Exercises — Boundary layer thickness, displacement thickness, momentum thickness

KYA: $y=0$ par, $u=0$. KYUN: Ek real fluid mein viscosity hoti hai, isliye wall se touch karne wala fluid ka layer uske upar slide nahi kar sakta — woh wall ke saath hi move karta hai. Ek stationary wall isliye $u(0)=0$ deti hai. Yahi no-slip condition hai. Answer: $u(0)=0$, no-slip condition se force hota hai.

Order: $\theta <{\delta }^{\ast }<\delta$. $\theta$ sabse chhota KYUN: iska integrand $\frac{u}{U}(1-\frac{u}{U})$ 0 aur 1 ke beech ke do fractions ka product hai, isliye yeh har jagah us single fraction $(1-\frac{u}{U})$ se chhota hai jo ${\delta }^{\ast }$ banata hai. Chhota integrand ek chhota integral deta hai.

Sirf $0\to \delta$ integrate karo. KYUN: $\delta$ ke baad fluid free-stream speed par move karta hai, $u=U$, isliye $1-\frac{u}{U}=0$ aur $\frac{u}{U}(1-\frac{u}{U})=0$ hota hai. Integrand vanish ho jaata hai, kuch contribute nahi karta — $\infty$ ek bookkeeping hai "jahan bhi deficit rehta hai" ke liye.

KYA: $\frac{u}{U}=\frac{y}{\delta }$ ko ${\delta }^{\ast }={\int }_0^{\delta }(1-\frac{u}{U})dy$ mein substitute karo. KYUN substitute karo: ${\delta }^{\ast }$ fractional velocity deficit $1-u/U$ ko height par stack karke count karta hai; use add karne ke liye humein woh deficit $y$ ke function ke roop mein chahiye, jo profile humein deti hai. ${\delta }^{\ast }={\int }_0^{\delta }\left(1-\frac{y}{\delta }\right)dy={\left[y-\frac{y^2}{2\delta }\right]}_0^{\delta }=\delta -\frac{\delta }{2}=\frac{\delta }{2}$ KYUN $\delta /2$ aata hai: deficit $1-y/\delta$ seedha 1 (wall par full deficit) se 0 (edge par koi deficit nahi) tak girta hai. Ek base $\delta$ par 1 se 0 tak seedhi line ke neeche ka area ek triangle hai — box ka bilkul aadha, isliye $\delta /2$. Answer: ${\delta }^{\ast }=\frac{\delta }{2}=0.5\delta$.

KYUN integrand badalta hai: momentum deficit velocity gap $(1-u/U)$ ko is baat se weight karta hai ki wahan actually kitna mass move kar raha hai, $u/U$ — isliye hum dono ko multiply karte hain, jo $\frac{y}{\delta }(1-\frac{y}{\delta })$ deta hai. $\theta ={\int }_0^{\delta }\frac{y}{\delta }\left(1-\frac{y}{\delta }\right)dy={\int }_0^{\delta }\left(\frac{y}{\delta }-\frac{y^2}{{\delta }^2}\right)dy=\frac{\delta }{2}-\frac{\delta }{3}=\frac{\delta }{6}$ KYUN $\theta <{\delta }^{\ast }$ yahan: extra factor $y/\delta$ ek fraction $\le 1$ hai, isliye yeh deficit ko har jagah shrink karta hai — integral chhota hi aana chahiye ($\delta /6<\delta /2$), jo Exercise 1.2 se consistent hai. $H=\frac{{\delta }^{\ast }}{\theta }=\frac{\delta /2}{\delta /6}=3$ Answer: $\theta =\frac{\delta }{6}$, $H=3$.

KYUN pehle $R{e}_x$: Blasius law $\delta =5x/\sqrt{R{e}_x}$ ko $R{e}_x$ chahiye, aur $R{e}_x$ measure karta hai ki inertia ($Ux$) viscous diffusion ($\nu$) ko kitna strongly beat karta hai — bada value matlab patli layer. $R{e}_x=\frac{Ux}{\nu }=\frac{30\times 0.5}{1.5\times 10^{-5}}=\frac{15}{1.5\times 10^{-5}}=10^6$ KYUN Blasius law mein plug karo: yeh exact laminar flat-plate result hai; $\sqrt{R{e}_x}=\sqrt{10^6}=1000$ shrink factor hai. $\delta =\frac{5x}{\sqrt{R{e}_x}}=\frac{5\times 0.5}{1000}=\frac{2.5}{1000}=2.5\times 10^{-3}\text{m}=2.5\text{mm}$ Answer: $R{e}_x=10^6$, $\delta =2.5\text{mm}$.

