A confined aquifer receives water from a lake (recharge zone) anl discharge into

Question

A confined aquifer receives water from a lake (recharge zone) anl discharge into a river adjacent located 6 miles from Lake. The elevation level (Head) in the lake is 200 ft msl and 100 ft river msl. A well located 2 miles downstream from the lake with an elevation of 230 ft msl and water depth of 60 ft. A geological fault truncates the aquifer at a distance "L" upstream from the river. The hydraulic conductivity of the area between the fault and the river is 50 ft day and hydraulic conductivity between the lake and the fault is 100 ft day. Since the thickness of the aquifer is 50 ft, calculate a-The Now rate from the lake and into the aquifer b-distance corresponding to the length L of low conductivity of the aquifer c-potentiometric load at a distance L. d trip time from the lake to the river, assuming a porosity of 30% mls above mean sea level 200 ft msl 230 ft msl Lake 100 ft msl K 100 ft/d River le 2 millas

Explanation / Answer

Total Head loss from lake to well = 60 - 30 = 30 fts over the distance of 2 miles = 10560 fts

Hydralic gradient = 30/10560 = 0.002841 ;

Flow rate in ft3/day Q = K * Hydralic Gradient * Area of lake

= 100 * 0.002841*(50*1) = 14.20 ft3/day = 0.0738 Gpm

Water level at C must be given to calculate the Length of aquifer connecting the river as it has different K value.

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