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Unformatted text preview: t, which is obtained by
material balance, Eq. (1587), which after rearrangement is:
qout = Q cF − cout
1(0.324 − 0.002)
=
= 1118 mg / g
.
S
0.288 Then, from a rearrangement of the given Freundlich adsorption isotherm,
c* = qout
32 1/ 0.428 = 1118
.
32 2 .336 = 0.000395 mg / L Substitution into Eq. (4) gives, 0.324 − 0.002
= 6, 640, 000 s = 110,700 minutes = 1,845 h = 76.9 days.
3.02 × 10 −5 ( 0.002 − 0.000395 )
This very large residence time is due to the very low exit concentration of benzene in the
solution, making the continuous mode impractical.
tres = (d) Semicontinuous mode.
Average residence time of the solution in the tank = 1.5(110,700) = 166,100 min.
Flow rate of the solution = 50 gal/min or 50(3.785) = 189 L/min.
The volume of solution in the tank = 189(166,100) = 31,400,000 L or 31,400 m3.
For a cylindrical tank with height = diameter, this gives a tank diameter of 34.2 m, which is
probably impractical, although many smaller tanks in parallel could be used.
In the tank, where the solid adsorbent charge resides, the variation of the loading, q, with run
time is given by Eq. (1591), which for time in minutes is,
dq
S
= k L a cout − c * t resQ = 3.02 × 10 −5 (60) cout − c * (166,100)(189) = 56,900 cout − c *
(5)
dt
where S is in g, q is in mg B/g, and t is in min. Exercise 15.22 (continued)
Analysis: (d) (continued)
From Eq. (1586),
c + k L at resc * 0.324 + 3.02 × 10−5 (60)(166,100)c *
cout = F
=
= 0.00107 + 0.99669c *
1 + k L at res
1 + 3.02 × 10−5 (60)(166,100)
From the given Freundlich equation, c* = (q/32)2.336
= 0.000305 q2.336
(7)
Combining Eqs. (5), (6), and (7), gives,
dq
S
= 56,900 0.00107 + 0.99669c * − c * = 60.88 − 188.34c* = 60.88 − 0.0574q 2.336
dt (6) (8) Let the volume of solids in the tank = 2.5 vol%. The solidsfree volume of liquid in the tank =
31,400 m3. Therefore, the solids volume = (2.5/97.5)(31,400) =805 m3. Assume a particle
density = 850 kg/m3. Therefore, S = 805 (850) = 684,000 kg = 6.84 x 108 g and Eq. (8) becomes,
dq
= 8.90 × 10 −8 − 8.39 × 10 −11 q 2.336
(9)
dt
Now, compute the run time to achieve a cumulative cout = 0.002 mg/L, where,
t ccum = 0.002 = cout dt (10)
t
Eq. (9) represents an initial value problem in ODEs, which can be solved by the Euler method
using a spreadsheet, if a small time step is taken. The Euler form, with j as the iteration index, is,
0 q ( j +1) = q ( j ) + ( ∆t ) [ 8.90 × 10−8 − 8.39 × 10 −11 q 2.336 ( j) (11) The total run time is specified to be not less than 10 times the average liquid residence time; thus,
10(166,100) = 1,661,000 minutes. Try a ∆t of 500,000 minutes. From the spreadsheet, the first
6 and last 6 values are:
t, min. q, mg/g c*, mg/L
c out
c cum
0
0
0 0.00154
0
500000 0.04450 2.12E07 0.00154 0.00154
1000000 0.08900 1.07E06 0.00154 0.00154
1500000 0.13500 2.76E06 0.00155 0.00154
2000000 0.17800 5.41E06 0.00155 0.00154
2500000 0.22250 9.11E06 0.00155 0.00154
20000000
20500000
21000000
21500000
22000000
22500000 1.77815
1.82249
1.86682
1.91114
1.95545
1.99974...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.
 Spring '11
 Levicky
 The Land

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