Worked examples — Boundary layer thickness, displacement thickness, momentum thickness

This page is a drill floor. The parent note built the three thicknesses $\delta$, ${\delta }^{\ast }$, $\theta$ from first principles. Here we throw every kind of input at those formulas so no exam question can surprise you.

Before we start, one reminder of the three tools, in plain words:

Recall The three definitions (know these cold)

• $\delta$ ::: distance from wall where speed reaches $99\%$ of the free-stream $U$.
• ${\delta }^{\ast }={\int }_0^{\infty }\left(1-\frac{u}{U}\right)dy$ ::: thickness of the "missing slab" of mass flow.
• $\theta ={\int }_0^{\infty }\frac{u}{U}\left(1-\frac{u}{U}\right)dy$ ::: thickness of the "missing slab" of momentum.

Here $u=u(y)$ is the fluid speed at height $y$ above the wall, $U$ is the free-stream speed, $\nu$ is kinematic viscosity, and everything obeys the No-slip condition ($u=0$ at $y=0$).

---

The scenario matrix

Every question about these thicknesses lands in one of these boxes. We will hit every box with a worked example.

Cell	Case class	What makes it tricky	Hit by
A	Linear profile	simplest polynomial, warm-up	Ex 1
B	Polynomial (cubic) profile	matching slope + curvature at edge	Ex 2
C	Trigonometric (sine) profile	integrating $\sin ,{\sin }^2$	Ex 3
D	Degenerate: plug/uniform flow ($u=U$ everywhere)	zero deficit — do the limits break?	Ex 4
E	Degenerate: stagnant limit ($u\to 0$)	maximal deficit, sanity ceiling	Ex 4
F	Profile defined only to $\delta$ (finite cutoff)	the ${\int }_0^{\infty }$ trap	Ex 5
G	Real-world word problem (numbers, units)	Reynolds number, mm-scale answer	Ex 6
H	Downstream growth / scaling with $x$	how $\delta ,{\delta }^{\ast },\theta$ all scale	Ex 7
I	Exam twist: given ${\delta }^{\ast }$ and $H$, back out $\theta$	reverse the Blasius solution shape factor	Ex 8
J	Sign/limiting sanity: ordering $\theta <{\delta }^{\ast }<\delta$	why the inequality can never flip	Ex 9

A recurring trick: for a profile that only depends on the shape, write $\eta =y/\delta$ (a pure number from $0$ to $1$). Then $dy=\delta d\eta$ and every thickness comes out as (a pure number) × $\delta$. That pure number is all that changes between profiles.

The figure below plots the three model profiles we use (linear, cubic, sine) on the same axes, so you can see why they give different thicknesses. The horizontal axis is $u/U$, the vertical axis is $\eta =y/\delta$. Look at how each curve leaves the wall (bottom-left) and reaches $u=U$ (the dashed navy line) at the top. The shaded magenta band between the linear curve and $u=U$ is exactly the deficit $(1-u/U)$ that ${\delta }^{\ast }$ integrates — a curve that hugs the dashed line more closely (like the orange sine) leaves a thinner band, hence a smaller ${\delta }^{\ast }$.

Keep this picture in mind: the further a profile sits from the dashed $u=U$ line, the fatter its deficit, the bigger its ${\delta }^{\ast }$.

---

Example 1 — Linear profile (Cell A)

Forecast: guess before computing — will ${\delta }^{\ast }$ be bigger or smaller than $\delta /2$? (The velocity is below $U$ across the whole layer, but the deficit shrinks linearly to zero...)

1. Switch to $\eta =y/\delta$. So $u/U=\eta$, $dy=\delta d\eta$, limits $0\to 1$. Why this step? It strips out $\delta$ so the integral is just a number. This is the red magenta line in the figure above.
2. Displacement: ${\delta }^{\ast }={\int }_0^1(1-\eta )\delta d\eta =\delta {\left[\eta -\frac{{\eta }^2}{2}\right]}_0^1=\delta \left(1-\frac{1}{2}\right)=\frac{\delta }{2}.$ Why this step? $1-u/U=1-\eta$ is the fractional deficit — the shaded band in the figure.
3. Momentum: $\theta ={\int }_0^1\eta (1-\eta )\delta d\eta =\delta {\left[\frac{{\eta }^2}{2}-\frac{{\eta }^3}{3}\right]}_0^1=\delta \left(\frac{1}{2}-\frac{1}{3}\right)=\frac{\delta }{6}.$ Why this step? The extra factor $\eta =u/U$ weights the deficit by how much mass is actually there.
4. Shape factor: $H=\frac{{\delta }^{\ast }}{\theta }=\frac{\delta /2}{\delta /6}=3.$ (Recall $H$ is defined in the box above.)

