05. Effective Medium Theories.pdf

Effective Medium Models

127Stanford Rock Physics Laboratory - Gary Mavko & Tapan Mukerji

Are real rocks hopelessly complex?


Theoretical Models

• Some are mathematically elegant (complicated).

• Most are extreme idealizations of the complexity of real

During the last four decades, many theoretical models

have appeared which try to describe the elastic and

transport properties of rocks.


• Most are extreme idealizations of the complexity of real rocks

Models with Idealized Geometries

Contact Theories Inclusion Models


?

Bounding Methods for Estimating Effective

Elastic Moduli

For many reasons we would like to be able to model or estimate the effective

elastic moduli of rocks in terms of the properties of the various constituent

minerals and pore fluids. To do it precisely one must incorporate

• the individual elastic moduli of the constituents

• the volume fractions of the constituents

• geometric details of how the various constituents


• geometric details of how the various constituents

are arranged

The geometric details are the most difficult to know or measure. If we ignore (or

don’t know) the details of geometry, then the best we can do is estimate upper

and lower bounds on the moduli or velocities.

The bounds are powerful and robust tools. They give rigorous upper and lower

limits on the moduli, given the composition. If you find that your measurements

fall outside the bounds, then you have made a mistake - in velocity, volume

fractions, or composition!

Voigt and Reuss BoundsOn a strictly empirical basis one can imagine defining a power law average of

the constituents

where

Special cases are the Voigt average (an upper bound):

= the effective modulus of the composite

= the modulus of the ith constituent

= the volume fraction of the ith constituent

α = a constant, generally between -1 and +1

M

M i

f i

Mα

= f1M1

α+ f2M2

α+ f3M3

α+ ...


Special cases are the Voigt average (an upper bound):

and the Reuss average (a lower bound):

Since these are upper and lower bounds, an estimate of the actual value is

sometimes taken as the average of the two, known as the Voigt-Reuss-Hill

average:MVRH =

MV + MR

2

KV = fQKQ + fF KF + fCKC ... + fW KW + fO KO + fGKG

µV = fQ µQ + fFµ F + fC µC ... + fW µW + fO µO + fGµG

K R

−1

= fQKQ

−1 + fFKF

−1 + fCKC

−1...+ fWKW

−1 + fOKO

−1 + fGKG

−1

µR

−1

= fQµQ

−1+ fFµF

−1+ fC µC

−1...+ fWµW

−1+ fOµO

−1+ fGµG

−1

The Voigt and Reuss averages are interpreted as the ratio of average stress

and average strain within the composite.

The stress and strain are generally unknown in the composite and are expected

to be nonuniform. The upper bound (Voigt) is found assuming that the strain is

everywhere uniform. The lower bound (Reuss) is found assuming that the

stress is everywhere uniform.

Geometric interpretations:


Voigt iso-strain model Reuss iso-stress model

Since the Reuss average describes an isostress situation,

it applies perfectly to suspensions and fluid mixtures.

ˆ E =ˆ σ ˆ ε

=f iσ i∑ˆ ε

=f i( ˆ ε E i)∑ˆ ε

ˆ E =ˆ σ ˆ ε

=ˆ σ

f iεi∑=

ˆ σ

f i(ˆ σ

E i

)∑1ˆ E

=f i

E i

∑ˆ E = f i∑ E i

Example: Representing a soft water-bottom sediment as a suspension.

A true suspension consists of particles suspended in a liquid. A suspension will act like a fluid, having zero shear modulus, and a bulk modulus given by the Reuss average. A suspension is isostress: both liquid and particles feel the same pressure.

We often approximate a soft-water bottom as a suspension, since it is composed of particles that are completely wet, with only slight grain-to-grain contacts.

