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According to Poiseuille's law, the effective resistance of a tube is inversely proportional to the fourth power of its radius (as given by the following equation).

$$R = \frac{ 8 \eta \Delta x}{\pi r^4} $$

I can go through the derivation of the law but is there an intuitive explanation for why it is the fourth power of radius, or a thought experiment that might make it more obvious?

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No-slip condition requires that the velocity is $0$ at the boundaries, thus you can see intuitively that the velocity profile has to behave in $a^2-r^2$ (at least as leading terms in its expansion), with $a$ the radius of the tube.

To get the hydraulic resistance, you need the mass flux: integrating $a^2$ and $r^2$ on a section (circle) you'll get terms in $a^4$ : the mass flux is in pressure drop times $a^4$, hence the dependence of the hydraulic resistance.

You can maybe conceptualize it as $a^{2+d}$, where 2 comes from the quadratic profile and $d$ is the dimension of space : in a Hele-Shaw flow you'll have a flux in $h^3$.

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Non-turbulent flow through a pipe of radius $r$ is proportional to the cross section $\pi r^2$ times the average flow velocity. The latter is proportional to the time $t$ it takes for the momentum in the fluid to diffuse to the wall: $t \approx (1/\nu) r^2$. Here $\nu$ is the kinematic viscosity ($\eta/\rho$). It follows that the flux (area times velocity) is proportional to $r^4$.

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