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

Every formula used here is built in the parent note Boundary Layer parent. The three we lean on:

Formula Toolkit for this page

First, the two material properties that describe "how sticky" and "how heavy" a fluid is:

$μ$ = dynamic viscosity, how strongly the fluid resists shearing (see Viscosity and Newton's law of viscosity), units $Pa\cdots = kg / (m\cdots)$ .
$ρ$ = density, mass per unit volume, units $kg/m^{3}$ .

Two velocities appear throughout, so pin them down now:

$U$ = free-stream speed, the steady speed of the fluid far from the plate (m/s).
$u$ = local flow speed inside the boundary layer, the speed of the fluid at a given height $y$ above the wall (m/s). At the wall $u = 0$ (no-slip); far out $u \to U$ .

From the material properties:

Local Reynolds number: $R e_{x} = \frac{U x}{ν}$ where $ν = \frac{μ}{ρ}$ is the kinematic viscosity (units $\frac{Pa\cdots}{kg/m ^{3}} = m^{2} / s$ ).
Laminar layer thickness (Blasius): $δ (x) = \frac{5.0 x}{R e _{x}} = 5.0 \frac{ν x}{U}$ .
Wall shear (order of magnitude): $τ_{w} \sim μ \frac{U}{δ}$ .

Here $x$ = distance from the leading edge (the front tip of the plate), $δ$ = thickness where $u = 0.99 U$ .

See also: Reynolds number, Skin friction drag, Laminar vs Turbulent flow.

Level 1 — Recognition

L1.1 In the boundary-layer formula $δ = 5.0 ν x / U$ , name what each symbol means and give the SI unit of each.

Recall Solution

$δ$ = boundary-layer thickness (distance from wall where the local speed $u$ reaches $0.99 U$ ), unit m.
$ν$ = kinematic viscosity $= μ / ρ$ , unit $m^{2} / s$ .
$x$ = distance downstream from the leading edge, unit m.
$U$ = free-stream speed, unit m/s.
The constant $5.0$ is dimensionless (from the exact Blasius solution). Sanity: inside the root, $\frac{( m ^{2} / s ) ( m )}{m/s} = m^{2}$ , and $m^{2} = m$ . ✔

L1.2 True or false, with one line of reasoning each: (a) At the leading edge $δ = 0$ . (b) A faster free stream makes the layer thicker. (c) The layer thickness grows linearly with $x$ .

Recall Solution

(a) True. At $x = 0$ , $δ = 5.0 ν \cdot 0/ U = 0$ . No fluid has yet been dragged. (b) False. $δ \propto 1/ U$ , so faster stream → thinner layer (less time for viscosity to diffuse sideways). (c) False. $δ \propto x$ , growth that slows down (see the figure below).

Walk through Figure s01. The horizontal axis is distance $x$ from the leading edge (in metres, $0$ to $2$ ); the vertical axis is $δ$ in millimetres, computed for air at $U = 15 m/s$ , $ν = 1.5 \times 1 0^{- 5} m^{2} / s$ . The thick grey line at $δ = 0$ is the plate surface. The blue arrows at the top are the uniform free stream $U$ arriving from the left. The red curve is $δ (x) = 5 ν x / U$ : notice it leaves the origin vertically (steep) at the leading edge, then bends over and flattens — that bending-over is the visual signature of the $x$ law. (Formally, squaring $δ = 5 ν x / U$ gives $x \propto δ^{2}$ , i.e. $x$ is a parabola in $δ$ ; plotted with $x$ horizontal the curve is that parabola turned on its side. The faint mirror curve below the plate just completes that sideways parabola so you can recognise the shape.)

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

Level 2 — Application

L2.1 Air ( $ν = 1.5 \times 1 0^{- 5} m^{2} / s$ ) flows at $U = 30 m/s$ over a flat plate. Find $R e_{x}$ and $δ$ at $x = 0.20 m$ .

