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Under consideration for publication in J. Fluid Mech. 1

Flow Past Periodic Arrays of Spheres atLow Reynolds Number

By HONGWEI CHENG1AND

GEORGE PAPANICOLAOU2

1Department of Civil Engineering and Operations Research, Princeton University,Princeton, NJ 08544, USA. Internet: [email protected].

2Department of Mathematics, Stanford University, Stanford CA 94305, USA.Internet: [email protected].

(Received ?? and in revised form ??)

We calculate the force on a periodic array of spheres in a viscous ow at small Reynoldsnumber and for small volume fraction. This generalizes the known results for the forceon a periodic array due to Stokes ow (zero Reynolds number) and the Oseen correctionto the Stokes formula for the force on a single sphere (zero volume fraction). We use ageneralization of Hasimoto's approach that is based on an analysis of periodic Green'sfunctions. We compare our results to the phenomenological ones of Kaneda for viscous ow past a random arrays of spheres.

1. Introduction

Inertial e�ects for particle motion in low Reynolds number ow are of interest in manyapplications but their theoretical analysis is rather complicated, even for a single particleas shown by Lovalenti & Brady (1993). In this paper we calculate inertial correctionsto the hydrodynamic force on a �xed periodic array of spheres in steady, viscous andincompressible ow. All spheres have the same radius a and their centers are placed ona cubic lattice of span L. The volume fraction

c = 4�a3=3L3 (1:1)

occupied by the spheres is assumed to be small as is the Reynolds number Re = U0a=�,which is based on the average ow rate U0 of the uid past the spheres, with U0=jU0jand � the kinematic viscosity.When both the volume fraction and the Reynolds number are in�nitesimal, we have

viscous ow past a single sphere with no inertial e�ects. The force is then given by theStokes formula (Batchelor 1967)

F = 6��aU0: (1:2)

When the Reynolds number is small, inertial e�ects appear with the Oseen correction

F = 6��aU0(1 +3

8Re): (1:3)

This was analyzed in detail by Proudman & Pearson (1957) using matched asymptoticexpansions.When the Reynolds number is zero and the volume fraction c occupied by the periodic

2 H. Cheng and G. Papanicolaou

array of spheres is small, the force on the array was studied by Hasimoto (1959) and hisresult for the simple cubic lattice is

F = 6��aU0(1 + 1:7601 3pc): (1:4)

This is one of several results concerning dilute suspension of small spheres in a viscous uid (Batchelor 1972; Brinkman 1947; Childress 1972; Sa�man 1973; Zick & Homsy1982).In this paper we calculate from �rst principles the small inertial corrections to Hasi-

moto's formula (1.4) when the Reynolds number is small but not zero, and the volumefraction c is also small but not zero. Kaneda (1986) studied this problem for a randomarray of �xed spheres. He started with the Brinkman's equation (Brinkman 1947) thatdescribes ow in a �xed random suspension of spheres, an e�ective equation, and addedto it inertial e�ects just as in the Oseen calculation. The Brinkman equation is rea-sonably well understood as an e�ective equation (Hinch 1977; Rubinstein 1986), but amathematical justi�cation for it, especially with inertial e�ects, is hard and unavailable.Kaneda (1986) obtained the following formula for the drag

F = 6��aU0[1 +ReF (S)]; (1:5)

where

F (S) =3

8

"(2S + 1)(4S + 1)

1

2 � 4S2 ln(4S + 1)

1

2 + 1

(4S + 1)1

2 � 1

#; (1:6)

and S = 9c=(2Re2). This result is consistent with Oseen's formula (1.3) when c = 0 andwith Brinkman's formula

F = 6��aU0(1 +3pcp2) (1:7)

when Re = 0. Note the characteristic di�erence between �xed periodic and randomarrays where the force depends on c1=3 in the periodic case and on c1=2 in the randomcase, for c small.Inertial e�ects and interacting sphere e�ects are not additive, even when the Reynolds

number and volume fraction are small, because the equations are nonlinear. This isclearly seen in Kaneda's result (1.5) although it is not discussed in detail in Kaneda (1986).In �gure 1 we plot the relative di�erence between the additive e�ects of inertia and par-ticle ow interaction, and Kaneda's formula (1.5). This relative di�erence is de�nedas

E(Re; c) =(38Re+

q92c)� ReF (S)

ReF (S); (1:8)

which can also be written in the form

E(Re; c) =(38 + S1=2)

F (S)� 1:

This implies that E(Re; c) is a constant along S = constant, i.e. Re=c1=2 = constant inthe (Re; c1=2) plane. This is not so clear from the black and white �gure 1, but can beseen easily from a colored picture in which the color shows the height of E(Re; c). Wenote from the �gure that E is positive, so that uid-particle interaction reduces drag, andthat it can be as large as 40%. A similar `screening' e�ect is observed in the calculationof heat transfer in a dilute �xed bed of spheres at a �xed temperature (Acrivos, Hinch& Je�rey 1980).

Flow Past Periodic Arrays of Spheres 3

0

0.1

0.2

0.3

0.4

0

0.1

0.2

0.3

0.40

0.1

0.2

0.3

0.4

0.5

Rel

ativ

e E

rror

Reynolds Number Volume Fraction1/2

Figure 1. Surface plot of the relative di�erence between the additive e�ects of inertia andparticle ow interaction, and Kaneda's results (1.5). The axis on the right is the square root of

the volume fraction, c1=2, while the axis on the left is the Reynolds number, Re.

We analyze here the periodic version of Kaneda's problem starting from the steadyNavier-Stokes equations and using matched asymptotic expansions (Lagerstrom 1988),combined with a generalized form of Hasimoto's method of periodic Green's functions.For simplicity we consider only the case of a simple cubic lattice of spheres with centersat lattice points xn = L(n1e1 + n2e2 + n3e3) for n = (n1; n2; n3) integers. Our mainresult is the following formula for the force on the array

F = 6��aU0

�I+

3

8ReI +

3

2Re[C(�)I +M(�)]

�+ ::: (1:9)

where C(�) and M(�) = (Cij(�))3�3 are given by (6.22) to (6.24) and I is the identitymatrix, and � = LRe=a, in terms of volume fraction c, it is � = Re=(3c=4�)1=3. Wheninertia is negligible and c is small we show that (1.9) reduces to Hasimoto's formula (1.4).In the opposite limit, where inertial e�ects dominate particle interaction, (1.9) reducesto the Oseen formula (1.3). A table of values for (1.9) is provided in section 8.In the intermediate regime where both inertial and particle interaction e�ects are

important, (1.9) is not the simple addition of the two e�ects. This is shown in �gure2 which is qualitatively similar to �gure 1. We plot the relative di�erence between theadditive Oseen-Hasimoto e�ects and our result (1.9) as a function of particle radius aand Re, de�ned similarly to (1.8) and can be written as

E(Re; c) =1:1735(4�=3)1=3

�[C(�) + C11(�)]� 1:

The error depends on Re and c through parameter �. From �gure 2, we see clearly thedrop in the relative drag correction, the screening e�ect that is due to the uid-particleinteraction. The wiggles in the �gure are due to numerical errors in calculating C(�) andC11(�).The paper is organized as follows. In the next section we formulate the problem.


