Environmental Engineering Reference
In-Depth Information
M
M
(
)
=
c x y x t
,
,
,
(
)
=
c x y z t
,
,
,
2 2
4
π
t
4
π
D D D t
1
+
φ
t
2 2
4
π
t
4
π
D D D t
1
+
φ
t
x
y
z
x
y
z
2
1
2
2
t
1
2
t
∫
(
)
( )
∫
(
)
x
−
V
τ τ
d
−
λ
y
+
λ
t
z
x
−
V
( )
τ
d
τ
−
λ
y
+
λ
t
z
0
y
z
0
y
z
0
0
exp
−
exp
−
(
)
(
)
4
D t
1
+
φ
2 2
t
4
D t
1
+
φ
2 2
t
x
x
2
2
y
D t
z
D t
−
−
−
kt
y
D t
2
z
D t
2
4
4
y
z
−
−
−
kt
4
4
100
y
z
(
)
=
c
100 0 0 600
,
,
,
(
)
(
)( )(
)(
)
4
π
600
4
π
10 5 0 1 600
.
1 063
.
(3.143)
−
(
)(
)
100
0 2 600
.
where
k
is the first-order decay coefficient, and
ϕ
is
defined by the relation
2
1
2
)
(
( )
+
( )
)(
−
0 1 0
.
0 05 0
.
600
exp
−
(
)(
)(
)
4 10 600 1 063
.
D
D
1
12
D
D
y
z
φ
2
=
λ
2
+
λ
2
(3.144)
y
z
0
2
x
x
−
( )(
)
4 5
600
0
4 0 1 600
Application of this equation is illustrated by the follow-
ing example.
2
)
)
−
(
)(
−
0 000167 600
.
(
)(
.
=
2 40 10
.
×
−
5
kg/m
3
=
2
4 . µg/L
EXAMPLE 3.14
Therefore, the concentration 100 m downstream of the
release location after 10 minutes is estimated as
24.0
μ
g/L.
one hundred kilograms of a tracer is released into the
deep ocean where the mean current is 20 cm/s, the
velocity shears in the horizontal-transverse and vertical-
transverse directions are 0.1 and 0.05 s
−1
, respectively,
and the diffusion coefficients in the longitudinal,
horizontal-transverse, and vertical-transverse directions
are 10, 5, and 0.1, respectively. The tracer has a first-
order decay constant of 0.01 min
−1
. Estimate the con-
centration at a point 100 m downstream of the release
location after 10 minutes.
3.3.3.3 Continuous Point Source with Constant Dif-
fusion Coefficient.
Consider the case of a steady con-
tinuous point source releasing a tracer into an ambient
environment flowing with a spatially uniform velocity,
V
. If we further assume that in the downstream direc-
tion the advective flux,
Vc
, is much greater than the
diffusive flux,
D
x
∂
c
/∂
x
, (i.e., Pe
x
>> 1), then the advection-
diffusion equation under steady-state conditions (i.e.,
∂
c
/∂
t
= 0) simplifies to
Solution
From the given data:
M
= 100 kg,
V
0
= 20 cm/s = 0.20 m/s,
λ
y
= 0.1 s
−1
,
λ
z
= 0.05 s
−1
,
D
x
= 10 m
2
/s,
D
y
= 5 m
2
/s,
D
z
= 0.1 m
2
/s,
k
= 0.01 min
−1
= 0.000167 s
−1
,
x
= 100 m,
y
=
z
= 0 m, and
t
= 10 minutes = 600 s. using these
data, gives the following:
∂
∂
c
x
∂
∂
2
c
y
∂
∂
c
z
2
V
=
D
+
D
(3.145)
y
z
2
2
where
c
is the tracer concentration,
x
is the coordinate
in the downstream direction,
y
and
z
are the transverse
coordinates, and
D
y
and
D
z
are the
x
and
y
components
of the diffusion coefficient. The solution to Equation
(3.145) is facilitated by changing coordinates from
(
x
,
y
,
z
) to (
τ
,
y
,
z
) where
1
12
D
D
D
D
1
12
5
10
0 1
10
.
=
y
2
2
2
z
2
2
φ
=
λ
+
λ
0 1
.
+
0 05
.
y
z
x
x
s
−
2
=
0
.000419
1
+
φ
t
2 2
= +
1
( .
0 000419
) (
2
600
)
2
=
1 063
.
τ=
x
V
Substituting the values of the given and derived param-
eters into Equation (3.143) gives
(3.146)
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