K +4 µ

0

P-wave velocity


VP =

K +3 µ

ρ

ρ = 1− φ( )ρsolid + φρbrine

K =1− φ

Ksolid

+φ

Kbrine

Gas-charged sediment: ρ = 1− φ( )ρsolid + φSwρbrine + φ 1− Sw( )ρgas

K = 1− φ( )Ksolid + φSwKbrine + φ 1− Sw( )Kgas

Wet sediment:

P-wave velocity

K = 5.05GPa

ρ =1.99g /cc

V =1.59km /s

K = 0.74 GPa

ρ =1.88 g /cc

V = 0.63 km /s

φ = 0.4; Sw = 0.7; Kqtz = 37; Kw = 2.2; Kgas = 0.1 GPa; ρqtz = 2.65; ρw =1.0; ρgas = 0.1 g /cc

Voigt


Velocity-porosity relationship in clastic sediments compared with the Voigt and Reuss bounds. Virtually all of the points indeed fall between the bounds. Furthermore, the suspensions, which are isostress materials (points with porosity > 40%) fall very close to the Reuss bound.

Data from Hamilton (1956), Yin et al. (1988), Han et al. (1986). Compiled by Marion, D., 1990, Ph.D.

dissertation, Stanford Univ.

Reuss water


Velocity-porosity relationship in clastic sediments compared with the Voigt and Reuss bounds. Virtually all of the points indeed fall between the bounds. Furthermore, the suspensions, which are isostress materials (points with porosity > 40%) fall very close to the Reuss bound.

Data from Hamilton (1956), Yin et al. (1988), Han et al. (1986). Compiled by Marion, D., 1990, Ph.D.

dissertation, Stanford Univ.

Reuss water5% gas

Hashin-Shtrikman Bounds

The narrowest possible bounds on moduli that we can estimate for an isotropic

material, knowing only the volume fractions of the constituents, are the Hashin-

Shtrikman bounds. (The Voigt-Reuss bounds are wider.) For a mixture of 2

materials:K HS ± = K 1 +

f 2

K 2 − K1( )−1

+ f1 K 1 +4

3µ 1

−1

µ HS ±= µ1 +

f 2

µ 2 − µ 1( )−1+

2 f1(K1 + 2µ1)

5µ 1 K 1 +4

3µ 1


Interpretation of bulk modulus:

where subscript 1 = shell, 2 = sphere. f1 and f2 are volume fractions.

These give upper bounds when stiff material is K1, µ1 (shell) and lower bounds

when soft material is K1, µ1.

3

Hashin-Shtrikman Bounds

A more general form that applies when more than two phases are being mixed (Berryman, 1993):

where

K HS + = Λ(µmax), K HS− = Λ(µmin)

µ HS+ = Γ ζ Kmax,,µmax( )( ), µHS − = Γ ζ Kmin ,µmin( )( )


where

indicates volume average over the spatially varying K(r), µ(r) of the constituents.

Λ(z) =1

K( r) +4

3z

−1

−4

3z

Γ( z) =1

µ(r) + z

−1

− z

ζ (K ,µ ) =µ

6

9K + 8µ

K + 2µ

Here we see that a mixture of calcite and water gives widely spacedbounds, but a mixture of calcite and dolomite gives very narrow bounds.

Distance between bounds depends on similarity/difference of end-member constituents.


G13

Wyllie Time Average

1

2

3

d1

d2

d3

D

Wyllie et al. (1956, 1958, 1962) found that travel time through water saturated

consolidated rocks could be approximately described as the volume weighted

average of the travel time through the constituents:

t =D

V


3d3

t = t1 + t2 + t3

D

V=

d1

V1

+d2

V2

+d3

V3

1

V=

d1 / D

V1

+d2 / D

V2

+d3 / D

V3

1

V=

f1

V1

+f2

V2

+f3

V3

Limitations:

• rock is isotropic• rock must be fluid-saturated• rock should be at high effective pressure• works best with primary porosity

Wyllie’s generally works best for

• water-saturated rocks

• consolidated rocks

• high effective pressures


• works best with primary porosity• works best at intermediate porosity• must be careful of mixed mineralogy (clay)

The time-average equation is heuristic and cannot be justified theoretically. It is based on ray theory which requires that (1) the wavelength is smaller than the grain and pore size, and (2) the minerals and pores are arranged in flat layers.

Note the problem for shear waves where one of the phases is a fluid, Vs-fluid → 0!

Modification of Wyllie's proposed by Raymer

V = (1−φ)2Vmineral+φVfluid

1

ρV 2 =φ

ρfluidVfluid

2 +1−φ

ρmineralVmineral

2

1 0.47−φ 1 φ −0.37 1

φ < 37%

φ > 47%


Still a strictly empirical relation.

