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

# The results of applying eq 1 are as follows solvent b

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Unformatted text preview: ble 3.5, and are listed below. The parachors for ethyl acetate, obtained by structural contributions from Table 3.6, is 2(55.5) + 40.0 + 63.8 = 214.8. Molecular volumes are obtained from Table 3.3, and are listed below. Solvent viscosities are obtained from Perry's Handbook. The results of applying Eq. (1) are as follows Solvent, B Benzene Acetone Ethyl acetate PB 205.3 161.5 214.4 υB 96 74 106.1 µB , cP 0.60 0.32 0.45 DA,B (H-M) 1.5 x 10-5 2.5 x 10-5 1.93 x 10-5 DA,B (Expt.) 2.09 x 10-5 2.92 x 10-5 2.18 x 10-5 The predictions by the Hayduk-Minhas equation are low, but quite good for ethyl acetate. For water, the Wilke-Chang equation, Eq. (3-39), is most applicable, DA,B 7.4 × 10 −8 φ B M B = 0 µ Bυ A.6 1/ 2 T (2) From Table 3.3, υA = (2)14.8 + 4(3.7) + 7.4 + 12 = 63.8 cm3/mol. Solvent viscosity = 0.95 cP, MB = 18, and φB = 2.6. Substitution into Eq. (1) gives, 7.4 × 10−8 ( 2.6 × 18 ) = 0.95(63.8)0.6 1/ 2 DA,B 298 = 1.3 × 10 −5 cm 2 /s This is close to the experimental value of 1.19 x 10-5 cm2/s. Exercise 3.14 Subject: Vapor diffusion through an effective film thickness. Given: Water (A) at 11oC (284 K) evaporating into dry air (B) at 25oC (298 K) at the rate of 0.04 g/h-cm2. Assumptions: 1 atm pressure. Ideal gas law. Raoult's law for water vapor mole fraction. Find: Effective stagnant air film thickness, assuming molecular diffusion of the water vapor through the air film. Analysis: Apply Eq. (3-31) for Fick's law with the bulk flow effect. Solving for film thickness, cDA,B ln ∆z = 1 − yA b 1 − yA I (1) NA From the ideal gas law, using T = (284 + 298)/2 = 291 K, c = P/RT = 1/(82.06)(284) = 4.29 x 10-5 mol/cm3. Molar flux of water vapor = NA = 0.04/MA = 0.04/(18)(3600) = 6.17 x 10-7 mol/s-cm2. Mole fraction of water vapor in the bulk dry air = yAb = 0 Mole fraction of water vapor at the interface, yA I = Ps at 11oC/P = 0.191/14.7 = 0.013. Estimate diffusivity of water vapor in air at 291 K and 1 atm from Eq. (3-36) of Fuller, Schettler, and Giddings, with, 2 M A,B = = 22.2 1 1 + 18 29 From Table 3.1, VA = 131, . VB = 19.7 DA,B = 0.00143(291)1.75 = 0.24 cm2 / s (1)(22.2)1/ 2 [1311/ 3 + 19.71/ 3 ]2 . Substitution into Eq. (1), ∆z = 4.29 × 10−5 ( 0.24 ) ln 6.17 × 10−7 Note that the bulk flow has little effect here. 1− 0 1 − 0.013 = 0.218 cm Exercise 3.15 Subject: Mass transfer of isopropyl alcohol (A) by molecular diffusion through liquid water and gaseous nitrogen at 35oC (308 K) and 2 atm Given: Critical conditions for nitrogen and isopropyl alcohol and molar liquid volume of isopropyl alcohol. Assumptions: Ideal gas law. Find: (a) (b) (c) (d) (e) (f) (g) Diffusivity of A in liquid water by Wilke-Chang equation. Diffusivity of A in gaseous nitrogen by the Fuller, Schettler, Giddings equation. The product, DABρΜ = DAB c for part (a). The product, DABρΜ = DAB c for part (b). Comparison of diffusivities in parts (a) and (b). Comparison of results from parts (c) and (d). Conclusions about diffusion in the liquid versus the...
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