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

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Unformatted text preview: nd p = partial pressure of water vapor in atm. Effective diffusivity of water vapor in the adsorbent particle = 0.05 cm2/s. Exit humidity of water vapor to be 0.0009 lb H2O/lb dry air. Assumptions: Isothermal and isobaric operation. Find: Time to breakthrough for the equilibrium model. Time to breakthrough for the mass-transfer model of Klinkenberg. Set of breakthrough curves for the mass-transfer model of Klinkenberg. Approximate width of the MTZ. Average loading of the bed at breakthrough. Analysis: First determine the moisture content of the entering air. From a psychrometric chart, the entering humidity for 80% R.H., 80oF, and 1 atm is 0.0177 lb H2O/lb dry air or a water vapor content of 0.0177/1.0177 x 100% = 1.739 wt%. By Dalton's law, partial pressure of water vapor in the entering air = (1)[0.0177/18.02]/[0.0177/18.02 + 1/28.97] = 0.0277 atm. The mole fraction of water vapor in the entering air = 0.0277. Assume a bed cross-sectional area of 1 ft2. Then, entering gas rate = 100 ft3/min. Bed volume = (1)(5) = 5 ft3. Bed contains 5(39) = 195 lb silica gel. The MW of the entering gas = 18.02(0.0277) + 28.97(0.9723) = 28.67. From the ideal gas law, density of entering gas = PM/RT = (1)(28.67)/[(0.7302)(540)] = 0.0727 lb/ft3. Therefore, the entering gas flow rate = 100(0.0727) = 7.27 lb/min with a water vapor flow rate of 0.01739(7.27) = 0.1264 lb H2O/min. For the dilute conditions, the desired c/cF = 0.0009/0.0177 = 0.05. So use this as the breakthrough value. Equilibrium model: Equilibrium loading based on entering feed = 15.9(0.0277) = 0.440 lb H2O/lb silica gel. Therefore, can adsorb 0.440(195) = 85.8 lb H2O. The ideal time for breakthrough = 85.8/0.1264 = 679 min = 11.3 h. q S 0.440(195) Alternatively, can apply Eq. (15-92), with Lideal/LB = 1, so tideal = F = = 679 min. QF cF 0.1264 Mass-transfer model using Klinkenberg equations: 2 Rp Rp 1 From Eq. (15-106), = + (1) kK 3k c 15De Rp = 0.28/2 = 0.14 cm = 0.0014 m, De = 0.05 cm2/s = 5 x 10-6 m2/s Exercise 15.26 (continued) Analysis: (continued) Dp G D From Eq. (15-65), k c = i 2 + 11 . Dp µ 0 .6 µ ρDi 1/ 3 (2) 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. Therefore, Di = 0.22[(80 + 460)/(32 + 460)]1.75 = 0.26 cm2/s = 0.26 x 10-4 m2/s. From above, ρ = 0.0727 lb/ft3 = 1.16 kg/m3. From a handbook, µ of air at 80oF = 1.75 x 10-5 kg/m-s. The mass velocity = G = 7.29/1.0 = 7.27 lb/min-ft2 = 0.592 kg/m2-s µ 1.75 × 10 −5 = 0.580 N Sc = = ρDi 116(0.26 × 10 −4 ) . N Re = Dp G µ = 0.0028(0.592) = 94.7 175 × 10 −5 . 0.26 × 10 −4 0 .6 1/ 3 2 + 11 94.7 . 0.58 = 0149 m/s . 0.0028 1 0.0014 (0.0014) 2 From Eq. (1), = + = 0.0031 + 0.0261 = 0.0292 s kK 3(0149) 15(0.05 × 10 −4 ) . The equilibrium constant, K, the given value is 15 lb H2O/lb gel-atm. We need to convert this to units of (lb H2O/ft3 gel)/(lb H2O/ft3 gas). From Table 15.2, assume small-pore silica gel with εb = 0....
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