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

8 40 applies diffusivity of a in c dd 13 x 10 9 m2s

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Unformatted text preview: ity is, µ 15µ D φ D . 1.0 15(3.0)(0.22) . µ M = C 1+ = 1+ = 16 cP or 3.87 lb/ft-h . φC µC + µ D 0.78 10 + 3.0 . Exercise 8.30 (continued) Analysis: (c) (continued) Compute minimum rpm from Eq. (8-25), where, far-right dimensionless group is, µ2 σ 387 2 (388,000) . M = = 3.42 × 10 −16 5 2 2 5 Di ρ M g ( ∆ρ) (143) (49.3)(4.17 × 108 )(18.2) 2 . 2 N min ρ M Di D = 1.03 T g∆ρ Di 2 .76 2 Therefore, N min = 20.88 φ 0.106 D µ2 σ M 5 Di ρ M g 2 ( ∆ρ) 2 0.084 = 1.03 4.3 1.43 2 .76 (0.22) 0.106 (3.42 × 10−16 ) 0.084 = 20.88 g∆ρ (4.17 × 108 )(18.2) = 20.88 = 9.9 × 107 (rph2) ρ M Di (49.3)(143) . and Nmin = 9,950 rph or 166 rpm (d) Determine the mixing vessel agitator Hp from Fig. 8.36 (b), which requires NRe. Assume N = Nmin = 9,950 rph From Eq. (8-22), N Re = Di2 Nρ M (143) 2 (9,950)(49.3) . = = 2.6 × 105 µM 387 . 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, P= N Po N 3 Di5ρ M (5.7)(9,950) 3 (143) 5 (49.3) . = = 4.0 × 106 ft-lbf/h or 2 Hp 8 gc 4.17 × 10 Compare this to the rule of thumb of 4 Hp/1,000 gallons. Vessel volume = 62 ft3 or 464 gal. Therefore, Hp = 4(464)/1,000 = 1.9 Hp, which is in very good agreement. Exercise 8.31 Subject: Droplet characteristics for extraction of acetic acid (A) from water (C) by isopropyl ether (S) in a mixer unit. Given: Conditions in Exercise 8.27. Find: (a) Sauter mean drop size. (b) Range of drop sizes. (c) Interfacial area of emulsion. Analysis: (a) 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 143 (9,950) 2 (45.3) . Di3 N 2ρC = = = 33,800 σi 388,000 Since NWe > 10,000, Eq. (8-39) applies. d vs = 0.39 Di N We (b) −0 . 6 = 0.39(143)(33,900) −0.6 = 0.00107 ft or 0.32 mm . dmin = dvs /3 = 0.32/3 = 0.107 mm dmax = 3dvs = 3(0.32) = 0.96 mm (c) From Eq. (8-36), the interfacial surface area per unit volume of emulsion is, a= 6φ D 6(0.22) = = 1, 234 ft2/ft3 d vs 0.00107 Exercise 8.32 Subject: Mass transfer characteristics for extraction of acetic acid (A) from water (C) by isopropyl ether (S) in a mixer unit. Given: Conditions in Exercises 8.27 and 8.28. Diffusivity of acetic acid in each phase. Distribution coefficient of acetic acid. Find: (a) (b) (c) (d) Dispersed-phase mass-transfer coefficient. Continuous-phase mass-transfer coefficient. Murphree dispersed-phase efficiency. Fraction of A extracted. Analysis: (a) For the raffinate, which is the dispersed phase, Eq. (8-40 applies. Diffusivity of A in C = DD = 1.3 x 10-9 m2/s and from Exercise 8.28, dvs = 0.32 mm = 0.00032 m DD 13 × 10−9 . = 6.6 = 26.8 × 10 −6 m/s or 0.32 ft/h From a rearrangement of Eq. (8-40), k D = 6.6 d vs 0.00032 (b) For the extract, which is the continuous phase, Eq. (8-50) applies. Diffusivity of A in S = DC = 2.0 x 10-9 m2/s or 7.75 x 10-5 ft2/h (10)(2.42) . µC In Eq. (8-44), N Sc = = = 690 ρC DC (45.3)(7.75 × 10 −5 ) In Eq. (8-50), N Re = In Eq....
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