Horizontal Infiltration
using Richards
Equation.
The Bruce and Klute
approach for horizontal
infiltration
1
Richards Eq: let’s derive it again for kicks
Richards Equation is easy to derive, so let's do it here
for one-dimensional horizontal flow.
For horizontal flow Darcy’s law says:
h
q = -K
z
The conservation of mass tells us:
the time rate of change in stored water is equal to the
negative of the change in flux with distance (i.e., an
increase or decrease in flux with distance results in
respective depletion or accumulation of stored water)

q
=
t
z
2

q
recall
and - =
t
z
Taking the first derivative of Darcy’s law with
respect to position,
h
q = -K
z
q

=z
z


K

h
z




and substituting the result from the
conservation of mass for the left side

  h 
=
K  

t
z   z 




Richards Equation for horizontal flow.
3
Richards Eq:

  h 
=
K  

t
z   z 




Using the definition of soil-water diffusivity
D  Kh
this may be written in the form of the diffusion
equation

  
=
D  

t
z  z 




Our goal: to solve this rascal for horizontal
infiltration into dry soil
4
So what are the rules here?
• If we find a solution to Richards Equation which
satisfies the boundary and initial conditions,
then we have the unique solution.
• The Green and Ampt solution had a one-to-one
relationship between the square root of time
and position.
With that motivation, let’s introduce a “similarity
variable" referred to as either the Boltzman or
Buckingham transform
z
=
t
5
Now putting the equation in terms of 
We are going to write Richards Eq. with  in
place of t and z. We need to calculate the
substitutions for the derivatives. For z:

1
=
z
t
or
z = 
t
For t:

-
-z
= 2 t 3/2 = 2 t
t
or
-2t
t =
 

6
-2t
  and z = 
We have t =


  h 
t and
=
K
t
z   z 




Using the expressions for dz and dt, we see
that the right side of Richards Eq is

t
 
d -
=
=
 t
d  2t
NOTE: The partial derivatives are now simple
derivatives since there is only one variable in
the similarity version of the equation.
The left side can be put in terms of :
  
D 

z  z 
    
=
D

z   z 
7
Finishing the substitution
  
D 

z  z 

1
=
z
t
    
=
D

z   z 
1 d   1 d 
=
D

t d    t d 
1 d  d 
= t
D

d   d 
Putting this all together we find:
- d
1 d  d  multiply -  d 
d  d 
=
D
=
D


2t d 
t d   d  by t:
2 d
d   d 
8
We have in terms of 
- d
d  d 
=
D

2 d
d   d 
Multiplying each side by d and
integrating from  =  i to  we obtain

- 
d

d ' = D( )

2
d

i
where ’ is the dummy variable of
integration. This may be rearranged:

The
Bruce
and
d

-1

D() =
  d '
Klute Eq.!
2 d 
i
9
Wait! That integral is constant!
-1 d 
D() =
2 d




 d '
i
If i is zero (initially dry soil), the integral is
identified as a soil-water parameter

S( ) 



d 
0
which is referred to as the soil “sorptivity.”
For infiltration into initially dry soil the B&K Eq. is
-S( ) d 
D() = 2
d
Pretty Simple! Clearly solutions to will depend on
the form of D() and S().
10
Now what? Need D and
-S( ) d 
S! D() = 2 d 
 What forms of the function D() allow for
analytical solutions?
 Philip (1960 a, b) developed a broad set of forms
of D() which produce exact solutions.
 Brutsaert (1968) then provided an expression for
diffusivity which fit well to natural soils and allow
solution
n 


D0 (n+1)
n 1 
DB =

n

n+1 
11
Using the Brutsaert
DB
n 
D0 (n+1) n 

DB =

1n

n+1 
• n and Do determined experimentally for soil.
• 1< n < 10, depending on pore size distribution
• Do is the diffusivity at saturation
Using this for D(), B&K is generally solved by
n
 = (1-  )
2D o (n+1) 



n2

1/2
12
THE SOLUTION!!!
n
 = (1-  )
2D o (n+1) 


