Soil Water Interactions With Other Soil Characteristics ... Water Interactions With Other Soil...

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 (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

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 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!

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 Θv

Fick’s 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) 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 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 (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

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