85 and 000777 on the 45o line the result is

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Unformatted text preview: 0.55, using Eq. (6-21), Na = Nt/Eo = 10/0.55 = 18.2 minimum plates. Analysis: (b and c) (continued) Exercise 7.29 (continued) Analysis: (b and c) (continued) Exercise 7.29 (continued) Analysis: (continued) (d) For an operating reflux ratio = L/V = 0.8, the reflux ratio, R = L/D = 0.8/(1-0.8) = 4. On the McCabe-Thiele diagram below, the rectifying section operating line has a slope of 0.8 and passes through the point, y=0.85, x=0.85. the stripping section operating line passes through the point, y=0.00777, x=0.0777 and intersects the vertical q-line at the point where the rectifying section operating line intersects the q-line. As seen, the equilibrium stages are stepped off starting at the top, with a switch from the rectifying section to the stripping section to minimize the number of stages and, thus, locating the optimal feed stage. The result is just less than 15 equilibrium stages. Call it 14 equilibrium stages plus a partial reboiler. Applying Eq. (6-21), Na = 14/0.55 = 25.5 or 26 actual plates plus...
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This note was uploaded on 09/08/2008 for the course CHE 244 taught by Professor Selebi during the Spring '06 term at Lehigh University .

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