1

Permeability Permeability and Thetaand Theta

K varies as a function of moisture content in the

vadose zone

K varies as a function of moisture content in the

vadose zoneWilliams, 2002 http://www.its.uidaho.edu/AgE558

Modified after Selker, 2000 http://bioe.orst.edu/vzp/

2

Extending Darcy to Unsaturated MediaExtending Darcy to Unsaturated Media1907 Buckingham saw Darcy could describe unsaturated flow

q = - K() H [2.102]

K(): 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 alone. To express K as a function of pressure you must employ the hysteretic functional relationship between and H.

1907 Buckingham saw Darcy could describe unsaturated flow

q = - K() H [2.102]

K(): 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 alone. To express K as a function of pressure you must employ the hysteretic functional relationship between and H.

3

How does K vary with How does K vary with ??It decreases precipitously!

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 volume of porous medium, 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.

It decreases precipitously!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 volume of porous medium, 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.

4

Permeability with theta...Permeability with theta...To get a feel for how fast K drops, let’s consider the effect of tortuosity alone (item 2 above). Particle flowpath and velocity must be distinguished

from the “Darcian” flow length and the “Darcian” velocity.

To get a feel for how fast K drops, let’s consider the effect of tortuosity alone (item 2 above). Particle flowpath and velocity must be distinguished

from the “Darcian” flow length and the “Darcian” velocity.

Actual flow length, L e

"Darcian" flow length, L

"Darcian" velcoity q = flow/area

q

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.

5

A few illustrative calculations…A few illustrative calculations…So writing the true pressure gradient that acts on the fluid we have

Le/L ratio of the true path length to the Darcian length.

Next write equation for capillary velocity, vf (Hagen-Poiseuille)

Translating into the Darcian velocity, q, we find

So writing the true pressure gradient that acts on the fluid we have

Le/L ratio of the true path length to the Darcian length.

Next write equation for capillary velocity, vf (Hagen-Poiseuille)

Translating into the Darcian velocity, q, we find

P

L =

Le

L P

Le[2.104]

vf = -

8  r 2 P

Le[2.105]

vf = qn

Le

L = -

8  r 2

P

L LLe

[2.106]

6

Conclusions on tortuosity...Conclusions on tortuosity...Solving for q we find

or, after comparing to Darcy’s law we see that

where C is some constant. K goes down with (tortuosity)2 and

(radius)2

K also hit by big pores emptying firstConductivity will drop precipitously

as decreases

Solving for q we find

or, after comparing to Darcy’s law we see that

where C is some constant. K goes down with (tortuosity)2 and

(radius)2

K also hit by big pores emptying firstConductivity will drop precipitously

as decreases

q = -  n r 2

8  P

L

L

L e

2[2.107]

K = C r 2

L

Le

2[2.108]

sr0

Linea

r

K

K = Ks

K = 023

1

Characteristics of K() 1: at saturation K= Ks. 2-3: K() = 0: pendular water

7

Adding Conservation of Mass: Richards EquationAdding 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 of porous medium. Keep track of fluid going into and out of this volume.

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 of porous medium. Keep track of fluid going into and out of this volume.

n

Direction of Flow

q

dSt

volume  dV = -

surfaceq n dS [2.109]

8

volume

t dV = -

surfaceq   n dS [2.110]

Volume is independent of time, bring the time derivative inside

Apply the divergence theorem to the right side

Combining [2.110] and [2.111]

Volume is independent of time, bring the time derivative inside

Apply the divergence theorem to the right side

Combining [2.110] and [2.111]

t

volume dV = -

surfaceq n dS [2.109]

surfaceq n dS =

volumeq dV [2.111]

volume

t  +  q  dV = 0 [2.112]

9

Since [2.112] is true for any volume element that the integrand must be zero for all points

Replace q using Darcy’s law to obtain Richards equation

where H is the total potential. This is also referred to as the Fokker-Plank equation

Since [2.112] is true for any volume element that the integrand must be zero for all points

Replace q using Darcy’s law to obtain Richards equation

where H is the total potential. This is also referred to as the Fokker-Plank equation

volume

t  +  q  dV = 0 [2.112]

t + q = 0 [2.113]

t = (K( )H) [2.114]

10

H = h + z [2.115]

z = zx i + zy j + zz k = 0i + 0j + 1k = 1k [2.116]

[K()H] = [K()(h1 k)]

= [K()(h)] - [K()1 k] [2.117]

[K( )1 k ] =

 z

x ’ 

z

y ’ 

z

z 

0 ’0 ’K( )

In terms of elevation and pressureIn terms of elevation and pressureTotal potential is sum of gravity potential and pressure, H = h + z

butSo we obtain

noticing that

Total potential is sum of gravity potential and pressure, H = h + z

butSo we obtain

noticing that

=

K( )

z [2.118]

11

And finally...And finally...So Richards Equation may be written (drum roll please....)

Before we can get anywhere we need the relationships, K() and h{}.

First order in time and second order in space; require

1 initial condition and 2 boundary conditions

So Richards Equation may be written (drum roll please....)

Before we can get anywhere we need the relationships, K() and h{}.

First order in time and second order in space; require

1 initial condition and 2 boundary conditions

t = [K()h] +

K()

z [2.119]

12

The diffusion form of R’s eq.The diffusion form of R’s eq.

