Hagen-Poiseuille flow from the Navier-Stokes equations

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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:

The flow is steady ( $\partial(...)/\partial t = 0$ ).
The radial and swirl components of the fluid velocity are zero ( $u r = u θ = 0$ ).
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 $u z$ 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:

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

The Hagen-Poiseuille equation relates the pressure drop $Δ p$ across a circular pipe of length $L$ to the average flow velocity in the pipe ${u_z}_{avg}$ 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 ${u_z}_{max}$ into the expression for ${u_z}_{avg}$ , and noting that the pipe diameter $D = 2 R$ , we get: