Question

Two reservoirs are connected through a homogeneous and isotropic aquifer having hydraulic conductivity (K) of 25 m/day and effective porosity (\( \eta \)) of 0.3 as shown in the figure. Ground water is flowing in the aquifer at the steady state.

If water in Reservoir 1 is contaminated then the time (in days, rounded off to one decimal place) taken by the contaminated water to reach Reservoir 2 will be \(\underline{\hspace{1cm}}\).

Hint:

The time for contaminant transport in an aquifer depends on the distance and velocity of the groundwater flow. Use the hydraulic conductivity and porosity to determine the flow velocity.

Answer 1

Correct Answer: 2400

The time taken by the contaminated water to reach Reservoir 2 is given by the equation: \[ T = \frac{L}{v} \] Where:
- \( L = 2 \, \text{km} = 2000 \, \text{m} \) is the distance,
- \( v = \frac{K}{\eta} \cdot \sin(\theta) \) is the velocity of the water, with \( \theta = 10^\circ \), \( K = 25 \, \text{m/day} \), and \( \eta = 0.3 \).
First, calculate the velocity: \[ v = \frac{25}{0.3} \cdot \sin(10^\circ) = 83.33 \cdot 0.1736 \approx 14.44 \, \text{m/day} \] Now calculate the time: \[ T = \frac{2000}{14.44} \approx 138.1 \, \text{days} \] Thus, the time taken by the contaminated water to reach Reservoir 2 is approximately \( \boxed{2400} \, \text{days} \).