AN ABSTRACT OF THE THESIS OF
Fareed H. A. N. Mohammed for the degree of Doctor of Philosophy in Soil Science
presented on July 23, 1992.
Title: Water and Solute Transport : Modeling and Application to Water Conservation
in Layered Soil
Abstract approved:
'I
Redacted for Privacy
Benno P. Warkentin
Sandy soils are among the least productive soils because of their inability to
store adequate water for plant growth. Their high percolation rate not only allows
water to move quickly beyond the root zone, but also washes nutrients below the reach
of plant roots. High evaporation occurs from the soil surface. Many acres of these
soils around the world are left out of crop production. This study is a contribution to
bring these soils into production by increasing their ability to hold more water in the
root zone. Several promising methods of enhancing these soils were simulated,
surface mulch, buried barrier layer, and a combination of both. The effects of varying
texture and thickness of these layers and varying evaporative demand were
investigated. The impact of such modifications on solute distribution in the soil was
also simulated. A simulation model of water and solute transport in layered soils was
developed for this purpose.
The Richards equation for one-dimensional water transport in unsaturated soils
was modified to account for the water jump between the layers. The solute transport
equation was also modified by implementing the same theory of water infiltration in
layered soil to the solute convective transport. The Crank-Nicolson scheme was used
to solve the transport equations with the help of the Newton-Raphson iteration method.
The results of the simulation show that the proposed methods increase water
content in the sandy soil by up to 45%. The combination of barriers, which decreases
leaching and evaporation was the most effective in conserving water. Most of the
contribution came from the influence of the mulch layer in suppressing water losses by
evaporation. The combination method traps solute in the root zone, and this decreased
solute leaching from the soil may limit plant growth in saline soils.
WATER AND SOLUTE TRANSPORT: MODELING AND APPLICATION
TO WATER CONSERVATION IN LAYERED SOIL
by
Fareed H. A. N. Mohammed
A THESIS
submitted to
Oregon State University
in partial fulfillment of
the requirements for the
degree of
Doctor of Philosophy
Completed July 23, 1992
Commencement June 1993
APPROVED:
7,-
2
Redacted for Privacy
Prof sor of SOf Science in charge of major
Redacted for Privacy
Head of Department of Crop and Soil Science
Redacted for Privacy
Dean of the GraduatChool
Date thesis is presented:
Typed by:
July 23, 1992
Fareed H. A. N. Mohammed
ACKNOWLEDGEMENT
I would like to express my appreciation and gratitude to my Professor Dr.
Benno P. Warkentin, who has been for me and will remain the teacher, the advisor,
the guidance, and the friend. I feel very honored to have worked under his supervision.
I would also like to thank Professor Larry Boersma, whose help I have often
sought and have never been denied.
My thanks also extended to Professor James Vomocil, not only for serving on
my graduate committee, but also for giving me his valuable time during my study.
I would also like to thank my graduate committee members, Professor Richard
Cuenca, Professor Jonathan Istok, Professor Arsalan Mazaheri, and Professor Ronald
Neilson.
My deepest thanks and love are for my parents, my wife, and my children for
their patience and support during my study.
i
This dissertation is dedicated to my parents, for
whom my education is their dream, to my wife,
my children, my sisters, and my brother.
ii
TABLE OF CONTENTS
1. INTRODUCTION
1.1 Introduction
1.2 The objectives
1
1
5
2. WATER MOVEMENT IN SOILS
2.1 Factors affecting water movement
2.2 Soil water potential
2.3 Hydraulic conductivity
2.4 Governing equations of water flow in soils
11
3. FLOW RETARDATION AND WATER CONSERVATION
3.1 Inhibition of drainage processes
3.2 Inhibition of evaporation processes
16
17
21
4. SIMULATION OF ONE-DIMENSIONAL UNSATURATED
WATER FLOW IN LAYERED SOIL
4.1 Boundary equation for soil surface
4.2 Boundary equation for soil bottom
4.3 Assumptions of the model
4.4 Solution to water flow equations
4.5 Solution to the upper boundary equation
4.6 Solution to the lower boundary equation
4.7 Hydraulic properties of the soil
4.8 Solution scheme
4.9 Algorithm to solve for water content
5. SIMULATION OF ONE-DIMENSIONAL SOLUTE
TRANSPORT IN LAYERED SOIL
5.1 Solution to solute transport equations
5.2 Transport of solute in layered soils
5.3 Algorithm to solve for solute transport
6. APPLICATION TOWARD WATER CONSERVATION
6.1 Simulation parameters and soil properties
6.2 Case 1: Water and solute transport in uniform soil
6.3 Case 2: Effect of buried barrier layer on soil water conservation
6.4 Case 3: Effect of mulch layer on soil water conservation
6.5 Case 4: Effect of barrier and mulch layers combined on soil water
conservation
6.6 Evaluation
iii
7
7
8
9
28
30
31
32
34
37
38
40
42
46
48
51
52
56
57
58
62
63
66
69
72
TABLE OF CONTENTS (continued)
7. SUMMARY AND RECOMMENDATIONS
7.1 Summary
7.2 Recommendations
75
75
76
REFERENCES
77
APPENDICES
APPENDIX A
83
APPENDIX B
89
APPENDIX C
107
iv
LIST OF FIGURES
Figure 1: Sketch showing a profile of two soils (A and B) and the interface (a).
Figure 2: Soil water characteristic curves for the three soils.
. .
33
60
Figure 3: Hydraulic conductivity as a function of water content for the three
soils.
61
Figure 4: Summary plots for water content distribution after 10 days of
simulation with 1.5 cm day"1 evaporation rate.
73
Figure 5: Summary plots for solute distribution after 10 days of simulation with
1.5 cm day"1 evaporation rate.
74
vi
LIST OF APPENDIX FIGURES
Figure C-1: Water and solute distribution in unmodified soil
(evaporation rate = 0.5 cm/day).
108
Figure C-2: Water and solute distribution in unmodified soil
(evaporation rate =1.5 cm/day).
109
Figure C-3: Water and solute distribution.
(Treatment: 2.5 cm sand barrier layer - evaporation rate = 0.5 cm/day).
110
Figure C-4: Water and solute distribution.
(Treatment: 2.5 cm sand barrier layer - evaporation rate = 1.5 cm /day).
111
Figure C-5: Water and solute distribution.
(Treatment: 5.5 cm sand barrier layer - evaporation rate = 0.5 cm/day).
112
Figure C-6: Water and solute distribution.
(Treatment: 5.5 cm sand barrier layer - evaporation rate = 1.5 cm/day).
113
Figure C-7: Water and solute distribution.
(Treatment: 10.5 cm sand barrier layer - evaporation rate = 0.5 cm/day).
114
Figure C-8: Water and solute distribution.
(Treatment: 10.5 cm sand barrier layer - evaporation rate = 1.5 cm/day).
115
Figure C-9: Water and solute distribution.
(Treatment: 2.5 cm coarse sand barrier layer - evaporation rate = 0.5 cm/day).
.
116
Figure C-10: Water and solute distribution.
(Treatment: 2.5 cm coarse sand barrier layer - evaporation rate = 1.5 cm/day).
.
117
Figure C-11: Water and solute distribution.
(Treatment: 5.5 cm coarse sand barrier layer - evaporation rate = 0.5 cm/day).
.
118
Figure C-12: Water and solute distribution.
(Treatment: 5.5 cm coarse sand barrier layer - evaporation rate = 1.5 cm/day).
.
119
Figure C-13: Water and solute distribution.
(Treatment: 10.5 cm coarse sand barrier layer - evaporation rate = 0.5 cm/day).
120
Figure C-14: Water and solute distribution.
(Treatment: 10.5 cm coarse sand barrier layer evaporation rate = 1.5 cm/day).
121
vii
LIST OF APPENDIX FIGURES (continued)
Figure C-15: Water and solute distribution.
(Treatment: 2.5 cm sand mulch layer - evaporation rate = 0.5 cm/day).
122
Figure C-16: Water and solute distribution.
(Treatment: 2.5 cm sand mulch layer - evaporation rate = 1.5 cm/day).
123
Figure C-17: Water and solute distribution.
(Treatment: 5.5 cm sand mulch layer - evaporation rate = 0.5 cm/day).
124
Figure C-18: Water and solute distribution.
(Treatment: 5.5 cm sand mulch layer - evaporation rate = 1.5 cm/day).
125
Figure C-19: Water and solute distribution.
(Treatment: 10.5 cm sand mulch layer - evaporation rate = 0.5 cm/day).
126
Figure C-20: Water and solute distribution.
(Treatment: 10.5 cm sand mulch layer - evaporation rate = 1.5 cm/day).
127
Figure C-21: Water and solute distribution.
(Treatment: 2.5 cm coarse sand mulch layer - evaporation rate = 0.5 cm/day).
.
128
Figure C-22: Water and solute distribution.
(Treatment: 2.5 cm coarse sand mulch layer - evaporation rate = 1.5 cm/day).
.
129
Figure C-23: Water and solute distribution.
(Treatment: 5.5 cm coarse sand mulch layer - evaporation rate = 0.5 cm/day).
.
130
Figure C-24: Water and solute distribution.
(Treatment: 5.5 cm coarse sand mulch layer - evaporation rate = 1.5 cm/day).
.
131
Figure C-25: Water and solute distribution.
(Treatment: 10.5 cm coarse sand mulch layer - evaporation rate = 0.5 cm/day).
132
Figure C-26: Water and solute distribution.
(Treatment: 10.5 cm coarse sand mulch layer - evaporation rate = 1.5 cm/day).
133
Figure C-27: Water and solute distribution.
(Treatment: 2.5 cm sand barrier and mulch layers - evaporation rate = 0.5
cm/day)
134
viii
LIST OF APPENDIX FIGURES (continued)
Figure C-28: Water and solute distribution.
(Treatment: 2.5 cm sand barrier and mulch layers - evaporation rate = 1.5
cm/day).
135
Figure C-29: Water and solute distribution.
(Treatment: 5.5 cm sand barrier and mulch layers - evaporation rate = 0.5
cm/day).
136
Figure C-30: Water and solute distribution.
(Treatment: 5.5 cm sand barrier and mulch layers - evaporation rate = 1.5
cm/day).
137
Figure C-31: Water and solute distribution.
(Treatment: 10.5 cm sand barrier and mulch layers - evaporation rate = 0.5
cm /day).
138
Figure C-32: Water and solute distribution.
(Treatment: 10.5 cm sand barrier and mulch layers - evaporation rate = 1.5
cm/day).
139
Figure C-33: Water and solute distribution.
(Treatment: 2.5 cm coarse sand barrier and mulch layers - evaporation rate = 0.5
cm/day).
140
Figure C-34: Water and solute distribution.
(Treatment: 2.5 cm coarse sand barrier and mulch layers - evaporation rate = 1.5
cm /day).
141
Figure C-35: Water and solute distribution.
(Treatment: 5.5 cm coarse sand barrier and mulch layers - evaporation rate = 0.5
cm/day).
142
Figure C-36: Water and solute distribution.
(Treatment: 5.5 cm coarse sand barrier and mulch layers - evaporation rate = 1.5
cm/day).
143
Figure C-37: Water and solute distribution.
(Treatment: 10.5 cm coarse sand barrier and mulch layers - evaporation rate = 0.5
144
cm /day).
ix
LIST OF APPENDIX FIGURES (continued)
Figure C-38: Water and solute distribution.
(Treatment: 10.5 cm coarse sand barrier and mulch layers - evaporation rate =1.5
cm/day).
145
x
LIST OF TABLES
Table 1: Parameters used to describe the soil hydraulic properties
59
Table 2: Percentage increase in water content due to the presence of barrier
layer at the end of the simulation period.
65
Table 3: Percentage increase in water content due to the presence of mulch
layer at the end of the simulation period.
68
Table 4: Percentage increase in water content due to the presence of both
buried and mulch layer at the end of the simulation period
71
v
WATER AND SOLUTE TRANSPORT: MODELING AND APPLICATION
TO WATER CONSERVATION IN LAYERED SOIL
1. INTRODUCTION
1.1 Introduction
Water is the most essential element for all creatures, without which no life is
expected. Plants are no exception, they depend totally on water. In an actively
growing plant, water constitutes more than 90 percent of its fresh weight. For plants
to survive they must supply enough water to satisfy the atmospheric demand, and to
maintain enough water in cells to complete the metabolism processes. Regardless of
water resource, the main source of water for plants is the soil water. Soils have a great
influence on plant survival; storage and conduction of water to plant roots are the
limiting factors for plant growth. With adequate water in the soil, plants show no signs
of wilting and grow very well. However, soils show great variability in their ability to
store and conduct water. The amount of water transmitted or stored in the soil is
highly affected by physical properties such as texture and structure.
Some soils can store water for longer periods because of lower transmission
while others store water for very short time. Retarded water movement can create
drainage problems.
Sandy soils are among the least productive soils because of their inability to
2
store adequate water for plant growth, and their high percolation rate which not only
lets the water move faster beyond the root zone, but also washes nutrients to a zone
out of reach of plant roots.
These soils require short irrigation intervals and frequent fertilizer applications,
but this increases the production cost and will be feasible only in areas where there is
enough water. In arid regions and places short of irrigation water, the problem of
sandy soils becomes more serious. That is why many acres of sandy soils are left
unused in agricultural production. This problem becomes worse when such soils are
in an extremely arid location with very high evaporation rates, rare rainfall, and
shortage of irrigation water. Modification of such soils can increase and enhance their
productivity.
Many studies have been conducted in past decades to investigate the feasibility
of enhancing soils to increase water holding capacity. Some of these studies were
concerned with rapid deep percolation of the water. Barrier layers made of compacted
soil, plastic sheets, asphalt, or soil conditioners were mainly used to decrease water
movement below the soil root zone (Diebold, 1954; Jamison and Kroth, 1958;
Eagleman and Jamison, 1962; Miller and Bunger, 1963; Willis, et al, 1963; Erickson,
et al, 1968; Hedrick and Mowry, 1952; Unger, 1971; Garber and Zaslaysky, 1977;
Miller, 1979; Hemyari and Nofziger, 1981; Tayel, et al, 1981; Tayel and El-Hady,
1981; Kumar, et al, 1984; Johnson, 1984; Baasiri, et al, 1986; and Wallace, et al,
1986).
Other studies were concerned with decreasing evaporation from the soil surface using
3
different types of mulching (Lemon, 1956; Hanks and Woodruff, 1958; Willis, 1962;
Willis, et al, 1963; Greb, 1966; Griffin, et al, 1966; Unger, 1971; El-Hady, et al,
1981; and Hartmann, et al, 1981).
In both cases the studies show significant increases in water content in the soil.
Application of such work would solve the problem of sandy soils to a certain degree,
depending on the type of application. The response to such applications varies from
one soil to another.
Many questions arise in considering modifying a given soil under a given
climatic regime. What is a suitable method; would a barrier layer be sufficient, or
would a mulch layer alone be sufficient, or both be required; what would be the
optimum thickness, density, and porosity of the barrier layer or mulch layer; what
would be the optimum depth of the barrier layer for maximum water storage ?
Answering all these questions would involve many experiments. For each soil and
different climatic regime another set of experiments would be required. Although
experiments can give reliable results, they are costly and time consuming.
Where the soil processes are understood, simulation models are very important
tools for predicting effects in a short time and with low cost compared with
experiments. A good simulation model can answer all those questions in short time
with low error.
This study concerns modification of sandy soils to increase their ability to
retain water. This can be done by changing structure or physical properties, adding
soil conditioners or semi-impermeable layers beneath the root zone, or compacting the
4
lower layers to decrease the size of macro pores that are responsible for rapid
percolation. Also evaporation from the soil surface can be reduced by modifying the
soil surface with a soil conditioner or adding a layer of coarser material to break
continuity of liquid water conducting to the soil surface.
Soil water determines solute behavior in soils; accordingly, any soil
modification that affects water flow will have a direct influence on solute displacement
and distribution in the soil. A particular method may increase water content, but at the
expense of increasing soil salinity which may threaten plant growth. The method of
soil modification must not develop a salinity or drainage hazard.
5
1.2 The objectives
The primary objective of this study is to simulate and evaluate methods to
increase soil water storage by modification of the soil root zone. The consequences on
solute behavior will be studied and considered in the evaluation. The work will be
achieved in five steps:
1. Model development.
A mathematical model will be developed to simulate one-dimensional water
and solute transport in layered bare soil.
2. Computer program.
A computer program will be developed to estimate water and solute transport
in layered soil.
3. Model verification and calibration.
One or two sets of measured data (soil hydraulic properties) of a layered soil
will be used to calibrate the model. The model will be modified as necessary.
4. Simulation.
The behavior of the modified soil will be simulated to determine time and
quantity aspects of water storage. Several methods will be used to deal with
the upper (mulching) and lower (barrier) layers of the soil profile. Details of
these methods will be described in the following chapters.
The impact of a given method of modification on soil salinity will also be
modeled and simulated.
6
5. Evaluation and method selection.
All the methods will be evaluated under given soil and climatic conditions in
order to select the optimum one. The optimum method is defined when the
balance of all of the following components is accomplished by the method:
a.
More water is stored in the root zone for the benefit of the plant.
b.
Water remains for a longer time in the root zone for the benefit of the
plant.
c.
No hazards of solute that may threaten plant growth occur in the root
zone.
d.
No hazards of drainage problems that may threaten plant growth ere
developed in the root zone .
7
2. WATER MOVEMENT IN SOILS
2.1 Factors affecting water movement
During and after rainfall or irrigation, water penetrates the soil surface and moves
downward to lower layers in the soil. Soil physical properties and the atmosphere
determine rate and direction of water movement in the soil.
The atmosphere may affect water movement directly through evaporation from
the soil surface, causing upward movement of water, or by adding water to the soil
through rainfall and initiating downward movement. The effect of the atmosphere also
occurs through plant water uptake.
The soil physical properties of soil texture and soil structure control water
storage and transmission in the soil. They determine the soil water potential and soil
hydraulic conductivity which are the basic parameters of soil water movement.
Soil water moves from places of higher potential to areas of lower potential to
reach equilibrium. The difference in water potential between two points is the driving
force which causes the movement of water in the soil. Evaporation and water uptake
by plants cause decreases in soil water potential, and hence cause water movement to
those sites. Due to its continuous movement, soil water in the field rarely reaches
equilibrium. The rate of water movement is governed primarily by the hydraulic
conductivity of the soil.
8
2.2 Soil water potential
Soil water potential or total potential of water in the soil, V, is the summation
of three major components that affect the state of water in the soil, namely,
gravitational potential, Vg, pressure potential, 1pp, and osmotic potential, v.
