Learn: Colebrook–White friction factor
In turbulent pipe flow, the friction factor depends on both Reynolds number and pipe-wall roughness. The Colebrook equation contains the unknown factor on both sides, so ordinary one-step substitution cannot isolate it.
What the formula is saying
Calculate relative roughness ε/D, start with a trial Darcy factor f = 0.02, and repeatedly evaluate fnew = [−2 log10(ε/(3.7D) + 2.51/(Re√f))]⁻². Stop only when successive values agree and the original-equation residual is small.
Read the symbols in plain language
- Re
- Reynolds number
Inertial-to-viscous flow parameter based on mean velocity and internal pipe diameter; choose the flow-regime equation accordingly.
Reynolds number (no unit)Use Reynolds number as the base unit shown here. Absolute roughness ε and internal diameter D use m; divide them to obtain a dimensionless ratio. Re and the Darcy factor f are dimensionless. log10 means base-ten logarithm, not the natural logarithm.
- ε
- Absolute roughness
Absolute height scale of the inner-wall roughness, not ε/D. Divide by internal diameter only once.
mMetres measure length; 1 m = 1000 mm.
- D
- Pipe diameter
Pipe diameter. Divide roughness height by internal diameter using matching length units.
mMetres measure length; 1 m = 1000 mm.
- f
- Result to find
Colebrook–White friction factor. The iteration table replaces this initial trial with converged Darcy factors and checks the residual.
ratio / no unit
Sort out the units first
Absolute roughness ε and internal diameter D use m; divide them to obtain a dimensionless ratio. Re and the Darcy factor f are dimensionless. log10 means base-ten logarithm, not the natural logarithm.
Assumptions before calculating
Assume steady, fully developed turbulent flow in a circular full pipe and use a representative equivalent sand roughness. The calculator requires Re ≥ 4000, D > 0 and 0 ≤ ε < D, excluding the transition interval.
Let’s solve one together
Read the given values, follow each operation, then check what the result means.
Use the following study data and find the requested result. Follow the calculation before trying the second case.
- Re · Reynolds number
- 100000 Reynolds number (no unit)
- ε · Absolute roughness
- 0.000045 m
- D · Pipe diameter
- 0.2 m
See the convergence: start with f = 0.02
Let u = 1/√f. Each row uses u(next) = −2 log10[ε/(3.7D) + 2.51u(previous)/Re]. The residual checks the original equation, not only the change between trials.
| Iteration | u(previous) | u(next) | f | Residual |
|---|---|---|---|---|
| 1 | 7.071067812 | 7.245771551 | 0.01904718173 | 0.01583836679 |
| 2 | 7.245771551 | 7.229933184 | 0.01913072516 | -0.001424036029 |
| 3 | 7.229933184 | 7.23135722 | 0.01912319126 | 0.0001281314034 |
| 4 | 7.23135722 | 7.231229089 | 0.01912386896 | -0.00001152818767 |
| 5 | 7.231229089 | 7.231240617 | 0.01912380798 | 0.000001037215781 |
| 6 | 7.231240617 | 7.23123958 | 0.01912381347 | -9.332047846e-8 |
| 7 | 7.23123958 | 7.231239673 | 0.01912381298 | 8.396239259e-9 |
| 8 | 7.231239673 | 7.231239665 | 0.01912381302 | -7.554277204e-10 |
| 9 | 7.231239665 | 7.231239666 | 0.01912381302 | 6.796785357e-11 |
| 10 | 7.231239666 | 7.231239666 | 0.01912381302 | -6.11510842e-12 |
| 11 | 7.231239666 | 7.231239666 | 0.01912381302 | 5.497824418e-13 |
Accepted only when the absolute equation residual is at most 10⁻¹².
Form the relative roughness
Divide roughness height by internal diameter using matching length units.
(0.000045) ÷ (0.2) = 0.000225Iterate until the original equation is satisfied
The iteration table replaces this initial trial with converged Darcy factors and checks the residual.
1 ÷ (7.231239666)^2 ≈ 0.01912381302
Does this worked answer make sense?
Substitute the final f into 1/√f + 2log10(ε/(3.7D) + 2.51/(Re√f)); the residual should be close to zero. At fixed Re, increasing equivalent roughness normally increases the Darcy factor.
A second worked example — different values
A second case uses different data. Predict which way the answer will change, then calculate it without reusing the first answer.
- Re · Reynolds number
- 250000 Reynolds number (no unit)
- ε · Absolute roughness
- 0.0001 m
- D · Pipe diameter
- 0.3 m
See the convergence: start with f = 0.02
Let u = 1/√f. Each row uses u(next) = −2 log10[ε/(3.7D) + 2.51u(previous)/Re]. The residual checks the original equation, not only the change between trials.
