Rotating flow of a second-order fluid on a porous plate

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International Journal of Non-Linear Mechanics 39 (2004) 767 – 777

Rotating ow of a second-order uid on a porous plate T. Hayat∗;1 , K. Hutter Institut fur Mechanik, Technische Universitat Darmstadt, Hochschulstr. 1, Darmstadt 64289, Germany Received 20 May 2002; accepted 21 January 2003

Abstract A study is made of the unsteady ow engendered in a second-order incompressible, rotating uid by an in0nite porous plate exhibiting non-torsional oscillation of a given frequency. The porous character of the plate and the non-Newtonian e3ect of the uid increase the order of the partial di3erential equation (it increases up to third order). The solution of the initial value problem is obtained by the method of Laplace transform. The e3ect of material parameters on the ow is given explicitly and several limiting cases are deduced. It is found that a non-Newtonian e3ect is present in the velocity 0eld for both the unsteady and steady-state cases. Once again for a second-order uid, it is also found that except for the resonant case the asymptotic steady solution exists for blowing. Furthermore, the structure of the associated boundary layers is determined. ? 2003 Elsevier Ltd. All rights reserved. Keywords: Analytic solution; Second order uid; Rotating system; Unsteady ow; Boundary layers

1. Introduction The study of non-Newtonian uid dynamics is important in connection with plastics manufacture, performance of lubricants, application of paints, processing of food, and movement of biological uids. Most biologically important uids contain higher molecular weight components and are, therefore, non-Newtonian. The unusual properties of polymer melts and solutions, together with the desirable attributes of many polymeric solids, have given rise to the world-wide industry of polymer processing. In recent years, considerable e3orts have been usefully devoted to the study of ow on non-Newtonian uids because of their practical and fundamental ∗

Corresponding author. E-mail address: t [email protected] (T. Hayat). 1 Permanent address: Department of Mathematics, Quaid-i-Azam University, Islamabad 45320, Pakistan.

importance associated with many technological applications. A vast amount of literature is now available for various kinds of geometries and for a variety of non-Newtonian uids. In order to increase the basic understanding of second-order uids, several authors including Rajagopal [1,2], Rajagopal and Gupta [3], Rajagopal et al. [4], Bandelli and Rajagopal [5], Garg [6], Bandelli [7], Ariel [8], Labropulu [9], FetecHau and FetecHau [10], Siddiqui and Kaloni [11], Kaloni and Huschilt [12], Benharbit and Siddiqui [13], Pop et al. [14], Chandna and Oku-Ukpong [15] and Hayat et al. [16–18] have studied the theory of non-Newtonian uids in various geometrical con0gurations. However, very few studies which illustrate the rotating ows of non-Newtonian uids have been reported, see [19– 21]. In general, it is not easy to study rotating ows of non-Newtonian uids. The main reason is that additional non-linear terms appear in the equations of motion rendering the problem more diIcult to solve. Another reason is that a universal non-Newtonian

0020-7462/04/$ - see front matter ? 2003 Elsevier Ltd. All rights reserved. doi:10.1016/S0020-7462(03)00040-4

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T. Hayat, K. Hutter / International Journal of Non-Linear Mechanics 39 (2004) 767 – 777

