{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

# D h packing hydraulic diameter 4 reynolds number n rel

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online