Environmental Engineering Reference
In-Depth Information
where
V
is a bulk volume and
c
is the contaminant
concentration in the pore water. According to Equation
(5.56), the inverse of the retardation factor represents
the fraction of contaminant that is present in the water.
For example, a tracer with retardation factor of 5 has
20% of its mass in the aqueous phase and 80% of its
mass sorbed onto the aquifer matrix. To illustrate the
application of the retardation factor in this context, con-
sider the case where a mass
M
of contaminant is con-
tained within the aquifer and the contaminated water is
to be extracted using a pumping well. After the first pore
volume is extracted, the mass of contaminant remaining
in the aquifer is
M
of the contaminant mass is given by Equation (5.57),
where
(
)
j
−
1
F
= −
1
1
−
R
j
d
(
)
j
−
1
0 90
.
= −
1
1 5
−
which yields
j
= 10 .
and
(
)
−
d
. Similarly, after the second
pore volume is extracted, the mass remaining is
M
1
1
−
R
pore water extraction
=
jnV
(
=
10 3 0 2 100
. )( . )(
m
3
)
(
)
1
2
=
206
m
3
d
, and after
j
pore volumes have been
extracted, the contaminant mass remaining is equal to
M
1
−
R
−
(
)
j
Therefore, 10.3 pore volumes or 206 m
3
must be
extracted to yield a 90% reduction in contaminant mass
in the aquifer.
d
. Therefore, the fraction of initial mass,
M
,
flushed after
j
pore volumes,
F
j
, is given by
1
−
R
−
1
−
1
j
M M R
M
−
(
1
−
)
d
It is important to keep in mind that the retardation
factor is only a constant if the sorption isotherm is linear
and can be described by a single parameter, the dis-
tribution coefficient. In cases where the sorption iso-
therm is nonlinear, the retardation factor will depend
on the aqueous concentration, with increasing retar-
dation factors corresponding to increasing aqueous
concentrations.
F
=
= −
1
(
1
−
R
−
1
)
j
(5.57)
j
d
This relation is particularly useful is determining the
number of pore volumes that must be removed from an
aquifer to obtain a given level of site remediation. Equa-
tion (5.57) assumes that soil water sorption-desorption
equilibrium occurs instantaneously. When pollutant
desorption from soil occurs much slower than the flow
rate of water through the contaminated zone, contami-
nant concentrations will be lower than the predicted
equilibrium, resulting in even lower mass fractions
removed. In cases where the sorption-desorption
process is relatively slow, contaminant concentration
distributions in groundwater show a heavy tail, in con-
trast to a near-Gaussian distribution.
5.5.2 First-Order Decay
many chemical compounds in the environment decom-
pose into other compounds, usually through chemical
reactions, such as hydrolysis or biodegradation. Biodeg-
radation rates in groundwater are much less than bio-
degradation rates in soils, due primarily to the much
lower density of microorganisms in groundwater. The
most frequently used model of decomposition is the
following first-order decay model:
EXAMPLE 5.10
A contaminated aquifer is estimated to contain 30 kg of
contaminant spread over a 100-m
3
volume of aquifer.
The porosity of the aquifer is 0.2, and the retardation
factor of the contaminant is 5. Estimate the volume of
pore water that must be removed to reduce the mass of
contaminant in the aquifer by 90%.
S
m
= −λ
c
(5.58)
where
S
m
is the rate at which tracer mass is added to
groundwater per unit volume of groundwater (mL−3T−1),
−3
T
−1
),
c
is the concentration of tracer in the groundwater
(mL
−3
), and
λ
is the first-order decay coefficient (T
−1
).
Substituting this decay model into the advection-
diffusion equation, Equation (5.17), gives
Solution
From the data given,
M
= 30 kg,
V
= 100 m
3
,
n
= 0.2,
R
d
= 5, and
F
j
= 0.9. The number of pore volumes,
j
,
that must be removed from the aquifer to extract 90%
3
3
∂
∂
c
t
∂
∂
c
x
∂
∂
2
c
∑
∑
+
V
=
D
−
λ
c
(5.59)
i
i
2
x
i
i
i
=
1
i
=
1
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