Boundary layer thickness, displacement thickness, momentum thickness

1. Boundary layer thickness $\delta$

WHY 99% and not 100%? Because $u \to U$ only as $y\to\infty$ (asymptotic). We need a finite cutoff, so we pick a conventional threshold. 99% is the standard.

WHAT it represents: a geometric edge of the slowed-down region.

HOW it grows: For laminar flow over a flat plate (Blasius), $\delta$ grows with distance $x$ : $\delta \approx \frac{5x}{\sqrt{Re_x}}, \qquad Re_x = \frac{Ux}{\nu}$ So the layer thickens downstream — there's more slowed fluid the longer the wall has acted.

Figure — Boundary layer thickness, displacement thickness, momentum thickness

2. Displacement thickness $\delta^*$ — derive it

Derivation from first principles (mass conservation):

Mass flux per unit width if the whole region moved at $U$ (ideal): $\dot m_{ideal} = \int_0^{\infty} \rho\, U \, dy$

Actual mass flux (real, slowed): $\dot m_{real} = \int_0^{\infty} \rho\, u \, dy$

The deficit in mass flux: $\Delta \dot m = \int_0^{\infty} \rho\,(U - u)\, dy$

We define $\delta^*$ as the thickness of an ideal stream (speed $U$ ) carrying this same deficit: $\rho U \,\delta^* = \int_0^{\infty}\rho\,(U-u)\,dy$

3. Momentum thickness $\theta$ — derive it

Derivation: The momentum actually carried by the slowed fluid is $\int \rho u \cdot u\, dy$ . The momentum that this same mass flow $\rho u$ would carry if moving at $U$ is $\int \rho u \cdot U\, dy$ . The deficit: $\Delta(\text{momentum}) = \int_0^\infty \rho\, u\,(U - u)\, dy$

Set this equal to the momentum of an ideal slab of thickness $\theta$ at speed $U$ , i.e. $\rho U^2 \theta$ : $\rho U^2 \theta = \int_0^\infty \rho\,u\,(U-u)\,dy$

4. Worked Examples

Worked example Example 1 — Linear velocity profile

Assume $\dfrac{u}{U} = \dfrac{y}{\delta}$ for $0\le y\le\delta$ (and $u=U$ beyond).

Find $\delta^*$ : $\delta^* = \int_0^\delta\left(1-\frac{y}{\delta}\right)dy = \left[y - \frac{y^2}{2\delta}\right]_0^\delta = \delta - \frac{\delta}{2} = \frac{\delta}{2}$ Why this step? Beyond $\delta$ , $1-u/U=0$ , so the integral stops at $\delta$ .

Find $\theta$ : $\theta = \int_0^\delta \frac{y}{\delta}\left(1-\frac{y}{\delta}\right)dy = \int_0^\delta\left(\frac{y}{\delta}-\frac{y^2}{\delta^2}\right)dy = \frac{\delta}{2}-\frac{\delta}{3} = \frac{\delta}{6}$ Shape factor: $H = \delta^*/\theta = (\delta/2)/(\delta/6) = 3$ .

Worked example Example 2 — Sine profile

Assume $\dfrac{u}{U} = \sin\!\left(\dfrac{\pi y}{2\delta}\right)$ .

$\delta^* = \int_0^\delta\left(1 - \sin\tfrac{\pi y}{2\delta}\right)dy = \delta - \frac{2\delta}{\pi} = \delta\left(1-\frac{2}{\pi}\right)\approx 0.363\,\delta$ Why this step? $\int_0^\delta \sin(\tfrac{\pi y}{2\delta})dy = \big[-\tfrac{2\delta}{\pi}\cos\big]_0^\delta = \tfrac{2\delta}{\pi}$ .

$\theta = \int_0^\delta \sin u_r(1-\sin u_r)\,dy = \frac{2\delta}{\pi} - \frac{\delta}{2} \approx 0.137\,\delta$ (using $\int_0^\delta \sin^2 = \delta/2$ ). So $H \approx 0.363/0.137 \approx 2.66$ — close to Blasius!

5. Common Mistakes

6. Active Recall Flashcards

What is the no-slip condition?

At a solid wall, fluid velocity equals the wall velocity (zero for a stationary wall), so

u(0)=0

Define boundary layer thickness

\delta

Distance from the wall where

u = 0.99U

(99% of free-stream speed).

Why 99% and not 100%?

Because

u \to U

only asymptotically; a finite cutoff is needed, 99% is the convention.

