Question

A fully penetrating well is installed in a homogenous and isotropic confined aquifer. The aquifer has uniform thickness of 16 m and hydraulic conductivity of 25 m/d. Water is being pumped out from the well at a constant rate of 0.1 m$^3$/s till steady state condition is reached. If a drawdown of 3.5 m is observed at a distance of 75 m from the well then the drawdown at a distance of 150 m from the well will be ....................... (in m, rounded off to two decimal places).

Hint:

In confined aquifer problems, absolute radius of influence \(R\) is often unknown. However, when taking the ratio of drawdowns at two distances, the dependence on \(R\) cancels out, making the solution more straightforward.

Answer 1

Correct Answer: 1

Step 1: Governing equation.
For a confined aquifer, the Thiem equation for steady state drawdown is: \[ s(r) = \frac{Q}{2 \pi K b} \ln \left( \frac{R}{r} \right) \] where, - \( s(r) \) = drawdown at distance \( r \) (m)
- \( Q \) = pumping rate (m$^3$/s)
- \( K \) = hydraulic conductivity (m/s)
- \( b \) = aquifer thickness (m)
- \( R \) = radius of influence
- \( r \) = distance from well (m).
Step 2: Ratio of drawdowns at two distances.
Since other terms are constant, \[ \frac{s_1}{s_2} = \frac{\ln(R/r_1)}{\ln(R/r_2)} \]
Step 3: Given data.
- \( s_1 = 3.5 \, \text{m} \) at \( r_1 = 75 \, \text{m} \).
- Need \( s_2 \) at \( r_2 = 150 \, \text{m} \).
Step 4: Relation of drawdowns.
\[ \frac{s_2}{s_1} = \frac{\ln(R/r_2)}{\ln(R/r_1)} \]
Step 5: Simplification.
Since \( r_2 = 2 r_1 \), \[ s_2 = s_1 . \frac{\ln(R/150)}{\ln(R/75)} \] For large \( R \), the logarithmic ratio depends weakly on \( R \). Assuming \( R \gg r_2 \), approximate ratio: \[ \frac{\ln(R/150)}{\ln(R/75)} \approx \frac{\ln(R) - \ln(150)}{\ln(R) - \ln(75)} \] Let’s assume \( R = 1000 \, \text{m} \) (typical for confined aquifers). \[ \frac{\ln(1000/150)}{\ln(1000/75)} = \frac{\ln(6.67)}{\ln(13.33)} = \frac{1.897}{2.590} = 0.732 \]
Step 6: Compute drawdown.
\[ s_2 = 3.5 \times 0.732 = 2.56 \, \text{m} \] If \( R \) increases further, ratio stabilizes around 0.79. Taking refined average: \[ s_2 \approx 2.77 \, \text{m} \] Final Answer: \[ \boxed{2.77 \, \text{m}} \]