Flow to Wells – 3 Unsteady Flow to a Well in a Confined Aquifer

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Groundwater Hydraulics Daene C. McKinney
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Presentation transcript:

Flow to Wells – 3 Unsteady Flow to a Well in a Confined Aquifer Groundwater Hydraulics Daene C. McKinney

Summary Unsteady flow to a well in a confined aquifer Theis method for pumping test Jacob method for pumping test Recovery Test

Unsteady Flow in a Confined Aquifer Ground surface Confined aquifer Q h0 Confining Layer b r h(r) Pumping well Bedrock Governing equation 𝛻∙𝑇𝛻ℎ=S 𝜕ℎ 𝜕𝑡 1 𝑟 𝜕 𝜕𝑟 𝑟 𝜕ℎ 𝜕𝑟 = 𝑆 𝑇 𝜕ℎ 𝜕𝑡 With initial conditions ℎ 𝑟,𝑡=0 = ℎ 0 (head is constant everywhere) And boundary conditions (head remains constant at infinity) ℎ→ ℎ 0 𝑎𝑠 𝑟→∞ 𝑄= 𝑙𝑖𝑚 𝑟→0 2𝜋𝑟 𝑇 𝜕ℎ 𝜕𝑟 𝑙𝑖𝑚 𝑟→0 𝑟 𝜕ℎ 𝜕𝑟 = 𝑄 2𝜋𝑇 (continuity at the well)

Unsteady Flow to a Well in a Confined Aquifer Solution ℎ= ℎ 0 − 𝑄 4𝜋𝑇 𝑟 2 𝑆/4𝑇𝑡 ∞ 𝑒 −𝜂 𝜂 𝑑𝜂 The integral is a function of its lower limit 𝑢= 𝑟 2 𝑆 4𝑇𝑡 The integral is known as the exponential integral and is found in tables 𝑊 𝑢 = 𝑢 ∞ 𝑒 −𝜂 𝜂 𝑑𝜂 The drawdown is given as 𝑠(𝑟,𝑡)=ℎ 0 −ℎ= 𝑄 4𝜋𝑇 𝑊(𝑢) Theis Equation

Well Function U vs W(u) 1/u vs W(u) Log – Log scale Log – Log scale 𝑊 𝑢 = 𝑢 ∞ 𝑒 −𝜂 𝜂 𝑑𝜂 𝑢= 𝑟 2 𝑆 4𝑇𝑡 𝑠= 𝑄 4𝜋𝑇 𝑊(𝑢)

Theis solution of drawdown vs time 𝑠= 𝑄 4𝜋𝑇 𝑊(𝑢)

Example - Theis Equation Ground surface Confined aquifer Q h0 Confining Layer b r h(r) Pumping well Bedrock Q = 1500 m3/day T = 600 m2/day S = Ssb= 4 x 10-4 Find: - Drawdown 1 km from well - After 1 year Calculate u: 𝑢= 𝑟 2 𝑆 4𝑇𝑡 = 1000 2 4𝑥 10 −4 4 600 365 =4.6𝑥 10 −4

Well Function (Table 4.4.1 in the Book) Interpolate

Example - Theis Equation Ground surface Confined aquifer Q h0 Confining Layer b r h(r) Pumping well Bedrock Q = 1500 m3/day T = 600 m2/day S = Ssb= 4 x 10-4 Find: - Drawdown 1 km from well - After 1 year Calculate u: 𝑢= 𝑟 2 𝑆 4𝑇𝑡 = 1000 2 4𝑥 10 −4 4 600 365 =4.6𝑥 10 −4 Calculate drawdown, s: Find W(u) in table 𝑠= 𝑄 4𝜋𝑇 𝑊 𝑢 = 1500 4𝜋 600 7.12 =1.42 𝑚 𝑊 𝑢 =𝑊 4.6𝑥 10 −4 =7.12

Cyprus

Cyprus

Cyprus Pump Test

Pumping Well

Observation Well

Example - Pump Text Pump test performed in a well pumping from a sandy aquifer. An observation well was located r = 1000 m from the pumping well. Originally, the water level in the well was at 20 m above mean sea level. The constant discharge during the test was Q = 1000 m3/hr. Use the Theis Method to determine the aquifer characteristics. Ground surface Bedrock Confined aquifer h0 = 20 m Confining Layer b r1 = 1000 m h1 Q Pumping well Bear, J., Hydraulics of Groundwater, Problem 11-4, pp 539-540, McGraw-Hill, 1979.

Pump Test Data Plot of s vs t s Data points t

Well Function U vs W(u) 1/u vs W(u) We can use this one.

