Page 87 - Water Engineering Hydraulics, Distribution and Treatment

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Drawdown constant = (114.6 Q)∕T = (114.6 × 700)∕(3.2 × 10 ) = 2.51 ft (0.765 m).
From Eq. (3.15), s = (114.6Q∕T)W(u).
a. Drawdown at face of pumping well:
s = 2.51 × 21.89 = 54.9ft (16.73 m).
1
Drawdown in central well:
s = 2.51 × 8.08 = 20.3ft (6.19 m).
2
Drawdown in the other outside well:
s = 2.51 × 6.69 = 16.8ft (5.12 m). 3.12 Multiple-Well Systems 65
3
b. Drawdown in outside wells:
s + s + s = 54.9 + 20.3 + 16.8 = 92 ft (28.04 m).
1
2
3
Drawdown in central well:
s + s + s = 54.9 + 20.3 + 20.3 = 95.5ft (29.11 m).
1
2
2
Solution 2 (SI System):
From Eq. (3.13),
1ft = 0.3048 m, 1,000 ft = 304.8 m, 2,000 ft = 609.6m.
2
u = r S∕(4Tt)
−5
2
u 0.3048 m = (0.3048) (3 × 10 )∕(4 × 397.4 × 10) = 1.75 × 10 −10
W(u) = 21.89
2
−5
u 304.8m = (304.8) (3 × 10 )∕(4 × 397.4 × 10) = 1.75 × 10 −4
W(u) = 8.08
2
−5
u = (609.6) (3 × 10 )∕(4 × 397.4 × 10) = 7.01 × 10 −4
609.6m
W(u) = 6.69.
From Eq. (3.12),
S = (Q∕4 T)W(u)
= (3,815∕4 × 3.14 × 397.4)W(u)
= 0.765W(u).
a. Drawdown at face of pumping well:
s = 0.765 × 21.89 = 16.73 m.
1
Drawdown in central well:
s = 0.765 × 8.08 = 6.18 m.
2
Drawdown in other outside well:
s = 0.765 × 6.69 = 5.12 m.
3
b. Drawdown in outside wells:
s + s + s = 16.73 + 6.18 + 5.12 = 28.03 m.
3
1
2
Drawdown in central well:
s + s + s = 16.73 + 6.18 + 6.18 = 29.09 m.
2
2
1