0

0.1

0.2

0.3

0.4

0

0.1

0.2

0.3

0.40

0.1

0.2

0.3

0.4

Rel

ativ

e E

rror

Particle RadiusReynolds Number

Figure 2. Surface plot of the relative di�erence between the additive Hasimoto-Oseen e�ectsand our result (1.9). The axis on the right is the radius of the spheres, a = (3c=4�)1=3. Theaxis on the left is the Reynolds number, Re.

In section 3 we review brie y Hasimoto's method. Sections 4 to 7 are devoted to thederivation of our results. In section 8 we present some numerical results that illustratehow the force and the average velocity are related by the new formula (1.9). Sometechnical mathematical calculations are presented in the appendices.

2. Formulation of the Problem

We consider a periodic array of identical rigid spherical particles of radius a in aNewtonian uid of viscosity � and density �, driven by an average pressure gradient.We wish to �nd the average uid ow that results when a no-slip boundary condition issatis�ed on the surface of the spherical particles. The ow satis�es the steady Navier-Stokes equations outside the particle array

[I] � � � � � � � � �8<:

�r2u�rp = �(u �r)ur � u = 0 for j x� xn j> a; 8nu = 0 for j x� xn j= a; 8n

with the requirements that u andrp be periodic. Here r2 is the Laplace operator. Ourgoal is to calculate the inertial correction to Hasimoto's formula (1.4), which gives therelation between the average ow rate U0 and the drag force per particle F

U0 = jV j�1ZV

u(x)dx; (2.1)

F =

Zjxj=a

� � nds:

Here V = [�L=2; L=2]3 � fjxj < ag is the cube of side L minus the sphere of radius aand � is the viscous stress tensor

�ij = �p�ij + 2�eij;


with

eij =1

2(@ui@xj

+@uj@xi

)

the rate of strain tensor. It is convenient to consider F as given and to attempt todetermine U0. In the near linear regime, where inertial e�ects are small, this relation isinvertible.

Let us consider the ow as a perturbation of a uniform one due to the presence of theparticle array. If the Renolds number Re = U0a=�, � = �=�, is not zero, then the owis not correctly described by the Stokes equations (Batchelor 1972; Lovalenti & Brady1993; Proudman & Pearson 1957) in the wake behind the particles, the Oseen region.The local inertia term is

�(u �r)u = ��(U0 �r)(U0 � u) + �((U0 � u) �r)(U0 � u); (2:2)

and when we have a dilute suspension, U0 � u may be approximated by a Stokesletplaced at a sphere center. Then, the �rst of the two terms on the right side of (2.2)behaves like �U2

0 a=r2, whereas the second behaves like �U2

0 a2=r3. Compared with the

viscous force, �U0a=r3, the second term is always negligible if Re � 1. However, theratio of the �rst term and the viscous force is

�U20 a

r2

��U0a

r3=r

aRe;

and is not small when r � O(a=Re), which is called the Oseen distance. In our case,the distance between particles is L and so when Re � O(a=L) = O(c1=3) the ow is notcorrectly described by the Stokes equations. We have to deal with the Navier-Stokesequations. We will use matched asymptotic expansions and Hasimoto's periodic Green'sfunctions for this purpose. His work is based on the Stokes equations

[II] � � � � � � � � �8<:

�r2u�rp = 0r �u = 0 for j x� xn j> a; 8nu = 0 for j x� xn j= a; 8n

and is reviewed brie y in the next section.

In the rest of the paper we will deal with the dimensionless form of the Navier-Stokesequations

[III] � � � � � � � � �8<:

�r2�u� �rp = Re(�u � �r)�u�r � �u = 0 for j �x � �xn j> 1; 8n�u = 0 for j �x � �xn j= 1; 8n

where the dimensionless quantities are de�ned by

�x =x

a; �u =

u

U0; �p =

p

�U0=a; Re =

�aU0�

;

and

�xn =L

a(n1; n2; n3):

The average ow velocity is now e0 = U0=U0 which is a unit vector. We will omit thebars in the sequel.


3. Brief Review of Hasimoto's Method

Hasimoto (1959) uses the periodic fundamental solutions of problem [II] de�ned by

[IV ] � � � � � � � � �8<: �r2u =rp+ F

Xn�(x� xn);

r � u = 0;

This may be used to construct a systematic expansion for the ow for small volumefraction (as in Sangani & Acrivos (1983) for the analogous di�usion problem). It is alsothe lowest order term in this expansion and it is then called the point force approximation,since the spherical inclusions are replaced by point forces [IV ]. Let

u(x) =Xk

uke�2�ix�k; (3.1)

�rp =Xk

pke�2�ix�k; (3.2)

where k has integer components. Then, from [IV ]

�4�2� j k j2 uk = �pk + F;

k � uk = 0;

k� pk = 0:

Therefore, u and �rp have the representations

�(rp)j = Fj � (4�)�1Fl@2S1@xl@xj

;

uj = U0j � (4��)�1(FjS1 � Fl@2S2@xl@xj

);

for j = 1; 2; 3. Here

S1(x) = ��1Xk6=0

j k j�2 e�2�ix�k;

S2(x) = �(4�3)�1Xk6=0

j k j�4 e�2�ix�k:

In order to apply this fundamental solution to problem [II], it is necessary to getexpansions for S1 and S2 for jxj � L. If we scale coordinates by L, or set L = 1, then

S1 =1

j x j �C + O(j x j2) (3.3)

@2S2@xl@xj

= � xjxl2 j x j3 + (

1

2 j x j �C

3)�jl + O(jxj2); (3.4)

uj = U0j� 1

4��

�Fj(

1

2 j x j �2C

3) +

F � x2 j x j3xj

�+O(jxj2): (3.5)

For a simple cubic lattice the constant C is equal to 1:7601 3

p4�=3. Moreover, the

boundary conditions in [II] are to lowest order in a small volume fraction expansionreplaced by

< u >=1

4�a2

Zjxj=a

uds = 0; (3:6)


as noted by Hasimoto (1959) and carried out in detail for the analogous di�usion problemin Sangani & Acrivos (1983). Substituting (3.5) in (3.6) gives

U0 = (4��)�1F(2

3a� 2C

3) (3:7)

and this determines F as in (1.4).