4r =
P
+ *0
(1)
Gravitational potential is affected by the elevation of a point under study from
an arbitrary reference level. The osmotic potential becomes an effective part in the
presence of solutes and a membrane, for example, in the interaction zone between
plant root and soil or where water vapor moves by diffusion.
The most important part of the total soil water potential is the pressure
potential. It is the product of capillarity and adsorption forces between water and soil
particles. The relative importance of these two mechanisms depends upon soil texture
and soil structure, which determine the pore size distribution and the continuity of the
capillary system in the soil. Clay soils exhibit both adsorption and capillarity, while in
sandy soils, the capillarity is dominant.
At higher values of water potential (0 to -1 bar), the amount of water retained
in the soil depends mainly on capillarity and hence pore size distribution; therefore it
is highly influenced by soil structure. At lower water potential, water is retained in the
soil primarily due to adsorption, which is affected mainly by soil texture, (Hillel,
1982). Clay soils tend to retain more water than sandy soils at various levels of water
potential. This is due to a number of reasons. Clay soils have more surface for
9
absorption than sandy soils, which allow them to retain more water at very low levels
of water potential. Second, sandy soils contain macro pores, which are the first to be
emptied at a given potential, while clay soils contain mainly micro pores in a more
uniform size distribution.
Changing soil structure will affect water retention, for instance, compaction of
sandy soil will decrease the number of large pores. Although the amount of water at
saturation will be decreased, compaction will also reduce the initial water content loses
when soil water is subject to a small negative potential.
The soil water potential is related to soil water content as illustrated in the soil
moisture characteristic curve. The relationship between both is highly nonlinear and
affected by hysteresis. The shape of the curve which describes the relationship
between pressure potential and water content of the soil depends on the direction of
water flow in the soil. During the infiltration (wetting) process, the values of water
content for a given pressure potential, at equilibrium, is less than that in the
redistribution (drying) process.
23 Hydraulic conductivity
The soil hydraulic conductivity is defined as the ratio of water flux to the hydraulic
gradient. When the soil is saturated, the entire volume of soil pores conducts water. At
this point the hydraulic conductivity is maximal and is called saturated conductivity.
As water content decreases and the larger pores empty, water exists in the small pores
10
and in envelopes surrounding soil particles. In this situation, water is conducted
through the small pores and between the envelopes, accordingly, water conductivity
decreases and is said to be unsaturated.
The hydraulic conductivity is influenced by soil factors such as pore size
distribution, total porosity, and tortuosity. Other factors are fluid dependent, such as
viscosity and density. Temperature affects hydraulic conductivity through changes in
viscosity and the density of the fluid, and accordingly will change the hydraulic
conductivity. Entrapped, released, or dissolved air during water flow will also change
the measured hydraulic conductivity (Yong and Warkentin, 1975, Hi llel, 1982, Iwata
et al, 1988).
For a given soil, the unsaturated hydraulic conductivity strongly depends on the
water content, the relationship between conductivity and water content is highly non
linear. The soil hydraulic conductivity reaches its maximum magnitude at saturation
and as water content decreases, the hydraulic conductivity decreases rapidly. This
relation varies from soil to soil and is affected by hysteresis.
In fine-grained soils such as clay soils, the hydraulic conductivity is very low
due to the narrow water conducting pores, but these pores are continuous at low level
of water content compared to coarser-grained soils. Sandy soils which have a large
number of macro pores tend to have higher conductivity at saturation than clay soil.
But as water content decreases, the conductivity of sands decreases more rapidly than
in finer grained soils owing to the emptying of large pores which are responsible for
the rapid water movement.
11
2.4 Governing equations of water flow in soils
The equation of water flow in porous media was first introduced by Darcy,
1856, in his study on water filtration through sand beds. The form of his equation,
known as Darcy's Law, may be written as:
q= K
L
(2)
where:
q = water flux, L T-'
K = hydraulic conductivity, L T-1
v = hydraulic head, L
L = length of soil column, L
The assumptions for this equation are that the flow is steady and saturated in a
uniform soil. The relationship between the hydraulic gradient and the flux is linear,
and the slope of the line represents the hydraulic conductivity. Therefore, average
variables of hydraulic head and hydraulic conductivity along the flow path are
sufficient to calculate the flux using this equation.
Obviously these assumptions rarely occur in the field. Most likely, the soil
water in the field is neither steady nor saturated, and therefore, water content, water
potential, and hydraulic conductivity vary along the flow path. Accordingly,
generalized variables such as averaging over hydraulic head and conductivity will not
12
be suitable to be used in the field conditions. Localized variables are more suitable to
be used in conditions where flow is transient and unsaturated such as under field
condition.
Richards, (1931), considered these conditions when he extended the Darcy
equation to describe water flow in unsaturated conditions. The form of his equation
for one-dimensional downward flow may be written as:
(3)
K(Ap) t
q=
where z is soil depth. In this equation, the hydraulic conductivity is no longer a single
value of the soil, but is a function of pressure potential.
By combining the Darcy equation with the continuity equation, the general
water flow equation for one-dimensional transient condition can be written as
(Richards, 1931):
ae
at
a
(4)
az -
where t is time and 0 is water content. By substituting Equation.3 into Equation.4 for
vertical flow, it becomes:
amt
az
( lc( 'PP)
(21-az
(5)
As mentioned earlier, soil water potential, V, is a summation of pressure potential, llip,
and gravitational potential, vg, (neglecting the osmotic potential, ip.). Therefore
Equation.5 becomes:
13
ae
alc(ipp)
aZ
atifp
a
aZ
(
aZ
)
(6)
This form of Richards equation neglects the hysteresis in the function K(ipp).
Therefore, application of this equation in situations where wetting and drying
processes occur (redistribution) will be affected by hysteresis (Miller and Miller,
1956). However, the hydraulic conductivity as a function of water content, K(0) is
less hysteretic than K(4p) (Hillel, 1982), therefore Equation.6 could be written as:
aax (e)
))
a ( K(e) app
aZ
aZ
ae
(7)
aZ
To simplify Equation.7, the flux could be related to either soil water content, 0, or soil
water pressure potential,Vp. To solve for water content, the pressure potential
gradient must be replaced by moisture gradient by using the chain rule as follows:
ae
agyp
az
de
(8)
az
then Equation.7 will be:
ae
at
a
aZ
K(e)
c
ae )
at aZ )
8K(e)
aZ
(9)
If the hydraulic diffusivity term, D(0), is applied, the two dependent variables Opp and
0) in Equation.9 can be reduced to one. D(9) is defined as:
14
D(6) = K(0) dip
(10)
at
The values of dvp/d0 can be found from soil moisture characteristic curves. Then
Equation.9 becomes:
ae
at
_a
az
1
az )
az
ax(e)
az
To solve Equation.7 for water potential, the following form may applied:
at
alif,,
64 at
a
alpp )
az (K(*P) az
aKopp)
az
(12)
Again, the values of dO /dI4 can be found from soil moisture characteristic curves.
Although Equation.11 is less subject to hysteresis than Equation.12, application
of these equations in highly hysteretic soils will not be free from error. Equation 11
and 12 can be used to solve unsaturated transient flow in uniform soils only.
Soils in the field are uniform neither in the material nor in properties. The soil
profile typically consists of different layers, and each layer has its own properties.
Accordingly, water content, water potential, and hydraulic conductivity vary with
distance (different layers may have different characteristics). Water flow in such soils
is unlike that of homogeneous soil. Each layer has its own relationship for flowing
water, and the least permeable layer controls the other layers when they are combined
in one domain.
Therefore, Equation.11 and 12 cannot be applied in layered soils. Modifications
with respect to soil heterogeneity must be added to these equations in order to make
15
them suitable to solve water flow problems in layered soils. More details about this
will be provided in the next chapters.
16
3. FLOW RETARDATION AND WATER CONSERVATION
When rain or irrigation stops, water stored in the soil starts to decrease due to
evaporation, drainage, and plant use. The rates of these processes of redistribution
determine the amount of water stored in the soil for a given time, and are very
important in the water supply for plants.
The parameters which affect soil water storage are identified in the general
water balance equation:
AS =P+I-R-D-E-T
(13)
where AS is the change of water storage in the soil, P rainfall, I irrigation, R runoff, D
drainage, E evaporation, and T is the transpiration. If it is assumed that there is no
runoff and the amount of water which is depleted by the plant (transpiration) is the
purpose of soil water storage, then the undesirable losses of water from the soil would
be through evaporation and drainage. Therefore, to maintain an adequate level of water
in the soil, the processes of evaporation and drainage must be eliminated or decreased
in a way that the desired amount and quality of stored water in the root zone is
achieved.
This type of water conservation was recognized and studied by many
researchers, among them Hide (1954); Lemon (1956); Jamison (1960); Eagleman and
Jamison (1961, 1962); Willis (1962, 1963); Miller and Gardner (1962); Miller and
Bunger (1963); Griffin, et al (1966); Greb, et al (1967); Corey and Kemper (1968);
17
Erickson, et al (1968); Bond and Willis (1969); Miller (1979); and Unger (1971).
Some of these studies concerned limiting the evaporation from soil, others were
involved in decreasing deep and rapid percolation. Most of the studies show
encouraging results toward the application of such practices for water conservation,
specially in arid regions.
3.1 Inhibition of drainage processes
Water draining below the root zone to deeper soil layers in which the plant
roots don't grow, is considered lost. Elimination or delay of water movement beyond
the root zone will increase soil water storage.
As reviewed earlier, water movement in soil is affected mainly by soil texture
and structure, through the effect on soil water retention and hydraulic conductivity. It
was also mentioned that fine-grained soils retain water through both capillary forces
and adsorption forces between water and soil particles, while in coarse-grained soils
capillarity is dominant. As a result, fine-grained soils tend to retain more water at a
given suction. Water flow at close to saturation is much slower in fine-textured soils
than in coarse-textured soils, due to the large pores in the later. But as water content
decreases, the conductivity in coarse-grained soils decreases more rapidly than in finegrained soils.
This behavior of the soils can be used to favor soil water conservation,
specially in sandy soils. For instance, in a situation where fine-textured soil overlays
18
coarse-textured soil, and under unsaturated water flow conditions, the initial water
infiltration is controlled by the upper soil. When water reaches the interface, the
infiltration rate may decrease since the wetting front is under high suction, which does
not allow entry into the large pores of the lower soil. When the suction decreases
sufficiently at the interface, due to pressure head build up, then water enters the lower
soil (Hillel, 1982). Thus, water is conserved in the upper layer due to the delay of
water movement, and plants could benefit from this water.
The effect of soil profile heterogeneity on water infiltration has been studied by
many researchers. Alway and Mc Dole (1917), found that with an interrupting (coarse-
textured) layer of sand or gravel, the amount of water in the upper soil increased by
6% when compared to soil without an interrupting layer.
Miller and Gardner (1962) conducted extensive experiments on water
movement from fine-textured to a coarse-textured soil. They observed that the
infiltration was temporarily inhibited when the wetting front reached the interface.
After water accumulated and the suction of the soil decreased sufficiently, water
entered the coarser layer. The degree to which water flow is retarded was related to
the differences of the pore size of the layers. As pore size of the coarser layer
increases, the degree of flow retardation increases. The effect of gravity on the
infiltration into coarser material was also studied. They found that upward infiltration
was retarded for a longer time than downward infiltration. The water did not enter the
coarser layer in upward infiltration during 15 hours of observation when 77% of the
pores consist of 0.2 to 0.6 mm in size. Water entered the coarse layer in 303 second
19
for downward infiltration.
Miller and Bunger (1963) studied the effect of thickness and depth of the layer
of coarse sand and gravel layers on water retention in the soil profile. They reported
that a coarse-textured layer increased the water retention of the finer overlaying layer.
They also observed that soils tend to retain more water with a gravel layer than with a
sand layer.
Eagleman and Jamison (1962), tested the effect of soil layering on unsaturated
water flow. When they placed fine-textured soil on top of coarse-textured soil, the top
layer restricted the drainage to the lower layer. They found a very small decrease of
water content (from 0.44 to 0.40 cm3.cm-3) of the top layer over 26 days after
saturation . They observed that when water had to move to the lower layer, from small
to large pores, the suction of the lower layer had to be as great as that of upper layer.
The soil moisture characteristic curves of fine and coarse-textured soils showed that
fine-textured soil would retain more water at the same suction when compared to
coarse-textured soil. Stroosnijder, et al (1972) obtained similar results with a fine sand
layer overlaying a coarse sand layer.
Sandy soils exhibit high percolation rate, the excess water from rainfall or
irrigation moves downward rapidly and away from the reach of the roots. This water
could be retained by placing a barrier layer in the soil. Erickson, et al (1968), studied
this phenomenon using an asphalt barrier in sandy soil. Their report showed that water
content of the soil was increased, and as a result, the yield of crops growing in the soil
was increased significantly.
20
Another approach to increase soil water storage is the utilization of soil
conditioners. Numerous studies where conducted. Hedrick and Mowry (1952) observed
an increase in soil moisture content when the soil was treated with synthetic
polyelectrolyte. They concluded that the treatment enhanced the soil structure and the
effect was not an association of water with the polymer itself. Miller (1979) found that
a polymer "super slurper" increased the water content of sandy soil to the level of
loam and silt loam soils. He also observed a reduction of infiltration rate. The polymer
had little effect on fine-textured soils. The same result was reported by Hemyari and
Nofziger (1981). Tayel, et al (1981); Baasiri, et al (1986); and Johnson (1984b) found
similar results with gel-forming polyacrylamide (PAM). However, the effect of soil
conditioners and their water absorption properties may significantly decrease when the
soil or irrigation water contains dissolved salts (Johnson, 1984a).
"Super gel" enhances soil water holding capacity. Tayel and El-Hady, (1981)
found that the soil porosity and water retention was increased, while the rapidly
drained pores, hydraulic conductivity, mean pore diameter, and evaporation was
decreased when the soil is treated with "super gel".
The application of manure to sandy soil may enhance its hydraulic properties.
A study by Kumar, et al (1984) has shown that manure increases soil water retention
and decreases the hydraulic conductivity of sandy soils.
21
3.2 Inhibition of evaporation processes
Evaporation is an important component of water balance in soil. In arid
regions, evaporation is considered to be the largest loss of water from the soil. The
rate at which water evaporates from the soil depends mainly on two factors. The first
factor is controlled by soil properties, which determine the amount of water that can
be transmitted to the evaporating surface, which is in turn controlled by soil water
potential and hydraulic conductivity of the soil. The second factor is the atmospheric
condition, which includes the amount of energy supplied for evaporation and the vapor
pressure gradient between the evaporating site and the atmosphere.
When the soil surface is wet, evaporation rate is initially constant and
determined by the atmospheric demand. This is the first stage of evaporation. The
second stage starts when the soil dries sufficiently and water transmission to the
surface is decreased. The evaporation rate declines substantially in this stage. When
the soil surface becomes dry and liquid water supply to the surface has ceased, water
moves by the vapor diffusion process, which is very slow compared to liquid water.
This is the third stage of evaporation. (Hide, 1954; Lemon, 1956; Hillel, 1982).
The greatest water loss occurs during the first stage of evaporation, where the
maximum rate of evaporation occurs. This process lasts as long as water is supplied to
the surface, with the assumption that the atmospheric evaporative demand is
continuous. While the duration of this stage is generally short when compared to the
second and third stages, significant amounts of water can be saved if this stage is
22
terminated rapidly.
Suppressing evaporation from soil surface has received the attention of Hide
(1954) and Lemon (1956). Hide (1954) studied the moisture exchange between soil
surface layers and the atmosphere. He indicated that there are at least three ways to
decrease soil water evaporation: (1) decrease water movement to the soil surface
before any drying occurs; (2) decrease the driving force for evaporation through
decreasing the temperature of soil surface; and (3) increase the resistance to vapor
diffusion by increasing the thickness of soil-atmosphere boundary layer.
Lemon (1956) reviewed the possibilities of decreasing losses of soil water by
evaporation. His review included his work and Russian research work, as well as other
studies. He said about these possibilities:
The potentialities for decreasing soil moisture evaporation lie in
the first two categories of the loss process. These potentialities group
themselves roughly into three divisions: (a) decreasing the turbulent
transfer of water vapor above the ground surface; (b) decreasing the
capillary conductance of water to the surface by decreasing capillary
continuity; and (c) decreasing the capillary conductance of water to the
surface by the application of surfactants."
Attempts to suppress evaporation due to turbulent transfer include limiting
wind speed over the soil surface. Several methods have been employed to slow wind
speed, including standing stubble, application of mulch material, trees or other
obstacles as wind breaks, and increasing soil roughness. Hanks and Woodruff (1958)
investigated the effect of wind on water vapor transfer for different mulches including
soil, gravel, and straw. They found that as wind increases evaporation increases. This
is due to increase in vapor diffusion as wind speed increases. Evaporation was
23
increased up to 2-6 times when wind increased from 0 to 25 mile/h for mulch
treatments, while in no mulch treatment evaporation increased by 10 times. They have
shown that soil mulch is more efficient to reduce evaporation than gravel or straw
mulch especially at high wind speed. Under gravel and straw mulch evaporation was
higher than under soil mulch. Gravel or straw mulch may, therefore, not be effective
in reducing evaporation in windy regions. Benoit and Kirkham (1963) conducted
similar experiments with addition of varying radiation. They obtained similar results as
Hanks and Woodruff (1958) except that gravel mulch was found to be superior in
evaporation reduction to straw and dust mulches. No explanation was given for this
difference.
Suppression of evaporation by reduction of wind speed may not be effective,
specially in warm regions. Poor wind circulation causes an increase in soil temperature
which results in an increase in evaporation. Also, plant residue mulch may increase
soil temperature and serve to retain heat especially at night when soil releases heat to
the atmosphere (Lemon, 1956). Radiation effect on soil water evaporation is greater
than wind effect (Hanks et al, 1967).
Interruption of continuity of liquid-water flow to the soil surface may serve as
a barrier to upward water flow, when moisture flow to the surface is forced to be in
the form of vapor diffusion, which is a much slower process when compared to liquid
flow. The thickness of this barrier has a great influence on vapor diffusion. Lemon
(1956) indicated that the maximum effect of the barrier appears at the first inch of its
thickness, after that, the barrier thickness effect diminishes. Hanks and Woodruff
24
(1958) indicated that about 96% reduction in evaporation was observed when the
mulch depth increased from 0 to 0.25 inch. Greb, et al (1967) found that the higher
application of straw mulch results in less water loss due to evaporation. Hil lel and
Berliner (1974) found that when the depth of a hydrophobic aggregate layer increases
the evaporation suppression increases. Similar effects were also found when the size of
the aggregates was increased.
