Chap 1 and 2. Introduction

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Soil Physics I. Introduction

1. INTRODUCTION TO SOIL PHYSICS

1.1. Definition of soil physics

Soil physics is the study of the physical properties of the soil and the

relation of these soil physical properties to agricultural, environmental and

engineering uses.

Other definitions include:-

the study of the state and transport of matter and energy in the soil.

the branch of soil science which deals with the physical properties of the

soil, as well as with measurement, prediction, and control of the physical

processes taking place in and through the soil.

The definition implies that:-

it is quantitative and mathematical in nature, and

it is primarily concerned with the fundamental properties of soils.

1.2. Areas of concern in soil physics

Over the years, the primary motivation of soil physics has been to obtain a

greater understanding of the physical processes of soil to ultimately aid in

agriculture, particularly in the production crops. Lately, this emphasis has

been expanded and supplemented to develop a greater understanding of :-

i) transport and retention of water soil or porous media

ii) transport of solutes, and

iii) heat flow in soil,

iv) hydrology and what flow,

v) various environmental and pollution concerns,

vi) spatial and temporal variability of soil properties in the

landscape, etc.

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AgEn 2002: Soil Physics II. Soil Physical Properties

2. SOIL PHYSICAL PROPERTIES

2.1 Definition and Physical Constituents of Soil

Soil refers to the weathered and fragmented outer layer of the earthsland surface. It is formed initially from disintegration and decomposition of

rocks by physical and chemical processes, and is influenced by the activity

and accumulated residues of numerous biological species.

Soil constitutes:-

Mineral matter quartz, feldspar, silicates,

Organic matter plant and animal residues,

Soil-water solution giving the liquid phase,

Soil air N2, O2, water vapor, CO2, etc

Soil is a heterogeneous, polyphasic, particulate, disperse, and porous

system, in which the interfacial area per unit volume can be very large. The

disperse nature of the soil and its consequent interfacial activity gives rise

to such phenomena as adsorption of water and chemicals, ion

exchange, adhesion and cohesion, swelling and shrinking,

dispersion and flocculation, andcapillarity.

States of soil: three phases of ordinary nature exist in the soil as well:

The solid phase- consists of the soil matrix (skeleton)

The liquid phase- consists of soil water, which always contains

dissolved substances so that it should properly called the soil

solution, and

The gaseous phase- the soil atmosphere.

The most important physical properties of soil are texture, structure,

surface area, thermal capacity, aeration, etc.

Q2.1

a) Suggest some situations where only two of the three phases occur

in soil, and

b) Is it possible for soil to consist of one phase?

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The spaces between the solid particles are called soil pores. Soil pores are

cavities of different shape and size determined by the shape, size and

arrangement of the solid particles. Normally, the pores form a continuous

space through out the soil. Although larger pores are connected by narrowchannels, there are strictly speaking no discrete pores.

Q2.2

A surface soil in good conditions for plant growth may have the following

volume fractions: mineral matter (0.45), organic matter 0.05, soil solution

0.2-0.3 and soil air 0.3 0.2.

a)- what is the volume fraction of the pores in this surface soil?

b) what is the volume fraction of liquid phase required to saturate the

soil?

Q2.3

In the above problem, consider 1 m3 of the surface soil.

a. What is the mass of water required to saturate it?

b. What is the mass of air in it when completely dry?

c. What is the mass of the solid phase?

d. What is the mass of the soil when it is saturated? And when dry?

e. What is the mass of the soil when the volume fractions of liquid

phase is only 0.25?

2.2. Soil texture

This is an expression of the predominant size or size ranges, of the

particles. Quantitatively, soil texture refers to the relative proportions of the

various sizes of particles in a given soil. It is static soil physical property,

The traditional method of characterizing particle sizes in soil is to divide the

particles into size ranges known as separates, and soil separate consists of

mineral particles between designated maximum and minimum diameters.

