Categories: Fluid dynamics, Fluid mechanics, Physics.

# Hagen-Poiseuille equation

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}

## References

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