Hydraulic Conductivity Calculation Setup

This solution details the calculation of hydraulic conductivity ($K$) for an unconfined aquifer using data from a pumping test.

Given Parameters:

• Initial water level (head), $h_{initial} = 25$ m
• Pumping rate, $Q = 0.05$ m$^3$/s
• Distance 1, $r_1 = 50$ m
• Drawdown at $r_1$, $\Delta s_1 = 3$ m
• Distance 2, $r_2 = 150$ m
• Drawdown at $r_2$, $\Delta s_2 = 1.2$ m
• Value of $\pi = 3.14$

Calculate Heads at Observation Points:

The head at each distance is calculated by subtracting the observed drawdown from the initial head.

• Head at $r_1$: $h_1 = h_{initial} - \Delta s_1 = 25 \, \text{m} - 3 \, \text{m} = 22$ m
• Head at $r_2$: $h_2 = h_{initial} - \Delta s_2 = 25 \, \text{m} - 1.2 \, \text{m} = 23.8$ m

Note: Since $r_1 < r_2$, we expect $h_1 < h_2$, which is true ($22 < 23.8$). This indicates larger drawdown closer to the well.

Governing Equation for Unconfined Aquifer

For steady-state radial flow to a well in an unconfined aquifer, assuming Dupuit's approximations, the flow rate ($Q$) is related to the hydraulic conductivity ($K$) and the heads ($h$) at different radial distances ($r$) by the following formula:

$Q = \pi K \frac{h_{outer}^2 - h_{inner}^2}{\ln(r_{outer}/r_{inner})}$

Where:

• $Q$ is the pumping rate.
• $K$ is the hydraulic conductivity.
• $h_{outer}$ is the head at the outer radius $r_{outer}$.
• $h_{inner}$ is the head at the inner radius $r_{inner}$.
• $\ln$ is the natural logarithm.

Hydraulic Conductivity Calculation

Rearrange the formula to solve for $K$:

$K = \frac{Q \ln(r_{outer}/r_{inner})}{\pi (h_{outer}^2 - h_{inner}^2)}$

Substitute the values:

• $Q = 0.05$ m$^3$/s
• $r_{inner} = r_1 = 50$ m, $h_{inner} = h_1 = 22$ m
• $r_{outer} = r_2 = 150$ m, $h_{outer} = h_2 = 23.8$ m
• $\pi = 3.14$

Perform Calculations:

• Ratio of radii: $\frac{r_{outer}}{r_{inner}} = \frac{150 \, \text{m}}{50 \, \text{m}} = 3$
• Natural logarithm of the ratio: $\ln(3) \approx 1.0986$
• Square of outer head: $h_{outer}^2 = (23.8 \, \text{m})^2 = 566.44 \, \text{m}^2$
• Square of inner head: $h_{inner}^2 = (22 \, \text{m})^2 = 484 \, \text{m}^2$
• Difference in squared heads: $h_{outer}^2 - h_{inner}^2 = 566.44 \, \text{m}^2 - 484 \, \text{m}^2 = 82.44 \, \text{m}^2$

Calculate K:

$K = \frac{(0.05 \, \text{m}^3/\text{s}) \times 1.0986}{3.14 \times (82.44 \, \text{m}^2)}$

$K = \frac{0.05493}{258.9096}$

$K \approx 0.00021216 \, \text{m/s}$

$K \approx 2.1216 \times 10^{-4} \, \text{m/s}$

Rounded Result:

Rounding to two decimal places for the value before $\times 10^{-4}$:

$K \approx 2.12 \times 10^{-4} \, \text{m/s}$

Result Verification

The calculated hydraulic conductivity is approximately $2.12 \times 10^{-4}$ m/s. This value lies within the given correct answer range of 2 to 2.4 (i.e., $2 \times 10^{-4}$ m/s to $2.4 \times 10^{-4}$ m/s).