# Why is the Reynolds number “the way it is?” Why is its order the way it is?

Why is the Reynolds number “the way it is?” Why is its order the way it is?

I'm not sure if this is an appropriate question for this context, but I would like more intuition on this matter and so I'm using a "naive" approach. What goes on in the transition from laminar to turbulent flow?

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• – Kyle Kanos Aug 5 '15 at 15:39

The Reynolds number, with $\rho$ the density, $u$ the velocity magnitude, $\mu$ the viscosity and $L$ some characteristic length scale (e.g. channel height or pipe diameter) is given by $$\text{Re}=\frac{\rho~u~L}{\mu}.$$ This is a dimensionless relation of the ratio of inertial forces ($\rho u u$) to viscous forces ($\mu\frac{u}{L}$). It therefore signifies the relative importance of inertial forces to viscous forces.

In the laminar regime, viscous forces are dominant (i.e. $\text{Re}\ll 1$) while in the turbulent regime, inertial forces are dominant (i.e. $\text{Re}\gg 1$). In the transition from laminar to turbulent flow, inertial forces start to overtake viscous forces which simply means that viscosity can no longer smooth out velocity gradients into smooth laminar flow (except for near a boundary where they are still important) and inertia of the flow causes it to 'trip' over itself causing vortices and in general chaotic behaviour associated with turbulence.

The Reynolds number is the way it is by a dimensional analysis of the hydrodynamic equations which govern the flow (i.e. the Navier-Stokes equations). Lets assume a steady flow (i.e. $\partial_t\mathbf{u}=0$) $$\rho~\mathbf{u}\cdot\mathbf{\nabla}\mathbf{u}=-\mathbf{\nabla}p + \mu~\mathbf{\nabla}^2\mathbf{u}.$$

Non-dimensionalizing this by defining $\bar{x}=\frac{x}{L}$, $\bar{\mathbf{u}}=\frac{\mathbf{u}}{U}$ and $\bar{p}=\frac{p}{P}$ where $U$ and $P$ are characteristic velocity and pressure scales respectively, we get:

$$\rho~\frac{U^2}{L}~\bar{\mathbf{u}}\cdot\bar{\mathbf{\nabla}}\bar{\mathbf{u}}=-\frac{P}{L}~\bar{\mathbf{\nabla}}\bar{p} + \mu \frac{U}{L^2}~\bar{\mathbf{\nabla}}^2\bar{\mathbf{u}}$$

we can simplify this by dividing through by $\mu\frac{U}{L^2}$ and defining $P=\mu\frac{U}{L}$ to get:

$$\text{Re}~\bar{\mathbf{u}}\cdot\bar{\mathbf{\nabla}}\bar{\mathbf{u}}=-\bar{\mathbf{\nabla}}\bar{p} + \bar{\mathbf{\nabla}}^2\bar{\mathbf{u}}$$

which reveals the Reynolds number. For $\text{Re}\ll 1$, where viscosity dominates, we see that the convective term on the left becomes negligible compared to the pressure gradient and viscous stress tensor on the right.

For $\text{Re}\gg 1$ we can do the same except we then need to divide by $\rho\frac{U^2}{L}$ and define $P=\rho U^2$ to get:

$$\bar{\mathbf{u}}\cdot\bar{\mathbf{\nabla}}\bar{\mathbf{u}}=-\bar{\mathbf{\nabla}}\bar{p} + \frac{1}{\text{Re}}\bar{\mathbf{\nabla}}^2\bar{\mathbf{u}}$$

Now the viscous stress tensor on the right becomes negligible compared to the pressure gradient and the convection term on the left.

Note that the characteristic pressure scale $P$ was defined in a viscous and inertial scale depending on which regime we were in. This is necessary as it is required that the dimensionless pressure gradient is of the same order as at least one other term.

Note also that real turbulence is inherently unsteady, my treatment above of the steady Navier-Stokes equations for different regimes was to focus on the role of the Reynolds number and simply to keep it as short as possible.

• I'm curious what steady state turbulent flow looks like :) – Bernhard Aug 5 '15 at 15:48
• @Bernhard I knew someone would comment on that :)... You are right, turbulence is inherently unsteady of course. But for the sake of not having to define a characteristic time scale as well and keeping it as simple as possible, I have invented steady turbulence. – nluigi Aug 5 '15 at 15:53
• @Bernhard Actually, steady state turbulence is an important concept. It means statistically steady, e.g. the energy spectrum or any other statistical measure of the flow is steady. It does, however, require that there be some forcing so the energy being put in can be balanced by the energy dissipated by the viscosity, so nlooije's equation needs to be supplemented with another term representing that force. (Sorry, nlooije, you didn't invent it! ;) ) – pwf Aug 5 '15 at 17:09

The question you ask is actually the central question of a huge sub-discipline of fluid dynamics. Some have even referred to it as "the last great unsolved problem in classical physics." If you get a complete answer, please let me know! (And don't tell anyone else. Just keep between us, eh?)

Generally, there are always small fluctuations in any flow, even if the flow is very laminar. (If nothing else, there are at least thermal fluctuations.) In laminar flow regimes, which generally means at low Re, those fluctuations dampen out quickly. At higher Re, viscosity becomes relatively less important, and those fluctuations can become unstable. For example, the sheared flow near the wall of a pipe can roll up into vortex tubes, and if they don't dampen out, they can get stretched into horseshoe and then hairpin shapes, and now you have a small structure with high local shear that has moved out away from the wall a bit. The process can repeat in the neighborhood of those structures, and if it grows and multiplies quickly, you've got a turbulent flow. That's just one example, but in general the path from laminar flow to turbulence involves one or more instabilities that grow, interact, multiply, and expand faster than viscosity can dampen them out or than the mean flow can wash them away.

There is (yet) no unified theory of instability, and so no universal answer to when or why turbulence is generated. It's a problem that many great minds, including Heisenberg and Feynman, pondered, but it's still open. Even in a particular situation, like pipe flow, the details are still poorly understood. But you can see why, in general, higher Re tends to make turbulence more likely - it's because at higher Re there's less damping, and so more tendency for instability. You can also see why the transition to turbulence depends sensitively on the details of the flow; for example, if the walls of a pipe are even a little bit rough, it's easier for instabilities to get excited and for turbulence to form than if they are extremely smooth.

• I want in on that prize money too :)! – nluigi Aug 5 '15 at 19:40