Environmental Engineering Reference
In-Depth Information
Hence, the concentration 100 m downstream of the
source is 62.7 mg/L.
10
c
(
100 0 0
,
, )
=
4
π
(
100 0 1 0 1
0 25 0
4 0 1 100
) ( . )( . )
2
2
( . )( )
( . )(
25 0
4 0 1 100
(
. )( )
( . )(
0
A similar analysis can be used in cases where the con-
tinuous source is well mixed over one of the lateral
dimensions, and diffusion is unlimited in the other
lateral dimension. Such a case is described by the fun-
damental solution to the one-dimensional diffusion
equation, and if the tracer is well mixed over a depth
Z
in the vertical dimension, then
exp
−
−
)
)
=
0 0796
.
kg/m
3
=
79 6
.
mg/L
Hence, the concentration 100 m downstream of the
source is 79.6 mg/L.
(b) If the tracer undergoes first-order decay with a
decay constant of
k
= 0.1 min
−1
= 0.001667 s
−1
, then
the concentration 100 m downstream of the source
is given by
m
w
V
y
D
2
c y
( , )
τ
=
exp
−
(3.153)
wZ
4
π τ
D
4
τ
y
y
k
x
V
which is simplified by replacing
τ
by
x
/
V
and yields
c
(
100 0 0
,
, )
=
(
79 6
. )exp
−
m
Vy
D x
2
100
0 25
c x y
( ,
)
=
exp
−
(3.154)
=
(
79 6
. )exp
−
( .
0 001667
)
4
.
Z
4
π
D xV
y
y
= 40 . mg/L
As a practical example, this equation could represent
the concentration in a wide stream downstream of a
tracer source where the tracer is well mixed over the
depth, but lateral diffusion is yet to be inhibited by the
the river banks.
Hence, the concentration 100 m downstream of the
source is 40.9 mg/L.
(c) If the tracer undergoes first-order decay and is 5 m
below the ocean surface, then an image source must
be added 5 m above the ocean surface to account
for the zero-flux boundary condition at the ocean
surface. The concentration 100 m downstream of
the source caused by the image is given by Equation
(3.152) by taking
y
= 0 m and
z
= 10 m, hence the
image concentrations,
c
i
, is given by
EXAMPLE 3.16
A contaminant is released at a rate of 1 kg/s from the
center of a wide stream, where the depth of the stream
is 1.0 m, the flow velocity is 30 cm/s, and the transverse
diffusion coefficient is 0.5 m
2
is If the contaminant is
initially well mixed over the depth, estimate the con-
taminant concentration at an instream intake that is
located 100 m downstream of the discharge and 2 m
from the center of the stream.
2
2
m
x D D
Vy
D x
Vz
D x
c x y z
( ,
, )
=
exp
−
−
i
4
4
4
π
y
z
y
z
10
c
i
(
100 0 10
,
,
)
=
4
π
(
100 0 1 0 1
0 25 0
4 0 1 100
) ( . )( . )
Solution
From the given data:
m
= 1 kg/s
,
Z
= 1.0 m,
V
= 0.30 m/s,
D
y
= 0.5 m
2
/s,
x
= 100 m, and
y
= 2 m. Neglecting the
effect of the lateral boundaries of the stream, the con-
centration at the downstream intake location is given
by Equation (3.154) as
( . )( )
( . )(
2
( .
0 25 10
4 0 1 100
)(
)
2
exp
−
−
)
( . )(
)
=
=
0 0426
42 6
.
kg/m
mg/L
3
.
Superposition of the source and image concentra-
tions gives a predicted concentration of 79.6 mg/L +
42.6 mg/L = 122.2 mg/L for a conservative tracer.
For a nonconservative tracer with
k
= 0.1 min
−1
=
0.001667 s
−1
, the concentration is given by
2
m
Vy
D x
c x y
( ,
)
=
exp
−
4
Z
4
π
D xV
y
y
2
1
0 30 2
4 0 5 100
( . )( )
( . )(
c
(
100 2
, )
=
)
exp
−
)
( )
1
4
π
( . )(
0 5 100 0 30
)( .
k
x
V
3
c
(
100 0 0
,
, )
=
(
122 2
. )exp
−
=
=
0 0733
73 3
.
kg/m
mg/L
.
100
0
=
(
122 2
. )exp
−
( .
0 001667
)
.
25
Hence, the concentration at the downstream intake is
73.3 mg/L.
=
62 7
. mg/L
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