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 soilwater 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 )=
Ms/V
Total porosity or pore space (TP; %)=
Pd = particle density; assume
2.65 g/cm3 or 2,650 kg/m3)
Gravimetric water content (Θg; % or g/g)=
Volumetric water content (Θv; % or cm3/cm3)=
(water by volume)/(volume of soil) or Bd*Θg
Air-filled porosity (AFP; % or cm3/cm3)= TP - Θv
Water-filled pore space (WFPS; % or cm3/cm3)=
water by volume/total porosity
Example Problem 1
If Bd = 1.3 g/cm3 and the Θg = 26%, what is the Θv, the
AFP, and how much of the TP space is filled with
water?
Step 1: calculate TP by
Step 2: calculate Θv by Bd*Θg
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: Θv = 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: Θv = 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!
So, surface area and matric forces
lead to this…
http://www.extension.umn.edu/distribution/cropsystems/DC7644.html
Soil Gas Exchange
Flux and Efflux of Soil Gases
• Driven by concentration gradients
• Driven by the AFP
– Which again depends on the Bd and Θv
Fick’s Law of Diffusion for the “Gradient Approach”
dC
C
F  DS
 DS
dz
z
-3
Molar Concentration, C (mol m )
Depth Below Surface, Z (m)
0
5e+4 1e+5 2e+5 2e+5 3e+5 3e+5 4e+5 4e+5
0.00
A1
0.05
A2) linear regression can be fit through the
[CO2] concentration with depth data and
CO2 gradient is the slope
A3
0.10
A2
0.15
c
Fitted model, C = a + bZ
Actual data
0.20
A1) fit a function to the [CO2] w/depth and
take first derivative at z = i
A3) finite-difference (concentration data
from discrete depth increments below the
surface)
= CO2 Sensor
-1 cm
-2 cm
-6 cm
-10 cm
DP
ML2x
-1 cm
-2 cm
-6 cm
-10 cm
Tom DeSutter ver.5
30 Oct 2006
DP
ML2x
Diffusivity of Soil Gases
Diffusivity in Air and Water
Gas
CO2 in air
CO2 in water
O2 in air
O2 in water
Diffusion Coefficient
(cm2/sec)
1.64 x 10-1
1.6 x 10-5 10,000X slower in water
1.98 x 10-1
1.9 x 10-5 10,000X slower in water
From Coyne and Thompson, 2006
Example Problem 3, find flux
Soil Bd = 1.3 g/cm3 and
1) Θv = 9% (almost “air-dry”)
2) Θv = 40% (nearly saturated)
Example Problem 3 cont.
Step 1: Solve for AFP (=TP – Θv)
Step 2: Solve for Ds (diffusivity for CO2 in soil)
Step 3: Solve for ΔC/Δz
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) = 533µL/L cm
conversion: 8.9µg CO2/cm4
Step 4: F = 0.04 * 8.9 = 0.36µg CO2/cm2 sec
For a nearly saturated soil, F = 0.09µg 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
Quartz
Organic
matter
Water
Air
Density
(g cm-3)
2.66
1.30
Heat Capacity
(cal cm-3 oC-1)
0.48
0.60
1.00
1.00
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 (Θv = 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 Θv was 51%, so
one-half saturation is 25.5%.
-So, Θv 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
Heat Capacity (cal/cm 3 oC)
0.8
0.7
0.6
0.5
0.4
0.3
Y intercept = 0.26; C of dry soil
0.2
0.0
0.1
0.2
0.3
0.4
0.5
Volumetric Water Content (cm3/cm3)
0.6
Heat Capacity (cal/cm3 oC)
0.8
Bd = 1.3 g/cm3
Bd = 1.5 g/cm3
Bd = 1.1 g/cm3
0.7
0.6
0.5
Bd
1.1
1.3
1.5
0.4
C of solids only
0.20
0.235
0.27
0.3
0.2
0.0
0.1
0.2
0.3
0.4
0.5
Volumetric Water Content (cm3/cm3)
0.6
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