# 8637 15 96 dl dab 155 108 t 129 pb05 pa05

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Unformatted text preview: ydraulic diameter = 4 0.977 =4 = 0.0423 m = 0.139 ft a 92.3 Reynolds number = N ReL , h = u L d h L (0.0168)(0.139)(50.6) = = 567 [(0.31)(0.000672)] L Take the surface tension of benzene at 180oF as = 21 dynes/cm = 0.00144 lbf/ft or 0.00144(32.2) = 0.0464 lbm/s2 Weber number = N WeL ,h ( 0.0168 ) (50.6)(0.139) = 0.0428 u 2 d = L L h = 0.0464 2 2 Froude number = N FrL , h ( 0.0168) = 0.0000631 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 ( 567 ) ( 0.0428 ) ( 0.0000631) -0.2 0.75 -0.45 = 1.56 HL = 0.24 1 = 0.154 m 1.56 Exercise 7.53 (continued) Analysis: (c) at the top of the column (continued) Now estimate the value of HG from Eq. (6-133). Estimate the gas diffusivity from Eq. (3-36), where from Table 3.1, . V = 6(15.9 + 2.31) - 18.3 = 90.96 for benzene, and V = 7(15.9 + 2.31) + 2.31 - 18.3 = 1115 for toluene and molecular weights are 787 for benzene and 92 for toluene. 0.00143T 1.75 DG = DAB = 1/ 2...
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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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