Exercises — Boundary layer thickness, displacement thickness, momentum thickness

WHAT: At $y=0$, $u=0$. WHY: A real fluid has viscosity, so the layer of fluid touching the wall cannot slide over it — it moves with the wall. A stationary wall therefore has $u(0)=0$. This is the no-slip condition. Answer: $u(0)=0$, forced by the no-slip condition.

Order: $\theta <{\delta }^{\ast }<\delta$. WHY $\theta$ is smallest: its integrand $\frac{u}{U}(1-\frac{u}{U})$ is the product of two fractions each between 0 and 1, so it is smaller everywhere than the single fraction $(1-\frac{u}{U})$ that builds ${\delta }^{\ast }$. A smaller integrand gives a smaller integral.

Integrate only $0\to \delta$. WHY: beyond $\delta$ the fluid moves at free-stream speed, $u=U$, so $1-\frac{u}{U}=0$ and $\frac{u}{U}(1-\frac{u}{U})=0$. The integrand vanishes, contributing nothing — the $\infty$ is bookkeeping for "wherever the deficit lives."

WHAT: substitute $\frac{u}{U}=\frac{y}{\delta }$ into ${\delta }^{\ast }={\int }_0^{\delta }(1-\frac{u}{U})dy$. WHY substitute: ${\delta }^{\ast }$ counts the fractional velocity deficit $1-u/U$ stacked up over height; to add it up we need that deficit written as a function of $y$, which the profile hands us. ${\delta }^{\ast }={\int }_0^{\delta }\left(1-\frac{y}{\delta }\right)dy={\left[y-\frac{y^2}{2\delta }\right]}_0^{\delta }=\delta -\frac{\delta }{2}=\frac{\delta }{2}$ WHY it comes out $\delta /2$: the deficit $1-y/\delta$ falls straight from 1 (full deficit at the wall) to 0 (no deficit at the edge). The area under a straight line from 1 to 0 over a base $\delta$ is a triangle — exactly half the box, hence $\delta /2$. Answer: ${\delta }^{\ast }=\frac{\delta }{2}=0.5\delta$.

WHY the integrand changes: momentum deficit weights the velocity gap $(1-u/U)$ by how much mass is actually moving there, $u/U$ — so we multiply the two, giving $\frac{y}{\delta }(1-\frac{y}{\delta })$. $\theta ={\int }_0^{\delta }\frac{y}{\delta }\left(1-\frac{y}{\delta }\right)dy={\int }_0^{\delta }\left(\frac{y}{\delta }-\frac{y^2}{{\delta }^2}\right)dy=\frac{\delta }{2}-\frac{\delta }{3}=\frac{\delta }{6}$ WHY $\theta <{\delta }^{\ast }$ here: the extra factor $y/\delta$ is a fraction $\le 1$, so it shrinks the deficit everywhere — the integral must come out smaller ($\delta /6<\delta /2$), consistent with Exercise 1.2. $H=\frac{{\delta }^{\ast }}{\theta }=\frac{\delta /2}{\delta /6}=3$ Answer: $\theta =\frac{\delta }{6}$, $H=3$.

WHY $R{e}_x$ first: the Blasius law $\delta =5x/\sqrt{R{e}_x}$ needs $R{e}_x$, and $R{e}_x$ measures how strongly inertia ($Ux$) beats viscous diffusion ($\nu$) — a big value means a thin layer. $R{e}_x=\frac{Ux}{\nu }=\frac{30\times 0.5}{1.5\times 10^{-5}}=\frac{15}{1.5\times 10^{-5}}=10^6$ WHY plug into the Blasius law: it is the exact laminar flat-plate result; $\sqrt{R{e}_x}=\sqrt{10^6}=1000$ is the shrink factor. $\delta =\frac{5x}{\sqrt{R{e}_x}}=\frac{5\times 0.5}{1000}=\frac{2.5}{1000}=2.5\times 10^{-3}\text{m}=2.5\text{mm}$ Answer: $R{e}_x=10^6$, $\delta =2.5\text{mm}$.

