Periodic flow due to oscillations of eccentric rotating porous disks

Abstract

This article is concerned with the periodic flow induced by non-torsional oscillations of two porous disks rotating about distinct axes. While the porous disks are initially rotating with the same angular velocity about non-coincident axes, they start to execute non-torsional oscillations in their own planes and in the opposite directions. An analytical solution corresponding to the velocity field in the periodic state is obtained. The variations in the components of the horizontal force per unit area exerted by the fluid on the top and bottom disks with the time are investigated for the suction/injection velocity parameter, the Reynolds number, the ratio of the frequency of oscillation to the angular velocity of the disks, and the dimensionless velocity amplitudes of oscillation. It is shown that the horizontal force on the upper disk is not equal to that on the lower disk.

Keywords

Newtonian fluid orthogonal rheometer porous disk non-torsional oscillation periodic flow

Introduction

Maxwell and Chartoff¹ claimed that it is possible to determine the material moduli of non-Newtonian fluids if an instrument called the orthogonal rheometer consisting of two parallel disks rotating with the same angular velocity about two different axes normal to the disks is used. Abbott and Walters² obtained an exact solution corresponding to the flow in this instrument for a Newtonian fluid. Rajagopal³ showed that this motion is one with constant principal relative stretch history.

Unsteady flows of Newtonian fluids between the eccentric rotating disks have also received considerable attention. Erdoğan⁴ studied the unsteady flow induced by the rotation about non-coincident axes while the disks are initially rotating with the same angular velocity about a common axis. Ersoy⁵ investigated the unsteady flow due to a sudden pull with constant velocities of the disks rotating eccentrically at the same speed. Ersoy⁶ examined the unsteady flow produced by the axes that are made coincident while the disks are initially rotating eccentrically. Mohyuddin⁷ studied the unsteady flow due to a sudden pull with constant velocities of the porous disks in the presence of a magnetic field. Das et al.⁸ examined the unsteady hydromagnetic flow between two disks rotating about non-coincident axes which are suddenly made coincident.

Time-dependent flows of Newtonian fluids caused by non-torsional oscillations of eccentric rotating disks have also attracted the attention of researchers. Erdoğan^9,10 studied the non-symmetrical unsteady flows due to the non-torsional oscillations of the disks initially rotating about a common axis. In the first article,⁹ he thought that the disks start to rotate eccentrically and the lower disk executes oscillations. In the second article,¹⁰ he took into account that the disks start to rotate eccentrically and both the disks execute oscillations in the same direction. Ersoy¹¹ studied the symmetrical unsteady flow caused by their oscillations in their own planes and in the opposite directions while the disks are initially rotating eccentrically.

Unsteady flows due to the oscillation of a porous disk have also attracted the interest of many investigators. The reader may consult Kasiviswanathan and Rao,¹² Hayat et al.,^13–16 Guria et al.,¹⁷ and Hayat et al.^18,19 for more details.

In this article, the periodic flow due to the non-torsional oscillations in their own planes and in the opposite directions of the porous disks initially rotating at the same speed about distinct axes is studied. In order to obtain a general solution, the oscillating disk velocity has two components. An analytical solution of the velocity field corresponding to the flow is found. Since the study of the horizontal force applied by the fluid on the disks in an orthogonal rheometer is important, a special attention is given to analyze the shear stresses. The main idea in this article is to study the effect of porous disks. It is shown that the horizontal force on the upper disk is not the same as that on the lower disk in the case of porous disks whereas they are equal to each other for non-porous disks. Moreover, it should be emphasized that the main difference between x- and y-components of the horizontal force is due to the eccentricity defined along the y-axis. The influences of the parameters controlling the periodic flow on the shear stresses corresponding to the components of the horizontal force are examined with the help of the figures.