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The International Soil Science Society Classification System

mm m

Coarse sand 2-0.2 2000- 200

Fine sand 0.2- 0.02 200 20Silt 0.02- 0.002 20 2

Clay < 0. 002 < 2

The USDA classification

Gravel > 2 mm

Sand 2-0.05 mm

Very coarse sand 2 1.0 mm

Coarse sand 1.0 0.5 mm

Medium sand 0.5 0.25 mm

Fine sand 0.25 0.10 mm

Very fine sand 0.10 0.05 mm

Silt 0.05 0.002 mm

Clay < 0.002 mm

Clay and humus form finer particles of the soil. They form the seat of

soil activity. Because of their extremely small particle size, clay and humus

have a large surface area per unit weight. Also they exhibit surface charges

that attract negatively and positively charged ions and water.

Since soils are a mixture of different size particles soils are classified using

the so-called soil textural triangle.

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The horizontal lines mark percentages of clay (by mass). The lines angle

upwards to the right mark percentages of silt. The lines angled upwards to

the left mark percentages of sand. So the central point in the loam range is

about 20% clay, 40% silt and 40% sand.

Measuring soil texture textural analysis of soil

There are two stages of particle size analysis. First stage is the mechanical

analysis using sieves. This is used to separate soil particles whose diameter

is greater than 0.05 mm. The second stage is the process of separating

particles whose diameter is less than 0.05 mm. This based on the principles

of Stokes law (sedimentation).

According to Stokes law, for a soil dispersed, and mixed with water (soil

suspension), settling velocity of individual particles depends on particle

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diameter, and the forces acting on soil particle are gravitation, buoyancy

and drag forces, all are functions of particle size.

Particle size determination can be made using either the pipette method orthe hydrometer method.

Pipette method is considered to be the most accurate it is based on

direct sampling of the density of the solution. Using a pipette, samples of

the suspension (usually 20 cm3) are withdrawn at a given depth after

various periods have elapsed after initiation of sedimentation.

At a depth L below the surface of the suspension and at time t, all

particles whose terminal velocity v is greater than L t will have passed

below this level e.g silt passes through but clay remains.

Stokes law of Sedimentation

The concept of mechanical equilibrium in the gravitational force field isused in particle size analysis by sedimentation. Forces acting on soil

particle are gravitation, buoyancy and drag forces. When a soil particle is

suspended in water, it first accelerates due to the influence of gravity. As

the velocity increases, the drag force exerted by the water on the particle

due to the viscosity of the water also increases. The net force (weight

minus buoyant force), however remains constant when the drag force has

become equal to the net static force, the resultant force is zero and thus

also the acceleration. Then particle continues to move at a constant

velocity also called terminal velocity.

When a solid spherical particle of radius r and density , moves through a

viscous liquid, the net static force acting downward is

( )grF lss

=3

3

4

Where l is the density of the liquid.

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The drag force on a spherical particle under laminar conditions was shown

by Stokes to be

vrFd 6=

Where is the viscosity of the liquid.

v

is the terminal velocity of the particle the ve sign indicates that dF is

directed oppositely to v

(and of course also to sF ). Laminar conditions will

persist as long as the Reynolds number which is calculated as

12 =vr

Re remains smaller than about 0.1, which is true for particles

less than 50 m

The sum of forces is zero when the sedimentation velocity has reached the

value for which

( )34

6 03

s d

s lF F r g r v

= = = r r

The terminal velocity, also known as settling velocity or sedimentation

velocity, is then

( )

9

2 12

= srg

v

Since the particles attain a terminal velocity almost instantly, the radius of

particles that travel a distance (S) during a time interval t is then

2

1

1 )(2

9

=

tg

sr

s

This sedimentation equation is based on a number of simplifying

assumption. When the equation is used for particle- size analysis, the

obtained radius is not actual radius but only an equivalent radius of the

particle. For non-spherical particles such as clays, r is the radius of a

hypothetical spherical particle with the same theoretical settling velocity.

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Viscosity of water is strongly dependent on temperature and has units of

poises or milliPascal- seconds. Values of viscosity of water at varies

temperatures is presented as shown below.

