Visual walkthrough — Blasius solution — exact laminar boundary layer solution

2.2.22 · D2 · Physics › Fluid Mechanics › Blasius solution — exact laminar boundary layer solution

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Step 1 — Woh picture jo hum explain karne ki koshish kar rahe hain

KYA. Ek flat, patli plate apni edge par hawa mein rakkhi hai. Door se saari hawa ek steady speed par chalti hai jise hum $U_{\infty }$ ("U-infinity" — free-stream speed, woh speed jo hawa ki hoti hai jab plate kahin paas nahi hoti) kehte hain. Surface par, hawa ke molecules plate se chipak jaate hain aur hil nahi sakte — yeh no-slip rule hai.

KYUN. "Wall par atak gaye" aur "$U_{\infty }$ par saans leke daud rahe hain" ke beech, hawa ki speed smoothly badhni chahiye. Woh patli badhti hui region hi boundary layer hai. Hamara poora kaam hai iska shape dhundhna (jaise jaise hum wall se door jaate hain speed kaise badhti hai) aur iska thickness pata karna.

PICTURE. Neeche: free stream (upar ke saare arrows barabar length ke), wall (neeche, arrows zero), aur woh layer jahan arrows kuch nahi se puri length tak badhte hain.

Do coordinates har jagah use hote hain:

• $x$ ::: plate ke saath saath distance, iske leading (aage ke) edge se naapa hua.
• $y$ ::: plate se door, seedha hawa mein.

$x$-direction mein hawa ki speed $u(x,y)$ likhi jaati hai — yeh depend karti hai kahan khade ho tum. Bahut choti sideways speed ($y$-direction mein) $v(x,y)$ hai. $v$ se Step 5 mein theek se milenge.

Fluid ki density ka naam bhi abhi fix karte hain, kyunki yeh drag stage par wapas aata hai:

• $\rho$ (Greek "rho") ::: fluid ki density — uska mass per unit volume, ${\text{kg/m}}^3$ mein (hawa ke liye, lagbhag $1.2{\text{kg/m}}^3$).

Is picture ki physical origin ke liye Prandtl Boundary Layer Theory dekho.

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Step 2 — Do laws jinhe hawa ko follow karna hai

KYA. Do physical statements layer ke andar flow ko control karti hain.

Law A — mass conserved hai (continuity): $\underset{\text{how }u\text{ changes along plate}}{\underbrace{\frac{\partial u}{\partial x}}}+\underset{\text{how }v\text{ changes away from wall}}{\underbrace{\frac{\partial v}{\partial y}}}=0$

Symbol $\partial u/\partial x$ (padho "partial-u-partial-x") ka matlab sirf itna hai: "$u$ kitni tezi se change hoti hai jab $x$ mein ek chota sa step lete ho, $y$ ko fixed rakh ke." Equation yeh kehti hai: agar along-flow $u$ ek direction mein speed up ho, toh sideways flow $v$ ko adjust karna hoga taaki koi fluid create ya destroy na ho.

Law B — fluid blob ke liye force = mass × acceleration (Navier–Stokes $x$-momentum equation): $\underset{\text{inertia: the blob’s acceleration}}{\underbrace{u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}}}=\underset{\text{viscous friction across the layer}}{\underbrace{\nu \frac{{\partial }^{2}u}{\partial y^2}}}$

KYUN. Left side inertia hai — ek moving blob apna momentum carry kar raha hai. Right side viscosity hai — hawa ki neighbouring layers ek doosre ko rub aur drag kar rahi hain. Yahan $\nu$ (Greek "nu") kinematic viscosity hai: fluid kitna shearing resist karta hai, ${\text{m}}^2/\text{s}$ mein measure hota hai. Double symbol ${\partial }^{2}u/\partial y^2$ speed profile ki curvature hai — velocity curve kitni bent hai.

PICTURE. Ek akela fluid blob jisme inertia arrows (use aage carry karte hain) viscous arrows (neeche ki slower layer se chipchipa drag) se lad rahe hain.

