Boundary layer — Prandtl's concept, growth along flat plate: Worked examples

This page is the drill floor. The parent note built the ideas; here we throw every kind of question at them and solve each one from the ground up. Nothing new is assumed — every symbol used below was earned in the parent, and we re-anchor each one as it appears.

Before touching numbers, let's list the master formulas we will keep reusing. Read each right-hand side as a sentence, not a spell.

Formula The tools we lean on

Local Reynolds number — a pure number comparing "push" (inertia) to "stickiness" (viscosity) at distance $x$ from the leading edge: $R e_{x} = \frac{U x}{ν}$ Here $U$ = free-stream speed (m/s), $x$ = distance downstream from the plate's front tip (m), $ν$ = kinematic viscosity ( $m^{2} / s$ , the "how fast momentum diffuses" number). See Reynolds number.
Laminar layer thickness (Blasius, exact constant), valid only while $R e_{x} < 5 \times 1 0^{5}$ : $δ (x) = \frac{5.0 x}{R e _{x}} = 5.0 \frac{ν x}{U}$
Turbulent layer thickness (empirical $1/5$ -power law, valid for $R e_{x} ≳ 5 \times 1 0^{5}$ ): $δ (x) = \frac{0.37 x}{R e _{x}^{1/5}} = 0.37 x (\frac{ν}{U x})^{1/5}$
Wall shear stress — the friction the fluid drags on the plate per unit area, $N/m^{2}$ : $τ_{w} \sim μ \frac{U}{δ}, μ = ρ ν$ where $μ$ = dynamic viscosity ( $Pa\cdots$ ) and $ρ$ = fluid density ( $kg/m^{3}$ ). See Viscosity and Newton's law of viscosity.

Two coordinate symbols will recur, so let's anchor them once here (both are inherited from the parent's picture of the plate): $x$ is the distance measured along the plate from the leading edge, and $y$ is the height measured straight up from the wall (so $y = 0$ is the plate surface, $y = δ$ is the top of the boundary layer). The symbol $u$ means the fluid's horizontal speed at a given height $y$ — it climbs from $u = 0$ at the wall (no-slip) up to $u = U$ at the layer's edge. The expression $\partial u / \partial y$ is simply the slope of that speed-versus-height curve: how fast the speed changes as you rise a little off the wall.

One extra convention we need repeatedly: the flow is laminar (smooth layers) roughly while $R e_{x} < 5 \times 1 0^{5}$ , and beyond that it goes turbulent (mixed-up eddies). See Laminar vs Turbulent flow. This threshold decides which formula is even allowed — Blasius below it, the $1/5$ -power law above it.

The scenario matrix

Every fluid-boundary-layer question is really one of these cells. The examples below tick each box.

Cell	Case class	What's special about it	Example
A	Direct forward calc	given $U, ν, x$ → find $δ$	Ex 1
B	Degenerate input: leading edge	$x \to 0$	Ex 2
C	Limiting behaviour: $U \to \infty$	very fast stream, layer $\to 0$	Ex 2
D	Scaling / ratio (only one variable changes)	use proportionality, no full numbers	Ex 3
E	Inverse problem	given $δ$ → find $x$ (or $U$ )	Ex 4
F	Transition check ( $R e_{x}$ vs $5 \times 1 0^{5}$ )	is laminar formula even legal?	Ex 5
G	Wall shear / skin friction	$τ_{w} \propto 1/ x$ , biggest at nose	Ex 6
H	Real-world word problem (units mixed, air)	translate story → symbols	Ex 7
I	Exam twist: two fluids / combine effects	change $ν$ AND $U$ together	Ex 8
J	Past transition: turbulent growth law	$R e_{x} > 5 \times 1 0^{5}$ , use $δ \propto x / R e_{x}^{1/5}$	Ex 9

Example 1 — Cell A: the plain forward calculation

Worked example Water over a plate

Water ( $ν = 1.0 \times 1 0^{- 6} m^{2} / s$ ) streams at $U = 2 m/s$ past a flat plate. Find $δ$ at $x = 0.2 m$ .

Forecast: Water is only mildly sticky and the plate is short — guess a millimetre-scale layer, not centimetres.

Step 1 — Compute $R e_{x}$ and check it is in range. $R e_{x} = \frac{U x}{ν} = \frac{2 \times 0.2}{1.0 \times 1 0 ^{- 6}} = 4.0 \times 1 0^{5}$ Why this step? Before choosing a thickness formula we must check which regime we are in. Here $R e_{x} = 4.0 \times 1 0^{5} < 5 \times 1 0^{5}$ , so the flow is laminar and the Blasius formula is legally applicable. (Had it exceeded $5 \times 1 0^{5}$ we would be forced to the turbulent law of Ex 9.)

