I'm trying to derive a scaling law for the Reynolds number, to get a better understanding of how it changes for microsystem applications, but I'm getting stuck. From textbook tables I should end up with $l^2$, but can't get there. The goal is to find a simplistic relation of Reynolds number and the system dimension.
This is my reasoning so far:
Considering a tube
$Re = \frac{\rho v 2R}{\mu}$$$Re = \frac{\rho v 2R}{\mu}$$
$v = \frac{Q}{A} = \frac{Q}{\pi R^2}$$$v = \frac{Q}{A} = \frac{Q}{\pi R^2}$$
From Hagen-Poiseuille: $Q = \frac{\delta P \pi R^4}{8 \mu L}$,$$Q = \frac{\delta P \pi R^4}{8 \mu L},$$ therefore:
$v = \frac{\frac{\delta P \pi R^4}{8 \mu L}}{\pi R^2} = \frac{\delta P \pi R^4}{8 \mu L \pi R^2} = \frac{\delta P R^2}{8 \mu L}$$$v = \frac{\frac{\delta P \pi R^4}{8 \mu L}}{\pi R^2} = \frac{\delta P \pi R^4}{8 \mu L \pi R^2} = \frac{\delta P R^2}{8 \mu L}$$
Replacing $v$ on $Re$:
$Re = \frac{\frac{\rho \delta P R^2 2R}{8 \mu L}}{\mu} = \frac{\rho \delta P R^3 2}{8 \mu^2 L}$$$Re = \frac{\frac{\rho \delta P R^2 2R}{8 \mu L}}{\mu} = \frac{\rho \delta P R^3 2}{8 \mu^2 L}$$
From this, if I consider the $\delta P$ as constant with the size scaling, $\rho$ and $\mu$ are material properties that are constant at different scales, I get:
$Re \sim \frac{R^3}{L}$ $$Re \sim \frac{R^3}{L}$$
I can only think that I could consider it as $\frac{l^3}{l} = l^2$ but doesn't seem right. What
What am I missing here?
