Reynolds Number & Flow Regime

Pick a fluid, a speed and a diameter to see whether the flow is laminar or turbulent, and how little it takes to cross over.

Worked example — Reynolds Number & Flow Regime

Kinematic viscosity	1.00 × 10⁻⁶ m²/s
Reynolds number	9,966
Regime	Turbulent
Laminar below	0.2308 m/s
Turbulent at and above	0.4014 m/s
Laminar in pipes narrower than	2.31 × 10⁻³ m
Turbulent in pipes this wide or wider	4.01 × 10⁻³ m

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

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Lesson

The theory — Reynolds Number & Flow Regime

The Reynolds number compares the two things that can happen to a moving fluid: inertia carrying it on, and viscosity smearing it out. Re = ρvD/η, and the units cancel — it is a pure number, which is the point. Two flows with the same Reynolds number behave the same way whatever their size, speed or fluid, which is why a model in a water tank can stand in for an aircraft.

What each symbol means

Re: the ratio itself, with no units at all
ν: kinematic viscosity, η/ρ, the only combination that matters here

Assumes: A smooth, straight, circular pipe with fully developed flow. Roughness, bends and the entrance length all move the transition, which is the honest reason the thresholds on this page are conventions rather than constants.

Two fluids, one pipe, one speed, opposite verdicts 🖖

Water in a 10 mm pipe at 1 m/s reads Re = 9,966, and the verdict is turbulent. Switch the fluid picker to honey, touching neither slider, and it reads 1.420, about as laminar as a flow gets. Nothing about the pipe changed and nothing about the speed changed, so neither of them decided the verdict.

The kinematic viscosity row says what did: 1.00 × 10⁻⁶ m²/s for water against 7.04 × 10⁻³ m²/s for honey. Since Re = vD/ν, the two Reynolds numbers stand in the ratio of those two properties and in nothing else. That ratio is 7018, and it survives every setting the tool allows: drag velocity across its six decades and diameter across its five, and both Reynolds numbers move together with 7018 still between them.

Watch the regime map while you switch fluids. The marked point does not move. It changes colour, because the boundary has slid almost four decades along the velocity axis to get past it.

Almost nothing in your plumbing is laminar 🖖

Water in a 10 mm pipe: the tool prints 0.2308 m/s as the laminar limit, and below that the verdict turns green. Push past 0.4014 m/s and it is fully turbulent. A kitchen tap runs faster than both.

Laminar pipe flow is what textbooks spend chapters on, because it can be solved in closed form. The velocity profile is an exact parabola, the pressure drop is proportional to the flow rate, and nothing crosses between the layers. Almost none of the water in a house behaves that way. What is in the walls mixes, and it has to be measured or correlated rather than derived.

Narrowing the pipe raises the limit rather than lowering it. Set the diameter to 5 mm and the laminar ceiling becomes 0.4616 m/s, because Re = vD/ν only cares about the product: halve D and you must double v to reach 2300 again. That is why a capillary stays laminar and a water main does not.

2300 and 4000 are conventions, and they are about pipes 🖖

The number the tool prints is exact arithmetic on the four values you set. The word beside it is a convention. Drag the velocity to 0.3 m/s in the 10 mm pipe: the Reynolds number reads 2,990 and the verdict reads Transitional, which is the tool declining to answer, correctly. 2300 and 4000 describe fully developed flow in a smooth round pipe, and neither the roughness of that pipe nor the stillness of its inlet reaches the Reynolds number. Hence an amber strip where a line would be tidier.

What the lower number means was settled in 2011, by Avila, Moxey, de Lozar, Avila, Barkley and Hof in Science, and their method is the part worth knowing. They injected puffs of turbulence into laminar pipe flow and timed how long each one lived. Below a critical point a puff decays; above it, turbulence spreads faster than it dies. The threshold is not the speed at which a pipe stops being laminar. It is the speed above which turbulence, once something has put it there, no longer goes away on its own.

Change the geometry and the comparison changes with it. A sphere settling through this same water is judged near Re = 1 and Re = 1000; a boundary layer along a flat plate transitions near Re = 5 × 10⁵.

Problem solved in full

A duct in a water tunnel 7 steps
A round ventilation duct 0.40 m across carries air at 3.0 m/s. You want to see what the flow inside it is doing, so you plan to run a 1:10 model in a water tunnel. How fast must the water go for the model's flow to be the same flow as the duct's?

Open this problem in the tool
1. Air's density and its viscosity never appear separately in the answer. Divide the second by the first and you have the kinematic viscosity, 1.813 × 10⁻⁵ over 1.204, which is 1.5058 × 10⁻⁵ m²/s. The Kinematic viscosity row rounds it to 1.51 × 10⁻⁵. That one number is everything the Reynolds number wants to know about the fluid.
  
