Environmental Engineering Reference
In-Depth Information
from the right-hand boundary. To simulate the presence
of the right-hand boundary, an image source is placed
at point 1 where
x
= 2
d
2
, and to simulate the presence
of the left-hand boundary an image is placed at point 1′
where
x
= −2
d
1
. However, the solution does not end
here because the image placed at point 1 creates an
imbalance of images relative to the left-hand boundary
(thereby violating the impermeable condition on the
left-hand boundary), and similarly, the image placed at
point 1' creates an imbalance of images relative to the
right-hand boundary. To correct these imbalances, addi-
tional images must be placed at points 2 and 2′, respec-
tively, where
x
= 2(
d
1
+
d
2
) at point 2 and
x
= −2(
d
1
+
d
2
)
at point 2′. These additional images again create imbal-
ances, creating the need for additional images ad infini-
tum and so the resulting concentration between the
boundaries is given as
0 00892
.
62 5
.
kg/m
c
(
50
, )
t
=
exp
−
3
t
t
(b) The concentration 25 m upstream from the spill
(
x
= 25 m) is given by
1
c
(
25
, )
t
=
20 4
π
(
10
25
4 10
)
t
2
2
+
[
25 2 50
4 10
−
(
)]
exp
−
exp
−
(
)
t
(
)
t
0 00446
.
15 6
.
141
+
=
exp
−
exp
−
t
t
t
and the concentration 25 m downstream from the
spill (
x
= −25 m) is given by
1
c
(
−
25
, )
t
=
20 4
π
(
10
25
4 10
)
t
M
x
D t
2
c x t
( , )
=
exp
−
4
2
−
−
[
25 2 50
4 1
−
(
)]
2
A
4
π
D t
+
x
x
exp
−
exp
(
)
t
(
0
)
t
∞
∑
(
x
−
x
D t
)
2
(
x
− ′
x
D t
)
2
+
n
n
+
exp
−
ex
p
−
0 00446
.
15 6
.
391
+
4
4
=
exp
−
exp
−
x
x
n
=
1
t
t
t
(3.91)
When
c
(25,
t
) is 10% higher than
c
(−25,
t
), then
where
x
n
and
x
n
are the locations of the image sources
′
given by
c
(
25
25
, )
t
=
1 1
.
c
(
−
, )
t
n d
(
+
d
)
+
(
d
−
d
)
when odd
when even
n
1
2
2
1
x
n
=
(3.92)
n d
(
+
d
)
n
or
1
2
15 6
.
141
+
and
exp
−
exp
−
t
t
= 1 .
15 6
.
391
+
−
n d
(
+
d
)
+
(
d
−
d
)
when odd
when even
n
exp
−
exp
−
1
2
2
1
(3.93)
x
′
=
t
t
n
−
n d
(
+
d
)
n
1
2
Solving this equation for
t
yields
t
= 55 seconds.
Therefore, after 55 seconds, the concentrations 25 m
upstream and downstream of the spill differ by 10%.
Since Equation (3.91) is based on the superposition of
the concentration distribution resulting from a single
real source and an infinite number of images, it is some-
times convenient to express Equation (3.91) as
In some cases of practical interest, the source is located
between two impermeable boundaries as shown in
Figure 3.12. In this case, the source is located at a dis-
tance
d
1
from the left-hand boundary and a distance
d
2
∞
∑
c x t
( , )
=
c x t
( , )
+
[
c x
(
−
x t
, )
+
c x
(
− ′
x t
, )]
(3.94)
0
0
n
0
n
n
=
1
d
1
d
2
boundary
boundary
source
image
2'
1'
1
2
-
2(2
d
1
+
d
2
)
x
-2
d
1
x
= 0
2
d
2
2(
d
1
+
d
2
)
Figure 3.12.
Two impermeable boundaries.
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