WHY sine? It satisfies $u=0$ at the wall and $u=U$ at $y=\delta$ smoothly (its slope flattens at the edge) — a realistic stand-in for the true profile, unlike the kinked linear one. WHY integrate the sine: ${\delta }^{\ast }$ still sums the deficit $1-u/U$; we just have a curved deficit now, so we need the antiderivative of $\sin$. ${\delta }^{\ast }={\int }_0^{\delta }\left(1-\sin \frac{\pi y}{2\delta }\right)dy=\delta -{\left[-\frac{2\delta }{\pi }\cos \frac{\pi y}{2\delta }\right]}_0^{\delta }$ WHY the $2\delta /\pi$ factor: the chain rule on $\cos (\frac{\pi y}{2\delta })$ brings out $\frac{\pi }{2\delta }$, so its reciprocal $\frac{2\delta }{\pi }$ appears when we antidifferentiate. The cosine term: at $y=\delta$, $\cos \frac{\pi }{2}=0$; at $y=0$, $\cos 0=1$. So the bracket is $-\frac{2\delta }{\pi }(0-1)=\frac{2\delta }{\pi }$. ${\delta }^{\ast }=\delta -\frac{2\delta }{\pi }=\delta \left(1-\frac{2}{\pi }\right)\approx 0.3634\delta$ Answer: ${\delta }^{\ast }=\delta (1-\frac{2}{\pi })\approx 0.363\delta$.

WHY split into two integrals: expanding $\sin (1-\sin )=\sin -{\sin }^2$ separates a piece we already know ($\int \sin$) from a new one ($\int {\sin }^2$), so each can be handled with a known result. $\theta ={\int }_0^{\delta }\sin \frac{\pi y}{2\delta }\left(1-\sin \frac{\pi y}{2\delta }\right)dy=\underset{=2\delta /\pi }{\underbrace{{\int }_0^{\delta }\sin \frac{\pi y}{2\delta }dy}}-\underset{=\delta /2}{\underbrace{{\int }_0^{\delta }{\sin }^2\frac{\pi y}{2\delta }dy}}$ WHY the identity is needed: ${\sin }^2$ has no elementary antiderivative on its own, so we rewrite it with ${\sin }^2\varphi =\frac{1-\cos 2\varphi }{2}$, whose average over this quarter-period is $\frac{1}{2}$, giving $\delta /2$. $\theta =\frac{2\delta }{\pi }-\frac{\delta }{2}=\delta \left(\frac{2}{\pi }-\frac{1}{2}\right)\approx 0.1366\delta$ $H=\frac{{\delta }^{\ast }}{\theta }=\frac{0.3634}{0.1366}\approx 2.66$ WHY compare to 2.59: $H$ is a shape fingerprint independent of $\delta$; the sine's $2.66$ landing near the exact Blasius $2.59$ is evidence the sine mimics the real profile well. Answer: $\theta \approx 0.137\delta$, $H\approx 2.66$ — very close to the true Blasius $2.59$, which is why the sine profile is a good approximation.

WHY check the ends first: a candidate profile is only physical if it starts at no-slip ($u=0$) and meets the free stream ($u=U$) — otherwise the deficit integrals are meaningless. Check ends. At $y=0$: $u/U=0$ ✓ (no-slip). At $y=\delta$: $u/U=2-1=1$ ✓. WHY the substitution $\eta =y/\delta$: it turns a $\delta$-cluttered integral into a clean number-integral from 0 to 1, and pulls one factor of $\delta$ out front (from $dy=\delta d\eta$) — so the answer must be a pure fraction times $\delta$. ${\delta }^{\ast }=\delta {\int }_0^1\left(1-2\eta +{\eta }^2\right)d\eta =\delta {\left[\eta -{\eta }^2+\frac{{\eta }^3}{3}\right]}_0^1=\delta \left(1-1+\frac{1}{3}\right)=\frac{\delta }{3}$ Answer: ${\delta }^{\ast }=\frac{\delta }{3}\approx 0.333\delta$.

