An Analysis of the Flow of a Newtonian Fluid between Two Moving Parallel Plates

We consider flow of an incompressible Newtonian fluid produced by two parallel plates, moving towards and away from each other with constant velocity. Inverse solutions of the equations of motion are obtained by assuming certain forms of the stream function. Analytical expressions for the stream function, fluid velocity components, and fluid pressure are derived.


Introduction
Owing to the nonlinear nature of the Navier-Stokes equations, their exact solutions are far and few in number. Importance of the exact solutions lies in the fact that they serve as standards for validating the corresponding solutions obtained by numerical methods and other approximate techniques. The inverse or the indirect method, see for example, Neményi [1], is often used to compute these exact solutions.
Finding exact solution using the inverse method consists of making an assumption on the general form of the stream function , involving certain unknown functions, without considering the shape of boundaries of the solution domain occupied by the fluid. We then substitute this assumed form of in the compatibility equation for the stream function to find the unknown functions involved in . This provides the stream function , and subsequently, the fluid velocity components. Once the fluid velocity components are available, then the second step is to compute the fluid pressure field using the component form of the Navier-Stokes equations. This kind of methods with applications in various fields of continuum mechanics are given in an article by Neményi [1]. Moreover, a number of reviews on the exact solutions for Navier-Stokes equations have been published, for example, Berker [2], Dryden et al. [3], Whitham [4], Schlichting [5], Wang [6], and Wang [7].
In this paper, we apply the technique described above to analyze the flow of a viscous incompressible fluid induced by the motion of two parallel plates. These plates are moving towards each other and in the opposite direction with a constant velocity , when size of the plates is much larger than distance between them. A large class of the processes such as the motion of liquid through a hydraulic pump and that of the underground water may mathematically be considered from this point of view, see Aristov and Gitman [8] and Siddiqui et al. [9,10]. We apply the inverse method to solve this problem for Berker type flow, Riabouchinsky type flow and the potential flow with perturbation. The results obtained are compared with the known viscous solutions by setting the relative velocity of the disks equal to zero.

Basic Equations
If u denotes the fluid velocity, f the total of body forces, T the stress tensor, the constant fluid density, and / = ( / ) + u ⋅ ∇ the material derivative, then the basic The constitutive equation, relating the stress tensor to the rate of strain tensor, for an incompressible Newtonian fluid is, T = − I + ∇u. Thus, in the absence of body forces, the basic set of equations governing the flow of an incompressible fluid becomes where denotes the fluid pressure, the coefficient of viscosity, and ] = / the kinematic viscosity.

Problem Formulation
Consider the motion of an incompressible Newtonian fluid produced by the motion of two parallel plates moving towards each other or in the opposite direction with constant velocity . The size of the disks is much larger than the distance between them. We assume that the horizontal velocity components and V are independent of vertical coordinate , whereas the vertical velocity component depends linearly on the distance between the disks, see Aristov and Gitman [8] and Siddiqui et al. [9,10].
Using (3) into (2), we obtain where * = ( , , )/ − 2 2 2 . The equation for the vertical component is identically satisfied. Equations (4)-(6) are three partial differential equations in three unknowns, namely, , V, and * . Once the velocity components and V are determined, the pressure field can be found through (5) and (6). Eliminating * between (5) and (6), we obtain where denotes magnitude of the vorticity vector. We now introduce the stream function defining the horizontal components of the fluid velocity through the relations [8,9]: Then, the continuity equation (4) is identically satisfied and momentum equations (5) and (6) yield the following compatibility equation: where The stream function is determined by solving the compatibility equation (9), and the velocity components and V can be determined through (8). One solution of the compatibility equation (9) is the trivial solution = 0 that corresponds to the potential motion, known as the motion near stagnation point. Other stationary solutions are examined in the following section.

