Visual walkthrough — Potential flow — irrotational, inviscid; superposition of basic flows

2.2.28 · D2 · Physics › Fluid Mechanics › Potential flow — irrotational, inviscid; superposition of ba

Step 1 — "Flow picture" hota kya hai

KYA HAI. Kisi bhi algebra se pehle, picture fix karo. Ek 2D flow ek flat sheet par rahti hai. Har point par hum ek chhota arrow draw karte hain: fluid wahan kis direction mein move karti hai, aur kitni fast (arrow ki length). Saare arrows ko collect karo aur tumhare paas velocity field aa jaata hai, jo likhte hain $\vec{v}=(u,v)$ jahan $u$ sideways (rightward, $+x$) speed hai aur $v$ upar ($+y$) ki speed hai.

KYUN. Niche sab kuch sirf arrow-fields ko stack karna hai. Agar tum ek arrow diagram padh sakte ho, tum poori derivation follow kar sakte ho.

PICTURE. Ek hi point ko describe karne ke do tarike:

• Cartesian $(x,y)$ — grid coordinates.
• Polar $(r,\theta )$ — origin se distance $r$, aur angle $\theta$ jo $+x$ axis se counter-clockwise measure hota hai. Hume polar chahiye kyunki jo flows hum add karte hain woh circular / radial hain, aur circles $x,y$ mein ugly hain lekin $r,\theta$ mein simple hain.

Step 2 — Seedhi nadi (uniform flow)

KYA HAI. Sabse simple flow: har arrow right point karta hai, sab ek hi length $U$ ke. To $u=U$ har jagah, $v=0$. Symbols mein stream function hai $\psi =Uy$.

Stream function $\psi$ kyun? Parent se yaad karo: streamlines woh curves hain jinpar fluid actually travel karti hai, aur woh exactly level curves $\psi =\text{const}$ hoti hain. Nadi ke liye, $\psi =Uy=\text{const}$ matlab $y=\text{const}$ — flat horizontal lines. Perfect: horizontal nadi, horizontal streamlines.

Term by term: $\underset{\text{streamline label}}{\underbrace{\psi }}=\underset{\begin{array}{c}\text{nadi ki speed}\\ \text{(ek chosen number)}\end{array}}{\underbrace{U}}\cdot \underset{\begin{array}{c}\text{point ki}\\ \text{height}\end{array}}{\underbrace{y}}$

Polar mein (baad mein chahiye hoga), $y=r\sin \theta$, to ${\psi }_{\text{uniform}}=Ur\sin \theta .$

PICTURE. Equally spaced horizontal lines ki ek family, sab arrows ek hi length ke, right point karte hue.

Step 3 — Magic dipole (the doublet)

KYA HAI. Ek doublet woh hai jo tab milta hai jab tum ek tiny source (fluid bahar nikalta hai) ko infinitesimally close ek sink ke paas rakhte ho (fluid andar jaati hai) aur unhe merge karne dete ho, unki combined "strength" $\kappa$ ko finite rakhte hue. Iska stream function hai ${\psi }_{\text{doublet}}=-\frac{\kappa }{2\pi }\frac{\sin \theta }{r}.$

HUM IS flow ko kyun choose karte hain, koi aur source kyun nahi? Hume kuch aisa chahiye jiske streamlines apne aap mein curl back hon — loops — kyunki ek solid round body ke aas-paas woh flow honi chahiye jo uske front se back tak jaaye. Ek akela source sirf outward push karta hai hamesha (no loops). Ek doublet ki streamlines closed circles hoti hain, exactly woh shape jo ek cylinder ko wrap kar sake. Isliye doublet sahi Lego brick hai.

Term by term: ${\psi }_{\text{doublet}}=-\underset{\begin{array}{c}\text{strength}\\ \text{(how strong}\\ \text{the dipole is)}\end{array}}{\underbrace{\frac{\kappa }{2\pi }}}\cdot \underset{\begin{array}{c}\text{shape: strong near}\\ \text{origin, tilts with }\theta \end{array}}{\underbrace{\frac{\sin \theta }{r}}}$

$\frac{1}{r}$ yeh kehta hai ki doublet ka influence door jaane par fade hota hai — origin ke paas yeh dominate karta hai, door jaake die ho jaata hai. Woh "local push, global fade" exactly wahi hai jo river ko door dominate karne deta hai aur doublet ko paas dominate karne deta hai.

