---
title: "Hagen-Poiseuille equation"
sort_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.