1

Horizontal Infiltration Horizontal Infiltration using Richards using Richards

EquationEquationThe Bruce and Klute The Bruce and Klute

approach for horizontal approach for horizontal infiltrationinfiltration

The Bruce and Klute The Bruce and Klute approach for horizontal approach for horizontal

infiltrationinfiltrationWilliams, 2002 http://www.its.uidaho.edu/AgE558

Modified after Selker, 2000 http://bioe.orst.edu/vzp

2

Richards Eq: let’s derive it againRichards Eq: let’s derive it again

Richards Equation is easy to derive, so let's do it here Richards Equation is easy to derive, so let's do it here for one-dimensional horizontal flow. for one-dimensional horizontal flow.

For For horizontalhorizontal flow Darcy’s law says: flow Darcy’s law says:

The The conservation of massconservation of mass tells us: tells us:

the time rate of change in stored water is equal to the the time rate of change in stored water is equal to the negative of the change in flux with distance (i.e., an negative of the change in flux with distance (i.e., an increase or decrease in flux with distance results in increase or decrease in flux with distance results in respective depletion or accumulation of stored water)respective depletion or accumulation of stored water)

Richards Equation is easy to derive, so let's do it here Richards Equation is easy to derive, so let's do it here for one-dimensional horizontal flow. for one-dimensional horizontal flow.

For For horizontalhorizontal flow Darcy’s law says: flow Darcy’s law says:

The The conservation of massconservation of mass tells us: tells us:

the time rate of change in stored water is equal to the the time rate of change in stored water is equal to the negative of the change in flux with distance (i.e., an negative of the change in flux with distance (i.e., an increase or decrease in flux with distance results in increase or decrease in flux with distance results in respective depletion or accumulation of stored water)respective depletion or accumulation of stored water)

-t =

qz

3

Taking the first derivative of Darcy’s law with respect to Taking the first derivative of Darcy’s law with respect to position, position,

and substituting the result from the conservation of mass and substituting the result from the conservation of mass for the left sidefor the left side

Richards Equation for horizontal flow.Richards Equation for horizontal flow.

Taking the first derivative of Darcy’s law with respect to Taking the first derivative of Darcy’s law with respect to position, position,

and substituting the result from the conservation of mass and substituting the result from the conservation of mass for the left sidefor the left side

Richards Equation for horizontal flow.Richards Equation for horizontal flow.

recall and

t =

z

K 

h

z  

4

Richards Eq:Richards Eq:

Using the definition of soil-water diffusivity Using the definition of soil-water diffusivity

this may be written in the form of the diffusion equationthis may be written in the form of the diffusion equation

Our goalOur goal: to solve this using : to solve this using boundary conditionsboundary conditionsand and initial conditionsinitial conditions for horizontal infiltration into dry soil for horizontal infiltration into dry soil

Using the definition of soil-water diffusivity Using the definition of soil-water diffusivity

this may be written in the form of the diffusion equationthis may be written in the form of the diffusion equation

Our goalOur goal: to solve this using : to solve this using boundary conditionsboundary conditionsand and initial conditionsinitial conditions for horizontal infiltration into dry soil for horizontal infiltration into dry soil

t =

z

K 

h

z  

hKD

5

So what are the rules here?So what are the rules here?

•• If we find a solution to Richards Equation which If we find a solution to Richards Equation which satisfies the boundary and initial conditions, then satisfies the boundary and initial conditions, then we have we have thethe unique solution. unique solution.

•• The Green and Ampt solution had a one-to-one The Green and Ampt solution had a one-to-one relationship between the square root of time and relationship between the square root of time and position. position.

With that motivation, let’s introduce a “similarity With that motivation, let’s introduce a “similarity variable" referred to as either the Boltzman or variable" referred to as either the Boltzman or Buckingham transformBuckingham transform

•• If we find a solution to Richards Equation which If we find a solution to Richards Equation which satisfies the boundary and initial conditions, then satisfies the boundary and initial conditions, then we have we have thethe unique solution. unique solution.

