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MHD non-Newtonian flow due to non-coaxial rotations of an accelerated disk and a fluid at infinity
Communications in Nonlinear Science and Numerical Simulation 12 (2007) 465–485 www.elsevier.com/locate/cnsns
MHD non-Newtonian flow due to non-coaxial rotations of an accelerated disk and a fluid at infinity S. Asghar a, K. Hanif b, T. Hayat b, C.M. Khalique
c,*
a
Department of Mathematics, COMSATS Institute of Information Technology, H-8 Islamabad, Pakistan b Department of Mathematics, Quaid-i-Azam University, Islamabad, Pakistan c Department of Mathematical Sciences, International Institute for Symmetry Analysis and Mathematical Modelling, University of North-West, Private Bag X 2046, Mmabatho 2735, South Africa Received 16 September 2004; received in revised form 19 April 2005; accepted 21 April 2005 Available online 5 July 2005
Abstract The magnetohydrodynamic (MHD) flow due to non-coaxial rotations of a porous disk, moving with uniform acceleration in its own plane and a second grade fluid at infinity is examined. The fluid is electrically conducting in the presence of a uniform magnetic field. The effects of non-Newtonian fluid characteristics and uniform acceleration of the disk on the velocity field are presented both analytically and numerically. A very good accuracy has been seen. Ó 2005 Elsevier B.V. All rights reserved. PACS: 76A05 Keywords: Second grade fluid; MHD flow; Uniformly accelerated disk; Numerical solutions
1. Introduction An investigation of MHD boundary layers under the influence of viscous forces is important in understanding a variety of geophysical, astrophysical and engineering phenomena such as those *
Corresponding author. Tel.: +27 18 389 2354 2319; fax: +27 18 389 2292. E-mail address: [email protected] (C.M. Khalique).
1007-5704/$ - see front matter Ó 2005 Elsevier B.V. All rights reserved. doi:10.1016/j.cnsns.2005.04.006
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that occur at the coremantle interface of the earth. Much progress has been made during the previous years in the development of Newtonian flows due to non-coaxial rotations of a disk and fluid at infinity. Mention may be made to the interesting work of Erdogan [1–3], Kasiviswanathan and Rao [4], Pop [5] and Hayat et al. [6]. Despite the above studies, a little work seems to have been done for flows of non-Newtonian fluids due to non-coaxial rotation of a disk and fluid at infinity. Few studies illustrating the non-Newtonian flows have been reported, see Erdogan [7], Ersoy [8], Siddiqui et al. [9] and Hayat et al. [10]. The study of non-Newtonian fluid dynamics is important in connection with plastic manufacture, performance of lubricants, processing of food, and movements of biological and geophysical fluids. Most geological fluids contains higher molecular weight components and are, therefore non-Newtonian. Alternatively, geophysical applications concern ice flow, magna flow which are based on non-Newtonian constitutive behaviours. In the present work, the unsteady flow of an incompressible and non-Newtonian fluid due to non-coaxial rotations of a porous moving disk and a fluid at infinity is studied with magnetic field. A uniform suction or injection is applied at the surface of the disk. The disk is non-conducting. The analytic solution is obtained by perturbation method in terms of small material parameter. The numerical solution of the problem is obtained using Crank–Nicolson method with some modification for entire range of all the parameters. The graphical results of the analytic solution are presented and discussed. A comparison between analytic and numerical solutions is made in the tabular form. An excellent accuracy has been seen in the two solutions.
2. Formulation of the problem Let us consider the disk at z = 0, i.e., it is the xy-plane or more precisely 1 < x < 1, 1 < y < 1. The fluid occupies the space z > 0. Thus, above the disk, the fluid can be prescribed as (x, y, z) such that 1 < x < 1, 1 < y < 1, 0 < z < 1. Naturally, the fluid occupies a 3-D space. Now the z and z 0 -axes are such that z-axis cuts the plane of disk at (0, 0, 0) and the z 0 -axis cuts the disk at (0, l, 0). Now, consider the disk at z = 0 and the fluid at z = 1. The disk and the fluid at infinity are initially rotating about z 0 -axis and suddenly the disk starts rotating about zaxis and moving along x-axis and the fluid at infinity continues to rotate about z 0 -axis. This is termed as non-coaxial rotation of the disk and the fluid at infinity [1–10]. This give rise to the following boundary and initial conditions u ¼ Xy þ ct;
v ¼ Xx;
at z ¼ 0;
u ¼ Xðy lÞ;
v ¼ Xx;
as z ! 1
u ¼ Xðy lÞ;
v ¼ Xx;
at t ¼ 0;
for t > 0; for all t > 0;
ð1Þ
for z > 0
in which u and v are the velocity components in the x and y directions respectively, t is the time and X is the angular velocity. We seek a solution of the form u ¼ Xy þ f ðz; tÞ;
v ¼ Xx þ gðz; tÞ.
ð2Þ
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The governing equations for fluid flow are the continuity and momentum equations which are respectively given by div v ¼ 0; ov þ ðv $Þv ¼ div T þ J B. q ot
ð3Þ ð4Þ
Since we are considering electrically conducting fluids so there is second term on the right hand side of above equation. The following equation gives the origin of B (OhmÕs law): J ¼ rðE þ v BÞ
ð5Þ
in which E is the electric field. In the last term of equation (4) on the right-hand side B = B0 + b is the total magnetic field, r is the electrical conductivity, v is the velocity vector, q is the density and J is the electric current density. For small magnetic Reynolds number the induced magnetic field is neglected and hence we can easily write 1 rB2 ðJ BÞ ¼ 0 v. q q
ð6Þ
From Eqs. (2) and (3) we have ð7Þ
w ¼ w0 ;
where w is z-component of velocity and w0 > 0 for suction and w0 < 0 for blowing or injection. The Cauchy stress T in an incompressible fluid of second grade is given by T ¼ pI þ lA1 þ a1 A2 þ a2 A21 ;
ð8Þ
where A1 and A2 are first two kinematical tensors given by A1 ¼ ðgrad vÞ þ ðgrad vÞ ; o þ v r A1 þ A1 ðgrad vÞ þ ðgrad vÞ A1 ; A2 ¼ ot where p is the pressure, I is the identity tensor, l is the dynamic viscosity and (*) denotes transpose of the matrix. If an incompressible fluid of second grade is to have motion which are compatible with thermodynamics in the sense of Clausius-Duhem inequality and the condition that the Helmholtz free energy be a minimum when the fluid is at rest, then the following conditions must be satisfied: l P 0;
a1 P 0;
a1 þ a2 ¼ 0.
In view of Eqs. (4) and (6)–(8), we can write 1 oP o2 f of of r ¼ t 2 þ w0 þ Xg þ X2 x B20 ½f ðz; tÞ Xy q ox oz oz ot q 3 3 2 a1 o f of og þ w0 3 þ X 2 ; 2 oz oz q otoz
ð9Þ
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1 oP o2 g og og r ¼ t 2 þ w0 Xf þ X2 y B20 ½gðz; tÞ þ Xx q oy oz oz ot q 3 3 2 a1 o g og of w0 3 þ X 2 ; þ 2 oz oz q otoz 1 oP rB20 ¼ w0 ; q oz q
ð10Þ ð11Þ
where ( P ¼ p ð2a1 þ a2 Þ
of oz
2
og þ oz
2 )
and the boundary and initial conditions in terms of f and g become f ð0; tÞ ¼ ct;
gð0; tÞ ¼ 0;
f ð1; tÞ ¼ Xl;
gð1; tÞ ¼ 0;
f ðz; 0Þ ¼ Xl;
gðz; 0Þ ¼ 0.
ð12Þ
Eliminating P from Eqs. (9) and (10) by differentiating with respect to z and combining them along with Eq. (11) and boundary conditions (12), we have a1 o3 F o3 F ia1 X o2 F oF oF rB20 w0 3 þ t þ w0 ð13Þ X iþ F ¼ 0; q q otoz2 oz oz2 oz ot qX F ðz; tÞ ¼ f þ ig Xl;
ð14Þ
where F ð0; tÞ ¼ ct Xl;
F ð1; tÞ ¼ 0;
F ðz; 0Þ ¼ 0.
ð15Þ
Introducing F ¼
N¼
F 1
;
ðmcÞ3
W ¼
c 13 rB2 0
m2
q
;
1 1
ðcmÞ3
w0 ;
g¼
1 a1 c2 3 a¼ ; q m4
c 13 m2
2 13 c s¼t ; m
z;
X1 ¼
m 13 c2
X;
ð16Þ
c 13 l1 ¼ l 2 m the problem (Eqs. (13)–(15)) reduces to o3 F o3 F o2 F oF oF a aW 3 þ ð1 iX1 aÞ 2 þ W ðiX1 þ N ÞF ¼ 0; osog2 og og og os F ð0; sÞ ¼ s X1 l1 ; F ð1; sÞ ¼ 0; F ðg; 0Þ ¼ 0.
