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A Note on Squeezing Flow Between Two Infinite Parallel Plates with Slip Boundary Conditions

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A Note on Squeezing Flow Between Two Infinite Parallel Plates with Slip Boundary Conditions International Journal of the Physical Sciences Vol. 6(14), pp. 3296-3301, 18 July, 2011 Available online at http://www.academicjournals.org/IJPS DOI: 10.5897/IJPS11.499 ISSN 1992 - 1950 ©2011 Academic Journals Full Length Research Paper A note on squeezing flow between two infinite parallel plates with slip boundary conditions Murad Ullah 1*, S. Islam 2, Gul Zaman 3 and Shazia Naeem Khalid 1 1Islamia College Peshawar (Charted University), Peshawar, NWFP, Pakistan. 2Mathematics, COMSATS Institute of Information technology, Park Road, Chakshazad, Islamabad, Pakistan. 3Centre for Advanced Mathematics and Physics, NUST, H-12, Islamabad, Pakistan. Accepted 04 May, 2011 The aim of this letter is to investigate an axisymmetric squeezing flow of an incompressible fluid generated by two large parallel plates, including fluid inertial effects. The governing equations have been transformed into a nonlinear ordinary differential equation using integribility condition. Solution to the problem is obtained by using an optimal homotopy asymptotic method (OHAM). The results reveal that the new method is very effective and simple. Key words: Axisymmetric squeezing flow, slip boundary conditions, optimal homotopy asymptotic method (OHAM). INTRODUCTION The study of squeezing flows has been published in a with slip boundary conditions, taking into account the wide variety of journals spanning a century or more due inertia effects. The optimal homotopy asymptotic method to its practical applications in chemical engineering and is applied to solve the title problem (Herisanu et al., 2008; food industry. The basic research in this field was carried Idrees et al., 2010a, b; Islam et al., 2010; Marinca and out by Stefan (1874). Of more recent origin is the interest Herisanu, 2010a,b; Marinca et al., 2008, 2009; in squeezing flows spurred by problems encountered in Hesameddini and Latifzadeh, 2009; Shah et al., 2010). lubrications and dusty fluids (Chatraei et al., 1981; Thien Navier assumed that the velocity u , at a solid surface is and Tanner, 1984; Debnath and Ghosh, 1988; Hamdan x and Baron, 1992; Kompani and Venerus, 2000). Also, the pro-portional to the shear rate at the surface analytical and experimental study of these flows together ux = β ∂ux ∂y , where β is the slip length Navier (1823). with the inertial term between the rotating cylinders and If β = 0 , we will get the general no-slip boundary parallel plates has resulted in increasing interest due to its importance in thin film of lubricants (Debbaut, 2001; condition. If β is finite, fluid slip occurs at the wall, but its Wang and Watson, 1979; Denn and Marrucci, 1999; effect depends upon the length scale of the flow. Estelle Hoffner et al., 2001; Lee et al., 1984). The mathematical and Lanos (2007), Tretheway and Meinhart (2002), Laun studies of these flows are concerned primarily with the et al. (1999) and Zhu and Granick (2001) have studied non linear partial differential equations which arise from this fact in more detail. the Navier-Stokes equations. These equations have no general solutions and only a few of number of exact solutions have been attained (He, 2006). To solve BASIC EQUATION practical problems, different perturbation and analytical techniques have been widely used in fluid mechanics and Here, we determine the basic equations which are used engineering (Ali et al., 2010). in the rest of this paper. In the absence of body forces, Our purpose in this contribution is to study the basic equations governing the flow in vorticity form axisymmetric fluid flow between two large parallel plates are given by: ∂ur u r ∂ u z + + = 0 (1) *Corresponding author. E-mail: [email protected]. ∂r r ∂ z Ullah et al. 3297 Figure 1. A steady squeezing axisymmetric fluid flow between two large parallel plates . E2ψ E 2 ψ ρ 2 ∂ ∂ 2 2 (4) −ρ()u% × ω% +∇ up % + =−∇× µ u % ∂ψr r ∂ ψ µ 2 2 (2) −ρ − = ()E 2 ψ ∂∂rz ∂∂ rzr where u% =( urztr( ,,) ,0, urzt z ( ,, )) is the velocity vector, ρ is the density of the fluid, p is the pressure, with the slip boundary conditions ω% = ∇× u% is the vorticity vector and µ is the dynamic ∂ur ∂ u r zd=, then ur=β , uz z =−= U, 0, then u z = 0, = 0, (5) viscosity