WHY expand the product: $\theta$ needs $\frac{u}{U}(1-\frac{u}{U})$, and with $u/U=2\eta -{\eta }^2$ this is a polynomial — multiplying it out gives terms we can integrate one power at a time. With $\eta =y/\delta$ and $u/U=2\eta -{\eta }^2$: $(2\eta -{\eta }^2)(1-2\eta +{\eta }^2)=2\eta -5{\eta }^2+4{\eta }^3-{\eta }^4$ $\theta =\delta {\int }_0^1(2\eta -5{\eta }^2+4{\eta }^3-{\eta }^4)d\eta =\delta {\left[{\eta }^2-\frac{5{\eta }^3}{3}+{\eta }^4-\frac{{\eta }^5}{5}\right]}_0^1$ WHY collect over 15: the denominators $1,3,1,5$ share least common multiple 15, so a common denominator makes the sum exact rather than decimal-rounded. $=\delta \left(1-\frac{5}{3}+1-\frac{1}{5}\right)=\delta \cdot \frac{15-25+15-3}{15}=\frac{2\delta }{15}$ $H=\frac{{\delta }^{\ast }}{\theta }=\frac{\delta /3}{2\delta /15}=\frac{15}{6}=2.5$ WHY $H=2.5$ makes sense: the parabola is "fuller" than the linear line, so it slows less fluid — a smaller shape factor than the linear $H=3$, exactly as the opening figure predicted. Answer: $\theta =\frac{2\delta }{15}\approx 0.133\delta$, $H=2.5$.

Values: linear $H=3.00$; sine $H\approx 2.66$; parabolic $H=2.50$. True Blasius $H\approx 2.59$. WHY $H$ is the right yardstick: it strips away $\delta$ and captures only the shape of the deficit, so comparing models to Blasius' $2.59$ is a fair, size-independent test. Ranking (closest to Blasius first): sine ($2.66$) $\approx$ parabolic ($2.50$) $>$ linear ($3.00$). Worst model: the linear profile, $H=3.00$, is farthest from $2.59$. WHY: a straight line has a sharp kink at $y=\delta$ (nonzero slope meeting the flat free stream) and constant shear — real profiles curve over and flatten smoothly. Sine and parabola both flatten near $\delta$, matching reality better.

WHY reuse ${\delta }^{\ast }=\delta /2$, $\theta =\delta /6$: those ratios came purely from the shape of the linear profile, so they hold at every $x$ — we just feed in the $x$-dependent $\delta$. From Exercise 2.1–2.2: ${\delta }^{\ast }=\delta /2$, $\theta =\delta /6$. So ${\delta }^{\ast }=\frac{1}{2}\cdot \frac{5x}{\sqrt{R{e}_x}}=\frac{2.5x}{\sqrt{R{e}_x}},\theta =\frac{1}{6}\cdot \frac{5x}{\sqrt{R{e}_x}}=\frac{0.8333x}{\sqrt{R{e}_x}}$ WHY the numbers follow instantly: at $x=0.5$ we already found $\delta =2.5\text{mm}$ (Exercise 2.3), so halving and sixthing it is all that remains. At $x=0.5$: $R{e}_x=10^6$, $\delta =2.5\text{mm}$. ${\delta }^{\ast }=\frac{2.5}{2}=1.25\text{mm},\theta =\frac{2.5}{6}=0.4167\text{mm}$ Answer: ${\delta }^{\ast }=1.25\text{mm}$, $\theta \approx 0.417\text{mm}$ (confirming $\theta <{\delta }^{\ast }<\delta$).

