Soil Water SOIL 206 – Soil Ecosystem Lab
After completing this laboratory the student should be able to:
1. Explain the terms polarity, hydrogen bonding, cohesion and adhesion as they apply to soil water
2. List and describe the three components of the total soil water potential
3. Describe soil water in terms of water potential and water content
4. Interpret a soil moisture-energy curve with respect to water retention, pore space, pore size
distribution and plant available water
5. Perform calculations involving gravimetric water content and soil water potential
6. Describe the influence of soil texture on plant available water
7. Describe the movement of water in saturated and unsaturated soils using the concepts of soil water
This lab continues your investigation of soil physical properties. Prior labs have taken a detailed look at
soil texture, porosity and density. In this lab you will examine the characteristics, behavior, measurement,
and quantification of soil water. It is virtually impossible to overstate the importance of water in soils.
Water is an important consideration in all aspects of soil science including parent material deposition,
weathering, horizon development, microbial activity, plant growth, and soil erosion.
One of the most important properties of the water molecule is its polarity; the hydrogen atoms in a water
molecule (H2O) exhibit partial positive charges while the oxygen atom exhibits a partial negative charge.
Polarity helps explain how water molecules interact with one another and with other atoms through the
process of hydrogen bonding. Hydrogen bonding, in turn, accounts for two basic forces responsible for
water retention and movement in soils. Cohesion is the force of attraction of water molecules for each
other and adhesion is the attraction of water molecules for solid surfaces.
Adhesive-cohesive forces give rise to surface tension and capillarity. Capillary rise is defined by the
h = (2 * T * cos / (r * * g)
h = the height of rise (cm)
T = surface tension of water
cos = the contact angle of water and capillary wall, assumed to be 1 for water
r = radius of the capillary
= density of liquid, assumed to be water at 1g cm-3
g = force of gravity, a constant of 981 cm sec-2
For water calculations, the surface tension, the contact angle, the density and the force of gravity can be
combined and simplified to:
h = 0.15/ r
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This indicates that the height of rise of water in a capillary tube or a soil pore is inversely proportional to
the tube or pore radius.
Capillary forces are at work in all moist soils. The extent of capillary rise depends somewhat on soil
texture. Capillary rise is usually greater with fine-textured (small pore size) soils as predicted by the
capillary rise equation. It is important to remember that capillarity can cause soil water movement in
horizontal as well as vertical directions. The significance of capillarity in controlling water movement
should be kept in mind as the topic of soil water potential is discussed.
Soil Water Potential
The fundamental principles controlling the behavior of soil water are its energy relationships. The
retention and movement of water in soil, its uptake and translocation in plants, and its loss to the
atmosphere through evapotranspiration can be explained in terms of free energy. Soil water, like other
bodies in nature, contains free energy in different forms and quantities.
Two principal forms of free energy are commonly described: kinetic and potential. Kinetic energy is
proportional to the square of the velocity of a moving body. Since the velocity of water moving in soil is
quite slow, its kinetic energy is generally considered negligible. Potential energy, on the other hand, is
a measure of the amount of work a body can perform by virtue of stored energy. The potential energy of
soil water is governed by its relative position within the soil profile, the arrangement of the solid matrix,
and the presence of solutes. Potential energy is of primary importance in determining the movement of
water in soil. The total soil water potential is the sum of three component potentials:
1. Gravitational potential (g):a function of the height of water above a reference point
2. Matric potential (m): a function of adhesive and cohesive forces, and
3. Osmotic potential (o): a function of the attraction of water molecules for solutes
The free energy of water at a given elevation is greater than the free energy of water at a lower elevation.
Thus the gravitational potential increases the free energy of water and is always positive in sign. The
adhesive and cohesive forces that comprise the matric potential significantly reduce the free energy of
soil water making the matric potential negative in sign. The interaction of water molecules with solutes
also lowers the free energy of soil water and hence, the osmotic potential is also negative in sign. The
combined effects of gravitational, matric and osmotic potentials cause the free energy of soil water to be
less than that of pure water and thus, in most cases, the total soil water potential is negative in sign.
Soil water potential, like other forms of energy, may be expressed using several different units. The most
common unit of water potential is the bar. One bar equals the pressure exerted by a column of water
1023 cm in height. One bar also equals one atmosphere, 14.7 lbs/sq. in. and 760 mm Hg. The SI unit of
water potential is the kilopascal (kPa). One hundred kPa is equal to 1 bar.
