water movement in soil and rocks. two principles to remember:

Water Movement in Soil and Rocks


Two Principles to Remember:


1. Darcy’s Law



1. Darcy’s Law

2. Continuity Equation:mass in = mass out + change in storage


“my name’s Bubba!”

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I. Critical in Engineering and Environmental GeologyA. Dams, Reservoirs, Levees, etc.

“ Pore Pressure”


I. Critical in Engineering and Environmental GeologyA. Dams, Reservoirs, Levees, etc.B. Groundwater Contamination

Landfills Leaking UndergroundStorage Tanks

SurfaceSpills


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




Porosity PermeabilityPermeability (def) the ease at which water can move through rock or soil

Porosity (def) % of total rock that isoccupied by voids.


B. The Bernoulli EquationA Demonstration:

http://home.earthlink.net/~mmc1919/venturi.html


B. The Bernoulli EquationA Demonstration:

Bernoulli's Principle states that as the speed of a moving fluid increases, the pressure within the fluid decreases.

http://home.earthlink.net/~mmc1919/venturi.html


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 = v2/2g + z + P/ρg

Where: h = total hydraulic head (units of length)v = velocityg = gravitational constantz = elevation above some datumP = pressure (where P = ρg*Δh) ρ = fluid density






h = v2/2g + z + P/ρg

Where: h = total hydraulic head (units of length)v = velocityg = gravitational constantz = elevation above some datumP = pressure (where P = ρg*Δh) ρ = fluid density

A quick problem……

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= v2/2g + z + P/g

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= v2/2g + z + P/g

(1*10-6 m/s)2 + +0.75 m + 1500 {(kg-m)/s2}m2

2 * 9.80 m/s2 9.80 m/s2 * 1.02 103 kg/m3

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= v2/2g + z + P/g

(1*10-6 m/s)2 + +0.75 m + 1500 {(kg-m)/s2}m2

2 * 9.80 m/s2 9.80 m/s2 * 1.02 103 kg/m3

5.10 * 10-14 m + 0.75 m + 0.15 m = 0.90 m = hydraulic head






h = v2/2g + z + P/ρg Total Head = velocity head + elevation head + pressure head h = zero + z + Ψ

Where: h = total hydraulic headz = elevation headΨ = pressure head


C. Darcy‘s LawHenri Darcy (1856)

Developed an empirical relationship of the discharge of water through porous mediums.


C. Darcy‘s Law

1. The experiment

K

II. Water Flow in a Porous MediumC. Darcy‘s Law

2. The results• unit discharge α permeability• unit discharge α head loss• unit discharge α hydraulic gradient

Also…..


2. The equation

v = Ki


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






v = K dh

dl

Darcy’s Law:

v = Ki






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


3. The Limits

Equation assumes ‘Laminar Flow’; which is usually the case for flow through soils.

C. Darcy‘s Law4. Some Representative Values for Hydraulic Conductivity


D. 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 = 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

Example Problem:

Q* dl= K

A dh



2. Falling Head Permeameter

More common for fine grained soils



2. Falling Head Permeameter

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. “


1. Double Ring Infiltrometer


2. Johnson Permeameter


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

0.01

0.1

1

0 1 2 3 4 5 6 7 8 9 10

Time (s)

h/h

oHzorslev Method

Example Problem:

K = r2 ln(L/R) 2LT0.37


initial change


Example Problem:

K = r2 ln(L/R) 2LT0.37


initial change


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


initial change


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


4. Pump Test also referred to as the Thiem Method

K = Q* ln(r1/r2) π(h1

2 – h22)

K = Q* ln(r1/r2) π(h1

2 – h22)

III. Flow Nets

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.

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

d2h + d2h = 0 gives the rate of change of dx2 dy2 h in 2 dimensions

• 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.

d2h + d2h = 0 gives the rate of change of dx2 dy2 h in 2 dimensions

FLOW NETS

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.

Q’ = Kph f

Where:Q’ = Discharge per unit depth of

flow net (L3/t/L)K = Hydraulic Conductivity (L/t)p = number of flow tubesh = head loss (L)f = number of equipotential drops

Q’ = Kph f

Where:Q’ = Discharge per unit depth of flow net (L3/t/L)K = Hydraulic Conductivity (L/t) = 1 * 10-4 m/sp = number of flow tubesh = head loss (L)f = number of equipotential drops

50 m32 m

Q’ = (1 * 10-4 m/s)(5)(18m) = Kph = 1 * 10-3 m3/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/sp = number of flow tubes = 5h = head loss (L) = 18 mf = number of equipotential drops = 9

50 m32 m

Derivations

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

K = Q* ln(r2/r1) π*(h2

2 – h12)

water movement in soil and rocks. two principles to remember:

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