constitutive relation that can be used for all uids and ows is not available. The investigations of rotating ows of a viscous uid have been examined by various authors including Loper [22–24], Gupta [25], Sarojamma and Krishna [26], Soundalgekar and Pop [27], Mazumder [28], Ganapathy [29], Singh [30], Thornley [31], Debnath [32–34], Deka et al. [35] and Acheson [36]. The instrument called orthogonal rheometer was considered to determine the material moduli of non-Newtonian uids. It consists of two parallel disks which rotate with the same angular velocity about two coincident axes perpendicular to the disks. In [37–40], the authors examined the ow in the orthogonal rheometer under the assumptions of negligible uid inertia. Abbott and Walters [41] who included the inertial e3ects both obtained an exact solution for a Newtonian uid between eccentric rotating disks and examined the ow of a viscoelastic uid assuming that the distance between the axes is small. Later, Berker [42] considered the possibility of solutions that are not necessarily axially symmetric and established a one-parameter family of solutions for the ow of classical linearly viscous uid between two plane parallel disks rotating about a common axis with the same angular speed. The only axially symmetric solution in this family is the rigid-body motion. However, Berker [42] did not investigate the implications of his study when the angular speed of two disks are distinct in which case there is also a ow in the axial direction. More importantly, the boundary-value problem studied by Berker is linear while the problem governing the rotation of the two disks with distinct angular speeds is non-linear. In the light of Berker’s work, Parter and Rajagopal [43] re-examined the problem of ow of the classical linearly viscous uid between parallel disks rotating about a common axis with di3ering angular speeds. Parter and Rajagopal [43] rigorously proved that the problem admits solutions that are lack axial symmetry and that the axial symmetric solutions are never isolated when considered within the full scope of the Navier–Stokes equation. Similar results apply to the case of the ow due to a single rotating disk and ow due to rotating disks subject to suction or injection at the disks. Based on the existence theorems of Parter and Rajagopal [43], extensive numerical computations have been carried out by Lai et al. [44,45]. In

continuation Huilgol and Rajagopal [46] and Ji et al. [47] constructed the possibility of asymmetric solutions for the ow of viscoelastic uids of the rate type, between rotating parallel plates. The ows of Newtonian and non-Newtonian uids between parallel disks rotating about a common axis has been reviewed by Rajagopal [48]. In this paper, we consider the unsteady ow of an incompressible, rotating second-order uid. The uid is bounded by a porous plate. Both the plate and the uid are in a state of solid body rotation with constant angular velocity about the z-axis normal to the plate. In addition, a non-torsional oscillation of a given frequency is superimposed on the plate for the generation of unsteady ow in the rotating system. Initially, both the plate and the uid are assumed to be at rest. It is shown that the equations of motion admit an exact solution. Since the equations governing the ow of non-Newtonian uids are more complicated than the Navier–Stokes equations, to solve the equations of motion of a non-Newtonian rotating uid is important. As the governing equations are linear, the Laplace transform technique is particularly well studied for the solutions (for small times). However, it is not a trivial matter to invert the Laplace transform. Bandelli et al. [49] already showed that the Laplace transform technique does not work for the Rayleigh problem (the obtained solution does not satisfy the initial condition). They showed that this is due to an incompatibility between the prescribed data. A comprehensive discussion on this issue has been given in great length by Bandelli [50]. Here, the solutions for small and large times are obtained. Applying the Laplace transform technique, the expression for the velocity 0eld has been obtained. The diIculty that arises is the solution of the integrals occurring in the inverse Laplace transform. These integrals are normally diIcult to handle. The analytic solution of these integrals is thus obtained and the velocity 0eld is presented. Finally, the results obtained are compared to the viscous uid case [51]. 2. Basic equations Let an in0nite porous plate at z = 0 bound a semi-in0nite expanse (z ¿ 0) of a second-order uid which is assumed to be incompressible. Both the

T. Hayat, K. Hutter / International Journal of Non-Linear Mechanics 39 (2004) 767 – 777

plate and the uid rotate as a solid body with constant angular velocity  about an axis normal to the plate. Initially, (t 6 0), both the uid and plate are assumed to be at rest. At t ¿ 0 the plate performs non-torsional oscillations. We take Cartesian axes (x; y; z) such that the z-axis is parallel to the common axis of rotation of the uid and the plate, and the x-, y-axis lie in the plane of the plate and rotate with it. Since the plate is in0nite in extent, all the physical quantities, except the pressure, depend on z and t only. Denoting the velocity components in the x, y, z directions by u, v, w, respectively. We de0ne the velocity vector as V = [u; (z; t); v(z; t); w(z; t)];

(1)

which together with the equation of continuity ∇ · V =0 gives w =−W0 (a constant). Obviously, W0 ¿ 0 for suction and W0 ¡ 0 for blowing. The governing equation of motion is
 @V + (V · ∇)V + 2 × V +  × ( × r) @t =∇ · T;

