# 8d 0480b 11 solving eqs 10 and 11 d 688 mols or

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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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## 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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