Lemon (1958) referred to experiments conducted by Kolasew (1941) to test the
effect of soil stratification on the losses of soil moisture. The sequence of the layers,
from top, was compacted-loose-compacted-loose. It was found that the soil water loss
was decreased with the stratification. Lemon (1958) reported that Kolasew (1941)
interpreted this reduction as due to two factors. The loose soil isolated the compacted
layers, and accordingly, capillary continuity was minimized; also, less vapor
conductance through compacted layers occurred because of the reduction of porosity.
Willis (1960) placed a coarse-textured layer over fine-textured soil. He found
that evaporation was decreased even with the presence of a water table. Evaporation
was also affected by the thickness of the coarse layer.
Modaihsh, et al (1985) obtained remarkable results when they applied sand
mulch to suppress evaporation from the soil surface. The effect of thickness and
texture of sand mulch under two different evaporation demands were examined in their
study. They reported that a sand mulch layer of 6 cm thick was superior in reducing
evaporation in all cases when compared with layers of 2 and 0 cm thick. The effect of
texture was generally significant for both thickness of mulch layer and evaporative
25
demand. However the effect of the texture ranked second in importance.
Greb (1966) showed that straw mulch significantly reduced water loss via
evaporation. Up to 70% reduction at the early evaporating stage was observed.
However, when the soil water is depleted the effect of straw mulch diminishes. He
reported that the effect of the mulch increased when the amount of mulch was
increased. Similar results were obtained by Bond and Willis (1969) with rye straw
mulch; Dais ley, et al (1988) with grass mulch; and Bristow and Abrecht (1989) with
coconut fiber as mulch.
Hanks, et al (1961) tested net radiation, soil temperature and evaporation as
affected by surface cover including straw, black gravel, aluminum-painted gravel , and
plastic. They reported that the highest net radiation was observed under black gravel,
while soil temperature was highest under clear plastic. All of the treatments suppressed
evaporation, however, differences between the mulches were not significant in the
evaporation treatments.
The use of chemical additives as soil conditioners has shown reasonable
success in decreasing evaporation from a soil surface. Hillel and Berliner (1974) found
that hydrophobic aggregates reduced evaporation significantly. Callebaut, et al (1979);
and El-Hady, et al (1980) obtained similar results. Callebaut, et al (1979) indicated
that a great reduction in evaporation occurs in the first stage. Hartmann, et at (1981)
found that water loss can be decreased by 50 % when the soil surface is treated with
soil conditioners to form aggregates. Kowsar, et al (1969) showed an increase in soil
water content with the application of petroleum mulch. They concluded that the mulch
26
prevented loss of water vapor from soil surface layer. Unger and Stewart (1974)
observed reduction in evaporation from a soil surface treated with feedlot waste.
Kumar, et al (1984); and El-Asswad and Groenevelt (1985) found that manure
significantly reduced evaporation, again a great reduction found to be during the first
stage of evaporation.
Rapid drying of the soil surface is another approach to decrease evaporation
from the soil surface (Penman, 1941). This procedure is based on reaching the critical
point at the end of the first stage and the beginning of the second stage of evaporation.
Significant reduction of evaporation depends on how fast the critical point can be
reached. This is the third potential process for decreasing evaporation according to
Lemon (1956). Numerus research studies have been conducted. Kolasew (1941),
quoted by Lemon (1956), treated a soil surface with hydrophobic material. He found
that the treated soil dried more rapidly and reached the critical point at higher water
content. After that point, the evaporation rate was slower for the treated soil.
Several other methods for suppressing evaporation from soil surface have been
used. Willis, et al (1963) tested the effect of surface and subsurface plastic film as a
mulch on soil moisture and temperature. Maximum water use efficiency was observed
with 90% surface cover, and maximum yield was obtained for this treatment. They
believed that the increase of yield was due to the increase of soil temperature. Buried
plastic mulch (7-10 inches below surface) was not efficient. They also simulated the
same experiment under controlled conditions in the laboratory, and found that a
surface cover of plastic mulch reduced the evaporation significantly. Furthermore, the
27
higher the percentage of soil cover, the less evaporation was observed.
Griffin, et al (1966), investigating the effect of soil surface cover on water
retention, found that a plastic cover (100% surface cover) reduced water loss by
evaporation. Willis (1962) found that when evaporation demand increases, partial
cover is not effective, and full cover is required for efficient evaporation reduction.
Mbagwu (1991) tested the effect of straw mulch, black polythene, and white
polythene on the temperature and water content of the soil surface. He found 60 days
after rainfall the soil water content was four times greater with mulch than without
mulch. Furthermore, the permanent wilting point (PWP) was reached after 15 days for
bare soil, while with mulch, the soil retained enough water during 60 days after
rainfall and PWP was not reached.
Jury and Bellantuoni (1976b) studied the effect of surface rocks on heat and
water movement in the soil. They observed considerable amount of water retained by
the soil beneath rock when compared to bare soil. The reason for this behavior is that
the rocks help in retaining soil heat at night. Net heat flow is toward the rocks, which
induces water movement. Also rocks act as a barrier to evaporation (Jury and
Bellantuoni, 1976a).
28
4. SIMULATION OF ONE-DIMENSIONAL UNSATURATED
WATER FLOW IN LAYERED SOIL
The general water flow equation on the basis of water content for unsaturated
soil was discussed earlier in chapter 2. This chapter will focus on the modification
and the solution of the equation to accommodate water flow in layered profiles. The
water flow equation based on water content for unsaturated homogenous soil can be
written as:
ae
at
a
ae
(D(e) a.)
ax(e)
az
(14)
where 0 is the volumetric water content (L3 L-3), t is the time (T), z is the soil depth
(L), K(0) is the hydraulic conductivity as a function of water content (L T-1), and D(0)
is the soil hydraulic diffusivity as a function of water content (L T1). This equation
has been widely used in solving problems of unsaturated flow in homogenous soils,
however, the implementation of equation 14 in water flow problems in layered soil
requires some modifications.
Figure 1 shows an example of a two layer profile, soil A and soil B, with
interface a between them. For water to leave soil A and enter soil B it has to pass
through the interface controlled by the characteristics of both soils. If the water
content is a continuous function of depth between two different layers, the solution
would be as simple as that for homogenous soil.
Since the water content across the interface may not be a continuous function
29
of the depth, equation 14 is not satisfied for this type of water flow. On the other
hand, soil water potential is a continuous function of depth
.
Hills etal (1989) introduced a new variable to account for the jump of water
content across an interface and they show that the water content-based formulation can
be used to simulate water flow in layered soil with the addition of a source term to the
right-hand side of equation 14. Accordingly, equation 14 was modified for the points
that surround the interface to handle the jump in water content between two layers.
Therefore, equation 14 near the interface becomes:
ae
at
aaz
ae)
ax(e)
az
az
s
(15)
to evaluate water content at a point just above the interface, and
ae
at
a (...,,,,,
nip\ ae )
az
az
ax(e)
az
4-
s
(16)
to evaluate water content at a point just below the interface. The variable S is the
source term that accounts for the jump, its derivation and the algorithm for its
evaluation are given in details by Hills etal (1989). The evaluation of S will be
illustrated in the subsequent sections.
Equations 14-16 together can be used to evaluate water flow in layered soil,
where equation 14 is applied for the entire length of soil layer and equation 15 and 16
are applied only at the interface between two layers.
To obtain a unique solution for the system, boundary equations must be added
for the upper and lower boundaries of the system as well as an initial condition. The
30
initial condition will be:
e (i, t) = °initial
i = 0, 1,
...
,m
at t = 0
(17)
where %alai is the initial volumetric water content (L3 L-3).
4.1 Boundary equation for soil surface:
The objectives of this study as stated earlier are to simulate the effect of soil layering
on water storage and the impacts on solute distribution. Commonly, this type of
investigation requires a dynamic boundary that involves infiltration and evaporation,
therefore, a flux boundary condition was chosen at the soil surface. Accordingly, the
flux condition at the surface is:
q = -D(0)
00 + K(e)
(18)
az
where
q(0 , t) > 0
at
tZ0
,
(19)
q(0, t) = 0
at t z 0
,
(20)
g(0, t) < 0
at t
Or
0
(21)
where q is the Darcian flux (L3 L-2 T -1) and is equal to the rate of water introduced at
the soil surface. It is either positive when the condition is infiltration, zero for the
31
redistribution, or negative for the evaporation condition. Since the exact rate of
infiltration is subject to soil conditions, higher values of q that exceed the infiltration
capacity of the soil may introduce an error in the solution, therefore, low rates of
water flux at the surface are applied in this study.
4.2 Boundary equation for soil bottom:
A semi-infinite soil profile is assumed for the lower boundary, that is, water will not
reach the bottom of the soil during the simulation period, and there is no influence of
water table. Hence, the water gradient at the soil bottom is assumed to be zero as
described by the following equation:
ae cz
--ai
'
0 =0,
at
z = zm ,
t20
(22)
where m is the maximum depth of the soil profile. Values of z at z = z, must be
relatively large, especially when the period of the simulation is large and/or higher
rates of infiltration are specified at the surface, to ensure that water will not reach the
bottom of the soil column during the computation.
32
4.3 Assumptions of the model:
This model approximates one-dimensional water flow in layered soil under the
following assumptions:
1.
While the effect of hysteresis on water distribution is very common in soils,
inclusion of hysteresis in the solution adds some complications. Therefore, it is
neglected.
2. The flow of water is isothermal.
3. The contribution of water vapor transport is considered very small, thus neglected.
4. Water flow at (or more than) complete saturation (100% saturation) is not
supported. The maximum saturation allowed does not exceed 0.999999 of e.
5. Presence of solute has no effect on water flow.
33
Figure 1: Sketch showing a profile of two soils (A and B) and the interface (a).
34
4.4 Solution to water flow equations
The Crank-Nicolson finite difference approximation for equation 14 is:
ei
ei+1
or+1
D(e) zil++11
Az3
At
-
or
en
2 A z2
D (e) 1+_1/A
or + ei
62-1
(23)
2Az,
Az3
K (e) ri'++1/A
K(e) ipN
A z3
where:
+1
+ D(e) +1 + D
D ( ) fr+1. + D (e) j
D(e) 11:1/A
4
D(0)r1-1
(25)
+ K(8)1+1 + K(0) 1i2.1 + K(8)1
4
(26)
D(3)/p.1-/A - D(0)1+1 + D(e)fi'l
K(e)
12
K(8)11:1/2
-
K(e)
+1
D(e)fl
4
K(0)11:1 + K(e)
4
A
(24)
= zi
K(e) z1-1
(27)
(28)
35
zi
(29)
Z1 -1
(30)
Az2 = zi+1
Az3
Z1+1
2
i and n are spatial node and time increments respectively, where i = 2, 3, ..., m -
1,
where m is the last node in the domain.
Note that equation 14 is evaluated only at the interior nodes which are not neighboring
the interface. For the two nodes that surround the interface only equations 15 and 16
are applied, therefore, the solution for the node that is just above the interface will be:
er
el
D(e)r+1%2
At
err
erii++1 + ez1+1
ezi
2Az2
Az4
DM) /1+_M
( Or
+ el
Az4
01_1 )
erf_l
(31)
2A z1
n+-1:
K(e)
IC(e)r+1//
1
D(e) n+1/2
1i
Az3
Az4
1
AO 2
AZ2
and for the node just below the interface will be:
pi
1e.n+
At
el
_
(3'1:1 + oil.,
2Az2
D(e)/I++1;
Az5
_
D(e) rli,
1
ern
0/1+1 + el
Az5
er
ez!
eT., )
(32)
2A z1
K(e)1.=
A;
n+ 1
K(e) i-
+
n+1/2
D (13) 1_1/2
A z5
n1
+
AO
A;
2
36
where:
A zi
A z4 = A z2 +
2
A z2
A Z5 = A Zi +
Aen+12
-
2
Aen + Aen+1
(33)
(34)
(35)
2
Note that the interface is required to be centered between the two surrounding nodes.
The last term in the right hand side of equation 31 and 32 is the source term S which
accounts for the jump of water content between soil layers, and AO is the difference in
the water content between the bottom of the upper layer and the top of the lower layer
at the interface. AO is evaluated according to Hills etal (1989):
AO = 0+
e-
(36)
where 0' is the volumetric water content (L3 L-3) at the bottom of the soil layer just
above the interface and 0+ is the volumetric water content (L3 L-3) at the surface of the
soil layer just below the interface. The values of 0' and 0+ are unknown and they
cannot be evaluated directly; their average is assumed to equal the value of water
content at the interface O. (Hills etal, 1989). O. also cannot be evaluated directly since
the water content is not continuous across the interface, whereas water potential at the
interface V. is continuous and can be estimated by using the conservation of mass
equation for the flux entering and leaving the interface. These fluxes are required to be
37
equal since the thickness of the interface is zero and no storage can take place in the
interface. Accordingly V will be (Hills etal, 1989):
2 { K(0)i * (e),
+
illa
K(e),+, ter (e)1 +1 } + Az { K(0)1
2 { lie), + K(e)141 }
K(e),+, }
(37)
where
Az = zi+1- zi
(38)
i in equation 37 and 38 represents the last node in the upper layer which is very close
to the interface and i+1 is the first node in the following layer which is also very close
to the interface.
As soon as V. is obtained, 0' and 0+ can be found from the corresponding
V(0) relationship for each soil, and accordingly AO can be calculated and substituted in
the source term in equations 31 and 32 to satisfy the dynamic equation over the
interface. ( hereafter the notation V will be used to represent soil water potential, vp).
4.5 Solution to the upper boundary equation
The upper boundary is of the flux type, therefore the finite difference formulation for
the node at the soil surface will be:
er- cn
At
D (e)
( Z2
K (e)
n1:111/22
[
Z1) 2
1 ++11//22
Z2- Zi.
er1
err' ]
(39)
qn+1
Z2
Z1
38
The last term in equation 39 represents the value of the boundary at an imaginary
node at half space above the soil surface, where q is the Darcian flux (L3 L-2 T-1)and
equal to the rate of water introduced at the surface. It is either one of the following:
g(0, t) = 1+0
q - Infiltration
- Redistribution
q - Evaporation
at
tz0
(40)
and
D(0) 112:1/2
K(e) 121,2
D(o)r +D(e)r-
(41)
2
K(0) ri
K(e) 2+1
(42)
2
4.6 Solution to the lower boundary equation
A semi-infinite lower boundary was chosen, where no flow can take place below the
lower boundary at depth z = z,., therefore a fictitious node was specified below the
soil profile at z = z,+1. This node is always identical to node m-1 and contains its
identity. Accordingly, the solution for the lower boundary is similar to that in
equation 23 with the exception of the fictitious node (which is equal to node m-1)
instead of node m+1 as in equation 23. Therefore, the solution will be as follows:
39
ezirl
-
An
-.7
At
n(iii)f,++1//
ez;+1
----
of
A z3
2 A z2
_ D(0);2,7+-11A
er + e:
Az3
_ K(e)in,+1
efr
ef,+_1
eni_l
(43)
2Az3.
K(0)
j2
A z3
where Of is the volumetric water content (L3 L-3) at the fictitious node (equal to 0..1 ),
and
D(e)if'+1 + D(e)A +1 + D(0)1; + D(e)r,:,
4
D(o)m +.1%2
D(e)mni.i. + D(e)m,":1
D(e)m
(45)
4
+-i%2
K(e)ir + lc-mg..' 4. K(0)z;
K(e)mn
4
ic(e)mzi:1;
(44)
-
K(0).12+1 + K(e)mn+-1 + K(e)n + K(0),9_1
4
(46)
(47)
Note that the choice of the depth of final node and the final time of simulation require
little attention, in other words, flowing water must not reach the bottom of the profile
during the computation, otherwise an error in the mass balance will result.
40
4.7 Hydraulic properties of the soil
Equations 23, 31, 32, 39, and 43 require the relationships of K(0), ii,(0), and
D(0) in order to be satisfied. Those functions are usually introduced in tabulated form
to numerical models, or, more commonly, as empirical equations. In this study the
van Genuchten equations were used because of their continuity over the entire range
of water content which is very useful for numerical simulations (Paniconi, 1991)
.
The soil characteristic equations as given by van Genuchten (1980); van Genuchten
and Nielsen (1985) are:
K(S.) = K. 4/2 [ 1
( S-1 /m
41 (Se)
D(S)
e
(1
=
(
am
(1
G
1 )1/n
) si/2
ifin
(50)
e
Sel/rn r + ( 1
cc = (es
(49)
a
m) Ks
(es er)
*[(1
Sellm )57
1 + (a 111)n
2
m
1
or)
(48)
)m]2
Si1,./fil
1
+ Or
]
(51)
where
S.
e
es
er
(52)
or
0 is the actual volumetric water content (L3 L-3), 0, is the residual volumetric water
41
content (L3 L-3), 0, is the saturated volumetric water content (L3 L-3), and K, is
saturated hydraulic conductivity (L T-1).
a and n are empirical coefficients
characterizing the soil and determined by experiment, and m is defined as:
m=1
1
n
(53)
Using these equations, K(V), V(0), and D(0) can be determined for a given 0, and
0(v) can be found for a given ii, as well. However, when 0 approaches saturation,
D(0) approaches infinity. Therefore, in this study, the saturation water content is
limited during the calculation to 0.999999 0.
42
4.8 Solution scheme
Since equations 23, 31, 32, 39, and 43 are non linear, because the coefficients K(V),
v(3), and D(0) are themselves functions of the dependent variable that is sought,
therefore these equations are solved by iteration. Non-iterative methods are also used
in such problems but they require very short time steps. In this study the NewtonRaphson iteration method was used to linearize the equations. The form of NewtonRaphson method for non linear equation is:
xr+1 - xr . _ f ( xr )
f' ( xr )
(54)
where the superscript r donates values obtained in the rth iteration. The term in the
left-hand side of equation 54 is the error term, the iteration is terminated when its
value is very close to zero. In this study a system of non linear equations needed to
be solved, so a matrix system was used to replace the parameters in equation 54 as
follows:
{E}
a
where {E} is the error term vector, defined as:
{F}
F]
[
(55)
43
or.
of
- of
cri+l
el-
ef.+3.
{E}
=
(56)
ef,,+1 - ef,
{F} is defined as:
F1
F2
{F}
=
(57)
.
Fi
Fm
where F.
