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<br />Part (3) <br /> <br />If pumping had continued for the entire 366-day period, !J/!/'E. = <br /> <br />" <br /> <br />1.53, and (see tabulation in (2) above), ~/g <br /> <br />= 0.568, and v/Qt = 0.366 <br />- - <br /> <br />- <br /> <br />1 = (0.568)(1.1 acre-ft/day) = 0.62 acre-ft/day <br />~ = (0.366)(1.1 acre-ft/day)(366 days) <br /> <br />= 147 acre-ft <br /> <br />~ during last 216 days = (147-31) acre-ft = 116 acre-ft. <br /> <br />Problem IV <br /> <br />A municipal well is to be drilled in an alluvial aquifer <br /> <br />near a stream. Downstream water uses require that depletion of the <br /> <br />stream be limited to no more than 5,000 cubic meters during the <br /> <br />. <br /> <br />dry season, which commonly is about 200 days long. The well will <br />be pumped continuously at the rate of 0.03 m3/sec (cubic meters per <br /> <br />second). !. = 30 cma/sec (square centimeters per second), and ~ = 0.20. <br /> <br />What is the minimum allowable distance between the well and the <br /> <br />stream? <br /> <br />Qt = (0.03 m3/sec) (200 days) (86400 see/day) = 5.184 (10)5m3 <br /> <br />- <br /> <br />v/Qt = 5000 m3 /5.184 (10tm3 = 0.01. <br /> <br />From Curve B: <br /> <br />-' <br /> <br />tT/aaS = 0.12 <br />-- <br />= (200 days) (86400 see/day) (30 ems /sec) <br />l (0.20) <br /> <br />aa = <br /> <br />(200) (86400) (30) <br />(0.12) (0.20) <br /> <br />a <br />em <br /> <br />= 2.16 (10)10 <br /> <br />a <br />em <br /> <br />"'0> <br />~ <br /> <br />a = 1.47 (10)5 em <br /> <br />- <br /> <br />. 1470 meters. <br /> <br />33 <br />