Page 126 - Fair, Geyer, and Okun's Water and wastewater engineering : water supply and wastewater removal

JWCL344_ch03_061-117.qxd 8/17/10 7:48 PM Page 88

88 Chapter 3 Water Sources: Groundwater

EXAMPLE 3.6 DRAWDOWN IN THREE WELLS
Three 24-in. (610-mm) wells are located on a straight line 1,000 ft (304.8 m) apart in an artesian
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3
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aquifer with T 3.2 10 gpd/ft (397.4 m /d/m) and S 3 l0 . Compute the drawdown at
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each well when (a) one of the outside wells is pumped at a rate of 700 gpm (3,815 m /d) for 10 days
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and (b) the three wells are pumped at 700 gpm (3,815 m /d) for 10 days.
Solution 1 (U.S. Customary System):
From Eq. 3.16
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u 1.87 r S>Tt
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4
u 1ft (1.87 1 3.0 l0 )>(3.2 l0 10) 1.75 l0 10 , W(u) 21.89
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4
u 1,000 ft 10 u 1 1.75 l0 , W(u) 8.08
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u 2,000 ft 4 l0 u 1 7 l0 , W(u) 6.69
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Drawdown constant (114.6 Q)>T (114.6 700)>(3.2 l0 ) 2.51 ft (0.765 m)
From Eq. 3.15 s (114.6 Q>T) W(u)
(a) Drawdown at face of pumping well:
s 1 2.51 21.89 54.9 ft (16.73 m)
Drawdown in central well:
s 2 2.51 8.08 20.3 ft (6.19 m)
Drawdown in the other outside well:
s 3 2.51 6.69 16.8 ft (5.12 m)
(b) Drawdown in outside wells:
s 1 s 2 s 3 54.9 20.3 16.8 92 ft (28.04 m)
Drawdown in central well:

s 1 s 2 s 3 54.9 20.3 20.3 95.5 ft (29.11 m)

Solution 2 (SI System):
From Eq. 3.13
1 ft 0.3048 m, 1,000 ft 304.8 m, 2,000 ft 609.6 m
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u = r S>(4Tt)
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u 0.3048 m = (0.3048) (3 * 10 )>(4 * 397.4 * 10) = 1.75 * 10 -10
W(u) = 21.89
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2
u 304.8 m = (304.8) (3 * 10 )>(4 * 397.4 * 10) = 1.75 * 10 -4
W(u) = 8.08
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2
u 609.6 m = (609.6) (3 * 10 )>(4 * 397.4 * 10) = 7.01 * 10 -4
W(u) = 6.69
From Eq. 3.12
S = (Q>4pT) W(u)
= (3,815>4 * 3.14 * 397.4) W(u)
= 0.765 W(u)