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

Both consist of a fixed bed column packed with

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: (124.08)/[(82.06)(294)] = 3.72 x 10-6 g/cm3 at z = 0. The cycle calculations are conveniently carried out by a modification of the FORTRAN program in Table 15.9, based on using the method of lines to obtain a set of ODEs, similar to the equations for TSA, with a stiff integrator, such as LSODES. The program in Table 15.9, which is for TSA, consists of: Main program to: Set initial parameters. Call LSODE Perform cycles for: Adsorption step Desorption step Write cycle results Subroutine FEX to: Compute derivatives of the set of ODEs For PSA, the program must add: Depressurization step Repressurization step Exercise 15.31 Subject: Three cycles of vacuum-swing adsorption. Given: Design basis in Example 15.13, except that PL = 0.12 atm and interstitial velocity during the desorption step corresponds to 44.5% of the product gas from the adsorption step. Assumptions: Plug flow with a constant interstitial velocity equal to the given inlet value. Negligible axial dispersion. Find: Result of the third VSA cycle starting from a clean bed. 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 44.5% of that for the adsorption step or (0.445)(0.000542)(60)(20) = 0.289 mol. This will be used for 20 min at a flow rate of 0.289/[(20)(60)] = 0.000241 mol/s At the conditions of desorption, 0.12 atm and 294 K, Q = nRT/P = 0.000241(82.06)(294)/0.12 = 48.45 cm3/s Superficial velocity for desorption = 48.45/0.95 = 51.0 cm/s Interstitial velocity for desorption = 51.0/0.43 = 118.6 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 15.31 (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...
View Full Document

Ask a homework question - tutors are online