Environmental Engineering Reference
In-Depth Information
t = 2 minutes = 120 seconds. The maximum concentra-
tion 20 m downstream of the source occurs at location
(20, 0, 0 m). Adapting Equation (3.169) to the case of a
nonzero current and first-order decay gives
tion as a function of time at a location 50 m north, 50 m
east, and 10 m above the centroid of the initial mass
release.
Solution
M L
t D D
/
(
x Vt
D t
)
2
y
D t
2
c x y z t
( ,
,
, )
=
exp
kt
From the given data: M = 10 kg, V = (1 m)(2 m)(2 m)
= 4 m 3 , D x = 10 m 2 /s, D y = 5 m 2 /s, D z = 0.05 m 2 /s,
x = 50 m, y = 50 m, and z = 10 m. The concentration dis-
tribution is given by Equation (3.170) as
4
4
8
π
x
y
x
y
z L
+
/
2
z L
D t
/
2
erf
erf
4
D t
4
z
z
100 5
/
c (
20 0 0 120
,
,
,
)
=
M V
/
8
π
(
120
) ( )( . )
1 0 1
20 0 10 120
4 1 1
c x y z t
( ,
,
, )
=
3
2
(
)
4
π
t
D D D
(
.
×
)
2
x
y
z
exp
(
)
2
( )(
20
)
x
ξ
4
D t
y
0
4 0 1 120
2
x
( .
0 000167 120
)(
)
(
)
2
( . )(
)
η
x
y
z
2
2
2
exp
d d d
ξ η ζ
4
D t
0 5 2
4 0 1 120
/
( . )(
+
x
y
z
1
1
1
y
erf
(
ζ 2
)
)
z
4
D t
0 5 2
4 0 1 120
/
( . )(
z
erf
10 4
/
)
c
(50, 50, 10,
t
) =
3
2
(
)
4
π
t
(10)(5) 0 05
50
4 10
50
4 5
( .
)
=
=
0 0140
14 0
.
kg/m
3
.
m
g/L.
(
)
2
ξ
(
)
t
Therefore, the maximum concentration 20 m down-
stream of the source after 2 minutes is 14.0 mg/L.
(
)
2
η
0 5
.
1
1
exp
d d d
ξ η ζ
( )
t
0 5
.
1
1
(
)
2
3.3.3.6  Instantaneous  Volume  Source.  The general
solution for an instantaneous release of a tracer mass
over a finite volume can be derived directly from Equa-
tion (3.139) by taking the mass per unit volume, denoted
by g ( x , y , z ), as a constant and equal to M / V , where M
is the tracer mass released over a volume V . This gives
10
4 0 05
ζ
( .
)
t
= 0 0355
.
3
2
t
(
)
2
50
40
50
20
10
0 2
ξ
t
M V
/
c x y z t
( ,
,
, )
=
2
(
)
η
0 5
.
1
1
3
2
(
)
exp
d d d
ξ η ζ
kg/m
3
4
π
t
D D D
x
y
z
t
0 5
.
1
1
2
2
ζ 2
(
)
(
)
(
)
x
ξ
y
η
z
x
y
z
2
2
2
2
(
)
ζ
exp
d d d
ξ η ζ
4
D t
4
D t
4
D t
x
y
z
.
t
1
1
1
x
y
z
(3.170)
This integral can be evaluated numerically to determine
the values of c (50, 50, 10, t ) for specified values of t .
Application of this equation is illustrated by the follow-
ing example.
3.4 TRANSPORT OF SUSPENDED PARTICLES
EXAMPLE 3.20
The advection-diffusion equation is appropriate for
describing the fate and transport of dissolved contami-
nants that are advected with the same velocity as the
ambient water. In the case of suspended particles, the
settling of the particles is influenced by the size, shape,
Ten kilograms of a contaminant in the form of a
1 × 2 × 2-m parallelepiped is released into the deep
ocean. The N-S, E-W, and vertical diffusion coefficients
are 10, 5, and 0.05 m 2 /s, respectively. Find the concentra-
 
Search WWH ::




Custom Search