ð17Þ ð18Þ
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3. Analytic solution Before proceeding with the solution of the above problem it would be interesting to remark here that in the classical viscous case (a = 0), we encounter a differential equation of order two [11]. The analysis of the flow of the second grade fluids, in particular, and the viscoelastic fluids, in general, is more challenging mathematically and computationally, because of a peculiarity in the equations governing the fluid motion; namely the order of the differential equation(s) characterizing the flow of these fluids is more than the number of the available boundary conditions. The difficulty is further accentuated by the fact that a non-Newtonian parameter of fluid (for example a, for a second grade fluid) usually occurs in the coefficient of the highest derivative. The usual attempts to resolve this difficulty centered around seeking a perturbation solution assuming the non-Newtonian fluid parameter to be small; the classical paper begin by Beard and Walters [12], who considered the two-dimensional stagnation point flow of the WalterÕs B fluid. One may also refer, for example, to the works of Shrestha [13], Misra and Mohapatra [14], Rajagopal et al. [15], Verma et al. [16] and Erdogan [17] for other problems in various geometries. In the present analysis, the difficulty is also removed by seeking a solution of the following form: F ðg; sÞ ¼ F 0 ðg; sÞ þ aF 1 ðg; sÞ þ Oða2 Þ;
ð19Þ
which is valid for small values of a only. Now, substituting Eq. (19) in Eq. (17) and conditions (18) and then collecting terms of like powers of a, one obtains the following systems of differential equations along with the boundary conditions: 3.1. System of order zero o2 F 0 oF 0 oF 0 þW ðiX1 þ NÞF 0 ¼ 0; 2 og og os F 0 ð0; sÞ ¼ s X1 l1 ; F 0 ð1; sÞ ¼ 0; F 0 ðg; 0Þ ¼ 0.
ð20Þ ð21Þ
3.2. System of order one o2 F 1 oF 1 oF 1 o3 F 0 o3 F 0 o2 F 0 ðiX þ W þ NÞF ¼ þ W þ iX ; 1 1 1 og2 og os osog2 og3 og2 F 1 ð0; sÞ ¼ 0; F 1 ð1; sÞ ¼ 0; F 1 ðg; 0Þ ¼ 0.
ð22Þ ð23Þ
3.3. Zeroth-order solution After taking Laplace transform, the zeroth order system becomes 00
0
F 0 þ W F 0 ðs þ N þ iX1 ÞF 0 ¼ 0;
ð24Þ
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1 X1 l1 ; s2 s F 0 ð1; sÞ ¼ 0; F 0 ð0; sÞ ¼
where primes indicate differentiation with respect to g. The solution of equation (24) satisfying conditions (25) is 1 X1 l1 gm F 0 ðg; sÞ ¼ 2 e s s
ð25Þ
ð26Þ
in which W pffiffiffiffiffiffiffiffiffiffiffi W2 þ s þ A; A ¼ þ N þ iX1 . 2 4 Laplace inversion of Eq. (26) yields pffiffi pffiffi pffiffiffiffiffiffi pffiffiffiffiffiffi s g g s g g w F 0 ðg; sÞ ¼ e 2 g pffiffiffi eg A erfc pffiffiffi As þ þ pffiffiffi eg A erfc pffiffiffi þ As 2 4 A 2 4 A 2 s 2 s pffiffi
pffiffi pffiffiffiffiffiffi pffiffiffiffiffiffi X1 l 1 g g eg A erfc pffiffiffi As þ eg A erfc pffiffiffi þ As . ð27Þ 2 2 s 2 s m¼
3.4. First-order solution Taking Laplace transform of Eq. (22) and then using Eq. (26) in the resulting equation, we have 1 X1 l 1 00 0 F 1 þ W F 1 ðs þ N þ iX1 ÞF 1 ¼ 2 ð28Þ X emg ; s s where X ¼
W4 3 2 þ W N þ X21 þ iX1 ðW 2 þ N Þ þ sf2W 2 þ N g 2 2 pffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi þ 2Ws A þ s þ s2 þ W fW 2 þ Ng A þ s.
ð29Þ
The boundary conditions now are F 1 ð0; sÞ ¼ 0;
F 1 ð1; sÞ ¼ 0.
Solving Eq. (28) along with the boundary conditions (30), we obtain pffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi g 1 X1 l 1 F 1 ðg; sÞ ¼ pffiffiffiffiffiffiffiffiffiffiffi 2 ½b1 þ sb2 þ 2Ws A þ s þ s2 þ b3 W A þ semg s 2 Aþs s
ð30Þ
ð31Þ
in which W4 3 2 þ W N þ X21 þ iX1 ðW 2 þ N Þ; 2 2 b2 ¼ 2W 2 þ N; b1 ¼
b3 ¼ W 2 þ N .
ð32Þ
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After lengthy calculations, the Laplace inversion of Eq. (31) gives pffiffiffiffiffiffi b1 g W g 1 s g gpffiffiA g 2 pffiffiffiffiffi pffiffiffi F 1 ðg; sÞ ¼ e e erfc pffiffiffi þ As 4 2 s A 2A 2 A3 rffiffiffi pffiffi pffiffiffiffiffiffi 1 s g g A g 2 s Asg2 4s erfc pffiffiffi As þ e e þ pffiffiffiffiffi3 þ pffiffiffi A p 2 s A 2A 2 A pffiffi pffiffiffiffiffiffi pffiffiffiffiffiffi gb2 W g gpffiffiA g g g A erfc pffiffiffi As e erfc pffiffiffi þ As þ pffiffiffi e 2 e 2 s 2 s 4 A pffiffi pffiffi pffiffiffiffiffiffi pffiffiffiffiffiffi gW W g g A g g g A 2 e e erfc pffiffiffi As þ e erfc pffiffiffi þ As þ 2 2 s 2 s pffiffi pffiffiffiffiffiffi g2 g b3 gW W g s g g W e 2 pffiffiffi eg A erfc pffiffiffi As þ pffiffiffiffiffi e 2 gAs 4s þ 2 2 4 A 2 ps 2 s pffiffi pffiffiffiffiffiffi s g g gX1 l1 b1 W þ pffiffiffi eg A erfc pffiffiffi þ As pffiffiffi e 2 g þ 2 4 A 2 s 4 A pffiffi pffiffi pffiffiffiffiffiffi pffiffiffiffiffiffi g2 g g gX1 l1 b2 W g A g A e erfc pffiffiffi As e erfc pffiffiffi þ As pffiffiffiffiffi e 2 gAs 4s 2 ps 2 s 2 s g2 g2 g2 X1 l1 W W gX1 l1 W g2 1 gX1 l1 Wb3 W g pffiffiffiffiffiffiffi e 2 gAs 4s pffiffiffiffiffi e 2 gAs 4s A e 2 2 3 2s 4s 4 2 ps 2 ps pffiffi pffiffi pffiffiffiffiffiffi pffiffiffiffiffiffi g g g A g A e erfc pffiffiffi As þ e erfc pffiffiffi þ As . 2 s 2 s
471
ð33Þ
With the help of Eqs. (19), (27) and (33) we write pffiffi pffiffiffiffiffiffi s g g X1 l1 b1 g 1 s g W2 g g A pffiffiffiffiffi pffiffiffi þ erfc pffiffiffi As pffiffiffi F ðg; sÞ ¼ e e a 2 4 A 4 2 A3 2 2 s A 2A b2 g gW b3 gW s g X1 l1 gb1 X1 l1 gWb3 þa pffiffiffi a pffiffiffi a þ a pffiffiffi þ a 2 2 2 4 A 4 4 A 4 A pffiffi pffiffiffiffiffiffi s g g X1 l1 b1 g 1 s g W2 g g A pffiffiffiffiffi pffiffiffi þ pffiffiffi þa þ e e erfc pffiffiffi þ As 2 4 A 4 2 A3 2 2 s A 2A b2 g gW b3 gW s g X1 l1 gb1 X1 l1 gWb3 þa þ pffiffiffi þ a pffiffiffi a a pffiffiffi þ a 2 2 2 4 A 4 4 A 4 A g2 W g2 g2 b1 g s g gX1 l1 b2 W W pffiffiffiffiffi eAs 4s 2 g þ a pffiffiffiffiffi e 2 gAs 4s a pffiffiffiffiffi e 2 gAs 4s A 2 ps 2 ps 2 ps g2 X1 l1 W W gAsg2 gX1 l1 W gAsg2 g2 1 4s a pffiffiffiffiffi e 2 4s a pffiffiffiffiffiffiffi e 2 A . 4s2 2s 2 ps 2 ps3
þa
ð34Þ
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The above equation after using Eq. (15) becomes pffiffi pffiffiffiffiffiffi f þ ig g W2 g g A e erfc pffiffiffi As 1 ¼ X1 l1 þ e 2 s ðmcÞ3 s g X1 l1 b1 g 1 s g b2 g gW pffiffiffiffiffi pffiffiffi þ pffiffiffi a þ a pffiffiffi þ a 2 4 A 4 2 A3 2A 2 2 A 4 A ffiffi p pffiffiffiffiffiffi b3 gW s g X1 l1 gb1 X1 l1 gWb3 g W2 g g A pffiffiffi a pffiffiffi a þ e e erfc pffiffiffi þ As þa 2 2 4 A 4 2 s 4 A s g X1 l1 b1 g 1 s g b2 g gW pffiffiffiffiffi pffiffiffi þ pffiffiffi þa a pffiffiffi þ a 3 2 4 A 4 2 A 2 2 4 A A 2A b3 gW s g X1 l1 gb1 X1 l1 gWb3 b1 g s Asg2 W g 4s 2 pffiffiffiffiffi e þ pffiffiffi þ a pffiffiffi a þa þa 2 2 4 A 2A ps 4 4 A g2 g2 g2 g gX1 l1 b2 W g2 X1 l1 W W W þ a pffiffiffiffiffi e 2 gAs 4s a pffiffiffiffiffi e 2 gAs 4s a pffiffiffiffiffiffiffi e 2 gAs 4s 2 ps 2 ps 2 ps3 2 2 g gX1 l1 W g 1 ð35Þ a pffiffiffiffiffi e 2 gAs 4s A . 2 2s 4s 2 ps It is worth emphasising to mention here that for a1 = c = 0, the results of Erdogan [2] can be recovered. This provides useful mathematical check.
