effective viscosity and permeability of porous media

Colloids and Surfaces

A: Physicochemical and Engineering Aspects 192 (2001) 363–375

Effective viscosity and permeability of porous media

Victor M. Starov *, Vjacheslav G. ZhdanovDepartment of Chemical Engineering, Loughborough Uni�ersity, Loughborough, Leicestershire, LE11 3TU, UK

Abstract

The effective properties of porous media composed of equally sized spherical particles are investigated. Flow insideporous media is modelled using Brinkman’s equations, which include two semi-empirical coefficients: an effectiveviscosity and a resistance coefficient (1/permeability). The differential method is used to deduce the dependencies ofthese coefficients on both porosity and the particle size. The deduced dependency of permeability on porosity is foundto agree reasonably well with known dependencies obtained by computer simulations of flow in random porousmedia. © 2001 Elsevier Science B.V. All rights reserved.

Keywords: Permeability; Porous bodies; Modelling

Nomenclature

pressurePu� velocityUb constant velocity far from the particle

hydrodynamic resistanceKN number of particlesL characteristic length scale

volumeVa radius of particles

Z(�)=KBa2

�0A� function defined by Eq. (8)function defined by Eq. (8)AK

distance from the fixed particle centreRx� Cartesian co-ordinates

ni=xi

r, nj=

xj

r, nk=

xk

rF force

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* Corresponding author. Tel.: +44-01509-222508; fax: +44-01509-223923.E-mail addresses: [email protected] (V.M. Starov), [email protected] (V.G. Zhdanov).

0927-7757/01/$ - see front matter © 2001 Elsevier Science B.V. All rights reserved.

PII: S0927 -7757 (01 )00737 -3

V.M. Staro�, V.G. Zhdano� / Colloids and Surfaces A: Physicochem. Eng. Aspects 192 (2001) 363–375364

relative permeabilityK=1/Zabsolute value of the velocityU

G, Q, N, M, C integration constants

Greek letters� volume concentration of particles�=1−� porosity

symmetric unit tensor�ij

small value�

Laplacian�small concentration of particles in PM4��

� viscosityanti-symmetric unit tensor� ijk

constant tensor�ij

� stress tensorspherical surface�

SubscriptsDarcyD

B Brinkmanconcentrated suspensionS

L liquid phaseparticlesp

0 pure liquidindexes with values 1, 2, 3i, j, k, m, n

1. Introduction

Viscous flow of a liquid in a porous medium isfrequently modelled using Darcy’s equations:

grad p= −KD(�)u� (1)

where KD is the porous medium hydrodynamicresistance (1/KD is the porous medium permeabil-ity); � is the particle volume concentration (�=1−� is porosity); p and u� are the pressure andaverage velocity, respectively. Eq. (1) should alsobe coupled with the incompressibility equation

di� u� =0. (2)

This model suffers from a well-known disad-vantage that it is impossible to match flows insideand outside the porous medium. To overcome this

difficulty two different approaches have been sug-gested: a slippage at the boundary between flowsinside and outside the porous medium approach[1] and Brinkman’s equations [2]. In the formerapproach, an empirical slippage coefficient isdefined, whereas in Brinkman’s approach two,physically meaningful, coefficients are introduced:an effective viscosity, �B (which is obviously dif-ferent from the liquid viscosity �0) and the porousmedium resistance coefficient, KB. Brinkman’sequations have the following form

grad p=�B(�)�u� −KB(�)u� . (3)

Eq. (2) should also be considered.Let LB be a characteristic length scale for Eq.

(3). This length scale can be determined in thefollowing manner: both terms in the right-hand

V.M. Staro�, V.G. Zhdano� / Colloids and Surfaces A: Physicochem. Eng. Aspects 192 (2001) 363–375 365

side of Eq. (3) have to be of the same order ofmagnitude. This requirement gives: �

B/LB

2 =KB,or LB=��B/KB,.where LB is referred to as the‘Brinkman’s length’. If the distance L in thedepth of the porous medium and remote from theporous medium-liquid interface satisfiesthe following inequality L�LB, then the viscosityterm in the right hand side of Eq. (3) can beneglected and Brinkman’s equations coincidewith Darcy’s equations (1). This considerationshows that hydrodynamic resistance in both equa-tions are equal; that is, KD=KB. Kozeny–Car-man and Happel–Brenner equations [3] arefrequently used to describe the dependency ofhydrodynamic resistance on porosity and particleradius.

