Difficulty: Medium
Correct Answer: 0.001 cm/s
Explanation:
Introduction / Context:
Pumping tests allow estimation of aquifer properties by relating discharge, drawdown, and geometry. For confined aquifers under steady conditions, the Thiem equation connects discharge to transmissivity and observed heads (or drawdowns) at different radii.
Given Data / Assumptions:
Concept / Approach:
Thiem (confined) equation in natural log form: Q = 2 * π * k * b * (s1 - s2) / ln(r2 / r1). Solve for k and convert to cm/s. The larger drawdown near the well and the logarithmic radial term govern the estimate.
Step-by-Step Solution:
Compute ln(r2/r1) = ln(6.184 / 3) ≈ ln(2.0613) ≈ 0.724.Compute numerator factor: 2 * π * b * (s1 - s2) = 6.283 * 16 * (2.6 - 0.3) = 6.283 * 16 * 2.3 ≈ 231.2.Rearrange for k: k = Q * ln(r2/r1) / [2 * π * b * (s1 - s2)] = 0.00291 * 0.724 / 231.2 ≈ 9.1 × 10^-6 m/s.Convert to cm/s: 9.1 × 10^-6 m/s × 100 = 9.1 × 10^-4 cm/s ≈ 0.001 cm/s.
Verification / Alternative check:
Check order of magnitude: Typical sandy aquifers have k from 10^-4 to 10^-2 cm/s; the computed 10^-3 cm/s is plausible for fine sand/silty sand, validating the result.
Why Other Options Are Wrong:
Common Pitfalls:
Using (s1/s2) instead of (s1 − s2) for confined Thiem, or forgetting to convert litres/s to m^3/s and m/s to cm/s.
Final Answer:
0.001 cm/s
Discussion & Comments