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

2 mm 002 cm e the initial rate of heat transfer to the

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Unformatted text preview: 1 Subject: Evaporation of a spherical water (A) drop into still, dry air (B). Given: Initial drop diameter = dp = 1 mm = 0.1 cm. Air temperature = T = 38oC. Drop temperature = 14.4oC. Pressure = P = 1 atm. Assumptions: Ideal gas law. Quasi-steady-state evaporation of water into air. That is, the rate of accumulation of water vapor in the air is negligible compared to the rate of diffusion of the water through the air. Find: (a) A Sherwood number of 2 for mass transfer from the drop surface to the air. (b) The initial mass of the drop. (c) The initial rate of evaporation of the drop in g/s. (d) The time in seconds for the drop diameter to be reduced to 0.2 mm = 0.02 cm. (e) The initial rate of heat transfer to the drop. If insufficient to supply the necessary heat, what will happen? Analysis: (a) With a pure water drop, the only resistance to mass transfer is in the air by ordinary molecular diffusion from the drop surface out to infinity. With the quasi-steady-state assumption, Eq. (3-74) for diffusion through a spherical region becomes, ∂cA DAB ∂ 2 ∂cA =2 r =0 ∂t r ∂r ∂r (1) Therefore, d 2 dcA r =0 dr dr (2) Integrating Eq. (2) twice, C1 + C2 with boundary conditions of: cA = cA s at r = rs and cA = 0 at r = ∞ r With these boundary conditions, the two constants of integration are evaluated to give for the region outside the drop, r (3) cA = cA s s r cA = − The rate of mass transfer from the drop surface to the bulk air in terms of a mass-transfer coefficient, Eq. (3-105), can be equated to the diffusion rate from the drop surface, as given by Fick's first law, as in Eq. (3-63), to give the following equation, which is similar to Eq. (3-106), Exercise 3.31 (continued) Analysis: (a) (continued) nA = k c A cA s − cA ∞ = − DAB A Differentiating Eq. (3), dcA dr r = rs = cA s rs − 1 r2 dcA dr =− r = rs (4) r = rs cA s rs (5) Substituting Eq. (5) into Eq. (4), k c A cA s − 0 = − DAB A − cA s (6) rs Simplifying and rearranging Eq. (6) into the form of a Sherwood number, like Eq. (3-112), with a characteristic length of the drop diameter, NSh = kc (2rs ) kc d p = =2 DAB DAB (b) Initial mass of a spherical drop = m = ρV = ρ π (7) d3 p 6 = (1.0) (3.14) (0.1)3 = 5.23 × 10−4 g 6 (c) Initial rate of evaporation can be obtained for Fick's law or from Eq. (3-105) with a masstransfer coefficient. From Eq. (3-105), nA = k c A cA s − cA ∞ = k c Ac yA s − 0 (8) The binary gas diffusivity is obtained from Eq. (3-36) using T = 38oC = 311 K and P = 1 atm. 2 M A,B = = 22.2 1 1 + 18 29 From Table 3.1, V A = 131, . DAB = V B = 19.7 0.00143(311)1.75 = 0.273 cm2 / s (1)(22.2)1/ 2 [(131)1/ 3 + (19.7)1/ 3 ]2 . Exercise 3.31 (continued) Analysis: (c) (continued) From Eq. (7), kc = 2 DAB (2)(0.273) = = 5.46 cm/s dp 0.1 2 Area for mass transfer = area of the drop = A = π d p = (3.14)(0.1)2 = 0.0314 cm2 From the ideal gas law, the total gas concentration is, c= P 1 = = 3.92 × 10−5 mol/cm3 RT (82.06)(311) The mole fraction of water in the air at the surface of the drop...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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