h and V' = V = 413
kmol/h. Therefore, the middle section operating line has a slope of L'/V' = 389.3/413 = 0.943
and intersects the q-line for x = 0.75 at y = 0.819. It intersects the q-line for Feed 2 at y = 0.543.
For the lower section, L" = L' + 0.5F2 = 389.3 + 50 = 439.3 kmol/h and V"=V' - 0.5F2=
413 - 50 = 363 kmol/h. Therefore, the lower section operating line has a slope of L"/V" =
439.3/363 = 1.210 and intersects the q-line for Feed 2 at y = 0.543 and the 45o line at xB = 0.05.
In the McCabe-Thiele diagrams below and on the next page for the high, middle, and low mole
fraction regions, the three operating lines are drawn and the equilibrium stages are stepped off so
as to place the two feeds at their optimal locations. As seen, the number of equilibrium stages
required = just less than 33 equilibrium stages. Call it 32 equilibrium stages in the column and a
partial reboiler. Optimal feed stages are located at Stages 17 and 27 from the top. Analysis: (continued) Exercise 7.35 (continued) Analysis: (continued) Exercise 7.35 (continued) Analysis: (continued) Exercise 7.35 (continued) Analysis: (continued) Exercise 7.35 (continued) Exercise 7.36
Distillation at 1 atm of a mixture of methanol (M) and ethanol (E) to obtain a
distillate, bottoms, and a liquid sidestream.
Given: 100 kmol/h of a 25 mol% vaporized mixture of 75 mol% methanol in ethanol.
Distillate is 96 mol% methanol and bottoms is 5 mol% methanol. Unit consists of a total
condenser, plate column, and partial reboiler. Sidestream is 15 kmol/h of 20 mol% methanol.
Reflux ratio, R = 1.2 times minimum.
Assumptions: Constant molar overflow. Raoult's law K-values.
Find: Number of theoretical stages and optimal locations of feed and sidestream.
Analysis: First compute the material balance.
Overall total material balance:
F = 100 = D + B + S = D + B + 15
Overall methanol material balance: 75 = 0.96D + 0.05B + 0.20(15)
Solving Eqs. (1) and (2), D = 74.45 kmol/h and B = 10.55 kmol/h
Now determine the equilibrium curve. At 1 atm (14.696 psia), methanol boils at 64.7oC
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.
- Spring '11
- The Land