Environmental Engineering Reference
In-Depth Information
EXAMPLE 3.2
Hence, the maximum contaminant concentration
observed at the recreation area is expected to be
2.58 mg/L.
(c) When the concentration at the recreation area is
10
μ
g/L = 10
−5
kg/m
3
, Equation (3.66) requires that
one hundred kilograms of a contaminant is spilled into
a small river and instantaneously mixes across the entire
cross section of the river. The cross section of the river
is approximately trapezoidal in shape, with a bottom
width of 5 m, side slopes of 2:1 (H :v), and a depth of
flow of 3 m. The discharge in the river is estimated as
30 m
3
/s, and the dispersion coefficient for mixing along
the river is estimated as 10 m
2
is Estimate (a) when the
maximum contaminant concentration will be observed
at a park recreation area 10 km downstream of the spill,
and (b) the maximum concentration expected at the
park. (c) If a safe level of this contaminant in recre-
ational waters is 10
μ
g/L, how long after the spill can
the park expect to resume normal operations? (d) If the
contaminant undergoes first-order decay and has a
decay rate of 7.85 d
−1
, express the concentration at the
park as a function of time?
M
(
x Vt
D t
−
)
2
c x t
( , )
=
exp
−
4
A
4
π
D t
x
x
100
33 4
(
10 000 0 909
4 10
,
−
.
t
)
2
10
−
=
5
exp
−
(
)
t
π(
10
)
t
which yields
t
= 9400 seconds and 12,850 seconds.
Clearly, the concentration is above 10
μ
g/L from
t
= 9400 to 12,850 seconds, and the park water is
expected to be safe when
t
> 12,850 seconds = 3.57
hours after the spill.
(d) If the contaminant undergoes first-order decay with
k
= 7.85 d
−1
= 9.09 × 10
−5
s
−1
, then Equation (3.67)
gives the concentration at the park as a function of
time as
Solution
Me
−
kt
(
x Vt
D t
−
)
2
c x t
( , )
=
exp
−
(a) From the data given,
M
= 100 kg,
D
x
= 10 m
2
/s, and
the flow rate,
Q
, in the river is 30 m
3
/s. The cross-
sectional area,
A
, of the river is given by
4
A
4
π
D t
x
x
×
−
5
−
9 09 10
.
t
2
100
33 4
e
(
10 000 0 909
4 10
,
−
.
t
)
c
park
=
exp
−
(
)
t
π
(
10
)
t
A by my
=
+
2
0 270
.
(
10 000 0 909
40
,
−
.
t
)
2
−
5
=
exp
−
−
9 09 10
.
×
t
where
b
= 5 m,
y
= 3 m, and
m
= 2; hence,
t
t
A
=
5 3
( )
+
2 3 2
( )
=
33
m
2
Although these calculations provide useful results,
water-quality measurements would be required to
ensure the safety of the water at the recreational area
prior to resuming normal operations after the spill.
Q
A
30
33
V
=
=
=
0 909
.
m/s
The distance,
x
m
, of the maximum concentration
from the spill location at any time,
t
, is given by
3.3.1.1 Spatially and Temporally Distributed Sources.
The fundamental solution to the one-dimensional advec-
tion-diffusion equation can be used to derive other solu-
tions to the one-dimensional advection-diffusion equation
for tracer sources that are distributed in space and/or
time. These derivations are based on the principle of
superposition, which states that the sum of multiple
solutions to a linear differential equation (such as the
advection-diffusion equation) is also a solution to the
differential equation, and the corresponding boundary
and initial condition is the sum of the multiple boundary
and initial conditions, respectively.
x
m
=
Vt
Therefore, for
x
m
= 10 km = 10,000 m,
x
V
10 000
0 909
,
.
m
t
=
=
=
11 000
,
seconds
=
3 06
.
hours
Hence, the park can expect to see the peak contami-
nant concentration 3.06 hours after the spill occurs.
(b) The maximum contaminant concentration at any
location (
x
) occurs at
t
=
x/V
and is given by Equa-
tion (3.66) as
Spatially Distributed Sources.
Consider the initial con-
centration distribution of a contaminant illustrated in
Figure 3.8, where the initial one-dimensional concentra-
tion distribution is defined by
M
100
c x t
( , )
=
=
A
4
π
D t
x
33 4
π
(
10 11 000
)(
,
)
−
3
3
=
2 58 10
.
×
kg/m
=
2 58
.
mg/L
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