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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 /...
View Full Document

Ask a homework question - tutors are online