Environmental Engineering Reference
In-Depth Information
The solutions for a steady continuous point source
derived using the “slab” approximation (i.e., Eqs. 3.152
and 3.154) are applicable for high Péclet number flows
in which the longitudinal advective flux is much greater
than the longitudinal diffusive flux. In cases where this
approximation is not valid, the concentration distribu-
tion downstream of a continuous point source is given
by (rubin and Atkinson, 2001)
environment has a small current of 1 cm/s and an esti-
mated diffusion coefficient of 1.0 m
2
/s in all three coor-
dinate directions. The tracer is estimated to have a
first-order decay coefficient of 0.1 d
−1
. (a) Estimate the
concentration after 10 minutes at two points of interest
50 m downstream of the release point, one point is on
the plume centerline and the other point is 25 m offset
from the centerline. (b) If the tracer release is continu-
ous at 10 kg/s, what is the steady-state concentration at
the points of interest, and how do these concentrations
compare with those calculated using the approximate
slab solution?
d
π
{
[
(
)
c x y z t
( ,
,
, )
=
exp(
2
ab
)
erf
F
+
F
1
2
2
a
(3.155)
(
)
]
+
−
erf
erf
F
+
F
exp(
−
2
ab
)
3
4
[
(
)
−
(
)
]
}
F
−
F
erf
F
−
F
1
2
3
4
Solution
where the parameters in Equation (3.155) are
From the given data:
m
= 10 kg/s
,
V
= 1 cm/s = 0.01 m/s,
D
x
=
D
y
=
D
z
= 1.0 m
2
/s,
k
= 0.1 d
−1
= 1.157 × 10
−6
s
−1
, and
x
1
=
y
1
=
z
1
= 0 m. The two points of interest (
x
,
y
,
z
) are
the centerline point (50 m, 0 m, 0 m) and the offset point
(50 m, 25 m, 0 m).
2
2
2
(
x
−
x
D
)
(
y y
D
−
)
(
z z
D
−
)
1
1
1
a
=
+
+
(3.156)
4
4
4
x
y
z
V
D
2
b
=
+
k
(3.157)
4
(a) For a finite release time,
t
1
= 1 minute = 60 seconds,
t
= 10 minutes = 600 seconds, and the derived
parameters for the centerline point (50 m, 0 m, 0 m)
are given by Equations (3.156-3.162) as
x
m
(
x
−
x V
D
)
1
d
=
exp
(3.158)
3
2
2
x
(
4
π
)
D D D
x
y
z
a
(3.159)
F
=
1
(
x x
D
−
)
2
(
y y
D
−
)
2
(
z z
D
−
)
2
t
−
t
1
1
1
a
=
+
+
1
4
4
4
x
y
z
(3.160)
F
=
b t
(
−
t
)
2
1
(
50 0
4 1 0
)
( . )
−
2
(
0 0
4 1. )
−
)
2
(
0 0
4 1 0
)
( . )
−
2
=
+
+
=
625
a
t
(
0
(3.161)
F
3
=
V
D
2
0 01
4 1 0
.
( . )
2
−
6
−
5
b
=
+ =
k
+
1 157 10
.
×
=
2 62 10
.
×
(3.162)
F
4
=
bt
4
x
m
(
x x V
D
−
)
1
d
=
exp
where
x
1
,
y
1
, and
z
1
are the coordinates of the source
location,
t
1
is the duration of the continuous release, and
k
is the first-order decay constant. Equation (3.155) was
derived from the integration (superposition) of instan-
taneous releases over time. For a continuous injection,
the injection time,
t
1
, is set equal to the injection time,
t
.
The steady-state solution is derived from Equation
(3.155) as
t
→ ∞, which gives
3
2
2
x
(
4
π
)
D D D
x
y
z
10
(
50 0 0 01
2 1 0
−
)( .
)
=
=
exp
0 288
.
3
2
( . )
(
4
π
)
( . )( . )( .
1 0 1 0 1 0
)
a
625
600 60
F
=
=
= .
1 076
1
t
−
t
−
1
F
=
b t
(
−
t
)
=
2 62 10
.
×
−
5
(
600 60
−
)
=
0 119
.
2
1
d
π
(
)
a
t
625
600
(3.163)
c x y z t
( ,
,
, )
=
exp
−
2
ab
F
=
=
= .
1 021
a
3
F
=
bt
=
( .
2 62 10
×
−
5
(
600
)
=
0 125
.
4
EXAMPLE 3.17
Substituting these derived parameters into Equa-
tion (3.155) gives the concentration,
c
0
, on the cen-
terline point (50 m, 0 m, 0 m) after 10 minutes as
A tracer is released into the deep ocean from a subma-
rine at a rate of 10 kg/s for 1 minute, where the ocean
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