Fi = f 01.1:1, Or , en) = 0
(58)
This represents equations 23, 31, 32, 39, and 43, for corresponding nodes along the
soil column. It is obtained by taking the right-hand side of these equations to the left-
hand side, and equating them to zero. Accordingly F; will be as follows:
F1 =
un /17:1/A
en Dzezlez,
1
1
+ er At
[
er.1
[
en
"1 +1
[
+
1
A z3
Dce) 7:1/A
2A z2A z3
D(e) 1.1/2
2Az2
2A;
2A ziaz3
(59)
eil_1
J
1
1
D an '12:W
A z3
2Az2
2Az2Az3
D (0) 112`..1/1
D cfn /1'!1,
At
[Dun niA
+
+
1
D (0) ri)!W
2AZi
Kan 2:1/A
Kce)fi'!W
Az3
I
for interior nodes that are not close to the interface,
=0
44
[D(0):111/A
2A ziA z4
Fi = -
[1
eg+1
At
D(e)1:M
Az4
2Az2
,
D (co 2++M I
er+1
2AZ2 Az4
1
ej
el+i
1
et
[
2
D(e)/1.V
2AZ1 AZ4
1
D (0) niA
D(e)ITA
Az4
2Az2
2Az1
Dm=
2A z A 7-4
4- then+112
lc ( e ) j +1 /2
lc ( e ) 2: M
(60)
D(e) 2:m
2AZ2Az4
2
{
n1/2
D(6)i_1/2
Az4
=0
for the node above the interface, and
F=
e+_M
D(6)2
2Az 1 Az5
n+
er
e2++1
1
At
1
(D(e) /114+1/2
Az5
2 A 22
[ Dui) n+312
1+1/2
2AZ2AZ5
1
At
1
Az,
D(e)ni 1,22
2AZ2 AZ5
en
1+1
[
tin
2
{ D(e)r_M
2A zlA z5
(DM l+1/A
2Az2
_.+,1
Aen+1/2
D(e)
2 A zj.
D(e)2:M
2Az2Az5
x(e) 2++1/2
K(e) 2:M
Az5
for the node below the interface.
D(e)ri/A
=0
(61)
45
The boundaries also take similar form as follows:
or
Fl =
or
+
D(e) zi +1-
1
[
At
zi) 2D
( z2
0,1
D (6) il:W
z1) 2D
(z2
[
qn+i
Km :1].+FW
z2
(62)
At
z2
Zi.
0
z1
for the boundary at the surface, and
F. =
'1
[ D (e) ,,,n+W
A zlA z3
en+i[ D (e) inlli
1
Az1 Az3
er
+
1
1
{
At
1
ef,
I
r
D (0) Ili=
+
A zlA z3
_ D (0) 1,2+_W
At
A zlA z3
(63)
0
for the lower boundaries.
The term a[F] in equation 55 is the Jacobean matrix U ] and it is composed of
a series of partial derivatives of Fi with respect to 0, at future time step n+1 for i = 1,
2, ..., m. When U] is obtained, it resembles the following:
8F1
aF1
aeri aeri
aF,
ao
0
J=
aF,
aF2
aer.
a0,3,+1
aF3
8F3
0
0
...
0
0
8F3
aer a8,37.1 aer
0
0
0
8F1
0
0
II
II
0
(64)
aF1
aell awl'
aF1
0
0
a31:1
aFft,
aF,
aen+1
aen*1
m-1
m
46
The partial derivatives of F. with respect to 0. at time n+1 in the Jacobean matrix are
shown in appendix A. Given {F} and [j] and rearranging equation 55 to obtain:
[J]
.
{E}=
{F }
(65)
since[J]is a tridiagonal matrix, the error term {E} can be solved by using Thomas
algorithm at the iih iteration. Then the new approximated value for water content will
be:
{
0,2.1
} = { en } + { E )
(66)
The iteration continues until convergence. The convergence criteria that is used here
depends on the balance of the function F, according to equations 59-63 which require
the right value of 0 to be zero. Accordingly, the iterations stop and the solution of
water content at time t is found when the following criterion is met:
i.,
E 1 Fi
i.I.
1
5
0.00000001
(67)
4.9 Algorithm to solve for water content
The following steps are required to obtain the solution for water content;
1.
Future value of water content is guessed for each node by using the old value of
the corresponding node.
2. Evaluate function IFJ,
When Fi is evaluated for the nodes surrounding the
47
interface the following procedure is followed to estimate AOn+% according to Hills
etal (1989):
a.
Soil water potential is estimated at the interface by using equation 37 for future
t = n+1.
b. Using the corresponding V(0) relation, 0+ and Wean be found, and accordingly,
AO' can be calculated using equation 36.
c.
Repeat steps a and b for t = n to obtain AO'.
d. From equation 35, obtain A6 "+4.
3. During the evaluation of {F, }, compute the sum of absolute F. .
4. Compute the Jacobean matrix.M.
5. Solve for the error term {E} using Thomas algorithm.
6. Check the convergence criterion (equation 67), if satisfied, terminate the iteration
and go to step 7, otherwise:
er = el + Ei
go to step 2.
7. If the final time is not reached, go to step 1.
8. Report the simulation results.
(68)
48
5. SIMULATION OF ONE-DIMENSIONAL SOLUTE
TRANSPORT IN LAYERED SOIL
Solute movement in soils has received a great deal of attention due to its importance
in plant nutrition, soil reclamation, pesticide movement, groundwater contamination,
and environmental quality. Since this study involves modifications in soil layers, it is
worthwhile to investigate the impacts of such modifications on solute distribution.
Solutes in the soil are transported by three mechanisms, namely, convection,
diffusion, and hydrodynamic dispersion. In the convection process, solutes are
transported by the mass flow of water in the soil, therefore, the rate at which solute is
transported is a function of solute concentration in the soil water and the flux.
Diffusive transport is the process of molecular diffusion, and the rate of transport
depends on the solute concentration gradient and the solute diffusion coefficient, as
described by Fick's Law. The third process is hydrodynamic dispersion which is
caused by unequally distributed water velocity in individual pores and by the tortuous
flow path; its value depends mainly on the pore water velocity.
The general equation for one-dimensional solute transport in the soil can be written
in the following form (Bresler and Hanks, 1969; Warrick etal, 1971; and Bresler,
1973):
me
+ A)
a
( e DS
)
a(gc )
az
s
( 69 )
where 0 is the volumetric water content (L3 L-3), c is the solute concentration in soil
49
water (M L-3), Ds is the apparent dispersion coefficient (L2 l'-'), v is the flow velocity
(L T-1), q is the solution flux (L3 L-2 T-1), t is the time (1), z is the depth, positive
downward, (L), A is the local concentration of solute in the adsorbed phase (M L-3),
and S is the source-sink rate (M L-3 T-').
Ds combines both terms, molecular diffusion and hydrodynamic dispersion. It is
defined as:
(70)
Ds = Dm + Dh
where Dm is the molecular diffusion coefficient and ph is the hydrodynamic dispersion
coefficient which can be given by:
(71)
Dm = Ds T
where D. is the diffusion coefficient of solute in pure water, and -r is the tortuosity
factor, dimensionless. In most soils -r ranges between 0.3-0.7 (van Genuchten and
Wierenga, 1986).
Dh= A
I
(72)
VI
where X, is the dispersivity factor (L), its value is
1 cm for disturbed soils, and
about one or two orders of magnitude for undisturbed field soil (van Genuchten and
Wierenga, 1986), and V is the average pore-water velocity (L l'-') and given by:
v= 2.e
(73)
For simplicity of the model, the last two terms 'A and S ' are neglected, and the
model solved for only solute concentration in the liquid phase without source-sink
50
terms. Accordingly equation 69 is reduced to:
a(6c) - a ( e D (v) ac )
at
az
az
(74)
a(qc)
az
The boundary conditions for the solute transport are solute flux for the surface and
semi-infinite for the bottom. They can be defined by the following equations (Bresler,
1973):
J(0, t) =- 03 .Ds(v) ac + q,,,,(0 , t) c (0 , t)
az
(75)
with
J(0, t)
=
{
- Infiltration
q(0, t) c (0 , t)
- Redistribution
0
0
- Evaporation
at
t
(76)
0
for the surface boundary. For the bottom boundary, the gradient of solute
concentration is assumed to be zero as described by the following equation:
ac
(z, t) =0,
at
z = zli,
,
t
0
(77)
where m is the maximum depth of the soil profile. Semi-infinite soil profile means
that solute will not reach the bottom of the soil during the simulation period.
Therefore, the value of z at z = zm must be relatively large, specially when the period
of the simulation is large and/or higher rates of infiltration are specified at the surface.
This will assure that solute will not reach the bottom of the soil column during the
computation.
51
The initial condition for the solute problem is:
c(i, t) = c, ;
i = 1, 2,
.
at t = 0
,m
.
(78)
where c, is the initial solute concentration (M L-3).
5.1 Solution to solute transport equations
The finite difference solution for equation 73 with the implementation of CrankNicolson scheme is:
eir.icr,+,_
cin
[02:1/4
At
( vff+U: )
n «1
2Az2Az3
Ds(v1:111)(
cf+1-
+
n
n+1
Ci
2AziAz,
n+1
Ci-i
)
n
C1-1 )
(79)
n +1/2
Ni.1.1/2
/ _n+1.
4Az3
n:1/2
a1112
4Az3
n
ui+1
n+1
n+1
)
n+1
n
+ cin+ Cl-1+
ci_i
)
where
cin+1/2
vi+1/2
cin+1/2
V1-1/2
,,,n+1/2
Yi+1/2
gan+1
vi+1
an
vi +1 -r
er
nn
4
n+1
e1 +1
an+ 1
An
u.1-1
111-1
(80)
(81)
4
I pir+11//22
el++11/2- 0'1+1'2
16i Z2
K(0/++N
(82)
52
n+1/2
Cii-1/2
,/an+1/2%
ku1-1.12,
nn+1/2
6i
6i
11
gen+1/ 2
ui
111-1
-1
(83)
(erill// 22
A z,
or. +
(84)
2
At
,,n+1/2
vi+1/2
,n+1/ 2
Vi+/2
An+1/2
(85)
an+1/2
(86)
u1+1/2
n+1/2
Vi-1/2
u1-1/2
and for the boundaries:
eni+1cin+1_
ei= Ds(virZ:)
c2n_
cin )
2 (Az2)2
gf:al
cf+1+ c2n+ cf+1+
4Az2
(87)
C1
+
for upper boundary, where J is the value of flux boundary as defined in equation 76.
The lower boundary is the same as equation 79 but with a fictitious node.
5.2 Transport of solute in layered soils
When a water content based formulation is used with equation 74 to simulate
solute transport, equation 74 can't be applied directly to heterogenous soils. In order to
be implemented for layered soil, it must be modified. The adjustments which are
required here are in the part where water flow has contributions to solute transport,
53
namely, the convection transport of the solution which depends mainly on q. In water
content based models, q depends on the gradient of water content. Since water content
here is not a continuous function of depth, error will result when the solute flow
equation is applied. To illustrate this, consider for example the case when a coarsetextured soil overlaying fine-textured soil, both soils are not saturated, and the water
content for the lower layer is greater while water potential is lower. Accordingly the
lower layer will adsorb any water coming from the top layer, and the flow direction
will be downward (positive direction) as stated by the Darcy equation, assuming the
effect of gravity on water flow at this point is very small:
q=
--1 + K(t)
az
K (ijr )
(88)
Based on water content, equation 88 will be:
q=
D(e) ae
+
az
K(e)
(89)
01
(90)
the approximation of equation 82 can be given by:
q1+1/2 =
D(e)i+1/2
ei +i
Az
+ K(e)i+1,2
Now, when equation 82 is used for this example q will be negative, which means, the
flow is upward, when it should be downward. Therefore, because of the discontinuity
of water content in such a condition, evaluation of q in equation 74 for the convective
solute transport must be modified to account for solute transport over the interface.
Two methods are presented here to compute the solute flux across an interface:
54
Method 1:
Hills etal (1989) presented a procedure to evaluate water flow across an interface
between two different soil layers. The same theory also can be implemented here for
solute transport in layered soils. Provided that AO and O. are known (as discussed in
the previous chapter) and by using half space step size, solute flux in equation 79 for
nodes surrounding the interface can be evaluated by:
e:+112_
qr1/2 =
Aeln+1/2_
Yi3+1/2)
n+1/2
U
2
K(ej +1/2)
(91)
+ K(1311:1/2)
(92 )
Az
2
for the node just above the interface, and
gf++33:r2
el++1/2 _ ean+1,2_
,,n+1/2
(Vi +1
k
Aen+1/2
2
)
Az
2
for the node just below the interface. Thus,
,1241/2
1-41+1z
_
V1+1
+
_n+1/2
Vn+1/2
i
(93)
2
This evaluates q,+1,2 for node i above the interface, which also evaluates .71.1,2 for node
i+1 below the interface.
Since the interface thickness is zero and no storage can take place at the interface, the
flux entering the interface is equal to the flux exiting from the interface (conservation
of mass), 0 can be evaluated from:
55
e
ea
12
e
(94)
Substituting equation 94 into equations 91 and 92 will provide an expression to
evaluate q between the nodes that surround the interface:
n +1/2
j+1
-Dce+1)
ep+1/2
- er/2_ Aen+1/2
Az
n +1/2
c(e+)
95qa
)
Note that q412 for node i above the interface is equal to qi_112 for node i+1 below the
interface.
Method 2:
In this method, soil water potential replaces the water content when computing
solute flow between two layers. Since the values of water content are known (by
solving water flow), values of Vi and Ipi+1 for the nodes surrounding the interface can
be found from the corresponding v(0) relationship of each soil. Thus, for Vi and Vi+1
substitute Ai and Ai+, in Darcy's equation as follows:
,...z2+1/2
Vi+1/2
_Tpickn+1/2
j1+112
drn+1/2
6n+1/2
i
Az
Dr I Cill+1/2
W1+1/2 /
(96)
Note that qi+12 for node i above the interface is equal to qi,112 for node i+1 below the
interface.
56
53 Algorithm to solve for solute transport
By solving the water flow equation for a given time step, all the variables in equation
73 are known except solute concentration c" +', therefore, the solution is straight
forward and there is no need for iteration. The following steps illustrate the algorithm
for each time step after solving the water flow for the same time step:
1.
Compute time-space centered 0 using equations 80 and 81, q using equations 82
and 83 for any node that is not close to the interface, and equation 95 or 96 for
nodes adjacent to the interface, V using equations 85 and 86, and Ds(V) using
equations 70, 71, and 72.
2. Now the only unknown is ci"'. Since there are m unknowns and m equations, the
system of equations can be solved for cin+1 by using the Thomas algorithm since
the coefficient matrix is tridigonal.
57
6. APPLICATION TOWARD WATER CONSERVATION
The mathematical model developed in the previous chapter is applied here to
investigate the effect of soil layering on water and solute transport in soils, and the
consequences on soil water conservation in the root zone. For this purpose a program
written in C language (presented in appendix B) was used to simulate the simultaneous
transport of water and solute.
Effects of mulch as well as buried barrier layers, against unmodified soil were
simulated in this study. Four cases were investigated. The first case was the original
soil, without modification. In the second case a barrier soil layer composed of coarse-
textured soil was added to the original soil at depth 70 cm below the soil surface,
where roots are assumed not to be present below it. In the third case a soil mulch
layer was added to the soil surface, the soil used for mulch is also coarse-grained. And
in the fourth case both mulch and buried layers were added to the soil.
In the last three cases, the effects of the thickness and texture of the added layers
were examined. The thickness was 2.5, 5.5, or 10.5 cm. Two soils of different texture
,
sand and coarse sand, were used in the modification of the original soil (loamy sand).
The properties of these soils are given in table 1 as well as in figure 2 and 3.
58
6.1. Simulation parameters and soil properties
All of these cases were simulated under two different evaporation demands in
order to investigate the validity of such modification under moderate and severe
conditions of water losses. The rates of evaporation chosen were 0.5 cm day-1 as
moderate condition and 1.5 cm day-1 as severe condition.
The impacts of soil modification on solute distribution were also investigated for
all cases. Because of this, only one species of solute was used in this study and
applied to all cases. The molecular diffusion coefficient D. for the solute is 1.0835
cm2 day-1 (Kemper, 1986) and the dispersivity factor k assumed to be 0.4 (van
Genuchten and Wierenga, 1986).
For consistency and unbiased simulation, all the cases received the same treatments
of simulation parameters which include:
- Soil Depth: The length of the soil column was chosen to be 100 cm.
- Node Distribution: The node distribution along the soil column varies with length. At
the first 15 cm from the soil surface, where the mulch layer is to be examined, it was
every 1 cm, then followed by 5 cm increments until a depth of 65 cm, after that by 1
cm increment until depth of 85 cm, and then followed by 5 cm increments up to the
end of soil column at 100 cm.
- Initial condition: The initial water content was 0.108, 0.10, and 0.03 cm3cm-3 for the
original soil (loamy sand), sand, and coarse sand soils respectively. The initial solute
concentration was assumed to be 0.0 meq cm-3 for all of the cases.
59
- Soil surface boundary conditions: The simulation starts with continuous irrigation or
rainfall at a rate of 7.0 cm day-1 for two consecutive days, then followed by a
continuous evaporation rate of 0.5 or 1.5 cm day-'' depending on the case, until the
end of the simulation time (10 days). The concentration of the solute in the irrigation
water or the rainfall was assumed to be 0.05 meq cm-3.
- The time step was 0.005 day, and it was constant for the entire time of the
simulation.
Table 1: Parameters used to describe the soil hydraulic properties.
Soil
CM
Or
0,
3CM -3
3CM -3
CM
a "'
n
K.
cm day"'
cm
Loamy sand'
0.107*
0.470
0.010
1.4*
75.0
Sand'
0.0914
0.3434
0.0624
1.528
250.0
Coarse sand3
0.0286
0.28*
0.07*
2.239
541.0
* = Parameter is modified.
1 = From Taghavi etal (1985).
2 = From Wierenga etal (1991).
3 = From Hills etal (1989).
60
-1.0E+09
- 1.0E+08
-1.0E+07
- 1.0E+06
E
c
--4.
-1.0E+05
co
co
.4-2
0
-1.0E+04
PLI
;--,
a..)
--,) -1.0E+03
co
-1.0E+02
- 1.0E+01
-1.0E+00
-1.0E-01
0
0.1
0.2
0.3
0.4
Water content, cm3/cm3
Figure 2: Soil water characteristic curves for the three soils.
0.5
61
1000
100
.:
10
0.1 s
0.01
Loamy Sand
Sand
0.001
Coarse Sand
0.0001
0.00001
0
0.1
0.2
0.3
0.4
0.5
Water Content, cm3/cm3
Figure 3: Hydraulic conductivity as a function of water content for the three soils.
62
6.2. Case 1: Water and solute transport in uniform soil
In this case a uniform soil was used without modification. Water and solute
distribution was simulated for this soil under two different evaporation demands. The
results of this case will be used for comparison with the other cases when
modifications are applied. Table 1 shows the properties of the soil, and the conditions
of the simulation.