T = −p1 I + A1 + 1 A2 + 2 A12 :

A1 = (grad V) + (grad V)T ;

1 ¿ 0;

 ¿ 0;

In the above equations grad denotes the gradient operator and d=dt the material time derivative. Coleman and Noll showed model (3) to be a second-order approximation to a simple uid in the sense of retardation [53]. However, since model (3) is invariant, it has been considered as an exact model for some uid. According to Dunn and Fosdick [54], the second-order uid

(5)

@u @u − W0 − 2v @t @z 1 @P @2 u + 2 @x @z   3 @3 u @u − W0 3 ; + @z 2 @t @z

=−

(6)

@v @v − W0 + 2u @t @z 1 @P @2 v + 2 @y @z   3 @3 v @v ; + − W 0 @z 2 @t @z 3

=−

where P = p1 − r 2 2 − (21 + 2 ) 2  = = ;

(4)

1 + 2 = 0:

Since we are dealing with second-order uid ow, the strict inequality holds true. Fosdick and Rajagopal [55] showed that when 1 ¡ 0, the uid exhibits anomalous behavior that is incompatible with any uid of rheological interest, and so results in a uid that is unstable. In view of Eqs. (1), (3) and (4), Eq. (2) can be rewritten in the component form

(3)

Here, p1 is the dynamic pressure function, I the identity tensor,  the dynamic viscosity and 1 , 2 the normal stress moduli. A1 and A2 are the Rivlin–Ericksen tensors [52] and are given by

dA1 A2 = + (grad V)T A1 + A1 (grad V): dt

model is compatible with thermodynamics when the Helmholtz free energy of the uid is a minimum for the uid in equilibrium. The uid model then has general and pleasant boundedness and stability properties. The aforementioned and the Clausius–Duhem inequality imply that the coeIcients , 1 and 2 must satisfy

(2)

where is the density, r is the radial coordinate given by r 2 = x2 + y2 ,  = k, k is a unit vector parallel to the z-axis, T is the Cauchy stress tensor and the constitutive equation for an incompressible homogeneous uid of second order is

769

(7) 

@u @z

2

 +

@v @z

2  ;

 = 1 = :

The initial and boundary conditions for the present problem are u=v=0

for all z and t 6 0;

(8)

u = −U + U ∗ (aei!t + be−i!t ); v=0

at z = 0; t ¿ 0;

u → 0; v → 0

as z → ∞; t ¿ 0;

(9) (10)

770

T. Hayat, K. Hutter / International Journal of Non-Linear Mechanics 39 (2004) 767 – 777

where U and U ∗ are real constants with the dimension of velocity, ! is the imposed frequency, and a, b, are complex constants. Condition (9) is a no-slip condition at the plate and (10) shows that there is no disturbance at in0nity. A general mathematical problem arises in uids of grade n. Since the equations are of higher order than the corresponding Newtonian ow, additional boundary conditions are needed. The issue is discussed in [56]. Later a detailed discussion is present by Rajagopal [57] and relevant references are given in that study. Introducing the complex velocity, q = u + iv, Eqs. (6) and (7) with no pressure gradient can be combined to form @q @q − W0 + 2iq @t @z   3 @2 q @3 q @ q : = 2 +  − W 0 @z @z 2 @t @z 3

(11)

The initial and boundary conditions now become q=0

for all z and t 6 0;

(12)

q→0

(13)

as z → ∞; t ¿ 0:

(14)

Introducing zˆ = zU ∗ =;

tˆ = t;

qˆ = q=U ∗ ;

Eq. (11) and conditions (12)–(14) on dropping the hats reduce to  3  @ q @q @2 q 2 @3 q +S +  S − 1 @z 2 @z 2 @t 1 @z 3 @z = q=0

E @q + iEq; 2 @t for all z and t 6 0;

(15) (16)

q = −U=U ∗ + (aei t + be−i t ) at z = 0; t ¿ 0; q → 0 as z → ∞; t ¿ 0;

S = W0 =U ∗ ; E = 2=U ∗2 ;

(17) (18)

= !=; 1 = =;

