KEY to EXERCISE:

EXPLORING APPLICATION OF THE SOLUTION TO A SYSTEM FOR WHICH IT WAS NOT ORIGINALLY INTENDED

To represent the half space, simply set h₂ equal to h₁ and use the results from the left hand side. The symmetry will represent a divide in the middle of the land mass which is the same as a no-flow boundary.


All of the conditions can be solved
1. K = 1x10^-4 m/day
2. K = 1x10^-1 m/day
3. ET = 0.85 m/yr
4. W = 0 m/yr

Diagrams are shown below for each case. Visit the applet to explore the response of the system and get specific values of flow and head for each case.

Evaluate and check your results as you proceed.

1. for K = 1x10^-4 m/day and all other parameters as in the base case except that h₂ equals h₁:









2. for K = 1x10^-1 m/day and all other parameters as in the base case except that h₂ equals h₁:









3. for ET 1.85 m/yr, that is W = -0.05 = -1.37 x 10^-4 m/day and all other parameters as in the base case except that h₂ equals h₁:







4. for W = 0 and all other parameters as in the base case except that h₂ equals h₁:







Do all of the results make physical sense to you?