4. Numerical solution Eq. (17) is a third order partial differential equation, it is difficult to obtain an exact analytical solution. Therefore, we have used perturbation technique to solve the problem, which gives an approximate solution. Numerical solution of the complete problem is now presented, which besides its own importance, to check the accuracy of analytic solution obtained by perturbation method. In this problem, we use modified Crank–Nicolson implicit formulation except for the term involving third order partial derivative. The governing equation (17) is transformed into algebraic equation by substituting the approximations to derivatives as a ½ðF iþ1;jþ1 2F i;jþ1 þ F i1;jþ1 Þ ðF iþ1;j 2F i;j þ F i1;j Þ kh2 aW 1 3 ½F iþ2;j 3F iþ1;j þ 3F i;j F i1;j þ 2 ð1 iX1 aÞ½ðF iþ1;jþ1 2F i;jþ1 þ F i1;jþ1 Þ h 2h W þ ðF iþ1;j 2F i;j þ F i1;j Þ þ ½ðF iþ1;jþ1 F i1;jþ1 Þ þ ðF iþ1;j F i1;j Þ 4h 1 1 ð36Þ ½F i;jþ1 F i;j ðiX1 þ N Þ ½F i;jþ1 þ F i;j ¼ 0. k 2 The unknown Fi,j + 1 cannot be expressed in term of known quantities at the time level j, namely Fi1,j, Fi,j, and Fi + 1,j. Therefore Eq. (36) represents an equation in three unknowns namely, Fi1,j + 1, Fi,j + 1 and Fi + 1,j + 1. Hence Eq. (36), when applied at a given mesh point P(ih, jk), can-
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473
not by itself result in a solution for Fi,j + 1. Therefore the equation must be written at all interior mesh points. Thus we obtain a system of algebraic equations. For that, multiplying Eq. (36) by k, we get Ai F i1;jþ1 þ Bi F i;jþ1 þ C i F iþ1;jþ1 ¼ Di
ð37Þ
with the following initial and boundary conditions: F 0;j ¼ jk X1 l1 ;
F M;j ¼ 0;
F i;0 ¼ 0; i ¼ 0; 1; 2; . . . ; M;
where a kð1 iX1 aÞ kW þ ; 4h h2 2h2 2a kð1 iX1 aÞ kðiX1 þ N Þ Bi ¼ 2 1 ; 2 2 h h a kð1 iX1 aÞ kW Ci ¼ 2 þ þ ; 4h h 2h2 a kaW Di ¼ 2 ½F iþ1;j 2F i;j þ F i1;j þ 3 ½F iþ2;j 3F iþ1;j þ 3F i;j F i1;j ð38Þ h h kð1 iX1 aÞ kW k ½F iþ1;j 2F i;j þ F i1;j ½F iþ1;j F i1;j F i;j þ ðiX1 þ N ÞF i;j 2 4h 2 2h and M is chosen large enough such that Mh approximates g at infinity. To find the value of DM1 at each time level j, we need the value of FM + 1,j. Since our governing equation (17) is of order three while given boundary conditions are two, therefore we introduce an augmented boundary condition Ai ¼
oF ð1; sÞ ¼0 og and consequently the problem becomes well-posed. This boundary condition is discretized to give F Mþ1;j ¼ F M;j : For i = 1 B1 F 1;jþ1 þ C 1 F 2;jþ1 ¼ D01 ; where D01 ¼ D1 A1 F 0;jþ1 . For 2 6 i 6 M 2 Ai F i1;jþ1 þ Bi F i;jþ1 þ C i F iþ1;jþ1 ¼ Di . For i = M 1 AM1 F M2;jþ1 þ BM1 F M1;jþ1 ¼ D0M1 ; where D0M1 ¼ DM1 C M1 F M;jþ1 .
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The above 2 B1 6A 6 2 6 6 0 6 6 6 6 6 0 6 6 6 6 6 0 6 6 4 0 0
set of M 1 equations can be written in the matrix form as 32 3 2 0 3 D1 F1 C1 0 0 0 0 0 0 0 7 7 6 6 B2 C 2 0 0 0 0 0 0 76 F 2 7 6 D2 7 7 76 7 6 7 7 7 6 6 A3 B3 C 3 0 0 0 0 0 76 F 3 7 6 D3 7 7 76 7 6 7 76 7 6 7 76 7 6 7 7 6 7 6 0 0 Ai Bi C i 0 0 0 7 76 F i 7 ¼ 6 Di 7. 76 7 6 7 76 7 6 7 76 7 6 7 7 6 7 6 0 0 0 0 AM3 BM3 C M3 0 7 76 F M3 7 6 DM3 7 76 7 6 7 0 0 0 0 0 AM2 BM2 C M2 54 F M2 5 4 DM2 5 D0M1 0 0 0 0 0 0 AM1 BM1 F M1
The above system of equations has been solved and the results are given in the tables.
5. Comparison of analytical and numerical solutions A comparison between numerical results and analytical results have been shown in the following tables. A good accuracy is obtained between the two results which supports the approximate results obtained by each method. It further justifies the correctness of the two methods and the results obtained thereof. This further suggests that the major contribution in the perturbation series comes from the first two terms which is the essence of the perturbation method. 6. Graphs and discussions In order to show the flow pattern of velocities for second grade fluid, we present the velocity field (f, g) in the form of graphs by separating F (Eq. (35)) into real and imaginary parts. Figs. 1 and 2 illustrate the graphs of f and g respectively, when s = 1, N = W = l1 = 0 and X1 = 1. It is seen that f increases for large values of a where as g decreases much as compared
Fig. 1. The variation of the velocity with distance from the disk for various values of second grade parameter a when s = 1, l1 = 0, W = N = 0 and X = 1.
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Fig. 2. The variation of the velocity with distance from the disk for various values of second grade parameter a when s = 1, l1 = 0, W = N = 0 and X = 1.
to the graph of f. The boundary layer thickness is found to increase in both cases. Now we add non-coaxial parameter l1 = 0.5. It is pertinent to mention that the values of f and g and boundary layer thickness in the Figs. 3 and 4 are larger when compared with the graphs when l1 = 0.
Fig. 3. The variation of the velocity with distance from the disk for various values of second grade parameter a when s = 1, l1 = 0.5, W = N = 0 and X = 1.
Fig. 4. The variation of the velocity with distance from the disk for various values of second grade parameter a when s = 1, l1 = 0.5, W = N = 0 and X = 1.
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However, the value of f stables at X1l1 and g decreases near the disk and then increases and become positive. The variation in angular velocity on the velocity profiles is depicted in Figs. 5 and 6. In these figures, we choose s = 1, N = W = l1 = 0 and a = 0.2. It is clearly seen that f and g decrease for large X1 as expected but the layer thickness in the case of f decreases whereas for g it increases. Now we plot Figs. 7 and 8 for l1 = 0.5. These show that the values of fand g and layer thickness increases. The graphs of f stables at the product X1l1. The rotational effects on g are larger when compared with f. Now we show the effects of time on the velocity profile in Figs. 9 and 10 when N = W = l1 = 0, a = 0.2 and X1 = 1. It is seen that f increases and g decreases with the increase in time. However, the layer thickness in both cases increases. In Figs. 11 and 12, we take l1 = 0.5. Here, it is revealed that f increases near the plate but g decreases. The layer thickness for f and g decreases and increases respectively. Also it is noteworthy to mention that the value of g in the present situation becomes negative from positive after some time which is quite distinct to the case l1 = 0.
