Soil Physics with HYDRUS:
Chapter 4

4.1 – 4.5
R n
 
)
S

R sky

R earth
R n
S

ET

J
H
ET
0

1



 
R n

G

   
1
 r / r c a


 c
  p a

 e

1

 e r d
/

/ r c a r
 a


 

T
4098
 e a
237.3

2
  c P p



1 o

1
1013J kg C
0.622
P


1629
P


4.6 – 4.9


J
H
   e dT dz
J
H z t


2 t
)
      t
J
H
( z
 
,


2 t
)
      t

 z
, t
  
(
2

 z
2
, )


      z s
H
    t

4.10 – 4.13
J
H z t


2 t
) x y t J
H
( z
 
,


2 t
)
      t





 z
2
, t
  
(

 z
2
, ) 


      z
 s
H
        t
J ( ,
H
 z

 t
)
2 
J
H
( z
 
,

 z
 t
2
) (


 z
, t
  
(

2
 t
 z
2
0=
J
H
( z
  z t

 t

2
 z
) J
H z t

 t
2
) (


 z
, t
  
(

2
 t
 z
2
 s
H
 s
H

4.14 – 4.18


J z
H
 lim
 
0


J
H
( z
 
, )

J
H
( , )
 z




H t
 lim
 
0


( , t ) H z t
 t




H t



J z
H
 s
H

0
H
 p
(

T ref
)


H t

C p

( T

T ref
 t
)

C p

T
 t

4.19 – 4.23
C p


T t


 z
 e

T
 z
C p

T
 t
  e


2
T z
2

T
 t

K
T
K
T

 e
C p


2
T z
2
T (0, )

T
0
T
 
0

4.24 – 4.28
( , 0)

0
( , ) T
0 erfc


2 z
K t
T


( )

T
A

A sin(
 
)
 
2


T (0, )

T
A

A sin(
 
)

4.29 – 4.33
( , )

T
A

Ae
 d z sin d

2 K
T


K
T



T
 z

T
 z z
 d

T i

T i
 z

1 i

1
2 i

1
2

T i

1

 z
T i z d

4.34 – 4.35


2
T z
2 i

 
T

 z
  z




T z
 i

1
2
 z


T z i

1
2


2
T z
2 i


T i

1

 z
T i
 
 z
T i

T i
 z

1


T i

1

2 T i

T i
 z
2

1

4.36 – 4.39

T
 t i j

1
2 
T i j

1 
T i
 t j
T i j

1 
T i
 t j

K
T
T i

1

2 T i

T i
 z
2

1
T i j

1 
T i j
 t

K
T
T i
 j
1

2 T i j 
T i
 z
2
 j
1
T i j

1 

 t
 z
2
 t
 z
2
K

T
T i
K T
T i
 j
1 j

1


2 T i j 
T i
 j
1


T i j
 t
 z
2
K T
T  i j 
 t
 z
2
K T
T i
 j
1

4.40 – 4.44
r

 t
 z
2
T i j

1  rK T
T i j

1
 rK

T
T i j  rK T
T i j

1
T i
1  rK T
T i
0

1
 rK

T
T i
0  rK T
T i
0

1
T
1
1  rK T
T 2
0
 rK
T

T
1
0  rK T
T 0
0
T i
2  rK T
T i
1

1
 rK

T
T i
1  rK T
T i
1

1

4.45 – 4.48
T i j

1 
0.001
T i
 j
1
 
T i j 
0.001
T i
 j
1
T i j

1 

1
1000
T i
 j
1

998
1000
T i j 
1
1000
T i
 j
1
0.563

T i j

1

1 1.126
 
T i j 
0.563

T i j

1

563
1000

T i j

1

126
1000

T i j 
563
1000

T i
 j
1
T i j

1 
T i
 t j

K
T
T i j


1
1 
2 T i j

1 
T i
 z
2
 j

1
1
 rK T
T i j


1
1
 rK T
T i
 j

1
1

2 rK T
T i j

1  rK T
T i j


1
1 
T i j

1 
 rK

T
T i j

1  rK T
T i j


1
1 
T i j
T i j

4.49 – 4.51
aT i j

1

1
 bT i j

1  cT i j

1

1

T i j a
   rK
T b
  rK
T




1 a
0 b
0 c
0
0




T
1 j

1

1
T
T
0
1 j
   

T
0 j j

1
  
  
0
0 a
0 b
0 c
1
 
T
T
2
3 j

1
   j

4.52 – 4.55
T
0 j

1 
T
0 j aT
0 j

1  bT
1 j

1  cT
2 j

1 
T
1 j aT
1 j

1  bT
2 j

1  cT
3 j

1 
T
2 j
T
3 j

1 
T
3 j
     j

1    j
  
1 
A

T j

1 
A

1
[ ] { } j
T j

1 
A

1
[ ] { } j

C p
4.56 – 4.60



T t


 z


T
 z

C J w w


T z
C p
 
C s
(1
 
)

C
 o o

C w
 
C a a
     e
 
C J t w w
 
1 b
2
  b
3
 e
T z t

T t
0
( ) at z

0 or z

L

4.61 – 4.62
 

T
 z

TC J w w

T t C J
0
( ) w w z
at z

0 or z

L
T
0
 
A sin


2
 p t t

7

12



f(z,t)
(t)
1 t N exp( -at ) sin( at )/ a cos( at )
Az /(2 t )
A
B
2 tA-zB
( C + D )/2
( C D )/(2 a )
A-aD
D
L
1
1/ s
N !/ s N +1
1/( s + a )
1/( s 2 + a 2 ) s /( s 2 + a 2 ) exp(z s 1/2 )
[exp(zs 1/2 )]/ s 1/2
[exp(zs 1/2 )]/ s
[exp(zs 1/2 )]/( s 3/2 )
[exp(zs 1/2 )]/( s a 2 )
[exp(zs 1/2 )]/[ s 1/2 ( s a 2 )]
[exp(zs 1/2 )]/( s 1/2 + a )
[exp(zs 1/2 )]/[ s 1/2 ( s 1/2 + a )]
6
7
8
1*
2
3**
4
5
9
10
11
12
13
14

Figure 4.1 Typical partitioning of the extraterrestrial radiation as it passes through the atmosphere
(Jury and Horton, 2004)

Figure 4.2 Typical partitioning of the global solar radiation as it reaches the land surface: (a) contributions to the net radiation; (b) net radiation partitioned into its components (Jury and
Horton, 2004).

