Answer:
T = 0.0088 m²/s
Explanation:
given,
initial piezometric elevation = 12.5 m
thickness of aquifer = 14 m
discharge = 28.24 L/s = 0.02824 m³/s
we know
[tex]k = \dfrac{qln(\dfrac{R_2}{R_1})}{2\pi D(H_2-H_1)}[/tex]
[tex]k = \dfrac{0.02824 \times ln(\dfrac{75}{25})}{2\pi \times 14 (11.45-10.89)}[/tex]
k = 0.629 mm/sec
Transmissibilty
T = k × H
T = 0.629 × 14 × 10⁻³
T = 0.0088 m²/s