Exercises — Kelvin's circulation theorem

Teeno assumptions hain: (1) inviscid (koi friction nahi), (2) conservative body force ($\mathbf{f}=-\nabla \Phi$), (3) barotropic ($\rho =\rho (p)$). (a) Satisfied. Incompressible $\Rightarrow \rho =$ const $\Rightarrow$ trivially barotropic; gravity conservative hai ($\Phi =gz$); assumption se inviscid. ✓ (b) (1) inviscid fail karta hai. Honey bahut zyada viscous hai — friction vorticity ko diffuse karta hai, isliye $D\Gamma /Dt\ne 0$. (c) (3) barotropic fail karta hai. Jab $\rho =\rho (p,T)$, toh term $\oint \frac{1}{\rho }\nabla p\cdot d\mathbf{\ell }$ ab vanish nahi hoti — yeh baroclinic generation hai.

Uniform $\omega$ area $A$ par $\Gamma =\omega A$ deta hai ($\Gamma ={\iint }_S\mathbf{\omega }\cdot d\mathbf{A}$ se). Toh $\omega =\frac{\Gamma }{A}=\frac{6}{3}=2{\text{s}}^{-1}.$

Kelvin kyun? Loop material hai, isliye uski circulation $\Gamma =\omega A$ conserved hai. ${\omega }_1{A}_1={\omega }_2{A}_2\Rightarrow {\omega }_2={\omega }_1\frac{A_1}{A_2}=5\cdot \frac{8}{2}=20{\text{s}}^{-1}.$ Area $4\times$ ghata, toh spin $4\times$ badha — skater-pulls-arms-in wala effect. Stretching term $(\mathbf{\omega }\cdot \nabla )\mathbf{u}$ ke liye Vorticity and the vorticity equation dekho jo yeh karta hai.

(a) Motion se pehle hawa still thi: wing aur wake ko enclose karne wale bade material loop mein $\Gamma =0$ tha. Kelvin us total ko $0$ par rakhta hai: ${\Gamma }_{\text{bound}}+{\Gamma }_{\text{starting}}=0\Rightarrow {\Gamma }_{\text{starting}}=-8{\text{m}}^2/\text{s}.$ (b) $L=\rho U{\Gamma }_{\text{bound}}=1.2\times 50\times 8=480\text{N/m}.$

Circulation algebraically add hoti hai (flux additive hai): $\Gamma ={\Gamma }_a+{\Gamma }_b=3+(-1)=2{\text{m}}^2/\text{s}.$ Kelvin's theorem se, jab tak dono tubes material loop se threaded rehti hain, $\Gamma$ $2{\text{m}}^2/\text{s}$ rehti hai — tubes move aur deform ho sakti hain lekin material loop cross nahi kar sakti (yeh Helmholtz consequence hai). $\Gamma$ nahi badlega.

Dekho $\mathbf{u}\cdot d\mathbf{u}=d\left(\frac{1}{2}\mathbf{u}\cdot \mathbf{u}\right)=d\left(\frac{1}{2}|\mathbf{u}{|}^2\right)$, kyunki $\frac{1}{2}|\mathbf{u}{|}^2$ ko differentiate karne par $\mathbf{u}\cdot d\mathbf{u}$ milta hai (dot product par product rule). Toh ${\oint }_C\mathbf{u}\cdot d\mathbf{u}={\oint }_{C}d\left(\frac{1}{2}|\mathbf{u}{|}^2\right)={\left[\frac{1}{2}|\mathbf{u}{|}^2\right]}_{\text{start}}^{\text{end}}=0.$ Zero kyun: yeh ek perfect differential (pure $d(\text{kuch bhi})$) ka integral hai. Ek closed loop ke around tum start par wapas aate ho, isliye end par value start par value ke barabar hoti hai — unka difference $0$ hai. Loop ka closed hona yeh karta hai.

Kinetic aur potential terms chale gaye lekin pressure term survive karti hai, $\frac{D\Gamma }{Dt}=-{\oint }_C\frac{1}{\rho }\nabla p\cdot d\mathbf{\ell }.$ Stokes' theorem se surface integral mein convert karne par baroclinic torque milta hai $\frac{D\Gamma }{Dt}={\iint }_S\frac{\nabla \rho \times \nabla p}{{\rho }^2}\cdot d\mathbf{A}.$ Mechanism hai baroclinic vorticity generation: jab constant density aur constant pressure ki surfaces misaligned hoti hain ($\nabla \rho \times \nabla p\ne 0$), toh circulation create hoti hai. Yahi se sea breezes aur weather fronts spin up hoti hain.

