Geoscience Reference
In-Depth Information
1000
Half-life time
10 d
50 d
Adsorption:
R
= 2
20 d
inf.
100
10
1
0.1
0
100
200
Time (days)
Figure 5.11
Breakthrough curves at retardation factor
R
= 2 for different rates of
decomposition. The experimental data of
Figure 5.2
apply.
where
μ
is the irst-order decomposition parameter (d
-1
). Suppose we have the amount
M
=
M
0
at
t
= 0. Integration of Eq. (
5.19
) results in an exponential decline of
M
0
with time:
0
e
µ
−
t
Mt
()=
M
(5.20)
This means that 50% of the mass is left at
t
= (ln 2)/
μ
, which is called the half-life
T
50
(d).
Question 5.10:
Derive the expression for half-life
T
50
= (ln 2)/
μ
with Eq. (
5.20
).
A mobile chemical with irst-order decay can be simulated with the convection-
dispersion equation (Eq. (
5.8
)), putting
S
s
=
μ θ C
l
. Therefore, for steady low condi-
tions (
θ
= constant) the transport equation may be written as:
∂
∂
=
C
t
∂
∂
2
C
z
−
∂
C
z
l
D
l
v
∂
−
l
µ
C
(5.21)
e
2
l
Proceeding with the leaching experiment of
Figure 5.2
,
Figure 5.11
shows the break-
through curves in case of convection, dispersion, adsorption (
R
= 2) and different
half-lives
T
50
. In case of
T
50
= ininity, no decay occurs and the breakthrough curve
is similar to the one in
Figure 5.10
with
R
= 2. At the soil surface we started with
solute concentrations equal to 1000 μg L
-1
. Note how effective the solute concentra-
tion decrease is owing to the combined effect of adsorption and decomposition: from
1000 μg L
-1
to less than 1 μg L
-1
!
Figure 5.12
shows that in case of higher adsorption,
the decomposition is much more effective. By increasing the retardation factor from
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