Basic equations and solution

Let us consider an incompressible Newtonian fluid between two porous disks located at $z = \pm h$ . The top and bottom disks are initially rotating about the $z'$ and $z ″$ axes with the same angular velocity Ω, respectively. The two distinct axes are separated by a distance $2 ℓ$ . A uniform suction and injection are applied perpendicular to the top and bottom disks, respectively. After this initial state, the top and bottom disks start to make non-torsional oscillations in their own planes with the velocities $U$ and $- U$ , respectively, where $U = (U_{x} \sin nt, U_{y} \sin nt, 0)$ . The aim of this article is to examine the periodic flow occurred in the fluid after the initiation of the motion. It should also be noted that the distance between the axes of rotation is fixed during the motion. The geometry of problem is shown in Figure 1.

Figure 1.

Flow geometry.

Therefore, the appropriate initial and boundary conditions for the velocity field are

\begin{matrix} u = - Ω y + \hat{f} (z), v = Ω x + \hat{g} (z), \\ w = w_{0} at t = 0 for - h \leq z \leq h \end{matrix}

(1)

u = - Ω (y - ℓ) + U_{x} \sin nt, v = Ω x + U_{y} \sin nt, w = w_{0} at z = h for t \geq 0

(2)

u = - Ω y, v = Ω x, w = w_{0} at z = 0 for t \geq 0

(3)

\begin{matrix} u = - Ω (y + ℓ) - U_{x} \sin nt, v = Ω x - U_{y} \sin nt, \\ w = w_{0} at z = - h for t \geq 0 \end{matrix}

(4)

The functions $\hat{f} (z)$ and $\hat{g} (z)$ represent the translational velocity components corresponding to the steady flow between eccentric rotating porous disks for a Newtonian fluid and are given by

\hat{f} (z) + i \hat{g} (z) = Ω ℓ \frac{{\hat{A}}_{1} \exp (K_{1} z) + {\hat{A}}_{2} \exp (K_{2} z) + {\hat{A}}_{3}}{{\hat{A}}_{4}}

(5)

where

\begin{matrix} K_{1} & = \frac{w_{0} + \sqrt{w_{0}^{2} + i 4 Ω ν}}{2 ν}, K_{2} = \frac{w_{0} - \sqrt{w_{0}^{2} + i 4 Ω ν}}{2 ν} \\ {\hat{A}}_{1} & = 1 - \cosh K_{2} h, {\hat{A}}_{2} = \cosh K_{1} h - 1 \\ {\hat{A}}_{3} & = \cosh K_{2} h - \cosh K_{1} h \\ {\hat{A}}_{4} & = \sinh K_{1} h - \sinh K_{2} h - \sinh (K_{1} - K_{2}) h \end{matrix}

(6)

It is the result obtained in Ersoy²⁰ (for $λ = 0$ ). In the light of these conditions, we seek a solution of the form

u = - Ω y + f (z, t), v = Ω x + g (z, t), w = w_{0}

(7)

Substituting equation (7) into the Navier–Stokes equations, one has

\begin{matrix} ν \frac{\partial^{2} f}{\partial z^{2}} - w_{0} \frac{\partial f}{\partial z} - \frac{\partial f}{\partial t} + Ω g = C_{1} (t) \\ ν \frac{\partial^{2} g}{\partial z^{2}} - w_{0} \frac{\partial g}{\partial z} - \frac{\partial g}{\partial t} - Ω f = C_{2} (t) \end{matrix}

(8)

Introducing $F (z, t) = f (z, t) + ig (z, t)$ , $K (t) = C_{1} (t) + {iC}_{2} (t)$ , and using equation (8), we obtain

ν \frac{\partial^{2} F}{\partial z^{2}} - w_{0} \frac{\partial F}{\partial z} - \frac{\partial F}{\partial t} - i Ω F = K (t)

(9)

For the periodic solution in the fluid, it is natural to assume a solution of the form

\begin{matrix} F (z, t) & = F_{0} (z) + F_{1} (z) \cos nt + F_{2} (z) \sin nt \\ K (t) & = K_{0} + K_{1} \cos nt + K_{2} \sin nt \end{matrix}

(10)

where $K_{0}$ , $K_{1}$ , and $K_{2}$ are constants, and $F_{0} (z)$ and $K_{0}$ correspond to the case of $n = 0$ . Substituting equation (10) into equation (9), one obtains

