Reynolds Number and Flow Transition, Through Visualization

The same air sits still in a nostril and churns in a vent. Nobody swapped the fluid. Velocity, length, and viscosity got packed into one number. That number is the Reynolds number.

Drag the Re slider to the right and the velocity vectors, starting from the default of 100, start to lose their direction behind the obstacle.

Hold the slider under 50 and the arrows wrap the body, then reattach at once. Push past 1000 and wake cells point every which way.

What Is the Reynolds Number?#

The definition is short. Inertia divided by viscosity.

$Re = \frac{\rho U L}{\mu} = \frac{U L}{\nu}$

The symbols stay fixed.

• $\rho$: fluid density $[\text{kg/m}^3]$
• $U$: characteristic velocity $[\text{m/s}]$
• $L$: characteristic length $[\text{m}]$
• $\mu$: dynamic viscosity $[\text{Pa·s}]$
• $\nu = \mu/\rho$: kinematic viscosity $[\text{m}^2/\text{s}]$

Flow Regimes#

The pipe value 2300 is a different threshold from the cylinder picture. The critical value where a jet breaks, and the layered structure of that boundary, continues in the critical Reynolds number where a jet breaks up.

The Physics of Transition#

The first crack from laminar flow into turbulence is the Kelvin–Helmholtz instability. Two layers at different speeds touch, a small wave rolls up, and a vortex is born. That vortex hands energy to smaller ones.

$\frac{\partial \mathbf{u}}{\partial t} + (\mathbf{u} \cdot \nabla)\mathbf{u} = -\frac{1}{\rho}\nabla p + \nu \nabla^2 \mathbf{u}$

In the Navier–Stokes equations, the nonlinear term $(\mathbf{u} \cdot \nabla)\mathbf{u}$ on the left carries inertia, while $\nu \nabla^2 \mathbf{u}$ on the right damps via viscosity. The Reynolds number measures the relative size of these two.

Velocity-Field Visualization: How Flow Changes with Re#

The picture above splits into three bands. At the low end the viscous term erases a disturbance where it appears. In the middle, symmetry behind the obstacle breaks first. Past 1000 the vortices stay in the cells. Do not put the pipe-flow 2300 and this obstacle picture on the same tick marks. Change the shape and the threshold moves.

Streamlines Past a Cylinder: the Kármán Vortex Street#

A periodic vortex release behind a cylinder is the Kármán vortex street. The frequency is packed into the Strouhal number.

The shedding frequency is normalized by the Strouhal number $St$:

$St = \frac{f D}{U_\infty}$

Where $f$ is shedding frequency, $D$ is cylinder diameter, $U_\infty$ is freestream velocity. For a cylinder, $St \approx 0.2$ stays roughly constant over $100 < Re < 10^5$.

Raise the speed slider and the same cylinder wake rocks left and right faster.

Watch the front stagnation point first, where the streamlines split. Then watch the rear symmetry break, and vortices drop off on alternating sides.

Reynolds Number in Numerics: Grid Requirements#

DNS has to put cells down to the smallest length, the Kolmogorov scale.

$\eta = \left(\frac{\nu^3}{\varepsilon}\right)^{1/4}$

The cell count grows with Re like this.

$N \sim Re^{9/4}$

So a 10× increase in $Re$ demands roughly 178× more grid points. This is why practical turbulence simulations rely on RANS or LES models.

The 9/4 Exponent and a Five-Diameter Cycle#

If St sits at 0.2, one cycle carries the fluid a diameter divided by 0.2. One over 0.2 is 5: five diameters. From Re 100 to 10 to the fifth that ratio barely moves, so a ruler reads the wake spacing. A pattern that repeats once every five diameters is already in this band.

Doubling Re scales the grid by 2 to the 9/4. Two squared is 4. Two to the 1/4 is about 1.19. Four times 1.19 is about 4.8, so cells grow about fivefold, not twofold. A factor of ten is steeper. Ten to the 9/4 is 100 times 10 to the 0.25. Ten to the 0.25 is about 1.78, and 100 times 1.78 is 178. That product is the 178-fold mesh.

Halving the Kolmogorov length uses the same power. Viscosity cubed over dissipation must fall to one sixteenth, because 2 to the fourth is 16. With viscosity fixed, dissipation rises by 16. One step toward smaller eddies costs that much. The 9/4 looks gentle only until a tenfold Re is on the table.

Is a 178× Mesh the Re You Can Afford?#

If the run that raises Re by a decade is a DNS, the mesh grows as $Re^{9/4}$. A cluster that cannot take the 178-fold increase should move to RANS or LES. On a cylinder, check the wake spacing against five diameters first. Off that spacing, the case sits outside the St 0.2 band.

The pipe threshold is not the 1000 on the obstacle picture. The band from 2300 to 4000 is 1700 wide. Before paying for 178 times the mesh, look at what happens when the model coefficient itself comes out negative. The averaging step where 49.9% of dynamic coefficients came out negative is the page for that decision.