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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  • Soil Water Interactions With Other Soil Characteristics With and

    Without Tile

    Soil and Soil Water Workshop

    15 January 2013

    Tom DeSutter

    Assistant Professor of Soil Science

    Program Leader for Soil Science

  • Why use subsurface drains?

    Decrease soil water content.

    Increase gas exchange

    Increase trafficability.

    Allow for improved soil warming.

    Remove excess soluble salts.

    Overall, improve production.

  • Soil Water Basics

    Only drainable water leaves via drains

    Total soil porosity minus soil moisture at field capacity

    This is generally thought of as gravitational water

    Drainable water depends on texture (and bulk density)

    Soil Texture Drainable Porosity (% by vol.)

    clays, clay loams, silty clays

    3-11%

    well structured loams

    10-15%

    sandy 18-35%

    http://www.extension.umn.edu/distribution/cropsystems/DC7644.html

  • Why does one care?

    Bulk density and water content drive most soil-water relationships.

    AFP influences gas exchange

    WFPS influences microbial activity

    Water content influences heat capacity which influences warming of the soil

  • Basic Soil-Water Concepts Bulk density (Bd; g/cm3 or kg/m3 )=

    Total porosity or pore space (TP; %)=

    Gravimetric water content (g; % or g/g)=

    Volumetric water content (v; % or cm3/cm3)=

    (water by volume)/(volume of soil) or Bd*g

    Air-filled porosity (AFP; % or cm3/cm3)= TP - v

    Water-filled pore space (WFPS; % or cm3/cm3)=

    water by volume/total porosity

    Ms/V

    Pd = particle density; assume 2.65 g/cm3 or 2,650 kg/m3)

  • 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?

    Step 1: calculate TP by

    Step 2: calculate v by Bd*g

    Step 3: calculate AFP by

    Step 4: calculate WFPS by water by volume/total porosity

    Next page

  • Step 1: TP = 1-(1.3/2.65)*100 = 51%

    Step 2: v = 1.3 * 26 = 33.8%

    Step 3: AFP = 51% - 33.8% = 17.2%

    Step 4: WFPS = (33.8%/51%)*100 = 66%

    What if the Bd changed to 1.5 g/cm3?

    Step 1: TP = 43%

    Step 2: v = 39%

    Step 3: AFP = 4%

    Step 4: WFPS = 91%

  • Clay vs Sand: water retention Sand and clay soils can have the same TP

    as long as their Bd and Pd are the same.

    But the sand soil has larger pores and less surface area (SA), and thus lower water retention than the clay soil

    SA of sand, silt, and clay are 30, 1,500, and 3,000,000 cm2/g, respectively

    A SA example

  • 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?

    Sand:14g*30cm2/g = 420 cm2

    Silt: 20g*1,500cm2/g = 30,000 cm2

    Clay: 15g*3,000,000cm2/g=45,000,000 cm2

    45,030,420 cm2

    5X the surface area as a 10,000 ft2 lot!

    10X the surface area as a basketball court!

  • http://www.extension.umn.edu/distribution/cropsystems/DC7644.html

    So, surface area and matric forces lead to this

  • Soil Gas Exchange

  • Flux and Efflux of Soil Gases

    Driven by concentration gradients

    Driven by the AFP

    Which again depends on the Bd and v

    Ficks Law of Diffusion for the Gradient Approach

  • Molar Concentration, C (mol m-3

    )

    0 5e+4 1e+5 2e+5 2e+5 3e+5 3e+5 4e+5 4e+5

    Depth

    Belo

    w S

    urf

    ace,

    Z (

    m)

    0.00

    0.05

    0.10

    0.15

    0.20

    Fitted model, C = a + bZc

    Actual data

    A1

    A2

    A3

    A1) fit a function to the [CO2] w/depth and take first derivative at z = i

    A2) linear regression can be fit through the [CO2] concentration with depth data and CO2 gradient is the slope

    A3) finite-difference (concentration data from discrete depth increments below the surface)

    z

    CD

    dz

    dCDF SS

  • Tom DeSutter ver.5

    30 Oct 2006

    = CO2 Sensor

    -2 cm

    -10 cm

    DP-1 cm

    ML2x-6 cm

    -2 cm

    -10 cm

    DP-1 cm

    ML2x-6 cm

  • Diffusivity of Soil Gases

  • Diffusivity in Air and Water Gas Diffusion Coefficient

    (cm2/sec)

    CO2 in air 1.64 x 10-1

    CO2 in water 1.6 x 10-5

    O2 in air 1.98 x 10-1

    O2 in water 1.9 x 10-5

    10,000X slower in water

    10,000X slower in water

    From Coyne and Thompson, 2006

  • Example Problem 3, find flux

    Soil Bd = 1.3 g/cm3 and 1) v = 9% (almost air-dry)

    2) v = 40% (nearly saturated)

  • Example Problem 3 cont.

