Flow around a Triaxial Ellipsoid in a Long Circular Tube

Flow around a Triaxial Ellipsoid in a Long Circular Tube

Doo-Sung Lee

Department of Mathematics, College of Education, Konkuk University, 1, Hwayang-Dong, Kwangjin-Gu, Seoul, Korea

Reprint requests to D.-S. L.; E-mail:dslee@konkuk.ac.kr

Z. Naturforsch.66a,481 – 488 (2011) / DOI: 10.5560/ZNA.2011-0011 Received November 25, 2010 / revised April 18, 2011

This paper deals with the three-dimensional analysis of viscous fluid flow in a long circular cylinder containing an ellipsoidal obstacle. The center of the ellipsoid coincides with that of the cylinder, and the flow is confined to the space between the ellipsoid and the cylinder when the fluid velocity at the large distance from the ellipsoid is uniform. The equations of the classical theory of fluid dynamics are solved in terms of an unknown function which is then shown to be the solution of a boundary integro-differential equation.

A numerical solution of the integro-differential equation is obtained and the pressure on the surface of the ellipsoid is presented in graphical forms for various values of the radius of the circular tube.

Key words:Triaxial Ellipsoid; Cylinder; Integro-Differential Equations; Reduction Formula.

1. Introduction

It has been long time since the study on the flow around a spherical and non-spherical object in a tube began. The investigations varied from the vortical to irrotational flow and from the inviscid to viscous flow.

The problem of determining the distribution of a vector potential in a long circular cylinder containing a spher- ical or a spheroidal obstacle has been investigated by Smythe [1,2]. The problem of flow around a sphere in a tube has been also investigated by others [3,4].

However, relatively sparse attention has been paid to the solution concerning a triaxial ellipsoidal obstacle, as a special case of which the analysis on spheres or spheroids can be dealt. Motion of a viscous liquid past an ellipsoid in an unbounded space was however inves- tigated by Venkates [5].

In more recent years, numerical studies on the motion of an ellipsoid in a circular tube have appeared.

Sugihara-Seki [6] studied numerically the motions of an ellipsoidal particle in a tube flow. She used a finite- element method to solve the Stokes equations for flow around a spheroid placed at various positions in the tube. The instantaneous velocity was used to compute the particle trajectories. Swaminathan et al. [7] have used direct numerical simulations to investigate the motion of an ellipsoid settling in an infinitely long

circular tube, under the influence of gravity, at low and intermediate Reynolds numbers. They examined the is- sue of damping of the oscillatory motion for different cases of particle inertia.

Information on the potential flow around an ellip- soid will be of value to the circumstances that occur in a wind tunnel, to a circular cylindrical flow with bubbles or to an electrical flow in a circular cylindri- cal conductor with defects that can be approximated by a triaxial ellipsoid.

Applications of the study on such flow can be made in a broad range of biological and engineering fields;

examples include flow due to the motion of proteins in various biomedical applications and transport of en- capsuled solid matter in pipelines.

In this paper, we derive the solution of the problem determining the distribution of the potential in a long circular cylinder containing a triaxial ellipsoid whose center coincides with that of the cylinder when the flow is uniform at a large distance from the ellipsoid. We assume that the fluid is incompressible and viscous. In more recent years, the present author has considered the same problem for the spheroid [8].

In Section2, by the use of the field equations and employing Fourier transform, the boundary integro- differential equation is derived in which the un- known function is subsequently solved by the Galerkin

c

2011 Verlag der Zeitschrift f¨ur Naturforschung, T¨ubingen·http://znaturforsch.com

method. In Section3, some numerical examples are given.

2. Derivation of the Integro-Differential Equation In this section we consider the Stokes problem when the fluid is viscous. Consider a circular cylinder of the radiushhaving an ellipsoid whose semi-axes area,b, andc.We take the center of the ellipsoid and the cylin- der as the origin of the Cartesian coordinates, and the x,y- andz-axis along the semi-axes of the ellipsoid, respectively, then the ellipsoid occupies the regionV which is governed by the equation ^x2

a²+^y2

b² +^z2

c² ≤1.

The surface of the ellipsoid is denoted byS. We shall also use cylindrical coordinates(r,θ,z)which are con- nected to the Cartesian coordinates by

x=rcosθ, y=rsinθ, z=z.

Let the velocity of the flow at large distances from the ellipsoid bev₀.

