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

By material balance around the condenser or because

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Unformatted text preview: molar overflow to give straight operating lines on a y-x diagram. Find: (a) (b) (c) (d) Compositions of V4 and L1 for 4 stages in cascade (a). Number of equilibrium stages for 85 mol% alcohol in exit vapor of cascade (a). Compositions of D and L1 for 4 stages in cascade (b). Number of equilibrium stages for 50 mol alcohol in D of cascade (b). Analysis: From the given vapor-liquid equilibrium data, in the composition range of interest, ethyl alcohol is more volatile than water. Therefore, the y and x coordinates in a y-x plot pertain to ethyl alcohol. (a) Since L = 100 mol and V = 100 mol, the slope of the operating line from Eqs. (7-6) or (7-11) = L/V = 100/100 = 1. The terminal points on the operating line as (y, x) are: (?, 0.7) at the top and (0.3, ?) at the bottom. To determine the compositions of V4 and L1 for 4 stages, this operating line is located so that exactly 4 stages are stepped off in a y-x diagram, as shown below. From the diagram, the ethanol compositions are 76 mol% in V4 and 24 mol% in L1. Exercise 7.8 (continued) Analysis: (continued) (b) It is impossible to obtain an overhead vapor with 85 mol% ethanol. With an infinite number of stages, the highest concentration of ethanol in the overhead vapor corresponds to that in equilibrium with the top liquid feed containing 70 mol% ethanol. From the given vapor-liquid equilibrium data, the highest concentration is an ethanol mole fraction of 0.82. (c) Since the bottom vapor feed, V0 = 100 mol and D = 50 mol, by overall material balance, L1 = V0 - D = 100 - 50 = 50 mol. Because of the assumption of constant molar overflow, L = LR = L1 = 50 mol. By material balance around the condenser or because of constant molar overflow, V = V4 = LR + D = 50 + 50 = 100 mol. The slope of the operating line from Eqs. (7-6) or (7-11) = L/V = 50/100 = 0.5. To determine the compositions of D and L1 for 4 stages, an operating line of this slope is located so that exactly 4 stages are stepped off in a y-x diagram, as shown below. From the diagram, the ethanol compositions are 45 mol% in D and 16 mol% in L1. (d) Since the distillate is 50 mol% ethanol, 25 moles of ethanol and 25 moles of water leave in the distillate. Because the feed is 30 moles of ethanol and 70 moles of water, L1 , the leaving liquid, contains 5 moles of ethanol and 45 moles of water. Thus, the terminal points on the operating line, because of the total condenser, as (y, x), are: (0.5, 0.5) at the top and (0.3, 0.1 ) at the bottom. However, point (0.3, 0.1) above the equilibrium line is impossible. Exercise 7.9 Subject: Separation of air in a reboiled stripper Given: Reboiled stripper with total reboiler operating at 1 atm. Liquid air (79.1 mol% N2 and 20.9 mol% O2) fed to top stage. 60% of O2 in the feed is drawn off in vapor product from the reboiler. Bottoms vapor product contains 0.2 mol% N2. Vapor-liquid equilibria data are given. Assumptions: Feed is a saturated liquid. Find: (a) Mol% N2 in vapor from top stage. (b) Moles of vapor generated in reboiler per 100 moles of feed. (...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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