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This paper is concerned with the steady flow of a second-grade fluid between two porous disks rotating eccentrically under the effect of a magnetic field. A perturbation solution for the velocity field is presented under the assumption that the second-grade fluid parameter
*β* is small. It is also studied the effect of all the parameters on the horizontal force per unit area exerted by the fluid on the disks. It is found that the
*x*- and
*y*-components of the force increase and decrease, respectively, when the second-grade fluid parameter
*β* and the Hartmann number
*M* increase. It is seen that the forces in the
*x*- and
*y*-directions on the top disk increase with the increase of the suction/injection velocity parameter
*P* but those on the bottom disk decrease. It is shown that the force acting on the top disk is greater than that acting on the bottom disk in view of the axial velocity in the positive
*z*-direction. It is observed that the increase in the Reynolds number
*R* leads to a rise in the horizontal force.

The orthogonal rheometer consisting of two parallel disks rotating with the same angular velocity about distinct axes was developed by Maxwell and Chartoff [

The steady flow between eccentric rotating disks under the effect of a magnetic field has been studied by many researchers. Mohanty [

The aim of this paper is to study the steady flow of an incompressible second-grade fluid between two insulated and porous disks rotating with the same angular velocity about two parallel axes under the effect of a magnetic field. It is a well-known fact that the governing equations of non-Newtonian fluids are in general of higher order than the Navier-Stokes equations. Even so, it is possible to find an exact solution for a second-grade fluid in this geometry when the disks are non-porous [

Let us consider an incompressible second-grade fluid between two porous and insulated disks located at z = ± h . The top and bottom disks rotate about the z ′ - and z ″ -axes with the same angular velocity Ω , respectively. The distance between the axes of rotation is defined by 2ℓ in the y-direction. A uniform magnetic induction B 0 is applied in the positive z-direction. It is assumed that the magnetic Reynolds number for the flow is very small so that the induced magnetic field is negligible in comparison with the applied magnetic field. A uniform suction and injection are applied perpendicular to the top and bottom disks, respectively. The schematic configuration of the problem is illustrated in

u = − Ω ( y − l ) , v = Ω x , w = w 0 at z = h , (1)

u = − Ω ( y + l ) , v = Ω x , w = w 0 at z = − h , (2)

u = − Ω y , v = Ω x , w = w 0 at z = 0 , (3)

where u, v, w denote the velocity components along the x, y, z-directions, respectively. We look for a solution of the velocity field of the form

u = − Ω y + f ( z ) , v = Ω x + g ( z ) , w = w 0 . (4)

The equations governing the flow are

ρ v ˙ = ∇ ⋅ T + J × B , (5)

∇ ⋅ v = 0 , (6)

where ρ is the density, v the velocity vector, J = ( J x , J y , J z ) the current density, B the magnetic induction and the overdot denotes material time differentiation. The velocity field in Equation (4) is compatible with the continuity Equation (6).

The stress T in a homogeneous second-grade fluid is given by [

T = − p I + μ A 1 + α 1 A 2 + α 2 A 1 2 , (7)

where p is the pressure, I the identity tensor, μ the dynamic viscosity coefficient, α 1 and α 2 the normal stress moduli. A 1 and A 2 stand for the first two Rivlin-Ericksen tensors defined through

A 1 = grad v + ( grad v ) T , (8)

and

A 2 = A ˙ 1 + A 1 ( grad v ) + ( grad v ) T A 1 . (9)

The coefficients μ, α 1 , and α 2 must satisfy

μ ≥ 0 , α 1 ≥ 0 , α 1 + α 2 = 0 . (10)

We refer the reader to [

Inserting Equation (4) into Equations (5), (7)-(9), we obtain

∂ p ∂ x = α 1 ( w 0 f ‴ + Ω g ″ ) + μ f ″ − ρ w 0 f ′ + ρ Ω ( Ω x + g ) + B 0 J y , (11)

∂ p ∂ y = α 1 ( w 0 g ‴ − Ω f ″ ) + μ g ″ − ρ w 0 g ′ + ρ Ω ( Ω y − f ) − B 0 J x , (12)

∂ p ∂ z = ( 2 α 1 + α 2 ) ( 2 f ′ f ″ + 2 g ′ g ″ ) , (13)

where a prime denotes differentiation with respect to z. Using J = σ ( E + v × B ) , we have

