Exercises — Vorticity — ω = ∇ × v, circulation Γ

KYA karte hain: spin-counter mein plug in karo. KYUN: definition sabse tezi se test karne ka tarika hai. Dono $v_x=3$ aur $v_y=-5$ constants hain — woh position ke saath change nahi hote, isliye har partial derivative $0$ hai: ${\omega }_z=\frac{\partial (-5)}{\partial x}-\frac{\partial (3)}{\partial y}=0-0=0.$ Answer: ${\omega }_z=0$. Ek uniform flow ek paddle-wheel ko saath le jaata hai lekin kabhi spin nahi karta — sense banta hai, kyunki kahin koi speed ka difference hi nahi hai jo use twist kare.

KYA: memorised relation ${\omega }_z=2\Omega$ apply karo. KYUN: parent note ne prove kiya ki is field ka curl exactly angular velocity ka double hai, isliye hum algebra skip karte hain. ${\omega }_z=2\Omega =2\times 4=8{\text{s}}^{-1}.$ Answer: ${\omega }_z=8{\text{s}}^{-1}$ (har jagah constant — ek rigid spin mein har point par same vorticity hoti hai).

(a) ${\omega }_z=\partial (0)/\partial x-\partial (y)/\partial y=0-1=-1\ne 0$. Rotational. (b) ${\omega }_z=\partial (y)/\partial x-\partial (x)/\partial y=0-0=0$. Irrotational (yeh ek pure stretch flow hai — fluid ko baahir spread karta hai, spin nahi karta). Answer: sirf (a) mein vorticity hai.

KYA: $v_x=\alpha y^2$ ko $y$ ke respect mein differentiate karo. KYUN: sirf $v_x$ position par depend karta hai, aur sirf $y$ ke through, isliye wahi ek surviving term hai. ${\omega }_z=\frac{\partial v_y}{\partial x}-\frac{\partial v_x}{\partial y}=0-\frac{\partial (\alpha y^2)}{\partial y}=-2\alpha y.$ $y=3$ par: ${\omega }_z=-2(2)(3)=-12{\text{s}}^{-1}$. Answer: ${\omega }_z=-12{\text{s}}^{-1}$. Note karo ki vorticity ab $y$ ke saath vary karti hai — simple shear se alag, yeh "curved shear" fluid ko jitna door, utna zyaada spin karta hai.

KYA: radius $r$ ka circle walk karo; har point par flow speed $v_{\theta }$ entirely walking direction ke along hai (yeh swirling part hai, $\hat{\theta }$ ki taraf pointing), isliye $\mathbf{v}\cdot d\mathbf{l}=v_{\theta }(rd\theta )$. KYUN $d\mathbf{l}=rd\theta$ likhte hain: jaise upar introduce kiya, circle par arc length radius times swept angle hoti hai. $\Gamma =\oint v_{\theta }rd\theta ={\int }_0^{2\pi }\frac{k}{r}rd\theta =k{\int }_0^{2\pi }d\theta =2\pi k=2\pi (6)=12\pi \approx 37.70{\text{m}}^2/\text{s}.$ Answer: $\Gamma =12\pi \approx 37.70{\text{m}}^2/\text{s}$, har radius ke liye same — free vortex ki pehchaan. Saari swirl singular centre par chhipi hoti hai.

Neeche figure dekho: dashed circles teen alag radius ke teen loops hain; teal (inner) arrows lambe hain orange (outer) arrows se kyunki $v_{\theta }=k/r$ radius ke saath ghatta hai. Phir bhi bade loop par chhote arrows exactly lamba walk karne se compensate ho jaate hain, isliye walk-around total $\Gamma =2\pi k$ har dashed circle par identical aata hai — yahi picture hai "$r$ se independent" ke peeche. Purple star singular centre mark karta hai jahan saari spin chhipi hoti hai.

KYA: disc par (constant) vorticity integrate karo. KYUN yahan Stokes use karein: vorticity uniform hai, isliye area integral bas ${\omega }_z$ times area hai — line integral se kaafi aasaan. $\Gamma ={\iint }_S{\omega }_{z}dA={\omega }_z\cdot \pi R^2=8\cdot \pi (2{)}^2=32\pi \approx 100.53{\text{m}}^2/\text{s}.$ Answer: $\Gamma =32\pi \approx 100.53{\text{m}}^2/\text{s}$.

