CE 473/573 Groundwater
Fall 2010
Comments on homeworks 4 and 5
24. As you have seen on problem 8 and exam 1, you can determine parameters such as
the radius of influence and transmissivity by fitting the theoretical equation to the
data and calculating the parameters from the coefficients. In this case, the steady
drawdown caused by the well in the confined aquifer is
Qw
ln
s=
2πT
or
s=−
R
r
,
(1)
Qw
Qw
ln r +
ln R.
2πT
2πT
(2)
In Excel, fiting a logarithmic curve to equation (2) will give you the coefficient of ln r
as well as the intercept. Then calculate T from the slope and R from the intercept.
25. 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).
27. In the PMWIN simulation I find a flow of 102.5 m3 /d in the left boundary and a flow
of 72.5 m3 /d out the right boundary. Excel gives 104 m3 /d in the left boundary and
71 m3 /d out the right boundary (an error of about 3%). My plots are below
400
350
300
y (m)
250
200
150
100
50
0
0
100
200
300
x (m)
400
500
600
100
200
300
x (m)
400
500
600
400
350
300
y (m)
250
200
150
100
50
0
0
28. Most groups computed the correct hydraulic conductivity. Some groups used the
result I stated in class to get the formula in the problem statement. No group started
from Darcy’s law.
29. No major problems.
31. Remember that ymax is half the maximum width. I computed the downgradient area
by plotting the curve x = y/ tan(y/x0 ) and integrating with the trapezoid rule. I find
an area of 233 m2 .
32. 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
R
Q0 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.
33. No major problems.
35. A few isolated problems. Remember that storativity should probably be less than
5 × 10−3 . Much higher values probably indicate an error. Be sure you can compute drawdown in unsteady flow to a well and apply the Jacob approximation. The
spreadsheet does it automatically.
36. Most groups computed the curve properly. I get 0.105 m for part a and 1.8 hours for
part b.
37. As Γ increases, the vertical flow becomes more important. Because vertical flow
can supply more of the water needed for the well, the drawdown will be less. This
behavior is illustrated in the plot of the unconfined well function since W decreases
as Γ increases. For part b, when the vertical flow is zero (i.e., Kv = 0 or Γ = 0),
the aquifer will behave as a confined aquifer because the only source of water is
the compressibility of the water and the soil matrix. As Γ goes to zero, the curve
approaches the solution for a confined aquifer.