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---
title: "Hagen-Poiseuille equation"
date: 2021-04-13
categories:
- Physics
- Fluid mechanics
- Fluid dynamics
layout: "concept"
---
The **Hagen-Poiseuille equation**, or simply the **Poiseuille equation**,
describes the flow of a fluid with nonzero [viscosity](/know/concept/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](/know/concept/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:
$$\begin{aligned}
(\va{v} \cdot \nabla) \va{v}
= - \frac{\nabla p}{\rho} + \nu \nabla^2 \va{v}
\qquad \quad
\nabla \cdot \va{v} = 0
\end{aligned}$$
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](/know/concept/cauchy-stress-tensor/)
$-\sigma_{zr}$ 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.
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