Categories: Fluid dynamics, Fluid mechanics, Physics.

The **Hagen-Poiseuille equation**, or simply the **Poiseuille equation**, describes the flow of a fluid with nonzero viscosity through a cylindrical pipe. Due to its viscosity, the fluid clings to the sides, limiting the amount that can pass through, for a pipe with radius \(R\).

Consider the Navier-Stokes equations of an incompressible fluid with spatially uniform density \(\rho\). Assuming that the flow is steady \(\pdv*{\va{v}}{t} = 0\), and that gravity is negligible \(\va{g} = 0\), we get:

\[\begin{aligned} (\va{v} \cdot \nabla) \va{v} = - \frac{\nabla p}{\rho} + \nu \nabla^2 \va{v} \qquad \quad \nabla \cdot \va{v} = 0 \end{aligned}\]

Into this, we insert the ansatz \(\va{v} = \vu{e}_z \: v_z(r)\), where \(\vu{e}_z\) is the \(z\)-axis’ unit vector. In other words, we assume that the flow velocity depends only on \(r\); not on \(\phi\) or \(z\). Plugging this into the Navier-Stokes equations, \(\nabla \cdot \va{v}\) is trivially zero, and in the other equation we multiply out \(\rho\), yielding this, where \(\eta = \rho \nu\) is the dynamic viscosity:

\[\begin{aligned} \nabla p = \vu{e}_z \: \eta \nabla^2 v_z \end{aligned}\]

Because only \(\vu{e}_z\) appears on the right-hand side, only the \(z\)-component of \(\nabla p\) can be nonzero. However, \(v_z(r)\) is a function of \(r\), not \(z\)! The left thus only depends on \(z\), and the right only on \(r\), meaning that both sides must equal a constant, which we call \(-G\):

\[\begin{aligned} \dv{p}{z} = -G \qquad \quad \eta \frac{1}{r} \dv{r} \Big( r \dv{v_z}{r} \Big) = - G \end{aligned}\]

The former equation, for \(p(z)\), is easy to solve. We get an integration constant \(p(0)\):

\[\begin{aligned} p(z) = p(0) - G z \end{aligned}\]

This gives meaning to the **pressure gradient** \(G\): for a pipe of length \(L\), it describes the pressure difference \(\Delta p = p(0) - p(L)\) that is driving the fluid, i.e. \(G = \Delta p / L\)

As for the latter equation, for \(v_z(r)\), we start by integrating it once, introducing a constant \(A\):

\[\begin{aligned} \dv{r} \Big( r \dv{v_z}{r} \Big) = - \frac{G}{\eta} r \quad \implies \quad \dv{v_z}{r} = - \frac{G}{2 \eta} r + \frac{A}{r} \end{aligned}\]

Integrating this one more time, thereby introducing another constant \(B\), we arrive at:

\[\begin{aligned} v_z = - \frac{G}{4 \eta} r^2 + A \ln{r} + B \end{aligned}\]

The velocity must be finite at \(r = 0\), so we set \(A = 0\). Furthermore, the Navier-Stokes equation’s *no-slip* condition demands that \(v_z = 0\) at the boundary \(r = R\), so \(B = G R^2 / (4 \eta)\). This brings us to the **Poiseuille solution** for \(v_z(r)\):

\[\begin{aligned} \boxed{ v_z(r) = \frac{G}{4 \eta} (R^2 - r^2) } \end{aligned}\]

How much fluid can pass through the pipe per unit time? This is denoted by the **volumetric flow rate** \(Q\), which is the integral of \(v_z\) over the circular cross-section:

\[\begin{aligned} Q = 2 \pi \int_0^R v_z(r) \: r \dd{r} = \frac{\pi G}{2 \eta} \int_0^R R^2 r - r^3 \dd{r} = \frac{\pi G}{2 \eta} \bigg[ \frac{R^2 r^2}{2} - \frac{r^4}{4} \bigg]_0^R \end{aligned}\]

We thus arrive at the main Hagen-Poiseuille equation, which predicts \(Q\) for a given setup:

\[\begin{aligned} \boxed{ Q = \frac{\pi G R^4}{8 \eta} } \end{aligned}\]

Consequently, the average flow velocity \(\expval{v_z}\) is simply \(Q\) divided by the cross-sectional area:

\[\begin{aligned} \expval{v_z} = \frac{Q}{\pi R^2} = \frac{G R^2}{8 \eta} \end{aligned}\]

The fluid’s viscous stickiness means it exerts a drag force \(D\) on the pipe as it flows. For a pipe of length \(L\) and radius \(R\), we calculate \(D\) by multiplying the internal area \(2 \pi R L\) by the shear stress \(-\sigma_{zr}\) on the wall (i.e. the wall applies \(\sigma_{zr}\), the fluid responds with \(- \sigma_{zr}\)):

\[\begin{aligned} D = - 2 \pi R L \: \sigma_{zr} \big|_{r = R} = - 2 \pi R L \eta \dv{v_z}{r}\Big|_{r = R} = 2 \pi R L \eta \frac{G R}{2 \eta} = \pi R^2 L G \end{aligned}\]

We would like to get rid of \(G\) for being impractical, so we substitute \(R^2 G = 8 \eta \expval{v_z}\), yielding:

\[\begin{aligned} \boxed{ D = 8 \pi \eta L \expval{v_z} } \end{aligned}\]

Due to this drag, the pressure difference \(\Delta p = p(0) - p(L)\) does work on the fluid, at a rate \(P\), since power equals force (i.e. pressure times area) times velocity:

\[\begin{aligned} P = 2 \pi \int_0^R \Delta p \: v_z(r) \: r \dd{r} \end{aligned}\]

Because \(\Delta p\) is independent of \(r\), we get the same integral we used to calculate \(Q\). Then, thanks to the fact that \(\Delta p = G L\) and \(Q = \pi R^2 \expval{v_z}\), it follows that:

\[\begin{aligned} P = \Delta p \: Q = G L \pi R^2 \expval{v_z} = D \expval{v_z} \end{aligned}\]

In conclusion, the power \(P\), needed to drive a fluid through the pipe at a rate \(Q\), is given by:

\[\begin{aligned} \boxed{ P = 8 \pi \eta L \expval{v_z}^2 } \end{aligned}\]

- B. Lautrup,
*Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition, CRC Press.

© "Prefetch". Licensed under CC BY-SA 4.0.