Soil Water Properties Lecture 1

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<ul><li><p>1BAEN 464 BAEN 464 Irrigation &amp; Drainage EngineeringIrrigation &amp; Drainage Engineering</p><p>SoilSoil--PlantPlant--Water Water RelationshipsRelationshipsRelationshipsRelationships</p><p>CHAPTER 2 OF TEXTCHAPTER 2 OF TEXT&amp;&amp;</p><p>Special HandoutsSpecial Handouts</p><p>Soil PropertiesSoil Properties</p><p>SOIL WATER RELATIONS</p><p> NEED TO KNOW HOW MUCH WATER NEED TO KNOW HOW MUCH WATER IS IN THE SOIL IS IN THE SOIL </p><p> AND THE AVAILABILITY OF THE AND THE AVAILABILITY OF THE WATER TO PLANTSWATER TO PLANTS</p><p> LEADS TO DEFINTIONS OF:LEADS TO DEFINTIONS OF:</p><p> WATER CONTENT, AND WATER CONTENT, AND </p><p> SOIL WATER POTENTIAL (TENSION)SOIL WATER POTENTIAL (TENSION)</p><p>SOIL WATER PROPERTIESSOIL WATER PROPERTIES COMPOSITION OF AN UNSATURATED SOIL SAMPLECOMPOSITION OF AN UNSATURATED SOIL SAMPLE</p><p>SOIL SOIL PARTICLEPARTICLE</p><p>WATERWATER</p><p>PORE SPACEPORE SPACE</p><p>AIRAIR AIRAIR</p><p>SOLID MATERIALSOLID MATERIAL</p><p>PORE SPACEPORE SPACE</p><p>MINERAL</p><p>WATERWATER ORGANICORGANICMATTERMATTER</p><p>AIRAIR</p><p>WATERWATER</p><p>SOILSOIL</p><p>SOIL PHYSICAL PROPERTIESSOIL PHYSICAL PROPERTIES</p><p>VolumesVolumesVVv v = Volume Voids= Volume VoidsVVaa = Volume Air= Volume AirVVww = Volume Water= Volume WaterVVss = Volume Solids= Volume Solids</p><p>VVvv = V= Vaa + V+ Vww</p><p>MassMass</p><p>MMaa = Mass of air (zero)= Mass of air (zero)MMss = Mass of solids= Mass of solidsMMww = Mass of water= Mass of water</p><p>AIRAIR</p><p>WATERWATER</p><p>SOILSOIL</p><p>SOIL PHYSICAL PROPERTIESSOIL PHYSICAL PROPERTIES</p><p>VolumeVolume</p><p>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</p><p>VVbb = V= Vvv + V+ Vss</p><p>VVbb = V= Vaa + V+ Vww + V+ Vss</p></li><li><p>2AIRAIR</p><p>WATERWATER</p><p>SOILSOIL</p><p>SOIL PHYSICAL PROPERTIESSOIL PHYSICAL PROPERTIES</p><p>Mass and DensityMass and Density</p><p>Specific Gravity of Solids (p)p = Ms/(Vs w)</p><p>SoilSoil ppMineralMineral 2.652.65ClayClay 2.702.70OrganicOrganic 2.602.60</p><p>AIRAIR</p><p>WATERWATER</p><p>SOILSOIL</p><p>SOIL PHYSICAL PROPERTIESSOIL PHYSICAL PROPERTIES</p><p>Mass and DensityMass and Density</p><p>Bulk Density (b) Apparent Specific Gravity (As)b = Ms/(Vb); As = b/w = Ms/(Vb w)</p><p>ONE CAN SHOW THAT = (1 As/p) PROVE THISPROVE THIS</p><p>SAND: 0.05 SAND: 0.05 -- 2.0 mm2.0 mm</p><p>RELATIVE SIZES OF SOIL PARTICLESRELATIVE SIZES OF SOIL PARTICLES</p><p>SILT: 0.002 - 0.05 mm</p><p>CLAY: CLAY: &lt; 0.002 mm&lt; 0.002 mm</p><p>SOIL TEXTURAL TRIANGLESOIL TEXTURAL TRIANGLE</p><p>MASS WATER CONTENTMASS WATER CONTENT</p><p>WET SOILWET SOIL</p><p>m = Mass of WaterMass of Dry Soil Weight of WaterWeight of Dry Soil=</p><p>DRY SOILDRY SOILWET SOILWET SOILSAMPLESAMPLE</p><p>WET WEIGHTWET WEIGHTOF SOILOF SOIL</p><p>DRY WEIGHTDRY WEIGHTOF SOILOF SOIL</p><p>-- ==</p><p>WEIGHT OF WATER</p><p>DRY