# Overall total mass balance 30 d b 1 2 overall benzene

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Unformatted text preview: tal mass balance: 30 = D + B (1) (2) Overall benzene mass balance: 0.40(30) = 12 = 0.97D + 0.02B D = 12.1 kg/h B = 17.9 kg/h Solving Eqs. (1) and (2): Converting to moles with molecular weights of 78.11 for benzene and 92.13 for toluene, Benzene Toluene Product kmol/h Mass fraction Mole fraction Mass fraction Mole fraction Distillate 0.154 0.97 0.974 0.03 0.0235 Bottoms 0.196 0.02 0.026 0.98 0.9765 1.00 1.000 1.00 1.0000 Total: 0.350 (b) Because benzene is the more volatile component of the feed, the x and y coordinates will be those of benzene in the diagram on the next page. . In moles, the feed consists of: Component kmol/h Mole fraction Benzene 0.154 0.44 Toluene 0.196 0.56 Total: 0.350 1.00 For a saturated liquid feed, the q-line is vertical and passes through x = 0.44. The slope of the rectifying operating line, L/V, is obtained from Eq. (7-7), using the specified reflux ratio = 3.5, L/V = R/(1 + R) = 3.5/4.5 = 0.778 For saturated liquid reflux, the rectifying operating line passes through the point {0.974, 0.974}...
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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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