Environmental Engineering Reference
In-Depth Information
Combining the finite difference form of Equation
(4.30) with Equations (4.31) and (4.32) gives
Solution
The longitudinal dispersion coefficient is given by Equa-
tion (4.34) as
2
2
2
2
2
2
1
2
σ
( )
t
−
−
σ
( )
t
1
2
V V
x V x V
σ
−
−
σ
x
2
x
1
t
t
K
=
=
(4.33)
2
1
L
t
t
/
/
2
1
2
1
3
2
2
V
σ
−
−
σ
t
t
K
L
=
2
x
x
which simplifies to the following useful expression
for
K
L
:
2
1
where
x
1
= 400 m and
x
2
= 700 m. The mean velocity,
V
,
and the temporal variances at
x
1
and
x
2
,
σ
t
2
and
σ
t
2
, must
be estimated from the measured data. The mean veloc-
ity,
V
, can be estimated by
3
2
2
V
σ
−
−
σ
t
t
K
L
=
(4.34)
2
x
x
2
1
x
−
−
x
where
σ
t
2
and
σ
t
2
are the temporal variances of the
concentration distributions measured at distances
x
1
and
x
2
downstream of the spill, respectively. Equation
(4.34) is useful in estimating the longitudinal dispersion
coefficient in open channels based on measurements of
concentration versus time at two locations along the
channel.
2
1
V
=
t
t
2
1
where
t
1
and
t
2
are the mean travel times to
x
1
and
x
2
,
respectively. Hence,
27
1
∑
t
=
t c
1
i
1
i
∑
27
c
i
=
1
1
i
i
=
1
EXAMPLE 4.5
where
t
i
and
c
1
i
are the measurement times and concen-
trations at
x
= 400 m. Since
A dye study is conducted to estimate the longitudinal
dispersion coefficient in a river. The dye is released
instantaneously from a bridge across the entire width of
the river, and the dye concentrations as a function of
time at locations 400 and 700 m downstream of the
bridge are as follows:
27
∑
=
c
i
35.9
mg/L
1
i
=
1
and
27
Concentration
(mg/L)
Concentration
(mg/L)
∑
=
t c
i
326
mg min/L
⋅
1
i
Time
(min)
Time
(min)
i
=
1
At 400 m At 700 m
At 400 m At 700 m
therefore
0
0
0
14
0.11
2.3
1
0
0
15
0.03
3.3
2
0
0
16
0
5.9
1
35.9
(326)
t
1
=
=
9.1
minutes
3
0.10
0
17
0
8.4
4
0.17
0
18
0
6.1
5
0.39
0
19
0
3.5
At
x
= 700 m,
6
1.4
0
20
0
2.1
7
2.9
0
21
0
1.3
8
7.1
0
22
0
0.65
27
1
∑
t
=
t c
9
10.5
0.01
23
0
0.21
2
i
2
i
∑
27
c
10
7.3
0.06
24
0
0.09
i
=
1
2
i
i
=
1
11
3.6
0.14
25
0
0.04
12
1.8
0.69
26
0
0
where
t
i
and
c
2
i
are the measurement times and concen-
trations at
x
= 700 m. Since
13
0.53
1.1
27
0
0
Estimate the longitudinal dispersion coefficient in the
river and verify that mass is conserved between the
upstream and downstream locations.
27
∑
=
c
i
35.9
mg/L
2
i
=
1
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