Verify: $\theta =\delta /6<{\delta }^{\ast }=\delta /2<\delta$ ✓ (ordering holds). Units: every term is $\times \delta$, a length ✓. Forecast check: ${\delta }^{\ast }=\delta /2$ exactly — the linear deficit averages to $1/2$.

---

Example 2 — Cubic profile (Cell B)

Forecast: this profile is steeper at the wall (bigger slope at $\eta =0$) than the linear one — it is the violet curve in the figure. Will its $H$ be above or below $3$?

1. Displacement: $1-\frac{u}{U}=1-\frac{3}{2}\eta +\frac{1}{2}{\eta }^3$. ${\delta }^{\ast }=\delta {\int }_0^1\left(1-\frac{3}{2}\eta +\frac{1}{2}{\eta }^3\right)d\eta =\delta \left[1-\frac{3}{4}+\frac{1}{8}\right]=\frac{3\delta }{8}.$ Why this step? Just integrate term by term; each power ${\eta }^n$ gives $1/(n+1)$.

2. Momentum — expand the integrand first. Multiply out $\frac{u}{U}\left(1-\frac{u}{U}\right)$: $\left(\frac{3}{2}\eta -\frac{1}{2}{\eta }^3\right)\left(1-\frac{3}{2}\eta +\frac{1}{2}{\eta }^3\right).$ Distribute term by term: $=\underset{\frac{3}{2}\eta \cdot 1}{\underbrace{\frac{3}{2}\eta }}-\underset{\frac{3}{2}\eta \cdot (-\frac{3}{2}\eta )}{\underbrace{\frac{9}{4}{\eta }^2}}+\underset{\frac{3}{2}\eta \cdot \frac{1}{2}{\eta }^3}{\underbrace{\frac{3}{4}{\eta }^4}}-\underset{-\frac{1}{2}{\eta }^3\cdot 1}{\underbrace{\frac{1}{2}{\eta }^3}}+\underset{-\frac{1}{2}{\eta }^3\cdot (-\frac{3}{2}\eta )}{\underbrace{\frac{3}{4}{\eta }^4}}-\underset{-\frac{1}{2}{\eta }^3\cdot \frac{1}{2}{\eta }^3}{\underbrace{\frac{1}{4}{\eta }^6}}.$ Collect like powers (the two $\frac{3}{4}{\eta }^4$ terms add to $\frac{3}{2}{\eta }^4$): $=\frac{3}{2}\eta -\frac{9}{4}{\eta }^2-\frac{1}{2}{\eta }^3+\frac{3}{2}{\eta }^4-\frac{1}{4}{\eta }^6.$ Why this step? Momentum needs $u(1-u)$; you cannot integrate the product without expanding it into powers first.

3. Integrate each power using ${\int }_0^1{\eta }^{n}d\eta =\frac{1}{n+1}$, then put every term over the common denominator $280$:

   term	value	$\times 280$
   $+\frac{3}{2}\cdot \frac{1}{2}$	$+\frac{3}{4}$	$+210$
   $-\frac{9}{4}\cdot \frac{1}{3}$	$-\frac{3}{4}$	$-210$
   $-\frac{1}{2}\cdot \frac{1}{4}$	$-\frac{1}{8}$	$-35$
   $+\frac{3}{2}\cdot \frac{1}{5}$	$+\frac{3}{10}$	$+84$
   $-\frac{1}{4}\cdot \frac{1}{7}$	$-\frac{1}{28}$	$-10$

   Sum of the last column: $210-210-35+84-10=39$. Hence $\theta =\frac{39}{280}\delta \approx 0.139\delta .$ Why this step? The table forces every fraction onto one denominator ($280$) so no term is mis-scaled — the safe way to add many fractions.

