Environmental Engineering Reference
In-Depth Information
(c) The maximum concentration 1 km downstream
from the release point occurs at time t 1 , where
A wd
=
=
(15)(3)
=
45
m
2
The concentration distribution resulting from an
instantaneous spill of a conservative contaminant
( k = 0) is given by Equation (4.37) as
x
V
1
t
=
1
In this case, x 1 = 1000 m and V = 0.35 m/s, and
therefore
M
(
x Vt
K t
)
2
c x t
( , )
=
exp
4
A
4
π
K t
L
L
1000
0.35
t 1
=
=
2860
seconds
In this case, V = 35 cm/s = 0.35 m/s, M = 10 kg, and
x = 500 m; hence, the concentration as a function of
time 500 m downstream of the spill site is given by
The maximum concentration, c 1 , occurring 1 km
downstream from the release point is given by
10
(45) 4
(500 0.35 )
4
t
2
c
(500, )
t
=
exp
0.0627
0.0627
(0.172)(2860)
K t
π
K t
3
c
=
=
=
0.00283
kg/m
L
L
1
K t
2
L
1
0.0627
(500 0.35 )
4
t
=
exp −
kg/m
3
=
2.83
mg/L
K t
L
K t
L
(d) The concentration 1 km downstream of the release
point as a function of time is given by
(b) The maximum concentration at any x occurs
(approximately) at a time t 0 given by
0.0627
(1000 0.35 )
4
t
2
x
V
3
c
(1000, )
t
=
exp
kg/m
t
0 =
K t
K t
L
L
When c (1000, t ) = 1 µ g/ L = 10 −6 kg/m 3 and K L =
0.172 m 2 /s,
In this case, x = 500 m and V = 0.35 m/s, which gives
500
0.35
t 0
=
=
1430
seconds
0.0627
0.172
(1000 0.35 )
4(0.172)
t
2
6
10
=
exp
t
t
The (maximum) concentration, c 0 , at time t 0 is given
by
which gives t = 2520 seconds. This time is relatively
close to the time that the maximum concentration
occurs (2860 seconds, for a difference of 340
seconds = 5.7 minutes). It should be noted that
there are two times when the concentration at
x = 1000 m is equal to 1 µ g/L: once before and once
after the arrival of the peak concentration.
M
0.0627
3
c
=
=
kg/m
0
A
4
π
K t
K t
L
0
L
0
When c 0 = 4 mg/L = 0.004 kg/m 3 and t 0 = 1430 s,
then
The dispersion problem consists primarily of predict-
ing downstream contaminant concentrations resulting
from a known or assumed spill. In some cases, the
(inverse) problem might be posed to determine the spill
characteristics that resulted in measured downstream
concentrations. However, this practical problem is
mathematically ill-posed and does not have a unique
solution.
0.0627
(1430)
0.004
=
K L
which gives
K L
= 0.172
m /s
This gives Pe = Vx / K L = (0.35)(500)/0.172 = 1017 at
x = 500 m, and Pe = 2035 at x = 1000 m. Since
Pe >> 1 at both x = 500 m and x = 1000 m, the
assumption that the maximum concentration occurs
at t 0 = x / V is valid.
4.3.1.2  Continuous  Release.  For the continuous
release of a tracer with concentration c 0 for a duration
δ , the following initial and boundary conditions are
appropriate:
 
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