Poiseuille’s law describes steady laminar flow of a Newtonian fluid through a long, circular pipe.
It connects the volume flow rate to the pressure drop, pipe radius, pipe length, and fluid viscosity.
1. Poiseuille’s law
For fully developed laminar flow in a circular pipe,
\[
\begin{aligned}
Q &= \frac{\pi R^4\Delta P}{8\eta L}.
\end{aligned}
\]
Here:
- \(Q\) is the volume flow rate.
- \(R\) is the pipe radius.
- \(\Delta P=P_1-P_2\) is the pressure drop.
- \(\eta\) is dynamic viscosity.
- \(L\) is pipe length.
2. Radius to the fourth power
The radius appears as \(R^4\), so the flow rate is extremely sensitive to pipe size.
If the radius is doubled while all other quantities stay the same,
\[
\begin{aligned}
Q_{\mathrm{new}} &= 2^4 Q = 16Q.
\end{aligned}
\]
This is why small changes in a tube, capillary, or blood vessel radius can produce large changes in flow.
3. Hydraulic resistance
Poiseuille’s law can be written in a resistance form:
\[
\begin{aligned}
\Delta P &= R_h Q.
\end{aligned}
\]
The hydraulic resistance of a circular pipe is
\[
\begin{aligned}
R_h &= \frac{8\eta L}{\pi R^4}.
\end{aligned}
\]
A larger viscosity or longer pipe increases resistance.
A larger radius greatly decreases resistance.
4. Pressure gradient
The pressure drop per unit length is
\[
\begin{aligned}
\frac{\Delta P}{L}.
\end{aligned}
\]
This quantity tells how strongly pressure is falling along the pipe.
5. Average speed
The cross-sectional area of the pipe is
\[
\begin{aligned}
A &= \pi R^2.
\end{aligned}
\]
The average flow speed is
\[
\begin{aligned}
\bar v &= \frac{Q}{A}
= \frac{Q}{\pi R^2}.
\end{aligned}
\]
6. Parabolic velocity profile
In laminar pipe flow, the velocity is not the same everywhere.
It is zero at the pipe wall and maximum at the center.
The profile is parabolic:
\[
\begin{aligned}
v(r) &= v_{\max}\left(1-\frac{r^2}{R^2}\right).
\end{aligned}
\]
Here \(r\) is the radial distance from the pipe centerline.
At the center, \(r=0\), so \(v=v_{\max}\).
At the wall, \(r=R\), so \(v=0\).
For Poiseuille flow,
\[
\begin{aligned}
v_{\max} &= 2\bar v.
\end{aligned}
\]
7. Wall shear stress
The pipe wall resists the motion of the fluid.
The wall shear stress for Poiseuille flow is
\[
\begin{aligned}
\tau_w &= \frac{\Delta P R}{2L}.
\end{aligned}
\]
It is larger when the pressure gradient is larger and when the pipe radius is larger.
8. Reynolds number check
Poiseuille’s law assumes laminar flow.
A common check is the Reynolds number:
\[
\begin{aligned}
Re &= \frac{\rho \bar v D}{\eta}.
\end{aligned}
\]
Here \(D=2R\) is the pipe diameter.
Flow in a pipe is usually laminar when \(Re\lesssim 2000\), transitional over a middle range, and turbulent at higher values.
The exact transition depends on pipe roughness, entrance conditions, and disturbances.
9. Solving for pressure drop
If the desired flow rate is known, rearrange Poiseuille’s law:
\[
\begin{aligned}
\Delta P
&= \frac{8\eta LQ}{\pi R^4}.
\end{aligned}
\]
This is useful when designing a pipe system to deliver a required flow.
10. Solving for radius
If the needed radius is unknown,
\[
\begin{aligned}
R
&= \left(\frac{8\eta LQ}{\pi\Delta P}\right)^{1/4}.
\end{aligned}
\]
Notice the fourth root. Even a large flow-rate change may require a less dramatic radius change than expected.
11. Solving for viscosity
Poiseuille flow can also be used to estimate viscosity from measured flow rate:
\[
\begin{aligned}
\eta
&= \frac{\pi R^4\Delta P}{8LQ}.
\end{aligned}
\]
This is the principle behind capillary viscometers.
12. Worked example
Suppose oil flows through a pipe with
\[
\begin{aligned}
\eta &= 0.50\ \mathrm{Pa\,s},\\
R &= 1.0\ \mathrm{cm}=0.010\ \mathrm{m},\\
L &= 3.0\ \mathrm{m},\\
\Delta P &= 10\ \mathrm{kPa}=10000\ \mathrm{Pa}.
\end{aligned}
\]
Use Poiseuille’s law:
\[
\begin{aligned}
Q
&= \frac{\pi R^4\Delta P}{8\eta L}\\
&= \frac{\pi(0.010)^4(10000)}{8(0.50)(3.0)}\\
&\approx 2.62\times10^{-5}\ \mathrm{m^3/s}.
\end{aligned}
\]
The pipe area is
\[
\begin{aligned}
A &= \pi(0.010)^2
\approx 3.14\times10^{-4}\ \mathrm{m^2}.
\end{aligned}
\]
Therefore the average speed is
\[
\begin{aligned}
\bar v
&= \frac{Q}{A}
\approx \frac{2.62\times10^{-5}}{3.14\times10^{-4}}
\approx 0.0833\ \mathrm{m/s}.
\end{aligned}
\]
The centerline maximum speed is
\[
\begin{aligned}
v_{\max}
&= 2\bar v
\approx 0.167\ \mathrm{m/s}.
\end{aligned}
\]
13. Formula summary
| Goal |
Formula |
Use |
| Flow rate |
\(Q=\frac{\pi R^4\Delta P}{8\eta L}\) |
Find volume flow rate through a circular pipe |
| Hydraulic resistance |
\(R_h=\frac{8\eta L}{\pi R^4}\) |
Use \(\Delta P=R_hQ\) |
| Average speed |
\(\bar v=Q/(\pi R^2)\) |
Find cross-sectional average velocity |
| Maximum speed |
\(v_{\max}=2\bar v\) |
Find centerline speed |
| Velocity profile |
\(v(r)=v_{\max}(1-r^2/R^2)\) |
Describe parabolic laminar flow |
| Pressure drop |
\(\Delta P=8\eta LQ/(\pi R^4)\) |
Find required pressure difference |
| Radius |
\(R=(8\eta LQ/(\pi\Delta P))^{1/4}\) |
Design pipe size for desired flow |
| Wall shear stress |
\(\tau_w=\Delta PR/(2L)\) |
Estimate wall stress in the pipe |
| Reynolds number |
\(Re=\rho\bar vD/\eta\) |
Check laminar-flow validity |
14. Assumptions and limitations
- The flow is steady.
- The flow is laminar.
- The fluid is Newtonian.
- The pipe is circular, rigid, and long.
- The velocity profile is fully developed.
- The pipe wall has no slip, so fluid speed at the wall is zero.
- Entrance effects, bends, valves, roughness, and turbulence are not included.
Key idea: Poiseuille flow has a parabolic velocity profile, and the volume flow rate depends on the fourth power of pipe radius.
