CE 473/573 Groundwater
Fall 2012
Comments on homework 4
16. Cindy graded the problem, but I will treat it as optional because we did not discuss
the contours around a river in detail.
18. All groups calculated a range of travel times approximately equal to what I found.
The range was very large. Predicting contaminant transport at this site would require
further information to narrow the range to something manageable.
19. Only three groups derived the effective conductivity correctly. Write Q1 = K1 (h0 −
hm )A/L1 and Q2 = K2 (hm − hL )A/L2 . Then set Q1 = Q2 to find hm . Most groups
did that in part d (but not in part a). To get Kef f , use the result for hm in Q1 and
use Q = Kef f (h0 − hL )A/(L1 + L2 ). The result is
Kef f =
K1 K2 (L1 + L2 )
.
K 1 L2 + K 2 L1
This result, unlike those produced by some groups, shows that the low conductivity
soil controls the flow. For example, if K1 = 0, then Kef f = 0; that is, if water cannot
pass through soil 1, then the flow is zero.