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Water Movement in Soil and Rocks

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Two Principles to Remember:

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

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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!”

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

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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

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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

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

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II. Water Flow in a Porous Medium A. Goal: Determine the permeability of the engineering material

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II. Water Flow in a Porous Medium A. Goal: Determine the permeability of the engineering material Porosity Permeability

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

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

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

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

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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

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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

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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……

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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

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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

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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

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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

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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.

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

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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

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Also…..

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II. Water Flow in a Porous Medium C. Darcy‘s Law 2. The equation v = Ki

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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

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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)

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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)

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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

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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

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

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Darcy’s Law: The exposed truth: these are only APPARENT velocities and discharges Q L = A K dh n e dl v L = 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%)

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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

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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 =-KdhV =-Kdh n e dl dl V = 1 * 10 -6 cm/sec V L = 4.55 * 10 -6 cm/sec 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

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II. Water Flow in a Porous Medium C. Darcy‘s Law 3. The Limits Equation assumes ‘Laminar Flow’; which is usually the case for flow through soils.

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

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

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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

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Example Problem: Q = A K dh dl Q* dl= K A dh 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

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Example Problem: Q* dl= K A dh

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Example Problem: Q* dl= K A dh

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

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

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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. “

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

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E. Field Methods for Determining Permeability 2. Johnson Permeameter

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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

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Example Problem: 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 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.

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Time since Injecti on (sec)H (ft)h/ho 00.881.000 10.60.682 20.380.432 30.210.239 40.120.136 50.060.068 60.040.045 70.020.023 80.010.011 900.000 Hzorslev Method

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Example Problem: 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 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.

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Example Problem: 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 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 = (0.083 ft) 2 ln(10 ft/ (0.083 ft) 2(10ft)(2.3 sec)

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Example Problem: 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 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 = (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

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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 1 2 – h 2 2 )

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

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III. Flow Nets

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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.

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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. Continuity Equation: mass in = mass out + change in storage d 2 h + d 2 h = 0 gives the rate of change of dx 2 dy 2 h in 2 dimensions

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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

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FLOW NETS

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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.

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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

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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 50 m32 m

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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 50 m32 m

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

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

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