# Ch07

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Unformatted text preview: 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. (7-11), y = (L"/V")x - BxB/V"= 1.524x - (61.1)(0.05)/116.7 = 1.524x - 0.026 (5) (b) A McCabe-Thiele 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 drawn from Eqs. (3), (4), and (5). The q-line is vertical, passing through x = 0.4. Note that the upper and middle section operating lines both pass through the point {0.95, 0.95}. The theoretical stages are stepped off starting from the top, switching to the middle section operating line after stage 2, and switching to the stripping section so as to locate the feed stage optimally. The result is just slightly less than 5 equilibrium stages or, say, 4 sta...
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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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