Page 88 - Water Engineering Hydraulics, Distribution and Treatment

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Water Sources: Groundwater
Chapter 3
EXAMPLE 3.7 DISCHARGE FROM THREE WELLS AT A GIVEN DRAWDOWN
Suppose we want to restrict the drawdown in each of the wells from Example 3.6 to 60 ft (18.3 m). What will be the corresponding
discharges for the individual wells?
Solution 1 (US Customary System):
Eq. (3.15) is s = (114.6Q∕T)W(u)
Well 1: [114.6∕T][Q 1 W(u 1 ) + Q 2 W(u 1,000 ) + Q 3 W(u 2,000 )] = 60 ft.
Well 2: [114.6∕T][Q 1 W(u 1,000 ) + Q 2 W(u 1 ) + Q 3 W(u 1,000 )] = 60 ft.
Well 3: [114.6∕T][Q 1 W(u 2,000 ) + Q 2 W(u 1,000 ) + Q 3 W(u 1 )] = 60 ft.
4
114.6∕(3.2 × 10 )[21.89Q 1 + 8.08Q 2 + 6.69Q 3 ] = 60 ft.
4
114.6∕(3.2 × 10 )[8.08Q 1 + 21.89Q 2 + 8.08Q 3 ] = 60 ft.
4
114.6∕(3.2 × 10 )[6.69Q 1 + 8.08Q 2 + 21.89Q 3 ] = 60 ft.
21.89Q 1 + 8.08Q 2 + 6.69Q 3 = 16, 800.
8.08Q 1 + 21.89Q 2 + 8.08Q 3 = 16, 800.
6.69Q 1 + 8.08Q 2 + 21.89Q 3 = 16, 800.
Solving the three equations for the three unknown discharges Q , Q ,and Q :
1 2 3
3
Q = Q = 468 gpm (2,550 m ∕d).
1
3
3
Q = 420 gpm (2,289 m ∕d).
2
Solution 2 (SI System):
Eq. (3.12) is s = (Q∕4 T)W(u).
Well 1: [1∕4 T][Q W(u 0.3048 m ) + Q W(u 304.8m )
2
1
+ Q W(u 609.6m )] = 18.3m.
3
Well 2: [1∕4 T][Q W(u 304.8m ) + Q W(u 0.3048 m )
2
1
+ Q W(u 304.8m )] = 18.3m.
3
Well 3: [1∕4 T][Q W(u 609.6m ) + Q W(u 304.8m )
2
1
+ Q W(u 0.3048 m )] = 18.3m.
3
Then,
[1∕(4 × 3.14 × 397.4)][21.89Q + 8.08Q + 6.69Q ] = 18.3m.
3
1
2
[1∕(4 × 3.14 × 397.4)][8.08Q + 21.89Q + 8.08Q ] = 18.3m.
1 2 3
[1∕(4 × 3.14 × 397.4)][6.69Q + 8.08Q + 21.89Q ] = 18.3m.
1 2 3
Then,
21.89Q + 8.08Q + 6.69Q = 91, 341.6.
3
2
1
8.08Q + 21.89Q + 8.08Q = 91, 341.6.
3
1
2
6.69Q + 8.08Q + 21.89Q = 91, 341.6.
1
3
2
Solving the three equations for the three unknowns, Q , Q ,and Q :
3
1
2
3
Q = Q = 2,550 m ∕d.
3
1
3
Q = 2,289 m ∕d.
2