Step 2 — Feed it into Blasius. $δ = \frac{5.0 x}{R e _{x}} = \frac{5.0 \times 0.2}{4.0 \times 1 0 ^{5}} = \frac{1.0}{632.5} = 1.58 \times 1 0^{- 3} m$ Why this step? The $R e_{x}$ in the denominator is exactly what turns "how thin" into a length. Large denominator → tiny $δ$ .

Answer: $δ \approx 1.58 mm$ .

Verify: Units — $x$ is metres, $R e_{x}$ is dimensionless, so $δ$ is metres. ✓ Magnitude ~1.6 mm matches the forecast, and $R e_{x}$ sits safely below transition so the formula was legal. ✓

Example 2 — Cells B & C: degenerate leading edge and infinite speed

Worked example Two limits at once

For the same water flow ( $ν = 1 0^{- 6}$ , $U = 2$ ): (a) what is $δ$ at the leading edge $x = 0$ ? (b) What happens to $δ$ at fixed $x = 0.2 m$ as $U \to \infty$ ?

Forecast: (a) The plate has just started grabbing fluid — guess zero. (b) A hurricane-fast stream gives the wall no time to slow anything — guess $δ \to 0$ .

Step 1 — Leading edge. Put $x = 0$ into $δ = 5.0 ν x / U$ : $δ (0) = 5.0 \frac{ν \cdot 0}{U} = 0$ Why this step? At the very front tip, no fluid has yet spent any time near the wall, so nothing has been slowed. The formula must (and does) give zero.

Step 2 — Infinite speed. Hold $x$ fixed, let $U \to \infty$ : $δ = 5.0 \frac{ν x}{U} U \to \infty 5.0 ν x \cdot \frac{1}{U} \to 0$ Why this step? $δ \propto 1/ U$ . Faster stream = less "exposure time" for viscosity to diffuse sideways, so the slowed zone can't grow. In the ideal limit ( $ν \to 0$ or $U \to \infty$ ) the layer vanishes — recovering the frictionless picture that gave d'Alembert's paradox.

Verify: Both limits push $δ$ toward $0$ , consistent with "high $R e_{x}$ ⇒ thin layer." Keeping the full Blasius constant, $δ = 5.0 x / R e_{x}$ ; since $R e_{x} = U x / ν \to \infty$ as $U \to \infty$ , the denominator blows up and $δ \to 0$ . ✓

Example 3 — Cell D: pure scaling (no calculator marathon)

Worked example Where is the layer twice as thick?

On a plate, the layer has thickness $δ_{1}$ at $x_{1} = 0.5 m$ . At what station $x_{2}$ is it twice as thick, everything else fixed (still laminar)?

Forecast: Because $δ \propto x$ , doubling thickness needs more than double the distance. Guess ~ $4 \times$ .

Step 1 — Write the ratio. With $U, ν$ fixed, only $x$ survives: $\frac{δ _{2}}{δ _{1}} = \frac{x _{2}}{x _{1}}$ Why this step? Constants cancel in a ratio, so we never need their values — the shape of the growth law is all that matters.

Step 2 — Set the ratio to 2 and solve. $2 = \frac{x _{2}}{x _{1}} \Rightarrow \frac{x _{2}}{x _{1}} = 4 \Rightarrow x_{2} = 4 (0.5) = 2.0 m$ Why this step? Squaring undoes the square root. The is exactly why quadrupling distance only doubles thickness — growth slows downstream.

Verify: $2.0/0.5 = 4 = 2$ . ✓ Matches the forecast " $4 \times$ distance." ✓

Example 4 — Cell E: the inverse problem

Worked example Read the thickness, recover the speed

Air ( $ν = 1.5 \times 1 0^{- 5} m^{2} / s$ ) flows over a plate. At $x = 0.3 m$ the laminar layer is measured to be $δ = 3.0 mm$ . What is the free-stream speed $U$ ?

Forecast: A fairly thick 3 mm layer in low-viscosity air over 30 cm suggests a slow stream (fast stream would give a thinner sliver). Guess a few m/s.

Step 1 — Invert Blasius for $U$ . Start from $δ = 5.0 ν x / U$ , square both sides: $δ^{2} = 25 \frac{ν x}{U} \Rightarrow U = \frac{25 ν x}{δ ^{2}}$ Why this step? We want $U$ alone; squaring removes the root and lets us isolate it algebraically.

