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: Pressureswing 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
Crosssectional 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. (1598) and (15105), 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 min1
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 = (10.43)(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)...
View
Full Document
 Spring '11
 Levicky
 The Land

Click to edit the document details