KYUN sine? Yeh $u=0$ wall par aur $u=U$ $y=\delta$ par smoothly satisfy karta hai (iska slope edge par flat ho jaata hai) — true profile ka ek realistic stand-in, unlike kinked linear wale ke. KYUN sine integrate karo: ${\delta }^{\ast }$ abhi bhi deficit $1-u/U$ sum karta hai; humein bas ab curved deficit mila hai, isliye humein $\sin$ ka antiderivative chahiye. ${\delta }^{\ast }={\int }_0^{\delta }\left(1-\sin \frac{\pi y}{2\delta }\right)dy=\delta -{\left[-\frac{2\delta }{\pi }\cos \frac{\pi y}{2\delta }\right]}_0^{\delta }$ KYUN $2\delta /\pi$ factor: $\cos (\frac{\pi y}{2\delta })$ par chain rule $\frac{\pi }{2\delta }$ bahar laata hai, isliye jab hum antidifferentiate karte hain toh iska reciprocal $\frac{2\delta }{\pi }$ appear hota hai. Cosine term: $y=\delta$ par, $\cos \frac{\pi }{2}=0$; $y=0$ par, $\cos 0=1$. Toh bracket hai $-\frac{2\delta }{\pi }(0-1)=\frac{2\delta }{\pi }$. ${\delta }^{\ast }=\delta -\frac{2\delta }{\pi }=\delta \left(1-\frac{2}{\pi }\right)\approx 0.3634\delta$ Answer: ${\delta }^{\ast }=\delta (1-\frac{2}{\pi })\approx 0.363\delta$.

KYUN do integrals mein split karo: $\sin (1-\sin )=\sin -{\sin }^2$ expand karna ek aisa piece alag karta hai jo humein pehle se pata hai ($\int \sin$) aur ek naya ($\int {\sin }^2$), toh har ek ko ek jaane-maane result se handle kiya ja sakta hai. $\theta ={\int }_0^{\delta }\sin \frac{\pi y}{2\delta }\left(1-\sin \frac{\pi y}{2\delta }\right)dy=\underset{=2\delta /\pi }{\underbrace{{\int }_0^{\delta }\sin \frac{\pi y}{2\delta }dy}}-\underset{=\delta /2}{\underbrace{{\int }_0^{\delta }{\sin }^2\frac{\pi y}{2\delta }dy}}$ KYUN identity ki zarurat hai: ${\sin }^2$ ka apna koi elementary antiderivative nahi hai, isliye hum ise ${\sin }^2\varphi =\frac{1-\cos 2\varphi }{2}$ se rewrite karte hain, jiska is quarter-period par average $\frac{1}{2}$ hai, jo $\delta /2$ deta hai. $\theta =\frac{2\delta }{\pi }-\frac{\delta }{2}=\delta \left(\frac{2}{\pi }-\frac{1}{2}\right)\approx 0.1366\delta$ $H=\frac{{\delta }^{\ast }}{\theta }=\frac{0.3634}{0.1366}\approx 2.66$ KYUN 2.59 se compare karo: $H$ ek shape fingerprint hai jo $\delta$ se independent hai; sine ka $2.66$ exact Blasius $2.59$ ke paas land karna iska evidence hai ki sine real profile ko achhi tarah mimic karta hai. Answer: $\theta \approx 0.137\delta$, $H\approx 2.66$ — true Blasius $2.59$ ke bahut kareeb, isliye sine profile ek achha approximation hai.

KYUN pehle ends check karo: ek candidate profile tab hi physical hai jab woh no-slip se start kare ($u=0$) aur free stream se mile ($u=U$) — warna deficit integrals meaningless hain. Ends check karo. $y=0$ par: $u/U=0$ ✓ (no-slip). $y=\delta$ par: $u/U=2-1=1$ ✓. KYUN substitution $\eta =y/\delta$: yeh ek $\delta$-cluttered integral ko 0 se 1 tak ek clean number-integral mein badal deta hai, aur $\delta$ ka ek factor front mein bahar nikaalta hai ($dy=\delta d\eta$ se) — isliye answer ek pure fraction times $\delta$ hona chahiye. ${\delta }^{\ast }=\delta {\int }_0^1\left(1-2\eta +{\eta }^2\right)d\eta =\delta {\left[\eta -{\eta }^2+\frac{{\eta }^3}{3}\right]}_0^1=\delta \left(1-1+\frac{1}{3}\right)=\frac{\delta }{3}$ Answer: ${\delta }^{\ast }=\frac{\delta }{3}\approx 0.333\delta$.

