### Groundwater Flow to Wells

```Groundwater Flow to Wells
I. Overview
A. Water well uses
1. Extraction
2. Injection
I. Overview
A. Water well uses
B. Terms
1. Cone of depression
2. Drawdown
I. Overview
A. Water well uses
B. Terms
C. Goals
1. Compute dh/dt given knowledge
of the properties of the aquifer
2. Determine the properties of the
aquifer based on the rate of dh/dt.
I. Overview
A. Water well uses
B. Terms
C. Goals
D. General Assumptions
General Assumptions
General Assumptions (continued)
I. Overview
A. Water well uses
B. Terms
C. Goals
D. General Assumptions
II. Theis Method
II. Theis Method
The aquifer is confined on the top and bottom
There is no source of recharge to the aquifer
The aquifer is compressible, and water is released instantaneously
from the aquifer as the hydraulic head is lowered.
The well is pumped at a constant rate.
II. Theis Method
B. The Equations
II. Theis Method
B. The Equations
ho -ht = Q* * wu
4πT
u = r2*S
4Tt
s = ho -ht
THEIS CURVE
II. Theis Method
C. Examples (with known values)
A well is located in an aquifer with a hydraulic conductivity of 15 m/d, storativity
is 0.005, aquifer thickness is 20 m, and the pumping of the water well is occurring
at a rate of 2725 m3/d. What is the drawdown at a distance of 7 m from the well
after 1 day of pumping?
ho -ht = Q* * wu
4πT
u = r2*S
4Tt
II. Theis Method
B. The Equations
C. Examples (with known values)
D. Examples (with unknown values)
THEIS CURVE
DRAWDOWN DATA
Problem: A well in a confined aquifer was pumped at a rate of 42,400 ft3/d
for 500 minutes. The aquifer is 48 ft. thick. Time drawdown data from an
observation well located 824 ft away yields the following data (see previous
slide of drawdown data).
Find T, K, and S.
III. Jacob Straight Line Method
A. Overview
B. Conditions
C. The Equation
D. Example
T = 2.3Q
4πΔh
S = 2.25T*t0
r2
III. Jacob Straight Line Method
D. Example
T = 2.3Q
4πΔh
S = 2.25T*t0
r2
Problem: A well in a confined aquifer was pumped at a rate of 42,400 ft3/d
for 500 minutes. The aquifer is 48 ft. thick. Time drawdown data from an
observation well located 824 ft away yields the following data (see previous
slide of drawdown data).
Find T, K, and S.
IV.
A.
B.
C.
Distance Drawdown Method
Overview
Equations
Example
T = 2.3Q
2πΔh
S = 2.25T*t
r02
D is ta n c e D ra w d o w n M e th o d
16
14
12
D ra w d o w n (ft)
10
8
6
4
2
0
1
10
100
D is ta n c e (ft)
1000
IV. Distance Drawdown Method
C. Example
D is ta n c e D ra w d o w n M e th o d
16
14
T = 2.3Q
2πΔh
12
S = 2.25T*t
r02
D ra w d o w n (ft)
10
8
6
4
2
0
1
10
100
D is ta n c e (ft)
A well is pumping 77,000 ft3/d, and has observational wells located 10, 40, 150,
300, and 400 ft away from the pumping well. After 0.14 days of pumping, the
Following drawdowns were observed in the observation wells (see graph).
Determine T (ft2/d) and S of the aquifer.
1000
V.
Hzorslev Method
(Slug or Bail Test)
K = r2*ln(L/R)
2Lt0.37
Hzorslev Method
Time
since
injection
h
h/ho
0
0.88
1.000
1
0.6
0.682
2
0.38
0.432
3
0.21
0.239
4
0.12
0.136
5
0.06
0.068
6
0.04
0.045
7
0.02
0.023
8
0.01
0.011
9
0
0.000
Hzorslev Method
h/ho
1
0.1
0.01
0
1
2
3
4
5
Tim e (s)
6
7
8
9
10
V.
Hzorslev Method
(Slug or Bail Test)
K = r2*ln(L/R)
2Lt0.37
h/ho
1
0.1
0.01
0
1
2
3
4
5
6
7
8
9
Tim e (s)
A slug test is performed by lowering a metal cylinder into a piezometer that is
screened in coarse sand. The inside of the bore hole has a radius of 0.500 ft, and
the inside radius of the piezometer is 0.083 ft. The screened section of the well is
10 ft. The well recovery data is shown via tables and the respective graph.
Determine the Hydraulic Conductivity of the aquifer.
10
VI. Intersecting Pumping Cones and Well
Interference
A. General Example
A.
Bounded Aquifers
1. Impermeable boundary
A.
Bounded Aquifers
VII. Recovery of Pumping Wells
A. Purpose
B. Example
VII. Recovery of Pumping Wells
Time(min)
Elevation (ft)
0
520
0.2
520.2
0.4
520.3
0.8
520.6
1
521
2
522
3
523
4
524
5
525
VII. Recovery of Pumping Wells
1 ft3 = 7.48 gallons
Well Recovery
Time(min)
Elevation (ft)
526
0
520
0.2
520.2
0.4
520.3
0.8
520.6
1
521
2
522
3
523
4
524
5
525
y = 1.0134x + 519.94
525
524
523
522
521
520
519
0
1
2
3
4
5
6
```