J x = σ ( E x + v B 0 ) , J y = σ ( E y − u B 0 ) , J z = σ E z , (14)

where σ is the electrical conductivity of the fluid, E = ( E x , E y , E z ) the electric

field and B 0 the magnitude of the applied magnetic induction B 0 . Since the disks are insulated, we obtain J z = 0 and E z = 0 . Using ∇ × E = 0 , we have

α 1 ( w 0 f ‴ + Ω g ″ ) + μ f ″ − ρ w 0 f ′ + ρ Ω g − σ B 0 2 f = C 1 , (15)

α 1 ( w 0 g ‴ − Ω f ″ ) + μ g ″ − ρ w 0 g ′ − ρ Ω f − σ B 0 2 g = C 2 , (16)

where C 1 and C 2 are constants. Combining Equations (15)-(16) and using F ( z ) = f ( z ) + i g ( z ) , we get

α 1 w 0 F ‴ + ( μ − i α 1 Ω ) F ″ − ρ w 0 F ′ − ( σ B 0 2 + i ρ Ω ) F = C , (17)

where C is a constant. The boundary conditions for F ( z ) are as follows:

F ( h ) = Ω l , F ( − h ) = − Ω l , F ( 0 ) = 0 . (18)

Introducing the dimensionless quantities,

F ¯ ( ζ ) = F ( z ) Ω l , ζ = z h , M = σ μ B 0 h , β = α 1 Ω μ ,

P = ρ w 0 h μ , R = ρ Ω h 2 μ , (19)

the governing Equation (17) and the conditions (18) reduce to the following dimensionless forms:

β P F ¯ ‴ + ( 1 − i β ) R F ¯ ″ − P R F ¯ ′ − ( M 2 + i R ) R F ¯ = K ¯ , (20)

F ¯ ( 1 ) = 1 , F ¯ ( 0 ) = 0 , F ¯ ( − 1 ) = − 1 , (21)

where a prime denotes differentiation with respect to ζ, and K ¯ is a constant.

We look for a perturbation solution for small values of the second-grade fluid parameter β as follows:

F ¯ ( ζ ) = F ¯ 0 ( ζ ) + β F ¯ 1 ( ζ ) + β 2 F ¯ 2 ( ζ ) + O ( β 3 ) , (22)

K ¯ = K ¯ 0 + β K ¯ 1 + β 2 K ¯ 2 + O ( β 3 ) , (23)

where K ¯ 0 , K ¯ 1 and K ¯ 2 are constants. Substituting Equations (22)-(23) into Equations (20)-(21) and equating the coefficients of different powers of β, we obtain

R F ¯ ″ 0 − P R F ¯ ′ 0 − ( M 2 + i R ) R F ¯ 0 = K ¯ 0 , (24)

P F ¯ ‴ 0 + R F ¯ ″ 1 − i R F ¯ ″ 0 − P R F ¯ ′ 1 − ( M 2 + i R ) R F ¯ 1 = K ¯ 1 , (25)

P F ¯ ‴ 1 + R F ¯ ″ 2 − i R F ¯ ″ 1 − P R F ¯ ′ 2 − ( M 2 + i R ) R F ¯ 2 = K ¯ 2 , (26)

and

F ¯ 0 ( 1 ) = 1 , F ¯ 0 ( 0 ) = 0 , F ¯ 0 ( − 1 ) = − 1 , (27)

F ¯ 1 ( 1 ) = 0 , F ¯ 1 ( 0 ) = 0 , F ¯ 1 ( − 1 ) = 0 , (28)

F ¯ 2 ( 1 ) = 0 , F ¯ 2 ( 0 ) = 0 , F ¯ 2 ( − 1 ) = 0 . (29)

The solution of Equation (24) with Equation (27) is

F ¯ 0 ( ζ ) = A ¯ 0 exp ( a ζ ) + B ¯ 0 exp ( b ζ ) + C ¯ 0 , (30)

where

a = P + P 2 + 4 ( M 2 + i R ) 2 , b = P − P 2 + 4 ( M 2 + i R ) 2 ,

A ¯ 0 = 1 − cosh b sinh a − sinh b − sinh ( a − b ) , B ¯ 0 = cosh a − 1 sinh a − sinh b − sinh ( a − b ) ,

C ¯ 0 = cosh b − cosh a sinh a − sinh b − sinh ( a − b ) . (31)

The solution of Equation (25) with Equation (28) is

F ¯ 1 = ( A ¯ 1 + B ¯ 1 ζ ) exp ( a ζ ) + ( C ¯ 1 + D ¯ 1 ζ ) exp ( b ζ ) + E ¯ 1 , (32)

where

A ¯ 1 = A ¯ 0 D ¯ 1 + E ¯ 0 B ¯ 1 , E ¯ 0 = cosh ( a − b ) − cosh a sinh a − sinh b − sinh ( a − b ) ,

