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6. A pharmaceutical industry owns a discharge well that completely penetrates a confined
aquifer and is pumped at a constant rate of 2,150 m3/day. The aquifer transmissivity is 880
m2/day. The steady state drawdown measured at a distance 80 m from the pumped well is
2.72 m. What is the drawdown impact of the industrial well on a domestic well that is 100 m
away? If a second industrial well is installed with the same diameter and pumping rate at a
distance of 140 m from the domestic well, what is the drawdown impact from both wells?
7. An industrial manufacturer owns a 12-in-diameter well that completely penetrates a 130–ft–
thick, unconfined aquifer with a coefficient of permeability of 0.00055 ft/sec. The well
pumping rate is 3.5 cfs and the radius of influence is 500 feet. Another industry plans to
move into an adjacent property and install a well with the same characteristics. If the new
well is 250 feet away, what is the impact in additional drawdown at the existing well? (Hint:
Consider using the radius of influence as an observation well with zero drawdown.)
8. A well is located in the middle of a circular island, as depicted below. If the well is pumped
at the rate of 7.5 gpm, the water table height in the well above the impervious layer is 25 ft as
shown. What would the water table height in the well be for a pump flow of 9.0 gpm?
Assume
.sec/ft 1002.1 5
K
Also determine the water table height 150 ft from the well.
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9. A well penetrating a confined aquifer will be pumped at a constant rate of 0.579 cfs. How
long can the well be pumped at this rate if the drawdown at a point 300 ft from the well can’t
exceed 3.66 ft. The aquifer transmissivity is 12,000 ft2/day and storage coefficient is 0.0003.
10. An industrial well that fully penetrates a confined aquifer is pumped at a constant rate of 300
m3/hr when needed. A second well is needed with a pumping rate of 200 m3/hr. However, the
drawdown of the aquifer at a nearby residential well (a distance of 100 m from the first well)
cannot exceed 10.5 m. If both wells are turned on at the same time and run for four days, how
close can the second well be placed to the residential well before the drawdown limitation is
violated? The aquifer transmissivity is 25.0 m2/hr and the storage coefficient is 0.00025.
11. A confined aquifer has a storage coefficient of 0.0005 and a transmissivity of 400 m2/day.
Wells 1 and 2 completely penetrate this aquifer and are pumped at constant rates of 600
m3/day and 1400 m3/day respectively. An observation well is located 400 m from well 1 and
490 m from well 2. Determine the drawdown in the observation well after 2.5 days of
pumping if the pumps in both wells start at the same time..
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12. A 30-m-thick confined aquifer has a piezometric surface 75 m above the top of the confining
layer. A 40-cm-diameter well draws water from the aquifer at the rate of 0.1 m3/sec, and the
drawdowns have stabilized. If the drawdown at the well is 30 m and the drawdown in an
observation well 50 m away is 10 m, determine the transmissivity and the radius of influence.
13. A field test is conducted in an unconfined aquifer by pumping 1,300 ft3/hr from an 8-inch-
diameter well penetrating the aquifer. The undisturbed aquifer thickness is 46 ft. The
drawdowns, measured at steady state at various locations, are tabulated below along with the
associated semi-log plot of h2 vs.log r. Determine the coefficient of permeability and how far
away from the well you must be before the drawdown becomes less than 2.5 ft.
r (ft)
0.33
40
125
350
s (ft)
6.05
4.05
3.57
3.05
Now use Eq’n (7.16) w/any observation well: s = 2.5 ft; h = ho – s = 46 – 2.5 = 43.5 ft.
Let rob = 40 ft, hob= 46 – 4.05 = 41.95 ft: h2 = hob2 – (Q/πK)[ln (rob/r)];
43.52 = 41.952 – (1300/π ∙11.9)[ln (40/r)]; r = 1800 ft (or approximate from the plot)
1550
1600
1650
1700
1750
1800
1850
1900
0.10 1.00 10.00 100.00 1000.00
h2(ft2)
Radial Distance, r (ft)
Unconfined Aquifer Test Data
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14. A pumping test was conducted in a 40-m-thick confined aquifer. Drawdown data were
collected in an observation well 20 m away, and is plotted below (drawdown = s = ho – h).
The test well was pumped at the constant rate of 8.50 m3/hr. Determine the aquifer
transmissivity, the storage coefficient, and the drawdown after a year if pumping continues.
Ans. From the best fit line, ∆os = 0.90 m. Then Equation (7.42) yields the transmissivity.
