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

10 kmolh almost equal to the m feed rate distillate

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Unformatted text preview: se above for the equilibrium-stage calculations. However, the % absorption of propane is now 39.2%. Five passes were needed to handle the high liquid flow rate. The back-calculated Murphree vapor tray efficiencies were mainly in the following ranges: Methane: 54 to 80% Ethane: 55 to 75% Propane: 46 to 77% n-Butane: 65 to 77% n-Pentane: 74 to 80% n-Dodecane: 73 to 78% (c) The tray efficiencies were higher than anticipated, probably due to somewhat higher temperatures at the elevated pressure, which reduces the liquid viscosity. The rate-based method has the advantage of predicting the efficiency. However, in this exercise, as pointed out above, the near 40% absorption of propane is independent of the number of equilibrium stages above three. A column diameter of 13 m is probably too large. Two or more columns in parallel should be considered. Exercise 12.18 Subject: Separation of methanol (M) from a mixture with ethanol (E) and water (W) by distillation, using equilibrium-stage and rate-based methods with a trayed column. Given: A feed at 1.3 atm and 316 K of 142.46 kmol/h of 65.36 mol% M, 3.51 mol% E, and 31.13 mol% W. Column is equipped with a total condenser (1.1 atm in and 1.0 atm out) and a partial reboiler. Reflux ratio = 1.2 and distillate rate = 93.10 kmol/h (almost equal to the M feed rate). Distillate is to have a mole fraction of W less than or equal to 0.0001. UNIFAC method for activity coefficients. Sieve-tray efficiency of 85%. Find: (a) Number of equilibrium stages, optimal feed stage location, and the split of all components. (b) Actual number of sieve trays, the split of all components, and the Murphree vapor tray efficiencies. Comparison and discussion of results. Analysis: The ChemSep method was applied to both parts, using the ideal gas law, Antoine vapor pressure, and excess enthalpy with UNIFAC. A column bottoms pressure of 1.3 atm was assumed. The calculations used automatic initialization with Newton's method and defaults for step limits on flows, temperature, and composition. The bottoms flow rate was specified as 142.46 - 93.10 = 49.36 kmol/h. (a) For the equilibrium-stage case, the total number of stages and the optimal feed stage were varied, starting with 42 stages (including the condenser and reboiler) and a feed stage of 21 from the condenser, until the water mol% specification in the distillate was met. A satisfactory result was achieved for 24 stages (including the condenser and reboiler) with the feed to stage 20 from the condenser. The splits were as follows: Component Feed Flow, kmol/h: Methanol 93.11 Ethanol 5.00 Water 44.35 Total: 142.46 Mole fraction: Methanol 0.6536 Ethanol 0.0351 Water 0.3113 Distillate Bottoms 89.49 3.60 0.01 93.10 3.62 1.40 44.34 49.36 0.9612 0.0387 0.000087 0.0734 0.0283 0.8983 Exercise 12.18 (continued) Analysis: (continued) (b) For the rate-based case, a sieve-tray column was used, with default properties and the Wilke-Chang method for liquid diffusivities. The Chan-Fair method was used to estimate masstransfer coefficients and interfacial area. Plug flow was used for the vapor and the mixed flow option was applied to the liquid. Initial...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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