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

8 02 kmolh a with a partial reboiler no other water

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Unformatted text preview: m reflux in terms of L/V is obtained from the slope of the rectifying section operating line, which passes through the point, y = xD = 0.85 and is tangent to the equilibrium curve, rather than being drawn through the intersection of the q-line and the equilibrium curve because that would cause the operating line to mistakenly cross over the equilibrium curve. The slope of the operating line = (L/V)min = 0.667. From Eq. (7-27), Rmin = (L/D)min = 0.667/(1-0.667) = 2.0. Exercise 7.32 (continued) Analysis: (b, c, and d) (continued) Exercise 7.32 (continued) Analysis: (b, c, and d) (continued) Exercise 7.32 (continued) Analysis: (b, c, and d) (continued) Exercise 7.33 Subject: Distillation of a mixture of isopropyl alcohol and water at 1 atm using either a partial reboiler or open steam. Given: Bubble-point feed containing 10 mol% isopropyl alcohol in water. Distillate to contain 67.5 mol% isopropyl alcohol with a 98% recovery. Vapor-liquid equilibrium data , with an azeotrope at 68.54 mol% alcohol. Assumptions: Constant molar overflow. Total condenser. Find: For a reflux ratio, R = L/D = 1.5 times minimum, determine number of stages, if: (a) Partial reboiler is used. (b) Open saturated steam is used. and (c) Minimum number of equilibrium stages. Analysis: In the composition region of operation, the alcohol is the most volatile component. First, compute the distribution of the alcohol. Take a basis of 100 kmol/h of feed. Then, the feed contains 10 kmol/h alcohol and 90 kmol/h of water. For a recovery of 98 mol%, alcohol, distillate contains 9.8 kmol/h of alcohol. For an alcohol purity of 67.5 mol%, the distillate rate = D = 9.8/0.675 = 14.52 kmol/h. Water in the distillate = 14.52 - 9.8 = 4.72 kmol/h. Alcohol in the bottoms = 10 - 9.8 = 0.2 kmol/h. (a) With a partial reboiler, no other water enters the system. Therefore, water in the bottoms = 90 - 4.72 = 85.28 kmol/h. Total bottoms rate = B = 85.28 + 0.2 = 85.48 kmol/h. Mole fraction of alcohol in bottoms = 0.2/85.48 = 0.0023. The minimum reflux is determined from the McCabe-Thiele diagram on the next page, where the equilibrium curve is drawn from the given data and the q-line is vertical, passing through x = 0.10. The rectifying section operating line for minimum reflux usually is a straight line that connects the distillate mole fraction on the 45o line to the intersection of the equilibrium curve and the q-line as shown by the dashed line on the diagram. However, in this case the line mistakenly crosses over the equilibrium curve. Therefore, instead, the operating line is drawn tangent to the equilibrium curve from the point x = xD = 0.675 as shown on the diagram by a solid line. The slope of the operating line = L/V = 0.467. From Eq. (7-27), Rmin = (L/D)min = 0.467/(1-0.467) = 0.876. The operating reflux ratio = R = 1.5Rmin = 1.5(0.876) = 1.314. From Eq. (7-7), L/V = R/(1+R) = 1.314/(1+1.314) = 0.568. On a set of three McCabe-Thiele diagrams on the next page, the rectifying section operating line has t...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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