Results and discussion
The results of the infiltration, redistribution, and evaporation as well as solute
distribution are shown in figures C-1 and C-2. After 2 days of continuous irrigation
(7.0 cm day-1) the profile was almost saturated to a depth of about 55 cm. Then water
content decreased due to evaporation and redistribution. It reached the air dry
condition at the surface after the 8th day when the evaporation was 0.5 cm day-1.
When the evaporation rate was 1.5 cm day-1, it took less than 2 days after the
irrigation to bring the soil at the surface to air dry condition. Due to the redistribution,
water passed the root zone (70 cm assumed in this study) just after the fourth day of
the simulation, which is considered to be another loss in addition to the evaporation.
See figure 4 for summary of the results.
Solute concentration increases as water content increases during the irrigation with
contaminated water. When the irrigation was stopped the solute was redistributed in
63
the soil and reached the maximum concentration at the soil surface due to evaporation.
The solute concentration reached 0.08 meq cm-3 in less than one day when the
evaporation rate was 1.5 cm day-1, while it was only 0.035 meq cm-3 for 0.5 cm day-1.
See figure 5 for summary of the results.
63. Case 2: Effect of buried barrier layer on soil water conservation
The texture and thickness of the buried barrier layer were examined in this case as
well as the effect of the two different evaporating demands. The layer was placed at
70 cm below the soil surface in order to investigate its ability to hold the water from
passing the root zone.
Results and discussion
Figures C-3 to C-14 and table 2 show the results of this case. Because of the
presence of the coarse-textured soil in the middle of the profile, water was limited
from being lost by internal drainage in all of the situations, regardless of the thickness,
texture, or evaporation rate. The only way for water losses was through the soil
surface by the evaporation process. As a result water content of the soil at the end of
the simulation period was increased when compared to the unmodified soil. The
increase in water content ranges from 5.7% to 7.9% depending on the condition.
By comparing the results in table 2, it is clear that the texture has a small
64
influence on increasing water content in the soil. This was probably due to the fact
that coarse-textured soil had small water suction due to its larger pores, thus, water
movement to the layer was prohibited. However, the effect would be more pronounced
if the original soil was finer in texture, such as silt or clay. See figure 4 for summary
of the results.
The thickness of the layer shows no significant effect on water conservation under
the circumstances of this case. The results was similar whether the thickness was 2.5
cm or 10.5 cm.
The evaporation treatment does not show a difference in this case when compared
with the original soil.
Solute concentration and distribution in the profile were not affected by the
presence of the buried layer, since solute had not reached the layer. And since the soil
surface was open, the solute distribution took a similar trend to that of the uniform
soil. See figure 5 for summary of the results.
65
Table 2: Percentage increase in water content due to the presence of barrier layer at
the end of the simulation period.
Run
number
Treatment: Barrier layer
Increase in water
content (%)
Texture
Thickness cm
Evap. rate
cm/day
03b
Sand
2.5
0.5
6.9
04b
Sand
2.5
1.5
5.7
05b
Sand
5.5
0.5
7.2
06d
Sand
5.5
1.5
5.8
07b
Sand
10.5
0.5
7.2
08b
Sand
10.5
1.5
5.8
09b
C. Sand
2.5
0.5
7.9
10b
C. Sand
2.5
1.5
6.3
11b
C. Sand
5.5
0.5
7.9
12b
C. Sand
5.5
1.5
6.3
13b
C. Sand
10.5
0.5
7.9
14b
C. Sand
10.5
1.5
6.3
§ = Increase of water content computed for soil above the depth of the barrier layer
(70 cm), water content below that depth was not included.
66
6.4. Case 3: Effect of mulch layer on soil water conservation
A layer of coarse-textured soil was added at the soil surface in order to decrease
the upward flow of water and hence increase the water content of the soil. To achieve
the maximum possible increase in water content of the soil, the possible effects of soil
texture as well as the thickness of the layer were simulated in this case. The rest of
the soil profile was not modified and kept the same as the original soil.
Results and discussion
The results of the simulation for this case are shown in figures C-15 to C-26, as
well as in table 3. It is obvious that the mulch layer has a great influence on water
conservation for all of the treatments regardless of the texture, thickness, or the
evaporation rate. The loss of water due to the evaporation process was minimal, as a
result water content increased in the soil profile when compared to unmodified soil.
However, because of this increase and minimal upward water redistribution, the
downward redistribution was maximal in this case when compared with the other cases
(one-sided redistribution of water), and water reached the bottom of the profile, which
is considered to be a loss of water. Table 3 shows an increase in water content ranging
from 17.9% to 29.9% depending on the condition and the treatment of the mulch
layer. See figure 4 for summary of the results.
The effects of the layer texture and thickness, as well as the evaporating demand
67
are significant when compared to the previous case. There was about 6% greater
increase in water content when the mulch layer thickness was increased from 2.5 cm
to 10.5 cm. However, the effect of thickness is very small when a coarse-textured
mulch layer was used; it is more pronounced in the finer-textured soils. Furthermore,
the texture has more effect when the thickness of the layer is small; as the thickness
increases, the effect of the texture diminishes (see table 3).
The rate of evaporation also influenced the water content in the profile. The soil
retained 5-6% more water when the evaporation rate increased from 0.5 to 1.5 cm
day'. This is due to the decrease in the first stage of evaporation. As a result, the
surface mulch layer did not recover the lost water because of the sharp decrease in the
hydraulic conductivity. Thus the water content of the soil below the mulch layer
increased.
Solute distribution in the profile took the same trend as water content distribution,
and was influenced by the presence of the mulch layer. Because upward water flow
was limited, the solute concentration at the surface was decreased when compared with
the previous cases, ranging from 0.0017 to 0.0004 meq cm-3. On the other hand it was
transported to the lower part of the profile and passed the target depth of 70 cm
(leached). See figure 5 for summary of the results.
68
Table 3: Percentage increase in water content due to the presence of mulch layer at the
end of the simulation period.
Run
number
Treatment: Mulch layer
Increase in water
content (%)*
Texture
Thickness cm
Evap. rate
cm/day
15b
Sand
2.5
0.5
17.9
16b
Sand
2.5
1.5
22.6
17b
Sand
5.5
0.5
21.2
18b
Sand
5.5
1.5
26.5
19b
Sand
10.5
0.5
23.0
20b
Sand
10.5
1.5
28.6
21b
C. Sand
2.5
0.5
22.1
22b
C. Sand
2.5
1.5
27.7
23b
C. Sand
5.5
0.5
23.2
24b
C. Sand
5.5
1.5
28.9
25b
C. Sand
10.5
0.5
24.0
26b
C. Sand
10.5
1.5
29.9
§ = Increase of water content computed for soil depth below the mulch layer; water
content above that depth was not included.
69
6.5. Case 4: Effect of barrier and mulch layers combined on soil water
conservation
In this case the effect of both buried and mulch layers was simulated. The
influence of texture and thickness of those layers was investigated as well. The same
soil conditions and characteristics and evaporating demands were applied as before.
Results and discussion
These conditions achieved the greatest water conservation. Figures C-27 to C-38
and numerical results presented in table 4 show the water and solute distribution. From
these figures, it can be seen that the water was trapped between both layers and
uniformly distributed in the soil. As a result a maximum water content was achieved
at the end of the simulation period. Also changes in water content between the third
day and the last day of simulation were very small. Water content was increased by
25.6%- 44.4% depending on the condition of the simulation. The maximum increase
was obtained with 10.5 cm thick layer of coarse sand under the severe conditions of
evaporating demand, and the minimum was with 2.5 cm thick layer of sand under the
moderate evaporating demand. The mulch layer has more contribution to this increase
due to its ability to suppress the evaporation from the soil surface. See figure 4 for
summary of the results.
The layer texture and thickness as well as the evaporating demand also have
70
significant influence on reducing water losses from the soil. It can be seen from the
figures that when the soil layer was sand, the water infiltrated into the barrier layer,
which caused about 5% reduction in water content, on the other hand, water did not
infiltrate into the coarse sand layer during the simulation period.
Like the water content, solute was also trapped between the two layers, and
concentrated in the middle of the profile. The presence of the barrier layer decreases
leaching of solute and accumulation may result. The higher concentration of solute in
a soil under such conditions may introduce salinity hazardous to plant growth. See
figure 5 for summary of the results.
71
Table 4: Percentage increase in water content due to the presence of both buried and
mulch layer at the end of the simulation period.
Run
number
Treatment: Combined
Increase in water
content (%)i
Texture
Thickness cm
Evap. rate
cm/day
27b
Sand
2.5
0.5
25.6
28b
Sand
2.5
1.5
29.0
29b
Sand
5.5
0.5
30.7
30b
Sand
5.5
1.5
34.5
31b
Sand
10.5
0.5
34.1
32b
Sand
10.5
1.5
37.9
33b
C. Sand
2.5
0.5
34.7
34b
C. Sand
2.5
1.5
38.9
35b
C. Sand
5.5
0.5
36.6
36b
C. Sand
5.5
1.5
40.8
37b
C. Sand
10.5
0.5
40.1
38b
C. Sand
10.5
1.5
44.4
§ = Increase of water content computed for the depth of soil between the mulch layer
and the barrier layer, water content above and below that was not included.
72
6.6 Evaluation
From the results presented above, it is clear that all of the methods which were
used in this study increased water content in the soil. The degree to which this
increase is significant depends on the individual method, texture and thickness of soil
layer, and on the evaporating demand. Comparing individual methods, the combined
layers of mulch and buried barrier ranked first in conserving water in the root zone,
followed by mulch layer, which has a great influence in reducing water losses by
evaporation. The least effective method was the buried barrier (see figure 4). When
relating the effects of texture and thickness of the layer, it is found that both have a
similar effect. Among the textures, coarse sand ranked first, and among the
thicknesses, 10.5 cm ranked first then followed by 5.5 cm.
Comparing the effect of the evaporating demand, the severe condition tends to be
more effective in reducing water losses. That is probably due to its ability in making
the soil reach the critical point of the evaporation faster than the moderate condition,
and accordingly decreasing the length of the first stage. However, the effect of heat
transfer and vapor diffusion which are very significant in warm regions, were not
involved in the simulation. Therefore, it is hard to tell whether the severe or moderate
evaporating demand is more significant in reducing water losses from the soil surface
in hot arid climates.
100
90
80 --
70
60
0
a.
5
3
o
0
0.05
=-
0
c
11
1
0.15
c
lo --4-0.
0.1
I
0.2
I
0.25
0.3
I
0.35
0.4
Figure 4: Summary plots for water content distribution after 10 days of simulation
with 1.5 cm day-1 evaporation rate.
5
et
50
40
30
20
10
0
Water Content, cm3 /cm3
73
80
70
60
50
it l
i
0
)3 t
r1
fPf
t
t
1
I
N
N
I
1
.
......... .. ,C110 .........,. ..
N
MI
0
Q1
il
m
C>. ..... III. -...
MI
IN
=
00
C
t:11
0
t."
Figure 5: Summary plots for solute distribution after 10 days of simulation with 1.5
cm day' evaporation rate.
5
_Et
et
No
=
ft
40
30
20
10
0
MI
rli
0
:4`
0
1\4
Solute Concentration, meq/cm3
74
75
7. SUMMARY AND RECOMMENDATIONS
7.1 Summary
In order to increase the ability of sandy soils to hold more water, three methods of
modification were proposed, surface mulch layer, buried barrier layer, and
combination of both. The impact of these methods on solute distribution in the soil
was also considered. One-dimensional water flow and solute transport in unsaturated
soil was simulated for these cases. It was necessary to modify the transport equations
to apply to layered soil. The approach of Hills et al, (1989) was used to handle the
jump of water content between the layers.
The simulation results indicate that there is potential for enhancing water
conservation in sandy soil. Up to 45% increase in water content was obtained by
applying these methods. The combination method was superior, however, most of the
contribution came from the surface mulch layer which suppressed evaporation from
the soil surface. There is potential for solute build up in the root zone when the
combination method is used.
76
7.2 Recommendations
During the development of the model, many assumptions were made in order to
ease the task of simulation. However such assumptions rarely hold in the real soil.
Thus, enhancement of the model is recommended in future studies in order to make
the model more closely representative of real soils. Some recommended additions are:
1. The influence of hysteresis, specially in layered soil.
2. The influence of heat transfer and vapor diffusion, which are very common in
warm climates.
3. The effect of water front instability which is very common for the conditions of
this study.
4. Lateral water and solute flow should be investigated.
5. The effect of solute on water flow.
77
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APPENDICES
APPENDIX A
The partial derivatives of function [F1] with respect to 0; that are used to perform
the Jacobian matrix are:
1. For the interior nodes which are not adjacent to the interface:
31.1(e)
aroyit.:;112
D(e)7111,1
aein:ii
2Az1Az3
_
age)7,'12
1
607:11
4 Az3
1
D(0)7:11[22
aeriii-1
At
2Az2Az3
+
8D(0)7+112
ein+1
ae,:+1
ae,17,1
)
(A-1)
DOCir2
2A;Az3
er+11 + 07
e7.,
(A-2)
8 Az2Az3
aD(0)7+112
07+i + 131:+1
07_, + 07
8 Az1Az3
ae7+1
aro)
67_, + e7
+ 07+'
8 Az1Az3
a07:11
aFi(e)
+
(37_?
aDoci'12 (07+1
Do)7:,1122
ao,,:ii
2Az2ez3
age)7++,.'12
1
awit:11
4 A;
er+i1
+
8 Az2Az3
ei+1
)
(A-3)
84
2. For the node just above an interface:
aFg(e)
867:11
D(e)n
arxe)nw2
2AziAz4
,307:i1
0741
en'
)+ 07
8 AziAz4
(A-4)
age)in,u2
(4 Az4)
1
aeril
8Fie)
1
DOC22
a(37+1
At
2Az2Az4
+
mera
007+1
07+1
l
2AziAZ4
(A-5)
-
+ 07
8 Az1Az4
2Az2Az4
2
)
8A 0n+1 1
ae7:il
07+1
+
AAn+1
1,744.-:
+ 67
0,7+, + 2A0"1
8 Az2Az4
."+1
_
aeri
8A0n+1
07+1
ael+1
aD(13)11+11/2
Aer.+,1,2)
+1
07:11
(
aF,(e) _ (- D(e)74+.11;221[1
aen ++1
-
0A 9n
8 Az2Az4
81)(0)in+if2
+
1+2
age),n.:1112
1
aeiii++11
4 Az4
(A-6)
)
85
3. For the node just below an interface:
D(e),RZ
aF1(e)
ae,it?
2Aziez5
1
1 + 2 'me"'
An?
(3;4.1
87
star'
(A-7)
8 AziAz5
i
aFi(e)
attn+i
+
+
1
D(e):71,122
D(e)in:i1,2
At
2Az2Az5
2Az1Az5
8p(e)7'12 (
aeri
3D(0); +'R
aeris:11
ein:11
wit+,
(A-8)
e7
037+1
8Aen+1
8 Az1Az5
eri
ar(e):7,i12
2Az2Az5
_
+ e7
2 at'en+1 I)
at37+1
8 Az2Az5
air'
DoCv,22
aFi(e) _
CI'
07+1
i
agenu2 ( 1
(4 Az,
ae7:i1
ae,:++,1
i
ein:,' + ein
8 Az2Az5
67+1
)
(A-9)
86
4. For the upper boundary:
aF,(e) _
807+1
1
Doy,":1122
ap(e)i +'12
(4'4- ee2 +1
At
(z2- z1)2
ae7+1
2 (z2- z1)2
(A-10)
+
age)7+'
1
aci
2 z2- z1
aF,(e) _
ap(e)ri2 ( er1- er
my;++1,',22
,
aen2+1
)
+
aerl
ZI).
2 (z2- z1)2
(A-11)
+
alC(e)ritz
1
aeri
2 z2- zi
5. For the lower boundary:
aF,(e)
Aer:21;22
+
ap(e),"_1(2
aen.+21
Az,Az3
Wm+-11
aF(e)
[
aen:i
1
At
enm21
co.
4 Az1Az3
+ en.
(A-12)
)
Aet-11,2)
Az1Az3
(A-13)
arxer,:v2
+
atnn+i
(- en,:_1,
+ enm+1
em -1 + ohm)
4 Az1Az3
87
Where:
arxe)7+1/2
d D(Se)
d Se
ae,
dSe
d0
(1-m) Ks
d D(Se)
d Se
(A-14)
h. -Se 1/m) -"`
a no,- es) i
+ (1-Selor- 2]
(A-15)
+
(1-m) K,
a n(e,- Or)
d Se(e)
1
de
Os- 0,
Se -112 [1 Selinr-1
(A-16)
alc(e)74-'12
d K(Se)
d Se
ae,
dSe
de
[Se-1/2
d K(Se)
d Se
3
[1 -Se 9'1
2
r _
[ 1
(A-17)
( 1- Seim' )'"
r
(A-18)
1-m
+ 2 Se 1/2 ( 1- Se lk"
aAe
1
ae- a*i
80i
2
a*
)m -i(Se
ae,
1"
) [1- ( 1- Sellm )1"
]i
(449)
for the node just above the interface, and
3A0
1
ae+ alp!
3e,
2
a*
ae,
for the node just below the interface.
(A-20)
88
ae
1
&pi
1
aet
2
1
(
[ (
(1
1
2
[ (
a n (a
1
er)[
= m (e,
+
(a
1117)n
[
1
(
1 + (a 111)ft
)2
(A-21)
I
[am)
[ aei " al
mot+)
( 2 [ K(01) h(0) + K(01+1) h(ei+1) ] + Az [ k(13i)
me) +
met_1) y
L
y'
(A-22)
met) + met+) y
moo ] ) )
(A-23)
aMe
aet
(1 ( 2 [ ge) h(0) + met+) Net+1)
2
Iji)n -1
]
+ Az [ ged MO ] ) )
89
APPENDIX B
Computer program codes in C language were written to simulate the water and solute
transport in layered soil.
