2.2.30 · D3Kelvin's circulation theorem
Yeh note Kelvin's circulation theorem ki drill-yard hai. Parent note ne bataya tha ki theorem kya kehta hai aur kyun sach hai. Yahaan hum isko har angle se dekhenge: shrinking loops, stretching loops, zero-swirl loops, loops jahan theorem fail karta hai, aur exam ke woh twists jo logon ko trap karte hain. Har worked example se pehle aapko apna forecast dena hoga — ek guess commit karo, phir khud check karo.
Sabse pehle hum theorem ke har word aur symbol ko pin down karte hain, taaki aapko kabhi parent note pe wapas jaana na pare.
Definition Woh words aur symbols jinpar hum build karte hain
Material derivative D t D fluid particle ke saath chalte hue , fixed point pe nahin. Likha jaata hai D t D ≡ ∂ t ∂ + u ⋅ ∇ carry ho raha hai. Jab hum "D Γ/ D t Line element d ℓ C saath us direction mein point karta hai jis direction mein tum chalo; in sab chhote arrows ko head-to-tail joodo aur woh poora loop trace kar denge. Iska length woh tiny distance hai jo tum step karte ho, iska direction local tangent hai.Vorticity ω ω = ∇ × u ω double hai. Dekho Vorticity and the vorticity equation .Scalar area A d A ::: scalar A ek chhote loop ka total flat area hai (pocket formula Γ = ω A d A ∬ S ω ⋅ d A vector area element hai — ek infinitesimal surface patch jo apne normal n ^ d A ω A Barotropic ::: density sirf pressure par akele depend karti hai, ρ = ρ ( p ) ρ baroclinic hai aur Kelvin fail ho sakta hai.Conservative body force ::: ek force jo f = − ∇Φ Φ
Yahan sab kuch do facts par tikaa hai parent note se:
Recall Woh do facts jo hum har jagah reuse karte hainFact A — Circulation ek material loop par conserved hoti hai. Ek inviscid, barotropic flow mein conservative body forces ke saath, material derivative D / D t ≡ ∂ / ∂ t + u ⋅ ∇ D t D Γ = 0 , Γ = ∮ C u ⋅ d ℓ . ∮ C u ⋅ d ℓ u d ℓ saath kitna point karti hai.
Fact B — Circulation vorticity flux ke barabar hai (Stokes' theorem ):
Γ = ∬ S ω ⋅ d A . scalar area A ω Γ = ω A
Har problem jo Kelvin's theorem phek sakta hai woh in cells mein se ek mein aati hai. Neeche ke examples us cell ke saath label hain jo woh hit karte hain, aur milake woh har cell fill karte hain.
#
Cell (case class)
Kya change hota hai
Theorem ka verdict
Example
C1
Area shrinks , positive ω
A ↓ ω ↑ Ex 1
C2
Area grows
A ↑ ω ↓ Ex 2
C3
Zero circulation startΓ = 0 hamesha 0
Ex 3
C4
Sign flip of enclosed swirlloop turns inside-out
Γ Ex 4
C5
Degenerate loop (area → 0 A → 0 ω → ∞ Γ Ex 5
C6
Real-world word problem (aerofoil)starting vortex shed
Γ total = 0 Ex 6
C7
Theorem FAILS — baroclinic∇ ρ × ∇ p = 0 D Γ/ D t = 0 Ex 7
C8
Theorem FAILS — viscousfriction ω
D Γ/ D t = 0 Ex 8
C9
Exam twist — fixed vs material loopgalat loop choose kiya
trap!
Ex 9
Pehle hum woh sign convention define karte hain jo hum har figure mein use karenge, kyunki galat sign sabse common mistake hai yahaan.
Definition Sign convention (right-hand rule)Ek direction choose karo loop C chalne ka . Apne right hand ki ungliyan usi tarah curl karo; tumhara thumb positive normal n ^ Γ > 0 usi tarah flow karta hai jis tarah tum chalte ho — ek counter-clockwise swirl n ^ Γ < 0 signed convention hai: circulation ek sign carry karta hai, sirf size nahin.