  $ν_{a} = \frac{η}{ρ} = \frac{1.813 \times 1 0 ^{- 5}}{1.204} = 1.5058 \times 1 0^{- 5} m^{2} /s$
2. Re = vD/ν for the duct: 3.0 × 0.40 over 1.5058 × 10⁻⁵ is 79,691. The Regime row reads Turbulent, which it would for anything past 4000, and this is twenty times past it.
  
  $Re_{d} = \frac{v _{d} D _{d}}{ν _{a}} = \frac{3.0 \cdot 0.40}{1.5058 \times 1 0 ^{- 5}} = 79, 691 \geq 4000$
3. Two flows are the same flow when the geometry is similar and the Reynolds numbers match. Re is the only dimensionless group in the problem, so matching it is the whole condition. Nothing in that condition says the model has to use the same fluid, and the fluid is the freedom worth spending.
  
  $Re_{m} = Re_{d} ⟺ \frac{v _{m} D _{m}}{ν _{w}} = \frac{v _{d} D _{d}}{ν _{a}}$
4. Water at 20 °C: 1.0016 × 10⁻³ Pa·s over 998.2 kg/m³ gives 1.0034 × 10⁻⁶ m²/s. Water is roughly eight hundred times denser than air and about fifty-five times more viscous, and the two nearly cancel. What is left is a factor of 15.007, and it runs the way beginners do not expect: per unit of inertia, air is the runnier of the two.
  
  $ν_{w} = \frac{1.0016 \times 1 0 ^{- 3}}{998.2} = 1.0034 \times 1 0^{- 6} m^{2} /s, \frac{ν _{a}}{ν _{w}} = 15.007$
5. Set the two Reynolds numbers equal and solve for the speed the tunnel needs. Shrinking the pipe by ten multiplies the required speed by ten; swapping air for water divides it by 15.007. The fluid wins, and the answer lands at 1.999 m/s. A model a tenth the size, and the water in it runs slower than the air it stands for.
  
  $v_{m} = v_{d} \cdot \frac{D _{d}}{D _{m}} \cdot \frac{ν _{w}}{ν _{a}} = 3.0 \cdot 10 \cdot \frac{1}{15.007} = 1.999 m/s$
6. Set the tool to water, 1.999 m/s, 0.040 m. The Reynolds number row reads 79,689 against the duct's 79,691, the two counts being the fourth digit of a speed quoted to four. The Laminar below row is a second way to see the match: the duct's air would have to drop under 0.0866 m/s to go laminar and the model's water under 0.0577 m/s, and each flow is running at 34.6 times its own limit.
  
  $Re_{m} = \frac{1.999 \cdot 0.040}{1.0034 \times 1 0 ^{- 6}} = 79, 689, \frac{3.0}{0.0866} \approx \frac{1.999}{0.0577} \approx 34.6$
7. Nobody sets a tunnel to 1.999. Round the dial to 2.0 m/s and the row reads 79,728, which is 0.05% high, and 0.05% of a Reynolds number is below anything the instruments in the tunnel will resolve.
  
  $v_{m} = 2.0 \Rightarrow Re_{m} = 79, 728, \frac{79 , 728}{79 , 691} - 1 = 0.05%$
Answer

About 2.0 m/s, through a 40 mm model: the exact requirement is 1.999 m/s. The page prints one Reynolds number for one setting, and the use of the number is comparing two, which is why 79,689 beside 79,691 is the answer rather than an accident. What crosses back from the model to the duct is dimensionless. The friction factor and the pressure coefficient transfer; pascals and newtons do not. A pressure difference measured in the water converts by the ratio of ρv², 998.2 × 2.0² against 1.204 × 3.0², so the tunnel reads 368 times the pressure the duct will.

$v_{m} = v_{d} \frac{D _{d}}{D _{m}} \frac{ν _{w}}{ν _{a}} = \frac{3.0 \cdot 10}{15.007} = 1.999 m/s$

References (2)

Where the number and the threshold come from: O. Reynolds, "An Experimental Investigation of the Circumstances Which Determine Whether the Motion of Water Shall Be Direct or Sinuous, and of the Law of Resistance in Parallel Channels." Philosophical Transactions of the Royal Society of London 174, 1883, pp. 935–982.
What the lower threshold actually marks, measured by timing whether turbulence survives: K. Avila, D. Moxey, A. de Lozar, M. Avila, D. Barkley & B. Hof, "The Onset of Turbulence in Pipe Flow." Science 333(6039), 2011, pp. 192–196.

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