SOILDRY SOILSAMPLESAMPLE WATERWATER</p><p>VOLUMETRIC WATER CONTENTVOLUMETRIC WATER CONTENT</p><p>WET SOILWET SOIL</p><p>vv = = Volume of WaterVolume of WaterBulk VolumeBulk Volume</p><p>DRY SOILDRY SOILWET SOILWET SOILSAMPLESAMPLE</p><p>WET WEIGHTWET WEIGHTOF SOILOF SOIL</p><p>-- ==</p><p>DRY WEIGHTDRY WEIGHTOF SOILOF SOIL</p><p>DRY SOILDRY SOILSAMPLESAMPLE WATERWATER</p><p>VOLUMEVOLUMEOF WaterOF Water</p><p>BULKBULKVOLUMEVOLUME</p></li><li><p>3AIRAIR</p><p>WATERWATERd = v L</p><p>DEPTH OF WATER (d) IN A SOIL LAYERDEPTH OF WATER (d) IN A SOIL LAYER</p><p>LL</p><p>Bulk Volume (VBulk Volume (Vbb))VVb b = = rr22 LL</p><p>SOILSOIL</p><p>rrWater Volume (VWater Volume (Vww))</p><p>VVw w = = rr22 d d </p><p>VOLUMETRIC WATER CONTENTVOLUMETRIC WATER CONTENT</p><p>vv = = Volume of WaterVolume of WaterBulk VolumeBulk Volume</p><p> = M= M /(V/(V ))vv = V= Vww/(V/(Vbb) ) </p><p>bb = M= Mss/(V/(Vbb) ) ww M Mww/(V/(Vww) ) </p><p>vv = V= Vww/(V/(Vbb) = (M) = (Mww//ww) / (M) / (Mss//bb))vv = (M= (Mww//ww) / (M) / (Mss//bb) = ) = mm bb // ww</p><p>VOLUMETRIC WATER CONTENTVOLUMETRIC WATER CONTENT</p><p>vv = = Volume of WaterVolume of WaterBulk VolumeBulk Volume</p><p>vv = (M= (Mww//ww) / (M) / (Mss//bb) = ) = mm bb // ww</p><p>vv = = mm bb // w w = = mm AAss</p><p>EXAMPLE OF SOIL WATER PROPERTIES</p><p>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?</p><p>Given:Soil sample volume (Vb) = 80 cm3</p><p>Dry weight of soil sample (Ms) = 100 gWet weight of soil sample (Ms + Mw) = 120 g</p><p>Find:Find:</p><p>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</p><p>EXAMPLE OF SOIL WATER PROPERTIES</p><p>d depth of water d depth of water </p></li><li><p>4EXAMPLE OF SOIL WATER PROPERTIES</p><p>Solution:Solution:</p><p>m m = M= Mww / M/ MssMMww = 120 = 120 100 = 20 g.100 = 20 g.ww gg</p><p>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 </p><p>EXAMPLE OF SOIL WATER PROPERTIES</p><p>Solution:Solution:</p><p>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</p><p>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</p><p>EXAMPLE OF SOIL WATER PROPERTIES</p><p>Solution:Solution:</p><p>Current depth of water in one foot of soilCurrent depth of water in one foot of soil</p><p>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 </p><p>minus 3.0, or 0.6 inchesminus 3.0, or 0.6 inches</p><p>GRAVITATIONAL POTENTIAL</p><p>g = DISTANCE ABOVE AN ARBITRARY ELEVATION</p><p>FLOWFROM </p><p> g2 = Z2</p><p>DATUM</p><p> g1 = Z1 </p><p>FROM HIGH TOLOWPOTENTIAL</p><p>UPWARD FORCE ( F1)</p><p>F1 = (2 r) COS() </p><p>FORCE BALANCE</p><p>CAPILLARY FORCES</p><p>Contact Angle</p><p>H i ht f</p><p>rRadius ofCapillary Tube</p><p>PRESSURE = - h</p><p>DOWNWARD FORCE (F2)</p><p>F2 = - g ( r2 ) h</p><p>F1 - F2 = 0</p><p>(2 r) COS() = g ( r2 ) hh = 2 COS() / g rh = CONSTANT/ r</p><p>hHeight ofCapillary Rise</p><p>PRESSURE = 0</p><p>CAPILLARY FORCES</p><p>Contact Angle</p><p>r</p><p>Radius ofCapillary Tube</p><p>PRESSURE = - h</p><p>Attraction of glass and water causes water molecules to adhere to the glass.