Environmental Engineering Reference
In-Depth Information
the initial condition of the advection-diffusion equation
is
500
2
erf
1800 1 5 0 02 1 86 400
4 2 1 86 400
+
.
−
( .
)( )(
,
)
c
(
1800 1
, )
=
( )( )(
,
)
M
1800 1 5 0 02 1 86 400
4 2 1 86 400
−
.
−
( .
)( )(
,
)
c
=
,
x
≤ ≤
x
x
−
erf
e
( )( )
1 1
0
L
R
c x
( , 0
)
=
f x
( )
=
A x
(
−
x
)
( )( )(
,
)
R
L
0
,
x
<
x
or
x
>
x
=
0 372 mg/L
.
L
R
(3.83)
Therefore, the concentration 1800 m downstream of the
spill after 1 day is 0.372 mg/L.
Substituting this initial condition into Equation (3.70)
yields the following solution for the concentration
distribution in a flowing environment with first-order
decay
Transient Source.
Suppose that a contaminant source is
located at a single point along the
x
-axis, and the source
injects the contaminant with a time-varying mass flux,
m
( )
, uniformly over a cross-sectional area
A
in a stag-
nant fluid. This scenario is equivalent to a mass of
m t dt
c
erf
(
x x
−
)
−
Vt
−
(
x x
−
)
−
Vt
0
L
R
c x t
( , )
=
erf
e
kt
( )
being released during every consecutive time interval
dt
. The concentration distribution,
dc
(
x
,
t
), resulting
from an instantaneous mass of
m d
2
4
D t
4
D t
x
x
(3.84)
( τ τ
released from
where
k
is the first-order decay constant, and
V
is the
fluid velocity. Cases where the ambient environment is
stagnant and the tracer is conservative can be accom-
modated by taking
V
= 0 and
k
= 0, respectively. In the
special case of a semi-infinite source, taking
x
L
→ −∞
and
x
r
= 0 yields Equation (3.81), which was derived
previously.
x
= 0 at time
τ
, is given by
m d
τ τ
π
( )
x
D t
2
dc x t
( , )
=
exp
−
(3.85)
4
(
−
τ
)
A
4
D t
(
−
τ
)
x
x
and superimposing all of the resulting concentration
distributions yields
EXAMPLE 3.4
m d
τ τ
π
( )
x
D t
2
t
∫
c x t
( , )
=
exp
−
(3.86)
Thirty kilograms of a toxic substance is spilled over a
3-m length of stream. The spilled substance is initially
well mixed across the stream, and the substance has a
first-order decay constant of 1 d
−1
. The stream is 10 m
wide and 2 m deep with an average velocity of 2 cm/s
and a longitudinal diffusion coefficient of 2 m
2
/s. Esti-
mate the concentration in the stream 1800 m down-
stream of the center of the spill after 1 day.
A
4
D t
(
−
τ
)
4
(
−
τ
)
0
x
x
This equation describes the concentration distribution
resulting from a transient source at
x
= 0. In the case of
a spatially distributed transient source with mass flux
m x t
( , )
, the principle of superposition indicates that the
resulting concentration distribution is given by
ξ τ ξ τ
π
(
)
(
)
−
2
m
,
d d
x
D t
−
ξ
t
x
Solution
R
∫
∫
c x t
( ,
)
=
exp
−
(
)
(
)
4
τ
A
4
D t
−
τ
0
x
L
x
x
From the given data:
M
= 30 kg,
L
= 3 m,
k
= 1 d
−1
,
W
= 10 m,
H
= 2 m,
V
= 2 cm/s = 0.02 m/s,
D
x
= 2 m
2
/s,
x
= 1800 m, and
t
= 1 day. using Equation (3.83), the
initial concentration,
c
0
, is given by
(3.87)
where
x
L
and
x
r
are the upper and lower bounds of the
tracer source location.
M
M
W H L
30
10 2 3
c
=
=
=
3.3.1.2 Impermeable Boundaries.
In the superposi-
tion examples cited so far, the boundary conditions have
required that the tracer concentration approaches zero
as
x
approaches infinity. Therefore, the superimposed
concentration distributions all have boundary condi-
tions in which the tracer concentration approaches zero
as
x
approaches infinity. In cases where impermeable
boundaries exist in relatively close proximity to the
tracer source, the diffusion equation must satisfy
0
A x
(
−
x
)
(
×
)
(
×
)( )
R
L
=
0.5
kg/m
3
=
500
mg/L
Taking
x
L
= −
L
/2 = −1.5 m and
x
r
= +
L
/2 = +1.5 m, the
required concentration is given by Equation (3.84) as
c
(
x
−
x
)
−
Vt
−
(
x
−
x
)
−
Vt
0
L
R
e
kt
c x t
( , )
=
2
erf
erf
4
D t
4
D t
x
x
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