CE 473/573 Groundwater
Fall 2009
Comments on homework 3
14. Be sure to explain your special cases: If K1 = K2 = K0 (say), then Kef f should equal
K. Also, if one of the conductivities is zero, then the flow will be zero; that means the
effective conductivity will also be zero, as seen by writing the effective conductivity
as
L
K1 K2 L
Kef f = L1
=
.
L2
K
L
+
K
L
+
1
2
2
1
K1
K2
15. The point of this problem is that three values of head are not enough to specify the
conductivities (or the flow) with any certainty. All groups divided the domain in half,
but when I computed the heads, one-quarter had low conductivity soil and the rest had
high-conductivity soil. If the soil is in fact divided into two sections of unknown length,
how many values of head would you need to specify the conductivities uniquely?
16. Table 5-1 of the report shows that the conductivities are not uniform (i.e., inhomogeneous) and the vertical and horizontal conductivities are different (i.e., anisotropic). I
computed a flow of about 4000 ft3 /d. Remember to use the distance from the bottom
of the aquifer to the water table—not the water table elevation—in Darcy’s law.
17. A few groups said that doubling K or ∆h would double the total volume out even
though the volume did not double in the simulation. Which two principles should you
have used to examine this problem? Also, be sure to explain your answers fully.
18. No problems among those who solved this optional exercise.
19. No problems.