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Theis Well Drawdown Calculator

Drawdown in a confined aquifer at any distance and time from a well pumping at a constant rate, from the Theis non-equilibrium equation — with the well function W(u), the Cooper-Jacob straight-line check and the radius of influence.

gpm
gpd/ft
ft
days
ft
ft
ft

For drawdown in the pumping well itself, enter the effective well radius as r — the result excludes well losses, so measured drawdown in a real well will be higher. Storativity is typically 10⁻⁵ to 10⁻³ for confined aquifers; for an unconfined aquifer use specific yield (≈0.05–0.30) only while drawdown is small compared with saturated thickness.

$$ s = \frac{Q}{4\pi T}\,W(u) \qquad u = \frac{r^2 S}{4 T t} \qquad W(u) = -0.5772 - \ln u + u - \frac{u^2}{2\cdot 2!} + \frac{u^3}{3\cdot 3!} - \cdots $$
$$ s_{CJ} = \frac{2.303\,Q}{4\pi T}\log_{10}\!\frac{2.25\,T\,t}{r^2 S} \qquad r_0 = \sqrt{\frac{2.25\,T\,t}{S}} $$
s drawdown · Q pumping rate · T transmissivity (= K·b) · S storativity · r radial distance from the pumping well · t time since pumping started · W(u) Theis well function · r₀ distance at which Cooper-Jacob drawdown is zero.

Well function W(u) table

Selected values of the Theis well function
uW(u)uW(u)
1×10⁻⁴8.6330.11.823
1×10⁻³6.3320.50.5598
0.014.03810.2194
0.052.46850.001148

Worked examples

Example 1 — Interference at a neighbouring well (US)

Given: Q = 500 gpm, T = 10,000 gpd/ft, S = 0.0005. Neighbouring well 100 ft away. Pumping for 1 day.
Convert: Q = 500 × 1,440 / 7.4805 = 96,250 ft³/day; T = 10,000 / 7.4805 = 1,336.8 ft²/day
u = (100² × 0.0005) / (4 × 1,336.8 × 1) = 9.35×10⁻⁴
W(u) = −0.5772 − ln(9.35×10⁻⁴) + 9.35×10⁻⁴ − … = 6.399
s = 96,250 / (4π × 1,336.8) × 6.399 = 36.66 ft  (check: 114.6 × 500 × 6.399 / 10,000 = 36.66 ft)
s = 36.7 ft · Cooper-Jacob gives 36.67 ft (u < 0.01, valid) · r₀ ≈ 2,450 ft after one day

Example 2 — The cone keeps growing

Same well, same observation point, after 10 days instead of 1.
u = 9.35×10⁻⁵ → W(u) = 8.700
s = 49.8 ft — ten times the pumping time adds 2.303·Q/(4πT) ≈ 13.2 ft, as the Cooper-Jacob straight line predicts

Example 3 — SI units

Given: Q = 2,000 m³/day, T = 500 m²/day, S = 1×10⁻⁴, r = 50 m, t = 0.5 day.
u = (50² × 10⁻⁴) / (4 × 500 × 0.5) = 2.5×10⁻⁴ → W(u) = 7.717
s = 2,000 / (4π × 500) × 7.717 = 2.46 m

Drawdown versus distance

For Example 1 after one day, drawdown is 89.4 ft at r = 1 ft, 63.0 ft at 10 ft, 36.7 ft at 100 ft, 18.4 ft at 500 ft and 10.8 ft at 1,000 ft. Each tenfold increase in distance removes about the same drawdown while u stays small — the logarithmic shape that makes semi-log distance-drawdown plots straight lines.

References: Theis, C.V. (1935). "The relation between the lowering of the piezometric surface and the rate and duration of discharge of a well using ground-water storage." Trans. AGU 16, 519–524. Cooper, H.H., Jacob, C.E. (1946). Trans. AGU 27, 526–534. Abramowitz, M., Stegun, I.A. (1964). Handbook of Mathematical Functions, eq. 5.1.11 and 5.1.54. Driscoll, F.G. (1986). Groundwater and Wells, 2nd ed.

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