Fluid flow and effective permeability of an infinite matrix containing disc-shaped cracks

Fluid flow and effective permeability of an infinite matrix containing disc-shaped cracks

Ahmad Pouya

^a,⇑

, Minh-Ngoc Vu

^b,c

aUniversité Paris-Est, Laboratoire Navier (ENPC/IFSTTAR/CNRS), IFSTTAR, 58 bd. Lefebvre, 75732 Paris, France

bBRGM/RNSC, F-45060 Orléans, Cedex 2, France

cLe Quy Don Technical University, 100 Hoang Quoc Viet, Hanoi, Viet Nam

a r t i c l e i n f o

Article history:

Received 16 August 2011

Received in revised form 7 March 2012 Accepted 8 March 2012

Available online 4 April 2012

Keywords:

Porous media Crack

Steady-state flow Poiseuille’s law Effective permeability Upscaling

a b s t r a c t

A basic equation governing the steady-state flow around a single crack in an infinite porous body is given.

The flow through the crack obeys to the Poiseuille’s law and the matrix has an anisotropic permeability. A semi-analytical solution is established for this equation in the case of elliptical disc-shaped crack. This solution takes a closed-form expression for the case of superconducting circular cracks. The results are compared to those obtained for flattened ellipsoidal inclusions obeying to the Darcy’s flow law, which are in some works supposed to represent the cracks. It is shown that the flow solution for an elliptical disc-shaped crack obeying the Poiseuille’s law is different from that obtained as the limiting case of flat- tened ellipsoidal inclusions. The results are then used to establish dilute Mori–Tanaka and self-consistent estimates of the effective permeability of porous media containing Poiseuille’s type elliptical cracks.

Ó2012 Elsevier Ltd. All rights reserved.

1. Introduction

The effective permeability of fractured rocks and micro-cracked porous materials is a great interest for many environmental and industrial applications such as hydrology, petroleum engineering, nuclear waste disposal, geothermal expansion etc. Numerous studies dealt with permeability of fracture network in three- dimensional media. Koudina et al. [1] presented a method to generate a triangular mesh on three-dimensional network of polygonal fracture surfaces, and used the finite volume method to compute the permeability of fractured media. The numerical tool developed in this way was employed to study the effective permeability of fractures network with statistical distribution[2].

However, the rock matrix was supposed to be impervious in these works which does not correspond to some real cases. Therefore, other theoretical and numerical investigations have focused on this problem[3–8]. A fundamental problem for modelling the effective permeability of cracked materials is the determination of the stea- dy state flow around a single crack in an infinite homogeneous ma- trix. In this work, we are interested in the case of cracks and fractures that can be represented geometrically by zero thickness surfaces with non pressure jump between their two faces, which

corresponds to an assumption of zero cross section resistance for the fracture. Therefore, we do not consider the cases of non-equi- librium flow[9]or of impervious fracture[10]. In many physical studies, the Poiseuille’s law is chosen preferentially to represent the flow in the cracks. The paper devotes to determine flow solu- tion around a single crack obeying to the Poiseuille’s law that is the key issue for determining the effective permeability.

The flow in cracks, embedded in an infinite matrix, has been determined in some works by using a model of ellipsoidal inclu- sion, in which the flow is governed by Darcy’s law [5,11]. It is implicitly supposed in these works that the flow in the crack with the Poiseuille’s law is obtained as the limiting case of a flattened ellipsoidal inclusion with thickness tending to zero. In this study, we show that this is not true, the solution obtained for the flow in and around an elliptical disc-shaped crack obeying to the Poiseuille’s law is different from that obtained as a limit of ellipsoi- dal inclusions obeying to Darcy’s law. Only some partial results, for instance, the pressure field in the case of superconducting cracks, are similar in the two problems.

To determine the solution of flow in a matrix containing a sin- gle elliptical disc-shaped crack, we use recent theoretical results established for the more general problem of multiple intersecting fracture surfaces [12]. The potential solution obtained for this general problem, as an extension of preceding works for the 2D case [13,14], is based on singular integral equations allowing to reduce the dimension of the problem from 3D to 2D and, hence, 0309-1708/$ - see front matterÓ2012 Elsevier Ltd. All rights reserved.

http://dx.doi.org/10.1016/j.advwatres.2012.03.005

⇑ Corresponding author.

E-mail address:ahmad.pouya@enpc.fr(A. Pouya).

Contents lists available atSciVerse ScienceDirect

Advances in Water Resources

j o u r n a l h o m e p a g e : w w w . e l s e v i e r . c o m / l o c a t e / a d v w a t r e s

particularly interesting for numerical methods. However, it will be seen in the sequel that for the case of a single elliptical disc- shaped crack, it allows also deriving semi-analytical solutions as an alternative way to the use of harmonic functions [15,16] or of dual integral equations methods[17,18].

The solution obtained for the single crack will be then used to estimate the effective permeability of the cracked porous materials by some classical upscaling methods.

1.1. Notations

In the sequel, light-face (Greek or Latin) letters denote scalars;

underlined letters denote vectors, bold-face letters designate sec- ond rank tensors or double-index matrices. The scalar product of two vectorsaandbis labelled asab. For second rank tensors, the tensor transposed fromAis denotedA^T, the matrix product is labelled asABand the determinant asjAj. The operation ofAon ais labelled asAa. The convention of summation on repeated indices isnotused for Latin indices (i,j,k. . .) that number surfaces, lines, etc., and are noted indifferently as subscript or upperscript.r represents the gradient andDthe Laplace operator for a scalar field and (r) the divergence for a vector field.

2. Governing equations

An infinite porous 3D bodyXcontaining an elliptical disc-frac- ture surfaceCis considered (Fig. 1). The points ofXare denoted by x= (x1,x2,x3) and, when they are located on the crack surface, by z= (z1,z2,z3). The surface C is defined by a function z(s) from R² ! R³parameterized bys= (s1,s2). Fluid flow through the ma- trix is governed by Darcy’s law:

8x2XC;

v

^{ðxÞ ¼ k} rpðxÞ ð1Þ where,kis the permeability tensor of porous matrix supposed to be uniform;

v

^{(x) and}p(x) the local velocity and pressure, respectively.