4. Inner Expansion

We will use the method of matched asymptotic expansions to solve [III]. We startwith an inner expansion of the form

u = u0 + Reu1 + o(Re); (4.1)

p = p0 +Rep1 + o(Re): (4.2)

Inserting these into (III) yields to O(Re0)

[V ] � � � � � � � � �8<:r2u0 �rp0 = 0r � u0 = 0 for j x � xn j> 1; 8nu0 = 0 for j x � xn j= 1; 8n;

and to O(Re1)

[V I] � � � � � � � � �8<:r2u1 �rp1 = (u0 �r)u0r � u1 = 0 for j x� xn j> 1; 8nu1 = 0 for j x� xn j= 1; 8n:

The conditions imposed on (u0; p0); (u1; p1) do not determine them uniquely. Addi-tional conditions are provided by matching them to the outer expansion. Speci�cally,we know that u0 must agree with the leading term of the outer expansion for jxj large.Since the ow is a perturbation of a uniform one, condition (2.1) in dimensionless formis

u0 �! e0 as jxj �!1:

Equations [V ] with this condition are just the Stokes equations for ow past a sphere.The solution is

u0 = e0 � 3

4(e0

jxj +e0 � xjxj3 x) +O(jxj�3) (4:3)

for jxj large. This will now provide a matching condition for the �rst-order outer expan-sion and it will give the leading term contribution to the force, which is exactly the Stokesforce. The �rst-order outer expansion will, moreover, provide boundary conditions forthe �rst-order inner approximation by matching, and this will determine it. We can thencalculate the �rst-order inertial force e�ect in Hasimoto's formula.

5. Outer Expansion

From the scaling analysis we carried out in section 2, we know that inertial e�ects areimportant when Re � O(a=L). Let us set L = 1 and assume that Re = �a with � oforder one. The outer expansion is, therefore, an expansion of solutions of [III] for smallRe and small volume fraction.The small volume fraction part of the outer expansion leads to the point force approx-

imation, as noted in section 3. We will not work this out in detail here (cf. Sangani &Acrivos 1983). We will begin instead with the point force approximation. This means


that the dimensionless equations [III] are to hold throughout space and the particles arereplaced by a distribution of forces at the sphere centers xn

[V II] � � � � � � � � �8<: r2u�rp = Re(u �r)u+F

Xn

�(x � xn) + � � �r �u = 0:

This is consistent to O(Re) with the outer expansion that follows, keeping in mind thatRe = �a with � of order one. To construct the outer expansion we need to introduceouter variables

~x = Rex; ~F = ReF; ~p =p

Re:

Equation [V II] is then

[V III] � � � � � � � � �8<:

~r2u� ~r~p = (u � ~r)u+ ~FXn�(~x � ~xn) + � � �

~r � u = 0:

where ~xn is �(n1; n2; n3) with (n1; n2; n3) integers (we have set L = 1).We now turn to the form of the outer expansion. We consider �rst the Oseen analysis

for a single particle. We have to divide the ow into several regions. One is the Stokes

Stokes Region

"Near" Oseen Region

"Far" Oseen Region

Figure 3. The Oseen ow region for a single particle.

and the other the Oseen region, which we must divide further into the Oseen `near' and`far' regions, as shown in �gure 3. There are two reasons for doing this. The �rst is that`far' Oseen region is di�erent in the periodic case since the other particles will be felt.The second is that the average ow velocity of the `near' Oseen region is di�erent fromthe one in the `far' Oseen region. The average ow velocity over the full ow domain,and the `far' Oseen region, is the uniform ow e0. But we know from the solution of theOseen equation (Brenner and Cox 1963) that the average ow velocity over the `near'Oseen region is (1 + 3

8Re)e0. This, in fact, is a convenient way to de�ne what we meanby `near' and `far' Oseen regions. We therefore consider the following outer expansionfor the velocity

u = e0 +3

8Ree0 + ReU1 + o(Re): (5:1)

We could combine the second and third terms on the right but with this de�nition U1

has average zero. For the pressure we consider the expansion

� ~r~p = (1

�)3 ~F� Re ~rP1 + o(Re):


Substituting these expansions into the outer equation [V III], we �nd that (U1; P1)satisfy 8<

:~r2U1 � ~rP1 = (e0 � ~r)U1 + F[

Xn�(~x � ~xn) � (1

�)3];

~r �U1 = 0:

Since the leading term for the force F is 6�e0 and additional terms do not contribute tothe velocity U1 to O(Re) in (5.1), we simplify further to

[IX] � � � � � � � � �8<:

~r2U1 � ~rP1 = (e0 � ~r)U1 + 6�e0[Xn�(~x � ~xn)� (1� )

3];

~r �U1 = 0:

This is exactly the Oseen version of equations [IV ], with zero average �elds. We solvefor (U1; P1) in the next section by generalizing Hasimoto's method.

6. The Generalized Periodic Fundamental Solution

We will deal with [IX] in the form

[X] � � � � � � � � �8<:

~r2U1 � ~rP1 = (e0 � ~r)U1 +FXn�(~x � ~xn);

~r �U1 = 0:

where F now stands for 6�e0. This is just [IX] with the mean pressure gradient includedin the new P1.As in Hasimoto (1959), we expand in Fourier series

U1(~x) =Xk

uke�2�i~x�k; (6.1)

� ~rP1 =Xk

pke�2�i~x�k: (6.2)

Here

k =1

�(k1; k2; k3);

with ki integers. In the rest of this section we will drop the tilde to simplify writing.We will pick it up again in the next section. Using the Fourier expansion of the periodicdelta functions X

n�(x � xn) =

1

�3

Xk

e�2�ix�k; (6:3)

in [X] we have

�4�2jkj2uk + pk �F

�3= �2�i(e0 � k)uk; (6.4)

k � uk = 0; (6.5)

k� pk = 0: (6.6)

