Blasius solution — exact laminar boundary layer solution: Worked examples

This page is a worked-example gym for the Blasius solution. The parent note built the theory; here we drill it against every kind of situation the exam (or reality) can hand you. If a symbol here feels new, it was defined in the parent — but we re-anchor the essentials as we go so you never stall.

Before we compute anything, pin down the four workhorses from the parent, with their plain meaning:

A quick word on the Blasius profile symbols ( $η$ , $f$ , $f^{'}$ )

We need one more idea from the parent before Example 7 (which peeks inside the layer). The parent showed that the velocity profile looks identical at every $x$ once you measure height the clever way:

The scenario matrix

Every Blasius problem lands in one of these cells. The examples below are labelled with the cell(s) they hit — together they cover the whole grid.

Cell	What varies / degenerate input	Question type	Example
A	Standard interior point, laminar	Find $δ$	Ex 1
B	Same point	Find $τ_{w}$ (local force)	Ex 2
C	Whole plate	Find total drag $F_{D}$	Ex 3
D	Limit $x \to 0$ (leading edge)	Degenerate: $δ \to 0$ , $τ_{w} \to \infty$	Ex 4
E	Ratio / scaling — change $x$ or $U_{\infty}$ , no numbers	How does $δ$ , $τ_{w}$ respond?	Ex 5
F	Boundary of validity: $R e_{x}$ near $5 \times 1 0^{5}$	Is Blasius even allowed? Find transition $x$	Ex 6
G	Inside the layer — a specific $y$ , or find $v$	Profile / vertical velocity ≠ 0	Ex 7
H	Real-world word problem (different fluid: water)	Full pipeline	Ex 8
I	Exam twist: work backwards from a measured drag	Inverse problem	Ex 9

We'll use one recurring "house" flow so numbers stay familiar: air, $ν = 1.5 \times 1 0^{- 5} m^{2} / s$ , $ρ = 1.2 kg/m^{3}$ , unless a problem says otherwise.

Example 1 — Cell A: thickness at an interior point

Worked example Statement

Air ( $ν = 1.5 \times 1 0^{- 5} m^{2} / s$ ) at $U_{\infty} = 10 m/s$ over a flat plate. Find $δ$ at $x = 0.5 m$ .

Forecast: Guess the layer thickness — millimetres, centimetres, or metres? (Boundary layers are famously thin.)

Step 1 — Reynolds number. $R e_{x} = \frac{U _{\infty} x}{ν} = \frac{10 \times 0.5}{1.5 \times 1 0 ^{- 5}} = 3.33 \times 1 0^{5}$ . Why this step? $R e_{x}$ is the master dial; nothing else can be computed without it, and it also tells us whether Blasius is even legal (need $R e_{x} ≲ 5 \times 1 0^{5}$ ). Here $3.33 \times 1 0^{5} < 5 \times 1 0^{5}$ ✓ laminar.

Step 2 — plug into $δ$ . $δ = \frac{5.0 x}{R e _{x}} = \frac{5.0 \times 0.5}{3.33 \times 1 0 ^{5}} = \frac{2.5}{577} = 4.3 \times 1 0^{- 3} m$ . Why this step? The boxed formula converts the master dial into a physical length.

Verify: $4.3 mm$ over half a metre of plate — a slow-zone one part in 116 of the distance travelled. Thin, as forecast ✓. Units: $\frac{m}{dimensionless} = m$ ✓.

Example 2 — Cell B: wall shear at a point

Worked example Statement

Same air, same $U_{\infty} = 10 m/s$ . Find the wall shear stress $τ_{w}$ at $x = 0.5 m$ .

Forecast: Air is light. Will the drag pressure be a big fraction of atmospheric ( $1 0^{5}$ Pa) or tiny?

