Unsteady Flow in Leaky Aquifers
Explore the dynamics of unsteady flow to wells in leaky aquifers through governing equations and solutions for confined and unconfined aquifers. Understand the implications of leakage on drawdown during pumping tests. Examples and calculations provided for comprehensive learning.
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Flow to Wells 4 Unsteady flow to a well in a leaky aquifer Groundwater Hydraulics Daene C. McKinney 1
Unsteady Flow in a Confined Aquifer Q ground surface Governing equation s(r) unconfined aquifer 0 ? = S? initial head ? ?? + ? ?? Cone of Depression leakage Cylindrical coordinates aquitard confined aquifer 1 ? Chain rule ? ?? ?? ?? +? 0 ? =? ? ?? R h(r) ? ? h0 b r K ?2 ??2+1 ? ??+ 0 ?2 =? ? ?? bedrock ? ? Well ? ? ? Leakage factor ? = 2
Radial Flow in a Leaky Aquifer Solution Q ? ?+?2 ground surface 4?2? ? s(r) unconfined aquifer ? = 4?? ?? initial head ? ? Cone of Depression leakage aquitard confined aquifer 4??? ?,? ? R ?(?,?) = h(r) h0 ? b r K bedrock ? ?+?2 4?2? ? ?,? Well = ?? ? ? ? When there is leakage from other layers, the drawdown from a pumping test will be less than the fully confined case. ? =?2? 4?? 3
? ?+?2 4?2? ? ?,? Leaky Well Function = ?? ? ? ? r/B = 0.01 r/B = 3 4 cleveland1.cive.uh.edu/software/spreadsheets/ssgwhydro/MODEL6.XLS
Leaky Aquifer Example Given: Well pumping in a confined aquifer Confining layer b = 14 ft. thick Observation well r = 96 ft. form well Well Q = 25 gal/min Find: T, S, and K t (min) s (ft) 5 28 41 60 75 244 493 669 958 1129 1185 0.76 3.3 3.59 4.08 4.39 5.47 5.96 6.11 6.27 6.4 6.42 Q ground surface s(r) unconfined aquifer initial head Cone of Depression leakage aquitard confined aquifer R h(r) h0 b r K bedrock Well 5 From: Fetter, Example, pg. 179
r/B = 0.15 = 0.20 = 0.30 = 0.40 Match Point W(u, r/B) = 1, 1/u = 10 s = 1.6 ft, t = 26 min, r/B = 0.15 6
Leaky Aquifer Example Match Point Wmp = 1, (1/u)mp = 10 smp = 1.6 ft, tmp = 26 min, r/Bmp = 0.15 Q = 25 gal/min * 1/7.48 ft3/gal*1440 min/d = 4800 ft3/d t = 26 min*1/1440 d/min = 0.01806 d Wmp=4800 ft3/d Q 4p(1.6 ft)*1= 238.7ft2/d T = 4psmp =4(238.7 ft2/d)(0.1)(0.01806) (96 ft)2 4Tump r2 t =1.87x10-4 S = mp b (r/B)2 r2 =(238.7 ft2/d)(14 ft)(0.15)2 (96 ft)2 K =T = 0.0081ft/d 7
Summary Unsteady flow to a well in a leaky aquifer 8
