# Assumptions constant molar overflow partial reboiler

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Unformatted text preview: diameter based on conditions at the top of the column, where T = 180oF and P = 14.7 psia = 1 atm. Use Figs. 6.36a, b, and c to compute flooding velocity, following Example 6.13. From the ideal gas law, V = PMV/RT = (1)(78.8)/[0.730(180 + 460)] = 0.171 lb/ft3 For the liquid, which is mainly benzene, use L = 50.6 lb/ft3 and L = 0.31 cP. LM L The abscissa in Fig. 6.36a = X = VM V V L 1/ 2 437(79.4) 0.171 = 712(78.8) 50.6 1/ 2 = 0.0376 From Fig. 6.36a, Y at flooding = 0.17; from Fig. 6.36b, for water/ L = 62.4/50.6 = 1.23, f{L } = 1.5; and from Fig. 6.36c, for L = 0.31 cP, f{L } = 0.8. From Table 6.8, FP = 16 for 50-mm Hiflow rings. From a rearrangement of the ordinate of Fig. 6.36a, g H2 O (L) 1 32.2 62.4 1 2 uo = Y = 0.17 = 127 (ft/s)2 and uo = 11.2 ft/s FP V f { L } f { L } 16 0171 123(0.8) . . For fraction of flooding = f = 0.7, uV = uo f = 11.2(0.7) = 7.84 ft/s. From Eq. (6-103), column diameter is, 4VM G DT = fu0 G 1/ 2 4VM G = uG G 1/ 2 4(712 / 3600)(78.8) = 7.84(3.14)(0.171) 1/ 2 = 3.8 ft....
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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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