Step 2 — Plug numbers ( $δ = 3.0 \times 1 0^{- 3} m$ ): $U = \frac{25 \times ( 1.5 \times 1 0 ^{- 5} ) \times 0.3}{( 3.0 \times 1 0 ^{- 3} ) ^{2}} = \frac{1.125 \times 1 0 ^{- 4}}{9.0 \times 1 0 ^{- 6}} = 12.5 m/s$ Why this step? Direct substitution once $U$ is isolated.

Answer: $U = 12.5 m/s$ .

Verify (plug back): $R e_{x} = \frac{12.5 \times 0.3}{1.5 \times 1 0 ^{- 5}} = 2.5 \times 1 0^{5}$ ; then $δ = \frac{5 \times 0.3}{2.5 \times 1 0 ^{5}} = \frac{1.5}{500} = 3.0 \times 1 0^{- 3} m$ . ✓ Exactly the given 3 mm. And $R e_{x} < 5 \times 1 0^{5}$ so laminar was legal. ✓

Example 5 — Cell F: is the laminar formula even allowed?

Worked example Transition check

Water ( $ν = 1.0 \times 1 0^{- 6}$ ) at $U = 2 m/s$ flows over a long plate. Up to what distance $x_{c}$ is the flow laminar, and what is $δ$ there?

Forecast: Transition is set by $R e_{x} = 5 \times 1 0^{5}$ . Since water diffuses momentum slowly, guess the laminar zone ends within tens of centimetres.

Step 1 — Set $R e_{x}$ to the critical value and solve for $x_{c}$ . $R e_{x, crit} = 5 \times 1 0^{5} = \frac{U x _{c}}{ν} \Rightarrow x_{c} = \frac{5 \times 1 0 ^{5} ν}{U} = \frac{5 \times 1 0 ^{5} \times 1 0 ^{- 6}}{2} = 0.25 m$ Why this step? Beyond $x_{c}$ the boundary layer trips into turbulence, and the smooth-layer Blasius formula no longer describes it. We must locate that frontier first. See Laminar vs Turbulent flow.

Step 2 — Thickness at transition. $δ (x_{c}) = \frac{5.0 \times 0.25}{5 \times 1 0 ^{5}} = \frac{1.25}{707.1} = 1.77 \times 1 0^{- 3} m$ Why this step? This is the thickest the laminar layer ever gets; past here we'd switch to a turbulent growth law (see Ex 9).

Answer: laminar up to $x_{c} = 0.25 m$ , where $δ \approx 1.77 mm$ .

Verify: $R e_{x_{c}} = 2 \times 0.25/1 0^{- 6} = 5 \times 1 0^{5}$ ✓ exactly critical. Units of $x_{c}$ : $[ν] / [U] = (m^{2} / s) / (m/s) = m$ . ✓

Example 6 — Cell G: wall shear stress, biggest at the nose

Figure 6.1 below is the visual backbone of this example — read the caption first, then follow how the wall slope shrinks downstream.

Figure — Boundary layer — Prandtl's concept, growth along flat plate

Figure 6.1 — Velocity profiles $u (y)$ at two stations along the plate. Blue arrows on the left are the uniform free stream $U$ . At each station a coloured curve shows how the fluid speed $u$ climbs from $0$ at the wall ( $y = 0$ ) up to $U$ at the layer edge $y = δ$ (dashed yellow curve $δ \propto x$ ). Near the nose (pink) the curve is squeezed into a thin layer, so its wall slope $\partial u / \partial y$ — the short pink arrow — is steep; downstream (blue) the layer is fatter and the wall slope is gentle. Because $τ_{w} \sim μ U / δ$ and $δ$ is smallest at the nose, friction is fiercest at the leading edge and fades as $1/ x$ .

Worked example Where does the plate feel the most friction?

Air ( $ρ = 1.2 kg/m^{3}$ , $ν = 1.5 \times 1 0^{- 5} m^{2} / s$ , so $μ = ρ ν = 1.8 \times 1 0^{- 5} Pa\cdots$ ) flows at $U = 10 m/s$ . Estimate the wall shear stress $τ_{w}$ at $x = 0.1 m$ , and say where along the plate friction is strongest. (Recall from the master list: $u$ is the horizontal speed at height $y$ above the wall, and $\partial u / \partial y$ is the slope of that speed profile.)

Forecast: In Figure 6.1 the velocity profile $u (y)$ is steepest right at the leading edge (thinnest layer). Steepest slope → biggest stress. Guess: friction is largest at the nose and fades as $1/ x$ .

