-------
Boundary Conditions for Chemical: Two types of boundary conditions can be imposed at the soil
surfaces. These conditions can be imposed at any time. They can also be modified at any time so
complex flow problems can be simulated.
The following boundary conditions are supported at x = 0:
1. Constant Concentration of Inflowing Solution: This boundary condition is used to
simulate movement of chemicals when the solution entering the soil has a known and
constant concentration, cs. The amount of chemical entering the soil depends upon the
flux of water entering the soil. Moreover, if water is moving out of the soil at x = 0 (as in
evaporation), no chemical moves with it. Mathematically, this boundary condition takes
the form
=q(0,t)cs ifq(0,t)>0
t,— 1\
(14)
-9D — + q(0,t)c =0 ifq(0,t)£0
x=0
2. Constant Concentration in Surface Soil (x = 0): This boundary condition specifies that the
concentration c(x, t) at x = 0 is a specified value c0. That is
c(0,t) = c0 (15)
Note that equation 15 approximates a system in which the concentration in the soil is
abruptly forced to take on a certain value and to remain at that value. This would likely
be difficult to carry out experimentally. Equation 14 will likely be a better approximation
to real soil systems.
The boundary conditions supported at x = L are described below.
1. Convective Flow Only: This boundary condition is used to simulate soil systems in which
the chemical moves out of the soil with the moving soil water, but dispersion and
diffusion do not contribute to this movement. This condition is equivalent to the
requirement that the gradient of the concentration is zero at x = L. That is,
(16)
2. Constant Concentration at Soil Surface (x = L): This boundary condition specifies that
the concentration c(x,t) at x = L is a specified value CL. That is
c(L,t) = cL (17)
34
-------
Additional Equations Related to Chemical Transport:
• Dispersion coefficient: The dispersion coefficient D at position x and time t is
approximated by the equation
D(x,t) = D0T+X|q(x,t)/0(x,t)| (18)
where A, is the dispersivity of the soil-chemical system, DO is the molecular diffusion
coefficient of the chemical in free solution and T is the tortuosity of the soil. The
tortuosity is estimated using the equation of Millington and Quirk (1961) where
(19)
e|
and 9s is the saturated water content of the soil.
• Flux density of chemical: The flux density of chemical at position x is the mass of
chemical passing that position in the soil per unit cross-sectional area per unit time. It is
positive in the direction of the x-axis, as explained for the flux density of water. The flux
of chemical, f(x,t), at location x and time t, is given by
f(x,t) = -0D^+qc (20)
• The cumulative flux of chemical is the mass of chemical moving past the position of
interest per unit cross-sectional area of soil from time t = 0 to the time of interest. That is,
the cumulative flux of chemical, F(x,t) is given by
F(x,t) = J*f(x,t)dt
(21)
where f(x,t) is the flux density of chemical.
• Total mass of chemical: The total mass of chemical in the soil at time t is the sum of the
mass of chemical in the liquid and solid phases. (Partitioning of the chemical to the vapor
phase is ignored in this model.) That is,
mT(t) = m|(t) + ms(t) (22)
where the mass of chemical in the liquid phase, nii(t), is
) c(x,t) dx (23)
and the mass of chemical in the solid phase, ms(t), is
-------
ms(t) = J0Lp(x)k(x)c(x,t)dx
(24)
• Total concentration of chemical: The total concentration, Cx(x,t) of chemical in a soil is
the sum of the mass of chemical in the soil solution per unit volume of soil plus the mass
of chemical adsorbed on the soil solids per unit volume of soil. That is
cT(x,t) = [9(x,t) + p(x)k(x)]c(x,t)
(25)
Computational Methods
Flow and transport equations are solved using finite difference methods. That is, difference
equations are used to approximate the governing differential equations. To do this, a set of mesh
points is defined in the soil. Initial conditions specified by the user determine the values of the
matric potential and concentration at these mesh points at time zero. Inserting these values into
the difference equations for water movement produces a system of equations (one equation for
each mesh point) that are then solved simultaneously to determine the matric potentials at each
point at time ti, a short time later. If chemical movement is being simulated, the initial values and
the solution to the water flow equations are used to define a second set of difference equations
that are solved for concentration at time ti. These solutions are then used to redefine the water
and chemical equations to obtain solutions at a later time, t^. This process is repeated until tj is
equal to the time of interest. The following pages contain the details of the equations used.
Water Movement: The computational method used to solve the Richards equation is based on the
work of Celia et al. (1990). This iterative method has the advantage of maintaining mass balance
of water.
The backward Euler approximation of equation 1 can be written as
9(xi,tj+i)-e(xi,tj)^ a
(tj+1-tp ~ax
fah(xj,tj+1) ^
K(Xj,tj+1) —J sin(A)
(26)
where x;, i = 0, 1,2, ..., N represent mesh points in space, and tj, j = 0, 1, 2, ... represent mesh
points in time. This is a non-linear problem in that the hydraulic conductivity depends upon the
matric potential or water content at time tj+i which is unknown when this equation is applied.
Following the work of Celia et al. (1990), this problem is solved using a Picard iteration scheme.
In that case, the above equation takes the form
(tj+1-tj)
ax
Km(Xj,tj+1)
ax
- sin(A)
(27)
36
-------
where m represents the iteration number at the current time step. Note that K on the right-hand
side is evaluated using iteration m when solving for iteration m+1. The iterations at a single time
step are continued until differences between iterations are "sufficiently small."
Expanding 9m+1 with respect to h by Taylor series yields
+ 0
I2}
ci,tj+1)-hm(xj,tj+1)J
Inserting equation 28 into 27 and ignoring second order and higher terms yields
(28)
-
m
9(Xi.tj+1)-9(Xi.tj)
rm
Km(xj,tj+1)
9hm+1(Xj,tj+1)
- sin(A)
(29)
where
r t. ,_
( " J+l)~
im
IX|,t
(30)
The last term in equation 29 was estimated by means of the following difference equation
9x
Km(Xj,tj+i)
Sx
- sin(A)
2^Km(xi+1/2,tj+1)-Km(Xi_1/2,tj+1)jSin(A)
(31)
where X;+i/2 = (x;+i + X;)/2 and X;.i/2 = (X; + Xi_i)/2.
37
-------
Inserting equation 31 into equation 29 and simplifying yields
Cm(xhtj+1) + 2
(tj+l-tj)
(Xj+l-xj)
= hm(Xj,tj+1)
-2
m
C(Xj,tj+1)
(tj+l-tj)
(Xj+i-xj)
V J
em(xi,tj+1)-9(xh
(tj+l-tj)
(32)
2^Km(xi+1/2,tj+1)-Km(Xi_1/2,tj+1)JSin(A)
fori= 1,2,3, ...,N-l;j = 0, 1, 2, ...; and m = 0, 1, 2, ....
Equation 32, along with two additional equations for the two ends of the soil system to be
developed later, define the system of N+l equations to be solved simultaneously. Because this
solution is an iterative one, we need to have additional equations to enable us to determine when
the difference between solutions for iteration m and m+1 are sufficiently small to allow the
process to stop. Celia et al. (1990) derived the needed equations. Equation 29 can be rewritten as
^ffl/v.
