Environmental Engineering Reference
In-Depth Information
=
=
/
m
3
on
basis (23.24 g on a dry basis), moisture
45%, bulk density of resin
750 kg
moist basis, and liquid flowrate
=
1.0 L
/
day. The design column will have a flowrate
of 400,000 L
day, the allowable breakthrough time is seven days of flow, and the resin
depth is approximately ten times the column diameter. Using the kinetic approach to
column design, determine:
/
(a) The mass of resin required;
(b) the column diameter and depth.
Data from breakthrough test.
Volume (L)
C
(mg
/
L)
C
(meq
/
L)
15.9
4.5
0.14
18.1
17.2
0.54
19.5
40.0
1.26
20.7
62.9
1.98
22.0
86.4
2.72
23.4
98.2
3.09
Solution:
A plot of ln(
C
0
/
C
−
1) vs
V
gives a straight line with a slope of
−
k
1
C
0
/
Q
and an
intercept of
k
1
q
0
M
/
Q
. The plot of the data is shown in Figure 8.9.
k
1
=
(slope)(
Q
/
C
0
)
=
(0
.
76
/
L)(1.0 L
/
day)(L
/
3.2 meq)(1000 meq
/
eq) = 240 L
/
day
·
eq.
q
0
=
(intercept)
Q
/
(
k
1
M
)
10
−
3
eq
=
(15.3)(1.0 L
/
day)(day
·
eq
/
240 L)(1
/
23.24 g)
=
2
.
74
×
/
g
.
The mass of the resin in the design column can now be calculated:
ln
10
−
3
eq
C
0
(240 L
/
day
·
eq)(2
.
74
×
/
g)
M
=
0
.
05
C
0
10
5
(4
.
0
×
L
/
day)
10
3
meq)(7 days)
−
(240 L
/
day
·
eq)(3.2 meq
/
L)(1 eq
/
.
10
6
10
3
So, (a)
M
=
5
.
1
×
g
=
5
.
1
×
kg resin (dry weight).
This amount of resin has a volume of:
moist density
→
dry
moist
55)(m
3
10
6
4m
3
(5
.
1
×
kg)(1
/
0
.
/
750 kg)
=
12
.
.
4)(
D
2
)(10
D
)
4m
3
The diameter,
D
, is calculated from (
π/
=
12
.
.
So, (b)
D
=
1
.
2 m, and the depth
Z
=
10
D
=
12 m.
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