We usually work at 20 oC in water and assume p = 2.65 g cm-3

Substituting thus values, check the SI units of sedimentation equation

sm

smkg

skg

smkg

mkgmsm

s

m===

11

2

11

322

..

.

..

....

Q2.4

Calculate the rate of fall in water at 20 oC of silt particle with a particle

diameter of 0.02 mm (i.e., 20 10-6) g. cm-3). Assume the density of the

particle is 2.65 g.cm-3 and the density of water is 1.0 g .cm-3

Q2.5

Calculate the terminal velocity corresponding to the upper size limit of theclay fraction 2 m , if the temperature is 20 oC and the particle density is

2650 kg-m-3.

Comparing this example with previous example the terminal velocity of clay

is 100 times lower than the medium sized silt particle having a diameter 10

times larger. Assuming that the terminal velocity (v) is attained almost

instantly, we can calculate the time needed for the particle to fall thorough

a distance L in the cylinder to be

Temperature

(o

C)

g/m.s.

(mPa-s)

0 1.78710 1.30820 1.00230 0.79840 0.65350 0.547

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g. What is the textural classification of the original soil sample

h. Justify the use of stakes law in the sedimentation method.

Q2.7Determine the sedimentation time for a soil particle having a diameter of 2

m and a particle density of 2.60 g/cm3 ( 2.60 g cm-3). Assume that the

sampling was made at the 0.10 m depth and at temperature of 20 0C.

Hydrometer method is less accurate but often used because of its

simplicity for taking measurements. It also depends on Stokes law.

Principle: Depth to the hydrometers center of buoyancy varies with the

density of the suspension and also with the particle size distribution. The

concentration, C, of the soil in suspension, in g/L, can be calculated from

the following equation:

+=

p

wws C

1

p - density of particle , w -density of water, s - density of suspension

The stem of the hydrometer is read in density units (g/L) on a scale of 0 to

60) or percentagesoC

C100when the initial concentration, C0, is equal to

0.04 Mg/m3 and p is 2.65 g cm-3

The sampling time is arbitrary, but geometric progression of time intervalsprovides a spread of data. For example, at 20 0C with the Bouyoucos

hydrometer, sampling times of 40 seconds and 8 hour result in

concentration of silt + clay and clay, respectively.

2.3. Specific surface area of soil (s, m2/g)

Surface area of soil particles is one important physical property soilaffecting soil for water and solute retention movement. Specific surface

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area depends of on shape of soil particles. For instance assuming spherical

shape (r=radius and =density). surface area (a) of soil is 24 r and its

mass (m) is [ ]34 3rV = . Thus the specific surface area of soil

rmas 3== . It shows that it is inversely proportional to radius. Surface

area of soil affects its physical and chemical properties and is largely

determined by amount of clay present in soil:

Q2.8. Find the mass of 2 mm soil particle

The surface area of the soil is estimated assuming:-

A spherical shape of diameter (d) is:

The ratio of surface area to volume is2

3/ 6

6v

da

d d

= =

The ratio of surface area to mass6 2.3

m

s

ad d

= =

Particle

Effective

Diameter

[cm] Mass (g)

Area

(cm2)

Specific

Surface area

(cm2 g-1)

Gravel 210-1 1.1310-2 1.310-1 11.1Sand 510-3 1.7710-7 7.910-5 444.4Silt 210-4 1.1310-11 1.310-7 11.1104Clay* 210-4 8.4810-15 6.310-8 7.4106

*Thickness = 10-7 cm

2.4. Soil Structure

Thisrefers to the arrangement and organization of soil particles in the soil,and the tendency of individual soil particles to bind together in aggregates.

The aggregation creates intra-aggregate and inter-aggregate pore space,

thereby changing flow paths for water, gases, solutes and pollutants. The

effects on plant growth operate through: aeration, soil compaction, water

relations, soil temperature.

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Soil texture determines the pore size distribution within peds or aggregates

while structure determines pore size distribution between peds or

aggregates.