(33)
^[h'"^(Xj,tj+1)-h'"(X
's[hm+1(xi,tj+1)-hm(xi,tj+1)J
Km(xj,tj+1)
Km(xj,tj+1)
Shm(xhtj+1)
- sin(A)
(tj+l-tj)
Note that if the right-hand side of this equation is zero, this implies that the solution at iteration
m solves equation 27. Therefore, these terms can be used to evaluate the residual r and to
determine when sufficient iterations have been carried out. Replacing the derivatives on the
right-hand side of equation 33 in a manner similar to that in equation 31 and simplifying yields
38
-------
n=
2Km(Xi+1/2,tj+1)^hm(xi+1,tj+1)-hm(Xi,tj+1)J
2(Km(Xj+1/2,tj+1)-Km(Xj_1/2,tj+1))Sin(A)
(34)
for i = 1, 2, 3, ..., N-l. Equation 34 plus two additional equations for the ends of the soil column
provide equations for determining the residual at each mesh point.
Boundary Conditions at x = 0: Equations for i = 0 must be developed based on the boundary
condition imposed. Specified matric potential, flux, and mixed type boundary conditions are
supported in this software. Mixed type is simply a combination of the other two, so no additional
equations are needed for it.
For a specified matric potential of h0 at x = 0, we have
(35)
for all j and all m. The residual equation in this case becomes
r0=0
(36)
To develop the equations for a specified flux q0 at x = 0, equation 1 is written in the form
59__aq
dt ~ dx
or
(37)
dt 9x
where q is the flux density of water. The iterative numerical equation for this becomes
9m+1(x0,tj+1)-9(XQ,
(tj+1-tj)
qm+1(xi/2,tj+i)-q0]
(xi-x0)
(38)
where
39
-------
-K (xi/2,tj+i)
-x0)
-sin(A)
(39)
m+1
Expanding 9m+ in a Taylor series as done previously and combining with equations 38 and 39
yields
»m,
C-(xo,tj+i).m+
(tj+i-tj) I J+1'
m
-.m
0m(xo,tj+i)-0(x0,tj)
m
2qo 2K(x1/2,tj+1)
-x0)
-x0)
-x0)
- sin(A)
(40)
or
hm+1(x0,tj+1)
»m
m.
C(x0,tj+1) 2K(x1/2,tj+1)
m
2K(x1/2,tj+1)
= hm(x0,tj+i)
cm(x0,tj+i) em(x0,tj+i)-e(x0,tj)
(tj+1-tj)
(tj+1-tj)
(41)
The residual equation to be used with equation 41 is obtained by rearranging equation 40. It is
2Km(xi/2,tj+i)
(XI-XQ)
hm(xi,tj+i)-hm(x0,tj+i)
«inrA^
(XI-XQ)
rvlTI f\
(42)
-x0)
(tj+1-tj)
Boundary Conditions at x = L: Equations for i = N must be developed based on the boundary
condition imposed. Specified matric potential, flux, and free drainage boundary conditions are
supported in this software. The steps involved in deriving these equations are similar to those
used at x = 0, so many details will be omitted.
For a specified matric potential of !IL at x = L, we have
hm+1(xN,tj+i) =
(43)
40
-------
for all j and all m. The residual equation in this case becomes
rN=0
For a specified flux qL at x = L, we obtain
2Km(xN_1/2,tj+1)
+ hm+1(xN,tj+1)
Cm(X|\|,tj+i) 2Km(xN_1/2,tj+1)
(tj+l-tj)
-+
Cm(xN,tj+1) 0m(xN,tj+1)-0(xN,tj)
= hm(xi\|,tj+i) —
J+1' (tj+1-tj) (tj+l-tj)
2fKm(xN_1/2,tj+1)Sin(A)-qL]
_I__J- *_
and
rN= -
2Km(xN_1/2,tj+1)
(XN-XN-I)
2qL 9mO
hm(xN,tj+1)-hm(xN_1,tj+1) 1
sin(A)
(XN-XN-I)
\ J _
«N'tj+i)-0(xN,tj)
XN-I) (tj+i-tj)
For free drainage at x = L, we again begin with equation 37. In this case we have
0m+1(xN,tj+1)-e(xN,tj) _ 2[qm+1(xN,tj+1) -
where
(tj+l-tj)
qm+1(xN,tj+i) = Km(xN,tj+i)Sin(A)
and
qm+1(xN_1/2,tj+1) =
rfT\
-Km(xN_1/2,tj+1)
hm+1(xN,tj+1)-hm+1(xN_1,tj+1) .
-sin(A)
m+l
Inserting the Taylor expansion for 9m and simplifying leads to
(44)
(45)
(46)
(47)
(48)
(49)
41
-------
2K (xN_1/2,tj+1)
+ hm+1(xN,tj+1)
= h (xN,tj+1)
Cm(xN,tj+1) 2Km(xN_1/2,tj+1)
(tj+1-tj)
(xN-xN_!)
(50)
Cm(xN,tj+1) 9m(xN,tj+1)-9(xN,tj)
(tj+1-tj) (tj+i-tj)
2^Km(xN_1/2,tj+1)-Km(xN,tj+1)Jsin(A)
The residual equation in this case becomes
rN=
-2Km(xN_1/2,tj+1)
(XN-XN-I)
'hm(xN,tj+1)-hm(xN_1,tj+1)
(XN-XN-I)
V
- sin(A)
(51)
2Km(xN,tj+1)Sin(A) 9m(xN,tj+1)-9(xN,tj)
xN-l) (tj+1-tj)
Calculations and Convergence: Equation 32 and the appropriate equations for the boundary
conditions define a set of N+l linear equations to be solved for each iteration at each time step.
The equations can be represented in matrix form where the coefficient matrix is tri-diagonal. As
a result, the solution can be obtained quite rapidly and accurately.
The solution process involves the following steps (ignoring logic for cases where convergence
fails):
1. Utilize the specified initial conditions to initialize h(x;, to)
2. Set time step index j = 1
3. Define ti
4. While tj is less than or equal to the time of interest
a. Set h°(xi, tj+i) = h(xi, tj)
b. Repeat the following steps until a solution is obtained (Boolean variable called
SolutionObtained is true):
i. Set up and solve the system of equations for the current iteration
ii. Set SolutionObtained to true
iii. For each mesh point in the system do the following
1. Calculate the residual using equation 34 (or the alternate equations
for nodes 0 and N)
2. If the residual exceeds the critical residual
a. Set SolutionObtained to false
b. Break out of residual calculation loop
c. Save last solution obtained as h(x;, tj)
d. Increment] by 1
e. Define tj
42
-------
In this program, the solution has converged when
R
max
(tj+i-tj)
(52)
for all values of i. Rmax is a number representing the convergence criterion and can be modified
by the user (see Mesh Size/Convergence option). Interpreting the physical meaning of Rmax is
not straightforward. The default value is 0.0001. It is recommended that users who are concerned
about this value examine the sensitivity of the results of interest to different values of this
parameter.
