Characterization of Soil Moisture Statusand the Movement of Water in Soils

Example:

You collect a 200 cm3 soil sample. Its moist weight is150 g. After drying, the dry weight is 100 g.

Gravimetric water content:

Moist weight – Dry weight

Dry weight=

Water weight

Dry weight

150 g - 100g 100g

= 50 g = 0.5 or 50%100g

Example:

You collect a 200 cm3 soil sample. Its moist weight is150 g. After drying the dry weight is 100 g.

Volumetric water content:

150 g - 100g 200 cm3 = = 50 cm3 water = 0.25 or 25%

200 cm3 soil

Volume Water

Volume SoilDensity of water

1 g/cm3

50 g200 cm3

Energy-Based

Characterizing Soil Moisture Status

Relating water content and matric potential (suction)

Characterizing Soil WaterCharacterizing Soil Water

Suction applied in discrete increments.

Water RemainingIn soil

Suction applied (cm)0 10,000

One soil

saturated

*

Soil Core

Moisture Release Curve

Texture, DensityTexture, Density

Water RemainingIn soil

Suction applied (cm)0 10,000

saturated

*A

B

Two Soils

coarser

finer

Pore Size Distribution

Water RemainingIn soil

Suction applied (cm) 10,000

saturated

*

SoilWaterContent

Suction applied (cm)

0 -15,000saturation -330F.C.

Soil Water Energy

PWP

Saturation: Water content of soil when all pores are filled Suction equivalent: 0 cm water

Field Capacity: Water content after drainage from saturation by gravity Suction equivalent: - 330 cm water

Permanent: Water can no longer be accessed by plantsWilting point Suction equivalent: - 15,000 cm water

Plant available

Hydraulic ConductivityHydraulic Conductivity

Strongly responsible for water distributionwithin the soil volume.

Determines the rate of water movement in soil.

TextureDensity

StructureWater content

The ease with which water moves through soils

TextureDensity

StructureWater content

Texture – small particles = small pores = poor conductivity

Density – high density suggests low porosity and small pores

Structure – inter-aggregate macropores improve conductivity

Water content – water leaves large pores first. At lower water contents, smaller pores conduct water, reducing conductivity. Maximum conductivity is under saturated conditions.

Coarseuncompacted

Finecompacted

Hydraulic ConductivityHydraulic Conductivity

Clay

SandHigh conductivity

low conductivity

Ponded Water

h

L

A

Flow Volume

Volume time

h * AL

SoIL

WATER

Measuring Saturated Hydraulic Measuring Saturated Hydraulic ConductivityConductivity

Volume time

h * A L

Volume time

= K h * A L

A

soil

h

L Vwater

AK = V * L h * A * t

Approximate Ksat and UsesApproximate Ksat and Uses

Ksat (cm/h)Ksat (cm/h) CommentsComments

3636 Beach sand/Golf Course GreensBeach sand/Golf Course Greens

1818 Very sandy soils, cannot filter Very sandy soils, cannot filter pollutantspollutants

1.81.8 Suitable for most agricultural, Suitable for most agricultural, recreational, and urban usesrecreational, and urban uses

0.180.18 Clayey, Too slow for most usesClayey, Too slow for most uses

<3.6 x 10<3.6 x 10-5-5 Extremely slow; good if Extremely slow; good if compacted material is neededcompacted material is needed

Saturated hydraulic conductivity

Determining Saturated FlowUsing Ksat and the Gradient

Difference in total potential between points Distance between the points

Gradient =

Determining Saturated Flow

Darcy’s Equation

Volume flowArea * time

= Q

A

= Ksat * gradient

Reference levelΨg = 0

Height (cm)

50

20

a

b10

ΨTa = -20 cm

ΨTb =-100 cm

Difference in total potential = 80 cm = 2 Distance between the points 40 cm=

Gradient

Difference in potential energy = -20 cm – (-100 cm) = 80 cm

Gradient =

Distance between points A and B = 40 cm

Darcy’s Equation

Volume flowArea * time

= Q = Ksat * gradient

(Q) = 5 cm/hr * 2

= 10 cm/hr

Difference in total potential = 80 cm = 2 Distance between the points 40 cm=Gradient =

If Ksat = 5 cm/hr, calculate Q

Distance (cm)0

Height (cm)

50

20

a b

10

Difference in total potential -100 - (-200) = 100 cm = 5 Distance between the points 20 cm 20 cm=

5 25

Ψma = -100 cm

Ψga = 0 cm

Ψmb = -200 cm

Ψgb = 0 cmRef.

If Ksat = 5 cm/hr, then the flow (Q) = 5 cm/hr * 5 = 25 cm/hr