For k = 0, (6.4) gives

p0 =F

�3; (6:7)


which is the mean pressure gradient that balances the point forces. For k 6= 0 we have

k � pk =k �F�3

; (6:8)

and from (6.4) and (6.6)

pk =(k �F)k�3jkj2 ; (6.9)

uk =F� (k�F)k

jkj2

2�i�3(u0 � k) � 4�2�3 j k j2 : (6.10)

With these Fourier coe�cients we have the representations

�(rP1)j = Fj�3� (4�)�1Fl

@2S1@xl@xj

; (6.11)

U1j = e0j � (4� )�1(Fj ~S1 � Fl@2 ~S2@xl@xj

); (6.12)

where:

S1 = ��1��3Xk6=0

j k j�2 e�2�ix�k; (6.13)

~S1 = ��1��3Xk6=0

[j k j2 � ie0 � k2�

]�1e�2�ix�k; (6.14)

~S2 = �(4�3)�1��3Xk 6=0

j k j�2 [j k j2 � ie0 � k2�

]�1e�2�ix�k: (6.15)

These expressions are similar to (6.11)-(6.15) for Stokes equations. The lattice sumfor ~S2 is absolutely convergent everywhere, but S1 and ~S1 are only weakly convergent, inthe sense of distributions. To continue with Hasimoto's approach for calculating the owpast small spheres we must estimate S1, ~S1 and ~S2 for small j x j� 1. Most of the work

in estimating ~S1 and @2 ~S2@xl@xj

as j x j! 0 goes in extracting their singular part, which is

the key point that makes Hasimoto's method successful. As in Hasimoto (1959), we will

use the Ewald summation technique to evaluate ~S1 and@2 ~S2@xl@xj

.

To carry out the small j x j expansion of the lattice sums we use the identity

1

(jkj2 � 12�ie0 � k)m

=�m

�(m)

Z 1

0

�m�1e��(jkj2� 1

2�ie0�k)�d�; (6:16)

where m is an integer. We now introduce a more general lattice sum �m and transformit as follows

�m(e0; �;x) =Xk6=0

e�2�ix�k

(j k j2 � 12� ie0 � k)m

=�m

�(m)

Xk6=0

Z 1

0

�m�1e��jkj2��2�ik�(x� �e0

4�)d�

=�m

�(m)

Z 1

0

�m�1[Xk

e��jkj2��2�ik�(x� �e0

4�) � 1]d�: (6.17)


Note that �m depends on � through k implicitly. We will also use a slight variant of thePoisson summation formula (Courant & Hilbert 1953)

Xk

e��jkj2��2�ik�(x� �e0

4�) = �3��

3

2

Xne�

�jx�xn��e04�

j2� ; (6:18)

in three dimensional space. We now split the integral in (6.17) into two parts. One isfrom 0 to � and we use (6.18) to it, and the other is from � to 1, where � is a suitablychosen constant. After some changes of variables we get

�m(e0; �;x) =�m�m

�(m)

��

3

2 �3Z 1

1��m+ 1

2 e��jx��

4��e0j2 �

� d�

+ �3��3

2

Xn6=0

Z 1

1

��m+ 1

2 e��jx�xn��4��

e0j2 �

� d�

� 1

m

+Xk6=0

e�2�ix�kZ 1

1

�m�1e��jkj2�+ 1

2�i(k�e0)�d�

35 : (6.19)

We can now estimate ~S1 and@2 ~S2@xl@xj

as j x j! 0. This is much more complicated than

in Hasimoto's case. We give the main results here and carry out the detailed calculationsin Appendix A.

~S1 =1

j x j +x � e02 j x j �C(�; e0) + O(j x j) (6.20)

@2 ~S2@xl@xj

= � xjxl2 j x j3 +

1 + 14x � e0

2 j x j �jl

�1

4

�xjxl(x � e0)2 j x j3 � xje0l + xle0j

2 j x j�

+Clj(�; e0) + O(j x j): (6.21)

Here

C =2p�e�

�16� +

�

�3� 1p

�

Xn 6=0

Z 1

1

��1

2 e��jxn+ �4��e0j2 �

� d�

� �

�3

Xk6=0

Z 1

1

e��jkj2�+ 1

2�i(k�e0)�d�; (6.22)

Cjj = � 1

2p�(1 + e�

�16� )�

p�

32�e20j

Z 1

1

��3

2 e��

16�� d�

� �

2�p�

Xn6=0

Z 1

1

�1

2

�x2nje

��jxnj2 �

� + (xnj +�

4��e0j)

2e��jxn+ �4��e0j2 �

�

�d�

+1

4p�

Xn6=0

Z 1

1

�1

2

he��jxnj

2 �

� + e��jxn+ �4��

e0j2 ��

id�

+��2

2�3

Xk6=0

k2j

Z 1

1

�e��jkj2�(1 + e

1

2�i(k�e0)�)d�


+1

8�3�3

Xk 6=0

k2j (k � e0)2j k j4 (j k j2 � 1

2� ik � e0)2; (j = 1; 2; 3) (6.23)

Clj = �p�

32�e0le0j

Z 1

1

��3

2 e��

16�� d�

� �

2�p�

Xn6=0

Z 1

1

�1

2 (xnj +�

4��e0j)(xnj +

�

4��e0l)e

��jxn+ �4��e0j2 �

� d�

+��2

2�3

Xk6=0

kjkl

Z 1

1

�e��jkj2�+ 1

2�i(k�e0)�d�

+1

8�3�3

Xk 6=0

kjkl(k � e0)2j k j4 (j k j2 � 1

2� ik � e0)2; (j 6= l): (6.24)

We note that the last term in (6.23) or (6.24) is new and is needed in calculating@2 ~S2@xl@xj

. Also, since the sum for ~S2 is

�� =Xk6=0

e�2�ix�k

j k j2 (j k j2 � 12� ie0 � k)

:

it is not possible to extract the singular part from it directly. However, we can write itin the form

�� =1

2

24Xk6=0

(1

j k j4 +1

(j k j2 � 12� ie0 � k)2

)e�2�ik�x

+1

4�2

Xk6=0

(e0 � k)2j k j4 (j k j2 � 1

2� ie0 � k)2e�2�ik�x

35 ;

and in terms of �m

�� =1

2(�2(0; �;x) + �2(e0; �;x)) +

1

8�2

Xk6=0

(e0 � k)2j k j4 (j k j2 � 1

2� ie0 � k)2e�2�ik�x: (6:25)

Now we can estimate �2(0; �;x) and �2(e0; �;x) together with their second order deriva-tives by the formulas above. Moreover, the last term and its second derivatives are allabsolutely convergent.

