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

A for a reflux ratio of 3 l 3d 3127 381 kmolh

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Unformatted text preview: n from Eqs. (3), (4), and (5). The q-line is vertical, passing through x = 0.4. Note that the upper and middle section operating lines both pass through the point {0.95, 0.95}. The theoretical stages are stepped off starting from the top, switching to the middle section operating line after stage 2, and switching to the stripping section so as to locate the feed stage optimally. The result is just slightly less than 5 equilibrium stages or, say, 4 stages plus a partial reboiler. Analysis: (continued) Exercise 7.31 (continued) Exercise 7.32 Subject: Distillation of a mixture of ethyl alcohol and water at 1 atm using open steam instead of a reboiler. Given: 100 kmol/h of a saturated liquid feed containing 12 mol% ethyl alcohol in water. Distillate to contain 85 mol% alcohol with a recovery of 90%. Reflux ratio, L/D = 3 with saturated liquid reflux. Feed stage located optimally. Vapor-liquid equilibrium data in Exercise 7.29. Assumptions: Constant molar overflow. Total condenser. Find: (a) (b) (c) (d) Open steam requirement, kmol/h Number of equilibrium stages Optimal feed stage location. Minimum reflux ratio. Analysis: First compute material balance. Because the ethanol mole fraction in the distillate is less than that of the azeotrope (89.43 mol% in Exercise 7.29), the ethanol is always the more volatile component. The feed contains 12 kmol/h of ethanol and 88 kmol/h of water. For a 90% recovery, the distillate contains 0.9(12) = 10.8 kmol/h of ethanol. Since the distillate is 85 mol% ethanol, the total distillate rate = D = 10.8/0.85 = 12.7 kmol/h. The bottoms contains 12 - 10.8 = 1.2 kmol/h of ethanol. The distillate contains 12.7 - 10.8 = 1.9 kmol/h of water. The bottoms contains 88 - 1.9 + open steam = 89.9 + open steam in kmol/h. (a) For a reflux ratio of 3, L = 3D = 3(12.7) = 38.1 kmol/h. Overhead vapor rate = V = L + D = 38.1 + 12.7 = 50.8 kmol/h. Below the feed stage, L' = L + F = 38.1 + 100 = 138.1 kmol/h. Boilup rate = V' = V = 50.8 kmol/h = flow rate of open steam. (b) The bottoms rate = 138.1 kmol/h. The bottoms consists of 1.2 kmol/h of ethanol and 138.1 - 1.2 = 136.9 kmol/h of water. The mole fraction of ethanol in the bottoms = 1.2/138.1 = 0.0087. The McCabe-Thiele diagram is given on the next page, where the equilibrium curve is obtained from Exercise 7.29 and the q-line is vertical at x = 0.12. The rectifying section operating line passes through the point {0.85, 0.85}and has a slope, L/V = 38.1/50.8 = 0.75. The stripping section operating line has a slope, L'/V' = 138.1/50.8 = 2.72 and, as shown in Fig. 7.27(c), passes through the point x = xB = 0.0087 at y = 0. Because the stages are so crowded at the high mole fraction end, a second McCabe-Thiele diagram is shown for the region above y = x = 0.7. As shown, with the use of the two diagrams, just less than 20 equilibrium stages are needed. (c) From the first McCabe-Thiele plot, the optimal feed stage is Stage 18 from the top. (d) From the third McCabe-Thiele diagram on the next page, the minimu...
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