WATER IN SOILS
Water – A unique substance A. Polar vs. Nonpolar Molecules
“water is a polar substance” Strong Surface Tension Strong Capillary Action
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 Resistance Block Potentiometer
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 b. 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
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+ + e- b. 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 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 Leaking Underground Storage Tanks Landfills 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 Permeability 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. 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 unit discharge α permeability unit discharge α head loss unit discharge α 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
The exposed truth: these are only APPARENT velocities and discharges 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.
The exposed truth: these are only APPARENT velocities and discharges Darcy’s Law: The exposed truth: these are only APPARENT velocities and discharges QL = A K dh ne dl vL = 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 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 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: Q = A K dh Q* dl= K dl A dh = 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. “
1. Double Ring Infiltrometer D. Field Methods for Determining Permeability 1. Double Ring Infiltrometer
D. Field Methods for Determining Permeability 2. Johnson Permeameter
1. Slug Test (Bail Test) D. Field Methods for Determining Permeability 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: K = r2 ln(L/R) 2LT0.37 Where: r = radius of well 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 Injection (sec) H (ft) h/ho 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 0.000
Hzorslev Method
Example Problem: K = r2 ln(L/R) 2LT0.37 Where: r = radius of well 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: K = r2 ln(L/R) 2LT0.37 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: K = r2 ln(L/R) 2LT0.37 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
4. Pump Test E. Field Methods for Determining Permeability also referred to as the Thiem Method K = Q* ln(r1/r2) π(h12 – h22)
K = Q* ln(r1/r2) π(h12 – h22)