8d 0495b 7 solving eqs 6 and 7 d 672 mols or 672

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Unformatted text preview: ng Eqs. (6) and (7), D = 67.2 mol/s or 67.2 mol / mol feed, and B = 32.8 mol/s. Therefore, vapor generated = V = 1.5D = 1.5(67.2) = 100.8 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 3: The slope and top point of the operating line are the same as for Procedures 1 and 2. We just have to extend Procedure 2 by stepping off a second equilibrium stage. From above, the results for the condenser and stage 1 are: yC = 0.80 xC = 0.615 y1 = 0.738 x1 = 0.530 y2 = 0.710 x2 = 0.495 Benzene material balance around Stage 2, y RV + x1 L = y2V + x2 L (8) L 1 Solving for yR, y R = y2 + ( x2 - x1 ) = 0.710 + (0.495 - 0.530) = 0.698 V 3 The vapor from the reboiler is in equilibrium with the liquid bottoms (residue). From the lefthand part of Eq. (2), yR 0.698 xR = = = 0.480 yR + (1 - yR ) 0.698 + 2.5(1 - 0.698) Overall total material balance, F = 100 = D + B (10) Overall benzene material balance, xFF = yCD + xRB or 70 = 0.8D + 0.480B (11) So...
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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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