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

2.2.21 · D2 · Physics › Fluid Mechanics › Boundary layer thickness, displacement thickness, momentum t

Hum yeh words poore walkthrough mein use karenge, isliye inhe ek baar, saaf language mein, kisi bhi symbol se pehle, pin kar lete hain:

• Wall — ek flat solid surface jiske upar fluid flow karta hai. Hum height seedha upar se measure karte hain.
• Free stream — wall se door ka fast fluid jo abhi tak wall ko "notice" nahi kar paya.
• Slow layer — wall ke paas ka fluid, jo surface se chipak jaata hai isliye peechhe kheencha jaata hai (no-slip condition).

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Step 1 — Picture set up karo: height, speed, aur do symbols $y$ aur $u$

KYA. Wall ko ek horizontal floor ki tarah draw karo. Uske upar bahut saari thin horizontal fluid layers stack karo. Har layer kisi height $y$ par baithe hai aur apni khud ki speed $u(y)$ se aage slide karti hai — downstream point karte hue arrow se draw ki gayi. Chhota arrow = slow, bada arrow = fast.

KYU. Boundary layers ki poori baat yeh hai ki "speed kaise change hoti hai jab tum upar jaate ho?" Toh naam dene layak do cheezein exactly height ($y$) aur wahan speed ($u$) hain. Abhi aur kuch nahi chahiye.

PICTURE. Sabse neeche ka arrow zero length ka hai (wall se chipka hua). Jaise tum upar chadhte ho, arrows bade hote jaate hain, jab tak upar ke paas sab ek hi poori length tak nahi pahunch jaate — free-stream speed, jise hum aage naam denge.

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Step 2 — Free-stream speed $U$ aur edge $\delta$ ko naam do

KYA. Upar free-stream arrow length $U$ mark karo. Woh height mark karo jahan arrows almost us poori length tak pahunch chuke hain; us height ko $\delta$ kaho.

$99\%$ kyun, $100\%$ kyun nahi? Speed sirf $U$ ke paas pahunchti hai; mathematically finite height par kabhi equal nahi hoti (yeh asymptotic hai — flatline hoti hai par kabhi bilkul touch nahi karti). Ek real edge draw karne ke liye ek cutoff chahiye, aur $99\%$ agreed convention hai. Yeh number kahan se aati hai, yeh dekhne ke liye Blasius dekho.

PICTURE. Height $\delta$ par ek vertical dashed line "slow, changing" region (neeche) aur "everyone at $U$" region (upar) ko alag karti hai.

Ek key fact note karo jis par hum baar baar rely karenge:

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Step 3 — "Velocity deficit": har layer kitni slow hai

KYA. Har layer ke liye, uske chhote arrow aur poore-length $U$ arrow ke beech ka gap draw karo. Woh red gap deficit $U-u$ hai.

YEH QUANTITY KYU. Boundary layer ka poora matlab yeh hai ki fluid missing speed hai. "Har layer mein kitni speed missing hai?" woh deficit hai. Un gaps ko wall ke upar stack karna wahi kaam hai jo Steps 5 aur 6 ke integrals karte hain.

PICTURE. Har layer ke bagal mein red gap-bars: neeche bade (bada deficit), $\delta$ par zero tak shrink hote hue.

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Step 4 — Ek thin layer mass flow ka ek thin slice carry karti hai

Sab kuch add karne se pehle, ek layer ko samjho jiska thickness $dy$ hai.

$\rho udy$ KYU? Ek second mein, is layer ka fluid $u$ distance aage slide karta hai. Woh height $dy$, length $u$ ka ek rectangle sweep karta hai, toh area $udy$; times density $\rho$ se woh mass milta hai jo cross kiya. Speed × height × density = mass flow rate.

PICTURE. Ek akela highlighted layer; ek horizontal arrow jisi length $u$ dikhata hai ki ek second mein fluid kitni door jaata hai; swept rectangle $udy$ shaded hai.

Ab hum sab layers ko do alag tarike se add kar sakte hain: mass ke liye (Step 5) aur momentum ke liye (Step 6).

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Step 5 — Mass deficit stack karo ⇒ displacement thickness ${\delta }^{\ast }$

KYA — deficit integral seedha banao. Do poore-column mass flows ko compare karne ki jagah (jिनमें से har ek, infinity tak integrate karne par, infinite hoga aur isliye term-by-term subtract nahi ho sakta), hum layer by layer kaam karte hain aur sirf difference add karte hain, jo finite hai. Height $y$ par ek layer ke liye:

• agar woh poore $U$ par chalti to mass flow $\rho Udy$ carry karti;
• woh actually $\rho udy$ carry karti hai (Step 4);
• toh uski missing mass flow $\rho (U-u)dy$ hai — aur yeh exactly $\delta$ ke upar zero hai (Step 2 ka fact).

Woh finite, well-behaved shortfall sab layers par sum karne par missing mass flow rate milta hai:

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

Yeh single integral converge karta hai kyunki $\delta$ ke baad integrand zero ho jaata hai — koi $\infty -\infty$ kabhi nahi aata.

$\rho U$ SE DIVIDE KYU. Ab hum poochhte hain: us missing mass flow ko ek saaf free-stream fluid slab se replace karo — density $\rho$, speed $U$, height ${\delta }^{\ast }$. Uska mass flow rate $\rho U{\delta }^{\ast }$ hai. Dono ko equal karo:

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

Dono sides ko $\rho U$ se divide karo (density aur $U$ constants hain, bahar aa jaate hain):

PICTURE. Red deficit region (sab gap-bars ka area) ek single solid rectangle mein "collapse" ho jaata hai jिसकी width $U$ aur height ${\delta }^{\ast }$ hai — same area, ek clean slab.

