Laminar, turbulent & the Reynolds number

Is the oil flowing in smooth layers, or churning chaotically? One number — the Reynolds number — tells you, and it decides how much pressure friction will cost.

Source: Rabie, Fluid Power Engineering, Ch. 3.

Before you start

What you need first

Kinematic viscosity $\nu=\mu/\rho$ (Topic 9).
Oil velocity in a line, $v=4Q/\pi D^2$ (Topic 15).

What you'll be able to do

Tell laminar from turbulent flow.
Compute $\mathrm{Re}=\dfrac{vD}{\nu}=\dfrac{\rho v D}{\mu}$ and classify.
Find the critical velocity at the threshold.

Start here · the picture

Two kinds of flow

Laminar flow moves in smooth, orderly layers that slide over each other — low loss, quiet. Turbulent flow churns and mixes with eddies — noisier, and it wastes much more pressure to friction.

Which one happens depends on a tug-of-war: the oil's inertia (which wants to break into eddies) versus its viscosity (which damps them out).

Smooth layers vs churning eddies — the Reynolds number says which.

The Reynolds number

That tug-of-war (inertia ÷ viscous) is captured by a single dimensionless number:

$$\mathrm{Re} = \dfrac{vD}{\nu} = \dfrac{\rho v D}{\mu}$$

where:
Symbol	Meaning	SI unit
Re	Reynolds number	— (none)
v	mean velocity	m/s
D	pipe bore	m
$\nu$	kinematic viscosity	m²/s
$\rho,\ \mu$	density, dynamic viscosity	kg/m³, Pa·s

The two forms are identical because $\nu=\mu/\rho$. Use $vD/\nu$ when you have $\nu$ (in cSt → m²/s); use $\rho vD/\mu$ when you have $\mu$ and $\rho$ — but never mix them.

The threshold

Reynolds number	Flow
$\mathrm{Re} < 2000$	Laminar (smooth)
$2000 < \mathrm{Re} < 4000$	Transition (unstable)
$\mathrm{Re} > 4000$	Turbulent

For hydraulics we take a single working threshold of $\mathrm{Re} \approx 2300$: below it, treat the flow as laminar; above, as turbulent. The critical velocity is the speed that makes $\mathrm{Re} = 2300$: $v_{\text{crit}} = \dfrac{2300\,\nu}{D}$.

✏️ Try it yourself — no numbers needed

The same line carries the same oil at the same speed, but on a cold morning the oil is thick and by midday it is hot and thin. In which condition is the flow more likely to be turbulent, and why?

Hot and thin. Thinner oil has a lower $\nu$, and $\mathrm{Re}=vD/\nu$ — a smaller $\nu$ makes $\mathrm{Re}$ larger, pushing past the turbulent threshold. Why: viscosity is what damps eddies; when the oil thins, there is less damping, so the inertia wins and the flow churns. Same pipe and speed, opposite regime.

Common mistakes to avoid

Mistake	Fix
Mixing the two forms (using $\mu$ where $\nu$ belongs)	$vD/\nu$ or $\rho vD/\mu$ — never $vD/\mu$.
Leaving $\nu$ in cSt	Convert: $1~\text{cSt}=10^{-6}~\text{m}^2/\text{s}$.
Forgetting $\mathrm{Re}$ has no units	If your answer has units, a conversion slipped.

Recap — the whole topic on one screen

$$\mathrm{Re}=\dfrac{vD}{\nu}=\dfrac{\rho vD}{\mu} \qquad \mathrm{Re}\approx 2300\ \text{threshold}\qquad v_{\text{crit}}=\dfrac{2300\,\nu}{D}$$

Idea	What you own now
Two regimes	Laminar (smooth) vs turbulent (eddies)
Reynolds number	Inertia ÷ viscous; $\mathrm{Re}=vD/\nu$, dimensionless
Threshold	≈ 2300; thin/hot oil → higher Re → turbulent

Next topic

Friction (major) losses

The regime feeds straight into the loss formula: next we use $\mathrm{Re}$ to get the friction factor $\lambda$, then the Darcy equation for the pressure lost along the pipe.

→ Friction (major) losses

Mistake	Fix
Mixing the two forms (using \(\mu\) where \(\nu\) belongs)	\(vD/\nu\) or \(\rho vD/\mu\) — never \(vD/\mu\).
Leaving \(\nu\) in cSt	Convert: \(1~\text{cSt}=10^{-6}~\text{m}^2/\text{s}\).
Forgetting \(\mathrm{Re}\) has no units	If your answer has units, a conversion slipped.

Reynolds number	Flow
\(\mathrm{Re} < 2000\)	Laminar (smooth)
\(2000 < \mathrm{Re} < 4000\)	Transition (unstable)
\(\mathrm{Re} > 4000\)	Turbulent