Step 1 — local friction coefficient. $C_{f} = \frac{0.664}{R e _{x}} = \frac{0.664}{577} = 1.15 \times 1 0^{- 3}$ . Why this step? $C_{f}$ is dimensionless and depends only on the master dial — compute it first, un-normalize later. This separates "shape of the physics" from "size of the numbers."

Step 2 — un-normalize with dynamic pressure. $τ_{w} = C_{f} \cdot \frac{1}{2} ρ U_{\infty}^{2} = 1.15 \times 1 0^{- 3} \times \frac{1}{2} \times 1.2 \times 1 0^{2} = 0.069 Pa$ . Why this step? $\frac{1}{2} ρ U_{\infty}^{2}$ is the natural pressure scale of the flow; multiplying restores real units of force-per-area.

Verify: $0.069 Pa$ vs atmospheric $\sim 1 0^{5} Pa$ — utterly tiny, as forecast (air drag is gentle) ✓. Units: dimensionless $\times kg/m^{3} \times (m/s)^{2} = kg / (m\cdots^{2}) = Pa$ ✓.

Example 3 — Cell C: total drag on the plate

Worked example Statement

Plate length $L = 0.5 m$ , width $b = 0.2 m$ , same air, $U_{\infty} = 10 m/s$ . Find the drag on one side.

Forecast: A tiny stress over a small area — will the force be newtons, or a small fraction of a newton?

Step 1 — averaged drag coefficient. $R e_{L} = \frac{10 \times 0.5}{1.5 \times 1 0 ^{- 5}} = 3.33 \times 1 0^{5}$ , so $C_{D} = \frac{1.328}{R e _{L}} = \frac{1.328}{577} = 2.30 \times 1 0^{- 3}$ . Why this step? Local $τ_{w}$ is largest at the leading edge and decays as $1/ x$ , so we can't just multiply one point's stress by area. $C_{D}$ is the average of $C_{f}$ over the plate. Where does the factor $2$ come from? Average $C_{f}$ means $\frac{1}{L} \int_{0}^{L} C_{f} d x$ , and since $C_{f} \propto x^{- 1/2}$ , the integral $\int_{0}^{L} x^{- 1/2} d x = 2 L$ , so the average is $\frac{2 L}{L} \times (value at L) / L^{- 1}$ — carrying the constants through gives exactly $\overset{ˉ}{C}_{f} = 2 C_{f} (L)$ . Hence $C_{D} = 2 C_{f} (L)$ .

Step 2 — un-normalize over the wetted area. $F_{D} = C_{D} \cdot \frac{1}{2} ρ U_{\infty}^{2} \cdot (b L) = 2.30 \times 1 0^{- 3} \times 60 \times 0.1 = 0.0138 N$ . Why this step? Same pressure scale as before, now multiplied by the plate area $b L = 0.1 m^{2}$ that the fluid actually rubs.

Verify: $0.0138 N \approx 1.4$ grams-force — feather-light, matching the tiny $τ_{w}$ ✓. Cross-check: $C_{D} = 2 C_{f} = 2 \times 1.15 \times 1 0^{- 3} = 2.30 \times 1 0^{- 3}$ ✓.

Example 4 — Cell D: the degenerate leading edge, $x \to 0$

Worked example Statement

Same air and speed. What happens to $δ$ and $τ_{w}$ as $x \to 0$ (right at the leading edge)? Evaluate the trend at $x = 0.5 m$ , $x = 0.05 m$ , $x = 0.005 m$ . (Recall $μ = ρ ν = 1.8 \times 1 0^{- 5} Pa\cdots$ , defined above.)

Forecast: The plate just started dragging the fluid. Is the layer thickest or thinnest here — and is the drag mildest or fiercest?