KYUN product expand karo: $\theta$ ko $\frac{u}{U}(1-\frac{u}{U})$ chahiye, aur $u/U=2\eta -{\eta }^2$ ke saath yeh ek polynomial hai — ise multiply out karne se aisi terms milti hain jo hum ek power at a time integrate kar sakte hain. $\eta =y/\delta$ aur $u/U=2\eta -{\eta }^2$ ke saath: $(2\eta -{\eta }^2)(1-2\eta +{\eta }^2)=2\eta -5{\eta }^2+4{\eta }^3-{\eta }^4$ $\theta =\delta {\int }_0^1(2\eta -5{\eta }^2+4{\eta }^3-{\eta }^4)d\eta =\delta {\left[{\eta }^2-\frac{5{\eta }^3}{3}+{\eta }^4-\frac{{\eta }^5}{5}\right]}_0^1$ KYUN 15 par collect karo: denominators $1,3,1,5$ ka least common multiple 15 hai, isliye common denominator sum ko exact banata hai decimal-rounded ki jagah. $=\delta \left(1-\frac{5}{3}+1-\frac{1}{5}\right)=\delta \cdot \frac{15-25+15-3}{15}=\frac{2\delta }{15}$ $H=\frac{{\delta }^{\ast }}{\theta }=\frac{\delta /3}{2\delta /15}=\frac{15}{6}=2.5$ KYUN $H=2.5$ sense banata hai: parabola linear line se "fuller" hai, isliye woh kam fluid slow karti hai — linear $H=3$ se chhota shape factor, bilkul waise jaise opening figure ne predict kiya tha. 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 $H\approx 2.59$. KYUN $H$ sahi yardstick hai: yeh $\delta$ ko strip away karta hai aur sirf deficit ki shape capture karta hai, isliye models ko Blasius' $2.59$ se compare karna ek fair, size-independent test hai. Ranking (Blasius ke sabse kareeb pehle): sine ($2.66$) $\approx$ parabolic ($2.50$) $>$ linear ($3.00$). Sabse bekar model: linear profile, $H=3.00$, $2.59$ se sabse door hai. KYUN: ek seedhi line $y=\delta$ par ek sharp kink rakhti hai (nonzero slope flat free stream se milti hui) aur constant shear — real profiles curve over hoti hain aur smoothly flatten karti hain. Sine aur parabola dono $\delta$ ke paas flatten hote hain, jo reality se zyada match karta hai.

KYUN ${\delta }^{\ast }=\delta /2$, $\theta =\delta /6$ reuse karo: woh ratios purely linear profile ki shape se aayi theen, isliye woh har $x$ par hold karti hain — hum bas $x$-dependent $\delta$ feed karte hain. Exercise 2.1–2.2 se: ${\delta }^{\ast }=\delta /2$, $\theta =\delta /6$. Toh ${\delta }^{\ast }=\frac{1}{2}\cdot \frac{5x}{\sqrt{R{e}_x}}=\frac{2.5x}{\sqrt{R{e}_x}},\theta =\frac{1}{6}\cdot \frac{5x}{\sqrt{R{e}_x}}=\frac{0.8333x}{\sqrt{R{e}_x}}$ KYUN numbers instantly follow karte hain: $x=0.5$ par humne pehle hi $\delta =2.5\text{mm}$ find kar liya tha (Exercise 2.3), isliye sirf use half aur sixth karna baaki hai. $x=0.5$ par: $R{e}_x=10^6$, $\delta =2.5\text{mm}$. ${\delta }^{\ast }=\frac{2.5}{2}=1.25\text{mm},\theta =\frac{2.5}{6}=0.4167\text{mm}$ Answer: ${\delta }^{\ast }=1.25\text{mm}$, $\theta \approx 0.417\text{mm}$ ($\theta <{\delta }^{\ast }<\delta$ confirm hota hai).