Formula for displacement thickness.

\delta^* = \int_0^\infty (1 - u/U)\,dy

Physical meaning of

\delta^*

Distance the outer streamlines are pushed away from the wall = thickness of an ideal

U

-stream carrying the missing mass flux.

Formula for momentum thickness.

\theta = \int_0^\infty \frac{u}{U}(1 - u/U)\,dy

Physical meaning of

\theta

Thickness of an ideal

U

-stream carrying the momentum lost to the wall; related to wall drag.

Why does

\theta

have the extra

u/U

factor?

Momentum deficit

= \rho u (U-u)

: needs the actual mass-flow

u

AND the velocity gap

(U-u)

Order of the three thicknesses.

\theta < \delta^* < \delta

Define shape factor

H

and its laminar value.

H = \delta^*/\theta

; Blasius laminar

H \approx 2.59

; rising

H

signals impending separation.

For a linear profile

u/U=y/\delta

, give

\delta^*

\theta

H

\delta^*=\delta/2

\theta=\delta/6

H=3

Blasius growth law for

\delta

\delta \approx 5x/\sqrt{Re_x}

, with

Re_x = Ux/\nu

Recall Feynman: explain to a 12-year-old

Imagine running your hand flat through water. The water touching your hand drags along with it, but water far away barely cares. There's a thin "sticky layer" near your hand where water goes from "stuck" to "free." That layer's height is $\delta$ . Now, because some water near the hand is moving slower, less water slips past than you'd expect — it's as if the hand were a tiny bit fatter and blocked a sliver of water. That fattening is the displacement thickness $\delta^*$ . And the hand also steals momentum (push) from the water, which is what makes drag — the amount of stolen push, written as a thickness, is the momentum thickness $\theta$ . Three rulers, three different things they measure.

Connections

No-slip condition — the physical origin of the boundary layer.
Reynolds number — governs whether the layer is laminar or turbulent.
Blasius solution — exact flat-plate laminar profile giving $\delta = 5x/\sqrt{Re_x}$ .
Skin friction drag — directly proportional to $\theta$ via the momentum-integral equation.
Boundary layer separation — predicted by rising shape factor $H$ .
Viscosity — the property that creates the velocity gradient at the wall.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, jab koi real fluid (viscous) kisi wall ke paas se behta hai, to wall ko touch karne wala fluid bilkul ruk jaata hai — isko no-slip condition kehte hain. Wall pe speed zero, aur door free-stream speed $U$ . Beech ki jo patli si layer hai jahan velocity $0$ se $U$ tak chadhti hai, usko boundary layer bolte hain. Problem yeh hai ki yeh layer kahan "khatam" hoti hai yeh clearly nahi pata, kyunki $u$ dheere-dheere $U$ tak pahunchti hai. Isliye hum 99% wali jagah ko $\delta$ keh dete hain — yeh bas geometry ka ruler hai.

Ab do aur important measures hain. Displacement thickness $\delta^*$ batata hai ki wall ke paas fluid slow hone ki wajah se kitna mass flow kam ho gaya — jaise wall thoda mota ho gaya ho aur streamlines ko bahar push kar raha ho. Formula: $\delta^*=\int(1-u/U)\,dy$ . Momentum thickness $\theta$ batata hai ki wall ne kitna momentum (push) chura liya — yahi cheez drag banati hai. Formula: $\theta=\int \frac{u}{U}(1-u/U)\,dy$ . Yaad rakho: momentum mein ek extra $u/U$ factor isliye aata hai kyunki momentum $\sim u^2$ hota hai — ek $u$ mass-flow ke liye, doosra velocity gap ke liye.

Trick samajh lo: $\delta^*$ mein ek deficit factor (mass), $\theta$ mein deficit ko $u/U$ se multiply (momentum). Aur hamesha order yeh: $\theta < \delta^* < \delta$ . Inka ratio $H=\delta^*/\theta$ shape factor hai — agar yeh badhne lage to flow separate hone wala hai, yeh engineering mein bahut kaam aata hai. Exam mein aksar linear ya sine profile dekar $\delta^*,\theta$ nikalne ko aata hai — bas integral lagao, ho gaya!

Boundary layer thickness, displacement thickness, momentum thickness

1. Boundary layer thickness $\delta$

2. Displacement thickness $\delta^*$ — derive it

3. Momentum thickness $\theta$ — derive it

4. Worked Examples

5. Common Mistakes

6. Active Recall Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Fluid Mechanics

Connections