The porous medium resistance, KB, vanishes intwo cases: either when concentration of the parti-cles is zero (i.e. for a pure liquid, when �B=�0,where �0 is the pure liquid viscosity) or when theparticles move with the same average velocity asthe liquid (concentrated suspension). In the lattercase �B=�S, where �S is the viscosity of theconcentrated suspension. The last observationgives a very important hint, namely, the depen-dency of �B on particle concentration should beclose to the viscosity of concentrated suspensionwith the same particle concentration in the casewhen KB is small enough. It is well known thatthe dependency of �S on particle volume fractionis an increasing function of particle volume frac-tion, �. Consequently, it is reasonable to assumethat �B dependency should also be an increasingfunction of particle volume fraction. In contrastto this assumption, a decrease of effective viscos-ity with particle volume fraction � has been pre-dicted [4].

2. Calculation of viscosity and hydrodynamicresistance in Brinkman’s equations using thedifferential model

Let us define and consider four different media(see Fig. 1) as follows:

(1) Porous medium 1, denoted as PM1, with avolume fraction of solid particles, � :

�=Vp

Vl+Vp

(4)

where Vl and Vp are the volumes of the liquid andparticles, respectively. As the particles are allspherical, Vp=4/3 �a3n, where a and n are, re-spectively, the particle radius and the number ofparticles in this volume. The liquid flow in PM1 isassumed to obey Eqs. (2) and (3). The dependen-cies of effective viscosity �B(�) and the resistancecoefficient KB(�) on � have yet to be determined.

(2) A homogeneous Brinkman’s medium, de-noted as BM2, where the liquid flow is describedby the same Eqs. (2) and (3) with exactly the sameviscosity �B(�), resistance coefficient KB(�) andvolume as for PM1. In contrast to PM1, BM2 issupposed to be a homogeneous medium withoutany solid particles at all.

Let a small number of particles, �n, be addedto both porous medium 1 (PM1) and Brinkman’smedium 2 (BM2). This addition results in a newporous medium, denoted as PM3 and a porousmedium from BM2, denoted as PM4 (Fig. 1).Media PM3 and PM4 have a greater volume,Vl+Vp+�Vp, as compared with PM1 and BM2,where �Vp=4/3 �a3�n.

Fig. 1. Differential method. Porous medium 1 (PM1) withparticle concentration � ; viscosity �B(�), resistance coefficientKB(�). Homogeneous Brinkman medium 2 (BM2) with thesame viscosity, resistance coefficient, and volume as PM1.Porous medium 3 (PM3) with added particles, concentration�+�� ; viscosity �B(�+��), resistance coefficient KB(�+��).Porous medium 4 (PM4) with the same added amount of newparticles as in PM3; small particle concentration, ��.


In the development that follows only smallquantities of the first order are retained.

The particle volume fraction in PM3, �+��, is

�+��=Vp+�Vp

Vl+Vp+�Vp

=Vp�

1+�Vp

Vp

�(Vl+Vp)

�1+

�Vp

Vl+Vp

��

Vp

Vl+Vp

�1+

�Vp

Vp

−�Vp

Vl+Vp

�=�+ (1−�)��,

(5)

where ��= (�Vp)/(Vl+Vp) is the particle volumefraction inside the PM4.

The liquid flow in PM3 is described by Eqs. (2)and (3) but at concentration �+�� instead of �,i.e. Eq. (3) transforms into

grad p=�B(�+��)�u� −KB(�+��) u� . (6)

The particle concentration in PM4, ��, accord-ing to Eq. (5) can be expressed as

��=��

1−�. (7)

It is recalled that BM2 is supposed not to haveany particles at all and to be a homogeneousmedium. Addition of a small number of particleswith concentration �� results in a small changeboth viscosity, ��B, and resistance coefficient,�KB, in PM4. The latter small changes are obvi-ously proportional to ��, i.e. we can write

��B=A�(�B, KB)��, �KB=AK(�B, KB)�� (8)

where A�(�B, KB) and AK(�B, KB) are two un-known functions. The main objective becomesone of calculating these two unknown functions.

Let us suppose that these two functions havebeen already determined. In this case the liquidflow in PM4 would be described by the followingequations

grad p= [�B(�)+A�(�B,KB)�� ]�u�

− [KB(�)+AK(�B, KB)�� ]u� . (9)

The main assumption of the differential methodis as follows: The �iscosity and resistance coeffi-cient in porous medium 3 and porous medium 4 areequal. Using this condition from Eqs. (6) and (9)we obtain:

�B(�+��)=�B(�)+A�(�B, KB)��

KB(�+��)=KB(�)+AK(�B, KB)��.

