Soil Water Properties Lecture 1

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  • 1BAEN 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

    SOIL PHYSICAL PROPERTIESSOIL PHYSICAL PROPERTIES

    VolumeVolume

    Porosity (Porosity () = V) = Vvv/V/VbbVoid Ratio (Void Ratio () = V) = Vbb/V/VssSaturation (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

  • 2AIRAIR

    WATERWATER

    SOILSOIL

    SOIL PHYSICAL PROPERTIESSOIL PHYSICAL PROPERTIES

    Mass and DensityMass and Density

    Specific Gravity of Solids (p)p = Ms/(Vs w)

    SoilSoil ppMineralMineral 2.652.65ClayClay 2.702.70OrganicOrganic 2.602.60

    AIRAIR

    WATERWATER

    SOILSOIL

    SOIL PHYSICAL PROPERTIESSOIL PHYSICAL PROPERTIES

    Mass and DensityMass and Density

    Bulk Density (b) Apparent Specific Gravity (As)b = Ms/(Vb); As = b/w = Ms/(Vb w)

    ONE CAN SHOW THAT = (1 As/p) 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

    m = Mass of WaterMass 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

    vv = = Volume of WaterVolume of WaterBulk 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

  • 3AIRAIR

    WATERWATERd = v L

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

    LL

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

    SOILSOIL

    rrWater Volume (VWater Volume (Vww))

    VVw w = = rr22 d d

    VOLUMETRIC WATER CONTENTVOLUMETRIC WATER CONTENT

    vv = = Volume of WaterVolume of WaterBulk VolumeBulk Volume

    = M= M /(V/(V ))vv = V= Vww/(V/(Vbb) )

    bb = M= Mss/(V/(Vbb) ) ww M Mww/(V/(Vww) )

    vv = V= Vww/(V/(Vbb) = (M) = (Mww//ww) / (M) / (Mss//bb))vv = (M= (Mww//ww) / (M) / (Mss//bb) = ) = mm bb // ww

    VOLUMETRIC WATER CONTENTVOLUMETRIC WATER CONTENT

    vv = = Volume of WaterVolume of WaterBulk VolumeBulk Volume

    vv = (M= (Mww//ww) / (M) / (Mss//bb) = ) = mm bb // ww

    vv = = mm bb // w w = = mm 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 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 gWet weight of soil sample (Ms + Mw) = 120 g

    Find:Find:

    m m water content on a dry weight basiswater content on a dry weight basisvv water content on a volume basiswater content on a volume basis

    EXAMPLE OF SOIL WATER PROPERTIES

    d depth of water d depth of water

  • 4EXAMPLE OF SOIL WATER PROPERTIES

    Solution:Solution:

    m m = M= Mww / M/ MssMMww = 120 = 120 100 = 20 g.100 = 20 g.ww gg

    mm = 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

    EXAMPLE OF SOIL WATER PROPERTIES

    Solution:Solution:

    vv = V= Vww / V/ Vbband and v v = = bb/ / ww ((mm) ) b b = M= Mss/V/Vbb = {100g/80 cm= {100g/80 cm33} = 1.25 g/cm} = 1.25 g/cm33

    vv = = bb/ / ww ((mm) = 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

    EXAMPLE OF SOIL WATER PROPERTIES

    Solution:Solution:

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

    d = d = vv L L d = d = vv L = 0.25 (12 in) = 3 inches of waterL = 0.25 (12 in) = 3 inches of waterThe depth of water in the soil when The depth of water in the soil when v v = 0.30 is = 0.30 is d = d = vv L = 0.30 (12 in) = 3.60 inches of L = 0.30 (12 in) = 3.60 inches of waterwaterThus, 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

    g = DISTANCE ABOVE AN ARBITRARY ELEVATION

    FLOWFROM

    g2 = Z2

    DATUM

    g1 = Z1

    FROM HIGH TOLOWPOTENTIAL

    UPWARD FORCE ( F1)

    F1 = (2 r) COS()

    FORCE BALANCE

    CAPILLARY FORCES

    Contact Angle

    H i ht f

    rRadius ofCapillary Tube

    PRESSURE = - h

    DOWNWARD FORCE (F2)

    F2 = - g ( r2 ) h

    F1 - F2 = 0

    (2 r) COS() = g ( r2 ) hh = 2 COS() / g rh = 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.

