Navier–Stokes: Flow Past a Cylinder

Set the Reynolds number and the page solves the Navier–Stokes equations for flow past a cylinder while you watch, coloured by which way the fluid spins. The trace under the picture is the sideways velocity three diameters downstream. It tells you what the picture cannot near the threshold: whether the wake is settling, dying away or swinging for good.

Worked example — Navier–Stokes: Flow Past a Cylinder

Viscosity the solver uses, in lattice units	0.01000
Relaxation time τ	0.5300
Swing the wake settles at	0.636 U
Strouhal number, from the swing's period	0.1690
Drag coefficient, time mean	1.489
Of which pressure	1.157
Share of the drag that is skin friction	22%
Strouhal number in Roshko's wind tunnel	0.1671

These are the values the calculator shows when the page opens. Change any input above to recompute them.

Loading interactive simulation...

Lesson

The theory — Navier–Stokes: Flow Past a Cylinder

The Navier–Stokes equations are Newton's second law for a fluid: each small parcel accelerates under the pressure pushing on it and the viscous drag of its neighbours. For a fluid that cannot be compressed they read ∂u/∂t + (u·∇)u = −∇p/ρ + ν∇²u and ∇·u = 0, where u is the velocity, p the pressure, ρ the density and ν the kinematic viscosity.

What each symbol means

Re: the Reynolds number UD/ν, from the flow speed U, the cylinder's diameter D and the kinematic viscosity ν. Step 1 shows it is the only constant left in the equations
St: the Strouhal number fD/U, the shedding frequency f in units of U/D
C_D: the drag coefficient: the drag on a unit length of cylinder divided by ½ρU²D
A: the swing's amplitude: half the peak-to-peak sideways velocity three diameters behind the cylinder, as a fraction of U
ω: the vorticity ∂v/∂x − ∂u/∂y, twice the rate at which a parcel of fluid spins. It is the colour in the picture

Where the formula comes from

Why one slider is enough. Measure lengths in diameters, speeds in units of U and time in units of D/U: put u = U u′, x = D x′, t = (D/U) t′ and p = ρU² p′. Every term on the left of the momentum equation then carries a factor U²/D, and so does the pressure term; the viscous term carries νU/D². Divide through by U²/D and the equation is the same as before, written in the primed quantities, except that ν has become ν/(UD) = 1/Re. Two flows with the same Re are the same flow at different sizes and speeds, so the page needs one control, not four.
How the page tells settling from dying. Call the swing's amplitude A. While it is small it grows or shrinks exponentially, dA/dt = σA, with σ positive above the threshold and negative below it, and close to the threshold σ is proportional to Re − Re_c. Growth cannot go on for ever, and the first term that can stop it, for a swing that looks the same flipped top to bottom, is cubic: dA/dt = σA − ℓA³, Landau's equation. Divided by A it is a straight line in A², d ln A/dt = σ − ℓA². The page fits that line to each run's cycles and reads σ off it, which is how it knows at Re 48 that a shrinking swing is settling rather than dying. Setting dA/dt = 0 gives the swing the wake settles at, A∞² = σ/ℓ: a straight line in Re that reaches zero at Re_c. That is the line in the second panel.
Zero drag without viscosity. Without viscosity the flow round a cylinder of radius R is ψ = U(r − R²/r) sin θ, and on the surface its speed is 2U sin θ: zero at the front and back, twice the free stream at the sides. Bernoulli, p + ½ρv² constant along the flow, turns that into a pressure, C_p = (p − p∞)/(½ρU²) = 1 − 4 sin²θ. The drag is that pressure pushing on the surface, added up all the way round: C_D = ∫₀^π C_p cos θ dθ = ∫₀^π cos θ dθ − 4∫₀^π sin²θ cos θ dθ = 0 − [4 sin³θ/3]₀^π = 0. The front half and the back half cancel exactly. The page integrates the solver's own pressure the same way and prints it as the pressure part of the drag.
The frequency. Scaled as in step 1, the shedding frequency f becomes St = fD/U, and like everything else in the scaled equations it can depend on Re and nothing else. Once St is known at a Reynolds number, f = St·U/D for any fluid, speed and diameter that share it. Roshko fitted St = 0.212(1 − 21.2/Re) to his wind-tunnel measurements between Re 50 and 150; this solver's St is 4% above it at Re 55, closing to 1% at Re 100.

Assumes: Two dimensions: an infinitely long cylinder with nothing varying along it. A fluid that cannot be compressed, which the solver approximates by running at about a sixth of its own speed of sound. A channel twenty diameters wide with its top joined to its bottom, so the cylinder has neighbours twenty diameters away, which adds a few percent to the drag and the frequency.
Breaks when: Above Re 188.5 a real wake grows waves along the cylinder (Barkley and Henderson, 1996) and the 2D answer stops describing it. Near the threshold everything slows down: at Re 47 the swing is still dying after 300 D/U, and would need more than 2,000 D/U more to fade out, which is why the page reads the trend with Landau's equation instead of waiting.

Below about Re 47 the wake settles; above it, the same equations swing for ever 🖖

Press Re 44 and then Re 48. The cylinder is symmetric, the flow arriving at it is symmetric, and at both Reynolds numbers the equations have a symmetric answer: two mirror-image eddies, sitting still. At 44 the kicked wake goes back to it. At 48 it never does. The symmetric answer still exists, but it has become unstable, and the flow leaves it at the smallest excuse. Here the excuse is the page's own: a sideways nudge in the inflow after every change, and a cylinder set a third of a cell off the grid's centre line. In a water channel it is any vibration at all. Camichel's group in Toulouse measured the threshold at 47. This solver's growth rate changes sign between 47 and 48, at about 47.4, and the fit in the second panel gives 47.5.

