Transcript Hydrodynamics in Porous Media

Hydrodynamics in
Porous Media
We will cover:
How fluids respond to local potential
gradients (Darcy’s Law)
Add the conservation of mass to
obtain Richard’s equation
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Darcy’s Law for saturated media
In 1856 Darcy hired to figure out the water supply to the
town’s central fountain.
Experimentally found that flux of water porous media could
be expressed as the product of the resistance to flow
which characterized the media, and forces acting to “push”
the fluid through the media.
 Q - The rate of flow (L3/T) as the volume of water passed through a
column per unit time.
 hi - The fluid potential in the media at position i, measured in
standing head equivalent. Under saturated conditions this is
composed of gravitational potential (elevation), and static pressure
potential (L: force per unit area divided by g).
 K - The hydraulic conductivity of the media. The proportionality
between specific flux and imposed gradient for a given medium (L/T).
 L - The length of media through which flow passes (L).
 A - The cross-sectional area of the column (L2).
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Darcy’s Law
Darcy then observed that the flow of water in a vertical
column was well described by the equation
Q= K
A
L
(H
1
-
H
0
)
[ 2 .6 8 ]
Darcy’s expression is written in a general form for
isotropic media as
q = -K H
[2.69]
q is the specific flux vector (L/T; volume of water per
unit area per unit time),
K is the saturated hydraulic conductivity tensor
(second rank) of the media (L/T), and
H is the gradient in hydraulic head (dimensionless) 3
Aside on calculus ...
What is this up-side-down triangle all about?
The “dell” operator: short hand for 3-d derivative




i,
j,
k
x y z
The result of “operating” on a scalar function (like potential) with 
is the slope of the function
F points directly towards the steepest direction of up hill with a
length proportional to the slope of the hill.
Later we’ll use •F. The dot just tells us to take the dell and
calculate the dot product of that and the function F (which needs to
be a vector for this to make sense).
“dell-dot-F” is the “divergence” of F.
If F were local flux (with magnitude and direction), •F would be
the amount of water leaving the point x,y,z. This is a scalar result!
F takes a scalar function F and gives a vector slope
•F uses a vector function F and gives a scalar result.
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Now, about those parameters...
Gradient in head is dimensionless, being length per length
H1  H 0
H =
L
Q = Aq
[ 2 .7 0 ]
Q has units volume per unit time
Specific flux, q, has units of length per time, or
velocity.
For vertical flow: speed at which the height of a pond
of fluid would drop
CAREFUL: q is not the velocity of particles of water
The specific flux is a vector (magnitude and direction).
Potential expressed as the height of a column of
water, has units of length.
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About those vectors...
q = -K  H
[2 .7 0 ]
Is the right side of Darcy’s law indeed a vector?
h is a scalar, but H is a vector
Since K is a tensor (yikes), KH is a vector
So all is well on the right hand side
Notes on K:
we could also obtain a vector on the right hand
side by selecting K to be a scalar, which is often
done (i.e., assuming that conductivity is
independent of direction).
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A few words about the K tensor
q=-
=-
 K xx K xy K xz   h

 K yx K yy K yz   x

K
K
K
 zx
zy
zz 
h

K xx
+K
x

h
xy
y
+K
h
xz
flux in x-direction
z
;K
h
y
h
yx
x
h 
z 
+K
h
yy
y
+K
h
yz
flux in y-direction
z
;K
h
zx
x
+K
h
zy
y
+K
h
zz

z 
flux in z-direction
Kab relates gradients in potential in the b-direction to
flux that results in the a-direction.
In anisotropic media, gradients not aligned with
bedding give flux not parallel with potential gradients. If
the coordinate system is aligned with directions of
anisotropy the "off diagonal” terms will be zero (i.e.,
Kab=0 where ab). If, in addition, these are all equal,
then the tensor collapses to a scalar.
The reason to use the tensor form is to capture the
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effects of anisotropy.
Looking holistically
Check out the intuitively aspects of Darcy’s result.
The rate of flow is:
Directly related to the area of flow (e.g., put two
columns in parallel and you get twice the flow);
Inversely related to the length of flow (e.g., flow
through twice the length with the same potential
drop gives half the flux);
Directly related to the potential energy drop
across the system (e.g., double the energy
expended to obtain twice the flow).
The expression is patently linear; all properties
scale linearly with changes in system forces and
dimensions.
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Why is Darcy Linear?
Because non-turbulent?
No.
Far before turbulence, there will be large local
accelerations: it is the lack of local acceleration which
makes the relationship linear.
Consider the Navier Stokes Equation for fluid flow.
The x-component of flow in a velocity field with
velocities u, v, and w in the x, y, and z (vertical)
directions, may be written
u
t
u
+ u
x
u
+ v
y
u
+ w
z
=
-1
P
 x
z
-g
x

+

 u
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Creeping flow
u
t
u
+ u
x
u
+ v
y
u
+ w
z
=
-1
P

x
z
-g
x

+

 u
Now impose the conditions needed for which Darcy’s Law
“Creeping flow”; acceleration (du/dx) terms small
compared to the viscous and gravitational terms
 u   u   u  0 [2.69]
x y z
Similarly changes in velocity with time are small
u  0
t
[2.70]
 P   gz     2 u
so N-S is: x 

