# But use the constant molar overflow values a column

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Unformatted text preview: 0 + 26,230 = 36,640 lbmol/h In Fig. 6.24, FLV LM L V = VMV L 1/ 2 = 36,640(18.2) 0.0994 26,230(18.29) 57.9 1/ 2 = 0.058 From Fig. 6.24, for 24-inch plate spacing, CF = 0.27 ft/s. For Eqs. (6-40) and (6-42), use FHA = 1.0 and FF = 1.0. For water, = 51 dyne/cm Therefore, FST = (51/20)0.2 = 1.21. From Eq. (6.42), C = 1.21(1)(1)0.37 = 0.45 ft/s. Analysis: (a) (continued) Exercise 7.47 (continued) 1/ 2 - V From Eq. (6-40), U f = C L V 57.9 - 0.0994 = 0.45 0.0994 1/ 2 1/ 2 = 10.9 ft/s Since FLV &lt; 0.1, from below Eq. (6-44), Ad /A = 0.1. Assume f = 0.80. 4VM V From Eq. (6-44), DT = fU f (1 - Ad / A)V 4(26, 230 / 3600)(18.2) = 0.80(10.9)(3.14)(1 - 0.1)(0.0994) 1/ 2 = 14.7 ft Column diameter at the top: From a simulation program, at 189oF and 33 psia, V = 0.1566 lb/ft3 and L =45.3 lb/ft3. V = 26,230 lbmol/h and L = 11,740 lbmol/h In Fig. 6.24, FLV LM L V = VMV L 1/ 2 = 11,740(34) 01566 . 26,230(34) 45.3 1/ 2 = 0.026 From Fig. 6.24, for 24-inch plate spacing, CF = 0.38 ft/s For Eqs. (6-40) and (6...
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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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