The fluid velocity v and the pressure p satisfy the Stokes equation and the continuity equation:

∇p=µ∇²v, (1)

∇·v=0, (2)

where µ is the coefficient of viscosity. Let v = (u_x,u_y,u_z)be the velocity components in Cartesian co- ordinates. If we choose the velocity components as ux=2B_x−∂ Φ

∂x +∂ Ω

∂y, uy=2B_y−∂ Φ

∂y −∂ Ω

∂x, uz=2B_z−∂ Φ

∂z,

(3)

whereΦis defined by

Φ=B₀+xB_x+yB_y+zB_z

with B₀,B_x,B_y,B_z, and Ω being the harmonic func- tions, we see that (1) and (2) are satisfied by

p=2µ ∂B_x

∂x +∂B_y

∂y +∂B_z

∂z

. (4)

The suitable functions for the problem are B_x=

∞ m=0

∑

Z ∞

−∞

A_m(ξ)I_m+1(ξr)e^−iξzdξcos(m+1)θ, (5) B_y=

∞ m=0

∑

Z ∞

−∞

A_m(ξ)I_m+1(ξr)e^−iξzdξsin(m+1)θ, (6)

B_z=

1−1 2

x∂

∂x+y∂

∂y+z∂

∂z Z

V

a(u)dv R(x−u), (7) B0= ∂

∂z Z

V

b(u)dv

R(x−u) (8)

+

∞

∑

m=0 Z ∞

−∞B_m(ξ)hI_m(ξr)e^−iξzdξcosmθ−v₀z 2 , Ω=

∞ m=1

∑

Z ∞

−∞

C_m(ξ)hI_m(ξr)e^−iξzdξsinmθ +1

h Z

V

c(u)dv

R(x−u)(1−δ_m,0),

(9)

where

R(x−u) = q

(x−u)²+ (y−v)²+ (z−w)² andI_m(x)is the modified Bessel function of the first kind,u= (u,v,w),dv is used for dudvdw, andδm,0is a Kronecker delta.

Let(u_r,u_θ,u_z)be the velocity components in cylin- drical coordinates. The velocity at the tube wall is zero:

u_r=0, (10a)

uθ=0, (10b)

uz=0. (10c)

Boundary condition (10a) can be written in an alterna- tive form as

F[u_r(h,θ,z);z→ξ] =0, whereFmeans the Fourier transform.

We make use of the known integral in Erd´elyi et al. [9]

Z ∞

0

cos(ξz)dz p(x−u)²+ (y−v)²+z²

=K₀ ξ

q

(x−u)²+ (y−v)²

(11)

and K₀

ξ q

(x−u)²+ (y−v)²

=K₀ ξ{r²+r⁰²−2rr⁰cos(θ−θ⁰)}¹²

=I₀(ξr_<)K₀(ξr_>) +2

∞

∑

m=1

cos{m(θ−θ⁰)}I_m(ξr<)K_m(ξr>),

(12)

wherer<=min(r,r⁰),r>=max(r,r⁰),K0,Km are the modified Bessel functions of the second kind, and we have set

u=r⁰cosθ⁰, v=r⁰sinθ⁰. (13) We obtain following relation for solving unknown Am(ξ),Bm(ξ), andCm(ξ):

{I_m+1(ξh)−ξhI_m+1⁰ (ξh)}A_m(ξ)

−ξhI_m⁰(ξh)B_m(ξ) +mIm(ξh)C_m(ξ)

= 1 πiζ

Z

V

a(u)e^iwξ

Km(|ζ|) I_m(ζ)

F(ξ,r⁰,w)[ζIm+1(ζ) +mI_m(ζ)] +f(ξ,r⁰,w)[−(m+2)ζI_m+1(ζ) + (ζ²−2m)I_m(ζ)]

(14)

−F(ξ,r⁰,w)−(2+m)f(ξ,r⁰,w) I_m(|ζ|)

cosmθ⁰dvε_m + 1

πiξ Z

V

b(u)e^iwξ|ξ|K_m⁰(|ζ|)I_m(|ξ|r⁰)cosmθ⁰dvε_m

− m πh²

Z

V

c(u)e^iwξK_m(|ζ|)I_m(|ξ|r⁰)sinmθ⁰dvε_m, whereζ=ξhand

F(ξ,r⁰,w) =1 2

{2(3+m)(1+m)

+ξ²(r⁰²−w²)}I_m(|ξ|r⁰) + (5+m)|ξ|r⁰I_m+1(|ξ|r⁰) +iwξ{3(2+m)I_m(|ξ|r⁰) +2I_m+1(|ξ|r⁰)|ξ|r⁰}

f(ξ,r⁰,w) =1

2{3I_m(|ξ|r⁰) +|ξ|r⁰I_m⁰(|ξ|r⁰) +iwξI_m(|ξ|r⁰)},

εm=

(1, if m=0 2, otherwise.

