Page 165 - Water and wastewater engineering

4-20 WATER AND WASTEWATER ENGINEERING

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4
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Example 4-2. If the storage coefficient is 2.74 10 and the transmissivity is 2.63 10 m /s,
calculate the drawdown that will result at the end of 100 days of pumping a 0.61 m diameter well
2
3
at a rate of 2.21 10 m /s.
Solution. Begin by computing u. The radius is 0.305 m and
2
(0 305 m ) (274 10 4 ) 10
.
.
u 280 10
.
2
.
0
( 42 63 10 3 m /s )(100 d)( 86 400 s/d)
,

From the series expansion, find
2 ( ..80 10 10 2
)
10 10 . . .
.
.
W(u) 0 577216 ln( 2 80 10 ) 2 80 10 21 .42
.

22!
and compute s as
2 3
221 10 m /s
.
s ( 21.442)
3
)(
.
.
(
4 3 14 2 63 10 3 m /s)
.
14 32 or 14 m
Comment. In this example, the transmissivity and storage coefficient are given. Obviously,
they need to be determined to use Equation 4-2. The method of estimating T and S are given in
the next section.
Determining the Hydraulic Properties of a Confined Aquifer. The estimation of the trans-
missivity and storage coefficient of an aquifer is based on the results of a pumping test. The
preferred situation is one in which one or more observation wells located at a distance from
the pumping well are used to gather the data. In the transient state, we cannot solve for T and S
directly. The Cooper and Jacob method has been selected from the several indirect methods that
are available (Cooper and Jacob, 1946). For values of u less than 0.01, they found that Equa-
tion 4-1 could be rewritten as:
.
Q 225 Tt
s ln (4-3)
2
T rS
4p
A semilogarithmic plot of s versus t (log scale) from the results of a pumping test ( Figure 4-9 )
enables a direct calculation of T from the slope of the line. From Equation 4-3, the difference in
drawdown at two points in time may be shown to be
Q t 2
s 2 s 1 ln
4p T t 1
Solving for T , we find
Q t 2
T ln (4-5)
4p( s 1 ) t 1
s
2