Water Resources and Environmental Hydraulics · Groundwater Resources · 2 marks · Numerical answer
An unconfined aquifer has an initial head of 25 m. After long-term pumping at $.05\,\mathrm{m^3/s}$,drawdowns are 3 m at radius 50 m and 1.2 m at radius 150 m. Find hydraulic conductivity as $n\times10^{-4}$ m/s, reporting n to two decimals. Use $\pi=3.14$.
The steady unconfined Thiem relation is $Q=\pi K(h_2^2-h_1^2)/\ln(r_2/r_1)$. Remaining saturated heads are $h_1=25-3=22$ m and $h_2=25-1.2=23.8$ m. Therefore $$K=\frac{.05\ln(150/50)}{3.14(23.8^2-22^2)}=.0002122007\ \mathrm{m/s}.$$ Hence **n=2.12**. Squared heads, not just drawdown difference, are required for unconfined flow.
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