Step 1 — Confirm laminar, then find thickness at $x = 0.1$ m. $R e_{x} = \frac{10 \times 0.1}{1.5 \times 1 0 ^{- 5}} = 6.67 \times 1 0^{4} < 5 \times 1 0^{5} (laminar), δ = \frac{5 \times 0.1}{6.67 \times 1 0 ^{4}} = \frac{0.5}{258.2} = 1.94 \times 1 0^{- 3} m$ Why this step? $τ_{w}$ needs $δ$ : the wall slope $\partial u / \partial y$ is roughly (full speed $U$ ) ÷ (layer thickness $δ$ ). Thin layer → steep slope → big stress. We check $R e_{x}$ first so we know Blasius is legal.

Step 2 — Estimate $τ_{w} \sim μU / δ$ . $τ_{w} \sim μ \frac{\partial u}{\partial y}_{y = 0} \sim \frac{μU}{δ} = \frac{( 1.8 \times 1 0 ^{- 5} ) ( 10 )}{1.94 \times 1 0 ^{- 3}} = 9.3 \times 1 0^{- 2} Pa$ Why this step? Newton's law of viscosity says stress = $μ \times$ velocity slope $\partial u / \partial y$ . We approximate that slope at the wall with the order-of-magnitude ratio $U / δ$ (speed change $U$ over height $δ$ ).

Step 3 — Where is it largest? Since $δ \propto x$ , we get $τ_{w} \propto μU / ν x / U \propto 1/ x$ . As $x \to 0$ (the leading edge), $τ_{w} \to \infty$ in this idealisation. Why this step? The layer is thinnest at the nose, so the gradient $\partial u / \partial y$ — and thus the friction — is fiercest there. This is the origin of Skin friction drag.

Verify: Units of $τ_{w}$ : $\frac{Pa\cdots \cdot m/s}{m} = Pa$ . ✓ Direction of the trend matches the steep pink wall-slope arrow at the nose in Figure 6.1. ✓

Example 7 — Cell H: real-world word problem

Worked example Wind over a solar panel

A wind of $6 km/h$ blows along the top of a flat rooftop solar panel. Take air $ν = 1.5 \times 1 0^{- 5} m^{2} / s$ . How thick is the boundary layer $0.4 m$ from the panel's leading edge?

Forecast: A gentle breeze over half a metre of air — expect a chunky layer, several millimetres, because slow + low-viscosity-but-not-tiny-plate.

Step 1 — Translate the story into SI symbols. Speed must be m/s: $U = 6 km/h = \frac{6000 m}{3600 s} = 1.667 m/s, x = 0.4 m, ν = 1.5 \times 1 0^{- 5} m^{2} / s$ Why this step? Every formula is in SI; mixing km/h into $R e_{x}$ silently corrupts the answer. Convert first, always.

Step 2 — Reynolds number and regime check. $R e_{x} = \frac{1.667 \times 0.4}{1.5 \times 1 0 ^{- 5}} = 4.44 \times 1 0^{4}$ Why this step? Below $5 \times 1 0^{5}$ , so the laminar assumption is genuinely valid here — no need for the turbulent law.

Step 3 — Thickness. $δ = \frac{5 \times 0.4}{4.44 \times 1 0 ^{4}} = \frac{2.0}{210.8} = 9.49 \times 1 0^{- 3} m \approx 9.5 mm$ Why this step? Direct substitution once $R e_{x}$ is known and laminar is confirmed.

Answer: $δ \approx 9.5 mm$ .

Verify: Slow speed gave a thicker (~1 cm) layer than the fast-water cases (~1–2 mm), exactly as $δ \propto 1/ U$ predicts. ✓ Matches forecast "several mm." ✓

Example 8 — Cell I: exam twist, two effects at once

Worked example Change the fluid

and the speed A plate is tested first in water ( $ν_{1} = 1.0 \times 1 0^{- 6}$ ) at $U_{1} = 4 m/s$ , then in a light oil ( $ν_{2} = 4.0 \times 1 0^{- 5}$ ) at $U_{2} = 1 m/s$ , at the same station $x$ (both laminar). By what factor does $δ$ change from water to oil?

Forecast: Oil is stickier (bigger $ν$ → thicker) and slower (smaller $U$ → thicker). Both push the same way, so expect $δ$ to grow by a large factor.

