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

0765 57 624 00765 1 2 020 fts from eq 6 52

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Unformatted text preview: = 0.0765 lb/ft3 Use entrainment flooding correlation of Fig. 6.24, where the abscissa is, FLV LM L ρV = VMV ρ L 1/ 2 27,800(18) 0.0765 = 1,077(29) 62.4 1/ 2 = 0.56 From Fig. 6.24, for 24-inch tray spacing, CF = 0.18 ft/s. Because 0.1 < FLV > 1, from below Eq. (6-44), Ad /A = 0.1 + (FLV - 0.1)/9 = 0.15 FHA = 1.0, FF = 0.9, and since σ = 80 dynes/cm, FST = (80/20)0.2 = 1.32 From Eq. (6-24), C = FSTFFFHACF = (1.32)(0.9)(1)(0.18) = 0.21 ft/s From Eq. (6-40), U f = C ρ L − ρV / ρV From Eq. (6-44), for f = 0.80, 4VM V DT = fU f π (1 − Ad / A ) ρV 1/ 2 1/ 2 = 0.21 62.4 − 0.0765 / 0.0765 1/ 2 = 6.0 ft / s 4(1, 077 / 3, 600)(29) = 0.80(6.0)(3.14)(1 − 0.15)(0.0765) (c) Vapor pressure drop per tray is given by Eq. (6-49), ht = hd + hl + hσ From the continuity equation, m = uAρ . Hole velocity for 10% hole area = uo = (1,077/3,600)(29)/(3.14/4)(6)2(0.0765)(0.1) = 40 ft/s Superficial velocity = (0.1)(40) = 4 ft/s 1/ 2 = 6.0 ft (1) Exercise 6.23 (continued) Analysis: (c) (continued) 2 uo ρV 402 0.0765 = 0186 . = 0.68 in. of liquid 2 2 Co ρ L 0.73 62.4 Active bubbling area for Ad /A = 0.15 is Aa = A - 2Ad = 0.7 A. Therefore, Ua = 4/0.7 = 5.7 ft/s From Eq. (6-50), hd = 0186 . ρV ρ L − ρV From Eq. (6-53), Ks = U a 1/ 2 0.0765 = 5.7 62.4 − 0.0765 1/ 2 = 0.20 ft/s From Eq. (6-52), φe = exp(-4.257Ks0.91) = exp[-4.257(0.20)0.91] = 0.37 Assume a 2-inch weir height = hw . . From Eq. (6-54), C = 0.362 + 0.317 exp( −35hw ) = 0.362 + 0.317 exp[ −35(2.0)] = 0.362 Take weir length, Lw = 0.73DT = 0.73(6)(12) = 53 in. qL From Eq. (6-51), hl = φe hw + C Lw φe 2/3 1, 000 = 0.37 2 + 0.362 ( 53) 0.37 2/3 = 2. 58 in. of liquid From Eq. (6-55), with maximum bubble size of 3/16 inch = 0.00476 m, hσ = 6σ / gρ L DBmax = 6(80 / 1,000) / (9.8)(1,000)(0.00476) = 0.0103 m = 0.40 in. of liquid From Eq. (2), ht = hd + hl + hσ =0.68 + 2.58 + 0.40 = 3.66 in. liquid = 0.13 psi/tray (d) DT = 6 ft, A = 28.3 ft2, Aa = 0.7(28.3) = 19.8 ft2 = 18,390 cm2 φe = 0.37, hl = 2.58 in. = 6.55 cm, ρG = 0.0765 lb/ft3 = 1.23 kg/m3 Ua = 5.7 ft/s = 1.74 m/s , Uf =6.0 ft/s, f = Ua/Uf = 5.7/6.0 = 0.95 F = UaρV0..5 =1.74(1.23)0.5 = 1.93 (kg/m)0.5/s , qL = 1,000 gpm = 63,100 cm3/s hA 6.55(18,390) From Eq. (6-64), t L = l a = = 19 s . qL 63,100 1 − φ e hl (1 − 0.37)6.55 = = 0.064 s φ eU a 0.37(174) From the Wilke-Chang Eq. (3-39), DL = 0.96 x 10-5 cm2/s 0 From Eq. (6-67), k L a = 78.8 DL.5 ( F + 0.425) = 78.8(0.96 × 10−5 )(193 + 0.425) = 0.58 s-1 . From Perry's Handbook, with T and P corrections based on Eq. (3-36), DV = 0.086 cm2/s From Eq. (6-65), t G = From Eq. (6-66), kG a = 0.5 1, 030 DV ( f − 0.842 f 2 ) 0.5 l h = 1, 030(0.086)0.5 0.95 − 0.842 ( 0.95 ) From Eq. (6-63), N L = k L at L = 0.58(1.9) = 1.1 6.55 0.5 2 = 2. 56 s-1 Analysis: (continued) Exercise 6.23 (continued) From Eq. (6-62), N G = kG atG = 2.56(0.064) = 0.16 (d) From Example 6.2, for benzene, S = KV/L = 9.89 1 1 1 From Eq. (6-61), N OG = = = = 0.066 1 KV / L 1 9.89 6.25 + 8.99 + + NG NL 016 11 . . From a rearrangement of...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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