CE 473/573 Groundwater Fall 2011 Comments on homework 5

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CE 473/573 Groundwater
Fall 2011
Comments on homework 5
29. No problems.
31. All groups computed the flow correctly. To compute the area affected downgradient
of the well, I plotted the curve and used the trapezoid rule to integrate under it. Two
other groups fit a curve and integrated that.
32. Most groups computed the drawdown correctly, but few explained 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.
33. 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
ln
,
h =H −
πK i=1
ri
2
2
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 safe approach is to compute the drawdown at all points in the construction area.
34. Most groups solved this problem without trouble. Remember that when rw /λ is large,
the product (rw /λ)K1 (rw /λ) in the denominator of the drawdown formula approaches
a value of 1.
36. All groups compute both sets of storativity and transmissivity, but a few forgot to
compare the values. The Jacob approximation gives a transmissivity that is 20% low
and a storativity that is low by a factor of 2. The Jacob approximation works best
for large time, or small u. In particular, it is quite accurate for u < 0.01. A few of the
points have larger values of u.
37. Two out of seven groups computed the correct answers. Some groups read the graphs
inaccurately, while others used the wrong values of W (uA , uB , Γ).
38. Most groups had the correct answers, but I wanted physical reasons too. When Γ
increases, more water comes from vertical flow (if Kh , r, and b are constant), and less
drawdown is needed to achieve the required flow. If Γ = 0, then for constant Kh , r,
and b, Kv = 0; the flow is then entirely horizontal, as in a confined aquifer.
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