Download Next adventure: The Flow of Water in the Vadose Zone

Chapter 3:
Flow of Water in the
Vadose Zone
The classic solutions for
infiltration and evaporation by
Green and Ampt, Bruce and
Klute, and Gardner
Williams, 2002
Modified after Selker, 2000
http://www.its.uidaho.edu/AgE558
http://bioe.orst.edu/vzp/
1
The Green & Ampt model for Infiltration
Conceived of in the
framework of a capillary
tube bundle concept:
Media is modeled as a
bundle of capillary tubes
oriented in the direction of
infiltration.
Tubes fill in parallel to the
same depth (i.e., the
wetting front is at a welldefined position, the
same in all tubes).
d
L
2
Goals
May be employed in calculating the infiltration into soils
which proceed at any angle relative to vertical by simply
adjusting the magnitude of gravity.
We will carry out some 1-dimensional calculations
Calculate the horizontal rate of infiltration into a dry
soil.
Solve for ponded vertical infiltration.
Solve for falling head vertical infiltration.
Note: This conceptual approach can be
applied to multidimensional problems.
3
The basic idea:
Infiltration is driven by
two forces:
Force of gravity acting on
the water.
d
L
Pressure drop produced
by a sharp wetting front
Assumptions:
Wetting front is completely “sharp”
Soil is saturated behind wetting front
4
The wetting front potential term
Where does this term come from and
what is its magnitude?
Let’s take the case of soil made up
capillary tubes of N sizes,
Each of the sizes makes up a fraction ai of
the total cross-sectional pore area. The
porosity n is
N
a
i 1
i
= n
5
Continuing on the wetting front potential
Part of the potential gradient is generated by the
capillary contacts.
Pressure due to capillary contacts will result in a
force per unit area.
The force is proportional to the length of solid/liquid
contact per unit area.
For a unit of cross-sectional area, total length,
la, of water/solid contact is

N
la =
area of tu be i w all len gth of i
u n it area
area of tu be i
[3.1]
i =1

N
la =
i =1
ai
2  ri
 r i2

N
=
i =1
2
ai
ri
[3.3]
6

N
la =
ai
2  ri
 r i2

N
=
ai
2
ri
[3.3]
i =1
i =1
•Assume a contact angle of f
•The pressure per unit area, Pf, at the water/solid
interface is
N
Pf = l a cos =  cos

2
ai
ri
[3.5]
i =1
•Resistance to flow? Poiseuille capillary flow! For the
ith tube, we have the flow per unit area, qi, of
qi
ri 2 DP
=
8L
[3.6]
Where P is the pressure drop across a capillary tube
of length L with a fluid flowing with viscosity .
7
Computing flux
qi
ri 2 DP
=
8 L
[3
For our capillary bundle the total flux would be

N
q=
ai
r i 2 P
P N
=
  a i r i 2
8L
8L
[3.7]
i =1
i =1
If the flow is horizontal, then the driving force is simply
the surface tension, so the pressure drop in the tubes is
Pf and the flux into the system is given by
q =
Pf
8L
N
  a i r i 2
[3.8]
i =1
For “convenience,” let’s define the system constant

N
K =
i =1
g a i r i 2
8
 [3.9]
8

N
K =
g a i r i 2
8
 [3.9]
i =1
We recognize K as the saturated hydraulic conductivity.
In the customary units of pressure head ghf =Pf [3.8] is
Khf
q
L
[3.10]
Now hang on there! That looks like Darcy’s Law! If we
have a soil with wetting front potential hf and conductivity
K, then the flux would be:
h
f
q   KH   K
L
So we have just found Darcy and seen how the terms
relate to capillary properties.
9
….. Solving for infiltration
Khf
q
L
The soil is going from dry to saturation, we can
relate the flux to the rate of movement of the
sharp wetting front
d
q=
nL
dt
[3.11]
Combining [3.10] and [3.11] we obtain a
differential equation for the position of the
wetting front in time
dL Khf
n