</p><p>hHeight ofCapillary Rise</p><p>PRESSURE = 0</p><p>Cohesive force between water molecules causes the tube to fill with water.</p><p>Water in the tube is belowatmospheric pressure. </p></li><li><p>5h = k / radius of tube</p><p>CAPILLARY FORCES</p><p>WATER RISES HIGHER FOR SMALL DIAMETER TUBES</p><p>WATER IN SMALL TUBES IS UNDER MORE TENSION</p><p>HOW DOES THAT RELATE TO SOIL-WATER RELATIONS</p><p>h</p><p>h</p><p>Capillary pressure = pw pa = k/r</p><p>NEARLY SATURATED SOIL</p><p>LARGE RADIUS (r)</p><p>SOIL PARTICLE</p><p>ATTRACTION OF SOIL PARTICLES FOR WATER --- (ADHESION)SURFACE TENSION OF WATER --- (COHESION)</p><p>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.</p><p>SMALL MATRIC POTENTIAL(i.e. little capillary action)</p><p>WATER</p><p>DRY SOIL</p><p>SMALL RADIUS (r)</p><p>SOIL PARTICLE</p><p>WATER</p><p>LARGE MATRIC POTENTIAL(i.e. large capillary action)</p><p>SMALL RADIUS OF CURVATURE AS THE SOIL DRIES AND THE WATER MIGRATES INTO THE CORNERS OF THE PORE SPACE</p><p>H</p><p>SOIL</p><p>POROUSPLATE / MEMBRANE</p><p>1 2</p><p>DEVELOPING MOISTURE RELEASE CURVESUSING A HANGING WATER COLUMN</p><p>INITIALLY: SOIL IS SATURATED WITH A POSITIVE HEAD CAUSING DRAINAGE THROUGH THE SOIL</p><p>t2 &gt; t1</p><p>MEMBRANE</p><p>SOIL</p><p>POROUSPLATE / MEMBRANE</p><p>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.</p><p>t2 = t1</p><p>HANGING WATER COLUMN FOR UNSATURATED CONDITIONS</p><p>POROUS PLATE REMAINS SATURATED AND PREVENTS AIR FROM ENTERING THE TUBE. </p><p>WATER FLOWS FROM THE SOIL SO THE SAMPLE IS </p><p>SOIL</p><p>POROUSPLATE / MEMBRANE</p><p>1</p><p>H = 100 cm</p><p>t1 = 0</p><p>UNSATURATED</p><p>THE SOIL WATER IS UNDER A TENSION</p><p>WHEN FLOW STOPS THE TENSION IN THE SOIL EQUALS THE DIFFERENCE IN ELEVATION BETWEEN POINT 1 AND 2.</p><p>2</p><p>t2 = 0</p><p>HANGING WATER COLUMN FOR UNSATURATED CONDITIONS</p><p>WHAT IS THE MATRIC POTENTIAL </p><p>AT POINT 1?</p><p>t = g + m</p><p>SOIL</p><p>POROUSPLATE / </p><p>1t1 = 0</p><p>WHEN NO FLOW</p><p>t = 0g = - 100 cmm = - 100 cm</p><p>PLATE / MEMBRANE</p><p>2</p><p>t2 = 0</p><p>H = 100 cm</p><p>DATUM</p></li><li><p>6HANGING WATER COLUMN FOR MOISTURE RELEASE CURVES</p><p>SOIL</p><p>1</p><p>m1 = - 100 cm DETERMINE THE </p><p>VOLUMETRIC WATER CONTENT:</p><p>MATCHING THE VOLUMETRIC WATER CONTENT TO THE MATRIC POTENTIA</p><p>POROUSPLATE / MEMBRANE</p><p>2</p><p>H = 100 cm</p><p>CONTENT:</p><p>WET WEIGHT = 500 g</p><p>DRY WEIGHT = 400 g</p><p>BULK VOLUME = 300 cm3</p><p>v = (500-400)/300 = 0.33</p><p>0.2</p><p>0.3</p><p>0.4</p><p>C W</p><p>ATER</p><p> CO</p><p>NTE</p><p>NT MOISTURE RELEASE CURVE</p><p>Field Capacity</p><p>PermanentWilting Point</p><p>50 100 200 400 600 1000 2000 4000 6000 104 15,0000</p><p>0.1</p><p>SOIL WATER TENSION, cm of water</p><p>VOLU</p><p>MET</p><p>RIC Wilting Point</p><p>SOIL WATER