PX285 - H6b - example of a flow

considering a flow between two parallel plates at $a$ and $- a$

the assumptions:
- laminar (1D, $∥ \hat{x}$ , no vortices)
- steady (time independent)
- viscous ( $μ$ is important)

\vec{u} = u_{x} (y) \hat{x}

the boundary conditions:
$\vec{u} (y = - a, a) = 0$
the steady motion requires a force which counteracts the viscous force
let it be the pressure difference, $P_{1} > P_{2}$

the viscous force:

F_{v} = μ A \frac{\partial u}{\partial y}

where, $A = Δ x Δ z$

the net force applied at the fluid element:

\begin{aligned} F_{v} & = A μ {\frac{\partial u_{x}}{\partial y} |_{y + Δ y} - \frac{\partial u_{x}}{\partial y} |_{y}} \\ = A μ \frac{Δ (\partial u_{x} / \partial y)}{Δ y} Δ y \\ F_{p} & = (P (x) - P (x + Δ x)) Δ y Δ z \\ \approx - \frac{\partial P}{\partial x} Δ x Δ y Δ z \end{aligned}

as the flow is staedy, there is no acceleration:

\begin{matrix} \frac{\partial P}{\partial x} Δ x Δ y Δ z = μ \frac{\partial^{2} u_{x}}{\partial y^{2}} Δ x Δ y Δ z \\ ⟹ \frac{\partial P}{\partial x} = μ \frac{\partial^{2} u_{x}}{\partial y^{2}} = - Q \end{matrix}

note: the negative sign vanishes as pressure is decreasing with $x$ , ie. $\frac{\partial P}{\partial x} < 0$
from the RHS:

u_{x} (y) = - \frac{Q}{2 μ} y^{2} + A y + B

using the boundary conditions

\begin{matrix} u_{x} (y = - a) = 0 = - \frac{Q a^{2}}{2 μ} - A a + B \\ u_{x} (y = + a) = 0 = - \frac{Q a^{2}}{2 μ} + A a + B \\ ⟹ A = 0 and B = \frac{Q a^{2}}{2 μ} \end{matrix}

∴ u_{x} (y) = \frac{Q}{2 μ} (a^{2} - y^{2})

constant $Q$ comes from the pressure difference
using the LHS:

Q = - \frac{\partial P}{\partial x} \approx \frac{P_{1} - P_{2}}{L}

∴ u_{x} (y) = \frac{P_{1} - P_{2}}{2 L μ} (a^{2} - y^{2})

therefore, the form of the flow is parabolic, called the poisseuille flow