Reynolds number Re = ρvL - μ — laminar vs turbulent criterion: Worked examples

Before anything, one reminder of the tools we lean on:

The scenario matrix

Every problem about $R e$ falls into one of these cells. The examples below are each tagged with the cell they cover.

#	Scenario class	What makes it tricky	Example
A	Plain pipe — plug and classify	Getting SI units right	Ex 1
B	High-viscosity limit ( $μ$ huge)	$R e \to$ tiny, always laminar	Ex 2
C	High-speed / large-body limit	$R e \to$ huge, always turbulent	Ex 3
D	Solve for the threshold (find $v_{c}$ , or $L_{c}$ )	Rearranging the formula	Ex 4
E	Degenerate input: $v = 0$ or $L = 0$	$R e \to 0$ ; is "no flow" laminar?	Ex 5
F	Radius-vs-diameter trap	Which $L$ does the table assume?	Ex 6
G	Unit-conversion word problem (cm, mm, cP, air)	Convert before plugging	Ex 7
H	Kinematic-viscosity form ( $ν$ given, not $μ, ρ$ )	Use $R e = v L / ν$ directly	Ex 8
I	Exam twist: scaling / "what if we double it?"	Reason with ratios, no calculator	Ex 9
J	Two fluids, same $R e$ (dynamic similarity)	Match $R e$ across a model & real object	Ex 10
K	Right in the middle ( $2000 < R e < 4000$ )	Reading the transitional band	Ex 11

The regime rule we classify against (pipe convention, $L =$ diameter):

$R e$	Regime
$< 2000$	laminar
$2000$ – $4000$	transitional
$> 4000$	turbulent

Figure — Reynolds number Re = ρvL - μ — laminar vs turbulent criterion

The worked examples

Worked example Example 1 — Cell A: plain pipe, classify

Water ( $ρ = 1000 kg/m^{3}$ , $μ = 1.0 \times 1 0^{- 3} Pa\cdots$ ) flows at $v = 2 m/s$ in a pipe of diameter $L = 0.05 m$ . Laminar or turbulent?

Forecast: water, moderately fast, a wide pipe — before computing, guess: turbulent or laminar?

Write the formula: $R e = \frac{ρ v L}{μ}$ . Why this step? We need a single number to place on the number line; this is the only quantity that decides the regime.
Substitute SI values: $R e = \frac{1000 \times 2 \times 0.05}{1.0 \times 1 0 ^{- 3}}$ . Why this step? Everything is already in SI, so the units cancel automatically and we get a pure number — no unit bookkeeping needed.
Simplify the top: $1000 \times 2 \times 0.05 = 100$ . Then $R e = \frac{100}{1 0 ^{- 3}} = 100000$ . Why this step? Do the numerator first so the division is a clean "divide by a small number = multiply up."
Classify: $100000 ≫ 4000 \Rightarrow$ turbulent.

Verify: Units check — $\frac{( kg/m ^{3} ) ( m/s ) ( m )}{kg/(m\cdots)} = \frac{kg/(m\cdots)}{kg/(m\cdots)} = 1$ , dimensionless. ✓ The forecast (turbulent) matches. On the figure this dot sits deep in the red band.

Worked example Example 2 — Cell B: high-viscosity limit

Glycerine ( $ρ = 1260 kg/m^{3}$ , $μ = 1.5 Pa\cdots$ ) flows at $v = 2 m/s$ in the same $L = 0.05 m$ pipe. Compare to Example 1.

Forecast: same speed, same pipe, but $1500 \times$ more viscous than water. Which cell of the number line?

$R e = \frac{1260 \times 2 \times 0.05}{1.5}$ . Why this step? Only $ρ$ and $μ$ changed vs Example 1, so we expect the answer to move by roughly $\frac{1260/1.5}{1000/1 0 ^{- 3}}$ — a huge shrink.
Numerator: $1260 \times 2 \times 0.05 = 126$ . So $R e = \frac{126}{1.5} = 84$ . Why this step? The big $μ$ sits in the denominator; dividing by $1.5$ instead of $0.001$ crushes $R e$ by a factor $\approx 1500$ .
Classify: $84 < 2000 \Rightarrow$ laminar — in fact deeply laminar.

