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

2600703 105 fts and co 073 1052 00703 from eq 3

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Unformatted text preview: with densities from a simulation program, where the abscissa is, FLV LM L ρV = VM G ρ L 1/ 2 = 945(30.9) 0.0703 1,944(30.9) 47.2 1/ 2 = 0.0192 (1) From Fig. 6.24, for 24-inch tray spacing, CF = 0.38 ft/s. Because FLV < 1, Ad /A = 0.1. FHA = 1.0, FF = 1.0, and since σ = 20 dynes/cm, FST = 1.0. From Eq. (6-24), C = FSTFFFHACF = (1)(1)(1)(0.38) = 0.38 ft/s From Eq. (6-40), U f = C ρ L − ρV / ρV 1/ 2 = 0.38 47.2 − 0.0703 / 0.0703 1/ 2 = 9.84 ft / s From a rearrangement of Eq. (6-44), solving for the fraction of flooding, 4VMV 4(1,944 / 3,600)(30.9) f= 2 =2 = 0.95 or 95% Unacceptable! D U f π 1 − Ad / A ρV 6 (9.84)(314)(1 − 01)(0.0703) . . At the bottom of the column, the boilup rate from Exercise 7.41 is 1.3(1.138)(1,209) = 1,789 lbmol/h. The liquid rate leaving the bottom tray = 1.3(1,376 + 1,201.9) = 3,351 lbmol/h. Use the entrainment flooding correlation of Fig. 6.24, with densities and molecular weights from a simulation program, where the abscissa is, FLV LM L ρV = VM G ρ L 1/ 2 = 3,351(181) 0.0408 . 1,789(18.8) 59.5 1/ 2 = 0.0473 From Fig. 6.24, for 24-inch tray spacing, CF = 0.37 ft/s. Because FLV < 1, Ad /A = 0.1. FHA = 1.0, FF = 1.0, and since σ = 20 dynes/cm, FST =(58/20)0.2 = 1.24 From Eq. (6-24), C = FSTFFFHACF = (1.24)(1)(1)(0.37) = 0.46 ft/s From Eq. (6-40), U f = C ρ L − ρV / ρV 1/ 2 = 0.46 59.5 − 0.04 / 0.0408 1/ 2 = 17.6 ft / s From a rearrangement of Eq. (6-44), solving for the fraction of flooding, 4VMV 4(1,789 / 3,600)(18.8) f= 2 =2 = 0.51 or 51% D U f π 1 − Ad / A ρV 6 (17.6)(314)(1 − 01)(0.0408) . . Exercise 7.49 (continued) Analysis: (continued) (b) Vapor pressure drop per tray is given by Eq. (6-49), ht = hd + hl + hσ u 2 ρV From Eq. (6-50), hd = 0.186 o2 Co ρ L (2) (3) Column area = A = πD2/4 = 3.14(6)2/4 = 28.3 ft2 Take downcomer area = Ad = 0.1(28.3) = 2.83 ft2 Bubbling area = Aa = A - 2Ad = 28.3 - 2(2.83) = 22.6 ft2 Hole area = Ah = 0.1(22.6) = 2.26 ft2 Consider first, the conditions at the top of the column. From the continuity equation, uo = m/AhρV uo = (1,944/3,600)(30.9)/[(2.26)(0.0703)] = 105 ft/s and Co = 0.73 1052 0.0703 From Eq. (3), hd = 0186 . = 5.73 inches of liquid 0.732 47.2 (4) Superficial velocity based on bubbling area = Ua = 105(2.26/22.6) = 10.4 ft/s From Eq. (6-53), Ks = U a ρV ρ L − ρV 1/ 2 = 10.4 0.0703 47.2 − 0.0703 1/ 2 = 0.40 ft/s From Eq. (6-52), φe = exp(-4.257Ks0.91) = exp[-4.257(0.40)0.91] = 0.16 From Eq. (6-54), Cl = 0.362 + 0.317 exp(−3.5hw ) = 0.362 + 0.317 exp[−3.5(2.0)] = 0.362 Weir length = Lw = 42.5 in. Volumetric liquid rate = qL = 945(30.9)/[60(8.33)(47.2/62.4)] = 77.2 gpm From Eq. (6-51), hl = φe hw + Cl qL Lw φe 2/3 77.2 = 0.16 2 + 0.362 ( 42.5 ) 0.16 2/3 = 0.61 in. liquid From Eq. (6-55), with maximum bubble size = DH = 3/16 inch = 0.00476 m and liquid density = 47.2 lb/ft3 or 756 kg/m3, hσ = 6σ / gρ L DBmax = 6(20 / 1000) / (9.8)(756)(0.00476) = 0.0034 m = 0.13 in. liquid From Eq. (2), ht = hd + hl + hσ =5.73 + 0.61 + 0.13 = 6.4...
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