Step 1 — write both as pure powers of $x$ . Start from the boxed $δ = 5.0 x / R e_{x}$ and substitute $R e_{x} = U_{\infty} x / ν$ : $δ = \frac{5.0 x}{U _{\infty} x / ν} = 5.0 x \frac{ν}{U _{\infty} x} = 5.0 \frac{ν x ^{2}}{U _{\infty} x} = 5.0 \frac{ν x}{U _{\infty}} \propto x^{1/2} .$ Likewise, from the $μ$ -form derived above, $τ_{w} = 0.332 μ U_{\infty} U_{\infty} / (ν x) \propto x^{- 1/2}$ . Why this step? Pushing the lone $x$ inside the square-root exposes the degenerate behaviour: one grows from zero, the other blows up.

Step 2 — tabulate the trend.

$x$ (m)	$δ = 5.0 ν x / U_{\infty}$ (mm)	$τ_{w} = 0.332 μ U_{\infty} U_{\infty} / ν x$ (Pa)
$0.5$	$4.33$	$0.069$
$0.05$	$1.37$	$0.218$
$0.005$	$0.433$	$0.690$

Why this step? Shrinking $x$ tenfold shrinks $δ$ by $10 \approx 3.16$ and grows $τ_{w}$ by the same factor — the two mirror each other.

Step 3 — take the limit. As $x \to 0^{+}$ : $δ \to 0$ and $τ_{w} \to + \infty$ . Why this step? This is the leading-edge singularity: the model predicts infinite shear at $x = 0$ . It's not physical — the thin-layer assumption $δ ≪ x$ breaks when $δ \sim x$ — but it correctly warns you the leading edge is where drag concentrates.

Verify: $δ$ scales as $x$ (÷10 in $x$ ⇒ ÷3.16 in $δ$ : $4.33/3.16 = 1.37$ ✓), $τ_{w}$ as $1/ x$ (÷10 in $x$ ⇒ ×3.16 in $τ_{w}$ : $0.069 \times 3.16 = 0.218$ ✓). The singularity is a genuine feature of the boxed formulas.

The figure below plots exactly this table as two curves so you can see the mirror-image trends and the blow-up at the origin.

Figure — Blasius solution — exact laminar boundary layer solution

Example 5 — Cell E: pure scaling, no numbers

Worked example Statement

A plate is tested at $U_{\infty}$ . Then the speed is quadrupled to $4 U_{\infty}$ , same fluid, same point $x$ . By what factor does (a) $δ$ change, (b) $τ_{w}$ change, (c) total drag $F_{D}$ change?

Forecast: Faster flow — thinner or thicker layer? More or less drag, and by how much?

Step 1 — read the exponents. $δ \propto U_{\infty}^{- 1/2}$ (from $δ = 5.0 ν x / U_{\infty}$ ), so $δ$ changes by $4^{- 1/2} = \frac{1}{2}$ . Why this step? We only need how each quantity depends on $U_{\infty}$ ; the constants cancel in a ratio. Faster free stream ⇒ inertia wins more ⇒ thinner layer.

Step 2 — shear stress. $τ_{w} = 0.332 μ U_{\infty} U_{\infty} / (ν x) \propto U_{\infty}^{3/2}$ , so it changes by $4^{3/2} = 8$ . Why this step? One power of $U_{\infty}$ from momentum flux, half a power from the steeper wall gradient — together $U_{\infty}^{3/2}$ .

Step 3 — total drag. First note $R e_{L} = U_{\infty} L / ν \propto U_{\infty}$ (from the definition, with $L, ν$ fixed). So $C_{D} = 1.328/ R e_{L} \propto R e_{L}^{- 1/2} \propto U_{\infty}^{- 1/2}$ . Then $F_{D} \propto C_{D} U_{\infty}^{2} \propto U_{\infty}^{- 1/2} U_{\infty}^{2} = U_{\infty}^{3/2}$ , so $F_{D}$ changes by $4^{3/2} = 8$ . Why this step? The hidden speed-dependence lives inside $C_{D}$ through $R e_{L} \propto U_{\infty}$ ; only after exposing it do the powers add correctly. $F_{D}$ and $τ_{w}$ end up sharing the same $U_{\infty}^{3/2}$ law.

