Hagen-Poiseuille flow

Here, I am going to go over the solution to fully developed laminar pipe flow. This is a canonical problem and provides an exact solution to the Navier-Stokes equations. It is often referred to as the Hagen-Poiseuille flow problem.

The problem is depicted in the figure below:

Slide2
Fig: Hagen-Poiseuille Flow Problem Definition

In this problem, we are examining laminar flow through a pipe. The problem states that the flow in the pipe is being driven by a constant pressure gradient in the axial direction (dP/dz=constant). We assume that the flow is purely axial (v_r = v_\theta =0), steady state (\partial / \partial t  =0), incompressible (ρ=constant), axisymmetric (\partial / \partial \theta  =0). We also neglect gravity.

First we start with the axial Navier-Stokes momentum equation in cylindrical coordinates:

\rho v_z \frac{\partial v_z}{\partial z} = -\frac{dp}{dz}+\frac{\mu}{r} \frac{d}{dr}\left(r \frac{dv_z}{dr}\right)

By using our assumptions we can reduce this equation to give us:

\frac{dp}{dz}=\frac{\mu}{r} \frac{d}{dr}\left(r \frac{dv_z}{dr}\right)

Because we know that dP/dz is a constant, this function is easily integrated twice. The first integration:

r\frac{dv_z}{dr}=\frac{r^2}{2\mu}\frac{dP}{dz}+A

And the second integration:

v_z=\frac{r^2}{4\mu}\frac{dP}{dz}+A \ln(r)+B

Now we apply our boundary conditions!

No slip condition at r=R : v_z=0=\frac{dP}{dz}\frac{R^2}{4\mu}+A\ln(r)+B

Finite axial velocity at r=0: v_z=0+A \ln(0)+B

Since ln(0) is a discontinuity, we know that in order for these equations to be satisfied, A must be equal to zero (ie A=0). Then combining these two equations, we get that:

B=\left(-\frac{dP}{dz}\right)\left(\frac{R^2}{4\mu}\right)

Thus our final solution is:

v_z=\left(\frac{dP}{dz}\right)\left(\frac{1}{4\mu}\right)\left(r^2-R^2\right)

And there we have our answer!

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