# CE 473/573 Groundwater Fall 2011 Comments on homework 4

```CE 473/573 Groundwater
Fall 2011
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 &lt; x &lt; 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 .
```