Download - Soil Water Properties Lecture 1

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BAEN 464 BAEN 464 Irrigation & Drainage EngineeringIrrigation & Drainage Engineering

SoilSoil--PlantPlant--Water Water RelationshipsRelationshipsRelationshipsRelationships

CHAPTER 2 OF TEXTCHAPTER 2 OF TEXT&&

Special HandoutsSpecial Handouts

Soil PropertiesSoil Properties

SOIL WATER RELATIONS

•• NEED TO KNOW HOW MUCH WATER NEED TO KNOW HOW MUCH WATER IS IN THE SOIL IS IN THE SOIL

•• AND THE AVAILABILITY OF THE AND THE AVAILABILITY OF THE WATER TO PLANTSWATER TO PLANTS

•• LEADS TO DEFINTIONS OF:LEADS TO DEFINTIONS OF:

–– WATER CONTENT, AND WATER CONTENT, AND

–– SOIL WATER POTENTIAL (TENSION)SOIL WATER POTENTIAL (TENSION)

SOIL WATER PROPERTIESSOIL WATER PROPERTIES COMPOSITION OF AN UNSATURATED SOIL SAMPLECOMPOSITION OF AN UNSATURATED SOIL SAMPLE

SOIL SOIL PARTICLEPARTICLE

WATERWATER

PORE SPACEPORE SPACE

AIRAIR AIRAIR

SOLID MATERIALSOLID MATERIAL

PORE SPACEPORE SPACE

MINERAL

WATERWATER ORGANICORGANICMATTERMATTER

AIRAIR

WATERWATER

SOILSOIL

SOIL PHYSICAL PROPERTIESSOIL PHYSICAL PROPERTIES

VolumesVolumesVVv v = Volume Voids= Volume VoidsVVaa = Volume Air= Volume AirVVww = Volume Water= Volume WaterVVss = Volume Solids= Volume Solids

VVvv = V= Vaa + V+ Vww

MassMass

MMaa = Mass of air (zero)= Mass of air (zero)MMss = Mass of solids= Mass of solidsMMww = Mass of water= Mass of water

AIRAIR

WATERWATER

SOILSOIL


VolumeVolume

Porosity (Porosity (ΦΦ) = V) = Vvv/V/Vbb

Void Ratio (Void Ratio (ηη) = V) = Vbb/V/Vss

Saturation (S) = VSaturation (S) = Vww/V/Vvv; 0 ≤ S ≤ 1.0; 0 ≤ S ≤ 1.0

VVbb = V= Vvv + V+ Vss

VVbb = V= Vaa + V+ Vww + V+ Vss

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AIRAIR

WATERWATER

SOILSOIL


Mass and DensityMass and Density

Specific Gravity of Solids (ρp)

ρp = Ms/(Vs ρw)

SoilSoil ρρpp

MineralMineral 2.652.65ClayClay 2.702.70OrganicOrganic 2.602.60

AIRAIR

WATERWATER

SOILSOIL


Mass and DensityMass and Density

Bulk Density (ρb) Apparent Specific Gravity (As)

ρb = Ms/(Vb); As = ρb/ρw = Ms/(Vb ρw)

