# Therefore at the top of the column l rd 0947768 727

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Unformatted text preview: Column diameter based on conditions at the top of the column, where T = 154oF 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)(30.9)/[0.730(154 + 460)] = 0.0689 lb/ft3 For the liquid, which is mainly methanol, use L = 0.77 g/cm3 = 48 lb/ft3 and L = 0.35 cP. LM L The abscissa in Fig. 6.36a = X = VM V V L 1/ 2 727(30.9) 0.0689 = 1495(30.9) 48 1/ 2 = 0.0184 From Fig. 6.36a, Y at flooding = 0.21; from Fig. 6.36b, for water/ L = 1/0.77 = 1.3, f{L } = 1.6; and from Fig. 6.36c, for L = 0.35 cP, f{L } = 0.8. From Table 6.8, FP = 14 for NOR PAC and FP = 33 for Montz. Analysis: (b) (continued) Exercise 7.52 (continued) g H2 O (L) FP V 1 f { L } f { L } 2 From a rearrangement of the ordinate of Fig. 6.36a, uo = Y 32.2 62.4 1 = 341 (ft/s)2 and uo = 18.5 ft/s 14 0.0689 16(0.8) . 32.2 62.4 1 2 For Montz, uo = 0.21 = 145 (ft/s)2 and uo = 12.0 ft/s 33 0.0689 16(0.8) . For fraction of flooding = f = 0.7, uV = uo f = 18.5(0.7) = 13.0...
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