[2.71]
Linear in gradient of hydraulic potential on left, proportional
to velocity and viscosity on right (same as Darcy).
Proof of Darcy’s Law? No! Shows that the creeping flow
assumption is sufficient to obtain correct form.
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Capillary tube model for flow
Widely used model for flow through porous media is a group
of cylindrical capillary tubes (e.g.,. Green and Ampt, 1911
and many more).
Let’s derive the equation for steady flow through a
capillary of radius ro
Consider forces on cylindrical control volume shown
F=0
[2.75]
 s
s
r
V
ro
0
Cylindrical Control Volume
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Force Balance on Control Volume
 s
s
r
V
ro
0
Cylindrical Control Volume
end pressures:
at S = 0
F1 = Pr2
at S = S
F2 = (P + S dP/dS) r2
shear force:
Fs = 2rS
where is the local shear stress
Putting these in the force balance gives
Pr2 - (P + S dP/dS) r2 - 2rS = 0
[2.76]
where we remember that dP/dS is negative in sign (pressure
drops along the direction of flow)
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continuing the force balance
Pr2 - (P + S dP/dS) r2 - 2rS = 0
[2.76]
With some algebra, this simplifies to
 = -
r dP
[2 .7 7 ]
2 dS
dP/dS is constant: shear stress varies
linearly with radius
  
From the definition of viscosity
Using this [2.77] says
Multiply both sides

by dr, and integrate

vv(r )
dv
dr
dv
dr
dP
=
2  dS
r
 dv = 
v0

r  ro
r
r dP
2  dS
dr
[2.78]
[2 .7 9 ]
[2.80]
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Computing the flux through the pipe...
Carrying out the integration we find
( r 2 -r 0 2 ) dP
v(r) =
4
[2.81]
dS
which gives the velocity profile in a cylindrical pipe
To calculate the flux integrate over the area
 vdA
Q =
[ 2 .8 2 ]
Area
in cylindrical coordinates, dA = r d dr, thus
 
Q =




= 
r
= r



r
0
r 2 
4
r
0

2 
dP
dS
r
dr
d
[2 . 8 0 ]
= 0
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Rearranging terms...
The integral is easy to compute, giving
 r 0 4 dP
Q= 8 dS
[2.84]
(fourth power!!)
which is the well known Hagen-Poiseuille Equation.
We are interested in the flow per unit area (flux), for
which we use the symbol q = Q/r2
1 r 0 2 dP
q=  8 dS
[2.85]
(second power)
We commonly measure pressure in terms of hydraulic
head, so we may substitute gh = P, to obtain
r 0 2  dh
q= 8  dS
[2.86]
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q= -
r
0
8
2

dh

dS
[2 . 8 3 ]
r02/8 is a geometric term: function of the media.
referred to as the intrinsic permeability, denoted by .
  is a function of the fluid alone
NOTICE:
Recovered Darcy’s law!
See why by pulling  out of the hydraulic
conductivity we obtain an intrinsic property of the solid
which can be applied to a range of fluids.
SO if K is the saturated hydraulic conductivity, K=   .
This way we can calculate the effective conductivity for
any fluid. This is very useful when dealing with oil spills
... boiling water spills ..... etc.
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Darcy's Law at Re# > 1
Often noted that Darcy's Law breaks down at Re# > 1.
Laminar flow holds capillaries for Re < 2000; HagenPoiseuille law still valid
Why does Darcy's law break down so soon?
Laminar ends for natural media at Re#>100 due to the
tortuosity of the flow paths (see Bear, 1972, pg 178).
Still far above the value required for the break down of
Darcy's law.
Real Reason: due to forces in acceleration of fluids
passing particles at the microscopic level being as
large as viscous forces: increased resistance to flow,
so flux responds less to applied pressure gradients.
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A few more words about Re#>1
Can get a feel for this
through a simple
calculation of the
relative magnitudes of
the viscous and inertial
forces.
FI  Fv when Re#  10.
Since FI go with v2,
while Fv goes with v,
at Re# 1 FI  Fv/10,
a reasonable cut-off for
creeping flow
approximation
Isometric View
d2
d1
d1
d2
Flow
Cross-Section
v1
v2
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L
aw
Deviations from Darcy’s law
D
ar
cy
's
Re=100
(a) The effect of inertial
terms becoming significant
at Re>1.
1
K
Re=10
(a)
q
Re=1
Re=0
h
10
100
L
cy
's
ar
D
(b) At very low flow there
may be a threshold
gradient required to be
overcome before any flow
occurs at all due to
hydrogen bonding of water.
1
aw
0
1
(b)
K
q
0
0
Threshold
Pressu re
h
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How does this apply to Vadose?
Consider typical water flow where v and d are maximized
Gravity driven flow near saturation in a coarse media.
maximum neck diameter will be about 1 mm,
vertical flux may be as high as 1 cm/min (14 meters/day).
d
R
=
1

= 0.167
v1
=
1 gr/cm
[2.100]
3
x 0.1 cm x 1 cm/min
0.01 gr/cm-sec
[2.101]
Typically Darcy's OK for vadose zone.
Can have problems around wells
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What about Soil Vapor Extraction?
Does Darcy's law apply?
Air velocities can exceed 30 m/day (0.035 cm/sec).
The Reynolds number for this air flow rate in the
coarse soil used in the example considered above is
d
R=
=
1
v1

0.001 gr/cm
= 0.02
3
x 0.1 cm x 0.035 cm/sec
-4 gr/cm-sec
1.8 x 10
[2.101]
[2.102]
again, no problem, although flow could be higher
than the average bulk flow about inlets and outlets
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Summary of Darcy and Poiseuille
For SATURATED MEDIA
Flow is linear with permeability and gradient
in potential (driving force)
At high flow rates becomes non-linear due to
local acceleration
Permeability is due to geometric properties of
the media (intrinsic permeability) and fluid
properties (viscosity and specific density)
Permeability drops with the square of pore
size
Assumed no slip solid-liquid boundary:
doesn't work with gas.
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