ν F ″_{0} - w_{0} F'_{0} - i Ω F_{0} = K_{0}

(11)

ν F ″_{1} - {nF}_{2} - w_{0} F'_{1} - i Ω F_{1} = K_{1}

(12)

ν F ″_{2} + {nF}_{1} - w_{0} F'_{2} - i Ω F_{2} = K_{2}

(13)

where a prime denotes differentiation with respect to $z$ . It is convenient to introduce the following dimensionless variables

\begin{matrix} Γ (ζ, τ) & = \frac{F (z, t)}{Ω ℓ}, Γ_{0} (ζ) = \frac{\hat{f} (z) + i \hat{g} (z)}{Ω ℓ}, ζ = \frac{z}{h}, τ = Ω t \\ R & = \frac{Ω h^{2}}{ν}, α = \frac{w_{0}}{\sqrt{Ω ν}}, k = \frac{n}{Ω}, V_{x} = \frac{U_{x}}{Ω ℓ}, V_{y} = \frac{U_{y}}{Ω ℓ} \end{matrix}

(14)

with the conditions

\begin{matrix} Γ (ζ, 0) = Γ_{0} (ζ), Γ (1, τ) = 1 + V_{x} \sin k τ + {iV}_{y} \sin k τ \\ Γ (- 1, τ) = - 1 - V_{x} \sin k τ - {iV}_{y} \sin k τ, Γ (0, τ) = 0 \end{matrix}

(15)

Thus, the solution is

\begin{matrix} Γ (ζ, τ) \\ = Γ_{0} (ζ) + \frac{V_{y} - {iV}_{x}}{2} [\frac{H_{1} \exp (m_{1} ζ) + H_{2} \exp (m_{2} ζ) + H_{3}}{H_{4}} \\ - \frac{G_{1} \exp (r_{1} ζ) + G_{2} \exp (r_{2} ζ) + G_{3}}{G_{4}}] \cos k τ \\ + \frac{V_{x} + {iV}_{y}}{2} [\frac{H_{1} \exp (m_{1} ζ) + H_{2} \exp (m_{2} ζ) + H_{3}}{H_{4}} \\ + \frac{G_{1} \exp (r_{1} ζ) + G_{2} \exp (r_{2} ζ) + G_{3}}{G_{4}}] \sin k τ \end{matrix}

(16)

where

\begin{matrix} Γ_{0} (ζ) & = \frac{A_{1} \exp (a ζ) + A_{2} \exp (b ζ) + A_{3}}{A_{4}} \\ a & = 0.5 \sqrt{R} (α + \sqrt{α^{2} + 4 i}) \\ b & = 0.5 \sqrt{R} (α - \sqrt{α^{2} + 4 i}) \\ A_{1} & = 1 - \cosh b \\ A_{2} & = \cosh a - 1 \\ A_{3} & = \cosh b - \cosh a \\ A_{4} & = \sinh a - \sinh b - \sinh (a - b) \\ m_{1} & = 0.5 \sqrt{R} (α + \sqrt{α^{2} + 4 i (1 + k)}) \\ m_{2} & = 0.5 \sqrt{R} (α - \sqrt{α^{2} + 4 i (1 + k)}) \\ H_{1} & = 1 - \cosh m_{2} \\ H_{2} & = \cosh m_{1} - 1 \\ H_{3} & = \cosh m_{2} - \cosh m_{1} \\ H_{4} & = \sinh m_{1} - \sinh m_{2} - \sinh (m_{1} - m_{2}) \\ r_{1} & = 0.5 \sqrt{R} (α + \sqrt{α^{2} + 4 i (1 - k)}) \\ r_{2} & = 0.5 \sqrt{R} (α - \sqrt{α^{2} + 4 i (1 - k)}) \\ G_{1} & = 1 - \cosh r_{2} \\ G_{2} & = \cosh r_{1} - 1 \\ G_{3} & = \cosh r_{2} - \cosh r_{1} \\ G_{4} & = \sinh r_{1} - \sinh r_{2} - \sinh (r_{1} - r_{2}) \end{matrix}

(17)

Here, $Γ_{0} (ζ)$ is equal to the solution obtained in Ersoy²⁰ (for $D = 0$ ).

