Soil Water Interactions With Other Soil Characteristics ... Water Interactions With Other Soil Characteristics With and Without Tile Soil and Soil Water Workshop 15 January 2013 Tom DeSutter Assistant Professor of Soil

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<ul><li><p>Soil Water Interactions With Other Soil Characteristics With and </p><p>Without Tile </p><p>Soil and Soil Water Workshop </p><p>15 January 2013 </p><p>Tom DeSutter </p><p>Assistant Professor of Soil Science </p><p>Program Leader for Soil Science </p></li><li><p>Why use subsurface drains? </p><p> Decrease soil water content. </p><p> Increase gas exchange </p><p> Increase trafficability. </p><p> Allow for improved soil warming. </p><p> Remove excess soluble salts. </p><p> Overall, improve production. </p></li><li><p>Soil Water Basics </p><p> Only drainable water leaves via drains </p><p> Total soil porosity minus soil moisture at field capacity </p><p> This is generally thought of as gravitational water </p><p> Drainable water depends on texture (and bulk density) </p><p>Soil Texture Drainable Porosity (% by vol.) </p><p>clays, clay loams, silty clays </p><p>3-11% </p><p>well structured loams </p><p>10-15% </p><p>sandy 18-35% </p><p>http://www.extension.umn.edu/distribution/cropsystems/DC7644.html </p></li><li><p>Why does one care? </p><p> Bulk density and water content drive most soil-water relationships. </p><p> AFP influences gas exchange </p><p> WFPS influences microbial activity </p><p> Water content influences heat capacity which influences warming of the soil </p></li><li><p>Basic Soil-Water Concepts Bulk density (Bd; g/cm3 or kg/m3 )= </p><p>Total porosity or pore space (TP; %)= </p><p>Gravimetric water content (g; % or g/g)= </p><p>Volumetric water content (v; % or cm3/cm3)= </p><p> (water by volume)/(volume of soil) or Bd*g </p><p>Air-filled porosity (AFP; % or cm3/cm3)= TP - v </p><p>Water-filled pore space (WFPS; % or cm3/cm3)= </p><p> water by volume/total porosity </p><p>Ms/V </p><p>Pd = particle density; assume 2.65 g/cm3 or 2,650 kg/m3) </p></li><li><p>Example Problem 1 If Bd = 1.3 g/cm3 and the g = 26%, what is the v, the AFP, and how much of the TP space is filled with water? </p><p>Step 1: calculate TP by </p><p>Step 2: calculate v by Bd*g </p><p>Step 3: calculate AFP by </p><p>Step 4: calculate WFPS by water by volume/total porosity </p><p>Next page </p></li><li><p>Step 1: TP = 1-(1.3/2.65)*100 = 51% </p><p>Step 2: v = 1.3 * 26 = 33.8% </p><p>Step 3: AFP = 51% - 33.8% = 17.2% </p><p>Step 4: WFPS = (33.8%/51%)*100 = 66% </p><p>What if the Bd changed to 1.5 g/cm3? </p><p>Step 1: TP = 43% </p><p>Step 2: v = 39% </p><p>Step 3: AFP = 4% </p><p>Step 4: WFPS = 91% </p></li><li><p>Clay vs Sand: water retention Sand and clay soils can have the same TP </p><p>as long as their Bd and Pd are the same. </p><p> But the sand soil has larger pores and less surface area (SA), and thus lower water retention than the clay soil </p><p> SA of sand, silt, and clay are 30, 1,500, and 3,000,000 cm2/g, respectively </p><p> A SA example </p></li><li><p>Example Problem 2 A 49 g sample of clay loam has 29, 41, and 31% sand, silt, and clay, respectively. What is its total surface area? </p><p>Sand:14g*30cm2/g = 420 cm2 </p><p>Silt: 20g*1,500cm2/g = 30,000 cm2 </p><p>Clay: 15g*3,000,000cm2/g=45,000,000 cm2 </p><p> 45,030,420 cm2 </p><p>5X the surface area as a 10,000 ft2 lot! </p><p>10X the surface area as a basketball court! </p></li><li><p>http://www.extension.umn.edu/distribution/cropsystems/DC7644.html </p><p>So, surface area and matric forces lead to this </p></li><li><p>Soil Gas Exchange </p></li><li><p>Flux and Efflux of Soil Gases </p><p> Driven by concentration gradients </p><p> Driven by the AFP </p><p> Which again depends on the Bd and v </p><p>Ficks Law of Diffusion for the Gradient Approach </p></li><li><p>Molar Concentration, C (mol m-3</p><p>)</p><p>0 5e+4 1e+5 2e+5 2e+5 3e+5 3e+5 4e+5 4e+5</p><p>Depth</p><p> Belo</p><p>w S</p><p>urf</p><p>ace, </p><p>Z (</p><p>m)</p><p>0.00</p><p>0.05</p><p>0.10</p><p>0.15</p><p>0.20</p><p>Fitted model, C = a + bZc</p><p>Actual data</p><p>A1</p><p>A2</p><p>A3</p><p>A1) fit a function to the [CO2] w/depth and take first derivative at z = i </p><p>A2) linear regression can be fit through the [CO2] concentration with depth data and CO2 gradient is the slope </p><p>A3) finite-difference (concentration data from discrete depth increments below the surface) </p><p>z</p><p>CD</p><p>dz</p><p>dCDF SS</p></li><li><p>Tom DeSutter ver.5</p><p>30 Oct 2006</p><p> = CO2 Sensor</p><p>-2 cm</p><p>-10 cm</p><p>DP-1 cm</p><p>ML2x-6 cm</p><p>-2 cm</p><p>-10 cm</p><p>DP-1 cm</p><p>ML2x-6 cm</p></li><li><p>Diffusivity of Soil Gases </p></li><li><p>Diffusivity in Air and Water Gas Diffusion Coefficient </p><p>(cm2/sec) </p><p>CO2 in air 1.64 x 10-1 </p><p>CO2 in water 1.6 x 10-5 </p><p>O2 in air 1.98 x 10-1 </p><p>O2 in water 1.9 x 10-5 </p><p>10,000X slower in water </p><p>10,000X slower in water </p><p>From Coyne and Thompson, 2006 </p></li><li><p>Example Problem 3, find flux </p><p>Soil Bd = 1.3 g/cm3 and 1) v = 9% (almost air-dry) </p><p> 2) v = 40% (nearly saturated) </p></li><li><p>Example Problem 3 cont. </p><p>Step 1: Solve for AFP (=TP v) </p><p>Step 2: Solve for Ds (diffusivity for CO2 in soil) </p><p>Step 3: Solve for C/z </p><p>Step 4: Solve for F </p><p>For the air-dry soil </p><p>Step 1: AFP = (1-1.3/2.65) 0.09 = 0.4 cm3/cm3 </p><p>Step 2: Ds = Dco2(0.66)(AFP) </p><p>Ds = 1.64 x 10-1(0.66)(0.4) = 0.04 cm2/sec </p><p>Penman, 1940 </p></li><li><p>Example Problem 3 cont. </p><p>Step 3: (10,000-2,000)/(20-5) = 533L/L cm </p><p> conversion: 8.9g CO2/cm4 </p><p>Step 4: F = 0.04 * 8.9 = 0.36g CO2/cm2 sec </p><p>For a nearly saturated soil, F = 0.09g CO2/cm2 sec, </p><p>which is 4X slower than an air-dry soil </p></li><li><p>Heat Capacity (C) Amount of energy that an object must absorb or </p><p>lose for its temperature