Visual walkthrough — Boundary layer thickness, displacement thickness, momentum thickness

We will use these words throughout, so let us pin them down once, in plain language, before any symbol appears:

• Wall — a flat solid surface the fluid flows over. We measure height straight up from it.
• Free stream — the fast fluid far from the wall that hasn't noticed the wall yet.
• Slow layer — fluid near the wall, dragged back because it sticks to the surface (the no-slip condition).

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Step 1 — Set up the picture: height, speed, and the two symbols $y$ and $u$

WHAT. Draw the wall as a horizontal floor. Stack many thin horizontal fluid layers on top of it. Each layer sits at some height $y$ and slides forward at its own speed $u(y)$ — drawn as an arrow pointing downstream. Short arrow = slow, long arrow = fast.

WHY. Everything about boundary layers is "how does speed change as you go up?" So the two things worth naming are exactly height ($y$) and speed there ($u$). Nothing else is needed yet.

PICTURE. The bottom arrow has length zero (stuck to the wall). As you climb, arrows get longer, until near the top they all reach the same full length — the free-stream speed, which we will name next.

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Step 2 — Name the free-stream speed $U$ and the edge $\delta$

WHAT. Mark the free-stream arrow length $U$ at the top. Mark the height where the arrows have almost reached that full length; call that height $\delta$.

WHY $99\%$ and not $100\%$? The speed only approaches $U$; it never mathematically equals it at a finite height (it's asymptotic — it flattens out but never quite touches). To draw a real edge we need a cutoff, and $99\%$ is the agreed convention. See Blasius for where the number comes from.

PICTURE. A vertical dashed line at height $\delta$ separates the "slow, changing" region below from the "everyone at $U$" region above.

Notice a key fact we will lean on repeatedly:

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Step 3 — The "velocity deficit": how much slower each layer is

WHAT. For each layer, draw the gap between its short arrow and the full-length $U$ arrow. That red gap is the deficit $U-u$.

WHY this quantity. The whole point of a boundary layer is that fluid is missing speed. "How much speed is missing, layer by layer?" is the deficit. Stacking those gaps up the wall is what the integrals in Steps 5 and 6 do.

PICTURE. Red gap-bars beside each layer: tall at the bottom (big deficit), shrinking to nothing at $\delta$.

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Step 4 — One thin layer carries a thin slice of mass flow

Before adding things up, understand one layer of tiny thickness $dy$.

WHY $\rho udy$? In one second, the fluid in this layer slides forward a distance $u$. That sweeps a rectangle of height $dy$, length $u$, so area $udy$; times density $\rho$ gives the mass that crossed. Speed × height × density = mass flow rate.

PICTURE. A single highlighted layer; a horizontal arrow of length $u$ shows how far its fluid travels in one second; the swept rectangle $udy$ is shaded.

Now we can add up all the layers two different ways: for mass (Step 5) and for momentum (Step 6).

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Step 5 — Stack the mass deficit ⇒ displacement thickness ${\delta }^{\ast }$

WHAT — build the deficit integral directly. Rather than compare two whole-column mass flows (each of which, integrated all the way to infinity, would be infinite and so cannot be subtracted term by term), we work layer by layer and add up only the difference, which is finite. For one layer at height $y$:

• if it moved at the full $U$ it would carry mass flow $\rho Udy$;
• it really carries $\rho udy$ (Step 4);
• so its missing mass flow is $\rho (U-u)dy$ — and this is exactly zero above $\delta$ (Step 2 fact).

Summing that finite, well-behaved shortfall over all layers gives the missing mass flow rate:

$\Delta \dot{m}={\int }_0^{\infty }\underset{\text{density}}{\underbrace{\rho }}\underset{\text{speed gap}}{\underbrace{(U-u)}}\underset{\text{slice height}}{\underbrace{dy}}$

This single integral converges because the integrand vanishes beyond $\delta$ — no $\infty -\infty$ ever appears.

WHY divide by $\rho U$. We now ask: replace that missing mass flow by a clean slab of free-stream fluid — density $\rho$, speed $U$, height ${\delta }^{\ast }$. Its mass flow rate is $\rho U{\delta }^{\ast }$. Set them equal:

$\rho U{\delta }^{\ast }={\int }_0^{\infty }\rho (U-u)dy$

Divide both sides by $\rho U$ (density and $U$ are constants, they slide out):

PICTURE. The red deficit region (area of all the gap-bars) gets "collapsed" into a single solid rectangle of width $U$ and height ${\delta }^{\ast }$ — same area, one clean slab.