A simple rearrangement of the latter equationsusing Eq. (7) results in the following system oftwo differential equations:

d�B

d�=

A�(�B, KB)(1−�)

dKB

d�=

AK(�B, KB)(1−�)

(10)

with the obvious boundary conditions

�B(0)=�0, KB(0)=0. (11)

This system of two ordinary differential Eqs.(10) with two boundary conditions (11) can besolved directly.

3. Determination of two unknown functionsA�(�B, KB) and AK(�B, KB)

According to its definition, BM2 is a homoge-neous medium and Eqs. (2) and (3) describe theliquid flow in this medium. After addition of asmall number of particles, �n, to BM2 it trans-forms into PM4. The most important observationis that the volume fraction of particles in PM4,��, is very small and, hence, particles are locatedfar from each other and their hydrodynamic inter-action can be neglected. In other words, in PM4one may consider flow in homogeneous mediumaround an isolated particle.

Let us consider an isolated particle placed inBM2 subjected to two different flow regimes: (a)homogeneous flow with constant velocity U, and(b) extensional flow (in both cases remote fromthe particle). Consideration of both flows is closeto that in [5]. Consideration of flow (a) gives usthe unknown function AK(�B, KB). Considerationof flow under condition (b) above gives a viscositychange, which determines the unknown functionA�(�B, KB). Detailed derivations are presented inAppendix A (Eq. (A16)) and Appendix B (Eq.(B23)), respectively. Substitution of these depen-dencies into Eq. (10) results in


dKB

d�=

9�B

2a2

1+a�KB

�B

+a2 KB

9�B

1−�(12)

d�B

d�=

KBa2

6

a�KB

�B

+6+15

a�KB

�B

+15

a2 KB

�B

(1−�)�

1+a�KB

�B

� (13)

with boundary conditions (11).Let us introduce two dimensionless functions

Z(�)=KBa2/�0, �(�)=�B/�0. Using these func-tions the system (12–13) with boundary condi-tions (11) can be rewritten as

dZd�

=9�

2

1+�Z

�+

19

Z�

1−�(14)

d�

d�=

Z6

�Z�

+6+15�Z

�

+15Z�

(1−�)�

1+�Z

�

� (15)

with boundary conditions

Z(0)=0

�(0)=1. (16)

4. Discussion

At low particle fraction, �, the dimensionlessresistance coefficient Z(�) becomes small and Eqs.(14) and (15) can be rewritten as

dZd�

=9�

21

1−�(17)

d�

d�=

52

�

(1−�). (18)

The latter system of equations with boundaryconditions (11) has the following solutions (revert-ing to dimensional variables)

�B(�)=�0(1−�)−5/2 (19)

KB(�)=95

�0

a2((1−�)−5/2−1). (20)

Eqs. (19) and (20) determine the dependence ofeffective viscosity and resistance coefficient ofBrinkman’s medium on the volume fraction ofparticles, �, in the case of low particle volumefraction. Both dependencies are increasing func-tions of particle concentration, �.

Dependency (19) coincides with the dependencyof the viscosity of concentrated suspensions [6] asit has been predicted in Section 1.

It is easy to see that Eq. (20) can be rewritten as

KB(�)=9

5a2[�B(�)−�0]. (21)

The latter equation gives the dependency of thehydrodynamic resistance on the particle volumefraction, �, if � is small. According to Eq. (21) thehydrodynamic resistance becomes negative if theeffective viscosity is lower than the liquid viscos-ity, which is obviously impossible.

Let us introduce the dimensionless permeabilityk=1/Z=�0/KBa2and investigate its dependenceon porosity �.