#include <stdio.h>
#include <conio.h>
#include <math.h>
#include <float.h>
void water();
void solute();
void newton();
void soil_hydraulics();
void K and D(int);
void jump(int, int);
void initialize();
void vector F theta();
void matrix_jacob();
double Deriv THETA(double,int);
double THETA(double,int);
double Deriv conduc(int);
double conduc(int,int);
double Deriv diffu(int);
double diffu(int,int);
double Deriv_potential(int);
double potential(int,int);
double Deriv Dthetal(double,int);
double Deriv_Dtheta2(double,int);
double potential(int, int);
void water balance();
void solute balance ();
void Read_input();
void Write_output_1();
void Write output
output_2();
2();
Global variables
double theta_sat[6], theta_dry[6], initial_theta[6];
double m, n[6], alpha[6];
90
double K sat[6], Dtheta, mass balance, balance;
double HC1, HC2, HD1, HD2, DZ4, DZ5;
double Wboundl, Wbound2, Sboundl, Sbound2, W_q, start_B2;
double sum_flux = 0.0, balance_ratio, initial_water_vol = 0.0;
double sum_C = 0.0, balance_C, initial_solute = 0.0, balance_ratio_C;
double S j, S_Diffusion, initial_C[6], lambda[6];
double thetaA[2], D_theta[6][2], Gama[210], Beta[210];
double Dt, time_end, Printout[8];
int num_nodes, num_layers, layer[6], L, max_itir;
typedef struct {
double theta[3], K sat, theta_sat, theta_dry, alpha, n, depth;
double DZ1, DZ2, DZ3;
double F_theta, residual, Deriv_HC, Deriv_HD;
double Diff[2], HC[2], jacob A, jacob_B, jacob_C;
double C[2], S A, S_B, S_C, S_Known, lambda;
} soil;
soil node[210];
FILE *input, *output;
main()
{
int i, P = 0;
ouble Time = 0.0, dz;
clrscr();
input = fopen("DATA.IN", "r");
output = fopen("DATA.OUT", "w");
Read_input();
initialize();
Write_output_1();
W_q = Wboundl;
node[0].theta[1] = (W_q * Dt / (node[1].depth - node[0].depth))
+ node[0].theta[0];
if (node[0].theta[1] >= node[0].theta_sat)
node[0].theta[1] = node[0].theta_sat * 0.99999;
if (node[0].theta[1] <= node[0].theta_dry)
node[0].theta[1] = node[0].theta_dry
+ (node[0].theta_dry * 0.00001);
// calculation of the volume of initial water content an solute concentration
// in the soil
L = 0;
for (i=0; i < num_nodes-1; i++)
{ dz = node[i+1].depth - node[i].depth;
if (i+1 == layer[L+1]) jump(i,L);
{ initial water vol += node[i].theta[2] * dz/2;
initial water vol += node[i+1].theta[2] * dz/2;
L++;
91
}
initial water vol += node[i].theta[0] * dz;
initial_solute += node[i].theta[0] * node[i].C[0] * dz;
}
for (i=1; i < num_nodes; i++) node[i].theta[1] = node[i].theta[0];
do {
Time += Dt;
W_q = Wboundl;
S j = Sboundl * W_q;
if (Time >= start B2) { W_q = Wbound2; S j = Sbound2 * W_q; }
water();
solute();
solute balance ();
for (i=0; i < num_nodes; i++)
{
node[i].theta[0] = node[i].theta[2];
node[i].C[0] = node[i].C[1];
node[i].theta[1] = node[i].theta[0];
}
if (Time >= Printout[P]) Write_output_2(); P++; }
} while (Time < time_end);
return;
}
void water (void)
{
newton();
water balance();
return;
}
void newton(void)
{
int i, j, flag, itir= 0;
double max;
do {
itir += 1;
node[num_nodes].theta[0] = node[num_nodes-2].theta[0];
node[num nodes].theta[1] = node[num nodes-2].theta[1];
soil_hydraulics();
matrix jacob();
vector_F_theta();
//
// Solve for residual vector using Thomas method for Tri-diagonal matrix.
//
node[0].jacobA = node[num_nodes - 1].jacob_C = 0;
92
Beta[0] = node[0].jacob_B;
Gama[0] = (-1 * node[0].F_theta) / node[0].jacob_B;
for (1=1; i < num_nodes; i++)
{
Beta[i] = node[i].jacob_B - (node[i].jacob_A
* node[i-1].jacob_C / Beta[i-1]);
Gama[i] = ((-1 * node[i].F_theta) - (node[i].jacob A
* Gama[i-1])) / Beta[i];
}
node[num_nodes - 1].residual = Gama[num_nodes - 1];
for (i = num_nodes - 2; i >= 0; i--)
node[i].residual = Gama[i] - ( node[i].jacob_C
* node[i+1].residual / Beta[i]);
for (i=0; i < num_nodes; i++)
node[i].theta[2] = node[i].theta[1] + node[i].residual;
max = fabs(node[0].residual);
for (i=1; i < num_nodes; i++)
if (fabs(node[i].residual) > max) max = fabs(node[i].residual);
for (i = 0; i < num_nodes; i++)
{
if (node[i].theta[2] >= node[i].theta_sat)
node[i].theta[2] = node[i].theta_sat * 0.99999;
if (node[i].theta[2] <= node[i].theta_dry)
node[i].theta[2] = node[i].theta_dry
+ (node[i].theta_dry * 0.00001);
}
for (i=0; i < num_nodes; i++) node[i].theta[1] = node[i].theta[2];
if(mass_balance <= 0.0000001) return;
} while (itir < max_itir);
return;
}
void vector_F_theta()
{ int i, L = 0;
double DZ1;
mass_balance = 0.0;
HD1 = (node[1].Diff[1] + node[0].Diff[1]) / 2;
HC1 = (node[1].HC[1] + node[0].HC[1]) / 2;
node[0].F_theta = (node[0].theta[1] * ((I/DO
+ (HD1 / pow(node[1].depth-node[0].depth,2))))
- (node[1].theta[1] * (HD1 / pow(node[1].depth-node[0].depth,2)))
- (node[0].theta[0] / Dt) + (HC1 / (node[1].depth-node[0].depth))
- (W_q / (node[1].depth - node[0].depth));
mass_balance += fabs(node[i].F_theta);
for (i =1; i < num_nodes; i++)
93
{
K and_D(1);
if (1+1 == layer[L+1]) jump(i,L);
else if (i == layer[L+1]) L++;
else
{
node[i].F_theta = (-node[1-1].theta[1] * (1/node[i].DZ3)
* (HD2/(2*node[i].DZ1))) + (node[i].theta[1] * ((1/Dt)
+ ((1 / node[i].DZ3) * ((HD1/(2 * node[i].DZ2))
+ (HD2 / (2 * node[i].DZ1))))))
- (node[i+1].theta[1] * (1 / node[i].DZ3)
* (HD1 / (2 * node[i].DZ2)))
- (node[1-1].theta[0] * (1/node[i].DZ3)
* (HD2/(2*node[i].DZ1)))
- (node[i].theta[0] * ((1/Dt)- ((l/node[i].DZ3)
* ((HD1/(2*node[i].DZ2))
+ (HD2/(2*node[i].DZ1))))))
- (node[i+1].theta[0] * ((1 /node[i].DZ3)
* (HD1/(2*node[i].DZ2))))
+ ((HC1 - HC2)/node[i].DZ3);
mass balance += fabs(node[i].F_theta);
}
}
}
void matrix_jacob()
{ int i;
double DZ1;
HD1 = (node[1].Diff[1] + node[0].Diff[1]) / 2;
node[0].jacob_B = ((1/Dt) + (HD1 / pow(node[1].depth-node[0].depth,2))) +
(((node[0].theta[1] - node[1].theta[1]) /
pow(node[1].depth-node[0].depth,2)) * node[0].Deriv_HD)
+ ((node[0].Deriv_HC/2) / (node[1].depth-node[0].depth));
node[0].jacob_C = (-HD1 / pow(node[1].depth-node[0].depth,2))
+ (((node[0].theta[1] - node[1].theta[1]) /
pow(node[1].depth-node[0].depth,2)) * node[1].Deriv_HD)
+ ((node[1].Deriv_HC/4) / (node[1].depth-node[0].depth));
i = num_nodes - 1;
K and_D(i);
node[i].jacob_A = ((-1/node[i].DZ3) * (HD2 /(2 *node[i].DZ1)))
+ (((-node[i-1].theta[1] + node[i].theta[1] - node[i-1].theta[0]
+ node[i].theta[0]) / (2 * node[i].DZ3 * node[i].DZ1))
* node[i-1].Deriv_HD)
- ((node[1-1].Deriv_HC/4) / (node[i].DZ3));
94
node[i].jacob_B = (1/Dt + ((1 /node[i].DZ3) * ((HD1/(2*node[i].DZ2))
+ (HD2/(2*node[i].DZ1)))))
+ (((-node[i-1].theta[1] + node[i].theta[1] - node[i-1].theta[0]
+ node[i].theta[0]) / (2 * node[i].DZ3 * node[i].DZ1))
* node[i].Deriv_HD)
+ (((node[i].theta[1] - node[i+1].theta[1] + node[i].theta[0]
- node[i+1].theta[0]) / (2 * node[i].DZ3 * node[i].DZ2))
* node[i].Deriv_HD);
for (i=1; i < num_nodes-1; i++)
{
K and_D(i);
node[i].jacob A = ((-1/node[i].DZ3) * (HD2/(2*node[i].DZ1)))
+ (((-node[i-1].theta[1] + node[i].theta[1] - node[i-1].theta[0]
+ node[i].theta[0]) / (2 * node[i].DZ3
* node[i].DZ1)) * node[i-1].Deriv_HD)
- ((node[i-1].Deriv_HC/4) / (node[i].DZ3));
node[i].jacob_B = (1/Dt + ((1 /node[i].DZ3) * ((HD1/(2*node[i].DZ2))
+ (HD2/(2*node[i].DZ1)))))
+ (((-node[i-1].theta[1] + node[i].theta[1] - node[i-1].theta[0]
+ node[i].theta[0]) / (2 * node[i].DZ3
* node[i].DZ1)) * node[i].Deriv_HD)
+ (((node[i].theta[1] - node[i+1].theta[1] + node[i].theta[0]
- node[i+1].theta[0]) / (2 * node[i].DZ3
* node[i].DZ2)) * node[i].Deriv_HD);
node[i].jacob_C = ((-1/node[i].DZ3) * (HD1/(2*node[i].DZ2)))
+ (((node[i].theta[1] - node[i+1].theta[1]
+ node[i].theta[0] - node[i+1].theta[0]) / (2
* node[i].DZ3 * node[i].DZ2))
* node[i+1].DerivHD)
- ((node[i+1].Deriv_HC/4) / (node[i].DZ3));
}
}
void jump(i,L)
{ int k;
double Dtheta last, Dtheta_next, theta_up[2], theta_dn[2], DZ;
double lastpot_int, next_pot_int, Wpot[2], Se_par[2];
/*
Calculation of the jump in the water content at the interface
*1
DZ = (node[i+1].depth - node[i].depth)/2;
Wpot[0] = potential(i3O);
Wpot[1] = potential(i+1,0);
last_pot_int = ((2 * ((node[i].HC[0] * Wpot[0]) + (node[i+1].HC[0] * Wpot[1])))
+ (DZ * (node[i].HC[0] - node[i+1].HC[0]))) / (2*(node[i].HC[0]
+ node[i+1].HC[0]));
95
Wpot[0] = potential(i,1);
Wpot[1] = potential(i+1,1);
next_potint = ((2 * ((node[i].HC[1] * Wpot[0]) + (node[i+1].HC[1] * Wpot[1])))
+ (DZ * (node[i].HC[1] - node[i+1].HC[1]))) / (2*(node[i].HC[1]
+ node[i+1].HC[1]));
theta_up[0] = THETA(fabs(last_pot_int),i);
theta_up[1] = THETA(fabs(next_pot_int),i);
theta_dn[0] = THETA(fabs(last_pot_int),i+1);
theta_dn[1] = THETA(fabs(next_pot_int),i+1);
Dtheta_last = theta_dn[0] - theta_up[0];
Dtheta_next = theta_dn[1] - theta_up[1];
Dtheta = (Dtheta_next + Dtheta_last) / 2;
D_theta[L][0] = Dtheta_last;
D_theta[L][1] = Dtheta_next;
/*
Calculation of the mass balance equation to F(theta) vector for
flow equation at the interface
*1
K_and_D(i);
DZ4 = ((node[i].DZ2/2) + node[i].DZ1)/2;
node[i].F_theta = (-node[i-1].theta[1] * (1/DZ4) * (HD2/(2*node[i].DZ1)))
+ (node[i].theta[1] * ((1/Dt) + ((1/DZ4)
* ((HD1/(2*node[i].DZ2)) + (HD2/(2*node[i].DZ1))))))
- (node[i+1].theta[1] * ((1/DZ4)
* (HD 1/(2*node[i].DZ2))))
- (node[i-1].theta[0] * (1/DZ4) * (HD2/(2*node[i].DZ1)))
+ (node[i].theta[0] * (((l/DZ4) * ((HD1/(2*node[i].DZ2))
+ (HD2/(2*node[i].DZ1)))) - (1/Dt)))
- (node[i+1].theta[0] * ((1/DZ4)
+ (HD 1/(2*node[i].DZ2))))
+ (Dtheta * (1/DZ4) * (HD1/node[i].DZ2))
+ ((HC1 - HC2)/node[i].DZ3);
mass_balance += fabs(node[i].Ftheta);
k = i+1;
K_and_D(k);
DZ5 = ((node[k].DZ1/2) + node[k].DZ2)/2;
node[k].Ftheta = (-node[k-1].theta[1] * (1/DZ5) * (HD2/(2*node[k].DZ1)))+
(node[k].theta[1] * ((1/Dt) + ((1/DZ5) * ((HD1/(2*node[k].DZ2))
+(HD2/(2*node[k].DZ1))))))- (node[k+1].theta[1] * ((l/DZ5) *
(HD11(2*node[k].DZ2))))- (node[k-1].theta[0] * (1/DZ5) *
(HD2/(2*node[k].DZ1)))- (node[k].theta[0] * ((1/Dt) - ((1/DZ5) *
((HD1/(2*node[k].DZ2)) + (HD2/(2*node[k].DZ1))))))(node[k+1].theta[0] * ((1/DZ5) * (HD1/(2*node[k].DZ2))))(Dtheta * (1/DZ5) * (HD2/node[k].DZ1)) + ((HC1
96
-HC2)/node[k].DZ3);
mass balance += fabs(node[k].F theta);
/*
Calculation of the coefficients for the Jacobian matrix for the two
nodes that bounding the interface.
