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

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

3. 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)

4. 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

5. 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)

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

7. 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 = 𝑥 10 − =4.6𝑥 10 −4

8. Well Function (Table 4.4.1 in the Book)
Interpolate

9. 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 = 𝑥 10 − =4.6𝑥 10 −4
Calculate drawdown, s:
Find W(u) in table
𝑠= 𝑄 4𝜋𝑇 𝑊 𝑢 = 𝜋 =1.42 𝑚
𝑊 𝑢 =𝑊 4.6𝑥 10 −4 =7.12

10. Cyprus

11. Cyprus

12. Cyprus Pump Test

13. Pumping Well

14. Observation Well

15. 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 , McGraw-Hill, 1979.

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

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

19. 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 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

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

21. 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 = 𝑚 3 /ℎ𝑟 4𝜋 𝑚 𝑚𝑝 =79.6 𝑚 2 ℎ𝑟
𝑆= 4𝑇 𝑟 2 𝑡𝑢 𝑚𝑝 = 𝑚 2 /ℎ𝑟 𝑚𝑖𝑛 60 𝑚𝑖𝑛/ℎ𝑟 1 =2.7𝑥 10 −5

22. 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 =− − ln 𝑢 +𝑢− 𝑢 2 2𝑥2! + 𝑢 3 3𝑥3! − 𝑢 4 4𝑥4! +⋯
𝑊 𝑢 ≈− − ln 𝑢
Good for small u (u < 0.001)
𝑠≈ 𝑄 4𝜋𝑇 − − ln 𝑢
𝑠≈ 𝑄 4𝜋𝑇 − − 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 𝑇𝑡 𝑟 2 𝑆

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

24. 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

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

26. 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 = 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)
𝑆=2.7𝑥 10 −5

27. 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.

28. Head Recovery After Pumping

29. 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 𝑇𝑡 𝑟 2 𝑆 − 𝑙𝑜𝑔 𝑇𝑡′ 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)

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

31. 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

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

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