2 = U ∗2 =2 ; 1 = 1 : 3. Method of solution Taking the Laplace transform (transformed functions carry on overhead bar) of Eq. (15) on using condition (16), takes the form (1 + p1 )

2 d 3 qQ d 2 qQ d qQ − 1 S 3 + S 2 dz 1 d z dz

E pqQ + iE q; Q 2 where  ∞ q(z; Q p) = q(z; t) e−pt dt; =

0

(19)

(p ¿ 0):

The transformed boundary conditions are U a b qQ = − ∗ + + at z = 0; U p p−i p+i qQ → 0

q = −U + U ∗ (aei!t + be−i!t ) at z = 0; t ¿ 0;

where

as z → ∞:

(20)

Before proceeding with the solution of the above problem it would be interesting to remark here that in the classical viscous case (1 = 0), we encounter a di3erential equation of order two [51]. The analysis of the ow of the second grade uids, in particular, and the viscoelastic uids, in general, is more challenging mathematically and computationally, because of a peculiarity in the equations governing the uid motion; namely, the order of the di3erential equation(s) characterizing the ow of these uids is more than the number of the available boundary conditions. The diIculty is further accentuated by the fact that a non-Newtonian parameter of the uid (for example 1 , for a second grade uid) usually occurs in the coeIcient of the highest derivative. The usual attempts to resolve this diIculty centered around seeking a perturbation solution assuming the non-Newtonian uid parameter to be small; the classical paper begin by Beard and Walters [58], who considered the two-dimensional stagnation point ow of the Walter’s B uid. One may also refer, for example, to the works of Shrestha [59], Misra and Mohapatra [60], Rajagopal et al. [61], Verma et al. [62] and Erdogan [63] for other problems

T. Hayat, K. Hutter / International Journal of Non-Linear Mechanics 39 (2004) 767 – 777

in various geometries. In the present analysis, the dif0culty is also removed by seeking a solution of the following form: qQ = qQ0 + 1 qQ1 + O(12 );

(21)

which is valid for small values of 1 only. Substituting expression (21) into Eq. (19) and boundary conditions (20), and then collecting terms of like powers of 1 , one obtains the following systems of di3erential equations along with the boundary conditions:

×

3.1. System of order zero   d qQ0 d 2 qQ0 E + S − p + iE qQ0 = 0; d z2 dz 2 a b U qQ0 = − ∗ + + at z = 0; U p p−i p+i qQ0 → 0

as z → ∞:

(22)

d qQ1 d 2 qQ1 − +S 2 dz dz =−p qQ1 = 0 qQ1 → 0



−

(24)

at z = 0; as z → ∞:

(25)

3.3. Zeroth-order solution Here, the problem is basically the same as that for classical viscous uids. The solution of the problem is known [51] and is given by   a b U qQ0 = + − ∗ p−i p+i U p √ 2 ×e−(S+ S +4iE+2Ep)z=2 ; (26) which, after inverse Laplace transform, yields a = ei 2

t−Sz=2

√   S 2 +4iE   −z E=2(i + 2E )1=2   +e erf c             

1=2 √   2   z E S +4iE  − i + t   × 

×

2t

2E

b −i t−Sz=2 e 2  √  S 2 +4iE   z E=2( 2E −i )1=2   e erf c           

2 1=2 √       −i t   × 2z 2tE + S +4iE  2E   √ S 2 +4iE     −z E=2( 2E −i )1=2   +e erf c           

 1=2   √ 2   z E S +4iE  − −i t   ×  2

 E p + iE qQ1 2

d 2 qQ0 2 d 3 qQ0 + S ; d z2 1 d z 3

q0 (z; t)

×

(23)

3.2. System of order one

 √  S 2 +4iE   z E=2(i + 2E )1=2   e erf c            

1=2   √   z E S 2 +4iE   × + i + t   2 2t 2E  

2

+

771

2t

2E

U −Sz=2 e 2U ∗  √  S 2 +4iE   z E=2( 2E )1=2   erf c   e          

1=2   √ 2   z E S +4iE   × + t   2 2t 2E  

; √ S 2 +4iE 1=2     −z E=2( )   2E +e erf c            1=2 √ 

2     z E S +4iE  × − t    2

2t

(27)