ONE CAN SHOW THAT ΦΦ = (1 – As/ρp) PROVE THISPROVE THIS

SAND: 0.05 SAND: 0.05 -- 2.0 mm2.0 mm

RELATIVE SIZES OF SOIL PARTICLESRELATIVE SIZES OF SOIL PARTICLES

SILT: 0.002 - 0.05 mm

CLAY: CLAY: < 0.002 mm< 0.002 mm

SOIL TEXTURAL TRIANGLESOIL TEXTURAL TRIANGLE

MASS WATER CONTENTMASS WATER CONTENT

WET SOILWET SOIL

θm = Mass of Water

Mass of Dry Soil

Weight of WaterWeight of Dry Soil

=

DRY SOILDRY SOILWET SOILWET SOILSAMPLESAMPLE

WET WEIGHTWET WEIGHTOF SOILOF SOIL

DRY WEIGHTDRY WEIGHTOF SOILOF SOIL

-- ==

WEIGHT OF WATER

DRY SOILDRY SOILSAMPLESAMPLE WATERWATER

VOLUMETRIC WATER CONTENTVOLUMETRIC WATER CONTENT

WET SOILWET SOIL

θθvv = = Volume of WaterVolume of Water

Bulk VolumeBulk Volume

DRY SOILDRY SOILWET SOILWET SOILSAMPLESAMPLE

WET WEIGHTWET WEIGHTOF SOILOF SOIL

-- ==

DRY WEIGHTDRY WEIGHTOF SOILOF SOIL

DRY SOILDRY SOILSAMPLESAMPLE WATERWATER

VOLUMEVOLUMEOF WaterOF Water

BULKBULKVOLUMEVOLUME

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AIRAIR

WATERWATERd = θv L

DEPTH OF WATER (d) IN A SOIL LAYERDEPTH OF WATER (d) IN A SOIL LAYER

LL

Bulk Volume (VBulk Volume (Vbb))VVb b = = ππ rr22 LL

SOILSOIL

rrWater Volume (VWater Volume (Vww))

VVw w = = ππ rr22 d d




ρρ = M= M /(V/(V ))

θθvv = V= Vww/(V/(Vbb) )

ρρbb = M= Mss/(V/(Vbb) )

ρρww M Mww/(V/(Vww) )

θθvv = V= Vww/(V/(Vbb) = (M) = (Mww//ρρww) / (M) / (Mss//ρρbb))

θθvv = (M= (Mww//ρρww) / (M) / (Mss//ρρbb) = ) = θθmm ρρbb // ρρww




θθvv = (M= (Mww//ρρww) / (M) / (Mss//ρρbb) = ) = θθmm ρρbb // ρρww

θθvv = = θθmm ρρbb // ρρw w = = θθmm AAss

EXAMPLE OF SOIL WATER PROPERTIES

A field soil sample prior to being disturbed has a volume of 80 cm3. The sample weighed 120 grams. After drying at 105 º C, the dry soil weighs 100 grams. What is the water content by weight? What is volumetric water content? What depth of water must be added to increase pthe volumetric water content of the top one-foot of soil to 0.30?

Given:Soil sample volume (Vb) = 80 cm3

Dry weight of soil sample (Ms) = 100 g

Wet weight of soil sample (Ms + Mw) = 120 g

Find:Find:

θθm m water content on a dry weight basiswater content on a dry weight basis

θθvv water content on a volume basiswater content on a volume basis


d depth of water d depth of water

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Solution:Solution:

θθm m = M= Mww / M/ Mss

MMww = 120 = 120 –– 100 = 20 g.100 = 20 g.ww gg

θθmm = M= Mww / M/ Ms s = {20g/100g} = 0.20 g of water/g = {20g/100g} = 0.20 g of water/g of soil of soil


Solution:Solution:

θθvv = V= Vww / V/ Vbb

and θand θv v = ρ= ρbb/ ρ/ ρww (θ(θmm) )

ρρb b = M= Mss/V/Vbb = {100g/80 cm= {100g/80 cm33} = 1.25 g/cm} = 1.25 g/cm33

θθvv = ρ= ρbb/ ρ/ ρww (θ(θmm) = 1.25/1.00 (0.20) ) = 1.25/1.00 (0.20)

= 0.25 cm= 0.25 cm3 3 of water per cmof water per cm33 of soilof soil


Solution:Solution:

Current depth of water in one foot of soilCurrent depth of water in one foot of soil

d = θd = θvv L L

d = θd = θvv L = 0.25 (12 in) = 3 inches of waterL = 0.25 (12 in) = 3 inches of water

The depth of water in the soil when θThe depth of water in the soil when θv v = 0.30 is = 0.30 is

d = θd = θvv L = 0.30 (12 in) = 3.60 inches of L = 0.30 (12 in) = 3.60 inches of waterwater