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Step 3 — Scaling guess: layer kitni thick honi chahiye?

KYA. Kuch bhi solve karne se pehle, hum thickness $\delta$ (Greek "delta") estimate karte hain sirf size mein Law B ke dono sides ko balance karke.

KYUN. Layer mein, inertia aur viscosity comparable hote hain (yahi layer ki definition hai). Toh unke rough sizes ko barabar set karo: $\underset{\text{inertia size}}{\underbrace{\frac{U_{\infty }^2}{x}}}\sim \underset{\text{viscous size}}{\underbrace{\frac{\nu U_{\infty }}{{\delta }^2}}}$ Yahan $U_{\infty }^2/x$ "speed² per length travelled" hai aur $\nu U_{\infty }/{\delta }^2$ "viscosity × speed ÷ thickness²" hai. $\delta$ ke liye solve karo: $\boxed{\delta \sim \sqrt{\frac{\nu x}{U_{\infty }}}}$

PICTURE. Layer edge $\sqrt{x}$ ki tarah badhte hue ek curve traced hai — pehle steep, phir flatten hoti hai. Ek (galat) straight line se contrast karo.

Yeh akela estimate poore solution ka seed hai: yeh humein batata hai layer ke andar height measure karne ka natural yardstick. Reynolds Number se connection notice karo: $\delta /x\sim 1/\sqrt{U_{\infty }x/\nu }$, aur $U_{\infty }x/\nu$ exactly $R{e}_x$ hai.

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Step 4 — Ruler ko stretch karo: similarity variable $\eta$

KYA. Raw height $y$ ki jagah, Step 3 ki natural thickness ke units mein height measure karo. Is stretched height ko $\eta$ (Greek "eta") kaho: $\boxed{\eta =y\sqrt{\frac{U_{\infty }}{\nu x}}}$

• $y$ ::: wall se raw distance.
• $\sqrt{U_{\infty }/(\nu x)}$ ::: stretch factor — leading edge ke paas bada (patli layer), downstream door (moti layer).

KYUN. Step 3 mein humne dekha ki layer $\sqrt{x}$ ki tarah thick hoti hai. Agar hum height ko usi $\sqrt{x}$ se divide karein, toh har $x$ par ke profiles line up ho jaate hain. Yeh ek similarity solution hai: ek universal curve, koi bhi dekho plate par kahin bhi. (Dekho Similarity Solutions in PDEs.)

PICTURE. Left panel: teen values of $x$ par raw profiles $u(y)$ — alag alag widths, ek fan. Right panel: wohi data $\eta$ ke against plot kiya — sab ek single curve par collapse ho jaate hain.

Law A aur Law B dono ko ek saath handle karne ke liye, hum stream function $\psi$ (Greek "psi") introduce karte hain. Definition se $u=\partial \psi /\partial y$ aur $v=-\partial \psi /\partial x$; is tarah velocities likhne se Law A (continuity) automatically true ho jaata hai — dekho Stream Function and Vorticity. Hum ise likhte hain $\psi =\underset{\text{amplitude}}{\underbrace{\sqrt{\nu x{U}_{\infty }}}}f(\eta )$

jahan $f(\eta )$ ek dimensionless unknown function hai — woh shape jo humein dhundhni hai. Sab kuch ab is ek function $f$ par depend karta hai.

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Step 5 — $f$ se actual velocities padhna

KYA. Stream function ko actual velocities mein convert karo.

Along-flow speed ke liye: $u=\frac{\partial \psi }{\partial y}=\sqrt{\nu x{U}_{\infty }}{f}^{\prime }(\eta )\cdot \underset{=\sqrt{U_{\infty }/(\nu x)}}{\underbrace{\frac{\partial \eta }{\partial y}}}=U_{\infty }{f}^{\prime }(\eta )$ toh $\boxed{u=U_{\infty }{f}^{\prime }(\eta )}$

KYUN yeh beautiful hai. Prime ka matlab derivative hai: $f^{\prime }(\eta )=\mathrm{d}f/\mathrm{d}\eta$. Toh $f^{\prime }$ literally velocity ratio $u/U_{\infty }$ hai — ek number $0$ (wall par) aur $1$ (free stream mein) ke beech. Boundary layer ka shape hai $f^{\prime }$ ka graph.