2E

where erf c(x) denotes the complementary error function. 3.4. First-order solution The solution of Eq. (24) together with Eq. (25), after inverse Laplace transform, has the form  c+i∞ 1 q1 (z; t) = qQ1 (z; p) ept dp; (28) 2#i c−i∞ where qQ1 (z; p) =ze−Sz=2

772

T. Hayat, K. Hutter / International Journal of Non-Linear Mechanics 39 (2004) 767 – 777

 S 2 +4iE  p +  +  p p + 3 4 10 2E     −1      − p S 2 +4iE p + 2E   11     −1    2   F − 12 p2 p + S +4iE × 2E         2   S +4iE   +13 p + 2E        −1   2 4iE −14 p + S 2E 

(29)

and  F=

a b U + − ∗ p−i p+i U p

×e−z S 3 = 2



√

E=2(p+

E2 1+ 21



S 2 2 6 = E i + 41 8 =



S 2 +4iE 1=2 2E ) ;



S2 2 1 (S + iE); 5 = ; 21 2

; 4 = 

+ zei t−Sz=2   √ i +S 2 +4iE 1=2   −z E=2( )  2E (H2 + H3 ) e erf c              

  1=2 √   2   z E S +4iE   × − i + t   2 2t 2E   ×

+ ze−i t−Sz=2   √ S 2 +4iE   −z E=2( 2E −i )1=2  (H4 + H5 ) e erf c            

2 1=2 √       +4iE z E S   −i t   × 2 2t −  2E   × √ S 2 +4iE     z E=2( 2E −i )1=2    + (H − H ) e erf c  4 5            

 1=2 √   2   z E S +4iE   − i t  × 2 2t +  2E −

E ; 7 = ; 2

S 2 2 iS 2 E2 ; 9 = ; 21 21

√ √ √ 10 = 5 2E; 11 = 6 ( 2E)−1 ; 12 = 7 ( 2E)−1 ;

×

√ √ 13 = 8 2E; 14 = 9 ( 2E)−1 :

=ze

 −Sz=2

×

e−Ez

2

3 (a + b)z 2t

=8t−t(

√



S 2 +4iE 1=2 2E )

#t

E + H1 2

Uz −Sz=2 e U∗  √ S 2 +4iE 1=2   (H6 + H7 ) e−z E=2( 2E ) erf c       1=2 √ 

2   z E S +4iE  × − t  2 2t 2E 



            

√ S 2 +4iE     z E=2( 2E )1=2  + (H − H ) e erf c    6 7           

1=2   √ 2   z E S +4iE   + t  ×  2

The integrals appearing in Eq. (28) have been obtained analytically. For that we use the transformation  p + (S 2 + 4iE)=2E = % and the resulting expression, after lengthy calculations, is given by q1 (z; t)

√   S 2 +4iE   z E=2(i + 2E )1=2    + (H − H ) e erf c  2 3             

1=2 √   2   z E S +4iE  t   × 2 2t + i + 2E 

2t

: (30)

2E

In the above equation 3 U i 10 (a − b) + ∗ U 4   U − (11 − 13 ) a + b + − ∗ U

H1 = −

 

2  

2   S +4iE a i − S +4iE − b i + 2E 2E   −12 

2  ; U S +4iE + U∗ 2E

T. Hayat, K. Hutter / International Journal of Non-Linear Mechanics 39 (2004) 767 – 777

a (i 3 + 4 ); 2 a H3 = 2  

(31) reduces to

H2 =

qs (z; t) ∼ [a + 21 z(H2 + H3 )] ei

  i 10 S 2 +4iE +  i + 13 4 2E   −1   × ;  2 + ( 2 12 − i 11 − 14 ) i + S +4iE 2E

b 2 

 13 − i

10 4

 S 2 +4iE

−i



2E   ;   × −1   2 S 2 +4iE + ( 12 + i 11 − 14 ) 2E − i

4 ; 2   −1   2 2 S + 4iE S + 4iE 1 : − 14 H7 = 13 2 2E 2E

H6 =

4. Discussion of the ow !eld The starting solution for the case of suction is q(z; t) = q0 (z; t) + 1 q1 (z; t):

(31)