Thus, the depth of water to be added is 3.6 Thus, the depth of water to be added is 3.6 minus 3.0, or 0.6 inchesminus 3.0, or 0.6 inches

GRAVITATIONAL POTENTIAL

ψg = DISTANCE ABOVE AN ARBITRARY ELEVATION

FLOWFROM

Ψ g2 = Z2

DATUM

Ψ g1 = Z1

FROM HIGH TOLOWPOTENTIAL

UPWARD FORCE ( F1)

F1 = (2 π r) σ COS(β)

FORCE BALANCE

CAPILLARY FORCES

βContact Angle

H i ht f

rRadius ofCapillary Tube

PRESSURE = - h

DOWNWARD FORCE (F2)

F2 = - ρg (π r2 ) h

F1 - F2 = 0

(2 π r) σ COS(β) = ρ g (π r2 ) h

h = 2 σ COS(β) / ρg r

h = CONSTANT/ r

hHeight ofCapillary Rise

PRESSURE = 0

CAPILLARY FORCES

Contact Angle

r

Radius ofCapillary Tube

PRESSURE = - h

Attraction of glass and water causes water molecules to adhere to the glass.

hHeight ofCapillary Rise

PRESSURE = 0

Cohesive force between water molecules causes the tube to fill with water.

Water in the tube is belowatmospheric pressure.

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h = k / radius of tube

CAPILLARY FORCES

•WATER RISES HIGHER FOR SMALL DIAMETER TUBES

•WATER IN SMALL TUBES IS UNDER MORE “TENSION”

•HOW DOES THAT RELATE TO SOIL-WATER RELATIONS

h

h

Capillary pressure = pw – pa = k/r

NEARLY SATURATED SOIL

LARGE RADIUS (r)

SOIL PARTICLE

ATTRACTION OF SOIL PARTICLES FOR WATER --- (ADHESION)SURFACE TENSION OF WATER --- (COHESION)

RESULTS: WATER FILL PORES BY FORMING SATRUATED ZONES AROUND THE PARTICLES. AS SOIL DRIES AND MATRIC POTENTIAL INCREASES A CURVED WATER SURFACE DEVELOPS ON THE WATER BETWEEN THE SOIL PARTICLES.

SMALL MATRIC POTENTIAL(i.e. little capillary action)

WATER

DRY SOIL

SMALL RADIUS (r)

SOIL PARTICLE

WATER

LARGE MATRIC POTENTIAL(i.e. large capillary action)

SMALL RADIUS OF CURVATURE AS THE SOIL DRIES AND THE WATER MIGRATES INTO THE CORNERS OF THE PORE SPACE

H

SOIL

POROUSPLATE / MEMBRANE

1 2

DEVELOPING MOISTURE RELEASE CURVESUSING A HANGING WATER COLUMN

INITIALLY: SOIL IS SATURATED WITH A POSITIVE HEAD CAUSING DRAINAGE THROUGH THE SOIL

ψt2 > ψt1

MEMBRANE

SOIL


1 2 ψt = 0AT EQUILIBRIUM: FLOW STOPS BECAUSE OF THE CAPILLARY ACTION OF THE SOIL. FOR THIS CASE THE AVERAGE MATRIC POTENTIAL IS ZERO.

ψt2 = ψt1

HANGING WATER COLUMN FOR UNSATURATED CONDITIONS

POROUS PLATE REMAINS SATURATED AND PREVENTS AIR FROM ENTERING THE TUBE.

WATER FLOWS FROM THE SOIL SO THE SAMPLE IS

SOIL


1

H = 100 cm

ψt1 = 0

UNSATURATED

THE SOIL WATER IS UNDER A “TENSION”

WHEN FLOW STOPS THE TENSION IN THE SOIL EQUALS THE DIFFERENCE IN ELEVATION BETWEEN POINT 1 AND 2.