Sideways speed, step by step. Definition se $v=-\partial \psi /\partial x$. Tricky part: $\psi =\sqrt{\nu x{U}_{\infty }}f(\eta )$ $x$ par do jagah depend karta hai — amplitude $\sqrt{\nu x{U}_{\infty }}$ aur $\eta$ ke through (kyunki $\eta =y\sqrt{U_{\infty }/(\nu x)}$ mein khud $x$ hai). Isliye hume product rule chahiye:

= -\underbrace{\frac{\partial\sqrt{\nu x U_\infty}}{\partial x}}_{\text{amplitude changes}}f \; -\; \sqrt{\nu x U_\infty}\,f'\underbrace{\frac{\partial\eta}{\partial x}}_{\eta\text{ changes}}$$ Ab do pieces: - $\dfrac{\partial}{\partial x}\sqrt{\nu x U_\infty} = \dfrac12\sqrt{\dfrac{\nu U_\infty}{x}}$ (kyunki $\sqrt{x}$ differentiate hokar $\tfrac{1}{2\sqrt{x}}$ banta hai — **yahan se $\tfrac12$ paida hota hai**). - $\dfrac{\partial\eta}{\partial x} = y\sqrt{\dfrac{U_\infty}{\nu}}\cdot\big(-\tfrac12\big)x^{-3/2} = -\dfrac{\eta}{2x}$ (phir denominator mein $\sqrt{x}$ ne $-\tfrac12$ diya). Dono substitute karo aur simplify karo (doosra term $+\tfrac12\sqrt{\nu U_\infty/x}\,\eta f'$ ban jaata hai): $$\boxed{\;v = \frac{1}{2}\sqrt{\frac{\nu U_\infty}{x}}\,\big(\eta f' - f\big)\;}$$ - Bracket $(\eta f' - f)$ generally **zero nahi** hota — toh layer ke andar $v \neq 0$! Jaise jaise layer thick hoti hai, mass conservation *force* karta hai ek gentle upward drift. **PICTURE.** Layer mein bade horizontal $u$-arrows ($f'$ se) aur chote upward $v$-arrows (bracket se), dikhate hue fluid gently bahar dhaka ja raha hai jaise layer grow karti hai. ![[deepdives/dd-physics-2.2.22-d2-s05.png]] --- ## Step 6 — Miracle: sab kuch ek ODE mein collapse ho jaata hai **KYA.** $u = U_\infty f'$ aur $v = \tfrac12\sqrt{\nu U_\infty/x}(\eta f' - f)$, unke $x$- aur $y$-derivatives ke saath, Law B (momentum equation) mein substitute karo. **Inertia side build karo, term by term.** Humein char derivatives chahiye. $\partial\eta/\partial x = -\eta/(2x)$ aur $\partial\eta/\partial y = \sqrt{U_\infty/(\nu x)}$ use karke: $$\frac{\partial u}{\partial x} = U_\infty f''\frac{\partial\eta}{\partial x} = -\frac{U_\infty}{2x}\,\eta f'',\qquad \frac{\partial u}{\partial y} = U_\infty f''\sqrt{\frac{U_\infty}{\nu x}}$$ Ab do inertia pieces banao: $$u\frac{\partial u}{\partial x} = U_\infty f'\cdot\Big(-\frac{U_\infty}{2x}\eta f''\Big) = -\frac{U_\infty^2}{2x}\,\eta f' f''$$ $$v\frac{\partial u}{\partial y} = \tfrac12\sqrt{\tfrac{\nu U_\infty}{x}}(\eta f'-f)\cdot U_\infty f''\sqrt{\tfrac{U_\infty}{\nu x}} = \frac{U_\infty^2}{2x}f''(\eta f'-f)$$ **Inhe add karo** — dekho $\eta f' f''$ terms exactly cancel ho jaate hain: $$u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y} = \frac{U_\infty^2}{2x}\big(-\eta f' f'' + \eta f' f'' - f f''\big) = -\frac{U_\infty^2}{2x}\,f f''$$ Woh surviving $-\tfrac12 f f''$ **exactly wahin se aata hai jahan Blasius equation mein $\tfrac12$ coefficient hai** — yeh woh leftover hai jo $\eta f' f''$ pieces ke annihilate hone ke baad bachta hai. **Viscous side:** $$\nu\frac{\partial^2 u}{\partial y^2} = \nu U_\infty f'''\Big(\frac{U_\infty}{\nu x}\Big) = \frac{U_\infty^2}{x}f'''$$ **Inertia = viscous set karo** aur $x/U_\infty^2$ se multiply karo: $$-\tfrac12 f f'' = f''' \;\Longrightarrow\; f''' + \tfrac12 f f'' = 0$$ **KYUN yeh kaam karta hai.