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

34 continued analysis open steam case exercise 735

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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-...
View Full Document

This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

Ask a homework question - tutors are online