Reynolds Number Calculator
Compute Re for a pipe or open-channel flow and read the regime: laminar, transitional, or turbulent.
ft/s
ft
ft²/s
——
——
Defaults are water at 60°F (ν = 1.08 × 10⁻⁵ ft²/s) in a 6-inch pipe at 5 ft/s.
$$ Re = \frac{V L}{\nu} $$
V mean velocity · L characteristic length (= pipe diameter D, or 4× hydraulic radius for open channel) · ν kinematic viscosity of fluid.
Regime thresholds — by geometry
| Geometry | Laminar | Transitional | Turbulent |
|---|---|---|---|
| Round pipe (internal flow) | < 2,300 | 2,300–4,000 | > 4,000 |
| Non-circular duct (use Dh) | < 2,300 | 2,300–4,000 | > 4,000 |
| Open channel (use 4Rh) | < 500 | 500–2,000 | > 2,000 |
| Flat plate boundary layer | < 5×10⁵ | 5×10⁵ | > 5×10⁵ |
| Sphere drag (transition to wake) | < 1 | 1–10⁵ | > 10⁵ |
| Cylinder cross-flow (vortex shed.) | < 40 | 40–10⁵ | > 3×10⁵ |
Open channels go turbulent at lower Re because the free surface adds instability. Engineering practice usually treats pipe flow above Re = 2,300 as effectively turbulent and uses rough-pipe friction formulas conservatively in the transitional band.
Kinematic viscosity ν of common fluids
| Fluid / temperature | ν (US ft²/s) | ν (SI m²/s) |
|---|---|---|
| Water, 32°F (0°C) | 1.92×10⁻⁵ | 1.79×10⁻⁶ |
| Water, 40°F (4°C) | 1.66×10⁻⁵ | 1.55×10⁻⁶ |
| Water, 50°F (10°C) | 1.41×10⁻⁵ | 1.31×10⁻⁶ |
| Water, 60°F (16°C) | 1.21×10⁻⁵ | 1.13×10⁻⁶ |
| Water, 70°F (21°C) | 1.06×10⁻⁵ | 9.84×10⁻⁷ |
| Water, 80°F (27°C) | 9.30×10⁻⁶ | 8.64×10⁻⁷ |
| Water, 100°F (38°C) | 7.39×10⁻⁶ | 6.87×10⁻⁷ |
| Air, 60°F | 1.58×10⁻⁴ | 1.47×10⁻⁵ |
| SAE 10W oil, 60°F | 1.0×10⁻³ | 9.3×10⁻⁵ |
| SAE 30 oil, 60°F | 4.1×10⁻³ | 3.8×10⁻⁴ |
| Glycerin, 70°F | 6.5×10⁻³ | 6.0×10⁻⁴ |
| Mercury, 60°F | 1.21×10⁻⁶ | 1.13×10⁻⁷ |
Worked examples
Example 1 — 6-inch water main at 5 ft/s
Given: 6-inch (D = 0.5 ft) pipe, V = 5 ft/s, water at 60°F (ν = 1.21×10⁻⁵ ft²/s).
Find: Reynolds number and regime.
Re = V·D/ν = 5·0.5/(1.21×10⁻⁵)
Re = 2.07×10⁵
Re ≈ 207,000 — fully turbulent. Use Swamee-Jain for friction factor.
Example 2 — Hydraulic-oil cooler line, small diameter
Given: ½-inch line, V = 2 ft/s, SAE 30 oil at 60°F (ν = 4.1×10⁻³ ft²/s).
Find: Re and regime.
D = 0.5/12 = 0.0417 ft
Re = 2·0.0417/(4.1×10⁻³) = 0.0833/0.0041 = 20.3
Re = 20 — strongly laminar. Use f = 64/Re = 3.15 in Darcy-Weisbach.
Reference: White, F.M. (2011). Fluid Mechanics (7th ed.). McGraw-Hill. Original concept: Reynolds, O. (1883). "An experimental investigation of the circumstances which determine whether the motion of water shall be direct or sinuous." Phil. Trans. Royal Soc. London, 174, 935–982.