V 26230 lbmolh and l 11740 lbmolh in fig 624 flv

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Unformatted text preview: -42), use FHA = 1.0 and FF = 1.0. For methanol, = 17 dyne/cm Therefore, FST = (17/20)0.2 = 0.97 From Eq. (6.42), C = 0.97(1)(1)0.38 = 0.37 ft/s. From Eq. (6-40), - V Uf =C L V 1/ 2 = 0.37 45.3 - 01566 . 0.1566 1/ 2 = 6.28 ft/s Since FLV < 0.1, from below Eq. (6-44), Ad /A = 0.1. Assume f = 0.80. From Eq. (6-44), 4VM V DT = fU f (1 - Ad / A)V 1/ 2 4(26, 230 / 3600)(34) = 0.80(6.28)(3.14)(1 - 0.1)(0.1556) 1/ 2 = 21.2 ft Therefore, column would be swaged. (b) Sizing of reflux drum. Assume a liquid residence time in the reflux drum of 5 minutes (0.0833 h), half full. From Eq. (7-44), VV = From Eq. (7-46), V DT = V 1/ 3 2 LM L t 2(26,230)(34)(0.0833) = = 3,280 ft3 L 45.3 3, 280 = 3.14 1/ 3 = 10.1 ft From Eq. (7-46), H = 4DT = 4(10.1) = 40.4 ft Exercise 7.48 Subject: Given: Tray hydraulics for methanol-water separation' Data from Exercise 7.41. Percent of flooding. Tray pressure drop in psi. Weeping potential. Entrainment rate. Downcomer froth height. Find: (a) (b) (c) (d) (e) Analysis: Use the material balance from Exercise 7.41. Make calculations at the top and...
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This note was uploaded on 09/08/2008 for the course CHE 244 taught by Professor Selebi during the Spring '06 term at Lehigh University .

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