Confined (artesian) aquifer pumping test – time–distance equivalence In an artesian aquifer, drawdown at r = 100 m equals the drawdown at r = 200 m after x hours. If the first equality occurs at t = 1 hour at 100 m, find x (hours) for the 200 m well, assuming similar conditions.

Introduction / Context:
Pumping-test interpretation in confined (artesian) aquifers often uses similarity via the Theis solution. Equal drawdown at two radii occurs when a particular nondimensional group is the same at the two observation wells.

Given Data / Assumptions:

• Confined aquifer; constant-rate pumping.
• Drawdown s is compared at r1 = 100 m after t1 = 1 hour, and at r2 = 200 m after t2 = x hours.
• Same aquifer properties (transmissivity T and storage S) and same pumping rate Q.

Concept / Approach:
The Theis variable u = r^2 S / (4 T t). For equal drawdown, u must be equal. Hence r1^2 / t1 = r2^2 / t2 when S and T cancel. Therefore, t ∝ r^2.

Step-by-Step Solution:
Use similarity: r1^2 / t1 = r2^2 / t2.Rearrange: t2 = t1 * (r2^2 / r1^2).Substitute: t2 = 1 hour * (200^2 / 100^2) = 1 * (40000 / 10000) = 4 hours.

Verification / Alternative check:
Since t scales with r^2, doubling r requires four times the duration for the same drawdown; result matches 4 hours.

Why Other Options Are Wrong:

• 2 hours underestimates; 9 and 16 hours overestimate given the r^2 proportionality.

Common Pitfalls:
Applying steady-state Thiem assumptions to early-time transient data; or forgetting that equality of drawdown requires the same u value, not the same r/t ratio.

Final Answer:
4 hours