Question

A 30 cm diameter well fully penetrates an unconfined aquifer of saturated thickness 20 m with hydraulic conductivity of 10 m/day. Under the steady pumping rate for a long time, the drawdowns in two observation wells located at 10 m and 100 m from the pumping well are 5 m and 1 m, respectively. The corresponding pumping rate (in m$^3$/day) from the well is \_\_\_\_\_\_\_\_\_\_ \textit{(rounded off to 2 decimal places).}

Hint:

When using the Thiem equation, ensure the units of hydraulic conductivity, saturated thickness, and drawdowns are consistent to avoid errors in calculation.

Answer 1

Step 1: Apply the Thiem equation for unconfined aquifers. The Thiem equation for steady-state pumping in an unconfined aquifer is: $$Q = 2 \pi K \cdot H \cdot \frac{\Delta h}{\ln(r_2/r_1)}$$ where: $Q$ = pumping rate (m$^3$/day), $K$ = hydraulic conductivity (10 m/day), $H$ = saturated thickness of the aquifer (20 m), $\Delta h$ = difference in drawdowns between two observation wells, $r_1$ and $r_2$ = radial distances of the observation wells from the pumping well (10 m and 100 m, respectively). Step 2: Substitute the given values. From the problem: $$\Delta h = (5 - 1) = 4 \, \text{m}, \quad r_1 = 10 \, \text{m}, \quad r_2 = 100 \, \text{m}, \quad K = 10 \, \text{m/day}, \quad H = 20 \, \text{m}.$$ Substitute into the Thiem equation: $$Q = 2 \pi (10) (20) \cdot \frac{4}{\ln(100/10)}.$$ Step 3: Simplify the equation. First, calculate the logarithmic term: $$\ln\left(\frac{100}{10}\right) = \ln(10) \approx 2.3026.$$ Now, substitute this value: $$Q = 2 \pi (10) (20) \cdot \frac{4}{2.3026}.$$ Simplify further: $$Q = 2 \pi (200) \cdot \frac{4}{2.3026}.$$ $$Q \approx 400 \pi \cdot 1.737 \approx 400 \cdot 3.1416 \cdot 1.737.$$ Step 4: Final calculation. $$Q \approx 400 \cdot 5.459 \approx 1858.00 \, \text{m\(^3\)/day}.$$ Conclusion: The corresponding pumping rate is approximately $1858.00 \, \text{m\(^3\)/day}$.