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

14 21 using 18 for the molecular weight of water and

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Unformatted text preview: Analysis: Dp = 3.3 mm = 0.0033 m. Provided that the dimensionless groups are in the specified ranges of the correlations, Eqs. (15-65) and (15-66) can be used to estimate the coefficients: Dp G D k c = i 2 + 11 . µ Dp Dp G D h = i 2 + 11 . µ Dp 0 .6 0.6 µ ρDi 1/ 3 CP µ k 1/ 3 (1) (2) At the dew-point temperature, the vapor pressure of water = 0.192 psi or 1.32 kPa. Therefore, at 653.3 kPa, the mole fraction of water in the air is only 1.32/653.3 = 0.002. Because the gas is so dilute in water vapor, use the properties of air in the coefficient correlations. At 21oC = 70oF, µ = 183 micropoise = 1.83 x 10-5 kg/m-s k = 0.0256 J/m-s-K CP = 1.09 kJ/kg-K = 1090 J/kg-K From Perry's Handbook, at 32oF, the diffusivity of water vapor in air at 1 atm = 0.22 cm2/s. From Eq. (3-36), the diffusivity is proportional to T to the 1.75 power and inversely proportional to the pressure. Therefore, Di = 0.22(101.3/653.3)[(70 + 460)/(32 + 460)]1.75 = 0.0388 cm2/s = 0.0388x10-4 m2/s. At the high pressure of 653.3 kPa, the compressibility factor of air = 0.998. From the generalized PM 653.3(29) gas law, ρ = = = 7.75 kg / m3 ZRT 0.998(8.314)(530 / 18) . CP µ 1090(183 × 10 −5 ) . = = 0.779 k 0.0256 µ 183 × 10 −5 . N Sc = = = 0.609 ρDi 7.75(0.0388 × 10 −4 ) N Pr = Exercise 15.17 (continued) Analysis (continued) In the Reynolds number, G = mass flow rate/bed cross-sectional area Bed cross-sectional area = Ab = πD2/4 = 3.14(0.1206)2/4 = 0.0114 m2 Gas flow rate = 1.327 kg/min = 1.327/60 = 0.0221 kg/s Gas mass velocity = G = 0.0221/0.0114 = 1.94 kg/m2-s NRe = 0.0033(1.94)/1.83 x 10-5 = 350 All dimensionless groups and the particle diameter are in the specified ranges of applicability of the correlations of Eqs (1) and (2). Therefore, using those two equations, kc = h= 0.0388 × 10−4 0.6 1/ 3 2 + 1.1( 350 ) ( 0.609 ) = 0.0392 m/s 0.0033 0.0256 0.6 1/ 3 = 280 J/m 2 -s-K 2 + 1.1( 350 ) ( 0.779 ) 0.0033 Exercise 15.18 Subject: Effective diffusivity of acetone in the pores of activated carbon. Given: Activated carbon with ρp = 0.85 g/cm3, εp = 0.48, dp = 25 Angstroms = 25 x 10-8 cm, and tortuosity = τ = 2.75. Diffusion of acetone vapor in nitrogen, where from Example 15.7, temperature = 297 K, pressure = 136 kPa, and molecular (bulk) diffusivity = Di = 0.085 x 10-4 m2/s or 0.085 cm2/s. Assumptions: Negligible surface diffusion. Find: Effective diffusivity of acetone vapor through nitrogen in the pores, accounting for bulk (molecular) diffusion and Knudsen diffusion. Analysis: From a modification of Eq. (15-75), omitting surface diffusion, De = εp 1 τ 1 1 + Di DK = 0.48 2.75 1 1 1 + 0.085 DK = 0.175 1 11.76 + DK (1) From Eq. (14-21), using 58 for the molecular weight of acetone, DK = 4850 dp(T/Mi)1/2 = 4850(25 x 10-8)(297/58)1/2 = 0.00274 cm2/s Substituting this value into Eq. (1) gives: 0.175 0.175 De = = = 4.66 × 10−4 cm 2 /s 1 11.76 + 364 11.76 + 0.00274 It is noted that Knudsen diffusion is very important and largely controls diffusion. Exercise 15.19 Subject...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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