Reynolds Number

R e = \frac{ρ V D}{μ} = \frac{V D}{ν}

Dimensionless ratio of inertia to viscous forces. For pipes, laminar if Re ≲ 2300 and turbulent if Re ≳ 4000. Kinematic viscosity ν = μ/ρ. Hydraulic diameter 4A/P is used for noncircular ducts.

$R e$ (—): Reynolds number
$ρ$ (kg/m³): Density
$V$ (m/s): Characteristic speed
$D$ (m): Characteristic length (diameter)
$μ, ν$ (Pa·s, m²/s): Dynamic and kinematic viscosity

Step-by-step solved example

Water μ = 1.0×10⁻³ Pa·s, ρ = 1000 kg/m³, V = 2.0 m/s, D = 50 mm. Re?

Re = ρVD/μ decides laminar vs turbulent in the pipe.

1. Compute Re
Re = ρVD/μ = 1000×2.0×0.050 / 0.0010 = 1.00×10⁵.
$R e = 1.00 \times 1 0^{5}$
2. Regime
Re ≫ 4000 → turbulent pipe flow.

Answer: 1.00×10⁵ (turbulent)

10 practice questions

0/10 correct

1.Reynolds number is

2.Kinematic viscosity ν is

3.Pipe flow is typically laminar if Re

4.Turbulent pipe flow is typically Re

5.ρ = 1000, V = 1.0 m/s, D = 0.10 m, μ = 0.001 Pa·s. Re =

6.Doubling V (other quantities fixed)

7.Air vs water at the same V and D: air typically has

8.Hydraulic diameter for a noncircular duct is

9.μ of water at 20°C is about

10.Re is used to