The local mass balance in the matrix reads:

8x2XC; r

v

^{ðxÞ ¼0 ð2Þ}

The flow in the fracture surface is characterised by a pressure field p(s) and an infiltration fieldq(s) in this surface. The total infiltration q(s) through a fracture section is determined by the integral of fluid velocity

v

over this physical section of the thickness 2eas follows:

8s2C; qðsÞ ¼ Z e

e

v

^{ðs;yÞdy ð3Þ}

The crack-matrix mass exchange at a point on the fracture surface excluded its boundary is established by considering the mass bal- ance in a small volume surrounding this point and reads[12]:

8x2CL;

v

^{ðxÞ nðsÞ þ}rsqðsÞ ¼0 ð4Þ where,rsq(s) is the surface divergence andLthe set of points on the crack boundaries. In addition, analysis of mass balance around a point on the crack boundary leads to the following mass balance condition at this point:

8zðsÞ 2L; qðsÞ mðsÞ ¼0 ð5Þ

In this equation,qis the infiltration vector in the crack andmis the outward unitnormalvector of the boundary line that is tangent to the crack plane.

The body is submitted at its infinite boundary to a pressure field p1(x):

kxk!1lim½pðxÞ p₁ðxÞ ¼0 ð6Þ p1satisfiesDp1= 0. For permeability upscaling, a constant pres- sure gradient is supposed to be prescribed to the matrix at its infi- nite boundary. Designating byAthe far-field pressure gradient, we have:

p₁ðxÞ ¼A x ð7Þ

The equations to be satisfied bypare the boundary condition(6), mass balances(2), (4), (5)and the constitutive Eq.(1). The solution obtained by Pouya[12]expresses the pressure field in the whole matrix as a function of the infiltration vector in the fracture as follows:

pðxÞ ¼p1ðxÞ þ 1 4

p

p^{ffiffiffiffiffiffi}jkj

Z

C

qðsÞ k¹ ðxzðsÞÞ

½ðxzðsÞÞ:k¹ ðxzðsÞÞ³⁼²ds ð8Þ In this equation,dsrepresents the surface element onC. This solu- tion satisfies well the far-field condition(6). Also, noting that the integrant function in the right-hand side integral is equal to q(s)rx(1/r) with r= [(xz)k¹(xz)]^1/2, the mass balance (2)can be shown forxnon situated on the fracture. To establish that (4)is well satisfied by(8)is little more technical and we refer to [12]for its demonstration.

It should be noted that the only significant assumptions limiting the validity of this solution are that matrix permeabilitykis as- sumed to be uniform and that there is no pressure gap between the two sides of the crack surface. Although we consider in the se- quel only the case of a plane crack surfaces, the solution(8)re- mains valid for curved-surface fractures, and, in its general form given in[12], for multiple intersecting fractures.

Also, in Eq.(8), the constitutive flow law in the fracture is not constrained. A linear relation between the infiltration vector and the pressure gradient, as an extension of the Poiseuille’s law for the viscous fluid flow between two parallel plates, is widely used in the literature to represent the flow in the crack:

qðsÞ ¼ cðsÞ rspðsÞ ð9Þ

The conductivity tensorcdepends on the fluid dynamic viscosity

l

, the hydraulic aperture of the fracturee, and also the roughness of fracture surfaces. In most cases,cis supposed isotropic (in the frac- ture surface) and replaced by a scalarc. In particular, the Poiseuille’s law, widely used for flow through rock fractures[19], determinesc from a laminar flow between two parallel plates separated by a constant distancee and leads to c=e³/(12

l

). However,c can be anisotropic in some cases as shown experimentally by Gentier et al.[20]. Introducing (9) into (8) leads to an integral equation involving only the unknown pressure field. A simple analysis shows that the solution for a constant pressure gradient along the disc Fig. 1.An infinite porous medium containing a flat elliptical disc shaped crack.

surface, corresponding to the Eshelby-type ellipsoidal inclusion with Dracy’s law, does not satisfy this equation. The comparison be- tween the Posieulle-type crack and the flattened ellipsoidal inclu- sion with Darcy’s law will be discussed further in this paper.

However, in this paper we are interested in the case of super- conducting cracks corresponding to the limiting casec?1. This is an important limiting case, currently addressed in the literature because it allows determining the upper bound of the effective per- meability for fixed crack density and geometry.

3. Superconductive elliptical crack-inclusion

In the case of superconducting fracture, the pressure remains constant along the fracture surface. We consider an elliptical disc shaped crackDand denote byd1andd2the half-diameters of the ellipse (SeeFig. 2) by the unit vectorse1ande2its principal direc- tions and bye3the unit normal to the crack plane. A coordinate system is defined with its origin located at the crack centre and its axes parallel toe₁,e₂ande₃. The equation of the ellipse reads xBx= 1 forxe3= 0. We note:

k1¼1=d²₁; k2¼1=d²₂; B¼k1e1e1þk2e2e2 ð10Þ The general result needed for permeability upscaling is the integral of the infiltration over the disc surface. This integral is a linear func- tion of the far-field gradientAand can be written asWA, where the tensorWis a function ofc,kandB. By application of the linear transformation[12], the problem can be changed into that of an ellipsoidal disc in an isotropic matrix. Nevertheless, a numerical method is required to solve the general problem for the case of fi- nite and anisotropic conductivity tensorc. In the following, we con- sider the particular case of superconducting crack, c?1, that allows us to derive an analytical or semi-analytical solution.

Eq.(8) written for a single elliptical crackDembedded in an infinite porous matrix with isotropic permeabilitykyields:

pðxÞ ¼A x 1 4

p

k

Z

D

qðnÞ ðnxÞ

knxk³ dn ð11Þ

In this equation,dnrepresents the surface element onD. The gen- eral solution can be obtained by superposition method. It requires three elementary solutions forAparallel toe1,e2ande3. A simple analysis shows that forAparallel toe3the solution is a uniform gra- dient pressure not perturbed by the presence of the crack, i.e., p(x) =Axin the whole body. To build the complete solution, it is sufficient to determine the solution for A parallel to the x-axis.

