Vorticity — ω = ∇ × v, circulation Γ: Worked examples

2.2.29 · D3 · Physics › Fluid Mechanics › Vorticity — ω = ∇ × v, circulation Γ

Scenario matrix

Numbers touch karne se pehle, chalte hain map karte hain ki vorticity/circulation problems kis tarah ke situations mein aate hain. Neeche har worked example us cell ke saath tagged hai jise woh fill karta hai.

Cell	Case class	Tricky kya hai	Example
A	Solid-body rotation, $ω_{z} > 0$	curl ka sign, factor of 2	Ex 1
B	Clockwise rotation, $ω_{z} < 0$	negative vorticity, direction	Ex 2
C	Free vortex, $r \neq = 0$	circles karta hai par irrotational hai	Ex 3
D	Free vortex, singular point $r = 0$	degenerate input, delta of vorticity	Ex 3 (part b)
E	Pure shear, straight streamlines	vorticity bina visible curving ke	Ex 4
F	Superposition / cancelling vorticity	$Γ = 0$ par $ω \neq = 0$ andar	Ex 5
G	Real-world word problem	words ko field mein translate karo, units	Ex 6 (draining sink)
H	Exam twist — lift from circulation	$Γ$ ko ek force se connect karo	Ex 7
I	Limiting behaviour	kya hota hai jab $r \to 0$ ya $R \to \infty$	Ex 8

Notation warm-up (ek baar padho)

Neeche har field ek velocity field $v = (v_{x}, v_{y}, v_{z})$ hai: space ke har point par yeh tumhe woh arrow batata hai jo dikhata hai paani kis direction mein move karta hai aur kitni tezi se. Hum 2D mein kaam karte hain (flat flow), isliye ek hi spin axis hai $z$ (page se seedha upar), aur hume sirf ek hi vorticity component chahiye:

$ω_{z} = \frac{\partial v _{y}}{\partial x} - \frac{\partial v _{x}}{\partial y} .$

Symbol $\partial v_{y} / \partial x$ ka matlab hai " $x$ direction mein step karne par up-down velocity $v_{y}$ kitni tezi se change hoti hai." Yeh rate of change hai — ek slope. Isliye derivative sahi tool hai: rotation ka matlab hai neighbours alag alag move karna, aur derivative exactly "do neighbours ke beech quantity kaise differ karti hai" hai.
$ω_{z} > 0$ ka matlab hai counter-clockwise local spin (standard positive sense, jaise ulta chalta clock). $ω_{z} < 0$ ka matlab hai clockwise.

Parent ka paddle-wheel picture yaad rakho: flow mein ek choti si cross daalo; $ω_{z}$ hai (do guna) woh speed jis par cross khud pivot karta hai.

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

Ex 1 — Cell A: solid-body rotation, positive spin

Step 1. Components padho: $v_{x} = - 3 y$ , $v_{y} = 3 x$ . Yeh step kyun? Curl formula ko do partial derivatives chahiye; components ko clearly identify karna sign slips se bachata hai.

Step 2. Differentiate karo: $\frac{\partial v _{y}}{\partial x} = \frac{\partial ( 3 x )}{\partial x} = 3, \frac{\partial v _{x}}{\partial y} = \frac{\partial ( - 3 y )}{\partial y} = - 3.$ Yeh step kyun? $ω_{z}$ precisely inhi do "neighbour differences" ka difference hai.

Step 3. Combine karo: $ω_{z} = 3 - (- 3) = 6 = 2Ω.$ Yeh step kyun? Yeh parent ke headline result $ω = 2 Ω$ ko confirm karta hai — factor 2 real hai, typo nahi.

Step 4. Stokes ke zariye Circulation: $ω_{z}$ constant hai, aur hamare convention se hum circle counter-clockwise chalte hain, isliye $Γ = ω_{z} \cdot (area) = 6 \cdot π R^{2} = 6 π (2)^{2} = 24 π \approx 75.40 m^{2} / s .$ Yeh step kyun? Jab vorticity uniform ho, surface integral sirf vorticity times area hai — koi calculus nahi chahiye.

Ex 2 — Cell B: clockwise rotation, negative vorticity

Step 1. Components: $v_{x} = 5 y$ , $v_{y} = - 5 x$ . Yeh step kyun? Wahi discipline — differentiate karne se pehle pieces ka naam lo.

Step 2. Derivatives: $\frac{\partial v _{y}}{\partial x} = - 5$ , $\frac{\partial v _{x}}{\partial y} = 5$ . Yeh step kyun? Yeh ab Ex 1 se opposite signs carry karte hain, jo is case ka poora point hai.

Step 3. $ω_{z} = - 5 - 5 = - 10 s^{- 1}$ . Yeh step kyun? Negative value clockwise spin encode karti hai — direction sign mein rehti hai, aur ise drop karna physics ko reverse kar deta.

