Question

An unconfined aquifer having a hydraulic conductivity of 12 m.day\(^{-1}\) covers an area of 1.0 ha. When this aquifer is pumped, it releases 6000 m\(^3\) of water and the water table drops from 3 m to 7 m below the ground level. The specific yield of the aquifer is _________%. (Answer in integer)

Hint:

To calculate the specific yield of an aquifer, use the volume of water released, the area of the aquifer, and the change in the water table depth.

Answer 1

The specific yield (Sy) is calculated using the formula: \[ Sy = \frac{V}{A \cdot \Delta h} \] where:
\(V\) is the volume of water released, which is 6000 m\(^3\).
\(A\) is the area of the aquifer, which is 1.0 ha = 10,000 m\(^2\).
\(\Delta h\) is the change in the water table, which is 7 m - 3 m = 4 m.
Now, substitute the values into the formula: \[ Sy = \frac{6000}{10000 \times 4} = \frac{6000}{40000} = 0.15 \] To convert this to a percentage, multiply by 100: \[ Sy = 0.15 \times 100 = 15% \] Thus, the specific yield of the aquifer is 15%.