Fig. 5. The variation of the velocity with distance from the disk for various values of rotation parameter X when s = 1, l1 = 0, W = N = 0, and a = 0.2.
Fig. 6. The variation of the velocity with distance from the disk for various values of rotation parameter X when s = 1, l1 = 0, W = N = 0, and a = 0.2.
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Fig. 7. The variation of the velocity with distance from the disk for various values of rotation parameter X when s = 1, l1 = 0.5, W = N = 0, and a = 0.2.
Fig. 8. The variation of the velocity with distance from the disk for various values of rotation parameter X when s = 1, l1 = 0.5, W = N = 0, and a = 0.2.
Fig. 9. The variation of the velocity with distance from the disk for various values of time s when X = 1, l1 = 0, W = N = 0, and a = 0.2.
The variations in non-coaxial parameter l1 is presented in Figs. 13 and 14. From these figures, it is observed that with the increase in l1, f stables at the product X1l1. At the same time the value of g increases and becomes positive from negative only in the presence of non-coaxial parameter.
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Fig. 10. The variation of the velocity with distance from the disk for various values of time s when X = 1, l1 = 0, W = N = 0, and a = 0.2.
Fig. 11. The variation of the velocity with distance from the disk for various values of time s when X = 1, l1 = 0.5, W = N = 0, and a = 0.2.
Fig. 12. The variation of the velocity with distance from the disk for various values of time s when X = 1, l1 = 0.5, W = N = 0, and a = 0.2.
The effects of magnetic field N are given in the Figs. 15 and 16. It is noted that with the increase in magnetic parameter N the values of velocity fields f and g decrease and increase respectively but
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Fig. 13. The variation of the velocity with distance from the disk for various values of noncoaxial parameter l1 when s = 1, X = 1, W = N = 0, and a = 0.2.
Fig. 14. The variation of the velocity with distance from the disk for various values of noncoaxial parameter l1 when s = 1, X = 1, W = N = 0, and a = 0.2.
Fig. 15. The variation of the velocity with distance from the disk for various values of magnetic field parameter N when s = 1, l1 = 0, W = 0, and a = 0.2.
the layer thickness in both the cases decreases. These figures are for W = l1 = 0, a = 0.2 and s = X1 = 1. In the presence of non-coaxial parameter l1 = 0.5 the variation of magnetic field
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Fig. 16. The variation of the velocity with distance from the disk for various values of magnetic field parameter N when s = 1, l1 = 0, W = 0, and a = 0.2.
Fig. 17. The variation of the velocity with distance from the disk for various values of magnetic field parameter N when s = 1, l1 = 0.5, W = 0, and a = 0.2.
Fig. 18. The variation of the velocity with distance from the disk for various values of magnetic field parameter N when s = 1, l1 = 0.5, W = 0, X = 1 and a = 0.2.
are shown in Figs. 17 and 18. It is noticed that with the increase in N, the change in the graph of f is very small and stables at the product X1l1 but the variation in g in the present situation is much. The value of g increases but the layer thickness decreases Tables 1–3.
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Table 1 s = 1, X = 1, a = 0.1, W = 0.5, N = 0 g 0.1 0.5 1.0 1.5 2.0 2.5 3.0
Real velocity f
Imaginary velocity g
Analytic solution
Numerical solution
Difference
Analytic solution
Numerical solution
Difference
0.87847 0.50934 0.24285 0.10690 0.04158 0.01290 0.00218
0.87546 0.50104 0.23645 0.10581 0.04511 0.01838 0.00717
0.00301 0.00831 0.00640 0.00109 0.00353 0.00548 0.00499
0.03793 0.10263 0.08619 0.04517 0.01385 0.00158 0.00573
0.03352 0.09251 0.08510 0.05673 0.03258 0.01705 0.00834
0.00441 0.01012 0.00109 0.01156 0.01873 0.01863 0.01407
Table 2 s = 1, X = 1, a = 0.1, W = 0, N = 0.5 g 0.1 0.5 1.0 1.5 2.0 2.5 3.0
Real velocity f
Imaginary velocity g
Analytic solution
Numerical solution
Difference
Analytic solution
Numerical solution
Difference
0.88384 0.52837 0.26684 0.12912 0.05885 0.02418 0.00822
0.87917 0.51204 0.24701 0.11289 0.04908 0.02036 0.00807
0.00467 0.01633 0.01983 0.01623 0.00977 0.00382 0.00015
0.03360 0.09682 0.09164 0.05893 0.02853 0.00869 0.00052
0.03250 0.09175 0.08667 0.05918 0.03473 0.01853 0.00923
0.00110 0.00507 0.00497 0.00025 0.00620 0.00984 0.00975
Table 3 s = 1, X = 1, a = 0.1, W = 0.5, N = 0.5 g 0.1 0.5 1.0 1.5 2.0 2.5 3.0
Real velocity f
Imaginary velocity g
Analytic solution
Numerical solution
Difference
Analytic solution
Numerical solution
Difference
0.86939 0.48638 0.22426 0.09681 0.03783 0.01250 0.00293
0.86260 0.46741 0.20770 0.08813 0.03581 0.01396 0.00522
0.00679 0.01897 0.01656 0.00868 0.00202 0.00146 0.00229
0.03230 0.08579 0.07167 0.03882 0.01405 0.00148 0.00253
0.03116 0.08224 0.07192 0.04580 0.02523 0.01270 0.00600
0.00114 0.00355 0.00025 0.00698 0.01118 0.01122 0.00853
The variation in suction velocity for N = 0, a = 0.2 and s = X1 = 1 can be seen from Figs. 19– 22. The layer thickness for the case l1 = 0 and l1 = 0.5 decreases. The effects of injection velocity are presented in Figs. 23–26. The value of f and g as well as layer thickness increases in the absence of non-coaxial parameter l1. For l1 = 0.5 the variation within the layer is very large. In this situation it is noted that for the increment in injection velocity the value of f decreases and g increases much when compared with the case of l1 = 0.
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Fig. 19. The variation of the velocity with distance from the disk for various values of suction parameter W when s = 1, l1 = 0, N = 0, X = 1 and a = 0.2.
Fig. 20. The variation of the velocity with distance from the disk for various values of suction parameter W when s = 1, l1 = 0, N = 0, X = 1 and a = 0.2.
Fig. 21. The variation of the velocity with distance from the disk for various values of suction parameter W when s = 1, l1 = 0.5, N = 0, X = 1 and a = 0.2.
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Fig. 22. The variation of the velocity with distance from the disk for various values of suction parameter W when s = 1, l1 = 0.5, N = 0, X = 1 and a = 0.2.
Fig. 23. The variation of the velocity with distance from the disk for various values of injection parameter W when s = 1, l1 = 0, N = 0, X = 1 and a = 0.2.
Fig. 24. The variation of the velocity with distance from the disk for various values of injection parameter W when s = 1, l1 = 0, N = 0, X = 1 and a = 0.2.
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Fig. 25. The variation of the velocity with distance from the disk for various values of injection parameter W when s = 1, l1 = 0.5, N = 0, X = 1 and a = 0.2.
Fig. 26. The variation of the velocity with distance from the disk for various values of injection parameter W when s = 1, l1 = 0.5, N = 0, X = 1 and a = 0.2.
References [1] Erdogan ME. Flow due to eccentric rotating a porous disk and a fluid at infinity. Trans ASME J Appl Mech 1976;43:203–4. [2] Erdogan ME. Unsteady flow of a viscous fluid due to non-coaxial rotations of a disk and a fluid at infinity. Int J Non-linear Mech 1997;32:285–90. [3] Erdogan ME. Flow induced by non-coaxial rotations of a disk executing non-torsional oscillations and a fluid at infinity. Int J Eng Sci 2000;38:175–96. [4] Kasiviswanathan SR, Rao AR. An unsteady flow due to eccentrically rotating porous disk and a fluid at infinity. Int J Eng Sci 1987;25:1419–25. [5] Pop I. Unsteady flow due to non-coaxially rotating a disk and a fluid at infinity. Bull Tech Uni 1st 1979;32:14–8. [6] Hayat T, Asghar S, Siddiqui AM, Haroon T. Unsteady MHD flow due to non-coaxial rotations of a porous disk and a fluid at infinity. Acta Mech 2001;151:127–34. [7] Erdogan ME. Non-Newtonian flow due to non-coaxially rotations of a disk and a fluid at infinity. Z Angew Math Mech (ZAMM) 1976;56:141–6. [8] Ersoy HV. MHD flow of an Oldroyd-B fluid due to non-coaxially rotations of a porous disk and a fluid at infinity. Int J Eng Sci 2000;38:1837–50.
