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# 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 ($partial\left(...\right)/partial t = 0$ ).
2. The radial and swirl components of the fluid velocity are zero ($u_r = u_theta = 0$ ).
3. The flow is axisymmetric ($partial\left(...\right)/partial theta = 0$ ) 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:

$frac\left\{1\right\}\left\{r\right\}frac\left\{partial\right\}\left\{partial r\right\}left\left(r frac\left\{partial u_z\right\}\left\{partial r\right\}right\right)= frac\left\{1\right\}\left\{mu\right\} frac\left\{partial p\right\}\left\{partial z\right\}$
The solution is
$u_z = frac\left\{1\right\}\left\{4mu\right\} frac\left\{partial p\right\}\left\{partial z\right\}r^2 + c_1 ln r + c_2$
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

$c_2 = -frac\left\{1\right\}\left\{4mu\right\} frac\left\{partial p\right\}\left\{partial z\right\}R^2.$

Thus we have finally the following parabolic velocity profile:

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

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

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

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

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

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 $\left\{u_z\right\}_mathrm\left\{avg\right\}$ and other parameters. Assuming that the pressure decreases linearly across the length of the pipe, we have $- frac\left\{partial p\right\}\left\{partial z\right\} = frac\left\{Delta p\right\}\left\{L\right\}$ (constant). Substituting this and the expression for $\left\{u_z\right\}_mathrm\left\{max\right\}$ into the expression for $\left\{u_z\right\}_mathrm\left\{avg\right\}$, and noting that the pipe diameter $D = 2R$, we get:

$\left\{u_z\right\}_\left\{avg\right\} = frac\left\{D^2\right\}\left\{32 mu\right\} frac\left\{Delta P\right\}\left\{L\right\}.$
Rearrangement of this gives the Hagen-Poiseuille equation:
$Delta P = frac\left\{32 mu L \left\{u_z\right\}_mathrm\left\{avg\right\}\right\}\left\{D^2\right\}.$