Amount and type of clay, as well as the exchangeable ions on the clay (also

water acts as bridge between clay particles). Amount and type of organic

matter, since it provides food for soil fungi and bacteria and their secretion

of cementing agents (polysaccharides); presence of iron and aluminum

oxides (cementing agents); binding between organic and inorganic

compounds (aluminum oxides, cations, clays), vegetation: produces OM,

roots act as holding soil together, and protects soil surface.

Types of soil structure: single grained (windblown particles such as silt;

sand) highly erodable; massive (heavy clays); aggregated (ideal soil

structure). Characterization of soil structure: ( mostly qualitative since it is a

function of time).

Size : particles (particle size distribution); aggregates ( dry-

sieving; water stability test by wet sieving); porosity

Morphological : blocky, plately, prismatic

Physical pore size distribution water desorption method.

Stable aggregate is caused by OM (arid soils) or oxides (humid soils).

Structural stability or resistance of soil to disintegrative forces (rain,

cultivation, soil swelling), depends on type of organic matter and its

changing from low to high C/N as a result of microbial breakdown of OM.

Soil structure deterioration causes soil compaction, reduced gaseous

exchange between atmosphere and soil (aeration), and reduction in

infiltration

The aggregate stability of a soil is the resistance of the soil structure

against mechanical or physico-chemical destructive forces. Soil structure is

one of the main factors controlling plant growth by its influence on root

penetration. Soil temperature and gas diffusion water transport and

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seedling emergence and therefore it is an important soil characteristics to

farmers

2.5. Soil Layers

Soil is generated by the weathering of rock in layers that are often very

distinct. The bottom layer is the bedrock, or regolith. As this weathers,

usually by the action of wind and water, it breaks into smaller chunks in the

unconsolidated or C layer. This can be the gravel ground up by glaciers or

by simple freezing and thawing. Often the soil overlaying bedrock has been

deposited by wind, ice or water from bedrock much farther away. The B

layer is composed of particles that are mostly sand-sized or smaller. In this

layer weathering proceeds at a bit faster pace because smaller particles

have a much higher relative surface area upon which chemical can act.

Most of the plant root and biological activity occurs in the A layer above it.

Here weathering is faster yet, accelerated by the chemical and physical

action of life. It is here that the fine clay particles are produced in the

largest quantity. Because they are so small, they tend, along with

chemically soluble products such as carbonates, to be leached out of this

zone into the B layer. If the soil is biologically productive, as in a tall-grass

prairie, the top layer is almost entirely organic, consisting of root mats and

wilted leaves, some of which are decomposing

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O1 Undecomposed litterO Horizon

Organic Plant

Residues O2 Partly decomposed debris

A1 Zone of humus accumulation

A2 Zone of strongest leaching

A Horizon

Zone of

eluviaiton(leaching) A3 Transition to B horizon

B1 Transition to A horizon

B2 Zone of strongest depositionSolum,TrueSoil

B Horizon

Zone of

illuviation

(deposition) B3 Transition to C horizonRegolith,Weathered

Material

C Horizon

Parent MaterialC Unconsolidated rock

R Layer - Bedrock R Consolidated rock

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Sand Silt Clay

Property 0.05-2 mm 0.002-0.05


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are called micropores, whereas the much larger voids made by roots and

worms and cracking soil are called macropores.

Storage, availability and transport of soil solution and soil air are not nearlyas much dependent on the porosity as such, as on how the total pore space

is partitioned, the so-called pore size distribution. Soil pores differ in size

and shape as a result of textural and structural arrangements. Based on the

diameter at the narrowest point, pores may be classified as follows:

Macropores > 100 m

Mesopores 30 1000 m

Micropores < 30 m

Although these limits are arbitrary, the functioning of these classes is

roughly the following.

Macropores:- conduct water only during flooding pending rain, etc. They

soon drain after cessation of such water supply. These they affect

aeration and drainage,

Mesopores:- are effective in conducting water also after the macropores

have become empty, such as during non- ponding rain, redistribution

of water, etc.

