Transcript Darcy meets theta - Oregon State Univ

Darcy meets
theta
The plummeting fortunes
of Permeability and other
distress caused by
inadequate fluids
1
Extending Darcy to Unsaturated Media
1907 Buckingham saw Darcy could describe
unsaturated flow
q = - K(q) H
[2.102]
K(q): a function of the moisture content.
Conductivity is not a function of pressure: the
geometry of the water filled pores is all that
matters, which is dictated by q alone. To
express K as a function of pressure you must
employ the hysteretic functional relationship
between q and H.
2
How does K vary with q?
Drops like a rock.
Three factors are responsible for this behavior:
1. Large pores empty first. These are the pores with least
resistance to flow, since they have the largest diameters
(recall the 1/r4 dependence of hydraulic resistance in the in
the Hagen-Poiseuille equation).
2. Flow paths increase in length. Instead of proceeding
straight through a chunk of media, the flow must avoid all the
empty pores, making the path more tortuous.
3. There is less cross-section of flow. In any given area
normal to flow, all the fluid must pass through a smaller
portion of this area; for a given aerial flux the pore velocity
must be higher.
3
Permeability with theta...
To get a feel for how fast K drops, let’s consider the
effect of tortuosity alone (item 2 above).
Must be distinguished from the “Darcian” flow length
and the “Darcian” velocity.
Difference between the
true microscopic fluid
flow path length and flow
velocity in comparison
with the “Darcian” values,
which are based on a
macroscopic picture of
the system.
Actual flow lengt h, Le
q
"Da rc ian" ve lc oity
q = flow/a rea
"Da rc ian" flow le ngth, L
4
A few illustrative calculations…
So writing the true pressure gradient that acts on the fluid
we have
P L  P
L
=
 e
L 


[2.104]
L e
Le/L ratio of the true path length to the Darcian length.
Next write equation for capillary velocity, vf (HagenPoiseuille)

P
vf = r2
[2.105]
8 
L e
Translating into the Darcian velocity, q, we find
q Le
vf =
=nL

8 
r
2
P L
L L e
[2.106]
5
Conclusions on tortuosity...
vf =
q Le
=n L
Solving for q we find
2
P  L 2
L L e 
K=Cr
2

P L
L L e
2
1
K = Ks
[2.107]
Li
ne
or, after comparing to Darcy’s law
we see that
 L 2


L 
 e
8
r
ar
q=-
nr
8 

K
[2.108]
where C is some constant.
K goes down with (tortuosity)2
and (radius)2
K also hit by big pores
emptying first
Conductivity will drop
precipitously as q decreases
3
2
0
qr
K=0
qs
Characteristics of K(q)
1: at saturation K= Ks.
2-3: K(q) = 0: pendular water
6
Adding Conservation of Mass: Richards Equation
 Need to add the constraint imposed by the
conservation of mass (L.A. Richards 1931).
 There are many ways to obtain this result (we’ll do a
couple).
Consider arbitrary volume media. Keep track of fluid
going into and out of this volume.

t
dS



volume
q dV = -
q n dS


surface
[2.109]
n
n of
o
i
t
c
Dire
q
Flow
7

t



volume
q dV


surface
=-
q n dS
[2.109]
Volume is independent of time, bring the time derivative
inside
 
 q dV = t
volume
q


surface

n dS
[2.110]
Apply the divergence theorem to the right side



 q n
surface
dS =



volume
q dV
[2.111]
Combining [2.110] and [2.111]




 q +
 t



volume

q  dV

=0
[2.112]
8
    q
 
+

    t

  q  dV

=0
[2.112]
vol ume
Since [2.112] is true for any volume element that the
integrand must be zero for all points
q
+  q = 0
t
[2.113]
Replace q using Darcy’s law to obtain Richards
equation
q
=  (K( q) H)
t
[2.114]
where H is the total potential. This is also referred to
as the Fokker-Plank equation
9
In terms of elevation and pressure
Total potential is sum of gravity
potential and pressure, H = h + z
H = h + z
z
z =
x
but
So we obtain
i
z
+
y
[2.115]
j
z
+
z
k
= 0i + 0j + 1k = 1k
[2 .1 1 6 ]
 [K(q)H] = [K(q)( h  1 k )]
=  [K(q)( h)] - [K(q)1 k ] [2.117]
noticing that
  
K (q )
  K (q )1k    , ,   (0,0, K (q )) 
[2.118]
z
 x y z 
10
And finally...
So Richards Equation may be written (drum
roll please....)
q
K( q)
= [K( q)h ] +
t
z
[2.119]
Before we can get anywhere we need
the relationships, K(q) and h{q}.
First order in time and second order in
space; require
1 initial condition and
2 boundary conditions (top and bottom)
11
The diffusion form of R’s eq.
Can put in more familiar form by introducing
h
D( q) = K( q)
q
[2.120]
the soil diffusivity. Note that
K h = K
 h h h 


x
y
z
’ ’ 

 h q h q h q 
=K 

q
x
q
y
q
z
’ ’ 

