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

# 826 0842 0826 277 05 2 554 s 1 from eq 6 63

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Unformatted text preview: n. Volumetric liquid rate = qL = m/ρL = (1,943/60)(18)/(1,000)(0.003785 m3/gal) = 154 gpm 2/3 qL From Eq. (6-51), hl = φ e hw + C Lw φ e 154 = 0.30 2 + 0.362 54.3 0.30 2/3 = 1.09 in. of liquid From Eq. (6-55), with maximum bubble size of 3/16 inch = 0.00476 m, hσ = 6σ / gρ L DBmax = 6(70 / 1000) / (9.8)(1,000)(0.00476) = 0.009 m = 0.35 in. of liquid From Eq. (2), ht = hd + hl + hσ =1.42 + 1.09 + 0.35 = 2.86 in. liquid = 0.103 psi/tray = 0.7 kPa/tray (c) DT = 1.89 m, A = 2.804 m2 = 28,040 cm2, Aa = 0.8(2.804) = 2.24 m2 = 22,400 cm2 φe = 0.30, hl = 1.09 in. = 2.77 cm Ua = 7.8 ft/s = 238 cm/s, Uf =9.44 ft/s, f = Ua/Uf = 7.8/9.44 = 0.826 F = UaρV0..5 = 2.38(1.05)0.5 = 2.44 (kg/m)0.5/s , qL = 154 gpm = 9,716 cm3/s hA 2.77(22,400) From Eq. (6-64), t L = l a = = 6.39 s qL 9,716 1 − φ e hl (1 − 0.30)2.77 = = 0.027 s φ eU a 0.30(238) From the Wilke-Chang Eq. (3-39), DL = 1.12 x 10-5 cm2/s 0 . From Eq. (6-67), k L a = 78.8 DL.5 ( F + 0.425) = 78.8(112 × 10−5 )(2.44 + 0.425) = 0.756 s-1 From Perry's Handbook, with a temperature correction based on Eq. (3-36), DV = 0.127 cm2/s From Eq. (6-66), From Eq. (6-65), t G = kG a = 0 1,030 DV .5 f − 0.842 f 2 hl0.5 = 1,030(0127) 0.5 0.826 − 0.842 0.826 . 2.77 0.5 2 = 55.4 s-1 From Eq. (6-63), N L = k L at L = 0.756(6.39) = 4.83 From Eq. (6-62), N G = kG atG = 55.4(0.027) = 1.5 (d) From p. 271, for acetone, A = L/KV = 1.38. Therefore, KV/L = 1/1.38 = 0.725 1 1 1 From Eq. (6-61), N OG = = = = 122 . 1 KV / L 1 0.725 0.667 + 0150 . + + NG NL 15 4.83 . (e) Mass transfer is controlled by the vapor phase. (f) From a rearrangement of Eq. (6-56), E OV = 1 − exp( − N OG ) = 1 − exp( −122) = 0.705 or 70 % . Now, the separation requires 10 equilibrium stages. Assume the liquid on a tray is well mixed. Then, from Eq. (6-31), EMV = EOV = 0.70. From below Eq. (6-33), take λ=KV/L=0.725. log 1 + E MV ( λ − 1) log 1 + 0.70(0.725 − 1) From Eq. (6-37), Eo = = = 0.66 (worst case) log λ log 0.725 Therefore, at most, we need from Eq. (6-21), Na =Nt /Eo = 10/0.66 = 15. Therefore, 30 okay. Exercise 6.23 Subject: Design of a column to strip VOCs from water by air at 15 psia and 70oF. Given: Conditions in Example 6.2 except that entering flow rates are twice as much. Assumptions: Dilute system such that changes in vapor and liquid rates in the column are negligible. Sieve trays on 24-inch spacing, with 10 % hole area of 3/16-inch holes, FF = 0.9, FHA = 1.0 and f = 0.80. Find: (a) (b) (c) (d) (e) Number of equilibrium stages required. Column diameter for sieve trays. Vapor pressure drop per tray. Murphree vapor-point efficiency from Chan-Fair method. Number of actual trays. Analysis: (a) Need 3 equilibrium stages, as determined in Example 6.2 and which is unchanged when flow rates are doubled. (b) liquid rate = 2(500) = 1,000 gpm or 1,000(60)(8.33)/18.02 = 27,800 lbmol/h = L vapor rate = 2(3,400) = 6,800 scfm or 6,800(60)/379 = 1,077 lbmol/h = V ρL = 62.4 lb/ft3. From ideal gas law, ρV = PM/RT = (15)(29)/(10.73)(530)...
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