to change by 1 oC. (Coyne and Thompson, 2006) </p><p> Most commonly reported on a volume basis: cal/cm3 oC or J/cm3 oC </p><p> Overall, </p><p>Ctotal = fmineralsCminerals + fomCom + fwaterCwater + fairCair </p><p>where C is the heat capacity and f is the volumetric fraction </p><p> C is strongly affected by water </p></li><li><p>Heat Capacity of Soil Components </p><p>Constituent Density (g cm-3) </p><p>Heat Capacity (cal cm-3 oC-1) </p><p>Quartz 2.66 0.48 </p><p>Organic matter </p><p>1.30 0.60 </p><p>Water 1.00 1.00 </p><p>Air 0.00125 0.003 </p><p>From Hillel, 1998 </p><p>Ctotal = fmineralsCminerals + fomCom + fwaterCwater + fairCair </p><p>Ctotal = fminerals0.48 + fom0.6 + fwater1.0 + fair0.003 </p></li><li><p>Example Problem 4 What is the C of a soil that has a Bd of 1.3 g/cm3 , is saturated with water (v = TP), and has 2% OM? Note: it is okay to neglect the properties of air. Assume Pd is 2.65 g/cm3. </p><p>Step 1: determine TP; = (1 1.3/2.65)*100 = 51% </p><p>Step 2: volume of solids = 100 51 2 = 47% </p><p>Step 3: solve for Ctotal Ctotal = fminerals0.48 + fom0.6 + fwater1.0 </p><p>Ctotal = 0.47*0.48+ 0.02*0.6 + 0.51*1.0 </p><p>Ctotal = 0.75 cal/cm3 oC </p></li><li><p>Continued What if the same soil was only half-saturated with water? </p><p>Step 1: Since at full saturation v was 51%, so one-half saturation is 25.5%. </p><p> -So, v is 25.5%, AFP is 25.5% </p><p>Step 2: solve for Ctotal Ctotal = 0.51*0.48+ 0.02*0.6 + 0.255*1.0</p><p>Ctotal = 0.5 cal/cm3 oC </p><p>Compared to 0.75 cal/cm3 oC at full saturation </p><p> -less energy needed to warm dryer soil </p></li><li><p>Volumetric Water Content (cm3/cm</p><p>3)</p><p>0.0 0.1 0.2 0.3 0.4 0.5 0.6</p><p>Heat </p><p>Cap</p><p>acit</p><p>y (</p><p>cal/</p><p>cm</p><p>3 o</p><p>C)</p><p>0.2</p><p>0.3</p><p>0.4</p><p>0.5</p><p>0.6</p><p>0.7</p><p>0.8</p><p>Y intercept = 0.26; C of dry soil</p></li><li><p>Volumetric Water Content (cm3/cm</p><p>3)</p><p>0.0 0.1 0.2 0.3 0.4 0.5 0.6</p><p>Heat </p><p>Cap</p><p>acit</p><p>y (</p><p>cal/</p><p>cm</p><p>3 o</p><p>C)</p><p>0.2</p><p>0.3</p><p>0.4</p><p>0.5</p><p>0.6</p><p>0.7</p><p>0.8</p><p>Bd = 1.3 g/cm3</p><p>Bd = 1.5 g/cm3</p><p>Bd = 1.1 g/cm3</p><p>Bd C of solids only1.1 0.201.3 0.2351.5 0.27</p></li><li><p>Summary </p><p> Bd and water drive many of the heat, biological, and gas exchange processes </p><p> Plus chemical reactions </p><p> So, soil and water management can too </p><p> compaction; tile vs no tile; surface drainage </p><p> Basic equations are simple, helpful tools </p><p> Easy to put actual numbers to what you are observing in the field </p><p> Quantitative vs Qualitative </p></li><li><p>Soil Water Interactions With Other Soil Characteristics With and </p><p>Without Tile </p><p>Soil and Soil Water Workshop </p><p>15 January 2013 </p><p>Tom DeSutter </p><p>Assistant Professor of Soil Science </p><p>Program Leader for Soil Science </p></li></ul>

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