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Step 6 — Momentum deficit stack karo ⇒ momentum thickness $\theta$

KYA — momentum mein speed do baar kyun hoti hai. Momentum = mass × velocity. Ek layer ke liye:

• actually flowing mass $\rho udy$ hai (Step 4) — note karo yeh real speed $u$ use karta hai,
• aur us mass ka har bit speed mein $(U-u)$ se short hai.

Toh ek layer ka momentum shortfall mass-flow × speed-gap hai:

$\rho \underset{\begin{array}{c}\text{kitna mass}\\ \text{move karta hai}\end{array}}{\underbrace{u}}\underset{\begin{array}{c}\text{kitni speed}\\ \text{usse chahiye}\end{array}}{\underbrace{(U-u)}}dy$

Sab layers par sum karo (phir finite — integrand $\delta$ ke upar zero ho jaata hai):

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

$\rho U^2$ SE DIVIDE KYU. Isse ek clean free-stream fluid slab se match karo jo momentum flow $\rho U^2\theta$ carry karta hai — woh momentum hai jo per second per unit span cross karta hai (mass flow $\rho U$ times speed $U$, height $\theta$ ke upar), units $\text{kg}\cdot {\text{m/s}}^2$ per metre of span (force per unit span). Dono equal karo:

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

$\rho U^2$ se divide karo:

PICTURE. Wall ke upar do curves — mass-weight $u/U$ upar jaata hua, deficit $1-u/U$ neecha aata hua — aur unka product beech mein bulge karta hua. Woh bulge, collapse hoke, height $\theta$ ban jaata hai.

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Step 7 — Order hamesha $\theta <{\delta }^{\ast }<\delta$ kyun hota hai

KYA. Teeno heights ko ek hi picture par line up karo.

YEH ORDERING FORCED KYU HAI.

• ${\delta }^{\ast }$ integrate karta hai $(1-\frac{u}{U})$ ko.
• $\theta$ integrate karta hai usi cheez ko lekin $\frac{u}{U}$ se multiply karke, jo 0 aur 1 ke beech ek number hai. Ek area ko $\le 1$ cheez se multiply karna use sirf shrink kar sakta hai — toh $\theta \le {\delta }^{\ast }$.
• Aur ${\delta }^{\ast }$ sirf poori height $\delta$ ka ek hissa hai (deficit kabhi poori layer nahi bharti), toh ${\delta }^{\ast }<\delta$.

Isliye, kisi bhi sensible profile ke liye: $\theta <{\delta }^{\ast }<\delta .$

PICTURE. Ek wall par teen stacked brackets: tall $\delta$, medium ${\delta }^{\ast }$, short $\theta$.

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Step 8 — Edge & degenerate cases (kabhi surprise mat ho)

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Step 9 — Ek real profile (linear) plug karo aur answers padho

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Ek-picture summary

Ek hi wall par sab kuch: velocity arrows $0$ se $U$ tak badhte hue; red deficit ${\delta }^{\ast }$ slab mein collapse hota hua; beech mein bulge karta product $\theta$ slab mein collapse hota hua; $99\%$ edge $\delta$ mark karta hua — aur brackets dikhate hue $\theta <{\delta }^{\ast }<\delta$.

Recall Feynman: poora walkthrough saadhe alfazon mein

Paani wall se chipakta hai, isliye wall ke paas uska ek thin sheet slow chalti hai. Har slow layer ko khada karo aur uske speed-arrow ko poore free-stream arrow $U$ se compare karo; gap yeh hai ki woh layer kitni speed miss kar rahi hai. Wall ke upar sab gaps add karo aur tumne missing mass flow count kar li — ise poori-speed fluid ke ek single clean slab mein squash karo aur uski height ${\delta }^{\ast }$ hai, woh distance jितna wall bahari flow ko bahar dhakelta hai slow fluid ke liye jagah banana ke liye. Ab har gap ko weight karo ki wahan actually kitna fluid move ho raha hai ($u/U$) — woh missing momentum deta hai, jo drag hai; us ko ek slab mein squash karo aur uski height $\theta$ hai. Kyunki momentum sum ek extra factor use karta hai jo ek se chhhota hai, $\theta$ hamesha sabse chhota hota hai, phir ${\delta }^{\ast }$, phir geometric edge $\delta$. Unka ratio ${\delta }^{\ast }/\theta$ ek warning gauge hai: jaise jaise yeh badhta hai, near-wall flow wall se peel off hone ki taraf weak hoti jaati hai.

Recall

Teeno thicknesses ka order kya hai? ::: $\theta <{\delta }^{\ast }<\delta$. $\theta$ mein extra $u/U$ factor kyun hota hai jo ${\delta }^{\ast }$ mein nahi hota? ::: Momentum lost $=\rho u(U-u)$: tumhe moving mass ($\propto u$) aur speed gap $(U-u)$ chahiye; mass loss ke liye sirf gap chahiye. Linear profile ke liye ${\delta }^{\ast },\theta ,H$ batao. ::: ${\delta }^{\ast }=\delta /2$, $\theta =\delta /6$, $H=3$. Tum sirf $\delta$ tak integrate kyun kar sakte ho? ::: $\delta$ ke upar, $u=U$ hai toh dono integrands zero hain.