Therefore the holdup is independent of the gas rate

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Unformatted text preview: top of the column. By the continuity equation, superficial liquid velocity is, uL = LML /LS = (727/3600)(30.9)/ [48(3.14)(DT)2/4] = 0.166/(DT)2 ft/s 2 0166 / DT 48 . 33900 u L L From Eq. (6-98), N Re L = = = 2 L a (0.35)(0.000672)a DT a From Eq. (6-99), N FrL 2 2 u L a ( 0.166 / DT ) a a = = = 0.000856 4 g 32.2 DT 2 Packing DT, ft uL, ft/s a, ft2/ft3 NOR PAC 4.3 0.00898 26.5 69.2 6.63 x 10-5 Montz 5.3 0.00591 91.4 13.2 9.91 x 10-5 0. 0 Since both values of N Re L > 5, use Eq. (6-101), ah / a = 0.85Ch N Re25 N Fr.1 L L For NOR PAC, ah / a = 0.85(0.651)(69.2) 0.25 (6.63 10 -5 ) 0.1 = 0.61 For Montz, ah / a = 0.85(0.482)(13.2) 0.25 (9.91 10 -5 ) 0.1 = 0.31 From Eq. (6-97), the fractional liquid holdup is hL = 12 6.63 10 -5 For NOR PAC, hL = (12) 69.2 1/ 3 N Re L N FrL N FrL N Re L 1/ 3 ah a 2/3 ( 0.61) 2/3 = 0.0162 m3/m3 9.91 10-5 2/3 For Montz, hL = (12) ( 0.31) = 0.0241 m3/m3 13.2 These liquid holdups at the top of the column are quite small. 1/ 3 Exercise 7.52 (continued) Analysis: (a) (conti...
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