Aquifer Area Unit Conversion

To ensure consistent units for calculations, the aquifer's surface area is converted from square kilometers (km$^2$) to square meters (m$^2$).

• Surface Area ($A$) = 100 km$^2$
• Since $1 \text{ km} = 1000 \text{ m}$, then $1 \text{ km}^2 = (1000 \text{ m})^2 = 10^6 \text{ m}^2$.
• $A = 100 \times 10^6 \text{ m}^2 = 1 \times 10^8 \text{ m}^2$

Pumping Volume and Specific Yield Calculation

The volume of groundwater pumped ($V$) over the aquifer's surface area ($A$) causing a water table drop ($\Delta h$) is directly related to the specific yield ($S_y$). The relationship is:

$ V = A \times \Delta h \times S_y $

Rearranging to solve for $S_y$:

$ S_y = \frac{V}{A \times \Delta h} $

Using the provided values:

• $V = 2 \times 10^7 \text{ m}^3$
• $A = 1 \times 10^8 \text{ m}^2$
• $\Delta h = 6 \text{ m}$

Substituting these values:

$ S_y = \frac{2 \times 10^7 \text{ m}^3}{1 \times 10^8 \text{ m}^2 \times 6 \text{ m}} = \frac{2 \times 10^7}{6 \times 10^8} = \frac{1}{30} $

Calculating the decimal value:

$ S_y \approx 0.0333 $

Specific Retention and Porosity Calculation

The porosity ($\phi$) of an aquifer is the sum of its specific yield ($S_y$) and specific retention ($S_r$).

$ \phi = S_y + S_r $

Given Specific Retention ($S_r$) = 15%, which is 0.15 in decimal form.

$ \phi = 0.0333 + 0.15 $

$ \phi = 0.1833 $

Converting to percentage and rounding to one decimal place:

• Porosity in percentage: $\phi (\%) = 0.1833 \times 100 \%$
• $\phi (\%) = 18.33 \%$

The final porosity is 18.3 %.