•• The Green and Ampt solution had a one-to-one The Green and Ampt solution had a one-to-one relationship between the square root of time and relationship between the square root of time and position. position.

With that motivation, let’s introduce a “similarity With that motivation, let’s introduce a “similarity variable" referred to as either the Boltzman or variable" referred to as either the Boltzman or Buckingham transformBuckingham transform

6

Now putting the equation in terms of Now putting the equation in terms of

We are going to write Richards Eq. with We are going to write Richards Eq. with in in place of t and z. We need to calculate the place of t and z. We need to calculate the substitutions for the derivatives. For z:substitutions for the derivatives. For z:

For t:For t:

We are going to write Richards Eq. with We are going to write Richards Eq. with in in place of t and z. We need to calculate the place of t and z. We need to calculate the substitutions for the derivatives. For z:substitutions for the derivatives. For z:

For t:For t:

t =

-z2 t 3/2 =

-2 t or t =

-2t 

7

We have and andWe have and and

Using the expressions for Using the expressions for z and z and t, we see that t, we see that the right side of Richards Eq isthe right side of Richards Eq is

NOTE: The partial derivatives are now simple NOTE: The partial derivatives are now simple derivatives since there is only one variable in the derivatives since there is only one variable in the similarity version of the equation.similarity version of the equation.

The left side can be put in terms of The left side can be put in terms of ::

Using the expressions for Using the expressions for z and z and t, we see that t, we see that the right side of Richards Eq isthe right side of Richards Eq is

NOTE: The partial derivatives are now simple NOTE: The partial derivatives are now simple derivatives since there is only one variable in the derivatives since there is only one variable in the similarity version of the equation.similarity version of the equation.

The left side can be put in terms of The left side can be put in terms of ::

t =

z

K 

h

z   t = -2t 

t =

t =

dd

-2t

8

Finishing the substitutionFinishing the substitution

Putting this all together we find:Putting this all together we find:

multiplymultiply

by t:by t:

Putting this all together we find:Putting this all together we find:

multiplymultiply

by t:by t:

z =

1t

-2t

dd =

1t

dd

D dd

-2

dd =

dd

D dd

9

We have in terms of We have in terms of

Multiplying each side by dMultiplying each side by d and integrating from and integrating from = = ii to to we obtain we obtain

where where ’’ is the dummy variable of integration. is the dummy variable of integration. This may be rearranged:This may be rearranged:

The Bruce and The Bruce and

Klute Eq.!Klute Eq.!

Multiplying each side by dMultiplying each side by d and integrating from and integrating from = = ii to to we obtain we obtain

where where ’’ is the dummy variable of integration. is the dummy variable of integration. This may be rearranged:This may be rearranged:

The Bruce and The Bruce and

Klute Eq.!Klute Eq.!

-2

dd =

dd

D dd

i

-2  d ' = D()

dd

D() = -12

dd

i

   d '

10

Wait!Wait! That integral is constant! That integral is constant!

If If ii is zero (initially dry soil), the integral is identified as a soil- is zero (initially dry soil), the integral is identified as a soil-

water parameterwater parameter

which is referred to as the soil “sorptivity.”which is referred to as the soil “sorptivity.”

For infiltration into initially dry soil the B&K Eq. isFor infiltration into initially dry soil the B&K Eq. is

Pretty Simple! Clearly solutions to will depend on the form of Pretty Simple! Clearly solutions to will depend on the form of D(D() and S() and S().).

If If ii is zero (initially dry soil), the integral is identified as a soil- is zero (initially dry soil), the integral is identified as a soil-

water parameterwater parameter

which is referred to as the soil “sorptivity.”which is referred to as the soil “sorptivity.”

For infiltration into initially dry soil the B&K Eq. isFor infiltration into initially dry soil the B&K Eq. is

Pretty Simple! Clearly solutions to will depend on the form of Pretty Simple! Clearly solutions to will depend on the form of D(D() and S() and S().).