The solution for $k = 1$ is

\begin{matrix} Γ (ζ, τ) = Γ_{0} (ζ) + \frac{V_{y} - {iV}_{x}}{2} [\frac{H_{1} \exp (m_{1} ζ) + H_{2} \exp (m_{2} ζ) + H_{3}}{H_{4}} - ζ] \cos τ + \frac{V_{x} + {iV}_{y}}{2} [\frac{H_{1} \exp (m_{1} ζ) + H_{2} \exp (m_{2} ζ) + H_{3}}{H_{4}} + ζ] \sin τ \end{matrix}

(18)

It is worth noting that the results obtained for large times by Ersoy¹¹ can be obtained as the special case of the present analysis by taking the suction/injection velocity parameter to be 0. The shear stress components $T_{xz}$ and $T_{yz}$ in the fluid are found as follows:

\begin{matrix} {\bar{T}}_{xz} (ζ, τ) + i {\bar{T}}_{yz} (ζ, τ) \\ = \frac{A_{1} a \exp (a ζ) + A_{2} b \exp (b ζ)}{A_{4}} \\ + \frac{V_{y} - {iV}_{x}}{2} [\frac{H_{1} m_{1} \exp (m_{1} ζ) + H_{2} m_{2} \exp (m_{2} ζ)}{H_{4}} - \frac{G_{1} r_{1} \exp (r_{1} ζ) + G_{2} r_{2} \exp (r_{2} ζ)}{G_{4}}] \cos k τ \\ + \frac{V_{x} + {iV}_{y}}{2} [\frac{H_{1} m_{1} \exp (m_{1} ζ) + H_{2} m_{2} \exp (m_{2} ζ)}{H_{4}} + \frac{G_{1} r_{1} \exp (r_{1} ζ) + G_{2} r_{2} \exp (r_{2} ζ)}{G_{4}}] \sin k τ \end{matrix}

(19)

where

{\bar{T}}_{xz} (ζ, τ) = \frac{T_{xz} (z, t)}{μ Ω ℓ / h}, {\bar{T}}_{yz} (ζ, τ) = \frac{T_{yz} (z, t)}{μ Ω ℓ / h}

(20)

The solution corresponding to $k = 1$ is

\begin{matrix} {\bar{T}}_{xz} (ζ, τ) + i {\bar{T}}_{yz} (ζ, τ) \\ = \frac{A_{1} a \exp (a ζ) + A_{2} b \exp (b ζ)}{A_{4}} \\ + \frac{V_{y} - {iV}_{x}}{2} [\frac{H_{1} m_{1} \exp (m_{1} ζ) + H_{2} m_{2} \exp (m_{2} ζ)}{H_{4}} - 1] \cos τ \\ + \frac{V_{x} + {iV}_{y}}{2} [\frac{H_{1} m_{1} \exp (m_{1} ζ) + H_{2} m_{2} \exp (m_{2} ζ)}{H_{4}} + 1] \sin τ \end{matrix}

(21)

The variations in ${({\bar{T}}_{xz})}_{ζ = \pm 1}$ and ${({\bar{T}}_{yz})}_{ζ = \pm 1}$ representing the x- and y-components of the dimensionless horizontal force per unit area exerted by the top and bottom disks on the fluid with the time are shown in Figures 2 –6. The solid and dotted lines show the variations on the top and bottom disks, respectively. Figure 2 shows the effect of the ratio of the frequency of oscillation to the angular velocity of the disks. The periodic time intervals in which the curves are drawn are $1.6 π$ , $2 π$ , and $4 π$ for $k = 1.25$ , $k = 1$ , and $k = 0.5$ , respectively. When the frequency of oscillation increases, it is observed that the x-component of the force has larger values when compared with the y-component of the force. Figure 3 displays the influence of suction/injection velocity parameter based on the axial velocity of fluid. In general, its increase causes an increase in the horizontal force on the upper disk. On the other hand, an opposite behavior is observed for the lower disk. In addition, it is seen that the y-component of the force changes its direction. The behavior of the Reynolds number related to the magnitude of the angular velocity of disks is plotted in Figure 4. With the increase in the Reynolds number, the horizontal force increases because of the effect of boundary layer near the top and bottom disks. Moreover, a change is observed in the direction of the y-component. Figures 5 and 6 depict the variations in the dimensionless velocity amplitude in the x- and y-directions, respectively. The increase in amplitudes results in the larger values of the horizontal force. As expected, the larger amplitude values lead to the change in direction of the force components. It is important to emphasize that the y-component has larger changes when compared with the x-component.

