1. Water Movement in Soil and Rocks

2. Two Principles to Remember:

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

4. 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!”

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

6. 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

7. 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

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

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

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

11. 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.

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

13. 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.

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

15. 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

16. 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

17. 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……

18. 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

19. 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

20. 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

21. 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

22. 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.

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

24. 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

25. Also…..

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

27. 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

28. 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)

29. 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)

30. 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

31. 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

32. 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

33. 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%)

34. 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

35. 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

36. 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.

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

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

39. 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

40. 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

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

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

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

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

48. 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. “

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

50. E. Field Methods for Determining Permeability 2. Johnson Permeameter

51. 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

52. 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.

53. 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

55. 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.

56. 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)

57. 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

58. 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 )

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

62. III. Flow Nets

63. 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.

64. 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

65. 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

66. FLOW NETS

67. 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.

68. 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

69. 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

70. 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

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

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