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

Froth height in downcomer analysis from exercise 616 v

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

Ask a homework question - tutors are online