Plane Poiseuille flow from Navier-Stokes

Derivation

Assumptions: steady, incompressible, fully developed ($\partial u/\partial x=0$), $v=w=0$, no body force effect along x, infinitely wide plates ($\partial/\partial z=0$).

   y=h  ======================  fixed plate
         ->    ->     ->
         ->   -->    --->   u(y)
         ->    ->     ->
   y=0  ======================  fixed plate

The x-momentum equation reduces to

$$0=-\frac{dp}{dx}+\mu\frac{d^2u}{dy^2}\ \Rightarrow\ \frac{d^2u}{dy^2}=\frac1\mu\frac{dp}{dx}$$

(the y-momentum equation shows that $p$ depends only on $x$). Integrate twice:

$$u=\frac{1}{2\mu}\frac{dp}{dx}y^2+C_1y+C_2$$

No-slip: $u=0$ at $y=0$ gives $C_2=0$; $u=0$ at $y=h$ gives $C_1=-\dfrac{1}{2\mu}\dfrac{dp}{dx}h$. So

$$u=-\frac{1}{2\mu}\frac{dp}{dx}\,y\,(h-y)$$

The profile is parabolic. Maximum velocity at $y=h/2$:

$$u_{max}=-\frac{h^2}{8\mu}\frac{dp}{dx}$$

Flow rate per unit width $q=\int_0^hu\,dy=-\dfrac{h^3}{12\mu}\dfrac{dp}{dx}$, so mean velocity $\bar u=q/h=\tfrac23u_{max}$. Wall shear stress $\tau_w=\mu\,\dfrac{du}{dy}\Big|_{0}=-\dfrac h2\dfrac{dp}{dx}$ in magnitude.

Numerical

Data: $\mu=0.08$ Pa·s, $\rho=870$ kg/m³, $h=0.006$ m, $G=-dp/dx=1500$ Pa/m.

$$u_{max}=\frac{G h^2}{8\mu}=\frac{1500\times(0.006)^2}{8\times0.08}=0.0844\ \text{m/s}$$ $$q=\frac{Gh^3}{12\mu}=\frac{1500\times(0.006)^3}{12\times0.08}=3.375\times10^{-4}\ \text{m}^2/\text{s per m width}$$ $$\bar u=\frac qh=\frac{3.375\times10^{-4}}{0.006}=0.05625\ \text{m/s}\quad(=\tfrac23\times0.0844)$$ $$\tau_w=\frac{Gh}{2}=\frac{1500\times0.006}{2}=4.5\ \text{Pa}$$ $$Re=\frac{\rho\bar uh}{\mu}=\frac{870\times0.05625\times0.006}{0.08}=3.67$$

$Re=3.67$ is far below the transition value for channel flow (of the order of 1000), so the flow is laminar and the assumption is valid.

Answer: $u_{max}=0.0844$ m/s; $q=3.375\times10^{-4}$ m²/s per m; $\bar u=0.0563$ m/s; $\tau_w=4.5$ Pa; $Re=3.67$.