2

ψt2 = 0

HANGING WATER COLUMN FOR UNSATURATED CONDITIONS

WHAT IS THE MATRIC POTENTIAL

AT POINT 1?

ψt = ψg + ψm

SOIL

POROUSPLATE /

1ψt1 = 0

WHEN NO FLOW

ψt = 0

ψg = - 100 cm

ψm = - 100 cm

PLATE / MEMBRANE

2

ψt2 = 0

H = 100 cm

DATUM

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HANGING WATER COLUMN FOR MOISTURE RELEASE CURVES

SOIL

1

ψm1 = - 100 cm

DETERMINE THE VOLUMETRIC WATER

CONTENT:

MATCHING THE VOLUMETRIC WATER CONTENT TO THE MATRIC POTENTIA


2

H = 100 cm

CONTENT:

WET WEIGHT = 500 g

DRY WEIGHT = 400 g

BULK VOLUME = 300 cm3

θv = (500-400)/300 = 0.33

0.2

0.3

0.4

C W

AT

ER

CO

NT

EN

T MOISTURE RELEASE CURVE

Field Capacity

Permanent

Wilting Point

50 100 200 400 600 1000 2000 4000 6000 104 15,0000

0.1

SOIL WATER TENSION, cm of water

VO

LU

ME

TR

IC Wilting Point

SOIL WATER POTENTIALMEASURE OF WATER AVAILABILITY

COMMON UNITS OF PRESSURE AND HEAD AND THEIR EQUIVALENTS

Unit Pressure Equivalent

Water Head Equivalent

1 Atmosphere 101.3 kPa 1034 cm H2O

34 ft H2O

1.013 bar 76 cm Hg

101.3 bar 29.9 in Hg

101.3 cb

14.7 psi (lb/in2)

1 psi (lb/in2) 6.69 kPa 2.31 ft H2O

MOISTURE RELEASE CURVES

0.3

0.4

0.5

0.6 W

AT

ER

CO

NT

EN

T

PERMANENT

FREE /GRAVITATIONAL

WATER 0.34

FIELDCAPACITY

0.0

0.1

0.2

10 100 1000 10000 105

SOIL WATER TENSION, cm

VO

LUM

ET

RIC

0.16

WILTINGPOINTAVAILABLE WATER

UNAVAILABLE WATER

FIELD CAPACITY & SATURATION FIELD CAPACITY & SATURATION -- RESERVOIR ANALOGY RESERVOIR ANALOGY

SATURATIONSATURATION FIELD CAPACITYFIELD CAPACITYDRAIN TUBEDRAIN TUBE

GRAVITATIONAL WATERGRAVITATIONAL WATER

RESERVOIRRESERVOIR

Spillway to protect the reservoir is like the water that drains due to gravity.

AVAILABLE WATER -- RESERVOIR ANALOGY

SATURATIONSATURATION FIELD CAPACITYFIELD CAPACITY

DRAIN TUBEDRAIN TUBE

PUMPPUMP

FLOWFLOW

FLOWFLOW

When reservoir is full it is easy to extract water, so we get large flows

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AVAILABLE WATER -- RESERVOIR ANALOGY


PUMPPUMP

FLOWFLOW

As water level drops it is more difficult to extract water, so flow is smaller

UNAVAILABLE WATER -- RESERVOIR ANALOGY


PUMPPUMP

PERMANENTPERMANENT

UNAVAILABLE WATERUNAVAILABLE WATER

No matter what we do there is some water that we can not extract.

