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    Calculates locally in your browser · Colebrook friction factor
    Fluid flow / regime classification

    Reynolds number

    Classify a pipe flow as laminar, transitional or turbulent, and get the Darcy friction factor that goes with it. Laminar uses the exact 64/Re law; turbulent solves Colebrook iteratively.

    RegimeTurbulentVelocity head0.1147 mKinematic ν1.0040 mm²/s
    Reynolds number
    7.470e+4
    ρVD/μ — the regime indicator
    Flow regime
    Turbulent
    1.50 m/s in a 50 mm bore
    Darcy friction factor
    0.02254
    Colebrook solution
    Head loss per metre
    0.0517 m/m
    f·(1/D)·V²/2g
    Flow regime thresholds
    Where this flow sits relative to the standard pipe-flow bands.
    Re = 7.470e+4
    Flow regime thresholds
    RegimeRe rangeFriction factorBehaviour
    Laminar< 2,30064/ReViscous forces dominate; parabolic profile
    Transitional2,300 – 4,000UnstableIntermittent turbulence; avoid designing here
    Turbulent smooth> 4,000ColebrookBlunted profile; f falls slowly with Re
    Fully roughHigh Re1/(2log(3.7D/ε))²f depends only on relative roughness

    Relative roughness here is 0.000900 (ε = 0.045 mm, drawn tubing). At this Re the friction factor is 0.02254.

    Inspect the model
    Choose a quick result view, a scaled comparison, or the complete diagnostic set.
    Friction factor vs Reynolds number
    The laminar 64/Re line gives way to the flatter Colebrook curve above 2,300.
    Friction factor vs Reynolds number0.0000.1430.2870.4300.5730.7171001005005001k1k2k2k3k3k5k5k10k10k50k50k100k100k500k500k1000k1000kf
    Reynolds number vs velocity
    Re is linear in velocity, so the regime boundary is a velocity threshold.
    Reynolds number vs velocity0.0005.876e+41.175e+51.763e+52.351e+52.938e+50.1 m/s0.1 m/s0.3 m/s0.3 m/s0.5 m/s0.5 m/s1 m/s1 m/s1.5 m/s1.5 m/s2 m/s2 m/s3 m/s3 m/s5 m/s5 m/sRe
    Canonical equations
    The exact relationships implemented by this engine.

    Re = ρVD/μ · dimensionless

    f = 64/Re · valid to Re ≈ 2,300

    1/√f = −2log(ε/3.7D + 2.51/(Re√f)) · implicit; solved by iteration

    f → [−2log(ε/3.7D)]⁻² · the high-Re asymptote

    engineering methodology

    A clear path from inputs to results.

    The Reynolds number is the ratio of inertial to viscous forces, and it decides which fluid-mechanics regime you are in. Below 2,300 the flow is laminar and the friction law is exact; above 4,000 it is turbulent and the friction factor must be solved implicitly.

    Step-by-step methodWorked exampleAssumptions explained

    How to use

    1. Enter the mean flow velocity.
    2. Set the internal pipe diameter.
    3. Enter the fluid density.
    4. Enter the dynamic viscosity in mPa·s.
    5. Read the Reynolds number and the regime classification.
    6. Note the friction factor: laminar uses 64/Re exactly, turbulent solves Colebrook.

    The formula

    Re=ρVDμ,f=64Re (laminar),1f=−2log⁡10 ⁣(ε3.7D+2.51Ref)Re = \frac{\rho V D}{\mu}, \qquad f = \frac{64}{Re}\ (\text{laminar}), \qquad \frac{1}{\sqrt{f}} = -2\log_{10}\!\left(\frac{\varepsilon}{3.7D} + \frac{2.51}{Re\sqrt{f}}\right)

    Reference: Reynolds (1883); Colebrook (1939); Moody (1944) chart.

    Worked example

    Water at 1.5 m/s in a 50 mm pipe (default inputs)

    Inputs
    • Velocity 1.5 m/s.
    • Diameter 50 mm.
    • Density 998 kg/m³.
    • Viscosity 1.002 mPa·s.
    Calculation path
    1. 1

      Convert: D = 0.05 m, μ = 1.002×10⁻³ Pa·s.

    2. 2

      Re = 998 × 1.5 × 0.05 / 1.002×10⁻³ = 74,701.

    3. 3

      Re is well above 4,000, so the flow is fully turbulent.

    4. 4

      Relative roughness ε/D = 0.045 mm / 50 mm = 0.0009.

    5. 5

      Colebrook iteration gives f = 0.0214.

    Result

    Re 74,701 · turbulent · f = 0.0214 · head-loss gradient 0.14 m per metre.

    Interpreting the result

    Almost every practical water system is turbulent: even 0.05 m/s in a 50 mm pipe exceeds 2,300. Laminar flow appears in viscous fluids, microchannels and very slow flows. The transitional band between 2,300 and 4,000 is unstable and should be avoided in design, because the friction factor is not well defined there.

    Assumptions & limitations

    • The pipe flows full and the cross-section is circular.
    • The fluid is Newtonian, with viscosity independent of shear rate.
    • Flow is steady and fully developed, so the velocity profile is established.
    • Roughness is uniform and evenly distributed; a single corroded patch behaves differently.
    • No fittings, valves or bends are included — those add minor losses separately.

    Frequently asked questions

    What are the exact regime boundaries?

    Laminar below about 2,300, transitional from 2,300 to 4,000, turbulent above 4,000. The boundaries are not sharp, which is why the transitional band is treated as unstable rather than as a regime.

    What is relative roughness?

    Absolute roughness divided by diameter. Only the ratio matters: a 0.045 mm scratch in a 50 mm pipe is hydraulically smooth-ish, but the same scratch in a 5 mm tube is very rough.

    How do I get viscosity for other fluids?

    Water at 20 °C is 1.002 mPa·s, air is about 0.018, motor oil ranges 50–2,000 depending on grade. Viscosity falls sharply with temperature for liquids and rises for gases.

    Why is the friction factor implicit for turbulent flow?

    Because Colebrook’s equation has f on both sides. It has no closed-form solution, so it is solved iteratively or approximated by the Swamee-Jain or Haaland correlations.

    Does Reynolds number apply to open channels?

    Yes, with the hydraulic radius substituted for diameter. The regime boundaries shift, and the Froude number becomes the more important parameter for surface behaviour.

    Why does my calculated f differ from a Moody chart reading?

    Chart reading error is typically ±5%. The iterative Colebrook solution is the reference the chart was drawn from, so trust the computed value.

    Methodology last reviewed 2026-10-18. Results are informational and not financial or engineering advice — see our disclaimer.