W2, Soil-water relations

Question 1

How do you calculate

Volumetric water content

Answer 1

θ_v = V_water/V_total

Question 2

True or False?

Expressing soil water content as volumetric water content is the best. Explain.

Answer 2

False.

The volumetric water content varies with soil compaction, making it unsuitable for lab measurement and especially for swelling soils because the volume of soil varies with water content. (Using the ratio of volume of water to soil mass is better for such soils).

Question 3

Convert a volumetric water content of 0.21 (21%) to a depth equivalent

Answer 3

21% = 210mm/m

Generally expressed in mm.

e.g. if a topsoil is 10 cm thick and has a volumetric water content 0.21, then the soil contains an equivalent depth of 21mm

Question 4

How do you calculate

Bulk density

Answer 4

ρ_b = m_{dry soil} / V_total

Question 5

True or False?

Expressing soil water content as gravimetric water content is the best. Explain.

Answer 5

True.

• Easier to determine accurately in a lab
• Independant of bulk density

Question 6

Explain the relationship between density and volume (in terms of calculations)

Answer 6

density = mass / volume

volume = mass / density

Question 7

How do you calculate

Porosity

Answer 7

Porosity, ℇ

= V_pores / V_total
= (V_total - V_solids) / V_total
= 1 - (V_solids / V_total)
= 1 - ρ_b / ρ_s

Question 8

What factors determine how much water is plant available?

Answer 8

• The water content of the soil
• The depth of the rooting zone
  
  • think about what limits the rooting depth:
    
    • soil depth
    • plant species and growth stage
    • limitations to growth
      
      • inpenetrable soil layers
      • toxicity
      • waterlogging
• Matric potential (how strongly the water is held by the soil)
• Osmotic potential (salt concentration)
  
  • not just salt, can be other species too
• Soil strength
  
  • can be limiting at SWC greater than the wilting point
• Aeration
  
  • can be limiting at SWC greater than the wilting point

Question 9

Define matric suction and describe how they influence a soil’s ability to provide water to plants

Answer 9

The negative of matric potential.

The suction force that an organism has to exert to extract water from the soil.

Plants are generally considered to be able to draw water from soil at matric suctions between 10 and 1500 kPa (i.e. 1 and 150 m head).

10 kPa = field capacity = drained upper limit (DUL)

1500 kPa = Permanent wilting point = lower limit (LL)

Question 10

Define field capacity and describe how it influences a soil’s ability to provide water to plants

Answer 10

• Also known as the drained upper limit (DUL)
• 10 kPa or 1 m (matric suction)
• The maximum amount of water that a soil can hold by capillary forces before it is drained by the force due to gravity.
• The official definition (by the Soil Science Glossary Terms Committee (2008)) is: the content of water, on a mass or volume basis, remaining in a soil 2-3 days after having been wetted with water and after free drainage is negligable

Question 11

Define pan evaporation and describe how it influences a soil’s ability to provide water to plants

Answer 11

• A measure of evaporation from an open body of water of fixed dimensions. The pan refers to the vessel.
• Estimates the amount of evapotranspiration
  
  • however, plants differ from pans in several ways
    
    • Less “free” water available
    • More surface area
    • Different radiation absorption characteristics (heat reflection, etc)
    • Different response to wind
  • In some situations (big, healthy crop, moist soil), evapotranspiration > pan evaporation
• Influenced by:
  
  • Temperature
  • Sunshine
  • Wind
  • Humidity
  • Materials that the pan is made out of

Question 12

Define infiltration rate and describe how it influences a soil’s ability to provide water to plants

Answer 12

*

Question 13

How does the plant available water vary depending on soil texture, assuming the same matric

Answer 13

• ↑ fine particles (sand → silt → clay)
  
  • = push curve upward
  • = ↑ water content at a given suction
• PAW_sand = θ_{1 m} - θ_{150 m} = 0.08 - 0.00 = 0.08 m³/m³
• PAW_loam = θ_{1 m} - θ_{150 m} = 0.29 - 0.06 = 0.23 m³/m³
• If you get any rain on a sandy soil, it is immediately plant available. If a clay soil is really dry, even if you get rain it might not be enough to reach the permanent wilting point, meaning it’s not plant available.