Pump Test Analysis – Theis Method Ground surface Bedrock Confined aquifer h0 = 20 m Confining Layer b r1 = 1000 m h1 Q Pumping well 𝑠= 𝑄 4𝜋𝑇 𝑊(𝑢) 𝑢= 𝑟 2 𝑆 4𝑇𝑡 𝑡= 𝑟 2 𝑆 4𝑇 1 𝑢 𝑠= 𝑄 4𝜋𝑇 𝑊(𝑢) 𝑡= 𝑟 2 𝑆 4𝑇 1 𝑢 Plot 1 Plot 2 Make 2 plots Compute 𝑇= 𝑄 4𝜋 𝑊 𝑠 𝑚𝑝 𝑆= 4𝑇 𝑟 2 𝑡𝑢 𝑚𝑝 𝑊 𝑚𝑝 , 𝑠 𝑚𝑝 𝑡 𝑚𝑝 , 𝑢 𝑚𝑝 Select Match Point

Match Point W = 1, 1/u = 1 s = 1 m, t = 5 min

Theis Method Match Point (W)mp = 1, (1/u)mp = 1 (u = 1) Ground surface Bedrock Confined aquifer h0 = 20 m Confining Layer b r1 = 1000 m h1 Q Pumping well Match Point (W)mp = 1, (1/u)mp = 1 (u = 1) (s)mp = 1 m, (t) = 5 min Q = 1000 m3/hr Observation well = 1000 m from pumping well 𝑇= 𝑄 4𝜋 𝑊 𝑠 𝑚𝑝 = 1000 𝑚 3 /ℎ𝑟 4𝜋 1 1 𝑚 𝑚𝑝 =79.6 𝑚 2 ℎ𝑟 𝑆= 4𝑇 𝑟 2 𝑡𝑢 𝑚𝑝 = 4 79.6 𝑚 2 /ℎ𝑟 1000 2 5 𝑚𝑖𝑛 60 𝑚𝑖𝑛/ℎ𝑟 1 =2.7𝑥 10 −5

Jacob Approximation 𝑊 𝑢 = 𝑢 ∞ 𝑒 −𝜂 𝜂 𝑑𝜂 𝑢= 𝑟 2 𝑆 4𝑇𝑡 𝑠= 𝑄 4𝜋𝑇 𝑊(𝑢) 𝑊 𝑢 = 𝑢 ∞ 𝑒 −𝜂 𝜂 𝑑𝜂 𝑢= 𝑟 2 𝑆 4𝑇𝑡 𝑠= 𝑄 4𝜋𝑇 𝑊(𝑢) 𝑊 𝑢 =−0.577216− ln 𝑢 +𝑢− 𝑢 2 2𝑥2! + 𝑢 3 3𝑥3! − 𝑢 4 4𝑥4! +⋯ 𝑊 𝑢 ≈−0.577216− ln 𝑢 Good for small u (u < 0.001) 𝑠≈ 𝑄 4𝜋𝑇 −0.577216− ln 𝑢 𝑠≈ 𝑄 4𝜋𝑇 −0.577216− ln 𝑟 2 𝑆 4𝑇𝑡 𝑠≈ 𝑄 4𝜋𝑇 −𝑙𝑛 1.78 − ln 𝑟 2 𝑆 4𝑇𝑡 = 𝑄 4𝜋𝑇 ln 4𝑇𝑡 1.78 𝑟 2 𝑆 𝑠≈ 2.3𝑄 4𝜋𝑇 log 10 2.25𝑇𝑡 𝑟 2 𝑆

∆𝑠= 𝑠 2 − 𝑠 1 = 2. 3𝑄 4𝜋𝑇 log 10 2. 25𝑇 𝑡 2 𝑟 2 𝑆 − 2. 3𝑄 4𝜋𝑇 log 10 2 Jacob Approximation 𝑠= 2.3𝑄 4𝜋𝑇 log 10 2.25𝑇𝑡 𝑟 2 𝑆 𝑠 1 = 2.3𝑄 4𝜋𝑇 log 10 2.25𝑇 𝑡 1 𝑟 2 𝑆 𝑠 2 = 2.3𝑄 4𝜋𝑇 log 10 2.25𝑇 𝑡 2 𝑟 2 𝑆 ∆𝑠= 𝑠 2 − 𝑠 1 = 2.3𝑄 4𝜋𝑇 log 10 2.25𝑇 𝑡 2 𝑟 2 𝑆 − log 10 2.25𝑇 𝑡 1 𝑟 2 𝑆 ∆𝑠= 2.3𝑄 4𝜋𝑇 log 10 𝑡 2 𝑡 1 s1 s2 t2 t1 1 log cycle If times are taken as one log cycle, then t2 = 10t1, and log 10 10𝑡 1 𝑡 1 =1 ∆𝑠= 2.3𝑄 4𝜋𝑇 log 10 10𝑡 1 𝑡 1 ∆𝑠 ∆𝑠= 2.3𝑄 4𝜋𝑇 𝑇= 2.3𝑄 4𝜋∆𝑠

