7
$\begingroup$

I'm studying fluid dynamics and recently the formula $Re=\frac{\rho vd}{\eta}$ was presented to me. I'm curious to know how Reynolds came up with this relations between this different variables.

Did $Re=\frac{\rho vd}{\eta}$ result from the formula $Re = \frac{\text{Inertial Forces}}{\text{Viscous Forces}}$ or did this last equation came up as an intuition/ physical interpretation after the Reynolds number was first discovered?

I tried to find the history behind Reynolds "scientific procedure", how he found the number, but I wasn't successful.

$\endgroup$
  • 1
    $\begingroup$ I know this is dusting off an old topic, but today the Philosophical Transactions of the Royal Society A published a review (open access) of Reynolds 1895 paper that first looked at the transition to turbulence occurring when a dimensionless number (now Reynolds number) exceeded a particular value. $\endgroup$ – tpg2114 Mar 6 '15 at 12:46
12
$\begingroup$

There's no magic behind it. It was done by non-dimensionalizing the momentum equation in the Navier-Stokes equations.

Starting with:

$$\frac{\partial u_i}{\partial t} + u_j\frac{\partial u_i}{\partial x_i} = -\frac{1}{\rho}\frac{\partial P}{\partial x_i} + \nu \frac{\partial^2 u_i}{\partial x_i x_j}$$

which is the momentum equation for an incompressible flow. Now you non-dimensionalize things by choosing some appropriate scaling values. Let's look at just the X-direction equation and assume it's 1D for simplicity. Introduce $\overline{x} = x/L$, $\overline{u} = u/U_\infty$, $\tau = tU_\infty/L$, $\overline{P} = P/(\rho U_\infty^2)$ and then substitute those into the equation. You get:

$$ \frac{\partial U_\infty \overline{u}}{\partial \tau L/U_\infty} + U_\infty\overline{u}\frac{\partial U_\infty \overline{u}}{\partial L \overline{x}} = - \frac{1}{\rho}\frac{\partial \overline{P}\rho U_\infty^2}{\partial L\overline{x}} + \nu \frac{\partial^2 U_\infty \overline{u}}{\partial L^2 \overline{x}^2} $$

So now, you collect terms and divide both sides by $U_\infty^2/L$ and you get:

$$ \frac{\partial \overline{u}}{\partial \tau} + \overline{u}\frac{\partial \overline{u}}{\partial \overline{x}} = -\frac{\partial \overline{P}}{\partial \overline{x}} + \frac{\nu}{U_\infty L}\frac{\partial^2 \overline{u}}{\partial \overline{x}^2}$$

Where now you should see that the parameter on the viscous term is $\frac{1}{Re}$. Therefore, it falls out naturally from the definitions of the non-dimensional parameters.

The intuition

There's some other ways to come up with it. The Buckingham Pi theorem is a popular way (demonstrated in Floris' answer) where you collect all of the units in your problem in this case $L, T, M$ and find a way to combine them into a number without dimension. There is one way to do that, which ends up being the Reynolds number.

The interpretation of inertial to viscous forces comes from looking at the non-dimensional equation. If you inspect the magnitude of the terms, namely the convective (or inertial term) and the viscous term, the role of the number should be obvious. As $Re \rightarrow 0$, the magnitude of the viscous term $\rightarrow \infty$, meaning the viscous term dominates. As $Re \rightarrow \infty$, the viscous term $\rightarrow 0$ and so the inertial terms dominates. Therefore, one can say that the Reynolds number is a measure of the ratio of inertial forces to viscous forces in a flow.

$\endgroup$
6
$\begingroup$

The way it was explained to me: you start by thinking of all the possible factors that could play in drag (size, velocity, density, viscosity, ...); then you do dimensional analysis and find dimensionless combinations - these tend to be "special" since they remain constant over different scales of time and space.

Reynolds number is one such combination. The interpretation follows from inspection.

Here's how that is done:

size: L
velocity: L/T
density: M/L^3
viscosity: M/LT

Now we look for a combination that is dimensionless. First we eliminate T by taking the ratio of velocity and viscosity:

velocity / viscosity = vv = L/T / (M/LT) = L^2/M

Next we eliminate M:

density * vv = dvv = M/L^3 * L^2 / M = 1/L

Finally we eliminate L:

dvv * size = 1/L * L = 1

So the final dimensionless expression is

density * velocity * size / viscosity

which is Reynold's number...

It was pointed out by tpg2114 that this is an application of the Buckingham Pi theorem . I use it all the time, and never knew it had a name...

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.