Exact solution of micropolar fluid for Poiseuille flow

Question

I am new here, so i wondered if you could help me with a sytem of ordinary equations with constant coefficients.

This model of equations refer to the well known Navier Stokes equations of fluid mechanics, but with a small discrimination. The original problem refers to a special category of fluids known as micropolar fluids. It considers the flow of a micropolar fluid between two parallel plates.

The general equations are these of conservation of momentum, mass, as well as one more equation of conservation of angular momentum.

The original probelm refers to the following (from book of 'Theory and applications of Micropolar Fluids' - by Grzegorz Lukaszewicz). To be more specific at page 24 of that book the general model is the following one: angular momentum and momentum conservation equations - modified Navier Stokes:

$$\rho(\frac{d\ u}{dt}+(u*\nabla)u)=-\nabla p+(\mu+μ_r)\nabla u+2\mu_r*rot\ ω+ρ*f ,\\ \rho I(\frac{d\ ω}{dt}+(u*\nabla)ω)=2\mu_r(rot\ u-2ω)+(c_0 + c_d - c_a)\nabla\ div\ ω+ (c_d + c_a)\Deltaω+ρg .$$

After the assumptions we made that $u=u(y) - ω=ω(y) $ and $p=p(x)$, from the equations above we finally obtain: -final system of ODE-

$$(\nu+ν_r)\frac{d^2\ u}{dy^2}+2ν_r\frac{d\ ω}{dy}=\frac{d\ p}{dx},\\ (c_d + c_a)\frac{d^2\ ω}{dy^2}-2ν_r(2ω+\frac{d\ u}{dy})=0. $$

Thats also what i am trying to solve and were i struggle most. Note that the derivative $dp/dx$ is constant. Also at page 217, where an exact solution to this problem has been introduced the author considers that $p$ should be such that so:

$$ u(y) = 1-(\frac{y}{h})^2,$$

but what i need to do, is to have $\frac{dp}{dx}$ inside my final form of $u(y)$.

Please, i need your help. Really anything would be much appreciated. Thank you again!

Answer 1

We wish to find the general solution of $$ (\nu+ν_r)\frac{d^2\ u}{dy^2}+2ν_r\frac{d\ ω}{dy}=\frac{d\ p}{dx} \\ (c_d + c_a)\frac{d^2\ ω}{dy^2}-2ν_r(2ω+\frac{d\ u}{dy})=0 $$ What I will present here is a method for solution.

Firstly, to simplify, define $$ a=\frac{2\nu_r}{\nu+\nu_r} \\ A=\frac{1} {\nu+\nu_r} \frac{dp}{dx} \\ b=\frac{2\nu_r} {c_d+c_a} \\ B=\frac{4\nu_r} {c_d+c_a} $$ thus $$ \frac{d^2u} {dy^2} + a\frac{d\omega} {dy} = A \\ \frac{d^2\omega} {dy^2} - b\frac{du} {dy} - B\omega =0 $$ Differentiating the second equation wrt $y$ yields $$ \frac{d^3\omega} {dy^3} - b\frac{d^2u} {dy^2} - B\frac{d\omega}{dy}=0 $$ and substituting from the first equation $$ \frac{d^3\omega} {dy^3} - b\left[A-a\frac{d\omega} {dy} \right] - B\frac{d\omega}{dy}=0 $$ which is a linear homogeneous ODE which can be solved by normal methods to obtain $\omega(y) $. This solution can then be substituted into $$ \frac{d^2u} {dy^2} = A - a\frac{d\omega} {dy} $$ which implies $$ u = \frac{Ay^2} {2} + const \cdot y + const - a\int \omega dy $$ and solving will give $u(y)$. Finally, substitute both solutions into $$ \frac{d^2\omega} {dy^2} - b\frac{du} {dy} - B\omega =0 $$ to find any remaining conditions on the constants of integration.

To specify the constants of integration, boundary conditions must be provided.