  • 5h = 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

    t2 > t1

    MEMBRANE

    SOIL

    POROUSPLATE / MEMBRANE

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

    t2 = t1

    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

    POROUSPLATE / MEMBRANE

    1

    H = 100 cm

    t1 = 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

    t2 = 0

    HANGING WATER COLUMN FOR UNSATURATED CONDITIONS

    WHAT IS THE MATRIC POTENTIAL

    AT POINT 1?

    t = g + m

    SOIL

    POROUSPLATE /

    1t1 = 0

    WHEN NO FLOW

    t = 0g = - 100 cmm = - 100 cm

    PLATE / MEMBRANE

    2

    t2 = 0

    H = 100 cm

    DATUM

  • 6HANGING WATER COLUMN FOR MOISTURE RELEASE CURVES

    SOIL

    1

    m1 = - 100 cm DETERMINE THE

    VOLUMETRIC WATER CONTENT:

    MATCHING THE VOLUMETRIC WATER CONTENT TO THE MATRIC POTENTIA

    POROUSPLATE / MEMBRANE

    2

    H = 100 cm

    CONTENT:

    WET WEIGHT = 500 g

    DRY WEIGHT = 400 g

    BULK VOLUME = 300 cm3

    v = (500-400)/300 = 0.33

    0.2

    0.3

    0.4

    C W

    ATER

    CO

    NTE

    NT MOISTURE RELEASE CURVE

    Field Capacity

    PermanentWilting Point

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

    0.1

    SOIL WATER TENSION, cm of water

    VOLU

    MET

    RIC 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

  • 7AVAILABLE WATER -- RESERVOIR ANALOGY

    SATURATIONSATURATION FIELD CAPACITYFIELD CAPACITYDRAIN TUBEDRAIN TUBE

    PUMPPUMP

    FLOWFLOW

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

    UNAVAILABLE WATER -- RESERVOIR ANALOGY

    SATURATIONSATURATION FIELD CAPACITYFIELD CAPACITYDRAIN TUBEDRAIN TUBE

    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

    SATURATIONSATURATION FIELD CAPACITYFIELD CAPACITYDRAIN TUBEDRAIN TUBE

    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

    ATER

    CO

    NTE

    NT

    WAT

    ER C

    ON

    TEN

    T

    PERMANENTPERMANENTWILTINGWILTINGPOINTPOINT

    0.34

    FIELDFIELDCAPACITYCAPACITY

    0.53

    0.420.36

    SATURATIONSATURATION

    0.0

    0.1

    0.2

    10 100 1000 10000 105

    SOIL WATER TENSION, cmSOIL WATER TENSION, cm

    VOLU

    MET

    RIC

    WVO

    LUM

    ETR

    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 = ((fcfc -- pwppwp))AWC is primarily a function of soil texture.AWC is primarily a function of soil texture.

    Soil Texture fc wp 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.

  • 8Soil Types VARIATION IN SOIL WATER BY SOIL TEXTUREVARIATION IN SOIL WATER BY SOIL TEXTURE

    5

    6

    7

    8

    OF

    SOIL

    , inc

    hes

    GRAVITATIONAL WATERAVAILABLE -- NO STRESSAVAILABLE -- SOME STRESSUNAVAILABLE

    5.25.8

    6.1

    4.4

    0

    1

    2

    3

    4

    Sand Loam Silty Clay Loam

    WA

    TER

    IN O

    NE

    FOO

    T 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

  • 9Plants 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 = (= (fcfc -- vv) / () / (fcfc -- wpwp) ) (9)dd (( fcfc vv) () ( fcfc wpwp)) ( )Fraction of available water remaining = Fraction of available water remaining = ffrr

    ffr r = (= (vv -- wpwp) / () / (fc fc -- wpwp)) (10)(10)AlsoAlso ffr r = 1 = 1 -- ffdd (10A)(10A)

  • 10

    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 = (= (fcfc -- vv) L) L (12)(12)

    EXAMPLE OF SOIL WATER PROPERTIES

    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 0 26 0 26Given:Given: vv = 0.26= 0.26fcfc = 0.34= 0.34wp wp = 0.16= 0.16Find:Find: ffdd

    ffrrAWC AWC Available Water Holding CapacityAvailable Water Holding CapacitySWD SWD Depth of Soil Water DepletedDepth of Soil Water Depleted

    EXAMPLE OF SOIL WATER PROPERTIES

    Solution:Solution:

    ffdd = (= (fcfc -- vv) / () / (fc fc -- wpwp) ) 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 = ((fcfc -- wpwp) = 0.34 ) = 0.34 0.16 = 0.18 in/in = 2.16 in/ft0.16 = 0.18 in/in = 2.16 in/ftSWD = 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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