Page 90 - Water Engineering Hydraulics, Distribution and Treatment

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Chapter 3
Water Sources: Groundwater
s, at any distance, r, from the pumping well and r from the
i
image well is the algebraic sum of the drawdowns due to the
tion and the pumping well, ft; r = the distance between the
i
real well, s , and buildup due to the image well, s :
drawdown location and the image well, ft; r = the radius
r
i
w
of the pumping well, ft; Q = pumping discharge rate, gpm;
(US customary units)
s = s + s = (528Q∕T) log(r ∕r)
r
i
i
T = transmissivity, gpd/ft; and a = the distance between the
stream and the pumping well, ft.
(3.36a)
The following are two equivalent equations using the SI
s = (528Q∕T) log(2a∕r )
(US customary units)
w
w
or metric units of s (m), s (m), s (m), s (m), r (m), r (m),
r
i
w
i
3
3
r (m), Q (m /d), T (m /d/m), and a (m):
(3.37a)
w
where s = the drawdown at any distance, r, from a pump-
(SI units) (3.36b)
s = s + s = (0.366Q∕T) log(r ∕r)
i
r
i
ing well, ft; s = the drawdown due to real well, ft; s =
r
i
s = (0.366Q∕T) log(2a∕r )
(3.37b)
(SI units)
the drawdown due to image well, ft; s = the drawdown at the well, ft; r = any distance between the drawdown loca-
w
w
w
EXAMPLE 3.8 DETERMINATION OF A WELL DRAWDOWN AND CONE OF DEPRESSION PROFILE
4
A gravel-packed well with an effective diameter of 24 in. (610 mm) pumps water from an artesian aquifer having T = 3.2 × 10 gpd∕ft
3
−5
(397.4 m /d/m) and S = 3.4 × 10 . The well lies at a distance of 1,000 ft (304.8 m) from a stream that can supply water fast enough
3
to maintain a constant head. Find the drawdown in the well after 10 days of pumping at 700 gpm (3,815 m /d). Determine the profile
of the cone of depression with a vertical plane through the well normal to the stream.
Solution 1 (US Customary System):
In the region of interest, the use of a semilogarithmic approximation is valid:
s = (528 Q∕T) log(2a∕r )
w
w
s = (528Q∕T) log(r ∕r)
i
4
528 Q∕T = (528 × 700)∕(3.2 × 10 ) = 11.55 ft.
Drawdown at the well:
a = 1,000 ft
r = 24 in./2 = 1ft
w
s = (528 Q∕T) log(2a∕r )
w
w
s = 11.55 log 2,000 = 38.0ft.
w
Drawdown at 500 ft from the stream:
r = 500 ft
r = a + r = 1,000 + 500 = 1,500 ft
i
s = 11.55 log(1,500∕500) = 5.5ft.
Drawdown at 1,500 ft from the stream:
r = 1,500 ft
r = a + r = 1,000 + 1,500 = 2,500 ft
i
s = 11.55 log(2,500∕1,500) = 2.56 ft.
Drawdowns at other points can be calculated in a similar manner.
The results are shown in Fig. 3.10.
Solution 2 (SI System):
In the region of interest, the use of semilogarithmic approximation is valid:
s = (0.366 Q∕T) log(r ∕r)
i
s = (0.366 Q∕T) log(2a∕r )
w
w
0.366 Q∕T = 0.366 × 3,815∕397.4 = 3.5m.