Blasius solution — exact laminar boundary layer solution: Exercises

This is the exercise companion to the Blasius topic note. Everything you need is repeated in the toolbox below — you never have to leave this page.

Definition Two symbols you will use constantly:

$η$ and $ψ$

$η$ (Greek "eta") is the similarity coordinate — a cleverly stretched wall-distance: $η = y \frac{U _{\infty}}{ν x}$ Here $y$ is the plain physical height above the plate (metres), $U_{\infty}$ the free-stream speed, $ν$ the kinematic viscosity, $x$ the distance downstream. Why stretch $y$ this way? Because the raw velocity profile at $x = 0.1$ m and at $x = 1$ m look different in $y$ , but identical when plotted against $η$ — one universal curve. $η$ is dimensionless; $η = 0$ is at the wall, $η \approx 5$ is the edge of the layer.
$ψ$ (Greek "psi") is the stream function. It is a single scalar field whose slopes give both velocity components: $u = \partial ψ / \partial y$ (horizontal speed) and $v = - \partial ψ / \partial x$ (vertical speed). Why bother? Defining velocity this way makes continuity $\partial u / \partial x + \partial v / \partial y = 0$ automatically true — one function replaces two, and mass is conserved for free. See Stream Function and Vorticity. In Blasius form $ψ = ν x U_{\infty} f (η)$ , which is where $f$ comes from.

Formula The toolbox (everything you need on this page)

Similarity coordinate: $η = y \frac{U _{\infty}}{ν x}$ (dimensionless stretched height; $η = 0$ at wall).
Stream function: $u = \partial ψ / \partial y, v = - \partial ψ / \partial x$ (guarantees continuity).
Reynolds number at distance $x$ : $R e_{x} = \frac{U _{\infty} x}{ν}$ — dimensionless ratio of inertia to viscosity. See Reynolds Number.
Boundary-layer thickness (where $u = 0.99 U_{\infty}$ ): $δ = \frac{5.0 x}{R e _{x}}$
Local skin-friction coefficient: $C_{f} = \frac{0.664}{R e _{x}}$
Wall shear stress: $τ_{w} = C_{f} \cdot \frac{1}{2} ρ U_{\infty}^{2} = 0.332 μ U_{\infty} \frac{U _{\infty}}{ν x}$
Total drag coefficient (one side, plate length $L$ ): $C_{D} = \frac{1.328}{R e _{L}}$
Blasius ODE: $f^{'''} + \frac{1}{2} f f^{''} = 0$ with $f (0) = 0, f^{'} (0) = 0, f^{'} (\infty) = 1$ , and the pinned number $f^{''} (0) = 0.332$ .
Here $μ = ρ ν$ (dynamic $=$ density $\times$ kinematic viscosity), and $f^{'} = u / U_{\infty}$ is the velocity ratio.

Level 1 — Recognition

Recall Solution

The group is the local Reynolds number $R e_{x} = U_{\infty} x / ν$ . Blasius (laminar) is valid up to roughly $R e_{x} \approx 5 \times 1 0^{5}$ ; above that we cross to the Turbulent Boundary Layer. What we did: recalled that inertia-vs-viscosity ratio is the only knob. Why: the Blasius ODE came out with all $x$ and $U_{\infty}$ absorbed into $η$ — the only thing left that "knows scale" is $R e_{x}$ .

Recall Solution

$f^{'} (η) = u / U_{\infty}$ , the velocity ratio at similarity height $η$ (recall $η = y U_{\infty} / (ν x)$ , the stretched wall-distance). At the wall $η = 0$ we get $f^{'} (0) = 0$ (no-slip: fluid frozen to the plate). Far out $f^{'} (\infty) = 1$ (matches free stream $U_{\infty}$ ). Why: $u = U_{\infty} f^{'}$ came straight from $u = \partial ψ / \partial y$ (the stream-function definition), so $f^{'}$ is literally "what fraction of full speed."