POTENTIALMEASURE OF WATER AVAILABILITY</p><p>COMMON UNITS OF PRESSURE AND HEAD AND THEIR EQUIVALENTS </p><p>Unit Pressure Equivalent Water Head Equivalent </p><p>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 </p><p>MOISTURE RELEASE CURVES</p><p>0.3</p><p>0.4</p><p>0.5</p><p>0.6 W</p><p>AT</p><p>ER</p><p> CO</p><p>NT</p><p>EN</p><p>T</p><p>PERMANENT</p><p>FREE /GRAVITATIONAL</p><p>WATER 0.34</p><p>FIELDCAPACITY</p><p>0.0</p><p>0.1</p><p>0.2</p><p>10 100 1000 10000 105</p><p>SOIL WATER TENSION, cm</p><p>VO</p><p>LUM</p><p>ET</p><p>RIC</p><p>0.16</p><p>WILTINGPOINTAVAILABLE WATER</p><p>UNAVAILABLE WATER</p><p>FIELD CAPACITY &amp; SATURATION FIELD CAPACITY &amp; SATURATION -- RESERVOIR ANALOGY RESERVOIR ANALOGY </p><p>SATURATIONSATURATION FIELD CAPACITYFIELD CAPACITYDRAIN TUBEDRAIN TUBE</p><p>GRAVITATIONAL WATERGRAVITATIONAL WATER</p><p>RESERVOIRRESERVOIR</p><p>Spillway to protect the reservoir is like the water that drains due to gravity.</p><p>AVAILABLE WATER -- RESERVOIR ANALOGY </p><p>SATURATIONSATURATION FIELD CAPACITYFIELD CAPACITY</p><p>DRAIN TUBEDRAIN TUBE</p><p>PUMPPUMP</p><p>FLOWFLOW</p><p>FLOWFLOW</p><p>When reservoir is full it is easy to extract water, so we get large flows</p></li><li><p>7AVAILABLE WATER -- RESERVOIR ANALOGY </p><p>SATURATIONSATURATION FIELD CAPACITYFIELD CAPACITYDRAIN TUBEDRAIN TUBE</p><p>PUMPPUMP</p><p>FLOWFLOW</p><p>As water level drops it is more difficult to extract water, so flow is smaller</p><p>UNAVAILABLE WATER -- RESERVOIR ANALOGY </p><p>SATURATIONSATURATION FIELD CAPACITYFIELD CAPACITYDRAIN TUBEDRAIN TUBE</p><p>PUMPPUMP</p><p>PERMANENTPERMANENT</p><p>UNAVAILABLE WATERUNAVAILABLE WATER</p><p>No matter what we do there is some water that we can not extract.</p><p>PERMANENTPERMANENTWILTING POINTWILTING POINT</p><p>WATER STATUS -- RESERVOIR ANALOGY </p><p>SATURATIONSATURATION FIELD CAPACITYFIELD CAPACITYDRAIN TUBEDRAIN TUBE</p><p>GRAVITATIONAL WATERGRAVITATIONAL WATER</p><p>PUMPPUMP</p><p>AVAILABLE WATERAVAILABLE WATERAVAILABLE WATERAVAILABLE WATER</p><p>UNAVAILABLE WATERUNAVAILABLE WATER</p><p>PERMANENT WILTINGPERMANENT WILTINGPOINTPOINT</p><p>MOISTURE RELEASE CURVESMOISTURE RELEASE CURVES</p><p>0.3</p><p>0.4</p><p>0.5</p><p>0.6W</p><p>ATER</p><p> CO</p><p>NTE</p><p>NT</p><p>WAT</p><p>ER C</p><p>ON</p><p>TEN</p><p>T</p><p>PERMANENTPERMANENTWILTINGWILTINGPOINTPOINT</p><p>0.34</p><p>FIELDFIELDCAPACITYCAPACITY</p><p>0.53</p><p>0.420.36</p><p>SATURATIONSATURATION</p><p>0.0</p><p>0.1</p><p>0.2</p><p>10 100 1000 10000 105</p><p>SOIL WATER TENSION, cmSOIL WATER TENSION, cm</p><p>VOLU</p><p>MET</p><p>RIC</p><p> WVO</p><p>LUM</p><p>ETR</p><p>IC W</p><p>0.16</p><p>0.10</p><p>0.06</p><p>0.23</p><p>0.15</p><p>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 </p><p>it Th AWC i l l t d bit Th AWC i l l t d b</p><p>AVAILABLE WATER &amp; SOIL WATER RESERVOIR</p><p>capacity. The AWC is calculated by:capacity. The AWC is calculated by:</p><p>AWC = AWC = ((fcfc -- pwppwp))AWC is primarily a function of soil texture.AWC is primarily a function of soil texture.