Separation Process Principles- 2n - Seader & Henley - Solutions Manual

D h packing hydraulic diameter 4 reynolds number n rel

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Unformatted text preview: = L L = = 145 µ L a (0.31)(0.000672)281 . u 2 a ( 0.0168 ) 28.1 From Eq. (6-99), N FrL = L = = 2.46 ×10 −4 g 32.2 Since NRe > 5, use Eq. (6-101), 2 0. 0 ah / a = 0.85Ch N Re25 N Fr.1 = 0.85(0.876)(145) 0.25 (2.46 × 10−4 ) 0.1 = 112 . L L Exercise 7.53 (continued) Analysis: (a) top of the column (continued) From Eq. (6-97), the fractional liquid holdup is, hL = 12 N FrL N Re L 1/ 3 2/3 ah a 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 /ρLS = (612/3600)(90.6)/ [50.6(3.14)(3.6)2/4] =0.0299 ft/s 0.0299 48.7 uρ From Eq. (6-98), N Re L = L L = = 308 µ 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, ε = 0.977, and uL = 0.00512 m/s. We need an estimate of the diffusivity of benzene in toluene at high concentrations of benzene. Because these two components form a nearly ideal solution, assume that this diffusivity is equal to that of toluene (A) at infinite dilution in benzene (B), as estimated from the Eq. (3-42), using Table 3.3 and Table 3.2. In this equation, T = 180oF = 356 K, B = 205.3, A = 245.5, µΒ = 0.32 cP, and υΒ = 6(14.8)+6(3.7) -15 = 96 DL = DAB = 155 × 10−8 . T 1.29 (PB0.5 / PA0.5 ) 3551.29 (205.30.5 / 24550.42 ) . = 155 × 10−8 . = 4.28 × 10−5 cm2/s 0.92 0.23 0.92 0.23 µB υ B 0.32 96 Exercise 7.53 (continued) Analysis: (c) top of the column (continued) From Eq. (6-132), 11 HL = CL 12 = 0.24 a aPh 1/ 6 4hL ε DL au L 1/ 2 uL a a 1 1 = aPh 1.168 12 1/ 6 4(0.0294)(0.977) (4.28 × 10 −9 )(92.3)(0.00512) 1/ 2 0.00512 92.3 m To compute (aPh /a), we need for the liquid phase, the Reynolds, Weber, and Froude numbers from equations (6-138), (6-139), and (6-140), respectively, based on the packing hydraulic diameter from (6-137). d h = packing hydraulic diameter = 4 Reynolds number = N ReL , h = ε 0.977 =4 = 0.0423 m = 0.139 ft a 92.3 u L d h ρ L (0.0168)(0.139)(50.6) = = 567 [(0.31)(0.000672)] µL Take the surface tension of benzene at 180oF as σ = 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 2ρ d = L L h= σ 0.0464 Froude number = N FrL , h ( 0.0168) = 0.0000631 u2 = L= gd h 32.2(0.139) 2 2 From (6-136), ( ) (N aPh −1/ 2 = 1.5 ( ad h ) N Re L ,h a −0.2 = 1.5 [ (92.3)(0.0423) ] −1/ 2 ) (N ) 0.75 We L , h −0.45 FrL , h ( 567 ) ( 0.0428 ) ( 0...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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