Verify: Ratio sanity: Ex 1 was $1 0^{5}$ ; multiplying $μ$ by $1500$ and $ρ$ by $1.26$ predicts $R e \approx 1 0^{5} \times \frac{1.26}{1500} \approx 84$ . ✓ Confirms the "viscosity is the peacemaker" rule: it lives on the bottom. On the figure this dot sits far left in the green band.

Worked example Example 3 — Cell C: high-speed / large-body limit

A submarine of length $L = 60 m$ cruises at $v = 10 m/s$ through seawater ( $ρ = 1025 kg/m^{3}$ , $μ = 1.1 \times 1 0^{- 3} Pa\cdots$ ). Estimate $R e$ .

Forecast: huge body, decent speed, thin water — do you expect a small or an astronomically large number?

$R e = \frac{1025 \times 10 \times 60}{1.1 \times 1 0 ^{- 3}}$ . Why this step? For a body (not a pipe) $L$ is the body's characteristic length; the same formula still applies because it is a ratio of the same two forces.
Numerator: $1025 \times 10 \times 60 = 615000$ . Then $R e = \frac{615000}{1.1 \times 1 0 ^{- 3}} \approx 5.59 \times 1 0^{8}$ . Why this step? Big $L$ and big $v$ both multiply the top; the tiny $μ$ divides — three effects all push $R e$ up.
Classify: $5.6 \times 1 0^{8} ⋙ 4000 \Rightarrow$ strongly turbulent.

Verify: Order-of-magnitude: $\frac{1 0 ^{3} \cdot 10 \cdot 60}{1 0 ^{- 3}} = \frac{6 \times 1 0 ^{5}}{1 0 ^{- 3}} = 6 \times 1 0^{8}$ . ✓ Any real vehicle in water/air lives at $R e ≫ 1 0^{5}$ — the everyday world is turbulent.

Worked example Example 4 — Cell D: solve for the threshold speed

For the Example-1 water in the $L = 0.05 m$ pipe, at what speed $v_{c}$ does the flow first become transitional ( $R e = 2000$ )?

Forecast: Example 1 had $v = 2 m/s$ giving $R e = 1 0^{5}$ . To get down to $R e = 2000$ we need $v$ smaller by roughly $50 \times$ . Guess $v_{c}$ .

Rearrange $R e = \frac{ρ v L}{μ}$ for $v$ : $v_{c} = \frac{R e μ}{ρ L}$ . Why this step? The unknown is speed, so we isolate $v$ — divide both sides by $ρ L / μ$ .
Substitute: $v_{c} = \frac{2000 \times 1 0 ^{- 3}}{1000 \times 0.05} = \frac{2}{50}$ . Why this step? Put the target $R e = 2000$ in the numerator; everything else is the geometry/fluid we're holding fixed.
Compute: $v_{c} = 0.04 m/s$ . At exactly $R e = 2000$ this is the edge of laminar / start of transition (see the endpoint box above).

Verify: Plug back: $R e = \frac{1000 \times 0.04 \times 0.05}{1 0 ^{- 3}} = \frac{2}{1 0 ^{- 3}} = 2000$ . ✓ And the ratio check: $2 m/s$ gave $1 0^{5}$ ; $2000$ is $50 \times$ smaller than $1 0^{5}$ , so speed must be $50 \times$ smaller: $2/50 = 0.04 m/s$ . ✓

Worked example Example 5 — Cell E: degenerate input,

$v \to 0$ The pump in the Example-1 pipe is switched off, so $v = 0$ . What is $R e$ , and what does that mean physically?

Forecast: Is "no flow" laminar, turbulent, or neither?

$R e = \frac{ρ \cdot 0 \cdot L}{μ} = 0$ . Why this step? $v$ is a factor in the numerator; a zero factor makes the whole product zero — no algebra tricks needed.
Interpret: $R e = 0 < 2000$ , so it is (trivially) in the laminar range. Why this step? The number line has no lower cutoff; every $R e$ below $2000$ counts as laminar, including the degenerate value $0$ .
Physical caveat: with $v = 0$ there is no flow at all — nothing is moving, so "laminar" here just means "no chaos," which is true but empty. Likewise $L = 0$ (a pinched-shut pipe) gives $R e = 0$ : no cross-section, no flow. Why this step? Extreme inputs must be checked for meaning, not just for arithmetic — $R e = 0$ is mathematically fine but describes a static fluid.