Can put in more familiar form by introducing

the soil diffusivity. Note that

Can put in more familiar form by introducing

the soil diffusivity. Note that

D() = K() h [2.120]

K h = K

 h

x ’ 

h

y ’ 

h

z 

= K

 h

 x ’ 

h

 y ’ 

h

 z

 

= K h

 x ’

 y ’ 

z

 

= D [2.121]

13

The diffusion form of R’s eq.The diffusion form of R’s eq.D() gives us a diffusion equation in

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() 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 and

D() gives us a diffusion equation in

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() 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 and

t = [D()] +

K()

z [2.122]

14

Gas Flow in Porous MediaGas Flow in Porous MediaMight expect movement of gases is a simple extension of liquids use gas density & viscosity with intrinsic permeability to get Kuse Darcy's law.

It isn’t that simple because of the shear forces.

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.

Might expect movement of gases is a simple extension of liquids use gas density & viscosity with intrinsic permeability to get Kuse Darcy's law.

It isn’t that simple because of the shear forces.

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

Gas flow in porous mediaGas flow in porous mediaGases mean free length of travel on the order of medium pore

size. No-slip condition does not applyNo-slip results for liquids inapplicable.

Taken together, results in "Klinkenberg effect" (experimental/theoretical 1941 paper by L.J. Klinkenberg).

Gas permeability, g, a function of gas pressure (dictates mean free path length)

Gases mean free length of travel on the order of medium pore

size. No-slip condition does not applyNo-slip results for liquids inapplicable.

Taken together, results in "Klinkenberg effect" (experimental/theoretical 1941 paper by L.J. Klinkenberg).

Gas permeability, g, a function of gas pressure (dictates mean free path length)

g =

1 + 4 c 

r[2.108b]

16

Parameters intrinsic permeability to liquid flow r characteristic radius apertures of the medium mean free path length of the gas moleculesc 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.

Parameters intrinsic permeability to liquid flow r characteristic radius apertures of the medium mean free path length of the gas moleculesc 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.

g =

1 + 4 c 

r[2.108b]

17

Final notes on gas permeabilityFinal notes on gas permeabilityg/ measure of size of conducting pathways,

can be used as a diagnostic parameter (Reeve, 1953).

As P the mean free path length r limit for g 10 Klinkenberg's found max g 5 - 20 times

Practical purposescorrection for vicinity of 1 bar 20 to 80% over the liquid permeability, depending upon the medium (Klinkenberg, 1941).

Methods of measurement of gas permeability: Corey (1986), New methods: Moore and Attenborough, 1992.

g depends on the liquid content, analogous to hydraulic conductivity with moisture content (see Corey, 1986).

g/ measure of size of conducting pathways, can be used as a diagnostic parameter (Reeve, 1953).

As P the mean free path length r limit for g 10 Klinkenberg's found max g 5 - 20 times

Practical purposescorrection for vicinity of 1 bar 20 to 80% over the liquid permeability, depending upon the medium (Klinkenberg, 1941).

Methods of measurement of gas permeability: Corey (1986), New methods: Moore and Attenborough, 1992.

g depends on the liquid content, analogous to hydraulic conductivity with moisture content (see Corey, 1986).

18

Expressions for Conductivity & RetentionExpressions for Conductivity & RetentionTo 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.

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.

D

  -   o

s   -   o =

Ks   | |h c ( 1 - )

(s   -   r )  

1   -  

  -   o

s   -   o

2

w it h 0 < < 1 Fu j it a ( 1 9 5 2 )

K( h ) = Ks e x p (h ) Ga r d n e r ( 1 9 5 8 )

K( h ) = Ks

h n   +  b w it h 1 < n < 4 Ga r d n e r ( 1 9 5 8 )

  -   o

s   -   o =

1

1   +   

h

h b

b

  B r u t s a e r t ( 1 9 6 6 )

D() = Do ( n +1 )n

n

1 -  n

n +1 w h e r e Do = D(s ) Br u t s a e r t ( 1 9 6 8 )

K( h ) = Ks e x p [h - h 0 ) ] fo r h < h o Rit g e m a ( 1 9 6 7 )K( h ) = Ks fo r h h o

  -  o

s  -  o =

hcr

h for h hcr with = s for h hcr

K() = Ks

  -  o

s  -  o

for h hcr

K() = Ks for h hcr

where = 2 + 3. Brooks and Corey (1964)

  -  o

s  -  o =

1

1  +   

h

hg

n m with hg < 0

and

K() = Ks

  -  o

s  -  o

1  - 

1-

  -  o

s  -  o

1 / m m

2

where m = 1 - 1/n with n >1. Van Genuchten (1980)

K() = Ks

  -  o

s  -  o

1  -   

s

D()  d

s

D()  d

where 01

Parlange et al. (1985)

= o +

s  -  o

2 erfc

ln{(c  -  )/(c  -  0)}  -  2

21/2 for < c

= s for c

Kosugi (1994)

wh ich can be pu t in to Mu alem ’s equ ation to yield a con du ctiv ity fu n ction

K(h ) = Ks

 - o

s - o

l

1

2 erfc

1

21/ 2

ln

h

h 0

2

Das an d Klu iten berg (1996)