The above solution describes the general features of the unsteady boundary layer ow in a rotating uid bounded by a plate including the e3ects of suction. This solution clearly brings out the contribution due to the material parameter of the second grade uid. The results for the corresponding Ekman problem can be obtained directly from Eq. (31) with = 0, S = 0 and 1 = 0. It should be noted that solution (31) includes the Thornley solution [31] as a special case for U = 0, S = 0 and 1 = 0. We also note that for 1 = 0, the velocity 0eld (31) is identical with that of Debnath and Mukherjee [51]. This provides a useful mathematical check. For large times we must recover the steady-state solution qs . Indeed, when t goes to in0nity, solution

t−((1 +i)1 )z

+ [b + 21 z(H4 + H5 )] e−i − [1 + 21 z(H6 + H7 )]

t−((2 +i)2 )z

U −((3 +i)3 )z e ; (32) U∗

where  1 S + √ [S 2 + S 4 + 4( + 2)2 E 2 ]1=2 ; 2 2 2  S 1 (2 = + √ [S 2 + S 4 + 4(2 − )2 E 2 ]1=2 ; 2 2 2  S 1 (3 = + √ [S 2 + S 4 + 16E 2 ]1=2 ; 2 2 2  1 )1 = √ [ − S 2 + S 4 + 4( + 2)2 E 2 ]1=2 ; 2 2  1 )2 = √ [ − S 2 + S 4 + 4(2 − )2 E 2 ]1=2 ; 2 2  1 )3 = √ [ − S 2 + S 4 + 16E 2 ]1=2 : 2 2 (1 =

b H4 = (4 − i 3 ); 2 H5 =

773

(33) (34) (35) (36) (37) (38)

It is clearly seen, that since the steady-state solution is to be valid for large values of time only, it is independent of the initial condition given by Eq. (16) and it is periodic in time. For some time after the initiation of the motion, the velocity 0eld contains transients and they gradually disappear in time. The transient solution is obtained by the subtraction of Eq. (32) from Eq. (31) i.e. qt (z; t) ∼ q(z; t) − qs (z; t);

(39)

where qt shows the transient solution. It is clearly seen that for a long time the transient solution given by Eq. (39) disappears. Further, expression (32) shows the existence of three distinct boundary layers of thicknesses of order =U ∗ (r (r = 1; 2; 3). These thicknesses decrease with an increase of the suction parameter and the rotation. It is also interesting to note that these thicknesses remain bounded for all values of the frequency of the imposed oscillations. When a = b = 1 = 0 then from expression (32) we have qs (z; t) ∼ −

U −((3 +i)3 )z e ; U∗

774

T. Hayat, K. Hutter / International Journal of Non-Linear Mechanics 39 (2004) 767 – 777

which in dimensional form becomes

In the limits yields

qs (z; t) ∼ U (1 − e−((3 +i)3 )z ): This is in accord with the result of Gupta [25] for viscous uid. In the limits t → ∞, → 0, solution (31) with S = 0 takes the following form: q(z; t) ∼ a[1 + 21 z(H6 + H7 )] −Sz=2−z

×e

√

 E=2

×e

√

 E=2

S 2 +4iE 2E

√

 E=2

S 2 +4iE 2E :

→2 t→∞

= lim lim q(z; t)

(40)

where 


 E i√ 4 : 4i + √ 2 2 4i → 2, S → 0, result (31)

We note that results (40) and (41) are di3erent. Further, result (40) does not even qualify for the solution because it fails to satisfy the boundary condition at in0nity unless b = 0. Result (41) satis0es all the conditions and hence is a correct solution. For uniform blowing S ¡ 0, and we take S = −S1 so that S1 ¿ 0. The solution in this case is given by q(z; ˜ t) = q˜0 (z; t) + 1 q˜1 (z; t);

t→∞ →2

2it−Sz=2−z ∼ [a + 21 z(Hˆ 2 + Hˆ 3 )]e

√

 E=2 2i+

S 2 +4iE 2E

+ [b + 21 z(Hˆ 4 + Hˆ 5 )] −2it−Sz=2−z

U −z√iE e ; U∗

(41)

lim lim q(z; t)

