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

# 34 continued analysis open steam case exercise 735

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Unformatted text preview: contain 40 mol% alcohol. Boilup, V/F = 0.246. Vapor-liquid equilibrium data from Exercise 7.33. Assumptions: Constant molar overflow. Find: Determine the number of equilibrium stages for: (1) Partial reboiler, (2) Open steam. Analysis: Take a basis of F = 100 kmol/h. Vapor rate leaving top of column = V = 0.246F = D = 0.246(100) = 24.6 kmol/h. Alcohol in overhead vapor = 0.4(24.6) = 9.84 kmol/h. Water in the overhead vapor = 24.6 - 9.84 = 14.76 kmol/h. (1) With a partial reboiler, bottoms rate = B = F - D = 100 - 24.6 = 75.4 kmol/h. Alcohol in bottoms = 10 - 9.84 = 0.16 kmol/h. Mole fraction of alcohol in bottoms = xB = 0.16/75.4 = 0.0021. With isopropanol as the most volatile component, the McCabe-Thiele diagram is given below, where the equilibrium curve is obtained from the data in Exercise 7.33 and the q-line is vertical through x = 0.10. The stripping section operating line passes through the point {x=0.0021, y=0.0021}with a slope = L/V = F/V =100/24.6 = 4.065. It also passes through the point {x=0.1, y=0.4}. From the plot, the number of equilibrium stages = just less than 3. Call it 2 equilibrium stages in the column + partial reboiler. (2) The open steam rate = V = 24.6 kmol/h. The liquid rate = L = 100 kmol/h. Therefore, the slope of the stripping section operating line is the same as for part (1), i.e. L./V = 100/24.6 = 4.065. Now the mole fraction of alcohol in the bottoms = xB = 0.16/100 = 0.0016. Thus, as shown in the McCabe-Thiele diagram below, the operating line passes through the points {x=0.0016, y=0}and {x=0.10, y=0.40}, with the slope of 4.065. Now, The number of equilibrium stages is equal to 3, all of them in the column. Exercise 7.34 (continued) Analysis: Partial Reboiler Case: Exercise 7.34 (continued) Analysis: Open Steam Case: Exercise 7.35 Subject: Distillation of two feeds of mixtures of water and acetic acid at 1 atm. Given: Feed 1 is a bubble-point liquid of 100 kmol/h containing 75 mol% water. Feed 2 is 50 mol% vaporized of 100 kmol/h containing 50 mol% water. Unit consists of a plate column, total condenser, and partial reboiler. Distillate is to contain 98 mol% water. Bottoms is to contain 5 mol% water. Reflux ratio, L/D = R = 1.2 times minimum. Vapor-liquid equilibrium data. Assumptions: Constant molar overflow. Find: Optimal feed stage locations and number of equilibrium stages. Analysis: Water is the more volatile component. Compute flow rates of distillate and bottoms. Overall total material balance: F1 + F2 = 100 + 100 = 200 = D + B (1) Overall water balance: (0.75)(100) + 0.5(100) = 125 = xDD + xBB = 0.98D + 0.05B (2) Solving Eqs. (1) and (2), D = 123.66 kmol/h and B = 76.34 kmol/h. Assume the minimum reflux is controlled by the upper feed. This is verified in the McCabe-Thiele diagram below, where the equilibrium curve is plotted from the data, the q-line for Feed 1 is vertical through the point, x = 0.75, the q-line for Feed 2 has a slope of -1 starting from x = 0.50, and the operating line for the upper section between Feed 1 and the condenser is drawn through the two points, {x=0.98, y=0.98} and the intersection of the equilibrium curve and the q-line for Feed 1. From the plot, for the upper section, L/V = (0.98-...
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## This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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