Environmental Engineering Reference
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function of time at a location 400 m downstream of
the gate. (b) How long after the gate is opened will the
concentration at the downstream location be equal to
1 mg/L?
c(x,t)
initial condition:
t = 0
Solution
c
0
t = t
3
t = t
2
t = t
1
(a) From the data given,
c
0
= 5 mg/L = 0.005 kg/m
3
,
V
= 20 cm/s = 0.20 m/s, and
D
x
= 5 m
2
/s. At
x
= 400 m, Equation (3.82) gives the concentration
as a function of time as
0
x
c
x Vt
D t
−
Figure 3.10.
Diffusion from a step-function initial condition.
o
c x t
( , )
=
erfc
2
4
x
0.005
2
400 0 20
4 5
−
.
( )
t
c
(400, )
t
=
erfc
A further simplification of Equation (3.79) comes from
the definition of the
complementary error function
,
erfc(
z
), where
t
89 4 0 0
.
−
. 447
t
kg/m
=
0.0025
erfc
3
t
erfc
( )
z
= −
1
erf
( )
z
(3.80)
(b) When the concentration 400 m downstream of the
gate is equal to 1 mg/L = 0.001 kg/m
3
,
The solution given by Equation (3.79) can therefore be
written in the form
89 4 0 0447
.
−
.
t
0.001
=
0.0025
erfc
t
c
x
D t
x
0
c x t
( , )
=
erfc
(3.81)
2
4
which leads to
89.4 0.0447
−
t
=
and the concentration distribution,
c
(
x
,
t
), is illustrated
in Figure 3.10. The corresponding solution for a fluid
moving with a velocity
V
is given by
erfc
0 4
.
t
or
c
x Vt
D t
−
89 4 0 0447
.
−
.
t
o
=
c x t
( , )
=
erfc
(3.82)
erf
0 6
.
2
4
t
x
This concentration distribution is identical to that illus-
trated in Figure 3.10 when viewed relative to an origin
moving with velocity
V
.
using the error function tabulated in Appendix D.1,
89 4 0 0447
.
−
.
t
=
0 595
.
t
EXAMPLE 3.3
which leads to
A long drainage canal is gated at the downstream
end and is designed to retain the runoff from an agri-
cultural area. The runoff into the canal is expected to
infiltrate into the groundwater. After a severe storm, the
concentration of a toxic pesticide in the channel rises to
5 mg/L and is distributed uniformly throughout the
channel. Because of the threat of flooding, the gate at
the downstream end of the canal is opened, and water
in the canal flows downstream at a velocity of 20 cm/s.
(a) If the longitudinal dispersion coefficient in the canal
is 5 m
2
is give an expression for the concentration as a
t
=
1487
seconds
=
24 8
.
minutes
Therefore, the concentration 400 m downstream
of the gate will reach 1 mg/L approximately 24.8
minutes after the gate is opened.
FINITE VOLUME SOURCE. This case occurs when a
tracer of mass
M
is released instantaneously and uni-
formly over a cross-sectional area
A
and over a longi-
tudinal interval
x
L
<
x
<
x
r
. under these circumstances,
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