Figure s01 yeh draw karta hai: chalk loop jisme blue walk-arrows counter-clockwise chalte hain, pale-yellow normal n ^ Γ < 0
Worked example Example 1 — Contracting vortex tube (cell C1)Ek chhota material loop ek vortex tube ke perpendicular baitha hai jiska area A 1 = 8 cm 2 ω 1 = 5 s − 1 A 2 = 2 cm 2 ω 2
Forecast: area quarter ho jaata hai — kya aap expect karte ho ki ω 2
Conserved quantity likho. Γ = ω A Yeh step kyun? Fact A kehta hai Γ ω A Start ko end ke barabar karo. ω 1 A 1 = ω 2 A 2 Yeh step kyun? Same loop, same conserved Γ Solve karo. ω 2 = ω 1 A 2 A 1 = 5 ⋅ 2 8 = 20 s − 1 Yeh step kyun? Pure algebra — koi physical decision baaki nahin.
Verify: Γ 1 = ω 1 A 1 = 5 ⋅ 8 × 1 0 − 4 = 4.0 × 1 0 − 3 m 2 / s Γ 2 = 20 ⋅ 2 × 1 0 − 4 = 4.0 × 1 0 − 3 m 2 / s quadruple hua kyunki area quarter hua — woh skater jo apni baahein andar kheenchti hai.
Figure s02 do states ko side by side dikhata hai: ek bada loop (light blue fill, kuch yellow spin-arrows, ω 1 = 5 ω 2 = 20 density vorticity hai — chhote loop mein denser arrows spin-up ko visible banate hain.
Worked example Example 2 — Loop ek diverging region par drag kiya gaya (cell C2)Example 1 se bilkul wahi material loop continue karo. Woh Example 1 ke end mein area 2 cm 2 Γ = ω 1 A 1 = 5 ⋅ 8 × 1 0 − 4 = 4.0 × 1 0 − 3 m 2 / s Γ A 3 = 10 cm 2 ω 3
Forecast: wider loop — zyada tez ya dheela spin?
Conserved Γ Γ = ω 3 A 3 Γ = 4.0 × 1 0 − 3 m 2 / s Yeh step kyun? Yeh ek continuous material loop hai; Kelvin iska Γ ω 3 ω 3 = A 3 Γ = 10 × 1 0 − 4 4.0 × 1 0 − 3 = 4.0 s − 1 Yeh step kyun? ω ∝ 1/ A Γ = ω A
Verify: ω 3 A 3 = 4.0 ⋅ 10 × 1 0 − 4 = 4.0 × 1 0 − 3 ⇒ dheela spin. Ex 1 aur Ex 2 compare karo: same tube, same loop, ω 5 → 20 → 4 Γ
Worked example Example 3 — Flow from rest (cell C3)t = 0 u = 0 har material loop Γ = 0
Forecast: kya ek smooth ideal flow kabhi stillness se swirl "spontaneously" grow kar sakta hai?
Initial value. Rest mein, u = 0 ⇒ Γ ( 0 ) = ∮ 0 ⋅ d ℓ = 0 Yeh step kyun? Koi velocity nahin matlab koi line integral nahin, kisi bhi loop par.Kelvin apply karo. D Γ/ D t = 0 Γ ( t ) = Γ ( 0 ) = 0 t Yeh step kyun? Fact A: value har material loop ke liye lock hai.Vorticity mein convert karo. Agar ∬ S ω ⋅ d A = 0 har surface ke liye, tab ω = 0 Yeh step kyun? Ek flux jo sabhi surfaces ke through vanish ho jaata hai woh integrand ko har jagah vanish karne par majboor karta hai.
Verify: Conclusion — ek ideal flow jo rest se start hua woh hamesha irrotational rehta hai — Kelvin's theorem ka ek standard corollary hai (kabhi kabhi persistence of irrotationality kaha jaata hai). Yeh velocity potential use karna justify karta hai. Ex 6 ke saath sanity check: air mein jo rest mein thi ek aerofoil ko is rule ka paalan karna chahiye — total Γ 0
Worked example Example 4 — Ek loop mein do counter-rotating vortices (cell C4)Ek material loop do vortex patches enclose karta hai: patch + Γ + = + 6 m 2 / s − Γ − = − 6 m 2 / s
Forecast: do equal-and-opposite swirls ek loop ke andar — total zero, ya total twelve?
Signed contributions add karo. Γ = Γ + + Γ − = 6 + ( − 6 ) = 0 Yeh step kyun? Circulation ek signed flux hai (Fact B); opposite swirls cancel hote hain, woh magnitude mein add nahin hote.Outer loop par Kelvin apply karo. Γ = 0 Yeh step kyun? Fact A internal structure se regardless hold karta hai — sirf total signed flux lock hoti hai.*Kya change ho sakta hai? Do patches stretch, migrate, aur individually tez ya dheela spin kar sakte hain — lekin material loop ke through unka signed sum 0 Yeh step kyun? Kelvin net ko constrain karta hai, pieces ko nahin; har sub-loop jo ek patch ko hug karta hai woh apna Γ
Verify: Patch + + 6 − − 6 = 0 pair bahut door ja sakta hai jabki far-field loop zero circulation read karta hai.