Verify: As $v \to 0$ , $R e \to 0$ continuously; there is no sudden jump, so the "laminar at rest" reading is self-consistent. ✓ Turbulence needs energy input; with $v = 0$ there is none.

Worked example Example 6 — Cell F: the radius-vs-diameter trap

A student is told a pipe has radius $r = 0.01 m$ , water at $v = 0.2 m/s$ ( $ρ = 1000$ , $μ = 1 0^{- 3}$ ). They plug $L = r$ and get one answer; the correct convention uses $L = D$ (diameter). Show both and the wrong conclusion.

Forecast: using radius instead of diameter changes $R e$ by what factor — and could it flip the regime?

Wrong (radius): $R e_{wrong} = \frac{1000 \times 0.2 \times 0.01}{1 0 ^{- 3}} = \frac{2}{1 0 ^{- 3}} = 2000$ . Why this step? Show exactly what the mistaken plug-in produces so we can see the error's size.
Right (diameter $D = 2 r = 0.02 m$ ): $R e = \frac{1000 \times 0.2 \times 0.02}{1 0 ^{- 3}} = \frac{4}{1 0 ^{- 3}} = 4000$ . Why this step? The standard pipe table is calibrated to diameter; using $D$ doubles $L$ and therefore doubles $R e$ .
Consequence, read with the endpoint box: the radius plug gives $R e = 2000$ = edge of laminar / start of transition; the correct diameter gives $R e = 4000$ = edge of transition / start of turbulent. So the trap shifts the flow across the entire transitional band — from "just leaving laminar" to "just entering turbulent." Why this step? This is why the mistake matters — it is not a rounding error; both endpoints are borderline, but they sit at opposite ends of the transition zone, so the practical verdict changes.

Verify: $R e_{D} / R e_{r} = D / r = 2$ exactly, since $R e \propto L$ . ✓ Always confirm which $L$ your table assumes before trusting the regime.

Worked example Example 7 — Cell G: unit-conversion word problem (air, with cP defined)

Air blows through a ventilation duct of diameter $50 cm$ at $v = 3 m/s$ . Air has $ρ = 1.2 kg/m^{3}$ and a viscosity quoted as $μ = 18 cP$ . Classify.

What is a cP? The centipoise (cP) is a common non-SI viscosity unit. The conversion is $1 cP = 1 0^{- 3} Pa\cdots$ (handily, pure water at room temperature is $\approx 1 cP$ ). So $18 cP = 18 \times 1 0^{- 3} cP \dots$ — no: $18 cP = 18 \times 1 0^{- 3} Pa\cdots$ ? Careful: $18 cP = 18 \times 1 0^{- 3} Pa\cdots = 0.018 Pa\cdots$ would be far too thick for air. Air is $\approx 0.018 cP$ , i.e. $μ_{air} = 0.018 cP = 1.8 \times 1 0^{- 5} Pa\cdots$ . We use this correct value below.

Forecast: air is light and thin — does the light $ρ$ (small numerator) or the thin $μ$ (small denominator) win?

Convert viscosity to SI: $μ = 0.018 cP \times \frac{1 0 ^{- 3} Pa\cdots}{1 cP} = 1.8 \times 1 0^{- 5} Pa\cdots$ . Why this step? The formula needs Pa·s; leaving $μ$ in cP would make $R e$ wrong by $1 0^{3}$ . Convert every non-SI quantity before plugging.
Convert length: $50 cm = 0.50 m$ . Why this step? Mixing cm and m would multiply $R e$ by $100$ and ruin the answer.
Substitute: $R e = \frac{1.2 \times 3 \times 0.50}{1.8 \times 1 0 ^{- 5}}$ . Why this step? Now every quantity is in SI, so units cancel and we get a pure number.
Numerator: $1.2 \times 3 \times 0.50 = 1.8$ . Then $R e = \frac{1.8}{1.8 \times 1 0 ^{- 5}} = 100000$ . Why this step? The tiny denominator ( $1 0^{- 5}$ ) wins decisively over the small numerator — air's thinness dominates.
Classify: $100000 > 4000 \Rightarrow$ turbulent. Real ventilation ducts are indeed turbulent (that's why they mix air well).

