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In a groundwater field test, a tracer took 8 hours to travel between two observation wells which are 56 m apart. The difference in water table elevations in these wells was 0.70 m. The volume of the void of the aquifer is 30% of the total volume of the aquifer. What is the hydraulic conductivity of the aquifer, if the dynamic viscosity of water is 0.995 × 10.8 Ns/m2 ?

In a groundwater field test, a tracer took 8 hours to travel between two observation wells which are 56 m apart. The difference in water table elevations in these wells was 0.70 m. The volume of the void of the aquifer is 30% of the total volume of the aquifer. What is the hydraulic conductivity of the aquifer, if the dynamic viscosity of water is 0.995 × 10.8 Ns/m2 ? Correct Answer 4.664 cm/s

Explanation:

Given,

Distance between wells = 56 m, Time took by tracer = 8 hours,

Difference in water table elevation = 0.70 m

Volume of void of aquifer = 30%, Dynamic viscosity = 0.995 × 10.8 Ns/m2

Hydraullic conductivity (k) = ?

We know, from darcy law V = ki

Where,

V is the actual velocity, k is hydraulic conductivity, i is hydraulic conductivity

Hydraulic gradient (i) = h/L = 0.70/56 = 1.25 × 10-2

We know the relation between Actual velocity (V) and porosity (η), Seepage velocity (Vs)

⇒ Vs = V/n 

Seepage Velocity (Vs) = distance/time = 56 m /8 hrs = 0.001945 m/sec

∴ Actual Velocity (V) = Vs × η = 0.001945 × 0.3 = 5.83 × 10-4 m/sec

Now , V = ki ⇒ k = V/i ⇒ k = 5.83 × 10-4/1.25 ×10-2  m/sec

 = 4.664 × 10-2 m/sec = 4.664 cm/sec

∴ Hydraullic conductivity (k) = 4.664 cm/sec

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