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

# 5 dynec from exercise 830 hets 4 with hets and dt in

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Unformatted text preview: lbf/ft uµ ρ Assume o C C = 0.01 σ i ∆ρ (0.000924)(63.5 − 45.3) σ ∆ρ Therefore, uo = 0.01 i = 0.01 = 0.177 ft/s (0.000021)(45.3) µ C ρC U D mD / ρ D 21,000 / 63.5 = = = 0.288 The flooding correlation of Fig. 8.39 is too difficult U C mC / ρC 52,000 / 45.3 to read for low values of UD/UC . Instead, use Eqs. (8-62) and (8-59). (1) From Eq. (8-62), with UC/UD = 1/0.289 = 3.46, φ Df = 1+ 8 UC / U D 1/ 2 4 UC / U D − 1 −3 = 1 + 8 3.46 1/ 2 4 3.46 − 1 −3 = 0.239 From Eq. (8-59), using φD = (φD)f = 0.239, UD UC UD UC + = + = uo 1 − φ D = 0.177(1 − 0.239) φ D 1 − φ D 0.239 1 − 0.239 which simplifies to, 4.184U D + 1314U C = 0.1347 . (2) Solving Eqs. (1) and (2) simultaneously, UC = 0.0534 ft/s and UD = 0.0154 ft/s Therefore, at flooding, (UD + UC) = 0.0154 + 0.0534 = 0.0688 ft/s Size the column for 50% of flooding. Therefore, (UD + UC) = 0.0688/2 = 0.0344 ft/s Total volumetric flow rate = Q = 21,000/63.5 + 52,000/45.3 = 1,480 ft3/h = 0.411 ft3/s Column cross-sectional area = AC = Q/(UD + UC) = 0.411/0.0344 = 11.9 ft2 Column diameter = DC = (4AC/π)1/2 = [4(11.9)/3.14]1/2 = 3.9 ft Exercise 8.35 Subject: Diameter of a Karr column for extraction of benzoic acid (A) from water (C) by toluene (S). Given: Conditions in Exercise 8.33. Find: Column diameter. Analysis: Volumetric flow rate of feed/raffinate = 500 gal/min Volumetric flow rate of solvent/extract = 750 gal/min Extract is dispersed. The following properties are given in Exercise 8.30. ρR = ρD = 0.860(62.4) = 53.7 lb/ft3 and ρE = ρC = 0.995(62.4) = 62.1 lb/ft3 µC = 0.95 cP or 0.000020 lbf-s/ft2 Interfacial tension = σI = 22 dyne/cm or 0.0015 lbf/ft uµ ρ Assume o C C = 0.01 σ i ∆ρ (0.0015)(62.1 − 53.7) σ ∆ρ = 0.01 = 0101 ft/s . Therefore, uo = 0.01 i µ C ρC (0.000020)(62.1) U D QD 750 = = = 150 . U C QC 500 From Fig. 8.39, U D + UC f uo = 0.27 Therefore, (UD +UC)f = 0.27 uo = 0.27(0.101) = 0.0273 ft/s Size the column for 50% of flooding. Therefore, (UD + UC) = 0.0273/2 = 0.0137 ft/s Total volumetric flow rate = Q =(500 + 750)(60)/7.48 = 10,030 ft3/h = 2.79 ft3/s Column cross-sectional area = AC = Q/(UD + UC) = 2.79/0.0137 = 204 ft2 Column diameter = DC = (4AC/π)1/2 = [4(204)/3.14]1/2 = 16.1 ft From Table 8.2, maximum Karr diameter is 1.5 m or 4.9 ft. Therefore, need multiple Karr πD 2 314(4.9) 2 . columns in parallel of maximum cross-sectional area = Ac max = = = 18.8 ft2 4 4 Therefore, need 204/18.8 = 11 units in parallel of 1.5 m diameter each. Would probably be better to specify one RDC column of 16.1 ft diameter or 5 m , which is less than the maximum diameter of 8 m. Exercise 8.36 Subject: HETS of an RDC column for extraction of acetic acid (A) from water (C) by isopropyl ether (S). Given: Conditions in Exercises 8.28, 8.30, and 8.34. Find: HETS. Analysis: From Fig. 8.40, for an interfacial tension of 13.5 dyne/c from Exercise 8.30, HETS = 4 with HETS and DT in inches. 1 DT/ 3 From Exercise 8.34, for the conditions of Exercise 8.28, DT = 2.8 ft or 33.6 inches. Therefore, HETS = 4(33.6)1/3 = 12.9 in...
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