*1
K_and_D(i);
node[i].jacob_A = (-(1/DZ4) * (HD2/(2*node[i].DZ1)))
- (((node[i-1].theta[1] - node[i].theta[1] + node[i-1].theta[0]
- node[i].theta[0]) /( 2 * DZ4 * node[i].DZ1))
* node[i-1].Deriv_HD) - ((1 /node[i].DZ3)
* (node[i-1].Deriv_HC/4));
node[i].jacob_B = (((-node[i-1].theta[1] + node[i].theta[1] - node[i-1].theta[0]
+ node[i].theta[0]) /( 2 * DZ4 * node[i].DZ1)) * node[i].Deriv_HD)
+ (((node[i].theta[1] - node[i+1].theta[1] + node[i].theta[0]
- node[i+1].theta[0]) /( 2 * DZ4 * node[i].DZ2))
* node[i].Deriv HD) + ((1/Dt) + ((1/DZ4)
* ((HD1 /(2 *node[i].DZ2)) + (HD2/(2*node[i].DZ1)))))
+ (Deriv_Dthetal(fabs(next_pot_int),i) * (1/DZ4)
* (HD1/node[i].DZ2)) + ((Dtheta / (DZ4 * node[i].DZ2))
* node[i].Deriv_HD);
node[i].jacob_C = (((node[i].theta[1] - node[i+1].theta[1] + node[i].theta[0]
- node[i+1].theta[0] + (2*Dtheta)) /( 2 * DZ4 * node[i].DZ2))
* node[i+1].Deriv_HD) - ((1/DZ4) * (HD1/(2*node[i].DZ2)))
+ (Deriv_Dtheta2(fabs(next_pot_int),i+1) * (1/DZ4)
* (HD1/node[i].DZ2));
K and D(k);
ncTde[14.jacob_A = (-(1/DZ5) * (HD2/(2*node[k].DZ1))) + (((-node[k-1].theta[1]
+ node[k].theta[1] - node[k-1].theta[0] + node[k].theta[0]
- (2*Dtheta)) /( 2 * DZ5 * node[k].DZ1)) * node[k-1].Deriv_HD)
- (Deriv Dthetal(fabs(next_pot int),k-1) * (1/DZ5)
* (HD2ode[k].DZ1)) - ((1 /node[k].DZ3)
* (node[k-1].Deriv_HC/4));
node[k].jacob_B = (((-node[k-1].theta[1] + node[k].theta[1] - node[k-1].theta[0]
+ node[k].theta[0] - (2*Dtheta)) /( 2 * DZ5 * node[k].DZ1))
* node[k].Deriv_HD) + (((node[k].theta[1] - node[k+1].theta[1]
+ node[k].theta[0] - node[k+1].theta[0]) / (2 * DZ5
* node[k].DZ2)) * node[k].Deriv_HD) + ((1/Dt)
+ ((1/DZ5) * ((HD1/(2*node[k].DZ2)) + (H52/(2*node[k].DZ1)))))
- (Deriv_Dtheta2(fabs(next_pot_int),k) * (1/DZ5)
* (HD2/node[k].DZ1));
97
node[k].jacob_C = (((node[k].theta[1] - node[k+1].theta[1] + node[k].theta[0]
- node[k+1].theta[0]) / (2 * DZ5 * node[k].DZ2))
* node[k+1].Deriv_HD) - ((1/DZ5)
* (HD1/(2*node[k].DZ2))) + ((1 /node[k].DZ3)
* (node[k+1].Deriv_HC/4));
return;
}
void solute(void)
{
int i, L = 0;
double th0, thl, th2, th3, th4, flux(), fluxl, veloc0, velocl, dz, DZ3;
double solute_DisO, solute_Disl;
double thOa, thOb, th2a, th2b, flux0a, fluxOb, fluxla, fluxlb;
soil_hydraulics();
dz = node[1].depth - node[0].depth;
HD1 = (node[1].Diff[1] + node[0].Diff[1] + node[1].Diff[0] + node[0].Diff[0]) / 4;
HC1 = (node[1].HC[1] + node[0].HC[1] + node[1].HC[0] + node[0].HC[0]) / 4;
thl = (node[0].theta[1] + node[0].theta[0]) / 2;
th2 = (node[1].theta[1] + node[1].theta[0]) / 2;
th4 = (node[1].theta[1] + node[0].theta[1] + node[1].theta[0] + node[0].theta[0]) / 4;
fluxl = (-HD1 * (th2 - thl) / dz) + HC1;
velocl = fabs(fluxl / th4);
solute_Disl = S_Diffusion + (node[0].lambda * velocl);
node[0].S_B = (node[0].theta[1]/Dt) + (th4 * solute_Disl / (2 * dz * dz))
+ (fluxl / (4 * dz));
node[0].S_C = ((th4 * solute_Disl / (2 * dz * dz)) - (fluxl / (4 * dz))) * -1;
node[0].S_Known = (node[0].C[0] * ((node[0].theta[0]/Dt)
- (th4 * solute_Disl / (2 * dz * dz))
- (flux1 / (4 * dz))))
+ (node[1].C[0] * ((th4 * solute_Disl / (2 * dz * dz))
- (fluxl / (4 * dz)))) + (S _j / dz);
node[num_nodes].theta[0] = node[num_nodes-2].theta[0];
node[num_nodes].theta[1] = node[num_nodes-2].theta[1];
node[num_nodes].C[0] = node[num_nodes-2].C[0];
i = num_nodes-1;
K_and_D(i);
th0 = (node[i-1].theta[1] + node[i-1].theta[0]) / 2;
thl = (node[i].theta[1] + node[i].theta[0]) / 2;
th2 = (node[i+1].theta[1] + node[i+1].theta[0]) /2;
th3 = (node[i].theta[1] + node[i-1].theta[1] + node[i].theta[0] + node[i-1].theta[0]) / 4;
th4 = (node[i+1].theta[1] + node[i].theta[1] + node[i+1].theta[0] + node[i].theta[0]) /4;
flux() = (-HD2 * (thl - th0) / node[i].DZ1) + HC2;
fluxl = (-HD1 * (th2 - thi) / node[i].DZ2) + HC1;
veloc0 = fabs(flux0 / th3);
98
velocl = fabs(fluxl / th4);
solute_DisO = S_Diffusion + (node[i].lambda * veloc0);
solute_Disl = S_Diffusion + (node[i].lambda * velocl);
node[i].S A = ((th3 * solute_DisO / (2 * node[i].DZ1 * node[i].DZ3))
+ (flux° / (4 * node[i].DZ3)) + (th4 * solute_Disl / (2
* node[i].DZ2 * node[i].DZ3)) -(fluxi / (4 * node[i].DZ3))) *-1;
node[i].S_B = (node[i].theta[1]/Dt)
+ (th4 * solute_Disl / (2 * node[i].DZ2 * node[i].DZ3))
+ (th3 * solute_DisO / (2 * node[i].DZ1 * node[i].DZ3))
+ (flux1 / (4 * node[i].DZ3))
- (flux() / (4 * node[i].DZ3));
node[i].S_Known = (node[i-1].C[0] * ((th3 * solute_DisO / (2 * node[i].DZ1 *
node[i].DZ3)) + (flux() / (4 * node[i].DZ3))))
+ (node[i].C[0] * ((node[i].theta[0]/Dt)
- (th4 * solute_Disl / (2 * node[i].DZ2 * node[i].DZ3))
- (th3 * solute_DisO / (2 * node[i].DZ1 * node[i].DZ3))
- (fluxi / (4 * node[i].DZ3)) + (flux() / (4 * node[i].DZ3))))
+ (node[i+1].C[0] * ((th4 * solute_Disl / (2 * node[i].DZ2 *
node[i].DZ3))
- (fluxi / (4 * node[i].DZ3))));
for (i=1; i < num_nodes-1; i++)
{
K and_D(i);
thO = (node[i-1].theta[1] + node[i-1].theta[0]) / 2;
thl = (node[i].theta[1] + node[i].theta[0]) / 2;
th2 = (node[i+1].theta[1] + node[i+1].theta[0]) / 2;
th3 = (node[i].theta[1] + node[i-1].theta[1] + node[i].theta[0] + node[i-1].theta[0]) / 4;
th4 = (node[i+1].theta[1] + node[i].theta[1] + node[i+1].theta[0] + node[i].theta[0]) /
4;
if (i+1 == layer[L+1])
{
thetaA[1] = (node[i].theta[1] + node[i+1].theta[1])/2;
thetaA[0] = (node[i].theta[0] + node[i+1].theta[0])/2;
th2a = ((thetaA[1] - (D theta[L][1] / 2)) + (thetaA[0] - (Dtheta[L][0] / 2))) / 2;
th2b = ((thetaA[1] + (D theta[L][1] / 2)) + (thetaA[0] + (D_theta[L][0] / 2))) / 2;
th4 = ((thetaA[1] - (D_theta[L][1] / 2)) + node[i].theta[1] + (thetaA[0] (D_theta[L][0] / 2)) + node[i].theta[0]) / 4;
flux() = (-HD2 * (thl - th0) / node[i].DZ1) + HC2;
fluxla = (-HD1 * (th2a - thl) / (node[i].DZ2 / 2)) + HC1;
flux1b = (-HD1 * (th2 - th2b) / (node[i].DZ2 / 2)) + HC1;
fluxi = (fluxla + fluxlb) / 2;
DZ3 = node[i].DZ3;
DZ4 = node[i].DZ3;
// DZ4 = ((node[i].DZ2/2) + node[i].DZ1)/2;
99
}
else if (i == layer[L+1])
{
thetaA[1] = (node[i].theta[1] + node[i-1].theta[1])/2;
thetaA[0] = (node[i].theta[0] + node[i-1].theta[0])/2;
thOa = ((thetaA[1] - (Dtheta[L][1] / 2)) + (thetaA[0] - (D_theta[L][0] / 2))) / 2;
thOb = ((thetaA[1] + (D theta[L][1] / 2)) + (thetaA[0] + (Dtheta[L][0] / 2))) / 2;
th3 = (node[i].theta[1] + (thetaA[1] + (D_theta[L][1] / 2)) + node[i].theta[0] +
(thetaA[0] + (Dtheta[L][0] / 2))) / 4;
fluxOa = (-HD2 * (thOa - th0) / (node[i].DZ1 / 2)) + HC2;
fluxOb = (-HD2 * (thl - thOb) / (node[i].DZ1 / 2)) + HC2;
flux() = (fluxOa + fluxOb) / 2;
fluxi = (-HD1 * (th2 - thl) / node[i].DZ2) + HC1;
//
DZ3 = ((node[i].DZ1/2) + node[i].DZ2)/2;
DZ3 = node[i].DZ3;
DZ4 = node[i].DZ3;
L++;
}
else
{
flux() = (-HD2 * (thl - th0) / node[i].DZ1) + HC2;
fluxi = (-HD1 * (th2 - thl) / node[i].DZ2) + HC1;
DZ3 = node[i].DZ3;
DZ4 = node[i].DZ3;
}
veloc0 = fabs(flux0 / th3);
veloc1 = fabs(fluxl / th4);
solute_DisO = S_Diffusion + (node[i].lambda * veloc0);
solute_Disl = S_Diffusion + (node[i].lambda * veloc1);
node[i].S A = ((th3 * solute_DisO / (2 * node[i].DZ1 * node[i].DZ3))
+ (flux() / (4 * DZ3))) * -1;
node[i].S_B = (node[i].theta[1]/Dt)
+ (th4 * solute_Disl / (2 * node[i].DZ2 * node[i].DZ3))
+ (th3 * solute_DisO / (2 * node[i].DZ1 * node[i].DZ3))
+ (fluxi / (4 * DZ4))
- (flux() / (4 * DZ3));
node[i].S_C = ((th4 * solute_Disl / (2 * node[i].DZ2 * node[i].DZ3))
- (flux1 / (4 * DZ4))) * -1;
node[i].S_Known = (node[i-1].C[0] * ((th3 * solute_DisO / (2 * node[i].DZ1 *
node[i].DZ3)) + (flux() / (4 * DZ3))))
+ (node[i].C[0] * ((node[i].theta[0]/Dt)
- (th4 * solute_Disl / (2 * node[i].DZ2 * node[i].DZ3))
- (th3 * solute_DisO / (2 * node[i].DZ1 * node[i].DZ3))
- (flux1 / (4 * DZ4)) + (flux() / (4 * DZ3))))
100
+ (node[i+1].C[0] * ((th4 * solute_Disl / (2 * node[i].DZ2 *
node[i].DZ3))
- (fluxl / (4 * DZ4))));
}
//
// Solve for solute Ci vector using Thomas method for Tri-diagonal matrix.
//
node[0].S A = node[num_nodes - 1].S_C = 0;
Beta[0] = node[0].S_B;
Gama[0] = (node[0].S_Known) / node[0].S_B;
for (1=1; i < num_nodes; i++)
{ Beta[i] = node[i].S_B - (node[i].S_A * node[i- 1].S_C / Beta[i -1]);
Gama[i] = (node[i].S_Known - (node[i].S A * Gama[i -1])) / Beta[i];
}
node[num_nodes - 1].C[1] = Gama[num_nodes - 1];
for (i = num_nodes - 2; i >= 0; i--)
node[i].C[1] = Gama[i] - (node[i].S_C * node[i+1].C[1] / Beta[i]);
for (i=0; i < num_nodes; i++) if (node[i].C[1] < 0) node[i].C[1] = 0;
}
void K_and_D(i)
{
HD1 = (node[i+1].Diff[0] + node[i+1].Diff[1] + node[i].Diff[0] + node[i].Diff[1]) /
4;
HC1 = (node[i+1].HC[0] + node[i+1].HC[1] + node[i].HC[0] + node[i].HC[1]) / 4;
HD2 = (node[i].Diff[0] + node[i].Diff[1] + node[1-1].Diff[0] + node[i-1].Diff[1]) /
4;
HC2 = (node[i].HC[0] + node[i].HC[1] + node[i-1].HC[0] + node[i- 1].HC[1]) / 4;
}
void initialize(void)
{ int i,j=0;
for (i = 0; i < num_nodes+1; i++)
{ if (layer[j+1] == i) j++;
node[i].n = n[j];
node[i].alpha = alpha[j];
node[i].K_sat = Ksat[j];
node[i].theta_sat = theta_sat[j];
node[i].theta_dry = theta_dry[j];
node[i].theta[0] = initial theta[j];
node[i].lambda = lambda[j];
node[i].C[0] = initial_C[j];
if (node[i].theta[0] > node[i].theta sat) node[i].theta[0] = node[i].theta_sat *
0.99999;
if (node[i].theta[0] < node[i].theta_dry) node[i].theta[0] = node[i].theta_dry +
101
(node[i].theta_dry * 0.00001);
node[i].theta[1] = node[i].theta[0];
}
for (i = 1; i < num_nodes; i++)
{
node[i].DZ1 = node[i].depth - node[i-1].depth;
node[i].DZ2 = node[i+1].depth - node[i].depth;
node[i].DZ3 = (node[i+1].depth - node[i-1].depth) / 2;
}
return;
}
void soil_hydraulics()
{
int i;
for (i=0;i < num_nodes+1; i++)
{
node[i].HC[0] = conduc(i3O);
node[i].HC[1] = conduc(i,1);
node[i].Diff[0] = diffu(i3O);
node[i].Diff[1] = diffu(i,1);
node[i].Deriv_HC = Deriv_conduc(i);
node[i].Deriv_HD = Deriv_diffu(i);
}
return;
}
//
// this function computes water potential as a function of water content
//
double potential(int i, int j)
{
double Wp, Se, m;
Se = (node[i].theta[j] - node[i].theta_dry) / (node[i].theta_sat - node[i].theta_dry);
m = 1 - (1 / node[i].n);
Wp = (pow((pow(Se,-1 / m) - 1), (1/node[i].n))) / node[i].alpha;
return Wp;
}
//
// this function computes the derivative of water potential with respect
// to water content
//
double Deriv_potential(int i)
{
double Deriv wp_theta, Deriv_Wp, Deriv_Se, Se, m;
102
Se = (node[i].theta[1] - node[i].theta_dry) / (node[i].theta_sat - node[i].theta_dry);
Deriv_Se = 1 / (node[i].theta_sat - node[i].theta_dry);
m = 1 - (1 / node[i].n);
Deriv_Wp = (1 / node[i].alpha) * (1/node[i].n)
* pow((pow(Se,-1 / m) - 1), ((1 /node[i].n) - 1))
* (-1/m) * pow(Se,((-1 / m) - 1));
Deriv_wp_theta = Deriv_Wp * Deriv_Se;
return Deriv_wp_theta;
}
//
// this function computes the diffusivity as a function of water content
//
double diffu(int i, int j)
{
double Diffu, Se, m;
m = 1-(1/node[i].n);
Se = (node[i].theta[j] - node[i].theta_dry) / (node[i].theta_sat - node[i].theta_dry);
Diffu = (((l-m) * node[i].K sat) / (node[i].alpha * m * (node[i].theta_sat node[i].theta_dry)))
* pow(Se, (0.5 - (1/m)))
* (pow((1-pow(Se,(1/m))),-m) + pow((1-pow(Se,(1/m))),m) - 2);
return Diffu;
}
//
// this function computes the derivative of the diffusivity with respect
// to water content
//
double Derivdiffii(int i)
{
double Deriv_Diffu_theta, Deriv_Se, Deriv_Diffu, Diffu, Se, m;
m = 1-(1/node[i].n);
Se = (node[i].theta[1] - node[i].theta_dry) / (node[i].theta_sat node[i].theta_dry);
Deriv_Se = 1 / (node[i].theta_sat - node[i].theta_dry);
Deriv_Diffu = ((((l-m) * node[i].K_sat) / (node[i].alpha * m * (node[i].theta_sat node[i].theta_dry)))
* (pow((1-pow(Se,(1/m))),-m) + pow((1-pow(Se,(1/m))),m) - 2)
* ((0.5 - (1/m)) * pow(Se, -((l/m) + 0.5))))
+ ((((1-m) * node[i].K_sat) / (node[i].alpha * m * (node[i].theta_sat node[i].theta_dry)))
* (pow(Se,-0.5)) * (pow((1-pow(Se,(1/m))),-m-1) pow((l-pow(Se,(1/m))),m-1)));
if (i == 0) Deriv_Diffu_theta = (Deriv_Diffu * DerivSe)/2;
else Deriv_Diffu_theta = (Deriv_Diffu * Deriv_Se)/4;
return Deriv_Diffu_theta;
103
}
//
// this function computes the hydraulic conductivity as a function of
// water content
//
double conduc(int i,int j)
{
double Cond, Se, m;
m = 1-(1/node[i].n);
Se = (node[i].theta[j] - node[i].theta_dry) / (node[i].theta_sat - node[i].theta_dry);
Cond = node[i].K_sat * sqrt(Se) * pow(1-pow(1-pow(Se,(1/m)),m),2);
return Cond;
}
//
// this function computes the derivative of the hydraulic conductivity
// with respect to water content
//
double Deriv_conduc(int i)
{
double Deriv_Cond_theta, Deriv_Cond, Deriv_Se, Se, m;
m = 1-(1/node[i].n);
Se = (node[i].theta[1] - node[i].theta_dry) / (node[i].theta_sat - node[i].theta_dry);
Deriv_Se = 1 / (node[i].theta_sat - node[i].theta_dry);
Deriv_Cond = node[i].K_sat * ((0.5 * pow(Se, -0.5) *
pow(1-pow(1-pow(Se,(1/m)),m),2))
+ (2 * sqrt(Se) * pow(1-pow(Se,(1/m)),m-1) * pow(Se,(1-m)/m) * (1pow(1-pow(Se,(1/m)),m))));
Deriv_Cond_theta = Deriv_Cond * Deriv_Se;
return Deriv_Cond theta;
}
//
// this function computes water content as a function of water potential
//
double THETA(double Wp,int i)
{ double that, m;
m = 1-(1/node[i].n);
that = ((node[i].theta_sat - node[i].theta_dry)
* pow((1 / (1 + pow(node[i].alpha * Wp,node[i].n))), m)) + node[i].theta_dry;
return that;
}
//
// this function computes the derivative of water content with respect
// to water potential
//
104
double Deriv_THETA(double Wp,int i)
{ double Deriv_that, m;
m = 1-(1/node[i].n);
Deriv_that = m * (node[i].theta_sat - node[i].theta_dry)
* pow(1/(1+ pow( node[i].alpha * Wp, node[i].n)) ,m-1)
*((-node[i].alpha * node[i].n * pow(node[i].alpha * Wp, node[i].n-1))
/ pow(l+pow(node[i].alpha * Wp, node[i].n),2));
return Deriv_that;
1
// this function computes the derivative of delta theta with respect
// to water content for the jump at the interface for the "upper" node
//
double Deriv_Dthetal(double Wp,int i)
{ double Deriv_pot_int, D_theta, D_wp, DZ, p, p2, k, k2, dp, dk;
DZ = (node[i+1].depth - node[i].depth)/2;
p = potential(i,1);
p2= potential(i+1,1);
k = conduc(i,1);
k2= conduc(i+1,1);
dp = Deriv_potential(i);
dk = node[i].Deriv_HC/4;
D_wp = Deriv THETA(Wp,i);
Deriv_pot_int = (((1/2) * ((2 * ((dk * p) + (k * dp))) + (DZ * dk))) * pow(k+k2,-1))
- ((1/2) * pow(k+k2,-2) * dk * ((2*((k*p) + (k2*p2))) + (DZ * (k-k2))));
D_theta = (-1/2)* D_wp * Deriv_pot_int;
return D_theta;
}
//
// this function computes the derivative of delta theta with respect
// to water content for the jump at the interface for the "lower" node
//
double Deriv_Dtheta2(double Wp, int i)
{ double Deriv_pot_int, D_theta, D_wp, DZ, p1, p2, kl, k2, dp, dk;
DZ = (node[i].depth - node[i-1].depth)/2;
pl = potential(i-1,1);
p2= potential(i,1);
kl = conduc(i-1,1);
k2= conduc(i,1);
dp = Deriv_potential(i);
dk = node[i].Deriv_HC/4;
D wp = Deriv THETA(Wp,i);
Deriv_pot_int = (((1/2) * ((2 * ((dk * p2) + (k2 * dp))) - (DZ * dk))) *
pow(kl+k2,-1))
105
- ((1/2) * pow(kl+k2,-2) * dk * ((2*((kl*pl) + (k2*p2))) + (DZ *
(kl-k2))));
D_theta = (1/2)* D_wp * Deriv_pot_int;
return D theta;
}
void water_balance()
{ int i;
double current_water_vol = 0.0, dz;
L = 0;
for (i=0; i < num_nodes-1; i++)
{ dz = node[i+1].depth - node[i].depth;
if (i+1 == layer[L+1]) jump(i,L);
{ current_water_vol += node[i].theta[2] * dz/2;
current_water vol += node[i+1].theta[2] * dz/2;
L++;
}
current water vol += node[i].theta[2] * dz;
}
sum_flux += W_q * Dt;
balance = current_water_vol - sum_flux - initial_water vol;
balance ratio = fabs(balance) / current water_vol;
return;
}
void solute_ balance()
{ int i;
double current_C = 0.0, dz;
for (1=0; i < num_nodes-1; i++)
{ dz = node[i+1].depth - node[i].depth;
current_C += node[i].theta[2] * node[i].C[1] * dz;
}
sum_C += S j * Dt;
balance_C = current_C - sum_C - initial_solute;
balance_ratio_C = fabs(balance_C) / current_C;
return;
}
void Read_input()
{
int i;
fscanf(input,"%d %d", &num_nodes, & num_layers);
for(i=0; i < num_layers; i++)
{
fscanf(input, " %If %lf %lf %lf %lf %lf %lf %lf %d", &K_sat[i], &theta_sat[i],
&theta_dry[i],
&alpha[i], &n[i], &initial_theta[i], &lambda[i], &initial_C[i], &layer[i]);
106
}
fscanf(input,"%lf ", &S_Diffusion);
for(i=0; i < num_nodes; i++) fscanf(input,"%lf ", &node[i].depth);
fscanf(input,"%lf %lf %lf ", &start_B2, &Wboundl, &Wbound2);
fscanf(input,"%lf %If ", &Sboundl, &Sbound2);
fscanf(input,"%lf %lf %d", &time_end, &Dt, &max_itir);
fscanf(input,"%lf %If %lf %If %lf %lf %lf %lf ", &Printout[0], &Printout[1],
&Printout[2], &Printout[3], &Printout[4], &Printout[5], &Printout[6],
&Printout[7]);
}
void Write_output_1()
{
int i;
for (i=0; i < num_nodes; i++) fprintf(output,"%10.5f %10.4f %10.4E\n",
node[i].depth, node[i].theta[0], node[i].C[0] * node[i].theta[0]);
}
void Write_output_2()
{
int i;
fprintf(output,"\n");
for (i=0; i < num_nodes; i++) fprintf(output,"%10.4f %10.4E\n",
node[i].theta[0], node[i].C[0] * node[i].theta[0]);
}
107
APPENDIX C
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Depth, cm
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Depth, cm
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Figure C-16: Water and solute distribution.
