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

# 1 904000 c 095242 di n 2 2634840 2 0148 g 417

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Unformatted text preview: = 750/1,250 = 0.60 and φC = 0.40. (a) We are given the following properties, Phase Dispersed (extract) Continuous (raffinate) Density, lb/ft3 53.7 62.1 Viscosity, cP 0.59 0.95 Diffusivity of A, cm/s 1.5 x 10-5 2.2 x 10-5 Also, interfacial tension = σI = 22.0 dyne/cm = 633,000 lb/h2 Density difference = ∆ρ =62.1 - 53.7 = 8.4 lb/ft3 Phase mixture density from Eq. (8-23) = ρM = 53.7(0.60) + 62.1(0.40) = 57.0 lb/ft3 From Eq. (8-24), mixture viscosity is, µ 15µ D φ D . 0.95 15(0.59)(0.60) . µ M = C 1+ = 1+ = 2.65 cP or 6.41 lb/ft-h φC µC + µ D 0.40 0.95 + 0.59 Compute minimum rpm from Eq. (8-25), where, far-right dimensionless group is, µ2 σ 6.412 (633,000) M = = 2.96 × 10 −19 Di5ρ M g 2 ( ∆ρ) 2 (2.63) 5 (57.0)(4.17 × 108 )(8.4) 2 2 N min ρ M Di D = 103 T . g∆ρ Di 2 .76 φ 0.106 D µ2 σ M Di5ρ M g 2 ( ∆ρ) 2 0.084 = 103 . 7.9 2.63 2 .76 (0.60) 0.106 (2.96 × 10−19 ) 0.084 = 1004 . Exercise 8.33 (continued) Analysis: (a) (continued) g∆ρ (4.17 × 108 )(8.4) 2 Therefore, N min = 1004 . = 1004 . = 235 × 106 (rph2) . ρ M Di (57.0)(2.63) and Nmin =4,840 rph or 81 rpm (b) Determine the mixing vessel agitator Hp from Fig. 8.36 (b), which requires NRe. Assume N = Nmin =4,840 rph From Eq. (8-22), N Re = Di2 Nρ M (2.63) 2 (4,840)(57.0) = = 2.98 × 105 µM 6.41 From Fig. 8.36 (b), for a mixing vessel with baffles (thus, no vortex), Npo = 5.7 From Eq. (8-21), the power is given by, N Po N 3 Di5ρ M (5.7)(4,840) 3 (2.63) 5 (57.0) . P= = = 1110 × 106 ft-lbf/h or 5.6 Hp 8 4.17 × 10 gc Compare this to the rule of thumb of 4 Hp/1,000 gallons. Vessel volume = 334 ft3 or 2,500 gal. Therefore, Hp = 4(2,500)/1,000 = 10 Hp, which is 80% higher. (c) The Sauter (surface-mean) diameter, dvs , is given by Eq. (8-38) or (8-39), depending on the Weber number. From Eq. (8-37), 3 N We = 2.63 (4,840) 2 (57.0) Di3 N 2ρC = = 41,800 633,000 σi Since NWe > 10,000, Eq. (8-39) applies. d vs = 0.39 Di N We −0 . 6 = 0.39(2.63)(41,800) −0.6 = 0.00173 ft or 0.53 mm (d) From Eq. (8-36), the interfacial surface area per unit volume of emulsion is, a= 6φ D 6(0.60) = = 2, 080 ft2/ft3 d vs 0.00173 Exercise 8.33 (continued) Analysis: (continued) 1 (e) Eq. (8-28) applies, where, KOD = 1 1 + k D mk C For the extract, which is the dispersed phase, Eq. (8-40 applies. D 15 × 10 −5 . = 188 × 10−3 cm/s or 0.22 ft/h . From a rearrangement of Eq. (8-40), k D = 6.6 D = 6.6 d vs 0.0527 For the raffinate, which is the continuous phase, Eq. (8-50) applies, with DC = 2.2 x 10-5 cm2/s or 8.53 x 10-5 ft2/h. µC (0.95)(2.42) = = 434 In Eq. (8-44), N Sc = ρC DC (62.1)(8.53 × 10−5 ) In Eq. (8-50), N Re = In Eq. (8-50), N Fr = Di2 NρC (2.63) 2 (4,840)(62.1) = = 904,000 µC (0.95)(2.42) Di N 2 (2.63)(4,840) 2 = = 0.148 g 4.17 × 108 In Eq. (8-50), Di /dvs = 2.63/(0.00173 = 1,520 In Eq. (8-50, dvs/DT = 0.00173/7.9 = 2.19 x 10-4 In Eq. (8-50, N Eo = 2 ρ D d vs g (53.7)(0.00173) 2 (4.17 × 108 ) = = 0.1058 633,000 σ From Eq. (8-50), N Sh C kd = C vs = 1237 ×...
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## This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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