** $x$ ki, $U_\infty$ ki, aur $\nu$ ki har appearance **cancel** ho jaati hai — kyunki amplitude $\sqrt{\nu x U_\infty}$ (Step 4) *choose* kiya gaya tha taaki common factor $U_\infty^2/x$ **dono** sides par aaye. Jo bachta hai woh purely $f$ aur $\eta$ mein equation hai: > [!formula] Blasius equation > $$\boxed{\;f''' + \tfrac{1}{2}\,f\,f'' = 0\;}$$ > - $f'''$ ::: third derivative — viscous term $\nu\,\partial^2 u/\partial y^2$ se aata hai. > - $\tfrac12 f f''$ ::: inertia terms ka leftover jab $\eta f' f''$ cancel hota hai. > > **Boundary conditions** (woh physics jo $f$ ko pin down karti hai): > $$f(0)=0,\qquad f'(0)=0,\qquad f'(\infty)=1$$ > - $f(0)=0$ ::: koi fluid wall se *guzarta* nahi ($v=0$ at $y=0$). > - $f'(0)=0$ ::: no-slip — wall par speed zero hai. > - $f'(\infty)=1$ ::: bahut door, $u \to U_\infty$, toh $u/U_\infty \to 1$. **PICTURE.** Ek "funnel" diagram: do messy PDEs (Laws A + B) plus similarity transform upar se daalo aur ek clean ODE neeche nikalta hai. ![[deepdives/dd-physics-2.2.22-d2-s06.png]] > [!mistake] $f$ ke liye ek neat formula expect karna > **Kyun sahi lagta hai:** zyaadatar physics problems ek neat expression mein khatam hoti hain. > **Fix:** $f'''+\tfrac12 f f''=0$ **nonlinear** hai aur **koi closed form nahi** hai. Ise numerically solve kiya jaata hai (*shooting method*: $f''(0)$ guess karo, integrate karo, adjust karo jab tak $f'(\infty)=1$ na mile). Magic guess jo kaam karta hai woh hai $f''(0)=0.332$. --- ## Step 7 — Universal shape aur uska ek magic number **KYA.** Numerically integrate karne par curve $f'(\eta)$ milta hai — velocity profile — aur crucial slope wall par, $f''(0)=0.332$. **KYUN $f''(0)$ matter karta hai.** Wall par velocity curve ki slope control karti hai ki hawa plate ko kitna drag karti hai (Step 8). Aur profile $f'=0.99$ tak $\eta \approx 5.0$ par pahunchti hai — wahin hum kehte hain layer "khatam" ho gayi. **PICTURE.** Blasius curve: $\eta$ side par, $f'=u/U_\infty$ across. $f'(0)=0$ (no-slip) mark karo, initial slope $0.332$, aur woh point $\eta=5.0$ jahan $f'=0.99$ hai. ![[deepdives/dd-physics-2.2.22-d2-s07.png]] Layer edge par $\eta \approx 5.0$ se: $$\boxed{\;\delta \approx 5.0\sqrt{\frac{\nu x}{U_\infty}} = \frac{5.0\,x}{\sqrt{Re_x}}\;},\qquad Re_x = \frac{U_\infty x}{\nu}$$ Exactly wahi $\sqrt{x}$ growth Step 3 mein guess ki gayi thi, ab precise number $5.0$ ke saath. --- ## Step 8 — Slope se drag tak **KYA.** Wall shear stress $\tau_w$ ("tau-w", plate par unit area par frictional force) wall par velocity slope ke proportional hai: $$\tau_w = \mu\left.