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

# 00274 cm2s substituting this value into eq 1 gives

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Unformatted text preview: ter packed with 4 x 6 mesh silica gel with an external void fraction of 0.5. Gas is benzene vapor in air, which flows through the bed at a benzene-free flow rate of 25 lb/min. Find: External gas-to-particle mass-transfer and heat-transfer coefficients at a location in the bed where the pressure is 1 atm, temperature is 70oF, and benzene bulk mole fraction = 0.005. Analysis: From Perry's Handbook, 4-mesh and 6-mesh screens have openings of 0.476 cm and 0.336 cm, respectively, with an average of (0.476 + 0.336)/2 = 0.406 cm. Aasssuming the silica gel is crushed particles, the sphericity = ψ = 0.65. Therefore, the effective particle diameter = 0.65(0.406) = 0.264 cm = 0.00264 m = Dp. 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) Because the gas is dilute in benzene, use the properties of air in the coefficient correlations. At 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 benzene in air at 1 atm = 0.077 cm2/s. From Eq. (3-36), the diffusivity is proportional to T to the 1.75 power. Therefore, Di = 0.077[(70 + 460)/(32 + 460)]1.75 = 0.0877 cm2/s = 0.0877 x 10-4 m2/s. PM 1013(29) . From the ideal gas law, ρ = = = 120 kg / m3 . RT 8.314(530 / 18) . CP µ 1090(183 × 10 −5 ) . = = 0.779 k 0.0256 µ 183 × 10−5 . N Sc = = = 174 . ρDi 1.20(0.0877 × 10−4 ) N Pr = Exercise 15.16 (continued) Analysis (continued) In the Reynolds number, G = mass flow rate/bed cross-sectional area bed cross-sectional area = Ab = πD2/4 = 3.14(2)2/4 = 3.14 ft2 or 0.292 m2 Air flow rate = 25 lb/min = 25/60 = 0.417 lb/s or 0.417/29 = 0.0144 lbmol/s Benzene flow rate = (0.005/0.995)0.0144 = 0.000072 lbmol/s or 0.000072(78.11) = 0.0056 lb/s Total gas mass flow = 0.417 + 0.0056 = 0.423 lb/s or 0.192 kg/s Gas mass velocity = G = 0.192/0.292 = 0.658 kg/m2-s NRe = 0.00264(0.658)/1.83 x 10-5 = 94.9 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, 0.0877 × 10 −4 0.6 1/ 3 2 + 1.1( 94.9 ) (1.74 ) kc = = 0.0741 m/s 0.00264 h= 0.0256 0.6 1/ 3 = 170.1 J/m 2 -s-K 2 + 1.1( 94.9 ) ( 0.779 ) 0.00264 Exercise 15.17 Estimation of external gas-to-particle mass-transfer and heat-transfer coefficients in a fixed-bed adsorption column. Given: Fixed bed of 12.06-cm inside diameter packed with 3.3-mm-diameter Alcoa F-200 activated alumina beads with an external void fraction of 0.442. Gas is air containing water vapor, which flows through the bed at a flow rate of 1.327 kg/min. Find: External gas-to-particle mass-transfer and heat-transfer coefficients at a location in the bed where the pressure is 653.3 kPa, temperature is 21oC , and the dew-point temperature = 11.2oC....
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