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CE 473/573 Groundwater Fall 2010 Comments on homework 2 10. Most groups ﬁgured out how to compute the travel time. To get the range, you need the smallest and largest travel times, which correspond to the maximum and minimum average linear velocities, respectively. The maximum velocity is computed with the largest gradient and the smallest porosity. Similarly, the minimum velocity is computed with the smallest gradient and the largest porosity. 11. Most groups realized that conservation of mass requires Q1 = Q2 = Q and used Darcy’s law to write −K1 hm − h0 hL − hm bw = −K2 bw, L1 L2 where hm is the head at x = L1 . Solve for hm , eliminate hm , and solve for the eﬀective conductivity, recognizing that Q = −Kef f h0 − hL bw. L Several groups used Δh = 0 as a special case; that works, but it does not tell you if your expression for Kef f is correct. Better cases are (a) equal conductivities, (b) L1 = 0 (or L2 = 0), and (c) K1 = 0 (or K2 = 0). In the numerical example, the head drop is almost entirely in soil 1. Therefore, the low conductivity soil controls the eﬀective conductivity. 14. The point of this problem is that three values of head are not enough to specify the conductivities (or the ﬂow) with any certainty. Darcy’s law can give you the ratio of conductivities; if you pick one (in the silt range), you can compute the other—and check to see that it is in the silt range. Notice that a source of uncertainty is that you do not know where soil 1 ends and soil 2 begins. 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? 15. To solve for the ﬂow, write down Darcy’s law, integrate, and apply the boundary conditions: h(0) = h0 and h(L) = hL . The result is Q = Kbw0 α h0 − hL eαL − 1 . Then eliminate the ﬂow to compute an expression for the head: h = h0 − (h0 − hL ) eαx − 1 . eαL − 1 To show that the result for a constant-width aquifer is recovered when α → 0 requires taking a limit or using Taylor series on the last term. The latter gives (1 + αx + . . .) − 1 x eαx − 1 ≈ = , eαL − 1 (1 + αL + . . .) − 1 L so that h = h0 + hL − h0 x, L as expected. The magnitude of the gradient should increase with x since the decreasing ﬂow area requires more driving force to maintain the same ﬂow.