KYA (spin): spin-counter se field differentiate karo. KYUN hum differentiate karte hain: poora sawaal yeh hai ki yahan paddle-wheel spin karta hai ya nahi, aur vorticity — curl — wahi ek quantity hai jo iska pointwise jawab deti hai, isliye hume seedha $\partial v_y/\partial x-\partial v_x/\partial y$ compute karna hoga. Maano $s=x^2+y^2$. $\frac{\partial v_y}{\partial x}=\frac{\partial }{\partial x}\left(\frac{x}{s}\right)=\frac{s-x(2x)}{s^2}=\frac{y^2-x^2}{s^2},$ $\frac{\partial v_x}{\partial y}=\frac{\partial }{\partial y}\left(\frac{-y}{s}\right)=\frac{-s+y(2y)}{s^2}=\frac{y^2-x^2}{s^2}.$ Subtract karne par: ${\omega }_z=\frac{y^2-x^2}{s^2}-\frac{y^2-x^2}{s^2}=0$ ($s\ne 0$ ke liye). KYUN origin par fail karta hai: wahan $s=0$ hai, field blow up hoti hai, derivatives undefined hain — saari vorticity us ek point par ek spike mein chhipi hoti hai. KYA (circulation): polar form $v_{\theta }=1/r$ mein convert karo aur walk-around integral reuse karo. KYUN yahan polar par switch karte hain: loop ek circle hai, aur polar language mein swirling speed $v_{\theta }$ us circle ke along constant hai, jo ek messy Cartesian line integral ko L2.2 ke one-line result mein badal deta hai. $k=1$ ke saath, origin ke around kisi bhi loop ke liye $\Gamma =2\pi$. Paradox resolved: Stokes phir bhi hold karta hai — disc mein singular point hai, isliye "area" integral zero nahi hai; woh exactly $2\pi$ worth ka concentrated spike pick up karta hai.

Line integral (counter-clockwise loop, standard positive direction). Parametrise karo $x=\cos \theta ,y=\sin \theta$, $d\mathbf{l}=(-\sin \theta ,\cos \theta )d\theta$. Phir $\mathbf{v}=(\sin \theta ,-\cos \theta )$, isliye $\mathbf{v}\cdot d\mathbf{l}=(\sin \theta )(-\sin \theta )+(-\cos \theta )(\cos \theta )=-{\sin }^2\theta -{\cos }^2\theta =-1.$ $\Gamma ={\int }_0^{2\pi }(-1)d\theta =-2\pi .$ Stokes. ${\omega }_z=\partial (-x)/\partial x-\partial (y)/\partial y=-1-1=-2$. Area $\pi (1{)}^2$ par: $\Gamma =-2\cdot \pi =-2\pi$. ✓ Answer: $\Gamma =-2\pi$. Minus sign honest hai: flow clockwise spin karta hai jabki humne counter-clockwise chalne ka choose kiya, isliye flow hamare chalne ka virodh karta hai — negative circulation.

KYA: do flux contributions add karo. Har half ka area $1\times 2=2{\text{m}}^2$ hai. $\Gamma =(+5)(2)+(-5)(2)=10-10=0.$ Answer: $\Gamma =0$, phir bhi flow irrotational nahi hai — vorticity lagbhag har jagah $\pm 5$ hai. Positives aur negatives sirf sum mein cancel ho jaate hain. Left par paddle-wheel ek taraf spin karta hai, right par doosri taraf; global loop yeh nahi dekh sakta.

KYA: lift law ko $\Gamma$ ke liye solve karo. KYUN yeh law: Kutta–Joukowski theorem kehta hai lift per unit span directly fluid density times flight speed times woh circulation hai jo wing apne around "bind" karta hai — circulation wing ka secret swirl hai. $\Gamma =\frac{L}{\rho U}=\frac{900}{1.2\times 50}=\frac{900}{60}=15{\text{m}}^2/\text{s}.$ Answer: $\Gamma =15{\text{m}}^2/\text{s}$. Zyaada bound circulation wala wing same speed par zyaada lift karta hai.

