Water Movement in Soil and
Rocks

Water Movement in Soil and
Rocks
Two Principles to Remember:

Water Movement in Soil and
Rocks
Two Principles to Remember:
1. Darcy’s Law

Water Movement in Soil and
Rocks
Two Principles to Remember:
1. Darcy’s Law
2. Continuity Equation: mass in = mass out + change in storage
“my name’s
Bubba!”

Water Movement in Soil and Rocks
I. Critical in Engineering and Environmental Geology
A. Dams, Reservoirs, Levees, etc.
“ Pore Pressure”

Water Movement in Soil and Rocks
I. Critical in Engineering and Environmental Geology
A. Dams, Reservoirs, Levees, etc.
B. Groundwater Contamination
Landfills
Leaking Underground
Storage Tanks
Surface
Spills

Water Movement in Soil and Rocks
I. Critical in Engineering and Environmental Geology
A. Dams, Reservoirs, Levees, etc.
B. Groundwater Contamination
C. Foundations
- Strength and Stability

I. Critical in Engineering and Environmental Geology
A. Dams, Reservoirs, Levees, etc.
B. Groundwater Contamination
C. Foundations
- Strength and Stability

II. Water Flow in a Porous Medium

II. Water Flow in a Porous Medium

II. Water Flow in a Porous Medium
Porosity (def) % of total rock that is occupied by voids.
Permeability (def) the ease at which water can move through rock or soil

II. Water Flow in a Porous Medium
B. The Bernoulli Equation
A Demonstration :

II. Water Flow in a Porous Medium
B. The Bernoulli Equation
A Demonstration :
Bernoulli's Principle states that as the speed of a moving fluid increases, the pressure within the fluid decreases.

II. Water Flow in a Porous Medium
B. The Bernoulli Equation
1. Components of Bernoulli
Total Energy = velocity energy + potential energy + pressure energy

II. Water Flow in a Porous Medium
B. The Bernoulli Equation
1. Components of Bernoulli
Total Energy = velocity energy + potential energy + pressure energy
Total Head = velocity head + elevation head + pressure head

II. Water Flow in a Porous Medium
B. The Bernoulli Equation
1. Components of Bernoulli
Total Energy = velocity energy + potential energy + pressure energy
Total Head = velocity head + elevation head + pressure head h = v 2 /2g + z + P/ ρg
Where: h = total hydraulic head (units of length) v = velocity g = gravitational constant z = elevation above some datum
P = pressure (where P = ρg*Δh)
ρ = fluid density

II. Water Flow in a Porous Medium
B. The Bernoulli Equation
1. Components of Bernoulli
Total Energy = velocity energy + potential energy + pressure energy
Total Head = velocity head + elevation head + pressure head h = v 2 /2g + z + P/ ρg
Where: h = total hydraulic head (units of length) v = velocity g = gravitational constant z = elevation above some datum
P = pressure (where P = ρg*Δh)
ρ = fluid density
A quick problem……

At a place where g = 9.80 m/s 2 , the fluid pressure is 1500 N/m 2 , the distance above a reference elevation is 0.75 m, and the fluid density is 1.02 103 kg/m 3 .
The fluid is moving at a velocity of 1* 10 -6 m/s.
Find the hydraulic head at this point.
h= v 2 /2g + z + P/
 g

At a place where g = 9.80 m/s2, the fluid pressure is 1500 N/m2, the distance above a reference elevation is 0.75 m, and the fluid density is 1.02 103 kg/m3 .
The fluid is moving at a velocity of 1* 10-6 m/s.
Find the hydraulic head at this point.
h= v 2 /2g + z + P/
 g
(1* 10 -6 m/s) 2 + +0.75 m + 1500 {(kg-m)/s 2 }m 2
2 * 9.80 m/s2 9.80 m/s2 * 1.02 10 3 kg/m 3

At a place where g = 9.80 m/s2, the fluid pressure is 1500 N/m2, the distance above a reference elevation is 0.75 m, and the fluid density is 1.02 103 kg/m3 .
The fluid is moving at a velocity of 1* 10-6 m/s. Gravity is 9.8 m/s2.
Find the hydraulic head at this point.
h= v 2 /2g + z + P/
 g
(1* 10 -6 m/s) 2 + +0.75 m + 1500 {(kg-m)/s 2 }m 2
2 * 9.80 m/s2 9.80 m/s2 * 1.02 10 3 kg/m 3
5.10 * 10 -14 m + 0.75 m + 0.15 m = 0.90 m = hydraulic head

II. Water Flow in a Porous Medium
B. The Bernoulli Equation
1. Components of Bernoulli
Total Energy = velocity energy + potential energy + pressure energy
Total Head = velocity head + elevation head + pressure head h = v 2 /2g + z + P/ ρg
Total Head = velocity head + elevation head + pressure head h = zero + z + Ψ
Where: h = total hydraulic head z = elevation head
Ψ = pressure head

II. Water Flow in a Porous Medium
C. Darcy‘s Law
Henri Darcy (1856)
Developed an empirical relationship of the discharge of water through porous mediums.