From (6.20) and (6.21), we get the expansion for U1 as jxj �! 0

U1j = e0j � (4�)�1�

Fj2 j x j (1 +

3

4x � e0)� FjC

+F � x2 j x j3xj � FlClj

+1

4

�(F � x)(x � e0)

2 j x j3 xj � F � e02 j x jxj �

F � x2 j x je0j

��: (6.26)


7. Calculation of the Force

The expression (6.26) for the principal term of the outer expansion for the velocitygives the solution of [IX] for j~xj small. Using the tilde variables again we have

U1 = � 3

4

�e0

j~xj +e0 � ~xj~xj3 ~x

�

+3

2(Ce0 + e0 �M)

� 3

16

�3~x � e0j~xj e0 +

(e0 � ~x)2j~xj3 ~x� ~x

j~xj �e0 � ~xj~xj e0

�+ o(1); (7.1)

as j~xj �! 0, where M = (Clj)3�3 is a matrix. Changing to the inner variables we obtainfor the outer expansion of u:

u = e0 � 3

4

�e0

jxj +e0 � xjxj3 x

�

+3

2Re (Ce0 + e0 �M) +

3

8Ree0

� 3

16Re

�3x � e0jxj e0 +

(e0 � x)2jxj3 x � x

jxj �e0 � xjxj e0

�+ O(Re2): (7.2)

The main term, on the �rst line, has been matched with the zero-order inner solutionalready. By the matching principle, the remaining terms give the following condition atin�nity for the �rst-order inner approximation (u1; p1)

u1 ;3

2(Ce0 + e0 �M) +

3

8e0

� 3

16

�3x � e0jxj e0 +

(e0 � x)2jxj3 x � x

jxj �e0 � xjxj e0

�; (7.3)

as jxj �! 1. Equation [V I] together with this condition determines (u1; p1) uniquely.We will not solve for (u1; p1) explicitly here. Instead, we will use the result of Brennerand Cox (1963) for the force on the particle due to (u1; p1), as determined by the aboveconditions: 9�(Ce0+e0 �M)+ 9

4�e0. This means that we do not have to get an improvedversion of formula (3.6), which is a considerable simpli�cation. We note that the termsinside the second parentheses of (7.3) are odd in x and thus do not contribute to theforce. Adding this force to the zero-order dimensionless force we get the drag

F = 6�e0

�I+

3

8ReI+

3

2Re(CI +M)

�+ ::: (7:4)

where C and M = (Cij)3�3 are giving by (6.22)-(6.24) and I is the identity matrix.This is the main result of our paper. We see that F and e0 are related in a nonlinear

way now, as expected. Furthermore, the drag depends on the volume fraction (througha) and on the Reynolds number. It is thus much more complicated than both Hasimoto's(1.4) and Oseen's (1.3) formulas. In Appendix B we show that our formula for the dragreduces to the Hasimoto and Oseen formulas, in the appropriate limit.

8. Numerical Calculations

We now calculate numerically the coe�cients C and Cij, given by (6.22)-(6.24), andthen the drag (7.4), for various values of the volume fraction and the Reynolds number.To simplify the computations we choose e0 = (1; 0; 0). This means that we do not discuss


� = Re=a Formula (8.1) Formula (8.2)

0.05 57.46 56.750.1 28.91 28.370.2 14.63 14.190.3 9.869 9.4580.6 5.090 4.7291.0 3.159 2.8372.0 1.690 1.4193.0 1.199 0.9465.0 0.807 0.56710.0 0.519 0.28415.0 0.428 0.189100.0 0.396 0.028

Table 1. Typical results for �F from our formula (8.1) and from Hasimoto's formula (8.2)

directional properties of the force, which are, however, important in distinguishing Stokesfrom Oseen ows.The numerical results are stated in terms of the di�erential drag coe�cient �F, de�ned

by

F = 6�e0(I +�FRe):

This is usually a 3�3 matrix, but with our choice of e0 we only need (�F)11, so we keeponly the one-one component

F = 6�e0(1 +�FRe):

From (7.4) we have

�F =3

8+3

2(C +C11); (8:1)

and C, C11 are given by (6.22) and (6.23). Thus, �F depends on e0 explicitly and onRe and a implicitly through their ratio � = Re=a.The results of our numerical calculations for �F in (8.1) are summarized in table 1.

For comparison, we also list the corresponding data from Hasimoto's formula as � varies

�F = 1:76013

r4�

3��1: (8:2)

For Oseen's formula the di�erential drag coe�cient is equal to 38 for all �.

Let us examine our results in three di�erent cases, depending on the magnitude of �.Case 1: � small

In this case our result should tend to the Hasimoto's formula (1.4). This is shown in�gure 4. The analytical behavior of �F for � small is discussed in Appendix B.Case 2: � largeWe expect that our results tend to Oseen's formula in this case. This is shown in �gure

5 where the dotted line is Oseen's formula. We do not show �F for � very large becausethe numerical calculation of the lattice sums converges very slowly. But the agreementobserved in �gure 5 is good.Case 3: � in between

Our result is shown in �gure 6. We see that when the Reynolds number Re and thedimensionless particle radius a, or 3

pc with c the volume fraction, are comparable in


0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.80

5

10

15

20

25

30

Figure 4. �F versus � for � small. The solid line is for �F computed from our formula andthe dotted line is from Hasimoto's formula.

8 9 10 11 12 13 14 15 160.35

0.4

0.45

0.5

0.55

0.6

Figure 5. �F versus � for � large. The solid line is for �F computed from our formula andthe dotted line is from Oseen's formula.

