---

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: the fluid clings to the sides, limiting the amount that can pass through per unit time.

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

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

Let the pipe have radius $R$, and be infinitely long and parallel to the $z$-axis. We insert the ansatz $\va{v} = \vu{e}_z \: v_z(r)$, where $\vu{e}_z$ is the $z$-axis’ unit vector, and we are assuming that the flow depends only on $r$, not on $\phi$ or $z$. With this, $\nabla \cdot \va{v}$ trivially vanishes, and in the main equation multiplying out $\rho$ yields 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 $G$: it is the pressure gradient, which for a pipe of length $L$ 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 $- \sigma_{zr}$):

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

$G$ is an inconvenient quantity here, so we remove it by substituting $R^2 G = 8 \eta \Expval{v_z}$:

$$\begin{aligned} \boxed{ D = 8 \pi \eta L \Expval{v_z} } \end{aligned}$$

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:

$$\begin{aligned} \boxed{ P = 8 \pi \eta L \Expval{v_z}^2 = \frac{8 \eta L}{\pi R^4} Q^2 } \end{aligned}$$

References

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

---