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[9] Siddiqui AM, Haroon T, Hayat T, Asghar S. Unsteady MHD flow of a non-Newtonian fluid due to eccentric rotations of a porous disk and a fluid at infinity. Acta Mech 2001;147:99–109. [10] Hayat T, Haroon T, Asghar S, Siddiqui AM. MHD flow of a third-grade fluid due to eccentric rotations of a porous disk and a fluid at infinity. Int J Non-linear Mech 2003;38:501–11. [11] Debnath L, Mukherjee S. Unsteady multiple boundary layers on a porous plate in a rotating system. Phys Fluids 1973;16:1418–21. [12] Beard DW, Walters K. Elastico-viscous boundary layer flow, l. Two-dimensional flow near a stagnation point. Proc Camb Philos Soc 1964;64:667–74. [13] Shrestha GM. Laminar elastico-viscous flow through channels with porous walls with different permeability. Appl Sci Res 1969;20:289–305. [14] Mishra SP, Mohapatra U. Elastico-viscous flow between a rotating and a stationary disk with uniform suction at the stationary disk. J Appl Phys 1977;48:1515–21. [15] Rajagopal KR, Na TY, Gupta AS. Flow of visco-elastic fluid over a stretching sheet. Rheol Acta 1984;23:213–5. [16] Verma PD, Sharma PR, Ariel PD. Applying quasilinearization to the problem of steady laminar flow of a second grade fluid between two rotating porous disk. J Tribol Trans ASME 1984;106:448–555. [17] Erdogan ME. On the flow of a non-Newtonian fluid past a porous flat plate. Z Angew Math Mech 1975;55:79–83.
MHD non-Newtonian flow due to non-coaxial rotations of an accelerated disk and a fluid at infinity S. Asghar a, K. Hanif b, T. Hayat b, C.M. Khalique
c,*
a
Department of Mathematics, COMSATS Institute of Information Technology, H-8 Islamabad, Pakistan b Department of Mathematics, Quaid-i-Azam University, Islamabad, Pakistan c Department of Mathematical Sciences, International Institute for Symmetry Analysis and Mathematical Modelling, University of North-West, Private Bag X 2046, Mmabatho 2735, South Africa Received 16 September 2004; received in revised form 19 April 2005; accepted 21 April 2005 Available online 5 July 2005
Abstract The magnetohydrodynamic (MHD) flow due to non-coaxial rotations of a porous disk, moving with uniform acceleration in its own plane and a second grade fluid at infinity is examined. The fluid is electrically conducting in the presence of a uniform magnetic field. The effects of non-Newtonian fluid characteristics and uniform acceleration of the disk on the velocity field are presented both analytically and numerically. A very good accuracy has been seen. Ó 2005 Elsevier B.V. All rights reserved. PACS: 76A05 Keywords: Second grade fluid; MHD flow; Uniformly accelerated disk; Numerical solutions
1. Introduction An investigation of MHD boundary layers under the influence of viscous forces is important in understanding a variety of geophysical, astrophysical and engineering phenomena such as those *
Corresponding author. Tel.: +27 18 389 2354 2319; fax: +27 18 389 2292. E-mail address: [email protected] (C.M. Khalique).
1007-5704/$ - see front matter Ó 2005 Elsevier B.V. All rights reserved. doi:10.1016/j.cnsns.2005.04.006
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that occur at the coremantle interface of the earth. Much progress has been made during the previous years in the development of Newtonian flows due to non-coaxial rotations of a disk and fluid at infinity. Mention may be made to the interesting work of Erdogan [1–3], Kasiviswanathan and Rao [4], Pop [5] and Hayat et al. [6]. Despite the above studies, a little work seems to have been done for flows of non-Newtonian fluids due to non-coaxial rotation of a disk and fluid at infinity. Few studies illustrating the non-Newtonian flows have been reported, see Erdogan [7], Ersoy [8], Siddiqui et al. [9] and Hayat et al. [10]. The study of non-Newtonian fluid dynamics is important in connection with plastic manufacture, performance of lubricants, processing of food, and movements of biological and geophysical fluids. Most geological fluids contains higher molecular weight components and are, therefore non-Newtonian. Alternatively, geophysical applications concern ice flow, magna flow which are based on non-Newtonian constitutive behaviours. In the present work, the unsteady flow of an incompressible and non-Newtonian fluid due to non-coaxial rotations of a porous moving disk and a fluid at infinity is studied with magnetic field. A uniform suction or injection is applied at the surface of the disk. The disk is non-conducting. The analytic solution is obtained by perturbation method in terms of small material parameter. The numerical solution of the problem is obtained using Crank–Nicolson method with some modification for entire range of all the parameters. The graphical results of the analytic solution are presented and discussed. A comparison between analytic and numerical solutions is made in the tabular form. An excellent accuracy has been seen in the two solutions.
2. Formulation of the problem Let us consider the disk at z = 0, i.e., it is the xy-plane or more precisely 1 < x < 1, 1 < y < 1. The fluid occupies the space z > 0. Thus, above the disk, the fluid can be prescribed as (x, y, z) such that 1 < x < 1, 1 < y < 1, 0 < z < 1. Naturally, the fluid occupies a 3-D space. Now the z and z 0 -axes are such that z-axis cuts the plane of disk at (0, 0, 0) and the z 0 -axis cuts the disk at (0, l, 0). Now, consider the disk at z = 0 and the fluid at z = 1. The disk and the fluid at infinity are initially rotating about z 0 -axis and suddenly the disk starts rotating about zaxis and moving along x-axis and the fluid at infinity continues to rotate about z 0 -axis. This is termed as non-coaxial rotation of the disk and the fluid at infinity [1–10]. This give rise to the following boundary and initial conditions u ¼ Xy þ ct;
v ¼ Xx;
at z ¼ 0;
u ¼ Xðy lÞ;
v ¼ Xx;
as z ! 1
u ¼ Xðy lÞ;
v ¼ Xx;
at t ¼ 0;
for t > 0; for all t > 0;
ð1Þ
for z > 0
in which u and v are the velocity components in the x and y directions respectively, t is the time and X is the angular velocity. We seek a solution of the form u ¼ Xy þ f ðz; tÞ;
v ¼ Xx þ gðz; tÞ.
ð2Þ
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467
The governing equations for fluid flow are the continuity and momentum equations which are respectively given by div v ¼ 0; ov þ ðv $Þv ¼ div T þ J B. q ot
ð3Þ ð4Þ
Since we are considering electrically conducting fluids so there is second term on the right hand side of above equation. The following equation gives the origin of B (OhmÕs law): J ¼ rðE þ v BÞ
ð5Þ
in which E is the electric field. In the last term of equation (4) on the right-hand side B = B0 + b is the total magnetic field, r is the electrical conductivity, v is the velocity vector, q is the density and J is the electric current density. For small magnetic Reynolds number the induced magnetic field is neglected and hence we can easily write 1 rB2 ðJ BÞ ¼ 0 v. q q
ð6Þ
From Eqs. (2) and (3) we have ð7Þ
w ¼ w0 ;
where w is z-component of velocity and w0 > 0 for suction and w0 < 0 for blowing or injection. The Cauchy stress T in an incompressible fluid of second grade is given by T ¼ pI þ lA1 þ a1 A2 þ a2 A21 ;
ð8Þ
where A1 and A2 are first two kinematical tensors given by A1 ¼ ðgrad vÞ þ ðgrad vÞ ; o þ v r A1 þ A1 ðgrad vÞ þ ðgrad vÞ A1 ; A2 ¼ ot where p is the pressure, I is the identity tensor, l is the dynamic viscosity and (*) denotes transpose of the matrix. If an incompressible fluid of second grade is to have motion which are compatible with thermodynamics in the sense of Clausius-Duhem inequality and the condition that the Helmholtz free energy be a minimum when the fluid is at rest, then the following conditions must be satisfied: l P 0;
a1 P 0;
a1 þ a2 ¼ 0.
In view of Eqs. (4) and (6)–(8), we can write 1 oP o2 f of of r ¼ t 2 þ w0 þ Xg þ X2 x B20 ½f ðz; tÞ Xy q ox oz oz ot q 3 3 2 a1 o f of og þ w0 3 þ X 2 ; 2 oz oz q otoz
ð9Þ
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1 oP o2 g og og r ¼ t 2 þ w0 Xf þ X2 y B20 ½gðz; tÞ þ Xx q oy oz oz ot q 3 3 2 a1 o g og of w0 3 þ X 2 ; þ 2 oz oz q otoz 1 oP rB20 ¼ w0 ; q oz q
ð10Þ ð11Þ
where ( P ¼ p ð2a1 þ a2 Þ
of oz
2
og þ oz
2 )
and the boundary and initial conditions in terms of f and g become f ð0; tÞ ¼ ct;
gð0; tÞ ¼ 0;
f ð1; tÞ ¼ Xl;
gð1; tÞ ¼ 0;
f ðz; 0Þ ¼ Xl;
gðz; 0Þ ¼ 0.