Verify: If we had forgotten to convert and used $L = 50$ (as cm), we'd get $R e = 1 0^{7}$ — obviously absurd for a duct, a red flag that conversion was skipped. Correct value $1 0^{5}$ . ✓

Worked example Example 8 — Cell H: kinematic-viscosity form

Oil flows in a pipe with $L = 0.03 m$ at $v = 0.5 m/s$ . You're given only its kinematic viscosity $ν = 1.0 \times 1 0^{- 4} m^{2} / s$ (density and $μ$ not given). Find $R e$ .

Forecast: $ν$ already bundles $μ / ρ$ together — do you even need $ρ$ ?

Use the second form: $R e = \frac{v L}{ν}$ . Why this step? When $ν$ is supplied directly, this form skips $ρ$ and $μ$ entirely — fewer numbers, fewer errors.
Substitute: $R e = \frac{0.5 \times 0.03}{1.0 \times 1 0 ^{- 4}} = \frac{0.015}{1 0 ^{- 4}}$ . Why this step? Compute the numerator $v L = 0.015 m^{2} / s$ ; dividing by $ν$ (also $m^{2} / s$ ) cancels units to a pure number.
Compute: $R e = 150$ . Why this step? Divide $0.015$ by $0.0001$ = multiply by $1 0^{4}$ then by $1.5$ → $150$ .
Classify: $150 < 2000 \Rightarrow$ laminar.

Verify: Both forms agree because $ν = μ / ρ$ , so $\frac{v L}{ν} = \frac{ρ v L}{μ}$ . Units: $\frac{( m/s ) ( m )}{m ^{2} / s} = 1$ . ✓

Worked example Example 9 — Cell I: scaling twist (no calculator)

A flow is laminar at $R e = 1800$ . The engineers double the pipe diameter and triple the flow speed, keeping the same fluid. New $R e$ ? New regime?

Forecast: guess the multiplier on $R e$ before doing any arithmetic.

Note $R e \propto v L$ (with $ρ, μ$ fixed). Doubling $L$ and tripling $v$ multiplies $R e$ by $2 \times 3 = 6$ . Why this step? Because $R e$ is linear in both $v$ and $L$ , changes just multiply — no need for the actual $ρ, μ$ values.
New $R e = 6 \times 1800 = 10800$ . Why this step? Apply the multiplier to the known starting value.
Classify: $10800 > 4000 \Rightarrow$ turbulent. The flow flipped from laminar all the way through transition to turbulent. Why this step? Report the regime change — that's the point of the twist.

Verify: Linearity check: if instead we halved both, factor $\frac{1}{4}$ , giving $450$ — still laminar, consistent with " $R e$ scales as $v L$ ." ✓

Worked example Example 10 — Cell J: dynamic similarity (two fluids, same

$R e$ ) A model car ( $L_{m} = 0.4 m$ ) is tested in a water tunnel ( $ν_{water} = 1.0 \times 1 0^{- 6} m^{2} / s$ ) to mimic a real car ( $L_{r} = 4 m$ ) driving in air ( $ν_{air} = 1.5 \times 1 0^{- 5} m^{2} / s$ ) at $v_{r} = 30 m/s$ . What test speed $v_{m}$ makes the two flows match ( $R e_{m} = R e_{r}$ )?

Forecast: water is $15 \times$ less viscous (kinematically) but the model is $10 \times$ smaller — do these effects add up to a fast or slow test speed?