Soil Water Content
Another means of describing soil moisture is the water content. Soil water content is simply the weight or
volume of water present in a given weight or volume of soil. The water content of a soil is most
commonly expressed as a percent of the oven-dry weight of the soil. This value, known as the mass
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water content (m) can be determined gravimetrically by drying a moist soil sample at 105 ºC to a
constant weight. The weight lost upon drying represents the water content and thus, the mass water
content (v) can be calculated as follows:
m = Ww/ Ws
Ww = weight of water (g)
Ws = weight of dry soil (g)
Another method of expressing water content is the volumetric water content (v), defined as the
volume of water associated with a given volume of soil. The volumetric water content (v) is usually
calculated from the mass water content (m) and the soil bulk density (Db):
v = m * Db
Volumetric water content is typically expressed in units of cm3 water per cm3 soil. These units allow easy
conversion to acre-inch of water per acre-foot of soil- terminology commonly used to express the volume
of water needed for irrigation. Sample calculations involving mass and volumetric water content are
shown in the following example.
Determination of mass and volumetric water content
Determination of both mass and volumetric water content involves a simple gravimetric (weight)
measurement. A moist soil is first weighed, dried at 105 ºC for 24 hrs and then reweighed. The difference
between the moist and oven-dried soil weights represents the weight of the water. The soil water content
can be expressed on a mass (m) or a volume basis (v). Remember that water content is always
expressed on a dry soil weight basis (i.e. the amount of water associated with 1 g of dry soil, not 1 g of
A soil sample weighs 126 g when moist and 98 g following oven drying. Calculate the mass and
volumetric water content. Assume a bulk density of 1.30 Mg/m3.
Moist soil weight = 126 g Dry soil weight = 98 g
Water weight = 126 g – 98 g = 28 g
m = Ww/ Ws = 28 g water/98 g soil = 0.29 g water/g soil = 29%
v = m * Db = 0.29 g water * 1 cm3 water * 1.3 g soil = 0.38 cm3 water = 38%
1 g soil 1 g water cm3 soil cm3 soil
Soil Moisture-Energy Curves and Plant Available Moisture
The relationship between water content and water potential is illustrated by the soil moisture-energy
curve. A soil moisture-energy curve indicates the manner in which a particular soil retains and loses
water, and also the proportion of stored water that is actually plant available.
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Consider the situation that occurs when a soil becomes waterlogged or saturated. Under these
conditions, the entire pore space (macropores plus micropores) is completely water-filled and the soil is
said to be at its maximum retentive capacity. Under normal field conditions, much of the water held in the
macropores soon drains in response to gravitational force. After gravity has drained the macropores, a
moisture content known as field capacity is attained. The soil at this point is unsaturated and water is
held by capillary (matric) action at potential values ranging from –10 to –30 kPa. Evaporation and plant
use further deplete the soil water supply to a moisture content known as the permanent wilting point.
Soil water present at the permanent wilting point is held at a potential value of approximately –1500 kPa.
Soil water held between field capacity and permanent wilting point is often termed plant available water.
The following example summarizes the calculations involved in quantifying plant available water.
-1 -10 -102 -103 -104 -105 -106 kPa
-0.01 -0.1 -1 -10 -102 -103 -104 bars
Soil water potential () (log scale)
The above graph illustrates the influence of texture on water content-water potential relationships. Note
that the clay-textured soil exhibits the greatest water content of the three soils at all water potential
values. This reflects both high total pore space of the clay, which gives rise to a large storage capacity,
and the extremely small pore size, which results in a low matric potential and strong water retention
force. An inverse relationship exists between the water content of a soil and the energy with which the
water is held. The radius of the pore containing the water determines the energy with which water is held.
The smaller the pore radius, the lower (more negative) the value of the matric potential of the water
contained in the pore and, hence the greater the energy required to remove the water.
Determination of available water
Plant available water in soil is defined as the water held between field capacity (-33 kPa) and the wilting
point (-1500 kPa). A soil sample has a mass water content of 34% at –33 kPa and 18% at –1500 kPa.
Calculate the grams of available water held in 10 kg of this soil.
v at –33 kP = 34%
v at –1500 kPa = 18%
Available Water = 16%
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Movement of Water in Saturated and Unsaturated Soils
The differences in water potential between one point and another give rise to the tendency of water to
flow in soils. Thermodynamic laws dictate that all matter in nature will move spontaneously from a region
of high free energy to a region of lower free energy, and for each parcel of matter to equilibrate with its
surroundings. Soil water, governed by the same laws of equilibria, constantly moves in the direction of
decreasing free energy. Thus soil water flows in response to a free energy gradient; that is from a region
of high (less negative) water potential to a region of low (more negative) water potential.