This relation recognizes that at large porosities (φ > 47%) the sediment behaves

as a suspension, with the Reuss average of the P-wave modulus, M = ρVp2.

1

V=

0.47−φ

0.10

1

V37

+φ −0.37

0.10

1

V47

37% < φ < 47%


Comparison of Wyllie's time average equationand the Raymer equations with Marion's compilation of shaly-sand velocities from Hamilton (1956), Yin et al. (1988), Han et al. 1986).

Backus Average for Thinly Layered Media

Backus (1962) showed that in the long wavelength limit a stratified medium

made up of thin layers is effectively anisotropic. It becomes transversely

isotropic, with symmetry axis normal to the strata. The elastic constants (see

next page) are given by:

whereA =

4µ(λ + µ)+

1−1

λ2

M =1

2A − B( )

A B F 0 0 0

B A F 0 0 0

F F C 0 0 0

0 0 0 D 0 0

0 0 0 0 D 0

0 0 0 0 0 M


where

are the isotropic elastic constants of the individual layers. The brackets

indicate averages of the enclosed properties, weighted by their volumetric

proportions. This is often called the Backus average.

A =4µ(λ + µ)

λ + 2µ+

1

λ + 2µ

λ

λ + 2µ

B =2µλ

λ + 2µ+

1

λ + 2µ

−1

λ

λ + 2µ

2

C =1

λ + 2µ

−1

F =1

λ + 2µ

−1

λ

λ + 2µ

D =1

µ

−1

M = µ

λλλλ, µµµµ

Hooke’s law relating stress and strain in a linear elastic

medium can be written as

elastic stiffnesses (moduli) elastic compliances

A standard shorthand is to write the stress and strain as

vectors:

σ ij = c ijkl εklΣkl

ε ij = S ijkl σ klΣkl

T =

σ 1= σ 11

σ 2= σ 22

σ 3= σ 33

σ 4= σ 23

σ 5= σ 13

E =

e1= ε11

e2= ε22

e3= ε33

e4=2ε23

e =2ε

Note the factor of 2 in the

definition of strains.


The elastic constants are similarly written in

abreviated form, and the Backus average constants shown on the previous

page now have the meaning:

σ 5= σ 13

σ 6= σ 12

e5=2ε13

e6=2ε12

σ 1

σ 2

σ 3

σ 4

σ 5

σ 6

=

A B F 0 0 0B A F 0 0 0F F C 0 0 00 0 0 D 0 00 0 0 0 D 00 0 0 0 0 M

e1

e2

e3

e4

e5

e6

Seismic Fluid Substitution

Pore fluids, pore stiffness,

and their interaction


Typical Problem: Analyze how rock properties, logs, and seismic change, when pore fluids change.


Example: We observe Vp, Vs, and density at a well and compute a synthetic seismic trace, as usual. Predict how the seismic will change if the fluid changes -- either over time at the same position, or if we move laterally away from the well and encounter different fluids in roughly the same rocks.

Gassmann to predict 4D changes

Initial

??

After

production

Can we predict signature of

saturation changes?


Gassmann to predict lateral variations

??

Can we predict

signature of oil?

Well penetrates wet

sand

Effective moduli for specific pore and grain geometries

Imagine a single linear elastic body. We do two separate experiments--apply

stresses σ1 and observe displacements u1, then apply stresses σ2 and observe

displacements u2.

The Betti-Rayleigh reciprocity theorem states that the work done by the first set

of forces acting through the second set of displacements is equal to the work

done by the second set of forces acting through the first set of displacements.

σij(1), u(1) σij

(2), u(2)


∆σ ∆σ

Estimate of Dry Compressibility

∆σ

∆σ

∆σ∆σ∆σ∆σ




Applying the reciprocity theorem we can write:

Assumptions• minerals behave elastically• friction and viscosity not important• assumes a single average mineral

limit as

∆σ∆σVbulk

K dry

− ∆σ∆Vpore =∆σ∆σVbulk

K mineral

1

Kdry

=1

Kmineral

+1

Vbulk

∂Vpore

∂σ

∆σ → 0

Relation of Rock Moduli to Pore Space Compressibility -- Dry Rock

A fairly general and rigorous relation between dry rock

bulk modulus and porosity is

is the pore space stiffness. This is a new concept

that quantifies the stiffness of a pore shape.

where 1

K dry

= 1K mineral

+φ

K φ

K φ

σφ ∂

∂=

pore

pore

v

vK

11

151Stanford Rock Physics Laboratory - Gary Mavko & Tapan Mukerji G.4

What is a “Dry Rock”?

