Question

An aquifer has a surface area of 100 km². With no recharge, assume that pumping out \(2 \times 10^{7}\) m³ of ground water resulted in a drop of the water table by 6 m. If the aquifer has a specific retention of 15%, then the porosity of the aquifer is % (rounded off to one decimal place).

Hint:

Specific yield = pumped volume / (area x drop). Porosity = specific yield + specific retention.

Answer 1

Correct Answer: 18.3

Step 1: Recall specific yield.
When a water table drops in an unconfined aquifer, only part of the drained pore volume actually comes out as water. That fraction is called specific yield, \(S_y\), defined as
\[ S_y = \frac{\text{Volume of water pumped}}{\text{Area} \times \text{drop in water table}} \]

Step 2: Convert the area to square meters.
The surface area is 100 km\(^2\). Since 1 km\(^2\) = \(10^6\) m\(^2\),
\[ A = 100 \times 10^6 = 1 \times 10^{8} \text{ m}^2 \]

Step 3: Substitute the given numbers.
The volume pumped is \(2\times10^{7}\) m\(^3\) and the drop is 6 m.
\[ S_y = \frac{2\times10^{7}}{(1\times10^{8})\times6} = \frac{2\times10^{7}}{6\times10^{8}} \]
\[ S_y = 0.03333 = 3.333\% \]

Step 4: Recall the link between porosity, specific yield and specific retention.
Total porosity is made up of the water that drains freely (specific yield) plus the water that stays held in the pores by surface tension (specific retention, \(S_r\)).
\[ n = S_y + S_r \]

Step 5: Add the specific retention.
The specific retention is given as 15%.
\[ n = 3.333\% + 15\% = 18.333\% \]

Final Answer:
Rounded to one decimal place, the porosity of the aquifer is
\[ \boxed{18.3\%} \]