water in soils. i.water – a unique substance a. polar vs. nonpolar molecules
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B. How Water is Held in Soils 1. Cohesion
a. Forces Bonding Water to Itself 2. Adhesion
a. Bonds Water to Soil Grainsb. Measured in Bars
1 Bar = 1 Atmosphere~15 psi
Positive end of the water molecule bonds with the negatively charged clay particle (hydrogen bonding)
ppt----cohesion-----adhesion
HygroscopicWater--
--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
to2. Field Capacity: -1/3 Bar
D. Hygroscopic Water
1. Held by Adhesion a. Greater than -31 Bars
saturation0 bar
F. Water Use in USA1. 83% Agriculture 2. Irrigation
a. Great Benefits – Great Problemsb. Mining Groundwater
ex: Ogallala Aquiferc. Salinizationd. Waterlogging of Soil
G. Soil Drainage1. Color
a. Oxidation State of IronFe 2+ <> Fe 3+ + e-
b. Organic MatterWet Soil Preserves Organics
c. Gleyingd. Mottling
2. Fragipan Soils
a. Can Cause Wetness
H. Vegetation1. Hydrophilic Plants
a. Cyprusb. Cattailsc. Willowsd. Reeds
2. Plants Requiring Good Drainage
a. Oak-Hickory Biomeb. Pinesc. Most Grasses
Water Movement in Soil and Rocks
1. Darcy’s Law
2. Continuity Equation:mass in = mass out + change in storage
Two Principles to Remember:
“my name’s Bubba!”
Water Movement in Soil and Rocks
I. Critical in Engineering and Environmental GeologyA. Dams, Reservoirs, Levees, etc.
“ Pore Pressure”
Water Movement in Soil and Rocks
I. Critical in Engineering and Environmental GeologyA. Dams, Reservoirs, Levees, etc.B. Groundwater Contamination
Landfills Leaking UndergroundStorage Tanks
SurfaceSpills
Water Movement in Soil and Rocks
I. Critical in Engineering and Environmental GeologyA. Dams, Reservoirs, Levees, etc.B. Groundwater ContaminationC. Foundations
- Strength and Stability
I. Critical in Engineering and Environmental GeologyA. Dams, Reservoirs, Levees, etc.B. Groundwater ContaminationC. Foundations
- Strength and Stability
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 PermeabilityPermeability (def) the ease at which water can move through rock or soil
Porosity (def) % of total rock that isoccupied by voids.
II. Water Flow in a Porous Medium
B. Darcy‘s LawHenri Darcy (1856)
Developed an empirical relationship of the discharge of water through porous mediums.
II. Water Flow in a Porous MediumB. Darcy‘s Law
2. The results• unit discharge α permeability• unit discharge α head loss• unit discharge α hydraulic gradient
II. Water Flow in a Porous MediumB. Darcy‘s Law
2. The equation
v = Kiwhere 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
Q = A K dh
dl
Vs.
v = K dh
dlQ = VA
Darcy’s Law:
The exposed truth: these are only APPARENT velocities and discharges
QL = A K dh
ne dlvL = K dh
ne dl
Where ne effective porosity VL = ave linear velocity (seepage velocity) QL = ave linear discharge (seepage discharge)
Both of these variablestake 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/sdh = 1.0dl = 100Area = 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/sdh = 1.0dl = 100Area = 75 cm2
Effective Porosity = 0.22
VL =-Kdh V =-Kdh nedl 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 MediumB. 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
1. Constant Head Permeameter
Q = A K dh
dlQ* dl= K
A dh
Example Problem:
Q = A K dh
dlQ* dl= K
A dh
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
= 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. Slug Test (Bail Test) also referred to as the Hzorslev Method
K = r2 ln(L/R) 2LT0.37
Where:r = radius of wellR = radius of bore holeL = length of screened sectionT0.37 = the time it take for the water level to rise or fall to 37% of the
initial change
Example Problem:
K = r2 ln(L/R) 2LT0.37
Where:r = radius of wellR = radius of bore hole (well casing)L = length of screened sectionT0.37 = the time it take for the water level to rise or fall to 37% of the
initial change
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 wellscreen is 10 feet in length. The data of the well recovery is shown below. Determine K from this test.
Time since
Injection
(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
0.01
0.1
1
0 1 2 3 4 5 6 7 8 9 10
Time (s)
h/h
o
Example Problem:
K = r2 ln(L/R) 2LT0.37
Where:r = radius of wellR = radius of bore hole (well casing)L = length of screened sectionT0.37 = the time it take for the water level to rise or fall to 37% of the
initial change
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 wellscreen is 10 feet in length. The data of the well recovery is shown below. Determine K from this test.
Example Problem:
K = r2 ln(L/R) 2LT0.37
Where:r = radius of wellR = radius of bore hole (well casing)L = length of screened sectionT0.37 = the time it take for the water level to rise or fall to 37% of the
initial change
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 wellscreen is 10 feet in length. The data of the well recovery is shown below. Determine K from this test.
K = (0.083 ft)2 ln(10 ft/ (0.083 ft) 2(10ft)(2.3 sec)
Example Problem:
K = r2 ln(L/R) 2LT0.37
Where:r = radius of wellR = radius of bore hole (well casing)L = length of screened sectionT0.37 = the time it take for the water level to rise or fall to 37% of the
initial change
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 wellscreen is 10 feet in length. The data of the well recovery is shown below. Determine K from this test.
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
E. Field Methods for Determining Permeability
4. Pump Test also referred to as the Thiem Method
K = Q* ln(r1/r2) π(h1
2 – h22)