In some cases, the system converges slowly or not at all. The algorithm used reduces the time
step if convergence fails after a predetermined number of iterations. The system may later
increase the time step again if convergence becomes rapid; however, the time step is never
increased beyond the value specified by the user. If reducing the time step several times still does
not result in convergence, the user may be notified and the calculation stops.
If the value of Rmax is not sufficiently small, the iterative process may converge based on the
criteria above, but the solution may be inaccurate. A mass balance calculation is incorporated to
detect such problems. This mass balance compares the difference between the quantity of water
entering the soil system and that leaving the soil system with the change in amount stored in the
soil. Conservation of mass implies that these amounts must be equal. If they are not nearly equal,
a warning is issued to the user who can then choose to use a smaller value of Rmax and / or
smaller mesh sizes.
43
-------
Chemical Transport : The preceding section presents the equations and logic for solving the
water flow equation. If chemical transport is being simulated, the system sets up and solves the
transport equations at each time step (before incrementing j and defining a new tj in the previous
outline of calculations). The flux density and water content at each point are used in the solution
to equation 9 for that time step. This alternating solution process continues until the time of
interest has been reached.
The numerical solution to equation 9 is based on that of van Genuchten (1978). In that work, he
derived a correction for numerical dispersion. The equations used in that work are outlined
below. An equation is derived for each mesh point in position. Values of concentration for time
t0 are determined from the initial conditions. This system of equations is solved simultaneously
to obtain the concentration at all points for time ti. Since no more than three mesh points are
involved in any single equation, the coefficients of the unknown concentrations form another tri-
diagonal matrix that can be readily solved. The equations are linear in this case so no iteration is
required.
The following equations define the system of equations that are solved for each time step.
Equations 53-59 define the equations for the interior mesh points or for x; for i = 1, 2, 3, ..., N-l.
{0Rc}(i,j- = ^^
(53)
1 I a I , Sc I I
+—4— 0D qc -(a0 + ppk)c+70Hi,j)
where
{0Rc}(i, j) = 0(x j ,t j) R(x j ,t j) c(x j ,t j) (54)
D_=D_q2(tj+1-tj)
60 R (55)
D+=D+q2(tj+1-tj)
662R
(56)
44
-------
q(xj,tj) + 2q(xi+1,tj)
x 1
~xi-l
q(Xj,tj) + 2q(Xj_i,tj)
{(a9 + ppk)c}(i,j) = —
(58)
1)k(xj_1)]
tj) + p(Xj)P(Xj)k(xi)
1)k(xi_1)]}
(59)
and x; for i = 0, 1,2, . . . , N are mesh points in position and tj for j =0, 1,2, ... are mesh points in
time. Equation 53 can be applied to mesh points x;for i = 1, 2, 3, . . ., N-l and all values of j.
Additional equations needed at the boundaries of the soil system (x0 = 0 and XN = L) are given
below.
Boundary condition #1 at x = 0 - Specified concentration of inflowing solution: Equation 53 can
be written as
» (60)
21 ax
where
df ,„ .,
1/2 ~
v,j,- — (61)
Sx (X!-x0)/2 v '
[9(x1,tj)D(x1,tj)+9(x0,tj)D(xo,tj)](c(x1,tj)-c(x0,tj)
2(xi-x0)(62)
_ .. „ t. . for q(x0,tj)>0
°~{o for q(x0,tj)^0 (63)
45
-------
+ a(x j )0(x !, t j) + P(x j )p(x i )k(x j)]
+ C(x1,tj)[a(xo)0(xo,tj) + P(xo)p(xo)k(x0)
+a(x1)6(x1,tj) + p(x1)p(x1)k(x1)]}
(65)
Equation 60 in combination with equations 61-65 provides the equation for the first mesh point
with this boundary condition.
Boundary condition #2 at x = 0 - Specified concentration of soil solution: This condition given
in equation 15 implies
c(x0,tj) = c0 (66)
where x0 = 0, j =0, 1,2, ..., and to = 0. This provides the relationship for the first equation with
this boundary condition.
Boundary Condition #1 at x = L - Convective Flow Only: Again, we can write equation 53 as
^ = -1-—-(a0 + ppk)c + y0j(N,j + l)
At 21 Sx ; (67)
fN~fN-l/2 (6g)
5x (xN-xN_!)/2
[9(x|\|,tj)D(x|\|,tj)+6(x|\|_1,tj)D(x|\|_1,tj)](c(x|\|,tj)-c(x|\|_1,tj)
N~1/2 ~
+ [2q(xN,tj)c(xN,tj) + q(xN,tj)c(xN_1,tj) (69)
+ q(X|\|_1,tj)c(X|\|,tj) + 2q(X|\|_1,tj)c(X|\|_1,tj)]/6
=q(X|\|,tj)c(X|\|,tj) (70)
j) = i{3c(xN,tj)[a(xN)0(xN,tj)+p(xN)p(xN)k(xN)
+ c(x|\|_1,tj)[a(x|\|)0(x|\|,tj
46
-------
j) = Y(xN)0(xN,tj) (72)
Equation 67, in combination with equations 68-72, provides the equation for the first mesh point
with this boundary condition.
Boundary Condition #2 as x = L - Specified concentration of soil solution: This condition given
in equation 17 implies
c(xN,tj) = cL (73)
The equations above form a system of N+l simultaneous equations that are solved for the
solution concentration at the mesh points, x; and time tj+i . The equations take to form
(74)
fori= 1,2,3, ...,N-1 where
f a(i + l,j
9 J T \.) — s
2 --.- --
a(xj+1)e(xj+1,tj+1) + p(xj+1)p(xj+1)k(xj+1)+a(xj)e(xj,tj+1) + |3(xj)p(xj)k(xj)
24
-+-
i+i -xj) (xi+1 -XJ^KXJ -xj_
1 a(l — 1 j _
W0,j + l) = --l—-i ^ ' V'J'^ +
(76)
"1-1. J (7?)
H
24
and
47
-------
Xj)k(Xj)]
(78)
i, j) = 0(xj ,
6[e(xj,tj)+p(xj)k(xj)]
For i = 0 and a specified concentration of the inflowing solution, the equation is
-c(x1,tj)u(0,j)-c(xo,tj)[v(0,j)-0(xo,tj)R(x0,tj)/(tj+1-tj)] (79)
Y(x0)[(0(xo,tj+1)+0(x0,tj)] cs[q(xo,tj+1) + q(x0,tj)]
2 (xi-x0)
where
-[a(l,j + l) + a(0,
u(0,j
-XQ)
2(X!-x0)
12
v(0,j
-x0)
2(x!-x0)
12
and
q(x0,tj+1)2(tj+1-tj)
= 0(x0,tj+1)D(x0,tj+1)- J+ J+ J
(80)
(81)
6[0(x0,tj+1) + p(xo)k(x0)]
q(x0,tj)2(tj+1-tj)
a(0,j) = 0(xo,tj)D(x0,tj)+ — —
J J 6[0(x0,tj) + p(xo)k(x0)]
Note: If q(xo,t) is zero or less than zero, the last term in equation 79 drops out.
For i = 0 and the specified concentration at x = 0, the equation is
c(x0,tj+i) = c0 (83)
For the convective flow boundary condition at x = L, we obtain the following equation for i = N.
