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Flow to a Well Importance of Wells • • • • • Extraction Control of salt-water intrusion Pump and treat Injection and pumping Disposal of hazardous waste Fig. 7.22, Manning Drawdown depends on: • K • T • S Basic Setup Fetter, Figure 5.2 Assumptions • • • • • • • • • • Confining layer on bottom Geologic formations are horizontal and infinite Potentiometric surface: horizontal, static before test Changes to potentiometric surface due to pumping Homogeneous, isotropic Flow is horizontal Darcy’s law is OK Constant density and viscosity Fully penetrating wells Well has infinitesimal diameter Two Ways to Go • Forward solution – Use equations to predict drawdown – K, T, S are known • Backward solution – Measure heads – Calculate K, T, S Radial Symmetry • Replace x and y with r and q • q doesn’t matter: drops out of equation Steady Radial Flow in a Confined Aquifer h2 dh Q 2rbK dr dh 2rT dr h1 b r2 r1 Steady Radial Flow in a Confined Aquifer dh Q 2rT dr Q dr dh 2T r h2 h1 Q dh 2T r2 r1 dr r Q ln r2 ln r1 h2 h1 2T h2 h1 Q r2 h2 h1 ln 2T r1 b r 2 r1 Q r2 T ln 2 h2 h1 r1 Steady Radial Flow in an Unconfined Aquifer dh Q 2rhK dr h2 h1 r2 r1 Steady Radial Flow in a Confined Aquifer dh Q 2rhK dr h2 Q dr hdh 2K r h1 r2 r1 h2 h1 Q hdh 2K r2 r1 dr r 1 2 Q r2 2 h2 h1 ln 2 2K r1 Q r2 K ln 2 2 h2 h1 r1 Unsteady Flow in a Confined Aquifer h h S h 2 2 x y T t 2 2 In polar coordinates: 2 h 1 h S h 2 r r r T t Boundary Conditions: h(r,0) = h0 dh At r=0, Q 2rT dr Unsteady Flow in a Confined Aquifer 2 r S u 4Tt Q h0 h 4T Q h0 h 4T u u e u u2 u3 u4 ... 0.5772 ln u u 2 2! 3 3! 4 4! We call u u e u the well function. W (u ) u u e u The Well Function Fetter, Figure 5.6 The Well Function 2 Q h0 h W (u ) 4T r S u 4Tt Q T W (u ) 4 h0 h 4Ttu S 2 r The Well Function Q h0 h W (u ) 4T Q logh0 h log logW (u ) 4T r 2S u 4Tt r 2S 1 t 4T u r 2S 1 log log t log u 4T Drawdown vs. Time Fetter, Figure 5.7 Curve Matching Fetter, Figure 5.8 Jacob Drawdown Method Q h0 h 4T u u u ... 0.5772 ln u u 2 2! 3 3! 4 4! 2 3 4 When u gets small, Q 0.5772 ln u h0 h 4T 2 Q r S h0 h ln1.78 ln 4T 4Tt Q h0 h 4T 4Tt Q ln 1.78r 2 S 4T 2.25Tt ln r 2 S Jacob Drawdown Method Q h0 h 4T 2.25Tt ln r 2 S ln x 2.3 log x 2.3Q 2.25Tt h0 h log 2 4T r S 2.3Q 2.25T 2.3Q h0 h log 2 logt 4T r S 4T Jacob Drawdown Method 2.3Q 2.25T 2.3Q h0 h log 2 logt 4T r S 4T 0 1 Drawdown 2 D(h0-h) 3 4 5 6 0 1 2 log Time 3 Jacob Drawdown At 0 drawdown, t = t0 2.3Q 2.25T 2.3Q 0 log 2 logt0 4T r S 4T 2.25T log r 2 S logt0 2.25T 1 2 r S t0 2.25Tt0 S r2 Fetter, Figure 5.9 Distance-Drawdown Method 2.3Q 2.25Tt h0 h log 2 4T r S 2.3Q 2.25Tt 2.3Q 1 h0 h log log 2 4T S 4T r 2.3Q 2.25Tt 2.3Q log r h0 h log 2 4T S 4T Fetter, Figure 5.10 Leaking Confining Beds Fetter, Figure 5.3 I. Nonleaky. Straight line: typical decay. As pumping affects larger and larger area, the rate of drawdown goes down. Fetter, Figure 5.4 II. Leaky, w/o storage from confining layer. As drawdown increases in confined unit, head difference increases between unconfined and confined units. Eventually, drawdown stops. Aquifer may appear nonleaky for hours or days. Fetter, Figure 5.4 IV. Like II, but with additional water from storage in the confining unit. Drawdown is less at early times, but storage effect ceases as system reaches steady state. Fetter, Figure 5.4 Fetter, Figure 5.3 h 1 h e S h 2 r r y T T t h0 h e K' b' 2 Fetter, Figure 5.3 Assume: •Water table elevation initially equal to head in aquifer •Water table doesn’t drop during test •No loss in storage from confining unit Hantush-Jacob Formula Q r h0 h W u, 4T B 2 r S u 4Tt Tb' B K' 0.5 Fetter, Figure 5.11 Also see Appendix 3 Hantush-Jacob Formula W(u, r/B) Tb' B K' Fetter, Figure 5.11 0.5 Curve Matching to Hantush-Jacob Fetter, Figure 5.12 Hantush-Jacob Formula Y-axis displacement gives you W(u, r/B) and (h0-h) Calculate T X-axis displacement gives you u and t Calculate S r / B r /Tb' / K ' 0.5 Tb' r / B K' r2 2 Fetter, Figure 5.3 Assume: •Water table elevation initially equal to head in aquifer •Water table doesn’t drop during test •Water initially comes from storage in the confining unit Hantush Solution Q h0 h H u , 4T 2 r S u 4Tt r S' 4B S 0.5 Tb' B K' Fetter, Figure 5.14 0.5 Hantush Solution Y-axis displacement gives you W(u, ) and (h0-h) Calculate T X-axis displacement gives you u and t Calculate S r S' 4B S 0.5 16 2Tb' S K' S' r2