ð12Þ
Eliminating P from Eqs. (9) and (10) by differentiating with respect to z and combining them along with Eq. (11) and boundary conditions (12), we have a1 o3 F o3 F ia1 X o2 F oF oF rB20 w0 3 þ t þ w0 ð13Þ X iþ F ¼ 0; q q otoz2 oz oz2 oz ot qX F ðz; tÞ ¼ f þ ig Xl;
ð14Þ
where F ð0; tÞ ¼ ct Xl;
F ð1; tÞ ¼ 0;
F ðz; 0Þ ¼ 0.
ð15Þ
Introducing F ¼
N¼
F 1
;
ðmcÞ3
W ¼
c 13 rB2 0
m2
q
;
1 1
ðcmÞ3
w0 ;
g¼
1 a1 c2 3 a¼ ; q m4
c 13 m2
2 13 c s¼t ; m
z;
X1 ¼
m 13 c2
X;
ð16Þ
c 13 l1 ¼ l 2 m the problem (Eqs. (13)–(15)) reduces to o3 F o3 F o2 F oF oF a aW 3 þ ð1 iX1 aÞ 2 þ W ðiX1 þ N ÞF ¼ 0; osog2 og og og os F ð0; sÞ ¼ s X1 l1 ; F ð1; sÞ ¼ 0; F ðg; 0Þ ¼ 0.
ð17Þ ð18Þ
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3. Analytic solution Before proceeding with the solution of the above problem it would be interesting to remark here that in the classical viscous case (a = 0), we encounter a differential equation of order two [11]. The analysis of the flow of the second grade fluids, in particular, and the viscoelastic fluids, in general, is more challenging mathematically and computationally, because of a peculiarity in the equations governing the fluid motion; namely the order of the differential equation(s) characterizing the flow of these fluids is more than the number of the available boundary conditions. The difficulty is further accentuated by the fact that a non-Newtonian parameter of fluid (for example a, for a second grade fluid) usually occurs in the coefficient of the highest derivative. The usual attempts to resolve this difficulty centered around seeking a perturbation solution assuming the non-Newtonian fluid parameter to be small; the classical paper begin by Beard and Walters [12], who considered the two-dimensional stagnation point flow of the WalterÕs B fluid. One may also refer, for example, to the works of Shrestha [13], Misra and Mohapatra [14], Rajagopal et al. [15], Verma et al. [16] and Erdogan [17] for other problems in various geometries. In the present analysis, the difficulty is also removed by seeking a solution of the following form: F ðg; sÞ ¼ F 0 ðg; sÞ þ aF 1 ðg; sÞ þ Oða2 Þ;
ð19Þ
which is valid for small values of a only. Now, substituting Eq. (19) in Eq. (17) and conditions (18) and then collecting terms of like powers of a, one obtains the following systems of differential equations along with the boundary conditions: 3.1. System of order zero o2 F 0 oF 0 oF 0 þW ðiX1 þ NÞF 0 ¼ 0; 2 og og os F 0 ð0; sÞ ¼ s X1 l1 ; F 0 ð1; sÞ ¼ 0; F 0 ðg; 0Þ ¼ 0.
ð20Þ ð21Þ
3.2. System of order one o2 F 1 oF 1 oF 1 o3 F 0 o3 F 0 o2 F 0 ðiX þ W þ NÞF ¼ þ W þ iX ; 1 1 1 og2 og os osog2 og3 og2 F 1 ð0; sÞ ¼ 0; F 1 ð1; sÞ ¼ 0; F 1 ðg; 0Þ ¼ 0.
ð22Þ ð23Þ
3.3. Zeroth-order solution After taking Laplace transform, the zeroth order system becomes 00
0
F 0 þ W F 0 ðs þ N þ iX1 ÞF 0 ¼ 0;
ð24Þ
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1 X1 l1 ; s2 s F 0 ð1; sÞ ¼ 0; F 0 ð0; sÞ ¼
where primes indicate differentiation with respect to g. The solution of equation (24) satisfying conditions (25) is 1 X1 l1 gm F 0 ðg; sÞ ¼ 2 e s s
ð25Þ
ð26Þ
in which W pffiffiffiffiffiffiffiffiffiffiffi W2 þ s þ A; A ¼ þ N þ iX1 . 2 4 Laplace inversion of Eq. (26) yields pffiffi pffiffi pffiffiffiffiffiffi pffiffiffiffiffiffi s g g s g g w F 0 ðg; sÞ ¼ e 2 g pffiffiffi eg A erfc pffiffiffi As þ þ pffiffiffi eg A erfc pffiffiffi þ As 2 4 A 2 4 A 2 s 2 s pffiffi
pffiffi pffiffiffiffiffiffi pffiffiffiffiffiffi X1 l 1 g g eg A erfc pffiffiffi As þ eg A erfc pffiffiffi þ As . ð27Þ 2 2 s 2 s m¼
3.4. First-order solution Taking Laplace transform of Eq. (22) and then using Eq. (26) in the resulting equation, we have 1 X1 l 1 00 0 F 1 þ W F 1 ðs þ N þ iX1 ÞF 1 ¼ 2 ð28Þ X emg ; s s where X ¼
W4 3 2 þ W N þ X21 þ iX1 ðW 2 þ N Þ þ sf2W 2 þ N g 2 2 pffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi þ 2Ws A þ s þ s2 þ W fW 2 þ Ng A þ s.
ð29Þ
The boundary conditions now are F 1 ð0; sÞ ¼ 0;
F 1 ð1; sÞ ¼ 0.
Solving Eq. (28) along with the boundary conditions (30), we obtain pffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi g 1 X1 l 1 F 1 ðg; sÞ ¼ pffiffiffiffiffiffiffiffiffiffiffi 2 ½b1 þ sb2 þ 2Ws A þ s þ s2 þ b3 W A þ semg s 2 Aþs s
ð30Þ
ð31Þ
in which W4 3 2 þ W N þ X21 þ iX1 ðW 2 þ N Þ; 2 2 b2 ¼ 2W 2 þ N; b1 ¼
b3 ¼ W 2 þ N .
