Definitions

Poiseuille Flow

Hagen-Poiseuille flow from the Navier-Stokes equations

The flow of fluid through a pipe of uniform (circular) cross-section is known as Hagen-Poiseuille flow. The Hagen-Poiseuille flow is an exact solution of the Navier-Stokes equations in fluid mechanics. The equations governing the Hagen-Poiseuille flow can be derived from the Navier-Stokes equation 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}
The solution is
u_z = frac{1}{4mu} 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}{4mu} frac{partial p}{partial z}R^2.

Thus we have finally the following parabolic velocity profile:

u_z = -frac{1}{4mu} 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}{4mu} 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 2pi 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}.

See also

Search another word or see Poiseuille Flowon Dictionary | Thesaurus |Spanish
  • Please Login or Sign Up to use the Recent Searches feature
FAVORITES
RECENT