Poiseuille Flow

In Poiseuille flow, we have steady, axisymmetric flow of a viscous, incompressible fluid through a long, straight pipe of circular cross-section (radius $R$).

In this case, we use cylindrical coordinates. We assume:

$$V_r=0,V_{\theta }=0,V_z=V_z$$

The continuity equation for incompressible fluids in cylindrical coordinates reduces to:

$$\frac{1}{r}\frac{\partial }{\partial r}(r{V}_r)+\frac{1}{r}\frac{\partial V_{\theta }}{\partial \theta }+\frac{\partial V_z}{\partial z}=0$$

Given our assumptions, this becomes

$$\frac{\partial V_z}{\partial z}=0\Rightarrow V_z=V_z(r)$$

Applying the Navier-Stokes in cylindrical coordinates:

The $x$-direction gives:

$$p=-\rho gr\sin \theta +f_1(\theta ,z)$$

The $y$-direction gives:

$$\begin{array}{rl}p& =-\rho gr\sin \theta +f_{2(r,z)}\end{array}$$

Thus, in general we have

$$p=-\rho gr\sin \theta +f(z)$$

The $z$-direction (axial) gives:

$$0=-\frac{\partial p}{\partial z}+\mu \left[\frac{1}{r}\frac{d}{dr}\left(r\frac{d{V}_z}{dr}\right)\right]$$

Solving the axial momentum equation gives:

$$V_z(r)=\frac{1}{4\mu }\frac{dp}{dz}{r}^2+C_1\ln r+C_2$$

Boundary conditions:

• No-slip at the wall: $V_z=0$ at $r=R$ gives $C_2=-\frac{1}{4\mu }\frac{dp}{dz}{R}^2$
• Velocity finite at centerline: $V_z$ finite as $r\to 0$ gives $C_1=0$

Thus, we have:

$$V_z(r)=\frac{1}{4\mu }\frac{dp}{dz}(r^2-R^2)=V_{\text{max}}\left(1-\frac{r^2}{R^2}\right)$$

Volumetric flow rate:

$$Q={\int }_0^R{V}_z(r)\cdot 2\pi rdr=\frac{\pi R^4}{8\mu }\left(-\frac{dp}{dz}\right)$$

If the pressure drop over length $l$ is $\Delta p$, then

$$\frac{\partial p}{\partial z}=-\frac{\Delta p}{l}$$

Substituting into the equation for $Q$:

$$Q=\frac{\pi R^4\Delta p}{8\mu l}$$

which is called Poiseuille’s Law.