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Biot Theory (Almost) ForDummies

Tad Patzek, Civil & Environmental Engineering, U.C. Berkeley

December 5, 2005, Seminar at the University of Houston


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Rock Classification

Isotropic Fractured Nonisotropic

Inhomogeneous

Homogeneous

Gassmann, Biot Biot?

Biot? Biot?

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Rock Types

Homogeneous, isotropic Heterogeneous, isotropic

GASSMANNs theory works only for the microscopically homogeneous rock (e.g., uniform

spheres)

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Rock Types

It is impossible to use equivalent homoge-

neous rock to explain heterogeneous rocks.

This is especially true for clay-rich rocks,

ZOBACK & BEYERLEE (1975), BERRYMAN,

(1992)

A new theory must be developed for

fractured, heterogeneous rocks

(In)homogeneous, anisotropic

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Porous Rock

Porous rock = Solid Skeleton + Pore Space

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Porous Rock Characterization

Partially saturated

Gas + Liquid

Unsaturated

Gas

Saturated

Liquid(s)

Pore Space

Gassmann, Biot? Gassmann, Biot

Bulk density

=mass of solid skeleton + mass of pore space fluids

bulk volume of rock

= (1 )s + f = skeleton + f

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Compressibility Measurements

The vertical stress, S1, is applied to a hol-

low piston. The tube in the piston is used

to regulate the pore pressure, p. The lat-

eral stresses, S2 = S3, are applied to

the copper-jacketed specimen by injecting oil

through the side tube. The confining pres-

sure is defined as

pc =

=

1

3 (S1 + S2 + S3)

The jacketed or drained triaxial rock com-

pressibility:

:= 1

V

V

pc

p,T

=1

K

p

S1S1

S2 =

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Compressibility Measurements

The unjacketed triaxial rock compressibility

measurement. The confining pressure,

pc = =1

3(S1 + S2 + S3),

is applied to all sides of the sample. The

tube in the piston is used to regulate the porepressure, p. Both the confining pressure and

the fluid pressure are changed at the same

time, so that their difference, pd = pc p,

remains constant.

s := 1

V

V

p

pd,T

=1

Ks

p

S1S1

S2 =

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Porous Rock Compressibilities

We can measure the following three compressibilities:

:= 1VVpc

p,T

= 1K

0@Biot : + p=0

1K1A

s := 1

V

V

p

pd,T

=1

Ks

Biot : +

p

=0

1

H

!

:= 1

V

V

p

pd,T

=1

K

Biot : +

p

=0

1

R= S

!

where V is the bulk volume of the sample, V is the pore space volume

A fourth compressibility may be defined as

p := 1

V

V

pc

p,T

=1

Kp

0

@Biot : +

p=0

1

H

1

Abut it depends on the porosity and the first two compressibilities above

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Porous Rock

At the reference state, we imagine a colored rock grain sample, in blue, filled with

colored water, in red. First, we remove the red water into a beaker and fill the pore space

with ordinary water. Second, we change the stress on the solid and the pore pressure,

and measure the new pore volume, V. Third, we measure the new red water volume

under the new pore pressure, Vf. In general, the new pore volume and water volume willnot be equal to each other, and water will have to flow in/out of the blue rock volume.

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Biots Increment of Fluid Mass

pc

V, V,mf

Final

State

p

f

V0, V0 mf0, Vf0, f0

Initial

State

pc = 0

p = 0

Initially Vf0 = V0; the pore space is fully saturated with red fluid

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Biots Increment of Fluid Mass At the final state

mf = mf0VVf

After Biot, I will introduce the increment of fluid mass perunit initial bulk volume V0, normalized by the initial fluiddensity mf0/Vf0:

:=mf/f0

V0=

Vf0V0

VVf

=

Vf0V0

VVf0 VfV0V2f0

=1

V0(V Vf) = 0( f)

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Talk Outline. . .

Refresher of Biots static poroelasticity model

Biots dynamic poroelastic model from thenon-equilibrium filtration theory

Low frequency reflections from a plane interfacebetween an elastic and an elastic fluid-saturated

layers

Different asymptotic regimes of the low-frequencyreflections

Conclusions

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Biot Theory. . .The isotropic, permeable porous rock, and thepore-filling fluid are in mechanical equilibrium

The stress is positive when it is tensile

The fluid pressure is positive

The state of rock and the fluid is described by the

total stress on the bulk material, ij, and the fluidpressure field p (ij is the total force in direction i,

acting on the surface element whose normal is indirection j)

Following BIOT, in one spatial dimension, the smallfluctuations of the total stress tensor, , and of thefluid pressure, p, will be called and p

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Biot Theory. . .

V

V0=

1

K +

1

Hp volumetric strain

mfV0f0

= 1H

+ 1R

p fluid volume per unit volume

p=0 1

Kdrained material compressibility

p=0

=

p

=0

1

H

poroelastic expansion coefficient

p

=0

1

R= S unconstrained specific storage

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Biot Theory. . .

p

=0

B =R

HSKEMPTONs coefficient

p

=0

1

M= S constrained specific storage

S = S KH2

K

H

BIOT-WILLIS coefficient

= +1

Mp

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Biot Theory. . .

