CE 473/573 Groundwater
Fall 2012
Comments on homeworks 6-9
39. The solution I gave you uses the result for a homogeneous unconfined aquifer instead
of a layered unconfined aquifer. The result for the latter gives x as a function of h,
which poses no problem when one wants a plot of the head profile.
40. The key to part b is to realize that the total flow is simply the recharge rate multiplied
by the horizontal area of the aquifer. For part c, if there is a groundwater divide, the
flow at x = 0 is negative. Therefore, the limiting case for no groundwater divide is
that the flow at x = 0 is zero. Use that information to find the recharge rate.
42. By applying conservation of mass and using the flow between two points in an unconfined aquifer, you should find
h2j =
Kj−1/2 h2j−1 + Kj+1/2 h2j+1
2ij (Δx)2
+
,
Kj−1/2 + Kj+1/2
Kj−1/2 + Kj+1/2
where ij is the recharge rate for the jth cell and Kj−1/2 = 2Kj−1 Kj /(Kj + Kj−1 ).
Apply this formula in the interior of the domain. You will find a groundwater divide,
which means that some water will flow to the drain on the left. To compute the
travel time, realize that the average linear velocity will vary because the head does
not decrease linearly. Therefore, compute the travel time between two cells as
Tj =
where
Kj−1/2
v =
n
Δx
,
v
hj−1 − hj
Δx
,
and add the times over the specified range (in this case, 13.5 m < x < 20 m). Some
groups did not account for the change in conductivity at the edge of the field. Your
plot of the water table should show a change in slope at the edge of the field because
the flow from the field should equal the flow to the river, but the conductivities are
different.
44. My Excel simulations shows that flow into the model includes flow through the west
boundary of 788 m3 /d and recharge of 480 m3 /d. Flow out includes pumping by the
well of 30 m3 /d and flow through the east boundary of 1242 m3 /d. The simulation
conserves mass to better than 1%. Some groups used a large cell size that yielded
inaccurate results; the usual guideline is to reduce the cell size till the answers change
by less than a specified amount (e.g., 1%, 5%, etc.) Other groups had errors in the
formulation, but they were difficult to find. Be sure to write out the equations you
use.
45. This problem is another example of determining parameters by fitting a theoretical
curve (here, a logarithmic function) to measured data.
47. The key to a letter to the editor is to be concise—e.g., no more than 150 words.
Therefore, there’s not much room for citing sources, developing arguments, citing
counterarguments in detail, etc.
48. No major problems.
50. Remember that the total maximum width is 2ymax . Also, as we discussed in class, the
well must be sufficiently far from the source for the maximum width to be achieved.
To check, use the equation for the boundary of the capture zone:
y
.
x=−
tan(y/x0 )
51. Most groups computed the drawdown correctly, but a few groups did not explain
correctly why the drawdown is larger in the case with the no flow boundary. The
no-flow boundary reduces the area from which the well can draw water; therefore,
more drawdown is needed to get the required flow.
52. I used six wells pumping at 450 m3 /d on each side and checked the point on the
centerline halfway up the short side since that point is farthest from the wells. I
computed the head with
N
Q0 R
2
2
,
ln
h =H −
πK i=1
ri
where N is the number of wells and ri is the distance from the ith well to the point in
question. For configurations in which the point of minimum drawdown is unclear, the
safest approach is to compute the drawdown at all points in the construction area.
53. The equation for drawdown
s=
K0 (r/λ)
Q0
2πT (rw /λ)K1 (rw /λ)
can be used to write
K0 (r1 /λ)
s1
= 0.
−
s2
K0 (r2 /λ)
Once λ is determined from the latter equation, it can be used in the former to find T .
Then K can be determine from the definition of λ.
54. Remember that drawdown is proportional to flowrate; therefore, if the flowrate increases by a factor of 1.5, so does the drawdown.
55. Some groups confused H, H0 , and h. H0 is the initial head, H is the head after
the rain event, and h is the head at any time. Here you are solving for the time for
h = 30.45 m.
56. No problems.