II. Water Flow in a Porous Medium
C. Darcy‘s Law
1. The experiment
K

II. Water Flow in a Porous Medium
C. Darcy‘s Law
2. The results
• unit discharge α permeability
• unit discharge α head loss
• unit discharge α hydraulic gradient

Also…..

II. Water Flow in a Porous Medium
C. Darcy‘s Law
2. The equation v = Ki where v = specific discharge (discharge per cross sectional area) (L/T)
* also called the Darcy Velocity
* function of the porous medium and fluid

Darcy’s Law: v = Ki where v = specific discharge (discharge per unit area)
(L/T)
K = hydraulic conductivity (L/T); also referred to as coefficient of permeability i = hydraulic gradient, where i = dh/dl (unitless variable)

Darcy’s Law: v = Ki where v = specific discharge (discharge per unit area)
(L/T)
K = hydraulic conductivity (L/T); also referred to as coefficient of permeability i = hydraulic gradient, where i = dh/dl (unitless variable)

Darcy’s Law: v = Ki where v = specific discharge (discharge per unit area)
(L/T)
K = hydraulic conductivity (L/T); also referred to as coefficient of permeability i = hydraulic gradient, where i = dh/dl (unitless variable) v = K dh dl

Darcy’s Law: v = Ki where v = specific discharge (discharge per unit area)
(L/T)
K = hydraulic conductivity (L/T); also referred to as coefficient of permeability i = hydraulic gradient, where i = dh/dl (unitless variable) v = K dh dl
If Q = VA, then
Q = A K dh dl

Darcy’s Law:
The exposed truth: these are only APPARENT velocities and discharges v = K dh dl
Q = A K dh dl
Q = VA
Vs.

Darcy’s Law:
The exposed truth: these are only APPARENT velocities and discharges v
L
= K dh n e dl
Q
L
= A K dh n e dl
Where n e effective porosity
V
L
= ave linear velocity (seepage velocity)
Q
L
= ave linear discharge (seepage discharge)
Both of these variables take into account that not all of the area is available for fluid flow
(porosity is less than 100%)

Find the specific discharge and average linear velocity of a pipe filled with sand with the following measurements.
K = 1* 10 -4 cm/s dh = 1.0
dl = 100
Area = 75 cm 2
Effective Porosity = 0.22

Find the specific discharge and average linear velocity of a pipe filled with sand with the following measurements.
K = 1* 10 -4 cm/s dh = 1.0
dl = 100
Area = 75 cm 2
Effective Porosity = 0.22
V
L
=-Kdh n e dl
V = 1 * 10 -6 cm/sec
V
L
= 4.55 * 10 -6 cm/sec
V =-Kdh dl
How much would it move in one year?
4.55 * 10 -6 cm * 3.15 * 10 7 sec * 1 meter = 1.43 meters for V
L sec year 100 cm
0.315 m for V

Equation assumes ‘Laminar Flow’; which is usually the case for flow through soils.

C. Darcy‘s Law
4. Some Representative Values for Hydraulic Conductivity

II. Water Flow in a Porous Medium
D. Laboratory Determination of Permeability

II. Water Flow in a Porous Medium
D. Laboratory Determination of Permeability
1. Constant Head Permeameter
Q = A K dh dl
Q* dl= K
A dh

Example Problem:
Given:
•Soil 6 inches diameter, 8 inches thick.
•Hydraulic head = 16 inches
•Flow of water = 766 lbs for 4 hrs, 15 minutes
•Unit weight of water = 62.4 lbs/ft3
Find the hydraulic conductivity in units of ft per minute
Q = A K dh dl
Q* dl= K
A dh

Example Problem:
Q* dl= K
A dh

Example Problem:
Q* dl= K
A dh

II. Water Flow in a Porous Medium
D. Laboratory Determination of Permeability
2. Falling Head Permeameter
More common for fine grained soils

II. Water Flow in a Porous Medium
D. Laboratory Determination of Permeability
2. Falling Head Permeameter

E. Field Methods for Determining Permeability
In one locality: “ Perk rates that are less than 15 minutes per inch or greater than
105 are unacceptable measurements. “

E. Field Methods for Determining Permeability
1. Double Ring Infiltrometer

E. Field Methods for Determining Permeability
2. Johnson Permeameter

E. Field Methods for Determining Permeability
1. Slug Test (Bail Test) also referred to as the Hzorslev Method
K = r 2 ln(L/R)
2LT
0.37
Where: r = radius of well
R = radius of bore hole
L = length of screened section
T
0.37
= the time it take for the water level to rise or fall to 37% of the initial change