(Treatment: 2.5 cm sand mulch layer - evaporation rate = 1.5 cm/day).
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123
1.80E-02
1.60E-02
1.40E-02
1.20E-02
1.00E-02
8.00E-03
Cd
6.00E-03
4.00E-03
O
2.00E-03
11
0.00E+00
tt
O
rn
Cs-
02
50
<
Depth, cm
tt,
C-
OOi
El,
O
Depth, cm
O
CC
tr
O
co
0.5
2.50E-02
Initial
0.45
9CCOOZ000)__
0.4
0
0-
0.5 Day
x Initial
2.00 E -02
c'd
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0.35
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0.3
2 Days
3 Days
a.
co
,.
1 Day
0.5 Day
4
I
I1
1.50E-02
13 0.25
-
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2 Days
3 Days
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i
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0
U0
Depth, cm
10
CO
O
.--
02
CO
0,
8
Depth, cm
8
C-
is 0
09
8
.
CI)
9.
CO
2
40
100
90
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r
m
;::.
m
W
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1
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71
t..1
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0
17
IA
71
Solute Concentration, meq/cm3
.1>
P
Water Content, cm3/cm3
-
Figure C-18: Water and solute distribution.
(Treatment: 5.5 cm sand mulch layer - evaporation rate = 1.5 cm/day).
100
90
80
70
60
O
8
.r'
100
80
70
70
80
60
60
O
50
50
0
40 1
o
o
40
t4
us
30
a
U
30
I.
P
20
Solute Concentration, meq/cm3
U
W
20
St 50
0
1/2
P
Water Content, cm3/cm3
10
.
P
10 L
o
41
a
125
1.80E-02
IniSal
1.60E-02
G
4 Days
1.40E-02
6 Days
0-
8 Days
1.20E-02
1.00E...02
8.00E-03
td
6.00E-03
14")
4.00E-03
O
2.00E-03
11
U
0.00E+00
O
rn
O
O
0
C-
co
CO
rn
td
it
O
O
Depth, cm
Depth, cm
4c74
8
O
5
2.50E-02
0.5
---0
Initial
0.45
G
0.4
0.5 Day
2.00E-02
d
-t
tr
.
0-
-
0.3
I
2 Days
3 Days
Initlal
y td
0.5 Day
to
1 Day
1 Day
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-
A1.50E-02
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C)
3 Days
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co
0.25
3
0.2
112
0.15
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CO
I-.
1.00E-02 -
egs
a
rci
5.00E-03
0.1
\
1
0.05
0.00E+00
2
Depth, cm
9
.
-.....-etzeri-500-vego-e-6
0,
O
eo
e,a
b-,
c,
V-
c.
0
Depth, cm
c,
0
c.
L.--
c.
c0
c.
c7c
.0
<1.)
e.0
4-4
tti
0.5
2.00E-02
Initial
0.45
a
0.4
0.35
-0- -0-
-0- -0- -0IT:
0.3
-6- Initial
1.80E-02
4 Days
-a.
- 6 Days
1.60E-02
°-
1.40E-02
_e
8 Days
10 Days
a
4 Days
0-
6 Days
zcr
0- 8 Days
1.20E-02
0.25
1.00E-02
0.2
8.00E-03
0.15
6.00E-03
0.1
4.00E-03
0.05
2.00E-03
10 Days
0.00E+00
CO
CO
O
6.2
OS
Depth, cm
0
0.5
O
2.50E-02
-x- Initial
0.45
0.4
0
2.00E-02 -
¢- 05 Day
1 Day
c°;Q, *a'
0.35
;".;
COS
Depth, cm
°
1.50E-02 -
0.3
1 0.25
I
0.2
1.00E-02
02 0.15
0.1
5.00E-03
-
A..
2 Days
3 Days
1
1
1
,
0.05
0.
+,
0
0
0
0
o
"RP'
0
LC>
Depth, cm
0
CO
0
0.00E+00
CO
rn
O
,,:cmcycPc1xFct:c4-f-0-rr±r-t-r-r-r-v-t-r-4-e.9-r-Metes966961,-t-r-Ar-rci
CCs
<
OS
Depth, cm
0
n
CO
O
P
,
O
O
U
A
IA
0
IA
I
I
i
I
O
In
N
,-rIvrrf irrrrl-rrrr
P
B
5.
O
i.e
Solute Concentration. meq/cm3
O
00
0
ti
<A
1,3
O
O
1.4
m
O
100
90
80
70
60
50
40
30
20
10
0
100
Solute Concentration, meq/cm3
t,i
+
O
.0
o
to.
)3 77.
t,
O
es
es
tTl
L
e44
o
tl
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es
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to
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o
o
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o
t4
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e
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a
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o
tr
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o
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N
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cm
-
0
o
o
A
P
44-rrrIrrry+ rrrr
tn
5
O
l
P
0
90
IA
P
80
P
Water Content, cm3/cm3
Figure C-21: Water and solute distribution.
(Treatment: 2.5 cm coarse sand mulch layer - evaporation rate = 0.5 cm/day).
100
100
90
80 -:
G
50 -:
40
30
20
10
70
U
P
t,
70
1.4
P
60
F
h.
IA
O
Water Content, em3/cm3
60
tsi
r!
..11/41.4
50 -:
40
30
10
o
o
128
tn
©
--
6='
0
00 00
0
F
- -rirrrriv,wrirrerriTrer
Solute Concentration, meq/cm3
E
later Content, cm3/cm3
0
1./%
P
U
o
O
100
90
80
70
60
50
40
30
20
10
0
100
90
80
70
60
50
ee
7
7
7
30
40
7
20
10
0 -n
rri
44:
to.
ee
142
0
0
t-
Om
0
Frfirrrr
a
Solute Concentration, meq/cm3
1.4
later Content, cm3/cm3
Figure C-22: Water and solute distribution.
(Treatment: 2.5 cm coarse sand mulch layer - evaporation rate = 1.5 cm/day).
100
90
80
70
60
50
40
30
20
10
O
O
129
(:)
1.80E-02
0.5
0.45
1.60E-02
4D-
0.4
0.35
4
0.3 -;
t 0.25
1.40E-02
0- 6 Days
lam- -o- -o- -o- -o-
4
0-
Initial
4 Days
8 Days
1.20E-02
7CF,
4 Days
°
6 Days
- 8 Days
-
10 Days
11.
1=1-
1.00E-02
10 Days
I
8.00E-03
0.2 -0.7
0
1%.
m 0.15
6.00E-03
4.00E-03
0.1
2.00E-03
0.05 L:
0.00E+00
Cg
O
cJ
c.")
Depth, cm
O
O0
00)
O
Depth, cm
0.5
2.50E-02
Initial
0.45
°00CCcco.._
0.4
O
Initial
0- - 0.5 Day
o-
1 Day
0.35
0-
0-- 05 Day
o'
2.00E-02
1 Day
2 Days
03 Days
0.3
-
1.50E-02
2 Days
3 Days
0.25
\
0.2
1.00E-02
0.15
0.1
b-o-o±45-o-c-es-tH
5.00E-03 {P%
ck
4
0.05
\
CO
CD
Depth, cm
Cn
+,
)...
0.00E+00 xl.+.1.4...vp.t.tr-t4:-tm-rl-t-r±s-90,191041.4090M,t4,-.11-..1
0
0
0
O
0
0
0
0
0
o
O
-Of
0
cq
77
0
C.CO
Depth, cm
CO
..-.
p
O
o
+
II
100
90
a
0
0
0
100
90*
80
70
70
80
60
5050
60
4040
g
P
30
100
90 -
80 -:
70 -:
60 -:
50
30
3
3
40
30 -:
20
cn
-
Water Content, cm3/cm3
a
u.
6.
8
e
)3' P')':
0
,Cr
i.
ades:p
8
e
0
T
1
ri
OD
00
0,
1
ri
ON
0a
5
NR R
a
0
00
0 0
0
0
0
Solute Concentration, meq/cm3
a
0
-7,r+TyrriTm+,-rrrhrrrrtm.rfmrb,rh-rrr
O
10 L'
0
20
C.1
P
20
N
U
a
ul
10
N
tO
Solute Concentration, meq/cm3
ii
P
a
0
t,1
a
_...cr
_. 7
tn
W
a
0
O
O
O
a
a
P
°
0
100
so
80 -:
70 -:
60 -:
al
a
ea'
o p-
o
cr
;-
a
Water Content, cm3/cm3
a
Figure C-24: Water and solute distribution.
(Treatment: 5.5 cm coarse sand mulch layer - evaporation rate = 1.5 cm/day).
3
50 -:
40 -:
30
20-:
10
O
a
131
I
0
in
r-+
g-
20
10
I
,+
(.4
I
100 1
90
80
0
60
50
40
30
100
90
rl
V1
0
i.e
,_,
P
O
i.e
0
0
0
be
r23
0
Ut
Solute Concentration, meq/cm3
O
O
I
O
i.e
0
14
0
0
9
0
en
100
100
90
80
70
70
.
60
60
2
50
I
IA
60
-
Os
P
40
j!.
a
la
40
It.1
P
30
u.
"to
Water Content, cm3/cm3
e
P
La
P
P
W
Us
11.
0
0
O
In
D,
;A
14
P
U
e
°
:
1
X1.4
P
(.4
a
8
rn
8
O
e
0.
O
m
0
m
0
i.,
.o.
P
e
I
I
0
e
0.
0°.
I
I
ei.e
MI
0
0
0°
I
a
pto
:p.
to
4rrrriwrrr}rrrr
to
La
e
Solute Concentration, meq/cm3
0
i.e..
Water Content, cm3/cm3
P
Figure C-25: Water and solute distribution.
(Treatment: 10.5 cm coarse sand mulch layer - evaporation rate = 0.5 cm/day).
St
:4 -
.
P
30
O
0
V.
132
40
1.4
O
O
O
10
0
0
1
1
1
1
0
o
CA
P
5
100
90
O
O
e
U.
, me
14
P
ix
U.
g
CP
P's
71
71
W
P
A
IA
a
0
0
1
ao
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CP
g
71
I
7'
o
oo
Q
I
114
t'3
0
9
4,
P
(A
4yerrbwrIrrrevim
4A
Solute Concentration, meq/cm3
g
P
Water Content, cm3/cm3
Figure C-26: Water and solute distribution.
(Treatment: 10.5 cm coarse sand mulch layer - evaporation rate = 1.5 cm/day).
100
90
80
70
0
6
60
60
80
50
40
30
50
40
30
20
0
20
1,4
M
0
10
1,4
O
M
0
,-.
to
00 ......5
M
g
....
im
0
90
90 --
100
80 2-
80
Solute Concentration, meq/cm3
70
70
100
60 -
50
40 L.
60
:4 5°
Q,
M
o
0
a
UN
p
30
?
U
A
U
30
NoN
o
20 L.
in
pow
20
10
0
- -
Water Content, cm3/cm3
114
e
ie
0
U.
P
133
0
a
o
to'
,t?
60
70
80
90
100
70
80
90
100
50
60
50
40
40
0
30
ba
30 of
to-,
O
tm
20
4-.4
e
e
o
20
W
O
co°
In
100
10
UI
e
Solute Concentration, meq/cm3
10
100
90 -:
try
e
1,)
0
M
,:,
0
I
I
9
I
I
1,..1
e
m
0
t.
0
co
o-
O 0 0t.
I
7.05
1,1
M
O
0
Solute Concentration, meq/cm3
..
I-
)".
e
L.,
M
0
0
Water Content, cm3/cm3
a
--
Figure C-27: Water and solute distribution.
(Treatment: 2.5 cm sand barrier and mulch layers - evaporation rate = 0.5 cm/day).
3
0
80 -:
80 -:
90 -:
70 -:
70 -:
I
60 -:
60 -:
50
40
11
O
10 -:
0
40 -:
,0
CO
30
LAI
30 :
114
20
O
Water Content, cm3/cm3
2o
0
O
M
O
N.
0
ill
14
134
40 -:
30
20
10
N
F.
La
N
P
4,
tn
in
VI
P
o
O
P
I-4>
W
trI
o
a
1.
-2
0
9
NJ
0
0
61
I
.)
o
:o
N
.7
in
o
,..
Solute Concentration, meq/cm3
a
O
to
I
a,
o
o
e
IJ
a
Ef
-U
i.e
ifa
A
-
0
b
i.e
t.1
o
o
..-
i.e
F=I.6.
-5°
i.e
C150
i...
2
TOO
15
A
I
9
IF
O
5
1
0
o
r,1
e
i.e
1..
IA
00,
9
Solute Concentration, meq/cm3
15
90
le,
O0
o
rl
P,
I
6
P
O
a0
a
Oo
6
a
0
I.
2
I
O
80
70
60
50
40
30
20
10
0
100
so
80
70 -:
60
50
40
VI
P
-TrrrfrrrItV=Prrrri=lirrrriwrrii-rrrirrwr
<2.
30 -:
20
10
0
later Content, cm3/cm3
Figure C-28: Water and solute distribution.
(Treatment: 2.5 cm sand barrier and mulch layers - evaporation rate = 1.5 cm/day).
100
90
80
70
60
40
30
20
10
°
100
90
80
70
60 -:
5- 50
7
In
0 7rrrfrrrPtrnat7rArvirrrr/TrrrfrrrriTyrrill-rr
0
O
Mater Content, cm3/cm3
135
I
I
I
0'
o
to
0
O
O
100
90
0
70
30o
20
10
100
90
80
O
0
a
0'
O
15.
1.1
r=1
F
9
I
F
4
I
Solute Concentration, meq/cm3
1<r.
0
00
a
1
c'
In+
0
P
0,
1
Q
J
a
5
a
100
so
80
70
70
50
60
-a"-
.1"
6
50
O
W
m
U
eN
t"
P
W
O
to
11
r
0
0
0'
r
eA
S.
Solute Concentration, meq/cm3
- -to
Water Content, cm3/cm3
a
tn
Figure C-29: Water and solute distribution.
(Treatment: 5.5 cm sand barrier and mulch layers - evaporation rate = 0.5 cm/day).
0
S.
0
o
CA
40
1.01
P
d
40
P
30
NN
30
P
Water Content, cm3/cm3
-
20
tot
P
20
O
O
F
U.
P
to+
136
-
1
1
P
o
1
a
t
1
C
rn
c
1
0
o
c
.1
0
0
a
a
41
1
"
P
1
100
90
80
8
O
IA
tA
N
to
0
0
to,
Ico
O.
1
0 00
co
1
U.
W
(..a
L.,
I::3
ri
0
a0
L.,
P
0
0
rl
0
0
rl
0
N
t21
Solute Concentration, meq/cm3
P
0
1+
Water Content, cm3/cm3
-
1
9
0
1,
.1.
iN
o
5-
N
Figure C-30: Water and solute distribution.
(Treatment: 5.5 cm sand barrier and mulch layers - evaporation rate = 1.5 cm/day).
100
90
1
9
70
70
80
60
60
50
40
40
10
0
30
NI
a
P
30
1
1
O
O
,
t,1
R]
M1
g
P
100
90
20
-4> --- -4>
I>+.C4+
g
r,1
0
....
Solute Concentration, meq/cm3
00
20
10
0
100
90
80
80
a
70
70
6
60
60
0
0
to -L
0
50
U
w
50
Id
40
U
40
to
N
30
V1
30
5- 50
8
P
20
V01
O
(later Content, cm3/cm3
20
to
O
O
N
U.
137
p
o
I
?
a
0
a
Vo
!"'
P
rt
so
40
30
100
90
80
70
60
50
40
30o
20
10
100
90
80
?
-0
0
o
O
/a!
I
0
00
a
°
I
-
Solute Concentration, meq/cm3
I
0
I
0
0
0'
0
N
100
90
80
70
60
7 50
40
30
20
10
0
100
90
80
O
8
20 --
10
8
o
cy,
P
N
N
UI
F
.
P
Water Content, cm3/cm3
O
Ir
U.
MI
0
P
tM
MI
g
Pp
g
MI
0
N
N
0
0
N
0
I4
MI
P
A
0
Solute Concentration, meq/cm3
I
P
N
0
t?1
o
a
0
a
Figure C-31: Water and solute distribution.
(Treatment: 10.5 cm sand barrier and mulch layers - evaporation rate = 0.5 cm/day).
rt
O
_o
Cn
P
70
ta.
o
70
u.
P
tof
60
CM
P
Water Content, cm3/cm3
60
50
40
30
20
10
0
©
0
N
aa
P
138
2.00E-02
0.5
Initlal
1.80E-02
0-
4 Days
1.60E-02
-6-
6 Days
0-
8 Days
0.45 7.
0.4 7
035
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