ð32Þ
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After lengthy calculations, the Laplace inversion of Eq. (31) gives pffiffiffiffiffiffi b1 g W g 1 s g gpffiffiA g 2 pffiffiffiffiffi pffiffiffi F 1 ðg; sÞ ¼ e e erfc pffiffiffi þ As 4 2 s A 2A 2 A3 rffiffiffi pffiffi pffiffiffiffiffiffi 1 s g g A g 2 s Asg2 4s erfc pffiffiffi As þ e e þ pffiffiffiffiffi3 þ pffiffiffi A p 2 s A 2A 2 A pffiffi pffiffiffiffiffiffi pffiffiffiffiffiffi gb2 W g gpffiffiA g g g A erfc pffiffiffi As e erfc pffiffiffi þ As þ pffiffiffi e 2 e 2 s 2 s 4 A pffiffi pffiffi pffiffiffiffiffiffi pffiffiffiffiffiffi gW W g g A g g g A 2 e e erfc pffiffiffi As þ e erfc pffiffiffi þ As þ 2 2 s 2 s pffiffi pffiffiffiffiffiffi g2 g b3 gW W g s g g W e 2 pffiffiffi eg A erfc pffiffiffi As þ pffiffiffiffiffi e 2 gAs 4s þ 2 2 4 A 2 ps 2 s pffiffi pffiffiffiffiffiffi s g g gX1 l1 b1 W þ pffiffiffi eg A erfc pffiffiffi þ As pffiffiffi e 2 g þ 2 4 A 2 s 4 A pffiffi pffiffi pffiffiffiffiffiffi pffiffiffiffiffiffi g2 g g gX1 l1 b2 W g A g A e erfc pffiffiffi As e erfc pffiffiffi þ As pffiffiffiffiffi e 2 gAs 4s 2 ps 2 s 2 s g2 g2 g2 X1 l1 W W gX1 l1 W g2 1 gX1 l1 Wb3 W g pffiffiffiffiffiffiffi e 2 gAs 4s pffiffiffiffiffi e 2 gAs 4s A e 2 2 3 2s 4s 4 2 ps 2 ps pffiffi pffiffi pffiffiffiffiffiffi pffiffiffiffiffiffi g g g A g A e erfc pffiffiffi As þ e erfc pffiffiffi þ As . 2 s 2 s
471
ð33Þ
With the help of Eqs. (19), (27) and (33) we write pffiffi pffiffiffiffiffiffi s g g X1 l1 b1 g 1 s g W2 g g A pffiffiffiffiffi pffiffiffi þ erfc pffiffiffi As pffiffiffi F ðg; sÞ ¼ e e a 2 4 A 4 2 A3 2 2 s A 2A b2 g gW b3 gW s g X1 l1 gb1 X1 l1 gWb3 þa pffiffiffi a pffiffiffi a þ a pffiffiffi þ a 2 2 2 4 A 4 4 A 4 A pffiffi pffiffiffiffiffiffi s g g X1 l1 b1 g 1 s g W2 g g A pffiffiffiffiffi pffiffiffi þ pffiffiffi þa þ e e erfc pffiffiffi þ As 2 4 A 4 2 A3 2 2 s A 2A b2 g gW b3 gW s g X1 l1 gb1 X1 l1 gWb3 þa þ pffiffiffi þ a pffiffiffi a a pffiffiffi þ a 2 2 2 4 A 4 4 A 4 A g2 W g2 g2 b1 g s g gX1 l1 b2 W W pffiffiffiffiffi eAs 4s 2 g þ a pffiffiffiffiffi e 2 gAs 4s a pffiffiffiffiffi e 2 gAs 4s A 2 ps 2 ps 2 ps g2 X1 l1 W W gAsg2 gX1 l1 W gAsg2 g2 1 4s a pffiffiffiffiffi e 2 4s a pffiffiffiffiffiffiffi e 2 A . 4s2 2s 2 ps 2 ps3
þa
ð34Þ
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The above equation after using Eq. (15) becomes pffiffi pffiffiffiffiffiffi f þ ig g W2 g g A e erfc pffiffiffi As 1 ¼ X1 l1 þ e 2 s ðmcÞ3 s g X1 l1 b1 g 1 s g b2 g gW pffiffiffiffiffi pffiffiffi þ pffiffiffi a þ a pffiffiffi þ a 2 4 A 4 2 A3 2A 2 2 A 4 A ffiffi p pffiffiffiffiffiffi b3 gW s g X1 l1 gb1 X1 l1 gWb3 g W2 g g A pffiffiffi a pffiffiffi a þ e e erfc pffiffiffi þ As þa 2 2 4 A 4 2 s 4 A s g X1 l1 b1 g 1 s g b2 g gW pffiffiffiffiffi pffiffiffi þ pffiffiffi þa a pffiffiffi þ a 3 2 4 A 4 2 A 2 2 4 A A 2A b3 gW s g X1 l1 gb1 X1 l1 gWb3 b1 g s Asg2 W g 4s 2 pffiffiffiffiffi e þ pffiffiffi þ a pffiffiffi a þa þa 2 2 4 A 2A ps 4 4 A g2 g2 g2 g gX1 l1 b2 W g2 X1 l1 W W W þ a pffiffiffiffiffi e 2 gAs 4s a pffiffiffiffiffi e 2 gAs 4s a pffiffiffiffiffiffiffi e 2 gAs 4s 2 ps 2 ps 2 ps3 2 2 g gX1 l1 W g 1 ð35Þ a pffiffiffiffiffi e 2 gAs 4s A . 2 2s 4s 2 ps It is worth emphasising to mention here that for a1 = c = 0, the results of Erdogan [2] can be recovered. This provides useful mathematical check.
4. Numerical solution Eq. (17) is a third order partial differential equation, it is difficult to obtain an exact analytical solution. Therefore, we have used perturbation technique to solve the problem, which gives an approximate solution. Numerical solution of the complete problem is now presented, which besides its own importance, to check the accuracy of analytic solution obtained by perturbation method. In this problem, we use modified Crank–Nicolson implicit formulation except for the term involving third order partial derivative. The governing equation (17) is transformed into algebraic equation by substituting the approximations to derivatives as a ½ðF iþ1;jþ1 2F i;jþ1 þ F i1;jþ1 Þ ðF iþ1;j 2F i;j þ F i1;j Þ kh2 aW 1 3 ½F iþ2;j 3F iþ1;j þ 3F i;j F i1;j þ 2 ð1 iX1 aÞ½ðF iþ1;jþ1 2F i;jþ1 þ F i1;jþ1 Þ h 2h W þ ðF iþ1;j 2F i;j þ F i1;j Þ þ ½ðF iþ1;jþ1 F i1;jþ1 Þ þ ðF iþ1;j F i1;j Þ 4h 1 1 ð36Þ ½F i;jþ1 F i;j ðiX1 þ N Þ ½F i;jþ1 þ F i;j ¼ 0. k 2 The unknown Fi,j + 1 cannot be expressed in term of known quantities at the time level j, namely Fi1,j, Fi,j, and Fi + 1,j. Therefore Eq. (36) represents an equation in three unknowns namely, Fi1,j + 1, Fi,j + 1 and Fi + 1,j + 1. Hence Eq. (36), when applied at a given mesh point P(ih, jk), can-
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not by itself result in a solution for Fi,j + 1. Therefore the equation must be written at all interior mesh points. Thus we obtain a system of algebraic equations. For that, multiplying Eq. (36) by k, we get Ai F i1;jþ1 þ Bi F i;jþ1 þ C i F iþ1;jþ1 ¼ Di
ð37Þ
with the following initial and boundary conditions: F 0;j ¼ jk X1 l1 ;
F M;j ¼ 0;
F i;0 ¼ 0; i ¼ 0; 1; 2; . . . ; M;
where a kð1 iX1 aÞ kW þ ; 4h h2 2h2 2a kð1 iX1 aÞ kðiX1 þ N Þ Bi ¼ 2 1 ; 2 2 h h a kð1 iX1 aÞ kW Ci ¼ 2 þ þ ; 4h h 2h2 a kaW Di ¼ 2 ½F iþ1;j 2F i;j þ F i1;j þ 3 ½F iþ2;j 3F iþ1;j þ 3F i;j F i1;j ð38Þ h h kð1 iX1 aÞ kW k ½F iþ1;j 2F i;j þ F i1;j ½F iþ1;j F i1;j F i;j þ ðiX1 þ N ÞF i;j 2 4h 2 2h and M is chosen large enough such that Mh approximates g at infinity. To find the value of DM1 at each time level j, we need the value of FM + 1,j. Since our governing equation (17) is of order three while given boundary conditions are two, therefore we introduce an augmented boundary condition Ai ¼
oF ð1; sÞ ¼0 og and consequently the problem becomes well-posed. This boundary condition is discretized to give F Mþ1;j ¼ F M;j : For i = 1 B1 F 1;jþ1 þ C 1 F 2;jþ1 ¼ D01 ; where D01 ¼ D1 A1 F 0;jþ1 . For 2 6 i 6 M 2 Ai F i1;jþ1 þ Bi F i;jþ1 þ C i F iþ1;jþ1 ¼ Di . For i = M 1 AM1 F M2;jþ1 þ BM1 F M1;jþ1 ¼ D0M1 ; where D0M1 ¼ DM1 C M1 F M;jþ1 .
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The above 2 B1 6A 6 2 6 6 0 6 6 6 6 6 0 6 6 6 6 6 0 6 6 4 0 0
set of M 1 equations can be written in the matrix form as 32 3 2 0 3 D1 F1 C1 0 0 0 0 0 0 0 7 7 6 6 B2 C 2 0 0 0 0 0 0 76 F 2 7 6 D2 7 7 76 7 6 7 7 7 6 6 A3 B3 C 3 0 0 0 0 0 76 F 3 7 6 D3 7 7 76 7 6 7 76 7 6 7 76 7 6 7 7 6 7 6 0 0 Ai Bi C i 0 0 0 7 76 F i 7 ¼ 6 Di 7. 76 7 6 7 76 7 6 7 76 7 6 7 7 6 7 6 0 0 0 0 AM3 BM3 C M3 0 7 76 F M3 7 6 DM3 7 76 7 6 7 0 0 0 0 0 AM2 BM2 C M2 54 F M2 5 4 DM2 5 D0M1 0 0 0 0 0 0 AM1 BM1 F M1
The above system of equations has been solved and the results are given in the tables.
5. Comparison of analytical and numerical solutions A comparison between numerical results and analytical results have been shown in the following tables. A good accuracy is obtained between the two results which supports the approximate results obtained by each method. It further justifies the correctness of the two methods and the results obtained thereof. This further suggests that the major contribution in the perturbation series comes from the first two terms which is the essence of the perturbation method. 6. Graphs and discussions In order to show the flow pattern of velocities for second grade fluid, we present the velocity field (f, g) in the form of graphs by separating F (Eq. (35)) into real and imaginary parts. Figs. 1 and 2 illustrate the graphs of f and g respectively, when s = 1, N = W = l1 = 0 and X1 = 1. It is seen that f increases for large values of a where as g decreases much as compared
Fig. 1. The variation of the velocity with distance from the disk for various values of second grade parameter a when s = 1, l1 = 0, W = N = 0 and X = 1.