Hence, we try to solve the problem forp1(x) =A1x1.

The condition of superconducting crack imposes that the pres- surepmust be constant in the crack surface, and because of sym- metry considerations, this constant is zero. Therefore, designating, for a fieldq(n), by:

IðqÞ ¼ Z

D

qðnÞ:ðnxÞ

knxk³ dn ð12Þ

the solutionqmust satisfy:

IðqÞ ¼4

p

kA1x1 ð13Þ

To find a solution for this problem, we try first the following expres- sion which is inspired by the 2D solution given by Pouya & Ghabez- loo[14]:

q⁰ðnÞ ¼h1

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1n B n

q e1 ð14Þ

As shown in Appendix A, the expression found for I is a linear function ofx1forq=q⁰which allows us to build the solution for the Eq. (13). The general expression found for this solution has

recourse to thecomplete elliptic integralsof first and second kind K(k) andE(k) defined by:

KðkÞ ¼ Z p=2

h¼0

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffidh 1k²sin²h

p ; EðkÞ ¼

Z p=2 h¼0

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1k²sin²h

p dh; ð06k61Þ

ð15Þ The semi-analytical or iterative numerical methods used to calcu- late thesecomplete elliptic integrals K(k) andE(k) can be found in mathematical handbooks[21].

Moreover, we define:

/1ðk1;k2Þ ¼Rp=2 h¼0

k1cos²h

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

k1cos²hþk2sin²h

p dh

/2ðk1;k2Þ ¼Rp=2 h¼0

k2sin²h

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

k1cos²hþk2sin²h

p dh

8>

<

>: ð16Þ

It can be noticed that, if d1Pd2, denoting byk¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 ^d_d²

1

r 2

, we have:

/₁ðk1;k2Þ þ/₂ðk1;k2Þ ¼_d¹₂EðkÞ d²₁/₁ðk1;k2Þ þd²₂/₂ðk1;k2Þ ¼d2KðkÞ (

ð17Þ

This system of equations allows the determination of/1and/2as function ofK(k) andE(k). Then, forq=q⁰, it is shown inAppendix Athat:

I¼ 2h1

p

/1ðk1;k2Þx1 ð18Þ

As a result,(13)is verified if we takeh1¼^2k_/

1A1, and therefore:

q⁰ðnÞ ¼ 2k /1

A1

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1n B n

q e1 ð19Þ

According to this expression, the fluxq⁰in the crack has a constant direction parallel to the far-field gradient, as in the case of ellipsoi- dal Darcy type inclusions. However,q⁰is not the real solution of the flow in the disc. As a matter of fact, it must be noted that the solu- tion of the equation(13)is not unique. If the vectorx(n) satisfies rx= 0 in the disc andxn= 0 on its boundary wherenis out- ward unit normal vector to the boundary, then we calculate:

Ið

x

Þ ¼ Z

C

x

nÞ rðknxk¹ dn

¼ Z

@C

knxk¹

x

ðnÞ n dsþ Z

C

knxk¹r

x

dn¼0 ð20Þ Consequently,q=q⁰+xalso satisfies(13). This degeneracy of the solution can be removed if we suppose that the superconducting crack is the limit case of a crack with the conductivityc=

r

c0where c0is constant and

r

?1. This implies thatc¹₀ qmust be a gradient field (pressure gradient) and satisfy the condition of symmetry of crossed derivatives. This condition allows us to determine a unique qsaved thatc0is symmetric and positive definite. In the case of a superconducting crack considered as the limit case of an isotropic conductivity crack, this condition implies thatqhas to be a gradient field. It can be checked thatq⁰given by(19)is not a gradient field. A complementary partx(n) should be added toq⁰to make it a gradient field satisfying(13). Thus, the solution of the infiltration fieldqreads:

qðnÞ ¼q⁰ðnÞ þ

x

ðnÞ ð21Þ

where, the complementary part x has to satisfy the following conditions:

8n2D; r

x

¼0 ð22Þ

8n2@D;

x

n¼0 ð23Þ

8n2D; @1 q⁰₂þ

x

2

¼@2 q⁰₁þ

x

1

ð24Þ The last condition assures thatqis a gradient field. Mathematically, Eq.(22)is satisfied if we take:

x

1¼@₂

u

;

x

2¼ @₁

u

ð25Þ with

u

an arbitrary function. Then, the condition(24)yields:

Ds

u

¼2kA1

/1

@₂ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1nBn

q ð26Þ

where,D_s

u

¼@²₁þ@²₂

u

. Substituting(25)into(23)and rearrang- ing the terms, we find:

8n2@D; r

u

t¼0 ð27Þ

wheretis the unit tangent vector to the crack boundary@D. This equation implies that

u

is constant on the boundary, hence:

8n2@D;

u

¼

u

0 ð28Þ

with,

u

0a constant. A solution for the Poisson’s equation(26)over the elliptical domain exists for every Dirichlet boundary condition and in particular(28)with constant

u

0. We can take

u

0= 0 because the addition of a constant value to

u

does not change the solution x. The solution of Eq.(26)with

u

= 0 on the boundary can be deter- mined numerically in the general case and analytically in some spe- cial cases. Two particular cases in which the solution can be derived analytically are the following:

Case 1: Limit of infinitely long elliptical disc

We suppose the limit case of elliptical disc with infinite exten- sion in the direction 2. Geometrically, this corresponds to an infi- nite band of width 2d1 parallel to the direction 2. The solution, for a far-field gradientA1parallel toe1corresponds to that of the 2D case with a crack of length 2D in the plane (x1,x2) parallel to x1. The solution can therefore be compared to that given for this case by Pouya and Ghabezloo[14]. In the 3D problem,d2?1 andk2= 0 andq⁰ðnÞ ¼ 2k

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi d²₁n²₁ q

A1e1is a gradient field. Thus, x= 0. The exact solution is then qðnÞ ¼ 2k

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi d²₁n²₁ q

A1e1 which is exactly the solution found in 2D case by Pouya and Ghabezloo [14].