Verify: $δ$ : $\times 0.5$ ; $τ_{w}$ : $\times 8$ ; $F_{D}$ : $\times 8$ . Sanity: $δ$ down means gradient steeper ⇒ more shear ⇒ consistent that $τ_{w}$ rose ✓. Numerically $4^{- 0.5} = 0.5$ , $4^{1.5} = 8$ ✓.

Example 6 — Cell F: is Blasius even allowed? Find transition.

Worked example Statement

Same air, $U_{\infty} = 10 m/s$ . At what distance $x_{tr}$ does the flow reach the transition Reynolds number $R e_{x, crit} = 5 \times 1 0^{5}$ ? Beyond it, is $δ = 5.0 x / R e_{x}$ valid?

Forecast: Guess: a few centimetres, tens of centimetres, or metres before turbulence sets in?

Step 1 — invert the Reynolds definition. $R e_{x} = U_{\infty} x / ν \Rightarrow x_{tr} = \frac{R e _{x, crit} ν}{U _{\infty}} = \frac{5 \times 1 0 ^{5} \times 1.5 \times 1 0 ^{- 5}}{10} = 0.75 m$ . Why this step? The validity boundary is stated as a Reynolds number, so convert it into a distance you can measure on the plate.

Step 2 — verdict on the formula. For $x > 0.75 m$ the flow is turbulent; the Blasius $δ = 5.0 x / R e_{x}$ no longer applies. Use the turbulent estimate $δ / x \approx 0.37 R e_{x}^{- 1/5}$ instead (see Turbulent Boundary Layer). Why this step? Blasius is a laminar solution — applying it past transition over-predicts the wrong power law ( $x$ vs $x^{4/5}$ ).

Verify: At $x = 0.5 m$ (Ex 1) we had $R e_{x} = 3.33 \times 1 0^{5} < 5 \times 1 0^{5}$ , safely laminar — consistent, since $0.5 < 0.75 m$ ✓. Recompute: $0.75 \times 10/1.5 \times 1 0^{- 5} = 5 \times 1 0^{5}$ ✓.

Example 7 — Cell G: inside the layer — where is $u = U_{\infty} /2$ , and is $v$ really nonzero?

Worked example Statement

Same air, $U_{\infty} = 10 m/s$ , at $x = 0.5 m$ . (a) At what height $y$ is the velocity half the free stream, $u = 5 m/s$ ? (b) Show the vertical velocity $v$ there is not zero. Use the Blasius read-offs defined above: $f^{'} = 0.5$ at $η \approx 1.6$ , and there $f \approx 0.65$ .

Forecast: Half-speed height — is it near the wall, mid-layer, or near the top of the $4.3 mm$ layer?

Step 1 — convert $η$ back to $y$ . Recall (from the definition block) $η = y U_{\infty} / (ν x)$ , so $y = η ν x / U_{\infty}$ . The scale $ν x / U_{\infty} = 1.5 \times 1 0^{- 5} \times 0.5/10 = 8.66 \times 1 0^{- 4} m$ . Then $y = 1.6 \times 8.66 \times 1 0^{- 4} = 1.39 \times 1 0^{- 3} m$ . Why this step? $η$ is the "clever stretched height" that makes the profile universal; we must un-stretch it to get a real millimetre reading. Because $f^{'} (η) = u / U_{\infty}$ , the value $f^{'} = 0.5$ marks exactly where $u = 5 m/s$ .

Step 2 — locate it in the layer. $δ = 4.33 mm$ (Ex 1), and $y = 1.39 mm$ , so half-speed sits at $y / δ = 0.32$ — about a third of the way up. Why this step? Confirms the profile is steep near the wall: you reach half speed after only a third of the height, then crawl the rest of the way to $0.99 U_{\infty}$ .

