# Liquid holdup based on conditions at the top of the

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Unformatted text preview: Re L 1/ 3 ah a 2/3 2.46 10-4 = (12) 145 1/ 3 (1.12 ) 2/3 = 0.0294 m3/m3 Liquid holdup based on conditions at the bottom of the column. By the continuity equation, superficial liquid velocity is, uL = LML /LS = (612/3600)(90.6)/ [50.6(3.14)(3.6)2/4] =0.0299 ft/s 0.0299 48.7 u From Eq. (6-98), N Re L = L L = = 308 L a (0.25)(0.000672)281 . u 2 a ( 0.0299 ) 28.1 = 7.80 10 -4 From Eq. (6-99), N FrL = L = 32.2 g Since NRe > 5, use Eq. (6-101), 0. 0 ah / a = 0.85Ch N Re25 N Fr.1 = 0.85(0.876)(308) 0.25 (7.80 10 -4 ) 0.1 = 152 . L L 2 From Eq. (6-97), the fractional liquid holdup is, hL = 12 N FrL N ReL 1/ 3 ah a 2/3 7.80 10-4 = (12) 308 1/ 3 (1.52 ) 2/3 = 0.0412 m3/m3 mV H L , where we must use the slope, m, of the L equilibrium curve instead of a K-value because the equilibrium curve is curved. At the top of the column: (c) From Eq. (7.51), HOG = HG + Estimate HL from Eq. (6-132), using the following properties and parameters: CL = 1.168, hL = 0.0294 m3/m3, = 0.977, and uL = 0.00512 m/s. We need an estim...
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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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