Hillel, pp. 449 - 465 & 620 - 623
6
The Hydrological Cycle and Water Balance
Evaporation
Infiltration
Deep
Percolation
Root
Extraction
Water
Table
Groundwater
Recharge
CE/ENVE 320 – Vadose Zone Hydrology/Soil Physics
Spring 2004
Copyright © Markus Tuller and Dani Or 20022002-2004
The Hydrological Cycle
● The hydrological cycle describes
terrestrial pathways and
transformation of water.
● Evaporation, precipitation, and
infiltration replenish soil water storage
and recharge groundwater - key life
supporting processes.
● The hydrologic cycle is “driven” by
solar radiation.
Volume
(km3 x108)
% of Total
1370
97.25
Ice Caps & Glaciers
29
2.05
Groundwater
9.5
0.68
Lakes
0.125
0.01
Soil Moisture
0.065
0.005
Atmosphere
0.013
0.001
Streams & Rivers
0.0017
0.0001
Biosphere
0.0006
0.00004
Reservoir
Oceans
● More than 97% of earth’s water
is found in oceans; soil water
storage is less than 0.005% !
● Residence times in various
“hydrological compartments”
vary considerably: >10,000yr
ice caps; >1000yr deep ground
water; <1 yr surface soil water.
Copyright© Markus Tuller and Dani Or2002-2004
The Water Balance Equation
The primary use of soil water content information is for evaluation of the
water balance equation given as:
P + I = ET + DR + RO − ∆W
P ............ Precipitation
I............. Irrigation
ET ......... Evapotranspiration (Soil and Plant)
DR......... Drainage
RO ........ Surface Runoff
∆W ........ Change in Water Storage within the Soil Profile
(Soil Water Depletion)
Evaporation
Infiltration
Deep
Percolation
Root
Extraction
Water
Table
Groundwater
Recharge
• The concept is based on conservation of mass, balancing inputs and outputs
from a soil profile (inputs are taken as positive, outputs negative sign).
• Soil water storage W is defined as the equivalent depth of water stored in a
certain soil depth.
• Changes in storage are calculated for a given time interval (day, year):
∆W=Winitial-Wfinal.
• Under typical conditions ∆W is fairly significant over short period of times
(weeks to months), but generally evens out to zero over one to several years.
Copyright© Markus Tuller and Dani Or2002-2004
The Water Balance Equation
Soil water measurements taken over time are coupled with other
climatic information, such as evapotranspiration or precipitation.
The resulting information may be used for:
- Irrigation Scheduling (time and amount of water)
- Estimation of Evaporation or Drainage
- Determination of Groundwater Recharge
Water balance calculations
may be applied to different
scale
Snow
Rain
s:
Solar
Radiation
Infiltration
Ice
Transpiration
Evaporation
River
Lake
Unsaturated
zone
Water
table
Ocean
- A single profile
- A field
- Watershed scale
- Basin to continental scale
- Global circulation models
Copyright© Markus Tuller and Dani Or2002-2004
Changes in Soil Water Storage
Final 2
)W i= )Z i*)2i
Initial 2
Soil Depth, Z
Soil Depth, Z
2
)W = j)Wi
i
Copyright© Markus Tuller and Dani Or2002-2004
The Water Balance Equation: Example
Example: Soil Water Balance
Problem Statement:
Estimate the drainage of water below a 2.0 m monitoring depth for the nonvegetated location of a recent chemical spill. Assume that the volumetric
soil water content was monitored with a neutron probe having the soilspecific calibration equation:
θ v = −0.0130278 + 0.207666 × CR
The measured neutron count ratios (CR) for 20 cm depth increments at the
beginning and the end of the monitoring period are given as:
Depth [m]
0.0-0.2
0.2-0.4
0.4-0.6
0.6-0.8
0.8-1.0
1.0-1.2
1.2-1.4
1.4-1.6
1.6-1.8
1.8-2.0
CR-Start
1.50
1.60
1.63
1.72
1.55
1.49
1.28
0.97
0.87
0.96
CR-End
0.90
1.10
1.62
1.68
1.56
1.47
1.30
1.05
0.90
1.10
The cumulative precipitation during the measurement interval was 42 cm,
the surface runoff was 3 cm, and the cumulative Evapotranspiration (ET)
was 36 cm.