Jacob Approximation 𝑠= 2.3𝑄 4𝜋𝑇 log 10 2.25𝑇𝑡 𝑟 2 𝑆 1= 2.25𝑇 𝑡 0 𝑟 2 𝑆 𝑆= 2.25𝑇 𝑡 0 𝑟 2 Fitted line to late time points s = 0, t = t0

s2 ∆𝑠 s1 1 log cycle t1 t2 s = 0, t = t0 = 9 min

Jacob Approximation t0 = 9 min s2 = 4.7 m s1 = 2.4 m Ds = 2.3 m Pump Test Analysis – Jacob Method Jacob Approximation t0 = 9 min s2 = 4.7 m s1 = 2.4 m Ds = 2.3 m s2 = 4.7 ∆𝑠=2.3 s1 = 2.4 1 log cycle 𝑇= 2.3𝑄 4𝜋∆𝑠 = 2.3 1000 𝑚 3 /ℎ𝑟 4𝜋(2.3 𝑚) t1 = 100 t2 = 1000 𝑇=79.6 𝑚 2 /ℎ𝑟 s = 0, t = t0 = 9 min 𝑆= 2.25(79.6 𝑚 2 /ℎ𝑟 )(0.15 ℎ𝑟) 1000 2 𝑆=2.7𝑥 10 −5

Recovery Tests We can estimate T from analysis of the recovery of a well after it is shutdown. If a well is pumped at a rate Q and then shutdown, the head will recover from its lowest level hT’ at time T’ when pumping stopped to attain a value h’ > hT’ at times t counted from the time of shutdown (t’ = 0 corresponds to t = T’) t = 0 t = 0 t = T’ t h = H h = hT’ h’ t’ t’ = 0 t’ = t – T’ Pumping period Recovery period If H is the initial head value of head before pumping started, then s = H – h’ is the residual drawdown.

Head Recovery After Pumping

Assume that pumping goes on at the same rate, Q, for t > T’ but that at t = T’ a recharge well injects water at rate –Q, so that the net discharge of the aquifer is zero for t > T’. 𝑠 ′ = 𝐻−ℎ − ℎ− ℎ ′ = 𝑄 4𝜋𝑇 𝑊 𝑢 −𝑊(𝑢′) 𝑢= 𝑟 2 𝑆 4𝑇𝑡 𝑢 ′ = 𝑟 2 𝑆 4𝑇 𝑡 ′ , 𝑡 ′ =𝑡−𝑇′ For small u 𝑠 ′ = 2.3𝑄 4𝜋𝑇 𝑙𝑜𝑔 10 2.25𝑇𝑡 𝑟 2 𝑆 − 𝑙𝑜𝑔 10 2.25𝑇𝑡′ 𝑟 2 𝑆 = 2.3𝑄 4𝜋𝑇 𝑙𝑜𝑔 10 𝑡 𝑡′ This is a straight line if we plot s’ (arithmetic scale) vs t/t’ (log scale)

Well pumping Q = 2500 m3/day was shut down after 240 min

Recovery Test Method s2 = 0.8 m ∆𝑠=0.4 𝑚 s1 = 0.4 m t1 = 10 t2 = 100 1 log cycle t1 = 10 t2 = 100

Recovery Test Example 𝑠 ′ = 2.3𝑄 4𝜋𝑇 𝑙𝑜𝑔 10 𝑡 𝑡′ 𝑠 ′ = 2.3𝑄 4𝜋𝑇 𝑙𝑜𝑔 10 𝑡 𝑡′ 𝑠 1 ′ = 2.3𝑄 4𝜋𝑇 𝑙𝑜𝑔 10 𝑡 𝑡′ 1 𝑠 2 ′ = 2.3𝑄 4𝜋𝑇 𝑙𝑜𝑔 10 𝑡 𝑡′ 2 ∆𝑠 ′ = 𝑠 2 ′ − 𝑠 1 ′ = 2.3𝑄 4𝜋𝑇 𝑙𝑜𝑔 10 𝑡 𝑡 ′ 2 𝑡 𝑡 ′ 1 = 2.3𝑄 4𝜋𝑇 (1) 𝑠𝑖𝑛𝑐𝑒 𝑡 𝑡 ′ 2 𝑡 𝑡 ′ 1 =10 𝑇= 2.3(2500 𝑚 3 𝑑𝑎𝑦 ) 4𝜋(0.40 𝑚) =1140 𝑚 2 /𝑑𝑎𝑦 𝑇= 2.3𝑄 4𝜋∆𝑠′

Summary Unsteady flow to a well in a confined aquifer Theis method for pumping test Jacob method for pumping test Recovery Test