D() = -12

dd

i

   d '

S()

0

   d

D() = -S()

2 dd

11

Now what? Need D and S!Now what? Need D and S!

What forms of the function D(What forms of the function D() allow for ) allow for analytical solutions? analytical solutions?

Philip (1960 a, b) developed a broad set of forms Philip (1960 a, b) developed a broad set of forms of D(of D() which produce exact solutions. ) which produce exact solutions.

Brutsaert (1968) then provided an expression for Brutsaert (1968) then provided an expression for diffusivity which diffusivity which fit well to natural soilsfit well to natural soils and allow and allow solutionsolution

What forms of the function D(What forms of the function D() allow for ) allow for analytical solutions? analytical solutions?

Philip (1960 a, b) developed a broad set of forms Philip (1960 a, b) developed a broad set of forms of D(of D() which produce exact solutions. ) which produce exact solutions.

Brutsaert (1968) then provided an expression for Brutsaert (1968) then provided an expression for diffusivity which diffusivity which fit well to natural soilsfit well to natural soils and allow and allow solutionsolution

D() = -S()

2 dd

12

Using the Brutsaert DUsing the Brutsaert DBB

• n and Dn and Doo determined experimentally for soil. determined experimentally for soil.

• 1< n < 10, depending on pore size distribution 1< n < 10, depending on pore size distribution

• DDoo is the diffusivity at saturation is the diffusivity at saturation

Using this for D(Using this for D(), B&K is ), B&K is generallygenerally solved by solved by

• n and Dn and Doo determined experimentally for soil. determined experimentally for soil.

• 1< n < 10, depending on pore size distribution 1< n < 10, depending on pore size distribution

• DDoo is the diffusivity at saturation is the diffusivity at saturation

Using this for D(Using this for D(), B&K is ), B&K is generallygenerally solved by solved by

DB = D0(n+1)

n n

1-n

n+1

= (1-n)

2Do(n+1)

n2 1/2

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THE SOLUTION!THE SOLUTION!

• This may be easily checked by putting the This may be easily checked by putting the solution into B&K and turning the crank. solution into B&K and turning the crank.

• This equation gives the exact solution for the This equation gives the exact solution for the shape of the wetting frontshape of the wetting front as a function of as a function of time. time.

• This is information that This is information that we could notwe could not get out get out of the Green and Ampt approach.of the Green and Ampt approach.

• This may be easily checked by putting the This may be easily checked by putting the solution into B&K and turning the crank. solution into B&K and turning the crank.

• This equation gives the exact solution for the This equation gives the exact solution for the shape of the wetting frontshape of the wetting front as a function of as a function of time. time.

• This is information that This is information that we could notwe could not get out get out of the Green and Ampt approach.of the Green and Ampt approach.

= (1-n)

2Do(n+1)

n2 1/2

Horizontal infiltration as a function of n for a Brutsaert Horizontal infiltration as a function of n for a Brutsaert soil with Do = 1, and n = 2, 5, and 10. The wetting soil with Do = 1, and n = 2, 5, and 10. The wetting front becomes increasingly sharp as n increases, front becomes increasingly sharp as n increases, making the pore size distribution narrower.making the pore size distribution narrower.

Horizontal infiltration as a function of n for a Brutsaert Horizontal infiltration as a function of n for a Brutsaert soil with Do = 1, and n = 2, 5, and 10. The wetting soil with Do = 1, and n = 2, 5, and 10. The wetting front becomes increasingly sharp as n increases, front becomes increasingly sharp as n increases, making the pore size distribution narrower.making the pore size distribution narrower.

15

HYDRUS-2D Simulations of Horiz. Infil.HYDRUS-2D Simulations of Horiz. Infil.Plotting moisture content Plotting moisture content vs position above and vs position above and moisture content vs x/tmoisture content vs x/t1/21/2 on the lower plot on the lower plot

They fit the Boltzman They fit the Boltzman transform!transform!

