Aquifer Transmissivity Determination from Pumping Test Data

The problem asks us to determine the transmissivity of a confined aquifer. We are provided with data from a pumping test, including the pumping rate and the observed drawdowns at two different distances from the center of the pumping well. To solve this, we will use Thiem's equation, which is suitable for calculating transmissivity under steady-state flow conditions in a confined aquifer.

Pumping Rate Conversion

The given pumping rate (Q) is in litres per minute, but for calculations in groundwater hydrology, it is standard to use cubic meters per day. Therefore, the first step is to convert the pumping rate to consistent units.

• Given pumping rate, Q = 2720 litres per minute
• We know that 1 litre is equivalent to \(0.001 \text{ m}^3\).
• We also know that 1 day consists of 24 hours, and 1 hour consists of 60 minutes. So, 1 day = \(24 \times 60 = 1440\) minutes.

Using these conversion factors, the pumping rate in cubic meters per day is calculated as follows:

\[ Q = 2720 \text{ litres/min} \times \left(\frac{0.001 \text{ m}^3}{1 \text{ litre}}\right) \times \left(\frac{1440 \text{ min}}{1 \text{ day}}\right) \]

\[ Q = 2.72 \text{ m}^3/\text{min} \times 1440 \text{ min/day} \]

\[ Q = 3916.8 \text{ m}^3/\text{day} \]

Thiem's Equation for Confined Aquifers

Thiem's equation is a key formula in groundwater hydrology used to calculate the transmissivity (T) of a confined aquifer. It uses data from a steady-state pumping test with two observation wells. The equation relates the pumping rate to the drawdowns observed at different distances from the pumping well.

The formula for Thiem's equation is:

\[ T = \frac{Q}{2\pi(s_1 - s_2)} \ln\left(\frac{r_2}{r_1}\right) \]

Where:

• \(T\) is the transmissivity of the aquifer (typically in \(\text{m}^2/\text{day}\)).
• \(Q\) is the constant pumping rate from the well (\(\text{m}^3/\text{day}\)).
• \(s_1\) is the drawdown observed at distance \(r_1\) from the pumping well (m).
• \(s_2\) is the drawdown observed at distance \(r_2\) from the pumping well (m).
• \(r_1\) is the radial distance of the first observation well from the pumping well (m).
• \(r_2\) is the radial distance of the second observation well from the pumping well (m).
• \(\ln\) represents the natural logarithm.

Given Data for Calculation

Let's summarize the given parameters from the pumping test, along with the converted pumping rate:

Calculation of Aquifer Transmissivity

Now, we will substitute these values into Thiem's equation to calculate the transmissivity.

• First, calculate the difference in drawdown: \(s_1 - s_2 = 3 \text{ m} - 0.5 \text{ m} = 2.5 \text{ m}\)
• Next, calculate the ratio of the distances: \(\frac{r_2}{r_1} = \frac{100 \text{ m}}{10 \text{ m}} = 10\)

Substitute these calculated and given values into the Thiem's equation:

\[ T = \frac{3916.8 \text{ m}^3/\text{day}}{2 \times \pi \times (2.5 \text{ m})} \ln(10) \]

\[ T = \frac{3916.8}{5\pi} \times \ln(10) \]

Using the approximate values of \(\pi \approx 3.14159\) and \(\ln(10) \approx 2.302585\):

\[ T = \frac{3916.8}{15.70796} \times 2.302585 \]

\[ T \approx 249.3503 \times 2.302585 \]

\[ T \approx 574.05 \text{ m}^2/\text{day} \]

The calculated transmissivity is approximately \(574.05 \text{ m}^2/\text{day}\). When comparing this value to the given options, \(576 \text{ m}^2/\text{day}\) is the closest available choice. Small differences in the final answer can occur due to rounding constants like \(\pi\) or \(\ln(10)\) during the problem formulation.

Therefore, the transmissivity of the aquifer is approximately \(576 \text{ m}^2/\text{day}\).