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

# 3 kpa and 25oc 298 k that is flowing past the plate

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Unformatted text preview: , y = L, from the top of the film, cCO 2 y / cCO 2 i = 0.5 . If we use Eqs. (3-99) and (3-100), we do not have to compute the average mass-transfer coefficient. Compute the value of η from (3-99): 1 − 0.5 = 0.5 = 0.7857 exp ( −5.1213η ) + 0.09726 exp ( −39.661η ) + 0.036093exp ( -106.25η ) Solving this nonlinear equation for η with a spreadsheet, η = 0.0894 From a rearrangement of Eq. (3-100), L= 3δ2 u y η 2 DAB 3 (1.15 × 10−2 ) (4.86)(0.0894) 2 = 2 (1.96 ×10 −5 ) = 4.4 cm Exercise 3.28 Subject: Evaporation of water (A) from a film on a flat plate into air (B) at 1 atm (101.3 kPa) and 25oC (298 K) that is flowing past the plate. Given: Free-stream air velocity = uo = 2 m/s. Plate length = L = 2 inches = 0.0508 m. Diffusivity of water vapor in air = DA,B = 0.25 cm2/s = 0.25 x 10-4 m2/s Assumptions: Ideal gas law. Laminar boundary-layer flow of the air. Negligible absorption of air by the water film. Equilibrium at the air-water interface. Find: Evaporation flux at 50% of L = x = 0.0254 m from the leading edge of the plate. Analysis: First check to see if the boundary layer is laminar. From Perry's Handbook, for air at 25oC, µ = 0.018 cP = 1.8 x 10-5 kg/m-s. From the ideal gas law, ρ = PM/RT = (101.3)(29)/(8.314)(298) = 1.186 kg/m3 From Eq. (3-127), at x = 0.0254 m, xu ρ (0.0254)(2)(1186) . N Re x = o = = 3,350 −5 µ 18 × 10 . Therefore, the boundary layer is laminar, since the criterion is less than 5 x 105. Since the liquid film is pure water, the only mass-transfer resistance for evaporation of water into air is in the air boundary layer. From Eq. (3-105), the molar mass-transfer flux of water is, nH 2 O = k cx cH 2 O − cH 2 O (1) i o A For a laminar boundary layer, the local mass-transfer coefficient can be obtained from Eq. (3132), which can be rearranged to give: k cx = 0.332 DA,B N Re x x 1/ 2 N Sc 1/ 3 (2) µ 18 × 10 −5 . = = 0.607 ρDA,B (1186) 0.25 × 10 −4 . From Eq. (3-101), N Sc = From Eq. (2), k cx = 0.332 0.25 × 10 −4 0.0254 3,350 1/ 2 0.607 1/ 3 = 0.16 m / s At 25oC, vapor pressure of water = 0.46 psia, Therefore, mole fraction of water at the interface = yH 2 O = 0.46/14.7 = 0.0313. cH 2 O = yH 2 O ρ/Mair = 0.0313(1.186)/29 = 0.0128 i kmol/m3.From Eq. (1), i nH 2O A i = (0.016)[0.00128 − 0] = 2.05 × 10−5 kmol/s-m 2 Exercise 3.29 Subject: Sublimation of a thin, flat plate of naphthalene (A) into air (B) at 100oC (373 K) and 1 atm (101.3 kPa) that is flowing past the plate. Given: Length of plate, L, is 1m. Reynolds number at the trailing edge of the plate = N Re L = 5 × 105 = limit of laminar boundary-layer flow. Find: (a) Average rate of sublimation. (b) Local rate of sublimation at x = 0.5 m from leading edge of plate. Analysis: This exercise is similar to Example 3.14, which also deals with the sublimation of naphthalene into air at the same temperature and pressure. Therefore, the physical properties used in Example 3.14 apply here. The only mass-transfer resistance is in the gas phase. (a)...
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