Thiem Steady-State Well Drawdown

Pump a well in a confined aquifer long enough for the cone of depression to stop growing, and the drawdown at any radius follows Dupuit and Thiem's equilibrium solution: $s = (Q/2\pi T)\ln(R/r)$. Pumping 1000 m³/d from an aquifer of transmissivity 500 m²/d, with an influence radius of 300 m and a well radius of 0.3 m, gives $0.318 \times \ln(1000) = 2.20$ m of drawdown at the well. The logarithm is what makes the cone the shape it is, steep and deep right at the well and almost flat a hundred metres out.

That logarithm has a consequence people find counter-intuitive: doubling the well diameter buys almost nothing. Going from a 300 mm to a 600 mm borehole changes $\ln(R/r)$ from 6.91 to 6.21, a ten percent reduction in drawdown for four times the drilling volume. Well capacity is bought with aquifer, screen length and development, essentially never with diameter. In the other direction, halving the transmissivity exactly doubles the drawdown, which is why a well in a tight formation is in trouble no matter how good the pump is.

The equation's weakness is the radius of influence, which is not a real physical boundary. True steady state requires a recharge source somewhere, and in its absence R keeps expanding slowly forever. Practitioners either estimate it from Sichardt's rule or, far better, sidestep it entirely by using two observation wells and taking the difference in drawdown, which cancels R out of the arithmetic. Also note this is the confined form, with T constant. In an unconfined aquifer the saturated thickness itself shrinks as you draw the water table down, the correct solution is written in terms of the squares of the heads, and using the confined equation on an unconfined aquifer under-predicts drawdown exactly when the well is already in difficulty.