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Colebrook-White equation for smooth and rough 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.

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.

Colebrook-White equation for turbulent flow in commercial pipes:

1f=−2log⁡10(k3.7D+2.51Ref)\frac{1}{\sqrt f} = -2\log_{10}\left(\frac{k}{3.7D} + \frac{2.51}{Re\sqrt f}\right)

where ff is the Darcy friction factor, kk the equivalent roughness, DD the diameter and ReRe the Reynolds number.

Smooth pipes (k→0k \to 0, so k/D→0k/D \to 0)

The roughness term vanishes:

1f=−2log⁡102.51Ref=2log⁡10Ref2.51=2log⁡10(Ref)−2log⁡102.51\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 2log⁡102.51=0.802\log_{10}2.51 = 0.80:

1f=2log⁡10(Ref)−0.8\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 ReRe, fully rough)

As Re→∞Re \to \infty the second term 2.51Ref→0\dfrac{2.51}{Re\sqrt f} \to 0:

1f=−2log⁡10k3.7D=2log⁡10Dk+2log⁡103.7\frac{1}{\sqrt f} = -2\log_{10}\frac{k}{3.7D} = 2\log_{10}\frac{D}{k} + 2\log_{10}3.7

Since 2log⁡103.7=1.142\log_{10}3.7 = 1.14:

1f=2log⁡10Dk+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 ff is independent of ReRe.

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.

✓ Answer checked by Claude Sonnet 5.5

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