Flow Analysis of Multilayer Gravity-Driven Sisko Fluid over a Flat Inclined Plane

Abstract

Gravity-driven thin film flow of two immiscible non-Newtonian Sisko fluids over a flat inclined plane is investigated. The equation, describing the flow, is nonlinear in nature and does not have a closed form of solution in general. In this study, the nonlinear governing differential equations are solved analytically by employing homotopy perturbation method (HPM). Besides, an exact solution for a special case of the non-Newtonian index is also obtained. The results of the solution based on HPM are compared with the exact solution for the special case, and it shows close agreement between the two results. This demonstrates the efficacy of HPM for solving nonlinear problems. The expression for the velocity is obtained, and the effects of various parameters like Sisko fluid parameters, density ratio, and viscosity ratio are also analyzed and discussed. The results indicate that with increase in the density ratio, it causes a significant increase in the velocity of fluids in both the layers and the velocities in both the layers are decreasing significantly with an increase in the Sisko fluid parameters. This study has got wider application in various industrial activities such as protective and decorative coating, painting.

Appendix A

Equation (12) is solved by substituting

$$ \frac{{{\text{d}}u_{1} }}{{{\text{d}}y}} = z_{1} $$

(A.1)

which gives:

$$ \frac{{{\text{d}}z_{1} }}{{{\text{d}}y}} + 2b_{1} z_{1} \frac{{{\text{d}}z_{1} }}{{{\text{d}}y}} + 1 = 0 $$

(A.2)

$$ \Rightarrow \frac{{{\text{d}}z_{1} }}{{{\text{d}}y}}(1 + 2b_{1} z_{1} ) = - {\text{d}}y $$

(A.3)

Integrating Eq. (A.3) and solving, we get: $z_{1} = \frac{{ - 1 + \sqrt {1 - 4b_{1} (y - c_{1} )} }}{{2b_{1} }}$ (taking the +ve value for $\frac{{{\text{d}}u_{1} }}{{{\text{d}}y}}$, as $u_{1}$ increases with y)

$$ = > \frac{{{\text{d}}u_{1} }}{{{\text{d}}y}} = \frac{{ - 1 + \sqrt {1 - 4b_{1} (y - c_{1} )} }}{{2b_{1} }} $$

(A.4)

$$ = > u_{1} = \frac{0.5}{{b_{1} }}\left[ { - y - \frac{1}{{6b_{1} }}\left\{ {1 - 4b_{1} (y - c_{1} )} \right\}^{3/2} } \right] + c_{2} $$

(A.5)

Similar procedure is used to solve $u_{2}$ from Eq. (13) which gives the solution as follows:

$$ u_{2} = \frac{0.5}{{b_{2} }}\left[ { - y - \frac{1}{{6b_{2} \left( {\frac{{k_{1} }}{{k_{2} }}} \right)}}\left\{ {1 - 4b_{2} \left( {\frac{{k_{1} }}{{k_{2} }}y - c_{1} } \right)} \right\}^{3/2} } \right] + c_{4} $$

(A.6)

Now, using the boundary conditions given by Eqs. (17.1, 17.2), we get the following equations:

$$ \frac{0.5}{{b_{1} }}\left[ { - \frac{1}{{6b_{1} }}\left\{ {1 + 4b_{1} c_{1} } \right\}^{3/2} } \right] + c_{2} = 0 $$

(A.7)

$$ \frac{0.5}{{b_{1} }}\left[ { - r - \frac{1}{{6b_{1} }}\left\{ {1 - 4b_{1} (r - c_{1} )} \right\}^{3/2} } \right] + c_{2} = \frac{0.5}{{b_{2} }}\left[ { - r - \frac{1}{{6b_{2} \left( {\frac{{k_{1} }}{{k_{2} }}} \right)}}\left\{ {1 - 4b_{2} \left( {\frac{{k_{1} }}{{k_{2} }}r - c_{3} } \right)} \right\}^{3/2} } \right] + c_{4} $$

(A.8)

$$ \begin{aligned} & \frac{0.5}{{b_{1} }}\left[ { - 1 + \sqrt {1 - 4b_{1} (r - c_{1} )} } \right] \\ &\quad + \frac{(0.5)}{{b_{1}^{{}} }}^{2} \left[ { - 1 + \sqrt {1 - 4b_{1}^{{}} (r - c_{1} )} } \right]^{2} \\ & \quad = \frac{0.5}{{b_{2}^{{}} }}k_{2} \left[ { - 1 + \sqrt {1 - 4b_{2}^{{}} \left( {\frac{{k_{1} }}{{k_{2} }}r - c_{3} } \right)} } \right] \\ &\quad + \frac{{(0.5)^{2} }}{{(b_{2}^{{}} )^{2} }}b_{1}^{{}} k_{3} \left[ { - 1 + \sqrt {1 - 4b_{2}^{{}} \left( {\frac{{k_{1} }}{{k_{2} }}r - c_{3} } \right)} } \right]^{2} \\ \end{aligned} $$

(A.9)

$$ - 1 + \sqrt {1 - 4b_{2}^{{}} \left( {\frac{{k_{1} }}{{k_{2} }} - c_{3} } \right)} = 0 $$

(A.10)

$c_{3} = \frac{{k_{1} }}{{k_{2} }}$. These four equations are solved for c₁, c₂, c₃, and c₄, and velocity in both layers is found as follows:

$$ c_{3} = \frac{{k_{1} }}{{k_{2} }} $$

(A.11)

$$ c_{1} = r - \frac{1}{{4b_{1}^{{}} }}\left[ {1 - \left\{ {0.5 + \sqrt {0.25 + b_{1} \left\{ \begin{aligned} 0.5k_{2} \frac{{b_{1}^{{}} }}{{b_{2}^{{}} }}\left( { - 1 + \sqrt {1 - 4b_{2}^{{}} \left( {\frac{{k_{1} }}{{k_{2} }}} \right)\left( {r - 1} \right)} } \right) \hfill \\ + 0.25b_{1}^{{}} k_{3} \left( { - 1 + \sqrt {1 - 4b_{2}^{{}} \left( {\frac{{k_{1} }}{{k_{2} }}} \right)\left( {r - 1} \right)} } \right)^{2} \hfill \\ \end{aligned} \right\}} } \right\}} \right] $$

(A.12)

$$ c_{2} = \frac{ - 0.5}{{6b_{1}^{{}} }}\left\{ {1 + 4b_{1}^{{}} c_{1} } \right\} $$

(A.13)

$$ c_{4} = \frac{0.5}{{b_{1}^{{}} }}\left[ { - r - \frac{1}{{6b_{1}^{{}} }}\left\{ {1 - 4b_{1}^{{}} (r - c_{1} )} \right\}^{3/2} } \right] + c_{2} - \frac{0.5}{{b_{2}^{{}} }}\left[ { - r - \frac{1}{{6b_{2}^{{}} \left( {\frac{{k_{1} }}{{k_{2} }}} \right)}}\left\{ {1 - 4b_{2}^{{}} \left( {\frac{{k_{1} }}{{k_{2} }}r - c_{3} } \right)} \right\}^{3/2} } \right] $$

(A.14)