# 8d 0530b 4 solving eqs 3 and 4 d 629 mols or 629

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Unformatted text preview: ng Eqs. (3) and (4), D = 62.9 mol/s or 62.9 mol/100 mol feed, and B = 37.1 mol/s. Therefore, vapor generated = V = 1.5D = 1.5(62.9) = 94.4 mol/s. The operating line for the y-x diagram passes through the (y, x) point (0.8, 0.8) with a slope, L/V = 1/3, as shown in the diagram below. Analysis: (continued) Procedure 2: The slope and top point of the operating line are the same as for Procedure 1. We just have to step off one more stage. Therefore from the results above, we have: yC = 0.80 xC = 0.615 y1 = 0.738 x1 = 0.530 Benzene material balance around Stage 1, y RV + xC L = y1V + x1 L (5) Solving for yR, yR = y1 + ( x1 - xC ) Exercise 7.11 (continued) L 1 = 0.738 + (0.530 - 0.615) = 0.710 V 3 The vapor from the reboiler is in equilibrium with the liquid bottoms (residue). From the lefthand part of Eq. (2), yR 0.710 xR = = = 0.495 yR + (1 - yR ) 0.710 + 2.5(1 - 0.710) Overall total material balance, F = 100 = D + B (6) Overall benzene material balance, xFF = yCD + xRB or 70 = 0.8D + 0.495B (7) Solvi...
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