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# Hagen-Poiseuille flow

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 ( ur = uθ = 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
$u_z = \frac{1}{4\mu} \frac{\partial p}{\partial z}r^2 + c_1 \ln r + c_2$

Since uz needs to be finite at r = 0, c1 = 0. The no slip boundary condition at the pipe wall requires that uz = 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:

${u_z}_{avg}=\frac{1}{\pi R^2} \int_0^R u_z \cdot 2\pi r dr = 0.5 {u_z}_{max}.$