Steady Evaporation from a Water Table

1
Steady Evaporation from a Water Table

• Following Gardner
• Soil Sci., 85228-232, 1958

2
Why pick on this solution?

• Of interest for several reasons
• it is instructive in how to solve simple
  unsaturated flow problems
• it provides very handy, informative results
• introduced widely used conductivity function.

3
The set-up

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

4
Getting down to business ..

• Richards equation is the governing equation
• At steady state the moisture content is constant
  in time, thus d?/ dt 0, and Richards equation
  becomes a differential equation in z alone

5
Simplifying further ...

• Since both sides are first derivatives in z, this
  may be integrated to recover the Buckingham-Darcy
  Law for unsaturated flow
• or
• where the constant of integration q is the
  vertical flux through the system. Notice that q
  can be either positive or negative corresponding
  to evaporation or infiltration.

6
Solving for pressure vs. elevation

• We would like to solve for the pressure as a
  function of elevation. Solving for dz we find
• which may be integrated to obtain
• h' is the dummy variable of integration
• h(z), or h is the pressure at the elevation
  z
• lower bound of this integral is taken at the
  water table where h(0) 0.

7
What next? Functional forms!

• To solve need a relationship between
  conductivity and pressure.
• Gardner introduced several conductivity
  functions which can be used to solve this
  equation, including the exponential relationship
• Simple, non-hysteretic, doesnt deal with hae,
• is only accurate over small pressure ranges

8
Now just plug and chug

• which may be re-arranged as
• To solve this we change variables and let
• or

9
Moving right along ...

• Our integral becomes
• which may be integrated to obtain

10
All Right!

• Solution for pressure vs elevation for steady
  evaporation (or infiltration)from the water table
  for a soil with exponential conductivity.
• Gardner (1958) notes that the problem may also
  be solved in closed form for conductivitys of
  the form K a/(hn b) for n 1, 3/2, 2, 3, and
  4

11
Rearranging makes it more intuitive

• We can put this into a more easily understood
  form through some simple manipulations. Note
  that we may write h (1/?)Lnexp(? h), so
  adding and subtracting h
• gives us a useful form

12
We now see ...

• Contributions of pressure and flux separately.
• As the flux increases, the argument of Ln
  gets larger, indicating that at a given
  elevation, the pressure potential becomes more
  negative (i.e., the soil gets drier), as expected
  for increasing evaporative flux.
• If q0, the second term on the right hand side
  goes to zero, and the pressure is simply the
  elevation above the water table (i.e.,
  hydrostatic, as expected).

13
Another useful form

• May also solve for pressure profile
• Although primarily interested in upward flux,
  note that if the flux is -Ks that the pressure is
  zero everywhere, which is as we would expect for
  steady infiltration at Ks.

14
(No Transcript)
15
The Maximum Evaporative Flux

• At the maximum flux, the pressure at the soil
  surface is -infinity, so the argument of the
  logarithm must go to zero. This implies
• solving for qmax, for a water table at depth z

16
So what does this tell us?

• Considering successive depths of
• z 1/?, 2/? , 3/?
• we find that
• qmax(z)/Ks 0.58, 0.16, and 0.05,
• very rapid decrease in evaporative flux as the
  depth to the water table increases.

17
(No Transcript)