A pumping test is conducted in an unconfined aquifer: a well penetrates 30 m below the static water table and is pumped at 31.40 L/min for 24 hours. Drawdown is 1.0 m at r = 20 m and 0.5 m at r = 80 m in observation wells. Using the steady-state Thiem equation, what is the transmissibility T of the aquifer (in m^2/min)?

Introduction:
This problem tests steady-state well hydraulics for a water-table (unconfined) aquifer using the Thiem equation. The goal is to compute transmissibility T from a constant-rate pumping test with drawdowns observed at two different radial distances.

Given Data / Assumptions:

• Pumping discharge Q = 31.40 L/min = 0.03140 m^3/min.
• Observation wells: r1 = 20 m with s1 = 1.0 m; r2 = 80 m with s2 = 0.5 m.
• Steady-state conditions are assumed; aquifer is laterally extensive and homogeneous in the test zone.
• Use natural logarithm in Thiem equation; flow is radial and horizontal toward the pumped well.

Concept / Approach:
For steady radial flow in a confined/unconfined aquifer of thickness large enough relative to drawdown, the Thiem equation between two observation wells is: T = (Q / (2 * pi * (s1 - s2))) * ln(r2 / r1) where T is transmissibility (m^2/min), Q is discharge, s are drawdowns (m), and r are radial distances (m). The formula exploits the difference in head at two radii to eliminate unknowns like the effective radius of the well.

Step-by-Step Solution:
Step 1: Convert discharge: 31.40 L/min = 0.03140 m^3/min. Step 2: Compute drawdown difference: s1 - s2 = 1.0 - 0.5 = 0.5 m. Step 3: Compute ln(r2 / r1) = ln(80 / 20) = ln(4) ≈ 1.3863. Step 4: Apply Thiem: T = (0.03140 / (2 * pi * 0.5)) * 1.3863. Step 5: Numerical value: T ≈ 0.01386 m^2/min.

Verification / Alternative check:
Check units: Q (m^3/min) divided by length (m) yields m^2/min after multiplying by the dimensionless ln term, which is consistent for T. Magnitude is reasonable for a moderately permeable aquifer under the given low discharge.

Why Other Options Are Wrong:

• 0.01285 m^2/min, 0.01185 m^2/min, 0.01485 m^2/min, 0.01585 m^2/min: These are near-miss distractors obtained by slightly perturbing ln(4) or the denominator 2 * pi * 0.5, but they do not match the correct computation.

Common Pitfalls:

• Forgetting to convert litres to cubic metres or minutes to consistent time units.
• Using log base 10 instead of natural log; if log10 is used, replace ln with 2.303 * log10.
• Using the wrong order for s1 and s2 or r1 and r2; only the difference s1 - s2 matters with the matching r2 / r1 inside the logarithm.

Final Answer:
0.01386 m^2/min.