Environmental Engineering Reference
In-Depth Information
D x
( ) =
a x
α
(3.165)
1 1
.
5 0
2
+
0 5
2
.
1 1 15 1 0
0 000206 0 0001
(
+
.
)(
+
)
−
1
+
y
y
c
(
100 0 0
,
, )
=
)
(
100
)
π
( .
)( .
D x
( ) =
a x
β
(3.166)
z
z
=
=
0 576
576
.
kg/m
mg/L
3
where
a
y
,
a
z
,
α
, and
β
are constants, yields the following
concentration distribution resulting from a continuous
point source (Walters, 1962),
Therefore, the maximum concentration 100 m down-
stream of the source is 576 mg/L.
α β
+
M
(
1
+
α
)(
1
+
β
)
− +
1
c x y z
( ,
, )
=
x
2
2
π
a a
3.3.3.5 Instantaneous Line Source.
In the case of an
instantaneous release of tracer mass uniformly over a
finite line along the
z
-axis, the concentration distribu-
tion is given by superposition as
y z
(3.167)
(
1
4
+
α
)
V
y
2
(
1
4
+
β
β
)
V
z
a
2
exp
−
−
x
1
+
α
a
x
1
+
y
z
where
M
is the mass release rate (MT
−1
). Application
of Equation (3.167) is illustrated by the following
example.
M L
/
x
D t
2
y
D t
2
c x y z t
( ,
,
, )
=
exp
−
−
3
2
4
4
(
)
4
π
t
D D D
x
y
x
y
z
2
(
ζ
)
z
−
z
2
∫
exp
−
d
ζ
4
D t
EXAMPLE 3.18
z
1
z
(3.168)
A toxic substance is released at a rate of 0.5 kg/s from
a point source in the open ocean where field research
has shown that the horizontal diffusion coefficient of a
tracer can be estimated by (okubo, 1971)
where
M
is the mass released and
L
is the length of the
source. If the line source is in the range −
L
/2 ≤
z
≤
L
/2,
then Equation (3.168) can be expressed in the more
convenient analytical form
D
= 2 06
.
L
1 15
.
M L
t D D
/
x
D t
2
y
D t
2
where
D
is the diffusion coefficient in the horizontal
plane (cm
2
/s) and
L
is the distance traveled by the tracer
plume (m). At the discharge location, the vertical diffu-
sion coefficient is estimated as 1.0 cm
2
/s, and the average
current speed is 20 cm/s. Estimate the maximum con-
centration of the discharged substance 100 m down-
stream of the source.
c x y z t
( ,
,
, )
=
exp
−
−
4
4
8
π
x
y
x
y
(3.169)
+
2
−
z L
D t
z
z L
D t
/
−
/ 2
erf
erf
4
4
z
Application of this equation is illustrated by the follow-
ing example.
Solution
From the given data:
M
= 0 . kg/s
,
V
= 20 cm/s =
0.20 m/s, and
x
= 100 m. The diffusion coefficients can
be put in the form
EXAMPLE 3.19
one hundred kilograms of a tracer is released instanta-
neously over a vertical depth of 5 m in a deep ocean.
The ocean current is 10 cm/s, and the diffusion
coefficients are 1, 0.1, and 0.1 m
2
/s in the longitudinal,
horizontal-transverse, and vertical-transverse direc-
tions, respectively. The first-order decay coefficient of
the tracer is 0.01 min
−1
. Estimate the maximum concen-
tration of the tracer 20 m downstream of the release
location after 2 minutes.
D
y
=
2 06
.
x
1 15
.
cm /s
2
=
0 000206
.
x
1 15
.
m /s
2
D
z
=
1 0
.
cm /s
2
=
0 0001
.
m /s
2
In terms of the parameters in Equations (3.165) and
(3.166):
a
y
= 0.000206 m
2
/s,
α
= 1.15,
a
z
= 0.0001 m
2
/s,
and
β
= 0. To estimate the maximum concentration
100 m downstream of the source using Equation (3.167),
set
y
=
z
= 0 m, which makes the exponential term equal
to unity, and Equation (3.167) yields
Solution
From the given data:
M
= 100 kg,
L
= 5 m,
V
= 10 cm/s = 0.10 m/s,
D
x
= 1 m
2
/s,
D
y
= 0.1 m
2
/s,
D
z
=
0.1 m
2
/s,
k
= 0.01 min
−1
= 0.000167 s
−1
,
x
= 20 m, and
π
α β
+
M
(
1
+
α
)(
1
+
β
)
−
1
+
c x
( ,
0 0
, )
=
x
2
2
a a
y z
Search WWH ::
Custom Search