0 1 2 3 4 5 6 7 80

2

4

6

8

Figure 6. �F versus � for intermediate values of �

magnitude and small, the change in the drag F depends on the ratio � = Re=a in acomplicated way. As explained in the introduction, the inertial and particle interactione�ects are not simply additive. We note also that �F is a full matrix for most e0 becausethere is directional sensitivity when inertial e�ects are included. This is to be expectedsince we approximate the ow around each particle by an Oseen ow which is not fullysymmetric.It is well known that in expressions generated by the Ewald summation technique, such

as (6.22) and (6.23), the parameter � does not a�ect the value of C and Cjj. This has


been veri�ed in several cases in our numerical calculations. However, di�erent choicesof � a�ect the numerical convergence of C and Cjj greatly. The optimal � for fastconvergence is hard to determine in advance. Our analysis of the limiting behavior of Cand Cjj in appendix B suggests that when � is small, � should be comparable to �2 andwhen � is large, � should be comparable to �. In between, we simply take (�2 + �)=2.As noted in the introduction, our results are in agreement with the phenomenological

analysis of Kaneda (1986). The main qualitative di�erence is that the inertia to parti-cle interaction parameter is � = Re=a (or Re=c1=3) in this paper and is S = c=Re2 (orRe=c1=2) for �xed random arrays in Kaneda (1986). This characteristic di�erence occurs,however, even without inertial e�ects. For Stokes ow (Sa�man 1973), the relation be-tween average force and average velocity for a uid-particle system is a di�erent functionof concentration for di�erent types of suspensions. In terms of �U = (U0 � U )=U0, it isknown that:

(i) For �xed periodic arrays (Hasimoto's case for a simple cubic lattice)

�U � 1:7601c1

3 ;

(ii) For �xed random arrays (Brinkman's case)

�U � 3p2c1

2 ;

(iii) For free random arrays (Batchelor's sedimentation case)

�U � 6:55c:

These relations show that uid-particle interaction is stronger for �xed periodic arraysthan for �xed random arrays, which is in turn stronger than that for random suspen-sions, at small volume fraction. We may expect, therefore, that inertial corrections toBatchelor's formula for sedimentation should depend on the parameter � = Re=c. As faras we know, this problem has not been analyzed.

We would like to acknowledge the referee's useful comments. H.C. is supported by USDepartment of Energy Grant DE-FG02-92ER14275. G.P. is supported by NSF GrantDMS 96-22854 and AFOSR F49620-94-1-0436.

Appendix A. Asymptotics for Generalized Lattice Sums

Part A: Asymptotics for ~S1

From its de�nition (6.4) we have

~S1 = ��1��3�1

=1p�

Z 1

1

��1

2 e��jx��

4��e0j2 �

� d�

+1p�

Xn6=0

Z 1

1

��1

2 e��j�x+xn+ �4��

e0j2 �

� d�

� �

�3

� �

�3

Xk6=0

Z 1

1

e��jkj2�+ 1

2�i(k�e0)�d�; (A 1)


The last three terms tend to �nite limits as jxj ! 0. So we need to analyze the �rstterm.Let

f(e0;x) =1p�

Z 1

1

��1

2 e��jx��

4��e0j2 �

� d�:

Then by a change of variables

f(e0;x) =1

jxje1

2x�e0 � 2

Z 1

jxjp�

e��(s2+

jxj216�2s2

)ds;

If we let

g(t) = 2

Z 1

tp�

e��(s2+ t2

16�2s2)ds;

then

g(t) = 2(

Z 1

0

�Z tp

�

0

)e��(s2+ t2

16�2s2)ds = g1(t)� g2(t):

Clearly g1(t) is an even function of t and g2(t) is odd. Thus

g(0) = g1(0) = 2

Z 1

0

e��s2

ds = 1;

g0(0) = �g02(0) = � limt!0

g2(t)

t= � 2p

�e�

�16� ;

g00(0) = g001 (0) 6= 0:

Expanding in a Taylor series we have

g(t) = 1� 2p�e�

�16� t+O(t2);

Collecting all these estimates we �nd that as jxj ! 0

f(e0;x) =1

jxj(1 +1

2x � e0 + O(jxj2))(1 � 2p

�e�

�16� jxj+O(jxj2))

=1

jxj +x � e02jxj �

2p�e�

�16� +O(jxj): (A 2)

This together with (A 1) gives the estimate (6.20) for ~S1.

Part B. Asymptotics for @2 ~S2@xl@xj

:

We use (6.25) so that the derivatives of ~S2 are

@2 ~S2@xl@xj

��x=0

= �(4�3)�1��3 12

limjxj!0

@2

@xl@xj(�2(0; �;x) + �2(e0; �;x))

+1

8�3�3

Xk 6=0

kjkl(k � e0)2j k j4 (j k j2 � 1

2� ik � e0)2: (A 3)

The last term is convergent. To estimate @2 ~S2@xl@xj

we have to estimate @2�2(e0;�;x)@xl@xj

.

There are two cases, one when j 6= l and the other when j = l. We analyze the �rst


case. The analysis of the second is similar. For j 6= l

@2�2(e0; �;x)

@xl@xj=

@2

@xl@xj

(�2�2

�(2)

"��

3

2 �3Xn

Z 1

1

��3

2 e��jx�xn��4��

e0j2 �

� d�

+Xk6=0

e�2�ix�kZ 1

1

�e��jkj2�+ 1

2�i(k�e0)�d�

359=;

= 4�4

"��

3

2 �3Xn

Z 1

1

�1

2 (xj � xnj � �

4��e0j)(xl � xnj � �

4��e0l)

�e��jx�xn� �4��

e0j2 �

� d�

��2Xk6=0

kjkle�2�ik�x

Z 1

1

�e��jkj2�+ 1

2�i(k�e0)�d�

35 : (A 4)

Letting jxj ! 0 we get

limjxj!0

@2�2(e0; �;x)

@xl@xj= 4�4��

3

2 �3�

limjxj!0

xjxl

Z 1

1

�1

2 e��jx��4��

e0j2 �

� d�

� �

4�limjxj!0

(xje0l + xle0j)

Z 1

1

��1

2 e��jx��4��

e0j2 �

� d�

�+ C0lj (A 5)

where C0lj is a constant related to the �rst three terms of (6.24). To estimate the rest ofthe above expression let

'(e0;x) = ��3

2

Z 1

1

�1

2 e��jx��4��e0j2 �

� d�:

After some transformations similar to the ones for f(e0;x), we have

'(e0;x) =e1

2x�e0

2�jxj32p�

Z 1

�jxj2�

pte�t�

jxj216�t dt:

We note that

(s) =2p�

Z 1

s2

pte�t�

s2

16�t dt

is an even function of s and (0) = 1. Thus

(s) = 1 + O(s2);

and hence

'(e0;x) =1

2�jxj3 (1 +1

2x � e0 +O(jxj2)):

From this last estimate and (A 2) we arrive at

@2�2(e0; �;x)

@xl@xj= 4�3�3

�xjxl2jxj3 +

xjxl(x � e0)4jxj3

�

��3 xjeol + xle0jjxj +C0lj +O(jxj); (A 6)

where j 6= l. This expression and the analogous one for j = l give us (6.21).


Appendix B. Consistency Calculations

Case 1. First we analyze (7.4) when � = Re=a �! 0. This means that inertial e�ectsare negligible so our result should tend to the Hasimoto's formula (1.4), and this is whatwe show here.