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Step 6 — Stack the momentum deficit ⇒ momentum thickness $\theta$

WHAT — why momentum needs the speed twice. Momentum = mass × velocity. For one layer:

• the mass actually flowing is $\rho udy$ (Step 4) — note it uses the real speed $u$,
• and each bit of that mass is short by $(U-u)$ in speed.

So the momentum shortfall of one layer is mass-flow × speed-gap:

$\rho \underset{\begin{array}{c}\text{how much}\\ \text{mass moves}\end{array}}{\underbrace{u}}\underset{\begin{array}{c}\text{how much}\\ \text{speed it lacks}\end{array}}{\underbrace{(U-u)}}dy$

Sum over all layers (again finite — the integrand vanishes above $\delta$):

$\Delta (\text{momentum})={\int }_0^{\infty }\rho u(U-u)dy$

WHY divide by $\rho U^2$. Match it to a clean slab of free-stream fluid carrying momentum flow $\rho U^2\theta$ — that is the momentum crossing per second per unit span (mass flow $\rho U$ times speed $U$, over height $\theta$), with units $\text{kg}\cdot {\text{m/s}}^2$ per metre of span (a force per unit span). Setting the two equal:

$\rho U^2\theta ={\int }_0^{\infty }\rho u(U-u)dy$

Divide by $\rho U^2$:

PICTURE. Two curves up the wall — the mass-weight $u/U$ rising, the deficit $1-u/U$ falling — and their product bulging in the middle. That bulge, collapsed, is the height $\theta$.

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Step 7 — Why the order is always $\theta <{\delta }^{\ast }<\delta$

WHAT. Line up all three heights on the same picture.

WHY the ordering is forced.

• ${\delta }^{\ast }$ integrates $(1-\frac{u}{U})$.
• $\theta$ integrates the same thing but multiplied by $\frac{u}{U}$, a number between 0 and 1. Multiplying an area by something $\le 1$ can only shrink it — so $\theta \le {\delta }^{\ast }$.
• And ${\delta }^{\ast }$ is only part of the full height $\delta$ (the deficit never fills the whole layer), so ${\delta }^{\ast }<\delta$.

Hence, for any sensible profile: $\theta <{\delta }^{\ast }<\delta .$

PICTURE. Three stacked brackets on one wall: tall $\delta$, medium ${\delta }^{\ast }$, short $\theta$.

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Step 8 — Edge & degenerate cases (never get surprised)

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Step 9 — Plug in a real profile (linear) and read the answers off

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The one-picture summary

Everything on one wall: the velocity arrows growing from $0$ to $U$; the red deficit collapsing into the ${\delta }^{\ast }$ slab; the middle-bulging product collapsing into the $\theta$ slab; the $99\%$ edge marking $\delta$ — and the brackets showing $\theta <{\delta }^{\ast }<\delta$.

Recall Feynman: the whole walkthrough in plain words

Water sticks to a wall, so a thin sheet of it near the wall runs slow. Stand each slow layer up and compare its speed-arrow to the full free-stream arrow $U$; the gap is how much speed that layer is missing. Add all the gaps up the wall and you've counted the missing mass flow — squash that into a single clean slab of full-speed fluid and its height is ${\delta }^{\ast }$, the distance the wall shoves the outside flow outward. Now weight each gap by how much fluid is actually moving there ($u/U$) — that gives the missing momentum, which is the drag; squash that into a slab and its height is $\theta$. Because the momentum sum uses an extra factor smaller than one, $\theta$ is always the smallest, then ${\delta }^{\ast }$, then the geometric edge $\delta$. Their ratio ${\delta }^{\ast }/\theta$ is a warning gauge: as it swells the near-wall flow is weakening toward peeling off the wall.

Recall

Order of the three thicknesses? ::: $\theta <{\delta }^{\ast }<\delta$. Why does $\theta$ carry an extra $u/U$ factor ${\delta }^{\ast }$ does not? ::: Momentum lost $=\rho u(U-u)$: you need the moving mass ($\propto u$) and the speed gap $(U-u)$; mass loss only needs the gap. For the linear profile, give ${\delta }^{\ast },\theta ,H$. ::: ${\delta }^{\ast }=\delta /2$, $\theta =\delta /6$, $H=3$. Why can you always integrate only up to $\delta$? ::: Above $\delta$, $u=U$ so both integrands are zero.