B ¯ 1 = a 2 A ¯ 0 P − 2 a ( P a R − i ) , C ¯ 1 = B ¯ 0 B ¯ 1 + H ¯ 0 D ¯ 1 ,

H ¯ 0 = cosh b − cosh ( a − b ) sinh a − sinh b − sinh ( a − b ) , D ¯ 1 = b 2 B ¯ 0 P − 2 b ( P b R − i ) ,

E ¯ 1 = I ¯ 0 ( B ¯ 1 − D ¯ 1 ) , I ¯ 0 = 1 − cosh ( a − b ) sinh a − sinh b − sinh ( a − b ) . (33)

Finally, the solution of Equation (26) subject to Equation (29) is

F ¯ 2 = ( A ¯ 2 + B ¯ 2 ζ + C ¯ 2 ζ 2 ) exp ( a ζ ) + ( D ¯ 2 + E ¯ 2 ζ + H ¯ 2 ζ 2 ) exp ( b ζ ) + I ¯ 2 , (34)

where

A ¯ 2 = A ¯ 0 E ¯ 2 + E ¯ 0 B ¯ 2 + N ¯ 0 C ¯ 2 − P ¯ 0 H ¯ 2 , N ¯ 0 = sinh ( a − b ) − sinh a sinh a − sinh b − sinh ( a − b ) ,

P ¯ 0 = sinh b sinh a − sinh b − sinh ( a − b ) , B ¯ 2 = L ¯ 1 P − 2 a + L ¯ 2 ( P − 2 a ) 2 ,

L ¯ 1 = P a 2 R ( a A ¯ 1 + 3 B ¯ 1 ) − i a ( a A ¯ 1 + 2 B ¯ 1 ) , L ¯ 2 = a 2 B ¯ 1 ( P a R − i ) ,

C ¯ 2 = L ¯ 2 2 ( P − 2 a ) , D ¯ 2 = B ¯ 0 B ¯ 2 + H ¯ 0 E ¯ 2 + R ¯ 0 H ¯ 2 + V ¯ 0 C ¯ 2 ,

R ¯ 0 = sinh b + sinh ( a − b ) sinh a − sinh b − sinh ( a − b ) , V ¯ 0 = sinh a sinh a − sinh b − sinh ( a − b ) ,

E ¯ 2 = L ¯ 3 P − 2 b + L ¯ 4 ( P − 2 b ) 2 , L ¯ 3 = P b 2 R ( b C ¯ 1 + 3 D ¯ 1 ) − i b ( b C ¯ 1 + 2 D ¯ 1 ) ,

L ¯ 4 = b 2 D ¯ 1 ( P b R − i ) , H ¯ 2 = L ¯ 4 2 ( P − 2 b ) ,

I ¯ 2 = I ¯ 0 ( B ¯ 2 − E ¯ 2 ) − U ¯ 0 ( C ¯ 2 + H ¯ 2 ) , U ¯ 0 = sinh ( a − b ) sinh a − sinh b − sinh ( a − b ) . (35)

The shear stress components T x z and T y z in the fluid are presented in the dimensionless complex form as follows:

T ¯ x z + i T ¯ y z = ( 1 − i β ) F ¯ ′ + β P R F ¯ ″ , (36)

where

T ¯ x z = T x z μ Ω l / h , T ¯ y z = T y z μ Ω l / h . (37)

For the small values of the second-grade fluid parameter, we obtain

T ¯ x z + i T ¯ y z = G ¯ 0 ( ζ ) + β G ¯ 1 ( ζ ) + β 2 G ¯ 2 ( ζ ) + O ( β 3 ) , (38)

where

G ¯ 0 ( ζ ) = a A ¯ 0 exp ( a ζ ) + b B ¯ 0 exp ( b ζ ) , G ¯ 1 ( ζ ) = [ G ¯ 11 + G ¯ 12 ζ ] exp ( a ζ ) + [ G ¯ 13 + G ¯ 14 ζ ] exp ( b ζ ) ,