Copyright© Markus Tuller and Dani Or2002-2004
The Water Balance Equation: Example
We can solve this problem by applying the climatic
water balance equation:
Solution:
P + I = ET + DR + RO − ∆ W
(1) Change in soil water storage
Depth
θvS
θvE
0.0-0.2
0.298
0.174
0.124
24.92
0.2-0.4
0.319
0.215
0.104
20.77
where P is precipitation, I is irrigation, ET is
evapotranspiration, DR is drainage, RO is surface
runoff, and DW is change in soil water storage. We
rearrange the equation to solve for drainage:
0.4-0.6
0.325
0.323
0.002
0.42
DR = P + I − ET − RO + ∆ W
0.6-0.8
0.344
0.336
0.008
1.66
0.8-1.0
0.309
0.311
-0.002
-0.42
1.0-1.2
0.296
0.292
0.004
0.83
1.2-1.4
0.253
0.257
-0.004
-0.83
1.4-1.6
0.188
0.205
-0.017
-3.32
1.6-1.8
0.168
0.174
-0.006
-1.25
1.8-2.0
0.186
0.215
-0.029
-5.81
θvS- θvE
SUM
(2) Drainage below 2.0 m
DR = 420 − 30 − 360 + 37 = 67
∆Wi
36.96
Since there was no irrigation during the
measurement interval, the only unknown is the
change in soil water storage DW.
The change in storage can be derived from neutron
probe measurements. We first calculate the
volumetric water content for each depth increment
at the beginning and at the end of the
measurement period using the calibration equation.
We then take the difference between these θv
values and calculate the change in storage (it is
equal to the change in the equivalent depth of
water) for each increment.
∆ Wi = ∆ θi × Di
[mm] The summation over the entire monitoring depth
gives the change in soil water storage. [make sure
you use the proper sign]
Copyright© Markus Tuller and Dani Or2002-2004
Equivalent Depth of Soil Water De
In the context of the water balance equation it is useful to recall the
concept of equivalent depth of soil water De relating volumetric water
content to water depth similar to climatic quantities (precipitation,
irrigation, evapotranspiration) commonly expressed in equivalent
units of water volume per unit soil surface area (or length).
De = θ v ⋅ D
Where D is the soil depth increment having uniform water content θv
De is the volumetric water content in a given depth increment
expressed as soil water storage (Length)
De is very useful in water balance calculations.
D
θv
De
Copyright© Markus Tuller and Dani Or2002-2004
The Water Balance Equation: Example
Example: Equivalent Water Depth and Redistribution of Rainfall
Problem Statement:
Calculate the depth that a soil profile, initially at uniform wetness of
θv=0.17m3m-3, would be wet to saturated water content of θs=0.49 m3m-3
following 120 mm rainfall. Next, calculate depth of soil profile wetting to field
capacity water content of θFC=0.24 m3m-3 following internal redistribution of
the added water.
Solution:
We employ the relationship D = De/∆θv, solving for the soil depth (D) wetted
from initial θv=0.17 m3m-3 to saturation water content θv=0.49 m3m-3 by 120
mm depth equivalent water (De). We must consider the initial water content
because all pore space is not available for occupancy by the invading
water.
120 mm
D1 =
= 37.5 cm
(0.49 − 0.17)
Next, we calculate redistribution to field capacity using the same
relationship:
D2 =
120 mm
= 171.4 cm
(0.24 − 0.17)
Copyright© Markus Tuller and Dani Or2002-2004
Plant Available Soil Water - Field Capacity
PLANT AVAILABLE SOIL WATER – The amount of soil water
between “field capacity” and “wilting point”.
The concept of “field capacity” is based on observations that very
wet soils tend to drain to a nearly constant value of water content
within a day or two after irrigation or rainfall.
FIELD CAPACITY θvFC is defined as the water content at which
internal drainage becomes negligible. The attainment of field
capacity is not always assured. It is dependent on:
S
iltL
o
m
a
- Initial soil water content and
depth of wetting
- The presence of impeding
layers or a water table effect
extent and rate of distribution
Copyright© Markus Tuller and Dani Or2002-2004
Equivalent Depth of Soil Water De
Conceptual sketch illustrating internal drainage and redistribution as
related to De
Copyright© Markus Tuller and Dani Or2002-2004
Plant Available Water - Permanent Wilting Point
θvWP (WILTING POINT) is the water content at which plants can no longer extract
soil water at a rate sufficient to meet evaporative demand, hence irreversible wilt
and die.