Figure 4.3 Components of the surface energy balance during daytime (left) and nighttime (right)
(Jury and Horton, 2004).

Figure 4.4 Effective soil thermal conductivity (a) and diffusivity (b) as a function of water content for various soil types. Numbers in parentheses refer to porosity (Jury and Horton, 2004).

Figure 4.5 Elementary volume of soil for calculating the heat conservation equation.

Figure 4.6 When the second derivative in T with respect to z is positive in the region of a point, the temperature increases at that point.

PDE
Laplace transform
ODE
Solve ODE
Solution to ODE
Inverse Laplace transform
Solution to PDE
Figure 4.7 Using the Laplace transform to solve a PDE.

-10
-15
-20
-25
-30
-35
-40
-45
0
-5 t = 1,000 s t = 10,000 s t = 30,000 s t = 100,000 s
-50
0 10 20 30
Temperature (C)
Figure 4.8 Temperature ( o
C) as a function of depth (cm) for times of 1, 000; 10, 000; 30, 000; and
100,000 seconds based on Equation 4.25.

Figure 4.9 Excel spreadsheet for Example 4.3. The formula bar shows the equation for temperatures
(Equation 4.25).

20
15
10
5
0
0 50000 100000
Time (s)
150000 200000
Figure 4.10 Temperature as a function of time at a depth of 20 cm using Equation 4.25 and a thermal diffusivity of 4.5∙10
-3
cm
2
s
-1
.

Figure 4.11 Long-term mean monthly temperature at a 10-cm depth measured at Davis, CA (Jury and Horton, 2004).

Figure 4.12 Temperature at depths of 0, 100, 200, and 600 cm as a function of time, calculated using
Equation 4.29 with K
T
= 388.8 cm
2
d
-1
, τ = 365 days, T
A
= 18 o
C, and A = 11 o
C.

Soil Surface
 z i = 0; z = 0 i = 1; z =
 z i = 2; z =
 z
Nodes i-1 i i+1 i = N; z =
 z
Figure 4.13 Dividing the soil profile into N discrete layers of the same thickness Δz results in N + 1 nodes.

Figure 4.14 The numerical solution is a matrix where the first column (column E) represents the initial temperatures at each node. The next column (column F) represents the temperatures at the next time level. The rows represent nodes with the surface node in the first row (row 5) (boundary condition of T = 30 o
C). The formula bar shows Equation 4.41.

Figure 4.15 Temperature as a function of depth after 5,000 seconds based on a numerical solution
(symbols) and an analytical solution (smooth curve) for the heat flow problem with a constant surface boundary condition of T = 30 o
C and a thermal diffusivity of 4.5 x 10
-3
cm
2
s
-1
. For the numerical solution, the time step is Δt = 1 s so that r∙K
T
= 0.001.

Figure 4.16 Temperature ( o
C) as a function of depth for t = 5,000 s based on an analytical solution
(curve) and a numerical solution (symbols) for the heat flow problem in Figure 4.15. For the numerical solution, the time step is Δt = 500 s so that r∙K
T
= 0.563.

Figure 4.17 Temperature as a function of depth after 5,000 seconds in a uniform wet soil (K
10
-3
cm
2
s
-1
) compared to a nonuniform soil with a wet layer in the top 5 cm (K
T
= 4.5 ∙ 10
-3 underlain by a dry layer (K
T
= 2.0 ∙ 10
-3
cm
2
s
-1
T
= 4.5 ∙
cm
2
s
-1
),
). An analytical solution was used for the uniform soil and a numerical solution was used for the nonuniform soil.

Figure 4.18 HYDRUS-1D overall display window.

Figure 4.19 Profile Information window showing the Material distribution for a uniform soil.

Figure 4.20 Profile Information window showing three observation points.

Figure 4.21 DOS window showing that the execution of the numerical solution is complete.

Figure 4.22 Overall display window showing results posted in the Post-processing panel.

Figure 4.23 Temperatures as a function of time at three observation nodes (N1 = 0, N2 = -10, and N3
= -50 cm).

Figure 4.24 Temperatures as a function of depth at six time intervals (T0 = 0, T1 = 8, T2 = 16, and T3
= 24 hr).

1.2
1.0
0.8
0.6
0.4
0.2
0.0
-4 -2 0 x
Figure 4.25 Plot of Error! Reference source not found.
as a function of x.
2 4

 
 
Figure 4.26 Error function and complementary error function of x = 0.5.

Figure 4.27 Excel spreadsheet for explicit finite difference numerical solution to heat flow equation.

Figure 4.28 Excel spreadsheet for solving heat flow equation using matrix algebra in problem 12. The coefficient matrix A is on the left in cells B10-L20.
The formula bar shows the equation for finding the inverse of the coefficient matrix A. The temperatures at the known time level T j
are in cells Z10-
Z20.

Figure 4.29 Excel spreadsheet for solving heat flow equation using matrix algebra in problem 12. The formula bar shows the equation for finding the product of the inverse of the coefficient matrix A
-1
and the vector of temperatures at the known time level T j
.