Recall Solution

Step 1 — Reynolds number. Why first? It tells us both which formula applies and appears inside $δ$ . $R e_{x} = \frac{U x}{ν} = \frac{30 \times 0.20}{1.5 \times 1 0 ^{- 5}} = \frac{6}{1.5 \times 1 0 ^{- 5}} = 4.0 \times 1 0^{5} .$ This is below the usual flat-plate transition ( $\approx 5 \times 1 0^{5}$ ), so laminar Blasius is valid. Step 2 — Thickness. Why? With $R e_{x}$ known, direct substitution into $δ = 5.0 x / R e_{x}$ is the only remaining move — it converts the dimensionless $R e_{x}$ back into a physical length. $δ = \frac{5.0 x}{R e _{x}} = \frac{5.0 \times 0.20}{4.0 \times 1 0 ^{5}} = \frac{1.0}{632.46} = 1.58 \times 1 0^{- 3} m \approx 1.58 mm .$ A 20 cm plate → 1.6 mm layer: genuinely thin. ✔

L2.2 For the same air flow, at what distance $x$ does the layer become $δ = 3.0 mm$ thick?

Recall Solution

Step 1 — Isolate $x$ . Why? We know $δ$ , want $x$ . Write $δ = 5.0 ν x / U$ and square: $δ^{2} = 25 \frac{ν x}{U} \Rightarrow x = \frac{δ ^{2} U}{25 ν} .$ Step 2 — Substitute ( $δ = 3.0 \times 1 0^{- 3} m$ ). Why? All three inputs ( $δ$ , $U$ , $ν$ ) are known, so plugging them into the rearranged formula gives the number directly. $x = \frac{( 3.0 \times 1 0 ^{- 3} ) ^{2} \times 30}{25 \times 1.5 \times 1 0 ^{- 5}} = \frac{9 \times 1 0 ^{- 6} \times 30}{3.75 \times 1 0 ^{- 4}} = \frac{2.7 \times 1 0 ^{- 4}}{3.75 \times 1 0 ^{- 4}} = 0.72 m .$ Check: at $x = 0.72$ , is flow still laminar? $R e_{x} = 30 \times 0.72/1.5 \times 1 0^{- 5} = 1.44 \times 1 0^{6}$ — above transition, so in reality it would be turbulent. We report $0.72 m$ as the laminar-formula value.

Level 3 — Analysis

L3.1 On one plate, water flows steadily. Compare $δ$ at $x_{2} = 1.6 m$ to $δ$ at $x_{1} = 0.1 m$ . By what factor is it thicker?

Recall Solution

Step 1 — Use the ratio, kill the constants. Why? $U, ν$ and the $5.0$ are identical at both stations, so form the ratio and they cancel — no numbers for $ν$ needed. Step 2 — Evaluate. Why? Only $x$ survives in the ratio, so the answer reduces to a single square root of the $x$ -ratio. $\frac{δ _{2}}{δ _{1}} = \frac{x _{2}}{x _{1}} = \frac{1.6}{0.1} = 16 = 4.$ The layer is 4× thicker at $x_{2}$ , even though the distance is 16× greater — the square root tames it.

L3.2 Wall shear stress scales as $τ_{w} \propto x^{- 1/2}$ . If $τ_{w} = 0.80 Pa$ at $x = 0.25 m$ , find $τ_{w}$ at $x = 1.0 m$ .

Recall Solution

Step 1 — Set up the scaling ratio. Why? Same fluid and speed, only $x$ changes, so use $τ_{w} \propto x^{- 1/2}$ directly. $\frac{τ _{w, 2}}{τ _{w, 1}} = (\frac{x _{2}}{x _{1}})^{- 1/2} = (\frac{1.0}{0.25})^{- 1/2} = 4^{- 1/2} = \frac{1}{2} .$ Step 2 — Multiply. Why? The ratio is the factor connecting the two stations, so the known $τ_{w, 1}$ times that factor is the unknown $τ_{w, 2}$ . $τ_{w, 2} = 0.80 \times \frac{1}{2} = 0.40 Pa .$ Friction is strongest at the leading edge (thinnest layer → steepest velocity gradient) and halves as we go 4× downstream.

