Geoscience Reference
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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 ) 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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