Irrigation and Drainage Engineering · Groundwater Hydrology · 1 mark · Numerical answer
A 300 mm well penetrates25 m below the static water table and pumps5400 L/min for 24 h. Observation drawdowns are1.11 m at 30 m and .53 m at 90 m. Assuming steady flow, find main-well drawdown in m, to two decimals. Use π=3.14.
For the intended unconfined Dupuit model, h₁=25−1.11=23.89 m, h₂=24.47 m. From $h_w^2=h_1^2-(h_2^2-h_1^2)\ln(30/.15)/\ln(90/30)$,h_w≈20.8677 m, so drawdown≈**4.13 m**. A confined constant-transmissivity approximation instead gives 3.91 m;the supplied saturated-depth datum supports the squared-head unconfined calculation.
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