Unformatted text preview: he bottoms contains 290.9 kmol/h of water. But, the flow rate of water entering in the two
feeds = 125 + 83.3  100 = 108.3 kmol/h.
Therefore, the open steam flow rate = 290.9 + 5  108.3 = 187.6 kmol/h
In the McCabeThiele diagram on the next page, 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 = Nt = 14. From Eq. (621), for a plate
efficiency of 70%, the actual number of trays = Na = Nt /Eo = 14/0.7 = 20 plates. Analysis: (continued) Exercise 7.30 (continued) Exercise 7.31
Subject:
intercooler. Distillation of a mixture of nhexane and noctane in a column with an Given: Saturated liquid feed of 40 mol% hexane in octane. Intercooler at second stage from the
top removes heat so as to condense 50 mol% of the vapor rising from the third. Distillate is to
contain 95 mol% of hexane and bottoms is to contain 5 mol% of hexane. Reflux ratio, L/D, at
the top, is equal to 0.5. Vaporliquid equilibrium data for 1 atm is plotted in Fig. 4.4.
Assumptions: Constant molar overflow. Total condenser and partial reboiler. Operating
pressure of 1 atm.
Find: (a) Equations to locate operating lines.
(b) Number of equilibrium stages if optimal feed stage location is used.
Analysis: First compute overall material balance. Take a basis of F = 100 kmol/h.
Overall total mole balance:
F = 100 = D + B
(1)
Overall hexane mole balance: FxF = 40 = DxD + BxB = 0.95D + 0.05B (2)
Solving Eqs. (1) and (2), D = 38.9 kmol/h and B = 61.1 kmol/h
(a) For a reflux ratio of 0.5, in the section above the intercooler, L = 0.5D = 19.45
kmol/h. The overhead vapor rate is V = L + D = 19.45 + 38.9 = 58.35 kmol/h. The slope of the
operating line = L/V = 19.45/58.35 = 0.333. Using Eq. (76), the equation for the operating line
y = 0.333x + DxD/V = 0.333x + (38.9)(0.95)/(58.35) = 0.333x + 0.633
(3)
is,
Now consider the section of stages between the intercooler at stage 2 from the top and the feed
stage. Because 50 mol% of the vapor from this section is condensed at stage 2 by the intercooler,
the vapor rate in this section = V' = 2V = 2(58.35) = 116.7 kmol/h. The liquid rate in this section
is L' = V'  D = 116.7  38.9 = 77.8 kmol/h. The slope of the operating line = L'/V' = 77.8/116.7
= 0.667. In this section, by hexane material balance, yV' = xL' + xDD or,
y = (L'/V')x + DxD/V' = 0.667x + (38.9)(0.95)/116.7 = 0.667x + 0.317
(4)
In the section below the feed stage, for a saturated liquid feed, L"= L' + F = 77.8 + 100 = 177.8
kmol/h. The vapor rate = V" = V' = 116.7 kmol/h. The slope of the operating line = L"/V" =
177.8/116.7 = 1.524. From Eq. (711),
y = (L"/V")x  BxB/V”= 1.524x  (61.1)(0.05)/116.7 = 1.524x  0.026
(5)
(b) A McCabeThiele diagram in terms of hexane, the more volatile component, is
shown on the next page, where the equilibrium curve is obtained from Fig. 4.4 and the operating
lines for the three sections are draw...
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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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