Transcript Water in Soil

WATER IN SOILS
I.
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Water – A unique substance
High Surface Tension
High Heat of Vaporization
High Boiling Point
Strong Capillary Action
Expansion During Freezing
High Freezing Point
I.
Water – A unique substance
A. Nonpolar vs. Polar Molecules
“water is a
polar substance”
-
+
“Electrical charge of molecule
is uniform in all directions”
Strong Surface Tension
Strong Capillary Action
“Electrical charge of molecule
is not uniform in all directions”
II. Soil Porosity
II. Soil Porosity
A. Varies with Texture
1. Approximately 50% for Undisturbed
Soils
III. Nature of Soil Water
A. Water Table
1. Zone of Aeration
2. Zone of Saturation
Nuclear
Gage
Potentiometer
Resistance
Block
A thought experiment……
A thought experiment……
B. How Water is Held in Soils
1. Cohesion
a. Forces Bonding Water to Itself
2. Adhesion
a. Bonds Water to Soil Grains
(Positive end of the water molecule bonds with the negatively
charged clay particle (hydrogen bonding)
b. Measured in Bars
1 Bar = 1 Atmosphere
~15 psi
Hygroscopic
Water--
--water that is tightly bound to the soil
particle and requires large expenditure of
energy to remove it.
C. Water Available to Plants
1. Wilting Point: -15 Bars
to
2. Field Capacity: -1/3 Bar
D. Hygroscopic Water
1. Held by Adhesion
a. Greater than -31 Bars
saturation
0 bar
E. Water Availability vs.Texture
1. Greatest in Loamy Soils
2. Least in Sandy and Clayey Soils
F. Water Use in USA
1. 83% Agriculture
2. Irrigation
a. Great Benefits – Great Problems
b. Mining Groundwater
ex: Ogallala Aquifer
c. Salinization
d. Waterlogging of Soil
G. Soil Drainage
1. Color
a. Oxidation State of Iron
Fe 2+ <> Fe 3+ + eb. Organic Matter
Wet Soil Preserves Organics
c. Gleying
d. Mottling
2. Fragipan Soils
a. Can Cause Wetness
Well drained soil,
Ferric iron
High organic matter
Gleyed soil
Mottled soil
H. Vegetation
1. Hydrophilic Plants
a. Cyprus
b. Cattails
c. Willows
d. Reeds
2. Plants Requiring Good Drainage
a. Oak-Hickory Biome
b. Pines
c. Most Grasses
Part II
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
A. Goal: Determine the permeability of the
engineering material
II. Water Flow in a Porous Medium
A. Goal: Determine the permeability of the
engineering material
Porosity
Permeability
II. Water Flow in a Porous Medium
A. Goal: Determine the permeability of the
engineering material
Porosity
Porosity (def) % of
total rock that is
occupied by voids.
Permeability
Permeability (def) the ease at
which water can move through
rock or soil
II. Water Flow in a Porous Medium
B. 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
B. Darcy‘s Law
1. The experiment
K
II. Water Flow in a Porous Medium
B. Darcy‘s Law
2. The results
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unit discharge α permeability
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unit discharge α head loss
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unit discharge α 1 / hydraulic
gradient
Also…..
II. Water Flow in a Porous Medium
B. Darcy‘s Law
2. The equation
v = Ki
II. Water Flow in a Porous Medium
B. 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
B. Darcy‘s Law
4. Some Representative Values for Hydraulic Conductivity
Darcy’s Law:
The exposed truth: these are only APPARENT
velocities and discharges
Q = A K dh
dl
v = K dh
dl
Q = VA
Vs.
Darcy’s Law:
The exposed truth: these are only APPARENT
velocities and discharges
vL = K dh
ne dl
QL = A K dh
ne dl
Where ne effective porosity
VL = ave linear velocity (seepage velocity)
QL = 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 cm2
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 cm2
Effective Porosity = 0.22
VL =-Kdh
nedl
V =-Kdh
dl
V = 1 * 10-6 cm/sec
VL = 4.55 * 10-6 cm/sec
How much would it move in one year?
4.55 * 10-6 cm * 3.15 * 107 sec * 1 meter = 1.43 meters for VL
sec
year 100 cm
0.315 m for V
II. Water Flow in a Porous Medium
B. Darcy‘s Law
3. The Limits
Equation assumes ‘Laminar Flow’; which is usually the case for flow
through soils.
II. Water Flow in a Porous Medium
C. Laboratory Determination of Permeability
II. Water Flow in a Porous Medium
C. Laboratory Determination of Permeability
1. Constant Head Permeameter
Q = A K dh
dl
Q* dl= K
A dh
Example Problem:
= 0.0481 ft3/min
Given:
•Soil 6 inches diameter, 8 inches thick.
•Hydraulic head = 16 inches
•Flow of water = 12.276 ft3 for 255 minutes
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
0.0481 ft3/min
Example Problem:
Q* dl= K
A dh
0.0481 ft3/min
II. Water Flow in a Porous Medium
C. Laboratory Determination of Permeability
2. Falling Head Permeameter
More common for fine grained soils
II. Water Flow in a Porous Medium
C. Laboratory Determination of Permeability
2. Falling Head Permeameter
D. 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. “
D. Field Methods for Determining Permeability
1. Double Ring Infiltrometer
D. Field Methods for Determining Permeability
2. Johnson Permeameter
D. Field Methods for Determining Permeability
1. Slug Test (Bail Test)
also referred to as the Hzorslev Method
K = r2 ln(L/R)
2LT0.37
Where:
r = radius of well
R = radius of bore hole
L = length of screened section
T0.37 = 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 = r2 ln(L/R)
2LT0.37
Where:
r = radius of well
R = radius of bore hole (well casing)
L = length of screened section
T0.37 = 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
Hzorslev Method
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 = r2 ln(L/R)
2LT0.37
Where:
r = radius of well
R = radius of bore hole (well casing)
L = length of screened section
T0.37 = 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 = r2 ln(L/R)
2LT0.37
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
T0.37 = 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 = r2 ln(L/R)
2LT0.37
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
T0.37 = 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(r1/r2)
π(h12 – h22)
K = Q* ln(r1/r2)
π(h12 – h22)