Pumping Test Interpretation and Well Interference
A confined sand-and-gravel aquifer is 18 m thick. A pumping well is pumped at 25 L/s for 24 hours, and observation wells at 15 m and 60 m show drawdowns of 4.2 m and 1.1 m respectively at the end of pumping. The storativity is estimated at 5 × 10⁻⁴. A second production well is proposed 40 m from the first, both pumping at 25 L/s continuously. The nearest property boundary is 150 m away, and the regulator requires that drawdown at the boundary not exceed 0.5 m.
1. Estimate the transmissivity from the observation-well data and check whether the two drawdowns are mutually consistent with a confined aquifer response.
Read the worked answer
Use the Thiem steady-state form as an approximation (valid for the late-time data):
T = Q·ln(r2/r1) / (2π·(s1 − s2))
Q = 25 L/s = 0.025 m³/s; r1 = 15 m, r2 = 60 m; s1 = 4.2 m, s2 = 1.1 m.
ln(60/15) = ln 4 = 1.386.
T = 0.025 × 1.386 / (2π × 3.1) = 0.03465 / 19.48 ≈ 1.78 × 10⁻³ m²/s (≈ 154 m²/day).
Check consistency: for a confined aquifer, drawdown should follow s ∝ ln(r), so s1 − s2 = 3.1 m over one log cycle of r (15→60 is a factor of 4, i.e. 0.60 of a decade). The implied slope per decade is 3.1/0.60 ≈ 5.2 m, giving T ≈ 0.025 × 2.303 / (2π × 5.2) ≈ 1.76 × 10⁻³ m²/s - consistent with the Thiem result.
Caveat: Thiem assumes steady state; 24 h may not be fully steady, so treat T as approximate. Storativity is not used in Thiem, which is fine for a confined aquifer at late time.
2. Estimate the combined drawdown at the 150 m property boundary from both wells and judge compliance with the 0.5 m limit.
Read the worked answer
Use the Theis/Thiem log approximation for each well at the boundary. For well 1 (r = 150 m) and well 2 (r = 110 m from the boundary, since it sits 40 m closer).
Reference drawdown at 60 m is 1.1 m. Scale by ln ratios:
Well 1 at 150 m: s = 1.1 × ln(150/15) / ln(60/15) = 1.1 × ln10 / ln4 = 1.1 × 2.303/1.386 = 1.1 × 1.661 ≈ 1.83 m.
Well 2 at 110 m: s = 1.1 × ln(110/15)/ln4 = 1.1 × ln(7.33)/1.386 = 1.1 × 1.992/1.386 ≈ 1.58 m.
Combined ≈ 1.83 + 1.58 = 3.4 m, far above the 0.5 m limit.
Judgement: non-compliant by a wide margin. Even a single well at 25 L/s gives ~1.8 m at the boundary. To meet 0.5 m, either move the wells farther from the boundary, reduce pumping rates, or increase spacing. Note the estimate is conservative (superposition ignores any recharge or boundary effects) but the exceedance is so large that the conclusion is robust.
3. The regulator also asks whether the two wells at 40 m spacing will interfere significantly. Quantify the interference and state its practical implication.
Read the worked answer
Interference drawdown at well 1 caused by well 2 (40 m apart), using the same log scaling from the 60 m reference:
s = 1.1 × ln(40/15)/ln4 = 1.1 × ln(2.667)/1.386 = 1.1 × 0.981/1.386 ≈ 0.78 m.
So each well experiences roughly 0.78 m of extra drawdown from its neighbour - about 19% of the 4.2 m drawdown observed at 15 m, and comparable to the entire 0.5 m boundary allowance.
Practical implication: the interference is significant but not catastrophic at the wells themselves; the real problem is cumulative - the second well roughly doubles the drawdown footprint at distance. Combined with the boundary calculation, this means the pair cannot both run at 25 L/s under the current layout. Options: increase spacing (interference falls only logarithmically, so this helps slowly), stagger pumping schedules, or reduce rates. Report the interference figure and recommend spacing/rate revision rather than claiming the wells are independent.