Step 4. $Γ = ω_{z} \cdot π R^{2} = - 10 π (1)^{2} = - 10 π \approx - 31.42 m^{2} / s .$ Yeh step kyun? Hamare counter-clockwise convention se, ek negative $Γ$ ka matlab hai flow tumhe apne chalne ki direction ke against push karta hai — exactly wahi jo ek clockwise swirl karta hai.

Ex 3 — Cells C & D: free vortex aur uska singular heart

Step 1. Circulation, circle ke around counter-clockwise walked, ka line integral hai. Circle par tangential velocity $d l$ ke saath aligned hai, aur $d l = r d θ$ : $Γ = \oint v_{θ} r d θ = \int_{0}^{2 π} \frac{k}{r} r d θ = k \int_{0}^{2 π} d θ = 2 π k .$ Yeh step kyun? Hum tangential component directly use karte hain kyunki circular loop ke liye, walking direction aur velocity parallel hain — dot product sirf magnitudes ka product hai.

Step 2. Plug in karo: $Γ = 2 π \cdot 4 = 8 π \approx 25.13 m^{2} / s$ , $r$ se independent. Yeh step kyun? Centre ke around har loop same $Γ$ deta hai. Woh constancy free vortex ki fingerprint hai.

Step 3 (Cell C, $r \neq = 0$ ). Do nested loops lo. Stokes ke zariye, unke beech vorticity flux hai $Γ_{outer} - Γ_{inner} = 8 π - 8 π = 0$ . Kyunki yeh kisi bhi annulus ke liye hold karta hai, axis se hatkar $ω_{z} = 0$ everywhere hai. Yeh step kyun? Yeh "same circulation ⇒ beech mein koi swirl nahi" ka rigorous version hai. Yahan paddle-wheel revolves karta hai par spin nahi karta: inner edge (fast) aur outer edge (slow) ise opposite senses mein torque karte hain aur cancel ho jaate hain.

Step 4 (Cell D, $r = 0$ ). Saari circulation singular centre par trapped hai. Formally vorticity origin par ek spike (delta function) hai jiska total flux $8 π$ ke equal hai. Formula $v_{θ} = k / r$ jab $r \to 0$ tab blow up karta hai, isliye yeh point ek degenerate input hai — tum wahan $ω_{z}$ pointwise evaluate nahi kar sakte. Yeh step kyun? Har case-complete analysis ko yeh naam dena chahiye ki excluded point par kya hota hai, na ki ise wave away karo.

Ex 4 — Cell E: pure shear, straight streamlines ke saath vorticity

Step 1. Components: $v_{x} = α y = 7 y$ , $v_{y} = 0$ . Yeh step kyun? $v_{y} = 0$ everywhere ek derivative ko khatam kar deta hai, shear ke effect ko isolate karta hai.

Step 2. $\frac{\partial v _{y}}{\partial x} = 0$ , $\frac{\partial v _{x}}{\partial y} = α = 7$ . Yeh step kyun? Nonzero piece velocity ke height ke saath change se aata hai — fluid element ka top uske bottom se faster move karta hai.

Step 3. $ω_{z} = 0 - 7 = - 7 s^{- 1}$ . Yeh step kyun? Straight streamlines ke bawajood nonzero — yahi is cell ka poora lesson hai. Ek choti si cross shear hoti hai, aur uski diagonal clockwise tilt karti hai, isliye $ω_{z} < 0$ .

Ex 5 — Cell F: cancelling vorticity, $Γ = 0$ par $ω \neq = 0$ andar

Step 1. Upper half ( $y > 0$ ): $v_{x} = β y$ , isliye $\frac{\partial v _{x}}{\partial y} = β = 2$ , deta hai $ω_{z} = - 2$ . Yeh step kyun? Ex 4 jaisa hi shear computation, piecewise us region par jahan formula apply hota hai.

Step 2. Lower half ( $y < 0$ ): $v_{x} = - β y$ , isliye $\frac{\partial v _{x}}{\partial y} = - β = - 2$ , deta hai $ω_{z} = + 2$ . Yeh step kyun? Sign flip hota hai, isliye do halves opposite spin karte hain — yeh engineered cancellation hai jo hum test karna chahte hain.

Step 3. Full rectangle ke around circulation, counter-clockwise walked. Top aur bottom edges (length 2) $v_{x}$ carry karti hain. Top par ( $y = 1$ ): $v_{x} = 2$ , $- x$ walk karna contribute karta hai $- 2 \cdot 2 = - 4$ . Bottom par ( $y = - 1$ ): $v_{x} = - β (- 1) = + 2$ , $+ x$ walk karna contribute karta hai $+ 2 \cdot 2 = + 4$ . Side edges mein $v_{x} ⊥ d l$ hai isliye contribute zero hai. $Γ = - 4 + 4 = 0.$ Yeh step kyun? Equal-and-opposite vorticities total flux mein cancel ho jaati hain.