48
-------
-c(xN_1,tj)w(N,j)-c(xN,tj)[v(N,
+Y(xN)[(6(xN,tj+1)+e(xN,tj)]/2
(84)
where
v(N,j + l) =
(XN-XN-I)|_ J 2(xN-xN_1) 6
tj+1) + 3p(X|\|)p(X|\|)k(X|\|)+a(X|\|_1)0(X|\|_1,tj+1) + p(X|\|_
12
(85)
w(N,
and
-1
a(N,j
2(x|M-X|M-l) 6
ltj+1) + p(xN)p(xN)k(xN)+g(xN_1)9(xN_1,tj+1) + p(xN_1)p(xN_1)k(xN_1)
12
(86)
j + l) = 0(xN,tj+1)D(xN,ti+1) *- *— *
N J+1 N J+1 6[0(xN,tj+1)+p(xN)k(xN)]
q(x|\|,tj)2(tj+1-tj)
l,J) = e(XN9ti)D(xN,tj)+ '- *— '-
J J 6[0(xN,tj) + p(xN)k(xN)]
For a specified soil solution concentration at x = L, the equation is
c(xN,tj+1) = cL
' ^
(88)
49
-------
Table 5. List of Symbols, Descriptions and Units Used in Text
Symbol
A
C(h)
c(x,t)
cT(x,t)
CinitialvV
cs
Co
CL
D(x,t)
D", D+
Do
df
F(x,t)
f(x,t)
H(x,t)
h(x,t)
hinitial(x)
hinitial
ho
hL
i
j
K(h)
Ks
k(x)
L
m
mi(t)
ms(t)
mT(t)
Description
- angle between the direction of flow and the horizontal direction
- specific water capacity, d9/dh
- concentration of chemical in the soil solution at position x and time t
- total concentration of chemical in the soil at position x and time t
- initial concentration of chemical in solution at t = 0
- concentration of inflowing solution at x = 0 for constant
concentration of inflowing solution boundary condition
- concentration of soil solution at x = 0 for constant concentration
boundary condition
- concentration of soil solution at x = L for constant concentration
boundary condition
- dispersion coefficient of chemical in soil at position x and time t
- dispersion coefficients corrected for numeric dispersion
- molecular diffusion coefficient of chemical in free solution
- driving force for water, -dWdx
- cumulative flux of chemical passing position x at time t
- flux density of chemical at position x and time t
- total soilwater potential (or total hydraulic head) at position x and
time t
- matric potential at position x and time t
- initial matric potential function at t = 0 for finite length soils
- initial matric potential at t = 0 for semi-infinite soil
- matric potential at x = 0 for constant potential or mixed type
boundary condition
- matric potential at x = L for constant potential boundary condition
- index of mesh points in position
- index of mesh points in time
- soil hydraulic conductivity at matric potential h
- saturated hydraulic conductivity
- partition coefficient of chemical in soil at position x
- length of soil column
- index of iteration number in solving flow equation
- mass of chemical in soil solution per unit cross-sectional area
- mass of chemical adsorbed on soil solids per unit cross-sectional
area
- total mass of chemical in soil system per unit cross-sectional area
Units
degrees
cm"1
g m"3 (water)
g m"3(soil)
g m"3 (water)
g m"3 (water)
g m"3 (water)
g m"3 (water)
cm2 hr"1
cm2 hf1
cm2 hf1
-
gm"2
g m"2 hr"1
cm
cm
cm
cm
cm
cm
-
-
cm hr"1
cm hr"1
m3 Mg"1
cm
-
gm"2
gm"2
gm"2
50
-------
Table 5. (continued)
Symbol
Q(x,t)
q(x,t)
qo
QL
R
l^-max
n
S(x,t)
t
tc
X
-------
Numerical Experiments for Water Movement
1. Infiltration of Water Ponded on the Soil Surface
Objective: 1. To determine the nature of the dynamic infiltration process for water ponded on the soil
surface.
2. To compare the infiltration rate at different times with the saturated hydraulic
conductivity of the soil.
Situation: Water is applied to a loam soil by flooding the soil to a depth of 2 cm. Water is continually
applied to the soil surface to maintain this height of water. The soil was somewhat dry
throughout before the water was applied. The objective is to observe the rate at which water
enters the soil and the total amount entering the soil.
Simulation: Simulate water movement into a vertical, semi-infinite loam with an initial matric potential
of -2000 cm. Apply water to the soil at a constant potential of 2 cm. Simulate movement for
12 hours. Display the flux of water at the soil surface and the cumulative infiltration of water
as functions of time.
1. What was the infiltration rate or the flux of water at the soil surface at 1, 2, 4, 6, 8, 10,
and 12 hours? Was the infiltration rate increasing or decreasing?
2. Look at the data for the loam soil. What is the saturated conductivity for the soil?
Compare the infiltration rates observed above with the saturated conductivity. Which is
larger? Why? What infiltration rate would be expected if the wetting process continued
for two days? Why?
3. How much water entered the soil profile in the first hour? How much in 2, 4, 6, 8, 10,
and 12 hours?
Additional Repeat the above exercises for different soils. Compare the infiltration rates and the
Work: saturated hydraulic conductivities. Are the results consistent with those above?
52
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Numerical Experiments for Water Movement (continued)
2. Infiltration of Water from Rainfall or Sprinkler Irrigation
Objective: 1. To determine the nature of the dynamic infiltration process for rainfall or sprinkler
irrigation.
2. To compare the infiltration rate at different times with the saturated hydraulic
conductivity of the soil.
3. To determine the total amount of infiltration, the total amount of runoff, and the time at
which runoff occurs.
Situation: A farmer recently irrigated his field so that the soil was relatively wet. Unexpectedly, a
rainstorm occurred. The storm lasted for six hours. The rainfall rate was 0.75 cm/hr. The
field was a silt loam soil. How much of the rain water entered the soil? Did any runoff
occur? If so, how much? When did runoff begin?
Simulation: Simulate water movement into a vertical, semi-infinite silt loam soil with an initial matric
potential of -100 cm. Apply water to the soil at a rainfall rate of 0.75 cm/hr. Simulate
movement for six hours. Display the flux of water at the soil surface and the cumulative
infiltration of water as functions of time.
1. Describe the curve for the infiltration rate as a function of time. It is initially constant.
What is the value of the infiltration rate during this constant phase? Compare this rate
to the rainfall rate.
2. Eventually the infiltration rate decreases. At this time the soil is no longer able to
transport water from its surface as fast as it is applied. This is the beginning of the
runoff phase. At what time does runoff begin?
3. How much water entered the soil surface by the end of the six-hour period? How much
water was applied as rainfall during this period? How much runoff occurred? (The
amount of runoff is the difference between the amount applied and the amount entering
the soil. This calculation assumes that there is no surface storage of water.)
4. Compare the rainfall rate to the saturated hydraulic conductivity for the soil. Compare
the final infiltration rate to the saturated conductivity of the soil.