In terms of a, F is

F = 6�e0

�I+ a � 3

8�I + a � 3

2�(CI +M)

�+ :::

To show that this tends to Hasimoto's formula as � �! 0, we note that the Oseen term38�I disappears and the other term is handled by the Lemma.

Lemma 1. As � �! 0, �C �! 1:7601 3

p4�=3 and �M �! �1

3 � 1:7601 3

p4�=3I

Proof. We give the details only for the �rst part since the other is similar. From(6.22), we have

�C =2�p�e�

�16� +

�

�2� �p

�

Xl 6=0

Z 1

1

��1

2 e��jl+�

4��e0j2 ��

2

� d�

� �

�2

Xl 6=0

Z 1

1

e��

�2jlj2�+ 1

2

��i(l�e0)�d�;

where l is in the integer lattice now. Note that � is a constant which we can changewithout a�ecting this Thus, we may take � = �2 and then

�C = 2e��2

16� + 1�Xl6=0

Z 1

1

��1

2 e��jl+�

4��e0j2�d�

�Xl6=0

Z 1

1

e��jlj2�+ 1

2�i(l�e0)�d�;

After some manipulation this goes to

�C = 3�Xl6=0

Z 1

1

��1

2 e��jlj2�d� �

Xl 6=0

Z 1

1

e��jlj2�d�;

as � �! 0 and the right hand side is approximately 1:7601 3

p4�=3, as shown by Hasimoto

(1959).Case 2. We analyze (7.4) when � �!1. Since we require Re to be small, this means

that the particle radius a is in�nitesimal and so particle interaction is negligible. Ourresult (7.4) should reduce to Oseen's formula (1.3), and this is what we show now.

Lemma 2. As � �!1, C �! 0 and M �! 0.Before proving this lemma we will look again at the the way we analyzed the outer

problem in section 5. When we apply Hasimoto's method to the Oseen region(the outerregion), we consider the whole Stokes region around each particle as an è�ective' particle(�gure 7). It is clear from the �gure that the radius of this è�ective' particle is the Oseendistance �Re. In our analysis, we � to be of order one, so the dimensionless radius issmall as long as Re � 1. But when we let � �! +1, the radius a may also large.This must be controlled so as not to the approximations in Hasimoto's method whichrequire that the radius be small. So we let � �! +1 to recover the Oseen's case fromour results, but we keep the è�ective' particle size small, i.e., �Re� 1.


Stokes Region

Re

Rea = θ

Rea

2

= θ Ree

0

Oseen’s Region

Figure 7. The ow domain in a period cell in terms of outer variables

We take these remarks into account by choosing new outer variables

~x =x

�; ~F =

F

�; ~p = �p:

Equation [X] take the form

[XI] � � � � � � � � �8<:

~r2U1 � ~rP1 = �Re(e0 � ~r)U1 + F

Xn�(~x� ~xn);

~r �U1 = 0:

We now repeat the previous analysis for this equation. We get exactly the same result(7.2), except that in formula (6.22)|(6.24) for C and M, � is replaced by 1=Re, and e0is replaced by �Ree0. Now �Re enters as a parameter. To keep the notation simple wedenote 1=Re by � again, and �Ree0 by �e0=�, where � � � �! +1. Instead of provingLemma 2 to recover Oseen's formula, we now prove the following.Lemma 3. C �! 0 and M �! 0 as � � � �! +1, where in the formula (6.22)|

(6.24) for C and M, e0 is substituted by �e0=�.Proof. First we show that C �! 0. We have

C =2p�e�

�16� +

�

�3� 1p

�

Xl 6=0

Z 1

1

��1

2 e��jl+ ��

4��2e0j2 ��

2

� d�

� �

�3

Xl6=0

Z 1

1

e��

�2jlj2�+ 1

2

��

�2i(l�e0)�d�; (B 1)

where l is an integer valued vector. We choose � = �. Then

C =2p�e�

�16� +

1

�2� 1p

�

Xl 6=0

Z 1

1

��1

2 e��jl+e0�4��

j2��d�

� 1

�2

Xl6=0

Z 1

1

e��1

�jlj2�+ �

2�i(l�e0)�d�: (B 2)


We can now show C �! 0 as � �! 1. The �rst two terms go to zero, clearly. Weassume that � > 1 and estimate the �rst sum as follows

Sum1 =

��1p�

Xl6=0

Z 1

1

��1

2 e��jl+�e04��

j2��d�

��<

1p�

Xl6=0

Z 1

1

��1

2 e��jl+e04� �

j2�d� =Kp�; ( �! +1)

for some constant K which does not depend on �. So this term vanishes like 1=p� as

� �!1.The last term is more di�cult. We begin with the estimate

Sum2 =

��1

�2

Xl6=0

Z 1

1

e��1

�jlj2�+ �

2�i(l�e0)�d�

�� 1

�2

Xl 6=0

Z 1

1

e��1

�jlj2�d�

=1

��

Xl6=0

e��1

�jlj2

jlj2

Since f(t) = e��t=t > 0, for t > 0, and decreases with t, we can estimate the last sum

by comparison with an integral

Xl6=0

e��1

�jlj2

jlj2 �Z Z Z

R3

e��1

�jxj2

jxj2 dV

=

Z 2�

0

Z �

0

Z 1

0

e��1

�r2

r2r2 sin�drd�d�

= Kp�;

where K is again a constant not depending on �. Therefore

Sum2 � 1

��Kp� �! 0

as � �!1 and C �! 0.To show that M �! 0, it is enough to show that Cjj �! 0. We replace in (6.23) e0

by �e0=�, let � = �, and remove the � dependence of xn and k. Then

Cjj = � 1

2p�(1 + e�

�16� )

�p�

32�(�

�)2e20j

Z 1

1

��3

2 e��

16�� d�

� �

2�p�

Xl6=0

Z 1

1

�1

2

��2l2j e

��jlj2� + �2(lj +�

4��e0j)

2e��jl+�

4��e0j2��

�d�

+1

4p�

Xl6=0

Z 1

1

�1

2

he��jlj

2� + e��jl+�

4��e0j2�

id�


+�

2�3

Xl6=0

l2j

Z 1

1

�e��jlj2�

�(1 + e�

2�i(l�e0)�)d�

+1

8�3�

Xl 6=0

l2j (�l � e0)2j l j4 (j l j2 � 1

2� i�l � e0)2; (j = 1; 2; 3)

Only the second and last terms need analysis. The other terms can be handled in thesame way as in showing C �! 0.For the second term, we note that

p�

Z 1

1

��3

2 e��

16�� d� =

Z 1

1

(�

�)�

3

2 e��

16�� d(�

�)

=

Z 1

1

�

s�3

2 e�1

16�s ds

�Z 1

0

s�3

2 e�1

16�s ds = K

where K is some constant not depending on �. So this term will vanish as �=� �!0 (� � �).