T = [2.30∙Qw/(4π ∙Δos)] = [2.30∙8.50/(4π ∙0.90)] = 1.73 m2/hr. Then from Eq’n (7.43):
S = (2.25∙T∙to)/(r2) = (2.25∙1.73∙1.72 x 10-2)/(202) = 1.67 x 10-4. To determine the
drawdown after a year, use Eq’n (7.40): s = [2.30∙Qw/(4πT)]log[2.25Tt/(r2S)]
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16. An industrial well taps into an 80–ft-thick confined aquifer that has a transmissivity of 0.0455
ft2/sec. The well is located 600 feet from a completely penetrating stream. A farmer’s
irrigation well is located halfway between the stream and the industrial well. What is the
maximum flow rate that can be pumped from the industrial well to limit the drawdown
impact on the irrigation well to 5 feet.
17. A homeowner has a well that has been used for their domestic water supply for years. The
well taps into a shallow, confined aquifer that has enough pressure to deliver their water
without a pump. An industry purchases an adjacent property and installs a high capacity well
in the same aquifer. To protect the homeowner’s domestic well from being impacted,
industrial representatives propose to drive sheet piling into the ground all along the border of
the two properties and deep enough to reach the aquifer’s underlying impermeable bottom.
Will the pressure (piezometric surface) in the homeowner’s well be affected by any of these
activities. How would go about analyzing the impacts?
18. A fully-penetrating, 12-in. diameter well pumps groundwater from a 25-ft thick confined
aquifer. An impermeable rock stratum is located in the aquifer 105 meters away. Will the
impermeable boundary impact the performance of the well when equilibrium conditions are
achieved if the pump rate is 20,000 gpd? The aquifer permeability is 20 gpd/ft2 and the
drawdown at the well is 30 feet. Prove your answer.
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19. A permeable stratum (coefficient of permeability of 0.12 m/hr, porosity of 0.45, and void
ratio of 0.82) lies under a dam and over an impermeable rock layer. A sheet pile is driven
into the stratum near the heal of the dam to reduce seepage, but doesn’t extend to the
impermeable rock strata. Based on the flow net below, determine the rate of seepage under
the dam (m3/day per meter width of dam), the pressure head at point “3” in meters, and the
seepage velocity at point “3” in m/day. Use the impervious rock layer as the datum for all
energy head determinations. The upstream water surface is 30 m above the impervious rock
layer (datum). The upstream water depth is 10 m, and this dimension can be used as a scale.
20. Determine the quantity of seepage (in gallons per minute) that can be expected to flow under
1,000 feet of sheetpile depicted in the figure below. The permeability of the soil under the
sheetpile is 1.50 x 10-5 ft/sec and the porosity is 0.35. Also determine the approximate exit
velocity of the water next to the sheetpile in ft/sec. (Note: The drawing below is roughly to
scale.)
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Ans. A flow net must be constructed. If three flow channels are drawn, there are
approximately six equipotential drops. However, the (m/n) ratio is roughly 3:6,
regardless of the number of streamlines used. Now applying Eq’n 7.56:
q = K(m/n)H = (1.5 x 10-5)(3/6)(5) = 3.75 x 10-5 cfs/ft
Q = q∙L = (3.75 x 10-5 cfs/ft)(1000 ft) = 0.0375 cfs = 16.8 gpm
Δh = H/n = (5 ft)/6 = 0.833 ft. The velocity at the exit next to the sheetpile:
VA= K(Δh/Δs)A= (1.5 x 10-5)[(0.833)/(1.5)] = 8.33 x 10-6 ft/sec, where Δs is the
distance measured across the last cell next to the sheet pile. The seepage velocity is
VSA = VA/α = (8.33 x 10-6)/0.35 = 2.38 x 10-5 ft/sec
21. Determine the seepage in m3/day through the homogeneous earth dam depicted in the figure
below. Also determine the seepage velocity when the water begins to surface on the
downstream embankment at point “D.” The upstream headwater is 7.0 m, the dam is 80 m
long, and soil used to construct the dam is silt. Assume this is a scale drawing with the
upstream depth of 7.0 m being the scale.
Ans. From Table 7.2, silt has a mid-range permeability of 5.0 x 10-8 m/sec. The flow net
provides the (m/n) ratio with n ≈ 17 (half way from D to F). Applying Eq’n 7.56:
q = K(m/n)H = (5.0 x 10-8)(4/17)(7.0) = 8.24 x 10-8 m3/s-m
Q = (8.24 x 10-8 m3/s-m)(80m)(86,400s/day) = 0.570 m3/day
The head drop (loss) between equipotential lines is: Δh = H/n = (7.0 m)/17 = 0.412 m.
At point “D”: H = 7.0 – 14.5(0.412) = 1.03 m. Using Darcy’s eq’n:
V= K(Δh/Δs)= (5.0 x 10-8)[(0.412)/1.0] = 2.06 x 10–8 m/s
where Δh is the equipotential drop from the 14th to the 15th equipotential line and the
distance (Δs = 1.0 m) is measured from the scale drawing.
Thus, VS = V/α = 4.58 x 10–8 m/s where the porosity is 0.45 from Table 7.1.