KAHIBARO
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Steady Flow Through a Narrow Tube
Poiseuille's law describes how a viscous fluid flows through a long, narrow cylindrical tube when the flow is smooth and steady. It gives the volume of fluid passing through the tube per unit time in terms of the pressure difference, the tube size, and the fluid's viscosity.
This law is very important for understanding flow in pipes, small channels, needles, and blood vessels. It shows that fluid flow can be very sensitive to the radius of the tube.
The Physical Situation
Imagine a liquid moving through a horizontal tube of radius $r$ and length $L$. A higher pressure at one end pushes the fluid toward the lower pressure at the other end. Because the fluid is viscous, layers of fluid rub against each other and against the tube wall.
The fluid right at the wall is effectively at rest, while fluid nearer the center moves faster. This produces a velocity profile across the tube.
Statement of Poiseuille's Law
For laminar flow of an incompressible Newtonian fluid in a cylindrical tube, Poiseuille's law is
$$
Q = \frac{\pi r^4 \Delta P}{8 \eta L}
$$
where $Q$ is the volume flow rate, $\Delta P = P_1 - P_2$ is the pressure difference, $r$ is the tube radius, $L$ is the tube length, and $\eta$ is the dynamic viscosity.
Poiseuille's law:
$$
Q = \frac{\pi r^4 \Delta P}{8 \eta L}
$$
Important consequences:
$$
Q \propto \Delta P
$$
$$
Q \propto r^4
$$
$$
Q \propto \frac{1}{\eta}
$$
$$
Q \propto \frac{1}{L}
$$
Meaning of the Formula
The law tells us several important things. If the pressure difference doubles, the flow rate doubles. If the fluid is more viscous, the flow rate becomes smaller. If the tube is longer, the fluid has more resistance and the flow rate decreases.
The most striking feature is the dependence on the fourth power of the radius. A small change in radius causes a large change in flow rate. For example, if the radius is doubled, the flow rate becomes
$$
2^4 = 16
$$
times larger, if everything else stays the same.
Average Speed Form
The volume flow rate is related to the average speed $v_{\text{avg}}$ by
$$
Q = A v_{\text{avg}} = \pi r^2 v_{\text{avg}}
$$
Combining this with Poiseuille's law gives
$$
v_{\text{avg}} = \frac{r^2 \Delta P}{8 \eta L}
$$
This form is useful when we want to know how fast the fluid moves on average along the tube.
Average speed in Poiseuille flow:
$$
v_{\text{avg}} = \frac{r^2 \Delta P}{8 \eta L}
$$
Flow Resistance
Poiseuille's law can be written in a form similar to Ohm's law in electricity. Rearranging gives
$$
\Delta P = R Q
$$
where the flow resistance is
$$
R = \frac{8 \eta L}{\pi r^4}
$$
This resistance becomes very large for narrow tubes and very small for wide tubes.
Velocity Profile in the Tube
The fluid does not move with the same speed everywhere in the cross section. It moves fastest at the center and slows down toward the walls. The velocity profile is parabolic.
If $\rho$ is the distance from the center axis, the speed is
$$
v(\rho) = \frac{\Delta P}{4 \eta L}\left(r^2 - \rho^2\right)
$$
At the center, where $\rho = 0$, the speed is maximum:
$$
v_{\max} = \frac{\Delta P \, r^2}{4 \eta L}
$$
At the wall, where $\rho = r$, the speed is zero.
Also,
$$
v_{\max} = 2 v_{\text{avg}}
$$
for Poiseuille flow in a circular tube.
Conditions for Using the Law
Poiseuille's law does not apply to every fluid flow. It works only under certain conditions. The flow must be laminar, not turbulent. The fluid should be incompressible and Newtonian. The tube should be cylindrical and have a constant radius. The flow should be steady, and entrance or edge effects should be small compared with the tube length.
If the flow becomes turbulent, this law no longer gives the correct result.
Poiseuille's law is valid only when the flow is smooth and laminar in a long, narrow cylindrical tube.
Units
The SI units in Poiseuille's law are shown below.
| Quantity | Symbol | SI unit |
|---|---|---|
| Volume flow rate | $Q$ | $\text{m}^3/\text{s}$ |
| Pressure difference | $\Delta P$ | $\text{Pa}$ |
| Radius | $r$ | $\text{m}$ |
| Length | $L$ | $\text{m}$ |
| Dynamic viscosity | $\eta$ | $\text{Pa}\cdot\text{s}$ |
A unit check confirms consistency:
$$
Q = \frac{\pi r^4 \Delta P}{8 \eta L}
\sim
\frac{\text{m}^4 \cdot \text{Pa}}{(\text{Pa}\cdot\text{s})\cdot \text{m}}
=
\frac{\text{m}^3}{\text{s}}
$$
Example
Suppose water flows through a tube with radius $r = 1.0 \times 10^{-3}\,\text{m}$, length $L = 0.50\,\text{m}$, viscosity $\eta = 1.0 \times 10^{-3}\,\text{Pa}\cdot\text{s}$, and pressure difference $\Delta P = 2000\,\text{Pa}$.
Using Poiseuille's law,
$$
Q = \frac{\pi (1.0 \times 10^{-3})^4 (2000)}{8 (1.0 \times 10^{-3})(0.50)}
$$
Since
$$
(1.0 \times 10^{-3})^4 = 1.0 \times 10^{-12}
$$
we get
$$
Q = \frac{\pi \times 10^{-12} \times 2000}{4.0 \times 10^{-3}}
= \pi \times 5.0 \times 10^{-7}
$$
so
$$
Q \approx 1.57 \times 10^{-6}\,\text{m}^3/\text{s}
$$
This is about $1.57\,\text{cm}^3/\text{s}$.
Why the Radius Matters So Much
The factor $r^4$ is one of the most important results in fluid mechanics. If a passage narrows slightly, the flow rate can drop dramatically. This is why clogged pipes, narrow needles, and constricted blood vessels can strongly affect fluid transport.
For comparison:
| Radius change | Flow rate change |
|---|---|
| $r \to 2r$ | $Q \to 16Q$ |
| $r \to 3r$ | $Q \to 81Q$ |
| $r \to r/2$ | $Q \to Q/16$ |
A small decrease in tube radius causes a very large decrease in flow rate because
$$
Q \propto r^4
$$
Summary
Poiseuille's law gives the flow rate of a viscous fluid through a cylindrical tube under laminar conditions. The central formula is
$$
Q = \frac{\pi r^4 \Delta P}{8 \eta L}
$$
It shows that flow increases with pressure difference, decreases with viscosity and tube length, and depends very strongly on tube radius. This makes it one of the most useful laws for understanding slow viscous flow in narrow tubes.