Many rock models incorporate the concept of a dry rock or the dry rock frame.

This includes the work by Biot, Gassmann, Kuster and Toksoz, etc, etc.

Caution: “Dry rock” is not the same as gas-saturated rock. The dry frame

modulus in these models refers to the incremental bulk deformation resulting from

an increment of applied confining pressure, with pore pressure held constant.

This corresponds to a “drained” experiment in which pore fluids can flow freely in


This corresponds to a “drained” experiment in which pore fluids can flow freely in

or out of the sample to insure constant pore pressure. Alternatively, it can

correspond to an undrained experiment in which the pore fluid has zero bulk

modulus, so that pore compressions do not induce changes in pore pressure –

this is approximately the case for an air-filled sample at standard temperature and

pressure. However, at reservoir conditions (high pore pressure), gas takes on a

non-negligible bulk modulus, and should be treated as a saturating fluid.

Relation of Rock Moduli to Pore Space Compressibility -- Saturated Rock

A similar general relation between saturated rock bulk modulus and porosity is

1K sat

= 1K mineral

+φ

K φ

K φ = K φ +

K mineral K fluid


where Pore spacecompressibilitymodified by fluids.

So we see that changing the pore fluid has the effect of changing the pore space

compressibility of the rock. The fluid modulus term is always just added to

K φ = K φ + mineral fluid

K mineral – K fluid

≈ K φ + K fluid

When we have a stiff rock with high velocity, then its value of is large, and

changes in do not have much effect. But a soft rock with small velocity will

have a small and changes in will have a much larger effect.

K φ

K φ K fluid

K fluid

K φ

Gassmann's Relations

These are Transformations! Pore space geometry and stiffness are incorporated automatically by measurements of Vp, Vs. Gassmann (1951) derived this general relation between the dry rock moduli and the saturated rock moduli. It is quite general and valid for all pore

Ksat

Kmineral − Ksat

=Kdry

Kmineral − Kdry

+K fluid

φ Kmineral − Kfluid( )

1

µsat

=1

µdry


the saturated rock moduli. It is quite general and valid for all pore geometries, but there are several important assumptions:

• the rock is isotropic

• the mineral moduli are homogeneous

• the frequency is low

“Dry rock” is not the same as gas saturated rock.

Be careful of high frequencies, high viscosity, clay.

Useful for Fluid Substitution problem:

gas

oilwater

Some Other Forms of Gassmann

K sat =

φ 1Kmin

– 1K fluid

+ 1K min

– 1Kdry

φKdry

1Kmin

– 1K fluid

+ 1Kmin

1K min

– 1Kdry

K sat = K dry +

1 –K dry

Kmin

2

φK fluid

+1 – φK min

–K dry

Kmin2


Kdry Kmin K fluid Kmin K min Kdry

1K sat

= 1Kmin

+φ

K φ +K minK fluid

K min – K fluid

K dry =

K sat

φK min

K fluid

+ 1 – φ – Kmin

φKmin

K fluid

+K sat

K min

– 1 – φ

1. Begin with measured velocities and density

2. Extract Moduli from Velocities measured with fluid 1:

3. Transform the bulk modulus using Gassmann

where K1, K2 are dynamic rock moduli with fluids 1, 2

Fluid Substitution Recipe

K2

Kmin − K2

−K fl 2

φ K min − K fl 2( )=

K1

Kmin − K1

−K fl 1

φ Kmin − Kfl 1( )

VP−1, VS−1, ρ1

K1 = ρ1 VP−1

2 −4

3VS−1

2

µ1 = ρ1VS−1

2


where K1, K2 are dynamic rock moduli with fluids 1, 2

bulk moduli of fluids 1, 2

density of rock with fluids 1, 2

mineral modulus and porosity

density of fluids 1, 2

4. µ2 = µ1 shear modulus stays the same

5. Transform density

6. Reassemble the velocities VP =K2 +

4

3µ 2

ρ2

VS =µ2

ρ2

K fl 1,K fl 2

ρ1,ρ2

Kmin,φ

ρfl 1,ρfl 2

ρ2 = 1− φ( )ρmin +φρfl 2 = ρ1 +φ ρfl 2 − ρfl 1( )