×e

2iE+2it

√

In the presence of suction (S = 0), the steady-state solution in the resonant case ( → 2) does not depend on the order of the double limit operation t → ∞, → 2. In fact, the double limiting procedure with S = 0 gives

√

√

q(z; t) ∼ a(1 + 21 zL) e−z 2iE+2it    √ U E z −2it − ∗ e−z iE : + be erf c 2 2t U

− [1 + 21 z(H6 + H7 )] U −Sz=2−z × ∗e U

+ be−2it −

For the limits t → ∞, reduces to

+ b[1 + 21 z(H6 + H7 )] −Sz=2−z

q(z; t) ∼ a(1 + 21 zL) e−z

a L= 2

S 2 +4iE 2E

→ 2, t → ∞, S → 0, solution (31)

 E=2

2

S +4iE 2E −2i

U − [1 + 21 z(Hˆ 6 + Hˆ 7 )] ∗ U  2 √ S +4iE −Sz=2−z E=2 2E ; ×e where Hˆ r (r = 2–7) are the values of Hr as → 2. However, in the resonant case with S = 0 the double limit does not commute. Therefore, some extra care is necessary to 0nd a meaningful correct solution.

(42)

where q˜0 (z; t) a = ei t+S1 z=2 2  √  S 2 +4iE   z E=2(i + 1 2E )1=2   e erf c            

1=2 √   2   S1 +4iE z E   × + i + t   2 2t 2E   ×

√   S 2 +4iE   −z E=2(i + 1 2E )1=2     erf c + e           

  1=2 √ 2   S1 +4iE  z E  ×  − i + t  2

2t

2E

T. Hayat, K. Hutter / International Journal of Non-Linear Mechanics 39 (2004) 767 – 777

+

×

−

×

b −i t+S1 z=2 e 2  √ S 2 +4iE    z E=2( 1 2E −i )1=2   e erf c            

   1=2 √ 2   S1 +4iE   z E   × + − i t   2 2t 2E   √ S 2 +4iE       −z E=2( 1 2E −i )1=2   erf c + e            

   1=2 √ 2   S +4iE  × z E − 1    − i t 2 2t 2E U S1 z=2 e 2U ∗  √ S 2 +4iE    z E=2( 1 2E )1=2   e erf c             

  1=2 √ 2   S1 +4iE   z E   × + t   2E 2 2t   √ S 2 +4iE       −z E=2( 1 2E )1=2   erf c + e             

  1=2 √ 2   S1 +4iE   z E  × 2 2t −  t 2E

 =zeS1 z=2

S1 H˜ 1 − 4t

−Ez 2 =8t−t(

e

√



E2 1+ 21



×

√ S 2 +4iE       z E=2( 1 2E −i )1=2  ˜ ˜ − H ) e erf c  + ( H   4 5         

   1=2 √ 2      × 2z 2tE + S1 +4iE  − i t 2E Uz S1 z=2 e U∗  √ S12 +4iE 1=2   ˜ 6 + H˜ 7 ) e−z E=2( 2E ) erf c ( H       1=2 √ 

2   S1 +4iE z E  t  2E  × 2 2t −

S12 +4iE 1=2 2E )

#t

√   S 2 +4iE     z E=2(i + 1 2E )1=2  ˜ ˜ − H ) e erf c  + ( H   2 3          

  1=2 √ 2      × 2z 2tE + i + S1 +4iE  t 2E

+ ze−i

t+S1 z=2

            

√ S12 +4iE 1=2      ˜ 6 − H˜ 7 ) e−z E=2( 2E ) erf c    +( H           

  1=2 √ 2   S +4iE   × z E + 1   t 2 2t 2E

;

where H˜ r (r = 2–7) are, respectively, given by H˜ r with S replaced by −S1 . The steady-state solution for blowing is of the form

