Water Resources and Environmental Hydraulics · Groundwater Resources · 1 mark · Numerical answer
An unconfined aquifer covers 20 km×20 km, has conductivity 4 m/day, porosity .32,specific yield .18 and initial saturated thickness 30 m. Extract $4\times10^8$ m³ of water. Find decline in saturated thickness in m, to two decimals.
Drainable storage change is $\Delta V=S_yA\Delta h$. Area is $400$ km²=$4\times10^8$ m². Therefore $\Delta h=(4\times10^8)/[.18(4\times10^8)]=5.55556$ m, rounded **5.56 m**. Use specific yield rather than total porosity; conductivity governs rates but is not needed for this storage calculation.
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