Poiseuille's law states that the rate of flow of water is proportionate to $r^4$ where $r$ is the radius of the pipe. I don't see why.

Intuitively I would expect rate of flow of fluid to vary with $r^2$ as the volume of a cylinder varies with $r^2$ for a constant length. The volume that flows past a point is equal to rate of flow fluid, and thus it should vary with $r^2$.

I cannot understand the mathematical proof completely. Is there an intuitive explanation for this?

  • $\begingroup$ The Hagen Pousseuille equation (if I've spelt that right) will answer your question in the laminar flow regime. $\endgroup$ Sep 13, 2016 at 20:18

3 Answers 3


The flow rate is the average velocity times the area.

If the velocity was constant, you would get a flow rate that scaled with $r^2$ (the area). But the velocity goes up for larger pipes - in fact, velocity scales with the square of the radius. And the product of these two squares gives us the 4th power relationship.

Let's break this into a few steps:

You may know that the velocity profile is a parabola - that is, if the velocity at the center is $v_0$, then the velocity at a distance $r$ from the center of a pipe with radius $R$ is given by

$$v(r) = v_0\left(1-\left(\frac{r}{R}\right)^2\right)$$

The mean velocity is exactly half the maximum velocity - you can see the proof for this in my earlier answer. So we just need to figure out how the maximum velocity scales with $r^2$. Once we know the velocity profile is quadratic, this is easy - because if we make the pipe a little bit bigger, the profile continues to follow the same parabolic shape with the same curvature.

It remains to prove for ourselves that the parabolic velocity profile is correct. This follows from the fact that the shear stress in a fluid is proportional to the viscosity times the velocity gradient. Looking at an annulus of liquid at a distance $r$ from the center, if there is a velocity gradient $\frac{dv}{dr}$ we know that the total force on the liquid inside the annulus is the pressure times the area, or $F = P\cdot A = P \cdot \pi r^2$. We also know this must equal the force due to the shear, which is the force per unit length of the annulus multiplied by the length of the circumference, so

$$\pi r^2 P = 2\pi r \frac{dv}{dr}\eta$$

It's easy to see this is a differential equation in $v$, with a parabolic solution.

$$\frac{P}{2\eta} r dr = dv\\ v = \frac{Pr^2}{4\eta} + C $$

For the boundary condition $v=0$ at $r=R$, we find that $C=-\frac{PR^2}{4\eta}$. With simple manipulation we find the same expression we had before with

$$v_0 = \frac{PR^2}{4\eta}$$.

So there you have it. Mean velocity is half the peak velocity, and peak velocity goes as the square of the radius; multiplied by the area we have the $R^4$ relationship.

  • $\begingroup$ Can you elaborate more on how $v_0 \propto R^2$? $\endgroup$
    – Ali
    Oct 10, 2017 at 15:03
  • $\begingroup$ @Ali I am not sure what I can add beyond the paragraph "It remains to prove..." which shows the equation that allows you to get velocity as a function of radius. Since the curvature is given by $P$ and $\eta$, the shape of the parabola will be the same for all $r$ - but if you double $r$, you will quadruple the height of the parabola (when the ends are kept at 0). Do you see it now or do I need to actually write the math into the answer? $\endgroup$
    – Floris
    Oct 10, 2017 at 15:13
  • $\begingroup$ I do indeed see it. Hopefully you don't think I'm nitpicking, I just found it missing from your nice answer. Maybe an equation like $v_0 = \frac{P}{\eta}R^2$ following the differential equation would be useful. Another comment, parabolas don't have a constant curvature along them, unless you mean the curvature at its vertex! $\endgroup$
    – Ali
    Oct 10, 2017 at 15:35
  • $\begingroup$ @Ali I used "curvature" when I really meant "second derivative" - sloppy of me. I have added the details as requested. $\endgroup$
    – Floris
    Oct 10, 2017 at 17:21

Intuitively, the rate goes up by faster than r squared because for laminar flow, there is no flow at the walls of the pipe. So the extra r squared term is due to a larger pipe having more area away from its walls.


Here is a really intuitive approach, not a rigorously correct one; but you might like it as a help to your memory.

Start with a square pipe with side one unit. Now put four of these pipes in a bundle. This gives you a pipe with side two units, a horizontal partition, and a vertical partition, a cross sectional area of four units, and a flow rate four times that of the smaller pipe.

Next, take out the horizontal partition. This makes the velocity profile more like the letter V on its side, instead of two little v's. This doubles the flow rate. Last, take out the vertical partition. This adds another dimension to the velocity profile, doubling the flow once more.

So the new pipe has twice the diameter of the old one, and the flow rate is 4 x 2 x 2 = 2^4 that of the old one.


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