Case 2: Circular crack with radius R

In this case, d1=d2=R. We find /₁¼/₂¼_4R^p and q⁰ðnÞ ¼

8kA1

p

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi R²r²

p e1 with r²=nn. In order to find the expression of the complementary functionx, we look for a solution of(26) in the following form:

u

ðnÞ ¼

v

ðnÞ þ@2fðrÞ ð29Þ

where,f(r) satisfies:

Dsf¼ 8kA1

p

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi R²r²

p ð30Þ

Then,vhas to satisfyDv= 0. The general solution of(30)is:

fðrÞ ¼ 8kA1

p

1

3R² ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi R²r²

p 1

6R³LnR ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi R²r² p Rþ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

R²r² p

"

1

9ðR²r²Þ³⁼²þC1LnrþC2

#

ð31Þ

where,C1,C2are two constants. Thus,

@2f¼ 8kA1

p

n2

r² 1

3ðR²r²Þ³⁼²C1

ð32Þ Substituting this expression in(29) and taking into account the boundary condition

u

(R) = 0, we find that

v

¼ C18kA₁

p n₂ R² on the boundary. Hence, the solution of the equation Dv= 0 is

v

ðnÞ ¼ C18kA1

p n₂

R²and therefore the solution for

u

is:

u

¼ 8kA1

p

n2

r² 1

3ðR²r²Þ³⁼²C1

C1

8kA1

p

n2

R² ð33Þ

The constantC1is determined by the condition thatqhas to be fi- nite in the crack or that

u

is not singular whenr?0. This imposes C1¼^R₃³. Substituting by this expression in(25) and (21)we find the solution of infiltration in the circular disc under the far-field pres- sure gradientA1as follows:

qðnÞ ¼ 8kA1

p

n²₁ r²

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi R²r²

p n²₁n²₂

3r⁴ R³ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi R²r² p

³

þR 3

" #

e1

8kA1

p

2n1n2

3r⁴ R³ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi R²r²

p ³

þn1n2

r² ðR²r²Þ¹⁼²

e2

ð34Þ The two components q₁(n), parallel to the far-field gradient and q2(n), normal to this gradient, are represented in the Fig. 3 for A1= 1 andk= 1.

The elementary solution for a far-field pressure gradientA2par- allel toe2is obtained from(34)by index changing 1M2. As men- tioned above, the solution for the general case of a far-field pressure gradientAis obtained by superposition of the three ele- mentary solutions. By introducing the notation:

r¼ ffiffiffiffiffiffiffiffiffiffiffi n n q

; ^n¼1

rn ð35Þ

T¼dnn ð36Þ

wherenis the unit normal vector to the crack plane, the general solution obtained forqhas the following expression:

qðnÞ ¼8k

3

p

r² Rr²R³þ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi R²r²

p ³

d

þ 3r² ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi R²r² p

þ2R³2 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi R²r²

p 3

ð^n^nÞ

TA ð37Þ It is important to note here a fundamental difference between this result and what could be obtained as the limiting case of flow solu- tion in a flattened ellipsoidal inclusion. The comparison between the two models is presented farther in the paper. But it is already possi- ble to note the flow in the elliptical crack given by(37), as seen also in theFig. 3,is not uniform and parallel to the farfield gradientcontrary to the case of ellipsoidal inclusions in which the fluid velocity is found to be uniform and parallel, at least if the matrix and inclusion are isotropic, to the farfield gradient. More precisely, for instance for a far-filed pressure gradient parallel to thex-axis, the solution given by(37)has anon vanishing component in the y-axis directionas shown by the expression(34)(see theFig. 3in the paper). However, for the same problem modelled by a flattened spheroidal inclusion,the fluid velocity in the inclusion would be parallel to the x-axis, so with non component in they-axis direction. This shows the fundamental dif- ference between the two types of flow and the fact that the solution for the Poiseuille crack cannot be obtained as the limiting case of ellipsoidal inclusions with Darcy’s law.

Fig. 2.Geometrical parameters defining the collocation pointxand the integration pointnin the elliptical crack surface.

The solution(37)could be probably obtained by other mathe- matical methods for instance based on the use of harmonics func- tions instead of the potential solution(11). But for the difference underlined here-above, it cannot be obtained as a limiting case of solutions presented in the literature for ellipsoidal inclusions obey- ing Darcy’s law.

The solution(37)represents an original and fundamental result that opens the way for further theoretical investigations concern- ing flow in cracked porous materials.

4. Pressure field in the matrix

The solution of the infiltrationqthrough the crack in(37)allows the determination of the pressure field in the matrix by using Eq.

(11).

For the discussions carried on farther in this paper, it is useful to decomposeqinq⁰andxparts according to(21)and to write:

pðxÞ ¼Ax 1 4

p

k

Z

D

½q⁰ðnÞ þ

x

ðnÞ ðnxÞ

knxk³ dn ð38Þ

The Eq.(20)shows that the contribution ofxto the integral van- ishes. Besides, we can deduce from(19), by superposition and by noting that a far-field gradient parallel toe3does not induce flow in the fracture, that, for a general far-field gradientA,q⁰is given by:

q⁰ðnÞ ¼ 2k ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1nBn q

UA ð39Þ

where,

U¼ ð/1Þ¹e1e1þ ð/2Þ¹e2e2 ð40Þ Substituting(40)into(38)yields:

pðxÞ ¼ x 1 2

p

^U

Z

D

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1nBn p

knxk³ ðnxÞdn

!

A ð41Þ

The pressure field solution given in this way is an explicit integral function of the geometrical parameters of the elliptical disc con- tained in the tensorB. It is interesting to notice that this solution is independent of the matrix permeabilityk. This is due in particular to the assumption of superconducting crack: the pressure solution would depend onkfor crack with finite conductivity. Anyway, the Eq.(41)allows us to determine at least numerically the pressure field around an elliptical crack by direct integration.

