WATER IN SOILS I Water A unique substance

WATER IN SOILS

I. Water – A unique substance A. Nonpolar vs. Polar Molecules - + “Electrical charge of molecule is uniform in all directions” “water is a polar substance” Strong Surface Tension Strong Capillary Action “Electrical charge of molecule is not uniform in all directions”

I. • • • Water – A unique substance High Surface Tension High Heat of Vaporization High Boiling Point Strong Capillary Action Expansion During Freezing High Freezing Point

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……

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. Soil Drainage 1. Color a. Oxidation State of Iron Fe 2+ <> Fe 3+ + eb. Organic Matter Wet Soil Preserves Organics c. Gleying d. Mottling

Well drained soil, Ferric iron

High organic matter

Gleyed soil

Mottled 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

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 (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 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) 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 v. L = 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 cm dl = 100 cm 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 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 “dh” “dl” Q = A K dh dl Q* dl= K A dh

Example Problem: = 0. 0481 ft 3/min Given: • Soil 6 inches diameter, 8 inches thick. • Hydraulic head = 16 inches • Flow of water = 12. 276 ft 3 for 255 minutes Find the hydraulic conductivity in units of ft per minute “dh” “dl” Q = A K dh dl Q* dl= K A dh

Example Problem: “dh” “dl” Q* dl= K A dh 0. 0481 ft 3/min

Example Problem: Q* dl= K A dh 0. 0481 ft 3/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 minutes per inch 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 = r 2 ln(L/R) 2 LT 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) 2 LT 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. 682 2 0. 38 0. 432 3 0. 21 0. 239 4 0. 12 0. 136 5 0. 068 6 0. 045 7 0. 023 8 0. 011 9 0 0. 000

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) 2 LT 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) 2 LT 0. 37 K = (0. 083 ft)2 ln(10 ft/ (0. 083 ft) 2(10 ft)(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) 2 LT 0. 37 K = (0. 083 ft)2 ln(10 ft/ (0. 083 ft) 2(10 ft)(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/r 2) π(h 12 – h 22)

K = Q* ln(r 1/r 2) π(h 12 – h 22)