Recall Solution

$δ = 5.0 \frac{ν x}{U _{\infty}}$ , so $δ \propto x$ (grows like square-root of distance) and $δ \propto 1/ U_{\infty}$ (faster flow ⇒ thinner layer). Why square-root: balancing inertia $U_{\infty}^{2} / x$ against viscosity $ν U_{\infty} / δ^{2}$ forces $δ^{2} \propto ν x / U_{\infty}$ .

Level 2 — Application

Recall Solution

Step 1 — $R e_{x}$ : $R e_{x} = \frac{0.8 \times 0.10}{1.0 \times 1 0 ^{- 6}} = 8.0 \times 1 0^{4}$ . Why first? It is $< 5 \times 1 0^{5}$ , so laminar Blasius is legal. Step 2 — $δ$ : $δ = \frac{5.0 \times 0.10}{8.0 \times 1 0 ^{4}} = \frac{0.5}{282.8} = 1.77 \times 1 0^{- 3} m \approx 1.8 mm$ . What it looks like: a sub-2 mm sheet of "slow water" hugging the plate.

Recall Solution

Step 1 — $C_{f}$ : $C_{f} = \frac{0.664}{8.0 \times 1 0 ^{4}} = \frac{0.664}{282.8} = 2.348 \times 1 0^{- 3}$ . Step 2 — un-normalise: $τ_{w} = C_{f} \cdot \frac{1}{2} ρ U_{\infty}^{2} = 2.348 \times 1 0^{- 3} \times \frac{1}{2} \times 1000 \times 0. 8^{2}$ $= 2.348 \times 1 0^{- 3} \times 320 = 0.751 Pa$ . Why: $\frac{1}{2} ρ U_{\infty}^{2}$ is the dynamic pressure — the natural "stress scale" you multiply the dimensionless $C_{f}$ by. See Skin Friction Drag.

Recall Solution

Step 1 — $R e_{L}$ : $R e_{L} = \frac{20 \times 0.30}{1.5 \times 1 0 ^{- 5}} = 4.0 \times 1 0^{5}$ (laminar, just below the limit). Step 2 — $C_{D}$ : $C_{D} = \frac{1.328}{4.0 \times 1 0 ^{5}} = \frac{1.328}{632.5} = 2.100 \times 1 0^{- 3}$ . Step 3 — force: $F_{D} = C_{D} \cdot \frac{1}{2} ρ U_{\infty}^{2} \cdot (b L)$ $= 2.100 \times 1 0^{- 3} \times \frac{1}{2} \times 1.2 \times 400 \times (0.15 \times 0.30)$ $= 2.100 \times 1 0^{- 3} \times 240 \times 0.045 = 0.0227 N$ . Why $C_{D}$ not $C_{f}$ : $C_{D}$ is the average friction over the whole plate; $C_{f}$ is local at a point.

Level 3 — Analysis

Recall Solution

Idea: $δ \propto x$ at fixed $U_{\infty}, ν$ , so $\frac{δ _{2}}{δ _{1}} = \frac{x _{2}}{x _{1}}$ . Compute: $x_{2} / x_{1} = 0.80/0.20 = 4$ , so $δ_{2} = δ_{1} 4 = 2 δ_{1} = 4.0 mm$ . What it looks like: quadruple the distance, only double the thickness — the square-root growth is decelerating (see figure below).

The figure below plots the whole curve $δ (x)$ for this flow. Look at the two dashed markers: the pale-yellow one sits at $(x_{1}, δ_{1}) = (0.20 m, 2.0 mm)$ , the pink one at $(x_{2}, δ_{2}) = (0.80 m, 4.0 mm)$ . Trace horizontally between them: $x$ jumps by a factor of 4 but the height only doubles. The curve is steep near the leading edge and flattening downstream — that visual concavity is the $x$ law. A straight line (the common wrong guess) would keep rising at the same rate and badly over-shoot far downstream.

