To calculate **friction factor (f)** using the **Colebrook formula**, solve for $f$ by taking the logarithm of the sum of two terms: the pipe roughness divided by the diameter and the Reynolds number multiplied by the reciprocal of the friction factor. This equation is implicit and typically solved iteratively.

The **Colebrook formula calculator** is actually designed to compute the **friction factor (f)** for turbulent flow in pipes. The **Colebrook equation** is widely used in fluid mechanics to estimate the head loss due to friction in pipe systems. It considers the pipe's roughness and the flow's Reynolds number to deliver accurate results.

Solving this equation manually can be challenging due to its iterative nature, which is why calculators are invaluable for simplifying the process.

**Formula:**

`$f = -2 \log_{10}\left(\frac{\varepsilon}{3.7D} + \frac{2.51}{Re\sqrt{f}}\righ$`

Variable |
Description |
---|---|

f |
Darcy-Weisbach friction factor |

ε |
Roughness of the pipe surface (in meters) |

D |
Diameter of the pipe (in meters) |

Re |
Reynolds number, indicating the type of flow |

**Solved Calculation:**

**Example 1:**

Step |
Calculation |
---|---|

Determine values | ε = 0.00005 m, D = 0.5 m, Re = 10,000 |

Calculate the first term (ε / 3.7D) | 0.00005 / (3.7 × 0.5) = 0.000027 |

Assume initial f (guess) | f = 0.02 |

Calculate the second term | 2.51 / (Re × √f) = 2.51 / (10,000 × √0.02) = 0.00395 |

Add the two terms | 0.000027 + 0.00395 = 0.003977 |

Logarithm calculation | -2 log₁₀(0.003977) ≈ 1.9 |

Adjust the guess for f | Refine f through iteration |

Result |
f ≈ 0.02 |

**Answer:** The friction factor (f) is approximately **0.02**.

**Example 2:**

Step |
Calculation |
---|---|

Determine values | ε = 0.0001 m, D = 1 m, Re = 20,000 |

Calculate the first term (ε / 3.7D) | 0.0001 / (3.7 × 1) = 0.000027 |

Assume initial f (guess) | f = 0.03 |

Calculate the second term | 2.51 / (20,000 × √0.03) = 0.00288 |

Add the two terms | 0.000027 + 0.00288 = 0.002907 |

Logarithm calculation | -2 log₁₀(0.002907) ≈ 2.0 |

Adjust the guess for f | Refine f through iteration |

Result |
f ≈ 0.03 |

**Answer:** The friction factor (f) is approximately **0.03**.