Unformatted text preview: L + 100 From above, V = 1.5D and L/V = 1/3. Also, L / V = L / V + 100 / V = 1 / 3 + 100 / V Therefore, Eq. (13) becomes, 1 100 1 100 17 11.33 (14) y R = 0.738 + 0.530 +  0.615  0.70 = 0.710  = 0.710  3 V 3 V V D The vapor from the reboiler is in equilibrium with the liquid bottoms (residue). From the lefthand part of Eq. (2), yR xR = (15) y R + 2.5(1  y R ) Overall total material balance, F = 100 = D + B (16) (17) Overall benzene material balance, xFF = yCD + xRB or 70 = 0.8D + xRB Solving Eqs. (14), (15), (16), and (17), yR = 0.565, xR = 0.342, D = 78.3 mol/s or 78.3 mol/100 mol feed, B = 21.7 mol/s Therefore, vapor generated = V = 1.5D = 1.5(78.3) = 117.5 mol/s The rectifying section operating line for the yx diagram passes through the (y, x) point (0.8, 0.8) with a slope, L/V = 1/3, the qline (feed line) is a vertical line, and the stripping section operating line passes throught the (y, x) point (0.565, 0.342) and the intersection of the other two lines, as shown in the diagram on the following page. The results fro...
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 Spring '06
 selebi
 Distillation, pH, Mass Transfer, Vaporliquid equilibrium, section operating line, reflux, feed, constant molar overflow

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