Environmental Engineering Reference
In-Depth Information
where
c
(
x
,
y
) is given by Equation (7.91). Typically, it is
not necessary to take
n
up to ∞, since
n
only has to be
large enough that the calculated value of
C
(
x
,
y
) does
not change significantly with the inclusion of additional
images, in practice a “significant change” usually means
changes within four significant digits.
( . )(
0 1 30
8 64 10
)
2
4
K
0
.
×
100
=
12
0 1
8 64 10
( . )
.
r
2
K
0
4
×
or
EXAMPLE 7.20
K
K
0 0323
0 00108
( .
)
0
A pollutant is released at a rate of 1.0 kg/s on the side
of a lake that is 3.0 m deep and 45 m wide. The lake is
estimated to have a dispersion coefficient of 1.0 m
2
/s
and an alongshore velocity of 15 cm/s. The first-order
decay constant of the pollutant is estimated as 0.15 min
−1
.
What is the steady-state concentration 30 m down-
stream of the source?
=
8 33
.
( .
r
)
0
Solving for
r
gives
r
= 918 m. Hence, a dilution of 100 is
achieved 918 m from the outfall. It is somewhat doubt-
ful that this formulation would be applicable out to
918 m from the shoreline.
The previous example uses a model in which the only
fate and transport processes are first-order decay and
turbulent diffusion. In cases where there is also an
alongshore current, the governing mass-balance equa-
tion is
Solution
From the given data:
M
= 1 . kg/s
,
H
= 3.0 m,
W
= 45 m,
D
= 1.0 m
2
/s,
U
= 15 cm/s = 0.15 m/s, and
k
= 0.15
minute
−1
= 0.0025 s
−1
. The situation is illustrated in
Figure 7.15, where the origin of the (
x
,
y
) coordinates is
at the source location, and the point P is where is con-
centration is to be estimated (30 m, 0 m). For a real or
image source located at (
x
i
,
y
i
), Equation (7.91) gives the
resulting concentration,
c
i
, at P as
∂
∂
c
x
∂
∂
2
c
+
∂
∂
2
c
y
−
(7.90)
U
=
D
kc
x
2
2
where
U
is the alongshore current (LT
−1
). For a vertical
line source of strength
M
(MT
−1
) at the origin and neg-
ligible concentration infinitely far from the source, the
solution of Equation (7.90) is given by (Boyce and
Hamblin, 1975)
2
M
HD
Ux
D
K
kr
D
2
U
D
+
c
=
exp
0
π
2
2
.
( . )(
1 0
3 0
( .
0 15 30
2 1 0
)(
)
c
i
=
. )
exp
π
1 0
( . )
2
M
HD
Ux
D
K
kr
D
2
U
D
+
c
=
exp
(7.91)
2
2
2
0
( .
0 0025
)[(
30
1 0
−
x
)
+
y
]
+
0 15
2 1 0
.
( . )
π
2
2
i
i
K
0
( .
)
where
r
is the radial coordinate (L) given by
c
=
1 007
.
K
( .
0 0025
)[(
30
−
x
)
2
+
y
2
]
+
0 005625
.
kg/m
3
i
0
i
i
2
(7.92)
mg/L
r
=
x
2
+
y
c
=
1007
K
( .
0 0025
)[(
30
−
x
)
2
+
y
2
]
+
0 005
.
625
i
0
i
i
(7.94)
As expected, when
U
= 0, Equation (7.91) reduces to
Equation (7.89). In cases where the width of the lake is
such that the calculated concentration is not negligible
on the side of the lake opposite the discharge, then
image sources can be used to ensure that the zero-flux
boundary condition is met on the opposite side of the
lake. If the width of the lake is
W
, then superposition of
images to meet the zero-flux boundary condition yields
the concentration distribution,
C
(
x
,
y
), as
lake boundary
15 cm/s
45 m
y
30 m
P
∞
∑
C x y
( ,
)
=
c x y
( ,
)
+
[ ( ,
c x y
+
2
nW c x y
)
+
( ,
−
2
nW
)]
x
lake boundary
source
n
=
1
(7.93)
Figure 7.15.
Pollutant source on the side of a lake.
Search WWH ::
Custom Search