The Hagen-Poiseuille equation, or simply the Poiseuille equation, describes the flow of a fluid with nonzero viscosity through a cylindrical pipe: the fluid clings to the sides, limiting the amount that can pass through per unit time.
Consider the Navier-Stokes equations of an incompressible fluid with spatially uniform density . Assuming that the flow is steady , and that gravity is negligible , we get:
Let the pipe have radius , and be infinitely long and parallel to the -axis. We insert the ansatz , where is the -axis’ unit vector, and we are assuming that the flow depends only on , not on or . With this, trivially vanishes, and in the main equation multiplying out yields this, where is the dynamic viscosity:
Because only appears on the right-hand side, only the -component of can be nonzero. However, is a function of , not ! The left thus only depends on , and the right only on , meaning that both sides must equal a constant, which we call :
The former equation for is easy to solve. We get an integration constant :
This gives meaning to : it is the pressure gradient, which for a pipe of length describes the pressure difference that is driving the fluid, i.e. .
As for the latter equation for , we start by integrating it once, introducing a constant :
Integrating this one more time, thereby introducing another constant , we arrive at:
The velocity must be finite at , so we set . Furthermore, the Navier-Stokes equation’s no-slip condition demands that at the boundary , so . This brings us to the Poiseuille solution for :
How much fluid can pass through the pipe per unit time? This is denoted by the volumetric flow rate , which is the integral of over the circular cross-section:
We thus arrive at the main Hagen-Poiseuille equation, which predicts for a given setup:
Consequently, the average flow velocity is simply divided by the cross-sectional area:
The fluid’s viscous stickiness means it exerts a drag force on the pipe as it flows. For a pipe of length and radius , we calculate by multiplying the internal area by the shear stress on the wall (i.e. the wall applies , the fluid responds with ):
is an inconvenient quantity here, so we remove it by substituting :
Due to this drag, the pressure difference does work on the fluid at a rate . Since power equals force (i.e. pressure times area) times velocity:
Because is independent of , we get the same integral we used to calculate . Then, thanks to the fact that and , it follows that:
In conclusion, the power needed to drive a fluid through the pipe at a rate is given by:
- B. Lautrup, Physics of continuous matter: exotic and everyday phenomena in the macroscopic world, 2nd edition, CRC Press.