1

Steady Steady Evaporation from Evaporation from

a Water Tablea Water Table

Following GardnerFollowing GardnerSoil Sci., 85:228-232, 1958Soil Sci., 85:228-232, 1958

Following GardnerFollowing GardnerSoil Sci., 85:228-232, 1958Soil Sci., 85:228-232, 1958

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

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

2

Why pick on this solution?Why pick on this solution?

Of interest for several reasons: Of interest for several reasons: •• it is instructive in how to solve simple it is instructive in how to solve simple

unsaturated flow problems; unsaturated flow problems;

•• it provides very handy, informative results;it provides very handy, informative results;

•• introduces a widely used conductivity introduces a widely used conductivity function.function.

Of interest for several reasons: Of interest for several reasons: •• it is instructive in how to solve simple it is instructive in how to solve simple

unsaturated flow problems; unsaturated flow problems;

•• it provides very handy, informative results;it provides very handy, informative results;

•• introduces a widely used conductivity introduces a widely used conductivity function.function.

3

The set-upThe set-up

The problem we will consider is that of evaporation from a broad The problem we will consider is that of evaporation from a broad land surface with a water table near by. land surface with a water table near by. Assume:Assume:• • The soil is uniform, The soil is uniform, • • The process is one-The process is one- dimensional (vertical). dimensional (vertical). • • The system is at The system is at steady state steady state

Notice: that since the system is at steady state, Notice: that since the system is at steady state, the flux must be constant with elevation, i.e. q(z) = q the flux must be constant with elevation, i.e. q(z) = q (conservation of mass).(conservation of mass).

The problem we will consider is that of evaporation from a broad The problem we will consider is that of evaporation from a broad land surface with a water table near by. land surface with a water table near by. Assume:Assume:• • The soil is uniform, The soil is uniform, • • The process is one-The process is one- dimensional (vertical). dimensional (vertical). • • The system is at The system is at steady state steady state

Notice: that since the system is at steady state, Notice: that since the system is at steady state, the flux must be constant with elevation, i.e. q(z) = q the flux must be constant with elevation, i.e. q(z) = q (conservation of mass).(conservation of mass).

4

Getting down to business ..Getting down to business ..

Richards equation is the governing equationRichards equation is the governing equation

At steady state the moisture content is constant At steady state the moisture content is constant in time, thus in time, thus / / t = 0, and Richards equation t = 0, and Richards equation becomes a differential equation in z alonebecomes a differential equation in z alone

Richards equation is the governing equationRichards equation is the governing equation

At steady state the moisture content is constant At steady state the moisture content is constant in time, thus in time, thus / / t = 0, and Richards equation t = 0, and Richards equation becomes a differential equation in z alonebecomes a differential equation in z alone

t =

z

K 

h

z   + Kz

- dKdz =

ddz

K 

dh

dz  

5

Simplifying further ...Simplifying further ...

Since both sides are first derivatives in z, this may Since both sides are first derivatives in z, this may be integrated to recover the Buckingham-Darcy Law be integrated to recover the Buckingham-Darcy Law for unsaturated flowfor unsaturated flow

oror

where the constant of integration q is the vertical flux where the constant of integration q is the vertical flux through the system. Notice that q can be either through the system. Notice that q can be either positive or negative corresponding to evaporation or positive or negative corresponding to evaporation or infiltration.infiltration.

Since both sides are first derivatives in z, this may Since both sides are first derivatives in z, this may be integrated to recover the Buckingham-Darcy Law be integrated to recover the Buckingham-Darcy Law for unsaturated flowfor unsaturated flow

oror

where the constant of integration q is the vertical flux where the constant of integration q is the vertical flux through the system. Notice that q can be either through the system. Notice that q can be either positive or negative corresponding to evaporation or positive or negative corresponding to evaporation or infiltration.infiltration.

- dKdz =

ddz

K 

dh

dz  

q = - K

dh

dz  + 1

6

Solving for pressure vs. elevationSolving for pressure vs. elevation

We would like to solve for the pressure as a function of We would like to solve for the pressure as a function of elevation. Solving for dz we find:elevation. Solving for dz we find:

which may be integrated to obtainwhich may be integrated to obtain••h' is the dummy variable of integration;h' is the dummy variable of integration;

• • h(z), or h is the pressure ath(z), or h is the pressure at the elevation z; the elevation z;

••lower bound of this integral is taken lower bound of this integral is taken at the water table where h(0) = 0. at the water table where h(0) = 0.

