WATER IN SOILS. I.Water – A unique substance A. Polar vs. Nonpolar Molecules.

WATER IN SOILS

I. 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

Potentiometer ResistanceBlock

NuclearGage

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 Grainsb. 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

HygroscopicWater--

--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

to2. Field Capacity: -1/3 Bar

D. Hygroscopic Water

1. Held by Adhesion a. Greater than -31 Bars

saturation0 bar

E. Water Availability vs. Texture

1. Greatest in Loamy Soils

2. Least in Sandy and Clayey Soils

F. Water Use in USA1. 83% Agriculture 2. Irrigation

a. Great Benefits – Great Problemsb. Mining Groundwater

ex: Ogallala Aquiferc. Salinizationd. Waterlogging of Soil

G. Soil Drainage1. Color

a. Oxidation State of IronFe 2+ <> Fe 3+ + e-

b. Organic MatterWet Soil Preserves Organics

c. Gleyingd. Mottling

2. Fragipan Soils

a. Can Cause Wetness

Well drained soil,Ferric iron

High organic matter

Gleyed soil

Mottled soil

H. Vegetation1. Hydrophilic Plants

a. Cyprusb. Cattailsc. Willowsd. Reeds

2. Plants Requiring Good Drainage

a. Oak-Hickory Biomeb. Pinesc. Most Grasses

Part II

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

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. Darcy‘s LawHenri Darcy (1856)

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


B. Darcy‘s Law

1. The experiment

K

II. Water Flow in a Porous MediumB. 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






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

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

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. 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 = 12.276 ft3 for 255 minutes

Find the hydraulic conductivity in units of ft per minute

= 0.0481 ft3/min

Example Problem:

Q* dl= K

A dh

0.0481 ft3/min

Example Problem:

Q* dl= K

A dh

0.0481 ft3/min



2. Falling Head Permeameter

More common for fine grained soils



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


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

E. Field Methods for Determining Permeability

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)

WATER IN SOILS. I.Water – A unique substance A. Polar vs. Nonpolar Molecules.

Documents

bar slide

adhesion slide

stability slide

pressure slide

wetness slide

gleyed soil slide

mottled soil slide

waterlogging of soil