Skip to content # 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:

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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Basic Fluids

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