# Number of transfer units above and below the feed

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Unformatted text preview: nzene 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 McCabe-Thiele diagram showing the equilibrium curve from the data of Exercise 7.29, a vertical q-line 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. (7-7), 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. (7-7), L/V = 1.59/(1 + 1.59) = 0.614. In the McCabe-Thiele 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 q-line 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 t...
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