G ¯ 11 = ( P a 2 R − i a ) A ¯ 0 + a A ¯ 1 + B ¯ 1 ,

G ¯ 12 = a B ¯ 1 ,

G ¯ 13 = ( P b 2 R − i b ) B ¯ 0 + b C ¯ 1 + D ¯ 1 ,

G ¯ 14 = b D ¯ 1 ,

G ¯ 2 ( ζ ) = [ G ¯ 21 + G ¯ 22 ζ + G ¯ 23 ζ 2 ] exp ( a ζ ) + [ G ¯ 24 + G ¯ 25 ζ + G ¯ 26 ζ 2 ] exp ( b ζ ) ,

G ¯ 21 = ( P a 2 R − i a ) A ¯ 1 + ( 2 P a R − i ) B ¯ 1 + a A ¯ 2 + B ¯ 2 ,

G ¯ 22 = ( P a 2 R − i a ) B ¯ 1 + a B ¯ 2 + 2 C ¯ 2 ,

G ¯ 23 = a C ¯ 2 ,

G ¯ 24 = ( P b 2 R − i b ) C ¯ 1 + ( 2 P b R − i ) D ¯ 1 + b D ¯ 2 + E ¯ 2 ,

G ¯ 25 = ( P b 2 R − i b ) D ¯ 1 + b E ¯ 2 + 2 H ¯ 2 ,

G ¯ 26 = b H ¯ 2 . (39)

The present flow is defined by the superposition of a helical motion and a rigid body translation that is different from plane to plane in each z = constant plane. The axial component of velocity chosen in the positive 𝑧-direction is constant as a result of the continuity equation. Thus, the top and bottom disks are subjected to uniform suction and injection, respectively. The velocity field is presented by obtaining the functions f ¯ ( ζ ) and g ¯ ( ζ ) that represent the dimensionless x- and y-components of the translational velocity. The variations of f ¯ ( ζ ) and g ¯ ( ζ ) with the Hartmann number M, the second-grade fluid parameter β, the suction/injection velocity parameter P, the Reynolds number R are drawn in Figures 2-5.

The determination of the horizontal force acting on the disks is also important.

( T ¯ x z ) ζ = ± 1 and ( T ¯ y z ) ζ = ± 1 represent the x- and y-components of the dimensionless

force per unit area exerted by the top and bottom disks on the fluid, respectively. When the disks are non-porous, it is clear that the x- and y-components of the

force on the top disk are equal to those on the bottom disk (i.e., ( T ¯ x z ) ζ = 1 = ( T ¯ x z ) ζ = − 1 and ( T ¯ y z ) ζ = 1 = ( T ¯ y z ) ζ = − 1 ). In view of the fact that the axial velocity of the fluid is chosen in the pozitive 𝑧-direction, the components of the horizontal force on the top disk are larger than those on the bottom disk (i.e., ( T ¯ x z ) ζ = 1 > ( T ¯ x z ) ζ = − 1 and ( T ¯ y z ) ζ = 1 > ( T ¯ y z ) ζ = − 1 ). The effects of all the parameters on ( T ¯ x z ) ζ = ± 1 and ( T ¯ y z ) ζ = ± 1 are illustrated in Figures 6-9.

The conclusions that can be drawn from the performed analysis are pointed out as follows:

・ When the Hartmann number M increases, the curves about which the fluid layers rotate start to get closer to the z-axis. These curves are closer to the plane x = 0 than the plane y = 0 . The dimensionless force in the x-direction increases but that in the y-direction decreases.

・ When the second-grade fluid parameter β increases, the curves mentioned tend to move away from the z-axis. The effect of β on the translational velocity in the y-direction is greater than that in the x-direction. The x-component of the dimensionless force increases but the y-component decreases. It may be noticed that the change is almost linear.

・ The functions f ¯ ( ζ ) and g ¯ ( ζ ) representing the dimensionless x- and y-components of the translational velocity vector are anti-symmetric for non-porous disks; however, they are not anti-symmetric for porous disks. When the suction/injection velocity parameter P increases, the space curves get closer to the plane y = 0 in the region between the top disk and the plane z = 0 , but move away from the plane x = 0 . An adverse effect is observed in the region between the bottom disk and the plane z = 0 . Both the x- and y-components of the force on the top disk increase but those on the bottom disk decrease. It is observed that the change is almost linear.

・ When the Reynolds number R increases, the space curves in the core region become closer to the z-axis, and thus the boundary layers developing on the disks lead to an increase in the horizontal force on the top and bottom disks.

Ersoy, H.V. (2018) Flow of a Second-Grade Fluid between Eccentric Rotating Porous Disks in the Presence of a Magnetic Field. Open Journal of Applied Sciences, 8, 159-169. https://doi.org/10.4236/ojapps.2018.85013