Reynolds number Re = ρvL - μ — laminar vs turbulent criterion: Question bank

True or false — justify

Whether an item is true or false, the reason is what matters. Say the reason, then reveal.

Raising the fluid's viscosity, all else fixed, makes the flow more turbulent.

False.

μ

sits in the denominator, so larger

μ

gives a smaller

R e

— viscosity is the peacemaker that damps disturbances, pushing flow toward laminar.

The Reynolds number carries units of newtons because we built it from forces.

False. It is a ratio of two forces with the same units, so the units cancel —

R e

is dimensionless. See Dimensional analysis.

If two flows (a toy pipe and an oil pipeline) share the same

R e

, they behave in the same regime.

True. Because

R e

is dimensionless, matching it means the inertia-to-viscosity balance is identical — this is the whole point of a universal number.

For pipe flow the correct characteristic length

L

is the pipe radius.

False. The standard convention uses the diameter

D

. Using radius halves

R e

and can mis-predict the regime.

R e_{c} \approx 2000

is a fundamental constant of nature like

c

g

False. The critical

R e

is geometry-dependent and fuzzy — flat plates, spheres and pipes all differ, and very smooth pipes stay laminar to

R e \sim 1 0^{5}

. It is a practical guideline.

Doubling the flow speed doubles the Reynolds number.

True.

R e = ρ v L / μ

is linear in

v

(the

v^{2}

was the inertial force; the extra

v

cancels against the viscous

v

), so

R e

scales in direct proportion to

v

Doubling the pipe diameter doubles the inertial force

ρ v^{2} L^{2}

False. Inertial force

\propto L^{2}

, so doubling

L

multiplies it by four. But

R e \propto L

(not

L^{2}

) because viscous force

\propto L

too, and one

L

cancels.

Bernoulli's principle is a good model for a very high-

R e

flow near a solid wall.

False. Bernoulli's principle assumes inviscid (ideal) flow — the

R e \to \infty

idealisation away from walls. Right at a wall viscosity always dominates in a thin boundary layer, no matter how large

R e

is.

Poiseuille's law can be safely applied at

R e = 8000

in a pipe.

False. Poiseuille's law assumes laminar flow (

R e < 2000

). At

R e = 8000

the flow is turbulent, so the parabolic velocity profile and the law's pressure-flow relation break down.

Kinematic viscosity

ν = μ / ρ

can replace

μ

and

ρ

together in the formula.

True.

R e = ρ v L / μ = v L / (μ / ρ) = v L / ν

. Bundling density and viscosity into

ν

is exact and often more convenient.

Spot the error

Each statement contains one flaw. Name it.

"Honey is thick and complicated, so it must flow turbulently."

The error is equating "thick" with "chaotic". Honey's huge

μ

makes

R e

tiny (often

\sim 1

), so it is deeply laminar and pours in smooth ribbons.

R e = ρ v L^{2} / μ

because inertial force is

ρ v^{2} L^{2}

The student forgot to divide by the viscous force

μv L

. The ratio is

ρ v^{2} L^{2} / (μv L) = ρ v L / μ

; one factor of

v

and one of

L

cancel.

"Since

R e

is dimensionless, its value doesn't depend on the units I plug in."

The value is unit-independent only if you use consistent units for every quantity (all SI, say). Mixing cm with m or poise with Pa·s gives a wrong number even though the true

R e

is a pure number.

"The velocity gradient is

d v / d y \approx v / L

where

L

is the pipe's length."

L

here is the transverse length across which speed falls from

v

0

(roughly the diameter), not the pipe's downstream length. See Viscosity and Newton's law of viscosity.

"A flow at

R e = 3000

is definitely laminar since it's below 4000."

2000 < 3000 < 4000

is the transitional band — the flow is unstable and intermittent, not cleanly laminar. Only

R e < 2000

is reliably laminar.

"Stokes' law drag

F = 6 π μ r v

holds for a fast cannonball in air."

