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Reynolds Number
100,000
dimensionless
Flow Regime Turbulent
Laminar Re < 2300
Transitional 2300 – 4000
Turbulent Re > 4000

What Is the Reynolds Number?

The Reynolds number (Re) is a dimensionless quantity that predicts whether fluid flow in a pipe will be smooth and orderly (laminar) or chaotic and mixing (turbulent). It represents the ratio of inertial forces to viscous forces within the moving fluid. Engineers use it constantly when designing piping, pumps, heat exchangers and HVAC systems because the flow regime determines pressure drop, heat transfer and mixing behaviour.

Comparison of laminar and turbulent flow streamlines inside a pipe
Laminar flow has smooth parallel streamlines, while turbulent flow is chaotic and mixing.

How to Use This Calculator

Enter four values: the fluid density ρ (kg/m³), the average flow velocity v (m/s), the inside pipe diameter D (m), and the dynamic viscosity μ (Pa·s). The calculator returns the Reynolds number and classifies the flow regime. For water at room temperature, \(\rho \approx 1000\) kg/m³ and \(\mu \approx 0.001\) Pa·s.

The Formula Explained

The governing equation is $$Re = \frac{\text{Density }\rho \cdot \text{Velocity }v \cdot \text{Diameter }D}{\text{Viscosity }\mu}$$ The numerator captures inertial momentum (heavier, faster fluid in a wider pipe resists deflection), while the denominator captures viscous damping. The conventional regime boundaries for flow in a circular pipe are: \(Re < 2300\) laminar, 2300–4000 transitional, and \(Re > 4000\) turbulent.

Diagram of pipe cross section showing diameter, fluid velocity arrow, density and viscosity symbols
The Reynolds number combines density (ρ), velocity (v), pipe diameter (D) and viscosity (μ).

Worked Example

Water (\(\rho = 1000\) kg/m³, \(\mu = 0.001\) Pa·s) flows at \(v = 2\) m/s through a pipe of \(D = 0.05\) m. Then $$Re = \frac{1000 \times 2 \times 0.05}{0.001} = \frac{100}{0.001} = 100{,}000$$ Since \(100{,}000 > 4000\), the flow is firmly turbulent.

FAQ

Why is the Reynolds number dimensionless? The units of \(\rho v D\) (kg/m³ × m/s × m) cancel exactly against μ (Pa·s = kg/m·s), leaving a pure number.

What diameter should I use for a non-circular duct? Use the hydraulic diameter, \(D_h = \frac{4A}{P}\), where A is the cross-sectional area and P the wetted perimeter.

Can I use kinematic viscosity instead? Yes. If you know \(\nu = \mu/\rho\), use \(Re = \frac{vD}{\nu}\), which gives the same result.

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