Calculate friction factor with the Colebrook equation from absolute roughness, diameter, and Reynolds number to determine the missing value.
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Colebrook Equation Formula
The Colebrook equation estimates the Darcy friction factor for turbulent flow in a pipe. The calculator solves for the friction factor by iteration using the pipe roughness, pipe diameter, and Reynolds number.
\frac{1}{\sqrt{f}} = -2\log_{10}\left(\frac{\epsilon}{3.7D}+\frac{2.51}{Re\sqrt{f}}\right)For numerical solving, the equation can be rearranged into a root-finding form:
F(f) = \frac{1}{\sqrt{f}} + 2\log_{10}\left(\frac{\epsilon}{3.7D}+\frac{2.51}{Re\sqrt{f}}\right)- f = Darcy friction factor, dimensionless
- ε = absolute roughness of the pipe wall
- D = inside pipe diameter
- Re = Reynolds number, dimensionless
- log10 = base-10 logarithm
The calculator converts the roughness and diameter inputs to meters before solving. Reynolds number and friction factor are dimensionless, so they do not need unit conversion.
To use the calculator, enter absolute roughness, diameter, and Reynolds number. Leave the friction factor field blank. The calculator starts with an initial friction factor guess and repeatedly updates it until the Colebrook equation is satisfied within a small tolerance.
Typical Pipe Roughness Values
Absolute roughness depends on pipe material, age, and condition. The values below are common reference ranges for clean commercial pipe.
| Pipe material | Typical absolute roughness ε | Meters |
|---|---|---|
| Drawn tubing, copper, brass, plastic | 0.0015 mm | 0.0000015 m |
| Commercial steel | 0.045 mm | 0.000045 m |
| Galvanized iron | 0.15 mm | 0.00015 m |
| Cast iron | 0.26 mm | 0.00026 m |
| Concrete | 0.3 mm to 3.0 mm | 0.0003 m to 0.003 m |
Flow Regime and Friction Factor Use
The Colebrook equation is intended for turbulent pipe flow. Check the Reynolds number before relying on the result.
| Reynolds number range | Flow regime | Friction factor note |
|---|---|---|
| Re < 2,300 | Laminar | Use f = 64 / Re instead of Colebrook. |
| 2,300 to 4,000 | Transitional | Results can be uncertain because flow may switch behavior. |
| Re > 4,000 | Turbulent | Colebrook is commonly used for Darcy friction factor. |
Example Problems
Example 1: Commercial steel pipe
Suppose you have a commercial steel pipe with:
- Absolute roughness: ε = 0.045 mm
- Diameter: D = 100 mm
- Reynolds number: Re = 100,000
Substitute into the Colebrook equation:
\frac{1}{\sqrt{f}} = -2\log_{10}\left(\frac{0.000045}{3.7(0.1)}+\frac{2.51}{100000\sqrt{f}}\right)Solving iteratively gives:
f ≈ 0.0185
Example 2: Smoother pipe at high Reynolds number
Suppose you have:
- Absolute roughness: ε = 0.0015 mm
- Diameter: D = 50 mm
- Reynolds number: Re = 200,000
Convert the length values to meters:
- ε = 0.0000015 m
- D = 0.05 m
Then solve:
\frac{1}{\sqrt{f}} = -2\log_{10}\left(\frac{0.0000015}{3.7(0.05)}+\frac{2.51}{200000\sqrt{f}}\right)The resulting Darcy friction factor is approximately:
f ≈ 0.0158
FAQ
Is the Colebrook friction factor the Darcy or Fanning friction factor?
The Colebrook equation gives the Darcy friction factor. The Fanning friction factor is one-fourth of the Darcy friction factor:
f_{Fanning} = \frac{f_{Darcy}}{4}If another equation or reference uses Fanning friction factor, do not mix it directly with a Darcy friction factor result.
Why does the Colebrook equation need iteration?
The friction factor appears on both sides of the equation, including inside a square root and a logarithm. Because of that, it cannot be solved directly with simple algebra. An iterative method starts with a guess for f, checks the equation, updates the guess, and repeats until the value changes very little.
Can you use the Colebrook equation for laminar flow?
No. For laminar pipe flow, use:
f = \frac{64}{Re}The Colebrook equation is used for turbulent flow, usually when Re is greater than about 4,000. For transitional flow, the result is less reliable because the flow is not fully laminar or fully turbulent.