According to Eq. (21) at high porosity, ��1:

k(�)=59

�5/2

1−�5/2, ��1. (22)

This dependency is shown by curve 4 in Fig. 2. Atlow porosity, ��0, according to Appendix C,k=1/t�

2 � and �=C��−4.130, hence:

k=�4.130

t�2 C�

. (23)

This dependency is shown by curve 5 in Fig. 2.Curve 3 in the same figure represents the perme-ability dependency on porosity calculated accord-ing to Eqs. (14) and (15). Curves 1 and 2 in Fig.2 give the permeability calculated according toHappel–Brenner and Kozeny–Carman equations[3], respectively. Points and squares show perme-ability obtained by direct computer simulations ofliquid flow in a random three-dimensional [7] anda two-dimensional porous medium [8], corre-spondingly. In both papers [7] and [8] simulatedporous medium consist of penetrating particles,which form clusters at high concentrations. It isrecalled that according to the model adopted herethat particles can not penetrate into each otherand do not form clusters. It appears that mutual


Fig. 2. Dependence of relative permeability, k=�0/KBa2 of theporous medium on porosity �=1−�. (1) — , according toHappel–Brener equation [3], (2) -----, according to Kozeny–Carman equation [3], (3) — , calculated according to Eqs. (14)and (15), (4) – – – , high porosity asymptote, ��1, accordingto Eq. (22), (5) �–�–�, low porosity asymptote, ��0,according to Eq. (23), �, from Cancelliere et al. [8], �, fromKoponen et al. [7].

Acknowledgements

This work was supported by the Royal Society(grant ESEP/JP/JEB/11159). Authors would liketo express their gratitude to Dr M Meireles foruseful discussions.

Appendix A. Uniform flow

Force applied to a single spherical particle inBrinkman’s medium and calculation of the func-tion AK(�B, KB).

Let us consider a single spherical particle withradius a placed in a uniform flow inside homoge-neous Brinkman’s homogeneous medium, de-noted previously as BM2. Far from the particlesthe following boundary conditions should besatisfied

u� �r��=Ub = (0,0,U) (A1)

that is, U1=U2=0, U3=U.In the case under consideration a force similar

to the Stock’s force is exerted on the particle andour aim is to calculate this force.

The flow obeys Brinkman’s Eq. (3) and theincompressibility condition (2).

If we apply a rot operation to Eq. (3), then weobtain

particle penetration and clustering is more impor-tant in the case of three-dimensional porous me-dia [7]. This results in a close agreement of ourprediction (curve 3, Fig. 2) with permeability de-pendency in the case of two-dimensional porousmedium (squares [8]).

The relative viscosity at high porosity (i.e. at��1) according to Eq. (19) can be rewritten as

log10 �= −2.5 log10 �, ��1 (24)

The latter dependency is shown as asymptote 3in Fig. 3. At low porosity, ��0, according toAppendix C the relative viscosity is

log10 �= −4.130 log10 �−0.2249, ��0. (25)

The latter dependency is shown by asymptote 3in Fig. 3. Curve 1 represents the relative viscositydependence on porosity calculated according toEqs. (14) and (15).

Fig. 3. Dependence of relative viscosity �=�B/�0 of porousmedium on the porosity �=1−� (log– log co-ordinates). 1,Solution of Eqs. (14) and (15); 2, high porosity asymptote,��1, according to Eq. (24); 3, low porosity asymptote, ��0,according to Eq. (25).


�B�rot u� −KBrot u� =0. (A2)

In the same way as in ref. [5] the velocity vectorcan be presented in the following form

u� =Ub +rot rot( f(r)Ub ) (A3)

where f(r) is a new unknown function, r= �r� � isthe radial distance from the origin (the origin isselected in the particle centre). Hence,

rotu� =rot rot rot( fUb )

=grad di� [rot( fUb )]−�[rot( fUb )]

= −�[rot( fUb )]

or

rotiu� = �3

j,k=1

� ijk�kuj= − �3

j,k=1

� ijkUk��j f=

−�f � �3

j,k=1

� ijkUk

xj

r

where � ijk is the anti-symmetric unit tensor(�123=1, � ijk= − � jik= − � ikj).

Substitution of the latter expression in Eq. (A2)gives the following equation for determination ofunknown function f(r):

0=��2f �(r)−

KB

�B

�f �(r)� �

3

j,k=1

� ijkUk

xj

r. (A4)

It is obvious, that �3j,k=1� ijkUk

xj

r�0 everywhere,

hence Eq. (A4) yields

0=��2f �−

KB

�B

�f ��

. (A5)

After substitution f=/r, where (r) is a newunknown function, and integration Eq. (A5) canbe written in the following form

’’’’(r)r

−KB

�B

’’(r)r

=const. (A6)

An integration constant should be set to zerobecause the fluid velocity remains finite far fromthe particle. Eq. (A6) has the following solution:

=G+Qr+M exp��KB

�B

r�

+N

exp−��KB

�B

r�

(A7)

where G, Q, M and N are integration constants.Taking into account boundary condition (A1) thelatter equation gives:

f=Gr+Q+N

�B

KB

�exp−��KB

�B

r�

r−

1r+�KB

�B

.