Step 1 — Ratio at fixed $x$ . From $δ = 5.0 ν x / U$ , with $x$ common: $\frac{δ _{2}}{δ _{1}} = \frac{ν _{2} / U _{2}}{ν _{1} / U _{1}} = \frac{ν _{2}}{ν _{1}} \cdot \frac{U _{1}}{U _{2}}$ Why this step? Only the group $ν / U$ varies between the two runs; forming the ratio cancels the constant $5.0 x$ , so no station value is even needed.

Step 2 — Insert numbers. $\frac{δ _{2}}{δ _{1}} = \frac{4.0 \times 1 0 ^{- 5}}{1.0 \times 1 0 ^{- 6}} \cdot \frac{4}{1} = 40 \times 4 = 160 = 12.65$ Why this step? Both factors ( $40$ from viscosity, $4$ from the slower speed) multiply, so their effects compound rather than partly cancel.

Answer: the oil layer is about $12.6 \times$ thicker.

Verify: Both changes fatten the layer, and the product $40 \cdot 4 = 160 \approx 12.65$ . Sanity: $160$ lies between $144 = 12$ and $169 = 13$ . ✓ Matches forecast "large factor." ✓

Example 9 — Cell J: past transition, the turbulent growth law

Worked example The next logical step into turbulence

Air ( $ν = 1.5 \times 1 0^{- 5} m^{2} / s$ ) flows at $U = 30 m/s$ over a long wing panel. Find $δ$ at $x = 2.0 m$ — but first check whether the flow is laminar or turbulent, and use the correct law.

Forecast: At high speed over 2 m, $R e_{x}$ will be many millions, far past transition. So we must switch to the turbulent law. Turbulent layers mix vigorously and grow faster than laminar ones, so expect a centimetre-scale layer, much fatter than the millimetre laminar cases.

Step 1 — Reynolds number and transition test. $R e_{x} = \frac{U x}{ν} = \frac{30 \times 2.0}{1.5 \times 1 0 ^{- 5}} = 4.0 \times 1 0^{6}$ Why this step? Compare with $5 \times 1 0^{5}$ : since $4.0 \times 1 0^{6} > 5 \times 1 0^{5}$ , the flow is turbulent here, so Blasius ( $δ = 5 x / R e_{x}$ ) is not allowed — using it would badly under-predict the thickness. This is the very cell (J) the earlier examples deliberately stayed below.

Step 2 — Apply the turbulent $1/5$ -power law. $δ = \frac{0.37 x}{R e _{x}^{1/5}} = \frac{0.37 \times 2.0}{( 4.0 \times 1 0 ^{6} ) ^{1/5}}$ Why this step? In turbulent flow, mixing eddies spread the wall's influence with a weaker $R e_{x}^{- 1/5}$ (not $R e_{x}^{- 1/2}$ ) dependence, so the layer thickens more quickly downstream. This is the empirical replacement for Blasius.

Step 3 — Crunch the fifth root. $(4.0 \times 1 0^{6})^{1/5} = 20.91, δ = \frac{0.74}{20.91} = 3.54 \times 1 0^{- 2} m \approx 35.4 mm$ Why this step? The fifth root of $4 \times 1 0^{6}$ is a much smaller divisor than the square root ( $\approx 2000$ ) would be, which is why turbulent layers come out thicker.

Answer: turbulent, $δ \approx 35 mm$ (about 3.5 cm).

Verify: A quick contrast — if we had wrongly used Blasius: $δ_{lam} = 5 (2) / 4 \times 1 0^{6} = 10/2000 = 5.0 mm$ , seven times thinner. The turbulent answer being centimetre-scale matches the forecast. ✓ Units: $x$ (m) $\times$ dimensionless $=$ m. ✓

Active recall

Recall Cover the answers and test yourself

The layer thickness at the leading edge $x = 0$ is ::: exactly zero (no fluid has been slowed yet). To go from $δ$ to $x$ you ::: invert Blasius: $x = δ^{2} U / (25 ν)$ (square and rearrange). The laminar formula stops being valid once ::: $R e_{x}$ exceeds about $5 \times 1 0^{5}$ (transition to turbulence). Past transition, the layer thickness follows ::: the turbulent law $δ = 0.37 x / R e_{x}^{1/5}$ (grows faster than laminar). Doubling the free-stream speed $U$ multiplies $δ$ by ::: $1/ 2 \approx 0.707$ (thinner, laminar). Wall shear stress varies with distance as ::: $τ_{w} \propto 1/ x$ , largest at the leading edge. If you switch to a fluid with $4 \times$ the viscosity and half the speed, $δ$ grows by ::: $4/0.5 = 8 \approx 2.83$ .