Environmental Engineering Reference
In-Depth Information
horizontal-transverse, and vertical-transverse disper-
sivities are 3, 0.3, and 0.03 m, respectively. (a) Assuming
that the contaminant is conservative, determine the
downstream location at which the contaminant plume
will be fully mixed over the depth of the aquifer.
(b) Estimate the contaminant concentrations at the
water table at locations 10, 100, and 1000 m downstream
of the source after 10 years. (c) If the contaminant
undergoes biodegradation with a decay rate of
0.01 day
−1
, estimate the effect on the concentrations
downstream of the source.
Since the contaminant becomes well mixed at
x
0
= 833 m, this formulation can only be used for
calculating the concentrations at
x
≤ 833 m. At
x
= 10 m and
x
= 100 m, the equation yields the fol-
lowing results:
x
c
(
x
, 0, 0, 3650)
c
(
x
, 0, 0, 3650)
(m)
(kg/m
3
)
(mg/L)
10
0.046
46
100
0.0090
9.0
At
x
= 1000 m, the plume is well mixed over
the vertical, and the contaminant concentration is
calculated by replacing
x
by
x
0
(= 833 m) in the
denominator of the error function in the
z
term to
yield
Solution
(a) From the given data,
Y
= 3 m,
Z
= 2 m,
c
0
= 100 mg/L = 0.1 kg/m
3
,
V
= 0.4 m/d,
H
= 7 m,
α
x
= 3 m,
α
y
= 0.3 m, and
α
z
= 0.03 m. The contami-
nant plume becomes well mixed at a distance
x
0
downstream, where
x
0
is given by Equation
x
−
1460
132
1 37
.
5 77
.
c x
( ,
0 0 3650
,
,
)
=
0 05
.
erfc
erf
erf
(5.27) as
x
833
x
−
1460
132
1 37
.
=
0 0111
.
erfc
erf
(
H Z
−
)
2
(
7 2
0 03
−
)
2
x
=
=
=
833
m
x
0
α
.
z
which gives
c
(1000, 0, 0, 3650) = 0.0011 kg/m
3
=
1.1 mg/L.
(c) If the contaminant undergoes first-order decay with
λ
= 0.01 d
−1
, the concentration profile along the line
y
= 0 m,
z
= 0 m, is given by Equation (5.28) as
(b) The concentration along the line
y
= 0 m,
z
= 0 m is
given by Equation (5.25) as
c
(
x Vt
Vt
−
)
=
0
c x
( ,
0 0
,
, )
t
erfc
8
2
2
(
α
)
1 2
/
x
−
(
Y
/
)
(
−
Y
x
/
2
)
erf
erf
c x
( ,
0 0 3650
0 1
8
,
,
)
2
(
α
x
)
1 2
/
2
(
α
)
1 2
/
y
y
1 2
/
.
x
4 0 01 3
0 4
( .
)( )
=
(
Z
x
)
−
(
−
Z
x
)
exp
1
−
1
+
erf
erf
2 3
( )
.
2
(
α
)
1 2
/
2
(
α
)
/
1 2
z
z
1 2
/
x
−
0 4 3650 1 4 0 01 3 0 4
2 3 0
. (
)(
+ ×
.
×
/
. )
erfc
1 2
/
and therefore at
t
= 10 years = 3650 days,
(
×
.
4 3650
×
)
3 2
2 0 3
(
/
)
−
3 2
2 0 3
(
−
/
)
erf
erf
0 1
8
.
0 4 3650
2 3 0 4 3650
1 2
(
x
−
.
×
)
=
( .
x
)
1 2
/
( .
x
)
1 2
/
c x
( ,
0 0 3650
,
,
)
erfc
(
×
.
×
)
/
2
2 0 03
( )
( .
2
2 0 03
(
−
)
erf
−
erf
3 2
2 0 3
(
/
)
3 2
2 0 3
(
−
/
)
erf
−
1 2
/
x
)
1 2
erf
x
)
( .
1 2
/
1 2
/
( .
x
)
( .
x
)
2
2 0 03
( )
( .
−
2
2 0 03
(
−
)
erf
erf
which simplifies to
x
)
1 2
/
( .
x
)
1 2
/
x
−
1665
132
which simplifies to*
c x
( ,
0 0 3650
,
,
)
=
0 05
.
exp(
−
0 0234
.
x
)
erfc
1 37
.
5 77
.
x
−
1460
132
1 37
.
5 77
.
erf
erf
c x
( ,
0 0 3650
,
,
)
=
0 05
.
erfc
erf
erf
x
x
x
x
and at
x
= 10 m and
x
= 100 m yields the following
results:
* This simplification uses the identity
erf
(
−
x
)
= −
erf
( )
x
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