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

# As seen the equilibrium stages are stepped off

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Unformatted text preview: 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 = 0.65. From Eq. (7-27), Rmin = (L/D)min = 0.65/(1-0.65) = 1.86. The liquid rate in the rectifying section = L = 1.86D = 1.86(22.82) = 42.44 kmol/h. Below the feed plate, L = L + F = 42.44 + 100 = 142.44 kmol/h. Vapor rate from the reboiler = VB = L - B = 142.44 - 77.18 = 65.26 kmol/h. Therefore, boilup ratio = VB/B = 65.26/77.18 = 0.846. (c) In the second McCabe-Thiele diagram on the next page, the minimum number of stages is determined by stepping off stages between the equilibrium curve and the 45o line (total reflux) from the points 0.85 and 0.00777 on the 45o line. The result is approximately 10 minimum equilibrium stages. For a stage efficiency of 0.55, using Eq. (6-21), Na = Nt/Eo = 10/0.55 = 18.2 minimum plates. Exercise 7.29 (continued) Analysis: (b and c) (continued) Analysis: (b and c) (continued) Exercise 7.29 (continued) Exercise 7.29 (continued) Analysis: (continued) (d) For an operating reflux ratio = L/V = 0.8, the reflux ratio, R = L/D = 0.8/(1-0.8) = 4. On the McCabe-Thiele diagram below, the rectifying section operating line has a slope of 0.8 and passes through the point, y=0.85, x=0.85. the stripping section operating line passes through the point, y=0.00777, x=0.0777 and intersects the vertical q-line at the point where the rectifying section operating line intersects the q-line. As seen, the equilibrium stages are stepped off starting at the top, with a switch from the rectifying section to the stripping section to minimize the number of stages and, thus, locating the optimal feed stage. The result is just less than 15 equilibrium stages. Call it 14 equilibrium stages plus a partial reboiler. Applying Eq. (6-21), Na = 14/0.55 = 25.5 or 26 actual plates plus the partial reboiler as an equilibrium stage. Exercise 7.30 Subject: Recovery by distillation with open steam of solvent A from water in two feeds. Given: Two saturated liquid feeds, each containing 50 kmol/h of A. Feed 1 contains 40 mol% A and Feed 2 contains 60 mol% A. Unit consists of a column and a total condenser. Open steam is used in lieu of a partial reboiler. Relative volatility = 3.0 for A with respect to water. Distillate is to contain 95 mol% A with a 95% recovery. Assumptions: Constant molar overflow. Open steam enters bottom stage as saturated vapor. Both feeds enter at optimal locations. Find: For an overall plate efficiency of 70% and an R = L/D = 1.33 times minimum, determine the number of actual plates. Compute the bottoms composition. Determine analytically the location of all three operatiing lines. Analysis: The total flow rate of Feed 1 = 50/0.4 = 125 kmol/h. The...
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