Poiseuille's Law

Equivalent forms

\Delta P = \frac{8 \eta L Q}{\pi r^4}

v_{max} = \frac{\Delta P \cdot r^2}{4 \eta L}

The r⁴ scaling — small changes in radius dominate everything.

Symbol	Name	SI unit	Dimension	Range
$r$	Pipe Radius Internal radius of the pipe	m	L	0.001 – 0.05
$\Delta P$	Pressure Drop Pressure difference along the pipe	Pa	M·L⁻¹·T⁻²	1 – 10000
$\eta$	Dynamic Viscosity Fluid viscosity	Pa·s	M·L⁻¹·T⁻¹	0.0001 – 1
$L$	Pipe Length Length of the pipe	m	L	0.01 – 10
$Q$	Volumetric Flow Rateoutput Volume of fluid per unit time	m³/s	L³·T⁻¹	0 – 0.001

Unit systems

Symbol	SI	CGS	Imperial
$r$	m	cm	ft
$\Delta P$	Pa	dyn/cm²	psi
$\eta$	Pa·s	P	lb/(ft·s)
$L$	m	cm	ft
$Q$	m³/s	cm³/s	ft³/s

Where it holds

Laminar (Re < 2300), Newtonian, incompressible, fully developed flow in a straight cylindrical pipe with no-slip walls. Fails for turbulent flow, non-Newtonian fluids, short pipes (entrance effects), and arteries (pulsatile flow).

Dimensional analysis

$[r^{4}][\Delta P]/([\eta ][L]) = (L^{4})(M\cdot L^{-1}\cdot T^{-2})/((M\cdot L^{-1}\cdot T^{-1})(L)) = L^{3}\cdot T^{-1} \checkmark$ (volume flow rate)

Discovery

Jean Léonard Marie Poiseuille · 1838

Poiseuille, a French physician, measured blood flow through capillaries and empirically discovered the r⁴ dependence. Gotthilf Hagen independently derived it in 1839.

Try this

Why does halving an artery's radius reduce blood flow by 16x?

Blood (viscosity 0.0035 Pa·s) flows through a 0.005 m radius artery, 0.1 m long, under a pressure drop of 100 Pa. What is the volumetric flow rate?

Research status: stable

Real-world applications

IV drip rate calculation in medicine
Capillary viscometers
Microfluidic chip design
Understanding stenosis (artery narrowing) and aneurysm hemodynamics

Common misconceptions

Poiseuille's law does NOT apply to blood in large arteries — they're pulsatile and turbulent at high Re
The flow profile is parabolic, not flat — velocity at the wall is zero, maximum at center
The $r^{4}$ dependence means a 20% reduction in radius (e.g., plaque) cuts flow $by \sim 60$ %

Experimental verification

Poiseuille's original 1838 measurements on capillaries

14-650 \mu m

in diameter showed the

r^{4}

scaling to high precision. Modern microfluidic experiments routinely confirm the law for water and oils with <2% deviation in the laminar regime.

Derivation

Start with the Navier-Stokes equation in cylindrical coordinates for steady, axisymmetric, fully developed flow.

The only non-trivial component reduces to

(1/r)d/dr(r\cdot dv/dr) = (1/\eta )\cdot dP/dz

= constant.

Integrating twice with no-slip

(v=0

at

r=R)

and finite velocity at center gives the parabolic profile

v(r) = (\Delta P/4\eta L)(R^{2}-r^{2})

.

Integrating v over the cross-section:

Q = \int v\cdot 2\pi r\cdot dr = \pi R^{4}\Delta P/(8\eta L)

.

Limiting cases

r \to 0

⟶

Q \to 0

A closed pipe carries no flow.

r doubles⟶

Q \times 16

The

r^{4}

dependence means a small dilation produces massive flow gains.

\eta \to \infty

⟶

Q \to 0

Infinitely viscous fluid cannot flow under finite pressure.

What if…

What if the pipe radius doubles?

Flow increases by $16\times$ . This is why aortic dilation dramatically increases cardiac output.

What if blood thickens (high hematocrit)?

Doubling viscosity halves flow rate. This is why dehydration strains the heart.

What if flow becomes turbulent?

Poiseuille's law fails. Use the Darcy-Weisbach equation with an empirical friction factor instead.

1

Blood flow through an artery

Given ·

r:: 0.005
\Delta P:: 100
\eta:: 0.0035
L:: 0.1

Find · Q

Steps

$Q = \pi r^{4}\Delta P/(8\eta L)$
Numerator: $\pi \times 6.25\times 10^{-10} \times 100 = 1.96\times 10^{-7}$
Denominator: $8 \times 0.0035 \times 0.1 = 0.0028$
$Q \approx 7.0 \times 10^{-5} m^{3}/s \approx 70\,\mathrm{mL/s}$

Result ·

Q = \pi (0.005)^{4}(100)/(8 \times 0.0035 \times 0.1) \approx 7.0 \times 10^{-5} m^{3}/s

2

Halving radius cuts flow 16-fold

Given ·

r:: 0.0025
\Delta P:: 100
\eta:: 0.0035
L:: 0.1

Find · Q

Steps

Same formula with r halved
Q scales as $r^{4}$ , so flow drops by $2^{4} = 16$
$Q \approx 70/16 \approx 4.4\,\mathrm{mL/s}$

Result ·

Q = \pi (0.0025)^{4} \times 100/(8 \times 0.0035 \times 0.1) \approx 4.4 \times 10^{-6} m^{3}/s

Continuity Equation→Reynolds Number→Darcy-Weisbach Equation→