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

# Find column diameter for 80 of flooding analysis use

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Unformatted text preview: te. For the 3 ft diameter column of Examples 6.1 and 6.4, liquid rate = 151.5 kmol/h of water. Assume the liquid rate is proportional to the column cross-sectional area. Thus, for the 11 ft diameter, the liquid rate = 151.5(11/3)2 = 2,037 kmol/h = 162 gpm. From Fig. 6.16, a single-pass tower is adequate. Assume liquid flow path = ZL = 7 ft. Eq. (6-34) is used to compute EMV, which requires the Peclet number, given by Eq. (6-36), Zu N Pe = L = 7(6) = 42 DE N From Eq. (6-35), η = Pe 2 4 λEOV 1+ N Pe 1/ 2 −1 = 42 2 1+ 4(0.68)(0.55) 42 1/ 2 − 1 = 0.37 From Eq. (6-34), EMV/EOV = 1.20. Therefore, EMV = 1.20(0.55) = 0.66 From Eq. (6-37), log [1 + EMV (λ − 1) ] log [1 + 0.66(0.68 − 1)] Eo = = 0.62 = log λ log 0.68 From these results, we see that the column is operating closely to plug flow for the liquid. Exercise 6.18 Subject: Estimation of column diameter based on conditions at bottom tray of a reboiled stripper. Given: Vapor and liquid conditions at bottom tray given in Fig. 6.48. Valve trays with 24-inch spacing. Find: Column diameter for 80% of flooding. Analysis: Use entrainment flooding correlation of Fig. 6.24, where the abscissa is, 1/ 2 LM L ρG FLV = VM G ρ L From Fig. 6.48, V = 546.2 lbmol/h and L = 621.3 lbmol/h (1) Average molecular weights of the gas and liquid are computed from M = C i =1 zi Mi , Component M y My x Mx Ethane 30.07 0.000006 0.0 0.000010 0.0 Propane 44.10 0.004817 0.2 0.001448 0.1 n-Butane 58.12 0.602573 35.0 0.391389 22.7 n-Pentane 72.15 0.325874 23.5 0.430599 31.1 n-Hexane 86.18 0.066730 5.8 0.176563 15.2 Total 64.5 69.1 Therefore, MV = 64.5 and ML = 69.1 Gas volumetric flow rate = QV = 6.192 ft3/s. Therefore, ρV = VMV /QV = (546.2)(64.5)/(6.192)(3,600) = 1.58 lb/ft3 Liquid volumetric flow rate = 171.1 gpm = (171.1)(60)/7.48 = 1,372 ft3/h Therefore, ρL = LML /QL = (621.3)(69.1)/1,372 = 31.3 lb/ft3 1/ 2 621.3(69.1) 158 . From Eq. (1), FLV = = 0.274 546.2(64.5) 31.3 Assume 24-inch tray spacing. From Fig. 6.24, CF = 0.24 ft/s From relations below Eq. (6-44), since FLV is between 0.1 and 1.0, F − 0.1 0.274 − 01 . Ad = 0.1 + LV = 01 + . = 0.119 A 9 9 From Perry's Handbook, surface tension = σ = 7 dyne/cm From relation below Eq. (6-42), FST = (σ/20)0.2 = (7/20)0.2 = 0.8. FF = 1.0 , FHA = 1.0 From Eq. (6-42), C = FST FF FHA CF = (0.8)(1.0)(1.0)(0.24) = 0.192 ft/s From Eq. (6-40), U f = C ( ρ L − ρV ) / ρV 1/ 2 = 0.192 ( 31.3 − 1.58 ) /1.58 1/ 2 = 0.833 ft/s Exercise 6.18 (continued) From Eq. (6-44), 4VM V DT = fU f π (1 − Ad / A ) ρV 1/ 2 4(546.2 / 3, 600)(64.5) = 0.80(0.833)(3.14)(1 − 0.119)(1.58) 1/ 2 = 3. 56 ft Exercise 6.19 Subject: Estimation of column diameter based on conditions at the top tray of an absorber. Given: Vapor and liquid conditions at the top tray. Valve trays with 24-inch spacing. Find: Flooding velocity and column diameter for 85% of flooding (f = 0.85) Analysis: Use entrainment flooding correlation of Fig. 6.24, where the abscissa is, FLV LM L ρV = VM V ρ L 1/ 2 889(109) 1.924 =...
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