Step 4. Conclude karo: poore loop mein $Γ = 0$ , phir bhi $ω_{z} = \pm 2 \neq = 0$ andar. Yeh step kyun? Yeh exactly parent ki warning hai: net flux zero ka matlab irrotational nahi hota. Positive aur negative vorticity balanced out ho gayi.

Ex 6 — Cell G: real-world word problem (draining bathtub)

Step 1. Model: $v_{θ} = k / r$ . Measurement use karke $k$ solve karo: $k = v_{θ} r = 0.5 \times 0.10 = 0.05 m^{2} / s$ . Yeh step kyun? Ek data point model ke single unknown ko fix karta hai.

Step 2. Circulation: $Γ = 2 π k = 2 π (0.05) = 0.1 π \approx 0.3142 m^{2} / s$ . Yeh step kyun? Ex 3 jaisa wahi free-vortex formula, ab physical $k$ ke saath.

Step 3. $r = 0.02$ par speed: $v_{θ} = k / r = 0.05/0.02 = 2.5 m/s$ . Yeh step kyun? $1/ r$ law ka matlab hai smaller radius ⇒ faster flow — paani drain mein accelerate karta hai, centre par visible fast whirl se match karta hai.

Ex 7 — Cell H: exam twist, lift from circulation

Step 1. Lift and the Kutta–Joukowski Theorem se relation likho: lift per unit span $L = ρ U Γ$ . Yeh step kyun? Yeh circulation (ek purely kinematic swirl) se ek real force tak ka bridge hai — wajah ki engineers $Γ$ ki parwah karte hain.

Step 2. Substitute karo: $L = 1.2 \times 60 \times 30 = 2160 N/m$ . Yeh step kyun? Formula choose hone ke baad direct plug-in.

Step 3. Forecast ka jawab do: $L \propto Γ$ linearly hai, isliye $Γ$ double karna lift double kar deta hai. Yeh step kyun? Proportionality recognize karna woh "twist" hai jo examiners test karte hain — arithmetic nahi.

Ex 8 — Cell I: limiting behaviour

Step 1 (a, $r \to \infty$ ). Free vortex: $v_{θ} = k / r \to 0$ . Door swirl die out ho jaati hai — flow calm hai. Yeh step kyun? Limits "background" state reveal karte hain; ek free vortex ek localized disturbance hai jo distance ke saath fade ho jaata hai.

Step 2 (a, $r \to 0$ ). Free vortex: $v_{θ} = k / r \to \infty$ (Cell D ka singular core). Physically, real fluids ise viscous core se cap karte hain, par ideal model diverge karta hai. Yeh step kyun? Blow-up ko naam dena degenerate-input picture complete karta hai — tumhe yeh kehna chahiye ki model us point par kya karta hai jise woh handle nahi kar sakta, aur reality ise kaise tame karti hai.

Step 3 (b, $r \to 0$ ). Solid body: $v_{θ} = Ω r \to 0$ . Rigidly rotating body ka centre rest mein hota hai — vortex se bilkul opposite behaviour, jo wahan blow up karta hai. Yeh step kyun? Do limits ko contrast karna cement karta hai ki "rotation" ek single flow nahi hai: free vortex centre par fastest hai, solid body door fastest hai.

Step 4 (c, matching speeds). Do speeds equal set karo aur crossover radius ke liye solve karo: $Ω r = \frac{k}{r} \Rightarrow r^{2} = \frac{k}{Ω} = \frac{4}{3} \Rightarrow r = \frac{4}{3} = \frac{2}{3} \approx 1.155 m .$ Yeh step kyun? Yeh crossover radius exactly Rankine vortex boundary hai — andar solid-body core, bahar free vortex — woh two-piece model jo real tornadoes aur drain vortices ke liye use hota hai.

Quick self-test

Kaun sa cell? Ek flow

v = (- 2 y, 2 x, 0)

ω_{z}

kya hai?

ω_{z} = 2 - (- 2) = 4

; solid-body, Cell A.

Free vortex

v_{θ} = k / r

: kya yeh

r > 0

ke liye rotational hai?

Nahi — vorticity axis se hatkar zero hai; saari swirl

r = 0

par baithi hai (Cell C/D).

Straight shear streamlines mein phir bhi vorticity ho sakti hai — true ya false?

True (Cell E);

ω_{z} = - \partial v_{x} / \partial y \neq = 0

Agar kisi loop ke around

Γ = 0

hai, kya interior irrotational hai?

Zaruri nahi —

+

aur

-

vorticity cancel ho sakti hai (Cell F).

Kutta–Joukowski lift per span formula?

L = ρ U Γ

(Cell H).

Hamare convention mein

Γ

ko positive banane ke liye kaun si walking direction?

Counter-clockwise (right-hand rule,

ω_{z}

page se bahar).

Connections

Parent: Vorticity & Circulation
Curl and Divergence (vector calculus)
Stokes' Theorem
Irrotational Flow and Velocity Potential
Kelvin's Circulation Theorem
Bernoulli's Equation
Lift and the Kutta–Joukowski Theorem
Angular Velocity of Rigid Bodies