Without viscosity a cylinder would feel no drag 🖖

Take viscosity out of the equations and the flow round a cylinder can be solved with a pencil (step 3 below). The pressure on the back half comes out as the mirror image of the pressure on the front, push and pull cancel, and the drag is exactly zero. d'Alembert published that in 1752 and it has been called a paradox since, because every real body feels drag. The first panel shows where it goes wrong. At Re 100 the frictionless pressure climbs back to +1 at the back of the cylinder; the solver's stops at −0.94, because the flow leaves the surface before it can recover. That missing pressure is more than three quarters of the drag. The rest is the fluid's grip on the surface: 22% of the drag at Re 100, 40% at Re 20.

The note a wire sings is not a fixed number 🖖

St = fD/U is the shedding frequency in units of the time the flow takes to pass one diameter. It is often rounded to 0.2. On this page it climbs from 0.128 at Re 50 to 0.169 at 100 and 0.192 at 180, and Roshko's wind-tunnel fit, the green line, climbs with it: 0.167 at Re 100. The 0.2 belongs to faster flows. Roshko's fit for Re 300 to 2,000 gives 0.203 to 0.211, and there the wake is three-dimensional, which a 2D solver cannot follow. A real wake goes three-dimensional at Re 188.5, which is why the slider stops at 180.

Problem solved in full

What note does a wire sing in the wind? 4 steps
A taut wire 3 mm thick stands in an 8 m/s wind. Vortices shed from it alternately on each side, and the wire hums at the shedding frequency. What frequency? Air at 20 °C has a viscosity of 1.813×10⁻⁵ Pa·s and a density of 1.204 kg/m³.

Open this problem in the tool
1. First the Reynolds number, from the kinematic viscosity ν = η/ρ:
  
  $ν = \frac{η}{ρ} = \frac{1 . 813 \times 1 0 ^{- 5} Pa s}{1 . 204 kg/ m ^{3}} = 1.506 \times 1 0^{- 5} m^{2} /s, Re = \frac{U D}{ν} = \frac{8 m/s \times 0 . 003 m}{1 . 506 \times 1 0 ^{- 5} m ^{2} /s} = 1, 594$
2. That is far past 188.5, where a real wake goes three-dimensional, so this page's solver cannot answer it. Roshko measured the shedding frequency in exactly this range and fitted it:
  
  $St = 0.212 (1 - \frac{12 . 7}{Re}) = 0.212 (1 - \frac{12 . 7}{1 , 594}) = 0.2103 (300 < Re < 2, 000)$
3. Then f follows from the definition of St:
  
  $f = \frac{St U}{D} = \frac{0 . 2103 \times 8 m/s}{0 . 003 m} = 561 Hz$
4. The solver earns its place below Re 150, where it and the same wind tunnel can be compared directly. Press Re 100 and read both off the results table:
  
  $Re = 100 : St_{solver} = 0.1690, St_{Roshko} = 0.212 (1 - \frac{21 . 2}{100}) = 0.1671$
Answer

The wire sings at about 561 Hz. Halve its thickness and the frequency goes to 1,113 Hz: a little short of doubling, because Re halves too and St falls from 0.2103 to 0.2086.

$f \approx 561 Hz; D = 1.5 mm : Re = 797, St = 0.2086, f = 1, 113 Hz$

References (5)

Wind-tunnel shedding frequencies behind circular cylinders from Re 40 to 10,000, and the two fits this page plots and the worked problem uses: S = 0.212(1 − 21.2/R) for 50 < R < 150 and S = 0.212(1 − 12.7/R) for 300 < R < 2,000: A. Roshko, "On the Development of Turbulent Wakes from Vortex Streets", NACA Report 1191, National Advisory Committee for Aeronautics, 1954.
Where a two-dimensional solver stops describing a real wake: the periodic 2D wake becomes unstable to three-dimensional disturbances at Re = 188.5 ± 1.0, which is where this page's slider stops: D. Barkley and R. D. Henderson, "Three-dimensional Floquet stability analysis of the wake of a circular cylinder", Journal of Fluid Mechanics 322, 215–241 (1996).
The history of the instability: Bénard's experiments, the threshold Re_c = 47 that Camichel's group in Toulouse measured with circular cylinders, and Mathis, Provansal and Boyer's 1984 measurement that the oscillation's amplitude grows as (Re − Re_c)^½, the Landau law this page fits: J. E. Wesfreid, "Henri Bénard: Thermal convection and vortex shedding", Comptes Rendus Mécanique 345, 446–466 (2017).
The method this page solves the equations with: a lattice-Boltzmann fluid with a curved wall placed on the true circle by interpolating along each lattice link, and the drag from the momentum each link exchanges with the wall: M. Bouzidi, M. Firdaouss and P. Lallemand, "Momentum transfer of a Boltzmann-lattice fluid with boundaries", Physics of Fluids 13, 3452–3459 (2001).
Steady flow past a cylinder computed up to Re 100, with the drag, the surface pressure and the separation bubble, which first appears just below Re 7: S. C. R. Dennis and G.-Z. Chang, "Numerical solutions for steady flow past a circular cylinder at Reynolds numbers up to 100", Journal of Fluid Mechanics 42, 471–489 (1970).

By the 42Math team · Reviewed against published sources. Corrections welcome.