Stokes' law and terminal velocity is a low- $R e$ result (

R e ≪ 1

). A cannonball has enormous

R e

, so its drag is inertial (pressure) drag, not viscous Stokes drag.

Why questions

Explain the mechanism, don't just state the fact.

Why does a critical Reynolds number exist at all?

Disturbances always exist; viscosity damps them while inertia amplifies them. At a threshold ratio the amplification overtakes the damping, and above it disturbances grow into turbulence.

Why is

R e

built as a ratio rather than a difference of the two forces?

A ratio is dimensionless and scale-free, so it compares the relative strength that governs the regime. A difference keeps units and depends on absolute size, so it wouldn't be universal.

Why does the inertial force scale as

ρ v^{2} L^{2}

Mass flowing per second through area

L^{2}

\overset{m}{˙} = ρ v L^{2}

; each unit mass carries velocity

v

, so momentum delivered per second (a force) is

\overset{m}{˙} v = ρ v^{2} L^{2}

Why does the viscous force scale as

μv L

Shear stress

τ = μ d v / d y \approx μv / L

(speed drops by

v

over thickness

L

); multiplying stress by area

L^{2}

gives

μ (v / L) L^{2} = μv L

Why does the drag coefficient

C_{d}

depend on

R e

rather than being a fixed constant?

[[Drag force and drag coefficient|

C_{d}

]] captures how the flow structure (attached vs separated, laminar vs turbulent wake) changes, and that structure is set by the inertia-to-viscosity balance — i.e. by

R e

Why can a very smooth pipe stay laminar up to

R e \sim 1 0^{5}

With no roughness to seed disturbances, there is almost nothing for inertia to amplify, so viscosity keeps the flow ordered far past the usual

\sim 2000

threshold. This shows

R e_{c}

is empirical, not fundamental.

Edge cases

Push the formula to its limits.

What is

R e

for a fluid at rest (

v = 0

R e = 0

. With no motion there is no inertial force, so the flow is trivially laminar (in fact motionless) — viscosity has nothing to fight.

What happens to

R e

as viscosity

μ \to 0

(an ideal/inviscid fluid)?

R e \to \infty

. This is the idealisation behind Bernoulli's principle: inertia utterly dominates, and there is no viscous mechanism to keep flow laminar.

What does

R e

do as the characteristic size

L \to 0

(e.g. a bacterium swimming)?

R e \to 0

. Micro-scale swimmers live in a world where viscosity dominates completely — inertia is negligible, so motion stops the instant propulsion stops.

Two pipes carry the same fluid at the same speed but one has twice the diameter — which has the larger

R e

The wider pipe.

R e \propto L

, so doubling the diameter doubles

R e

, making the larger pipe more prone to turbulence.

If you keep

R e

fixed but scale the whole system up (larger

L

), what must change to compensate?

Speed or kinematic viscosity must adjust: since

R e = v L / ν

, a larger

L

needs a proportionally smaller

v

(or larger

ν

) to hold

R e

constant — the basis of scale-model testing.

Is turbulence possible at low

R e

if you shake the fluid violently?

Not sustained turbulence. Strong forcing can create momentary disorder, but at low

R e

viscosity re-damps the disturbances quickly, so the flow relaxes back to laminar.

Active Recall

Recall One-line self-check

Q: Which force is on the bottom of $R e$ ? → viscous ( $μ$ ).
Q: Units of $R e$ ? → none, dimensionless.
Q: $R e = 3000$ regime? → transitional.
Q: $L$ for a pipe? → diameter $D$ .
Q: $μ \to 0$ gives $R e \to$ ? → $\infty$ (inviscid limit).

Connections

Parent topic (Hinglish)
Viscosity and Newton's law of viscosity — the $τ = μ d v / d y$ behind the viscous force.
Poiseuille's law — laminar-only regime.
Stokes' law and terminal velocity — low- $R e$ drag.
Bernoulli's principle — the $R e \to \infty$ idealisation.
Dimensional analysis — why dimensionless ratios are universal.
Drag force and drag coefficient — $C_{d}$ as a function of $R e$ .