(A8)

Eq. (A3) can be rewritten as

ui=Ui+grad di� fUi−�fUi, i=1,2,3. (A9)

From Eqs. (A8) and (A9) we can conclude

ui=Ub i+� 3

r3 ni �3

j=1

njUb j−1r3 Ub i

�G+N

�B

KB

�KB

�B

exp−��KB

�B

r�

rni �

3

j=1

njUb j−KB

�B

exp−��KB

�B

r�

rUb i

−�KB

�B

exp−��KB

�B

r�

r2 Ub i+3�KB

�B

exp−��KB

�B

r�

r2 ni �3

j=1

nj Ub j

−exp−

��KB

�B

r�

r3 Ub i+Ub i

r3 +3 exp−

��KB

�B

r�

r3 ni �3

j=1

njUb j−3r3 ni �

3

j=1

njUb j�

. (A10)


The latter equations and non-slip conditions atthe particle surface r=a determine value of con-stants G and N :

N=32

exp��KB

�B

a�

a, G=

−a3

2

�1+

3a��B

KB

+3a2

�B

KB

−3 exp

��KB

�B

a�

a2

�B

KB

.

(A11)All functions under consideration below (compo-

nents of velocity vector, viscose stress tensor andpressure) are presented as F=F (0)+F (1), whereF (0) is caused by the flow far from the particle andF (1) is caused by the presence of the particle. Belowwe are interested only in values caused by theparticle and, hence, in first order terms. Superscript(1) is omitted.

The pressure can be calculated after substitutionin Eq. (3) the velocity expression via function f(r)(Eq (A3)):

grad p=grad(�B�di� fUb −KB di� fUb )

− (�BUb �2f−KBUb �f)

=grad (�B�di� fUb −KB di� fUb ).

And after integration:

p=�B�di� fUb −KBdi� fUb .

Using Eq. (A10) the latter equation gives:

p=�

KB

Gr3−�B

Nr3

� �3

j=1

xjUb j (A12)

where constants G and N are given by Eq. (A11).Let us rewrite the velocity vector (A10) and

pressure (A12) using spherical co-ordinate systemwith polar axe along the direction of the velocityvector, Ub . This gives:

Components of the stress tensor at the particlesurface, r=a can be deduced using the latterexpressions as

�rr=�

KB

a2+

32

��BKB+32

�B

a�

U cos

�r= −�3

2��BKB+

32

�B

a�

U sin (A13)

where �rr, �r are components of the stress tensorin polar co-ordinate system. The force, FB, exertedto the spherical particle according to [3] can becalculated as:

FB=

�(�rr cos −�r sin ) d�, (A14)

where � is the particle surface, d� is the surfaceelement. Eqs. (A13) and (A14) yield

FB=6��BaU�

1+a�KB

�B

+19

a2 KB

�B

�. (A15)

Additional resistance cause by the presence of �nparticles inside Brinkman’s medium 4 is

�KB=FB�n

V=

3FB��

4�a3 =3FB��

4�a3(1−�).

After substitution of Eq. (A15) into the latterequation the result is

�KB=3FB��

4�a3(1−�)

=6��BaU�

1+a�KB

�B

+19

a2 KB

�B

� 3��

4�a3(1−�)

=9

2a2(1−�)�BU

�1+a

�KB

�B

+19

a2 KB

�B

��.

The latter equation shows that

u� r=U cos

�N

�B

KB

�2

exp−��KB

�B

r�

r2

�KB

�B

+2exp−

��KB

�B

r�

r3 −2r3

+

2Gr3

u� =−U sin

�N

�B

KB

�−�KB

�B

exp−��KB

�B

r�

r2 −exp−

��KB

�B

r�

r3 +1r3−

KB

�B

exp−��KB

�B

r�

r

−

Gr3

p=

�KB

Gr2−�B

Nr2

�U cos .


AK=9�B

2a2

�1+a

�KB

�B

+19

a2 KB

�B

�. (A16)

Appendix B. Extensional flow, calculation of func-tion A�(�B, KB).

An extensional flow in the Brinkman’s medium2 around a single sphere of radius a is consideredbelow. Origin coincides with the particle centre.Far from the particle the flow is given by

�3

j=1

�ijxj �r��=Ub i, (B1)

where �ij is a symmetric tensor, �3k=1 �kk=0.

