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 flow of fluid 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 equations in cylindrical coordinates by making the following set of assumptions:

  1. The flow is steady ( \( \partial(...)/\partial t = 0 \) ).
  2. The radial and swirl components of the fluid velocity are zero ( \( u_r = u_\theta = 0 \) ).
  3. The flow is axisymmetric ( \( \partial(...)/\partial \theta = 0 \) ) and fully developed (\( \partial u_z/\partial z = 0 \) ).

Then the second of the three Navier–Stokes momentum equations and the continuity equation are identically satisfied. The first momentum equation reduces to \( \partial p/\partial r = 0 \), i.e., the pressure \( p \) is a function of the axial coordinate \( z \) only. The third momentum equation reduces to:

\[ \frac{1}{r}\frac{\partial}{\partial r}\left(r \frac{\partial u_z}{\partial r}\right)= \frac{1}{\mu} \frac{\partial p}{\partial z}\] where \(\mu\) is the dynamic viscosity of the fluid.

The solution is

\[ u_z = \frac{1}{4\mu} \frac{\partial p}{\partial z}r^2 + c_1 \ln r + c_2 \] Since \( u_z \) needs to be finite at \( r = 0 \), \( c_1 = 0 \). The no slip boundary condition at the pipe wall requires that \( u_z = 0\) at \( r = R \) (radius of the pipe), which yields

\[ c_2 = -\frac{1}{4\mu} \frac{\partial p}{\partial z}R^2.\]

Thus we have finally the following parabolic velocity profile:

\[ u_z = -\frac{1}{4\mu} \frac{\partial p}{\partial z} (R^2 - r^2). \]

The maximum velocity occurs at the pipe centerline (\( r=0 \)):

\[ {u_z}_{max}=\frac{R^2}{4\mu} \left(-\frac{\partial p}{\partial z}\right). \]

The average velocity can be obtained by integrating over the pipe cross section: \[ {u_z}_\mathrm{avg}=\frac{1}{\pi R^2} \int_0^R u_z \cdot 2\pi r dr = 0.5 {u_z}_\mathrm{max}. \]

The Hagen–Poiseuille equation relates the pressure drop \( \Delta p\) across a circular pipe of length \( L \) to the average flow velocity in the pipe \( {u_z}_\mathrm{avg} \) and other parameters. Assuming that the pressure decreases linearly across the length of the pipe, we have \( - \frac{\partial p}{\partial z} = \frac{\Delta p}{L} \) (constant). Substituting this and the expression for \( {u_z}_\mathrm{max} \) into the expression for \( {u_z}_\mathrm{avg} \), and noting that the pipe diameter \( D = 2R \), we get: \[ {u_z}_{avg} = \frac{D^2}{32 \mu} \frac{\Delta P}{L}. \] Rearrangement of this gives the Hagen–Poiseuille equation: \[ \Delta P = \frac{32 \mu L ~{u_z}_\mathrm{avg}}{D^2}. \]

References

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See also