Steady Evaporation from a Water Table

1. Steady Evaporation from a Water Table Following Gardner Soil Sci., 85:228-232, 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 /t = 0, and Richards equation becomes a differential equation in z alone (e.g., straight-backed d’s)

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, doesn’t 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 conductivities 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/)Ln[exp( h)], so adding and subtracting h (yes, the a is an alpha…) • 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 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., 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.

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.