2

n

1/2
• This may be easily checked by putting the
solution into B&K and turning the crank.
• This equation gives the exact solution for
the shape of the wetting front as a function
of time.
• Exactly the information that we could not
get out of the Green and Ampt approach.
13
1
n=2
Moisture Content
0.8
n=5
n = 10
0.6
0.4
0.2
0
0
0.2
0.4
0.6
0.8
1
1.2
1.4
x/[t^1/2]
Horizontal infiltration as a function of n for a Brutsaert
soil with Do = 1, and n = 2, 5, and 10. The wetting
front becomes increasingly sharp as n increases,
making the pore size distribution narrower.
HYDRUS-2D Simulations of Horz. Infil.
Horizontal Infiltration
Moisture Content (vol/vol)
Plotting moisture
content vs position
above and moisture
content vs x/t1/2 on
the lower plot
0.5
0.4
0.3
0.2
0.1
0
0
20
40
60
80
100
120
X/t^.5
Also recall that in
Miller similarity time
scales with the
square root of
macroscopic length
scale…not bad!
t = 0.2
t = 0.4
t = 1.6
t = 2.56
t = 6.55
t = 42.9
t = 200
Horizontal Infiltration
Moisture Content (vol/vol)
They fit the
Boltzman transform!
t = 0.1
0.5
0.45
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
t = 0.1
t = 0.2
t = 0.4
t = 1.6
t = 2.56
t = 6.55
t = 42.9
0
5
10
15
t = 200
X/t^.5
15
Now about that sorptivity ...
Can also solve for the S. Using Brutsaert’s
equation and the definition of S

S() =



d ’
0

=




(1- ' n )
2D o (n+1) 1/2


2

n

d '
0
Let’s pull out the constant
C
=
2D o (n+1) 


2

n

1/2
16
Computing the Sorptivity ...
So we have the result

S() = C  (1- ' n )d '
0
Which is easy to integrate to obtain
S() = C
n+1

 
- n+1





Most often sorptivity is reported for
saturated soil,  = 1
 2D o 
nC
S(1) = (n+1) = n+1 
1/2
17
Why bother (with S)?
S(1) =
 2D o 


n+1 
1/2
Suppose we want to calculate the infiltration
Integrating the moisture content over all positions at a
given time

I=



 dx

0
x
We can evaluate the same integral by switching the
bounds of integration so that we integrate all positions
over the moisture content

I =   x d
0

or in
of 

I=t
1/2



d 
0
18

Computing cumulative infiltration
I=t
1/2



d 
0
Which is just
I=St
1/2
which is exactly the form obtained by Green
and Ampt! (i.e. square root of time)
Can calculate the rate of infiltration
dI
q=
dt
1
=
St
2
-1/2
Again, identical Green and Ampt!
19
OK, but what is sorptivity?
A parameter which expresses the macroscopic
balance between the capillary forces and the
hydraulic conductivity.
Recall From the discussion of the Green and
Ampt results that
SGA =
2 Ksat f n
Ksat goes up with 2 and f goes with 1/ ( is the
characteristic microscopic length scale, for
instance d50), then we can guess that sorptivity
will get larger for coarser soils, but only with 1/2 20
Miller Scaling Big Time
From the definition of S() we know that
d
S( ) = 2 D( )
d
dh
= 2 K( )
d
dh
=2 K( )  
x
d  
t 
where S is the sorptivity, D is the diffusivity, K is the
conductivity,  is the moisture content and  is the
Boltzman transform variable,  = xt-1/2
21
S( ) =2 K( )
Miller Scaling S
dh
x 
d  
t 
To derive the scaled value of sorptivity we must
replace the variables with the appropriate
t
scaled quantities
x
x• =
t• =

L
L
(L is macro h 
h
K• =  K 
h• =
scopic scale)





We find:

d 


S=
K•


d


=

h


•

Lx • 
L 2 t • 

 

dh •
2K •
=

d •

S•

22
Wrapping this up ...
S=

S•

S scales with 1/2, (Just as we saw in the
Green and Ampt Sorptivity).
As a little bonus, we see what the effect
of changing fluid properties would
produce in the value of sorptivity
23