θvWP is dependent on soil texture (specific surface area), and on soil ability to
transmit water, and to a lesser extent on plant’s ability to withstand drought. It is
commonly considered as the water content at -15 bar matric potential.
FIELD CAPACITY and WILTING POINT enable determination of plant available
soil water (more realistic than considering all soil water available to plants!) which
varies with soil texture.
Rule of Thumb
θ vFC ≅
θ v −Saturation
2
θ vWP ≅
θ vFC
2
Copyright© Markus Tuller and Dani Or2002-2004
Hillel, pp. 142 - 154
The Energy State of Soil Water
CE/ENVE 320 – Vadose Zone Hydrology/Soil Physics
Spring 2004
Copyright © Markus Tuller and Dani Or 20022002-2004
The Energy State of Soil Water
The liquid phase content alone is insufficient to
characterize soil water status.
Phenomena such as water exchange between two
soils with identical water contents but different
textures; or water accumulation at bottom of initially
uniform vertical soil column – require examination
of the energy state of the liquid phase.
Like all matter, soil water contains various
amounts and forms of energy, the most
important for hydrological applications are
kinetic and potential energy.
E kin
m ⋅ v2
=
2
E pot = m ⋅ g ⋅ h
Kinetic energy is acquired by virtue of motion – however, because water
moves very slowly through soils we often neglect this form of energy.
Potential energy results from position within a force field or internal
conditions is the primary form of energy that determines movement of
soil water.
Copyright© Markus Tuller and Dani Or2002-2004
The Energy State of Soil Water
Like with all other forms of matter, water flows from locations with
high potential energy to locations of lower potential energy in pursuit
of equilibrium state.
High potential energy
∆ψ1= ψ1- ψ2
i=−
ψ1
ψ1 − ψ2
∆ψ
=−
L
L
L
ψ2
Low potential energy
The driving force for flow is a potential (energy) gradient, or the
difference in potentials between different spatial locations.
Copyright© Markus Tuller and Dani Or2002-2004
The Energy State of Soil Water
Soil water is subject to several forces. Their combined effects result
in a deviation of potential energy relative to a predefined reference
state. This potential difference is termed Total Soil Water Potential ψT
and is expressed as the sum of the various components as:
ψT = ψz + ψm + ψp + ψs +....
ψz
ψm
ψp
ψs
gravitational potential
matric potential
pressure potential
solute or osmotic potential
• DEFINITION - The amount of work that an infinitesimal unit
quantity of water at equilibrium is capable of doing when it moves
to a standard (reference) state.
• As a conventional reference state we consider a hypothetical
reservoir of pure water (no solutes and no external forces other
than gravity) at a reference (atmospheric) pressure, temperature
and elevation.
Copyright© Markus Tuller and Dani Or2002-2004
The Energy State of Soil Water
The difference in chemical and mechanical potentials between
soil water and water at reference state is known as Soil Water
Potential ψw :
ψw = ψ m + ψp + ψs
ψP matric potential
ψS pressure potential
ψV solute or osmotic potential
Soil water potential is thus the result of inherent properties of soil
water itself, and its physical and chemical interactions with its
surroundings, while total potential includes the effects of gravity
which is an external force.
Copyright© Markus Tuller and Dani Or2002-2004
The Energy State of Soil Water
The potential energy of soil water can be expressed in terms of
chemical potential µ (energy/mass), soil water potential ψ
(energy/volume), or soil water head H (energy/ weight).