Figure — Blasius solution — exact laminar boundary layer solution

Recall Solution

Idea: $C_{f} = 0.664/ R e_{x} \propto 1/ x$ . Halving $C_{f}$ means multiplying $x$ by $2$ , i.e. multiplying $x$ by $4$ . Compute: $x = 4 \times 0.10 = 0.40 m$ . Why: $C_{f} \propto x^{- 1/2}$ , so $C_{f} (4 x) = C_{f} (x) /2$ . Friction fades downstream — the profile flattens, wall slope drops.

Recall Solution

Step 1 — invert $η$ : since $η = y U_{\infty} / (ν x)$ , we get $y = η ν x / U_{\infty}$ . Step 2 — plug in: $ν x / U_{\infty} = \frac{1.5 \times 1 0 ^{- 5} \times 0.25}{15} = 2.5 \times 1 0^{- 7} = 5.0 \times 1 0^{- 4} m$ . Step 3: $y = 1.84 \times 5.0 \times 1 0^{- 4} = 9.2 \times 1 0^{- 4} m = 0.92 mm$ . Why this works: $η$ is the "clever height." A fixed shape point (half-speed) always sits at fixed $η$ ; convert once to physical $y$ .

Level 4 — Synthesis

Recall Solution

Step 1 — total drag per unit width: integrate wall shear along the plate: $\frac{F _{D}}{b} = \int_{0}^{L} τ_{w} d x = 0.332 μ U_{\infty} \frac{U _{\infty}}{ν} \int_{0}^{L} x^{- 1/2} d x .$ Step 2 — do the integral: $\int_{0}^{L} x^{- 1/2} d x = 2 L$ , so $\frac{F _{D}}{b} = 0.332 μ U_{\infty} \frac{U _{\infty}}{ν} \cdot 2 L = 0.664 μ U_{\infty} \frac{U _{\infty} L}{ν} .$ Step 3 — normalise into $C_{D}$ : $C_{D} = \frac{F _{D} / b}{\frac{1}{2} ρ U _{\infty}^{2} L}$ . Using $μ = ρ ν$ : $C_{D} = \frac{0.664 ρ ν U _{\infty} U _{\infty} L / ν}{\frac{1}{2} ρ U _{\infty}^{2} L} = \frac{1.328}{U _{\infty} L / ν} = \frac{1.328}{R e _{L}} .$ Step 4 — compare: $C_{f} (L) = 0.664/ R e_{L}$ , hence $C_{D} (L) = 2 C_{f} (L)$ . ✓ Physical why: near the leading edge ( $x \to 0$ ) the wall slope — and so $τ_{w}$ — is huge (blows up as $x^{- 1/2}$ ). Averaging that big upstream friction with the small trailing-edge friction gives a mean that is exactly twice the trailing-edge local value. The integral $\int x^{- 1/2}$ is dominated by the front.

Recall Solution

At the wall: $η = 0$ gives $η f^{'} = 0$ , and the BC $f (0) = 0$ kills the other term, so $v (x, 0) = \frac{1}{2} ν U_{\infty} / x (0 - 0) = 0$ . ✓ (No fluid crosses the solid plate — impermeability.) Above the wall: because the layer thickens downstream, $\partial u / \partial x < 0$ inside it (fluid decelerates as $x$ grows at fixed $y$ ). Continuity $\partial u / \partial x + \partial v / \partial y = 0$ then forces $\partial v / \partial y > 0$ , so $v$ climbs from $0$ upward — a gentle outflow. This is the L4 synthesis of continuity + Stream Function and Vorticity: the stream function $ψ$ was chosen precisely so continuity holds automatically, and it still predicts this non-zero $v$ . Mistake it kills: "flow is purely parallel" — false; $v \neq = 0$ carries momentum into the layer.

