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

# Froth height in downcomer analysis from exercise 616 v

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Unformatted text preview: 530(26.6) 41.1 1/ 2 = 1.487 (1) For 24-inch tray spacing, from Fig. 6.24, CF = 0.082 ft/s From relations below Eq. (6-44), since FLV is < 0.1, Ad/A = 0.1 From relation below Eq. (6-42), FST = (σ/20)0.2 = (18.4/20)0.2 =0.98. FF = 0.75 , FHA = 1.0 From Eq. (6-42), C = FST FF FHA CF = (0.98)(0.75)(1.0)(0.082) = 0.0603 ft/s From Eq. (6-40), U f = C ( ρ L − ρV ) / ρV 1/ 2 4VM V From Eq. (6-44), DT = fU f π (1 − Ad / A ) ρV = 0.0603 ( 41.1 − 1.924 ) /1.924 1/ 2 1/ 2 = 0.272 ft/s 4(530 / 3, 600)(26.6) = 0.85(0.272)(3.14)(1 − 0.1)(1.924) 1/ 2 = 3. 53 ft Exercise 6.20 Subject: Hydraulic calculations for bottom tray of a sieve-tray column Given: Absorber of Exercise 6.16 with an absorbent flow rate of 40,000 gpm of nC8. Sieve trays on 24-inch spacing , with a weir height of 2.5 inches and 1/4-inch holes. Foaming factor = 0.80 and fraction flooding = 0.70. Find: (a) (b) (c) (d) (e) (f) Column diameter at bottom. Vapor pressure drop/tray. Whether weeping will occur. Entrainment rate. Fractional decrease in Murphree efficiency due to entrainment. Froth height in downcomer. Analysis: From Exercise 6.16, V = 800,000/359 = 2,230 lbmol/min of entering gas. Average MW of gas = 0.725(2.016) + 0.25(16.04) + 0.025(30.05) = 6.22 Assuming ideal gas law for gas at 400 psia and 100oF, ρV = PM/RT = (400)(6.22)/(10.73)(560) = 0.414 lb/ft3 For the liquid, neglect absorbed components. ρL = 43.9 lb/ft3 = 5.86 lb/gal MW of nC8 = 114. Therefore, L = 40,000(5.86)/114 = 2,056 lbmol/min (a) Use entrainment flooding correlation of Fig. 6.24, where the abscissa is, FLV LM L ρV = VM V ρ L 1/ 2 = 2, 056(114) 0.414 2, 230(6.22) 43.9 1/ 2 = 1.64 (1) From Fig. 6.24, for 24-inch tray spacing, CF = 0.08. Because FLV > 1, Ad /A = 0.2. FHA = 1.0, FF = 0.8, and since σ = 20 dynes/cm, FST = 1.0. From Eq. (6-24), C = FSTFFFHACF = (1)(0.8)(1)(0.08) = 0.064 ft/s From Eq. (6-40), U f = C ρ L − ρV / ρV 1/ 2 = 0.064 43.9 − 0.414 / 0.414 4VM V From Eq. (6-44), DT = fU f π (1 − Ad / A ) ρV 1/ 2 1/ 2 = 0.66 ft / s 4(2, 230 / 60)(6.22) = 0.70(0.66)(3.14)(1 − 0.2)(0.414) 1/ 2 = 44 ft This is a very large diameter. Would probably use two parallel columns of 31 ft each, but continue with 44 ft diameter. (b) Vapor pressure drop per tray is given by Eq. (6-49), ht = hd + hl + hσ (2) From the continuity equation, m = uAρ , hole velocity for 10% hole area = uo = (2,230/60)(6.22)/(3.14/4)(44)2(0.414)(0.1) = 3.68 ft/s and superficial velocity = 0.368 ft/s u 2 ρV 3.682 0.414 From Eq. (6-50), hd = 0186 o2 . = 0186 . = 0.045 in. of nC8 Co ρ L 0.732 43.9 Analysis: (continued) Exercise 6.20 (continued) Active bubbling area for Ad /A = 0.2 is Aa = A - 2Ad = 0.6 A. So, Ua = 0.368/0.6 = 0.61 ft/s From Eq. (6-53), Ks = U a ρV ρ L − ρV 1/ 2 1/ 2 0.414 = 0.61 43.9 − 0.414 = 0.06 ft/s From Eq. (6-52), φe = exp(-4.257Ks0.91) = exp[-4.257(0.06)0.91] = 0.72 From Eq. (6-54), C = 0.362 + 0.317 exp( −35hw ) = 0.362 + 0.317 exp[ −3.5(2.5)] = 0.362 . Take weir length, Lw = 0.73DT = 0.73(44) = 32.1 ft = 385 i...
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