For the case of circular disc (B=R²T) a closed-form expression for the integral(41)can be derived for a point located on the plane of the crack, i.e.x3= 0. We find (seeAppendix B):

Z

D

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1nBn p

knxk³ ðnxÞdn¼

p

2RðWsinWÞx ð42Þ where,Wis, as shown inFig. 4, the angular diameter of the disc seen from the positionx.

Due to superconducting crack,p(x) = 0 whenxis in the crack.

We also haveU= (4R/

p

)Tand the expression ofpfor the points exterior to the disc is given by the following expression:

x e3¼0; pðxÞ ¼ 11

p

^ðWsinWÞ

Ax ð43Þ

It can be checked that whenx?1thenW?0 and sop(x)?Ax according to(43). The infinite boundary condition is recovered in this way. Also, whenxis on the boundary of the disc,W=

p

/2 and we find the pressure condition in the disc,p(x= 0).Fig. 5displays the variation of pressure field in the plan containing the supercon- ducting fracture with a far-field pressure gradientA= (1, 0, 0).

It could be noted that this pressure solution could be, in princi- ple, obtained as the limiting case of the pressure field in an infinite matrix containing a superconducting ellipsoidal inclusion. The two problems obey, as a matter of fact, to the same equations and boundary conditions: constant pressure in the crack or inclusion, constant farfield pressure gradient and mass conservation equation in the matrix. The solution could be then obtained as a classical application of potential theories using harmonics functions. How- ever, we have not found in the literature any explicit expression of this solution as given by(43).

5. Comparison between Darcy-ellipsoidal inclusions and Poiseuille-crack

The crack inclusion has been sometimes considered as the lim- iting case of a penny-shaped inclusion with aspect ratio (ratio be- tween its thickness and diameter) tending to zero. However, a further investigation shows this similarity holds only for some Fig. 3.Closed-form solution of infiltrationqin fracture of unit radius in an infinite isotropic matrix under a far-field pressure gradient inx1-direction:q1component parallel to the far-field gradient,q2orthogonal to the far-field gradient.

Fig. 4.Parameters in polar coordinate system.

partial results and in restricted cases and this difference seems to be ignored in the literature.

To show this difference, let a circular penny-shaped inclusion be considered with the revolution axis taken as the third axis of coor- dinates and with semi-diametersd1=d2=d,d3d(SeeFig. 6). The inclusion has an isotropic permeabilityk^⁄embedded in the matrix of isotropic permeabilityk. The fluid velocityvin the inclusion is found to be uniform and its value can be deduced, for instance, from the results given by Shafiro and Kachanov[11]. We consider the limit when

e

=d3/d?0 and k^⁄?1. The contribution of the inclusion to the equivalent permeability of the material is made through the integral of the fluid velocity in the inclusion:

Q¼ Z

X

v

^{ðnÞdn ð44Þ}

This integral is a linear function of the far-field gradientA:

Q¼ WA ð45Þ

Shafiro and Kachanov[11]gave the following general expression of Wfor an arbitrary valuek^⁄of the inclusion and

e

:

W¼X X^ðk

kÞ 1þkk k

pe

4

¹

ðdnnÞ (

þ 1þkk k 1

pe

2

¹

nn )

ð46Þ

where,XandX^⁄represent respectively the matrix reference vol- ume and the inclusion volume.

Considering the limiting case of a superconducting crack by tak- ing the limit

e

?0 andk^⁄?1Shafiro and Kachanov[11]proposed the following expression for this limit:

W¼16

3

m

R³kðdnnÞ ð47Þ

where,

m

is the crack density. Nevertheless, a careful study shows that the limit is not unique and depends on the value of

e

k^⁄which is degenerate when

e

?0 andk^⁄?1. Let us suppose thatk^⁄=bk/

e

where b is a fixed ratio, hence,

e

k^⁄=bk remains constant when

e

?0 andk^⁄?1. The limit ofWfrom the Shafiro and Kachanov [11], is obtained to be:

W_¼¹⁶

3

m

R³k b

bþ ð4=

p

ÞðInnÞ ð48Þ The limit expression (47) corresponds then to the assumption b?1.

Let us consider now the ellipsoidal inclusion havingk^⁄tends to infinity andd3 tends to zero. At the boundary, we note

v

^{m the}

velocity in the matrix and

v

ⁱ the velocity in the inclusion. The velocity in the inclusion is constant for ellipsoidal inclusion. We note

v

ⁱ⁼

v

^e1. At the boundary, the mass balance yields

v

ⁱⁿ⁼

v

^mn,

where n is the unit normal vector to the ellipsoid, and then,

v

^mn=vn1wheren1¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi^ex1

d²r²ð1e²Þ

p .

The general value of velocity

v

in the inclusion, for small ratio

e

=d3/d, can be deduced from the expression given by Shafiro and Kachanov[11]for the integral(44)since

v

is constant in the inclusion. We find, in first order development with respect to

e

:

v

¼ kk

1þ^pe₄ ^k_k^k ð49Þ

The equivalent infiltration vector in the crack section is deduced from the relation(3)where the thicknesseis given, for the ellipsoi- dal inclusion, bye¼

e

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

d²r²

p and

v

is constant and given by(49).

The limit of the equivalent infiltrationq= 2e

v

^for

e

?0 is found to be:

lime!0q¼ 8kc

p

cþ8dk

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi d²r² q

ð50Þ where,c= 2d3k^⁄. We note that for the limit casec?1, the Eq.(50) yields a solution different from that given by(37). As a matter of fact, it yields only the expression ofq⁰in the Eq.(19)for circular disc (/1=

p

/4d), which is not the complete solution for infiltration in the Poiseuille’s crack since it misses the complementary partx. It can also be checked that the expression(50)does not satisfy the Eq.(11). It is then deduced that, in the general case of finite conductivityc, the solution of flow in the crack with Poiseuille’s law is not given as the limit of thickness vanishing ellipsoidal inclusions with Darcy’s law. This implies also that the pressure fields in the inclusion and the matrix are different for the two problems.