The condition (10b) can be alternatively written as F[u_θ(h,θ,z);z→ξ] =0

from which we obtain another relation to solve the un- knownsA_m(ξ),B_m(ξ), andC_m(ξ):

A_m(ξ)I_m(ζ)(m+2) +mB_m(ξ)I_m(ζ)−ζI_m⁰(ζ)C_m(ξ)

=− m πiζ

Z

V

a(u)e^iwξ

K_m(|ζ|)F(ξ,r⁰,w) +f(ξ,r⁰,w)K_m(|ζ|)

ζIm+1(ζ) I_m(ζ) −2

(15)

−f(ξ,r⁰,w) I_m(|ζ|)

cosmθ⁰dvε_m

− m πih

Z

V

b(u)e^iwξξK_m(|ζ|)I_m(|ξ|r⁰)cosmθ⁰dvε_m + 1

πh Z

V

c(u)e^iwξ|ξ|K_m⁰(|ζ|)I_m(|ξ|r⁰)sinmθ⁰dvε_m.

The condition (10c) can be alternatively written as F[u_z(h,θ,z);z→ξ] =0

from which we obtain another relation to solve un- knownA_m(ξ)andB_m(ξ):

ζIm+1(ζ)A_m(ξ) +ζIm(ζ)B_m(ξ)

=−1 πi

Z

V

a(u)e^iwξ

K_m(|ζ|)F(ξ,r⁰,w) +f(ξ,r⁰,w)

I_m+1(ζ)

Im(ζ) ζK_m(|ζ|)− 1 Im(|ζ|)

·cosmθ⁰dvε_m− 1 πiξ²

Z

V

b(u)e^iwξKm(|ζ|)I_m(|ξ|r⁰)

·cosmθ⁰dvε_m+iv₀

2δ(ξ), (16)

whereδ(ξ)is the Dirac delta function, and we have used the known relation

1 π

Z ∞

0

cosξzdz=δ(ξ).

Therefore if we solve (14), (15), and (16) simultane- ously forA_m(ξ),B_m(ξ), andC_m(ξ), we obtain follow-

ing equations:

A_m(ξ) =−1 πi

_Z

V

a(u)

F(ξ,r⁰,w)

∆_m(ζ)I_m(|ζ|)ζI_m(ζ)I_m⁰(ζ) +f(ξ,r⁰,w)

·

G_m(ζ)−2(m+1)ζI_m(ζ)I_m⁰(ζ)

∆m(ζ)I_m(|ζ|)

e^iξwcosmθ⁰dv +ξ²ζI_m⁰(ζ)

∆m(ζ) Z

V

b(u)I_m(ξr⁰)e^iξwcosmθ⁰dv

εm

− mζI_m(ζ) π∆_m(ζ)h²

Z

V

c(u)I_m(ξr⁰)e^iξwsinmθ⁰dvε_m

−δ(ξ)ζ²I₁²(ζ)v₀

2i∆₀(ζ) δm,0, (17)

where

∆_m(ζ) =2mζI_m²(ζ)I_m+1(ζ) +ζ²I_m⁰(ζ)

· {I_m(ζ)I_m+1(ζ)2(m+1)−ζ(I_m²(ζ)−I_m+1² (ζ))}, G_m(ζ) =K_m(|ζ|)

I_m(ζ) −ζI_m+1(ζ)

∆_mI_m(|ζ|){ζI_m+1(ζ) +mI_m(ζ)}, and

B_m(ξ) =−1 πi

_Z

V

a(u) 1

ζG_m(ζ)F(ξ,r⁰,w)

− f(ξ,r⁰,w)

∆m(ζ)I_m(|ζ|)[2mI_m+1(ζ)

−ζ²I_m⁰(ζ)]I_m(ζ)

e^iξwcosmθ⁰dvε_m +G_m(ζ)ξ

h Z

V

b(u)I_m(|ξ|r⁰)e^iξwcosmθ⁰dvε_m

+ ζm

πh²I_m+1(ζ) Z

V

c(u)I_m(ξr⁰)e^iξwsinmθ⁰dvε_m +δ(ξ)ζ²I₁(ζ)I₂(ζ)v₀

2i∆₀(ζ) δ_m,0,

(18)

and

Cm(ξ) =− 2 πiζ

Z

V

a(u)

