A pump test that checks itself

Well hydraulics · Cooper–Jacob with its own audit

A production well pumps a steady 30 m³/h. The hydrogeology report from the original drilling gives the aquifer a transmissivity of 500 m²/d and a storativity of 2 × 10⁻⁴, and two observation wells stand at 25 m and 100 m. Twelve hours in, the tape finds 0.96 m of drawdown at the near well — right on the old report's prediction — and 0.646 m at the far one. Predict the near-well drawdown from the report's numbers, then let the two wells speak for themselves: the transmissivity the distance-drawdown pair actually measures, and the u-value that says whether Cooper–Jacob was ever valid here.

Qr₁r₂s₁s₂T · Su

Every number in this problem is editable — change any value below and the whole chain recalculates.

Given
  • Q = 30 m³/hPumping rate, held steady
  • T = 500 m²/dTransmissivity, from the old report
  • S = 0.0002 Storativity, from the old report
  • t = 12 hTime since the pump started
  • r₁ = 25 mNear observation well
  • r₂ = 100 mFar observation well
  • s₂ = 0.646 mMeasured drawdown at the far well
Determine
  1. (a)the drawdown the old report predicts at 25 m
  2. (b)the transmissivity the two wells actually measure
  3. (c)the validity parameter u at the near well
Step 1 of 3(a) · solve for Drawdown

This is the prediction, made entirely from the old report: 500 m²/d and 2 × 10⁻⁴ say the near well should stand 0.964 m down after twelve hours of 30 m³/h. The tape agrees — which feels like confirmation, and is only half of one.

QsrT, St
Rearranged for s
s=2.303Q4πTlog10 ⁣(2.25Ttr2S)s = \frac{2.303 \, Q}{4 \pi T} \log_{10}\!\left(\frac{2.25 \, T \, t}{r^{2} S}\right)
Your values, in your units
s=2.303(30 m3/h)4π(500 m2/d)log10 ⁣(2.25(500 m2/d)(12 h)((25 m))2(0.0002))s = \frac{2.303 \cdot \left( 30\ \text{m}^{3}\text{/h} \right)}{4 \pi \cdot \left( 500\ \text{m}^{2}\text{/d} \right)} \log_{10}\!\left(\frac{2.25 \cdot \left( 500\ \text{m}^{2}\text{/d} \right) \cdot \left( 12\ \text{h} \right)}{\left(\left( 25\ \text{m} \right)\right)^{2} \left( 0.0002 \right)}\right)
Answer
s=963.93 mms = 963.93\ \text{mm}

Carried onward at full precision, not this rounded figure.

Open the Cooper-Jacob Drawdown solver →

Step 2 of 3(b) · solve for Transmissivity (m²/d)

Two wells at one moment beat one well over time, because the slope of drawdown against log-distance is set by transmissivity alone — storativity drops out entirely. The 0.318 m of difference across the 4:1 distance ratio gives T = 499.7 m²/d: the old report holds.

Qs1s2r1r2
Rearranged for T
T=2.303Q2π(s1s2)log10 ⁣r2r1T = \frac{2.303 \, Q}{2 \pi (s_1 - s_2)} \log_{10}\!\frac{r_2}{r_1}
963.93 mmcarried from step 1
Your values, in your units
T=2.303(30 m3/h)2π((0.963925 m)(0.646 m))log10 ⁣(100 m)(25 m)T = \frac{2.303 \cdot \left( 30\ \text{m}^{3}\text{/h} \right)}{2 \pi \left(\left( 0.963925\ \text{m} \right) - \left( 0.646\ \text{m} \right)\right)} \log_{10}\!\frac{\left( 100\ \text{m} \right)}{\left( 25\ \text{m} \right)}
Answer
T=499.67 m2/dT = 499.67\ \text{m}^{2}\text{/d}

Carried onward at full precision, not this rounded figure.

Open the Jacob Distance-Drawdown Transmissivity solver →

Step 3 of 3(c) · solve for Dimensionless time u

Cooper–Jacob is a truncated series, and u is the fine print: the approximation holds where u < 0.01. With the measured transmissivity, u = 1.25 × 10⁻⁴ at the near well — two orders inside the limit, so every number above was computed on solid ground.

QtrST
Rearranged for u
u=r2S4Ttu = \frac{r^{2} S}{4 T t}
499.67 carried from step 2
Your values, in your units
u=((25 m))2(0.0002)4(499.67 m2/d)(12 h)u = \frac{\left(\left( 25\ \text{m} \right)\right)^{2} \left( 0.0002 \right)}{4 \cdot \left( 499.67\ \text{m}^{2}\text{/d} \right) \cdot \left( 12\ \text{h} \right)}
Answer
u=125.08 ppmu = 125.08\ \text{ppm}

Carried onward at full precision, not this rounded figure.

Open the Cooper-Jacob Validity Parameter u solver →

Answer

Therefore the report predicts 0.964 m at the near well and the tape agrees; the two observation wells measure T = 499.7 m²/d, confirming the report rather than merely repeating it; and u = 1.25 × 10⁻⁴ says the straight-line method was valid all along.

Why this order

The chain is shaped like good field practice: predict, measure, and then audit the method itself. Step (b) is the quiet masterstroke of well hydraulics — taking two wells at the same instant, the storativity that plagues every time-based analysis cancels out of the slope, and transmissivity comes out of nothing but a pumping rate, two distances and two water levels. That is why a proper test has observation wells at all: the production well's own drawdown is polluted by well losses and tells you about the well, where the observation pair tells you about the aquifer.

Step (c) is the one practitioners skip and shouldn't. Cooper–Jacob drops every term of the well function past the logarithm, which is only honest once u = r²S/4Tt is small — the standard line is u < 0.01. Early times, distant wells, tight aquifers all push u up; use the straight line there and the numbers look fine while being quietly wrong. A method that publishes its own validity check deserves to have it run.

Carried values move at full precision, not the rounded figure shown — chaining rounded numbers compounds error.