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 \(\underline{\hspace{1cm}}\).
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 \(\underline{\hspace{1cm}}\).
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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.
Updated On: Jan 2, 2026
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Verified By Collegedunia
Correct Answer: 2400
Solution and Explanation
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 \( \boxed{2400} \, \text{days} \).
- \( L = 2 \, \text{km} = 2000 \, \text{m} \) 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 \( \boxed{2400} \, \text{days} \).
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