Example Problem:
A slug test is performed by injecting water into a piezometer finished in coarse sand. The inside diameter of both the well screen and well casing is 2 inches. The well screen is 10 feet in length. The data of the well recovery is shown below. Determine K from this test.
K = r 2 ln(L/R)
2LT
0.37
Where: r = radius of well
R = radius of bore hole (well casing)
L = length of screened section
T
0.37
= the time it take for the water level to rise or fall to 37% of the initial change

Hzorslev Method
Time since
Injecti on
(sec) H (ft) 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
1
0.1
0.01
0 1 2 3 4 5
Tim e (s)
6 7 8 9 10

Hzorslev Method
1
0.1
0.01
0 1 2 3 4 5
Tim e (s)
6 7 8 9 10

Example Problem:
A slug test is performed by injecting water into a piezometer finished in coarse sand. The inside diameter of both the well screen and well casing is 2 inches. The well screen is 10 feet in length. The data of the well recovery is shown below. Determine K from this test.
K = r 2 ln(L/R)
2LT
0.37
Where: r = radius of well
R = radius of bore hole (well casing)
L = length of screened section
T
0.37
= the time it take for the water level to rise or fall to 37% of the initial change

Example Problem:
A slug test is performed by injecting water into a piezometer finished in coarse sand. The inside diameter of both the well screen and well casing is 2 inches. The well screen is 10 feet in length. The data of the well recovery is shown below. Determine K from this test.
K = r 2 ln(L/R)
2LT
0.37
K = (0.083 ft) 2 ln(10 ft/ (0.083 ft)
2(10ft)(2.3 sec)
Where: r = radius of well
R = radius of bore hole (well casing)
L = length of screened section
T
0.37
= the time it take for the water level to rise or fall to 37% of the initial change

Example Problem:
A slug test is performed by injecting water into a piezometer finished in coarse sand. The inside diameter of both the well screen and well casing is 2 inches. The well screen is 10 feet in length. The data of the well recovery is shown below. Determine K from this test.
K = r 2 ln(L/R)
2LT
0.37
K = (0.083 ft) 2 ln(10 ft/ (0.083 ft)
2(10ft)(2.3 sec)
K = 7.18 * 10 -4 ft/s
K = 62.0 ft/day
Where: r = radius of well
R = radius of bore hole (well casing)
L = length of screened section
T
0.37
= the time it take for the water level to rise or fall to 37% of the initial change

E. Field Methods for Determining Permeability
4. Pump Test also referred to as the Thiem Method
K = Q* ln(r
1
π(h
1
2
/r
– h
2
)
2
2 )

K = Q* ln(r
1
π(h
1
2
/r
– h
2
)
2
2 )

III. Flow Nets

III. Flow Nets
A. Overview
• one of the most powerful tools for the analysis of groundwater flow.
• provides a solution to the Continuity Equation for 2-D, steady state, boundary value problem.

III. Flow Nets
A. Overview
• one of the most powerful tools for the analysis of groundwater flow.
• provides a solution to the Continuity Equation for 2-D, steady state, boundary value problem.
d 2 h + d 2 h = 0 gives the rate of change of dx 2 dy 2 h in 2 dimensions
Continuity Equation: mass in = mass out + change in storage

• Composed of 2 sets of lines
– equipotential lines (connect points of equal hydraulic head)
– flow lines (pathways of water as it moves through the aquifer.
d 2 h + d 2 h = 0 gives the rate of change of dx 2 dy 2 h in 2 dimensions

III. Flow Nets
B. To solve, need to know:
– have knowledge of the region of flow
– boundary conditions along the perimeter of the region
– spatial distribution of hydraulic head in region.

Q’ = Kph f
Where:
Q’ = Discharge per unit depth of flow net (L3/t/L)
K = Hydraulic Conductivity (L/t) p = number of flow tubes h = head loss (L) f = number of equipotential drops

32 m 50 m
Q’ = Kph f
Where:
Q’ = Discharge per unit depth of flow net (L3/t/L)
K = Hydraulic Conductivity (L/t) = 1 * 10 -4 m/s p = number of flow tubes h = head loss (L) f = number of equipotential drops

32 m 50 m
Q’ = ( 1 * 10 -4 m/s)(5)(18m) = Kph = 1 * 10 -3 m 3 /s/m thickness
9 f
Where:
Q’ = Discharge per unit depth of flow net (L3/t/L)
K = Hydraulic Conductivity (L/t) = 1 * 10 -4 m/s p = number of flow tubes = 5 h = head loss (L) = 18 m f = number of equipotential drops = 9

Derivations
• These are extra slides in the case you want to see how the equations are created, or derived…..

K = Q* ln(r
2
π*(h
2
2
/r
1
)
– h
1
2 )