</p><p>Soil Texture fc wp AWC AWCin/in or </p><p>m/min/in or </p><p>m/min/in or </p><p>m/min/ft</p><p>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</p><p>Table 2.3 Example values of soil water characteristics for various soil </p><p>textures. * TEXT</p><p>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</p><p>* These are examples only. Considerable variation exits from these values within each soil texture.</p></li><li><p>8Soil Types VARIATION IN SOIL WATER BY SOIL TEXTUREVARIATION IN SOIL WATER BY SOIL TEXTURE</p><p>5</p><p>6</p><p>7</p><p>8</p><p>OF </p><p>SOIL</p><p>, inc</p><p>hes</p><p>GRAVITATIONAL WATERAVAILABLE -- NO STRESSAVAILABLE -- SOME STRESSUNAVAILABLE</p><p>5.25.8</p><p>6.1</p><p>4.4</p><p>0</p><p>1</p><p>2</p><p>3</p><p>4</p><p>Sand Loam Silty Clay Loam</p><p>WA</p><p>TER</p><p> IN O</p><p>NE </p><p>FOO</p><p>T O</p><p>1.0</p><p>2.0</p><p>1.8</p><p>2.1</p><p>3.8</p><p>1.1</p><p>1.8</p><p>2.6</p><p>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:</p><p>TAW = (AWC) (RTAW = (AWC) (Rdd))</p><p>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</p><p>RRd d = depth of the plant root zone.= depth of the plant root zone.</p><p>Soil Types</p></li><li><p>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. </p><p>RAW = (AWC) (MAD)RAW = (AWC) (MAD)</p><p>where where MAD is the management allowed MAD is the management allowed deficiency (decimal) which can be deficiency (decimal) which can be removed. removed. </p><p>EXAMPLE PROBLEMEXAMPLE PROBLEM</p><p>Cotton crop with root zone of 3 feet.</p><p>FIND: READILY AVAILABLE WATER (RAW)</p><p>TAW = (AWC) (RTAW = (AWC) (Rdd))</p><p>From TABLE 14.1, AWC = 1.4 inches/footFrom TABLE 14.1, AWC = 1.4 inches/foot</p><p>Rd = 3 feetRd = 3 feet</p><p>TAW = 1.4 (3 feet) = 4.2 inchesTAW = 1.4 (3 feet) = 4.2 inches****************************************************************************************************</p><p>RAW = (AWC) (MAD)RAW = (AWC) (MAD)</p><p>From TABLE 14.2, MAD = 0.6From TABLE 14.2, MAD = 0.6</p><p>RAW = 4.2 (0.6) = 2.5 inchesRAW = 4.2 (0.6) = 2.5 inches</p><p>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.</p><p>AVAILABLE WATER &amp; SOIL WATER RESERVOIRAVAILABLE WATER &amp; SOIL WATER RESERVOIR</p><p>Fraction of available water depleted = Fraction of available water depleted = ffdd</p><p>ffdd = (= (fcfc -- vv) / () / (fcfc -- wpwp) ) (9)dd (( fcfc vv) () ( fcfc wpwp)) ( )Fraction of available water remaining = Fraction of available water remaining = ffrr</p><p>ffr r = (= (vv -- wpwp) / () / (fc fc -- wpwp)) (10)(10)AlsoAlso ffr r = 1 = 1 -- ffdd (10A)(10A)</p></li><li><p>10</p><p>It is very useful in irrigation management It is very useful in irrigation management to know the depth of water required to fil...</p></li></ul>

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