Darcy Friction Factor (Moody / Colebrook)

\frac{1}{f} = - 2 lo g_{10} (\frac{ε / D}{3.7} + \frac{2.51}{Re f})

Colebrook is implicit; Moody chart is the exam tool. Laminar f=64/Re (smooth or rough). Fully rough turbulent: 1/√f ≈ −2 log10((ε/D)/3.7). Then hf=f (L/D) V²/2g.

$f$: Darcy friction factor
$ε$ (mm): Absolute roughness
$R e$: Reynolds number ρVD/μ

Step-by-step solved example

Laminar Re=800. f? If instead fully rough ε/D=0.01, f nearest?

f from Moody/Colebrook then hf = f (L/D) V²/2g.

1. Laminar
f=64/800=0.080.
2. Rough
1/√f≈−2 log10(0.01/3.7)=−2 log10(0.00270)=5.14, f≈0.038.

Answer: laminar f=0.080; rough f≈0.038

10 practice questions

0/10 correct

1.Laminar f =

2.Transition Re is nearest

3.Blasius smooth turbulent f ≈

4.Smoother pipe (smaller ε) in turbulent flow

5.Swamee–Jain is

6.hf doubles if L doubles at fixed V,D,f

7.ε for commercial steel is on the order of

8.If Re→∞, f depends only on

9.f is dimensionless

10.Haaland equation is another