Solutions of the Problem
Berker Type Flows. In this type of flow we look for the stream function in the following general form where ( ) and ( ) are arbitrary functions of and , respectively. This type of flow was also studied in the context of second grade fluid by Siddiqui et al. [9,10] using certain forms of ( ) and ( ). Here, we consider more general cases. Using (11) in the compatibility equation (9), we obtain For = 0 (12) reduces to Berker's case. Now following Berker [2], we plan to study the following particular cases. If ( ) = 0, ( ) ̸ = 0. Then, in this case the vorticity = −∇ 2 is a function of only and (12) reduces to (4) where ( ) = 1 , 1 being an arbitrary constant. Now, integrate (13) twice to get where , are arbitrary constants. For = = 0, the solution of (14) is given by ] and 1 denotes an arbitrary constant. Therefore, the stream function is given by where 1 is an arbitrary constant. Although the solution (16) for the stream function is a power series, we were able to identify the series solution for . Thus, the velocity components are Consequently, the analytical expression for the fluid pressure distribution in this case is the following: where 1 and 1 are arbitrary constants. Similarly, if ( ) ̸ = 0, ( ) = 0, then the stream function in the form of a power series is given by where 2 = exp[ 2 − 2 2 /(2] )], 2 = ( ). Here, 2 , 2 , and 2 are arbitrary constants. Again, we were able to identify the power series for V, and components of the fluid velocity are Correspondingly, the analytical expression for the fluid pressure distribution in this case is where 2 and 2 are arbitrary constants.
If ̸ = 0. Then, we have to solve (12) to find the functions ( ) and ( ), which is both nonlinear and nonseparable. However, by differentiating first w.r.to and then w.r.to this equation becomes (4) which is in the separable form. Therefore, separating variables, (22) yields the following two linear ordinary differential equations with constant coefficients: where is an arbitrary constant.

ISRN Mathematical Analysis
(c) If > 0, let = 2 , being constant then solution to (23) is given by where 1 , 2 , 3 , 4 , 1 , 2 , 3 , and 4 are arbitrary constants. Therefore, the stream function is given by Consequently, the velocity components of the fluid are the following: The pressure distribution in all the three cases (a), (b), and (c) is given by We remark that all the solutions given in (26), (29), and (32) are free from viscosity coefficient and are known as universal solutions. However, pressure field in each case involves the viscosity coefficient . Further note that the solutions are essentially the same as that of the Berker [2] in 2-dimensional case.

Riabouchinsky Type Flow.
Here, we assume that the stream function is linear in and has the following general form: where ( ) and ( ) are arbitrary functions of . Using (34), Aristov and Gitman [8] obtained an analytical solution of the compatibility equation by taking ( ) as a linear function of . Siddiqui et al. [9,10] studied the same in the context of second grade fluid. Here, we consider some more particular cases of ( ).
On substituting from (34) the compatibility equation (9) yields the following two differential equations: and (35) admits the following particular solutions: where , are arbitrary constants. Therefore, the stream function and the corresponding fluid velocity components are which is a second order differential equation with variable coefficients. To solve, we assume where is a constant to be determined. We obtain the following solution of (45) Therefore, the function ( ) finally becomes Hence, the stream function and the fluid velocity components are given by ] + 2 1 , It is not easy to obtain the general solution of (50). Therefore, we consider a special case when 2 = 0 and obtain In order to solve this equation we reduce its order by letting = 2 / 2 . Then (51) becomes We notice that for 2 = 0, the particular solution (39) falls back to the solution given in (37) with = and = ] − 4 / . This is why (52) is similar to (41) having solution similar to (42), that is, where 1 and 2 are arbitrary constants. Thus, Remark. If we set = 0 in (37), we can recover the case ( ) = studied by Aristov and Gitman [8]. Similarly, letting = 0 in the cases (38) and (39) additionally studied in this paper, Riabouchinsky type 2-dimensional solutions can be obtained, see Berker [2], Polyanin and Zaitsev [11]. Moreover, the case (39) has also been considered by Siddiqui et al. [9,10] in the context of second grade fluid. However, the special cases considered for evaluation of ( ) they choose