Well Drawdown Calculator: Theis and Cooper–Jacob Cone of Depression

How it works

s = Q ÷ (4πT) × W(u), u = r²S ÷ (4Tt), W(u) = −0.5772 − ln u + u − u²/(2·2!) + u³/(3·3!) − … · for u < 0.01: s ≈ 2.303Q ÷ (4πT) × log₁₀(2.25Tt ÷ r²S) · R = √(2.25Tt ÷ S) · unconfined: s′ = b − √(b² − 2b·s) · superposition: s_total = Σ s from each well

When a well pumps a confined aquifer, water is released from storage as the pressure drops, and a cone of depression deepens and spreads with time. Theis (1935) solved the flow equation by analogy with heat conduction. Drawdown depends on the pumping rate, the aquifer’s transmissivity (how easily it carries water sideways) and its storativity (how much water it releases per metre of head drop), through the exponential-integral "well function" W(u). This calculator evaluates W(u) accurately for any u, using the series for small u and a rational approximation (Abramowitz & Stegun 5.1.56) for large u, so drawdown far from the well and at short times is also correct. Cooper and Jacob noticed that for small u (long times or short distances) the series reduces to a logarithm. That is why drawdown plotted against log time is a straight line, and why every tenfold increase in pumping time adds the same drawdown, Δs = 2.303Q/(4πT). The approximation is shown next to the Theis value with its validity (u < 0.01 is the strict rule; up to 0.05 is often accepted). The radius of influence is where the Cooper–Jacob line reaches zero drawdown; it grows with the square root of time. Because the equations are linear, the cones of several wells simply add up (superposition): enter the neighbour’s pumping rate to see the combined drawdown. For a water-table (unconfined) aquifer, use specific yield as S and choose Unconfined. The Jacob correction converts the Theis result to the larger real drawdown as long as it stays under about a quarter of the saturated thickness. Real pumping wells also have well losses in the screen and gravel pack; a well efficiency below 100% adds them to the drawdown inside the well.

Worked example

With the defaults: Q = 1,200 m³/day, T = 250 m²/day, S = 0.0003, at the well face r = 0.15 m after 30 days. u = 0.15² × 0.0003 ÷ (4 × 250 × 30) = 2.25 × 10⁻¹⁰ and W(u) = 21.64, so s = 1,200 ÷ (4π × 250) × 21.64 = 8.26 m (27.1 ft). Cooper–Jacob gives 0.880 × log₁₀(2.5 × 10⁹) = 8.27 m, almost the same because u is tiny. The radius of influence is √(2.25 × 250 × 30 ÷ 0.0003) = 7,500 m. A neighbour 300 m away sees 2.46 m of extra drawdown after 30 days (3.41 m after a year), and each tenfold increase in time adds 0.88 m. Specific capacity is 1,200 ÷ 8.26 = 145.2 m³/day per metre (8.12 gpm per ft).

Sources: Heath (1983), Basic Ground-Water Hydrology, USGS Water-Supply Paper 2220 · Aquifer test: Theis solution and well function — Wikipedia

Frequently asked questions