Only in the case of superconducting crack, the pressure fields in the two problems are the same since the pressure is constant in the crack or the inclusion. In this case, the integral ofqover the inclu- sion (determining the contribution to the effective permeability, see the next section) is also the same for the two models since it depends only onq⁰and not on the complementary partx.

In conclusion, the limit of the flattened inclusion with Darcy’s flow does not give in the general case the same result of pressure and infiltration fields that the model of Poiseuille-type crack.

6. Application to the effective permeability of cracked porous materials

The effective permeabilityk^ of the heterogeneous domain is calculated from the average velocity and average pressure gradient in this domain. For a domainXcontaining a distribution of zero- thickness fracturesC^j, the equivalent average velocity is defined by:

V1 X

Z

X

v

d

x

þP

j

Z

C^j

qds

" #

ð51Þ

It can be shown that if a pressure condition corresponding to a uni- form macroscopic gradient is applied on the boundary ofX, i.e., p(x) =AxonoX, the average pressure gradientGis equal toA:

Fig. 5.Pressure field around a single superconducting circular disc crack in an infinite matrix prescribed by a constant farfield pressure gradient parallel toX direction.

G¼1 X Z

X

rp dX¼A ð52Þ

We assume that the permeability tensor of the porous matrix is constant and equal tok. Introducing(1) and (52)into(51)and using V¼ ^kGyield:

^k¼kþk^f ð53Þ

where,k^fis the contribution of the cracks to the effective perme- ability obtained by the following relationship:

1 X

P

j

Z

C^j

qds¼ k^fA ð54Þ

The dilute Mori–Tanaka scheme estimates the left-hand side of(54) as the sum of the integral ofqover each individual crack by neglect- ing the effects of cracks interactions. As seen in Section3, the infil- tration in the crack is expressed byq=q⁰+x. According to(25),x can be written as:x=e3 r

u

. Using this relation and(28), we calculate:

Z

D

x

ðnÞds¼ e3^ Z

D

r

u

ds¼ e3^ Z

@D

u

n dl¼

u

₀

Z

@D

t dl¼0 ð55Þ Therefore, onlyq⁰contributes to the integral over the crack surface.

From the results given in Section3, we deduce:

Q¼ Z

C

qðnÞds¼ Z

C

q⁰ðnÞds¼ 4k

3 SCUA ð56Þ

where, the tensorUdefined by(40)andSC=

p

d1d2=

p

/jBj^1/2repre- sents the area of the ellipseC. Comparing this relation to(54), the Mori–Tanaka estimate of k^f for an isotropic matrix containing a family of elliptical cracks is found as follows:

k^f ¼4

3

m

khSCUi ð57Þ

where,

m

is the crack density, i.e: number of cracks (centres) per unit volume andhSCUiis the average value ofSCUcalculated over the cracks population. For an isotropic distribution of circular cracks with radiusR, we find:

k^f ¼32

9 k

m

R³d ð58Þ

and the following expression for the dilute Mori–Tanaka effective permeability:

^k^MT¼k 1þ32 9

m

R³

ð59Þ The self-consistent estimate is obtained by replacing, when calcu- lating the contribution of cracks to the effective permeability in equation(58), the matrix permeability by the unknown effective permeability. Therefore, Eq.(59) leads to the following equation for the self-consistent estimation^k^sc:

^k^sc¼kþ32 9

^k^sc

m

R³ ð60Þ

Furthermore,

^k^sc¼ k

1³²₉

m

R³ ð61Þ

The dilute Mori–Tanaka(59)estimation is found to give the first or- der development for small

m

of the self-consistent estimation(61) (Fig. 7). By introducing the dimensionless crack density

m

¼⁴₃

pm

R³, the self-consistent estimation becomes singular for a critical value

m

¼³_p1:18. The model is valid only for small values of

m

and we do not think a physical signification can be attached to this singu- larity that appears for relatively high values of crack density.

It is worth noting that occasionally this singularity has been re- lated to the percolation limit of the cracks network[5]. However, Dormieux and Kondo [5] obtained a different critical value for self-consistent model,

m

¼³₄^p2:36. The percolation threshold has sometimes been studied numerically. When the domain size is finite, some fracture network configuration percolate and some do not for a given density. It can be only defined a percolation threshold

m

cfor the limit of infinite size of domain. Studying the hexagonal fracture networks with isotropic distribution, Bogdanov et al.[3]found that there is rarely percolation for finite fracture network with

m

⁶0:42; percolating and nonpercolating fracture network coexist for the range of densities 0:426

m

⁶1:70 and al- most all the fracture are connected when

m

P1:70. Huseby et al.

[22] gave a more accurate result for the percolation threshold

m

P1:00, for infinite matrix, which is very close to the critical va- lue obtained in this work by the self-consistent scheme.

7. Extension to anisotropic matrix

The linear transformation presented in Pouya[12]can be used to extend the solutions obtained in the previous sections for ellip- tical crack in an isotropic matrix to the case of anisotropic matri- ces. Considering a reversible tensorM, we define thetransformed variables~x;

m

~ð~xÞandpð~~xÞas follows:

~x¼M ~x; ~

m

ð~xÞ ¼M

m

ðxÞ; ~pð~xÞ ¼pðxÞ ð62Þ Therefore,~x¼Mxtransforms the domainXto a domainXe, the cracks Cto a new geometry Ce The flow problem is then trans- formed to a new problem corresponding to matrix permeability.

Let us consider an infinite homogeneous porous medium embedding an elliptical crack with a geometry defined by two principal axese₁ande₂and half-diametersd₁andd₂. A pressure gradientAis imposed at the infinite boundary of the domain. We refer to Pouya[12]for the complete set of transformation relations x2

x1

d d3

x3

Fig. 6.Darcy ellipsoidal inclusion.

