Question

A pumping test pumps 5400 L/min for 24 h from a well. Observation wells at 30 m and 90 m show drawdowns of 1.11 m and 0.53 m. Well diameter = 30 cm. Estimate drawdown in the pumped well (round off to 2 decimals), using steady–state flow and \(\pi = 3.14\).

Hint:

Use the Thiem equation: well drawdown increases logarithmically with radius ratio.

Answer 1

Correct Answer: 4

Flow rate: \[ Q = 5400\ \text{L/min} = 5.4\ \text{m}^3/\text{min} = 0.09\ \text{m}^3/\text{s} \] Use Thiem equation for confined steady flow:
\[ s(r) = s_1 + \frac{Q}{2\pi K}\ln\left(\frac{r}{r_1}\right) \] Difference between observation wells: \[ 1.11 - 0.53 = \frac{Q}{2\pi K}\ln\left(\frac{90}{30}\right) \] \[ 0.58 = \frac{0.09}{2\pi K}\ln 3 \] Solve for hydraulic conductivity term: \[ \frac{Q}{2\pi K} = \frac{0.58}{\ln 3} = \frac{0.58}{1.099} = 0.528 \] Drawdown at pumped well (well radius \(r_w = 0.15\ \text{m}\)): \[ s_w = s_1 + 0.528 \ln\left(\frac{30}{0.15}\right) \] \[ s_w = 1.11 + 0.528 \ln(200) \] \[ s_w = 1.11 + 0.528 (5.298) \] \[ s_w = 1.11 + 2.80 = 3.91\ \text{m} \] Rounded:
\[ s_w = 3.91\ \text{m} \approx 4.00\ \text{m} \]