F(ξ,r⁰,w)

∆_m(ζ)I_m(|ζ|)

·ζIm+1(ζ)I_m(ζ) +f(ξ,r⁰,w)

·

G_m(ζ)−(m+2)ζIm+1(ζ)I_m(ζ)

∆m(ζ)I_m(|ζ|)

e^iξwcosmθ⁰dvε_m

−2ξ²I_m+1(ζ)

∆_m(ζ)πi Z

V

b(u)I_m(ξr⁰)e^iξwcosmθ⁰dvε_m

−

G_m(ζ)−ζ{(m+2)I_m+1(ζ)−ζIm(ζ)}I_m(ζ)

∆_m(ζ)I_m(|ζ|)

1 πh²

· Z

V

c(u)I_m(|ξ|r⁰)e^iξwsinmθ⁰dvε_m. (19)

The velocity components (u_x,uy,uz) are zero on the surface of the ellipsoid. Thus if we substitute the value ofA_m(ξ),B_m(ξ), andC_m(ξ)given by (17), (18), and (19) intoB_x,B_y,andB₀in (5), (6), (8), and (9), we ob- tain following three conditions:

u_x=2

∞ m=0

∑

B_mcos(m+1)θ−∂ Φ

∂x +∂ Ω

∂y =0, (x,y,z)∈S,

(20)

u_y=2

∞

∑

m=0

B_msin(m+1)θ−∂ Φ

∂y −∂ Ω

∂x =0, (x,y,z)∈S,

(21)

u_z=2L Z

V

a(u)dv R(x−u)−∂ Φ

∂z +v₀

1−r² h²

=0, (x,y,z)∈S,

(22)

where B_m=−2

πεm

Z

V

a(u)cosmθ⁰ Z ∞

0

F₁(ξ,r⁰,w,z)

·ζIm+1(ζ) +mI_m(ζ)

∆m(ζ) +F₂(ξ,r⁰,w,z)

·

Gm(ζ)−2(m+1)ζI_m⁰(ζ)

∆_m(ζ)

Im+1(ξr)dξdv +

Z

V

b(u)cosmθ⁰ Z ∞

0

ζI_m+1(ζ) +mI_m(ζ)

∆m(ζ)

·I_m(ξr⁰)I_m+1(ξr)ξ²sinξ(w−z)dξdv +m

h² Z

V

c(u)sinmθ⁰ Z _∞

0

ζI_m(ζ)

∆_m(ζ)

·I_m(ξr⁰)I_m+1(ξr)cosξ(w−z)dξdv

(23)

with

F₁(ξ,r⁰,w,z) =ℜF(ξ,r⁰,w)sinξ(w−z) +IF(ξ,r⁰,w)cosξ(w−z), F2(ξ,r⁰,w,z) =ℜf(ξ,r⁰,w)sinξ(w−z)

+If(ξ,r⁰,w)cosξ(w−z), L=1−1

2

x∂

∂x+y∂

∂y+z∂

∂z

,

and Φ=−2

π

∞ m=0

∑

cosmθ ε_m Z

V

a(u)cosmθ⁰

· Z ∞

0

1 ξ

F₁(ξ,r⁰,w,z)

G_m(ζ)I_m(ξr) +ξrIm+1(ξr)

∆m(ζ)

·(ζI_m+1(ζ) +mI_m(ζ))

+F₂(ξ,r⁰,w,z)

·

Gm(ζ)−2(m+1)ζI_m⁰(ζ)

∆_m(ζ)

ξrIm+1(ξr)− ζ

∆_m(ζ)

·[(2m−ζ²)I_m+1(ζ)−mζI_m(ζ)]I_m(ξr)

dξdv +

Z

V

b(u)cosmθ⁰ Z ∞

0

G_m(ζ)I_m(ξr) +ξrIm+1(ξr)

∆_m(ζ) ζI_m⁰(ζ)

Im(ξr⁰)ξsinξ(w−z)dξdv +m

h Z

V

c(u)sinmθ⁰ Z ∞

0

{I_m(ζ)ξrI_m+1(ξr)