| Iteration | u(previous) | u(next) | f | Residual |
|---|---|---|---|---|
| 1 | 7.071067812 | 7.585897287 | 0.01737745145 | 0.02743365059 |
| 2 | 7.585897287 | 7.558463637 | 0.01750382424 | -0.001440202107 |
| 3 | 7.558463637 | 7.559903839 | 0.01749715573 | 0.00007566664257 |
| 4 | 7.559903839 | 7.559828172 | 0.017497506 | -0.000003975278536 |
| 5 | 7.559828172 | 7.559832147 | 0.01749748759 | 2.088486175e-7 |
| 6 | 7.559832147 | 7.559831939 | 0.01749748856 | -1.0972248e-8 |
| 7 | 7.559831939 | 7.55983195 | 0.01749748851 | 5.764473343e-10 |
| 8 | 7.55983195 | 7.559831949 | 0.01749748851 | -3.028421958e-11 |
| 9 | 7.559831949 | 7.559831949 | 0.01749748851 | 1.59072755e-12 |
| 10 | 7.559831949 | 7.559831949 | 0.01749748851 | -8.348877145e-14 |
Accepted only when the absolute equation residual is at most 10⁻¹².
Form the relative roughness
Divide roughness height by internal diameter using matching length units.
(0.0001) ÷ (0.3) ≈ 0.0003333333333Iterate until the original equation is satisfied
The iteration table replaces this initial trial with converged Darcy factors and checks the residual.
1 ÷ (7.559831949)^2 ≈ 0.01749748851
Now try your own values
Change a value or its unit. The same method will show your calculation, step by step.
Results update only when you calculate. The lesson example above stays unchanged.
Your turn — check your understanding
Solve this separate case yourself. Use only the values below; the two worked examples use different data. Give the requested result in the selected unit.
- Re · Reynolds number
- 80000 Reynolds number (no unit)
- ε · Absolute roughness
- 0.00002 m
- D · Pipe diameter
- 0.15 m
Find: Learn: Colebrook–White friction factor
A hint, not the answer
Calculate relative roughness ε/D, start with a trial Darcy factor f = 0.02, and repeatedly evaluate fnew = [−2 log10(ε/(3.7D) + 2.51/(Re√f))]⁻². Stop only when successive values agree and the original-equation residual is small.
Absolute roughness ε and internal diameter D use m; divide them to obtain a dimensionless ratio. Re and the Darcy factor f are dimensionless. log10 means base-ten logarithm, not the natural logarithm.
Show the full practice solution
Compare the steps with your work; revealing a solution does not mark the lesson complete.
See the convergence: start with f = 0.02
Let u = 1/√f. Each row uses u(next) = −2 log10[ε/(3.7D) + 2.51u(previous)/Re]. The residual checks the original equation, not only the change between trials.
| Iteration | u(previous) | u(next) | f | Residual |
|---|---|---|---|---|
| 1 | 7.071067812 | 7.17712834 | 0.01941326467 | 0.01113599485 |
| 2 | 7.17712834 | 7.165992345 | 0.01947364821 | -0.001162555893 |
| 3 | 7.165992345 | 7.167154901 | 0.01946733124 | 0.0001214392212 |
| 4 | 7.167154901 | 7.167033461 | 0.01946799096 | -0.00001268460409 |
| 5 | 7.167033461 | 7.167046146 | 0.01946792204 | 0.00000132494454 |
| 6 | 7.167046146 | 7.167044821 | 0.01946792924 | -1.383942942e-7 |
| 7 | 7.167044821 | 7.167044959 | 0.01946792849 | 1.445568554e-8 |
| 8 | 7.167044959 | 7.167044945 | 0.01946792857 | -1.509937952e-9 |
| 9 | 7.167044945 | 7.167044946 | 0.01946792856 | 1.577173947e-10 |
| 10 | 7.167044946 | 7.167044946 | 0.01946792856 | -1.647482151e-11 |
| 11 | 7.167044946 | 7.167044946 | 0.01946792856 | 1.721289777e-12 |
| 12 | 7.167044946 | 7.167044946 | 0.01946792856 | -1.794120408e-13 |
Accepted only when the absolute equation residual is at most 10⁻¹².
Form the relative roughness
Divide roughness height by internal diameter using matching length units.
(0.00002) ÷ (0.15) ≈ 0.0001333333333Iterate until the original equation is satisfied
The iteration table replaces this initial trial with converged Darcy factors and checks the residual.
1 ÷ (7.167044946)^2 ≈ 0.01946792856
Avoid the common trap
Do not mix mm of roughness with m of diameter. Do not use a Fanning factor in the iteration, replace log10 with ln, or present the starting guess as the final solution.
When this method applies — and when it does not
Colebrook is an empirical turbulent-pipe relation, not a laminar or transitional-flow model. Roughness must suit the actual pipe condition. The arithmetic may converge for an unusual roughness, but convergence alone does not establish physical applicability.
For study and understanding, not approval of a real structure, site operation or design. Apply the correct standard, National Annex and professional review to actual engineering work.
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Sources & further reading
References open in a new tab and explain the underlying principles. The teaching text and examples here are SimpleFlick’s own; the source organisations have not endorsed this calculator.
Reading focus: Colebrook–White friction factor. Fundamentals, hydrostatics, energy, pipe flow and open-channel flow. The lesson identifies whether the model is uniform, critical, pressurized or rapidly varied flow.
Lesson updated: · Both examples and the separate practice case are checked against an independent high-precision numerical implementation. This verifies arithmetic for the stated model, not engineering certification.