Environmental Engineering Reference
In-Depth Information
where indices 1-3 refer to soil, roots and leaves, respectively;
C
(mg kg
−
1
) is con-
centration;
k
1
,
k
2
and
k
3
are the sum of all first-order loss processes in compartment
1, 2 and 3, respectively, and
k
12
and
k
23
are the transfer rates from compartment
1 to 2 and 2 to 3, respectively.
I
i
(mg d
−
1
) describes the constant input to the
compartments, e.g., from air, and
M
i
(kg) is the mass of compartment
i
,
i
1, 2,
3. The matrix elements
k
and
I
can be derived from the differential equations above
(Eqs.
9.15
,
9.18
and
9.20
).
Linear differential equations approach steady state for
t
=
→∞
, i.e. the change
of concentration with time is zero,
dC/dt
0. The steady-state solutions for matrix
equations 1 (soil), 2 (roots) and 3 (leaves) with continuous input are as follows:
=
I
1
k
1
M
1
C
1
(
t
→∞
)
=
(9.29)
I
2
k
2
M
2
+
k
12
k
2
C
2
(
t
→∞
)
=
C
1
(
t
→∞
)
(9.30)
I
3
k
3
M
3
+
k
23
k
3
C
3
(
t
→∞
)
=
C
2
(
t
→∞
)
(9.31)
The steady-state solution follows the general scheme:
I
n
k
n
M
n
+
k
n
−
1,
n
k
n
C
n
(
t
→∞
)
=
×
C
n
−
1
(
t
→∞
)
(9.32)
where
n
is the compartment number.
The analytical solutions for the differential equations 1 (soil), 2 (roots) and
3 (leaves) for a pulse input is the same as for initial concentrations
C(0)
=
0:
e
−
k
1
t
C
1
(
t
)
=
C
1
(0)
×
(9.33)
e
−
k
1
t
(
k
2
−
e
−
k
2
t
(
k
1
−
e
−
k
2
t
C
2
(
t
)
=
k
12
C
1
(0)
×
k
1
)
+
+
C
2
(0)
×
(9.34)
k
2
)
e
−
k
1
t
e
−
k
2
t
e
−
k
3
t
C
3
(
t
)
=
k
12
k
23
C
1
(0)
×
k
3
)
+
k
3
)
+
(
k
1
−
k
2
)(
k
1
−
(
k
2
−
k
1
)(
k
2
−
(
k
3
−
k
1
)(
k
3
−
k
2
)
(9.35)
e
−
k
2
t
(
k
3
−
e
−
k
3
t
(
k
2
−
+
C
3
(0)
×
e
−
k
3
t
k
2
)
+
+
k
23
C
2
(0)
×
k
3
)
The general solution scheme for pulse input to soil only, i.e.
C
1
(0)
=
0 and
C
n
(0)
=
0 with
n
>
2 is as follows:
⎧
⎨
⎫
⎬
n
−
1
n
e
−
k
j
t
C
n
(
t
)
=
k
i
,
i
+
1
C
1
(0)
×
(9.36)
⎩
n
⎭
(
k
k
−
k
j
)
i
−
1
j
=
1
k
=
1,
k
=
j
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