−ζI_m+1(ζ)I_m(ξr)} 1

∆m(ζ)I_m(ξr⁰)cosξ(w−z)dξdv

+zL Z

V

a(u)dv R(x−u)+ ∂

∂z Z

V

b(u)dv

R(x−u), (24) and

Ω =−2 π

∞ m=1

∑

ε_msinmθ _Z

V

a(u)cosmθ⁰

· Z ∞

0

2 ξ

F₁(ξ,r⁰,w,z)ζIm+1(ζ)

∆m(ζ) +F₂(ξ,r⁰,w,z)

·

G_m(ζ)−(m+2)I_m+1(ζ)

∆m(ζ)

I_m(ξr)dξdv +

Z

V

b(u)cosmθ⁰ Z ∞

0

2ζI_m+1(ζ)

∆_m(ζ)

·I_m(ξr⁰)I_m(ξr)ξsinξ(w−z)dξdv+1 h Z

V

c(u)sinmθ⁰

· Z ∞

0

G_m(ζ)−ζ{(m+2)I_m+1(ζ)−ζI_m(ζ)}

∆m(ζ)

·I_m(ξr⁰)I_m(ξr)cosξ(w−z)dξdv

+1 h Z

V

c(u)dv

R(x−u)(1−δ_m,0). (25) Of interest is the case when the radius of the cylinder tends to infinity. All terms involvinghvanish, and per- tinent functions for the solution are constants. So a(u) =a1, b(u) =b1 (say).

The following formula is useful:

Z

V

dv

R(x−u) (26)

=πabc Z _∞

λ

1− x²

a²+s− y²

b²+s− z² c²+s

ds

∆(s), where∆(s) =p

(a²+s)(b²+s)(c²+s)andλ is the greatest root of

1− x²

a²+λ − y²

b²+λ− z² c²+λ =0.

Then from (26) we find that Φ=πza1χ−2πzb₁γ, where

χ=abc Z ∞

λ

ds

∆(s), γ=abc Z ∞

λ

ds

(c²+s)∆(s). (27) Then conditions (20) and (21) require

−a₁dχ

dλ +2b₁dγ dλ

λ=0

=0, or −a₁+2b₁ c² =0.

With the help of this relation, the conditionu_z=0 re- duces to

v0+ (a₁χ0+2b1γ0)π=0, (28) where the suffix denotes that the lower limit in the in- tegrals (27) is to be replaced by zero. Hence,

b₁=1

2a₁c², a₁=− v₀ π(χ₀+γ₀c²). This is in agreement with Lamb [10].

Equations (20)–(22) are solved by the Galerkin method. For this, following formulae [11] along with (26) are useful:

Z

V

vwdv

R(x−u)=b²c²πabcyz

· Z ∞

λ

1− x²

a²+s− y²

b²+s− z² c²+s

· ds

p(a²+s)(b²+s)³(c²+s)³, Z

V

w²dv

R(x−u)=πabc

· Z _∞

λ

c²s

4(c²+s)ω²(s)− c⁴z² (c²+s)²ω(s)

ds

∆(s), (29)

where

ω(s) = x²

a²+s+ y²

b²+s+ z² c²+s−1.

We also need to evaluate following integral:

I_`,m,n= Z ∞

0

1

(a²+s)^`(b²+s)^m(c²+s)ⁿ ds

∆(s). (30) To evaluate (30), we let

s= (a²−c²)sn⁻²u

and use the following identities for the Jacobian elliptic functions:

k²sn²u+dn²u=1, sn²u+cn²u=1, and

k= s

a²−b²

a²−c², k⁰²=1−k². So

Z ∞

0

1

(a²+s)^`(b²+s)^m(c²+s)ⁿ ds

∆(s)

= 2

(a²−c²)^`+m+n+12 Z F

0

sn^2`+2m+2nudu dn^2mucn²ⁿu ,

(31)

where F=

Z θ

0

(1−k²sin²u)⁻¹²du, θ=sin⁻¹

√ a²−c²

a

.

(32)

The integral on the right-hand side of (31) is L_`,m,n=

Z F 0

sn^2`+2m+2nudu dn^2mucn²ⁿu

= 1

k^02`+2m+2n

`+m+n

∑

(−1)^j

`+n+m j

· Z F

0

dn^2`+2n−2junc^2n−2judu,

(33)

where we used

sn²u=dn²u−cn²u

k⁰² , ncu= 1 cnu.