We would like to solve for the pressure as a function of We would like to solve for the pressure as a function of elevation. Solving for dz we find:elevation. Solving for dz we find:

which may be integrated to obtainwhich may be integrated to obtain••h' is the dummy variable of integration;h' is the dummy variable of integration;

• • h(z), or h is the pressure ath(z), or h is the pressure at the elevation z; the elevation z;

••lower bound of this integral is taken lower bound of this integral is taken at the water table where h(0) = 0. at the water table where h(0) = 0.

q = - K

dh

dz  + 1

dz = - dh

qK + 1

7

What next? Functional forms!What next? Functional forms!

••To solve need a relationship between conductivity and To solve need a relationship between conductivity and pressure. pressure.

••Gardner introduced several conductivity functions which Gardner introduced several conductivity functions which can be used to solve this equation, including the can be used to solve this equation, including the exponential relationshipexponential relationship

Simple, non-hysteretic, doesn’t deal with hSimple, non-hysteretic, doesn’t deal with haeae,,

is only accurate over small pressure rangesis only accurate over small pressure ranges

••To solve need a relationship between conductivity and To solve need a relationship between conductivity and pressure. pressure.

••Gardner introduced several conductivity functions which Gardner introduced several conductivity functions which can be used to solve this equation, including the can be used to solve this equation, including the exponential relationshipexponential relationship

Simple, non-hysteretic, doesn’t deal with hSimple, non-hysteretic, doesn’t deal with haeae,,

is only accurate over small pressure rangesis only accurate over small pressure ranges

z =

0

h(z)- dh'qK + 1

K(h) = K sexp(h)

8dx = K sexp(h) dh dx = x dh

Now just plug and chugNow just plug and chug

which may be re-arranged aswhich may be re-arranged as

To solve this we change variables and letTo solve this we change variables and let

oror

which may be re-arranged aswhich may be re-arranged as

To solve this we change variables and letTo solve this we change variables and let

oror

z =

0

h- dh'q

K sexp(h') + 1

z =

0

h(z)- dh'qK + 1

K(h) = K sexp(h)

z =

0

h- K sexp(h') dh' q + K sexp(h')

9

Moving right along ...Moving right along ...

Our integral becomesOur integral becomes

which may be integrated to obtainwhich may be integrated to obtain

Our integral becomesOur integral becomes

which may be integrated to obtainwhich may be integrated to obtain

z = -

0

hx dh'

 q + x = - 1

x(0)

x(h)dx

 q + x

dx = x dh

z = - 1 Ln

q + K sexp(h)

 q + K s

10

All Right!All Right!

•• Solution for pressure vs elevation for steady Solution for pressure vs elevation for steady evaporation (or infiltration)from the water table evaporation (or infiltration)from the water table for a soil with exponential conductivity. for a soil with exponential conductivity.

•• Gardner (1958) notes that the problem may also Gardner (1958) notes that the problem may also be solved in closed form for conductivity’s of the be solved in closed form for conductivity’s of the form: K = a/(hform: K = a/(hnn +b) for n = 1, 3/2, 2, 3, and 4 +b) for n = 1, 3/2, 2, 3, and 4

•• Solution for pressure vs elevation for steady Solution for pressure vs elevation for steady evaporation (or infiltration)from the water table evaporation (or infiltration)from the water table for a soil with exponential conductivity. for a soil with exponential conductivity.

•• Gardner (1958) notes that the problem may also Gardner (1958) notes that the problem may also be solved in closed form for conductivity’s of the be solved in closed form for conductivity’s of the form: K = a/(hform: K = a/(hnn +b) for n = 1, 3/2, 2, 3, and 4 +b) for n = 1, 3/2, 2, 3, and 4

z = - 1 Ln

q + K sexp(h)

 q + K s

11

Rearranging makes it more intuitiveRearranging makes it more intuitive

We can put this into a more easily understood We can put this into a more easily understood form through some simple manipulations. form through some simple manipulations. Note that we may write: h = (1/Note that we may write: h = (1/)Ln[exp()Ln[exp( h)], h)], so adding and subtracting hso adding and subtracting h

gives us a useful formgives us a useful form

We can put this into a more easily understood We can put this into a more easily understood form through some simple manipulations. form through some simple manipulations. Note that we may write: h = (1/Note that we may write: h = (1/)Ln[exp()Ln[exp( h)], h)], so adding and subtracting hso adding and subtracting h

gives us a useful formgives us a useful form

z = - h + 1 Ln[exp(ah)] -

1 Ln

q + K sexp(h)

 q + K s

z = - h - 1 Ln

q exp(- h) + K s

 q + K s

12

We now see ...We now see ...