KYA: circulation ka conservation. KYUN: Kelvin's theorem kehta hai ideal fluid ke liye ek loop ke around circulation jo fluid ke saath move karti hai, woh frozen hai — apni initial value $0$ se change nahi ho sakti. Agar loop ab wing ke bound vortex ($+15$) aur shed starting vortex (${\Gamma }_s$) dono ko enclose karta hai: ${\Gamma }_b+{\Gamma }_s=0\Rightarrow {\Gamma }_s=-{\Gamma }_b=-15{\text{m}}^2/\text{s}.$ Answer: ${\Gamma }_s=-15{\text{m}}^2/\text{s}$ — ek equal aur opposite vortex trailing edge se shed hota hai, jo exactly wahi hai jo wind-tunnel smoke dikhata hai.

(a) Vorticity. Purely swirling flow ke liye (sirf $v_{\theta }$, $v_r=0$ ke saath) axial vorticity hai ${\omega }_z=\frac{1}{r}\frac{d}{dr}\left(r{v}_{\theta }\right).$ Yeh kahan se aata hai (KYUN, magic nahi): radius $r$ ke chhote circle par walk-around total $\Gamma$ lo; kyunki $v_{\theta }$ us par constant hai, $\Gamma (r)=v_{\theta }\cdot 2\pi r=2\pi (r{v}_{\theta })$. Stokes kehta hai $\Gamma$ disc ke through vorticity flux hai, isliye radius $r$ tak flux hai $\Phi (r)=2\pi (r{v}_{\theta })$. Vorticity woh flux density hai: $r$ aur $r+dr$ ke beech thin ring mein extra flux ${\Phi }^{\prime }(r)dr$ hai, aur ring ka area $2\pi rdr$ hai, isliye ${\omega }_z=\frac{{\Phi }^{\prime }(r)dr}{2\pi rdr}=\frac{1}{2\pi r}\frac{d}{dr}\left(2\pi r{v}_{\theta }\right)=\frac{1}{r}\frac{d}{dr}\left(r{v}_{\theta }\right).$ Yahi toolbox ka 2D curl hai, bas circular coordinates mein likha. Ab apply karo:

• Andar ($r\le a$): $r{v}_{\theta }=\Omega r^2$, isliye $\frac{d}{dr}(\Omega r^2)=2\Omega r$, jo deta hai ${\omega }_z=\frac{1}{r}(2\Omega r)=2\Omega$. Ek uniform core spin — solid-body, exactly ${\omega }_z=2\Omega =6{\text{s}}^{-1}$.
• Baahir ($r>a$): $r{v}_{\theta }=\Omega a^2$ (constant), isliye uska derivative $0$ hai, jo deta hai ${\omega }_z=0$. Irrotational halo — free vortex ki tarah.

(b) Circulation.

• Andar: $\Gamma (r)=\oint v_{\theta }rd\theta =2\pi r\cdot \Omega r=2\pi \Omega r^2$. (Equivalent roop se ${\omega }_z\cdot \pi r^2=2\Omega \pi r^2$. ✓)
• Baahir: $\Gamma (r)=2\pi r\cdot \frac{\Omega a^2}{r}=2\pi \Omega a^2$ — $r$ cancel ho jaata hai.

(c) Baahir constant. $\Gamma (r>a)=2\pi \Omega a^2=2\pi (3)(2{)}^2=24\pi \approx 75.40{\text{m}}^2/\text{s}$, $r$ se independent. Boundary $r=a$ par continuity check karo: andar ka formula deta hai $\Gamma (a)=2\pi \Omega a^2$, jo baahir ki value se exactly match karta hai — dono pieces smoothly join hote hain. Conclusion: saari vorticity core ke andar bottled hai (${\omega }_z=2\Omega$ wahan, baahir $0$), isliye $r=a$ ke baad tumhe phir bhi full accumulated swirl $\Gamma =24\pi {\text{m}}^2/\text{s}$ feel hoti hai lekin koi local spin nahi — wahan ka paddle-wheel bas revolve karta hai, apne pin par turn nahi karta.

$r=4\text{m}$ par numbers (baahir): ${\omega }_z=0$, $\Gamma =24\pi \approx 75.40{\text{m}}^2/\text{s}$.

Kya tum, bina notes ke: (1) 2D field se ${\omega }_z$ compute kar sakte ho, (2) $rd\theta$ arc length including circulation line integral kar sakte ho, (3) explain kar sakte ho kyun $\Gamma =0$ ≠ irrotational, (4) $\Gamma$ ko lift se relate kar sakte ho, aur (5) Rankine vortex ko rotational core + irrotational halo mein split kar sakte ho? Agar koi bhi shaky lage, toh uss level ko dobara dekho.