Figure s03 bade chalk loop ko draw karta hai jisme left par blue counter-clockwise patch (+ 6 − 6 + 6 + ( − 6 ) = 0 directions — unki count nahin — signs carry karte hain yahaan.
Worked example Example 5 — Kya ek point vortex allowed hai? (cell C5)Ex 1 ka loop lo aur imagine karo ki iska area zero ki taraf squeeze ho raha hai, A → 0 ω
Forecast: jab A → 0 Γ
Γ ω Γ = ω A 4.0 × 1 0 − 3 m 2 / s Yeh step kyun? Kelvin Γ ω A Limit lo. Jab A → 0 Γ ω = Γ/ A → ∞ Yeh step kyun? Ek finite numerator ek vanishing denominator par diverge karta hai.Interpret karo. Vorticity ek delta spike ban jaati hai ek line par concentrated — ek point vortex (2D mein) — jo finite circulation Γ ω Yeh step kyun? Circulation physically finite, measurable, conserved quantity hai; local vorticity infinity tak idealized hai.
Verify: ω A = 2 × 1 0 − 4 , 2 × 1 0 − 6 , 2 × 1 0 − 8 m 2 ω = 20 , 2000 , 2 × 1 0 5 s − 1 Γ = ω A = 4.0 × 1 0 − 3 Moral: Kelvin Γ ω
Worked example Example 6 — Ek aeroplane wing start karna (cell C6)Ek wing U = 50 m/s ρ = 1.2 kg/m 3 L = 1800 N per metre of span Kutta–Joukowski L = ρ U Γ bound
Forecast: agar wing clockwise bound swirl gain karta hai, to wake ko kya karna chahiye — kuch nahin, ya ek opposite vortex shed karna?
Bound circulation ke liye Kutta–Joukowski solve karo. Γ bound = ρ U L = 1.2 ⋅ 50 1800 = 30 m 2 / s Yeh step kyun? Lift bound circulation ke proportional hai; Γ bound Pre-motion state yaad karo. Takeoff se pehle air rest mein hai, isliye koi bhi material loop jo wing + wake enclose kare Γ = 0 Yeh step kyun? Kelvin woh 0 Books balance karo. Γ bound + Γ starting = 0 ⇒ Γ starting = − 30 m 2 / s Yeh step kyun? Conserved total ko 0 + 30 − 30
Verify: Total = 30 + ( − 30 ) = 0 [ Γ ] = m 2 / s ρ U Γ = 1.2 ⋅ 50 ⋅ 30 = 1800 kg m − 3 ⋅ m/s ⋅ m 2 / s = 1800 N/m isliye exist karta hai kyunki Kelvin ek shed starting vortex allow karta hai.
Worked example Example 7 — Ek sea breeze spin up karta hai (cell C7)Warm land air aur cool sea air side by side baithti hain. Pressure height ke saath decrease karta hai, lekin density horizontally vary karti hai (warm=light left par, cool=dense right par). Explain karo, surviving term ke saath, ki D Γ/ D t = 0
Forecast: agar ρ p
Derivation mein pressure term yaad karo. Kelvin ko chahiye tha ∮ ρ 1 ∇ p ⋅ d ℓ = 0 ρ 1 ∇ p = ∇ P pure gradient tha (barotropic).
Yeh step kyun? Woh gradient property hi poori wajah thi ki loop integral vanish hua.Barotropicity todo. Jab ρ = ρ ( p , T ) ρ 1 ∇ p nahin hai. Iska curl baroclinic term hai:
D t D Γ = − ∮ C ρ ∇ p ⋅ d ℓ = − ∬ S ρ 2 ∇ ρ × ∇ p ⋅ d A = 0. Yeh step kyun? Stokes' theorem surviving loop integral ko ∇ ρ × ∇ p Geometry padho. Ek gradient increasing values ki taraf point karta hai. Pressure ground par sabse zyada hai, isliye ∇ p downward point karta hai (neeche zyada pressure ki taraf). Density right ki taraf cool sea ki taraf zyada hai, isliye ∇ ρ right ki taraf point karta hai. Unka cross product ∇ ρ × ∇ p × page ke bahar point karta hai ⇒ Yeh step kyun? Cross product ki direction wahi hai jo circulation ka sign hai jo pump in ho raha hai, aur gradient directions sahi karna woh sign fix karta hai.
