Transcript Miller Similarity and Scaling of Capillary Properties

Miller Similarity and
Scaling of Capillary
Properties
How to get the most out of
your lab dollar by cheating
with physics
1
A little orientation...
How can we take information about the hydraulic
properties of one media to make quantitative
predictions of the properties of another media?
1956 Miller and Miller presented a comprehensive
methodology.
Bounds on our expectations:
can’t expect to make measurements in a sand
and hope to learn about the behavior of clays:
fundamentally differing in their chemical properties and
pore-scale geometric configuration.
To extrapolate from one media to another, the
two systems must be similar in the geometric
sense (akin to similar triangles).
2
Characterizing each medium
We need “characteristic microscopic length scales”
of each media with a consistent definition.
Need ’s in some dimension which can be
identified readily and reflects the typical
dimensions at the grain scale.
In practice people often use the d50 as their  as it
is easy to measure and therefore widely reported.
Any other measure would be fine so long as you
are consistent in using the same measure for each
of the media.


3
Enough generalities, let’s see how this works!
Assumptions Need For Similarity:
Media: Uniform with regard to position, orientation and
time (homogeneous, isotropic and permanent).
Liquid: Uniform, constant surface tension, contact angle,
viscosity and density. Contact angle must be the same in
the two systems, although surface tension and viscosity
may differ.
Gas: Move freely in comparison to the liquid phase, and
is assumed to be at a uniform pressure.
Connectivity: Funicular states in both the air and water:
no isolated bubbles or droplets.
What about hysteresis? No problem, usual bookkeeping.
4
Rigorous definition of geometric similarity
Necessary and sufficient conditions:
Media 1 and 2 are similar if, and only if, there exists
a constant  = 1/2 such that if all length
dimensions in medium 1 are multiplied by  , the
probability of any given geometric shape to be seen
in the scaled media1 and medium 2 are identical.
Most convenient to define a “scaled medium”
which can then be compared to any other
similar medium.
Scaled quantity noted with dot suffix: K• is
scaled conductivity.
5
Getting to some math..
Consider moisture content of two media with
geometrically similar emplacement of water.
Volumetric moisture content will be the same in the
two systems.
For any particular gas/liquid interface Laplace’s
equation gives the pressure in terms of the
reduced radius R
2cos
p=
R
[2.123]
where  is the contact angle,  is the surface
tension, and p is the difference in pressure
between the gas and liquid.
6
2cos
p=
R
[2.123]
Multiplying both sides by 
2cos

R
=
p

[2.124]
Stuff on left side is the same for any similar media,
THUS
Stuff on the right side must be constant as well.
We have obtained a method for scaling the
pressure!
p• =
p

[2.125]
7
Example Application
Let’s calculate the pressure in media 1 at some
moisture content  given that we know the pressure
in media 2 at . From above, we note that the
pressures are related simply as
.
h () =
λ1h1 ( ) λ 2h2 ( )
=
σ1
σ2
[2.126]
so we may obtain the pressure of the second media
as
1 2
h2() =
2 1
h 1()
[2.127]
8
Some data from our lab
Scaling of the characteristic curves for four
similar sands. Sizes indicated by mesh
bounds.
60
70
12/20
60
20/30
12/20
50
50
20/30
S caled P otential
Matric Potential (cm H2O)
80
30/40
40
40/50
30
20
40
30/40
30
40/50
20
10
10
0
0
0
0.2
0.4
0.6
0.8
1
0
0 .2
0 .4
0 .6
0 .8
1
Degree of Saturation
Degree of Saturation
9
How about scaling hydraulic conductivity
Need to go back to the underlying physical equations
to derive the correct expression for scaling.
Identify the terms which make up K in Darcy’s law in
the Navier-Stokes equation for creeping flow
(g-p) = 2v
[2.128]
Compared to Darcy’s law
v = K(p)(g-p)
[2.129]
which can be written
v
(g-p) =
K(p)
[2.130]
10
(g-p) =
2v
v
(g-p) =
K(p)
[2.128]
Solving for K we find
v
K(p) =
2v
[2.130]
[2.131]
Now looking at this for a one-dimensional case for the
l
l•
moment we see that
v=
t
=
t

[2.132]
where l is a microscopic unit of length, and
2 l
2 l• 1
2v =
=
= •2v•
l2t 2l•2 t

[2.133]
so equation [2.131] may be rewritten as
K(p) =
v•
•2v•
2
[2.134]
Putting the unscaled variables on the left we see that
K(p)
v•
= 2
2

• v•
[2.135]
11
From last slide:
K(p)
v•
=
2
•2v•
[2.135]
The right-hand side of [2.135] is only
dependent on the properties of the scaled
media, implying that the left-hand side must be
as well
K(p)
K•(p) =
2
[2.136]
Careful: p is not the scaled pressure! Should
write
K•(p•) =
p
K(  )
2
K()
=
2
[2.137]
The scaling relationship for permeability!
12
Example
Two similar media at moisture content : Scaled
conductivities will be identical
1K1() 2K2()
K•(p•) =
=
2
1
22
[2.138]
or, solving for K2 in terms of K1 we find
122
K2() = K1()
212
[2.139]
which can also be written in terms of pressure
12  122
K2() = K1

21  212
[2.140
13
Last scaling parameter required: time
By looking to the macroscopic properties of the
system, we can obtain the scaling relationship for
time
Consider Darcy’s law and the conservation of
mass.
In the absence of gravity Darcy’s law states
v = -K(p)p
[2.141]
Multiplying both sides by , we find
p


v = 2 K(p) 
[2.142]



14
Scaling time...
p


v = 2 K(p) 



[2.142]
From a macroscopic viewpoint, both v and  are
functions of the macroscopic length scale, say L.
The product L is the reduced form of the
gradient operator. So we can multiply both sides
by L to put the right side of this equation in the
reduced form
p
L

v = 2 K(p)L  = K••p•



[2.143]
Since right side is, then left side is in reduced
form, thus the reduced macroscopic velocity is
given by
L
v• =
v
[2.144]

15
Finishing up t scaling
Would like to obtain the scaling parameter for
time, say , such that  t = t•. Using the definition
of velocity we can write
dx•
v• =
dt•
[2.145]
and using the fact that x•=x/L and t•=t, v• can be
rewritten
dx/L
1 dx
1
v• =
dt
=
=
v
L dt
L
[2.146]
Now solving for  we find
and solving for t•
JOB DONE!!

v
= v L = 2
•
L

t• = 2 t
L
[2.147]
[2.148]
16
Squared
scaling of K
w.r.t.
particle size
Data from Warrick
et al. demonstrating
scaling of
saturation permeability
relationship
Warrick et al.
demonstration
of scaled
pressure saturation
relationship