Also recall that in Miller Also recall that in Miller similarity time scales similarity time scales with the square root of with the square root of macroscopic length macroscopic length scale…scale…

Plotting moisture content Plotting moisture content vs position above and vs position above and moisture content vs x/tmoisture content vs x/t1/21/2 on the lower plot on the lower plot

They fit the Boltzman They fit the Boltzman transform!transform!

Also recall that in Miller Also recall that in Miller similarity time scales similarity time scales with the square root of with the square root of macroscopic length macroscopic length scale…scale…

Horizontal Infiltration

0

0.1

0.2

0.3

0.4

0.5

0 20 40 60 80 100 120

X/t^.5

Mo

istu

re C

on

ten

t (v

ol/v

ol)

t = 0.1 t = 0.2 t = 0.4 t = 1.6 t = 2.56 t = 6.55 t = 42.9 t = 200

Horizontal Infiltration

00.050.1

0.150.2

0.250.3

0.350.4

0.450.5

0 5 10 15

X/t^.5

Mo

istu

re C

on

ten

t (v

ol/v

ol)

t = 0.1

t = 0.2

t = 0.4

t = 1.6

t = 2.56

t = 6.55

t = 42.9

t = 200

16

Now for the sorptivity ...Now for the sorptivity ...

Can also solve for sorptivity (S). Using Brutsaert’s Can also solve for sorptivity (S). Using Brutsaert’s equation and the definition of Sequation and the definition of S

Let’s pull out the constant Let’s pull out the constant

Can also solve for sorptivity (S). Using Brutsaert’s Can also solve for sorptivity (S). Using Brutsaert’s equation and the definition of Sequation and the definition of S

Let’s pull out the constant Let’s pull out the constant

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Computing the Sorptivity ...Computing the Sorptivity ...

So we have the resultSo we have the result

Which is easy to integrate to obtainWhich is easy to integrate to obtain

Most often sorptivity is reported for Most often sorptivity is reported for saturated soil, saturated soil, = 1 = 1

So we have the resultSo we have the result

Which is easy to integrate to obtainWhich is easy to integrate to obtain

Most often sorptivity is reported for Most often sorptivity is reported for saturated soil, saturated soil, = 1 = 1

S() =C

0

 (1- 'n)d'

S() = C

-n+1

n+1

S(1) = n C

(n+1) =

2Do

n+1 1/2

18

Why bother (with S)?Why bother (with S)?

Suppose we want to calculate the infiltrationSuppose we want to calculate the infiltration

Integrating the moisture content over all positions at a given timeIntegrating the moisture content over all positions at a given time

We can evaluate the same integral by switching the bounds of We can evaluate the same integral by switching the bounds of integration so that we integrate all positions over the moisture integration so that we integrate all positions over the moisture contentcontent

or in termsor in termsof of

Suppose we want to calculate the infiltrationSuppose we want to calculate the infiltration

Integrating the moisture content over all positions at a given timeIntegrating the moisture content over all positions at a given time

We can evaluate the same integral by switching the bounds of We can evaluate the same integral by switching the bounds of integration so that we integrate all positions over the moisture integration so that we integrate all positions over the moisture contentcontent

or in termsor in termsof of

S(1) =

2Do

n+1 1/2

I =

0

   dx

x

I =

0

 x d I = t1/2

0

   d

19

Computing cumulative infiltrationComputing cumulative infiltration

Which is justWhich is just

which is exactly the form obtained by Green and which is exactly the form obtained by Green and Ampt! (i.e. square root of time)Ampt! (i.e. square root of time)

Can calculate the rate of infiltration Can calculate the rate of infiltration

Again, identical Green and Ampt! Again, identical Green and Ampt!

