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Darcy · Colebrook · Swamee-Jain

Friction Factor
Calculator

Compute the Darcy friction factor for laminar and turbulent pipe flow, with full step-by-step solutions.

Formula

f=64Ref = \dfrac{64}{Re}

Quick Answer

The Darcy friction factor quantifies frictional pressure loss in pipe flow. For laminar flow (Re < 2100), f = 64/Re exactly. For turbulent flow, f is found from the implicit Colebrook equation (solved iteratively) or approximated directly with the explicit Swamee-Jain equation, both using Reynolds number and relative pipe roughness ε/D as inputs. This calculator computes both and shows every step.

Friction Factor Calculator

f=64/Ref = 64/Re (laminar) · 1f=2log10(ε/D3.7+2.51Ref)\dfrac{1}{\sqrt f} = -2\log_{10}\left(\dfrac{\varepsilon/D}{3.7}+\dfrac{2.51}{Re\sqrt f}\right) (Colebrook)

How to Use the Friction Factor Calculator

  1. 1

    Enter the Reynolds number

    Enter the flow's Reynolds number Re (from the Reynolds Number Calculator if you don't already have it).

  2. 2

    Enter the relative roughness

    Enter the pipe's relative roughness ε/D (roughness height divided by pipe diameter), or pick a common material preset.

  3. 3

    Click Calculate

    Click Calculate to determine the flow regime and compute the Darcy friction factor.

  4. 4

    Read the friction factor result

    For laminar flow, read the exact f = 64/Re result. For turbulent flow, compare the iterative Colebrook solution against the explicit Swamee-Jain approximation.

Need the Reynolds number first? Use the Reynolds Number Calculator, or see the full Fluid Mechanics topic guide.

What Is the Darcy Friction Factor?

The Darcy friction factor f quantifies how much a pipe's internal surface resists flow, feeding directly into the Darcy-Weisbach equation ΔP=f(LD)(ρv22)\Delta P = f\left(\dfrac{L}{D}\right)\left(\dfrac{\rho v^2}{2}\right)for frictional pressure drop. For laminar flow it has an exact analytical value; for turbulent flow it depends on both the Reynolds number and the pipe's relative roughness, and must be found from the implicit Colebrook equation or an explicit approximation like Swamee-Jain.

This calculator computes both, the exact laminar result when Re < 2100, and both the Colebrook (iterative) and Swamee-Jain (explicit) turbulent results when Re > 4000, so you can see how closely the fast approximation tracks the more rigorous iterative solution.

The corroded, pitted interior surface of an old metal pipe
This is what relative roughness ε/D is actually measuring — real surface texture, not an abstract number picked from a table. NIOSH, Public domain, via Wikimedia Commons.

Derivation: From Force Balance to Colebrook

In laminar flow, the friction factor comes directly from the exact Hagen-Poiseuille solution to the Navier-Stokes equations for fully developed pipe flow. Equating the Hagen-Poiseuille pressure drop to the Darcy-Weisbach definition of f gives a closed-form result with no empirical fitting at all:

ΔP=32μLvD2  =!  f(LD)(ρv22)    f=64μρvD=64Re\Delta P = \dfrac{32\mu L v}{D^2} \;\overset{!}{=}\; f\left(\dfrac{L}{D}\right)\left(\dfrac{\rho v^2}{2}\right) \;\Rightarrow\; f = \dfrac{64\mu}{\rho v D} = \dfrac{64}{Re}

Turbulent flow has no equivalent closed-form solution, the velocity profile is chaotic and not analytically tractable, so the Colebrook equation is instead a semi-empirical fit combining the smooth-pipe turbulent correlation with a roughness term, validated against a huge body of experimental friction-factor data (the same data underlying the Moody chart):

1f=2log10(ε/D3.7+2.51Ref)\dfrac{1}{\sqrt f} = -2\log_{10}\left(\dfrac{\varepsilon/D}{3.7}+\dfrac{2.51}{Re\sqrt f}\right)

Because f appears on both sides, once outside the logarithm, once inside it, this equation has no algebraic solution for f. It's solved by fixed-point iteration: guess an initial f (commonly from Swamee-Jain), substitute it into the right-hand side to get an improved f, and repeat until the value stops changing meaningfully, usually only 3–5 iterations for engineering accuracy. The Swamee-Jain equation itself was derived by algebraically rearranging Colebrook into an explicit approximate form, trading a small amount of accuracy for the convenience of a direct formula.

