# CE 473/573 Groundwater Fall 2012 Comments on homework 4

```CE 473/573 Groundwater
Fall 2012
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.
```