$t=0$ par, har material loop $C$ mein $\Gamma ={\iint }_S\mathbf{\omega }\cdot d\mathbf{A}=0$ hai. Kelvin se, har material loop par $\frac{D\Gamma }{Dt}=0$ hai, isliye $\Gamma (t)=0$ sabhi $t$ ke liye, sabhi loops ke liye. Agar ${\iint }_S\mathbf{\omega }\cdot d\mathbf{A}=0$ har surface ke liye hai (loop ko kisi bhi orientation mein ek point tak shrink karo), toh $\mathbf{\omega }=0$ everywhere hai sabhi time ke liye. Yeh Lagrange–Cauchy theorem hai. Shortcut justify hua: ek irrotational velocity field ko $\mathbf{u}=\nabla \varphi$ likh sakte hain ek scalar velocity potential $\varphi$ ke liye, jo vector problem ko ek scalar equation tak collapse kar deta hai (agar incompressible toh ${\nabla }^2\varphi =0$).

(a) ${\omega }_1={\Gamma }_0/A_1=4/0.5=8{\text{s}}^{-1}.$ (b) Volume $V=A\ell$ fixed hai aur $\ell \to 2\ell$, isliye $A_2=A_1/2=0.25{\text{m}}^2$. Kelvin ${\Gamma }_0$ ko material loop par fixed rakhta hai, toh ${\omega }_2=\frac{{\Gamma }_0}{A_2}=\frac{4}{0.25}=16{\text{s}}^{-1}.$ Double length tak stretching ne spin double kar di. (c) Ek fluid tube jo $\Gamma$ conserve karte hue narrow hoti hai, woh ek spinning mass ki tarah hai jo andar khichti hai: moment of inertia kam, spin fast, angular momentum same. Kelvin ⟷ angular momentum conservation; vortex stretching ⟷ skater ka arms tucking.

Circulation conserve karo: $\Gamma =\omega A=10\times 0.2=2{\text{m}}^2/\text{s}$. ${\omega }^{\prime }=\frac{\Gamma }{A^{\prime }}=\frac{2}{0.05}=40{\text{s}}^{-1}.$ Ratio $\frac{{\omega }^{\prime 2}}{{\omega }^2}={\left(\frac{40}{10}\right)}^2=16.$ Kyunki $\omega \propto 1/A$, area ko $4\times$ squeeze karne se $\omega$ $4\times$ badhta hai aur spin-energy proxy $4^2=16\times$ — concentrated swirl locally bahut zyada energetic hoti hai, jo Bernoulli's principle ke tight vortex cores mein pressure drops ke saath consistent hai.

$\nabla p=(\partial p/\partial x)\hat{\mathbf{x}}=(-2)\hat{\mathbf{x}}$ aur $\nabla \rho =(\partial \rho /\partial y)\hat{\mathbf{y}}=(3\times 10^{-4})\hat{\mathbf{y}}$ ke saath, $\nabla \rho \times \nabla p=(3\times 10^{-4}\hat{\mathbf{y}})\times (-2\hat{\mathbf{x}})=(3\times 10^{-4})(-2)(\hat{\mathbf{y}}\times \hat{\mathbf{x}})=-6\times 10^{-4}(-\hat{\mathbf{z}})=6\times 10^{-4}\hat{\mathbf{z}}.$ Integrand ka magnitude: $\left|\frac{\nabla \rho \times \nabla p}{{\rho }^2}\right|=\frac{6\times 10^{-4}}{(1.2{)}^2}=\frac{6\times 10^{-4}}{1.44}\approx 4.17\times 10^{-4}{\text{s}}^{-2}.$ Non-zero hai kyunki gradients misaligned hain — yeh purely baroclinic spin-up hai, jo Kelvin's theorem ke barotropic form ko invisible hai.

1. Motion se pehle: still air, isliye wing + wake ko enclose karne wale kisi bhi bade material loop mein $\Gamma =0$ tha.
2. Impulsive start par: potential (ideal) outer flow sharp trailing edge ke around infinite velocity ke saath slip karta — physically impossible.
3. Viscosity act karti hai thin boundary layer mein (Kelvin ki inviscid assumption locally fail hoti hai), vorticity generate karti hai jo roll up hoti hai aur circulation $-{\Gamma }_{\text{bound}}$ ke starting vortex ke roop mein shed hoti hai.
4. Kelvin globally bade loop par hold karta rehta hai: ${\Gamma }_{\text{bound}}+{\Gamma }_{\text{starting}}=0$, isliye total $0$ rehta hai jabki wing bound circulation ${\Gamma }_{\text{bound}}$ rakhti hai.
5. Kutta–Joukowski phir steady lift deta hai $L=\rho U{\Gamma }_{\text{bound}}$. Indispensable failure: boundary layer mein inviscid assumption ka local breakdown. Viscosity ke bina starting vortex shed karne ke liye, Kutta condition kabhi meet nahi ho sakti aur koi net bound circulation (aur isliye koi lift) arise nahi ho sakti. Kelvin's theorem wahi hai jo guarantee karta hai ki shed vortex exactly $-{\Gamma }_{\text{bound}}$ hona chahiye.

Teeno assumptions pehchano ⟶ $\Gamma =\omega A$ conservation apply karo ⟶ un terms ko analyse karo jo vanish hoti hain (aur woh jo zaroori nahi) ⟶ stretching + energy synthesise karo ⟶ real baroclinic aur viscous exceptions master karo.