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

# 7226226 797 fts from eq 6 53 ks u a v l v 1

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Unformatted text preview: .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(1,495 / 3,600)(30.9) f= 2 =2 = 0.73 or 73% . . 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 1.138(1,209) = 1,376 lbmol/h. The liquid rate leaving the bottom tray = 1,376 + 1,201.9 = 2,578 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 = 2,578(181) 0.0408 . 1,376(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(1,376 / 3,600)(18.8) f= 2 =2 = 0.39 or 39% D U f π 1 − Ad / A ρV 6 (17.6)(314)(1 − 01)(0.0408) . . Exercise 7.48 (continued) Analysis: (continued) (b) Vapor pressure drop per tray is given by Eq. (6-49), ht = hd + hl + hσ 2 uo ρV From Eq. (6-50), hd = 0.186 2 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 = (1,495/3,600)(30.9)/[(2.26)(0.0703)] = 80.7 ft/s and Co = 0.73 80.7 2 0.0703 From Eq. (3), hd = 0186 . = 3.39 inches of liquid 0.732 47.2 (4) Superficial velocity based on bubbling area = Ua = 80.7(2.26/22.6) = 7.97 ft/s From Eq. (6-53), Ks = U a ρV ρ L − ρV 1/ 2 = 7.97 0.0703 47.2 − 0.0703 1/ 2 = 0.31 ft/s From Eq. (6-52), φe = exp(-4.257Ks0.91) = exp[-4.257(0.31)0.91] = 0.23 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 = 727(30.9)/[60(8.33)(47.2/62.4)] = 59.4 gpm From Eq. (6-51), hl = φe hw + Cl qL Lw φe 2/3 59.4 = 0.23 2 + 0.362 ( 42.5 ) 0.23 2/3 = 0.74 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 = 3.39 + 0.74 + 0.13 = 4.26 inches liquid = 0.115 psi/tray. Consider next, the conditions at the bottom tray of the column. From the continuity equation, uo = m/AhρV (4) uo = (1,376/3,600)(18.1)/[(2.26)(0.0408)] = 75.0 ft/s and Co = 0.73 75.02 0.0408 From Eq. (3), hd = 0186...
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## This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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