# CE 473/573 Groundwater Fall 2010 Comments on homework 2 ```CE 473/573 Groundwater
Fall 2010
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.
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