PICTURE. Nested closed loops origin ko hug karte hue, arrows unke around sweep karte hue.

Step 4 — Unhe stack karna (superposition)

WHAT. Laplace's equation linear hai, to hum simply do stream functions ko add kar sakte hain: $\psi =\underset{\text{river}}{\underbrace{Ur\sin \theta }}-\underset{\text{doublet}}{\underbrace{\frac{\kappa }{2\pi }\frac{\sin \theta }{r}}}.$

Yeh legal kyun hai. Dono $\psi$ ke values ${\nabla }^2\psi =0$ satisfy karte hain. Ek linear equation ke do solutions ka sum phir bhi ek solution hota hai, to combined field ek genuine potential flow hai — koi cheating nahi.

Key algebra — $\sin \theta$ factor out karo: $\psi =U\sin \theta \left(r-\frac{\kappa }{2\pi Ur}\right).$

Dekho humne kya kiya: humne common $\sin \theta$ ko front par le aaya. Yeh ek bracket isolate karta hai jo sirf $r$ par depend karta hai. Woh bracket humein abhi ek circle dene wala hai.

PICTURE. Left: nadi ke arrows. Middle: doublet loops. Right (added): arrows origin ke paas ek blob ke around split aur jaane lagte hain.

Step 5 — Kahin se ek circle appear hoti hai

KYA HAI. Poochho: kya $\psi =0$ ka ek streamline hai jo ek closed curve hai? Set karo $\psi =0$: $U\sin \theta \left(r-\frac{\kappa }{2\pi Ur}\right)=0.$

Ek product tab zero hoti hai jab koi bhi factor zero ho:

• $\sin \theta =0$ → $x$-axis (woh flow line jo seedhi andar aur bahar jaati hai).
• $r-\frac{\kappa }{2\pi Ur}=0$ → $r^2=\frac{\kappa }{2\pi U}$, yaani ek fixed radius jo $\theta$ se independent hai.

Yeh kyun poora trick hai. Ek fixed radius, har angle ke liye same, ek circle hai. Ise naam do $a$: $\boxed{a^2=\frac{\kappa }{2\pi U}}⟺\kappa =2\pi U{a}^2.$

To agar hum doublet strength ko $\kappa =2\pi U{a}^2$ choose karein, to streamline $\psi =0$ mein circle $r=a$ aa jaata hai. Fluid kabhi ek streamline cross nahi karti — to flow exactly aise behave karti hai jaise wahan radius $a$ ka ek solid cylinder khada ho. Humne ek solid body ko ek number choose karke conjure kar liya.

Boxed result mein term by term: $a^2=\frac{\kappa }{2\pi U}:a(\text{woh radius jo hum chahte hain}),\kappa (\text{doublet strength jo hum tune karte hain}),U(\text{nadi ki speed}).$

PICTURE. Step 4 ka blob snap karke radius $a$ ka perfect circle ban jaata hai; streamlines bahar smoothly uske around bend karti hain.

Step 6 — Surface par speed padhna

KYA HAI. $\kappa =2\pi U{a}^2$ ke saath, rewrite karo: $\psi =U\sin \theta \left(r-\frac{a^2}{r}\right).$ Surface par fluid circle ke along move karti hai, isliye wahan sirf tangential speed $u_{\theta }$ matter karti hai. Polar mein, tangential velocity $\psi$ se recover hoti hai $u_{\theta }=-\frac{\partial \psi }{\partial r}.$

Woh formula kyun hai (aur minus sign kyun)? Definition se $\psi$ is tarah bana hai ki fluid uski level lines ke along flow kare; tangential speed yeh hai ki $\psi$ kitni fast change hoti hai jab tum radially outward step karte ho. Sign convention $u_{\theta }=-\partial \psi /\partial r$ counter-clockwise ko positive banata hai, match karta hai jaise humne $\theta$ draw kiya. Yeh $u=\partial \psi /\partial y,v=-\partial \psi /\partial x$ ka polar twin hai parent se.

