# Hagen–Poiseuille flow from the Navier–Stokes equations

In fluid dynamics, the derivation of the Hagen–Poiseuille flow from the Navier–Stokes equations shows how this flow is an exact solution to the Navier–Stokes equations.[1][2]

## Derivation

The laminar flow through a pipe of uniform (circular) cross-section is known as Hagen–Poiseuille flow. The equations governing the Hagen–Poiseuille flow can be derived directly from the Navier–Stokes momentum equations in 3D cylindrical coordinates by making the following set of assumptions:

1. The flow is steady ( ${\displaystyle \partial (...)/\partial t=0}$ ).
2. The radial and swirl components of the fluid velocity are zero ( ${\displaystyle u_{r}=u_{\theta }=0}$ ).
3. The flow is axisymmetric ( ${\displaystyle \partial (...)/\partial \theta =0}$ ).
4. The flow is fully developed (${\displaystyle \partial u_{z}/\partial z=0}$ ). However, this can be proved via mass conservation, and the above assumptions.

Then the angular equation in the momentum equations and the continuity equation are identically satisfied. The first momentum equation reduces to ${\displaystyle \partial p/\partial r=0}$, i.e., the pressure ${\displaystyle p}$ is a function of the axial coordinate ${\displaystyle z}$ only. The third momentum equation reduces to:

${\displaystyle {\frac {1}{r}}{\frac {\partial }{\partial r}}\left(r{\frac {\partial u_{z}}{\partial r}}\right)={\frac {1}{\mu }}{\frac {\partial p}{\partial z}}}$ where ${\displaystyle \mu }$ is the dynamic viscosity of the fluid.
The solution is
${\displaystyle u_{z}={\frac {1}{4\mu }}{\frac {\partial p}{\partial z}}r^{2}+c_{1}\ln r+c_{2}}$

Since ${\displaystyle u_{z}}$ needs to be finite at ${\displaystyle r=0}$, ${\displaystyle c_{1}=0}$. The no slip boundary condition at the pipe wall requires that ${\displaystyle u_{z}=0}$ at ${\displaystyle r=R}$ (radius of the pipe), which yields

${\displaystyle c_{2}=-{\frac {1}{4\mu }}{\frac {\partial p}{\partial z}}R^{2}.}$

Thus we have finally the following parabolic velocity profile:

${\displaystyle u_{z}=-{\frac {1}{4\mu }}{\frac {\partial p}{\partial z}}(R^{2}-r^{2}).}$

The maximum velocity occurs at the pipe centerline (${\displaystyle r=0}$):

${\displaystyle {u_{z}}_{max}={\frac {R^{2}}{4\mu }}\left(-{\frac {\partial p}{\partial z}}\right).}$

The average velocity can be obtained by integrating over the pipe cross section:

${\displaystyle {u_{z}}_{\mathrm {avg} }={\frac {1}{\pi R^{2}}}\int _{0}^{R}u_{z}\cdot 2\pi rdr=0.5{u_{z}}_{\mathrm {max} }.}$

The Hagen–Poiseuille equation relates the pressure drop ${\displaystyle \Delta p}$ across a circular pipe of length ${\displaystyle L}$ to the average flow velocity in the pipe ${\displaystyle {u_{z}}_{\mathrm {avg} }}$ and other parameters. Assuming that the pressure decreases linearly across the length of the pipe, we have ${\displaystyle -{\frac {\partial p}{\partial z}}={\frac {\Delta p}{L}}}$ (constant). Substituting this and the expression for ${\displaystyle {u_{z}}_{\mathrm {max} }}$ into the expression for ${\displaystyle {u_{z}}_{\mathrm {avg} }}$, and noting that the pipe diameter ${\displaystyle D=2R}$, we get:

${\displaystyle {u_{z}}_{avg}={\frac {D^{2}}{32\mu }}{\frac {\Delta p}{L}}.}$

Rearrangement of this gives the Hagen–Poiseuille equation:

${\displaystyle \Delta p={\frac {32\mu L~{u_{z}}_{\mathrm {avg} }}{D^{2}}}.}$

## References

1. ^ White, Frank M. (2003). "6". Fluid Mechanics (5 ed.). McGraw-Hill.
2. ^ Bird, Stewart, Lightfoot (1960). Transport Phenomena.CS1 maint: Multiple names: authors list (link)