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Unformatted text preview: = πD2L/4 = 3.14(2)2(30)/4 = 94.2 ft3
Volume of SG particles = (1εb)V = (10.5)(94.2 = 47.1 ft3
During adsorption at 70oF, from Example 15.10, the concentration of benzene in the entering airbenzene mixture = 0.00181 lb benzene/ft3 = cF. The adsorbent loading in equilibrium with the
feed, as given in Example 15.10, is qF* = 5,120 cF = 5,120(0.00181) = 9.27 lb benzene/ft3 of
silica gel particles. If the bed were completely saturated with the feed, the benzene in the bed
would be 9.27(47.1) = 437 lb benzene. If the breakthrough occurred for c just greater than zero
at the given 641 minutes for adsorption (i.e. all of the benzene in the feed were adsorbed), then
with a feed flow rate, from Example 15.10, of 310 ft3/min, the benzene loading would be:
310(641)(0.00181) = 360 lb benzene, which, as expected, is less than complete saturation.
However, the breakthrough at 641 minutes corresponds to φ = c/cF = 0.05 and not at 0.
Therefore, it is best to integrate the given loading data at the end of adsorption to obtain a more
accurate value of benzene loading. Using the given data on the next page, compute the average
value of ψ by numerical integration. Note that because the flow of desorption agent is opposite
to the feed gas, the bed distances below are given as z' = 30 ft  z. ψ avg = 0.5ψ z '= 0 + 29
z ' =1 ψ at z ' + 0.5ψ z '= 30
30 = 0.022 + 24.079 + 0.500
= 0.820
30 Therefore, the amount of benzene on the adsorbent in the bed at the end of the adsorption period
is: 0.820(9.27)(47.1) = 358.0 lb benzene, which is very close to the above rough estimate of
360. The desorption period must be long enough to reduce the loading to (10.90)(358) = 35.8 lb
benzene. Exercise 15.29 (continued)
Analysis: (continued)
Given conditions at the beginning of the desorption period:
z, ft
z' = 30  z, ft φ = c/cF ψ = q/ qF*
30
0
0.050
0.044
29
1
0.090
0.081
28
2
0.150
0.137
27
3
0.235
0.217
26
4
0.343
0.321
25
5
0.468
0.444
24
6
0.599
0.575
23
7
0.722
0.701
22
8
0.825
0.808
21
9
0.901
0.890
20
10
0.951
0.944
19
11
0.978
0.975
18
12
0.992
0.990
17
13
0.997
0.997
16
14
0.999
0.999
15
15
1.000
1.000
14 to 0
16 to 30
1.000
1.000
The desorption calculations utilize the following forms of Eqs. (15123) and (15124):
1 − εb
1 − 0.5
∂φ
∂φ
∂φ
= −u
−
(0.206)(1,000) φ − ψ
kK φ − ψ = −98.5 −
0.5
∂t
∂z '
εb
∂z '
= −98.5 ∂φ
− 206 φ − ψ
∂z ' ∂ψ
= 0.206 φ − ψ
∂t (1)
(2) Conditions for φ and ψ at t = 0 are given as a function of z' in the above table.
Condition for φ at z' = 0 is 0 because the desorption gas contains no benzene.
Equations (1) and (2) can be solved for φ{t,z'} and ψ{t,z'} for z' = 0 to 30 ft and t > 0 by various
means, including:
(1) A FORTRAN code similar to that on for Example 1513, with the method of lines and a stiff
integrator in connection with Eqs. (15127) to (15133).
(2) Use of the windows code PDESOL (www.pdesol.com), which also utilizes the method lines
∂φ
in connection with a number of different integrators an...
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 Spring '11
 Levicky
 The Land

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