Verify: Agar ∇ ρ ∥ ∇ p ∇ ρ × ∇ p = 0 D Γ/ D t = 0 misalignment angle se measure hoti hai ∇ ρ ∇ p
Figure s04 ∇ p down point karte hue draw karta hai (ground par zyada pressure ki taraf), ∇ ρ right point karte hue (dense sea air ki taraf), aur resulting yellow counter-clockwise circulation loop; caption state karta hai ki ∇ ρ × ∇ p Γ
Worked example Example 8 — Vorticity ek material loop se leak ho rahi hai (cell C8)Ek real (viscous) fluid ki kinematic viscosity ν Vorticity equation mein ek diffusion term ν ∇ 2 ω Γ ν = 1 0 − 6 m 2 / s δ = 1 mm
Forecast: friction — kya yeh swirl lock in karta hai, ya tumhare loop se seep hone deta hai?
Euler mein friction add karo. Navier–Stokes momentum balance clean Euler right-hand side − ρ 1 ∇ p − ∇Φ ν ∇ 2 u Yeh step kyun? Kelvin ke Step 3 ne assume kiya tha D u / D t = − ρ 1 ∇ p − ∇Φ Dikhao ki extra term loop integral se survive karta hai. Ise D Γ/ D t = ∮ C D t D u ⋅ d ℓ ∮ C ν ∇ 2 u ⋅ d ℓ = ν ∮ C ∇ 2 u ⋅ d ℓ nahin hai aur generally nahin vanish karta. Isliye D Γ/ D t = 0 Yeh step kyun? Pressure aur potential terms proof mein isliye khatam hue kyunki har ek ek perfect differential tha; viscous term nahin hai, isliye wahi ek term hai jo circulation create ya destroy kar sakta hai.Diffusion time estimate karo. Vorticity ek distance δ τ ∼ δ 2 / ν ν t ∼ δ τ ∼ δ 2 / ν = 1 0 − 6 ( 1 0 − 3 ) 2 = 1 s Yeh step kyun? Yeh concrete timescale deta hai jis par viscosity vorticity ko ek millimetre loop se carry karta hai aur visibly Kelvin violate karta hai.
Verify: Units: δ 2 / ν = m 2 / ( m 2 / s ) = s τ = 1 s D Γ/ D t = 0
Worked example Example 9 — Woh loop jo Kelvin "violate" karta lagta hai (cell C9)Ek wing ke past steady flow mein, space mein ek fixed rectangular loop (fluid ke saath nahin chalte) ke through circulation 0 30 m 2 / s
Forecast: kya Kelvin kisi bhi loop par conservation promise karta hai, ya sirf ek special par?
Loop type identify karo. Student ne ek loop use kiya jo space mein fixed hai; fluid particles usse through pass karte hain.
Yeh step kyun? Kelvin ka D / D t material derivative hai (upar define kiya: ∂ / ∂ t + u ⋅ ∇ Batao Kelvin actually kya claim karta hai. Sirf ek material loop ke liye (jo hamesha same fluid particles se bana ho) D Γ/ D t = 0 alag equation follow karti hai aur freely change ho sakti hai jab vortices uske boundary se advect hote hain.
Yeh step kyun? Galat tarah ke loop par theorem apply karna parent note ki second listed mistake hai — poora trap yahaan rehta hai.Paradox resolve karo. Starting vortex (Ex 6, circulation − 30 + 30 Yeh step kyun? Jabki material loop jo sab kuch enclose karta tha abhi bhi 0
Verify: Consistency chain: material loop total = Γ bound + Γ start = 30 + ( − 30 ) = 0 + 30 − 30
Recall Main kis cell mein hoon? (decision check)Loop shrinks, positive ω ω ω Γ = 0 → 0 ω → ∞ Γ − Γ bound ∇ ρ × ∇ p = 0 material loop chahiye
"Same water, no stickiness, one-to-one density" — material loop, inviscid, barotropic. Teeno mein se ek miss karo aur Γ Helmholtz vortex theorems aur Bernoulli's principle neighboring ideal-flow results ke liye jo exactly yahi hypotheses share karte hain).