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Fig. 2. The variation of the velocity with distance from the disk for various values of second grade parameter a when s = 1, l1 = 0, W = N = 0 and X = 1.
to the graph of f. The boundary layer thickness is found to increase in both cases. Now we add non-coaxial parameter l1 = 0.5. It is pertinent to mention that the values of f and g and boundary layer thickness in the Figs. 3 and 4 are larger when compared with the graphs when l1 = 0.
Fig. 3. The variation of the velocity with distance from the disk for various values of second grade parameter a when s = 1, l1 = 0.5, W = N = 0 and X = 1.
Fig. 4. The variation of the velocity with distance from the disk for various values of second grade parameter a when s = 1, l1 = 0.5, W = N = 0 and X = 1.
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However, the value of f stables at X1l1 and g decreases near the disk and then increases and become positive. The variation in angular velocity on the velocity profiles is depicted in Figs. 5 and 6. In these figures, we choose s = 1, N = W = l1 = 0 and a = 0.2. It is clearly seen that f and g decrease for large X1 as expected but the layer thickness in the case of f decreases whereas for g it increases. Now we plot Figs. 7 and 8 for l1 = 0.5. These show that the values of fand g and layer thickness increases. The graphs of f stables at the product X1l1. The rotational effects on g are larger when compared with f. Now we show the effects of time on the velocity profile in Figs. 9 and 10 when N = W = l1 = 0, a = 0.2 and X1 = 1. It is seen that f increases and g decreases with the increase in time. However, the layer thickness in both cases increases. In Figs. 11 and 12, we take l1 = 0.5. Here, it is revealed that f increases near the plate but g decreases. The layer thickness for f and g decreases and increases respectively. Also it is noteworthy to mention that the value of g in the present situation becomes negative from positive after some time which is quite distinct to the case l1 = 0.
Fig. 5. The variation of the velocity with distance from the disk for various values of rotation parameter X when s = 1, l1 = 0, W = N = 0, and a = 0.2.
Fig. 6. The variation of the velocity with distance from the disk for various values of rotation parameter X when s = 1, l1 = 0, W = N = 0, and a = 0.2.
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Fig. 7. The variation of the velocity with distance from the disk for various values of rotation parameter X when s = 1, l1 = 0.5, W = N = 0, and a = 0.2.
Fig. 8. The variation of the velocity with distance from the disk for various values of rotation parameter X when s = 1, l1 = 0.5, W = N = 0, and a = 0.2.
Fig. 9. The variation of the velocity with distance from the disk for various values of time s when X = 1, l1 = 0, W = N = 0, and a = 0.2.
The variations in non-coaxial parameter l1 is presented in Figs. 13 and 14. From these figures, it is observed that with the increase in l1, f stables at the product X1l1. At the same time the value of g increases and becomes positive from negative only in the presence of non-coaxial parameter.
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Fig. 10. The variation of the velocity with distance from the disk for various values of time s when X = 1, l1 = 0, W = N = 0, and a = 0.2.
Fig. 11. The variation of the velocity with distance from the disk for various values of time s when X = 1, l1 = 0.5, W = N = 0, and a = 0.2.
Fig. 12. The variation of the velocity with distance from the disk for various values of time s when X = 1, l1 = 0.5, W = N = 0, and a = 0.2.
The effects of magnetic field N are given in the Figs. 15 and 16. It is noted that with the increase in magnetic parameter N the values of velocity fields f and g decrease and increase respectively but
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Fig. 13. The variation of the velocity with distance from the disk for various values of noncoaxial parameter l1 when s = 1, X = 1, W = N = 0, and a = 0.2.
Fig. 14. The variation of the velocity with distance from the disk for various values of noncoaxial parameter l1 when s = 1, X = 1, W = N = 0, and a = 0.2.
Fig. 15. The variation of the velocity with distance from the disk for various values of magnetic field parameter N when s = 1, l1 = 0, W = 0, and a = 0.2.
the layer thickness in both the cases decreases. These figures are for W = l1 = 0, a = 0.2 and s = X1 = 1. In the presence of non-coaxial parameter l1 = 0.5 the variation of magnetic field
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Fig. 16. The variation of the velocity with distance from the disk for various values of magnetic field parameter N when s = 1, l1 = 0, W = 0, and a = 0.2.
Fig. 17. The variation of the velocity with distance from the disk for various values of magnetic field parameter N when s = 1, l1 = 0.5, W = 0, and a = 0.2.
Fig. 18. The variation of the velocity with distance from the disk for various values of magnetic field parameter N when s = 1, l1 = 0.5, W = 0, X = 1 and a = 0.2.
are shown in Figs. 17 and 18. It is noticed that with the increase in N, the change in the graph of f is very small and stables at the product X1l1 but the variation in g in the present situation is much. The value of g increases but the layer thickness decreases Tables 1–3.
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Table 1 s = 1, X = 1, a = 0.1, W = 0.5, N = 0 g 0.1 0.5 1.0 1.5 2.0 2.5 3.0
Real velocity f
Imaginary velocity g
Analytic solution
Numerical solution
Difference
Analytic solution
Numerical solution
Difference
0.87847 0.50934 0.24285 0.10690 0.04158 0.01290 0.00218
0.87546 0.50104 0.23645 0.10581 0.04511 0.01838 0.00717
0.00301 0.00831 0.00640 0.00109 0.00353 0.00548 0.00499
0.03793 0.10263 0.08619 0.04517 0.01385 0.00158 0.00573
0.03352 0.09251 0.08510 0.05673 0.03258 0.01705 0.00834
0.00441 0.01012 0.00109 0.01156 0.01873 0.01863 0.01407
Table 2 s = 1, X = 1, a = 0.1, W = 0, N = 0.5 g 0.1 0.5 1.0 1.5 2.0 2.5 3.0
Real velocity f
Imaginary velocity g
Analytic solution
Numerical solution
Difference
Analytic solution
Numerical solution
Difference
0.88384 0.52837 0.26684 0.12912 0.05885 0.02418 0.00822
0.87917 0.51204 0.24701 0.11289 0.04908 0.02036 0.00807
0.00467 0.01633 0.01983 0.01623 0.00977 0.00382 0.00015
0.03360 0.09682 0.09164 0.05893 0.02853 0.00869 0.00052
0.03250 0.09175 0.08667 0.05918 0.03473 0.01853 0.00923
0.00110 0.00507 0.00497 0.00025 0.00620 0.00984 0.00975
Table 3 s = 1, X = 1, a = 0.1, W = 0.5, N = 0.5 g 0.1 0.5 1.0 1.5 2.0 2.5 3.0
Real velocity f
Imaginary velocity g
Analytic solution
Numerical solution
Difference
Analytic solution
Numerical solution
Difference
0.86939 0.48638 0.22426 0.09681 0.03783 0.01250 0.00293
0.86260 0.46741 0.20770 0.08813 0.03581 0.01396 0.00522
0.00679 0.01897 0.01656 0.00868 0.00202 0.00146 0.00229
0.03230 0.08579 0.07167 0.03882 0.01405 0.00148 0.00253
0.03116 0.08224 0.07192 0.04580 0.02523 0.01270 0.00600
0.00114 0.00355 0.00025 0.00698 0.01118 0.01122 0.00853
The variation in suction velocity for N = 0, a = 0.2 and s = X1 = 1 can be seen from Figs. 19– 22. The layer thickness for the case l1 = 0 and l1 = 0.5 decreases. The effects of injection velocity are presented in Figs. 23–26. The value of f and g as well as layer thickness increases in the absence of non-coaxial parameter l1. For l1 = 0.5 the variation within the layer is very large. In this situation it is noted that for the increment in injection velocity the value of f decreases and g increases much when compared with the case of l1 = 0.
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Fig. 19. The variation of the velocity with distance from the disk for various values of suction parameter W when s = 1, l1 = 0, N = 0, X = 1 and a = 0.2.
Fig. 20. The variation of the velocity with distance from the disk for various values of suction parameter W when s = 1, l1 = 0, N = 0, X = 1 and a = 0.2.
Fig. 21. The variation of the velocity with distance from the disk for various values of suction parameter W when s = 1, l1 = 0.5, N = 0, X = 1 and a = 0.2.
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Fig. 22. The variation of the velocity with distance from the disk for various values of suction parameter W when s = 1, l1 = 0.5, N = 0, X = 1 and a = 0.2.
Fig. 23. The variation of the velocity with distance from the disk for various values of injection parameter W when s = 1, l1 = 0, N = 0, X = 1 and a = 0.2.
Fig. 24. The variation of the velocity with distance from the disk for various values of injection parameter W when s = 1, l1 = 0, N = 0, X = 1 and a = 0.2.
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Fig. 25. The variation of the velocity with distance from the disk for various values of injection parameter W when s = 1, l1 = 0.5, N = 0, X = 1 and a = 0.2.
Fig. 26. The variation of the velocity with distance from the disk for various values of injection parameter W when s = 1, l1 = 0.5, N = 0, X = 1 and a = 0.2.
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