Leaky Aquifers. log s drawdown. log time. will the red observation well look more like?.. A infinite aquifer. D none of the above

Transcription

1 Leaky Aquifers b K S s unpumped aquifer b K S s b 1 K 1 S s1 aquitard pumped aquifer previously K was zero (i.e. no leakage) subscript 1 = pumped zone subscript = unpumped aquifer prime = aquitard = pumping rate b = thickness K = hydraulic conductivity S s = specific storage assume: no head change in shallow aquifer horizontal flow in aquifers vertical flow in aquitards aquifer extends far enough to intercept enough leakage to satisfy Theis assumptions (other than impermeable aquitard) apply A What sufficiently conditions permeable justify assuming shallow aquifer no head such change that flow in the to shallow the disk aquifer? above Solution the drawdown is similar cone to is Theis low enough Solution to but require well function nearly zero is more gradient. complex water level in shallow aquifer Greatest Drawdown What S Vertical & T will controls of aquifer leakage, we be happen will expect the the be maximum where first limited rate that from of did to regarding head extent the drawdown AND not storage radius occur difference of in drawdown? at in cone the which the aquitard, is completely largest, the then the decreasing entire development? S confined from and magnitude the Kv leaks of shallow aquifer? away the from of aquitard from leakage? aquifer. the the well. upper auifer to the lower. will the red observation well look more like?.. log s drawdown B A infinite aquifer C D none of the above log time 1

2 The system is hydrostatic at a head of 100 when pumping begins. The shallow aquifer head remains 100. The pumped aquifer is laterally infinite Work with a partner to draw the head distribution at early, middle and late time. K= 1x10-1 /day K =1x10-6 /day S & S = 1x10-5 b = 10 b =5 = 0. GPM head in shallow aquifer= Hantush & Jacob 1955 assumed no storage in the aquitard and expressed this solution in terms of a dimensionless parameter r/b What does no storage in the aquitard mean relative to the last transient animation we watched? How does it change?

3 Curve match W(u, r/b) r/b 1/u matched with s t Solve for T 1 by rearranging and using s and W from curve match Solve for K 1 from Solve for K from Solve for S 1 from =0.004 m 3 /sec r=55m b 1 =30.5m b =3.05m 3

4 W(u,r/B)=0.6 (r/b=1.0) s=0.08m 1/u= sec =0.004 m 3 /sec r=55m b 1 =30.5m b =3.05m Calculate K 1 S s1 K Work with a neighbor =0.004 m 3 /sec r=55m W(u,r/b)=0.6 (r/b=1.0) s=0.08m b 1 =30.5m b =3.05m 1/u=0.8 u= sec T = m 3 /sec * * 0.08m =.4x10-3 m /sec K 1 = T/b = 7.8x10-5 m/sec K = (1/55m) *7.8x10-5 m/sec*30.5m*3.05m =.4x10-6 m/sec S = (1.5*4*.4x10-3 m /sec * 150sec)/(55m) = 6x10-4 S s = S/b = x10-5 m -1 4

5 The Hantush & Jacob 1955 solution was based on restrictive assumptions 1. hydraulic head in unpumped aquifer remains constant. rate of leakage into pumped aquifer is proportional to gradient across aquitard Hantush 1960 added concept of S in the aquitard to equations Neuman & Witherspoon 1969 presented complete solution including release from aquitard storage and head decrease in unpumped aquifer Allows us to evaluate properties of both aquifers and the aquitard in aquifers s(r,t) in aquitard s(r,t,z) Is this graph and legend correct? = 0. = 0.8 = 0.5 = 0.5 = 0. = 0.8 Click here to see visualization of different conceptual models for pumping in IE 5

6 Unconfined - aquifer is dewatered, not only depressurized aquifer thickness decreases and vertical components of flow exist Two mechanisms for water delivery 1. first elastic storage. second actual dewatering Click here to visualize pumping in an uncofined aquifer 14wh3LeakyUnconf/PumpingUnconfinedAquifer.html Three distinct phases of time-drawdown curves 1. shortly aer start of pumping, water from elastic storage, horizontal flow. water table begins to decline, water primarily from gravity drainage, horizontal & vertical flow 3. at later times, rate of drawdown decreases, essentially horizontal flow Type Curves based on: Valid for: S y >> S s << b Fully penetrating pumping and observation wells 6

7 Example Problem: Data on previous sheet corrected drawdowns to adjust unconfined conditions for confined equations: s < 10% b OK s 10% - 5% b: s > 5% don't trust Plot data, curve match and read values of: comes from selected type curve and is the same for all given r it may be easier to match early time & shi horizontally to later time curves solve for 7

8 u A =.5 x 10 - (1/4u A = 10) t = 6 min = 0.06 W (u, ) = 1 s = 0.55 SLIDE LATERALLY DO NOT SHIFT UP AND DOWN T DOES NOT CHANGE WITH TIME BUT S DOES = 0.06 W (u, ) = 1 s = 0.55 t = 53 min u B = 0.5 (1/4u B =1) 8

9 Match early time = 0.06 W (u, ) = 1 u A =.5 x 10 - (1/4u A = 10) t = 6 min s = 0.55 = /min r = 73 b = 100 late time same slide horizontally same s = 0.55 t = 53 min u B = 0.5 (1/4u B =1) Calculate T S K v K h S y Early time match results: T = W(u, Γ) = min (1) 4πs 4π Late time match results: 0.9 ( 0.55) min A = (.5x10 ) min 4u min ATt S = = = x10 r ( 73) T. Same (match by sliding horizontally) 4 4uBTt = = r ( 0.5) min min Sy = K K V H = T = x10 b Γb K = r H 0.06 = 1 ( 73) min ( 100) 0. min = x ( 73) min Use distributed data and type curve to estimate aquifer properties Notice the curve can be used for a confined or unconfined aquifer Think about what parameters you can get from the data you have the exam may only ask you to report aquifer parameters Distance of fully penetrating observation well from pumping well = 190 Initial saturated thickness = 88 Pumping rate = 35 GPM 9