dt
L
[3.12]
10
Solve for the position of the wetting front in time
dL Kh f
n
=
dt
L
[3.12]
2
integrate to find that
n L
= K hft
2
[3.13]
or, solving for the length of infiltration
L=
2K h ft
n
[3.14]
the wetting front advances with the square root of
time.
the total infiltration is also proportional to the square
11
root of time.
We may now compute flux
Recall
d
q=
nL
dt
L=
2K h ft
n
[3.11]
So we may put in the result for L and take the
derivative
d 2 Kh f t
q= n
q =
dt
n
Kh fn
-1/ 2
= 1/ 2 SGA t
2t
[3.15]
[3.16]
where SGA (referred to as the sorptivity) is a constant
dependent on the media, and equal to (2nKhf)1/2.
Key observation:
Horizontal flux decreases proportionally to
one over the square root of time.
12
What are effects of soil texture
Plugging in our previous results
2
SGA = 

 N a i 
n cos  

 ga r 2 

i i
i 1 gr i 
2 
[3.17]
To help “see” how SGA relates to soil characteristics,
make the vastly simplifying assumption that the soil has
only one pore radius. SGA becomes
2
SGA
 cos r
= 
 =Cr
2 
[3.18]
where C is a constant which depends only on the fluid.
This then tells us that for our simplified soil that the
infiltrating flux will be given by
q=
C r
t
[3.19]
13
Infiltrating flux varies as r
r
1
0.9
0.8
0.7
0.6
0.5
0.4
K
q
r squared
1
0.81
0.64
0.49
0.36
0.25
0.16
sqrt r
1
0.95
0.89
0.84
0.77
0.71
0.63
14
Let’s look at this...
q=
C r
t
[3.19]
Infiltrating flux goes down with the square root
of the radius.
Compared to conductivity which decreases
with r2.
Why the difference?
The larger surface area of the particles in the
finer soil provides a large capillary “pull” into
the soil, which is almost enough to overcome
the lower hydraulic conductivity.
15
Next stop: Vertical infiltration
Essentially as easy as solving for horizontal infiltration.
Ponded infiltration: Darcy's law for the system is
simplified by the fact that the wetting front is assumed to
be sharp, with saturation behind the front
head loss 

q = -Ksat
[3.20]
length of travel
The head loss across the system is the sum of the
potential at the wetting front, hf, the depth of ponding, d,
plus the depth of infiltration, L, while the length of travel
is L
d
h f+ d + L

n L = Ksat
[3.21]
dt
L


16
d
h f+ d + L

n L = Ksa t
dt
L


[3.21]
Vertical,
constant head
infiltration
Separating variables
n  L dL 
dt =
Ksa t h f + d + L
hf + d 
n 
=
1dL
Ksa t  h f + d + L
[3.22]
which may be integrated to obtain
n 
h f+d+L

t=
L - (h f+d)ln
 [3.23]
Ksat 
 h f+d 
Square root of time behavior for horizontal
infiltration replaced with logarithmic relationship.
Note: taking Taylor expansion at short time, the
horizontal and vertical results become identical
17
Vertical falling head...
Same procedure can solve for infiltration
under falling head.
In this case, we replace the depth of
ponding, d, with this depth minus the depth
of water which has infiltrated, nL (where
nL<d, after which the pond is gone). Using
(3.20) as before, we obtain
n
h f+d+L(1-n ) 


t=

2 L(1-n ) - (h f+d)ln 
h
+d

Ksa t (1-n ) 
f
[3.28]
18
Typical Results
35
Depth of Infiltration (cm)
30
25
20
15
Falling Head
10
Constant Head
Horizontal
5
0
0
50
100
150
200
250
Time (seconds)
Green and Ampt model predictions for infiltration in a
soil with 30% pore space, Ks = 0.03 cm/sec, hf = 25 cm
and d = 20 cm.
19
Field Measurement of G&A Parameters
Bouwer Infiltrometer
Get Ks by falling head
Get hf by shutting valve
and measuring the
vacuum created by the
wetting front pressure
Measured Ks about 1/2
fully saturated Ks
Measured hf really the
air entry pressure
which is 2 times water
entry pressure
2r
Volumetricly
Graduated
Reservoir
Water Flood Valve
Vacuum
Gauge
Air Purge Valve
H
O-Ring Seal
G
SteelWool
2
R
L
Approx
10 cm
20