# 53 continued 11 hl cl 12 1 6 4hl dl au l 1 2 ul a a

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Unformatted text preview: r = N WeL ,h = 2 uL L d h ( 0.0299 ) (48.7)(0.139) = = 0.152 0.0397 2 2 Froude number = N FrL ,h ( 0.0299 ) = 0.000200 u2 = L = gd h 32.2(0.139) From (6-136), aPh -1/ 2 = 1.5 ( ad h ) N Re L ,h a = 1.5 [ (92.3)(0.0423) ] ( ) (N -0.2 -1/ 2 We L , h ) (N ) 0.75 FrL , h -0.45 (1205 ) ( 0.152 ) ( 0.000200 ) -0.2 0.75 -0.45 = 2.07 HL = 0.26 1 = 0.126 m 2.07 Analysis: (c) bottom of the column (continued) Exercise 7.53 (continued) Now estimate the value of HG from Eq. (6-133). Estimate the gas diffusivity from Eq. (3-36), by ratio from the value above for the top of the column, DV = 0.052(382/354)1.75 = 0.059 cm2/s Use the following additional properties and parameters, together with V = 0.0092 cP and V = 0.179 lb/ft3 , where from Eq. (6-134), N ReV = From (6-135),) NScV = uV V (5.9)(0.181) = = 6,150 aV (28.1)(0.0092)(0.000672) V (0.0092)(0.000672) = = 0.54 V DV 0.059 (0.181) (2.54) 2 (12) 2 From Eq. (6-133), 1 1/ 2 4 HG = ( - hL ) CV a4 1/ 2 (N ) (N ) -3/ 4 ReV ScV -1/ 3 uV a DV aPh -3/ 4 1 1/...
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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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