Which is justWhich is just

which is exactly the form obtained by Green and which is exactly the form obtained by Green and Ampt! (i.e. square root of time)Ampt! (i.e. square root of time)

Can calculate the rate of infiltration Can calculate the rate of infiltration

Again, identical Green and Ampt! Again, identical Green and Ampt!

I = t1/2

0

   d

q = dIdt =

12 S t-1/2

I = S t1/2

20

OK, but what is sorptivity? OK, but what is sorptivity?

A parameter which expresses the macroscopic balance A parameter which expresses the macroscopic balance between the capillary forces and the hydraulic conductivity.between the capillary forces and the hydraulic conductivity.

Recall From the discussion of the Green and Ampt results Recall From the discussion of the Green and Ampt results that that

KKsatsat goes up with goes up with 22 and and ff goes with 1/ goes with 1/ ( ( is the is the

characteristic microscopic length scale, for instance dcharacteristic microscopic length scale, for instance d5050), ),

then we can guess that sorptivity will get larger for coarser then we can guess that sorptivity will get larger for coarser soils, but only with soils, but only with 1/21/2

A parameter which expresses the macroscopic balance A parameter which expresses the macroscopic balance between the capillary forces and the hydraulic conductivity.between the capillary forces and the hydraulic conductivity.

Recall From the discussion of the Green and Ampt results Recall From the discussion of the Green and Ampt results that that

KKsatsat goes up with goes up with 22 and and ff goes with 1/ goes with 1/ ( ( is the is the

characteristic microscopic length scale, for instance dcharacteristic microscopic length scale, for instance d5050), ),

then we can guess that sorptivity will get larger for coarser then we can guess that sorptivity will get larger for coarser soils, but only with soils, but only with 1/21/2

SGA = 2  Ksat f  n

21

Miller Scaling Big TimeMiller Scaling Big Time

From the definition of S(From the definition of S() we know that) we know that

where S is the sorptivity, D is the diffusivity, K is the conductivity, where S is the sorptivity, D is the diffusivity, K is the conductivity, is is

the moisture content and the moisture content and is the Boltzman transform variable, is the Boltzman transform variable, = xt = xt-1/2-1/2

From the definition of S(From the definition of S() we know that) we know that

where S is the sorptivity, D is the diffusivity, K is the conductivity, where S is the sorptivity, D is the diffusivity, K is the conductivity, is is

the moisture content and the moisture content and is the Boltzman transform variable, is the Boltzman transform variable, = xt = xt-1/2-1/2

S() = 2 D() dd

= 2 K() dhd

=2 K() dh

d

x

t

22

Miller Scaling SMiller Scaling S

To derive the scaled value of sorptivity we To derive the scaled value of sorptivity we must replace the variables with the appropriate must replace the variables with the appropriate scaled quantitiesscaled quantities

(L is macro-(L is macro-

scopic scale)scopic scale)

We find:We find:

To derive the scaled value of sorptivity we To derive the scaled value of sorptivity we must replace the variables with the appropriate must replace the variables with the appropriate scaled quantitiesscaled quantities

(L is macro-(L is macro-

scopic scale)scopic scale)

We find:We find:

S() =2 K() dh

d

x

t

x• = xL t• =

tL

K• = K

h

h• = h

S =

K•

d

 h •

d

Lx •

L2t•

= 2K•

dh•

d• =

S•

23

Wrapping this up ...Wrapping this up ...

S scales with S scales with 1/21/2, (Just as we saw in the , (Just as we saw in the Green and Ampt Sorptivity). Green and Ampt Sorptivity).

As a little bonus, we see what the effect As a little bonus, we see what the effect of changing fluid properties would of changing fluid properties would produce in the value of sorptivityproduce in the value of sorptivity

S scales with S scales with 1/21/2, (Just as we saw in the , (Just as we saw in the Green and Ampt Sorptivity). Green and Ampt Sorptivity).

As a little bonus, we see what the effect As a little bonus, we see what the effect of changing fluid properties would of changing fluid properties would produce in the value of sorptivityproduce in the value of sorptivity

S = S•