Step 3 — vertical velocity is nonzero. The parent's formula is $v = \frac{1}{2} ν U_{\infty} / x (η f^{'} - f)$ . Plug in: the root is $1.5 \times 1 0^{- 5} \times 10/0.5 = 3 \times 1 0^{- 4} = 1.73 \times 1 0^{- 2}$ , and $(η f^{'} - f) = (1.6 \times 0.5 - 0.65) = (0.8 - 0.65) = 0.15$ . So $v = 0.5 \times 1.73 \times 1 0^{- 2} \times 0.15 = 1.30 \times 1 0^{- 3} m/s$ . Why this step? Continuity forces fluid upward as the layer thickens — a real, if tiny, $v > 0$ . Ignoring it (a common mistake) breaks mass conservation.

Verify: $y = 1.39 mm$ lies inside $δ = 4.33 mm$ ✓ (as it must). $v = 1.30 \times 1 0^{- 3} m/s$ is $\sim 1 0^{- 4}$ of $U_{\infty}$ — small but strictly positive ✓. Units of $v$ : $m^{2} / s^{2} = m/s$ ✓.

The figure shows the full universal profile so you can see why half-speed lands so low.

Example 8 — Cell H: real-world, different fluid (water)

Worked example Statement

A ship model panel runs through water ( $ν = 1.0 \times 1 0^{- 6} m^{2} / s$ , $ρ = 1000 kg/m^{3}$ ) at $U_{\infty} = 2 m/s$ . Panel: $L = 0.3 m$ , width $b = 0.5 m$ . Find (a) $δ$ at the trailing edge, (b) drag on one side. Check laminar validity.

Forecast: Water is 15× less viscous (kinematically) than our air and much denser. Thinner or thicker layer than air? Bigger or smaller drag?

Step 1 — Reynolds & validity. $R e_{L} = \frac{2 \times 0.3}{1.0 \times 1 0 ^{- 6}} = 6.0 \times 1 0^{5}$ . Why this step? Master dial first; and $6.0 \times 1 0^{5} > 5 \times 1 0^{5}$ , so strictly it's just past laminar transition — Blasius gives an estimate only. We proceed as a first approximation and flag it.

Step 2 — thickness. $δ = \frac{5.0 \times 0.3}{6.0 \times 1 0 ^{5}} = \frac{1.5}{775} = 1.94 \times 1 0^{- 3} m$ . Why this step? Same boxed formula; low $ν$ ⇒ high $R e$ ⇒ thin layer (≈1.9 mm, thinner than air's 4.3 mm relative to length).

Step 3 — drag. $C_{D} = \frac{1.328}{6.0 \times 1 0 ^{5}} = 1.71 \times 1 0^{- 3}$ . Then $F_{D} = C_{D} \cdot \frac{1}{2} ρ U_{\infty}^{2} \cdot b L = 1.71 \times 1 0^{- 3} \times \frac{1}{2} \times 1000 \times 4 \times 0.15 = 0.514 N$ . Why this step? Water's density is ~800× air's, so despite the small $C_{D}$ , the force is now sizeable (half a newton, not milli-newtons).

Verify: $\frac{1}{2} ρ U_{\infty}^{2} = 2000 Pa$ ; $b L = 0.15 m^{2}$ ; $2000 \times 0.15 = 300 N \times 1.71 \times 1 0^{- 3} = 0.514 N$ ✓. Dense fluid ⇒ big drag, as forecast ✓. Caveat noted: $R e_{L}$ slightly above $5 \times 1 0^{5}$ , so treat as an upper-ballpark laminar estimate.

Example 9 — Cell I: exam twist, work backwards from a measured drag

Worked example Statement

In a wind tunnel, a plate ( $L = 1.0 m$ , $b = 0.4 m$ , air $ν = 1.5 \times 1 0^{- 5}$ , $ρ = 1.2$ ) records a one-sided drag of $F_{D} = 0.02 N$ . Find the flow speed $U_{\infty}$ . (Assume laminar Blasius.)

