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Unformatted text preview: (45o line for total
reflux condition) are plotted and equilibrium stages are stepped off between x = 0.298 and 0.653.
The number of stages = Nt = 14.4. From Eq. (673), HETP = 10/14.4 = 0.7 ft. This value must
be used with caution for design, because the HETP is a function of the L/V ratio. The value of
0.7 ft applies only to L/V =1. Exercise 7.51
Subject:
atm. Design of a packed column for separation of a mixture of ethanol and water at 1 Given: Bubblepoint feed of 10 mol% ethanol. Bottoms of 1 mol% ethanol and distillate of 80
mol% ethanol. Reflux ratio at 1.5 times minimum. Equilibrium data from Exercise 7.29.
Assumptions: Constant molar overflow.
Find: (a)
(b)
(c)
(d)
(e) Equilibrium stages above and below feed.
Number of transfer units above and below the feed.
Height of plate column for 18inch tray spacing and 80% tray efficiency.
Height of packing for HOG = 1.2 ft.
HTU from HTU for benzenetoluene system. Analysis: By material balance, with a basis of 100 moles of feed,
Total material balance, F = 100 = D + B
Benzene material balance, 0.10F = 10 = 0.80D + 0.01B
Solving these two equations, D = 11.39 moles and B = 88.61 moles.
(a) Below is the McCabeThiele diagram showing the equilibrium curve from the data of
Exercise 7.29, a vertical qline for the saturated liquid feed at xF = 0.10, and the construction of
the rectification section operating line for determining the minimum reflux ratio. The slope of
that line is, (0.80  0.44)/(0.80  0.10) = 0.514 = Lmin/V. From a rearrangement of Eq. (77),
Rmin = Lmin / V
0.514
=
= 1.058 and R = 1.5 Rmin = 1.5(1.058) = 1.59
1 − Lmin / V
1 − 0.514 Slope of the rectification section operating line is, from Eq. (77), L/V = 1.59/(1 + 1.59) = 0.614.
In the McCabeThiele diagrams below, one for the higher concentration region and one
for the lower concentration region, a rectification section operating line of this slope is drawn
that passes through the 45o line at x = xD = 0.8. A stripping section operating line is drawn that
extends from the intersection of the rectifying operating line and the qline to the 45o line at x =
xB = 0.01. The stages are stepped off, with the optimal feed stage location. The result is 3
equilibrium stages in the stripping section, including one for the partial reboiler, and 13.8
equilibrium stages in the rectifying section. Analysis: (a) (continued) Exercise 7.51 (continued) Exercise 7.51 (continued)
Analysis: (a) (continued) Exercise 7.51 (continued)
Analysis: (a) (continued) Exercise 7.51 (continued)
Analysis: (continued)
(b) Because the value of HOG is given in part (d), calculate NOG for each section, using from
Table 6.7 for EM diffusion, N OG = dy
y −y
* For the stripping section, the first stage is the partial reboiler. Therefore, from the McCabeThiele diagram, starting from Stage 2, the limits on y are y = 0.09 at x = 0.03 to y = 0.37 at x =
0.10.
Using the trapezoidal method to solve the integral, ( NOG )SS = ∆y
(0.23 − 0.09) (0.37 − 0.23)
=
+
= 2.4
0.10
0.14
( y *−...
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 Spring '11
 Levicky
 The Land

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