75 xd 075 x1 0545 benzene material balance around

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Unformatted text preview: .545 Benzene material balance around Stage 1, which now includes the feed, xFF + y BV + x D L = y1V + x1 L (6) Solving for yB, V L L F V L L 100 y B = y1 + x1 - xD - xF = 0.750 + 0.545 - 0.750 - 0.50 (7) V V V V V V V V Because the feed is a saturated liquid, , V =V and L = L + 100 From above, V = 4D and L/V = 3/4. Also, L / V = L / V + 100 / V = 3 / 4 + 100 / V Therefore, Eq. (7) becomes, 3 100 3 100 4.5 1125 . (8) y B = 0.750 + 0.545 + - 0.750 - 0.50 = 0.596 - = 0.596 - 4 V 4 V V D The vapor from the reboiler is in equilibrium with the liquid bottoms (residue). From the lefthand part of Eq. (2), yB xB = (9) y B + 2.5(1 - y B ) Overall total material balance, F = 100 = D + B (10) Overall benzene material balance, xFF = xDD + xBB or 50 = 0.75D + xBB (11) Solving Eqs. (8), (9), (10), and (11), yB = 0.647, xB = 0.423, D = 23.5 moles or 23.5 mol/100 mol feed, B = 76.5 moles (d) With a partial condenser, the mole fraction of the liquid reflux is that in equilibrium with the vapor distillate. Therefore,...
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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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