Categories: Fluid dynamics, Fluid mechanics, Physics.

# Reynolds number

The Navier-Stokes equations are infamously tricky to solve, so we would like a way to qualitatively predict the behaviour of a fluid without needing the flow $\va{v}$. Consider the main equation:

\begin{aligned} \pdv{\va{v}}{t} + (\va{v} \cdot \nabla) \va{v} = - \frac{\nabla p}{\rho} + \nu \nabla^2 \va{v} \end{aligned}

In this case, the gravity term $\va{g}$ has been absorbed into the pressure term: $p \to p\!+\!\rho \Phi$, where $\Phi$ is the gravitational scalar potential, i.e. $\va{g} = - \nabla \Phi$.

Let us introduce the dimensionless variables $\va{v}'$, $\va{r}'$, $t'$ and $p'$, where $U$ and $L$ are respectively a characteristic velocity and length of the system at hand:

\begin{aligned} \va{v} = U \va{v}' \qquad \va{r} = L \va{r}' \qquad t = \frac{L}{U} t' \qquad p = \rho U^2 p' \end{aligned}

In this non-dimenionsalization, the differential operators are scaled as follows:

\begin{aligned} \pdv{}{t} = \frac{U}{L} \pdv{}{t'} \qquad \quad \nabla = \frac{1}{L} \nabla' \end{aligned}

Putting everything into the main Navier-Stokes equation then yields:

\begin{aligned} \frac{U^2}{L} \pdv{\va{v}}{t'} + \frac{U^2}{L} (\va{v}' \cdot \nabla') \va{v}' = - \frac{U^2}{L} \nabla' p' + \frac{U \nu}{L^2} \nabla'^2 \va{v}' \end{aligned}

After dividing out $U^2/L$, we arrive at the form of the original equation again:

\begin{aligned} \pdv{\va{v}}{t'} + (\va{v}' \cdot \nabla') \va{v}' = - \nabla' p' + \frac{\nu}{U L} \nabla'^2 \va{v}' \end{aligned}

The constant factor of the last term leads to the definition of the Reynolds number $\mathrm{Re}$:

\begin{aligned} \boxed{ \mathrm{Re} \equiv \frac{U L}{\nu} } \end{aligned}

If we choose $U$ and $L$ appropriately for a given system, the Reynolds number allows us to predict the general trends. It can be regarded as the inverse of an “effective viscosity”: when $\mathrm{Re}$ is large, viscosity only has a minor role, but when $\mathrm{Re}$ is small, it dominates the dynamics.

Another way is thus to see the Reynolds number as the characteristic ratio between the advective term (see material derivative) to the viscosity term, since $\va{v} \sim U$:

\begin{aligned} \mathrm{Re} \approx \frac{\big| (\va{v} \cdot \nabla) \va{v} \big|}{\big| \nu \nabla^2 \va{v} \big|} \approx \frac{U^2 / L}{\nu U / L^2} = \frac{U L}{\nu} \end{aligned}

In other words, $\mathrm{Re}$ describes the relative strength of intertial and viscous forces. Returning to the dimensionless Navier-Stokes equation:

\begin{aligned} \pdv{\va{v}}{t'} + (\va{v}' \cdot \nabla') \va{v}' = - \nabla' p' + \frac{1}{\mathrm{Re}} \nabla'^2 \va{v}' \end{aligned}

For large $\mathrm{Re} \gg 1$, we can neglect the latter term, such that redimensionalizing yields:

\begin{aligned} \pdv{\va{v}}{t} + (\va{v} \cdot \nabla) \va{v} = - \frac{\nabla p}{\rho} \end{aligned}

Which is simply the main Euler equation for an ideal fluid, i.e. a fluid without viscosity.

## Stokes flow

A notable case is so-called Stokes flow or creeping flow, meaning flow at $\mathrm{Re} \ll 1$. In this limit, the Navier-Stokes equations can be linearized: since $\mathrm{Re}$ is the advective-to-viscous ratio, $\mathrm{Re} \ll 1$ implies that we can ignore the advective term, leaving:

\begin{aligned} \boxed{ \pdv{\va{v}}{t} = - \frac{\nabla p}{\rho} + \nu \nabla^2 \va{v} } \end{aligned}

This equation is called the unsteady Stokes equation. Usually, however, such flows are assumed to be steady (i.e. time-invariant), leading to the steady Stokes equation, with $\eta = \rho \nu$:

\begin{aligned} \boxed{ \nabla p = \eta \nabla^2 \va{v} } \end{aligned}

This equation is much easier to solve than the full Navier-Stokes equation thanks to being linear, and has some interesting properties, such as time-reversibility.

1. B. Lautrup, Physics of continuous matter: exotic and everyday phenomena in the macroscopic world, 2nd edition, CRC Press.
2. R. Fitzpatrick, Dimensionless numbers in incompressible flow, University of Texas.