•• Contributions of pressure and flux separately.Contributions of pressure and flux separately.

•• As the flux increases, the argument of Ln[ ] gets larger, indicating that As the flux increases, the argument of Ln[ ] gets larger, indicating that at a given elevation, the pressure potential becomes more negative at a given elevation, the pressure potential becomes more negative (i.e., the soil gets drier), as expected for increasing evaporative flux. (i.e., the soil gets drier), as expected for increasing evaporative flux.

•• If q=0, the second term on the right hand side goes to zero, and the If q=0, the second term on the right hand side goes to zero, and the pressure is simply the elevation above the water table (i.e., pressure is simply the elevation above the water table (i.e., hydrostatic, as expected). hydrostatic, as expected).

•• Contributions of pressure and flux separately.Contributions of pressure and flux separately.

•• As the flux increases, the argument of Ln[ ] gets larger, indicating that As the flux increases, the argument of Ln[ ] gets larger, indicating that at a given elevation, the pressure potential becomes more negative at a given elevation, the pressure potential becomes more negative (i.e., the soil gets drier), as expected for increasing evaporative flux. (i.e., the soil gets drier), as expected for increasing evaporative flux.

•• If q=0, the second term on the right hand side goes to zero, and the If q=0, the second term on the right hand side goes to zero, and the pressure is simply the elevation above the water table (i.e., pressure is simply the elevation above the water table (i.e., hydrostatic, as expected). hydrostatic, as expected).

z = - h - 1 Ln

q exp(- h) + K s

 q + K s

13

Another useful formAnother useful form

May also solve for pressure profileMay also solve for pressure profile

• • Although primarily interested in upward flux, Although primarily interested in upward flux, note that if the flux is -Knote that if the flux is -Kss that the pressure is that the pressure is

zero everywhere, which is as we would expect zero everywhere, which is as we would expect for steady infiltration at Kfor steady infiltration at Kss..

May also solve for pressure profileMay also solve for pressure profile

• • Although primarily interested in upward flux, Although primarily interested in upward flux, note that if the flux is -Knote that if the flux is -Kss that the pressure is that the pressure is

zero everywhere, which is as we would expect zero everywhere, which is as we would expect for steady infiltration at Kfor steady infiltration at Kss..

h = 1 Ln

exp(- z) 

q

K s + 1  - 

qK s

15

The Maximum Evaporative FluxThe Maximum Evaporative Flux

At the maximum flux, the pressure at the soil At the maximum flux, the pressure at the soil surface is -infinity, so the argument of the surface is -infinity, so the argument of the logarithm must go to zero. This implieslogarithm must go to zero. This implies

solving for qsolving for qmaxmax, for a water table at depth z, for a water table at depth z

At the maximum flux, the pressure at the soil At the maximum flux, the pressure at the soil surface is -infinity, so the argument of the surface is -infinity, so the argument of the logarithm must go to zero. This implieslogarithm must go to zero. This implies

solving for qsolving for qmaxmax, for a water table at depth z, for a water table at depth z

qmax

K s= exp(-z)

qmax

K s + 1

qmax = K s 

exp(z)-1

16

So what does this tell us?So what does this tell us?

Considering successive depths of Considering successive depths of

z = 1/z = 1/, 2/, 2/ , 3/ , 3/

we find thatwe find that

qqmaxmax(z)/K(z)/Kss = 0.58, 0.16, and 0.05, = 0.58, 0.16, and 0.05,

very rapid decrease in evaporative flux as very rapid decrease in evaporative flux as the depth to the water table increases.the depth to the water table increases.

Considering successive depths of Considering successive depths of

z = 1/z = 1/, 2/, 2/ , 3/ , 3/

we find thatwe find that

qqmaxmax(z)/K(z)/Kss = 0.58, 0.16, and 0.05, = 0.58, 0.16, and 0.05,

very rapid decrease in evaporative flux as very rapid decrease in evaporative flux as the depth to the water table increases.the depth to the water table increases.

qmax = K s 

exp(z)-1