Separation Process Principles- 2n - Seader & Henley - Solutions Manual

8 08 with a slope lv 13 as shown in the diagram below

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Unformatted text preview: 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 + α (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) Solving 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. Solving for yR, yR = y1 + ( x1 − xC ) Exercise 7.11 (continued) 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 + α (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) Solving 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, L/V = 1/3, as shown in the diagram below. Exercise 7.11 (continued) 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 L = 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 − 3V 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 =...
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