Question

A 50 cm diameter pumping well and site details are provided on the next page. The hydraulic conductivity of the unconfined aquifer is estimated to be about 1.2 m/day. The bottom of the aquifer is approximately horizontal at a depth of 100 m below the ground surface. Under natural conditions, the water table is nearly horizontal at a depth of 3 m below the ground surface. What is the minimum discharge that can be used that has no effect on-site if you know the water table of the pumping well below the ground level was equal to 8 m? Then, what is the drawdown at point A? Assuming steady-state conditions.

A. Minimum discharge: 1.2 m³/day; Drawdown at point A: 5 m

B. Minimum discharge: 0.8 m³/day; Drawdown at point A: 8 m

C. Minimum discharge: 1.5 m³/day; Drawdown at point A: 3 m

D. Minimum discharge: 2.0 m³/day; Drawdown at point A: 2 m

Answer 1

The minimum discharge that has no effect on the site is 2.56335 m³/day and the drawdown at point A is approximately 2.148 meters.

Step-by-step explanation:

To determine the minimum discharge that has no effect on the site, we need to calculate the rate at which water is flowing out of the pumping well. The flow rate can be calculated using the equation Q = K * A * h, where Q is the discharge, K is the hydraulic conductivity, A is the cross-sectional area of the pumping well, and h is the difference in water level between the pumping well and the water table. In this case, the diameter of the pumping well is 50 cm, so the cross-sectional area can be calculated as (pi * r^2) = (pi * (25 cm)^2) = 1963.5 cm^2. The difference in water level between the pumping well and the water table is 8 m - (-3 m) = 11 m. Plugging in the values, we get Q = (1.2 m/day) * (1963.5 cm^2) * (11 m) = 256,335 cm^3/day. Converting to m^3/day, we get Q = 2.56335 m^3/day. Therefore, the minimum discharge that can be used without having an effect on the site is 2.56335 m^3/day.

To calculate the drawdown at point A, we need to use the equation s = Q / (4 * pi * T * B), where s is the drawdown, Q is the discharge, T is the transmissivity, and B is the aquifer thickness. The transmissivity can be calculated as T = K * B, where K is the hydraulic conductivity and B is the aquifer thickness. In this case, the hydraulic conductivity is 1.2 m/day and the aquifer thickness is 100 m. Therefore, the transmissivity is T = (1.2 m/day) * (100 m) = 120 m²/day. Plugging in the values, we get s = (2.56335 m^3/day) / (4 * pi * (120 m²/day) * (100 m)) ≈ 2.148 m. Therefore, the drawdown at point A is approximately 2.148 meters.