WHY do it in $\eta$: with $\eta =y/\delta$ the profile is just ${\eta }^{1/n}$ and every integral runs 0→1, so we get formulas valid for all $n$ in one stroke — a synthesis of every earlier special case. $\frac{{\delta }^{\ast }}{\delta }={\int }_0^1\left(1-{\eta }^{1/n}\right)d\eta =1-\frac{1}{\frac{1}{n}+1}=1-\frac{n}{n+1}=\frac{1}{n+1}$ WHY the power-rule step: ${\int }_0^1{\eta }^{1/n}d\eta =\frac{1}{\frac{1}{n}+1}=\frac{n}{n+1}$; subtracting from 1 leaves $\frac{1}{n+1}$. $\frac{\theta }{\delta }={\int }_0^1{\eta }^{1/n}\left(1-{\eta }^{1/n}\right)d\eta ={\int }_0^1\left({\eta }^{1/n}-{\eta }^{2/n}\right)d\eta =\frac{n}{n+1}-\frac{n}{n+2}=\frac{n}{(n+1)(n+2)}$ WHY the extra ${\eta }^{2/n}$ term: squaring ${\eta }^{1/n}$ doubles the exponent, giving the second integral $\frac{n}{n+2}$; the difference collapses to a single fraction. Shape factor: $H=\frac{{\delta }^{\ast }/\delta }{\theta /\delta }=\frac{1/(n+1)}{n/[(n+1)(n+2)]}=\frac{n+2}{n}$.

For $n=7$: $\frac{{\delta }^{\ast }}{\delta }=\frac{1}{8}=0.125,\frac{\theta }{\delta }=\frac{7}{8\cdot 9}=\frac{7}{72}\approx 0.0972,H=\frac{9}{7}\approx 1.286$ WHY $H\approx 1.29$ is "fuller": large $n$ makes ${\eta }^{1/n}$ shoot up fast near the wall — most fluid is already near $U$, tiny deficit, so $H$ sits far below the laminar values. Answer: ${\delta }^{\ast }=\delta /(n+1)$, $\theta =\frac{n\delta }{(n+1)(n+2)}$, $H=\frac{n+2}{n}$; at $n=7$: ${\delta }^{\ast }=0.125\delta$, $\theta \approx 0.0972\delta$, $H\approx 1.29$.

WHY take the limit: pushing a parameter to an extreme is the cheapest way to test whether a formula is physically sane — if it breaks here, it is wrong. As $n\to \infty$, ${\eta }^{1/n}\to 1$ for every $\eta >0$: the fluid is at full speed $U$ everywhere except an infinitely thin sliver at the wall — a plug (slug) profile. $\frac{{\delta }^{\ast }}{\delta }=\frac{1}{n+1}\to 0,\frac{\theta }{\delta }=\frac{n}{(n+1)(n+2)}\to 0,H=\frac{n+2}{n}\to 1$ Physical sense: with almost no slowed fluid there is almost no mass or momentum deficit, so ${\delta }^{\ast }\to 0$ and $\theta \to 0$. And $H\to 1$ is the theoretical minimum shape factor — the flow is as "full" as possible, the opposite extreme from separation. ✓

WHY compare integrands, not integrals: if one integrand is $\le$ the other at every height, the integral inherits the inequality — this is the cleanest way to prove an ordering without computing either integral. Write both as integrals with the same $dy$ and let $f=\frac{u}{U}$, which satisfies $0\le f\le 1$: ${\delta }^{\ast }={\int }_0^{\infty }(1-f)dy,\theta ={\int }_0^{\infty }f(1-f)dy$ WHY the pointwise bound holds: since $0\le f\le 1$, the factor $(1-f)\ge 0$, and multiplying a non-negative number by $f\le 1$ can only shrink (or hold) it: $f(1-f)\le 1\cdot (1-f)=(1-f)$ As this holds at every height $y$, integrating preserves the inequality: $\theta =\int f(1-f)dy\le \int (1-f)dy={\delta }^{\ast }$ Hence $\theta \le {\delta }^{\ast }$, i.e. $H={\delta }^{\ast }/\theta \ge 1$ always. Combined with ${\delta }^{\ast }<\delta$ (the deficit slab is thinner than the whole layer), we recover $\theta <{\delta }^{\ast }<\delta$. $■$

Linear $H=3$ ::: parabolic $H=2.5$ ::: sine $H\approx 2.66$ ::: Blasius $H\approx 2.59$ ::: $\frac{1}{7}$-power $H\approx 1.29$ ::: and always $H\ge 1$.