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 ρ\rho. Assuming that the flow is steady v/t=0\ipdv{\va{v}}{t} = 0, and that gravity is negligible g=0\va{g} = 0, we get:

(v)v=pρ+ν2vv=0\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 RR, and be infinitely long and parallel to the zz-axis. We insert the ansatz v=e^zvz(r)\va{v} = \vu{e}_z \: v_z(r), where e^z\vu{e}_z is the zz-axis’ unit vector, and we are assuming that the flow depends only on rr, not on ϕ\phi or zz. With this, v\nabla \cdot \va{v} trivially vanishes, and in the main equation multiplying out ρ\rho yields this, where η=ρν\eta = \rho \nu is the dynamic viscosity:

p=e^zη2vz\begin{aligned} \nabla p = \vu{e}_z \: \eta \nabla^2 v_z \end{aligned}

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

dpdz=Gη1rddr(rdvzdr)=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)p(z) is easy to solve. We get an integration constant p(0)p(0):

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

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

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

ddr(rdvzdr)=Gηr    dvzdr=G2ηr+Ar\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 BB, we arrive at:

vz=G4ηr2+Alnr+B\begin{aligned} v_z = - \frac{G}{4 \eta} r^2 + A \ln{r} + B \end{aligned}

The velocity must be finite at r=0r = 0, so we set A=0A = 0. Furthermore, the Navier-Stokes equation’s no-slip condition demands that vz=0v_z = 0 at the boundary r=Rr = R, so B=GR2/(4η)B = G R^2 / (4 \eta). This brings us to the Poiseuille solution for vz(r)v_z(r):

vz(r)=G4η(R2r2)\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 QQ, which is the integral of vzv_z over the circular cross-section:

Q=2π0Rvz(r)rdr=πG2η0RR2rr3dr=πG2η[R2r22r44]0R\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 QQ for a given setup:

Q=πGR48η\begin{aligned} \boxed{ Q = \frac{\pi G R^4}{8 \eta} } \end{aligned}

Consequently, the average flow velocity vz\Expval{v_z} is simply QQ divided by the cross-sectional area:

vz=QπR2=GR28η\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 DD on the pipe as it flows. For a pipe of length LL and radius RR, we calculate DD by multiplying the internal area 2πRL2 \pi R L by the shear stress σzr-\sigma_{zr} on the wall (i.e. the wall applies σzr\sigma_{zr}, the fluid responds with σzr- \sigma_{zr}):

D=2πRLσzrr=R=2πRLηdvzdrr=R=2πRLηGR2η=πR2LG\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}

GG is an inconvenient quantity here, so we remove it by substituting R2G=8ηvzR^2 G = 8 \eta \Expval{v_z}:

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

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

P=2π0RΔpvz(r)rdr\begin{aligned} P = 2 \pi \int_0^R \Delta p \: v_z(r) \: r \dd{r} \end{aligned}

Because Δp\Delta p is independent of rr, we get the same integral we used to calculate QQ. Then, thanks to the fact that Δp=GL\Delta p = G L and Q=πR2vzQ = \pi R^2 \Expval{v_z}, it follows that:

P=ΔpQ=GLπR2vz=Dvz\begin{aligned} P = \Delta p \: Q = G L \pi R^2 \Expval{v_z} = D \Expval{v_z} \end{aligned}

In conclusion, the power PP needed to drive a fluid through the pipe at a rate QQ is given by:

P=8πηLvz2=8ηLπR4Q2\begin{aligned} \boxed{ P = 8 \pi \eta L \Expval{v_z}^2 = \frac{8 \eta L}{\pi R^4} Q^2 } \end{aligned}


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