2.25

2.3

2.35

2.4

2.45V

p (

km

/s)

sandstoneporosity = 30%

patchy

homogeneous


2.15

2.2

2.25

0 0.2 0.4 0.6 0.8 1

Vp

(k

m/s

)

Oil Saturation

homogeneous

KVoigt = SwKw + SoKo + SgKg

Summary of Mixing Rules

K


1 / KReuss = Sw / Kw + So / Ko + Sg / Kg

KBrie = K liquid − Kg( )1− Sg( )e

+ Kg

Brie, et al.SPE�30595

K

Why is the shear modulus unaffected by fluids in

Gassmann’s relations?


Imagine first an isotropic sample of rock with a hypothetical spherical pore. Under “pure shear” loading there is no volume change of the rock sample or the pore -- only shape changes. Since it is easy tochange the shape of a fluid, the rock stiffness is not affected by the type of fluid in the pore.

Why do the Gassmann relations

only work at low frequencies?

This crack decreases involume. Its pore pressurelocally increases if the fluid cannot flow out of the crack.

This crack increases in

+∆Pp

-∆Pp


Imagine an isotropic sample of rock with cracks at all orientations. Under “pure shear” loading there is no volume change of the rock sample or the pore space, because some cracks open while others close. If the frequency is too high, there is a tendency for local pore pressures to increase in some pores and decrease in others: hence the rock stiffness depends on the fluid compressibility.However, if the frequency is low enough, the fluid has time to flow and adjust: there is no net pore volume change and therefore the rock stiffness is independent of the fluids.

This crack increases involume. Its pore pressurelocally decreases if the fluidcannot flow into the crack.

Graphical Interpretation of Gassmann's Relations

1. Plot known effective modulus K, with initial fluid.

2. Compute change in fluid term:

3. Jump vertically up or down that number of contours.

Example: for quartz and water ~ 3 contours.

∆KmineralK fluid

Kmineral − K fluid

≈ ∆Kfluid

∆Kfluid

Kmineral

= 0.6

161Stanford Rock Physics Laboratory - Gary Mavko & Tapan Mukerji G.6

C

C‘

Approximate Gassmann Relationwhen Shear Velocity is Unknown

Normally, to apply Gassmann's relations, we need to know both Vp and Vs so that we can extract the bulk and shear moduli:

and then compute the change of bulk modulus with fluids using the usual expression:

K1 = ρ VP

2−

4

3VS

2

µ1 = ρVS

2

Ksat

Kmineral − Ksat

=Kdry

Kmineral − Kdry

+K fluid

φ Kmineral − Kfluid( )


The problem is that we usually don't know Vs.

One approach is to guess Vs, and then proceed.

We have also found that a reasonably good approximation to Gassmann is

where M is the P-wave modulus:

mineral sat mineral dry mineral fluid( )

M = ρVp

2

Msat

Mmineral− Msat

≈Mdry

Mmineral − Mdry

+Mfluid

φ Mmineral − Mfluid( )

Approximate Gassmann RelationWhen Shear Velocity is Unknown

satu

rate

d,

Fro

m A

ppro

xim

ate

Gassm

ann


Predictions of saturated rock Vp from dry rock Vp are virtually the same for the approximate and exact forms of Gassmann’s relations.

Vp-saturated, From Gassmann Vp-s

atu

rate

d,

Fro

m A

ppro

xim

ate

Gassm

ann

Gassmann's is a Low Frequency Relation


It is important to remember that Gassmann’s relations assume low frequencies.

Measured ultrasonic Vp in saturated rocks is almost always faster than

saturated Vp predicted from dry rock Vp using Gassmann. Data here are for

shaly sandstones (Han, 1986).

