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Unformatted text preview: and t is in seconds. This equation 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, where j is the iteration
number, is,
c ( j +1) = c ( j ) − ( ∆t ) 6.46 × 10 −5 c ( j ) − 0.223 − c ( j )
5.72 1.773 (3) If a time step, ∆t, of 500 s is used to find the time when c becomes 0.01 mg/L, the first 10 values
of c are as follows:
Time, s
c of C, mg/L
0
0.223
500
0.216
1000
0.209
1500
0.202
2000
0.196
2500
0.189
3000
0.183
3500
0.177
4000
0.172
4500
0.166
5000
0.161
The time required is found to be 51,500 s = 858 minutes = 14.3 h. A plot of c vs. t is given on
the next page, where semilog coordinates gives almost a straight line. Exercise 15.23 (continued)
Analysis: (b) (continued) C o nc entratio n o f C hloro fo rm in so lution , m g/L 1 Slurry Batch Adsorption 0.1 0.01
0 10000 20000 30000 40000 Time, seconds 50000 60000 Exercise 15.23 (continued)
Analysis: (continued)
(c) For continuous slurry adsorption, assume a solution feed rate, Q, of 1 L/h. Therefore, the
adsorbent feed rate, from Part (a) = S = 2(0.286) = 0.572 g carbon/h. A rearrangement of Eq.
(1586) gives the residence time as:
t res = 0.223 − 0.01
cF − cout
=
*
6.46 × 10 −5 0.01 − c*
k L a cout − c (4) The quantity c* is the ficticious concentration in equilibrium with qout, which is obtained by
material balance, Eq. (1587), which after rearrangement is:
qout = Q cF − cout
1(0.223 − 0.01)
=
= 0.372 mg / g
S
0.572 Then, from a rearrangement of the given Freundlich adsorption isotherm,
q
c* = out
10 1/ 0.564 0.372
=
10 1.773 = 0.00292 mg / L Substitution into Eq. (4) gives, tres = 0.223 − 0.01
= 466, 000 s = 7,770 minutes = 130 h = 5.4 days.
6.46 × 10−5 ( 0.01 − 0.00293) This is a large residence time compared to the batch time in Part (b).
(d) Semicontinuous mode.
Average residence time of the solution in the tank = 1.5(7,700) = 11,600 min.
Flow rate of the solution = 50 gal/min or 50(3.785) = 189 L/min.
The volume of solution in the tank = 189(11,600) = 2,190,000 L or 2,200 m3.
For a cylindrical tank with height = diameter, this gives a tank diameter of 14.1 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 = 6.46 × 10−5 (60) cout − c * (11,600)(189) = 8,500 cout − c *
(5)
dt
where S is in g, q is in mg C/g, and t is in min.
From Eq. (1586),
cF + k L at resc * 0.223 + 6.46 × 10 −5 (60)(11,600)c *
cout =
=
= 0.00485 + 0.97826c *
(6)
1 + k L at res
1 + 6.46 × 10 −5 (60)(11,600) Exercise 15.23 (continued)
Analysis: (d) (continued)
From the given Freundlich equation, c* = (q/10)1.773
(7) = 0.01687 q1.773 Combining Eqs. (5), (6), and (7), gives,
dq
S
= 8,500 0.00485 + 0.97826c * − c * = 412 − 184.79c* = 412 − 31...
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 Spring '11
 Levicky
 The Land

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