# 0376 from fig 636a y at flooding 017 from fig 636b

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Unformatted text preview: Analysis: (b) (continued) Column diameter based on conditions at the bottom of the column, where T = 227oF 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)(90.6)/[0.730(227 + 460)] = 0.181 lb/ft3 For the liquid, which is mainly benzene, use L = 48.7 lb/ft3 and L = 0.25 cP. LM L The abscissa in Fig. 6.36a = X = VM V V L 1/ 2 Exercise 7.53 (continued) 612(90.6) 0.181 = 437(91.5) 48.7 1/ 2 = 0.085 From Fig. 6.36a, Y at flooding = 0.12; from Fig. 6.36b, for water/ L = 62.4/48.7 = 1.28, f{L } = 1.5; and from Fig. 6.36c, for L = 0.25 cP, f{L } = 0.78. 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 = 0.12 = 71 (ft/s)2 and uo = 8.43 ft/s . . V 16 0181 150(0.78) FP f { L } f { L } For fraction of flooding = f = 0.7, uV = uo f = 8.43(0.7) = 5.9 ft/s. From Eq. (6-103), column diameter is, 4VM V DT = fu0 V 1/ 2 4VM V = uV V 1/ 2 4(437 /...
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