Page 86 - Water Engineering Hydraulics, Distribution and Treatment

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Chapter 3
Water Sources: Groundwater
Similarly, for the second period:
′
log(t∕t ) =log(1,200∕1,080) = 0.046.
For the third period:
log(t∕t ) =log(960∕780) = 0.090.
3
Total residual drawdown:
′
Solution 2 (SI System): s = (264Q∕T) [ ∑ 2 ′ log t∕t ′ ] = 2.89 × 0.174 = 0.5ft(0.15 m).
The problem can be decomposed into three pumping and recovery periods, with Eq. (3.26b) applied to each of the subproblems:
′
′
s = (0.1833Q∕T) log(t∕t )
′
= (0.1833 × 1907.5∕397.38) log(t∕t )
′
= 0.8798 log(t∕t ).
For the first period of pumping:
t = time since pumping started = 1,440 min.
′
t = time since pumping stopped = 1,320 min.
′
log(t∕t ) =log(1,440∕1,320) = 0.038.
1
For the second period of pumping:
′
log(t∕t ) =log(1,200∕1,080) = 0.046.
2
For the third period of pumping:
′
log(t∕t ) =log(960∕780) = 0.090.
3
Total residual drawdown:
]
′
[ ∑
s = (0.1833Q∕T) log t∕t ′
= 0.8798(0.038 + 0.046 + 0.090)
= 0.15 m.

EXAMPLE 3.6 DRAWDOWN IN THREE WELLS
4
Three 24 in. (610 mm) wells are located on a straight line 1,000 ft (304.8 m) apart in an artesian aquifer with T = 3.2 × 10 gpd∕ft
3
−5
(397.4 m /d/m) and S = 3 × 10 . Compute the drawdown at each well when (a) one of the outside wells is pumped at a rate of
3
3
700 gpm (3,815 m /d) for 10 days and (b) the three wells are pumped at 700 gpm (3,815 m /d) for 10 days.
Solution 1 (US Customary System):
From Eq. (3.16),
2
u = 1.87r S∕Tt
4
−5
2
u 1ft = (1.87 × 1 × 3.0 × 10 )∕(3.2 × 10 × 10) = 1.75 × 10 −10 , W(u) = 21.89
−4
6
u 1,000 ft = 10 u = 1.75 × 10 , W(u) = 8.08
1
−4
6
u 2,000 ft = 4 × 10 u = 7 × 10 , W(u) = 6.69.
1