fluid dynamics - 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?

Answer

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.