Further,

L_`,m,n= 1

k<sup>02`+2m+2n</sup> `+n

∑

j=0

(−1)^j

`+n+m j

· Z F

0

(k⁰²+k²cn²u)^`+n−jnc^2n−2judu +

`+m+n

∑

j=`+n+1

(−1)^j

`+n+m j

· Z F

0

nd^2j−2`−2nu

dn²u −k⁰² k²

j−n

du

,

(34)

where we used

dn²u=k⁰²+k²cn²u, ndu= 1 dnu.

Expanding the powered terms by using the binomial expansion, we find

L_`,m,n= 1

k^02`+2m+2n

"_`+n

∑

j=0

`+n−j

∑

i=0

(−1)^j

`+n+m j

·

`+n−j i

k⁰²ⁱk2`+2n−2j−2i

C2`−2i

+

`+m+n j=`+n+1

∑

j−n i=0

∑

(−1)^j+i

`+n+m j

j−n i

·k⁰²ⁱk^2n−2kG2`−2i

#

, (35)

where C_2n=

Z F 0

cn²ⁿudu, G_2n= Z F

0

dn²ⁿudu.

We find the following reduction formula forC_2n in Byrd and Friedman [12, p. 194]:

C_2n+2= (36)

2n(2k²−1)C_2n+ (2n−1)k⁰²C2n−2+snFdnFcn²ⁿ⁻¹F

(2n+1)k² .

If`−i<0,we findC−2n=D_2n, where

D2n+2= (37)

(2n−1)k²D2n−2+2n(1−2k²)D_2n+tnFdnFnc²ⁿF

(2n+1)k⁰² .

Also we have following reduction formula forG2n:

G_2n+2= (38)

k²dn²ⁿ⁻¹FsnFcnF+(1−2n)k⁰²G_2n−2+2n(2−k²)G_2n

(2n+1) .

If`−i<0,we findG−2n=I_2n, where

I_2n+2= (39)

2n(2−k²)I_2n+ (1−2n)I2n−2−k²snFcnFnd²ⁿ⁺¹F

(2n+1)k⁰² .

pressure on the spheroid (553) -0,3 -0,2 -0,1 0 0,1 0,2 0,3

3 1,8 0,6 -0,6 -1,8 3

z (cm) pressure p/2µ (sec-1)

h=10cm h=11cm h=12cm h=infinity

Fig. 1. Variation of the pressure on the sur- face of the spheroid; a=b=5 cm, c= 3 cm.

pressure on the ellipsoid (543) -0,3 -0,2 -0,1 0 0,1 0,2 0,3

3 1,8 0,6 -0,6 -1,8 3

z (cm) pressure p/2µ (sec-1)

h=10 h=11 h=12 h=infinity

Fig. 2. Variation of the pressure on the sur- face of the ellipsoid;a=5 cm,b=4 cm, c=3 cm.

Thus, finally we see that one needs the following starting values for evaluating the general terms of C_2n,D_2n,G_2n,andI_2n:

C₀=D₀=F, C₂= 1

k²[E−k⁰²F], D₂= 1

k⁰²[k⁰²F−E+dnFtnF],

whereE is the elliptic integral of the second kind of moduluskand argumentθgiven in (32) and

G0=I0=F,G2=E, I2= 1

k⁰²[E−k²snFcdF]

variation of pressure w.r.t. h -3,5

-3 -2,5 -2 -1,5 -1 -0,5

0 ^{13 11 9,9 9,7 9,5 9,3 9,1 8,9 8,7}

h (cm)

pressure p/2µ (sec-1)

Fig. 3. Variation of the pressure on the sur- face of the ellipsoid atz=−3 cm with re- spect to h;a=5 cm,b=4 cm,c=3 cm.

and

snF=

√ a²−c²

a , cnF=c

a, dnF=b a. 3. Numerical Examples

In this section, we present some numerical exam- ples. In Figure1we present the variation ofp/2µwith respect tozon the surface of the spheroid whenx=0 witha=b=5 cm andc=3 cm, and on Figure2the pressure on the surface of a triaxial ellipsoid whena=

5 cm,b=4 cm, andc=3 cm is computed. We choose v₀=1 cm/s. On Figure3we present the variation of p/2µon the surface of the ellipsoid with respect toh.

4. Conclusion

We have presented the solution of a viscous fluid flow around a triaxial ellipsoid in a circular tube. This solution agrees with the published accounts when the radius of the cylinder approaches to infinity. Judging from the results for the infinite medium, the solution appears to be correct.

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