The quantity of water that moves through a soil profile is determined by two factors:
1. The hydraulic conductivity and
2. The driving force (potential energy gradient).
This relationship is expressed by Darcy’s Law, which for a soil column of fixed dimensions, is:
Q = Ksat *P
= Ksat * [(H2-H1)/ (Z2-Z1)]
Q = quantity of water flowing through the soil
Ksat = saturated hydraulic conductivity, a soil property
P = driving force (potential energy gradient) or
H = the change in hydraulic head
Z = the change in distance.
In saturated soils, the driving force is due primarily to the gravitational potential gradient. The saturated
hydraulic conductivity, or ease with which soil water moves, is dependent on the size and configuration
of individual soil pores. The rate of water movement through a pore is proportional to the fourth power of
the radius of the pore. Thus, as pore size increases, water flow increases exponentially. This explains
why water moves more rapidly through sand than clay. It follows then that water moving in response to a
gravitational gradient through macropores accounts for the majority of water flow in a saturated soil.
Normally, a soil is not at saturation. If drainage is not restricted, gravity will remove a certain amount of
water leaving many of the macropores filled with air. Water in unsaturated soils is governed by the same
general principle discussed above; that is a potential gradient moving water through soil exhibiting a
given hydraulic conductivity. However, both the driving force and the hydraulic conductivity change
significantly as a soil shifts from saturated to unsaturated conditions. Under unsaturated conditions,
water movement is confined to micropores and the driving force is due to a matric potential gradient. This
matric potential gradient is due to the difference in matric potential or moisture content between two
adjacent regions in the soil profile. Water, under unsaturated conditions, will move toward an area with a
lower, more negative matric potential value. For example, water will flow from an area with a matric
potential of –0.1 bar (wet soil) to an area with a matric potential of –15 bars (drier soil).
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Most soils exhibit stratification with respect to texture. That is, strata or layers with varying textures and
pore sizes exist in the profile. In a saturated soil, the size of individual pores profoundly affects water
flow. Water moving through a coarse-textured soil encounters many large pores and the rate of
movement is very rapid. If the advancing wetting front encounters a fine-textured layer, the resistance
caused by the micropores drastically slows water movement.
A more interesting phenomenon occurs when water moving through a fine-textured layer comes in
contact with a coarse-textured layer. As the wetting front encounters the macropores of the coarse
textured layer, downward water movement stops. This is because water movement in an unsaturated soil
is governed by matric potential. The capillary force exerted by the micropores of the fine-textured soil
exceeds the gravitational potential moving the water in a downward direction. Water will move laterally
through the micropores by capillary action, gradually increasing the water content of the fine-textured
soil. Downward movement of water will begin when the fine-textured soil nears saturation, and the
gravitational potential again exceeds the matric potential.
Exercise 1: Field capacity, permanent wilting point and plant available water
From the soil water potential graph above, estimate the water holding capacity, the permanent wilting
point and the plant available water of your soils. Use the particle size line on the graph from the textural
class you calculated in the Particle Size Analysis Lab.
Soil A Soil B
v at FC
v at PWP
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Exercise 2: Flow through saturated soil
If we know Q, the amount of water that will flow through a soil column of given dimensions, we can
calculate Ksat for a given soil, and then apply that value to water movement experiments in the field. In
this exercise, we will measure Q and use Darcy’s Law to calculate Ksat.
1. Obtain a cylinder containing the cores taken during the field trip. These have been soaking to obtain
saturated conditions. When you remove the cylinder from the water, handle with care. Let them drip
for a few seconds.
2. Measure, in cm, the inside diameter of the cylinder and the height of the soil within it. Record these
values. Mark the cylinder 5 cm above the soil level. Calculate the area of the cylinder. Record this
3. Place the cylinders in a funnels. Adjust the funnel height so it will extend into a numbered centrifuge
4. Pour tap water into the core cylinder using care to pour onto the Kimwipe, not the soil surface.
Maintain a level of 5 cm above the top of the soil. We will begin timing when everyone is ready to
begin. Keep the water level constant at 5 cm above the soil surface throughout the timing interval.
We will catch samples until we have three samples + or – 10%.
H2 = Z2 + 5 cm
Z2 = height of soil surface
H1 = 0 Z1 = 0
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Soil A Soil B
Diameter of cylinder
Area of soil surface = * r2
H1 0 0
H2 = Z2 + 5 cm
Z1 0 0
Z2 = height of soil
Volume of water/ 10 minutes
Q = cm3 of water/ hour
Ksat = cm/ hr
Ksat = cm/ sec
Darcy’s Law Q = Ksat * A * [(H2-H1)/ (Z2-Z1)]
Ksat = Q/ (A * [(H2-H1)/ (Z2-Z1)])
= cm3 = cm/ t
t * [cm * (cm/ cm)]
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