We consider now the last term and take e0 = (1; 0; 0) for simplicity. We have, therefore

Sum3 =1

8�3�

Xl6=0

l2j (�l1)2

j l j4 (j l j2 � 12� i�l1)

2;

to be estimated for � � � �! +1. We show that��1

�

Xl 6=0

l2j (�l1)2

j l j4 (j l j2 � 12� i�l1)

2

�� K (B 3)

as � �! +1, for some constant K that does not depend on �.With this estimate, it is clear that jSum3j = O(�=�) �! 0 as � � � �! +1 and the

proof of Lemma 3 is complete.To prove (B 3), we will prove the slightly stronger estimate

Sum =

��1

�

Xl1>0

Xl2;l3

(�l1)2

j l j2 (j l j2 �i�l1)2

�� K

as � �! +1 and again K does not depend on �. We split the sum into two parts

Sum � 1

�

Xl1>0

Xjlj2>�l1

(�l1)2

j l j2 j j l j2 �i�l1 j2

+1

�

Xl1>0

Xjlj2��l1

(�l1)2

j l j2 j j l j2 �i�l1j2

= (I) + (II):

Consider (II) �rst. We have

�l1 � l21 � l22 + l23 � 0;


so that l1 � �. Thus

(II) � 1

�

X0<l1��

Xl22+l2

3��l1�l21

1

jlj2

� K

�

X0<l1��

Z �l1�l2

1

0

dr2

l21 + r2

� K

�

X0<l1��

log�

l1

� K

�

Z �

1

log�

xdx

� K

Z 1

0

log1

tdt

= K:

Here K stands for di�erent constants that do not depend on �.Consider (I). We have

(I) � 1

�

Xl1>0

Xl22+l2

3>�l1�l21

(�l1)2

jlj6

� 1

�

Xl1��

Xl2;l3

(�l1)2

jlj6 +1

�

Xl1<�

Xl22+l2

3>�l1�l21

(�l1)2

jlj6

� K

�

Xl1��

(�l1)2

Z 1

0

dr2

(l21 + r2)3+K

�

Xl1<�

(�l1)2

Z 1

�l1�l21

dr2

(l21 + r2)3

� K

�

Xl1��

(�l1)2 1

l41+K

�

Xl1<�

(�l1)2 1

(�l1)2

� K�Xl1��

1

l21+K

� K�

Z 1

�

dx

x2+K

= K;

and again K denotes constants independent of �. So this term is also bounded and theproof of (B 3) complete.

REFERENCES

Acrivos, A., Hinch, E.J. & Jeffrey, D.J. 1980 Heat transfer to a slowly moving uid froma dilute �xed bed of heated spheres. J. Fluid Mech. 101, 403-421.

Batchelor, G.K. 1972 Sedimentation in a dilute suspension of spheres. J. Fluid Mech. 52,245-268.

Batchelor, G.K. 1967 An introduction to uid dynamics. Cambridge University Press, Ch 4.Brenner, H. & Cox, R.G. 1963 The Resistance to a Particle of Arbitrary Shape in Transla-

tional Motion at Small Reynolds Numbers. J. Fluid Mech. 17, 561-595.Brinkman, H.C. 1947 A calculation of the viscous force exerted by a owing uid on a dense

swarm of particles. Appl. Sci. Res. A1, 27-34.


Caflisch, R. & Rubinstein, J. 1986 Lectures on the mathematical theory of multiphase ow.Lecture Notes at Courant Institute, NYU.

Childress, S. 1972 Viscous ow past a random array of spheres. J. Chem. Phys. 56, 2527-2539.Courant, R. & Hilbert, D. 1953 Methods of mathematical physics, Vol. 1, New York, Inter-

science.Happel, J. & Brenner, H. 1965 Low Reynolds number hydrodynamics, Prentice-Hall, Ch 8.Hasimoto, H. 1959 On the periodic fundamental solutions of the Stokes equation and their

application to viscous ow past a cubic array of spheres. J. Fluid Mech. 5, 317-328.Hinch, E.J. 1977 An averaged-equation approach to particle interactions in a uid suspension.

J. Fluid Mech. 83, 695-720.Kaneda, Y. 1986 The drag on a sparse random array of �xed spheres in ow at small but

�nite Reynolds number. J. Fluid Mech. 167, 455-463.Kevorkian, J. & Cole, J.D. 1981 Perturbation Methods in Applied Mathematics. Springer-

Verlag, New York, Ch 4.Lagerstrom, P.A. 1988 Matched Asymptotic Expansions. Springer-Verlag, New York.Leal, L. 1980 Particle motions in a viscous uid. Ann. Rev. Fluid Mech. 12, 435-476.Lovalenti, P. & Brady, J. 1993 The hydrodynamic force on a rigid particle undergoing

arbitrary time-dependent motion at small Reynolds number J. Fluid Mech. 256, 561-589.Proudman, I & Pearson, J. R. A. 1957 Expansions at small Reynolds numbers for the ow

past a sphere and a circular cylinder. J. Fluid Mech. 2, 237-262.Rubinstein, J. 1986 On the Macroscopic Description of slow viscous ow past a random array

of spheres. J. Stat. Phys. 44, 849-863.Sangani, A. & Acrivos, A. 1982 On the thermal conductivity and permeability of regular

arrays of spheres. In Macroscopic Properties of Disordered Media (ed. Burridge, Childress& Papanicolaou). Lecture Notes in Physics, vol. 159, pp. 216 -225, Springer.

Sangani, A. & Acrivos, A. 1983 The e�ective conductivity of a periodic array of spheresProc. R. Soc. Lond. A 386, 263-275.

Saffman, P. 1973 On the settling speed of free and �xed suspensions. Stud. Appl. Math. 52,115-127.

Van Dyke, M. 1975 Perturbation methods in uid mechanics. The Parabolic Press, Ch 8.Zick A. A. & Homsy G. M. 1982 Stokes ow through periodic arrays of spheres. J. Fluid

Mech. 115, 13-26.

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