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

Analysis a base column diameter on the top tray where

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Unformatted text preview: 0.37)2.24 = = 0.028 s φ eU a 0.37(137) 0 From Eq. (6-67), k L a = 78.8 DL.5 ( F + 0.425) = 78.8(181 × 10−5 )(190 + 0.425) = 0.78 s-1 . . From Eq. (6-66), kG a = 0 1,030 DV .5 f − 0.842 f 2 0.5 l h = 1,030(7.86 × 10 −2 ) 0.5 0.44 − 0.842 0.44 2.24 0.5 2 = 53.4 s-1 From Eq. (6-63), N L = k L at L = 0.78(28.6) = 22.3 From Eq. (6-62), N G = k G at G = 53.4(0.28) = 15 . From Example 6.7, KV/L = 0.662 1 1 1 From Eq. (6-61), N OG = = = = 144 . 1 KV / L 1 0.662 0.667 + 0.030 + + NG NL 15 22.3 . Mass transfer is controlled by the vapor phase. From a rearrangement of Eq. (6-52), EOV = 1 − exp(− N OG ) = 1 − exp(−1.44) = 0.76 or 76%. Exercise 6.22 Subject: Sizing, hydraulics, and mass transfer for an acetone absorber Given: Acetone absorber of Fig. 6.1, using sieve trays with 10% hole area, 3/16-inch holes, and an 18-inch tray spacing. Foaming factor = 0.85. Find: (a) Column diameter for 75% of flooding. (b) Vapor pressure drop per tray. (c) Number of transfer units. (d) Number of overall transfer units. (e) Controlling resistance to mass transfer. (f) Murphree point vapor efficiency If 30 trays are adequate. Analysis: (a) Base column diameter on the top tray, where pressure is the lowest and gas flow rate is the highest. Use data in Fig. 6.1. L = 1,943 kmol/h, ML = 18, ρL = 1,000 kg/m3 V = 6.9 + 144.3 + 536.0 + 22.0 + 0.1 = 709.3 kmol/h, MV = 28.8, P = 90 kPa, T = 25oC From the ideal gas law, ρV = PM/RT = (90)(28.8)/(8.314)(298) = 1.05 kg/m3 Use entrainment flooding correlation of Fig. 6.24, where the abscissa is, 1/ 2 1/ 2 LM L ρV 1,943(18) 105 . FLV = = = 0.056 VMV ρ L 709.2(28.8) 1,000 From Fig. 6.24, for 18-inch tray spacing, CF = 0.28 ft/s. Because FLV < 1, Ad /A = 0.1. FHA = 1.0, FF = 0.85, and since σ = 70 dynes/cm, FST = (70/20)0.2 = 1.285 From Eq. (6-24), C = FSTFFFHACF = (1.285)(0.85)(1)(0.28) = 0.306 ft/s From Eq. (6-40), U f = C ρ L − ρV / ρV From Eq. (6-44), 4VM V DT = fU f π (1 − Ad / A ) ρV 1/ 2 1/ 2 = 0.306 1,000 − 105 / 105 . . 1/ 2 4(709.2 / 3, 600)(28.8) = 0.75(9.44 / 3.28)(3.14)(1 − 0.1)(1.05) = 9.44 ft / s 1/ 2 = 1.89 m = 6.2 ft (b) Vapor pressure drop per tray is given by Eq. (6-49), ht = hd + hl + hσ (1) From the continuity equation, m = uAρ , hole velocity for 10% hole area = uo = (709.2/3,600)(28.8)/(3.14/4)(1.9)2(1.05)(0.1) = 19.0 m/s =62.3 ft/s Superficial velocity = (0.1)(19.0) = 1.90 m/s = 6.23 ft/s u 2 ρV 62.32 105 . From Eq. (6-50), hd = 0186 o2 . = 0186 . = 142 in. of liquid . 2 Co ρ L 0.73 1,000 Active bubbling area for Ad /A = 0.1 is Aa = A - 2Ad = 0.8 A. So, Ua = 6.23/0.8 = 7.8 ft/s From Eq. (6-53), Ks = U a ρV ρ L − ρV 1/ 2 1.05 = 7.8 1,000 − 105 . 1/ 2 = 0.253 ft/s From Eq. (6-52), φe = exp(-4.257Ks0.91) = exp[-4.257(0.253)0.91] = 0.30 Assume a 2-inch weir height = hw Exercise 6.22 (continued) Analysis: (b) (continued) . . 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(6.2) = 4.52 ft = 54.3 i...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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