CE 473/573 Groundwater
Fall 2011
Comments on homework 4
24. 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.
26. 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 (or eastern) boundary. I suspect that a few groups did not
specify the boundary conditions (or boundary types) correctly. 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
27. Most groups determined 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
R
Qw
ln
s=
,
(1)
2πT
r
or
Qw
Qw
ln r +
ln R.
(2)
2πT
2πT
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.
s=−
28. Most groups got this problem. Just remember that s2 = H 2 − h2 .