Environmental Engineering Reference
In-Depth Information
impermeable boundary
c
(
x
,
t
)
L
L
c
1
(
x
,
y
) +
c
2
(
x
,
y
):
(source and image)
c
1
(
x
,
y
):
(from source)
c
2
(
x
,
y
)
(from image)
source
image
x
= 0
x
=
L
x
= 2
L
x
Figure 3.11.
Impermeable boundary condition.
which is a Gaussian distribution in which mass
M
is
contained in one-half the distribution (i.e., where
x
> 0).
boundary conditions that require zero mass flux across
the impermeable boundary. This scenario is illustrated
in Figure 3.11. Since mass flux,
M
x
, is governed by the
Fickian diffusion equation
EXAMPLE 3.5
∂
∂
c
x
one kilogram of a contaminant is spilled into an open
channel at a location 50 m from the end of the channel.
The channel has a rectangular cross section 10 m wide
and 2 m deep, and the diffusion coefficient along the
channel is estimated as 10 m
2
is (a) Assuming that the
contaminant is initially well mixed across the channel,
express the concentration as a function of time at the
end of the channel. (b) How long will it take for
the contaminant concentrations 25 m upstream (in the
direction of the channel end) to be 10% higher than
the concentration 25 m downstream of the spill (in the
direction away from the channel end)?
M
(3.88)
= −
D
x
x
an impermeable boundary (across which the mass flux is
equal to zero) requires that the concentration gradient,
∂
c
/∂
x
, be equal to zero at the boundary. referring to
Figure 3.11, if the tracer source is located at
x
= 0 and the
impermeable boundary is located at
x
=
L
, then if an
identical tracer source is superimposed at
x
= 2
L
, the
resulting (superimposed) concentration distribution has
the following properties within the domain
x
∈ [−∞,
L
]:
(1) satisfies the diffusion equation (via linearity);
(2) satisfies the initial condition of an instantaneous mass
release at
x
= 0; and (3) ∂
c
/∂
x
= 0 at
x
=
L
, due to the sym-
metry of the superimposed solutions around
x
=
L
. These
results demonstrate that the symmetrical placement of
an
image source
produces a solution that satisfies the dif-
fusion equation, as well as the required initial and bound-
ary conditions. Applying this result to the case of an
instantaneous source of mass,
M
, located at a distance
L
from an impermeable boundary, indicates that the result-
ing concentration distribution is given by
Solution
(a) The concentration distribution is given by
M
x
D t
2
(
x
−
L
D t
2 )
2
+
c x t
( , )
=
exp
−
exp
−
4
4
A
4
π
D t
x
x
x
where the
x
-coordinate is measured from the
spill location in the direction of the end of the
channel. From the data given,
M
= 1 kg,
A
= 10 m ×
2 m = 20 m
2
,
D
x
= 10 m
2
/s,
L
= 50 m, and
x
= 50 m
(at the end of the channel). The concentration as a
function of time at the end of the channel is there-
fore given by
M
x
D t
2
(
x
−
L
D t
2
)
2
+
c x t
( , )
=
exp
−
exp
−
4
4
A
4
π
D t
x
x
x
(3.89)
1
It is interesting to note that if a source is located on the
boundary (i.e.,
L
→ 0), then Equation (3.89) gives the
concentration distribution as
c
(
50
, )
t
=
20 4
π
(
10
50
4 10
)
t
2
[
50 2 50
4 10
−
(
)]
2
+
exp
−
exp
−
(
)
t
(
)
t
2
4
M
x
D t
2
c x t
( , )
=
exp
−
(3.90)
4
A
π
D t
which reduces to
x
x
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