Micropores:- The remaining soil water solution is retuned or moves very

slowly within the micropores. Part of this water can be taken by plant

roots.

Soil porosity consists of gas filled porosity or aeration porosity g and liquid

filled porosity, l , which is equal to the volume fraction of liquid, , For

optimum plant growth the aeration porosity should be at least 0.10 0.12

within two to three days after irrigation or heavy rainfall.

2.7. Clay minerals

Primary and secondary minerals

Minerals that have persisted with little change in composition since they

were extruded in molten lava (e.g. quartz, micas and feldspars) are knownas primary minerals. They are most prominent in the sand and silt fractions.

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Tetrahedron

oxygen atoms

silicon inside

Octahedron

hydroxyls

aluminum, iron or

magnesium inside


Other minerals, such as the silicates, clays and iron oxides, have been

formed by the breakdown and weathering of less resistant minerals as soil

formation progressed. These minerals are called secondary minerals. They

tend to dominate the clay and in some cases the silt fraction of soil.

Soil particles range in size from slightly larger than molecular size to

stones. The larger particles form the framework or skeleton of the soil,

while the smaller ones fill the space around the contact points and cover

the surfaces of the larger particles. The liquid phase in the soil pores is in

physical contact with the solid phase. At the contact surface, forces of

electrical nature cause phenomena such as swelling, shrinkage,

aggregation, flocculation and dispersion, which in turn influence transport

of air and soil solution. They are typical of soils containing clays and humus.

Clay minerals are formed of sheets of tetrahedral or octahedral basic

molecules, about 5 (10-10) m thick. In the 1:1 clays, one tetrahedral sheet

and one octahedral sheet bind tightly together in alternate layers to form a

larger particle. Water cannot penetrate between these sheets and can only

interact with the platelets at their outer edges and planes. In 2:1 clays,

each octahedral sheet is sandwiched between two tetrahedral sheets.

Water can be drawn in between the adjacent tetrahedral sheets, causing

the clay to swell and have an extremely large specific surface, Sv. The

orientations of the sheets from one clay particle to the next is generally

random.

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1:1 clay

tetrahedral sheet

octahedral sheet

tetrahedral sheet

octahedral sheet


The specific surface area of clay minerals, which governs many soil

properties, varies from one mineral to another. It is

Montmorillonite 700 800 m2 . g-1

Vermiculite 3000 500 m2 . g-1

Mica - 100 300 m2

. g-1

Kaolinite - 5 100 m2 . g-1

The thickness of clay mineral layers varies from mineral to mineral. It is

about 1 nm for Montmorillonite, illite and mica 0.7 nm for kaolinite. The

planar extensions (i.e. length and width) of clay minerals layers varies from

100 nm to 2 m. The ratio thickness/planar extension varies between about

0.5 10-3 and 10 10-3.

2.8 Surface charges of clay minerals

During the formation of clay minerals the tetrahedral and octahedral sheets

do not meet as already finished sheets. Rather, they grow together, unit by

unit. During growth, Si, Al and Mg are hardly ever present in the ideal ratios

required for the clay mineral. Therefore, if A13+- ions are present in excess,

they may occupy Si 4+ - sites in the crystal lattice. Similarly, Mg2+ - ions

may occupy sites of A13+ - ions if there is a shortage of Al3+ ions. This

tetrahedral sheet

octahedral sheet

tetrahedral sheet

tetrahedral sheetoctahedral sheet

tetrahedral sheet

water2:1 clay

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process leads to an excess negative charge in the crystal lattice, which

manifests itself as specific surface charge density.

Since overall electrical neutrality is always maintained in nature, the net

negative charge of a clay platelet is compensated by an equivalent positive

charge of cations, ions, such as Na+, K+, Ca2+ Mg2+, H+ which are either

adsorbed at the platelet surface or swarm in the surrounding solution.

These cations counteract the negative electrical charge of the clay platelet

towards its surroundings, and are, therefore, also called counter ions.