The poroelastic expansion coefficient 1/H has noanalog in elasticity

It describes how much a change of pore pressurealso changes the bulk volume, while the appliedstress is held constant

1/H, and two other constants, K drained bulkmodulus, and the unconstrained storage coefficientS, completely describe the linear, poroelasticresponse to volumetric deformation

Other constants, such as SKEMPTONs coefficient, orBIOT-WILLIS coefficient can be derived from thethree fundamental BIOT constants

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Definitions. . .p pressure increment, Pa stress increment, Pau displacement of skeleton grains, m

ut velocity of displacement of skeleton grains, m/sw superficial displacement of fluid relative to solid, mW wt Darcy velocity of fluid relative to solid, m/s

isothermal compressibility, Pa

1

(1 )g, dry bulk density, kgm3

b (1 )g + f, bulk density, kgm3

V /V , increment of volumetric strain

small parameter in series expansions mf/f0/V0, increment of fluid content per unit volume

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The Bulk Momentum Balance. . .

d

dt

V

but

solid+liquidmomentum+ fW

relative liquidmomentum

dV

=

V

totalstressn dA +

V

Fb

bodyforcedV

Small perturbation from equilibrium

Incremental body force is zero

t but + fW =

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The Bulk Momentum Balance. . .

Almost incompressible grains ( 1)

Poroelastic effective stress , and Terzaghi effective

stress are equal1D normal deformations, = xx

t but + fW = xxx =

xx

x p

x

xx Ku

x=

1

u

x

K is the drained bulk modulus

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Force Balance. . .

The second Newtons law for the bulk solid is

bttu + ftW =1

xxu xp (1)

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Darcys Law. . .

Consider steady state, single-phase flow of analmost incompressible fluid

The superficial fluid velocity relative to the solid

W =

x

In horizontal flow, viewed from a non-inertialcoordinate system moving with the solid, thedifferential of the flow potential is

d

Mechanical

energy

= dp

Viscousdissipation+ fttu dx

Inertial force12.05.05 p.21/


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Extended Darcys Law. . .In time-dependent, single-phase flow, we can write

W

t

Wfuture W

where Wfuture is a future value of Darcys velocity, and is a characteristic time of transition

At constant position x, and constant value of Wfuture,we can integrate

Wfuture W expt Therefore, is a characteristic relaxation time for

transient flow, e.g., JAMES C. MAXWELL, 1867

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Extended Darcys Law. . .

In time-dependent, single-phase flow, we now write

Wfuture W +W

t + =

This is the essence of ALISHAEVs, and BARENBLATT& VINNICHENKOs extension of DARCYs law

Dimensional analysis suggests that

= fF(/L2)

where L is the characteristic length scale of REV

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Extended Darcys Law. . .

We characterize the dynamics of horizontal fluid flow in anon-inertial coordinate system as follows

W + Wt

=

px f

2

ut2

(2)

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Mass Balances & Isothermal EOSs. .Slightly compressible fluid

( f)

t=

x

fW +

f

u

t

d f

f= fdp

Almost incompressible solid grains

[ g(1 )]

t=

x g(1 )u

t 1

gd g

= gsdx + gfdp

gs

and gf

f

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Reduced Mass Balances. . .

With almost incompressible grains, the bulkdeformation occurs only through the porosity change

With some algebra, the mass balance equationsreduce to

2

uxt

+ fpt

= Wx

(3)

Note that we now have three unknowns u, p and W,and three balance equations: (1) Force balance ofbulk solid, (2) Force balance in viscous-dominatedfluid flow, and (3) Combined mass balance of fluid

and solid

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The Governing Equations. . .

For a linearly compressible rock skeleton and fluid, andsmall perturbations from thermodynamic equilibrium:

Force balance of bulk material

bttu + ftW = 1

xxu xp (1)

Force balance of viscous fluid

W + tW =

xp fttu

(2)

F/S mass balances + EOSs

ftp = x (W + tu) (3)

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Biots Theory. . .

We define the superficial fluid displacement

W := tw (4)

and insert it into mass balance equation (3)

ftp = xt(w + u)

By integration in t and differentiation in x, we obtain

xp =

1

fxx (u + w) (5)

Now we substitute the displacement (4) and the finalresult (5) into the governing equations

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Biots Theory. . .Our equations

b2u

t2

+ f2w

t2

= 1

+1

f2u

x2

+1

f

2w

x2

f2u

t2+

2w

t2=

1

f

2u

x2+

1

f

2w

x2

w

t

Biots 1962 equations

2

t2 bu + fw =

x A11u

x + M11

w

x2

t2 fu + mw

=

x

M11

u

x+ M

w

x

w

t

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Biots Theory. . .We have assumed an isotropic porous medium andincompressible grains

The Biot-Willis coefficient = K/H

1The undrained bulk modulus Ku = K+ Kf/

The Biot coefficients are then constant and equal to

A11 = Ku 1

+

1

fand M11 = M = KuB

1

f

where B = R/H is Skemptons coefficient, 1/H beingthe poroelastic expansion coefficient, and 1/R theunconstrained specific storage coefficient

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Biots Theory. . .The dynamic coupling coefficient in Biots theory, m,is equal to the inverse fluid mobility, /

The dynamic coupling coefficient is often expressed

through the tortuosity factor T: m = T f/

Hence, for the tortuosity and relaxation time, weobtain the following relationship:

T =

f

Inv. kinematicmobility

or = T f

(6)

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