HYD-34

Colebrook snapshot (Haaland f)

1/√f = −1.8 log10[(ε/D/3.7)^{1.11} + 6.9/Re]. Explicit Haaland fit to Colebrook.

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Pipe flowColebrook–WhiteHaaland

Governing equation

\dfrac1{\sqrt f}=-1.8\log_{10}\!\left[\left(\dfrac{\varepsilon/D}{3.7}\right)^{1.11}+\dfrac{6.9}{\mathrm{Re}}\right]

$\mathrm{Re} \in \left[4000, 1\times 10^{7}\right]$

where

\varepsilon: Roughness (mm)
D: Diameter (mm)
\mathrm{Re}: Reynolds (—)
f: Friction factor (—)

The figure, part by part

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①

Governing law — red

\dfrac1{\sqrt f}=-1.8\log_{10}\!\left[\left(\dfrac{\varepsilon/D}{3.7}\right)^{1.11}+\dfrac{6.9}{\mathrm{Re}}\right]

$\mathrm{Re} \in \left[4000, 1\times 10^{7}\right]$

$\mathrm{Re} = 4000$ $f = 0.041$

Along the curve

②

Domain of the variable — blue

\mathrm{Re}_{\min} = 4000,\quad \mathrm{Re}_{\max} = 1\times 10^{7}

$\mathrm{Re} \in \left[4000, 1\times 10^{7}\right]$

$\mathrm{Re} = 4000$ $\mathrm{Re} = 4000$

Along the curve

③

Response curve — green

\dfrac1{\sqrt f}=-1.8\log_{10}\!\left[\left(\dfrac{\varepsilon/D}{3.7}\right)^{1.11}+\dfrac{6.9}{\mathrm{Re}}\right]

$\mathrm{Re} \in \left[4000, 1\times 10^{7}\right]$

$\mathrm{Re} = 4000$ $f = 0.041$

Along the curve

④

Rate of change — amber

\frac{\mathrm{d}f}{\mathrm{d}\mathrm{Re}}

$\mathrm{Re} \in \left[4000, 1\times 10^{7}\right]$

$\mathrm{Re} = 66867.9$ $f^{\prime} = -1.7\times 10^{-7}$

Along the curve

Lecture brief

Historical brief

Open-channel and pipe flow were written by Chezy, Manning, Darcy and Weisbach in the nineteenth century, then Bakhmeteff and Bélanger on the hydraulic jump. The sheets compute conveyance, head loss and gradually varied profiles. This sheet (HYD-34 — Colebrook snapshot (Haaland f)) is the form associated with Colebrook–White · Haaland. Working symbols: $\varepsilon$ , $D$ , $\mathrm{Re}$ $\rightarrow$ $f$ . Colebrook matches the log-law overlap to Nikuradse's sand grain. Implicit in f; Haaland is within ~2 %.

Purpose

Purpose: compute

f

from

\varepsilon

D

\mathrm{Re}

in Hydraulics via

\dfrac1{\sqrt f}=-1.8\log_{10}\!\left[\left(\dfrac{\varepsilon/D}{3.7}\right)^{1.11}+\dfrac{6.9}{\mathrm{Re}}\right]

1/√f = −1.8 log10[(ε/D/3.7)^{1.11} + 6.9/Re]. Explicit Haaland fit to Colebrook. Use it when a real hydraulics question must be answered in SI before a code check.

Live realistic example

In symbols

Live case. Given

\varepsilon = 0.150\,\mathrm{mm}

D = 300.000\,\mathrm{mm}

\mathrm{Re} = 200000.000\,\mathrm{—}

, the governing relation

\dfrac1{\sqrt f}=-1.8\log_{10}\!\left[\left(\dfrac{\varepsilon/D}{3.7}\right)^{1.11}+\dfrac{6.9}{\mathrm{Re}}\right]

yields

f = 0.0186\,\mathrm{—}

. A Moody point, a roughness, a friction factor. Move a slider: the numbers are this situation, not a canned story.

Calculator

Inputs

Roughness \varepsilon0.15 mmDiameter D300 mmReynolds \mathrm{Re}200000 —

Outputs

Friction factor f0.0186 —

Reading speed

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Narration of this film

A Moody point, a roughness, a friction factor.

Colebrook matches the log-law overlap to Nikuradse's sand grain. Implicit in f; Haaland is within ~2 %.

Reading speed

Watch on YouTube