The charged surface of a clay platelet with its swarm of counter ions is

called a diffuse electrical double layer. The term double layer refers to the

spatial separation of the negatively charged clay platelet and positively

charged counter ions. The later can be divides into two layers:

a fixed layer, also called stern layer, and

a diffuse layer

The fixed layer consists of cations strongly adsorbed at the surface of the

clay platelet, where as the diffuse layer is the adjacent zone of more loosely

bound cations. The thickness of the fixed layer is little more than the

diameter of a single hydrated cation.

The extent of the double layer it is defined as the distance from the clay

surface at which the cation concentration reaches a unit form value or a

minimum. It is also the distance over which the electrical influence of the

clay platelet on its surroundings vanishes.

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Since the bulk solution is electrically neutral, it contains positive and

negative ions in equal concentration of ionic equivalents. In the double

layer, however, the anion concentration decreases towards the clay

surface. The difference between the total positive charge of the cations and

the total negative charge of the anions in the double layer is equal to the

negative charge of the clay platelet.

2.9. Particle density

Particle density is defined as the mass of soil particles divided by the

volume occupied by the solids, (i.e. excluding voids and water)

typical values for soils range from 2.5 - 2.8 g cm-3 with 2.65 g cm-3

being representative of many soils.

quartz the dominant soil mineral has a value of about 2.65 g cm-3

which is why this value is frequently given as representing all soils

(1.0 g cm-3)

organic matter (1.0 g cm-3) (is removed in determining the particle

density)

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Q2.10To determine s a surveyor took 82 g of powdered dry soil at random, put it

in a glass bottle slightly larger than 100 cm3, and added deaerated water at

room temperature. After ascertaining that all the air in the soil was driven

out, he filled the bottle to exactly 100 cm3. A bottle with which this can be

done with great accuracy is called a pychnometer. The mass of the water

and the soil was 151 g.

a. Calculate s

b. Why was it necessary to drive out all the air?

Characteristics of bulk soil

Many of the important transport and retention processes in the soil are

influenced strongly by the composite properties of the soil matrix, which

are sometimes called bulk soil properties. These properties are

commonly characterized with samples that contain many individual soilparticles, void spaces, and water films. Hence, the bulk soil properties are

sold to be volume averaged.

2.10. Volume and mass relationships in soil.

Solid phase - under field conditions, the solid phase occupies from 30

60% of the total soil volume. Ideally the solid phase occupies 50% of thesoil by volume.

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Liquid phase dynamic rather than stable, volume varies between 0 -

50%, and liquid phase varies in amount and chemical composition.

Gaseous phase is also dynamic and not stable

Mt = Ms +Mw + Mg = total mass of the soil

Vt = Vs + Vw + Vg = total volume of the soil

For instance an individual soil particle has a diameter of 2 mm. What is its

mass?

Mass = volume density

Volume =3933 10189.4001.0

3

4

3

4mr ==

Density = 2650 kg m-3

Mass of the particle (assuming spherical) = 1.11 X 10-2 g.

Dry bulk density, b , is calculated as

t

sb

V

M=

Dry bulk density varies from 1000 to 1800 kg m-3 but cannot be considered

static particular at the surface due to effects tillage, compaction, etc.

Organic soil bulk density varies between 800- 1000 kg m-3 .

Particle density this is calculated as

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s

sp

V

M=

For mineral soil it varies between 2600 2700 kg m-3. For most mineral

soil, it is taken as 2650 kg m

-3

(2.6 g cm

-3

). This is considered to be a staticproperty of soil.

Total porosity is an index of the relative volume of pores in the soil and

is calculated as:

Porosity ( )V

Vpores= where V is bulk volume of soil (total soil volume).

Porosity in soil varies between 0.3 (sand, silt) to 0.45 (clays) to 0.7 (peat),

and is largely determined by the soil bulk density.

t

gw

V

VV += = m3 of voids/m3 of soil

For mineral soils, its value generally is between 0.3 and 0.6 m3 m-3. For

coarse- textured soils, value of tend to be less than for fine- textured

soils, even though the average sizes of the pores are larger in the coarse-

textured soils. Total porosity tends to decrease with depth in the profile

due to compaction. Total porosity gives no information about the pore size

distribution.

