Question

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.

Hint:

Use the Thiem equation for steady radial flow in an unconfined aquifer, which relates the pumping rate to the heads (not the drawdowns directly) at the two observation distances.

Answer 1

Correct Answer: 2.12

Step 1: Write the Thiem equation for steady radial flow to a well in an unconfined aquifer.
For a fully penetrating well pumped at a constant rate \(Q\) from an unconfined aquifer, the discharge is related to the heads (saturated thicknesses) at two observation points by \[ Q = \frac{\pi K (h_2^2 - h_1^2)}{\ln(r_2/r_1)} \] where \(h_1\) and \(h_2\) are the heads above the aquifer base at radial distances \(r_1\) and \(r_2\) from the well, and \(K\) is the hydraulic conductivity.

Step 2: Convert the given drawdowns into actual heads.
The static head before pumping is \(H = 25\) m. The drawdown at \(r_1 = 50\) m is \(s_1 = 3\) m, so \[ h_1 = H - s_1 = 25 - 3 = 22 \text{ m} \] The drawdown at \(r_2 = 150\) m is \(s_2 = 1.2\) m, so \[ h_2 = H - s_2 = 25 - 1.2 = 23.8 \text{ m} \]

Step 3: Compute \(h_2^2 - h_1^2\) and \(\ln(r_2/r_1)\).
\[ h_2^2 - h_1^2 = (23.8)^2 - (22)^2 = 566.44 - 484 = 82.44 \text{ m}^2 \] \[ \ln\left(\frac{r_2}{r_1}\right) = \ln\left(\frac{150}{50}\right) = \ln(3) = 1.0986 \]

Step 4: Rearrange the Thiem equation and solve for K.
\[ K = \frac{Q \ln(r_2/r_1)}{\pi (h_2^2 - h_1^2)} = \frac{0.05 \times 1.0986}{3.14 \times 82.44} \] \[ K = \frac{0.05493}{258.86} = 2.122 \times 10^{-4} \text{ m/s} \]

Step 5: State the answer.
The hydraulic conductivity of the unconfined aquifer works out to \(2.12 \times 10^{-4}\) m/s, so the required value is \(2.12\).