Page 167 - Water and wastewater engineering

4-22 WATER AND WASTEWATER ENGINEERING

From Figure 4-9 , we find that the extrapolation of the straight portion of the graph to a drawdown
of 0.0 m yields t 0 0.30 min. Using the distance between the pumping well and the observation
( r 68.58 m), we find
2
.

.
(225 )(4 3110 3 m /s )(0 30 min )(60 s/min) )
.
S 2
(68 58 m )
.
3 7 . 10 − 5
Now we should check to see if our implicit assumption that u is less than 0.01 was true. We use
t 100.0 min for the check.
2
.
.
(68 58 m ) (3 7 10 5 )
u − 3 2
.
( 44 31 10 m /s )(100..0 min)( 60 s/min)
. 168 10 3
Thus, the assumption is acceptable. In this example, the pumping test curve plots as a straight line
on semilog paper. In most instances a more complex curve results. Analysis of the more complex
case is left for more advanced texts.
Calculating Interference. As was mentioned earlier, the cones of depression of wells located
close together will overlap. This interference will reduce the potential yield of the wells. In severe
circumstances, well interference can cause drawdowns that lower the piezometric surface below
the bottom of the aquiclude and, thus, cause structural failure of the well and surrounding struc-
tures due to settlement of the ground surface.
The method of superposition is used to estimate the total drawdown due to interference. This
method assumes that the drawdown at a particular location is equal to the sum of the drawdowns
from all of the influencing wells. Mathematically this can be represented as follows:
n
s resultant ∑ s (4-7)
i
i =1
where s i individual drawdown caused by well i at location r.
This method is illustrated in the next example for the case of three identical wells located
symmetrically in a line.

Example 4-4. Three wells are located at 75-m intervals along a straight line. Each well is 0.50
3
2
m in diameter. The coefficient of transmissivity is 2.63 10 m /s, and the storage coefficient
4
2
3
is 2.74 10 . Determine the drawdown at each well if each well is pumped at 4.42 10 m /s
for 10 days.
Solution. The drawdown at each well will be the sum of the drawdown of each well pumping
by itself plus the interference from each of the other two wells. Because each well is the same
diameter and pumps at the same rate, we may compute one value of the term Q /(4 T ) and apply
it to each well.
3
Q 442 10 2 m/s
.
1134. m
2
(
T 4 314 2 63 10 3 m/s)
.
.
)(
4p