Additional Retain the lines for the graphs used above. Then decrease the saturated conductivity of the
Work: soil by 0.1 cm/hr and repeat the simulation. Compare the sets of curves. Did the reduction
of saturated conductivity have much impact upon the time at which runoff began or on the
total amount entering the soil?
Repeat the exercise with different rainfall rates. How does rainfall rate influence the time at
which runoff begins and the total amount entering the soil in the six-hour period?
53
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Numerical Experiments for Water Movement (continued)
3. Water Content and Matric Potential Distributions during Infiltration
Objective: 1. To observe water content and matric potential distributions during infiltration at
selected times.
2. To observe changes in water content and matric potential with time at selected distances
from the inlet.
3. To compare the distributions for infiltration due to ponding with those due to
infiltration of rainfall.
Situation: The settings are similar to those described in the preceding exercises. However, in this case
there is an interest in the behavior of the water within the soil profile, not just the rate at
which it enters the soil.
Simulation: Simulate water movement into a vertical, semi-infinite silt loam soil with an initial matric
potential of -2000 cm.
Distributions at selected times:
1. Apply water to the soil at a constant potential of 2 cm, as done in Exercise 1.
A. Set the time of interest to 0 and press the calculate button. Display graphs of water
content vs. distance and matric potential vs. distance. These graphs represent the
water content and matric potential at time zero or before flow begins.
B. Increase the time of interest by one hour, retain the line, and calculate again.
Observe the change in water content and matric potential graphs. Note that the soil
water content and matric potential change only to a certain depth. Below that
depth, the parameters retain their initial value. This indicates that water movement
has penetrated through only part of the soil. Record the depth of wetting for each
hour for 12 hours. Does the wetting depth increase uniformly with time or does it
appear to slow down? Why?
C. Observe the water content at the soil surface (distance = 0). What is its value? Does
the value change with time when the water is applied by ponding?
D. Compare the shape of the water content profiles with those of the matric potential
profiles. Remember that these curves are related to each other by means of the
water release or water characteristic curve.
2. Repeat the simulation for the same soil with the same initial condition but with water
applied as rainfall at an intensity of 0.5 cm/hr. Compare these results with those found
in Part 1. Take special note of changes in water content near the soil surface.
54
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3. Water Content and Matric Potential Distributions during Infiltration (Continued)
Simulation Distributions at specific locations:
Continued: 1. Using the same initial conditions as earlier in this experiment, simulate movement for
12 hours with position of interest set to zero. Select graphs of water content vs. time
and matric potential vs. time. Press the button to retain these lines. Then repeat the
simulation for a position of interest equal to 5 cm. Retain these lines as well. Finally,
simulate a third time for a position of interest equal to 10 cm. These sets of curves
illustrate the way the water content and matric potential change with time at each
location.
A. Observe the curves for the three depths. What similarities and differences are
found?
B. In general, the water content at a specific location remains at its initial value until
water reaches that depth. It then increases. At what time does the water content
begin to increase at 0, 5, and 10 cm? Does it take longer for the water to move from
5 cm to 10 cm than it took for water to move from 0 to 5 cm?
C. Compare the slope of the water content vs. time graphs for the three depths. Which
depth shows the most rapid change? Which changes least rapidly?
2. Repeat the simulation for the same soil with the same initial condition but with water
applied as rainfall at an intensity of 0.5 cm/hr. Compare the times at which water
content changes occur and the slopes of the curves for water applied by ponding with
those values for water applied as rainfall. What differences are observed? Why?
Additional Determine the time at which runoff began for the simulation with water applied as rainfall.
Work: What was the value of the water content at a depth of zero at that time? What was the value
of the matric potential at that time? Formulate an hypothesis about the relationship between
these parameters and test that hypothesis.
55
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Numerical Experiments for Water Movement (continued)
4. Comparison of Horizontal and Vertical Water Movement in Unsaturated Soils
Objective: 1. To compare the water movement in horizontal and vertical soil systems.
2. To assess the significance of the force of gravity in unsaturated water movement.
Situation: Water is applied to a silt loam soil by means of furrow irrigation. The field was irrigated
for ten hours. The farmer noticed that the soil surface between the furrows became wet
rather quickly. The farmer wanted to compare the distance the water moved horizontally
with the distance it moved vertically downward.
Simulation: Simulate water movement for one hour into a vertical semi-infinite soil with an initial
matric potential of -2000 cm. Apply water to the soil at a constant potential of 1 cm.
Observe the water content as a function of distance from the inlet and the cumulative
flux of water passing the inlet. Retain these lines. Change the orientation of the column
so the soil is oriented horizontally. Calculate the flow for this system.
1. How far did the water move in each direction? (Choose a water content in the middle
of the range and determine the depth to which that water content penetrated.)
2. Compare the flux of water at the soil surface (the infiltration rate) for the two cases.
3. Compare the cumulative infiltration in both cases.
4. Compare the driving forces for the two systems.
5. Repeat the simulation for 2, 4, 6, 8, and 10 hours and make the observations
suggested above. What trends are observed as time increases?
6. Why does water move horizontally nearly as fast as vertically?
Additional Repeat the above exercises for a sandy clay loam and a sandy loam soil. What
Work: similarities and differences are observed?
Repeat the experiments for soils that are initially wetter. For example, choose an initial
matric potential of -200 cm and repeat the experiments. Are the results consistent with
those above?
These simulations compare vertically downward and horizontal water movement.
Compare the vertically downward, horizontal, and vertically upward flow. What can be
seen? Do the simulations support expectations?
56
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Numerical Experiments for Water Movement (continued)
5. Redistribution of Soil Water Following Infiltration
Objective: To observe movement of water within a soil profile after infiltration has taken place but
in the absence of evaporation.
Situation: A scientist flooded a plot of loam soil by maintaining 5 cm of water on the surface for
six hours. At that time, borders were opened and the water quickly flowed off the soil
surface. Then the surface was covered with plastic and insulated to prevent evaporation.
If the soil had an initial matric potential of -1000 cm at all depths, what will be the
distribution of water at the time infiltration stopped? How will this distribution of water
change during the next five days while the plot is covered?
Simulation: Simulate water infiltration into a vertical, semi-infinite loam soil with an initial matric
potential of -1000 cm. Water was applied by flooding the soil surface to a depth of 5 cm
for six hours. At that time, the surface was covered so no infiltration or evaporation took
place. This is simulated by specifying a constant flux boundary condition of zero at the
soil surface beginning at six hours. This zero flux boundary condition should continue
until the simulated time reaches 120 hours. Answer the following questions using graphs
of water content profiles, matric potential profiles, and water content vs. time.
1. Describe the water content profiles during the redistribution process. In what ways
do they differ from profiles observed during infiltration?
2. What was the depth of wetting when infiltration stopped? What was the depth of
wetting at 12, 24, 48, 72, 96, and 120 hours?
3. What changes in water content and matric potential occurred in the soil near the
surface?
4. Does movement continue beyond two days? When will movement stop?
5. Why does the rate of water movement decrease as redistribution time increases?
Additional Conduct additional experiments to determine if the observations made above can be
Work: generalized for other soils, wetting times, and initial conditions.