4. Shape factor: $H=\frac{3/8}{39/280}=\frac{3}{8}\cdot \frac{280}{39}=\frac{105}{39}=\frac{35}{13}\approx \mathrm{2.69.}$

Verify: $H\approx 2.69$ is much closer to the true Blasius solution value $2.59$ than the linear profile's $3$ — as expected, the more realistic curved profile predicts drag better. Ordering: $0.139\delta <0.375\delta <\delta$ ✓.

---

Example 3 — Sine profile (Cell C)

Forecast: the sine bulges above the linear line for most of the layer (it rises fast then flattens) — the orange curve in the figure, hugging the dashed $u=U$ line. So its deficit is smaller — expect ${\delta }^{\ast }<\delta /2$.

1. Displacement: ${\delta }^{\ast }=\delta {\int }_0^1\left(1-\sin \frac{\pi \eta }{2}\right)d\eta =\delta {\left[\eta +\frac{2}{\pi }\cos \frac{\pi \eta }{2}\right]}_0^1=\delta \left(1-\frac{2}{\pi }\right)\approx 0.363\delta .$ Why this step? ${\int }_0^1\sin \frac{\pi \eta }{2}d\eta =[-\frac{2}{\pi }\cos \frac{\pi \eta }{2}{]}_0^1=\frac{2}{\pi }$. We use a trig tool here (not polynomials) because the profile itself is trig — the antiderivative of $\sin$ is a known clean function.
2. Momentum: $\frac{u}{U}\left(1-\frac{u}{U}\right)=\sin \frac{\pi \eta }{2}-{\sin }^2\frac{\pi \eta }{2}$. Using ${\int }_0^1\sin \frac{\pi \eta }{2}d\eta =\frac{2}{\pi }$ and ${\int }_0^1{\sin }^2\frac{\pi \eta }{2}d\eta =\frac{1}{2}$: $\theta =\delta \left(\frac{2}{\pi }-\frac{1}{2}\right)\approx 0.137\delta .$ Why this step? We used the half-angle identity ${\sin }^{2}x=\frac{1-\cos 2x}{2}$ to integrate ${\sin }^2$ — that's why we reach for it, because you can't integrate ${\sin }^2$ directly.
3. Shape factor: $H=\frac{0.363}{0.137}\approx \mathrm{2.66.}$

Verify: $H\approx 2.66$, again close to Blasius $2.59$ ✓. Forecast confirmed: ${\delta }^{\ast }\approx 0.363\delta <0.5\delta$ ✓.

---

Example 4 — Degenerate limits: plug flow and stagnant flow (Cells D, E)

Forecast: with no slowdown, the deficit is zero — both thicknesses should vanish. With maximal slowdown, the deficit should approach its ceiling: the whole layer $\delta$.

1. (a) Plug flow: $1-u/U=0$ everywhere, so ${\delta }^{\ast }={\int }_0^{\infty }0dy=0,\theta ={\int }_0^{\infty }0dy=0.$ Why this step? If nothing is slowed, no mass and no momentum are "missing." This is the sanity floor.
2. (b) Power-law $u/U={\eta }^n$: $\frac{{\delta }^{\ast }}{\delta }={\int }_0^1(1-{\eta }^n)d\eta =1-\frac{1}{n+1}=\frac{n}{n+1}.$ Why this step? One clean integral covers a whole family of profiles.
3. Take $n\to \infty$: $\frac{{\delta }^{\ast }}{\delta }=\frac{n}{n+1}\to 1.$ Why this step? As $n\to \infty$, the fluid is near-zero for almost all $\eta <1$ then jumps to $U$ — the deficit fills the whole layer.

Verify: For $n=1$ (linear) this gives $\frac{1}{2}$, matching Example 1 ✓. The limit ${\delta }^{\ast }\to \delta$ is the theoretical ceiling: ${\delta }^{\ast }$ can never exceed $\delta$ because $1-u/U\le 1$. Plug flow gives $0$, the floor ✓.

---

Example 5 — The "${\int }_0^{\infty }$" trap (Cell F)

Forecast: the piece from $\delta$ to $\infty$ contributes... what?