h  q q q 
=K


q  x ’ y ’ z 
= D q
[2.121]
12
The diffusion form of R’s eq.
D(q) gives us a diffusion equation in q
q
K( q)
= [D( q) q ] +
t
z
[2.122]
Favorite trick in solving diffusion problems is to
assume that D is constant over space and pull it
outside the derivative: it will find no place here!
D(q) strongly non-linear function of q and varies
drastically as a function of space.
This makes Richards equation quite an interesting
challenge in terms of finding tidy analytical
solutions, and even makes numerical modelers
wince a bit due to the very rapid changes in both D
13
and q.
Briefly: how this is applied
to gas
What about that no-slip
boundary?
14
Gas Flow in Porous Media
Might expect movement of gases is a simple
extension of liquids
use gas density & viscosity with intrinsic permeability
to get K
use Darcy's law.
Well, it ain't quite that simple.
Recall the "no-slip" boundary condition
 Idea: collisions between molecules in liquid so frequent that
near a fixed surface the closest molecules will be constantly
loosing all of their wall-parallel energy, rendering them
motionless from the macroscopic perspective.
 Requisite: mean free path of travel short compared to the
aperture through which the liquid is moving.
15
… about Gases
Gases
mean free length of travel on the order of media
pore size.
No-slip condition for liquids does not apply
Taken together, results in "Klinkenberg effect"
(experimental/theoretical 1941 paper by L.J.
Klinkenberg).
Gas permeability, Kg, a function of gas pressure
(dictates mean free path length)
16
kg = k

1 +

4c
l

r 
[2.108b]
k intrinsic permeability to liquid flow
r characteristic radius apertures of the media
l mean free path length of the gas molecules
c proportionality factor between the mean free path for
the free gas compared to the mean free path for gas
which just collided with the capillary wall (c is just
slightly less than 1, Klinkenberg, 1941).
Coarse media: capillaries larger than the mean free path
length, the intrinsic permeability for liquids is recovered
Fine media: gas permeability exceeds liquid permeability.
17
Final notes on gas permeability
kg/k measure of size of conducting pathways,
 can be used as a diagnostic parameter (Reeve, 1953).
As P the mean free path length lr
 limit for kg 10 k
 Klinkenberg's found max kg 5 - 20 times k
Practical purposes
correction for vicinity of 1 bar 20 to 80% over the liquid
permeability, depending upon the media (Klinkenberg, 1941).
Methods of measurement of gas permeability:
Corey (1986),
New methods: Moore and Attenborough, 1992.
kg depends on the liquid content, analogous to hydraulic
conductivity with moisture content (see Corey, 1986).
18
Relative permeabilities of wetting and nonwetting phases
Ksnw
0
Ksw
0
0
qsnw
qsnw
0
19
Expressions for Conductivity & Retention
To solve Richards equation we need the
mathematical relationships between pressure,
moisture content and conductivity.
Conductivity is a non-hysteretic function of
moisture content
Moisture content and pore pressure are related
through a hysteretic functional (often hysteretic
ignored; gives expressions of conductivity in terms
of matric potential).
For analytical solution, must use analytical
expressions.
For numerical solutions may use constructed as
interpolations between successive laboratory
data.
20
 q - qo 
 =
D
q
q
o
 s
Ks | h c | (1 - )
w it h 0 <  < 1 Fu jit a (1 9 5 2 )

 q - qo  2 
 
- qr ) 1 - 
q
q
o 

 s
(qs
K(h ) = Ks e xp (h )
K(h ) =
Ga r d n e r (1 9 5 8 )
Ks
hn + b
q - qo
qs - qo
w it h 1 < n < 4
Ga r d n e r (1 9 5 8 )
 h b
 
h b 
Br u t s a e r t (1 9 6 6 )
1
=
1 +
n
Do (n +1 )q
D(q) =
n

1 
qn 
n +1 
w h e r e Do = D(qs )
K(h ) = Ks e xp [h - h 0 )]
for h < h o
K(h ) = Ks
for h 
  h o
Br u t s a e r t (1 9 6 8 )
Rit ge m a (1 9 6 7 )
q - qo
qs - qo
 h cr l
=  
h
for h  h cr with q = qs for h 
  h cr
 q - qo  
K(q) = K s  q - q  for h  h cr
 s
o
for h 
  h cr
K(q) = K s
wher e =
q - qo
qs - qo
2
+ 3.
l
=
Br ooks and Cor ey (1964)
1

1 +

n m
h
  
 h g 
with h g < 0
and
  q - qo  1 / m  m  2
 q - qo   
K(q) = K s  q - q 
 1 -  1- 

 
q
q
 s
o

s
o



 
wher e m = 1 - 1/n with n >1.
Van Genuchten (1980)



q
q

o
K(q) = Ksq - q  1  s
o


qs


 where 01
dq 


D() d
q

qs

D(q)

Parlange et al. (1985)
 qs - qo
 ln{(c - )/(c - 0 )} - 2
q = qo +  2  erfc
 for  < c
1/2




2 
q = qs
for   c
Kosugi (1994)
w h ich can be pu t in to M u alem ’s equ ation to yield a con du ctiv ity fu n ction
h  
 q - q o  l 1
 1

K(h ) = Ks 
erfc
ln
 
 1/ 2

h

0
q s - q o  2
2



2
Das an d Klu iten berg (1996)