57
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Numerical Experiments for Water Movement (continued)
6. Redistribution and Evaporation of Soil Water Following Infiltration
Objective: To observe the simultaneous downward redistribution and upward evaporation of water
within a soil profile after infiltration.
Situation: A field of a loam soil was flooded by maintaining 5 cm of water on the surface for six
hours. At that time, the water was removed from the soil surface. If the soil had an initial
matric potential of -1000 cm at all depths, what was the distribution of water at the time
infiltration stopped? How will this distribution of water change during the next five days.
Simulation: Simulate water infiltration into a vertical, semi-infinite loam soil with an initial matric
potential of -1000 cm. Water is applied by flooding the soil surface to a depth of 5 cm for
six hours. At that time, evaporation and redistribution will take place. Suppose the
atmospheric conditions are capable of evaporating 0.05 cm of water per hour as long as
water is available at the surface for evaporation. After that time, the soil limits the
evaporation rate. This scenario is simulated by specifying a mixed type boundary
condition at the surface beginning six hours after flooding. Set the flux to -0.05 cm/hr
and the critical matric potential at the soil surface to -5000 cm. This mixed type
boundary condition should continue until 120 hrs have been simulated. Answer the
following questions using graphs of water content profiles, matric potential profiles, and
water content vs. time.
1. Describe the water content profiles during the evaporation and redistribution
process. In what ways do they differ from profiles observed during infiltration? How
does the evaporation process change the curves?
2. What was the depth of wetting when infiltration stopped? What was the depth of
wetting at 12, 24, 48, 72, 96, and 120 hours?
3. What changes in water content and matric potential occurred in the soil near the
surface?
4. What is the evaporation rate from the soil surface? Does it change with time?
Explain. (The evaporation rate is the flux of water at a depth of zero. This number
will be negative since it is upward.)
5. What is the total amount of water lost to evaporation during the 120 hours?
6. Why does the rate of water movement decrease as time increases?
Additional Design and carry out additional experiments to determine the influence of the initial
Work: evaporation rate upon the duration of the "constant rate" phase of the evaporation process
and upon the total amount of water lost to evaporation during the 120-hour period.
Conduct additional experiments to determine if the observations made above can be
generalized for other soils, wetting times, and initial conditions.
58
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Numerical Experiments for Water Movement (continued)
7. Influence of Rainfall Rate upon Infiltration and Depth Wetting
Objective: To determine the influence of rainfall rate upon the infiltration rate, time to runoff, and
depth of wetting
Situation: A farmer has the option of applying water by sprinkler irrigation at different rates. He
would like to know if the rate of application affects the way the soil wets and the amount
of water entering the soil. He wants to try irrigating the sandy loam soil at 0.25, 0.5, 1.0,
and 2.0 cm/hr. In each case he will apply 4 cm of water. The soil before irrigation has a
matric potential of -1000 cm.
Simulation: Simulate water infiltration into the vertical sandy loam soil for a rainfall rate of 0.25
cm/hr for 16 hrs. Observe the infiltration rate, depth of wetting, and total infiltration.
Calculate the amount of runoff, if any. Repeat the simulation for a rainfall rate of 0.5
cm/hr applied for eight hrs, 1.0 cm/hr applied for four hrs, and 2 cm/hr applied for two
hrs. Answer the following questions:
1. What was the total amount of water entering the soil at each rate?
2. What was the amount of runoff?
3. Did the application rate affect the form of the wetting process and the depth of
wetting in the soil? If so, how?
4. Should the farmer be concerned about the rate of irrigation? Explain.
Additional Allow water to redistribute from the end of the irrigation until the time equals 16 hrs.
Work: (This is the number of hours required to apply 4 cm of water at the 0.25 cm/hr rate so
this allows us to compare water content profiles at the same time after irrigation began.)
Now compare the shapes of the water content profiles and the depth of wetting.
Repeat the experiment for other soils.
59
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Numerical Experiments for Water Movement (continued)
8. Influence of Initial Water Content upon Water Movement
Objective: To observe the influence of initial soil wetness upon infiltration, runoff, and depth of
wetting for a fixed rainfall rate.
Situation: A person wanting to assess the runoff potential of a certain field, applied water to it by
sprinkling at an intensity of 2.5 cm/hr for six hours. Runoff began at 5.3 hours. He
concluded that runoff would not occur unless storms of 2.5 cm/hr intensity exceeded five
hour in duration. Another person stated that the time to runoff would depend upon the
initial wetness of the soil. Simulate infiltration for different initial water contents to
inform them of the importance of initial water content upon wetting. Make comparisons
of the total amount infiltrating the soil in six hours and the depth of wetting.
Simulation: Simulate water movement into four vertical, semi-infinite columns of the default soil
with initial matric potentials of -5000 cm, -1000 cm, -500 cm, and -100 cm. Apply water
at a rate of 2.5 cm/hr for six hours. Use graphs of water content profiles, infiltration
rates, and cumulative infiltration to answer the questions.
1. Compare the infiltration rates each hour for the different soil systems. How does the
infiltration rate depend upon the initial soil wetness? Explain why this change
occurs.
2. Compare the cumulative infiltration amounts during the entire application.
3. Does the time to the beginning of runoff depend upon the initial soil wetness? How
much does it change for this soil?
4. Compare the final infiltration rates for the different initial conditions. How do they
compare with the saturated hydraulic conductivity of the soil?
5. What is the depth of wetting for each case? Do the water content profiles change?
6. Does the time to runoff depend upon the initial wetness? Does this answer depend
upon the soil? Does it depend upon the rainfall rate? Explain.
Additional Reduce the saturated hydraulic conductivity of the soil used by 25% and repeat the
Work: experiment. How much does that change the answers obtained?
60
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Numerical Experiments for Water Movement (continued)
9. Steady-State Water Movement in Finite Unsaturated Soils
Objective: 1. To determine the rate of water movement through an unsaturated soil when the
system is at steady-state.
2. To determine the distribution of matric potential, water content, driving force,
conductivity, and flux throughout the soil system.
Note: A soil is said to be at steady-state when water properties of the soil do not change
with time. That is, at steady-state, the matric potential, water content, driving force,
conductivity, and flux at a particular location do not change with time.
Situation: A bare soil has a water table at 50 cm. It has not rained for a long time, but water is
evaporating from the surface. The surface has become nearly air-dry with a matric
potential of -5,000 cm. What is the steady-state rate of water movement through the
soil?
Simulation: Simulate water movement in a 50-cm vertical soil (the default soil can be used). The
boundary condition at the top should be a constant matric potential of-5,000 cm. The
boundary condition at the water table is a constant potential of zero. Use the option for
non-uniform initial conditions to enter a series of matric potentials that will be close to
the distribution at steady-state. Simulate movement for 50 hrs, 100 hrs, or until the
parameters no longer change with time. Observe graphs of water content and matric
potential vs. distance as well as other parameters of interest. Print tables of the calculated
quantities, also, and then answer the following questions:
1. What is the flux of water at the top of the soil? What is the flux at the bottom? How
does it vary through the soil?
2. Describe the final matric potential and water content distributions in the soil. Are
they what was expected?
3. Describe the hydraulic conductivity and driving force distributions. Why does the
hydraulic conductivity increase with depth?