    Step 1: Solve for AFP (=TP v)

    Step 2: Solve for Ds (diffusivity for CO2 in soil)

    Step 3: Solve for C/z

    Step 4: Solve for F

    For the air-dry soil

    Step 1: AFP = (1-1.3/2.65) 0.09 = 0.4 cm3/cm3

    Step 2: Ds = Dco2(0.66)(AFP)

    Ds = 1.64 x 10-1(0.66)(0.4) = 0.04 cm2/sec

    Penman, 1940

  • Example Problem 3 cont.

    Step 3: (10,000-2,000)/(20-5) = 533L/L cm

    conversion: 8.9g CO2/cm4

    Step 4: F = 0.04 * 8.9 = 0.36g CO2/cm2 sec

    For a nearly saturated soil, F = 0.09g CO2/cm2 sec,

    which is 4X slower than an air-dry soil

  • Heat Capacity (C) Amount of energy that an object must absorb or

    lose for its temperature to change by 1 oC. (Coyne and Thompson, 2006)

    Most commonly reported on a volume basis: cal/cm3 oC or J/cm3 oC

    Overall,

    Ctotal = fmineralsCminerals + fomCom + fwaterCwater + fairCair

    where C is the heat capacity and f is the volumetric fraction

    C is strongly affected by water

  • Heat Capacity of Soil Components

    Constituent Density (g cm-3)

    Heat Capacity (cal cm-3 oC-1)

    Quartz 2.66 0.48

    Organic matter

    1.30 0.60

    Water 1.00 1.00

    Air 0.00125 0.003

    From Hillel, 1998

    Ctotal = fmineralsCminerals + fomCom + fwaterCwater + fairCair

    Ctotal = fminerals0.48 + fom0.6 + fwater1.0 + fair0.003

  • 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.

    Step 1: determine TP; = (1 1.3/2.65)*100 = 51%

    Step 2: volume of solids = 100 51 2 = 47%

    Step 3: solve for Ctotal Ctotal = fminerals0.48 + fom0.6 + fwater1.0

    Ctotal = 0.47*0.48+ 0.02*0.6 + 0.51*1.0

    Ctotal = 0.75 cal/cm3 oC

  • Continued What if the same soil was only half-saturated with water?

    Step 1: Since at full saturation v was 51%, so one-half saturation is 25.5%.

    -So, v is 25.5%, AFP is 25.5%

    Step 2: solve for Ctotal Ctotal = 0.51*0.48+ 0.02*0.6 + 0.255*1.0

    Ctotal = 0.5 cal/cm3 oC

    Compared to 0.75 cal/cm3 oC at full saturation

    -less energy needed to warm dryer soil

  • Volumetric Water Content (cm3/cm

    3)

    0.0 0.1 0.2 0.3 0.4 0.5 0.6

    Heat

    Cap

    acit

    y (

    cal/

    cm

    3 o

    C)

    0.2

    0.3

    0.4

    0.5

    0.6

    0.7

    0.8

    Y intercept = 0.26; C of dry soil

  • Volumetric Water Content (cm3/cm

    3)

    0.0 0.1 0.2 0.3 0.4 0.5 0.6

    Heat

    Cap

    acit

    y (

    cal/

    cm

    3 o

    C)

    0.2

    0.3

    0.4

    0.5

    0.6

    0.7

    0.8

    Bd = 1.3 g/cm3

    Bd = 1.5 g/cm3

    Bd = 1.1 g/cm3

    Bd C of solids only1.1 0.201.3 0.2351.5 0.27

  • Summary

    Bd and water drive many of the heat, biological, and gas exchange processes

    Plus chemical reactions

    So, soil and water management can too

    compaction; tile vs no tile; surface drainage

    Basic equations are simple, helpful tools

    Easy to put actual numbers to what you are observing in the field

    Quantitative vs Qualitative

  • Soil Water Interactions With Other Soil Characteristics With and

    Without Tile

    Soil and Soil Water Workshop

    15 January 2013

    Tom DeSutter

    Assistant Professor of Soil Science

    Program Leader for Soil Science

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