Permeability and transmissibility from pumping test

Given: unconfined aquifer, saturated depth $H = 25$ m; well diameter 30 cm ($r_w = 0.15$ m); $Q = 2100$ lpm $= 0.035$ m³/s; observation wells $r_1 = 30$ m, $s_1 = 5$ m and $r_2 = 90$ m, $s_2 = 4$ m.

Water-table depths: $h_1 = 25 - 5 = 20$ m, $h_2 = 25 - 4 = 21$ m.

(a) Coefficient of permeability

For a steady unconfined flow (Dupuit):

$$Q = \frac{\pi K (h_2^2 - h_1^2)}{\ln(r_2/r_1)}$$ $$K = \frac{Q\ln(r_2/r_1)}{\pi (h_2^2-h_1^2)} = \frac{0.035 \times \ln(3)}{\pi (441 - 400)} = \frac{0.035\times 1.0986}{128.8}$$ $$K = 2.985\times10^{-4}\ \text{m/s} \;(= 25.8\ \text{m/day})$$

(b) Transmissibility

$$T = K H = 2.985\times10^{-4} \times 25 = 7.46\times10^{-3}\ \text{m}^2/\text{s} \;(\approx 645\ \text{m}^2/\text{day})$$

(c) Drawdown at the pumping well

Apply the same equation between $r_w$ and $r_1$:

$$h_1^2 - h_w^2 = \frac{Q \ln(r_1/r_w)}{\pi K} = \frac{0.035 \times \ln(200)}{\pi \times 2.985\times10^{-4}} = 197.7\ \text{m}^2$$ $$h_w^2 = 400 - 197.7 = 202.3 \Rightarrow h_w = 14.22\ \text{m}$$ $$s_w = 25 - 14.22 = 10.78\ \text{m}$$

Answer: K = 2.985 × 10⁻⁴ m/s (25.8 m/day); T = 7.46 × 10⁻³ m²/s (about 645 m²/day); drawdown at the well = 10.78 m.