4. Why does the matric potential change slowly near the water table and then more
rapidly near the top?
Additional It is possible to have an unsaturated soil at steady-state with uniform water content
Work: throughout and a non-zero flux. What will be the orientation and boundary conditions of
such a soil?
61
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Numerical Experiments for Water Movement (continued)
10. Sensitivity of Depth of Wetting and Cumulative Infiltration to Changes in Hydraulic
Conductivity
Objective: To observe the impact of hydraulic conductivity upon water content profiles and
cumulative infiltration under low intensity rainfall or irrigation
Situation: A scientist is interested in predicting the depth of wetting and the total amount of water
entering a soil when water is applied by a sprinkler irrigation system capable of applying
water at 1.0 cm/hr. The scientist has measured the saturated hydraulic conductivity of the
soil and found values of 0.8, 1.5, 2.0 cm/hr. He is concerned about the impact of these
differences upon the predictions.
Simulation: Simulate water movement into a vertical, semi-infinite sandy clay loam having an initial
matric potential of-1000 cm. The boundary condition at the top can be approximated by
a rainfall boundary condition with intensity of 1.0 cm/hr. Modify the sandy clay loam
soil in CHEMFLO so that the saturated hydraulic conductivity is 0.8 cm/hr. Simulate
water movement for five hours with a position of interest equal to the inlet or zero. Select
graphs for water content vs. distance and cumulative flux vs. time. Retain these lines.
Change the soil saturated conductivity to 1.5 cm/hr. Repeat the simulation and retain the
lines. Repeat these steps once more for the saturated conductivity of 2.0 cm/hr. The
screen will now contain three sets of lines, one for each saturated conductivity.
1. How deep did the water penetrate into the soil in each case?
2. What was the water content near the inlet in each case?
3. How much water entered the soil in each case?
4. How significant were these differences in conductivity for these flow conditions?
5. How large would the differences be if the sprinkler intensity were 2.0 cm/hr? Would
the answer to question 4 change in this case?
6. Repeat the simulations and compare results when water is applied by ponding water
on the soil surface to a depth of 2 cm.
7. What conclusions can be drawn from these results?
Additional Simulate infiltration for these three conductivity values when water is applied by
Work: ponding water to a depth of 2 cm on the soil surface. Determine the time required for 5
cm of water to enter each soil. Compare the water content distributions for the three soils
after the same amount of water has entered each soil.
62
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Numerical Experiments for Water and Chemical Movement
1. Chemical Movement During Infiltration Due to Rainfall
Objective: To observe the movement of a chemical applied with irrigation water.
Situation: A farmer decided to apply nitrate nitrogen to his field with his irrigation water. He
irrigated for six hours at 1 cm/hr with a solution of 20 g m "3 nitrate. The soil contained
no nitrate nitrogen before irrigation. What will be the distribution of the chemical
immediately after irrigation?
Simulation: Simulate movement into a loam soil with an initial matric potential of -500 cm and a
length of 100 cm. The lower boundary condition for water is a constant potential of -500
cm. The upper boundary condition is a constant rainfall rate of 1 cm/hr. The upper
boundary condition for chemical is that the inflowing solution has a concentration of 20
g m "3. The chemical leaves the bottom by mass flow only. The loam has a bulk density
of 1.55 Mg m"3. The partition coefficient, degradation rate constants, and zero-order rate
constants are all zero. Simulate movement for six hours. Then answer the following
questions:
1. Compare the shape of the water content vs. distance graph with the concentration vs.
depth. What similarities are observed? What are the differences? Does one seem to
be ahead of the other?
2. Compare graphs of the flux of chemical at selected depths with the flux of water at
those depths.
Additional Imagine that an unexpected storm came up after the irrigation. During the 12-hour storm,
Work: rain fell at a rate of .5 cm/hr. The concentration of the chemical in the rainfall was zero.
1. What was the distribution of chemical after the rainstorm? Compare the water
content and concentration distributions.
2. How far has the chemical moved into the soil?
63
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Numerical Experiments for Water and Chemical Movement (continued)
2. Influence of Initial Soil Wetness upon Depth of Chemical Movement
Objective: To determine the influence of the initial water content of a soil upon the depth of
movement of surface applied chemicals.
Situation: An alert environmental consultant recognizes the importance of initial soil water content
upon the depth of wetting. He is now concerned about the depth of penetration of a
chemical applied with the infiltrating water.
Simulation: A column of loam soil, 50 cm long, is oriented vertically. Water is applied at the top at a
constant potential of 1 cm. Make a series of simulations for initial matric potential values
of -1000, -500, -100, and -50 cm. In each case, the potential at the lower boundary
should be the same as the initial matric potential. The concentration of the inflowing
solution should be 10 g m"3. The chemical leaves the lower surface by mass flow only.
The partition coefficient and rate constants are zero. Simulate movement until the water
content at 40 cm begins to change. Discuss the following using graphs of water content
vs. distance and concentration vs. distance.
1. Construct a table showing the depth that has a concentration of 5g m"3 at times when
the wetting front is at 10, 20, and 30 cm.
Initial Matric Potential (cm)
-1000 -500 -100 -50
Initial Water Content
Depth of Chemical for:
• Wet front of 10 cm
• Wet front of 20 cm
• Wet front of 30 cm
2. Discuss the influence of the initial water content of the soil upon the depth of
penetration of the chemical? What explanations can be given for this behavior?
Additional Repeat the experiment for other soils. Are the depths of penetration similar to those for
Work: the loam? Can the same conclusions be made that were made for the loam soil?
64
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Numerical Experiments for Water and Chemical Movement (continued)
3. Influence of Adsorption on Chemical Movement
Objective: To determine the impact of adsorption of chemicals upon the chemical movement during
infiltration.
Situation: The farmer described in Exercise 1 is considering applying other chemicals to his soil.
These soils are adsorbed on the soil solids. How will this affect the movement of the
chemicals?
Simulation: Define the flow problem as in the case of Exercise 1 with one exception. In this case,
specify a partition coefficient of 0.5 m3 Mg"1 for one experiment and 5 m3 Mg"1 for a
second experiment.
1. Compare the shape of the water content vs. distance graph with the concentration vs.
depth. What similarities are observed? What are the differences? Does one seem to
be ahead of the other?
2. Compare the graphs of concentration vs. time at the soil surface for the different
partition coefficients with those for a partition coefficient of zero. Explain the
differences.
65
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Numerical Experiments for Water and Chemical Movement (continued)
4. Chemical Movement without Water Movement
Objective:
Situation:
Simulation:
Additional
Work:
To observe the movement of a chemical due to diffusion only.
A loam soil has a uniform matric potential of 0 cm. The soil is oriented horizontally so
no water movement occurs. A concentration of 100 g nr
one end. How will the chemical move in this soil?
of a chemical is maintained at
Simulate water and chemical movement into a horizontal loam soil with an initial matric
potential of 0 cm and a length of 50 cm. The initial concentration of chemical is zero
throughout. The concentration in the soil solution at one end is zero and at the other end
is 100 g m"3. The water content at each end of the soil is maintained at a matric potential
of 0 cm. The partition coefficient and rate constants are zero. The diffusion coefficient is
5 cm2 / hr. Simulate movement for 16 hours.