Question 14

Define pan evaporation and describe how it influences a soil’s ability to provide water to plants

Answer 14

• A measure of evaporation from an open body of water of specific dimensions to estimate the maximum evapotranspiration from a crop.
• Several factors influence pan evaporation
  
  • Temperature
  • Wind
  • Humidity
  • Sunshine
  • The contruction of the pan (material, colour, etc.)
• Not the best measure of evapotranspiration because a plant is not a pan
  
  • less ‘free’ water available
  • more surface area
  • different radiation absorption characteristics (heat reflection, etc.)
  • different response to wind
  • in some situations (e.g. when you’ve got a big, healthy crop and moist soil), evapotranspiration can exceed pan evaporation.
  • evapotranspiration varies depending on the crop and its growth stage
  • several equations exist to include these factors with the pan evaporation measurements to provide a more accurate representation of crop transpiration (ET_crop).
    
    • To get ET_crop from ET, you need to multiply it by a ‘crop factor’, K_c, which have been developed by organisations like the FAO.

Question 15

True or False?

Evapotranspiration (ET) is dependant on plant available water (PAW).

Answer 15

Yes and no.

Below a certain critical level, which varies between soils, ET has a linear relationship with PAW. (In other words: up to a certain point, ET and PAW have a positive linear relationship; but after that point the relationship disappears).

Question 16

Describe the Philip Equation

Answer 16

Describes the rate of water infiltration into soil

I = St^1/2 + At

Where:

• I = Infiltration
• S = Sorptivity
• t = Time
• A = Steady-state infiltration rate

Sorptivity is high for dry and fine-textured soils

• For dry clays, it takes so long to reach the steady-state that A is negligible.
  
  • I ≈ St^1/2
• For wet sands, the steady state is reached almost instantaneously because there is no sorptive phase, so S is negligible
  
  • I ≈ At

Question 17

True or False?

For dry clays, the infiltration equation can be simplified to

I ≈ St^1/2

Answer 17

True.

The it takes so long to reach the steady-state phase that A is negligible.

Question 18

True or False?

For dry clays, the infiltration equation can be simplified to

I ≈ At

Answer 18

False.

The sorptive phase is the most significant when it comes to dry and fine-textured soils.

Question 19

True or False?

For wet sands, the infiltration equation can be simplified to

I ≈ At

Answer 19

True.

The steady-state phase is reached almost instantaneously, so S is negligible.

Question 20

True or False?

Movement of water through unsaturated soils is faster than through saturated ones.

Answer 20

False.

Water moves through unsaturated soils slower because not all of the pores are filled with water (and water only moves through water-filled pores).

Question 21

Explain what tortuosity refers to

Answer 21

The length of the path that water travels from point A to point B.

Tortuosity increases as soil moisture decreases (water fills the smallest pores first and ‘hugs’ the edge of solid particles).

Question 22

What’s the difference between K and K_s?

Answer 22

K = unsaturated hydraulic conductivity

K_s = saturated hydraulic conductivity

Question 23

True or False?

For very sandy soils, drainage is slow at water contents higher than field capacity, therefore increasing the plant-available water.

Answer 23

True

Question 24

True or False?

For very sandy soils, low unsaturated hydraulic conductivity (K) reduces water accession (↓ PAW) at water contents higher than wilting point.

Answer 24

True.

Question 25

Explain this diagram.

Answer 25

• Sandy soils hold very little water because of their large pores, meaning water drains quickly (i.e. saturated hydraulic conductivity, K_s, is high).
• After the large pores are drained, very little water remains, so the unsaturated hydraulic conductivity decreases rapidly as a sandy soil dries out.
• Because loams and clays have a higher proportion of small pores, they retain more water at any given matric suction, and the rate of K decreasing is much less.
• Even at field capacity, K is so low that the water moves very slowly
  
  • its still not held very tightly, it just doesn’t move.