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.