Question

A 0.30 m well penetrates an unconfined aquifer (saturated depth = 40 m). Pumping 0.03 m\(^3\)/s for 8 hours. Drawdowns in two observation wells: 3 m at 20 m, 2 m at 50 m. Find drawdown in pumping well in m.

Hint:

Use Thiem's equation with two observation wells to find aquifer properties, then compute drawdown at the pumping well.

Answer 1

Step 1: Use Thiem's equation for unconfined aquifer. \[ Q = \frac{\pi K (h_1^2 - h_2^2)}{\ln(r_2/r_1)} \] Given drawdowns: \(s_1 = 3, \, r_1 = 20\), \(s_2 = 2, \, r_2 = 50\).

Step 2: Initial saturated head. \[ H = 40 \, m \] \[ h_1 = H - s_1 = 37, h_2 = H - s_2 = 38 \]

Step 3: Solve for hydraulic conductivity. \[ Q = 0.03 = \frac{\pi K (38^2 - 37^2)}{\ln(50/20)} \] \[ 0.03 = \frac{3.14 K (1444 - 1369)}{\ln(2.5)} = \frac{3.14 K (75)}{0.916} \] \[ 0.03 = 257.2 K \Rightarrow K = 1.17 \times 10^{-4} \, m/s \]

Step 4: Drawdown in pumping well. \[ Q = \frac{\pi K (H^2 - h_w^2)}{\ln(R/r_w)} \] where \(r_w = 0.15 \, m\). Effective radius \(R\) (using r\(_1\), r\(_2\)): \[ R = \sqrt{r_1 r_2} = \sqrt{20 \times 50} = 31.62 \, m \] \[ 0.03 = \frac{\pi (1.17 \times 10^{-4})(1600 - h_w^2)}{\ln(31.62/0.15)} \] \[ 0.03 = \frac{3.67 \times 10^{-4} (1600 - h_w^2)}{5.35} \] \[ 0.03 = 6.87 \times 10^{-5} (1600 - h_w^2) \] \[ 1600 - h_w^2 = 437 \Rightarrow h_w^2 = 1163 \] \[ h_w = 34.1 \Rightarrow s_w = H - h_w = 40 - 34.1 = 5.9 \, m \] Rounded: \[ \boxed{5.9 \, m} \]