Brinkman’s Eq. (3) and the equation of incom-pressibility (2) describe the flow around thesphere.On the particle surface non-slip conditionsare

u� �r=a=0. (B2)

In the same way as in Appendix A the velocityvector can be presented in the following form

u� =�� x� +rot [rot(��f)], (B3)

where an unknown function f(r) depends on ra-dial distance r only, �� x� �i=�3

j=1 �ijxj. Applyingrot to the both sides of Eq. (3) yields

�B�rot u� −KBrot u� =0. (B4)

From Eq. (B3) we find

rot u� =rot [rot [rot(��f)]]

=grad [di� [rot(��f)]]−�rot(��f)

= −�rot(��f).

Substitution of the latter equation into Eq. (B4)gives

−�B�2rot(��f)+KB�rot(��f)=0. (B5)

Using the antisymetric unit tensor � ijk (�123=1,� ijk= − � jik= − � ikj) in Eq. (B5) we arrive (inthe same way as in Appendix A) to the followingequation for f(r) determination:

�B�2f �−KB�f �=0,, (B6)

where � means a derivative in respect to r= �r� �.After substitution f=/r (where (r) is a newunknown function) we obtain from Eq. (B6):�

�B

’’’’r

−KB� ’’r��

=0. (B7)

Eq. (B7) has the following solution

f=(r)

r=

Qr

+G+Mexp

��KB

�B

r�

r

+Nexp−

��KB

�B

r�

r+Ar4+Br2

where Q, G, M and N are integration constants.In order to satisfy boundary conditions far fromthe particle (B1) constants M, A and be B must beset to zero and constant G can be easily deter-mined, this gives:

f=Qr

+N�exp−

��KB

�B

r�

r−

1r+�KB

�B

�B

KB

. (B8)

Eq. (B3) can be rewritten as

u=��x� +grad [di�(��f)]−��f (B9)

Using latter two equations we obtain

u� k=�−15

Qr4+N

�−�KB

�B

exp−��KB

�B

r�

r−6

exp−��KB

�B

r�

r2 −15��B

KB

exp−��KB

�B

r�

r3 −15�B

KB

exp−��KB

�B

r�

r4 +15�B

KB

1r4

��3

i, j=1�ijninjnk

+�

r+6Qr4 +N

��KB

�B

exp−��KB

�B

r�

r+3

exp−��KB

�B

r�

r2 +6��B

KB

exp−��KB

�B

r�

r3 +6�B

KB

exp−��KB

�B

r�

r4 −6�B

KB

1r4

��3

i=1�ki ni (B10)


Non-slip conditions at the particle surface resultin the following values of integration constants Nand Q

Now function f(r) is completely determined. Ac-cording to Eqs. (3) and (B9) the pressure gradientcan be expressed as

�p=grad di� (�B�� f−KB�� f)−KB�� x.

After integration of the latter equation

p= −12

KB �3

i, j=1

�ijxixj+3(N�B−QKB)�3

i, j=1

�ijxixj

r5

(B12)

Using definition of the stress tensor�ij= −p�ij+

�B��u� i

�xj

+�u� j�xi

�and Eqs. (B10)– (B12) gives the fol-

lowing expression for the viscos stress tensor:

Let us calculate a correction to the viscous stresstensor at low concentration of particles. For anyfunction g we introduce an average value as

�g�=1/V Vg dV, where V is the volume of thebig sphere with centre coincides with the particlecentre. The volume V does not include any otherparticles (low particles concentration). Integrationof the viscous stress tensor over the volume Vresults in

��ij�= −�p��ij+�B��u� j

�xi

�+��u� i

�xj

��+

1V

V

��ij+p�ij−�B

��u� i�xj

+�u� j�xi

��dV

(B14)

The expression under the integral in the right

N= −53

exp��KB

�B

a�

a3

1+�KB

�B

a

Q=a5

9�

1+�KB

�B

a��

a�KB

�B

+6+15��B

KB

1a+15

�B

KB

1a2−15

�B

KB

exp��B

KB

a�

a2

, (B11)

