Transcript Steady Evaporation from a Water Table.

Steady
Evaporation from
a Water Table
Following Gardner
Soil Sci., 85:228-232, 1958
1
Why pick on this solution?
Of interest for several reasons:
• it is instructive in how to solve
simple unsaturated flow problems;
• it provides very handy, informative
results;
• introduced widely used
conductivity function.
2
The set-up
The problem we will consider is that of evaporation
from a broad land surface with a water table near by.
Assume:
• The soil is uniform,
• The process is oneEvaporative f lux q
K
=
saturated
Conductivity
dimensional (vertical).
positive upw ard
z positive upw ard
• The system is at
z = 0 at the
w ater table
steady state
s
Notice: that since the system is at steady state,
the flux must be constant with elevation, i.e. q(z) = q.
3
Getting down to business ..
Richards equation is the governing equation
   h   K
K    +
=
t z   z   z
At steady state the moisture content is
constant in time, thus d/ dt = 0, and Richards
equation becomes a differential equation in z
alone
dK
d 
=
K
dz
dz 
dh

 dz









4
Simplifying further ...
dK
d 
=
K
dz
dz 
 dh

 dz









Since both sides are first derivatives in z, this
may be integrated to recover the
Buckingham-Darcy Law for unsaturated flow
dh
- K =K
+q
dz
or
q =- K
 dh


 dz
+1




where the constant of integration q is the
vertical flux through the system. Notice that
q can be either positive or negative
corresponding to evaporation or infiltration.
5
Solving for pressure vs. elevation q = - K
 dh


 dz




+1
We would like to solve for the pressure as a
function of elevation. Solving for dz we find:
- dh
dz =
q
+1
K
which may be integrated to obtain
•h' is the dummy variable of integration;
• h(z), or h is the pressure at
the elevation z;
•lower bound of this integral is taken
at the water table where h(0) = 0.
h(z)
z
- dh'

dz = z = q
0
 +1
K
0
6
What next? Functional forms!
•To solve need a relationship between
conductivity and pressure.
h(z)
- dh'
z= 
q
 +1
K
0
•Gardner introduced several conductivity
functions which can be used to solve this
equation, including the exponential
relationship
K(h) = K s exp( h)
Simple, non-hysteretic, doesn’t deal with hae,
is only accurate over small pressure ranges 7
Now just plug and chug
h(z)
- dh'
z= 
q
 +1
K
K(h) = K s exp( h)
h
- dh'

z= 
q
+1

K
exp(
h')
 s
0
0
which may be re-arranged as
h
- K s exp( h') dh'
z= 
 q + K s exp( h')
0
To solve this we change variables and let
x = K s exp( h)
dx = Ks exp( h) dh
or dx =  x dh
8
dx =  x dh
Moving right along ...
Our integral becomes
h
z= -




x(h)
x dh'
q+x
1
=
0




dx
q+x
x(0)
which may be integrated to obtain
1
z= Ln ( q + x

1
z = - Ln

q



 x=x(h)

)
 x= x(0)
+ K s exp( h) 

q+K s 
9
All Right!
1
z = - Ln

q + K
s exp( h)


q+K s





• Solution for pressure vs elevation for steady
evaporation (or infiltration)from the water table
for a soil with exponential conductivity.
• Gardner (1958) notes that the problem may also
be solved in closed form for conductivity’s of the
form: K = a/(hn +b) for n = 1, 3/2, 2, 3, and 4
10
Rearranging makes it more intuitive
We can put this into a more easily understood
form through some simple manipulations.
Note that we may write: h = (1/)Ln[exp( h)],
so adding and subtracting h
1
1 q + K s exp( h) 
z= - h +
Ln[exp(ah)] Ln 

q
+
K

 

s
gives us a useful form
z= - h -
1 q exp(- h) + K
Ln 
q+K s
 

s


11
We now see ...
z= - h -
1 q exp(- h) + K
Ln 
q+K s
 

s


• Contributions of pressure and flux separately.
• As the flux increases, the argument of Ln[] gets
larger, indicating that at a given elevation, the
pressure potential becomes more negative (i.e., the
soil gets drier), as expected for increasing
evaporative flux.
• If q=0, the second term on the right hand side goes to
zero, and the pressure is simply the elevation above
the water table (i.e., hydrostatic, as expected).
12
Another useful form
May also solve for pressure profile
h=

1

Ln exp(- z)


q

K s

+1




-
q 
K s 
• Although primarily interested in upward flux,
note that if the flux is -Ks that the pressure is
zero everywhere, which is as we would
expect for steady infiltration at Ks.
13
Evaporation Rate Dependence on Upper Boundary
Wat er T able at 2 meters dept h. Exponential conduct ivit y
function with alpha = 0.01 (aft er Gardner, 1958)
0.2
Relative Flux (q/Ksat)
0.175
0.15
0.125
0.1
0.075
0.05
0.025
0
0
250
500
750
Mat ric Pot ent ial at Surface (-cm)
1000
The Maximum Evaporative Flux
h=

1

Ln exp(- z)


q

K s

+1




-
q 
K s 
At the maximum flux, the pressure at the soil
surface is -infinity, so the argument of the
logarithm must go to zero. This implies
q max
Ks
= exp(- z)
q max

 Ks

+1




solving for qmax, for a water table at depth z
q max
Ks
=
exp( z)-1
15
So what does this tell us?
q max
Ks
=
exp( z)-1
Considering successive depths of
z = 1/, 2/ , 3/
we find that
qmax(z)/Ks = 0.58, 0.16, and 0.05,
 very rapid decrease in evaporative flux as
the depth to the water table increases.
16
Maxim um Evaporation From a Water Table
Following Gardner (1958), using exponential conductiv ity . Alpha is taken as
0.01, 0.03 and 0.1 cm^-1 and Ks as 1, 10 and 50 cm/day .
1000
100
10
Flux (cm/day)
1
0.1
0.01
Clay Soil
0.001
Sandy Soil
0.0001
Silty Soil
0.00001
0.000001
1
10
100
Depth to Water Table (cm)
1000
10000