PERMANENTPERMANENTWILTING POINTWILTING POINT

WATER STATUS -- RESERVOIR ANALOGY


GRAVITATIONAL WATERGRAVITATIONAL WATER

PUMPPUMP

AVAILABLE WATERAVAILABLE WATERAVAILABLE WATERAVAILABLE WATER

UNAVAILABLE WATERUNAVAILABLE WATER

PERMANENT WILTINGPERMANENT WILTINGPOINTPOINT

MOISTURE RELEASE CURVESMOISTURE RELEASE CURVES

0.3

0.4

0.5

0.6W

AT

ER

CO

NT

EN

TW

AT

ER

CO

NT

EN

T

PERMANENTPERMANENTWILTINGWILTINGPOINTPOINT

0.34

FIELDFIELDCAPACITYCAPACITY

0.53

0.42

0.36

SATURATIONSATURATION

0.0

0.1

0.2

10 100 1000 10000 105

SOIL WATER TENSION, cmSOIL WATER TENSION, cm

VO

LU

ME

TR

IC W

VO

LU

ME

TR

IC W

0.16

0.10

0.06

0.23

0.15

The water held between field capacity and The water held between field capacity and permanent wilting point is called the available permanent wilting point is called the available water or the available water capacity (AWC) and is water or the available water capacity (AWC) and is sometimes called the available water holding sometimes called the available water holding

it Th AWC i l l t d bit Th AWC i l l t d b

AVAILABLE WATER & SOIL WATER RESERVOIR

capacity. The AWC is calculated by:capacity. The AWC is calculated by:

AWC = AWC = (θ(θfcfc -- θθpwppwp))

AWC is primarily a function of soil texture.AWC is primarily a function of soil texture.

Soil Texture θfc θwp AWC AWCin/in or

m/min/in or

m/min/in or

m/min/ft

Coarse Sand 0.10 0.05 0.05 0.60Sand 0.15 0.07 0.08 0.96Loamy Sand 0.18 0.07 0.11 1.32

Table 2.3 Example values of soil water characteristics for various soil

textures. * TEXT

Sandy Loam 0.20 0.08 0.12 1.44Loam 0.25 0.10 0.15 1.80Silt Loam 0.30 0.12 0.18 2.16Silty Clay Loam 0.38 0.22 0.16 1.92Clay Loam 0.40 0.25 0.15 1.80Silty Clay 0.40 0.27 0.13 1.56Clay 0.40 0.28 0.12 1.44

* These are examples only. Considerable variation exits from these values

within each soil texture.

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Soil Types VARIATION IN SOIL WATER BY SOIL TEXTUREVARIATION IN SOIL WATER BY SOIL TEXTURE

5

6

7

8

OF

SO

IL,

inc

he

s

GRAVITATIONAL WATERAVAILABLE -- NO STRESSAVAILABLE -- SOME STRESSUNAVAILABLE

5.2”

5.8”

6.1”

4.4”

0

1

2

3

4

Sand Loam Silty Clay Loam

WA

TE

R I

N O

NE

FO

OT

O

1.0”

2.0”

1.8”

2.1”

3.8”

1.1”

1.8”

2.6”

The capacity of the available soil water The capacity of the available soil water reservoir (TAW), Depends on both the AWC and reservoir (TAW), Depends on both the AWC and the depth that the plant roots have penetrated. the depth that the plant roots have penetrated. This relationship is given by:This relationship is given by:

TAW = (AWC) (RTAW = (AWC) (Rdd))

where where TAW = the total available water TAW = the total available water capacity in the plant root zone, andcapacity in the plant root zone, and

RRd d = depth of the plant root zone.= depth of the plant root zone.

Soil Types

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Plants can remove only a portion of the available Plants can remove only a portion of the available soil water reservoir (TAW) before growth and soil water reservoir (TAW) before growth and ultimately yield are affected. This portion is ultimately yield are affected. This portion is termed readily available water (termed readily available water (RAWRAW), and for ), and for most crops it ranges between 40 and 65 percent most crops it ranges between 40 and 65 percent of the available water in the crop root zone. of the available water in the crop root zone.

RAW = (AWC) (MAD)RAW = (AWC) (MAD)

where where MAD is the management allowed MAD is the management allowed deficiency (decimal) which can be deficiency (decimal) which can be removed. removed.

EXAMPLE PROBLEMEXAMPLE PROBLEM

Cotton crop with root zone of 3 feet.