�km=�1

2KBr2−

3(N�B−QKB)r3

n�km �

3

i, j=1�ijninj

+�B

�−

30r5 Q+N exp−

��KB

�B

r��

−�KB

�B

2r2−

12r3 −

��B

KB

30r4 −

�B

KB

30r5 +30

�B

KB

exp��KB

�B

r�

r5

��km �

3

i, j=1�ijninj

+�B�210Q

r5 +N exp−��KB

�B

r��KB

�B

2r+20

�KB

�B

1r2+

90r3 +

210r4

��B

KB

+210r5

�B

KB

−210r5

�B

KB

exp��KB

�B

��n �3

i, j=1�ijninjnknm

+�B�−60

Qr5+N exp−

��KB

�B

r��

−KB

�B

1r−�KB

�B

7r2−

27r3 −

��B

KB

60r4 −

�B

KB

60r5 +

60�B

KB

1r5 exp−

��KB

�B

r��n� �

3

i=1�kininm+ �

3

i=1�minink

�

+�B

�12r5 Q+N exp−

��KB

�B

r��KB

�B

2r2+

6r3+

��B

KB

12r4+

�B

KB

12r5−

�B

KB

12 exp−��KB

�B

r�

r5

+2

��km (B13)


hand side of the latter equation is zero outside theparticle and differs from zero inside the particles.A direct calculation of such integral inside theparticle requires investigations of the stress tensorinside the particle. However, it is possible to avoidthis problem by a transformation of the integralover volume to the integral over comprising sur-faces: inner surface of the volume V, �, whichdoes not go through any other particle.

Using Eqs. (A1)– (A3) and definition of thestress tensor we can rewrite the stress tensor as

�ij=12

�3

k=1

��(�ikxj)�xk

+�(�jkxi)

�xk

−KB

�(u� kxixj)�xk

�.

(B15)

Substitution of Eq. (B15) into (B14) gives

��ij�=��u� i

�xj

�+��u� j

�xi

��+

12V

�� 3

k=1

(�ikxj dfk+�jkxi dfk

−KBu� kxixj dfk)−2�Bu� i dfj−2�Bu� j dfi�

(B16)

The value �p� disappears because the averagepressure is equal to zero: �p� is a scalar, whichmust be determined by a linear combination ofcomponents of the tensor �ik, the latter scalar isequal to �3

i=1�ii=0.In order to calculate the inte-gral in (B16) the following relations which arevalid at integration over spherical surface, areused

1��

�ninj d�=

13

�ij

1��

�ninjnknl d�

=1

15(�ij�kl+�ik�jl+�il�jk) (B17)

where ni=xi/r, d� is the differential on thespherical surface.Eq. (B16) may be rewritten inthe following way

��ij�=�B��u� i

�xj

�+��u� j

�xi

��+

12V

�� 3

k=1

(�iknjnkr+�jkninkr

−KBu� kninjnkr2)−2�Bu� inj−2�Bu� jni�

d�.

(B18)

For calculation of the integral (B18) we will usestress tensor given by Eq. (B15) and the velocityvector given by Eq. (B10). In the final equationwe keep terms proportional to 1/r2 and omit thosetends faster to zero:

��ij�=�B��u� i

�xj

�+��u� j

�xi

��+

12V

��

−15(N�B−QKB)r2

�3

p,q=1

(�pqnpnq)ninj d�. (B19)

Where��u� i�xj

�+��u� j

�xi

�=2�ij. (B20)

After substitution of integration constants N andQ from Eq. (B11) we find using Eqs. (B13), (B18),(B19)

12V

��

−15(N�B−QKB)r2 �

3

p,q=1

(�pqnpnq)ninj d�

=KB�

a3�KB

�B

+6a2+15��B

KB

a+15�B

KB

�3�

1+�KB

�B

a�

43

�a3

V�ij.

(B21)In the case under consideration, ��= ((4/

3) �a3)/V, hence, from Eqs. (B18)– (B21) weconclude

��ij�=�B2�ij

+KB�

a3�KB

�B

+6a2+15��B

KB

a+15�B

KB

�3�

1+�KB

�B

a� �ij��.

The latter equation can be rewritten as

��ij�/2�ij=�B


+KB�

a3�KB

�B

+6a2+15��B

KB

a+15�B

KB

�6�

1+�KB

�B

a� ��.

(B22)

Comparison of Eqs. (B22) and (9) gives

A�=KB�

a3�KB

�B

+6a2+15��B

KB

a+15�B

KB

�6�

1+�KB

�B

a� .

(B23)

If �(KB)/(�B)a�0 then the latter equation givesA�=5/2, which coincides with Einstein’sequation.

Appendix C. Solution of the system of the differen-tial Eqs. (14)– (15).

