Question:
Two reservoirs differ by 20 m in water levels and are connected through a confined aquifer
(thickness = 5 m, length = 2 km, hydraulic conductivity $K = 3\times10^{-3}$ m/s,
porosity $\eta = 0.3$). If Reservoir 1 is contaminated, the time (in days, rounded off to one decimal place)
taken by the pollutant to reach Reservoir 2 by advection is __________.
Two reservoirs differ by 20 m in water levels and are connected through a confined aquifer
(thickness = 5 m, length = 2 km, hydraulic conductivity $K = 3\times10^{-3}$ m/s,
porosity $\eta = 0.3$). If Reservoir 1 is contaminated, the time (in days, rounded off to one decimal place)
taken by the pollutant to reach Reservoir 2 by advection is __________.
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Seepage velocity = Darcy velocity divided by porosity.
Updated On: Dec 17, 2025
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Verified By Collegedunia
Correct Answer: 230
Solution and Explanation
Hydraulic gradient:
\[ i = \frac{\Delta h}{L} = \frac{20}{2000} = 0.01. \]
Darcy velocity:
\[ v_D = K\, i = (3\times10^{-3})(0.01) = 3\times10^{-5}\ \text{m/s}. \]
Seepage (pore water) velocity:
\[ v = \frac{v_D}{\eta} = \frac{3\times10^{-5}}{0.3} = 1\times10^{-4}\ \text{m/s}. \]
Travel time:
\[ t = \frac{L}{v} = \frac{2000}{1\times10^{-4}} = 2\times10^{7}\ \text{s}. \]
Convert to days:
\[ t = \frac{2\times10^{7}}{86400} \approx 231.5\ \text{days}. \]
Thus,
\[ \boxed{231.5\ \text{days}} \quad (\text{acceptable range: } 230.0\text{–}233.0) \]
\[ i = \frac{\Delta h}{L} = \frac{20}{2000} = 0.01. \]
Darcy velocity:
\[ v_D = K\, i = (3\times10^{-3})(0.01) = 3\times10^{-5}\ \text{m/s}. \]
Seepage (pore water) velocity:
\[ v = \frac{v_D}{\eta} = \frac{3\times10^{-5}}{0.3} = 1\times10^{-4}\ \text{m/s}. \]
Travel time:
\[ t = \frac{L}{v} = \frac{2000}{1\times10^{-4}} = 2\times10^{7}\ \text{s}. \]
Convert to days:
\[ t = \frac{2\times10^{7}}{86400} \approx 231.5\ \text{days}. \]
Thus,
\[ \boxed{231.5\ \text{days}} \quad (\text{acceptable range: } 230.0\text{–}233.0) \]
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