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

# Flow rate of solventextract 52000 lbh raffinate is

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Unformatted text preview: 10−5 N Sc . DC 1/ 3 N Re 2/3 φ −1/ 2 D N Fr 5/12 Di d vs 2 d vs DT 1/ 2 N Eo 5/ 4 . . = 1.237 × 10 −5 (434)1/ 3 (904,000) 2 / 3 (0.60) −1/ 2 (0148) 5/12 (1,520) 2 (2.19 × 10−4 )1/ 2 01058 N Sh DC 5/ 4 = 1,052 1,052(8.53 × 10−5 ) = 518 ft/h . d vs 0.00173 The slope of the equilibrium curve for dilute conditions is given by Eq. (8-29) as m = dcC/dcD = cC/cD = 1/21 = 0.0476 1 1 1 From Eq. (8-28), K OD = = = = 0.202 ft/h 1 1 1 1 4.55 + 0.405 + + k D mkC 0.22 0.0476(51.8) where, it is seen that the dispersed phase controls the rate of mass transfer. kC = C = Exercise 8.33 (continued) Analysis: (continued) (f) From Eq. (8-32, N OD = K OD aV (0.202)(2, 080)(334) = = 26.3 QD (750)(60) / 7.48 (g) From Eq. (8-33), E MD = N OD 26.3 = = 0.963 or 96.3% 1 + N OD 1 + 26.3 (h) Because the extract is the dispersed phase, Eq. (3) in Example 8.7 applies, f extracted 750(0.963) QD E MD 1 500 QC m 21 = = = 0.968 or 96.8% extracted QD E MD 750(0.963) 1+ 1+ 1 QC m 500 21 Exercise 8.34 Subject: Diameter of an RDC column for extraction of acetic acid (A) from water (C) by isopropyl ether (S). Given: Conditions in Exercises 8.28 and 8.30. Find: Column diameter. Analysis: Conditions of Exercise 8.28: Flow rate of feed/raffinate = 12,400 lb/h. Flow rate of solvent/extract = 24,000 lb/h Assume raffinate is dispersed. The following properties are given in Exercise 8.30. ρR = ρD = 63.5 lb/ft3 and ρE = ρC = 45.3 lb/ft3 µC = 1.0 cP or 0.000021 lbf-s/ft2 Interfacial tension = σI = 13.5 dyne/cm or 0.000924 lbf/ft uµ ρ Assume o C C = 0.01 σ i ∆ρ σ ∆ρ (0.000924)(63.5 − 45.3) Therefore, uo = 0.01 i = 0.01 = 0.177 ft/s µ C ρC (0.000021)(45.3) U D mD / ρ D 12,400 / 635 . = = = 0.369 The flooding correlation of Fig. 8.39 is too difficult U C mC / ρC 24,000 / 45.3 to read for low values of UD/UC . Instead, use Eqs. (8-62) and (8-59). From Eq. (8-62), with UC/UD = 1/0.369 = 2.71, (1) φ Df = 1+ 8 UC / U D 1/ 2 4 UC / U D − 1 −3 = 1 + 8 2.71 1/ 2 4 2.71 − 1 −3 = 0.257 From Eq. (8-59), using φD = (φD)f = 0.257, UD UC UD UC + = + = uo 1 − φ D = 0.177(1 − 0.257) φ D 1 − φ D 0.257 1 − 0.257 which simplifies to, 3891U D + 1346U C = 01315 . . . (2) Solving Eqs. (1) and (2) simultaneously, UC = 0.0473 ft/s and UD = 0.0175 ft/s Therefore, at flooding, (UD + UC) = 0.0175 + 0.0473 = 0.0648 ft/s Size the column for 50% of flooding. Therefore, (UD + UC) = 0.0648/2 = 0.0324 ft/s Total volumetric flow rate = Q = 12,400/63.5 + 24,000/45.3 = 725 ft3/h = 0.201 ft3/s Column cross-sectional area = AC = Q/(UD + UC) = 0.201/0.0324 = 6.2 ft2 Column diameter = DC = (4AC/π)1/2 = [4(6.2)/3.14]1/2 = 2.8 ft Exercise 8.34 (continued) Analysis: (continued) Conditions of Exercise 8.30: Flow rate of feed/raffinate = 21,000 lb/h. Flow rate of solvent/extract = 52,000 lb/h Raffinate is dispersed. The following properties are given in Exercise 8.30. ρR = ρD = 63.5 lb/ft3 and ρE = ρC = 45.3 lb/ft3 µC = 1.0 cP or 0.000021 lbf-s/ft2 Interfacial tension = σI = 13.5 dyne/cm or 0.000924...
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