Unsteady Flow to Wells in Aquifers
Analyzing unsteady flow to a well in an unconfined aquifer, this content covers topics such as dewatering, water table decline, Neuman solution, and determining aquifer properties like transmissivity and storativity. Learn about early, middle, and late water release patterns and how to calculate hydraulic conductivities in unconfined aquifers.
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Flow to Wells 5 Unsteady flow to a well in an unconfined aquifer Groundwater Hydraulics Daene C. McKinney 1
Summary Unsteady flow to a well in an unconfined aquifer 2
Unsteady Flow to Wells in Unconfined Aquifers Unsteady Flow to a Well in an Unconfined Aquifer Water is produced by Dewatering of unconfined aquifer Compressibility factors as in a confined aquifer Lateral movement from other formations Q Ground surface Prepumping Water level Pumping well Water Table Observation wells h0 r1 hw h 2 h1 Q Unconfined aquifer r2 Bedrock rw 3
Unsteady Flow to Wells in Unconfined Aquifers Analyzing Drawdown in An Unconfined Aquifer Early Release of water is from compaction of aquifer and expansion of water like confined aquifer. Water table doesn t drop significantly Middle Release of water is from gravity drainage Decrease in slope of time- drawdown curve relative to Theis curve Late Release of water is due to drainage of formation over large area Water table decline slows and flow is essentially horizontal 4
Unsteady Flow to Wells in Unconfined Aquifers Unconfined Aquifer (Neuman Solution) ? ? = 4??? ??,??,? 2 ? =?? ? ? ??= ???????? ???????? ???????????? ?? = horizontal hydraulic conductivity ?? ? = 0 ua=r2S Early time (a) 4Tt uy=r2Sy Late time (y) 4Tt 5
Unconfined Well Function ? ??,??,? ? Late (y) Early (a) 6
Example Well pumping Q = 144.4 ft3/min in an unconfined aquifer Determine transmissivity, storativity, specific yield, and horizontal and vertical hydraulic conductivities Observation well is 73 ft away from pumping well Initial saturated thickness = 25 ft 7
Early Match point: t = 0.17 min, s = 0.57 ft and 1/ua = 1.0, W(ua, uy, n) = 1.0 8
? ? = 4??? ??,??,? ? 144.4 4?(0.57) 1 = 20.16 ??2/??? 4??? ??,??,? = ua=r2S 4Tt ? = ? =4???? =4 20.16 0.17 (1.0) (73)2 = 0.00257 ?2 9
Late Match point: t = 13 min, s = 0.57 ft and 1/uy = 0.1, W(ua, uy, n) = 1.0 10
? ? = 4??? ??,??,? ? = 20.16 ??2/??? ? = 0.00257 uy=r2Sy ??=4???? =4 20.16 13 (0.1) (73)2 = 0.02 4Tt ?2 ??=? ?=? =20.16 25 ?? ???= 1160 ??/??? = 0.806 0 ??=??2? 0.06 252(1160 ??/???) (73)2 = = 8.2 ??/??? ?2 11
Summary Unsteady flow to a well in an unconfined aquifer 12