Real-car Reynolds number: $R e_{r} = \frac{v _{r} L _{r}}{ν _{air}} = \frac{30 \times 4}{1.5 \times 1 0 ^{- 5}} = \frac{120}{1.5 \times 1 0 ^{- 5}} = 8 \times 1 0^{6}$ . Why this step? Dynamic similarity means the dimensionless number must be equal; first find the target $R e$ to match.
Set $R e_{m} = R e_{r}$ and solve for the model speed: $v_{m} = \frac{R e _{r} ν _{water}}{L _{m}} = \frac{8 \times 1 0 ^{6} \times 1.0 \times 1 0 ^{- 6}}{0.4}$ . Why this step? Rearranging $R e_{m} = v_{m} L_{m} / ν_{water}$ isolates the only unknown, $v_{m}$ .
Compute: numerator $8 \times 1 0^{6} \times 1 0^{- 6} = 8$ ; then $v_{m} = \frac{8}{0.4} = 20 m/s$ . Why this step? This test speed gives the model the same $R e$ as the real car, so the flow pattern is physically identical.
Classify: both flows share $R e = 8 \times 1 0^{6} > 4000 \Rightarrow$ turbulent for model and real car — that is the whole point: they are in the same regime. Why this step? Equal $R e$ ⇒ same regime, same wake, same drag coefficient.

Verify: Check both sides: $R e_{m} = \frac{20 \times 0.4}{1 0 ^{- 6}} = \frac{8}{1 0 ^{- 6}} = 8 \times 1 0^{6} = R e_{r}$ . ✓ Same $R e$ ⇒ same regime — this is exactly why wind-tunnel testing works. See Drag force and drag coefficient.

Worked example Example 11 — Cell K: right in the middle (transitional band)

Water ( $ρ = 1000 kg/m^{3}$ , $μ = 1.0 \times 1 0^{- 3} Pa\cdots$ ) flows at $v = 0.15 m/s$ in a pipe of diameter $L = 0.02 m$ . Which regime?

Forecast: slow-ish water in a narrow-ish pipe — will it land cleanly in laminar, cleanly in turbulent, or awkwardly between?

$R e = \frac{1000 \times 0.15 \times 0.02}{1.0 \times 1 0 ^{- 3}}$ . Why this step? Same plug-and-classify as Ex 1, but chosen so the answer falls between the two thresholds — the case the table's middle row describes.
Numerator: $1000 \times 0.15 \times 0.02 = 3$ . Then $R e = \frac{3}{1 0 ^{- 3}} = 3000$ . Why this step? Compute the top first, then divide by the small $μ$ .
Classify with the endpoint box: $2000 < 3000 < 4000 \Rightarrow$ transitional. Why this step? $3000$ is strictly inside the yellow band — neither $\leq 2000$ nor $\geq 4000$ — so it is genuinely in the middle regime.
What "transitional" means physically: the flow flickers — patches of smooth laminar motion appear and break into turbulent bursts intermittently, and the exact behaviour is sensitive to pipe roughness and vibration. Engineers avoid designing in this band because it is unpredictable. Why this step? Naming the regime is not enough; the transitional band is defined by intermittency, and that is the exam-relevant physical fact.

Verify: Plug back and locate on the figure: $R e = 3000$ sits squarely in the yellow band, between the $2000$ and $4000$ ticks. To push it laminar you'd need $v < 0.1 m/s$ ( $R e < 2000$ ); to push it turbulent, $v > 0.2 m/s$ ( $R e > 4000$ ). ✓

Recall Quick self-test across all cells

Doubling only the speed does what to $R e$ ? ::: Doubles it (linear in $v$ ). Switching from radius to diameter as $L$ does what? ::: Doubles $R e$ . $v = 0$ gives $R e =$ ? and which regime? ::: $R e = 0$ , trivially laminar (but no flow). Given $ν$ instead of $μ, ρ$ , which formula? ::: $R e = v L / ν$ . Two flows with equal $R e$ are called? ::: Dynamically similar (same regime & pattern). $R e = 3000$ is which regime? ::: Transitional (between 2000 and 4000). How do you read the exact endpoint $R e = 2000$ ? ::: Borderline — edge of laminar / start of transition.

Connections

Parent topic — the derivation these examples apply.
Viscosity and Newton's law of viscosity — where $μ$ and $ν = μ / ρ$ come from.
Poiseuille's law — the laminar-only law (valid for Ex 2, 5, 8).
Stokes' law and terminal velocity — the low- $R e$ sphere case (relatives of Ex 2).
Bernoulli's principle — the $R e \to \infty$ inviscid idealisation (Ex 3).
Dimensional analysis — why matching $R e$ (Ex 10) guarantees similarity.
Drag force and drag coefficient — $C_{d} (R e)$ , the payoff of Ex 10's similarity.