Walk through Figure s02. The horizontal axis is again distance $x$ (m, from $0.05$ to $2.0$ ); the vertical axis is wall shear $τ_{w}$ in pascals. The orange curve is the $τ_{w} \propto x^{- 1/2}$ law, normalised so that $τ_{w} = 0.40 Pa$ at $x = 1.0 m$ . The two marked dots are exactly this problem's stations: the red dot at $(0.25, 0.80)$ and the green dot at $(1.0, 0.40)$ — the value halves as $x$ quadruples. Notice how steeply the curve climbs toward the left edge: this is the leading-edge blow-up flagged in the toolkit ( $τ_{w} \to \infty$ as $x \to 0$ ), which is why the plot starts at $x = 0.05$ rather than $0$ .

Level 4 — Synthesis

L4.1 A plate of length $L = 1.0 m$ and width $b = 0.5 m$ sits in air ( $ρ = 1.2 kg/m^{3}$ , $ν = 1.5 \times 1 0^{- 5} m^{2} / s$ ) at $U = 15 m/s$ . Using the laminar drag-coefficient formula for one side of the plate, $C_{D} = \frac{1.328}{R e _{L}}, F_{D} = C_{D} \cdot \frac{1}{2} ρ U^{2} \cdot (b L),$ find (a) $R e_{L}$ , (b) $C_{D}$ , (c) the friction drag force $F_{D}$ on one side.

Recall Solution

Step 1 — Length-based Reynolds number. Why? $C_{D}$ integrates the shear over the whole plate, so it depends on $R e$ at the trailing edge, $R e_{L}$ . $R e_{L} = \frac{U L}{ν} = \frac{15 \times 1.0}{1.5 \times 1 0 ^{- 5}} = 1.0 \times 1 0^{6} .$ Step 2 — Drag coefficient. Why? The formula $C_{D} = 1.328/ R e_{L}$ turns the just-computed $R e_{L}$ into the dimensionless friction number. $C_{D} = \frac{1.328}{1.0 \times 1 0 ^{6}} = \frac{1.328}{1000} = 1.328 \times 1 0^{- 3} .$ Step 3 — Force. Why this form? Dynamic pressure $\frac{1}{2} ρ U^{2}$ times area $b L$ times the dimensionless $C_{D}$ gives a force. $q = \frac{1}{2} ρ U^{2} = \frac{1}{2} (1.2) (1 5^{2}) = 0.5 \times 1.2 \times 225 = 135 Pa .$ $A = b L = 0.5 \times 1.0 = 0.5 m^{2} .$ $F_{D} = C_{D} q A = 1.328 \times 1 0^{- 3} \times 135 \times 0.5 = 0.089640 N .$ So one side of the plate feels about $0.0896 N$ of skin-friction drag — small, as expected from a thin laminar layer. See Skin friction drag.

L4.2 Same plate. What is the layer thickness $δ$ at the trailing edge ( $x = L = 1.0 m$ ), and what fraction of the plate length is it?

Recall Solution

Step 1 — Reuse $R e_{L}$ . Why? $δ$ at the trailing edge needs the local $R e_{x}$ evaluated at $x = L$ , which is exactly the $R e_{L} = 1.0 \times 1 0^{6}$ we already computed in L4.1 — same $U$ , $ν$ , and same $x = L$ . No new arithmetic needed. Step 2. Why? Substituting $R e_{L}$ into $δ = 5.0 L / R e_{L}$ converts the whole-plate Reynolds number into the trailing-edge thickness. $δ (L) = \frac{5.0 L}{R e _{L}} = \frac{5.0 \times 1.0}{1000} = 5.0 \times 1 0^{- 3} m = 5.0 mm .$ Step 3 — Fraction. Why? Dividing $δ$ by $L$ expresses the thickness as a fraction of the plate, the quantity that judges whether Prandtl's thin-layer split is valid. $\frac{δ}{L} = \frac{5.0 \times 1 0 ^{- 3}}{1.0} = 0.5%.$ Half a percent of the length: this tiny ratio is exactly why Prandtl's two-region split works.