Water Flood Example: Pore PressureIncrease and Change From Oil to Brine

1250

Brine Flood into Oil

de

pth

(m

)

Pressure

oil to water

One typical depth point

(laboratory)


Calculated using Gassmann via dry lab data from Troll (Blangy, 1992). Virgin condition taken as low frequency, oil saturated at Peff=30 Mpa Pressure drop to Peff=10 MPa, then fluid substitution to brine.Koil = 1., Kbrine = 2.2

G.12

2 2.5

1300

1350

Vp (km/s)

de

pth

(m

)

oil to water

original oil

oil atincreased Pp

brine atincreased Pp • effect of pressure on frame

• effect of pressure on fluids

• frame+fluid: fluid substitution

Gas Flood Example: Pore PressureIncrease and Change From Oil to Gas

1250

1300

Brine Flood into Oil

dep

th (

m)

Pressure

oil to water



Calculated using Gassmann from dry lab data from Troll (Blangy, 1992). Virgin condition taken as low frequency, oil saturated at Peff=30 MPa. Pressure drop to Peff=10 MPa, then fluid substitution to gas.Koil = 1., Kbrine = 2.2

• effect of pressure on frame

• effect of pressure on fluids

• frame+fluid: fluid substitution

2 2.5

1300

1350

Vp (km/s)

dep

th (

m)

original oil

oil atincreased Pp

brine atincreased Pp

G.12

Brine Flood Example: Pore PressureDecrease and Change From Oil to Brine

1250

1300

Brine Flood with Pressure Decline

dep

th (

m)



Calculated using Gassmann from dry lab data from Troll (Blangy, 1992). Virgin condition taken as low frequency, oil saturated at Peff=25 MPa. Pore pressure drop to Peff=30 MPa, then fluid substitution to brine. Koil = 1., Kbrine = 2.2

1.8 2.4 3

1300

1350

Vp (km/s)

dep

th (

m)

original oil

oil atdecreased Pp

brine atdecreased Pp

frame effectdecreased Peff

Stiff, Turbidite Sand, Heavy Oil, Water Flood with Pp Increase

168Stanford Rock Physics Laboratory - Gary Mavko & Tapan Mukerji Stiff,deepwater sand, heavy oil (API20,GOR=15, T=75,Pp=18->25, Sw=.3->.8)

Fluid Substitution in Anisotropic Rocks: Brown and

Korringa’s Relations

where

effective elastic compliance tensor of dry rock

effective elastic compliance tensor of rock saturated with pore fluid

Sijkl

(dry)− Sijkl

(sat )=

Sijαα(dry) − Sijαα

0( )Sklαα(dry) − Sklαα

0( )Sααββ

(dry)− Sααββ

0( )+ βfl − β 0( )φ

Sijkl

(dry )

Sijkl

(sat )


effective elastic compliance tensor of mineral

compressibility of pore fluid

compressibility of mineral material =

porosity

This is analogous to Gassmann’s relations. To apply it, one must measure

enough velocities to extract the full tensor of elastic constants. Then invert these

for the compliances, and apply the relation as shown.

Sijkl

0

β fl

β0

φ

Sααββ

0

Challenge of Reactive Fluid Substitution

Images show changes in the rock frame after CO2 injection. Gassmann’s equations assume that the frame remains unchanged.


Noiriel, 2005

Laboratory Data vs. Gassmann’s Predicted Moduli

Pure Elastic Systems, Low Frequency, no rock-fluid reactions


Factors Influencing Reactions in Carbonates

a) The properties of the sediment itself

- Mineralogy

- Grain size

- Texture

- Permeability

- Surface area

2CaCO3 + Mg2+ CaMg(CO3)2 + Ca2+

[Mg2+] [CaCO ]2


- Surface area

b) The properties of the pore-fluid

- salinity

- Mg/Ca ratio

- SO4-2

- HCO3-

Kcd=[Mg2+]

[Ca2+]

[CaCO3]2

[CaMg(CO3)2]

Kcd=[Mg2+]

[Ca2+]


a) The properties of the sediment itself

- Mineralogy

- Grain size

- Texture

- Permeability

- Surface area


- Surface area

b) The properties of the pore-fluid

- salinity

- Mg/Ca ratio

- SO4-2

- HCO3-

Kcd= 0.67 - SPT conditions

The properties of the pore-fluid

- Mg/Ca ratio

- SO4-2

- HCO3-

Case A



CO2 injection Calcite dissolution

Case A

Case C

Gypsum dissolution dolomite dissolution

de-dolomitation driven by Gypsum dissolution

Case B

Transformation Anhydrite ⇒ Gypsum

Tucker et al., 1990



Dissolution and mineral transformation affect the elastic moduli in a way that opposes the

high-frequency dispersion mechanisms. Thus, Gassmann fluid substitution may either

overestimate or fit high-frequency, saturated velocities, depending on the balancing of

chemical processes against dispersion mechanisms.

