Well hydraulics – time–distance relation for equal drawdown in an artesian aquifer Two observation wells at distances r1 and r2 from a pumped well in a confined (artesian) aquifer show the same drawdown at times t1 and t2 hours after pumping starts. Which relation between the times and distances holds good (assume homogeneous aquifer and identical conditions)?

Difficulty: Medium

Correct Answer: t2 / t1 = (r2 / r1)^2

Explanation:


Introduction / Context:
Pumping-test analysis in confined aquifers often uses type curves or simplifying relations. A useful heuristic is that the time at which a given drawdown is observed at an observation well scales approximately with the square of the distance from the pumped well under homogeneous conditions.



Given Data / Assumptions:

  • Confined (artesian) aquifer of uniform transmissivity and storativity.
  • Constant-rate pumping begins at t = 0.
  • Equal drawdown observed at r1 at time t1 and at r2 at time t2.



Concept / Approach:
Solutions based on the Theis equation (exponential integral) and its Cooper–Jacob approximation indicate that the “propagation” of a given drawdown contour outward from the pumped well scales with r^2/t approximately constant for a fixed drawdown level. Therefore, for the same drawdown, time is proportional to the square of distance.



Step-by-Step Solution:
For equal drawdown s: r^2 / t ≈ constant → t ∝ r^2.Hence, t2 / t1 = (r2^2) / (r1^2) = (r2 / r1)^2.Choose the option that exactly matches this proportionality.



Verification / Alternative check:
Cooper–Jacob straight-line method in semi-log plots shows that for a given s, the “time–distance” shift corresponds to r^2 scaling, consistent with the relation given.



Why Other Options Are Wrong:

  • Linear scaling with r or inverse relations contradict the diffusion character of confined aquifers.
  • t1 = t2 ignores distance effects entirely.



Common Pitfalls:
Applying the relation to unconfined conditions without caution; delayed yield and nonlinearity in unconfined aquifers modify the simple scaling.



Final Answer:
t2 / t1 = (r2 / r1)^2

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