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Unformatted text preview: lving Eqs. (10) and (11), D = 68.8 mol/s or 68.8 mol / 100 mol feed, and B = 31.2 mol/s. Therefore, vapor generated = V = 1.5D = 1.5(68.8) = 103.2 mol/s The operating line for the y-x diagram passes through the (y, x) point (0.8, 0.8) with a slope, Exercise 7.11 (continued) L/V = 1/3, as shown in the diagram below. Analysis: (continued) Procedure 4: If the reflux bypasses the top stage, the vapor and liquid pass through that stage without change. Therefore, this procedure is the same as Procedure 2, i.e. just one stage in the column.
Procedure 5: The slope and top point of the operating line are the same as for Procedure 1. We just have to add the feed to the stage in the column. 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, which now includes the feed, xFF + y RV + xC L = y1V + x1 L (12) Solving for yR, V L L F V L L 100 y R = y1 + x1 - xC - xF = 0.738 + 0.530 - 0.615 - 0.70 (13) V V V V V V V V Because the feed is a saturated liquid, , V =V and Exercise 7.11 (continued) L =...
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