of the fluid. ∂z ∂ z We consider viscous incompressible fluid, squeezed 2 between two large planar and parallel plates, separated where the differential operator E is defined by by a distance 2 d. The plates are moving towards each 2 2 2 ∂ 1 ∂ ∂ . other with velocity U, as shown in the Figure 1. The E = − + 2 2 surfaces of both plates are covered by special material ∂ r r ∂ r ∂ z with slip length (slip coefficient) β . For small values of U, Equation (4) admits a solution of the form (Stefan, 1874): the gap distance 2 d between the plates varies slowly with the time t, so that it can be taken as constant and the flow 2 as quasi-steady (Zhu and Granick, 2001; Papanastasiou, ψ ()()rz, = rFz . (6) 2000; Siddiqui et al., 2007; Idrees et al., 2010; Ran et al., 2009). The velocity field u% is given as follows. By virtue of Equation (6), the compatibility Equation (4) and boundary conditions (5) becomes u= urzt,, ,0, urzt ,, (3) 4 3 % ( r( ) z ( )) dFρ dF 2 0, 4+F 3 = (7) dzµ dz It is easily shown that the stream function ψ (r, z ) , 1∂ψ 1 ∂ ψ subjecting to the boundary conditions defined by ur= −, u z = satisfies the rz∂ rz ∂ F(00) = ,F′′ ( 00) = , . (8) continuity equation identically. Substituting u and u U r z Fd() = − ,Fd′()() = γ Fd ′′ into the z- and r-components of the Navier-Stokes 2 equation, and eliminating the pressure lead to the following equation To deal with the problem, we introduce dimensionless 3298 Int. J. Phys. Sci. parameters given by Now we solve the problems (14) to (16) in succession and obtain the series solutions with unknown constants. F z β ρ d U F*= ,* z = , γ = , R = , U/ 2 d d µ Zeroth order solution and dropping ‘*’ for simplicity, the boundary value problem (7) becomes, 1 3 Fz0 () =() z +()6γ − 3 z . (17) 2() 3γ − 1 dF4ρ dF 3 (9) 4+2F 3 = 0, dzµ dz First order solution which satisfies the boundary conditions 3 5 7 19 RzC 1−39 RzC 1 +21 RzC 11 −− RzC 1 F00= ,F′′ 00111 = ,F = ,F ′ = γ F ′′ 1 . (10) 3 5 ( ) ( ) ( ) ( ) ( ) F1 =3 171 RzC γ 1 +−+273 RzC γ 1105 RzC γ 1 . 560 (− 1 + 3 γ ) 7 2 32 52 3RzCγ1+ 294 RzC γ 1 -420 Rz γ C 1 + 126 Rz γ C 1 SOLUTION BY OPTIMAL HOMOTOPY ASYMPTOTIC METHOD (18) Here we apply the basic idea of OHAM (Idrees et al., 2010) to Equations (9) and 10). Defining linear and non linear operators Second order solution respectively as: R z (-1+ z 2 ) (4620 (1-3 γ) 2 (-19 + 3 (57-98 γ) γ 4 z, p 4 2 2 ∂ φ ( ) +++z (-1 3 γ) 2 z (10 -51 γ +63 γ )) C + (4620 ’ (11) 1 L()φ () z, p = 4 2 4 2 ∂z (1-3 γ) (-19+ 3 (57-98 γγ) +++z (-1 3 γ ) 2 z (10-51 γγ+632 )) +R (63 z 8(1-3 γ) 262 + 7 z (1-3 γ ) 2 1 4 F2 = 5 (-21 1+++ 440 γ)z (-1 3 γ)(-10205 + 7 γ (10609 + . ∂ φ ( z, p ) 2587200 (-1+ 3 γ ) , (12) 2 N()φ() zp,= R φ () zp , 2 660 γγ(-37+ 27 ))) -z (-1 + 3 γ)(-5503 ++ 7 γ (9587 ∂z 264γ (-107+ 90 γ))) -12(274 +++ γ (-4384 7 γ (2405 99γγ (-38+ 25 )))))) + 4620(1-3 γ )2 (-19 + 3(57-98 γγ ) g z = 0 . (13) 4 ( 1 3 ) 2 2 (10 51 63 2 )) ) ( ) +−+zγ + z − γ + γ C 2 Equating the coefficients of like powers of p, we get the following (19) problems of different orders. Thus the optimal solution up to second order is given by: Zeroth order problem z(1293600 (1− 3 γ)4 ( −++ 3 zRz 2 6 γ ) +−+ ( 1 2 )(9240 (4 ) 2 4 2 F = 0 , (1− 3 γ)(19 −+ 3 (57 − 98 γγ) +−++z ( 1 3) γ 2 z (10 0 −+51γγ 632 ))C + (4620 (1 − 3 γ)( 2 −+− 19 3(57 98 γ ) ′ ′ ″ 1 F00= ,F 00111 = ,F = ,F = γ F 1 (14) γ+−++z4(1 3) γ 2 z 2 (10 −+ 51 γγ 63 282 )) + Rz (63 (1 − 3) γ 0()()()()() 0 0 0 0 1 6 2 4 F= +7 z (1 − 3)(γ−+ 211 440 γ ) +−+z ( 1 3 γ )( − 10205 + 7 γ . 2587200 (-1+ 3 γ ) 5 (10609+ 660γγ ( −+ 37 27 ))) −−+−z2 ( 1 3 γ )( 5503 + 7 γ First order problem (9587+ 264γ ( −+ 107 90 γ ))) − 12(274 +−+ γ ( 4384 7 γ (2405+−+ 99γ ( 38 25 γ))))))C 2 + 4620(1 −−+ 3 γ ) 2 ( 19 ()4 () 4 ″ 1 F z,C=1 + CF z + RCFzF z, (15) 1()()()()() 1 11 100 3(57 98)z4 (1 3) 2 z 2 (10 51 63 2 )) C )) −γγ +−++ γ −+ γγ 2 (20) ′ ′ ″ F00= ,F 00101 = ,F = ,F = γ F, 1 1()()()()() 1 1 11 Here we used the method of least squares for finding the values of C1 and C 2 . Thus for γ =1 and R=1 , we have Second order problem C1 =-0.9453415636834306, C2 =0.00397319798144 2457 In view of the values of C and C , the simplified form of our 1+CF(4) zC , + CF (4) z + RCFzF′′′ zC , 1 2 (4) ()()()()()11 120 10 1 1 F2() zCC,, 1 2 = , solution is; ′′′ ′′′ +RCFzF 11 ()()()() 1 zC, 1 + RCFz 20 F 1 zC, 1 (16) 20697600 (3++−+z2 ) ( 1 z 2 )( − 34866.4( −+ 142 44 z 2 z (21) Fz() = ++2 z4 ) 0.893671( −− 44592 60380 zz2 + 35716 4 .
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