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

000236306124088206294 372 x 10 6 gcm3 at z

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Unformatted text preview: n the bed, the following values for benzene loading are obtained: Desorption time, minutes 0 50 100 150 200 250 300 350 400 Lb benzene not desorbed 358.0 286.5 214.9 144.0 79.4 33.0 9.8 2.1 0.3 % of benzene desorbed 0.0 20.0 40.0 59.8 22.2 90.8 97.3 99.4 99.9 From the above results, the required desorption time is almost 250 minutes. Exercise 15.30 Subject: Pressure-swing adsorption cycles to approach a cyclic steady state Given: Design basis in Example 15.13. Assumptions: Plug flow with a constant interstitial velocity equal to the given inlet value. Negligible axial dispersion. Find: PSA cycle results that approach the cyclic steady state starting from: (a) a clean bed. (b) a bed saturated with the feed. Are the two cyclic steady states essentially the same? Analysis: Mole fraction of DMMP in the feed gas = 236/1,000,000 = 0.000236 Feed gas superficial velocity =us = uεβ = 10.465(0.43) = 4.5 cm/s Cross-sectional area of bed = A = 3.14(1.1)2/4 = 0.95 cm2 Volumetric feed gas flow rate = Q = us A = 4.5(0.95) = 4.275 cm3/s From the ideal gas law at 3.06 atm and 294 K, molar feed gas flow rate = n = PQ/RT = (3.06)(4.275)/[(82.06)(294)] = 0.000542 mol/s Because the feed is so dilute, the product gas is almost equal to the feed gas. Therefore, for the desorption step, the gas used for 20 minutes will be pure air at 41.6% of that for the adsorption step or (0.416)( 0.000542)(60)(20) = 0.270 mol. This will be used for 20 min at a flow rate of 0.270/[(20)(60)] = 0.000225 mol/s At the conditions of desorption, 1.07 atm and 294 K, Q = nRT/P = 0.000225(82.06)(294)/1.07 = 5.073 cm3/s Superficial velocity for desorption = 5.073/0.95 = 5.34 cm/s Interstitial velocity for desorption = 5.34/0.43 = 12.42 cm/s Convert the adsorption isotherm into the form consistent with the following equations, which are modifications of Eqs. (15-98) and (15-105), in terms of the gas concentration and loading of DMMP: ∂c ∂c 1 − ε b = −u − k q * −q ∂t ∂z εb ∂q = k q * −q ∂t where, the units are: c in g/cm3 of gas t in minutes u in cm/min k in min-1 q in g/cm3 of particles Exercise 15.30 (continued) Analysis: (continued) The given Langmuir isotherm is: 48,360 p , where q* is in g/g of adsorbent and p is partial pressure in atm. 1 + 98,700 p This isotherm must be converted to units of g/cm3 of particles for q* and the variable c in g/cm3 of gas must be substituted for the partial pressure in atm. First consider q*. The bed volume = 0.95(12.80) = 12.16 cm3 The particle volume = (1-0.43)(12.16) = 6.93 cm3 Particle density = 5.25/6.93 = 0.758 g/cm3 of particles The molecular weight of DMMP, dimethyl methylphosphonate, CH3PO(OCH3)2 = M = 124.08 Assuming the ideal gas law, p = cRT/M = c (82.06)(294)/124.08 = 194.4 c Therefore, the Langmuir isotherm becomes: q* = q* = 0.738 g/cm3 48,360(194.4)c 6.94 × 106 c = , where q* is in g/cm3 of particles and c is 6 1 + 98,700(194.4)c 1 + 19.19 × 10 c The initial conditions for adsorption step of the first cycle are: c = 0 for t = 0 and q = 0 at t = 0 The boundary condition for all adsorption steps is: cF = yFPM/RT = (0.000236)(3.06)...
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