1. Split the integral at $y=\delta$ (i.e. $\eta =1$): ${\delta }^{\ast }={\int }_0^{\delta }\left(1-\frac{u}{U}\right)dy+{\int }_{\delta }^{\infty }\left(1-\frac{u}{U}\right)dy.$ Why this step? The profile is defined piecewise, so split where the pieces meet.
2. Beyond $\delta$, $u=U$, so $1-u/U=0$ and the second integral is $0$. Why this step? This is the whole resolution — the tail contributes nothing. That's why ${\int }_0^{\infty }$ and ${\int }_0^{\delta }$ agree for any profile that reaches $U$ at $\delta$.
3. Compute the finite part with $u/U=2\eta -{\eta }^2$, so $1-u/U=1-2\eta +{\eta }^2=(1-\eta {)}^2$: ${\delta }^{\ast }=\delta {\int }_0^1(1-\eta {)}^{2}d\eta =\delta {\left[-\frac{(1-\eta {)}^3}{3}\right]}_0^1=\frac{\delta }{3}.$

Verify: $(1-\eta {)}^2\ge 0$ so ${\delta }^{\ast }>0$ ✓. Value $\delta /3\approx 0.333\delta <\delta$ ✓. Sanity: this parabola sits above the linear line, smaller deficit than $\delta /2$, and indeed $\frac{1}{3}<\frac{1}{2}$ ✓.

---

Example 6 — Real-world word problem (Cell G)

Forecast: boundary layers on everyday objects are surprisingly thin — guess $\delta$ in mm.

1. Reynolds number: $R{e}_x=\frac{Ux}{\nu }=\frac{30\times 0.5}{1.5\times 10^{-5}}=10^6.$ Why this step? $R{e}_x$ tells us the flow is laminar (well below the transition band) and it drives the growth law.
2. Boundary layer thickness (Blasius solution growth law): $\delta =\frac{5x}{\sqrt{R{e}_x}}=\frac{5\times 0.5}{\sqrt{10^6}}=\frac{2.5}{1000}=2.5\times 10^{-3}\text{m}=2.5\text{mm}.$ Why this step? $5/\sqrt{R{e}_x}$ is the fraction of $x$ occupied by slowed fluid.
3. Choose a profile model. We only measured $\delta$, not the full velocity field $u(y)$. To get ${\delta }^{\ast }$ and $\theta$ we must assume a profile shape; the simplest is the linear one from Example 1, which gave the fixed ratios ${\delta }^{\ast }=\delta /2$ and $\theta =\delta /6$. Why this step? ${\delta }^{\ast }$ and $\theta$ are integrals of $u(y)$ — without knowing the profile they are undefined, so a model profile is mandatory. The linear model is the crudest estimate but shows the correct scale.
4. Displacement & momentum from those ratios: ${\delta }^{\ast }=\frac{\delta }{2}=1.25\text{mm},\theta =\frac{\delta }{6}=0.417\text{mm}.$
5. Shape factor — for the linear profile it is a pure number independent of the physical size: $H=\frac{{\delta }^{\ast }}{\theta }=\frac{1.25}{0.417}=3.$ Why this step? $H$ depends only on the profile shape, not on $\delta$, so it equals the Example 1 value $3$.

Verify: Units: $\frac{(\text{m/s})(\text{m})}{{\text{m}}^2/\text{s}}=$ dimensionless ✓ for $R{e}_x$. $\delta =2.5\text{mm}$ — thin, as forecast ✓. Ordering $0.417<1.25<2.5\text{mm}$ ✓, and $H=3$ matches Example 1 ✓.

---

Example 7 — Downstream scaling (Cell H)

Forecast: more wall means more slowed fluid — all three grow. But by how much? Linearly with $x$? Or slower?

1. Growth law: $\delta =\frac{5x}{\sqrt{R{e}_x}}$ with $R{e}_x=\frac{Ux}{\nu }$, so $\sqrt{R{e}_x}\propto \sqrt{x}$ and $\delta \propto \frac{x}{\sqrt{x}}=\sqrt{x}.$ Why this step? Substitute $R{e}_x$ into the law — the $x$ in the numerator partly cancels the $\sqrt{x}$ downstairs.
2. Since ${\delta }^{\ast }$ and $\theta$ are fixed fractions of $\delta$ (for a self-similar laminar profile), they also scale as $\sqrt{x}$. Why this step? Self-similarity means the profile shape is unchanged with $x$; only $\delta$ stretches, so $H$ stays constant while all three lengths grow together.
3. Factor for 4× distance: $\sqrt{4}=2$. All three double.

