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.
Perform cycles for:
Write cycle results
Subroutine FEX to:
Compute derivatives of the set of ODEs
For PSA, the program must add:
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
∂c 1 − ε b
= −u −
k q * −q
= k q * −q
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)
The given Langmuir isotherm is:
, 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...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.
- Spring '11
- The Land