Fig. 7.Self-consistent and Mori-Tanaka estimations of effective permeability.

for different variables in which the tensor of permeability is taken the following expression:

k~¼MkM^T ð63Þ

The transformation to the problem studied in previous section is applied by takingM¼ ffiffiffiffiffiffiffiffi

k¹ p

. In the transformed problem, we find an isotropic matrix with unit permeability:~k¼d. The equation of the elliptical cracknBn= 1 is equivalent to ~nM^1;TBM¹~n¼ 1 or~neB~n¼1 with:

eB¼ ffiffiffi pk

B ffiffiffi pk

ð64Þ The tensorBedefines the geometry of the crack in the transformed problem. In order to calculate the half-diameters~d1;d~2and the prin- cipal directions~e1;~e2(unit vectors) of the transformed ellipse, we need to write:

eB¼~k1~e1~e1þ~k2~e2~e2 ð65Þ

The expressions~k1and~k2, and of~/1and/~1 that can be deduced from~k1and~k2by the same relations(16)are required for writing the solutions of pressure and velocity in the transformed problem.

However, if only the pressure field in the matrix in Eq.(41)or the application to effective permeability in Eq.(56)is envisaged, it is easier to express directlyUas function ofB(seeAppendix C):

U¼fdþgB ð66Þ

where,fandgare scalar functions ofB=B:dandb=jBj.

As a matter of fact, the pressure transformation ~pð~xÞ ¼pðxÞ leads toeA~x¼Axat infinity; hence,Ae¼ ffiffiffi

pk

A. Moreover,~q⁰ð~nÞ in the transformed problem is given by(39)such as:

~q⁰ð~nÞ ¼ 2~k

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1~nB~n

q UeeA ð67Þ

where~k¼1 and:

e

U¼fðeB;~bÞdþgðeB;bÞ~ eB ð68Þ The general transformation rule for infiltration vectors is (see in Pouya[12]):

~q⁰¼ 1 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi nkn p

ffiffiffiffiffiffiffiffi k¹ q

q⁰ ð69Þ

Replacing by these values function of initial variables yields:

q⁰ðnÞ ¼ 2 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi nkn

p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1nBn

q ffiffiffi

pk e U ffiffiffi

pk A

witheBgiven byBe(64),eB¼eB:d¼k:Band~b¼ jBj ¼ jkkBje and the expressions the functionsfandggiven inAppendix C. In this way, q⁰(n) is calculated explicitly for arbitrary elliptical shaped crack in a matrix with arbitrary anisotropic permeabilityk. The pressure field in the matrix is deduced from:

pðxÞ ¼Axþ 1 4

p

p^{ffiffiffiffiffiffi}jkj

Z

D

q⁰ðnÞ k¹ðnxÞ

kðnxÞ k¹ ðnxÞk³⁼²dn ð70Þ For extension of the results concerning the effective permeability, it is sufficient to write(57)for the transformed problem:

k~^f ¼4

3~

m

kh^~eSCUie ð71Þ

Noting thatk^¼1;

m

~¼

m

=jMj ¼

m

p^{ffiffiffiffiffiffi}jkj

; eS_C¼

p

= ffiffiffiffiffiffiffi jeBj q

; k^f¼ ffiffiffi pkk~^f ffiffiffi

pk , and using the expressions obtained here above for UandB, we find:

k^f ¼4 3

m

p^ffiffiffik

hSCUie ffiffiffi pk

ð72Þ

Substituting(72)into(53)allows us to determine the effective per- meability of a porous body with anisotropic matrix permeabilityk containing a distribution of elliptical superconducting cracks by using the Mori–Tanaka or the self-consistent estimate.

In practice, field observation permits us to determine the cracks geometries (size, aperture and orientation) and also their density [23]. An estimation of the crack conductivity is also required that can be obtained by a Poiseuille’s model. However, in our work, only superconducting disc-shaped cracks are considered. The crack net- work is assumed to be randomly distributed. With these data, the models given in Section6are suitable to estimating the effective permeability.

8. Conclusions and perspectives

The equation governing the steady state flow around a crack surface in an infinite porous body, deduced from previous investi- gations[12], was considered in this paper. An analytical solution was derived for this equation in the case of a superconducting and elliptical disc shaped crack.

Cracks have been very often modelled as flattened ellipsoidal inclusions obeying Darcy’s law because for this problem analytical results are available. The question raises then if this model can rep- resent well a crack with zero thickness and obeying to the Poiseu- ille’s law. In the literature, it is often implicitly responded positively to this question: it is assumed that the flow field in and around a crack with Poiseuille’s law can be obtained as the limiting case of the flow field in and around a flattened ellipsoidal inclusion with Darcy’s law when its thickness tends to zero. The re- sults obtained in the present work allowed the comparison. We could show that the equivalence between the two models does not hold in the general case. Only some partial results in restricted cases, mainly the case of superconducting cracks, can be obtained as a limit case. In general, for cracks with finite conductivity, the pressure gradient through the mean crack surface is constant in ellipsoidal inclusion model but not in Poiseuille’s type crack. These results change fundamentally the vision we can have for the flow in a Poiseuille’s type crack.

The results obtained for a single crack have been used to esti- mate the effective permeability of cracked porous materials. The dilute Mori–Tanaka model is a very simple model that does not take into account the interaction between cracks. The expression of the effective permeability is a linear function of the crack den- sity according to this model, for anisotropic matrix permeability and elliptical disc cracks. The crack interaction is taken into ac- count implicitly in the self consistent model. We established the exact expression of the effective permeability according to this model for circular cracks in a matrix with isotropic permeability.

This non linear model presents a singularity for a critical value of crack density that is sometimes interpreted as representing the percolation threshold of crack network. This critical value is then compared to analogous values given in the literature, and found to be very close to that obtained by numerical simulation for the percolation threshold of crack networks.

The analytical solutions given in this paper open the way for dee- per theoretical investigations. The effective permeability of micro- cracked materials can take benefit from these results as well as the study of crack interactions within a porous body. Also it is worth reminding that a fundamental assumption underlying the present work was the infinite transversal conductivity of cracks that re- moves the possibility of pressure jump between to sides of the crack.

