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and thus:
d
I
d
d
s
t
= −
( )
I
=
cum
θθ
f
(4.30)
t
i
d
t
Equations (
4.28
) and (
4.30
) yield:
h
s
d
d
s
t
=
( )
k
1−
f
θθ
f
(4.31)
t
t
i
f
When gravity can be neglected (early stage of iniltration or horizontal iniltration),
the term ∂
z
/∂
s
in Eq. (
4.28
) vanishes and Eq. (
4.31
) becomes:
h
s
d
d
s
t
−=−
( )
k
f
θθ
f
(4.32)
t
t
i
f
With rearranging and integrating Eq. (
4.32
) we obtain:
−= −
( )
+
kht
½
θθ
2
s
C
(4.33)
tf
t
i
f
Because
s
f
= 0 at
t
= 0, the integration constant is zero, and thus Eq. (
4.33
) becomes:
½
=
−
2
θθ
kh
½
(4.34)
s
tf
t
f
−
t
i
Because
k
t
,
θ
t
,
θ
I
and
h
f
remain constant during the low process, Eq. (
4.34
) can be
read as
s
f
/ √
t
= constant, that is, the depth of the wetting front is proportional to the
square root of time.
Combination of Eqs. (
4.29
) and (
4.34
) leads to (Koorevaar et al.,
1983
):
2
θθ
½
−
( )
½
½
I
=−
k h
t
=
St
(4.35)
cum
t
f
t
i
where the sorptivity
S
(m d
-½
) is:
θθ
½
−
( )
S
=−
2
tf
k h
(4.36)
t
i
Question 4.12:
Does sorptivity
S
remain constant during the iniltration process?
Question 4.13:
Two horizontal columns of the same soil, with initial volume fractions
of water
θ
1
and
θ
2
, respectively, are iniltrated with water. Assume
θ
t
and
h
f
indepen-
dent of
θ
i
. Denote parameters pertaining to sample 1 by subscript 1 and to sample 2 by
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