Derivative karo: $u_{\theta }=-\frac{\partial }{\partial r}\left[U\sin \theta \left(r-\frac{a^2}{r}\right)\right]=-U\sin \theta \left(1+\frac{a^2}{r^2}\right).$ Ab surface par khade ho, $r=a$, to $a^2/r^2=1$: $\boxed{u_{\theta }{|}_{r=a}=-2U\sin \theta }$

Term by term: $u_{\theta }=-\underset{\begin{array}{c}\text{from }1+1\end{array}}{\underbrace{2}}\underset{\text{nadi ki speed}}{\underbrace{U}}\underset{\begin{array}{c}\text{circle par}\\ \text{position}\end{array}}{\underbrace{\sin \theta }}$

PICTURE. Circle ke tangent arrows, top/bottom par longest, front/back par zero tak shrink karte hue.

Step 7 — Surface par har case (koi bhi mat chhodhna!)

KYA HAI. $\theta$ ko poore around walk karo aur har landmark par $u_{\theta }=-2U\sin \theta$ padho.

Sabhi chaar kyun cover karein. Ek reader kabhi cylinder par kisi aisa spot par na pahunche jise humne explain na kiya ho. Do stagnation points (front aur back) woh hain jahan incoming nadi poori tarah rest mein aa jaati hai — flow wahan split hoti hai. Do speed maxima ($|u_{\theta }|=2U$, free-stream speed se do guna) top aur bottom par hain. Top aur bottom ke beech sign flip sirf yeh record karta hai ki fluid top ke upar ek taraf jaati hai aur bottom ke neeche doosri taraf — symmetric.

Degenerate check — $U\to 0$: agar nadi ruk jaaye, $u_{\theta }=-2U\sin \theta \to 0$ har jagah. Koi stream nahi, koi cheez ke aas-paas flow nahi. Consistent. Isi tarah $a\to 0$ (ya $\kappa \to 0$) doublet ko hata deta hai aur hum bare nadi $\psi =Ur\sin \theta$ ke saath reh jaate hain. Har limit well behaved hai.

PICTURE. Circle apne chaar landmark arrows ke saath labelled: do zero-length (stagnation) $0,\pi$ par; do full-length $\pm 90^{\circ }$ par.

Step 8 — Speed se pressure tak (ek honest warning)

KYA HAI. Jahan fluid fast hoti hai, pressure low hoti hai — woh hai Bernoulli's equation: $p+\frac{1}{2}\rho v^2=\text{const}.$ Kyunki $|u_{\theta }|$ top aur bottom par sabse bada hai, pressure wahan sabse kam hoti hai aur front aur back stagnation points par sabse zyada hoti hai.

Warning kyun. Bernoulli $v$ mein nonlinear hai ($v^2$ hai). To tum velocities ki tarah pressures add nahi kar sakte. Recipe strict hai: stream functions add karo → total velocity lo → woh single total velocity Bernoulli mein ek baar daalo. Plain (non-spinning) cylinder ke liye pressure front-to-back perfectly symmetric hai, isliye net push zero hai — yeh d'Alembert's paradox hai parent se. Ek vortex add karo aur top/bottom symmetry break ho jaati hai, jisse Kutta–Joukowski theorem ke zariye lift milti hai — Magnus effect ka seed.

PICTURE. Circle par color map: hot (low pressure) top aur bottom, cool (high pressure) front aur back.

Ek-picture summary

Ise left to right padho: nadi $\psi =Ur\sin \theta$ $+$ doublet $\psi =-\frac{\kappa }{2\pi }\frac{\sin \theta }{r}$ $=$ cylinder flow, jahan $\kappa =2\pi U{a}^2$ choose karna streamline $\psi =0$ ko circle $r=a$ mein turn kar deta hai, aur surface speed padhti hai $u_{\theta }=-2U\sin \theta$.

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