Figure 2.

Variations in ${({\bar{T}}_{xz})}_{ζ = \pm 1}$ and ${({\bar{T}}_{yz})}_{ζ = \pm 1}$ with $τ$ for $k = 0.5, 1, 1.25$ ( $R = 10$ , $α = 0.1$ , $V_{x} = 1$ , $V_{y} = 1$ ).

Figure 3.

Variations in ${({\bar{T}}_{xz})}_{ζ = \pm 1}$ and ${({\bar{T}}_{yz})}_{ζ = \pm 1}$ with $τ$ for $α = 0.1, 0.2, 0.3$ ( $R = 10$ , $k = 0.5$ , $V_{x} = 1$ , $V_{y} = 1$ ).

Figure 4.

Variations in ${({\bar{T}}_{xz})}_{ζ = \pm 1}$ and ${({\bar{T}}_{yz})}_{ζ = \pm 1}$ with $τ$ for $R = 10, 20, 40$ ( $α = 0.1$ , $k = 0.5$ , $V_{x} = 1$ , $V_{y} = 1$ ).

Figure 5.

Variations in ${({\bar{T}}_{xz})}_{ζ = \pm 1}$ and ${({\bar{T}}_{yz})}_{ζ = \pm 1}$ with $τ$ for $V_{x} = 1, 2, 3$ ( $R = 10$ , $α = 0.1$ , $k = 0.5$ , $V_{y} = 1$ ).

Figure 6.

Variations in ${({\bar{T}}_{xz})}_{ζ = \pm 1}$ and ${({\bar{T}}_{yz})}_{ζ = \pm 1}$ with $τ$ for $V_{y} = 1, 2, 3$ ( $R = 10$ , $α = 0.1$ , $k = 0.5$ , $V_{x} = 1$ ).

Results and discussion

This article deals with the periodic flow due to non-torsional oscillations of eccentric rotating porous disks. The two porous disks are initially rotating with the same angular velocity about non-coincident axes. The direction of constant axial velocity is upward. Hence, the initial condition is taken as the solution corresponding to $λ = 0$ in Ersoy’s²⁰ article. The porous disks suddenly start to execute non-torsional oscillations in their own planes and in the opposite directions. The sine oscillations of the disks are preferred since the cosine oscillations do not satisfy the initial condition. After the start of oscillation motion, the flow contains transients for small times and then attains the periodic state. The periodic flow starts at a specific time that depends on the values of the parameters. For example, the periodic flow occurs at the dimensionless time $τ = 7 - 8$ for $R = 10$ , $V_{x} = 1$ , $V_{y} = 1$ , $k = 0.5$ , and $α = 0$ (see Ersoy¹¹). In this article, the examination is made after $τ = 8 π$ at which the flow already attains its periodic state as a reasonable estimation for the selected values of the parameters.

When the disks are non-porous, the Poiseuille-type pressure gradient is 0. In other words, we have $C_{1} (t) = C_{2} (t) = 0$ as a consequence of the symmetrical condition.¹¹ However, $C_{1} (t)$ and $C_{2} (t)$ are not equal to 0 for the porous disks.

Although the shear stresses $T_{xz}$ and $T_{yz}$ on the top and bottom disks for the non-porous disks have the same values, that is, ${(T_{xz})}_{z = h} = {(T_{xz})}_{z = - h}$ and ${(T_{yz})}_{z = h} = {(T_{yz})}_{z = - h}$ , they have different values when the disks are porous.

