A well penetrates an unconfined aquifer. The water level (i.e., head) in the well prior to pumping is 25 m. After a long period of pumping at a constant rate of 0.05 m³/s, the drawdowns at distances of 50 m and 150 m from the well are observed to be 3 m and 1.2 m, respectively.
The hydraulic conductivity of the unconfined aquifer is ______ × 10^-4 m/s (rounded off to two decimal places).
Consider π = 3.14.

Question

Which one of the following gases is primarily responsible for the formation of photochemical smog?

Answer 1

We can arrive at the same result by deriving the governing equation from Darcy's law instead of directly quoting the Thiem formula. Under the Dupuit assumption, flow in an unconfined aquifer toward a pumped well is essentially horizontal, so at any radius $r$ the discharge crossing a cylindrical surface of height $h$ (the local saturated thickness) is

\[ Q = 2\pi r K h \frac{dh}{dr} \]

since the flow area is $2\pi r h$ and the hydraulic gradient is $dh/dr$. Separating variables gives

\[ Q\,\frac{dr}{r} = 2\pi K h\, dh \]

Before integrating, note the two observation points: at $r_1 = 50$ m the drawdown is 3 m, so the head is $h_1 = 25 - 3 = 22$ m; at $r_2 = 150$ m the drawdown is 1.2 m, so $h_2 = 25 - 1.2 = 23.8$ m. Integrating the left side from $r_1$ to $r_2$ and the right side from $h_1$ to $h_2$,

\[ Q \ln\left(\frac{r_2}{r_1}\right) = 2\pi K \cdot \frac{h_2^2 - h_1^2}{2} = \pi K (h_2^2 - h_1^2) \]

which is exactly the Thiem relation. Now substitute the numbers. The head-squared difference is $h_2^2 - h_1^2 = 23.8^2 - 22^2 = 566.44 - 484 = 82.44\ \text{m}^2$, and the log term is $\ln(150/50) = \ln 3 = 1.0986$. With $Q = 0.05\ \text{m}^3/\text{s}$ and $\pi = 3.14$,

\[ K = \frac{Q\ln(r_2/r_1)}{\pi(h_2^2-h_1^2)} = \frac{0.05 \times 1.0986}{3.14 \times 82.44} = \frac{0.05493}{258.86} = 2.122\times10^{-4}\ \text{m/s} \]

Rounding to two decimal places,

\[\boxed{K = 2.12 \times 10^{-4}\ \text{m/s}}\]

Answer 2