To solve this problem, we must calculate the decline in the saturated thickness of an unconfined aquifer after a specific volume of water is extracted. Here's the step-by-step approach:
Given:
The change in water storage is related to the specific yield and the change in saturated thickness (\(\Delta b\)) by the equation:
\(V = S \cdot A \cdot \Delta b\)
Where:
- \(A\) is the area of the aquifer (in m\(^2\))
- \(\Delta b\) is the decline in saturated thickness (in meters)
Convert the areal extent from km\(^2\) to m\(^2\):
\(400 \text{ km}^2 = 400 \times 10^6 \text{ m}^2\)
Rearrange the equation to solve for \(\Delta b\):
\(\Delta b = \frac{V}{S \cdot A}\)
Substitute the known values:
\(\Delta b = \frac{4 \times 10^8 \text{ m}^3}{0.18 \times 400 \times 10^6 \text{ m}^2}\)
\(\Delta b = \frac{4 \times 10^8}{72 \times 10^6}\)
\(\Delta b = \frac{4 \times 10^8}{72 \times 10^6} = \frac{400}{72} \approx 5.56 \text{ m}\)
Verify the computed value falls within the provided range:
The computed decline in saturated thickness, 5.56 m, slightly exceeds the expected range of 5.5 m. This could be due to rounding differences or specific interpretations of the aquifer's characteristics.
Thus, the decline in the saturated thickness of the aquifer, after 4×10\(^8\) m\(^3\) of water is extracted, is approximately 5.56 meters.