  E (a + b)z 2

+ zei t+S1 z=2   √ S 2 +4iE   −z E=2(i + 1 2E )1=2  ˜ ˜ (H 2 + H 3 )e erf c             

  1=2 √ 2   S1 +4iE   z E   × − i + t   2 2t 2E   ×

−

  √ S 2 +4iE   −z E=2( 1 2E −i )1=2  ˜ ˜ erf c   (H 4 + H 5 ) e          

   1=2 √ 2      × 2z 2tE − S1 +4iE  − i t   2E  

q˜s (z; t) ∼ [a + 21 z (H˜ 2 + H˜ 3 )] ei

q˜1 (z; t)

×

×

775

t+((˜1 −i)˜1 )z

+ b + 21 z (H˜ 4 + H˜ 5 )] e−i

t+((˜2 −i)˜2 )z

z U ˜ − [1 + 21 z (H˜ 6 + H˜ 7 )] ∗ e((3 −i)˜3 ) ; (43) U

where S1 1 (˜1 = − √ [S12 + 2 2 2 S1 1 (˜2 = − √ [S12 + 2 2 2 S1 1 (˜3 = − √ [S12 + 2 2 2



S14 + 4( + 2)2 E 2 ]1=2 ; (44)



S14 + 4(2 − )2 E 2 ]1=2 ; (45) S14 + 16E 2 ]1=2

(46)

and )˜r (r = 1; 2; 3) can be obtained from )r be replacing S with −S1 . It is well known that in an inertial frame, no steady asymptotic solution is possible for ow past a porous plate subjected to uniform blowing. The physical mechanism of non-existence of the solution is that the blowing causes thickening

776

T. Hayat, K. Hutter / International Journal of Non-Linear Mechanics 39 (2004) 767 – 777

of the boundary layer. At suIciently large distance from the leading edge the boundary layer becomes so thick that it becomes turbulent. In a rotating frame the boundary layer thickness decreases with the increase in rotation and for blowing not too large, the thinning e3ect of rotation may just counterbalance the thickening e3ect of blowing and thus similar to a viscous uid, a steady asymptotic solution is possible in case of second-order uid. In the case of blowing and resonance, solution (43) does not satisfy the boundary condition at in0nity. Physically, the blowing promotes the spreading of the shear oscillations far away from the boundary and hence the depth of penetration of the oscillations tends to in0nity as → 2 (one of the three boundary layers becomes in0nitely thick). In other words, in case of blowing and resonance, the oscillatory boundary layer ow of a second grade uid con0ned to well-de0ned boundary layers is hardly possible which is also in agreement with the Newtonian uid. Acknowledgements We are grateful to Professor K.R. Rajagopal for his valuable guidance and suggestions. Dr. Hayat is thankful to the Alexander Von Humboldt Foundation for the 0nancial support. References [1] K.R. Rajagopal, A note on unsteady unidirectional ows of a non-Newtonian uid, Int. J. Non-Linear Mech. 17 (1982) 369–373. [2] K.R. Rajagopal, On the creeping ow of the second order uid, J. Non-Newtonian Fluid Mech. 15 (1984) 239–246. [3] K.R. Rajagopal, A.S. Gupta, An exact solution for the ow of a non-Newtonian uid past an in0nite porous plate, Meccanica 19 (1984) 158–160. [4] K.R. Rajagopal, A.S. Gupta, T.Y. Na, A note on the Falkner– Skan ows of a non-Newtonian uid, Int. J. Non-Linear Mech. 18 (1983) 313–320. [5] R. Bandelli, K.R. Rajagopal, Start-up ows of second grade uids in domains with one 0nite dimension, Int. J. Non-Linear Mech. 50 (1995) 817–839. [6] V.K. Garg, Heat transfer due to stagnation point ow of a non-Newtonian uid, Acta Mech. 104 (1994) 159–171. [7] R. Bandelli, Unsteady unidirectional ows of second grade uids in domains with heated boundaries, Int. J. Non-Linear Mech. 30 (1995) 263–269.

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