Question

An unconfined aquifer of areal extent 20 km × 20 km has hydraulic conductivity of 4 m/day, porosity of 0.32, and storage coefficient (specific yield) of 0.18. If the initial saturated thickness of the aquifer is 30 m, and 4×10^8 m\(^3\) of water is extracted from the aquifer, then the decline in the saturated thickness is ____________ m.

Hint:

N/A

Answer 1

Step 1: Given values.

• Area of the aquifer = \( 20 \, \text{km} \times 20 \, \text{km} = 400 \, \text{km}^2 = 4 \times 10^8 \, \text{m}^2 \)
• Hydraulic conductivity (\(K\)) = \( 4 \, \text{m/day} \)
• Porosity (\( \phi \)) = \( 0.32 \)
• Storage coefficient (specific yield) (\(S_y\)) = \( 0.18 \)
• Initial saturated thickness (\(h_0\)) = \( 30 \, \text{m} \)
• Water extracted = \( 4 \times 10^8 \, \text{m}^3 \)

Step 2: Calculate the volume of water extracted from the aquifer.
\[ \text{Volume of water} = \text{Area} \times \text{Decline in saturated thickness} \times \text{Specific yield} \] \[ 4 \times 10^8 \, \text{m}^3 = (4 \times 10^8 \, \text{m}^2) \times \Delta h \times 0.18 \] Step 3: Rearranging the equation to find the decline in saturated thickness (\( \Delta h \)).
\[ \Delta h = \frac{4 \times 10^8 \, \text{m}^3}{4 \times 10^8 \, \text{m}^2 \times 0.18} = \frac{4 \times 10^8}{7.2 \times 10^7} = 5.56 \, \text{m} \] Conclusion:
The decline in the saturated thickness is approximately \( 5.50 \, \text{m} \) (rounded to two decimal places).