\frac{\partial u}{\partial y}\right|_{y=0} = \mu U_\infty\sqrt{\frac{U_\infty}{\nu x}}\;\underbrace{f''(0)}_{=\,0.332} = 0.332\,\mu U_\infty\sqrt{\frac{U_\infty}{\nu x}}$$ Yahan $\mu = \rho\nu$ *dynamic* viscosity hai — density $\rho$ (Step 1 mein define hui) times kinematic viscosity $\nu$. **KYUN.** Fluid se friction proportional hai ki surface par speed kitni sharply change hoti hai — steeper profile ka matlab hai neighbouring layers zyada tezi se slide karti hain, zyada drag hoti hai. Woh steepness *hai* $f''(0)$. Dynamic pressure $\tfrac12\rho U_\infty^2$ (natural "flow-energy per volume", density $\rho$ aur speed $U_\infty$ se bana) se divide karne par dimensionless local [[Skin Friction Drag|skin-friction coefficient]] milta hai: $$\boxed{\;C_f = \frac{0.664}{\sqrt{Re_x}}\;}\qquad(0.664 = 2\times 0.332)$$ aur $\tau_w$ ko total length $L$ ki plate par integrate karne par total-drag coefficient milta hai. Yahan: - $L$ ::: plate ki **puri length** leading se trailing edge tak (metres mein). - $Re_L$ ::: us puri length par built Reynolds number, $Re_L = \dfrac{U_\infty L}{\nu}$ — same $Re_x$ formula lekin $x=L$ par evaluate kiya. $$\boxed{\;C_D = \frac{1.328}{\sqrt{Re_L}}\;}\qquad(1.328 = 2\times 0.664)$$ **PICTURE.** Wall jisme $y=0$ par profile ki tangent line hai; uski slope ($0.332$ marked) plate par drag arrow ko feed karti hai. Downstream, slope soft ho jaata hai aur drag arrow chhhota ho jaata hai. ![[deepdives/dd-physics-2.2.22-d2-s08.png]] --- ## Step 9 — Edge cases (reader ko kabhi stranded mat chhhodo) **KYA & KYUN.** - **Leading edge, $x\to 0$.** Formula deta hai $\delta\to 0$ aur $\tau_w\to\infty$. Yeh *leading-edge singularity* hai: thin-layer assumption ($\delta \ll x$) tip par hi break hoti hai, toh Blasius ek tiny nose region mein invalid hai. Wahan se downstream har jagah excellent hai. - **Far downstream / high $Re_x$.** Jab $Re_x \gtrsim 5\times10^5$ ho jaata hai flow **turbulent** ho jaata hai (dekho [[Turbulent Boundary Layer]]); phir $\delta$ $\sqrt{x}$ ki tarah nahi balki $x^{4/5}$ ki tarah grow karta hai, aur Blasius apply nahi hota. - **Zero viscosity, $\nu\to 0$.** Layer vanish ho jaati hai ($\delta\to 0$) — ek ideal *inviscid* flow plate ke paas freely slip karta hai. Viscosity exactly wahi hai jo layer create karti hai. **PICTURE.** Plate ke saath ek strip: ek shaded *invalid* nose region ($x\to0$), wide *valid laminar Blasius* band, aur transition line ke baad *turbulent* zone. ![[deepdives/dd-physics-2.2.22-d2-s09.png]] --- ## Ek-picture summary ![[deepdives/dd-physics-2.2.22-d2-s10.png]] Poori derivation ek canvas par: kai $x$ par raw profiles (top-left) → stretch $\eta = y\sqrt{U_\infty/\nu x}$ se collapsed hokar **ek** curve $f'$ mein (centre), governed by single ODE $f'''+\tfrac12 f f''=0$, teen headline numbers $5.0,\;0.332,\;0.664$ deliver karta hai aur drag law $C_D=1.328/\sqrt{Re_L}$. > [!recall]- Poore walkthrough