Bounding Average Method (BAM)


Marion (1990) discovered a simple, semi-empirical way to solve the fluid substitution problem. The Hashin-Shtrikman bounds define the range of velocities possible for a given volume mix of two phases, either liquid or solid. The vertical position within the bounds, d/D, is a measure of the relative geometry of the two phases. For a given rock, the bounds can be computed for any two pore phases, 0 and 1. If we assume that d/D remains constant with a change of fluids, then a measured velocity with one fluid will determine d/D, which can be used to predict the velocity relative to the bounds for any other pore phase.

An Example of the Bam Method. The wax saturated

velocities are predicted from the dry rock velocities.

3600

3800

4000

4200

Massillon Light Sandstone

P-V

elo

cit

y (

m/s

)

measured parowax


Velocity in Massilon sandstone saturated with parowax. Data from Wang (1988). Wax saturated velocities were predicted using BAM, from Wang's measured velocities in the dry rock and in wax (from Marion, 1990)

G.82800

3000

3200

3400

0 20 40 60 80 100 120 140

P-V

elo

cit

y (

m/s

)

Temperature ( °°°°C)

BAMcalculatedparowax

measured dry


Velocity in dry and saturated Westerly granite. Data from Nur and Simmons (1969). Saturated velocities were predicted using BAM, from measured velocities in the dry rock (from Marion, 1990)

G.9

Ellipsoidal Models for Pore Deformation

Most deterministic models for effective moduli assume a specific

Recall the general expression for the dry rock modulus:

Gassmann’s relation is a transformation, allowing us to predict how measured velocities are perturbed by changing the pore fluid. Now wediscuss a different approach in which we try to model the moduli “from scratch”.

1K dry

= 1K mineral

+φ

K φ


Most deterministic models for effective moduli assume a specific idealized pore geometry in order to estimate the pore space compressibility:

The usual one is a 2-dimensional or 3-dimensional ellipsoidal inclusion or pore.

The quantity α = b/c is called the aspect ratio.

bc

σφ ∂

∂=

pore

pore

v

vK

11

Estimating the Dry Rock Modulus

An externally applied compression tends to narrow the crack, with the faces displacing toward each other.

bc

σ


Mathematicians have worked out in great detail the 3-D deformation field U, of an oblate spheroid (penny-shaped crack) under applied stress. For example, the displacement of the crack face is:

We can easily integrate to get the pore volume change and the dry modulus:

σ

1

K dry

= 1K mineral

+16 1 – ν 2

9 1 – 2ν1

K mineral

Nc 3

Vbulk

U(r) = σc

K mineral

4 1 – ν 2

3π 1 – 2ν1 – r

c2

Dry Rock Bulk Modulus

1

Kdry

=1

Kmineral

+16 1− v

2( )9 1− 2v( )

1

Kmineral

Nc 3

Vbulk

1

Kdry

=1

Kmineral

1 +16 1− v2( )9 1− 2v( )

Nc3

Vbulk

1

Kdry

=1

Kmineral

1 +16 1− v

2( )9 1− 2v( )

∈


"Crack density parameter"

Modulus depends directly on crack density. Crack geometry or stiffness must be specified to get a dependence on porosity.

∈=N

Vbulk

c3

≈φ

α

3

4π

Crack Density Parameter

In these and other theories we often encounter the quantity:

This is called the Crack Density Parameter, and has the interpretation of the

number of cracks per unit volume.

Example: 2 cracks per small cell. Each crack about 2/3 the length of a cell.