Void ratio , (e) this is calculated as

s

wg

V

VVe

+= =

solidsm

voidsm3

3

=[-]

Void ratio expresses the relationship between the volumes occupied by

solids and by voids. Therefore, the volume of voids in a soil volume is the

sum of the volumes of the liquid and gaseous phases. It is used where the

soil is undergoing compaction, shrinking, or swelling and mostly in soil

engineering and mechanics.

Values ofe vary between 0.3 and 2.0 m3. m-3. Compacted soils tend to have

values less than one.

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Relation between void ratio and porosity is

The relation between volumetric water content( )v , saturation ratio ( )s , and

total porosity( ( )f is

fsv =

gw

ws

VV

V

+= = m3 of water/m3 voids. The ratio ranges from 0 1 m3 m-

3.

The relation between total porosity (f), bulk density b and particle density

m

is

w g t s

t t

V V V Vf

V V

+ = =

=

t

s

V

V1 =

s

s

t

s

M

M

V

V1 =

s

s

t

s

V

M

V

M1

1 b

p

f

=

The relation between volumetric water content, water content by weight,

bulk density and water density is

w

bwv

=

Soil wetness

Mass wetness, m also called

gravimetric water content at 105 0C dried at temperature little above

boiling point.

s

wm

mm= - maximum value of m ranges from 25 to 60%

Volume wetness sand 40-50%fs

w

t

w

VV

V

V

V

+==

Degree of saturation (S),sat

V

wa

wv

VV

VS

=

+==

This ranges from zero in dry soil to unity (or 100%) for a saturated soil.

Air filled porosity - this measures the relative air content of the soil, andas such is an important criterion of soil aeration

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was

a

t

aa

VVV

V

V

V

++==

Q2.11.

A soil having a bulk density of 1.2 g . cm-3 and a particle density of 2.65

g.cm-3 weigh 0.1 kg when sampled and 80 g after oven drying. Find values

of l , a , w and v . Calculate the equivalent depth (in meters) of water in

each meter of soil when sampled.

Volume of water in the soil is also expressed by equivalent depth

of water, De:

soil

e

soil

ele

DD

ADAD

VV

areaSurfaceofVolumewaterVolumeD ====

Equivalent depth of water per depth of soil

The quantity De is depth of water = De = zvw =

Where is the depth interval

Where n is the number of layers.

Q2.12

A 100 cm3 undisturbed soil sample has a mass of 162.5g. After drying at

105 oC, its mass is reduced to 132. 5g.

a. Find w , and d

b. Find gls and ,,

c. Is the soil well aerated?

I.24 BG. 2012

Water content

Soild

epth

Z

( ) ( )=

=

==ni

oi

ivi

z

veZdzzD

0


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d. Are the obtained values of gls and ,, true for every part of the

soil sample?

Q2.13.

An undisturbed soil core is 10 cm in diameter and 10 cm in length. The wetsoil mass is 1320 g. After oven drying the core, the dry soil mass is 1100 g.

The mineral density of the soil is 2.6 g cm-3. Calculate: a. Dry soil bulk

density, b. water content on a mass basis, c. water content on volume

basis, d. soil porosity, e. equivalent depth of water (cm) contained in a 1 m

soil profile, if the undisturbed core is representative of the 1 m soil depth.

Q2.14

Consider a 1.2 m depth of soil profile with 3 layers. The dry bulk density of

each layer (top, center, bottom) is 1.20, 1.35, and 1.48 g/cm3. The top 30

cm layer has a water content of 0.12 g/g, the center 50 cm layer has awater content of 0.18 g/g, and the bottom 40 cm layer has a water content

of 0.22 g/g.

a. what is the total amount of water I the whole profile in mm

b. How much water (mm) do you need to apply to bring the 1.2 m soil

profile to a volumetric water content of 0.35 cm3 cm-3.

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Documents

Chap 1 and 2. Introduction