ki Feynman retelling > Hawa ek flat plate ke paas se guzarti hai. Plate par hi hawa chipak jaati hai (Step 1). Do rules apply hote hain: kuch bhi create nahi hota (mass conserved hai) aur pushes equal accelerations (Newton) — Step 2. Kyunki chipchipa "slow zone" patla hai, sirf speed ki sabse sharp bending matter karti hai, jo hume ek hard term throw away karne deti hai. Humne zone ki thickness *guess* ki thi dekh ke ki hawa ka momentum kitna zor se dhakelta hai aur chippak kitna drag karti hai — yeh distance ke square root ki tarah grow karta hai (Step 3). Toh humne height *stretched* units $\eta$ mein measure ki jo ussi tarah grow karti hai, aur — miracle — profile plate par har jagah identical lagti hai (Step 4). Flow ko ek clever helper $f$ ke through likha jiska amplitude humne *choose* kiya taaki $x$ cancel ho jaayein, horizontal speed simply $U_\infty$ times $f$ ka slope ban gayi, aur product rule se born ek small lekin real upward drift $v$ bhi tha (Step 5). Sab plug in karne par, extra $\eta f' f''$ pieces cancel ho gaye aur ek clean equation bacha, $f'''+\tfrac12 f f''=0$, teen common-sense conditions ke saath (Step 6). Ek computer ise solve karta hai: profile $99\%$ speed par stretched height $5.0$ par hit karti hai, aur wall par steepness $0.332$ hai (Step 7). Wahi steepness drag hai: friction coefficient $0.664/\sqrt{Re}$, total drag $1.328/\sqrt{Re}$ (Step 8). Yeh sirf bilkul tip par aur jab flow turbulent ho jaata hai tabhi fail hota hai (Step 9). Ek curve, teen numbers, poora flat-plate boundary layer. > [!mnemonic] Numbers double hote hain > **Five thick · three-thirty steep · six-six-four friction · one-three-two-eight total.** > $0.664 = 2\times0.332$, aur $1.328 = 2\times0.664$ — har step mein double hota hai. --- ## #flashcards/physics Momentum equation mein sirf $\partial^2 u/\partial y^2$ kyun rakhte hain? ::: Layer patli hai, isliye speed across it (in $y$) sharply bend hoti hai lekin along it (in $x$) gently; $y$-curvature dominate karta hai. Inertia aur viscosity balance karne par kaunsi natural thickness milti hai? ::: $\delta \sim \sqrt{\nu x/U_\infty}$, yaani yeh $\sqrt{x}$ ki tarah grow karti hai. $f'(\eta)$ physically kya represent karta hai? ::: Velocity ratio $u/U_\infty$ — boundary-layer profile ki shape. Stream-function amplitude $\sqrt{\nu x U_\infty}$ kyun choose kiya jaata hai? ::: Taaki $u=\partial\psi/\partial y$ exactly $U_\infty f'$ nikle bina kisi leftover $x$ ke; amplitude forced hai, guessed nahi. $f'''+\tfrac12 f f''=0$ mein $\tfrac12$ kahan se aata hai? ::: Product rule mein $\sqrt{x}$ (aur $x^{-1/2}$) differentiate karne se; inertia terms add karne par $\eta f' f''$ parts cancel ho jaate hain, $-\tfrac12 f f''$ bachta hai. Drag number $0.664$ kahan se aata hai? ::: Wall slope $f''(0)=0.332$ se, $\tau_w$ ko dynamic pressure $\tfrac12\rho U_\infty^2$ se divide karne par double ho jaata hai.