Units
Symbol
Name
Dimensions
SI Units
Energy/Mass
µ
Chemical Potential
L2/t2
J/kg
Energy/Volume
Ψ
Soil Water Potential
Suction
Tension
M/(L2t2)
N/m2 (Pa)
Energy/Weight
H
Soil Water Head
L
m
ψ
µ=
= g⋅H
ρw
g acceleration of gravity
ρw density of water
Copyright© Markus Tuller and Dani Or2002-2004
Water Potential Conversions - Example
Convert a soil water head of -0.01 m to soil water potential in kPa
and to chemical potential in J/kg
1. From m to kPa
g = 9.81 m/s2
ρw= 1000 kg/m3
ψ = ρw ⋅ g ⋅ H
⎡ kg ⎤ ⎡ m ⎤
ψ = 1000⋅ 9.81⋅ (−0.01) ⎢ 3 ⎥ ⋅ ⎢ 2 ⎥ ⋅ [m]
⎣m ⎦ ⎣ s ⎦
⎡kg ⋅ m⎤ ⎡ 1 ⎤ ⎡ N ⎤
= [Pa]
⋅
=
2 ⎥ ⎢ 2⎥ ⎢ 2⎥
⎣ s ⎦ ⎣m ⎦ ⎣m ⎦
ψ = −98.1 ⎢
− 98.1
ψ=
= −0.098
1000
[kPa]
Copyright© Markus Tuller and Dani Or2002-2004
Water Potential Conversions - Example
2. From m to J/kg
g = 9.81 m/s2
µ = g⋅H
µ = 9.81⋅ (− 0.01)
⎡m⎤
⎢ 2 ⎥ ⋅ [m]
⎣s ⎦
⎡kg⎤ ⎡ m ⎤
⎡kg ⋅ m⎤
⎡1⎤
⎡N ⋅ m ⎤
⎡J⎤
=⎢ ⎥
µ = − 0.098 ⎢ ⎥ ⋅ ⎢ 2 ⎥ ⋅ [m] = ⎢ 2 ⎥ ⋅ [m] ⋅ ⎢ ⎥ = ⎢
⎥
⎣ s ⎦
⎣kg⎦ ⎣ s ⎦
⎣kg⎦ ⎣ kg ⎦ ⎣kg⎦
Copyright© Markus Tuller and Dani Or2002-2004
The Gravitational Potential
ψT = ψz + ψm + ψp + ψs +....
A body on the earth‘s surface is attracted towards the earth‘s center by
a gravitational force that is equal to the weight of the body (w=mg).
To raise a body against the gravitational force work has to be
expended, and this work is stored by the raised body in form of
gravitational potential energy (conservation of energy).
Copyright© Markus Tuller and Dani Or2002-2004
Gravitational Potential
When potentials are expressed as energy per unit weight, the
gravitational potential is simply the vertical distance from a reference
level to the point of interest.
The following examples illustrate that the reference level may be set
at an arbitrary location.
Case 2
Case 1
CASE 1:
Point A is 100 mm above the reference
and point B 200 mm below.
+
Reference
A
+
A
100
400
200
B
100
B
∆ψ z = ψ zA −ψ zB = 100 − (− 200 ) = 300 mm
CASE 2:
Point A is 100 mm below the reference
and point B 400 mm below.
∆ψ z = ψ zA −ψ zB = −100 − (− 400 ) = 300 mm
Copyright© Markus Tuller and Dani Or2002-2004
Matric Potential
ψT = ψz + ψm + ψp + ψs +....
The matric potential results from interactive capillary and adsorptive
forces between the water and the soil matrix, which in effect binds
water in the soil resulting in lower potential energy relative to that of
bulk water.
The value of ψm ranges from zero, when the soil is saturated to often
Tensiometer
very low negative numbers when the soil is dry.
The matric potential per
unit of weight is defined as
the vertical distance
between a porous cup in
contact with the soil and
the water level in a
manometer connected to
the cup [Hanks, 1992]
Copyright© Markus Tuller and Dani Or2002-2004
Pressure Potential
ψT = ψz + ψm + ψp + ψs +....
The pressure potential is the hydrostatic pressure exerted by
unsupported water saturating the soil above a point of interest.
When expressed as energy per unit weight it is equal to the vertical
distance from the point of interest to the free water surface, or
unconfined water table elevation.
The pressure potential is
always positive
(superatmospheric) below the
water table and zero if the
point of interest is exactly at,
or above the water table.
ψp and ψm are
“mutually exclusive“.
Copyright© Markus Tuller and Dani Or2002-2004
Solute Potential
ψT = ψz + ψm + ψp + ψs +....
The presence of solutes in soil water lowers its potential energy and
its vapor pressure.
The effect of ψs is important when (1) there are appreciable amounts
of solutes in the soil; and (2) in the presence of a selectively
permeable membrane or a diffusion barrier which transmits water
more readily than salts.
The solute potential, also called the osmotic pressure, is
proportional to the solute concentration and temperature according
to the van‘t Hoff relationship:
ψs = -R T Cs [kPa]
where R is the universal gas constant [8,314x10-3 kPa m3/(mol K)], T
is the absolute temperature in Kelvin, and Cs is the solute
concentration [mol/m3].
Copyright© Markus Tuller and Dani Or2002-2004