Jacob Distance-Drawdown Transmissivity

Transmissivity cannot be solved out of the single-well Cooper-Jacob equation, but it falls out immediately if you read two observation wells at the same instant. Write the equation twice, once for each radius, and subtract. Everything inside the logarithm that does not depend on $r$ cancels: the 2.25, the transmissivity, the time, and the storativity all disappear together, leaving $s_1 - s_2 = \frac{2.303\,Q}{2\pi T}\log_{10}\frac{r_2}{r_1}$. Rearranged for $T$, that is a clean, closed-form, single-step answer, and it is what hydrogeologists actually do.

The neatest way to use it is to pick two wells one log cycle apart in distance, say 10 m and 100 m, because then $\log_{10}(r_2/r_1) = 1$ and the whole equation reduces to $T = 2.303\,Q/(2\pi\,\Delta s)$, where $\Delta s$ is simply the drawdown change per log cycle. On a semi-log plot of drawdown against distance the data fall on a straight line, and its slope per log cycle is all you need. Two metres per cycle at 100 m³/h gives $T = 2.303\times2400/(2\pi\times2) = 440$ m²/d. Three wells are better than two, because three points reveal whether the line is actually straight, and a curve tells you the aquifer is not what the equation assumes.

Here is the honest part, and it is worth sitting with rather than glossing over. This algebra is identical to the steady-state Thiem equation, which lives on its own page. That is not a coincidence and it is not duplication for its own sake. What it means is that at any single instant, the shape of the cone of depression is already the steady-state shape, even though the entire cone is still deepening with time. Storage controls how fast the cone drops; transmissivity controls how steeply it slopes. Subtracting two simultaneous readings removes the dropping and leaves only the sloping. Which is also why this method measures transmissivity and tells you absolutely nothing about storativity. For $S$ you need the time-drawdown form and its zero-drawdown intercept.

Both wells must satisfy u < 0.01 at the moment of the reading, and it is always the far well that fails first, because $u$ grows with the square of distance. A common error is reading the two wells at different times, which quietly reintroduces the storage term you thought you had cancelled. Another is putting the far well outside the cone entirely, where its drawdown is zero and the slope becomes meaningless. If the near well shows less drawdown than the far one, something other than this pump is moving the water table, or the labels have been swapped.