85 085and has a slope lv 381508 075 the stripping

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Unformatted text preview: ing 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 McCabe-Thiele 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 McCabe-Thiele plot, the optimal feed stage is Stage 18 from the top. (d) From the third McCabe-Thiele diagram on the next page, the minimum reflux in terms of L/V is obtained from the slope of the rectifying section operating line, which passes through the point, y = xD = 0.85 and is tangent to the equilibrium curve, rather than being drawn through the intersection of the q-line and the equilibrium curve because that would cause the operating line to mistakenly cross over the equilibrium curve. The slope of the operating line = (L/V)min = 0.667. From Eq. (7-27), Rmin = (L/D)min = 0.667/(1-0.667) = 2.0. Analysis: (b, c...
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