Therefore from the above results yd 075 xr 0545 y1

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Unformatted text preview: from the above results, yD = 0.75 xR = 0.545 y1 = 0.596 x1 = 0.371 Benzene material balance around the theoretical plate, which includes the feed, xFF + y BV + x R L = y1V + x1 L (12) Solving for yB, V L L F V L L 100 y B = y1 + x1 - xR - xF = 0.596 + 0.371 - 0.545 - 0.50 (13) V V V V V V V V Analysis: (d) (continued) Because the feed is a saturated liquid, From above, V = 4D and L/V = 3/4. Therefore, Eq. (13) becomes, 3 100 y B = 0.596 + 0.371 + - 0.545 4 V Exercise 7.12 (continued) , V =V and L = L + 100 Also, L / V = L / V + 100 / V = 3 / 4 + 100 / V 3 100 12.9 3.23 - 0.50 = 0.466 - = 0.466 - 4 V V D (14) The vapor from the reboiler is in equilibrium with the liquid bottoms (residue). From the lefthand part of Eq. (2), yB xB = (15) y B + 2.5(1 - y B ) Overall total material balance, F = 100 = D + B Overall benzene material balance, xFF = yDD + xBB or 50 = 0.75D + xBB Solving Eqs. (14), (15), (16), and (17), yB = 0.405, xB = 0.214, D = 53.4 moles or 53.4/100 mol feed, (16) (17) B = 46.6 moles...
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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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