Environmental Engineering Reference
In-Depth Information
f
(
x
)
f
(
x
)
concentration distribution at
t
= 0
f
(
e
)
c
0
d
e
e
x
Figure 3.8.
Initial concentration distribution.
0
x
Figure 3.9.
Step-function initial condition.
c x
( , )
0 =
f x
( )
(3.68)
This initial concentration distribution is equivalent to
an infinite number of adjacent instantaneous sources of
mass
f
(
x
)
Adx
located along the
x
-axis between
x
L
and
x
r
, where
A
is the cross-sectional area over which the
contaminant is well mixed. For each of these incremen-
tal sources located a distance
ξ
away from the origin,
the fundamental solution of the one-dimensional diffu-
sion equation is applicable, and the resulting concentra-
tion distribution is given by
Changing variables from
x
to
u
, where
x
− ξ
u
=
(3.73)
4
D t
x
Equation (3.72) becomes
c
∞
∫
2
o
c x t
( , ) =
e
−
u
du
(3.74)
x
D t
x
π
2
f
( )
ξ ξ
π
d
D t
(
x
−
ξ
)
4
c x t
( , )
=
exp
−
(3.69)
4
D t
4
x
x
This integral cannot be evaluated analytically, but is
similar to a special function in mathematics called the
error function
, erf(
z
), which is defined as
using the principle of superposition to sum the solu-
tions for all the incremental sources, results in the total
solution
2
z
∫
2
−
ξ
(3.75)
erf(
z
=
e
d
ξ
π
2
f
( )
ξ ξ
π
d
D t
(
x
−
ξ
)
0
x
∫
R
c x t
( , )
=
exp
−
(3.70)
4
D t
4
x
L
x
x
and values of this function are tabulated in Appendix
D.1. It is useful to note the following properties
Application of this equation to derive solutions to the
advection-diffusion equation for semi-infinite and finite
volume sources are illustrated below.
erf
(
−
z
)
= −
erf
( )
z
(3.76)
erf( 0
=
(3.77)
SEMI-INFINITE VOLUME SOURCE. A frequently encoun-
tered case is where the initial concentration equal to a
fixed nonzero value for
x
≤ 0 and equal to zero for
x
> 0.
This initial concentration distribution is the step func-
tion shown in Figure 3.9, which is described by
erf( ∞ = 1
(3.78)
Comparing the solution of the diffusion equation, Equa-
tion (3.74), with the definition of the error function,
Equation (3.75), gives
c
,
,
x
x
≤
>
0
o
x
D t
c x
( , )
0
=
f x
( )
=
(3.71)
c
∞
∫
∫
2
2
0
0
o
c x t
( , ) =
e
−
u
du
−
4
e
−
u
du
x
π
0
0
and substituting this initial condition into Equation
(3.70) yields
c
π
π
erf
x
D t
(3.79)
o
=
−
2
2
π
4
x
2
c d
D t
ξ
(
x
−
ξ
)
c
x
D t
0
∫
o
o
c x t
( ,
)
=
exp
−
(3.72)
=
1
−
erf
4
D t
2
4
π
4
−∞
x
x
x
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