Level 5 — Mastery

Recall Solution

Step 1 — assemble the drag law: from L4.1, $F_{D} = 0.664 b μ U_{\infty} U_{\infty} L / ν \propto U_{\infty}^{3/2} L^{1/2}$ (since $U_{\infty} \cdot U_{\infty} = U_{\infty}^{3/2}$ and $L$ ). Step 2 — form the ratio: $\frac{F _{B}}{F _{A}} = (\frac{U _{B}}{U _{A}})^{3/2} (\frac{L _{B}}{L _{A}})^{1/2} = 2^{3/2} \cdot 2^{1/2} = 2^{2} = 4.$ Step 3 — sanity: speed matters more (exponent $3/2$ ) than length (exponent $1/2$ ). Doubling both gives a clean $4 \times$ . ✓ What it looks like: drag climbs steeply with speed — the $U_{\infty}^{3/2}$ term dominates the growth.

Recall Solution

Step 1 — read $f^{'''} (0)$ from the ODE: $f^{'''} = - \frac{1}{2} f f^{''}$ . At $η = 0$ , $f (0) = 0$ , so $f^{'''} (0) = - \frac{1}{2} f (0) f^{''} (0) = - \frac{1}{2} (0) (0.332) = 0.$ Step 2 — interpret: $f^{'''}$ is the curvature of the velocity gradient. $f^{'''} (0) = 0$ means the profile has an inflection-free, zero-jerk start — the shear $f^{''}$ is momentarily flat at the wall. The velocity near the wall behaves like $f^{'} \approx f^{''} (0) η = 0.332 η$ (linear) with the next correction only at order $η^{4}$ (since the $η^{2}$ and $η^{3}$ coefficients, $\propto f^{'''} (0)$ , vanish). Step 3 — why a parabola fails: a parabola $u / U_{\infty} = a η + b η^{2}$ has a non-zero $η^{2}$ term, i.e. non-zero curvature $f^{''}$ that stays constant, which contradicts $f^{'''} (0) = 0$ and the true flattening. The parabola is only a crude integral-method approximation; the exact Blasius wall region is linear to remarkably high order. This is mastery: the ODE itself hands you $f^{'''} (0) = 0$ for free — no numerics needed — and that pins the profile shape.

Recall Solution

As $x \to 0^{+}$ (leading edge):

$δ = 5.0 ν x / U_{\infty} \to 0$ (layer has zero thickness right at the edge). ✓ sensible.
$τ_{w} = 0.332 μ U_{\infty} U_{\infty} / (ν x) \to \infty$ (wall shear diverges). This is the leading-edge singularity.
$C_{f} = 0.664/ R e_{x} \to \infty$ too (since $R e_{x} \to 0$ ). Which prediction breaks: the infinite $τ_{w}$ is unphysical — no wall feels infinite drag. Why the theory forbids it: the Prandtl reduction assumed $δ ≪ x$ (that is what let us drop $\partial^{2} u / \partial x^{2}$ ). At $x \to 0$ , $δ$ and $x$ become comparable, so that founding approximation dies and the boundary-layer equations are simply invalid there. The true tip flow needs the full Navier–Stokes Equations (and see Prandtl Boundary Layer Theory for where the approximation holds). As $x \to \infty$ : $δ \to \infty$ but slowly ( $\propto x$ ), while $τ_{w} \to 0$ and $C_{f} \to 0$ — friction fades toward nothing. But $R e_{x} = U_{\infty} x / ν$ keeps growing and eventually exceeds $\sim 5 \times 1 0^{5}$ , so the flow transitions to turbulent long before "infinity." Blasius is therefore a finite laminar window, not a formula valid for all $x$ — beyond the window use the Turbulent Boundary Layer scaling $δ / x \approx 0.37 R e_{x}^{- 1/5}$ .

Recall One-line recap of the scaling exponents (the spine of every problem above)

$δ \propto x^{1/2} U_{\infty}^{- 1/2}$ · $τ_{w} \propto x^{- 1/2} U_{\infty}^{3/2}$ · $C_{f} \propto R e_{x}^{- 1/2}$ · $F_{D} \propto L^{1/2} U_{\infty}^{3/2}$ · and the constants double: $0.332 \to 0.664 \to 1.328$ .