But in some physical cases, cracks act as an impervious membrane with a pressure discontinuity across the crack surface[3,10,24].

The approach used by Martin[24]could be used in this case to ex- tend some of the results of the present paper to these types of cracks.

Analytical infiltration solution through a single crack obtained here was restricted to the case of superconducting cracks. For the case of Poiseuille-type cracks with finite conductivity, semi-analyt- ical and numerical calculations are required that we have investi- gated in an ongoing work and we hope present the results in a future paper. Moreover, numerical study allows us to take into consideration explicitly the crack interaction and the crack inter- section. The three-dimensional effective permeability model that can be obtained in this way would allow extending the two-dimen- sional application to CO2 storage [25] to a three-dimensional modelling.

Appendix A. Integral calculation for a point inside the elliptical disc

Replacing by the(14) in the main text in (12), the following equation(A.1) is found in whichBis diagonal with eigenvalues k1,k2, andnBn¼k1n²₁þk2n²₂. We use fornthe polar coordinates (

q

,h) in the local coordinate system havingxas origin (seeFig. 2).

We haven1=x1+

q

cosh,n2=x2+

q

sinh. Thus, we have:

I1¼ Z

D

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1nBn p

knxk³ ðn1x1Þdn

¼ Z 2p

h¼0

Z qm

q¼0

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1nBn p

q

^d

q

coshdh ðA:1Þ

It is obvious that this integral must be considered as the Cauchy principal value:

I1¼lim

e!0I1e; I1e¼ Z 2p

h¼0

J_eðhÞcoshdh ðA:2Þ where:

J_eðhÞ ¼ Z qm

e

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1nBn p

q

^d

q

¼

c

e Z qm

e

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 ðe

q

þgÞ²

q d

q

q

^ðA:3Þ

with:

c

¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi k1cos²hþk2sin²h q

; H¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

c

²k1k2ðx1sinhþx2coshÞ² q

e¼

c

²

H; g¼k1x1coshþk2x2sinh H

ðA:4Þ We take the variable

a

:

sin

a

¼e

q

þg ðA:5Þ

When

q

varies between

e

and

q

m,

a

varies between

a

e=Arcsin (

qe

+g) to

p

/2. In this interval, cos

a

P0, and variable change

q

?

a

allows calculating:

J_eðhÞ ¼

c

e Z p=2

a¼ae

1g²

sin

a

g ðsin

a

þgÞ

d

a

ðA:6Þ

and find:

J_eðhÞ ¼

c

ð1g²Þ

e FeðhÞ

c

e½cos

a

eþgð

p

=2

a

eÞ ðA:7Þ where:

FeðhÞ ¼ Z p=2

a¼ae

1

sin

a

gd

a

ðA:8Þ

Now we can write:

I1;e¼

Z p

h¼0

½J_eðhÞ J_eðhþ

p

Þcoshdh ðA:9Þ

When changinghtoh+

p

, the functions

c

,e,Hremain unchanged, but g is changed to g,

a

e to

a

⁰_e¼Arcsinð

qe

gÞ and the ue to u⁰_e¼tan

a

⁰_e=2

. It can be shown that:

lime!0½FeðhÞ Feðhþ

p

Þ ¼ Z p=2

p=2

1

sin

a

gd

a

ðA:10Þ where the integral in the right-hand side must be considered as the Cauchy principal value and can be shown to take the following values:

Rp=2 p=2

1

sinagd

a

¼ ^p

g ffiffiffiffiffiffiffiffi

1¹ g2

q if jgj>1

¼0 if jgj<1 8>

<

>: ðA:11Þ

By using this result (with jgj< 1) and the limits lime!0

a

e¼ lim_e!0

a

⁰_e¼ArcsinðgÞ, we find:

I1¼

p

Z p

h¼0

c

g

e coshdh ðA:12Þ

Replacing by(A.4)for

c

,gandein this expression and after integra- tion we obtain:

I1¼ 2

p

k1x1

Z p=2 h¼0

cos²h ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi k1cos²hþk2sin²h

q dh¼ 2

p

/1ðk1;k2Þx1

ðA:13Þ where the function/1(k1,k2) is defined by the Eq.(16)in the main text.

Furthermore, using the same notation withx1=x2= 0, and thus, g= 0 ande=

c

, we have:

Z 2p h¼0

Z q_m q¼0

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1nMn

q

q

d

q

dh¼

Z 2p h¼0

Z 1=c q¼0

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1

c

²

q

²

q

q

d

q

dh

¼ Z 2p

h¼0

dh 3

c

^2¼

2

p

3 ffiffiffiffiffiffiffiffiffiffi k1k2

p ðA:14Þ

Appendix B. Integral calculation for a point outside the circular disc

As seen inFig. 4in the text, the integral in the left-hand side of (42)can be re-written in the polar coordinate system as follows:

I¼ Z

D

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1nBn p

knxk³ ðnxÞdn

¼ Z W⁰

W⁰

Zq₂

q1

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1nn=R² q

ðucos

u

þ

v

^sin

^u

^Þ

q

^d

q

dh ðB:1Þ

where:

W⁰¼W=2 ðB:2Þ

Also, we can write ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1nn=R² q

¼_eR¹

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 ðe

q

þgÞ² q

with: e¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1

R²kxk²sin²u

p andg=ekxkcos

u

. Then the symmetries with respect to

u

?

u

imply that we can write: I¼ RW⁰

W⁰ 1

eRJð

u

Þcos

u

d

u

u where:Jð

u

Þ ¼Rq2

q1

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

1ðeqþgÞ²

p

q d

q

For the same variable change(A.5),

a

varies between

p

/2 to

p

/ 2 when

q

varies between

q

1and

q

2. By the same method than in the previous case, and using (A.11) with, this time, jgj> 1, we calculate:

1

eRJð

u

Þcos

u

¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffi 11

g² s

1

!g

e

p

¼

p

sinW^0ðcosW⁰cos

u

Þcos

u

ðB:3Þ