Level 5 — Mastery

L5.1 (Derive the scaling of total drag with speed.) For a fixed laminar plate, show how the total friction force $F_{D}$ scales with the free-stream speed $U$ . Then predict the factor by which $F_{D}$ grows if $U$ triples.

Recall Solution

Step 1 — Assemble the pieces symbolically. Why symbolic? We want a scaling law, so keep letters and track powers of $U$ only. $C_{D} = \frac{1.328}{R e _{L}} = 1.328 (U L / ν)^{- 1/2} \propto U^{- 1/2} .$ $q = \frac{1}{2} ρ U^{2} \propto U^{2} .$ $F_{D} = C_{D} q A \propto U^{- 1/2} \cdot U^{2} = U^{3/2} .$ Step 2 — State the law. Laminar skin-friction drag grows as $F_{D} \propto U^{3/2} .$ Step 3 — Triple the speed. $F_{D}$ multiplies by $3^{3/2} = 33 \approx 5.196$ . Physical read: the dynamic pressure ( $U^{2}$ ) pushes drag up, but the thinning layer ( $C_{D} \propto U^{- 1/2}$ ) claws back half a power of $U$ — net $U^{3/2}$ , weaker than the $U^{2}$ you might naively guess.

L5.2 (Design problem.) You need the trailing-edge boundary layer on a $2.0 m$ plate in water ( $ν = 1.0 \times 1 0^{- 6} m^{2} / s$ ) to be exactly $δ = 4.0 mm$ at $x = L = 2.0 m$ , staying laminar. What free-stream speed $U$ achieves this? Then check the flow is genuinely laminar there.

Recall Solution

Step 1 — Solve $δ = 5.0 ν L / U$ for $U$ . Why? $δ$ and $x = L$ are fixed targets; $U$ is the unknown dial. Square: $δ^{2} = 25 \frac{ν L}{U} \Rightarrow U = \frac{25 ν L}{δ ^{2}} .$ Step 2 — Substitute ( $δ = 4.0 \times 1 0^{- 3} m$ ). Why? All inputs are known, so the rearranged formula yields $U$ directly. $U = \frac{25 \times ( 1.0 \times 1 0 ^{- 6} ) \times 2.0}{( 4.0 \times 1 0 ^{- 3} ) ^{2}} = \frac{5.0 \times 1 0 ^{- 5}}{1.6 \times 1 0 ^{- 5}} = 3.125 m/s .$ Step 3 — Laminarity check. Why? Blasius is only valid below transition, so we must confirm $R e_{L}$ against the $5 \times 1 0^{5}$ threshold before trusting the design. $R e_{L} = \frac{U L}{ν} = \frac{3.125 \times 2.0}{1.0 \times 1 0 ^{- 6}} = 6.25 \times 1 0^{6} .$ This is well above the transition $R e \approx 5 \times 1 0^{5}$ — so in reality the flow would be turbulent long before $x = 2 m$ . The laminar target is not physically achievable on this plate; you would need a shorter plate or higher $ν$ (thicker fluid) to keep it laminar. Reporting the honest verdict is the mastery step. See Laminar vs Turbulent flow.

Active Recall

Recall One-line answers (cover them)

$δ$ scales with $x$ as ::: $x$ (grows, but decelerating) $δ$ scales with $U$ as ::: $1/ U$ (faster stream → thinner layer) To find $x$ for a target $δ$ you must ::: square the relation, since $x \propto δ^{2}$ Wall shear $τ_{w}$ scales with $x$ as ::: $x^{- 1/2}$ (largest at the leading edge) What happens to $τ_{w}$ as $x \to 0$ ::: it blows up ( $\to \infty$ ); Blasius breaks down at the sharp tip Laminar total drag scales with speed as ::: $F_{D} \propto U^{3/2}$ The two velocities on this page, $U$ and $u$ , are ::: free-stream speed and local in-layer speed Before trusting any Blasius result, check that ::: $R e_{x} ≲ 5 \times 1 0^{5}$ (still laminar)