ε =Nc

3

Vbulk


L2c

v = L3

ε =c

L

3

≈ 0.07

Distribution of Aspect Ratios

Modulus depends on the number of cracks and their average lengths

An idealized ellipsoidal crack will close when the amount of deformation equals the original crack width:

1

Kdry

=1

Kmineral

+16 1− v2( )

9Kmineral 1− 2v( )Nc

3

Vbulk

U = b


solving gives:

We generally model rocks as having a distribution of cracks with different aspect ratios. As the pressure is increased, more and more of them close, causing the rock to become stiffer.

U = b

σclose ≈αKmineral

3π

4

1− 2v( )1− v

2( )≈αKmineral

Kuster and Toksöz (1974)

fmorulation based on long-

wavelength, first order scattering

theory (non self-consistent)

KKT

*− Km( )

Km +4

3µm

KKT

* +4

3µm

= xi

i =1

N

∑ Ki − Km( )Pmi

µKT

*− µm( )

µm +ζ m( )µKT

*+ζ m( )

= xi

i=1

N

∑ µ i − µm( )Qmi

ζ =µ

6

9K + 8µ( )K + 2µ( )


Self-Consistent Embedding Approximation

Walsh's expression for the moduli in terms of the pore compressibility is fairly general. However attempts to estimate the actual pore compressibility are often based on single, isolated pores.

1

Kdry

=1

Kmineral

+16 1− v2( )

9Kmineral 1− 2v( )Nc

3

Vbulk


The self-consistent approach uses a single porein a medium with the effective modulus.

Solving for Kdry gives:

Kdry Kmineral 9Kmineral 1− 2v( ) Vbulk

1

Kdry

=1

Kmineral

+16 1− v 2( )

9Kdry 1− 2v( )Nc

3

Vbulk

Kdry = Kmineral 1−16 1− v2( )9 1− 2v( )

Nc3

Vbulk

Self-Consistent Approximations

O’Connell and Budiansky (1974) model for medium with randomly oriented thin dry cracks

KSC

*

K=1−

16

9

1− vSC

*2

1− 2vSC

*

ε

µSC

*

µ= 1−

32

45

1− vSC

*( ) 5 − vSC

*( )2 − vSC

*( )ε

45 v − v*( ) 2 − v

*( )


K and µ are the bulk and shear moduli of the uncracked medium, ν is the Poisson’s ratio, and ε is the crack density parameter. The calculations are simplified by the approximation:

Assumes small aspect ratios (α → 0).

ε =45

16

v − vSC

*( ) 2 − vSC

*( )1− vSC

*2( )10v − 3vvSC

* − vSC

*( )

vSC

*≈ v 1−

16

9ε

Self-Consistent Approximations

Berryman’s (1980) model for N-phase composites

coupled equations solved by simultaneous iteration

xi K i − K*( )P

*i

i=1

N

∑ = 0

xi

i=1

N

∑ µ i − µ*( )Q*i= 0



Comparison of Han's (1986) sandstone data with models of idealized pore shapes. At high pressure (40-50 MPa), there seems to be some equivalent pore shape that is more compliant than any of the convex circular or spherical models.


Comparison of self-consistent elliptical crack models with carbonate data. The rocks with stiffer pore shapes are fit best by spherical pore models, while the rocks with thinner, more crack-like pores are fitbest by lower aspect ratio ellipsoids.

Data from Anselmetti and Eberli., 1997, in Carbonate Seismology, SEG.

Differential Effective Medium

ModelThe differential effective medium (DEM) theory models two-phase composites by incrementally adding inclusions of one phase (phase 2) to the matrix phase

d


(*2)

2

(*2)

2

(1 ) [ *( )] ( *) ( )

(1 ) [ *( )] ( *) ( )

dy K y K K P y

dy

dy y Q y

dyµ µ µ

− = −

− = −

Coupled differential equations with initial conditions K*(0) = K1 and µ*(0) = µ1, where

K1, µ1 = bulk and shear moduli of the initial host material (phase 1)

K2, µ2 = bulk and shear moduli of the incrementally added inclusions (phase 2)

y = concentration of phase 2


ModelThe differential effective medium (DEM) theory models two-phase composites by incrementally adding inclusions of one phase (phase 2) to the matrix phase


25

30

35

40

K (Bulk m

odulus, G

Pa)


Model


0 0.2 0.4 0.6 0.8 10

5

10

15

20

porosity

K (Bulk m

odulus, G

Pa)

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