FIND: READILY AVAILABLE WATER (RAW)

TAW = (AWC) (RTAW = (AWC) (Rdd))

From TABLE 14.1, AWC = 1.4 inches/footFrom TABLE 14.1, AWC = 1.4 inches/foot

Rd = 3 feetRd = 3 feet

TAW = 1.4 (3 feet) = 4.2 inchesTAW = 1.4 (3 feet) = 4.2 inches****************************************************************************************************

RAW = (AWC) (MAD)RAW = (AWC) (MAD)

From TABLE 14.2, MAD = 0.6From TABLE 14.2, MAD = 0.6

RAW = 4.2 (0.6) = 2.5 inchesRAW = 4.2 (0.6) = 2.5 inches

Define depleted and remaining water as Define depleted and remaining water as fraction of available water depleted or fraction of available water depleted or fraction of available water remaining.fraction of available water remaining.

AVAILABLE WATER & SOIL WATER RESERVOIRAVAILABLE WATER & SOIL WATER RESERVOIR

Fraction of available water depleted = Fraction of available water depleted = ffdd

ffdd = (= (θθfcfc -- θθvv) / () / (θθfcfc -- θθwpwp) ) (9)dd (( fcfc vv) () ( fcfc wpwp)) ( )

Fraction of available water remaining = Fraction of available water remaining = ffrr

ffr r = (= (θθvv -- θθwpwp) / () / (θθfc fc -- θθwpwp)) (10)(10)

AlsoAlso ffr r = 1 = 1 -- ffdd (10A)(10A)

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It is very useful in irrigation management It is very useful in irrigation management to know the depth of water required to fill a to know the depth of water required to fill a layer of soil to field capacity. This depth is layer of soil to field capacity. This depth is equal to equal to SWD.SWD.

AVAILABLE WATER & SOIL WATER RESERVOIRAVAILABLE WATER & SOIL WATER RESERVOIR

SWD = fSWD = f (AWC) L(AWC) L (11) (11) SWD = fSWD = fdd (AWC) L(AWC) L (11) (11)

by through a bit of algebra you will by through a bit of algebra you will find the equivalent to:find the equivalent to:

SWDSWD = (= (θθfcfc -- θθvv) L) L (12)(12)


A SAMPLE OF A SILT LOAM SOIL HAS A A SAMPLE OF A SILT LOAM SOIL HAS A VOLUMETRIC WATER CONTENT OF 0.26. VOLUMETRIC WATER CONTENT OF 0.26. CALCULATE fCALCULATE fdd, f, frr, AWC, AND SWD. ASSUME , AWC, AND SWD. ASSUME THE SOIL IS 36 INCHES DEEP.THE SOIL IS 36 INCHES DEEP.

GiGi θθ 0 26 0 26Given:Given: θθvv = 0.26= 0.26θθfcfc = 0.34= 0.34θθwp wp = 0.16= 0.16

Find:Find: ffddffrrAWC AWC –– Available Water Holding CapacityAvailable Water Holding CapacitySWD SWD –– Depth of Soil Water DepletedDepth of Soil Water Depleted


Solution:Solution:

ffdd = (= (θθfcfc -- θθvv) / () / (θθfc fc -- θθwpwp) ) Equation 10Equation 10

= (0.34 = (0.34 –– 0.26) / (0.34 0.26) / (0.34 –– 0.16) = 0.440.16) = 0.44

ffr r = 1 = 1 –– ffdd Equation 10AEquation 10Affr r 1 1 ffdd Equation 10AEquation 10A

= 1.0 = 1.0 –– 0.44 = 0.560.44 = 0.56

AWC = AWC = ((θθfcfc -- θθwpwp) = 0.34 ) = 0.34 –– 0.16 = 0.18 in/in = 2.16 in/ft0.16 = 0.18 in/in = 2.16 in/ft

SWD = fSWD = fd d (AWC) L(AWC) L Equation 11Equation 11

= 0.44 (0.18 in/in) 36 in = 2.85 in= 0.44 (0.18 in/in) 36 in = 2.85 in

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