---
title: "Colebrook-White equation for smooth and rough pipes"
subject: "Hydraulics"
chapter: 1
key: "semester_4/hydraulics:colebrook-white-equation-validity"
source: https://www.bctengineer.com/q/semester_4/hydraulics/colebrook-white-equation-validity
site: BCT Engineering
---

# Colebrook-White equation for smooth and rough pipes

Hydraulics · Chapter 1: Pipe flow

- Asked in: 2069 Poush, 4 marks
- Video: [Cole Brook White equation - Hydraulics 3rd year](https://www.youtube.com/watch?v=NtUctM_ysg0) — Eng. Ahmed Metwally
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/colebrook-white-equation-validity
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-1

## Question

Write down Colebrook and White equation. Show that this equation is also valid for variation of friction factor for turbulent rough as well as smooth pipes.

## Short answer

The Colebrook-White equation is 1/√f = -2 log₁₀(k/(3.7D) + 2.51/(Re√f)). For smooth pipes (k→0) it reduces to 1/√f = 2 log₁₀(Re√f) - 0.8, Prandtl's smooth law. For fully rough pipes (Re→∞) it becomes 1/√f = 2 log₁₀(D/k) + 1.14, the von Karman-Prandtl rough law, covering both limits and the transition.

## Answer

**Colebrook-White equation** for turbulent flow in commercial pipes:
$$
\frac{1}{\sqrt f} = -2\log_{10}\left(\frac{k}{3.7D} + \frac{2.51}{Re\sqrt f}\right)
$$
where $f$ is the Darcy friction factor, $k$ the equivalent roughness, $D$ the diameter and $Re$ the Reynolds number.

#### Smooth pipes ($k \to 0$, so $k/D \to 0$)
The roughness term vanishes:
$$
\frac{1}{\sqrt f} = -2\log_{10}\frac{2.51}{Re\sqrt f} = 2\log_{10}\frac{Re\sqrt f}{2.51} = 2\log_{10}(Re\sqrt f) - 2\log_{10}2.51
$$
Since $2\log_{10}2.51 = 0.80$:
$$
\frac{1}{\sqrt f} = 2\log_{10}(Re\sqrt f) - 0.8
$$
This is **Prandtl's universal resistance law for smooth pipes**.

#### Rough pipes (large $Re$, fully rough)
As $Re \to \infty$ the second term $\dfrac{2.51}{Re\sqrt f} \to 0$:
$$
\frac{1}{\sqrt f} = -2\log_{10}\frac{k}{3.7D} = 2\log_{10}\frac{D}{k} + 2\log_{10}3.7
$$
Since $2\log_{10}3.7 = 1.14$:
$$
\frac{1}{\sqrt f} = 2\log_{10}\frac{D}{k} + 1.14
$$
This is the **von Karman-Prandtl law for rough pipes**, where $f$ is independent of $Re$.

For intermediate values both terms in the bracket matter, so the equation describes the transition zone, and it covers the whole turbulent range with one formula.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What happens to the Colebrook-White equation when k tends to 0?
   A. The Re term vanishes and 1/sqrt f = 2 log(D/k) + 1.14
   B. The roughness term vanishes, giving 1/sqrt f = 2 log(Re sqrt f) - 0.8
   C. f becomes 64/Re
   D. f becomes constant
   Answer: B. With k/(3.7D) = 0 only the Reynolds term remains, which is Prandtl's smooth-pipe law. (Key point: Smooth pipe limit)

2. What happens at very large Re?
   A. The roughness term vanishes, leaving f = 64/Re
   B. Both terms vanish and f = 0
   C. 2.51/(Re sqrt f) tends to 0, giving 1/sqrt f = 2 log(D/k) + 1.14, independent of Re
   D. f depends on Re only
   Answer: C. The Reynolds term disappears, leaving the von Karman-Prandtl rough-pipe law. (Key point: Rough pipe limit)

3. Why does one equation cover the whole turbulent range?
   A. It ignores roughness
   B. It is valid only at the two extremes
   C. It includes only the viscous term
   D. Both terms in the bracket matter at intermediate Re, giving the transition, and each vanishes at its extreme
   Answer: D. The two terms represent roughness and viscous effects; their relative sizes decide the regime. (Key point: Transition zone)

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