Forecast: Inverse problem — we know the force, want the cause. Will $U_{\infty}$ come out a few m/s or tens of m/s?

Step 1 — write $F_{D}$ fully in terms of $U_{\infty}$ . $F_{D} = C_{D} \cdot \frac{1}{2} ρ U_{\infty}^{2} \cdot b L = \frac{1.328}{U _{\infty} L / ν} \cdot \frac{1}{2} ρ U_{\infty}^{2} b L$ . Why this step? We can't isolate $U_{\infty}$ until every hidden $U_{\infty}$ (inside $C_{D}$ too, via $R e_{L} = U_{\infty} L / ν$ !) is on the page — the classic trap is forgetting $C_{D}$ depends on speed.

Step 2 — collect the power of $U_{\infty}$ . $C_{D} \propto U_{\infty}^{- 1/2}$ and the dynamic pressure $\propto U_{\infty}^{2}$ , so $F_{D} \propto U_{\infty}^{3/2}$ . Explicitly: $F_{D} = 1.328 \frac{ν}{L} \cdot \frac{1}{2} ρ b L U_{\infty}^{3/2} = 0.664 ρ b ν L U_{\infty}^{3/2} .$ Why this step? Reducing to a single power $U_{\infty}^{3/2}$ makes the inversion a clean root-type solve.

Step 3 — solve for $U_{\infty}$ . The prefactor $K = 0.664 \times 1.2 \times 0.4 \times 1.5 \times 1 0^{- 5} \times 1.0 = 0.664 \times 1.2 \times 0.4 \times 3.873 \times 1 0^{- 3} = 1.234 \times 1 0^{- 3}$ . Then $U_{\infty}^{3/2} = F_{D} / K = 0.02/1.234 \times 1 0^{- 3} = 16.21$ , so $U_{\infty} = 16.2 1^{2/3} = 6.42 m/s$ . Why this step? Raise both sides to the $2/3$ power to undo the $3/2$ exponent — the algebraic inverse of the physics.

Verify (plug forward): $R e_{L} = 6.42 \times 1.0/1.5 \times 1 0^{- 5} = 4.28 \times 1 0^{5} < 5 \times 1 0^{5}$ laminar ✓. $C_{D} = 1.328/ 4.28 \times 1 0^{5} = 2.03 \times 1 0^{- 3}$ ; $F_{D} = 2.03 \times 1 0^{- 3} \times \frac{1}{2} \times 1.2 \times 6.4 2^{2} \times 0.4 = 2.03 \times 1 0^{- 3} \times \frac{1}{2} \times 1.2 \times 41.2 \times 0.4 = 0.0201 N \approx 0.02 N$ ✓. Round-trip closes.

Recall Quick self-test

Which cell blows up at the leading edge, and which quantity? ::: Cell D — $τ_{w} \to \infty$ as $x \to 0$ , while $δ \to 0$ . If you double $U_{\infty}$ , by what factor does total drag change? ::: $2^{3/2} \approx 2.83$ (drag $\propto U_{\infty}^{3/2}$ ). Why must you recompute $R e$ before trusting any Blasius answer? ::: To confirm $R e < 5 \times 1 0^{5}$ (laminar); above that the $x$ laws fail — see Turbulent Boundary Layer. In the inverse problem (Cell I), what is the trap? ::: Forgetting that $C_{D}$ itself depends on $U_{\infty}$ (through $R e_{L} \propto U_{\infty}$ ), giving $F_{D} \propto U_{\infty}^{3/2}$ , not $U_{\infty}^{2}$ . How do $μ$ and $ν$ relate? ::: $μ = ρ ν$ (dynamic = density × kinematic).

See also: Prandtl Boundary Layer Theory · Navier–Stokes Equations · Skin Friction Drag · Similarity Solutions in PDEs · Stream Function and Vorticity.