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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