The resulting system of the differential equa-tions is (14)– (15) with boundary conditions (16) isunder consideration below. Dividing Eq. (14) byEq. (15) gives

dZd�

=27�

Z

�1+

�Z�

+19

Z�

��1+

�Z�

��Z

�+6+

15�Z�

+15Z�

. (C1)

After substitution Z= t2(�)×�, where t(�) is anew unknown function the differential Eq. (C1)becomes

1�

d�

dt=

2t(t3+6t2+15t+15)(− t5−6t4−12t3+15t2+54t+27)

.

(C2)

Eq. (C2) can be directly integrated

ln �+C= 2t(t3+6t2+15t+15)(− t5−6t4−12t3+15t2+54t+27)

dt (C3)

where C is an integration constant.Polynomial function in the denominator can be

presented as

Fig. 4. Dependence of ��4.130 on porosity � ‘(calculatedaccording to Eqs. (14) and 15)). Broken horizontal line corre-sponds to 0.5957.

− t5−6t4−12t3+15t2+54t+27

= − (t+1.63209)

× (t+0.67371)(t−1.89389)

× (− t2−5.58797t−12.96558)

Using the latter representation the integral inthe right hand side of Eq. (C3) can be directlycalculated. The result is: 2t(t3+6t2+15t+15)

(− t5−6t4−9t3+18t2+54t+27)dt

=0.0457026 arctg(0.220129(5.58797+2t))

−1.10585 ln�1.89389− t �−0.41469 ln�t+0.67371�+0.319579 ln�t+1.63209�−0.39952 ln(12.9656+5.58797t+ t2).

In derivation of the latter two equationsMatematica symbolic manipulation is used. Thelatter equation and Eq. (C3) results in

ln �+C

=0.0457026 arctg(0.220129(5.58797+2t))

−1.10585 ln�1.89389− t �−0.41469ln�t+0.67371�+0.319579 ln�t+1.63209�−0.39952 ln(12.9656+5.58797t+ t2). (C4)

The integration constant C can be calculatedusing condition

t ��=1=0 (C5)


which can be directly deduced from boundaryconditions (16). Eq. (C4) and boundary condition(C4), which gives: C= −1.368995.

Below the case when the concentration of theparticles, �, tends to 1 is investigated.

It is clearly seen from Eq. (C4) that t rangesbetween 0 and value 1.89389. It happens becauset starts from 0 at �=1 and can not pass valuet�=1.89389 (this value arises in the third term−1.10585 ln�1.89389− t �, Eq. (C4)). Conse-quently, at � tends to infinity �Z/�� t�. Eq.(15) after substitution Z= t�

2 � yields

d�

d�=�

�

(1−�), (C6)

where �= (t�3 +6t�

2 +15t�+15)/6(1+ t�)=4.130. Eq. (C6) corresponds to the limiting case��1. Solution of this equation is

�=C�(1−�)−4.130, (C7)

where C� is an integration constant. According toEq. (C7) exact solution should satisfy the follow-ing requirement

��4.130�C�, at ��0. (C8)

Constant C� is calculated using direct integrationof system (14)– (15) (see Fig. 4 for details). Theresult is C�=0.5957 and log100.5957= −0.2249.The asymptotic dependence log10 �= −4.130 log10 �−0.2249 of the viscosity on porosityis presented in Fig. 3 (asymptote 3).

References

[1] G. Beaverse, D. Johns, J. Fluid Mech. 30 (1) (1967) 197.[2] H. Brinkman, J. Chem. Phys., 20 (1952) 571; H. Brinkman,

Appl. Sci. Res., A1 (1947) 27.[3] J. Happel, H. Brenner, Low Reynolds Number Hydrody-

namics, Prentice-Hall, Englewood Cliffs, NJ, 1965.[4] H. Koplik, J. Levin, A. Zee, Phys. Fluids 26 (10) (1983)

2864.[5] L. Landau, E. Lifshiz, Fluid Mechanics, Pergamon, 1959.[6] V. Starov, V. Zhdanov, M. Meireles, C. Molle, J. Colloid

Interface Sci., (2001) accepted.[7] A. Koponen, M. Kataja, M. Timonen, J. Phy. Rev. E 56

(3) (1997) 3319.[8] A. Cancelliere, C. Chang, E. Foti, D. Rothman, S. Succi,

Phys. Fluids A2 (12) (1990) 2085.

effective viscosity and permeability of porous media

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