In this article, an exact solution corresponding to the velocity field is obtained. However, the attention is focused on the variations in the dimensionless shear stresses on the top and bottom disks with the time. The results are obtained by means of Figures 2 –6.

Conclusion

Since the dimensionless shear stresses ${\bar{T}}_{xz}$ and ${\bar{T}}_{yz}$ do not depend on $x$ and $y$ , ${({\bar{T}}_{xz})}_{ζ = \pm 1}$ and ${({\bar{T}}_{yz})}_{ζ = \pm 1}$ are equal to the x- and y-components of the dimensionless force per unit area exerted by the top and bottom disks on the fluid, respectively. If ${\bar{X}}_{t} = {({\bar{T}}_{xz})}_{ζ = 1} > 0$ , the upper disk applies a force to the fluid in the positive x-direction, but the direction becomes opposite if ${({\bar{T}}_{xz})}_{ζ = 1} < 0$ . If ${\bar{X}}_{b} = {({\bar{T}}_{xz})}_{ζ = - 1} > 0$ , the force the lower disk applies to the fluid is in the negative x-direction, but the direction is in the positive x-direction if ${({\bar{T}}_{xz})}_{ζ = - 1} < 0$ . In a similar manner, the dimensionless force is in the positive y-direction if ${\bar{Y}}_{t} = {({\bar{T}}_{yz})}_{ζ = 1} > 0$ , and it is in the negative y-direction if ${\bar{Y}}_{b} = {({\bar{T}}_{yz})}_{ζ = - 1} > 0$ .

The direction of the force exerted by the fluid is opposite to that exerted by the disk. Hence, it is sufficient to carry out the analysis of the magnitudes of the dimensionless horizontal force exerted by the fluid. The main conclusions are pointed out as follows:

The direction of the y-component of force changes more often in comparison with that of x-component in the periodic time interval since the eccentricity is defined along the y-axis. The changes in $| {\bar{Y}}_{t} |$ and $| {\bar{Y}}_{b} |$ are larger than those in $| {\bar{X}}_{t} |$ and $| {\bar{X}}_{b} |$ .

In general, it is observed that $| {\bar{X}}_{t} | > | {\bar{X}}_{b} |$ and $| {\bar{Y}}_{t} | > | {\bar{Y}}_{b} |$ at a specific time. In other words, the force on the top disk is greater than that on the bottom disk.

When $k$ increases, the changes in $| {\bar{X}}_{t} |$ and $| {\bar{X}}_{b} |$ become larger. Conversely, the changes in $| {\bar{Y}}_{t} |$ and $| {\bar{Y}}_{b} |$ become smaller.

An increase in $α$ leads to an increase in $| {\bar{X}}_{t} |$ but a decrease in $| {\bar{X}}_{b} |$ . In general, it is observed that $| {\bar{Y}}_{t} |$ increases but $| {\bar{Y}}_{b} |$ decreases. For $α = 0$ , the force component on the top disk is equal to that on the bottom disk, that is, ${\bar{X}}_{t} = {\bar{X}}_{b}$ and ${\bar{Y}}_{t} = {\bar{Y}}_{b}$ .

When $R$ increases, the magnitudes of all the forces increase. In other words, $| {\bar{X}}_{t} |$ , $| {\bar{X}}_{b} |$ , $| {\bar{Y}}_{t} |$ , and $| {\bar{Y}}_{b} |$ have larger values.

The increase in $V_{x}$ or $V_{y}$ causes larger changes in the values of $| {\bar{X}}_{t} |$ , $| {\bar{X}}_{b} |$ , $| {\bar{Y}}_{t} |$ , and $| {\bar{Y}}_{b} |$ .

Footnotes

Appendix 1 Acknowledgements

The author would like to express his sincere thanks to the referees for their valuable comments and suggestions.

Academic Editor: Thirumalisai S Dhanasekaran

Declaration of conflicting interests

The author declared no potential conflicts of interest with respect to the research, authorship, and/or publication of this article.

Funding

The author received no financial support for the research, authorship, and/or publication of this article.

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