KYUN $\eta$ mein karo: $\eta =y/\delta$ ke saath profile bas ${\eta }^{1/n}$ hai aur har integral 0→1 run karta hai, isliye hum ek hi stroke mein sab $n$ ke liye valid formulas paate hain — har pehle special case ki ek synthesis. $\frac{{\delta }^{\ast }}{\delta }={\int }_0^1\left(1-{\eta }^{1/n}\right)d\eta =1-\frac{1}{\frac{1}{n}+1}=1-\frac{n}{n+1}=\frac{1}{n+1}$ KYUN power-rule step: ${\int }_0^1{\eta }^{1/n}d\eta =\frac{1}{\frac{1}{n}+1}=\frac{n}{n+1}$; 1 se subtract karne par $\frac{1}{n+1}$ bachta hai. $\frac{\theta }{\delta }={\int }_0^1{\eta }^{1/n}\left(1-{\eta }^{1/n}\right)d\eta ={\int }_0^1\left({\eta }^{1/n}-{\eta }^{2/n}\right)d\eta =\frac{n}{n+1}-\frac{n}{n+2}=\frac{n}{(n+1)(n+2)}$ KYUN extra ${\eta }^{2/n}$ term: ${\eta }^{1/n}$ ko square karne se exponent double hota hai, jo second integral $\frac{n}{n+2}$ deta hai; difference ek single fraction mein collapse ho jaata hai. Shape factor: $H=\frac{{\delta }^{\ast }/\delta }{\theta /\delta }=\frac{1/(n+1)}{n/[(n+1)(n+2)]}=\frac{n+2}{n}$.

$n=7$ ke liye: $\frac{{\delta }^{\ast }}{\delta }=\frac{1}{8}=0.125,\frac{\theta }{\delta }=\frac{7}{8\cdot 9}=\frac{7}{72}\approx 0.0972,H=\frac{9}{7}\approx 1.286$ KYUN $H\approx 1.29$ "fuller" hai: bada $n$ ${\eta }^{1/n}$ ko wall ke paas fast upar le jaata hai — zyada fluid already $U$ ke paas hai, tiny deficit, isliye $H$ laminar values se bahut neeche baithta hai. Answer: ${\delta }^{\ast }=\delta /(n+1)$, $\theta =\frac{n\delta }{(n+1)(n+2)}$, $H=\frac{n+2}{n}$; $n=7$ par: ${\delta }^{\ast }=0.125\delta$, $\theta \approx 0.0972\delta$, $H\approx 1.29$.

KYUN limit lo: ek parameter ko extreme par push karna ek formula ko physically sane hai ya nahi test karne ka sabse sasta tarika hai — agar yah yahan toota, toh galat hai. Jaise $n\to \infty$, har $\eta >0$ ke liye ${\eta }^{1/n}\to 1$: fluid wall par ek infinitely thin sliver ko chhodkar har jagah full speed $U$ par hai — yeh ek plug (slug) profile hai. $\frac{{\delta }^{\ast }}{\delta }=\frac{1}{n+1}\to 0,\frac{\theta }{\delta }=\frac{n}{(n+1)(n+2)}\to 0,H=\frac{n+2}{n}\to 1$ Physical sense: jab almost koi slowed fluid nahi hai toh almost koi mass ya momentum deficit nahi hai, isliye ${\delta }^{\ast }\to 0$ aur $\theta \to 0$. Aur $H\to 1$ theoretical minimum shape factor hai — flow jitna ho sake utna "full" hai, separation se bilkul ulta extreme. ✓

KYUN integrands compare karo, integrals nahi: agar ek integrand har height par $\le$ doosre ka hai, toh integral woh inequality inherit karta hai — yeh kisi bhi integral compute kiye bina ordering prove karne ka sabse clean tarika hai. Dono ko same $dy$ ke saath integrals likhein aur $f=\frac{u}{U}$ lo, jo $0\le f\le 1$ satisfy karta hai: ${\delta }^{\ast }={\int }_0^{\infty }(1-f)dy,\theta ={\int }_0^{\infty }f(1-f)dy$ KYUN pointwise bound hold karta hai: kyunki $0\le f\le 1$, factor $(1-f)\ge 0$ hai, aur ek non-negative number ko $f\le 1$ se multiply karna use sirf shrink kar sakta hai (ya hold): $f(1-f)\le 1\cdot (1-f)=(1-f)$ Kyunki yeh har height $y$ par hold karta hai, integrate karna inequality preserve karta hai: $\theta =\int f(1-f)dy\le \int (1-f)dy={\delta }^{\ast }$ Isliye $\theta \le {\delta }^{\ast }$, yaani $H={\delta }^{\ast }/\theta \ge 1$ hamesha. ${\delta }^{\ast }<\delta$ ke saath combine karke (deficit slab poori layer se pateeli hai), hum $\theta <{\delta }^{\ast }<\delta$ recover karte hain. $■$

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