The Moody Chart

The laminar line and the Colebrook turbulent curves below are computed live from the same equations derived above, not traced from a scanned image, so the two amber-marked points line up exactly with the worked examples further down this page.

10³10⁴10⁵10⁶10⁷10⁸Reynolds number, Re0.0060.0080.010.020.030.050.070.1Darcy friction factor, flaminar, f = 64/Reε/D=0.05ε/D=0.01ε/D=0.001ε/D=1.5×10⁻⁴ε/D=smoothExample 1Example 2
Curves computed live from f = 64/Re (laminar) and the Colebrook-White equation (turbulent) — the same equations derived above, not a scanned chart. Amber dots mark the two worked examples.

When You'll Need It in GATE Chemical Engineering

Friction factor rarely stands alone, it's the missing piece a pipe-flow pressure-drop or pumping-power problem needs:

  • Pressure drop / pumping power problems, any question asking for frictional pressure loss or the power needed to overcome it requires f from the Darcy-Weisbach equation, after first checking the flow regime via Reynolds number.
  • Moody chart reading questions, GATE sometimes gives a Moody chart directly and asks you to read off f for a stated Re and ε/D, which is exactly what the Colebrook equation reproduces numerically.
  • Pipe network / sizing problems, selecting a pipe diameter for a target pressure drop is an iterative problem that repeatedly calls the friction-factor calculation at each trial diameter.
  • Fanning-vs-Darcy conversion checks, some reference correlations use the Fanning friction factor; recognizing and converting between the two conventions is a common trap (see Common Mistakes below).

Two Fully Worked Examples

Example 1: Laminar Flow in a Small-Bore Pipe

Problem: Oil flows through a pipe at Re = 1200. Find the Darcy friction factor.

Re=1200<2100laminar flowRe = 1200 < 2100 \Rightarrow \text{laminar flow}
f=64Re=641200f = \dfrac{64}{Re} = \dfrac{64}{1200}

Answer: f = 0.0533 (exact, no roughness dependence)

Example 2: Turbulent Flow in Commercial Steel Pipe

Problem: Water flows through a commercial steel pipe (ε/D = 0.00015) at Re = 100,000. Estimate f using Swamee-Jain, then refine with one Colebrook iteration.

fSJ=0.25[log10(0.000153.7+5.74100,0000.9)]20.01798f_{SJ} = \dfrac{0.25}{\left[\log_{10}\left(\frac{0.00015}{3.7}+\frac{5.74}{100{,}000^{0.9}}\right)\right]^2} \approx 0.01798
1f=2log10(0.000153.7+2.51100,0000.01798)\dfrac{1}{\sqrt{f}} = -2\log_{10}\left(\dfrac{0.00015}{3.7}+\dfrac{2.51}{100{,}000\sqrt{0.01798}}\right)
fColebrook0.01791 (after iterating)f_{Colebrook} \approx 0.01791\ \text{(after iterating)}

Answer: f ≈ 0.0179, Swamee-Jain and Colebrook agree to within about 0.4% here.

Common Mistakes GATE Students Make

  • Using f = 64/Re for turbulent flow.That formula is exact only for laminar flow (Re < 2100), using it for turbulent flow ignores roughness entirely and gives a badly wrong answer.
  • Confusing Darcy and Fanning friction factors. fDarcy = 4 × fFanning, always check which convention a given correlation, textbook, or Moody chart uses before plugging into Darcy-Weisbach.
  • Forgetting relative roughness entirely. In the turbulent regime, f depends on both Re and ε/D, using a smooth-pipe (ε/D = 0) assumption for a rough pipe underestimates pressure drop.
  • Starting Colebrook iteration from a poor initial guess. The iteration converges quickly from almost any reasonable starting point, but a wildly wrong guess (like f = 1) wastes iterations, starting from the Swamee-Jain estimate is standard practice.

Key Takeaways

  • Laminar flow (Re < 2100): f = 64/Re exactly, no roughness dependence.
  • Turbulent flow (Re > 4000): f depends on both Re and relative roughness ε/D.
  • Colebrook is implicit (solved iteratively); Swamee-Jain is explicit (direct formula, ~1% accurate).
  • Darcy friction factor is 4× the Fanning friction factor, check which convention applies.
  • Avoid designing for steady operation in the transitional regime (2100 < Re < 4000).

Frequently Asked Questions

Almost always as an input to something else, not the final answer, it feeds the pressure-drop and pump-power calculations that actually answer the question a problem is asking. See the derivation below for where the Darcy-Weisbach relationship it plugs into comes from.

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