1. Describe the concentration profile after 1, 2, 4, 8, and 16 hrs. How are the
concentration profiles similar to the profiles in exercise 1? How do they differ?
2. What similarities and differences are seen between the concentration and water
content profiles found in the water movement exercises?
Repeat the above exercise with diffusion coefficients of 1 and 10 cm2 / hr. What impact
does this have on the shape and position of the curves?
Repeat the above exercise for initial matric potentials of-100 cm and -1000 cm. In each
case the matric potential at each end of the column should be equal to the initial matric
potential used in the remainder of the column so that no water movement occurs. Does
the initial matric potential (or initial water content) affect the movement of the chemical?
Why?
66
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Numerical Experiments for Water and Chemical Movement (continued)
5. Degradation and Production of Chemicals
Objective: To observe the effect of first order and zero order terms upon the concentration of
chemical in the soil.
Simulation: Define the soil system to be 20 cm long and oriented horizontally. The initial matric
potential should be -10 cm. A constant potential of -10 cm is imposed at each end of the
soil. The initial concentration of chemical in the system is 100 g/m3. The inflowing
solution has a concentration of 0. Outflow is by convective flow only. The partition
coefficient is zero.
1. Will water move in this soil system? If so, in what direction? Explain.
2. Simulate movement as needed to complete the table below.
3. The degradation process is considered a "first-order" process. What differences are
observed in the manner in which the concentration changes for this process when
compared with a zero-order process?
4. The last column in the table represents a system in which the chemical is degrading
by a first-order process and is being produced by a zero-order process. What impact
does the production have on the shape of the curves? What will be the concentration
at 100 hrs? 1000 hrs? Explain.
Data Table:
Parameters:
Degradation Rate, liquid (1/hr)
Degradation Rate, solid (1/hr)
Zero-order Rate Const(g/m3/hr)
Observations:
Concentration at 1 hr
Concentration at 2 hrs
Concentration at 3 hrs
Concentration at 4 hrs
Concentration at 5 hrs
Concentration at 10 hrs
Concentration at 20 hrs
Experiment
#1
0.1386
0.1386
0.0
#2
0.0693
0.0693
0.0
#3
0.0
0.0
-4.0
#4
0.0693
0.0693
5.0
67
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Related Software
A collection of software has been developed for use in understanding water movement and the fate of
chemicals in soils. These are available as applets usable from most browsers, as well as more fully
functional applications for use in a stand-alone computer system. They can be found at
http://soilphysics.okstate.edu. An overview of available tools is provided below.
Steady-State Water Movement in Homogeneous and Layered Soils: This software solves the Darcy
(or Buckingham-Darcy) equation for steady-state flow in one dimension. It is useful in understanding
basic flow principles and forms a good foundation for understanding transient flow as calculated in
CHEMFLO.
Water Balance Calculations: This software deals with long-term (on the order of weeks and months)
flow in soils using daily time steps. It is best suited to estimating the amount of water stored in the soil,
the amount of water passing through the root zone, and the amount of water lost by plant uptake and
evaporation. Weather data are provided for several years so the user can gain an appreciation for the range
of values these parameters can take on from year to year. Water balance methods are often used for
making management decisions.
Soil Temperature Changes with Time and Depth: Many soil processes are influenced by soil
temperature. This software provides information on the seasonal changes in soil temperature with time of
year and with depth in the soil.
Degradation of Chemicals: Two programs are available to illustrate degradation of chemicals in soils.
One simply calculates first-order degradation based on user-specified degradation rates or half-life values.
The second incorporates corrections for temperature and changes in temperature over time upon
degradation rate.
Diffusion of Chemicals in Soils: Diffusion of chemicals from regions of higher to lower concentrations
is one means by which chemicals move in soils. This software demonstrates the way chemical
concentrations change in soils due to diffusion only. It provides an appreciation for contribution of
diffusion upon chemical transport.
Chemical Movement Under Conditions of Steady-State Water Flow: This software solves the
convection-dispersion equation for conditions when water movement does not change with time. This is
the same equation for chemical movement as used here in CHEMFLO. However, in this case, the
transient nature of water movement is ignored. These solutions would be most applicable to laboratory
experiments where flow conditions are controlled. The software helps users get acquainted with soil and
chemical parameters influencing transport and their importance in determining the fate of chemicals.
Aquifer Mixing: This program uses a simple mixing model to estimate the rate at which the
concentration of a chemical in the aquifer will change with time as a result of user-specified recharge
rates and concentrations in water entering the aquifer.
CMIS Chemical Movement in Soils Educational Model: This simplified chemical transport model
enables users to define two pesticide-soil-management systems and visually compare the chemical
movement in the two systems. Dr. Art Hornsby at the University of Florida initially designed the software
for use in public schools.
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References
Brooks, R.H., and A.T. Corey. 1964. Hydraulic Properties of Porous Media. Hydrology Paper
No. 3. Colorado State Univ., Fort Collins, CO. 27 pp.
Buckingham, E. 1907. Studies on the Movement of Soil Moisture. USDA Bur. Soils. Bull. 38.
U.S. Gov. Print. Office, Washington, DC.
Celia, M.A., E.T. Bouloutas, and R.L. Zarba. 1990. A general mass-conservative numerical
solution for the unsaturated flow equation. Water Re sour. Res. 26:1483-1496.
Gardner, W.R. 1958. Some steady state solutions of the unsaturated moisture flow equation with
application to evaporation from a water table. Soil Sci. 85:228-232.
Millington, R.J., and J.M. Quirk. 1961. Permeability of porous solids. Trans. Faraday Soc.
57:1200-1207.
Nofziger, D.L., K. Rajender, Sivaram K. Nayudu, and Pei-Yao Su. 1989a. CHEMFLO: One-
Dimensional Water and Chemical Movement in Soil. Computer Software Series CSS-38.
Oklahoma Agricultural Exp. Sta., Oklahoma State Univ., Stillwater, OK. 109 pp.
Nofziger, D.L., K. Rajender, Sivaram K. Nayudu, and Pei-Yao Su. 1989b. CHEMFLO: One-
Dimensional Water and Chemical Movement in Unsaturated Soil. EPA/600/8-89/076,
August 1989.
Richards, L.A. 1931. Capillary conduction of liquids through porous mediums. Physics 1:318-
O •"> O
333.
Simmons, C.S., D.R. Nielsen, and J.W. Biggar. 1979. Scaling of field-measured soil water
properties. Hilgardia 47:77-173.
van Genuchten, M. Th. 1978. Mass Transport in Saturated-Unsaturated Media: One-Dimensional
Solutions. Water Resources Program, Dept. of Civil Engineering, Princeton Univ.,
Princeton, NJ. 118 pp.
van Genuchten, M. Th. 1980. A closed-form equation for predicting the hydraulic conductivity
of unsaturated soils. Soil Sci. Soc. Am. J. 44:892-898.
69
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&EPA
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Environmental Protection
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Research Laboratory
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Official Business
Penalty for Private Use
$300
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March 2003
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