Verify: $\delta (2.0)=\frac{5\times 2.0}{\sqrt{30\times 2.0/1.5\times 10^{-5}}}=\frac{10}{\sqrt{4\times 10^6}}=\frac{10}{2000}=5\times 10^{-3}\text{m}=5\text{mm}$. That is exactly $2\times 2.5\text{mm}$ ✓. The layer thickens like $\sqrt{x}$, not linearly — a key laminar fact behind Skin friction drag.

---

Example 8 — Exam twist: reverse the shape factor (Cell I)

Forecast: rising $H$ means ${\delta }^{\ast }$ grows relative to $\theta$ — the profile is getting "emptier" near the wall. What physical event does that precede?

1. Definition of shape factor: $H=\frac{{\delta }^{\ast }}{\theta }\Rightarrow \theta =\frac{{\delta }^{\ast }}{H}.$ Why this step? Rearranging the single definition (see the definition box) is all we need — no integrals.
2. (a) $\theta =\frac{1.8}{2.59}\approx 0.695\text{mm}.$
3. (b) Rising $H$ signals impending Boundary layer separation — the near-wall fluid is slowing so much it is about to reverse. New $\theta =\frac{1.8}{3.5}\approx 0.514\text{mm}.$ Why this step? Same formula; larger $H$ (with ${\delta }^{\ast }$ fixed) means smaller $\theta$ — the momentum-carrying capacity has collapsed near the wall.

Verify: $H\theta =2.59\times 0.695\approx 1.80\text{mm}={\delta }^{\ast }$ ✓. And $3.5\times 0.514\approx 1.80\text{mm}$ ✓. Both reproduce the measured ${\delta }^{\ast }$.

---

Example 9 — Why $\theta <{\delta }^{\ast }<\delta$ can never flip (Cell J)

Forecast: compare the two integrands term-by-term — one always sits below the other. Why?

1. Write both integrands: ${\delta }^{\ast }$ has $\left(1-\frac{u}{U}\right)$; $\theta$ has $\frac{u}{U}\left(1-\frac{u}{U}\right)$. Why this step? Same factor $\left(1-\frac{u}{U}\right)\ge 0$ appears in both; only the extra $\frac{u}{U}$ differs. Comparing them reduces the whole ordering to comparing this one extra factor against $1$.
2. The extra multiplier is $\le 1$: since $0\le u\le U$, we have $0\le \frac{u}{U}\le 1$. Multiplying a non-negative quantity by something in $[0,1]$ can only shrink it (or leave it unchanged): $\frac{u}{U}\left(1-\frac{u}{U}\right)\le \left(1-\frac{u}{U}\right).$ Why this step? This is the core of the whole ordering — the momentum deficit carries the extra mass factor $u/U$, and a mass fraction never exceeds $1$.
3. Integrate both sides over $y:0\to \infty$. Integration preserves the $\le$ sign, so $\theta ={\int }_0^{\infty }\frac{u}{U}\left(1-\frac{u}{U}\right)dy\le {\int }_0^{\infty }\left(1-\frac{u}{U}\right)dy={\delta }^{\ast }.$ Why this step? If $f(y)\le g(y)$ everywhere and both are integrable, then $\int f\le \int g$.
4. Strictness. Equality in step 2 needs $\frac{u}{U}=1$ (or $\frac{u}{U}=0$) at every height. A real profile climbs strictly from $0$ to $U$, so it is not at those extremes across the whole layer — hence the inequality is strict: $\theta <{\delta }^{\ast }$.
5. Top end. Likewise ${\delta }^{\ast }={\int }_0^{\delta }(1-u/U)dy<{\int }_0^{\delta }1dy=\delta$, since $1-u/U<1$ wherever $u>0$. Chaining gives the full guarantee: $\boxed{\theta <{\delta }^{\ast }<\delta }$ This single argument is why every earlier example obeyed the ordering — it is not a coincidence, it follows from $0\le u\le U$ alone.

Verify: Check every earlier example: linear $(\frac{1}{6}<\frac{1}{2}<1)$✓, cubic $(0.139<0.375<1)$✓, sine $(0.137<0.363<1)$✓, parabola ${\delta }^{\ast }=\frac{1}{3}<1$✓ — all obey $\theta <{\delta }^{\ast }<\delta$ exactly as the proof demands.

---